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There are two hyperbolic closed 3-manifolds, but I don't know whether they are homeomorphic or not. The only thing I know is that the Cayley graphs of their fundamental groups are quasi-isometric.

My question is:

Will the manifolds be homeomorphic?

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If the manifolds are compact then their Cayley graphs are quasi-isometric to hyperbolic 3-space; so you get no info this way. –  Anton Petrunin Aug 17 '13 at 1:35
    
@Anto: Will you explain it ? –  yanqing Aug 17 '13 at 7:25
    
@Anton: Also, please consider posting your comment (and any further explanation) as an answer. Thank you. –  Ricardo Andrade Aug 17 '13 at 7:40
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@RicardoAndrade I conjecture that Anton does not think that this is a research level discussion -- I know I think it more appropriate for stackexchange. –  Igor Rivin Aug 17 '13 at 15:17
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1 Answer

up vote 10 down vote accepted

$\newcommand{\HH}{\mathbb{H}}$Here is an expansion of what Anton is saying.

Suppose that $M$ is a closed hyperbolic three-manifold. It follows that the universal cover of $M$ is $\HH^3$: hyperbolic space. The covering map of $M$ comes with a deck group - namely there is an action of $\pi_1(M)$ on $\HH^3$ so that the quotient $\pi_1(M) \backslash \HH^3$ is homeomorphic to $M$.

Now fix a set $S$ of generators for $\pi_1(M)$. Let $\Gamma = \Gamma(M, S)$ be the Cayley graph for $\pi_1(M)$ relative to $S$. Also, fix a point $x$ of $\HH^3$. Let $\rho_x{:}\,\Gamma \to \HH^3$ be the orbit map: namely $\rho_x(g) = g \cdot x$ for vertices of $\Gamma$, and all edges of $\Gamma$ are sent to geodesics. The Švarc-Milnor lemma says that $\rho_x$ is a quasi-isometry.

It follows that the Cayley graphs for any pair of closed hyperbolic three-manifolds are quasi-isometric. Thus homeomorphism types cannot be distinguished this way.

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