[proofplan] The proof compares positivity of curvature with the cohomological degree of the line bundle. One direction follows by integrating the positive Chern form, using that every Chern form of a Hermitian metric represents the first Chern class and has total integral equal to $\deg L$. For the converse, we choose a positive smooth area form with the prescribed total integral and use the compact-surface Hodge theorem to modify an arbitrary Hermitian metric by a smooth conformal factor so that its Chern form becomes exactly that area form. The divisor statement then follows from the identity $\deg \mathcal{O}_C(D)=\deg D$. [/proofplan]

[step:Relate the degree of $L$ to the integral of any Chern curvature form]Let $h$ be a smooth Hermitian metric on the holomorphic line bundle $L \to C$. Its normalized Chern curvature form is the real smooth $(1,1)$-form $c_1(L,h)$ representing the first Chern class $c_1(L) \in H^2(C;\mathbb{R})$. Fix a smooth positive area form $\Omega$ on $C$, and let $\mu_\Omega$ denote the positive measure induced by $\Omega$. For every smooth real $2$-form $\beta$ on $C$, define its density with respect to $\Omega$ to be the smooth map \begin{align*} f_\beta:C &\to \mathbb{R} \end{align*} characterized by $\beta=f_\beta\,\Omega$. By the [Chern-Weil degree formula](/page/Chern-Weil%20Theory) for holomorphic line bundles on compact Riemann surfaces, applied with the complex orientation on $C$, \begin{align*} \int_C f_{c_1(L,h)}\,d\mu_\Omega = \deg L. \end{align*} This density integral is independent of the auxiliary positive area form $\Omega$ and is the de Rham integral of the top-degree form $c_1(L,h)$. The left-hand side in the degree formula is independent of the metric because two Chern forms for two smooth Hermitian metrics differ by an exact real $(1,1)$-form.[/step]

[guided]Fix a smooth Hermitian metric $h$ on $L$. The associated normalized Chern curvature form $c_1(L,h)$ is a real smooth $(1,1)$-form on $C$. Choose a positive smooth area form $\Omega$ on $C$, and let $\mu_\Omega$ be the positive measure induced by $\Omega$. For every smooth real $2$-form $\beta$ on $C$, define the density map \begin{align*} f_\beta:C &\to \mathbb{R} \end{align*} by the identity $\beta=f_\beta\,\Omega$. The [Chern-Weil degree formula](/page/Chern-Weil%20Theory) for holomorphic line bundles on compact Riemann surfaces says that $c_1(L,h)$ represents the first Chern class of $L$ and that its density integral computes the degree: \begin{align*} \int_C f_{c_1(L,h)}\,d\mu_\Omega = \deg L. \end{align*} This is the bridge between the analytic condition, namely positivity of the normalized Chern curvature form, and the algebro-geometric invariant $\deg L$. We also record why the metric chosen here is irrelevant. If $h_1$ and $h_2$ are two smooth Hermitian metrics on $L$, then locally their ratio has the form $h_2 = h_1 e^{-u}$ for a smooth real-valued function $u$, and the corresponding Chern forms differ by an exact real $(1,1)$-form. Since $C$ is compact without boundary, the integral of an exact top-degree form over $C$ is zero by [Stokes' theorem](/theorems/1530). Hence every smooth Hermitian metric gives the same integral, namely $\deg L$.[/guided]

[step:Integrate a positive curvature form to prove positive degree]Assume that $L$ admits a smooth Hermitian metric $h$ such that $c_1(L,h)$ is positive at every point of $C$. Since $C$ is connected and compact, a positive smooth $(1,1)$-form on $C$ is a positive smooth area form. With respect to the fixed auxiliary area form $\Omega$, its density $f_{c_1(L,h)}:C\to(0,\infty)$ is smooth and positive, so its total measure integral is strictly positive: \begin{align*} \int_C f_{c_1(L,h)}\,d\mu_\Omega > 0. \end{align*} Using the degree formula from the previous step gives \begin{align*} \deg L = \int_C f_{c_1(L,h)}\,d\mu_\Omega > 0. \end{align*}[/step]

[guided]Suppose $h$ is a smooth Hermitian metric whose Chern form $c_1(L,h)$ is positive everywhere. On a Riemann surface, a real $(1,1)$-form is a top-degree real $2$-form. Positivity means that in every holomorphic coordinate chart $(U,z)$, with $z=x+iy$, the form is locally a positive smooth multiple of the standard orientation form: \begin{align*} c_1(L,h)|_U = a_U\, d\mathcal{L}^2(x,y), \end{align*} where $a_U:U \to (0,\infty)$ is smooth and $d\mathcal{L}^2(x,y)$ denotes the local Lebesgue area element in the coordinate chart. Since $C$ is compact and the local density is everywhere positive, the resulting global area form has strictly positive total measure integral. Using the auxiliary area form $\Omega$ and the density map $f_{c_1(L,h)}:C\to(0,\infty)$ defined by $c_1(L,h)=f_{c_1(L,h)}\Omega$, this is \begin{align*} \int_C f_{c_1(L,h)}\,d\mu_\Omega > 0. \end{align*} The degree formula then converts this analytic positivity into the numerical statement \begin{align*} \deg L = \int_C f_{c_1(L,h)}\,d\mu_\Omega > 0. \end{align*} This proves the necessary direction.[/guided]

