The following question originates from analyzing the modular $S$-formation of a modular differential equation that annihilates the character of a chiral algebra. A term pops up that is proportional to the $M_{ij}$ matrix described below which should be diagonalized (it is by observation, but I want to prove it in general).
Consider a rank $r$ simple Lie algebra $\mathfrak{g}$. The roots are $\alpha$, the simple roots $\alpha_i$. Any root can be expanded as $$ \alpha = \sum_{i = 1}^r m^\alpha_i \alpha_i \ . $$ The Killing form $K(X, Y) = \frac{1}{2h^\vee}\operatorname{tr}_\mathbf{adj}XY$ induces an inner product $(\mu, \nu)$ for weights, which satisfy $$ \sum_{\alpha} (\alpha, \mu) (\alpha, \nu) = 2h^\vee (\mu, \nu) \ . $$ The norm-squared $|\alpha|^2:= (\alpha, \alpha)$ is such that long roots have $|\alpha|^2 = 2$. And $\alpha_i^\vee$ denotes the simple coroots.
By some direct computations, we notice the following: the matrix $$ M_{ij} := \sum_{\alpha} m^\alpha_i |\alpha_i|^2 (\alpha, \alpha^\vee_j)^{2n + 1} \ , \qquad n = 0, 1, 2, 3, ... $$ is always diagonal. I have checked cases like $A_{1, 2, 3}, B_{4, 5}, C_6, D_5, F_4, G_2, E_{6,7,8}$, for $n = 0, 1,2, 3, ... 9$
I hope to prove the above observation in general for all $\mathfrak{g}$ and all $n \in \mathbb{N}$.
The case with $n = 0$ is simple to prove. It just says $$ \sum_{\alpha} (\alpha, \omega_i^\vee) |\alpha_i|^2 (\alpha, \alpha_j^\vee) = 2h^\vee (\omega_i^\vee, \alpha_j^\vee) |\alpha_i|^2 = 2h^\vee (\omega_i^\vee, \alpha_j) \frac{2}{|\alpha_j|^2} |\alpha_i|^2 \\ = 2h^\vee \delta_{ij} \frac{2|\alpha_i|^2}{|\alpha_j|^2} \ . $$ Here the first equality makes use of the previously mentioned property of the inner product. For $n > 0$, I don't known if similar formula exists (the question itself suggests some formula exists). If one were to write the inner product using Cartan-Weyl basis $H^I, E^\alpha$, then the equality seems to be related to some property of the trace $$ \operatorname{tr}( H^I H^J H^K ...) $$
Another way to approach the problem is to do it for ABCDEFG separately, utilizing the explicit root structure. But that's not what I want.
It feels like a simple thing to understand but at the moment I can't find a way. I wonder how to prove that $M_{ij}$ is always (for all $\mathfrak{g}$ and $n \in \mathbb{N}$) diagonal? Or, if it is not always true, what is the criteria for it to be true?
A snapshot of some checked cases
