[proofplan] We prove the equivalence by connecting the three conditions through first Chern classes. A projective embedding pulls back the hyperplane bundle and the Fubini-Study form, giving both a positive line bundle and an integral Kähler class. Conversely, an integral Kähler class determines a holomorphic line bundle through the exponential sequence and the Hodge type condition; the $\partial\bar{\partial}$-lemma then adjusts a Hermitian metric so that its Chern form is the prescribed Kähler form. Finally, a positive line bundle gives a projective embedding by Kodaira's embedding theorem for positive line bundles. [/proofplan]

[step:Pull back the hyperplane bundle from a projective embedding] Assume that $X$ is projective. Choose $N \in \mathbb{N}$ and a holomorphic embedding \begin{align*} j: X &\to \mathbb{P}^N. \end{align*} Let $\mathcal{O}_{\mathbb{P}^N}(1) \to \mathbb{P}^N$ denote the hyperplane line bundle, equipped with its Fubini-Study Hermitian metric $h_{\mathrm{FS}}$. Let $\omega_{\mathrm{FS}}$ denote the corresponding Fubini-Study Kähler form, normalized so that \begin{align*} c_1(\mathcal{O}_{\mathbb{P}^N}(1),h_{\mathrm{FS}})=\omega_{\mathrm{FS}}. \end{align*} The cohomology class $[\omega_{\mathrm{FS}}] \in H^2(\mathbb{P}^N,\mathbb{R})$ is the image of the integral first Chern class $c_1(\mathcal{O}_{\mathbb{P}^N}(1)) \in H^2(\mathbb{P}^N,\mathbb{Z})$. Define the holomorphic line bundle \begin{align*} L := j^*\mathcal{O}_{\mathbb{P}^N}(1) \to X \end{align*} and equip it with the pulled-back Hermitian metric \begin{align*} h := j^*h_{\mathrm{FS}}. \end{align*} By functoriality of the Chern connection and its curvature, \begin{align*} c_1(L,h)=j^*c_1(\mathcal{O}_{\mathbb{P}^N}(1),h_{\mathrm{FS}})=j^*\omega_{\mathrm{FS}}. \end{align*} Because $j$ is a holomorphic embedding, its differential \begin{align*} dj_x: T_xX &\to T_{j(x)}\mathbb{P}^N \end{align*} is injective for every $x \in X$. Since $\omega_{\mathrm{FS}}$ is positive on complex tangent vectors in $\mathbb{P}^N$, the pullback $j^*\omega_{\mathrm{FS}}$ is positive on nonzero complex tangent vectors in $X$. Thus $j^*\omega_{\mathrm{FS}}$ is a Kähler form on $X$. Therefore $L$ is a positive holomorphic line bundle, and its Chern class \begin{align*} c_1(L)=j^*c_1(\mathcal{O}_{\mathbb{P}^N}(1)) \in H^2(X,\mathbb{Z}) \end{align*} maps to the Kähler class $[j^*\omega_{\mathrm{FS}}] \in H^2(X,\mathbb{R})$. Hence $X$ admits an integral Kähler class and a positive holomorphic line bundle. [/step]

[step:Recover an integral Kähler class from a positive line bundle] Assume that $X$ admits a positive holomorphic line bundle. Thus there are a holomorphic line bundle \begin{align*} L \to X \end{align*} and a smooth Hermitian metric $h$ on $L$ such that \begin{align*} \omega := c_1(L,h) \end{align*} is a Kähler form on $X$. The Chern-Weil representative $c_1(L,h)$ represents the real image of the integral first Chern class $c_1(L) \in H^2(X,\mathbb{Z})$. Therefore \begin{align*} [\omega] = c_1(L)_{\mathbb{R}} \in H^2(X,\mathbb{R}), \end{align*} where $c_1(L)_{\mathbb{R}}$ denotes the image of $c_1(L)$ under $H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{R})$. Since $\omega$ is Kähler, this proves that $X$ admits an integral Kähler class. [/step]

