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Update: Originally, I formulated this question for finite abelian hypergroups, but in a discussion with Geoff Robinson below I realized that the abelian hypergroups defined below are equivalent to table algebras up to a simple normalization [reference]. Hence, I am equally interested on the table algebra case.


Question

A collaborator and I are investigating finite abelian hypergroups (see a short definition below or the following surveys [survey1],[survey2], in a quantum computing context. Our interest is more focused on finite hypergroups that arise from finite groups (namey, conjugacy class and character hypergroups) but we would also be interested on knowing about the arbitrary finite abelian hypergroup case. The question is:

Question. Is there a classification of finite conjugacy class / character hypergroups? If not, is this problem considered to be "impossible" (or, more precisely, wild)?

Of course, this question is very easy for finite abelian groups. Yet, the case for finite abelian hypergroups seems actually quite hard: as formulated, our question is probably equivalent to asking whether classifying finite groups "up to conjugation" and their characters is believed to be as hard as the problem of classifying finite groups. The latter problem, in turn, is wild.

Still, I have not rigorous way to turn this intuition into an argument. My belief is not fully rigorous because classifying groups "up to conjugation" and their characters without classifying groups could (in principle) be a lot easier than classifying groups. So, can something more rigorous be said?

Definition. We are using the following standard definition: a finite abelian hypergroup is a set $H = \{ C_0 , C_1 , . . . , C_n\}$ together with an associative unital abelian algebra structure on $\mathbb{R}H$ $$C_iC_j=\sum_{k}n_{ij}^{k}C_k$$

and an involution $*: H \rightarrow H$ such that

  1. $n_{ij}^{k}\geq 0$
  2. $\sum_{k} n_{ij}^{k}=1$
  3. $C_0$ is the identity
  4. $n_{ij}^{0}> 0$ if and only if $C_i^*=C_j$

Example 1. The irreducible characters $\widehat{G}$ of any finite group $G$ define a hypergroup using the tensor product of a characters the multiplication of the algebra and the complex conjugation as involution. This follows from the formula $$\chi_\mu \otimes \chi_\nu = \sum_{\gamma\in\widehat{G}} m_{\mu,\nu}^\gamma \chi_\gamma.$$ The coefficients $m_{\mu\nu}^\gamma$ in that expansion are not normalized, but one can easily normalize characters so that condition 1 is met [2].

Example 2. The conjugacy classes of any finite group $G$ also have a natural hypergroup structure (this is what we call conjugacy class hypergroup). The simplest way to define this hypergroup is to identify every class $C_g\subset G$ witht he following element of the group algebra $\mathbb{R}G$:

$$C_g = \frac{1}{|C_g|}\sum_{aga^{-1}\in C_g} aga^{-1}$$.

It is shown in [1] that the classes with the product in $\mathbb{R}G$ define a hypergroup.

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    $\begingroup$ These seem rather similar to table algebras, and there is a lot of literature about these- see, eg work of H. Blau. But in any case, it seems to me that a complete classification is likely to be out of reach- is there a way to define simple objects in this context? $\endgroup$ Aug 17, 2015 at 11:12
  • $\begingroup$ According to this paper, a hypergroups seems to be just a normalized table algebra. This difference is probably not too important for our purposes. The simplest example of a hypergroup (likely also a table algebra) that we have in mind would be irreducible characters of finite groups with the tensor product as product, via the formula $\chi_\mu \otimes \chi_\nu = \sum_{\gamma\in\widehat{G}} m_{\mu,\nu}^\gamma \chi_\gamma$, which can be normalized to define a hypergroup. I will update the question to include the example. $\endgroup$ Aug 17, 2015 at 11:21
  • $\begingroup$ So, in short, I am (partially) asking to what extend characters and conjugacy classes of finite groups can be classified without classifying groups. This seems to be likely out of reach to me but I would like to know if anything rigorous is known. $\endgroup$ Aug 17, 2015 at 11:24
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    $\begingroup$ Yes, the irreducible characters of a finite group form a table algebra under pointwise multiplication. For the last question, I can only suggest you check out the existing literature on table algebras- this is not really my own field. $\endgroup$ Aug 17, 2015 at 11:36

1 Answer 1

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I had never heard of table algebras before, but it seems to me that a subset of the table algebras is the abelian unital based rings in the sense of Section 2 of Ostrik's "Module categories, weak Hopf algebras and modular invariants" (MR1976459). Classifying abelian unital based rings is extremely difficult. For example, there are infinite families of rank $n$ abelian unital based rings for all $n\geq 2$ (e.g., near group fusion rings for $|G|=n-1$), and we don't know which rank 4 unital based rings (which are automatically abelian) are categorifiable.

