Basic geometry of curves

7/25/2019 Basic geometry of curves

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Basic geometry of curves and

surfaces

by

RABAHS OUAM

The purpose of these notes is to give a short introduction to the geometry of curves in R2

and surfaces in R3.

1 Introduction

We will introduce some basic facts about the geometry of curves in R2 and surfaces in R3.Needless to say, this short text will be far from being exhaustive. It is intended for readers

not familiar with differential geometry. We have selected some topics in order to give an idea

about the subject so that the interested reader can continue by himself this study. In particular

we have chosen to skip the study of curves in R3. We give at the end of these notes somereferences. In all what follows the word smooth will mean of classC. All the objects wewill consider (curves, surfaces, mappings...) will be assumed smooth unless otherwise stated.

2 Geometry of curves in R2

2. 1 Local theory of curves in R2

The Euclidean plane R2 will be endowed with its usual orientation, i.e the counter-clockwise

sense.{e1, e2} will denote the canonical positive orthonormal basis.Definition 2.1(parametrized curves) A parametrized curve in R2 is a smooth map: IR2 of an open interval

IR into R2 such that the derivative vector

(t)does not vanish for

everytI .For eachtI ,there is a well-defined straight line containing(t)and directedby(t).It is called the tangent line to att.

Remark 2.1. The reader should be warned that our definition of a parametrized curve coin-

cides with the defintion of a regular parametrized curve used in many standard textbooks.

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44 GEO METRIES ETDYNAMIQUES

The length of a curve: I R2 is the positive (possibly infinite) real number

L() =I|(t)|dt,

where|(t)|is the Euclidean norm of(t).It follows from the change of variable formulathat the length is invariant under reparametrizations, i.e if : I Jis a diffeomorphism,thenL( ) = L().Givent0 I, the arc length function of a parametrized curve : I R2, measured fromt0,is the function defined by

s(t) =

tt0

|(t)|dt.

Since(t)= 0, the arc lengths is a smooth function oft and ds/dt =|(t)|. Given thecurve parametrized bys (a, b), we may consider the curve defined on(b, a), by(s) =(s),which has the same trace (i.e same image) as the first one but is traversed inthe opposite direction. We say that these two curves differ by a change of orientation.The arc length functions has a differentiable inverse since ds/dt=|(t)| = 0.We denoteby abuse of notation its inverse function byt = t(s), which is then defined on the intervals(I) = J.Setting = t : J R2, we obtain a curve which is parametrized by arclength, since|(s)| =|(t).(dt/ds)| = 1,and which has the same trace as. We refer toas areparametrization ofby arc length.Given a parametrized curve : I R2,for eachtI ,we denote byT(t)the unique unitvector tangent to the curve and compatible with its orientation. There is a unique unit vector

Nnormal to Tsuch that the basis {T, N} has the same orientation as the standard orientationofR2. Nis called the unit normal vector of. So Ndefines a differentiable map from I tothe unit circleS1 centered at the origin in R2.Suppose is parametrized by arc length. Then differentiating the equation|T(s)|2 = 1,wesee that the vector dT/ds(s) is orthogonal toT(s),and so is parallel to N(s). We then set

the definiton:

Definition 2.2 (curvature) The curvaturek of the curve parameterized by arc lengths isdefined by

dT

ds(s) = k(s)N(s),

or equivalently by

k(s) =dTds

(s), N(s).For a better understanding of the curvature, we now see how to compute it in different

situations.

For a general curve, not necessarily parametrized by arc length, we have

dT

ds(s) =

dt

ds (t)

dT

dt(t) =

1

|(t)|dT

dt(t)

and so setting:v(t) =|(t)|,we get

k(t) = 1

v(t)dT

dt(t), N(t).


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R. SOUAM 45

Note that the sign ofk changes when we change either the orientation of or the orien-tation ofR2.

Let (x, y) denote the canonical coordinates in R2. Let then (t) = (x(t), y(t)) be aparametrized curve in these coordinates. Then

v(t) = (x2 + y2)1/2, T(t) = (x, y)

(x2 + y2)1/2 and N(t) =

(y, x)(x2 + y2)1/2

and so

k(t) = xy xy(x2 + y2)3/2

.

Suppose is parameterized by arc length. Denote by (s) the oriented angle betweenthe positivex direction andT(s). This is a locally well defined function (up to an additiveconstant multiple of2). We have

T(s) = cos (s)e1+ sin (s)e2 and N(s) =sin (s)e1+ cos (s)e2.We deduce the following formula which may also be taken as a definition of the curvature:

k(s) =d

ds(s).

This formula shows the curvature at s measures the rate of change of the angle which theneighboring tangents make with the tangent at s and so:

The curvature ats measures how rapidly the curve pulls away from the tangent lineats, in the neighborhood ofs.

Consider now a curve which is given as a graph (x) = (x, (x)) of a differentiablefunction. Note that the orientation of the curve is in the direction of increasing x. Then,

with the previous notations:

tan (x) = (x) and ds

dx= (1 + 2(x))1/2.

Thus

k(x) =d

ds =

d

dx dx

ds =

(x)

(1 + 2(x))3/2.

Given a parametrized curve : I R2, andt0 I, there exists an interval I0 Icontainingt0 such that the image(I0) may be represented as a graph over its tangent lineat t0. Choosing a coordinate system (x, y) so that (t0) coincides with the origin, T(t0)coincides withe1andN(t0)coincides withe2,we may assume(I0)is a graph(x, (x))ofa functionsatisfying(0) =(0) = 0.Then, the previous formula shows the curvature att0 is given by the second derivative

(0).In particular ifk(t0)= 0,the image of the curve

is situated on one side of its tangent line at t0 in a neighborhood oft0.Ifk(t0) > 0 it issituated in the half-plane into which N(t0)points and in the other half-plane ifk(t0)< 0.Ifk(t0) = 0 then, in a neighborhood oft0,the curve may stay on one side of its tangent line att0 or may traverse it.


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Examples. (1) A line segment has curvature zero.

