# Root Systems: existence of a root give two other roots

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.

Thanks!

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 $$F_4$$.