INTRODUCTION TO RIEMANNIAN GEOMETRY

Transcription

1 INTRODUCTION TO RIEMANNIAN GEOMETRY In this course we shall discuss the following notions: smooth manifolds, tangent spaces, fiber bundles, Lie groups and Lie algebras, topics on curvature and geodesics. Recommended books: - Sigmundur Gudmunsson, Riemannian Geometry, - Juergen Jost, Riemannian Geometry and Geometric Analysis, Springer

2 2 Contents 1. Manifolds, differentiable maps, Lie groups Topological manifolds and smooth manifolds Differentiable maps Lie groups Manifolds with boundary 6 2. Tangent bundle and tangent map Tangent space and tangent map Tangent bundle and cotangent bundle Vector fields Lie algebras Submanifolds and fiber bundles Submanifolds and a regular value of a smooth map Fiber bundles Tensors and Riemannian metrics Tensors and Riemannian metrics The existence of a Riemannian metric Levi-Civita connection Linear connections and metric connection The existence and uniqueness of the Levi-Civita connection Parallel transport and geodesics Geodesic and variation of the length of curves The second fundamental form and totally geodesic submanifolds The Riemannian curvature tensor Manifolds, differentiable maps, Lie groups In this section we introduce the notion of differentiable manifolds, which are generalization of domains in Euclidean spaces with good enough properties to apply tools of differential calculus. Morphisms between differentiable manifolds are differentiable maps. Most important examples of differentiable manifolds are Lie groups Topological manifolds and smooth manifolds. I suppose that you know what are curves and surfaces. Curves and surfaces are subjects of differential geometry at the very beginning of its development. In analytic geometry you already learn geometry of some curves and some surfaces. In differential geometry you shall use extensively the tool of differential calculus, taking derivatives, taking integrations, solving differential equations related with geometric objects. Differential geometry has many applications in physics, economy, computer graphic, information theory, where we recognize pattern of differentiable manifolds and laws governed by differentiable mappings. Let me now introduce you to geometric objects in differential geometry. In classical analysis we are concerned with calculus on domains Ω of Euclidean spaces R m : we can differentiate a smooth function f(x 1,, x n ) of n-variables on Ω, take

3 INTRODUCTION TO RIEMANNIAN GEOMETRY 3 integration of f etc. So any domain in R n is an elementary geometric object in differential geometry. To enlarge the class of differential geometric objects we glue together domains of the same dimension. The obtained geometric objects look locally like open subsets in spaces R n. Definition (Dold, chapter VIII, Def.1.1) Let M be a Hausdorff space 1 with a countable basis for its topology. We call M an n-dimensional topological manifold, if for each point p M there is an open neighborhood U of p in M such that U is homeomorphic to an open set in R n by some homeomorphism φ, which is also called coordinate map of U. Such a pair (U, φ) is called a local coordinates system or a chart on M, and the number n is called the dimension of M n. Clearly every open set in R n is a topological n-dimensional space. But there are plenty of closed subsets of R n which are topological manifolds. Example The sphere S n = {x R n+1, x = 1} is a n-dimensional manifold. We write S n as the union of two open sets {S n \ N} {S n \ S} where N and S are the north pole and the south pole of S n. Clearly each point of S n belongs to one of those two open sets. And these open sets are homeomorphic to R n by stereographic projections π. The pair ({S n \ N, π N }, {S n \ S, π S } is a chart on S n. We are now interested in the class of topological manifolds with a good gluing, that is a good agreement between different homeomorphisms φ i and φ j on the common domain U i U j. Definition A (smooth) atlas on a topological manifold M is a collection A = {(U i, φ i )} of charts such that U i form an open covering of M and for each pair (U i, φ i ) and (U j, φ j ) in A the transition map Φ ij = φ j φ 1 i : φ i (U i U j ) φ j (U i U j ) is a smooth map between the open sets in Euclidean space. Two atlas are equivalent, if their union is also an atlas. A smooth structure on a topological manifold is an equivalence class of smooth atlases, and a smooth manifold is a topological manifold with a specified smooth structure. Example Let us consider a circle S 1 as the subset {z C z = 1}. Write z = re iθ we identify S 1 with the interval [0, 2π] after gluing points 0 and 2π together. We write S 1 as a union of two open intervals I 1 := S 1 \ {0} and I 2 := S 1 \ {2π}. These open intervals are charts on S 1 with φ i : I i R being the identity map for i = 1, 2. Thus S 1 is a topological manifold of dimension 1. Since the transition map φ 1 φ 1 2 on I 1 I 2 is the identity map, S 1 is a smooth manifold. A simple example of atlas of S 2 is the world atlas everybody knows. It was not easy in the ancient time for atlas makers to patch charts together. It may happen that after gluing a chart we will have our earth of the shape of a torus. 1 an equivalent definition is the existence of unique limit for any sequence, a nice property of a Hausdorff space is that the compactness implies the closedness,

4 4 Exercise (i) Show that if M m and N n are topological manifolds, then the Cartesian product M m N n is a topological manifold. If {(U i, φ i )} is a smooth structure on M m and {(V j, ψ j )} is a smooth structure on N n, then {(U i U j, φ i ψ j )} is a smooth structure on M m N n. (ii) Prove that the charts on S n in Example form a smooth atlas. (iii) Let CP n := {[z 0 :, z n ]} be the set of all complex lines in C n+1. Prove that the set of chart U i := CP n {[z 0 :, z i 0, z n ]} C n, [z 0 :, z i 0, z n ] ( z 0 zi,, 1 i, zn z i ) forms a smooth atlas on CP m. Hint. i) Show that π N (p 1,, p n+1 ) = 1 1 p 1 (p 2,, p n+1 ) and π S (p 1,, p n+1 ) = 1 1+p 1 (p 2,, p n+1 ). It is known that there are topological manifolds which have no smooth structure and there are topological manifolds which have more than one smooth structure. Simon Donaldson (1983) proved that a positive definite intersection form of a simply connected smooth manifold of dimension 4 is diagonalisable to the identity matrix. Michael Freedman (1982) had shown that any positive definite unimodular symmetric bilinear form is realized as the intersection form of some four-manifold; combining his and Donaldson s result, any non-diagonalizable intersection form gives rise to a fourdimensional topological manifold with no differentiable structure. An exotic sphere is a differentiable manifold that is homeomorphic to the standard Euclidean n-sphere S n, but not diffeomorphic to S n. That means that such a manifold M is a sphere from a topological point of view, but not from the point of view of its differential structure. The first exotic spheres were constructed by John Milnor (1956) in dimension n = 7 as S 3 -bundles over S 4. He showed that the oriented exotic 7-spheres are the non-trivial elements of a cyclic group of order 28 under the operation of connected sum. In any dimension Milnor (1959) showed that the diffeomorphism classes of oriented exotic spheres form the non-trivial elements of an abelian monoid under connected sum, which is a finite abelian group if the dimension is not 4. Remark The concept of a differentiable manifold was implicitly introduced in the habilitation address of Bernhard Riemmann in Göttingen. The first clear formulation of this concept was given by Herman Weyl later in Differentiable maps. Let M be a differentiable manifold. To study M we need to understand all social relations of M with other differentiable manifolds N, i.e. to study all maps from N to M and all maps from M to N. A mapping from M to N = R n is also called a R n -valued function. Definition A continuous map f : M 1 M 2 is said to be C k -differentiable at a point p M 1, if there is a chart (U, φ) on M 1 with p U and a chart (V, ψ) on M 2 with f(p) V such that the composition ψ f φ 1 is C k -differentiable. If f is C k -differentiable at every point p M 1, we say that f is a C k -map. If f is a homeomorphism such that f and f 1 are smooth, we call f a diffeomorphism. If M 1 is an interval or the whole real line R and k =, we say f is a smooth curve on M 2.

