12
$\begingroup$

In "Topological Gauge Theories and Group Cohomology", Dijkgraaf and Witten construct a 3-manifold invariant from a finite group $G$ and 3-cocycle $\omega$. I would think there is also an associated knot invariant, coming in particular from the twisted quantum double $D^\omega(G)$, but haven't been able to find a place where it is explicitly described or computed.

If it exists, could someone point me to the right reference? I would be interested in both a general construction and a specific example, for instance computing it in the case of $\mathbb{Z}_n$.

$\endgroup$

2 Answers 2

11
$\begingroup$

Whenever you have a 2+1-dimensional TQFT you also have a link invariant by taking the path integral of the complement of the link.

More specifically, let $L\subset S^3$ be a framed, oriented link and let $N$ be the complement of a tubular neighborhood of $L$. The path integral $Z(N)$ lives in the vector space $Z(\def\bd{\partial}\bd N)$ associated to $\bd N$, which is a disjoint union of tori. The orientation and framing of $L$ determine a diffeomorphism of each torus in $\bd N$ to the standard torus. The standard torus has an orthonormal basis indexed by irreducible representations of the semisimple category associated to a circle by the TQFT. Labeling each component of the link by an irrep, we get a numerical invariant of oriented, framed, labeled links in $S^3$ (or in any 3-manifold, for that matter).

In the specific case of DW theory, the category of the circle has objects {maps of the circle into BG} and morphisms linear combinations of maps of $S^1\times I$ into BG, twisted by the cocycle $\omega$. In the case where $\omega$ is zero, the irreps of this category are indexed by pairs $(x, \rho)$, where $x\in G$ runs through representatives of the conjugacy classes of $G$, and $\rho$ runs through irreps of the centralizer of $x$.


ADDED LATER:

Here's a sketch of an argument that for $G$ abelian, and for any twisting cocycle, the link invariant depends only on the linking matrix of the link.

Step 1. The irreducible representations of the annular category mentioned above are all 1-dimensional. (This is the category whose morphisms are maps of an annulus into BG.)

Step 2. If $\alpha$ and $\beta$ are two such irreps, then $\alpha\otimes\beta$ is again a 1-dimensional irrep (as opposed to a direct sum of several irreps). Here the tensor product is induced by a pair of pants.

Step 3. It follows that the Hilbert space of a 4-punctured sphere, with punctures labeled by $\alpha, \beta, \beta^*, \alpha^*$, is 1-dimensional. (cf. Witten's famous argument that in the case of $SU(2)$ Chern-Simons theory the analogous Hilbert space is 2-dimensional.)

Step 4. It follows that there is a skein relation saying that the path integral of a overcrossing is a scalar multiple of the path integral of an undercrossing. (Again cf. Witten in the SU(2) case, where one gets a three term skein relation.) The over- and undercrossings live in a 3-ball whose boundary is the sphere of Step 3.

Step 5. Using the skein relation of Step 4, it follows that the link invariant depends only on the linking matrix and the labels of the components of the link.

$\endgroup$
5
  • $\begingroup$ @Kevin In Chern-Simons theory Witten suggested invariants of links as "Wilson loops" (i.e. expectation values of traces of monodromies of connections) is it equivalent to " taking the path integral of the complement of the link" ? $\endgroup$ Jan 25, 2012 at 19:04
  • $\begingroup$ Yes, it's equivalent. Think of the solid torus $D^2\times S^1$ with a Wilson loop at $\{0\}\times S^1$. The path integral (with Wilson loop) of this 3-manifold determines a vector in the Hilbert space of a torus, and this vector is the same as the one I refer to above. $\endgroup$ Jan 25, 2012 at 19:18
  • 3
    $\begingroup$ @Kevin: A key word in Iva's question is "explicit". $\mathbb Z_m$ we all know, it's $H^3$ is also known, so someone must have figured out explicitly what are the corresponding DW invariants. Are they powerful or silly? Are they evaluations of the Alexander polynomial? If you draw a random 10 crossings knot, can you compute its DW invariants by staring at it and marking this strand with $A$ and the other one with $B^{-1}$ and multiplying everything in some this or that simple-minded algebra? I.e., is there a description of the answer that is as simple-minded and as explicit as the Kauffman br $\endgroup$ Jan 25, 2012 at 19:32
  • $\begingroup$ @Dror: I guess I misunderstood the emphasis in the question. I was just pointing out that there is, indeed, a well-defined invariant of labeled links, for very general reasons. I'm not aware of explicit computations in the literature, especially in the case of a non-trivial twisting. For untwisted $\mathbb{Z}/m$, it will boil down to a homology computation (with $\mathbb{Z}/m$ coefficients), so I doubt there's anything interesting in that case. $\endgroup$ Jan 25, 2012 at 19:57
  • $\begingroup$ @Kevin: Thank you for the answer, it gives a nice overview of the general mechanism. I am in fact looking for a specific formula for the knot invariants in the DW setting, as Dror points out, that would allow me to evaluate them on given knots. By "general" I meant for any $G$ and $\omega$. Sorry if that wasn't very clear in the question. $\endgroup$ Jan 25, 2012 at 22:43
5
$\begingroup$

You may be interested in this paper by Chen.

One of the simplest nontrivial knot invariants is the number of "Fox n-colorings", which count homomorphisms of the knot group to a dihedral group $D_{2n}$ sending the meridians to reflections. This count can be interpreted as a value in a Dijkgraaf-Witten theory on the knot complement. Chen's paper computes those invariants in the nontwisted case in terms of the representation category of the quantum double $D(D_{2n})$ and gives a nice formula for the number of colorings of a rational link.

I tried computing values of the twisted invariant, but it turns out that these are equal to the untwisted invariants times a power of $i$ depending on the writhe.

It may be possible that something more intersting happens with a more complicated group: $D_{2n} = \mathbb{Z}_2 \ltimes \mathbb{Z}_n$ is the simplest possible case where you get interesting homomorphism counts, and I'd be curious what happens in the twisted (and untwisted) case for other semidirect products. In particular, the twisting that actually affects the DW invariants seems to mostly involve the $\mathbb{Z}_2$ part, which has few interesting cocycles.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.