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 $g\cdot \phi (g^{\prime })=\phi (g^{-1}{g}^{\prime })$. 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 $\phi :V\to V/W^{\prime }\cong k$. Now consider the characteristic function of the identity $e\in G$

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

$$\eta =\sum _{g\in G}\eta (g){\delta }_g$$

for all $\eta \in V$,

$$W^{\prime }=\phi (\eta )=\mathrm{\ell }\left(\sum _{g\in G}\eta (g)\right).$$

Thus,

$$\ker \phi =\{\eta \in V|\sum _{\in G}\eta (g)=0\}.$$

But since the characteristic of the field $k$ divides the order of $G$, $W\le W^{\prime }$, 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
Canonical name	ARepresentationWhichIsNotCompletelyReducible
Date of creation	2013-03-22 13:31:47
Last modified on	2013-03-22 13:31:47
Owner	bwebste (988)
Last modified by	bwebste (988)
Numerical id	5
Author	bwebste (988)
Entry type	Example
Classification	msc 20C15