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By a threefold, I mean a compact complex manifold of dimension three.

My question is a simple one:

Are there known INFINITELY many non-homeomorphic threefolds that have the same Betti numbers and the same Chern numbers?

I would like to know answers to cases of other dimensions.

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1 Answer 1

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The complex parallelizable (hence, all Chern classes are trivial) Iwasawa manifold is constructed by taking the complex Lie group of matrices of the form $$\begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix}$$ and modding out by the subgroup of matrices with entries in the Gaussian integers $\mathbb{Z}[i]$. The Betti numbers are $1,4,8,10,8,4,1$, calculated by forming the corresponding cdga generated by global 1-forms, calculating cohomology there, and applying Nomizu's theorem which tells us the inclusion of that cdga into all forms on the quotient manifold is a quasi-isomorphism.

Now, notice that we can quotient the above Lie group by different lattices, e.g. fixing an integer $n$, by those where $x,y$ are Gaussian integers, but $z$ is allowed to be of the form $a + i \frac{b}{n}$ for $a,b$ integers. The corresponding quotient manifold is also complex parallelizable, with the same Betti numbers as the Iwasawa manifold above (both following from the same arguments that apply there), but has the Iwasawa manifold as an $n$-sheeted covering. Varying $n$, this gives you infinitely many compact threefolds with pairwise distinct fundamental groups. Cross with any fixed compact complex manifold to get examples in higher dimension.

In dimension four, you can use the Kodaira-Thurston manifold, which is constructed by a similar procedure as above, except one starts with just a real Lie group, namely real matrices of the form $$\begin{pmatrix} 1 & x & z & 0 & 0 \\ 0 & 1 & y & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & u \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}.$$ The quotient by the integer matrices (or scaled versions as above, where we take the $z$ entry to be an integer divided by a fixed $n$) admit complex structures (the Kodaira-Thurston manifold is famous for admitting complex and symplectic structures, but not admitting a Kähler metric). Since we are starting with a real Lie group, complex parallelizability is not a priori guaranteed, but since you only ask for equality of Chern numbers, this follows from the Euler characteristic being trivial (hence $c_2 = 0$) and ordinary parallelizability (giving $0 = p_1 = c_1^2 - 2c_2$ so $c_1^2 = 0$). Again this property and the Betti numbers are shared by all instances of this construction for varying $n$, but the fundamental groups are pairwise non-isomorphic. We could have also constructed examples for your question in all higher dimensions by starting with this family of complex surfaces.

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  • $\begingroup$ Great answer! What if simply-connectedness is added to the conditions? Maybe it's asking too much? $\endgroup$
    – Basics
    Commented Jan 2, 2023 at 16:48
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    $\begingroup$ You might look at Example 14, due to Oguiso, in "Cubic forms and complex 3-folds" by Okonek-Van de Ven. Namely, they write Oguiso "proves that for every integer $d\geq 1$ there exists a smooth complete intersection $X_{d'}$ of type (2,4) in $\mathbb{P}^5$ which contains a non-singular rational curve $C_d$ of degree $d$ with normal bundle $\mathcal{O}(-1)^{\oplus 2}$." Then taking the flop of $X_{d'}$ along this curve gives simply connected 3-folds which, if I am understanding correctly, have equal Betti numbers, and have trivial canonical (hence only $c_3$ is non-trivial, and equal for all). $\endgroup$ Commented Jan 2, 2023 at 18:03
  • $\begingroup$ However, the ring structures on integral cohomology differ, so they are not homotopy equivalent. I do not know what exactly happens in the flop construction, so I am not sure if my interpretation above is correct. $\endgroup$ Commented Jan 2, 2023 at 18:04
  • $\begingroup$ Also I suspect easier simply connected examples can be found in the Okonek-Van de Ven paper. $\endgroup$ Commented Jan 2, 2023 at 18:19

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