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Let $\Gamma$ be an arithmetic group and $X$ its symmetric space. Borel-Serre constructed a space $\bar{X} \supset X$ such that $\bar{X}/\Gamma$ is a compactification of $X/\Gamma$ [Corners and Arithmetic Groups, Comm. Math. Helv. 48(1973), 436-491, §7].

Moreover $\bar{X}$ is a contractible, finite-dimensional CW-complex and $\Gamma$ operates properly and cellularly on $\bar{X}$. In particular, if $H \le \Gamma$ is a finite subgroup, then the fixed point space $\bar{X}^H$ is non-empty.

Is $\bar{X}^H$ contractible or at least path-connected ?

Background: If so, it would follow that the non-abelian cohomology $H^1(G;\Gamma)$ is finite for $\Gamma$ arithmetic and $G \subseteq \operatorname{Aut}(\Gamma)$ finite. See also Finiteness of non-abelian cohomology

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It appears to me that $\bar{X}^H$ is disconnected in general, will always contain the connected contractible piece $X^H$, and have other components corresponding to intersections of $H$ with rational parabolics of $G$. For example, any $\mathbb{Q}$-irreducible finite subgroup $F$ of $G_{\mathbb{Z}}$ will only have trivial intersection with a rational parabolic, and hence $\bar{X}^F=X^F$ is contractible (by the usual CAT(0) arguments). In case $F$ meets some rational parabolics $\{P_\nu\}_\nu$, then $F< G_{\mathbb{Z}}$ is a $\mathbb{Q}$-reducible representation, and stabilizes some finite set $\{ \nu\}$ of vertices in the rational Tits building. It does not seem possible to describe in general the possible adjacency relations of these $F$-invariant vertices, i.e. $\bar{X}^F \setminus X^F$ may or may not be a simplex (e.g. for integral subgroups of rational Weyl group representations) and may perhaps be a pair of adjacent edges o--o--o.

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