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?

  • $\begingroup$ I can move this question to MathOverflow, if you want. $\endgroup$ Jul 1, 2014 at 7:22
  • $\begingroup$ @MarianoSuárez-Alvarez, I considered asking it there, but I though that it is likely that I am just not seeing something obvious here. If you think it's an Ok question for MathOverflow, I would be grateful if you do move. $\endgroup$ Jul 1, 2014 at 7:25
  • 2
    $\begingroup$ @Peter: Can you clarify the formulation? For instance, your $g^\vee$ is called $h^\vee$ by Kac and coincides with the usual Coxeter number $h$ in simply-laced cases. Does your question have an analogue for other irreducible root systems? Also, what does "mark 1" mean? (And note that the old-fashioned symbol \varpi for pi, rather than \omega, was used for weights to suggest the French "poids".) Do you have a specific normalization of the inner product in mind (though it doesn't seem to matter here)? $\endgroup$ Jul 1, 2014 at 15:06
  • $\begingroup$ P.S. To be more specific, are you looking at just the fundamental weights corresponding to minuscule representations? (These have just a single orbit of weights under the Weyl group, but don't exist for some types including $E_8$.) $\endgroup$ Jul 1, 2014 at 19:17
  • $\begingroup$ @JimHumphreys, of course notation isn't the important thing, but maybe $g$ in place of $h$ comes from Suter's paper (the only other detailed discussion of dual Coxeter numbers that I've seen)? ams.org/mathscinet-getitem?mr=1600666 $\endgroup$
    – LSpice
    Jul 1, 2014 at 20:10

2 Answers 2


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 $k\omega$.

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 $\langle\omega,\frac\rho{h^\vee}\rangle=\langle\omega,\frac\omega2\rangle=\frac12\|\omega\|^2$.

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

  • 1
    $\begingroup$ (In the remark, "right" should be "left".) $\endgroup$
    – S. Carnahan
    Sep 24, 2014 at 12:46
  • $\begingroup$ Ah!, `right'! ;-) $\endgroup$ Sep 24, 2014 at 20:39
  • $\begingroup$ Wow, thanks! That's a very usefull generalization and a very nice proof. $\endgroup$ Sep 28, 2014 at 7:26

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.)

  • $\begingroup$ Thanks! The minuscule property seems to be indeed very important here. I'll try finish your idea.. $\endgroup$ Jul 3, 2014 at 6:04
  • $\begingroup$ I tried using Freudenthal's formula, it indeed does simplify a lot for minuscule representations. It gives information about $(\varpi,\rho)$,however, I have not yet been able to relate it to squared norm of $\varpi$. Though I get some weird relations like $(\varpi,4\rho)=|\Delta|-|\Delta/\varpi|$, where $\Delta$ is the set of all roots, and $\Delta/\varpi$ is the set of roots obtained by deleting the node corresponding to $\varpi$ from Dynkin diagram, which is at least fun. $\endgroup$ Jul 4, 2014 at 9:47
  • $\begingroup$ A typo in your answer? the dual Coxeter number is defined using the highest short coroot right? $\endgroup$
    – brunoh
    Jun 8, 2016 at 0:13
  • $\begingroup$ @brunch: Yes, I've edited accordingly in both places where I misquoted the definition. $\endgroup$ Jun 9, 2016 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.