OK, now we’ve gotten some of the basic facts about the root space decomposition down. So, as usual {\mathfrak{g}} is a semisimple Lie algebra and {\mathfrak{h}} a Cartan subalgebra; we have the decomposition {\mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}}, where {\Phi \subset \mathfrak{h}^{\vee}} is the root system. For each {\alpha \in \Phi}, we can choose a pair of vectors {X_{\alpha} \in \mathfrak{g}_{\alpha}< Y_{\alpha} \in \mathfrak{g}_{-\alpha}, H_{\alpha} \in \mathfrak{h}}. Then {X_{\alpha}, Y_{\alpha}, H_{\alpha}} generate a subalgebra {\mathfrak{s}_{\alpha} \subset \mathfrak{g}} which is isomorphic to {\mathfrak{sl}_2}. Here {\alpha(H_{\alpha})=2} and {H_{\alpha}} is a multiple of {T_{\alpha}}, which in turn is the dual to {\alpha} under the Killing form that identifies {\mathfrak{h} \simeq \mathfrak{h}^{\vee}}.

That was a lightning review of where we are; if you’ve missed something, check back at this post.

The notation {\mathfrak{s}_{\alpha}} suggests that the algebra should only depend on {\alpha} and not on the particular choice of {X_{\alpha}, Y_{\alpha}} (but {H_{\alpha}} is uniquely determined from {\alpha(H_{\alpha})=2} and {H_{\alpha} \in \mathbb{C} T_{\alpha}}). Indeed, this is the case, and it follows from

Proposition 1 When {\alpha \in \Phi}, {\mathfrak{g}_{\alpha}} is one-dimensional.

 

Choose any {\mathfrak{s}_{\alpha}} coming from suitable {X_{\alpha}, Y_{\alpha}} and {H_{\alpha}}. We have a representation of {\mathfrak{s}_{\alpha}} on

\displaystyle V := \bigoplus_{\mathbb{Z} \alpha} \mathfrak{g}_{\alpha}

(recall {\mathfrak{g}_0 = \mathfrak{h}}) and we can apply the representation theory of {\mathfrak{sl}_2} to it. (more…)

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I talked about the Lie algebra {\mathfrak{sl}_2} a while back.  Now I’m going to do it more properly, and using the tools developed.  This is going to feature prominently in some of the proofs in the sequel.

Now, let’s see how all this works for the familiar case of {\mathfrak{sl}_2}, with its usual generators {H,X,Y}. This is a simple Lie algebra in fact. To see this, let’s consider the ideal {I} of {\mathfrak{sl}_2} generated by some nonzero vector {aX + bH + cY}; I claim it is all of {\mathfrak{sl}_2}.

Consider the three cases {a \neq 0, b \neq 0, c \neq 0}:

First, assume {a} or {c} is nonzero. Bracketing with {H}, and again, gives

\displaystyle -2aX + 2 c Y \in I , \ (-2)^2 a X + 2^2 cY \in I, \ (-2)^3 a X + 2^3 cY \in I.

Using a vanderMonde invertibility of this system of linear equations, we find that either {X} or {Y} belongs to {I}. Say {X} does, for definiteness; then {H = [X,Y] \in I} too; from this, {Y = -\frac{1}{2} [H,Y] \in I} as well. Thus {I = \mathfrak{sl}_2}.

If {a=c=0}, then from {b \neq 0}, we find {H \in I}, which implies {X = \frac{1}{2}[H,X] \in I} and similarly for {Y}. Thus {I= \mathfrak{sl}_2}.

I claim now that the algebra {\mathbb{C} H} is in fact a Cartan subalgebra. Indeed, it is easily checked to be maximal abelian. Moreover, since {H} acts by a diagonalizable operator on the faithful representation on {\mathbb{C}^2}, it follows that {H \in \mathfrak{sl}_2} is (abstractly) semisimple. (more…)

Recall that in the representation theory of {\mathfrak{sl}_2}, one considered an element {H} and its action on a representation {V}. We looked for its largest eigenvalue and the corresponding highest weight vector.

There is something along the same lines to be done here for arbitrary semisimple Lie algebras, though it is much more complicated (and interesting).   I’m only going to scratch the surface today.

Let {\mathfrak{g}} be a semisimple Lie algebra and {\mathfrak{h}} a Cartan subalgebra. Then {\mathfrak{h}} is to play the role of {H} in {\mathfrak{sl}_2}; the {X,Y} matrices in {\mathfrak{sl}_2} are now replaced by the root space decomposition

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

We know that {\mathfrak{h}} acts on a representation {V} of {\mathfrak{g}} by commuting semisimple transformations, so we can write

\displaystyle \mathfrak{h} = \bigoplus_{\beta \in \mathfrak{h}^{\vee}} V_{\beta}

where {V_{\beta} := \{ v \in V: hv = \beta(h) v \ \forall h \in \mathfrak{h} \}}. These are called the weight spaces, and the {\beta} are called weights.

Now

\displaystyle g_{\alpha} V_{\beta} \subset V_{\alpha + \beta }

by an analog of the “fundamental calculation,” proved as follows. Let {h \in \mathfrak{h}, x \in \mathfrak{g}_{\alpha}, v \in V_{\beta}}. Then

\displaystyle h (x v) =xh(v) + [h,x] v = x (\alpha(h)) v + \beta(h) x v = (\alpha + \beta)(h) xv. (more…)

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…)