# Monday, February 15 :::{.remark} Last time: we saw that \( \chi(\PP^1, \OO) = 1 \), and we'd like to generalize to holomorphic line bundles on a Riemann surface. This will be the main ingredient for Riemann-Roch. ::: :::{.theorem title="Euler characteristic and homological vanishing for holomorphic vector bundles"} Let $X \in \Mfd_\CC$ be compact and let \( \bundle{F} \) be a holomorphic vector bundle on \( X \) [^coh_sheaf_general] Then $\chi$ is well-defined and \[ h^{> \dim_\CC X}(X; \bundle{F} ) = 0.\] [^coh_sheaf_general]: Or more generally a finitely-generated \( \OO\dash \)module, i.e. a coherent sheaf. ::: :::{.remark} The locally constant sheaf \( \constantsheaf{\CC} \) is not an \( \OO\dash \)module, i.e. \( \constantsheaf{\CC}(U) \not\in \mods{\OO(U)} \). In fact, $h^{2i}(X, \constantsheaf{\CC}) = \CC$ for all $i$. ::: :::{.proof title="of theorem"} We'll can resolve $\bundle{F}$ as a sheaf by first mapping to its smooth sections and continuing in the following way: \[ 0 \to \bundle{F} \to C^{\infty } \bundle{F} \mapsvia{\delbar} F \tensor A^{0, 1} \to \cdots ,\] where $\delbar f = \sum_i \dd{f}{\conjugate{z}_i} \, d\conjugate{z}_i$. Suppose we have a holomorphic trivialization of \( \ro {\bundle{F} }{U} \cong \OO_{U}^{\oplus r} \) and we have sections $(s_1, \cdots, s_r) \in C^{\infty } \bundle{F}(U)$, which are smooth functions on $U$. In local coordinates we have \[ \delbar s \da (\delbar s_1, \cdots, \delbar s_r) ,\] but is this well-defined globally? Given a different trivialization over $V \subseteq X$, the $s_i$ are related by transition functions, so the new sections are $t_{UV}(s_1, \cdots, s_r)$ where $t_{UV}: U \intersect V \to \GL_r(\CC)$. Since $t_{UV}$ are holomorphic, we have \[ \delbar( t_{UV} (s_1, \cdots, s_r)) = t_{UV} \delbar(s_1, \cdots, s_r) .\] This makes $\delbar: C^{\infty } \bundle{F} \to F\tensor A^{0, 1}$ a well-defined (but not $\OO\dash$linear) map. We can thus continue this resolution using the Leibniz rule: \[ 0 \to \bundle{F} \to C^{\infty } \bundle{F} \mapsvia{\delbar} F \tensor A^{0, 1} \mapsvia{\delbar} \cdots F \tensor A^{0, 2} \mapsvia{\delbar} \cdots ,\] which is an exact sequence of sheaves since $(A^{0, \wait}, \delbar)$ is exact. \todo[inline]{Why? Split into line bundles?} We can identify $C^{\infty }\sheaf{F} = \sheaf{F} \tensor A^{0, 0}$, and $\sheaf{F} \tensor A^{0, q}$ is a smooth vector bundle on $X$. Using partitions of unity, we have that $\sheaf{F} \tensor A^{0, q}$ is acyclic, so its higher cohomology vanishes, and \[ H^i(X; \sheaf{F} ) \cong \frac { \ker ( \delbar: \sheaf{F}\tensor A^{0, i} \to \sheaf{F} \tensor A^{0, i+1} } { \im ( \delbar: \sheaf{F}\tensor A^{0, i-1} \to \sheaf{F} \tensor A^{0, i} } .\] However, we know that $A^{0, p} = 0$ for all $p> n \da \dim_\CC X$, since any wedge of $p>n$ forms necessarily vanishes since there are only $n$ complex coordinates. ::: :::{.warnings} This only applies to holomorphic vector bundles or $\OO\dash$modules! ::: ## Riemann-Roch :::{.theorem title="Riemann-Roch"} Let $C$ be a compact connected Riemann surface, i.e. $C\in \Mfd_\CC$ with $\dim_\CC(C) = 1$, and let $\bundle{L}\to C$ be a holomorphic line bundle. Then \[ \chi(C, \bundle{L}) = \deg(L) + (1-g) && \text{where } \deg(L) \da \int_C c_1(\bundle{L}) \] and $g$ is the genus of $C$. ::: :::{.proof title="of Riemann-Roch"} We'll introduce the notion of a "point bundle", which are particularly nice line bundles, denoted $\OO(p)$ for $p\in \CC$. \begin{tikzpicture} \fontsize{34pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-15_14-16.pdf_tex} }; \end{tikzpicture} Taking $\DD$ to be a disc of radius $1/2$ and $V$ to be its complement, we have $t_{uv}(z) = z\inv \in \OO^*(U \intersect V)$. We can take a holomorphic section $s_p \in H^0( C, \OO(p) )$, where $\ro{s_p}{U} = z$ and $\ro{s_p}{V} = 1$. Then $t_{uv}( \ro{s_p}{U} ) = \ro{s_p}{V}$ on the overlaps. We have a function which precisely vanishes to first order at $p$. Recall that $c_1( \OO(p) )$ is represented by $[ V(s) ] = [p]$, and moreover $\int_C c_1 ( \OO(p) ) = 1$. We now want to generalize this to a **divisor**: a formal $\ZZ\dash$linear combination of points. :::{.example title="?"} Take $p, q,r\in C$, then a divisor can be defined as something like $D \da 2[p] - [q] + 3[r]$. ::: Define $\OO(D) \da \bigotimes_{i} \OO(p_i)^{\tensor n_i}$ for any $D = \sum n_i [p_i]$. Here tensoring by negatives means taking duals, i.e. $\OO( -[p] ) \da \OO^{\tensor_{-1}} \da \OO(p)\dual$, the line bundle with inverted transition functions. $\OO(D)$ has a meromorphic section given by \[ s_D \da \prod s_{p_i}^{n_i} \in \Mero(C, \OO(D) ) \] where we take the sections coming from point bundles. We can compute \[ \int_C c_1 ( \OO(D) ) = \sum n_i \da \deg(D) .\]. :::{.example title="?"} \[ \deg( 2[p] -[q] + 3[r]) = 4 .\] ::: :::{.remark} Assume our line bundle $L$ is $\OO(D)$, we'll prove Riemann-Roch in this case by induction on $\sum \abs{n_i}$. The base case is $\OO$, which corresponds to taking an empty divisor. Then either - Take $D = D_0 + [p]$ with $\deg(D_0) < \sum \abs{n_i}$ (for which we need some positive coefficient), or - Take $D_0 = D + [p]$. ::: :::{.claim} There is an exact sequence \[ 0 \to \OO(D_0) &\to \OO(D) \to \CC_p \to 0\\ s\in \OO(D_0)(U) &\mapsto s \cdot s_p \in \OO( D_0 + [p] ) (U) ,\] where the last term is the skyscraper sheaf at $p$. ::: :::{.proof title="of claim"} The given map is $\OO\dash$linear and injective, since $s_p\neq 0$ and $s s_p=0$ forces $s=0$. Recall that we looked at \( \OO \mapsvia{\cdot z} \OO \) on $\CC$, and this section only vanishes at $p$ (and to first order). The same situation is happening here. ::: Thus there is a LES \begin{tikzcd} &&&& 0 \\ \\ {H^0( \OO(D_0) )} && {H^0( \OO(D) )} && {H^0( \OO(\CC_p) )} \\ \\ {H^1( \OO(D_0) )} && {H^1( \OO(D) )} && {H^1( \OO(\CC_p) ) = 0} \\ \\ 0 \arrow[from=3-5, to=5-1, out=0, in=180] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow[from=1-5, to=3-1, out=0, in=180] \arrow[from=5-5, to=7-1, out=0, in=180] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwyLCJIXjAoIFxcT08oRF8wKSApIl0sWzIsMiwiSF4wKCBcXE9PKEQpICkiXSxbNCwyLCJIXjAoIFxcT08oXFxDQ19wKSApIl0sWzAsNCwiSF4xKCBcXE9PKERfMCkgKSJdLFsyLDQsIkheMSggXFxPTyhEKSApIl0sWzQsNCwiSF4xKCBcXE9PKFxcQ0NfcCkgKSA9IDAiXSxbNCwwLCIwIl0sWzAsNiwiMCJdLFsyLDNdLFswLDFdLFsxLDJdLFszLDRdLFs0LDVdLFs2LDBdLFs1LDddXQ==) We also have $h^1(\CC_p) = 0$ by taking a sufficiently fine open cover where $p$ is only in one open set. So just checking Čech cocycles yields $C_U^1(C, \CC_p) \da \prod_{i