37

Your question presupposes that people were excited about the Jones polynomial because it would help them to classify/distinguish knots. In fact, I suspect the interest came from the fact that this knot invariant was originally defined using operator algebras (rather than in the more combinatorial way people usually define it now). Operator algebras had grown ...

36

To strengthen Sam Nead's answer, note that it is trivial to compute the Jones polynomial from the Khovanov homology. It is known that computing (or even approximating) the Jones polynomial is #P-hard:
Kuperberg, Greg, How hard is it to approximate the Jones polynomial?, Theory Comput. 11 (2015), 183–219. http://arxiv.org/pdf/0908.0512.pdf
Vertigan, Dirk, ...

35

Yes, forcing can add fundamentally new knots, not equivalent to any ground model knot. Indeed, whenever you extend the set-theoretic universe to add new reals, then you must also have added fundamentally new knots.
Theorem. If $V\subset W$ are two models of set theory with the same ordinals and different reals (for example, any forcing extension with new ...

answered May 25 '18 at 19:01

Joel David Hamkins

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30

A quick skim of the paper you linked to finds (on paragraph two of page two) comments by Bar-Natan suggesting that his improvement should take time $\exp({\sqrt{n}})$, beating the naive algorithm (taking time $\exp(n)$). That is, the improvement hopefully makes an exponential algorithm subexponential. He is not claiming a sublinear algorithm.

25

The evaluation of the Jones polynomial at $e^{i\pi/3}$ is related to the number of 3-colourings $tri(K)$ of $K$ (see also here) as well as to the topology of the branched double cover $\Sigma(K)$:
$$tri(K) = 3\left|V^2_K(e^{i\pi/3})\right| = 3^{\dim H_1(\Sigma(K);\mathbb{Z}/3\mathbb{Z})+1}$$
This was proved by Przytycki in this paper (Theorem 1.13) and ...

25

As a historical note (others may have had a different perspective - I was a graduate student when the Jones polynomial made its appearance), when it came out there was some mild excitement because the Jones polynomial could distinguish some pairs of knots which the previous invariants (Alexander polynomial) could not. There was some excitement that the ...

23

This is a question that I remember worrying about when I first started learning about knot theory. Older books have a tendency to skim over this point rather lightly, perhaps because the resolution of the question seems to involve techniques that have little to do with standard knot theory. One book that doesn't avoid treating this issue squarely is Burde ...

23

No, there cannot exist infinitely many alternating knots with the same Alexander polynomial.
To see why, suppose for the contrary that $K$ belongs to an infinite family $\{K_n\}_{n\in\mathbb{Z}}$ of alternating knots with $\Delta_{K_n}(t)=\Delta_K(t)$. Immediately we have
$$ \det(K_n ) = |\Delta_{K_n} (−1)| = |\Delta_K (−1)| = \det(K ). $$
for all $n$. ...

21

This is a result of Lickorish, in his paper "A representation of orientable combinatorial 3-manifolds". The paper is only eleven pages, and is very readable. In his proof, Lickorish rediscovers some ideas first investigated by Max Dehn 40 years earlier.
Lickorish's theorem was unexpected at the time; I don't think that Witten's remark is really ...

21

This isn't directly what you ask, but it's also worth noting that unknot detection is in $\text{NP} \cap \text{co-NP}$, that is, there are polynomial-checkable certificates that will show that either a knot is the unknot or that the knot is not the unknot. The $\text{NP}$ certificate uses normal surface theory:
Ian Agol, Joel Hass, William Thurston, The ...

21

The statement that for arbitrary K in $S^3$, if for some $n \ge 2$, the n-fold cyclic branched cover is $S^3$ (or in some versions, a homotopy 3-sphere) then K is the unknot, was known as the Smith conjecture. It was proved around 1979, combining work of many authors: Thurston, Meeks-Yau, Gordon-Litherland, Bass, Shalen, and perhaps others. Its proof was ...

21

Yes, this is done in
Penney, D.E., Generalized Brunnian links, Duke Math. J. 36, 31-32 (1969). ZBL0176.22201.
Call a link $(n,k)$-Brunnian if it has $n$ components, and every sublink with $m$ components: does not split when $k<m\le n$; completely splits when $1\le m\le k$. Then it is shown in the above paper how to use iterated commutators to build an $...

20

I'm very curious where this came up. In any case, the answer to the first question is yes, it does distinguish these trefoils; you found the minimal representatives.
Let $a_0,\dots,a_{N-1}$ be the roots of unity that are visited along the knot, in (cyclic) order. Suppose we have a minimal representative for some non-trivial knot. Then we cannot have $|...

20

It wasn't the properties of knots, but rather the hydrodynamical properties of closed fluids. It stems from the most basic facts in fluid mechanics:
Kelvin proved (assuming inviscid flows) that a closed curve $C$ of fluid particles (velocity field $u$) has its circulation $\oint_Cu\cdot dl$ independent of time. His theorem isn't true if the curve is fixed in ...

20

There have been some topological applications of the Jones polynomial and its various generalizations. I believe that these applications increased the interest in these invariants by topologists.
One application was to the Tait conjectures. Jones used his polynomial to give lower bounds on the bridge number of links; see Proposition 15.6 of
Jones, V. F. R....

