Differential Geometry

James Emery

5/29/2015

Contents

1 Introduction 2

2 Curvature and Elementary Differential Geometry 2

3 Frame Vectors 6

4 The Serret-Frenet Formulas 6

5 The Reflection of a Ray From a Surface Tangent Plane 7

6 Curve Examples 8

6.1 The Circular Involute . . . . . . . . . . . . . . . . . . . . . . . 8

7 The Covariant Derivative in Rn 13

8 The Weingarten Map, The Shape Operator 14

9 Normal Curvature 16

10 Bilinear Forms 17

11 Quadratic Approximation 20

12 The Tangent Vector as a Derivation 20

13 Envelope Surfaces 23

1

14 The Covariant Derivative On a Surface 25

15 Christoffel Symbols 25

16 The Intrinsic Covariant Derivative 27

17 The Covariant Derivative in Riemmanian Geometry 28

18 The Bracket 30

19 Differential Manifolds 31

20 Riemannian Geometry 32

21 The Unique Riemannian Connexion Defined by the Riemannian

or semi-Riemannian Metric 33

22 Bibliography 33

1 Introduction

Differential Geometry is the study of curves and surfaces and their abstractgeneralization: the differential manifold. More generally it is the study ofthe calculus of curves and surfaces and involves definitions of curve tangents,normals, and curvature. In higher dimensions we study surfaces and man-ifolds and various kinds of curvature and so on. Some related subjects areVector Analysis, Algebraic Geometry, Tensor Analysis, Differential Topology,Metrics and metric spaces, Riemanian Geometry, Differential Manifolds, andGeneral Manifold Theory, and so on. Areas of physics that use differentialgeometry include General Relativity Theory, various subjects in TheoreticalPhysics, Mechanics, Particle theory, and String Theory.

2 Curvature and Elementary Differential Geom-

etry

Curvature is the ratio of the change in turning to the distance traveled.Consider the circular path. As a point on the circle moves through an angle

2

change ∆θ, it moves a distance ∆s = r∆θ. The ratio is a measure of thecurvature

∆θ

∆s=

∆θ

r∆θ=

1

r.

The angle change of the tangent ∆φ is here equal to the angle change ∆θ, sowe can use the tangent angle in our definition of the curvature. So supposewe are given a general curve in the plane

r = x(t)i + y(t)j.

Suppose dx/dt 6= 0, then the angle of the tangent is

φ = tan−1

[

dy/dt

dx/dt

]

.

Let us writedx

dt= x,

dy

dt= y,

d2x

dt2= x,

d2y

dt2= y.

Then we havedφ

dt=

1

1 + (y/x)2

yx− yx

x2

=yx− yx

x2 + y2.

We haveds

dt=

√

(dx/dt)2 + (dy/dt)2 =√

x2 + y2

So the curvature κ is

κ =dφ

ds=dφ/dt

ds/dt

=yx− yx

x2 + y2

1√x2 + y2

.

3

=yx− yx

(x2 + y2)3/2.

We excluded the case where dx/dt = 0. However, we can just as welldefine the tangent angle as

φ = cot−1

[

dx/dt

dy/dt

]

.

If we carry out a similure derivation to the one above we shall find that weget the same formula for the curvature. Hence as long as at least one ofdx/dt or dy/dt is not zero, the formula holds.

Suppose we have a function y = f(x). This is a curve with t = x as theparameter. Then x = 1, x = 0, and the curvature formula reduces to

κ =dφ

ds=

y

(1 + y2)3/2,

where

y =d2y

dx2,

and

y =dy

dx.

Notice that in this formula the curvature can be either positive or negative. Inthe case of a 1-dimensional function the curvature determines if the functionis concave up or down. In the more general definition in three space, thecurvature is always positive. So the new curvature will be defined to be theabsolute value of the 2d curvature defined above.

In three dimensional space there is no obvious tangent angle. So we mustdefine curvature using another approch. Let

r = x(t)i + y(t)j + z(t)k.

The velocity is

v =dr

dt= dx/dt(t)i + dy/dt(t)j + dz/dt(t)k.

The magnitude of the velocity is

v =√

x2 + y2 + z2 =ds

dt.

4

The unit tangent vector T is defined as

T =v

v=dr/dt

ds/dt=dr

ds.

T is a unit vector soT · T = 1.

We haved(T · T)

ds= dT/ds · T + T · dT/ds = 0,

which implies that2T · dT/ds = 0.

So T and it derivative are orthogonal. Thus dT/ds is a vector normal to thecurve. The unit normal vector N is defined as

N =dT/ds

‖dT/ds‖ .

