Character formula of Verma module over a Kac-Moody algebra

Question

The character formula of a Verma module over a Kac-Moody algebra is given by

$$\textrm{ch}\ M(\Lambda)=\frac{e(\Lambda)}{\prod_{\alpha\in\Phi+}(1-e(-\alpha))^{\textrm{mult}(\alpha)}}$$

Here $\Phi_+$ are the positive root system and $e(\lambda)$ is the formal exponential.

When proving this we first consider a basis $\{e_{-\alpha,i}\}_{1\leq i\leq \textrm{mult}(\alpha)}$ of $(\mathfrak{n}^-)_{-\alpha}$. Using PBW theorem gives a basis for $U(\mathfrak{n}^-)$ consisting of products $$\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$$

For $h\in U(\mathfrak{g})$ then

$$h\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}=\left(\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha\right)(h)\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$$

Thus the weight space $U(\mathfrak{n}^-)_{-\lambda}$ has a basis consisting of elements satisfying $$\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha=\lambda$$

Then

$$\textrm{ch}\ U(\mathfrak{n}_-)=\sum_{\alpha\in\Phi_+}\dim( (U(\mathfrak{n}^-))_{-\alpha}) e(-\alpha)=\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$$ Kac states that this holds since the number of times $e^{-\lambda}$ appears on the right side is the number of sets of positive integers $k_{\alpha,i}$ st. $\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}{k_{\alpha,i}}\right)\alpha=\lambda$.

This leads to my questions:

1. Why is $\left(\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha\right)$ the corresponding weight of the weight vector $\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$?
2. I'm not sure I understand Kac's argument that $$\sum_{\alpha\in\Phi_+}\dim( (U(\mathfrak{n}^-))_{-\alpha}) e(-\alpha)=\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$$ Why is this the case?

Answer 1

$$h\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}=\left(\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha\right)(h)\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$$

The actual weight should have $-$ in front because $[h,e_{-\alpha,i}]=-\alpha(h)e_{-\alpha,i}$. With this, it will then be true that $U(\mathfrak{n}^-)_{\color{red}{-}\lambda}$ has a basis consisting of elements satisfying $$\tag{1} \sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha=\lambda.$$

2. I'm not sure I understand Kac's argument that $$\sum_{\alpha\in\Phi_+}\dim( (U(\mathfrak{n}^-))_{-\alpha}) e(-\alpha)=\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$$

From previous argument, we know $\dim( (U(\mathfrak{n}^-))_{-\lambda})$ is number of $(k_{\alpha,i})_{\lambda\in \Phi_+, 0\le i\le \text{mult}(\lambda)}$ such that (1) holds. Each such $(k_{\alpha,i})$ corresponds to a way to write $e(-\lambda)=\prod_{\alpha\in \Phi_+}e(-\alpha)^{k_{\alpha,i}}$, which is a term in the product $\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$, hence the equality.