Riemannian Geometry

it is a draft of Lecture Notes of H.M. Khudaverdian.Manchester, 28 April 2017

Contents

1 Riemannian manifolds 11.1 Manifolds. Tensors. (Recalling) . . . . . . . . . . . . . . . . . 1

1.1.1 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Tensors on Manifold . . . . . . . . . . . . . . . . . . . 31.1.3 Push-forward and pull-back . . . . . . . . . . . . . . . 5

1.2 Riemannian manifold . . . . . . . . . . . . . . . . . . . . . . . 61.2.1 Riemannian manifold— manifold equipped with Rie-

mannian metric . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Pull-back of Riemannian metric . . . . . . . . . . . . . 91.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.4 ∗ Pseudoriemannian manifold . . . . . . . . . . . . . . 111.2.5 Scalar product→ Length of tangent vectors and angle

between them . . . . . . . . . . . . . . . . . . . . . . 121.2.6 Length of curves . . . . . . . . . . . . . . . . . . . . . 131.2.7 Conformally Euclidean metric . . . . . . . . . . . . . . 15

1.3 Riemannian structure on the surfaces embedded in Euclideanspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.3.1 Internal and external coordinates of tangent vector . . 171.3.2 Formulae for induced metric . . . . . . . . . . . . . . . 191.3.3 Induced Riemannian metrics. Examples. . . . . . . . . 211.3.4 Inversion and metric on circle and sphere in stero-

graphic cooridnates . . . . . . . . . . . . . . . . . . . . 28

1

1.3.5 ∗Induced metric on two-sheeted hyperboloid embeddedin pseudo-Euclidean space. . . . . . . . . . . . . . . . . 29

1.4 Isometries of Riemannian manifolds. . . . . . . . . . . . . . . 311.4.1 Isometries of Riemannian manifold (on itself) . . . . . 32

1.5 ∗Infinitesimal isometries of Riemannian manifold (Killing vec-tor fields) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.5.1 Locally Euclidean Riemannian manifolds . . . . . . . 35

1.6 Volume element in Riemannian manifold . . . . . . . . . . . . 371.6.1 Motivation: Gramm formula for volume of parallelepiped 381.6.2 ∗Invariance of volume element under changing of co-

ordinates . . . . . . . . . . . . . . . . . . . . . . . . . 391.6.3 Examples of calculating volume element . . . . . . . . 40

2 Covariant differentiaion. Connection. Levi Civita Connec-tion on Riemannian manifold 432.1 Differentiation of vector field along the vector field.—Affine

connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.1.1 Definition of connection. Christoffel symbols of con-

nection . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.1.2 Transformation of Christoffel symbols for an arbitrary

connection . . . . . . . . . . . . . . . . . . . . . . . . . 462.1.3 Canonical flat affine connection . . . . . . . . . . . . . 472.1.4 ∗ Global aspects of existence of connection . . . . . . . 50

2.2 Connection induced on the surfaces . . . . . . . . . . . . . . . 512.2.1 Calculation of induced connection on surfaces in E3. . . 52

2.3 Levi-Civita connection . . . . . . . . . . . . . . . . . . . . . . 542.3.1 Symmetric connection . . . . . . . . . . . . . . . . . . 542.3.2 Levi-Civita connection. Theorem and Explicit formulae 542.3.3 Levi-Civita connection of En . . . . . . . . . . . . . . . 562.3.4 Levi-Civita connection on 2-dimensional Riemannian

manifold with metric G = adu2 + bdv2. . . . . . . . . . 562.3.5 Example of the sphere again . . . . . . . . . . . . . . . 57

2.4 Levi-Civita connection = induced connection on surfaces in E3 572.5 ∗ Killing vectors, antisymmetric operator and antisymmetric

bilinear form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3 Parallel transport and geodesics 613.1 Parallel transport . . . . . . . . . . . . . . . . . . . . . . . . . 61

2

3.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 613.1.2 Parallel transport is a linear map. Parallel transport

with respect to Levi-Civita connection . . . . . . . . . 623.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.2.1 Definition. Geodesic on Riemannian manifold . . . . . 643.2.2 Geodesics and Lagrangians of ”free” particle on Rie-

mannian manifold. . . . . . . . . . . . . . . . . . . . . 653.2.3 Calculations of Christoffel symbols and geodesics us-

ing the Lagrangians of a free particle. . . . . . . . . . . 673.2.4 † Magnitudes preserved along geodesics—Integrals of

motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.2.5 ∗ Variational principe and Euler-Lagrange equations . 723.2.6 Un-parameterised geodesic . . . . . . . . . . . . . . . . 733.2.7 Parallel transport of vectors along geodesics . . . . . . 743.2.8 Geodesics on surfaces in E3 . . . . . . . . . . . . . . . 743.2.9 ∗ Geodesics and shortest distance. . . . . . . . . . . . . 763.2.10 ∗ Again geodesics for sphere and Lobachevsky plane . . 78

4 Surfaces in E3 804.1 Parallel transport of the vector. Formulation of result. . . . . 80

4.1.1 Theorem of parallel transport over closed curve . . . . 804.1.2 Gauß Theorema Egregium . . . . . . . . . . . . . . . . 834.1.3 †Formula for Gaussian curvature in isothermal (confor-

mal) coordinates . . . . . . . . . . . . . . . . . . . . . 844.2 Derivation formula . . . . . . . . . . . . . . . . . . . . . . . . 85

4.2.1 Gauss condition (structure equations) . . . . . . . . . . 884.2.2 Geometrical meaning of derivation formula. Weingarten

operator (shape oeprator) in terms of derivation for-mula. . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

4.2.3 Gaussian and mean curvature in terms of derivationformula . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4.3 Calculations with use of derivation formulae . . . . . . . . . . 914.3.1 Examples of calculations of derivation formulae and

curvatures for cylinder, cone and sphere . . . . . . . . 914.3.2 †Curvatures for surface z = F (x, y) . . . . . . . . . . . 964.3.3 ∗Proof of the Theorem of parallel transport along

closed curve . . . . . . . . . . . . . . . . . . . . . . . . 99

3

4.3.4 † Proof of the Theorem on curvature of surfaces givenin conformal coordinates using derivation formulae . . 102

5 Curvature tensor 1055.1 Curvature tensor for connection . . . . . . . . . . . . . . . . . 105

5.1.1 Properties of curvature tensor . . . . . . . . . . . . . . 1065.2 Riemann curvature tensor of Riemannian manifolds. . . . . . . 107

5.2.1 Curvature of surfaces in E3.. Theorema Egregium again 1095.2.2 Relation between Gaussian curvature and Riemann

curvature tensor and Theorema Egregium . . . . . . . 1105.2.3 † Straightforward proof of the Proposition (5.27) . . . . 113

5.3 †Gauss Bonnet Theorem . . . . . . . . . . . . . . . . . . . . . 116

6 Appendices 1186.1 ∗Integrals of motions and geodesics. . . . . . . . . . . . . . . . 118

6.1.1 ∗Integral of motion for arbitrary Lagrangian L(x, x) . . 1186.1.2 ∗Basic examples of Integrals of motion: Generalised

momentum and Energy . . . . . . . . . . . . . . . . . . 1196.1.3 ∗Integrals of motion for geodesics . . . . . . . . . . . . 1206.1.4 ∗Using integral of motions to calculate geodesics . . . . 1226.1.5 ∗Killing vectors of Lobachevsky plane and geodesics . . 122

6.2 Induced metric on surfaces. . . . . . . . . . . . . . . . . . . . 1246.2.1 Recalling Weingarten operator . . . . . . . . . . . . . . 1246.2.2 Recalling Weingarten operator . . . . . . . . . . . . . . 1256.2.3 Second quadratic form . . . . . . . . . . . . . . . . . . 1266.2.4 Gaussian and mean curvatures . . . . . . . . . . . . . . 1276.2.5 Examples of calculation of Weingarten operator, Sec-

ond quadratic forms, curvatures for cylinder, cone andsphere. . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

i

1 Riemannian manifolds

1.1 Manifolds. Tensors. (Recalling)

1.1.1 Manifolds

I recall briefly basics of manifolds and tensor fields on manifolds.An n-dimensional manifold is a space such that in a vicinity of any point

one can consider local coordinates x1, . . . , xn (charts). One can considerdifferent local coordinates. If both coordinates x1, . . . , xn, x1′ , . . . , xn

′are defined in a vicinity of the given point then they are related by bijectivetransition functions (functions defined on domains in Rn and taking valuesin Rn).

x1′ = x1′(x1, . . . , xn)

x2′ = x2′(x1, . . . , xn)

. . .

xn−1′ = xn−1′(x1, . . . , xn)

xn′= xn

′(x1, . . . , xn)

(1.1)

We say that n-dimensional manifold is differentiable or smooth if transitionfunctions are diffeomorphisms, i.e. they are smooth and rank of Jacobian isequal to k, i.e.

det

∂x1′

∂x1∂x1′

∂x2 . . .∂x1′

∂xn

∂x2′

∂x1∂x2′

∂x2 . . .∂x2′

∂xn

. . .∂xn

′

∂x1∂xn

′

∂x2 . . .∂xn

′

∂xn

6= 0 . (1.2)

A good example of manifold is an open domain D in n-dimensional vectorspace Rn. Cartesian coordinates on Rn define global coordinates on D. Onthe other hand one can consider an arbitrary local coordinates in differentdomains in Rn. E.g. one can consider polar coordinates r, ϕ in a domainD = x, y : y > 0 of R2 defined by standard formulae:

x = r cosϕ

y = r sinϕ, (1.3)

det

(∂x∂r

∂x∂ϕ

∂y∂r

∂y∂ϕ

)= det

(cosϕ −r sinϕsinϕ r cosϕ

)= r (1.4)

1

or one can consider spherical coordinates r, θ, ϕ in a domainD = x, y, z : x >0, y > 0, z > 0 of R3 (or in other domain of R3) defined by standard for-mulae

x = r sin θ cosϕ

y = r sin θ sinϕ

z = r cos θ

,

,

det

∂x∂r

∂x∂θ

∂x∂ϕ

∂y∂r

∂y∂θ

∂y∂ϕ

∂z∂r

∂z∂θ

∂z∂ϕ

= det

sin θ cosϕ r cos θ cosϕ −r sin θ sinϕsin θ sinϕ r cos θ sinϕ r sin θ cosϕ

cos θ −r sin θ 0

= r2 sin θ

(1.5)Choosing domain where polar (spherical) coordinates are well-defined we

have to be award that coordinates have to be well-defined and transitionfunctions (1.1) have to obey condition (1.2), i.e. they have to be diffeomor-phisms. E.g. for domain D in example (1.3) Jacobian (1.4) does not vanishif and only if r > 0 in D.

Examples of manifolds: Rn, Circle S1, Sphere S2, in general sphere Sn,torus S1 × S1, cylinder, cone, . . . .

Example Consider in detail circle S1. Suppose it is given as x2 + y2 = 1.One can consider two local polar coordinates: 1) ϕ, 0 < ϕ < 2π which coversall points except the point (1, 0) and ϕ′ : −π < ϕ′ < π which covers all pointsexcept point (−1, 0). In a vicinity of any point M on the circle (except thesetwo exceptional points) one can consider both coordinates ϕ and ϕ′

We come to another very useful coordinates on a circle using stereographicprojection. Take north pole of the circle: the point N = (0, 1). Assign toevery point M = (x, y) on the circle the point (t, 0) on the x-axis such thatthe point (t, 0), the point M and the north pole N are on the one line. Thiscan be done for every point of circle except the north pole (0, 1) itself. Wecome to stereographic projection of circle S1 without North pole on the lineR. In the same way we can define stereographic projection of circle withoutsouth pole (the point (0,−1)) on the x-axis. We come to coordinate t′. Onecan see that these coordinates are related by the following simple formula:

t′ =1

t,

† One very important property of stereographic projetion which we do not use in

this course but it is too beautiful not to mention it: under stereographic projection

2

all points on the circle x2 + y2 = 1 with rational coordinates x and y and only

these points transform to rational points on line. Thus we come to Pythagorean

triples a2 + b2 = c2.

1.1.2 Tensors on Manifold

Recall briefly what are tensors on manifold. For every point p on manifoldM one can consider tangent vector space TpM— the space of vectors tangentto the manifold at the point M .

Tangent vector A(x) = Ai(x) ∂∂xi

. Under changing of coordinates it trans-forms as follows:

A = Ai(x)∂

∂xi= Ai(x)

∂xm′(x)

∂xi∂

∂xm′ = Am′(x′(x))

∂

∂xm′ .

Hence

Ai′(x′) =

∂xi′(x)

∂xiAi(x) . (1.6)

Consider also cotangent space T ∗pM (for every point p on manifold M)—space of linear functions on tangent vectors, i.e. space of 1-forms whichsometimes are called covectors.:

One-form (covector) ω = ωi(x)dxi transforms as follows

ω = ωm(x)dxm = ωm∂xm(x′)

∂xm′ dxm′= ωm′(x′)dxm

′.

Hence

ωm′(x′) =∂xm(x′)

∂xm′ ωm(x) . (1.7)

Differential form sometimes is called covector.Tensors:

One can consider contravariant tensors of the rank p

T = T i1i2...ip(x)∂

∂xi1⊗ ∂

∂xi2⊗ · · · ⊗ ∂

∂xik

with components T i1i2...ik(x). Under changing of coordinates (x1, . . . , xn)→(x1′ , . . . , xn

′) (see (1.1)) they transform as follows:

T i′1i

′2...i

′p(x′) =

∂xi′1

∂xi1∂xi

′2

∂xi2. . .

∂xi′p

∂xipT i1i2...ip(x) . (1.8)

3

One can consider covariant tensors of the rank q

S = Sj1j2...jqdxj1 ⊗ dxj2 ⊗ . . . dxjq

with components Sj1j2...jq. Under changing of coordinates (x1, . . . , xn) →(x1′ , . . . , xn

′) they transform as follows:

Sj′1j′2...j′q(x′) =

∂xi1

∂xi′1

∂xi2

∂xi′2. . .

∂xip

∂xi′pSj1j2...jq(x) . (1.9)

One can also consider mixed tensors:

Q = Qi1i2...ipj1j2...jq

∂

∂xi1⊗ ∂

∂xi2⊗ · · · ⊗ ∂

∂xik⊗ dxj1 ⊗ dxj2 ⊗ . . . dxjq

with components Qi1i2...ipj1j2...jq

. We call these tensors tensors of the type

(pq

).

Tensors of the type

(p0

)are called contravariant tensors of the rank p. They

have p upper indices.

Tensors of the type

(0q

)are called covariant tensors of the rank q. They

have q lower indices.Having in mind (1.6), (1.7),(1.8) and (1.9) we come to the rule of trans-

formation for tensors which have p upper and q lover indices, tensors of type(pq

):

Qi′1i

′2...i

′p

j′1j′2...j

′q(x′) =

∂xi′1

∂xi1∂xi

′2

∂xi2. . .

∂xi′p

∂xip∂xj1

∂xj′1

∂xj2

∂xj′2. . .

∂xjq

∂xj′qQi1i2...ipj1j2...jq

(x) .

E.g. if Sik is a covariant tensor of rank 2 then

Si′k′(x′) =

∂xi(x′)

∂xi′∂xk(x′)

∂xk′Sik(x) . (1.10)

If Aik is a tensor of rank

(11

)(linear operator on TpM) then

Ai′

k′(x′) =

∂xi′(x)

∂xi∂xk(x′)

∂xk′Aik(x) .

4

If Smik is a tensor of the type

(12

)) then

Sm′

i′k′ =∂xm

′

∂xm∂xi

∂xi′∂xk

∂xk′Smik (x) . (1.11)

Transformations formulae (1.6)—(1.11) define vectors, covectors and ingenerally any tensor fields in components. E.g. covariant tensor (covari-ant tensor field) of the rank 2 can be defined as matrix Sik (matrix valuedfunction Sik(x)) such that under changing of coordinates x1, x2, . . . , xn 7→x1′ , x2′ , . . . , xn

′, (1.1) Sik change by the rule (1.10).Remark Einstein summation rulesIn our lectures we always use so called Einstein summation convention. it

implies that when an index occurs twice in the same expression in upper andin lower postitions, then the expression is implicitly summed over all possiblevalues for that index. Sometimes it is called dummy indices summation rule.

1.1.3 Push-forward and pull-back

Let F : M → N be a smooth map from a manifold M to manifold N . Wehave “zoo” of geometrical objects on both manifolds: functions, functions ontangent space1 vectors tangent at given points, vector fields, tensors attachedat the point, and tensor fields, e.t.c. How they transform under the map FIt turns out that even in the case if F is not diffeomorphism (e.g. if F is amap of m dimensional manifold on n-dimensional, m 6= n) a map F inducescanonical transformations of some objects on N to objects on M or viceversa.

Consider important examples.Example 1 Let x0 be a point on manifold M and y0 be its image under

the map F : y0 = F (x0). Let xi be local coordinates in the vicinity of thepoint x0, and let ya be local coordinates in the vicinity of the point y0,(i = 1, . . .m, a = 1, . . . , n in general we do not suppose that m = n.) Wesuppose that in these local coordinates map F has appearance

ya = ya(xi) = F a(xi) .

An arbitrary vector A tangent to M at the point x0, A ∈ Tx0M , A = Ai ∂∂i

defines a vector B = F∗A tangent to manifold N at the point y0, B = F∗A ∈1i.e. Lagrangians

5

Tx0 such that

Ba∣∣y0=F (x0)

= (F∗A)a∣∣ =

∂ya(x)∂

xi∣∣x0Ai(x) .

Remark Push-forward F∗ if tangent vector is the action of differentialdF of the map F on the vector A. Sometimes push forward F∗ is denotedjust dF .

Remark Note that in general one cannot define push-forward Fp of vectorfield A = A(x) on M . since map F in general may map different points inthe same point.

We do not go in detail of defining a smooth map of smooth manifolds.

Now consider operation pull backAs usual Let F : M → N be a smooth map from a manifold M to mani-

fold N . One can see that a map F defines pull-back F ∗ of functions, forms,Lagrangians from N to M

An arbitrary function g = g(y) on N defines a function f = F ∗g on M :

f : = (F ∗g)(x) = g(F (x)) (1.12)

An arbitrary differential form ω on N defines a form F ∗ω on M . Forexample if ω is 1-form ω = Ra(y)dya then

(F ∗ω) = ωa(ya(x))dya

∣∣ya=ya(x)

= (1.13)

For example if F : R2 → R, t = t(x, y) = xey, then this map assigns tothe every 1=form ω = f(t)dt the 1=form on R2:

(F ∗f(t)dt) = f(t(x, y))

(∂t

∂xdx+

∂t

∂ydy

)= f(xey)(eydx+ xeydy) .

(See discussion of pull-back also in ?? below.)

1.2 Riemannian manifold

1.2.1 Riemannian manifold— manifold equipped with Rieman-nian metric

Definition The Riemannian manifold is a manifold equipped with a Rie-mannian metric.

6

The Riemannian metric on the manifold M defines the length of thetangent vectors and the length of the curves.

Definition Riemannian metric G on n-dimensional manifold Mn definesfor every point p ∈ M the scalar product of tangent vectors in the tangentspace TpM smoothly depending on the point p.

It means that in every coordinate system (x1, . . . , xn) a metric G =gikdx

idxk is defined by a matrix valued smooth function gik(x) (i = 1, . . . , n; k =1, . . . n) such that for any two vectors

A = Ai(x)∂

∂xi, B = Bi(x)

∂

∂xi,

tangent to the manifoldM at the point p with coordinates x = (x1, x2, . . . , xn)(A,B ∈ TpM) the scalar product is equal to:

〈A,B〉G∣∣p

= G(A,B)∣∣p

= Ai(x)gik(x)Bk(x) =

(A1 . . . An

)g11(x) . . . g1n(x). . . . . . . . .

gn1(x) . . . gnn(x)

B1

···Bn

(1.14)

where

• G(A,B) = G(B,A), i.e. gik(x) = gki(x) (symmetricity condition)

• G(A,A) > 0 if A 6= 0, i.e.

gik(x)uiuk ≥ 0, gik(x)uiuk = 0 iff u1 = · · · = un = 0 (positive-definiteness)

• G(A,B)∣∣p=x

, i.e. gik(x) are smooth functions.

The matrix ||gik|| of components of the metric G we also sometimes denoteby G.

Now we establish rule of transformation for entries of matrix gik(x), ofmetric G.

Notice that an arbitrary matrix entry gik is nothing but scalar productof vectors ∂i, ∂k at the given point:

gik(x) =

⟨∂

∂xi,∂

∂xk

⟩, in coordinates (x1, . . . , xn) (1.15)

7

Use this formula for establishing rule of transformations of gik(x). In the newcoordinates xi

′= (x1′ , . . . , xn

′) according this formula we have that

gi′k′(x′) =

⟨∂

∂xi′,∂

∂xk′

⟩, in coordinates (x1, . . . , xn) .

Now using chain rule, linearity of scalar product and formula (??) we seethat

gi′k′(x′) =

⟨∂

∂xi′,∂

∂xk′

⟩=

⟨∂xi

∂xi′∂

∂xi,∂xk

∂xk′∂

∂xk

⟩=∂xi

∂xi′

⟨∂

∂xi,∂

∂xk

⟩︸ ︷︷ ︸

gik(x)

∂xk

∂xk′=

∂xi

∂xi′gik(x)

∂xk

∂xk′(1.16)

This transformation law justifies that gik entries of matrix ||gik|| are compo-nents of covariant tensor field G = gikdx

idxk of rank 2(see equation (1.10)).One can say that Riemannian metric is defined by symmetric covariant

smooth tensor field G of the rank 2 which defines scalar product in the tangentspaces TpM smoothly depending on the point p. Components of tensor fieldG in coordinate system are matrix valued functions gik(x):

G = gik(x)dxi ⊗ dxk . (1.17)

In practice it is more convenient to perform transformation of metric G underchanging of coordinates in the following way:

G = gikdxi ⊗ dxk = gik

(∂xi

∂xi′dxi

′)⊗(∂xk

∂xk′dxk

′)

=

∂xi

∂xi′gik

∂xk

∂xk′dxi

′ ⊗ dxk′ = gi′k′dxi′ ⊗ dxk′

Hence

gi′k′ =∂xi

∂xi′gik

∂xk

∂xk′. (1.18)

We come to transfomation rule (1.16).Later by some abuse of notations we sometimes omit the sign of tensor

product and write a metric just as

G = gik(x)dxidxk .

8

1.2.2 Pull-back of Riemannian metric

Return again to pull-back (see 1.74).Let M be a Riemannian manifold M equipped with Riemannina metric

GM , and N be an arbitrary manifold.One can see that an arbitrary map F : N →M induces pull-back of GM

on Riemannian metric GN = (F ∗GM). The Riemannian metric GN is definedby the following relation: if A,B ∈ TxN are two vectors tangent to manifoldN at an arbitrary point x ∈ N then

GN(A,B) = (F ∗GM)(A,B) = GM(F∗A, F∗B) . (1.19)

In particular if A = ∂∂xi

, B = ∂∂xj

are local coordinate tangent vectors, then

GN(∂i, ∂j) = (F ∗GM)(∂i, ∂j) = GM

(∂ya

∂xi∂

∂ya,∂yb

∂xj∂

∂yb

)=∂ya(x)

∂xi(gM)ab

∂yb(x)

∂xj,

i.e.

GN(∂i, ∂j) = (F ∗GM)(∂i, ∂j) = GM

(∂ya

∂xi∂

∂ya,∂yb

∂xj∂

∂yb

)=∂ya(x)

∂xiGM

(∂

∂ya,∂

∂yb

)∂yb(x)

∂xj,

Now using (1.15) we see that in local coordinates

(gN)ik(x) = (F ∗gM)ik =∂ya(x)

∂xigM

(∂

∂ya,∂

∂yb

)∂yb(x)

∂xj, (1.20)

1.2.3 Examples

• Rn with canonical coordinates xi and with metric

G = (dx1)2 + (dx2)2 + · · ·+ (dxn)2

G = ||gik|| = diag [1, 1, . . . , 1]

Recall that this is a basis example of n-dimensional Euclidean space,where scalar product is defined by the formula:

G(X,Y) = 〈X,Y〉 = gikXiY k = X1Y 1 +X2Y 2 + · · ·+XnY n .

In the general case if G = ||gik|| is an arbitrary symmetric positive-definite metric then G(X,Y) = 〈X,Y〉 = gikX

iY k. One can show that

9

there exists a new basis ei such that in this basis G(ei, ek) = δik. Thisbasis is called orthonormal basis. (See the Lecture notes in Geometry)

Scalar product in vector space defines the same scalar product at allthe points. In general case for Riemannian manifold scalar productdepends on a point. In Riemannian manifold we consider arbitrarytransformations from local coordinates to new local coordinates.

• R2 with polar coordinates in the domain y > 0 (x = r cosϕ, y =r sinϕ):

dx = cosϕdr − r sinϕdϕ, dy = sinϕdr + r cosϕdϕ. In new coordi-nates the Riemannian metric G = dx2 + dy2 will have the followingappearance:

G = (dx)2+(dy)2 = (cosϕdr−r sinϕdϕ)2+(sinϕdr+r cosϕdϕ)2 = dr2+r2(dϕ)2

We see that for matrix G = ||gik||

G =

(gxx gxygyx gyy

)=

(1 00 1

)︸ ︷︷ ︸

in Cartesian coordinates

, G =

(grr grϕgϕr gϕϕ

)=

(1 00 r2

)︸ ︷︷ ︸

in polar coordinates

• Circle

Interval [0, 2π) in the line 0 ≤ x < 2π with Riemannian metric

G = a2dx2 (1.21)

Renaming x 7→ ϕ we come to habitual formula for metric for circle ofthe radius a: x2 + y2 = a2 embedded in the Euclidean space E2:

G = a2dϕ2

x = a cosϕ

y = a sinϕ, 0 ≤ ϕ < 2π, (1.22)

• Domain in R2 with metric G = du2 + u2dv2

(Compare with R2 with polar coordinates).

• Cylinder surface

10

Domain in R2 D = (x, y) : , 0 ≤ x < 2π with Riemannian metric

G = a2dx2 + dy2 (1.23)

We see that renaming variables x 7→ ϕ, y 7→ h we come to habitual,familiar formulae for metric in standard polar coordinates for cylindersurface of the radius a embedded in the Euclidean space E3:

G = a2dϕ2 + dh2

x = a cosϕ

y = a sinϕ

z = h

, 0 ≤ ϕ < 2π,−∞ < h <∞

(1.24)

• Sphere

Domain in R2, 0 < x < 2π, 0 < y < π with metric G = dy2 + sin2 ydx2

We see that renaming variables x 7→ ϕ, y 7→ h we come to habitual,familiar formulae for metric in standard spherical coordinates for spherex2 + y2 + z2 = a2 of the radius a embedded in the Euclidean space E3:

G = a2dθ2+a2 sin2 θdϕ2

x = a sin θ cosϕ

y = a sin θ sinϕz = a cos θ, 0 ≤ ϕ < 2π,−∞ < h <∞

(1.25)

(See examples also in the Homeworks.)

1.2.4 ∗ Pseudoriemannian manifold

If we omit the condition of positive-definiteness for Riemannian metric wecome to so called Pseudoriemannian metric. Manifold equipped with pseu-doriemannian metric is called pseudoriemannian manifold. Pseudorieman-nian manifolds appear in applications in the special and general relativitytheory.

In pseudoriemanninan space scalar product (X,X) may take an arbitraryreal values: it can be positive, negative, it can be equal to zero. Vectors Xsuch that (X,X) = 0 are called null-vectors. (See the problem 6 in Homework1).

11

Example Consider n+1-dimensional linear space Rn+1 with pseudomet-ric 2

G = (dx0)2 − (dx1)2 − (dx2)2 − · · · − (dxn)2 .

For an arbitrary vector X = (a0, a1, a2, . . . , an) scalar product (X,X) is pos-itive if (a0)2 > (a1)2 + (a2)2 + · · · + (an)2, it is negative if (a0)2 < (a1)2 +(a2)2 + · · ·+ (an)2, and X is null-vector if (a0)2 = (a1)2 + (a2)2 + · · ·+ (an)2.

1.2.5 Scalar product → Length of tangent vectors and angle be-tween them

The Riemannian metric defines scalar product of tangent vectors attachedat the given point. Hence it defines the length of tangent vectors and anglebetween them. If X = Xm ∂

∂xm,Y = Y m ∂

∂xmare two tangent vectors at the

given point p of Riemannian manifold with coordinates x1, . . . , xn, then wehave that lengths of these vectors equal to

|X| =√〈X,X〉 =

√gik(x)X iXk, |Y| =

√〈Y,Y〉 =

√gik(x)Y iY k,

(1.26)and an ‘angle’ θ between these vectors is defined by the relation

cos θ =〈X,Y〉|X| · |Y|

=gikX

iY k√gik(x)X iXk

√gik(x)Y iY k

(1.27)

Remark We say ‘angle’ but we calculate just cosinus of angle.Example Let M be 3-dimensional Riemannian manifold. Consider the

vectors X = 2∂x+2∂y−∂z and Y = ∂x−2∂y−2∂z attached at the point p ofM with local coordinates (x, y, z), where x = y = 1, z = 0. Find the lengthsof these vectors and angle between them if the expression of Riemannianmetric in these coordinates is

G =dx2 + dy2 + dz2

(1 + x2 + y2 + z2)2. (1.28)

We see that matrix of Riemannian metric gik = σ(x, y, z)δik, where σ(x, y, z) =1

(1+x2+y2+z2)2 is a scalar function, i.e. matrix G = ||gik|| is proportional tounity matrix. According to formulae above

|X| =√〈X,X〉 =

√gik(x, y, z)X iXk =

√σ(x, y, z)

√X iX i = 3

√σ(x, y, z) = 1 .

2In the case n = 3 it is so called Minkovski space. The coordinate x0 plays a role ofthe time: x0 = ct, where c is the value of the speed of the light. Vectors X such that(X,X) > 0 are called time-like vectors and they called space-like vectors if (X,X) < 0

12

The same answer for |Y|. The scalar product between vectors X,Y equal tozero:

〈X,Y〉 = σ(x, y, z)δikXiY k = 0

Hence these vectors are unit vectors which are orthogonal to each other.(δik is Kronecker symbol: δik = 1 if i = k and it vanishes otherwise.)

This example is related with the notion of so called conformally euclideanmetric (see 1.2.7)

1.2.6 Length of curves

Let γ : xi = xi(t), (i = 1, . . . , n)) (a ≤ t ≤ b) be a curve on the Riemannianmanifold (M,G).

At the every point of the curve the velocity vector (tangent vector) isdefined:

v(t) =

x1(t)···

xn(t)

The length of velocity vector v ∈ TxM (vector v is tangent to the manifold

M at the point x) equals to

|v|x =√〈v,v〉G

∣∣x

=√gikvivk

∣∣x

=

√gikdxi(t)

dt

dxk(t)

dt x

∣∣x.

For an arbitrary curve its length is equal to the integral of the length ofvelocity vector:

Lγ =

∫ b

a

√〈v,v〉G

∣∣x(t)dt =

∫ b

a

√gik(x(t))xi(t)xk(t)dt . (1.29)

Bearing in mind that metric (1.17) defines the length we often write metricin the following form

ds2 = gikdxidxk

For example consider 2-dimensional Riemannian manifold with metric

||gik(u, v)|| =(g11(u, v) g12(u, v)g21(u, v) g22(u, v)

).

13

Then

G = ds2 = gikduidvk = g11(u, v)du2 + 2g12(u, v)dudv + g22(u, v)dv2 .

The length of the curve γ : u = u(t), v = v(t), where t0 ≤ t ≤ t1 according to(1.29) is equal to Lγ =

∫ t1t0

√〈v,v〉 =

∫ t1t0

√gik(x)xixk =∫ t1

t0

√g11 (u (t) , v (t))u2

t + 2g12 (u (t) , v (t))utvt + g22 (u (t) , v (t)) v2t dt .

