There should be a category $3\text{CobTang}$ whose

  • objects are some kind of surfaces with a finite set of marked points
  • morphisms $M : S \to T$ are some kind of $3$-dimensional cobordisms $M$ between surfaces together with a choice of (framed) tangle in $M$ beginning at the marked points in $S$ and ending at the marked points in $T$.

Composition is given by gluing. Taking free $\mathbb{Z}[q, q^{-1}]$-modules on the morphisms in this category and quotienting by the skein relations defining the Kauffman bracket should give a category (the "Kauffman skein bracket category") $K$ in which

  • the Temperley-Lieb category,
  • skein algebras of surfaces, and
  • skein modules of $3$-manifolds

all sit as subcategories. The first bullet is the subcategory where $S, T$ are both, say, the open disc with some marked points and $M$ is a cylinder. The second bullet is the subcategory where $S, T$ are both a fixed surface with no marked points and $M$ is a cylinder. The third bullet is the subcategory where $S, T$ have no marked points.

There is also a forgetful functor $3\text{CobTang} \to 3\text{Cob}$ given by forgetting the marked points and the tangle, and this gives a span of categories $3\text{Cob} \leftarrow 3 \text{CobTang} \rightarrow K$.

(Where) does this construction appear in the literature?

It seems related to a TQFT except that we only get a span and not a functor. In addition, the Temperley-Lieb category quantizes the representation theory of $\text{SU}(2)$ while skein algebras quantize character varieties. So:

What does $K$ quantize?

  • 1
    $\begingroup$ If you're interested in doing explicit calculations here, Kauffman-Lins's book is invaluable. $\endgroup$ Commented Oct 9, 2012 at 15:08

1 Answer 1


I'm not sure where in the literature this exact construction occurs, but it is certainly well-known. I think some skein-module-centric papers consider non-empty boundary conditions on the skein modules (which would be equivalent to what you describe), so you might try looking for that. You could also have a look at the (incomplete, a little out of date) "TQFTs" notes on my web page.

There is a standard way to view this as a functor $\mathrm{3Cob}\to K$ instead of a span. Assign to a surface $Y$ the category $A(Y)$ whose objects are finite collections of framed points in $Y$ and whose morphisms are (finite linear combinations of) framed tangles in $Y\times I$ modulo Kauffman bracket relations. Then given a 3-dimensional bordism between surfaces $Y_1$ and $Y_2$ we can construct an $A(Y_1)$-$A(Y_2)$ bimodule. These assignments give a functor from $\mathrm{3Cob}$ to the category whose objects are linear categories and whose morphisms are bimodules.

The above construction can be extended "all the way down", assigning 2-categories to 1-manifolds and 3-categories to 0-manifolds, with gluing of 1- or 2-manifolds corresponding to the appropriate categorified tensor product over a 3- or 2-category.

To extend the construction "all the way up", assigning an intertwiner of bimodules to a 4-dimensional bordism, one needs to assume that the deformation parameter $q$ is a root of unity and take the usual quotient. The resulting fully extended 3+1-dimensional TQFT is a disguised version of the Witten-Reshetikhin-Turaev TQFT based on the Kauffman bracket at that root of unity.

Finally, I'll make a small but important quibble with your statement that the Kauffman bracket quantizes the representation theory of $SU(2)$. As it's name/notation suggests, $Rep(U_q(SU(2)))$ quantizes the representation theory of $SU(2)$. The Kauffman bracket differs from $Rep(U_q(SU(2)))$ by a sprinkling of $\pm 1$'s. By tweaking the signs like this, one gains the convenience of a trivial Frobenius-Schur indicator. (This means that one can work with unoriented tangles.) But one loses some positivity properties. In the short run the Kauffman bracket is more convenient, but in the long run $Rep(U_q(SU(2)))$ is the nicer object. For example, $Rep(U_q(SU(2)))$ can be categorified (Khovanov homology), while the Kauffman bracket can only be categorified up to ambiguous $\pm$ signs.


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.