180˚ vs 360˚ Twists in String Diagrams for Ribbon Categories - MathOverflow

180˚ vs 360˚ Twists in String Diagrams for Ribbon Categories

2010-06-14T17:31:48Z

Ribbon categories are braided monoidal categories with a twist or balance, $\theta_B:B\to B$, which is a natural transformation from the identity functor to itself. In the string diagram calculus for ribbon categories, the ribbon is represented as a 360˚ twist in a ribbon (op. cit.). (See for example Street's Quantum Groups or Kassel's Quantum Groups for details.)

My questions are:

Is there work describing what happens if we consider a 180˚ twist?
If not, what goes wrong if we take this half twist one of the operations of interest on ribbons?
How are ribbons with a 180˚ twist axiomatised? What if loops are possible and we have twisted tangles? (I believe that Traced Monoidal Categories by Verity, Street and Joyal doesn't cover this case.)

Answer by Noah Snyder for 180˚ vs 360˚ Twists in String Diagrams for Ribbon Categories

2010-06-15T03:40:02Z

As Scott points out, Peter Tingley and I wrote a paper about this question. For the sake of concreteness (and because it included our main example) we only deal with the case of Hopf algebras whose representation theory has a ribbon half-twist, but the whole theory carries over to general monoidal categories. I'll sketch this generalization below, but you should probably read my paper with Peter first which is more accessible. The main result we use are formulas for the braiding given (independently) by Kirillov-Reshetikhin and Levendorskii-Soibleman which can be interpreted as given a formula for the half-twist. Certainly Reshetikhin (and presumably some of the other authors) were aware that these formulas could be interpreted in terms of half-twists, but it didn't explicitly appear until Peter and my paper.

One important warning that applies to everything though, in this theory the front and the back of the ribbon correspond to a priori different objects, so you still can't talk about Mobius bands.

Recall that a monoidal functor (these are often called "weak monoidal functors" in the quantum groups literature to distinguish them from strict monoidal functors, while they're called "strong monoidal functors" in the category theory literature to distinguish them from "lax monoidal functors) is a pair a functor F: C->D together with a binatural isomorphism $F(X\otimes Y) \rightarrow F(X) \otimes F(Y)$ which plays well with the associator.

Let's define a commutor to be a monoidal functor from C to C' (which will denote C with the opposite tensor product) whose underlying functor is the identity. The natural transformation is thus a map $X \otimes Y \rightarrow Y \otimes X$. The consistency condition says that there's a well-defined map $X \otimes Y \otimes Z \rightarrow Z \otimes Y \otimes X$. This definition of a commutor is the common generalization of braidings (which additionally satisfy the Yang-Baxter equation) and cactus commutors (which additionally square to 1). It's a natural condition that is satisfied by all known interesting "commutivity constraints."

There is a common way to produce commutors which comes up in a paper of Kamnitzer and Henriques (which is a very beautiful paper) which is closely related to half-twists.

First let's define a "dark side functor" to be any monoidal functor from C to C'. The name comes from the fact that this functor is what lets us talk about the "dark side" of the ribbon. If F is a dark side functor, then a half-twist for F is a natural transformation between F and the identity functor from C to C'.

Certainly you can use a half-twist for a dark side functor to produce a commutor, just compose the natural transformation with the dark side functor to get a commutor! If you work through what this tautological explanation means, you'll see that you get the commutor by first applying the half-twist to each object seperately, and then the inverse of the half-twist to the tensor product (or maybe visa-versa).

A ribbon half-twist is just a half-twist for a dark side functor whose resulting commutor is a braiding.

Answer by supercooldave for 180˚ vs 360˚ Twists in String Diagrams for Ribbon Categories

2010-06-23T17:14:16Z

I stumbled across Autonomous categories in which $A \cong A^∗$ by Peter Selinger. It states that the graphical representation of the self-duality $h_A:A\to A^*$ is represented by a half-twist of a ribbon. A coherence result is conjectured, but not proven (of course).

Answer by Peter Selinger for 180˚ vs 360˚ Twists in String Diagrams for Ribbon Categories

2010-06-24T12:58:09Z

The completeness result, which I conjectured in "Autonomous categories in which A is isomorphic to A*" (as cited by Dave above), has been proven last month. I talked about this at QPL 2010 in May, but it is not yet written. It is actually relatively easy to prove, although it took me over a month to realize that this is so. Essentially it is a reduction to the known result for ribbon categories. The absence of Moebius strips is one of the things that makes this possible.

What must be shown is: given two terms (in the half-twist language) that have the same diagram, then the terms can be proved equal by the axioms.

In a nutshell: first, it suffices to show this for terms that use the half-twist map only at object generators (half twists on A tensor B, on I, and on A* can be immediately reduced using the axioms). Now given two terms t and s that have the same diagram, there are two possibilities:

(1) each ribbon in the diagram has an even number of half-twists on it. In this case, they can all be moved next to each other and replaced by full twists, using the axioms. Then one can simply use coherence for ribbon categories to show that s and t are provably equal.

(2) some ribbon in the diagram has an odd number of half-twists on it. Since there are no Moebius strips, this can only happen if the ribbon has two ends, each of which is either connected to a box or to a source/sink of the diagram. W.l.o.g. assume one side of the ribbon is connected to an output of the box f in the diagram. Using the above trick, we can use the axioms to replace all but one of the half-twists by full twists, and to move the remaining half-twist adjacent to the box f. The key point is that this happens in both terms s and t. In both s and t, now replace this particular occurrence of the half-twist by a new variable H:A->A*. Note that this is no longer a half-twist graphically, but simply a new box. Call the modified terms s' and t'. Since both s' and t' have the H in the same place, s' and t' still have isomorphic diagrams. But they have one less ribbon with an odd number of twists, so the result follows by induction.

This proof is annoyingly simple, but it's correct.