We will discuss more about sheaf cohomology on curves and its applications in this post, including the cohomology version for Riemann-Roch's theorem, and the genus degree formula. We always fix $X$ an irreducible projective smooth curve. Then $X$ is covered by two affine open subsets $ U_0, U_1$. As we have discuss, the $n$-th cohomology group of $X$ vanish, for $n\ge 2$. And hence, for all sheaf $\mathscr{F}$ on $X$, we just need to care about the first cohomology group, which is defined by the $\mathscr{F}(U_0\cap U_1)/d^0(\mathscr{F}(U_0)\times \mathscr{F}(U_1))$.
1. Some examples and the Kodarai's vanishing theorem.
Example 1.1. Let us compute $H^1(\mathbb{P}^1, \mathscr{O})$. Here, $U_0=\{x_0\ne 0\}, U_1=\{x_1\ne 0\}$. Then $\mathscr{O}(U_0)=\langle\frac{x_1^a}{x_0^a}|a\in\mathbb{Z}_{\ge 0}\rangle$, similarly for $\mathscr{O}(U_1)=\langle \frac{x_0^a}{x_1^a}|a\in\mathbb{Z}_{\ge 0}\rangle$. Also, $\mathscr{O}(U_0\cap U_1)=\langle\frac{x_0^ax_1^b}{x_0^cx_1^d}|a+b=c+d; a,b,c,d\in\mathbb{Z}_{\ge 0}\rangle$. Because $a+b=c+d$, either $a\ge c$, or $b\ge d$. If $a\ge c$, then $b\le d$, and hence, $\frac{x_0^ax_1^b}{x_0^cx_1^d}=\frac{x_0^{a-c}}{x_1^{d-b}}$, which is in $\mathscr{O}(U_0)$. Similarly, when $b\ge d$, we will have $\frac{x_0^ax_1^b}{x_0^ax_1^b}$ is in $\mathscr{O}(U_1)$. And hence, the map $d^0: \mathscr{O}(U_0)\oplus \mathscr{O}(U_1)\to \mathscr{O}(U_0\cap U_1)$ sending $(f,g)$ to $f-g|_{U_0\cap U_1}$ is a surjective map. This yields $H^1(\mathbb{P}^1,\mathscr{O})=0$.
Example 1.2. Let us compute $H^1(\mathbb{P}^1,\mathscr{O}[-P])$, where $-P\in Div(\mathbb{P}^1)$. Choose $U_0, U_1$ as above. If $P\in U_0\cap U_1$, then we have $\mathscr{O}[-D](U_0\cap U_1)=\{f\in k(X)^\times|div(f) - P\ge 0 \text{ on }U_0\cap U_1\}\cup\{0\}$. This will yield $\mathscr{O}[-D](U_0\cap U_1)=0$, since $\deg(div(f))=0$, and hence, $div(f)-P$ cannot be an effective divisor. If $P\in U_0\setminus(U_0\cap U_1)$. Then for any $f\in \mathscr{O}[-P](U_0\cap U_1)$, we have $(div(f)-P)|_{U_1}\ge 0$ since $P\notin U_1$. Therefore, the map $d^0$ is again surjective. We conclude that $H^1(X,\mathscr{O}[-P])=0$.
We are now ready for the Kodarai's vanishing theorem.
Theorem 1.3 (Kodarai's vanishing theorem). Let $D\in Div(X)$ be a divisor such that $\deg(D)\ge 2g-1$. Then $H^1(X,\mathscr{O}[-D])=0$.
Proof. If the genus of $X$ is 0, we know that $X\cong \mathbb{P}^1$, and any divisor of zero degree is in $Pic(X)$. Then if $D\in Div(X)$, and $\deg D = -1$, we can assume $D=-P$. Then by Example 1.2 above, we can see $H^1(X,\mathscr{O}[D])=0$. If $\deg D\ge 0$, then by Riemann-Roch's theorem, $l(D)=1+\deg D\ge 1$. This yields, there exists $f\in L(D)$. Therefore, we can assume that $D$ is effective. If the genus of $X\ge 1$, then for all $D$, such that $\deg(D)\ge 2g-1$, by Riemann-Roch's theorem again, we obtain $l(D)\ge g\ge 1$. And we can assume $D$ is effective in all cases.
