Riemannian Geometry

Home , Riemannian manifold, Volume element

it is a draft of Lecture Notes of H.M. Khudaverdian. Manchester, 20-th May 2011

Contents

1 Riemannian manifolds 4 1.1 Manifolds. Tensors. (Recalling) ...... 4 1.2 Riemannian manifold—manifold equipped with Riemannian metric ...... 7 1.2.1 ∗ Pseudoriemannian manifold ...... 11 1.3 Scalar product. Length of tangent vectors and angle between vectors. Length of the curve ...... 11 1.3.1 Length of the curve ...... 12 1.4 Riemannian structure on the surfaces embedded in Euclidean space ...... 15 1.4.1 Internal and external coordinates of tangent vector . . 15 1.4.2 Explicit formulae for induced Riemannian metric (First Quadratic form) ...... 17 1.4.3 Induced Riemannian metrics. Examples...... 20 1.4.4 ∗Induced metric on two-sheeted hyperboloid embedded in pseudo-Euclidean space...... 27 1.5 Isometries of Riemanian manifolds...... 28 1.5.1 Examples of local isometries ...... 29 1.6 Volume element in Riemannian manifold ...... 31 1.6.1 Volume of parallelepiped ...... 31 1.6.2 Invariance of volume element under changing of coordinates ...... 33 1.6.3 Examples of calculating volume element ...... 34

1 2 Covariant differentiaion. Connection. Levi Civita Connec- tion on Riemannian manifold 36 2.1 Differentiation of vector field along the vector field.—Affine connection ...... 36 2.1.1 Definition of connection. Christoffel symbols of connection ...... 37 2.1.2 Transformation of Christoffel symbols for an arbitrary connection ...... 39 2.1.3 Canonical flat affine connection ...... 40 2.1.4 ∗ Global aspects of existence of connection ...... 43 2.2 Connection induced on the surfaces ...... 44 2.2.1 Calculation of induced connection on surfaces in E3. . . 45 2.3 Levi-Civita connection ...... 47 2.3.1 Symmetric connection ...... 47 2.3.2 Levi-Civita connection. Theorem and Explicit formulae 47 2.3.3 Levi-Civita connection on 2-dimensional Riemannian manifold with metric G = adu2 + bdv2...... 49 2.3.4 Example of the sphere again ...... 49 2.4 Levi-Civita connection = induced connection on surfaces in E3 50

3 Parallel transport and geodesics 52 3.1 Parallel transport ...... 52 3.1.1 Definition ...... 52 3.1.2 ∗Parallel transport is a linear map ...... 53 3.1.3 Parallel transport with respect to Levi-Civita connection 54 3.2 Geodesics ...... 54 3.2.1 Definition. Geodesic on Riemannian manifold...... 54 3.2.2 Un-parameterised geodesic ...... 55 3.2.3 Geodesics on surfaces in E3 ...... 56 3.3 Geodesics and Lagrangians of ”free” particle on Riemannian manifold...... 58 3.3.1 Lagrangian and Euler-Lagrange equations ...... 58 3.3.2 Lagrangian of ”free” particle ...... 58 3.3.3 Equations of geodesics and Euler-Lagrange equations . 59 3.3.4 Examples of calculations of Christoffel symbols and geodesics using Lagrangians...... 60 3.3.5 Variational principe and Euler-Lagrange equations . . 62 3.4 Geodesics and shortest distance...... 63

2 3.4.1 Again geodesics for sphere and Lobachevsky plane . . . 65

4 Surfaces in E3 67 4.1 Formulation of the main result. Theorem of parallel transport over closed curve and Theorema Egregium ...... 67 4.1.1 GaußTheorema Egregium ...... 69 4.2 Derivation formulae ...... 70 4.2.1 ∗Gauss condition (structure equations) ...... 72 4.3 Geometrical meaning of derivation formulae. Weingarten operator and second quadratic form in terms of derivation formulae...... 73 4.3.1 Gaussian and mean curvature in terms of derivation formulae ...... 75 4.4 Examples of calculations of derivation formulae and curvatures for cylinder, cone and sphere ...... 76 4.5 ∗Proof of the Theorem of parallel transport along closed curve. 80

5 Curvtature tensor 84 5.1 Deﬁnition ...... 84 5.1.1 Properties of curvature tensor ...... 85 5.2 Riemann curvature tensor of Riemannian manifolds...... 86 5.3 †Curvature of surfaces in E3.. Theorema Egregium again . . . 87 5.4 Relation between Gaussian curvature and Riemann curvature tensor. Straightforward proof of Theorema Egregium . . . . . 88 5.4.1 ∗Proof of the Proposition (5.25) ...... 90 5.5 Gauss Bonnet Theorem ...... 94

6 Appendices 97 6.1 ∗Integrals of motions and geodesics...... 97 6.1.1 ∗Integral of motion for arbitrary Lagrangian L(x, x˙) . . 97 6.1.2 ∗Basic examples of Integrals of motion: Generalised momentum and Energy ...... 97 6.1.3 ∗Integrals of motion for geodesics ...... 98 6.1.4 ∗Using integral of motions to calculate geodesics . . . . 100 6.2 Induced metric on surfaces...... 101 6.2.1 Recalling Weingarten operator ...... 101 6.2.2 Second quadratic form ...... 102 6.2.3 Gaussian and mean curvatures ...... 103

3 6.2.4 Examples of calculation of Weingarten operator, Sec- ond quadratic forms, curvatures for cylinder, cone and sphere...... 103

1 Riemannian manifolds

1.1 Manifolds. Tensors. (Recalling) 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 consider different local coordinates. If both coordinates {x1, . . . , xn}, {y1, . . . , yn} are defined in a vicinity of the given point then they are related by bijective transition functions (functions defined on domains in Rn and taking values in Rn).   1′ 1′ 1 n x = x (x , . . . , x )  ′ ′ x2 = x2 (x1, . . . , xn) ...  − ′ − ′ xn 1 = xn 1 (x1, . . . , xn)  ′ ′ xn = xn (x1, . . . , xn) We say that manifold is differentiable or smooth if transition functions are diffeomorphisms, i.e. they are smooth and rank of Jacobian is equal to k, i.e.  ′ ′ ′  ∂x1 ∂x1 ∂x1 1 2 ... n  ∂x ′ ∂x ′ ∂x ′   ∂x2 ∂x2 ... ∂x2  det  ∂x1 ∂x2 ∂xn  ≠ 0 (1.1)  ...  ′ ′ ′ ∂xn ∂xn ∂xn ∂x1 ∂x2 ... ∂xn A good example of manifold is an open domain D in n-dimensional vector space Rn. Cartesian coordinates on Rn define global coordinates on D. On the other hand one can consider an arbitrary local coordinates in different domains in Rn. E.g. one can consider polar coordinates {r, φ} in a domain D = {x, y : y > 0} of R2 (or in other domain of R2) defined by standard formulae: { x = r cos φ , y = r sin φ

4 ( ) ( ) ∂x ∂x − ∂r ∂φ cos φ r sin φ det ∂y ∂y = det = r (1.2) ∂r ∂φ sin φ r cos φ or one can consider spherical coordinates {r, θ, φ} in a domain D = {x, y, z : x > 0, y > 0, z > 0} of R3 (or in other domain of R3) defined by standard formulae  x = r sin θ cos φ y = r sin θ sin φ ,  z = r cos θ ,     ∂x ∂x ∂x −  ∂r ∂θ ∂φ  sin θ cos φ r cos θ cos φ r sin θ sin φ  ∂y ∂y ∂y    2 det ∂r ∂θ ∂φ = det sin θ sin φ r cos θ sin φ r sin θ cos φ = r sin θ ∂z ∂z ∂z cos θ −r sin θ 0 ∂r ∂θ ∂φ (1.3) Choosing domain where polar (spherical) coordinates are well-defined we have to be award that coordinates have to be well-defined and transition functions (1.1) have to be diffeomorphisms. Examples of manifolds: Rn, Circle S1, Sphere S2, in general sphere Sn, torus S1 × S1, cylinder, cone, . . . . We also have to recall briefly what are tensors on manifold. Tensors on Manifold For every point p on manifold M one can consider tangent vector space TpM—the space of vectors tangent to the manifold at the point M. i ∂ Tangent vector A = A ∂xi . Under changing of coordinates it transforms as follows: m′ ∂ ∂x ∂ ′ ∂ A = Ai = Ai = Am ∂xi ∂xi ∂xm′ ∂xm′ Hence

i′ ′ ∂x Ai = Ai (1.4) ∂xi ∗ Consider also cotangent space TpM (for every point p on manifold M)— space of linear functions on tangent vectors, i.e. space of 1-forms which sometimes are called covectors.:

5 i One-form (covector) ω = ωidx transforms as follows m m ∂x m′ m′ ω = ω dx = ω dx = ω ′ dx . m m ∂xm′ m Hence ∂xm ω ′ = ω . (1.5) m ∂xm′ m Tensors: One can consider contravariant tensors of the rank p ∂ ∂ ∂ T = T i1i2...ip ⊗ ⊗ · · · ⊗ ∂xi1 ∂xi2 ∂xik with components {T i1i2...ik }. One can consider covariant tensors of the rank q

j1 ⊗ j2 ⊗ jq S = Sj1j2...jq dx dx . . . dx { } with components Sj1j2...jq . One can also consider mixed tensors:

i i ...i ∂ ∂ ∂ 1 2 p ⊗ ⊗ · · · ⊗ ⊗ j1 ⊗ j2 ⊗ jq Q = Qj j ...j dx dx . . . dx 1 2 q ∂xi1 ∂xi2 ∂xik ( ) p { i1i2...ip } with components Qj j ...j . We call these tensors tensors of the type . (1 )2 q q p Tensors of the type are called contravariant tensors of the rank p. 0 ( ) 0 Tensors of the type are called covariant tensors of the rank q. q Having in mind( (1.4)) and (1.5) we come to the rule of transformation for p tensors of the type : q

′ ′ ′ ′ ′ ′ i1 i2 ip j1 j2 jq i1i2...ip ∂x ∂x ∂x ∂x ∂x ∂x i1i2...ip Q ′ ′ ′ = ... ′ ′ ... ′ Q (1.6) j j ...j i i i j j j j1j2...jq 1 2 q ∂x 1 ∂x 2 ∂x p ∂x 1 ∂x 2 ∂x q ( ) p E.g. if S is a covariant tensor of rank 2 (tensor of the type ) then ik q

∂xi ∂xk S ′ ′ = S . (1.7) i k ∂xi′ ∂xk′ ik

6 ( ) 1 If Ai is a tensor of rank (linear operator on T M) then k 1 p i′ k i′ ∂x ∂x i A ′ = A k ∂xi ∂xk′ k Remark Transformations formulae (1.4)—(1.7) define vectors, covectors and in generally any tensor fields in components. E.g. covariant tensor (covariant tensor field) of the rank 2 can be defined as matrix Sik (matrix valued 1 2 n function Sik(x)) such that under changing of coordinates {x , x , . . . , x } 7→ 1′ 2′ n′ {x , x , . . . , x } Sik change by the rule (1.7). Remark Einstein summation rules In our lectures we always use so called Einstein summation convention. it implies that when an index occurs more than once in the same expression in upper and in.. postitions.., the expression is implicitly summed over all pos- sible values for that index. Sometimes it is called dummy indices summation rule.

1.2 Riemannian manifold—manifold equipped with Rie- mannian metric Definition The Riemannian manifold is a manifold equipped with a Rie- mannian metric. The Riemannian metric on the manifold M defines the length of the tangent vectors and the length of the curves. Definition Riemannian metric G on n-dimensional manifold M n defines for every point p ∈ M the scalar product of tangent vectors in the tangent space TpM smoothly depending on the point p. It means that in every coordinate system (x1, . . . , xn) a metric G = i k gikdx dx 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) , B = Bi(x) , ∂xi ∂xi tangent to the manifold M at the point p with coordinates x = (x1, x2, . . . , xn) (A, B ∈ TpM) the scalar product is equal to:

⟨ ⟩ i k A, B G p = G(A, B) p = A (x)gik(x)B (x) =

7     B1 g11(x) . . . g1n(x)   ( )    ·   ......    A1 ...An    ·  (1.8) gn1(x) . . . gnn(x)  ·  Bn where

• G(A, B) = G(B, A), i.e. gik(x) = gki(x) (symmetricity condition) • G(A, A) > 0 if A ≠ 0, i.e. i k i k 1 n gik(x)u u ≥ 0, gik(x)u u = 0 iﬀ u = ··· = u = 0 (positive- deﬁniteness)

• G(A, B) p=x, i.e. gik(x) are smooth functions. 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 tangent spaces TpM smoothly depending on the point p. Components of tensor field G in coordinate system are matrix valued functions gik(x):

i k G = gik(x)dx ⊗ dx . (1.9)

The matrix ||gik|| of components of the metric G we also sometimes denote by G. Rule of transformation for entries of matrix gik(x) gik(x)-entries of the matrix ||gik|| are components of tensor ﬁeld G in a given coordinate system. How do these components transform under transformation of coordinates ′ {xi} 7→ {xi }? ( ) ( ) i k ∂x ′ ∂x ′ G = g dxi ⊗ dxk = g dxi ⊗ dxk = ik ik ∂xi′ ∂xk′

i k ∂x ∂x i′ k′ i′ k′ g dx ⊗ dx = g ′ ′ dx ⊗ dx ∂xi′ ik ∂xk′ i k Hence ∂xi ∂xk g ′ ′ = g . (1.10) i k ∂xi′ ik ∂xk′

8 One can derive transformations formulae also using general formulae (1.7) for tensors. Important remark ⟨ ⟩ ∂ ∂ g = , (1.11) ik ∂xi ∂xk Later by some abuse of notations we sometimes omit the sign of tensor product and write a metric just as

i k G = gik(x)dx dx .

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:

i k 1 1 2 2 n n G(X, Y) = ⟨X, Y⟩ = gikX Y = X Y + X Y + ··· + X Y .

In the general case if G = ||gik|| is an arbitrary symmetric positive- deﬁnite metric then

i k G(X, Y) = ⟨X, Y⟩ = gikX Y .

One can show that there exist a new basis {ei} such that in this basis

G(ei, ek) = δik .

This basis is called orthonormal basis. (See the Lecture notes in Ge- ometry) Scalar product in vector space deﬁnes the same scalar product at all the points. In general case for Riemannian manifold scalar product depends on the point.

9 • R2 with polar coordinates in the domain y > 0 (x = r cos φ, y = r sin φ): dx = cos φdr − r sin φdφ, dy = sin φdr + r cos φdφ. In new coordinates the Riemannian metric G = dx2 + dy2 will have the following appearance:

G = (dx)2+(dy)2 = (cos φdr−r sin φdφ)2+(sin φdr+r cos φdφ)2 = dr2+r2(dφ)2

We see that for matrix G = ||gik|| ( ) ( ) ( ) ( ) g g 1 0 g g 1 0 G = xx xy = ,G = rr rφ = g g 0 1 g g 0 r | yx {zyy } | φr {zφφ } in cartesian coordinates in polar coordinates

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

G = a2dx2 (1.12)

Renaming x 7→ φ we come to habitual formula for metric for circle of the radius a: x2 + y2 = a2 embedded in the Euclidean space E2: { x = a cos φ G = a2dφ2 , 0 ≤ φ < 2π, (1.13) y = a sin φ

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

G = a2dx2 + dy2 (1.14)

We see that renaming variables x 7→ φ, y 7→ h we come to habitual, familiar formulae for metric in standard polar coordinates for cylinder surface of the radius a embedded in the Euclidean space E3:  x = a cos φ G = a2dφ2 + dh2 y = a sin φ , 0 ≤ φ < 2π, −∞ < h < ∞  z = h (1.15)

10 • 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 sphere x2 + y2 + z2 = a2 of the radius a embedded in the Euclidean space E3: { x = a sin θ cos φ G = a2dθ2+a2 sin2 θdφ2 , 0 ≤ φ < 2π, −∞ < h < ∞ y = a sin θ sin φz = a cos θ (1.16)

1.2.1 ∗ Pseudoriemannian manifold If we omit the condition of positive-deﬁniteness for Riemannian metric we come to so called Pseudorimannian metric. Manifodl equipped with pseu- doriemannan metric is called pseudoriemannian manifold. Pseudoriemannian manifolds appear in applications in the special and general relativity theory. Example Consider n+1-dimensional linear space Rn+1 with pseudomet- ric (dx0)2 − (dx1)2 − (dx2)2 − · · · − (dxn)2 in coordinates x0, x1, . . . , xn. In the case n = 3 it is so called Minkovski space. The coordinate x0 the role of the time: x0 = ct, where c is the value of the speed of the light.

1.3 Scalar product. Length of tangent vectors and angle between vectors. Length of the curve The Riemannian metric defines scalar product of tangent vectors attached at the given point. Hence it defines the length of tangent vectors and angle m ∂ m ∂ between them. If X = X ∂xm , Y = Y ∂xm are two tangent vectors at the given point p of Riemannian manifold with coordinates x1, . . . , xn, then we have that lengths of these vectors equal to √ √ √ √ i k i k |X| = ⟨X, X⟩ = gik(x)X X , |Y| = ⟨Y, Y⟩ = gik(x)Y Y , (1.17) and he angle θ between these vectors is defined by the relation ⟨X, Y⟩ g XiY k cos θ = = √ ik √ (1.18) | | · | | i k i k X Y gik(x)X X gik(x)Y Y

11 Example Let M be 3-dimensional Riemanian manifold. Consider the vectors X = 2∂x +2∂y −∂z and Y = ∂x −2∂y −2∂z attached at the point p of M with local coordinates (x, y, z), where x = y = 1, z = 0. Find the lengths of these vectors and angle between them if the expression of Riemannian dx2+dy2+dz2 metric in these coordinates is coordinates (1+x2+y2)2 . We see that matrix of Riemannian gik = σ(x, y, z)δik, where σ(x, y, z) = 1 ̸ || || (1+x2+y2+z2)2 δik = 1 if i = k and δik = 0 if i = k, i.e. matrix gik is proportional to unity matrix. According to formulae above √ √ √ √ √ √ i k i i |X| = ⟨X, X⟩ = gik(x)X X = σ(x, y, z) X X = 3 σ(x, y, z) = 3 .

The same answer for |Y|. The scalar product between vectors X, Y equal to zero: i k ⟨X, Y⟩ = σ(x, y, z)δikX Y = 0 Hence these vectors are orthogonal to each other.

1.3.1 Length of the curve Let γ : xi = xi(t), (i = 1, . . . , n)) (a ≤ t ≤ b) be a curve on the Riemannian manifold (M,G). At the every point of the curve the velocity vector (tangent vector) is deﬁned:   x˙ 1(t)    ·    v(t) =  ·  (1.19)  ·  x˙ n(t)

The length of velocity vector v ∈ TxM (vector v is tangent to the manifold M at the point x) equals to √ √ √ dxi(t) dxk(t) | | ⟨ ⟩ i k v x = v, v G x = gikv v x = gik x (1.20) dt dt x The length of the curve is deﬁned by the integral of the length of velocity vector: ∫ ∫ b √ b √ ⟨ ⟩ i k Lγ = v, v G x(t)dt = gik(x(t))x ˙ (t)x ˙ (t)dt (1.21) a a

12 Bearing in mind that metric (1.9) deﬁnes the length we often write metric in the following form 2 i k ds = gikdx dx (1.22)

For example consider 2-dimensional Riemannian manifold with metric ( ) g11(u, v) g12(u, v) ||gik(u, v)|| = . g21(u, v) g22(u, v) Then

2 i k 2 2 G = ds = gikdu dv = g11(u, v)du + 2g12(u, v)dudv + g22(u, v)dv

The length of the curve γ : u = u(t), v = v(t), where t0 ≤ t ≤ t1 according to (1.21) is equal to ∫ ∫ t1 √ t1 √ i k Lγ = ⟨v, v⟩ = gik(x)x ˙ x˙ = (1.23) t0 t0 ∫ √ t1 2 2 g11 (u (t) , v (t)) ut + 2g12 (u (t) , v (t)) utvt + g22 (u (t) , v (t)) vt dt t0 (1.24) The length of the curve deﬁned by the formula(1.21) obeys the following natural conditions

• It coincides with the usual length in the Euclidean space En (Rn with standard metric G = (dx1)2 + ··· + (dxn)2 in cartesian coordinates). E.g. for 3-dimensional Euclidean space ∫ ∫ b √ b √ i k 1 2 2 2 3 2 Lγ = gik(x(t))x ˙ (t)x ˙ (t)dt = (x ˙ (t)) + (x ˙ (t)) + (x ˙ (t)) dt a a (1.25)

• It does not depend on parameterisation of the curve

∫ ∫ b √ b′ √ i k i k Lγ = gik(x(t))x ˙ (t)x ˙ (t)dt = gik(x(τ))x ˙ (τ)x ˙ (τ)dτ, a a′ (1.26) where xi(τ) = xi(t(τ)), a′ ≤ τ ≤ b′ while a ≤ t ≤ b.

