space of simple loops in the plane Consider the space $S$ of smooth free simple (non-self-intersecting) loops in the plane $\mathbb{R}^2.$ By Grayson's theorem, $S$ is connected, but is more known about its topology? Is it known to be contractible (or known NOT to be contractible)? By contrast, the set of self-intersecting loops in not connected, but of course one can ask the same question about the connected components.
EDIT To answer @RyanBudney's question, parametrized or not is fine, and oriented or not (as @Ryan asserts, the answers are different depending on the setting, but I would be curious to know the whole set).
 A: The space of smooth embeddings $S^1 \to \mathbb R^2$ has the homotopy-type of $O_2$.  Denote this embedding space by $Emb(S^1,\mathbb R^2)$.
The proof goes like this.   Let $Emb(D^2, \mathbb R^2)$ be the space of smooth embeddings of the 2-disc in $\mathbb R^2$.  There is a locally trivial fibre bundle (due to Palais) given by restricting the embedding to the boundary circle
$$Emb(D^2, \mathbb R^2) \to Emb(S^1, \mathbb R^2)$$
It is onto by the Schoenflies theorem, and the fibre is the group of diffeomorphisms of $D^2$ which restrict to the identity on the boundary, $Diff(D^2)$.  This group is contractible by Smale.  So $Emb(S^1,\mathbb R^2)$ and $Emb(D^2,\mathbb R^2)$ have the same homotopy-type.  But by the linearization process (equivalently, the with-parameters version of the tubular neighbourhood theorem), $Emb(D^2,\mathbb R^2)$ has the homotopy-type of the linear subspace, which is $O_2$. 
If you think through this argument you'll see everything is equivariant enough so that you can mod out by parametrizations to show $Emb(S^1,\mathbb R^2) / Diff(S^1)$ is contractible, as $Diff(S^1)$ also has the homotopy-type of $O_2$ (Smale uses this in his proof). 
So other than ODEs, the main theorems we're using is Schoenflies and Smale's paper on the homotopy-type of $Diff(S^2)$. 
