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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)?

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    $\begingroup$ I think M. Geck has done quite a bit of work on this question for groups of Lie type $\endgroup$
    – Geoff Robinson
    Commented Sep 19, 2020 at 10:34
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    $\begingroup$ 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. $\endgroup$
    – Geoff Robinson
    Commented Sep 19, 2020 at 10:40
  • $\begingroup$ 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 $+1$. $\endgroup$
    – spin
    Commented Sep 20, 2020 at 13:01
  • $\begingroup$ @spin Yes: mathoverflow.net/q/54800/34538 $\endgroup$
    – Sebastien Palcoux
    Commented Sep 20, 2020 at 14:58
  • $\begingroup$ 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. $\endgroup$
    – Geoff Robinson
    Commented Nov 13, 2020 at 12:52

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