Extending braidings to tensor powers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:14:50Z http://mathoverflow.net/feeds/question/85753 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85753/extending-braidings-to-tensor-powers Extending braidings to tensor powers Ago Szekeres 2012-01-15T18:03:10Z 2012-01-15T19:29:19Z <p>Given a braiding $\Psi: X \otimes Y \to Y \otimes X$ for two objects $X,Y$ in a monoidal category, it seems reasonable to assume that $\Psi$ extends uniquely to a braiding $X^k \otimes Y^l \to Y^l \otimes X^k$, for all $k,l \in {\mathbb N}$. The proof would surely be based upon the Yang--Baxter property of $\Psi$ and the fact that one can express any permutation as a product of transpositions. However, I can't seem to write it down exactly.</p> http://mathoverflow.net/questions/85753/extending-braidings-to-tensor-powers/85757#85757 Answer by MTS for Extending braidings to tensor powers MTS 2012-01-15T19:29:19Z 2012-01-15T19:29:19Z <p>$\newcommand{\id}{\mathrm{id}} \newcommand{\ot}{\otimes}$ Your assumption is correct. $\Psi$ does extend uniquely to the map that you want. By drawing string diagrams and playing with them, you can intuitively see what to do, namely: every time you see an $X$ strand to the left of a $Y$ strand, use $\Psi$ to braid $X$ over $Y$. However, there may be many ways to do this. For example if $k=l=2$, you can do $$(\id_Y \ot \Psi \ot \id_X) \circ (\id_Y \ot \id_X \ot \Psi) \circ (\Psi \ot \id_X \ot \id_Y) \circ (\id_X \ot \Psi \ot \id_Y),$$ or you can do $$(\id_Y \ot \Psi \ot \id_X) \circ (\Psi \ot \id_Y \ot \id_X) \circ (\id_X \ot \id_Y \ot \Psi) \circ (\id_X \ot \Psi \ot \id_Y),$$ and the braid relations show that those are the same map. Obviously this is easier to see if you draw a picture.</p> <p>The challenge is how to efficiently prove that, given any $k$ and $l$ and any possible choice of way you build your braiding, that you get the same map. This is the sort of thing that is known as a "coherence theorem." These are often stated in the fashion: Every diagram in (some category) commutes, where the category is a sort of "free category" whose morphisms are the structural ones that you are talking about.</p> <p>There are coherence theorems for the associativity morphisms in a monoidal category (see MacLane's book, Chapter VII, Section 2) which is what allows you to drop parentheses when doing iterated tensor products. And similarly there is a coherence theorem for morphisms built from the braidings in a braided monoidal category. This is spelled out in detail in Joyal and Street's 1993 article <em>Braided Tensor Categories.</em></p>