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I want to show that the space of smooth $(1,0)$ forms on a compact Riemann surface $X$ has the natural splitting: $\mathcal{E}^{1,0}(X)=\Omega(X) \oplus \partial\mathcal{E}^{0}(X)$, where $\Omega(X)$ is the space of holomorphic 1-forms on $X$, $\mathcal{E}^{0}(X)$ is the space of smooth 0-forms on $X$ (i.e. smooth functions), and $\partial$ comes from the splitting $d=\partial + \bar{\partial}$.

How can I show this? I am a little confused because it seems that $\Omega(X)$ is identically $\mathcal{E}^{1,0}(X)$ since the condition for a form to be holomorphic on $X$ is for it to be type $(1,0)$. So what exactly does $\partial\mathcal{E}^{0}(X)$ do?

Using this result, I want to show that $H^{1}(X,\mathcal{H}) \cong H^{1,1}(X)$ where $\mathcal{H}$ is the sheaf of complex valued harmonic functions on $X$. I can show this directly, but I'm not sure how to use the above result. Any help would be appreciated!

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Here's the point to get you started: $\Omega(X)$ is a proper subset of $\mathcal E^{1,0}(X)$, as the latter includes all forms that look like $f dz$ for any $C^\infty$ function $f$. If you start with an arbitary smooth function $g$, $dg = \partial g + \overline\partial g = \dfrac{\partial g}{\partial z}dz+\dfrac{\partial g}{\partial \bar z}d\bar z$. Of course, $\overline\partial g = 0$ if and only if $g$ is holomorphic.

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  • $\begingroup$ Thank you for the start! But after proving this, how do we show the desired isomorphism? I know that there is a long exact sequence of sheaves $0 \rightarrow \mathcal{H}(X) \rightarrow \mathcal{E}^{0} \rightarrow \mathcal{E}^{2} \rightarrow H^{1}(X,\mathcal{H}) \rightarrow 0$ where the third and fourth arrows are given by $\partial\bar{\partial}$ and $\Delta$ respectively. This gives an isomorphism $\mathcal{E}^{2}/\partial\bar{\partial}(\mathcal{E}^{0}) \cong H^{1}(X,\mathcal{H})$. Where does the splitting fit into this isomorphism? $\endgroup$ – user 3462 Mar 31 '14 at 19:28
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    $\begingroup$ In this comment you're confusing sheaves and global sections. So, what tools do we have at our disposal? $\endgroup$ – Ted Shifrin Mar 31 '14 at 20:47
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Regarding the difference between $\mathcal E^{1,0}$ and $\Omega$:

$\mathcal E^{1,0}(X)$ consists of forms forms locally of the shape $f(z)dz,$ where $z$ is a holomorphic local coord., and $f$ is a smooth function of $z$.

$\Omega(X)$ consists of holomorphic $1$ forms, so of the shape $f(z) dz$ where $f$ is a holomorphic function.

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