currently studying algebraic surfaces over the complex numbers. Before i did some algebraic geometry (I,II,start of III of Hartshorne) and a course on Riemann surfaces.

Now i understood that by GAGA, a lot of results transfer from complex analytic geometry to algebraic geometry over $\mathbb{C}$. Question: in my RS surface course the genus for a compact RS was defined as the number of holes, formally half of the dimension of the first de rham cohomology group (or singular with coefficients in some field of characteristic zero). Now i encountered the algebraic definition: the dimension of the first cohomology group of the structure sheaf.

I expect them to be the same, but did not find a quick proof. Can anyone show me how this works? I learned that de rham cohomology with coefficients in $\mathbb{C}$ corresponds to sheaf cohomology of the constant sheaf $\mathbb{C}$. However i'm not sure, can someone either confirm this or tell me why it's not true? If it is true, i find it hard to work with, since up till now i've been basically using Serre duality and Riemann Roch for bundles to reduce computing homology to computing spaces of global sections, however since the constant sheaf is not locally free i cannot apply this trick (or can i?). Also, the constant sheaf makes sense both algebraically and analytically, so which one should i take in the de rham <-> constant sheaf correspondence? Or do both work? (with GAGA in mind, i expect them to have the same cohomology, but this might not be true)

So summarizing

  • Do both definitions of genus agree? (of course assuming the algebraic curve to be smooth, so it is a RS)
  • If they do, can you show me a proof? (preferably using some sheaf cohomology)
  • Is it true that de rham cohomology corresponds to sheaf cohomology of the constant sheaf, if so is it the the analytic one, the algebraic one, or both?
  • Am i limiting myself unnecessarily by using Riemann Roch and Serre duality just for bundles, i.e. can i use them for all sheaves?

Lastly, when answering my questions, it would be immensely appreciated if you could elaborate a little on how to use GAGA in general.

Hope this makes sense, i suspect them to be silly questions once i understand them, but right now it's a fuzz to me..


Edit: i just found a result from Hodge theory for surfaces in Beauville, stating $h^{0}(S,\Omega^1_S) = \frac{1}{2}h^{1}(S,\mathbb{R})$, all cohomology being analytic. So assuming (1) the same thing holds for curves and (2) GAGA identifies the algebraic and analytic cotangent bundle, Serre duality states that the left hand side is equal to $h^{1}(C,\mathcal{O}_C)$, i.e. the genus found in my AG textbook and i proved one thingwhat i wanted to proof. I was trying to find a reference on Hodge theory so i could verify (1) but a quick scan yielded no results. Does anyone have good idea on this?

Also i was wondering about (2) for a longer time, i'd be really happy with any comments on this.

  • $\begingroup$ (1) Yes, both definitions of genus agree. Probably the fastest way to check this is to use Riemann–Roch or Serre duality... (3) De Rham cohomology agrees with the cohomology of the constant sheaf $\mathbb{C}$, but only in the so-called classical topology $\endgroup$
    – Zhen Lin
    Commented Jul 26, 2012 at 12:21
  • $\begingroup$ Thanks @Zhen! Concerning (1), since in Riemann Roch the genus is already part of the statement, i would find it more enlightening to see a proof that has a result like $$ h^{1}(C,\mathcal{O}_C) = \frac{1}{2} h^{1}_{\text{de Rham}}(\Sigma)$$ where i mean $\Sigma = C$ but $\Sigma$ is the Riemann surface one, and $C$ is the algebraic curve one.. $\endgroup$
    – Joachim
    Commented Jul 26, 2012 at 12:29
  • $\begingroup$ Or is that impossible without using those theorems? $\endgroup$
    – Joachim
    Commented Jul 26, 2012 at 12:35

1 Answer 1


First let me emphasize what Serre did : in 1956 he wrote an article Géométrie algébrique et géométrie analytique, which everybody calls GAGA (I heard him explain that he deliberately chose the title so that this would happen!).
This article proves many results, the gist of which is that calculations for coherent sheaves on a projective complex variety $X$ give the same (or isomorphic) results as those done on the underlying holomorphic variety $X^h$.

However he did not consider non coherent sheaves and for good reason: if for example you consider the constant sheaf $\mathbb C_X$ on $X$, it is flabby ("flasque") so that all its cohomology groups vanish: $H^i(X,\mathbb C_X)=0$, whereas of course the cohomology vector spaces $H^i(X^h,\mathbb C_{X^h})$ are not zero in general:
For example if $X$ is a smooth projective curve of genus $g$, then $X^h$ is a Riemann surface and $dim _\mathbb C H^1(X^h,\mathbb C_{X^h})=2g $.

Fortunately thanks to the rich techniques known for holomorphic manifolds many apparently transcendental invariants of a projective smooth manifold can be computed using only coherent sheaves.
A splendid example is Dolbeault's 1953 result for Betti numbers $$b_r(X^h)=\sum_{p+q=r} dim_\mathbb C H^q (X^h,\Omega_{X^h}^p)$$ Since the right hand size is cohomology of coherent sheaves,it can be calculated algebraically thanks to GAGA: $$ dim_\mathbb C H^q (X^h,\Omega_{X^h}^p)=dim_\mathbb C H^q (X,\Omega_{X}^p) $$ In particular, for a smooth projective curve $X$, the genus of the associated Riemann surface $X^h$ is given algebraically by $$g(X^h)=\frac {1}{2}(b_1(X^h))=\frac {1}{2}(dim_\mathbb C H^1 (X,\mathcal O_X)+dim_\mathbb C H^0 (X,\Omega_{X}^1))$$ where the two summands on the right are equal by Serre duality (him again!).

Warning This answer is meant to address your request "it would be immensely appreciated if you could elaborate a little on how to use GAGA in general ", not to show the shortest or most elementary road to technically answer your questions on Riemann surfaces.

  • $\begingroup$ thanks a lot for your nice answer! You did answer my main question in a really nice way too, i hoped that there would a proof like this. Also from your answer i understand that in fact under GAGA, the analytic and algebraic cotangent bundle (sheaf) are identified (as i expected). That settles everything, thanks again! $\endgroup$
    – Joachim
    Commented Jul 26, 2012 at 18:32
  • $\begingroup$ My pleasure, Joachim. $\endgroup$ Commented Jul 26, 2012 at 18:49

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