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let $W$ be a weyl group and $\alpha\in R$ we have $s_{w(\alpha)}=ws_{\alpha}w^{-1}$, to prove this the author says $ws_{\alpha}w^{-1}$ acts as identity on $wL_{\alpha}=L_{w(\alpha)}$ , and $ws_{\alpha}w^{-1}(w(\alpha))=-w(\alpha)$ , I really dont understand why, could any one help me?

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$ws_\alpha w^{-1}\cdot wL_\alpha=w s_\alpha L_\alpha=w L_\alpha$ and $w s_\alpha w^{-1}(w(\alpha))= w s_\alpha (\alpha) = w(-\alpha)=-w(\alpha)$.

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thank you but I need to show $s_{w(\alpha)}=ws_{\alpha}w^{-1}$ –  Une Femme Douce Oct 1 '12 at 11:15
    
Yes, and the above calculations show that $ws_\alpha w^{-1}$ does exactly the same thing we would expect from $s_{w(\alpha)}$. –  Hagen von Eitzen Oct 1 '12 at 11:22

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