(1.30)The length of curves defined by the formula(1.29) obeys the following

natural conditions

• It coincides with the usual length in the Euclidean space En (Rn withstandard metric G = (dx1)2 + · · · + (dxn)2 in Cartesian coordinates).E.g. for 3-dimensional Euclidean space

Lγ =

∫ b

a

√gik(x(t))xi(t)xk(t)dt =

∫ b

a

√(x1(t))2 + (x2(t))2 + (x3(t))2dt

• It does not depend on parameterisation of the curve

Lγ =

∫ b

a

√gik(x(t))xi(t)xk(t)dt =

∫ b′

a′

√gik(x(τ))xi(τ)xk(τ)dτ ,

(xi(τ) = xi(t(τ)), a′ ≤ τ ≤ b′ while a ≤ t ≤ b) since under changing ofparameterisation

xi(τ) =dx(t(τ))

dτ=dx(t(τ))

dt

dt

dτ= xi(t)

dt

dτ.

• It does not depend on coordinates on Riemannian manifold M

Lγ =

∫ b

a

√gik(x(t))xi(t)xk(t)dt =

∫ b

a

√gi′k′(x′(t))xi

′(t)xk′(t)dt .

This immediately follows from transformation rule (1.75) for Rieman-nian metric:

gi′k′xi′(t)xk

′(t) = gik

(∂xi

∂xi′(t)xi

′(t)

)(∂xk

∂xk′xk

′(t)

)gik x

i(t)xk(t) .

14

• It is additive: length of the sum of two curves is equal to the sumof their lengths. If a curve γ = γ1 + γ, i.e. γ : xi(t), a ≤ t ≤ b,γ1 : xi(t), a ≤ t ≤ c and γ2 : xi(t), c ≤ t ≤ b where a point c belongs tothe interval (a, b) then Lγ = Lγ1 + Lγ2 .

One can show that formula (1.29) for length is defined uniquely by these conditions.More precisely one can show under some technical conditions one may show that any localadditive functional on curves which does not depend on coordinates and parameterisation,and depends on derivatives of curves of order ≤ 1 is equal to (1.29) up to a constantmultiplier. To feel the taste of this statement you may do the following exercise:

Exercise Let A = A(x(t), y(t), dx(t)

dt , dy(t)dt

)be a function such that an integral

L =∫A(x(t), y(t), dx(t)

dt , dy(t)dt

)dt over an arbitrary curve γ in E2 does not change under

reparameterisation of this curve and under an arbitrary isometry, i.e. translation androtation of the curve. Then one can easy show (show it!) that

A

(x(t), y(t),

dx(t)

dt,dy(t)

dt

)= c

√(dx(t)

dt

)2

+

(dy(t)

dt

)2

,

where c is a constant, i.e. it is a usual length up to a multiplier

1.2.7 Conformally Euclidean metric

Definition We say that metric G is locally conformally Euclidean in a vicin-ity of the point p if in a vicinity of this point there exist local coordinatesxi such that in these coordinates metric has an appearance

G = σ(x)δikdxidxk = σ(x)

((dx1)2 + · · ·+ (dxn)2

), (1.31)

i.e. it is proportional to ‘Euclidean metric’. We call coordinates ui copnfor-mall coordinates or isothermic coordinates if condition (1.31) holds.

We say that metric is conformally Euclidean if it is locally conformallyEuclidean in the vicinity of every point. In the example above we consid-ered just conformally Euclidean metric. One can see that for conformallyEuclidean metric the angle between vectors (more precisely the cosinus ofthe angle) is the same as for Eucldean vectors.

It is evident that coefficient σ(x) in (1.31) has to be positive. It is conve-nient sometimes to denote it as σ(x) = eλ(x),

G = σ(x)δikdxidxk = eλ(x)δikdx

idxk , (1.32)

Let G be conformally Euclidean metric and let xi be local coordinates suchthat the metric has an appearnace (1.31) in these coordinates. Then it is easy

15

to see that the ‘angle’ between two vectors (1.27) (more exactly the cosinusof the angle) has the same appearance as for Euclidean case. Namely letwe have two non-vanishing vectors X = Xm(x) ∂

∂xm,Y = Y m(x) ∂

∂xm(|X| 6=

0, |Y| 6= 0). Then

cos θ =〈X,Y〉|X| · |Y|

=gikX

iY k√gik(x)X iXk

√gik(x)Y iY k

=

λ(x)δikXiY k√

λ(x)δik(x)X iXk√λ(x)δik(x)Y iY k

=

∑kX

kY k√∑k(x)XkXk

√∑k Y

kY k.

(1.33)(see also the example above)

We say that Riemannian manifold (M.G) is conformally Euclidean if themetric G on it is conformally Euclidean.

Later we cosnider many interesting examples of confromally EuclideanRiemannian manifolds.

1.3 Riemannian structure on the surfaces embeddedin Euclidean space

Let M be a surface embedded in Euclidean space. Let G be Riemannianstructure on the manifold M .

Let X,Y be two vectors tangent to the surface M at a point p ∈M . AnExternal Observer calculate this scalar product viewing these two vectors asvectors in E3 attached at the point p ∈ E3 using scalar product in E3. AnInternal Observer will calculate the scalar product viewing these two vectorsas vectors tangent to the surface M using the Riemannian metric G (see theformula (1.38)). Respectively

If L is a curve in M then an External Observer consider this curve as acurve in E3, calculate the modulus of velocity vector (speed) and the lengthof the curve using Euclidean scalar product of ambient space. An InternalObserver (”an ant”) will define the modulus of the velocity vector and thelength of the curve using Riemannian metric.

Definition Let M be a surface embedded in the Euclidean space. Wesay that metric GM on the surface is induced by the Euclidean metric if thescalar product of arbitrary two vectors A,B ∈ TpM calculated in terms ofthe metric G equals to Euclidean scalar product of these two vectors:

〈A,B〉GM = 〈A,B〉GEuclidean(1.34)

16

In other words we say that Riemannian metric on the embedded surface isinduced by the Euclidean structure of the ambient space if External andInternal Observers come to the same results calculating scalar product ofvectors tangent to the surface.

In this case modulus of velocity vector (speed) and the length of the curveis the same for External and Internal Observer.

Note that surface M embedded in Euclidean space is defined by embed-ding ι : M → E3. One can express equation (1.34) for induced metric in thefollowing way:

GM = ι∗GEuclidean (1.35)

Using this formula we will write down in 1.3.2 explicit formulae for inducedmetric on surfaces in Euclidean space

1.3.1 Internal and external coordinates of tangent vector

In this subsection we will just recall the material from the lectures of secondyear course Introduction to Geometry. In particular we recall the conceptionof Internal and External Observers when dealing with surfaces in Euclideanspace, and conception of First Quadratic form for surfaces in E3.

It will be useful to read this paragraph for all students.Tangent plane

Here we recall basic notions from the course of Geometry which we willneed here.

Let r = r(u, v) be parameterisation of the surface M embedded in theEuclidean space:

r(u, v) =

x(u, v)y(u, v)z(u, v)

Here as always x, y, z are Cartesian coordinates in E3.

Let p be an arbitrary point on the surface M . Consider the plane formedby the vectors which are adjusted to the point p and tangent to the surfaceM . We call this plane plane tangent to M at the point p and denote it byTpM .

For a point p ∈ M one can consider a basis in the tangent plane TpMadjusted to the parameters u, v. Tangent basis vectors at any point (u, v)

17

are

ru =∂r(u, v)

∂u=

∂x(u,v)∂u

∂y(u,v)∂u

∂z(u,v)∂u

=∂x(u, v)

∂u

∂

∂x+∂y(u, v)

∂u

∂

∂y+∂z(u, v)

∂u

∂

∂z

Every vector X ∈ TpM can be expanded over this basis:

X = Xuru +Xvrv,

where Xu, Xv are coefficients, components of the vector X.Internal Observer views the basis vector ru ∈ TpM , as a velocity vector

for the curve u = u0 + t, v = v0, where (u0, v0) are coordinates of the point p.Respectively the basis vector rv ∈ TpM for an Internal Observer, is velocityvector for the curve u = u0, v = v0 + t, where (u0, v0) are coordinates of thepoint p.

Explicit formulae for induced Riemannian metric (First Quadratic form)Now we are ready to write down the explicit formulae for the Rieman-

nian metric on the surface induced by metric (scalar product) in ambientEuclidean space (see the Definition (1.34) as we learnt it in the course of Ge-ometry. In fact this definition coicides with (1.35). We will return to inducedmetric again in 1.3.2.

Let M : r = r(u, v) be a surface embedded in E3.The formula (??) means that scalar products of basic vectors ru = ∂u, rv =

∂v has to be the same calculated in the ambient space and on the surface:For example scalar product 〈∂u, ∂v〉M = guv calculated by the Internal Ob-server is the same as a scalar product 〈ru, rv〉E3 calculated by the ExternalObserver, scalar product 〈∂v, ∂v〉M = guv calculated by the Internal Observeris the same as a scalar product 〈rv, rv〉E3 calculated by the External Observerand so on:

G =

(guu guvgvu gvv

)=

(〈∂u, ∂u〉 〈∂u, ∂v〉〈∂v, ∂u〉 〈∂v, ∂v〉

)=

(〈ru, ru〉E3 〈ru, rv〉E3

〈rv, ru〉E3 〈rv, rv〉E3

)(1.36)

where as usual we denote by 〈 , 〉E3 the scalar product in the ambient Eu-clidean space. (Here see also the important remark (1.15))

Remark It is convenient sometimes to denote parameters (u, v) as (u1, u2)or uα (α = 1, 2) and to write r = r(u1, u2) or r = r(uα) (α = 1, 2) insteadr = r(u, v)

18

In these notations:

GM =

(g11 g12

g12 g22

)=

(〈ru, ru〉E3 〈ru, rv〉E3

〈ru, rv〉E3 〈rv, rv〉E3

), gαβ = 〈rα, rβ〉 ,

GM = gαβduαduβ = g11du

2 + 2g12dudv + g22dv2 (1.37)

where ( , ) is a scalar product in Euclidean space.The formula (1.37) is the formula for induced Riemannian metric on the

surface r = r(u, v). It is First Quadratic Form of this surface.If X,Y are two tangent vectors in the tangent plane TpC then G(X,Y)

at the point p is equal to scalar product of vectors X,Y:

(X,Y) = (X1r1 +X2r2, Y1r1 + Y 2r2) = (1.38)

X1(r1, r1)Y 1 +X1(r1, r2)Y 2 +X2(r2, r1)Y 1 +X2(r2, r2)Y 2 =

Xα(rα, rβ)Y β = XαgαβYβ = G(X,Y)

1.3.2 Formulae for induced metric

Now after recalling the material of the scourse of Geometry return again tothe formula (1.35).

The Riemannian structure of Euclidean space—standard Euclidean met-ric in Euclidean coordinates is given by

GE3 = (dx)2 + (dy)2 + (dz)2 . (1.39)

Then the induced metric (1.35) on the surface M defined by equation r =r(u, v) is equal to

GM = ι∗GE3 = GE3

∣∣r=r(u,v)

=((dx)2 + (dy)2 + (dz)2

) ∣∣r=r(u,v)

= GM = gαβduαduβ

(1.40)i.e. ((dx)2 + (dy)2 + (dz)2)

∣∣r=r(u,v)

=(∂x(u, v)

∂udu+

∂x(u, v)

∂udu

)2

+

(∂x(u, v)

∂udu+

∂x(u, v)

∂udu

)2

+

(∂x(u, v)

∂udu+

∂x(u, v)

∂udu

)2

=

(x2u + y2

u + z2u)du

2 + 2(xuxv + yuyv + zuzv)dudv + (x2v + y2

v + z2v)dv

2

We see that

GM = gαβduαduβ = g11du

2 + 2g12dudv + g22dv2, (1.41)

19

where g11 = guu = (x2u + y2

u + z2u) = 〈ru, ru〉E3 , g12 = g21 = guv = gvu =

(xuxv + yuyv + zuzv) = 〈ru, rv〉E3 , g22 = gvv = (x2v + y2

v + z2v) = 〈rv, rv〉E3 . We

come to same formula (1.37).(See the examples of calculations in the next subsection.)Remark Sometimes it is convenient to denote Cartesian coordinates of

Euclidean space by xi, (i = 1, 2, 3). Let surface M be given in local param-eterisation xi = xi(uα). Riemannian metric of Euclidean space (1.39) hasappearance

GE = dxiδikdxk . (1.42)

and induced metric (1.40) has appearance

GM = dxiδikdxk∣∣xi=xi(uα)

=∂xi(u)

∂uαδik∂xk(u)

∂uβduαduβ = gαβ(u)duαduβ

(1.43)Why this representation is useful? it is easy to see that formulae (1.42) ,(1.43) work forarbitrary dimensions, i.e. if we have m-dimensional manifold embedded in n-dimensionalEuclidean space. We just have to suppose that in this case i = 1, . . . , n and α = 1, . . . ,m;manifold is given by parameterisation xi = xi(uα) (α = 1, . . . ,m). Moreover in the face ifmanifold is embedded not in Euclidean space but in an arbitrary Riemannian space thenone can see comparing formulae (??) and (1.42) we come to the induced metric

GM = dxigik ((x(u))) dxk∣∣xi=xi(uα)

=∂xi(u)

∂uαgik ((x(u)))

∂xk(u)

∂uβduαduβ = gαβ(x(u))duαduβ

Check explicitly again that length of the tangent vectors and curves onthe surface calculating by External observer (i.e. using Euclidean metric(1.39)) is the same as calculating by Internal Observer, ant (i.e. using theinduced Riemannian metric (1.37), (1.41)). Let X = Xαrα = aru + brv be avector tangent to the surface M . The square of the length |X| of this vectorcalculated by External observer (he calculates using the scalar product inE3) equals to

|X|2 = 〈X,X〉 = 〈ru + brv, aru + brv〉 = a2〈ru, ru〉+ 2ab〈ru, rv〉+ b2〈rv, rv〉(1.44)

where 〈 , 〉 is a scalar product in E3. The internal observer will calculate thelength using Riemannian metric (1.37) (1.41):

G(X,X) =(a, b

)·(g11 g12

g21 G22

)·(ab

)= g11a

2 + 2g12ab+ g22b2 (1.45)

20

External observer (person living in ambient space E3) calculate the lengthof the tangent vector using formula (1.44). An ant living on the surfacecalculate length of this vector in internal coordinates using formula (1.45).External observer deals with external coordinates of the vector, ant on thesurface with internal coordinates. They come to the same answer.

Let r(t) = r(u(t), v(t)) a ≤ t ≤ b be a curve on the surface.Velocity of this curve at the point r(u(t), v(t)) is equal to

v = X = ξru + ηrvwhere ξ = ut, η = vt : v = dr(t)dt

= utru + vtrv .

The length of the curve is equal to

L =

∫ b

a

|v(t)|dt =

∫ b

a

√〈v(t),v(t)〉E3dt =

∫ b

a

√〈utru + vtrv, utru + vtrv〉E3dt =

(1.46)∫ b

a

√〈ru, ru〉E3u2

t + 2〈ru, rv〉E3utvt + 〈rv, rv〉E3v2t dτ =∫ b

a

√g11u2

t + 2g12utvt + g22v2t dt (1.47)

An external observer will calculate the length of the curve using (1.46).An ant living on the surface calculate length of the curve using (1.47) usingRiemannian metric on the surface. They will come to the same answer.

1.3.3 Induced Riemannian metrics. Examples.

We consider here examples of calculating induced Riemannian metric on somequadratic surfaces in E3. using calculations for tangent vectors (see (1.37))or explicitly in terms of differentials (see (1.40) and (1.41)).

First of all consider the general case when a surface M is defined by theequation z − F (x, y) = 0. One can consider the following parameterisationof this surface:

r(u, v) :

x = u

y = v

z = F (u, v)

(1.48)

Then

21

ru =

10Fu

rv =

01Fv

(1.49)

,(ru, ru) = 1 + F 2

u , (ru, rv) = FuFv, (rv, rv) = 1 + F 2v

and induced Riemannian metric (first quadratic form) (1.37) is equal to

||gαβ|| =(g11 g12

g12 g22

)=

((ru, ru) (ru, rv)(ru, rv) (rv, rv)

)=

(1 + F 2

u FuFvFuFv 1 + F 2

v

)(1.50)

GM = ds2 = (1 + F 2u )du2 + 2FuFvdudv + (1 + F 2

v )dv2 (1.51)

and the length of the curve r(t) = r(u(t), v(t)) on C (a ≤ t ≤ b) can becalculated by the formula:

L =

∫ ∫ b

a

√(1 + F 2

u )u2t + 2FuFvutvt + (1 + Fv)2v2

t dt

One can calculate (1.51) explicitly using (1.40):

GM =(dx2 + dy2 + dz2

) ∣∣x=u,y=v,z=F (u,v)

= (du)2 + (dv)2 + (Fudu+Fvdv)2 =

= (1 + F 2u )du2 + 2FuFvdudv + (1 + F 2

v )dv2 . (1.52)

CylinderCylinder is given by the equation x2 + y2 = a2. One can consider the

following parameterisation of this surface:

r(h, ϕ) :

x = a cosϕ

y = a sinϕ

z = h

(1.53)

We have Gcylinder = ι∗GE3 = (dx2 + dy2 + dz2)∣∣x=a cosϕ,y=a sinϕ,z=h

=

= (−a sinϕdϕ)2 + (a cosϕdϕ)2 + dh2 = a2dϕ2 + dh2 (1.54)

The same formula in terms of scalar product of tangent vectors:

22

rh =

001

rϕ =

−a sinϕa cosϕ

0

(1.55)

,(rh, rh) = 1, (rh, rϕ) = 0, (rϕ, rϕ) = a2

and

||gαβ|| =(

(ru, ru) (ru, rv)(ru, rv) (rv, rv)

)=

(1 00 a2

),

G = dh2 + a2dϕ2 (1.56)

and the length of the curve r(t) = r(h(t), ϕ(t)) on the cylinder (a ≤ t ≤ b)can be calculated by the formula:

L =

∫ b

a

√h2t + a2ϕtdt (1.57)

ConeCone is given by the equation x2 + y2 − k2z2 = 0. One can consider the

following parameterisation of this surface:

r(h, ϕ) :

x = kh cosϕ

y = kh sinϕ

z = h

(1.58)

Calculate induced Riemannian metric:We have

Gconus = ι∗GE3 =(dx2 + dy2 + dz2

) ∣∣x=kh cosϕ,y=kh sinϕ,z=h

=

(k cosϕdh− kh sinϕdϕ)2 + (k sinϕdh+ kh cosϕdϕ)2 + dh2

Gconus = k2h2dϕ2 + (1 + k2)dh2, ||gαβ|| =(

1 + k2 00 k2h2

)(1.59)

The length of the curve r(t) = r(h(t), ϕ(t)) on the cone (a ≤ t ≤ b) can becalculated by the formula:

L =

∫ b

a

√(1 + k2)h2

t + k2h2ϕ2tdt (1.60)

23

CircleCircle of radius R is given by the equation x2 + y2 = R2. Consider

standard parameterisation ϕ of this surface:

r(ϕ) :

x = R cosϕ

y = R sinϕ

Calculate induced Riemannian metric (first quadratic form)

GS1 = ι∗GE3 =(dx2 + dy2

) ∣∣x=R cosϕ,y=R sinϕ

=

(−R sinϕdϕ)2 + (a cosϕdϕ)2 = (R2 cos2 ϕ+R2 sin2 ϕ)dϕ2 = R2dϕ2 .

One can consider stereographic coordinates on the circle (see Example inthe subsection 1.1) A point x, y : x2 + y2 = R2 has stereographic coordinatet if points (0, 1) (north pole), the point (x, y) and the point (t, 0) belong tothe same line, i.e. x

t= R−y

R, i.e.

t =Rx

R− y,

x = 2tR2

R2+t2

y = t2−R2

t2+R2R. since x2 + y2 = R2 .

Induced metric in coordinate t is

G = (dx2 + dy2)∣∣x=x(t),y=y(t)

=

(d

(2tR2

R2 + t2

))2

+

(d

(t2 −R2

R2 + t2R

))2

=

(2R2dt

R2 + t2− 4t2R2dt

(R2 + t2)2

)2

+

(− 4R2tdt

(t2 +R2)2

)2

=4R4dt2

(R2 + t2)2. (1.61)

(See for detail Homework 2. Another beatiful deduction of this formula seein 1.3.4)

Remark Stereographic coordinates very often are preferable since theydefine birational equivalence between circle and line.

SphereSphere of radius R is given by the equation x2 + y2 + z2 = R2. Consider

the following (standard ) parameterisation of this surface:

r(θ, ϕ) :

x = R sin θ cosϕ

y = R sin θ sinϕ

z = R cos θ

(1.62)

24

Calculate induced Riemannian metric (first quadratic form)

GS2 = ι∗GE3 =(dx2 + dy2 + dz2

) ∣∣x=R sin θ cosϕ,y=R sin θ sinϕ,z=R cos θ

=

(R cos θ cosϕdθ−R sin θ sinϕdϕ)2+(R cos θ sinϕdθ+R sin θ cosϕdϕ)2+(−R sin θdθ)2 =

R2 cos2 θdθ2 +R2 sin2 θdϕ2 +R2 sin2 θdθ2 =

, = R2dθ2 +R2 sin2 θdϕ2 , ||gαβ|| =(R2 00 R2 sin2 θ

)(1.63)

One comes to the same answer calculating scalar product of tangent vec-tors:

rθ =

R cos θ cosϕR cos θ sinϕ−R sin θ

rϕ =

−R sin θ sinϕR sin θ cosϕ

0

,

(rθ, rθ) = R2, (rh, rϕ) = 0, (rϕ, rϕ) = R2 sin2 θ

and

||g|| =(

(ru, ru) (ru, rv)(ru, rv) (rv, rv)

)=(

R2 00 R2 sin2 θ

), GS2 = ds2 = R2dθ2 +R2 sin2 θdϕ2

The length of the curve r(t) = r(θ(t), ϕ(t)) on the sphere of the radius a(a ≤ t ≤ b) can be calculated by the formula:

L =

∫ b

a

R

√θ2t + sin2 θ · ϕ2

tdt (1.64)

One can consider on sphere as well as on a circle stereographic coordi-nates:

u = RxR−z

v = RyR−z

,

x = 2uR2

R2+u2+v2

y = 2vR2

R2+u2+v2

z = u2+v2−R2

u2+v2+R2R

(1.65)

In these coordinates Riemannian metric is

G = (dx2 + dy2 + dz2)∣∣x=x(u,v),y=y(u,v),z=z(u,v)

=(d

(2uR2

R2 + u2 + v2

))2

+

(d

(2vR2

R2 + u2 + v2

))2

+

(d

(1− 2R2

R2 + u2 + v2

)R

)2

=

25

=4R4(du2 + dv2)

(R2 + u2 + v2)2. (1.66)

(See for detail Homework 2. Another beatiful deduction of this formulasee in 1.3.4)

Notice that we showed that metric on sphere is conformally Euclidean.Saddle (paraboloid)3

Consider paraboloid z = x2 − y2. It can be rewritten as z = axy andit is called sometimes “saddle” (rotation on the angle ϕ = π/4 transformsz = x2 − y2 onto z = 2xy.) Paraboloid and saddle they are ruled surfaces which are

formed by lines. We considered this surface in the course of Geometry.Consider the following (standard ) parameterisation of this surface:

r(u, v) :

x = u

y = v

z = uv

(1.67)

Calculate induced metric:

Gsaddle =(dx2 + dy2 + dz2

) ∣∣x=u cosϕ,y=v sinϕ,z=uv

= du2 +dv2 +(udv+vdu)2 =

Gsaddle = (1 + v2)du2 + 2uvdudv + (1 + u2)dv2 .

One-sheeted and two-sheeted hyperboloids.These examples were not considered on lectures, but they are interesting

for learning purposes.Consider surface given by the equation

x2 + y2 − z2 = c

If c = 0 it is a cone. We considered it already above.If c > 0 it is one-sheeted hyperboloid—connected surface in E3.If c < 0 it is two-sheeted hyperboloid— a surface with two sheets: upper

sheet z > 0 and another sheet: z < 0.Consider these cases separately.

1) One-sheeted hyperboloid: x2 + y2 − z2 = a2. It is ruled surface.Exercise† Find the lines on two-sheeted hyperboloid

3This example was not considered on lectures. It could be useful for learning purposes.

26

One-sheeted hyperboloid is given by the equation x2 + y2− z2 = a2. It isconvenient to choose parameterisation:

r(θ, ϕ) :

x = a cosh θ cosϕ

y = a cosh θ sinϕ

z = a sinh θ

(1.68)

x2 + y2 − z2 = a2 cosh2 θ − a2 sinh2 θ = a2 .

(Compare the calculations with calculations for sphere! We changed func-tions cos, sin on cosh, sinh.)

Induced Riemannian metric (first quadratic form) is equal to

GHyperbolI =(dx2 + dy2 + dz2

) ∣∣x=a cosh θ cosϕ,y=a cosh θ sinϕ,z=a sinh θ

=

(a sinh θ cosϕdθ−a cosh θ sinϕdϕ)2+(a sinh θ sinϕdθ+a cosh θ cosϕdϕ)2+(a cosh θdθ)2 =

a2 sinh2 θdθ2 + a2 cosh2 θdϕ2 + a2 cosh2 θdθ2 =

, = a2(1+2 sinh2 θ)dθ2+a2 cosh2 θdϕ2 , ||gαβ|| =(a2(1 + 2 sinh2 θ) 0

0 a2 cosh2 θ

)

2) Two-sheeted hyperboloid: z2 − x2 − y2 = a2. It is not ruled surface!For two-sheeted hyperboloid calculations will be very similar.In the same way as for one-sheeted hyperboloid (see equation (1.68)) it

is convenient to choose parameterisation:

r(θ, ϕ) :

x = a sinh θ cosϕ

y = a sinh θ sinϕ

z = a cosh θ

(1.69)

z2 − x2 − y2 = a2 cosh2 θ − a2 sinh2 θ = a2

(Compare the calculations with calculations for sphere and one-sheeted hy-perboloid.

Induced Riemannian metric (first quadratic form) is equal to

GHyperbolI =(dx2 + dy2 + dz2

) ∣∣x=a sinh θ cosϕ,y=a sinh θ sinϕ,z=a cosh θ

=

(a cosh θ cosϕdθ−a sinh θ sinϕdϕ)2+(a cosh θ sinϕdθ+a sinh θ cosϕdϕ)2+(a sinh θdθ)2 =

27

a2 cosh2 θdθ2 + a2 sinh2 θdϕ2 + a2 sinh2 θdθ2 =

, = a2(1+2 sinh2 θ)dθ2+a2 sinh2 θdϕ2 , ||gαβ|| =(a2(1 + 2 sinh2 θ) 0

0 a2 sinh2 θ

)(1.70)

We calculated examples of induced Riemannian structure embedded in Euclidean spacealmost for all quadratic surfaces.

Quadratic surface is a surface defined by the equation

Ax2 +By2 + Cz2 + 2Dxy + 2Exz + 2Fyz + ex+ fy + dz + c = 0

One can see that any quadratic surface by affine transformation can be transformed toone of these surfaces

• cylinder (elliptic cylinder) x2 + y2 = 1

• hyperbolic cylinder: x2 − y2 = 1)

• parabolic cylinder z = x2

• paraboloid x2 + y2 = z

• hyperbolic paraboloid x2 − y2 = z

• cone x2 + y2 − z2 = 0

• sphere x2 + y2 + z2 = 1

• one-sheeted hyperboloid x2 + y2 − z2 = 1

• two-sheeted hyperboloid z2 − x2 − y2 = 1

(We exclude degenerate cases such as ”point” x2 + y2 + z2 = 0, planes, e.t.c.)

1.3.4 Inversion and metric on circle and sphere in sterographiccooridnates

Formulae (1.61) and (1.66) for metric in stereographic coordinates are veryimportant, look very nice, but everyone who tried to calculate them wasforced to do difficult calculations. In this paragraph we will explain howthese formulae can be derived almost automatically with use of inversion

Let O be an arbitrary point in Euclidean space En (Here we consider justthe case n = 2, 3 4)

Let Sa be a sphere of radius a with centre at the point 0.If ni are coordinates of the point O, then points of the shere are defined

by equation (∑n

i=1(xi − ni)2 = a2. We call this sphere base of inversion.

4These considerations can be generalised for arbitrary n

28

We define inversion of En with respect to the sphere Sa as a map whichmaps an arbitrary point P 6= 0 in En to the point P ′ such

• point P ′ belongs to the ray OP

•|0P | · ||OP ′| = a2 (1.71)

We see that in particular points of the inversion sphere remian fixed underinversion.

It can be proved that transforms lines, k-dimensional, planes, circles,.spheresto lines, planes, circles, spheres, and that the inversion does not change angelbetween tangent vectors.

Stereographic projection is restriction of inversion.Hence stereographi projection is conformal map. This is why in stereo-

graphic coordinates Riemannian metric has confprmal appearance.

1.3.5 ∗Induced metric on two-sheeted hyperboloid embedded inpseudo-Euclidean space.

Consider the same two-sheeted hyperboloid z2 − x2 − y2 = 1 embedded R3 (Seeequation (1.69). For simplicity we assume now that a = 1.) Now we consider theambient space R3 not as Euclidean space but as pseudo-Euclidean space, i.e. inR3 instead standard scalar product

〈X,Y〉 = X1Y 1 +X2Y 2 +X3Y 3

we consider pseudo-scalar product defined by bilinear form

〈X,Y〉pseud = X1Y 1 +X2Y 2 −X3Y 3

The ”pseudoscalar” product is bilinear, symmetric. It is defined by non-degeneratematrix. But it is not positive-definite. E.g. The ”pseudo-length” of vectors X =(a cosϕ, a sinϕ,±a) is equals to zero (such vectors are called null vectors):

X = (a cosϕ, a sinϕ,±a)⇒ 〈X,X〉pseudo = 0,

The corresponding pseudo-Riemannian metric is:

Gpseudo = dx2 + dy2 − dz2 (1.72)

It turns out that the following remarkable fact occurs:

29

Proposition The pseudo-Riemannian metric (1.72) in the ambient 3-dimensionalpseudo-Euclidean space induces Riemannian metric on two-sheeted hyperboloidx2 + y2 − z2 = 1.

Remark This is not the fact for one-sheeted hyperboloid (see problem 7 inHomework 2)

Show it. (See also problems 5 and 6 in Homework 2. ) Repeat the calculationsabove for two-sheeted hyperboloid changing in the ambient space Riemannianmetric G = dx2 + dy2 + dz2 on pseudo-Riemannian dx2 + dy2 − dz2:

Using (1.69) and (1.72) we come now to

G =(dx2 + dy2 − dz2

) ∣∣x=a sinh θ cosϕ,y=a sinh θ sinϕ,z=a cosh θ

=

(a cosh θ cosϕdθ−a sinh θ sinϕdϕ)2+(a cosh θ sinϕdθ+a sinh θ cosϕdϕ)2−(a sinh θdθ)2 =

a2 cosh2 θdθ2 + a2 sinh2 θdϕ2 − a2 sinh2 θdθ2

, GL = a2dθ2 + a2 sinh2 θdϕ2 , ||gαβ|| =(

1 00 sinh2 θ

)(1.73)

The two-sheeted hyperboloid equipped with this metric is called hyperbolic orLobachevsky plane.

Now express Riemannian metric in stereographic coordinates. (We did it indetail in homework 2)

Calculations are very similar to the case of stereographic coordinates of 2-sphere x2 + y2 + z2 = 1. (See homework 1). Centre of projection (0, 0,−1): Forstereographic coordinates u, v we have u

x = yv = 1

1+z . We come to

u = x

1+z

v = y1+z

,

x = 2u

1−u2−v2

y = 2v1−u2−v2

z = u2+v2+11−u2−v2

(4)

The image of upper-sheet is an open disc u2 + v2 = 1 since u2 + v2 = x2+y2

(1+z)2 =

z2−1(1+z)2 = z−1

z+1 . Since for upper sheet z > 1 then 0 ≤ z−1z+1 < 1.

