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Consider a two-component tame link in 3-space, consisting of an arc from $(-1,1,0)$ to $(1,1,0)$ and an arc from $(-1,-1,0)$ to $(1,-1,0)$, confined to the slab $-1 \leq x \leq 1$. Call such a link trivial if it can be deformed so that the arcs admit parameterizations in which the $x$-coordinate is strictly increasing. (Note that this notion of triviality is rather forgiving, in that it permits the arcs to twist around one another.) There is a natural way to add two such links by sticking them side-by-side. (More formally, one compresses the links so that the $x$-coordinate in the arcs goes from $-1$ to 0 and from 0 to 1, respectively, and then applies the usual way of composing paths to the respective arcs.) For an example of what such a sum can look like, see the top figure at http://en.wikipedia.org/wiki/File:Fisherman's_knot.png . [I can't get the link to embed properly; can anyone fix this?]

Can the sum of two non-trivial links of this kind be trivial?

Note the confinement property that prevents us from pulling the arcs outside the slab.

It is a classical result that the analogously-defined sum of two non-trivial 1-component links cannot be trivial. (Can anyone provide a definitive reference for this result? I learned about it from a Martin Gardner column that presented a proof due to Conway.)

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What you are talking about are called string links. String links have a stacking monoid operation. It's non-commutative (provided you have two or more strands). The invertible elements are precisely the string links that are braids. There are a few references on this topic.

I forget who first proved the result about invertible string links, but you should be able to find most of the references in this Blair-Burke-Koytcheff paper. They cite Krebes, but it might go back further.

In the one-stranded case your monoid is free commutative with countably-infinite number of generators. That's Schubert's theorem (1949 Die eindeutige Zerlegbarkeit eines Knotens in Primeknoten).

http://arxiv.org/abs/1308.1594

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  • $\begingroup$ Definition 2.1 defines two such links to be equivalent iff there an isotopy of the slab (denoted by $M$ in the paper) that fixes the boundary and maps one link to the other. This is a priori stronger than the notion of isotopy that I had in mind, namely a way of deforming one link into the other that pays no attention to "how the rest of the slab gets out of the way" during the course of the deformation. But maybe the two notions of equivalence coincide. Do they? $\endgroup$
    – James Propp
    Commented Oct 9, 2015 at 16:32
  • $\begingroup$ @JamesPropp: I do not understand what kind of isotopy relation you are using. I had assumed you were using a standard one that allowed for the definition of the stacking operation. If your isotopy relation is too weak, then stacking isn't well-defined. $\endgroup$
    – Ryan Budney
    Commented Oct 9, 2015 at 17:12
  • $\begingroup$ My kind of isotopy between links is given by a continuous map $f: [0,1] \times \{1,2\} \times [0,1] \rightarrow M$ such that $f(0,\cdot,\cdot)$ parametrizes one link and $f(1,\cdot,\cdot)$ parametrizes the other (where a link parametrization is a continuous injection from $\{1,2\} \times [0,1]$ to $M$; the discrete coordinate indexes the 2 components of the 2-component link and the continuous coordinate parametrizes each link from one endpoint to the other). Are you saying I should call this homotopy rather than isotopy? Anyway, it's the kind of equivalence relation I'm trying to ask about. $\endgroup$
    – James Propp
    Commented Oct 9, 2015 at 19:02
  • $\begingroup$ It is easy to show that my kind of deformation gives an equivalence relation that allows for the definition of the stacking operation (as illustrated in Figure 1 on page 3 of arxiv.org/abs/1308.1594) at the level of equivalence classes. $\endgroup$
    – James Propp
    Commented Oct 9, 2015 at 19:08
  • $\begingroup$ Your definition appears to be equivalent to the Blair-Burke-Koytcheff definition via the isotopy extension theorem. $\endgroup$
    – Ryan Budney
    Commented Oct 11, 2015 at 3:57

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