13 • It does not depend on coordinates on Riemannian manifold M ∫ ∫ b √ b √ i k ′ i′ k′ Lγ = gik(x(t))x ˙ (t)x ˙ (t)dt = gi′k′ (x (t))x ˙ (t)x ˙ (t)dt a a (1.27)

i i • It is additive: if a curve γ = γ1+γ, i.e. γ : x (t), a ≤ t ≤ b, γ1 : x (t), a ≤ i t ≤ c and γ2 : x (t), c ≤ t ≤ b where a point c belongs to the interval (a, b) then Lγ = Lγ1 + Lγ2 , i.e. ∫ b √ i k gik(x(t))x ˙ (t)x ˙ (t)dt = a ∫ ∫ c √ b √ i k i k gik(x(t))x ˙ (t)x ˙ (t)dt + gik(x(t))x ˙ (t)x ˙ (t)dt (1.28) a c Conditions (1.25) and (1.28) evidently are obeyed. Condition (1.26) follows from the fact that

dx(t(τ)) dx(t(τ)) dt dt x˙ i(τ) = = =x ˙ i(t) . dτ dt dτ dτ Bearing in mind the above formula we have ∫ ∫ √ ( ) b′ √ b′ 2 i k i k dt gik(x(τ))x ˙ (τ)x ˙ (τ)dτ = gik(x(t(τ)))x ˙ (t)x ˙ (t) dτ = a′ a′ dτ ∫ ∫ b′ √ b √ i k dt i k gik(x(t(τ)))x ˙ (t)x ˙ (t) dτ = gik(x(t))x ˙ (t)x ˙ (t)dt = Lγ a′ dτ a Condition (1.27) follows from the condition (1.76): ∫ ∫ √ b √ b i k ′ ′ ∂x ∂x ′ ′ ′ ′ ′ i k i k gi k (x (t))x ˙ (t)x ˙ (t)dt = gik(x(t)) i′ k′ x˙ (t)x ˙ (t)dt = a a ∂x ∂x ∫ √ ( )( ) ∫ b i k b √ ∂x i′ ∂x k′ i k gik(x(t)) i′ x˙ (t) k′ x˙ (t) dt = gik(x(t))x ˙ (t)x ˙ (t)dt . a ∂x ∂x a (1.29)

14 1.4 Riemannian structure on the surfaces embedded in Euclidean space Let M be a surface embedded in Euclidean space. Let G be Riemannian structure on the manifold M. Let X, Y be two vectors tangent to the surface M at a point p ∈ M. An External Observer calculate this scalar product viewing these two vectors as vectors in E3 attached at the point p ∈ E3 using scalar product in E3. An Internal Observer will calculate the scalar product viewing these two vectors as vectors tangent to the surface M using the Riemannian metric G (see the formula (1.39)). Respectively If L is a curve in M then an External Observer consider this curve as a curve in E3, calculate the modulus of velocity vector (speed) and the length of the curve using Euclidean scalar product of ambient space. An Internal Observer (”an ant”) will define the modulus of the velocity vector and the length of the curve using Riemannian metric. Definition Let M be a surface embedded in the Euclidean space. We say that metric GM on the surface is induced by the Euclidean metric if the scalar product of arbitrary two vectors A, B ∈ TpM calculated in terms of the metric G equals to Euclidean scalar product of these two vectors: ⟨ ⟩ ⟨ ⟩ A, B GM = A, B GEuclidean (1.30) In other words we say that Riemannian metric on the embedded surface is induced by the Euclidean structure of the ambient space if External and Internal Observers come to the same results calculating scalar product of vectors tangent to the surface. In this case modulus of velocity vector (speed) and the length of the curve is the same for External and Internal Observer. Before going in details of this definition recall the conception of Internal and External Observers when dealing with surfaces in Euclidean space:

1.4.1 Internal and external coordinates of tangent vector Tangent plane Let r = r(u, v) be parameterisation of the surface M embedded in the Eu- clidean space:   x(u, v) r(u, v) = y(u, v) z(u, v)

15 Here as always x, y, z are Cartesian coordinates in E3. Let p be an arbitrary point on the surface M. Consider the plane formed by the vectors which are adjusted to the point p and tangent to the surface M. We call this plane plane tangent to M at the point p and denote it by TpM. For a point p ∈ M one can consider a basis in the tangent plane TpM adjusted to the parameters u, v. Tangent basis vectors at any point (u, v) are   ∂x(u,v) ∂r(u, v)  ∂u  ∂x(u, v) ∂ ∂y(u, v) ∂ ∂z(u, v) ∂ r = =  ∂y(u,v)  = + + u ∂u ∂u ∂z(u,v) ∂u ∂x ∂u ∂y ∂u ∂z ∂u

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

X = Xuru + Xvrv, (1.31) 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 velocity vector for the curve u = u0, v = v0 + t, where (u0, v0) are coordinates of the point p. Let r = r(t) be a curve belonging to the surface C, which passes through the point p, r(t) = r(u(t), v(t)) and p = r(t0). Then vector dr dr(u(t), v(t)) r = = (1.32) t dt dt belongs to the tangent plane TpM. Note that for the vector (1.32) components Xu,Xv are equal to Xu = ut,Xv = vt because dr dr(u(t), v(t)) r = = = u r + v r (1.33) t dt dt t u t v 3 An External Observer describes the vector rt as a vector in E attached at the point p. The Internal Observer describes this vector as a vector which has components (ut, vt) in the basis ru, rv according to the formula (1.33). In general consider an arbitrary tangent vector X ∈ TpM. Denote Xu = a, Xv = b

16       xu(u, v) xv(u, v) axu(u, v) + bxv(u, v)       X = Xuru+Xvrv = aru+brv = a yu(u, v) +b xv(u, v) = ayu(u, v) + byv(u, v) zu(u, v) xv(u, v) azu(u, v) + bzv(u, v) (1.34) The last column in this formula represents the three components of the vector X in the ambient space. The pair (a, b) can be considered as internal coordinates of the tangent vector X. An Internal Observer, Ant living on the surface, deals with the vector X in terms of coordinates (a, b). External observer which contemplates the surface embedded in three-dimensional ambient space deals with vector X as with vector with three external coordinates (see the last right column in the formula (1.34).) In condensed notation instead denoting coordinates by (u, v) we often denote them by uα = (u1, u2). Respectively we denote by

dr r = , r = r , r = r α duα u 1 v 2 The formula (1.34) for tangent vector ﬁeld will have the following appearance: α 1 2 1 2 X = X rα = X r1 + X r2, (X = Xu,X = Xv) (1.35)

When using condensed notations∑ we usually omit explicit summation α 2 α 1 2 symbols. E.g. we write u rα instead i=1 u rα or u r1 + u r2 One can consider also diﬀerentials duα = (du1, du2):

α α 1 2 1 2 du (rβ) = δβ : du (r1) = du (r2) = 1, du (r2) = du (r1) = 0 (1.36)

1.4.2 Explicit formulae for induced Riemannian metric (First Quadratic form) Now we are ready to write down the explicit formuale for the Riemannian metric on the surface induced by metric (scalar product) in ambient Eu- clidean space (see the Deﬁnition (1.30)). Let M : r = r(u, v) be a surface embedded in E3. The formula (1.30) means that scalar products of basic vectors ru = ∂u, rv = ∂v has to be the same calculated in the ambient space and on

17 the surface: For example scalar product ⟨∂u, ∂v⟩M = guv calculated by the Internal Observer is the same as a scalar product ⟨ru, rv⟩E3 calculated by the External Observer, scalar product ⟨∂v, ∂v⟩M = guv calculated by the Internal Observer is the same as a scalar product ⟨rv, rv⟩E3 calculated by the External Observer and so on: ( ) ( ) ( ) g g ⟨∂ , ∂ ⟩⟨∂ , ∂ ⟩ ⟨r , r ⟩ 3 ⟨r , r ⟩ 3 G = uu uv = u u u v = u u E u v E (1.37) gvu gvv ⟨∂v, ∂u⟩⟨∂v, ∂v⟩ ⟨rv, ru⟩E3 ⟨rv, rv⟩E3 where as usual we denote by ⟨ , ⟩E3 the scalar product in the ambient Eu- clidean space. (Here see also the important remark (1.11)) 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) instead r = r(u, v) In these notations: ( ) ( ) g11 g12 ⟨ru, ru⟩E3 ⟨ru, rv⟩E3 GM = = , gαβ = ⟨rα, rβ⟩ , g12 g22 ⟨ru, rv⟩E3 ⟨rv, rv⟩E3

α β 2 2 GM = gαβdu du = g11du + 2g12dudv + g22dv (1.38) where ( , ) is a scalar product in Euclidean space. The formula (1.38) is the formula for induced Riemannian metric on the surface—First Quadratic Form. 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:

1 2 1 2 (X, Y) = (X r1 + X r2,Y r1 + Y r2) = (1.39)

1 1 1 2 2 1 2 2 X (r1, r1)Y + X (r1, r2)Y + X (r2, r1)Y + X (r2, r2)Y = α β α β X (rα, rβ)Y = X gαβY = G(X, Y) We can come to this formula just transforming diﬀerentials. In carteisan coordinates ⟨X, Y⟩ = X1Y 1 + X2Y 2 + X3Y 3, i.e. the Euclidean metric in cartesian coordinates is given by

2 2 2 GE3 = (dx) + (dy) + (dz) . (1.40)

The condition that Riemannian metric (1.38) is induced by Euclidean scalar product means that

18 ( ) 2 2 2 α β 3 GE r=r(u,v) = (dx) + (dy) + (dz) r=r(u,v) = GM = gαβdu du (1.41)

2 2 2 i.e. ((dx) + (dy) + (dz) ) r=r(u,v) = ( ) ( ) ( ) ∂x(u, v) ∂x(u, v) 2 ∂x(u, v) ∂x(u, v) 2 ∂x(u, v) ∂x(u, v) 2 du + du + du + du + du + du = ∂u ∂u ∂u ∂u ∂u ∂u

2 2 2 2 2 2 2 2 (xu + yu + zu)du + 2(xuxv + yuyv + zuzv)dudv + (xv + yv + zv )dv We see that

α β 2 2 GM = gαβdu du = g11du + 2g12dudv + g22dv , (1.42)

2 2 2 ⟨ ⟩ where g11 = guu = (xu + yu + zu) = ru, ru E3 , g12 = g21 = guv = gvu = ⟨ ⟩ 2 2 2 ⟨ ⟩ (xuxv + yuyv + zuzv) = ru, rv E3 , g22 = gvv = (xv + yv + zv ) = rv, rv E3 . We come to the same formula. (See the examples of calculations in the next subsection.) Check explicitly again that length of the tangent vectors and of the curves calculating by External observer (i.e. using Euclidean metric (1.40)) is the same as calculating by Internal Observer, ant (i.e. using the induced Rie- mannian metric (1.38)) α Consider a vector X = X rα = aru + brv tangent to the surface M. Calculate its length by External and Internal Observer. The square of the length |X| of this vector calculated by External observer (he calculates using the scalar product in E3) equals to

2 2 2 |X| = ⟨X, X⟩ = ⟨ru + brv, aru + brv⟩ = a ⟨ru, ru⟩ + 2ab⟨ru, rv⟩ + b ⟨rv, rv⟩ (1.43) where ⟨ , ⟩ is a scalar product in E3. The internal observer will calculate the length using Riemannian metric (1.38): ( ) ( ) ( ) g11 g12 a 2 2 G(X, X) = a, b · · = g11a + 2g12ab + g22b (1.44) g21 G22 b

External observer (person living in ambient space E3) calculate the length of the tangent vector using formula (1.43). An ant living on the surface calculate length of this vector in internal coordinates using formula (1.44).

19 External observer deals with external coordinates of the vector, ant on the surface 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

dr(t) v = X = ξru + ηrvwhere ξ = ut, η = vt : v = dt = utru + vtrv . The length of the curve is equal to ∫ ∫ ∫ b b √ b √ L = |v(t)|dt = ⟨v(t), v(t)⟩E3 dt = ⟨utru + vtrv, utru + vtrv⟩E3 dt = a a a ∫ (1.45) b √ ⟨ ⟩ 2 ⟨ ⟩ ⟨ ⟩ 2 ru, ru E3 ut + 2 ru, rv E3 utvt + rv, rv E3 vt dτ = a ∫ b √ 2 2 g11ut + 2g12utvt + g22vt dt (1.46) a An external observer will calculate the length of the curve using (1.45). An ant living on the surface calculate length of the curve using (1.46) using Riemannian metric on the surface:

2 i k 2 2 ds = gikdu du = g11du + 2g12dudv + g22dv (1.47)

They will come to the same answer.

1.4.3 Induced Riemannian metrics. Examples. We consider here examples of calculating induced Riemanian metric on some quadratic surfaces in E3. using calculations for tangent vectors (see (1.38)) or explicitly in terms of diﬀerentials (see (1.41) and (1.42)). First of all consider the general case when a surface M is deﬁned by the equation z − F (x, y) = 0. One can consider the following parameterisation of this surface:  x = u r(u, v): y = v (1.48)  z = F (u, v) Then

20     1 0     ru = 0 rv = 1 (1.49) Fu Fv , 2 2 (ru, ru) = 1 + Fu , (ru, rv) = FuFv, (rv, rv) = 1 + Fv and induced Riemannina metric (ﬁrst quadratic form) (1.38) is equal to ( ) ( ) ( ) 2 || || g11 g12 (ru, ru)(ru, rv) 1 + Fu FuFv gαβ = = = 2 (1.50) g12 g22 (ru, rv)(rv, rv) FuFv 1 + Fv

2 2 2 2 2 GM = ds = (1 + Fu )du + 2FuFvdudv + (1 + Fv )dv (1.51) and the length of the curve r(t) = r(u(t), v(t)) on C (a ≤ t ≤ b) can be calculated by the formula: ∫ ∫ b √ 2 2 2 2 L = (1 + Fu )ut + 2FuFvutvt + (1 + Fv) vt dt a One can calculate (1.51) explicitly using (1.41): ( ) 2 2 2 2 2 2 GM = dx + dy + dz x=u,y=v,z=F (u,v) = (du) + (dv) + (Fudu + Fvdv) =

2 2 2 2 = (1 + Fu )du + 2FuFvdudv + (1 + Fv )dv . (1.52)

Cylinder Cylinder is given by the equation x2 + y2 = a2. One can consider the following parameterisation of this surface:  x = a cos φ r(h, φ): y = a sin φ (1.53)  z = h

2 2 2 We have Gcylinder = (dx + dy + dz ) x=a cos φ,y=a sin φ,z=h =

= (−a sin φdφ)2 + (a cos φdφ)2 + dh2 = a2dφ2 + dh2 (1.54)

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

21     0 −a sin φ     rh = 0 rφ = a cos φ (1.55) 1 0 , 2 (rh, rh) = 1, (rh, rφ) = 0, (rφ, rφ) = a and ( ) ( ) || || (ru, ru)(ru, rv) 1 0 gαβ = = 2 , (ru, rv)(rv, rv) 0 a 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: ∫ b √ 2 2 L = ht + a φtdt (1.57) a

Cone Cone is given by the equation x2 + y2 − k2z2 = 0. One can consider the following parameterisation of this surface:  x = kh cos φ r(h, φ): y = kh sin φ (1.58)  z = h

Calculate induced Riemannian metric: We have ( ) 2 2 2 Gconus = dx + dy + dz x=kh cos φ,y=kh sin φ,z=h =

(k cos φdh − kh sin φdφ)2 + (k sin φdh + kh cos φdφ)2 + dh2 ( ) 1 + k2 0 G = k2h2dφ2 + (1 + k2)dh2, ||g || = (1.59) conus αβ 0 k2h2 The length of the curve r(t) = r(h(t), φ(t)) on the cone (a ≤ t ≤ b) can be calculated by the formula: ∫ b √ 2 2 2 2 2 L = (1 + k )ht + k h φt dt (1.60) a

22 Sphere Sphere is given by the equation x2 + y2 + z2 = a2. Consider the following (standard ) parameterisation of this surface:  x = a sin θ cos φ r(θ, φ): y = a sin θ sin φ (1.61)  z = a cos θ

Calculate induced Riemannian metric (ﬁrst quadratic form) ( ) 2 2 2 2 GS = dx + dy + dz x=a sin θ cos φ,y=a sin θ sin φ,z=a cos θ =

(a cos θ cos φdθ−a sin θ sin φdφ)2+(a cos θ sin φdθ+a sin θ cos φdφ)2+(−a sin θdθ)2 = a2 cos2 θdθ2 + a2 sin2 θdφ2 + a2 sin2 θdθ2 = ( ) a2 0 , = a2dθ2 + a2 sin2 θdφ2 , ||g || = (1.62) αβ 0 a2 sin2 θ One comes to the same answer calculating scalar product of tangent vectors:     a cos θ cos φ −a sin θ sin φ     rθ = a cos θ sin φ rφ = a sin θ cos φ (1.63) −a sin θ 0 , 2 2 2 (rθ, rθ) = a , (rh, rφ) = 0, (rφ, rφ) = a sin θ and ( ) (r , r )(r , r ) ||g|| = u u u v = (ru, rv)(rv, rv) ( ) 2 a 0 2 2 2 2 2 2 ,G 2 = ds = a dθ + a sin θdφ 0 a2 sin2 θ S 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: ∫ b √ 2 2 · 2 L = a θt + sin θ φt dt (1.64) a We considered all quadratic surfaces except paraboloid z = x2 − y2. It can be rewritten as z = xy () and is call sometimes ”saddle””

23 Saddle (paraboloid) Saddle is given by the equation z − xy = 0. Why paraboloid? Exercise Show that the equation of saddle can be rewritten as z = x2−y2 (This surface is a ruled surface containing lines...) Consider the following (standard ) parameterisation of this surface:  x = u r(u, v): y = v (1.65)  z = uv

Calculate induced metric: ( ) 2 2 2 Gsaddle = dx + dy + dz x=u cos φ,y=v sin φ,z=uv =

du2 + dv2 + (udv + vdu)2 = 2 2 2 2 Gsaddle = (1 + v )du + 2uvdudv + (1 + u )dv . (1.66) The length of the curve r(t) = r(u(t), v(t)) on the sphere of the radius a (a ≤ t ≤ b) can be calculated by the formula: ∫ b √ 2 2 2 2 L = a (1 + v )ut + 2uvutvt + (1 + u )vt dt (1.67) a

One-sheeted and two-sheeted hyperboloids. 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 z > 0 and z < 0 L Consider these cases separately. 1) One-sheeted hyperboloid: x2 + y2 − z2 = a2. It is ruled surface. It contains two family of lines Exercise† Find the lines on two-sheeted hyperboloid

24 One-sheeted hyperboloid is given by the equation x2 + y2 − z2 = a2. it is convenient to choose parameterisation:  x = a cosh θ cos φ r(θ, φ): y = a cosh θ sin φ (1.68)  z = a sinh θ

x2 + y2 − z2 = a2 cosh2 θ − a2 sinh2 θ = a2 Compare the calculations with calculations for sphere! We changed functions cos, sin on cosh, sinh Induced Riemannian metric (ﬁrst quadratic form).

( ) 2 2 2 GHyperbolI = dx + dy + dz x=a cosh θ cos φ,y=a cosh θ sin φ,z=a sinh θ =

(a sinh θ cos φdθ−a cosh θ sin φdφ)2+(a sinh θ sin φdθ+a cosh θ cos φdφ)2+(a cosh θdθ)2 = a2 sinh2 θdθ2 + a2 cosh2 θdφ2 + a2 cosh2 θdθ2 = ( ) 1 + 2 sinh2 θ 0 , = a2(1+2 sinh2 θ)dθ2+a2 cosh2 θdφ2 , ||g || = αβ 0 cosh2 θ (1.69) For two-sheeted hyperboloid calculatiosn will be very similar. 1) Two-sheeted hyperboloid: z2 − x2 − y2 = a2. It is not ruled surface! it is convenient to choose parameterisation:  x = a sinh θ cos φ r(θ, φ): y = a sinh θ sin φ (1.70)  z = a cosh θ

z2 − x2 − y2 = a2 coshθ −a2 sinh2 θ = a2 Compare the calculations with calculations for sphere! We changed functions cos, sin on cosh, sinh Induced Riemannian metric (ﬁrst quadratic form).

( ) 2 2 2 GHyperbolI = dx + dy + dz x=a sinh θ cos φ,y=a sinh θ sin φ,z=a cosh θ =

25 (a cosh θ cos φdθ−a sinh θ sin φdφ)2+(a cosh θ sin φdθ+a sinh θ cos φdφ)2+(a sinh θdθ)2 = a2 cosh2 θdθ2 + a2 sinh2 θdφ2 + a2 sinh2 θdθ2 = ( ) 1 + 2 sinh2 θ 0 , = a2(1+2 sinh2 θ)dθ2+a2 sinh2 θdφ2 , ||g || = αβ 0 sinh2 θ (1.71)

We calculated examples of induced Riemannian structure embedded in Euclidean space almost for all quadratic surfaces. Quadratic surface is a surface deﬁned by the equation

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

One can see that any quadratic surface by aﬃne transformation can be trans- formed to one 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

• two-sheeted hyperboloid

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

26 1.4.4 ∗Induced metric on two-sheeted hyperboloid embedded in pseudo-Euclidean space. Consider two-sheeted hyperboloid embedded in pseudo-Euclidean space with pseudoscalar product defined by bilinear form 1 1 2 2 3 3 ⟨X, Y⟩pseudo = X Y + X Y − X Y (1.72) The ”pseudoscalar” product is bilinear, symmetric. It is defined by non-degenerate matrix. But it is not positive-definite: The ”pseudo-length” of vectors X = (a cos φ, a sin φ, ±a) equals to zero:

X = (a cos φ, a sin φ, ±a) ⇒ ⟨X, X⟩pseudo = 0, (1.73) This is not scalar product. The pseudo-Riemannian metric is: 2 2 2 Gpseudo = dx + dy − dz (1.74) it turns out that the following remarkable fact occurs: Proposition The pseudo-Riemannian metric (1.74) in the ambient 3-dimensional pseudo-Euclidean space induces Riemannian metric on two-sheeted hyperboloid x2 + y2 − z2 = 1. Show it. repeat the calculations above for two-sheeted hyperboloid changing in the ambient space Riemannian metric G = dx2 + dy2 + dz2 on pseudo-Riemannian dx2 + dy2 − dz2: Using (1.70) and (1.74) we come now to ( ) 2 2 − 2 G = dx + dy dz x=a sinh θ cos φ,y=a sinh θ sin φ,z=a cosh θ = (a cosh θ cos φdθ−a sinh θ sin φdφ)2+(a cosh θ sin φdθ+a sinh θ cos φdφ)2−(a sinh θdθ)2 = a2 cosh2 θdθ2 + a2 sinh2 θdφ2 − a2 sinh2 θdθ2 ( ) 1 0 ,G = a2dθ2 + a2 sinh2 θdφ2 , ||g || = (1.75) L αβ 0 sinh2 θ The two-sheeted hyperboloid equipped with this metric is called hyperbolic or Lobachevsky plane. Now express Riemannian metric in stereographic coordinates. 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): For u y 1 stereographic coordinates u, v we have x = v = 1+z . We come to  {  2u x = 2 2 u = x 1−u −v 1+z , 2v (4) y y = 1−u2−v2 v = 1+z  u2+v2+1 z = 1−u2−v2

27 2 2 2 2 x2+y2 The image of upper-sheet is an open disc u + v = 1 since u + v = (1+z)2 = z2−1 z−1 ≤ z−1 (1+z)2 = z+1 . Since for upper sheet z > 1 then 0 z+1 < 1.