G = (dx2 + dy2 − dz2)∣∣x=x(u,v),y=y(u,v),z=z(u,v)

=

(d

(2u

1− u2 − v2

))2

+

(d

(2v

1− u2 − v2

))2

−(d

(u2 + v2 + 1

1− u2 − v2

))2

=4(du)2 + 4(dv)2

(1− u2 − v2)2.

These coordinates are very illuminating. One can show that we come to so calledhyperbolic plane (see in detail Homework 2)

30

1.4 Isometries of Riemannian manifolds.

Let (M1, G(1)), (M2, G(2)) be two Riemannian manifolds— manifolds equippedwith Riemannian metric G(1) and G(2) respectively.

Loosely speaking isometry is the diffeomorphism of Riemannian manifoldswhich preserves the distance.

Definition Let F be a diffeomorphims (one-one smooth map with smoothinverse) of manifold M1 on manifold M2.

We say that diffeomorphism F is an isometry of Riemannian manifolds (M1, G(1))and (M2, G(2)) if it preserves the metrics, i.e. G(1) is pull-back of G(2):

F ∗G(2) = G(1) . (1.74)

In local coordinates this means the following: Let p1 be an arbitrary point onmanifold M1 and p2 ∈ M2 be its image:F (p1) = p2. Let xi be arbitraryoordinates in a vicinity of a point p1 ∈ M1 and ya be arbitrary coordinatesin a vicinity of a point p2 ∈ M2. Let Riemannian metrics G(1) on M1 has local

expression G(1) = g(1)ik(x)dxidxk in coordinates xi and respectively Riemannian

metrics G(2) has local expression G(2) = g(2)ab(y)dyadyb in coordinates ya on M2.Then the formula (1.74) has the following appearance in these local coordinates:

F ∗(g

(2)ab(y)dyadyb)

= g(2)ab(y)dyadyb

∣∣y=y(x)

=

g(2)ab(y(x))

∂ya(x)

∂xidxi

∂yb(x)

∂xkdxk = g

(1)ik(x)dxidxk , (1.75)

i.e.

g(1)ik(x) =

∂ya(x)

∂xig

(2)ab(y(x))∂yb(x)

∂xk, (1.76)

where ya = ya(x) is local expression for diffeomorphism F . We say that diffeo-morphism F is isometry of Riemanian manifolds (M1, G(1)) and M2, G(2).

Diffeomorphism F establishes one-one correspondence between local coordi-nates on manifolds M1 and M2. The left hand side of equation (1.75) can beconsidered as a local expression of metric G(2) in coordinates xi on M2 and theright hand side of this equation is local expression of metric G(1) in coordinates xi

on M1. Diffeomorphism F identifies manifolds M1 and M2 and it can be consideredas changing of coordinates.

Example Consider surface of cylinder C, x2 + y2 = a2 in E3 with inducedRiemannian metric GC = a2dϕ2 + dh2 (see equations (1.53) and (1.54)). If weremove the line l : x = a, y = 0 from the cylinder surface C we come to surface

31

C ′ = C\l. Consider a map F of this surface in Euclidean space E2 with Cartesiancoordinates u, v (with standard Euclidean metric GEucl = du2 + dv2):

F :

u = aϕ

v = h0 < ϕ < 2π . (1.77)

One can see that F is the diffeomorphism of C ′ on the domain 0 < u < 2πain E2 and this diffeomorphism is an isometry: it transforms the metric GEucl onEuclidean space in metric GC on cylinder, i.e. pull-back condition (1.74) is obeyed:

F ∗GEucl = F ∗(du2 + dv2

)=(du2 + dv2

) ∣∣u=aϕ,v=h

= a2dϕ2 + dh2 = G1 .

We see that cylinder surface with removed line is isometric to domain in E2 andthe map F establishes this isometry.

Remark Let F be diffeomorphism of manifold M1 on a manifold M2. Leta manifold M2 be equipped with Riemannian metric G(2). Then consider thepull-back of this metric, Riemannian metric G(1) = F ∗G(2) on M1. We see thatdiffeomorphism F is an isometry of Riemannian manifold (M1, G(1)) on Rieman-nian manifold (M2, G(2)).

1.4.1 Isometries of Riemannian manifold (on itself)

Definition Let (M,G) be a Riemannian manifold. We say that a diffeomorphismF is an isometry of Riemannian manifold on itself if it preserves the metric, i.e.F ∗G = G. In local coordinates this means that

gik(x) = gpq(x′(x))

∂xp(x′)

∂xi∂xq(x′)

∂xk, (1.78)

where x′ = x′(x) is a local expression for diffeomorphism F . Example Let E2 beEuclidean plane with metric dx2 +dy2 in Cartesian coordinates x, y. Consider thetransformation

x′ = p+ ax+ by

y′ = q + cx+ dy

is isometry if and only if the matrix A =

(a bc d

)is an orthogonal matrix, i.e. if

the trasformation above is combination of translation, rotation and reflection.

32

1.5 ∗Infinitesimal isometries of Riemannian manifold(Killing vector fields)

Let X be an arbitrary vector field on Riemannian manifold M . It induces infinites-imal diffeomorphism

F : xi′

= xi + εXi(x), where ε2 = 0 .

(the condition ε2 = 0 reflects the fact that we ignore terms of order ≥ 2 overε.) Find a condition which guarantees that infinitesimal diffeomorphism is anisometry. If xi

′= xi + εXi(x), then one can see that the inverse infinitesimal

diffeomorphism is defined by the equation xi = xi′ − εXi(x′) and equation (1.78)

implies that

gik(x) = gpq(x′(x))

∂xp(x′)

∂xi∂xq(x′)

∂xk= gpq(x

i+εXi)

(δpi + ε

pXp(x)

∂xi

)(δqk + ε

pXq(x)

∂xk

)=

gik(x) + ε

[Xp(x)

∂gik(x)

∂xp+ giq(x)

∂Xq(x)

∂xk+ gpk(x)

∂Xp(x)

∂xi,

]Here we consider only terms of first and zero order over ε since ε2 = 0 (thisis related with the fact that transformation is infinitesimal). The last relationimplies that

Xp(x)∂gik(x)

∂xp+ giq(x)

∂Xq(x)

∂xk+ gpk(x)

∂Xp(x)

∂xi= 0 . (1.79)

Left hand side of this relation we denote LXG— Lie derivative of Riemannianmetric along vector field X. Vector field X induces isometry if Lie derivative ofmetric along this vector field vanishes. We come to

Proposition Vector field X on Riemannian manifold (M,G) induces infinites-imal isometry if LXG = 0:

LXG = Xp(x)∂gik(x)

∂xp+ giq(x)

∂Xq(x)

∂xk+ gpk(x)

∂Xp(x)

∂xi= 0 . (1.80)

Definition) We call vector field X Killing vector field) if it preserves the metric,i.e. if equation (1.80) is obeyed.

Example Consider plane (x, y) with Riemannian metric G = σ(x, y)(dx2 +dy2). Find differential equation for infinitesimal isometries of this metric, i.e. writedown equations (1.80) for this metric.

We have ||gik(x, y)|| =(σ(x, y) 0

0 σ(x, y)

).

33

Let X = A(x, y)∂x + B(x, y)∂y. Write down equations (1.80) for componentsg11, g12, g21 and g22: We will have the following three equations

A(x, y)∂σ∂x +B(x, y)∂σ∂y + 2∂A(x,y)∂x σ = 0 for component g11

A(x, y)∂σ∂x +B(x, y)∂σ∂y + 2∂B(x,y)∂y σ = 0 for component g22

∂B(x,y)∂x + ∂A(x,y

∂y = 0 for components g12 and g21

(1.81)

Practically for sphere, Lobachevsky plane, e.t.c. it is much easier to find theKilling fields not solving these equations, but considering the usual isometries (seeexamples in solutions of Coursework and in the Appendix about Killing vectorfields for Lobachevsky plane.))

Another simple and interesting exercise: How look Killing vectors for Euclideanspace En. In this case we come from (1.80) to equation

LKG = δiq(x)∂Kq(x)

∂xk+ δpk(x)

∂Kp(x)

∂xi= 0 ,

i.e.∂Ki(x)

∂xk+∂Kk(x)

∂xi= 0 . (1.82)

Solve this equation. Differentiating by x we come to

∂2Ki(x)

∂xm∂xk+∂Kk(x)

∂xm∂xi= 0

Consider tensor field

T imk =∂2Ki

∂xm∂xk(1.83)

It follows from equation (1.82) that

T imk = T ikm = −Tmik . (1.84)

It is easy to see that this implies that T imk ≡ 0!!!:

T imk = −Tmik = −Tmki = T kmi = T kim = −T ikm = −T imk ⇒ T imk = −T imk ,

i.e. T imk = ∂2Ki(x)∂xm∂xk

= 0. This implies that

Ki(x) = Ci +Bikx

k

We come toTheorem All infinitesimal isometries of En are translations and infinites-

imal rotations.What happens in general case?

34

1.5.1 Locally Euclidean Riemannian manifolds

It is useful to formulate the local isometry condition between Riemannianmanifold and Euclidean space. A neighbourhood of every point of n-dimensionalmanifold is diffeomorphic to Rn. Let as usual En be n-dimensional Eu-clidean space, i.e. Rn with standard Riemannian metric G = dxiδikdx

k =(dx1)2 + · · ·+ (dxn)2 in Cartesian coordinates (x1, . . . , xn).

Definition We say that n-dimensional Riemannian manifold (M,G) islocally isometric to Euclidean space En, i.e. it is locally Euclidean Rieman-nian manifold, if for every point p ∈M there exists an open neighboorhoodD (domain) containing this point, p ∈ D such that D is isometric to a do-main in Euclidean plane. In other words in a vicinity of every point p thereexist local coordinates u1, . . . , un such that Riemannian metric G in thesecoordinates has an appearance

G = duiδikduk = (du1)2 + · · ·+ (dun)2 . (1.85)

Consider examples.Example Consider again cylinder surface..We know that cylinder is not diffeomorphic to plane (cylinder surface

is S1 × R, E2 = R × R, and circle is not diffeomorphic to line ). In theprevious subsection we cutted the line from cylindre. Thus we came tosurface diffeomorphic to plane. We established that this surface is isometricto Euclidean plane. (See equation (1.77) and considerations above.) Localisometry of cylinder to the Euclidean plane, i.e. the fact that it is locallyEuclidean Riemannian surface immediately follows from the fact that underchanging of local coordinates u = aϕ, v = h in equation (1.77), the standardEuclidean metric du2 + dv2 transforms to the metric Gcylinder = a2dϕ2 + dh2

on cylinder.Remark Strictly speaking we consider all the points except the points

on the cutted line (with coordinate ϕ − 0). On the other hand for thepoints on cuttng line we can consider instead coordinate ϕ another coordinateϕ′ = ϕ− π, −π < ϕ′ < π, and we will come to the same answer. In this casethe cutted line will be the line ϕ/ = π.

Example Now show that cone is locally Euclidean Riemannian surface,i.e, it is locally isometric to the Euclidean plane. This means that we have tofind local coordinates u, v on the cone such that in these coordinates induced

35

metric G|c on cone would have the appearance G|c = du2 + dv2. Recallcalculations of the metric on cone in coordinates h, ϕ where

r(h, ϕ) :

x = kh cosϕ

y = kh sinϕ

z = h

,

x2 + y2−k2z2 = k2h2 cos2 ϕ+k2h2 sin2 ϕ−k2h2 = k2h2−k2h2 = 0. We havethat metric Gc on the cone in coordinates h, ϕ induced with the Euclideanmetric G = dx2 + dy2 + dz2 is equal to

Gc =(dx2 + dy2 + dz2

) ∣∣x=kh cosϕ,y=kh sinϕ,z=h

= (k cosϕdh− kh sinϕdϕ)2+

(k sinϕdh+ kh cosϕdϕ)2 + dh2 = (k2 + 1)dh2 + k2h2dϕ2 .

In analogy with polar coordinates try to find new local coordinates u, v such

that

u = αh cos βϕ

v = αh sin βϕ, where α, β are parameters. We come to du2 +dv2 =

(α cos βϕdh− αβh sin βϕdϕ)2+(α sin βϕdh+ αβh cos βϕdϕ)2 = α2dh2+α2β2h2dϕ2.

Comparing with the metric on the cone Gc = (1 + k2)dh2 + k2h2dϕ2 we seethat if we put α = k and β = k√

1+k2 then du2 + dv2 = α2dh2 + α2β2h2dϕ2 =

(1 + k2)dh2 + k2h2dϕ2.Thus in new local coordinates

u =√k2 + 1h cos k√

k2+1ϕ

v =√k2 + 1h sin k√

k2+1ϕ

induced metric on the cone becomes G|c = du2 + dv2, i.e. cone locally isisometric to the Euclidean plane

Of course these coordinates are local.— Cone and plane are not homeo-morphic, thus they are not globally isometric.

Example and counterexampleConsider domain D in Euclidean plane with two metrics:

G(1) = du2 + sin2 vdv2 , and G(2) = du2 + sin2 udv2 (1.86)

36

Thus we have two different Rimeannian manifolds (D,G(1)) and (D,G(2)).Metrics in (1.86) look similar. But.... It is easy to see that the first one islocally isometric to Euclidean plane, i.e. it is locally Euclidean Riemannianmanifold since sin2 vdv2 = d(− cos v)2: in new coordinates u′ = u, v′ = cos vRiemannian metric G(1) has appearance of standard Euclidean metric:

(du′)2 + (dv′)2 = (du)2 + (d(cos v))2 = du2 + sin2 vdv2 = G(1) .

This is not the case for second metric G(2). If we change notations u 7→ θ,v 7→ ϕ then G(2) = dθ2 + sinθ2dϕ2. This is local expression for Riemannianmetric induced on the sphere of radius R = 1. Suppose that there existcoordinates u′ = u′(θ, ϕ) v′ = v′(θ, ϕ) such that in these coordinates metrichas Eucldean appearance. This means that locally geometry of sphere is as ageometry of Euclidean plane. On the other hand we know from the course ofGeometry that this is not the case: sum of angles of triangels on the sphereis not equal to π, sphere cannot be bended without shrinking. Later in thiscourse we will return to this question....

There are plenty other examples:

2) Plane with metric 4R4(dx2+dy2)(R2+x2+y2)2 is isometric to the sphere with radius R.

3) Disc with metric du2+dv2

(1−u2−v2)2 is isometric to half plane with metricdx2+dy2

4y2 .

(see also exercises in Homeworks and Coursework.)

1.6 Volume element in Riemannian manifold

The volume element in n-dimensional Riemannian manifold with metric G =gikdx

idxk is defined by the formula√det gik dx

1dx2 . . . dxn . (1.87)

If D is a domain in the n-dimensional Riemannian manifold with metricG = gikdx

i then its volume is equal to to the integral of volume element overthis domain.

V (D) =

∫D

√det gik dx

1dx2 . . . dxn . (1.88)

Remark Students who know the concept of exterior forms can read thevolume element as n-form√

det gik dx1 ∧ dx2 ∧ · · · ∧ dxn .

37

Note that in the case of n = 1 volume is just the length, in the case ifn = 2 it is area.

1.6.1 Motivation: Gramm formula for volume of parallelepiped

Explain how formulae (1.87), (1.88) are related with basic formulae of linearalgebra.

Recall the formulae for calculation of volume of n-dimensional paral-lelepiped. Let En be Euclidean vector space equipped with orthonormalbasis ei. i

Let ai = v1, . . . , an be an arbitrary basis in this vector space. Con-sider n-dimensional parallelepiped formed by vectors ai:

Πai : r = tivi, 0 ≤ ti ≤ 1.

We know that the volume of this parallelepiped is equal to

V ol(Πai) = det ||ami || , (1.89)

where matrix A is defined by expansion of vectors ai over orthonormalbasis ei ai = ema

im. (We know this formula at least for n = 1, 2–length of

interval, are of parallelogram).(Volume vanishes ⇔ if ai is not a basis.)Now consider the scalar product (Riemannian metric) in En in the basis

a1, . . . , an:gik = 〈ai, ak〉 , (1.90)

where 〈 , 〉 is scalar product in En: 〈ei, ej〉 = δij. We see that in (1.90)

gij = 〈ai, ak〉 = 〈∑m

ami em,∑n

anj en〉 = ami δmnani = (AT ·A)ij ⇒ detG = (detA)2 ,

where G = ||gij||. Comparing with formula (1.89) we come to Gramm for-mula:

V ol(Πai) =√

det gik (1.91)

The matrix G = ||gik|| is called Gramm matrix for the vectors ai. Grammformula states that volume of parallelogram formed by the vectors a1, . . . , αmisequal to the square root of the determinant of Gramm matrix.

Remark One can easy see that formula (1.91) works for arbitrary n-dimensionalparallelogramm in m-dimensional space. Indeed if α1, . . . , αn are just arbitrary n vectors

38

in m-dimensional Euclidean space then if n < m, the formula (1.89) s failed (matrix A ism × n matrix), but formula (1.91) works. For example the area of parallelogram formedby arbitrary vectors a1,a2 in En is equal to√

det

(g11 g12

g21 g22

)=

√det

(〈a1,a1〉 〈a1,a1〉〈a1,a1〉 〈a1,a1〉

).

One can say that n-dimensional parallelepiped Πai in new coordinatesti corresponding to the basis ai becomes n-dimensional cube, StandardEuclidean metric G = dxiδikdx

k (in orthonormal basis ei) transforms to

G = dxiδikdxk = (aimdt

m)δik(akndt

n) = (ATA)mndtmdtn

and

VolumeΠai =

∫x∈Π

dx1 . . . dxn =

∫0≤ti≤1

√Gdt1 . . . dtn =

√detG =

√detATA .

1.6.2 ∗Invariance of volume element under changing of coordi-nates

Check straighforwardly that volume element is invariant under coordinatetransformations, i.e. if y1, . . . , yn are new coordinates: x1 = x1(y1, . . . , yn), x2 =x2(y1, . . . , yn)...,

xi = xi(yp), i = 1, . . . , n , p = 1, . . . , n

and gpq(y) matrix of the metric in new coordinates:

gpq(y) =∂xi

∂ypgik(x(y))

∂xk

∂yq. (1.92)

Then √det gik(x) dx1dx2 . . . dxn =

√det gpq(y) dy1dy2 . . . dyn (1.93)

This follows from (1.92). Namely

√det gik(y) dy1dy2 . . . dyn =

√det

(∂xi

∂ypgik(x(y))

∂xk

∂yq

)dy1dy2 . . . dyn

39

Using the fact that det(ABC) = detA · detB · detC and det(∂xi

∂yp

)=

det(∂xk

∂yq

)5 we see that from the formula above follows:

√det gik(y) dy1dy2 . . . dyn =

√det

(∂xi

∂ypgik(x(y))

∂xk

∂yq

)dy1dy2 . . . dyn =

√(det

(∂xi

∂yp

))2√det gik(x(y))dy1dy2 . . . dyn =

√det gik(x(y)) det

(∂xi

∂yp

)dy1dy2 . . . dyn = (1.94)

Now note that

det

(∂xi

∂yp

)dy1dy2 . . . dyn = dx1 . . . dxn

according to the formula for changing coordinates in n-dimensional integral6. Hence√

det gik(x(y)) det

(∂xi

∂yp

)dy1dy2 . . . dyn =

√det gik(x(y))dx1dx2 . . . dxn

(1.95)Thus we come to (1.93).

1.6.3 Examples of calculating volume element

Consider first very simple example: Volume element of plane in Cartesiancoordinates, metric g = dx2 + dy2. Volume element is equal to

√det gdxdy =

√det

(1 00 1

)dxdy = dxdy

5determinant of matrix does not change if we change the matrix on the adjoint, i.e.change columns on rows.

6Determinant of the matrix(∂xi

∂yp

)of changing of coordinates is called sometimes Ja-

cobian. Here we consider the case if Jacobian is positive. If Jacobian is negative thenformulae above remain valid just the symbol of modulus appears.

40

Volume of the domain D is equal to

V (D) =

∫D

√det gdxdy =

∫D

dxdy

If we go to polar coordinates:

x = r cosϕ, y = r sinϕ (1.96)

Then we have for metric:G = dr2 + r2dϕ2

because

dx2 + dy2 = (dr cosϕ− r sinϕdϕ)2 + (dr sinϕ+ r cosϕdϕ)2 = dr2 + r2dϕ2

(1.97)Volume element in polar coordinates is equal to

√det gdrdϕ =

√det

(1 00 r2

)drdϕ = rdrdϕ .

Lobachesvky plane.In coordinates x, y (y > 0) metric G = dx2+dy2

y2 , the corresponding matrix

G =

(1/y2

0

0 1/y2

). Volume element is equal to

√det gdxdy = dxdy

y2 .

Sphere in stereographic coordinates In stereographic coordinates

G =4R4(du2 + dv2)

(R + u2 + v2)2(1.98)

(It is isometric to the sphere of the radius R without North pole in stere-ographic coordinates (see the Homeworks.))

Calculate its volume element and volume. It is easy to see that:

G =

(4R4

(R2+u2+v2)2 0

0 4R4

(R2+u2+v2)2

)det g =

16R8

(R2 + u2 + v2)4(1.99)

and volume element is equal to√

det gdudv = 4R4dudv(R2+u2+v2)2

41

One can calculate volume in coordinates u, v but it is better to considerhomothety u → Ru, v → Rv and polar coordinates: u = Rr cosϕ, v =Rr sinϕ. Then volume form is equal to

√det gdudv = 4R4dudv

(R2+u2+v2)2 = 4R2rdrdϕ(1+r2)2 .

Now calculation of integral becomes easy:

V =

∫4R2rdrdϕ

(1 + r2)2= 8πR2

∫ ∞0

rdr

(1 + r2)2= 4πR2

∫ ∞0

du

(1 + u)2= 4πR2 .

Domain in Lobachevsky plane.Consider in Lobachevsky plane the domain Da such that

Da = x, y : x2 + y2 ≥ 1, |y| ≤ a , (|a| ≤ 1) . (1.100)

Remark Note that vertical lines and half-circle are geodesics. One can seethat the distance between these lines tends to zero. (We will studly it later).If we denote by A a point (a,

√1− a2) and by B the point (−a,

√1− a2),

then the domain Da can be cinsidered as a ‘triangle’ with vertices at thepoint A,B,C where C is a point at infinity. The meaning of this remark wewill study later.

One can calculate the area of this domain, using area form on Lobachevskyplane

V (Da) =

∫−a≤y≤a ,x2+y2≥1

dxdy

y2= 2 arcsin a (1.101)

(See in detail Homework) We will discuss later the geometrical meaning ofthis formula.

Segment of the sphere.Consider sphere of the radius a in Euclidean space with standard Rie-

mannian metrica2dθ2 + a2 sin2 θdϕ2

This metric is nothing but first quadratic form on the sphere (see (1.3.3)).The volume element is√

det gdθdϕ =

√det

(a2 00 a2 sin θ

)dθdϕ = a2 sin θdθdϕ

Now calculate the volume of the segment of the sphere between two parallelplanes, i.e. domain restricted by parallels θ1 ≤ θ ≤ θ0: Denote by h be theheight of this segment. One can see that

h = a cos θ0 − a cos θ1 = a(cos θ0 − a cos θ1)

42

There is remarkable formula which express the area of segment via the heighth:

V =

∫θ1≤θ≤θ0

(a2 sin θ

)dθdϕ =

∫ θ1

θ0

(∫ 2π

0

(a2 sin θ

)dϕ

)dθ =

∫ θ0

θ1

2πa2 sin θdθ = 2πa2(cos θ0 − cos θ1) = 2πa(a cos θ0 − acosθ1) = 2πah

(1.102)E.g. for all the sphere h = 2a. We come to S = 4πa2. It is remarkableformula: area of the segment is a polynomial function of radius of the sphereand height (Compare with formula for length of the arc of the circle)

2 Covariant differentiaion. Connection. Levi

Civita Connection on Riemannian mani-

fold

2.1 Differentiation of vector field along the vector field.—Affine connection

How to differentiate vector fields on a (smooth )manifold M?Recall the differentiation of functions on a (smooth )manifold M .Let X = Xi(x)ei(x) = ∂

∂xi be a vector field on M . Recall that vectorfield 7 X = Xiei defines at the every point x0 an infinitesimal curve: xi(t) =xi0 + tX i (More exactly the equivalence class [γ(t)]X of curves xi(t) = xi0 +tX i + . . . ).

Let f be an arbitrary (smooth) function on M and X = X i ∂∂xi

. Thenderivative of function f along vector field X = X i ∂

∂xiis equal to

∂Xf = ∇Xf = X i ∂f

∂xi

The geometrical meaning of this definition is following: If X is a velocityvector of the curve xi(t) at the point xi0 = xi(t) at the ”time” t = 0 thenthe value of the derivative ∇Xf at the point xi0 = xi(0) is equal just to the

7here like always we suppose by default the summation over repeated indices. E.g.X =Xiei is nothing but X =

∑ni=1X

iei

43

derivative by t of the function f(xi(t)) at the ”time” t = 0:

if X i(x)∣∣x0=x(0)

=dxi(t)

dt

∣∣t=0, then ∇Xf

∣∣xi=xi(0)

=d

dtf(xi (t)

) ∣∣t=0

(2.1)Remark In the course of Geometry and Differentiable Manifolds the

operator of taking derivation of function along the vector field was denotedby ”∂Xf”. In this course we prefer to denote it by ”∇Xf” to have the uniformnotation for both operators of taking derivation of functions and vector fieldsalong the vector field.

One can see that the operation∇X on the space C∞(M) (space of smoothfunctions on the manifold) satisfies the following conditions:

• ∇X (λf + µg) = λ∇Xf+µ∇Xg where λ, µ ∈ R (linearity over numbers)

• ∇hX+gY(f) = h∇X(f) + g∇Y(f) (linearity over the space of functions)

• ∇X(λfg) = f∇X(λg) + g∇X(λf) (Leibnitz rule)

(2.2)

Remark One can prove that these properties characterize vector fields:operatoron smooth functions obeying the conditions above is a vector field. (You willhave a detailed analysis of this statement in the course of Differentiable Man-ifolds.)

How to define differentiation of vector fields along vector fields.The formula (2.1) cannot be generalised straightforwardly because vec-

tors at the point x0 and x0 + tX are vectors from different vector spaces.(We cannot substract the vector from one vector space from the vector fromthe another vector space, because apriori we cannot compare vectors fromdifferent vector space. One have to define an operation of transport of vec-tors from the space Tx0M to the point Tx0+tXM defining the transport fromthe point Tx0M to the point Tx0+tXM).

Try to define the operation ∇ on vector fields such that conditions (2.2)above be satisfied.

44

2.1.1 Definition of connection. Christoffel symbols of connection

Definition Affine connection on M is the operation ∇ which assigns to everyvector field X a linear map, (but not necessarily C(M)-linear map!) (i.e. amap which is linear over numbers not necessarily over functions) ∇X on thespace of vector fields on M :

∇X (λY + µZ) = λ∇XY + µ∇XZ, for every λ, µ ∈ R (2.3)

(Compare the first condition in (2.2)).which satisfies the following conditions:

• for arbitrary (smooth) functions f, g on M

∇fX+gY (Z) = f∇X (Z) + g∇Y (Z) (C(M)-linearity) (2.4)

(compare with second condition in (2.2))

• for arbitrary function f

∇X (fY) = (∇Xf) Y + f∇X (Y) (Leibnitz rule) (2.5)

Recall that ∇Xf is just usual derivative of a function f along vectorfield: ∇Xf = ∂Xf .

(Compare with Leibnitz rule in (2.2)).

The operation ∇XY is called covariant derivative of vector field Y alongthe vector field X.

Write down explicit formulae in a given local coordinates xi (i =1, 2, . . . , n) on manifold M .

Let

X = X iei = X i ∂

∂xiY = Y iei = Y i ∂

∂xi

The basis vector fields ∂xi

we denote sometimes by ∂i sometimes by eiUsing properties above one can see that

∇XY = ∇Xi∂iYk∂k = X i

(∇i

(Y k∂k

)), where ∇i = ∇∂i (2.6)

Then according to (2.4)

∇i

(Y k∂k

)= ∇i

(Y k)∂k + Y k∇i∂k

45

Decompose the vector field ∇i∂k over the basis ∂i:

∇i∂k = Γmik∂m (2.7)

and

∇i

(Y k∂k

)=∂Y k(x)

∂xi∂k + Y kΓmik∂m, (2.8)

∇XY = X i∂Ym(x)

∂xi∂m +X iY kΓmik∂m, (2.9)

In components

(∇XY)m = X i

(∂Y m(x)

∂xi+ Y kΓmik

)(2.10)

Coefficients Γmik are called Christoffel symbols in coordinates xi. Thesecoefficients define covariant derivative—connection.

If operation of taking covariant derivative is given we say that the con-nection is given on the manifold. Later it will be explained why we us theword ”connection”

We see from the formula above that to define covariant derivative of vectorfields, connection, we have to define Christoffel symbols in local coordinates.

2.1.2 Transformation of Christoffel symbols for an arbitrary con-nection

Let ∇ be a connection on manifold M . Let Γikm be Christoffel symbolsof this connection in given local coordinates xi. Then according (2.7) and(2.8) we have

∇XY = Xm ∂Yi

∂xm∂

∂xi+XmΓimkY

k ∂

∂xi,

and in particularlyΓimk∂i = ∇∂m∂k

Use this relation to calculate Christoffel symbols in new coordinates xi′

Γi′

m′k′∂i′ = ∇∂′m∂k′

We have that ∂m′ = ∂∂xm′ = ∂xm

∂xm′∂

∂xm= ∂xm

∂xm′ ∂m. Hence due to properties(2.4), (2.5) we have

Γi′

m′k′∂i′ = ∇∂m′∂k′ = ∇∂′m

(∂xk

∂xk′∂k

)=

(∂xk

∂xk′

)∇∂′m∂k +

∂

∂xm′

(∂xk

∂xk′

)∂k =

46

(∂xk

∂xk′

)∇ ∂xm

∂xm′ ∂m

∂k +∂2xk

∂xm′∂xk′∂k =

∂xk

∂xk′∂xm

∂xm′∇∂m∂k +∂2xk

∂xm′∂xk′∂k

∂xk

∂xk′∂xm

∂xm′ Γimk∂i +

∂2xk

∂xm′∂xk′∂k =

∂xk

∂xk′∂xm

∂xm′ Γimk

∂xi′

∂xi∂i′ +

∂2xk

∂xm′∂xk′∂xi

′

∂xk∂i′

Comparing the first and the last term in this formula we come to the trans-formation law:

If Γikm are Christoffel symbols of the connection ∇ in local coordinatesxi and Γi′k′m′ are Christoffel symbols of this connection in new localcoordinates xi′ then

Γi′

k′m′ =∂xk

∂xk′∂xm

∂xm′

∂xi′

∂xiΓikm +

∂2xr

∂xk′∂xm′

∂xi′

∂xr(2.11)

Remark Christoffel symbols do not transform as tensor. If the secondterm is equal to zero, i.e. transformation of coordinates are linear (see theProposition on flat connections) then the transformation rule above is the the

same as a transformation rule for tensors of the type

(12

)(see the formula

(1.1.2)). In general case this is not true. Christoffel symbols do not trans-form as tensor under arbitrary non-linear coordinate transformation: see thesecond term in the formula above.

Remark On the other hand note that difference of two arbitrary connections is atensor. If Γikm and Γikm are corresponding Chrstoffel symbols then it follows from (1.1.2))

that their difference Sikm = Γikm − Γikm transforms as a tensor:

Si′

k′m′ = Γi′

k′m′ − Γi′

k′m′ =∂xk

∂xk′∂xm

∂xm′

∂xi′

∂xi

(Γikm − Γikm

)=

∂xk

∂xk′∂xm

∂xm′

∂xi′

∂xiSikm

(See for detail the Homework.)

2.1.3 Canonical flat affine connection

It follows from the properties of connection that it is suffice to define con-nection at vector fields which form basis at the every point using (2.7), i.e.to define Christoffel symbols of this connection.

Example Consider n-dimensional Euclidean space En with Cartesiancoordinates x1, . . . , xn.

Define connection such that all Christoffel symbols are equal to zero inthese Cartesian coordinates xi.

