A property forcing the Frobenius-Schur indicator to be positive

Question

Let $G$ be a finite group. Two irreducible complex representations $V,V'$ of $G$ are called dual to each other if $V \otimes V'$ admits a trivial component, i.e. $\hom_G(V \otimes V',V_0)$ is positive dimensional (thus one-dimensional) with $V_0$ the trivial representation. Then the representation $V'$ is denoted $V^*$ (and $V \simeq V^{**}$).

Let $V, W$ be irreducible complex representations of $G$ such that:

• $W^* \simeq W$,
• $\hom_G(V \otimes V^*,W)$ is odd-dimensional.

Question: Is it true that the Frobenius-Schur indicator of $W$ is equal to $1$?

The odd-dimensional assumption cannot be extended to even (see counterexamples in Appendix).

It is conjectured to be true for every pivotal fusion category, see [a, Conjecture 4.26]. See also [b] and [c].

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References

[a] Z. Wang, Topological quantum computation, CBMS Reg. Conf. Ser. Math. (112) xiii + 115pp, (2010).

[b] G. Mason, A brief history of the positivity conjecture in tensor category theory. Bull. Inst. Math. Acad. Sin. (N.S.) 14 (2019), no. 2, 149--153

[c] J. Fuchs; I. Runkel; C. Schweigert. A reason for fusion rules to be even. J. Phys. A 35 (2002), no. 19, L255--L259.

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Appendix

Smallest counterexample among the finite groups:

gap> G:=PSU(3,2);; Order(G);
72
gap> Indicator(CharacterTable(G),2);
[ 1, 1, 1, 1, -1, 1 ]
gap> M:=RepGroupFusionRing(G);;
gap> M[6][6];     # FS(V)=1
[ 1, 1, 1, 1, 2, 7 ]

Smallest counterexample among the finite simple groups:

gap> G:=PSU(3,5);; Order(G);
126000
gap> Indicator(CharacterTable(G),2);
[ 1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0 ]  
gap> M:=RepGroupFusionRing(G);;  
gap> M[8][8];     # FS(V)=1
[ 1, 2, 2, 3, 3, 3, 9, 9, 11, 11, 11, 11, 12, 12 ]
gap> M[11][12];   # FS(V)=0
[ 1, 2, 3, 4, 4, 4, 10, 13, 17, 17, 15, 15, 18, 18 ]

with the following code:

RepGroupFusionRing:=function(g)
    local irr,n,M;
    irr:=Irr(g);
    n:=Size(irr);
    M:=List([1..n],i->List([1..n],j->List([1..n],k->ScalarProduct(irr[i]*irr[j],irr[k]))));
    return M;
end;;

Answer 1

Here is a more general statement, see also Lemma 1.2 in [1].

Lemma: Let $Z$ be a self-dual $kG$-module which admits a non-degenerate $G$-invariant symmetric (alternating) bilinear form $b$. Suppose that $W$ is a self-dual irreducible $kG$-module. If $W$ occurs in $Z$ as a composition factor of odd multiplicity, then $W$ admits a non-degenerate $G$-invariant symmetric (alternating) bilinear form.

Proof: Induction on $\dim Z$. Let $Q \subset Z$ be an irreducible $G$-submodule of $Z$, so either $Q$ is non-degenerate or totally singular.

If $Q$ is non-degenerate, then $Z = Q \oplus Q^\perp$. If $Q \cong W$ the lemma follows, and if $Q \not\cong W$ the lemma follows by applying induction on $Q^\perp$.

If $Q$ is totally singular, then $b$ induces a non-degenerate bilinear form on $Q^\perp/Q$, which is of the same type as $b$. So we can apply induction on $Q^\perp / Q$ since $Z/Q^\perp \cong Q^*$.

We can apply the lemma for $Z = V \otimes V^*$, since it always admits a non-degenerate $G$-invariant symmetric bilinear form. Identifying $V \otimes V^* \cong \operatorname{End}(V)$, one such form is given by $b(x,y) = \operatorname{Tr}(xy)$ for all $x, y \in \operatorname{End}(V)$.

So in your setting: if $W$ is irreducible, $W \cong W^*$, and $\operatorname{Hom}_G(V \otimes V^*, W)$ is one-dimensional, it follows from the lemma that $W$ admits a nondegenerate $G$-invariant symmetric bilinear form. In other words, the Frobenius-Schur indicator of $W$ is $1$.

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[1] R. Gow, W. Willems, Methods to decide if simple self-dual modules over fields of characteristic 2 are of quadratic type, J. Algebra 175 (1995) 1067–1081.