[proofplan] We first construct the function $\psi$ pointwise as the negative logarithm of the ratio of the two Hermitian norms, and verify that this ratio is independent of the chosen nonzero vector in each line. Smoothness follows by computing the ratio in a local holomorphic frame. We then compare the local Chern potentials for $h_0$ and $h_1$: multiplying the metric by $e^{-\psi}$ adds $\psi$ to the local potential, so applying $\frac{i}{2\pi}\partial\bar\partial$ gives the asserted difference of Chern forms. Finally, the difference form is $d$-exact because $d=\partial+\bar\partial$ on complex-valued forms and $\bar\partial^2=0$. [/proofplan]

[step:Construct the global metric ratio and define $\psi$]For each point $p \in M$, choose any nonzero vector $v \in L_p$. Define \begin{align*} \rho(p) := \frac{h_{1,p}(v,v)}{h_{0,p}(v,v)}. \end{align*} This number is independent of the chosen nonzero vector $v$. Indeed, since $L_p$ is a one-dimensional complex [vector space](/page/Vector%20Space), every other nonzero vector has the form $w=\lambda v$ for some $\lambda \in \mathbb{C}\setminus\{0\}$. Hermitian sesquilinearity gives \begin{align*} \frac{h_{1,p}(w,w)}{h_{0,p}(w,w)} = \frac{h_{1,p}(\lambda v,\lambda v)}{h_{0,p}(\lambda v,\lambda v)} = \frac{|\lambda|^2 h_{1,p}(v,v)}{|\lambda|^2 h_{0,p}(v,v)} = \frac{h_{1,p}(v,v)}{h_{0,p}(v,v)}. \end{align*} Since both Hermitian metrics are positive definite on every fiber, $\rho(p)>0$ for every $p \in M$. Define the function \begin{align*} \psi: M &\to \mathbb{R} \\ p &\mapsto -\log \rho(p). \end{align*} Then $\rho=e^{-\psi}$, so for every $p \in M$ and every $v \in L_p$, \begin{align*} h_{1,p}(v,v)=e^{-\psi(p)}h_{0,p}(v,v). \end{align*} By sesquilinearity and polarization for Hermitian forms, this equality of quadratic forms implies \begin{align*} h_{1,p}=e^{-\psi(p)}h_{0,p} \end{align*} as Hermitian forms on $L_p$. Hence $h_1=e^{-\psi}h_0$ globally.[/step]

[guided]At a point $p \in M$, the two Hermitian metrics $h_{0,p}$ and $h_{1,p}$ are positive definite Hermitian forms on the same one-dimensional complex vector space $L_p$. Because the fiber is one-dimensional, the ratio of the two squared norms is forced to be a single positive scalar. Choose a nonzero vector $v \in L_p$ and define \begin{align*} \rho(p) := \frac{h_{1,p}(v,v)}{h_{0,p}(v,v)}. \end{align*} We must check that this does not depend on $v$. If $w \in L_p\setminus\{0\}$ is another choice, then $w=\lambda v$ for some $\lambda \in \mathbb{C}\setminus\{0\}$. Hermitian metrics satisfy $h_{j,p}(\lambda v,\lambda v)=|\lambda|^2h_{j,p}(v,v)$ for $j \in \{0,1\}$. Therefore \begin{align*} \frac{h_{1,p}(w,w)}{h_{0,p}(w,w)} = \frac{|\lambda|^2h_{1,p}(v,v)}{|\lambda|^2h_{0,p}(v,v)} = \frac{h_{1,p}(v,v)}{h_{0,p}(v,v)}. \end{align*} Thus $\rho: M \to (0,\infty)$ is well-defined. Now define \begin{align*} \psi: M &\to \mathbb{R} \\ p &\mapsto -\log \rho(p). \end{align*} The choice of the negative logarithm is exactly the choice that makes $\rho=e^{-\psi}$. Hence, for every $p \in M$ and every $v \in L_p$, \begin{align*} h_{1,p}(v,v)=\rho(p)h_{0,p}(v,v)=e^{-\psi(p)}h_{0,p}(v,v). \end{align*} Since Hermitian forms are determined by their values $h(v,v)$ through the polarization identity, equality on all vectors gives \begin{align*} h_{1,p}=e^{-\psi(p)}h_{0,p}. \end{align*} Since $p$ was arbitrary, $h_1=e^{-\psi}h_0$ on all of $L$.[/guided]

