Topological factors of complex projective manifolds

Question

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 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 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.

Answer 1

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 $h^{2,0}=h^{0,2}=0$.