# Weyl group reflects root to a base?

I am reading Humphreys' Introduction to Lie Algebras and Representation Theory. In Theorem 10.3(c), it is stated that if $\alpha$ is a root then there exists a Weyl group reflection $\sigma$ such that $\sigma(\alpha) \in \Delta$, where here $\Delta$ is a base for the root system. The proof begins by selecting $\gamma' \in E$ ($E$ is the vector space for our root system) such that $(\gamma',\alpha) = \epsilon > 0$ and $|(\gamma',\beta)| > \epsilon$ for all other roots $\beta \neq \pm \alpha$. I can follow the proof up to this point. From there, it is then "evident" that $\alpha$ lies in some base $\Delta$ (specifically, in the notation of Humphreys', in $\Delta(\gamma')$). I don't understand why this is "evident".

• Try to draw a diagram of some semisimple Lie algebras and convince yourself that it works. The proof may not be very illuminating. – Bombyx mori Jun 17 '14 at 7:40

For $\gamma' \in E$ a regular point, Humphreys' defines $$\Phi_+(\gamma') = \{\beta \in \Phi:(\beta,\gamma')>0\}.$$ Thus $(\alpha,\gamma')>0$ shows that $\alpha \in \Phi_+(\gamma')$. By Theorem 10.1 of Humphreys' book, we know that there exists a base $\Delta(\gamma')$ consisting of roots in $\Phi_+(\gamma')$. Suppose $\alpha \notin \Delta(\gamma')$. Thus $$\Delta(\gamma') = \{\beta_1, \ldots, \beta_r\}$$ with $\beta_i \neq \pm \alpha$ (we know $\beta_i \neq -\alpha$ since $-\alpha \notin \Phi_+(\gamma')$). Since $\beta_i \in \Phi_+(\gamma')$, it follows from the statement that $|(\gamma',\beta)|>\epsilon$ for all $\beta \in \Phi$ that we have $(\gamma',\beta_i)>\epsilon$ for all $i$. Since $\Delta(\gamma')$ is a base, it follows that there exists $k_i \in \mathbb{Z}_{\geq 0}$ such that $\alpha = \sum k_i \beta_i$ with at least one $k_j >0$. But then $$\epsilon = (\gamma',\alpha) \geq (\gamma',\beta_j) > \epsilon,$$
a contradiction. Thus $\alpha \in \Delta(\gamma')$.
• Correct. Same idea, maybe slightly shorter: If $\alpha = \beta_1 + \beta_2$ for some $\beta_i \in \Phi(\gamma)^+$, then $\beta_i \neq \pm \alpha$, so $(\beta_i, \gamma) > \epsilon$. This implies $(\alpha, \gamma) > 2\epsilon$ which is a contradiction. Hence $\alpha$ is indecomposable, ie. $\alpha \in \Delta(\gamma')$. – spin Jun 18 '14 at 20:00