Does one of the hexagon identities imply the other one? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:23:39Z http://mathoverflow.net/feeds/question/51900 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51900/does-one-of-the-hexagon-identities-imply-the-other-one Does one of the hexagon identities imply the other one? Dmitri Pavlov 2011-01-13T02:22:57Z 2011-01-13T06:19:56Z <p>Suppose we have a monoidal category equipped with additional data that almost makes it a braided monoidal category except that only one of the hexagon identities is satisfied. Can we then prove the other hexagon identity? If not, is there an explicit counterexample and can we prove the other identity under the additional condition that the braiding is symmetric?</p> http://mathoverflow.net/questions/51900/does-one-of-the-hexagon-identities-imply-the-other-one/51904#51904 Answer by David Roberts for Does one of the hexagon identities imply the other one? David Roberts 2011-01-13T03:26:17Z 2011-01-13T03:26:17Z <p>In the presence of a symmetric braiding on a monoidal category, one of the hexagon identities does imply the other (it doesn't matter which one you start from). I can offer no other reference other than the <a href="http://ncatlab.org/nlab/show/symmetric+monoidal+category" rel="nofollow">nLab page on symmetric monoidal categories</a>. But I would imagine that this result goes back to early work on coherence (Kelly or MacLane, maybe). As far as the two hexagon identities, I would guess that you need both. These have interpretations as string diagrams (see <a href="http://math.ucr.edu/home/baez/rosetta.pdf" rel="nofollow">Baez-Stay</a> definition 11/page 21 for pictures) and I think it very unlikely that you can recover one from the other without extra structure/operations on your 'prebraided' monoidal category. It would be worth checking the original Joyal-Street report <a href="http://rutherglen.ics.mq.edu.au/~street/JS86.pdf" rel="nofollow">Braided monoidal categories</a> to see if they give a counterexample.</p> http://mathoverflow.net/questions/51900/does-one-of-the-hexagon-identities-imply-the-other-one/51908#51908 Answer by Theo Johnson-Freyd for Does one of the hexagon identities imply the other one? Theo Johnson-Freyd 2011-01-13T04:04:36Z 2011-01-13T04:04:36Z <p>Recall first that the free braided monoidal category on one generator is the category of braids. Recall also that to every braid you can associate a "crossing number": every overcrossing counts as +1, and every undercrossing as -1. Given a braid $B$, let me write $\chi(B)$ for this crossing number, since I spell is $\chi$rossing. Anyway, there's another braided category where I multiply every braid $B$ by $q^{\chi(B)}$, for $q \neq 0$ some "variable". What I mean by this is that I'm taking the same monoidal category, but a different braiding, where the basic crossing is rescaled by $q$. Then the crossing of $n$ strands over $m$ strands is rescaled by $q^{nm}$. I guess implicitly I'm taking my category over $k[q,q^{-1}]$, for $k$ a field. An undercrossing $n$ under $m$ is rescaled by $q^{-nm}$.</p> <p>Now here's what I'd like to do. I'd like to build a "prebraided category", which is the same monoidal category, but where the braiding of $n$ strands over $m$ strands is rescaled by $q^n$ (except when $m=0$, in which case don't rescale). I think you'll find that this satisfies one of the hexagons but not the other.</p> http://mathoverflow.net/questions/51900/does-one-of-the-hexagon-identities-imply-the-other-one/51923#51923 Answer by Victor Ostrik for Does one of the hexagon identities imply the other one? Victor Ostrik 2011-01-13T06:19:56Z 2011-01-13T06:19:56Z <p>Consider the category of $A-$graded vector spaces (here $A$ is an abelian group) with obvious tensor product and trivial associator. Then each isomorphism $a\otimes b\to b\otimes a$ can be specified as a nonzero complex number $B(a,b)$. Now the first hexagon axiom says that the function $B(a,b)$ is linear in the first variable and the second axiom says that $B(a,b)$ is linear in the second variable. Clearly these conditions are independent. The symmetric braiding would correspond to skew-symmetric function $B(a,b)$ and linearity in one variable would imply bilinearity.</p>