Let $(M,g)$ be a Riemannian manifold, $N$ a smooth manifold and $$\pi:M\to N$$ a surjective smooth submersion. Then, each level set $M_q=\pi^{-1}(q)$ is a properly embedded submanifold of $M$ so we can define the vertical space $$V_p=T_pM_{\pi(p)}=\ker d\pi_p\subseteq T_pM$$ for each $p\in M$. Let $H_p\subseteq T_pM$ be the orthogonal complement of $V_p$ with respect to $g$, called the horizontal space, so that $T_pM=V_p\oplus H_p$, $\forall p\in M$.

Problem: Show that if $Y$ is a smooth vector field on $N$, then there is a unique smooth vector field $X$ on $M$ such that $X_p\in H_p$ and $d\pi_p(X_p)=Y_{\pi(p)}$ for all $p\in M$.

The only problem I have is with smoothness of $X$. Indeed, $$d\pi_p:V_p\oplus H_p\to T_{\pi(p)}N$$ is surjective by assumption. But, $V_p$ is mapped to zero, and $$\dim H_p=\dim M-\dim V_p=\dim N=\dim T_{\pi(p)}N,$$ so we have a linear isomorphism $$H_p\to T_{\pi(p)}N.$$ Thus, for each $p\in M$, there is a unique $X_p\in H_p$ such that $d\pi_p(X_p)=Y_{\pi(p)}$. It remains to show that $X:M\to TM$ is smooth. How?


Recall that the problem of smoothness is local. So around any point $p \in M$ you can find a coordinate system $x_1,\cdots,x_k,\cdots,x_n$ centered at $p$ and a coordinate system $y_1,\cdots,y_k$ of $N$ centered at $\pi(p)$ such that the map $\pi$ is given by $$\pi(x_1,\cdots,x_k,\cdots,x_n)=(x_1,\cdots,x_k)$$

Your vector field $Y$ is writen (in such coordinates) as $$Y = \sum_{i=1}^k f_i(y_1,\dots,y_k) \frac{\partial}{\partial y_i} $$

So you get an smooth vector field $W$ on $M$ such that $$d\pi_p (W_p) = Y_{\pi(p)} .$$ just by defining $$W := \sum_{i=1}^k f_i(x_1,\dots,x_k) \frac{\partial}{\partial x_i} $$

Observe that $W$ is smooth but in general is not horizontal. To get (or to show) that your horizontal $X$ is smooth just observe that $$X = \sum_{i=k}^n \langle W,e_i \rangle e_i$$ where $e_k,\cdots,e_n$ is a smooth orthonormal frame of the horizontal distribution $H$.

  • $\begingroup$ Why $d\pi(\sum \langle W,e_i\rangle e_i)=Y$? $\endgroup$ – user253278 Jul 8 '15 at 20:19
  • $\begingroup$ Because is the orthogonal projection of $W$ onto the horizontal subspace. So $d \pi (W) = d \pi (X) = Y$ $\endgroup$ – Holonomia Jul 8 '15 at 20:22
  • $\begingroup$ I think you misread my comment. $\endgroup$ – user253278 Jul 8 '15 at 20:26
  • $\begingroup$ If you agree that $d \pi (W) = Y$ then also $d \pi (W + V) = Y$ for any vertical vector field $V$. Now $W$ is smooth and also $\sum \langle W,e_i \rangle e_i$ is smooth and $W - \sum \langle W,e_i \rangle e_i$ is vertical. So $d \pi (W) = d \pi(\sum \langle W,e_i \rangle e_i) = Y$. Thus, $X = \sum \langle W,e_i \rangle e_i$ is smooth as you wanted. Did you agree? $\endgroup$ – Holonomia Jul 8 '15 at 20:49
  • $\begingroup$ Yes I get it.${}$ $\endgroup$ – user253278 Jul 8 '15 at 21:25

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