Let $B$ be the open unit ball in $\mathbb C^n$. Consider the space $\mathcal F$ of holomorphic embeddings of $B$ in $\mathbb C^n$ equipped with the compact-open topology. (A holomorphic embedding of $B$ in $\mathbb C^n$ is a holomorphic map $f: B\to \mathbb C^n$ such that $f(B)$ is open and there is a holomorphic inverse $f^{-1}: f(B)\to B$). Is it known whether the space $\mathcal F$ is connected?
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$\begingroup$ Do you know the answer for $n=1$? $\endgroup$– QfwfqCommented Sep 25, 2019 at 11:29
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$\begingroup$ @Qfwfq For $n=1$ it seems to be true and should follow from a result claimed in Kirillov, A. A.; Golenishcheva-Kutuzova, M. I., The geometry of moments for groups of diffeomorphisms. (Russian) Akad. Nauk SSSR Inst. Prikl. Mat. Preprint 1986, no. 101, 25 pp. - also see Kirillov, A. A.; Yurʹev, D. V. Kähler geometry of the infinite-dimensional homogeneous manifold M=Diff+(S1)/Rot(S1). (Russian) Funktsional. Anal. i Prilozhen. 20 (1986), no. 4, 79–80. However I am not sure whether the results above deal with an open or a closed unit disk. $\endgroup$– user102829Commented Sep 25, 2019 at 14:04
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1$\begingroup$ I believe the space of holomorphic embeddings with Jacobi matrix = 1 at zero is contractible due to the standard construction $f_t = (1/t)f(zt)$, and $f_0$ defined as a limit will be equal to the identity map ($t$ changes from 1 to 0). Am I missing something? The similar proof also works in the smooth category... $\endgroup$– Lev SoukhanovCommented Sep 25, 2019 at 16:43
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$\begingroup$ @LevSoukhanov You are right! I missed it because of a wrong recollection that some gluing appears in this argument in the smooth case. Would you post it as an answer? $\endgroup$– user102829Commented Sep 25, 2019 at 20:00
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$\begingroup$ @MichaelEntov ok, sure $\endgroup$– Lev SoukhanovCommented Oct 2, 2019 at 14:12
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The space of holomorphic embeddings with Jacobi matrix = 1 at zero is contractible due to the standard construction $f_t=(1/t)f(zt)$, and $f_0$ defined as a limit will be equal to the identity map ($t$ changes from 1 to 0). The similar proof also works in the smooth category.
