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I want to show this fact:

Let $M, N$ be two manifolds, $\pi : M \to N$ a surjective submersion and $X$ a vector field over $M$. If $d \pi_q(X_q) = d \pi_p(X_p)$ whenever $\pi(p) = \pi(q)$ then there exists a unique vector field $Y$ over $N$ such that $Y_{\pi(p)} = d \pi_{p}(X_p)$ for every $p \in M$.

I don't know where to start. I could define $Y : N \to TN$ such that $Y_p = d \pi_q (X_q)$ where $q$ is an element of $\pi^{-1}(p)$, but that seems to be too... ugly and not much rigorous, even if it's the right path I can't see how $Y$ can be smooth. Since $\pi$ is a submersion I tried to use its normal local form but following this road is even worse: I can't even see how to build properly $Y$. Can you give me a hint please?

Thanks. English is not my mother tongue, please excuse any errors on my part.

Ker
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2 Answers2

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You are in the right track. Given $y \in N$, set $Y_y = {\rm d}\pi_x(X_x)$, where $x \in \pi^{-1}(y)$ is any point. This is well defined since ${\rm d}\pi_{x_1}(X_{x_1}) = {\rm d}\pi_{x_2}(X_{x_2})$ whenever $x_1,x_2 \in M$ satisfy $\pi(x_1)=\pi(x_2)$. As a matter of fact, this also ensures the uniqueness of such $Y$, as another $\tilde{Y}$ would have to satisfy $\tilde{Y}_y = {\rm d}\pi_x(X_x)$ for some $x \in \pi^{-1}(y)$, which is precisely how we defined $Y_y$.

So what is left to do is showing that such $Y$ is smooth. This follows from the "universal property" of surjective submersions (Proposition 7.17 in John Lee's Introduction to Smooth Manifolds, 2nd edition): if $P$ is a smooth manifold and $F\colon N \to P$ is a given map, then $F$ is smooth if and only if $F\circ \pi$ is smooth. We take $P = TN$ and $F = Y$. That is to say, smoothness of $Y$ follows from smoothness of $Y\circ \pi = {\rm d}\pi \circ X$.

Ivo Terek
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  • Thanks for the reply. This is a very elegant approach. Thank you! – Ker Apr 29 '19 at 21:04
  • Glad to help. I feel that coordinates are useful for explicit computations, but it is better if we try to avoid them for theoretical results like this one. There's even this funny quote by Hermann Weyl: "the introduction of numbers as coordinates is an act of violence" :P – Ivo Terek Apr 29 '19 at 21:07
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I would argue as follows:

1) First of all, the relation $Y_{\pi(p)}=(d\pi)_{p}(X_{p})$ determines $Y$ uniquely on all of $N$ since $\pi$ is surjective. As a consequence, it suffices to construct $Y$ locally on an arbitrary coordinate chart $U_{\alpha}$. Indeed, on the overlap of such charts $U_{\alpha}$ and $U_{\beta}$, these locally defined vector fields will agree by the uniqueness, hence they patch together to a smooth vector field $Y$.

2) Let us now locally construct the vector field $Y$. Since $\pi$ is a submersion, it is locally given by $$\pi:M\rightarrow N: (x_{1},\ldots,x_{k},y_{1},\ldots,y_{l})\mapsto (x_{1},\ldots,x_{k})$$ Let the vector field $X$ be given by $$ X=\sum_{i=1}^{k}f_{i}(x,y)\partial_{x_{i}}+\sum_{j=1}^{l}g_{j}(x,y)\partial_{y_{j}}. $$ The given condition on $X$ means exactly that the $f_{i}$ don't depend on the $y$-coordinates, i.e. $f_{i}(x,y)=f_{i}(x)$. Consequently, we should define $Y$ locally by $$ Y=\sum_{i=1}^{k}f_{i}(x)\partial_{x_{i}}. $$

studiosus
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  • Thanks for the reply. I was missing that $f_i$ don't depend on $y$, now it's all clear! Thank you. – Ker Apr 29 '19 at 21:00