I am having some trouble with a paper by Livingston (Infinite Order Amphicheiral Knots, Algebraic and Geometric Topology 1, 2001, 231-241). He starts with a knot $K$, and constructs a new knot "replacing the neighborhood of an unknotted circle $L$ in the complement of a Seifert surface for $K$ with the complement of a knot $J$. The identification map switches longitude and meridian so that the resulting manifold is still $S^3$. The effect of this construction is to tie that portion of $K$ that passes through $L$ into a knot".

I see what happens to $K$, but I cannot understand how the construction goes. For example, I think that the neighborhood of $L$ is a tubular neighborhood, but then why replacing it affects the original knot $K$?


1 Answer 1


In the background of this construction is JSJ theory, which deals with decomposing $3$-manifolds along embedded tori. Let's analyze what happens with tori in knot complements so we understand the particular synthesis.

In $S^3$, a key fact about embedded tori is that one side is a solid torus and the other is a knot exterior (possibly also a solid torus if it's an unknot exterior), which is a nice consequence of the loop theorem. Suppose $T$ is an embedded torus in a knot exterior $S^3-\nu(K)$, where $\nu(K)$ is a tubular neighborhood of a knot $K$. By the above consideration, $K$ is either (1) on the solid torus side or (2) on the knot exterior side of $T$. In case (2), $T$ bounds a solid torus disjoint from $K$, so it is simply the boundary of a regular neighborhood of a knot in the exterior of $K$; we won't say anything more about this case.

In case (1), then the solid torus has a loop $L\subset T$ that bounds a disk in it. The solid torus also has a curve $M\subset T$ so that when gluing in another solid torus with its meridian glued to $M$ and its longitude glued to $L$, one obtains an $S^3$. Let's refer to this $S^3$ by $\Sigma^3$ since we want to consider $K$ both in the original $S^3$ and in this $\Sigma^3$. In $\Sigma^3$, $L$ is an unknot in the complement of $K$. It is best to think of $L$ as being core of the solid torus that was glued in, and we can think of that solid torus as being a tubular neighborhood $\nu(L)$ of $L$. Hence, $\Sigma^3-\nu(L)$ is the solid torus on the $K$ side of $T$.

Going in reverse, if we take $\Sigma^3-\nu(L)$ and glue in the knot complement that was on the other side of $T$ from $K$, we return to $K\subset S^3$. In the following, the left-hand side is $K$ and $L$ in $\Sigma^3$, where $T$ is the boundary of the closure of $\nu(L)$, and the right-hand side is $K$ where the $L$ side of $T$ has been replaced by a knot exterior of a knot $J$. The longitude of this knot exterior corresponds to a meridian loop of $L$ (i.e., $M$ from before), and the meridian of this knot exterior corresponds to $L$ back when we thought of it as being on $T$ itself -- or, we may think of the meridian as being $L$ pushed slightly into the disk it bounds in the solid torus on the $K$ side of $T$.

Satellite operation

This is essentially the definition of a satellite knot, with $J$ being the companion and $(K,L)$ being the pattern.

Notice we made no use of the fact that $L$ is disjoint from a Seifert surface for $K$ in $\Sigma^3$. This is a separate condition, and it is equivalent to saying that $L$ has linking number $0$ with $K$ in $\Sigma^3$, or that $K$ is nullhomotopic in the complement of $L$.

Roughly speaking, $K$ in $S^3$ has a Seifert surface that is "knotted up" by $J$. $K$ always has a Seifert surface where only thin bands pass through $L$. Then this "bundle" of bands is knotted up by $J$.

One example of this construction is a Whitehead double. It's illuminating to think about a Whitehead double's genus-$1$ Seifert surface and how it relates to all of this.

  • $\begingroup$ Thank you so much for your help. My knowledge of 3-manifolds is still very low, but I am working on it! Thanks again. $\endgroup$ May 5, 2020 at 21:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .