7
$\begingroup$

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".

$\endgroup$
  • 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 '14 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 '14 at 9:34
  • 1
    $\begingroup$ @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 '14 at 10:22
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.