Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic...

13
Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic Space Mahmut Mak 1 and Baki KarlJLa 2 1 Department of Mathematics, Faculty of Arts and Sciences, Ahi Evran University, 40100 Kırs ¸ehir, Turkey 2 Department of Mathematics, Faculty of Arts and Sciences, Gazi University, 06500 Ankara, Turkey Correspondence should be addressed to Mahmut Mak; [email protected] Received 23 May 2014; Accepted 19 September 2014; Published 20 November 2014 Academic Editor: Henggui Zhang Copyright © 2014 M. Mak and B. Karlı˘ ga. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We consider hyperbolic rotation ( 0 ), hyperbolic translation ( 1 ), and horocyclic rotation ( 2 ) groups in H 3 , which is called Minkowski model of hyperbolic space. en, we investigate extrinsic differential geometry of invariant surfaces under subgroups of 0 in H 3 . Also, we give explicit parametrization of these invariant surfaces with respect to constant hyperbolic curvature of profile curves. Finally, we obtain some corollaries for flat and minimal invariant surfaces which are associated with de Sitter and hyperbolic shape operator in H 3 . 1. Introduction Hyperbolic space has five analytic models, which are isomet- rically equivalent to each other [1, 2]. In this study, we choose Minkowski model of hyperbolic space which is denoted by H 3 . In a different point of view, we may consider invariant surface as rotational surface. In this sense, rotational surfaces in different ambient spaces were studied by many authors. For instance, in [3], Carmo and Dajczer define rotational hyper- surfaces with constant mean curvature (cmc) in hyperbolic -space. ey also give a local parametrization of this surface in terms of the cmc under some special conditions. In [4], Mori studied elliptic, spherical, and parabolic type rotational surfaces with cmc in H 3 . In [5], the total classification of the timelike and spacelike hyperbolic rotation surfaces is given in terms of cmc in 3-dimensional de Sitter space S 3 1 . As a general form, explicit parametrizations of rotational surfaces with cmc are given in Minkowski -space by [6]. is paper is organized as follows. In Section 2, we give briefly the notions of H-point, H-line, H-plane, and H- distance in hyperbolic geometry of H 3 . roughout this work, the prefix “H-” is used is the sense of belonging to hyperbolic space. It is well known that H-isometry is a map which is preserved H-distance in H 3 . e set of H-isometries is a group which is identified with restriction of isometries of Minkowski 4-space R 4 1 to H 3 . Let the group of H-isometries be denoted by in H 3 . We consider subgroups 0 , 1 , and 2 of with respect to leaving fixed timelike, spacelike, and lightlike planes of R 4 1 , respectively. en, we give the notions of H-rotation, H-translation, and horocyclic rotation which are one-parameter actions of in H 3 . Moreover, we obtain some properties of H-isometries. ere exist three kinds of totally umbilical surfaces which are called H-sphere, equidistant surface, and horosphere in H 3 . We obtain a classification of H-isometries by the subgroups 0 , 1 , and 2 with respect to leaving fixed equidistant surfaces, H-spheres, and horospheres in H 3 , respectively. In Section 3, we give the basic theory of extrinsic differential geometry of curves and surfaces in H 3 . In Section 4, we investigate surfaces which are invariant under a subgroup of H-translations in H 3 . Moreover, in the sense of de Sitter and hyperbolic shape operator in H 3 , we study extrinsic differential geometry of these invariant surfaces by using notations in [7, 8]. We give a relation between one of the principal curvatures of the invariant surface and hyperbolic curvature of profile curve of the invariant surface in H 3 . In a different viewpoint, we obtain explicit parametrization of some invariant surfaces in terms Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 838564, 12 pages http://dx.doi.org/10.1155/2014/838564

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Research ArticleInvariant Surfaces under Hyperbolic Translations inHyperbolic Space

Mahmut Mak1 and Baki KarlJLa2

1 Department of Mathematics Faculty of Arts and Sciences Ahi Evran University 40100 Kırsehir Turkey2Department of Mathematics Faculty of Arts and Sciences Gazi University 06500 Ankara Turkey

Correspondence should be addressed to Mahmut Mak mahmutmakgmailcom

Received 23 May 2014 Accepted 19 September 2014 Published 20 November 2014

Academic Editor Henggui Zhang

Copyright copy 2014 M Mak and B Karlıga This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider hyperbolic rotation (1198660) hyperbolic translation (119866

1) and horocyclic rotation (119866

2) groups in H3 which is called

Minkowski model of hyperbolic spaceThen we investigate extrinsic differential geometry of invariant surfaces under subgroups of119866

0in H3 Also we give explicit parametrization of these invariant surfaces with respect to constant hyperbolic curvature of profile

curves Finally we obtain some corollaries for flat andminimal invariant surfaces which are associated with de Sitter and hyperbolicshape operator in H3

1 Introduction

Hyperbolic space has five analytic models which are isomet-rically equivalent to each other [1 2] In this study we chooseMinkowski model of hyperbolic space which is denoted byH3 In a different point of view we may consider invariantsurface as rotational surface In this sense rotational surfacesin different ambient spaceswere studied bymany authors Forinstance in [3] Carmo and Dajczer define rotational hyper-surfaces with constant mean curvature (cmc) in hyperbolic119899-spaceThey also give a local parametrization of this surfacein terms of the cmc under some special conditions In [4]Mori studied elliptic spherical and parabolic type rotationalsurfaces with cmc in H3 In [5] the total classification of thetimelike and spacelike hyperbolic rotation surfaces is givenin terms of cmc in 3-dimensional de Sitter space S3

1 As a

general form explicit parametrizations of rotational surfaceswith cmc are given in Minkowski 119899-space by [6]

This paper is organized as follows In Section 2 we givebriefly the notions of H-point H-line H-plane and H-distance in hyperbolic geometry ofH3Throughout this workthe prefix ldquoH-rdquo is used is the sense of belonging to hyperbolicspace It is well known that H-isometry is a map which ispreserved H-distance in H3 The set of H-isometries is a

group which is identified with restriction of isometries ofMinkowski 4-space R4

1to H3 Let the group of H-isometries

be denoted by 119866 in H3 We consider subgroups 1198660 119866

1 and

1198662of 119866 with respect to leaving fixed timelike spacelike

and lightlike planes of R4

1 respectively Then we give the

notions of H-rotation H-translation and horocyclic rotationwhich are one-parameter actions of 119866 in H3 Moreover weobtain some properties of H-isometries There exist threekinds of totally umbilical surfaces which are called H-sphereequidistant surface and horosphere in H3 We obtain aclassification ofH-isometries by the subgroups119866

0119866

1 and119866

2

with respect to leaving fixed equidistant surfaces H-spheresand horospheres in H3 respectively In Section 3 we givethe basic theory of extrinsic differential geometry of curvesand surfaces in H3 In Section 4 we investigate surfaceswhich are invariant under a subgroup of H-translations inH3 Moreover in the sense of de Sitter and hyperbolic shapeoperator in H3 we study extrinsic differential geometry ofthese invariant surfaces by using notations in [7 8] We givea relation between one of the principal curvatures of theinvariant surface and hyperbolic curvature of profile curve ofthe invariant surface inH3 In a different viewpoint we obtainexplicit parametrization of some invariant surfaces in terms

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 838564 12 pageshttpdxdoiorg1011552014838564

2 Journal of Applied Mathematics

of constant hyperbolic curvature of profile curve Moreoverwe give some geometric results with respect to constanthyperbolic curvature of profile curve for flat and minimalinvariant surfaces in H3 Finally we give a classificationtheorem for the totally umbilical invariant surfaces in H3

2 Isometries of H3

In [9] Reynold give a brief introduction to hyperbolicgeometry of hyperbolic plane H2 Also he described explicitdescriptions of the hyperbolic metric and the isometries ofthe hyperbolic plane In this section we consider hyperbolicgeometry in H3 We especially determine isometry groups ofH3with respect to causal character of hyperplanes ofR4

1 then

these isometry groups are classified in terms of leaving thosetotally umbilic surfaces of H3 fixed

Let R4

1denote the 4-dimensional Minkowski space that

is the real vector space R4 endowed with the scalar product

⟨x y⟩ = minus11990901199100

+

3

sum

119894=1

119909119894119910119894

(1)

for all x = (1199090 119909

1 119909

2 119909

3) y = (119910

0 119910

1 119910

2 119910

3) isin R4

Let e0 e

1 e

2 e

3 be pseudo-orthonormal basis forR4

1 Then

⟨e119894 e

119895⟩ = 120575

119894119895120576119895for signatures 120576

0= minus1 120576

1= 120576

2= 120576

3= 1 The

function

119902 R4

1997888rarr R 119902 (x) = ⟨x x⟩ (2)

is called the associated quadratic form of ⟨sdot sdot⟩A vector k isin R4

1is called spacelike timelike and lightlike

if ⟨k k⟩ gt 0 (or k = 0) ⟨k k⟩ lt 0 and ⟨k k⟩ = 0respectively The Lorentzian norm of a vector k is defined byk = radic|⟨k k⟩|

The sets

H3

= x isin R4

1⟨x x⟩ = minus1 119909

0ge 1

S3

1= x isin R

4

1⟨x x⟩ = 1

LC+

= x isin R4

1| ⟨x x⟩ = 0 119909

0gt 0

(3)

are calledMinkowskimodel of hyperbolic space de Sitter spaceand future light cone respectively

Let 119875 be a vector subspace of R4

1 Then 119875 is said to be

timelike spacelike and lightlike if and only if 119875 contains atimelike vector and every nonzero vector in 119875 is spacelikeotherwise respectively

Nowwe give basic notions for hyperbolic geometry inH3From now on we use the prefix ldquoH-rdquo instead of ldquohyperbolicrdquofor brevity

An H-point is intersection 1198800

cap H3 such that 1198800is 1-

dimensional timelike subspace of R4

1and is called 119860

1198800 An

H-line is intersection 1198801

cap H3 such that 1198801is 2-dimensional

timelike subspace of R4

1and is called 119897

1198801 An H-plane is

intersection 1198802

cap H3 such that 1198802is 3-dimensional timelike

subspace of R4

1and is called 119863

1198802

H-coordinate axes 1198970119895

are denoted by intersections 1198970119895

=

1198810119895

capH3 such that1198810119895

= Spe0 e

119895 for 119895 = 1 2 3H-coordinate

planes 119863119894119895are denoted by intersections 119863

119894119895= 119882

0119894119895cap H3 such

that 1198820119894119895

= Spe0 e

119894 e

119895 for 119894 119895 = 1 2 3 H-upper (H-lower)

half-spaces of 119863119894119895are defined by intersections H3 and upper

(lower) half-space of 1198820119894119895

A hyperplane in R4

1is defined by HP(k 119888) = x isin R4

1|

⟨x k⟩ = 119888 for a pseudo-normal k isin R4

1and a real number 119888 If

k is spacelike timelike or lightlike HP(k 119888) is called timelikespacelike or lightlike respectively

Three kinds of totally umbilic surfaces haveH3 which aregiven by intersections of H3 and hyperplanes HP(k 119888) in R4

1

A surface HP(k 119888)capH3 is calledH-sphere equidistant surfaceand horosphere if HP(k 119888) is spacelike timelike and lightlikerespectively

We now give the existence and uniqueness of any H-lineor H-plane in H3 Any given two distinct points determineunique 2-plane through origin inR4

1and three distinct points

determine unique 3-plane through origin in R4

1 So the

following propositions are clear

Proposition 1 Any given two distinct H-points lie on a uniqueH-line in H3

Proposition 2 Any given three distinct H-points lie on aunique H-plane in H3

Also we say that H-line segments 119897119860119861 H-ray 119897997888997888rarr

119860119861are

determined by two different H-points119860 and119861 in natural way

Definition 3 Let 120574 [119886 119887] sub R rarr 119897119860119861

sub H3 beparametrization of 119897

119860119861 Then H-length of 119897

119860119861is given by

119889H (119860 119861) = int

119887

119886

100381710038171003817100381710038171205741015840(119905)

10038171003817100381710038171003817119889119905 (4)

If we take any hyperbolic space curve instead of 119897119860119861

inDefinition 3 thenH-arc length of any hyperbolic space curveis calculated by formula (4) in the same way Moreover H-distance between H-points 119860 and 119861 is given by

119889H (119860 119861) = minusarccosh (⟨119860 119861⟩) (5)

Let 119879 R4

1rarr R4

1be a linear transformation Then 119879

is called linear isometry (with respect to 119902) if it satisfies thefollowing equation

119902 (119879 (119909)) = 119902 (119909) (6)

Letmatrix form of linear transformation119879 be denoted byT =

[119905119894119895] 0 ⩽ 119894 119895 ⩽ 3 with respect to pseudo-orthonormal basis

e0 e

1 e

2 e

3 The set of all linear isometries of R4

1is a group

under matrix multiplication and it is denoted by

1198741

(4) = T isin GL (4R) | T119905J0T = J

0 (7)

where signature matrix J0

= diag(minus1 1 1 1) It is also calledsemiorthogonal group of R4

1and so

det (T) = plusmn1 (8)

Journal of Applied Mathematics 3

The subgroup 1198781198741(4) = T isin 119874

1(4) | det(T) = 1 is

called special semiorthogonal group Let block matrix form ofT isin 119874

1(4) be T = [

T119905 119887

119888 T119904 ] Then T119905and T

119904are called timelike

and spacelike part of T respectively

Definition 4 (i) If det(T119905) gt 0 (det(T

119905) lt 0) thenTpreserves

(reverses) time orientation(ii) If det(T

119904) gt 0 (det(T

119904) lt 0) then T preserves

(reverses) space orientation

Thus1198741(4) is decomposed into four disjoint sets indexed

by the signs of det(T119905) and det(T

119904) in that order They are

called 119874++

1(4) 119874

+minus

1(4) 119874

minus+

1(4) and 119874

minusminus

1(4) We define the

group

119866 = 119879 isin 1198741

(4) | 119879H3 H3

997888rarr H3 (9)

Elements of 119866 preserve H-distance in H3 It is clear that

119889H (119879 (119860) 119879 (119861)) = 119889H (119860 119861) (10)

for every 119879 isin 119866 and 119860 119861 isin H3 Thus we are ready to give thefollowing definition

Definition 5 Every element of 119866 is an H-isometry in H3

Thus we say that 119866 is union of subgroup which preservestime orientation of 119874

1(4) That is 119866 = 119874

++

1(4) cup 119874

+minus

1(4)

We consider 1198660 119866

1 and 119866

2subgroups of 119866 which

leave fixed timelike spacelike and lightlike planes of R4

1

respectively Let matrix representation of H-isometries beT = [119905

119894119895] 0 ⩽ 119894 119895 ⩽ 3 and let H-isometries J

1 J

2 and J

3

be denoted by J1

= diag(1 minus1 1 1) J2

= diag(1 1 minus1 1) andJ3

= diag(1 1 1 minus1) respectivelyWe suppose that 119879 isin 119866

0 Then 119879 leaves fixed timelike

planes 1198810119895

= Spe0 e

119895 for 119895 = 1 2 3 of R4

1 So that 119879(119881

0119895) =

1198810119895If 119879(119881

01) = 119881

01for 119895 = 1 then entries of matrix T must

be 11990500

= 1 11990510

= 0 11990520

= 0 and 11990530

= 0 and 11990501

= 0 11990511

= 111990521

= 0 and 11990531

= 0 By using (7) and (8) we have 11990503

= 011990502

= 0 11990512

= 0 and 11990513

= 0 and the following equationsystem

1199052

22+ 119905

2

32= 1 119905

2

23+ 119905

2

33= 1

11990522

11990523

+ 11990532

11990533

= 0

11990522

11990533

minus 11990523

11990532

= plusmn1

(11)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed timelike plane 119881

01of R4

1is given by

T01

= R01

120579J1198981J1198993 119898 119899 = 0 1 (12)

such that

R01

120579=

[[[

[

1 0 0 0

0 1 0 0

0 0 cos 120579 minus sin 120579

0 0 sin 120579 cos 120579

]]]

]

(13)

for all 120579 isin R In other cases if 119879(11988102

) = 11988102and 119879(119881

03) = 119881

03

then general forms of H-isometries that leave fixed timelikeplanes 119881

02and 119881

03of R4

1are given by

T02

= R02

120579J1198981J1198992

T03

= R03

120579J1198982J1198993

119898 119899 = 0 1

(14)

such that

R02

120579=

[[[

[

1 0 0 0

0 cos 120579 0 sin 120579

0 0 1 0

0 minus sin 120579 0 cos 120579

]]]

]

R03

120579=

[[[

[

1 0 0 0

0 cos 120579 minus sin 120579 0

0 sin 120579 cos 120579 0

0 0 0 1

]]]

]

(15)

for all 120579 isin R respectively Thus we say that the group 1198660is

union of disjoint subgroups of 119866+

0and 119866

minus

0such that

119866+

0= R01

11989811205791R02

11989821205792R03

11989831205793| 119898

119894isin Z 120579

119895isin R

119866minus

0= R01

11989811205791R02

11989821205792R03

11989831205793J11989841J11989852J11989863

|

119898119894isin Z 120579

119895isin R 119898

4+ 119898

5+ 119898

6equiv 1 (mod 2)

(16)

that is 1198660

= 119866+

0cup 119866

minus

0

We suppose that 119879 isin 1198661 Then 119879 leaves fixed spacelike

planes 119881119894119895

= Spe119894 e

119895 for 119894 119895 = 1 2 3 of R4

1 That is 119879(119881

119894119895) =

119881119894119895If 119879(119881

23) = 119881

23for 119894 = 2 119895 = 3 then entries of T must

be 11990502

= 0 11990512

= 0 11990522

= 1 and 11990532

= 0 and 11990503

= 0 11990513

= 011990523

= 0 and 11990533

= 1 By using (7) and (8) we have 11990520

= 011990530

= 0 11990521

= 0 and 11990531

= 0 and the following equationsystem

minus1199052

00+ 119905

2

10= minus1 minus119905

2

01+ 119905

2

11= 1

minus11990500

11990501

+ 11990510

11990511

= 0 (11990500

11990511

minus 11990501

11990510

) 11990522

11990533

= plusmn1

(17)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed spacelike plane 119881

23of R4

1is given by

T23

= L01119904J1198981J1198992J1198963 119898 119899 119896 = 0 1 (18)

such that

L01119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

(19)

4 Journal of Applied Mathematics

for all 119904 isin R In other cases if 119879(11988113

) = 11988113and 119879(119881

12) = 119881

12

then general forms of H-isometries that leave fixed spacelikeplanes 119881

13and 119881

12of R4

1are given by

T13

= L02119904J1198981J1198992J1198963

T12

= L03119904J1198981J1198992J1198963

119898 119899 119896 = 0 1

(20)

such that

L02119904

=[[[

[

cosh 119904 0 sinh 119904 0

0 1 0 0

sinh 119904 0 cosh 119904 0

0 0 0 1

]]]

