Discrete subgroup of centralizer of transvections in isometries acts properly discontinuously

Question

My question will rely on a clarification of a proof, which I simply don't understand.

Let us denote by $X$ a pseudo-riemannian symmetric space and define
$$ Z_{\mathrm{Iso}\left(X\right)}G(X) = \{\, f \in \mathrm{Iso}\!\left(X\right) \mid gf = fg \text{ for all } g \in G(X) \,\} $$ as the centralizer of the transvection group $G(X)$ of the space $X$ and let $\mathrm{Iso}\left(X\right)$ be the isometry group of the regarding space.

In Cahen,Parker,"Pseudo-riemannian symmetric spaces" following assertion has been stated in Proposition 3.9:

Assertion: Any discrete subgroup $A$ of $Z_{\mathrm{Iso}\left(X\right)}G(X)$ acts properly discontinuously on $X$.

Proof: Let $C \subseteq X$ be a compact subset of $X$ with the property that $A_C = \{\, g \in A \mid gC \cap C \neq \emptyset \,\}$ is infinite, $\lvert A_C \rvert = \infty$. There exists a sequence $(a_n) \subseteq A_C$ with distinct elements and a sequence $(c_n) \subseteq C$ such that $a_n c_n \in C$ for all $n \in \mathbb{N}$. By compactness of $C$ and therefore taking sub-sequences, we may assume that $(c_n)$ converges towards $c \in C$ and that $(a_nc_n)$ converges towards $c^\prime \in C$. Let us choose $c$ as a basepoint of $X$. Since the transvection group $G(X)$ acts transitively on $X$ we can find a $g \in G(X)$ such that $gc = c^\prime$ and let $V_1 \subseteq U_1 \subseteq W_1$ be three neighborhoods of the identity in $G(X)$ such that $$ V_1=V_1^{-1}, g^{-1}V_1g\subseteq U_1 \text{ and } U_1^2 \subseteq W_1. $$ Define; $$ W_c = V_1 c, \\ \tilde{W}_c = W_1 c,\\ W_{c^\prime}=gW_c\\\tilde{W}_{c^\prime}=g\tilde{W}_c. $$

Question: Why does this define neighborhoods for the regarding points $c$ and $c^\prime$ and furthermore why does this define a neighborhoodbasis?

There exists an integer $N$ such that for all $n \geq N$ we have $c_n \in W_c$ and $a_nc_n \in W_{c^\prime}$. Indeed, there exists $g_n \in V_1$ such that $c_n = g_nc$ for all $n \geq N$. We want to show that $\lim_{n \to \infty} (a_nc) = c^\prime$. Consider therefore $$ a_n c = a_ng_n^{-1}c_n = g_n^{-1} a_n c_n \in g_n^{-1} g W_c \\ = gg^{-1}g_n^{-1}gW_c \subseteq gU_1V_1c \subseteq gU_1^2c \subseteq gW_1c = \tilde{W}_{c^\prime}, $$ which shows the claimed result.

The sequence $(b_n = a_{n+1}^{-1} a_n)$ in $A$ is clearly such that $\lim_{n \to \infty}b_n c = c$.

Question: Unfortunatley I have to redefine the term "clearly" into "highly unclearly". To me it is not obvious why this series converges towards $c$.

Answer 1

Question 1: This is a standard Baire category argument. Let $U$ be some small neighborhood of $c$. Since $G(X)$ acts continuously on $X$, there is a neighborhood $V$ of the identity in $G(X)$, such that $V^{-1} V c \subseteq U$. Choose a neighbourhood $V_1$ of the identity, such that the closure $\overline{V_1}$ is contained in $V$, and is compact. Since $G(X)$ is separable, it is covered by countably many translates $\{g_i V_1\}$ of $V_1$. Then $X$ is the union of $\{g_i V_1 c\}$, so the Baire Category Theorem implies that the closure of some $g V_1 c$ has nonempty interior. Since $g$ acts by a homeomorphism on $X$, this means that $\overline{V_1c}$ has nonempty interior. Since $\overline{V_1}$ is compact, we know that $\overline{V_1}c$ is closed, so $\overline{V_1c} \subseteq \overline{V_1}c \subseteq Vc$. So $Vc$ contains an open set. Then $V^{-1} V c$ contains a neighborhood of $c$. By assumption, this neighborhood is contained in $U$.

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Question 2: Since $a_n c \to c'$, we know that if $n$ is large, then there exist $w,w'$ close to the identity in $G(X)$, such that $a_n c = w' c'$ and $a_{n+1}c = w c'$. Since $a_{n+1}$ commutes with $w$ and $w'$, we have $$ a_{n+1}^{-1} (a_n c) = a_{n+1}^{-1} (w' c') = w' w^{-1} (a_{n+1}^{-1} w c') = w' w^{-1} c .$$ Since $w'$ and $w$ are both close to the identity, this is close to $c$. So $a_{n+1}^{-1} a_n c \to c$.