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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.

KJA
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    Have you tried answering these questions just for $\mathfrak{g} = \mathfrak{sl}2$? A highest-weight vector is a weight vector, and you can use the $\mathfrak{sl}_2$ commutation relations to see that any submodule generated by a weight vector is a weight module. The weight spaces are finite-dimensional because $U(\mathfrak{n}^-)$ has finite-dimensional weight spaces, and always moves things downward in weight. If there were an infinite dimensional weight space $V\lambda$, then ${x \in U(\mathfrak{n}^-) \mid x \cdot v_{\Lambda} \in V_\lambda}$ would be infinite-dimensional. – Joppy Dec 24 '20 at 00:02
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    Lastly, since $V = U(\mathfrak{n}^-) v_{\Lambda}$, we have that if $V_\lambda \neq 0$ then $\lambda \leq \Lambda$ in the usual partial order on the weight space $\mathfrak{h}^*$. Therefore the set ${\Lambda}$ upper bounds all the weights of $V$. – Joppy Dec 24 '20 at 00:04
  • Thank you! It makes sense that $U(\mathfrak{n}-)$ gives the desired decomposition. However I'm still having a hard time seeing why $V=U(\mathfrak{n}-)v_\Lambda$ gives $\lambda\leq \Lambda$ or equivalently for $\Lambda-\lambda$ to be a positive integral combination of the roots? – KJA Dec 24 '20 at 12:49
  • Also, why do we need to argue in perspective of $U(\mathfrak{n}_-)$ for the weight-space decomposition? Wouldn't $U(\mathfrak{g})$ give a weight-space decomposition with finite-dimensional spaces? – KJA Dec 24 '20 at 13:04
  • By the PBW theorem $U(\mathfrak{g}) = U(\mathfrak{n}^-) \otimes U(\mathfrak{h}) \otimes U(\mathfrak{n}+)$ (as a tensor product of vector spaces). Acting on $v_{\Lambda}$ we get $U(\mathfrak{n}^+) v_{\Lambda} = U(\mathfrak{h}) v_{\Lambda} = \mathbb{C}v_{\Lambda}$ by the conditions of a highest weight vector, so we are really only left with $U(\mathfrak{n}^-) v_{\Lambda}$. (It doesn't matter whether you look at this from the whole enveloping algebra or just the negative half, you can get finite dimensionality). For $\lambda \leq \Lambda$, recall $\mathfrak{n}^-$ is generated by $f_i$. – Joppy Dec 24 '20 at 21:18
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    I guess I want to show $h(f_{i_1}^{k_1}\cdots f_{i_d}^{k_d}v_\Lambda)=(\Lambda-\sum k_j\alpha_j)(h)(f_{i_1}^{k_1}\cdots f_{i_d}^{k_d}v_\Lambda)$. Is it correct to assume that the action between $h$ and $f_j$ works as the commutator since we are in $U(\mathfrak{g})$? Ie. that $h(f_{i_1}^{k_1}\cdots f_{i_d}^{k_d}v_\Lambda)=f_{i_1}h(f_{i_1}^{k_1-1}\cdots f_{i_d}^{k_d}v_\Lambda)-hf_{i_1}(f_{i_1}^{k_1-1}\cdots f_{i_d}^{k_d}v_\Lambda)$? – KJA Dec 27 '20 at 17:24

1 Answers1

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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 $v \in V_{\lambda}$.

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.

Joppy
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