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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 ?.)

Thanks for your time/help

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    $\begingroup$ 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. $\endgroup$ – Nick L Jun 27 '17 at 14:40
  • $\begingroup$ 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. $\endgroup$ – Bertie Jun 27 '17 at 15:35
  • $\begingroup$ @NickL : there is no obligation to obtain a general result. $\endgroup$ – R. Alexandre Jun 27 '17 at 15:38
  • $\begingroup$ @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 $\endgroup$ – R. Alexandre Jun 27 '17 at 15:40
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    $\begingroup$ 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. $\endgroup$ – Sándor Kovács Jun 27 '17 at 22:03
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OK, so here is a little more on the contents of my comment:

  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.
    In any case, it should be easy to find references for Hodge theory which nowadays goes way further than what I mentioned above. If you just google the phrase you'll see the abundance. A very good reference is a couple of books written by Voisin, though it is a bit tough, but it is just the nature of the subject.
  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.
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