( ( )) 2u 2 G = (dx2 + dy2 − dz2) = d + x=x(u,v),y=y(u,v),z=z(u,v) 1 − u2 − v2 ( ( )) ( ( )) 2v 2 u2 + v2 + 1 2 d − d = 1 − u2 − v2 1 − u2 − v2 (Compare with calculations for sphere x2 + y2 + z2 = 1). We have G = ( ) ( ) 2du 2u(2udu + 2vdv) 2 2dv 2v(2udu + 2vdv) 2 + + + 1 − u2 − v2 (1 − u2 − v2)2 1 − u2 − v2 (1 − u2 − v2)2 ( ) 2udu + 2vdv (u2 + v2 + 1)(2udu + 2vdv) 2 4(du)2 + 4(dv)2 − + = 1 − u2 − v2 (1 − u2 − v2)2 (1 + u2 + v2)2 (To perform these calculations it is convenient to denote by s = 1 − u2 − v2).

1.5 Isometries of Riemanian manifolds.

Let (M1,G(1)), (M2,G(2)) be two Riemannian manifolds—manifolds equipped with Riemannian metric G1 and G2 respectively. Definition We say that these Riemannian manifolds are isometric if there exists a diffeomorphism F (one-one smooth map) which preserves the distances. This means the following: We say that these Riemannian manifolds are isometric if there exists a diffeomorphism F (one-one smooth map) such that ∗ F G(2) = G(1) , which means the following: Let p1 be an arbitrary point on manifold M1 and p2 ∈ M2 be its image:F (p1) = i a p2. Let {x } be coordinates in a vicinity of a point p1 ∈ M1 and {y } be coordinates in a vicinity of a point p2 ∈ M2. Let Riemannian metrics G1 on i k i M1 has local expression G(1) = g(1)ik(x)dx dx in coordinates {x } and re- a b spectively Riemannian metrics G(2) has local expression G2 = g(2)ab(y)dy dy i in coordinates {y } on M2. Then ∂ya(x) ∂yb(x) g (x)dxidxk = g (y)dyadyb = g (y(x)) dxi dxk (1.76) (1)ik (2)ab (2)ab ∂xi ∂xk

28 i.e. ∂ya(x) ∂yb(x) g (x) = g (y(x)) . (1.77) (1)ik ∂xi (2)ab ∂xk where yi = yi(x) is local expression for diﬀeomorphism F .

Definition We say that two Riemannian manifolds (M1,G(1)), (M2,G(2)) are locally isometric if the following conditions hold: For arbitrary point p1 on the manifold M1 there exists a point p2 on the i manifold M2 such that there exist coordinates {x } in a vicinity of a point a p1 ∈ M1 and coordinates {y } in a vicinity of a point p2 ∈ M2 such that local i expression for metric G(1) on M1 in coordinates {x } and local expression for a metric G(2) on M2 in coordinates {y } are related via formuale (1.76), (1.29). Locally isometric Riemannian manifolds have not to be diffeomorphic. E.g. Euclidean plane is locally isometric to cylinder, but they are not diffeomorphic.

1.5.1 Examples of local isometries Consider examples. Example 1 Cylinder-Cone—Plane, 2 2 2 Riemannian metric on cylinder is Gcylinder = a dφ + dh and on the cone 2 2 2 2 2 Gconus = k h dφ + (1 + k )dh (see formulae (1.54) and (1.59)). To show that cylinder is isometric to Euclidean plan we have to ﬁnd new local coordinates u, v on the cylinder such that in these coordinates the metric on the cylinder equals to du2 + dv2. If we put u = aφ, v = h then

2 2 2 2 2 2 2 du + dv = d(aφ) + dh = a dφ + dh = Gcylinder . (1.78)

Thus we prove the local isometry. Of course the coordinate u = φ is not global coordinate on the surface of cylinder. It is evident that cylinder and plane are not globally isometric since there are no diffeomorphism of cylinder on the plane. (They are different non-homeomorphic topological spaces.) Now show that cone is locally isometric to the plane. This means that we have to find local coordinates u, v on the cone such that in these coordinates induced metric G|c on cone would have the appear- 2 2 ance G|c = du + dv .

29 First of all calculate the metric on cone in natural coordinates h, φ where  x = kh cos φ r(h, φ): y = kh sin φ .  z = h

(x2 + y2 − k2z2 = k2h2 cos2 φ + k2h2 sin2 φ − k2h2 = k2h2 − k2h2 = 0. Calculate metric Gc on the cone in coordinates h, φ induced with the Euclidean metric G = dx2 + dy2 + dz2: ( ) 2 2 2 − 2 Gc = dx + dy + dz x=kh cos φ,y=kh sin φ,z=h = (k cos φdh kh sin φdφ) +

(k sin φdh + kh cos φdφ)2 + dh2 = (k2 + 1)dh2 + k2h2dφ2 . In analogy{ with polar coordinates try to ﬁnd new local coordinates u, v such u = αh cos βφ that , where α, β are parameters. We come to du2 + dv2 = v = αh sin βφ

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

2 2 2 2 2 Comparing with the metric on the cone Gc = (1 + k )dh + k h dφ we see that if we put α = k and β = √ k then du2 + dv2 = α2dh2 + α2β2h2dφ2 = 1+k2 (1 + k2)dh2 + k2h2dφ2. Thus in new local coordinates { √ u = k2 + 1h cos √ k φ √ k2+1 v = k2 + 1h sin √ k φ k2+1

2 2 induced metric on the cone becomes G|c = du + dv , i.e. cone locally is isometric to the Euclidean plane Of course these coordinates are local.— Cone and plane are not homeomorphic, thus they are not globally isometric. One can show also that 4R2(dx2+dy2) 2) Plane with metric (1+x2+y2)2 is isometric to the sphere with radius R(a) = ... du2+dv2 3) Disc with metric (1−u2−v2)2 is isometric to half plane with metric dx2+dy2 4y2 . (see exercises in Homeworks and Coursework.)

30 1.6 Volume element in Riemannian manifold The volume element in n-dimensional Riemannian manifold with metric G = i k gikdx dx is deﬁned by the formula √ 1 2 n det gik dx dx . . . dx (1.79) If D is a domain in the n-dimensional Riemannian manifold with metric i G = gikdx then its volume is equal to to the integral of volume element over this domain. ∫ √ 1 2 n V (D) = det gik dx dx . . . dx (1.80) D Remark Students who know the concept of exterior forms can read the volume element as √ 1 2 n det gik dx ∧ dx ∧ · · · ∧ dx (1.81)

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

1.6.1 Volume of parallelepiped Note that the formula (1.79) gives the volume of n-dimensional parallelepiped. n Show this. Let E be Euclidean vector space with orthonormal basis {ei}. Let vi be an arbitrary basis in this vector space (vectors vi in general have not unit length and are not orthogonal to each other). Consider n-parallelepiped spanned by vectors {vi}: i ≤ i ≤ Πvi : r = t vi, 0 t 1. The volume of this parallelepiped equals to || m|| V ol(Πvi ) = det ai , (1.82) || m|| m where A = ai is transition matrix, vi = emai . On the other hand

i m i i m i r = x ei = t vm, hence x = amt , where vm = eiam.

1 2 n 2 i k Let G = (dx ) + ··· + (dx ) = gikdt dt be usual Euclidean metric in new coordinates ti. The

i′ k′ 1 2 n 2 i k i ∂x ∂x k G = (dx ) + ··· + (dx ) = dx δ dx = dt δ ′ ′ dt . ik ∂ti i k ∂tk

31 ′ ∂xi i′ Since ∂ti = ai then ∑ √ i′ i′ ⇒ 2 gik = ai ak det g = (det A) , det g = det A. i′ and according to the formula (1.80) ∫ √ 1 2 n V ol(Πvi ) = det gdt dt . . . dt = det A. 0≤ti≤1 We come to (1.82). Perform these calculations in detail for 3-dimensional case. E.g. if Euclidean space is 3-dimensional then the parallelepiped spanned by basis vectors {a, b, c}

1 2 3 1 2 3 Πa,b,c = t a + t b + t c , 0 ≤ t , t , t ≤ 1. Volume of parallelepiped equals to   ax bx cx   V olΠa,b,c = det ay by cy , (1.83) az bz cz and       1 x ax bx cx t      2 y = ay by cy t . 3 z az bz cz t The Riemannian metric in new coordinates (t1, t2, t3) equals to       1 ( ) ax ay az ax bx cx dt 2 2 2 1 2 3      2 dx + dy + dz = dt dt dt bx by bz ay by cy dt , 3 cx cy cz az bz cz dt i.e. in coordinates ti Riemannian metric G = g dtidtk where    ik  g11 g12 g13 ax ay az ax bx cx       g21 g22 g23 = bx by bz ay by cy g31 g32 g33 cx cy cz az bz cz i.e.   √ ax bx cx   det gik = det ay by cy = V olΠa,b,c . (1.84) az bz cz

32 1.6.2 Invariance of volume element under changing of coordinates

Prove that volume element is invariant under coordinate transformations, 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 andg ˜pq(y) matrix of the metric in new coordinates: ∂xi ∂xk g˜ (y) = g (x(y)) . (1.85) pq ∂yp ik ∂yq Then √ √ 1 2 n 1 2 n det gik(x) dx dx . . . dx = detg ˜pq(y) dy dy . . . dy (1.86) This follows from (1.85). Namely √ ( ) √ ∂xi ∂xk det g (y) dy1dy2 . . . dyn = det g (x(y)) dy1dy2 . . . dyn ik ∂yp ik ∂yq ( ) ( ) · · ∂xi ∂xk 1 Using the fact that det(ABC) = det A det B det C and det ∂yp = det ∂yq we see that from the formula above follows: √ ( ) √ ∂xi ∂xk det g (y) dy1dy2 . . . dyn = det g (x(y)) dy1dy2 . . . dyn = ik ∂yp ik ∂yq √( ( )) ∂xi 2√ det det g (x(y))dy1dy2 . . . dyn = ∂yp ik ( ) √ ∂xi det g (x(y)) det dy1dy2 . . . dyn = (1.87) ik ∂yp Now note that ( ) ∂xi det dy1dy2 . . . dyn = dx1 . . . dxn ∂yp according to the formula for changing coordinates in n-dimensional integral 2. Hence ( ) √ ∂xi √ det g (x(y)) det dy1dy2 . . . dyn = det g (x(y))dx1dx2 . . . dxn (1.88) ik ∂yp ik

1determinant of matrix does not change if we change the matrix on the adjoint, i.e. change columns on rows. ( ) 2 ∂xi Determinant of the matrix ∂yp of changing of coordinates is called sometimes Ja- cobian. Here we consider the case if Jacobian is positive. If Jacobian is negative then formulae above remain valid just the symbol of modulus appears.

33 Thus we come to (1.86).

1.6.3 Examples of calculating volume element Consider ﬁrst very simple example: Volume element of plane in cartesian coordinates, metric g = dx2 + dy2. Volume element is equal to √ ( ) √ 1 0 det gdxdy = det dxdy = dxdy 0 1

Volume of the domain D is equal to ∫ √ ∫ V (D) = det gdxdy = dxdy D D If we go to polar coordinates:

x = r cos φ, y = r sin φ (1.89)

Then we have for metric: G = dr2 + r2dφ2 because

dx2 + dy2 = (dr cos φ − r sin φdφ)2 + (dr sin φ + r cos φdφ)2 = dr2 + r2dφ2 (1.90) Volume element in polar coordinates is equal to √ ( ) √ 1 0 det gdrdφ = det drdφ = drdφ . 0 r2

Lobachesvky plane. 2 2 In coordinates x, y (y > 0) metric G = dx +dy , the corresponding matrix ( ) y2 2 √ 1/y 0 dxdy G = 2 . Volume element is equal to det gdxdy = y2 . 0 1/y Sphere in stereographic coordinates Consider the two dimensional plane with Riemannian metrics 4R2(du2 + dv2) G = (1.91) (1 + u2 + v2)2

34 (It is isometric to the sphere of the radius R without North pole in stereographic coordinates (see the Homeworks.)) Calculate its volume element and volume. It is easy to see that: ( ) 4R2 4 (1+u2+v2)2 0 16R G = 4R2 det g = 2 2 4 (1.92) 0 (1+u2+v2)2 (1 + u + v ) √ 4R2dudv and volume element is equal to det gdudv = (1+u2+v2)2 One can calculate volume in coordinates u, v but it is better to consider volume form in polar coordinates u = r cos φ, v = r sin φ. Then it is easy 2 2 2 to see that according to (1.90) we have for the metric G = R (du +dv ) = √ (1+u2+v2)2 R2(dr2+r2dφ2) 4R2rdrdφ (1+r2)2 and volume form is equal to det gdrdφ = (1+r2)2 Now calculation of integral becomes easy: ∫ ∫ ∫ 2 ∞ ∞ 4R rdrdφ 2 rdr 2 du 2 V = 2 2 = 8πR 2 2 = 4πR 2 = 4πR . (1 + r ) 0 (1 + r ) 0 (1 + u)

Segment of the sphere. Consider sphere of the radius a in Euclidean space with standard Riema- nian metric a2dθ2 + a2 sin2 θdφ2 This metric is nothing but ﬁrst quadratic form on the sphere (see (1.4.3)). The volume element is √ ( ) √ a2 0 det gdθdφ = det dθdφ = a2 sin θdθdφ 0 a2 sin θ

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

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

There is remarkable formula which express the area of segment via the height h: ∫ ∫ (∫ ) ( ) θ1 2π ( ) V = a2 sin θ dθdφ = a2 sin θ dφ dθ = 0 θ1≤θ≤θ0 θ 0

35 ∫ θ0 2 2 2πa sin θdθ = 2πa (cos θ0 − cos θ1) = 2πa(a cos θ0 − acosθ1) = 2πah θ1 (1.93) E.g. for all the sphere h = 2a. We come to S = 4πa2. It is remarkable formula: area of the segment is a polynomial function of radius of the sphere and height (Compare with formula for length of the arc of the circle)

2 Covariant diﬀerentiaion. Connection. Levi Civita Connection on Riemannian manifold

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. i ∂ Let X = X (x)ei(x) = ∂xi be a vector field on M. Recall that vector 3 i i field X = X ei defines at the every point x0 an infinitesimal curve: x (t) = i i i i x0 + tX (More exactly the equivalence class [γ(t)]X of curves x (t) = x0 + tXi + ... ). i ∂ Let f be an arbitrary (smooth) function on M and X = X ∂xi . Then i ∂ derivative of function f along vector field X = X ∂xi is equal to ∂f ∂ f = ∇ f = Xi X X ∂xi The geometrical meaning of this definition is following: If X is a velocity i i i vector of the curve x (t) at the point x0 = x (t) at the ”time” t = 0 then ∇ i i the value of the derivative Xf at the point x0 = x (0) is equal just to the derivative by t of the function f(xi(t)) at the ”time” t = 0: dxi(t) d ( ) i ∇ i if X (x) = , then Xf i i = f x (t) x0=x(0) dt t=0 x =x (0) dt 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 denoted

3 here like always we suppose∑ by default the summation over repeated indices. E.g.X = i n i X ei is nothing but X = i=1 X ei

36 by ”∂Xf”. In this course we prefer to denote it by ”∇Xf” to have the uniform notation for both operators of taking derivation of functions and vector fields along the vector field. ∞ One can see that the operation ∇X on the space C (M) (space of smooth functions on the manifold) satisfies the following conditions:

• ∇X (λf + µg) = a∇Xf +b∇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:operator on smooth functions obeying the conditions above is a vector field. (You will have 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 vectors 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 from the another vector space, because apriori we cannot compare vectors from different vector space. One have to define an operation of transport of vectors from the space Tx0 M to the point Tx0+tX M defining the transport from the point Tx0 M to the point Tx0+tX M). Try to define the operation ∇ on vector fields such that conditions (2.2) above be satisfied.

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

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

37 (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 vector ﬁeld: ∇Xf = ∂Xf. (Compare with Leibnitz rule in (2.2)).

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

Write down explicit formulae in a given local coordinates {xi} (i = 1, 2, . . . , n) on manifold M. Let ∂ ∂ X = Xie = Xi Y = Y ie = Y i i ∂xi i ∂xi ∂ The basis vector ﬁelds xi we denote sometimes by ∂i sometimes by ei Using properties above one can see that ( ( )) k i k ∇ ∇ i ∇ ∇ ∇ XY = X ∂i Y ∂k = X i Y ∂k , where i = ∂i (2.6) Then according to (2.4) ( ) ( ) k k k ∇i Y ∂k = ∇i Y ∂k + Y ∇i∂k

Decompose the vector ﬁeld ∇i∂k over the basis ∂i: ∇ m i∂k = Γik∂m (2.7) and ( ) ∂Y k(x) ∇ Y k∂ = ∂ + Y kΓm∂ , (2.8) i k ∂xi k ik m

38 ∂Y m(x) ∇ Y = Xi ∂ + XiY kΓm∂ , (2.9) X ∂xi m ik m In components ( ) ∂Y m(x) (∇ Y)m = Xi + Y kΓm (2.10) X ∂xi ik

{ m} { i} Coefficients Γik are called Christoffel symbols in coordinates x . These coefficients define covariant derivative—connection. If operation of taking covariant derivative is given we say that the connection is given on the manifold. Later it will be explained why we us the word ”connection” We see from the formula above that to define covariant derivative of vector fields, connection, we have to define Christoffel symbols in local coordinates.

2.1.2 Transformation of Christoffel symbols for an arbitrary connection ∇ { i } Let be a connection on manifold M. Let Γkm be Christoffel symbols of this connection in given local coordinates {xi}. Then according (2.7) and (2.8) we have ∂Y i ∂ ∂ ∇ Y = Xm + XmΓi Y k , X ∂xm ∂xi mk ∂xi and in particularly i ∇ Γmk∂i = ∂m ∂k ′ Use this relation to calculate Christoffel symbols in new coordinates xi

i′ ′ ′ ′ ∇ ′ ′ Γm k ∂i = ∂m ∂k

∂ ∂xm ∂ ∂xm We have that ∂ ′ = ′ = ′ = ′ ∂ . Hence due to properties m ∂xm ∂xm ∂xm ∂xm m (2.4), (2.5) we have ( ) ( ) ( ) k k k i′ ∂x ∂x ∂ ∂x Γ ′ ′ ∂ ′ = ∇ ∂ ′ = ∇ ′ ∂ = ∇ ′ ∂ + ∂ = m k i ∂m′ k ∂m ∂xk′ k ∂xk′ ∂m k ∂xm′ ∂xk′ k ( ) ∂xk ∂2xk ∂xk ∂xm ∂2xk ′ ∇ ∂xm ∂ + ′ ′ ∂ = ′ ′ ∇ ∂ + ′ ′ ∂ k ′ ∂m k m k k k m ∂m k m k k ∂x ∂xm ∂x ∂x ∂x ∂x ∂x ∂x k m 2 k k m i′ 2 k i′ ∂x ∂x i ∂ x ∂x ∂x i ∂x ∂ x ∂x Γ ∂ + ∂ = Γ ∂ ′ + ∂ ′ ∂xk′ ∂xm′ mk i ∂xm′ ∂xk′ k ∂xk′ ∂xm′ mk ∂xi i ∂xm′ ∂xk′ ∂xk i

39 Comparing the first and the last term in this formula we come to the transformation law: { i } ∇ If Γkm are Christoffel symbols of the connection in local coordinates i i′ {x } and {Γ ′ ′ } are Christoffel symbols of this connection in new local k m′ coordinates {xi } then

k m i′ 2 r i′ i′ ∂x ∂x ∂x i ∂ x ∂x Γ ′ ′ = Γ + (2.11) k m ∂xk′ ∂xm′ ∂xi km ∂xk′ ∂xm′ ∂xr Remark Christoffel symbols do not transform as tensor. If the second term is equal to zero, i.e. transformation of coordinates are linear (see the Proposition on flat connections) then the transformation( rule) above is the the 1 same as a transformation rule for tensors of the type (see the formula 2 (1.6)). In general case this is not true. Christoffel symbols does not transform as tensor under arbitrary non-linear coordinate transformation: see the second term in the formula above.