∇eiek = Γmikem = 0, Γmik = 0 (2.12)

47

Does this mean that Christoffel symbols are equal to zero in an arbitraryCartesian coordinates if they equal to zero in given Cartesian coordinates?

Does this mean that Christoffel symbols of this connection equal to zeroin arbitrary coordinates system?

it follows from transformation rules (??) for Christoffel symbols thatChristofel symbols vanish also in new coordinates xi

′if and only if

∂2xi

∂xm′∂xi′= 0, i.e. xi = bi + aikx

k (2.13)

i.e. the relations between new and old coordinates are linear. We come tosimple but very important

Proposition Let all Christoffel symbols of a given connection be equal tozero in a given coordinate system xi. Then all Christoffel symbols of thisconnection are equal to zero in an arbitrary coordinate system xi′ such thatthe relations between new and old coordinates are linear:

xi′= bi + aikx

k (2.14)

If transformation to new coordinate system is not linear, i.e. ∂2xi

∂xm′∂xi′6= 0

then Christoffel symbols of this connection in general are not equal to zero innew coordinate system xi′.

Definition We call connection ∇ flat if there exists coordinate systemsuch that all Christoffel symbols of this connection are equal to zero in agiven coordinate system.

In particular connection (2.12) has zero Christoffel symbols in arbitraryCartesian coordinates.

Corollary Connection has zero Christoffel symbols in arbitrary Cartesiancoordinates if it has zero Christoffel symbols in a given Cartesian coordinates.

Hence the following definition is correct:

Definition A connection on the Euclidean space En which Christoffelsymbols vanish in Cartesian coordinates is called canonical flat connection.

Remark Canonical flat connection in Euclidean space is uniquely defined,

since Cartesian coordinates are defined globally. On the other hand on arbitrary

manifold one can define flat connection locally just choosing any arbitrary local

coordinates and define locally flat connection by condition that Christoffel symbols

vanish in these local coordinates. This does not mean that one can define flat

48

connection globally. We will study this question after learning transformation law

for Christoffel symbols.

Remark One can see that flat connection is symmetric connection.

Example Consider a connection (2.12) in E2. It is a flat connection.Calculate Christoffel symbols of this connection in polar coordinates

x = r cosϕ

y = y sinϕ

r =

√x2 + y2

ϕ = arctan yx

(2.15)

Write down Jacobians of transformations—matrices of partial derivatives:

(xr yrxϕ yϕ

)=

(cosϕ sinϕ−r sinϕ r cosϕ

),

(rx ϕxry ϕy

)=

x√x2+y2

− yx2+y2

y√x2+y2

xx2+y2

(2.16)

According (2.11) and since Chrsitoffel symbols are equal to zero in Cartesiancoordinates (x, y) we have

Γi′

k′m′ =∂2xr

∂xk′∂xm′

∂xi′

∂xr, (2.17)

where (x1, x2) = (x, y) and (x1′ , x2′) = (r, ϕ). Now using (2.16) we have

Γrrr =∂2x

∂r∂r

∂r

∂x+

∂2y

∂r∂r

∂r

∂y= 0

Γrrϕ = Γrϕr =∂2x

∂r∂ϕ

∂r

∂x+

∂2y

∂r∂ϕ

∂r

∂y= − sinϕ cosϕ+ sinϕ cosϕ = 0 .

Γrϕϕ =∂2x

∂ϕ∂ϕ

∂r

∂x+

∂2y

∂r∂ϕ

∂r

∂y= −xx

r− yy

r= −r .

Γϕrr =∂2x

∂r∂r

∂ϕ

∂x+

∂2y

∂r∂r

∂ϕ

∂y= 0 .

Γϕϕr = Γϕrϕ =∂2x

∂r∂ϕ

∂ϕ

∂x+

∂2y

∂r∂ϕ

∂ϕ

∂y= − sinϕ

−yr2

+ cosϕx

r2=

1

r

Γϕϕϕ =∂2x

∂ϕ∂ϕ

∂ϕ

∂x+

∂2y

∂ϕ∂ϕ

∂ϕ

∂y= −x−y

r2− y x

r2= 0 . (2.18)

49

Hence we have that the covariant derivative (??) in polar coordinates hasthe following appearance

∇r∂r = Γrrr∂r + Γϕrr∂ϕ = 0 , , ∇r∂ϕ = Γrrϕ∂r + Γϕrϕ∂ϕ =∂ϕr

∇ϕ∂r = Γrϕr∂r + Γϕϕr∂ϕ =∂ϕr, ∇ϕ∂ϕ = Γrϕϕ∂r + Γϕϕϕ∂ϕ = −r∂r (2.19)

Remark Later when we study geodesics we will learn a very quick methodto calculate Christoffel symbols.

2.1.4 ∗ Global aspects of existence of connection

We defined connection as an operation on vector fields obeying the special axioms(see the subsubsection 2.1.1). Then we showed that in a given coordinates con-nection is defined by Christoffel symbols. On the other hand we know that ingeneral coordinates on manifold are not defined globally. (We had not this troublein Euclidean space where there are globally defined Cartesian coordinates.)

• How to define connection globally using local coordinates?

• Does there exist at least one globally defined connection?

• Does there exist globally defined flat connection?

These questions are not naive questions. Answer on first and second questionsis ”Yes”. It sounds bizzare but answer on the first question is not ”Yes” 8

Global definition of connectionThe formula (2.11) defines the transformation for Christoffer symbols if we go

from one coordinates to another.Let (xiα), Uα be an atlas of charts on the manifold M .If connection ∇ is defined on the manifold M then it defines in any chart (local

coordinates) (xiα) Christoffer symbols which we denote by (α)Γikm. If (xiα), (xi

′

(β))

are different local coordinates in a vicinity of a given point then according to (2.11)

(β)Γi′k′m′ =

∂xk(α)

∂xk′

(β)

∂xm(α)

∂xm′

(β)

∂xi′

(β)

∂xi(α) (β)

Γ(α)imk +

∂2xk(α)

∂xm′

(β)∂xk′(β)

∂xi′

(β)

∂xk(α)

(2.20)

Definition Let (xiα), Uα be an atlas of charts on the manifold M

8Topology of the manifold can be an obstruction to existence of global flat connection.E.g. it does not exist on sphere Sn if n > 1.

50

We say that the collection of Christoffel symbols Γ(α)ikm defines globally a

connection on the manifold M in this atlas if for every two local coordinates(xi(α)), (xi(β)) from this atlas the transformation rules (2.20) are obeyed.

Using partition of unity one can prove the existence of global connection con-structing it in explicit way. Let (xiα), Uα (α = 1, 2, . . . , N) be a finite atlas onthe manifold M and let ρα be a partition of unity adjusted to this atlas. Denoteby (α)Γikm local connection defined in domain Uα such that its components in these

coordinates are equal to zero. Denote by(α)(β)Γ

ikm Christoffel symbols of this local

connection in coordinates (xi(β)) ((α)(β)Γ

ikm = 0). Now one can define globally the

connection by the formula:

(β)Γikm(x) =

∑α

ρα(x)(α)(β)Γ

ikm(x) =

∑α

ρα(x)∂xi(β)

∂xi′

(α)

∂2xi′

(α)(x)

∂xk(β)∂xm(β)

. (2.21)

This connection in general is not flat connection9

2.2 Connection induced on the surfaces

Let M be a manifold embedded in Euclidean space10. Canonical flat connec-tion on EN induces the connection on surface in the following way.

Let X,Y be tangent vector fields to the surfaceM and∇can.flat a canonicalflat connection in EN . In general

Z = ∇can.flatX Y is not tangent to manifold M (2.22)

Consider its decomposition on two vector fields:

Z = Ztangent + Z⊥,∇can.flatX ,Y =

(∇can.flat

X Y)tangent

+(∇can.flat

X Y)⊥ , (2.23)

where Z⊥ is a component of vector which is orthogonal to the surface Mand Z|| is a component which is tangent to the surface. Define an inducedconnection ∇M on the surface M by the following formula

∇M : ∇MX Y : =

(∇can.flat

X Y)tangent

(2.24)

9See for detail the text: ”Global affine connection on manifold” ” in my homepage:”www.maths.mancheser.ac.uk/khudian” in subdirectory Etudes/Geometry

10We know that every n-dimensional manifold can be embedded in 2n+ 1-dimensionalEuclidean space

51

One can see that this formula really defines the connection on surface M ,i.e. the operation defined by this relation obeys all axioms of connection.Indeed it is easy to see that for arbitrary vector fields X and Y, the vectorfield ∇M

X Y is tangent vector field, and this operation obeys relations (2.3),(2.4) and (2.5). For example check Leibnitz rule:

∇MX (fY) =

(∇can.flat

X (fY))tangent

=(∂XfY + f∇can.flat

X Y)tangent

=

∂XfY + f∇can.flatX Ytangent = ∂XfY + f∇M

X Y .

We mainly apply this construction for 2-dimensional manifolds (surfaces)in E3.

2.2.1 Calculation of induced connection on surfaces in E3.

Let r = r(u, v) be a surface in E3. Let ∇can.flat be a flat connection in E3.Then

∇M : ∇MX Y : =

(∇can.flat

X Y)|| = ∇

can.flatX Y − n(∇can.flat

X Y,n), (2.25)

where n is normal unit vector field to M . Consider a special exampleExample (Induced connection on sphere) Consider a sphere of the radius

R in E3:

r(θ, ϕ) :

x = R sin θ cosϕ

y = R sin θ sinϕ

z = R cos θ

then

rθ =

R cos θ cosϕR cos θ sinϕ−R sin θ

, rϕ =

−R sin θ sinϕR sin θ cosϕ

0

,n =

sinθ cosϕsinθ sinϕ

cos θ

,

where rθ = ∂r(θ,ϕ)∂θ

, rϕ = ∂r(θ,ϕ)∂ϕ

are basic tangent vectors and n is normal unitvector.

Calculate an induced connection ∇ on the sphere.First calculate ∇∂θ∂θ.

∇∂θ∂θ =

(∂rθ∂θ

)tangent

= (rθθ)tangent .

52

On the other hand one can see that rθθ =

−Rsinθ cosϕ−Rsinθ sinϕ−R cos θ

= −Rn is

proportional to normal vector, i.e. (rθθ)tangent = 0. We come to

∇∂θ∂θ = (rθθ)tangent = 0⇒ Γθθθ = Γϕθθ = 0 . (2.26)

Now calculate ∇∂θ∂ϕ and ∇∂ϕ∂θ.

∇∂θ∂ϕ =

(∂rϕ∂θ

)tangent

= (rθϕ)tangent , ∇∂ϕ∂θ =

(∂rθ∂ϕ

)tangent

= (rϕθ)tangent

We have

∇∂θ∂ϕ = ∇∂ϕ∂θ = (rϕθ)tangent =

−R cos θ sinϕR cos θ cosϕ

0

tangent

.

We see that the vector rϕθ is orthogonal to n:

〈rϕθ,n〉 = −R cos θ sinϕ sin θ cosϕ+R cos θ cosϕ sin θ sinϕ = 0.

Hence

∇∂θ∂ϕ = ∇∂ϕ∂θ = (rϕθ)tangent = rϕθ =

−R cos θ sinϕR cos θ cosϕ

0

= cotan θrϕ .

We come to

∇∂θ∂ϕ = ∇∂ϕ∂θ = cotan θ∂ϕ ⇒ Γθθϕ = Γθϕθ = 0, Γϕθϕ = Γϕϕθ = cotan θ (2.27)

Finally calculate ∇∂ϕ∂ϕ

∇∂ϕ∂ϕ = (rϕϕ)tangent =

−R sin θ cosϕ−R sin θ sinϕ

0

tangent

Projecting on the tangent vectors to the sphere (see (2.25)) we have

∇∂ϕ∂ϕ = (rϕϕ)tangent = rϕϕ − n〈n, rϕϕ〉 =

53

−R sin θ cosϕ−R sin θ sinϕ

0

−sin θ cosϕ

sin θ sinϕcos θ

(−R sin θ cosϕ sin θ cosϕ−R sin θ sinϕ sin θ sinϕ) =

− sin θ cos θ

R cos θ cosϕR cos θ sinϕ−R sin θ

= − sin θ cos θrθ,

i.e.

∇∂ϕ∂ϕ = − sin θ cos θrθ ⇒ Γθϕϕ = − sin θ cos θ, Γϕϕϕ = Γϕϕϕ = 0 . (2.28)

2.3 Levi-Civita connection

2.3.1 Symmetric connection

Definition. We say that connection is symmetric if its Christoffel symbolsΓikm are symmetric with respect to lower indices

Γikm = Γimk (2.29)

The canonical flat connection and induced connections considered above aresymmetric connections.

Invariant definition of symmetric connectionA connection ∇ is symmetric if for an arbitrary vector fields X,Y

∇XY −∇YX− [X,Y] = 0 (2.30)

If we apply this definition to basic fields ∂k, ∂m which commute: [∂k, ∂m] = 0 wecome to the condition

∇∂k∂m −∇∂m∂k = Γimk∂i − Γikm∂i = 0

and this is the condition (2.29).

2.3.2 Levi-Civita connection. Theorem and Explicit formulae

Let (M,G) be a Riemannian manifold.Definition-TheoremA symmetric connection ∇ is called Levi-Civita connection if it is com-

patible with metric, i.e. if it preserves the scalar product:

∂X〈Y,Z〉 = 〈∇XY,Z〉+ 〈Y,∇XZ〉 (2.31)

54

for arbitrary vector fields X,Y,Z.There exists unique levi-Civita connection on the Riemannian manifold.In local coordinates Christoffel symbols of Levi-Civita connection are given

by the following formulae:

Γimk =1

2gij(∂gjm∂xk

+∂gjk∂xm

− ∂gmk∂xj

). (2.32)

where G = gikdxidxk is Riemannian metric in local coordinates and ||gik|| is

the matrix inverse to the matrix ||gik||.ProofSuppose that this connection exists and Γimk are its Christoffel symbols. Con-

sider vector fields X = ∂m,Y = ∂i and Z = ∂k in (2.31). We have that

∂mgik = 〈Γrmi∂r, ∂k〉+ 〈∂i,Γrmk∂r〉 = Γrmigrk + girΓrmk . (2.33)

for arbitrary indices m, i, k.Denote by Γmik = Γrmigrk we come to

∂mgik = Γmik + Γmki, i.e.

Now using the symmetricity Γmik = Γimk since Γkmi = Γkim we have

Γmik = ∂mgik − Γmki = ∂mgik − Γkmi = ∂mgik − (∂kgmi − Γkim) =

∂mgik−∂kgmi+ Γkim = ∂mgik−∂kgmi+ Γikm = ∂mgik−∂kgmi+ (∂igkm − Γimk) =

∂mgik − ∂kgmi + ∂igkm − Γmik .

Hence

Γmik =1

2(∂mgik + ∂igmk − ∂kgmi)⇒ Γkim =

1

2gkr (∂mgir + ∂igmr − ∂rgmi) (2.34)

We see that if this connection exists then it is given by the formula(2.32).On the other hand one can see that (2.32) obeys the condition (2.33). We

prove the uniqueness and existence.

since ∇∂i∂k = Γmik∂m.

Consider examples.

55

2.3.3 Levi-Civita connection of En

For Euclidean space En in standard Cartesian coordinates

GEucl

= (dx1)2 + · · ·+ (dxn)2 = δikdxidxk

Components of metric are constants (they are equal to 0 or 1). Hence obvi-ously Christoffel symbols of Levi-Civita connection in Cartesian coordinatesaccording formula (2.32) vanish:

ΓIkm = 0 in Cartesian coordiantes

Recalling canonical flat connection (see 2.1.3) we come to simple but impor-tant observation:

Observation Levi-Civita connection coincides with canonical flat con-nection in Euclidean space En. They have vanishing Cristoffel symbols inCartesian coordinates.

2.3.4 Levi-Civita connection on 2-dimensional Riemannian mani-fold with metric G = adu2 + bdv2.

Example Consider 2-dimensional manifold with Riemannian metrics

G = a(u, v)du2 + b(u, v)dv2, G =

(g11 g12

g21 g22

)=

(a(u, v) 0

0 b(u, v)

)Calculate Christoffel symbols of Levi Civita connection.

Using (2.34) we see that:

Γ111 = 12

(∂1g11 + ∂1g11 − ∂1g11) = 12∂1g11 = 1

2au

Γ211 = Γ121 = 12

(∂1g12 + ∂2g11 − ∂1g12) = 12∂2g11 = 1

2av

Γ221 = 12

(∂2g12 + ∂2g12 − ∂1g22) = −12∂1g22 = − 1

2bu

Γ112 = 12

(∂1g12 + ∂1g12 − ∂2g11) = −12∂2g11 = − 1

2av

Γ122 = Γ212 = 12

(∂2g21 + ∂1g22 − ∂2g21) = 12∂1g22 = 1

2bu

Γ222 = 12

(∂2g22 + ∂2g22 − ∂2g22) = 12∂2g22 = 1

2bv

(2.35)

56

To calculate Γikm = girΓkmr note that for the metric a(u, v)du2 + b(u, v)dv2

G−1 =

(g11 g12

g21 g22

)=

(1

a(u,v)0

0 1b(u,v)

)

Hence

Γ111 = g11Γ111 = au

2a, Γ1

21 = Γ112 = g11Γ121 = av

2a, Γ1

22 = g11Γ221 = −bu2a

Γ211 = g22Γ112 = −av

2b, Γ2

21 = Γ212 = g22Γ122 = bu

2b, Γ2

22 = g22Γ222 = bv2b

(2.36)

2.3.5 Example of the sphere again

Calculate Levi-Civita connection on the sphere.On the sphere first quadratic form (Riemannian metric) G = R2dθ2 +

R2 sin2 θdϕ2. Hence we use calculations from the previous example witha(θ, ϕ) = R2, b(θ, ϕ) = R2 sin2 θ (u = θ, v = ϕ). Note that aθ = aϕ = bϕ = 0.Hence only non-trivial components of Γ will be:

Γθϕϕ =−bθ2a

=− sin 2θ

2,

(Γϕϕθ =

−R2 sin 2θ

2

), (2.37)

Γϕθϕ = Γϕϕθ =bθ2b

=cos θ

sin θ

(Γθϕϕ =

R2 sin 2θ

2

)(2.38)

All other components are equal to zero:

Γθθθ = Γθθϕ = Γθϕθ = Γϕθθ = Γϕϕϕ = 0

Remark Note that Christoffel symbols of Levi-Civita connection on thesphere coincide with Christoffel symbols of induced connection calculated inthe subsection ”Connection induced on surfaces”. later we will understandthe geometrical meaning of this fact.

2.4 Levi-Civita connection = induced connection onsurfaces in E3

We know already that canonical flat connection of Euclidean space is the Levi-Civita connection of the standard metric on Euclidean space. (see section

57

2.3.3.) Now we show that Levi-Civita connection on surfaces in Euclideanspace coincides with the connection induced on the surfaces by canonical flatconnection. We perform our analysis for surfaces in E3.

Let M : r = r(u, v) be a surface in E3. Let G be induced Riemannianmetric on M and ∇ Levi–Civita connection of this metric.

We know that the induced connection ∇(M) is defined in the followingway: for arbitrary vector fields X,Y tangent to the surface M , ∇M

X Y equalsto the projection on the tangent space of the vector field ∇can.flat

X Y:

∇MX Y =

(∇can.flat

X Y)

tangent,

where ∇can.flat is canonical flat connection in E3 (its Christoffel symbolsvanish in Cartesian coordinates). We denote by Atangent a projection ofthe vector A attached at the point of the surface on the tangent space:A⊥ = A− n(A,n) , (n is normal unit vector field to the surface.)

Theorem Induced connection on the surface r = r(u, v) in E3 coincideswith Levi-Civita connection of Riemannian metric induced by the canonicalmetric on Euclidean space E3.

ProofLet ∇M be induced connection on a surface M in E3 given by equations

r = r(u, v). Considering this connection on the basic vectors rh, rv we seethat it is symmetric connection. Indeed

∇M∂u∂v = (ruv)tangent = (rvu)tangent = ∇M

∂v∂u .⇒ Γuuv = Γuvu,Γvuv = Γvvu .

Prove that this connection preserves scalar product on M . For arbitrarytangent vector fields X,Y,Z we have

∂X〈Y,Z〉E3 = 〈∇can. flatX Y,Z〉E3 + 〈Y,∇can.flat

X Z〉E3 .

since canonical flat connection in E3 preserves Euclidean metric in E3 (itis evident in Cartesian coordinates). Now project the equation above onthe surface M . If A is an arbitrary vector attached to the surface andAtangent is its projection on the tangent space to the surface, then for ev-ery tangent vector B scalar product 〈A,B〉E3 equals to the scalar product〈Atangent,B〉E3 = 〈Atangent,B〉M since vector A − Atangent is orthogonal tothe surface. Hence we deduce from (2) that ∂X〈Y,Z〉M =

〈(∇can. flat

X Y)

tangent,Z〉E3+〈Y,

(∇can.flat

X Z)

tangent〉E3 = 〈∇M

X Y,Z〉M+〈Y,∇MX Z〉M .

58

We see that induced connection is symmetric connection which preservesthe induced metric. Hence due to Levi-Civita Theorem it is unique and isexpressed as in the formula (2.32).

Remark One can easy to reformulate and prove more general statement: Let M be

a submanifold in Riemannian manifold (E,G). Then Levi-Civita connection of the metric

induced on this submanifold coincides with the connection induced on the manifold by

Levi-Civita connection of the metric G.

2.5 ∗ Killing vectors, antisymmetric operator and an-tisymmetric bilinear form

We return to Killing vectors.First consider the following construction.Let K be an arbitrary vector field, then consider bilinear form

SK(X,Y) = G (∇XK ,Y) = 〈∇XK ,Y〉 = 〈Y ,∇XK〉 , (2.39)

where ∇ is an arbitrary connection, G Riemannian metric, defining scalarproduct 〈 , 〉, and K,X,Y arbitrary vector fields. One can see that forarbitrary functions f, g

S(fX, gY) = fgS(X,Y)

This immediately follows from definition of connection (see condition (2.4)in 2.1.1). In local coordinates if X = Xm∂m, Y = Y k∂k then

SK(X,Y) = S(∇XK,Y) = S(∇Xm∂mKi∂i, Y

n∂n) = XmY nSmn ,

where

Smn = 〈(∂mKi+ΓimrKr)Kr∂i, ∂n〉 = 〈(∂mKi+ΓimrK

r)∂i, ∂n〉 =(∂mK

i + ΓimrKr)gin .

(2.40)We see that this construction defines covariant tensor field.

Theorem Let ∇ be Levi-Civita connection on Riemannian manifold(M,G). Then vector field K is a Killing vector field on M if and only if tensorfield SK(X,Y) is antisymmetric tensor field: SK(X,Y) = −SK(X,Y).

ProofFirst recall properties of Killing vector field.

59

Let M be Riemannian manifold with Riemannian metric G. Recall thata vector field K is Killing vector field, i.e. it defines infinitesimal isometry, if

LKG = 0 ,

Notice that for an arbitrary vector field Z and arbitrary vetor fields X,Y wehave

LZG(X,Y) = ∂ZG(X,Y) =

= (LZG)(X,Y)+G(LZX,Y)+G(X,LZY) = LZG(X,Y)+G([Z,X],Y)+G(X, [Z,Y]) .(2.41)

. In the case if Z = K is Killing vector field then condition (2.5) reads

∂KG(X,Y) = G(LKX,Y) +G(X,LKY) = G([K,X], Y ) +G(X, [K,Y]) .(2.42)

Now let ∇ be Levi-Civita connection of Riemannian metric, i.e.

∂ZG(X,Y) = G(∇ZX,Y) +G(X,∇ZY)

for arbitrary vector fields Z,X,Y (see levicivitaconnection1 ). In particular for Z = K we have

∂KG(X,Y) = G(∇KX,Y) +G(X,∇KY) .

Now transform the relation (2.42) and compare it with this relation. Per-forming this transformation we will use the symmetricity of Levi-Civita con-nection, i.e.

∇XY −∇YX = [X,Y] . (2.43)

(see symmetricconnectioninvariant and Homework 4.) We have:

∂KG(X,Y) = G(LKX,Y) +G(X,LKY) = G([K,X],Y) +G(X, [K,Y]) =

G(∇KX−∇XK,Y) +G(X,∇KY −∇YK) =

G(∇KX,Y) +G(X,∇KY)︸ ︷︷ ︸∂K(X,Y)

−G(∇XK,Y)−G(X,∇YK)

This implies that

G(∇XK,Y) +G(X,∇YK) = SK(X,Y) + SK(Y,X) = 0 .

, i.e. SK is antisymmetric. One can easy to see that converse implication isalso true.

Remark Note that the antisymmetric tensor field SK defines antisym-metric linear operator

A : X 7→ ∇XK .

60

3 Parallel transport and geodesics

3.1 Parallel transport

3.1.1 Definition

Let M be a manifold equipped with affine connection ∇.Definition Let C : x(t), t0 ≤ t ≤ t1 be a curve on the manifold M ,

starting at the point p0 = x(t0) and ending at the point p1 = x(1) ((withcoordinates xi = xi(t))). Let X = X(t0) be an arbitrary tangent vectorattached at the initial point p0 = x0 (with coordinates xi(t0)) of the curveC, i.e. X(t0) ∈ Tp0M is a vector tangent to the manifold M at the pointp0 with coordinates xi(t0). (The vector X is not necessarily tangent to thecurve C)

We say that X(t), t0 ≤ t ≤ t1 is a parallel transport of the vector X(t0) ∈T∂0M along the curve C : xi = xi(t), t0 ≤ t ≤ t1 if

• For an arbitrary t, t0 ≤ t ≤ t, vector X = X(t), (X(t)|t=t0 = X(t0)) isa vector attached at the point x(t) of the curve C, i.e. X(t) is a vectortangent to the manifold M at the point x(t) of the curve C.

• The covariant derivative of X(t) along the curve C equals to zero:

∇X

dt= ∇vX = 0 . (3.1)

In components: if Xm(t) are components of the vector field X(t) andvm(t) are components of the velocity vector v of the curve C ,

X(t) = Xm(t)∂

∂xm|x(t) , v =

dx(t)

dt=dxi

dt

∂

∂xm|x(t)

then the condition (3.1) can be rewritten as

dX i(t)

dt+ vk(t)Γikm(xi(t))Xm(t) ≡ 0 . (3.2)

Remark We say sometimes that X(t) is covariantly constant along thecurve C if X(t) is parallel transport of the vector X along the curve C. Ifwe consider Euclidean space with canonical flat connection then in Cartesian

61

coordinates Christoffel symbols vanish and parallel transport is nothing butdXdt

= ∇vX = 0, X(t) is constant vector.Remark Compare this definition of parallel transport with the defini-

tion which we consider in the course of ”Introduction to Geometry” wherewe consider parallel transport of the vector along the curve on the surfaceembedded in E3 and define parallel transport by the condition, that onlyorthogonal component of the vector changes during parallel transport, i.e.dX(t)dt

is a vector orthogonal to the surface

3.1.2 Parallel transport is a linear map. Parallel transport withrespect to Levi-Civita connection

We usualy consider parallel transport on Riemannian manifold with respectto Levi-Civita connection. If (M,G) is Riemannian manifold then we con-sider parallel transport with respect to connection ∇ which is Levi-Civitaconnection of the Riemannian metric G.

Consider again curve C : x = x(t), t0 ≤ t ≤ t1 on manifold M startingat the point p0 and ending at the point p1 (see above). Let X ∈ T∂0 be anarbitrary tangent vector at the point ∂0, and X(t) be parallel transport (3.1)of this vector along the curve C:

X(t)∣∣t=t0

= X ,∇X(t)

dt= 0 .

Taking value of X(t) at the final point p1 of the curve C we come to the newvector X′ = X(t)

∣∣t=t1

tangent to the manifold M at the point p1. Thus wedefine the map between tangent vectors at the initial point p0 of the curveC and tangent vectors at the ending point p1 of this curve:

PC : Tp0M 3 X −→ PC(X) = X′ ∈ Tp1M . (3.3)

Sure this map depends on the curve C which joins starting and ending points(if we are not in Euclidean space).

PropositionLet C be a an arbitrary curve with starting point p0 and ending point

p1. Then the map (3.3) defines linear operator PC which does not depend

62

on parameterisation of the curve11:

PC(λX1 + µX2) = λPC(X1) + µPC(X2) . (3.4)

In the case if connection ∇ is Levi-Civita connection, then PC is an orthog-onal operator: for two arbitrary vectors X,Y ∈ T∂0M

〈X,Y〉p0 = 〈X′,Y′〉p1 X′ = PC(X) ∈ Tp1M , Y′ = PC(Y) ∈ Tp1M ,(3.5)

where as usual 〈 , 〉p is the scalar product at the point p.In particular the length of the vector is preserved during parallel trans-

port.

Proof The fact that it is a linear map follows immediately from the factthat differential equations (3.2) are linear. If vector fields X(t),Y(t) arecovariantly constant along the curve C, i.e. they obey differential equation(3.2), then their linear combination λX(t) +µY(t) obeys this equation, also,This implies (3.4).

The fact that the map (3.3) does not depend on the parameterisation(if it does not change the orinetation) also follows from differential equation(3.2). Indeed let t = t(τ), τ0 ≤ τ ≤ τ1, t(τ0) = t0, t(τ1) = τ1 be anotherparameterisation of the curve C, which does not change parameterisation, i.e.initial and ending points of the curve do not intechange. Then multiplyingthe equation (3.2) on dt

dτand using the fact that velocity v′(τ) = tτv(t) we

come to differential equation:

dX i(t(τ))

dτ+ v′k(t(τ))Γikm(xi(t(τ)))Xm(t(τ)) ≡ 0 . (3.6)

The functions X(t(τ)) with the same initial conditions are the solutionsof this equation.

It remains to prove that PC is orthogonal operator.It follows immediately from the definition (3.1) of a parallel transport and

the definition (2.31) of Levi-Civita connection that during parallel transportthe scalar product 〈X(t),Y(t)〉x(t) is preserved:

d

dt〈X(t),Y(t)〉 = ∂v〈X(t),Y(t)〉 = 〈∇vX(t),Y(t)〉+〈X(t),∇vY(t)〉 = 〈0,Y(t)〉+〈X(t), 0〉 = 0 .

(3.7)This implies (3.5).

11we consider parameterisatons with the same initial and ending points, i.e. all repa-rameterisations do not change orientation. One can say that linear operator PC is anoperator defined for orineted curve, since we fix initial and ending poitns of the curve.

63

3.2 Geodesics

3.2.1 Definition. Geodesic on Riemannian manifold

Let M be manifold equipped with connection ∇.Definition A parameterised curve C : xi = xi(t) is called geodesic if

velocity vector v(t) : vi(t) = dxi(t)dt

is covariantly constant along this curve,i.e. parallel transport of velocity vector along the curve preserves the velocityvector:

∇vv =∇v

dt=dvi(t)

dt+ vk(t)Γikm(x(t))vm(t) = 0, i.e. (3.8)

d2xi(t)

dt2+dxk(t)

dtΓikm(x(t))

dxm(t)

dt= 0 . (3.9)

These are linear second order differential equations. One can prove thatthis equations have solution and it is unique12 for an arbitrary initial data(xi(t0) = xi0, x

i(t0) = xi0. )In other words the curve C : x(t) is a geodesic if parallel transport of

velocity vector along the curve is a velocity vector at any point of the curve.

Geodesics defined with Levi-Civita connection on the Riemannian mani-fold is called geodesic on Riemannian manifold. We mostly consider geodesicson Riemannian manifolds.

Since velocity vector of the geodesics on Riemannian manifold at anypoint is a parallel transport with the Levi-Civita connection, hence due toProposition above (see equation (3.4), (3.5) and (3.7)) the length of thevelocity vector remains constant:

Proposition If C : x(t) is a geodesics on Riemannian manifold then thelength of velocity vector is preserved along the geodesic.

Proof Since the connection is Levi-Civita connection then it preservesscalar product of tangent vectors, (see (2.31)) in particularly the length ofthe velocity vector v:

Example 1 Geodesics of Euclidean space. In Cartesian coordinatesChristoffel symbols of Levi-Civita connection vanish, and differential equa-tion (3.8), (3.9) are reduced to equation

d2xi(t)

dt2= 0 ,⇒ dxi(t)

dt= vi ⇒ xi = xi0 + vit . (3.10)

12this is true under additional technical conditions which we do not discuss here

64

We come to straight lines.

