Let $C$ be an algebraic curve over an arbitrary field $k$ and $D= \sum n_i[c_i]$ a divisor of $C$.
I am trying to understand what the global sections of $\Omega^1_{C/k}(D)$ are. I read (and this is also what for me makes seem to make sense from the definition $\mathcal{O}_C(D) \otimes_{ \mathcal{O}_C}\Omega^1_{C/k}$) that for a uniformizer $z_i$ for a point $c_i$ a section looks locally like $f(z_i)dz_i$ with $f \in k(c_i)$ having a pole of order at most $n_i$ if $n_i \geq 0$ and a zero of order at least $-n_i$ if $n_i < 0$.
But would this imply that $\Gamma \Omega^1_{C/k}(D) \subset$ $\Gamma\Omega^1_{C/k}(D+[n_i])$? This conclusion looks straight forward to me, but from what I found in literature it often looks like that if the degree of $D$ is not "correct" then there are no non-trivial global sections and the existence of such a section serves as a certificate for some nice properties of such a divisor.
this looks like an application of Riemann-Roch. Let $C$ be a complete non-singular curve over $k$ alg. closed. Let $K$ denote the divisor class of $\Omega^1_C$.
Then Riemann Roch tells you that $dim\: \Gamma(K+D)+dim \:\Gamma(K-(K+D))=deg(K)+deg(D)+1-g$, where $g$ is the genus of $C$ which is the dimension of $\Gamma(C,\Omega^1_C)$. Now on the LHS, the second term is just $dim \Gamma(C,\mathcal{O}_C(-D))$. If $D$ has positive degree, this vector space is zero, because a line bundle of negative degree can't have non-zero global sections. Considering the RHS, you see that if you twist $D$ further, i.e. increase the degree of $D$, the other terms remain the same and $\Gamma(K+D)$ becomes bigger.
Does this help?