Let $K_1$ the trefoil (left or right hopefully does not matter?) and let $K_2$ be the figure-eight knot in $S^3 = \partial B^4$. Are there any smooth properly embedded annulus $A$ in $B^4$ with $\partial A = K_1 \coprod K_2$?

I'm really interested in knowing some general obstructions for this sort of thing for more general knots $K_1,K_2$ so any ideas are welcome. I figured that I would ask the question with two specific knots just for fun.

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    When you say the boundary is the disjoint union of the two knots, are you insisting they are not geometrically linked? If so, as Andy Putman says, this is just concordance. However, if you allow the boundary of the annulus to be a nontrivial link of the two knots, then it is a different question. – Jim Conant Oct 12 at 3:30
  • @JimConant I had in mind initially that the two knots were geometrically unlinked. However, I am also interested in how you could address this if there was linking. – user101010 Oct 12 at 23:41
up vote 11 down vote accepted

The relationship you're asking for is called concordance. Determining if knots are concordant is quite difficult: there are many concordance invariants, but no kind of global picture of what it means.

One concordance invariant is the knot signature. The trefoil has signature $-2$ and the figure-8 knot has signature $0$, so they are not concordant. I got this data from the wonderful website Table of Knot Invariants.

As far as where to read about knot concordance, a good place to start is Livingston and Naik's book-in-progress here.


EDIT: Knots $K_0$ and $K_1$ are usually said to be (smoothly) concordant if there is a smoothly embedded annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$. In the comments, I was asked to sketch why this is the same as having an annulus connecting $K_0$ and $K_1$ in $B^4$ (assuming that $K_0$ and $K_1$ are unlinked).

There are a lot of ways to prove this. Here's a brief sketch of one of the easiest. If $D$ is a small open round ball in $S^3$, then

$$\left(S^3 \times [0,1]\right) \setminus \left(D \times [0,1]\right) \cong B^4.$$

There is a tiny issue in that the right hand side is a manifold with corners, but as always corners can be smoothed. Using this diffeomorphism, we immediately see that if there is a smoothly embedded annulus in $B^4$ connecting $K_0$ and $K_1$, then after possibly moving $K_0$ and $K_1$ around we can obtain a smoothly embedded annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$; this annulus avoids $D \times [0,1]$. Conversely, if we can find an annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$, then homotoping everything we can assume that this annulus avoids some $D \times [0,1]$, and then transport it to an annulus in $B^4$ connecting $K_0$ and $K_1$.

The above sketch elides various identification we are making, but once you understand the geometry (draw a picture!) you'll see that they are all pretty clear.

  • @AndyPutnam I have seen the definition of concordant - namely that there is an embedded annulus in $S^3 \times I$ with the boundary being the two knots, one in each component. I am a confused as to how this is the same as the existence of an annulus in $B^4$ though I imagine I am missing something simple. – user101010 Oct 12 at 23:39
  • @user101010: They are equivalent conditions. Which direction seems hard to you? I can sketch the argument. – Andy Putman Oct 13 at 3:26
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    (ps: I really wish people could spell my name...) – Andy Putman Oct 13 at 3:26
  • Whoops - I am sorry about the spelling. Both directions actually are a little unclear to me. In addition, I have been leaving the question open in the event that anyone knows how to tackle the question if there is some possible linking occurring between $K_1$ and $K_2$ - but thank you very much for your answer! – user101010 Oct 14 at 23:29
  • @user101010: I just added some details. – Andy Putman Oct 15 at 0:26

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