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I am looking for an example of a compact complex analytic space, reduced and irreducible, which does not admit any holomorphic embedding into any (smooth) complex manifold (possibly non-compact).

I am aware of many examples of a similar kind of algebraic varieties which cannot be embedded in any smooth scheme, see for example this question and references there (in particular, the paper of Roth-Vakil linked there, and the corrigendum and further comments at the bottom of Vakil's page). It is possible that the Roth-Vakil example also works for my question, but I don't quite see how to prove it.

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  • $\begingroup$ One approach is to find a proper algebraic space $X$ that is local complete intersection, everywhere nonreduced, and such that every locally free sheaf on $X$ is a direct sum of copies of the structure sheaf. That is not as far-fetched as it sounds: the plane with double structure from Hartshorne, Exercise III.5.9, p. 299 (at least in the Russian copy) has no invertible sheaves other than the structure sheaf. If $X$ is contained in a complex manifold $Y$, then for the associated ideal sheaf $\mathcal{I}$, the conormal sheaf $\mathcal{I}/\mathcal{I}^2$ is locally free . . . $\endgroup$
    – Jason Starr
    Commented Feb 21, 2018 at 1:49
  • $\begingroup$ . . . If $X$ is swept out by rational curves, then triviality of $\mathcal{I}/\mathcal{I}^2$ should imply that those rational curves deform out of $X$ to cover $Y$. It might be possible to choose a family of these curves that sweep out a hypersurface in $Y$ whose intersection with $X$ is a hypersurface in $X$. In the case of smooth schemes, every hypersurface on $Y$ gives a line bundle, and the restriction of this line bundle to $X$ is a nontrivial invertible sheaf, contradicting that $\text{Pic}(X)$ is trivial. $\endgroup$
    – Jason Starr
    Commented Feb 21, 2018 at 1:55
  • $\begingroup$ @JasonStarr Thank you for the comment; quick question: I was looking for $X$ reduced though. Also, if in the Roth-Vakil example one can show that divisors on $X$ can be deformed out into $Y$, then perhaps one can obtain a contradiction as in their paper, which in the end relies on intersection numbers calculations $\endgroup$
    – YangMills
    Commented Feb 21, 2018 at 2:01
  • $\begingroup$ Here is a variant that is reduced. It has been many years since I spoke with Roth and Vakil about these things -- this may be similar to examples in their paper. Let $f :S\to B$ be a non-projective, Kaehler K3 with an elliptic fibration. Let $i_0:F_0\hookrightarrow S$ and $i_\infty:F_\infty\hookrightarrow S$ be smooth fibers, and let $g:F_0\to F_\infty$ be an isogeny of positive degree. Form the cofiber coproduct $X$ of the two morphisms, $i_0\sqcup i_\infty: F_0\sqcup F_\infty \hookrightarrow S$ and $g\sqcup \text{Id}:F_0\sqcup F_\infty \to F_\infty$. This is reduced and LCI . . . $\endgroup$
    – Jason Starr
    Commented Feb 21, 2018 at 2:16

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