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

$S=\left\lbrace H\in \mathrm{End}_k\left(k^n\right) \mid \rho\left(g\right)^T H\rho\left(g\right)=H\text{ for any }g\in G\right\rbrace$

$=\left\lbrace \sum\limits_{g\in G}\rho\left(g\right)^T H\rho\left(g\right)\mid H\in \mathrm{End}_k\left(k^n\right)\right\rbrace$

and its subspace

$T=\left\lbrace H\in S\mid H\text{ is a symmetric matrix}\right\rbrace$.

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 $V\to V^{\ast}$ (namely, $H\in S$ corresponds to the homomorphism $v\mapsto\left(w\mapsto v^THw\right)$), while the elements of $T$ uniquely correspond to $G$-invariant quadratic forms on $V$ (namely, $H\in T$ corresponds to the quadratic form $v\mapsto v^THv$).

(1) In the case when $k=\mathbb C$, Schur's lemma yields $\dim S\leq 1$, with equality if and only if $V\cong V^{\ast}$ (which holds if and only if $V$ is a real or quaternionic representation). Thus, $\dim T\leq 1$, 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 $T\neq 0$ (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 $k\subseteq\mathbb R$, do we still have $\dim T=1$ as in the $\mathbb R$ case? In fact, $T\neq 0$ can be shown in the same way.

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up vote 5 down vote accepted

There are certainly examples over $k=\mathbb{Q}$ where $\dim T\ge2$. Let's take the cyclic group $G$ of order $5$ and the representation space $$V=\{(a_0,\ldots,a_4)\in\mathbb{Q}^5:a_0+\cdots +a_4=0\}$$ where $G$ acts by cyclic permutation. Two linearly independent elements of $T$ are given by $$\left(\begin{array}{rrrrr} 2&-1&0&0&-1\\\ -1&2&-1&0&0\\\ 0&-1&2&-1&0\\\ 0&0&-1&2&-1\\\ -1&0&0&-1&2\end{array}\right) $$ and $$\left(\begin{array}{rrrrr} 2&0&-1&-1&0\\\ 0&2&0&-1&-1\\\ -1&0&2&0&-1\\\ -1&-1&0&2&0\\\ 0&-1&-1&0&2\end{array}\right) $$ (these define quadratic forms on $V$ since they annihilate the all-one vector).

The point here is that this representation splits into two over $\mathbb{R}$. I think the dimension of $T$ in general will be the number of irreducible representations it splits into over $\mathbb{R}$.

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Thanks! I should have thought about that - another case when the irreducibility of cyclotomic polynomials over $\mathbb Q$ comes in ahndy. – darij grinberg Apr 4 '10 at 13:03
If $V$ is the representation space, then the space $T$ corresponds to $G$-maps from $\mathrm{Sym}^2(V)$ to $k$ considered as a trivial $G$-space. Thus $\dim T$ is the number of copies of the trivial representation inside $\mathrm{Sym}^2(V)$. This won't change when one passes from $k$ to an extension field. – Robin Chapman Apr 4 '10 at 15:59

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