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The question that I'd like to answer can be generalized to the following: if $M$ is an orientable 3-manifold and $F$ is a boundary component of $M$ (which may have other boundary components), can an arbitrary diffeomorphism of $F$ with itself be extended to a diffeomorphism of $M$ with itself? I would be very surprised if the answer to the question in this generality is yes, so I'll ask a more specific question.

If $M$ is a complement of a link with 1 or 2 components in $S^3$ and $\sigma:T^2 \to T^2$ is the antipodal map in each component, is there a diffeomorphism $\sigma':M \to M$ that restricts to $\sigma$?

I don't really know how to approach this, but here's a couple thoughts:

1) If I understand right, this question can be reduced to the question of whether the subgroup of the mapping class group of $M$ that preserves $F$ surjects onto the mapping class group of $F$ (since it's my impression that isotopies can be extended from submanifolds).

2) If we lower the dimension by 1 it seems like the answer is yes, at least if the last comment was right. To see this, let $C$ be the boundary component that is to be sent to itself, and write the surface as a quotient of a polygon in $\mathbb R^2$ that is symmetric about the origin. Also, center the boundary component $C$ around the origin, and put the other boundary components symmetrically around the origin. Then the antipodal map in $\mathbb R^2$ induces a diffeomorphism of $F$ with the desired properties.

(I'm not sure about tags - retag as appropriate.)

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2 Answers 2

The more general question has a negative answer for the following reason (which could probably be extended to the special case with more thought). The mapping class group of a finite volume hyperbolic 3-manifold is finite by Mostow rigidity. But finite volume hyperbolic 3-manifolds can have tori as boundary components, whose mapping class group is infinite. So in this case most diffeomorphisms of a boundary component cannot be lifted to the 3-manifold.

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I interprete the "antipodal map" $T^2 \to T^2$ as the elliptic involution, which acts as multiplication by $-1$ on homology. For many link complements the elliptic involution extends to the link complement, but not for all. This holds if and only if the link is strongly invertible: the Wikipedia page furnishes some counterexamples.

The elliptic involution in fact extends to the solid torus neighborhood of each knot component of the link via a map that inverts the orientation of the knot, and therefore you get a map $S^3 \to S^3$ which preserves the link but inverts the orientation of every knot component.

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Since the question did not require that the extension have order 2, it's equivalent to invertibility rather than strong invertibility. Most simple knots and links are invertible, but typical complicated knots and links are not. Jeff Weeks' program snappea will (usually) determine whether or not a given knot or link is invertible. –  Bill Thurston Mar 21 '11 at 13:04

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