Is it possible to reconstruct a finitely generated group from its category of representations?

Question

Suppose $G$ is a finitely generated group, and suppose $Rep_k(G)$ is its category of representations over some field (or maybe even a ring) $k$, endowed with whatever extra structure is needed --- monoidal structure, fiber functor etc. Is it possible to reconstruct $G$ from this category?

Please note that my question isn't about Tannakian reconstruction. Indeed, Tannakian reconstruction (in the forms familiar to me) requires the tensor category to be rigid (that is, it considers only finite dimensional representations) and produces not $G$ itself, but some proalgebraic group over $k$, called $k$-proalgebraic completion. Proalgebraic completion of a finitely generated group could be trivial: for example, Tarski monster doesn't have any finite dimensional representations.

Thanks in advance!

Answer 1

Let $F: Rep_k(G) \to k\mathsf{-mod}$ be the fibre functor. $G$ can be reconstructed from this functor and the monoidal structure as the group of automorphisms, because $t: G\to Aut(F,\otimes,k), g\mapsto (t_g^V)_{V\in Rep_k(G)}$ is an isomorphism, where $t_g^V$ is just the $k$-linear map $F(V)\to F(V), v\mapsto gv$. Because the morphisms in $Rep_k(G)$ are $k[G]$-linear maps, $t_g$ really is a natural transformation. It clearly satisfies $t_g^{-1}=t_{g^{-1}}$ so that it is a natural automorphism of $F$. By definition of the monodial structure it also satisfies $t_g^{V\otimes W} = t_g^V\otimes t_g^W$ and $t_g^{k} = id$.

$t$ is injective, because $g = t_g^{k[G]}(1)$. $t$ is surjective, because if $\tau=(\tau^V)$ is any natural automorphism of $F$ and $u:=\tau^{k[G]}(1)$, then $u$ is a unit of $k[G]$ because $\tau$ is invertible and by naturality of $\tau$ applied to the $k[G]$-linear map $k[G]\to V, 1\mapsto v$, one sees that $\tau^V = v\mapsto uv$. $u$ is not just any invertible element, it is an element of $G$, because $\tau^{V\otimes W} = \tau^V\otimes\tau^W$ and $\tau^k=id$: Write $u=\sum_g a_g g$ with coefficients $a_g\in k$ and consider how $\tau$ acts on $k[G]\otimes k[G]$. On the one hand, it is multiplication by $u$, so $\tau(x\otimes y) = u(x\otimes y) = \sum_g a_g g(x\otimes y) = \sum_g a_g (gx)\otimes(gy)$. On the other hand, it is $\tau^{k[G]}\otimes\tau^{k[G]}$, so that $\tau(x\otimes y) = (ux)\otimes(uy) = \sum_{g,h} a_g a_h (gx)\otimes(hy)$. Letting $x=y=1$ and comparing coefficients, we find $a_g a_h = 0$ for all $g\neq h$. This means that at most one of the coefficients is non-zero. The action on the trivial module $k$ is trivial so that $1=\tau(1) = u1 = \sum_g a_g$ from which we conclude that there is exactly one non-zero coefficient that it is equal to one, i.e. $u$ is a group element.

Therefore if you have the full category of all representations with its fibre functor and the monoidal structure, it is trivially easy to reconstruct $G$. The point of the Tannakian theory is that for certain groups knowledge of the finite-dimensional part of $Rep_k(G)$ is sufficient to reconstruct $G$.