[proofplan] We prove the forward implications by unpacking the definitions and using the Fubini--Study Kähler form on projective space. We then disprove the reverse implications in every dimension $n \geq 2$ by giving standard families of counterexamples: products of Hopf surfaces with projective space for non-Kähler compact complex manifolds, and generic complex tori for Kähler non-projective manifolds. Finally, in dimension $1$, we invoke the projectivity theorem for compact Riemann surfaces. [/proofplan] [step:Restrict the Fubini--Study form to prove that projective manifolds are Kähler] Let $X$ be a projective complex manifold. By definition, there exist an integer $N \geq 1$ and a holomorphic embedding \begin{align*} \iota: X &\to \mathbb{P}^N \end{align*} whose image $\iota(X)$ is a closed complex submanifold of complex projective space. Since $\mathbb{P}^N$ is compact and $\iota(X)$ is closed, $X$ is compact. The complex structure on $X$ is the one induced by the embedding, so $X$ is a compact complex manifold. Let $\omega_{\mathrm{FS}}$ denote the Fubini--Study form on $\mathbb{P}^N$. The form $\omega_{\mathrm{FS}}$ is a closed positive real $(1,1)$-form on $\mathbb{P}^N$ (citing a result not yet in the wiki: Fubini--Study form is Kähler). Define \begin{align*} \omega_X := \iota^*\omega_{\mathrm{FS}}. \end{align*} Because pullback by a holomorphic map preserves type, $\omega_X$ is a real $(1,1)$-form on $X$. Because exterior differentiation commutes with pullback, \begin{align*} d\omega_X = d(\iota^*\omega_{\mathrm{FS}}) = \iota^*(d\omega_{\mathrm{FS}}) = \iota^*0 = 0. \end{align*} Because $\iota$ is an embedding, $d\iota_x: T_xX \to T_{\iota(x)}\mathbb{P}^N$ is injective for every $x \in X$, and the positivity of $\omega_{\mathrm{FS}}$ implies positivity of $\omega_X$ on every nonzero complex tangent vector of $X$. Hence $\omega_X$ is a Kähler form on $X$, so $X$ is compact Kähler. [guided] We start from the definition of a projective complex manifold. It means that $X$ is realized holomorphically inside some projective space: there are an integer $N \geq 1$ and a holomorphic embedding \begin{align*} \iota: X &\to \mathbb{P}^N \end{align*} with $\iota(X)$ a closed complex submanifold. Since complex projective space $\mathbb{P}^N$ is compact, and closed subsets of compact spaces are compact, the image $\iota(X)$ is compact. Since $\iota$ is an embedding, $X$ is homeomorphic to this image, so $X$ is compact. Its complex manifold structure is part of the projective embedding data, hence $X$ is already a compact complex manifold. The Kähler structure comes from projective space. Let $\omega_{\mathrm{FS}}$ be the Fubini--Study form on $\mathbb{P}^N$. We use the standard fact that $\omega_{\mathrm{FS}}$ is a closed positive real $(1,1)$-form on $\mathbb{P}^N$ (citing a result not yet in the wiki: Fubini--Study form is Kähler). Pull it back along the embedding and define \begin{align*} \omega_X := \iota^*\omega_{\mathrm{FS}}. \end{align*} Because $\iota$ is holomorphic, pullback preserves the Hodge type $(1,1)$; therefore $\omega_X$ is a real $(1,1)$-form. Because exterior differentiation commutes with pullback, \begin{align*} d\omega_X = d(\iota^*\omega_{\mathrm{FS}}) = \iota^*(d\omega_{\mathrm{FS}}) = \iota^*0 = 0. \end{align*} It remains to check positivity. For every point $x \in X$ and every nonzero vector $v \in T_xX$, the differential \begin{align*} d\iota_x: T_xX &\to T_{\iota(x)}\mathbb{P}^N \end{align*} is injective because $\iota$ is an embedding. Hence $d\iota_x(v) \neq 0$. Positivity of $\omega_{\mathrm{FS}}$ gives positivity on $d\iota_x(v)$, and by definition of pullback this is exactly positivity of $\omega_X$ on $v$. Therefore $\omega_X$ is a Kähler form on $X$, so every projective complex manifold is compact