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While every algebraic manifold is complex analytic the converse is far from true.

If $M$ is a compact complex manifold then $a(M)$, the algebraic dimension of $M$, is defined as the transcendence degree of $\mathbb C(M)$ over $\mathbb C$, where $\mathbb C(M)$ is the field of meromorphic functions on $M$.

The algebraic dimension is at most the dimension of $M$ and when the equality holds $M$ is called a Moishezon space. In general what one has is a variety $\hat M$ with $\dim \hat M = \dim M$, a projective variety $N$ with $\dim N = a(M)$, and analytic morphisms $\pi : \hat M \to M$ and $\varphi : \hat M \to N$ such that

• $\pi$ is bimeromorphic;
• the generic fiber of $\varphi$ is irreducible;
• the field of meromorphic functions on $M$ is the same as the one of $N$. More precisely: $\varphi^* \mathbb C(N) = \pi^* \mathbb C ( M)$.

It also has to be noted that from a topological point of view there are many more compact complex manifolds than compact algebraic manifolds. To be more precise, there are strong restrictions on the fundamental group of compact Kähler manifolds, while there is a deep result of Taubes that implies that every finitely generated group is the fundamental group of a compact complex $3$-fold.

Motivated by comments on Kuperberg's answwer, let me also mention that there were a conjecture by Kodaira claiming that any compact Kähler manifold is deformation equivalent to projective manifolds. While the conjecture is true in dimension two, as Kodaira has himself shown, it has been recently disproved by Claire Voisin.

Compact complex manifolds which are not algebraic are not artificial beasts. Let me recall a very natural example of algebraic nature. Consider the quotient $X=SL(2,\mathbb C)/\Gamma$, where $\Gamma$ is a discrete cocompact subgroup. I think it was Mostow who proved that $X$ has no analytic hypersurfaces, which implies in particular $a(X)=0$. To see that $X$ is far from an algebraic variety, despite being algebraically defined, notice that there are $1$-forms on $SL(2,\mathbb C)$ which are invariant by right translations and which are not closed. So they induce global holomorphic $1$-forms on the quotient $X$ which are not closed. But on compact Kähler manifolds, and more generally on compact manifolds bimeromorphic to them, every holomorphic $1$-form is closed.

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While every algebraic manifold is complex analytic the converse is far from true.

If $M$ is a compact complex manifold then $a(M)$, the algebraic dimension of $M$, is defined as the transcendence degree of $\mathbb C(M)$ over $\mathbb C$, where $\mathbb C(M)$ is the field of meromorphic functions on $M$.

The algebraic dimension is at most the dimension of $M$ and when the equality holds $M$ is called a Moishezon space. In general what one has is a variety $\hat M$ with $\dim \hat M = \dim M$, a projective variety $N$ with $\dim N = a(M)$, and analytic morphisms $\pi : \hat M \to M$ and $\varphi : \hat M \to N$ such that

• $\pi$ is bimeromorphic;
• the generic fiber of $\varphi$ is irreducible;
• the field of meromorphic functions on $M$ is the same as the one of $N$. More precisely: $\varphi^* \mathbb C(N) = \pi^* \mathbb C ( \hat M)$.

It also has to be noted that from a topological point of view there are many more compact complex manifolds than compact algebraic manifolds. To be more precise, there are strong restrictions on the fundamental group of compact Kähler manifolds, while there is a deep result of Taubes that implies that every finitely generated group is the fundamental group of a compact complex $3$-fold.

Motivated by comments on Kuperberg's answwer, let me also mention that there were a conjecture by Kodaira claiming that any compact Kähler manifold is deformation equivalent to projective manifolds. While the conjecture is true in dimension two, as Kodaira has himself shown, it has been recently(?) recently disproved by Claire Voisin.

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While every algebraic manifold is complex analytic the converse is far from true.

If $M$ is a compact complex manifold then $a(M)$, the algebraic dimension of $M$, is defined as the transcendence degree of $\mathbb C(M)$ over $\mathbb C$, where $\mathbb C(M)$ is the field of meromorphic functions on $M$.

The algebraic dimension is at most the dimension of $M$ and when the equality holds $M$ is called a Moishezon space. In general what one has is a variety $\hat M$ with $\dim \hat M = \dim M$, a projective variety $N$ with $\dim N = a(M)$, and analytic morphisms $\pi : \hat M \to M$ and $\varphi : \hat M \to N$ such that

• $\pi$ is bimeromorphic;
• the generic fiber of $\varphi$ is irreducible;
• the field of meromorphic functions on $M$ is the same as the one of $N$. More precisely: $\varphi^* \mathbb C(N) = \pi^* \mathbb C ( \hat M)$.

It also has to be noted that from a topological point of view there are many more compact complex manifolds than compact algebraic manifolds. To be more precise, there are strong restrictions on the fundamental group of compact Kähler manifolds, while there is a deep result of Taubes that implies that every finitely generated group is the fundamental group of a compact complex $3$-fold.

Motivated by comments on Kuperberg's answwer, let me also mention that there were a conjecture by Kodaira claiming that any compact Kähler manifold is deformation equivalent to projective manifolds. While the conjecture is true in dimension two, as Kodaira has himself shown, it has been recently(?) disproved by Claire Voisin.

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