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Let $C$ be a finite 2-dimensional aspherical complex. The Michael Davis' trick is basically the following: One embeds $C$ to, say, ${\mathbb R}^5$, then takes a regular neighborhood in ${\mathbb R}^5$, producing a manifold $M$ with boundary. Then we triangulate the boundary and using a right angled Coxeter reflection group $G$ acting on ${\mathbb R}^5$, by gluing together all the images $GM'$, produce a manifold $M'$ on which $G$ acts properly by isometries. Factoring by a torsion-free finite index subgroup $H < G$, produce a closed aspherical manifold $M'/H$ whose fundamental group contains $\pi_1(C)$.

Question. Can we get a closed smooth (Riemannian) aspherical manifold where $C$ $\pi_1$-embeds?

I think, by reading Davis' Annals paper published in 1983 that the answer is "yes", but I would like to make sure.

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I'm pretty sure the answer is yes: you just have to make sure you model $M$ on a manifold with corners. Then the reflections should fit together to get a smooth manifold structure on M'. –  Ian Agol Mar 14 '11 at 19:31
    
@Ian: Yes, that is what I thought too. But I need something to refer to. It is needed for my paper. –  Mark Sapir Mar 14 '11 at 19:50
    
@Ian: Davis also writes that in dim 4 the situation is different. –  Mark Sapir Mar 14 '11 at 19:58
    
@Mark: I think, the reference is Davis's book, Proposition 11.1.3 (iv). That's what the number is in my electronic (and posssibly preliminary) copy; it is in the chapter on reflection group trick. The statement is that if the regular neighborhood and the triangulation of the boundary are smooth (which can always be arranged in your case) then the quotient aspherical manifold is smooth. –  Igor Belegradek Mar 14 '11 at 21:15
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Actually, the book is on his homepage: math.osu.edu/~mdavis/my%20book/My%20book.html –  Igor Belegradek Mar 14 '11 at 21:55
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