Holomorphic deformation of complex structure on the real plane

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.

• How about $M=\{|z w|<1 \}\subset D\times \mathbb{C}$ with projection to the first coordinate? – user_1789 Nov 29 '19 at 9:13
• Beautiful! I think it works. If you write it in an answer with a small argument it will be the accepted solution – Paul Nov 29 '19 at 13:25

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.

• 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 – Paul Nov 29 '19 at 5:01
• Your map $f$ is my family $M \to D$ ? – Paul Nov 29 '19 at 5:18
• $1$ is not satisfied in my situation. – Paul Nov 29 '19 at 5:40
• 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. – Alexandre Eremenko Nov 29 '19 at 13:39
• 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. – Paul Nov 29 '19 at 19:12