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? 



A question on Cayley graphs and hyperbolic 3-manifolds
$\newcommand{\HH}{\mathbb{H}}$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. 


