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Suppose $\mathfrak{g}$ is an (untwisted) affine Lie algebra with the normalized invariant form $(\cdot | \cdot)$. Let $\lambda \in \mathfrak{h}^\ast$ be a dominant integral weight such that $\lambda(d)=0$ for $d$ the derivation in the loop algebra construction; that is, $\lambda = c_0\Lambda_0 +c_1\Lambda_1+\cdots+c_n\Lambda_n$, where $c_i \in \mathbb{Z}_{\geq0}$, $\Lambda_i$ the fundamental weights. Let $l$ be the level of $\lambda$.

While doing some (albeit limited) explicit computations for various $\mathfrak{g}$ and $\lambda$, I am encountering an inequality of the form $$ 2l(\lambda|\rho)-h^\vee(\lambda|\lambda) \geq 0 $$ where $h^\vee$ is the dual Coxeter number of $\mathfrak{g}$ and $\rho$ is as usual a choice of weight satisfying $\rho(\alpha_i^\vee)=1$ for all simple coroots $\alpha_i^\vee$.

Question: Should this inequality hold for all such $\lambda$?

I have little intuition for whether this should be expected; this very well could be an artifact of the specific $\lambda$ that appear in my computations. For example, I would be somewhat satisfied if this were to hold under the stronger condition that each $c_i \in \{0,1\}$.

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  • $\begingroup$ What is $d$?... $\endgroup$
    – LSpice
    Commented Jul 2, 2020 at 15:05
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    $\begingroup$ @LSpice $d$ is the derivation coming from the loop algebra construction of $\mathfrak{g}$; really, I was just emphasizing that $\lambda$ has no $\delta$ term in the summand. I will add that. $\endgroup$
    – SamJeralds
    Commented Jul 2, 2020 at 15:09

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The answer to this is found as theorem 13.11 in Kac, "Infinite dimensional Lie Algebras". To be specific, we have $2k(\Lambda|\rho) \geq h^{\vee} (\Lambda| \Lambda)$ for all $\Lambda \in P^k_+$, with equality if and only if $\Lambda = k \Lambda_j$ mod $\mathbb{C} \delta$. Here $j \in J$, where $J$ is a set depending on the Kac labeling of the Dynkin diagram in question (J should correspond to the set of simple roots with Kac label 1, roughly).

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    $\begingroup$ This is actually fantastic! Thank you--I am still woefully unfamiliar with the content of the last two chapters of Kac's book, but I'm not completely surprised that this was "hidden" in there! $\endgroup$
    – SamJeralds
    Commented Jul 15, 2020 at 2:25

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