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I start by saying that I am not an expert in this field and I apologize if the question is too elementary.

Let $K$ be a knot in $S^3$. I denote by $\pi_1(K)$ the knot group, which is the fundamental group of its exterior: $$ \pi_1(K) = \pi_1(S^3 \smallsetminus K) .$$

The minimal number of generators of a knot $K$ is the minimal number of generators of $\pi_1(K)$.

I am searching for knots with an arbitrarily high minimal number of generators. In particular:

  1. I found in https://arxiv.org/abs/1007.3175 , Lemma 4.17, a reference: Goodrick, R. (1968). Non-simplicially collapsible triangulations of In. In this article, the author proves that the connected sum of $n$ copies of a two-bridge knot is a $m$-bridge knot with $m$>$n$. A sharper result should hold from Schultens, Jennifer (2003). Additivity of bridge numbers of knots. I cannot understand, though, how this should prove the statement. In particular, by Knot theory question: bridge number vs. min generators of fundamental group of complement , this should not imply the result. Maybe I am missing something in the article, I did not go through the details. However, the article is quite dated and there is, hopefully, a simple way to prove this fact nowadays. So the first question is: is there a simple way to prove that there are knots with an arbitrarily high minimal number of generators?

  2. The techniques used seem to rely on the connected sum. What if we search for prime knots with an arbitrarily high minimal number of generators?

  3. What if we search for hyperbolic knots with an arbitrarily high minimal number of generators?

Thank you in advance for the attention.

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If $\pi_1(S^3\setminus K)$ has a presentation with $n$ generators then its representation variety $\mathrm{Hom}(\pi_1(S^3\setminus K),SL_2(\mathbb{C}))$ is a subvariety of $(SL_2(\mathbb{C}))^n$, which has complex dimension $3n$, so any component will have complex dimension at most $3n$. So if you want the minimal number of generators to be arbitrarily large, you just have to find knots with high-dimensional representation varieties. (You can replace $SL_2(\mathbb{C})$ with other groups if you prefer.)

Cooper and Long ("Remarks on the A-polynomial of a knot", section 8) show how to construct hyperbolic knots where there are components of arbitrarily large dimension. Start with some $n$-fold connected sum, and realize it as the closure of a braid $\beta$ such that the union of the closure $\hat\beta$ with its braid axis is hyperbolic. Lift $\hat\beta$ to the $p$-fold cyclic branched cover of $S^3$, branched over the braid axis -- this will again be $S^3$, since the axis is unknotted -- to get a new knot, which is hyperbolic for all large enough $p$. (This knot will be the closure of the braid $\beta^p$.) Then the representation variety of this new knot turns out to have a component of dimension at least $n$ as well.

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I think the rank of the Alexander polynomial gives you a lower bound on the number of generators of your fundamental group. i.e. just compute the Alexander module by lifting a 2-complex for the knot exterior to the universal cover.

It's a very coarse bound, but it seems to answer your question, as there are hyperbolic knots with arbitrarily large Alexander polynomials.

edit: my estimation was too quick. See Kyle Miller's comment below. His argument gets a lower bound and carefully avoids my mistake. But the statement needs to be changed, i.e. he uses full Alexander modules rather than Alexander polynomials.

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    $\begingroup$ What is the "rank" of a Laurent polynomial in $\mathbf{Z}[t^{\pm 1}]$? $\endgroup$
    – YCor
    Commented Jul 14, 2021 at 8:56
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    $\begingroup$ I probably should have said degree. This is with consideration that the Alexander polynomial is only well-defined up to multiplication by a unit. $\endgroup$
    – Ryan Budney
    Commented Jul 14, 2021 at 9:25
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    $\begingroup$ @RyanBudney: torus knots have Alexander polynomials with arbitrarily large degrees, but their fundamental groups have two generators. Maybe the number of generators of the Alexander module is a lower bound? $\endgroup$
    – Marco Golla
    Commented Jul 14, 2021 at 10:53
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    $\begingroup$ The usual term is “span” - the difference in degree between the maximal and minimal terms. And, as noted, the span is not related to the minimal number of generators of the fundamental group. Instead it is a lower bound for (twice) the Seifert genus of the knot. $\endgroup$
    – Sam Nead
    Commented Jul 14, 2021 at 13:44
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    $\begingroup$ The sequence of higher Alexander polynomials gives some information on the number of generators. If $\Delta_K^i(t)$ is the GCD of the $i$th Fitting/elementary ideal of the Alexander module (so $\Delta_K^0(t)$ is the usual Alexander polynomial), then as a course bound if $\Delta_K^i(t)\neq 1$ then the fundamental group needs at least $i$ generators. (Also, the bridge number is an upper bound for the minimal number of generators, and for a given bridge number you can make Alexander polynomials of arbitrarily high breadth/span/rank.) $\endgroup$
    – Kyle Miller
    Commented Jul 14, 2021 at 17:36

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