How to show that $\alpha_{i_p}(s_{i_{p-1}} \cdots s_{i_1}(h)) = (s_{i_1} \cdots s_{i_{p-1}}(\alpha_{i_p}))(h)$?

Question

Let $i_1, \ldots, i_p$ be integers and $\alpha_i$ be simple roots and $s_i$ be simple reflections in a Weyl group of type A. I checked some examples and it seems that $$\alpha_{i_p}(s_{i_{p-1}} \cdots s_{i_1}(h)) = (s_{i_1} \cdots s_{i_{p-1}}(\alpha_{i_p}))(h).$$ For example, let $h=\operatorname{diag}(h_1,h_2,h_3)$ we have $$\alpha_1(s_2 s_1(h)) = \alpha_1( s_2 \operatorname{diag}(h_2,h_1,h_3) ) = \alpha_1( h_2, h_3, h_1 ) = h_2/h_3 $$ and $$(s_1 s_2 (\alpha_1))(h) = (s_1(\alpha_1 + \alpha_2))(h) = \alpha_2(h) = h_2/h_3.$$ How could we prove that $$\alpha_{i_p}(s_{i_{p-1}} \cdots s_{i_1}(h)) = (s_{i_1} \cdots s_{i_{p-1}}(\alpha_{i_p}))(h)$$ in general? Thank you very much.

Answer 1

It is easy to check that for a simple root $\alpha_j$, simple reflection $s_i$, $h \in T$, $(s_i(\alpha_j))(h) = \alpha_j( s_i(h) )$. Therefore for any root $\alpha$, we have $(s_i(\alpha))(h) = \alpha(s_i(h))$. It follows that \begin{align} & \alpha_{i_p}(s_{i_{p-1}} \cdots s_{i_1}(h)) \\ & = \alpha_{i_p}(s_{i_{p-1}} (s_{i_{p-2}} \cdots s_{i_1}(h))) \\ & = (s_{i_{p-1}} (\alpha_{i_p})) ((s_{i_{p-2}} \cdots s_{i_1}(h))) \\ & = ( s_{i_{p-2}} s_{i_{p-1}} (\alpha_{i_p})) ((s_{i_{p-3}} \cdots s_{i_1}(h))) \\ & = \cdots \\ & = (s_{i_1} \cdots s_{i_{p-1}}(\alpha_{i_p}))(h). \end{align}