]

(21)

L03119904

=[[[

[

cosh 119904 0 0 sinh 119904

0 1 0 0

0 0 1 0

sinh 119904 0 0 cosh 119904

]]]

]

(22)

for all 119904 isin R respectively Thus we say that the group 1198661is

union of disjoint subgroups of 119866+

1and 119866

minus

1such that

119866+

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 0 (mod 2)

(23)

119866minus

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 1 (mod 2)

(24)

that is 1198661

= 119866+

1cup 119866

minus

1

We suppose that 119879 isin 1198662 Then 119879 leaves fixed lightlike

planes D119894119895

= Spe0

+ e119894 e

119895 for 119895 = 1 2 3 of R4

1 So that

119879(D119894119895) = D

119894119895

If 119879(D12

) = D12for 119894 = 1 119895 = 2 then entries of matrix T

must be

11990500

+ 11990501

= 1 11990510

+ 11990511

= 1 11990520

+ 11990521

= 0

11990530

+ 11990531

= 0 11990502

= 11990512

= 11990532

= 0 11990522

= 1

(25)

By using (7) and (8) we obtain the following equation system

1 minus 211990500

+ 1199052

03= minus1 1 minus 2119905

10+ 119905

2

13= 1

1199052

23= 0 119905

2

33= 1

1 minus 11990500

minus 11990510

+ 11990503

11990513

= 0 minus11990530

+ 11990503

11990533

= 0

minus11990520

+ 11990513

11990523

= 0 minus11990530

+ 11990513

11990533

= 0

11990523

11990533

= 0 minus11990503

11990530

+ 11990513

11990530

+ (11990500

minus 11990510

) 11990533

= plusmn1

(26)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleaves fixed lightlike plane D

12of R4

1is given by

T012

= H012

120582J1198982J1198993 119898 119899 = 0 1 (27)

such that

H012

120582=

[[[[[[

[

1 +1205822

2minus

1205822

20 120582

1205822

21 minus

1205822

20 120582

0 0 1 0

120582 minus120582 0 1

]]]]]]

]

(28)

for all 120582 isin R In other cases we apply similar method Henceif H-isometry that leaves fixed lightlike plane D

119894119895of R4

1is

denoted byH0119894119895

120582 then we have the following H-isometries

H013

120582=

[[[[[[

[

1 +1205822

2minus

1205822

2120582 0

1205822

21 minus

1205822

2120582 0

120582 minus120582 1 0

0 0 0 1

]]]]]]

]

H021

120582=

[[[[[[

[

1 +1205822

20 minus

1205822

2120582

0 1 0 0

1205822

20 1 minus

1205822

2120582

120582 0 minus120582 1

]]]]]]

]

H023

120582=

[[[[[[

[

1 +1205822

2120582 minus

1205822

20

120582 1 minus120582 0

1205822

2120582 1 minus

1205822

20

0 0 0 1

]]]]]]

]

H031

120582=

[[[[[[

[

1 +1205822

20 120582 minus

1205822

20 1 0 0

120582 0 1 minus120582

1205822

20 120582 1 minus

1205822

2

]]]]]]

]

H032

120582=

[[[[[[

[

1 +1205822

2120582 0 minus

1205822

2120582 1 0 minus120582

0 0 1 0

1205822

2120582 0 1 minus

1205822

2

]]]]]]

]

(29)

and we obtained the following general forms

T013

= H013

120582J1198982J1198993

T021

= H021

120582J1198981J1198993

T023

= H023

120582J1198981J1198993

T031

= H031

120582J1198981J1198992

T032

= H032

120582J1198981J1198992

119898 119899 = 0 1

(30)

Journal of Applied Mathematics 5

So we say that the group 1198662is union of disjoint subgroups of

119866+

2and 119866

minus

2such that

119866+

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 0 (mod 2)

119866minus

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 1 (mod 2)

(31)

that is 1198662

= 119866+

2cup 119866

minus

2

Hence it is clear that

119866 = 1198660

cup 1198661

cup 1198662

1198660

cap 1198661

cap 1198662

= I4 J1 J2 J3 J1J2 J1J3 J2J3 J1J2J3

(32)

However we see that easily from matrix multiplication

R0119895

1205791R0119895

1205792= R0119895

1205791+1205792

L01198951199041L01198951199042

= L01198951199041+1199042

H0119894119895

1205821H0119894119895

1205822= H0119894119895

1205821+1205822

(33)

for any 1199041 119904

2 120579

1 120579

2 120582

1 120582

2isin R

Let parametrization of H-coordinate axes 11989701 119897

02 and 119897

03

be

11989701

[0 infin) 997888rarr 1198673 119897

01 (119903) = (cosh 119903 sinh 119903 0 0)

11989702

[0 infin) 997888rarr 1198673 119897

02(119903) = (cosh 119903 0 sinh 119903 0)

11989703

[0 infin) 997888rarr 1198673 119897

03(119903) = (cosh 119903 0 0 sinh 119903)

(34)

and let their matrix forms be

l01119903

=[[[

[

cosh 119903

sinh 119903

0

0

]]]

]

l02119903

=[[[

[

cosh 119903

0

sinh 119903

0

]]]

]

l03119903

=[[[

[

cosh 119903

0

0

sinh 119903

]]]

]

(35)

respectively After applying suitable H-isometry to H-coordinate axis l0119895

119903for 119895 = 1 2 3 we obtain the following

different parametrizations

H(11990312057911205792)

= R01

1205792R03

1205791l01119903

H(11990312057911205792)

= R02

1205792R01

1205791l02119903

H(11990312057911205792)

= R03

1205792R02

1205791l03119903

(36)

for hyperbolic polar coordinates 119903 gt 0 1205791

isin [0 120587] and 1205792

isin

[0 2120587] Thus we see roles of H-isometries R0119895

120579119894and L0119896

119904in H3

from equations

R01

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R02

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R03

120593

H(11990312057911205792)

=H

(1199031205791120593+1205792)

R03

120579H

(11990312057910)= H

(119903120579+1205791 0)

R01

120601H

(11990312057910)= H

(119903120601+1205791 0)

R02

120601

H(11990312057910)

=H

(119903120601+1205791 0)

L01119904H

(11990300)= H

(119904+11990300)

L02119904H

(11990300)= H

(119904+11990300)

L03119904

H(11990300)

=H

(119904+11990300)

(37)

Moreover L0119895119904and R0119895

120579leave fixed l0119895

119903 that is

L0119895119904l0119895119903

= l0119895119904+119903

R0119895

120579l0119895119903

= l0119895119903

(38)

Thus we are ready to give the following definitions by (33)and (38)

Definition 6 L0119895119904isH-translation by 119904 alongH-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 7 R0119895

120579is H-rotation by 120579 about H-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 8 H0119894119895

120582is horocyclic rotation by 120582 about lightlike

plane D119894119895for 119894 119895 = 1 2 3 (119894 = 119895) in H3

Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863

119894119895

Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is

R0119895

120579l01119903

=

l02119903

119895 = 3 120579 =120587

2

l03119903

119895 = 2 120579 =3120587

2

(39)

Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely

R0119895

120601l0119896119903

=

D23

(119903120601) 119895 = 1 119896 = 2

D13

(119903120601) 119895 = 2 119896 = 3

D12

(119903120601) 119895 = 3 119896 = 1

(40)

6 Journal of Applied Mathematics

Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is

R0119895

120579D12

(119903120601)=

D23

(119903120601) 119895 = 3 120579 =

3120587

2

D13

(119903120601) 119895 = 1 120579 =

120587

2

(41)

Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely

R0119897

minus120595R0119896

minus120601R0119895

minus120579H012

120582R0119895

120579R0119896

120601R0119897

120595

=

H013

120582 119895 = 1 120579 =

120587

2 120601 = 0 120595 = 0

H031

120582 119895 = 1 119896 = 2 120579 =

120587

2 120601 =

120587

2 120595 = 0

H032

120582 119895 = 1 119896 = 2 119897 = 3 120579 =

120587

2 120601 =

120587

2 120595 =

120587

2

H023

120582 119895 = 3 119896 = 2 120579 = minus

120587

2 120601 = minus

120587

2 120595 = 0

H021

120582 119895 = 3 120579 = minus

120587

2 120601 = 0 120595 = 0

(42)

After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed

Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878

1015840

for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3

Theorem 14 An H-sphere is invariant under H-translation inH3

Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3

cap HP(k minus119896) for 119896 gt 0 So

HP (k minus119896) = x isin R4

1|

minus V01199090

+ V11199091

+ V21199092

+ V31199093

= minus119896 119896 gt 0

(43)

Moreover for w = kk isin H3 and 119888 = 119896k we have

HP (w minus119888) = x isin R4

1| ⟨xw⟩ = minus119888 (44)

Since w isin H3

w = (cosh 1199040 sinh 119904

0cos120601

0 sinh 119904

0sin120601

0cos 120579

0

sinh 1199040sin120601

0sin 120579

0)

(45)

for any hyperbolic polar coordinates 1199040

isin [0 infin) 1206010

isin [0 120587]

and 1205790

isin [0 2120587] If we apply H-isometry T = L01minus1199040

R03

minus1206010R01

minus1205790isin

119866 then we have unit timelike vector e0such that

119879 (w) = e0 (46)

However unit timelike normal vector e0is invariant under

1198710119895

119904 That is

1198710119895

119904(e

0) = (cosh 119904) e

0 119895 = 1 2 3 (47)

For this reason if is an H-sphere which is generated fromspacelike hyperplane

HP (e0 minus119888) = x isin R

4

1| 119909

0= 119888 119888 ge 1 (48)

then we have

1198710119895

119904() = (49)

by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)

The following theorems also can be proved using similarmethod

Theorem 15 An equidistant surface is invariant under H-rotation in H3

Theorem 16 A horosphere is invariant under horocyclicrotation in H3

Finally we give the following corollary

Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866

0 119866

1 and 119866

2in H3

respectively

3 Differential Geometry of Curves andSurfaces in H3

In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]

TheLorentzian vector product of vectors x1 x2 x3 is givenby

x1 and x2 and x3 =

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

minuse0

e1

e2

e3

1199091

01199091

11199091

21199091

3

1199092

01199092

11199092

21199092

3

1199093

01199093

11199093

21199093

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(50)

where e0 e

1 e

2 e

3 is the canonical basis of R4

1and x119894 =

(119909119894

0 119909

119894

1 119909

119894

2 119909

119894

3) 119894 = 1 2 3 Also it is clear that

⟨x x1 and x2 and x3⟩ = det (x x1 x2 x3) (51)

for any x isin R4

1 Therefore x1 and x2 and x3 is pseudo-orthogonal

to any x119894 119894 = 1 2 3We recall the basic theory of curves inH3 Let120572 119868 rarr H3

be a unit speed regular curve for open subset 119868 sub R Since⟨120572

1015840(119904) 120572

1015840(119904)⟩ = 1 tangent vector of 120572 is given by t(119904) = 120572

1015840(119904)

The vector 12057210158401015840(119904) minus 120572(119904) is orthogonal to 120572(119904) and t(119904) We

suppose that 12057210158401015840(119904) minus 120572(119904) = 0 Then the normal vector of 120572

is given by n(119904) = (12057210158401015840(119904) minus 120572(119904))120572

10158401015840(119904) minus 120572(119904) However

the binormal vector of 120572 is given by e(119904) = 120572(119904) and t(119904) and n(119904)

Journal of Applied Mathematics 7

Hence we have a pseudo-orthonormal frame field120572(119904) t(119904)n(119904) e(119904) of R4

1along 120572 and the following Frenet-

Serret formulas

[[[

[

1205721015840(119904)

t1015840 (119904)

n1015840(119904)

e1015840 (119904)

]]]

]

=[[[

[

0 1 0 0

1 0 120581ℎ (119904) 0

0 minus120581ℎ

(119904) 0 120591ℎ

(119904)

0 0 minus120591ℎ

(119904) 0

]]]

]

[[[

[

120572 (119904)

t (119904)

n (119904)

e (119904)

]]]

]

(52)

where hyperbolic curvature and hyperbolic torsion of 120572

are given by 120581ℎ(119904) = 120572

10158401015840(119904) minus 120572(119904) and 120591

ℎ(119904) =

(minus det(120572(119904) 1205721015840(119904) 120572

10158401015840(119904) 120572

101584010158401015840(119904)))(120581

ℎ(119904))

2 under the assump-tion ⟨120572

10158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 respectively

Remark 18 The condition ⟨12057210158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 is equivalent

to 120581ℎ(119904) = 0 Moreover we see easily that 120581

ℎ(119904) = 0 if and only

if there exists a lightlike vector c such that 120572(119904) minus 119888 a geodesic(H-line)

If 120581ℎ(119904) = 1 and 120591

ℎ(119904) = 0 then 120572 in H3 is called a

horocycle We give a lemma about existence and uniquenessfor horocycles (cf [8 Proposition 43])

Lemma 19 For any a0

isin H3 and a1 a

2isin S3

1such that

⟨a119894 a

119895⟩ = 0 the unique horocycle with the initial conditions

120574(0) = a0 1205741015840(0) = a

1 and 120574

10158401015840(0) = a

0+ a

2is given by

120574 (119904) = a0

+ 119904a1

+1199042

2(a

0+ a

2) (53)

Now we recall the basic theory of surfaces in H3 Let x

119880 rarr H3 be embedding such that open subset 119880 sub R2 Wedenote that regular surface 119872 = x(119880) and identify 119872 and 119880

through the embedding x where 119894 119880 rarr 119872 is a local chartFor x(119906) = 119901 isin 119872 and 119906 = (119906

1 119906

2) isin 119880 if we define spacelike

unit normal vector

120578 (119906) =x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817

(54)

where x119906119894

= 120597x120597119906119894 119894 = 1 2 thenwehave ⟨x

119906119894 x⟩ = ⟨x

119906119894 120578⟩ =

0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3

1 Then the following maps 119864 119880 rarr 119878

3

1

119864(119906) = 120578(119906) and 119871plusmn

119880 rarr 119871119862+ 119871

plusmn(119906) = x(119906) plusmn 120578(119906)

are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906

0) can be identified with

identity mapping 119868119879119901119872

on the tangent space 119879119901119872 at x(119906

0) =

119901 isin 119872 We have that minus119889119871plusmn

= minus119868119879119901119872

plusmn (minus119889119864)For any given x(119906

0) = 119901 isin 119872 the linear transforms 119860

119901=

minus119889119864(1199060) 119879

119901119872 rarr 119879

119901119872 and 119878

plusmn

119901= minus119889119871

plusmn(119906

0) 119879

119901119872 rarr

119879119901119872 are called de Sitter shape operator and hyperbolic shape

operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and

119878plusmn

119901are denoted by 119896

119894(119901) and 119896

plusmn

119894(119901) for 119894 = 1 2 respectively

Obviously 119860119901and 119878

plusmn

119901have same eigenvectors Also the

eigenvalues satisfy

119896plusmn

119894(119901) = minus1 plusmn 119896

119894(119901) 119894 = 1 2 (55)

where 119896119894(119901) and 119896

plusmn

119894(119901) are called de Sitter principal curvature

and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872

respectivelyThe de Sitter Gauss curvature and the de Sitter mean

curvature of 119872 are given by

119870119889

(1199060) = det119860

119901= 119896

1(119901) 119896

2(119901)

119867119889

(1199060) =

1

2Tr119860

119901=

1198961

(119901) + 1198962

(119901)

2

(56)

at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss

curvature and the hyperbolic mean curvature of 119872 are givenby

119870plusmn

ℎ(119906

0) = det 119878

plusmn

119901= 119896

plusmn

1(119901) 119896

plusmn

2(119901)

119867plusmn

ℎ(119906

0) =

1

2Tr 119878

plusmn

119901=

119896plusmn

1(119901) + 119896

plusmn

2(119901)

2

(57)

at x(1199060) = 119901 respectively Evidently we have the following

relations119870

plusmn

ℎ= 1 ∓ 2119867

119889+ 119870

119889

119867plusmn

ℎ= minus1 plusmn 119867

119889

(58)

We say that a point x(1199060) = 119901 isin 119872 is an umbilical point

if 1198961(119901) = 119896

2(119901) Also 119872 is totally umbilical if all points

on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])

Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification

(1) Supposing that 1198962

= 1

(a) if 119896 = 0 and 1198962

lt 1 then M is a part of anequidistant surface

(b) if 119896 = 0 and 1198962

gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)

(2) If 1198962

= 1 then M is a part of horosphere

4 1198661-Invariant Surfaces in H3

In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces

Let 119872 = x(119880) be a regular surface via embedding x

119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860

the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4

1H3 and 119872 respectively Then the Gauss

and Weingarten explicit formulas for 119872 in H3 are given by

119863119883

119884 = 119863119883

119884 + ⟨119860 (119883) 119884⟩ 120578 + ⟨119883 119884⟩ x (59)

119860 (119883) = minus119863119883

120578 = minus119863119883

120578 (60)

for all tangent vector fields 119883 119884 isin X(119872) respectively

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Page 2: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

2 Journal of Applied Mathematics

of constant hyperbolic curvature of profile curve Moreoverwe give some geometric results with respect to constanthyperbolic curvature of profile curve for flat and minimalinvariant surfaces in H3 Finally we give a classificationtheorem for the totally umbilical invariant surfaces in H3

2 Isometries of H3

In [9] Reynold give a brief introduction to hyperbolicgeometry of hyperbolic plane H2 Also he described explicitdescriptions of the hyperbolic metric and the isometries ofthe hyperbolic plane In this section we consider hyperbolicgeometry in H3 We especially determine isometry groups ofH3with respect to causal character of hyperplanes ofR4

1 then

these isometry groups are classified in terms of leaving thosetotally umbilic surfaces of H3 fixed

Let R4

1denote the 4-dimensional Minkowski space that

is the real vector space R4 endowed with the scalar product

⟨x y⟩ = minus11990901199100

+

3

sum

119894=1

119909119894119910119894

(1)

for all x = (1199090 119909

1 119909

2 119909

3) y = (119910

0 119910

1 119910

2 119910

3) isin R4

Let e0 e

1 e

2 e

3 be pseudo-orthonormal basis forR4

1 Then

⟨e119894 e

119895⟩ = 120575

119894119895120576119895for signatures 120576

0= minus1 120576

1= 120576

2= 120576

3= 1 The

function

119902 R4

1997888rarr R 119902 (x) = ⟨x x⟩ (2)

is called the associated quadratic form of ⟨sdot sdot⟩A vector k isin R4

1is called spacelike timelike and lightlike

if ⟨k k⟩ gt 0 (or k = 0) ⟨k k⟩ lt 0 and ⟨k k⟩ = 0respectively The Lorentzian norm of a vector k is defined byk = radic|⟨k k⟩|

