A special property of the Lie bracket and general questions about vector fields

Question

Problem: Let $s: M \rightarrow N$ be a surjective submersion. Let $X_1, X_2$ be vector fields on $M$, and $Y_1, Y_2$ be vector fields on $N$. Suppose that $(T_p s)(X_i|_p) = Y_i|_{s(p)}$ for $i = 1,2$. Show: $$(T_p s)([X_1, X_2]|_p) = [Y_1, Y_2]|_{s(p)}$$ where $[\cdot,\cdot]$ is the Lie bracket.

Attempt: I am a beginner in differential geometry and am trying to wrap my head around the concepts of tangent spaces, tangent maps and vector fields right now so please point out any of my (potential) misunderstanding. The first approach that comes to mind is to work in coordinates, and eventually reach a stage that verifies that both sides agree, i.e. for chart $(U,\phi)$ around $p$, we consider $(T_p\phi) \circ ([X_1,X_2]) \circ \phi^{-1} = \sum_{i}\alpha_i(x_1,...,x_m) \partial_{x_i}$ in coordinates, and $(T_p s)([X_1, X_2])(f) = (T_p\phi) \circ ([X_1,X_2]) \circ \phi^{-1}|_p(\phi \circ f \circ s)$ for $f \in C^{\infty}(N)$ (is this correct?). Now it's a matter of actually writing out the expressions on both sides and rearranging them.

Firstly, I haven't gone through the entire computation as it is quite long, but am I on the right track? More importantly, is there a cleaner of proving the equality, e.g. a more "algebraic", abstract way? The issue is that, when I write, for example, $(X_1 \circ X_2)|_p(f \circ s)$, but what now? Any help is appreciated!

Answer 1

You have that $s$ is a submersion, so by the constant rank theorem there are charts $(U_1,\phi_1 = (x^i)) \in \Sigma(M)$ and $(U_2,\phi_2 = (y^i))\in \Sigma(N)$ around $p$ and $s(p)$ such that $$ \phi_2\circ s\circ\phi_1^{-1}(x^1,\ldots, x^m) = (x^1,\ldots, x^n),$$where $m = \dim M$ and $n = \dim N$. In this coordinate system, we have that $$X_1\big|_q = \sum_{i=1}^m a^i(q) \frac{\partial}{\partial x^i}\bigg|_q \implies Y_1\big|_{s(q)} ={\rm d}s_q(X_1\big|_q) = \sum_{i=1}^n a^i(q) \frac{\partial}{\partial y^i}\bigg|_{s(q)},$$and similarly for $X_2$ and $Y_2$ (with coefficients $b^i$, say), for $q$ near $p$. You can compute $$[X_1,X_2]\big|_q = \sum_{i=1}^m \left(\sum_{j=1}^ma^j(q)\frac{\partial b^i}{\partial x^j}(q) - b^j(q)\frac{\partial a^i}{\partial x^j}(q)\right)\frac{\partial}{\partial x^i}\bigg|_q$$and $$[Y_1, Y_2]\big|_{s(q)} = \sum_{i=1}^n \left(\sum_{j=1}^n a^j(q) \frac{\partial b^i}{\partial y^j}(s(q)) - b^j(q) \frac{\partial a^i}{\partial y^j}(s(q))\right)\frac{\partial}{\partial y^i}\bigg|_{s(q)},$$and the result follows, since basically the derivative of $s$ will kill the $\partial/\partial x^i$ terms for $n+1\leq i \leq m$, and this is the key for solving the problem. More explicitly, we have that $${\rm d}s_q\left(\frac{\partial}{\partial x_i}\bigg|_q\right) = \begin{cases} \dfrac{\partial}{\partial y^i}\bigg|_{s(q)}, \mbox{ for } 1 \leq i \leq n, \\[1em] 0, \mbox{ for }n+1 \leq i \leq m.\end{cases}$$

I should probably point there I am doing a single abuse of notation here, in the expression for the $Y$'s. What I'm differentiating with respect to $y^j$ is a function $\widetilde{b}^i$, which satisfies $\widetilde{b}^i \circ s= b^i$.

Answer 2

Take a test function $f \in C^{\infty} (N)$. If we can show that the vector fields $[Y_1, Y_2]$ and $(T_p s)([X_1, X_2])$ act give the same result when applied to $f$ (thinking of vector fields as differential operators), then the vector fields are equal.

We compute: $$[Y_1, Y_2](s(p))f = (Y_1 Y_2 - Y_2 Y_1) (s(p))f = Y_1 (s(p))(Y_2 f) - Y_2 (s(p)) (Y_1 f)$$ $$ = (T_p s)(X_1 (p))(Y_2 f) - (T_p s)(X_2 (p))(Y_1 f) = (X_1(p))(Y_2f \circ s) - (X_2(p))(Y_1f \circ s)$$ Now, note that the function $Y_2f \circ s$ is a function $M \to \mathbb{R}$ whose value at $p \in M$ is given by differentiating $f$ in the direction $Y_2 (s(p))$, so $$(Y_2 f \circ s)(p) = Y_2 (s(p)) f = (T_p s)(X_2 (p)) f = X_2 (p) (f \circ s) = X_2 (f\circ s)(p) \implies Y_2f \circ s = X_2 (f\circ s)$$ and similarly we have $Y_1 f \circ s = X_1 (f\circ s)$. Therefore, $$(X_1(p))(Y_2f \circ s) - X_2(p)(Y_1f \circ s) = X_1 (p) (X_2 ( f\circ s)) - X_2(p) (X_1 (f\circ s)) $$ $$= (X_1 X_2 - X_2 X_1)(p) (f\circ s) = (T_p s)([X_1, X_2](p))f$$ We conclude $[Y_1, Y_2] (s(p)) = (T_p s)([X_1, X_2](p))$ as desired.