# Wednesday, March 30 :::{.remark} Topics: - General vanishing (Serre 1 and 2) - Čech cohomology - Riemann-Roch and Serre duality - Advanced vanishing (e.g. Kodaira vanishing) ::: ## Čech Cohomology :::{.remark} Setup: $X$ and $\mcf\in \Sh(X; \Ab\Grp)$, an open cover $\mcu \covers X$. We defined the Čech complex: \[ \Cc^p(\mcu; \mcf) = \bigoplus _{i_1 < \cdots < i_p} \mcf(U_{i_1, \cdots, i_p}) ,\] which had certain differentials. ::: :::{.theorem title="?"} Suppose $X\in \Alg\Var$ or $X\in \Sch$ is separated (e.g. a quasiprojective scheme), $F\in \QCoh(X)$ an $\OO_X\dash$module, and let $\mcu \covers X$ be an affine open cover. Then \[ \Hc(\mcu; F) = \RR\Gamma(X; F) .\] ::: :::{.remark} More generally, we can just assume that all intersections of affines are affine, and instead there is a spectral sequence. This can fail if $X$ is not separated, e.g. $X \da \AA^2 \glue{\AA^2\smz} \AA^2$ where the intersection $\AA^2\smz$ is not affine. Recall that $X$ is separated iff $X \injectsvia{\Delta_X} X\fiberpower{X}{2}$ is closed. ::: :::{.example title="?"} Consider $X=\PP^1$ and $F = \bigoplus _{d\in \ZZ} \OO_X(d)$, we can compute $\Hc(X; \OO(d))$ for all $d$. Take a cover $U_i = \ts{x_i\neq 0}$ where $U_0$ has coordinate $x \da x_1/x_0$ and $U_1$ has coordinate $y= x_0/x_1$ which intersect at $U_{01} = \ts{x,y\neq 0}$ and are glued by $y=1/x$. The Čech resolution is \[ 0\to F(U_0) \oplus F(U_1) \mapsvia{f} F(U_{01}) \to 0 ,\] so $H^0 = \ker f$ and $H^1 = \coker f$. Recall that sections of $\OO(d)$ are locally ratios of polynomials with valuation $d$. We have $\ro{\OO_{\PP^1}(d)}{\AA_1} = x_0^d \OO_{\PP^1}$ by rewriting $p/q = x_0^d p'/q'$. We can thus write this sequence as \[ 0 \to \bigoplus _{d\in \ZZ} x_0^d k\adjoin{x= {x_1\over x_0}} = \bigoplus_d \gens{\text{degree $d$ monomials in } x_0^{\pm 1}, x_1} \oplus \bigoplus _{d} \gens{\text{degree $d$ monomials in } x_0, x_1^{\pm 1}} \to \bigoplus_d \gens{\text{degree $d$ monomials in } x_0^{\pm 1}, x_1^{\pm 1}} \to 0 .\] :::{.claim} \[ H^0(X; F) = k[x_0, x_1], \qquad H^1 = {1\over x_0 x_1} k\adjoin{{1\over x_0}, {1\over x_1}} .\] ::: Being in the kernel means $v_{x_0}(f)>0$ and $v_{x_1}(f) > 0$, which yields monomials $x_0^n x_1^m$ where $d=n+m$. For the cokernel, note $(p, 1) \mapsto p-q$, what's missing? Monomials where both powers are negative. ::: :::{.example title="?"} Similar computations work for $X=\PP^n$ and yield \[ {H}^0\qty{X; \bigoplus _{d\in \ZZ} \OO_{\PP^n}(d) } = \kxn, \quad {H}^n\qty{X; \bigoplus _{d\in \ZZ} \OO_{\PP^n}(d) } = {1\over \prod x_i} k\adjoin{{1\over x_0}, \cdots, {1\over x_n}} .\] Note that both sides are graded by degree. This can be done in affine opens $U_i = \ts{x_i\neq 0} \cong \AA^n$, $\ro{\OO_X(d)} = x_i^d \OO_X$, and similarly \[ 0 \to \bigoplus_d \gens{\text{degree $d$ monomials in } x_0^{\pm 1}, x_1, \cdots, x_n} \oplus \bigoplus _d \gens{\text{degree $d$ monomials in} x_0, x_1^{\pm 1}, \cdots, x_n} \oplus \cdots \to \cdots \to 0 .\] The kernel is again spanned by monomials $f$ with $v_{x_i}(f) \geq 0$ for all $i$. Which monomials don't come from the middle step? Those where $v_{x_i}(f) < 0$ for all $i$. ::: :::{.remark} A combinatorial device to keep track of monomials: let $X=\PP^2$, and build simplices which track which monomials are allowed to be negative. See Hartshorne for a description of how to encode this as a simplicial set: ![](figures/2022-03-30_11-07-34.png) ::: :::{.remark} As a result, we can compute \[ \dim H^0(\PP^n; \OO_{\PP^n}(d)) = {n+d\choose n} = {n+d\choose d} \] by counting monomials using a stars and bars argument. Moreover \[ \dim H^n(\PP^n; \OO(d)) = \dim H^0(\PP^n; \OO(n-1-d)) = \dim H^0(\PP^n; \OO(K) \tensor \OO(d)\inv ) \] where the canonical class of $\PP^n$ is given by $\OO(K_{\PP^n}) = \OO(-n-1)$. :::