What is the relationship between a complex manifold being Kähler, projective, nonprojective, and nonKähler? I was wondering if this implication is true. I read a few places that
$$\text{nonprojective} \Longrightarrow \text{nonKähler}$$
but I think I maybe have misunderstood. Equivalently, this is of course asking if $$\text{Kähler} \Longrightarrow \text{projective ?}$$
 A: I give the Campana-Moishezon criterion here which is less known.
Definition: A complex manifold $X$ is algebraically connected if for any general pair of points $x, y \in X,$
there exists a proper, connected and not necessarily irreducible curve in $X$ containing $x$ and $y$. 
Campana-Moishezon criterion: A compact complex manifold is projective if and only if it is Kähler and algebraically connected.
Note that by another result from Campana et al,
A compact hyper- Kahler manifold $X$ is projective if and only if the Néron-Severi group $NS(X)$ is hyperbolic, I.e., negative semi definite but not negative definite with respect to Beauville-Bogomolov-Fujiki's form
F. Campana. Coréduction algébrique d’un espace analytique faiblement kählérien compact. Invent. Math., 63(2):187–223, 1981.
https://projecteuclid.org/euclid.jdg/1292940689
A: Both of these implications are incorrect.
As $\mathbb{CP}^n$ is Kähler and complex submanifolds of Kähler manifolds are Kähler, all projective manifolds are Kähler. The converse however is not true. That is, there exist Kähler manifolds which are not projective. For example, all two (complex) dimensional tori $\mathbb{C}^2/\Lambda$ are Kähler, but many of them are not projective. In order to be projective, the lattice must satisfy the so-called Riemann conditions (see Griffiths & Harris, Chapter 2, Section 6).
In summary, projective implies Kähler. Taking the contrapositive, we see that non-Kähler implies non-projective.
