\ifx\streetdiagsloaded\relax\else
\documentclass{monog}
\usepackage{amsfonts}
\usepackage{xypic}
\xyoption{2cell}
%\documentstyle{monog}
\input streetdiags.mac 
%\dump
\fi
\bogustitlepage{11}{``Tensor categories''}

\chapter{Tensor categories}
%
It is clear that specific categories have
entered explicitly into the above discussion, 
but we have made little use of them as categories 
apart from diagrams and duality. 
For what follows it is hard to imagine how to
express the results without categories.

\medskip\noindent
%
A {\em tensor category\/}\index{tensor category}
is a category $\scrV$ together with functor 
$\map\XO\from{\!\scrV\times\!\scrV}\to{\!\scrV}$ 
called {\em tensor product},\index{tensor product}
an object $I$ of $\scrV$ called the {\em unit object\/},
\index{unit object}
and natural families of isomorphisms
\begin{eqnarray*}
&\relax\map a_{\!A,B,C}^{\phj}\from(A\XO B)\XO C\to{A\XO(B\XO C)}&\\
&\relax\map r_{\!\!A}^{\phj}\from A\XO I\to A\quad,\quad
\map l_{\!A}^{}\from I\XO A\to A
\end{eqnarray*}
called respectively 
the {\em associativity\/}\index{associativity } constraint, the
{\em right unit\/}\index{right unit constraint}%
\index{unit \subitem right} constraint and the
{\em left unit\/}\index{left unit constraint}%
\index{unit \subitem left} constraint,
subject to the two conditions:

$$
\spreaddiagramcolumns{-2pc}
\spreaddiagramrows{.5pc}
\diagram
&&\relax\llap{$(A\XO$}B)\XO(C\relax\rlap{$\XO D)$}
 \drrto^-{a_{\!A,B,C\xo D}^{\phj}}&&\\
\relax\llap{$((A\XO B)\XO C)$}\XO D)\urrto^-{a_{\!A\xo B,C,D}^{\phj}}
	\drto_-{a_{\!A,B,C}^{\phj}\xo1}&&&& A\XO\rlap{$(B\XO(C\XO D))$}\\
&\relax\llap{$(A\XO(B\XO C))$}\XO D\rrto_-{a_{\!A,B\xo C,D}^{\phj}\;}
 &\hskip3pc& 
 A\XO\rlap{$((B\XO C)\XO D)$}\urto_-{1\xo a_{\!B,C,D}^{\phj}} 
 & \\
\enddiagram
$$

$$\spreaddiagramcolumns{-.5pc}\spreaddiagramrows{.5pc}
\diagram
(A\XO I)\XO C\drto_-{r_{\!\!A}^{\phj}\xo1\quad}\rrto^-{a_{\!A,I,C}^{\phj}}&
 & A\XO(I\XO C)\dlto^-{\qquad1\xo l_{\!A}^{}}\\
& A\XO C \\
\enddiagram
$$
Define $A_{\!1}^{}\XO\dots\XO A_n$ to be the
object obtained by inserting brackets in some chosen
preassigned way, such as from the left
$((\dots(A_{\!1}^{}\XO A_2^{})\XO\dots)\XO A_n\,$.
It is an important fact (MacLane's coherence theorem)
\index{coherence theorem of MacLane}%
\index{MacLane, coherence theorem}%
that the only automorphisms of the form
$1\xO(x\xO 1)$ or $(1\xO x)\xO 1\,$,
where $x$ is a component of $a$, $r$, $l$ or their inverses, 
is the identity arrow of $A_{\!1}^{}\XO\dots\XO A_n\,$. 
This essentially allows one to work 
as if the $a$, $r$, $l$ are all identities. 
If all the $a$, $r$, $l$ are indeed identities, 
then the tensor category is called {\em strict\/}.
\index{tensor category \subitem strict}%
The {\em opposite\/}\index{tensor category \subitem opposite}% 
$\scrV^{\op}$ of a tensor category $\scrV$ 
consists of the opposite category of $\scrV$ 
(obtained by reversing the direction of arrows of $\scrV$) 
and the reverse tensor product, so that $A\XO B$ in $\scrV^{\op}$ 
is just $B\XO A$ in $\scrV$. 

\medskip\noindent
%
A {\em braiding\/}\index{braiding }
for a tensor category $\scrV$ is a
natural family of isomorphisms
\[
\map\comm_A\,B\from A\XO B\to B\XO A
\]
subject to the conditions

