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Is there an integral fusion category of global dimension $210$, such that the simple objects have dimensions $\{1,5,5,5,6,7,7\}$ and the following fusion matrices?

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

or also the same rules with a little $\color{purple}{\text{variation}}$ for the 7-dim. simple objects (and mult. 3 instead of 2):

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

Remark: these are the fusion matrices of the first two simple integral non-trivial fusion rings.

Note that $210 = 2.3.5.7$ and that these matrices are self-dual and irreducibles. They also commute.
By arXiv:0809.3031 proposition 9.11, if such integral fusion categories exist, they couldn't be "weakly group theoretical", and by arXiv:1208.0840 corollary 6.16, they would be abelian but not braided.
Thank you to Eric Rowell and Leonid Vainermann for these references.
Also thanks to Dave Penneys for asking Eric.


The proof that such a fusion category $\mathcal{C}$ can't be braided is the following completed argument:
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 the corollary 6.16 cited.
- Else if it's degenerated, then by simplicity $\mathcal{C}′=\mathcal{C}$, so $\mathcal{C}$ is symmetric, and by Deligne, $\mathcal{C}≃Rep(G)$ as fusion category (without considering the symmetric structures), with $G$ a finite simple group, contradiction (because there is no simple group of order $210$).

Edit about the original motivation (July 2013):
These matrices are naturally came from my will of classifying the cyclic subfactors:
The first case I consider is "depth 2, irreducible, finite index", i.e. finite dimensional C*-Hopf algebras (also called Kac algebras). The first question to answer is:
Are there non-trivial cyclic Kac algebras ? If so, the first example is certainly maximal.
Now a Kac algebra gives a unitary integral fusion category, so I have written an algorithm investigating all the integral fusion rings of a restrictive class containing necessarily those related to the non-trivial maximal Kac algebras. There are finitely many possibilities for each dimension.

Edit (June 2014):
I had also discovered eight fusion rings of global dimensions 360 and 660, with simple objects of dimensions $\{1,5,5,8,8,9,10\}$ and $\{1,5,5,10,10,11,12,12\}$. Two of them come from the simple groups $A_6$ and $A_1(11)$, the six others are new (see the fusion rules here).

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I am interested to know how you got those matrices: where do they come from? –  André Henriques Jul 4 '13 at 21:16
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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. –  César Galindo Jul 12 '13 at 16:18
    
@CésarGalindo : Do you confirm they have the same fusion category (not only the same fusion ring). Maybe the fusion categories of the Kac algebra and its dual, give completely the Kac algebras. It's true for the few twist deformation of group algebras I know, but I don't know if it's true in general, and you ? –  Sébastien Palcoux Sep 10 '13 at 14:58
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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. –  César Galindo Sep 12 '13 at 22:13
    
In the same flavor that Cérar's comment, see the paper of Etingof-Gelaki : Isocategorical Groups –  Sébastien Palcoux Jan 30 at 19:24

1 Answer 1

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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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. –  Scott Morrison Oct 27 at 23:24
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@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... –  Dave Penneys Oct 28 at 2:31
    
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 $\sum |d_i|^2$. –  Scott Morrison Oct 28 at 2:33
    
@DavePenneys: you're right because $M_{i^*}= M_i^* (= M_i)$. –  Sébastien Palcoux 2 days ago
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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. –  Sébastien Palcoux 2 days ago

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