Quadratic Casimir of fundamental irreps of simply-laced Lie algebras

Question

I have the following question, motivated by the expression for the character of level 1 highest weight integrable representations of simply-laced affine algebras (in terms of the string function). It follows from the character expression (by comparing the leading "conformal dimension") that for a fundamental weight $\omega$ with mark 1 the following identity should hold in simply-laced case: $$ \frac{(\omega,\omega+2\rho)}{2(h^\vee+1)}=\frac{1}{2}|\omega|^2=\frac{1}{2}(\omega,\omega), $$ where $h^\vee$ is the dual Coxeter number of the simply-laced simple $\mathfrak{g}$. The lhs is the quadratic Casimir divided by $2(h^\vee+1)$. I checked it for $A_n$ series and $E_6,\,E_7$ explicitly, but I feel there should be some simple general argument.

So the question is if there indeed is a simple argument why this should be true? Is there any generalization?

Answer 1

It turns out that it's easier to prove the following generalization:

Let $\mathfrak g$ be a simple Lie algebra (not necessarily simply laced), let $\omega$ be a fundamental weight whose Dynkin mark is 1, and let $k$ be any number. Then we have $$ \frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)} = \frac{k}{2}\|\omega\|^2. $$ (The original question is the case $k=1$)

Remark: The right left hand side $\frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)} $ is the minimal energy of the positive energy representation of the affine Lie algebra $\hat{\mathfrak g}$ of level $k$ and highest weight $k\omega$.

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Proof: We'll show that $$ \frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)} \,\,\stackrel{(1)}=\,\, \frac{\| k\omega+\rho\|^2-\|\rho\|^2}{2(h^\vee+k)} \,\,\stackrel{(2)}=\,\, \frac{k\langle\omega,\rho\rangle}{h^\vee} \,\,\stackrel{(3)}=\,\, \frac{k}{2}\|\omega\|^2. $$

(1) Easy. (2) By the lemma below, the numerator $\| k\omega+\rho\|^2-\|\rho\|^2$ vanishes when $k=-h^\vee$. The function $k\mapsto \frac{\| k\omega+\rho\|^2-\|\rho\|^2}{2(h^\vee+k)}$ is therefore linear. One easily checks that it vanishes at zero and that its derivative at zero is $\frac{\langle\omega,\rho\rangle}{h^\vee}$. (3) Again by the lemma, the point $\frac \rho {h^\vee}$ is equidistant to $0$ and to $\omega$. It is on the bisecting hyperplane of the segment $[0,\omega]$, and so $\langle\omega,\frac\rho{h^\vee}\rangle=\langle\omega,\frac\omega2\rangle=\frac12\|\omega\|^2$.
QED

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Lemma: Let $\omega$ be as above. Then $\| \rho - h^\vee\omega\| = \|\rho\|$.

Proof: Let $\mathcal A$ be the Weyl alcove. Its isometry group is the automorphism group of the extended Dynkin diagram $\Gamma^e=\Gamma\cup \{\circ\}$. The vertices of $\Gamma$ whose Dynkin mark is $1$ are exactly those in the $Aut(\Gamma^e)$ orbit of the extra vertex $\circ$. Recall that $\rho$ is the unique weight in the interior of $h^\vee \mathcal A$. To finish the argument, note that the vertices $0$ and and $h^\vee\omega$ are in the same orbit under the isometry group of $h^\vee \mathcal A$, and therefore equidistant to $\rho$. QED

Answer 2

Probably the best explanation will come from Freudenthal's method for recursive computation of weight multiplicities in any irreducible representation of a finite dimensional simple Lie algebra $\mathfrak{g}$ (say over an algebraically closed field of characteristic 0). The quadratic Casimir action is by the scalar $c$ indicated in the question, for any highest weight. But here only the minuscule highest weights are involved, as defined in Bourbaki's Lie Groups and Lie Algebras (Chap. 8, $\S7.3$). Such weights are among the fundamental weights $\varpi_1, \dots, \varpi_\ell$, characterized for example as those $\varpi_i$ for which the coefficient of the corresponding simple root $\alpha_i$ in the highest root is 1. The key property of the representation with this highest weight is that all weights are in a single Weyl group orbit and thus have multiplicity 1.

The case-by-case construction of irreducible root systems and lists of data including formulas for $2\rho$ are given in Bourbaki, Chap. 6. The Coxeter number $h$ (of the root system or its Weyl group) turns out to be 1 plus the sum of coefficients of the highest root when expressed in terms of simple roots, while the "dual Coxeter number" $h^\vee$ or $g$ defined by Kac substitutes the highest short root of the dual root system in cases where there are two root lengths. In the question here, $h = h^\vee$.

While all fundamental weights in type $A_\ell$ are minuscule, only 3, 2, 1 (resp.) are of this type for $D_\ell, E_6, E_7$. Note that the factor $1/2$ on both sides of the equation in the question can be cancelled, so the essential formulation for a minuscule $\varpi = \varpi_i$ reads: $$\frac{c}{(h+1)} = \frac{(\varpi,\varpi+2\rho)}{(h+1)} = (\varpi, \varpi).$$

For example, in type $A_1$, $2 \varpi = \alpha$ (the single simple root), $h=2$, and the equation is clear. For type $E_6$ and $\varpi = \varpi_1$ (highest weight of the natural $27$-dimensional representation), Bourbaki's tables make it easy to verify the equation; here $h=12$.

I think the reason for the term $h+1$ can be seen in Freudenthal's method for the minuscule case, but will have to check that more carefully. Similarly, it may take some care to get the right formulation for the other types $B_\ell, C_\ell, F_4, G_2$. But only classical finite dimensional representation theory should be needed here. [EDIT: I'm still undecided about the best way to get a case-free proof, but some further checking does indicate that for minuscule highest weights $\varpi$ only, the Casimir scalar $c$ is always given by $(h+1) (\varpi, \varpi)$. Only the extra cases $B_\ell, C_\ell$ occur here, with respective weights $\varpi_\ell, \varpi_1$ in Bourbaki numbering. The dual Coxeter number $h^\vee$ seems to play no role at all.]

ADDED: Freudenthal did not use the Coxeter number directly, but indirectly it helps one to compute the Casimir operator here due to the standard identity $(h+1) \ell = \dim \mathfrak{g}$. (This was first observed empirically, but later explained theoretically by Kostant.)