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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.)

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For the case of quantum groups, see arxiv.org/abs/0810.0084 –  S. Carnahan Jun 14 '10 at 17:42
    
@Scott: I should (re)read Noah and Peter's paper. Do you know if anyone's worked out a "Tannaka Krein theorem" form half-twist ribbon categories? –  Theo Johnson-Freyd Jun 14 '10 at 22:39
    
@Theo: I have no idea. You should ask an expert (and/or peruse Google Scholar). –  S. Carnahan Jun 15 '10 at 1:03
    
If you take a strip of paper, put a half twist in it and glue the ends together you get a möbius band. Do the same thing with a full twist and you get an annulus. It probably means that there will be a forking in resolutions that will keep you from having truly local skein relations for invariants derived from such a theory. –  Charlie Frohman Jun 15 '10 at 2:29
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As far as I know Peter and my paper is the only paper that really deals with any questions about half-ribbon categories. In particular, we only define half-ribbon Hopf algebras, not half-ribbon categories, so it's very unlikely that there's any Tannaka-Krein theorem in the literature. –  Noah Snyder Jun 15 '10 at 3:41

3 Answers 3

up vote 6 down vote accepted

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.

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If I understand correctly, the absence of Möbius strips follows from the fact that the objects corresponding to (using terminology above) the light and dark sides of the ribbon are different, so the light and dark strips cannot be plugged together. Do you have a reference for this result? –  supercooldave Jun 24 '10 at 13:38
    
Much of this argument is written up in Section 4 of my paper with Noah (cited above). See especially Prop 4.15 which argues that there is a unique functor incorporating the half twist. This argument actually requires almost no assumptions on the half twist (only that it is invertible), but does not imply all the properties one would want (for instance, it does not imply all caps/cups are what you expect). These extra properties are checked later with additional assumptions on t. We work with the quantum group case, but I think the argument is unchanged for more general ribbon categories. –  Peter Tingley Jun 24 '10 at 15:35
    
@dave: I don't have a reference; it seems to be just a consequence of soundness. By induction, the diagram corresponding to every term has the property that no "light" and "dark" sides are plugged together. @Peter: great, then I'll have another paper to cite! I agree that it is essentially a similar proof. The setting is slightly different, e.g. you don't identify the two duals of an object (the dual from the rigid structure and the light/dark dual). This actually simplifies the proof a bit. But yes, it's a very similar idea. –  Peter Selinger Jun 24 '10 at 18:45
    
@Peter I'm actually very interested to see the case where the dark side functor is duality worked out in detail. Among the things that confuse me about that case is that, the way Noah and I set things up, the dark side functor and the duality functor don't actually commute, so cannot be equal. See our Comment 4.12. They do commmute up to isomorphism, and one can set things up so that they commute exactly. But then other things get worse. In particular, our proposition 4.18 becomes messy (i.e. the half-twist on a tensor product is hard to describe). Do you see such issues in your work? –  Peter Tingley Jun 25 '10 at 15:19

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.

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@Noah. Thanks. I'll get to work on the two papers you recommend. –  supercooldave Jun 15 '10 at 14:23
    
Noah, you went to the trouble of clarifying the terminology for "monoidal functor", but then you didn't specify that the transformation $F(X\otimes Y)\to F(X) \otimes F(Y)$ should be invertible! That's exactly what distinguishes strong monoidal from (co)lax monoidal in the category theory literature, at least. –  Mike Shulman Jul 10 '11 at 7:14
    
Good point. Fixed. –  Noah Snyder Jul 10 '11 at 15:15

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).

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Thanks, I was not aware of this work. In Noah's language, perhaps the most obvious guess for a "dark side functor" would be duality. A half twist should then be a natural transformation from the identity functor $C \rightarrow C'$ and the duality functor, and it seems Selinger is exactly writing down the conditions to make this work. –  Peter Tingley Jun 23 '10 at 19:30
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I actually once started working out this story in the case of representations of quantum groups. There $V$ need not necessarily isomorphic to $V^*$, but one can define natural inner products on each irreducible $V_\lambda$. Equivalently, one has a chosen system of isomorphism of vector spaces $V_\lambda \rightarrow V_\lambda^*$ (at least up to scaling). These behave quite well with respect to the representation structure, but are not morphisms). I wanted to use this as the half-twist...but never worked out the details. I've very glad to see that other people are thinking these things through! –  Peter Tingley Jun 23 '10 at 19:38

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