Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.