# Getting the Weyl dimension formula geometrically

I'm reading Borel and Hirzebruch's "Characteristic Classes and Homogeneous Spaces, II" and at one point in the paper (Section 22.4) they obtain Weyl's dimension formula for an irrep of a semisimple compact Lie group $G$. This accomplished by some playing around with the Todd class of $G/T$. However, Weyl's character formula is also invoked---and my question is: can this be avoided?

Here's the same question, but in a more specific form: Let $\lambda$ be a dominant character and denote by $L_\lambda$ the associated line bundle over $X=G/T$. Then the Borel-Weil-Bott theorem implies that $H^0(X,L_\lambda)$ is an irrep of highest weight $\lambda$ and $\chi(X, L_\lambda) = \sum_i (-1)^i \dim H^i(X,L_\lambda) = \dim H^0(X,L_\lambda)$. So, with some straightforward manipulation (which can be found in the Borel-Hirzebruch paper), the Hirzebruch-Riemann-Roch theorem yields $$\dim H^0(X, L_\lambda) = \int_X e^{\lambda+\rho} \prod_{\alpha > 0} \frac{\alpha}{e^{\alpha/2}-e^{-\alpha/2}},$$ where $\rho$ is the half sum of positive roots. Is there some way to reduce the RHS to $$\prod_{\alpha>0} \frac{(\lambda+\rho,\alpha)}{(\rho,\alpha)}$$ without invoking the character formula?

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You can think of the Weyl Character Formula as geometric because it can be proved using Borel-Weil-Bott and Riemann-Roch. David Speyer wrote a nice answer in regards to this, found here mathoverflow.net/questions/11422/… – Mike Skirvin Jul 13 '10 at 3:36
Mike, that looks--at least to me--like the Borel-Hirzebruch argument. In particular, David only gets the dimension of the irrep after using the Weyl char formula. I've done some searching and have found a paper by Bernhard Koeck where he derives the Weyl char formula using some sophisticated version of Riemann-Roch. I was hoping something more elementary would suffice for the dimension formula. – Faisal Jul 13 '10 at 3:56
Can you, please, clarify where the RHS of your formula "lives", i.e. the meaning of the exponential functions $e^\alpha$ and $e^\lambda$ when $\alpha, \lambda\in\mathfrak{t}^*$ (together with the identifications made)? – Victor Protsak Jul 13 '10 at 7:12
If I remember correctly, you can get full Weyl character formula from equivariant $K$-theory and the dimension by "forgetting" the $T$-action. – Victor Protsak Jul 13 '10 at 7:15
Victor: The clarification you seek can be found in the link posted by Mike. (Apologies for the delayed response!) – Faisal Jul 21 '10 at 23:25

Like others I am skeptical about the possibility of finding an elementary proof of the dimension formula which is detached completely from the character formula. But of course I can't prove an impossibility theorem. The history of Weyl's character formula shows that it can be approached from a number of different directions in creative ways, though it is hard to declare one proof "easier" than another since all of them presuppose serious background. Weyl's original method used integration on a compact Lie group (a convenient real form of a complex semisimple Lie group), exploiting the fact that every element has a conjugate in a fixed maximal torus. The proof highlights the importance of what is now called the Weyl group (but was used by E. Cartan), in its realization as the quotient of the normalizer of such a torus by the torus itself.

Once the character formula is in hand, a formal calculation yields the dimension as well. (It seems unlikely that the dimension formula alone could be strong enough to recover the full character, however.) There have been other approaches to the character formula, including a more algebraic but opaque one found in older textbooks. The approach of Bernstein-Gelfand-Gelfand in 1971 including their BGG resolution provided a much more elegant algebraic proof, pointing the way toward the more general treatment of infinite dimensional highest weight modules (Kazhdan-Lusztig Conjecture, etc.) and the version by Victor Kac for certain Kac-Moody algebras.

