4
$\begingroup$

Let $V$ be a smooth manifold obtained by attaching the ``open collar'' $[0,1)\times \partial N$ to a compact smooth manifold $N$ along the boundary. Let $\mathrm{Emb}(N, V)$ be the space of embeddings of $N$ into $V$ isotopic to the inclusion. The identity component $\mathrm{Diff\,} N$ of the diffeomorphism group of $N$ embeds into $\mathrm{Emb}(N, V)$ by precomposing a diffeomorphism with the inclusion.

Question: Under what conditions the embedding $\mathrm{Diff\,} N\to\mathrm{Emb}(N, V)$ is not a homotopy equivalence?

It is relevant to consider the $\mathrm{Diff\,} V$-action on $\mathrm{Emb}(N, V)$ by postcomposing. The orbit map $\mathrm{Diff\,} V\to\mathrm{Emb}(N, V)$ is a homotopy equivalence: its homotopy fiber consists of diffeomorphisms that are identity on $N$, and the group of such diffeomorphisms is contractible by the Alexander trick towards infinity.

The case when $V=\mathbb R^n$ and $N=D^n$ is fairly well understood: both $\mathrm{Diff\,} \mathbb R^n$ and $\mathrm{Diff\,}\mathbb D^n$ deformation retract onto $O(n)$ [Edit: as Allen Hatcher explains in comments $\mathrm{Diff\,}\mathbb D^n$ does not deformation retract onto $O(n)$]. On the other hand, $\mathrm{Diff\,}(\mathbb D^n, rel\,\partial D^n)$ is homotopically more complicated when $n$ large (computing it involves pseudoisotopy and Waldhausen K-theory).

Naively I hope the case of a disk be used to produce elements of $\pi_i(\mathrm{Diff\,} N)$ that die in $\pi_i(\mathrm{Emb}(N,V))\cong\pi_i(\mathrm{Diff\,} V)$, e.g. by considering boundary connected sum of $N$ with $D^n$, and extending diffeomorphisms of $D^n$ that are identity on $\partial D^n$ to $N$ by the identity. Has this ever been done?

I am even more interested in ways to produce elements of $\pi_i(\mathrm{Emb}(N,V))$ that do not come from $\pi_i(\mathrm{Diff\,} N)$.

$\endgroup$
  • $\begingroup$ In the first paragraph, does $Diff(N)$ mean diffeomorphisms fixing the boundary pointwise or not? $\endgroup$ – Oscar Randal-Williams Dec 16 '13 at 18:37
  • $\begingroup$ @Oscar Randal-Williams: no, I do not assume that diffeomorphisms of $N$ are identity on the boundary. $\endgroup$ – Igor Belegradek Dec 16 '13 at 18:43
  • 1
    $\begingroup$ $Diff(D^n)$ is homotopy equivalent to $O(n)\times Diff(D^n\ rel\ D^{n-1})$ where $D^{n-1}$ is a disk in $\partial D^n$. The factor $Diff(D^n\ rel\ D^{n-1})$ can be identified with the pseudoisotopy space $Diff(D^{n-1}\times I\ rel\ D^{n-1} \times 0)$, so it has a complicated homotopy type when $n$ is large enough. $\endgroup$ – Allen Hatcher Dec 16 '13 at 21:00
  • 1
    $\begingroup$ As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\operatorname{Diff} N \to \operatorname{Emb}(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times \{0\}$ pointwise. $\endgroup$ – Ricardo Andrade Dec 16 '13 at 21:17
  • 1
    $\begingroup$ The paper of LaBach mentioned in an earlier comment (now deleted?) uses a nonstandard topology on $Diff(D^n)$. The restriction map $Diff(D^n)\to Diff(int(D^n))$ is injective so can be viewed as an inclusion, and LaBach uses the subspace topology on $Diff(D^n)$ induced from the compact-open topology on $Diff(int(D^n))$. In this topology one can do a sort of "reverse Alexander trick" and push all the complications of a diffeomorphism of $D^n$ out to $\partial D^n$ and make them disappear. $\endgroup$ – Allen Hatcher Dec 16 '13 at 21:46
1
$\begingroup$

[The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$

As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.

Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map described by Tom in his answer, $$ r = k^\ast : \Diff(N) \To \Emb(N_0,\interior N) $$ which is given by restriction along $k:N_0 \to N$.

On the other hand, consider, as in the question, a manifold $V$ obtained from $N$ by gluing an open collar along the boundary of $N$. Then we also have the map defined in the question $$ i = l_\ast : \Diff(N) \To \Emb(N,V) $$ given by composing (on the left) with the inclusion $l:N \to V$.

We only need to consider the following commutative square diagram $$ \begin{matrix} \Diff(N) & \overset{i}{\To} & \Emb(N,V) \\ \Big\downarrow\rlap{r} & & \Big\downarrow\rlap{s} \\ \Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V) \end{matrix} $$ where $s = k^\ast : \Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j = m_\ast : \Emb(N_0,\interior N) \to \Emb(N_0,V)$ is given by composition (on the left) with the inclusion $m:\interior N \to V$.

Finally, observe that both $s$ and $j$ are homotopy equivalences (one needs to use the collars to show this). Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Hence, the homotopy fibre of $r$ is equivalent to its fibre. The desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.

$\endgroup$
  • $\begingroup$ The map in Goodwillie's answer is post-composing with the inclusion--mine is pre-composing with the inclusion. $\endgroup$ – Igor Belegradek Dec 16 '13 at 22:07
  • $\begingroup$ @Igor Belegradek, to make things clearer, if $\iota$ denotes the inclusion, by "post-composing with the inclusion" do you mean $- \circ \iota$? $\endgroup$ – Ricardo Andrade Dec 16 '13 at 22:12
  • $\begingroup$ My map takes a diffeomorphism $\phi\in Diff(N)$ to $k\circ\phi$, where $k: N\to V$ is the inclusion. $\endgroup$ – Igor Belegradek Dec 16 '13 at 22:16
  • $\begingroup$ @Igor Belegradek, then I believe I correctly understood the definition of your map. So I am confused as to what issue you find with my explanation above. Perhaps it is related to how you have changed notation from your question to the comment above? In my answer, I described your map as taking $\phi\in \operatorname{Diff}(N)$ to $l\circ\phi$, where $l:N\to V$ is the inclusion. Can you please clarify where you think there is a problem with my answer? $\endgroup$ – Ricardo Andrade Dec 16 '13 at 22:22
  • 1
    $\begingroup$ @Igor Belegradek, to clarify, I have actually not changed the meaning of $N$ from your question. When I write $N$ in my answer, I mean the same $N$ you do in the question. Your map is called $i$ in my answer, and Tom's map is called $r$. $V$ is an open manifold (diffeomorphic to the interior of $N$) which contains $N$ as in your question. Moreover, $N_0$ is a compact manifold (diffeomorphic to $N$) which is contained in the interior of $N$. Does that make things clearer? $\endgroup$ – Ricardo Andrade Dec 16 '13 at 22:57

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.