# Space of embeddings of circle in a surface

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 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 at 21:28

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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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 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 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 at 4:47
@Don: Is Sam's argument purely topological? –  Will Sawin Feb 3 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 at 17:28
Reading a bit of that I learned that Kneser was the first to prove (in 1926!) that $\Homeo_0(S^2)$ deformation retracts to $\SO(3)$.