rt.representation theory - Invariant quadratic forms of irreducible representations

Let $G$ be a finite group, and $k$ be a field of characteristic zero (not necessarily algebraically closed!). Let $rho : G to mathrm{End}_k left(k^nright)$ be a irreducible representation of $G$ over $k$. Consider the vector space

$S=leftlbrace Hin mathrm{End}_kleft(k^nright) mid rholeft(gright)^T Hrholeft(gright)=Htext{ for any }gin Grightrbrace$

$=leftlbrace sumlimits_{gin G}rholeft(gright)^T Hrholeft(gright)mid Hin mathrm{End}_kleft(k^nright)rightrbrace$

and its subspace

$T=leftlbrace Hin Smid Htext{ is a symmetric matrix}rightrbrace$.

It is easy to show that, if we denote our representation of $G$ on $k^n$ by $V$, then the elements of $S$ uniquely correspond to homomorphisms of representations $Vto V^{ast}$ (namely, $Hin S$ corresponds to the homomorphism $vmapstoleft(wmapsto v^THwright)$), while the elements of $T$ uniquely correspond to $G$-invariant quadratic forms on $V$ (namely, $Hin T$ corresponds to the quadratic form $vmapsto v^THv$).

(1) In the case when $k=mathbb C$, Schur's lemma yields $dim Sleq 1$, with equality if and only if $Vcong V^{ast}$ (which holds if and only if $V$ is a real or quaternionic representation). Thus, $dim Tleq 1$, and it is known that this is an equality if and only if $V$ is a real representation. (Except of the equality parts, this all pertains to the more general case when $k$ is algebraically closed of zero characteristic).

(2) In the case when $k=mathbb R$, it is easily seen that $Tneq 0$ (that's the famous nondegenerate unitary form, which in the case $k=mathbb R$ is a quadratic form), and I think I can show (using the spectral theorem) that $dim T=1$. As for $S$, it can have dimension $>1$.

(3) I am wondering what can be said about other fields $k$; for instance, $k=mathbb Q$. If $ksubseteqmathbb R$, do we still have $dim T=1$ as in the $mathbb R$ case? In fact, $Tneq 0$ can be shown in the same way.

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