# Example $4.8$ - Chapter $0$ - Do Carmo's Riemannian Geometry

I have some doubts about the proof in example $$4.8$$ in chapter $$0$$ of Do Carmo's Riemannian Geometry book concerning to the part of the proof that the canonical projection of a differentiable manifold $$M$$ in it's quotient space $$M/G$$ given by the relation defined by an properly discontinuous action is a local diffeomorphism. Precisely, I didn't understand the following statement:

Let $$p_1 = (\pi_1^{-1} \circ \pi_2) (p_2)$$. Then $$p_1$$ and $$p_2$$ are equivalent in $$M$$, hence there is a a $$g \in G$$ such that $$gp_2 = p_1$$. It follows easily that the restriction $$(\pi_1^{-1} \circ \pi_2)|_{\textbf{x}_2(W)}$$ coincides with the diffeomorphism $$\varphi_g|_{\textbf{x}_2(W)}$$.

Why $$(\pi_1^{-1} \circ \pi_2)|_{\textbf{x}_2(W)}$$ coincides with the diffeomorphism $$\varphi_g|_{\textbf{x}_2(W)}$$? How the fact that the action is properly discontinuous is used here to ensure that the functions coincide on $$\textbf{x}_2(W)$$? Is $$G$$ the group of $$\textbf{all}$$ diffeomorphisms of $$M$$ or just a group of a $$\textbf{specific}$$ diffeomorphisms of $$M$$?

About my last question, I think $$G$$ is the group of $$\textbf{all}$$ diffeomorphisms of $$M$$, because he stated that $$p_1$$ and $$p_2$$ are equivalent.

Below, there is the example $$4.8$$.

$$4.8$$ Example. ($$\textit{Discontinuous action of a group}$$). We say that a group $$G$$ acts on a differentiable manifold $$M$$ if there exists a mapping $$\varphi: G \times M \rightarrow M$$ such that

(i) For each $$g \in G$$, the mapping $$\varphi_g: M \longrightarrow M$$ given by $$\varphi_g(p) = \varphi(g,p)$$, $$p \in M$$, is a diffeomorphism and $$\varphi_e = \text{identity}$$.

(ii) If $$g_1, g_2 \in G$$, $$\varphi_{g_1g_2} = \varphi_{g_1} \circ \varphi_{g_2}$$.

Frequently, when dealing with a single action, we set $$\varphi(g,p) = gp$$; in this notation, condition $$(ii)$$ can be interpreted as a formof associativity: $$(g_1g_2)p = g_1(g_2p)$$.

We say that the action is $$\textit{properly discontinuous}$$ if every $$p \in M$$ has a neighborhood $$U \subset M$$ such that $$U \cap g(U) = \emptyset$$ for all $$g \neq e$$.

When $$G$$ acts on $$M$$, the action determines an equivalence relation $$\sim$$ on $$M$$, in which $$p_1 \sim p_2$$ if and only if $$p_2 = gp_1$$, for some $$g \in G$$. Denote the quotient space of $$M$$ by this equivalence relation by $$M / G$$. The mapping $$\pi: M \longrightarrow M/G$$, given by

$$\pi(p) = \text{equiv. class of} \ p = Gp$$

will be called the $$\textit{projection}$$ of $$M$$ onto $$M/G$$.

Now let $$M$$ be a differentiable manifold and let $$\varphi: G \times M \longrightarrow M$$ be a properly discontinuous action of a group $$G$$ on $$M$$. We are going to show that $$M/G$$ has a differentiable structure with respect to which the projection $$\pi: M \longrightarrow M/G$$ is a local diffeomorphism.

For each $$p \in M$$ choose a parametrization $$\textbf{x}: V \longrightarrow M$$ at $$p$$ so that $$\textbf{x}(V) \subset U$$, where $$U \subset M$$ is a neighborhood og $$p$$ such that $$U \cap gU = \emptyset$$, $$g \neq e$$. Clearly $$\pi|_U$$ is injective, hence $$\textbf{y} = \pi \circ \textbf{x}: V \longrightarrow M/G$$ is injective. The family $$\{ (V,\textbf{y}) \}$$ clearly covers $$M/G$$; for such a family to be a differentiable structure, it suffices to show that given two mappings $$\textbf{y}_1 = \pi \circ \textbf{x}_1: V_1 \longrightarrow M/G$$ and $$\textbf{y}_2 = \pi \circ \textbf{x}_2: V_2 \longrightarrow M/G$$ with $$\textbf{y}_1(V_1) \cap \textbf{y}_2(V_2) \neq \emptyset$$, then $$\textbf{y}_1^{-1} \circ \textbf{y}_2$$ is differentiable.

