I'm studying algebraic geometry and I need some help to understand the Riemann-Roch theorem.
Let us consider a holomorphic line bundle $\xi$ over a Riemann surface $X$. The unique invariant of a holomorphic bundle is the equivalence class $[D]$ of the divisor $D$ of an arbitrary section, but it looks like this class is "complicated to calculate".
The most simplest invariant of line bundles is the degree $\text{deg}\,D$ of divisor $D$. In fact, it is just the Euler characteristic. This number is enough to classify all line bundles over $\mathbb{CP}^{1}$.
The second clear invariant is the count of linearly independent holomorphic sections.
This number is precisely $l(-D)$.
According to the Riemann-Roch theorem we have the formula
$$
l(D) – l'(D) = 1 – g – \text{deg}\,D,
$$
that we can use to calculate the $l(D)$. Am I understanding this theorem correctly?
Ok. Could you show me some example of a non-trivial holomorphic line bundle over torus that has degree zero.
Thanks.
Best Answer
Let $\mbox{Pic}^0(X)$ denote the group of all line bundles of degree 0 on $X$. Then it is a classical fact that $\mbox{Pic}^0(X)$ is an abelian variety (that is, a projective complex torus) of dimension $g$ called the Jacobian of $X$ (I'm assuming that $X$ is compact). So basically, for $g\geq1$ there are "plenty" of line bundles of degree 0 that are non-trivial. Actually, for $g\geq1$, $X$ is contained in $\mbox{Pic}^0(X)$; this can be seen by using the line bundle that Asal Beag Dubh described in the comments. Let $q\in X$ be fixed, and define the map $$\phi_q:X\to\mbox{Pic}^0(X)$$ where $p\mapsto\mathcal{O}_X(p)\otimes\mathcal{O}_X(-q)$. This is a holomorphic immersion.
Since we have the exact sequence $$0\to\mbox{Pic}^0(X)\to\mbox{Pic}(X)\stackrel{\deg}{\to}\mathbb{Z}\to0,$$ we see that the degree classifies line bundles on $\mathbb{P}^1$, but for higher genus this is no longer true. For example, for $g=1$, a divisor $n_1p_1+\cdots+n_rp_r$ is linearly equivalent to zero (that is, defines the trivial line bundle) if and only if it is of degree 0 and if the sum $n_1p_1+\cdots+n_rp_r$ in $X$ (remember that $X$ is a torus) is 0. This then generalizes to higher genus by using what I said above, and so a divisor $D$ on $X$ is linearly equivalent to 0 if and only if it is of degree 0 and $\alpha_p(D)=0$ in $\mbox{Pic}^0(X)$ (the map $\alpha_p$ extends linearly to $\mbox{Div}(X)$).
Your Riemann-Roch formula isn't quite right. The Riemann-Roch formula actually says that $$\chi(D):=h^0(D)-h^1(D)=\deg D+1-g.$$ Using Serre duality, $h^1(D)=h^0(K-D)$, where $K$ is a canonical divisor of $X$ (that is, the zero set of a differential form). Here $h^0(D)$ denotes the dimension of the global sections of $D$ and $h^1(D)$ denotes the dimension of the first cohomology group of $D$ (actually of $\mathcal{O}_X(D)$). Another thing you mentioned is $l(-D)$; if $D$ is effective, this is actually the dimension of all holomorphic functions that vanish on $D$, and so equals 0.