THEODORE VORONOV DIFFERENTIABLE MANIFOLDS. Fall Last updated: November 26, (Under construction.)

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1 4 Vector fields Last updated: November 26, (Under construction.) 4.1 Tangent vectors as derivations After we have introduced topological notions, we can come back to analysis on manifolds. Let M = M n be a manifold. Consider a vector v T x0 M. Suppose γ : ( ε, ε) M is a curve such that v = dx dt at t = 0. Definition 4.1. For an arbitrary function f : M R the number v f := d dt f(x(t)). (the derivative at t = 0) is called the derivative of f along v. Comparing it with the definition of the differential, we see that v f = df(x)(v). If x 1,..., x n are coordinates near x and v i, i = 1,..., n are the components of v, we have n f v f = x (x) i vi. i=1 Proposition 4.1. The operation v : C (M) R satisfies the following properties: linearity over R and the Leibniz rule Proof. Immediate. v (fg) = v f g(x) + f(x) v g. Note that the map sending a function f C (M) to the number f(x) R is a homomorphism, called the evaluation homomorphism at x. We denote it ev x. There are a fundamental algebraic notions. 1

2 Definition 4.2. For a given homomorphism of algebras α: A 1 linear map D : A 1 A 2 is called an derivation over α if A 2, a D(ab) = D(a) α(b) + α(a) D(b) for all a, b A 1. In the special case of a single algebra A 1 = A 2 = A and α = id (the identity map), a derivation over it is simply called a derivation of the algebra A. Hence for any v T x0 M the operation v : C (M) R is a derivation over the evaluation homomorphism at x 0 M. Remark 4.1. It is possible to consider v on C (V ) for any open set V M s.t. x 0 V, and it is a derivation C (V ) R over ev x0 as well. It turns out that all the derivations of the algebra of functions on a manifold to numbers are of the form v. Let us first explore the case of R n. Theorem 4.1. Let x 0 R n. For an arbitrary derivation D : C (R n ) R over the evaluation homomorphism ev x0 there is a vector v T x0 R n such that D = v. The proof uses the following simple but fundamental statement: Lemma 4.1 (Hadamard s Lemma). For any smooth function f C (R n ) and any x 0 R n there is an expansion f(x) = f(x 0 ) + n (x i x i 0)g i (x) i=1 where g i C (R n ) are some smooth functions. Proof. Consider the segment joining x and x 0 and write f(x) = f(x 0 ) d dt f(x 0 + t(x x 0 )) dt = f(x 0 ) + (x i x i 0) 1 0 f x i (x 0 + t(x x 0 )) dt 2

3 Corollary 4.1. There is an expansion f(x) = f(x 0 ) + n i=1 (x i x i 0) f x i (x 0) + n (x i x i 0)(x j x j 0)g ij (x) (1) i,j=1 where g ij C (R n ). Proof. By iterating the previous expansion we arrive at f(x) = f(x 0 ) + n n (x i x i 0)a i + (x i x i 0)(x j x j 0) g ij (x) i=1 i,j=1 where a i R are numbers and g ij C (R n ), functions. Apply partial derivative at x x i 0 and obtain a i = f (x x i 0 ). Now we can prove the main theorem. Proof of Theorem 4.1. Let D : C (R n ). Consider a point x 0 R n. (We are keeping x as a running point.) Apply D to the expansion (1). First note that derivations kill constants; indeed, D(1) = D(1 1) = D(1) 1+1 D(1) = 2D(1), hence D(1) = 0 and then D(c) = D(c 1) = 0 for any c R. Therefore we obtain D(f) = D(x i ) f x (x 0)+ i i ( ) D(x i )(x j 0 x j 0) g ij (x 0 )+(x i 0 x i 0)D(x j ) g ij (x 0 )+(x i x i 0)(x j x j 0) D(g ij ) = ij for v = (v 1,..., v n ) where v i = D(x i ). i D(x i ) f x i (x 0) = v f Theorem 4.1 can be immediately transferred to open domains of a manifold U M admitting a single coordinate system: any derivation D : C (U) R ev x0, where x 0 U M, is of the form D = v for some v T x0 M if there is a chart φ: V U, where V R n. 3