5 INTRODUCTION TO RIEMANNIAN GEOMETRY 5 Example We identify S 1 with [0, 2π]/({0} = {2π}). A map π k : S 1 S 1, θ kθ mod 2π is a smooth map for any k Z. 2. A map R k (t) U (f 1,,f n ) R n is a smooth map, iff f 1,, f n are smooth functions. Exercise (i) Prove that the embedding map S n R n+1 is a smooth map. (ii) Show that there is a smooth map sq : S 2 S 2 such that its restriction to the chart U 1 = {(1, z 1 )} has the form (1, z k 1 ). Clearly the C k -differentiability does not depend on the choice of local coordinates, since the composition of a smooth map (the change of local coordinates) with a map of class C k is of class C k. We denote by C k (M 1, M 2 ) the set of all C k -maps between M 1 and M 2. It follows immediately from the definition that the composition of two C k -maps is also a C k -map. In particular, for φ C k (M 1, M 2 ) (1.2.1) φ (C k (M 2 )) C k (M 1 ), where the induced map φ is defined as follows φ (f(x)) := f(φ(x)). In fact, we can take (1.2.1) for a definition of a C k -map, i.e. a continuous map φ is said to be of class C k, if (1.2.1) holds. A smooth map f is called diffeomorphism, if it is bijective with a smooth inverse. Now working on differentiable manifolds it is very useful to calculate every thing in local coordinates. Now let us see how a differentiable map ψ looks in local coordinates. Fix a chart (U 1, φ 1 ) on M 1. Suppose that ψ(u 1 ) U 2 where (U 2, φ 2 ) is a chart on M 2. The the composition γ : φ 2 ψ φ 1 1 is a map from an open set Ω 1 R n to Ω 2 R m. Since γ(x 1,, x n ) is a point (y 1,, y m ) R m, we can write where γ i is a function on R n. y i = γ i (x 1,, x n ), 1.3. Lie groups. We now introduce main players in the study of differential geometry. These are Lie groups studied by Sophus Lie, Felix Klein, Wilhelm Killing, Elie Cartan. The theory of Lie group remains an active domain of research nowadays. Definition A Lie group is a smooth manifold G with a smooth multiplication and a smooth inversion. Example The set GL n (R) = {x Mat n (R) det(x) 0} is a smooth manifold, since it is an open set in the space Mat n (R) = R n2. It is easy to check that the multiplication µ : GL n (R) GL n (R) GL n (R) and the inverse τ : GL n (R) GL n (R) is smooth. Definition A homomorphism between Lie groups G and H is a smooth map φ : G H which is also a homomorphism in the sense of group theory. Example The exceptional mapping exp : (R, +) (R +, ) is an isomorphism of Lie groups (i.e. it and its inverse are homomorphisms of Lie groups).

6 6 The idea of symmetry in differential geometry is expressed via the notion of an action of a Lie group on a (topological) differentiable manifold. For example we see that the round sphere S 2 is symmetric because we can rotate it along an axis, say x 3. Equivalently, there is an action of S 1 on S 2, i.e. we have a map S 1 S 2 S 2 : (θ, r 3, r 2 φ) (r 3, r 2, φ + θ) In general, we say that a group G acts on a manifold M if there exists a differentiable map χ : G M M such that χ(g 1 g 2, m) = χ(g 1, χ(g 2, m)). So the action looks like an associative multiplication with value in M. Exercise Prove that the product of two rotations in R 3 around v i with an angle θ i, i = 1, 2 is a rotation around some axis. Hint. Prove that the product of two rotations leaves some vector unchanged, using the fact that the characteristic polynomial of a linear transformation on R 3 has a real root. Exercise ) Prove that S 1 is a Lie group. 2) A quaternion is quadruple (a 1, ia 2, ja 3, ka 4 ), where a i R. We can add quaternions together as well to multiply two quaternions using the following recipe i 2 = j 2 = k 2 = 1, ij = ji = k, jk = kj = i, ki = ik = j. Show that the set S 3 of quaternions with norm 1 (i.e. a a2 2 + a2 3 + a2 4 = 1) has a structure of a Lie group w.r.t. the quaternion multiplication. We have seen that S 1 and S 3 have a structure of a Lie group. We can also ask, if S 2, or more general S k has a structure of a Lie group. There are many proofs showing that S 2 has no Lie group structure. One of the proof uses the property that any smooth map S 2 S 2 which is homotopic to identity has a fixed point. We also observe that the Lie structure on S 1 and S 3 are related to the normed algebra of complex numbers and the normed algebra of quaternions. Many important equations in modern physics are related to geometric structures generalizing the notion of complex numbers and quaternions Manifolds with boundary. Denote by E m the half-space of R m with x 1 < 0. Its completion Ēm is the closed subset in R m defined by x 1 0. Definition A topological space M is called a m-(differentiable) manifold with boundary, if M = M and (i) M is an open m-(differentiable) manifold, (ii) for all x M := M \ M there is an open neighborhood M U x and a homeomorphism φ x : U U Ēm. We require that the transition functions to be C k -differentiable for a C k -differentiable manifold M m. Remark It is an interesting question to define the notion of a C k -differentiable function on a half space E m. One way to define a function f to be differentiable at a boundary point x R m 1, if there is an open neighborhood of x in R m where f can

