In this journey of understanding some of differential geometry I started to do some proof of results. And now I on doubt about this one which is the commutativity between Pushforward and the Lie bracket without draw upon local coordinates:
Let $M$ and $N$ two differential manifolds, and
$$ \varphi: M \longrightarrow N $$
A diffeomorphism between the two.
So if I have a map $g: N \rightarrow \mathbb{R} $ then I can define a function $\varphi^*g: M \rightarrow \mathbb{R} $ the Pullback of $g$.
And with the same philosophy, given a vector field on $M$, $X: C^\infty(M) \rightarrow C^\infty(M)$ the Pushforward vector field on $N$ is defined as: $\varphi_*X(g) = X(\varphi^*g), \ g \in C^\infty(N)$
(I made all this intro because I'm not even 100% sure about if I got this concepts right, so any comment is welcome).
So my idea of proof is first start with the pushformward of the Lie bracket, with $f \in C^\infty(N)$:
$$ \varphi_*[X,Y](f) = [X,Y](\varphi^*f) = X(Y(\varphi^*f)) – Y(X(\varphi^*f))$$
Since the two terms are the same under a swaping of $X$ and $Y$ I will continue with the first one.
So, by the definition of pushforward $Y(\varphi^*f) = \varphi_*Y(f)$, but $\varphi_*Y(f) \in C^\infty(N)$ so it cannot be an argument for $X(\cdot{})$ therefore the expression
$$ X(Y(\varphi^*f)) = X(\varphi_*Y(f))$$
Cannot the be true.
My intuition tells me that the solution of this is that, since $\varphi_*Y(f)$ is another map on $N$ I can first "pull-it back" to $M$ before feed it into $X$. Having:
$$X(Y(\varphi^*f)) = X(\varphi^*(\varphi_*Y(f))) = \varphi_*X(\varphi_*Y(f))$$
Letting me completing my "proof" after a few straightforward steps.
Is this correct? Is so, what is the justification for incorporate that last pullback? How can I know that the pullback of $\varphi_*Y(f)$ if the map on $M$ that I'm looking for?
Sorry for the bad English and thanks in advance!
Best Answer
So far, the definitions and setup seem fine.
Yes, it is correct that $Y(\varphi^*f) = \varphi^*(\varphi_*Y(f))$.
Here is a rough proof.
I'm mostly copying this proof from Proposition 8.16 of John Lee's book "Smooth Manifolds"
To prove it, take a point $p \in M$. Then $Y(\varphi^*f)$ is a function on $M$, which at the point $p$ is: $$ Y(\varphi^*f)(p) = (\varphi_*Y)_{\varphi(p)}(f) $$ Here, $(\varphi_*Y)_{\varphi(p)}$ is the pushforward vector field evaluated at $\varphi(p)$. This last expression is equal to $\varphi_*Y(f)(\varphi(p))$. That is, $\varphi_*Y(f)$ is a function on $N$, and evaluating it at $\varphi(p)$ is equal to $(\varphi_*Y)_p(f)$. Now, this last expression, $\varphi_*Y(f)(\varphi(p))$, is the composition $\Big( \varphi_*Y(f) \Big) \circ \varphi$, evaluated at $p$. By definition of pullback of functions, this is equal to $\varphi^*(\varphi_*Y(f))$.