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Mar 30, 2019 at 16:03 comment added Sebastien Palcoux In the group case, these numbers would be the order of the centralizers for a representative of each conjugacy class (and so the indices of the conjugacy classes by the Orbit-Stabilizer Theorem, which should justify the name formal codegree). I edited the formal character table.
Oct 28, 2014 at 14:59 history edited Ryan T Johnson CC BY-SA 3.0
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Oct 28, 2014 at 13:42 comment added Sebastien Palcoux 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.
Oct 28, 2014 at 13:41 comment added Sebastien Palcoux @DavePenneys: you're right because $M_{i^*}= M_i^* (= M_i)$.
Oct 28, 2014 at 2:33 comment added Kim Morrison 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 $\sum |d_i|^2$.
Oct 28, 2014 at 2:31 comment added Dave Penneys @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...
Oct 27, 2014 at 23:24 comment added Kim Morrison 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.
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Oct 27, 2014 at 21:57 history answered Ryan T Johnson CC BY-SA 3.0