Weyl group and Character Formula of Lie Group

This article is one of Lie Group & Representation contents.

Weyl Group

The Weyl reflection $S_\alpha$ of weight $M$ along a root $\alpha$ is $$S_\alpha M = M-2\frac{\left\langle M,\alpha\right\rangle}{\left\langle \alpha,\alpha \right\rangle}=\alpha$$

Weyl reflections forms the Weyl group $W$, which is a subgroup for the root system.

Since any irreducible representation is invarant under weyl group, we can obtain all weight from one weight.

Character

By Peter-Weyl theorem, class function can characterize by character.

Given irreducible representation $\pi$, Carten subalgebra element $H$,

$$\chi_\pi(H)=\mbox{Tr}(e^{\pi(H)})=\sum_\mu e^{\mu(H)}$$

for weight $m$.

However, sum over all weight is not convenient.

Maximal Torus

Compact group's congugate class's coordiante is torus, it called maximal torus.

For example, diagonal compoent build maximal torus of $SU(N)$, $N-1$ dim torus.

$$T=\{ \mbox{diag} (e^{i\theta_1},e^{i\theta_2}, \cdots, e^{i\theta_n}): \forall j, \theta_j\in \mathbb{R}\}$$

This became maximal Abelian group, or maximal Carten Subalgebra.

Root system also defined from maximal torus.

The roots are the weights for the adjoint action of $T$ on the complexified Lie algebra of $G$. To be more explicit, let $\mathfrak{t}$ denote the Lie algebra of $T$, let $\mathfrak{g}$ denote the Lie algebra of $G$, and let $\mathfrak{g}_C:=\mathfrak{g} \oplus i \mathfrak{g}$ denote the complexification of $\mathfrak{g}$. Then we say that an element $\alpha \in \mathfrak{t}$ is a root for $G$ relatice to $T$ if $\alpha \ne 0$ and there exists a nonzero $X \in \mathfrak{g}_C$ such that $$\mbox{Ad}_{e^H} (X)=e^{i\left\langle \alpha, H \right\rangle} X$$ for all $H \in \mathfrak{t}$. Here $\left\langle \cdot, \cdot \right\rangle$ is a fixed inner product on $\mathfrak{g}$ that is invariant under the adjoint action of connected compact Lie groups.

Weyl group is defined differently.

If $T$ is a maximal torus in $K$, then the normalizer of $T$, denoted $N(T)$, is the group of elements $x\in K$ such that $xTx^{-1}=T$. The quotient group $$W:=N(T)/T$$ is the Weyl group of $T$.

Weyl Character Formula

Charater formula is invariant under Weyl group.

Then alternative sum over weyl group cancell every weight except highest weight.

$$\chi_\pi (H)=\frac{\sum_{w\in W} \epsilon(w) e^{w(\lambda+\rho)(H)}}{\sum_{w\in W} \epsilon(w) e^{w(\rho)(H)}}$$

where $W$ is the Weyl group, $\rho$ is the half-sum of the positive roots(Weyl vector), $\lambda$ is the highest weight, $\epsilon(w)$ is the determinant of the action of $w$ on the Cartan subalgebra (equal to $(-1)^{l(w)}$, where $l(w)$ is the length of the Weyl group element, defined to be the minimal number of reflections with respect to simple roots such that $w$ equals the product of those reflections.)

Haar Measure

Let's calculate the integral of a class function $f$ on $G$ as a suitable integral over the torus $T$. For this, consider the map $$\pi: G/T\times T \rightarrow G,\ \pi(xT,y)=xyx^{-1}$$

By what we said earlier, $\pi$ is a generically finite-sheeted covering, with $\left| \mathfrak{W}\right|$ sheets. If follows that $$\int_G f d\mu = \frac{1}{\left| \mathfrak{W}\right|} \int_{G/T\times T} \pi^*(f) \pi^* d\mu$$

