Interpolated simple integral fusion categories of Lie type

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

• 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. – Sebastien Palcoux 2 days ago