[step:Verify smoothness and uniqueness of $\psi$] Let $U \subset M$ be an [open set](/page/Open%20Set) over which $L$ admits a nowhere-vanishing holomorphic frame \begin{align*} e: U &\to L|_U. \end{align*} Define smooth positive functions \begin{align*} a_j: U &\to (0,\infty) \\ p &\mapsto h_{j,p}(e(p),e(p)), \end{align*} for $j \in \{0,1\}$. On $U$, \begin{align*} \rho(p)=\frac{a_1(p)}{a_0(p)}. \end{align*} Since $a_0$ and $a_1$ are smooth and $a_0>0$, $\rho|_U$ is smooth. Since the logarithm is smooth on $(0,\infty)$, $\psi|_U=-\log(\rho|_U)$ is smooth. These local frames cover $M$, so $\psi \in C^\infty(M;\mathbb{R})$. If $\widetilde{\psi}:M\to\mathbb{R}$ is another smooth function satisfying $h_1=e^{-\widetilde{\psi}}h_0$, then for every $p \in M$ and every nonzero $v \in L_p$, \begin{align*} e^{-\widetilde{\psi}(p)} = \frac{h_{1,p}(v,v)}{h_{0,p}(v,v)} = \rho(p) = e^{-\psi(p)}. \end{align*} The real exponential function is injective, so $\widetilde{\psi}(p)=\psi(p)$ for every $p \in M$. Thus $\psi$ is unique. [/step]

[step:Compare the local Chern potentials]Let $U \subset M$ be an open set with nowhere-vanishing holomorphic frame \begin{align*} e: U &\to L|_U. \end{align*} For $j \in \{0,1\}$, define the local Chern potential \begin{align*} \varphi_j: U &\to \mathbb{R} \\ p &\mapsto -\log h_{j,p}(e(p),e(p)). \end{align*} Equivalently, \begin{align*} h_{j,p}(e(p),e(p))=e^{-\varphi_j(p)}. \end{align*} Using $h_1=e^{-\psi}h_0$, we obtain for every $p \in U$, \begin{align*} e^{-\varphi_1(p)} = h_{1,p}(e(p),e(p)) = e^{-\psi(p)}h_{0,p}(e(p),e(p)) = e^{-\psi(p)}e^{-\varphi_0(p)} = e^{-(\varphi_0(p)+\psi(p))}. \end{align*} Taking the real logarithm gives \begin{align*} \varphi_1=\varphi_0+\psi|_U. \end{align*} We use the defining convention that, for a Hermitian holomorphic line bundle whose metric is written in a holomorphic frame as $h_{j,p}(e(p),e(p))=e^{-\varphi_j(p)}$, the first Chern form is \begin{align*} c_1(L,h_j)|_U=\frac{i}{2\pi}\partial\bar\partial\varphi_j \end{align*} for $j \in \{0,1\}$. Therefore, on $U$, \begin{align*} c_1(L,h_1)-c_1(L,h_0) &= \frac{i}{2\pi}\partial\bar\partial\varphi_1 - \frac{i}{2\pi}\partial\bar\partial\varphi_0 \\ &= \frac{i}{2\pi}\partial\bar\partial(\varphi_0+\psi) - \frac{i}{2\pi}\partial\bar\partial\varphi_0 \\ &= \frac{i}{2\pi}\partial\bar\partial\psi. \end{align*} Since $U$ was arbitrary and $\psi$ is globally defined, these local identities agree on overlaps and give the global identity \begin{align*} c_1(L,h_1)-c_1(L,h_0)=\frac{i}{2\pi}\partial\bar\partial\psi. \end{align*}[/step]

[guided]The first Chern form is computed locally from a holomorphic frame. Fix an open set $U \subset M$ on which there is a nowhere-vanishing holomorphic frame \begin{align*} e: U &\to L|_U. \end{align*} For each metric $h_j$, where $j \in \{0,1\}$, define the local potential \begin{align*} \varphi_j: U &\to \mathbb{R} \\ p &\mapsto -\log h_{j,p}(e(p),e(p)). \end{align*} This is the same as writing \begin{align*} h_{j,p}(e(p),e(p))=e^{-\varphi_j(p)}. \end{align*} The purpose of introducing $\varphi_j$ is that the Chern form is locally obtained by applying $\frac{i}{2\pi}\partial\bar\partial$ to this potential. We now compare the two potentials. Since $h_1=e^{-\psi}h_0$, evaluating both sides on the local frame $e$ gives \begin{align*} e^{-\varphi_1(p)} = h_{1,p}(e(p),e(p)) = e^{-\psi(p)}h_{0,p}(e(p),e(p)) = e^{-\psi(p)}e^{-\varphi_0(p)} = e^{-(\varphi_0(p)+\psi(p))} \end{align*} for every $p \in U$. The real exponential function is injective, so \begin{align*} \varphi_1=\varphi_0+\psi|_U. \end{align*} We now use the sign convention for the first Chern form. Under the convention used in this theorem, if a Hermitian metric is written in a holomorphic frame as $h_{j,p}(e(p),e(p))=e^{-\varphi_j(p)}$, then the associated first Chern form is defined locally by \begin{align*} c_1(L,h_j)|_U=\frac{i}{2\pi}\partial\bar\partial\varphi_j \end{align*} for $j \in \{0,1\}$. Therefore, \begin{align*} c_1(L,h_1)-c_1(L,h_0) &= \frac{i}{2\pi}\partial\bar\partial\varphi_1 - \frac{i}{2\pi}\partial\bar\partial\varphi_0 \\ &= \frac{i}{2\pi}\partial\bar\partial(\varphi_0+\psi) - \frac{i}{2\pi}\partial\bar\partial\varphi_0 \\ &= \frac{i}{2\pi}\partial\bar\partial\psi \end{align*} on $U$. Because $\psi$ is a globally defined smooth function, the form $\frac{i}{2\pi}\partial\bar\partial\psi$ is globally defined. Since the same computation holds in every holomorphic frame, the local identities patch to the global identity \begin{align*} c_1(L,h_1)-c_1(L,h_0)=\frac{i}{2\pi}\partial\bar\partial\psi. \end{align*}[/guided]

