# 'Non-algebraic Calabi-Yau' threefolds

By a Calabi-Yau threefold I mean a simply connected compact Kahler threefold with trivial canonical bundle.

By an algebraic compact complex manifold, I mean one that admits a closed immersion into a complex projective space.

What are examples of non-algebraic Calabi-Yau threefolds? Or is there none?

What I know: if the Calabi-Yau threefold $$M$$ has $$h^{2,0} (M) = 0$$, then it is algebraic. (Algebraicity holds for any compact Kahler manifold with the vanishing condition, btw). So any such example must have nonzero $$h^{2,0}$$.

I don't know the Ricci flow viewpoint to this subject, so I'm probably missing something but I'm eager to learn.

All Calabi-Yau threefolds as you defined them are algebraic. In fact any Calabi-Yau manifold of complex dimension $$\geq 3$$ is algebraic. An important property here is that your manifold is simply connected. Otherwise there are many examples like complex tori which are not algebraic. The only proof that I am aware of is due to D. Joyce and is published in [this paper][1] and also in
Gross, Mark (ed.); Huybrechts, Daniel (ed.); Joyce, Dominic (ed.), Calabi-Yau manifolds and related geometries. Lectures at a summer school in Nordfjordeid, Norway, June 2001, Universitext. Berlin: Springer. viii, 239 p. EUR 49.95/net; sFr. 86.00; \textsterling 35.00; $59.95 (2003). ZBL1001.00028. [1]: https://arxiv.org/abs/math/0108088 • And the reason is? I know about the complex tori, of course. Mar 17, 2022 at 12:46 • Always a Calabi-Yau manifold has$h^{2,0}=0\$. A nice proof can be found in link.springer.com/book/10.1007/978-3-642-19004-9. Mar 17, 2022 at 12:55
If $$X$$ is a CY3 as defined in the question, $$X$$ is simply connected so $$H^1(X, \mathbb{C} ) = 0$$ after abelianization of $$\pi_1$$ and Poincare duality. Next since $$X$$ is Kahler, $$H^1(X, \mathcal{O}_X) = 0$$ and finally using Serre duality alongwith the fact that $$\omega_X$$ is trivial implies $$H^2(X, \mathcal{O}_X ) = 0$$. Then the general fact stated in the original question implies that $$X$$ is projective.