7 INTRODUCTION TO RIEMANNIAN GEOMETRY 7 be extended to a differentiable function. There is another intrinsic way to define this notion suggested by Seeley in his PAMS paper (1994, 15, Extension of C -jets from a half space), see also the book by Adams, Sobolev spaces (p ). We need only to take the limit of k-derivatives of a function f from the lower half space and the existence of an extension of f to the upper space is guaranteed. Example A connected 1-dimensional manifold with boundary is either diffeomorphic to interval [0, 1) or interval [0, 1]. Exercise Show that if M m is not empty then it is a closed differentiable manifold of dimension (m 1) without boundary. 2. Tangent bundle and tangent map In this section, using the notion of a derivation of a function, we define the tangent bundle T M of a differentiable manifold M as a smooth manifold which is a Taylor expansion of M up to first order. We define the tangent map Df of a smooth map g : M N as a natural linear transformation T x M T f(x) N Tangent space and tangent map. Denote by C (M) the space of all smooth differentiable functions on a differentiable manifold M. Definition A tangent vector δ in a point x 0 M is a R-linear map δ : C (M) R satisfying the Leibniz rule δ(f g) = f(x 0 )δ(g) + g(x 0 )δ(f) for all f, g C (M). Such a map is called a derivation of C (M) at x 0. It is easy to see that the space of tangent vectors in x 0 is a vector space. We denote this space by T x0 M. Example i) The partial derivative x i defines a derivation at a point x by x i (f) x = ( x i f)(x). ii) Let γ : ( ε, ε) M be a smooth map (curve) such that γ(0) = p. Then γ defines a derivation of C (M) at point p by for all f C (M). D γ (f) := d dt t=0f(γ(t)) Remark (Existence of the differential of a smooth map). If g : (M m, x) (N n, y) is a smooth map, then g induces a linear map (Dg) x : T x M T y N defined by for any ψ C (M). ((Dg) x δ)ψ = δ(ψ g) The linear map Dg : T x M T f(x) N is called the differential of g at x M, also the tangent map of g at x M.

8 8 Let us compute the differential of the composition of two smooth maps. If g : (M, x) (N, y) and f : (N, y) (P, z) are smooth map, then for δ T x M and for ψ C (P ) we have (D(f g) x (δ))ψ = δ(ψ f g) = [Dg x (δ)](ψ f) = [Df f(x) Dg x (δ)]ψ. Thus we get the following chain rule (D(f g)) x = (Df) f(x) (Dg) x. Corollary If f : M N is a diffeomorphism in some neighborhood of x M then (Df ) x : T x M T f(x) M is an isomorphism. Corollary says that the tangent space T x M is determined by a coordinate neighborhood of x. Now we shall look at the tangent space of a point in R n. Exercise Show that D γ = Dγ( t), where t is the partial derivative on R. Theorem If U is an open subset of R n and D is a derivation at p U, then D = D(x i ). x i p 1 i n Proof. We will represent f(x) in Taylor expansion, and then use the Leibniz rule to conclude that only the first term in the Taylor expansion enters in the formula Df. Write (2.1.1) f(x) f(p) = 1 Integrating by parts we get (2.1.2) f(p + t(x p)) t dt = 1 i n f (p + t(x p)) dt = t f (p + t(x p)) 1 0 x i x i 1 (x i p i ) f x i (p + t(x p)) dt. t 2 f(p + t(x p)) x i t We write the second term in the RHS of (2.1.2) as (2.1.3) 1 t 2 f(p + t(x p)) dt = 1 2 f (p+t(x p))(x j p j ) tdt = 0 x i t 0 x i x j j 1 j n where b j (x) are smooth functions. From (2.1.2) and (2.1.3) we get (2.1.4) 1 0 f (p + t(x p)) dt = f (x) + a i (x), x i x i dt. (x j p j )b ij (x), where a i (x) = j (x j p j )b ij (x) are smooth and vanish at p. Now we get from (2.1.1) and (2.1.4) (2.1.5) f(x) = f(p) + (x i p i )( f (x) + a i (x)). x i 1 n 1 i n Applying D to both sides of (2.1.5) and using the Leibniz rule we get Df p = D(x i p i )( f (p) + a i (x)) = f D(x i ) p (p), x i x i 1 i n

9 INTRODUCTION TO RIEMANNIAN GEOMETRY 9 since D(x i p i ) = D(x i ) follows from the Leibniz rule and linearity of D D(λf) = λd(f) + D(λ)f for any constant function λ R. This completes the proof of Theorem Corollary The tangent space T x R n is spanned by D i = n. x i p. Hence dim T x M n = Exercise Assume that g = (g 1,, g n ) is a smooth map from R m R n. Prove that (Dg) p (D i ) = a ij D j where a ij = ( g j / x i )(p). Hint. Use (Dg)( x i (f(x)) = 1 j n x i (f(g(x)) = f g j x j x i Tangent bundle and cotangent bundle. We define the tangent bundle T M as the disjoint union x M T x M. We shall provide T M with a topology and a smooth structure. Denote by π the projection T M M sending the vector v T p (M) to the point p. Let U be a coordinate neighborhood of M. By Theorem we have a set isomorphism T U τ = U R n such that π 1 (τ(v)) = π(v), where π 1 is the projection to the first factor. The isomorphism τ supplies T U with the product topology. Finally, the open sets on T M are generated by the open sets on T U i, where {U i } is an open covering of M. Example T R n = R n R n, T S 1 = S 1 R. Proposition The space T M has a smooth manifold structure. Proof. By the above T M m has an open covering {T U i = U i R m }, where {U i } is an open covering on M m. We define a coordinate map φ : T U i R m R m by φ i (y, δ) = (φ i (y), (Dφ i ) y (δ)), where φ i : U i U R n is a coordinate map. We have to show that the transition functions φ ij = ( φ 1 j ) ( φ i ) = (φ j φ 1 i, Dφ j Dφ 1 i ) are smooth. Using the chain rule in Exercise?? we reduce the proof of the smoothness of φ ij to the smoothness of the tangent map D(φ j φ 1 i ), which is obvious by the definition of the smooth atlas. We define the cotangent space T x M m at x M m by T x M m := Hom(T M m, R). We define the cotangent bundle T M m as the disjoint union x M mt x M m. This space can be provided with a topology in the same way as we did for the tangent bundle T M m, since we have T U i = U i R m. Exercise Prove that the cotangent bundle T M m is a smooth manifold.

