I'm reading this [paper](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/s3-55.1.59) and at page 67, he states that for any line bundle $L$ over a Rieman surface there is a connection $A$ whose curvature is
$$
F(A)=(\deg L)\omega,
$$
where $\omega$ is a positive form. Does anyone know how can I prove it without using algebraic geometry arguments?



Existence of a connection $A$ on a holomorphic line bundle $L$, s.t $F(A)=(\deg L)\omega$
Let me expand a bit Henri's (hi Henri!) answer, even if this is completely standard. In general, given a compact Kähler manifold $X$ of any dimension, given a holomorphic line bundle $L\to X$, and given any $(1,1)$-form $\eta$ representing $c\_1(L)$, there exists a smooth hermitian metric $h\_\eta$ on $L$ such that its Chern curvature satisfies
$$
\frac i{2\pi}\Theta(L,h\_\eta)=\eta.
$$
The proof goes as Henri was saying, that is: fix any hermitian metric $h$ on $L$, take its Chern curvature form, then $\frac i{2\pi}\Theta(L,h)\in c\_1(L)$. But then, $\frac i{2\pi}\Theta(L,h)-\eta$ is an exact form, and the $\partial\bar\partial$-lemma tells us that there exists a real function $f$ on $X$ such that 
$$
\frac i{2\pi}\Theta(L,h)-\eta=\frac i{2\pi}\partial\bar\partial f.
$$
Now, make a conformal change of the metric $h$ using $f$ as follows: $h\_\eta:=e^fh$. The Chern curvature of this new metric satisfies 
$$
\frac i{2\pi}\Theta(L,e^fh)=-\frac i{2\pi}\partial\bar\partial f+\frac i{2\pi}\Theta(L,h)=\eta.
$$


To finish with, if your $X$ is a compact Riemann surface (you didn't mention explicitly compactness in your question, but I guess you just missed it? ) then $H^2(X,\mathbb Z)\simeq\mathbb Z$. Thus (I also also imagine that your Kähler class $[\omega]$ was chosen to be a positive generator of $H^2(X,\mathbb Z)$, otherwise what you say is trivially false...), $c\_1(L)=\deg L$ and $\deg L\,\omega$ is in the right cohomology class (namely $c\_1(L)$), so that in the preceding argument you can replace $\eta$ by $\deg L\,\omega$.


