Is $H^1(X,\mathbb{C})$ the cohomology group or the vector space of holomorphic differentials in a Riemann surface $X$? I am an undergraduate studying Riemann surfaces so I don't have a huge background in algebra. As far as I know, $H^1(X,\mathbb{C})$ is the cohomology group of $X$ with coefficients in $\mathbb{C}$. However, in some books I saw that people also call $H^1(X,\mathbb{C})$ the complex vector space of holomorphic differentials. I don't see the relationship between these objects.
Are they really the same object? If so, why?
 A: Edit As peter a g points out below, Hodge theory tells us that $H^1(X,\Bbb{C})=H^{(1,0)}(X,\Bbb{C})\oplus H^{(0,1)}(X,\Bbb{C})$, where $H^{(1,0)}$ are the holomorphic 1-forms and $H^{(0,1)}$ are the antiholomorphic 1-forms. I've got to figure out where exactly I lost the antiholomorphic 1-forms below. 
Edit 2 Oh, I figured it out. I was being silly and starting with the assumption that my 1-form was a multiple of $dz$ to begin with. So the story is a bit more complicated than I originally described, but the general idea is correct.
More or less fixed version:
By de Rham's theorem, we can regard $H^1(X,\Bbb{C})$ as being the complexification of $\newcommand{\dR}{\text{dR}}H^1_{\dR}(X)$, which is the set of closed differential 1-forms with coefficients in $\Bbb{C}$ modulo the set of exact differential 1-forms with coefficients in $\Bbb{C}$.
Working locally, we may assume that we have $U\subseteq \Bbb{C}$. When is a differential $1$-form $P\,dx + Q\, dy$ closed? Well, it's precisely when $Q_x-P_y=0$, since
$$d(P\,dx+Q\, dy)=(Q_x-P_y)\,dx\wedge dy.$$
Now since we are doing complex analysis, it's nicer to use the basis $dz = dx+i\,dy$ and $d\bar{z}=dx-i\,dy$ for the basis for our differential forms.
We can therefore write any complex differential form locally as $f \,dz + g\, d\bar{z}$ (where $f,g$ are smooth functions to $\Bbb{C}$).
So let's now consider what happens when $g=0$, so we just have $f=u+iv$.
Let's apply the condition for a differential form to be closed to the differential form 
$$f\,dz=(u+iv)(dx+i\,dy)=(u+iv)\, dx +(-v+ui)\, dy.$$
This differential form is closed when $(-v+ui)_x = (u+iv)_y$, i.e., when $u_y=-v_x$ and $u_x = v_y$.
However these equations are precisely the Cauchy-Riemann equations, so a differential 1-form of the form $f\,dz$ on $X$ is closed if and only if it is holomorphic.
Now it would be nice if we could always decompose our forms into $f\,dz+g\,d\bar{z}$ such that the differential form is closed if and only if both pieces are closed. This is not always possible, but it is possible when $X$ is a compact Kähler manifold. This is the content of the Hodge theorem.
I think I'll have to come back and check this last part, but it should be more or less correct
Then the explanation for why $H^{(1,0)}(X)$ is the space of holomorphic $1$-forms when $X$ is a compact Kähler manifold is more or less the same as my prior explanation, since holomorphic $1$-forms are the $1$-forms locally of the form $f\, dz$, with $f$ holomorphic.
Then the exact forms are the forms $$\frac{\partial f}{\partial z} \,dz + \frac{\partial f}{\partial \bar{z}}\, d\bar{z},$$
and when $\frac{\partial f}{partial \bar{z}}=0$, $f$ is holomorphic. However since $X$ is compact, global holomorphic functions are constants. Thus there are no nonzero exact holomorphic 1-forms.
Note
It's safe to write $\frac{\partial f}{\partial z}$ even though a priori we only know $f$ is $C^\infty$ because if $f=u+iv$, then 
$$df = f_x\,dx + f_y\,dy = f_x\,dx +(-if_y)i\,dy$$
$$= \frac{f_x-if_y}{2} \, dz + \frac{f_x+if_y}{2} \, d\bar{z}.$$
Then if we know $f_x+if_y=0$, then if $f=u+iv$, we get $u_x+iv_x = -iu_y + v_y$, which are once again the Cauchy-Riemann equations. Thus by the decomposition of closed forms, if an exact form $df$ ends up in $\Omega^{(1,0)}$, $f$ must have been holomorphic.
