Why is $f(\alpha) = \frac{1}{2}\langle \alpha, f(\alpha) \rangle \alpha^{\wedge}$?

Question

Let $(X,R,X^{\wedge},R^{\wedge})$ be a root datum (reference: 7.4, Springer Linear Algebraic Groups). Define a homomorphism $f: X \rightarrow X^{\wedge}$ by $$f(x) = \sum\limits_{\alpha \in R} \langle x,\alpha^{\wedge} \rangle \alpha^{\wedge}$$ I am trying to understand why for $\alpha \in R$, $$f(\alpha) = \frac{1}{2}\langle \alpha, f(\alpha) \rangle \alpha^{\wedge}$$ This should be easy, but I have been trying for over an hour and I still don't see it. We have $$ \langle \alpha, f(\alpha) \rangle \alpha^{\wedge} = \langle \alpha, \sum\limits_{\beta} \langle \alpha, \beta^{\wedge} \rangle \beta^{\wedge} \rangle \alpha^{\wedge} = (\sum\limits_{\beta} \langle \alpha, \beta^{\wedge} \rangle^2 )\alpha^{\wedge}$$ I was hoping I could use something like $$2f(\alpha) = \sum\limits_{\beta} 2\langle \alpha, \beta^{\wedge} \rangle \beta^{\wedge} = \sum\limits_{\beta} \langle\beta, \beta^{\wedge} \rangle \langle \alpha, \beta^{\wedge} \rangle \beta^{\wedge}$$

Answer 1

I found a solution in Casselman's notes. View $X, X^{\wedge}$ as embedded in vector spaces $V, V^{\wedge}$. For $\alpha \in \Sigma$, consider the maps $$s_{\alpha}: X \rightarrow X, v \mapsto v - \langle v, \alpha^{\wedge} \rangle \alpha$$ $$s_{\alpha^{\wedge}}: X^{\wedge} \rightarrow X^{\wedge}, w \mapsto w - \langle \alpha, w \rangle \alpha^{\wedge}$$

One can check that $\langle v, s_{\alpha^{\wedge}}(w) \rangle = \langle s_{\alpha}(v), w \rangle$.

For $\alpha, \beta \in \Sigma$, we have $\langle \alpha, \beta^{\wedge} \rangle \alpha^{\wedge} = \beta^{\wedge} - s_{\alpha^{\wedge}}(\beta^{\wedge})$, so $$\langle \alpha, \beta^{\wedge} \rangle^2 \alpha^{\wedge} = \langle \alpha, \beta^{\wedge} \rangle \beta^{\wedge} - \langle \alpha, \beta^{\wedge} \rangle s_{\alpha^{\wedge}}(\beta^{\wedge})$$ $$ = \langle \alpha, \beta^{\wedge} \rangle \beta^{\wedge} - \langle - s_{\alpha}(\alpha), \beta^{\wedge} \rangle s_{\alpha^{\wedge}}(\beta^{\wedge}) = \langle \alpha, \beta^{\wedge} \rangle \beta^{\wedge} + \langle s_{\alpha}(\alpha), \beta^{\wedge} \rangle s_{\alpha^{\wedge}}(\beta^{\wedge})$$ $$ = \langle \alpha, \beta^{\wedge} \rangle \beta^{\wedge} + \langle \alpha , s_{\alpha^{\wedge}}(\beta^{\wedge}) \rangle s_{\alpha^{\wedge}}(\beta^{\wedge})$$ You then sum over all $\beta$. Since $s_{\alpha^{\wedge}}$ is a bijection from $\Sigma^{\wedge}$ to itself, the right hand side is $2f(\alpha)$. The left hand side is $\langle \alpha, f(\alpha) \rangle \alpha^{\wedge}$ as I mentioned in my question.