As the title suggests, I have the following question:
Is there a compact complex manifold with $b_1(X)=b_2(X)=b_3(X)=b_4(X)=0$?
Clarification:
Denote by $b_k$ the $k$th Betti number of a compact complex manifold of positive dimension.
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4$\begingroup$ Do you mean Betti numbers? The point seems to work. $\endgroup$– Fernando MuroCommented Dec 31, 2021 at 1:55
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12$\begingroup$ I think the answer to mathoverflow.net/questions/117098/… applies here as well. $\endgroup$– dhyCommented Dec 31, 2021 at 2:05
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6$\begingroup$ In dimension 3 an example would be a homology $\Bbb{Q}$-sphere. This is very close to ask whether $\Bbb{S}^6$ admits a complex structure... $\endgroup$– abxCommented Dec 31, 2021 at 4:52
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4$\begingroup$ There are infinitely many six-dimensional rational homology spheres which admit almost complex structures (and infinitely many which do not), but as far as I know, it is unknown whether any of them admit a complex structure. Aside from dimension two, rational homology spheres in every other dimension cannot even admit an almost complex structure. $\endgroup$– Michael AlbaneseCommented Dec 31, 2021 at 12:46
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4$\begingroup$ @YCor Unless I'm mistaken, a compact Hermitian locally symmetric space would be Kähler, so $b_2$ would be positive. $\endgroup$– Donu ArapuraCommented Dec 31, 2021 at 13:27
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1 Answer
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There are complex manifolds with reduced cohomology vanishing in arbitrarily high degrees. Namely, product of two odd-dimensional spheres admits complex structure coming from representing it as a quotient of $(\Bbb C^n \setminus {0})\times (\Bbb C^m \setminus {0})$ by diagonal action of $\Bbb C$. They are known as Calabi-Eckmann manifolds.
Easiest example is $(\Bbb C^n \setminus {0})\times (\Bbb C^m \setminus {0}) / (e^At, e^At)$, where $A$ is a scalar matrix with non-real value $a$. It maps to a product of complex projective spaces by further quotiening, giving it structure of principal bundle with fiber being elliptic curve $\Bbb C / (\Bbb Z\cdot 1, \Bbb Z \cdot a)$. This map is the algebraic reduction of the manifold.
This easy example admits deformations of several types; for example, you can take different contracting flow instead of linear diagonal one. Most deformations do not have complex submanifolds at all, and have algebraic dimension zero as "most" generic complex manifolds.
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6$\begingroup$ This example was already brought up by dhy in the comments (by linking to Dmitri Panov's answer to another question) $\endgroup$ Commented Dec 31, 2021 at 14:18