10 Vector fields. A vector field on a smooth manifold M is a smooth section s of the tangent bundle T M, i.e. a smooth map M T M such that π s = Id, where π is the canonical projection T M M. Example Since T n R = Rn R n, a smooth section R n T R n is a smooth map R n (x 1,, x n ) R n ( x 1,, x n ). Thus any smooth vector field X on R n can be written as X(x) = i f i(x) x i, where f i (x) C (R n ). Exercise Show that a vector field X of M is smooth, if and only X(f) is smooth for all f C (M). Hint. Write X in local coordinates. Since any tangent vector in T x0 M m is a derivation of the algebra C (M) at x 0, a vector field X defines a derivation X : C (M) C (M), i.e. a R-linear map satisfying X(fg) = X(f)g + fx(g). Denote the set of all derivations C (M) C (M) by Der(C (M)). We define [X, Y ] := XY Y X. Is is easy to check that [X, Y ] is again a smooth vector field, i.e. for any ψ 1, ψ 2 C (M) we have [X, Y ]ψ 1 ψ 2 = ([X, Y ]ψ 1 )ψ 2 +ψ 1 [X, Y ]ψ 2. Furthermore we note that this bracket turns Der(C (M)) into a Lie algebra, i.e. the following Jacobi identity holds [[X, Y ], Z] + [[Y, Z], X] + [[Z, X], Y ] = 0. The following Proposition shows the naturality of Lie bracket under the smooth map Proposition Let f : M N is a smooth map, and ψc (M). Then for any smooth vector fields X, Y on M we have (2.3.1) [f X, f Y ] = f [X, Y ]. (2.3.2) [X, ψy ] = X(ψ)Y + ψ[x, Y ]. Proof. 1. Let f : C (N) C (M) be defined by ψ ψ f for any ψ C (N). Then f (X)ψ f(x) = X(f ψ) x. Hence we have [f X, f Y ]ψ f(x) = (f Xf Y f Y f X)ψ f(x) = = X(f (f Y ψ)) x Y (f (f Xψ)) x = XY (f ψ) x Y X(f ψ) x = f [X, Y ]ψ. This proves the first assertion of Proposition Let α C (M). We compute [X, ψy ]α = X(ψY α) ψy (Xα) = X(ψ)Y (α) ψ[xy Y X](α) This completes the proof of Proposition Exercise Compute [x x + y y, y x x y]. Exercise Every derivation of C (M) arises from a smooth vector field. Hint. Combine the definition of a tangent vector with Exercise Now we shall find a local normal form of a vector field.

11 INTRODUCTION TO RIEMANNIAN GEOMETRY 11 Theorem (Linearization of a vector field). Let X is a vector field on a smooth n-manifold M. Suppose that X(p) 0. Then there is a local coordinate system (U, φ) around p such that φ(u) = ( ε, ε) ( ε, ε) ( ε, ε) and φ (X) = / x 1. We prove this theorem by using the inverse function theorem, a very important theorem in analysis and in differential geometry. Theorem (Inverse function theorem). Let f be a C k -map from an open domain U R n to R n. Suppose that the tangent map Df p : R n R n is invertible. Then there exists a neighborhood U(p) p in U such that the restriction f to U(p) is bijective on the image f(u(p)), moreover, f 1 f(u(p)) is also a Ck -map. Proof of Theorem Since this is a local result we may assume that M is an open set U R n, p = 0, X never vanishes on U and after a change of coordinate X(0) = / x 1. Then we will use the fibration structure of R n whose base is R n 1 and whose fibers are smooth curves tangent to X to find a local chart (U, φ). We denote by R n 1 the vector space spanned on ( / x 2 (0),, / x n (0)). Now let us consider the ODE system for σ(t, x) : R(t) R n with initial value x R n 1 (2.3.3) with the initial condition dσ(t, x) dt = X(σ(t, x)) (2.3.4) σ(0, x) = x. We know that for any x R n 1 there is an ε > 0 such that this system has a unique smooth solution σ : ( ε, ε) R n. Let G : R n U R n be a map defined by G(t, x) := σ(t, x). The map G is smooth because the solution σ(t, x) depends smoothly on x. We shall show that for a small neighborhood U of 0 the map G 1 is the required diffeomorphism whose tangent map DG 1 sends X to / x 1. We observe that (DG) 0 R n 1 = Id, since G(0, x) = x. By (2.3.3) (DG) (t,x)( / x 1 ) = X(G(t, x)). Since X(0) = / x 1, the tangent map (DG) 0 is an invertible map. Hence G is a local diffeomorphism by inverse function theorem. By (2.3.3) (DG 1 ) y (X) = / x 1 (G 1 (y)). In the proof of Theorem we use the notion of integral curves. An integral curve for a smooth vector field X on a smooth manifold M is a parametrized curve σ : (( ε, ε), 0) (M, p) whose tangent at the point σ(t) is the vector X σ(t) : X σ(t) = σ(t) = Dσ( t t). Example Let M = S 3. For each S 3 (z 0, z 1 ) we consider the vector field X(z 0, z 1 ) = d exp 1t x = ( 1z 0, 1z 1 ). dt t=0 The integral curves of this vector field X are the fiber of the Hopf fibration. As we have seen in the proof of Theorem 2.3.6, the local existence and uniqueness of an integral curve of a smooth vector field X through a point p where X(p) 0 follows from the existence and uniqueness of the ODE (2.3.3) with the initial condition (2.3.4).

12 12 Now we shall examine the global existence of integral curves for a smooth vector field X. Assume that σ j ((a j, b j ), c j ) (M, p), j = 1, 2, are two integral curves through p. After a translation of the parameter for σ 2 we may assume that c 1 = c 2. Then by the uniqueness of the solution of ODE we conclude that σ 1 = σ 2 on (a 1, b 1 ) (a 2, b 2 ). Hence we can extend σ i on (a 1, b 1 ) (a 2, b 2 ). Applying Zorn s lemma we see that a maximal integral curve through a point p always exists. Up to a translation of parameters there are only four possible types of maximal integral curves with the following domains (0, a), (0, ), (, 0), (, ). An integral curve is called complete if its domain is of the last type. A vector field is called complete if all of its integral curves are complete. Proposition A vector field with compact support is complete. Proof. We consider the graph Y = (X, / t) of X on M R. The projection of the integral curve τ(t) := (σ(t), t) is an integral curve of X on M. Clearly τ(t) is complete if and only if σ(t) is complete. If the projection is not complete, then we can assume w.l.g. that it is of type (a, b) with b <. We can assume further that X never vanishes on (a, b). Using the compactness of sppt(x) we find a point y M such that lim tk σ(t k ) = y. Let us consider point (y, b) M R. Applying Theorem we find a flow of Y through (y, b) in some time interval ( ε, +ε). But then the integral curve τ(t) is defined on an interval (a, b + ε), so b is not maximal. So we arrive at a contradiction. Exercise Find integral curves of the following vector fields X on R 2 a) X = x x + y y, b) X = y x x y Lie algebras. Definition Let G be a Lie group. algebra of G. We denote T e G also by g. The tangent space T e G is called the Lie Example T e R = R, T e S 1 = R. Definition A vector field V on a Lie group G is called left invariant, if for any g G the left multiplication L g : G G preserves V : L g V = V. Lemma Any left invariant vector field V on G is defined uniquely by its value at T e G. The Lie bracket on left invariant vector fields induces a Lie bracket on the Lie algebra g. Proof. The first assertion follows from the fact that for any point g G there is unique left multiplication L g sending e to g. The second assertion follows from Proposition Thus T e G = g is a finite-dimensional Lie algebra.

