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Is there an integral fusion category of rank $7$, FPdim $210$ and type $(1,5,5,5,6,7,7)$ with the following fusion rules (or the little $\color{purple}{\text{variation}}$ below)?

$$\small{\begin{smallmatrix} 1 & 0 & 0 & 0& 0& 0& 0 \\ 0 & 1 & 0 & 0& 0& 0& 0 \\ 0 & 0 & 1 & 0& 0& 0& 0 \\ 0 & 0 & 0 & 1& 0& 0& 0 \\ 0 & 0 & 0 & 0& 1& 0& 0 \\ 0 & 0 & 0 & 0& 0& 1& 0 \\ 0 & 0 & 0 & 0& 0& 0& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 1 & 0 & 0& 0& 0& 0 \\ 1 & 1 & 0 & 1& 0& 1& 1 \\ 0 & 0 & 1 & 0& 1& 1& 1 \\ 0 & 1 & 0 & 0& 1& 1& 1 \\ 0 & 0 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 1 & 0& 0& 0& 0 \\ 0 & 0 & 1 & 0& 1& 1& 1 \\ 1 & 1 & 1 & 0& 0& 1& 1 \\ 0 & 0 & 0 & 1& 1& 1& 1 \\ 0 & 1 & 0 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 1& 0& 0& 0 \\ 0 & 1 & 0 & 0& 1& 1& 1 \\ 0 & 0 & 0 & 1& 1& 1& 1 \\ 1 & 0 & 1 & 1& 0& 1& 1 \\ 0 & 1 & 1 & 0& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0& 1& 0& 0 \\ 0 & 0 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 0 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 0& 1& 1& 1 \\ 1 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 2& 1 \\ 0 & 1 & 1 & 1& 1& 1& 2 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0& 0& 1& 0 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 2& 1 \\ 1 & 1 & 1 & 1& 2& \color{purple}{1}& \color{purple}{2} \\ 0 & 1 & 1 & 1& 1& \color{purple}{2}& \color{purple}{2} \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0& 0& 0& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 2 \\ 0 & 1 & 1 & 1& 1& \color{purple}{2}& \color{purple}{2} \\ 1 & 1 & 1 & 1& 2& \color{purple}{2}& \color{purple}{1} \end{smallmatrix}}$$

This is the first known non-trivial simple integral fusion ring. It was found by computation with SageMath (this code giving this output after 2 minutes). Note that there is no finite simple group of order $210$, because the order of a (non-trivial) perfect group is divisible by $4$ (but see this post).

Note that $210 = 2\cdot 3\cdot 5\cdot 7$ and that the above matrices are irreducible; they are also self-dual and commuting, so simultaneously diagonalizable. We deduce the following "formal" character table:

$$\scriptsize \begin{array}{c|c} \text{class}&C_1&C_2&C_3&C_4&C_5&C_6&C_7 \newline \text{dim}&1&35&30&30&30&42&42 \newline \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \newline \chi_2 & 5 & -1 & -\zeta_7 -\zeta_7^6 & -\zeta_7^5 - \zeta_7^2 & -\zeta_7^4 - \zeta_7^3 & 0 & 0 \newline \chi_3 & 5 & -1 & -\zeta_7^5 - \zeta_7^2 & -\zeta_7^4 - \zeta_7^3 & -\zeta_7 -\zeta_7^6 & 0 & 0 \newline \chi_4 & 5 & -1 & -\zeta_7^4 - \zeta_7^3 & -\zeta_7 -\zeta_7^6 & -\zeta_7^5 - \zeta_7^2 & 0 & 0 \newline \chi_5 & 6 & 0 & -1 & -1 & -1 & 1 & 1 \newline \chi_6 & 7 & 1 & 0 & 0 & 0 & \zeta_5+\zeta_5^4 & \zeta_5^2+\zeta_5^3 \newline \chi_7 & 7 & 1 & 0 & 0 & 0 & \zeta_5^2+\zeta_5^3 & \zeta_5+\zeta_5^4 \newline \end{array} $$

See Schur orthogonality relations and Orbit-Stabilizer theorem together with MR2863455.

Assume the existence of an integral fusion category $\mathcal{C}$ having the above fusion ring.

Theorem: The integral fusion category $\mathcal{C}$ would be not weakly-group-theoretical.
Proof: the fusion ring is simple, so the result follows by MR2735754 Proposition 9.11. because there is no simple group of order $210$. $\square$

Theorem: The integral fusion category $\mathcal{C}$ would be abelian but not braided.
Proof: it is abelian because the above matrices commute. Now, if it's braided, then it can be non-degenerated (i.e. $\mathcal{C}′=Vec$) or degenerated:

  • If it's non-degenerated then the contradiction follows by MR3077244 Corollary 6.16.
  • Else it's degenerated, then $Vec \subsetneq \mathcal{C}′$, so by simplicity $\mathcal{C}′=\mathcal{C}$, so $\mathcal{C}$ is symmetric, and by Deligne (see MR3077244 Example 4.6 citing MR1106898), $\mathcal{C}≃Rep(G)$ as fusion category (without considering the symmetric structure), with $G$ a finite simple group, contradiction. $\square$

Remark: A pointed fusion category is equivalent to $Vec_{G, \omega}$ with $G$ a finite group and $\omega$ a $3$-cocycle in $Z^3 (G, \mathbb{C}^*)$ as associator. In particular, the fusion ring of a pointed fusion category is trivial.

