In Chapter 15 Section 15.2.1 of Quantum Groups and Noncommutative Geometry, 2nd edition, the authors raised a question: can we reconstruct a finite group $G$ from its category of finite dimensional complex representations $\text{rep}_{\mathbb{C}}(G)$.

The answer depends on how we consider $\text{rep}_{\mathbb{C}}(G)$. Actually we cannot reconstruct $G$ if we only consider $\text{rep}_{\mathbb{C}}(G)$ as an abelian category or monoidal category, but we can reconstruct $G$ if we consider $\text{rep}_{\mathbb{C}}(G)$ as a symmetric monoidal category.

The proof outlined in the book is to consider the forgetful functor $$ F: \text{rep}_{\mathbb{C}}(G)\to \text{vect}_{\mathbb{C}} $$ to the symmetric monoidal category of finite dimensional $\mathbb{C}$-vector spaces. Then we show there is an isomorphism $$ G\to \text{Aut}^{\otimes}(F) $$ where $\text{Aut}^{\otimes}(F)$ denotes the group of natural isomorphisms of $F$ which are compatible with the tensor product on $\text{rep}_{\mathbb{C}}(G)$.

I want to see the impact $\text{Aut}^{\otimes}(F)$ when we add the symmetric structure on the monoidal categories. But it seems to me that there is no difference between the symmetric monoidal case and merely monoidal case. For example, symmetric means the following diagram commutes for $\alpha\in \text{Aut}^{\otimes}(F)$ $$ \require{AMScd} \begin{CD} F(V\otimes W)@>F(S_{V,W})>> F(W\otimes V)\\ @V \alpha_{V\otimes W} VV @VV \alpha_{W\otimes V} V\\ F(V\otimes W) @>>F(S_{W,V})> F(W\otimes V) \end{CD} $$ where $S_{V,W}$ is the symmetry functor in $\text{rep}_{\mathbb{C}}(G)$.

However since $F$ is the forgetful functor we automatically have $$ F(S_{V,W})\cong S_{F(V),F(W)} $$ and since $\alpha$ preserve the monoidal structure we have $$ \alpha_{V\otimes W}\cong \alpha_V\otimes \alpha_W. $$ Therefore since we have the commutative diagram $$ \begin{CD} F(V)\otimes F(W)@>S_{F(V),F(W)}>> F(W)\otimes F(V)\\ @V \alpha_{V} \otimes\alpha_{W} VV \circlearrowright @VV \alpha_{W}\otimes\alpha_{V} V\\ F(V)\otimes F(W) @>>S_{F(W),F(V)}> F(W)\otimes F(V) \end{CD} $$ the first diagram also commute. In conclusion there is no difference of $\text{Aut}^{\otimes}(F)$ whether we consider $\text{rep}_{\mathbb{C}}(G)$ as a symmetric monoidal category or just a monoidal category.

What did I do wrong in this argument?

  • 1
    $\begingroup$ Being symmetric is not a property but a structure on monoidal category specifying the commutativity isomorphism and you implicitly assumed that it is a "standard one". $\endgroup$ – Denis T. Oct 11 '19 at 14:28
  • $\begingroup$ Easy example is the category of graded vector spaces which is monoidally isomorphic to vector spaces, but not symmetrically. $\endgroup$ – Denis T. Oct 11 '19 at 14:35
  • $\begingroup$ @DenisT. That's right but it doesn't solve my problem: is there any difference on $\text{Aut}^{\otimes}(F)$ if we only consider $\text{rep}_{\mathbb{C}}(G)$ as a monoidal category? $\endgroup$ – Zhaoting Wei Oct 11 '19 at 14:43
  • $\begingroup$ You also need duals. $\endgroup$ – BWW Oct 12 '19 at 22:25
  • $\begingroup$ The issue is to construct a fibre functor. $\endgroup$ – BWW Oct 12 '19 at 22:26

I don't think you've made an error. There's a version of Tannakian reconstruction for merely monoidal categories with fiber functors. Namely, a finite dimensional Hopf algebra is the same thing as a finite tensor category together with a monoidal functor to vector spaces. See Theorem 5.3.12 of the Tensor Categories. This reconstruction is compatible with the one for groups, in the sense that the Hopf algebra you recover is $k[G]$. However, for the Hopf algebra one you look at all endomorphisms instead of tensor isomorphisms. If you start with H-mod for a Hopf algebra H and look at $\mathrm{Aut}^\otimes(F)$ you'll end up with the group G of grouplike elements of H, and Rep(G) will not be the same as H-mod because H is not spanned by its grouplike elements.

  • $\begingroup$ Just to clarify the last sentence, are you are saying that if $H$ is not isomorphic to a group ring, then Rep(G) will be different from H-mod? $\endgroup$ – S. Carnahan Oct 11 '19 at 15:59
  • $\begingroup$ Yes. I edited accordingly and added an additional clarification. $\endgroup$ – Noah Snyder Oct 11 '19 at 16:01
  • $\begingroup$ @NoahSnyder Thank you! Do you know then why $\text{rep}_{\mathbb{C}}(G)$ as a monoidal category cannot reconstruct the group $G$? $\endgroup$ – Zhaoting Wei Oct 11 '19 at 16:47
  • $\begingroup$ Corrected an error. H = End(F) is all endomorphisms, not just the tensor automorphisms. This is why the tensor automorphisms are the grouplike elements (and not just all invertible elements). $\endgroup$ – Noah Snyder Oct 11 '19 at 17:11
  • 4
    $\begingroup$ @ZhaotingWei: There's two separate issues here: Tannakian reconstruction which recovers an algebraic structure from a category with a fiber functor, and the relationship between symmetric tensor categories and fiber functors. Rep(G) might have many monoidal fiber functors, and each of them will give you a Hopf algebra such that Rep(G) = H-mod via reconstruction. But it only has one symmetric fiber functor, essentially because you can see what vector space you have to assign by looking at which wedge power vanishes. $\endgroup$ – Noah Snyder Oct 11 '19 at 17:16

Although you already have one sufficient answer, I thought I would add here that Etingof and Gelaki, "Isocategorical groups" considered exactly this question. For finite groups, they gave a complete characterization of when two non-isomorphic finite groups can have the same monoidal categories $rep(G)$ (ignoring the symmetric structure). One consequence of their characterization is that all finite groups of odd order, or twice odd order, and all simple finite groups are determined even by $rep(G)$ considered only as a monoidal category.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.