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Let $S$ be a compact oriented surface of genus at least $2$ (possibly with boundary). Let $X$ be a connected component of the space of embeddings of $S^1$ into $S$.

Question : what is the fundamental group of $X$? My guess is that the answer is $\mathbb{Z}$ with generator the loop of embeddings obtained by precomposing the base embedding with a sequence of rotations of $S^1$.

I'm also interested in the higher homotopy groups of $X$, which I would guess are trivial.

Edit: In response to Sam Nead's question, I'm most interested in the smooth category, but am also interested in the topological category. There are technical issues in giving an appropriate topology to mapping spaces in the PL category, so the question doesn't really make sense there.

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There is a result by Gramain (1973, Le type d'homotopie...) which says that the space $Emb((S^1,v),(M,w))$ of embeddings with prescribed behaviour on a fixed unit tangent vector has contractible components, for each compact surface. His proof is purely topological. This should imply what you are looking for, though I do not see it immediately. – Johannes Ebert Feb 2 '14 at 22:15
If I may ask: please edit your post to say which category you are working in? Top, PL, Diff? – Sam Nead Feb 4 '14 at 21:28
See the very similar question here:… – Sam Nead May 10 '14 at 18:33

Edit -

As pointed out by Igor below, my "proof" of the first bullet point is incomplete. I'll leave the rest of the post here: perhaps some kind soul will fix the gap.

Original -

Let's restrict attention to the case where all embeddings are smooth and where $S$ has negative Euler characteristic, or is an annulus or Möbius band.

  • If $X$ contains a trivial curve, then $X$ is homotopy equivalent to the unit tangent bundle to the surface $S$.
  • If $X$ contains an essential curve, then $X$ is homotopy equivalent to a circle.

Both of these results follow from Matt Grayson's curve-shortening flow (Annals, 1989). If $S$ is the torus or Klein bottle, then first bullet still holds, but the second does not. Instead $X$ is homotopy equivalent to (a cover of) $S$. Finally, if $S$ is the sphere or projective plane, then I believe that $X$ has nontrivial higher homotopy groups.

If the embeddings are only continuous, well, that seems tricky.

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Could you explain how Grayson's paper proves what you say it does? – Igor Belegradek May 9 '14 at 11:46
For the first bullet - Suppose that $\gamma$ is a trivial curve. Then curve shortening gives a one-parameter family $\gamma_t$ of curves so that $\gamma_t$ shrinks towards a point $x$, becomes round (and constant speed), and does not rotate infinitely about $x$. The tangent to the track of $\gamma_t(1)$ is the desired tangent vector at $x$. – Sam Nead May 10 '14 at 18:05
For the second bullet - The homeomorphism type of $X$ (the space of embeddings) is blind to the metric on $S$. So we may suppose that $S$ has a hyperbolic metric. Then $\gamma_t$ approaches the unique geodesic representative $\hat{\gamma}$, and $\gamma_t(1)$ selects a point of $\hat{\gamma}$. – Sam Nead May 10 '14 at 18:30
For the first bullet: does this construct a deformation retraction of $X$ to the unit tangent bundle to $S$? How does the latter embed into the former? – Igor Belegradek May 10 '14 at 19:19
The latter embeds, but not in a unique fashion. There is a copy of $T_1 S \times \mathbb{R}$ in $X$. To see this, fix a hyperbolic or Euclidean metric on $S$ (an oriented compact surface without boundary). Fix $\epsilon$ sufficiently small - say less than one-third of the injectivity radius of $S$. At any point $x$ we can draw the circle $C_x$ of radius $\epsilon$, equipped with a counter-clockwise orientation and with a marked point. This specifies an element of $X$. – Sam Nead May 10 '14 at 19:24

For $S$ the sphere, assuming smooth embeddings, any curve divides the sphere into two discs, hence is diffeomorphic to the equator. Then $X$ is a quotient space of the orientation-preserving diffeomorphism group of the sphere by the subgroup that preserves the equator. The orientation-preserving diffeomorphism group of the sphere is homotopic to $SO_3$. The subgroup that preserves the equator is a product of two copies of the group of diffeomorphisms of the disc that fix the boundary. This is known to be contractible.

Thus the space is diffeomorphic to $SO_3$, which is the unit tangent bundle on $S^2$.

So as Sam Nead suspected, there is a lot of higher homotopy.

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Slick. I think there may be a missing factor of $S^1$ someplace, as $X$ is the space of parametrized embedded loops? Anyway - can we use the fact that the components of $\operatorname{Homeo}(S)$ are contractible (for generic $S$) to answer the original question this way? – Sam Nead Feb 2 '14 at 23:12
The $S^1$ is the $SO_2$ inside $SO_3$, I believe. With regards to your second question, one would have to compute the kernel of the homomorphism from the mapping class group of the surface cut along the circle to the mapping class group of the surface. – Will Sawin Feb 3 '14 at 4:16
@Will Sawin : this kernel is Z, and is generated by the mapping class associated to the loop in question. This is Theorem 3.18 in Farb-Margalit's primer. You've actually discovered my secret motivation, which is to give a purely topological proof of this (running your argument backwards!) – Don Feb 3 '14 at 4:47
@Don: Is Sam's argument purely topological? – Will Sawin Feb 3 '14 at 16:28
@Will Sawin : The approach via Grayson's flow is lovely, but I was hoping for something more topological (say, using techniques like those in Gramain's paper that Ebert mentions in a comment). – Don Feb 3 '14 at 17:28

$ \newcommand{\Homeo}{\operatorname{Homeo}} \newcommand{\SO}{\operatorname{SO}} $Since you are interested in the topological category, then I think it will suffice to prove the necessary facts about $\Homeo_0(S)$ and about the curve stabilizer. Now, there is a topological proof that $\Homeo_0(S)$ is contractible -- see this mathoverflow question:

Homotopy type of set of self homotopy-equivalences of a surface

Here is the paper of Hamstrom that they are referring to.

Reading a bit of that I learned that Kneser was the first to prove (in 1926!) that $\Homeo_0(S^2)$ deformation retracts to $\SO(3)$.

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