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Nov 30, 2019 at 14:39 comment added Alexandre Eremenko My $f(\lambda,z)$ is what Winkler denotes by $\phi_t(z)$ on p. 3 of the paper you cite, my $\lambda$ is his $t$.
Nov 30, 2019 at 14:35 comment added Paul But you claimed that your theorem implies that the answer to my question is no. I say that I don’t know how to apply your theorem to my situation so I need to know what are your letters and maps in terms of my letters and maps.
Nov 30, 2019 at 14:32 comment added Alexandre Eremenko $f$ is what I would call "a deformation". But I do not insist. I just stated a theorem. You decide whether it is useful for you or not.
Nov 30, 2019 at 14:09 comment added Paul and what is the map f? The projection onto the first factor or the second factor? Again, what is f, D,A,C in terms of my M, arrow and D? (So that I can try to apply the lemma)
Nov 30, 2019 at 14:07 comment added Alexandre Eremenko Take $A=C$, for example. Then no other fiber can be a disk.
Nov 30, 2019 at 14:03 comment added Paul I still don’t know what is your possible definition of complex analytic family of complex structures. My definition agrees with your notion of “family of deformations deepending on a complex parameter”. What is your f,D,A,C in terms of my $M \to D$? Where is your central fiber biholomorphic to C and your generic fiber biholomorphic to $\Delta$ in order to arrive to a contradiction?
Nov 30, 2019 at 1:49 comment added Alexandre Eremenko @Paul: "What does it mean?" That you have now two answers, depending on what is meant by a "deformation". And you can choose the one that you need. Nothing more.
Nov 29, 2019 at 19:12 comment added Paul 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.
Nov 29, 2019 at 18:54 comment added Alexandre Eremenko But then you have a trivial example (in the comments).
Nov 29, 2019 at 14:39 comment added Paul In this context, the simplest definition that fits is: a complex analytic family of complex manifolds is the triple $(M,D,\pi)$ where $\pi:M \to D$ is a flat holomorphic submersion which is a $C^\infty$ locally trivial (and hence trivial) fibration. The usual context is that of analytic families of compact manifolds (in this case you only ask that $\pi$ is a holomorphic proper submersion). The case where fibers are open is much more subtle and less studied. By definition, the fibers are all diffeomorphic but not neccesarily biholomorphic.
Nov 29, 2019 at 13:39 comment added Alexandre Eremenko 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.
Nov 29, 2019 at 5:40 comment added Paul $1$ is not satisfied in my situation.
Nov 29, 2019 at 5:18 comment added Paul Your map $f$ is my family $M \to D$ ?
Nov 29, 2019 at 5:01 comment added Paul 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
Nov 29, 2019 at 4:43 history answered Alexandre Eremenko CC BY-SA 4.0