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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

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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

1 Answer 1

up vote 43 down vote accepted

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 negative 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.

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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 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 Speyer Jan 11 '10 at 22:01
    
To David: Thank you! –  Peter Lee Jan 12 '10 at 1:16

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