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?



Quadratic Casimir of fundamental irreps of simply-laced Lie algebras
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*](https://mathoverflow.net/questions/179695/minimal-energy-of-affine-lie-algebra-reps) 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 $\langle\omega,\frac\rho{h^\vee}\rangle=\langle\omega,\frac\omega2\rangle=\frac12\|\omega\|^2$.  
 QED




---


*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