(2) A circle of radius R (centered at the origin) and oriented in the counter-clockwisesense has as an arc length parameterization: (s) = R(cos(s/R), sin(s/R)), wheres[0, 2]and so its curvature is constant and equal to1/R.

Conversely, suppose : I R is a parametrized curve having constant curvature k0.Then differentiating the equation N(s), (s)= 0,we get

N(s), (s)=N(s), (s)=k0.Since N(s), N(s)= 0,we conclude that

N(s) =k0(s).Integrating, we see that N(s) + k0(s) is a constant vector, say v . Ifk0 = 0, the normal

N(s) = v would be constant and so also

(s)and integrating again we see that (I)lies ona line. Ifk0= 0,then(s) v

k0=

N(s)

k0

and thus (s) vk0 = 1|k0| ,

which means(I)lies on the circle of radius1/|k0| and centerv/k0.More generally, the arc length and the curvature function entirely determine the curve up

to isometries ofR2 :

Proposition 2.3 Given a smooth function k : s I k(s) R, the arc lengthparametrized curves : I

R

2 havingk as curvature function are given by:

(s) = (

cos (s)ds,

sin (s)ds) + (a, b),

where

(s) =

k(s)ds + ,

with R and(a, b)R2.Proof. Consider a curve : I R2 parametrized by arclength whose curvature functionis the given function k. Then the angle (s) that (s) makes with the positive x axis sat-isfiesk(s) = (s). So (s) =

k(s)ds+ , for some constant R. From (s) =

(cos (s), sin (s)),we get

(s) = (

cos (s)ds,

sin (s)ds) + (a, b), (a, b) R2.

Conversely, it is easy to check that a map : I R2 given by the previous formula definesa curve parametrized by arclength with curvature functionk.


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R. SOUAM 47

Definition 2.4(the osculating circle) Let : I R2 be a parameterized curve andt0Isuch thatk(t0)

= 0.The osculating circle of the curve att0 is the circle of radius 1/

|k(t0)

|and center(t0) + 1/k(t0)N(t0).It is the circle which approximates the best the curve att0(cf. exercise (3) below).

Exercises. (1) LetA be an isometry ofR2 and : I R2 a parametrized curve. Checkthat the curvaturekAofA satisfies:kA(t) = k(t)ifAis orientation preserv-ing andkA(t) =k(t)ifAis orientation reversing.

(2) Letbe a parameterized curve and >0 a positive number. Check that the curvaturekof and the curvaturek of the homothetic curve,are related by the formula:k=

1k.

(3) Let : I R2 be a curve parametrized by arc length and s0I , such that k(s0)= 0.Changing orientation of the curve if necessary, we may assumek(s0)> 0.Let Rand denoted byC the circle tangent to the curve at s0 with center the point (s0) +

N(s0).Check that if >1/k(s0)(respectively if


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Integrating by parts, we obtain

dLdt

(0) =12

l0

Ft

, 2

Fs2

|t=0ds=12

l0

Ft

|t=0, k(s)N(s)ds.

The mapping Ft|t=0 C([0, l],R2) is a vector field along and is called the variationvector field associated to the variation F. We may consider the spaceCof closed curves inR2.The tangent spaceTC at a curve : [0, l] R2 may be identified withC([0, l],R2)

and endowed with the inner product:

V1, V2= l0

V1(s), V2(s)ds, V 1, V2C([0, l],R2),

where we assumed the curve is parametrized by arclength. Thus the vector field 12kNalongmay be viewed as the gradientat of the length functional defined on the space C.

The Jordan curve theorem

The Jordan curve theorem asserts that a simple closed curve divides the plane into two con-

nected components, an unbounded one, called its exterior, and a bounded one called its inte-

rior, which moreover is homeomorphic to a disk. We shall refer to the latter as the domain

bounded by the curve. Although intuitive, this fact, which is also true forC0 maps, is not soeasy to prove. For a proof see [2].

Remark 2.2. Let : t [a, b] (x(t), y(t)) R2 be a simple closedcurve and let denote its interior domain. The curve is said to be positively oriented if its oriented unit

normal points towards.

We now derive formulae for the area A of the domainbounded by a simple closed curve.Using Stokes theorem, the area A can be written as follows:

A=

dx dy=

d(xdy) =

d(ydx) = b

a

x(t)y(t)dt= b

a

x(t)y(t)dt.

Another useful formula for the area is as follows. Denote by Xthe position vector fieldon R2, i.eX(x, y) = (x, y). The divergence ofX is divX = 2. So if : [0, l] R2 ispositively oriented and parametrized by arc length, using the divergence theorem we get:

A=1

2

(divX) dx dy =12

l0

(s), N(s)ds.

The isoperimetric inequality

The isoperimetric problem is stated as follows: among all domains having the same boundary

length, find those of maximum area. The solutions to this problem in the plane are the circular

disks. The problem can be formulated in all dimensions and the result extends: the solutionsare the round balls. There are many different proofs of this fact. We present one which is

maybe the cheapest one but has the inconvenient that, so far, it has not been generalized to

higher dimensions. The isoperimetry property of the circular disks is a consequence of the

following isoperimetric inequality:


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R. SOUAM 49

Theorem 2.6 LetCbe a simple closed curve in R2 with lengthL, and letAbe the area ofthe domain bounded byC. Then

L2 4A,and equality holds if and only ifCis a circle.

For the proof, we need the following:

Lemma 2.7(Wirtingers inequality) LetfC1(R)a2-periodic function with zero mean,that is

20

f(t)dt= 0.Then

20

f2(t)dt 20

f2(t)dt,

with equality if and only iff(t) = a cos(t) + b sin(t).

Proof. fand f are equal to their Fourier series in the L2[0, 2]sense. Since 20

f(t)dt= 0,the Fourier series expansion off reads:

n=1

ancos(nt) + bnsin(nt)

and that off readsn=1

nbncos(nt) nansin(nt)

By Parsevals identity we have:

1

20

f2(t)dt=

n=1

a2n+ b2n

1

20

f2(t)dt=

n=1

n2a2n+ n2b2n.

The result follows immediately.