If we choose two affine covers $U_0,U_1$ of $X$ such that $U_0=X\setminus\{P_1,...,P_m\}$, and $U_1$ contains all $P_i$, and $U_1=X\setminus\{Q_1,...,Q_n\}$. Then for any $f\in \mathscr{O}[D](U_0\cap U_1)$, we will construct $f_0\in \mathscr{O}[D](U_0)$, $f_1\in \mathscr{O}[D](U_1)$ such that $f=f_0+f_1$.
If $f$ has no pole in $\{P_1,...,P_m,Q_1,...,Q_n\}$ then one can easily choose $f_0=0$, $f_1=f$ (note that in this case $f$ is also in $\mathscr{O}[D](U_1)$).
Otherwise, assume that $f$ has pole at $P_1,...,P_n$ with order $a_1\ge 1,a_1,...,a_n$, respectively. We can see $\deg(D+\sum_{i}a_iP_i)\ge 2g$, and hence, applying the Riemann-Roch's theorem. we get
$$l(D+\sum_{i}a_iP_i)=\deg(D)+\sum_ia_i+1-g$$
And also
$$l(D+(a_1-1)P_1+a_2P_2+...+a_mP_m)=\deg(D)+\sum_ia_i-g$$
i.e. there exists $f'_0\in L(D+\sum_{i}a_iP_i)$ such that $f'_0\notin L(D+(a_1-1)P_1+a_2P_2+...+a_mP_m)$. Also, one can see that $f\in L(D+\sum_{i}a_iP_i)$, and not in $L(D+(a_1-1)P_1+a_2P_2+...+a_mP_m)$. This yields there exists $\lambda\in k^\times$ such that $f-\lambda f'_0\in L(D+(a_1-1)P_1+a_2P_2+...+a_mP_m)$. Note that $f'_0\in \mathscr{O}[D](U_0)$, since $P_i\notin U_0$, and the degree of poles of $f-\lambda f'$ decreases 1 compared to $f$. Continuing this process $\sum_{i}a_i$ times, we finally get $f_0\in \mathscr{O}[D](U_0)$, and $f_1:=f-f_0$ has no pole at $P_i$ anymore. Note that $(div(f-f_0)+D)|_{U_\cap U_1}\ge 0$, and $(div(f-f_0)+D)|_{\{P_1,...,P_m\}}\ge 0$ (becase $D$ is effective), and $U_1=(U_0\cap U_1)\cup\{P_1,...,P_m\}$, and hence $f-f_0=f_1\in \mathscr{O}[D](U_1)$. This yields $H^1(X,\mathscr{O}[D])=0$. (Q.E.D)
With the Kodarai's vanishing theorem at hand, we can easily prove the cohomology version of Riemann-Roch's theorem in the next section.
2. Skyscraper sheaf and Riemann-Roch's theorem revisited. Back to our previous note, we already constructed the locally constant sheaf. Now, we will construct another kind of sheaf, which is called skyscraper sheaf.
If we identify each point $p\in X$ with an abelian group $G_p$. We then define $\mathscr{S}(U):=\prod_{p\in U}G_p$. Then it is easy to see that $\mathscr{S}$ is a sheaf of abelian group. Such a sheaf $\mathscr{S}$ is called totally disconnected sheaf. And skyscraper sheaf is a special case of totally disconnected sheaf, where we fix a point $p\in X$, and for all $q\ne p$, define $G_q:=0$, i.e. $G_q$ is the trivial group, and $G_p=k$, the base field. Say another words, we have defined a sheaf $k_p(U)=0$ if $p\notin U$, and $k_p(U)=k$ if $p\in U$. This is called skyscraper sheaf, and denoted by $k_p$.
Lemma 2.1. $H^0(X,k_p)=k$, and $H^1(X,k_p)=0$.
