Let $M$ be a closed complex manifold that is not deformation equivalent to a complex projective manifold.

Can $M$ be orientedly diffeomorphic to a complex projective manifold? What if $M$ is moreover Kähler?

MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.

Sign up to join this community
Anybody can ask a question

Anybody can answer

The best answers are voted up and rise to the top

$\begingroup$
$\endgroup$

Let $M$ be a closed complex manifold that is not deformation equivalent to a complex projective manifold.

Can $M$ be orientedly diffeomorphic to a complex projective manifold? What if $M$ is moreover Kähler?

$\begingroup$
$\endgroup$

I believe the answer is yes and follows from the combination of Theorem 4.6 here https://arxiv.org/pdf/math/0111245.pdf and Theorem 1.3 here https://arxiv.org/pdf/math/0111245.pdf

The first result shows that deformations of standard complex tori are complex tori (i.e. $\mathbb C^n/\Gamma$ where $\Gamma\cong \mathbb Z^{2n}$). The second result shows that on $T^6$ there is an infinite dimensional family of complex structures.

**PS.** As for the second version of the question, where $M$ is asked additionally to be Kahler, I would guess that it can be safely counted as an open problem. Remember in https://link.springer.com/article/10.1007/s00222-003-0352-1 Voisin solved negatively *Kodaira problem* by constructing the first ever example of a Kahler manifold that is not deformation equivalent to a projective one. It seems to me that since then no new examples were found of such a phenomenon. And as you pointed out in a different post, it was proven recently that in dimension $3$ every Kahler manifold can be deformed to a projective one, but this is hard work (strongly relying on dim $3$). So in order to advance in your question one has to advance in one of these two directions - trying to extend the $3$-dimensional result to dimension $4$ and trying to look for new constructions of Kahler manifolds...

$\begingroup$
$\endgroup$

Yes.

As everyone knows, flat complex structures on $\Bbb R^4$ are parametrised by two copies of $\Bbb CP^1$. Take your favourite elliptic curve $E$; map $e \mapsto -e$ gives two-sheeted ramified covering $p: E \to \Bbb CP^1$. Consider a complex structure on $E \times T^4$ where complex structure on the fiber over $e \in E$ corresponds to the point $p(e)$. This almost complex structure is integrable (somewhat easy exercise) and non-Kahler (if it was, then corresponding Kahler metric would be flat and globally pluriclosed, but computation shows that such metric could not be compatible).

There is a paper by Catanese from early 00s where he shows that any deformation of a flat complex 3-dim torus is again a flat torus (and same result for products of curves of higer genus with torus $\Sigma \times T^4$). (UPD: it's the first link in Dmitri's answer)

I'm not aware of any Kahler examples.