If $latex V$ is a module of a Lie algebra $latex L$, there is one submodule that turns out to be rather interesting: the submodule $latex V^0$ of vectors $latex v\in V$ such that $latex x\cdot v=0$…| The Unapologetic Mathematician
There are a few constructions we can make, starting with the ones from last time and applying them in certain special cases. First off, if $latex V$ and $latex W$ are two finite-dimensional $latex …| The Unapologetic Mathematician
There are a few standard techniques we can use to generate new modules for a Lie algebra $latex L$ from old ones. We’ve seen direct sums already, but here are a few more. One way is to start …| The Unapologetic Mathematician
As might be surmised from irreducible modules, a reducible module $latex M$ for a Lie algebra $latex L$ is one that contains a nontrivial proper submodule — one other than $latex 0$ or $latex…| The Unapologetic Mathematician
Sorry for the delay; it’s getting crowded around here again. Anyway, an irreducible module for a Lie algebra $latex L$ is a pretty straightforward concept: it’s a module $latex M$ such …| The Unapologetic Mathematician
It should be little surprise that we’re interested in concrete actions of Lie algebras on vector spaces, like we were for groups. Given a Lie algebra $latex L$ we define an $latex L$-module t…| The Unapologetic Mathematician
It turns out that all the derivations on a semisimple Lie algebra $latex L$ are inner derivations. That is, they’re all of the form $latex \mathrm{ad}(x)$ for some $latex x\in L$. We know tha…| The Unapologetic Mathematician
We say that a Lie algebra $latex L$ is the direct sum of a collection of ideals $latex L=I_1\oplus\dots\oplus I_n$ if it’s the direct sum as a vector space. In particular, this implies that $…| The Unapologetic Mathematician
Let’s go back to our explicit example of $latex L=\mathfrak{sl}(2,\mathbb{F})$ and look at its Killing form. We first recall our usual basis: $latex \displaystyle\begin{aligned}x&=\begin{…| The Unapologetic Mathematician
The first and most important structural result using the Killing form regards its “radical”. We never really defined this before, but it’s not hard: the radical of a binary form $…| The Unapologetic Mathematician