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Let $M$ be a smooth manifold,$G$ a subgroup of diffeomorphisms on $M$.We require that $G$ acts properly discontinuous on $M$,that is:for any $p\in M$,there is a neighborhood $U$ s.t.$$gU\cap U=\emptyset,\forall g\in G,g\neq e,$$where $e$ is the identity.

There is an equivalent relation in $M$ given by :$$p\text{~}q\Leftrightarrow \exists g\in G ~s.t.gp=q.$$Let $M/G$ denote the quotient space of $M$ under this equivalent relation,we now introduce an atlas on $M/G$.Since $G$ acts properly discontinuous on $M$,there exist a coordinate cover $\{(U_i,\phi_i)\}$ on $M$ s.t. $gU_i\cap U_i = \emptyset,\forall g\in G,g\neq e$,let $\pi_i$ be the restriciton of the projection $\pi\colon M\to M/G$ on $U_i$,we see that $\pi_i:U_i\to \pi_i(U_i)$ is bijective.To show that $\{(\pi(U_i),\phi_i\pi_i^{-1})\}$ gives an atlas on $M/G$,we need to check the compatibility. In fact,on $\pi(U_i)\cap\pi(U_j)\neq \emptyset$,we have $$\phi_i\pi_i^{-1}(\phi_j\pi_j^{-1})^{-1}\mid_{\phi_j\pi_j^{-1}(\pi(U_i)\cap\pi(U_j))}=\phi_i\pi_i^{-1}(\phi_j\pi_j^{-1})^{-1}\mid_{\phi_j(U_i\cap U_j)}=\phi_i\pi_i^{-1}\pi_j\phi_j^{-1}\mid_{\phi_j(U_i\cap U_j)},$$ so it reduced to show $\pi_i^{-1}\pi_j$ is smooth on $U_i\cap U_j$.

let $p\in U_i\cap U_j$,and $q=\pi_i^{-1}\pi_j(p)$,since $\pi_i(q)=\pi_j(p)$,there exist $g\in G s.t.gp=q$.Since g is a diffeomorphism, $\pi_i^{-1}\pi_j$ is smooth.Now my question is :the diffeomorphism $g$ apparently depends on the particular point p,I think we need to show that $$\pi_i^{-1}\pi_j\mid_{ U_i\cap U_j}=g\mid_{ U_i\cap U_j},$$but how?

Here is my idea:note that $\pi_i^{-1}\pi_j:U_i\cap U_j\to U_i\cap U_j,$ so $q=gp\in g(U_i\cap U_j)\cap(U_i\cap U_j)$,by the choice of $U_i,U_j$,this happens only if $g=e$.Hence $\pi_i^{-1}\pi_j\mid_{ U_i\cap U_j}=i_d\mid_{ U_i\cap U_j}$.Is this argument correct?

Also there is an excercise I can't solve:show that $M/G$ is orientable iff there is an orientation on $M$ and this orientation is preserved by all $g\in G$.

Edit:I may have find what caused the problem.The thing is $$\pi_j^{-1}(\pi(U_i)\cap\pi(U_j))\neq U_i\cap U_j,$$ and it may well happen that $\pi(U_i)\cap\pi(U_j)\neq \emptyset ~\text{but}~ U_i\cap U_j=\emptyset$.Maybe it is not appropriate to use $\{(\pi(U_i),\phi_i\pi_i^{-1})\}$ as a atlas?

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  • $\begingroup$ It is not true that $\pi_i^{-1}\pi_j:U_i\cap U_j\to U_i\cap U_j$. $\endgroup$ – Amitai Yuval Dec 4 '14 at 9:13
  • $\begingroup$ @Amitai Yuval:Can you show me why? $\endgroup$ – Wei Xia Dec 4 '14 at 9:19
  • $\begingroup$ In fact, $U_i\cap U_j$ is not what you need to look at. It is $\pi_j^{-1}\pi_i(U_i)$. $\endgroup$ – Amitai Yuval Dec 4 '14 at 9:25
  • $\begingroup$ It's something wrong in your question: I suppose $\pi_i$ is just the restriction of $\pi : M \to M/G$ to $U_i$. So if $p \in U_i \cap U_j$, then $\pi_j^{-1}\pi_i(p) = p$. $\endgroup$ – user99914 Dec 4 '14 at 9:26
  • $\begingroup$ @ John:$\pi_i$ is exactly the restriction of $\pi$ to $U_i$.But what you have suggested is just the same thing as in my question,just the indices changed. $\endgroup$ – Wei Xia Dec 4 '14 at 14:11
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Suggestion: Let $p\in\pi_j^{-1}\pi_i(U_i)$, and take $g\in G$ such that $g(p)=\pi_i^{-1}\pi_j(p)$. Set $$V=g^{-1}(U_i)\cap\pi_j^{-1}\pi_i(U_i),$$and note that $V$ is an open neighborhood of $p$. Show that if some $q\in V,h\in G$ satisfy $h(q)=\pi_i^{-1}\pi_j(q),$ it follows that $h=g$, hence $\pi_i^{-1}\pi_j|_V=g$.

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