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A lecturer of mine once ``proved'' the existence of non-constant meromorphic functions on a compact Riemann surface $X$ by using analysis of the Laplacian to decompose the de Rham cohomology group as $$H^1_{\text{dR}}(X) \cong H^{0,1}(X) \oplus H^{1,0}(X)$$ where the groups on the right are the Dolbeault cohomology groups. My understanding is that this is a particular instance of Hodge theory, although I don't really know anything about that.

Once we have an inclusion $H^{0,1}(X) \hookrightarrow H^1_{\text{dR}}(X)$ then I am perfectly happy with the construction of these meromorphic functions: as the latter space is easily shown to be finite dimensional, the former is too and so we can find a relation among our obstructions and use that to build a global meromorphic function with at least one pole.

The issue I have with this proof is that I have never studied the Laplacian. Does anyone know of another way to prove such an inclusion, ideally more algebraically? I was toying around with using the isomorphism between Dolbeault and Cech cohomology, but could not come up with a satisfactory proof.

To clarify: I am aware that one can prove this inclusion using Hodge theory, and also that Deligne had an algebraic approach by way of going through characteristic $p$. I am hoping for an elementary argument that $H^{0,1}$ includes into $H^1_{\text{dR}}$. I am willing to accept that one doesn't exist, but I thought I'd ask.,

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    $\begingroup$ i'm not sure u can really expect an algebraic argument for this. one has to do analysis at some point to show that compact riemann surfaces are algebraic in the first place, so that one can at least hope for an algebraic argument from that point on $\endgroup$ – EBz Jun 22 at 12:41
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(This would a comment, but it's hard to squeeze all the notation into the comment box.)

If you are comfortable with sheaf theory, then you can use the exact sequence $$0\to \mathbb{C}\to \mathcal{O}_X\to \Omega_X^1\to 0$$ to get $$\to H^0(X,\Omega_X^1)\xrightarrow{\iota} H^1(X,\mathbb{C})\xrightarrow{\pi} H^1(X,\mathcal{O}_X)\to $$ There are various ways to argue that $\iota$ is injective, and $\pi$ is surjective*. This would give you a noncanonical decomposition $$H^1(X,\mathbb{C})\cong H^0(X,\Omega_X^1)\oplus H^1(X,\mathcal{O}_X)$$ However, if you want to get a natural splitting of $\pi$ (which is what you are asking for) then I don't know any approach that avoids talking about harmonic forms*. So it's a good thing to learn.

Added footnotes

  1. For $\iota$ look at the previous terms in the sequence and observe that global holomorphic functions are constant. For $\pi$, it's more involved, and I'd prefer not to get into to it here.
  2. In a nutshell, $H^1(X,\mathcal{O}_X)$ can be identified with the space of harmonic $(0,1)$-forms, or equivalently complex conjugates of holomorphic $1$-forms.
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    $\begingroup$ I am very happy with this, but my concern is that in proving surjectivity of $\pi$, we secretly need some harmonic analysis, the details of which I cannot prove. The claim that $\pi$ is surjective amounts to the statement that everything is in the kernel of the map $H^1(X, \mathcal{O}_X) \to H^1(X, \Omega^1_X)$ induced by the exterior derivative, and this is not at all clear to me. $\endgroup$ – Martin Skilleter Jun 22 at 14:36
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    $\begingroup$ Right, I didn't elaborate because it wouldn't be trivial. You could use Serre duality, for example, but this is a story in itself. $\endgroup$ – Donu Arapura Jun 22 at 14:46
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    $\begingroup$ I’m much more comfortable with line bundles, so Serre duality is a very acceptable tool for me. In this case, I think the map following $\pi$ in this long exact sequence becomes (after applying Serre duality) the exterior derivative from anti-holomorphic functions into holomorphic 1-forms, so everything is in the kernel. If so, I am fully satisfied. Thank you for all your help. $\endgroup$ – Martin Skilleter Jun 22 at 15:09
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    $\begingroup$ The argument I had in mind is that $H^1(X,\Omega)\to H^2(X,\mathbb{C})$ is surjective map of vector spaces of the same dimension (by duality), so.... $\endgroup$ – Donu Arapura Jun 22 at 15:22

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