Let ${\mathfrak{g}}$ be a semisimple Lie algebra over ${\mathbb{C}}$ and ${\mathfrak{h}}$ a Cartan subalgebra.

Given ${\alpha \in \mathfrak{h}^{\vee}}$, we can define a subspace of ${\mathfrak{g}}$

$\displaystyle \mathfrak{g}_{\alpha} = \{ x \in \mathfrak{g}: (\mathrm{ad} H)x = \alpha(H) x , \ \forall H \in \mathfrak{h} \}.$

The nonzero ${\alpha}$ that occur with ${\mathfrak{g}_{\alpha} \neq 0}$ are called roots, and they form a set ${\Phi}$. Because ${\mathfrak{h}}$ acts on ${\mathfrak{g}}$ by commuting diagonalizable operators (by semisimplicity of the elements of ${\mathfrak{h}}$), it follows by simultaneous diagonalization, that

$\displaystyle \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}.$

Recall that ${\mathfrak{g}_0 = \mathfrak{h}}$, because a Cartan subalgebra is maximal abelian.

This is called the root space decomposition. A simple but important property is that ${[\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}] \subset \mathfrak{g}_{\alpha + \beta}}$; this is checked because the ${\mathrm{ad} H}$ are derivations.

The root space decomposition is highly useful in studying simple representations of ${\mathfrak{g}}$.

I shall collect here a few facts about it.

Proposition 1 ${\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}}$ are orthogonal under the Killing form unless ${\alpha + \beta = 0}$.

This follows by a familiar argument, in view of ${[\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}] \subset \mathfrak{g}_{\alpha + \beta}}$. (more…)

So, we now know that in any representation of a semisimple Lie algebra ${\mathfrak{g}}$, the semisimple elements act by diagonalizable operators. Now if we were given an abelian subalgebra ${\mathfrak{h} \subset \mathfrak{g}}$ whose elements were semisimple, then every simple representation ${V}$ of ${\mathfrak{g}}$ would split into simple ${\mathfrak{h}}$-modules ${W}$, and perhaps we could study ${V}$ through the various ${W}$. After all, each ${W}$ is easy to describe: it is one-dimensional, and the action of ${\mathfrak{h}}$ can be described by a linear functional on it (i.e., a weight).

The ${\mathfrak{h}}$ in question that will be chosen are Cartan subalgebras.

More precisely, a Cartan subalgebra of the semisimple Lie algebra ${\mathfrak{g}}$ is a maximal abelian subalgebra that consists only of semisimple elements. The special linear algebra ${\mathfrak{sl}_n}$ of matrices of trace zero has the diagonal matrices of trace zero as a Cartan subalgebra, for instance.

In this post, I shall prove the existence of Cartan subalgebras. We will always be working over an algebraically closed field of characteristic zero. (more…)

Orthogonal complements of semisimple ideals

There is a general fact about semisimple ideals in arbitrary Lie algebras that we prove next; it will be an application of the material on complete reducibility.  We will use it to complete the picture of the abstract Jordan decomposition in a semisimple Lie algebra.

Proposition 1 Let ${\mathfrak{s} \subset \mathfrak{g}}$ be a semisimple ideal in the Lie algebra ${\mathfrak{g}}$. Then there is a unique ideal ${\mathfrak{a} \subset \mathfrak{g}}$ with ${\mathfrak{g} = \mathfrak{s} \oplus \mathfrak{a}}$.

The idea is that we can find a complementary ${\mathfrak{s}}$-module ${A \subset \mathfrak{g}}$ with

$\displaystyle \mathfrak{g} = \mathfrak{s} \oplus A$

as ${\mathfrak{s}}$-modules. Now ${[\mathfrak{s},A] \subset A \cap \mathfrak{s} = \{0\}}$ because ${\mathfrak{s}}$ is an ideal and because ${A}$ is stable under the action of ${\mathfrak{s}}$. The converse is true: ${A}$ is the centralizer of ${\mathfrak{s}}$, and thus an ideal, because anything commuting with ${\mathfrak{s}}$ can have no part in ${\mathfrak{s}}$ in the ${\mathfrak{s} \oplus A}$ decomposition (${\mathfrak{s}}$ being semisimple). Thus, I hereby anoint ${A}$ with the fraktur font and call it ${\mathfrak{a}}$ to recognize its Lie algebraness.