13 INTRODUCTION TO RIEMANNIAN GEOMETRY 13 Example The Lie algebra of Gl(R, n) is the space M at(r, n). For any X Mat(R, n) the vector field X L on G defined by X L (g) := d dt (g exp(tx)) is a leftinvariant vector field. It follows that the Lie bracket is defined by [X, Y ] = X Y Y X, t=0 where is the matrix multiplication of matrices. Exercise Compute the Lie bracket on the Lie algebra of S 3 = {x H x = 1}. 3. Submanifolds and fiber bundles In this section we introduce the notion of submanifolds and fiber bundles. We show that these objects arise from special classes of smooth mappings between smooth manifolds Submanifolds and a regular value of a smooth map. Originally, manifolds were regarded as subsets of Euclidean space. In many case a nice subset can be situated very complicated in a smooth manifold. Let us consider the simplest case, how a curve can be situated in a manifold. D o Let N be a subset in a smooth manifold M m. Definition A subset N n of M m is called a smooth n-dimensional submanifold if for any point p N n there exists a chart (U(p) p, φ p ) on M m such that φ p (U(p) N n ) = φ p (U(p)) (R n R m ) for some linear subspace R n R m. Example Let f : R(x) S 1 (θ 1 ) S 1 (θ 2 ) be defined by f(x) = (xθ 1, α xθ 2 ),

14 14 where α is some constant. The image f(x) is a submanifold in S 1 S 2, if and only if α is a rational number. Remark Let M m be a smooth manifold and N n a submanifold equipped with the induced topology. The restriction of the a chart (U p, φ p ) to N n provides a chart on N n. Thus N is a topological manifold and the induced charts (U(p) N n, φ p (U(p) N n), p N n provide a smooth atlas for N. How to describe a submanifold in a manifold? There are two ways to do it. In the first way we describe a submanifold explicitly (equivalently, parametrically) by giving some smooth map f : N M and we verify that the image f(n) is a smooth submanifold. In the second way we describe a submanifold implicitly, as a solution of a certain system of equations, more precisely as the preimage of a regular value of a smooth map f between smooth manifolds M m and L l, where m l. A point p M m is called a regular value of a smooth map f : M m L l, m l, if for any point x f 1 (p) the differential (tangent map) Df(x) has rank l. A smooth map f : N n M m whose tangent map is a linear embedding at every point p N n is said to be an immersion. An immersion is called an embedding, if it is 1-1 on its image. A map f is called proper, if the preimage f 1 (S) of any compact set is a compact set. The following theorem gives us a criterion for defining a submanifold parametrically. Theorem Let M m and N n be smooth manifolds and let f : N n M m be a proper embedding. Then f(n n ) is a submanifold of M m. Proof. We shall use the immersion condition to show that locally f looks like an embedding R n R n R m n. We shall use the properness and the embeddness to show the existence of a chart on M around a point f(p), p N n, satisfying the condition in the definition of a submanifold. Choose p N n and set q = f(p). We first show that there are coordinate systems (U, φ) and (V, ψ) around p and f(p) respectively such that the composition ψ f φ 1 is a restriction of the coordinate inclusion R n R n R m n, i.e. (3.1.1) π (ψ f φ 1 ) φ(u) = Id, where π is the projection R m R n. This shall imply that ψ(v f(u)) = ψ(v ) R n. After translation we may assume that φ(p) = 0 and ψ(f(p)) = 0. By assumption the derivative (ψ f φ 1 )(0) is an embedding R n R m, so in suitable coordinates on R n and R m we can write D(ψ f φ 1 ) 0 = ( Ir 0 Let D 0 := D(ψ f φ 1 ) 0. Then D 0 is an isomorphism from R n to D 0 (R n ) R m. Hence in the new chart (U, D 0 φ) the composition g = ψ f φ 1 D0 1 is a map from D 0 (φ(u)) R n to R m with g(d 0 (φ(u))) D 0 (R n ). Denote by π the projection from R m to D 0 (R n ). Then π g is a local diffeomorphism of D 0 (φ 0 (U)). Hence, by the inverse function theorem, there exists a local right inverse (ḡ) : D 0 (φ 0 (U 1 )) D 0 (φ 0 (U 1 )) of (π g) for some smaller neighborhood U 1 U of p, i.e. (π g) ḡ = Id. Thus on U 1 with the coordinates φ := ḡ 1 D 0 φ we get ). π (ψ f ( φ) 1 ) φ(u1 ) = π (g ḡ) = Id.