ThusdT/ds = ‖dT/ds‖N.

We can define the curvature κ as

κ = ‖dT/ds‖,

because we can show that for a two dimensional curve this agrees with thetwo dimensional curvature.

Indeed, in two dimensions the unit tangent vector T can be written as

T = cos(φ)i + sin(φ)j,

where the tangent angle φ is a function of the arc length s. Then

dT/ds = − sin(φ)(dφ/ds)i + cos(φ)(dφ/ds)j.

So‖dT/ds‖ = |dφ/ds|

√

sin2(φ) + cos2(φ) = |dφ/ds| = κ.

Notice that here the curvature is always nonnegative. We can thereforewrite

dT/ds = κN.

5

3 Frame Vectors

Let the curve R(s) in Euclidean 3-space be parameterized by arc length.Define the tangent vector by

T =dR

ds,

and the curvature by

κ =

∣

∣

∣

∣

∣

dT

ds

∣

∣

∣

∣

∣

.

The normal vector is defined to be

N =1

κ

dT

ds,

and the binormal vector is defined by

B = T × N.

The vectors T,N, and B, are called the frame vectors.

4 The Serret-Frenet Formulas

There are a set of formulas relating the frame vectors T,N, and B and theirderivatives.

The derivatives of the frame vectors with respect to arc length s are equalto linear combinations of the frame vectors themselves.

These are called the Serret-Frenet formulas. The Serret-Frenet formulasare derived from the facts that the frame vectors are mutually perpendicular,and that they have unit length. The dot product of any pair of frame vectorsis zero. So the derivative of their dot product is also zero. Unit vectors areperpendicular to their derivatives, and N is a unit vector. So dN/ds is per-pendicular to N. Consequently dN/ds can be written as a linear combinationof T and B only. Thus

dN

ds= a1T + a3B.

Because T is perpendicular to N,

a1 =dN

ds· T = −N · dT

ds.

6

By definition the right hand expression is equal to −κ. So we conclude thata1 is equal to the curvature κ.

Define the torsion τ to be a3. Thus

dN

ds= −κT + τB.

LetdB

ds= b1T + b2N.

b1 =dB

ds· T = −B · dT

ds= −B · N = 0.

ThendB

ds· N = −B · dN

ds= −B · (−κT + τB) = −τ.

ThusdB

ds= −τN.

Therefore we have the Serret-Frenet transformation,

T′

N′

B′

=

0 κ 0−κ 0 τ0 −τ 0

T

N

B

Notice that the matrix of the transformation is antisymmetric, and that thetop right element of the matrix is zero. An antisymmetric matrix has a zerodiagonal. The lower triangular part has negative elements. These facts mightaid one in remembering the formula.

5 The Reflection of a Ray From a Surface

Tangent Plane

Suppose the ray intersects the surface at a point p. Let n be a unit normalto the tangent plane at point p. Then −n is a second unit normal to theplane. It does not matter which of these normals we choose. Let the ray berepresented by vector a. The component of a in the direction of n is

b = (n · a)n.

7

Notice that if we use the other normal we get the same vector b,

(−n · a)(−n) = (n · a)n = b.

The vector b is actually the projection of vector a to the line orthogonal tothe tangent plane.

We subtract this component from a, getting

c = a − b,

which is the component of a orthogonal to n. So

a = b + c.

For the reflected ray, the component in the direction of n is the negative of theoriginal component, but the orthogonal component is in the same directionas the original orthogonal component. Therefore the reflected ray is

d = −b + c = c − b.

That isd = (a− b) − b = a − 2b.

As a check we see that

n · d = n · a− 2n · b = −n · a.

6 Curve Examples

6.1 The Circular Involute

The involute of a curve C is a curve generated by unwinding a string wrappedaround C.

In the case of a circle of radius a centered at the origin, let t be theangle at the string tangent point on the circle, as the string is unwrapped togenerate the involute. Then we find the parametric equation of the involuteto be

x = a(cos(t) + t sin(t))

y = a(sin(t) − t cos(t)).

8

Figure 1: Two caustics (envelope curves of reflected rays) from the involutecurve. The two curves are formed from two sets of parallel rays, the firstmaking a zero angle with the horizontal, and the second making a twentydegree angle with the horizontal. The smaller caustic comes from the zeroangle rays.

9

Figure 2: A caustic (envelope curve of reflected rays) from the involute curve.This curve is produced from reflections from a set of parallel rays making anangle of zero degrees with the horizontal axis. Notice that the cusp pointoccurs at a height equal to the height of the circle.

