How many definitions are there of the Jones polynomial? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:06:29Z http://mathoverflow.net/feeds/question/67712 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67712/how-many-definitions-are-there-of-the-jones-polynomial How many definitions are there of the Jones polynomial? unknown (google) 2011-06-13T23:37:50Z 2011-06-14T18:46:55Z <p>Even with the connection to quantum groups being made clearer (I believe it was not known when the Jones polynomial was first introduced), it seems to me that still we don't have the "right" definition of the Jones polynomial. It is certainly true though that we know a lot of different definitions, some more useful than others. I'm thinking of:</p> <p>1) Kauffman bracket (i.e. the skein relation). This defines the Jones polynomial by giving a straightforward algorithm to compute it, but leaves any other significance a mystery.</p> <p>2) Quantum groups. Take $U_q(\mathfrak s\mathfrak l_2)$, and observe via universal $R$-matrices that "its category of representations is a braided monoidal tensor category", so that in particular, $\overbrace{V\otimes\cdots\otimes V}^{n\text{ times}}$ gives a representation of $B_n$, for any given representation $V$ of $U_q(\mathfrak s\mathfrak l_2)$. The Jones polynomial is easily derived from this representation of $B_n$.</p> <p>3) KZ equations (closely related to (2)). Let $X_n$ be the configuration space of $n$ points $(z_1,\ldots,z_n)$ in $\mathbb C$. Now write down the one-form <code>$A=\hbar\cdot\sum_{i&lt;j}\Omega_{ij}d\log(z_i-z_j)$</code> (taking values in $U(\mathfrak s\mathfrak l_2)^{\otimes n}$), and observe that this gives a flat connection on a trivial bundle of $V^{\otimes n}$ over $X_n$, for a representation $V$ of $\mathfrak s\mathfrak l_2$. The monodromy of this connection gives a representation of $\pi_1(X_n)=B_n$ on $V^{\otimes n}$.</p> <p>Methods (2) &amp; (3) (especially method 3) are natural constructions for representations of $B_n$.</p> <p>Question: Are there any other constructions of the Jones polynomial that are not trivially (interpret as you wish) equivalent to the ones above?</p> <p>I am particularly interested in ones which seem natural for the case of <em>knots in $\mathbb R^3$</em> (note that (2) and (3) seem natural ways to get representations of $B_n$, but, at least to me, it seems that the extension to knots is sort of ad-hoc). I feel like there are a number of "moral" approaches which "should" give the Jones polynomial, but have yet to be made rigoruous, and I'd be interested to know how close they are to being so:</p> <p>A) [warning: this is kind of sketchy] Start with $M_K=\operatorname{Hom}(\pi_1(\mathbb S^3-K),G)/\!/G$ (where $G=\operatorname{SL}(2)$) and consider this as a left-module over $R=\operatorname{Hom}(\pi_1(\text{torus}),G)/\!/G$. Make a noncommutative deformation $R^q$ of $R$ to get the Kauffman bracket skein module of the torus, and observe that $M_T^q$, the Kauffman bracket skein module of the solid torus $D^2\times S^1$ is a right-module over $R^q$. Since the Kauffman bracket skein module of $\mathbb S^3$ is $\mathbb C$, this means $M_K^q\otimes_{R^q}M_T^q=\mathbb C$. Then take $1\in M_K^q$ and some canonical elements (Jones-Wenzel idempotents) in $M_T^q$ and take their tensor in $M_K^q\otimes_{R^q}M_T^q=\mathbb C$. This should give the colored Jones polynomial of the knot. The problem with this is that we don't know how to define the deformed left-module structure on $M_K$ to get $M_K^q$.</p> <p>B) Take an ideal triangulation of the knot complement. Apply some black magic "TQFT with corners" (perhaps just some explicit formulae) and get back the Jones polynomial of the knot. I thought that Dylan Thurston was working on this at one point (in relation to the volume conjecture), but that was a while ago, and as far as I know, there is still no definition of the Jones polynomial from an ideal triangulation of the complement (I'm thinking something along the lines of Turaev-Viro invariants of $3$-manifolds). (Please correct me if I'm wrong)</p> <p>(Certainly my question could be stated in a more general setting for general quantum knot invariants.)</p> http://mathoverflow.net/questions/67712/how-many-definitions-are-there-of-the-jones-polynomial/67718#67718 Answer by Joel Kamnitzer for How many definitions are there of the Jones polynomial? Joel Kamnitzer 2011-06-14T01:41:30Z 2011-06-14T01:41:30Z <p>In Witten's famous paper "Quantum Field Theory and the Jones Polynomial", <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.cmp/1104178138" rel="nofollow">http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.cmp/1104178138</a>, he shows that the Jones polynomial can be defined using Chern-Simons theory. This definition has the advantage of being directly defined from a knot, rather than having to pick a braid first (which I agree is an important distinction).</p> http://mathoverflow.net/questions/67712/how-many-definitions-are-there-of-the-jones-polynomial/67755#67755 Answer by Bruno Martelli for How many definitions are there of the Jones polynomial? Bruno Martelli 2011-06-14T12:50:52Z 2011-06-14T18:46:55Z <p>Concerning point B, you can calculate the value of the Jones polynomial at the roots of unity by taking a spine of the knot complement and a kind of "projection" of the knot on the spine, as <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=189053&amp;vfpref=html&amp;r=67&amp;mx-pid=1249172" rel="nofollow">shown by Turaev</a>.</p> <p>More generally, the quantum invariants of a pair $(M^3, Y)$ where $M^3$ is a closed 3-manifold and $Y\subset M^3$ is a ribbon 3-valent graph (for instance, a framed link) which is <i> admissibly coloured </i> can be calculated by taking a <i> shadow </i> $X$ of the pair $(M^3, Y)$. As defined by Turaev in <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=189053&amp;vfpref=html&amp;r=63&amp;mx-pid=1292673" rel="nofollow">his book</a>, a shadow is a two-dimensional simple polyhedron with boundary equal to $Y$, whose thickening is a 4-manifold whose boundary is the pair $(M^3, Y)$. It is a nice geometric object for defining and computing quantum invariants.</p> <p>The quantum invariants are calculated as a sum over all admissible extensions of the admissible colouring from $Y$ to the whole shadow $X$. In most cases the number of such extensions is infinite and in order for this state-sum to become finite you need to fix a root of unity.</p> <p>It turns out however that if $X$ is contractible (or <a href="http://arxiv.org/abs/math/0510048" rel="nofollow">more generally</a> collapses onto a 1-dimensional polyhedron) the number of admissible extensions is finite, and the resulting quantum invariant is then a rational function. One may always choose a contractible shadow $X$ when $M$ is the 3-sphere. </p> <p>As for Witten's approach, the fact that the resulting rational function is a Laurent polynomial for links in the 3-sphere is mysterious from this point of view.</p>