Proof of the isoperimetric inequality. Note that if we perform a homothety of magnitude ,then the length is multiplied by and the area is multiplied by2. So, up to a homothety,we may assume the curve has length 2. We may also assume it is positively oriented andparametrized by arc length, = (x, y) : [0, 2] R2. Up to a translation in R2, we mayassume the center of gravity of the curve is at the origin, i.e

2

0

xdt= 2

0

ydt = 0.

Since the curve is parametrized by arclength we have

20

(x2 + y2)dt= 2= L,


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whereas

A= 2

0 xy

dt.

So

2( A) = 20

(x2 + y2 2xy)dt= 20

(x2 x2)dt + 20

(x y)2dt.

The integral in the last term is nonnegative by Wirtingers inequality, so A .If equalityholds then x(t) = a cos t+ b sin t and y(t) = a sin tb cos t, and the curve is a circleparametrized by arc length.

Remark 2.3. It follows that the isoperimetric inequality is in fact true for any bounded regular

domain in R2.A domain is said to be regular if each of its boundary components is the trace(i.e the image) of either a one-to-one parametrized curve defined on R or a simple closed

curve. A compact regular domain is then bounded by a finite number of simple closed curves

and is contained in the interior of one of them, this follows easily from the Jordan curvetheorem. The isoperimetric inequality (and the characterization of the equality case) follows

directly from the result forby filling the holes.

The rotation index and total curvature of simple closed curves

The rotation index of a closed curve (not necessarily simple) is the algebraic number of times

the tangent vector rotates along the curve. We need some preparatory material before stating

some results about the rotation index. We refer to [1] for the proofs. In order to give a

rigourous definition of the rotation index we need to introduce the notion of degree of maps

S1 S1. Denote byp : R S1 = R/2Z the canonical projection. We then have thefollowing:

Lemma 2.8(Paths lifting property) LetfC0

([a, b], S1

),andxp1

(f(a)).Then thereexists a unique map fC0([a, b],R)such thatf=p f andf(a) = x.The map fis called a lift off.Any two lifts differ by an integer mutiple of2.

Proof. we leave the proof as an exercise. Hint: consider the set I={t[a, b], f : [0, t]R has a (continuous) lift f : [0, t] R, f(a) = x} and show that it is non empty, openand closed. See [1] for the details.

Definition 2.9(the degree of maps S1 S1) Letf :S1 S1 be a continuous map. Applythe above lemma to obtain a map f : [0, 2] R liftingfp.The degree offis the integer

deg(f) :=f(2) f(0)

2 .

Remark 2.4. (1) The degree is indeed an integer since f(2) = f(0), and does not dependon the lift since any two lifts differ by an additive constant. One can also replace 0 byanys R,and gets the same result setting

deg(f) =f(2+ s) f(s)

2 .


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R. SOUAM 51

(2) If one takes a periodic map having a periodLnot necessarily equal to2, that is a mapf : R/LZ

R/2Z,the degree is then given by:

deg(f) =f(L + s) f(s)

2 .

A mapf1 C0(S1, S1)is said to be homotopic to a map f2 C0(S1, S1)if there is acontinuous map:

H: [0, 1] S1 S1,such that

H(0, t) = f1(t),

H(1, t) = f2(t).

This defines an equivalence relation between maps in C0(S1, S1). A fundamental propertyof the degree is its invariance under homotopy:

Homotopic maps have the same degree.

See [1] for the details and further reading.

Definition 2.10(The rotation index of closed curves) Let : S1 R2 be a closed planecurve andT(t) = 1|(t)|

(t), its tangent unit vector att S1. Tmay be seen as a mapT : S1 S1,called the tangent map. The rotation index of the curve is by definition thedegree of its tangent map T.

Note that the rotation index depends on the orientation (and is independent on the choice

of orientation preserving parametrizations). An important fact about the rotation index is the

following:

Theorem 2.11(The turning tangent theorem) The rotation index of a simple closed curve

is 1,where the sign depends on the orientation of the curve.Proof. We outline the proof and refer to [1], [2] or [5] for the details and figures. The idea

of the proof is to construct a homotopy between the tangent map and a secant mapwhich has

degree one. Let : [0, l] R2 be a simple closed curve. We may easily find a lineL inthe plane which is tangent to ([0, l])at some pointp and such that the whole trace ([0, l])is located on one side ofL (take a line not touching the curve and move it parallel to itselfuntil it touches the curve for the first time). We may assume without loss of generality that

p= (0).Consider now the triangle

={(t1, t2)[0, l] [0, l]; 0t1t2l}

and define a secant map : S1 by

(t1, t2) = (t2) (t1)|(t2) (t1)| for t1=t2, (t1, t2) {(0, l)}

(t, t) = (t)

|(t)| , (0, l) =(0)

|(0)| .


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is well defined because the curve is simple. It is easily seen to be continuous. LetA =(0, 0), B = (0, l), C = (l, l) be the vertices of the triangle . Note that restricted to the

sideAC is the tangent map of. The tangent map is thus homotopic to the restriction ofto the remaining sidesAB and BC.It is thus enough to see the latter map has degree 1.

Assume the orientations of the plane and the curve are such that the oriented angle from

(0)to(0)is . The restriction of to AB covers half ofS1 in the positive direction,and the restriction oftoBCcovers the remaining half also in the positive direction. Thus,the degree of restricted toAB andBC is+1. If we reverse the rientation, we obtain1for this degree. This completes the proof.

Definition 2.12(Total curvature of closed curves) Let : [0, l] R2 be a closed curveparametrized by arc length. The total curvature ofis by definition the quantity:

l0

k(s)ds.

Note that the total curvature depends on the orientation of the curve. Let : [0, l]R2be a simple closed curve parametrized by arc length. A lift T : [0, l] R, of the tangent mapT : [0, l]S1 is a measure of the oriented angle between the positivex axis andT(s).Wehave seen thatk(s) = dT(s)/ds.So:

l0

k(s)ds= T(l) T(0) = 2deg(T).

It follows that:

The total curvature of a closed curve is equal to2times its rotation index.

A consequence of the Theorem 2.11 is thus:

Corollary 2.13 The total curvature of a simple closed curve is equal to 2.Remark 2.5. The total curvature is equal to2in case the curve is positively oriented.

