# An algebraic proof: A line bundle on a curve with a connection must be of degree 0

Let me state it using the language of $$D$$-modules. Let $$X$$ be a smooth projective curve and $$\mathcal{L}$$ a line bundle on it. Assume that $$\mathcal{L}$$ has a left action of $$\mathcal{D}_X$$. Then show that $$\mathrm{deg}\, \mathcal{L} = 0$$.

This is true for $$\mathcal{L}=\mathcal{O}_X$$ clearly. And this question is purely algebraic. Though it has a proof based in analytic method over $$\mathbb{C}$$, I would like to ask for a proof in algebraic method, maybe available in positive characteristic.

• If $F:C\to C$ is the absolute Frobenius morphism and $C$ is a proper smooth curve over the finite field ${\bf F}_p$, then for any line bundle $L$ on $C$, there is a connection on $F^*(L)=L^{\otimes p}$ but in general one does not have ${\rm deg}(F^*(L))=0$. I think one needs some form of the Riemann-Hilbert correspondence to prove the statement you consider in char. $0$, so analytic methods might be required. Oct 26, 2021 at 14:21
• see Biswas article dx.doi.org/10.1007/s10977-005-4708-8 which deals with vector bundles, also in positive characteristic Oct 26, 2021 at 14:25
• $c_1(L)\in H^1(X, \Omega^1)$ is $\pm$ the class of the $\Omega^1$-torsor of connections on $L$. So, this class is zero if, and only if, $L$ admits a connection. For a smooth projective curve $H^1(X, \Omega^1) = k$ and the image of $c_1(L)$ is the image of $\deg(L)$ under the map $\mathbb{Z}\rightarrow k$. In positive characteristic this means the degree has to be divisible by $p$, explaining Damian Rössler's example. Oct 26, 2021 at 14:27