The sets

H3

= x isin R4

1⟨x x⟩ = minus1 119909

0ge 1

S3

1= x isin R

4

1⟨x x⟩ = 1

LC+

= x isin R4

1| ⟨x x⟩ = 0 119909

0gt 0

(3)

are calledMinkowskimodel of hyperbolic space de Sitter spaceand future light cone respectively

Let 119875 be a vector subspace of R4

1 Then 119875 is said to be

timelike spacelike and lightlike if and only if 119875 contains atimelike vector and every nonzero vector in 119875 is spacelikeotherwise respectively

Nowwe give basic notions for hyperbolic geometry inH3From now on we use the prefix ldquoH-rdquo instead of ldquohyperbolicrdquofor brevity

An H-point is intersection 1198800

cap H3 such that 1198800is 1-

dimensional timelike subspace of R4

1and is called 119860

1198800 An

H-line is intersection 1198801

cap H3 such that 1198801is 2-dimensional

timelike subspace of R4

1and is called 119897

1198801 An H-plane is

intersection 1198802

cap H3 such that 1198802is 3-dimensional timelike

subspace of R4

1and is called 119863

1198802

H-coordinate axes 1198970119895

are denoted by intersections 1198970119895

=

1198810119895

capH3 such that1198810119895

= Spe0 e

119895 for 119895 = 1 2 3H-coordinate

planes 119863119894119895are denoted by intersections 119863

119894119895= 119882

0119894119895cap H3 such

that 1198820119894119895

= Spe0 e

119894 e

119895 for 119894 119895 = 1 2 3 H-upper (H-lower)

half-spaces of 119863119894119895are defined by intersections H3 and upper

(lower) half-space of 1198820119894119895

A hyperplane in R4

1is defined by HP(k 119888) = x isin R4

1|

⟨x k⟩ = 119888 for a pseudo-normal k isin R4

1and a real number 119888 If

k is spacelike timelike or lightlike HP(k 119888) is called timelikespacelike or lightlike respectively

Three kinds of totally umbilic surfaces haveH3 which aregiven by intersections of H3 and hyperplanes HP(k 119888) in R4

1

A surface HP(k 119888)capH3 is calledH-sphere equidistant surfaceand horosphere if HP(k 119888) is spacelike timelike and lightlikerespectively

We now give the existence and uniqueness of any H-lineor H-plane in H3 Any given two distinct points determineunique 2-plane through origin inR4

1and three distinct points

determine unique 3-plane through origin in R4

1 So the

following propositions are clear

Proposition 1 Any given two distinct H-points lie on a uniqueH-line in H3

Proposition 2 Any given three distinct H-points lie on aunique H-plane in H3

Also we say that H-line segments 119897119860119861 H-ray 119897997888997888rarr

119860119861are

determined by two different H-points119860 and119861 in natural way

Definition 3 Let 120574 [119886 119887] sub R rarr 119897119860119861

sub H3 beparametrization of 119897

119860119861 Then H-length of 119897

119860119861is given by

119889H (119860 119861) = int

119887

119886

100381710038171003817100381710038171205741015840(119905)

10038171003817100381710038171003817119889119905 (4)

If we take any hyperbolic space curve instead of 119897119860119861

inDefinition 3 thenH-arc length of any hyperbolic space curveis calculated by formula (4) in the same way Moreover H-distance between H-points 119860 and 119861 is given by

119889H (119860 119861) = minusarccosh (⟨119860 119861⟩) (5)

Let 119879 R4

1rarr R4

1be a linear transformation Then 119879

is called linear isometry (with respect to 119902) if it satisfies thefollowing equation

119902 (119879 (119909)) = 119902 (119909) (6)

Letmatrix form of linear transformation119879 be denoted byT =

[119905119894119895] 0 ⩽ 119894 119895 ⩽ 3 with respect to pseudo-orthonormal basis

e0 e

1 e

2 e

3 The set of all linear isometries of R4

1is a group

under matrix multiplication and it is denoted by

1198741

(4) = T isin GL (4R) | T119905J0T = J

0 (7)

where signature matrix J0

= diag(minus1 1 1 1) It is also calledsemiorthogonal group of R4

1and so

det (T) = plusmn1 (8)

Journal of Applied Mathematics 3

The subgroup 1198781198741(4) = T isin 119874

1(4) | det(T) = 1 is

called special semiorthogonal group Let block matrix form ofT isin 119874

1(4) be T = [

T119905 119887

119888 T119904 ] Then T119905and T

119904are called timelike

and spacelike part of T respectively

Definition 4 (i) If det(T119905) gt 0 (det(T

119905) lt 0) thenTpreserves

(reverses) time orientation(ii) If det(T

119904) gt 0 (det(T

119904) lt 0) then T preserves

(reverses) space orientation

Thus1198741(4) is decomposed into four disjoint sets indexed

by the signs of det(T119905) and det(T

119904) in that order They are

called 119874++

1(4) 119874

+minus

1(4) 119874

minus+

1(4) and 119874

minusminus

1(4) We define the

group

119866 = 119879 isin 1198741

(4) | 119879H3 H3

997888rarr H3 (9)

Elements of 119866 preserve H-distance in H3 It is clear that

119889H (119879 (119860) 119879 (119861)) = 119889H (119860 119861) (10)

for every 119879 isin 119866 and 119860 119861 isin H3 Thus we are ready to give thefollowing definition

Definition 5 Every element of 119866 is an H-isometry in H3

Thus we say that 119866 is union of subgroup which preservestime orientation of 119874

1(4) That is 119866 = 119874

++

1(4) cup 119874

+minus

1(4)

We consider 1198660 119866

1 and 119866

2subgroups of 119866 which

leave fixed timelike spacelike and lightlike planes of R4

1

respectively Let matrix representation of H-isometries beT = [119905

119894119895] 0 ⩽ 119894 119895 ⩽ 3 and let H-isometries J

1 J

2 and J

3

be denoted by J1

= diag(1 minus1 1 1) J2

= diag(1 1 minus1 1) andJ3

= diag(1 1 1 minus1) respectivelyWe suppose that 119879 isin 119866

0 Then 119879 leaves fixed timelike

planes 1198810119895

= Spe0 e

119895 for 119895 = 1 2 3 of R4

1 So that 119879(119881

0119895) =

1198810119895If 119879(119881

01) = 119881

01for 119895 = 1 then entries of matrix T must

be 11990500

= 1 11990510

= 0 11990520

= 0 and 11990530

= 0 and 11990501

= 0 11990511

= 111990521

= 0 and 11990531

= 0 By using (7) and (8) we have 11990503

= 011990502

= 0 11990512

= 0 and 11990513

= 0 and the following equationsystem

1199052

22+ 119905

2

32= 1 119905

2

23+ 119905

2

33= 1

11990522

11990523

+ 11990532

11990533

= 0

11990522

11990533

minus 11990523

11990532

= plusmn1

(11)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed timelike plane 119881

01of R4

1is given by

T01

= R01

120579J1198981J1198993 119898 119899 = 0 1 (12)

such that

R01

120579=

[[[

[

1 0 0 0

0 1 0 0

0 0 cos 120579 minus sin 120579

0 0 sin 120579 cos 120579

]]]

]

(13)

for all 120579 isin R In other cases if 119879(11988102

) = 11988102and 119879(119881

03) = 119881

03

then general forms of H-isometries that leave fixed timelikeplanes 119881

02and 119881

03of R4

1are given by

T02

= R02

120579J1198981J1198992

T03

= R03

120579J1198982J1198993

119898 119899 = 0 1

(14)

such that

R02

120579=

[[[

[

1 0 0 0

0 cos 120579 0 sin 120579

0 0 1 0

0 minus sin 120579 0 cos 120579

]]]

]

R03

120579=

[[[

[

1 0 0 0

0 cos 120579 minus sin 120579 0

0 sin 120579 cos 120579 0

0 0 0 1

]]]

]

(15)

for all 120579 isin R respectively Thus we say that the group 1198660is

union of disjoint subgroups of 119866+

0and 119866

minus

0such that

119866+

0= R01

11989811205791R02

11989821205792R03

11989831205793| 119898

119894isin Z 120579

119895isin R

119866minus

0= R01

11989811205791R02

11989821205792R03

11989831205793J11989841J11989852J11989863

|

119898119894isin Z 120579

119895isin R 119898

4+ 119898

5+ 119898

6equiv 1 (mod 2)

(16)

that is 1198660

= 119866+

0cup 119866

minus

0

We suppose that 119879 isin 1198661 Then 119879 leaves fixed spacelike

planes 119881119894119895

= Spe119894 e

119895 for 119894 119895 = 1 2 3 of R4

1 That is 119879(119881

119894119895) =

119881119894119895If 119879(119881

23) = 119881

23for 119894 = 2 119895 = 3 then entries of T must

be 11990502

= 0 11990512

= 0 11990522

= 1 and 11990532

= 0 and 11990503

= 0 11990513

= 011990523

= 0 and 11990533

= 1 By using (7) and (8) we have 11990520

= 011990530

= 0 11990521

= 0 and 11990531

= 0 and the following equationsystem

minus1199052

00+ 119905

2

10= minus1 minus119905

2

01+ 119905

2

11= 1

minus11990500

11990501

+ 11990510

11990511

= 0 (11990500

11990511

minus 11990501

11990510

) 11990522

11990533

= plusmn1

(17)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed spacelike plane 119881

23of R4

1is given by

T23

= L01119904J1198981J1198992J1198963 119898 119899 119896 = 0 1 (18)

such that

L01119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

(19)

4 Journal of Applied Mathematics

for all 119904 isin R In other cases if 119879(11988113

) = 11988113and 119879(119881

12) = 119881

12

then general forms of H-isometries that leave fixed spacelikeplanes 119881

13and 119881

12of R4

1are given by

T13

= L02119904J1198981J1198992J1198963

T12

= L03119904J1198981J1198992J1198963

119898 119899 119896 = 0 1

(20)

such that

L02119904

=[[[

[

cosh 119904 0 sinh 119904 0

0 1 0 0

sinh 119904 0 cosh 119904 0

0 0 0 1

]]]

]

(21)

L03119904

=[[[

[

cosh 119904 0 0 sinh 119904

0 1 0 0

0 0 1 0

sinh 119904 0 0 cosh 119904

]]]

]

(22)

for all 119904 isin R respectively Thus we say that the group 1198661is

union of disjoint subgroups of 119866+

1and 119866

minus

1such that

119866+

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 0 (mod 2)

(23)

119866minus

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 1 (mod 2)

(24)

that is 1198661

= 119866+

1cup 119866

minus

1

We suppose that 119879 isin 1198662 Then 119879 leaves fixed lightlike

planes D119894119895

= Spe0

+ e119894 e

119895 for 119895 = 1 2 3 of R4

1 So that

119879(D119894119895) = D

119894119895

If 119879(D12

) = D12for 119894 = 1 119895 = 2 then entries of matrix T

must be

11990500

+ 11990501

= 1 11990510

+ 11990511

= 1 11990520

+ 11990521

= 0

11990530

+ 11990531

= 0 11990502

= 11990512

= 11990532

= 0 11990522

= 1

(25)

By using (7) and (8) we obtain the following equation system

1 minus 211990500

+ 1199052

03= minus1 1 minus 2119905

10+ 119905

2

13= 1

1199052

23= 0 119905

2

33= 1

1 minus 11990500

minus 11990510

+ 11990503

11990513

= 0 minus11990530

+ 11990503

11990533

= 0

minus11990520

+ 11990513

11990523

= 0 minus11990530

+ 11990513

11990533

= 0

11990523

11990533

= 0 minus11990503

11990530

+ 11990513

11990530

+ (11990500

minus 11990510

) 11990533

= plusmn1

(26)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleaves fixed lightlike plane D

12of R4

1is given by

T012

= H012

120582J1198982J1198993 119898 119899 = 0 1 (27)

such that

H012

120582=

[[[[[[

[

1 +1205822

2minus

1205822

20 120582

1205822

21 minus

1205822

20 120582

0 0 1 0

120582 minus120582 0 1

]]]]]]

]

(28)

for all 120582 isin R In other cases we apply similar method Henceif H-isometry that leaves fixed lightlike plane D

119894119895of R4

1is

denoted byH0119894119895

120582 then we have the following H-isometries

H013

120582=

[[[[[[

[

1 +1205822

2minus

1205822

2120582 0

1205822

21 minus

1205822

2120582 0

120582 minus120582 1 0

0 0 0 1

]]]]]]

]

H021

120582=

[[[[[[

[

1 +1205822

20 minus

1205822

2120582

0 1 0 0

1205822

20 1 minus

1205822

2120582

120582 0 minus120582 1

]]]]]]

]

H023

120582=

[[[[[[

[

1 +1205822

2120582 minus

1205822

20

120582 1 minus120582 0

1205822

2120582 1 minus

1205822

20

0 0 0 1

]]]]]]

]

H031

120582=

[[[[[[

[

1 +1205822

20 120582 minus

1205822

20 1 0 0

120582 0 1 minus120582

1205822

20 120582 1 minus

1205822

2

]]]]]]

]

H032

120582=

[[[[[[

[

1 +1205822

2120582 0 minus

1205822

2120582 1 0 minus120582

0 0 1 0

1205822

2120582 0 1 minus

1205822

2

]]]]]]

]

(29)

and we obtained the following general forms

T013

= H013

120582J1198982J1198993

T021

= H021

120582J1198981J1198993

T023

= H023

120582J1198981J1198993

T031

= H031

120582J1198981J1198992

T032

= H032

120582J1198981J1198992

119898 119899 = 0 1

(30)

Journal of Applied Mathematics 5

So we say that the group 1198662is union of disjoint subgroups of

119866+

2and 119866

minus

2such that

119866+

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 0 (mod 2)

119866minus

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 1 (mod 2)

(31)

that is 1198662

= 119866+

2cup 119866

minus

2

Hence it is clear that

119866 = 1198660

cup 1198661

cup 1198662

1198660

cap 1198661

cap 1198662

= I4 J1 J2 J3 J1J2 J1J3 J2J3 J1J2J3

(32)

However we see that easily from matrix multiplication

R0119895

1205791R0119895

1205792= R0119895

1205791+1205792

L01198951199041L01198951199042

= L01198951199041+1199042

H0119894119895

1205821H0119894119895

1205822= H0119894119895

1205821+1205822

(33)

for any 1199041 119904

2 120579

1 120579

2 120582

1 120582

2isin R

Let parametrization of H-coordinate axes 11989701 119897

02 and 119897

03

be

11989701

[0 infin) 997888rarr 1198673 119897

01 (119903) = (cosh 119903 sinh 119903 0 0)

11989702

[0 infin) 997888rarr 1198673 119897

02(119903) = (cosh 119903 0 sinh 119903 0)

11989703

[0 infin) 997888rarr 1198673 119897

03(119903) = (cosh 119903 0 0 sinh 119903)

(34)

and let their matrix forms be

l01119903

=[[[

[

cosh 119903

sinh 119903

0

0

]]]

]

l02119903

=[[[

[

cosh 119903

0

sinh 119903

0

]]]

]

l03119903

=[[[

[

cosh 119903

0

0

sinh 119903

]]]

]

(35)

respectively After applying suitable H-isometry to H-coordinate axis l0119895

119903for 119895 = 1 2 3 we obtain the following

different parametrizations

H(11990312057911205792)

= R01

1205792R03

1205791l01119903

H(11990312057911205792)

= R02

1205792R01

1205791l02119903

H(11990312057911205792)

= R03

1205792R02

1205791l03119903

(36)

for hyperbolic polar coordinates 119903 gt 0 1205791

isin [0 120587] and 1205792

isin

[0 2120587] Thus we see roles of H-isometries R0119895

120579119894and L0119896

119904in H3

from equations

R01

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R02

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R03

120593

H(11990312057911205792)

=H

(1199031205791120593+1205792)

R03

120579H

(11990312057910)= H

(119903120579+1205791 0)

R01

120601H

(11990312057910)= H

(119903120601+1205791 0)

R02

120601

H(11990312057910)

=H

(119903120601+1205791 0)

L01119904H

(11990300)= H

(119904+11990300)

L02119904H

(11990300)= H

(119904+11990300)

L03119904

H(11990300)

=H

(119904+11990300)

(37)

Moreover L0119895119904and R0119895

120579leave fixed l0119895

119903 that is

L0119895119904l0119895119903

= l0119895119904+119903

R0119895

120579l0119895119903

= l0119895119903

(38)

Thus we are ready to give the following definitions by (33)and (38)

Definition 6 L0119895119904isH-translation by 119904 alongH-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 7 R0119895

120579is H-rotation by 120579 about H-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 8 H0119894119895

120582is horocyclic rotation by 120582 about lightlike

plane D119894119895for 119894 119895 = 1 2 3 (119894 = 119895) in H3

Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863

119894119895

Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is

R0119895

120579l01119903

=

l02119903

119895 = 3 120579 =120587

2

l03119903

119895 = 2 120579 =3120587

2

(39)

Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely

R0119895

120601l0119896119903

=

D23

(119903120601) 119895 = 1 119896 = 2

D13

(119903120601) 119895 = 2 119896 = 3

D12

(119903120601) 119895 = 3 119896 = 1

(40)

6 Journal of Applied Mathematics

Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is

R0119895

120579D12

(119903120601)=

D23

(119903120601) 119895 = 3 120579 =

3120587

2

D13

(119903120601) 119895 = 1 120579 =

120587

2

(41)

Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely

R0119897

minus120595R0119896

minus120601R0119895

minus120579H012

120582R0119895

120579R0119896

120601R0119897

120595

=

H013

120582 119895 = 1 120579 =

120587

2 120601 = 0 120595 = 0

H031

120582 119895 = 1 119896 = 2 120579 =

120587

2 120601 =

120587

2 120595 = 0

H032

120582 119895 = 1 119896 = 2 119897 = 3 120579 =

120587

2 120601 =

120587

2 120595 =

120587

2

H023

120582 119895 = 3 119896 = 2 120579 = minus

120587

2 120601 = minus

120587

2 120595 = 0

H021

120582 119895 = 3 120579 = minus

120587

2 120601 = 0 120595 = 0

(42)

After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed

Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878

1015840

for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3

Theorem 14 An H-sphere is invariant under H-translation inH3

Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3

cap HP(k minus119896) for 119896 gt 0 So

HP (k minus119896) = x isin R4

1|

minus V01199090

+ V11199091

+ V21199092

+ V31199093

= minus119896 119896 gt 0

(43)

Moreover for w = kk isin H3 and 119888 = 119896k we have

HP (w minus119888) = x isin R4

1| ⟨xw⟩ = minus119888 (44)

Since w isin H3

w = (cosh 1199040 sinh 119904

0cos120601

0 sinh 119904

0sin120601

0cos 120579

0

sinh 1199040sin120601

0sin 120579

0)

(45)

for any hyperbolic polar coordinates 1199040

isin [0 infin) 1206010

isin [0 120587]

and 1205790

isin [0 2120587] If we apply H-isometry T = L01minus1199040

R03

minus1206010R01

minus1205790isin

119866 then we have unit timelike vector e0such that

119879 (w) = e0 (46)

