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Is there a smooth (complex projective) 3-fold of general type which is homeomorphic (in the complex topology) to a rational $3$-fold?

I am aware of such examples in complex dimension $2$, for example the Barlow surface is homeomorphic to a blow up $\mathbb{P}^{2}$ in 8 points.

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    $\begingroup$ To the best of my knowledge, I do not remember any smooth threefold of general type with $h^{1,0}=h^{2,0}=h^{3,0}=0$. However, if you allow basically harmless (e.g. quotient) singularities you can find examples of canonically polarised threefolds where $|-K|$ is empty (and the other hodge number vanishes as well by Lefschetz). Go to grdb.co.uk/search/gt3 and simply pick the threefold where no ambient weight is equal to 1. (of course this is just a necessary condition, far from sufficient) $\endgroup$
    – Enrico
    Commented Dec 18, 2018 at 10:42
  • $\begingroup$ Maybe it's worth to ask the authors of arxiv.org/abs/1606.09237 ... $\endgroup$
    – Dmitri Panov
    Commented Dec 18, 2018 at 14:29
  • $\begingroup$ @ Enrico, thanks, how do you show that the homeomorphism preserves Hodge numbers? It will be true that $h^{1,0}=0$ since rational $3$-folds are simply connected, but I am not sure about $h^{2,0},h^{3,0}$. $\endgroup$
    – Nick L
    Commented Dec 18, 2018 at 16:02
  • $\begingroup$ For sure all $h^{p,0}$ are birational invariants. I am not how to prove the full statement though $\endgroup$
    – Enrico
    Commented Dec 18, 2018 at 23:43
  • $\begingroup$ Presumably topologists are well-aware of the following, but I just learned of the Manifold Atlas Project. There is a wonderful page about simply connected 6-manifolds and a complete list of homeomorphism invariants: map.mpim-bonn.mpg.de/6-manifolds:_1-connected I will try to see if the simply connected examples from the following answer might be homeomorphic to rational 3folds: mathoverflow.net/questions/318990/… $\endgroup$
    – Jason Starr
    Commented Dec 19, 2018 at 22:02

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This is not a complete solution. We prove that 69 out of the 105 smooth Fano $3$-folds are not homeomorphic to a 3-fold of general type.

The proof is similar to Sai-Kee Yeung's proof of the non-existence of a "fake $\mathbb{CP}^{3}$" (Theorem 3.1 of http://journals.math.ac.vn/acta/pdf/1001199.pdf), which I learned about is dhy's comment here 3-folds with "simple" Betti numbers and positive Kodaira dimension and relies on the Miyaoka-Yau inequality.

In particular, we prove the following: There is no smooth $3$-fold of general type with $b_{1}=0$ and $b_{3} \leq 2$.

Proof: By the assumption $b_{1}=0$, $h^{1,0}=0$. Then by applying the Hirzbruch Riemann-Roch theorem to $\mathcal{O}_{X}$ we obtain:$$\frac{c_{1}c_{2}(X)}{24} = \chi(\mathcal{O}_{X}) = 1 + h^{2,0} - h^{3,0}$$. Since $b_{3} \leq 2$, $h^{3,0} \leq 1$ and hence $c_{1}c_{2}(X) \geq 0$. Then by the Miyaoka-Yau inequality for smooth $3$-folds $(c_{1})^{3} \geq \frac{8 c_{1}c_{2}}{3} \geq 0$ contradicting the fact that $K_{X}$ is ample. (The exact form of the Yau inequality we use is that same as in Yeung's paper, namely that $8c_{2}(X)- 3c_{1}^{2}(X)$ is a pseudo-effective class).QED.

To obtain the statement about Fano's, it is well-known that smooth Fano's satisfy $b_{1}=0$ (they are even simply connected), and looking at the Ivskovskikh-Mori-Mukai list, we see that 69 families satisfy $b_{3} \leq 2$, in particular no smooth Fano $3$-fold with Picard rank at least $4$ is homeomorphic to a smooth $3$-fold of general type. By Miyaoka's generalisation of Yau's inequality the class $3c_{2}-c_{1}^{2}$ is pseudo-effective for $3$-folds with $K_{X}$ nef and big, so the above applies to this class also.

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