Today’s is going to be a long post, but an important one.  It tells us precisely what weights are allowed to occur as highest weights in finite-dimensional representations of a semisimple Lie algebra.

Dominant integral weights

Let ${V}$ be a finite-dimensional simple representation of a semisimple Lie algebra ${\mathfrak{g}}$, with Cartan subalgebra ${\mathfrak{h}}$, and root space decomposition ${\mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}}$. Suppose given a base ${\Delta}$ and a corresponding division ${\Phi = \Phi^+ \cup \Phi^-}$.

For each ${\alpha \in \Phi^+}$, choose ${X_{\alpha} \in \mathfrak{g}_{\alpha}, Y_{\alpha} \in \mathfrak{g}_{-\alpha}}$ such that ${[X_{\alpha},Y_{\alpha}] = H_{\alpha}}$ and ${X_{\alpha}, Y_{\alpha}, H_{\alpha}}$ generate a subalgebra ${\mathfrak{s}_{\alpha}}$ isomorphic to ${\mathfrak{sl}_2}$.

Consider the weight space decomposition

$\displaystyle V = \bigoplus_{\beta \in \Pi} V_{\beta}$

where ${\Pi}$ denotes the set of weights of ${V}$. Then if ${\beta \in \Pi}$ is the weight associated to a highest weight vector, ${\beta(H_{\alpha})}$ is necessarily a nonnegative integer by the representation theory of ${\mathfrak{sl}_2}$. In other words,

$\displaystyle <\beta, \alpha> := 2 \frac{ (\beta, \alpha)}{(\alpha, \alpha)} \in \mathbb{Z}_{\geq 0}.$

Any weight ${\beta}$ satisfying that identity for all ${\alpha \in \Phi^+}$ is called dominant integral. We have shown that the highest weight of a finite-dimensional simple ${\mathfrak{g}}$-representation is necessarily dominant integral. In fact, given a dominant integral weight, we can actually construct such a finite-dimensional simple module.

The set of merely integral weights—those ${\beta}$ with ${<\beta, \alpha> \in \mathbb{Z}}$ for ${\alpha \in \Phi}$—form a lattice, spanned by vectors ${\lambda_i}$ such that ${<\lambda_i, \delta_j> = \delta_{ij}}$, where the last ${\delta_{ij}}$ is the Kronecker delta.

Theorem 1 The unique simple quotient of the Verma module ${V(\beta)}$ is finite-dimensional if and only if ${\beta}$ is dominant integral.

We will show that if ${<\beta, \delta> \in \mathbb{Z}_{\geq 0}}$ for all ${\delta \in \Delta}$, then finite-dimensionality follows.

Let the simple quotient be denoted ${L(\beta)}$. We shall prove the theorem in a series of lemmas.

Lemma 2 Let ${v}$ be the maximal vector in ${L(\beta)}$. Then ${v}$ is annihilated by a power of ${Y_{\delta}}$.

To see this, we recall the commutation relations in ${U\mathfrak{g}}$:

$\displaystyle [X_{\delta}, Y_{\delta}^n] = -nY_{\delta}^{n-1} ( (n-1) - H_{\delta})$

and

$\displaystyle [X_{\alpha}, Y_{\delta}^n] = 0, \quad \alpha \in \Phi^+ - \{\delta\}.$

The first is just a fact about ${\mathfrak{sl}_2}$ and was proved earlier. For the second, it reduces (because bracketing is a derivation) to checking that ${[X_{\alpha}, Y_{\delta} ] \in \mathfrak{g}_{\alpha - \delta} =0}$.

We can now prove the lemma. It is already known that ${X_{\delta}}$ acts locally nilpotently (because it does on the Verma module). The hard part is to show that ${Y_\delta}$ does so. Let the highest weight vector be ${v}$. Then

$\displaystyle [X_{\delta}, Y_{\delta}^{\beta(\delta)+1}] v = 0$

by the first identity above. So it follows that ${Y_{\delta}^{\beta(\delta)+1}v}$ is actually annihilated by ${X_{\delta}}$. It is also annihilated by ${X_{\alpha}, \ \alpha \in \Phi^+ - \{ \delta\}}$ by the second commutation relation.

In other words, ${Y_{\delta}^{\beta(\delta)+1}v}$ is a highest weight vector—of a weight lower than ${\beta}$! This cannot happen unless ${Y_{\delta}^{\beta(\delta)+1}v = 0}$.

Lemma 3 ${L(\beta)}$ is the sum of finite-dimensional ${\mathfrak{s}_{\delta}}$-modules.

Indeed, let ${L' \subset L(\beta)}$ be the sum of all finite-dimensional ${\mathfrak{s}_\delta}$-modules contained in ${L(\beta)}$. Then ${v \in L'}$, so ${L' \neq 0}$. Next, I claim that ${\mathfrak{g} L' \subset L'}$, which will imply ${L' = L(\beta)}$ by irreducibility. Indeed, ${\mathfrak{g}L'}$ is the sum of finite-dimensional objects, and

$\displaystyle \mathfrak{s}_{\delta} \mathfrak{g} L \subset \mathfrak{g} \mathfrak{s}_{\delta} L + [\mathfrak{g}, \mathfrak{s}_{\delta}]L \subset L.$

By maximality of ${L'}$, we have ${\mathfrak{g}L' = L'}$.

Corollary 4 ${X_{\delta}, Y_{\delta}}$ act locally nilpotently on ${L(\beta)}$.

This is because local nilpotence is true for finite-dimensional representations (this is the ${\mathfrak{sl}_2}$ theory again).