However unit timelike normal vector e0is invariant under

1198710119895

119904 That is

1198710119895

119904(e

0) = (cosh 119904) e

0 119895 = 1 2 3 (47)

For this reason if is an H-sphere which is generated fromspacelike hyperplane

HP (e0 minus119888) = x isin R

4

1| 119909

0= 119888 119888 ge 1 (48)

then we have

1198710119895

119904() = (49)

by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)

The following theorems also can be proved using similarmethod

Theorem 15 An equidistant surface is invariant under H-rotation in H3

Theorem 16 A horosphere is invariant under horocyclicrotation in H3

Finally we give the following corollary

Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866

0 119866

1 and 119866

2in H3

respectively

3 Differential Geometry of Curves andSurfaces in H3

In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]

TheLorentzian vector product of vectors x1 x2 x3 is givenby

x1 and x2 and x3 =

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

minuse0

e1

e2

e3

1199091

01199091

11199091

21199091

3

1199092

01199092

11199092

21199092

3

1199093

01199093

11199093

21199093

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(50)

where e0 e

1 e

2 e

3 is the canonical basis of R4

1and x119894 =

(119909119894

0 119909

119894

1 119909

119894

2 119909

119894

3) 119894 = 1 2 3 Also it is clear that

⟨x x1 and x2 and x3⟩ = det (x x1 x2 x3) (51)

for any x isin R4

1 Therefore x1 and x2 and x3 is pseudo-orthogonal

to any x119894 119894 = 1 2 3We recall the basic theory of curves inH3 Let120572 119868 rarr H3

be a unit speed regular curve for open subset 119868 sub R Since⟨120572

1015840(119904) 120572

1015840(119904)⟩ = 1 tangent vector of 120572 is given by t(119904) = 120572

1015840(119904)

The vector 12057210158401015840(119904) minus 120572(119904) is orthogonal to 120572(119904) and t(119904) We

suppose that 12057210158401015840(119904) minus 120572(119904) = 0 Then the normal vector of 120572

is given by n(119904) = (12057210158401015840(119904) minus 120572(119904))120572

10158401015840(119904) minus 120572(119904) However

the binormal vector of 120572 is given by e(119904) = 120572(119904) and t(119904) and n(119904)

Journal of Applied Mathematics 7

Hence we have a pseudo-orthonormal frame field120572(119904) t(119904)n(119904) e(119904) of R4

1along 120572 and the following Frenet-

Serret formulas

[[[

[

1205721015840(119904)

t1015840 (119904)

n1015840(119904)

e1015840 (119904)

]]]

]

=[[[

[

0 1 0 0

1 0 120581ℎ (119904) 0

0 minus120581ℎ

(119904) 0 120591ℎ

(119904)

0 0 minus120591ℎ

(119904) 0

]]]

]

[[[

[

120572 (119904)

t (119904)

n (119904)

e (119904)

]]]

]

(52)

where hyperbolic curvature and hyperbolic torsion of 120572

are given by 120581ℎ(119904) = 120572

10158401015840(119904) minus 120572(119904) and 120591

ℎ(119904) =

(minus det(120572(119904) 1205721015840(119904) 120572

10158401015840(119904) 120572

101584010158401015840(119904)))(120581

ℎ(119904))

2 under the assump-tion ⟨120572

10158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 respectively

Remark 18 The condition ⟨12057210158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 is equivalent

to 120581ℎ(119904) = 0 Moreover we see easily that 120581

ℎ(119904) = 0 if and only

if there exists a lightlike vector c such that 120572(119904) minus 119888 a geodesic(H-line)

If 120581ℎ(119904) = 1 and 120591

ℎ(119904) = 0 then 120572 in H3 is called a

horocycle We give a lemma about existence and uniquenessfor horocycles (cf [8 Proposition 43])

Lemma 19 For any a0

isin H3 and a1 a

2isin S3

1such that

⟨a119894 a

119895⟩ = 0 the unique horocycle with the initial conditions

120574(0) = a0 1205741015840(0) = a

1 and 120574

10158401015840(0) = a

0+ a

2is given by

120574 (119904) = a0

+ 119904a1

+1199042

2(a

0+ a

2) (53)

Now we recall the basic theory of surfaces in H3 Let x

119880 rarr H3 be embedding such that open subset 119880 sub R2 Wedenote that regular surface 119872 = x(119880) and identify 119872 and 119880

through the embedding x where 119894 119880 rarr 119872 is a local chartFor x(119906) = 119901 isin 119872 and 119906 = (119906

1 119906

2) isin 119880 if we define spacelike

unit normal vector

120578 (119906) =x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817

(54)

where x119906119894

= 120597x120597119906119894 119894 = 1 2 thenwehave ⟨x

119906119894 x⟩ = ⟨x

119906119894 120578⟩ =

0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3

1 Then the following maps 119864 119880 rarr 119878

3

1

119864(119906) = 120578(119906) and 119871plusmn

119880 rarr 119871119862+ 119871

plusmn(119906) = x(119906) plusmn 120578(119906)

are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906

0) can be identified with

identity mapping 119868119879119901119872

on the tangent space 119879119901119872 at x(119906

0) =

119901 isin 119872 We have that minus119889119871plusmn

= minus119868119879119901119872

plusmn (minus119889119864)For any given x(119906

0) = 119901 isin 119872 the linear transforms 119860

119901=

minus119889119864(1199060) 119879

119901119872 rarr 119879

119901119872 and 119878

plusmn

119901= minus119889119871

plusmn(119906

0) 119879

119901119872 rarr

119879119901119872 are called de Sitter shape operator and hyperbolic shape

operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and

119878plusmn

119901are denoted by 119896

119894(119901) and 119896

plusmn

119894(119901) for 119894 = 1 2 respectively

Obviously 119860119901and 119878

plusmn

119901have same eigenvectors Also the

eigenvalues satisfy

119896plusmn

119894(119901) = minus1 plusmn 119896

119894(119901) 119894 = 1 2 (55)

where 119896119894(119901) and 119896

plusmn

119894(119901) are called de Sitter principal curvature

and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872

respectivelyThe de Sitter Gauss curvature and the de Sitter mean

curvature of 119872 are given by

119870119889

(1199060) = det119860

119901= 119896

1(119901) 119896

2(119901)

119867119889

(1199060) =

1

2Tr119860

119901=

1198961

(119901) + 1198962

(119901)

2

(56)

at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss

curvature and the hyperbolic mean curvature of 119872 are givenby

119870plusmn

ℎ(119906

0) = det 119878

plusmn

119901= 119896

plusmn

1(119901) 119896

plusmn

2(119901)

119867plusmn

ℎ(119906

0) =

1

2Tr 119878

plusmn

119901=

119896plusmn

1(119901) + 119896

plusmn

2(119901)

2

(57)

at x(1199060) = 119901 respectively Evidently we have the following

relations119870

plusmn

ℎ= 1 ∓ 2119867

119889+ 119870

119889

119867plusmn

ℎ= minus1 plusmn 119867

119889

(58)

We say that a point x(1199060) = 119901 isin 119872 is an umbilical point

if 1198961(119901) = 119896

2(119901) Also 119872 is totally umbilical if all points

on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])

Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification

(1) Supposing that 1198962

= 1

(a) if 119896 = 0 and 1198962

lt 1 then M is a part of anequidistant surface

(b) if 119896 = 0 and 1198962

gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)

(2) If 1198962

= 1 then M is a part of horosphere

4 1198661-Invariant Surfaces in H3

In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces

Let 119872 = x(119880) be a regular surface via embedding x

119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860

the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4

1H3 and 119872 respectively Then the Gauss

and Weingarten explicit formulas for 119872 in H3 are given by

119863119883

119884 = 119863119883

119884 + ⟨119860 (119883) 119884⟩ 120578 + ⟨119883 119884⟩ x (59)

119860 (119883) = minus119863119883

120578 = minus119863119883

120578 (60)

for all tangent vector fields 119883 119884 isin X(119872) respectively

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Page 3: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

Journal of Applied Mathematics 3

The subgroup 1198781198741(4) = T isin 119874

1(4) | det(T) = 1 is

called special semiorthogonal group Let block matrix form ofT isin 119874

1(4) be T = [

T119905 119887

119888 T119904 ] Then T119905and T

119904are called timelike

and spacelike part of T respectively

Definition 4 (i) If det(T119905) gt 0 (det(T

119905) lt 0) thenTpreserves

(reverses) time orientation(ii) If det(T

119904) gt 0 (det(T

119904) lt 0) then T preserves

(reverses) space orientation

Thus1198741(4) is decomposed into four disjoint sets indexed

by the signs of det(T119905) and det(T

119904) in that order They are

called 119874++

1(4) 119874

+minus

1(4) 119874

minus+

1(4) and 119874

minusminus

1(4) We define the

group

119866 = 119879 isin 1198741

(4) | 119879H3 H3

997888rarr H3 (9)

Elements of 119866 preserve H-distance in H3 It is clear that

119889H (119879 (119860) 119879 (119861)) = 119889H (119860 119861) (10)

for every 119879 isin 119866 and 119860 119861 isin H3 Thus we are ready to give thefollowing definition

Definition 5 Every element of 119866 is an H-isometry in H3

Thus we say that 119866 is union of subgroup which preservestime orientation of 119874

1(4) That is 119866 = 119874

++

1(4) cup 119874

+minus

1(4)

We consider 1198660 119866

1 and 119866

2subgroups of 119866 which

leave fixed timelike spacelike and lightlike planes of R4

1

respectively Let matrix representation of H-isometries beT = [119905

119894119895] 0 ⩽ 119894 119895 ⩽ 3 and let H-isometries J

1 J

2 and J

3

be denoted by J1

= diag(1 minus1 1 1) J2

= diag(1 1 minus1 1) andJ3

= diag(1 1 1 minus1) respectivelyWe suppose that 119879 isin 119866

0 Then 119879 leaves fixed timelike

planes 1198810119895

= Spe0 e

119895 for 119895 = 1 2 3 of R4

1 So that 119879(119881

0119895) =

1198810119895If 119879(119881

01) = 119881

01for 119895 = 1 then entries of matrix T must

be 11990500

= 1 11990510

= 0 11990520

= 0 and 11990530

= 0 and 11990501

= 0 11990511

= 111990521

= 0 and 11990531

= 0 By using (7) and (8) we have 11990503

= 011990502

= 0 11990512

= 0 and 11990513

= 0 and the following equationsystem

1199052

22+ 119905

2

32= 1 119905

2

23+ 119905

2

33= 1

11990522

11990523

+ 11990532

11990533

= 0

11990522

11990533

minus 11990523

11990532

= plusmn1

(11)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed timelike plane 119881

01of R4

1is given by

T01

= R01

120579J1198981J1198993 119898 119899 = 0 1 (12)

such that

R01

120579=

[[[

[

1 0 0 0

0 1 0 0

0 0 cos 120579 minus sin 120579

0 0 sin 120579 cos 120579

]]]

]

(13)

for all 120579 isin R In other cases if 119879(11988102

) = 11988102and 119879(119881

03) = 119881

03

then general forms of H-isometries that leave fixed timelikeplanes 119881

02and 119881

03of R4

1are given by

T02

= R02

120579J1198981J1198992

T03

= R03

120579J1198982J1198993

119898 119899 = 0 1

(14)

such that

R02

120579=

[[[

[

1 0 0 0

0 cos 120579 0 sin 120579

0 0 1 0

0 minus sin 120579 0 cos 120579

]]]

]

R03

120579=

[[[

[

1 0 0 0

0 cos 120579 minus sin 120579 0

0 sin 120579 cos 120579 0

0 0 0 1

]]]

]

(15)

for all 120579 isin R respectively Thus we say that the group 1198660is

union of disjoint subgroups of 119866+

0and 119866

minus

0such that

119866+

0= R01

11989811205791R02

11989821205792R03

11989831205793| 119898

119894isin Z 120579

119895isin R

119866minus

0= R01

11989811205791R02

11989821205792R03

11989831205793J11989841J11989852J11989863

|

119898119894isin Z 120579

119895isin R 119898

4+ 119898

5+ 119898

6equiv 1 (mod 2)

(16)

that is 1198660

= 119866+

0cup 119866

minus

0

We suppose that 119879 isin 1198661 Then 119879 leaves fixed spacelike

planes 119881119894119895

= Spe119894 e

119895 for 119894 119895 = 1 2 3 of R4

1 That is 119879(119881

119894119895) =

119881119894119895If 119879(119881

23) = 119881

23for 119894 = 2 119895 = 3 then entries of T must

be 11990502

= 0 11990512

= 0 11990522

= 1 and 11990532

= 0 and 11990503

= 0 11990513

= 011990523

= 0 and 11990533

= 1 By using (7) and (8) we have 11990520

= 011990530

= 0 11990521

= 0 and 11990531

= 0 and the following equationsystem

minus1199052

00+ 119905

2

10= minus1 minus119905

2

01+ 119905

2

11= 1

minus11990500

11990501

+ 11990510

11990511

= 0 (11990500

11990511

minus 11990501

11990510

) 11990522

11990533

= plusmn1

(17)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed spacelike plane 119881

23of R4

1is given by

T23

= L01119904J1198981J1198992J1198963 119898 119899 119896 = 0 1 (18)

such that

L01119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

(19)

4 Journal of Applied Mathematics

for all 119904 isin R In other cases if 119879(11988113

) = 11988113and 119879(119881

12) = 119881

12

then general forms of H-isometries that leave fixed spacelikeplanes 119881

13and 119881

12of R4

1are given by

T13

= L02119904J1198981J1198992J1198963

T12

= L03119904J1198981J1198992J1198963

119898 119899 119896 = 0 1

(20)

such that

L02119904

=[[[

[

cosh 119904 0 sinh 119904 0

0 1 0 0

sinh 119904 0 cosh 119904 0

0 0 0 1

]]]

]

(21)

L03119904

=[[[

[

cosh 119904 0 0 sinh 119904

0 1 0 0

0 0 1 0

sinh 119904 0 0 cosh 119904

]]]

]

(22)

for all 119904 isin R respectively Thus we say that the group 1198661is

union of disjoint subgroups of 119866+

1and 119866

minus

1such that

119866+

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 0 (mod 2)

(23)

119866minus

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 1 (mod 2)

(24)

that is 1198661

= 119866+

1cup 119866

minus

1

We suppose that 119879 isin 1198662 Then 119879 leaves fixed lightlike

planes D119894119895

= Spe0

+ e119894 e

119895 for 119895 = 1 2 3 of R4

1 So that

119879(D119894119895) = D

119894119895

If 119879(D12

) = D12for 119894 = 1 119895 = 2 then entries of matrix T

must be

11990500

+ 11990501

= 1 11990510

+ 11990511

= 1 11990520

+ 11990521

= 0

11990530

+ 11990531

= 0 11990502

= 11990512

= 11990532

= 0 11990522

= 1

(25)

By using (7) and (8) we obtain the following equation system

1 minus 211990500

+ 1199052

03= minus1 1 minus 2119905

10+ 119905

2

13= 1

1199052

23= 0 119905

2

33= 1

1 minus 11990500

minus 11990510

+ 11990503

11990513

= 0 minus11990530

+ 11990503

11990533

= 0

minus11990520

+ 11990513

11990523

= 0 minus11990530

+ 11990513

11990533

= 0

11990523

11990533

= 0 minus11990503

11990530

+ 11990513

11990530

+ (11990500

minus 11990510

) 11990533

= plusmn1

(26)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleaves fixed lightlike plane D

12of R4

1is given by

T012

= H012

120582J1198982J1198993 119898 119899 = 0 1 (27)

such that

H012

120582=

[[[[[[

[

1 +1205822

2minus

1205822

20 120582

1205822

21 minus

1205822

20 120582

0 0 1 0

120582 minus120582 0 1

]]]]]]

]

(28)

for all 120582 isin R In other cases we apply similar method Henceif H-isometry that leaves fixed lightlike plane D

119894119895of R4

1is

denoted byH0119894119895

120582 then we have the following H-isometries

H013

120582=

[[[[[[

[

1 +1205822

2minus

1205822

2120582 0

1205822

21 minus

1205822

2120582 0

120582 minus120582 1 0

0 0 0 1

]]]]]]

]

H021

120582=

[[[[[[

[

1 +1205822

20 minus

1205822

2120582

0 1 0 0

1205822

20 1 minus

1205822

2120582

120582 0 minus120582 1

]]]]]]

]

H023

120582=

[[[[[[

[

1 +1205822

2120582 minus

1205822

20

120582 1 minus120582 0

1205822

2120582 1 minus

1205822

20

0 0 0 1

]]]]]]

]

H031

120582=

[[[[[[

[

1 +1205822

20 120582 minus

1205822

20 1 0 0

120582 0 1 minus120582

1205822

20 120582 1 minus

1205822

2

]]]]]]

]

H032

120582=

[[[[[[

[

1 +1205822

2120582 0 minus

1205822

2120582 1 0 minus120582

0 0 1 0

1205822

2120582 0 1 minus

1205822

2

]]]]]]

]

(29)

and we obtained the following general forms

T013

= H013

120582J1198982J1198993

T021

= H021

120582J1198981J1198993

T023

= H023

120582J1198981J1198993

T031

= H031

120582J1198981J1198992

T032

= H032

120582J1198981J1198992

119898 119899 = 0 1

(30)

Journal of Applied Mathematics 5

So we say that the group 1198662is union of disjoint subgroups of

119866+

2and 119866

minus

2such that

119866+

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 0 (mod 2)

119866minus

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 1 (mod 2)

(31)

that is 1198662

= 119866+

2cup 119866

minus

2

Hence it is clear that

119866 = 1198660

cup 1198661

cup 1198662

1198660

cap 1198661

cap 1198662

= I4 J1 J2 J3 J1J2 J1J3 J2J3 J1J2J3

(32)

However we see that easily from matrix multiplication

R0119895

1205791R0119895

1205792= R0119895

1205791+1205792

L01198951199041L01198951199042

= L01198951199041+1199042

H0119894119895

1205821H0119894119895

1205822= H0119894119895

1205821+1205822

(33)

for any 1199041 119904

2 120579

1 120579

2 120582

1 120582

2isin R

Let parametrization of H-coordinate axes 11989701 119897

02 and 119897

03

be

11989701

[0 infin) 997888rarr 1198673 119897

01 (119903) = (cosh 119903 sinh 119903 0 0)

11989702

[0 infin) 997888rarr 1198673 119897

02(119903) = (cosh 119903 0 sinh 119903 0)

11989703

[0 infin) 997888rarr 1198673 119897

03(119903) = (cosh 119903 0 0 sinh 119903)

(34)

and let their matrix forms be

l01119903

=[[[

[

cosh 119903

sinh 119903

0

0

]]]

]

l02119903

=[[[

[

cosh 119903

0

sinh 119903

0

]]]

