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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 $R^N$?

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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.

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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 $R^3$). – William 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.

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