The next step is to prove that the weights of ${L(\beta)}$ are invariant under the Weyl group. This is nontrivial.

Another realization of reflections

Fix a positive root ${\alpha}$. We will now define an automorphism ${\tau_{\alpha}}$ of ${\mathfrak{g}}$ that preserves ${\mathfrak{h}}$ and induces the reflection ${s_{\alpha}}$ on ${\mathfrak{h}^{\vee}}$, via the isomorphism ${\mathfrak{h} \rightarrow\mathfrak{h}^{\vee}}$ from the Killing form.

This is

$\displaystyle \tau_{\alpha} := \exp( -\mathrm{ad} X_{\alpha}) \exp( \mathrm{ad} Y_{\alpha}) \exp(- \mathrm{ad} X_{\alpha}).$

The exponential of a derivation is an automorphism, so ${\tau_{\alpha}}$ is an automorphism too. Let us compute how it acts on ${\mathfrak{h}}$. If ${H \in \mathfrak{h}}$ is orthogonal to ${H_{\alpha}}$ (or equivalently, ${\alpha(H)=0}$), then

$\displaystyle \tau_{\alpha}(H) = H .$

It can also be checked directly, using the commutation relations for ${\mathfrak{sl}_2}$, that

$\displaystyle \tau_{\alpha}(H_{\alpha}) = -H_{\alpha} .$

It now follows that ${\tau_{\alpha}}$ indeed has the appropriate properties.

Now, consider a ${\mathfrak{g}}$-module ${V}$ (with the Lie algebra representation ${\phi: \mathfrak{g} \rightarrow End(V)}$) on which ${X_{\alpha}, Y_{\alpha}}$ act locally nilpotently, so that

$\displaystyle \psi := \exp( \phi(X_{\alpha})) \exp( \phi(-Y_{\alpha})) \exp( \phi(X_{\alpha})) \in Aut(V)$

is well-defined. We will use ${\psi}$ heavily below.

The weights are invariant under the Weyl group

Lemma 5 Suppose ${X_{\alpha}, Y_{\alpha}}$ act locally nilpotently on ${V}$. Then ${\psi(V_{\gamma}) = V_{s_{\alpha} \gamma}}$ for ${\gamma}$ any weight.

Let ${v \in V_{\gamma}}$.

First, we have some linear algebra. Let ${N \in End(V)}$ be nilpotent, ${T \in End(V)}$ arbitrary. Then:

${ T \exp(N) v = \exp(N) \left( \exp(-N) T \exp(N) \right) v}$

which is ${\exp(N) \left( \exp( -\mathrm{ad} N) T \right)}$

This is because ${\mathrm{ad} N}$ as a nilpotent operator on ${End(V)}$ is given by ${L_N- R_N}$ for ${L_N, R_N}$ left nad right translation. Then ${L_N, R_N}$ commute so

$\displaystyle \exp( \mathrm{ad} N) T = \exp(L_N) \exp(-R_N) T = \exp(N) T \exp(-N).$

As a consequence, we may prove the lemma:

$\displaystyle \phi(H) \psi(v) = \psi( \exp(- \mathrm{ad} \phi(X_{\alpha})) \exp( \mathrm{ad} \phi(Y_{\alpha})) \exp( -\mathrm{ad} \phi(X_{\alpha})) \phi(H) ) v$

(Whoops.  Looks like the end of this isn’t appearing.  It’s supposed to end $\phi(H)) v$.  I.e., the ad part acts on the $\phi(H)$, which itself is a transformation of $V$.)

In this huge mess, ${\phi}$ is a homomorphism of Lie algebras, so this becomes

$\displaystyle \psi( \phi( \tau_{\alpha} H) ) v = \gamma( \tau_{\alpha} H) \psi (v)$

which, finally, proves the lemma; we have the appropriate weight.

Proof of finite-dimensionality

So, we now know that ${L(\beta)}$‘s weights are invariant under the Weyl group, because the simple reflections generate it, and we have the previous lemma. I now claim there are only finitely many weights; this, when proved, will imply that ${L(\beta)}$ is finite-dimensional, since each weight space is finite.

Now recall that any weight ${\mu}$ of ${L(\beta)}$ must satisfy that ${\beta - \mu}$ is a sum (with nonnegative integral coefficients) of positive roots.

There is a closed Weyl chamber ${\{v \in E: (v, \delta) \geq 0 \ \forall \delta \in \Delta\}}$, of dominant weights. Any weight of ${L(\beta)}$ can be conjugated into this Weyl chamber, so it suffices to prove that the number of weights of ${L(\beta)}$ in this Weyl chamber is finite.

Suppose ${\mu}$ is a dominant weight such that ${\beta - \mu = \sum k_i \delta_i}$ for each ${k_i}$ nonegative. Then I claim ${(\mu, \mu) \leq (\beta, \beta)}$. Indeed,

$\displaystyle (\beta, \beta) - (\mu, \mu) = (\beta+\mu, \beta - \mu) \geq 0$

because ${\beta + \mu}$ is dominant, and ${\beta - \mu}$ is a sum ${\sum k_i \delta_i}$ with each ${k_i \geq 0}$.

There are thus only finitely many weights ${\mu}$ which are dominant, ${\beta - \mu}$ is integral, and such that ${\beta}$ is eligible to be weights of ${L(\beta)}$; this is because ${\mu}$ has to lie in a closed ball and in a discrete set.

Phew! Next time I’m going to aim for character formulas.