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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.

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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

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  • $\begingroup$ And the reason is? I know about the complex tori, of course. $\endgroup$ Mar 17, 2022 at 12:46
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    $\begingroup$ 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. $\endgroup$
    – adhalanay
    Mar 17, 2022 at 12:55
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    $\begingroup$ Please present a bit more detail of the argument in your post. (You also say dimension at least four in contrast to the op who asks for dimension 3 - perhaps this is a typo?) $\endgroup$
    – KReiser
    Mar 17, 2022 at 14:13
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This seems very simple, but I believe it is correct. Upon revisiting it again, it should never have been so complicated.

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.

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