2.1.3 Canonical flat affine connection It follows from the properties of connection that it is suffice to define connection 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 cartesian coordinates {x1, . . . , xn}. Define connection such that all Christoffel symbols are equal to zero in these cartesian coordinates {xi}. ∇ m m ei ek = Γikem = 0, Γik = 0 (2.12) Does this mean that Christoffel symbols are equal to zero in an arbitrary cartesian coordinates if they equal to zero in given cartesian coordinates? Does this mean that Christoffel symbols of this connection equal to zero in arbitrary coordinates system? To answer these questions note that the relations (2.12) mean that ∂Y i ∂ ∇ Y = Xm (2.13) X ∂xm ∂xi in coordinates {xi}

40 ′ ′ Consider an arbitrary new coordinates xi = xi (x1, . . . , xn). Recall the transformation rule for an arbitrary vector ﬁeld (see subsection 1.1)

m′ m′ m ∂ ∂x ∂ ′ ∂x ∂x ′ R = Rm = Rm , i.e.Rm = Rm , and ,Rm = Rm . ∂xm ∂xm ∂xm′ ∂xm ∂xm′ Hence we have from (2.13) that ( ) i m′ i ∂Y ∂ ∂ ( ) ∂ ∂x ∂ ∂x ′ ∂ ∇ Y = Xm = Xm Y i = Xm Y i = X ∂xm ∂xi ∂xm ∂xi ∂xm ∂xm′ ∂xi′ ∂xi ( ) i ( ) i 2 i ( ) ′ ∂ ∂x ′ ∂ ′ ∂ ′ ∂x ∂ ′ ∂ x ′ ∂ Xm Y i = Xm Y i +Xm Y i = ∂xm′ ∂xi′ ∂xi ∂xm′ ∂xi′ ∂xi ∂xm′ ∂xi′ ∂xi i′ 2 i m′ ∂Y ∂ m′ ∂ x i′ ∂ X ′ ′ + X ′ ′ Y = (2.14) ∂xm ∂xi | ∂xm {z∂xi ∂x}i an additional term We see that an additional term equals to zero for arbitrary vector fields X, Y if and only if the relations between new and old coordinates are linear: ∂2xi = 0, i.e. xi = bi + ai xk (2.15) ∂xm′ ∂xi′ k Comparing formulae (2.15) and (2.13) we come to simple but very important Proposition Let all Christoffel symbols of a given connection be equal to zero in a given coordinate system {xi}. Then all Christoffel symbols of this ′ connection are equal to zero in an arbitrary coordinate system {xi } such that the relations between new and old coordinates are linear:

i′ i i k x = b + akx (2.16)

∂2xi If transformation to new coordinate system is not linear, i.e. ′ ′ ≠ 0 ∂xm ∂xi then Christoffel symbols of this connection in general are not equal to zero in ′ new coordinate system {xi }. Definition We call connection ∇ flat if there exists coordinate system such that all Christoffel symbols of this connection are equal to zero in a given coordinate system. In particular connection (2.12) has zero Christoffel symbols in arbitrary cartesian coordinates.

41 Corollary Connection has zero Christoffel symbols in arbitrary Cartesian coordinates if it has zero Christoffel symbols in a given Cartesian coordinates. Hence the following definition is correct: Definition A connection on En which Christoffel symbols vanish in cartesian coordinates is called canonical flat connection. Remark Canonical flat connection in Euclidean space is uniquely defined, sincce 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 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 cos φ r = x2 + y2 y (2.17) y = y sin φ φ = arctan x Write down Jacobians of transformations—matrices of partial derivatives:   ( ) ( ) ( ) √ x − y 2 2 xr yr cos φ sin φ rx φx  x2+y2 x +y  = − , = √ y x xφ yφ r sin φ r cos φ ry φy 2 2 x2+y2 x +y (2.18) According (2.11) and since Chrsitoffel symbols are equal to zero in cartesian coordinates (x, y) we have

2 r i′ i′ ∂ x ∂x Γ ′ ′ = , (2.19) k m ∂xk′ ∂xm′ ∂xr

′ ′ where (x1, x2) = (x, y) and (x1 , x2 ) = (r, φ). Now using (2.18) we have

∂2x ∂r ∂2y ∂r Γr = + = 0 rr ∂r∂r ∂x ∂r∂r ∂y

∂2x ∂r ∂2y ∂r Γr = Γr = + = − sin φ cos φ + sin φ cos φ = 0 . rφ φr ∂r∂φ ∂x ∂r∂φ ∂y

42 ∂2x ∂r ∂2y ∂r x y Γr = + = −x − y = −r . φφ ∂φ∂φ ∂x ∂r∂φ ∂y r r ∂2x ∂φ ∂2y ∂φ Γφ = + = 0 . rr ∂r∂r ∂x ∂r∂r ∂y ∂2x ∂φ ∂2y ∂φ −y x 1 Γφ = Γφ = + = − sin φ + cos φ = φr rφ ∂r∂φ ∂x ∂r∂φ ∂y r2 r2 r ∂2x ∂φ ∂2y ∂φ −x y Γφ = + = −x − y = 0 . (2.20) φφ ∂φ∂φ ∂x ∂φ∂φ ∂y r2 r2 Hence we have that the covariant derivative (2.13) in polar coordinates has the following appearance ∂ ∇ ∂ = Γr ∂ + Γφ ∂ = 0 ,, ∇ ∂ = Γr ∂ + Γφ ∂ = φ r r rr r rr φ r φ rφ r rφ φ r ∂ ∇ ∂ = Γr ∂ + Γφ ∂ = φ , ∇ ∂ = Γr ∂ + Γφ ∂ = −r∂ (2.21) φ r φr r φr φ r φ φ φφ r φφ φ r Remark Later when we study geodesics we will learn a very quick method to 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 connection is defined by Christoffel symbols. On the other hand we know that in general coordinates on manifold are not defined globally. (We had not this trouble in 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 questions is ”Yes”. It sounds bizzare but answer on the first question is not ”Yes” 4 Global definition of connection 4Topology 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.

43 The formula (2.11) defines the transformation for Christoffer symbols if we go from one coordinates to another. { i } Let (xα),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 i i i i′ coordinates) (xα) Christoffer symbols which we denote by (α)Γkm. If (xα), (x(β)) are different local coordinates in a vicinity of a given point then according to (2.11)

′ ′ ∂xk ∂xm ∂xi ∂2xk ∂xi i′ (α) (α) (β) (α)i (α) (β) Γ ′ ′ = ′ ′ Γ + ′ ′ (2.22) (β) k m ∂xk ∂xm ∂xi mk ∂xm ∂xk ∂xk (β) (β) (α) (β) (β) (β) (α) { i } Definition Let (xα),Uα be an atlas of charts on the manifold M { (α)i} We say that the collection of Christoffel symbols Γkm defines globally a connection on the manifold M in this atlas if for every two local coordinates i i (x(α)), (x(β)) from this atlas the transformation rules (2.22) are obeyed. Using partition of unity one can prove the existence of global connection con- { i } structing it in explicit way. Let (xα),Uα (α = 1, 2,...,N) be a finite atlas on the manifold M and let {ρα} be a partition of unity adjusted to this atlas. Denote (α) i by Γkm local connection defined in domain Uα such that its components in these (α) i coordinates are equal to zero. Denote by (β)Γkm Christoffel symbols of this local i (α) i connection in coordinates (x(β))((β)Γkm = 0). Now one can define globally the connection by the formula:

′ ∑ ∑ ∂xi ∂2xi (x) i (α) i (β) (α) (β)Γkm(x) = ρα(x) (β)Γkm(x) = ρα(x) i′ k m . (2.23) α α ∂x(α) ∂x(β)∂x(β) This connection in general is not ﬂat connection5

2.2 Connection induced on the surfaces Let M be a manifold (surface) embedded in Euclidean space6. Canonical flat connection on EN induces the connection on surface in the following way. Let X, Y be tangent vector fields to the surface M and ∇can.flat a canonical flat connection in EN . In general ∇can.flat Z = X Y is not tangent to manifold M (2.24) 5See for detail the text: ”Global affine connection on manifold” ” in my homepage: ”www.maths.mancheser.ac.uk/khudian” in subdirectory Etudes/Geometry 6We know that every n-dimensional manifodl can be embedded in 2n + 1-dimensional Euclidean space

44 Consider its decomposition on two vector fields: ( ) ( ) can.flat can.flat can.flat ⊥ ∇ ∇ ∇ Z = Ztangent + Z , X , Y = X Y tangent + X Y ⊥ , (2.25) where Z⊥ is a component of vector which is orthogonal to the surface M and Z|| is a component which is tangent to the surface. Define an induced connection ∇M on the surface M by the following formula ( ) ∇M ∇M ∇can.flat : X Y : = X Y tangent (2.26) Remark One can imply this construction for an arbitrary connection in EN .

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 ∇M ∇can.flat ∇can.flat − ∇can.flat : X Y : = X Y || = X Y n( X Y, n), (2.27) where n is normal unit vector field to M. Consider a special example Example (Induced connection on sphere) Consider a sphere of the radius R in E3:  x = R sin θ cos φ r(θ, φ): y = R sin θ sin φ  z = R cos θ then       R cos θ cos φ −R sin θ sin φ sinθ cos φ       rθ = R cos θ sin φ , rφ = R sin θ cos φ , n = sinθ sin φ , −R sin θ 0 cos θ

∂r(θ,φ) ∂r(θ,φ) where rθ = ∂θ , rφ = ∂φ are basic tangent vectors and n is normal unit vector. Calculate an induced connection ∇ on the sphere. ∇ First calculate ∂θ ∂θ. ( ) ∇ ∂rθ ∂θ ∂θ = = (rθθ)tangent . ∂θ tangent

45   −Rsinθ cos φ   On the other hand one can see that rθθ = −Rsinθ sin φ = −Rn is −R cos θ proportional to normal vector, i.e. (rθθ)tangent = 0. We come to ∇ ⇒ θ φ ∂θ ∂θ = (rθθ)tangent = 0 Γθθ = Γθθ = 0 . (2.28) ∇ ∇ Now calculate ∂θ ∂φ and ∂φ ∂θ. ( ) ( ) ∇ ∂rφ ∇ ∂rθ ∂θ ∂φ = = (rθφ)tangent , ∂φ ∂θ = = (rφθ)tangent ∂θ tangent ∂φ tangent We have   −R cos θ sin φ ∇ ∇   ∂θ ∂φ = ∂φ ∂θ = (rφθ)tangent = R cos θ cos φ . 0 tangent

We see that the vector rφθ is orthogonal to n:

⟨rφθ, n⟩ = −R cos θ sin φ sin θ cos φ + R cos θ cos φ sin θ sin φ = 0.

Hence   −R cos θ sin φ ∇ ∇   ∂θ ∂φ = ∂φ ∂θ = (rφθ)tangent = rφθ = R cos θ cos φ = cotan θrφ . 0

We come to

∇ ∇ ⇒ θ θ φ φ ∂θ ∂φ = ∂φ ∂θ = cotan θ∂φ Γθφ = Γφθ = 0, Γθφ = Γφθ = cotan θ (2.29) ∇ Finally calculate ∂φ ∂φ   −R sin θ cos φ ∇ −  ∂φ ∂φ = (rφφ)tangent = R sin θ sin φ 0 tangent Projecting on the tangent vectors to the sphere (see (2.27)) we have ∇ − ⟨ ⟩ ∂φ ∂φ = (rφφ)tangent = rφφ n n, rφφ =

46     −R sin θ cos φ sin θ cos φ −R sin θ sin φ−sin θ sin φ (−R sin θ cos φ sin θ cos φ − R sin θ sin φ sin θ sin φ) = 0 cos θ   R cos θ cos φ   − sin θ cos θ R cos θ sin φ = − sin θ cos θrθ, −R sin θ i.e. ∇ − ⇒ θ − φ φ ∂φ ∂φ = sin θ cos θrθ Γφφ = sin θ cos θ, Γφφ = Γφφ = 0 . (2.30)

2.3 Levi-Civita connection 2.3.1 Symmetric connection Deﬁnition. We say that connection is symmetric if its Christoﬀel symbols i Γkm are symmetric with respect to lower indices

i i Γkm = Γmk (2.31) The canonical flat connection and induced connections considered above are symmetric connections. Invariant definition of symmetric connection A connection ∇ is symmetric if for an arbitrary vector fields X, Y

∇XY − ∇YX − [X, Y] = 0 (2.32)

If we apply this deﬁnition to basic ﬁelds ∂k, ∂m which commute: [∂k, ∂m] = 0 we come to the condition ∇ − ∇ i − i ∂k ∂m ∂m ∂k = Γmk∂i Γkm∂i = 0 and this is the condition (2.31).

2.3.2 Levi-Civita connection. Theorem and Explicit formulae Let (M,G) be a Riemannian manifold. Deﬁnition. Theorem A 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⟩ (2.33)

47 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: ( ) 1 ∂g ∂g ∂g Γi = gij jm + jk − mk . (2.34) mk 2 ∂xk ∂xm ∂xj

i k ik where G = gikdx dx is Riemannian metric in local coordinates and ||g || is the matrix inverse to the matrix ||gik||. Proof i Suppose that this connection exists and Γmk are its Christoﬀel symbols. Con- sider vector ﬁelds X = ∂m, Y = ∂i and Z = ∂k in (2.33). We have that ⟨ r ⟩ ⟨ r ⟩ r r ∂mgik = Γmi∂r, ∂k + ∂i, Γmk∂r = Γmigrk + girΓmk . (2.35) for arbitrary indices m, i, k. r Denote by Γmik = Γmigrk we come to

∂mgik = Γmik + Γmki, i.e.

k k Now using the symmetricity Γmik = Γimk since Γmi = Γim 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 1 1 Γ = (∂ g + ∂ g − ∂ g ) ⇒ Γk = gkr (∂ g + ∂ g − ∂ g ) (2.36) mik 2 m ik i mk k mi im 2 m ir i mr r mi We see that if this connection exists then it is given by the formula(2.34). On the other hand one can see that (2.34) obeys the condition (2.35). We prove the uniqueness and existence. ∇ m since ∂i ∂k = Γik∂m. Consider examples.

48 2.3.3 Levi-Civita connection on 2-dimensional Riemannian manifold with metric G = adu2 + bdv2. Example Consider 2-dimensional manifold with Riemannian metrics ( ) ( ) g g a(u, v) 0 G = a(u, v)du2 + b(u, v)dv2,G = 11 12 = g21 g22 0 b(u, v)

Calculate Christoﬀel symbols of Levi Civita connection. Using (2.36) we see that:

1 − 1 1 Γ111 = 2 (∂1g11 + ∂1g11 ∂1g11) = 2 ∂1g11 = 2 au

1 − 1 1 Γ211 = Γ121 = 2 (∂1g12 + ∂2g11 ∂1g12) = 2 ∂2g11 = 2 av

1 − − 1 − 1 Γ221 = 2 (∂2g12 + ∂2g12 ∂1g22) = 2 ∂1g22 = 2 bu (2.37) 1 − − 1 − 1 Γ112 = 2 (∂1g12 + ∂1g12 ∂2g11) = 2 ∂2g11 = 2 av

1 − 1 1 Γ122 = Γ212 = 2 (∂2g21 + ∂1g22 ∂2g21) = 2 ∂1g22 = 2 bu

1 − 1 1 Γ222 = 2 (∂2g22 + ∂2g22 ∂2g22) = 2 ∂2g22 = 2 bv

i ir 2 2 To calculate Γkm = g Γkmr note that for the metric a(u, v)du + b(u, v)dv ( ) ( ) 11 12 1 −1 g g a(u,v) 0 G = 21 22 = 1 g g 0 b(u,v)

Hence

1 11 au 1 1 11 av 1 11 −bu Γ11 = g Γ111 = 2a , Γ21 = Γ12 = g Γ121 = 2a , Γ22 = g Γ221 = 2a

2 22 −av 2 2 22 bu 2 22 bv Γ11 = g Γ112 = 2b , Γ21 = Γ12 = g Γ122 = 2b , Γ22 = g Γ222 = 2b (2.38)

2.3.4 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 with

49 2 2 2 a(θ, φ) = R , b(θ, φ) = R sin θ (u = θ, v = φ). Note that aθ = aφ = bφ = 0. Hence only non-trivial components of Γ will be: ( ) −b − sin 2θ −R2 sin 2θ Γθ = θ = , Γ = , (2.39) φφ 2a 2 φφθ 2 ( ) b cos θ R2 sin 2θ Γφ = Γφ = θ = Γ = (2.40) θφ φθ 2b sin θ θφφ 2 All other components are equal to zero:

θ θ θ φ φ Γθθ = Γθφ = Γφθ = Γθθ = Γφφ = 0 Remark Note that Christoﬀel symbols of Levi-Civita connection on the sphere coincide with Christoﬀel symbols of induced connection calculated in the subsection ”Connection induced on surfaces”. later we will understand the geometrical meaning of this fact.

2.4 Levi-Civita connection = induced connection on surfaces in E3 Christoffel symbols of canonical flat connection vanish in Cartesian coordinates. On the other hand standard Riemannian metric on Euclidean space i k has the appearance: G = dx δikdx . The matrix gik has constant entries. Hence according to Levi-Civita formula (2.34) the Christoffel symbols of Levi-Civita connection vanish also. Hence we proved very important fact: Canonical flat connection of Euclidean space is the Levi-Civita connection of the standard metric on Euclidean space. Now we show that Levi-Civita connection on surfaces in Euclidean space coincides with the connection induced on the surfaces by canonical flat connection. We perform our analysis for surfaces in E3. Let M : r = r(u, v) be a surface in E3. Let G be induced Riemannian metric on M and ∇ Levi–Civita connection of this metric. We know that the induced connection ∇(M) is defined in the following ∇M way: for arbitrary vector fields X, Y tangent to the surface M, X Y equals ∇can.flat to the projection on the tangent space of the vector field X Y: ( ) ∇M ∇can.flat X Y = X Y tangent ,

50 where ∇can.flat is canonical flat connection in E3 (its Christoffel symbols vanish in cartesian coordinates). We denote by Atangent a projection of the 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 coincides with Levi-Civita connection of Riemannian metric induced by the canonical metric on Euclidean space E3. Proof Let ∇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 see that it is symmetric connection. Indeed

∇M ∇M ⇒ u u v v ∂u ∂v = (ruv)tangent = (rvu)tangent = ∂v ∂u . Γuv = Γvu, Γuv = Γvu . Prove that this connection preserves scalar product on M. For arbitrary tangent vector ﬁelds X, Y, Z we have

⟨ ⟩ ⟨∇can. flat ⟩ ⟨ ∇can. flat ⟩ ∂X Y, Z E3 = X Y, Z E3 + Y, X Z E3 . since canonical flat connection in E3 preserves Euclidean metric in E3 (it is evident in Cartesian coordinates). Now project the equation above on the surface M. If A is an arbitrary vector attached to the surface and Atangent is its projection on the tangent space to the surface, then for every 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 to the surface. Hence we deduce from (2) that ∂X⟨Y, Z⟩M = ( ) ( ) can. flat can. flat M M ⟨ ∇ ⟩ 3 ⟨ ∇ ⟩ 3 ⟨∇ ⟩ ⟨ ∇ ⟩ X Y tangent , Z E + Y, X Z tangent E = X Y, Z M + Y, X Z M . We see that induced connection is symmetric connection which preserves the induced metric. Hence due to Levi-Civita Theorem it is unique and is expressed as in the formula (2.34). 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.

51 3 Parallel transport and geodesics

3.1 Parallel transport 3.1.1 Definition Let M be a manifold equipped with affine connection ∇. i i Definition Let C : x(t) with coordinates x = x (t), t0 ≤ t ≤ t1 be a curve on the manifold M. Let X = X(t0) be an arbitrary tangent vector i attached at the initial point x0 (with coordinates x (t0)) of the curve C, i.e. ∈ X(t0) Tx0 M is a vector tangent to the manifold M at the point x0 with i coordinates x (t0). (The vector X is not necessarily tangent to the curve C) We say that X(t), t0 ≤ t ≤ t1 is a parallel transport of the vector X(t0) ∈ i i ≤ ≤ Tx0 M along the curve C : x = x (t), t0 t t1 if • ≤ ≤ | For an arbitrary t, t0 t t, vector X = X(t), (X(t) t=t0 = X(t0)) is a vector attached at the point x(t) of the curve C, i.e. X(t) is a vector tangent 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 = ∇ X = 0 . (3.1) dt v In components: if Xm(t) are components of the vector ﬁeld X(t) and vm(t) are components of the velocity vector v of the curve C ,

∂ dx(t) dxi ∂ X(t) = Xm(t) | , v = = | ∂xm x(t) dt dt ∂xm x(t) then the condition (3.1) can be rewritten as

dXi(t) + vk(t)Γi (xi(t))Xm(t) ≡ 0 . (3.2) dt km

Remark We say sometimes that X(t) is covariantly constant along the curve C If X(t) is parallel transport of the vector X along the curve C. If we consider Euclidean space with canonical ﬂat connection then in Cartesian coordinates Christoﬀel symbols vanish and parallel transport is nothing but dX ∇ dt = vX = 0, X(t) is constant vector.