Example 2 Geodesics of cylindrical surface One can see that if Rieman-nian metric G = Gikdu

idvk have constant coefficients in coordinates ui thenChristoffel symbols of Levi-Civita connection vanish in these coordinates,(see formula (2.32)) and according to (3.10) geodesics are “straight lines” incoordinates ui. In particular this is a case for cylinder: If surface of cylin-

der is given by equation

x = a cosϕ

y = a sinϕ

z = h

then Riemannian metric is equal to

G = a2dϕ2 + dh2 and we come to equations:d2ϕ(t)dt2

= 0d2h(t)dt2

= 0⇒

dϕ(t)dt

= Ωdh(t)dt

= c⇒

ϕ(t)dt

= ϕ0 + Ωtdh(t)dt

= h0 + tc. (3.11)

In general case we come to helix:x = a cosϕ(t) = a cos (ϕ0 + Ωt)

y = a sinϕ(t) = a sin (ϕ0 + Ωt)

z = h(t) = h0 + ct

(3.12)

If c = 0 then geodesics are circles x2 + y2 = a2, z = h0. If angular velocityΩ = 0 then geodesics are vertical lines x = x0, y = y0, z = h0 + ct.

3.2.2 Geodesics and Lagrangians of ”free” particle on Riemannianmanifold.

Lagrangian and Euler-Lagrange equationsA function L = L(x, x) on points and velocity vectors on manifold M is aLagrangian on manifold M .

We assign to Lagrangian L = L(x, x) the following second order differen-tial equations

d

dt

(∂L

∂xi

)=∂L

∂xi(3.13)

In detaild

dt

(∂L

∂xi

)=

∂2L

∂xm∂xixm +

∂2L

∂xm∂xi··xm

=∂L

∂xi. (3.14)

65

These equations are called Euler-Lagrange equations of the LagrangianL. We will explain later the variational origin of these equations 13.

Lagrangian of ”free” particle

Let (M,G), G = gikdxidxk be a Riemannian manifold.

Definition We say that Lagrangian L = L(x, x) is the Lagrangian of a a‘free’ particle on the Riemannian manifold M if

L =gikx

ixk

2(3.15)

Example ”Free” particle in Euclidean space. Consider E3 with standardmetric G = dx2 + dy2 + dz2

L =gikx

ixk

2=x2 + y2 + z2

2(3.16)

Note that this is the Lagrangian that describes the dynamics of a free particle.

Example A ‘free’ particle on a sphere.The metric on the sphere of radius R is G = R2dθ2 + R2 sin2 θdϕ2. Re-

spectively for the Lagrangian of ”free” particle we have

L =gikx

ixk

2=R2θ2 +R2 sin2 θϕ2

2(3.17)

Equations of geodesics and Euler-Lagrange equations

Theorem. Euler-Lagrange equations of the Lagrangian of a free particleare equivalent to the second order differential equations for geodesics.

This Theorem makes very easy calculations for Christoffel indices.

This Theorem can be proved by direct calculations.Calculate Euler-Lagrange equations (3.13) for the Lagrangian (3.15):

d

dt

(∂L

∂xi

)=

d

dt

∂(gmkx

mxk

2

)∂xi

=d

dt

(gikx

k)

= gik··x k +

∂gik∂xm

xmxk

13To every mechanical system one can put in correspondence a Lagrangian on config-uration space. The dynamics of the system is described by Euler-Lagrange equations.The advantage of Lagrangian approach is that it works in an arbitrary coordinate system:Euler-Lagrange equations are invariant with respect to changing of coordinates since theyarise from variational principe.

66

and

∂L

∂xi=∂(gmkx

mxk

2

)∂xi

=1

2

∂gmk∂xi

xmxk.

Hence we have

d

dt

(∂L

∂xi

)= gik

··x k +

∂gik∂xm

xmxk =∂L

∂xi=

1

2

∂gmk∂xi

xmxk ,

i.e.

gik··x k + ∂mgikx

mxk =1

2∂igmkx

mxk .

Note that ∂mgikxmxk = 1

2

(∂mgikx

mxk + ∂kgimxmxk

). Hence we come to

equation:

gikd2xk

dt2+

1

2(∂mgik + ∂kgim − ∂igmk) xmxk

Multiplying on the inverse matrix gik we come

d2xi

dt2+

1

2gij(∂gjm∂xk

+∂gjk∂xm

− ∂gmk∂xj

)dxm

dt

dxk

dt= 0 . (3.18)

We recognize here Christoffel symbols of Levi-Civita connection (see (2.32))and we rewrite this equation as

d2xi

dt2+dxm

dtΓimk

dxk

dt= 0 . (3.19)

This is nothing but the equation (3.8).Applications of this Theorem: calculation of Christoffel symbols of Levi-

Civita connection.

3.2.3 Calculations of Christoffel symbols and geodesics using theLagrangians of a free particle.

It turns out that equation (3.19) is the very effective tool to caluclatieChristoffel symbols of Levi-Civita connection.

Examples in this subsection will be calculated in detail on tutorial (seeHomework 7)

Consider two examples: We calculate Levi-Civita connection on spherein E3 and on Lobachevsky plane using Lagrangians and find geodesics.

67

1) Sphere of the radius R in E3:Lagrangian of ”free” particle on the sphere is given by (3.17):

L =R2θ2 +R2 sin2 θϕ2

2

Euler-Lagrange equations defining geodesics are

d

dt

(∂L

∂θ

)− ∂L

∂θ=

d

dt

(R2·θ

)−R2 sin θ cos θϕ2 ⇒

··θ − sin θ cos θϕ2 = 0 ,

(3.20)d

dt

(∂L

∂ϕ

)− ∂L

∂ϕ=

d

dt

(R2sin2 θϕ

)= 0⇒ ··

ϕ+ 2cotan θθϕ = 0 .

Comparing Euler-Lagrange equations with equations for geodesic in terms ofChristoffel symbols:

··θ + Γθθθθ

2 + 2Γθθϕθϕ+ Γθϕϕϕ2 = 0,

··ϕ+ Γϕθθθ

2 + 2Γϕθϕθϕ+ Γϕϕϕϕ2 = 0

we come toΓθθθ = Γθθϕ = Γθϕθ = 0 ,Γθϕϕ = − sin θ cos θ , (3.21)

Γϕθθ = Γϕϕϕ = 0, Γϕθϕ = Γϕϕθ = cotan θ . (3.22)

(Compare with previous calculations for connection in subsections 2.2.1 and2.3.4)

We know already and we will prove later in a elegant way that geodesics on the sphereare great circles. (see subsection 3.2.8 below). Consider another technically more difficultbut straightforward proof of this fact. To find geodesics one have to solve second orderdifferential equations (3.19)

One can see that the great circles: ϕ = ϕ0, θ = θ0 + t are solutions of second orderdifferential equations (3.20) with initial conditions

θ(t)∣∣t=0

= θ0, θ(t)∣∣t=0

= 1, ϕ(t)∣∣t=0

= ϕ0, θ(t)∣∣t=0

= 0 . (3.23)

The rotation of the sphere is isometry, which does not change Levi-Civta connection.Hence an arbitrary great circle is geodesic.

Prove that an arbitrary geodesic is an arc of great circle. Let the curve θ = θ(t), ϕ =ϕ(t), 0 ≤ t ≤ t1 be geodesic. Rotating the sphere we can come to the curve θ = θ′(t), ϕ =ϕ′(t), 0 ≤ t ≤ t1 such that velocity vector at the initial time is direccted along meridian,i.e. initial conditions are

θ′(t)∣∣t=0

= θ0, θ′(t)∣∣t=0

= a, ϕ′(t)∣∣t=0

= ϕ0, ϕ′(t)∣∣t=0

= 0 . (3.24)

68

(Compare with initial conditions (3.23)) Second order differential equations with boundaryconditions for coordinates and velocities at t = 0 have unique solution. The solutions ofsecond order differential equations (3.20) with initial conditions (3.24) is a curve θ′(t) =θ0 + at, ϕ′(t) = ϕ0. It is great circle. Hence initial curve the geodesic θ = θ(t), ϕ = ϕ(t),0 ≤ t ≤ t1 is an arc of great circle too.

This is another proof that geodesics are great circles.

2) Lobachevsky plane.Lagrangian of ”free” particle on the Lobachevsky plane with metric G =

dx2+dy2

y2 is

L =1

2

x2 + y2

y2.

Euler-Lagrange equations are

∂L

∂x= 0 =

d

dt

∂L

∂x=

d

dt

(x

y2

)=

··x

y2− 2xy

y3, i.e.

··x− 2xy

y= 0 ,

∂L

∂y= − x

2 + y2

y3=

d

dt

∂L

∂y=

d

dt

(y

y2

)=

··y

y2− 2y2

y3, i.e.

··y +

x2

y− y2

y= 0 .

Comparing these equations with equations for geodesics:··xi− xkΓikmxm = 0

(i = 1, 2, x = x1, y = x2) we come to

Γxxx = 0,Γxxy = Γxyx = −1

y, Γxyy = 0, Γyxx =

1

y,Γyxy = Γyyx = 0,Γyyy = −1

y.

In a similar way as for a sphere one can find geodesics on Lobachevsky plane. First we

note that vertical rays are geodesics. Then using the inversions with centre on the absolute

one can see that arcs of the circles with centre at the absolute (y = 0) are geodesics too.

See another examples in Homework 6 and 7.

3.2.4 † Magnitudes preserved along geodesics—Integrals of mo-tion

It is very useful to find magnitudes which are preserved along geodesics, functions F =F (x, x) such that for geodesic C : xi = xi(t) the magnitude

I(t) = F (x, x)∣∣ xi = xi(t) is preserved along geodesic xi(t),

dI(t)

dt= 0 . (3.25)

69

Geodesics are solutions of equations of motions for the Lagrangian of a free particle L =gik(x)xixk

2 . One can consider such functions F = F (x, x) for an arbitrary Lagrangian L.In this case xi(t) is a solution of the Lagrangian L.

These magnitudes which are preserved along solutions of equations of motion (inparticular along geodesics in the case if L is the Lagrangian of a free particle) are calledintegrals of motion (See in detail about integrals of motion in Appendix to this lectures).

There is the following very useful criterion to find magnitudes, which are preserved onequations of motions, i.e. integrals fo motion.

Proposition Let Lagrangian L(xi, xi) in coordinates xi does not depend, say onthe coordinate x1. L = L(x2, . . . , xn, x1, x2, . . . , xn). Then the function

F1(x, x) =∂L

∂x1

is integral of motion. (In the case if L(xi, xi) does not depend on the coordinate xi. thefunction Fi(x, x) = ∂L

∂xiwill be integral of motion.)

Proof is simple. Check the condition (3.25): Euler-Lagrange equations of motion are:

d

dt

(∂L

∂xi

)− ∂L

∂xi= 0 (i = 1, 2, . . . , n)

In particular for first coordinate x1, ∂L∂x1 = 0 and

d

dt

(∂L

∂x1

)− ∂L

∂x1=

d

dt

(∂L

∂x1

)= 0 ,

i.e. the magnitude I(t) = F (x, x) is preserved if F = ∂L∂x1

. We see that exactly firstequation of motion is

d

dt

(∂L

∂x1

)=

d

dtF1(q, q) = 0 since ∂L

∂x1 = 0, .

(if L(xi, xi) does not depend on the coordinate xi then the function Fi(x, x) = ∂Lx1 is

integral of motion since i− th equation is exactly the condition Fi = 0.)The integral of motion Fi = ∂L

∂xiis called sometimes generalised momentum.

Consider examples of calculation of preserved mangnitudes along geodesics.Example (sphere)Sphere of the radius R in E3. Riemannian metric: G = Rdθ2 + R2 sin2 θdϕ2 and

Lfree = 12

(R2θ2 +R2 sin2 θϕ2

). Lagrangian does not depend explicitly on coordinate ϕ.

The integral of motion is

F =∂Lfree

∂ϕ= R2 sin2 θϕ .

It is preserved along geodesics, i.e. along great circles.

Example (cone)

70

Consider cone

x = kh cosϕ

y = kh sinϕ

z = h

. Riemannian metric:

G = d(kh cosϕ)2 + d(kh sinϕ)2 + (dh)2 = (k2 + 1)dh2 + k2h2dϕ2 .

and free Lagrangian

Lfree =(k2 + 1)h2 + k2h2ϕ2

2.

Lagrangian does not depend explicitly on coordinate h. The integral of motion is

F =∂Lfree

∂ϕ= k2h2ϕ.

It is preserved along geodesics.Remark One has to note that for the Lagrangian of a free particle F = L = gikx

ixk,kinetik energy, is integral of motion preserved along geodesic:it is nothing that square ofthe length of velocity vector which is preserved along the geodesic.

See these and other examples in Homework 6.

Using integral of motions to calculate geodesicsIntegrals of motions may be very useful to calculate geodesics. The equations for

geodesics are second order differential equations. If we know integrals of motions theyhelp us to solve these equations. Consider just an example.

For Lobachevsky plane the free Lagrangian L = x2+y2

2y2 . We already calculated geodesicsin the subsection 3.3.4. Geodesics are solutions of second order Euler-Lagrange equations

for the Lagrangian L = x2+y2

2y2 (see the subsection 3.3.4)··x− 2xy

y = 0··y + x2

y −y2

y = 0

It is not so easy to solve these differential equations.For Lobachevsky plane we know two integrals of motions:

E = L =x2 + y2

2y2, and F =

∂L

∂x=

x

y2. (3.26)

These both integrals are preserved in time: if x(t), y(t) is geodesics thenF = x(t)

y(t)2

E = x(t)2+y(t)2

2y(t)2 = C2

⇒

x = C1y

2

y = ±√

2C2y2 − C21y

4(3.27)

These are first order differential equations. It is much easier to solve these equations in

general case than initial second order differential equations.

71

3.2.5 ∗ Variational principe and Euler-Lagrange equations

Here very briefly we will explain how Euler-Lagrange equations follow fromvariational principe.

Let M be a manifold (not necessarily Riemannian) and L = L(xi, xi) bea Lagrangian on it.

Denote myMx2,t2

x1,t1the space of curves (paths) such that they start at the

point x1 at the ”time” t = t1 and end at the point x2 at the ”time” t = t2:

Mx2,t2

x1,t1= C : x(t), t1 ≤ t ≤ t2, x(t1) = x1,x(t2) = x2 . (3.28)

Consider the following functional S on the space Mx2,t2

x1,t1:

S [x(t)] =

∫ t2

t1

L(xi(t), xi(t)

)dt . (3.29)

for every curve x(t) ∈Mx2,t2

x1,t1.

This functional is called action functional.Theorem Let functional S attaints the minimal value on the path x0(t) ∈

Mx2,t2

x1,t1, i.e.

∀x(t) ∈Mx2,t2

x1,t1S[x0(t)] ≤ S[x(t)] . (3.30)

Then the path x0(t) is a solution of Euler-Lagrange equations of the La-grangian L:

d

dt

(∂L

∂xi

)=∂L

∂xiif x(t) = x0(t) . (3.31)

Remark The path x(t) sometimes is called extremal of the action func-tional (3.29).

We will use this Theorem to show that the geodesics are in some senseshortest curves 14.

14The statement of this Theorem is enough for our purposes. In fact in classical mechan-ics another more useful statement is used: the path x0(t) is a solution of Euler-Lagrangeequations of the Lagrangian L if and only if it is the stationary ”point” of the actionfunctional (3.29), i.e.

S[x0(t) + δx(t)]− S[x0(t) + δx(t)] = 0(δx(t)) (3.32)

for an arbitrary infinitesimal variation of the path x0(t): δx(t1) = δx(t2) = 0.

72

3.2.6 Un-parameterised geodesic

We defined a geodesic as a parameterised curve such that the velocity vectoris covariantly constant along the curve.

What happens if we change the parameterisation of the curve?

Another question: Suppose a tangent vector to the curve remains tangentto the curve during parallel transport. Is it true that this curve (in a suitableparameterisation) becomes geodesic?

Definition We call un-parameterised curve geodesic if under suitableparameterisation it obeys the equation (3.8) for geodesics.

Let C–be un-parameterised geodesic. Then the following statement isvalid.

Proposition A curve C (un-parameterised) is geodesic if an only if anon-zero vector tangent to the curve remains tangent to the curve duringparallel transport.

Proof. Let A be tangent vector at the point p ∈ C of the curve. Paralleltransport does not depend on parametersiation of the curve (see subsection3.1.2, equation (3.3)). Choose a suitable parameterisation xi = xi(t) suchthat xi(t) obeys the equations (3.8) for geodesics, i.e. the velocity vector v(t)is covariantly constant along the curve: ∇vv = 0. If A(t0) = cv(t0) at thegiven point p (c is a scalar coefficient) then due to linearity A(t) = cv(t) isa parallel transport of the vector A. The vector A(t) is tangent to the curvesince it is proportional to velocity vector. We proved that any tangent vectorremains tangent during parallel transport.

Now prove the converse: Let A(t) be a parallel transport of non-zerovector and it is proportional to velocity. If in a given parameterisation A(t) =

c(t)v(t) choose a reparameterisation t = t(τ) such that dt(τ)dτ

= c(t). In the

new parameterisation the velocity vector v′(τ) = dt(τ)dτ

v(t(τ)) = c(t)v(t) =A(t(τ)). We come to parameterisation such that velocity vector remainscovariantly constant. Thus we come to parameterised geodesic. Hence C isa geodesic.

Remark In particularly it follows from the Proposition above the follow-ing important observation:

Let C is un-parameterised geodesic, xi(t) be its arbitrary parameterisa-tion and v(t) be velocity vector in this parameterisation. Then the velocityvector remains parallel to the curve since it is a tangent vector.

73

In spite of the fact that velocity vector is not covariantly constant alongthe curve, i.e. it will not remain velocity vector during parallel transport,since it will be remain tangent to the curve during parallel transport.

Remark One can see that if xi = xi(t) is geodesic in an arbitrary pa-rameterisation and s = s(t) is a natural parameter (which defines the lengthof the curve) then xi(t(s)) is parameterised geodesic.

3.2.7 Parallel transport of vectors along geodesics

We already now that during parallel transport along curve with respect toLevi-Civita connection scalar product of vectors, i.e. lengths of vectors andangle between them does not change (see subsection 3.1.2, equations (3.3)and (3.5)). This remark makes easy to calculate parallel transport of vectorsalong geodesics in Riemannian manifold. Indeed let C a geodesic (in generalun-parameterised) and a vectors X(t) is attached to the point p1 ∈ C onthe curve C. In the special case if X is a tangent vector to geodesic C thenduring parallel transport it remains tangent, i.e. proportional to velocityvector:

X(t) = a(t)v(t) . (3.33)

Here v(t) = dr(t)dt

and r = r(t) is an arbitrary parameterisation of geodesic C.Note that in general t is not parameter such that r = r(t) is parameterisedgeodesic; t is an arbitrary parameter. In the special case if t is a parametersuch that r = r(t) is parameterised geodesic then velocity vector remainsvelocity vector during parallel transport, i.e. X(t) = av(t) where a is notdependent on t.

To calculate the dependence of coefficient a on t in (3.34) we note that thelength of the vector is not changed (see equation (3.5) in the section 3.1.2,i.e.

〈X(t),X(t)〉 = 〈a(t)v(t), a(t)v(t)〉 = a2(t)|v(t)|2 = constant (3.34)

3.2.8 Geodesics on surfaces in E3

Let M : r = r(u, v) be a surface in E3. Let GM be induced Riemannianmetric and ∇ a Levi-Civita connection on M . We consider on M Levi-Civitaconnection of the metric GM .

Let C be an arbitrary geodesic and v(t) = dr(t)dt

the velocity vector. Ac-cording to the definition of geodesic ∇vv = 0. On the other hand we know

74

that Levi-Civita connection coincides with the connection induced on thesurface by canonical flat connection in E3 (see the Theorem in subsection2.4). Hence

∇vv = 0 = ∇Mv v =

(∇can.flat

v v)

tangent(3.35)

In Cartesian coordinates ∇can.flatv v = ∂vv = d

dtv(u(t), v(t)) = d2r(t)

dt2= a.

Hence according to (3.35) the tangent component of acceleration equalsto zero.

Converse if for the curve r(t) = r(u(t), v(t)) the acceleration vector a(t)is orthogonal to the surface then due to (3.35) ∇vv = 0.

We come to very beautiful observation:Theorem The acceleration vector of an curve r = r(u(t), v(t)) on M is

orthogonal to the surface M if and only if this curve is geodesic.

In other words due to Newton second law particle moves along alonggeodesic on the surface if and only if the force is orthogonal to the surface.

One can very easy using this Proposition to calculate geodesics of cylinderand sphere.

Geodesic on the cylinder

Let r(h(t), ϕ(t)) be a geodesic on the cylinder

x = a cosϕ

y = a sinϕ

z = h

. We have

v = drdt

=

−aϕ sinϕaϕ cosϕ

h

and for acceleration:

a =dv

dt=

−a··ϕ sinϕ

a··ϕ cosϕ··h

︸ ︷︷ ︸

tangent acceleration

+

−aϕ2 cosϕ−aϕ2 sinϕ

0

︸ ︷︷ ︸

normal acceleration

Since tangential acceleration equals to zero hence d2hdt2

= 0 and h(t) = h0 + ctNormal acceleration is centripetal acceleration of the rotation over circlewith constant speed (projection on the plane OXY ). The geodesic is helix.(Compare these calculations with calculations of geodesics of cylinder in thelast example of section 3.2.1: see (3.12).)

75

Geodesics on sphere

Let r = r(θ(t), ϕ(t)) be a geodesic on the sphere of the radius a: r(θ, ϕ) :

x = a sin θ cosϕ

y = a sin θ sinϕ

z = a cos θ

Consider the vector product of the vectors r(t) and velocity vector v(t)M(t) = r(t)× v(t). Acceleration vector a(t) is proportional to the r(t) sincedue to Proposition it is orthogonal to the surface of the sphere. This impliesthat M(t) is constant vector:

d

dtM(t) =

d

dt(r(t)× v(t)) = (v(t)× v(t)) + (r(t)× a(t)) = 0 (3.36)

We have M(t) = M0. r(t) is orthogonal to M = r(t) × v(t). We seethat r(t) belongs to the sphere and to the plane orthogonal to the vectorM0 = r(t) × v(t). The intersection of this plane with sphere is a greatcircle. We proved that if r(t) is geodesic hence it belongs to great circle (asun-parameterised curve).

The converse is evident since if particle moves along the great circle withconstant velocity then obviously acceleration vector is orthogonal to the sur-face.

Remark The vector M = r(t) × v(t) is the torque. The torque is inte-gral of motion in isotropic space.—This is the core of the considerations forgeodesics on the sphere.

3.2.9 ∗ Geodesics and shortest distance.

Many of you know that geodesics are in some sense shortest curves. Wewill give an exact meaning to this statement and prove it using variationalprincipe:

Let M be a Riemannian manifold.Theorem Let x1 and x2 be two points on M . The shortest curve which

joins these points is an arc of geodesic.Let C be a geodesic on M and x1 ∈ C. Then for an arbitrary point x2 ∈ C

which is close to the point x1 the arc of geodesic joining the points x1,x2 isa shortest curve between these points15 .

15More precisely: for every point x1 ∈ C there exists a ball Bδ(x1) such that for anarbitrary point x2 ∈ C ∩Bδ(x1) the arc of geodesic joining the points x1,x2 is a shortestcurve between these points.

76

This Theorem makes a bridge between two different approach to geodesic:the shortest disntance and parallel transport of velocity vector.

Sketch a proof:Consider the following two Lagrangians: Lagrangian of a ”free ” particle Lfree =

gik(x)xixk

2 and the length Lagrangian

Llength(x, x) =√gik(x)xixk =

√2Lfree .

If C : xi(t), t1 ≤ t ≤ t2 is a curve on M then

Length of the curve C =∫ t2

t1

Llength(xi(t), xi(t))dt =

∫ t2

t1

√gik(x(t))xi(t)xk(t)dt . (3.37)

The proof of the Theorem follows from the following observation:Observation Euler-Lagrange equations for the length functional (3.37) are equivalent

to the Euler-Lagrange equations for action functional (3.29). This means that extremalsof the length functional and action functionals coincide.

Indeed it follows from this observation and the variational principe that the shortestcurves obey the Euler-Lagrange equations for the action functional. We showed beforethat Euler-Lagrange equations for action functional (3.29) define geodesics. Hence theshortest curves are geodesics.

One can check the observation by direct calculation: Calculate Euler-Lagrange equa-tions for the Lagrangian Llength =

√gik(x)xixk =

√2Lfree:

d

dt

(∂Llength

∂xi

)− ∂Llength

∂xi=

d

dt

(1√

gikxixkgikx

k

)− 1

2√gikxixk

∂gkmxkxm

∂xi

=d

dt

(1

Llength

∂Lfree

∂xi

)− 1

Llength

∂Lfree

∂xi= 0 . (3.38)

To facilitate calculations note that the length functional (3.37) is reparameterisationinvariant. Choose the natural parameter s(t) or a parameter proportional to the naturalparameter on the curve xi(t). We come to Llength = const and it follows from (3.38) that

d

dt

(∂Llength

∂xi

)− ∂Llength

∂xi=

1

Llength

(d

dt

(∂Lfree

∂xi

)− ∂Lfree

∂xi

)= 0 .

We prove that Euler-Lagrange equations for length and action Lagrangians coincide.

In the Euclidean space straight lines are the shortest distances betweentwo points. On the other hand their velocity vectors are constant. We realisenow that in general Riemannian manifold the role of geodesic is twofold also:they are locally shortest and have covariantly constant velocity vectors.

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3.2.10 ∗ Again geodesics for sphere and Lobachevsky plane

The fact that geodesics are shortest gives us another tool to calculate geodesics.Consider again examples of sphere and Lobachevsky plane and find geodesics

using the fact that they are shortest. The fact that geodesics are locally theshortest curves

Consider again sphere in E3 with the radius R: Coordinates θ, ϕ, inducedRiemannian metrics (first quadratic form):

G = R2(dθ2 + sin2 θdϕ2) . (3.39)

Consider two arbitrary points A and B on the sphere. Let (θ0, ϕ0) be coor-dinates of the point A and (θ1, ϕ1) be coordinates of the point B

Let CAB be a curve which connects these points: CAB : θ(t), ϕ(t) suchthat θ(t0) = θ0, θ(t1) = θ1, ϕ(t0) = θ0, θ(t1) = θ1 then:

LCAB =

∫R√θ2t + sin2 θ(t)ϕ2

tdt (3.40)

Suppose that points A and B have the same latitude, i.e. if (θ0, ϕ0) arecoordinates of the point A and (θ1, ϕ1) are coordinates of the point B thenϕ0 = ϕ1 (if it is not the fact then we can come to this condition rotating thesphere)

Now it is easy to see that an arc of meridian, the curve ϕ = ϕ0 is geodesics:Indeed consider an arbitrary curve θ(t), ϕ(t) which connects the points A,B:θ(t0) = θ(t1) = θ0, ϕ(t0) = ϕ(t1) = ϕ0. Compare its length with the lengthof the meridian which connects the points A, B:∫ t1

t0

R

√θ2t + sin2 θϕ2

tdt ≥ R

∫ t1

t0

√θ2t dt = R

∫ t1

t0

θtdt = R(θ1 − θ0) (3.41)

Thus we see that the great circle joining points A,B is the shortest. Thegreat circles on sphere are geodesics. It corresponds to geometrical intuition:The geodesics on the sphere are the circles of intersection of the sphere withthe plane which crosses the centre.

Geodesics on Lobachevsky planeRiemannian metric on Lobachevsky plane:

G =dx2 + dy2

y2(3.42)

78

The length of the curve γ : x = x(t, y = y(t)) is equal to

L =

∫ √x2t + y2

t

y2(t)dt

In particularly the length of the vertical interval [1, ε] tends to infinity ifε→ 0:

L =

∫ √x2t + y2

t

y2(t)dt =

∫ 1

ε

√1

t2dt = log

1

ε

One can see that the distance from every point to the line y = 0 is equal toinfinity. This motivates the fact that the line y = 0 is called absolute.

Consider two points A = (x0, y0), B = (x1, y1) on Lobachevsky plane.It is easy to see that vertical lines are geodesics of Lobachevsky plane.Namely let points A,B are on the ray x = x0. Let CAB be an arc of

the ray x = x0 which joins these points: CAB : x = x0, y = y0 + t Thenit is easy to see that the length of the curve CAB is less or equal than thelength of the arbitrary curve x = x(t), y = y(t) which joins these points:x(t)

∣∣t=0

= x0, y(t)∣∣t=0

= y0, x(t)∣∣t=t1

= x0, y(t)∣∣t=t1

= y1:

∫ t

0

√x2t + y2

t

y2(t)dt ≥

∫ t

0

√y2t

y2(t)dt =

∫ y1

y0

dt

tdt = log

y1

y0

= length of CAB

Hence CAB is shortest. We prove that vertical rays are geodesics.Consider now the case if x0 6= x1. Find geodesics which connects two

points A,B which are not on the same vertical ray. Consider semicirclewhich passes these two points such that its centre is on the absolute. Weprove that it is a geodesic.

Proof Let coordinates of the centre of the circle are (a, 0). Then consider polar coor-dinates (r, ϕ):

x = a+ r cosϕ, y = r sinϕ (3.43)

In these polar coordinates r-coordinate of the semicircle is constant.Find Lobachevsky metric in these coordinates: dx = −r sinϕdϕ + cosϕdr, dy =

r cosϕdϕ+ sinϕdr, dx2 + dy2 = dr2 + r2dϕ2. Hence:

G =dx2 + dy2

y2=dr2 + r2dϕ2

r2 sin2 ϕ==

dϕ2

sin2 ϕ+

dr2

r2 sin2 ϕ(3.44)

79

We see that the length of the arbitrary curve which connects points A,B is greater orequal to the length of the arc of the circle:

LAB =

∫ t1

t0

√ϕ2t

sin2 ϕ+

r2t

r2 sin2 ϕdt ≥

∫ t1

t0

√ϕ2t

sin2 ϕdt = (3.45)

∫ t1

t0

ϕtsinϕ

dt =

∫ ϕ1

ϕ0

dϕ

sinϕ= log

tanϕ1

tanϕ1

The proof is finished.

4 Surfaces in E3

Now equipped by the knowledge of Riemannian geometry we consider sur-faces in E3. We reconsider again conceptions of Shape (Weingarten) opera-tor, Gaussian and mean curvatures, focusing attention on the the fact whatproperties are internal and what properties are external. In particular weconsider again Gaussian curvature and derive its internal meaning. We willconsider again Theorema Egregium.

4.1 Parallel transport of the vector. Formulation ofresult.

We formulate here very important theorem about parallel transport of vectorsover closed curve and deduce Theorema Egregium form this theorem.

We will consider also the formula for Gaussian curvature in isothermal coordinates

(This year this is not compulsory).

4.1.1 Theorem of parallel transport over closed curve

Let M be a surface in Euclidean space E3. Consider a closed curve C on M ,M : r = r(u, v), C : r = r(u(t, v(t)), 0 ≤ t ≤ t1, x(0) = x(t1). (u(t), v(t) areinternal coordinates of the curve C.)

Consider the parallel transport of an arbitrary tangent X vector alongthe closed curve C: Recall that we did it in the section 3.1.2 where wealready considered parallel transport over curve for arbitrary connection andfor Levi-Civita connection for arbitrary Riemannian manifold (see equations(3.3) and (3.5)). Now we will repeat these considerations for this special

80

case.)

X(t) = Xα(t)∂

∂uα

∣∣uα(t)︸ ︷︷ ︸

Internal observer

= Xα(t)rα∣∣r(u(t),v(t))︸ ︷︷ ︸

External observer

,

(rα =

∂xi

∂uα∂

∂xi

).

