For the definition of Coxeter System, you can see: <http://en.wikipedia.org/wiki/Coxeter_group>


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



How to judge a manifold generated by Coxeter system smooth?
See:
Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Reflection groups on Riemannian manifolds. Annali di Matematica 186 (2006), 25-58.
[(pdf)](http://www.mat.univie.ac.at/~michor/reflec.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. 


