Ribbon categories are braided monoidal categories with a [twist](http://nlab.mathforge.org/nlab/show/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.)



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


