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Oct 3, 2019 at 7:48 vote accept user102829
Oct 2, 2019 at 14:12 comment added Lev Soukhanov @MichaelEntov ok, sure
Oct 2, 2019 at 14:12 answer added Lev Soukhanov timeline score: 4
Sep 25, 2019 at 20:00 comment added user102829 @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?
Sep 25, 2019 at 16:43 comment added Lev Soukhanov 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...
Sep 25, 2019 at 14:05 history edited user102829 CC BY-SA 4.0
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Sep 25, 2019 at 14:04 comment added user102829 @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.
Sep 25, 2019 at 11:29 comment added Qfwfq Do you know the answer for $n=1$?
Sep 25, 2019 at 10:55 history asked user102829 CC BY-SA 4.0