[step:Use the exponential sequence to build a holomorphic line bundle from an integral Kähler class]Assume that $X$ admits an integral Kähler class. Choose a Kähler form $\omega$ on $X$ and a class \begin{align*} \alpha \in H^2(X,\mathbb{Z}) \end{align*} whose image in $H^2(X,\mathbb{R})$ is $[\omega]$. Because $\omega$ is a real closed $(1,1)$-form, the real class $[\omega]$ lies in \begin{align*} H^{1,1}(X) \cap H^2(X,\mathbb{R}). \end{align*} Let $H^2(X,\mathcal{O}_X)$ denote the second sheaf cohomology group of the structure sheaf $\mathcal{O}_X$ of holomorphic functions on $X$. Under the Dolbeault identification \begin{align*} H^2(X,\mathcal{O}_X) \cong H^{0,2}_{\bar{\partial}}(X), \end{align*} the Hodge-theoretic description of the sheaf-cohomology map induced by $\mathbb{Z} \hookrightarrow \mathcal{O}_X$ is as follows: first map $H^2(X,\mathbb{Z})$ to $H^2(X,\mathbb{R})$, then complexify to $H^2(X,\mathbb{C})$, and finally project, using the [Hodge decomposition](/theorems/2745) of the compact Kähler manifold $X$, to the $(0,2)$-summand. Since the real image of $\alpha$ is $[\omega]$ and $[\omega]$ has type $(1,1)$, this $(0,2)$-component is zero. Thus the image of $\alpha$ in $H^2(X,\mathcal{O}_X)$ is zero. By the exponential sequence for holomorphic line bundles, applied to the compact complex manifold $X$, \begin{align*} 0 \to \mathbb{Z} \to \mathcal{O}_X \xrightarrow{\exp(2\pi i\,\cdot)} \mathcal{O}_X^* \to 0, \end{align*} the image of the first Chern class map \begin{align*} c_1: \operatorname{Pic}(X) &\to H^2(X,\mathbb{Z}) \end{align*} is precisely the kernel of the map $H^2(X,\mathbb{Z}) \to H^2(X,\mathcal{O}_X)$. Here $\operatorname{Pic}(X)$ denotes the group of holomorphic line bundles on $X$ up to holomorphic isomorphism. Hence there exists a holomorphic line bundle \begin{align*} L \to X \end{align*} such that \begin{align*} c_1(L)=\alpha. \end{align*}[/step]

[guided]The purpose of this step is to turn the integral cohomology class underlying the Kähler form into an actual holomorphic line bundle. We start with a Kähler form $\omega$ and an integral class $\alpha \in H^2(X,\mathbb{Z})$ whose real image is $[\omega] \in H^2(X,\mathbb{R})$. Since $\omega$ is a Kähler form, it is a real closed form of type $(1,1)$. Therefore its cohomology class belongs to the Hodge summand \begin{align*} H^{1,1}(X) \cap H^2(X,\mathbb{R}). \end{align*} Let $H^2(X,\mathcal{O}_X)$ denote the second sheaf cohomology group of the structure sheaf $\mathcal{O}_X$ of holomorphic functions on $X$. The obstruction to an integral class being the first Chern class of a holomorphic line bundle is its image in this group. More concretely, the Dolbeault identification gives \begin{align*} H^2(X,\mathcal{O}_X) \cong H^{0,2}_{\bar{\partial}}(X), \end{align*} and the Hodge-theoretic description of the sheaf-cohomology map induced by $\mathbb{Z} \hookrightarrow \mathcal{O}_X$ is this: first pass from $H^2(X,\mathbb{Z})$ to real cohomology, then complexify, and then project to the $(0,2)$-component in the Hodge decomposition of the compact Kähler manifold $X$. The real image of $\alpha$ is $[\omega]$, and $[\omega]$ has only type $(1,1)$. Hence the $(0,2)$-component of the image of $\alpha$ is zero, so $\alpha$ lies in the kernel of $H^2(X,\mathbb{Z}) \to H^2(X,\mathcal{O}_X)$. Now apply the exponential sequence \begin{align*} 0 \to \mathbb{Z} \to \mathcal{O}_X \xrightarrow{\exp(2\pi i\,\cdot)} \mathcal{O}_X^* \to 0. \end{align*} The associated long exact sequence in sheaf cohomology contains the connecting homomorphism from holomorphic line bundles to integral cohomology: \begin{align*} c_1: \operatorname{Pic}(X) &\to H^2(X,\mathbb{Z}), \end{align*} where $\operatorname{Pic}(X)=H^1(X,\mathcal{O}_X^*)$ is the group of holomorphic line bundles on $X$ up to holomorphic isomorphism. Exactness says that the image of this map is the kernel of \begin{align*} H^2(X,\mathbb{Z}) &\to H^2(X,\mathcal{O}_X). \end{align*} Since $\alpha$ lies in that kernel, there exists a holomorphic line bundle $L \to X$ with $c_1(L)=\alpha$.[/guided]

