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A fusion category is called noncommutative if its Grothendieck ring is noncommutative. Let us call a fusion category strongly noncommutative if every fusion category Morita equivalent to it (i.e. same Drinfeld center up to equiv.) is noncommutative.

Question: Is there a strongly noncommutative fusion category (say over $\mathbb{C}$)?
If so, what are the known examples?

Note that if $G$ is a finite group then $Vec(G)$ is not strongly noncommutative (even if $G$ is noncommutative) because it is Morita equivalent to $Rep(G)$ which has a commutative Grothendieck ring. Moreover, the Extended Haagerup fusion categories are also not strongly noncommutative because they form a Morita equivalent class and one of them is commutative.

This post is in the same spirit than this one about strongly simple fusion categories. The main difference is that I know examples of strongly simple fusion categories whereas I do not know a single strongly noncommutative fusion category.

The next step would be about fusion categories which are both strongly simple and strongly noncommutative.

Sebastien Palcoux
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  • Is not having same Drinfeld center too strong? E. g. for all I know there might be tons of very different fusion categories with "trivial" Drinfeld center (equivalent to the category of fd vector spaces), no? In any case, could you please refer to some introductory material about Morita equivalence in this context? – მამუკა ჯიბლაძე May 27 '22 at 10:06
  • @მამუკაჯიბლაძე: see Section 7.13 in this book. My guess is that if (Drinfeld center) $\mathcal{Z}(\mathcal{C}) \simeq Vec$ then $\mathcal{C} \simeq Vec$ (at least over $\mathbb{C}$), but I may be wrong, I need to check. Maybe you had in mind general tensor categories. – Sebastien Palcoux May 27 '22 at 10:36
  • Thank you for the link, but I could not find there that Morita equivalence in their sense (7.12.17, that one is equivalent to the category of bimodules over an algebra in the other) follows from having the same Drinfeld center? – მამუკა ჯიბლაძე May 27 '22 at 11:28
  • Also now I noticed that my "too strong" looks wrong. I meant not the condition but the resulting equivalence notion. Writing "too coarse" would be probably better. – მამუკა ჯიბლაძე May 27 '22 at 11:35
  • Oh I found it now, it is here. Spectacular. – მამუკა ჯიბლაძე May 27 '22 at 11:53
  • Sorry, no, I still cannot reconstruct it. What I understand is that if $\mathcal C$ and $\mathcal D$ have the same center $\mathcal Z$, then there are two commutative algebras $C,D\in\mathcal Z$ such that $\mathcal C\simeq C$-mod, $\mathcal D\simeq D$-mod. Why does it follow that there is an algebra $A$ in $\mathcal C$ such that $\mathcal D\simeq A$-bimod?? – მამუკა ჯიბლაძე May 27 '22 at 12:11
  • @მამუკაჯიბლაძე See Theorem 3.1 in Etingof, Pavel; Nikshych, Dmitri; Ostrik, Victor. Weakly group-theoretical and solvable fusion categories. Adv. Math. 226 (2011), no. 1, 176--205. MR2735754 – Sebastien Palcoux May 27 '22 at 12:37
  • That's what I was looking at. My problem is (sorry) in there. If I understand correctly, they take for $A\in\mathcal C$ the commutative algebra $D\in\mathcal Z$ viewed as an algebra in $\mathcal C$. They then claim that the forgetful functor $D$-mod $\to A$-bimod is an equivalence (there is a subtlety that $A$ may fail to be decomposable in $\mathcal C$ but let us suppose for simplicity that it is indecomposable). What I don't understand is that every $A$-bimodule $M$ in the image of that functor has the property that its right action is given by $M\otimes A\to A\otimes M\to M$ – მამუკა ჯიბლაძე May 27 '22 at 13:12
  • where $M\otimes A\to A\otimes M$ is the isomorphism coming from $\mathcal Z$ and $A\otimes M\to M$ is the left action. But in $\mathcal C$, although $A$ is equipped with natural isomorphisms $A\otimes X\to X\otimes A$ for all $X$, there might still easily exist $A$-bimodules whose left and right actions are not determined by each other in the above way, no? – მამუკა ჯიბლაძე May 27 '22 at 13:14
  • @მამუკაჯიბლაძე It may be helpful to read Definition 2.9 in Grossman, Pinhas; Snyder, Noah. Quantum subgroups of the Haagerup fusion categories. Comm. Math. Phys. 311 (2012), no. 3, 617--643. MR2909758 – Sebastien Palcoux May 27 '22 at 13:29
  • @მამუკაჯიბლაძე That theorem is also in the book mentioned above, see Theorem 8.12.3. – Sebastien Palcoux May 27 '22 at 14:43
  • Thank you! I still don't understand it well, but I will continue to try. – მამუკა ჯიბლაძე May 28 '22 at 05:33
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    @მამუკაჯიბლაძე I can at least provides a definite answer to your first comment. Theorem 7.16.6 in the book mentioned above (usually called EGNO book) states that for any finite tensor category $FPdim(\mathcal{Z}(\mathcal{C})) = FPdim(\mathcal{C})^2$. So if $\mathcal{Z}(\mathcal{C}) \simeq Vec$ then $FPdim(\mathcal{Z}(\mathcal{C})) =1$, so $FPdim(\mathcal{C}) = 1$, which means that $\mathcal{C} \simeq Vec$, as I guessed. – Sebastien Palcoux May 28 '22 at 18:07