19

(Since this is just a string of references, I do not believe this constitutes a 'real answer' but it is too long for a comment, so I'm placing it in the answer field. Editors, please feel free to correct my etiquette.)
As for a general introduction or survey article, you might also look at these:
"An introduction to Heegaard Floer homology" by Ozsvath and ...

18

I did not see any mention of this preprint by Marc Lackenby, probably because the question is quite old :
A polynomial upper bound on Reidemeister moves
http://arxiv.org/abs/1302.0180
Building on the techniques introduced by Dynnikov and adding some normal surface theory, he shows that any unknot can be unknotted using only a polynomial number of ...

18

I would recommend you to look at Jones' survey paper from 1986, entitled A New Knot Polynomial and Von Neumann Algebras. It is very readable. Let me try to make a brief summary, though.
The basic object you start with is a $II_1$ factor. This is a von Neumann algebra $M$ with trivial center $Z(M)\simeq \mathbb{C}$, possessing a faithful trace $\tau : M \...

18

This is not a definitive answer to your query, but you might be interested in the model
explored in the paper,
"Localization of Breakage Points in Knotted Strings," Piotr Pieranski, Sandor Kasas, Giovanni Dietler, Jacques Dubochet, and Andrzej Stasiak, New Journal of Physics, vol. 3, June 2001, p. 1-13 (journal link).
They model a knot by a curve with ...

answered Sep 18 '12 at 1:52

Joseph O'Rourke

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18

Here is a figure of a (4,2)-Brunnian link (in the terminology of Mark Grant's answer):
And here is an image of a (5,3)-Brunnian link:
These are taken from G.C. Shephard's 2006 article "Interlinked Loops". He was not aware of the work of Debrunner and Penney at that time as he states essentially this MO question as an open problem.
However, the 2009 ...

17

Such a knot would yield a counterexample to one of two important conjectures in the area. A preliminary definition: a slice knot is homotopically ribbon if the inclusion of the knot into the slice disk complement induces a surjection on the fundamental group. It's easy to see that a ribbon knot is homotopically ribbon; note that homotopically ribbon doesn't ...

17

Le and Murakami (HERE and HERE) discovered several previously unknown relations between multiple zeta values through the study of quantum invariants of knots. Further relations were later discovered through knot theory by Takamuki, and by Ihara and Takamuki.
These relations stem from the fact that the Kontsevich invariant extends to an invariant not of ...

answered Aug 12 '16 at 5:47

Daniel Moskovich

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17

A proof may be given along the lines of the proof of the Jordan Curve theorem by Doyle (see this answer). This uses the fundamental group and a variation on Van Kampen, but not homology. So this probably still won't be satisfying to you, but it doesn't use Alexander duality (which is the normal way to do this).
By removing a point from the knot, we may ...

16

The relationship between operator algebras and braids is fairly straightforward to explain, and is nicely written up in many places (e.g. in Kauffman's Knots and Physics). Jones studied representations of the braid group $B_n$ into the Temperley-Lieb algebra $TL_n$. The existence of such a representation is not so surprising (the following explanation is ...

answered May 24 '12 at 7:52

Daniel Moskovich

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16

Revision: For $n>6$, there is no embedding of $\mathcal{S}_n \hookrightarrow \mathcal{S}_{n+1}$.
First, recall that there is an extension $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}_n \to Mod(S_{0,n})$, where $Mod(S_{0,n})$ is the (orientation preserving) mapping class group of the $n$-punctured sphere.
Inside $Mod(S_{0,n})$, there is a subgroup isomorphic ...

16

Here is one reason not to expect such a relationship (although I'm not sure if it can be completed to a proof). The Jones polynomial $J_\sigma$ (roughly) comes from taking the trace of a linear map $A_\sigma$ associated to the braid $\sigma$, so the question (roughly) asks about relations between $Tr(A_\sigma)$, $Tr(A_\tau)$, and $Tr(A_\sigma A_\tau)$.
Let ...

16

This is just a comment. The same week (!) when Dylan asked this question, we received at our department a message from a non-professional mathematician who wrote a computer program that tries to simplify knots using level moves. (A "level move" is like an under move, but there can be more strands lying below the arc that you move.) He says that he tried all ...

16

This paper from earlier this year (Jan 18, to be precise) proves the existence of $\mathbb{Z}/n\mathbb{Z}$-torsion for $n\le 8$ and $\mathbb{Z}/2^s\mathbb{Z}$-torsion for $s\le23$. It also states at the beginning of Section 3.4:
Until now, no knot or link with torsion larger than $\mathbb{Z}/8\mathbb{Z}$ was known.
I believe this is the state of art.
...

15

A non-trivial (and much more general) result of Gabai implies that $g_n(K)=g_1(K)$ for all $n$. This is encapsulated in the phrase ``Gromov norm equals Thurston norm". Roughly, the Gromov norm represents the minimal genus of an immersed Seifert surface, whereas the Thurston norm represents the minimal genus embedded Seifert surface. It follows that the ...

15

If by the symmetry group of a knot you mean the group of isometries of $S^3$ leaving the knot invariant, then this can only be cyclic or dihedral, apart from the special case of torus knots which can have an $O(2)$ group of symmetries. By restricting symmetries to the knot itself one gets a homomorphism from the symmetry group of the knot to the symmetry ...

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