Convexity

We now give a converse to Corollary 2.13.

Proposition 2.14 Let : [0, l]R2 be a closed curve with l0

k(s)ds=2. Assume itscurvature does not change sign. Then is simple. Moreover, for eachs [0, l], the trace([0, l])lies in one of the closed half-planes determined by the tangent line ats.

Proof. Up to a change of orientation on the curve, we may assume k 0. Take any liftT : [0, l] R of the tangent map T : [0, l] S1. Since k does not change sign, T ismonotone. Lets0

[0, l], we will show the trace of lies in one of the closed half-planes

determined by the tangent lineat s0. Assume this is not the case, there exist then two pointsp, qon the curve lying on different sides of.LetN(s0)be the unit normal vector at s0,andconsider the function:

h(s) =(s) (s0), N(s0), s[0, l].


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Thenh takes positive and negative values (its values at p and qare of opposite sign by hy-pothesis). By compactness it then has a positive maximum at a points1

=s0 and a negative

minimum at a point s2= s0. Thus two of the unit tangent vectors at s0, s1, s2 coincide,say for instance at s0 ands1. Then T(s0) T(s1) 2Z. Since the rotation index is one,|T(s0)T(s1)|< 2, so necessarily T(s0) = T(s1)and then by monotonicity, Tis constanton the interval[s0, s1](assumings0 < s1). Thereforek vanishes identically on[s0, s1]andso([s0, s1]),which is a contradiction.

Let us show now is simple. Assume there are0 s0 < s1 < l, such that (s0) =(s1). If the tangent lines at s0 ands1 do not coincide, this clearly contradicts the aboveproperty about the position of the curve with respect to its tangent lines. Therefore either

(s0) = (s1) or

(s0) =(s1). In the first case, as before, the monotonicity ofTand the fact that the rotation index is one imply that ([s0, s1])is contained in a line. Since(s0) =

(s1), this implies the existence of a zero of, which is a contradiction. So

necessarily(s0) =(s1).Consider then, as before, the function

h(s) =(s) (s0), N(s0), s[s0, s1].

Then h(s0) = h(s1)and so h reaches its maximum on[s0, s1]at some point s2 , s0 < s2 0, such that(0) =pand(0) =w. The set of all such vectors is denoted by TpS.

IfX : U S is a parametrization with p X(U) then it is not difficult to checkthatTpS= (dX)X

1

(p)(R2

).In particularTpS is a linear subspace ofR

3

and is called thetangent planetoSat p. The affine subspacep + TpSofR3 is called theaffine tangent plane

toSat p.

The first fundamental form

Definition 3.3(The first fundamental form) LetSbe a surface in R3 andp S. Thetangent spaceTpS inherits an inner product, denoted by Ip, from the inner product on R

3.More precisely,

Ip(w1, w2) =w1, w2, for all w1, w2TpS.

Iis called the first fundamental form ofS.

I measures the norm of vectors tangent to Sand hence the length of curves on S. Letc: JSbe a parameterized curve onS(Jan interval ofR) and letc(t) = dcdt (t)Tc(t)Sdenote its tangent vector at tJ,then the length ofcis given by

L(c) =

J

Ic(t)(c(t), c(t))dt.

The length is invariant under reparametrizations.

IfX : U R2 X(U) is a local parametrization of an open set X(U) S, andw = Xu+ Xv a tangent vector then:

I(w, w) = 2Xu, Xu + 2Xu, Xv + 2Xv, Xv

LetE(u, v) =Xu, Xv, F(u, v) =Xu, Xv, and G(u, v) =Xv, Xv,denote the coef-ficients of the first fundamental form in the basis{Xu, Xv}.Usually one represents the firstfundamental form by thearclength elementds written as

ds2 =Edu2 + 2Fdudv+ Gdv2,


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R. SOUAM 55

which means that if : J X(U), (t) = X(u(t), v(t)) is a curve in X(U),then its arclength satisfies:

ds

dt =|(t)|=

Edu

dt

2+ 2F

dudt

dvdt

+ G

dvdt

2.

Its length is given by the integral:

J

Edu

dt

2+ 2F

dudt

dvdt

+ G

dvdt

2dt.

Definition 3.4 LetSandS be two surfaces in R3.A diffeomorphism: SS is said tobe an isometry if for eachpSand allw1, w2TpS

Ip(w1, w2) = I(p)(dp(w1), dp(w2)).

A map : V S of a neighborhoodV ofp Sis a local isometry atp if there exists aneighborhoodV of(p)inS such that: V V is an isometry. If there is a local isom-etry intoS at everypS, the surfaceSis said to be locally isometric to S.The surfacesSandS are locally isometric ifSis locally isometric toS andS is locally isometric toS.

Exercise. Suppose there are local parametrizations X : U X(U) ofS andX : UX(U)ofS such thatE= E, F = F andG = G.Check that the map: : X X1 :X(U)X(U)is an isometry.Examples. (1) A graph{(x, y, h(x, y))}of some functionh : U R3, U R2 has as a

parametrization the map:

(x, y)UX(x, y) = (x,y,h(x, y)).

The coefficients of the first fundamental form are:

E(x, y) =Xx, Xx= 1 + h2x,

F(x, y) =Xx, Xy= hxhy,G(x, y) =Xy, Xy= 1 + h2y.

So,

ds2 = (1 + h2x)dx2 + 2hxhydxdy+ (1 + h

2y)dy

2.

For instance, the planez = 0has as arclength element: ds2 =dx2 + dy2.

(2) Let : s(a, b)((x(s), y(s))R2 be a curve in R2 parametrized by arc length.The right cylinder with base the curve in R3 has the following parametrization:

X(s, t) = (x(s), y(s), t), s(a, b), tR.

Then

Xs= (x(s), y(s), 0), Xt = (0, 0, 1),


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so

E= 1, F= 0, G= 1.

We have thus found (local) coordinates on the cylinder where the first fundamental

form is the same as the first fundamental form of a plane. Thus the plane and a cylinder

are locally isometric.

