3
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Sam Hopkins
    Commented Dec 4, 2020 at 23:30
  • $\begingroup$ You normalise lengths in $\mathsf G_2$ so that the short roots have squared length $2/3$? $\endgroup$
    – LSpice
    Commented Dec 5, 2020 at 1:30
  • 1
    $\begingroup$ @LSpice: Yes indeed. In type $G_2$, the squared length of the short roots is $2/3$, according to my conventions. $\endgroup$
    – André Henriques
    Commented Dec 5, 2020 at 11:11

1 Answer 1

Reset to default
3
$\begingroup$

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 $h^\vee\alpha_0 = \sum_{\alpha \in \Phi_+} \langle\alpha, \alpha_0^\vee\rangle\alpha$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .