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What is the homotopy type of the space of (topological) emdeddings of $S^1$ in $\mathbb R^2$?

My conjecture: This space deformation retracts to $S^1\sqcup S^1$, and a retraction in each of orientation preserving and reversing components is given by "rotation number" in $S^1$. (I don't have an exact proof or even an exact definition of the conjectured retraction map)

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    $\begingroup$ What do you mean by embedding,exxactly? If the maps are injective then you have only two contractible components, no? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 13, 2015 at 14:51
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    $\begingroup$ @MarianoSuárez-Alvarez Yes, I mean injective and continuous. But I can't see why it's contractible? $\endgroup$
    – Mostafa - Free Palestine
    Commented May 13, 2015 at 15:01
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    $\begingroup$ My guess is because the group of self-homeos of the circle has contractible identity component (a homeo lifts to a map $\mathbb R\to\mathbb R$, and the latter is a strictly increasing or decreasing function which you can deform to a linear one) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 13, 2015 at 15:51
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    $\begingroup$ You can use the Riemann mapping theorem to prove your conjecture. $\endgroup$
    – John Pardon
    Commented May 13, 2015 at 16:45
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    $\begingroup$ @MarianoSuárez-Alvarez Self-homeos of circle are selfhomeos of $\mathbb{R}$ mod integer translates, so it's fundamental group is $\mathbb{Z}$. $\endgroup$
    – Mostafa - Free Palestine
    Commented May 13, 2015 at 17:47

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For $C^2$-embeddings you can follow weighted (length preserving) mean curvature flow until you have a circle, then move the center to 0, then use a homothethy to go to the unit circle, then use the contraction in $\{\phi\in Diff(S^1):\phi(1)=1\}$ to get arc-length parametrization, so you end up with a circle. The initial point of the parametrization is then the $S^1$ for your orientation.

Now, how to get from topological embedding to $C^2$-embedding? Use convolution with a smooth bump function? Does this destroy topological embedding?

Added:

Mean curvature flow preserves embeddings; see http://www.ams.org/journals/bull/2015-52-02/S0273-0979-2015-01468-0/S0273-0979-2015-01468-0.pdf

About John Pardons remark: Embedded simple closed curves (without parametrisation!) are diffeomorphic to $PSL(2,\mathbb R)\backslash Diff(S^1)$ and this diffeomorphism goes down to $H^{3/2}$-embeddings and corresponds to a subgroup of quasi symmetric homeomorphisms of the circle. The idea is to take a Riemann mapping $\phi_{int}$ from the unit disk to the interior of $\Gamma$ (= the embedded circle) which is unique up to a Möbius transformation, and a Riemann mapping $\phi_{ext}$ from the exterior of the disk to the exterior of $\Gamma$, fixing $\infty$ and the derivative at $\infty$, so that it becomes unique. Then $\phi_{ext}^{-1}\circ \phi_{int}$ is in $PSL(2,\mathbb R)\backslash Diff(S^1)$ which is contractible.

See Takhtaja and Teo, Memoirs of the AMS, or see the following papers for smooth embeddings.

  • E. Sharon and D. Mumford. 2d-shape analysis using conformal mapping. In Proc. IEEE Conf. Computer Vision and Pattern Recognition, pages 350–357, 2004.

  • E. Sharon and D. Mumford. 2d-shape analysis using conformal mapping. International Journal of Computer Vision, 70:55–75, 2006.

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