Is the Weyl denominator globally well defined on $T$?

Question

The Weyl denominator function on $T$, the maximal torus of a compact connected Lie group $G$ is given by (for $H \in \mathrm{Lie}(T)$) $$\delta(\exp(H)) = \sum_{w \in W} \det(w) e^{\rho(w(H))}$$ where $W = N(T)/T$ is the Weyl group $\rho = \frac{1}{2} \sum_{\alpha > 0} \alpha$ is half the sum of positive roots (with respect to some chosen Weyl chamber).

I tried computing this for $G = U(n)$ and I found the following. We take $T$ to be the diagonal matrices in $U(n)$ and $W$ simply becomes the group of permutations on the $n$ eigenvalues. Then for $\exp(H) = \mathrm{diag}(e^{i\theta_1}, \dots , e^{i\theta_n} )$ I find that $$\delta(\exp(H)) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) e^{i \sum_{k=0}^{n-1} \left( k - \frac{n-1}{2}\right) \theta_{k+1}}$$ For this function to be well defined on $T$ it needs to be unchanged when replacing $\theta_k \to \theta_k + 2\pi$. However it changes by a factor of $e^{- i(n-1) \pi}$. This is equal to $1$ if $n$ is odd but $-1$ if $n$ is even. Thus for $n$ even $\delta$ does not seem to be globally well defined on $T$. This is confusing me. It would be much appreciated if someone could clear up my misunderstanding.

UPDATE: In this paper by Peter Frenkel he says that the antisymmetric characters (of which the Weyl denominator is just one) are defined on the preimage of $T$ under the universal covering map $\tilde{G} \to G$. However I have not been able to find a textbook that discusses this and I would be grateful if anyone could suggest one.

Answer 1

Let us take $G = U(2)$. Then the Weyl denominator is: $$ \exp\left[ \frac{i}{2}\left(\theta_1 - \theta_2\right) \right] + \exp\left[ -\frac{i}{2}\left(\theta_1 - \theta_2\right) \right] = 2 \cos\left(\frac{\theta_1 - \theta_2}{2}\right).$$

Moreover, as you can see, if $\theta_1 \mapsto \theta_1 + 2\pi$, then the Weyl denominator picks up a minus sign. So i this case, the Weyl denominator is not globally defined on $T$ (I think it is up to a sign, but please check).

However, in Weyl's character formula, the numerator also picks up a minus sign (for similar reasons), so the ratio is well defined (as it should).

Note however, that if you consider $G = SU(2)$ instead, then $\theta_2 = -\theta_1 = \theta$, say (since the trace of an element in the Lie algebra of $SU(2)$ must vanish). So the Weyl denominator then becomes:

$$2 \cos\left( \frac{2\theta}{2} \right) = 2 \cos(\theta),$$

which is globally defined on $T$ in this case.

I believe that, if $G$ is a compact connected and simply connected Lie group, then also the Weyl denominator is globally defined on a maximal torus $T$ of $G$. I think this follows if one proves that

$\exp(w\rho(H)) = 1$, whenever $H$ belongs to the lattice $L$ in the universal cover of $T$ (with the universal cover being a copy of $\mathbb{R}^k$, with $k$ being the rank of $G$) consisting of the preimages of $1 \in T$. Thus, as a lattice, $L$ is isomorphic to $\mathbb{Z}^k$.

This in turn follows if

$w\rho(H) \in 2\pi i \mathbb{Z}$, whenever $H \in L$.

I believe this integrality condition is satisfied, and should be a standard fact. Some references may be Knapp's "Lie Groups beyond an introduction", or if needed, Bourbaki's reference volumes on Lie groups and Lie algebras.

Edit: in order to help clear the OP's possible confusion, note that if $G = SU(n)$ and $n$ is odd, then it is clear that $n-1$, $n-3$,$\ldots$, $-(n-1)$ are all even, so they can be evenly divided by $2$, and the integrality condition is clearly satisfied.

However, if $G = SU(n)$ and $n$ is even, then one has to use that the trace of $H$ is $0$. Note that $w \rho$ consists of $\frac{1}{2}\left(n-1,n-3,\ldots,-(n-3),-(n-1)\right)$ up to a permutation. We want to show that if you dot it with $(\alpha_1,\ldots,\alpha_n)$, where the $\alpha_i \in \mathbb{Z}$ such that $\alpha_1 + \cdots + \alpha_n = 0$, then the result is also an integer. After taking the previous dot product, replace $\alpha_n$ by $-(\alpha_1 + \cdots + \alpha_{n-1})$. By combining the coefficients of $\alpha_1$, say, one sees that the result is $1/2$ times the sum of two odd integers (since we are assuming $n$ is even), i.e. $1/2$ times an even integer, which is thus an integer. Similarly for the coefficients of $\alpha_i$, for $i = 1, \ldots, n-1$.