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Let $G$ be a finite group and $\pi$ an irreducible complex representation. The Frobenius-Schur indicator of $\pi$ is defined as:
$$ \nu_2(\pi):=\frac{1}{|G|} \sum_{g \in G} \chi_{\pi}(g^2) $$ with $\chi_{\pi}$ the character of $\pi$. Recall that $\nu_2(\pi) \in \{-1, 0,1\}$. The only finite simple groups $G$ with $|G|<10^7$ and having an irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$ are $\mathrm{PSU}(3,q)$ with $q=3,4,5,7,8$ and $\mathrm{PSU}(4,3)$.

gap> it:=SimpleGroupsIterator(10,10000000);; for g in it do if -1 in Indicator(CharacterTable(g),2) then Print([g]); fi; od;
[ PSU(3,3)][ PSU(3,4)][ PSU(3,5)][ PSU(4,3)][ PSU(3,8)][ PSU(3,7)]
gap>

Question: What are the finite simple groups known to have no irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$?
What are those known to have an irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$?

I am specifically interested in the groups $\mathrm{PSL}(2,q)$ for which I ckecked by GAP that for $q<500$, there is no irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$:

gap> for q in [2..500] do if -1 in Indicator(CharacterTable( "PSL", 2, q),2) then Print([q]); fi; od;
gap>

So I expect that it is true for all $q$. Is there a proof using the character table (including class type) of $\mathrm{PSL}(2,q)$ (see this post)?

Sebastien Palcoux
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    I think M. Geck has done quite a bit of work on this question for groups of Lie type – Geoff Robinson Sep 19 '20 at 10:34
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    In the particular case of PSL(2,q), I think it is true for odd q because the Sylow 2-subgroups are dihedral (including possibly Klein 4) when q is odd. When q is a power of 2, there is one conjugacy class of involutions, and the number of solutions of x^2 = 1 is q^2, which is also the sum of the irreducible character degrees of PSL(2,q), and the FS-indicator formula for the number of solutions of x^2 = 1 forces all indicators to be 1. – Geoff Robinson Sep 19 '20 at 10:40
  • For sporadic $G$, you can check the only case with indicator $-1$ is $G = McL$, which has two irreducible characters $\chi$ with $\nu_2(\chi) = -1$. When $G$ is an alternating group, you always have $\nu_2(\chi) \geq 0$ since all irreducible characters of $S_n$ have indicator $+1$. – spin Sep 20 '20 at 13:01
  • @spin Yes: https://mathoverflow.net/q/54800/34538 – Sebastien Palcoux Sep 20 '20 at 14:58
  • Further to my earlier comment, it is definitely true that when $q$ is a power of $2$, all indicators are $+1$, so you only need to check $q$ odd. When I find time, I will try to make the rest of my comment into a formal answer. – Geoff Robinson Nov 13 '20 at 12:52
  • @GeoffRobinson: when there is an irreducible character whose square is the sum of all the irreducible characters (with multiplicity one, or odd) then all the FS indicators are 1, by the result considered here. This applies to PSL(2,2^n) with the Steinberg character. – Sebastien Palcoux Nov 13 '20 at 17:15
  • But I already explained in my September comment why it worked for $PSL(2,2^{n})$ (for a different reason than that given in the answer to yesterday's question). – Geoff Robinson Nov 13 '20 at 17:56
  • @GeoffRobinson: Yes. This answer should help for $q$ odd also. – Sebastien Palcoux Nov 13 '20 at 18:01

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