(3) Consider a curve t(a, b)(x(t), z(t))in the(x, z)-plane in R3 and assume x(t)>0. Rotating this curve around the z-axis, one obtains a surface of revolution whichadmits the parametrization:

X(, t) = (x(t)cos , x(t)sin , z(t)), a < t < b, 02.The coefficients of the second fundamental form in these coordinates are then:

E= (x(t))2, F= 0, G= (x(t))2 + (z(t))2.

The arclength element is:

ds2 = (x(t))2d2 + ((x(t))2 + (z(t))2)dt2.

3. 2 The Gauss map

In order to study the way a surface curves into space one is laid, as in the case of curves in

the plane, to study the variation of the tangent plane or equivalently the variation of the unit

normal to the tangent plane. This leads to the notion of Gauss map which is fundamental in

the theory of surfaces in R3.We point out again that the following study is local and thus isvalid for immersed surfaces.

Definition 3.5 LetS R3 be an orientable surface in R3, oriented by the choice of a(differentiable) unit normal N . The map N : S

R

3 takes its values in the unit sphere

S2 ={(x, y, z) R3, x2 + y2 + z2 = 1}.It can thus be viewed as a mapN :S S2,

and is called the Gauss map ofS.

For anyq S2,the tangent planeTqS2 is the linear plane orthogonal in R3 to the vectorq. So forpS, the tangent space TN(p)S2 coincides withTpSand so the linear mapdNp :TpSTN(p)S2 is a linear endomorphism ofTpS,

dpN :TpSTpS.Definition 3.6 The linear endomorphism dNp: TpSTpSis called the shape operatoror Weingarten map ofSatp.

The linear mapdNp

:Tp

S

Tp

Soperates as follows. Letp

Sand w

Tp

S.If(t)is a parametrized curve inS with(0) = p and (0) = w, we consider the parametrizedcurveN (t) = N(t)in S2 R3.Then

dNp(w) = d

dt(N )(t)|t=0 = N(0).


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R. SOUAM 57

Examples. (1) For a plane, say the (x, y)-plane,N = (0, 0, 1) at each point and so theshape operator of the plane vanishes identically.

(2) Consider the sphereS2a(r) R3 of radiusr centered at a point a and oriented by theinterior unit normalN(p) =1r (p a), pS2a(r).ThendNp =1r Idp,whereIdpis the identity map ofTpS

2a(r).

(3) An important feature of the shape operator is the following:

Proposition 3.7 dNp: TpSTpSis symmetric.

Proof. Take any local parametrizationX : (u, v) U Saroundp S. Since for eachqU, {Xu(q), Xv(q)} is a basis ofTX(q)S, it is enough to check that:

dN(Xu), Xv=Xu, dN(Xv).Since N X, Xv= 0,differentiating with respect tou,we get

dN(Xu), Xv=(N X)u, Xv=(N X)uv, X,

and symmetrically,

dNp(Xv), Xu=(N X)v, Xu=(N X)vu, X.

The result now follows from the Schwarz lemma.

Being symmetric, the operatordNp has two real eigenvalues and can be diagonalizedin an orthonormal basis. We thus set the following

Definition 3.8 The eigenvalues 1and2of the symmetric linear map dNp: TpSTpSare called the principal curvatures ofS atp, and its eigenvectors are called the principaldirections ofSatp.A curve onSwhose tangent vector is a principal direction at each pointis called a line of curvature.

The half of the trace ofdNp,

H(p) =12

trdNp =1+ 2

2 ,

is called the mean curvature ofSatp.

The determinant ofdNp,K(p) = det dNp= 12,

is called the Gaussian curvature ofSatp.


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The second fundamental form

Definition 3.9 (The second fundamental form) The symmetric bilinear form on TpS atpSdefined by

IIp(w1, w2) =dNp(w1), w2=w1, dNp(w2)

is called the second fundamental form ofSatp.

Remark 3.1. If we change the orientation of the surface (i.e replace N byN), thenII, 1, 2and Hchange by sign but Kdoes not change.

Examples. (1) For a plane, at each pointII = 0, 1 = 2 = H= K= 0.

(2) The sphereS2a(r) R3 of radius r centered at a point a and oriented by the interior unitnormal has as second fundamental formII =

1

r I,and as curvatures1 = 2 = H= 1

randK= 1r2 everywhere.

Note that in these two examples, at each point, all the tangent directions are principal,

otherwise said the second fundamental form is a multiple of the first one. A point on a

surface where the second fundamental form is a multiple of the first one is called an umbilic

point. A surface is called totally umbilic if all its points are umbilic. So planes and spheres

are totally umbilic. Conversely, it can be shown that a totally umbilic surface is a part of a

plane or a sphere (cf. [2]).

Curvature of normal sections

The second fundamental form has the following interesting interpretation: letp

S and

wTpSa unit tangent vector and consider the affine plane w throughpspanned byw andN(p). Nearp, the intersection wS is a curve in the planew which is the graph of asmooth function over the axis determined by w. Orient the planewby the basis {w, N(p)}.Letbe a parametrization by arc length of the intersection curve with (0) =pand(0) =w.Differentiating the equality N((s)), (s)= 0,we get

dNp((0)), (0)=N((0)), (0)

that is

IIp(w, w) =N(p), (0).So: The second fundamental form calculated on the unit vectorw coincides with the curva-ture atpof the normal section defined byw.

Also, it is easily seen that on a two dimensional Euclidean space, the eigenvalues of a

symmetric operator are the extremums (the maximum and the minimum) on the unit sphere

of the associated quadratic form. It follows that the principal curvatures atp Sare theextremums of the curvatures of plane sections at pdefined by the normalN(p).


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R. SOUAM 59

The Gauss map in local coordinates

In this section, we consider a local parametrization X : U S. We can choose the unitnormal fieldNdetermined by the relation

(N X)(u, v) = Xu Xv|Xu Xv| .

Let(aij)be the matrix ofdNin the basis {Xu, Xv}.This means:dN(Xu) = (N X)u= a11Xu+ a21Xv,dN(Xv) = (N X)v = a12Xu+ a22Xv.

