Complex tori are not associated to projective varieties in general.

But can one find an open $U$ inside a complex torus $\mathbb C^g/L$ such that $U$ is the analytification of a quasi-projective variety?

What if we just ask $U$ to be the analytification of a (finite type separated) scheme?

I think that if $U$ comes from a scheme, then $U$ has a compactification which should map to the complex torus $\mathbb C^g/L$. This map should be birational proper and therefore $\mathbb C^g/L$ is Moishezon (and Kahler). Therefore, it is projective.

So, probably the answer is "you can find a quasi-projective open if and only if the torus is algebraic".

  • 1
    $\begingroup$ If $U$ is the analytification of a $g$-dimensional integral separated scheme, then that scheme has a set of $g$ algebraically independent elements in its function field. $\endgroup$
    – S. Carnahan
    Nov 18, 2014 at 9:31
  • $\begingroup$ There exist varieties which are not projective but birational to projective varieties, so a priori the answer could be yes. $\endgroup$
    – Giulio
    Nov 18, 2014 at 9:34
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    $\begingroup$ @S.Carnahan Thank you for your comment. If I understand correctly, it answers the question (very easily). The open subset $U$ has $g$ algebraicaly independent elements in its function field. Therefore, so does $\mathbb C^g/L$ (as $U$ and $\mathbb C^g/L$ have the same function field). But then we conclude that $\mathbb C^g/L$ is Moishezon and thus algebraic (projective). $\endgroup$
    – Steven
    Nov 18, 2014 at 10:22

1 Answer 1


The answer is negative if the complex torus is not algebraic (or equivalently, not Moishezon). More generally, if $U$ is the analytification of a separated scheme of finite type over $\mathbf{C}$ and $Y$ is a proper complex-analytic space admitting an open immersion $j:U \hookrightarrow Y$ onto the complement of a nowhere-dense analytic set then $Y$ must be Moishzeon. However, even if $Y$ is a manifold, without knowing the geometry of the complement of $Y-U$ in $Y$ perhaps the restriction map $M_Y\rightarrow j_{\ast}(M_U)$ with sheaves of meromorphic functions is not an equality on global sections (i.e., the Levi extension theorem may not apply). Thus, to prove that $Y$ must be Moishezon it is necessary to use more serious input from complex analysis as follows.

We may assume $Y$ is reduced and irreducible (by the very definition of "Moishzeon") and may shrink $U$ to be affine (if it weren't already quasi-projective) and even smooth (by reducedness). The normalization $Y'$ of $Y$ is connected and $Y' \rightarrow Y$ is an isomorphism over $U$, and $M(Y) = M(Y')$ by the theory of analytic normalization, so we may replace $Y$ with $Y'$ to arrange that $Y$ is normal. By resolution of singularities, $U$ is the complement of a strict normal crossings divisor in a connected projective manifold $X$. Borel's extension theorem extends $U\rightarrow Y$ to $\nu:X\rightarrow Y$ by normality of $Y$. But $M_Y\rightarrow\nu_{\ast}(M_X)$ is an isomorphism by 10.6.2 in "Coherent Analytic Sheaves" (or really Remark 8.1.3 in loc. cit.), which rests crucially on Grauert's coherence theorem, so $M_Y(Y) = M_X(X)$ has transcendence degree $\dim X = \dim Y$ over $\mathbf{C}$. Hence, $Y$ is Moishezon.


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