Kähler. [/guided] [/step] [step:Forget the Kähler form to obtain a compact complex manifold] Let $X$ be a compact Kähler manifold. By definition, $X$ is a compact complex manifold equipped with a Kähler form, namely a closed positive real $(1,1)$-form. Forgetting this additional differential-geometric structure leaves a compact complex manifold. Therefore \begin{align*} \text{compact Kähler} \implies \text{compact complex}. \end{align*} [/step] [step:Use Hopf surfaces and products to show that compact complex need not imply Kähler] We first use the standard Hopf surface counterexample. Let $H$ be a Hopf surface. Then $H$ is a compact connected complex manifold of complex dimension $2$ and its first Betti number is \begin{align*} b_1(H) = 1 \end{align*} (citing a result not yet in the wiki: Hopf surfaces are compact complex surfaces with first Betti number one). Every compact Kähler manifold has even first Betti number by [Hodge decomposition](/theorems/2745) (citing a result not yet in the wiki: compact Kähler manifolds have even first Betti number). Since $1$ is odd, $H$ cannot be Kähler. Thus the implication \begin{align*} \text{compact complex} \implies \text{compact Kähler} \end{align*} fails in dimension $2$. Now let $n \geq 3$. Define \begin{align*} X_n := H \times \mathbb{P}^{n-2}. \end{align*} The product $X_n$ is a compact connected complex manifold of dimension \begin{align*} \dim_{\mathbb{C}} X_n = 2 + (n-2) = n. \end{align*} Since $\mathbb{P}^{n-2}$ is simply connected, the Künneth formula gives \begin{align*} b_1(X_n) = b_1(H) + b_1(\mathbb{P}^{n-2}) = 1 + 0 = 1 \end{align*} (citing a result not yet in the wiki: Künneth formula for first Betti numbers of products). Again $b_1(X_n)$ is odd, so $X_n$ is not Kähler. Therefore, for every $n \geq 2$, there exists a compact connected complex manifold of dimension $n$ that is not Kähler. [guided] The obstruction we use is topological: compact Kähler manifolds have even first Betti number. More precisely, Hodge decomposition gives \begin{align*} b_1(X) = 2h^{1,0}(X) \end{align*} for every compact Kähler manifold $X$, and therefore $b_1(X)$ is even (citing a result not yet in the wiki: compact Kähler manifolds have even first Betti number). A Hopf surface $H$ is a compact connected complex manifold of complex dimension $2$ with \begin{align*} b_1(H) = 1 \end{align*} (citing a result not yet in the wiki: Hopf surfaces are compact complex surfaces with first Betti number one). Since $1$ is not even, $H$ cannot be Kähler. This gives a compact complex surface that is not compact Kähler, so the reverse implication fails in dimension $2$. To get examples in every larger dimension, we take products with projective space. Let $n \geq 3$ and define \begin{align*} X_n := H \times \mathbb{P}^{n-2}. \end{align*} The product of compact connected complex manifolds is again a compact connected complex manifold. Its complex dimension is additive: \begin{align*} \dim_{\mathbb{C}} X_n = \dim_{\mathbb{C}} H + \dim_{\mathbb{C}} \mathbb{P}^{n-2} = 2 + (n-2) = n. \end{align*} The first Betti number is also additive for products, by the Künneth formula in degree $1$: \begin{align*} b_1(X_n) = b_1(H) + b_1(\mathbb{P}^{n-2}). \end{align*} Since $\mathbb{P}^{n-2}$ is simply connected, $b_1(\mathbb{P}^{n-2}) = 0$, so \begin{align*} b_1(X_n) = 1 + 0 = 1. \end{align*} Thus $X_n$ has odd first Betti number. If $X_n$ were Kähler, its first Betti number would have to be even by the compact Kähler parity theorem. This contradiction proves that $X_n$ is not Kähler. Hence compact complex does not imply compact Kähler in any dimension $n \geq 2$. [/guided] [/step] [step:Use generic complex tori to show that compact Kähler need not imply projective] Let $n \geq 2$. Choose a complex torus \begin{align*} T := \mathbb{C}^n / \Lambda, \end{align*} where $\Lambda \subset \mathbb{C}^n$ is a lattice of rank $2n$, such that \begin{align*} H^{1,1}(T) \cap H^2(T,\mathbb{Z}) = \{0\}. \end{align*} Such complex tori exist for every $n \geq 2$ (citing a result not yet in the wiki: generic complex tori have no nonzero integral $(1,1)$-classes). The torus $T$ is compact and complex. It is Kähler because the standard translation-invariant Hermitian form on $\mathbb{C}^n$ descends to a closed positive real $(1,1)$-form on the quotient $T$. If $T$ were projective, then it would admit an ample line bundle, and the first Chern class of that line bundle would give a nonzero integral $(1,1)$-class on $T$ by the Lefschetz $(1,1)$ theorem and the positivity of ample line bundles (citing a result not yet in the wiki: projective complex tori have nonzero integral $(1,1)$-classes). This contradicts \begin{align*} H^{1,1}(T) \cap H^2(T,\mathbb{Z}) = \{0\}. \end{align*} Therefore $T$ is compact Kähler but not projective. Hence, for every $n \geq 2$, compact Kähler does not imply projective. [guided] We now need examples that have a Kähler form but are not algebraic. Complex tori provide exactly this distinction. Fix $n \geq 2$ and choose a lattice $\Lambda \subset \mathbb{C}^n$ of rank $2n$. The quotient \begin{align*} T := \mathbb{C}^n / \Lambda \end{align*} is a compact connected complex manifold of dimension $n$. A complex torus is Kähler: choose any positive Hermitian inner product on $\mathbb{C}^n$. Its associated constant real $(1,1)$-form is translation-invariant, hence invariant under the lattice action by translations. Therefore it descends to a real $(1,1)$-form on $T$. Since the form on $\mathbb{C}^n$ has constant coefficients, it is closed; since it is positive on nonzero tangent vectors upstairs, the descended form is positive on nonzero tangent vectors downstairs. Thus $T$ is compact Kähler. The non-projectivity comes from choosing the torus generically. For every $n \geq 2$, there exist complex tori satisfying \begin{align*} H^{1,1}(T) \cap H^2(T,\mathbb{Z}) = \{0\} \end{align*} (citing a result not yet in the wiki: generic complex tori have no nonzero integral $(1,1)$-classes). If such a torus were projective, it would carry an ample holomorphic line bundle. The first Chern class of an ample line bundle is an integral cohomology class of Hodge type $(1,1)$, and it is nonzero because an ample class has positive degree on suitable algebraic curves or, equivalently, gives a positive polarization (citing a result not yet in the wiki: projective complex tori have nonzero integral $(1,1)$-classes). This would produce a nonzero element of \begin{align*} H^{1,1}(T) \cap H^2(T,\mathbb{Z}), \end{align*} contradicting the chosen property of $T$. Therefore $T$ is Kähler but not projective. Since $n \geq 2$ was arbitrary, the reverse implication compact Kähler implies projective fails in every complex dimension at least $2$. [/guided] [/step] [step:Invoke the projectivity theorem for compact Riemann surfaces] Let $X$ be a compact connected complex manifold of complex dimension $1$. Then $X$ is a compact Riemann surface. By the projectivity theorem for compact Riemann surfaces, every compact Riemann surface admits a holomorphic embedding into some complex projective space (citing a result not yet in the wiki: every compact Riemann surface is projective). Hence $X$ is projective. Combining the first two steps gives \begin{align*} \text{projective} \implies \text{compact Kähler} \implies \text{compact complex}. \end{align*} The Hopf-surface and complex-torus examples show that neither reverse implication holds in any complex dimension $n \geq 2$, while the compact Riemann surface projectivity theorem proves the asserted exceptional behavior in dimension $1$. [/step]