Denote byMthe matrix associated to the first fundamental form in the coordintaes(u, v) :

M=E F

F G

Set

e=(N X)u, Xu=N X, Xuuf=(N X)v, Xu=N X, Xuv=N X, Xvu=(N X)u, Xv

g=Nv, Xv=N X, Xvv.These are called the coefficients of the second fundamental form in the coordinates (u, v).Denote bythe associated matrix:

=

e ff g

Then, it is easily checked that =M(aij)and so:(aij) =M1.

The mean and Gaussian curvatures are therefore given by:

2H=tr(aij) = tr(M1) = eG 2fF + gEEG F2 ,

K= det(aij) = det

detM=

eg f2EG F2 .

Example(Graphs). For a surface given as a graphS={(x, y, h(x, y), h: U R2 R},

Xx= (1, 0, hx), Xy = (0, 1, hy),

Xxx= (0, 0, hxx), Xxy= (0, 0, hxy), Xyy = (0, 0, hyy),

(N X)(x, y) = (hx, hy, 1)(1 + h2x+ h

2y)

1/2,


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e=N X, Xxx= hxx(1 + h2x+ h

2y)

1/2, g =N X, Xyy= hyy

(1 + h2x+ h2y)

1/2

f=N X, Xxy= hxy(1 + h2x+ h

2y)

1/2.

So

K=hxxhyy h2xy

(1 + h2x+ h2y)

2,

2H=(1 + h2x)hyy 2hxhyhxy+ (1 + h2y)hxx

(1 + h2x+ h2y)

3/2 .

Once known the values ofHandK, the principal curvatures are then the roots of the poly-nomial X2 2HX+ K,that is,i = H

H2 K, i= 1, 2.

Given a pointpS, up to a translation and a rotation in space, we may assume the pointpcoincides with the origin and the tangent plane TpScoincides with the plane{z = 0}.Wemay then represent Snear p as a graph of a function h with h(0, 0) = 0 and hx(0, 0) =hy(0, 0) = 0,then the second fundamental form atp calculated on a tangent vector (x, y)isgiven by:

IIp((x, y), (x, y)) = hxx(0, 0)x2 + 2hxy(0, 0)xy+ hyy(0, 0)y

2

and so it coincides with the hessiand2h(0, 0)ofhat (0, 0).In particular, ifK(p) = det d2h(0, 0) = hxx(0, 0)hyy(0, 0) h2xy(0, 0)is>0 thenhhas

either a local maximum or minimum at (0, 0)and soS lies on one side of its affine tangentplane in some neighborhood ofp, the same side asN(p) if1, 2 > 0,the opposite side if1, 2 < 0.In this case, the point pis called elliptic.

IfK(p)< 0,that is, if1, 2 have opposite signs, thenh changes sign in any neighbor-

hood of(0, 0)ans soStraverses its affine tangent plane in any neighborhood ofp.The pointpis called asaddlepoint or a hyperbolic point.

IfK(p) = 0but IIp= 0,which means one of the principal curvatures is 0 and the otheris non zero, thenp is called a parabolic point. IfIIp = 0,that is to sayk1 =k2 = 0,thenpis called a flat point.

Example(Surfaces of revolution). Let t(a, b)(x(t), z(t))be a curve in the(x, z)-planein R3 with x(t)> 0. Rotating this curve around the z-axis, one obtains a surface of revolutionwhich admits the parametrization:

X(, t) = (x(t)cos , x(t)sin , z(t)), a < t < b, 02.

We will now see that the meridians t X(, t) and the parallels X(, t) are lines ofcurvature. Since the surface is invariant under rotation around the z-axis, it is enough to check

that the curve t(x(t), 0, z(t))is a line of curvature. Indeed, by symmetry this will show allthe meridians are lines of curvature and also all the parallels since the two principal directions

are orthogonal at each point. Now, just notice that by symmetry the unit normal to the curve

t (x(t), 0, z(t))in the(x, z)-plane is normal to the surface and so its differential will beparallel to the tangent vector(x(t), 0, z(t)),which means the curve is a line of curvature.


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R. SOUAM 61

Let us compute the principal curvatures. Again, by symmetry, it is enough to do the

computation on the curve t

(x(t), 0, z(t)). This curve is a normal section, so the associated

principal curvature is equal to its curvature as a plane curve provided the plane is correctlyoriented .

Therefore (cf. 2. 1):

1 = xz xz(x2 + z2)3/2

.

As a unit normal to the curve, we may take:

N(0, t) = (z, 0, x)(x2 + z2)1/2

.

So the unit normal to the surface at the point X(, t)is obtained by rotating N(0, t)aroundthez-axis by the angle :

N(, t) =(z cos , z sin , x)

(x2 + z2)1/2 .

Fortfixed, put() = X(, t),then:

II((), ()) =N, ()= xz

(x2 + z2)1/2.

Hence:

2 =II((), ())

|()|2 = z

x(x2 + z2)1/2.

3. 3 The area element and the area form

Let Sbe a surface in R3 and X : U R3 a local parametrization, U R2, and asusual denote byE , F , G the coefficients of the first fundamental form in these coordinates.We consider the measure on Ugiven by

EG F2dudv.Iff : S Ris a continuous

function with compact support inX(U),its integral is by definition:

S

f=

U

(f X)(u, v)

EG F2dudv.

The value of the integral does not depend on the choice of paramaterization. Indeed ifX :U S is another parametrization with X(U)X(U)=, set = X1 X, V =X1(X(U) X(U)) and V = X1(X(U) X(U)). Iff is a function with compactsupport inX(U) X(U),then by the change of variables formula:V

(fX)(u, v)

EG F2dudv =V

(fX)((u, v))

EG F2 |detd|dudv

If we denote by M (resp. M) the matrix of the first fundamental form in the coordinates(u, v)(resp. in the coordinates(u, v)) and byPthe change of coordinates matrix from the


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basis{Xu, Xv}ofR2 to the basis{Xu, Xv},thenM = P MtP. The conclusion followssincePis also the matrix ofdin the bases

{ u ,

v

}and

{ u ,

v

}Using a partition of unity, one then extends this defintion to functions with compact sup-port inSand it can be checked this definition is independent on the choice of the partitionof unity. In this way we have defined a measure onScalled the area element or canonicalmeasure and usually denoted by dA. In particular the area of a domain in Sis by definitionthe integral of the function identically equal to 1:

area() =

dA.