[step:Show the difference form is exact in real de Rham cohomology]Let \begin{align*} \alpha := \frac{i}{2\pi}\bar\partial\psi, \end{align*} viewed as a complex-valued smooth one-form on $M$, and define the real-valued smooth one-form \begin{align*} \beta := \operatorname{Re}(\alpha). \end{align*} Since $d=\partial+\bar\partial$ on complex-valued differential forms and $\bar\partial^2=0$, we have \begin{align*} d\alpha = \frac{i}{2\pi}d(\bar\partial\psi) = \frac{i}{2\pi}(\partial\bar\partial\psi+\bar\partial^2\psi) = \frac{i}{2\pi}\partial\bar\partial\psi. \end{align*} The first Chern forms $c_1(L,h_0)$ and $c_1(L,h_1)$ are real-valued two-forms, so the identity already proved implies that $\frac{i}{2\pi}\partial\bar\partial\psi$ is real-valued. Therefore \begin{align*} d\beta = d\operatorname{Re}(\alpha) = \operatorname{Re}(d\alpha) = \operatorname{Re}\left(\frac{i}{2\pi}\partial\bar\partial\psi\right) = \frac{i}{2\pi}\partial\bar\partial\psi. \end{align*} Using the identity proved above, \begin{align*} c_1(L,h_1)-c_1(L,h_0)=d\beta. \end{align*} Thus the two first Chern forms differ by an exact real de Rham form. Consequently they determine the same real de Rham cohomology class: \begin{align*} [c_1(L,h_1)] = [c_1(L,h_0)]. \end{align*} This completes the proof.[/step]

[guided]The identity already proved gives the difference of Chern forms as \begin{align*} c_1(L,h_1)-c_1(L,h_0)=\frac{i}{2\pi}\partial\bar\partial\psi. \end{align*} To prove equality in ordinary de Rham cohomology, we need a real-valued primitive, not only a complex-valued one. First introduce the complex-valued smooth one-form \begin{align*} \alpha := \frac{i}{2\pi}\bar\partial\psi. \end{align*} On complex-valued differential forms, the [exterior derivative](/theorems/1525) decomposes as $d=\partial+\bar\partial$. Also $\bar\partial^2=0$. Applying these identities to the smooth real-valued function $\psi:M\to\mathbb{R}$ gives \begin{align*} d\alpha = \frac{i}{2\pi}d(\bar\partial\psi) = \frac{i}{2\pi}(\partial\bar\partial\psi+\bar\partial^2\psi) = \frac{i}{2\pi}\partial\bar\partial\psi. \end{align*} This shows exactness after complexifying the de Rham complex. To descend to real de Rham cohomology, take the real part. Define \begin{align*} \beta := \operatorname{Re}(\alpha), \end{align*} so $\beta$ is a real-valued smooth one-form on $M$. The first Chern forms $c_1(L,h_0)$ and $c_1(L,h_1)$ are real-valued two-forms, so their difference is real-valued. Since that difference equals $\frac{i}{2\pi}\partial\bar\partial\psi$, the form $\frac{i}{2\pi}\partial\bar\partial\psi$ is real-valued. Because $d$ commutes with taking real parts of complex-valued forms, \begin{align*} d\beta = d\operatorname{Re}(\alpha) = \operatorname{Re}(d\alpha) = \operatorname{Re}\left(\frac{i}{2\pi}\partial\bar\partial\psi\right) = \frac{i}{2\pi}\partial\bar\partial\psi. \end{align*} Combining this with the metric-change identity gives \begin{align*} c_1(L,h_1)-c_1(L,h_0)=d\beta. \end{align*} Thus the difference is exact in the real de Rham complex, and therefore \begin{align*} [c_1(L,h_1)] = [c_1(L,h_0)] \end{align*} in real de Rham cohomology.[/guided]