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Bounded non-symmetric domains covering a compact manifold

MathOverflow Asked by diverietti on November 3, 2021

This question is somewhat related to this other question of mine.

I was wondering which are the known examples of bounded domains $Omega$ in $mathbb C^n$ admitting a compact free quotient.

By a theorem of Siegel, such a domain must be holomorphically convex. Moreover, if the boundary is sufficiently regular, say $C^2$ (even if, by a recent theorem of A. Zimmer, $C^{1,1}$ suffices), by the classical theorem of Wong-Rosay, then $Omega$ must be biholomorphic to the unit ball.

Of course all bounded symmetric domains give such examples, by a classical theorem of E. Borel. But I am interested in more "exotic" examples, specifically non symmetric examples.

The only I am aware of live in $mathbb C^2$ and are the universal covers of Kodaira fibrations (see this question for more details).

  • Is it possibile for instance to construct higher dimensional analogous of the universal cover of a Kodaira fibration?

For dimension $nge 4$, there exists (many in fact) homogeneous bounded domains in $mathbb C^n$ which are non symmetric. In 1979 J. E. D’Atri proved that there exists bounded homogeneous non-symmetric domains, for $nge 6$, whose Bergman metric has positive holomorphic sectional curvature somewhere.

Unfortunately, they can never cover a compact manifold. Indeed, it was shown by J. Hano in 1957 that if a homogeneous bounded domain covers a manifold of finite volume, i.e. if its automorphism group admits a lattice so that it’s unimodular, than the domain is in fact symmetric.

Remark that I am really looking for compact, discrete, free quotients.

Thank you very much in advance.

One Answer

A silly generalization would be a direct product of Koadaira's surface and, say, a Riemann surface. A better construction is below.

I will use

Griffiths, Phillip A., Complex-analytic properties of certain Zariski open sets on algebraic varieties, Ann. Math. (2) 94, 21-51 (1971). ZBL0221.14008.

specifically, Lemma 6.2: Suppose that $Uto S$ is a (nonsingular) holomorphic family of compact Riemann surfaces such that $S$ is uniformed by a bounded contractible domain of holomorphic in ${mathbb C}^n$. Then the same holds for $U$.

Given this, one inducts: Take a compact Kodaira surface $Ksubset {mathcal M}_g$, let $xi: {mathcal M}_{g,1}to {mathcal M}_{g}$ be the "universal curve." Then the pull-back of $xi$ to $K$ is a holomorphic family of genus $g$ Riemann surfaces $Uto K$ as in Griffiths lemma. Hence, $U$ is compact and is again uniformed by a bounded domain in ${mathbb C}^3$. To continue, one needs a trick since $U$ lies in ${mathcal M}_{g,1}$ and the universal curve over that will have noncompact fibers. However, one can regard a puncture on a genus $g$ surface as an orbifold cone-point of order $2$, hence, ${mathcal M}_{g,1}$ (as an orbifold) holomorphically embeds in some ${mathcal M}_{h}$. To prove this, take a 2-dimensional oriented compact connected orbifold ${mathcal O}$ of genus $g$ with one cone point of order $2$. It admits a finite manifold-covering $S_hto {mathcal O}$. Hence, the moduli space of ${mathcal O}$ embeds holomorphically (as an orbifold) in ${mathcal M}_h$.

Thus, $U$ is embedded in ${mathcal M}_{h}$ and, so we can continue.

Edit. I do not know how to prove that in general these domains are non-symmetric (in dimension 2 this is understood). But, in all the examples I am aware of, the compact complex manifolds given by this construction are non-rigid and, hence, cannot be locally symmetric (except for trivial families which I ignore).

Answered by Misha on November 3, 2021

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