15 INTRODUCTION TO RIEMANNIAN GEOMETRY 15 Clearly φ := π g D 0 φ satisfies (3.1.1). Now f is proper and an embedding, so we may shrink U 1 and V so that f(u 1 ) = f(n n ) V, which implies that ψ(v f(n n )) = ψ f(u 1 ) = ψ(v ) R n, was is required to prove. Exercise Let a point M moves with a constant speed on a ray ON which rotates around the origin O with a constant speed. Write the equation describing the trajectory of M. On which time interval (t 0, t 1 ) R this equation defines a proper embedding from (t 0, t 1 ) to R 2? Now we turn to another very important class of submanifolds described implicitly. Theorem (Implicit function theorem). Let M m and N n be smooth manifold with m n, and let q be a regular value of a smooth map f : M m N n. Then the set f 1 (q) is a smooth submanifold of M m. Proof. Let p M q := f 1 (q). We first show that there are coordinate charts (U, φ) and (V, ψ) around p and q = f(p) such that φ(p) = 0, ψ(q) = 0, and moreover the composition ψ f φ 1 is the restriction of the canonical projection (3.1.2) π : R m = R n R m n R n (3.1.3) (x, y) x. First we can assume that ψ(v ) R n R m. Furthermore, by translation, we can assume the equalities φ(p) = 0 and ψ(q) = 0. Let us denote by D 0 the derivative D(ψ f φ 1 ) 0. Denote by R m n the kernel of D 0. Clearly R m = R m n R n. Applying a linear transformation B on R n we can assume that D 0 is the canonical projection π. Recall that φ 1 (U) is an open set in R n R m with coordinates (x 1,, x n, y 1,, y m ). Let us define a map g : φ 1 (U) ψ(v ) R m n by g(x, y) := (ψ f φ 1 (x, y), y). Then we have Dg (0,0) = Id. So by inverse function theorem there is a local diffeomorphism J from ψ(v ) R m n φ 1 (U) such that g J = Id. It follows that on the smaller open set φ 1 (U 1 ) φ 1 (U) we have ψ f φ 1 J = π g J = π. Clearly the map φ := J φ. Now let us assume that we have a covering U α on M m satisfying (3.1.2) and (3.1.3). Since ψ f φ 1 (φ(u f 1 (q)) = 0, using (3.1.3) we get φ(u f 1 (q)) R m n. Hence φ((u f 1 (q)) φ(u) R m n. Using (3.1.3) again we get φ(u) R m n φ((u f 1 (q)). Hence φ((u f 1 (q)) = φ((u f 1 (q)). Example Let F : R n R is given by F (x) = x, x. Then F x(v) = 2x, v is an immersion, if x 0. Hence the pre-image F 1 (1) = S n is a smooth manifold. Exercise Let us define O n (R) = {A Mat n (R) AA t = Id}. This is the group of orthogonal transformation on R n. Prove that O n (R) is a Lie group. Hint. Denote by Sym n (R) the space of symmetric bilinear forms on R. Show that n i=1 (ei ) is a regular value of a map f : Mat n (R) Sym n (R), f(a) = A A t.

16 Fiber bundles. Fiber bundles arise when we have a projection π from a differentiable manifold E m+n onto a differentiable manifold M m such that π is a differentiable map and all point q M m is a regular value of π. The implicit function theorem implies all the fiber π 1 (p) is a smooth submanifold in E m+n. Thus we can think of a fiber bundle as a family of smooth (sub)manifolds parametrized by the base space. When we require that the special structures on (sub)manifolds change smoothly we need to modify our definition of fiber bundles. Definition Let E be a topological space and π : E B a continuous map. We call the quadruple ξ = (E, F, π, B) a fiber bundle (or fibration), if for each point b B, there is an open set U B containing p such that π 1 (U) is homeomorphic to U F by a homeomorphism φ satisfying the following condition of commutativity diagram π 1 φ (U) U F π proj U π is called the projection, B is the base of the fibration, F is the fiber over B and E is the total space. The pair (U, φ) is called a chart, or a local bundle coordinate system. The map φ is also called a trivialization of E over U. Note that π 1 (b), the fiber over b, denoted by F b is homeomorphic to F for all b B. If the base, fiber and the total space are smooth manifold, π and φ are smooth maps then we have a smooth bundle. If the fiber is a vector space V and the transition bundle map φ j φ 1 i : (U i U j ) V (U i U j ) V restricted to each fiber p V is a linear transformation on V, the bundle F is called a vector bundle. Two bundles E 1 and E 2 over the same base B is called isomorphic, if there is a homeomorphism (diffeomophism) φ between the total spaces such that the following diagram commutes φ E 1 = π 1 π 2 B The map φ is called a bundle isomorphism. If E 1 = E 2 then φ is called a bundle automorphism. The set of all bundle isomorphisms forms a group. This group is called the gauge group of E. Example The simplest example of a bundle is the product bundle E = B F with π being the projection on the first factor. Any bundle which is isomorphic to a product bundle is called a trivial bundle. - Given two bundles (E 1, F 1, π 1, B 1 ) and (E 2, F 2, π 2, B 2 ) we shall define their product bundle by taking product of fibers and bases: (E 1 E 2, F 1 F 2, π 1 π 2, B 1 B 2 ). E 2

17 INTRODUCTION TO RIEMANNIAN GEOMETRY 17 -Tangent bundles and cotangent bundles are vector bundles, since the transition bundle map restricted to each fiber p has the form D(φ 1 j φ i ) or D(φ j φ 1 i ) which are linear maps. - Given two vector bundles (E 1, V 1, π, M) and (E 2, V 2, π, M) with local trivialization φ 1 i and φ2 i over a manifold M we construct the Whitney sum E 1 E 2 - a vector bundle over M with fiber V 1 V 2 by using the bundle transition functions (φ 1 j φ2 j ) (φ1 i φ2 i ) 1. We also construct the Whitney product E 1 E 2 - a vector bundle over M with fiber V 1 V 2 using the bundle transition functions (φ 1 j φ2 j ) (φ1 i φ2 i ) 1. Example (Möbius band). If we twist a paper thin bank in and glue the ends together in the following way then we get a Möbius band. It is a fiber bundle but it is not a trivial bundle, since if it were, then its boundary would consist of two components. It is often important to know, if a given bundle is trivial or not. As examples let us consider the Hopf bundle and the canonical line bundle over a real projective space. Example (Hopf bundle). Let us consider a map π : S 3 S 2 defined by π(e iθ 1 r 1, e iθ 2 r 2 ) = (r 1, e i(θ 2 θ 1 ) r 2 ) R C. It is is to see that π is a projection of S 3 onto S 2 and the fibers π 1 are integral curves in example which are circle. Clearly the Hopf bundle is not a trivial bundle, since S 3 is simply-connected. Example The space RP n of all lines in R n+1 is a differentiable manifold. Let [l] be a line in R n+1 through point (x 0,, x n ), so we write [l] = [x 0 : x n ]. Clearly RP n can be covered by (n + 1) open sets U i defined by the condition x i 0. We define coordinate map φ i : U i R n by φ i ([x 0 : x n ]) = ( x 0 x i,, x n x i ). It is easy to check that {U i, φ i } defines an atlas on RP n. There is a canonical line bundle l(n) over RP n consists of all pair ([l], x) where x is a point in [l]. If l(n) is trivial, then its sub-bundle l (n) = {([l], x x 0)} were trivial,