4 This can be made slightly more formal by introducing the so-called germs of functions at a given point. Consider functions defined on open neighborhoods U M of x 0. A function f defined on U and a function g defined on U are said to be equivalent if there is an open neighborhood U of x 0 such that f U = g U. The equivalence class of a (local) function f defined near x 0 is called its germ. Germs at x make an algebra, notation: F x ; and there an evaluation homomorphism ev x to R. We get Theorem 4.2. The derivations F R over ε = ev x correspondence with the tangent vectors v T x M. are in one-to-one It is common to identify vectors v with the corresponding derivations v. For example, the coordinate basis vectors e i = x are identified with the x i partial derivatives i =. x i Now what about global functions, i.e., the algebra C (M)? Theorem 4.3. Every derivation C (M) R over ev x0 a tangent vector v T x0 M. has the form v for Proof. Note that if a function vanishes on an open neighborhood U of x 0, it belongs to I 2 x 0. Indeed, consider h such that h = 0 on W U and h = 1 on M U. Then hf = f. Note that both f and h vanish at x 0, and f is the product of such functions. Therefore every derivation C (M) R over ev x0 annihilates h. It follows that all such derivations can be restricted to local functions (defined near x 0 ) or to germs at x 0, because the results of their application to global functions depend only on the behaviour of functions near x 0. Now we apply the previous Theorem. 4.2 Commutator of vector fields Everything what is said below applies to vector fields defined not necessarily on the whole manifold M, but on an open subset U M. We shall speak of vector fields on M for the simplicity of notation. Recall that a tangent vector v at a point x M defines a linear map v : C (M) R satisfying v (fg) = v f g(x) + f(x) v g 4

5 for all f, g C (M), and is frequently identified with this map. (Such linear maps C (M) R are called derivations at x.) Shortly: vectors at a given point are identified with derivations at this point. Now, if we consider a vector field instead of a single vector, that means that we allow x in the formula above to vary. For a vector field v X(M), we arrive at the linear map v : C (M) C (M) of the algebra C (M) to itself satisfying v (fg) = v f g + f v g. Such linear operators on algebras are known as derivations of algebras. Hence every vector field on M defines a derivation of the algebra C (M). Written in local coordinates, v = v i (x) x i if v = v i (x) x x. i Very frequently vector fields and the corresponding derivations are identified: v v, and one writes v = v i (x) x. i There is a convenient abbreviation i =, so one can write v = v i x i i. It is an important fact that not only we can identify vector fields on M with the corresponding derivations of the algebra of functions C (M), but all the derivations of it arise in this way: every derivation D of the algebra C (M) has the form D = v for some vector field v X(M). Recall the following notion: for two linear operators A and B on a vector space V, their commutator is denoted by [A, B] and defined as [A, B] = AB BA. (Here the product of operators is the composition: AB = A B.) Theorem 4.4. The commutator of vector fields considered as linear operators on the algebra of functions is again a vector field. Proof. Calculation in coordinates. Suppose we are given X, Y X(M) and in some local coordinates X = X i i, Y = Y i i. For an arbitrary function f we have X(Y f) = X i i (Y j j f) = X i i Y j j f + X i Y j 2 ijf 5

6 (where we identify vector fields with operators on functions). Here 2 ijf = 2 f x i x j. In the same way we have Y (Xf) = Y i i X j j f + Y i X j 2 ijf. We can rename indices in the second term to get Y (Xf) = Y i i X j j f + Y j X i 2 jif = Y i i X j j f + Y j X i 2 ijf, where we have used the commutativity of the second partial derivatives. Therefore we obtain (XY Y X)f = X(Y f) Y (Xf) = X i i Y j j f Y i i X j j f = (X i i Y j Y i i X j ) j f. We conclude that the commutator of vector fields X and Y is indeed a vector field given by ) [X, Y ] = (X i i Y j Y i i X j j (2) in local coordinates. We have obtained a binary operation on vector fields, called their commutator or (sometimes) the Lie bracket. An explicit formula is coordinates is equation (2). The following form of this formula is convenient for practical calculations: [X, Y ] = X(Y j ) j Y (X j ) j. (3) (It has the appearance of taking the derivative of the vector field Y along the vector field X minus the same with X and Y swapped. Here taking the derivative of one vector field along the other means taking the derivative of the components. Unfortunately, this makes sense only in a fixed coordinate system. However, the difference, which is the commutator, makes good sense independent of coordinates.) Remark 4.2. As one can see from the calculation, the product (composition) XY of two vector fields X, Y as operators on functions is no longer a vector field. (It is a differential operator of the second order.) Hence the significance of the fact that the commutator [X, Y ] is a vector field. Theorem 4.5. The commutator of vector fields has the following properties: 6

