3
$\begingroup$

Let $M$ and $N$ be smooth manifolds. Consider an isotopy of $M$ inside $N$. This means that we have a level preserving embedding $J\colon M\times [0,1] \to N \times [0,1]$. Put $J(x,t)=(\phi_t(x),t)$. Hence $\phi_t\colon M \to N$ is an embedding for each $t\in [0,1]$.

If $M$ is compact and $N$ has no boundary, then the classical isotopy extension theorem states that there exists a diffeotopy (aka ambient isotopy) $H\colon M \times [0,1] \to M$ (i.e. a 1-parameter family of diffeomorphisms), extending $J$, i.e. such that

a) $\phi_t(x)=H(\phi_0(x),t)$, for $x \in M$ and $t \in [0,1]$, and

b) $H(y,0)=y$, for each $y \in N$.

Question: how unique is such ambient isotopy $H$? Can any two ambient isotopies $H$ and $H'$ extending $J$ always be connected by some form of second order ambient isotopy?

(The main example I am thinking of is that of isotopies of a disjoint union of 1-spheres inside the 3-sphere.)

$\endgroup$
5
  • 2
    $\begingroup$ You're essentially asking about the homotopy type of $\text{Emb}(M,N)$ (k-simplices of embeddings, which restrict to some fixed embedding on the boundary, give maps from the k-sphere to this space.) In particular you want $\pi_1 = 0$. This is an interesting question. For links inside S^3, there is work of Hatcher and Budney calculating this space; if I recall, it was rarely simply connected. $\endgroup$
    – mme
    May 31, 2018 at 14:29
  • $\begingroup$ That is a very good point. $\endgroup$ May 31, 2018 at 14:36
  • $\begingroup$ Do you mean the homotopy type of the following subgroup of ${\rm diff(N)}$, $\{f \in {\rm diff}(N) \,\,|\,\, f_M=id_M\}$? Indeed if it is not simply connected then it appears that the answer to what I asked is negative. $\endgroup$ May 31, 2018 at 15:03
  • $\begingroup$ I did not read closely enough and missed the word 'ambient'. Your comment is correct. The work of Hatcher and Budney I mentioned is on embedding spaces, but you can probably exploit it using the fiber sequence $\text{Diff}(N \text{ rel } M) \to \text{Diff}(N) \to \text{Emb}(M, N)$. In particular $\text{Diff}^+(S^3) \simeq SO(4)$ (a different theorem of Hatcher) should be useful for the case of links. $\endgroup$
    – mme
    May 31, 2018 at 15:07
  • 1
    $\begingroup$ As Tom's answer below makes clear, my above comments are wrong for stupid reasons (your space is the path-space of Diff(N,M)). They are only relevant if you make demands of both endpoints of your isotopy. $\endgroup$
    – mme
    Jun 1, 2018 at 0:42

1 Answer 1

2
$\begingroup$

If I interpret the question correctly then the answer is "yes". You seem to be asking whether, if $H'$ is an isotopy satisfying the same conditions as $H$, there must be a one-parameter family of such isotopies joining $H$ to $H'$. I claim that the space of all such isotopies $H'$ is not only path-connected but contractible.

Write $H'(y,t)=H(K(y,t))$. This defines an isotopy $K:N\times I\to N\times I$, such that $K(y,0)=(y,0)$ for all $y$, and such that $K(\phi_0(x),t)=(\phi_0(x),t)$ for all $x$ and $t$. So $K$ is a one-parameter family, starting at $1_N$, in the space of all diffeomorphisms of $N$ that fix $\phi_0(M)$ pointwise. The space of all such $K$ is contractible because more generally for any space $X$ and any point $p\in X$ the space of all paths in $X$ starting at $p$ is contractible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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