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For which $3$-manifolds $M$ is the fundamental group $\pi_1(M)$ finitely generated and has positive rank gradient?

Recall that the rank gradient of a finitely generated group $G$ is defined to be $$\inf_{H} \frac{d(H) - 1}{[G : H]}$$ where the infimum is taken over all finite index subgroups $H$ of $G$, and $d(H)$ stands for the minimal cardinality of a generating set of $H$.

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  • $\begingroup$ Well, any virtually fibred 3-manifold must have zero rank-gradient, so that doesn't leave too many possibilities. $\endgroup$
    – HJRW
    Commented Nov 16, 2015 at 11:34
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    $\begingroup$ @HJRW right! I knew this. So I want to know in which cases the rank gradient is known to be $ > 0$. Is there some exact meaning in "too many possibilities" ? $\endgroup$
    – Pablo
    Commented Nov 16, 2015 at 11:37

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This is answered in the proof of Theorem 8.5 of this paper. This says that the rank gradient is zero iff $M$ is prime or $\mathbb{RP}^3\#\mathbb{RP}^3$.

Edit: There's a small step missing from the argument. The argument shows that if $G$ is an orientable connected sum 3-manifold group (not $\mathbb{RP}^3\#\mathbb{RP}^3$), then there exists $G' \lhd G$ of finite index so that $G'$ has positive corank gradient, hence positive rank gradient. Since $d(G')-1\leq [G:G'](d(G)-1)$, if $G$ has rank gradient $0$, so would $G'$ a contradiction (one applies this inequality to $G'\cap H \lhd H$). This applies as well to non-orientable 3-manifolds, where one has to consider the $\mathbb{RP}^2$-decomposition, or just posit that there is a prime cover.

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    $\begingroup$ You assume that $M$ is closed and orientable. What about the general case? $\endgroup$
    – Pablo
    Commented Nov 16, 2015 at 14:41
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    $\begingroup$ @Pablo: see my revision. $\endgroup$
    – Ian Agol
    Commented Nov 16, 2015 at 14:51
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    $\begingroup$ Correct, in the orientable case. $\endgroup$
    – Ian Agol
    Commented Nov 16, 2015 at 15:33
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    $\begingroup$ For non-orientable manifolds, one has to consider the $\mathbb{RP}^2$ decomposition. So, for example, if a hyperbolic 3-manifold $M$ admits an orientation-reversing isometry with two fixed points, the quotient will be an oribfold with two point which are cones over $\mathbb{RP}^2$. Remove neighborhoods of these and glue the $\mathbb{RP}^2$'s together. This manifold has orientable double cover $M\# (S^2\times S^1)$, so has positive rank gradient. You can imagine generalizations of this. See: plms.oxfordjournals.org/content/s3-11/1/469 $\endgroup$
    – Ian Agol
    Commented Nov 16, 2015 at 16:01
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    $\begingroup$ @Pablo: correct $\endgroup$
    – Ian Agol
    Commented Nov 16, 2015 at 16:16

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