An irreducible representation of a finite group whose symmetric square is also irreducible

Is there an irreducible representation $V$ (over $\mathbb{C}$) of a finite group $G$ such that $\dim V > 1$ and $\mathrm{Sym}^2V$ is again irreducible? (I believe that there are examples for compact $G$, such as the fundamental representation of $SU(2)$.)

Searching for examples:

Some obvious candidates don't work:

• $G$ cannot be dihedral, as all dihedral irreps are of dimension 1 or 2;
• $G$ cannot be $Q_8$ for analogous reasons;
• $G$ cannot be $S_n$ for $n \le 6$; dimension-counting rules out all irreps except for the two-dimensional irrep of $S_4$, which satisfies $\mathrm{Sym}^2V = 1 \oplus V$;
• $G$ cannot be $A_n$ for $n \le 6$ for similar dimension-counting reasons.

Obviously the Frobenius-Schur indicator of $V$ cannot be 1, but this is sort of tautological. If $G$ has odd order, then I think the Frobenius-Schur indicator of $V$ must be zero, so perhaps something like the subgroup of $\mathop{PGL}_2(\mathbb{F}_7)$ of order 21 is a good place to look.

A suitable GAP query might also produce examples; unfortunately this is beyond my present ability.

Trying to prove this is impossible:

I also haven't been able to make any headway on showing that there can be no such $G$. The obvious strategy would be to find a nice simplification of $$ \left\langle \chi_{\rho}, \chi_{\mathrm{Sym}^2V} \right \rangle, $$ where presumably the clever choice is to take $\rho = V$. (Taking $\rho = 1$ cannot prove this is impossible, since groups with Frobenius-Schur indicators other than 1 exist.)

Note that if $V$ is not a summand of $\mathrm{Sym}^2V$, we have $$(\dim V)^2 = \sum_{g \in G} \overline{\chi}_V(g)\chi_V(g^2).$$ This feels possibly helpful in special cases; for example using this formula I think I can (via a sort of "complex rearrangement inequality") rule out this case for $G$ of odd order.

It might be possible to use the hook-length formula to rule out most examples in $S_n$. The only candidate irreps $V$ of $S_n$ for $n \le 8$ are the quotient of the standard representation by the trivial representation and the tensor product of this representation with the sign representation. I haven't yet written this down to see if I can make it work; I'd also have to look up a lot about the representations of the symmetric groups to rule out the possibilities that would be left.

Solution 1:

Since you mention GAP, here is a way how one could test it using the character table functionality:

# test for property: Arguments: table, character
test:=function(t,c). 
local p;
  if c[1]=1 or ScalarProduct(c,c)<>1 then return false;fi;
  p:=SymmetricParts(t,[c],2)[1];
  return ScalarProduct(p,p)=1;
end;

# test for group -- any irreducible characetr has property?
gptest:=function ( g )
local t;
  t := CharacterTable( g );
  return ForAny( Irr( t ), x->test( t, x );end );
end;

Lets check group up to order 100:

gap> g:=OneSmallGroup(Size,[2..100],gptest,true);
<pc group of size 24 with 4 generators>
gap> StructureDescription(g); # only makes sense for small orders
"SL(2,3)"

and lets check:

gap> t:=CharacterTable(g);
CharacterTable( SL(2,3) )
gap> List(Irr(t),x->x[1]);
[ 1, 1, 1, 2, 2, 2, 3 ]
gap> List(Irr(t),x->test(t,x));
[ false, false, false, true, true, true, false ]

Thus the 2-dimensional representations of $SL_2(3)$ are an example of minimal order. $SL_2(5)$ is a minimal nonsolvable example (also degree 2), and $SL_3(2)$ is an exampel with character degree 3.

Solution 2:

Here's a strategy to find lots of two-dimensional examples!

Let $\rho\colon G\to \mathrm{GL}_2(\mathbb C)$ be an irreducible, two-dimensional representaiton representation and suppose that the three-dimensional representation $\mathrm{Sym}^2(\rho)$ is reducible. Then, in particular, $\mathrm{Sym}^2(\rho)$ contains a $1$-dimensional subrepresentation, $\chi$.

Claim: There exists an index two subgroup $H\le G$ and a character $\eta\colon H\to\mathbb C^\times$ such that $\rho = \mathrm{Ind}_H^G(\eta)$.

Proof: First note that $\rho \cong \rho^\vee\otimes \det\rho$. This is always true for two-dimensional representations. Hence, we have $$1=(\rho, \rho) = (\rho\otimes\rho^\vee, 1) = (\rho\otimes\rho, \det(\rho)) = (\mathrm{Sym}^2(\rho) + \det(\rho), \det(\rho)),$$ from which it follows that $(\mathrm{Sym}^2(\rho),\det(\rho)) = 0$. Thus $\det(\rho)\ne \chi$.

Now $$\begin{align*}(\rho, \rho\otimes\chi\det(\rho)^{-1}) &= (\rho, \rho^\vee\otimes\chi)\\&=(\rho\otimes\rho , \chi)\\&=(\mathrm{Sym}^2(\rho), \chi) = 1\end{align*}$$

Hence, $\rho\cong\rho\otimes\chi'$ for a non trivial character $\chi'$. Taking determinants, it's clear that $\chi'$ has order $2$. Hence, its kernel is a subgroup $H\le G$ of index $2$. It follows from Clifford theory that $\rho|_H$ contains a one-dimensional subrepresentation $\eta$. Finally, $$1\ge (\rho, \mathrm{Ind}_H^G\eta) = (\rho|_H, \eta) \ge 1$$ and the result follows.

So to find an example, just chose any group with a two-dimensional representation whose projective image is not dihedral. As pointed out in the other answer, the smallest example is the two-dimensional representation of $\mathrm{SL}_2(\mathbb F_3)$, whose image in $\mathrm{PGL}_2(\mathbb C)$ is $A_4$.