7 1. bilinearity over numbers: [cx, Y ] = c[x, Y ], [X + Y, Z] = [X, Z] + [Y, Z], (and the same w.r.t. the second argument); 2. antisymmetry: [X, Y ] = [Y, X] ; 3. Jacobi identity: for all X, Y, Z X(M) and c R. [X, [Y, Z]] + [Z, [X, Y ]] + [Y, [Z, X]] = 0, Proof. Bilinearity and antisymmetry are obvious from the definition. Let us check the Jacobi identity. By expanding the commutators we have [X, [Y, Z]] + [Z, [X, Y ]] + [Y, [Z, X]] = as claimed. X(Y Z ZY ) (Y Z ZY )X + Z(XY Y X) (XY Y X)Z + Y (ZX XZ) (ZX XZ)Y = XY Z XZY Y ZX + ZY X + ZXY ZY X XY Z + Y XZ + Y ZX Y XZ ZXY + XZY = 0 Definition 4.3. A vector space endowed with a bilinear antisymmetric operator satisfying the Jacobi identity is called a Lie algebra. Therefore the space X(M) with the operation of commutator is a Lie algebra. Lie algebras play fundamental role in many areas of mathematics. Typical examples of Lie algebras, besides the Lie algebras of vector fields X(M), include various matrix Lie algebras, where the operation is the matrix commutator. See examples below. Remark 4.3. Vector fields can also be multiplied by functions, but the commutator of vector fields is not bilinear w.r.t. this multiplication (one cannot take functions out). Check that [X, fy ] = f[x, Y ] + (Xf) Y, where Xf = X f = X i i f is the extra term. 7

8 Our proof of Theorem 4.4 was based on a direct coordinate calculation. A more elucidating approach is possible. For an arbitrary algebra A consider the space of all derivations of A, i.e., the linear operators D : A A satisfying the Leibniz identity D(ab) = D(a) b + a D(b) for all elements a, b A. It is denoted Der A. One can consider commutators of derivations (as linear operators). Theorem 4.6. The commutator of derivations is a derivation. Therefore the space Der A is a Lie algebra w.r.t. the commutator. Proof. Consider two derivations D 1 and D 2. We have D 1 (D 2 (ab)) = D 1 ( D2 (a) b + a D 2 (b) ) = D 1 (D 2 (a)) b + D 2 (a) D 1 (b)+ and similarly D 1 (a) D 2 (b) + a D 1 (D 2 (b)) D 2 (D 1 (ab)) = D 2 (D 1 (a)) b + D 1 (a) D 2 (b) + D 2 (a) D 1 (b) + a D 2 (D 1 (b)). After subtracting, the cross terms such as D 1 (a) D 2 (b) are cancelled and we arrive at [D 1, D 2 ](ab) = D 1 (D 2 (ab)) D 2 (D 1 (ab)) = D 1 (D 2 (a)) b D 2 (D 1 (a)) b + a D 1 (D 2 (b)) a D 2 (D 1 (b)) = Hence [D 1, D 2 ] is a derivation as claimed. [D 1, D 2 ](a) b + a [D 1, D 2 ](b). Now we see that Theorem 4.4 follows from Theorem 4.6 if we identify the space X(M) with Der A for the algebra A = C (M). Example 4.1. Check that the following subspaces of the space of all n n matrices are closed under the commutator: the space of all antisymmetric matrices; the space of all trace-free matrices; 8

9 (for matrices with complex entries) the space of all anti-hermitian matrices, i.e., satisfying A T = A; the space of all upper-triangular matrices (unlike the previous examples, it is already closed under the matrix product). Therefore all these spaces give examples of Lie algebras. Example 4.2. The ordinary Euclidean three-space becomes a Lie algebra w.r.t. the operation of the cross-product (or vector product) of vectors: (u, v) u v, defined, e.g., via a determinant. It is not easy to establish the Jacobi identity directly. However, one can check that the following linear transformation 0 u 3 u 2 u = (u 1, u 2, u 3 ) u 0 u 1 u 2 u 1 0 maps the vector product on R 3 to the commutator of matrices (it is sufficient to check this for the standard basis vectors i, j, k), and the Jacobi identity follows. The Jacobi identity for the matrix Lie algebras is checked in the same way as we did for vector fields. Since the check uses nothing but the associativity of the composition, it carries through to arbitrary associative algebras: each such an algebra gives rise to a Lie algebra by setting [a, b] = ab ba. 4.3 Flow of a vector field Vector fields on a manifold have the following geometrical interpretation. We have a manifold (or its open subset) and to each point of it a tangent vector is attached. If we imagine our manifold as sitting in some large R N, we have a picture of tangent arrows at each point. Interpreting a tangent vector as the velocity of a curve, we have a flow of some fluid flowing on our manifold so that the velocity of the flow at each point is given by the vector attached to that point. (In particular, this is a stationary flow, in the sense that the velocities of particles traveling through any given point are the same and do not depend on the time, so the velocity is a function of a point only.) Such a hydrodynamic interpretation of vector fields as the velocity fields is very 9