[step:Adjust the Hermitian metric so that its Chern form is the given Kähler form]Choose any smooth Hermitian metric $h_0$ on the holomorphic line bundle $L \to X$ obtained above, and define the real closed $(1,1)$-form \begin{align*} \theta_0 := c_1(L,h_0). \end{align*} Both $\theta_0$ and $\omega$ represent the same real cohomology class in $H^2(X,\mathbb{R})$, namely the real image of $\alpha$. Hence the real $(1,1)$-form $\omega-\theta_0$ is exact. By the $\partial\bar{\partial}$-lemma for compact Kähler manifolds, applied to the exact real $(1,1)$-form $\omega-\theta_0$, there exists a smooth real-valued function \begin{align*} \varphi: X &\to \mathbb{R} \end{align*} such that \begin{align*} \omega-\theta_0 = \frac{i}{2\pi}\partial\bar{\partial}\varphi. \end{align*} Here the constant and sign are fixed by replacing the real potential supplied by the $\partial\bar{\partial}$-lemma by the corresponding real scalar multiple, so the displayed normalization matches the Chern-form convention used for Hermitian metric changes. Define a new smooth Hermitian metric $h$ on $L$ by \begin{align*} h := e^{-\varphi}h_0. \end{align*} Under this conformal change of Hermitian metric on a holomorphic line bundle, the Chern form transforms as \begin{align*} c_1(L,h)=c_1(L,h_0)+\frac{i}{2\pi}\partial\bar{\partial}\varphi. \end{align*} Therefore \begin{align*} c_1(L,h)=\theta_0+\frac{i}{2\pi}\partial\bar{\partial}\varphi=\omega. \end{align*} Since $\omega$ is Kähler, the line bundle $L$ is positive.[/step]

[guided]At this point we have a holomorphic line bundle $L \to X$ with the correct topological first Chern class, but positivity is a statement about a Hermitian metric and its curvature form. We therefore choose an arbitrary smooth Hermitian metric $h_0$ on $L$ and set \begin{align*} \theta_0 := c_1(L,h_0). \end{align*} The Chern-Weil theorem identifies $[\theta_0]$ with the real image of $c_1(L)$, so \begin{align*} [\theta_0]=[\omega] \in H^2(X,\mathbb{R}). \end{align*} Thus $\omega-\theta_0$ is an exact real form. It is also of type $(1,1)$ because both $\omega$ and $\theta_0$ are real $(1,1)$-forms. Now we use the compact Kähler hypothesis through the $\partial\bar{\partial}$-lemma. The lemma says that an exact form of pure type on a compact Kähler manifold is $\partial\bar{\partial}$-exact. Applied to the exact real $(1,1)$-form $\omega-\theta_0$, it gives a smooth real-valued potential. We rescale that real potential, if necessary, so that it is a smooth real-valued function \begin{align*} \varphi: X &\to \mathbb{R} \end{align*} with the normalization \begin{align*} \omega-\theta_0 = \frac{i}{2\pi}\partial\bar{\partial}\varphi. \end{align*} This rescaling only absorbs the conventional constant and sign in the $\partial\bar{\partial}$-lemma, and it is chosen to match the Chern-form convention for the metric change below. We now modify the metric rather than the line bundle. Define \begin{align*} h := e^{-\varphi}h_0. \end{align*} For a holomorphic line bundle, this conformal change changes the Chern form by \begin{align*} c_1(L,h)-c_1(L,h_0)=\frac{i}{2\pi}\partial\bar{\partial}\varphi. \end{align*} Therefore \begin{align*} c_1(L,h)=\theta_0+\frac{i}{2\pi}\partial\bar{\partial}\varphi=\omega. \end{align*} Because $\omega$ is positive and closed, it is a Kähler form. Hence the Hermitian holomorphic line bundle $(L,h)$ is positive.[/guided]

