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 $c$.* 



Discrete subgroup of centralizer of transvections in isometries acts properly discontinuously
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 $U$.




---


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 $a\_{n+1}^{-1} a\_n c \to c$.


