Let $T$ be a solid torus and $K$ be a knot. Suppose $i: T \rightarrow S^3 $ be an embedding of $T$ such that it is homeomorphic to regular neighbourhood of the knot(i.e., thickened knot). Now, how does the complement of this embedding $iT$ look like?

I think it will also be a solid torus, say $T^\prime$, such that the meridian and longitude of $T^\prime$ are longitude and meridian of $T$ respectively.


2 Answers 2


The only time the complement of a (thickened) knot is a another torus is when the knot is trivial, or the unknot. So if the knot is just a circle, you are right. But for every other knot, there is a lot going on.

By the way, this space you describe is called the knot complement or the exterior of the knot.

What exactly is going on is a current area of research and has many interesting questions. In particular, these manifolds are often hyperbolic and methods from geometry and topology are used together to study them. See wikipedia for some more information.

  • $\begingroup$ I understand that gluing two solid tori along the boundary, such that meridian of one goes to the longitude of other would yield $S^3$. So, can't we just take homeomorphic images of these two tori and say the same? $\endgroup$ Sep 28, 2017 at 7:00
  • 1
    $\begingroup$ @AjayKumarNair No. The defining feature of a solid torus is a compressible disc, which you can still find in the inside of the torus that follows the knot. But if the exterior were a solid torus, there would be a compressible disk there too, and there is not one. Try reading "The Knot Book" by Colin Adams, which you can find online for free. Page 84. $\endgroup$
    – N. Owad
    Sep 28, 2017 at 8:37

A Seifert surface $\Sigma$ of a knot is an orientable surface in the complement of $i(T)$ whose boundary lies in $\partial i(T)$ and where $\partial\Sigma$ is a loop that is homologically trivial (it is a longitude). A non-trivial knot's Seifert surface has genus at least $1$, and only the unknot has a genus-$0$ Seifert surface. Using Kneser's lemma, one may assume $\pi_1(\Sigma)\to\pi_1(S^3-i(T)^\circ)$ is injective.

If the complement of $i(T)$ were a solid torus, then $\pi_1(S^3-i(T)^\circ)$ would be $\mathbb{Z}$, so $\pi_1(\Sigma)=1$, hence $\Sigma$ is a disk, implying the knot is trivial.

If the complement were a solid torus, then you could reassemble the space using Dehn filling. To get $S^3$, you will find that, up to Dehn twists, the meridian disk of one solid torus has to be glued along the longitude of the other solid torus, again implying the knot was the unknot because that disk is a Seifert surface.

If the complement were a solid torus, then the fundamental group of the knot complement (the knot group) would be $\mathbb{Z}$, however the trefoil knot has a knot group isomorphic to the braid group on three strands, for example.


You must log in to answer this question.

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