As the title suggests, I have the following question:



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> Is there a compact complex manifold with $b\_1(X)=b\_2(X)=b\_3(X)=b\_4(X)=0$?
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Clarification:


Denote by $b\_k$ the $k$th Betti number of a compact complex manifold of positive dimension.



Is there a compact complex manifold with $b_1(X)=b_2(X)=b_3(X)=b_4(X)=0$?
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.


