Could someone give an laymans explanation of the relation between QFT and knot theory? What are the central ideas in Wittens work on the Jones polynomial?


Solution 1:

I am a topologist who uses gauge theory. This answer reflects my taste and background. I make no claims of historical accuracy. It would be nice to see answers from people with other perspectives.

In the 80s, Donaldson used Yang-Mills theory to great effect in studying the topology of smooth 4-manifolds. In it, one (in some sense) counts solutions to the PDE $F_A^+ = 0$ defined on the space of connections on a smooth manifold $M$. I will call solutions to this instantons and the equation the instanton equation. Later, Seiberg and Witten introduced the (related) Seiberg-Witten equations, which led to much less technical proofs of many of the same theorems. Solutions to these equations are called monopoles, and the invariants you get from counting them are called the Seiberg-Witten invariants.

In studying the Thom conjecture and more general adjunction inequalities, Kronheimer and Mrowka were led to the idea of studying monopoles that were singular along an embedded surface $\Sigma$. This was an extraordinarily successful avenue of research.

Orthogonally, after Floer's original invention of Floer homology groups, it was realized that the instanton and monopole invariants fit into the tidy formalism of a TQFT (well, mostly). By using the technique of dimensional reduction, and following the idaes of Floer's construction, every connected orientable 3-manifold $Y$ gets homology groups $HF(Y)$ (different depending on whether you're using instanton or monopole invariants; there are other related constructions), and for every cobordism $W: Y \to Y'$ we get a map $HF(Y) \to HF(Y')$ satisfying certain criteria. Equivalently from the perspective of TQFT, every 4-manifold $W$ with boundary $Y$ has an (instanton or monopole) invariant living in $HF(Y)$; one obtains the original invariants by deleting a 4-ball and looking at the invariant in $HF(S^3) \cong \Bbb Z$. There are numerous technical difficulties I have either avoided or lied about in this paragraph.

One begins to think: well, that idea of studying monopoles singular along a surface seemed pretty good. Why don't I take the dimensional reduction of that idea? Do so, and you obtain $HF(Y,K)$, the "knot Floer homology" of a knot $K \subset Y$. This is functorial with respect to cobordisms $(W,\Sigma): (Y,K) \to (Y',K')$ and gets you a sort of knot TQFT.

I'm a little uncomfortable saying this with complete certainty, but in the monopole case, the (graded) Euler characteristic of $HF(S^3,K)$ should be the Alexander polynomial of $K$. (I don't remember what it is for the instanton knot Floer homology; I would not guess it's the Alexander polynomial) This is a well-known knot invariant. One might ask: well, what about the Jones polynomial? Does that come as the Euler characteristic of some group? Well, it does, but not one that's obviously defined via gauge theory: it's the Euler characteristic of Khovanov homology. Now, separately, in a story I don't know very well, Witten gave a construction of the Jones polynomial in a way defined by Chern-Simons theory; my understanding is that it was defined by studying solutions to an equation that are singular along a knot (but I really have not looked at this much at all).

This suggests that one should expect there to be a gauge-theoretically defined version of Khovanov homology, which Witten described here. This is not yet known to work mathematically, but to my understanding it should be some sort of "$SL_2(\Bbb C)$-Floer homology" as opposed to the $SU(2)$ or $SO(3)$ or $S^1$-based ones above. (Note that $SL_2(\Bbb C)$ is not compact. This leads to an extraordinary amount of technical trouble - which is why this Floer homology doesn't really exist yet.) Then the Jones polynomial for knots in arbitrary manifolds should be the Euler characteristic of this thing. This is an active area of research, and when completed, should be the natural place where Witten's work lives.

Solution 2:

This will not be by any means a "layman's" explanation, but here's something short that might be helpful. There also won't be much physics here, if that's what you're looking for.

Let's take for granted, following Witten, that if $G$ is a simply connected compact Lie group (for simplicity) and $k$ is an integer, then there is a 3d topological field theory called "Chern-Simons theory with gauge group $G$ and level $k$." Chern-Simons theory is so named because its heuristic physical description involves path integrals where the action has something to do with the Chern-Simons 3-form. I won't get into all of the structure that this TFT has; it's enough for now that it assigns (up to some subtleties I don't want to get into)

  • a finite-dimensional complex vector space $Z(\Sigma)$ to every closed surface $\Sigma$, and
  • a linear map $Z(\Sigma_1) \to Z(\Sigma_2)$ to every 3d cobordism $M$ between $\Sigma_1$ and $\Sigma_2$

and these assignments satisfy various compatibilities. Also, you'll need to know one more thing: the vector space $Z(T^2)$ assigned to the torus has a natural basis that can be identified with either the irreducible representations of the corresponding loop group ("at level $k$") or with the irreducible representations of a version of the corresponding quantum group (for some root of unity $q$ depending on $k$). In physics language these are "Wilson loop operators."

How do you get knot invariants out of this? Given a knot $K$ sitting inside $S^3$, consider a tubular neighborhood around it. The complement of this tubular neighborhood describes a cobordism from the torus $T^2$ to the empty manifold, and hence hitting it with Chern-Simons gives a linear map

$$Z(T^2) \to \mathbb{C}.$$

Now evaluating this linear map at an irreducible representation of, according to taste, either the loop group or the quantum group produces a knot invariant. When $G = SU(2)$ and the representation is the standard one, you get the Jones polynomial evaluated at some root of unity $q$ depending on $k$.