Now $\pi^*(f)(xT,y)=f(y)$ since $f$ is a class function. To calculate $\pi^*d\mu$, consider the induced map on tangent spaces $$\pi_*=d\pi:\ \mathfrak{g}/\mathfrak{h} \times \mathfrak{h} \rightarrow \mathfrak{g}.$$

At the point $(x_0 T,y_0) \in G/T \times T$, $$(x_0 e^{tx}T,y_0 e^{ty})\mapsto x_0 e^{tx} y_0 e^{ty} e^{-tx} x_0^{-1}.$$  

We want to calculate $$\frac{d}{dt}(x_0 e^{ix}y_0e^{ty}e^{-tx}x_0^{-1})|_{t=0}(x_0y_0x_0^{-1})^{-1},$$ which is $$x_0(xy_0+y_0y-y_0x)x_0^{-1}(x_0y_0^{-1}x_0^{-1})=x_0(x+y_0yy_0^{-1}-y_0xy_0^{-1})x_0^{-1}.$$

Now $y_0yy_0^{-1}=y$ since $y_0\in T$ and $y\in \mathfrak{h}$. To calculate the determinant of $\pi_*$ we can ignore the volume-preserving transformation $x_0(\ )x_0^{-1}$. If we identify $\mathfrak{g}$ with $\mathfrak{g}/\mathfrak{h}\times \mathfrak{h}$, the matrix becomes $$\begin{pmatrix}I-\mbox{Ad}(y_0)& 0 \\ 0 & I\end{pmatrix}.$$ (Becaue $y$ is same, $x$ becomes $x-y_0xy_0^{-1}$.

So the determinant of $\pi_*$ is $\det (I-\mbox{Ad}(y_0))$. Now $(\mathfrak{g}/\mathfrak{h})_C=\bigoplus \mathfrak{g}_\alpha$, and $\mbox{Ad}(y_0)$ act as $e^{2\pi i \alpha(y_0)}$ on $\mathfrak{g}_\alpha$. Hence $$\det (\pi_*)=\prod_{\alpha \in R} (1-e^{2\pi i \alpha}),$$ as a function on $T$ alonw, independent of the factor $G/T$. THis gives Weyl's integration formula: $$\int_G f d\mu_G=\frac{1}{\left| \mathfrak{W} \right|} \int_T f(y) \prod_{\alpha \in R} (1-e^{2\pi i \alpha (y)})d\mu_T.$$

This also give another proof of Weyl's Character formula.

In Weyl Integral Formula, we can see only positive root with no Weyl group normalization.

$$\int_G f d\mu_G=\frac{1}{\left| \mathfrak{W} \right|} \int_T f(y) \prod_{\alpha \in R} (1-e^{2\pi i \alpha (y)})d\mu_T=\frac{1}{\left| \mathfrak{W} \right|} \int_T f(y) \prod_{\alpha \in R^+} (1-e^{2\pi i \alpha (y)})(1-e^{-2\pi i \alpha (y)})d\mu_T=\frac{1}{\left| \mathfrak{W} \right|} \int_T f(y) \prod_{\alpha \in R^+} ([1-e^{2\pi i \alpha (y)}]+[1-e^{-2\pi i \alpha (y)}])d\mu_T=\frac{1}{\left| \mathfrak{W} \right|} \int_T f(y) \sum_{\forall \mbox{ half of }R}\prod_{\alpha \in \mbox{ half of }R} (1-e^{2\pi i \alpha (y)})d\mu_T=\int_T f(y) \prod_{\alpha \in R^+} (1-e^{2\pi i \alpha (y)})d\mu_T$$

...This fails.

see this paper B.2.

Reference

Matthew Foster - 9. Casimir operators, characters, dimension and strange formulae

Brian C. Hall - Lie Groups, Lie Algebras, and Representations

www.math.columbia.edu/~woit/LieGroups-2012/weylcharacter.pdf

Fulton, Harris - Representation Theory A First Course

Computational Invariant Theory

Introduction to Lie Algebras and Representation Theory