[step:Apply Kodaira embedding to a positive line bundle]Assume that $X$ admits a positive holomorphic line bundle $L \to X$. The manifold $X$ is compact complex because it is a compact Kähler manifold, and $L$ is positive by hypothesis. Therefore the hypotheses of Kodaira's embedding theorem for positive line bundles are satisfied. By that theorem (citing a result not yet in the wiki: [Kodaira Embedding Theorem](/theorems/3898)), there exists $k_0 \in \mathbb{N}$ such that for every integer $k \geq k_0$, the tensor power \begin{align*} L^k := L^{\otimes k} \end{align*} is globally generated and has enough global holomorphic sections to separate points and tangent directions. Choose such a $k \geq k_0$, and define the finite-dimensional complex [vector space](/page/Vector%20Space) \begin{align*} V := H^0(X,L^k). \end{align*} Global generation means that for every $x \in X$ there exists a section $s \in V$ with $s(x) \neq 0$, so the evaluation functional at $x$ is nonzero. Hence the evaluation construction defines a holomorphic map \begin{align*} \Phi_k: X &\to \mathbb{P}(V^*) \end{align*} by sending $x$ to the line in $V^*$ spanned by evaluation at $x$. The point-separation and tangent-separation properties imply that $\Phi_k$ is a holomorphic embedding. Since $\mathbb{P}(V^*)$ is a complex projective space, this realizes $X$ as a compact complex submanifold of projective space. Hence $X$ is projective.[/step]

[guided]The remaining implication is the analytic heart of the theorem. Positivity of $L$ means that some Hermitian metric on $L$ has positive curvature form. Since $X$ is a compact Kähler manifold, it is in particular a compact complex manifold, and therefore $X$ together with the positive holomorphic line bundle $L$ satisfies the hypotheses of Kodaira's embedding theorem for positive line bundles. The theorem says that this curvature positivity forces high tensor powers of $L$ to have many global holomorphic sections. More precisely, there is $k_0 \in \mathbb{N}$ such that for every integer $k \geq k_0$, the line bundle \begin{align*} L^k := L^{\otimes k} \end{align*} is globally generated and sufficiently generated by its global holomorphic sections: for every $x \in X$ at least one section of $L^k$ does not vanish at $x$, and the sections separate distinct points of $X$ and separate tangent directions at each point. Set \begin{align*} V := H^0(X,L^k), \end{align*} the finite-dimensional complex vector space of global holomorphic sections of $L^k$. The first property is base-point-freeness. It is needed before we can define the evaluation map: for each $x \in X$, the evaluation functional on $V$ is nonzero because some $s \in V$ satisfies $s(x) \neq 0$. Therefore the evaluation construction gives a holomorphic map \begin{align*} \Phi_k: X &\to \mathbb{P}(V^*). \end{align*} At a point $x \in X$, the map records the hyperplane of sections vanishing at $x$, equivalently the evaluation functional up to scalar. Because the sections separate points, $\Phi_k$ is injective. Because they separate tangent directions, the differential \begin{align*} d(\Phi_k)_x: T_xX &\to T_{\Phi_k(x)}\mathbb{P}(V^*) \end{align*} is injective for every $x \in X$. Therefore $\Phi_k$ is a holomorphic embedding. Since $\mathbb{P}(V^*)$ is projective space, this proves that $X$ is projective.[/guided]

[step:Combine the implications] The first step proves that projectivity implies both the existence of an integral Kähler class and the existence of a positive holomorphic line bundle. The second step proves that a positive holomorphic line bundle gives an integral Kähler class. The third and fourth steps prove that an integral Kähler class gives a positive holomorphic line bundle. The fifth step proves that a positive holomorphic line bundle implies projectivity. Hence all three stated conditions are equivalent. [/step]