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Let $X$ be a Riemann surface. According to Griffiths and Harris "Principles of Algebraic Geometry", the complexified tangent space of $X$ at $p\in X$ is \begin{equation} \begin{split} T_{\mathbb{C},p}X=\{\xi:\mathcal{C}^{\infty}(X,\mathbb{C})_p\rightarrow\mathbb{C}:\xi(\lambda f+\mu g)=\lambda \xi(f)+\mu\xi(g)\text{ and}\\ \xi(fg)=g(p)\xi(f)+f(p)\xi(g)\text{ }\forall f,g\in\mathcal{C}^\infty(X,\mathbb{C})_p,\text{ }\forall\lambda,\mu\in\mathbb{C}\}. \end{split} \end{equation} We can consider its dual $T^*_{\mathbb{C},p}X=\{\alpha:T_{\mathbb{C},p}X\rightarrow \mathbb{C}:\alpha\text{ is $\mathbb{C}$-linear}\}$. Then, $\mathcal{C}^\infty$ $1$-forms are the sections of the bundle, $$ T^*_\mathbb{C}X=\{(p,\alpha):p\in X, \alpha\in T^*_{\mathbb{C},p}X\}. $$ In a chart they have the form $$ f\cdot\mathrm{d}x+g\cdot\mathrm{d}y $$ for some complex-valued $\mathcal{C}^\infty$ functions $f,g$.

On the other hand, we have the subbundle $$ \{(p,\alpha):p\in X, \alpha\in\mathbb{C}\cdot (\mathrm{d}z)_p\subseteq T^*_{\mathbb{C},p}X\}, $$ where $(\mathrm{d}z)_p=(\mathrm{d}x)_p+i\cdot(\mathrm{d}y)_p$. If I am not wrong, type $(1,0)$ $\mathcal{C}^\infty$ $1$-forms are the sections of this subbundle. Hence, in a chart they have the form $$ f\cdot dz $$ for some complex-valued $\mathcal{C}^\infty$ function $f$.

Now, holomorphic $1$-forms are of the type $$ f\cdot dz $$ for some holomorphic function $f$.

I guess they are sections of some bundle (related with the previous one). What bundle are we talking about?

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    $\begingroup$ The complexified cotangent bundle; take a look in Griffiths and Harris "Principles of Algebraic Geometry". $\endgroup$ – Moishe Kohan Jul 2 '17 at 10:54
  • $\begingroup$ Dear @MoisheCohen, first of all I wanted to thank you for the suggestion. Now, I have been reading Griffiths and Harris "Principles of Algebraic Geometry", but it is not clear for me yet. I have edited the question with my new doubts. $\endgroup$ – Zé Pequeno Jul 3 '17 at 8:22
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    $\begingroup$ No, there is no reason for $f$ to be holomorphic for general $(1,0)$ forms $fdz$. $\endgroup$ – Moishe Kohan Jul 3 '17 at 10:19
  • $\begingroup$ @MoisheCohen edited again. $\endgroup$ – Zé Pequeno Jul 3 '17 at 10:57
  • $\begingroup$ @ZéPequeno: It's not clear to me from the edits what your confusion is. A $(1, 0)$-form is a smooth section of the $(1, 0)$-subbundle of the complexified cotangent bundle. A holomorphic $1$-form is a particular type of smooth $(1, 0)$-form, i.e., a section of the same complex line bundle. Since the $(1, 0)$-cotangent bundle is rank one, it has no proper, non-trivial (complex) subbundles. $\endgroup$ – Andrew D. Hwang Jul 3 '17 at 14:18

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