Why is the highest weight module in $\mathcal{O}$ category?

Question

The $\mathcal{O}$ category is defined as follows:

Let $\mathfrak{g}$ be a Kac-Moody algebra and $V$ a $\mathfrak{g}$-module. $V$ is an object in $\mathcal{O}$ if

• $V$ has decomposition $V=\bigoplus_{\lambda\in \mathfrak{h}^*}V_\lambda$ with $V_\lambda=\{v\in V\ |\ hv=\lambda(h)v\ for\ all\ h\in\mathfrak{h}\}$
• $\dim V_\lambda<\infty$
• There is a finite set $\lambda_1,\cdots,\lambda_k\subset\mathfrak{h}^*$ st. each $\lambda$ with $V_\lambda\neq 0$ satisfies $\lambda\leq \lambda_i$ for some $i$

A highest weight module is defined as follows

A $\mathfrak{g}$-module $V$ is a highest weight module with highest weight $\Lambda\in\mathfrak{h}^*$ if there exists a non-zero element $v_\Lambda\in V$ st.

• $x(v_\Lambda)=0$ for all $x\in\mathfrak{n}_+$.
• $h(v_\Lambda)=\Lambda(h)v_\Lambda$ for all $h\in\mathfrak{h}$.
• $U(\mathfrak{g})(v_\Lambda)=V$.

Kac states that $V$ is in the $\mathcal{O}$ category since $V=\bigoplus_{\lambda\leq \Lambda}V_\lambda$, $V_\Lambda=\mathbb{C}v_\Lambda$ and $\dim V_\lambda<\infty$, however I'm having trouble seeing why these holds. I've managed to show that $U(\mathfrak{n}_-)(v_\Lambda)=V$ by the Poincare Birkhoff Witt theorem.

So the second condition of a highest weight module gives a weight space $V_\Lambda$ of $V$, however why does this imply that we have a decomposition $V=\bigoplus_{\lambda\leq \Lambda}V_\lambda$? How do we know all the other weight space $V_\lambda$ exist and why are these finite dimensional?

The third criteria for something to be in the $\mathcal{O}$ category is that we have a finite set, which bounds the roots. I assume that the finite set is just $\Lambda$, however how do we know that $\Lambda$ is in fact the biggest root?

Since we know that $U(\mathfrak{n}_-)(v_\Lambda)=V$, then by PBW we would have to show that $$h(f_{i_1}^{k_1}\cdots f_{i_d}^{k_d}v_\Lambda)=(\Lambda-\sum k_j\alpha_j)(h)(f_{i_1}^{k_1}\cdots f_{i_d}^{k_d}v_\Lambda)$$ This would imply that $v_\Lambda$ is the highest root. I've tried shown in by induction on $k_1+\cdots+k_d$, however I'm having trouble with he inductive step.

Answer 1

Let $V$ be a Kac-Moody algebra, $V$ a representation. For $\lambda \in \mathfrak{h}^*$, let $V_\lambda$ be the $\lambda$ weight space. Then we have:

1. For all $h \in \mathfrak{h}$, $h(V_\lambda) \subseteq V_\lambda$.
2. For all $i$, $e_i(V_\lambda) \subseteq V_{\lambda + \alpha_i}$.
3. For all $i$, $f_i(V_\lambda) \subseteq V_{\lambda - \alpha_i}$.

(1) follows from the definition of a weight space. (2) and (3) are easy to prove and I'll give you a hint: rearrange the commutation relation $[h, f_i] = -\alpha_i(h)f_i$ inside $U(\mathfrak{g})$ to read $h f_i = f_i(h - \alpha_i(h))$, then try to compute $h_i f v$ for some $v \in V_{\lambda}$.

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Now, suppose that $(V, v_{\Lambda})$ is a highest-weight module. By this we mean that: $0 \neq v_{\Lambda} \in V_{\Lambda}$, $e_i v_{\Lambda} = 0$ for all $i$, and $V$ is generated by $v_{\Lambda}$. We want to show that $V$ is a weight module, with finite dimensional weight spaces, with all weights bounded above by $\Lambda$.

By the PBW theorem, $V = \sum \mathbb{C} f_{i_1}^{k_d} \cdots f_{i_d}^{k_d} v_{\Lambda}$, where the sum is over all possible choices of $d$, $i_1, \ldots, i_d$, and $k_1, \ldots, k_d$. Looking at point $3$ above, we can see that every such spanning vector is in a weight space, for example $f_1^2 f_3 v_{\Lambda} \in V_{\Lambda - 2 \alpha_1 - \alpha_3}$. Since $V$ is spanned by weight vectors, it is a weight representation.

We can already see that every weight of $V$ is bounded above by $\Lambda$, since $\lambda \geq \lambda - \alpha_i$ for all $\lambda$ and $\alpha_i$. In particular, if $V_{\lambda} \neq 0$ then $\lambda \leq \Lambda$.

As for finite-dimensionality, this comes from the fact that the simple roots are linearly independent, and there are only so many ways of writing a sequence $f_{i_1} \ldots f_{i_d} v_{\Lambda}$ so that the resulting vector is in the correct weight space. Take for example $\lambda = \Lambda - 2 \alpha_1 - \alpha_3$ Then $\lambda - \Lambda = 2 \alpha_1 + \alpha_3$ is an element of the root lattice, with "height" 3. There are only three sequences of $f_i$ that lead to vectors with this weight, namely $f_1^2 f_3 v_{\Lambda}$, $f_1 f_3 f_1 v_{\Lambda}$, and $f_3 f_1^2 v_{\Lambda}$. Therefore $\dim V_{\Lambda - 2 \alpha_1 - \alpha_3} \leq 3$. These inequalities do not in general give good bounds on the dimensions of weight spaces, but all the bounds they give show that the weight spaces are finite-dimensional, which is all that we need.