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A strong version of the loop theorem implies that if an essential closed curve on the boundary of a 3-manifold $M$ is nullhomotopic, then realizing its image in $\partial M$ as a 4-valent graph, you can draw a cycle with no edge repeats that is homotopic on $\partial M$ to the boundary of an essential embedded disk in $M$. (Compare with Bing's "The geometric topology of 3-manifolds", pg 205.)

Q: Is there a similarly strong version of the annulus theorem?

For instance, suppose you have a (say, hyperbolizable) 3-manifold M with incompressible boundary, and you have two closed curves on the boundary of M that bound an essential singular annulus in M. (E.g., they're homotopic in M but not within its boundary.) Homotoping them to be in general position, can you find two cycles with no edge repeats on the 4-valent graph that is the union of their images, such that the two corresponding curves are homotopic to the two boundary components of an essential embedded annulus in M?

The closest thing I've found is Thm 2 in Cannon-Feustel's "Essential embeddings of annuli...", but that requires the two original curves to be disjoint.

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2 Answers 2

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I think this might follow from JSJ theory. Assume that $M$ is irreducible with incompressible boundary. Then any essential annulus is homotopic into an $I$-bundle region or a Seifert-fibered region of the JSJ decomposition. In the Seifert case, the region meets the boundary in annuli, in which case the boundaries of the immersed annulus will be a multiple of the core of the annuli, and I think one can cut and paste to get a simple closed essential core of each annulus, which then cobound an embedded essential annulus. Otherwise, if the annulis lies in a product region, then your immersed annulus goes between two curves on the surface. So this amounts to asking: for a given homotopy class of curve on a surface, can one always cut and paste a subset of the curve to give an embedded curve in the same isotopy class? This seems plausible, e.g., if the curve is not filling (one should be able to cut and paste to get a boundary component of the subsurface that it fills). In any case, it seems that at least this gives a reformulation of the problem (although I'm not sure what happens for twisted $I$-bundles).

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  • $\begingroup$ We can isotope the annulus to be vertical. At that point we should be good. Um, assuming that Ian B. is happy with perhaps getting an embedded Mobius band, instead of an annulus. $\endgroup$
    – Sam Nead
    Commented Aug 9, 2016 at 0:20
  • $\begingroup$ My impression is that Ian doesn't want to homotope the boundary of the annulus. $\endgroup$
    – Ian Agol
    Commented Aug 9, 2016 at 4:11
  • $\begingroup$ Thanks for the answers! I'll wait till I find Peter Scott's book to accept one, for fairness. You're right that I don't really want to homotope the boundary of the annulus, although I realize that wasn't stated well in the question. Do you think you can still get an embedded annulus by cut and paste if the boundary is compressible? I admit this is my real interest, although it's a very good point that it should follow from JSJ in the incompressible case. (I consider the question above asked and answered, of course.) $\endgroup$
    – biringer
    Commented Aug 9, 2016 at 16:08
  • $\begingroup$ Hey @biringer, I'm still not actually sure that my answer resolves this in the case that there's a non-trivial JSJ decomposition. If the $I$-bundle is the entire manifold, then I think it reduces to a question about curves on surfaces like I indicated. Otherwise, if the annulus is homotopic into an $I$-bundle piece, one can pass to a covering space corresponding to the fundamental group of the $I$-bundle piece, and do cut and paste up there to get an embedded annulus upstairs. Now the issue is whether projecting down, this can be cut and paste as well. One could probably use a tower argument. $\endgroup$
    – Ian Agol
    Commented Aug 9, 2016 at 17:19
  • $\begingroup$ In the compressible boundary case, pass to the cover corresponding to the fundamental group of the annulus. This cover compactifies to a solid torus. Again, the solid torus case sounds doable, although I certainly haven't considered it carefully. If the manifold is a handlebody, then I suspect one can use 2-separability and a tower argument to embed it in a finite-sheeted cover. $\endgroup$
    – Ian Agol
    Commented Aug 9, 2016 at 17:28
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I think that Theorem 15.1.5 in Scott's unpublished book on three-manifolds (email him for a copy) is very close to what you want.

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