# Monday, April 04

## Riemann-Roch and Serre Duality

:::{.remark}
Let $X\in \Proj\Var\slice k$ and $F \in \Coh(\mods{\OO_X})$.
By Grothendieck, $\cocomplex{H}(X; F)$ is supported in degrees $0 \leq d \leq \dim X$ and $h^i = \dim_k H^d(X; F) < \infty$ for all $d$.
:::

:::{.proposition title="Riemann-Roch"}
If $X\in \smooth\Proj\Var\slice k$,
\[
\chi(X; F) \da \sum_{0\leq i \leq \dim X} (-1)^i h^i(F) = \int_X \chern(F) \Todd(\T_X)
.\]
:::

:::{.remark}
What this formula means: for $X$ smooth projective, there is a Chow ring $A^*(X) = \bigoplus _{0\leq i \leq \dim X} A^i(X)$ where $A^i$ is analogous to $H^{2i}_\sing(X; \CC)$.
These are often different, but sometimes coincide (which can only happen if odd cohomology vanishes).
For curves, these differ, and $A^1(X) \cong \Pic(X)$ which breaks up as a discrete part (degree) and continuous part (Jacobian).
Define $A^i(X) \da \ZZ[C_i]\sim$ where $C_i$ are codimension $i$ algebraic cycles (subvarieties) and we quotient by linear equivalence.
Recall that for divisors, $D_1\sim D_2$ if $D_1-D_2$ is the divisor of zeros/poles of a rational functions.
More generally, for $Z$ of codimension $i$ and $Z \mapsvia{f} X$, consider $f_* D_1 \sim f_* D_2$ in order to define linear equivalence.
:::

:::{.example title="?"}
Consider $X_4 \subseteq \PP^3$ a quartic, the easiest example of a K3 surface.
Then $A^0[X] = \ZZ[X]$, $A^1(X) = \Pic(X)$, so what is $A^2(X)$?
These are linear equivalence classes of points, and any two points are equivalent if they are equivalent in the image of a curve.
It's a fact that K3s are not covered by rational curves -- instead these form a countable discrete set, with finitely many in each degree.
There is a formula which says that the generating function of curve counts is modular, and
\[
\sum n_d x^d = {1\over x} {1\over \prod_{1\leq n\leq \infty} (1-x^n)^{24} } 
,\]
where $n_d$ is the number of rational curves of degree $2d$.
So $A^2(X)$ is not obvious!
A theorem of Mumford says that it's torsionfree and infinitely generated.
Note that $n_d = p_{24}(d+1)$ where $p_\ell(\wait)$ is the numbered of *colored* integer partitions
:::

:::{.remark}
The integration map:
\[
\int_X: A^{\dim X}(X) &\to \ZZ \\
\sum n_i p_i &\to \sum n_i
.\]

There are two non-homogeneous polynomials $\chern(F)$ and $\Todd(\T_X)$ in $A^*(X)\tensor_\ZZ \QQ$, and the formula for Riemann-Roch says to multiply and extract only the top-dimensional component, i.e. take $\deg(\chern(F) \Todd(\T_X))_{\dim X}$.
This is very computable!
:::

:::{.example title="?"}
A Chern class: if $F = \OO_X(D)$, then
\[
\chern(F) = e^D = \sum_{1\leq i\leq n} D^i/i!
\]
where
\[
\OO_X(D)(U) = \ts{ f\in \OO_X(U) \st (f) + D \geq 0}
\]
and $D^n = D \cupprod D \cupprod \cdots \cupprod D$ is the $n\dash$fold self-intersection of $D$.
Note that $c_1(F) = D$.
:::

:::{.remark}
The Chern character of $F$ is additive on SESs, i.e. $0\to A\to B\to C\to 0$ yields $\chern(B) = \chern(A) + \chern(C)$.
:::

:::{.proposition title="RR for curves"}
If $X$ is a smooth projective curve,
\[
h^0(X) - h^1(X) = \deg D - g(X) + 1
.\]
In this case, $\chern(F) = 1+D$ and $\Todd(\T_X) = 1 + (1-g)[\pt]$ where $[\pt]$ is a certain well-defined divisor in $A^1(X)$.
One can rewrite this as $\Todd_X = 1 + {1\over 2}c_1 = 1 - {1\over 2} K_X$ (the canonical class, where $\deg K_X = 2g-2$).
This uses that
\[
c_1 = c_1(\T_X) = -c_1(\Omega_X) = -K_X
.\]
:::

:::{.example title="?"}
For $X$ a smooth surface,

- $\chern(F) = 1 + D + {D\over 2}$
- $\Todd(\T_X) = 1 - {1\over 2}c_1 + {1\over 12}(c_1^2 + c_2)$,

thus
\[
\chi(X; \OO_X(D)) = {D(D-2) \over K} + \chi(X; \OO_X)
.\]
:::

:::{.example}
If $X$ is a K3 surface, then $K_X = 0$ and $h^0(\OO_X) = h^1(\OO_X) = 0$, so $\chi(X; \OO_X) = 2$ and
\[
\chi(X; \OO_X(D)) = {D^2\over 2} + 2
.\]
:::

:::{.example title="?"}
For $X = \PP^2$ with $F = \OO(d)$, note 

- $K_X = \OO(-3)$
- $h^0(\OO_X) = 1, h^1(\OO_X) = h^2(\OO_X) = 0$

So
\[
\chi(X; \OO(d)) = {d(d+3) \over 2} +1 = {d+2\choose 2}
.\]
As a corollary, for $d\geq 0$,
\[
h^0(\OO_{\PP^n}(d)) = {d+n\choose n}
.\]
:::