# What is the Hirzebruch-Riemann-Roch formula for the flag variety of a Lie algebra?

If we have a finite dimensional Lie algebra g, then the flag variety of g is a projective scheme.

My question is what is Hirzebruch-Riemann-Roch formula for this projective scheme? Are there any interesting results?

I wonder know whether Riemann-Roch in this setting have some beautiful representation theory explanations.

Thanks

• I have edited the question to say Hirzebruch-Riemann-Roch rather than Grothendieck-Riemann-Roch, as it is really about HRR, not GRR. But feel free to change it back if you think that I am wrong. – Kevin H. Lin Oct 27 '10 at 6:25

Riemann-Roch for the flag variety is the Weyl Character formula!

More specifically, let $$L$$ be an ample line bundle on $$G/B$$, corresponding to the weight $$\lambda$$. According to Borel-Weil-Bott, $$H^0(G/B,L)$$ is $$V_{\lambda}$$, the irreducible representation of $$G$$ with highest weight $$\lambda$$, and $$H^i(G/B,L)=0$$ for $$i>0$$. So the holomorphic Euler characteristic of $$L$$ is $$\mathrm{dim} \ V_{\lambda}$$.

As we will see, computing the holomorphic Euler characteristic of $$L$$ by Hirzebruch-Riemann-Roch gives the Weyl character formula for $$\mathrm{dim} \ V_{\lambda}$$.

## Notation:

$$G$$ is a simply-connected semi-simple algebraic group, $$B$$ a Borel and $$T$$ the maximal torus in $$B$$. The corresponding Lie algebras are $$\mathfrak{g}$$, $$\mathfrak{b}$$, $$\mathfrak{t}$$. The Weyl group is $$W$$, the length function on $$W$$ is $$\ell$$ and the positive roots are $$\Phi^{+}$$. It will simplify many signs later to take $$B$$ to be a lower Borel, so the weights of $$T$$ acting on $$\mathfrak{b}$$ are $$\Phi^{-}$$.

We will need notations for the following objects: $$\rho = (1/2) \sum_{\alpha \in \Phi^{+}} \alpha.$$ $$\Delta = \prod_{\alpha \in \Phi^{+}} \alpha.$$ $$\delta = \prod_{\alpha \in \Phi^{+}} (e^{\alpha/2}-e^{-\alpha/2}).$$

They respectively live in $$\mathfrak{t}^*$$, in the polynomial ring $$\mathbb{C}[\mathfrak{t}^*]$$ and in the power series ring $$\mathbb{C}[[\mathfrak{t}^*]]$$.

## Geometry of flag varieties

Every line bundle $$L$$ on $$G/B$$ can be made $$G$$-equivariant in a unique way. Writing $$x$$ for the point $$B/B$$, the Borel $$B$$ acts on the fiber $$L_x$$ by some character of $$T$$. This is a bijection between line bundles on $$G/B$$ and characters of $$T$$. Taking chern classes of line bundles gives classes in $$H^2(G/B)$$. This extends to an isomorphism $$\mathfrak{t}^* \to H^2(G/B, \mathbb{C})$$ and a surjection $$\mathbb{C}[[\mathfrak{t}^*]] \to H^*(G/B, \mathbb{C})$$. We will often abuse notation by identifiying a power series in $$\mathbb{C}[[\mathfrak{t}^*]]$$ with its image in $$H^*(G/B)$$.

We will need to know the Chern roots of the cotangent bundle to $$G/B$$. Again writing $$x$$ for the point $$B/B$$, the Borel $$B$$ acts on the tangent space $$T_x(G/B)$$ by the adjoint action of $$B$$ on $$\mathfrak{g}/\mathfrak{b}$$. As a $$T$$-representation, $$\mathfrak{g}/\mathfrak{b}$$ breaks into a sum of one dimensional representations, with characters the positive roots. We can order these summands to give a $$B$$-equivariant filtration of $$\mathfrak{g}/\mathfrak{b}$$ whose quotients are the corresponding characters of $$B$$. Translating this filtration around $$G/B$$, we get a filtration on the tangent bundle whose associated graded is the direct sum of line bundles indexed by the positive roots. So the Chern roots of the tangent bundle are $$\Phi^{+}$$. (The signs in this paragraph would be reversed if $$B$$ were an upper Borel.)

The Weyl group $$W$$ acts on $$\mathfrak{t}^*$$. This extends to an action of $$W$$ on $$H^*(G/B)$$. The easiest way to see this is to use the diffeomorphism between $$G/B$$ and $$K/(K \cap T)$$, where $$K$$ is a maximal compact subgroup of $$G$$; the Weyl group normalizes $$K$$ and $$T$$ so it gives an action on $$K/(K \cap T)$$.

