# Coxeter group action on the product of root systems

Let W be a finite Coxeter group and $$\Phi^+$$ the set of its positive roots. The Coxeter group acts on $$\Phi^+$$ by $$(w, \alpha) \mapsto w \cdot \alpha$$ if $$w \cdot \alpha \in \Phi^+$$ and $$(w, \alpha) \mapsto -w \cdot \alpha$$ if $$w \cdot \alpha \in \Phi^-$$, where $$\Phi^-$$ is the set of negative roots.

The group $$W$$ acts on $$(\Phi^+ \times \Phi^+) \backslash \{(\alpha, \alpha): \alpha \in \Phi^+\}$$ by $$w\cdot (\alpha, \beta) = (w \cdot \alpha, w \cdot \beta)$$. Is this action transitive? That is, for $$(\alpha, \beta)$$, $$(\alpha', \beta') \in (\Phi^+ \times \Phi^+) \backslash \{(\alpha, \alpha): \alpha \in \Phi^+\}$$, is there $$w \in W$$ such that $$w \cdot(\alpha, \beta) = (\alpha', \beta')$$?

In type $$A_2$$, this action is transitive: let $$s_1, s_2$$ be the simple reflections and $$\alpha_1, \alpha_2$$ be simple roots. Then \begin{align} & s_1 \cdot (\alpha_1, \alpha_2) = (\alpha_1, \alpha_1+\alpha_2), \ s_2 (\alpha_1, \alpha_1+\alpha_2)=(\alpha_1+\alpha_2, \alpha_1), \\ & s_2(\alpha_1, \alpha_2) = (\alpha_1+\alpha_2, \alpha_2), s_1(\alpha_1+\alpha_2, \alpha_2) = (\alpha_2, \alpha_1+\alpha_2), \\ & s_2(\alpha_2, \alpha_1+\alpha_2)=(\alpha_2, \alpha_1). \end{align} Is it true in general? Are there some references about this? Thank you very much.

• I don't think it can possibly be transitive: if $\alpha$ and $\beta$ are orthogonal, then $w\cdot \alpha$ and $w\cdot\beta$ will have to be orthogonal too. – Sam Hopkins Dec 7 at 15:15
• @SamHopkins, thank you very much. – Jianrong Li Dec 7 at 17:06
• @SamHopkins In G_2, two roots inclined at 2$\pi$/3 are not part of a simple system. – Richard Lyons Dec 7 at 23:38
• @RichardLyons: very good point. I think what I should’ve said is that if the roots form a simple system for the intersection of the root system with the subspace they span, then they are part of a simple system for the whole root system. – Sam Hopkins Dec 8 at 2:14