X(t) :∇X(t)

dt= 0, 0 ≤ t ≤ t1 ,

i.e.

dXα(t)

dt+Xβ(t)Γαβγ(u(t))

duγ(t)

dt= 0, 0 ≤ t ≤ t1 , α, β, γ = 1, 2 , (4.1)

where ∇ is the connection induced on the surface M by canonical flatconnection (see (2.25)), or (it is the same) the Levi-Civita connection (2.32)of the induced Riemannian metric on the surface M and Γαβγ its Christoffelsymbols:

Γγαβ =1

2gγπ(∂gπα∂uβ

+∂gπβ∂uα

− ∂gαβ∂uπ

),where gαβ = 〈rα, rβ〉 =

∂xi

∂uα∂xi

∂uβ

(4.2)are components of induced Riemannian metric GM = gαβdu

αduβ/Let r(0) = p be a starting (and ending) point of the curve C: r(0) =

r(t1) = p. The differential equation (4.1) defines the linear operator

PC : TpM −→ TpM (4.3)

For any vector X ∈ TpM , its image the vector RCX as the solution of thedifferential equation (4.1) with initial condition X(t)

∣∣t=0

= X. (See alsosection 3.1.2, equations (3.3) and (3.5)). )

On the other hand we know that parallel transport is orthogonal operator,it does not change the scalar product of two vectors, and it does not chnagelengths of vectors (see (3.5) in the subsection 3.1.2):

〈X,X〉 = 〈PCX, PCX〉 (4.4)

We see that PC is an orthogonal operator in the 2-dimensional vector spaceTpM . We know that orthogonal operator preserving orientation is the oper-ator of rotation on some angle φ.

81

One can see that PC preserves orientation 16 then the action of operatorPC on vectors is rotation on the angle, i.e. the result of parallel transportalong closed curve is rotation on the ∠φ. This angle depends depends onthe curve. The very beautiful question arises: How to calculate this angle∆Φ(C)

Theorem Let M be a surface in Euclidean space E3. Let C be a closedcurve C on M such that C is a boundary of a compact oriented domainD ⊂M . Consider the parallel transport of an arbitrary tangent vector alongthe closed curve C. As a result of parallel transport along this closed curveany tangent vector rotates through the angle

∠φ = ∠ (X, PCX) =

∫D

Kdσ , (4.5)

where K is the Gaussian curvature and dσ =√

det gdudv is the area elementinduced by the Riemannian metric on the surface M , i.e. dσ =

√det gdudv.

Remark One can show that the angle of rotation does not depend oninitial point of the curve.

Remark Here we assume that a curve C is smooth curve. E.g. if C is a smooth closedgeodesics then it follows from this Theorem and properties of geodesics that rotation angleis equal to 2πn (where n is integer). In general if C is piecewise smooth curve, then onecan see that rotation angle is equal to

∑αi− π(n− 1) where n is number of smooth arcs,

and αi angles between them.

Example Consider the closed curve, ”latitude” Cθ0 : θ = θ0 on the sphereof the radius R. Calculations show that

∠φ(Cθ0) = 2π(1− cos θ0) (4.6)

(see also the Homework 8). On the other hand the latitude Cθ0 is theboundary of the segment D with area 2πRH where H = R(1−cos θ0). Hence

∠ (X,RCX) =2πRH

R2=

1

R2· area of the segment =

∫D

Kdσ

since Gaussian curvature is equal to 1R2

To prove this Theorem we develop the technique of derivation formulae;we apply this technique to caclulate the Gaussain and mean curvature forsurfaces, and then in the section 4.3.3 we will prove this Theorem.

16We will consider mainly the case if the closed curve C is a boundary of a compactoriented domain D ⊂M . In this case one can see by continuity arguments that operatorRC preserves an orientation.

82

4.1.2 Gauß Theorema Egregium

We defined Gaussian curvature in terms of Weingarten (shape) operator asa product of principal curvatures. This definition was in terms of ExternalObserver.

The Theorem about transport over closed curve implies the remarkableCorollary:

Corollary Gauß Egregium TheoremaGaussian curvature of the surface can be expressed in terms of induced

Riemannian metric. It is invariant of isometries.

Indeed let D be a small domain around a given point p, let C its boundaryand ∠φ(D) be an angle of rotation. Denote by S(D) an area of this domain.Applying the Theorem for the case when area of the domain D tends to zerowe we come to the statement that

if S(D)→ 0 then ∠φ(D) =

∫D

Kdσ → K(p)S(D), i.e.

K(p) = limS(D)→0

∠φ(D)

S(D). (4.7)

Now notice that left hand side od this equation defining Gaussian cur-vature K(p) depends only on Riemannian metric on the surface C. Indeednumerator of LHS is defined by the solution of differential equation (4.1)which depends on Levi-Civita connection depending on the induced Rieman-nian metric, and denominator is an area depending on Riemannian metrictoo. Thus we come to GaußTheorema Egregium.

In next subsections we develop the technique which itself is very interest-ing. One of the applications of this technique is the proof of the Theorem(4.5). Thus we will prove Theorema Egregium too.

There are different other proofs of Theorema Egregium.E.g. it immediately follows from the formula (4.8) in the Theorem in subsection 4.1.3,

where the right hand side of the formula (4.8) depends on metric on the surface, i.e.Gaussian curvature maybe independently calculated by Internal Observer.

Later in the fifth section we will give also another proof of the Theorema Egregium,

calculating straightforwardly Gaussian curvature in terms of Riemannian curvature ten-

sor.

83

4.1.3 †Formula for Gaussian curvature in isothermal (conformal)coordinates

Now we will consider one very beautiful and illuminating formula to calculate Gaussiancurvature for surfaces.

Let M : r = r(u, v) be a surface in E3.Definition We say that coordinates (parameters) u, v are isothermal (or conformal)

if the induced Riemannian metric 1.3.1 is equal to

G = σ(u, v)(du2 + dv2)

Consider examples. For example if (u, v) are stereographic coordinates for sphere of radiusR we know that

G =4R4(du2 + dv2)

(R2 + u2 + v2)2)

i.e. stereographic coordiantes are conformal coordiantes:

G = σ(u, v)(du2 + dv2) with σ =4R4

(R2 + u2 + v2)2.

Another example: an arbitrary locally Euclidean surface, i.e. surface with induced Rie-mannian metric G = du2 + dv2 in some local coordinates u, v.

One can show that locally one can always find conformal (isothermal) coordinates onsurface in E3 17

Theorem Let surface M is given in conformal coordinates: r = r(u, v) such thatinduced Riemannian metric is equal to G = σ(u, v)(du2 + dv2). Then Gaussian curvatureK of the surface M is given by the formula

K = −1

2

∆(log σ)

σ, where ∆ = ∂2

∂u2 + ∂2

∂v2 . (4.8)

In particular this formula states that Gaussian curvature is expressed in terms of Rieman-nian metric. This implies GaußTheorema Egregium.

If u, v are conformal coordinates, G = σ(du2 + dv2), then it is often convenient tointroduce a function Φ(u, v) such that Φ = log σ, i.e. σ = eΦ(u,v). Then

G = eΦ(u,v)(du2 + dv2) . (4.9)

Then formula (4.8) will have the following appearance:

K = −1

2e−Φ∆(Φ) = −1

2e−Φ

(∂2Φ(u, v)

∂u2+∂2Φ(u, v)

∂v2

). (4.10)

What about existence of conformal (isothermal) coordinates? Proposition For sur-face M in E3

17The existence of local isothermal coordiantes is a part of famous Gauss theorem, whichcan be formulated in modern terms in the following way: every surface has a canonicalcomplex structure (z = u+ iv, z = u− iv). We will consider this question later.

84

• in a vicinity of an arbitrary point there exist isothermal coordinates i.e. coordinatessuch that induced metric G = eΦ(du2 + dv2) = eΦdzdz.

• If (u, v) and (u′, v′) are two arbitrary isothermal coordinates then the functionz = f(w) is holomorphic function or anti-holomorphic function,

We denote u+ iv = z, u− iv = z. Recall that if z = u+ iv then

Fz =∂F

∂z=

1

2

(∂

∂u− i ∂

∂v

)F, and Fz =

∂F

∂z=

1

2

(∂

∂u+ i

∂

∂v

)F (4.11)

Function F = f + ig is holomorphic ⇔ Fz = 0 ⇔ (fu + igu) + i(fv + igv) = 0 ⇔ fu = gvand fv = −gv (Cauchy Riemann conditions). Function F = f + ig is anti-holomorphic⇔ Fz = 0 ⇔ (fu + igu) − i(fv − igv) = 0. E.g. F = z2 = (u + iv)2 = u2 − v2 + 2iuv isholomorphic function, F = ez = eu−iv = eu(cos v − i sin v) is anti-holomorphic function(See also ??).

This Proposition immediately implies the following important corollary-Theorem:Two-dimensional surface in E3 has canonical complex structure, i.e. one can consider

a canonical atlas of local complex charts such that transition funtctions are analytic.Idea of Proof of this PropositionLet (ξ, η) be arbitrary parameters of surface and G = Adξ2 + 2Bdξdη + Cdη2 (g11 =

A, g12 = g21 = B, g22 = C). The positive-definiteness of the metric implies that G = ωωwhere ω = df + idg is 1-form. Use the fact that an arbitrary 1-form up to a mulitplierfunction is an exact form: ω = λdF . We come to isothermal coordinates: G = λλdFdF .To prove the second part of Proposition we just perform straightforward calculation. LetG = eΦ(du2 + dv2) = eΦdzdz in local coordinates z = u+ iv, and in new local coordinatesw = u′ + iv′ G = eΦ′

(du′2 + dv′2) = eΦ′dwdw, where Let w = F (z). Then

G = eΦdzdz = eΦ (Fwdw + Fwdw)(Fwdw + Fwdw

)=

eΦ(FwFwdw

2 +(|Fw|2 + |Fw|2|

)dwdw + FwFwdw

2)

(4.12)

The condition that new coordinates are isothermal too means that FwFw = 0 , i.e. Fw = 0,i.e. F is anti-holomorphic function or Fw = 0, i.e. F is holomorphic function.

Illustrate this idea on the example: Let G = dx2 + f2(x)dy2 be a metric on a domainof Riemannian manifold (e.g. for sphere x = θ, y = ϕ, f(x) = sin2 x, for cone x = h, y =ϕ, f(x) = x). Then G = (dx+ ifdy)()dx− if(x)dy). For 1-form ω = dx+ if(x)dy we havethat dx+ if(x)dy = f(x) (dG(x) + idy) = f(x)d (L(x) + iy), where L(x) is antiderivativeof a function 1

f and dx2 + f2(x)dy2 = f2(x)d (L(x) + iy) d (G(x)− iy) = eΦ(du2 + dv2),

where eΦ = f2(x), u = L(x), v = y 18

4.2 Derivation formula

Let M be a surface embedded in Euclidean space E3, M : r = r(u, v).

18in general case we use essentially the condition of analiticity. This proof was done byGauss. The general smooth case was proved only in the beginning of XX century.

85

Let e, f ,n be three vector fields defined on the points of this surface suchthat they form an orthonormal basis at any point, so that the vectors e, fare tangent to the surface and the vector n is orthogonal to the surface19.Vector fields e, f ,n are functions on the surface M :

e = e(u, v), f = f(u, v) ,n = n(u, v) .

Consider 1-forms de, df , dn:

de =∂e

∂udu+

∂e

∂vdv, df =

∂f

∂udu+

∂f

∂vdv , dn =

∂n

∂udu+

∂n

∂vdv

These 1-forms take values in the vectors in E3, i.e. they are vector valued1-forms. Any vector in E3 attached at an arbitrary point of the surface canbe expanded over the basis e, f ,n. Thus vector valued 1-forms de, df , dncan be expanded in a sum of 1-forms with values in basic vectors e, f ,n. E.g.for de = ∂e

∂udu+ ∂e

∂vdv expanding vectors ∂e

∂uand ∂e

∂vover basis vectors we come

to

∂e

∂u= A1(u, v)e+B1(u, v)f+C1(u, v)n,

∂e

∂v= A2(u, v)e+B2(u, v)f+C2(u, v)n

thus

de =∂e

∂udu+

∂e

∂vdv = (A1e +B1f + C1n) du+ (A2e +B2f + C2n) dv =

= (A1du+ A2dv)︸ ︷︷ ︸M11

e + (B1du+B2dv)︸ ︷︷ ︸M12

f + (C1du+ C2dv)︸ ︷︷ ︸M11

e, (4.13)

i.e.de = M11e +M12f +M13n,

where M11,M12 and M13 are 1-forms on the surface M defined by the relation(4.13).

In the same way we do the expansions of vector-valued 1-forms df anddn we come to

de = M11e +M12f +M13ndf = M21e +M22f +M23ndn = M31e +M32f +M33n

19One can say that e, f ,n is an orthonormal basis in TpE3 at every point of surfacep ∈M such that e, f is an orthonormal basis in TpE3 at every point of surface p ∈M .

86

This equation can be rewritten in the following way:

d

efn

=

M11 M12 M13

M21 M22 M23

M31 M32 M33

efn

(4.14)

Proposition The matrix M in the equation (4.14) is antisymmetrical ma-trix, i.e.

M11 = M22 = M33 = 0M12 = −M21 = aM13 = −M31 = bM23 = −M32 = −b

(4.15)

i.e.

d

efn

=

0 a b−a 0 c−b −c 0

efn

, (4.16)

where a, b, c are 1-forms on the surface M .

Formulae (4.16) are called derivation formula.Prove this Proposition. (Here I give the detailed proof, but later in Re-

mark, very short proof in condensed notations.)Recall that e, f ,n is orthonormal basis, i.e. at every point of the surface

〈e, e〉 = 〈f , f〉 = 〈n,n〉 = 1, and 〈e, f〉 = 〈e,n〉 = 〈f ,n〉 = 0

Now using (4.14) we have

〈e, e〉 = 1⇒ d〈e, e〉 = 0 = 2〈e, de〉 = 〈e,M11e +M12f +M13n〉 =

M11〈e, e〉+M12〈e, f〉+M13〈e,n〉 = M11 ⇒M11 = 0 .

Analogously

〈f , f〉 = 1⇒ d〈f , f〉 = 0 = 2〈f , df〉 = 〈f ,M21e+M22f+M23n〉 = M22 ⇒M22 = 0 ,

〈n,n〉 = 1⇒ d〈n,n〉 = 0 = 2〈n, dn〉 = 〈n,M31e+M32f+M33n〉 = M33.⇒M33 = 0 .

We proved already that M11 = M22 = M33 = 0. Now prove that M12 =−M21, M13 = −M31 and M13 = −M31.

〈e, f〉 = 0⇒ d〈e, f〉 = 0 = 〈e, df〉+ 〈de, f〉 =

87

〈e,M21e +M22f +M23n〉+ 〈M11e +M12f +M13n, f〉 = M21 +M12 = 0.

Analogously

〈e,n〉 = 0⇒ d〈e,n〉 = 0 = 〈e, dn〉+ 〈de,n〉 =

〈e,M31e +M32f +M33n〉+ 〈M11e +M12f +M13n,n〉 = M31 +M13 = 0

and〈f ,n〉 = 0⇒ d〈f ,n〉 = 0 = 〈f , dn〉+ 〈df ,n〉 =

〈f ,M31e +M32f +M33n〉+ 〈M21e +M22f +M23n,n〉 = M32 +M23 = 0.

Remark This proof may be be performed much more shortly in con-densed notations. Derivation formula (4.16) in condensed notations are

dei = Mikek (4.17)

Orthonormality condition means that 〈ei, ek〉 = δik. Hence

d〈ei, ek〉 = 0 = 〈dei, ek〉+〈ei, dek〉 = 〈Mimem, ek〉+〈ei,Mknen〉 = Mik+Mki = 0

(4.18)Much shorter, is not it?

4.2.1 Gauss condition (structure equations)

Derive the relations between 1-forms a, b and c in derivation formula.Recall that a, b, c are 1-forms, e, f,n are vector valued functions (0-forms)

and de, df , dn are vector valued 1-forms. (We use the simple identity thatddf = 0 and the fact that for 1-form ω ∧ ω = 0.) We have from derivationformula (4.16) that

d2e = 0 = d(af + bn) = daf − a ∧ df + dbn− b ∧ dn =

da f − a ∧ (−ae + cn) + dbn− b ∧ (−be− cf) =

(da+b∧c)f +(a∧a+b∧b)e+(db−a∧c)n = (da+b∧c)f +(db+c∧a)n = 0 .

We see that(da+ b ∧ c)f + (db+ c ∧ a)n = 0 (4.19)

Hence components of the left hand side equal to zero:

(da+ b ∧ c) = 0 (db+ c ∧ a) = 0 . (4.20)

88

Analogously

d2f = 0 = d(−ae + cn) = −dae + a ∧ de + dcn− c ∧ dn =

−dae + a ∧ (af + bn) + dcn− c ∧ (−be− cf) =

(−da+ c ∧ b)e + (dc+ a ∧ b)n = 0 .

Hence we come to structure equations:

da+ b ∧ c = 0db+ c ∧ a = 0dc+ a ∧ a = 0

(4.21)

4.2.2 Geometrical meaning of derivation formula. Weingarten op-erator (shape oeprator) in terms of derivation formula.

Let M be a surface in E3.Let e, f ,n be three vector fields defined on the points of this surface such

that they form an orthonormal basis at any point, so that the vectors e, f aretangent to the surface and the vector n is orthogonal to the surface. Notethat in generally these vectors are not coordinate vectors.

Describe Riemannian geometry on the surface M in terms of this basisand derivation formula (4.16).

Induced Riemannian metricIf G is the Riemannian metric induced on the surface M then since e, f

is orthonormal basis at every tangent space TpM then

G(e, e) = G(f , f) = 1, G(e, f) = G(f , e) = 0 (4.22)

The matrix of the Riemannian metric in the basis e, f is

G =

(1 00 1

)(4.23)

Induced connection Let ∇ be the connection induced by the canonical flat connectionon the surface M .

Then according equations (2.25) and derivation formula (4.16) for every tangent vectorX

∇Xe = (∂Xe)tangent = (de(X))tangent = (a(X)f + b(X)n)tangent = a(X)f . (4.24)

89

and

∇Xf = (∂Xf)tangent = (df(X))tangent = (−a(X)e + c(X)n)tangent = −a(X)e . (4.25)

In particular∇ee = a(e)f ∇fe = a(f)f∇ef = −a(e)e ∇f f = −a(f)e

(4.26)

We know that the connection ∇ is Levi-Civita connection of the induced Riemannianmetric (4.24) (see the subsection 4.2.1)20.

Second Quadratic form Second quadratic form is a bilinear symmetric functionA(X,Y)on tangent vectors which is well-defined by the condition A(X,Y)n = (∂XY)orthogonal (seee.g. subsection 6.4 in Appendices.)

Let A(X,Y) be second quadratic form. Then according to derivation formula (4.16)we have

A(e, e) = 〈∂ee,n〉 = 〈de(e),n〉 = 〈a(e)f + b(e)n,n〉 = b(e) ,

A(f , e) = 〈∂fe,n〉 = 〈de(f),n〉 = 〈a(f)f + b(f)n,n〉 = b(f) ,

A(e, f) = 〈∂ef ,n〉 = 〈df(e),n〉 = 〈−a(e)f + c(e)n,n〉 = c(e) ,

A(f , f) = 〈∂f f ,n〉 = 〈df(f),n〉 = 〈−a(f)f + c(f)n,n〉 = c(f) ,

The matrix of the second quadratic form in the basis e, f is

A =

(A(e, e) A(f , e)A(e, f) A(f , f)

)=

(b(e) b(f)c(e) c(f)

)(4.28)

This is symmetrical matrix (see the subsection 4.3.2):

A(f , e) = b(f) = A(e, f) = c(e) . (4.29)

Weingarten (Shape) operatorLet S be Weingarten operator: SX = −∂Xn (see the subsection 6.4 in

Appendix, or Geometry lectures). Then it follows from derivation formulathat

SX = −∂Xn = −dn(X) = − (−b(X)e− c(X)f) = b(X)e + c(X)f

In particular

S(e) = b(e)e + c(e)f , S(f) = b(f)e + c(f)f

20In particular this implies that this is symmetric connection, i.e.

∇fe−∇ef − [f , e] = a(f)f + a(e)e− [f , e] = 0 . (4.27)

90

and the matrix of the Weingarten operator in the basis e, f is

S =

(b(e) b(f)c(e) c(f)

)(4.30)

Remark According to the condition (4.29) the matrix S is symmetrical. The relations

A = GS, S = G−1A for Weingarten operator, Riemannian metric and second quadratic

form are evidently obeyed for matrices of these operators in the basis e, f where G = 1,

A = S.

4.2.3 Gaussian and mean curvature in terms of derivation formula

Now we are equipped to express Gaussian and mean curvatures in terms ofderivation formula. Using (4.30) we have for Gaussian curvature

K = detS = b(e)c(f)− c(e)b(f) = (b ∧ c)(e, f) (4.31)

and for mean curvature

H = TrS = b(e) + c(f) (4.32)

What next? We will study in more detail formula (4.31) later.Now consider some examples of calculation of Weingarten operator, e.t..c.

for using derivation formula.

4.3 Calculations with use of derivation formulae

4.3.1 Examples of calculations of derivation formulae and curva-tures for cylinder, cone and sphere

Last year we calculated Weingarten operator and curvatures for cylinder,cone and sphere (see also the subsection 6.4 in Appendices.). Now we do thesame but in terms of derivation formula.

Cylinder

We have to define three vector fields e, f ,n on the points of the cylindersurface x2 + y2 = a2:

r(h, ϕ) :

x = a cosϕ

y = a sinϕ

z = h

(4.33)

91

such that they form an orthonormal basis at any point, so that the vectorse, f are tangent to the surface and the vector n is orthogonal to the surface.We calculated many times coordinate vector fields rh, rϕ and normal unitvector field:

rh =

001

, rϕ =

−a sinϕa cosϕ

0

, n =

cosϕsinϕ

0

. (4.34)

Vectors rh, rϕ and n are orthogonal to each other but not all of them haveunit length. One can choose

e = rh =

001

, f =rϕa

=

− sinϕcosϕ

0

, n =

cosϕsinϕ

0

(4.35)

These vectors form an orthonormal basis and e, f form an orthonormal basisin tangent space.

Derive for this basis derivation formula (4.16). For vector fields e, f ,n in(4.35) we have

de = 0, df = d

− sinϕcosϕ

0

=

− cosϕ− sinϕ

0

dϕ = −ndϕ,

dn = d

cosϕsinϕ

0

=

− sinϕcosϕ

0

= fdϕ,

i.e.

d

efn

=

0 a b−a 0 c−b −c 0

efn

=

0 0 00 0 −dϕ0 dϕ 0

efn

, (4.36)

i.e. in derivation formula 1-forms a, b vanish a = b = 0 and c = −dϕ.

The matrix of Weingarten operator in the basis e, f is

S =

(b(e) c(e)b(f) c(f)

)=

(0 −dϕ(e)0 −dϕ(f)

)=

(0 00 − 1

R

)

92

According to (4.31) and(4.32) Gaussian curvatureK = b(e)c(f)−b(e)c(f) = 0and mean curvature

H = b(e) + c(f) = −dϕ(f) = −dϕ(rϕR

)= −1

a

Remark We denote by the same letter a the radius of the cylinder surface(4.33) and 1-form a in derivation formula. I hope that this will not lead tothe confusion. (May be it is better to denote the radius of the cylindricalsurface by the letter R.)

ConeFor cone:

r(h, ϕ) :

x = kh cosϕ

y = kh sinϕ

z = h

,

rh =

k cosϕk sinϕ

1

, rϕ =

−kh sinϕkh cosϕ

0

, n =1√

1 + k2

cosϕsinϕ−k

Tangent vectors rh, rϕ are orthogonal to each other. The length of the vector rh equals to√

1 + k2 and the length of the vector rϕ equals to kh. Hence we can choose orthonormalbasis e, f ,n such that vectors e, f are unit vectors in the directions of the vectors rh, rϕ:

e =rh√

1 + k2=

1√1 + k2

k cosϕk sinϕ

1

, f =rϕhk

=

− sinϕcosϕ

0

, n =1√

1 + k2

cosϕsinϕ−k

Calculate de, df and dn:

de = d

k cosϕk sinϕ

1

=kdϕ√1 + k2

− sinϕcosϕ

0

=kdϕ√1 + k2

f ,

df = d

− sinϕcosϕ

0

=

− cosϕ− sinϕ

0

dϕ =

−k1 + k2

k cosϕk sinϕ

1

dϕ− dϕ

1 + k2

cosϕsinϕ−k

=−kdϕ√1 + k2

e− dϕ√1 + k2

n ,

and

dn =1√

1 + k2d

cosϕsinϕ−k

=dϕ√

1 + k2

− sinϕcosϕ

0

.

93

We come to

d

efn

=

0 a b−a 0 c−b −c 0

efn

=

0 kdϕ√1+k2

0

− kdϕ√1+k2

0 −dϕ√1+k2

0 dϕ√1+k2

0

e

fn

, (4.37)

i.e. in derivation formula for 1-forms a = kdϕ√1+k2

, b = 0 and and c = − −dϕ√1+k2

.

Remark Note that calculation of df are little bit hard. On the other hand the answerfor df follows from answers for de and dn since the matrix in (4.37) is antisymmetric. Sowe can omit the straightforward calculations of df .

The matrix of Weingarten operator in the basis e, f is

S =

(b(e) c(e)b(f) c(f)

)= S =

(0 −dϕ(e)√

1+k2

0 −dϕ(f)√1+k2

)=

(0 00 −1

kh√

1+k2

).

since dϕ(f) = dϕ( rϕkh

)= 1

khdϕ(∂ϕ) = 1kh .

According to (4.31), (4.32) Gaussian curvature

K = b(e)c(f)− b(e)c(f) = 0

and mean curvature

H = b(e) + c(f) = −dϕ(f) = −dϕ(rϕR

)= − 1

kh√

1 + k2.

Sphere

For sphere

r(θ, ϕ) :

x = R sin θ cosϕ

y = R sin θ sinϕ

z = R cos θ

(4.38)

rθ(θ, ϕ) =∂r

∂θ=

R cos θ cosϕR cos θ sinϕ−R sin θ

, rϕ(θ, ϕ) =∂r

∂ϕ=

−R sin θ sinϕR sin θ cosϕ

0

,

n(θ, ϕ) =r

R=

sin θ cosϕsin θ sinϕ

cos θ

.

Tangent vectors rθ, rϕ are orthogonal to each other. The length of the vectorrθ equals to R and the length of the vector rϕ equals to R sin θ. Hence we

94

can choose orthonormal basis e, f ,n such that vectors e, f are unit vectorsin the directions of the vectors rθ, rϕ:

e(θ, ϕ) =rθR

=

cos θ cosϕcos θ sinϕ− sin θ

, f(θ, ϕ) =rϕ

R sin θ=

− sinϕcosϕ

0

, n(θ, ϕ) =r

R=

sin θ cosϕsin θ sinϕ

cos θ

.

Calculate de, df and dn:

de = d

cos θ cosϕcos θ sinϕ− sin θ

=

− cos θ sinϕcos θ cosϕ

0

dϕ−

sin θ cosϕsin θ sinϕ− cos θ

dθ = cos θdϕf − dθn,

df = d

− sinϕcosϕ

0

= −

cosϕsinϕ

0

dϕ =

− cos θdϕ

cos θ cosϕcos θ sinϕ− sin θ

− sin θdϕ

sin θ cosϕsin θ sinϕ

cos θ

= − cos θdϕe− sin θdϕn ,

dn = d

sin θ cosϕsin θ sinϕ

cos θ

=

cos θ cosϕcos θ sinϕ− sin θ

dθ +

− sin θ sinϕsin θ cosϕ

0

dϕ

= dθe + sin θdϕf .

i.e.

d

efn

=

0 a b−a 0 c−b −c 0

efn

=

0 cos θdϕ −dθ− cos θdϕ 0 − sin θdϕ

dθ sin θdϕ 0

efn

,

(4.39)i.e. in derivation formula a = cos θdϕ, b = −dθ, c = − sin θdϕ.

Remark The same remark as for cone: equipped by the properties ofderivation formula we do not need to calculate df . The calculation of de anddn and the property that the matrix in derivation formula is antisymmetricgives us the answer for df .

95

The matrix of Weingarten operator in the basis e, f is

S =

(b(e) c(e)b(f) c(f)

)=

(−dθ(e) − sin θdϕ(e)−dθ(f) − sin θdϕ(f)

)=

(−1R

00 − 1

R

)since dθ(e) = dθ

(∂θR

)= 1

Rdθ(∂θ) = 1

R, dϕ(e) = dϕ

(∂θR

)= 1

Rdϕ(∂θ) = 0.

According to (4.31) and(4.32) Gaussian curvature

K = b(e)c(f)− b(e)c(f) =1

R2

and mean curvature

H = b(e) + c(f) = − 2

RNotice that for calculation of Weingarten operator and curvatures we usedonly 1-forms b and c, i.e. the derivation equation for dn, (dn = dθe +sin θdϕf).

Mean curvature is define up to a sign. If we change n → −n meancurvature H → 1

Rand Gaussian curvature will not change.

We see that for the sphere Gaussian curvature is not equal to zero, whilstfor cylinder and cone Gaussian curvature equals to zero.

4.3.2 †Curvatures for surface z = F (x, y)

.Now using derivation formulae we calculate curvature for arbitrary surface

z = F (x, y) and later we will calculate curvatures for surfaces in conformal(isometric coordinates).

We will caclulate curvature not at an arbitrary point but only at thepoints of extrema of function F (x, y). (In fact this condition is not verydemanding.)

Derivation formulae become very useful tool for solving these questions21.

A surface z = F (x, y) can be defined by parameterisation

r(u, v) :

x = u

y = v

z = F (u, v)

21In the previous section we calculated curvatures of cylinder, and sphere using deriva-tion formulae. These calculations may be even easier to perform using just usual methodswhich we studied in the course of Geometry.

96

Conisder coordinate vector fields of the surface

∂

∂u− ru =

10Fu

,∂

∂v− rv =

01Fv

,

and unit normal vector field

n(u, v) =1√

1 + F 2u + F 2

v

−Fu−Fv1

.

It is obviously orthogonal to ru, rv and it has unbit length.

One can see that e = ru|ru| = 1√

1+F 2u

10Fu

is unit vector field tangent to

surface. The vector field rv|rv | = 1√

1+F 2v

01Fv

is also tangent to surface, it is

also unit, but in general it is not orthogonal to vector field e, 〈ru, rv〉 = FuFv.To find a second tangent vector field orthgonal to e we may consider thevector field f which is vector product f = n× e:

f = n× e =1√

1 + F 2u + F 2

v

−Fu−Fv1

× 1√1 + F 2

u

10Fu

=

1√(1 + F 2

u + F 2v )(1 + F 2

u )

−FuFv1 + F 2v

Fv

×Since vector field n is orthogonal to surface, and it is unit, hence vectorfields e, f ,n form orthonormal basis, e, f are tangent. Thus we constructedorthonormal basis e, f ,n attached to the surface. We want to calculatecurvatures at the origin a point p with coordinates u = v = 0 Put thefollowing condition: the surface z = F (x, y) has extremum at the origin, i.e.