Given a splitting as above, ${\mathfrak{a}}$ would have to be the centralizer of ${\mathfrak{s}}$, so uniqueness is evident. (more…)

The purpose of this post is to show that the category of finite-dimensional representations of a semismple Lie algebra is  a semisimple category; there is thus an analogy with Maschke’s theorem, except in this case the proofs are more complicated.  They can be simplified somewhat if one uses the cohomology of Lie algebras (i.e., appropriate Ext groups), which I may talk more about, but most likely only later.  Here we will give the proofs based on linear algebra.

The first step is to construct certain central elements in the enveloping algebra.

Casimir elements

Let ${B}$ be a nondegenerate invariant bilinear form on the Lie algebra ${\mathfrak{g}}$. (E.g. ${\mathfrak{g}}$ could be semisimple and ${B}$ the Killing form.) Given a basis ${e_i \in \mathfrak{g}}$, we can consider the dual basis ${f_j}$ with respect to it, i.e. such that ${B(e_i, f_j) = \delta_{ij}}$. Consider the Casimir element

$\displaystyle C := \sum e_i f_i \in U \mathfrak{g}.$

I claim that ${C}$ is independent of the choice of ${e_i}$ and is in the center of the enveloping algebra. First off, consider the isomorphisms of ${\mathfrak{g}}$-modules,

$\displaystyle \hom_k( \mathfrak{g}, \mathfrak{g}) \simeq \mathfrak{g} \otimes \mathfrak{g}^{\vee} \simeq \mathfrak{g} \otimes \mathfrak{g} .$

The last one is given by the form ${B}$. (more…)

I learned the material in this post from the book by Humphreys on Lie algebras and representation theory.

Recall that if ${A}$ is any algebra (not necessarily associative), then the derivations of ${A}$ form a Lie algebra ${Der(A)}$, and that if ${A}$ is actually a Lie algebra, then there is a homomorphism ${\mathrm{ad}: A \rightarrow Der(A)}$. In this case, the image of ${\mathrm{ad}}$ is said to consist of inner derivations.

Theorem 1 Any derivation of a semisimple Lie algebra ${\mathfrak{g}}$ is inner.

To see this, consider ${\mathrm{ad}: \mathfrak{g} \rightarrow D :=Der(\mathfrak{g})}$; by semisimplicity this is an injection. Let the image be ${D_i}$, the inner derivations. Next, I claim that ${[D, D_i] \subset D_i}$. Indeed, if ${\delta \in D}$ and ${\mathrm{ad} x \in D_i}$, we have

$\displaystyle [\delta, \mathrm{ad} x] y = \delta( [x,y]) - [x, \delta(y)] = [\delta(x),y] = (\mathrm{ad}(\delta(x)))y.$

In other words, ${[\delta, \mathrm{ad} x] = \mathrm{ad}(\delta(x))}$. This proves the claim.

Consider the Killing form ${B_D}$ on ${D}$ and the Killing form ${B_{D_i}}$ on ${D_i}$. The above claim and the definition as a trace shows that ${B_D|_{D_i \times D_i} = B_{D_i}}$. (more…)

A semisimple Lie algebra ${\mathfrak{g}}$ is one that has no nonzero abelian ideals. This is equivalent to the absence of solvable ideals. Indeed, if ${\mathfrak{g}}$ had a solvable ideal ${\mathfrak{z}}$, we could consider the derived series of ${\mathfrak{z}}$, i.e. ${D^1\mathfrak{z} = [\mathfrak{z},\mathfrak{z}], D^n\mathfrak{z} = [D^{n-1}\mathfrak{z}, D^{n-1}\mathfrak{z}]}$. These are ideals because, by the Jacobi identity, the Lie product of two ideals is an ideal. These ${D^n \mathfrak{z}}$ eventually become zero by the hypothesis of solvability, and the last nonzero one is abelian.

One justification for the epithet “semisimple” is that the category of finite-dimensional representations is in fact semisimple, i.e. that any exact sequence of ${\mathfrak{g}}$ representations for ${\mathfrak{g}}$ semisimple

$\displaystyle 0 \rightarrow V' \rightarrow V \rightarrow V'' \rightarrow 0$

splits. This is what happens for finite groups, because by Maschke’s theorem the group algebra of a finite group is semisimple. Nevertheless, the enveloping algebra ${U\mathfrak{g}}$ is not generally semisimple; we are restricting ourselves to finite-dimensional ${U \mathfrak{g}}$-modules.

Cartan’s criterion

Before getting there, we will prove a basic criterion for semisimplicity.

Theorem 1 (Cartan) The Lie algebra ${\mathfrak{g}}$ is semisimple if and only if its Killing form is nondegenerate.

The proof turns out to be a relatively easy consequence of Cartan’s criterion for solvability, which I’ve already given a twopost spiel on. (more…)