]

l03119903

=[[[

[

cosh 119903

0

0

sinh 119903

]]]

]

(35)

respectively After applying suitable H-isometry to H-coordinate axis l0119895

119903for 119895 = 1 2 3 we obtain the following

different parametrizations

H(11990312057911205792)

= R01

1205792R03

1205791l01119903

H(11990312057911205792)

= R02

1205792R01

1205791l02119903

H(11990312057911205792)

= R03

1205792R02

1205791l03119903

(36)

for hyperbolic polar coordinates 119903 gt 0 1205791

isin [0 120587] and 1205792

isin

[0 2120587] Thus we see roles of H-isometries R0119895

120579119894and L0119896

119904in H3

from equations

R01

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R02

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R03

120593

H(11990312057911205792)

=H

(1199031205791120593+1205792)

R03

120579H

(11990312057910)= H

(119903120579+1205791 0)

R01

120601H

(11990312057910)= H

(119903120601+1205791 0)

R02

120601

H(11990312057910)

=H

(119903120601+1205791 0)

L01119904H

(11990300)= H

(119904+11990300)

L02119904H

(11990300)= H

(119904+11990300)

L03119904

H(11990300)

=H

(119904+11990300)

(37)

Moreover L0119895119904and R0119895

120579leave fixed l0119895

119903 that is

L0119895119904l0119895119903

= l0119895119904+119903

R0119895

120579l0119895119903

= l0119895119903

(38)

Thus we are ready to give the following definitions by (33)and (38)

Definition 6 L0119895119904isH-translation by 119904 alongH-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 7 R0119895

120579is H-rotation by 120579 about H-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 8 H0119894119895

120582is horocyclic rotation by 120582 about lightlike

plane D119894119895for 119894 119895 = 1 2 3 (119894 = 119895) in H3

Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863

119894119895

Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is

R0119895

120579l01119903

=

l02119903

119895 = 3 120579 =120587

2

l03119903

119895 = 2 120579 =3120587

2

(39)

Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely

R0119895

120601l0119896119903

=

D23

(119903120601) 119895 = 1 119896 = 2

D13

(119903120601) 119895 = 2 119896 = 3

D12

(119903120601) 119895 = 3 119896 = 1

(40)

6 Journal of Applied Mathematics

Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is

R0119895

120579D12

(119903120601)=

D23

(119903120601) 119895 = 3 120579 =

3120587

2

D13

(119903120601) 119895 = 1 120579 =

120587

2

(41)

Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely

R0119897

minus120595R0119896

minus120601R0119895

minus120579H012

120582R0119895

120579R0119896

120601R0119897

120595

=

H013

120582 119895 = 1 120579 =

120587

2 120601 = 0 120595 = 0

H031

120582 119895 = 1 119896 = 2 120579 =

120587

2 120601 =

120587

2 120595 = 0

H032

120582 119895 = 1 119896 = 2 119897 = 3 120579 =

120587

2 120601 =

120587

2 120595 =

120587

2

H023

120582 119895 = 3 119896 = 2 120579 = minus

120587

2 120601 = minus

120587

2 120595 = 0

H021

120582 119895 = 3 120579 = minus

120587

2 120601 = 0 120595 = 0

(42)

After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed

Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878

1015840

for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3

Theorem 14 An H-sphere is invariant under H-translation inH3

Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3

cap HP(k minus119896) for 119896 gt 0 So

HP (k minus119896) = x isin R4

1|

minus V01199090

+ V11199091

+ V21199092

+ V31199093

= minus119896 119896 gt 0

(43)

Moreover for w = kk isin H3 and 119888 = 119896k we have

HP (w minus119888) = x isin R4

1| ⟨xw⟩ = minus119888 (44)

Since w isin H3

w = (cosh 1199040 sinh 119904

0cos120601

0 sinh 119904

0sin120601

0cos 120579

0

sinh 1199040sin120601

0sin 120579

0)

(45)

for any hyperbolic polar coordinates 1199040

isin [0 infin) 1206010

isin [0 120587]

and 1205790

isin [0 2120587] If we apply H-isometry T = L01minus1199040

R03

minus1206010R01

minus1205790isin

119866 then we have unit timelike vector e0such that

119879 (w) = e0 (46)

However unit timelike normal vector e0is invariant under

1198710119895

119904 That is

1198710119895

119904(e

0) = (cosh 119904) e

0 119895 = 1 2 3 (47)

For this reason if is an H-sphere which is generated fromspacelike hyperplane

HP (e0 minus119888) = x isin R

4

1| 119909

0= 119888 119888 ge 1 (48)

then we have

1198710119895

119904() = (49)

by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)

The following theorems also can be proved using similarmethod

Theorem 15 An equidistant surface is invariant under H-rotation in H3

Theorem 16 A horosphere is invariant under horocyclicrotation in H3

Finally we give the following corollary

Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866

0 119866

1 and 119866

2in H3

respectively

3 Differential Geometry of Curves andSurfaces in H3

In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]

TheLorentzian vector product of vectors x1 x2 x3 is givenby

x1 and x2 and x3 =

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

minuse0

e1

e2

e3

1199091

01199091

11199091

21199091

3

1199092

01199092

11199092

21199092

3

1199093

01199093

11199093

21199093

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(50)

where e0 e

1 e

2 e

3 is the canonical basis of R4

1and x119894 =

(119909119894

0 119909

119894

1 119909

119894

2 119909

119894

3) 119894 = 1 2 3 Also it is clear that

⟨x x1 and x2 and x3⟩ = det (x x1 x2 x3) (51)

for any x isin R4

1 Therefore x1 and x2 and x3 is pseudo-orthogonal

to any x119894 119894 = 1 2 3We recall the basic theory of curves inH3 Let120572 119868 rarr H3

be a unit speed regular curve for open subset 119868 sub R Since⟨120572

1015840(119904) 120572

1015840(119904)⟩ = 1 tangent vector of 120572 is given by t(119904) = 120572

1015840(119904)

The vector 12057210158401015840(119904) minus 120572(119904) is orthogonal to 120572(119904) and t(119904) We

suppose that 12057210158401015840(119904) minus 120572(119904) = 0 Then the normal vector of 120572

is given by n(119904) = (12057210158401015840(119904) minus 120572(119904))120572

10158401015840(119904) minus 120572(119904) However

the binormal vector of 120572 is given by e(119904) = 120572(119904) and t(119904) and n(119904)

Journal of Applied Mathematics 7

Hence we have a pseudo-orthonormal frame field120572(119904) t(119904)n(119904) e(119904) of R4

1along 120572 and the following Frenet-

Serret formulas

[[[

[

1205721015840(119904)

t1015840 (119904)

n1015840(119904)

e1015840 (119904)

]]]

]

=[[[

[

0 1 0 0

1 0 120581ℎ (119904) 0

0 minus120581ℎ

(119904) 0 120591ℎ

(119904)

0 0 minus120591ℎ

(119904) 0

]]]

]

[[[

[

120572 (119904)

t (119904)

n (119904)

e (119904)

]]]

]

(52)

where hyperbolic curvature and hyperbolic torsion of 120572

are given by 120581ℎ(119904) = 120572

10158401015840(119904) minus 120572(119904) and 120591

ℎ(119904) =

(minus det(120572(119904) 1205721015840(119904) 120572

10158401015840(119904) 120572

101584010158401015840(119904)))(120581

ℎ(119904))

2 under the assump-tion ⟨120572

10158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 respectively

Remark 18 The condition ⟨12057210158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 is equivalent

to 120581ℎ(119904) = 0 Moreover we see easily that 120581

ℎ(119904) = 0 if and only

if there exists a lightlike vector c such that 120572(119904) minus 119888 a geodesic(H-line)

If 120581ℎ(119904) = 1 and 120591

ℎ(119904) = 0 then 120572 in H3 is called a

horocycle We give a lemma about existence and uniquenessfor horocycles (cf [8 Proposition 43])

Lemma 19 For any a0

isin H3 and a1 a

2isin S3

1such that

⟨a119894 a

119895⟩ = 0 the unique horocycle with the initial conditions

120574(0) = a0 1205741015840(0) = a

1 and 120574

10158401015840(0) = a

0+ a

2is given by

120574 (119904) = a0

+ 119904a1

+1199042

2(a

0+ a

2) (53)

Now we recall the basic theory of surfaces in H3 Let x

119880 rarr H3 be embedding such that open subset 119880 sub R2 Wedenote that regular surface 119872 = x(119880) and identify 119872 and 119880

through the embedding x where 119894 119880 rarr 119872 is a local chartFor x(119906) = 119901 isin 119872 and 119906 = (119906

1 119906

2) isin 119880 if we define spacelike

unit normal vector

120578 (119906) =x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817

(54)

where x119906119894

= 120597x120597119906119894 119894 = 1 2 thenwehave ⟨x

119906119894 x⟩ = ⟨x

119906119894 120578⟩ =

0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3

1 Then the following maps 119864 119880 rarr 119878

3

1

119864(119906) = 120578(119906) and 119871plusmn

119880 rarr 119871119862+ 119871

plusmn(119906) = x(119906) plusmn 120578(119906)

are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906

0) can be identified with

identity mapping 119868119879119901119872

on the tangent space 119879119901119872 at x(119906

0) =

119901 isin 119872 We have that minus119889119871plusmn

= minus119868119879119901119872

plusmn (minus119889119864)For any given x(119906

0) = 119901 isin 119872 the linear transforms 119860

119901=

minus119889119864(1199060) 119879

119901119872 rarr 119879

119901119872 and 119878

plusmn

119901= minus119889119871

plusmn(119906

0) 119879

119901119872 rarr

119879119901119872 are called de Sitter shape operator and hyperbolic shape

operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and

119878plusmn

119901are denoted by 119896

119894(119901) and 119896

plusmn

119894(119901) for 119894 = 1 2 respectively

Obviously 119860119901and 119878

plusmn

119901have same eigenvectors Also the

eigenvalues satisfy

119896plusmn

119894(119901) = minus1 plusmn 119896

119894(119901) 119894 = 1 2 (55)

where 119896119894(119901) and 119896

plusmn

119894(119901) are called de Sitter principal curvature

and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872

respectivelyThe de Sitter Gauss curvature and the de Sitter mean

curvature of 119872 are given by

119870119889

(1199060) = det119860

119901= 119896

1(119901) 119896

2(119901)

119867119889

(1199060) =

1

2Tr119860

119901=

1198961

(119901) + 1198962

(119901)

2

(56)

at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss

curvature and the hyperbolic mean curvature of 119872 are givenby

119870plusmn

ℎ(119906

0) = det 119878

plusmn

119901= 119896

plusmn

1(119901) 119896

plusmn

2(119901)

119867plusmn

ℎ(119906

0) =

1

2Tr 119878

plusmn

119901=

119896plusmn

1(119901) + 119896

plusmn

2(119901)

2

(57)

at x(1199060) = 119901 respectively Evidently we have the following

relations119870

plusmn

ℎ= 1 ∓ 2119867

119889+ 119870

119889

119867plusmn

ℎ= minus1 plusmn 119867

119889

(58)

We say that a point x(1199060) = 119901 isin 119872 is an umbilical point

if 1198961(119901) = 119896

2(119901) Also 119872 is totally umbilical if all points

on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])

Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification

(1) Supposing that 1198962

= 1

(a) if 119896 = 0 and 1198962

lt 1 then M is a part of anequidistant surface

(b) if 119896 = 0 and 1198962

gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)

(2) If 1198962

= 1 then M is a part of horosphere

4 1198661-Invariant Surfaces in H3

In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces

Let 119872 = x(119880) be a regular surface via embedding x

119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860

the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4

1H3 and 119872 respectively Then the Gauss

and Weingarten explicit formulas for 119872 in H3 are given by

119863119883

119884 = 119863119883

119884 + ⟨119860 (119883) 119884⟩ 120578 + ⟨119883 119884⟩ x (59)

119860 (119883) = minus119863119883

120578 = minus119863119883

120578 (60)

for all tangent vector fields 119883 119884 isin X(119872) respectively

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Page 4: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

4 Journal of Applied Mathematics

for all 119904 isin R In other cases if 119879(11988113

) = 11988113and 119879(119881

12) = 119881

12

then general forms of H-isometries that leave fixed spacelikeplanes 119881

13and 119881

12of R4

1are given by

T13

= L02119904J1198981J1198992J1198963

T12

= L03119904J1198981J1198992J1198963

119898 119899 119896 = 0 1

(20)

such that

L02119904

=[[[

[

cosh 119904 0 sinh 119904 0

0 1 0 0

sinh 119904 0 cosh 119904 0

0 0 0 1

]]]

]

(21)

L03119904

=[[[

[

cosh 119904 0 0 sinh 119904

0 1 0 0

0 0 1 0

sinh 119904 0 0 cosh 119904

]]]

]

(22)

for all 119904 isin R respectively Thus we say that the group 1198661is

union of disjoint subgroups of 119866+

1and 119866

minus

1such that

119866+

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 0 (mod 2)

(23)

119866minus

1= L01

11989811199041L0211989821199042

L0311989831199043

J11989841J11989852J11989863

|

119898119894isin Z 119904

119895isin R 119898

4+ 119898

5+ 119898

6equiv 1 (mod 2)

(24)

that is 1198661

= 119866+

1cup 119866

minus

1

We suppose that 119879 isin 1198662 Then 119879 leaves fixed lightlike

planes D119894119895

= Spe0

+ e119894 e

119895 for 119895 = 1 2 3 of R4

1 So that

119879(D119894119895) = D

119894119895

If 119879(D12

) = D12for 119894 = 1 119895 = 2 then entries of matrix T

must be

11990500

+ 11990501

= 1 11990510

+ 11990511

= 1 11990520

+ 11990521

= 0

11990530

+ 11990531

= 0 11990502

= 11990512

= 11990532

= 0 11990522

= 1

(25)

By using (7) and (8) we obtain the following equation system

1 minus 211990500

+ 1199052

03= minus1 1 minus 2119905

10+ 119905

2

13= 1

1199052

23= 0 119905

2

33= 1

1 minus 11990500

minus 11990510

+ 11990503

11990513

= 0 minus11990530

+ 11990503

11990533

= 0

minus11990520

+ 11990513

11990523

= 0 minus11990530

+ 11990513

11990533

= 0

11990523

11990533

= 0 minus11990503

11990530

+ 11990513

11990530

+ (11990500

minus 11990510

) 11990533

= plusmn1

(26)

If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleaves fixed lightlike plane D

12of R4

1is given by

T012

= H012

120582J1198982J1198993 119898 119899 = 0 1 (27)

such that

H012

120582=

[[[[[[

[

1 +1205822

2minus

1205822

20 120582

1205822

21 minus

1205822

20 120582

0 0 1 0

120582 minus120582 0 1

]]]]]]

]

(28)

for all 120582 isin R In other cases we apply similar method Henceif H-isometry that leaves fixed lightlike plane D

119894119895of R4

1is

denoted byH0119894119895

120582 then we have the following H-isometries

H013

120582=

[[[[[[

[

1 +1205822

2minus

1205822

2120582 0

1205822

21 minus

1205822

2120582 0

120582 minus120582 1 0

0 0 0 1

]]]]]]

]

H021

120582=

[[[[[[

[

1 +1205822

20 minus

1205822

2120582

0 1 0 0

1205822

20 1 minus

1205822

2120582

120582 0 minus120582 1

]]]]]]

]

H023

120582=

[[[[[[

[

1 +1205822

2120582 minus

1205822

20

120582 1 minus120582 0

1205822

2120582 1 minus

1205822

20

0 0 0 1

]]]]]]

]

H031

120582=

[[[[[[

[

1 +1205822

20 120582 minus

1205822

20 1 0 0

120582 0 1 minus120582

1205822

20 120582 1 minus

1205822

2

]]]]]]

]

H032

120582=

[[[[[[

[

1 +1205822

2120582 0 minus

1205822

2120582 1 0 minus120582

0 0 1 0

1205822

2120582 0 1 minus

1205822

2

]]]]]]

]

(29)

and we obtained the following general forms

T013

= H013

120582J1198982J1198993

T021

= H021

120582J1198981J1198993

T023

= H023

120582J1198981J1198993

T031

= H031

120582J1198981J1198992

T032

= H032

120582J1198981J1198992

119898 119899 = 0 1

(30)

Journal of Applied Mathematics 5

So we say that the group 1198662is union of disjoint subgroups of

119866+

2and 119866

minus

2such that

119866+

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 0 (mod 2)

119866minus

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 1 (mod 2)

(31)

that is 1198662

= 119866+

2cup 119866

minus

2

Hence it is clear that

119866 = 1198660

cup 1198661

cup 1198662

1198660

cap 1198661

cap 1198662

= I4 J1 J2 J3 J1J2 J1J3 J2J3 J1J2J3

(32)

However we see that easily from matrix multiplication

R0119895

1205791R0119895

1205792= R0119895

1205791+1205792

L01198951199041L01198951199042

= L01198951199041+1199042

H0119894119895

1205821H0119894119895

1205822= H0119894119895

1205821+1205822

(33)

for any 1199041 119904

2 120579

1 120579

2 120582

1 120582

2isin R

Let parametrization of H-coordinate axes 11989701 119897

02 and 119897

03

be

11989701

[0 infin) 997888rarr 1198673 119897

01 (119903) = (cosh 119903 sinh 119903 0 0)

11989702

[0 infin) 997888rarr 1198673 119897

02(119903) = (cosh 119903 0 sinh 119903 0)

11989703

[0 infin) 997888rarr 1198673 119897

03(119903) = (cosh 119903 0 0 sinh 119903)

(34)

and let their matrix forms be

l01119903

=[[[

[

cosh 119903

sinh 119903

0

0

]]]

]

l02119903

=[[[

[

cosh 119903

0

sinh 119903

0

]]]

]

l03119903

=[[[

[

cosh 119903

0

0

sinh 119903

]]]

]

(35)

respectively After applying suitable H-isometry to H-coordinate axis l0119895

119903for 119895 = 1 2 3 we obtain the following

different parametrizations

H(11990312057911205792)

= R01

1205792R03

1205791l01119903

H(11990312057911205792)

= R02

1205792R01

1205791l02119903

H(11990312057911205792)

= R03

1205792R02

1205791l03119903

(36)

for hyperbolic polar coordinates 119903 gt 0 1205791

isin [0 120587] and 1205792

isin

[0 2120587] Thus we see roles of H-isometries R0119895

120579119894and L0119896

119904in H3

from equations

R01

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R02

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R03

120593

H(11990312057911205792)

=H

(1199031205791120593+1205792)

R03

120579H

(11990312057910)= H

(119903120579+1205791 0)

R01

120601H

(11990312057910)= H

(119903120601+1205791 0)

R02

120601

H(11990312057910)

=H

(119903120601+1205791 0)

L01119904H

(11990300)= H

(119904+11990300)

L02119904H

(11990300)= H

(119904+11990300)

