# Dimension of the linear system of a divisor and Chern class

Before exposing my question, I'd like to advise that I'm not a geometer. So, please, be explicit as much as possible. I first exposed this problem in math.stackexchange, but I didn't get any answer, so maybe this is the right place.

Given a divisor $D$ on a plane complex projective curve $X$, there is a nice way to build a line bundle, $[D]$, which first Chern class is almost $D$ : it's just the Poincaré dual of $D$ when understood as homology class. (This is a much more general fact.)

For example, if I take the canonical bundle $K_X$ of $X$, then $$K_X = c_1(T^*X) = -c_1(X).$$

And now there is two ways to compute $l(K_X)$, the dimension of the linear system of $K_X$. On one hand, we can do some explicit computation (classical algebraic geometry). On the other hand, if I admit that $l(K_X)$ is the genus of $X$, $g(X)$, then since : $$\langle c_1(X),[X]\rangle = \chi(X) = 2-2g(X)$$ we get $$l(K_X) = \frac {\langle K_X,[X]\rangle}2 - 1.$$

My question is the following :

Is there such a way to compute $l(D)$ for a general divisor $D$ and complex algebraic manifold $X$ (put here the assumptions you want), i.e. in terms of the topology of the line bundle associated ?

(I know we can tell that $l(D) = h^0(X,\mathcal O(D)) -1$. But this is not very explicit and useful for computations, is It ?.)

• Not in general. There a divisors of degree 0 on an elliptic curve such that the corresponding line bundle has no sections, whereas the trivial line bundle has a 1-dimensional space of sections. – Nick L Jun 27 '17 at 14:40
• The general picture is encapsulated in the Riemann-Roch theorem. That theorem tells you how to compute the Euler characteristic $\chi(D)$ in terms of "topological" information about $D$. But $\chi(D)$ may differ from $h^0(D)$ due to the presence of higher cohomology (which is difficult to deal with in general); often to obtain $h^0(D)$ itself, one invokes an appropriate vanishing theorem (if it exists!) for the bundle $D$. This is a huge subject. – Bertie Jun 27 '17 at 15:35
• @NickL : there is no obligation to obtain a general result. – R. Alexandre Jun 27 '17 at 15:38
• @Bertie : initially my motivation was to prove Riemann-Roch from a nice formula of this kind. So indeed, it's the difference between singular cohomology and "holomorphic" cohomology that interests me – R. Alexandre Jun 27 '17 at 15:40
• You can relate $h^i(X,\mathscr O_X)$ to singular cohomology via Hodge theory and relate $h^i(X,\mathscr O_X(D))$ to $h^i(Y,\mathscr O_Y)$ for an appropriate branched cover $Y\to X$ for any very ample $D$ (a "general" divisor should probably be very ample, although you'd have to say what exactly you mean by that). Then again, proving RR this way seems kind of like buying a coat to match a button. – Sándor Kovács Jun 27 '17 at 22:03

1. Regarding Hodge theory. The classical result is that if $X$ is a smooth projective variety (or if you like equivalently, a compact complex manifold that can be embedded into $\mathbb P_{\mathbb C}^n$ for some $n$), then for any $m\in\mathbb N$ one has a direct sum decomposition, $$H^m(X,\mathbb C) \simeq \bigoplus_{p+q=m} H^q(X,\Omega_X^p).$$ If $\dim X=1$, then the relevant $m$'s are $0,1,2$, and the only $p,q$ for which you get anything that's not obviously $0$ is $p,q=0,1$, so the only non-trivial information you get out of this is that $$H^1(X,\mathbb C) \simeq H^1(X,\mathscr O_X) \oplus H^0(X,\omega_X).$$ By Serre duality the two spaces on the right are dual to each other and that's why you get that singular cohomology is essentially equivalent to coherent cohomology.
If $\dim X>1$, then you have more terms and more independent data which accounts for the defect of this "equivalence" in that case.
2. Branched covers, or ramified covers. A recent, good reference for this is Kollár's book on singularities (see the link). It has a section called "ramified covers", but this is also discussed at many places. The main idea is that if you have a divisor $D$ on $X$ such that some large multiple $aD$ has an effective representative which itself is smooth (e.g., if $D$ is ample), then taking a cover $f:Y\to X$ ramified along this representative connects the cohomology of $Y$ to the cohmologies of the powers of $-D$. More precisely, what you get is that for any $i$ $$H^i(Y,\mathscr O_Y) \simeq \bigoplus_{j=0}^{a-1} H^i(X,\mathscr O_X(-jD)).$$ So ultimately you get that the coherent cohomology groups of $\mathscr O_X(-D)$ are direct summands of the singular cohomology of $Y$. So, there is indeed a link between these, but not as straightforward as in the curve case.