If we fix a complex structure $c$ on a closed oriented surface $S$, then the space of holomorphic 1-forms $\Omega^1(S,c)$ has complex dimension equal to the genus $g$ of $S$ and - since they are all closed and don't differ by an exact form - we can see $\Omega^1(S,c)$ as a subspace of the de Rham-cohomology group $H^1(S, \mathbb C)$, which has complex dimension $2g$.
If I consider a different complex structure $c'$ on $S$ (inducing the same orientation), would I get the same vector subset? In other words: for all complex structures $c, c'$ on $S$ and for any holomorphic 1-form for $(S,c)$, can I find a holomorphic 1-form for $(S,c')$ that differs from the former by an exact form?
I don't want to push it too far (you can stop here if you are not confident with the next part), but it seems to me that, by de Rham theorem, the vector subspace is the same if the complex structures differ by an isotopy, so one seems to have a map from the Teichmuller space $\mathcal T(S)$ to the Grassmanian of $g$-dimensional subspaces of $H^1(S, \mathbb C)$. Does anyone know what this map looks like?
Thanks a lot, any help would be great!