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Let $K, K'$ be knots in $S^3$, and $T, T'$ the boundaries of their tubular neighborhoods.

Recall that by theorems of Waldhausen, and Gordon and Luecke, one knows the following: an isomorphism $[\pi_1(T) \to \pi_1(S^3 \setminus K)] \cong [(\pi_1(T') \to \pi_1(S^3 \setminus K')]$ implies that $K$ and $K'$ are isotopic.

Suppose the category of linear representations of $[\pi_1(T) \to \pi_1(S^3 \setminus K)]$ is equivalent to the category of linear representations of $[\pi_1(T') \to \pi_1(S^3 \setminus K')]$. Are $K$ and $K'$ isotopic?

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Note: In a previous version of this question, I asked whether this is possible if you interpret the category above as a tensor category; in the comments it is pointed out that then, yes, at least if you allow big representations. But, now I realize that in fact I don't have the tensor product structure.

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  • $\begingroup$ What do you mean by linear representations of a morphism of groups? $\endgroup$
    – Qiaochu Yuan
    Commented Nov 7, 2015 at 0:14
  • $\begingroup$ representation of one, representation of the other, morphism between them compatible with the map of groups. $\endgroup$
    – Vivek Shende
    Commented Nov 7, 2015 at 0:40
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    $\begingroup$ Do you allow infinite-dimensional representations? If so, the answer should be trivially yes, since if you allow infinite dimensional representations, a (possibly infinite) discrete group $G$ is the group of symmetric monoidal endomorphisms of the unique (up to isomorphism) symmetric monoidal exact functor $\mathrm{Rep}_{\mathbb C}(G) \to \mathrm{Vect}_{\mathbb C}$, and homomorphisms $G \to H$ are uniquely determined by the corresponding symmetric monoidal functors $\mathrm{Rep}_{\mathbb C}(H) \to \mathrm{Rep}_{\mathbb C}(G)$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 7, 2015 at 1:59
  • $\begingroup$ If the knot is hyperbolic then the hyperbolic representation is faithful, so maybe infinite-dimensional representations aren't needed, at least for hyperbolic knots. $\endgroup$
    – Peter Samuelson
    Commented Nov 7, 2015 at 2:14
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    $\begingroup$ Every knot group admits a faithful finite-dimensional representation. Over the integers this is a theorem of Przytycki--Wise, though I have a feeling that over the reals it was proved by Kronheimer--Mrowka. $\endgroup$
    – HJRW
    Commented Nov 7, 2015 at 5:57

1 Answer 1

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So long as I allow myself infinite size representations, and writing $\mathbb{Z}[G]-mod$ for representations of $G$ in $\mathbb{Z}$-modules, then so long as I have the forgetful functor

$$\mathbb{Z}[G]-mod \to \mathbb{Z}-mod$$

I can recover $\mathbb{Z}[G]$ as endomorphisms of this functor.

In general, one cannot recover $G$ even from $\mathbb{Z}[G]$, even for finite groups $G$.

However, as it turns out, knot groups are known to be left orderable, and it is known that for left orderable groups, the only units in $\mathbb{Z}[G]$ are the obvious ones, from which $G$ can be recovered.

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