4
$\begingroup$

It is well known that not all graphs embed on the plane (e.g. the graph $K_{3,3}$). However, every finite graph embeds into a surface of some genus. One can think of this procedure as starting with a drawing of a graph on the plane and for each crossing one glues in a torus to resolve the intersection. So the takeaway is that adding 1-handles helps to resolve intersections. My question is:

Given a knot in the 3-ball, does it become trivial, i.e. it extends to an embedding of $D^2$, if we glue enough 1- and 2-handles? Can we use 2-handles to resolve crossings?

In my mind, these two problems, embedding graphs in $\mathbb{R}^2$ and embedding 2-disks with a given boundary knot in $\mathbb{R}^3$, are similar in the sense that both become solved when we stabilize by dimension, i.e. when we add an extra dimension to the target. In some sense, the obstructions to embedding graphs in the plane are local and so they vanish if we add "a small extra dimension" of room to resolve an intersection (much like blowing up resolves singularities.)

I am aware that this intuition is a bit sketchy but I would be curious to see any thoughts on this direction.

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ If you allow attachment of the $2$-handles after the $1$-handle attachments then you are basically asking about the map $\pi_0 Emb(S^1, D^3) \to \pi_0 Emb(S^1, M)$ where $D^3$ is a subset of a manifold $M$, i.e. attachment of $3$-handles does not really change the story. I think we have threads on this topic: mathoverflow.net/questions/389582/…, mathoverflow.net/questions/173235/…. Is your question not answered in these threads? $\endgroup$
    – Ryan Budney
    Commented May 8, 2023 at 8:56
  • $\begingroup$ @RyanBudney I don't really see how these threads answer this question. They seem to fix a specific $3$-manifold, while I "fix" a knot and ask whether there exists a $3$-manifold where this knot is trivial. $\endgroup$
    – João Lobo Fernandes
    Commented May 9, 2023 at 9:28
  • $\begingroup$ Your question is still unclear: My guess is that you are allowing to add only 3-dimensional handles (namely 1-handles and 2-handles) to $B^3$ (but Sam's answer interprets your question as allowing to add 4-dimensional handles). $\endgroup$
    – Moishe Kohan
    Commented May 9, 2023 at 14:02

2 Answers 2

Reset to default
5
$\begingroup$

If you mean adding 1 and 2 handles to the boundary of the ball, that will never change the (non)triviality of the knot in the 3-ball. On the other hand, perhaps the notion of tunnel number is what you are looking for. Find a handlebody H containing the knot as a core loop (i.e. the handlebody can be built by adding 1-handles to a solid torus neighborhood of the knot) such that the exterior of H is also a handlebody (that is: the 3-ball can be built by attaching 2-handles to the boundary of H.) The tunnel number of the knot is one less than the minimal genus of such a handlebody. Alternatively, it is the minimum number of 1-handles we needed to add to the solid torus neighborhood of the knot in order to create $H$. Since $H$ and its complement form a Heegaard splitting of the 3-ball they are (in some sense) knot-theoretically trivial. Indeed, $H$ will have a planar spine.

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ The trouble is that OP never defined what they mean by triviality of a knot $K$ in a general 3-manifold $M$. One natural definition is that there exists a 3-ball $B\subset M$ containing $K$ such that $K\subset B$ is unknotted. I am not sure your answer addresses this. $\endgroup$
    – Moishe Kohan
    Commented May 8, 2023 at 20:53
  • $\begingroup$ @MoisheKohan my definition is: a knot $f:S^1\to M$ is trivial if it extends to an embedding of $D^2$ to $M$. I will add this definition to the question. $\endgroup$
    – João Lobo Fernandes
    Commented May 8, 2023 at 22:05
  • 2
    $\begingroup$ @JoãoLoboFernandes: Your definition is equivalent to mine. $\endgroup$
    – Moishe Kohan
    Commented May 8, 2023 at 22:22
  • 1
    $\begingroup$ Sorry: here's why $K$ can't become trivial after the handle attachments. After attaching 1-handles and 2-handles to the boundary of $B$ you get a 3-manifold $M$. The boundary of $B$ is a sphere $S$. If $K$ bounds a (smooth) disc $D$ in $M$ it can be isotoped to be transverse to $S$. An innermost disc argument then produces a smooth disc with boundary $K$ disjoint from $S$ showing that $K$ was trivial in $B$. $\endgroup$
    – Scott Taylor
    Commented May 9, 2023 at 15:06
5
$\begingroup$

You example of graphs in the plane "becoming embedded via local operations" is not modelled by handle attachment. (For example, if you immerse $K_5$ in a disk $D^2$ and then attach one-handles to the boundary of $D^2$, then your $K_5$ does not become embedded.) Instead, you are performing a "surgery and re-embedding" supported in a small neighbourhood of a crossing. Now, knots (or more generally, links) in three-space do not have crossings. However, disks (and collections of disks) in four-space can. So you might rephrase as follows.

Question: Suppose that $K$ is a tame knot in the three-sphere $S^3$. Recall that $S^3$ bounds a standard four-ball $B^4$. Suppose that $K$ bounds a smooth immersed disk. That is, there is a proper immersion $(D^2, S^1) \to (B^4, S^3)$ with image $K$ in $S^3$. Can we perform a "surgery and re-embedding" in a small neighbourhood of each of the self-intersections (of the image of $D^2$) to obtain a four-manifold $X^4$ with boundary $S^3$ so that $(D^2, S^1)$ now embeds in $(X^4, S^3)$?

Improve this answer
$\endgroup$
3
  • $\begingroup$ This is a very nice idea, this is probably similar to what I had in mind in the first place, but I have some questions. What precisely do you mean by "surgery and re-embedding"? Also why do you consider immersions of $(D^2,S^1)\to (B^4,S^3)$ and not just $D^2\to D^3$, which is closer to the idea of showing that the knot $K$ is unknotted in a $3$-manifold ? $\endgroup$
    – João Lobo Fernandes
    Commented May 11, 2023 at 10:55
  • $\begingroup$ I should not have said "surgery and re-embed" - probably "surgery and re-immerse" is a better phrase. And, by this I mean that you have pair of spaces A and B and an immersion f of A into B. You seem to want to change the space B (in a local way, say surgery) and then change the immersion f (in a local way, say, re-immersion). $\endgroup$
    – Sam Nead
    Commented May 12, 2023 at 12:26
  • $\begingroup$ I am mapping into the four-ball instead of the three-ball because it is impossible (by definition) for a non-trivial knot to span an embedded disk in the three-ball. $\endgroup$
    – Sam Nead
    Commented May 12, 2023 at 12:27

You must log in to answer this question.

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