# $0\in \mathfrak h^*$ is a weight of $V(\lambda)$ if and only if $\lambda \in \Lambda^+$ is a sum of roots.

This question appears in Humphrey's book on Lie algebras (Introduction to lie algebras and representation theory) page 116. Let $$L$$ be a semisimple Lie algebra with root system $$\Phi$$ and base $$\Delta = \{\alpha_1,\cdots,\alpha_\ell\}$$. Denote by $$\Lambda^+$$ the set of dominant weights.

If $$\lambda \in \Lambda^+$$, then $$0$$ occurs as a weight of the irreducible highest weight $$L$$-module $$V(\lambda)$$ if and only if $$\lambda$$ is a sum of roots.

The direction $$\Rightarrow$$ is simple: since $$V(\lambda)$$ is generated by a maximal vector $$v$$ of weight $$\lambda$$, its weights $$\mu$$ all satisfy $$\mu\leq \lambda$$, so $$\lambda - \mu$$ is a sum of positive roots. In particular, $$\mu = 0$$ shows that $$\lambda$$ is a sum of (positive) roots.

I'm actually struggling to prove the direction $$\Leftarrow$$ . We may write $$\lambda = \sum \langle \lambda, \alpha_i\rangle \omega_i$$ in terms of the fundamental weights. Since the fundamental weights are a linear comb. of simple roots with rational positive coefficients and $$\lambda \in \Lambda^+$$, then $$\lambda$$ is a fortiori a sum of positive roots. Let then $$\lambda = \sum k_i \alpha_i$$ with $$k_i\in \mathbb Z_{\geq 0}.$$ Now the vectors $$y_{\alpha_1}^{i_1}\cdots y_{\alpha_\ell}^{i_\ell}v \in V(\lambda)$$ have weight $$\lambda - \sum i_j \alpha_j$$, so I must show that $$y_{\alpha_1}^{i_1}\cdots y_{\alpha_m}^{i_m}v \neq0$$ for some $$\{i_1,\cdots, i_\ell\} = \{k_1,\cdots, k_\ell\}$$. How can I prove that? I already know that $$y_{\alpha_j}^{m_j+1}v = 0$$ for $$m_i = \langle \lambda,\alpha_j \rangle$$, so why we cannot have $$k_j>m_i$$? If this situation can happen, how can I cancel out the factor $$k_j\alpha_j$$ of $$\lambda$$?

Any hint or help is very much appreciated.

Denote by $$\Pi(\lambda)$$ the set of weights of $$V(\lambda)$$. By proposition 21.3, it is saturated. We prove that if the set $$\{\mu \in \Pi(\lambda): \mu \mbox{ is a sum of positive roots}\}\neq \emptyset$$, then $$0 \in \Pi(\lambda)$$. In particular, $$\mu = \lambda$$ solves the exercise.
Proceed by induction on $$\mbox{ht}\,\mu$$, the case $$\mbox{ht}\,\mu = 0$$ being clear, since $$\mu = 0$$ if and only if $$\mbox{ht}\,\mu =0$$ and $$\mu \in \Pi(\lambda)$$ by hyphotesis.
The inductive hyphotesis states that if $$\{\nu \in \Pi(\lambda): \nu \mbox{ is a um of positive roots and } \mbox{ht}\,\nu < \mbox{ht}\,\mu\}\neq \emptyset$$, then $$0 \in \Pi(\lambda)$$.
Now, if $$\mu \neq 0,$$ let us write $$\mu = \sum k_i\alpha_i, k_i\in \mathbb Z_{\geq 0}$$. Then, $$0<(\mu,\mu) = \sum k_i(\mu,\alpha_i)$$ implies that $$(\mu,\alpha_i)>0$$ for some index $$i$$ with $$k_i\geq 1$$, so it follows that $$\langle \mu, \alpha_i \rangle\geq 1.$$ Since $$\Pi(\lambda)$$ is saturated, we have at least that $$\mu - \alpha_i \in \Pi(\lambda)$$, and $$k_i\geq 1$$ assures that $$\mu-\alpha_i$$ is still a sum of positive roots. The induction hyphotesis now gives us that $$0\in \Pi(\lambda)$$.
• Sorry I dont understand the part, why $0$ inside the sets of weights of $V(\lambda)$ May 8, 2023 at 7:55