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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.)

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    $\begingroup$ 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
    Commented May 31, 2018 at 14:29
  • $\begingroup$ That is a very good point. $\endgroup$
    – Joao Faria Martins
    Commented 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. $\endgroup$
    – Joao Faria Martins
    Commented 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
    Commented May 31, 2018 at 15:07
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    $\begingroup$ 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
    Commented Jun 1, 2018 at 0:42

1 Answer 1

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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.

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