When Sis orientable and oriented by a choice of a unit normal N, one can define a differential2-form on Sas follows:

(p)(w1, w2) = det(w1, w2, N(p)), pS, w1, w2TpS.

(p)(w1, w2)is the algebraic area in TpSof the quadrelateral determined by the ordered pair(w1, w2). The form is called thearea formon S. Integrating a function fon Swith respectto the area element amounts to the same to integrating the 2-formf .

Denote by the area form on the unit sphere oriented by its exterior normal. Then, thepull-back byNof is such that forw1, w2TS :

N(w1, w2) = (dN(w1), dN(w2))

=det(dN(w1), dN(w2), N)

=det(dN)det(w1, w2, N).

So we get the following relation:

N= K.

Npreserves the orientation at points whereK >0 and reverses it at points whereK


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R. SOUAM 63

Theorem 3.10(Existence of conformal coordinates) LetSbe an immersed surface in R3.Then around each point inSthere exists a conformal parametrization.

Remark 3.2. This theorem, which is of fundamental importance, is valid on any abstract

Riemannian surface. A Riemannian surface is a surface (not necessarily immersed in R3)

whose tangent plane at each point is endowed with an inner product which depends smoothly

on the point. This means that when written in any local chart, the coefficients of the inner

product are smooth functions.

Exercise(Riemann surface structure). LetSbe a surface in R3 (more generally any Rieman-nian surface). By the Theorem 3.10 there exists an atlas of conformal charts. The change of

coordinates maps may be viewed as functions between open sets in C.Prove that each suchfunction is either holomorphic or anti-holomorphic. In particular, ifS is orientable, we canget an atlas of conformal charts and the change of coordinates maps are holomorphic. The

surface Sadmits thus an underlying complex structure compatible with its metric. A complex

manifold with complex dimension 1 is by definition a Riemann surface.

In a conformal parametrization, the mean curvature has a particularily simple expression:

Proposition 3.11 LetX: UX(U)be a conformal parametrization of an open setX(U)in surfaceS. CallNa unit normal field to X(U)andHthe mean curvature computed withrespect to this orientation. Then:

Xuu+ Xvv = 22HN,

where2 =|Xu|2 =|Xv|2.

Proof. By hypothesis, we have:

Xu, Xu=Xv, Xv, Xu, Xv= 0.

We deduce that:

Xuu, Xu=Xvu, Xv=Xu, Xvv.So

Xuu+ Xvv , Xu= 0and in a symmetric way:

Xuu+ Xvv , Xv= 0.It follows that the vectorXuu+ Xvv is parallel toN.The formula for the mean curvature inconformal coordinates (cf.

3.2) takes the form:

2H=Xuu, N + Xvv , N

2

and the result follows.


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3. 5 Minimal surfaces

Definition 3.12 (Minimal surfaces) A surface Sin R3

is called minimal if its mean curvaturevanishes identically, i.eH= 0.

It can be proved that each point on a minimal surface has a neighborhood which mini-

mizes the area among all surfaces with the same boundary, whence the terminology mini-

mal.

Remark 3.3. Up to now we have considered oriented surfaces. However, note that the min-

imality condition is invariant under change of orientation. Therefore, it makes sense for non

orientable surfaces too: a (not necessarily orientable) surface is called minimal if each of its

points admits an orientable neighborhood which is minimal.

A consequence of the Proposition 3.11 is the following:

Corollary 3.13 A surfaceSin R3 is minimal if and only if for any conformal parametriza-

tion X : U X(U), the (vector valued) map X is harmonic (i.e its coordinate functionsare harmonic). In particular, a minimal surface is analytic.

The analyticity of the surface means that one may find a local analytic parametrization

around each point and follows immediately from the analyticity of harmonic maps.

Corollary 3.14 There is no compact minimal surface without boundary in R3.

Proof. Without loss of generality we may assume the surface connected. Fix any coordinate

function, by compactness its restriction toSreaches its maximum, saya.The set of points ofSwhere this coordinate function equalsa is clearly closed. It is also open. Indeed in a localparametrizationX : U S,around such a point, this coordinate function is harmonic andhas a local maximum and so is constant by the maximum principle for harmonic functions.

This coordinate function is thus constant andS lies on a plane, which is absurd sinceShas

no boundary.

Example(Minimal Surfaces of revolution). Let us determine all the surfaces of revolution

which are minimal. The planes are trivial examples. Consider the surface of revolution

obtained by rotating around thez-axis the curvet(x(t), 0, z(t)), x(t) > 0.Consider aninterval on which the tangent plane to the surface is never parallel to the plane {z= 0}.Onecan then takez as a parameter and using the formulae in example 2 of3.3, we have to solvethe differential equation:

x

(1 + x2)3/2 =

1

x(1 + x2)1/2.

Multiplying both sides byx(1 + x2)1/2 and integrating, we find:

log(1 + x2

) = logx2

a2 ,

for somea > 0. Taking the exponentials and integrating again we find the (maximal) solu-tions:

x= a coshz

a, zR


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R. SOUAM 65

where a > 0 is a parameter. The solution curves are thus homothetic to the catenary,x = cosh z, z

R. The minimal surfaces so obtained are called catenoids. The previous

computation shows that if the tangent plane of a minimal surface of revolution is not parallelto the plane {z= 0} at some point then the same happens at each point. So the catenoids arethe only minimal surfaces of revolution besides the planes.

Exercise(The Laplacian). Let = 2

x2 + 2

y2 = 4 2

zz denote the usual Laplacian in R2.

(1) LetU, Vbe open subsets ofC and : U V a holomorphic (resp. anti-holomor-phic) function and finally letfC2(V, R).Check that, for anyzU,

(f )(z) =|(z)|2(f)((z)) (resp. |(z)|2(f)((z))).

