# Bounded non-symmetric domains covering a compact manifold

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 $$n\ge 4$$, 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 $$n\ge 6$$, 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.

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 $$U\to S$$ 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 $$K\subset {\mathcal M}_g$$, 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 $$U\to K$$ 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 $$S_h\to {\mathcal O}$$. 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).

• Hey Misha! Long time no see! Hope you are fine. Thank you very much for your answer, I’ll read it carefully. Hope to meet you soon, best. Commented Jul 17, 2020 at 22:18