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I've been calculating some Jones polynomials lately and I was just curious if there was a "physical" (or, rather, geometric) meaning to evaluating the Jones polynomial at a particular value of $t$.

For example, if I take the Jones polynomial for the (right) Trefoil knot, I have

$J(t) = t + t^3 - t^4$.

Is there some way I can interpret $J(0)$? $J(1)$?

I understand that the Jones polynomial is a laurent polynomial, so I don't expect $J(0)$ to make sense for a lot of knots (for example the left trefoil has $J(t) = t^{-1} + t^{-3} - t^{-4}$), but I thought it was worth asking.

I also know that $J(t^{-1})$ gives the Jones polynomial of the mirror image knot. Is there a way to interpret $J(-t)$? $J(t^2)$? How about $J(t) = 0$?

Edit to clarify what I mean when I say "physical meaning": Since the Jones polynomial is a link invariant, $J(0)$ is also a link invariant (if it exists). Does this invariant correspond to a property of the knot that you can visualise, such as, say, the linking number or the crossing number?

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    $\begingroup$ The Jones polynomial evaluated at roots of unity is closely related to Chern-Simons theory. I don't know if that's the kind of physical meaning you're looking for though. The face that launched a thousand ships here is this paper of Witten's: projecteuclid.org/…. $\endgroup$ – Qiaochu Yuan Jun 30 '14 at 6:05
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    $\begingroup$ From the perspective of Khovanov homology the thing that really has meaning is not the evaluation of the Jones polynomial at some value but its coefficients; these are Euler characteristics of certain chain complexes, and also have a physical interpretation that I can't even vaguely describe. $\endgroup$ – Qiaochu Yuan Jun 30 '14 at 7:02
  • $\begingroup$ $V(1)$ is always zero on knots and a constant depending on the number of link components for links. You can see this by plugging $t=1$ into the skein relation. $\endgroup$ – Jim Conant Jun 30 '14 at 7:34
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    $\begingroup$ I think $V(-1)$ is the determinant of the knot up to sign. $\endgroup$ – Jim Conant Jun 30 '14 at 7:35
  • $\begingroup$ You mean $V(1)$ is always 1 for knots? $\endgroup$ – Mark B Jul 1 '14 at 4:06
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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 Lickorish-Millet here. I don't know whether similar relations hold for more general Fox colourings.

This is not really an answer to the precise questions you're asking, but it's a pretty result.


UPDATE (Aug 19, 2014): I have found some more references and some more info in this problem list: the third remark on page 383 (page 11 of the PDF) covers what was known in 2004. In particular, it says that computing $V_K(\omega)$ is $\#P$-hard (see Neil Hoffman's comment below) unless $\omega$ is a power of $e^{i\pi/3}$ or $\omega = \pm i$, and it gives the interpretation for $V_K(\omega)$ in the four remaining cases (the first two have been mentioned by Jim Conant in the comments above). If $L$ is a link, I will call $\ell$ the number of components, and $\Sigma(L)$ the double cover of $S^3$ branched over all components of $L$.

  • $V_L(1) = (-2)^{\ell - 1}$; for a knot, $V_K(1) = 1$;
  • $\left|V_L(-1)\right| = \left|H_1(\Sigma(L))\right|$ if $H_1(\Sigma(L))$ is torsion, and is 0 otherwise; for a knot, $\left|V_K(-1)\right| = \left|\det(K)\right|$;
  • $V_L(i) = (-\sqrt2)^{\ell-1}(-1)^{\mathrm{Arf}(L)}$ if $L$ is a proper link (i.e. ${\rm lk}(K,L\setminus K)$ is even for every component $K$ of $L$), and vanishes otherwise (Murakami); notice that the Arf invariant is defined only for proper links.
  • $V_L(e^{2i\pi/3}) = 1$.
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    $\begingroup$ For more on $\#$P-hard, see the wikipedia page: en.wikipedia.org/wiki/Sharp-P. Essentially, is an analog of NP, but these questions focus on counting "how many?" as opposed to just finding then verifying one solution. $\endgroup$ – Neil Hoffman Aug 19 '14 at 7:05
  • $\begingroup$ Does $Arf(K)$ always exist for a knot K? Is it only links for which the Arf invariant may not exist? $\endgroup$ – Mark B Mar 9 '15 at 7:32
  • $\begingroup$ Additionally, I think that the fourth dot point has a typo - it shouldn't be the magnitude of V(i). For instance, in a single knot with Arf(L) = 1, the RHS is negative. The formula in the paper doesn't have absolute value signs on it. $\endgroup$ – Mark B Mar 9 '15 at 7:46
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    $\begingroup$ @MarkB: thanks, now it's fixed. And yes, the Arf invariant is defined for all knots, but not for all links (not as far as I know). $\endgroup$ – Marco Golla Mar 9 '15 at 10:46
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The volume conjecture predicts the existence limit of (a certain normalization of) the colored Jones polynomials evaluated at roots of unity (which is not known to exist), and that this limit is equal to the hyperbolic volume of the knot complement. This volume is uniquely defined and in some sense is a "physical" quantity.

(Note: This doesn't literally answer the question, since the answer generalizes "Jones polynomial" to "colored Jones polynomial.")

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    $\begingroup$ Still absolutely in the spirit of what I was asking though! $\endgroup$ – Mark B Sep 7 '17 at 0:36

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