(2) The Proposition 3.11 suggests to define an operator acting on smooth functions defined

on a Riemannian surfaceSin the following way: Let f C2(S,R) and X : UX(U)

Sa conformal parametrization of an open subset X(U)ofSwhere the metric

takes the formds2 =2(du2 + dv2).Put

Sf(q) =(f X)(X1(q))

2 , qU

where denotes the usual Laplacian in the coordinates (u, v).Show that this definesa global operator onC2(S,R)by checking the invariance under conformal change ofcoordinates.Sis called the Laplacian on the (Riemannian) surface S. It is linear andelliptic and is of fundamental importance in Riemannian geometry. It can be defined

without using conformal coordinates but one needs to introduce the notion of covariant

derivative.

3. 6 The Gauss and Codazzi-Mainardi equations

We have seen in the Proposition 2.3 that defining a curve is equivalent, up to rigid motions,

to giving an arc length and a curvature function. The analogous problem for surfaces in R3 is

formulated as follows:

Problem. LetU R2 be an open set. LetE , F , G, e, f, g : U R be smooth functionssatisfying E >0, G > 0, EG F2 >0. Under which conditions is it possible to find aroundeach pointpU, a neighborhood Vand an embedding X :V R3 whose first fundamentalform isEdu2 + 2Fdudv+ Gdv2 and whose second fundamental form isedu2 + 2fdudv+gdv2?

The Gauss and Codazzi-Mainardi equations give necessary and sufficient compatibility

conditions on the functionsE,F,G,e,f andg for the problem to have a solution. The ideais to look first for the trihedron X

u, X

v, N :U

R

3.If a solution exists, then this trihedronsatisfies a linear differential system:

Xuu= 111Xu+

211Xv+ eN,

Xuv = 112Xu+

212Xv+ fN,


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Xvu = 121Xu+

221Xv+ fN,

Xvv =

1

22Xu+

2

22Xv+ gN,Nu= a11Xu+ a21Xv,

Nv =a12Xu+ a22Xv,

where we used the same notations as previously. The coefficientsijk are called theChristof-

fel symbols in the coordinates (u, v). By definition 111 is the first component ofXuu inthe basis{Xu, Xv, N} etc... It can be proved they depend only on the functionsE , F andG and their derivatives (cf. [2]). Note that Xuv = Xvu implies the symmetry relations:112 =

121,

212 =

221. The integrability conditions onE , F , G, e, f, g in the previous sys-

tem are derived from the identities:

(Xuu)v = (Xuv)u, (Xvv)u = (Xvu)v, Nuv = Nvu.

One thus obtains the following compatibility conditions:

The Gauss equation:

(212)u (211)u+ 112211+ 212212 211222 111212 =E eg f2EG F2 ,

The Codazzi-Mainardi equations:

ev fu= e112+ f(212 111) g211,fv gu= e122+ f(222 112) g212.

It turns out these conditions are also sufficient:

Theorem 3.15 If The Gauss and Codazzi-Mainardi equations are satisfied then for each

p U, there exists a neighborhoodV ofp in U and an embeddingX : V X(V) in R3such that the surface X(V)has Edu2 + 2Fdudv + Gdv2 andedu2 + 2fdudv + gdv2 as first

and second fundamental forms respectively. IfV is connected andX : V X(V) R3

is another solution to the problem, then there exists a translation Tand a rotationO(3)such that

X= T X.In fact one shows the system is integrable, which gives a trihedron Y1, Y2, Nand then one

findsXsuch thatXu = Y1 and Xv =Y2.The integrability condition for this new system isguaranteed by the symmetries of the Christoffel symbols. For a proof see [2] or [7].

Remark 3.4. One can define the notion of (sectional) curvature for an abstract Riemannian

surface and the Gauss equation expresses the fact that this curvature coincides with the Gauss

curvature when the surface is immersed in R3.This is the Gauss theorema egregium whichexpresses the remarkable fact that the Gaussian curvature, although defined using the extrinsic

unit normal, is in fact completely intrinsic, i.e depends only on the first fundamental form. In

conformal coordinates where the metric writes ds2 =2(du2 + dv2), the Gaussian curvatureis given by the formula:

K=(log )2

,

being the usual Laplacian in the coordinates (u, v).


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R. SOUAM 67

Further reading: We did not address the global theory of surfaces (immersed in the Eu-

clidean space or abstract ones). This a rich subject. We strongly recommend the standard

texbook [2] as well as the recent one [5]. More information on curves can be found in [1].Chapters 10 and 11 of [1] offer a nice guide with very interesting comments for the study of

local and global geometry of surfaces. [3] gives a nice and complete introduction to different

topics in Riemannian geometry with overview on some recent research works. For the notion

of curvature in Riemannian geometry and its meaning the delightful survey articles [6] and

[4] are strongly recommended.

References

[1] M. Berger and B. Gostiaux.: Geometrie differentielle: varietes, courbes et surfaces.

Deuxieme edition.Presses universitaires de France, (1992).

[2] M. P. do Carmo.: Differential Geometry of Curves and Surfaces. Prentice-Hall Inc.,Englewood Cliffs,N.J., (1976).

[3] S. Gallot, D. Hulin and J. Lafontaine.:Riemannian geometry. Third edition.Universi-

text. Springer-Verlag, Berlin, (2004).

[4] M. Gromov.: Sign and geometric meaning of curvature. Rend. Sem. Mat. Fis. Milano

61 (1991), 9123 (1994).

[5] S. Montiel and A.Ros: Curves and Surfaces. Graduate Studies in Mathematics, Vol.

69, American Mathematical Society, Providence, RI, (2005).

[6] R. Osserman.:Curvature in the eighties.Amer. Math. Monthly 97 (1990), no. 8, 731

756.

[7] J.J. Stoker.: Differential Geometry.Wiley Interscience, New York,(1969).

Rabah Souam

Institut de Mathematiques de Jussieu - CNRS UMR 7586

Universite Paris Diderot - Paris 7, Geometrie et Dynamique

Case 7012 75205 - Paris Cedex 13

e-mail: [email protected]

Basic geometry of curves

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