On connection between Knot theory and Operator algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:01:19Z http://mathoverflow.net/feeds/question/97788 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97788/on-connection-between-knot-theory-and-operator-algebra On connection between Knot theory and Operator algebra Arnab 2012-05-23T20:10:06Z 2012-12-29T02:50:32Z <p>What exactly is the connection between knots and operator algebra? I heard that Jones established such a connection while discovering the celebrated Jones Polynomial. </p> <p>Now Jones Polynomial is probably understood out of that context on its own, but what was it with Operator algebras in this space ? Can someone explain in plain English?</p> <p>As is probably very evident, I am a complete newbie to this area.</p> http://mathoverflow.net/questions/97788/on-connection-between-knot-theory-and-operator-algebra/97796#97796 Answer by MTS for On connection between Knot theory and Operator algebra MTS 2012-05-23T22:30:24Z 2012-05-23T22:30:24Z <p>I would recommend you to look at Jones' survey paper from 1986, entitled <em>A New Knot Polynomial and Von Neumann Algebras.</em> It is very readable. Let me try to make a brief summary, though.</p> <p>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 $\tau : M \to \mathbb{C}$ (i.e. a positive normal state such that $\tau(a^*a)=0$ implies $a=0$) and having no minimal projections (this excludes matrix algebras).</p> <p>For whatever reason, it is a good idea to study subfactors of $N \subseteq M$. It turns out that the most important thing about a subfactor is not so much its isomorphism type (it is very hard to tell if von Neumann algebras are isomorphic or not) but rather the way it sits inside the big factor.</p> <p>Using the trace $\tau$, you can perform the GNS construction for $M$. This means that you define an inner product on $M$ via $\langle x,y \rangle = \tau(y^*x)$ (which is positive definite since $\tau$ is faithful), and then take the completion to get a Hilbert space called $L^2(M,\tau)$. Then $M$ acts on this Hilbert space by left-multiplication.</p> <p>Now you do what is called Jones' basic construction. Inside the Hilbert space $L^2(M,\tau)$ is the subspace $L^2(N,\tau)$, so there is the orthogonal projection of $L^2(M,\tau)$ onto $L^2(N,\tau)$. Call that projection $e_1$. Then define a new algebra $M_1$ to be the von Neumann algebra generated by $M$ and $e_1$ (inside $\mathcal{L}(L^2(M,\tau))$). It is immediate that $e_1$ commutes with $N$.</p> <p>It turns out that if the inclusion $N \subseteq M$ was of finite index (which I won't get into here) then $M_1$ is also a $II_1$ factor (so it comes with a faithful trace also), and the inclusion $M \subseteq M_1$ has the same index as $N \subseteq M$.</p> <p>Then you just keep going! Repeat the basic construction for $M \subseteq M_1$ to get a projection $e_2$, then let $M_2$ be the von Neumann algebra generated by $M_1$ and $e_2$, etc. So you end up with a sequence of projections $e_1,e_2,\dots$ which satisfy the following relations:</p> <ol> <li>$e_i e_j = e_j e_i$ if $|i-j| > 1$.</li> <li>$e_i e_j e_i = \lambda e_i$ whenever $|i-j| = 1$, where $\lambda$ is the inverse of the index of $N$ in $M$.</li> </ol> <p>The projections $e_i$ give a representation of something called the Temperley-Lieb algebra. You can see that these relations are reminiscent of the relations in the braid group. That is where the connection comes in. Knots are connected to braids, braids are connected to the Temperley-Lieb algebra (and hence to these projections) and then you can use the trace in the von Neumann algebra to define invariants of knots.</p> <p>That is the gist of it. Read Jones' paper for more details.</p> http://mathoverflow.net/questions/97788/on-connection-between-knot-theory-and-operator-algebra/97824#97824 Answer by Daniel Moskovich for On connection between Knot theory and Operator algebra Daniel Moskovich 2012-05-24T07:52:45Z 2012-05-24T08:03:07Z <p>The relationship between operator algebras and <em>braids</em> is fairly straightforward to explain, and is nicely written up in many places (<i>e.g.</i> in Kauffman's <a href="http://www.amazon.com/Knots-Physics-Louis-H-Kauffman/dp/9810203446" rel="nofollow">Knots and Physics</a>). Jones studied representations of the braid group $B_n$ into the <a href="http://en.wikipedia.org/wiki/Temperley%E2%80%93Lieb_algebra" rel="nofollow">Temperley-Lieb algebra</a> $TL_n$. The existence of such a representation is not so surprising (the following explanation is with hindsight- historically, this isn't how it happened): A Temperley-Lieb element is a <a href="http://en.wikipedia.org/wiki/Transfer_matrix" rel="nofollow">transfer matrix</a> in a <a href="http://en.wikipedia.org/wiki/Potts_model" rel="nofollow">Potts model</a>, in which each $e_i$ implements one more interaction, and you can think of a braid as a motion of $n$ distinct points in the lattice, with the crossings of points as an interaction, so it's roughly sort of like a universal model for this type of lattice statistic mechanical setup. Taking the trace of the representation gives the <a href="http://en.wikipedia.org/wiki/Partition_function_(statistical_mechanics)" rel="nofollow">partition function</a> for the model, so it's a natural thing for a statistical mechanic to do. Diagrammatically, taking the trace may be visualized as closing the braid.</p> <p>The surprise is that the trace is invariant under <a href="http://mathworld.wolfram.com/MarkovMoves.html" rel="nofollow">Markov moves</a>, and by <a href="http://mathworld.wolfram.com/MarkovsTheorem.html" rel="nofollow">Markov's theorem</a>, any two braids whose closure gives the same knot are related by a finite sequence of Markov moves. Thus, the trace of the representation actually ends up giving an invariant of a knot!!!</p> <p>Birman said in a talk that Jones verified invariance under Markov moves by chance- basically it was good luck, and nobody could have anticipated such invariance, or that the Jones polynomial would be a knot invariant. So there is some significant element of mystery in the question of what knots have to do with operator algebras. It ties in to the biggest question in quantum topology, which is a curious one: What do quantum invariants mean topologically? Why is any quantum invariant a topological invariant?</p> <p>The braid group is a group (tautology alert) which mimics interacting point particles, so you could imagine that it might be related to operator algebras (or to subfactors), but a knot isn't an element of a group, and the algebraic structure which you can equip the set of knots with is much more complicated and mysterious.</p> http://mathoverflow.net/questions/97788/on-connection-between-knot-theory-and-operator-algebra/117470#117470 Answer by Koushik for On connection between Knot theory and Operator algebra Koushik 2012-12-29T02:50:32Z 2012-12-29T02:50:32Z <p>you may looks into David Evans monumental book on quantum symmetries on operator algebras</p>