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Math Forum / Mathematics / Research / April 2009



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Dehn's lemma in higher dimensions

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Blake Winter - 26 Mar 2009 08:30 GMT
Dehn's lemma can be phrased as saying that if we have an S^1 K in S^3
which bounds an immersion f:D^2->S^3 s.t. f^-1(K)=\partial D^2, then K
bounds an embedded D^2.
Is the analogous statement true for higher dimensions (in the smooth
category)? Specifically, is it known whether:
Given an S^n K in S^n+2 which bounds an immersion f:D^n+1->S^n+2 s.t.
f^-1(K)=\partial D^n+1, K bounds an embedded D^n+1?
Blake Winter - 03 Apr 2009 07:29 GMT
> Dehn's lemma can be phrased as saying that if we have an S^1 K in S^3
> which bounds an immersion f:D^2->S^3 s.t. f^-1(K)=\partial D^2, then K
[quoted text clipped - 3 lines]
> Given an S^n K in S^n+2 which bounds an immersion f:D^n+1->S^n+2 s.t.
> f^-1(K)=\partial D^n+1, K bounds an embedded D^n+1?

I seem to have found the answer to my question. Swarup, GA, 1975: An
unknotting criterion. J. Pure Appl. Algebra, 6 (1975) and Dicks, W.
and Dunwoody, M.J. Groups acting on Graphs, Cambridge studies in
advanced mathematics 17, Cambridge University Press, Cambridge-New
York-Melbourne (1989), are cited by Hillman, Four-manifolds,
Geometries, and Knots as providing the result that if a codimension-2
sphere bounds an immersed codimension-1 disk which has no double-
points on its boundary, then the complement of the sphere is homotopy
equivalent to the circle. The results of Levine et al. then show that
in dimensions five and up, there is an embedded disk, whereas in
dimension four this is true in the topological category by Freedman,
but unknown (to my knowledge) in the PL/smooth categories.

This does raise a related question. In dimension 3 and 4, the
existence of such an immersed disk is guaranteed if the fundamental
group vanishes. But in higher dimensions, is it possible to have a non-
trivially embedded sphere where the fundamental group of the
complement vanishes? Clearly by the result of Swarup it would have to
have some nontrivial homotopy group.
 
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