52 Remark Compare this definition of parallel transport with the definition which we consider in the course of ”Introduction to Geometry” where we consider parallel transport of the vector along the curve on the surface embedded in E3 and define parallel transport by the condition, that only orthogonal component of the vector changes during parallel transport, i.e. dX(t) dt is a vector orthogonal to the surface (see the Exercise in the Homework 7).

3.1.2 ∗Parallel transport is a linear map

Consider two different points x0, x1 on the manifold M with connection ∇. Let C be a curve x(t) joining these points. The parallel transport (3.1) X(t) defines the map between tangent vectors at the point x0 and tangent vectors at the point x1. This map depends on the curve C. Parallel transport along different curves joining the same points is in general different (if we are not in Euclidean space). On the other hand parallel transport is a linear map of tangent spaces which does not depend on the parameterisation of the curve joining these points.

Proposition Let X(t), t0 ≤ t ≤ t1 be a parallel transport of the vector ∈ ≤ ≤ X(t0) Tx0 M along the curve C : x = x(t), t0 t t1, joining the points x0 = x(t0) and x1 = x(t1). Then the map ∋ −→ ∈ τC : Tx0 M X(t0) X(t1) Tx1 M (3.3)

is a linear map from the vector space Tx0 M to the vector space Tx1 M which does not depend on the parametersiation of the curve C. The fact that the map (3.3) does not depend on the parameterisation follows from the diﬀerential equation (3.2) also. Indeed let t = t(τ), τ0 ≤ τ ≤ τ1, t(τ0) = t0, t(τ1) = τ1 be another parameterisation of the curve C. Then multiplying the equation (3.2) on dt ′ dτ and using the fact that velocity v (τ) = tτ v(t) we come to diﬀerential equation:

dXi(t(τ)) + v′k(t(τ))Γi (xi(t(τ)))Xm(t(τ)) ≡ 0 . (3.4) dτ km The functions X(t(τ)) with the same initial conditions are the solutions of this equation.

53 The fact that it is a linear map follows immediately from the fact that diﬀerential equations (3.2) are linear. E.g. let vector ﬁelds X(t), Y(t) be covariantly constant along the curve C. Then since linearity X(t) + Y(t) is a solution too.

3.1.3 Parallel transport with respect to Levi-Civita connection We mostly consider parallel transport on Riemannian manifold. If (M,G) is Riemannian manifold we mostly consider parallel transport with respect to connection ∇ which is Levi-Civita connection of the Riemannian metric G Proposition The length of the vector preserves during parallel transport with respect to Levi-Civita connection. Proof follows immediately from the deﬁnition (3.1) of aprallel traspotr and the deﬁnition (2.33) of Levi-Civita connection

d ⟨X(t), X(t)⟩ = ∂ ⟨X(t), X(t)⟩ = 2 ⟨∇ X(t), X(t)⟩ = 2 ⟨0, X(t)⟩ = 0 . dt v v (3.5) Hence |X(t)| is constant. Exercise Show that scalar product preserves during parallel transport.

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 i dxi(t) velocity vector v(t): v (t) = dt is covariantly constant along this curve, i.e. it remains parallel along the curve:

∇v dvi(t) ∇ v = = + vk(t)Γi (x(t))vm(t) = 0, i.e. (3.6) v dt dt km d2xi(t) dxk(t) dxm(t) + Γi (x(t)) = 0 . dt2 dt km dt These are linear second order diﬀerential equations. One can prove that this equations have solution and it is unique7 for an arbitrary initial data i i i i (x (t0) = x0, x˙ (t0) =x ˙ 0.) 7this is true under additional technical conditions which we do not discuss here

54 In other words the curve C : x(t) is a geodesic if parallel transport of velocity vector to along the curve is a velocity vector at any point of the curve. Geodesics deﬁned with Levi-Civita connection on the Riemannian manifold is called geodesic on Riemannian manifold. We mostly consider geodesics on Riemannian manifolds. Since velocity vector of the geodesics on Riemannian manifold at any point is a parallel transport with the Levi-Civita connection, hence due to Proposition (3.5) the length of the velocity vector remains constant: Proposition If C : x(t) is a geodesics on Riemannian manifold then the length of velocity vector is preserved along the geodesic. Proof Since the connection is Levi-Civita connection then it preserves scalar product of tangent vectors, (see (2.33)) in particularly the length of the velocity vector v:

3.2.2 Un-parameterised geodesic We deﬁned a geodesic as a parameterised curve such that the velocity vector is 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 tangent to the curve during parallel transport. Is it true that this curve (in a suitable parameterisation) becomes geodesic? Deﬁnition We call un-parameterised curve geodesic if under suitable parameterisation it obeys the equation (3.6) for geodesics. Let C–be un-parameterised geodesic. Then the following statement is valid. Proposition A curve C (un-parameterised) is geodesic if an only if a non-zero vector tangent to the curve remains tangent to the curve during parallel transport. Proof. Let A be tangent vector at the point p ∈ C of the curve. Paral- lel transport does not depend on parametersiation of the curve (see 3.1.2). Choose a suitable parameterisation xi = xi(t) such that xi(t) obeys the equations (3.6) for geodesics, i.e. the velocity vector v(t) is covariantly constant along the curve: ∇vv = 0. If A(t0) = cv(t0) at the given point p (c is a scalar

55 coeﬃcient) then due to linearity A(t) = cv(t) is a parallel transport of the vector A. The vector A(t) is tangent to the curve since it is proportional to velocity vector. We proved that any tangent vector remains tangent during parallel transport. Now prove the converse: Let A(t) be a parallel transport of non-zero vector and it is proportional to velocity. If in a given parameterisation A(t) = dt(τ) c(t)v(t) choose a reparameterisation t = t(τ) such that dτ = c(t). In the ′ dt(τ) new parameterisation the velocity vector v (τ) = dτ v(t(τ)) = c(t)v(t) = A(t(τ)). We come to parameterisation such that velocity vector remains covariantly constant. Thus we come to parameterised geodesic. Hence C is a geodesic. Remark In particularly it follows from the Proposition above the following important observation: Let C is un-parameterised geodesic, xi(t) be its arbitrary parameterisation and v(t) be velocity vector in this parameterisation. Then the velocity vector remains parallel to the curve since it is a tangent vector. In spite of the fact that velocity vector is not covariantly constant along the 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 parameterisation and s = s(t) is a natural parameter (which deﬁnes the length of the curve) then xi(t(s)) is parameterised geodesic.

3.2.3 Geodesics on surfaces in E3

3 Let M : r = r(u, v) be a surface in E . Let GM be induced Riemannian metric and ∇ a Levi-Civita connection on M. We consider on M Levi-Civita connection of the metric GM . dr(t) Let C be an arbitrary geodesic and v(t) = dt the velocity vector. Ac- cording to the definition of geodesic ∇vv = 0. On the other hand we know that Levi-Civita connection coincides with the connection induced on the surface by canonical flat connection in E3 (see the Theorem in subsection 2.4). Hence ( ) ∇ ∇M ∇can.flat vv = 0 = v v = v v tangent (3.7) ∇can.flat d d2r(t) In cartesian coordinates v v = ∂vv = dt v(u(t), v(t)) = dt2 = a. Hence according to (3.7) the tangent component of acceleration equals to zero.

56 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 (3.7) ∇vv = 0. We come to very beautiful observation: Proposition 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 along geodesic 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 cylinder and sphere. Geodesic on the cylinder  x = a cos φ Let r(h(t), φ(t)) be a geodesic on the cylinder y = a sin φ . We have    z = h −aφ˙ sin φ dr   v = dt = aφ˙ cos φ and for acceleration: h˙  ··    −aφ sin φ −aφ˙ 2 cos φ dv  ··  a = =  aφ cos φ  + −aφ˙ 2 sin φ dt ·· h 0 | {z } | {z } tangent acceleration normal acceleration

d2h Since tangential acceleration equals to zero hence dt2 = 0 and h(t) = h0 + ct Normal acceleration is centripetal acceleration of the rotation over circle with constant speed (projection on the plane OXY ). The geodesic is helix. Geodesics on sphere  x = a sin θ cos φ Let r = r(θ(t), φ(t)) be a geodesic on the sphere of the radius a: r(θ, φ): y = a sin θ sin φ  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) since due to Proposition it is orthogonal to the surface of the sphere. This implies that M(t) is constant vector:

d d M(t) = (r(t) × v(t)) = (v(t) × v(t)) + (r(t) × a(t)) = 0 (3.8) dt dt

57 We have M(t) = M0. r(t) is orthogonal to M = r(t) × v(t). We see that r(t) belongs to the sphere and to the plane orthogonal to the vector M0 = r(t) × v(t). The intersection of this plane with sphere is a great circle. We proved that if r(t) is geodesic hence it belongs to great circle (as un-parameterised curve). The converse is evident since if particle moves along the great circle with constant velocity then obviously acceleration vector is orthogonal to the surface. Remark The vector M = r(t) × v(t) is the torque. The torque is integral of motion in isotropic space.—This is the core of the considerations for geodesics on the sphere.

3.3 Geodesics and Lagrangians of ”free” particle on Riemannian manifold. 3.3.1 Lagrangian and Euler-Lagrange equations A function L = L(x, x˙) on points and velocity vectors on manifold M is a Lagrangian on manifold M. We assign to Lagrangian L = L(x, x˙) the following second order diﬀeren- tial equations ( ) d ∂L ∂L = (3.9) dt ∂x˙ i ∂xi In detail ( ) d ∂L ∂2L ∂2L ··m ∂L = x˙ m + x = . (3.10) dt ∂x˙ i ∂xm∂x˙ i ∂x˙ m∂x˙ i ∂xi These equations are called Euler-Lagrange equations of the Lagrangian L. We will explain later the variational origin of these equations 8.

3.3.2 Lagrangian of ”free” particle

i k Let (M,G), G = gikdx dx be a Riemannian manifold.

8To 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 they arise from variational principe.

58 Deﬁnition We say that Lagrangian L = L(x, x˙) is the Lagrangian of a ”free” particle on the Riemannian manifold M if g x˙ ix˙ k L = ik (3.11) 2 Example ”Free” particle in Euclidean space. Consider E3 with standard metric G = dx2 + dy2 + dz2 g x˙ ix˙ k x˙ 2 +y ˙2 +z ˙2 L = ik = (3.12) 2 2 Note that this is the Lagrangian that describes the dynamics od free particle. Example ”Free” particle on sphere. The metric on the sphere of radius R is G = R2dθ2 + R2 sin2 θdφ2. Re- spectively for the Lagrnagian of ”free” particle we have g x˙ ix˙ k R2θ˙2 + R2 sin2 θφ˙ 2 L = ik = (3.13) 2 2

3.3.3 Equations of geodesics and Euler-Lagrange equations Theorem. Euler-Lagrange equations of Lagrangian of free particle are 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 (3.9) for the Lagrangian (3.11):  ( ) m k ( ) gmkx˙ x˙ d ∂L d ∂ 2 d ( ) ·· ∂g =   = g x˙ k = g x k + ik x˙ mx˙ k dt ∂x˙ i dt ∂x˙ i dt ik ik ∂xm and ( ) m k gmkx˙ x˙ ∂L ∂ 2 1 ∂g = = mk x˙ mx˙ k. ∂xi ∂xi 2 ∂xi Hence we have ( ) d ∂L ·· ∂g ∂L 1 ∂g = g x k + ik x˙ mx˙ k = = mk x˙ mx˙ k , dt ∂x˙ i ik ∂xm ∂xi 2 ∂xi

59 i.e. ·· k m k 1 m k gikx + ∂mgikx˙ x˙ = ∂igmkx˙ x˙ . ( 2 ) m k 1 m k m k Note that ∂mgikx˙ x˙ = 2 ∂mgikx˙ x˙ + ∂kgimx˙ x˙ . Hence we come to equation: d2xk 1 g + (∂ g + ∂ g − ∂ g )x ˙ mx˙ k ik dt2 2 m ik k im i mk Multiplying on the inverse matrix gik we come ( ) d2xi 1 ∂g ∂g ∂g dxm dxk + gij jm + jk − mk = 0 . (3.14) dt2 2 ∂xk ∂xm ∂xj dt dt

We recognize here Christoﬀel symbols of Levi-Civita connection (see (2.34)) and we rewrite this equation as

d2xi dxm dxk + Γi = 0 . (3.15) dt2 dt mk dt This is nothing but the equation (3.6). Applications of this Theorem: calculation of Christoﬀel symbols of Levi- Civita connection.

3.3.4 Examples of calculations of Christoﬀel symbols and geodesics using Lagrangians. Consider two examples: We calculate Levi-Civita connection on sphere in E3 and on Lobachevsky plane using Lagrangians and ﬁnd geodesics. 1) Sphere of the radius R in E3: Lagrangian of ”free” particle on the sphere is given by (3.13):

R2θ˙2 + R2 sin2 θφ˙ 2 L = 2 Euler-Lagrange equations deﬁning geodesics are ( ) ( ) d ∂L ∂L d · ·· − = R2θ − R2 sin θ cos θφ˙ 2 ⇒ θ − sin θ cos θφ˙ 2 = 0 , dt ∂θ˙ ∂θ dt ( ) (3.16) d ∂L ∂L d ( ) ·· − = R2sin2 θφ˙ = 0 ⇒ φ + cotan θθ˙φ˙ = 0 . dt ∂φ˙ ∂φ dt

60 Comparing Euler-Lagrange equations with equations for geodesic in terms of Christoﬀel symbols:

·· θ ˙2 θ ˙ θ 2 θ + Γθθθ + 2Γθφθφ˙ + Γφφφ˙ = 0,

·· φ ˙2 φ ˙ φ 2 φ + Γθθθ + 2Γθφθφ˙ + Γφφφ˙ = 0 we come to θ θ θ θ − Γθθ = Γθφ = Γφθ = 0 , Γφφ = sin θ cos θ , (3.17) φ φ φ φ Γθθ = Γφφ = 0, Γθφ = Γφθ = cotan θ . (3.18) (Compare with previous calculations for connection in subsections 2.2.1 and 2.3.4) We proved that geodesics on the sphere are great circles. (see subsection 3.2.3 above). Consider another more straightforward proof of this fact. To find geodesics one have to solve second order differential equations (3.16): One can see that the great circles: φ = φ0, θ = θ0 + t are solutions of second order differential equations (3.16) with initial conditions

˙ ˙ θ(t) t=0 = θ0, θ(t) t=0 = 1, φ(t) t=0 = φ0, θ(t) t=0 = 0 . (3.19) 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 . (3.20) (Compare with initial conditions (3.19)) Second order diﬀerential equations with boundary conditions for coordinates and velocities at t = 0 have unique solution. The solutions of second order diﬀerential equations (3.16) with initial conditions (3.20) 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 1 x˙ 2 +y ˙2 L = . 2 y2 Euler-Lagrange equations are

61 ( ) ·· ∂L d ∂L d x˙ x 2x ˙y˙ ·· 2x ˙y˙ = 0 = = = − , i.e. x − = 0 , ∂x dt ∂x˙ dt y2 y2 y3 y

( ) ·· ∂L x˙ 2 +y ˙2 d ∂L d y˙ y 2y ˙2 ·· x˙ 2 y˙2 = − = = = − , i.e. y + − = 0 . ∂y y3 dt ∂y˙ dt y2 y2 y3 y y ··i − k i m Comparing these equations with equations for geodesics: x x˙ Γkmx˙ = 0 (i = 1, 2, x = x1, y = x2) we come to 1 1 1 Γx = 0, Γx = Γx = − , Γx = 0, Γy = , Γy = Γy = 0, Γy = − . xx xy yx y yy xx y xy yx yy 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.

3.3.5 Variational principe and Euler-Lagrange equations Here very brieﬂy we will explain how Euler-Lagrange equations follow from variational principe. Let M be a manifold (not necessarily Riemannian) and L = L(xi, x˙ i) be a Lagrangian on it. Denote my Mx2,t2 the space of curves (paths) such that they start at the x1,t1 point x1 at the ”time” t = t1 and end at the point x2 at the ”time” t = t2:

x2,t2 M = {C : x(t), t1 ≤ t ≤ t2, x(t1) = x1, x(t2) = x2} . (3.21) x1,t1

Consider the following functional S on the space Mx2,t2 : x1,t1 ∫ t2 ( ) S [x(t)] = L xi(t), x˙ i(t) dt . (3.22) t1 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 , i.e. x1,t1 x2,t2 ∀x(t) ∈ M S[x0(t)] ≤ S[x(t)] . (3.23) x1,t1

62 Then the path x0(t) is a solution of Euler-Lagrange equations of the La- grangian L: ( ) d ∂L ∂L = if x(t) = x (t) . (3.24) dt ∂x˙ i ∂xi 0

Remark The path x(t) sometimes is called extremal of the action functional (3.22). We will use this Theorem to show that the geodesics are in some sense shortest curves 9.

3.4 Geodesics and shortest distance. Many of you know that geodesics are in some sense shortest curves. We will give an exact meaning to this statement and prove it using variational principe: 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 is a shortest curve between these points10 . 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 i k gik(x)x ˙ x˙ Lfree = 2 and the length Lagrangian √ √ i k Llength(x, x˙) = gik(x)x ˙ x˙ = 2Lfree .

9The 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-Lagrange equations of the Lagrangian L if and only if it is the stationary ”point” of the action functional (3.22), i.e.

S[x0(t) + δx(t)] − S[x0(t) + δx(t)] = 0(δx(t)) (3.25) for an arbitrary inﬁnitesimal variation of the path x0(t): δx(t1) = δx(t2) = 0. 10 More precisely: for every point x1 ∈ C there exists a ball Bδ(x1) such that for an arbitrary point x2 ∈ C ∩ Bδ(x1) the arc of geodesic joining the points x1, x2 is a shortest curve between these points.

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

Length of the curve C = ∫ ∫ t2 t2 √ i i i k Llength(x (t), x˙ (t))dt = gik(x(t))x ˙ (t)x ˙ (t)dt . (3.26) t1 t1 The proof of the Theorem follows from the following observation: Observation Euler-Lagrange equations for the length functional (3.26) are equivalent to the Euler-Lagrange equations for action functional (3.22). This means that extremals of the length functional and action functionals coincide. Indeed it follows from this observation and the variational principe that the shortest curves obey the Euler-Lagrange equations for the action functional. We showed before that Euler-Lagrange equations for action functional (3.22) deﬁne geodesics. Hence the shortest curves are geodesics. One can check the observation by direct calculation:√ Calculate√ Euler- i k Lagrange equations for the Lagrangian Llength = gik(x)x ˙ x˙ = 2Lfree: ( ) ( ) k m d ∂Llength ∂Llength d 1 k 1 ∂gkmx˙ x˙ − = √ gikx˙ − √ i i i k i k i dt ∂x˙ ∂x dt gikx˙ x˙ 2 gikx˙ x˙ ∂x ( ) d 1 ∂Lfree − 1 ∂Lfree = i = 0 . (3.27) dt Llength ∂x˙ i Llength ∂x To facilitate calculations note that the length functional (3.26) is reparameterisation invariant. Choose the natural parameter s(t) or a parameter proportional to the natural parameter on the curve xi(t). We come to Llength = const and it follows from (3.27) that ( ) ( ( ) ) d ∂Llength − ∂Llength 1 d ∂Lfree − ∂Lfree i i = i = 0 . dt ∂x˙ ∂x Llength dt ∂x˙ i ∂x We prove that Euler-Lagrange equations for length and action Lagrangians coincide. In the Euclidean space straight lines are the shortest distances between two points. On the other hand their velocity vectors are constant. We realise now that in general Riemannian manifold the role of geodesic is twofold also: they are locally shortest and have covariantly constant velocity vectors.

64 3.4.1 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 the shortest curves Consider again sphere in E3 with the radius R: Coordinates θ, φ, induced Riemannian metrics (ﬁrst quadratic form):

G = R2(dθ2 + sin2 θdφ2) . (3.28)

Consider two arbitrary points A and B on the sphere. Let (θ0, φ0) be coordinates 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) such that θ(t0) = θ0, θ(t1) = θ1, φ(t0) = θ0, θ(t1) = θ1 then: ∫ √ 2 2 2 LCAB = R θt + sin θ(t)φt dt (3.29)

Suppose that points A and B have the same latitude, i.e. if (θ0, φ0) are coordinates of the point A and (θ1, φ1) are coordinates of the point B then φ0 = φ1 (if it is not the fact then we can come to this condition rotating the sphere) 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 length of the meridian which connects the points A, B: ∫ √ ∫ ∫ t1 t1 √ t1 2 2 2 ≥ 2 − R θt + sin θφt dt R θt dt = R θtdt = R(θ1 θ0) (3.30) t0 t0 t0 Thus we see that the great circle joining points A, B is the shortest. The great circles on sphere are geodesics. It corresponds to geometrical intuition: The geodesics on the sphere are the circles of intersection of the sphere with the plane which crosses the centre.

