holomorphic exoticness

A topological manifold is an exotic copy of another smooth manifold if it is homeomorphic to it, but not diffeomorphic (and when you switch diffeomorphic by homotopic, you get a fake copy, following some authors).

Reading Wild world of 4-manifolds (Scorpan), I wonder if there are topological manifolds $X^n$, such that $X^n$ is heomeorphic and diffeomorphic to an $M^n$, but not (complex-)biholomorphic to it (i.e. when you look to it from an analytic point of view, you get a different structure, but looking from differential point of view they are the same).

To put in other words: Does: "$X^n$ diffeomorphic to $Y^n$" if and only if "$X^n$ biholomorphic to $Y^n$" hold ? And if so, does it hold for all n ?

Solution 1:

What a lovely question! Let's state it again for reference:

"Is $X$ diffeomorphic to $Y$ if and only if $X$ is biholomorphic to $Y$?"

As biholomorphic maps are diffeomorphisms, one implication is clear; manifolds which are biholomorphic are diffeomorphic. The other does not hold, it does not hold at all, and is the starting point of a vast and diffcult subject called "deformation theory".

As Riemann was the first to remark, and others after him did until the foundational works of Kodaira and Spencer in the 60's, complex structures on manifolds come many at a time. Unlike discrete invariants such as dimension or "number of holes", complex structures on a fixed manifold form a continuous space.

Once you give the words in the last sentence the right definitions, you'll be able to prove that if a fixed manifold $X$ admits one complex structure, then in general it admits an entire family of complex structures. Thus you get infinitely many complex manifolds $X_s$, depending on some parameter $s$, which will all be diffeomorphic as smooth manifolds. The converse implication of your question thus fails dramatically.

A nice example where one sees this going in is the one of elliptic curves. Let $s$ denote a point in the Poincaré half-plane $\mathbb H = \{ s \in \mathbb C \,|\, \text{Im} s > 0\}$. Each such point determines an elliptic curve $X_s = \mathbb C / \Lambda_s$, where $\Lambda_s$ is the lattice $\mathbb Z \oplus s \mathbb Z$. When are two such curves biholomorphic?

By lifting a potential biholomorphism $f : X_s \to X_{s'}$ to a map $\tilde f : \mathbb C \to \mathbb C$, one can prove that $f$ must be induced by a linear map $\tilde f$ that sends the lattice $\Lambda_s$ to $\Lambda_{s'}$. Now these maps are very rare. For example, for $\epsilon$ small enough, the lattices $\mathbb Z \oplus i\mathbb Z$ and $\mathbb Z \oplus i(1 + \epsilon) \mathbb Z$ will never be isomorphic (for a heuristic reason: holomorphic maps cannot stretch the $y$-axis while keeping the $x$-axis fixed). Thus we get infinitely many non-biholomorphic elliptic curves $X_s$.

On the other hand, it is easy to see that any elliptic curve is diffeomorphic to $\mathbb R^2 / \mathbb Z^2$. Indeed, there is a unique $\mathbb R$-linear map of $\mathbb R^2$ to itself sending the lattice $\mathbb Z^2$ to any lattice of our choice. This map will induce a diffeomorphism of the corresponding elliptic curves.

I'm not sure what would be a good reference to point you to if you want to know more, as I don't know your background. Some keywords that you might find useful are "deformation theory", "moduli space", and "hilbert scheme". The books by Kodaira and Morrow-Kodaira might be useful, as maybe "Moduli of curves" by Harris and Morrison. Unfortunately there is a dearth of good introductions to this subject.

Solution 2:

The set of complex structures on an orientable genus $g$ surface is called Teichmüller space, which is a manifold. After modding out by diffeomorphisms, you still get an uncountable space called moduli space. It is well-known that there are no exotic surfaces, but this gives uncountably many complex structures.

Solution 3:

Consider just the case of 1 complex dimension.

The Poincare disk and the complex plane $\mathbb{C}$ are diffeomorphic but not biholomorphic (Riemann mapping theorem).

Solution 4:

The graph of a smooth bump function on $\mathbb{R}^2$ (such as this: http://en.wikipedia.org/wiki/File:Bump2D_illustration.png) should be a simple counterexample. It's diffeomorphic to the plane, but the only analytic functions that are constant on an open set are everywhere constant, so there can be no analytic bijection between them.