Let $S$ be a closed real surface having two complex structures $c\_1$ and $c\_2$ which are not biholomorphic (so $S$ is a Riemann surface with genus at least 1). Consider $\omega$ a 1-form on $S$ which is holomorphic for both $c\_1$ and $c\_2$. Can we conclude that $\omega=0$ ?


Intuitively I would proceed as follows. Since $\omega$ is holomorphic for $c\_1$, I would write $\omega= f(z) dz$ in conformal coordinates around a given point in $S$. Then, I would compare the Cauchy–Riemann equations for $c\_1$ and $c\_2$ : since $c\_1$ and $c\_2$ are not biholomorphic, the Cauchy–Riemann equations should take the form $\partial\_{\bar{z}} f = 0$ for $c\_1$ and some linear combination $\alpha \partial\_x f + \beta \partial\_y f=0$, with $\beta/\alpha \neq -i$ for $c\_2$. So the derivatives of $f$ vanish.  

So $\omega$ should be proportional to $dz$, in every choice of conformal coordinates for $c\_1$. But the same also holds for $c\_2$. So the only possibility should be $\omega=0$.



Common holomorphic forms for two distinct complex structures
Any $C^1$ function $g$ on a connected Riemann surface is holomorphic if and only if $dg\wedge \omega=0$, for $\omega$ holomorphic and not the zero form. Where $\omega\ne 0$ this is clear, expanding out in a holomorphic coordinate, say $\omega=f(z)\,dz$, and then integrating along curves to construct a local holomorphic coordinate $Z=\int f(z)\,dz$ so that $\omega=dZ$. Expand out to find that if $g=u+iv$ and $Z=x+iy$ then
$$dg\wedge\omega=du\wedge dx-dv\wedge dy+i(du\wedge dy+dv\wedge dx),\\=((-u\_y-v\_x)+i(u\_x-v\_y))dx\wedge dy,$$
vanishing just when the Cauchy--Riemann equations are satisfied. Elsewhere it follows by continuity. Hence the complex structures are identical.