L03119904

H(11990300)

=H

(119904+11990300)

(37)

Moreover L0119895119904and R0119895

120579leave fixed l0119895

119903 that is

L0119895119904l0119895119903

= l0119895119904+119903

R0119895

120579l0119895119903

= l0119895119903

(38)

Thus we are ready to give the following definitions by (33)and (38)

Definition 6 L0119895119904isH-translation by 119904 alongH-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 7 R0119895

120579is H-rotation by 120579 about H-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 8 H0119894119895

120582is horocyclic rotation by 120582 about lightlike

plane D119894119895for 119894 119895 = 1 2 3 (119894 = 119895) in H3

Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863

119894119895

Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is

R0119895

120579l01119903

=

l02119903

119895 = 3 120579 =120587

2

l03119903

119895 = 2 120579 =3120587

2

(39)

Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely

R0119895

120601l0119896119903

=

D23

(119903120601) 119895 = 1 119896 = 2

D13

(119903120601) 119895 = 2 119896 = 3

D12

(119903120601) 119895 = 3 119896 = 1

(40)

6 Journal of Applied Mathematics

Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is

R0119895

120579D12

(119903120601)=

D23

(119903120601) 119895 = 3 120579 =

3120587

2

D13

(119903120601) 119895 = 1 120579 =

120587

2

(41)

Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely

R0119897

minus120595R0119896

minus120601R0119895

minus120579H012

120582R0119895

120579R0119896

120601R0119897

120595

=

H013

120582 119895 = 1 120579 =

120587

2 120601 = 0 120595 = 0

H031

120582 119895 = 1 119896 = 2 120579 =

120587

2 120601 =

120587

2 120595 = 0

H032

120582 119895 = 1 119896 = 2 119897 = 3 120579 =

120587

2 120601 =

120587

2 120595 =

120587

2

H023

120582 119895 = 3 119896 = 2 120579 = minus

120587

2 120601 = minus

120587

2 120595 = 0

H021

120582 119895 = 3 120579 = minus

120587

2 120601 = 0 120595 = 0

(42)

After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed

Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878

1015840

for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3

Theorem 14 An H-sphere is invariant under H-translation inH3

Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3

cap HP(k minus119896) for 119896 gt 0 So

HP (k minus119896) = x isin R4

1|

minus V01199090

+ V11199091

+ V21199092

+ V31199093

= minus119896 119896 gt 0

(43)

Moreover for w = kk isin H3 and 119888 = 119896k we have

HP (w minus119888) = x isin R4

1| ⟨xw⟩ = minus119888 (44)

Since w isin H3

w = (cosh 1199040 sinh 119904

0cos120601

0 sinh 119904

0sin120601

0cos 120579

0

sinh 1199040sin120601

0sin 120579

0)

(45)

for any hyperbolic polar coordinates 1199040

isin [0 infin) 1206010

isin [0 120587]

and 1205790

isin [0 2120587] If we apply H-isometry T = L01minus1199040

R03

minus1206010R01

minus1205790isin

119866 then we have unit timelike vector e0such that

119879 (w) = e0 (46)

However unit timelike normal vector e0is invariant under

1198710119895

119904 That is

1198710119895

119904(e

0) = (cosh 119904) e

0 119895 = 1 2 3 (47)

For this reason if is an H-sphere which is generated fromspacelike hyperplane

HP (e0 minus119888) = x isin R

4

1| 119909

0= 119888 119888 ge 1 (48)

then we have

1198710119895

119904() = (49)

by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)

The following theorems also can be proved using similarmethod

Theorem 15 An equidistant surface is invariant under H-rotation in H3

Theorem 16 A horosphere is invariant under horocyclicrotation in H3

Finally we give the following corollary

Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866

0 119866

1 and 119866

2in H3

respectively

3 Differential Geometry of Curves andSurfaces in H3

In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]

TheLorentzian vector product of vectors x1 x2 x3 is givenby

x1 and x2 and x3 =

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

minuse0

e1

e2

e3

1199091

01199091

11199091

21199091

3

1199092

01199092

11199092

21199092

3

1199093

01199093

11199093

21199093

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(50)

where e0 e

1 e

2 e

3 is the canonical basis of R4

1and x119894 =

(119909119894

0 119909

119894

1 119909

119894

2 119909

119894

3) 119894 = 1 2 3 Also it is clear that

⟨x x1 and x2 and x3⟩ = det (x x1 x2 x3) (51)

for any x isin R4

1 Therefore x1 and x2 and x3 is pseudo-orthogonal

to any x119894 119894 = 1 2 3We recall the basic theory of curves inH3 Let120572 119868 rarr H3

be a unit speed regular curve for open subset 119868 sub R Since⟨120572

1015840(119904) 120572

1015840(119904)⟩ = 1 tangent vector of 120572 is given by t(119904) = 120572

1015840(119904)

The vector 12057210158401015840(119904) minus 120572(119904) is orthogonal to 120572(119904) and t(119904) We

suppose that 12057210158401015840(119904) minus 120572(119904) = 0 Then the normal vector of 120572

is given by n(119904) = (12057210158401015840(119904) minus 120572(119904))120572

10158401015840(119904) minus 120572(119904) However

the binormal vector of 120572 is given by e(119904) = 120572(119904) and t(119904) and n(119904)

Journal of Applied Mathematics 7

Hence we have a pseudo-orthonormal frame field120572(119904) t(119904)n(119904) e(119904) of R4

1along 120572 and the following Frenet-

Serret formulas

[[[

[

1205721015840(119904)

t1015840 (119904)

n1015840(119904)

e1015840 (119904)

]]]

]

=[[[

[

0 1 0 0

1 0 120581ℎ (119904) 0

0 minus120581ℎ

(119904) 0 120591ℎ

(119904)

0 0 minus120591ℎ

(119904) 0

]]]

]

[[[

[

120572 (119904)

t (119904)

n (119904)

e (119904)

]]]

]

(52)

where hyperbolic curvature and hyperbolic torsion of 120572

are given by 120581ℎ(119904) = 120572

10158401015840(119904) minus 120572(119904) and 120591

ℎ(119904) =

(minus det(120572(119904) 1205721015840(119904) 120572

10158401015840(119904) 120572

101584010158401015840(119904)))(120581

ℎ(119904))

2 under the assump-tion ⟨120572

10158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 respectively

Remark 18 The condition ⟨12057210158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 is equivalent

to 120581ℎ(119904) = 0 Moreover we see easily that 120581

ℎ(119904) = 0 if and only

if there exists a lightlike vector c such that 120572(119904) minus 119888 a geodesic(H-line)

If 120581ℎ(119904) = 1 and 120591

ℎ(119904) = 0 then 120572 in H3 is called a

horocycle We give a lemma about existence and uniquenessfor horocycles (cf [8 Proposition 43])

Lemma 19 For any a0

isin H3 and a1 a

2isin S3

1such that

⟨a119894 a

119895⟩ = 0 the unique horocycle with the initial conditions

120574(0) = a0 1205741015840(0) = a

1 and 120574

10158401015840(0) = a

0+ a

2is given by

120574 (119904) = a0

+ 119904a1

+1199042

2(a

0+ a

2) (53)

Now we recall the basic theory of surfaces in H3 Let x

119880 rarr H3 be embedding such that open subset 119880 sub R2 Wedenote that regular surface 119872 = x(119880) and identify 119872 and 119880

through the embedding x where 119894 119880 rarr 119872 is a local chartFor x(119906) = 119901 isin 119872 and 119906 = (119906

1 119906

2) isin 119880 if we define spacelike

unit normal vector

120578 (119906) =x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817

(54)

where x119906119894

= 120597x120597119906119894 119894 = 1 2 thenwehave ⟨x

119906119894 x⟩ = ⟨x

119906119894 120578⟩ =

0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3

1 Then the following maps 119864 119880 rarr 119878

3

1

119864(119906) = 120578(119906) and 119871plusmn

119880 rarr 119871119862+ 119871

plusmn(119906) = x(119906) plusmn 120578(119906)

are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906

0) can be identified with

identity mapping 119868119879119901119872

on the tangent space 119879119901119872 at x(119906

0) =

119901 isin 119872 We have that minus119889119871plusmn

= minus119868119879119901119872

plusmn (minus119889119864)For any given x(119906

0) = 119901 isin 119872 the linear transforms 119860

119901=

minus119889119864(1199060) 119879

119901119872 rarr 119879

119901119872 and 119878

plusmn

119901= minus119889119871

plusmn(119906

0) 119879

119901119872 rarr

119879119901119872 are called de Sitter shape operator and hyperbolic shape

operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and

119878plusmn

119901are denoted by 119896

119894(119901) and 119896

plusmn

119894(119901) for 119894 = 1 2 respectively

Obviously 119860119901and 119878

plusmn

119901have same eigenvectors Also the

eigenvalues satisfy

119896plusmn

119894(119901) = minus1 plusmn 119896

119894(119901) 119894 = 1 2 (55)

where 119896119894(119901) and 119896

plusmn

119894(119901) are called de Sitter principal curvature

and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872

respectivelyThe de Sitter Gauss curvature and the de Sitter mean

curvature of 119872 are given by

119870119889

(1199060) = det119860

119901= 119896

1(119901) 119896

2(119901)

119867119889

(1199060) =

1

2Tr119860

119901=

1198961

(119901) + 1198962

(119901)

2

(56)

at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss

curvature and the hyperbolic mean curvature of 119872 are givenby

119870plusmn

ℎ(119906

0) = det 119878

plusmn

119901= 119896

plusmn

1(119901) 119896

plusmn

2(119901)

119867plusmn

ℎ(119906

0) =

1

2Tr 119878

plusmn

119901=

119896plusmn

1(119901) + 119896

plusmn

2(119901)

2

(57)

at x(1199060) = 119901 respectively Evidently we have the following

relations119870

plusmn

ℎ= 1 ∓ 2119867

119889+ 119870

119889

119867plusmn

ℎ= minus1 plusmn 119867

119889

(58)

We say that a point x(1199060) = 119901 isin 119872 is an umbilical point

if 1198961(119901) = 119896

2(119901) Also 119872 is totally umbilical if all points

on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])

Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification

(1) Supposing that 1198962

= 1

(a) if 119896 = 0 and 1198962

lt 1 then M is a part of anequidistant surface

(b) if 119896 = 0 and 1198962

gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)

(2) If 1198962

= 1 then M is a part of horosphere

4 1198661-Invariant Surfaces in H3

In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces

Let 119872 = x(119880) be a regular surface via embedding x

119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860

the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4

1H3 and 119872 respectively Then the Gauss

and Weingarten explicit formulas for 119872 in H3 are given by

119863119883

119884 = 119863119883

119884 + ⟨119860 (119883) 119884⟩ 120578 + ⟨119883 119884⟩ x (59)

119860 (119883) = minus119863119883

120578 = minus119863119883

120578 (60)

for all tangent vector fields 119883 119884 isin X(119872) respectively

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Page 5: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

Journal of Applied Mathematics 5

So we say that the group 1198662is union of disjoint subgroups of

119866+

2and 119866

minus

2such that

119866+

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 0 (mod 2)

119866minus

2= H012

11989811205821H013

11989821205822H021

11989831205823H023

11989841205824H031

11989851205825H032

11989861205826J11989871J11989882J11989893

|

119898119894isin Z 120582

119895isin R 119898

7+ 119898

8+ 119898

9equiv 1 (mod 2)

(31)

that is 1198662

= 119866+

2cup 119866

minus

2

Hence it is clear that

119866 = 1198660

cup 1198661

cup 1198662

1198660

cap 1198661

cap 1198662

= I4 J1 J2 J3 J1J2 J1J3 J2J3 J1J2J3

(32)

However we see that easily from matrix multiplication

R0119895

1205791R0119895

1205792= R0119895

1205791+1205792

L01198951199041L01198951199042

= L01198951199041+1199042

H0119894119895

1205821H0119894119895

1205822= H0119894119895

1205821+1205822

(33)

for any 1199041 119904

2 120579

1 120579

2 120582

1 120582

2isin R

Let parametrization of H-coordinate axes 11989701 119897

02 and 119897

03

be

11989701

[0 infin) 997888rarr 1198673 119897

01 (119903) = (cosh 119903 sinh 119903 0 0)

11989702

[0 infin) 997888rarr 1198673 119897

02(119903) = (cosh 119903 0 sinh 119903 0)

11989703

[0 infin) 997888rarr 1198673 119897

03(119903) = (cosh 119903 0 0 sinh 119903)

(34)

and let their matrix forms be

l01119903

=[[[

[

cosh 119903

sinh 119903

0

0

]]]

]

l02119903

=[[[

[

cosh 119903

0

sinh 119903

0

]]]

]

l03119903

=[[[

[

cosh 119903

0

0

sinh 119903

]]]

]

(35)

respectively After applying suitable H-isometry to H-coordinate axis l0119895

119903for 119895 = 1 2 3 we obtain the following

different parametrizations

H(11990312057911205792)

= R01

1205792R03

1205791l01119903

H(11990312057911205792)

= R02

1205792R01

1205791l02119903

H(11990312057911205792)

= R03

1205792R02

1205791l03119903

(36)

for hyperbolic polar coordinates 119903 gt 0 1205791

isin [0 120587] and 1205792

isin

[0 2120587] Thus we see roles of H-isometries R0119895

120579119894and L0119896

119904in H3

from equations

R01

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R02

120593H

(11990312057911205792)= H

(1199031205791120593+1205792)

R03

120593

H(11990312057911205792)

=H

(1199031205791120593+1205792)

R03

120579H

(11990312057910)= H

(119903120579+1205791 0)

R01

120601H

(11990312057910)= H

(119903120601+1205791 0)

R02

120601

H(11990312057910)

=H

(119903120601+1205791 0)

L01119904H

(11990300)= H

(119904+11990300)

L02119904H

(11990300)= H

(119904+11990300)

L03119904

H(11990300)

=H

(119904+11990300)

(37)

Moreover L0119895119904and R0119895

120579leave fixed l0119895

119903 that is

L0119895119904l0119895119903

= l0119895119904+119903

R0119895

120579l0119895119903

= l0119895119903

(38)

Thus we are ready to give the following definitions by (33)and (38)

Definition 6 L0119895119904isH-translation by 119904 alongH-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 7 R0119895

120579is H-rotation by 120579 about H-coordinate axis

1198970119895for 119895 = 1 2 3 in H3

Definition 8 H0119894119895

120582is horocyclic rotation by 120582 about lightlike

plane D119894119895for 119894 119895 = 1 2 3 (119894 = 119895) in H3

Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863

119894119895

Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is

R0119895

120579l01119903

=

l02119903

119895 = 3 120579 =120587

2

l03119903

119895 = 2 120579 =3120587

2

(39)

Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely

R0119895

120601l0119896119903

=

D23

(119903120601) 119895 = 1 119896 = 2

D13

(119903120601) 119895 = 2 119896 = 3

D12

(119903120601) 119895 = 3 119896 = 1

(40)

6 Journal of Applied Mathematics

Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is

R0119895

120579D12

(119903120601)=

D23

(119903120601) 119895 = 3 120579 =

3120587

2

D13

(119903120601) 119895 = 1 120579 =

120587

2

(41)

Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely

R0119897

minus120595R0119896

minus120601R0119895

minus120579H012

120582R0119895

120579R0119896

120601R0119897

120595

=

H013

120582 119895 = 1 120579 =

120587

2 120601 = 0 120595 = 0

H031

120582 119895 = 1 119896 = 2 120579 =

120587

2 120601 =

120587

2 120595 = 0

H032

120582 119895 = 1 119896 = 2 119897 = 3 120579 =

120587

2 120601 =

120587

2 120595 =

120587

2

H023

120582 119895 = 3 119896 = 2 120579 = minus

120587

2 120601 = minus

120587

2 120595 = 0

H021

120582 119895 = 3 120579 = minus

120587

2 120601 = 0 120595 = 0

(42)

After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed

Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878

1015840

for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3

Theorem 14 An H-sphere is invariant under H-translation inH3

Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3

cap HP(k minus119896) for 119896 gt 0 So

HP (k minus119896) = x isin R4

1|

minus V01199090

+ V11199091

+ V21199092

+ V31199093

= minus119896 119896 gt 0

(43)

Moreover for w = kk isin H3 and 119888 = 119896k we have

HP (w minus119888) = x isin R4

1| ⟨xw⟩ = minus119888 (44)

Since w isin H3

w = (cosh 1199040 sinh 119904

0cos120601

0 sinh 119904

0sin120601

0cos 120579

0

sinh 1199040sin120601

0sin 120579

0)

(45)

for any hyperbolic polar coordinates 1199040

isin [0 infin) 1206010

isin [0 120587]

and 1205790

isin [0 2120587] If we apply H-isometry T = L01minus1199040

R03

minus1206010R01

minus1205790isin

119866 then we have unit timelike vector e0such that

119879 (w) = e0 (46)

However unit timelike normal vector e0is invariant under

1198710119895

119904 That is

1198710119895

119904(e

0) = (cosh 119904) e

0 119895 = 1 2 3 (47)

For this reason if is an H-sphere which is generated fromspacelike hyperplane

HP (e0 minus119888) = x isin R

4

1| 119909

0= 119888 119888 ge 1 (48)

then we have

1198710119895

119904() = (49)

by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)

The following theorems also can be proved using similarmethod

Theorem 15 An equidistant surface is invariant under H-rotation in H3

Theorem 16 A horosphere is invariant under horocyclicrotation in H3

Finally we give the following corollary

Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866

0 119866

1 and 119866

2in H3

respectively

3 Differential Geometry of Curves andSurfaces in H3

In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]

TheLorentzian vector product of vectors x1 x2 x3 is givenby

x1 and x2 and x3 =

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

minuse0

e1

e2

e3

1199091

01199091

11199091

21199091

3

1199092

01199092

11199092

21199092

3

1199093

01199093

11199093

21199093

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(50)

where e0 e

1 e

2 e

3 is the canonical basis of R4

1and x119894 =

(119909119894

0 119909

119894

1 119909

119894

2 119909

119894

3) 119894 = 1 2 3 Also it is clear that

⟨x x1 and x2 and x3⟩ = det (x x1 x2 x3) (51)

for any x isin R4

1 Therefore x1 and x2 and x3 is pseudo-orthogonal

to any x119894 119894 = 1 2 3We recall the basic theory of curves inH3 Let120572 119868 rarr H3

be a unit speed regular curve for open subset 119868 sub R Since⟨120572

1015840(119904) 120572

1015840(119904)⟩ = 1 tangent vector of 120572 is given by t(119904) = 120572

1015840(119904)

The vector 12057210158401015840(119904) minus 120572(119904) is orthogonal to 120572(119904) and t(119904) We

suppose that 12057210158401015840(119904) minus 120572(119904) = 0 Then the normal vector of 120572

is given by n(119904) = (12057210158401015840(119904) minus 120572(119904))120572