10

Figure 3: Reflections of parallel rays from the right at zero degrees from aninvolute curve.

11

The position vector of the curve is

R(t) = xi + yj.

We have the derivatives

x =dx

dt= r(− sin(t) + sin(t) + t cos(t)) = rt cos(t)

y =dy

dt= r(cos(t) − cos(t) + t sin(t)) = rt sin(t)

x =d2x

dt2= r(cos(t) − t sin(t))

y =d2y

dt2= r(sin(t) + t cos(t)),

and the derivative of the arclength

ds

dt=

√

x2 + y2 = rt

The tangent vector is

T =dR

ds=dR

dt

dt

ds

= cos(t)i + sin(t)j

The normal vector is

N = κdT

ds= κ

dT

dt

dt

ds

= − sin(t)i + cos(t)j.

c from involute.ftn

c+ function fx definition

real*8 function fx(t)

implicit real*8(a-h,o-z)

a=1.

c=1.

fx=a*(cos(t) + c*t*sin(t))

return

end

c+ function fy definition

real*8 function fy(t)

implicit real*8(a-h,o-z)

a=1.

c=1.

fy=a*(sin(t) - c*t*cos(t))

return

end

12

7 The Covariant Derivative in Rn

Given any curve α in Rn with P = α(0) and

dα

dt(0) = X,

the covariant derivative of the vector field Y in the direction X at the pointP is

∇XY (P ) =d

dtY (α)(0).

The chain rule may be used to show that the definition does not depend onthe curve α.

This definition may be applied to a two dimensional surface patch em-bedded in R3. Let φ be a surface patch mapping an open set in R2 to R3.Suppose β is a curve in the domain of φ so that α(t) = φ(β(t)). Then

∇XY (P ) =d

dtY (α) =

=dY (φ(β(t)))

dt

≈ Y (φ(β+)) − Y (φ(β−))

4t .

This is a central difference approximation where

β− = β(−4t/2),

andβ+ = β(4t/2).

The tangent vector X is

X =dφ(β(t))

dt

≈ φ(β+) − φ(β−)

4t .

Richardson extrapolation can be applied to these central difference approx-imations. An intrinsic covariant derivative may be defined on the patch,which is derived from the covariant derivative in R3. This is accomplishedby subtracting the normal component of the R3 covariant derivative (seeHicks, p26).

13

8 The Weingarten Map, The Shape Operator

Let N be a unit normal vector field on the surface. The Weingarten map,sometimes called the shape operator, is the mapping from the tangent spaceat P to itself, defined by

W (X) = ∇XN(P ),

where X is in the tangent space at P . The Weingarten map is linear and it issymmetric. To prove that it is symmetric, let φ(u, v) be a coordinate patch.Then φu and φv are linearly independent tangent vectors and so a basis forthe tangent space. We have

W (φu) · φv = ∇φuN(φ) · φv

=∂N(φ)

∂u· φv

=∂N(φ)

∂u· ∂φdv

= −N(φ) · ( ∂∂u

∂φ

dv)

= −N(φ) · ( ∂∂v

∂φ

du)

=∂N(φ)

∂v· ∂φdu

= W (φv) · φu.

We have used the fact that N is perpendicular to the tangent space andhence to tangent vectors

∂φ

∂v,

and∂φ

∂u.

We will present a method for determining W numerically when surfacepatch values are available, but derivatives values are not. In addition wesuppose that the surface normal is available (although it could be computednumerically from patch values).

14

Let β1 be a curve in the uv coordinate space defined by

β1(t) = (u0, v0) + tT1,

where T1 = (1, 0) is a tangent vector in the uv coordinate space. Let α1(t)be the curve φ(β1(t)). Let

X1 =dα1(0)

dt.

Similarly define

X2 =dα2(0)

dt,

where the uv tangent vector is T2 = (0, 1). If Φ has nonzero Jacobian thenX1 and X2 are independent vectors in the tangent space. X1 and X2 are notnecessarily perpendicular to one another.

Let Z1 and Z2 be orthogonal vectors in the tangent space, which meansthat Z1 and Z2 are in R3, and are perpendicular to N at P . Z1 and Z2 area basis of the tangent space, so there exists bij so that

X1 = b11Z1 + b12Z2

andX2 = b21Z1 + b22Z2.

We have

bij =Xi · Zj

‖Zi‖2.

Then using the linearity of W we get the vector equations

W (X1) = b11W (Z1) + b12W (Z2)

andW (X2) = b21W (Z1) + b22W (Z2).

