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Consider a finite $n$-element classical (real) link and the resulting link structure obtained by cutting each of the component elements (knots). Let us represent the resulting structures in a tableau, where each term gives the cut element and a list of the resulting links made from the remaining component elements.

For instance, in a $3$-element Borromean ring (in which cutting any component leads to separation of every element, $6_2^3$ in Rolfsen's table of links) the tableau would be:

  • $l_1 \to \{ l_2, l_3 \}$
  • $l_2 \to \{ l_1, l_3 \}$
  • $l_3 \to \{ l_1, l_2 \}$

A fully interlinked link ($6_3^3$ in Rolfsen's table of links) would have the tableau:

  • $l_1 \to \{ l_2 l_3 \}$
  • $l_2 \to \{ l_1 l_3 \}$
  • $l_3 \to \{ l_1 l_3 \}$

A simple $3$-element linear chain (link) would have the tableau:

  • $l_1 \to \{ l_2 l_3 \}$
  • $l_2 \to \{ l_1, l_3 \}$
  • $l_3 \to \{ l_1 l_2 \}$

where of course here multiplication denotes linked elements.

Call such a tableau "feasible" if there exists a corresponding $n$-element link.

Questions

  1. Is there an algebraic method to determine if an arbitrary tableau is feasible? For instance can one algebraically show whether the following tableau is feasible?

• $l_1 \to \{ l_2 l_3 \}$

• $l_2 \to \{ l_1 l_3 \}$

• $l_3 \to \{ l_1, l_2 \}$

  1. Assuming a tableau is feasible, is there an algorithm to construct a corresponding link? (Of course the corresponding link need not be unique, up to ambient isotopy.)
  2. What is the number of feasible tableaus (up to permutation symmetry of the labels of the component links) for a given $n$?
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If I'm reading your notation correctly, the symbol $\{l_1, l_2\}$ means the two element link with components labelled $l_1, l_2$ together with the additional information that they are unlinked? Similarly $\{l_1l_2\}$ would mean they are linked?

If this is what your notation means, you question was answered by H.Debrunner in 1964. What you are interested in Debrunner called "the splitting property of a link".

Given an n-component link $L$, Debrunner chose the language $\frak{U}$ as a subset of the power set of $\{1,2,\cdots, n\}$. Membership $I \in \frak{U}$ is equivalent to saying the sublink of $L$ whose components are indexed by elements of $I$ is not a split link, i.e. in your notation it would not be of the form $\{l_{i1}, l_{i2}, \cdots, l_{ik}\}$. You can have commas, but you can't have every component separated by commas. Debrunner also used the convention that $\frak{U}$ is not allowed to have singletons.

Debrunner observed the only rules on splitting properties was this: If $A, B \in \frak{U}$ and if $A \cap B \neq \emptyset$ then $A \cup B \in \frak{U}$. i.e. he provided examples for every splitting property.

The answers to your questions (1) and (2) are immediate.

(1) Yes.

(2) Yes.

For question (3) this reduces it to a counting problem. It's too early in the morning for me to count.

The Debrunner paper I don't have a link do, but Taizo Kanenobu went one step further than Debrunner and proved that every splitting property (for non-split links) is realizable by a hyperbolic link:

https://projecteuclid.org/download/pdf_1/euclid.jmsj/1230395010

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  • $\begingroup$ Debrunner's paper is available here eudml.org/doc/170292 . This paper also seems to be relevant arxiv.org/abs/0804.4323v1 $\endgroup$ – j.c. Jun 21 '18 at 15:08
  • $\begingroup$ @RyanBudney: Yes, you properly interpret my notation; and thanks for the citation (+1). I'm a bit surprised that the answer to question 2 is affirmative, as it is not obvious which link satisfies even the simple example I gave under Questions. I'll read the references (translating the German) and once I understand the full paper will likely accept. Thanks again. $\endgroup$ – David G. Stork Jun 21 '18 at 15:56

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