Finite simple groups and negative Frobenius-Schur indicator

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)?

• I think M. Geck has done quite a bit of work on this question for groups of Lie type – Geoff Robinson Sep 19 at 10:34
• 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 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 at 13:01
• @spin Yes: mathoverflow.net/q/54800/34538 – Sebastien Palcoux Sep 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 at 12:52