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It is known that each complex structures on $\mathbb{R}^2$ is biholomorphic to either $\mathbb{C}$ or the open unit disk $\Delta$.

One can continuously deform one complex structure to the other as is for example done in Winkelmann - Deformations of Riemann surfaces (page 3).

My question is

Can this deformation be taken to be holomorphic on the deformation parameter? That is, does there exist a non-trivial complex analytic family $M \to D$ where $D \subset \mathbb{C}$ is a small disk, the central fiber is biholomorphic to $\mathbb{C}$, and the generic fiber is biholomorphic to $\Delta$?

Note that all the theorems that assure complex analytic triviality of deformations when $H^1(X,TX)$ vanishes, use the hypothesis that $X$ is compact.

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    $\begingroup$ How about $M=\{|z w|<1 \}\subset D\times \mathbb{C}$ with projection to the first coordinate? $\endgroup$
    – user_1789
    Nov 29, 2019 at 9:13
  • $\begingroup$ Beautiful! I think it works. If you write it in an answer with a small argument it will be the accepted solution $\endgroup$
    – Paul
    Nov 29, 2019 at 13:25

1 Answer 1

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The answer is no. This follows from the so-called $\lambda$-lemma of Sullivan, Mane Sad and Lyubich: Let $D$ be a disk, $C$ the complex plane and $A$ any set in C. Let $f:D\times A\to C$ be a function with the following properties:

  1. $\lambda\mapsto f(\lambda,z)$ is holomorphic for every $z\in A$.

  2. $z\mapsto f(\lambda,z)$ is injective for every $\lambda\in D$,

  3. $z\mapsto f(0,z)=z$ for every $z\in A$.

Then for every $\lambda\in D$ the map $z\mapsto f(\lambda,z)$ is quasisymmetric (=quasiconformal if $A$ is open)

Since the plane is not conformally equivalent to a disk, this implies that the answer to your question is negative.

Ref. MR0732343 Mañé, R.; Sad, P.; Sullivan, D., On the dynamics of rational maps, Ann. Sci. École Norm. Sup. (4) 16 (1983), no. 2, 193–217.

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    $\begingroup$ I am sorry but I don't know how to apply your answer to my situation. You are looking at my $D$ to be embedded in your C? And then my family $M$ would be your product? Observe that my central fiber is not a disk by hypothesis. Also the conclusion of the $\lambda$-lemma is that the map admits a quasiconformal extension $\endgroup$
    – Paul
    Nov 29, 2019 at 5:01
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    $\begingroup$ Your map $f$ is my family $M \to D$ ? $\endgroup$
    – Paul
    Nov 29, 2019 at 5:18
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    $\begingroup$ $1$ is not satisfied in my situation. $\endgroup$
    – Paul
    Nov 29, 2019 at 5:40
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    $\begingroup$ I probably misunderstood what is a "complex analytic family". Can you define it? On my opinion "a deformation depending on a complex analytic parameter" is exactly what satisfies 1,2,3. $\endgroup$
    – Alexandre Eremenko
    Nov 29, 2019 at 13:39
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    $\begingroup$ What does that mean? That it was my fault that your answer is wrong because my question was too easy? I don’t mind admitting that my question was “trivial” (whatever that means) or that I didn’t know something (that I learned today) or that I’m more ignorant than a lot of people. I asked this question because I am no expert and I’m learning deformation theory on my own with no experts around me. $\endgroup$
    – Paul
    Nov 29, 2019 at 19:12

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