I just begin to learn about Hodge Theory. The following statement is heard from somewhere, which I know it is true. But I don't understand the exact detail of how and why.

Let $M$ be a closed Riemann Surface. Then $$H^1(M,\mathbb{C})=\mathcal H^{0,1}(M)\oplus \mathcal H^{1,0}(M),$$ where $\mathcal H$ is the space of harmonic forms.

The question is why then $\dim \mathcal H^{1,0}(M,\mathbb{C})=\operatorname{genus}(M)$? How to understand it intuitively, and how can I prove it?

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    $\begingroup$ Because it is well-known that $dim H^1(M)=2g$ and taking conjugation is an isomorphism between linear spaces $\mathcal H^{i,j}=\mathcal H^{j,i}$. $\endgroup$ – Akatsuki Jan 1 '18 at 23:52

There is an isomorphism $\mathcal H^{p,q}(M) \simeq H^{p,q}(M) \simeq H^q(M, \Omega_M^p)$ which you can find in any textbook deriving the Hodge decomposition for Kahler manifolds (Voisin's "Hodge Theory and Complex Algebraic Geometry", or Carlson-Muller Stach-Peters' "Period Mappings and Period Domains", for example).

Riemann surfaces are curves and therefore have the classic Riemann-Roch formula associated to them $$h^0(M, \mathcal O(D)) = \deg D + 1 - g + h^0(M, \Omega_M^1).$$ Notice that the last term on the right hand side is $\dim H^{1,0}(M)$. It is easy to see that $h^0(M, \Omega_M^1) = g$ by this formula when you look at $D = 0$.

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    $\begingroup$ Another good reference is Daniel Huybrecht's book "Complex Geometry". $\endgroup$ – user347489 Apr 15 '18 at 1:04

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