Lie derivatives, Lie brackets and vector fields

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1 Lie derivatives, Lie brackets and vector fields over curves Gonzalo E. Reyes Université de Montréal 27 October 2006 Although Lie derivatives can be defined for tensor fields quite generally, we shall limit ourselves to define this notion for vector fields and vector fields over curves. 0.1 Lie derivative of a vector field along a curve with respect to a vector field Let M be a (microlinear) space, Q : M M D a vector field, γ : R M an integral curve of Q and W : R M D a vector field along γ. We define L Q W the Lie derivative of W with respect to Q to be the new vector field along γ characterized by the property hl Q W (t) = Q D h(w (t + h)) W (t) = Q h W (t + h) W (t) for every h D. The fact that this equation characterizes L Q W is a consequence of Kock-Lawvere for the fiber M γ(t). Proposition 0.1 The following hold 1. L Q (W 1 + W 2 ) = L Q (W 1 ) + L Q (W 2 ) 2. L Q (fw ) = f W + fl Q W, whenever f : R R. Proof: This is a consequence of the following 1

2 Lemma 0.2 The map Q D d : M γ(t) M γ(t+d) is R-linear. Proof: Indeed, Q D d (v)(h) = Q(d, v(h)). Hence, Q D d (λv)(h) = Q(d, (λv)(h)) = Q(d, v(λh)) = = Q D d (v)(λh) λq D d (v)(h) The first is immediate by linearity of Q D d. The second is proved as in elementary calculus: hl Q (fw )(t) = Q D h(f(t + h)w (t + h)) f(t)w (t) = Q D h(f(t + h)w (t + h)) Q D h(f(t)w (t + h)) + Q D h(f(t)w (t + h)) f(t)w (t) Since f(t + h) = f(t) + hf (t), linearity of Q D h allows us to rewrite hl Q (fw )(t) = hf (t)q D h(w (t + h)) + f(t)[q D h(w (t + h)) W (t)] = hf (t)q D h(w (t + h)) + hf(t)l Q W (t) = hf (t)w (t) + hf(t)l Q W (t)) To justify the last equality, it is enough to check that hq D h(w (t + h)) = hw (t). But this is clear: indeed, define φ(h) = Q D h(w (t + h)) M γ(t) for all h D. Thus, φ Mγ(t) D. Since the Kock-Lawvere axiom holds for the fiber M γ(t), φ(h) = A + Bh for all h D, where A, B M γ(t). But A = φ(0) = W (t). Therefore φ(h) = W (t) + Bh and this implies that hφ(h) = hw (t), i.e. hq D h(w (t + h)) = hw (t). This justifies the last equality and finishes the proof, by cancellation of the universally quantified h. An alternative proof proceeds by showing the (local) existence of a W with W γ = W (cf. Axiomsfieldequations(final).pdf, proposition 3.7) and then using the corresponding result for [Q, W ]. Define W d : R M by the prescription W d (t) = W (t)(d) for d D. Proposition 0.3 The following are equivalent: 1. Every W d is an integral curve of Q 2. L Q W = 0 2

3 Proof: A simple computation: (L Q W )(t) = 0 iff h D Q D h(w (t + h)) = W (t) iff h D d D Q( h, W (t + h)(d)) = W (t)(d) iff h D d D Q( h, W d (t + h)) = W d (t) Since for all t (L Q W )(t) = 0, we may substitute first h by h and then t by t+h to obtain the equivalence between the RHS of last line and Q(h, W d (t)) = W d (t + h) (for all h, d D and all t,) i.e., for each d, W d is an integral curve of Q. Notice that W is completely determined by the family {W d } d and viceversa and both notions will be used interchangeably. 0.2 Lie derivative of a vector field relative to a vector field Let X, Y : M M D be vector fields. We define the Lie derivative of Y with respect to X by the formula h(l X Y ) x = (X h ) D (Y Xh (x)) Y x (Notice the abuse of language here: X h (x) = X x (h)) Proposition L X (Y 1 + Y 2 ) = L X (Y 1 ) + L X (Y 2 ) 2. L X (fy ) = fl X Y + X(f)Y Proof: Linearity of (X h ) D. The first is immediate. As for the second, h(l X fy ) x = (X h ) D ((fy ) Xh (x)) (fy ) x = (X h ) D (f(x h (x))y Xh (x)) f(x)y x = (X h ) D ((f(x) + hx(f)(x))y Xh (x) f(x)y x = f(x)(x h ) D (Y Xh (x)) + hx(f)(x)x h(y D Xh (x)) f(x)y x = f(x)[(x h ) D (Y Xh (x))) Y x ] + hx(f)(x)y Xh (x) = hf(x)(l X Y ) x + hx(f)(x)y x The last equality follows from Y Xh (x) = Y x + terms in h This Lie derivative is not new, since we knew it under another guise: 3

