Simply-connected complex manifolds are symplectic

Question

Could you suggest a reference for the proof of the following fact:

Complex surfaces with trivial fundamental group are symplectic.

UPD: Once again I found my own question by googling the conditions on when complex surfaces have symplectic structure so it is unfortunate that this question is closed. Maybe I briefly explain why I care about this to reopen.

I'm interested in classification of symplectic 4-manifolds. Many natural examples of 4-manifolds are complex surfaces, so it is important for me to know which of them have symplectic structure.

UPD2: Thanks to Michael's kind comment and explanation, I'd like to add one more detail about what I tried.

I know how to show that concrete manifolds are symplectic by writing down explicitly the symplectic form. Also, I know that algebraic, projective manifolds can be endowed with symplectic structure by restricting Fubini-Studi on $\mathbb{C}\mathbb{P}^n$. But the general statement, I didn't have a faintest idea and googling wasn't helpful at that time.

Answer 1

As I pointed out in the comments above, the statement is not true. For example, Calabi-Eckmann manifolds are complex manifolds which are diffeomorphic to $S^{2m-1}\times S^{2n-1}$ where $m, n > 1$. They are simply connected, but they can't be symplectic as $b_2 = 0$. Note that such manifolds have complex dimension $m + n - 1 \geq 3$. On the other hand, if $X$ is a simply connected complex manifold of complex dimension $2$, then it is necessarily symplectic. To show this, first note that as $X$ is simply connected, it is in particular connected. We must now consider the compact and non-compact cases separately.

Suppose $X$ is compact. If $\pi_1(X) = 0$, then $b_1(X) = 0$. A connected compact complex surface with even first Betti number admits a Kähler metric, and is therefore symplectic. This statement was originally conjectured by Kodaira, and was later proved using the classification of compact complex surfaces, with the final case of K3 surfaces completed by Siu in the paper Every K3 surface is Kähler. Later, a proof which does not rely on the classification was found independently by Buchdahl and Lamari, see On compact Kähler surfaces and Courants kählériens et surfaces compactes respectively. For a textbook reference, see Chapter IV, Theorem 3.1 of Compact Complex Surfaces by Barth, Hulek, Peters, and Van de Ven.

Suppose $X$ is non-compact (note, $X$ is connected and has no boundary, so $X$ is open). Gromov used the h-principle to show that an open manifold admits a symplectic form if and only if it admits an almost complex structure. See 4.3.2 of Stable Mappings of Foliations into Manifolds by Gromov, also Example 3 in Lectures on the Theorem of Gromov by Haefliger. Note that we didn't use the simply connected assumption here (only connectedness to ensure $X$ is open). One can show that an open four-manifold admits a symplectic form if and only if it is orientable!

We can summarise the above as follows:

Let $X$ be a connected complex surface.

• If $X$ is compact, then $X$ admits a symplectic form if and only if $b_1(X)$ is even.
• If $X$ is non-compact, then $X$ admits a symplectic form.

Finally, let me just point out that in complex dimension $1$, every complex manifold is Kähler and hence symplectic.