# a representation which is not completely reducible

If $G$ is a finite group, and $k$ is a field whose characteristic does divide the order of the group, then Maschke’s theorem fails. For example let $V$ be the regular representation of $G$, which can be thought of as functions from $G$ to $k$, with the $G$ action $g\cdot\varphi(g^{\prime})=\varphi(g^{-1}g^{\prime})$. Then this representation is not completely reducible.

There is an obvious trivial subrepresentation $W$ of $V$, consisting of the constant functions. I claim that there is no complementary invariant subspace to this one. If $W^{\prime}$ is such a subspace, then there is a homomorphism $\varphi:V\to V/W^{\prime}\cong k$. Now consider the characteristic function of the identity $e\in G$

 $\delta_{e}(g)=\begin{cases}1&g=e\\ 0&g\neq e\end{cases}$

and $\ell=\varphi(\delta_{e})$ in $V/W^{\prime}$. This is not zero since $\delta$ generates the representation $V$. By $G$-equivarience, $\varphi(\delta_{g})=\ell$ for all $g\in G$. Since

 $\eta=\sum_{g\in G}\eta(g)\delta_{g}$

for all $\eta\in V$,

 $W^{\prime}=\varphi(\eta)=\ell\left(\sum_{g\in G}\eta(g)\right).$

Thus,

 $\mathrm{ker}\,\varphi=\{\eta\in V|\sum_{\in G}\eta(g)=0\}.$

But since the characteristic of the field $k$ divides the order of $G$, $W\leq W^{\prime}$, and thus could not possibly be complementary to it.

For example, if $G=C_{2}=\{e,f\}$ then the invariant subspace of $V$ is spanned by $e+f$. For characteristics other than $2$, $e-f$ spans a complementary subspace, but over characteristic 2, these elements are the same.

Title a representation which is not completely reducible ARepresentationWhichIsNotCompletelyReducible 2013-03-22 13:31:47 2013-03-22 13:31:47 bwebste (988) bwebste (988) 5 bwebste (988) Example msc 20C15