Let $\Omega$ be a root system and fix a set of positive roots. Let $\gamma \in \Omega$ be a positive root, $\alpha \in \Omega$ a simple positive root, and $s_{\alpha}$ the assosociated reflection in the Weyl group. Suppose $s_{\alpha}(\gamma) = \gamma + 2\alpha \in \Omega$, can we conclude that $\gamma+ \alpha \in \Omega$?

Apparently if $\gamma$ is long and $\alpha$ is short $s_{\gamma}(\alpha)=\gamma +\alpha$. I Don't know if the other possible length combinations can occur and what happens in those cases.



1 Answer 1


In general one calls $n(\beta, \alpha) = \check{\alpha}(\beta) = \dfrac{2\langle \beta, \alpha \rangle}{\langle \alpha, \alpha \rangle}$ the integer such that $s_\alpha(\beta)= \beta-n(\beta, \alpha) \alpha$. So in your case, you have $$n(\gamma, \alpha)=-2.$$

The standard tables (cf. e.g. Bourbaki's volume on Lie groups and algebras, summary at the end of volume 6; or think through this) show there are only two possibilities for two roots $\gamma, \alpha$ to satisfy $n(\gamma, \alpha)=-2$:

  • either $\gamma = -\alpha$, which is excluded in your case because both $\alpha$ and $\gamma$ are supposed to be positive; or
  • $\lvert \lvert\gamma\rvert \rvert /\lvert\lvert \alpha\rvert\rvert =\sqrt2$, and the angle between $\alpha$ and $\gamma$ is $\frac{3\pi}{4}$ a.k.a. $135°$. That is the case you describe. Among irreducible root systems, it occurs exactly in systems of type $B_{n \ge 2}$, $C_{n\ge 2}$, and $F_4$.

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