# On the Weyl character formula

So let $G$ be a compact real Lie group. Let $\rho:G\rightarrow GL_n(\mathbb{C})$ be an irreducible representation of $G$ and let $\chi_{\rho}$ be the character associated to $\rho$. Let $\Lambda_{\rho}$ be the highest weight of $\rho$ (this of course depends on a choice of a labelling of the roots of $G$) then it seems that one has the following formula \begin{align}\label{a} dim(\rho)=\prod_{\alpha\in\Delta^+}\frac{(\Lambda_{\rho}+\delta,\alpha)}{(\delta,\alpha)} \end{align} where $\delta$ is half the sum of the positive roots $\Delta^+$. In order to prove this equality one has to show the following identity: $$\sum_{\sigma\in W}sign(\sigma)e^{2\pi i(\sigma(\alpha),t\delta)}= \prod_{\alpha\in\Delta^+}(e^{\pi i(\alpha,\delta)t}-e^{-\pi i(\alpha,\delta)t})$$ where $W$ is the Weyl group, $( , )$ is an $ad$-invariant inner product on $\mathfrak{h}$ (we think of the roots as living in $\mathfrak{h}$) and $t$ is a formal parameter.

Q: Is there a simple way to prove the equality above? Is there some geometry hidden behind it?

• The first formula is the Weyl dimension formula, the second the Weyl denominator formula. They both follow easily from the Weyl character formula, and I wrote up some notes that give the standard somewhat geometric proof of this using the Weyl dimension formula here: math.columbia.edu/~woit/notes12.pdf A more sophisticated geometrical way to get the denominator formula would be to apply the Atiyah-Bott fixed point formula to the index-theory version of Borel-Weil-Bott for the trivial representation, as an index of an operator on the flag manifold. Commented Jan 6, 2011 at 23:50
• Thnaks a lot for your nice set of notes :) Commented Jan 7, 2011 at 1:07
• Oops, in the above comment, the second occurrence of "Weyl denominator formula" should be "Weyl integral formula". Also, I just remembered that there's a nice discussion of the Atiyah-Bott fixed point calculation I mentioned, see section 14.2 of Pressley and Segal's Loop Groups. Commented Jan 7, 2011 at 1:25
• You can find reasonably straightforward algebraic proofs (that apply in the more general Kac-Moody case) in Kac's Infinite dimensional Lie algebras or Kumar's book on Kac-Moody groups. Commented Jan 7, 2011 at 4:44

## 3 Answers

The character formula should be viewed here as a purely formal statement about weight multiplicities in the irreducible representation, so the analytic-looking exponential notation for compact Lie groups doesn't really add anything significant to the combinatorics. (The roots and weights actually live in the dual of the Cartan, but Killing form duality allows your identification.) The "sum equals product" formula is at the heart of the Weyl formula, but isn't simple to prove in the original compact Lie group setting. It gets much easier to see in the algebraic setting pioneered by Bernstein-Gelfand-Gelfand, which recovers the Weyl formula as related to Kostant's later weight multiplicity formula (the two eventually being in fact equivalent). This is all done in the setting of Verma modules in their early 1970s papers, within the BGG category $$\mathcal{O}$$.

Anyway, it's important to be flexible about translating the compact Lie group formula as Weyl proved it into the setting of complex semisimple Lie algebras. This allows also for a beautiful generalization of the Weyl denominator formula to the infinite dimensional setting of Kac-Moody algebras. There you see connections with much older infinite "sum = product" identities going back to people like Jacobi and Euler.

Is there some geometry hidden here? Maybe yes, but that's a longer story.

P.S. Weyl's character formula can be approached from many directions, so what is "simple" depends heavily on what you already know and what your further interests are. The Weyl denominator formula amounts to the character formula for the trivial 1-dimensional module and is hard to make intuitive. The BGG proof and its extension to infinite dimensional Lie algebras have much to offer algebraically, while other references mentioned in the comments involve radically different ideas. Besides Weyl's approach through invariant integration on compact groups, there is an approach through complex algebraic geometry at the end of J.L. Taylor's 2002 AMS graduate text Several complex variables with connections to algebraic geometry and Lie groups, as well as different treatments of Weyl's formula using algebraic geometry due to Steve Donkin and Henning Andersen (A new proof of old character formulas (MSN)). Geometry enters via the flag variety of a semisimple algebraic group in characteristic 0, which is comparable to working with the flag variety of a corresponding compact Lie group as in Peter Woit's comment.

• Where is the proof by Donkin expounded? Commented Dec 27, 2018 at 21:33
• @LSpice: Probably I was thinking of this one mathscinet.ams.org/mathscinet-getitem?mr=1958906 Commented Dec 30, 2018 at 0:26
• Thanks! That seems at a glance to be specific to the modular case. Do the techniques in it also work in characteristic 0? Commented Dec 30, 2018 at 0:39
• @LSpice: I am presumably remembering wrong, In any case, Donkin has been mainly concerned with prime characteristic. Commented Dec 31, 2018 at 18:19

If I recall correctly, one can recover the character formula for the irrep $V_\lambda$ by using fixed-point localization to compute $f_{*}${\mathcal L}$$, where $f_{*}$ is the K-theoretic pushforward map

$f_*: K_G(G/T) \to K_G(pt)$

and $${\mathcal L}$$ is the K-theory class of the Borel-Weil line bundle ${\mathcal L}$ whose global sections on the flag variety $G/T$ are the irrep $V_\lambda$.

Is that the sort of geometry you had in mind?

• No, I used the word geometry in a more naive sense, i.e., understanding enough about the action of the Weyl group on the root system $\Delta$ in order to deduce the Weyl integral formula. Commented Jan 8, 2011 at 0:50

So one may prove the second equality of the question (the so-called Weyl integral formula) in the following way:

For every $H\in\mathfrak{h}=Lie(T)$ we denote $$Q(H):=\prod_{\alpha\in\Delta^+}(e^{\pi i(\alpha,H)}-e^{-\pi i(\alpha,H))}).$$ We think of the roots of $G$ as elements of a fixed Weyl chamber (which we denote by $C_0$) of $\mathfrak{h}$.

The Weyl group $W$ is generated by reflexions $r_j$. For every $r_j$ there is an associated simple root $\alpha_j$ such that $r_j(\Delta^+)=(\Delta^+-\{\alpha_j\})\cup\{-\alpha_j\}$. Using this observation one sees that $Q(r_jH)=-Q(H)$ and hence $Q(\sigma H)=sign(\sigma)Q(H)$. Now since $Q(H)$ is an alternating function with respect to the $W$-action one has that $$Q(H)=\sum_{\sigma\in W}sign(\sigma)e^{2\pi i(\sigma(\delta),H)}+\mbox{possible other terms}.$$ Now the key observation is to note that $\delta$ is the only vector of the form $\frac{1}{2}\sum \pm\alpha$ which is in $C_0$. It follows from this that there is in fact no other terms!