If two complex projective manifolds are homotopy equivalent are they homeomorphic?

1$\begingroup$ Just a remark: The simplest examples for homotopy equivalent, nonhomeomorphic manifolds are lens spaces. Since they are quotients of odd dimensional spheres, they won't have a complex structure, though. $\endgroup$– HenrikRüpingSep 3 '20 at 19:16

$\begingroup$ Idea to look for possible candidates, following an idea from my (wrong) deleted answer. Are there two nonhomeomorphic Kodaira fibered surfaces with isomorphic fundamental group? $\endgroup$– Francesco PolizziSep 3 '20 at 19:24

3$\begingroup$ @FrancescoPolizzi: Wouldn't such beasts give a counterexample to the Borel conjecture? $\endgroup$– Mark GrantSep 3 '20 at 19:52

2$\begingroup$ @Henrik: a minor remark but presumably you mean homotopy equivalent, nonhomeomorphic closed manifolds; for open manifolds much easier examples are possible, of course. $\endgroup$– Qiaochu YuanSep 3 '20 at 21:55

4$\begingroup$ I believe that there are holomorphic $\mathbb {CP}^2$ bundles over $\mathbb {CP}^3$ that are homotopy equivalent but have different Pontrjagin classes. Thus they are easily seen to be not diffeomorphic. Given the difficult theorem of topological invariance of Pontrjagin classes, they are in fact not homeomorphic. $\endgroup$– Ben WielandSep 3 '20 at 23:53
For curves this follows from the classification of (2dimensional topological) surfaces, and for simplyconnected surfaces this follows from Freedman's theorem.
My former colleagues Anatoly Libgober and John Wood found examples of pairs of 3folds which are complete intersections and are homotopy equivalent but not diffeomorphic, in fact have distinct Pontryagin classes. See Example 9.2. Since in this case $H^4(M;\mathbb{Z})\cong \mathbb{Z}$, this implies that the manifolds are not homeomorphic by the topological invariance of rational Pontryagin classes (see Ben Wieland's comment).
For the higher dimensional case see:
Fang, Fuquan, Topology of complete intersections, Comment. Math. Helv. 72, No. 3, 466480 (1997). ZBL0896.14028.

2$\begingroup$ A nice summary of what is known about the classification of complete intersections can be found here. $\endgroup$ Sep 5 '20 at 1:03

$\begingroup$ what about the surfaces with nontrivial $\pi_1$? $\endgroup$– user164740Sep 5 '20 at 13:34

$\begingroup$ @JoeT the only thing that I’m aware of is that for aspherical manifolds, the Borel conjecture predicts that they should be homotopy rigid (homotopy implies homeomorphism). This is known for products of curves and ball quotients (complex projective spaces), but is open in general. For manifolds with nontrivial π_1, π_2 or π_3, I’m not sure what to expect. Part of the difficulty is that surgery theory is not completely understood in 4D  this has to do with Freedman’s ABslice conjecture. $\endgroup$– Ian AgolSep 5 '20 at 13:46

$\begingroup$ @JoeT Freedman’s theory carries over to 4manifolds with sub exponential growth. But one would have to consult an expert for the classification of such manifolds. The only known finitely presented groups with sub exponential growth are polynomial growth, so virtually nilpotent. There may be interesting surfaces with nilpotent fundamental group, and one could try to see how the theory applies in that case. $\endgroup$– Ian AgolSep 5 '20 at 13:49
EDIT: Oops, I just remembered that you're asking the manifolds to be projective, which these aren't. Still, it's an example for just complex manifolds.
The CalabiEckman manifold (https://en.wikipedia.org/wiki/Calabi%E2%80%93Eckmann_manifold) is the quotient of $\mathbb{C}^m \setminus 0 \times \mathbb{C}^n \setminus 0$ by the holomorphic $\mathbb{C}$action $t(x,y) = (e^t x, e^{\alpha t}y)$ for some fixed nonreal $\alpha$. This quotient is a complex manifold diffeomorphic to $S^{2m1} \times S^{2n1}$. It is clear that the usual Lens space action on each of the factors commutes with this $\mathbb{C}$action, and so we obtain a complex structure on products of Lens spaces. As mentioned in the comments, there are examples of homotopyequivalent nondiffeomorphic Lens spaces, so this should furnish an example. (I believe Lens spaces are not so pathological that they could be nondiffeomorphic but become diffeomorphic after taking a product with e.g. $S^1$.)
Edit: I misread the question. The statement below explains only, that if a homotopy complex projective space other than $\mathbb{CP}^3$ supports a complex projective structure, then the answer would be no. As far as I know, it is not known if such spaces support even a symplectic structure.
Let us call a manifold which is homotopy equivalent to a complex projective space a homotopy complex projective space (HCP). In dimension 6 there are $\mathbb Z$ many manifolds (up to diffeomorphism) with homotopy type of $\mathbb{CP}^3$. They are distinguished by their first Pontryagin class. In dimension $6$ we have that (under certain conditions, which are fullfilled for HCPs) if a topological manifold admits a smooth structure, then this structure is unique. Hence if two HCPs would be homeomorphic, they would be diffeomorphic, hence they would have the same first Pontryagin class. But as I mentioned above there are $\mathbb Z$ many HCPs with pairwise different first Pontryagin classes.

3$\begingroup$ Are these manifolds the underlying topological manifolds of complex projective manifolds? $\endgroup$ Sep 3 '20 at 17:15

$\begingroup$ is there an easy way to rule out the existence of an almost complex or a complex structure on these manifolds? $\endgroup$– user164740Sep 3 '20 at 17:42

$\begingroup$ No, every closed, oriented $6$manifold with vanishing third integer cohomology supports an almost complex structure. $\endgroup$ Sep 3 '20 at 17:44