Geodesics on Lobachevsky plane Riemannian metric on Lobachevsky plane: dx2 + dy2 G = (3.31) y2

65 The length of the curve γ : x = x(t, y = y(t)) is equal to ∫ √ x2 + y2 L = t t dt y2(t)

In particularly the length of the vertical interval [1, ε] tends to inﬁnity if → ε 0: ∫ √ ∫ √ 2 2 1 xt + yt 1 1 L = 2 dt = 2 dt = log y (t) ε t ε One can see that the distance from every point to the line y = 0 is equal to inﬁnity. 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 Then it is easy to see that the length of the curve CAB is less or equal than the length of the arbitrary curve x = x(t), y = y (t) which joins these points: x(t) = x0, y(t) = y0, x(t) = x0, y(t) = y1: t=0 t=0 t=t1 t=t1 ∫ √ ∫ √ ∫ t 2 2 t 2 y1 xt + yt ≥ yt dt y1 2 dt 2 dt = dt = log = length of CAB 0 y (t) 0 y (t) y0 t y0

Hence CAB is shortest. We prove that vertical rays are geodesics. Consider now the case if x0 ≠ x1. Find geodesics which connects two points A, B which are not on the same vertical ray. Consider semicircle which passes these two points such that its centre is on the absolute. We prove that it is a geodesic. Proof Let coordinates of the centre of the circle are (a, 0). Then consider polar coordinates (r, φ): x = a + r cos φ, y = r sin φ (3.32) In these polar coordinates r-coordinate of the semicircle is constant. Find Lobachevsky metric in these coordinates: dx = −r sin φdφ + cos φdr, dy = r cos φdφ + sin φdr, dx2 + dy2 = dr2 + r2dφ2. Hence:

dx2 + dy2 dr2 + r2dφ2 dφ2 dr2 G = = == + (3.33) y2 r2 sin2 φ sin2 φ r2 sin2 φ

66 We see that the length of the arbitrary curve which connects points A, B is greater or equal to the length of the arc of the circle: ∫ √ ∫ √ t1 φ2 r2 t1 φ2 L = t + t dt ≥ t dt = (3.34) AB 2 2 2 2 t0 sin φ r sin φ t0 sin φ ∫ ∫ t1 φ φ1 dφ tan φ t dt = = log 1 t0 sin φ φ0 sin φ tan φ1 The proof is ﬁnished.

4 Surfaces in E3

Now equipped by the knowledge of Riemannian geometry we consider surfaces in E3. We reconsider again conceptions of Shape (Weingarten) operator, Gaussian and mean curvatures, focusing attention on the the fact what properties are internal and what properties are external.

4.1 Formulation of the main result. Theorem of parallel transport over closed curve and Theorema Egregium 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) are internal coordinates of the curve C.) Consider the parallel transport of an arbitrary tangent X vector along the closed curve C: ( ) i α ∂ α ∂x ∂ X(t) = X (t) uα(t) = X (t)rα r(u(t),v(t)) , rα = α i . | ∂u{z α } | {z } ∂u ∂x Internal observer External observer ∇X(t) X(t): = 0, 0 ≤ t ≤ t , dt 1 i.e. dXα(t) duγ(t) + Xβ(t)Γα (u(t)) = 0, 0 ≤ t ≤ t , (4.1) dt βγ dt 1 where ∇ is the connection induced on the surface M by canonical ﬂat connection (see (??)), i.e. the Levi-Civita connection of the induced Riemannian

67 α metric on the surface M and Γβγ its Christoffel symbols: ( ) 1 ∂g ∂g ∂g ∂xi ∂xi Γγ = gγπ πα + πβ − αβ , where g = ⟨r , r ⟩ = αβ 2 ∂uβ ∂uα ∂uπ αβ α β ∂uα ∂uβ (4.2) α β are components of induced Riemannian metric GM = gαβdu du / 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

RC : TpM −→ TpM (4.3)

For any vector X ∈ TpM, its image the vector RC X as the solution of the diﬀerential equation (4.1) with initial condition X(t) t=0 = X. On the other hand we know that parallel transport of the vector does not change its length (see (3.5) in the subsection 3.2.1):

⟨X, X⟩ = ⟨RC X,RC X⟩ (4.4)

We see that RC is an orthogonal operator in the 2-dimensional vector space TpM. We know that orthogonal operator preserving orientation is the operator of rotation on some angle Φ. 11 We see that if RC preserves orientation then the action of operator RC on vectors is rotation on the angle, i.e. the result of parallel transport along closed curve is rotation on the angle

∆Φ = ∆Φ(C) (4.5) which depends on the 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 closed curve C on M such that C is a boundary of a compact oriented domain D ⊂ M. Consider the parallel transport of an arbitrary tangent vector along the closed curve C. As a result of parallel transport along this closed curve any tangent vector rotates through the angle ∫

∆Φ = ∠ (X, RC X) = Kdσ , (4.6) D

11 We consider the case if the operator RC preserves orientation. In our considerations we consider only the case if the closed curve C is a boundary of a compact oriented domain D ⊂ M. In this case one can see that operator RC preserves an orientation.

68 √ where K is the Gaussian curvature and dσ = det gdudv is the area√ element induced 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 on initial point of the curve.

Example Consider the closed curve, ”latitude” Cθ0 : θ = θ0 on the sphere of the radius R. Calculations show that

− ∆Φ(Cθ0 ) = 2π(1 cos θ0) (4.7)

(see the Homework 6). On the other hand the latitude Cθ0 is the boundary of the segment D with area 2πRH where H = R(1 − cos θ0). Hence ∫ ∠ 2πRH 1 · (X, RC X) = 2 = 2 area of the segment = Kdσ R R D

1 since Gaussian curvature is equal to R2 The proof of this Theorem is the content of the subsections above.

4.1.1 GaußTheorema Egregium Show that this Theorem implies the remarkable corollary. Let D be a small domain around a given point p, let C its boundary and ∆Φ(D) the 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 zero we we come to the statement that ∫ if S(D) → 0 then ∆Φ(D) = Kdσ → K(p)S(D), i.e. D

∆Φ(D) K(p) = lim , i.e. (4.8) S(D)→0 S(D) Now notice that left hand side od this equation defining Gaussian curvature K(p) depends only on Riemannian metric on the surface C. Indeed numerator 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 metric too. Thus we come to very important

69 Corollary Gauß Egregium Theorema Gaussian curvature of the surface is invariant of isometries.

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 (4.6). Thus we will prove Theorema Egregium too. Later in the ﬁfth section we will give another proof of the Theorema Egregium.

4.2 Derivation formulae Let M be a surface embedded in Euclidean space E3, M : r = r(u, v). 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 are tangent to the surface and the vector n is orthogonal to the surface12. 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: ∂e ∂e ∂f ∂f ∂n ∂n de = du + dv, df = du + dv , dn = du + dv ∂u ∂v ∂u ∂v ∂u ∂v These 1-forms take values in the vectors in E3, i.e. they are vector valued 1-forms. Any vector in E3 attached at an arbitrary point of the surface can be expanded over the basis {e, f, n}. Thus vector valued 1-forms de, df, dn can be expanded in a sum of 1-forms with values in basic vectors e, f, n. E.g. ∂e ∂e ∂e ∂e for de = ∂u du+ ∂v dv expanding vectors ∂u and ∂v over basis vectors we come to ∂e ∂e = A (u, v)e+B (u, v)f+C (u, v)n, = A (u, v)e+B (u, v)f+C (u, v)n ∂u 1 1 1 ∂v 2 2 2 thus ∂e ∂e de = du + dv = (A e + B f + C n) du + (A e + B f + C n) dv = ∂u ∂v 1 1 1 2 2 2

12 3 One can say that {e, f, n} is an orthonormal basis in TpE at every point of surface 3 p ∈ M such that {e, f} is an orthonormal basis in TpE at every point of surface p ∈ M.

70 = (|A1du{z+ A2dv}) e + (|B1du{z+ B2dv}) f + (|C1du{z+ C2dv}) e, (4.9)

M11 M12 M11 i.e. de = M11e + M12f + M13n, where M11,M12 and M13 are 1-forms on the surface M deﬁned by the relation (4.9). In the same way we do the expansions of vector-valued 1-forms df and dn we come to de = M11e + M12f + M13n df = M21e + M22f + M23n dn = M31e + M32f + M33n This equation can be rewritten in the following way:       e M11 M12 M13 e       d f = M21 M22 M23 f (4.10) n M31 M32 M33 n

Proposition The matrix M in the equation (4.10) is antisymmetrical matrix, i.e. M11 = M22 = M33 = 0 M = −M = a 12 21 (4.11) M13 = −M31 = b M23 = −M32 = −b i.e.       e 0 a b e d f  = −a 0 c f  , (4.12) n −b −c 0 n where a, b, c are 1-forms on the surface M. Formulae (5.13) are called derivation formulae. Prove this Proposition. 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 (4.10) we have

⟨e, e⟩ = 1 ⇒ d⟨e, e⟩ = 0 = 2⟨e, de⟩ = ⟨e,M11e + M12f + M13n⟩ =

71 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⟩ =

⟨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 could be performed much more shortly in condensed notations. Derivation formulae (5.13) in condensed notations are

dei = Mikek (4.13)

Orthonormality condition means that ⟨ei, ek⟩ = δik. Hence d⟨ei, ek⟩ = 0 = ⟨dei, ek⟩+⟨ei, dek⟩ = ⟨Mimem, ek⟩+⟨ei,Mknen⟩ = Mik+Mki = 0 (4.14) Much shorter, is not it?

4.2.1 ∗Gauss condition (structure equations) Derive the relations between 1-forms a, b and c in derivation formulae. 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 that

72 ddf = 0 and the fact that for 1-form ω ∧ ω = 0.) We have from derivation formulae (5.13) that

d2e = 0 = d(af + bn) = daf − a ∧ df + dbn − b ∧ dn =

da f − a ∧ (−ae + cn) + db n − 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 (4.15) Hence components of the left hand side equal to zero:

(da + b ∧ c) = 0 (db + c ∧ a) = 0 . (4.16)

Analogously

d2f = 0 = d(−ae + cn) = −dae + a ∧ de + dcn − c ∧ dn =

−dae + a ∧ (af + bn) + dc n − c ∧ (−be − cf) = (−da + c ∧ b)e + (dc + a ∧ b)n = 0 . Hence we come to structure equations:

da + b ∧ c = 0 db + c ∧ a = 0 (4.17) dc + a ∧ a = 0

4.3 Geometrical meaning of derivation formulae. Wein- garten operator and second quadratic form in terms of derivation formulae. 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 are tangent to the surface and the vector n is orthogonal to the surface. Note that in generally these vectors are not coordinate vectors. Describe Riemannian geometry on the surface M in terms of this basis and derivation formulae (5.13). Induced Riemannian metric

73 If 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 (4.18)

The matrix of the Riemannian metric in the basis {e, f} is ( ) 1 0 G = (4.19) 0 1

Induced connection Let ∇ be the connection induced by the canonical ﬂat connection on the surface M. Then according equations (??) and derivation formulae (5.13) for every tangent vector X

∇ Xe = (∂Xe)tangent = (de(X))tangent = (a(X)f + b(X)n)tangent = a(X)f . (4.20) and

∇ − − Xf = (∂Xf)tangent = (df(X))tangent = ( a(X)e + c(X)n)tangent = a(X)e . (4.21) In particular ∇ e = a(e)f ∇ e = a(f)f e f (4.22) ∇ef = −a(e)e ∇f f = −a(f)e We know that the connection ∇ is Levi-Civita connection of the induced Riemannian metric (4.20) (see the subsection 4.2.1)13. Second Quadratic form Second quadratic form is a bilinear symmetric function A(X, Y) on tangent vectors which is well-deﬁned by the condition A(X, Y)n = (∂XY)orthogonal (see e.g. subsection 6.4 in Appendices.) Let A(X, Y) be second quadratic form. Then according to derivation formulae (5.13) we have A(e, e) = ⟨∂ee, n⟩ = ⟨de(e), n⟩ = ⟨a(e)f + b(e)n, n⟩ = b(e) ,

A(f, e) = ⟨∂f e, 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) ,

13In particular this implies that this is symmetric connection, i.e.

∇fe − ∇ef − [f, e] = a(f)f + a(e)e − [f, e] = 0 . (4.23)

74 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(e, e) A(f, e) b(e) b(f) A = = (4.24) A(e, f) A(f, f) c(e) c(f) This is symmetrical matrix (see the subsection 4.3.2): A(f, e) = b(f) = A(e, f) = c(e) . (4.25)

Weingarten operator Let S be Weingarten operator: SX = −∂Xn (see the subsection 6.4 in Appendix, or last year Geometry letures). Then it follows from derivation formulae that

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 and the matrix of the Weingarten operator in the basis {e, f} is ( ) b(e) c(e) S = (4.26) b(f) c(f) Remark According to the condition (4.25) 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.

4.3.1 Gaussian and mean curvature in terms of derivation formulae Now we are equipped to express Gaussian and mean curvatures in terms of derivation formulae. Using (4.26) we have for Gaussian curvature K = det S = b(e)c(f) − c(e)b(f) = (b ∧ c)(e, f) (4.27) and for mean curvature H = Tr S = b(e) + c(f) (4.28) What next? We will study in more detail formula (4.27) later. Now consider some examples of calculation derivation formulae, Wein- garten operator, e.t..c. for some examples using derivation formulae.

75 4.4 Examples of calculations of derivation formulae and curvatures for cylinder, cone and sphere Last year we calculated Weingarten oeprator, second quadratic form and curvatures for cylinder, cone and sphere (see also the subsection 6.4 in Ap- pendices.). Now we do the same but in terms of derivation formulae. Cylinder We have to define three vector fields e, f, n on the points of the cylinder x = R cos φ r(h, φ): y = R sin φ such that they form an orthonormal basis at any  z = h point, so that the vectors e, 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 unit vector field:       0 −R sin φ cos φ       rh = 0 , rφ = R cos φ , n = sin φ . (4.29) 1 0 0

Vectors rh, rφ and n are orthogonal to each other but not all of them have unit length. One can choose       0 − sin φ cos φ r e = r = 0 , f = φ =  cos φ  , n = sin φ (4.30) h R 1 0 0

These vectors form an orthonormal basis and e, f form an orthonormal basis in tangent space. Derive for this basis derivation formulae (5.13). For vector ﬁelds e, f, n in (4.30) we have     − sin φ − cos φ de = 0, df = d  cos φ  = − sin φ dφ = −ndφ, 0 0     cos φ − sin φ dn = d sin φ =  cos φ  = fdφ, 0 0

76 i.e.           e 0 a b e 0 0 0 e d f  = −a 0 c f  = 0 0 −dφ f  , (4.31) n −b −c 0 n 0 dφ 0 n i.e. in derivation formulae a = b = 0, c = −dφ. The matrix of Weingarten operator in the basis {e, f} is ( ) ( ) ( ) b(e) c(e) 0 −dφ(e) 0 0 S = = − = − 1 b(f) c(f) 0 dφ(f) 0 R According to (4.27) and(4.28) Gaussian curvature K = b(e)c(f)−b(e)c(f) = 0 and mean curvature ( ) r 1 H = b(e) + c(f) = −dφ(f) = −dφ φ = − R R (Compare with calculations in the subsection 4.3.4)

Cone For cone:  x = kh cos φ r(h, φ): y = kh sin φ ,  z = h       k cos φ −kh sin φ cos φ     1   rh = k sin φ , rφ = kh cos φ , n = √ sin φ 2 1 0 1 + k −k

Tangent vectors√ rh, rφ are orthogonal to each other. The length of the vector 2 rh equals to 1 + k and the length of the vector rφ equals to kh. Hence we can choose orthonormal basis {e, f, n} such that vectors e, f are unit vectors in the directions of the vectors rh, rφ:       k cos φ − sin φ cos φ r 1 r 1 e = √ h = √ k sin φ , f = φ =  cos φ  , n = √ sin φ 2 2 hk 2 1 + k 1 + k 1 0 1 + k −k Calculate de, df and dn:     k cos φ − sin φ kdφ kdφ de = d k sin φ = √  cos φ  = √ f, 2 2 1 1 + k 0 1 + k

77     − sin φ − cos φ df = d  cos φ  = − sin φ dφ = 0 0     k cos φ cos φ −k dφ −kdφ dφ k sin φ dφ − sin φ = √ e − √ n , 1 + k2 1 + k2 2 2 1 −k 1 + k 1 + k and     cos φ − sin φ 1 dφ dn = √ d sin φ = √  cos φ  . 2 2 1 + k −k 1 + k 0 We come to           0 √kdφ 0 e 0 a b e  1+k2  e   −    − √kdφ 0 √−dφ    d f = a 0 c f = 1+k2 1+k2 f , n −b −c 0 n 0 √ dφ 0 n 1+k2 (4.32) i.e. in derivation formulae for 1-forms a = √kdφ , b = 0 and and c = − √−dφ . 1+k2 1+k2 The matrix of Weingarten operator in the basis {e, f} is ( ) ( ) − ( ) 0 √dφ(e) b(e) c(e) 1+k2 0 0 S = = S = − = −1 . √dφ(f) √ b(f) c(f) 0 0 2 1+k2 kh 1+k ( ) rφ 1 1 since dφ(f) = dφ kh = kh dφ(∂φ) = kh . According to (4.27), (4.28) Gaussian curvature

K = b(e)c(f) − b(e)c(f) = 0 and mean curvature ( ) r 1 H = b(e) + c(f) = −dφ(f) = −dφ φ = − R R Remark Equipped by the properties of derivation formulae we do not need to calculate df. The calculation of de and dn and the property that the matrix in derivation formulae is antisymmetric gives us the answer for df.

Sphere

78 For sphere  x = R sin θ cos φ r(θ, φ): y = R sin θ sin φ (4.33)  z = R cos θ     R cos θ cos φ −R sin θ sin φ ∂r ∂r r (θ, φ) = = R cos θ sin φ , r (θ, φ) = =  R sin θ cos φ  , θ ∂θ φ ∂φ −R sin θ 0   sin θ cos φ r n(θ, φ) = = sin θ sin φ . R cos θ

Tangent vectors rθ, rφ are orthogonal to each other. The length of the vector rθ equals to R and the length of the vector rφ equals to R sin θ. Hence we can choose orthonormal basis {e, f, n} such that vectors e, f are unit vectors in the directions of the vectors rθ, rφ:       cos θ cos φ − sin φ sin θ cos φ r r r e(θ, φ) = θ = cos θ sin φ , f(θ, φ) = φ =  cos φ  , n(θ, φ) = = sin θ sin φ . R R sin θ R − sin θ 0 cos θ Calculate de, df and dn:   cos θ cos φ de = d cos θ sin φ = − sin θ     − cos θ sin φ cos θ cos φ  cos θ cos φ  dφ − cos θ sin φ = cos θdφf − dθn, 0 − sin θ     − sin φ cos φ df = d  cos φ  = − sin φ dφ = 0 0     cos θ cos φ sin θ cos φ − cos θdφ cos θ sin φ − sin θdφ sin θ sin φ = − cos θdφe − sin θdφn , − sin θ cos θ       sin θ cos φ cos θ cos φ − sin θ sin φ dn = d sin θ sin φ = cos θ sin φ dθ +  sin θ cos φ  dφ cos θ − sin θ 0

79 = dθe + sin θdφf . i.e.           e 0 a b e 0 cos θdφ −dθ e d f  = −a 0 c f  = − cos θdφ 0 − sin θdφ f  , n −b −c 0 n dθ sin θdφ 0 n (4.34) i.e. in derivation formulae a = cos θdφ, b = −dθ, c = − sin θdφ. The matrix of Weingarten operator in the basis {e, f} is ( ) ( ) ( ) − − −1 b(e) c(e) dθ(e) sin θdφ(e) R 0 S = = − − = − 1 b(f) c(f) dθ(f) sin θdφ(f) 0 R ( ) ( ) ∂θ 1 1 ∂θ 1 since dθ(e) = dθ R = R dθ(∂θ) = R , dφ(e) = dφ R = R dφ(∂θ) = 0. According to (4.27) and(4.28) Gaussian curvature 1 K = b(e)c(f) − b(e)c(f) = R2 and mean curvature 2 H = b(e) + c(f) = − R Notice that for calculation of Weingarten operator and curvatures we used only 1-forms b and c, i.e. the derivation equation for dn,(dn = dθe + sin θdφf). (Compare with calculations in the subsection 4.3.4) Mean curvature is deﬁne up to a sign. If we change n → −n mean → 1 curvature H R 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. Remark The same remark as for cone: equipped by the properties of derivation formulae we do not need to calculate df. The calculation of de and dn and the property that the matrix in derivation formulae is antisymmetric gives us the answer for df.

4.5 ∗Proof of the Theorem of parallel transport along closed curve. We are ready now to prove the Theorem. Recall that the Theorem states following:

80 If C is a closed curve on a surface M such that C is a boundary of a compact oriented domain D ⊂ M, then during the parallel transport of an arbitrary tangent vector along the closed curve C the vector rotates through the angle ∫

∆Φ = ∠ (X, RC X) = Kdσ , (4.35) √D where K is the Gaussian curvature and dσ = det gdudv is the area√ element induced by the Riemannian metric on the surface M, i.e. dσ = det gdudv. (see (4.6). Recall that for derivation formulae (5.13) we obtained structure equations

da + b ∧ c = 0 db + c ∧ a = 0 (4.36) dc + a ∧ a = 0

We need to use only one of these equations, the equation

da + b ∧ c = 0 . (4.37)

This condition sometimes is called Gauß condition. 3 Let as always {e, f, n} be an orthonormal basis in TpE at every point of surface p ∈ M such that {e, f} is an orthonormal basis in TpM at every point of surface p ∈ M. Then the Gauß condition (4.37) and equation (4.27) mean that for Gaussian curvature on the surface M can be expressed through the 2-form da and base vectors {e, f}:

K = b ∧ c(e, f) = −da(e, f) (4.38)

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) is tangent to the surface M at the point u(t) of the curve C and vector ﬁeld X(t) is covariantly constant along the curve: ∇X(t) = 0 dt

81 To write this equation in components we usually expanded the vector field in the coordinate basis {ru = ∂u, rv = ∂v} and used Christoffel symbols of α ∇ α the connection Γβγ : β∂γ = Γβγ∂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 the connection ∇ has the following appearance in this basis

∇ve = a(v)f,, ∇vf = −a(v)e (4.39)

(see the equations (4.20) and (4.21)) Let X = X(u(t)) = X1(t)e(u(t)) + X2(t)f(u(t)) Lbe an expansion of tangent vector ﬁeld X(t) over basis {e, f}. Let v be ve- ∇X(t) locity vector of the curve C. Then the equation of parallel transport dt = 0 will have the following appearance:

∇X(t) ( ) = 0 = ∇ X1(t)e(u(t)) + X2(t)f(u(t)) = dt v dX1(t) dX2(t) e(u(t)) + X1(t)∇ e(u(t)) + f(u(t)) + X2(t)∇ f(u(t)) = dt v dt v dX1(t) dX2(t) e(u(t)) + X1(t)a(v)f(u(t)) + f(u(t)) − X2(t)a(v)e(u(t)) = (dt ) ( dt ) dX1(t) dX2(t) − X2(t)a(v) e(u(t)) + + X1(t)a(v) f(u(t)) = 0. dt dt Thus we come to equation: { X˙ 1(t) − a(v(t))X2 = 0 X˙ 2(t) + a(v(t))X1 = 0

There are many ways to solve this equation. It is very convenient to consider complex variable Z(t) = X1(t) + iX2(t) We see that

Z˙ (t) = X˙ 1(t) + iX˙ 2(t) = a(v(t))X2 − ia(v(t)X1 = −ia(v)Z(t),

82 i.e. dZ(t) = −ia(v(t))Z(t) (4.40) dt The solution of this equation is: ∫ −i t a(v(τ))dτ Z(t) = Z(0)e 0 (4.41) ∫ t1 Calculate 0 a(v(τ))dτ for closed curve u(0) = u(t1). Due to Stokes Theo- rem: ∫ ∫ ∫ t1 a(v(t))dt = a = da 0 C D Hence using Gauss condition (4.37) we see that ∫ ∫ ∫ ∫ t1 a(v(t))dt = a = da = − b ∧ c 0 C D D Claim ∫ ∫ ∫ b ∧ c = − da = Kdσ . (4.42) D D Theorem follows from this claim: ∫ ∫ −i a i b∧C Z(t1) = Z(0)e C = Z(0)e D (4.43) ∫ ∫ ∧ ∧ Denote the integral i D b C by ∆Φ: ∆Φ = i D b C. We have ( ) 1 2 1 2 i∆Φ Z(t1) = X (t1) + iX (t1) = X (0) + iX (0) e = (4.44)

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 (4.45) √ (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 (4.38). 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 (4.42).

83 5 Curvtature tensor

5.1 Definition Let X, Y, Z be an arbitrary vector fields on the manifold equipped with affine connection ∇. Consider the following operation which assings to the vector fields X, Y and Z 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 and it is C∞(M) with respect to vector ﬁelds X, YZ, i.e. for an arbitrary functions f, g, h

R(fX, gY)(hZ) = fghR(X, Y)Z, (5.2) ( ) 1 i.e. it defines the tensor field of the type : If X = Xi∂ , X = Xi∂ , X = 3 i i i X ∂i then according to (5.2) R R m n r r i m n (X, Y)Z = (X ∂m,Y ∂n)(Z ∂r) = Z RrmnX Y i R where we denote by Rrmn the components of the tensor in the coordinate basis ∂i i R R rmn∂i = (∂m, ∂n)∂r (5.3) Express components of the curvature tensor in terms of Christoffel symbols ∇ r of the connection. If m∂n = Γmn∂r then according to the (5.1) we have: i R ∇ ∇ − ∇ ∇ R rmn∂i = (∂m, ∂n)∂r = ∂m ∂n ∂r ∂n ∂m ∂r, i ∇ p − ∇ p R rmn = ∂m (Γnr∂p) ∂n (Γmr∂p) = i i p − ↔ ∂mΓnr + ΓmpΓnr (m n) (5.4) The proof of the property (5.2) can be given just by straightforward calculations: 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−

84 ∂Y∂XhZ − ∂Xh∇YZ − ∂Yh∇XZ + h∇Y∇XZ− ∂ hZ − h∇ Z = [ [X,Y] ] [X,Y] ] 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

i Tensor R kmn is expressed trough derivatives of Christoffel symbols. In spite this fact it is is much more ”pleasant” object than Christoffel symbols, since the latter is not the tensor. i It follows from the definition that the tensor Rkmn is antisymmetrical with respect to indices m, n:

i − i Rkmn = Rknm . (5.5)

One can prove that for symmetric connection this tensor obeys the following identities:

i i i R kmn + Rmnk + Rnkm = 0 , (5.6) The curvature tensor corresponding to Levi-Civita connection obeys also another identities too (see the next subsection.) We know well that If Christoffel symbols vanish in a vicinity of a given point p in some chosen coordinate system then in general Christoffel symbols do not vanish in arbitrary coordinate systems. E.g. Christoffel symbols of canonical flat connection in E2 vanish in Cartesian coordinates but do not vanish in polar coordinates. This unpleasant property of Christoffel symbols is due to the fact that Christoffel symbols do not form a tensor. In particular if a tensor vanishes in some coordinate system, then it vanishes in arbitrary coordinate system too. This implies very simple but important i Proposition If curvature tensor R kmn vanishes in some coordinate system, then it vanishes in arbitrary coordinate systems. We see that if Christoffel symbols vanish in a vicinity of a given point p in some chosen coordinate system then it Riemannian curvature tensor vanishes at the point p (see the formula (5.4)) and hence it vanishes at the point p in arbitrary coordinate system.

85 5.2 Riemann curvature tensor of Riemannian manifolds. Let M be Riemannian manifold equipped with Riemannian metric G In 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 indices j Rikmn = gijR kmn (5.7) Due to identities (5.5) and (5.6) for curvature tensor Riemann tensor obeys the following identities:

Rikmn = −Riknm ,Rikmn + Rimnk + Rinkm = 0 (5.8)

Riemann curvature tensor which is curvature tensor for Levi-Civita connection obeys also the following identities:

Rikmn = −Rkimn ,Rikmn = Rmnki . (5.9)

These condition lead to the fact that for 2-dimensional Riemannian manifold the Riemann curvature tensor of Levi-Civita connection has essentially only one non-vanishing component: all components vanish or equal to component R1212up to a sign. Indeed consider for 2-dimensional Riemannian manifold Riemann tensor Rikmn, where i, k, m, n = 1, 2. Since antisymmetricity with respect to third and fourth indices (Rikmn = −Riknm), Rik11 = Rik22 = 0 and Rik12 = −Rik21. The same for ﬁrst and second indices: since antisymmetricity with respect to the the ﬁrst and second indices (R12mn = −R21mn), R11mn = R22mn = 0 and R12mn = −Rik21. If we denote R1212 = a then

R1212 = R2121 = a, R1221 = R2112 = −a and all other components vanish. For Riemann tensor one can consider Ricci tensor,

i Rmn = Rmin (5.10) which is symmetrical tensor: Rmn = Rnm.

86 One can consider scalar curvature:

i kn kn R = R king = g Rkn (5.11) where gkn is Riemannian metric with indices above (the matrix ||gik|| is inverse to the matrix ||gil||). Ricci tensor and scalar curvature play essential role for formulation of Ein- stein gravity equations. In particular the space without matter the Einsten equations have the following form: 1 R − g = 0 . (5.12) ik R ik 5.3 †Curvature of surfaces in E3.. Theorema Egregium again Express Riemannian curvature of surfaces in E3 in terms of derivation formulae (5.13). Consider derivation formulae for the orthonormal basis {e, f, n, } adjusted to the surface M:       e 0 a b e d f  = −a 0 c f  , (5.13) n −b −c 0 n where as usual e, f, n vector ﬁelds of unit length which are orthogonal to each other and n is orthogonal to the surface M. As we know the induced connection on the surface M is deﬁned by the formulae (4.20) and (4.21):

∇ ∇ − Ye = (de(Y))tangent = a(X)f , Yf = (df(Y))tangent = a(X)e , (5.14) According to the deﬁnition of curvature calculate

R(e, f)e = ∇e∇f e − ∇f ∇ee − ∇[e,f]e .

Note that since the induced connection is symmetrical connection then:

∇ef − ∇f e − [e, f] = 0 . (5.15) hence due to (5.14)

[e, f] = ∇ef − ∇f e = a(e)e + a(f)f (5.16) Thus we see that R(e, f)e =

∇e∇f e − ∇f ∇ee − ∇[e,f]e = ∇e (a(f)f) − ∇f (a(e)f) − ∇a(e)e+a(f)f e =

∂ea(f)f + a(f)∇ef − ∂f a(e)f − a(e)∇f f − a(e)∇ee − a(f)∇f e =

87 ∂ea(f)f − a(f)a(e)e − ∂f a(e)f + a(e)a(f)e − a(e)a(e)f − a(f)a(f)f =

[∂ea(f)f − ∂f a(e)f − a ((e)e − a(f)f)] f = da(e, f)f . Recall that we established in 4.38 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 . (5.17)

This means that 2 − R112 = K (in the basis e, f), i.e. the scalar curvature

R = 2R1212 = 2K

We come to fundamental relation which claims that the Gaussian curvature (the magnitude deﬁned in terms of External observer) equals to the scalar curvature, the magnitude deﬁned in terms of Internal observer. This gives us the straightforward proof of Theorema Egregium. Below we will give the proof of Theorema Egregium by straightforward calcultaions.

5.4 Relation between Gaussian curvature and Riemann curvature tensor. Straightforward proof of Theo- rema Egregium

3 i Let M be a surface in E and R kmp be Riemann tensor, Riemann curva- i ture tensor of Levi-Civita connection. Recall that this means that R kmp is curvature tensor of the connection ∇, which is Levi-Civita connection of the Riemannian metric gαβ induced on the surface M by standard Euclidean metric dx2 + dy2 + dz2. Recall that Riemann curvature tensor is expressed via Christoﬀel symbols of connection by the formula

i i i p − i − i p R kmn = ∂mΓnk + ΓmpΓnk ∂nΓmk ΓnpΓmk (5.18) (see he formula (5.4)) where Christoﬀel symbols of Levi-Civita connection are deﬁned by the formula ( ) 1 ∂g ∂g ∂g Γi = gij jm + jk − mk (5.19) mk 2 ∂xk ∂xm ∂xj

(see Levi-Civita Theorem)

88 i km Recall that scalar curvature R of Riemann tensor equals to R = Rkimg , where gkm is Riemannina metric tensor with upper indices (matrix ||gik|| is inverse to the matrix ||gik||). Note that as it was mentioned before the formula for scalar curvature becomes very simple in two-dimensional case (see formulae (5.8) and (5.8) above). Using these formulae calculate Ricci tensor and scalar curvature R. Let R1212 = a. We know that all other components of Riemann tensor equal to zero or equal to ±a (see (5.8) and (5.8)). One can show that scalar curvature R can be expressed via the component R1212 = a by the formula 2R R = 1212 (5.20) det g 2 where det g = det gik = g11g22 = g12. Show it. Using identities (5.8) and (5.8) we see that

i 2 22 21 22 22 R11 = R 1i1 = R 121 = g R2121 + g R1121 = g R1212 = g a (5.21)

i 1 11 12 11 11 R22 = R 2i2 = R 212 = g R1212 + g R2221 = g R1212 = g a (5.22) i 1 12 − 12 − 12 R12 = R21 = R 1i2 = R 112 = g R2112 = g R1212 = g a (5.23) Thus ( ) g22a −g12a R = ik −g21a g11a Now for scalar curvature we have

i km km 11 12 21 22 R = R kimg = Rkmg = g R11 + g R12 + g R21 + g R22 = (5.24) 2a 2R (g11g22 − g12g12) = 2a det g−1 = 1212 det g Now we formulate very important Proposition For an arbitrary point of the surface M

R = 2K (5.25)

i km where R = Rkimg 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 know the Gaussian curvature then we know all components of Riemann curvature tensor (since all components vanish or equal to ±a.): using (5.20) one can rewrite the formula (5.25) R 1212 = K (5.26) det g

89 This is nothing but Theorema Egregium! Indeed Theorema Egregium (see beginning of the section 4) immediately follows from this Proposition. Indeed according to the formulae (5.18) and (5.19) the left hand side of the relation R = 2K depends only on induced Riemannian metric. Hence Gaussian curvature K depends only on induced Riemannian metric. It remains to prove the Proposition. We do it by direct calculations in the next subsection. Example Let M = S2 be sphere of radius R in E3. Show that one cannot ﬁnd local coordinates u, v on the sphere such that induced Riemannian metric equals to du2 + dv2 in these coordinates. This immediately follows from the Proposition. Indeed suppose there exist local coordinates u, v on the sphere such that induced Riemannian metric equals to du2 + dv2, i.e. Riemannian metric is given by unity matrix. Then according to the formulae for Levi-Civita connection, the Christoﬀel symbols equal to zero in these coordinates. Hence Riemann curvature tensor equals to zero, and scalar curvature too. Due to Proposition this is in contradiction 1 with the fact that Gaussian curvature of the sphere equals to R2 .

5.4.1 ∗Proof of the Proposition (5.25) We prove this fact by direct calculations. The plan of calculations is following: Let M be a surface in E3 For an arbitrary point p of the surface M we consider cartesian coordinates x, y, z such that orgin coincides with the point p and coordinate plane OXY is the palne attached at the surface at the point p and the axis OZ is orthogonal to the surface. In these coordinates calculations become easy. The surface M in these Cartesian coordinates can be expressed by the equation  x = u y = v (5.27)  z = F (u, v) where F (u, v) has local extremum at the point u = v = 014.

14more in detail this is stationary point. It is local extremum if quadratic form corre- − 2 sponding to second diﬀerential is positive or negative deﬁnite, i.e. FuuFvv Fuv keeps the sign. This means that the Gaussian curvature is not negative

90 Calculate explicitly Gaussian curvature at the pointp. In a vicinity of 1 0     the point p basis vector ru = 0 , basis vector rv = 1 . One can see Fu Fv that   −F 1 u n(u, v) = −F  (5.28) 1 + F 2 + F 2 v u v 1 is unit normal vector. In particular in the origin at the point p, we have       1 0 0       ru = 0 , rv = 1 , and n = 0 (5.29) 0 0 1

Now calculate shape operator. We need this operator only at the origin, hence during calculations of all derivatives we have to note that we need the ﬁnal result only at the point u = v = 0. This will essentially simplify calculations since at the point p derivatives Fu,Fv equal to zero.    −F ∂ 1 u Sr = −∂ n(u, v)| = −  −F  | = u u p ∂u 1 + F 2 + F 2 v u=v=0 u v 1   Fuu   −Fvu |u=v=0 = Fuuru + Fuvrv 1 and    −F ∂ 1 u Sr = −∂ n(u, v)| = −  −F  | = v v p ∂v 1 + F 2 + F 2 v u=v=0 u v 1   Fuv   −Fvv |u=v=0 = Fuvru + Fvvrv 1 since Fu = Fv = 0. We see that at the origin the Shape (Weingarten) operator equals to ( ) F F S = uu uv (5.30) Fvu Fvv

91 (all the derivatives at the origin). Gaussian curvature at the point p equals to − 2 K = det S = FuuFvv Fuv . (5.31) (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 ( ) ( ) ⟨ ⟩⟨ ⟩ 2 ru, ru ru, rv 1 + Fu FuFv G = ⟨ ⟩⟨ ⟩ = 2 . (5.32) rv, ru rv, rv FvFu 1 + Fv

This immediately follows from the expression for basic vectors ru, rv Note that Riemannian metric gik at the point u = v = 0 is deﬁned by unity( matrix guu = g)vv = 1,(guv )= gvu = 0 since p is extremum point: 2 1 + Fu FuFv 1 0 G = 2 p = since p is stationary point, extremum FvFu 1 + Fv 0 1 i (Fu = Fv = 0). Hence the components of the tensor R kmn and Rikmn = j gijR kmn at the point p are the same. i Recall that the components of R kmn are deﬁned by the formula i i i p − i − i p R kmn = ∂mΓnk + ΓmpΓnk ∂nΓmk ΓnpΓmk .

Notice that at the point p not only the matrix of the metric gik equals to unity matrix, but more: Christoﬀel symbols vanish at this point in coordinates u, v since the derivatives of metric at this point vanish. (Why they vanish: this immediately follows from Levi-Civita formula applied to the metric (5.32), see also in detail the ﬁle ”The solution of the problem 5 in the coursework, i revisited”). Hence to calculate Rkmn at the point p one can consider more simple formula: i | i | − i | R kmn p = ∂mΓnk p ∂nΓmk p Try to calculate in a more ”economical” way. Due to Levi-Civita formula ( ) 1 ∂g ∂g ∂g Γi = gij jm + jk − mk mk 2 ∂xk ∂xm ∂xj ( ) 1 0 Since metric g equals to unity matrix g = at the point p hence gij ik 0 1 is unity matrix also: ( ) 1 0 gik| = = δik . p 0 1

92 (We denote δik the unity matrix: all diagonal components equal to 1, all other components equal to zero. (It is so called Kronecker symbols)) Moreover we know also that all the ﬁrst derivatives of the metric vanish at the point p: ∂g ik | = 0 . ∂xm p Hence it follows from the formulae above that for an arbitrary indices i, j, k, m, n ( ) ∂ ∂g ∂gkm ∂g ∂2g ∂2g gkm pr = pr + gkm pr = δkm pr . ∂xi ∂xj p ∂xi p ∂xj p p ∂xi∂xj p ∂xi∂xj p Now using the Levi-Civita formula for the Christoﬀel symbols of connection: ( ) 1 ∂g ∂g ∂g Γi = gij jm + jk − mk mk 2 ∂xk ∂xm ∂xj ( ( )) ∂g i | ∂ 1 ij ∂gjm jk − ∂gmk we come to ∂nΓmk p = ∂xn 2 g ∂xk + ∂xm ∂xj p = ( ) 1 ∂2g ∂2g ∂2g δij jm + jk − mk . (5.33) 2 ∂xn∂xk ∂xn∂xm ∂xn∂xj p Now using this formula we are ready to calculate Riemann curvature tensor i 1 1 R kmn. Remember that it is enough to calculate R 212 and R 212 = R1212 at 1 | the point p since gik = δik at the point p. We have that at p R 212 p = 1 | − 1 | ∂1Γ22 p ∂2Γ12 p and according to (5.33) ( ) 1 ∂2g ∂2g ∂2g ∂ Γ1 = δ1j j2 + j2 − 22 = 1 22 2 ∂x1∂x2 ∂x1∂x2 ∂x1∂xj p ( ) 1 ∂2g ∂2g ∂2g ∂2g 1 ∂2g 12 + 12 − 22 = 12 − 22 , 2 ∂x1∂x2 ∂x1∂x2 ∂x1∂x1 p ∂x1∂x2 2 ∂x1∂x1 p ( ) 1 ∂2g ∂2g ∂2g ∂ Γ1 = δ1j j1 + j2 − 12 = 2 12 2 ∂x2∂x2 ∂x2∂x1 ∂x2∂xj p ( ) 1 ∂2g ∂2g ∂2g 1 ∂2g 11 + 12 − 12 = 11 , 2 ∂x2∂x2 ∂x2∂x1 ∂x2∂x1 p 2 ∂x2∂x2 p Hence ( ) ∂2g 1 ∂2g 1 ∂2g R | = R1 | = ∂ Γ1 | −∂ Γ1 | = 12 − 22 − 11 1212 p 212 p 1 22 p 2 12 p ∂x1∂x2 2 ∂x1∂x1 2 ∂x2∂x2 p

93 2 2 ∂ g12 ∂ g11 Now using expression (5.32) for metric calculate second derivatives ∂x1∂x2 , ∂x2∂x2 2 ∂ g22 and ∂x1∂x1 :

∂2g ∂2(F F ) ∂ ( ) 12 | = u v = (F F + F F ) | = F F + F 2 | . ∂x1∂x2 p ∂u∂v p ∂u u vv uv v p uu vv uv p

1 2 since Fu = Fv = 0 at the extremum point p (x = u, x = v). Analogoulsy

∂2g ∂2(1 + F 2) ∂ 11 | = u = (2F F ) | = 2F 2 | ∂x2∂x2 p ∂v∂v p ∂v u uv p uv p and ∂2g ∂2(1 + F 2) ∂ 22 | = v = (2F F ) | = 2F 2 | , ∂x1∂x1 p ∂u∂u p ∂u v uv p uv p We have ﬁnally that ( ) ∂2g 1 ∂2g 1 ∂2g R | = R1 | = ∂ Γ1 | −∂ Γ1 | = 12 − 22 − 11 = F F −F 2 . 1212 p 212 p 1 22 p 2 12 p ∂x1∂x2 2 ∂x1∂x1 2 ∂x2∂x2 p uu vv uv

The scalar curvature according to (5.20) equals to R = 2R1212 = 2(FuuFvv − 2 − 2 Fuv). Comparing with the expression K = FuuFvv Fuv in (5.31) for Gaussian curvature we come to R = 2K .

5.5 Gauss Bonnet Theorem Consider the integral of curvature over whole closed surface M. According to the Theorem above the answer has to be equal to 0 (modulo 2π), i.e. 2πN where N is an integer, because this integral is a limit when we consider very small curve. We come to the formula: ∫ Kdσ = 2πN D (Compare this formula with formula (4.6)). What is the value of integer N? We present now one remarkable Theorem which answers this question and prove this Theorem using the formula (4.6).

