# Proof that $H^{0,1} \oplus H^{1,0} = H_{DR}^1$

I am struggling with a proof from Donaldson's Riemann Surfaces which he leaves as an exercise. we want to construct an isomorphism from the direct sum of $H^{1,0}(X)$, the set of holomorphic 1-forms on a riemann-surface $X$, direct summed with $H^{0,1}$, the cosets of $\overline{\partial}$-closed 1-forms modulo $\overline{\partial}$-exact forms. (alternatively, the first sheaf cohomology $H^1(\mathscr{O})$ of holomorphic functions.)

The isomorphism we construct is this: we know that conjugation induces a map (and isomorphism from $H^{1,0}$ to $H^{0,1}$ from a representative of a class in $H^{0,1}$ to a holomorphic 1-form and vice-versa. We define a map $\phi$ from holomorphic 1-forms to their de-rham cohomology classes. now we take the map $\Psi(\alpha, [\beta]) = \phi(\alpha) + \overline{\phi(\overline{\beta'})}$ where $\beta'$ is a representative of $\beta$.

I really am not sure how to prove this is an isomorphism, however.

• There is a natural inverse map. Hint: any form on a Riemann surface is a sum of the form $f dz + g d\overline{z}$. See for example the beginning of Griffiths & Harris. – Mose Wintner Aug 27 '15 at 0:46

This does follow from the Hodge decomposition, which lets you view the first de Rham cohomology as the set of harmonic 1-forms. The map you are discussing can be viewed as $\Omega^{1,0} \oplus \overline{\Omega^{1,0}} \rightarrow \mathscr{H}^{(1)}$. This is then a canonical map, and you can see it is an isomorphism pretty easily.

• Is there any way I can do this more directly without using the Hodge decomposition? – user264059 Aug 27 '15 at 0:56

In response to user264059's comment, the decomposition you presented in your original post is generally called the Hodge decomposition, so it is not really clear what you are looking for here. (A more direct proof of the Hodge decomposition than what?)

That said, there are several approaches.

The theory of elliptic PDE's gives a proof of the Hodge decomposition on any Kähler manifold, in particular a compact Riemann surface. See Griffiths-Harris's Principles of Algebraic Geometry or Voisin's Hodge Theory and Complex Algebraic Geometry, for example.

There are simpler proofs of the Hodge decomposition on a Riemann surface based on solving the equation $\Delta f = g$, where $\Delta$ is the Laplace operator. See e.g. Donaldson's Riemann Surfaces (as you have already mentioned) or Narasimhan's Compact Riemann Surfaces .

The fact that $\dim H^{1,0} = g$ implies easily the Hodge decomposition, and this follows from the Riemann-Roch theorem. Usually this fact is used in the proof of Riemann-Roch so this is circular reasoning, but it is possible to prove Riemann-Roch directly after first solving the Dirichlet problem on Riemann surfaces with boundary. Reyssat's Quelques Aspects des Surfaces de Riemann takes this approach.