Complete reducibility of integrable modules over symmetrizable Kac-Moody Lie algebras

Question

I am reading the book "Infinite-dimensional Lie algebras" by Victor G Kac. This is a long question regarding my understanding of the following theorem.

In Theorem 10.7 Kac proves the complete reducibility of the integrable modules over symmetrizable Kac-Moody lie algebras. we have the following proof where he proves that if $\lambda$ and $\mu$ be two primitive weights and $\lambda - \mu = \beta \in Q_+$ then $$2(\lambda+\rho \mid \beta) \ne (\beta,\beta).$$ To prove complete reproducibility, it is enough to prove the above condition in view of the following Proposition 9.9. But the proof of the above Proposition uses the following Lemma where it is claimed that to prove a module is completely reducible it is enough to prove that for any two primitive weights $\lambda$ and $\mu$ of $V$ the inequality $\lambda \ge \mu$ implies $\lambda = \mu$.

I want to understand the proof of Theorem 10.7. I have the following questions:

1. Let $V \in \mathcal O$ and $\lambda_1,\dots,\lambda_k$ be the associated weights such that every weights in V is of the form $\lambda_i - \beta$ for some $\beta \in Q_+$ (Root lattice). Then $V^0 = \oplus_{i=1}^k V_{\lambda_i}$? ($V^0$ is defined in Lemma 9.5 given above)

2. Can theorem 10.7 be proved directly using Lemma 9.5 by proving for any two primitive weights $\lambda$ and $\mu$ of $V$ the inequality $\lambda \ge \mu$ implies $\lambda = \mu$ in $V$? Here, it is proved in multiple steps using the equation $2(\lambda+\rho,\beta) = (\beta,\beta)$. What is the problem if there are two comparable primitive weights in the complete reducibility of $V$?

3. In the proof of Proposition 9.9(b). How to show that $\Omega$ is locally finite and there exists $a \in \Bbb C$ such that $\Omega-aI$ is locally impotent.

4. What is the role of the equations $|\lambda + \rho|^2 = |\mu + \rho|^2$ and $2(\lambda+\rho,\beta) = (\beta,\beta)$ (one implies the other) (where $\lambda - \mu = \beta \in Q_+$) in the entire proof?

5. I am finding the flow of the proof a little confusing in my first reading. Anywhere else can I find a direct or an alternate proof?

6. What is the analogue of Theorem 10.7 for the case of superalgebras? Kindly share some references. Thank you.

Answer 1

I managed to prove (3) :

Claim: The Casimir element $\Omega$ preserves the weight spaces of $V$.

Proof: Let $v \in V_{\lambda}$. Since $\Omega$ commutes with the action of $\operatorname{lie} g$ we have $$h \cdot \Omega(v) = \Omega (h \cdot v) = \Omega(\lambda(h)(v)) = \lambda(h)(\Omega(v)).$$ This proves the claim.

Claim: The Casimir element $\Omega$ acts locally finite on $V$

Proof: Let $v \in V$, then $v \in \oplus_{i=1}^k V_{\lambda_i}$ for some weights $\lambda_1,\lambda_2,\dots,\lambda_k$ of $V$. Since the weight spaces are preserved under the action of $\Omega$, $\oplus_{i=1}^k V_{\lambda_i}$ is the required finite-dimensional $\Omega$ invariant subspace containing $v$.

Claim: There exists $a \in \mathbb C$ such that $\Omega - a I$ is locally nilpotent on $V$.

Proof: Since $\Omega$ acts on $V$, we can write $V = \oplus_{\alpha \in \mathbb C}V_{\alpha}$ where the $V_{\alpha}$s are generalized eigenspaces of $\Omega$. If $v \in V_{\alpha}$ and $x \in \mathrm{lie} g$ then there exists $r > 0$ such the $(\Omega - \alpha I)^r(v) = 0$. We want to show that there exists $s>0$ such that $(\Omega - \alpha I)^s(x \cdot v) = 0$. If $v \in V$, $x,y \in \operatorname {lie} g$ and $\alpha,\beta \in \Bbb C$ then $$(y - (\alpha+\beta)I)^r(x\cdot v) = \sum\limits_{i=0}^r\binom{r}{i} = ((\text{ad }y - \beta I)^i(x))((y - \alpha I)^{r-i}(v)).$$ Since $\Omega$ is an operator on $V$, $y$ can be replaced with $\Omega$ and we take $\beta = 0$. Then the above equation becomes $$(\Omega - \alpha I)^r(x\cdot v) = \sum\limits_{i=0}^r\binom{r}{i} ((\text{ad }\Omega)^i(x))((\Omega - \alpha I)^{r-i}(v)).$$ Since $\Omega$ is an central element, we get $(\text{ad }\Omega)^i(x)) = 0$ whenever $i>0$. Hence the above equation becomes $$(\Omega - \alpha I)^r(x\cdot v) = (\Omega - \alpha I)^{r}(v).$$ This shows that the generalized eigenspaces $V_{\alpha}$s are $\operatorname{lie} g$-invariant. We have assumed that $V$ is indecomposable, therefore $V = V_{\alpha}$ for some $\alpha \in \mathbb C$. This proves the claim.