5
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$\DeclareMathOperator\PSL{PSL} \DeclareMathOperator\Rep{Rep}$The idea motivating this post is that there should exist a global understanding of the unitary fusion categories $\Rep(G(q))$, with $G(q)$ a finite group of Lie type (on the finite field $\mathbb{F}_q$, with $q$ a prime power), allowing to interpolate an extended family denoted $“\Rep(G(n))”$, for all (odd, at least) integer $n>1$. The integer $n$ corresponds to a virtual "finite field of order $n$", in the same flavor than the noncommutative geometry or the field with one element. In particular, for $G$ classical, it is NOT given by $G(\mathbb{Z}/n\mathbb{Z})$, because $G(q)$ is already different from $G(\mathbb{Z}/q\mathbb{Z})$ when $q=p^r$, with $p$ prime and $r>1$.

Such constructions would provide infinite families of non-group-like simple integral fusion categories, and so a lot of examples of non weakly-group-theoretical integral fusion categories, answering Question 2 in Etingof-Nikshych-Ostrik (2011), thanks to their Proposition 9.11.

A global understanding of the type of $\Rep(\PSL(2,q))$ is known:

  • if $q \equiv 0 \pmod2$, it is of type $[[1,1],[q-1,\frac{q}{2}],[q,1],[q+1,\frac{q-2}{2}]]$,
  • if $q \equiv 1 \pmod4$, it is of type $[[1,1],[\frac{q+1}{2},2],[q-1,\frac{q-1}{4}],[q,1],[q+1,\frac{q-5}{4}]]$,
  • if $q \equiv 3 \pmod4$, it is of type $[[1,1],[\frac{q-1}{2},2],[q-1,\frac{q-3}{4}],[q,1],[q+1,\frac{q-3}{4}]],$

and so is for the character table, see for example the webpage Character table of $\PSL(2,\mathbb{F}_q)$ by J. Adams (Warning: there are typos, see the comment below).

Question: Is there a global understanding of the F-symbols (also called 6j-symbols) for $\Rep(\PSL(2,q))$? If not (yet), how to compute them for $q$ small (it could be enough to guess for small $n$)?

Note that the knowledge of the F-symbols is exactly what is lost when we consider the Grothendieck ring of a fusion category.

The types and character tables mentionned above can immediately be interpolated, by replacing the prime power $q$ by an integer $n$. The problem here is the existence of a global understanding of the unitary fusion category $\Rep(\PSL(2,q))$, to find $“\Rep(\PSL(2,n))”$ by interpolation.

Note that by the Schur orthogonality relations we can already compute what would be their Grothendieck rings. For example, the Grothendieck ring of $“\Rep(\PSL(2,6))”$ would be the (first) fusion ring mentionned in this post, and the one for $“\Rep(\PSL(2,15))”$ would be the following of rank $10$, FPdim $1680$ and type $[[1,1],[7,2],[14,3],[15,1],[16,3]]$:

$$\scriptsize{\begin{smallmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\end{smallmatrix}, \ \begin{smallmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{smallmatrix}, \ \begin{smallmatrix}0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{smallmatrix}, \ \begin{smallmatrix}0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 2 & 2 & 2 & 1 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 1 & 1 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 1 & 1 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2\end{smallmatrix}, \ \begin{smallmatrix}0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 & 1 & 1 & 2 & 2 & 2 & 2 \\ 1 & 1 & 1 & 1 & 2 & 2 & 1 & 2 & 2 & 2 \\ 0 & 0 & 0 & 1 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 1 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2\end{smallmatrix}},$$ $$\scriptsize{\begin{smallmatrix}0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 & 1 & 1 & 2 & 2 & 2 & 2 \\ 0 & 0 & 0 & 1 & 2 & 2 & 2 & 2 & 2 & 2 \\ 1 & 1 & 1 & 1 & 2 & 2 & 1 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 1 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2\end{smallmatrix}, \ \begin{smallmatrix}0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 1 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 1 & 2 & 2 & 2 & 2 \\ 1 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 3 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 3\end{smallmatrix}, \ \begin{smallmatrix}0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 2 & 2 \\ 1 & 1 & 1 & 2 & 2 & 2 & 3 & 2 & 3 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 2 & 3 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 3 & 3\end{smallmatrix}, \ \begin{smallmatrix}0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 3 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 2 & 3 \\ 1 & 1 & 1 & 2 & 2 & 2 & 3 & 2 & 2 & 3 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 3 & 2\end{smallmatrix}, \ \begin{smallmatrix}0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 3 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 2 & 3 & 3 \\ 0 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 3 & 2 \\ 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 2 & 2\end{smallmatrix} \ } $$

Here is its character table: $$\scriptsize{\begin{array}{c|c} \text{class}&C_1&C_2&C_3&C_4&C_5&C_6&C_7&C_8&C_9&C_{10} \newline \text{size}&1&240 & 240& 240& 112& 112& 105& 210& 210& 210 \newline \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \newline \chi_2 & 7 & 0 & 0 & 0 & \frac{-1+i\sqrt{15}}{2} & \frac{-1-i\sqrt{15}}{2} & -1 & 1 & 1 & -1 \newline \chi_3 & 7 & 0 & 0 & 0 & \frac{-1-i\sqrt{15}}{2} & \frac{-1+i\sqrt{15}}{2} & -1 & 1 & 1 & -1 \newline \chi_4 & 14 & 0 & 0 & 0 & -1 & -1 & -2 & 0 & 0 & 2 \newline \chi_5 & 14 & 0 & 0 & 0 & -1 & -1 & 2 & \sqrt{2} & -\sqrt{2} & 0 \newline \chi_6 & 14 & 0 & 0 & 0 & -1 & -1 & 2 & -\sqrt{2} & \sqrt{2} & 0 \newline \chi_7 & 15 & 1 & 1 & 1 & 0 & 0 & -1 & -1 & -1 & -1 \newline \chi_8 & 16 & 2\cos(\frac{2\pi}{7}) & -2\cos(\frac{3\pi}{7})& -2\cos(\frac{\pi}{7}) & 1 & 1 & 0 & 0 & 0 & 0 \newline \chi_9 & 16 & -2\cos(\frac{3\pi}{7})& -2\cos(\frac{\pi}{7}) & 2\cos(\frac{2\pi}{7}) & 1 & 1 & 0 & 0 & 0 & 0 \newline \chi_{10}& 16 & -2\cos(\frac{\pi}{7}) & 2\cos(\frac{2\pi}{7}) & -2\cos(\frac{3\pi}{7}) & 1 & 1 & 0 & 0 & 0 & 0 \newline \end{array}}$$

$\endgroup$
  • $\begingroup$ There are two typos in the second table that J. Adams put here (I pointed out to him): first line: (q-7)/4 should be (q-3)/4; last column, middle line: 1 should be -1. I hope he will fix them soon. $\endgroup$ – Sebastien Palcoux 2 days ago

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