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

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.

• There are non-isomorphic (finite) groups with isomorphic group algebras over some field (Berman). So you have to cheat at some point to keep track of the group.
– YCor
Commented Mar 1, 2019 at 8:31
• There are groups $G_1,G_2$ such that $\mathbb{Z}G_1 \cong \mathbb{Z}G_2$, as described here: jstor.org/stable/3062112 Commented Mar 1, 2019 at 11:07
• You can then tensor the group rings by any ring $R$ you like to get an isomorphism of the corresponding $R$-algebras. However, finite groups are compact groups, so we can use the Tannakian theory for compact groups (equivalently affine proalgebraic groups over $\mathbb{R}$). The key point is that we need the symmetric monoidal structure to recover the group multiplication. This doesn't affect the question as asked, because the monoidal structure was explicitly included. Commented Mar 1, 2019 at 11:12
• You cannot recover the group from the group ring as a ring, but you can recover it from the group ring as a Hopf algebra. A Hopf algebra contains a subset of grouplike elements: $\{x\;|\;\Delta x=x\otimes x\}$. Over a field, the nonzero grouplike elements recover the group. Invertible grouplike elts for general nonzero ring? Alternately, Tannaka's reconstruction theorem identifies the group with the group of natural automorphisms of the fiber functor. Krein's hypotheses about dualizability are only necessary to identify a category of representations, but you assume that you have one. Commented Mar 1, 2019 at 17:47
• What am I missing here? Isn't $t: G\to Aut(F), g\mapsto (t_V^g)_{V\in Rep_k(G)}$ an isomorphism, where $F: Rep_k(G)\to k-mod$ is the fibre functor and $t_V^g$ is $F(V)\to F(V), v\mapsto gv$ ? $t$ is easily seen to be bijective by evaluation on the regular representation. Isn't that the whole point of the Tannakian theorem that one does not need the regular representation if one has all the finite-dimensional ones? Commented Mar 1, 2019 at 17:56

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 $$G$$.
• The answer as stated seems false to me: The fiber functor functor is representable (as a $k$-linear functor) by the regular representation. So the $k$-enriched natural transformation are exactly the automorphisms of the regular representations. And there are already a lot more of these than justs $G$. A more reasonable statement would be that $G$ identifies with the enriched & monoidal natural transformation, meaning that you also needs the monoidal structure. Commented Mar 3, 2019 at 12:56
• Oooh. So that was what I missing! $\tau^{k[G]}(1)$ can be any unit of $k[G]$, not just a group element. I should have seen that. Commented Mar 3, 2019 at 14:47
• @Johannes Hahn can you expand the last part of the proof please ? Why is $u$ a group element ? Commented May 10, 2021 at 11:28