\subsection{Braidings} \label{sec:braidings}

\subsubsection{Braidings for monoidal categories}
If you already know this stuff, skip it!

For a monoidal category, a braiding is a family of natural isomorphisms \[c_{X,Y}: X \otimes Y \rightarrow Y\otimes X\] satisfying the relations of braids.  For example, one requires that $$c_{X\otimes Y, Z} = (c_{X, Y} \otimes \Id) \circ (\Id \otimes c_{Y, Z})$$ so that the following equation of diagrams holds.

\[\mathfig{.5}{quasitriangular}\]


\subsubsection{Braidings for bi-oidal categories}

What is a braiding for a bi-oidal category?  Let's consider a typical two bimodules inside $\bimod{}{}$, say $X \in \bimod{N}{M}$ and $Y \in \bimod{M}{M}$.  Notice that  $X \tensorover{M} Y$ makes perfect sense, but there is no way to take the tensor product in the other order!  So a braiding can't just be a map $X \otimes Y \rightarrow Y \otimes X$.  So what are we meant to do?

Rather than guessing a new definition, we will return to the motivating example: the Jones polynomial.  There is a braiding on the bi-oidal category $\TL_d$ (see \ref{sec:temperley-lieb}) given by the usual Kauffman bracket formula.

\[\mathfig{.5}{kauffmanbracket}\]

This braiding is a map $$c_{M, \bar{M}}:\bimodule{N}{M}{M} \tensorover{M} \bimodule{M}{\overline{M}}{N} \rightarrow \bimodule{N}M{M} \tensorover{M} \bimodule{M}{\overline{M}}{N}.$$

What about using the Kauffman bracket to compute the braiding of more complicated bimodules?  For example where does $c_{M\otimes \bar{M}, M}$ live?  To decide this question we draw the diagram below and read off what it means.

$$\mathfig{.5}{biggerbraiding}$$

We read off that the braiding is a map $$c_{M\otimes \bar{M}, M}: \left(\bimodule{N}{M}{M} \tensorover{M} \bimodule{M}{\bar{M}}{N}\right) \tensorover{N} \bimodule{N}{M}{M} \rightarrow \bimodule{N}{M}{M} \tensorover{M} \left(\bimodule{M}{\bar{M}}{N} \tensorover{n} \bimodule{N}{\bar{M}}{M}\right).$$

In general, we see that if we braid some object across an odd object, then we need to reverse all the shadings.  If we braid some object across an even object, then we leave the shadings the same.  Since the braiding is natural, we will also need a way to reverse the shadings on morphisms.  For Temperley-Lieb, it is clear what shading reversing does.  But for a more general planar algebra a shading-reversing is an additional piece of data.

Let's think about how shading-reversing might work for $\dfour$.  We need to know how to shading-reverse the $2$-box $R$, the obvious picture doesn't make any sense.

PICTURE

Since we know how things in Temperley-Lieb shading-reverse, we know that the shading-reverse of $R$ must satisfy the defining relations of $R$.  Hence the shading-reverse of $R$ must be one of the following two options.

PICTURE

Since either possibility seems reasonable, we will allow ourselves the flexibility to use either shading change.  In the end this will be necessary in order to make the following to pictures work.

PICTURE

\begin{defn}
A shading reversing for a bi-oidal category is two involutions ${}^\star$ and $_{\star}$ (read "over" and "under") which reverse all the shadings.  We also require that $\Id^\star = \Id = \Id_\star$, that $( \otimes )_{\star} = {}_{\star} \otimes {}_{\star}$, and that $( \otimes )^{\star} = {}^{\star} \otimes {}^{\star}$.
\end{defn}

When something passes under something odd you apply the "under" involution to it, when something passes over something odd you apply the "over" involution to it.  We use notation like $X^{\star \deg Y}$ to mean "apply the over involution to $X$ if the degree of $Y$ is odd, but leave $X$ alone if the degree of $Y$ is even."

\begin{defn}
A braiding for a bi-oidal category is a family of isomorphisms $c_{X,Y}: X \otimes Y \rightarrow Y_{\star \deg X} \otimes X^{\star \deg Y}$ such that 
\begin{enumerate}
\item $c$ is natural:
	\begin{itemize}
	\item $c_{X,\Id} = \Id_X =  c_{\Id, X}$.
	\item For any $f \in \Hom{}{X}{Z}$, we have that $c_{Z,Y}\circ (f\otimes \Id) = (\Id \otimes f^{\star \deg Y}) \circ c_{X,Y}$.
	\item For any $f \in \Hom{}{Y}{Z}$, we have that $c_{X,Z}\circ (\Id \otimes f) = (f_{\star \deg X}\otimes \Id) \circ c_{X,Y}$.
	\end{itemize}
	
\item $c$ is natural with respect to color changing:
	\begin{itemize}
	\item $c_{X,Y}^\star = c_{X^\star, Y^\star}$
	\item $(c_{X,Y})_\star = c_{X_\star, Y_\star}$.
	\end{itemize}

\item $c$ is quasitriangular:
	\begin{itemize}
	\item $c_{X\otimes Y, Z} = (c_{X, Z_{\star \deg Y}} \otimes \Id) \circ (\Id \otimes c_{Y, Z})$
	\item $c_{X, Y \otimes Z} = (\Id \otimes c_{X^{\star \deg Y}, Z}) \circ (c_{X, Y} \otimes \Id).$
	\end{itemize}
\end{enumerate}
\end{defn}

INSERT PROOFS OF THE REIDEMEISTER RELATIONS

\subsubsection{Where might you look for a braiding?}

One nice thing about a braiding (as opposed to more general commutors) is that the quasi-triangularity condition combined with naturality, tells you that the entire braiding is determined by the braiding for $V \otimes V$ where $V$ is a tensor generator.  To see this notice that by naturality $c_{X, Y} = \restrict{c_{V^{\otimes m}, V^{\otimes n}}}{X \otimes Y}$, and that by quasitriangularity $c_{V^{\otimes m}, V^{\otimes n}}$ can be written in terms of only $c_{V,V}$.  In pictures this looks like:

PICTURE

Suppose we're looking at a subfactor planar algebra, all we need to know to determine a braiding is how to braid $M$ with $\bar{M}$.  But the Kauffman bracket formula already gives us a braiding for $M \otimes \bar{M}$.  In fact, if the principal graph starts out like $A_3$, you can show that this is the only possible braiding (for example, for $D_{2n}$ for $n>2$).  So rather than working hard to find a braiding, let's just look at the one we already have and see if it works.

What do we need to check to see that this braiding works?

PICTURE

