8
$\begingroup$

Let $M^{n}$ be a differential manifold and $S$, $N$ submanifolds of $M$ with same dimension $\lt n$, $\phi$ is a diffeomorphism from $S$ to $N$, can $\phi$ be extended to a self-diffeomorphism of $M$? Or does there exist such $\phi$, $\phi$ can be extended?

Special case is $M=\mathbb{R}^{n}$, $S$ is an $s$-dimensional submanifold $(s\le n)$,and $\phi$ is a diffeomorphism from $S$ to an open set in $R^{s}$, the question is can $\phi$ be extended to a self-diffeomorphism of $\mathbb{R}^{n}$? If not, does there exist such $\phi$, $\phi$ can be extended?

$\endgroup$
15
$\begingroup$

Consider a standard circle and a trefoil knot in $\mathbb{R}^3$. Any diffeomorphism of $\mathbb R^3$ to itself that took one to the other would have to be a diffeomorphism of the complements. But the fundamental groups of the complements are different, so such a diffeomorphism does not exist.

$\endgroup$
  • $\begingroup$ Or related, pick any embedded nullhomotopic simple loop and a non-nullhomotopic simple loop. (Such things always exist in non-simply connected $n$-manifolds with $n\geq 3$). $\endgroup$ – Jason DeVito Sep 26 '16 at 18:01
  • $\begingroup$ @JasonDeVito You'll find they exist in $n=2$ as well. $\endgroup$ – PVAL-inactive Sep 26 '16 at 20:11
  • $\begingroup$ @PVAL: What I was thinking in my head is that every homotopy class is represented by a simple closed curve. Is this always true when $n=2$? (It's certainly false when $n = 1$). For example, with $\pi_1(S^1\times S^1)\cong \mathbb{Z}^2$ with usual choice of generators, is the element $(2,0)$ represented by a simple closed curve? $\endgroup$ – Jason DeVito Sep 27 '16 at 0:44
  • $\begingroup$ No, it's not true for $n = 2$, and your example is about the simplest thing that shows it. (I guess you could look at the element 2 in $\pi_1(cylinder)$ and that'd be simpler...but it's also not compact...) $\endgroup$ – John Hughes Sep 27 '16 at 1:02
  • 1
    $\begingroup$ @JasonDeVito No but you can proof for instance that for any surface $\pi_1$ is generated by simple closed curves. $\endgroup$ – PVAL-inactive Sep 27 '16 at 3:56
14
$\begingroup$

Let $M = \mathbb{R}$ and $S = N = \{0, 1, 2\}$.

Consider the diffeomorphism $\varphi : \{0, 1, 2\} \to \{0, 1, 2\}$, given by $\varphi (0) = 0$, $\varphi(1) = 2$, $\varphi(2) = 1$. As every diffeomorphism $\mathbb{R} \to \mathbb{R}$ is monotonic, there is no diffeomorphism $\psi : \mathbb{R} \to \mathbb{R}$ such that $\psi|_{\{0, 1, 2\}} = \varphi$.

$\endgroup$
9
$\begingroup$

I know that this is a trivial counterexample with a specific $\phi$, but if you're not putting any constraints and let $s\leq n$ then you can also consider two disjoint open bounded intervals $I_1, I_2\subset\mathbb R$ and the diffeomorphism which is the identity on $I_1$ and flips $I_2$. This does not extend because for example the two "inner" extremes must be sent to the same connected component of $\mathbb R\backslash(I_1\cup I_2)$, which is impossible.

$\endgroup$
  • $\begingroup$ Nice! You've (almost) shown my conjecture about the "simplest possible" was wrong. I was looking for compact manifolds-without-boundary, and never thought of this. I say "almost" because OP asked that the dimension of the submanifold be strictly less than that of the embedding manifold. $\endgroup$ – John Hughes Sep 26 '16 at 20:50
  • 1
    $\begingroup$ This can, however, be remedied: Let $S$ be two copies of the discrete manifold $\{1, 2, 3, \ldots}$, and embed one part via $n \mapsto \frac{1}{n}$ and the other part by $n \mapsto -\frac{1}{n}$. Let $N$ be the same thing, but let the second map by $n \mapsto -1 + \frac{1}{n}$. For compact $S$ and $N$, I'm not sure you can do anything like this, however. :) $\endgroup$ – John Hughes Sep 26 '16 at 20:52
  • 3
    $\begingroup$ @JohnHughes Or even $\{1,2,3\}$ with the diffeomorphism being flipping $1$ and $2$ :) $\endgroup$ – Del Sep 26 '16 at 21:05
5
$\begingroup$

Consider $S = $ a union of two concentric circles, $N = $ two circles, side by side (e.g., unit circles centered at $(\pm 2, 0)$), and $M = \mathbb R^2$. The complement of $S$ contains two components whose fundamental groups are isomorphic to the integers (the annulus between the two circles, and the unbounded piece); the complement of $N$ contains two components (the disks bounded by the circles) whose fundamental groups are trivial. Since a self-homeomorphism of $M$ that carries $S$ to $N$ would have to carry the complements to the complements, there can be no such self-homeomorphism.

This is probably as simple an example as you're going to find, for in the line, no such example works.

$\endgroup$

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.