Aknowledgment: The above theorems were discussed during the workshop on fusion categories (May 2013, IMB Dijon, France). Thanks to Peter Schauenburg for his invitation, and thanks to Eric Rowell, Leonid Vainermann and David Penneys for above references and encouragements.


There is a second integral simple fusion ring of same FPdim, type and rules except a little $\color{purple}{\text{variation}}$ for the $7$-dimensional simple objects (and multiplicity $3$ instead of $2$):

$$\small{\begin{smallmatrix} 1 & 0 & 0 & 0& 0& 0& 0 \\ 0 & 1 & 0 & 0& 0& 0& 0 \\ 0 & 0 & 1 & 0& 0& 0& 0 \\ 0 & 0 & 0 & 1& 0& 0& 0 \\ 0 & 0 & 0 & 0& 1& 0& 0 \\ 0 & 0 & 0 & 0& 0& 1& 0 \\ 0 & 0 & 0 & 0& 0& 0& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 1 & 0 & 0& 0& 0& 0 \\ 1 & 1 & 0 & 1& 0& 1& 1 \\ 0 & 0 & 1 & 0& 1& 1& 1 \\ 0 & 1 & 0 & 0& 1& 1& 1 \\ 0 & 0 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 1 & 0& 0& 0& 0 \\ 0 & 0 & 1 & 0& 1& 1& 1 \\ 1 & 1 & 1 & 0& 0& 1& 1 \\ 0 & 0 & 0 & 1& 1& 1& 1 \\ 0 & 1 & 0 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 1& 0& 0& 0 \\ 0 & 1 & 0 & 0& 1& 1& 1 \\ 0 & 0 & 0 & 1& 1& 1& 1 \\ 1 & 0 & 1 & 1& 0& 1& 1 \\ 0 & 1 & 1 & 0& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0& 1& 0& 0 \\ 0 & 0 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 0 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 0& 1& 1& 1 \\ 1 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 2& 1 \\ 0 & 1 & 1 & 1& 1& 1& 2 \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0& 0& 1& 0 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 2& 1 \\ 1 & 1 & 1 & 1& 2& {\color{purple}{0}}& {\color{purple}{3}} \\ 0 & 1 & 1 & 1& 1& {\color{purple}{3}}& {\color{purple}{1}} \end{smallmatrix} , \ \begin{smallmatrix} 0 & 0 & 0 & 0& 0& 0& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 1 \\ 0 & 1 & 1 & 1& 1& 1& 2 \\ 0 & 1 & 1 & 1& 1& {\color{purple}{3}}& {\color{purple}{1}} \\ 1 & 1 & 1 & 1& 2& {\color{purple}{1}}& {\color{purple}{2}} \end{smallmatrix}}$$

All the properties above hold for this fusion ring. Here is the corresponding formal character table:

$$\scriptsize \begin{array}{c|c} \text{class}&C_1&C_2&C_3&C_4&C_5&C_6&C_7 \newline \text{dim}&1&35&30&30&30&70&14 \newline \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \newline \chi_2 & 5 & -1 & -\zeta_7 -\zeta_7^6 & -\zeta_7^5 - \zeta_7^2 & -\zeta_7^4 - \zeta_7^3 & 0 & 0 \newline \chi_3 & 5 & -1 & -\zeta_7^5 - \zeta_7^2 & -\zeta_7^4 - \zeta_7^3 & -\zeta_7 -\zeta_7^6 & 0 & 0 \newline \chi_4 & 5 & -1 & -\zeta_7^4 - \zeta_7^3 & -\zeta_7 -\zeta_7^6 & -\zeta_7^5 - \zeta_7^2 & 0 & 0 \newline \chi_5 & 6 & 0 & -1 & -1 & -1 & 1 & 1 \newline \chi_6 & 7 & 1 & 0 & 0 & 0 & 0 & -3 \newline \chi_7 & 7 & 1 & 0 & 0 & 0 & -1 & 2 \newline \end{array} $$


About the original motivation
These matrices came from my will of classifying the cyclic subfactors (see this post and MR3708264). I first wondered whether there is an irreducible finite index depth $2$ maximal subfactor which is not a group subfactor, in other words, a finite dimensional Hopf ${\rm C}^{\star}$-algebra (or Kac algebra) nontrivial and without left coideal $\star$-subalgebra (see this post). Such a Kac algebra must be simple. Now a Kac algebra gives a unitary integral fusion category. This algorithm investigates all the (non-trivial) simple integral fusion rings. There are finitely many possibilities for each dimension.

