# Can a diffeomorphism between submanifolds be extended to a self-diffeomorphism of total manifold?

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?

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.

• 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 $n\geq 3$). – Jason DeVito Sep 26 '16 at 18:01
• @JasonDeVito You'll find they exist in $n=2$ as well. – PVAL-inactive Sep 26 '16 at 20:11
• @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? – Jason DeVito Sep 27 '16 at 0:44
• 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...) – John Hughes Sep 27 '16 at 1:02
• @JasonDeVito No but you can proof for instance that for any surface $\pi_1$ is generated by simple closed curves. – PVAL-inactive Sep 27 '16 at 3:56

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 $\psi|_{\{0, 1, 2\}} = \varphi$.

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.

• 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. – John Hughes Sep 26 '16 at 20:50
• 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. :) – John Hughes Sep 26 '16 at 20:52
• @JohnHughes Or even $\{1,2,3\}$ with the diffeomorphism being flipping $1$ and $2$ :) – Del Sep 26 '16 at 21:05

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.