Fu∣∣p

= Fv∣∣p

= 0 , (p it has coordinates u = v = 0). (4.40)

This conidition is not demanding. For every point A on the surface onthe surface one can find adjusted Cartesian coordinates such that in these

97

coordinates this surface will have extremum at the point A. On the otherhand this condition drastically simplifies calculations. Note that if condition(4.40) is obeyed then at the point p vector fields e, f ,n look in a very simpleway:

e∣∣p

= e(u, v)u=v=0 =

100

, f∣∣p

= f(u, v)u=v=0 =

010

,n∣∣p

= n(u, v)u=v=0 =

001

,

Now obtain derivation formulae (at the point p) We will calculate everythingjust at the point p. Note that if condition (4.40) is obeyed then

de∣∣p

= d

1√1 + F 2

u

10Fu

p

=

00dFu

u=v=0

= (ndFu)p ,

df∣∣p

= d

1√(1 + F 2

u + F 2v )(1 + F 2

u )

−FuFv1 + F 2v

Fv

p

=

00dFu

u=v=0

= (ndFv)p ,

and

dn∣∣p

= d

1√1 + F 2

u + F 2v

−Fu−Fv1

p

=

00dFu

u=v=0

= (−edFu − fdFv)p ,

since all other terms vanish at u = v = 0. Comparing with derivation formula

d

efn

=

0 a b−a 0 c−b −c 0

d

efn

we see that forms a, b, c at origin are equal to

a∣∣p

= 0 , b∣∣p

= dFu∣∣p

= (Fuudu+ Fuvdv)p , c∣∣p

= dFv∣∣p

= (Fvudu+ Fvvdv)p ,

Now calculate the values of these forms on vectors e, f at origin. We havethat b(e) = dFu(e)p = (Fuudu + Fuvdv)(ru) = Fuu. Analogously b(f) =dFu(f)p = (Fuudu+Fuvdv)(rv) = Fuv, c(e) = dFv(e)p = (Fvudu+Fvvdv)(ru) =Fvu, and c(f) = dFv(f)p = (Fvudu+ Fvvdv)(rv) = Fvv,

98

Hence we have that matrix of Weintegarten (shape) operator at the originis equal to

S =

(b(e) c(e)b(f) c(f)

)=

(Fuu FuvFvu Fvv

)p

, K = detS = FuuFvv−F 2uv , H = TrS = Fuu+Fvv

Theorem For surface z = F (x, y), Wengarten (Shape) operator in extremum point isgiven by quadratic form (Hessian) of function.

ExampleConsider surface defined by equation z = Ax2 +2Bxy+y2, The point x =

y = 0 is extremum point. All derivatives Fu = 2Au+ 2Bv, Fv = 2Bu+ 2Cvvanish t origin.

Then Gaussian curvature at point x = y = 0 is equal to

K = FxxFyy − F 2xy

Gaussian curvature at arbitrary point of surface z = F (x, y) is equal to

K =FxxFyy − F 2

xy

(1 + F 2x + F 2

y )3/2

4.3.3 ∗Proof of the Theorem of parallel transport along closedcurve

We are ready now to prove the Theorem. Recall that the Theorem statesfollowing:

If C is a closed curve on a surface M such that C is a boundary of acompact oriented domain D ⊂ M , then during the parallel transport of anarbitrary tangent vector along the closed curve C the vector rotates throughthe angle

∆Φ = ∠ (X,RCX) =

∫D

Kdσ , (4.41)

where K is the Gaussian curvature and dσ =√

det gdudv is the area elementinduced by the Riemannian metric on the surface M , i.e. dσ =

√det gdudv.

(see (4.5).Recall that for derivation formula (4.16) we obtained structure equations

da+ b ∧ c = 0db+ c ∧ a = 0dc+ a ∧ a = 0

(4.42)

99

We need to use only one of these equations, the equation

da+ b ∧ c = 0 . (4.43)

This condition sometimes is called Gauß condition.Let as always e, f ,n be an orthonormal basis in TpE3 at every point

of surface p ∈ M such that e, f is an orthonormal basis in TpM at everypoint of surface p ∈M . Then the Gauß condition (4.43) and equation (4.31)mean that for Gaussian curvature on the surface M can be expressed throughthe 2-form da and base vectors e, f:

K = b ∧ c(e, f) = −da(e, f) (4.44)

We use this formula to prove the Theorem.Now calculate the parallel transport of an arbitrary tangent vector over

the closed curve C on the surface M .Let r = r(u, v) = r(uα) (α = 1, 2, (u, v) = (u1, v1)) be an equation of the

surface M .Let uα = uα(t) (α = 1, 2) be the equation of the curve C. Let X(t) be

the parallel transport of vector field along the closed curve C, i.e. X(t) istangent to the surface M at the point u(t) of the curve C and vector fieldX(t) is covariantly constant along the curve:

∇X(t)

dt= 0

To write this equation in components we usually expanded the vector fieldin the coordinate basis ru = ∂u, rv = ∂v and used Christoffel symbols ofthe connection Γαβγ : ∇β∂γ = Γαβγ∂α.

Now we will do it in different way: instead coordinate basis ru = ∂u, rv =∂v we will use the basis e, f. In the subsection 3.4.4 we obtained that theconnection ∇ has the following appearance in this basis

∇ve = a(v)f , ,∇vf = −a(v)e (4.45)

(see the equations (4.24) and (4.25))Let

X = X(u(t)) = X1(t)e(u(t)) +X2(t)f(u(t))

100

Lbe an expansion of tangent vector field X(t) over basis e, f. Let v be ve-

locity vector of the curve C. Then the equation of parallel transport ∇X(t)dt

= 0will have the following appearance:

∇X(t)

dt= 0 = ∇v

(X1(t)e(u(t)) +X2(t)f(u(t))

)=

dX1(t)

dte(u(t)) +X1(t)∇ve(u(t)) +

dX2(t)

dtf(u(t)) +X2(t)∇vf(u(t)) =

dX1(t)

dte(u(t)) +X1(t)a(v)f(u(t)) +

dX2(t)

dtf(u(t))−X2(t)a(v)e(u(t)) =(

dX1(t)

dt−X2(t)a(v)

)e(u(t)) +

(dX2(t)

dt+X1(t)a(v)

)f(u(t)) = 0.

Thus we come to equation:X1(t)− a(v(t))X2 = 0

X2(t) + a(v(t))X1 = 0

There are many ways to solve this equation. It is very convenient to considercomplex variable

Z(t) = X1(t) + iX2(t)

We see that

Z(t) = X1(t) + iX2(t) = a(v(t))X2 − ia(v(t)X1 = −ia(v)Z(t),

i.e.dZ(t)

dt= −ia(v(t))Z(t) (4.46)

The solution of this equation is:

Z(t) = Z(0)e−i∫ t0 a(v(τ))dτ (4.47)

Calculate∫ t1

0a(v(τ))dτ for closed curve u(0) = u(t1). Due to Stokes Theo-

rem: ∫ t1

0

a(v(t))dt =

∫C

a =

∫D

da

Hence using Gauss condition (4.43) we see that∫ t1

0

a(v(t))dt =

∫C

a =

∫D

da = −∫D

b ∧ c

101

Claim ∫D

b ∧ c = −∫D

da =

∫Kdσ . (4.48)

Theorem follows from this claim:

Z(t1) = Z(0)e−i∫C a = Z(0)ei

∫D b∧C (4.49)

Denote the integral i∫Db ∧ C by ∆Φ: ∆Φ = i

∫Db ∧ C. We have

Z(t1) = X1(t1) + iX2(t1) =(X1(0) + iX2(0)

)ei∆Φ = (4.50)

It remains to prove the claim. The induced volume form dσ is 2-form.Its value on two orthogonal unit vector e, f equals to 1:

dσ(e, f) = 1 (4.51)

(In coordinates u, v volume form dσ =√

det gdu ∧ dv).The value of the form b∧ c on vectors e, f equals to Gaussian curvature

according to (4.44). We see that

b ∧ c(e, f) = −da(e, f) = Kdσ(e, f)

Hence 2-forms b∧c, −da and volume form dσ coincide. Thus we prove (4.48).

4.3.4 † Proof of the Theorem on curvature of surfaces given inconformal coordinates using derivation formulae

We return here to subsection 4.1.3 where we formulated the Theorem aboutGaussian curvatureof surface r = r(u, v) in E3 in conformal coordinates(4.8) .Let (u, v) be local conformal coordinates, and metricG = σ(u, v)(du2+dv2) =eΦ(du2 + dv2).

Consider vectors

e = e−Φ2∂

∂u, f = e−

Φ2∂

∂v, n = e× f .

It is evident that e, f ,n form orthonoromal basis:

〈e, e〉 = 1 , 〈e, f〉 = 0 , 〈e,n〉 = 0, 〈f , f〉 = 1 〈f ,n〉 = 0 , 〈n,n〉 = 1 .

102

Consider derivation formula (4.16) for this basis:

d

efn

=

0 a b−a 0 c−b −c 0

efn

, (4.52)

To calculate Gaussian curvature we need to calculate 1-form a in this equa-tions since accordind equations (4.31) and (4.21) K = b ∧ c(e, f) and da +b ∧ c = 0, i.e. K = −da(e, f). Now calculate 1-form a. We have

de = d(e−

Φ2 ru

)= af + bn .

Taking scalar product of this equation of f we come to

a = 〈de, f〉 =⟨d(e−

Φ2 ru

), e−

Φ2 rv⟩. (4.53)

Calculate it. Since 〈ru, rv〉 = 0 then⟨d(e−

Φ2 ru

), e−

Φ2 rv⟩

= e−Φ〈dru, rv〉 = e−Φ〈ruu, rv〉du+ e−Φ〈ruv, rv〉dv .

Now using the fact that 〈rv, rv〉 = 〈ru, ru〉 = eΦ and 〈ru, rv〉 = 0 calculate〈ruv, rv〉 and 〈ruu, rv〉. We have

〈ruv, rv〉 =1

2

∂

∂u〈rv, rv〉 =

1

2

∂

∂u

(eΦ)

=1

2Φue

Φ

and

〈ruu, rv〉 =∂

∂u〈ru, rv〉−〈ru, ruv〉 = 0−〈ru, ruv〉 = −1

2

∂

∂v〈rv, rv〉 = −1

2

∂

∂v

(eΦ)

= −1

2Φve

Φ

Hence we see that 1-form a in (4.53) is equal to

a = 〈de, f〉 =⟨d(e−

Φ2 ru

),(e−

Φ2 rv

) ⟩= e−Φ〈ruu, rv〉du+e−Φ〈ruv, rv〉dv =

1

2(Φudv−Φvdu) ,

(4.54)and 2-form

da = d

(1

2(Φudv − Φvdu)

)=

1

2(Φuudu∧dv−Φvvdv∧du) =

1

2(Φuu + Φvv) dv∧du .

103

Now using Gauss formula (4.21) and (4.31) we come to

K = b ∧ c(e, f) = −da(e, f) = −1

2(Φuu + Φvv) du ∧ dv

(e−

Φ2∂

∂u, e−

Φ2∂

∂v

)=

−e−Φ

2

2(Φuu + Φvv) du ∧ dv

(∂

∂u,∂

∂v

)= −e

−Φ

2(Φuu + Φvv) = −e

−Φ

2∆Φ,

where Laplacian ∆ = ∂2

∂u2 + ∂2

∂v2 .It is useful to write down this fiormula in complex coordinates Write down

the formula in holomorphic coordinates: z = u + iv, z = u − iv. We havethat

K = −e−Φ

2

(∂2

∂u2+

∂2

∂v2

)Φ = −e

−Φ

2

(∂

∂u+ i

∂

∂v

)(∂

∂u− i ∂

∂v

)Φ = −2e−Φ ∂2Φ

∂z∂z,

(4.55)(for definition of ∂

∂zand ∂

∂zsee (4.11)). This expression is sometimes very

convenient for calculations.Example Consider sphere of radius 1 in stereographic coordinates. Then G =

4(du2+dv20(1+u2+v2)2) . In complex coordinates G = 4dzdz

(1+zz)2 = eΦdzdz with eΦ = 4(1+zz)2 , i.e.

Φ = log 4 − 2 log(1 + zz). We see that Φz = − 21+zz and Φzz = − 2

(1+zz)2 , i.e. K =

−2e−ΦΦzz = 1.Exercise Let z = f(w) be an holomorphic changing of complex coordinates. Due to

Theorem new coordinates u′, v′ (w = u′+ iv′, z = u+ iv ) are isothermal coordinates too:If

G = eΦ(du2 + dv2) = eΦdzdz = eΦ′dwdw = eΦ′

(du′2 + dv′2) .

It is very illuminating to check straightforwardly that calculating of Gaussian curvaturein new coordinates we will come to the same answer. Do it. According to (4.12) we seethat eΦdzdz = eΦfwfwdwdw, i.e. Φ′ = Φ + log fw + log fw. Hence

∂2Φ′

∂w∂w=

∂2Φ

∂w∂w+

∂2

∂w∂w

(log fw + log fw

).

Notice that the function log fw is holomorphic function ⇔ ∂∂w log fw = 0 and the function

log fw is anti-holomorphic function ⇔ ∂∂w log fw = 0 too. Hence

∂2

∂w∂w

(log fw + log fw

)= 0 .

This implies that∂2Φ′

∂w∂w=

∂2Φ

∂w∂w.

104

Again using the fact that functions z = f(w) and zw = fw are holomorphic functions wesee that

∂2Φ′

∂w∂w=

∂2Φ

∂w∂w=

∂

∂w(Φzfw) =

∂Φz∂w

fw = Φzzfwfw.

Finally we come to

K = −2e−Φ′ ∂2Φ′

∂w∂w= −2e−Φ−log fw−log fwΦzzfwfw = −2e−Φ ∂2Φ

∂z∂z.

Thus we check by straightforward calculations that Gaussian curvature remains the same.Inthese calculations In these calculations we used intensively properties (4.11) of holomorphicand anti-holomorphic functions.

5 Curvature tensor

5.1 Curvature tensor for connection

Definition-Proposition Let manifold M be equipped with connection ∇.Consider the following operation which assigns to arbitrary vector fields X,Yand Z on M the new vector field:

R(X,Y)Z =(∇X∇Y −∇Y∇X −∇[X,Y]

)Z (5.1)

This operation is obviously linear over the scalar coefficients.One can show that this operation is C∞(M)-linear with respect to vector

fields X,Y Z, i.e. for an arbitrary functions f, g, h

R(fX, gY)(hZ) = fghR(X,Y)Z . (5.2)

This means that the operation defines the tensor field of the type

(13

): If

X = X i∂i,X = X i∂i,X = X i∂i then according to (5.2)

R(X,Y)Z = R(Xm∂m, Yn∂n)(Zr∂r) = ZrRi

rmnXmY n

where we denote by Rirmn the components of the tensor R in the coordinate

basis ∂iRi

rmn∂i = R(∂m, ∂n)∂r (5.3)

This tensor is called curvature tensor of the connection ∇.

105

Express components of the curvature tensor in terms of Christoffel sym-bols of the connection. If ∇m∂n = Γrmn∂r then according to the (5.1) wehave:

Rirmn∂i = R(∂m, ∂n)∂r = ∇∂m∇∂n∂r −∇∂n∇∂m∂r,

Rirmn = ∇∂m (Γpnr∂p)−∇∂n (Γpmr∂p) =

∂mΓinr + ΓimpΓpnr − ∂nΓimr − ΓinpΓ

pmr . (5.4)

The proof of the property (5.2) can be given just by straightforwardcalculations: Consider e.g. the case f = g = 1, then

R(X,Y)(hZ) = ∇X∇Y(hZ)−∇Y∇X(hZ)−∇[X,Y](hZ) =

∇X (∂YhZ + h∇YZ)−∇Y (∂XhZ + h∇XZ)− ∂[X,Y]hZ− h∇[X,Y]Z =

∂X∂YhZ + ∂Yh∇XZ + ∂Xh∇YZ + h∇X∇YZ−∂Y∂XhZ− ∂Xh∇YZ− ∂Yh∇XZ + h∇Y∇XZ−

∂[X,Y]hZ− h∇[X,Y]Z =

h[∇X∇YZ−∇Y∇XZ)−∇[X,Y]Z

]+[∂X∂Yh− ∂Y∂Xh− ∂[X,Y]h

]Z =

h∇X∇YZ−∇Y∇XZ)−∇[X,Y]Z = hR(X,Y)Z ,

since ∂X∂Yh− ∂Y∂Xh− ∂[X,Y]h = 0.

5.1.1 Properties of curvature tensor

Tensor Rikmn is expressed trough derivatives of Christoffel symbols. In spite

this fact it is is much more ”pleasant” object than Christoffel symbols, sincethe latter is not the tensor.

It follows from the definition that the tensor Rikmn is antisymmetrical

with respect to indices m,n:

Rikmn = −Ri

knm . (5.5)

One can prove that for symmetric connection this tensor obeys the following identities:

Rikmn +Rimnk +Rinkm = 0 , (5.6)

The curvature tensor corresponding to Levi-Civita connection obeys alsoanother identities too (see the next subsection.)

We know well that If Christoffel symbols vanish in a vicinity of a givenpoint p in some chosen coordinate system then in general Christoffel symbols

106

do not vanish in arbitrary coordinate systems. E.g. Christoffel symbols ofcanonical flat connection in E2 vanish in Cartesian coordinates but do notvanish in polar coordinates. This unpleasant property of Christoffel symbolsis due to the fact that Christoffel symbols do not form a tensor.

In particular if a tensor vanishes in some coordinate system, then it van-ishes in arbitrary coordinate system too. This implies very simple but im-portant

Proposition If curvature tensor Rikmn vanishes in some coordinate sys-

tem, then it vanishes in arbitrary coordinate systems.

We see that if Christoffel symbols vanish in a vicinity of a given pointp in some chosen coordinate system then its Riemannian curvature tensorvanishes in a vicinity of the point p (see the formula (5.4)) and hence itvanishes locally (in a vicinity of point p) in arbitrary coordinate system.

In fact one can proveTheorem If a connection is symmetric then curvature tensor vanishes in

a vicinity of a point if and only there exist local Cartesian coordiantes,n avicinity of this point i.e. coordinates in which Christoffel symbol of connec-tion vanish.

5.2 Riemann curvature tensor of Riemannian mani-folds.

Let M be Riemannian manifold equipped with Riemannian metric GIn this section we will consider curvature tensor of Levi-Civita connection

∇ of Riemannian metric G.The curvature tensor for Levi-Civita connection will be called later Rie-

mann curvature tensor, or Riemann tensor.Using Riemannian metric one can consider Riemann tensor with all low

indicesRikmn = gijR

jkmn (5.7)

In the subsection above we formulated very important Theorem that van-ishing of curvature tensor means that connection is locally flat. For Rie-mann tensor one can formulate the analogous Theorem. If Riemannina man-ifold is locally Euclidean, i.e. there exist coordinates (x1, . . . , xn) such thatG = (dx1)2 + · · · + (dxn)2 then it is evident that Christoffel symbols ofLevi-Civita connection vanish in these coordinates, hence curvature tensorvanishes also. The converse implication is true also:

107

Theorem Riemann curvature tensor vanishes if and only if Riemannianmanifold is locally Euclidean, i.e. if Ri

kmn ≡ 0 in a vicinity of the pointp of Riemannian manifold, then in a vicinity of this point there exist localcoordinates (x1, . . . , xn) such that Riemannian metric G = (dx1)2 + · · · +(dxn)2.

For Riemann tensor one can consider Ricci tensor,

Rmn = Rimin (5.8)

which is symmetrical tensor: Rmn = Rnm.One can consider scalar curvature:

R = Riking

kn = gknRkn (5.9)

where gkn is Riemannian metric with indices above (the matrix ||gik|| isinverse to the matrix ||gil||).

Ricci tensor and scalar curvature also play essential role for formulationof famous Einstein gravity equations. In particular the space without matterthe Einstein equations have the following form:

Rik −1

2Rgik = 0 . (5.10)

Due to identities (5.5) and (5.6) for curvature tensor Riemann tensorobeys the following identities:

Rikmn = −Riknm , Rikmn +Rimnk +Rinkm = 0 (5.11)

Riemann curvature tensor which is curvature tensor for Levi-Civita con-nection obeys also the following identities:

Rikmn = −Rkimn , Rikmn = Rmnki . (5.12)

These condition lead to the fact that for 2-dimensional Riemannian man-ifold the Riemann curvature tensor of Levi-Civita connection has essentiallyonly one non-vanishing component: all components vanish or equal to com-ponent R1212up to a sign.

Indeed consider for 2-dimensional Riemannian manifold Riemann tensorRikmn, where i, k,m, n = 1, 2. Since antisymmetricity with respect to thirdand fourth indices (Rikmn = −Riknm), Rik11 = Rik22 = 0 and Rik12 = −Rik21.

108

The same for first and second indices: since antisymmetricity with respectto the the first and second indices (R12mn = −R21mn), R11mn = R22mn = 0and R12mn = −Rik21. If we denote R1212 = a then

R1212 = R2121 = a,R1221 = R2112 = −a (5.13)

and all other components vanish.

5.2.1 Curvature of surfaces in E3.. Theorema Egregium again

For surfaces in E3 Gaussian curvature is equal to half of scalar curvature:

K =R

2, (5.14)

where R = Riking

kn is scalar curvature of Riemann curvature tensor.Equation (5.17) is the fundamental relation which claims that the Gaus-

sian curvature (the magnitude defined in terms of External observer) equalsto the scalar curvature (up to a coefdficient), the magnitude defined in termsof Internal Observer. This gives us another proof of Theorema Egregium.

Prove this formula.Express Riemannian curvature of surfaces in E3 in terms of derivation formula (4.16).Consider derivation formula (4.16) for the orthonormal basis e, f ,n, adjusted to the

surface M :

d

efn

=

0 a b−a 0 c−b −c 0

efn

, (5.15)

where as usual e, f ,n vector fields of unit length which are orthogonal to each otherand n is orthogonal to the surface M . As we know the induced connection on the surfaceM is defined by the formula (4.24) and (4.25):

∇Ye = (de(Y))tangent = a(X)f ,∇Yf = (df(Y))tangent = −a(X)e , (5.16)

According to the definition of curvature calculate

R(e, f)e = ∇e∇fe−∇f∇ee−∇[e,f ]e .

Using these formulae one can calculate straightforwardly that for surfaces in E3 Gaussiancurvature is equal to half of the scalar curvature:

K =R

2(5.17)

Detailed calculations are following:

109

Note that since the induced connection is symmetrical connection then:

∇ef −∇fe− [e, f ] = 0 .

hence due to (5.16)

[e, f ] = ∇ef −∇fe = −a(e)e− a(f)f

Thus we see that R(e, f)e =

∇e∇fe−∇f∇ee−∇[e,f ]e = ∇e (a(f)f)−∇f (a(e)f) +∇a(e)e+a(f)fe =

∂ea(f)f + a(f)∇ef − ∂fa(e)f − a(e)∇f f + a(e)∇ee + a(f)∇fe =

∂ea(f)f − a(f)a(e)e− ∂fa(e)f + a(e)a(f)e + a(e)a(e)f + a(f)a(f)f =

[∂ea(f)f − ∂fa(e)f − a [−a(e)e− a(f)f ]] f =

= [∂ea(f)f − ∂fa(e)f − a ([e, f ])] f = da(e, f)f .

Recall that we established in 4.44 that for Gaussian curvature K

K = b ∧ c(e, f) = −da(e, f)

Hence we come to the relation:

R(e, f)e = da(e, f) = −Kf .

This means thatR2

112 = −K

(in the basis e, f), i.e. the scalar curvature

R = 2R1212 = 2K

Thus we come to equation (5.17).The proof of Theorema Egregium by straightforward calculations see in the last ap-

pendix.

5.2.2 Relation between Gaussian curvature and Riemann curva-ture tensor and Theorema Egregium

Let M be a surface in E3 and Rikmp be Riemann tensor, Riemann curva-

ture tensor of Levi-Civita connection. Recall that this means that Rikmp

is curvature tensor of the connection ∇, which is Levi-Civita connection ofthe Riemannian metric gαβ induced on the surface M by standard Euclideanmetric dx2 + dy2 + dz2. Recall that Riemann curvature tensor is expressedvia Christoffel symbols of connection by the formula

Rikmn = ∂mΓink + ΓimpΓ

pnk − ∂nΓimk − ΓinpΓ

pmk (5.18)

110

(see he formula (5.4)) where Christoffel symbols of Levi-Civita connectionare defined by the formula

Γimk =1

2gij(∂gjm∂xk

+∂gjk∂xm

− ∂gmk∂xj

)(5.19)

(see Levi-Civita Theorem)Recall that scalar curvature R of Riemann tensor equals to R = Ri

kimgkm,

where gkm is Riemannian metric tensor with upper indices (matrix ||gik|| isinverse to the matrix ||gik||).

Now consider 2-dimensional case. One can show that in this case scalarcurvature R can be expressed via the component R1212 = a by the formula

R =2R1212

det g(5.20)

where det g = det gik = g11g22− g212. This can be checked by simple straight-

forward calculations.To see it note that as it was mentioned in the subsection above the formula for scalar

curvature becomes very simple in two-dimensional case (see formulae (5.11) and (5.13)above) and in this case it is very easy to calculate Ricci tensor and scalar curvature R.Indeed let R1212 = a. For 2-dimensional Riemannian surface all other components ofRiemann tensor equal to zero or equal to ±a (see (5.11) and (5.13)). Show it. Usingidentities (5.11) and (5.11) we see that

R11 = Ri1i1 = R2121 = g22R2121 + g21R1121 = g22R1212 = g22a (5.21)

R22 = Ri2i2 = R1212 = g11R1212 + g12R2221 = g11R1212 = g11a (5.22)

R12 = R21 = Ri1i2 = R1112 = g12R2112 = −g12R1212 = −g12a (5.23)

Thus using the formula for inverse 2× 2 matrix we come to the relation

Rik =

(R11 R12

R21 R22

)=

(g22a −g12a−g21a g11a

)=

1

det g

(g11a g12ag21a g11a

),

i.e. for 2-dimensional Riemannian manifold

Rik =1

det gR1212gik , (5.24)

Hence for scalar curvatre of 2-dimension Riemannian manifold

R = Rikimgkm = Rkmg

km =1

det gR1212gikg

ik =2

det gR1212 . (5.25)

Note that the relations (5.24) and (5.25) imply thatt

Rik =1

2Rgik . (5.26)

111

One can say that gravity equation for n = 2 are trivial. The mathematical meaningof this formula is the following: equation (5.26) means that variation of functional S =∫R√

det gdσ vanishes and this is one of corollaries of Gauss-Bonet Theorem (see later).

Formula (5.20) expresses scalar curvature for surface in terms of non-trivial componentR1212. On the other hand Gaussian curvature of 2-dimensionalsurface is equal to the half of the scalar curvature (see equation (5.14)). Hencewe come to

Proposition For an arbitrary point of the surface M

K =R

2=R1212

det g. (5.27)

where R = Rikimg

km is scalar curvature and K is Gaussian curvature.We know also that for surface M the scalar curvature R is expressed

via Riemann curvature tensor by the formula (5.20). Hence if we knowthe Gaussian curvature then we know all components of Riemann curvaturetensor (since all components vanish or equal to ±a.). This is nothing butTheorema Egregium! Theorema Egregium (see beginning of the section 4)immediately follows from this Proposition which states that Gaussian cur-vature is equal (up to a coefficient) to scalar curvature whcih is expressed interms of Riemannian metric.

In the next section we will present straightforward calculations where wecheck equation (5.27) just by brute force calculating Riemannian metric andRiemanian curvature tensor.

Before going to calculations, just a simple example

Example Let M = S2 be sphere of radius R in E3. Show that one cannotfind local coordinates u, v on the sphere such that induced Riemannian metricequals to du2 + dv2 in these coordinates.

This immediately follows from the Proposition. Indeed suppose there ex-ist local coordinates u, v on the sphere such that induced Riemannian metricequals to du2 + dv2, i.e. Riemannian metric is given by unity matrix. Thenaccording to the formulae for Levi-Civita connection, the Christoffel symbolsequal to zero in these coordinates. Hence Riemann curvature tensor equalsto zero, and scalar curvature too. Due to Proposition this is in contradictionwith the fact that Gaussian curvature of the sphere equals to 1

R2 .(The straightforward proof see in the next paragraph)

112

5.2.3 † Straightforward proof of the Proposition (5.27)

(Content of this paragraph is similar to the content of the solution of exercise6 in Homework 6. It is useful to read them both.)

We prove this fact by direct calculations. The plan of calculations isfollowing:

Let M be a surface in E3 For an arbitrary point p of the surface Mwe consider Cartesian coordinates x, y, z such that origin coincides with thepoint p and coordinate plane OXY is the palne attached at the surface at thepoint p and the axis OZ is orthogonal to the surface. In these coordinatescalculations become easy. (See the subsection 4.3.2) The surface M in theseCartesian coordinates can be expressed by the equation

x = u

y = v

z = F (u, v)

(5.28)

where F (u, v) has local extremum at the point u = v = 0. We calculated insubsection 4.3.2 Gaussian curvature at this point: Gaussian curvature at thepoint p equals to

K = detS = FuuFvv − F 2uv . (5.29)

(all the derivatives at the origin).Now it is time to calculate the Riemann curvature tensor at the origin.First of all recall the expression for Riemannian metric for the surface M

in a vicinity of origin is

G =

(〈ru, ru〉 〈ru, rv〉〈rv, ru〉 〈rv, rv〉

)=

(1 + F 2

u FuFvFvFu 1 + F 2

v

). (5.30)

This immediately follows from the expression for basic vectors ru, rvNote that Riemannian metric gik at the point u = v = 0 is defined by

unity matrix guu = gvv = 1, guv = gvu = 0 since p is extremum point:

G =

(1 + F 2

u FuFvFvFu 1 + F 2

v

) ∣∣p

=

(1 00 1

)since p is stationary point, extremum

(Fu = Fv = 0). Hence the components of the tensor Rikmn and Rikmn =

gijRjkmn at the point p are the same.

For 2-dimensional surface Riemann curvature tensor has oessentially onlyone not-vanishing component R1

212. All other components vanish all are equal

113

up to a sign to this component:

R1212 = −R2112 = R2121R1112 = · · · = R2111 = 0 .

So in fact we need to calculate only one component, the component R1212.In our calculations we will use the fact that Riemannian metric at the

point p is defined by unity matrix, and that first derivatives of metric at pvanish, i.e.Christoffel symbols in coordinates u, v vanish at the point p.

Recall that the components of Rikmn are defined by the formula

Rikmn = ∂mΓink + ΓimpΓ

pnk − ∂nΓimk − ΓinpΓ

pmk .

(see equation (5.4)). Notice that at the point p not only the matrix of themetric gik equals to unity matrix, but more: Christoffel symbols vanish at thispoint in coordinates u, v since the derivatives of metric at this point vanish.(Why they vanish: this immediately follows from Levi-Civita formula appliedto the metric (5.30), see also in detail the file ”The solution of the problem inthe coursework”). Hence to calculate Ri

kmn at the point p one can considermuch more simple formula than formula (5.4):

Rikmn|p = ∂mΓink|p − ∂nΓimk|p

Try to continue calculations in a more ”economical” way. Due to Levi-Civitaformula

Γimk =1

2gij(∂gjm∂xk

+∂gjk∂xm

− ∂gmk∂xj

)Since metric gik equals to unity matrix g =

(1 00 1

)at the point p hence gij

is unity matrix also:

gik|p =

(1 00 1

)= δik .

(We denote δik the unity matrix: all diagonal components equal to 1, all othercomponents equal to zero. (It is so called Kronecker symbols)) Moreover weknow also that all the first derivatives of the metric vanish at the point p:

∂gik∂xm|p = 0 .

Hence it follows from the formulae above that for an arbitrary indices i, j, k,m, n

∂

∂xi

(gkm

∂gpr∂xj

) ∣∣p

=∂gkm

∂xi∣∣p

∂gpr∂xj

∣∣p

+ gkm∣∣p

∂2gpr∂xi∂xj

∣∣p

= δkm∂2gpr∂xi∂xj

∣∣p.

114

Now using the Levi-Civita formula for the Christoffel symbols of connection:

Γimk =1

2gij(∂gjm∂xk

+∂gjk∂xm

− ∂gmk∂xj

)we come to ∂nΓimk|p = ∂

∂xn

(12gij(∂gjm∂xk

+∂gjk∂xm− ∂gmk

∂xj

)) ∣∣p

=

1

2δij(∂2gjm∂xn∂xk

+∂2gjk∂xn∂xm

− ∂2gmk∂xn∂xj

) ∣∣p. (5.31)

since first derivatives of metric vanish at the point p.Now using this formula we are ready to calculate Riemann curvature

tensor Rikmn. Remember that it is enough to calculate R1

212 and R1212 =

R1212 at the point p since gik = δik at the point p. We have that at pR1

212|p = ∂1Γ122|p − ∂2Γ1

12|p. Now using equation (5.31) we come to

R1212|p = ∂1Γ1

22|p−∂2Γ112|p =

1

2

∂

∂x1

(2∂g12

∂x2− ∂g22

∂x1

)−1

2

∂

∂x2

(∂g11

∂x2+∂g12

∂x1− ∂g12

∂x1

)p =

∂2guv∂u∂v

|p −1

2

∂2gvv∂u2|p −

1

2

∂2guu∂v2|p (5.32)

Now return to the surface (5.28) We have that guu = 1 + F 2u , guv = FuFv

and gvv = 1 + F 2v ,hence

∂2guv∂u∂v

|p = (FuuFv + FuFuv)v = FuuFvv + F 2uv ,

∂2gvv∂u2|p = (2FvFvu)u = 2F 2

uv ,

∂2guu∂v2|p = (FuuFu)v = 2F 2

uv ,

Hence

R1212|p =((FuuFvv + F 2

uv)− 2F 2uv

)p

=(FuuFvv − F 2

uv

)p

= Kp .

