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By using the definition

$$\tag{1} \mathcal{L}_V U=\lim_{t\to 0}\frac{U_{\phi_t x}-\phi_{t*}U_x}{t} $$

For the Lie derivative of the vector field $U$ in the direction of $V$, where $\phi_t$ is the flow along $V$ and $\phi_*$ is the push-forward. I'd like to show that $\mathcal{L}_V U=[V,U]$ by using the components of $U$ and $V$.

Let $U_x=u^i(x)\frac{\partial}{\partial x^i}$, $V_x=v^i(x)\frac{\partial}{\partial x^i}$ be vector fields. Summation convention is used. We're on a smooth manifold in local co-ordinates $x^i$. We need:

A. $U_{\phi_t x}$ which is $U$ evaluated at the point $\phi_t x$. If we take the derivative at $x=x_a$, and define $\phi_t x_a=x_b$ then $x^i_b=x^i_a+t v^i(x_a)$ for sufficiently small $t$. Thus

$$\tag{2} U_{\phi_t x_a}=u^i(x)\frac{\partial}{\partial x^i}\bigg|_{x=x_b=x_a+t v} $$

B. $\phi_{t*}U_x$ which is (in this case) $U_{x_a}$ under the co-ordinate transformation $x^i_b=x^i_a+t v^i(x_a)$. Since $\frac{\partial}{\partial x_a^i}=\frac{\partial x_b^j}{\partial x_a^i}\frac{\partial}{\partial x_b^j}=\left(\delta_i^j+t\frac{\partial v^j}{\partial x^i_a}\right)\frac{\partial}{\partial x^j_b}$ (where $\delta$ is the Kronecker delta) we have

$$\tag{3} \phi_{t*}U_{x_a}=u^i(x_a)\frac{\partial}{\partial x_b^i}+tu^i(x_a) \frac{\partial v^j(x_a)}{\partial x^i_a}\frac{\partial}{\partial x^j_b} $$

C. To find the order $t$ difference between eq. (2) and eq. (3). The difference is

$$\tag{4} U_{\phi_t x_a}-\phi_{t*}U_{x_a}=u^i(x_b)\frac{\partial}{\partial x_b^i}-u^i(x_a)\frac{\partial}{\partial x_b^i}-tu^i(x_a) \frac{\partial v^j(x_a)}{\partial x^i_a}\frac{\partial}{\partial x^j_b} $$

Expanding the first term on the RHS around $t=0$

$$\tag{5} u^i(x_b)=u^i(x_a)+tv^j(x_a)\frac{\partial u^i(x_a)}{\partial x_a^j}+\mathcal{O}(t^2) $$

Substituting eq. (5) into eq. (4) yields

$$\tag{6} t^{-1}\left(U_{\phi_t x_a}-\phi_{t*}U_{x_a}\right)=v^j(x_a)\frac{\partial u^i(x_a)}{\partial x^j_a} \frac{\partial}{\partial x_b^i}-u^i(x_a)\frac{\partial v^j(x_a)}{\partial x_a^i}\frac{\partial}{\partial x_b^j} \stackrel{t\to 0}{=}[V,U]_{x_a} $$

Which appears to be the correct result. Question: do I make any conceptual (esp. part A or B) or algebraic mistakes?

I have seen a few slick derivations of 'Lie derivative equals commutator', and have looked at many similar questions on this site, but have not seen one in the spirit of the above (I admit it's not pretty), which is why I'd like to check for errors.

