Irreducible tensor product representations in finite simple groups

Question

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PSp{PSp}\DeclareMathOperator\PSU{PSU}$Background:

A representation $ \rho: G \to \GL(V) $ of a group $ G $ on a (complex) vector space $ V $ is said to be irreducible if there are no nontrivial invariant subspaces under the action of $ G $. A tensor product of two irreducible representations $ \rho_i: G \to \GL(V_i) $, $i=1,2$, is a representation on the tensor product space $ V_1 \otimes V_2 $, defined by $ \rho_1(g) \otimes \rho_2(g) $ for $ g \in G $.

In some cases, this tensor product representation $ \rho_1 \otimes \rho_2 $ can itself be irreducible.

Examples:

• If $V_1$ or $V_2$ is one-dimensional, then $ \rho_1 \otimes \rho_2 $ is irreducible.

• Let $G$ be the product $ G_1 \times G_2 $. If $ \rho_i $ is an irreducible representation of $ G_i $, then the representation $\rho_1 \boxtimes \rho_2 = (\rho_1 \boxtimes 1) \otimes (1 \boxtimes \rho_2)$ is irreducible for $ G_1 \times G_2 $.

To avoid such straightforward examples, let us assume that $G$ is a non-abelian finite simple group and $\rho_i \neq 1 $. Surprisingly, such phenomenon is still possible in this case, but quite rare.

It occurs just on $10$ (among $204$) such groups of order less than $2.7\times 10^8$ (see the computations in Appendix), namely,
$$\PSp(4,3), \ M_{12}, \ A_9, \ \PSp(6,2), \ M_{23}, \ \PSU(5,2), \ 2F(4,2)', \ O_+(8,2), \ ^3D(4,2), \ M_{24}.$$

More specifically, for the alternating groups $A_n$, with $5 \le n \le 21$, this phenomenon occurs just for $n=9,16$, with $(\dim(V_1),\dim(V_2))=(8,21), (15,12012)$, respectively.

For the sporadic groups, it occurs for $9$ ones (among $26$), namely,
$$M_{12}, \ M_{23}, \ M_{24}, \ Co_3, \ Co_2, \ Th, \ Co_1, \ F_{3+}, \ M.$$ Warning: This last list differs from (the complement of) the list in this review mentioned by Nick Gill in this comment.

Question: For which non-abelian finite simple groups this phenomenon occurs? more specifically, for which alternating groups $A_n$? if and only if $n$ is a square? and $\dim(V_1) = n-1$?

My laptop was not able to check up to $n=25$.

---

Computations

For general non-abelian finite simple groups (we know that we can avoid $\PSL(2,q)$, see here):

gap> simpnames:= AllCharacterTableNames( IsSimple, true, IsAbelian, false, IsDuplicateTable, false : OrderedBy:= Size);
gap> for nam in simpnames do if Size(nam)<3 or List([1..3],i->nam[i]) <> "L2(" then ct:=CharacterTable(nam); Print(nam," ",Size(ct)," ",c,"\n"); Phenomenon(ct); fi; od; 
...
U4(2), M12, A9, S6(2), M23, U5(2), 2F4(2)', O8+(2), 3D4(2), M24, ... the list should be incomplete after, because AllCharacterTableNames contains finitely many simple groups

For the alternating group $A_n$, for $5 \le n \le 21$:

gap> for n in [5..21] do Print(n,"\n");; ct:=CharacterTable("Alternating", n); Phenomenon(ct);; od;
...
A9, A16

For the sporadic groups:

gap> spornames:= AllCharacterTableNames(IsSporadicSimple, true, IsDuplicateTable, false : OrderedBy:= Size);;
gap> for nam in spornames do ct:=CharacterTable(nam); Print(nam," ",Size(ct),"\n"); Phenomenon(ct); od;    
M12, M23, M24, Co3, Co2, Th, Co1, F3+, M

Script

LoadPackage("ctbllib"); LoadPackage("atlasrep");
Phenomenon:=function(ct)
    local irr,r,L,i,j,k;
    irr:=Irr(ct);
    r:=Size(irr);
    for i in [2..r] do
        for j in [i..r] do
            if IsIrreducible(irr[i]*irr[j]) then
                Print([i,j],"\n");; 
            fi;
        od;
    od;
end;;

Answer 1

This is not an answer, but it reminded me of a beautiful general argument of R. Brauer (which I first saw in a paper of David Wales, (credited to Brauer, though I am not sure whether Brauer ever published it in his own right)). This result may be marginally relevant to the question.

If we have an irreducible character $\chi$ of a finite group $G$ such that $\chi \overline{\chi} = 1 + \theta$ where $\theta$ is a (necessarily non-trivial and real-valued) irreducible character, then either $\theta$ is rational valued, or else there is a Galois conjugation $\sigma$ such that $\chi \chi^{\sigma}$ is irreducible.

For suppose that $\theta$ is not rational valued. Then there is a Galois conjugation $\sigma$ (commuting with complex conjugation) such that $\theta^{\sigma} \neq \theta.$

Then $\chi \chi^{\sigma} \overline{\chi \chi^{\sigma}} = 1 + \theta + \theta^{\sigma} + \theta \theta^{\sigma}.$

Since $\theta = \overline{\theta}\neq \theta^{\sigma},$ the irreducible character $\theta^{\sigma}$ is different from $\overline{\theta}.$ Hence the trivial character only occurs once in $\chi \chi^{\sigma} \overline{\chi \chi^{\sigma}}$ and $\chi \chi^{\sigma}$ is irreducible.

Two examples where this occurs in practice are when $G ={\rm SL}(2,5)$ and $\chi$ is an irreducible character of degree $2$. Then $\chi \overline{\chi} = 1 +\theta,$ where $\theta$ is irreducible of degree three, but is not rational valued. Then $\chi \chi^{\sigma}$ is irreducible of degree $4$ where $\sigma$ is the Galois automorphism which squares odd order roots of unity and fixes $2$-power roots of unity. A similar example occurs for $G$ a triple cover of $A_{6}$, with $\chi$ irreducible of degree $3$ (with $\sigma$ cubing roots of unity of order prime to $3$ anf fixing roots of unity of $3$-power order).

Answer 2

Also not an answer, just a hint on how the GAP code can be made a bit faster, for future reference: GAP actually "knows" the character tables for the alternating groups. For some reason I don't know this is not used here, but we can ask GAP directly for that character table, bypassing all computations in groups, to get a faster computation, like so:

LoadPackage("ctbllib");
Phenomenon:=function(n)
    local ct, irr,r,L,i,j,k;
    ct:=CharacterTable("Alternating", n);
    irr:=Irr(ct);
    r:=Size(irr);
    for i in [2..r] do
        for j in [i..r] do
            if IsIrreducible(irr[i]*irr[j]) then
                Print([i,j],"\n");; 
            fi;
        od;
    od;
end;;

for n in [5..20] do Print(n,"\n");; Phenomenon(n);; od;