8
$\begingroup$

I'm pretty sure that both of the following spaces are contractible. However, I can't seem to find a proof or a reference. Can anyone provide one? Let $D^2$ be the unit disc in $\mathbb{R}^2$.

  1. The space of smooth embeddings $f\colon D^2 \rightarrow \mathbb{R}^2$ such that $f(0)=0$ and $(Df)_0 = \text{id}$, with the $C^{\infty}$ topology.

  2. The space of topological embeddings $f\colon D^2 \rightarrow \mathbb{R}^2$ such that $f(0)=0$, with the compact-open topology.

$\endgroup$
5
  • 1
    $\begingroup$ Please clarify what you mean by "proper". Usually a proper embedding is one which takes boundary to boundary, which in this case would force the embeddings to be diffeomorphisms or homeomorphisms, respectively, in the two cases. $\endgroup$ Sep 21 '14 at 3:29
  • $\begingroup$ @AllenHatcher : You're right, the word proper should not have been there. What I really wanted was for the image to lie in the interior; I fixed this by making the target be Euclidean space. Thanks! $\endgroup$
    – Lior
    Sep 21 '14 at 3:33
  • 14
    $\begingroup$ For question 2 the space is not contractible. There is a map from the space of topological embeddings to ${\mathbb R}^2 - \{0\}$ given by $f\mapsto f(1,0)$ and this map splits (i.e., has a section) by considering only embeddings which are orientation-preserving Euclidean similarities. So the space is not simply-connected. An extra condition analogous to the condition on the derivative in question 1 is needed, though it is perhaps not obvious what the best way to state this condition is. $\endgroup$ Sep 21 '14 at 3:44
  • $\begingroup$ Lior, you should perhaps download and read Jean Cerf's dissertation. It was published by IHES. $\endgroup$ Sep 21 '14 at 4:50
  • 1
    $\begingroup$ in question 2 you have also 2 connected components given by orientation preserving/reversing embeddings $\endgroup$ Sep 21 '14 at 6:43
10
$\begingroup$

For #1, you can take one parameter family of maps $f_t\colon D\to\mathbb R^2$ defined as $f_0=\mathrm{id}$ and $$f_t(x)=\tfrac1t{\cdot}f(t{\cdot}x).$$

$\endgroup$
3
  • 2
    $\begingroup$ I think, this only proves that the diffeomorphism group of $\mathbb R^2$ that fixes $0$ and acts as the identity at the tangent space at $0$ is contractible. $\endgroup$ Sep 21 '14 at 12:58
  • 7
    $\begingroup$ @IgorBelegradek: The formula seems to work equally well for embeddings $D^2\to {\mathbb R}^2$ and diffeomorphisms of ${\mathbb R}^2$. For embeddings the term $f(tx)$ is in effect restricting embeddings to smaller and smaller concentric disks, so one still gets embeddings, and then the $1/t$ factor re-expands to keep the derivative the identity at the origin. The limit exists, just as for diffeomorphisms, and is the identity. $\endgroup$ Sep 21 '14 at 21:44
  • 3
    $\begingroup$ I take my comment back. This does seem to work. $\endgroup$ Sep 22 '14 at 12:16
10
$\begingroup$

For details I recommend looking at the papers of Yagasaki on arXiv especially the paper

Let me answer 1. Consider be the diffeomorphism group of $\mathbb R^2$ that fixes $0$ and acts as the identity at the tangent space to $0$. The group acts on your space of embeddings by postcomposition. By the parametrized isotopy extension theorem the action is transitive on each path-component and the orbit map is a fiber bundle whose fiber is the subspace of diffeomorphisms that are identity on the inclusion of the disk. The latter space is contractible by the Alexander trick towards infinity. Note that so far the two-dimensionality has not been used and the argument generalizes to the embedding of a compact manifold onto an open manifold obtained by attaching a collar to the compact one.

We are left to understand the homotopy type of the above diffeomorphism group. The group is contractible as proved in the last mentioned paper of Yagasaki (or even easier, look at the Anton Petrunin's answer which really proves contractibility of the diffeomorphism group). Thus each component of the embedding space is contractible. In fact, the embedding space is path-connected essentially because any two embedded circles in $\mathbb R^2-\{0\}$ are isotopic via a compactly supported ambient isotopy.

The same strategy can be used to find the homotopy type of the space of embeddings in 2. As Allen Hatcher says in his comment, the space is 2 is not contractible, and one way to explain is that the homeomorphism group of $\mathbb R^2$ that fixes $0$ is not contractible. I think the group is homotopy equivalent to $O(2)$. See the paper above of Yagasaki.

$\endgroup$
0
$\begingroup$

I believe at least the second (and somewhat more tentatively, the first) of the questions is answered (in much greater generality) in:

Homotopy types of the components of spaces of embeddings of compact polyhedra into 2-manifolds Tatsuhiko Yagasaki

(Topology and applications, 2005)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.