Complex structures on $\mathbb{R}^{2n}$

Question

I am wondering what are the complex structures on $\mathbb{R}^{2n}$, up to diffeomorphism. I know that for $n = 1$, there is only one by the Riemann mapping theorem. However, there are more for $n >1$. For example, the unit polydisk $\{|z_1| < 1, \cdots, |z_n| < 1\}$ is not biholomorphic to the unit ball $\{|z_1|^2 + \cdots + |z_n|^2 < 1\}$; see https://mathoverflow.net/questions/154612/automorphism-groups-of-unit-disk-mathbfdn-and-unit-ball-bn. Since these are both diffeomorphic to $\mathbb{R}^{2n}$, I get different induced complex structures on $\mathbb{R}^{2n}$.

For example, what is the dimension of the moduli space of complex structures on $\mathbb{R}^{2n}$? Here I am not assuming anything about the complex structure at infinity but I could ask a similar question assuming that the complex structure at infinity is standard; what is the dimension in that case?

Note that it is known that there are infinitely many Stein structures on $\mathbb{C}^n$ of finite type up to Stein homotopy; this is by work of Mclean on Weinstein structures and work of Eliashberg/Cieliebak on equivalence of Weinstein and Stein.

Answer 1

Let me answer the part of your question which I find reasonable:

Suppose that $\sigma$ is a complex structure on ${\mathbb R}^{2n}$ which is "standard at infinity". Then $\sigma$ is biholomorphic to the standard complex structure on ${\mathbb R}^{2n}$. The only interesting case is that of $n\ge 2$. I will prove more:

Theorem. Suppose that $M$ is a contractible complex manifold of dimension $n\ge 2$, $K\subset M$ is compact, $f: M - K \to {\mathbb C}^n$ is a biholomorphic map to the complement of a compact subset of ${\mathbb C}^n$. Then $f$ extends to a biholomorphic map $F: M\to {\mathbb C}^n$.

Proof.

Lemma. $f$ extends to a holomorphic map $F: M\to {\mathbb C}^n$.

Note that if $M$ were biholomorphic to ${\mathbb C}^n$, this would be just the Hartogs extension theorem. A version of the extension theorem though holds in our case due to the following fact: The compact subset $K\subset M$ is contained in a larger compact subset $C\subset M$ with Levi-convex smooth boundary. This larger subset is bounded by the preimage $f^{-1}(S)$ of a round sphere $S\subset {\mathbb C}^n$ of sufficiently large radius. Now you use, for instance the main theorem in (sadly, behind the Jstor paywall)

J. Kohn and H Rossi, On the extension of holomorphic functions from the boundary of a complex manifold. Ann. Math., 81 (1965), 451-472.

or Corollary 4.3 in (freely downloadable)

A. Andreotti, C. D. Hill, E. E. Levi convexity and the Hans Lewy problem. I and II, Ann. Scuola Norm. Sup. Pisa 26 (1972), 325–363, 747–806.

to extend $f|\partial C$ to a holomorphic function $C\to {\mathbb C}^n$. But, again, in view of Levi-convexity of $\partial C$, we also have local uniqueness of extension, which, therefore, coincides with $f$ on $M-K$. Hence, $f$ extends to the entire $M$: Use $F$ to define the extension on $C$ and $f$ to define the extension on $M-C$. qed

By the maximum principle, $F$ is surjective. Since $M$ is contractible, $M$ cannot contain any compact analytic subvarieties of positive dimension (since each such subvariety has a "fundamental class"). Therefore, point preimages for $F$ are finite. The map $F$ then is a local homeomorphism (use the fact that $det(dF)$ can vanish only only on a compact analytic subvariety whose image under $F$ would then have to be finite). Since $F$ is a proper map, it has to be a covering map. But its target is simply-connected, hence, $F$ is a homeomorphism. It is a standard fact (see e.g. [Griffiths-Harris, Principles of Algebraic Geometry p. 19]) that a holomorphic homeomorphism is in fact biholomorphic. Hence, $M$ is biholomorphic to ${\mathbb C}^n$. qed