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I am looking for a proof of this fact, where $H^{1,0} = Ker(d: \mathscr{E}^{1,0} \rightarrow \mathscr{E}^{2})$ and $H^{0,1} = Coker(\overline{\partial}: \mathscr{E} \rightarrow \mathscr{E}^{0,1}$, (These are the dolbeault cohomology groups) where we are working on a riemann-surface $X$. This statement is equivalent to $H^0(\Omega) = H^1(\mathscr{O})$, where the latter two are the sheaf cohomology groups of the sheaf of holomorphic 1-forms and holomorphic differentials, respectively. I have read the proof in Donaldson, in which he first establishes an isomorphism $\alpha \rightarrow \overline{\alpha}$ that takes $H^{1,0}$ to $\overline{H^{0,1}}$, and then composes that with the inner product $\langle \alpha, \beta \rangle = \int_X \alpha \wedge \beta$, but I don't understand how we know that this original conjugation map produces an isomorphism. Another proof I have seen uses the theory of distributions, but was incomplete.

I would greatly appreciate a proof or a reference to a complete proof, preferably one that doesn't use too much heavy machinery (The proofs I've seen only seem to rely on a couple of facts form analysis.

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  • $\begingroup$ Doesn't this follow from Serre duality? $\endgroup$ – user40276 Aug 25 '15 at 7:07
  • $\begingroup$ Yes, I am trying to prove it in a more elementary manner. Sorry, I should have said. I am looking for a proof which hopefully constructs an explicit isomorphism in an analytical manner. $\endgroup$ – user264059 Aug 25 '15 at 7:28
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    $\begingroup$ I thought Donaldson's proof was fairly complete—is the issue that he uses the Main Theorem that isn't proved until later? $\endgroup$ – Takumi Murayama Aug 26 '15 at 2:27
  • $\begingroup$ Well, do you understand the statement of Theorem 5, the "Main Theorem"? Donaldson uses the $\Leftarrow$ direction of the statement to prove surjectivity of $\sigma$. Injectivity is shown by composing with the inner product, which is positive definite. The hard part is probably understanding the proof of Theorem 5, which is Chapter 9, and this uses "a couple of facts from analysis." $\endgroup$ – Takumi Murayama Aug 27 '15 at 8:13
  • $\begingroup$ I actually do understand it now. I was being silly while reading the proof at first. thanks! $\endgroup$ – user264059 Aug 28 '15 at 22:52

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