Dual Killing form is a $\mathbb{Q}$-bilinear form

Question

Given a semisimple Lie algebra $\mathfrak{g}$, with Cartan subalgebra $\mathfrak{h}$, s.t. $ \mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{\alpha \in R}\mathfrak{g}_\alpha $, I'm trying to show that the dual killing form on $\sum_{\alpha \in R} \mathbb{Q}\alpha$ the roots is a positive definite bilinear form. Since we have that $$n_{\alpha \beta}=2\frac{(\alpha,\beta)}{(\alpha,\alpha)} \in \mathbb{Z} \ \ \ \forall \ \alpha, \ \beta \ \in R$$
If I could show that $(\alpha,\alpha) \in \mathbb{Q} \ \forall \ \alpha \in R$, I would be able to conclude the proof quite easily, but I'm stuck at this passage, does anyone know how to prove this?

Answer 1

By definition, $K(h_\alpha, h_\alpha)$ is the trace of $ad(h_\alpha) \circ ad(h_\alpha) \in End(\mathfrak g)$.

Now you already have the root space decomposition which gives a very convenient basis w.r.t. which the endomorphism $ad(h_\alpha)$ (and hence its composition with itself) is diagonal. In particular, note that if $v \in \mathfrak g_\beta$ for any root $\beta$, we have $ad(h_\alpha) (v) = [h_\alpha, v] = \beta(h_\alpha) \cdot v = \check{\alpha}(\beta) \cdot v$.

See how this implies that the trace of $ad(h_\alpha) \circ ad(h_\alpha)$ is the sum of squares of certain rational numbers, which are not all zero?