Let $M$ be a closed orientable smooth 4-manifold. Assume $\pi\_1(M)=\{0\}$ and $b\_2(M)>0$.


Let $S$ be a closed orientable surface. Denote $P=M\times S$.


Can it so happen that there is no complex projective manifold homotopy equivalent to $P$?


Is it possible to rule out the existence of a closed symplectic 6-manifold homotopy equivalent to $P$? It seems unlikely, see [this](https://www.math.stonybrook.edu/%7Emilivojevic/symplecticalgebras.pdf) preprint.


Note that


* $P$ is formal
* the Betti numbers are even in odd degree
* there is a class $c\in H^2(P, \mathbb{R})$ satisfying hard Lefschetz
* $\pi\_1(P)$ is Kähler
* $P$ admits an almost complex structure.


A related example is constructed [here](https://arxiv.org/abs/1601.04337) but I don't think it decomposes as a direct product. It may be relevant that Hsueh-Yung Lin claims that every closed Kähler threefold is deformation equivalent to a complex projective manifold.



Topological factors of complex projective manifolds
Let $M=\mathbb CP^2\#\mathbb CP^2$ and let $S=T^2$ be the $2$-dimensional torus. I think this gives an example for the original question. As for the symplectic version of the question, I am sure it is an open problem.


*Proof.* Suppose by contradiction $P=M\times S=\mathbb CP^2\#\mathbb CP^2\times T^2$ is homotopic to a complex projective manifold. The contradiction will be derived from the Hodge index theorem, applied to $H^{1,1}(P)$.


Let us calculate the cubic intersection form on $H^2(P,\mathbb Z)$. First we choose the basis $e\_1,e\_2, e\_3$ in $H^2(P,\mathbb Z)$ as follows. Let $S\_1\subset \mathbb CP^2\#\mathbb CP^2$ be the sphere generating $H\_2$ of the first summand and $S\_1$ of the second summand. Then we set $e\_1$ to be Poincare dual to $S\_1\times T^2$, $e\_2$ dual to $S\_2\times T^2$, and $e\_3$ dual to the fibre $M$ in $M\times T^2$. It is easy to see then, that


$$(a\_1e\_1+a\_2e\_2+a\_3e\_3)^3=3(a\_1^2+a\_2^2)a\_3=Q$$


Let us now choose the any positive class $h$ in $H^{1,1}(P)$ with $h^3=3$. It is easy to see that applying a linear transformation to $\mathbb R^3$ that preserves $Q$ we can send $h$ to the vector $(1,0,1)$. Let us now apply Hodge index theorem to $H^{1,1}$. First, the class $h$ induces a quadratic from on $H^{1,1}$ and this form should be definite on the orthogonal $h^{\perp}$ to $h$. The quadratic form is $(a\_1e\_1+a\_2e\_2+a\_3e\_3)^2h=$


$$=(a\_1e\_1+a\_2e\_2+a\_3e\_3)^2(e\_1+e\_3)=a\_1^2+a\_2^2+2a\_1a\_3=(a\_1+a\_3)^2+a\_2^2-a\_3^2.$$


So, we see, its signature is $(2,1)$. The vector $h$ is positive by definition, so the orthogonal to it has signature $(1,1)$. This contradicts the Hodge index theorem. QED.


By the way, would $P$ be complex projective, one would be able to prove that $h^{1,1}(P)=3$. I'll give the argument, even if it is not needed. Indeed, since $H^1(P,\mathbb Z)\cong\mathbb Z^2$, we have the Albanese map $A:P\to Alb(P)$, where $Alb(P)$ is a 1-dimensional abelian variety, i.e. an elliptic curve. Take a regular point $x\in Alb(P)$, then the fiber $A^{-1}(x)$ is smooth $2$-dimensional divisor on $P$ with zero square. However, since $P$ is projective, there is a class in $H^{1,1}(P)$ with positive cube. I follows $h^{1,1}\ge 2$, and so $h^{2,0}=h^{0,2}=0$.