10158401015840(119904) minus 120572(119904) However

the binormal vector of 120572 is given by e(119904) = 120572(119904) and t(119904) and n(119904)

Journal of Applied Mathematics 7

Hence we have a pseudo-orthonormal frame field120572(119904) t(119904)n(119904) e(119904) of R4

1along 120572 and the following Frenet-

Serret formulas

[[[

[

1205721015840(119904)

t1015840 (119904)

n1015840(119904)

e1015840 (119904)

]]]

]

=[[[

[

0 1 0 0

1 0 120581ℎ (119904) 0

0 minus120581ℎ

(119904) 0 120591ℎ

(119904)

0 0 minus120591ℎ

(119904) 0

]]]

]

[[[

[

120572 (119904)

t (119904)

n (119904)

e (119904)

]]]

]

(52)

where hyperbolic curvature and hyperbolic torsion of 120572

are given by 120581ℎ(119904) = 120572

10158401015840(119904) minus 120572(119904) and 120591

ℎ(119904) =

(minus det(120572(119904) 1205721015840(119904) 120572

10158401015840(119904) 120572

101584010158401015840(119904)))(120581

ℎ(119904))

2 under the assump-tion ⟨120572

10158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 respectively

Remark 18 The condition ⟨12057210158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 is equivalent

to 120581ℎ(119904) = 0 Moreover we see easily that 120581

ℎ(119904) = 0 if and only

if there exists a lightlike vector c such that 120572(119904) minus 119888 a geodesic(H-line)

If 120581ℎ(119904) = 1 and 120591

ℎ(119904) = 0 then 120572 in H3 is called a

horocycle We give a lemma about existence and uniquenessfor horocycles (cf [8 Proposition 43])

Lemma 19 For any a0

isin H3 and a1 a

2isin S3

1such that

⟨a119894 a

119895⟩ = 0 the unique horocycle with the initial conditions

120574(0) = a0 1205741015840(0) = a

1 and 120574

10158401015840(0) = a

0+ a

2is given by

120574 (119904) = a0

+ 119904a1

+1199042

2(a

0+ a

2) (53)

Now we recall the basic theory of surfaces in H3 Let x

119880 rarr H3 be embedding such that open subset 119880 sub R2 Wedenote that regular surface 119872 = x(119880) and identify 119872 and 119880

through the embedding x where 119894 119880 rarr 119872 is a local chartFor x(119906) = 119901 isin 119872 and 119906 = (119906

1 119906

2) isin 119880 if we define spacelike

unit normal vector

120578 (119906) =x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817

(54)

where x119906119894

= 120597x120597119906119894 119894 = 1 2 thenwehave ⟨x

119906119894 x⟩ = ⟨x

119906119894 120578⟩ =

0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3

1 Then the following maps 119864 119880 rarr 119878

3

1

119864(119906) = 120578(119906) and 119871plusmn

119880 rarr 119871119862+ 119871

plusmn(119906) = x(119906) plusmn 120578(119906)

are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906

0) can be identified with

identity mapping 119868119879119901119872

on the tangent space 119879119901119872 at x(119906

0) =

119901 isin 119872 We have that minus119889119871plusmn

= minus119868119879119901119872

plusmn (minus119889119864)For any given x(119906

0) = 119901 isin 119872 the linear transforms 119860

119901=

minus119889119864(1199060) 119879

119901119872 rarr 119879

119901119872 and 119878

plusmn

119901= minus119889119871

plusmn(119906

0) 119879

119901119872 rarr

119879119901119872 are called de Sitter shape operator and hyperbolic shape

operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and

119878plusmn

119901are denoted by 119896

119894(119901) and 119896

plusmn

119894(119901) for 119894 = 1 2 respectively

Obviously 119860119901and 119878

plusmn

119901have same eigenvectors Also the

eigenvalues satisfy

119896plusmn

119894(119901) = minus1 plusmn 119896

119894(119901) 119894 = 1 2 (55)

where 119896119894(119901) and 119896

plusmn

119894(119901) are called de Sitter principal curvature

and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872

respectivelyThe de Sitter Gauss curvature and the de Sitter mean

curvature of 119872 are given by

119870119889

(1199060) = det119860

119901= 119896

1(119901) 119896

2(119901)

119867119889

(1199060) =

1

2Tr119860

119901=

1198961

(119901) + 1198962

(119901)

2

(56)

at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss

curvature and the hyperbolic mean curvature of 119872 are givenby

119870plusmn

ℎ(119906

0) = det 119878

plusmn

119901= 119896

plusmn

1(119901) 119896

plusmn

2(119901)

119867plusmn

ℎ(119906

0) =

1

2Tr 119878

plusmn

119901=

119896plusmn

1(119901) + 119896

plusmn

2(119901)

2

(57)

at x(1199060) = 119901 respectively Evidently we have the following

relations119870

plusmn

ℎ= 1 ∓ 2119867

119889+ 119870

119889

119867plusmn

ℎ= minus1 plusmn 119867

119889

(58)

We say that a point x(1199060) = 119901 isin 119872 is an umbilical point

if 1198961(119901) = 119896

2(119901) Also 119872 is totally umbilical if all points

on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])

Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification

(1) Supposing that 1198962

= 1

(a) if 119896 = 0 and 1198962

lt 1 then M is a part of anequidistant surface

(b) if 119896 = 0 and 1198962

gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)

(2) If 1198962

= 1 then M is a part of horosphere

4 1198661-Invariant Surfaces in H3

In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces

Let 119872 = x(119880) be a regular surface via embedding x

119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860

the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4

1H3 and 119872 respectively Then the Gauss

and Weingarten explicit formulas for 119872 in H3 are given by

119863119883

119884 = 119863119883

119884 + ⟨119860 (119883) 119884⟩ 120578 + ⟨119883 119884⟩ x (59)

119860 (119883) = minus119863119883

120578 = minus119863119883

120578 (60)

for all tangent vector fields 119883 119884 isin X(119872) respectively

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Stochastic AnalysisInternational Journal of

Page 6: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

6 Journal of Applied Mathematics

Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is

R0119895

120579D12

(119903120601)=

D23

(119903120601) 119895 = 3 120579 =

3120587

2

D13

(119903120601) 119895 = 1 120579 =

120587

2

(41)

Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely

R0119897

minus120595R0119896

minus120601R0119895

minus120579H012

120582R0119895

120579R0119896

120601R0119897

120595

=

H013

120582 119895 = 1 120579 =

120587

2 120601 = 0 120595 = 0

H031

120582 119895 = 1 119896 = 2 120579 =

120587

2 120601 =

120587

2 120595 = 0

H032

120582 119895 = 1 119896 = 2 119897 = 3 120579 =

120587

2 120601 =

120587

2 120595 =

120587

2

H023

120582 119895 = 3 119896 = 2 120579 = minus

120587

2 120601 = minus

120587

2 120595 = 0

H021

120582 119895 = 3 120579 = minus

120587

2 120601 = 0 120595 = 0

(42)

After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed

Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878

1015840

for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3

Theorem 14 An H-sphere is invariant under H-translation inH3

Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3

cap HP(k minus119896) for 119896 gt 0 So

HP (k minus119896) = x isin R4

1|

minus V01199090

+ V11199091

+ V21199092

+ V31199093

= minus119896 119896 gt 0

(43)

Moreover for w = kk isin H3 and 119888 = 119896k we have

HP (w minus119888) = x isin R4

1| ⟨xw⟩ = minus119888 (44)

Since w isin H3

w = (cosh 1199040 sinh 119904

0cos120601

0 sinh 119904

0sin120601

0cos 120579

0

sinh 1199040sin120601

0sin 120579

0)

(45)

for any hyperbolic polar coordinates 1199040

isin [0 infin) 1206010

isin [0 120587]

and 1205790

isin [0 2120587] If we apply H-isometry T = L01minus1199040

R03

minus1206010R01

minus1205790isin

119866 then we have unit timelike vector e0such that

119879 (w) = e0 (46)

However unit timelike normal vector e0is invariant under

1198710119895

119904 That is

1198710119895

119904(e

0) = (cosh 119904) e

0 119895 = 1 2 3 (47)

For this reason if is an H-sphere which is generated fromspacelike hyperplane

HP (e0 minus119888) = x isin R

4

1| 119909

0= 119888 119888 ge 1 (48)

then we have

1198710119895

119904() = (49)

by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)

The following theorems also can be proved using similarmethod

Theorem 15 An equidistant surface is invariant under H-rotation in H3

Theorem 16 A horosphere is invariant under horocyclicrotation in H3

Finally we give the following corollary

Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866

0 119866

1 and 119866

2in H3

respectively

3 Differential Geometry of Curves andSurfaces in H3

In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]

TheLorentzian vector product of vectors x1 x2 x3 is givenby

x1 and x2 and x3 =

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

minuse0

e1

e2

e3

1199091

01199091

11199091

21199091

3

1199092

01199092

11199092

21199092

3

1199093

01199093

11199093

21199093

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(50)

where e0 e

1 e

2 e

3 is the canonical basis of R4

1and x119894 =

(119909119894

0 119909

119894

1 119909

119894

2 119909

119894

3) 119894 = 1 2 3 Also it is clear that

⟨x x1 and x2 and x3⟩ = det (x x1 x2 x3) (51)

for any x isin R4

1 Therefore x1 and x2 and x3 is pseudo-orthogonal

to any x119894 119894 = 1 2 3We recall the basic theory of curves inH3 Let120572 119868 rarr H3

be a unit speed regular curve for open subset 119868 sub R Since⟨120572

1015840(119904) 120572

1015840(119904)⟩ = 1 tangent vector of 120572 is given by t(119904) = 120572

1015840(119904)

The vector 12057210158401015840(119904) minus 120572(119904) is orthogonal to 120572(119904) and t(119904) We

suppose that 12057210158401015840(119904) minus 120572(119904) = 0 Then the normal vector of 120572

is given by n(119904) = (12057210158401015840(119904) minus 120572(119904))120572

10158401015840(119904) minus 120572(119904) However

the binormal vector of 120572 is given by e(119904) = 120572(119904) and t(119904) and n(119904)

Journal of Applied Mathematics 7

Hence we have a pseudo-orthonormal frame field120572(119904) t(119904)n(119904) e(119904) of R4

1along 120572 and the following Frenet-

Serret formulas

[[[

[

1205721015840(119904)

t1015840 (119904)

n1015840(119904)

e1015840 (119904)

]]]

]

=[[[

[

0 1 0 0

1 0 120581ℎ (119904) 0

0 minus120581ℎ

(119904) 0 120591ℎ

(119904)

0 0 minus120591ℎ

(119904) 0

]]]

]

[[[

[

120572 (119904)

t (119904)

n (119904)

e (119904)

]]]

]

(52)

where hyperbolic curvature and hyperbolic torsion of 120572

are given by 120581ℎ(119904) = 120572

10158401015840(119904) minus 120572(119904) and 120591

ℎ(119904) =

(minus det(120572(119904) 1205721015840(119904) 120572

10158401015840(119904) 120572

101584010158401015840(119904)))(120581

ℎ(119904))

2 under the assump-tion ⟨120572

10158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 respectively

Remark 18 The condition ⟨12057210158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 is equivalent

to 120581ℎ(119904) = 0 Moreover we see easily that 120581

ℎ(119904) = 0 if and only

if there exists a lightlike vector c such that 120572(119904) minus 119888 a geodesic(H-line)

If 120581ℎ(119904) = 1 and 120591

ℎ(119904) = 0 then 120572 in H3 is called a

horocycle We give a lemma about existence and uniquenessfor horocycles (cf [8 Proposition 43])

Lemma 19 For any a0

isin H3 and a1 a

2isin S3

1such that

⟨a119894 a

119895⟩ = 0 the unique horocycle with the initial conditions

120574(0) = a0 1205741015840(0) = a

1 and 120574

10158401015840(0) = a

0+ a

2is given by

120574 (119904) = a0

+ 119904a1

+1199042

2(a

0+ a

2) (53)

Now we recall the basic theory of surfaces in H3 Let x

119880 rarr H3 be embedding such that open subset 119880 sub R2 Wedenote that regular surface 119872 = x(119880) and identify 119872 and 119880

through the embedding x where 119894 119880 rarr 119872 is a local chartFor x(119906) = 119901 isin 119872 and 119906 = (119906

1 119906

2) isin 119880 if we define spacelike

unit normal vector

120578 (119906) =x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817

(54)

where x119906119894

= 120597x120597119906119894 119894 = 1 2 thenwehave ⟨x

119906119894 x⟩ = ⟨x

119906119894 120578⟩ =

0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3

1 Then the following maps 119864 119880 rarr 119878

3

1

119864(119906) = 120578(119906) and 119871plusmn

119880 rarr 119871119862+ 119871

plusmn(119906) = x(119906) plusmn 120578(119906)

are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906

0) can be identified with

identity mapping 119868119879119901119872

on the tangent space 119879119901119872 at x(119906

0) =

119901 isin 119872 We have that minus119889119871plusmn

= minus119868119879119901119872

plusmn (minus119889119864)For any given x(119906

0) = 119901 isin 119872 the linear transforms 119860

119901=

minus119889119864(1199060) 119879

119901119872 rarr 119879

119901119872 and 119878

plusmn

119901= minus119889119871

plusmn(119906

0) 119879

119901119872 rarr

119879119901119872 are called de Sitter shape operator and hyperbolic shape

operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and

119878plusmn

119901are denoted by 119896

119894(119901) and 119896

plusmn

119894(119901) for 119894 = 1 2 respectively

Obviously 119860119901and 119878

plusmn

119901have same eigenvectors Also the

eigenvalues satisfy

119896plusmn

119894(119901) = minus1 plusmn 119896

119894(119901) 119894 = 1 2 (55)

where 119896119894(119901) and 119896

plusmn

119894(119901) are called de Sitter principal curvature

and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872

respectivelyThe de Sitter Gauss curvature and the de Sitter mean

curvature of 119872 are given by

119870119889

(1199060) = det119860

119901= 119896

1(119901) 119896

2(119901)

119867119889

(1199060) =

1

2Tr119860

119901=

1198961

(119901) + 1198962

(119901)

2

(56)

at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss

curvature and the hyperbolic mean curvature of 119872 are givenby

119870plusmn

ℎ(119906

0) = det 119878

plusmn

119901= 119896

plusmn

1(119901) 119896

plusmn

2(119901)

119867plusmn

ℎ(119906

0) =

1

2Tr 119878

plusmn

119901=

119896plusmn

1(119901) + 119896

plusmn

2(119901)

2

(57)

at x(1199060) = 119901 respectively Evidently we have the following

relations119870

plusmn

ℎ= 1 ∓ 2119867

119889+ 119870

119889

119867plusmn

ℎ= minus1 plusmn 119867

119889

(58)

We say that a point x(1199060) = 119901 isin 119872 is an umbilical point

if 1198961(119901) = 119896

2(119901) Also 119872 is totally umbilical if all points

on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])

Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification

(1) Supposing that 1198962

= 1

(a) if 119896 = 0 and 1198962

lt 1 then M is a part of anequidistant surface

(b) if 119896 = 0 and 1198962

gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)

(2) If 1198962

= 1 then M is a part of horosphere

4 1198661-Invariant Surfaces in H3

In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces

Let 119872 = x(119880) be a regular surface via embedding x

119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860

the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4

1H3 and 119872 respectively Then the Gauss

and Weingarten explicit formulas for 119872 in H3 are given by

119863119883

119884 = 119863119883

119884 + ⟨119860 (119883) 119884⟩ 120578 + ⟨119883 119884⟩ x (59)

119860 (119883) = minus119863119883

120578 = minus119863119883

120578 (60)

for all tangent vector fields 119883 119884 isin X(119872) respectively

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Stochastic AnalysisInternational Journal of

Page 7: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

Journal of Applied Mathematics 7

Hence we have a pseudo-orthonormal frame field120572(119904) t(119904)n(119904) e(119904) of R4

1along 120572 and the following Frenet-

Serret formulas

[[[

[

1205721015840(119904)

t1015840 (119904)

n1015840(119904)

e1015840 (119904)

]]]

]

=[[[

[

0 1 0 0

1 0 120581ℎ (119904) 0

0 minus120581ℎ

(119904) 0 120591ℎ

(119904)

0 0 minus120591ℎ

(119904) 0

]]]

]

[[[

[

120572 (119904)

t (119904)

n (119904)

e (119904)

]]]

]

(52)

where hyperbolic curvature and hyperbolic torsion of 120572

are given by 120581ℎ(119904) = 120572

10158401015840(119904) minus 120572(119904) and 120591

ℎ(119904) =

(minus det(120572(119904) 1205721015840(119904) 120572

10158401015840(119904) 120572

101584010158401015840(119904)))(120581

ℎ(119904))

2 under the assump-tion ⟨120572

10158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 respectively

Remark 18 The condition ⟨12057210158401015840(119904) 120572

10158401015840(119904)⟩ = minus1 is equivalent

to 120581ℎ(119904) = 0 Moreover we see easily that 120581

ℎ(119904) = 0 if and only

if there exists a lightlike vector c such that 120572(119904) minus 119888 a geodesic(H-line)

If 120581ℎ(119904) = 1 and 120591

ℎ(119904) = 0 then 120572 in H3 is called a

horocycle We give a lemma about existence and uniquenessfor horocycles (cf [8 Proposition 43])

Lemma 19 For any a0

isin H3 and a1 a

2isin S3

1such that

⟨a119894 a

119895⟩ = 0 the unique horocycle with the initial conditions

120574(0) = a0 1205741015840(0) = a

1 and 120574

10158401015840(0) = a

0+ a

2is given by

120574 (119904) = a0

+ 119904a1

+1199042

2(a

0+ a

2) (53)

Now we recall the basic theory of surfaces in H3 Let x

119880 rarr H3 be embedding such that open subset 119880 sub R2 Wedenote that regular surface 119872 = x(119880) and identify 119872 and 119880

through the embedding x where 119894 119880 rarr 119872 is a local chartFor x(119906) = 119901 isin 119872 and 119906 = (119906

1 119906

2) isin 119880 if we define spacelike

unit normal vector

120578 (119906) =x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817x (119906) and x

1199061(119906) and x

1199062(119906)

