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.

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