We refer to Chapter 8 of the book Tensor Categories for notions related to modular tensor categories and J.P. Serre for the basic theory of linear representations of finite groups over $\mathbb C$.

Let $G$ be a finite group, $\mathrm{Vec}_G^\omega$ be a category of finite dimensional $G$-graded vector spaces (potentially twisted by some non-trivial 3-cocycle $\omega$) and $\mathrm{Rep}(G)$ be the category of finite dimensional complex (for ease) semi-simple representations of $G$. The fusion rules of $\mathrm{Vec}_G^\omega$ (resp. $\mathrm{Rep}(G)$) are given by the product of elements (resp. irreducible characters) of $G$.

The number of conjugacy classes of $G$ (the class number) is equal to the number of its irreducible characters, but there is no "natural" bijection between these two sets (see this post), in particular, the character ring is not equivalent to the conjugacy class ring in general, but note that the equivalence holds in a specific case mentionned here, properly containing the abelian groups.

Two fusion categories are said to be 'Grothendieck equivalent' if their Grothendieck rings (i.e. the de-categorification of their monoidal structure) are equivalent as fusion rings. Let $A$ be a finite abelian group, then $\mathrm{Vec}_A^\omega$ and $\mathrm{Rep}(A)$ are Grothendieck equivalent.

Example 8.13.5 of 1 mentions way to make a modular tensor category using a finite abelian group $A$ and a non-degenerate quadratic form $q: A \rightarrow \mathbb C^*$. It is denoted $\mathcal C(A, q)$ and (see on page 205) is Grothendieck equivalent to $\mathrm{Rep}(A)$.

Thus, for every finite abelian group $A$ on which there exists a non-degenerate quadratic form, $\mathrm{Rep}(A)$ is Grothendieck equivalent to a modular tensor category. But it exists for everyone according to the answers of this post.

Question: Is there a classification or a group-theoretical characterization of the finite groups $G$ such that the tensor category $\mathrm{Rep}(G)$ is Grothendieck equivalent to a modular category? Is there a non-abelian one?

The paper On the classification of weakly integral modular categories shows that all integral modular categories of rank at most $7$ are pointed. It follows that for all non-abelian finite group $G$ of class number at most $7$ (as $S_3$ or $A_5$), $\mathrm{Rep}(G)$ is not Grothendieck equivalent to a modular category.

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    $\begingroup$ A non-trivial symmetric monoidal category cannot be modular (e.g. because its Müger center is obviously non-trivial in that case), so it seems to me it's never the case that $Rep(G)$ is modular, and this also applies if $G$ is abelian which seems to contradict your statement, am I missing something ? $\endgroup$
    – Adrien
    Oct 12, 2020 at 8:05
  • $\begingroup$ @Adrien: I am confused... I should ask up to Grothendieck equivalence... The correct question should be: For which finite groups G the Grothendieck ring of Rep(G) admits a modular categorification? $\endgroup$ Oct 12, 2020 at 9:27
  • $\begingroup$ @Adrien: For every finite abelian group $A$, the Grothendieck ring of $Rep(A)$ admits a modular categorification (Example 8.13.5 in cited book), right? $\endgroup$ Oct 12, 2020 at 9:40
  • $\begingroup$ @Adrien: Edited with "up to Grothendieck equivalence" $\endgroup$ Oct 12, 2020 at 11:08
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    $\begingroup$ @Adrien: the group algebra of an abelian group is a self-dual Hopf algebra (see the third line of the introduction of this paper by Andruskiewitsch-Natale). $\endgroup$ Oct 15, 2020 at 7:52

1 Answer 1


Here is a necessary condition for a group $G$ such that Rep($G$) is Grothendieck equivalent to a modular category:

there is a bijection between irreducible complex characters of $G$ and conjugacy classes of $G$ such that the size of a conjugacy class equals the square of dimension of the corresponding representation. In particular, the sizes of conjugacy classes are all squares, and the squares of degrees of irreducible characters divide the order of $G$.

Example: the Monster simple group $M$ has a character of degree 196,883; square of this degree does not divide the order of $M$. Thus there is no modular tensor category which is Grothendieck equivalent to Rep($M$).

In fact, I don't know a single non-abelian group $G$ satisfying the condition above.

One obtains the condition above as follows: it is well known that the columns of $S-$matrix of a modular tensor category ${\mathcal C}$ are proportional to various homomorphisms $K({\mathcal C})\to {\mathbb C}$ evaluated at basis elements; also the columns of the character table of $G$ are precisely all the homomorphisms $K({\mbox Rep}(G))\to {\mathbb C}$. Thus the $S-$matrix of a modular category Grothendieck equivalent to Rep($G$) can be obtained from the character table by normalizing and permuting the columns. Using the orthogonality relations for the characters, it is easy to compute that the normalization factors above are precisely square roots of the sizes of the conjugacy classes; since the $S-$matrix must be symmetric we get the condition.

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    $\begingroup$ Apparently this was asked about on MO before! The smallest group with this property has order $64$, no non-abelian finite simple group has this property, and conjecturally all finite groups with this property are nilpotent: mathoverflow.net/questions/20374/… $\endgroup$ Oct 16, 2020 at 17:49
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    $\begingroup$ @Qiaochu Yuan Thanks for pointing out this example! The same arguments as in my answer give an apparently much stronger (but more difficult to verify) necessary condition: after normalization and permutation of columns as above the character table must become symmetric matrix. It would be interesting to check whether it works for the group of order 64 from your comment. $\endgroup$ Oct 16, 2020 at 19:00
  • $\begingroup$ What about the groups with character ring equivalent to conjugacy class ring? According to this answer they are exactly the groups $p$-nilpotent with abelian Sylow $p$-subgroup. Perhaps this paper in Russian of Saksonov (cited as reference) contains a non-abelian example. I don’t know if the equivalence must keep the usual bases. $\endgroup$ Oct 16, 2020 at 20:02
  • $\begingroup$ @Sebastien Palcoux I think that my "stronger condition" from the answer to Qiaochu is equivalent to the existence of isomorphism between character ring and conjugacy class ring which send irreducible characters to class sums divided by square roots of class sizes (so it is almost isomorphism of based rings). This is much stronger than just isomorphism of rings. Also this is equivalent to group $G$ being self-dual in the sense of Bannai (see the answer by A.Stasinski in Qiaochu's link). For example paper by Andrus et al in the same answer proves that such groups are nilpotent. $\endgroup$ Oct 16, 2020 at 22:25
  • $\begingroup$ @VictorOstrik There is something unclear: the paper of Andrus et al provides examples of non-abelian self-transpose p-groups (called stem groups). As you wrote, the property to be self-transpose should be much stronger than the (unbased) ring isomorphism between character ring and conjugacy class ring. So, according to Saksonov (mentioned above) these groups should be p-nilpotent with abelian Sylow p-subgroup, but they already are p-groups, so they should be abelian... Anyway, do you believe that Rep(G) is Grothendieck equivalent to a modular category if and only if G is self-transpose? $\endgroup$ Oct 17, 2020 at 9:28

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