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Let $\mathfrak g$ be a semisimple Lie algebra generated by $x_i^+,x_i^-$, $1\leq i\leq n$, via the Chevalley-Serre relations and let $V(\mu)$ be a Verma module with highest-weight $\mu$. I gather that the ($\mathfrak g$-module) isomorphism $V(\mu)\cong U(\mathfrak n^-)\otimes_\mathbb K \mathbb Kw$ (where $w$ is the highest-weight vector of weight $\mu$) shows that the weight spaces $V_\lambda$ are finite-dimensional, because every $x_i^-$ lowers the weight in some way.

When $n=1$, the highest weight is, say, $\mu$ and every other weight can be written as $\mu-2k$. In particular, $V_{\mu-2k}$ is spanned by $(x^-)^kw$, where $w$ is again the highest weight vector.

I don't quite understand the higher-dimensional analogue. Is it correct to write, for $n\geq 1$ the highest weight as $\mu=(k_1,\dotsc,k_n)$ ($k_i\in\mathbb N$) and the weight of the weight space spanned by the collection of $(x_1^-)^{a_1}\dotsb (x_n^-)^{a_n}w$ ($\sum a_i=a$ some integer) as $\mu-\sum a_i\alpha_i$, where $\alpha_i$ are the simple roots?

Earthliŋ
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1 Answers1

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Yes. For any weight module (this includes Verma modules) you have that $x_i^-$ sends vectors of weight $\lambda$ to vectors of weight $\lambda-\alpha_i$ where $\alpha_i$ is the simple root associated with $x_i^+$.

That is -- if $w$ is a vector of weight $\lambda$ then $x_i^- \cdot w$ has weight $\lambda-\alpha_i$. Likewise, $x_i^+ \cdot w$ has weight $\lambda+\alpha_i$.

Your "$n=1$" is just a special case. In this case we identify $\alpha_1$ with "2". So if $w$ has weight $\ell$, then $x_1^- \cdot w$ has weight $\ell-\alpha_1=\ell-2$.

Thus in general a monomial $(x_1^-)^{a_1} \cdots (x_n^-)^{a_n}$ will send a vector of weight $\mu$ to a vector of weight $\mu-a_1\alpha_1-\cdots-a_n\alpha_n$ (as you surmised).

This all flows from the following: (suppose $h$ belongs to your Cartan subalgebra for $\mathfrak{g}$ and $w$ has weight $\lambda$)

$$h \cdot (x_i^- \cdot w) = x_i^- \cdot (h \cdot w) + [h,x_i^-] \cdot w = x_i^-\cdot(\lambda(h)w)-\alpha_i(h)x_i^- \cdot w = (\lambda-\alpha_i)(h)w$$

The first equality follows from the definition of "module". The second equality comes from the two facts: (1) $h \cdot w = \lambda(h)w$ since $w$ has weight $\lambda$ and (2) $[h,x_i^-]=-\alpha_i(h)x_i^-$ by the definition of the root vector $x_i^-$.

Bill Cook
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