Let $G$ be a Lie group acting isometrically, freely and properly on a Riemannian manifold $(M,g)$ and $\pi:M\to N:=M/G$ the natural projection. Show that there is a unique metric $h$ on $N$ which makes the projection $\pi:(M,g)\to (N,h)$ a Riemannian submersion.
For $x\in N$ and $p\in\pi^{-1}(x)$, we take the vertical/horizontal spaces $V_p:=\ker d\pi_p$ and $H_p:=V_p^\perp$, so that $T_pM=V_p\oplus H_p$. Since $d\pi_p$ is surjective, then $d\pi_p|_{H_p}:H_p\to T_{\pi(p)}N$ is bijective. For $v_1,v_2\in T_xN$, we define:
$$h_x(v_1,v_2)=g_p(d\pi|_{H_p}^{-1}v_1,d\pi|_{H_p}^{-1}v_2)$$
(in particular, this satisfies $|v|_g=|d\pi(v)|_h$)
I was able to prove $h_x$ doesn't depend on the choice of $p$ using the fact that $G$ acts isometrically. The uniqueness comes easy from the fact that another $h'$ would be such that $|v|_{h'}=|v|_h$ for all $v\in T_xN$, which is enough to conclude $h'=h$.
I'm having trouble to prove $h$ is smooth. I wish to argue that if $X,Y$ are a fields in $N$, we may find horizontal fields $X_H, Y_H$ in $M$ such that $d\pi(X_H)=X,d\pi(Y_H)=Y$, so $h(X,Y)=g(X_H,Y_H)$. I only need to show that such fields $X_H,Y_H$ exist (I believe they do), but I don't know how.
Any suggestions?
Best Answer
Let $y\in N$, and $x\in M$ such that $p(x)=y$. Let $U$ be an open subset containing $x$ whose adherence $V$ is compact. Since the action is proper, $Orb_x\cap V$ has a finite number of connected components. We can shrink $U$ such that $U$ contains a unique connected component of $Orb_x\cap V$, $C$, let $T$ be a submanifold transverse to $C$, the restriction $p$ of $\pi$ to $T$ is a diffeomorphism onto its image, for every $z\in T$, we can define $Y_H(z)=dp^{-1}_{\pi(z)}(X_H(\pi(z))$.