Invariant quadratic forms of irreducible representations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:05:00Z http://mathoverflow.net/feeds/question/20267 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20267/invariant-quadratic-forms-of-irreducible-representations Invariant quadratic forms of irreducible representations darij grinberg 2010-04-03T22:08:11Z 2010-04-04T10:04:48Z <p>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</p> <p>$S=\left\lbrace H\in \mathrm{End}_k\left(k^n\right) \mid \rho\left(g\right)^T H\rho\left(g\right)=H\text{ for any }g\in G\right\rbrace$</p> <p>$=\left\lbrace \sum\limits_{g\in G}\rho\left(g\right)^T H\rho\left(g\right)\mid H\in \mathrm{End}_k\left(k^n\right)\right\rbrace$</p> <p>and its subspace</p> <p>$T=\left\lbrace H\in S\mid H\text{ is a symmetric matrix}\right\rbrace$.</p> <p>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 $v\mapsto v^THv$).</p> <p><strong>(1)</strong> 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).</p> <p><strong>(2)</strong> 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 $>1$.</p> <p><strong>(3)</strong> 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.</p> http://mathoverflow.net/questions/20267/invariant-quadratic-forms-of-irreducible-representations/20296#20296 Answer by Robin Chapman for Invariant quadratic forms of irreducible representations Robin Chapman 2010-04-04T08:52:06Z 2010-04-04T08:52:06Z <p>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 $$\left(\begin{array}{rrrrr} 2&amp;-1&amp;0&amp;0&amp;-1\\ -1&amp;2&amp;-1&amp;0&amp;0\\ 0&amp;-1&amp;2&amp;-1&amp;0\\ 0&amp;0&amp;-1&amp;2&amp;-1\\ -1&amp;0&amp;0&amp;-1&amp;2\end{array}\right)$$ and $$\left(\begin{array}{rrrrr} 2&amp;0&amp;-1&amp;-1&amp;0\\ 0&amp;2&amp;0&amp;-1&amp;-1\\ -1&amp;0&amp;2&amp;0&amp;-1\\ -1&amp;-1&amp;0&amp;2&amp;0\\ 0&amp;-1&amp;-1&amp;0&amp;2\end{array}\right)$$ (these define quadratic forms on $V$ since they annihilate the all-one vector).</p> <p>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 $\mathbb{R}$.</p>