The proof is finished: we showed by straightforward calculations that R1212 is

equal to Gaussian curvature K. On the other hand at the point p, det g = 1.Thus we come to the statement of Proposition5.27.

115

Repeat again: all other components of Riemann curvature tensor Rikmn

are equal to R1212 up to a sign or vanish. Hence we calculated Riemanncurvature tensor at the point p and showed that it is essentially is definedby Gaussian curvature.

It is important to note that in our calculations of R1212 (see formula(5.32)) we used only the fact that Riemannian metric at the point p is de-fined by unity matrix, and all first derivatives at this point vanish.(see alsoStatement 1 in the solution of exercise 6 of Homework 9)

5.3 †Gauss Bonnet Theorem

Consider the integral of curvature over whole closed surface M . Accordingto the Theorem above the answer has to be equal to 0 (modulo 2π), i.e. 2πNwhere N is an integer, because this integral is a limit when we consider verysmall curve. We come to the formula:∫

D

Kdσ = 2πN

(Compare this formula with formula (4.5)).What is the value of integer N?We present now one remarkable Theorem which answers this question

and prove this Theorem using the formula (4.5).Let M be a closed orientable surface.22 All these surfaces can be clas-

sified up to a diffeomorphism. Namely arbitrary closed oriented surface Mis diffeomorphic either to sphere (zero holes), or torus (one hole), or pretzel(two holes),... ”Number k” of holes is intuitively evident characteristic of thesurface. It is related with very important characteristic—Euler characteristicχ(M) by the following formula:

χ(M) = 2(1− g(M)), where g is number of holes (5.33)

Remark What we have called here ”holes” in a surface is often referredto as ”handles” attached o the sphere, so that the sphere itself does not have

22Closed means compact surface without boundaries. Intuitively orientability meansthat one can define out and inner side of the surface. In terms of normal vectors ori-entability means that one can define the continuous field of normal vectors at all thepoints of M . The direction of normal vectors at any point defines outward direction.Orientable surface is called oriented if the direction of normal vector is chosen.

116

any handles, the torus has one handle, the pretzel has two handles and soon. The number of handles is also called genus.

Euler characteristic appears in many different way. The simplest appear-ance is the following:

Consider on the surface M an arbitrary set of points (vertices) connectedwith edges (graph on the surface) such that surface is divided on polygonswith (curvilinear sides)—plaquets. (”Map of world”)

Denote by P number of plaquets (countries of the map)Denote by E number of edges (boundaries between countries)Denote by V number of vertices.Then it turns out that

P − E + V = χ(M) (5.34)

It does not depend on the graph, it depends only on how much holes hassurface.

E.g. for every graph on M , P − E + V = 2 if M is diffeomorphic tosphere. For every graph on M P −E+V = 0 if M is diffeomorphic to torus.

Now we formulate Gauß -Bonnet Theorem.Let M be closed oriented surface in E3.Let K(p) be Gaussian curvature at any point p of this surface.

Theorem (Gauß -Bonnet) The integral of Gaussian curvature over theclosed compact oriented surface M is equal to 2π multiplied by the Eulercharacteristic of the surface M

1

2π

∫M

Kdσ = χ(M) = 2(1− number of holes) (5.35)

In particular for the surface M diffeomorphic to the sphere κ(M) = 2,for the surface diffeomorphic to the torus it is equal to 0.

The value of the integral does not change under continuous deformationsof surface! It is integer number (up to the factor π) which characterisestopology of the surface.

E.g. consider surface M which is diffeomorphic to the sphere. If it issphere of the radius R then curvature is equal to 1

R2 , area of the sphere isequal to 4πR2 and left hand side is equal to 4π

2π= 2.

117

If surface M is an arbitrary surface diffeomorphic to M then metrics andcurvature depend from point to the point, Gauß -Bonnet states that integralnevertheless remains unchanged.

Very simple but impressive corollary:Let M be surface diffeomorphic to sphere in E3. Then there exists at least

one point where Gaussian curvature is positive.Proof: Suppose it is not right. Then

∫MK√

det gdudv ≤ 0. On the otherhand according to the Theorem it is equal to 4π. Contradiction.

Proof of Gauß-Bonet TheoremConsider triangulation of the surface M . Suppose M is covered by N triangles. Then

number of edges will be 3N/over2. If V number of vertices then according to EulerTheorem

N − 3N

2+ V = V − N

2= χ(M).

Calculate the sum of the angles of all triangles. On the one hand it is equal to 2πV . Onthe other hand according the formula (4.5) it is equal to

N∑i=1

(π +

∫4iKdσ

)= πN +

N∑i=1

(∫4iKdσ

)= Nπ +

∫M

Kdσ

We see that 2πV = Nπ +∫MKdσ, i.e.∫

M

Kdσ = π

(2V − N

2

)= 2πχ(M)

6 Appendices

6.1 ∗Integrals of motions and geodesics.

We see how useful in Riemannian geometry to use the Lagrangian approach.To solve and study solutions of Lagrangian equations (in particular geodesics which

are solutions of Euler-Lagrange equations for Lagrangian of free particle) it is very usefulto use integrals of motion

6.1.1 ∗Integral of motion for arbitrary Lagrangian L(x, x)

Let L = L(x, x) be a Lagrangian, the function of point and velocity vectors on manifoldM (the function on tangent bundle TM).

118

Definition We say that the function F = F (q, q) on TM is integral of motion forLagrangian L = L(x, x) if for any curve q = q(t) which is the solution of Euler-Lagrangeequations of motions the magnitude I(t) = F (x(t), x(t)) is preserved along this curve:

F (x(t), x(t)) = const if x(t) is a solution of Euler-Lagrange equations(3.13). (6.1)

In other words

d

dt(F (x(t), x(t))) = 0 if xi(t) :

d

dt

(∂L

∂xi

)− ∂L

∂xi= 0 . (6.2)

6.1.2 ∗Basic examples of Integrals of motion: Generalised momen-tum and Energy

Let L(xi, xi) does not depend on the coordinate x1. L = L(x2, . . . , xn, x1, x2, . . . , xn).Then the function

F1(x, x) =∂L

∂x1

is integral of motion. (In the case if L(xi, xi) does not depend on the coordinate xi. thefunction Fi(x, x) = ∂L

∂xiwill be integral of motion.)

Proof is simple. Check the condition (6.2): Euler-Lagrange equations of motion are:

d

dt

(∂L

∂xi

)− ∂L

∂xi= 0 (i = 1, 2, . . . , n)

We see that exactly first equation of motion is

d

dt

(∂L

∂x1

)=

d

dtF1(q, q) = 0 since ∂L

∂x1 = 0, .

(if L(xi, xi) does not depend on the coordinate xi then the function Fi(x, x) = ∂Lx1 is

integral of motion since i− th equation is exactly the condition Fi = 0.)The integral of motion Fi = ∂L

∂xiis called sometimes generalised momentum.

Another very important example of integral of motion is: energy.

E(x, x) = xi∂L

∂xi− L . (6.3)

One can check by direct calculation that it is indeed integral of motion. Using Euler

Lagrange equations ddt

(∂L∂xi

)− ∂L

∂xi we have:

d

dtE(x(t), x(t)) =

d

dt

(xi∂L

∂xi− L

)=∂L

∂xidxi

dt+d

dt

(∂L

∂xi

)xi − dL

dt=

∂L

∂xidxi

dt+∂L

∂xidxi

dt− dL(x, x)

dt=dL(x, x)

dt− dL(x, x)

dt= 0 .

119

6.1.3 ∗Integrals of motion for geodesics

Apply the integral of motions for studying geodesics.

The Lagrangian of ”free” particle Lfree = gik(x)xixk

2 . For Lagrangian of free particlesolution of Euler-Lagrange equations of motions are geodesics.

If F = F (x, x) is the integral of motion of the free Lagrangian Lfree = gik(x)xixk

2 thenthe condition (6.1) means that the magnitude I(t) = F (xi(t), xi(t)) is preserved along thegeodesics:

I(t) = F (xi(t), xi(t)) = const, i.e.d

dtI(t) = 0 if xi(t) is geodesic. (6.4)

Consider examples of integrals of motion for free Lagrangian, i.e. magnitudes whichpreserve along the geodesics:

Example 1 Note that for an arbitrary ”free” Lagrangian Energy integral (6.3) is anintegral of motion:

E = xi∂L

∂xi− L = xi

∂(gpq(x)xpxq

2

)∂xi

− gik(x)xixk

2=

xigiq(x)xq − gik(x)xixk

2=gik(x)xixk

2. (6.5)

This is an integral of motion for an arbitrary Riemannian metric. It is preserved on anarbitrary geodesic

dE(t)

dt=

d

dt

(1

2gik(x(t))xi(t)xk(t)

)= 0 .

In fact we already know this integral of motion: Energy (6.5) is proportional to the squareof the length of velocity vector:

|v| =√gik(x)xixk =

√2E . (6.6)

We already proved that velocity vector is preserved along the geodesic (see the Propositionin the subsection 3.2.1 and its proof (??).)

Example 2 Consider Riemannian metric G = adu2 + bdv2 (see also calculations insubsection 2.3.3) in the case if a = a(u), b = b(u), i.e. coefficients do not depend on thesecond coordinate v:

G = a(u)du2 + b(u)dv2, Lfree =1

2

(a(u)u2 + b(u)v2

)(6.7)

We see that Lagrangian does not depend on the second coordinate v hence the magnitude

F =∂Lfree

∂v= b(u)v (6.8)

is preserved along geodesic. It is integral of motion because Euler-Lagrange equation forcoordinate v is

d

dt

∂Lfree

∂v− ∂Lfree

∂v=

d

dt

∂Lfree

∂v=

d

dtF = 0 since ∂Lfree

∂v = 0. .

120

In fact all revolution surfaces which we consider here (cylinder, cone, sphere,...) haveRiemannian metric of this type. Indeed consider further examples.

Example (sphere)Sphere of the radius R in E3. Riemannian metric: G = Rdθ2 + R2 sin2 θdϕ2 and

Lfree = 12

(R2θ2 +R2 sin2 θϕ2

)It is the case (6.7) for u = θ, v = ϕ, b(u) = R2 sin2 θ The

integral of motion is

F =∂Lfree

∂ϕ= R2 sin2 θϕ

Example (cone)

Consider cone

x = ah cosϕ

y = ah sinϕ

z = bh

. Riemannian metric:

G = d(ah cosϕ)2 + d(ah sinϕ)2 + (dbh)2 = (a2 + b2)dh2 + a2h2dϕ2 .

and free Lagrangian

Lfree =(a2 + b2)h2 + a2h2ϕ2

2.

The integral of motion is

F =∂Lfree

∂ϕ= a2h2ϕ.

Example (general surface of revolution)Consider a surface of revolution in E3:

r(h, ϕ) :

x = f(h) cosϕ

y = f(h) sinϕ

z = h

(f(h) > 0) (6.9)

(In the case f(h) = R it is cylinder, in the case f(h) = kh it is a cone, in the casef(h) =

√R2 − h2 it is a sphere, in the case f(h) =

√R2 + h2 it is one-sheeted hyperboloid,

in the case z = cosh it is catenoid,...)For the surface of revolution (6.9)

G = d(f(h) cosϕ)2 + d(f(h) sinϕ)2 + (dh)2 = (f ′(h) cosϕdh− f(h) sinϕdϕ)2+

(f ′(h) sinϕdh+ f(h) cosϕdϕ)2 + dh2 = (1 + f ′2(h))dh2 + f2(h)dϕ2 .

The ”free” Lagrangian of the surface of revolution is

Lfree =

(1 + f ′2(h)

)h2 + f2(h)ϕ2

2.

and the integral of motion is

F =∂Lfree

∂ϕ= f2(h)ϕ.

121

6.1.4 ∗Using integral of motions to calculate geodesics

Integrals of motions may be very useful to calculate geodesics. The equations for geodesicsare second order differential equations. If we know integrals of motions they help us tosolve these equations. Consider just an example.

For Lobachevsky plane the free Lagrangian L = x2+y2

2y2 . We already calculated geodesicsin the subsection 3.3.4. Geodesics are solutions of second order Euler-Lagrange equations

for the Lagrangian L = x2+y2

2y2 (see the subsection 3.3.4)··x− 2xy

y = 0··y + x2

y −y2

y = 0

It is not so easy to solve these differential equations.For Lobachevsky plane we know two integrals of motions:

E = L =x2 + y2

2y2, and F =

∂L

∂x=

x

y2. (6.10)

These both integrals preserve in time: if x(t), y(t) is geodesics thenF = x(t)

y(t)2

E = x(t)2+y(t)2

2y(t)2 = C2

⇒

x = C1y

2

y = ±√

2C2y2 − C21y

4

These are first order differential equations. It is much easier to solve these equations ingeneral case than initial second order differential equations.

6.1.5 ∗Killing vectors of Lobachevsky plane and geodesics

Killing vector field of Rimeannina manifold (M,G) is an infinitesimal isometry of theRimeannian metric G: under infinitesimal transform x → x + εK, xi → xi + εKI(x)(ε2 = 0) metric does not change:

gik(x)dxidxk = gik(xr + εKi)(dxi + ∂mKidxm)(dxk + ∂nK

kdxn) . (1)

Expanding this formula by ε and using the fact that ε2 = 0 we come to

Ki∂igkm + ∂kKrgrm + ∂mK

rgrk = 0 , (1a)

(i.e. Lie derivative LKG = 0.)

Examples: Killings of plane, sphere, cylindre, Lobachevsky plane.....

Theorem Let V be a vector space of all Killing vector fields of Riemannian manifold

M . Then the dimension of V is less or equal than n(n+12) .

It means that for surfaces the number of independent Killing vector fields is less orequal to 3.

122

One can prove that it is only for plane, sphere and Lobachevsky plane that numberof independent Killing vector fiels is equal to 3.

We calculate here Killing vector fields for Lobachevsky plane and use them for findinggeodesics.

Theorem Let K be Killing tor field on Riemannian manifold (M,G) , and L =gkpx

kxp

2Lagragian of ‘free’ particle on M . We know that geodesics are solutions of its equationsof motions.

The magnitude

I = IK = Ki(x)∂L(x, x)

∂xi

is an integral of motion, i.e. it is preserved along geodesics.

The proof of the Theorem is obvious. The condition that K is Killing vector fieldmeans that

L(xi + εKi, xi + εKi) , (2)

i.e.

Ki(x)∂L

∂xi+dKi

dt

∂L

∂xi= 0 . (2a)

Hence

d

dtIK =

d

dt

(Ki(x)

∂L(x, x)

∂xi

)=dKi

dt

∂L(x, x)

∂xi+Ki d

dt

(∂L(x, x)

∂xi

)=

Ki(x)∂L

∂xi+dKi

dt

∂L

∂xi︸ ︷︷ ︸condition that K is Killing

+Ki

(∂L(x, x)

∂xi− ∂L(x, x)

∂xi

)︸ ︷︷ ︸

equations of motion

= 0 .

Use this Theorem to find geodesics.First find Killing vector fields, i.e. infinitesimal isometries.Since the dimension is equal 2, the dimension of space of Killing vector fields is ≤ 3.

We will find three independent Killing vector fields.

There are two evident Killing vectors: Metric G = dx2+dy2

y2 is evidently invariant with

respect to translations x→ x+ a and homothety:

x→ λx

y 7→ λy: d(λx)2+d(λy)2

(λy)2 = dx2+dy2

y2 .

Infinitesimal translation is x′ = x + ε, y′ = y, the vector field D1 = ∂∂x . Infinitesimal

homothety is x′ = x+ εx, y′ = y + εy, the vector field D2 = x ∂∂x + y ∂

∂y .Now most interesting: find the third Killing vector field. Use the fact that inversion

O : (x, y) 7→(

xx2+y2 ,

yx2+y2

)preserves the metric. Consider the infinitesimal transforma-

tion Lε = O Tε O (L0 = id):

Lε :

(xy

)→( xx2+y2y

x2+y2

)→( xx2+y2 + ε

yx2+y2

)7→

x

x2+y2+ε(

xx2+y2

+ε)2

+(

y

x2+y2

)2

y

x2+y2(x

x2+y2+ε

)2+(

y

x2+y2

)2

.

123

6.2 Induced metric on surfaces.

Recall here again induced metric (see for detail subsection 1.4)If surface M : r = r(u, v)is embedded in E3 then induced Riemannian metric GM is

defined by the formulae

〈X,Y〉 = GM (X,Y) = G(X,Y) , (6.11)

where G is Euclidean metric in E3:

GM = dx2 + dy2 + dz2∣∣r=r(u,v)

=

3∑i=1

(dxi)2∣∣r=r(u,v)

=

3∑i=1

(∂xi

∂uαduα

)2

=∂xi

∂uα∂xi

∂uβduαduβ ,

i.e.

GM = gαβduα, where gαβ =

∂xi

∂uα∂xi

∂uβduαduβ .

We use notations x, y, z or xi (i = 1, 2, 3) for Cartesian coordinates in E3, u, v or uα

(α = 1, 2) for coordinates on the surface. We usually omit summation symbol over dummyindices. For coordinate tangent vectors

∂

∂uα︸︷︷︸Internal observer

= rα =∂xi

∂uα∂

∂xi︸ ︷︷ ︸External observer

We have already plenty examples in the subsection 1.4. In particular for scalar product

gαβ =

⟨∂

uα,∂

uβ

⟩= xiαxiβ .〈rα, rβ〉 . (6.12)

6.2.1 Recalling Weingarten operator

Continue to play with formulae 23.Recall the Weingarten (shape) operator which acts on tangent vectors:

SX = −∂Xn , (6.13)

where we denote by n-unit normal vector field at the points of the surface M : 〈n, rα〉 = 0,〈n, rα〉 = 1.

Now we realise that the derivative ∂XR of vector field with respect to another vectorfield is not a well-defined object: we need a connection. The formula ∂XR in Cartesiancoordinates, is nothing but the derivative with respect to flat canonical connection: If

23In some sense differential geometry it is when we write down the formulae expressingthe geometrical facts, differentiate these formulae then reveal the geometrical meaning ofthe new obtained formulae e.t.c.

124

we work only in Cartesian coordinates we do not need to distinguish between ∂XR and∇can.flat

X R. Sometimes with some abuse of notations we will use ∂XR instead ∇can.flatX R,

but never forget: this can be done only in Cartesian coordinates where Christoffel symbolsof flat canonical connection vanish:

∂XR = ∇can.flatX R in Cartesian coordinates .

So the rigorous definition of Weingarten operator is

SX = −∇can.flatX n , (6.14)

but we often use the former one (equation (6.16)) just remembering that this can be doneonly in Cartesian coordinates.

Recall that the fact that Weingarten operator S maps tangent vectors to tangentvectors follows from the property: 〈n,X〉 = 0⇒ X is tangent to the surface.

Indeed:

0 = ∂X〈n,n〉 = 2〈∂Xn,n〉 = −2〈SX,n〉 = 0⇒ SX is tangent to the surface

Recall also that normal unit vector is defined up to a sign, n → −n. On the other hand

if n is chosen then S is defined uniquely.We use notations x, y, z or xi (i = 1, 2, 3) for Cartesian coordinates in E3, u, v or uα

(α = 1, 2) for coordinates on the surface. We usually omit summation symbol over dummyindices. For coordinate tangent vectors

∂

∂uα︸︷︷︸Internal observer

= rα =∂xi

∂uα∂

∂xi︸ ︷︷ ︸External observer

We have already plenty examples in the subsection 1.4. In particular for scalar product

gαβ =

⟨∂

uα,∂

uβ

⟩= xiαxiβ .〈rα, rβ〉 . (6.15)

6.2.2 Recalling Weingarten operator

Continue to play with formulae 24.Recall the Weingarten (shape) operator which acts on tangent vectors:

SX = −∂Xn , (6.16)

where we denote by n-unit normal vector field at the points of the surface M : 〈n, rα〉 = 0,〈n, rα〉 = 1.

24In some sense differential geometry it is when we write down the formulae expressingthe geometrical facts, differentiate these formulae then reveal the geometrical meaning ofthe new obtained formulae e.t.c.

125

Now we realise that the derivative ∂XR of vector field with respect to another vectorfield is not a well-defined object: we need a connection. The formula ∂XR in Cartesiancoordinates, is nothing but the derivative with respect to flat canonical connection: Ifwe work only in Cartesian coordinates we do not need to distinguish between ∂XR and∇can.flat

X R. Sometimes with some abuse of notations we will use ∂XR instead ∇can.flatX R,

but never forget: this can be done only in Cartesian coordinates where Christoffel symbolsof flat canonical connection vanish:

∂XR = ∇can.flatX R in Cartesian coordinates .

So the rigorous definition of Weingarten operator is

SX = −∇can.flatX n , (6.17)

but we often use the former one (equation (6.16)) just remembering that this can be doneonly in Cartesian coordinates.

Recall that the fact that Weingarten operator S maps tangent vectors to tangentvectors follows from the property: 〈n,X〉 = 0⇒ X is tangent to the surface.

Indeed:

0 = ∂X〈n,n〉 = 2〈∂Xn,n〉 = −2〈SX,n〉 = 0⇒ SX is tangent to the surface

Recall also that normal unit vector is defined up to a sign, n → −n. On the other hand

if n is chosen then S is defined uniquely.

6.2.3 Second quadratic form

We define now the new object: second quadratic formDefinition. For two tangent vectors X,Y A(X,Y) is defined such that(

∇can.flatX Y

)⊥ = A(X,Y)n (6.18)

i.e. we take orthogonal component of the derivative of Y with respect to X.This definition seems to be very vague: to evaluate covariant derivative we have to

consider not a vector Y at a given point but the vector field. In fact one can see thatA(X,Y) does depend only on the value of Y at the given point.

Indeed it follows from the definition of second quadratic form and from the propertiesof Weingarten operator that

A(X,Y) =⟨(∇can.flat

X Y)⊥ ,n

⟩=⟨∇can.flat

X Y,n⟩

=

∂X〈Y,n〉 −⟨Y,∇can.flat

X n⟩

= 〈S(X),Y〉 (6.19)

We proved that second quadratic form depends only on value of vector field Y at thegiven poit and we established the relation between second quadratic form and Weingartenoperator.

Proposition The second quadratic form A(X,Y) is symmetric bilinear form on tan-gent vectors X,Y in a given point.

126

A : TpM ⊗ TpM → R, A(X,Y) = A(Y,X) = 〈SX,Y〉 . (6.20)

In components

A = Aαβduαduβ = 〈rαβ ,n〉 =

∂2xi

∂uα∂uβni . (6.21)

andSαβ = gαπAπβ = gαπxiπβn

i , (6.22)

i.e.A = GS, S = G−1A .

Remark The normal unit vector field is defined up to a sign.

6.2.4 Gaussian and mean curvatures

Recall that Gaussian curvatureK = detS

and mean curvatureH = TrS

It is easy to see that for Gaussian curvature

K = detS = det(G−1A) =detA

detG.

We know already the geometrical meaning of Gaussian and mean curvatures from thepoint of view of the External Observer:

Gaussian curvature K equals to the product of principal curvatures, and mean cur-vartures equals to the sum of principal curvatures.

Now we ask a principal question: what bout internal observer, ”aunt” living on thesurface?

We will show that Gaussian curvature can be expressed in terms of induced Rieman-nian metric, i.e. it is an internal characteristic of the surface, invariant of isometries.

It is not the case with mean curvature: cylinder is isometric to the plane but it havenon-zero mean curvature.

6.2.5 Examples of calculation of Weingarten operator, Secondquadratic forms, curvatures for cylinder, cone and sphere.

Cylinder

We already calculated induced Riemannian metric on the cylinder (see (1.56)).Cylinder is given by the equation x2 + y2 = R2. One can consider the following

parameterisation of this surface:

r(h, ϕ) :

x = R cosϕ

y = R sinϕ

z = h

, rh =

001

, rϕ =

−R sinϕR cosϕ

0

, (6.23)

127

Gcylinder =(dx2 + dy2 + dz2

) ∣∣x=a cosϕ,y=a sinϕ,z=h

=

= (−a sinϕdϕ)2 + (a cosϕdϕ)2 + dh2 = a2dϕ2 + dh2 , ||gαβ || =(

1 00 R2

).

Normal unit vector n = ±

cosϕsinϕ

0

. Choose n =

cosϕsinϕ

0

. Weingarten operator

S∂h = −∇can.flatrh

n = −∂rhn = −∂h

cosϕsinϕ

0

= 0 ,

S∂ϕ = −∇can.flatrϕ n = −∂rϕn = −∂ϕ

cosϕsinϕ

0

=

sinϕ− cosϕ

0

= −∂ϕR.

S

(∂h∂ϕ

)=

(0−∂ϕR

), S =

(0 00 −1

R

). (6.24)

Calculate second quadratic form: rhh = ∂hrh =

000

, rhϕ = rϕh =

∂h

−R sinϕR cosϕ

0

= 0, rϕϕ = ∂ϕ

−R sinϕR cosϕ

0

=

−R cosϕ−R sinϕ

0

= −Rn .

We have

Aαβ = 〈rαβ ,n〉, A =

(〈rhh,n〉 〈rhϕ,n〉〈rϕh,n〉 〈rϕϕ,n〉

)=

(0 00 −R

), (6.25)

A = GS =

(1 00 R2

)(0 00 − 1

R

)=

(0 00 −R

),

For Gaussian and mean curvatures we have

K = detS =detA

detG= det

(0 00 − 1

R

)= 0 , (6.26)

and mean curvature

H = TrS = Tr

(0 00 − 1

R

)= − 1

R. (6.27)

Mean curvature is define up to a sign. If we change n→ −n mean curvature H → 1R and

Gaussian curvature will not change.

ConeWe already calculated induced Riemannian metric on the cone (see (??).

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Cone is given by the equation x2 + y2 − k2z2 = 0. One can consider the followingparameterisation of this surface:

r(h, ϕ) :

x = kh cosϕ

y = kh sinϕ

z = h

, rh =

k cosϕk sinϕ

1

, rϕ =

−kh sinϕkh cosϕ

0

, (6.28)

Gcone =(dx2 + dy2 + dz2

) ∣∣x=kh cosϕ,y=kh sinϕ,z=h

=

= (−kh sinϕdϕ+ k cosϕdh)2 + (kh cosϕdϕ+ k sinϕdh)2 + dh2 =

k2h2dϕ2 + (k2 + 1)dh2, ||gαβ || =(k2 + 1 0

0 k2h2

).

One can see that N =

cosϕsinϕ−k

is orthogonal to the surface: N⊥rh, rϕ. Hence normal

unit vector n = ± 1√1+k2

cosϕsinϕ−k

. Choose n = 1√1+k2

cosϕsinϕ−k

. Weingarten operator

S∂h = −∇can.flatrh

n = −∂rhn = −∂h

1√1 + k2

cosϕsinϕ−k

= 0 ,

S∂ϕ = −∇can.flatrϕ n = −∂rϕn = −∂ϕ

1√1 + k2

cosϕsinϕ−k

=

1√1 + k2

sinϕ− cosϕ

0

= − ∂ϕ

kh√k2 + 1

.

S

(∂h∂ϕ

)=

(0

− ∂ϕkh√k2+1

), S =

(0 00 −1

kh√k2+1

). (6.29)

Calculate second quadratic form: rhh = ∂hrh =

000

, rhϕ = rϕh =

∂h

−kh sinϕkh cosϕ

0

=

−k sinϕk cosϕ

0

, rϕϕ = ∂ϕ

−kh sinϕkh cosϕ

0

=

−kh cosϕ−kh sinϕ

0

.

We have

Aαβ = 〈rαβ ,n〉, A =

(〈rhh,n〉 〈rhϕ,n〉〈rϕh,n〉 〈rϕϕ,n〉

)=

(0 00 − kh√

1+k2

), (6.30)

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A = GS =

(k2 + 1 0

0 k2h2

)(0 00 −1

kh√k2+1

)=

(0 00 −kh√

k2+1

),

For Gaussian and mean curvatures we have

K = detS =detA

detG= det

(0 00 −1

kh√k2+1

)= 0 , (6.31)

and mean curvature

H = TrS = Tr

(0 00 −1

kh√k2+1

)=

−1

kh√k2 + 1

. (6.32)

Mean curvature is define up to a sign. If we change n→ −n mean curvature H → 1R and

Gaussian curvature will not change.

Sphere

Sphere is given by the equation x2 + y2 + z2 = a2. Consider the parameterisation ofsphere in spherical coordinates

r(θ, ϕ) :

x = R sin θ cosϕ

y = R sin θ sinϕ

z = R cos θ

(6.33)

We already calculated induced Riemannian metric on the sphere (see (6.2.5)). Recallthat

rθ =

R cos θ cosϕR cos θ sinϕ−R sin θ

, rϕ =

−R sin θ sinϕR sin θ cosϕ

0

and

GS2 =(dx2 + dy2 + dz2

) ∣∣x=R sin θ cosϕ,y=R sin θ sinϕ,z=R cos θ

=

(R cos θ cosϕdθ −R sin θ sinϕdϕ)2 + (R cos θ sinϕdθ +R sin θ cosϕdϕ)2+

(−R sin θdθ)2 = R2 cos2 θdθ2 +R2 sin2 θdϕ2 +R2 sin2 θdθ2 =

= R2dθ2 +R2 sin2 θdϕ2 , ||gαβ || =(R2 00 R2 sin2 θ

).

For the sphere r(θ, ϕ) is orthogonal to the surface. Hence normal unit vector n(θ, ϕ) =

± r(θ,ϕ)R = ±

sin θ cosϕsin θ sinϕ

cos θ

. Choose n = rR =

sin θ cosϕsin θ sinϕ

cos θ

. Weingarten operator

S∂θ = −∇can.flatrθ

n = −∂θn = −∂θ( r

R

)= −rθ

R,

S∂ϕ = −∇can.flatrϕ n = −∂ϕn = −∂ϕ

( r

R

)= −rϕ

R.

130

S

(∂θ∂ϕ

)=

(−∂θR−∂ϕR

), S = −

(1R 00 1

R

). (6.34)

For second quadratic form: rθθ = ∂θrθ =

−R sin θ cosϕ−R sin θ sinϕ−R cos θ

, rθϕ = rϕθ =

∂θ

−R sin θ sinϕR sin θ cosϕ

0

=

−R cos θ sinϕR cos θ cosϕ

0

, rϕϕ = ∂ϕ

−R sin θ sinϕR sin θ cosϕ

0

=

−R sin θ cosϕ−R sin θ sinϕ

0

.

We have

Aαβ = 〈rαβ ,n〉, A =

(〈rθθ,n〉 〈rθϕ,n〉〈rϕθ,n〉 〈rϕϕ,n〉

)=

(−R 00 −R sin2 θ

), (6.35)

A = GS =

(R2 00 R2 sin2 θ

)(−1R 00 −1

R

)= −R

(1 00 sin2 θ

),

For Gaussian and mean curvatures we have

K = detS =detA

detG= det

(− 1R 0

0 − 1R

)=

1

R2, (6.36)

and mean curvature

H = TrS = Tr

(− 1R 0

0 − 1R

)= − 2

R, (6.37)

Mean curvature is define up to a sign. If we change n→ −n mean curvature H → 1R and

Gaussian curvature will not change.We see that for the sphere Gaussian curvature is not equal to zero, whilst for cylinder

and cone Gaussian curvature equals to zero.

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