Other examples
We also discovered $17$ fusion rings of FPdims $360, 660$, ranks $7, 8$, and types $(1,5,5,8,8,9,10)$, $(1,5,5,10,10,11,12,12)$. Two of them come from the simple groups $A_6$ and ${PSL}(2,11)$, the $15$ others are new (see the fusion rules here).

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    $\begingroup$ I am interested to know how you got those matrices: where do they come from? $\endgroup$ – André Henriques Jul 4 '13 at 21:16
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    $\begingroup$ Kac algebras are not completely given by a unitary fusion category. You can have two non isomorphic Kac algebras with the same tensor category of representation, for example, a Drinfeld twist deformation of a group algebra. $\endgroup$ – César Galindo Jul 12 '13 at 16:18
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    $\begingroup$ Yes, twist deformations of finite dimensional Hopf algebras have the same category of representation, look for example arxiv.org/pdf/math/0107167.pdf A finite dimensional Hopf algebra is completely determined by their category of representation and its fiber functor. $\endgroup$ – César Galindo Sep 12 '13 at 22:13
  • $\begingroup$ In the same flavor that Cérar's comment, see the paper of Etingof-Gelaki : Isocategorical Groups $\endgroup$ – Sebastien Palcoux Jan 30 '14 at 19:24
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I'm still new to this specific field, so I apologize in advance if I have this wrong.

EDIT: Neither example is categorifiable. [The answer is still unkown to me.]

If you let $M_i$ be the fusion matrices for $i=1...7$ and $A = \sum_{i=1}^7 M_i M_{i^*}$. Then the eigenvalues of $A$ are the formal codegrees arxiv:0810.3242v2. However, both of your cases yield a matrix $A$ with three eigenvalues equal to 0. [I did not compute your $A$ matrix correctly.]

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    $\begingroup$ Ryan, I don't think this is correct. Certainly Cor 1.7 of Victor's paper doesn't say that codegrees are non-zero (zero is a rational integer...). Moreover, there are counterexamples to this claim. The first one I came up with is $SU(3)_4$, discussed for example in arxiv.org/abs/1205.2742. $\endgroup$ – Scott Morrison Oct 27 '14 at 23:24
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    $\begingroup$ @Scott, I don't think the formal codegrees can be zero. For example, see Section 2.3 of arXiv:1309.4822. However, if A is the matrix in the answer above, then it's a positive definite operator which is $\geq I$, the identity, since $M_1=I$. Thus all it's eigenvalues are at least 1, regardless of whether they are fusion matrices or not. So there must be an error somewhere... $\endgroup$ – Dave Penneys Oct 28 '14 at 2:31
  • $\begingroup$ Yup, there's definitely an error in my computations (which were in the FusionAtlas mathematica package...). I was assuming that dimension functions were real, and computing $\sum d_i^2$ instead of $\sum |d_i|^2$. $\endgroup$ – Scott Morrison Oct 28 '14 at 2:33
  • $\begingroup$ @DavePenneys: you're right because $M_{i^*}= M_i^* (= M_i)$. $\endgroup$ – Sebastien Palcoux Oct 28 '14 at 13:41
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    $\begingroup$ I've computed the Jordan form of $A$ and I've obtained $diag(210, 6, 5, 5, 7, 7, 7)$ for the first fusion ring and $diag(210, 15, 6, 3, 7, 7, 7)$ for the second. I don't know how interpreting all these numbers and their multiplicities. $\endgroup$ – Sebastien Palcoux Oct 28 '14 at 13:42
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Partial answer

The second fusion ring admits no unitary categorification. It is proved in this paper, using Character Table and Fourier Analysis. Here are some explanations:

Let $\mathcal{F}$ be a commutative fusion ring. Then its fusion matrices $(M_i)$ are commuting and normal so simultaneously diagonalisable: there is an invertible matrix $P$ such that $P^{-1}M_iP = diag(\lambda_{i,j})$. Let us call $\Lambda=(\lambda_{i,j})$ the character table of $\mathcal{F}$. Note that if $G$ is a finite group and $\mathcal{F}$ the Grothendieck ring of $Rep(G)$ then $\Lambda$ is the usual character table of $G$.

If $\mathcal{F}$ is the Grothendieck ring of a unitary fusion category, and if we choose $\lambda_{i,1}$ to be $\Vert M_i \Vert$, then by Corollary 7.5 in this paper, for every triple $(j,k,\ell)$ $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ Note that for $Rep(G)$ this identity admits a combinatorial/probabilistic interpretation (see here).

Now observe the character table of the second fusion ring above, in particular its last column, then $$ \frac{1^3}{1} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{1^3}{6} + \frac{(-3)^3}{7} + \frac{2^3}{7} = -\frac{65}{42}<0.$$ It follows that the second fusion ring above admits no unitary categorification.

Note that the identity holds for the first fusion ring.

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