Riemannian manifolds $(M/\Gamma_1,g_1)$ and $(M/\Gamma_2,g_2)$ are isometric if and only if $\Gamma_1$ and $\Gamma_2$ are conjugate

Question

This is problem 3.21 of Lee's RM. Let $(M,g)$ be a simply connected Riemannian manifold, and suppose $\Gamma_1$ and $\Gamma_2$ are countable subgroups of $\mathrm{Iso}(M,g)$ acting smoothly, freely, and properly on $M$.

Prove that Riemannian manifolds $(M/\Gamma_1,g_1)$ and $(M/\Gamma_2,g_2)$ are isometric if and only if $\Gamma_1$ and $\Gamma_2$ are conjugate subgroups of $\mathrm{Iso}(M,g)$

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My attempt: if $\Gamma_1$ and $\Gamma_2$ are conjugate then there is an isometry $\varphi$ such that $\varphi\Gamma_1\varphi^{-1}=\Gamma_2$ or $\varphi\Gamma_1=\Gamma_2 \varphi$. One can see that orbit space $\Gamma_1.p$ (=points on $M/\Gamma_1$) map to orbit space $\Gamma_2.\varphi(p)$ (=points on $M/\Gamma_2$) by $\varphi$. call this map $\phi$. Thus $\pi_2\circ \varphi=\phi\circ \pi_1$ where $\pi_i\colon M\to M/\Gamma_i$ are Riemannian covering map. because $\pi_i$s are local isometry so $\phi$ is also a local isometry (Right?). How to proceed?

For converse, if $(M/\Gamma_1,g_1)$ and $(M/\Gamma_2,g_2)$ are isometric by $\psi$, because $M$ is simply connected therefore there is a unique lift $\widetilde{\psi}:M/\Gamma_1\to M$ such that $\pi_2\circ \widetilde{\psi}=\psi$. Now how to involve $\Gamma_i$s? $$\require{AMScd} \begin{CD} M @>>> M\\ @V \displaystyle \pi_1 V V @VV \displaystyle{\pi_2} V\\ M/\Gamma_1 @>>\displaystyle \psi> M/\Gamma_2 \end{CD}$$

Honestly I don't know how to reach to the goal.

Answer 1

Suppose $\Gamma_1 = g \Gamma_2 g^{-1}$. Then $g:M \to M$ induces an isometry $\bar{g}:M/\Gamma_1 \to M/\Gamma_2$ since for every $\gamma_1 \in \Gamma_1,$ there is a (unique) $\gamma_2 \in \Gamma_2$ such that $g\gamma_1 m= \gamma_2 g m$.

On the other hand, suppose $\phi: M/\Gamma_1 \to M/\Gamma_2$ is an isometry. Since $M$ is simply connected and the actions are free and proper, $\Gamma_i$ identifies with the group of deck transformations of $\pi_i:M \to M/\Gamma_i$.

To work with fundamental groups we should really pick a basepoint. Choose $x_0 \in M/\Gamma_1$ and a lift $\tilde{x_0} \in M$. Also choose a lift $\tilde{y_0} \in M$ of $y_0=\phi(x_0) \in M/\Gamma_2$. Then there is a unique lift $\tilde{\phi}:M \to M$ such that $\tilde{\phi}(\tilde{x_0})=\tilde{y_0}$ fitting into your commutative diagram. Now let $\gamma_1 \in \Gamma_1$. Then $\tilde{\phi} \circ \gamma_1 \circ \tilde{\phi}^{-1}$ maps $\tilde{y_0}$ to another lift of $y_0$, and is therefore an element of $\Gamma_2$.