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For the definition of Coxeter System, you can see:

Given a chamber Q, given a Coxeter System $(\Gamma,V)$, we can defined a set M by the following way: defined $M=\Gamma\times X/$~, the ~ is defined by $(g,x)$~$(h,y)$ if and only if x=y and $g^{-1}h\in \Gamma_{V(x)}$

V(x) is the element in V that the reflection generated by the element fixed x. $\Gamma_{(V(x))}$ is the group generated by $V(x)$

if the Coxeter system satisfies that all the $\Gamma_{V(x)}$ is finite, then Davis[1] has a theorem that: the manifold M generated by Q and $(\Gamma,V)$ is a topological manifold.

My question is in what condition can we judge that M is smooth?

If we assume the coxeter system is right angular, can we claim that M is smooth?

[1] Groups Generated by reflections and aspherical manifolds not covered by Euclidean space. Annals of Math. pp.293-324

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up vote 2 down vote accepted

See: Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Reflection groups on Riemannian manifolds. Annali di Matematica 186 (2006), 25-58. (pdf)

The idea there is to put a Riemannian metric on the chamber with the walls totally geodesic and with the right conditions on the angles betweens walls.

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That's great! At first I think that nobody has consider this! Thank you for your paper and answer! – Siqi He Oct 29 '12 at 9:28

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