MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question 1: Given a smooth Riemannian surface $M$ in $R^3$ (i.e., a smooth Riemannian 2-manifold embedded in $R^3$) and a diffeomorphism $f: M\rightarrow M$ of class $C^{k\geq 2}$, does $f$ admit a smooth extension $\tilde{f}$ to all of $R^3$? If not always, then are there sufficient conditions?

Question 2: If the answer to Q1 is affirmative, then given two diffeomorphisms $f,g: M\rightarrow M$ of class $C^{k\geq 2}$ which are close in the $C^2$-topology, can we find extensions $\tilde{f}, \tilde{g}$ which are also close in the $C^2$-topology?

Edit 1: I should add that $M$ carries the induced metric (from $R^3$).

Edit 2: We can ask a more general question. Say $M$ is a smooth Riemannian $m$-manifold. Embed $M$ in $R^N$ isometrically. Say $f: M\rightarrow M$ is a diffeomorphism of class $C^k$. Can we extend $f$ smoothly to $R^N$?

share|cite|improve this question
up vote 4 down vote accepted

Q1: Definately not always. More like "almost never". If the automorphism extends to $\mathbb R^3$, then the bundle $S^1 \ltimes_f M$ would embed in $S^4$. $S^1 \ltimes_f M$ is the bundle over $S^1$ with fiber $M$ and monodromy $f$. The most-commonly used obstructions to embedding in this case are things like the Alexander polynomial, and Milnor signatures.

I don't see where the metric on $M$ plays a role for this.

If you want to see automorphisms that extend (and do not extend) for your Q1, take a look at my arXiv paper. You'll also find some references to several Jonathan Hillman papers that explore such obstructions.

In the case that your surface is unknotted -- bounding handlebodies on both sides (thinking of $M \subset S^3$) then the automorphisms of $M$ that extend in this case are well-known. They're called the mapping class group of a Heegaard splitting of $S^3$. It's an infinite group. Generators are known for it (if I recall, they're the automorphisms induced by handle slides) but off the top of my head I'm not sure how much is known about the structure of the group. Do a little Googling on "mapping class group of a Heegaard splitting of S^3" and you should start finding relevant material.

To respond to your 2nd edit, if the co-dimension is high enough all automorphisms extend. This is a theorem of Hassler Whitney's. The basic idea is this, let $f : M \to M$ be an automorphism. Let $i : M \to \mathbb R^k$ be any embedding. So you have two embeddings $i \circ f$ and $i$ of $M$ in $\mathbb R^k$. Any two maps $M \to \mathbb R^k$ are isotopic provided the co-dimension is large enough $k \geq 2m+3$ suffices, for example. So isotope your standard inclusion to the one pre-composed with $f$. The Isotopy Extension Theorem gives you the result.

For example, if $\Sigma$ is a Heegaard splitting / the surface is unknotted, $\Sigma \subset \mathbb R^3$ (or $\subset \mathbb S^3$) and you have an automorphism $f : \Sigma \to \Sigma$ a neccessary and sufficient condition for $f$ to extend to $\mathbb R^3$ (or a side-preserving automorphism of the pair $(S^3,\Sigma)$ in the $S^3$ case) is that if $C \subset \Sigma$ is a curve that bounds a disc on either the inside or outside of $\Sigma$ respectively, then $f(C)$ bounds a disc on the inside or outside of $\Sigma$ respectively (here I'm using inside/outside re the Jordan-Brouwer separation theorem). Since the fundamental group of the complement is just a free product of infinite cyclic groups this is something that can be checked rather easily provided you know the map $f$ well enough.

share|cite|improve this answer
Thank you Ryan for your answers. While I am asking more general questions, my situation is rather specific. I have a 2-manifold which is the locus of zeros of some smooth (in fact analytic) S(x,y,z) = 0 in $R^3$. I have a diffeomorphism $f: M\rightarrow M$, and I wish to extend $f$ smoothly to $R^3$. Now, I presume I could embed $M$ in $R^k$ for large enough $k$ and extend there (since it isn't of principle importance to me in which space to do work - $R^k$ or $R^3$). – user12918723509187 Sep 7 '10 at 2:41
Yes, $\mathbb R^4$ suffices, since any closed surface in $\mathbb R^3$ unknots in $\mathbb R^4$, so you're in the "Whitney" situation I described above. I'll give you a neccessary and sufficient condition for extendability in $\mathbb R^3$ but I'll just edit it into my answer above. – Ryan Budney Sep 7 '10 at 2:47

If $f$ is the time $1$ of a vector field, then it is easy to extend (in a stable way ragarding the second part of the question) by extending the vector field. IF $f$ is isotopic to identity you should be able to do the same.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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