Let $\Phi$ a root system and $W$ the Weyl group relative to basis $\Delta$. Let $\mu \in \overline{C(\Delta)}$, where $C(\Delta)$ is the Weyl chamber fundamental, $w \in W$ and $\alpha \in \Delta$ such that $w\alpha \prec 0$. Show that $(\mu, w\alpha) \leq 0$.

Comments: I need this to understand a demonstration of the Lemma 10.3B of book J.E. Humphreys - Introduction to Lie algebras and representation theory. He states that this fact occurs because $\mu \in \overline{C(\Delta)}$. But I can not see why.

I'm assuming that $(\mu, w\alpha) > 0$. If $\mu \in C(\Delta)$ then $w\alpha \in \Delta \Rightarrow w \alpha \succ 0$ a contradiction. I can not solve the case where $\mu \in \overline{C(\Delta)} \setminus C(\Delta)$.

Another way is as follows: $(\mu, w \alpha) \leq 0 \Leftrightarrow -(\mu, w \alpha) \geq 0 \Leftrightarrow (\mu, w(- \alpha)) \geq 0 \Leftrightarrow w(- \alpha) \in \Delta$. Knowing that $\alpha \in \Delta$ and $w\alpha \prec 0$ I can conclude that $w(-\alpha) \in \Delta$?


Write $\alpha_1,\dots,\alpha_n$ for the given choice of simple roots. Suppose $\beta$ is a root with $\beta \prec 0$, that is $$\beta=\sum_{i=1}^n k_i \alpha_i,$$ with $k_i \leq 0$. We must show that $(\beta,\mu)$ is non-positive for all $\mu$ in the closure of the fundamental chamber. But this is an immediate consequence of the definition, which we now recall.

The fundamental chamber $C(\Delta)$ is defined by $$C(\Delta)=\{v \in \mathfrak{h} \ | \ (\alpha,v) > 0 \quad \hbox{for all positive roots $\alpha$} \}$$ and its closure is $$\overline{C(\Delta)}=\{v \in \mathfrak{h} \ | \ (\alpha,v) \geq 0 \quad \hbox{for all positive roots $\alpha$} \}.$$ Therefore by definition, for any $\mu \in \overline{C(\Delta)}$, $$(\beta,\mu)=\sum_{i=1}^n k_i (\alpha_i,\mu) \leq 0.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.