It is not difficult to build new manifolds out of old in the smooth category, for example

  • taking the direct product or constructing a fiber bundle,

  • taking the level set of a regular value of a smooth function,

  • quotienting by the free action of a cpt Lie group,

  • forming the connected sum and more generally gluing (in different ways) two manifolds

The connected sum construction is very flexible and important for the topological classification of manifolds. Unfortunately, in the complex category it doesn't behave well *: maybe the connected sum doesn't admit a complex structure at all. In my attempt to understand the topology of complex manifolds, it is natural to ask

What are the most common/useful constructions that allows us to form a new complex manifold $X = \mathcal{F}(M, N)$ given two complex manifolds $M,N$?

The constructions I am interested in should have a topological interpretation and when meaningful the complex structure on $X$ should be compatible with the one of $M,N$.

*see Does $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ ever support an almost complex structure? and https://math.stackexchange.com/questions/1411694/is-the-connected-sum-of-complex-manifolds-also-complex)

  • $\begingroup$ What do you mean by "most powerful"? I mean, direct product (for instance) makes perfect sense for complex manifolds. $\endgroup$ – Francesco Polizzi Feb 19 at 22:53
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    $\begingroup$ Recurring/useful to understanding the topology or complex structure of various class of manifolds. For example with the connected sum and just 3 models we can form all the compact (smooth) manifolds of (real) dimension 2. The question is of course deliberately vague. $\endgroup$ – Warlock of Firetop Mountain Feb 19 at 23:13
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    $\begingroup$ I guess that, if there exists a complex structure on $S^{6}$ then there is a natural connect sum operation for complex $3$-folds. This is definitely true in the almost complex category (see exercise 7.28 of "Introduction to symplectic topology" McDuff and Salamon), I would guess that the same idea would work if $S^{6}$ has a complex structure. $\endgroup$ – Nick L Feb 19 at 23:46

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