18 18 which implies that l (n) has two connected component. On the other hand, it is easy to see that l (n) = R 3 \ {0} is connected. Hence l(n) is not a trivial bundle. Exercise Denote by CP n the space of all complex lines [z] in C n+1. Prove that CP n is a differentiable manifold. The canonical line bundle L(n) over CP n consists of all pair ([z], y) where y is a point in the complex line [z]. Prove that L(n) is not a trivial vector bundle. Hint. Using the argument in Example Alternatively we can argue that if L(n) is trivial, then the restriction of L(n) to RP n is trivial which implies that the canonical line bundle over RP n is trivial. 4. Tensors and Riemannian metrics In this section we consider natural vector bundles over a differentiable manifolds and their sections which are called tensors. A Riemannian metric is a special type of tensor, with which we can measure the length of curves on a manifold. We prove the existence of a Riemmanian metric on any differentiable manifold Tensors and Riemannian metrics. Recall that a vector field is a section of the tangent bundle T M. In general, a field on a manifold M is a section s of some vector bundle π : E M, i.e. a smooth map M E such that π s = Id. Clearly fields are generalization of the notion of functions. Definition A tensor ω of type (r, p) on a differentiable manifold M is a section of a natural vector bundle T M r times T M T M p times T M. Clearly the space T (r,p) (M) of tensors of type (r, p) on M is a linear space. The ring C (M) acts on T (r,p) (M) by f(ω)(x) := f(x) ω(x). This action provides T (r,p) (M) with a C (M)-module structure. Let us look at a local expression of a tensor T of type (r, p). On a local coordinate neighborhood U i R n the space T U i r times T U i T U i p times T U i is a direct product U i (R n ) r times (R n ) (R n ) p times (R n ). Thus a tensor T at a point x has a value i 1,,i r,j 1,,j p f i1 j p x i1 dx jp. Example For any function f we define a tensor df of type (0, 1) as follows: df(x) = X (f). We note that any tensor type ω type (0, 1) defines a C -linear map ω : V ect(m) C (M) : ω(v )(x) := (ω(x), V (x)). Exercise Show that a linear function ω : V ect(m) C (M) is a tensor type (0, 1) if and only if for all f C (M) and V V ect(m) we have ω(fv ) = fω(v ). Definition A Riemmannian metric g is a tensor of type (1, 1) on M such that at each point x M the value g(x) is a symmetric positive definite bilinear form on T x M.

19 INTRODUCTION TO RIEMANNIAN GEOMETRY 19 Example Let M be a submanifold in R n. Then the restriction of the Euclidean metric to each tangent space T x M is a positive definite bilinear form. This restriction defines a Riemannian metric on M. - Let f : N n M m be an immersion. If g is a Riemannian metric on M m then f induces a Riemannian metric on N n by f (g)(x, Y ) x = g(df(x), Df(Y )). Exercise Let f : R 2 R 3 is defined by f(x, y) = (x, y, z(x, y)). Find the induced metric f (g), where g is the Euclidean metric. On a coordinate domain U R n a Riemannian metric g can be written as g(x) = g ij (x)dx i dx j. On other coordinate domain y(u) with y : R n R n the same metric g(y) has the following expression g(y(x)) = g ij (y(x))dỹ i dỹ j = g ij (y(x)) y i x k dx k y j x l dx l = g kl (y(x))dx k dx l. Hence ( g ij )(y(x)) = ( x k y i ) x(g kl (y(x)))( x l y j ). Exercise i) Compute the Riemmannian metric dx 2 + dy 2 in polar coordinates (r, φ) with x = r cos φ and y = r sin φ. ii ) Compute the induced Riemmanian metric on the sphere in polar coordinate z = cos θ, x = sin θ cos φ, y = sin θ sin φ. Remark If γ : [0, 1] (M m, g) is a curve on a Riemannian manifold, then we define the length L([γ]) by L([γ]) := 1 0 γ(t) dt. Clearly the length of a curve γ does not depend on a parametrization of γ, i.e. if s : [0, 1] [0, 1] is a monotone smooth function, then a new curve γ(t) := γ(s(t)) has the same length as γ(t), since 1 0 γ(t) dt = 1 0 d 1 ds γ ṡ dt = γ dt. Exercise i) Compute the length a a curve γ(t) : [0, 1] S 3 defined by θ(t) =, φ(t) = π t. π 4 Exercise Let (M, g) be a Riemannian manifold. Show that a linear function V : V ect(m) C is a tensor type (1,0) if and only if for all f C (M) and W V ect(m) we have g(v, fw ) = fg(v, W ). 0

20 The existence of a Riemannian metric. Theorem On each differentiable manifold M m there exists a Riemannian metric g. To prove this theorem we need the fact that every locally compact Hausdorff space M with countable basis is paracompact, i.e. any open covering on M possesses a locally finite refinement, see e.g. Kobayashi-Nomidzu, v.1. (In many textbooks ones requires that a differentiable manifold is a paracompact Hausdorff topological space). This fact implies the existence of partition of unity on a differentiable manifold. This fact is a fundamental Lemma in the theory of differentiable manifolds. Denote by sppt f the support of a function f on a differentiable manifold M, i.e. sppt f := {x M f(x) 0}. Lemma (Partition of unity). Let M be a differentiable manifold, (U α ) alpha A an open covering. Then there exists a partition of unity, subordinate to (U α ). This means that there exists a locally finite refinement (V β ) β B of (U α ) and smooth functions φ β : M R with (i) sppt φ β V β for all β B, (ii) 0 φ β 1 for all x M, β B, (iii) β B φ β(x) = 1 for all x M. Using a partition of unity we shall prove Theorem Proof. Let {(U α, φ α )} be an atlas on M. W.l.g. we can assume that (U α ) is locally finite. We define a Riemannian metric g on M by setting its value at a pair of tangent vectors v, w T p M as follows v, w = α A φ α (p)v i αw i α, where (vα, 1, vα) n and (wα 1, wα) n are representation of Dφ α (v) and Dφ α (w) in R n. It is easy to check that g is well-defined. Remark Let N n be a submanifold in a Riemannian manifold (M m, g). Then the restriction of g to the tangent bundle T N defines a Riemannian metric on N n which we called the induced metric. Thus any smooth submanifold in R n carries a Riemmanian metric which is induced from the Euclidean metric on R n. A celebrated theorem of Nash asserts that any a Riemannian metric on a manifold N n is induced from the Eulidean metric on some space R m via a smooth embedding f : N n R m. A submanifold N n in a Riemannian manifold (M m, g) provided with the induced metric ḡ is called a Riemannian submanifold of (M m, g). Exercise Let N n be a submanifold in a manifold M. Let ω be a tensor on N n. Show that ω can be extended to a tensor on M m. Hint. Unse partition of unity. 5. Levi-Civita connection In this section we introduce the notion of Levi-Civita connection on a Riemannian manifold which is a central notion in Riemannian geometry (a branch of differential geometry which studies Riemannian manifolds). The Levi-Civita connections give us