The Borel-Weil theorem using the geometry of line bundles on flag manifolds gave another interpretation of the character formula (as an Euler character), while Bott's further work clarified the behavior of sheaf cohomology for non-dominant weights. (But Bott's theorem isn't needed for the Weyl formula.) In the 1960s Demazure found a short, elegant proof of Borel-Weil-Bott in the setting of algebraic geometry and cohomology of line bundles on flag varieties; this proof uses only the most basic facts about algebraic groups and sheaf cohomology. Later Donkin showed how to streamline the derivation of Weyl's formula in this setting, while Andersen and others extended some of the ideas to prime characteristic: see Chapter II.5 in Jantzen's book Representations of Algebraic Groups. Offshoots of such thinking have permeated the study of Lie group representations, as well as representations of algebraic groups in prime characteristic or quantum groups at a root of unity.

While the Weyl character formula has been a catalyst for developments in representation theory of various flavors, the dimension formula is limited to the classical finite dimensional setting and seems to be mainly a byproduct of the character formula.

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1. There is an interesting $q$-analogue of the character of a finite-dimensional module that involves the notion of $q$-multiplicity of weight due to Lusztig, and it specializes into a natural $q$-analogue of the Weyl dimension formula. 2. Yet another proof of the Weyl character formula follows from Kostant's version of Borel-Weil-Bott theorem involving Lie algebra cohomology (this can also be used to derive the BGG resolution). – Victor Protsak Jul 14 '10 at 10:19

Here is a proof of the Weyl dimension formula, up to a constant factor, that does not go through the Weyl character formula. I leave up to you how geometric you consider this.

Set $\mu=\lambda+\rho$. It is clear that the answer will be a polynomial in $\mu$, of degree $\leq \dim_{\mathbb{C}} X$.

Lemma: This polynomial is $W$-antisymmetric.

Proof: To see this, it will be easier to work with the description of $X$ as $K/T$, where $K$ is the compact real form of $G$, and $T$ the maximal torus. There are two $W$ actions on $K/T$: the one from the left, which exists because $K/T$ is a coset space, and the one from the right, which exists because $W$ normalizes $T$. We will be concerned with the second action. Note that this action is not holomorphic, and is very hard to see from the $G/B$ description of $X$.

Under the identification of $H^2(X)$ with $\mathfrak{t}^*$, this $W$ action induces the standard reflection representation of $W$. On top cohomology, $W$ acts by the sign character. Using the latter fact, for any differential form $\eta$, we have $\int_X w^*(\eta) = (-1)^{\ell(w)} \int_X \eta$. Let's apply this to the integrand we care about.

We write $\Phi^{+}$ for the set of positive roots. $$w^* \left( e^{\mu} \prod_{\alpha \in \Phi^{+}} \frac{\alpha}{e^{\alpha/2} - e^{- \alpha/2}} \right) = e^{w(\mu)} \prod_{\alpha \in w \Phi^{+}} \frac{\alpha}{e^{\alpha/2} - e^{- \alpha/2}}$$ We can partition $w \Phi^{+}$ into $A := \Phi^{+} \cap w \Phi^{+}$ and $B = \Phi^{-} \cap w \Phi^{+}$. Then $A \sqcup (-B) = \Phi^{+}$. Abbreviate $T(\alpha) := \alpha/(e^{\alpha/2} - e^{- \alpha/2})$ and note that $T(\alpha) = T(- \alpha)$. So $$\prod_{\alpha \in w \Phi^{+}} T(\alpha) = \prod_{\alpha \in A } T(\alpha) \prod_{\alpha \in B} T(\alpha) = \prod_{\alpha \in A } T(\alpha) \prod_{\alpha \in B} T(-\alpha)= \prod_{\alpha \in \Phi^{+}} T(\alpha)$$

Putting it all together, we see that $$\int_X e^{w \mu} \prod_{\alpha \in \Phi^{+}} T(\alpha) = \int_X w^* \left( e^{\mu} \prod_{\alpha \in \Phi^{+}} T(\alpha) \right) = (-1)^{\ell(w)} \int_X e^{\mu} \prod_{\alpha \in \Phi^{+}} T(\alpha)$$

This proves the lemma. $\square$

Now, let $\beta$ be any root in $\Phi^{+}$. Suppose that $\langle \mu, \beta \rangle =0$. Let $t$ be the reflection which negates $\beta$. Then $t(\mu)=\mu$, and $(-1)^{\ell(t)} = -1$, so our polynomial must vanish at $\mu$. Thus, our polynomial is divisible by $\langle \cdot , \beta \rangle$. Since our polynomial has degree $\leq \dim_{\mathbb{C}} X = |\Phi^{+}|$, it must be $c \prod_{\beta \in \Phi^{+}} \langle \cdot , \beta \rangle$ for some constant $c$. I found a couple of ways to work out the constant, but none of them were as clean as I'd like, so I'll leave it there.

