When is the induced representation functor exact?

139 Views Asked by At

I only have superficial familiarity with concepts of homological algebra, and couldn't find this written down explicitly anywhere so I wanted to make sure. Here is my basic argument:

Let $G$ be a finite groups and $ H \subset G $ a subgroup, and suppose we have the following short exact sequence of complex representations of $H$: $$ 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 $$ so $ A,B,C $ are all finite-dimensional complex vector spaces on which $H$ acts by invertible linear maps, and the arrows corresponds to $H$-equivariant linear maps. Since all complex representations of $H$ are completely reducible, this short exact sequence splits. Now, the functor $ \mathrm{Ind} $ is additive, so it preserves split short exact sequences, hence we obtain a split short exact sequence of complex representations of $G$: $$ 0 \rightarrow \mathrm{Ind} (A) \rightarrow \mathrm{Ind} (B) \rightarrow \mathrm{Ind} (C) \rightarrow 0 $$ which implies that the map from $ \mathrm{Ind} (A) $ to $ \mathrm{Ind} (B) $ is injective, so the first left derived functor of $ \mathrm{Ind} $ vanishes on $C$. Since this short exact sequence can be continued to the left by an infinite sequence of zeros, it seems that all higher order derived functors vanish as well, for $A,B,C$ (and presumably any other finite-dimensional complex representation of $H$).

Question: is this argument correct? And if so, how can it be generalized? Is $ \mathrm{Ind} $ always exact for cases where all representations are completely reducible? When is it nontrivial?

2

There are 2 best solutions below

0
On BEST ANSWER

It seems that my argument was indeed correct: for any finite group $H$, the category of finite-dimensional complex representations of $H$ is a semi-simple abelian category. According to the first two answers to this MO post, every semi-simple category is split, i.e. has the property that every short exact sequence splits. Equivalently, every additive functor from this category to an abelian category is exact. In particular, $\mathrm{Ind}$ is exact, so its derived functor is zero.

0
On

A different approach: induction for $k$-linear representations ($k$ a field) can be defined as $\text{Ind}_H^G(M) := kG \otimes_{kH} M$. The group algebra $kG$ is free (hence flat) as a right $kH$-module, so the functor $kG \otimes_{kh} -$ is exact. This is true in general, including the semisimple case and the case where $\text{char}~k$ divides $|G|$.