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Part 1: The following picture is from Saveliev's book Lectures on Topology of 3-manifolds, page 130:

knot

He indicates that the knot drawn in the solid torus $S^1 \times D^2$ is homologous to $S^1 \times \{ 0\} \subset S^1 \times D^2$.

How can we prove such a claim? Do we know the classification of null-homologous knots in $S^1 \times D^2$?

Part 2: To obtain a Mazur manifold $W$, he says that we attach a $2$-handle to $S^1 \times D^3$ (the dotted circle) with framing $3$. Diagrammatically, we have:

enter image description here

If the knot is not null-homologous, where does the framing $3$ come from?

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    $\begingroup$ The knot shown is homologous to $S^1 \times \{0\}$ but it is not null-homologous. You might want to fix your question and its title... $\endgroup$
    – Sam Nead
    Mar 15 at 17:23
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    $\begingroup$ Why? If not, how can we talk about the framings of given knots? Because the next step is attaching a $2$-handle along this knot. $\endgroup$ Mar 15 at 17:30
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    $\begingroup$ I am super confused! I have added a second part to the question... $\endgroup$ Mar 15 at 17:43
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    $\begingroup$ You are using the terminology "null homologous" incorrectly. I suggest correcting your question as this will continue to throw people off. Also, why are you asking for a classification of these knots? Seems beside the point of your question. $\endgroup$ Mar 15 at 18:08
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    $\begingroup$ Correct, and your curve does not have a Seifert surface in $S^1 \times D^2$, that's why I asked you to try and construct one -- it does not exist. $\endgroup$ Mar 15 at 20:31

2 Answers 2

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Two knots in the solid torus $U = S^1 \times D^2$ are homologous if and only if they have the same (signed) winding number. Proving this boils down to computing the first homology group $H_1(U, \mathbb{Z})$.

For a given oriented knot $K$, its winding number can be computed by (a) finding an oriented meridian disk $D$ for $U$ which is transverse to $K$ and then (b) computing the algebraic intersection number of $K$ and $D$.

Finally you ask: "Do we know the classification of null-homologous knots in $S^1 \times D^2$?" Well, there is a (very complicated) algorithm that decides if two null-homologous knots in $U$ are isotopic. Using this we could produce the (infinite) list of all such knots (said list being complete and without repeats).

where does the framing 3 come from?

I'll guess that he means either (a) the blackboard framing for $K$ or (b) the framing coming from the surface that $K$ and $S^1 \times \{0\}$ cobound. Luckily, these are the same in this particular case.

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  • $\begingroup$ For the final part, do you suggest a reference? $\endgroup$ Mar 15 at 18:18
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    $\begingroup$ The blackboard framing of a knot $K$ (drawn in the plane) is the normal vector field to $K$, thought of as an immersed curve in the plane. A quick google search found some research articles as well as pictures of framed blackboards, as well as pictures of knots equipped with the blackboard framing. For example: researchgate.net/figure/… $\endgroup$
    – Sam Nead
    Mar 15 at 18:25
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    $\begingroup$ The framing comes from the fact that you drew the knot in $S^3$! This is the "magic" idea of Akbulut's carving notation, that allows you to label framings of homologically essential knots $\#^g S^1 \times S^2$ by integers, once you fix a representation of $(\#^g S^1 \times S^2,K)$ in "dotted form" (which is exactly what you have). I have learnt this from Gompf and Stipsicz's book. $\endgroup$ Mar 15 at 18:33
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    $\begingroup$ @MarcoGolla Thanks! I have two questions: What does homologically essential knot mean here? Do you suggest section 5.4 of Gompf and Stipsicz's book? $\endgroup$ Mar 15 at 19:37
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    $\begingroup$ "Homologically essential" means that its not null-homologous (sometimes with rational coefficients, but in this case it's the same, since there's no torsion). And, yes, I think it's Section 5.4. $\endgroup$ Mar 15 at 19:55
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As @MarcoGolla mentioned, the framing can be controlled due to Akbulut's carving technology: a dotted circle notation. It was introduced in the following article:

Akbulut, Selman. "On 2-dimensional homology classes of 4-manifolds." Mathematical Proceedings of the Cambridge Philosophical Society. Vol. 82. No. 1. Cambridge University Press, 1977.

Let $U$ be the unknot in $S^3$ and $D_U$ be the ribbon disk in $B^4$ with $\partial D_U = U$. Observe that $S^1 \times B^3$ is the ribbon disk exterior of $D_U$, i.e., it is diffeomorphic to $B^4 \setminus \nu(D_U)$ where $\nu(D_U) \approx D_U \times B^2$.

Consider a ribbon knot with a ribbon disk $(K,D) \subset (S^3,B^4)$. Similarly, one can try to understand the $4$-manifold $B^4 \setminus \nu(D)$. The procedure of the construction of the Kirby diagram was answered, for instance here.

Once it is understood, one can also put a dot on the ribbon disk exterior. The reference is again Akbulut's book Section 1.1 and 1.4 about carving ribbon disks:

Akbulut, Selman. 4-manifolds. Vol. 25. Oxford University Press, 2016.

It equivalently represents the ribbon disk exterior. See again Exercise 1.10 and Figure 1.22 in Akbulut's book.

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