We may solve these equations for the x, y and z components of W (Z1) andW (Z2). Then let

W (Z1) = c11Z1 + c12Z2

andW (Z2) = c21Z1 + c22Z2.

Z1 and Z2 are perpendicular, so we have

15

cij =Zi ·W (Zj)

‖Zi‖2.

The matrix representation of the Weingarten transformation in this basis is.

[

c11 c21c12 c22

]

9 Normal Curvature

Define the normal curvature of a surface in the direction of a curve α, at apoint p, by

κn = −W (α′) · α′,

where a(0) = p, and α′ = α′(0). This turns out to be the component of thecurvature of the curve α in the direction of the surface normal, when α isparameterized by arc length. Let N be the surface normal restricted to α.Then N is a curve, and

W (α′) = ∇α′N =dN(α)

ds= N ′.

We haveα′ ·N = 0.

Soα′′ ·N = −α′ ·N ′.

But

α′′ =dα′

ds=dt

ds= κn.

Soκn = −N ′ · α′ = α′′ ·N = κn ·N = κ cos(ψ),

where ψ is the angle between the surface normal N and the curve normal n.From the definition one sees that the normal curvature depends only uponthe direction of the unit vector t = α′. Hence notice that given two curvesthat pass through a given point p and have the same direction at that pointhave the same surface normal component of their curvature vector. Also ifthe curve lies in a plane which contains the surface normal, then the curve

16

normal coincides with the surface normal so the normal curvature is the sameas the curvature of the curve. If λ is an eigenvalue of W with eigenvector tthen the normal curvature is

κn = −W (t) · t = −λt · t = −λ.

The direction of t is called a principal direction. W is a symmetric trans-formation so the eigenvalues are real and the eigenvectors are orthogonal.The principal directions are well defined except at an umbilic point wherethe eigenvalues are equal and every vector is an eigenvector. The normalcurvature is a quadratic form and the maximum and minimum values of theform occur at the eigenvalues of W . Thus the principal curvatures are themaximum and the minimum normal curvatures.

10 Bilinear Forms

The classical fundamental bilinear forms defined on pairs of tangent vectorsinclude the first fundamental form

B1(T1, T2) = T1 · T2,

and the second fundamental form

B2(T1, T2) = W (T1) · T2.

The quadratic form derived from the first fundamental form is

Q1(T ) = T · T.

This is the metric form.In classical differential geometry a tangent vector is written as

T = ds =n

∑

i=1

∂X

∂uidui.

Then the first fundamental quadratic form would be written as

Q1(ds) = ds · ds =n

∑

i=1

n∑

j=1

gijduiduj.

17

g is called the metric tensor. Suppose X(u, v) is a coordinate patch, then wehave

ds2 = Q1(ds) = Xu ·Xudu2 + 2Xu ·Xvdudv +Xv ·Xvdv

2.

= Edu2 + 2Fdudv +Gdv2.

As an example, consider a spherical patch of radius a,

X(u, v) = (a cos(v) sin(u), a sin(v) sin(u), a cos(u)).

ThenXu(u, v) = (a cos(v) cos(u),−a sin(v) cos(u),−a sin(u)).

andXv(u, v) = (−a sin(v) sin(u),−a cos(v) sin(u), 0).

ThusE = Xu ·Xu = a2

F = Xu ·Xv = 0

andG = Xv ·Xv = a2 sin2(u).

The second fundamental quadratic form is

Q2(ds) · ds = W (Xu) ·Xudu2 + 2W (Xu) ·Xvdudv +W (Xv) ·Xvdv

2.

= edu2 + 2fdudv + gdv2.

Let N be a unit normal. Then

0 = Xu ·N = Xv ·N,

soe = W (Xu) ·Xu = Xu ·Nu = −Xuu ·N,f = W (Xv) ·Xu = Xu ·Nv = −Xuv ·N,

andg = W (Xv) ·Xv = Xv ·Nv = −Xvv ·N.

The normal curvature in the direction of T is

κn = −W (T/‖T‖) · T/‖T‖ = −W (T ) · TT · T = −Q2(T )

Q1(T ).

18

Proposition. The Gaussian curvature is given by

K(X) =eg − f 2

EG− F 2,

and the mean curvature by

H(X) =Ge−Eg − 2Ff

2(EG− F 2),

Proof. Let T1 and T2 be linearly independent tangent vectors. Let a be thecoefficient matrix of W with respect to this basis. Write

W (Ti) = ajiTj

ThenW (T1) ×W (T2) = Det(a)T1 × T2 = KT1 × T2.

and

W (T1) × T2 + T1 ×W (T2) = 2Trace(a)T1 × T2 = 2HT1 × T2.

