# Existence of Hermitian form .

I came across this question in my notes : If $G$ is finite group and $V$ is finite dimensional $CG$ module ( $C$ complex ) , how do i show that there exist a positive definite Hermitian Form $(.,.)$ for $($gv, gw) for all $g$ and $v$,$w$. And show that V is semisimple without using Maschke's theorem.

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The basic idea is that you know that $V$ admits some inner product $\langle -,-\rangle_0$. Then, just normalize it by definiing

$$\langle u,v\rangle=\frac{1}{|G|}\sum_{g\in G}\langle gu,gv\rangle_0$$

You can then prove Maschke's theorem which says that $\mathbb{C}[G]$ is semisimple by showing that every $\mathbb{C}[G]$-submodule $W\leqslant V$ has a complement by showing that $W^\perp$ (with respect to $\langle-,-\rangle$) is such a subspace--which of course proves the semisimplicity of $\mathbb{C}[G]$ since $V$ was arbitrary.

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Another interesting fact is the following. Let $V$ be an arbitrary representation of $G$ (it could be infinite-dimensional). A given $v\in V$ is contained in a finite-dimensional subrepresentation: $$V_v=\mbox{span}\{g\cdot v:g\in G\}$$ (it is finite dimensional since $G$ is a finite group). The answer of Alex Youcis proves that a finite-dimensional representation is semisimple. Hence, it follows that every $v\in V$ is contained in a semisimple subrepresentation, and so we can conclude that $V$ itself is semisimple. Therefore, all representations of a finite group (even infinite-dimensional) is semisimple.

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It would be nice if you could you give me a explicit example of a non reducible representation ( i.e not semisimple). – Theorem Apr 24 '12 at 17:21
Consider the following complex representation of $\mathbb{R}$: $$t\mapsto \left( \begin{array}{cc} 1 & t \\ 0 & 1 \end{array} \right).$$ – M Turgeon Apr 24 '12 at 17:34
The only way a vector $(a,b)$ can be fixed by all elements of $\mathbb{R}$ is if $b=0$. Since there is a unique 1-dimensional subrepresentation, the above representation is not reducible. – M Turgeon Apr 24 '12 at 17:37