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?


1 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?

  • $\begingroup$ $Tr(ad(h_\alpha) \circ ad(h_\alpha))= \sum_\beta \beta(h_\alpha)^2$, and $\alpha(h_\alpha) > 0$, but how do I know that $\beta(h_\alpha) \in \mathbb{Q}$? $\endgroup$
    – Alessandro
    Jan 7, 2021 at 23:16
  • $\begingroup$ Actually, all $\beta(h_\alpha) \in \mathbb Z$. That's a basic fact usually proved via representation theory of $\mathfrak{sl}_2$. Without that, I think one would have a hard time seeing that the roots form a root system to begin with. $\endgroup$ Jan 7, 2021 at 23:32
  • $\begingroup$ Are $\beta(h_\alpha)$ supposed to be the cartan numbers? $\endgroup$
    – Alessandro
    Jan 7, 2021 at 23:43
  • $\begingroup$ I guess so. For that matter, $\beta(h_\alpha)= \check{\alpha}(\beta)$ which is $=2(\alpha, \beta)/(\alpha, \alpha)$ for any invariant scalar product on the roots. I'm just not sure now what you have and don't have already at that point, of course you have to make sure your argument is not circular. $\endgroup$ Jan 7, 2021 at 23:50
  • $\begingroup$ All right, thanks a lot for the help! $\endgroup$
    – Alessandro
    Jan 7, 2021 at 23:56

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