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Consider the symmetric group group $G = S_3$ of order $6$. Then $\mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1)) \cong \mathbb Z/6\mathbb Z$. Choose a generator $\omega \in \mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1))$. Then $\omega$ restricts nontrivially to every nontrivial subgroup of $G$.

It follows that $\mathbf{Vec}^\omega[G]$ is not Morita-equivalent to any other fusion category. (Recall that, for any $G,\omega$, fusion categories equivalent to $\mathbf{Vec}^\omega[G]$ are indexed by pairs consisting of a subgroup $H \subset G$ together with a 2-cochain $\psi$ on $H$ solving $\mathrm{d}\psi = \omega|_H$.)

But $G$ is noncommutative.

Theo Johnson-Freyd
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  • What is the reference for what you recalled? – Sebastien Palcoux May 27 '22 at 13:47
  • I'm sure it's in EGNO somewhere. And I should didn't quite quote the indexing correctly --- the correct statement is that pairs (fusion category $\mathcal F$, morita equivalence $M : \mathcal F \simeq Vec^\omega[G]$) are indexed by pairs $(H,\psi)$, and that if you want to work with just sets (and not spaces) then you should take $(\mathcal F, M)$ up to appropriate equivalence, and you have to complement this by taking $\psi$ modulo $\psi \leadsto \psi+d\phi$. Anyway, if $\omega|_H$ is always nontrivial (except $H = {1}$), then there are no nontrivial Morita equivalents. – Theo Johnson-Freyd May 27 '22 at 14:19
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    One proof is roughly: Given fusion category $\mathcal C$, to give a Morita equivalence $(\mathcal{F},M)$ is to give the simple $\mathcal C$-module $M$ (at which point $\mathcal F = End_{\mathcal C}(M)$), and by Ostrik these are the same as simple algebra objects, and then you have to convince yourself that in $Vec^\omega[G]$ any simple algebra must be $A = \bigoplus_{h \in H} h$ for some subgroup $H \subset G$, at which point $\psi$ gives you the multiplication rule on $A$. – Theo Johnson-Freyd May 27 '22 at 14:22
  • Is it Corollary 7.12.20 in EGNO book? – Sebastien Palcoux May 27 '22 at 15:30
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    Technically no --- that corollary only does the case of trivial $\omega$. You want EGNO 9.7.2. – Theo Johnson-Freyd May 27 '22 at 15:59
  • What happens with the trivial subgroup? – მამუკა ჯიბლაძე May 28 '22 at 14:00
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    @მამუკაჯიბლაძე The trivial subgroup is the identity morita equivalence $Vec^\omega[G] \simeq Vec^\omega[G]$. – Theo Johnson-Freyd May 29 '22 at 15:11