Naturality of Lie bracket — alternate proof

Question

Let $M$ and $N$ be smooth manifolds, and let $F: M \to N$ be a smooth map. Let $X$ and $Y$ be vector fields on $M$, and let $\tilde{X}$ and $\tilde{Y}$ be vector fields on $N$. We say that $X$ and $\tilde{X}$ are $F$-related if $dF_p(X_p) = \tilde{X}_{F(p)}$ for every $p \in M$. There is a theorem (naturality of the Lie bracket):

If $X$ and $\tilde{X}$ are $F$-related and $Y$ and $\tilde{Y}$ are $F$-related, then $[X,Y]$ and $[\tilde{X}, \tilde{Y}]$ are $F$-related.

A very concise proof is as follows:

Let $f \in C^\infty(N)$. We only need to show that $[X,Y](f \circ F) = [\tilde{X}, \tilde{Y}]f$, which is $$(XY - YX)(f \circ F) = (\tilde{X}\tilde{Y} - \tilde{Y}\tilde{X})f.$$ This equality follows from $XY(f \circ F) = X(\tilde{Y}f \circ F) = \tilde{X}\tilde{Y}f$.

My question is, how can we prove this theorem without introducing the smooth function $f$?

When $F$ is a local diffeomorphism, we can prove the theorem directly using coordinates. Let the matrix $(b^i_j) = (\partial_jF^i)$ with inverse $(a^i_j)$. We compute: \begin{align*} \tilde{x}^j \tilde{\partial}_j \tilde{y}^i - \tilde{y}^j \tilde{\partial}_j \tilde{x}^i &= b^j_k x^k a^l_j \partial_l(b^i_my^m) - b^j_k y^k a^l_j \partial_l(b^i_mx^m)\\ &= x^l\partial_l(b^i_my^m) - y^l\partial_l(b^i_mx^m)\\ &= b^i_m(x^l \partial_ly^m - y^l \partial_lx^m) + \partial_l\partial_mF^i(x^ly^m - x^my^l)\\ &= b^i_m(x^l \partial_ly^m - y^l \partial_lx^m). \end{align*}

This proof is also in the accepted answer of this MSE question: Pushforward of Lie Bracket.

However, when $F$ is not locally invertible, then the matrix $(b^i_j)$ won't be invertible either. The matrix $(a^i_j)$ in the proof above would be undefined! This situation seems very strange. It appears that the only way to prove this theorem is to use the abstract definition of $[X,Y]f = (XY - YX)f$. The coordinate expression $[X,Y] = (x^j \partial_j y^i - y^j \partial_j x^i)$ seems inadequate.

Another abstract definition of $[X,Y]$ is the Lie derivative $\mathcal{L}_XY$, which is computed using the flow map of $X$. We can also discuss the naturality of the Lie derivative: are $\mathcal{L}_XY$ and $\mathcal{L}_{\tilde{X}}\tilde{Y}$ $F$-related? If $F$ is not locally invertible, then it seems we can't prove this only using the abstract notion of Lie derivative either. We have to use the fact that $\mathcal{L}_XY = [X,Y]$ and return to the Lie bracket case. This also feels strange because the definition $[X,Y]f = (XY - YX)f$ is so special here.

I also discussed the question in this article on Banana Space (a Chinese website dedicated to sharing math): Intuitive interpretation of Lie bracket. However, it is written in Chinese.

Update: I seem to find a way to prove $\mathcal{L}_XY$ and $\mathcal{L}_{\tilde{X}}\tilde{Y}$ are $F$-related only using the abstract notion of Lie derivative. I have written an anwser.

Answer 1

My favorite proof of this takes a step back.

If $M$ is a manifold, let $TM$ be its tangent bundle. The total space is itself a manifold, so you can form the tangent bundle of $TM$, i.e.

$$T(TM) = T^2 M.$$

The Clairaut theorem about mixed partial derivatives being equal tells you that there is an involution of $T^2 M$,

$$\iota_M : T^2 M \to T^2 M.$$

If you write $TM$ as the set of all pairs

$$TM = \{(p,v) : p \in M, v \in T_p M\}$$

then

$$\iota_M(p,v,w,y) = (p,w,v,y).$$

From this your Lie bracket of vector fields could be described this way. Let $X$ and $Y$ be vector fields, then they have the form

$$X,Y : M \to TM.$$

So you can form the directional derivatives

$$TX \circ Y : M \to T^2 M,\hskip 1cm TY \circ X : M \to T^2 M$$

$TX$ denotes the induced map of tangent bundles, i.e. the bundle-theoretic derivative. $TX(p,v) = (X(p), DX_p(v))$.

Now observe that $TX \circ Y - \iota_M \circ TY \circ X$ is not only well-defined but it takes values in $ker(T\pi : T^2 M \to TM)$, where $\pi : TM \to M$ is the tangent bundle projection map, $\pi(p,v) = p$. $T\pi$ evaluates as $T\pi(p,v,w,y) = (p,w)$, thus the kernel are elements in $T^2 M$ of the form $(p,v,0,y)$, i.e. $y \in T_p M$ thus the fibers are canonically identified with $T_p M$, so via this identification we have

$$TX \circ Y - \iota_M \circ TY \circ X : M \to TM$$

and this is the Lie bracket of $X$ and $Y$.

Your result then just pops out of the above.

Answer 2

The Lie bracket can somehow be computed from flows, and the flows intertwine. In more detail, if $X$ maps to $\bar{X}$, i.e. $F'(p)X(p)=\bar{X}(F(p))$ then clearly the flows of $X$ and $\bar{X}$ intertwine: just look at the differential equations of the flows. But then the Lie bracket is computed by switchbacks of flows: $$ [X,Y]:=\left.\frac{d}{dt}\right|_{t=0}e^{-\sqrt{t}Y}e^{-\sqrt{t}X}e^{\sqrt{t}Y}e^{\sqrt{t}X}. $$

Answer 3

Here I present a proof of the naturality of Lie derivative, only using the abstract definition, and without referencing the naturality of Lie bracket.

Let $\phi(t,p)=\phi^t(p)$ be the flow generated by $X$, and $\psi(t,p)=\psi^t(p)$ be the flow generated by $\tilde X$.

Note that $\psi(\cdot,F(p))$ and $F\circ\phi(\cdot,p)$ are both integral curves of $Y$ started from $F(p)$. From the uniqueness of integral curve, we deduce $F \circ \phi^t =\psi^t \circ F$. As a consequence, $$dF_{\phi^t(p)} \phi^t_* = \psi^t_* dF_p,$$ so $$(\psi^t_*)^{-1}dF_{\phi^t(p)}=dF_p(\phi^t_*)^{-1}.$$ Now we see

$$dF_p\mathcal L_XY =\frac d {dt} dF_p(\phi^t_*)^{-1}Y_{\phi^t(p)} =\frac d {dt} (\psi^t_*)^{-1}dF_{\phi^t(p)}Y_{\phi^t(p)} =\frac d {dt} (\psi^t_*)^{-1}\tilde Y_{\psi^t(F(p))} =\mathcal L_{\tilde X} \tilde Y.$$

This proof does not require $F$ to be a local diffeomorphism.