## Why is the Euler characteristic of powers of a line bundle a polynomial in the power?

Mumford's book Abelian Varieties asserts that for a line bundle L on a projective variety (if necessary, you can assume it is as nice as possible), the Euler characteristic $\chi(L^k)$ of tensor powers of $L$ is a polynomial in $k$. If L is very ample, this is just the Hilbert polynomial, and this can be proven by an induction argument twisting the short exact sequence $0 \to \mathcal O(-1) \to \mathcal O \to K \to 0$. More generally, if $L$ (or $L^*$) is an ideal sheaf, the same argument should work. Why does the result still hold for arbitrary $L$?

Edit: I'd be particularly interested in an elementary proof that does not involve proving an entire Riemann-Roch theorem--Mumford is using this result to prove Riemann-Roch for abelian varieties!

-

## 3 Answers

OK, here is another way to see it more in line with what you had in mind I think. Write your $L$ as $\mathcal O(D)$ for some divisor $D$ on $X$. Set $J_1$ to be the ideal sheaf defined by $\mathcal O(-D) \cap \mathcal O_X$ and $J_2$ to be the ideal sheaf defined by $\mathcal O(D) \cap \mathcal O_X$ (intersections taken inside of $K_X$). Let $Y_i$ be the closed subschemes of $X$ defined by these ideal sheaves (they have dimension smaller than that of $X$). Then we have the exact sequences

$$0 \to J_1(kD) \to \mathcal O(kD) \to \mathcal O_{Y_1}(kD) \to 0$$

$$0 \to J_2((k-1)D) \to \mathcal O((k-1)D) \to \mathcal O_{Y_2}((k-1)D) \to 0$$

The two left hand terms are equal by construction. Then by the induction hypothesis, and chasing the Euler characteristics, $\chi(kD) - \chi((k-1)D)$ is a numerical polynomial. This implies that that $\chi(kD)$ itself is a numerical polynomial (Section 1.7 of Harshorne's Algebraic Geometry).

(Here I swept something under the rug, because the subschemes $Y_i$ may not be as nice as $X$ was. But they are at least proper, and we should show that the result we want is that for a proper variety $W$, $\chi(kD)$ is polynomial for a divisor $D$. Then reduce this to the case where $W$ is reduced by looking at the inclusion of $W_\mathrm{red}$ into $W$. Then further reduce to the case where $W$ is integral.)

-
 Great. To avoid technicalities in the induction step, I believe you can first write your divisor as a difference of very ample divisors, so that by Bertini you can assume the Y_i are smooth. – Eric Wofsey Oct 18 2009 at 16:26 If X is not smooth then you can't expect the Y_i to be able to be chosen to be smooth. But the result is true by "soft" methods without needing smoothness. The point is to introduce an auxiliary coherent sheaf to carry along in the argument. See answer below. – BCnrd Feb 25 2010 at 6:24

In the case of a curve of genus $g$, this is the standard Riemann-Roch theorem, it says $\chi(L^k) = k \cdot deg(L) + 1 - g$. In higher dimensions, this is a result of the more general Grothendieck-Riemann-Roch theorem, though in the case I am about to state, it is commonly called the Hirzebruch-Riemann-Roch theorem. In the case of a line bundle on a $n$ dimensional projective variety, it says $\chi(L^k) = (exp(1 + k \cdot L) \cdot td(X))_n$. Here $td(X)$ means the Todd class of the tangent bundle of $X$ (a fixed cohomology class) and the subscript $n$ means we take the degree $n$ piece of the above expression. If you expand this out, you will exactly find a degree $n$ polynomial in $k$ (the Hilbert polynomial). A proof can be found, for example, in Fulton's book on Intersection Theory.

-
 Is there a simpler way to know that it's given by some polynomial without actually finding what the polynomial is? – Eric Wofsey Oct 17 2009 at 18:21

See EGA III$_1$ 2.5.3 and EGA IV$_2$ 5.3. The elegant generalization there which incorporates an auxiliary coherent sheaf opens the door to using Grothendieck's unscrewing lemma (EGA III$_1$ 3.1.2) to vary that and get the result in considerably more generality: any line bundle on any proper scheme over a field (or artin local ring, using length in place of dimension). The method is via Chow's Lemma to reduce to projective case and ultimately use slicing methods in the style of DeLand's answer to get back to the very ample case.

-
 By slicing method, do you mean cutting by hyperplane sections? What's Chow's lemma? – Fei YE Feb 25 2010 at 14:28 Fei, please speak with one of the many excellent algebraic geometry professors at your university to learn more in this direction. Iitaka's book on algebraic geometry may also have a good discussion, or SGA6 somewhere (article of Kleiman?); the relevant phrase is "Snapper polynomials". – BCnrd Feb 25 2010 at 14:59