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

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

So ${xv \in V_{\alpha + \beta}}$.

To talk about “highest weights,” we need an ordering on the roots ${\Phi}$. We can do this by choosing a suitable real functional on the vector space ${\mathfrak{h}^{\vee}}$ which is irrational with respect to the lattice generated by ${\Phi}$. In particular, with respect to such a functional, we can talk about positive roots ${\Phi^+}$ and negative roots ${\Phi^-}$, i.e. those that are mapped to positive (resp. negative) values under this functional.

So, let’s say that a highest weight vector in some not-necessarily-finite-dimensional representation ${V}$ is one that is annihilated by ${\mathfrak{g}_{\alpha}}$ for ${\alpha \in \Phi^+}$. One way to get a highest weight vector in a finite-dimensional representation is to choose any nonzero vector in ${V_{\gamma}}$, where ${\gamma \in \Phi}$ maximizes the linear functional on ${\mathfrak{h}^{\vee}}$ among the weights. It turns out that we can study the simple representations of ${\mathfrak{g}}$—and that is enough to study all representations of ${\mathfrak{g}}$, by complete reducibility—by using these highest weight vectors. In particular, there is a unique one in every simple ${\mathfrak{g}}$-module.

First, we define the Borel subalgebra ${\mathfrak{b}}$ as

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

(Borel subalgebra means maximal solvable.) It is clear that this is solvable, because the commutator ${[\mathfrak{b}, \mathfrak{b}] \subset \mathfrak{n} := \bigoplus_{\alpha \in \Phi^+} \mathfrak{g}_{\alpha}}$. Here ${\mathfrak{n}}$ is fact a nilpotent subalgebra, and ${\mathfrak{b} = \mathfrak{h} \oplus \mathfrak{n}}$. One should think of ${\mathfrak{h}}$ as being something like diagonal matrices (indeed, if we are working with ${\mathfrak{sl}_n}$, this is indeed a Cartan subalgebra), and ${\mathfrak{b}}$ something like upper-traingular matrices.

We can also define an opposite Borel subalgebra

$\displaystyle \mathfrak{b}^- = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak{g}_{-\alpha}$

and a corresponding nilpotent ${\mathfrak{n}^- = \bigoplus_{\alpha \in \Phi^+} \mathfrak{g}_{-\alpha} }$.

Proposition 1 If ${v \in V}$ is a highest weight vector, then ${U \mathfrak{b}^- v}$ is a ${\mathfrak{g}}$-submodule of ${V}$, and it is the smallest one containing ${v}$.

In detail, if ${q_1, \dots, q_n}$ is a basis for ${\mathfrak{b}^-}$, then ${U \mathfrak{b}^- v}$ is spanned by vectors of the form

$\displaystyle q_{i_1}^{j_1} \dots q_{i_k}^{j_k } v.$

(Here the ${j_r}$ may be zero.)

The reason for this proposition is that the Poincare-Birkhoff-Witt theorem implies that, as subrings of ${U\mathfrak{g}}$,

$\displaystyle U\mathfrak{g} = U \mathfrak{n}^- U \mathfrak{h} U \mathfrak{n} ,$

and since ${U\mathfrak{g}v}$ is obviously a ${\mathfrak{g}}$-subrepresentation and ${\mathfrak{n}}$ annihilates ${v}$, the result is clear.

Note that if ${v}$ has weight ${\lambda}$ and ${q_i \in \mathfrak{g}_{-\alpha_i}}$ (where ${\alpha_i \in \Phi^+}$), we have

$\displaystyle q_1 \dots q_n v \in V_{\lambda - \sum \alpha_i} \ (*)$

which is a lower weight vector.

Now, we are interested in the case where the highest weight vector generates all of ${V}$; in this case we will call ${V}$ a highest weight module, though it may not be finite-dimensional. Let the weight be ${\lambda}$. Of course, any simple finite-dimensional representation is a highest weight module. However, it is also possible to construct infinite-dimensional ones (as we will see). We will discuss some of the properties of highest weight modules.

First, note that the highest weight vector ${v}$ is necessarily unique. As we saw, when we multiply it by ${U\mathfrak{n}^-}$, its weight goes down. So ${\dim V_{\lambda}=1}$. Moreover, I claim that for all ${\mu, \dim V_{\mu} < \infty}$. This is because there are only finitely many ways the weights of the ${q}$‘s in (*) can add up to ${\lambda - \mu}$ by positivity. Thus, we do not necessarily have finite-dimensionality, but the spaces in question are not too bad either.

Proposition 2

If ${W \subset V}$ is a subrepresentation, then$\displaystyle W = \bigoplus_{\mu} W \cap V_{\mu}.$

This is proved by a familiar inductive trick. Suppose we have ${w = w_1 + \dots + w_n}$ where ${w \in W, w_i \in V_{\mu_i}}$ for ${\mu_i}$ a nonrendant collection of weights. Suppose ${n}$ is the smallest natural number with such a contradiction possible; then each ${w_i \notin W}$, and clearly ${n>1}$. Then we may choose ${h \in \mathfrak{h}}$ to take different values on all the ${\mu_i}$; in particular

$\displaystyle h w - \mu_1(h) w = \sum_{k=2}^n ( \mu_k(h) - \mu_2(h)) w_k.$

But this is an expression of the same form resolving ${hw - \mu_1(w) \in W}$ into its weight components, so by the inductive hypothesis ${w_i}$ for ${i \geq 2}$ belongs to ${W}$, contradiction.

So we have a few tools for dealing with highest weight modules. Next, we’re going to specialize this to the case of ${\mathfrak{sl}_2}$ and then use that to get a few more results about the root space decomposition.