# The Unapologetic Mathematician

## Irreducible Modules

Sorry for the delay; it’s getting crowded around here again.

Anyway, an irreducible module for a Lie algebra $L$ is a pretty straightforward concept: it’s a module $M$ such that its only submodules are $0$ and $M$. As usual, Schur’s lemma tells us that any morphism between two irreducible modules is either $0$ or an isomorphism. And, as we’ve seen in other examples involving linear transformations, all automorphisms of an irreducible module are scalars times the identity transformation. This, of course, doesn’t depend on any choice of basis.

A one-dimensional module will always be irreducible, if it exists. And a unique — up to isomorphism, of course — one-dimensional module will always exist for simple Lie algebras. Indeed, if $L$ is simple then we know that $[L,L]=L$. Any one-dimensional representation $\phi:L\to\mathfrak{gl}(1,\mathbb{F})$ must have its image in $[\mathfrak{gl}(1,\mathbb{F}),\mathfrak{gl}(1,\mathbb{F})]=\mathfrak{sl}(1,\mathbb{F})$. But the only traceless $1\times1$ matrix is the zero matrix. Setting $\phi(x)=0$ for all $x\in L$ does indeed give a valid representation of $L$.

September 15, 2012