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(Just to be precise, in this question, the word "knot" means "ambient isotopy class of a (EDIT: smooth) knot in $S^3$.") A knot in $S^3$ is called prime if it is not the connected sum of two other non-trivial knots in $S^3$. Clearly any knot is a sum of prime knots, and it is a theorem that this decomposition is unique. One downside of this is that there are infinitely many prime knots (for example, all non-trivial torus knots are prime). Here's a vague version of the question - Is there a way to trade off the uniqueness result (and add finitely many operations) in exchange for starting with a finite list of knots?

To try to make my question a little more precise, I'll define an "operation" as a function that takes a list of knots as input and outputs a finite list of knots. I'm not sure how to say this well, but I'd like to avoid very dumb operations like "fix an ordered list of all knots $K_1,K_2,\ldots$ and if $K_i$ is input, output $K_{i+1}$." However, I am interested in dumb answers, just not as dumb as that :-)

Is there finite list of knots $L$ and a finite list of operations $O$ on knots with the property that if $S$ is the smallest set of knots which contains $L$ and is closed under the operations $O$, then $S$ contains all knots?

For example, the first paragraph, phrased in this language, is
L = all prime knots
O = input two knots and output their connected sum
S = all knots

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    $\begingroup$ 1) There are infintely many prime integers. Why do you want there to be finitely many fundamental knots? If you insist on having finitely many fundamental integers, then you are ignoring the natural multiplicative structure. 2) You probably want to consider tame knots. 3) You can get every tame knot from the unknot if your operation is a crossing change. $\endgroup$
    – Douglas Zare
    Commented Aug 4, 2010 at 6:23
  • $\begingroup$ 1) $\mathbb{Z}[x]$ is not finitely generated as an abelian group (under addition) or as a monoid (under multiplication), but it is finitely generated as a ring, and I was wondering if there is a somewhat analogous situation for knots. 2) Definitely. 3) Do you mean take a diagram, perform a Reidmeister move that adds a crossing, and then change that crossing? That's a good point. Is it obvious this works? (I'm quite new to knot theory and have come at it kind of backwards.) $\endgroup$
    – Peter Samuelson
    Commented Aug 4, 2010 at 7:27
  • $\begingroup$ It's a good exercise, anyway. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 4, 2010 at 17:17

3 Answers 3

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If you allow yourself to consider knotted trivalent graphs, instead of just knots, then you can start with just the tetrahedron and Mobius bands, and use the operations unzip, bubbling, and connect sum to get all knotted trivalent graphs. This result is due to Dylan Thurston and Dror Bar-Natan and is written up by Dylan in a GT monograph. Their motivation was the lack of a good candidate for an answer to your question which stayed entirely in the language of knots. However for many situations KTGs are just as good as knots (e.g. click on a random link at Dror's wiki).

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  • $\begingroup$ This is very interesting! It looks like they were trying to answer almost the exact same question that I asked, and in addition to adding operations, they also expanded "all knots" to "all knotted trivalent graphs." (If I had thought of this possibility (expanding "knots" to "knots + other things") I would have written the question differently to include that possibility also.) $\endgroup$
    – Peter Samuelson
    Commented Aug 5, 2010 at 22:06
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There is a refinement of the connect-sum decomposition called 'satellite operation' or 'splicing'. Like the connect-sum operation, it's originally due to Schubert, although Schubert didn't prove a uniqueness theorem. That came much later with the JSJ-decomposition.

Unfortunately, this consists of infinitely-many operations. You could think of it as one operation that takes as input $n$ knots and an $(n+1)$-component link $L = (L_0, L_1, \cdots, L_n)$ such that the sublink $(L_1, \cdots, L_n)$ is trivial. You also demand that the complement of $L$ in $S^3$ is either Seifert-fibred or hyperbolic. It's also an essentially unique decomposition (like the connect-sum decomposition) -- the link is well defined but the indexing of the knots is only unique up to the symmetry group of the link $L$.

Whitehead double of figure-8

This is called the Whitehead double of a figure-8 knot. The link $W$ (in the $\mathbb G_K$ diagram) is the Whitehead link, the input knot (only one) is the figure-8 knot.

For these definitions there is a framing convention, twisted Whitehead doubles would decompose as:

Twisted double

This doesn't have a standard name. I like to call it a "(twisted) Borromean double". The input link is the Borromean rings (technically, a twisted version of them), and the input knots are two trefoils. Another example:

Another twisted example

This decomposition also applies to links in $S^3$. Here is a picture of the decomposition in an interesting case:

Link decomp

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  • $\begingroup$ These are very nice pictures, thanks! Are there finitely many knots that are "indecomposable" in this sense? $\endgroup$
    – Peter Samuelson
    Commented Aug 4, 2010 at 7:37
  • $\begingroup$ Hyperbolic knots and torus knots are precisely the indecomposables. There's infinitely many of both. $\endgroup$
    – Ryan Budney
    Commented Aug 4, 2010 at 7:50
  • $\begingroup$ Ah, I should have looked at Wikipedia before commenting. It looks like there's much more information here arxiv4.library.cornell.edu/abs/math/0506523 :-) And it looks like the answer to the question in the comment above is no... $\endgroup$
    – Peter Samuelson
    Commented Aug 4, 2010 at 7:59
  • $\begingroup$ Hyperbolic knots and links have further decompositions but I'm not aware of any that decompose them into knots and links. A classic and closely-related decomposition to the one above is sometimes called the "rational-tangle decomposition" or the "$\mathbb Z_2$-equivariant JSJ-decomposition." It is what you get when you look at the JSJ-decomposition of the $\mathbb Z_2$-branched cover of $S^3$, branched over the link. www-bcf.usc.edu/~fbonahon/Research/Preprints/BonSieb.pdf $\endgroup$
    – Ryan Budney
    Commented Aug 4, 2010 at 9:03
  • $\begingroup$ Again with this, there's infinitely many indecomposables. $\endgroup$
    – Ryan Budney
    Commented Aug 4, 2010 at 9:04
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The knot tabulations do something like this. First they generate all planar 4-valent graphs (with a given number of vertices). Then they replace each vertex with an over and an under crossing. This set contains all link diagrams with the given number of crossings. Then what you do next depends on what you are interested in.

http://pzacad.pitzer.edu/~jhoste/HosteWebPages/downloads/Enumeration.pdf

The procedure for generating planar 4-valent graphs is I think the type of procedure you are asking for. It is implemented in nauty

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