# Asymptotic Riemann Roch

I am reading Ravi Vakil's Foundation of Algebraic Geometry 20.1.I.

I want to show this:

Suppose $X$ is projective (over an infinite field), $\mathfrak{F}$ is a coherent sheaf on $X$ with support of dimension $= n$, and $\mathfrak{L}$ is a line bundle on $X$. Show that $\chi(\mathfrak{L}^{\otimes m}\otimes \mathfrak{F})$ is a polynomial with degree at most $n$, and the coefficient of $m^n$ is $(\mathfrak{L}^n\cdot \mathfrak{F})$

Hint: Expanding $(\mathfrak{L}^{n+1}\cdot (\mathfrak{L}^{\otimes i}\otimes \mathfrak{F}))=0$ to get a recursion for $\chi (\mathfrak{L}^m \otimes \mathfrak{F})$.

For the case that $\mathfrak{L}$ is very ample, I can obtain the result by induction on $n$ as follows:

The case $n=1$ follows from Riemman Roch for nonreduced curves (Exercise 18.4 S). And if $n>1$, then I can find a global section $s$ of $\mathfrak{L}$, such that $s$ does not passes through the assocaited points of $\mathfrak{F}$. By the short exact sequence $$0\to \mathfrak{L}^{\vee}\otimes \mathfrak{F}\xrightarrow{\times s}\mathfrak{F}\to \mathfrak{G}\to 0$$ where $\mathfrak{G}$ has support with dimension $n-1$. I can obtain the result from the fact that Euler characteristic is additive.

What I want to know is how to do the general case, I know that every line bundle on $X$ can be expressed as the difference of two very ample line bundles, but I don't know how to obtain the result from this. I also don't know how to use the hint, from the hint, I can obtain $$\chi(\mathfrak{L}^{\otimes m}\otimes \frak{F})=\sum_{i=1}^{n+1}(-1)^{j+1}\chi(\frak{L}^{\otimes (m-i)}\otimes \frak{F})$$ but I don't know why this formula is useful.

Any help or hints are appreciated, thank you.

I am also not sure how the recursive formula $$\mathcal{L}^{n+1} \cdot (\mathcal{L}^{\otimes i} \otimes \mathcal{F})=0$$ is supposed to help.

To prove polynomiality one can combine your idea for very ample line bundles with some techniques from this section of stacks project.

Write $$\mathcal{L}=\mathcal{A} \otimes \mathcal{B}^{\vee}$$ as the product of very ample bundles, $$\mathcal{A}=\mathcal{O}(D_1), \mathcal{B}=\mathcal{O}(D_2)$$. (16.6.F Vakil).

Consider $$0 \to \mathcal{O}(-D_1) \to \mathcal{O}_X \to i_{*} \mathcal{O}_{D_1} \to 0,$$ tensor it with $$\mathcal{A}^{\otimes n_1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F}$$ and apply additivity of $$\chi$$: $$\chi(\mathcal{A}^{\otimes n_1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F})-\chi(\mathcal{A}^{\otimes n_1 - 1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F}) = \chi(\mathcal{A}^{\otimes n_1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F} \otimes i_{*} \mathcal{O}_{D_1} )$$where now $$\mathcal{F} \otimes i_{*} \mathcal{O}_{D_1}$$ has lower dimensional support.

Analagously for $$D_2$$; thus we can inductively assume that $$P(n_1,n_2)-P(n_1 -1, n_2)$$ $$P(n_1, n_2) - P(n_1, n_2 -1)$$ are polynomials in $$(n_1,n_2)$$. Then, as the Stacks Project says, "A simple arithmetic argument shows that $$P$$ is a numerical polynomial of total degree at most dim (Supp ($$\mathcal{F}$$))". Then set $$p(m)=P(m,m)$$.

As far as the leading coefficient is concerned, what we want to do is to "take $$n$$-th derivative" (where $$n$$ is dim (Supp ($$\mathcal{F}$$))). Discretely, we could take "$$n$$-th order backward discrete difference" to obtain $$n! a_n = \sum_{i=0}^{n} (-1)^i{{n}\choose{i}} p(-i)=\sum_{i=0}^{n} (-1)^i{{n}\choose{i}} \chi((\mathcal{L}^{\vee})^{i} \otimes \mathcal{F})= \mathcal{L}^n \cdot \mathcal{F}.$$

• Why tensor $\mathcal{F}$ and dual are exact?
– Mike
Nov 21, 2019 at 19:39
• Thank you, I've edited the post. We can resolve $\mathcal{F}$ by locally free and therefore assume it is locally free to begin with. The dualization part is not necessary at all I think. Nov 22, 2019 at 0:53
• How could we resolve $\mathcal{F}$ by locally free sheaf? Also, I think that analagously for $D_{2}$, we could get that $P(n_{1},n_{2})-P(n_{1},n_{2}+1)$ is a polynomials, combine the first equality, could we get that $P(n_{1},n_{2})$ is a polynomial?
– Mike
Nov 22, 2019 at 1:17
• Yes, you are right about $D_2$. For the resolution I am invoking a standard fact that a coherent sheaf on a smooth variety can by resolved by a finite complex of locally free sheaves. This though requires smoothness, which I think is not assumed in this exercise. My solution is just bad, probably there is a better one. Nov 22, 2019 at 16:07

Here is my approach:

Let $$f(t) = \chi(X, L^{\otimes t} \otimes F)$$. Then, $$int_X(L^{int: n+1} , L^{\otimes i} \otimes F) = 0$$.

Expanding $$L^{\otimes i} \otimes F) = 0$$ using the definition, we note that there are $$\binom{n+1}{k}$$ subsets of cardinality $$k$$ of $$\{1, ... n\}$$. Therefore, $$int_X(L^{int: n+1} , L^{\otimes i} \otimes F) = \sum_{k=0}^{n+1} (-1)^k \binom{n+1} {k}\chi(X, L^{\otimes t-k} \otimes F)$$. Therefore, $$\sum_{k=0}^{n+1} (-1)^k \binom{n+1} {k} f(t-k) = 0$$.

Extracting the $$k=0$$ piece separately , we get $$f(t) = \sum_{k=0}^{n+1} (-1)^{k+1} \binom{n+1} {k} f(t-k) = 0$$.

This is a linear recurrence whose characteristic polynomial is $$(x-1)^{n+1}$$. Therefore, from the formula for linear recurrences, $$f(t)$$ is a polynomial of degree $$n$$.

• This was actually the approach I intended, so nicely done! I have now edited the exercise so as to more directly lay this out. Jun 4 at 4:42