In more detail, a unital based ring consists of:

  • An $\mathbb{Z}$-algebra $A$, which is free as a module over $\mathbb{Z}$, with fixed basis $B=\{b_i\}_{i=0}^n$ such that $b_i b_j = \sum_k N_{i,j}^k b_k$ with $N_{i,j}^k\in\mathbb{Z}_{\geq 0}$ for all $i,j,k$
  • $b_0=1_A$, the identity of $A$
  • There is an involution $*$ on $I$ which extends to a ring anti-isomorphism of $A$ such that $N_{i,j}^0=\delta_{i=j^*}$

These are also known as fusion rule algebras. An important question is given such an algebra, when does it arise as the Grothendieck ring of a fusion category? (I did not see the normalization condition 3 in your definition above in the definition of a table algebra, so it looks like these qualify. Please correct me if I'm mistaken.)

It turns out that given any unital based ring $(A,B)$, there is a unique ring homomorphism $d: A\to \mathbb{C}$ such that $d(b_i)>0$ for all $i\in I$. Also, $A\otimes_\mathbb{Z} \mathbb{C}$ is semisimple, so it is a direct sum of matrix algebras. Since $M_2(\mathbb{C})$ has no characters, we see that $A\otimes_\mathbb{Z} \mathbb{C}\cong \mathbb{C}^4$, which is abelian.

This argument appears in Larson's "Pseudo-unitary non-self-dual fusion categories of rank 4" (MR3229513), which treats the case of pseudo-unitary categorifiable rank 4 unital based rings such that $*$ is not the identity (there is a dual pair $b_1^*=b_2$ WLOG).

The case where $*$ is the identity seems much more difficult. Certainly you can write down the associativity equations for $(A,B)$ and look for integral solutions, but there are infinitely many solutions. (This is already the case in rank 2 and 3. In rank 2, there's only one possible family, and I'm not sure how many families there are for rank 3. See Ostrik's "Pivotal fusion categories of rank 3" arXiv:1309.4822). I know of several infinite families of possible rings for rank 4 with $*=Id$, but I do not know how many infinite families there are. One expects there should only be finitely many categorifiable rank 4 unital based rings.

One interesting infinite family of unital based rings is the near groups. The basis here consists of a finite group $G$ together with one additional symbol $\rho$ satisfying the following:

  • $g\cdot h = gh$ for all $g,h\in G$
  • $g^* = g^{-1}$ for all $g\in G$
  • $g\rho=\rho g=\rho$ for all $g\in G$
  • $\rho^*=\rho$
  • $\rho^2 = n\rho + \sum_{g\in G} g$.

It is not known in general when such fusion rings are categorifiable (see e.g. Evans and Gannon's "Near-group fusion categories and their doubles" MR3167494), but a necessary requirement is that $n=|G|-1$ or $n=k|G|$ for some $k\geq 0$.

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  • $\begingroup$ I am interested on your answer. One first thing I see is that your claims are relevant to at least the first half of my question because the conjugacy class hypergroup of any finite group $G$ is a special type of fusion algebra / unital based ring normalized by the ring homomorphism $d$ you define (this is explained eg [here]). Fusion algebras can always to be renormalized to be abelian hypergroups, and you are saying these are already hard to classify. $\endgroup$ Aug 17, 2015 at 16:58
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    $\begingroup$ Yes, my point is that abelian unital based rings are already very complicated, so you'll need a lot of extra structure to gain traction on your problem. In particular, you'll have to rely on extra properties/structure for conjugacy class hypergroups, since arbitrary hypergroups (even abelian ones) are difficult. I see I failed to explicitly mention this in my answer. $\endgroup$ Aug 17, 2015 at 17:10
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    $\begingroup$ Good point. Yes, I agree with you now. Just for completeness, it does seem hard to classify conjugacy class hypergroups as well because classifying arbitrary abelian hypergroups / fusion algebras is hard (as you point out) and, on top of that, classifying finite groups is also hard (you can always determine the structure constants $N_{ij}^{k}$ of these hypergroups / fusion algebras if you know everything about the finite group conjugacy classes, but then you know a lot of information about the group already...). $\endgroup$ Aug 17, 2015 at 17:19
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    $\begingroup$ Classifying conjugacy class HG might not be hopeless. There is a class of conjugacy class HG which correspond to finite simple group and whose classification happen to be exactly the classification of finite simple groups. An arbitrary group corresponds to a series of extension by simple groups... the problem of classifying such things is very difficult, but maybe understanding the conjugacy class HG of an extension is easier than understanding the actual extension... $\endgroup$ Aug 17, 2015 at 17:36
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    $\begingroup$ (But I still think it is very difficult, just maybe not as difficult as classifying finite group or arbitrary table algebra) $\endgroup$ Aug 17, 2015 at 20:30

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