# Degree of canonical divisor on curve, higher order

If $X$ is a smooth algebraic curve (over $\mathbb{C}$) with genus $g$, and $\omega\in \Omega^1(X)$ is a meromorphic 1-form, then it is a corollary of the Riemann-Roch theorem that $deg(div(\omega)) = 2g-2$.

My question is, if $\omega \in \Omega^{k}(X)$ is a higher order differential, what sort of estimates can we give to $deg(div(w))$ ?

-
The notation $\Omega^k$ generally means sheaf of regular $k$-forms on $X$ and for a curve it is zero for all $k\geq 2$.
What you probably mean is the tensor product $\Omega ^{\otimes k}$ of $\Omega ^{1}$ with itself $k$ times.
In that case you have for $w\in \Omega ^{\otimes k}(X)$ the formula $$\operatorname {deg}(\operatorname {div}(w))=\operatorname {deg} \Omega ^{\otimes k}=k(2g-2)$$
This formula is valid for all $k\in \mathbb Z$ and also for all $w\in \Omega _{rat} ^{\otimes k}(X)$ where $\Omega _{rat} ^{\otimes k}=\Omega^{\otimes k} \otimes _{\mathcal O_X}\mathcal K_X$.
Here $\mathcal K_X$ is the sheaf of rational functions and the pleasant thing is that there always exist non-zero $w\in \Omega _{rat} ^{\otimes k}(X)$ for which you can calculate $\operatorname {deg}(\operatorname {div}(w))$, whereas you might have $\Omega ^{\otimes k}(X) =0$