[step:Choose a positive area form with total integral equal to $\deg L$] Assume now that $\deg L>0$. Choose any smooth positive area form $\omega_0$ on $C$. Let $\mu_{\omega_0}$ denote the positive measure induced by the area form $\omega_0$. Define the positive real constant \begin{align*} A := \int_C 1\,d\mu_{\omega_0}. \end{align*} Since $\omega_0$ is positive and $C$ is compact and nonempty, $A>0$. Define \begin{align*} \omega := \frac{\deg L}{A}\,\omega_0. \end{align*} Then $\omega$ is a smooth positive real $(1,1)$-form on $C$. If $\mu_\omega$ is the positive measure associated to the area form $\omega$, then \begin{align*} \int_C 1\,d\mu_\omega = \frac{\deg L}{A}\int_C 1\,d\mu_{\omega_0} = \deg L. \end{align*} [/step]

[step:Modify an arbitrary Hermitian metric to prescribe its Chern form]Let $h_0$ be any smooth Hermitian metric on $L$. Define the smooth real $(1,1)$-form \begin{align*} \alpha := \omega - c_1(L,h_0). \end{align*} Define the smooth density map \begin{align*} f_\alpha:C &\to \mathbb{R} \end{align*} by $\alpha=f_\alpha\,\Omega$. By the degree formula and the construction of $\omega$, \begin{align*} \int_C f_\alpha\,d\mu_\Omega = \int_C 1\,d\mu_\omega - \int_C f_{c_1(L,h_0)}\,d\mu_\Omega = \deg L - \deg L = 0. \end{align*} The hypotheses for the [compact-surface Hodge solvability theorem](/page/Hodge%20Theory) are satisfied: $C$ is compact and connected, the complex structure orients $C$, and $\alpha$ is a smooth real top-degree form whose density has total measure integral zero. With the Chern-form normalization used here, that statement gives a smooth real-valued function $u:C\to\mathbb{R}$ such that $\frac{i}{2\pi}\partial\bar{\partial}u=\alpha$. Thus choose $u \in C^\infty(C;\mathbb{R})$ such that \begin{align*} \frac{i}{2\pi}\partial\bar{\partial}u = \alpha. \end{align*} Define a new smooth Hermitian metric $h$ on $L$ by \begin{align*} h := h_0 e^{-u}. \end{align*} The conformal transformation law for Chern forms gives \begin{align*} c_1(L,h) = c_1(L,h_0) + \frac{i}{2\pi}\partial\bar{\partial}u = c_1(L,h_0) + \alpha = \omega. \end{align*} Since $\omega$ is positive everywhere, $h$ has positive curvature.[/step]

[guided]We begin with an arbitrary smooth Hermitian metric $h_0$ on $L$. The goal is to change $h_0$ by multiplying it by a positive smooth function so that the new Chern form is the already chosen positive form $\omega$. Define \begin{align*} \alpha := \omega - c_1(L,h_0). \end{align*} This is a smooth real $(1,1)$-form on $C$. Define the smooth density map \begin{align*} f_\alpha:C &\to \mathbb{R} \end{align*} by $\alpha=f_\alpha\,\Omega$. Its total measure integral is zero, because both terms have the same total integral: \begin{align*} \int_C f_\alpha\,d\mu_\Omega = \int_C 1\,d\mu_\omega - \int_C f_{c_1(L,h_0)}\,d\mu_\Omega = \deg L - \deg L = 0. \end{align*} Here $\int_C 1\,d\mu_\omega=\deg L$ by construction, while $\int_C f_{c_1(L,h_0)}\,d\mu_\Omega=\deg L$ by the [Chern-Weil degree formula](/page/Chern-Weil%20Theory). Now we use the compactness of $C$. The [compact-surface Hodge solvability theorem](/page/Hodge%20Theory) says that a smooth real top-degree form has zero total measure integral exactly when it is $\frac{i}{2\pi}\partial\bar{\partial}$ of a smooth real-valued function, with this sign and normalization matching the Chern-form convention used in the conformal change formula below. The hypotheses are met: $C$ is compact and connected, its complex structure supplies the orientation, and $\alpha$ is a smooth real top-degree form with $\int_C f_\alpha\,d\mu_\Omega=0$. Therefore there exists a function $u \in C^\infty(C;\mathbb{R})$ such that \begin{align*} \frac{i}{2\pi}\partial\bar{\partial}u = \alpha. \end{align*} This is precisely the scalar Laplace equation hidden in the proof: on a Riemann surface, the operator $u \mapsto i\partial\bar{\partial}u$ is a constant multiple of the Laplacian times the area form. Define the new Hermitian metric \begin{align*} h := h_0 e^{-u}. \end{align*} This is again a smooth Hermitian metric because $e^{-u}:C\to(0,\infty)$ is a positive smooth function. The curvature transformation formula for a conformal change of Hermitian metric gives \begin{align*} c_1(L,h) = c_1(L,h_0) + \frac{i}{2\pi}\partial\bar{\partial}u. \end{align*} Substituting the equation satisfied by $u$ yields \begin{align*} c_1(L,h) = c_1(L,h_0) + \alpha = c_1(L,h_0) + \omega - c_1(L,h_0) = \omega. \end{align*} Since $\omega$ is positive everywhere, this metric $h$ has positive curvature.[/guided]

[step:Apply the criterion to effective divisors of positive degree] Let $D$ be an effective divisor on $C$ with $\deg D>0$. The associated holomorphic line bundle $\mathcal{O}_C(D)$ satisfies \begin{align*} \deg \mathcal{O}_C(D) = \deg D. \end{align*} Hence \begin{align*} \deg \mathcal{O}_C(D) > 0. \end{align*} By the equivalence just proved, $\mathcal{O}_C(D)$ admits a smooth Hermitian metric with positive curvature. Therefore $\mathcal{O}_C(D)$ is positive. [/step]