Sal
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  • Hmm interesting, it feels like the whole site just started asking about lie derivative voernight – Clemens Bartholdy Aug 19 '22 at 22:54
  • Looks conceptually right to me. Also you can see this – Clemens Bartholdy Aug 19 '22 at 22:55
  • @Beautifullyirrational Thank you – Sal Aug 19 '22 at 23:31
  • $x_b = \phi_t(x_a) = x_a + tV(x_a)$ is not true, even for $t$ small. It is true that $\phi_t(x_a) = x_a + tv(x_a) + O(t^2)$ by definition of derivative. – Mason Aug 20 '22 at 02:18
  • @Mason the second expression in your comment is what I use in OP (cf. above eq. (2)), nowhere do I assert the first – Sal Aug 20 '22 at 21:12
  • @Sal You assert the equality here: "If we take the derivative at $x=x_a$, and define $\phi_t x_a=x_b$ then $x^i_b=x^i_a+t v^i(x_a)$ for sufficiently small $t$." – Mason Aug 20 '22 at 21:17
  • @Mason I do not follow. What you quote is the expression which you claimed was true, if we understand the words: 'for sufficiently small $t$' to mean at $\mathcal{O}(t^2)$ – Sal Aug 20 '22 at 21:34
  • @Mason Maybe you object to my words: "if we take the derivative at $x_a$"? If so; yes, those were poorly chosen. I meant: "if we want to take the Lie derivative at $x_a$" – Sal Aug 20 '22 at 21:45
  • @Sal You cannot write $x^i_b=x^i_a+t v^i(x_a)$. sufficiently small $t$ means that there exists $\delta > 0$ such that it holds for all $t$ with $|t| < \delta$. Your analysis is probably correct though since you only used the $O(t^2)$ equality. – Mason Aug 20 '22 at 22:48
  • I don't follow the $\epsilon$-$\delta$ argument. It seems like you are saying that $\lim_{t \to 0} \phi_t x_a$ does not exist. Conversely, accepting that $\lim_{t \to 0} \phi_t x_a =x_a$ suggests that I will be able to find such a $\delta$ – Sal Aug 20 '22 at 23:35

1 Answers1

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You can do this simply without the use of indices. By definition, \begin{eqnarray*} L_V U (x) & = & \frac{d}{d t} (\varphi_t^{\ast} U (x)) |_{t = 0} . \end{eqnarray*} We have \begin{eqnarray*} \varphi_t^{\ast} U (x) & = & D \varphi_t (x)^{- 1} U (\varphi_t (x))\\ & = & D \varphi_{- t} (\varphi_t (x)) U (\varphi_t (x))\\ & = & D \varphi (- t, \varphi (t, x)) U (\varphi (t, x)) . \end{eqnarray*} Now first take the derivative with respect to $t$ by using the product rule and chain rule, then set $t = 0$. Using the definition $\frac{\partial \varphi}{\partial t} (t, x) = V (\varphi (t, x))$ and $\varphi(0, x) = x$, you will arrive at \begin{eqnarray*} L_V U (x) & = & - D V (x) U (x) + D U (x) V (x) . \end{eqnarray*} The right hand side is the coordinate formula for $[V, U] (x)$.