10 important. Now we shall elaborate it and in particular define the flow of a vector field as a precise mathematical notion. If u = u(x) is a vector field on M, we can associate with it the following ordinary differential equation on M, which will become a system of ODEs when written in coordinates: dx dt = u(x). (4) Here x = x(t) and the parameter t (the time) runs over some interval, e.g., t ( ε, ε). In coordinates, dx i dt = ui (x), (5) where i = 1,..., n. This is a system of non-linear (in general) ordinary differential equations, with the RHS not depending on the time explicitly. (Such systems are called autonomous.) We shall make use of the two main facts concerning such systems. Firstly, if an initial value x(t 0 ) = x 0 is fixed for some moment of time t 0, there is a unique solution x = x(t) with this initial value for t in some interval around t = t 0 ( the existence and uniqueness theorem for solutions of ODEs ). The solutions x = x(t) are called the trajectories or the integral curves of the vector field u. Secondly, for this unique solution of our ODE with a given initial value x(t 0 ) = x 0, it depends smoothly on x 0 M ( the smooth dependence on initial value theorem ). It follows that we have a well-defined smooth map g t : M M, g t : x 0 g t (x 0 ) = x(t), for each t, where x(t) is the solution of (4) with the initial value x(0) = x 0. Here t belongs to some interval around zero, depending, in principle, on x 0. To avoid complicated notation we shall neglect this fact and write all the formulas as if t can take any value. The uniqueness of solution implies that the family of maps g t : M M has the following properties: g 0 = id (6) (indeed, this simply restates that g 0 (x 0 ) = x(0) = x 0 for any x 0 M taken as the initial value for x(t) at t = 0); g t = g 1 t (7) 10

11 (indeed, this simply says that if we travel along a trajectory from any given point backward in time for the time interval t and then take the result as an initial value and travel in the forward direction for the same t, we shall return to the original point; and the same if we do it other way round: first moving forward and then, back); g t+s = g t g s (8) (this means that, starting from any point, if we travel along a trajectory for the time interval s and then, t, it is the same as to travel for the interval t + s; again follows from the uniqueness of solution). Any family of transformations of a manifold M satisfying (6), (7), (8), which are automatically invertible due to (7), is called a one-parameter group of transformations (or diffeomorphisms) of M or, shortly, a flow on M. We see that any vector field on M gives rise to a flow, which is called the flow of a vector field (or: generated by a vector field), with the vector field called as the generator of a flow. Finding the flow for a given vector field u is the same as solving equation (4) for all initial values. Example 4.3. Let M = R n and u(x) = a (a constant vector). Then (4) will be dx dt = a, which has the general solution x = at + x 0, where x 0 = x(0). Hence the flow g t consists of parallel shifts of all points in the direction of the constant vector a: g t : x x + at for any t R. The trajectories are the straight lines parallel to a. Example 4.4. Let M = R 2 and suppose a vector field X is given in Cartesian coordinates as X = ye x + xe y. To find its flow we have to solve the system { x = y y = x (the time derivative denoted by the dot). This can be written in the matrix form as ( ) ( ) ( ) d x 0 1 x =. dt y 1 0 y 11

12 The solution is given by the matrix exponential: where A = x = e At x 0 ( ) Since in our case A 2 = E, A 3 = A, A 4 = E, etc. (E is the identity matrix), we have ( ) cos t sin t e At =. sin t cos t Therefore the flow g t : R 2 R 2 is ( ) ( ) ( ) x cos t sin t x g t :, y sin t cos t y i.e., is the rotations around the origin through angle t. The trajectories are the circles with the center at the origin, and the origin itself (the whole trajectory is one point). 12

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