# What does the representation category of the knot group know?

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?



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.

• What do you mean by linear representations of a morphism of groups? – Qiaochu Yuan Nov 7 '15 at 0:14
• representation of one, representation of the other, morphism between them compatible with the map of groups. – Vivek Shende Nov 7 '15 at 0:40
• 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 $\mathrm{Rep}_{\mathbb C}(H) \to \mathrm{Rep}_{\mathbb C}(G)$. – Theo Johnson-Freyd Nov 7 '15 at 1:59
• If the knot is hyperbolic then the hyperbolic representation is faithful, so maybe infinite-dimensional representations aren't needed, at least for hyperbolic knots. – Peter Samuelson Nov 7 '15 at 2:14
• 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. – HJRW Nov 7 '15 at 5:57

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.