Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak{g}$. The group has a maximal torus $T$ with Lie algebra $\mathfrak{t}$. Let $\rho = \frac{1}{2}\sum_{\alpha > 0} \alpha$ be the half-sum of positive roots, the so-called "Weyl vector."

When one considers the classical compact groups $U(n)$,$Sp(n)$,$SO(2n)$,$SO(2n+1)$ one gets the following Weyl vectors $$\left(\frac{n-1}{2}, \frac{n-3}{2}, \dots, -\frac{n-1}{2} \right)$$ $$\left(n, n-1, \dots, 1 \right)$$ $$\left(n-1, n-2, \dots, 0 \right)$$ $$\left(n-\frac{1}{2}, n-\frac{3}{2}, \dots, \frac{1}{2} \right)$$ I have two questions: firstly I note that the difference between any two components is an integer, $\rho_i - \rho_j \in \mathbb{Z}$. Is there a general way of proving this?

Of course I am implicitly choosing a basis of $\mathfrak{t}$ and $\mathfrak{t}^\ast$ when I write the above vectors. And this is my second question. I am taking a basis $e_1, \dots, e_n \in \mathfrak{t}$, $e_1^\ast, \dots, e_n^\ast \in \mathfrak{t}^\ast$, such that $\exp(2\pi e_j) = \mathrm{Id}$ and such that $\langle e_i , e_j \rangle = \delta_{ij}$, where $\langle \cdot, \cdot \rangle$ is an Ad-invariant inner product. In the groups considered I don't see any systematic reason why such a basis (with these properties) should exist, though in each case a computation shows that it does. Is there some property that all these groups share that makes this possible?


The basis you have chosen to write $\rho$ in is the basis of fundamental weights, so for $U(n)$ for example you are saying that $$\rho = \frac{n - 1}{2} \varpi_1 + \frac{n - 3}{2} \varpi_2 + \cdots - \frac{n - 1}{2} \varpi_{n - 1}.$$ We can then rewrite your question: since the fundamental weights are dual to the simple coroots, why is it true for any two simple coroots $\alpha^\vee, \beta^\vee$ that $\langle \rho, \alpha^\vee - \beta^\vee \rangle$ is an integer? In fact, more is true: each of $\langle \rho, \alpha^\vee \rangle$ and $\langle \rho, \beta^\vee \rangle$ will be an integer.

Let $S \subseteq \Delta^+ \subseteq \Delta \subseteq \mathfrak{t}^*$ denote the simple roots, positive roots, and roots. For every simple root $\alpha \in S$, the simple reflection $s_\alpha$ permutes the other positive roots $s_\alpha(\Delta^+ - \{\alpha\}) = \Delta^+ - \{\alpha\}$ and of course acts on $\alpha$ as $s_\alpha(\alpha) = -\alpha$, hence we get that $s_\alpha(\rho) = \rho - \alpha$. By definition the reflection $s_\alpha(\rho) = \rho - \langle \rho, \alpha^\vee \rangle \alpha$ (where $\alpha^\vee \in \mathfrak{t}$ is the coroot corresponding to $\alpha$) and therefore we see that for $\alpha \in S$, $\langle \rho, \alpha^\vee \rangle = 1$.

If the Lie algebra $\mathfrak{g}$ is semisimple, then we can replace the fact "$\langle \rho, \alpha^\vee \rangle = 1$ for all simple coroots $\alpha^\vee$" with the equivalent (and easier to remember fact) $\rho = \varpi_1 + \cdots + \varpi_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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