Mason
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  • Valid points but I don't think this was OP's issue, they wanted verification if what they did was right – Clemens Bartholdy Aug 19 '22 at 23:06
  • As I say in OP, I'm asking specifically about the horrible derivation using components. I know the expression may be derived by more elegant means – Sal Aug 19 '22 at 23:17
  • @Sal This is the "horrible" derivation using components. All I say in my answer is that you can derive the formula by using the matrices like $DU(x)$ instead of expanding the matrix multiplications as you have done. In addition, probably more importantly, you can use the product rule instead of the big $O$ you use. – Mason Aug 19 '22 at 23:59
  • Where precisely do you propose I use the product rule? Also, you still start with a different definition than I ask for in OP – Sal Aug 20 '22 at 01:30
  • @Sal Your definition of $L_{V}U(x)$ looks wrong. The vector in your quotient is in a different vector space at each $t$, namely $T_{\phi_t(x)}M$, so the limit doesn't make sense. – Mason Aug 20 '22 at 02:31
  • The push-forward in the second term ensures that at each $t$ the difference is taken in $T_{\phi_t x}M$, because that's what the push forward does. Have a look at Frankel's book The geometry of physics to see a nice pictorial description and explanation of it – Sal Aug 20 '22 at 21:33
  • Yes you are taking a limit, but at each $t$, the quantity is in $T_{\phi_t(x)}M$, which is a different vector space for each $t$. The definition of limit requires that the thing you are taking a limit of lies in the same space for all $t$. In my definition, the space is $T_xM$, as it has to be in order that $L_V U$ defines a vector field. – Mason Aug 20 '22 at 22:53
  • Accepting this requirement for limits, I think you cannot compute things like $\lim_{x \to x_0}V_x$ for a vector field $V$... for each $x$, $V_x$ will live in a different vector space – Sal Aug 20 '22 at 23:32
  • @Sal Correct. That's why your limit doesn't make sense. The only way in which it kind of makes sense is in coordinates, where all tangent spaces can be identified with $\mathbb{R}^n$. – Mason Aug 20 '22 at 23:52
  • But the tangent spaces are always identifiable with $\mathbb{R}^n$, no? The purpose of my last comment was to point out that in this context the requirement is very steep. You also cannot compute eg. $\lim_{t\to t_0}\phi_{t*} V$. Where exactly does this requirement come from? – Sal Aug 22 '22 at 22:49
  • @Sal Actually isn't $(\phi_{t}){*}U(x) = D\phi_t(\phi_t^{-1}(x))U(\phi_t^{-1}(x))$? We have $U(\phi_t(x)) \in T{\phi_t(x)}M$ while $(\phi_{t}){}U(x) \in T_xM$. So the object you are taking the limit of only makes sense in coordinates. The point of the usual definition $L_V U(p) = \frac{d}{dt}(\phi_t^{}U(p))|{t = 0}$ for $p \in M$ is that the definition makes sense on the manifold $M$ immediately when you write it down. With your definition, you have to through make sure that $L_V U(x)$ is independent of choice of coordinates $x$. – Mason Aug 23 '22 at 03:29
  • @Sal I see that pushforward is defined in a nonstandard way in your geometry book. Nevertheless, your definition of the Lie derivative is equivalent to the standard one. See how the author of your book shows that your definition is equivalent to equation (4.2) on page 126 of https://fedika.com/wp-content/uploads/2019/02/The-Geometry-of-Physics.pdf . (4.2) is the definition I use in my answer. – Mason Aug 23 '22 at 03:45
  • If $U_x \in T_x M$ then $\phi_{t*}U_x$ is in $T_{\phi_t x}M$, where have you seen otherwise? I know the definition you begin with is equivalent- as I said in OP I'm asking about this particular derivation starting with the definition in eq 1 – Sal Aug 23 '22 at 04:33
  • @Sal For the definition of pushforward, see page 183 of "Introduction to Smooth Manifolds" by Lee. My thought is that the pushforward is supposed to be the inverse of the pullback. See exercise 12-10 on page 326. Another reason why your definition is weird is that using your definition, $(\phi_t)_{*}U$ is not a vector field. – Mason Aug 23 '22 at 04:51
  • Eq 13.2 is equivalent to my equation 1. The definition in Lee p62 of pushforward is precisely what I said: if $F:M \to N$ then $F_* Y_p \in T_{F(p)}N$ for vector field $Y$. What confuses you? – Sal Aug 24 '22 at 18:35
  • No, page 62-63 doesn't define the pushforward $F_{}X$ anywhere. Actually it remarks that some people use $F_{}$ to mean $DF(p)$, which seems to be how you use it. But on page 183, Lee defines the pushforward of a vector field $X$ on $M$ by a diffeomorphism $F : M \to N$ as a vector field $F_{}X$ on $N$, defined by $(F_{}X)(q) = DF(p)X(p)$, where $p = F^{-1}(q)$. – Mason Aug 24 '22 at 19:32
  • Your last comment is not in conflict with what I'm saying. It is in conflict with your prior assertion, specifically that $\phi_{t*}U_x \in T_x M$ – Sal Aug 24 '22 at 20:26
  • In the notation of my last comment, $(F_{}X)(q) \in T_qN$. So $((\phi_t)_{}U)(x) \in T_xM$. Your notation $(\phi_{t}){*}U_x$ corresponds to $D\phi_t(x)U(x)$ in my/Lee's notation. Yours is in $T{\phi_t(x)}M$ while mine is in $T_xM$. – Mason Aug 24 '22 at 21:24