These equations can be solved for the Gaussian curvature K and the meancurvature H by dotting each equation by

T1 × T2.

We have(W1 ×W2) · (T1 × T2) = K(T1 × T2) · (T1 × T2).

The left side is

(W1 ×W2) · (T1 × T2) = W1 · (W2 × (T1 × T2))

= W1 · (T1(W2 · T2) − T2(W2 · T1))

= (W1 · T1)(W2 · T2) − (W1 · T2)(W2 · T1)

= (W1 · T1)(W2 · T2) − (W1 · T2)2.

We have used the BAC − CAB rule. Similarly

(T1 × T2) · (T1 × T2) = (T1 · T1)(T2 · T2) − (T1 · T2)2.

19

Thus

K =(W1 · T1)(W2 · T2) − (W1 · T2)

2

(T1 · T1)(T2 · T2) − (T1 · T2)2

IfT1 = Xu

andT2 = Xv,

then

K(X) =eg − f 2

EG− F 2.

The mean curvature equation is proved similarly.

11 Quadratic Approximation

Let p be a point on the surface. We may assume a coordinate system inwhich p is the origin and the normal is in the z direction. Thus we may writethe surface as

z = f(x, y).

We expand f in a Taylor series. Keeping only second order terms we obtaina quadratic approximation to the surface. Now we could fit local data toa quadric of this form using least squares and then obtain approximationsto the principal curvatures. However, the least squares quadratic goes tothe quadratic approximation only as the data approaches p. In effect we areapproximating second derivatives, while the previous technique requires onlycovariant derivatives of the first order.

12 The Tangent Vector as a Derivation

Let c be a curve with parameter t. A tangent to the curve is v = dc/dt. Letf be a function. We define

Dv(f) =df(c(t))

dt= ∇f · dc

dt= ∇f · v.

This defines a functional Dv mapping functions to real numbers. It dependsonly on the tangent vector v, not specifically on the curve c. It does not

20

depend on the coordinate system because

df(c(t))

dt

is coordinate independent. From the theory of differential equations, given atangent vector v at a point p, there exists a curve c so that

dc(0)

dt= v,

and c(0) = p. Thus for every tangent vector there is a functional Dv. Thefunctional has the obvious properties

Dv(f + g) = Dv(f) +Dv(g)

andDv(ag) = aDv(f),

where a is a scalar. Dv is a linear functional. We have

Dv(fg) =df(c)g(c)

dt= f(c)

dg(c)

dt+df(c)

dtg(c) =

f(c)Dv(g) +Dv(f)g(c).

Functionals satisfying these properties are called derivations. Consider theexample of the x coordinate curve in Cartesian three space. We have c(t) =tux where ux is the unit coordinate vector in the x direction. Then

v =dc

dt= ux

and

Dv(f) = ∇f · ux =∂f

∂x.

We have shown that

Dv =∂

∂x.

Similarly if ψi is a local surface coordinate, then

∂

∂ψi,

21

is the derivation corresponding to its tangent vector. We have

∂

∂ψj(ψi) = δi

j .

Hence

A = { ∂

∂ψi

: 1 ≤ i ≤ n}

is a linearly independent set of derivations. We shall show that tangentvectors and derivations can be identified. So A is a basis of the tangentspace. Let v1 and v2 be tangent vectors ,then

Dv1+v2f = ∇f · (v1 + v2) =

∇f · v1 + ∇f · v2 = Dv1f +Dv2

f

If a is a scalar, thenDav1

= aDv1.

Therefore the mappingv 7−→ Dv

is a homomorphism. Since the tangent space of an n-dimensional manifoldis n-dimensional, and there are tangent vectors mapping to A, the mappingv 7−→ Dv is actually an isomorphism.

Now we no longer have to rely on a coordinate system in the embeddingspace, in which the surface lies, to define a tangent vector. Given any curvein the surface c, and a parameter point t0, there is a derivation defined by

Dcf =df(c)

dt(t0).

This is independent of the embedding. A local surface coordinate system canbe used to find the components of Dc with respect to the local tangent basis.

The covariant derivative ∇XY can be given as the action of the derivationX on the vector field Y . Indeed let

X = x1∂

∂u1+ x2

∂

∂u2+ x3

∂

∂u3

and

Y = y1∂

∂u1+ y2

∂

∂u2+ y3

∂

∂u3

22

be tangent vectors in R3 and let X act as a derivation. Then

∇XY = (X(y1), X(y2), X(y3)).

