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I am studying differentiable topology and I am facing the definition on vector field and Lie bracket. If $M$ is an $m-$manifold and $V,W:T\to TM$ are vector fields on $M$, we define the Lie bracket of $V$ and $W$ as the following vector field: \begin{equation*} [V,W]=VW-WV \end{equation*} (where $V$ and $W$ are intended as derivations). In particular, if $(A,\varphi)$ is a local chart on $M$, with coordinates $(x^1,\dots,x^m)$, then we can write: \begin{equation*} [V,W]=\sum_j\sum_i\left(v^i_x\frac{\partial w^j_x}{\partial x^i}-w_x^i\frac{\partial v_x^j}{\partial x^i}\right)\frac{\partial}{\partial x^j} \end{equation*} where $v^i_x, w^i_x:M\to \mathbb{R}$ are differentiable functions representing the components of $V$ and $W$ with respect to the basis $\left(\frac{\partial}{\partial x^1},\dots, \frac{\partial}{\partial x^m}\right)$.

Now, we have given this definition using a local chart of $M$, but we want to ensure that this definition does not depend on the coordinate system, that is: if $(B,\psi)$ is another local chart with coordinates $(y^1,\dots, y^m)$, then in $A\cap B$ we must have: \begin{equation*} \sum_j\sum_i\left(v^i_x\frac{\partial w^j_x}{\partial x^i}-w_x^i\frac{\partial v_x^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}=\sum_j\sum_i\left(v^i_y\frac{\partial w^j_y}{\partial y^i}-w_y^i\frac{\partial v_y^j}{\partial y^i}\right)\frac{\partial}{\partial y^j} \end{equation*}

I want to prove this fact using the jacobian matrix of the change of coordinates, but I am having many many difficulties with the indices and, moreover (maybe it is a silly question), if \begin{equation*} \frac{\partial }{\partial x^i}=\sum_k \frac{\partial y^k}{\partial x^i}\frac{\partial }{\partial y^k} \end{equation*} then what is $\frac{\partial w_x^j}{\partial x^i}$ ? Is it \begin{equation*} \frac{\partial w_x^j}{\partial x^i}=w^j_x\sum_k \frac{\partial y^k}{\partial x^i}\frac{\partial }{\partial y^k} \end{equation*} or \begin{equation*} \frac{\partial w_x^j}{\partial x^i}=\sum_k \frac{\partial y^k}{\partial x^i}\frac{\partial w^j_x}{\partial y^k} \end{equation*} or something else? In the second case, what would $\frac{\partial w^j_x}{\partial y^k}$ mean?

I know I am boring, but can you give me any suggestion, or a complete proof, or give me a reference where I can find the answers to my questions?

Thanks in advance for any help!

Edit: Tell me if my solution is correct or not (if not, please, tell me what is wrong). In what follows, I will use Einstein's convention of the summation over the repeated indices, in order to avoid the mess caused by too many sums. As I wrote above, we know that: \begin{equation*} \frac{\partial}{\partial x^i}=\frac{\partial y^k}{\partial x^i}\frac{\partial}{\partial y^k} \end{equation*} and, using the jacobian matrix of the change of coordinates, \begin{equation*} v^j_y=\frac{\partial y^j}{\partial x^i}v^i_x \end{equation*} (the same holds for $w^i_y$) Then \begin{equation*} \frac{\partial v^j_y}{\partial y^i}=\frac{\partial}{\partial y^i}\left(\frac{\partial y^j}{\partial x^k}v^k_x\right)=\frac{\partial y^j}{\partial x^k}\frac{\partial v^k_x}{\partial y^i} \end{equation*} (the same holds for $w^j_y$). Using these facts, we obtain: \begin{align*} \left(v^i_x\frac{\partial w^j_x}{\partial x^i}-w^i_x\frac{\partial v^j_x}{\partial x^i}\right)\frac{\partial}{\partial x^j}&=\left(v^i_x\frac{\partial y^k}{\partial x^i}\frac{\partial w^j_x}{\partial y^k}-w^i_x\frac{\partial y^k}{\partial x^i}\frac{\partial v^j_x}{\partial y^k}\right)\frac{\partial y^h}{\partial x^j}\frac{\partial}{\partial y^h}\\ &=\left(v^k_y\frac{\partial w^j_x}{\partial y^k}-w^k_y\frac{\partial v^j_x}{\partial y^k}\right)\frac{\partial y^h}{\partial x^j}\frac{\partial}{\partial y^h}\\ &=\left(v^k_y\frac{\partial y^h}{\partial x^j}\frac{\partial w^j_x}{\partial y^k}-w^k_y\frac{\partial y^h}{\partial x^j}\frac{\partial v^j_x}{\partial y^k}\right)\frac{\partial}{\partial y^h}\\ &=\left(v^k_y\frac{\partial w^h_y}{\partial y^k}-w^k_y\frac{\partial v^h_y}{\partial y^k}\right)\frac{\partial}{\partial y^h} \end{align*} which is what we wanted to prove.

