I am asking because the literature seems to contain some inconsistencies as to the definition of a braided monoidal category, and I'd like to get it straight. According Chari and Pressley's book ``A guide to quantum groups," a braided monoidal category is a monoidal category $\mathcal{C}$ along with a natural system of isomorphisms $\sigma_{U,V}: U \otimes V \rightarrow V \otimes U$ for all pairs of objects $U$ and $V$, such that

(i) The ``Hexagon" axioms (two commutative diagrams) hold.

(ii) The ``identity object" axioms: $\rho_V= \lambda_V \circ \sigma_{{\bf 1},V}: {\bf 1} \otimes V \rightarrow V$ and $\lambda_V= \rho_V \circ \sigma_{V, {\bf 1}}: {V} \otimes {\bf 1} \rightarrow V$, where $\lambda_V$ and $\rho_V$ are the isomorphisms of $V \otimes {\bf 1}$ and ${\bf 1} \otimes V$ with $V$ that are part of the definition of monoidal category. See Chari-Pressley Definitions 5.2.1 and 5.2.4. They use the term "quasitensor category," but note on p153 that the term "braided monoidal category" is equivalent.

However, in some references (ii) seems to have been dropped. I am thinking in particular of Definition 3.1 is this expository paper, and the wikipedia article. The wikipedia article goes further, and suggests that (ii) somehow follows from (i) and the axioms of a monoidal category. So, my questions are.

1) Is (ii) needed? That is if we do not impose (ii), does it follow from (i) and the axioms of a monoidal category?

2) If (ii) is needed, can someone provide an example demonstrating why? That is, provide an example of a monoidal category $\mathcal{C}$ along with maps $\sigma_{U,V}$ such that (i) holds but (ii) fails. Alternatively, if (ii) is not needed, I'd like a proof (or reference to a proof) that it follows from other axioms.

  • $\begingroup$ It's a good question, and I don't know the answer. However, here's a sociological observation: often people are sloppy about identities. For example, people sometimes speak as if the non-strictness of associativity were the only non-strict aspect of bicategories or monoidal categories, forgetting the non-strictness of the identity axioms. This may be related to a general vagueness in certain parts of topology and algebra about whether associative algebras are assumed to have a unit. So: on those grounds, I wouldn't be surprised if (ii) were needed but nevertheless ignored by many authors. $\endgroup$ – Tom Leinster Dec 18 '09 at 20:20
  • $\begingroup$ My impression is also that (ii) should be needed, and people have been sloppy. But I have been unable to construct a counter example, so I'm not sure. It would be nice to get this straight. $\endgroup$ – Peter Tingley Dec 18 '09 at 21:42

This is Proposition 1 in the seminal paper "Braided Monoidal Categories" by Joyal and Street. Relation (ii) is implied by the others.

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    $\begingroup$ I think the paper is "Braided Tensor Categories", although they use the words interchangeably despite current convention. Also, it is Proposition 2.1. In proposition 1.1 they prove a related claim about certain unit axioms being redundant in the definition of a monoidal category. Here is the link: sciencedirect.com.libproxy.mit.edu/… $\endgroup$ – David Jordan Dec 19 '09 at 1:28
  • $\begingroup$ Thanks for bringing this up, as it is confusing. There are several slightly different versions of this paper in circulation. The copy I have is definitely called "Braided Monoidal Categories" and it is also the first proposition in that paper that is relevant here. I know that the copy I have is NOT the published one, because the published one, I am told, omits the very useful section on classifying certain braided monoidal categories in terms of group cohomology. $\endgroup$ – Chris Schommer-Pries Dec 19 '09 at 2:57
  • $\begingroup$ Yes, this is a very good reference. Thanks! $\endgroup$ – Peter Tingley Dec 19 '09 at 14:43

There is a proof that ii) follows from the other axioms in Kassel's book (proposition X111.1.2 p.316).

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    $\begingroup$ I actually prefer the presentation of the proof in this reference. But I can only give one correct answer, and you can't beat ``the statement is in the seminal paper on the subject"! $\endgroup$ – Peter Tingley Dec 19 '09 at 14:44

I am not sure what is meant by the "hexagon identity." Is it an expression of the Yang-Baxter relation? If so, this follows from naturality and (ii) if I am understanding (ii) correctly.

I think that Turaev's book gives a precise and accurate definition. See also Baez and Langford's paper on braided moniodal 2-categories --- you have to understand the one to generalize to the other.

  • $\begingroup$ Yes, the hexagon identity" is the expression of the Yang-Baxter relation. I'm not quite sure I understand your answer though. It is certainly not true that (i) follows from (ii), since I know examples where (ii) holds but (i) does not, such as the coboundary categories" studied in e.g. Henriques and Kamnitzer's paper "crystals and coboundary categories". What I was asking is if (ii) follows from (i). I will take a look at the references you suggest. $\endgroup$ – Peter Tingley Dec 18 '09 at 23:56
  • $\begingroup$ (ii) and naturality give the YBE. You can see this by playing with a ribbon with a 1/2 twist and an arc over-crossing it. $\endgroup$ – Scott Carter Dec 19 '09 at 0:48
  • $\begingroup$ To clarify my previous comment, (i) is what is sometime called ``quasi-triangularity". Ignoring associators, it says two maps $A \otimes B \otimes C \rightarrow C \otimes A \otimes B$ are equal (and one other identity). The Yang-Baxter condition is an identity of maps $A \otimes B \otimes C \rightarrow C \otimes B \otimes A$, which follows from quasi-triangularity and naturality. So saying that the Hexagon axiom is the Yang-Baxter condition is not quite accurate. $\endgroup$ – Peter Tingley Dec 19 '09 at 14:41
  • $\begingroup$ The condition (ii) above is different. Notice that it only involves one object other than the identity. There are examples of monoidal categories with commutativity constraints that do not satisfy (i) but not (ii), like the coboundary categories I mentioned earlier. $\endgroup$ – Peter Tingley Dec 19 '09 at 19:10

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