To prove this let c be a curve so that c′ = X. Then

∇XY =∂Y (c(t))

dt=

(∂y1(c(t))

dt,∂y2(c(t))

dt,∂y3(c(t))

dt) = (X(y1), X(y2), X(y3)).

Reference Abraham Goetz, ”Introduction to Differential Geometry,” Addison-Wesley 1970.

13 Envelope Surfaces

Given a one parameter family of surfaces

f(x, y, z, t) = 0,

the envelope surface is defined by eliminating t from the equations

f(x, y, z, t) = 0

and∂f(x, y, z, t)

∂t= 0

(see Goetz p125). To establish this, let the envelope surface have parametricequation

r(u1, u2) = (x(u1, u2), y(u1, u2), z(u1, u2)).

Given the coordinates of a point u1u2 of a point p on the envelope, there isan element of the family of surfaces that is tangent to the envelope at thispoint. Let the corresponding parameter of this tangent surface be t(u1, u2).Now ∇f is normal to this tangent surface. Hence

∇f · drdu1

= 0.

But let

23

g(u1, u2) = f((x(u1, u2), y(u1, u2), z(u1, u2), t(u1, u2)).

Then g = 0, so

0 =dg

du1= ∇f · dr

du1+∂f(x, y, z, t)

∂t

dt

du1.

Thus∂f(x, y, z, t)

∂t

dt

du1= 0.

Similarly∂f(x, y, z, t)

∂t

dt

du2

= 0.

We shall assume thatdt

du1

anddt

du2

do not vanish simultaneously. Thus

∂f(x, y, z, t)

∂t= 0.

A canal surface is the envelope of a family of spheres. For example thetorus is a canal surface, and Dupin cyclides are also canal surfaces. A spheri-cal Milling cutter produces a canal surface, other cutter shapes produce otherenvelope surfaces. A rather general milling cutter may be considered the en-velope of a family of flat disk like ellipsoids. Then a milled surface may beconsidered the envelope of a two parameter family of surfaces

f(x, y, x, s, t) = 0.

The envelope surface is found by eliminating t and s from

f(x, y, z, s, t) = 0, ft(x, y, z, s, t) = 0, fs(x, y, z, s, t) = 0.

See Goetz.The milled internal screw thread is an envelope of surfaces generated by

swept disks. It can also be considered a the envelope of a two parameterfamily of algebraic surfaces and is thus algebraic.

An envelope of a one-parameter set of planes is called a developable sur-face.

24

14 The Covariant Derivative On a Surface

Define the covariant derivative on the surface by

DXY = ∇XY+ < ∇XN, Y > N.

We will show that this is the projection of the R3 covariant derivative to thetangent space. For if we take the dot product with the normal N , we get

DXY ·N =< ∇XY,N > + < ∇XN, Y >= X < Y,N >= 0.

X, as a derivation, is simply a differential operator. Thus the expression onthe right is a differential operator operating on an inner product, and Y isorthogonal to N , so it is zero. Then we have

< ∇XN, Y >= − < ∇X , N > .

Then we haveDXY = ∇XY− < ∇XY,N > N,

which shows that DXY is the projection of ∇XY to the tangent space.The covariant derivative has an explicit coordinate representation, given

in terms of the soon to be introduced Christoffel symbols.

15 Christoffel Symbols

Let φ(u1, u2) be a surface patch (an inverse coordinate system). Write

φij =∂2φ

∂ui∂uj.

We may express these derivatives in the moving basis

{φ1, φ2, N}

We haveφij = Γk

ijφk + βijN.

The Γkij are called the Christoffel symbols of the second kind. Taking inner

products we find that the βij are the coefficients bij of the second fundamentalform. Thus

βij =< φij, n >= bij , .

25

whereb11 = E, b12 = b21 = F, b22 = G.

Let us write the derivative of the normal as

ni = −bki φk.

Then we get< ni, φj >= −bki gkj.

By differentiating< n, φj >= 0,

we get< ni, φj >= − < n, φij >= −bij

Thusbij = bki gkj.

Thenbijg

kj = bki gkjgkj = bki δ

lk = bli,

where{gij} = {gij}−1.

The Christoffel symbols are symmetric. In fact φij and bij are symmetric,so

Γkijφk = φij − bijn = φji − bjin = Γk

jiφk.

ThenΓk

ij = Γkji,

because the φi are linearly independent.The Christoffel symbols of the first kind are defined by

Γijl =< φij , φl >=< Γkijφk + bijn, φl >= Γk

ij < φk, φl >= Γkijgkl.

They are symmetric in the first two indices because of the symmetry of φij .It follows that

Γkij = Γiljg

lk.