For this, let $$\pi_i$$ be the restriction of $$\pi$$ to $$\textbf{x}_i(V_i)$$, $$i = 1,2$$. Let $$q \in \textbf{y}_1(V_1) \cap \textbf{y}_2(V_2)$$ and let $$\textbf{r} = \textbf{x}_2^{-1} \circ \pi_2^{-1}(q)$$. Let $$W \subset V_2$$ be a neighborhood of $$\textbf{r}$$ such that $$(\pi_2 \circ \textbf{x}_2)(W) \subset \textbf{y}_1(V_1) \cap \textbf{y}_2(V_2)$$. Then, the rstriction to $$W$$ is given by

$$\textbf{y}_1^{-1} \circ \textbf{y}_2|_W = \textbf{x}_1^{-1} \circ \pi_1^{-1} \circ \pi_2 \circ \textbf{x}_2.$$

Therefore, it is enough to show that $$\pi_1^{-1} \circ \pi_2$$ is differentiable at $$p_2 = \pi_2^{-1}(q)$$. Let $$p_1 = \pi_1^{-1} \circ \pi_2 (p_2)$$. Then $$p_1$$ and $$p_2$$ are equivalent in $$M$$, hence there is a $$g \in G$$ such that $$gp_2 = p_1$$ . It follows easily that the restrinction $$\pi_1^{-1} \circ \pi_2|_{\textbf{x}_2(W)}$$ coincides with the diffeomorphism $$\varphi_g|_{\textbf{x}_2(W)}$$, which proves that $$\pi_1^{-1} \circ \pi_2$$ is differentiable at $$p_2$$, as stated.

$$G$$ is any group acting by diffeomorphisms on $$M$$. The points $$p_1, p_2$$ are equivalent simply because both are mapped by $$\pi$$ to $$q \in M/G$$.

The claim that $$(\pi_1^{-1} \circ \pi_2)|_{\textbf{x}_2(W)} = \varphi_g|_{\textbf{x}_2(W)}$$ is potentially false for the set $$W$$ considered by do Carmo. However, it is true if $$W$$ is chosen carefully.

Given $$q \in \textbf{y}_1(V_1) \cap \textbf{y}_2(V_2)$$, let $$p_i = \pi_i^{-1}(q) \in \textbf{x}_i(V_i)$$. Since $$\pi(p_i) = q$$, these two points are equivalent which means that there exists $$g \in G$$ such that $$gp_2 = p_1$$. Choose a neighborhood $$U_2$$ of $$p_2$$ in $$M$$ such that $$\varphi_g(U_2) \subset \textbf{x}_1(V_1)$$ and let $$V'_2 = \textbf{x}_2^{-1}(U_2)$$ which is an open subset of $$V_2$$ containing $$\textbf{r} = \textbf{x}_2^{-1}(p_2)$$. Now let $$W \subset V'_2$$ be an open neighborhood of $$\textbf{r}$$ such that $$(\pi_2 \circ \textbf{x}_2)(W) \subset \textbf{y}_1(V_1) \cap \textbf{y}_2(V_2)$$.

Then $$(\pi_1^{-1} \circ \pi_2)|_{\textbf{x}_2(W)} = \varphi_g|_{\textbf{x}_2(W)}$$ is equivalent to $$\pi_2|_{\textbf{x}_2(W)} = \pi_1 \circ \varphi_g|_{\textbf{x}_2(W)}$$. Note that the composition on the right hand side makes sense because we constructed $$W$$ such that $$\varphi_g(\textbf{x}_2(W)) \subset \varphi_g(\textbf{x}_2(V'_2)) = \varphi_g(U_2) \subset \textbf{x}_1(V_1)$$.

But now we have for $$p \in \textbf{x}_2(W)$$ $$(\pi_1 \circ \varphi_g)(p) = \pi(gp) = \pi(p) = \pi_2(p) .$$

• What is $\pi(g)$ and where it was used that the action is properly discontinuous? – George Apr 12 at 22:33
• $\pi(g)$ was a typo which I have corrected. That the action is properly discontinuous enters a the very beginning of the prroof: $U$ was chosen such that $U \cap gU = \emptyset$ for all $g \ne e$. – Paul Frost Apr 12 at 22:59
• Ok, sorry for this stupid question. How do you know that $\pi(gp) = \pi(p)$ if the $g$ is not arbritrary and, therefore you can choose and $g$, which acts on the diffeomorphisms of $M$ such that $g(\textbf{x}_2(W)) = \textbf{x}_2(W)$? – George Apr 13 at 1:31
• $\pi : M \to M/G$ identifies all points which are equivalent under the action of $G$. We have $p \sim p'$ if and only if there exists $g \in G$ such that $p' = gp$. This explains why $\pi(gp) = \pi(p)$, and it is true for any $g$. Since $\pi(p_1) = \pi(p_2) = q$, we know that $p_1 \sim p_2$, that is, there exists a $g \in G$ such that $gp_2 = p_1$. This $g$ is not necessarily uniquely determined, but it is not arbitrary. For this $g$, $W$ was constructed so that $g\mathbf{x}_2(W) \subset \mathbf{x}_1(V_1)$. Then in fact $\pi_2|_{\textbf{x}_2(W)} = \pi_1 \circ \varphi_g|_{\textbf{x}_2(W)}$. – Paul Frost Apr 13 at 12:49
• Ok, so if you ensure that $g(\textbf{x}_2(W)) \subset \textbf{x}_1(V_1)$ for some $g \in G$, then you ensure that there is $\tilde{p} \in \textbf{x}_1(V_1)$ such that $gp = \tilde{p}$ for some $p \in \textbf{x}_2(W)$, which confuse me is that you said that given $p \in \textbf{x}_2(W)$, exist some $g \in G$ such that $gp = p$, because $g(\textbf{x}_2(W)) \subset \textbf{x}_1(V_1)$, but this doesn't ensure that $gp = p$, but it ensures that $gp = \tilde{p}$ for some $\tilde{p} \in \textbf{x}_1(V_1)$. – George Apr 14 at 18:52