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I was wondering if there were any known resources containing information on two-torus embeddable knots? I have checked the knot atlas with no luck, and it seems that many other cites only reference the figure eight knot.

Is there anyway to compute or algorithmically generate two-torus embeddable knots? I am specifically interested in obtaining several of these and examining their respective Alexander polynomial.

Thanks

Broudy
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    This is sometimes known as Heegaard-genus-$2$ knots. There's a little bit of discussion on this question with some references in the second answer. https://math.stackexchange.com/a/3160324/172988 – Kyle Miller Dec 22 '20 at 00:00
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    According to that answer, if I understand correctly, Heegaard-genus-$2$ knots have tunnel number either $1$ or $2$, and the Knotinfo database at least contains tunnel numbers https://knotinfo.math.indiana.edu/ – Kyle Miller Dec 22 '20 at 00:03

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As noted in the other answer, any knot embeds into a genus 2 surface. However, as you clarify in the comments, it seems that you mean the notion of embedding into a genus 2 Heegaard surface, which is much more restrictive.

There are two possible cases: the knot is separating or non-separating on the genus 2 surface. If the knot is separating, then it has Seifert genus 1, and hence its Alexander polynomial will have degree 2. I’m not sure if this case has been investigated, but there exists an algorithm to determine if such a knot is of this type. Certainly at the very least it must have Seifert genus 1 which is computable via normal surface theory (Haken gave an algorithm). What is needed is a knot that has two genus 1 Seifert surfaces that decompose the knot complement into two handle bodies. One may modify Haken’s algorithm to handle this case as well, although I don’t know if it could be done in practice. The program Regina can handle normal surfaces well.

The second case is that the knot is non-separating. Special classes of these knots appear in the Dehn surgery literature, such as twisted torus knots. This knot will be strongly invertible: the hyperelliptic involution of the genus two surface extends over each handlebody and preserving the knot, flipping the knot over.

Now consider the Heegaard surface in the knot exterior. This will be a genus 1 surface with two boundary components of integral slope. If the surface is compressible in the knot complement, then compression gives a knot lying on a torus. Moreover this torus is a tubular neighborhood of a tunnel number one knot. Thus the knot is a cable of a tunnel number 1 knot (it might be tunnel number one itself). I think conversely that a cable of a tunnel number 1 knot will lie on a genus 2 Heegaard splitting. There are normal/almost normal surface algorithms to compute if a knot is a cable knot and to compute the tunnel number, so this case is algorithmic.

If the surface is incompressible in the knot complement, then there are normal surface algorithms to enumerate such surfaces. Moreover again one may determine if the complements are handlebodies. So in summary there exist algorithms, although I don’t know if these are written up in the literature.

Agol
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Take any knot diagram, choose two points encircled by the knot:

enter image description here

These represent the positions of the holes in the two-torus.

Imagine placing two cylinders vertically on the plane at the positions of your dots.

Return your knot to three dimensions, and onto the two-torus.

enter image description here QED

David G. Stork
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  • Right, but I am interested in examples beyond the figure eight. Are there any other examples? Are there any methods of determining whether a knot could be mapped onto a two-torus? – Broudy Dec 21 '20 at 23:32
  • I think every knot can be embedded onto the surface of a two-torus. (Surely the unknot can.) Can you think of even a single counter-example? – David G. Stork Dec 21 '20 at 23:34
  • I am uncertain of any counter-example; however, I have never seen such a proof. Would you be able to direct me to one? – Broudy Dec 21 '20 at 23:37
  • Take any knot. Project it to the plane. Find two separate points encircled by the projection. Let these correspond to the holes in the two-torus. Project the knot onto the two torus respecting the points and the holes in the two torus. QED. (Notice this even works for the unknot.) – David G. Stork Dec 21 '20 at 23:47
  • @DavidG.Stork A two-torus embeddable knot would be a knot $K\subset S^3$ with a genus-two surface $\Sigma$ such that $K\subset\Sigma\subset S^3$. It's true that every knot has such a $\Sigma$, but presumably Isaac Broudy meant a $\Sigma$ that's a Heegaard splitting surface, to be properly analogous to torus knots. – Kyle Miller Dec 21 '20 at 23:59
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    KyleMiller: It was by no means obvious to me that IsaacBoudy meant that... but who knows? – David G. Stork Dec 22 '20 at 00:03
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    I don't see why your argument gives an embedding of the knot on a 2-torus, by the way, since I'm only seeing that you're putting a knot diagram (with crossings) on a 2-torus. The simplest way to do it is to have a torus run along the knot and then add an extra genus hole somewhere -- but this is certainly not a Heegaard surface in general. ($n$-embeddability of knots is defined in some places like Colin Adams's book on knot theory.) – Kyle Miller Dec 22 '20 at 00:06
  • @Kyle Miller, I believe this is what I meant to get across. I have been following Adam's book on Knot Theory and he defines n-embeddable torus knots. I am interested in doing computations with these types of knots. – Broudy Dec 22 '20 at 03:39