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Here is one of the odd things to note here. The product $\prod_{\alpha \in \Phi^{+}} T(\alpha)$ is $W$-symmetric. By the standard presentation of the cohomology ring of $X$, this means that every high degree term is cohomologically trivial. So we are really just trying to compute $\int_X e^{\mu}$. This partially responds to Bugs' objection that the formula for the Todd genus is messy. (continued...) – David Speyer Jul 14 '10 at 14:45
Here is an attempt to explain why this happens. Let $X$ be an algebraic variety whose real tangent bundle is stably trivial. $K/T$ has this property, although I'm not sure what the quickest proof is. Writing $T_\mathbb{R}$ and $T_\mathbb{C}$ for the real and holomorphic tangent bundles, we have $T_\mathbb{R} \otimes \mathbb{C} \cong T_{\mathbb{C}} \oplus T_{\mathbb{C}}^*$ (topologically), so the Chern classes of $T_{\mathbb{C}} \oplus T_{\mathbb{C}}^*$ are trivial. (Continued...) – David Speyer Jul 14 '10 at 15:00
Let $\alpha_1$, $\alpha_2$, ..., $\alpha_N$ be the Chern roots of $T_{\mathbb{C}} X$. Set $\rho=(1/2) \sum \alpha_i$. So the Todd class is $e^{\rho} \prod T(\alpha_i)$. Now, $\prod T(\alpha_i)$ is a symmetric function of the $\alpha$'s, and is invariant under negating any subset of the $\alpha$'s. So it must be a power series in the elementary symmetric functions of $\alpha_i^2$. However, those are the chern classes of $T_{\mathbb{C}} \oplus T_{\mathbb{C}}^*$, which we just argued were trivial. So, in this setting, the Todd genus is simply $e^{\rho}$. – David Speyer Jul 14 '10 at 15:04
Any other cool examples of algebraic varieties whose real tangent bundles are stably trivial? The only one that comes to my mind is abelian varieties. – David Speyer Jul 14 '10 at 15:05
Thank you David. – Faisal Jul 21 '10 at 23:28

This is a comment which won't fit in the comment box.

Let me see if I understand the question. There are three equalities: $$\dim \ H^0(X, L_{\lambda}) = \int_X e^{\lambda+\rho} \prod \frac{\alpha}{e^{\alpha/2} - e^{-\alpha/2}} \quad (1)$$ $$\int_X e^{\lambda+\rho} \prod \frac{\alpha}{e^{\alpha/2} - e^{-\alpha/2}} = \left. \left( \sum_{w \in W} e^{w(\lambda+\rho)} {\LARGE /} \prod(e^{\alpha/2} - e^{- \alpha/2}) \right)\right|_{0} \quad (2)$$ $$\left. \left( \sum_{w \in W} e^{w(\lambda+\rho)} {\LARGE /} \prod(e^{\alpha/2} - e^{- \alpha/2}) \right)\right|_{0} = \prod \frac{\langle \lambda+\rho, \alpha \rangle}{\langle \rho, \alpha \rangle} \quad (3)$$

Here $|_{0}$ means that the formula, being a power series in the weights of $T$, is a function on the Cartan algebra $\mathfrak{h}$, and we are to evaluate this function at the origin.

Is your question how to go from the LHS of $(2)$ to the RHS of $(3)$ without passing through the RHS of $(2)$?

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Publish it if you can find one :-)) You have no convenient basis for cohomology with easily described multiplication. You have a clean formula for Chern character and not so clean formula for Todd class, which you are trying to multiply somehow explicitly! Since you know the answer, there may be a shortcut in calculation. Good Luck!! but you'd better be good at combinatorics.

BTW, the logic usually goes the other way. Weyl's Dimension Formula is easy and proved using algebra. Now you can use it to derive a clean formula for Todd class (I think it was done by Brion a while ago but I don't remember the reference).

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