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.


1 Answer 1


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 $0\in \Pi(\lambda)$.

  • $\begingroup$ Sorry I dont understand the part, why $0$ inside the sets of weights of $V(\lambda)$ $\endgroup$
    – James Chiu
    May 8, 2023 at 7:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .