f-related vector field

Question

Here is the question:

Let $M$ and $N$ be manifolds and let $f:M\rightarrow N$ be a smooth map. A vector field $X$ on $M$ and a vector field $Y$ on $N$ are said to be $f$-related if $$df_p(X_p)=Y_{f(p)},\forall p\in M.$$ This means that for every function $$h:N\rightarrow R$$ $$X_p[h\circ f]=Y_{f(p)} h.$$Suppose that $X$ is $f$-related to $Z$ and $Y$ is $f$-related to $W$. Show that $[X,Y]$ is $f$-related to $[Z,W]$.

I know I have to prove that $$df_p([x,y])=[df_p(x),df_p(y)]_{f(p)}$$But that I can not figure out this calculation.

Answer 1

Sorry for the delay, I have been away for a few days. This calculation is horrible and I don't guarantee that this is the best way to do this. I just did it in coordinates and it seems to work.

We want to show that

$$f_*[X,Y]=[f_*(X),f_*(Y)]$$

We will show this by showing that both sides act the same way on functions on $N$ and so must be the same vector field.

Let $h:N\to\mathbb R$, we have by definition

$$f_*([X,Y])(h)=[X,Y](h\circ f)$$

We use coordinates $x^i$ for $M$ and $y^m$ for $N$. We have \begin{equation} \begin{aligned}f_*([X,Y])(h)&=[X,Y](h\circ f)(x)\\ &=X^i\frac{\partial}{\partial x^i}\left(Y^j\frac{\partial}{\partial x^j}(h\circ f)\right)(x)-Y^i\frac{\partial}{\partial x^i}\left(X^j\frac{\partial}{\partial x^j}(h\circ f)\right)(x)\\ &=X^i\frac{\partial}{\partial x^i}\left(Y^j\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x)\right)-Y^i\frac{\partial}{\partial x^i}\left(X^j\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x)\right)\\ &=X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x)-Y^i\frac{\partial X^j}{\partial x^i}\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x) \end{aligned} \end{equation}

While going from the other end \begin{equation} \begin{aligned} \phantom a[f_*(X),f_*(Y)](h)&=f_*(X)(f_*(Y)(h))-f_*(X)(f_*(Y)(h))\\ &=f_*(X)\left(Y^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))\right)-f_*(Y)\left(X^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))\right)\\ &=X^iY^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial y^n}{\partial x^i}(x)\frac{\partial^2 h}{\partial y^m\partial y^n}(f(x))+X^i\frac{\partial Y^j}{\partial x^m}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial x^m}{\partial y^n}(f(x))\frac{\partial y^n}{\partial x^i}(x)\frac{\partial h}{\partial y^m}(f(x))\\ &\phantom=-Y^iX^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial y^n}{\partial x^i}(x)\frac{\partial^2 h}{\partial y^m\partial y^n}(f(x))-Y^i\frac{\partial X^j}{\partial x^m}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial x^m}{\partial y^n}(f(x))\frac{\partial y^n}{\partial x^i}(x)\frac{\partial h}{\partial y^m}(f(x))\\ &=X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))-Y^i\frac{\partial X^j}{\partial x^i}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))\\ \end{aligned} \end{equation}

We have now shown that both sides are the same. In this derivation I have used the chain rule as well as the Jacobian, some cancellations and the fact that $\frac{\partial y^m}{\partial x^j}(f(x))\frac{\partial x^k}{\partial y^m}(x)=\delta^k_j$.

Answer 2

I stumbled upon this question recently and decided to finally create an account. Your question relates to the commutation of the pushforward map and the Lie-bracket which is why the below proof together with the fact that the vector fields are $f$-related (or $\varphi$ in my case) should solve the problem.

Theorem: Given a smooth map $\varphi:M\rightarrow N$ between two manifolds $M$ and $N$ and two vector fields $X,Y \in \Gamma(M)$ the pushforward map $\varphi_*$ commutes with the Lie-bracket, namely $\varphi_* [X,Y] = [\varphi_*X, \varphi_*Y]$.

Proof: Let a smooth function $g\in C^{\infty}(N)$ and a point $p\in M$ be given. We first note that from $\varphi_* (X)_{\varphi(p)}(g) = X_p (\varphi^* g) = X_p (g \circ \varphi)$ we deduce that $\varphi_* (X)(g) \circ \varphi = X (g \circ \varphi)$. We use this fact to calculate \begin{align*} \varphi_* [X,Y]_{\varphi(p)}(g) &= [X,Y]_{p}(g \circ \varphi)\\ &= X_p \circ Y(g \circ \varphi) - Y_p \circ X(g \circ \varphi)\\ &= X_p (\varphi_*(Y)(g) \circ \varphi) - Y_p( \varphi_*(X)(g) \circ \varphi)\\ &= \varphi_* X_{\varphi(p)} (\varphi_*(Y)(g)) - \varphi_* Y_{\varphi(p)} ( \varphi_*(X)(g))\\ &= [\varphi_* X, \varphi_* Y]_{\varphi(p)}(g)\\ \end{align*} This implies that indeed $\varphi_* [X,Y] = [\varphi_* X, \varphi_* Y]$. While a proof using coordinates was given, some may prefer a coordinate free approach.