Theorem

Γijk =1

2(∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk),

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and

Γkij =

gkl

2(∂gjl

∂ui+∂gli

∂uj− ∂gij

∂ul).

Proof. We have

∂gij

∂uk=∂ < φi, φj >

∂uk=< φik, φj > + < φi, φjk >=

Γikj + Γjki.

The permutationi→ k, j → i, k → j

gives

∂gki

∂uk= Γkji + Γijk = Γjki + Γijk,

and the permutationi→ j, j → k, k → i

gives∂gjk

∂ui= Γjik + Γkij = Γijk + Γikj.

Adding the last two equations and subtracting the first, we obtain the resultwhich was to be proved.

16 The Intrinsic Covariant Derivative

Let φ(u1, u2) be a coordinate patch. Let X be the tangent to curve c(t) =(u1(t), u2(t)). Let

Y = y1φ1 + y2φ2 = yiφi

Then the covariant derivative is

∇XY =dY (c)

dt=dyi

dtφi + yiφij

duj

dt

We may substitute the expressions involving the Christoffel symbols for thesecond derivatives, obtaining a linear combination of the basis vectors

{φ1, φ2, N}.

27

Taking the projection into the tangent space, (deleting terms involving N),we get an expression for the surface covariant derivative

DXY =dyi

dtφi + yiΓ

kijφk

dyj

dt=

dyk

dtφk + yiΓk

ijφkduj

dt=

(dyk

dt+ yiΓk

ij

duj

dt)φk

which involves the basis vectors

{φ1, φ2},

and the Christoffel symbols. The Christoffel symbols involve only the coef-ficients of the first fundamental form and their derivatives. So the surfacecovariant derivative is intrinsic, and depends only on the metric gij.

Let X = φm and Y = φn. Then

dyk

dt= 0

andyi = δi

n,

andduj

dt= δj

m.

ThenDφm

φn = Γkmnφk.

17 The Covariant Derivative in Riemmanian

Geometry

In Hicks, the covariant derivative of vector field Y with respect to vector fieldX is written as

DXY.

28

We shall continue to write it as

∇XY.

Above we introduced the derivation corresponding to vector v. A deriva-tion D is a linear functional on a manifold that has the properties:

(1)D(f + g) = D(f) +D(g)

(2)D(af) = aD(f),

(3)D(fg) = f(c)D(g) +D(f)g(c).

The derivations play the roll of tangent vectors. The set of derivations con-stitute the tangent space. A derivation may also be defined as a parametricderivative along a curve.

In the Euclidean case suppose we have a vector

X = (a1, a2, ...., an)

Then a directional derivative in the direction of this vector is a derivationand the operation on a function f is given by

Xf =n

∑

i=1

ai∂f

∂ui.

In this case it is easy to see that property (3) of the derivation follows fromthe rule for diferentiating the product of functions. Further in this Euclideancase one can show that every derivation is of this form. If Y is a vector fieldin this Euclidean space with ith coordinate yi, then a covarient derivative isgiven by

∇XY = (Xy1, ...., Xyn).

Covariant derivatives have the following four characteristic properties:

(1)∇X(Y + Z) = ∇XY + ∇XZ

(2)∇X+WY = ∇XY + ∇WZ

(3)∇f(p)XY = f(p)∇XY

(4)∇Xf(p)Y = (Xf)Y + f(p)∇XY

Property (4) follows from the property (3) of derivations.

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Given a curve σ and a vector field Y defined along the curve, if T is thecurve tangent and

∇TY = 0,

then the vector Y (0) is said to be parallel translated along σ.If

∇TT = 0,

then the curve tangent is parallel translated along the curve σ and thecurve is a geodesic, a shortest path connecting the starting point and theending point.

Consider the case of translating a tangent vector around a path on asphere, say the earth. Say one starts at the north pole and suppose wetranslate a vector along the zero degree meridian. Let the vector Y be per-pendicular to the meridian as it is translated, that is it is always pointingeast, always making the same angle with the meridian, i.e. the geodesiccurve.

When we reach the equator the vector points east in the direction of theequator, which is another geodesic. We keep it in the direction of the equatoruntil we get to say the 30thth degree meridian. The vector makes a 90 degreeangle with the 30th degree meridian as we translate it back up to the northpole. When we reach the north pole we find that the translated vector ispointing in a different direction from the starting direction. So on the earthwe have traveled along a spherical triagle path, always keeping our vectorpointing in the same direction with respect to what we consider locally asstraight lines. But when we return to our starting point we see that thevector has been turned. So this tells us that our surface is curved. Becauseon a flat surface this would not occur.

