# A formula for the dual Coxeter number

Let $$\Phi$$ be the root system of a finite dimensional simple Lie algebra $$\mathfrak g$$, with dual Coxeter number $$h^\vee$$.

Let $$\alpha_0\in \Phi$$ be a long root
(if all the roots have the same length, then let $$\alpha_0\in \Phi$$ be any root).

Let $$\langle\cdot,\cdot\rangle$$ be the basic inner product (the inner product according to which $$\langle\alpha_0,\alpha_0\rangle=2$$).

Then I believe that the following formula is true:            $$\displaystyle\sum_{\alpha\in\Phi}\,\, \langle \alpha,\alpha_0\rangle^2=4h^\vee$$

How does one prove that formula?

• Are you sure you don't mean to sum in your sum over only the simple roots (not all the roots)? That would look more like what I've seen as the definition of the dual Coxeter number. Dec 4, 2020 at 23:30
• You normalise lengths in $\mathsf G_2$ so that the short roots have squared length $2/3$? Dec 5, 2020 at 1:30
• @LSpice: Yes indeed. In type $G_2$, the squared length of the short roots is $2/3$, according to my conventions. Dec 5, 2020 at 11:11

By Weyl invariance, it suffices to prove this for some long root. When $$\alpha_0$$ is the highest root, it is an immediate consequence of Lemma 4 of Suter - Coxeter and dual Coxeter numbers, which states that $$h^\vee\alpha_0 = \sum_{\alpha \in \Phi_+} \langle\alpha, \alpha_0^\vee\rangle\alpha$$.