94 Let M be a closed orientable surface.15 All these surfaces can be clas- sified up to a diffeomorphism. Namely arbitrary closed oriented surface M is diffeomorphic either to sphere (zero holes), or torus (one hole), or pretzel (two holes),... ”Number k” of holes is intuitively evident characteristic of the surface. It is related with very important characteristic—Euler characteristic χ(M) by the following formula:

χ(M) = 2(1 − g(M)), where g is number of holes (5.34)

Remark What we have called here ”holes” in a surface is often referred to as ”handles” attached o the sphere, so that the sphere itself does not have any handles, the torus has one handle, the pretzel has two handles and so on. The number of handles is also called genus. Euler characteristic appears in many diﬀerent way. The simplest appearance is the following: Consider on the surface M an arbitrary set of points (vertices) connected with edges (graph on the surface) such that surface is divided on polygons with (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) (5.35)

It does not depend on the graph, it depends only on how much holes has surface. E.g. for every graph on M, P − E + V = 2 if M is diﬀeomorphic to sphere. For every graph on MP − 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.

15Closed means compact surface without boundaries. Intuitively orientability means that one can define out and inner side of the surface. In terms of normal vectors orientability means that one can define the continuous field of normal vectors at all the points 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.

95 Theorem (Gauß -Bonnet) The integral of Gaussian curvature over the closed compact oriented surface M is equal to 2π multiplied by the Euler characteristic of the surface M ∫ 1 Kdσ = χ(M) = 2(1 − number of holes) (5.36) 2π M 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 deformations of surface! It is integer number (up to the factor π) which characterises topology of the surface. E.g. consider surface M which is diffeomorphic to the sphere. If it is 1 sphere of the radius R then curvature is equal to R2 , area of the sphere is 2 4π equal to 4πR and left hand side is equal to 2π = 2. If surface M is an arbitrary surface diffeomorphic to M then metrics and curvature depend from point to the point, Gauß -Bonnet states that integral nevertheless 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 M K det gdudv 0. On the other hand according to the Theorem it is equal to 4π. Contradiction.

Proof of Gauß-Bonet Theorem Consider 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 Euler Theorem 3N N N − + V = V − = χ(M). 2 2 Calculate the sum of the angles of all triangles. On the one hand it is equal to 2πV . On the other hand according the formula (4.6) it is equal to ( ) ( ) ∑N ∫ ∑N ∫ ∫ π + Kdσ = πN + Kdσ = Nπ + Kdσ △ △ i=1 i i=1 i M ∫ We see that 2πV = Nπ + M Kdσ, i.e. ∫ ( ) N Kdσ = π 2V − = 2πχ(M) M 2

96 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 useful to 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 manifold M (the function on tangent bundle TM). Deﬁnition We say that the function F = F (q, q˙) on TM is integral of motion for Lagrangian L = L(x, x˙) if for any curve q = q(t) which is the solution of Euler-Lagrange equations 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(3.9). (6.1)

In other words ( ) d d ∂L ∂L (F (x(t), x˙(t))) = 0 if xi(t): − = 0 . (6.2) dt dt ∂x˙ i ∂xi

6.1.2 ∗Basic examples of Integrals of motion: Generalised momentum and Energy Let L(xi, x˙ i) does not depend on the coordinate x1. L = L(x2, . . . , xn, x˙ 1, x˙ 2,..., x˙ n). Then the function ∂L F (x, x˙) = 1 ∂x˙ 1 is integral of motion. (In the case if L(xi, x˙ i) does not depend on the coordinate xi. the function F (x, x˙) = ∂L will be integral of motion.) i ∂x˙ i Proof is simple. Check the condition (6.2): Euler-Lagrange equations of motion are: ( ) d ∂L ∂L − = 0 (i = 1, 2, . . . , n) dt ∂x˙ i ∂xi We see that exactly ﬁrst equation of motion is ( ) d ∂L d ∂L = F1(q, q˙) = 0 since 1 = 0,. dt ∂x˙ 1 dt ∂x

i i i ∂L (if L(x , x˙ ) does not depend on the coordinate x then the function Fi(x, x˙) = x˙ 1 is integral of motion since i − th equation is exactly the condition F˙i = 0.) The integral of motion F = ∂L is called sometimes generalised momentum. i ∂x˙ i

97 Another very important example of integral of motion is: energy. ∂L E(x, x˙) =x ˙ i − L. (6.3) ∂x˙ i One can check by direct( calculation) that it is indeed integral of motion. Using Euler Lagrange equations d ∂L − ∂L we have: dt ∂x˙ i ∂xi ( ) ( ) d d ∂L ∂L dx˙ i d ∂L dL E(x(t), x˙(t)) = x˙ i − L = + x˙ i − = dt dt ∂x˙ i ∂x˙ i dt dt ∂x˙ i dt

∂L dx˙ i ∂L dxi dL(x, x˙) dL(x, x˙) dL(x, x˙) + − = − = 0 . ∂x˙ i dt ∂xi dt dt dt dt 6.1.3 ∗Integrals of motion for geodesics Apply the integral of motions for studying geodesics. i k gik(x)x ˙ x˙ The Lagrangian of ”free” particle Lfree = 2 . For Lagrangian of free particle solution of Euler-Lagrange equations of motions are geodesics. i k gik(x)x ˙ x˙ If F = F (x, x˙) is the integral of motion of the free Lagrangian Lfree = 2 then the condition (6.1) means that the magnitude I(t) = F (xi(t), x˙ i(t)) is preserved along the geodesics: d I(t) = F (xi(t), x˙ i(t)) = const, i.e. I(t) = 0 if xi(t) is geodesic. (6.4) dt Consider examples of integrals of motion for free Lagrangian, i.e. magnitudes which preserve along the geodesics: Example 1 Note that for an arbitrary ”free” Lagrangian Energy integral (6.3) is an integral of motion: ( ) p q gpq (x)x ˙ x˙ ∂L ∂ 2 g (x)x ˙ ix˙ k E =x ˙ i − L =x ˙ i − ik = ∂x˙ i ∂x˙ i 2 g (x)x ˙ ix˙ k g (x)x ˙ ix˙ k x˙ ig (x)x ˙ q − ik = ik . (6.5) iq 2 2 This is an integral of motion for an arbitrary Riemannian metric. It is preserved on an arbitrary geodesic ( ) dE(t) d 1 = g (x(t))x ˙ i(t)x ˙ k(t) = 0 . dt dt 2 ik In fact we already know this integral of motion: Energy (6.5) is proportional to the square of the length of velocity vector: √ √ i k |v| = gik(x)x ˙ x˙ = 2E. (6.6)

We already proved that velocity vector is preserved along the geodesic (see the Proposition in the subsection 3.2.1 and its proof (??).)

98 Example 2 Consider Riemannian metric G = adu2 + bdv2 (see also calculations in subsection 2.3.3) in the case if a = a(u), b = b(u), i.e. coeﬃcients do not depend on the second coordinate v: 1 ( ) G = a(u)du2 + b(u)dv2,L = a(u)u ˙ 2 + b(u)v ˙ 2 (6.7) free 2 We see that Lagrangian does not depend on the second coordinate v hence the magnitude ∂L F = free = b(u)v ˙ (6.8) ∂v˙ is preserved along geodesic. It is integral of motion because Euler-Lagrange equation for coordinate v is

d ∂Lfree ∂Lfree d ∂Lfree d − = = F = 0 since ∂Lfree = 0. . dt ∂v˙ ∂v dt ∂v˙ dt ∂v

In fact all revolution surfaces which we consider here (cylinder, cone, sphere,...) have Riemannian 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 1 2 ˙2 2 2 2 2 2 Lfree = 2 R θ + R sin θφ˙ It is the case (6.7) for u = θ, v = φ, b(u) = R sin θ The integral of motion is ∂L F = free = R2 sin2 θφ˙ ∂φ˙

Example (cone) x = ah cos φ Consider cone y = ah sin φ . Riemannian metric:  z = bh

G = d(ah cos φ)2 + d(ah sin φ)2 + (dbh)2 = (a2 + b2)dh2 + a2h2dφ2 . and free Lagrangian (a2 + b2)h˙ 2 + a2h2φ˙ 2 L = . free 2 The integral of motion is ∂L F = free = a2h2φ.˙ ∂φ˙

Example (general surface of revolution) Consider a surface of revolution in E3:  x = f(h) cos φ r(h, φ): y = f(h) sin φ (f(h) > 0) (6.9)  z = h

99 (In√ the case f(h) = R it is cylinder, in the case√ f(h) = kh it is a cone, in the case f(h) = R2 − h2 it is a sphere, in the case f(h) = R2 + h2 it is one-sheeted hyperboloid, in the case z = cos h it is catenoid,...) For the surface of revolution (6.9)

G = d(f(h) cos φ)2 + d(f(h) sin φ)2 + (dh)2 = (f ′(h) cos φdh − f(h) sin φdφ)2+

(f ′(h) sin φdh + f(h) cos φdφ)2 + dh2 = (1 + f ′2(h))dh2 + f 2(h)dφ2 . The ”free” Lagrangian of the surface of revolution is ( ) 1 + f ′2(h) h˙ 2 + f 2(h)φ ˙ 2 L = . free 2 and the integral of motion is ∂L F = free = f 2(h)φ. ˙ ∂φ˙

6.1.4 ∗Using integral of motions to calculate geodesics Integrals of motions may be very useful to calculate geodesics. The equations for geodesics are second order diﬀerential equations. If we know integrals of motions they help us to solve these equations. Consider just an example. x˙ 2+y ˙2 For Lobachevsky plane the free Lagrangian L = 2y2 . We already calculated geodesics in the subsection 3.3.4. Geodesics are solutions of second order Euler-Lagrange equations x˙ 2+y ˙2 for the Lagrangian L = 2y2 (see the subsection 3.3.4) { ·· − 2x ˙ y˙ x y = 0 ·· 2 x˙ 2 − y˙ y + y y = 0 It is not so easy to solve these diﬀerential equations. For Lobachevsky plane we know two integrals of motions:

x˙ 2 +y ˙2 ∂L x˙ E = L = , and F = = . (6.10) 2y2 ∂x˙ y2

These both integrals preserve in time: if x(t), y(t) is geodesics then { { x˙ (t) 2 F = y(t)2 x˙ = C1y 2 2 ⇒ √ x˙ (t) +y ˙(t) ± 2 − 2 4 E = 2y(t)2 = C2 y˙ = 2C2y C1 y

These are first order differential equations. It is much easier to solve these equations in general case than initial second order differential equations.

100 6.2 Induced metric on surfaces. Recall here again induced metric (see for detail subsection 1.4) 3 If surface M : r = r(u, v)is embedded in E 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: ( ) ∑3 ∑3 ∂xi 2 G = dx2 + dy2 + dz2 = (dxi)2 = duα M r=r(u,v) r=r(u,v) ∂uα i=1 i=1

∂xi ∂xi = duαduβ , ∂uα ∂uβ i.e. ∂xi ∂xi G = g duα, where g = duαduβ . M αβ αβ ∂uα ∂uβ 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 dummy indices. For coordinate tangent vectors

∂ ∂xi ∂ = rα = α i |∂u{zα} | ∂u{z ∂x} Internal observer External observer We have already plenty examples in the subsection 1.4. In particular for scalar product ⟨ ⟩ ∂ ∂ i i gαβ = , = x αx β .⟨rα, rβ⟩ . (6.12) uα uβ

6.2.1 Recalling Weingarten operator Continue to play with formulae 16. 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 vector field is not a well-defined object: we need a connection. The formula ∂XR in Cartesian coordinates, is nothing but the derivative with respect to flat canonical connection: If

16In some sense diﬀerential geometry it is when we write down the formulae expressing the geometrical facts, diﬀerentiate these formulae then reveal the geometrical meaning of the new obtained formulae e.t.c.

101 we work only in Cartesian coordinates we do not need to distinguish between ∂XR and ∇can.flat ∇can.flat X R. Sometimes with some abuse of notations we will use ∂XR instead X R, but never forget: this can be done only in Cartesian coordinates where Christoffel symbols of flat canonical connection vanish:

∇can.ﬂat ∂XR = X R in Cartesian coordinates . So the rigorous deﬁnition of Weingarten operator is

−∇can.ﬂat SX = X n , (6.14) but we often use the former one (equation (6.13)) just remembering that this can be done only in Cartesian coordinates. Recall that the fact that Weingarten operator S maps tangent vectors to tangent vectors 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.

6.2.2 Second quadratic form We define now the new object: second quadratic form Definition. For two tangent vectors X, Y A(X, Y) is defined such that ( ) ∇can.flat X Y ⊥ = A(X, Y)n (6.15) 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 that A(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 properties of Weingarten operator that ⟨( ) ⟩ ⟨ ⟩ ∇can.flat ∇can.flat A(X, Y) = X Y ⊥ , n = X Y, n = ⟨ ⟩ ⟨ ⟩ − ∇can.flat ⟨ ⟩ ∂X Y, n Y, X n = S(X), Y (6.16) We proved that second quadratic form depends only on value of vector field Y at the given poit and we established the relation between second quadratic form and Weingarten operator. Proposition The second quadratic form A(X, Y) is symmetric bilinear form on tangent vectors X, Y in a given point.

A: TpM ⊗ TpM → R,A(X, Y) = A(Y, X) = ⟨SX, Y⟩ . (6.17)

102 In components ∂2xi A = A duαduβ = ⟨r , n⟩ = ni . (6.18) αβ αβ ∂uα∂uβ and α απ απ i i Sβ = g Aπβ = g xπβn , (6.19) i.e. A = GS, S = G−1A. Remark The normal unit vector ﬁeld is deﬁned up to a sign.

6.2.3 Gaussian and mean curvatures Recall that Gaussian curvature K = det S and mean curvature H = Tr S It is easy to see that for Gaussian curvature det A K = det S = det(G−1A) = . det G We know already the geometrical meaning of Gaussian and mean curvatures from the point 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 the surface? 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 have non-zero mean curvature.

6.2.4 Examples of calculation of Weingarten operator, Second quadratic 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:      x = R cos φ 0 −R sin φ     r(h, φ): y = R sin φ , rh = 0 , rφ = R cos φ , (6.20)  z = h 1 0

103 ( ) 2 2 2 Gcylinder = dx + dy + dz x=a cos φ,y=a sin φ,z=h = ( ) 1 0 = (−a sin φdφ)2 + (a cos φdφ)2 + dh2 = a2dφ2 + dh2 , ||g || = . αβ 0 R2     cos φ cos φ Normal unit vector n = ± sin φ. Choose n = sin φ. Weingarten operator 0 0   cos φ S∂ = −∇can.ﬂatn = −∂ n = −∂ sin φ = 0 , h rh rh h 0     cos φ sin φ can.ﬂat     ∂φ S∂φ = −∇ n = −∂r n = −∂φ sin φ = − cos φ = − . rφ φ R 0 0 ( ) ( ) ( ) ∂h 0 0 0 S = −∂φ ,S = −1 . (6.21) ∂φ 0 R R   0   Calculate second quadratic form: rhh = ∂hrh = 0 , rhφ = rφh = 0       −R sin φ −R sin φ −R cos φ       ∂h R cos φ = 0, rφφ = ∂φ R cos φ = −R sin φ = −Rn . 0 0 0

We have ( ) ( ) ⟨rhh, n⟩⟨rhφ, n⟩ 0 0 Aαβ = ⟨rαβ, n⟩,A = = , (6.22) ⟨rφh, n⟩⟨rφφ, n⟩ 0 −R ( )( ) ( ) 1 0 0 0 0 0 A = GS = 2 − 1 = − , 0 R 0 R 0 R For Gaussian and mean curvatures we have ( ) det A 0 0 K = det S = = det − 1 = 0 , (6.23) det G 0 R and mean curvature ( ) 0 0 − 1 H = Tr S = Tr − 1 = . (6.24) 0 R R → − → 1 Mean curvature is deﬁne up to a sign. If we change n n mean curvature H R and Gaussian curvature will not change.

Cone We already calculated induced Riemannian metric on the cone (see (1.59).

104 Cone is given by the equation x2 + y2 − k2z2 = 0. One can consider the following parameterisation of this surface:      x = kh cos φ k cos φ −kh sin φ     r(h, φ): y = kh sin φ , rh = k sin φ , rφ = kh cos φ , (6.25)  z = h 1 0 ( ) 2 2 2 Gcone = dx + dy + dz x=kh cos φ,y=kh sin φ,z=h =

= (−kh sin φdφ + k cos φdh)2 + (kh cos φdφ + k sin φdh)2 + dh2 = ( ) k2 + 1 0 k2h2dφ2 + (k2 + 1)dh2, ||g || = . αβ 0 k2h2   cos φ   One can see that N = sin φ is orthogonal to the surface: N⊥rh, rφ. Hence normal  −k   cos φ cos φ unit vector n = ± √ 1 sin φ. Choose n = √ 1 sin φ. Weingarten operator 1+k2 1+k2 −k −k    cos φ can.ﬂat  1   S∂h = −∇ n = −∂r n = −∂h √ sin φ = 0 , rh h 2 1 + k −k    cos φ can.ﬂat  1   S∂φ = −∇ n = −∂r n = −∂φ √ sin φ = rφ φ 2 1 + k −k   sin φ 1 ∂ √ − cos φ = − √ φ . 2 2 1 + k 0 kh k + 1 ( ) ( ) ( ) ∂h 0 0 0 S = ,S = − . (6.26) ∂ − √∂φ 0 √ 1 φ kh k2+1 kh k2+1   0   Calculate second quadratic form: rhh = ∂hrh = 0 , rhφ = rφh = 0         −kh sin φ −k sin φ −kh sin φ −kh cos φ         ∂h kh cos φ = k cos φ , rφφ = ∂φ kh cos φ = −kh sin φ . 0 0 0 0

We have ( ) ( ) ⟨r , n⟩⟨r , n⟩ 0 0 A = ⟨r , n⟩,A = hh hφ = , (6.27) αβ αβ ⟨ ⟩⟨ ⟩ 0 − √ kh rφh, n rφφ, n 1+k2

105 ( )( ) ( ) k2 + 1 0 0 0 0 0 A = GS = − = − , 2 2 0 √ 1 0 √ kh 0 k h kh k2+1 k2+1 For Gaussian and mean curvatures we have ( ) det A 0 0 K = det S = = det 0 √−1 = 0 , (6.28) det G kh k2+1 and mean curvature ( ) 0 0 −1 H = Tr S = Tr −1 = √ . (6.29) 0 √ 2 kh k2+1 kh k + 1 → − → 1 Mean curvature is deﬁne up to a sign. If we change n n mean curvature H R and Gaussian curvature will not change.

Sphere Sphere is given by the equation x2 + y2 + z2 = a2. Consider the parameterisation of sphere in spherical coordinates  x = R sin θ cos φ r(θ, φ): y = R sin θ sin φ (6.30)  z = R cos θ

We already calculated induced Riemannian metric on the sphere (see (6.2.4)). Recall that     R cos θ cos φ −R sin θ sin φ     rθ = R cos θ sin φ , rφ = R sin θ cos φ −R sin θ 0 and ( ) 2 2 2 2 GS = dx + dy + dz x=R sin θ cos φ,y=R sin θ sin φ,z=R cos θ = (R cos θ cos φdθ − R sin θ sin φdφ)2 + (R cos θ sin φdθ + R sin θ cos φdφ)2+ (−R sin θdθ)2 = R2 cos2 θdθ2 + R2 sin2 θdφ2 + R2 sin2 θdθ2 = ( ) R2 0 = R2dθ2 + R2 sin2 θdφ2 , ||g || = . αβ 0 R2 sin2 θ For the sphere r(θ, φ) is orthogonal to the surface. Hence normal unit vector n(θ, φ) = sin θ cos φ sin θ cos φ ± r(θ,φ) ±   r   R = sin θ sin φ . Choose n = R = sin θ sin φ . Weingarten operator cos θ cos θ ( ) can.ﬂat r rθ S∂θ = −∇ n = −∂θn = −∂θ = − , rθ R R ( ) can.ﬂat r rφ S∂φ = −∇ n = −∂φn = −∂φ = − . rφ R R

106 ( ) ( ) ( ) ∂ − ∂θ 1 0 S θ = R ,S = − R . (6.31) − ∂φ 1 ∂φ 0 R R   −R sin θ cos φ   For second quadratic form: rθθ = ∂θrθ = −R sin θ sin φ , rθφ = rφθ = −R cos θ         −R sin θ sin φ −R cos θ sin φ −R sin θ sin φ −R sin θ cos φ         ∂θ R sin θ cos φ = R cos θ cos φ , rφφ = ∂φ R sin θ cos φ = −R sin θ sin φ . 0 0 0 0

We have ( ) ( ) ⟨rθθ, n⟩⟨rθφ, n⟩ −R 0 Aαβ = ⟨rαβ, n⟩,A = = 2 , (6.32) ⟨rφθ, n⟩⟨rφφ, n⟩ 0 −R sin θ ( )( ) ( ) R2 0 −1 0 1 0 R − A = GS = 2 2 −1 = R 2 , 0 R sin θ 0 R 0 sin θ For Gaussian and mean curvatures we have ( ) det A − 1 0 1 K = det S = = det R = , (6.33) − 1 2 det G 0 R R and mean curvature ( ) − 1 0 2 R − H = Tr S = Tr − 1 = , (6.34) 0 R R → − → 1 Mean curvature is deﬁne up to a sign. If we change n n mean curvature H R 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.

107