The Weyl group preserves inner product

Question

Show that the Weyl group $W$ preserves the inner product: $(w(\lambda)\,,\, w(\mu)) = (\lambda, \mu)$ for all $w\in W$ and $\lambda, \mu\in E$.

I know it suffices to check this on reflections $w_\alpha$ for $\alpha\in \Phi$, where $\Phi$ is my root system. So far I have the following straight-forward computation:

$$(w_\alpha(\lambda)\,,\, w_\alpha(\mu)) = (\lambda\,,\, \mu)-\langle \lambda,\alpha\rangle(\alpha\,,\,\mu)-\langle \mu,\alpha\rangle(\lambda\,,\,\alpha)+\langle \lambda,\alpha\rangle\langle\mu,\alpha\rangle(\alpha\,,\,\alpha)$$ where $\langle u,v\rangle =2\dfrac{(u,v)}{(v,v)}$. But how in the world do I show the three "extraneous" terms on the right-hand side are zero?

Thanks

Answer 1

It seems just expansion is enough.

$$ \begin{align*} (w_\alpha(\lambda), w_\alpha(\mu)) &= \Big(\lambda - 2\frac{(\alpha, \lambda)}{(\alpha, \alpha)}\alpha, \mu - 2\frac{(\alpha, \mu)}{(\alpha, \alpha)}\alpha\Big) \\ &= (\lambda, \mu) - 2\frac{(\alpha, \lambda)(\alpha, \mu)}{(\alpha, \alpha)} - 2\frac{(\alpha, \mu)(\lambda, \alpha)}{(\alpha, \alpha)} + 4\frac{(\alpha, \lambda)(\alpha, \mu)}{(\alpha, \alpha)} \\ &= (\lambda, \mu). \end{align*} $$