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  • $\begingroup$ You skipped the crucial step. Write down the equation relating $w^i_x$ and $w^j_y$ (and similarly for the $v$'s). In any event, neither of those two equations is correct. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ I don’t understand what you are trying to show. Your first definition is manifestly coordinate independent. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ @Malady If so, it is far from clear that it is yet again a vector field. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ I agree, but that seems to me a separate issue than showing coordinate independence. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ Yes. It should be clear that you need to use the product rule there, but it's subtle because you have to realize that you need the chain rule to differentiate $\partial y^j/\partial x^k$ with respect to $y^i$ and get functional combinations of $\partial^2 y^j/\partial x^k\partial x^\ell$. $\endgroup$ Commented 17 hours ago

2 Answers 2

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Note: what follows is not a verification of your argument. Instead, it is an alternative argument showing coordinate independence (an elaboration on the comments by Malady and Ted Shifrin).

Identifying vector fields $V,W$ with their corresponding derivations, we define $[V,W]$ as the differential operator acting on scalar functions: $$[V,W](f) = V(W(f))-W(V(f)).$$ No choice of coordinates was involved in this definition, case closed. However, in order for the definition to make sens, we need to make sure that this differential operator is indeed a derivation with respect to a (uniquely defined) vector field. There are at least two options for that:

1. It's possible that you define tangent vectors as derivations. In that case, the only thing to check is that $[V,W]$ is indeed a derivation, or in other words, it satisfies the Leibniz formula $$[V,W](fg)(p) = f(p)[V,W](g)(p) + g(p)[V,W](f)(p) \quad \text{for each point } p. $$ This follows immediately from the Leibniz formula for $V$ and $W$.

2. If that's not your definition, let's look at the coordinate expression you gave: $$[V,W]=\sum_{i,j}\left(v^i_x\frac{\partial w^j_x}{\partial x^i}-w_x^i\frac{\partial v_x^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}.$$ It is clear from this expression that there is a vector field $X$ whose derivation matches exactly the differential operator $[V,W]$ (this $X$ is given by $X^j = v^i_x\frac{\partial w^j_x}{\partial x^i}-w_x^i\frac{\partial v_x^j}{\partial x^i}$). One last question to ask is: OK, we constructed $X$ via a map, so maybe it depends on the choice of coordinates? And the answer is no - if two vector fields $X$, $Y$ (say, constructed via different maps) have the same derivation operator, they have to coincide.

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Here is the bit of the computation I've been referring to in my comments. Differentiating $v^j_y = \dfrac{\partial y^j}{\partial x^k}v^k_x$, we obtain $$\frac{\partial v^j_y}{\partial y^i}=\frac{\partial y^j}{\partial x^k}\frac{\partial v^k_x}{\partial y^i}+ v^k_x\frac{\partial^2 y^j}{\partial x^k\partial x^\ell}\frac{\partial x^\ell}{\partial y^i}.$$ The second derivative terms now contribute as follows to the Lie bracket calculation: \begin{align*} \big(v^i_yw^k_x-v^k_xw^i_y\big)\frac{\partial^2 y^j}{\partial x^k\partial x^\ell}\frac{\partial x^\ell}{\partial y^i} &= \frac{\partial y^i}{\partial x^m}\big(v^m_x w^k_x -v^k_xw^m_x\big)\frac{\partial^2 y^j}{\partial x^k\partial x^\ell}\frac{\partial x^\ell}{\partial y^i} \\ &= \big(v^m_x w^k_x -v^k_xw^m_x\big)\frac{\partial^2 y^j}{\partial x^k\partial x^\ell}\frac{\partial y^i}{\partial x^m}\frac{\partial x^\ell}{\partial y^i} \\ &= \big(v^m_x w^k_x -v^k_xw^m_x\big)\frac{\partial^2 y^j}{\partial x^k\partial x^\ell}\delta^\ell_m \\ &= \big(v^m_x w^k_x -v^k_xw^m_x\big)\frac{\partial^2 y^j}{\partial x^k\partial x^m}. \end{align*} This sum is $0$: We switch $k$ and $m$ in one of the terms and use symmetry of the second-order partial derivatives. This sort of thing is ubiquitous in differential topology/geometry.

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  • $\begingroup$ Thank you, now it is clear. I had forgotten to apply chain rule in this expression. $\endgroup$ Commented 7 hours ago

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