# Space of holomorphic embeddings of open unit ball in ${\mathbb C}^n$

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?

• Do you know the answer for $n=1$? – Qfwfq Sep 25 '19 at 11:29
• @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. – Michael Entov Sep 25 '19 at 14:04
• 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... – Lev Soukhanov Sep 25 '19 at 16:43
• @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? – Michael Entov Sep 25 '19 at 20:00
• @MichaelEntov ok, sure – Lev Soukhanov Oct 2 '19 at 14:12

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.