2
$\begingroup$

Let $M$ be a smooth manifold with boundary $\partial M$. Let $Diff_0(M)$ be the group of all diffeomorphisms homotopic to identity. According to this article (Page 6, section " Beyond mapping class group"), the restriction of a diffeomorphism to the boundary gives a well defined surjactive homomorphism $$\phi: Diff_0(M)\rightarrow Diff_0(\partial M).$$

I could not find a reference for the above two results. So I have two questions.

1) Why is $\phi$ well defined, i.e., why the restriction of a difeomorphism homotopic to identity in $M$ is homotopic to identity in $\partial M$.

2) Why every diffeomorphism homotopic to identity in $\partial M$ is a restriction of a diffeomorphism of $M$ which is homotopic to identity.

As I am not quite familiar with diffeomorphism groups, any suggestion/reference/comment will be extremely helpful. Also I would request you to improve the tags if possible.

Thanks in advance.

$\endgroup$
3
  • 1
    $\begingroup$ Both claims hold (with easy proofds) if you replace homotopy with isotopy; are you sure they use homotopy in the paper? The restriction map does not (in general, send diffeomorphisms homotopic to the identity to diffeomorphisms homotopic to the identity. $\endgroup$ May 9 '20 at 4:16
  • $\begingroup$ Good point. The paper states simply the "path component of identity". Also even for isotopy the first one is straightforward but how to prove the second one? Any reference for that. $\endgroup$
    – Cusp
    May 9 '20 at 4:19
  • 3
    $\begingroup$ Path component of the identity means isotopy. To get 1, just restrict the isotopy to the boundary, it remains isotopy. (The difference with homotopy is that boundary maps to boundary.) To get 2, use the existence of a collar neighborhood of the boundary which is the product $\partial M\times [0,1]$. $\endgroup$ May 9 '20 at 4:23
4
$\begingroup$

As discussed in comments, $Diff_0$ stands for the subgroup of the diffeomorphism group, consisting of diffeomorphisms isotopic (rather than homotopic) to the identity. With this in mind, the fact that the restriction map $\phi: Diff(M)\to Diff(\partial M)$ sends $Diff_0(M)$ to $Diff_0(\partial M)$ is clear. Let's prove surjectivity. First of all, $\partial M$ admits a "collar" $C$ in $M$, a closed neighborhood of $\partial M$ in $M$, $C$ is diffeomorphic to $\partial M\times [0,1]$. Now, given $h\in Diff_0(\partial M)$, let $H(x, t), t\in [0,1]$, denote the isotopy of $h= H(\cdot, 0)$ to $id_{\partial M}= H(\cdot, 1)$. I leave it to you to prove that $H$ can be chosen so that $H(x,t)=x$ for all $t\in [1/4, 1]$. Then, using the diffeomorphism $C\cong \partial M\times [0,1]$, extend $h$ first to $C$ and then, by identity, to the rest of $M$. Call the extension $\hat{h}$. Clearly, $\phi(\hat{h})=h$. It remains to prove that $\hat{h}\in Diff_0(M)$. To prove this, play the same game as before: Given an isotopy $H(x,t)$ from $h$ to $id_{\partial M}$, extend it to $C\cong \partial M\times [0,1]$ by $$ (x,t,s)\mapsto H(x, t+s), $$ and then by identity to the rest of $M$. This will be an isotopy $\hat{H}$ from $\hat{h}$ to $id_M$.

Here is an example to ponder: Let $M$ be the annulus $S^1\times [0,1]$. Consider the diffeomorphism $f(s,t)=(s, 1-t)$; $f: M\to M$ is homotopic to the identity, but its restriction to $\partial M$ is not.

$\endgroup$

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.