We need the following formula, valid for any $$h \in \mathbb{C}[[\mathfrak{t}^*]]$$: $$\int h = \ \mbox{constant term of}\left( (\sum_{w \in W} (-1)^{\ell(w)} w^*h)/\Delta \right). \quad (*)$$ Two comments: on the left hand side, we are considering $$h \in H^*(G/B)$$ and using the standard notation that $$\int$$ means "discard all components not in top degree and integrate." On the right hand side, we are working in $$\mathbb{C}[[\mathfrak{t}^*]]$$, as $$\Delta$$ is a zero divisor in $$H^*(G/B)$$.

Sketch of proof of (*): The action of $$w$$ is orientation reversing or preserving according to the sign of $$\ell(w)$$. So $$\int h = \int (\sum_{w \in W} (-1)^{\ell(w)} w^*h) / |W|$$. Since the power series $$\sum_{w \in W} (-1)^{\ell(w)} w^*h$$ is alternating, it is divisible by $$\Delta$$ and must be of the form $$\Delta(k + (\mbox{higher order terms}))$$ for some constant $$k$$. The higher order terms, multiplied by $$\Delta$$, all vanish in $$H^*(G/B)$$, so we have $$\int h = k \int \Delta/|W|$$. The right hand side of $$(*)$$ is just $$k$$.

By the Chern root computation above, the top chern class of the tangent bundle is $$\Delta$$. So $$\int \Delta$$ is the (topological) Euler characteristic of $$G/B$$. The Bruhat decomposition of $$G/B$$ has one even-dimensional cell for every element of $$W$$, and no odd cells, so $$\int \Delta = |W|$$ and we have proved formula $$(*)$$.

## The computation

We now have all the ingredients. Consider an ample line bundle $$L$$ on $$G/B$$, corresponding to the weight $$\lambda$$ of $$T$$. The Chern character is $$e^{\lambda}$$. HRR tells us that the holomorphic Euler characteristic of $$L$$ is $$\int e^{\lambda} \prod_{\alpha \in \Phi^{+}} \frac{\alpha}{1 - e^{- \alpha}}.$$ Elementary manipulations show that this is $$\int \frac{ e^{\lambda + \rho} \Delta}{\delta}.$$

Applying $$(*)$$, and noticing that $$\Delta/\delta$$ is fixed by $$W$$, this is $$\mbox{Constant term of} \left( \frac{1}{\Delta} \frac{\Delta}{\delta} \sum_{w \in W} (-1)^{\ell(w)} w^* e^{\lambda + \rho} \right)=$$ $$\mbox{Constant term of} \left( \frac{\sum_{w \in W} (-1)^{\ell(w)} e^{w(\lambda + \rho)}}{\delta} \right).$$

Let $$s_{\lambda}$$ be the character of the $$G$$-irrep with highest weight $$\lambda$$. By the Weyl character formula, the term in parentheses is $$s_{\lambda}$$ as an element of $$\mathbb{C}[[\mathfrak{t}^*]]$$. More precisely, a character is a function on $$G$$. Restrict to $$T$$, and pull back by the exponential to get an analytic function on $$\mathfrak{t}$$. The power series of this function is the expression in parentheses. Taking the constant term means evaluating this character at the origin, so we get $$\dim V_{\lambda}$$, as desired.

• This isn't that important, but I think in the second paragraph it should say that ${\rm H}^0(G/B, L)$ is $V_\lambda^*$ and not $V_\lambda$. For example, if $G/B = {\bf P}(E)$ for some vector space E, then ${\rm H}^0({\bf P}(E), \mathcal{O}(n))$ is ${\rm Sym}^n(E)^*$ and not ${\rm Sym}^n(E)$. – Steven Sam Jan 11 '10 at 19:09
• Hmm. But the formula I got was the formula for the character of $V_{\lambda}$. – David E Speyer Jan 11 '10 at 19:48
• But Riemann--Roch only gives a number, right? I don't think the character can be computed this way, and since $\dim V_\lambda = \dim V_\lambda^*$ there's no discrepancy. To get the Weyl character formula, we can use something like the Atiyah-Bott fixed point theorem on G/B. – Steven Sam Jan 11 '10 at 21:22
• First of all, I could certainly be wrong. The first draft of this post had several sign errors, and I may not have gotten them all. We should be able to get a character formula, rather than a number, out of this by using equivariant RR. See front.math.ucdavis.edu/9912.5088 Most of the parts will go through unchanged, but we'd need to figure out the equivariant version of (*). – David E Speyer Jan 11 '10 at 22:01
• arxiv.org/abs/math/0312454 – Vít Tuček Dec 17 '12 at 0:18