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Here are four remarks about the homology and homotopy type of a compact, complex manifold $M$:

  1. If $M$ is Kähler, then it is symplectic and thus $H^2(M,\mathbb{R}) \ne 0$. (Also, as explained in a blog posting by David Speyer, you still have $H^2(M,\mathbb{R}) \ne 0$ even if $M$ is non-projective but algebraic.)

  2. An interesting first example of a non-Kähler manifold is a Hopf manifold, by definition $(\mathbb{C}^n\setminus 0)/\Gamma_r$, where $\Gamma_r$ is a rescaling by $r$ with $|r| \ne 0,1$. This example has $H^1(M,\mathbb{R}) \ne 0$.

  3. On the other hand, even-dimensional, compact Lie groups have left-invariant complex structures. If $M$ is such a manifold and is simply connected, then it is also 2-connected. $H^1(M,\mathbb{Z}) = H^2(M,\mathbb{Z}) = 0$ and $M$ is manifestly not Kähler. On the other hand, no such example is 3-connected and you always have $H^3(M,\mathbb{R}) \ne 0$.

  4. There is (or was) a long-standing conjecture that no even-dimensional sphere other than $S^2$ has a complex structure.

So, question: Is there for each $n$, a compact, complex manifold $M$ which is $n$-connected?

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2 Answers 2

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E. Calabi, B. Eckmann, A class of compact, complex manifolds which are not algebraic. Ann. of Math. (2) 58, (1953). 494–500.

From Chern's MR review (MR0057539):

This paper defines on the topological product $S^{2p+1} \times S^{2q+1}$ of two spheres of dimensions $2p+1$ and $2q+1$ respectively, $p$ > 0, a complex analytic structure. The complex manifold so obtained ... admits a complex analytic fibering, with two-dimensional tori as fibers and having as base space the product $\mathbb{P}^p \times \mathbb{P}^q$ of complex projective spaces of (complex) dimensions $p$ and $q$ respectively.
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  • $\begingroup$ A slam dunk for the question! I did do a few Google searches (including Google Books and Google Scholar) and I just didn't see it. $\endgroup$ Commented Jun 13, 2010 at 21:36
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    $\begingroup$ There's also a Wikipedia page, with a construction that's very similar to that of a Hopf manifold. en.wikipedia.org/wiki/Calabi%2dEckmann_manifold $\endgroup$ Commented Jun 13, 2010 at 21:53
  • $\begingroup$ My googling was directed by your Lie group examples, among which there are certain products of spheres! (BTW, when you described the Hopf manifolds, I expect you meant to delete zero.) $\endgroup$
    – Tim Perutz
    Commented Jun 13, 2010 at 23:07
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As shown by Calabi and Eckmann, products of odd-dimensional spheres admit complex structures. See Anns of Maths 58, 1953, 494-500.

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