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. 



Do an unlinked trefoil and figure-eight cobound an annulus in $B^4$?
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](https://www.indiana.edu/~knotinfo/).


As far as where to read about knot concordance, a good place to start is Livingston and Naik's book-in-progress [here](http://pages.iu.edu/~livingst/Book.pdf).




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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.