Proof. By Proposition 1.2 of the previous note, we have $H^0(X,k_p)\cong k_p(X)=k$. Let $\{U_0,U_1\}$ be affine open cover of $X$. Then one can see if $p\notin U_0\cap U_1$, then the group $k_p(U_0\cap U_1)$ is trivial. Otherwise, $p\in U_0\cap U_1$, then $k_p(U_0)=k_p(U_1)=k_p(U_0\cap U_1)=k$. And for any $s\in k_p(U_0\cap U_1)$, we can choose $s_0:=-s\in k_p(U_0), s_1=0\in k_p(U_1)$. And $d^0(s_0,s_1)=s_1-s_0=s$. This also yields $H^1(X,k_p)=0$. (Q.E.D)
The skyscraper sheaf arises in a natural situation as follows. If we fix a divisor $D\in Div(X)$, and then $L(D)\subset L(D+p)$. For any $f\in L(D+p)$, we have $ord_p(f)\ge -D(p)-1$, and one can write locally at $p$, $f=z^{-D(p)}u_f$, where $ord_p(u_f)\ge 0$. This induce a $k$-linear map $\phi: L(D+p)\to k$ sending $f$ to $u_f(p)$, and $\ker\phi$ is exactly $L(D)$. It can be seen then for any open subset $U\subset X$, if $p\notin U$, then for all $f\in \mathscr{O}[D+p](U)$, we define $\phi(f)=0$, also in this case, $\mathscr{O}[D](U)=\mathscr{O}[D+p](U)$. If $p\in U$, then the map $\mathscr{O}[D+p](U)\to k$ defined as sending $f$ to $u_f(p)$, which has the kernel is exactly $\mathscr{O}[D](U)$. Hence, in any case, we get an exact sequence of sheaf
$$0\to \mathscr{O}[D]\to \mathscr{O}[D+p]\to k_p\to O$$
This exact sequence also play a major roles in the proof of the following
Theorem 2.2 (Riemann-Roch, cohomology version). Let $D\in Div(X)$ be any divisor, then $h^0(D)-h^1(D)=\deg(D)+1-g$, where $h^0(D)=l(D)=\dim L(D)$, and $h^1(D)=\dim_kH^1(X,\mathscr{O}[D])$.
Proof. For short, we can denote $H^i(X,\mathscr{O}[D])$ as $H^i(D)$. Let us consider the short exact sequence
$$0\to \mathscr{O}[D]\to \mathscr{O}[D+p]\to k_p\to O$$
This yields a long exact sequence (note that $H^n(X,\mathscr{F})$ vanishes for all $n\le 2$, and all sheaf $\mathscr{F}$, $H^0(X,k_p)=k$, and $H^1(X,k_P)=0$)
$$0\to H^0(D)\to H^0(D+p)\to $k$ \to H^1(D)\to H^1(D+p)\to 0$$
By Proposition 1.2 of the previous note, we have $h^0(D)=l(D)$. And taking the dimension, we get $h^0(D)-h^0(D+p)+1-h^1(D)+h^1(D+p)=0$. Let $\chi(D)=h^0(D)-h^1(D)$, we get $\chi(D+p)=\chi(D)+1$.
And hence, by induction, we have $\chi(D)=\deg(D)+c$, for some constant $c$, and for all $D\in Div(X)$. If we choose $D$ such that $\deg(D)\ge 2g-1$, then by the Kodarai's vanishing theorem, $h^0(D)=\chi(D)=\deg(D)+1-g$. Hence, $c=1-g$. This yields $\chi(D)=\deg(D)+1-g$, for all $D\in Div(X)$. Or equivalently, $h^0(D)-h^1(D)=\deg(D)+1-g$. (Q.E.D)
The very important corollary of the theorem below is that
Corollary 2.3. $g=h^1(0)$, where $g$ is the genus of $X$.
Proof. If we take $D=0$ in the Riemann-Roch's theorem, then $h^0(0)$ is obviously 1 (it consists of all rational function have poles nowhere, and hence, are constant), and $h^1(0)=g$. (Q.E.D)
A consequence of the Corollary above is the genus degree formula.
Theorem 2.4. Let $X\subset \mathbb{P}^2$ be a smooth irreducible projective curve, given by an equation of degree $d$, then $g_X=\frac{1}{2}(d-1)(d-2)$.
Proof. The genus can be obtained by $h^1(0)$. We assume that $(0:0:1)$ is not in $X$, so that the term $x_2^d$ is in the equation of $X$ (if $x_2^d$ is not in the equation of $X$, then $(0:0:1)$ is in $X$). Then $x_2^d$ can be represented in terms of $x_0,x_1$. Choose two open affine sets covering $X$, $U_0=\{x_0\ne 0\}$, $U_1=\{x_1\ne 0\}$. By very similar argument of Example 1.1, we get $h^1(0)=\frac{1}{2}(d-1)(d-2) = g$ (Q.E.D)
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