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It is well-known that if $\widetilde M\to M$ is a Galois cover of a compact Riemannian manifold $M$ with deck-transformation group $G$, then the growth of $G$ equals the volume growth of $\widetilde M$ (in the pullback metric).

Question. Is the same true when $M$ is a finite volume complete Riemannian manifold and $G$ is finitely generated?

The usual proof (a la Svarc-Milnor) argues that $G$ and $\widetilde M$ are quasiisometric (since $M$ is compact), and then uses that growth is a quasiisometry invariant. One could hope that the reasoning extends to the finite volume case when quasi-isometry is replaced by measure equivalence (ME), but growth type is not an invariant of ME. On the other hand, I do not have counterexamples for the above question.

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what do you mean by equal? is $\exp(2n)=\exp(3n)$? –  YCor May 19 '13 at 14:16
    
By "equal" I mean that the growth functions have the same growth type, i.e. they dominate each other, where $g$ dominates $f$ if and only if $f(t)\le Ag(At+B)+B$ for all $t$ and some constants $A, B$. –  Igor Belegradek May 19 '13 at 14:34
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up vote 8 down vote accepted

No, it's not true. A cusp is a counterexample (see below for an example with no boundary). Here by cusp I mean the subset $\{\text{Im}(z)\ge 1\}$ of the upper half-plane with hyperbolic metric modulo $z\mapsto z+1$. This has finite volume and the fundamental group has linear $\mathbf{Z}$ growth, but the universal covering, which is a horodisc, has exponential growth (because the $n$-ball contains, for some $c>0$, the rectangle $[0,e^{cn}]\times [1,2]$, which contains exponentially many disjoint balls $[m,m+1]\times [1,2]$, which all have the same volume).

If you want no boundary, take the cylinder $\mathbf{C}$ modulo $z\mapsto z+1$, endow it with the cusp Riemannian metric on both components of $\{|\text{Im}(z)|\ge 1\}$, and extend arbitrarily to a Riemannian metric in between.

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This is very helpful. I see now that I have had some silly confusions about growth. –  Igor Belegradek May 19 '13 at 14:50
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