The Unapologetic Mathematician

Mathematics for the interested outsider

Reducible Modules

As might be surmised from irreducible modules, a reducible module M for a Lie algebra L is one that contains a nontrivial proper submodule — one other than 0 or M itself.

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

\displaystyle0\to N\to M\to M/N\to 0

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

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

\displaystyle\left[phi_{A\oplus B}(x)\right](a,b)=\left(\left[\phi_a(x)\right](a),\left[\phi_B(x)\right](b)\right)

Clearly the usual subspace inclusions and projections between A, B, and A\oplus B intertwine these actions, so they’re the required module morphisms. Further, it’s clear that (A\oplus B)/A\cong B.

So, do all short exact sequences of representations split? no. Indeed, let \mathfrak{t}(n,\mathbb{F}) be the algebra of n\times n upper-triangular matrices, along with the obvious n-dimensional representation. If we let e_i be the basic column vector with a 1 in the ith row and 0 elsewhere, then the one-dimensional space spanned by e_1 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 M that any nontrivial proper submodule N has a complementary submodule N'\subseteq M with M=N\oplus N'. In this case, N is either irreducible or it’s not; if not, then any proper nontrivial submodule of N will also be a proper nontrivial submodule of M, and we can continue taking smaller submodules until we get to an irreducible one, so we may as well assume that N is irreducible. Now the same sort of argument works for N', 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 M. At each step, the complement gets smaller and smaller, until we have decomposed M entirely into a direct sum of irreducible submodules.

If M is decomposable into a direct sum of irreducible submodules, we say that M 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 M, and the complement will consist of the rest.

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September 16, 2012 - Posted by | Algebra, Lie Algebras, Representation Theory

1 Comment »

  1. [...] techniques we can use to generate new modules for a Lie algebra from old ones. We’ve seen direct sums already, but here are a few [...]

    Pingback by New Modules from Old « The Unapologetic Mathematician | September 17, 2012 | Reply


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