# A torus whose middle is tied in an overhand knot; how many holes does it have?

## 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.

## 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.

$$\dagger$$: . . . or some other operation . . .

• The surface is just a torus, like you said, so the number of holes of the surface is the same as the number of holes of the torus. Or are you asking about the holes of something else? May 31, 2023 at 0:49
• I don't recall saying that, @Frank; how is it a torus? I'm interested in the number of holes in shape depicted. I'm not sure what else you might think I mean. If you can prove it's a torus in an answer, then I would be likely to accept it. May 31, 2023 at 0:57
• You refer to the shape as a torus whose middle is "tied up." This allows you to define a homeomorphism (a bijection that is continuous both ways) from the torus to your space. Tying it up doesn't change the fact its a torus; this is analogous to the fact that every knot is homeomorphic to a circle even though it's tied up (in fact that's how knots are defined). May 31, 2023 at 0:59
• I believe if you consider the space "inside" the torus, you get the knot complement of the trefoil knot. Maybe someone with more knowledge can post an answer about that. May 31, 2023 at 1:01
• This whole thing about "counting holes" is just an imprecise intuition for other much more precisely defined mathematical invariants. So your question is not really a mathematical question, without more precision in its formulation. May 31, 2023 at 2:12

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.