The geodesics are the paths of particles when they do not experience anyforce. Their motion is the analogue of classical uniform motion.

18 The Bracket

If X and Y are C∞ vector fields, their bracket is a C∞ vector field (HicksP8) defined by

[X, Y ]f = X(Y f) − Y (Xf).

The bracket has the following properties:

[X, Y ] = −[Y,X].

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[X,X] = 0.

[X1 +X2, Y ] = [X1, Y ] + [X2, Y ]

and[fX, gY ] = f(Xg)Y − g(Y f)X + fg[X, Y ].

At a fixed point p on the manifold,

[X, Y ]pf

is a functional, mapping f to a real number. For a fixed p and a fixed f ,

[X, Y ]pf

maps two vectors from the product tangent space to a real number. If it weremultilinear with respect to X and Y , then it would be a second rank tensor.But property

[fX, gY ] = f(Xg)Y − g(Y f)X + fg[X, Y ],

shows that this is not the case.The bracket satisfies the Jacoby identity

[X, [X,Z]] + [Z, [X, Y ]] + [Y, [Z,X]] = 0.

The bracket comes from the Poisson bracket of mechanics, and was intro-duced by Sophus Lie in his work on Lie Groups (originally called continuousgroups of transformations) and Lie Algebras in solving partial differentialequations.

19 Differential Manifolds

A differential manifold is a topological space M and a set of charts, whereeach chart Φ is a one to one mapping from an open neighborhood of M toan open connected neighborhood in Rn. If Φ1 and Φ2 are overlapping charts,then Φ−1

1 Φ2 is a C∞ map. M is covered by the domains of the set of charts.The charts specify a set of local coordinates for the manifold. For exampleon the spherical earth, the mapping of a point on the earth to its latitudeand longitude is a chart. It takes more than one chart to cover the earth.Let us take the example of a 2d unit sphere embedded in R3. The use of

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spherical coordinates constitutes a chart. That is a point on the sphere ismapped to the φ, θ coordinates. The inverse of this chart is the mapping

x = sin(θ) cos(φ)

y = sin(θ) sin(φ)

z = cos(θ)

Tangent vectors in these coordinate directions are the functionals (deriva-tions)

∂

∂φ, and

∂

∂θ.

20 Riemannian Geometry

A Riemannian Space is a differential manifold with a Riemannian metric. ARiemannian Metric is a function < X, Y >, where X and Y are elements ofa tangent space. This is the distance between X and Y . In classical notationthis is written as g(X,Y), and has tensor components gij. The Riemannianmetric is real valued, bilinear, symmetric, and positive definite. The metric isa quadratic form and is represented by a symmetric matrix. Positive definitemeans that if X is not zero, then g(X,X) > 0. The eigenvalues of the matrixare all positive.

A Semi-Riemannian metric is real valued, bilinear, and symmetric. Butis not necessarily positive definite, but must be nonsingular. The eigenval-ues must be nonzero, but not necessarily positive. This is the case for theMinkowski metric of relativity theory. A positive definite matrix is clearlynonsinglular, so a Riemannian metric is a Semi-Riemannian metric. Clearly,there can be a nonsingular matrix that is not positive definite, namely a diag-onal matrix. A Riemannian metric tensor is given classically as a symmetricquadratic form

ds2 =n

∑

i=1

n∑

j=1

gijdxidxj .

where ds is the infinitesimal distance.

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21 The Unique Riemannian Connexion De-

fined by the Riemannian or semi-Riemannian

Metric

The properties 1-6 of the standard connexion on Rn (Hicks pp19-20) aresatisfied by this Riemannian connection properties 5 and 6 are additionalproperties not necessarily satisfied by a covariant derivative or connexionthat is not Riemannian.

There exists a unique Riemannian connexion (pp 69-70) on a Riemannianor semi-Riemannian manifold. The Christoffel symbols are given in terms ofthe metric tensor. We have by definition

∇Xk(Xj) =

n∑

i=1

ΓijkXi.

From the properties of the covariant derivative or connexion we have theChristoffel symbols in terms of the metric

Γkij =

1

2

∑

r

(g−1)kr(∂grj

∂xi+∂gri

∂xj− ∂gij

∂xr).

The symbol g−1 is the inverse of the symmetric matrix of the metric quadraticform. This is where the nonsingularity of the semi-Riemannian metric isneeded. This defines the connexion, because it has been defined for the basisvector fields Xi.

22 Bibliography

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and Lie Algebras with Applications and Computational Methods,1972, SIAM.

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