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Let $G$ be a torsionfree kleinian group. Is there necessary an sufficient conditions on $G$ to be a knot group ?

It seems that there is some necessary conditions:

  1. $H_{1}(BG) = \mathbb{Z}$
  2. $H_{2}(BG) =H_{3}(BG) = 0$
  3. $M= \mathbb{H}^3/G$ has a finite hyperbolic volume and $M$ has exactly one cusp.
  4. $\mathbb{Z}^2$ is isomorphic to a subgroup of $G$.
  5. (I am not sure about this one) there exists an element $x\in G$ , $ G/<x>$ is the trivial group, where $<x>$ is the smallest normal subgroup of $G$ generated by $x$.

Question1: are these conditions sufficient?

Question2: let $G$ and $H$ be knot groups of hyperbolic knots such that the hyperbolic volume of $\mathbb{H}^3/G$ and $\mathbb{H}^3/H$ are equal. Are $G$ and $H$ isomorphic?

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  • $\begingroup$ Your 5 doesn't hold for all knot groups, not even all hyperbolic ones. For instance, if K is a fibered knot and x is represented by a curve on the fiber surface, then <x> is contained in the fundamental group of the fiber, which is a normal subgroup with quotient Z. Maybe you mean that there is an element x (eg a meridian) with that property. $\endgroup$
    – Danny Ruberman
    Commented Nov 13, 2021 at 23:07
  • $\begingroup$ @DannyRuberman thanks I was not sure about item 5. I made an edit. $\endgroup$
    – GSM
    Commented Nov 13, 2021 at 23:09
  • $\begingroup$ Also, I think these are a bit redundant: 4 follows from 3, as does $H_3(BG)=0$. $\endgroup$
    – Danny Ruberman
    Commented Nov 13, 2021 at 23:13
  • $\begingroup$ Have you read the Kervaire conditions for knot groups? See either Kawauchi's survey book or arxiv.org/pdf/1409.0704.pdf $\endgroup$
    – Ryan Budney
    Commented Dec 11, 2023 at 20:51

2 Answers 2

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For question 2, the answer is no. There are plenty of distinct hyperbolic knots (and hence having distinct fundamental group) with the same volume. For example, the Conway and Kinoshita-Terasaka knots differ by a mutation, and hence their complements have the same volume. You can get arbitrarily many knots with the same volume by doing mutations on knots that contain many Conway spheres.

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  1. suffices if x is conjugate to a primitive element in the peripheral subgroup. But in general it won’t suffice. There are many cusped hyperbolic homology circles that have weight 1, eg if they have tunnel number 1, so are 2 generated, yet are not knot complements. Choose a generator that generates the first homology. Killing this element kills the group, since the quotient is 1-generator and perfect.
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