# The Unapologetic Mathematician

## 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 $i$th 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.

September 16, 2012 -

## 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 […]

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