[proofplan] We reduce the prescribed Ricci equation on the compact connected Kähler manifold $X$ to the normalized complex Monge-Ampère equation in the fixed Kähler class $[\omega]$. The cohomological hypothesis and the real $\partial\bar\partial$-lemma produce a real potential $F$ satisfying $\operatorname{Ric}(\omega)-\rho=i\partial\bar\partial F$, with the additive constant chosen to satisfy the volume compatibility condition. Yau's solution of the Calabi conjecture then gives a smooth real potential $\varphi$ with $\omega+i\partial\bar\partial\varphi>0$ and prescribed volume form. Finally, the uniqueness part of the same normalized Monge-Ampère theorem identifies any two potentials up to an additive constant, hence gives uniqueness of the Kähler form. [/proofplan] [step:Choose the Ricci potential with the correct volume normalization] Let, for any Kähler form $\eta$ on $X$, $dV_\eta$ denote the volume measure induced by $\eta^n$. Set \begin{align*} V:=\int_X 1\, dV_\omega \end{align*} to denote the total Kähler volume of $X$ with respect to $\omega$. Since $\rho$ represents $2\pi c_1(X)$ and the Ricci form $\operatorname{Ric}(\omega)$ also represents $2\pi c_1(X)$, the real closed $(1,1)$-form $\operatorname{Ric}(\omega)-\rho$ is cohomologous to zero. By the real $(1,1)$-form version of the $\partial\bar\partial$-lemma on compact Kähler manifolds, there exists a smooth real-valued function \begin{align*} F_0:X\to \mathbb R \end{align*} such that \begin{align*} \operatorname{Ric}(\omega)-\rho=i\partial\bar\partial F_0. \end{align*} The function $F_0$ is determined only up to an additive constant. Define the real constant \begin{align*} c:=\log\left(\frac{V}{\int_X e^{F_0}\, dV_\omega}\right) \end{align*} and define \begin{align*} F:X&\to\mathbb R \end{align*} by $F:=F_0+c$. Then $i\partial\bar\partial F=i\partial\bar\partial F_0$, so \begin{align*} \operatorname{Ric}(\omega)-\rho=i\partial\bar\partial F. \end{align*} Moreover, \begin{align*} \int_X e^F\, dV_\omega=\int_X e^{F_0+c}\, dV_\omega=e^c\int_X e^{F_0}\, dV_\omega=V. \end{align*} [guided] The first task is to turn the cohomological hypothesis into an actual global function. The Ricci form $\operatorname{Ric}(\omega)$ represents $2\pi c_1(X)$, and the hypothesis says that $\rho$ represents the same cohomology class. Hence the difference \begin{align*} \operatorname{Ric}(\omega)-\rho \end{align*} is an exact real closed $(1,1)$-form. Because $X$ is compact Kähler, the real $(1,1)$-form version of the $\partial\bar\partial$-lemma applies: every exact real closed $(1,1)$-form is of the form $i\partial\bar\partial$ of a smooth real-valued function. Therefore there is a smooth map \begin{align*} F_0:X\to\mathbb R \end{align*} such that \begin{align*} \operatorname{Ric}(\omega)-\rho=i\partial\bar\partial F_0. \end{align*} The function $F_0$ is not unique, because adding a constant does not change $i\partial\bar\partial F_0$. We use this freedom to impose the volume condition required by the Monge-Ampère equation. Define the measure $dV_\omega$ induced by $\omega^n$, set \begin{align*} V:=\int_X 1\, dV_\omega \end{align*} and \begin{align*} c:=\log\left(\frac{V}{\int_X e^{F_0}\, dV_\omega}\right). \end{align*} The denominator is positive because $e^{F_0}>0$ and $dV_\omega$ is a positive measure on the compact manifold $X$. Now set $F:=F_0+c$. Since constants vanish under $i\partial\bar\partial$, we still have \begin{align*} \operatorname{Ric}(\omega)-\rho=i\partial\bar\partial F. \end{align*} The choice of $c$ gives \begin{align*} \int_X e^F\, dV_\omega=\int_X e^{F_0+c}\, dV_\omega \end{align*} \begin{align*} \int_X e^{F_0+c}\, dV_\omega=e^c\int_X e^{F_0}\, dV_\omega \end{align*} \begin{align*} e^c\int_X e^{F_0}\, dV_\omega=V \end{align*} \begin{align*} V=\int_X 1\, dV_\omega \end{align*} This is exactly the compatibility condition for replacing $\omega^n$ by the volume form $e^F\omega^n$ inside the same Kähler class. [/guided] [/step] [step:Translate the prescribed Ricci equation into a Monge-Ampère equation] Let \begin{align*} \varphi:X&\to\mathbb R \end{align*} be a smooth real-valued function such that \begin{align*} \omega_\varphi:=\omega+i\partial\bar\partial\varphi \end{align*} is positive. Then $\omega_\varphi$ is a Kähler form and $[\omega_\varphi]=[\omega]$. For two Kähler forms in the same complex structure, the Ricci forms satisfy \begin{align*} \operatorname{Ric}(\omega_\varphi)-\operatorname{Ric}(\omega) =-i\partial\bar\partial\log\left(\frac{\omega_\varphi^n}{\omega^n}\right). \end{align*} Thus, if \begin{align*} \omega_\varphi^n=e^F\omega^n, \end{align*} then \begin{align*} \operatorname{Ric}(\omega_\varphi) =\operatorname{Ric}(\omega)-i\partial\bar\partial F =\rho. \end{align*} Conversely, if $\operatorname{Ric}(\omega_\varphi)=\rho$, then \begin{align*} i\partial\bar\partial\log\left(\frac{\omega_\varphi^n}{e^F\omega^n}\right)=0. \end{align*} Since $X$ is compact and connected, a real pluriharmonic function on $X$ is constant by applying the maximum principle locally to its [harmonic representatives](/theorems/2747). Hence \begin{align*} \omega_\varphi^n=C e^F\omega^n \end{align*} for some constant $C>0$. Since $\omega_\varphi$ and $\omega$ are cohomologous, their total volumes agree: \begin{align*} \int_X 1\, dV_{\omega_\varphi}=\int_X 1\, dV_\omega. \end{align*} Together with \begin{align*} \int_X e^F\, dV_\omega=\int_X 1\, dV_\omega, \end{align*} this forces $C=1$. Therefore the prescribed Ricci equation is equivalent to \begin{align*} (\omega+i\partial\bar\partial\varphi)^n=e^F\omega^n, \end{align*} with $\omega+i\partial\bar\partial\varphi>0$. [/step] [step:Apply Yau's complex Monge-Ampère theorem to solve the normalized equation] We now invoke Yau's complex Monge-Ampère theorem, also called Yau's solution of the Calabi conjecture, in the following form: if $(X,\omega)$ is a compact Kähler manifold and $F:X\to\mathbb R$ is smooth with \begin{align*} \int_X e^F\, dV_\omega=\int_X 1\, dV_\omega, \end{align*} then there exists a smooth real-valued function \begin{align*} \varphi:X&\to\mathbb R \end{align*} such that \begin{align*} \omega+i\partial\bar\partial\varphi>0 \end{align*} and \begin{align*} (\omega+i\partial\bar\partial\varphi)^n=e^F\omega^n. \end{align*} Moreover, $\varphi$ is unique up to addition of a constant. The hypotheses of this theorem are satisfied: $X$ is compact Kähler, $\omega$ is a Kähler form, $F$ is smooth, and the normalization was established in the first step. Hence such a potential $\varphi$ exists. Define \begin{align*} \omega':=\omega+i\partial\bar\partial\varphi. \end{align*} Then $\omega'$ is a Kähler form, $[\omega']=[\omega]$, and the previous step gives \begin{align*} \operatorname{Ric}(\omega')=\rho. \end{align*} The invoked result is the external analytic input Yau's complex Monge-Ampère theorem. [/step] [step:Compare two potentials to prove uniqueness] Suppose $\omega_1$ and $\omega_2$ are Kähler forms in $[\omega]$ satisfying \begin{align*} \operatorname{Ric}(\omega_1)=\rho \end{align*} and \begin{align*} \operatorname{Ric}(\omega_2)=\rho. \end{align*} Because both forms lie in $[\omega]$, the $\partial\bar\partial$-lemma gives smooth real-valued functions \begin{align*} \varphi_1:X&\to\mathbb R, \end{align*} and \begin{align*} \varphi_2:X&\to\mathbb R \end{align*} such that \begin{align*} \omega_1=\omega+i\partial\bar\partial\varphi_1 \end{align*} and \begin{align*} \omega_2=\omega+i\partial\bar\partial\varphi_2. \end{align*} Adding constants to $\varphi_1$ and $\varphi_2$ does not change $\omega_1$ or $\omega_2$, so normalize them by requiring \begin{align*} \int_X \varphi_1 \, dV_\omega=0 \end{align*} and \begin{align*} \int_X \varphi_2 \, dV_\omega=0. \end{align*} By the equivalence proved above, both potentials solve \begin{align*} (\omega+i\partial\bar\partial\varphi_j)^n=e^F\omega^n \end{align*} for $j\in\{1,2\}$. Set \begin{align*} u:X&\to\mathbb R \end{align*} by $u:=\varphi_1-\varphi_2$. At this point the hypotheses of the uniqueness part of Yau's complex Monge-Ampère theorem are satisfied: $\varphi_1$ and $\varphi_2$ are smooth real potentials, both forms $\omega+i\partial\bar\partial\varphi_j$ are positive, both potentials are measured relative to the same background Kähler form $\omega$, and both solve the normalized equation \begin{align*} (\omega+i\partial\bar\partial\varphi_j)^n=e^F\omega^n \end{align*} for $j\in\{1,2\}$. The uniqueness theorem gives that $\varphi_1-\varphi_2$ is constant. The imposed normalizations force that constant to be zero: \begin{align*} 0=\int_X (\varphi_1-\varphi_2)\, dV_\omega=\int_X u\, dV_\omega. \end{align*} Thus $\varphi_1=\varphi_2$, and consequently \begin{align*} \omega_1=\omega_2. \end{align*} This proves uniqueness of the Kähler form in the class $[\omega]$. [guided] The point of reducing both Ricci equations to the same Monge-Ampère equation is that uniqueness is an analytic statement about potentials, not a pointwise determinant comparison at a single maximum or minimum. We have already written \begin{align*} \omega_j=\omega+i\partial\bar\partial\varphi_j \end{align*} for $j\in\{1,2\}$, with each $\varphi_j:X\to\mathbb R$ smooth and each $\omega+i\partial\bar\partial\varphi_j$ positive. The equivalence step applied to each $\omega_j$ shows that both potentials solve \begin{align*} (\omega+i\partial\bar\partial\varphi_j)^n=e^F\omega^n \end{align*} for $j\in\{1,2\}$. These are exactly the hypotheses of the uniqueness part of Yau's normalized complex Monge-Ampère theorem: two smooth real potentials for the same background Kähler form $\omega$, with positive associated Kähler forms, and with the same right-hand side $e^F\omega^n$. Therefore the theorem gives that $u:=\varphi_1-\varphi_2$ is constant on the connected manifold $X$. Since the potentials were normalized by \begin{align*} \int_X \varphi_1\, dV_\omega=0 \end{align*} and \begin{align*} \int_X \varphi_2\, dV_\omega=0, \end{align*} we get \begin{align*} 0=\int_X (\varphi_1-\varphi_2)\, dV_\omega=\int_X u\, dV_\omega. \end{align*} If $u$ is the constant $a\in\mathbb R$, then the last equality says $a\int_X 1\,dV_\omega=0$. The volume $\int_X 1\,dV_\omega$ is positive, so $a=0$. Hence $\varphi_1=\varphi_2$, and therefore \begin{align*} \omega_1=\omega_2. \end{align*} This proves uniqueness of the Kähler form in the class $[\omega]$. [/guided] [/step]