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A Calabi-Yau threefold is a compact Kahler threefold which is simply connected and has trivial canonical bundle.

So my question is as in the title. What is the moduli space of such objects? I'm interested in general info about how its geometry looks like, its irreducible components etc.

I do know a few examples of CY3s but mostly from algebraic geometry - a non-singular quintic in $ \mathbb{P}^4 $, or more generally, appropriate complete intersections in higher dimensional projective spaces / products of projective spaces.

The points of this supposed moduli space should correspond to isomorphism classes of CY3s. How is this space constructed?

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    $\begingroup$ By the Tian-Todorov Theorem, the local Kuranishi spaces are all smooth. So connected components are irreducible. The main open conjecture about the components is the Kawamata-Morrison Conjecture. $\endgroup$
    – Jason Starr
    Commented Nov 29, 2021 at 23:46
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    $\begingroup$ There are better MO users than me to discuss this, but according to the Beauville-Bogomolov decomposition (closely related to the Berger-de Rham decomposition), such manifolds are finite group quotients of a complex torus, a product of a K3 surface and an elliptic curve, or an "irreducible Calabi-Yau manifold", i.e., one for which the Hodge numbers $h^{p,0}$ equal $0$ for $p=1,2$. For the last class, all of these are projective. So you can construct moduli stacks of these manifolds using Artin's algebraization theorems, cf. "Versal deformations and algebraic stacks". $\endgroup$
    – Jason Starr
    Commented Nov 30, 2021 at 0:43
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    $\begingroup$ Since the Hodge numbers $h^{1,0}$ and $h^{2,0}$ vanish, the stacks of Calabi-Yau threefolds can be glued together from "charts" that are of the form $H_{n,d}/\text{Aut}(\mathbb{P}^n)$, where $H_[n,d}$ is the open subscheme of the Hilbert scheme parameterizing closed subschemes of $\mathbb{P}^n$ that are smooth Calabi-Yau threefolds of degree $d$. So the stacks are locally quite close to the scheme $H_{n,d}$. By the Tian-Todorov Theorem, this scheme is smooth and quasi-projective. $\endgroup$
    – Jason Starr
    Commented Nov 30, 2021 at 20:27
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    $\begingroup$ These are projective because the second cohomology group $H^2(X,\mathbb{C})$ is purely $(1,1)$. Since $H^2(X,\mathbb{Q})$ is dense in $H^2(X,\mathbb{C})$, and since the Kaehler cone is open in $H^2(X,\mathbb{C})$, it follows that there are elements of $H^2(X,\mathbb{Q})$ that are in the Kaehler cone. Then, by the Kodaira embedding theorem, a suitably positive and divisible integer multiple of this class equals the first Chern class of a very ample invertible sheaf. This is how you can embed the Calabi-Yau threefold in $\mathbb{P}^n$, and thus get a point of $H_{n,d}$. $\endgroup$
    – Jason Starr
    Commented Nov 30, 2021 at 20:46
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    $\begingroup$ Yes, that is correct. None of these arguments is my proof -- I have seen this argument repeated in lectures many times (unfortunately I cannot immediately think of a written reference, but I am sure that there is a reference). $\endgroup$
    – Jason Starr
    Commented Nov 30, 2021 at 21:00

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