4
$\begingroup$

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 $H$.

$\endgroup$
2
  • $\begingroup$ Well, any virtually fibred 3-manifold must have zero rank-gradient, so that doesn't leave too many possibilities. $\endgroup$
    – HJRW
    Nov 16, 2015 at 11:34
  • 1
    $\begingroup$ @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
    Nov 16, 2015 at 11:37

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
9
  • 1
    $\begingroup$ You assume that $M$ is closed and orientable. What about the general case? $\endgroup$
    – Pablo
    Nov 16, 2015 at 14:41
  • 1
    $\begingroup$ @Pablo: see my revision. $\endgroup$
    – Ian Agol
    Nov 16, 2015 at 14:51
  • 1
    $\begingroup$ Correct, in the orientable case. $\endgroup$
    – Ian Agol
    Nov 16, 2015 at 15:33
  • 1
    $\begingroup$ 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
    Nov 16, 2015 at 16:01
  • 1
    $\begingroup$ @Pablo: correct $\endgroup$
    – Ian Agol
    Nov 16, 2015 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.