Reducible Modules

As might be surmised from irreducible modules, a reducible module for a Lie algebra is one that contains a nontrivial proper submodule — one other than or itself.

Now obviously if is a submodule we can form the quotient . This is the basic setup of a short exact sequence:

The question is, does this sequence split? That is, can we write as the direct sum of and some other submodule isomorphic to ?

First of all, let’s be clear that direct sums of modules do make sense. Indeed, if and are -modules then we can form an action on by defining it on each summand separately

Clearly the usual subspace inclusions and projections between , , and intertwine these actions, so they’re the required module morphisms. Further, it’s clear that .

So, do all short exact sequences of representations split? no. Indeed, let be the algebra of upper-triangular matrices, along with the obvious -dimensional representation. If we let be the basic column vector with a in the th row and elsewhere, then the one-dimensional space spanned by forms a one-dimensional submodule. Indeed, all upper-triangular matrices will send this column vector back to a multiple of itself.

On the other hand, it may be the case for a module that any nontrivial proper submodule has a complementary submodule with . In this case, is either irreducible or it’s not; if not, then any proper nontrivial submodule of will also be a proper nontrivial submodule of , and we can continue taking smaller submodules until we get to an irreducible one, so we may as well assume that is irreducible. Now the same sort of argument works for , showing that if it’s not irreducible it can be decomposed into the direct sum of some irreducible submodule and some complement, which is another nontrivial proper submodule of . At each step, the complement gets smaller and smaller, until we have decomposed entirely into a direct sum of irreducible submodules.

If is decomposable into a direct sum of irreducible submodules, we say that is “completely reducible”, or “decomposable”, as we did when we were working with groups. Any module where any nontrivial proper submodule has a complement is thus completely reducible; conversely, complete reducibility implies that any nontrivial proper submodule has a complement. Indeed, such a submodule must consist of some, but not all, of the summands of , and the complement will consist of the rest.