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For a simple Lie algebra $\mathfrak{g}$ we can define a Killing form $K(X,Y) \equiv \frac{1}{2 h^\vee}\operatorname{tr}(\mathfrak{ad}_X \mathfrak{ad}_Y)$, where $\mathfrak{ad}_X Y \equiv [X, Y]$ as usual, and $h^\vee$ is the dual Coxeter number.

If we consider a Cartan-Weyl basis $H^I, E^\alpha$, then it is easy to show that (where $\alpha$ are all roots) $$ \frac{1}{2h^\vee}\sum_{\alpha} \alpha^I \alpha^J = K^{IJ} = K(H^I, H^J), \qquad \Rightarrow \qquad \frac{1}{2h^\vee}\sum_\alpha (\lambda,\alpha)(\alpha, \mu) = (\lambda, \mu) \ , $$ where $(\lambda_1, \lambda_2) \equiv \sum_{I,J = 1}^r K_{IJ}\lambda_1^I\lambda_2^J $ is the usual inner product between weights. In particular, $$ \frac{1}{2h^\vee}\sum_\alpha (\lambda,\alpha)^2 = (\lambda, \lambda) \ . $$

Now, in a physics problem I need to solve the following equation, $$ N (\lambda, \lambda)^2 = \sum_\alpha (\alpha, \lambda)^4, \qquad \forall \lambda \ , $$ where $N$ is some real number depending only on $\mathfrak{g}$ but not on the specific weight $\lambda$. Not all Lie algebras admit a solution of $N$ by direct tests, since the expressions on the two sides depend on $\lambda$ rather differently.

I wonder: what are the Lie algebras which admit a solution of $N$?

After some testing, I think at least the famous Deligne series of Lie algebras $A_1, A_2, G_2, D_4, F_4, E_6, E_7, E_8$ admit solutions, $$ \sum_{\alpha} (\lambda, \alpha)^4 = (h^\vee + 6) (\lambda, \lambda)^2\ . $$ but I'm not sure if there are more, or how to prove that they are the only possibilities.

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    $\begingroup$ What is $\alpha^I$? $\endgroup$ Commented Dec 23, 2022 at 12:33
  • $\begingroup$ @VítTuček It's the eigenvalue $[H^I, E^\alpha] = \alpha^I E^\alpha$ corresponding to the Cartan-Weyl basis. $\endgroup$
    – Lelouch
    Commented Dec 23, 2022 at 12:58
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    $\begingroup$ OK. And your question is for which complex simple Lie algebras does there exists a natural number $N$ such that the equation holds? $\endgroup$ Commented Dec 23, 2022 at 23:02
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    $\begingroup$ @VítTuček Well any $N \in \mathbb{R}$ would do, not necessarily a natural number. The object on both sides of the equation depend on $\lambda$ rather differently in general, and I am looking for cases where they are proportional. $\endgroup$
    – Lelouch
    Commented Dec 24, 2022 at 1:35

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