a representation which is not completely reducible
If is a finite group, and is a field whose characteristic does divide the order of the group, then Maschke’s theorem fails. For example let be the regular representation of , which can be thought of as functions from to , with the action . Then this representation is not completely reducible.
There is an obvious trivial subrepresentation of , consisting of the constant functions. I claim that there is no complementary invariant subspace to this one. If is such a subspace, then there is a homomorphism . Now consider the characteristic function of the identity
and in . This is not zero since generates the representation . By -equivarience, for all . Since
for all ,
But since the characteristic of the field divides the order of , , and thus could not possibly be complementary to it.
For example, if then the invariant subspace of is spanned by . For characteristics other than , spans a complementary subspace, but over characteristic 2, these elements are the same.
|Title||a representation which is not completely reducible|
|Date of creation||2013-03-22 13:31:47|
|Last modified on||2013-03-22 13:31:47|
|Last modified by||bwebste (988)|