Motivation:
I stumbled upon the following image on Twitter recently:
Description: A see-through torus is depicted in which the "tube" that ordinarily forms its middle is tied in an overhand knot.
It had me wondering:
The Question:
How many holes does the shape have?
That is, if you ignore the gaps in its illustration and, instead, work with the surface of the shape then how many holes does it have in the sense of topology?
Context and Thoughts:
Topology is definitely not my forté. My guess, though, is that it would have something to do with the trefoil knot: if we take out the middle tube of the shape and glue its ends together trivially, then that's what you get, so maybe the number of holes is the number of holes of a torus "plus"${}^\dagger$ the number of holes of a trefoil knot.
I doubt this heuristic because it seems too naïve, even for someone who is uncomfortable with topology.
Given the surplus of facts out there concerning the trefoil knot, it's hard for me to track down which constant associated with it that I need; what is the number of holes of a trefoil knot?
I tried a Google search of the image above and it got me nowhere.
Due to my ignorance all round here, it is difficult for me to search for the right things, so please forgive me for not producing more accurate mathematics here.
This is not a question I think I can answer myself. The answers I'm looking for should be pitched at an introductory undergraduate level.
Please help 🙂
$\dagger$: . . . or some other operation . . .
Best Answer
The surface in the picture is still a torus, even though it is twisted in the middle. We can still define a homeomorphism from the torus to the surface in the picture, for instance by using the coordinate grid displayed.
In general, "twisting" a space in this way does not change its homeomorphism type. For example, the image of every knot is homeomorphic to the circle $S^1$, and in fact this is how knots are defined.
A more interesting space is the "inside" of the torus in your picture. This space is actually equal to the knot complement of the trefoil knot, i.e., the complement of the trefoil knot in the $3$-sphere $S^3$. We can think of $S^3$ as the one-point compactification of $\mathbb{R}^3$, which is just $\mathbb{R}^3$ plus an extra "point at infinity." We can also talk about knot complements in $\mathbb{R}^3$, but it turns out $S^3$ is more convenient (apparently for these reasons). To see that the inside of the torus is the knot complement of the trefoil, you can imagine "pulling" half of the boundary outward and through the point at infinity, so that it comes back around the other side. Then the outside of the torus becomes a (thick) trefoil knot, and the inside of the torus therefore becomes the knot complement.
The fundamental group of the knot complement of the trefoil is $\pi_1 = \langle x, y \mid x^3 = y^2 \rangle$ (this is proven here). The abelianization of $\pi_1$ is $\mathbb{Z}$, which is the first homology group $H_1$. In particular, the rank of $H_1$ is $1$, which can be interpreted as saying the knot complement has a single one-dimensional hole. According to this article, the remaining homology groups are all zero (except $H_0$), so there are no higher-dimensional holes.
Disclaimer: I know very little about knot theory, so if someone reading this does know more, please let me know if I have made any errors or missed any important details.