21 INTRODUCTION TO RIEMANNIAN GEOMETRY 21 tool to analyze global properties of a Riemmannian manifold M by using local invariants on M, since we can compare the tangent spaces at different points on M in a canonical way Linear connections and metric connection. We have observed in exercise??(ii) that for a p M the derivative v f of function f is equal to the speed of the change of f along a curve γ(t) at t = 0, if γ(0) = v. If M = R n we can also take derivative v W of any vector field W on R n by setting v W = d dt t=0w (γ(t)). This formula is possible, since the tangent space T x R n at any point x R n is canonically identified with R n. On a general differentiable manifold there is a no canonical way to identify T x M m and T y M m. Thus we can imagine that there are many ways to define a derivative v W of a vector field W on a differentiable manifold M for a tangent vector v T p M. A linear connection on T M is a method to define such a derivative which we also require to satisfy some additional properties. Denote by X V the (covariant) derivative of a vector field V at a tangent vector X depending on a connection. Then X V must be linear in variable V (5.1.1) X (V 1 + V 2 ) = X (V 1 ) + X (V 2 ). This derivative must also satisfy the Leibniz rule (5.1.2) X (f s) = df(x) s + f X (s). This derivative must be also linear on the variable X i.e. (5.1.3) X+Y (s) = X s + Y s. Now let (M, g) be a Riemannian manifold. We say that a connection is metric, if for all vector fields V, W and for all X T M we have (5.1.4) X g(v, W ) = g( X V, W ) + g(v, X W ). We shall prove that a metric connection on a Riemannian manifold always exists. Moreover, this connection is unique, if we pose a certain condition on. Remark The notion of connection can be defined for any vector bundle E over a smooth manifold M. We require that all conditions (5.1.1), (5.1.2), (5.1.3) holds for any tangent vector X T M and any sections V, W of the vector bundle E The existence and uniqueness of the Levi-Civita connection. For a connection on T M we define its torsion T ( ) C (T M T M) (a tensor of type (0, 2)) by T (V p, W p ) := V W W V [V, W ], for any vector fields V on M with V (p) = V p and W (p) = W p. Exercise Prove that T is well-defined, i.e. its value T (V p, W p ) does not depend on the choice of extensions V and W.

22 22 Hint. Compare with Exercise Show that T (fv, gw ) = fgt (V, W ) for all f, g C (M). In particular we have that T (V, W ) p = 0 if V p = 0 or W p = 0. Theorem On any Riemannian manifold there exists a unique metric torsion free connection. Proof. We set for X, Y, Z V ect(m) (5.2.1) X Y, Z := 1 {X Y, Z Z X, Y + Y Z, X X, [Y, Z] + Z, [X, Y ] + Y, [Z, X] } 2 First we show the uniqueness of a torsion free metric connection, i.e. we have to show that any torsion free metric connection satisfies (5.2.1). Since is metric it should satisfy X Y, Z = X Y, Z + Y, X Z, Using the torsion free condition we get Y Z, X = Y Z, X + Z, Y X, Z X, Y = Z X, Y + X, Z Y. X Y, Z + Y Z, X Z X, Y = 2 X Y, Z [X, Y ], Z + Y, [X, Z] + X, [Y, Z] which yields (5.2.1). Now we will show that (5.2.1) defines a torsion free metric connection, i.e. satisfies (5.1.1), (5.1.2), (5.1.3), 5.1.4) and T ( ) = 0. It will imply the existence of a torsion free metric connection. Then we shall show that any torsion free metric connection satisfies (5.2.1). First we note that for any fixed X, Y V ect(m) the value of X Y defined by RHS of (5.2.1) is a tensor field of type (1,0) on M. Clearly X Y defines a linear map V ect(m) R. By Exercise it suffices to show (5.2.2) X Y, (fz) = f X, Y. A straightforward calculation of the RHS of (5.2.1) gives X Y, (fz) = f X Y, Z + 1 [(Xf) Y, Z + (Y f) X, Z 2 (Xf) Y, Z (Y f) X, Z ] which is equal to the RHS of (5.2.2). Thus X defines a linear map V ect(m) V ect(m), so the first condition (5.1.1) holds. In the same manner we verify that properties (5.1.2) and (5.1.3) hold. So defines a linear connection. To check that is metric, we add the RHS of (5.2.1) associated to X, Y and to X Z, Y. Finally the torsion free condition follows directly from (5.2.1). X Y, Z Y X, Z = [X, Y ], Z

23 INTRODUCTION TO RIEMANNIAN GEOMETRY 23 A torsion free metric connection on a Riemannian manifold M m is called the Levi- Civita connection. Now let us look at the expression of Levi-Civita connection in local coordinate (x i ) with g(x) = g ij (x)dx i dx j. Using the Leibniz rule, it suffices to compute Di D j for i, j, = 1, m. Now we define function Γ k ij (x) as follows Di D j (x) = Γ k ij(x)d k. The functions Γ k ij are called Chritoffel symbol. By definition m Di D j, D l = Γ k ijd k, D l = Γ k ijg kl. Using (5.2.1) we get immediately Di D j, D l = 1 2 (D i D j, D l + D j D l, D i D l D i, D j ) k=1 Thus, letting g kl := (g) 1 kl, we have = 1 2 ( g jl x i + g li x j g ij x l ). (5.2.3) Γ k ij = 1 m 2 gkl ( g jl + g li g ij ). x i x j x l l= Parallel transport and geodesics. Let (M, g) be a Riemannian manifold, its Levi-Civita connection and γ(t) : ( ε, ε) M be an embedded curve. Note that the restriction of T M to γ(t) defines a vector bundle over submanifold γ(t) t ( ε, ε). Lemma For a section V (t) : γ(t) T M let Ṽ be a extension of V to M. We define for any t a linear map γ(t) : Γ(T M γ(t) ) Γ(T M γ(t) ) as follows γ(t) V (t) := γ(t) Ṽ. This linear map is well-defined. It is a connected on the vector bundle T M γ(t). From Lemma we get immediately Corollary The covariant derivative γ(t) V for any vector field V on M depends only on the restriction of V to γ(t). Proof of Lemma It suffices to show that if V (t) is a zero vector field, then γ(t) Ṽ = 0. Choose a local coordinate at γ(t). W.l.g. we assume that γ(t) = x 1. Using the theorem on linearization of a vector field we can assume that γ(t) = (t, 0,, 0) for a sufficient small t. Let Ṽ = f i(x)d i. Since V γ(t) = 0 we have f i (t, 0,, 0) = 0. Now we compute D1 V (t) 0 = D1 (f i D i ) 0 = f i ( D1 D i ) 0 + D 1 (f i )D i = 0.