4 Proposition 0.5 L X Y = [X, Y ] Proof: (Lavendhomme page 237) By unravelling the definition h(l X Y ) x (d) = [(X h ) (Y Xh (x)) Y x ](d) = X h Y d X h (x) Y d (x) Define l h (d, d ) = (X h Y d X h Y d ) x Then { lh (d, 0) = (X h Y d X h ) x l h (0, d) = (Y d ) x Thus, recalling the definition of addition on fibers, h(l X Y ) x (d) = l h (d, d) = (X h Y d X h Y d ) x = [Y, X] x (h( d)) = [X, Y ] x (hd) = h[x, Y ] x (d) (For the properties of [. ] see curvaturetorsion.pdf, page 9) The Lie bracket and the Lie derivative are connected as follows: Proposition 0.6 If Q X (M), γ : R M an integral curve of Q, W : R M D a vector field along γ and W X (M) is such that W γ = W, then L Q W = [Q, W ] γ Proof: This is practically a repetition of the proof that L X Y = [X, Y ]. Recall that L Q W is (uniquely) characterized by the equation hl Q W (t) = Q D h(w (t + h)) W (t) for all h D. A simple computation allows us to write h(l Q W )(t)) d = (Q h W (t + h) W (t)) d = (Q h W d Q h )(γ(t)) W d (γ(t) In fact, by hypothesis, W (t) = W γ and Q h (γ(t)) = γ(t + h), since γ is an integral curve of Q. 4

5 Define l h (d, d ) = (Q h W d Q h W d )(γ(t)). Then { lh (d, 0) = (Q h W d )(γ(t)) l h (0, d) = W d (γ(t)) = W d (γ(t)) Therefore, by definition of the addition on fibers h(l Q W )(t)(d) = (Q h W d Q h )(γ(t)) W d (γ(t) = l h (d, d) = (Q h W d Q h W d )(γ(t)) = [ W, Q] γ(t) (h.( d)) = [Q, W ] γ(t) (hd) = h([q, W ] γ)(t)(d) Corollary 0.7 Let γ : R M be an integral curve of Q. Assume, furthermore, that W X (M) is such that W = W γ. Then the following are equivalent 1. Every γ h is an integral curve of Q 2. L Q W = 0 3. [Q, W ] γ = Computations for M = R n Let W : R M D be a vector field over γ : R M and M = R n. Then M D = M R n and W is completely determined by n functions f i : R R by the equation (Kock-Lawvere) W (t) = [d γ(t) + d(f 1 (t),..., f n (t))] Alternatively, this equation may be re-written as n W (t) = f i (t)( / x i ) γ(t) with ( / x i ) x = [d x + d(0,..., 1,..., 0)] (the 1 in the i th place). Notice that the sequence {( / 1 ) x.... ( / n ) x } form a basis of M x. 5

6 Proposition 0.8 If then { X = n a i / x i Y b j / x i n n L X Y = [a i ( b j / x i ) / x j b j ( a i / x j ) / x i ] j=1 Proof: Straightforward computation from the properties of L X On the other hand, ( ) L X Y L X (b i / x i ) [X(b i ) / x i + b i L X / x i ] L X / x i = [ / x i, X] = n j=1 L / x i(a j / x j ) = n j=1 ( a j / x i ) / x j + a j [ / x i, / x j ] = n j=1 ( a j / x i ) / x j since [ / x i, / x j ] = 0. Finally, X(b i ) j=1 a j / x j (b i ) j=1 a j b i / x j Substituting in ( ), we obtain L X Y [ n j=1 a j ( b i / x j ) / x i b i n j=1 ( a j / x i ) / x j ] a j ( b i / x j ) / x i n b i n j=1 ( a j / x i ) / x j a j ( b i / x j ) / x i n b i ( a j / x i ) / x j [a j ( b i / x j ) / x i b i ( a j / x i ) / x j ] By interchanging i and j we obtain the desired result: n n L X Y = [a i ( b j / x i ) / x j b j ( a i / x j ) / x i ] j=1 By further interchanging i and j in the second double sum we obtain L X Y [a i ( b j / x i ) / x j b i ( a j / x i ) / x j ] j=1 [ n a i ( b j / x i ) b i ( a j / x i )] / x j which is the formula given by Spivak I, page 212. In particular, the j th -component of L X Y is n (L X Y ) j = [a i ( b j / x i ) b i ( a j / x i )] 6