10038171003817100381710038171003817

(54)

where x119906119894

= 120597x120597119906119894 119894 = 1 2 thenwehave ⟨x

119906119894 x⟩ = ⟨x

119906119894 120578⟩ =

0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3

1 Then the following maps 119864 119880 rarr 119878

3

1

119864(119906) = 120578(119906) and 119871plusmn

119880 rarr 119871119862+ 119871

plusmn(119906) = x(119906) plusmn 120578(119906)

are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906

0) can be identified with

identity mapping 119868119879119901119872

on the tangent space 119879119901119872 at x(119906

0) =

119901 isin 119872 We have that minus119889119871plusmn

= minus119868119879119901119872

plusmn (minus119889119864)For any given x(119906

0) = 119901 isin 119872 the linear transforms 119860

119901=

minus119889119864(1199060) 119879

119901119872 rarr 119879

119901119872 and 119878

plusmn

119901= minus119889119871

plusmn(119906

0) 119879

119901119872 rarr

119879119901119872 are called de Sitter shape operator and hyperbolic shape

operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and

119878plusmn

119901are denoted by 119896

119894(119901) and 119896

plusmn

119894(119901) for 119894 = 1 2 respectively

Obviously 119860119901and 119878

plusmn

119901have same eigenvectors Also the

eigenvalues satisfy

119896plusmn

119894(119901) = minus1 plusmn 119896

119894(119901) 119894 = 1 2 (55)

where 119896119894(119901) and 119896

plusmn

119894(119901) are called de Sitter principal curvature

and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872

respectivelyThe de Sitter Gauss curvature and the de Sitter mean

curvature of 119872 are given by

119870119889

(1199060) = det119860

119901= 119896

1(119901) 119896

2(119901)

119867119889

(1199060) =

1

2Tr119860

119901=

1198961

(119901) + 1198962

(119901)

2

(56)

at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss

curvature and the hyperbolic mean curvature of 119872 are givenby

119870plusmn

ℎ(119906

0) = det 119878

plusmn

119901= 119896

plusmn

1(119901) 119896

plusmn

2(119901)

119867plusmn

ℎ(119906

0) =

1

2Tr 119878

plusmn

119901=

119896plusmn

1(119901) + 119896

plusmn

2(119901)

2

(57)

at x(1199060) = 119901 respectively Evidently we have the following

relations119870

plusmn

ℎ= 1 ∓ 2119867

119889+ 119870

119889

119867plusmn

ℎ= minus1 plusmn 119867

119889

(58)

We say that a point x(1199060) = 119901 isin 119872 is an umbilical point

if 1198961(119901) = 119896

2(119901) Also 119872 is totally umbilical if all points

on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])

Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification

(1) Supposing that 1198962

= 1

(a) if 119896 = 0 and 1198962

lt 1 then M is a part of anequidistant surface

(b) if 119896 = 0 and 1198962

gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)

(2) If 1198962

= 1 then M is a part of horosphere

4 1198661-Invariant Surfaces in H3

In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces

Let 119872 = x(119880) be a regular surface via embedding x

119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860

the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4

1H3 and 119872 respectively Then the Gauss

and Weingarten explicit formulas for 119872 in H3 are given by

119863119883

119884 = 119863119883

119884 + ⟨119860 (119883) 119884⟩ 120578 + ⟨119883 119884⟩ x (59)

119860 (119883) = minus119863119883

120578 = minus119863119883

120578 (60)

for all tangent vector fields 119883 119884 isin X(119872) respectively

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Stochastic AnalysisInternational Journal of

Page 8: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

8 Journal of Applied Mathematics

Let 119868 be an open interval of R and let

120572 119868 997888rarr 11986323

sub 1198673 120572 (119905) = (120572

1(119905) 0 120572

3(119905) 120572

4(119905))

(61)

be a unit speed regular curve which is lying onH-plane11986323

=

(1199090 119909

1 119909

2 119909

3) isin H3

| 1199091

= 0 Without loss of generality weconsider the subgroup

1198661

=

L119904

=[[[

[

cosh 119904 sinh 119904 0 0

sinh 119904 cosh 119904 0 0

0 0 1 0

0 0 0 1

]]]

]

119904 isin R

sub 1198661

(62)

which isH-translation group alongH-coordinate axis 11989701

(119904) =

(cosh 119904 sinh 119904 0 0) in H3 Let 119872 = x(119880) be a regular surfacewhich is given by the embedding

x 119880 997888rarr H3 x (119904 119905) = 119871

119904(120572 (119905)) (63)

where 119880 = R times 119868 is open subset of R2 Since 119871119904(119872) = 119872

for all 119904 isin R 119872 is invariant under H-translation group 1198661

Hence we say that 119872 is a 1198661-invariant surface and 120572 is the

profile curve of 119872 in H3

Remark 21 From now on we will not use the parameter ldquotrdquoin case of necessity for brevity

By (61) and (62) the parametrization of 119872 is

x (119904 119905) = (1205721

(119905) cosh 119904 1205721

(119905) sinh 119904 1205723

(119905) 1205724

(119905)) (64)

for all (119904 119905) isin 119880 By (59)

120597119904

= 119863x119904x minus ⟨x119904 x⟩ x = 119863x119904x = x

119904

120597119905

= 119863x119905x minus ⟨x119905 x⟩ x = 119863x119905x = x

119905

(65)

where 120597119904

= 119863x119904x 120597119905

= 119863x119905x x119904 = 120597x120597119904 and x119905

= 120597x120597119905 Sowe have

120597119904 (119904 119905) = (120572

1 (119905) sinh 119904 1205721 (119905) cosh 119904 0 0)

120597119905 (119904 119905) = (120572

1015840

1(119905) cosh 119904 120572

1015840

1(119905) sinh 119904 120572

1015840

3(119905) 120572

1015840

4(119905))

⟨120597119904 (119904 119905) 120597

119905 (119904 119905)⟩ = 0

(66)

Hence 120595 = 120597119904 120597

119905 is orthogonal tangent frame of X(119872)

If 120596(119904 119905) = x(119904 119905) and 120597119904(119904 119905) and 120597

119905(119904 119905) then we have that

⟨120596(119904 119905) 120596(119904 119905)⟩ = 1205721(119905)

2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since

120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that

120578 (119904 119905) = ((12057231205721015840

4minus 120572

1015840

31205724) cosh 119904 (120572

31205721015840

4minus 120572

1015840

31205724) sinh 119904

12057211205721015840

4minus 120572

1015840

11205724 120572

1015840

11205723

minus 12057211205721015840

3)

(67)

and it is clear that

⟨120578 120597119904⟩ equiv ⟨120578 120597

119905⟩ equiv 0 (68)

for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A

119901= [

119886 119888

119887 119889] at any x(119904 119905) = 119901 isin 119872 where

119886 =

⟨minus119863120597119904

120578 120597119904⟩

⟨120597119904 120597

119904⟩

=

⟨119863119909119904

119909119904 120578⟩

⟨119909119904 119909

119904⟩

119887 =

⟨minus119863120597119904

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119904

119909119905 120578⟩

⟨119909119905 119909

119905⟩

119888 = 119887

119889 =

⟨minus119863120597119905

120578 120597119905⟩

⟨120597119905 120597

119905⟩

=

⟨119863119909119905

119909119905 120578⟩

⟨119909119905 119909

119905⟩

(69)

After basic calculations the de Sitter principal curvatures of119872 are

1198961

=1205721015840

31205724

minus 12057231205721015840

4

1205721

(70)

1198962

= 12057210158401015840

1(120572

1015840

31205724

minus 12057231205721015840

4) + 120572

10158401015840

3(120572

11205721015840

4minus 120572

1015840

11205724)

+ 12057210158401015840

4(120572

1015840

11205723

minus 12057211205721015840

3)

(71)

Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581

ℎ 120591

ℎ in H3

Proposition 22 Thebinormal vector of the profile curve of1198661-

invariant surface 119872 is constant in H3

Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591

ℎ= 0 Moreover by

(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)

119863te = 0 This completes the proof

From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582

0 120582

1 120582

2 120582

3) such that 120582

119894

is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872

Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896

2=

1205821120581ℎ

Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581

ℎe = 120572 and 120572

1015840and 120572

10158401015840 Also by (50)and (71) we obtain that 120572 and 120572

1015840and 120572

10158401015840= (0 119896

2 0 0) Thus it

follows that 120581ℎe = (0 119896

2 0 0) For this reason we have that

1198962

= 1205821120581ℎ

As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are

119870119889

(119901) =1205721015840

31205724

minus 12057231205721015840

4

1205721

1205821120581ℎ (72)

119867119889

(119901) =

(1205721015840

31205724

minus 12057231205721015840

4) + 120582

1120581ℎ1205721

21205721

(73)

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Page 9: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

Journal of Applied Mathematics 9

at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are

119870plusmn

ℎ(119901) =

(1205721

∓ (1205721015840

31205724

minus 12057231205721015840

4)) (1 ∓ 120582

1120581ℎ)

1205721

(74)

119867plusmn

ℎ(119901) =

1205721

(minus2 plusmn 1205821120581ℎ) plusmn (120572

1015840

31205724

minus 12057231205721015840

4)

21205721

(75)

at any x(119904 119905) = 119901 respectively

Proposition 24 Let 120572 119868 rarr 11986323

sub 1198673 120572(119905) =

(1205721(119905) 0 120572

3(119905) 120572

4(119905)) be unit speed regular profile curve of119866

1-

invariant surface 119872 Then its components are given by

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(76)

Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations

minus1205721

(119905)2

+ 1205723

(119905)2

+ 1205724

(119905)2

= minus1 (77)

minus1205721015840

1(119905)

2+ 120572

1015840

3(119905)

2+ 120572

1015840

4(119905)

2= 1 (78)

for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that

1205723 (119905) = radic120572

1 (119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

(79)

such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that

1205931015840(119905)

2=

1205721 (119905)

2minus 120572

1015840

1(119905)

2minus 1

(1205721

(119905)2

minus 1)2

(80)

Finally by (80) we have that

120593 (119905) = plusmn int

119905

0

radic1205721

(119906)2

minus 1205721015840

1(119906)

2minus 1

1205721 (119906)

2minus 1

119889119906 (81)

such that 1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1 gt 0 for all 119905 isin 119868 Without loss

of generality when we choose positive of signature of 120593 thiscompletes the proof

Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3

Now we will give some results which are obtained by (72)and (74)

Corollary 26 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205723

= 0 or 1205724

= 0 then 119872 is a part of de Sitter flatsurface

(ii) if 120581ℎ

= 0 then 119872 is a part of de Sitter flat surface

(iii) if 1205821

= 0 then 119872 is a part of de Sitter flat surface

Corollary 27 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 If 1205723

= 1205831205724such that 120583 isin R then 119872 is a

de Sitter flat surface

Theorem 28 Let 120572 be the profile curve of1198661-invariant surface

119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ

=

plusmn11205821

Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn

ℎ=

0 By (74) it follows that

1205721

(119905) ∓ (1205721015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (82)

or

1 ∓ 1205821120581ℎ

(119905) = 0 (83)

for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572

1(119905) ∓

(minusradic1205721(119905)

2minus 120572

1015840

1(119905)

2minus 1) = 0 Hence it follows that

1205721015840

1(119905)

2+ 1 = 0 (84)

There is no real solution of (84) This means that the onlyone solution is 120581

ℎ= plusmn1120582

1by (83) On the other hand if we

assume that 120581ℎ

= plusmn11205821 then the proof is clear

Corollary 29 Let 120572 be the profile curve of 1198661-invariant

surface 119872 in H3 Then

(i) if 1205821

= 1 and 120581ℎ

= 1 then 119872 is 119870+

ℎ-flat surface which

is generated from horocyle

(ii) if 1205821

= minus1 and 120581ℎ

= 1 then 119872 is 119870minus

ℎ-flat surface which

is generated from horocyle

Now we will give theorem and corollaries for 1198661-

invariant surface which satisfy minimal condition in H3 by(73) and (75)

Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 inH3 Then 119872 is de

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

10 Journal of Applied Mathematics

Sitter minimal surface if and only if the parametrization of 120572 isgiven by

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724 (119905) = radic120572

1 (119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721 (119906)

2minus 120572

1015840

1(119906)

2minus 1

1205721

(119906)2

minus 1119889119906

(85)

with the condition 120581ℎ

isin (minus11205821 1120582

1) such that 119888

1is an

arbitrary constant

Proof Suppose that 119872 is de Sitter minimal surface that is119867

119889equiv 0 By (73) it follows that

(1205721015840

3(119905) 120572

4 (119905) minus 1205723 (119905) 120572

1015840

4(119905)) + 120582

1120581ℎ1205721 (119905) = 0 (86)

for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation

1205721015840

1(119905)

2minus (1 minus 120582

2

11205812

ℎ) 120572

1 (119905)2

+ 1 = 0 (87)

There exists only one real solution of (87) under the condition1 minus 120582

2

11205812

ℎgt 0 Moreover 120582

1must not be zero by Corollary 26

So that we obtain

120581ℎ

isin (minus1

1205821

1

1205821

) (88)

Hence the solution is

1205721

(119905) = ((minus2 + 1205822

11205812

ℎ) cosh ((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ)

minus1205822

11205812

ℎsinh((119905 + 119888

1) radic1 minus 120582

2

11205812

ℎ))

times (minus2 (1 minus 1205822

11205812

ℎ))

minus1

(89)

where 1198881is an arbitrary constant under the condition (88)

Finally the parametrization of 120572 is given explicitly byProposition 24

On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867

119889equiv 0

Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866

1-invariant surface 119872 in H3 Then 119872 is

hyperbolic minimal surface if and only if the parametrizationof 120572 is given by

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

1205723

(119905) = radic1205721

(119905)2

minus 1 cos120593 (119905)

1205724

(119905) = radic1205721

(119905)2

minus 1 sin120593 (119905)

120593 (119905) = int

119905

0

radic1205721

(119905)2

minus 1205721015840

1(119905)

2minus 1

1205721 (119905)

2minus 1

119889119906

(90)

with the condition 1minus(2∓1205821120581ℎ)2

gt 0 such that 1198881is an arbitrary

constant

Proof Suppose that 119872 is hyperbolic minimal surface that is119867

plusmn

ℎequiv 0 By (75) it follows that

(minus2 plusmn 1205821120581ℎ) 120572

1(119905) plusmn (120572

1015840

3(119905) 120572

4(119905) minus 120572

3(119905) 120572

1015840

4(119905)) = 0 (91)

If Proposition 24 is applied to (91) we have that

1205721015840

1(119905)

2+ ((minus2 plusmn 120582

1120581ℎ)2

minus 1) 1205721

(119905)2

+ 1 = 0 (92)

There exists only one real solution of (92) under the condition

(minus2 plusmn 1205821120581ℎ)2

minus 1 lt 0 (93)

Thus the solution is

1205721

(119905)

= (((2 ∓ 1205821120581ℎ)2

minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582

1120581ℎ)2)

+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888

1) radic1 minus (2 ∓ 120582

1120581ℎ)2))

times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))

minus1

(94)

where 1198881is an arbitrary constant under the condition (93)

However 1205821must not be zero by Corollary 26 So that if 119872

is119867+

ℎ-minimal surface (119867minus

ℎ-minimal surface) thenwe obtain

120581ℎ

isin (11205821 3120582

1) (120581

ℎisin (minus3120582

1 minus1120582

1)) by (93) Finally the

parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of

119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867

plusmn

ℎequiv 0

Now we will give classification theorem for totally umbil-ical 119866

1-invariant surfaces

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

Journal of Applied Mathematics 11

Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-

invariant surface 119872 inH3 Then the hyperbolic curvature of 120572

is constant

Proof Let 119872 be totally umbilical 1198661-invariant surface By

Theorem 23 and (70) we may assume that 1198961(119901) = 119896

2(119901) =

1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following

equations

119860 (120597119904) = minus119863

120597119904120578 = minus119863x119904120578 = minus120582

1120581ℎ (119905) x119904

119860 (120597119905) = minus119863

120597119905120578 = minus119863x119905120578 = minus120582

1120581ℎ

(119905) x119905

119863120597119905

(119863120597119904

120578) = 119863x119905 (119863x119904120578) = 119863x119905120578119904

119863120597119904

(119863120597119905

120578) = 119863x119904 (119863x119905120578) = 119863x119904120578119905

(95)

By (95) we obtain that

minus119863x119905120578119904 = minus1205821

(1205811015840

ℎ(119905) x119904 + 120581

ℎ (119905) x119904119905)

minus119863x119904120578119905 = minus1205821120581ℎ

(119905) x119905119904

(96)

Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905

= x119905119904in

(96) then it follows that 12058211205811015840

ℎ(119905) = 0 for all 119905 isin 119868 Moreover

1205821must not be zero by Corollary 26Thus 120581

ℎis constant

Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following

classification

(1) Supposing that 1205852

= 1

(a) if 120585 = 0 and 1205852

lt 1 then 119872 is a part of anequidistant surface

(b) if 120585 = 0 and 1205852

gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane

(2) If 1205852

= 1 then 119872 is a part of horosphere

where 120585 = 1205821120581ℎis a constant

Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and

Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860

119901= 120585119868

2

where 1198682is identity matrix Finally the proof is complete by

Lemma 20

Now we will give some examples of 1198661-invariant surface

in H3 Let the Poincare ball model of hyperbolic space begiven by

B3

= (1199091 119909

2 119909

3) isin R

3|

3

sum

119894=1

1199092

119894lt 1 (97)

with the hyperbolic metric 1198891199042

= 4(1198891199092

1+119889119909

2

2+119889119909

2

3)(1minus119909

2

1minus

1199092

2minus 119909

2

3) Then it is well known that stereographic projection

of H3 is given by

Φ H3

997888rarr B3

Φ (1199090 119909

1 119909

2 119909

3) = (

1199091

1 + 1199090

1199092

1 + 1199090

1199093

1 + 1199090

)

(98)

We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3

Example 34 The 1198661-invariant surface which is generated

from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ

= radic2 is drawn in Figure 1(a)

Example 35 Let the profile curve of 119872 be given by

120572 (119905) = (radic2 +1

2(minus1 + radic2) 119905

2 0 1 minus

1

2(minus1 + radic2) 119905

2 119905)

(99)

such that hyperbolic curvature 120581ℎ

= 1 Then 119872 is hyperbolicflat 119866

1-invariant surface which is generated from horocycle

in H3 (see Figure 1(b))

Example 36 The 1198661-invariant surface which is generated

from

120572 (119905) = (1

3(minus1 + 4 cosh

radic3119905

2) 0

2

3(minus1 + cosh

radic3119905

2)

2

radic3sinh

radic3119905

2)

(100)

with hyperbolic curvature 120581ℎ

= 12 is drawn in Figure 1(c)

Example 37 Let the profile curve of 119872 be given by

120572 (119905)

= (2 cosh 119905

radic3minus sinh 119905

radic3 0 cosh 119905

radic3minus 2 sinh 119905

radic3 radic2)

(101)

such that hyperbolic curvature 120581ℎ

= radic2radic3 Then 119872 istotally umbilical 119866

1-invariant surface with 120585

2= 23 in H3

(see Figure 1(d))

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

12 Journal of Applied Mathematics

(a) (b)

(c) (d)

Figure 1 Examples of some 1198661-invariant surfaces

Acknowledgment

The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)

References

[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997

[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006

[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983

[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867

3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983

[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian

Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000

[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010

[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003

[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010

[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 13: Research Article Invariant Surfaces under Hyperbolic Translations in Hyperbolic …downloads.hindawi.com/journals/jam/2014/838564.pdf · 2019-07-31 · Research Article Invariant

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


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