What exactly is the connection between knots and operator algebra? I heard that Jones established such a connection while discovering the celebrated Jones Polynomial.

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?

As is probably very evident, I am a complete newbie to this area.

  • 11
    $\begingroup$ I am never sure how much mathematics can be explained "in plain English" without resorting to, well, lying. What exactly do you know about (i) braids (ii) Hilbert space (iii) von Neumann algebras? $\endgroup$
    – Yemon Choi
    May 23, 2012 at 21:10
  • $\begingroup$ Yemon, I agree with you. In fact, I probably used it a bit loosely here. I know decently about Hilbert Spaces and Von Neumann Algebras; just started learning braids. Let me amend my words - instead of 'plain english', let me ask - what are the concepts I need to learn to understand this connection? $\endgroup$
    – Arnab
    May 23, 2012 at 21:49
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    $\begingroup$ Also, I love how so many things are "celebrated" in mathematics. Every time I hear that, I picture a quiet, dignified celebration, maybe over some tea and cookies. $\endgroup$
    – MTS
    May 24, 2012 at 0:58
  • $\begingroup$ Yes, 'celebration' has a different meaning in mathematics, I suppose. Thanks very much for sketching the path to the connection. I read up a bit on Temperley-Lieb and it is slowly emerging for me. Daniel, thanks a lot for pointing out the invariance under Markkov moves. No surprise, but cannot be overstated that abstract math often becomes alive while representing physical systems. $\endgroup$
    – Arnab
    May 24, 2012 at 17:52

3 Answers 3


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 $\tau : M \to \mathbb{C}$ (i.e. a positive normal state such that $\tau(a^*a)=0$ implies $a=0$) and having no minimal projections (this excludes matrix algebras).

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.

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.

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 $N$.

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 $N \subseteq M$.

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:

  1. $e_i e_j = e_j e_i$ if $|i-j| > 1$.
  2. $e_i e_j e_i = \lambda e_i$ whenever $|i-j| = 1$, where $\lambda$ is the inverse of the index of $N$ in $M$.

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.

That is the gist of it. Read Jones' paper for more details.


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 with hindsight- historically, this isn't how it happened): A Temperley-Lieb element is a transfer matrix in a Potts model, 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 partition function 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.

The surprise is that the trace is invariant under Markov moves, and by Markov's theorem, 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!!!

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?

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.

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    $\begingroup$ Daniel, that is a very good point you make about the invariance under the Markov moves. In fact the first Markov move is clear - certainly the trace will be invariant under conjugation. But the fact that it is invariant under the second Markov move, that's the miracle. $\endgroup$
    – MTS
    May 24, 2012 at 16:03

you may looks into David Evans monumental book on quantum symmetries on operator algebras


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