Revision b1c287cd-692a-4dad-9261-872d8deec75a

$\DeclareMathOperator\PSL{PSL} \DeclareMathOperator\Rep{Rep}$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][1] (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][2] or the [field with one element][3]. 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)][4], 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 Section 3 in [arXiv.2102.01663][5].  Note that the generic type of many other families of Lie type is available in [this webpage][6] of Frank Lübeck.

> **Question**: Is there a global understanding of the F-symbols (or [6j-symbols][7]) 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][8] 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][9], 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: 
$$\scriptsize{\begin{array}{c|c}  
\text{class}&C_1&C_2&C_3&C_4&C_5&C_6&C_7&C_8&C_9&C_{10} \newline  
\text{size}&1&240 & 240& 240& 112& 112& 105& 210& 210& 210 \newline \hline
\chi_1   & 1  & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1  \newline 
\chi_2   & 7  & 0 & 0 & 0 & \frac{-1+i\sqrt{15}}{2} & \frac{-1-i\sqrt{15}}{2} & -1 & 1 & 1 & -1 \newline 
\chi_3   & 7  & 0 & 0 & 0 & \frac{-1-i\sqrt{15}}{2} & \frac{-1+i\sqrt{15}}{2} & -1 & 1 & 1 & -1  \newline 
\chi_4   & 14 & 0 & 0 & 0 & -1 & -1 & -2 & 0 & 0 & 2 \newline 
\chi_5   & 14 & 0 & 0 & 0 & -1 & -1 &  2 & \sqrt{2} & -\sqrt{2} & 0 \newline 
\chi_6   & 14 & 0 & 0 & 0 & -1 & -1 &  2 & -\sqrt{2} & \sqrt{2} & 0  \newline 
\chi_7   & 15 & 1 & 1 & 1 & 0 & 0 & -1 & -1 & -1 & -1  \newline 
\chi_8   & 16 & 2\cos(\frac{2\pi}{7}) & -2\cos(\frac{3\pi}{7})& -2\cos(\frac{\pi}{7}) & 1 & 1 & 0 & 0 & 0 & 0  \newline 
\chi_9   & 16 & -2\cos(\frac{3\pi}{7})& -2\cos(\frac{\pi}{7}) & 2\cos(\frac{2\pi}{7}) & 1 & 1 & 0 & 0 & 0 & 0  \newline 
\chi_{10}& 16 & -2\cos(\frac{\pi}{7}) & 2\cos(\frac{2\pi}{7}) & -2\cos(\frac{3\pi}{7}) & 1 & 1 & 0 & 0 & 0 & 0  \newline 
\end{array}}$$


  [1]: https://en.wikipedia.org/wiki/Group_of_Lie_type
  [2]: https://en.wikipedia.org/wiki/Noncommutative_geometry
  [3]: https://en.wikipedia.org/wiki/Field_with_one_element
  [4]: https://doi.org/10.1016/j.aim.2010.06.009
  [5]: https://doi.org/10.48550/arXiv.2102.01663
  [6]: http://www.math.rwth-aachen.de/~Frank.Luebeck/chev/DegMult/index.html
  [7]: https://en.wikipedia.org/wiki/6-j_symbol#Mathematical_interpretation
  [8]: https://en.wikipedia.org/wiki/Schur_orthogonality_relations
  [9]: https://mathoverflow.net/q/132866/34538