$$\spreaddiagramcolumns{-1pc}\spreaddiagramrows{.5pc}
\diagram
&\relax\llap{$A\XO(B\XO$}C)\rrto^-{\comm_\!A\,{B\xO C}\;}
 &&(B\XO\rlap{$C)\XO A$} 
 \drto^(.6){a_{\!B,C,A}^{\phj}}& \\
\relax\llap{$A\XO(B\XO$}C)\urto^(.4){a_{\!A,B,C}^{\phj}}
\drto_(.4){\comm_A\,B\xo1} &&&& B\rlap{$\XO(C\XO A)$}\\
&\relax\llap{$(B\XO A)$}\XO C \rrto^-{a_{\!B,A,C}^{\phj}\;}
 &\hskip.5pc & 
 B\XO\rlap{$(A\XO C)$}\urto_(.6){1\xo\comm_A\,C} & \\
\enddiagram
$$
%\unitlength=0.1pt
%\displaydiagram yy
%{\begin{picture}(2000,1000)(0,0)
%\hmap{500}{1000}{A\otimes (B\otimes C)}
%     {1500}{(B\otimes C)\otimes A}
%     {c_{A,B\otimes C}}u
%\hmap{500}{0}{(B\otimes A)\otimes C}
%     {1500}{B\otimes (A\otimes C)}
%     {a_{B,A,C}}d
%\put(0,500){\cent{(A\otimes B)\otimes C}}
%\put(2000,500){\cent{B\otimes (C\otimes A)}}
%\svector0{500}11{500}{a_{A,B,C}}54l
%\svector0{500}1{-1}{500}{c_{A,B}\otimes 1}54l
%\svector{1500}{1000}1{-1}{500}{a_{B,C,A}}34r
%\svector{1500}{0}1{1}{500}{1\otimes c_{A,C}}34r
%\end{picture}}

\smallskip

$$
\spreaddiagramcolumns{-1pc}
\spreaddiagramrows{.5pc}
\diagram
&\relax\llap{$(A\XO B)$}\XO C\rrto^-{\comm_\!A\xO B\,C\;}&
& C\XO\rlap{$(A\XO B)$}\drto^(.6){a^{-1}_{\!C,A,B}} &\\
\relax\llap{$A\XO(B\XO$}C)\urto^(.4){a^{-1}_{\!A,B,C}}
\drto_(.4){1\xo\comm_B\,C} &&&& (C\rlap{$\XO A)\XO B$}\\
&\relax\llap{$A\XO(C$}\XO B)\rrto^-{a^{-1}_{\!A,C,B}\;}
 & \hskip.5pc & 
 (A\XO\rlap{$C)\XO B$}\urto_(.6){\comm_A\,C\xo1}
\enddiagram
$$

%\unitlength=0.1pt
%\displaydiagram yy
%{\begin{picture}(2000,1000)(0,0)
%\hmap{500}{1000}{(A\otimes B)\otimes C}
%     {1500}{C\otimes (A\otimes B)}
%     {c_{A\otimes B,C}}u
%\hmap{500}{0}{A\otimes (C\otimes B)}
%     {1500}{(A\otimes C)\otimes B}
%     {a_{A,C,B}^{-1}}d
%\put(0,500){\cent{A\otimes (B\otimes C)}}
%\put(2000,500){\cent{(C\otimes A)\otimes B}}
%\svector0{500}11{500}{a_{A,B,C}^{-1}}54l
%\svector0{500}1{-1}{500}{1\otimes c_{B,C}}54l
%\svector{1500}{1000}1{-1}{500}{a_{C,A,B}^{-1}}34r
%\svector{1500}{0}1{1}{500}{c_{A,C}\otimes 1}34r
%\end{picture}}

\smallskip\noindent
%
A {\em braided tensor category\/}
\index{braided tensor category}%
\index{tensor category \subitem braided}%
is a tensor category with a chosen braiding.

\goodbreak\medskip\noindent
%
A {\em symmetry\/}\index{symmetry }
for a tensor category is a braiding
which satisfies the following extra condition

$$
\spreaddiagramcolumns{-1pc}\spreaddiagramrows{1pc}
\diagram
 A\xO B\drto_-{\comm_A\,B}\rrto^-{1\;} &
 & A\xO B\\
& B\xO A \urto_-{\comm_B\,A}& \\
\enddiagram
$$
%\unitlength=0.06pt
%\displaydiagram yn
%{\begin{picture}(1000,750)(0,0)
%\hmap{0}{750}{A\otimes B}
%     {1000}{A\otimes B}
%     {1}u
%\put(500,0){\cent{B\otimes A}}
%\svector0{750}2{-3}{500}{c_{A,B}}11l
%\svector{1000}{750}{-2}{-3}{500}{c_{B,A}}11r
%\end{picture}}

\medskip\noindent
%
A {\em symmetric tensor category\/}
\index{symmetric tensor category}%
\index{tensor category \subitem symmetric}%
is a tensor category with a chosen symmetry.

\bigskip

\begin{example} 
The {\em braid category\/}\index{braid category} 
$\Braid$ has as objects the natural numbers\/ $0,1,2,\dots$ 
and as arrows $\map\alpha\from n\to n$ the braids on\/ $n$ strings; 
there are no arrows\/ $\map\from m\to n$ for\/ $m\ne n\,$. 
A braid\/ $\alpha$

\MissingPicture

\noindent
on\/ $n$ strings can be regarded as an element of the
Artin braid group\/\index{braid group of Artin}
$\Braid_n$ with generators\/
$s_{\!1}^{},\,\dots\,, s_{n-1}^{}$ subject to the relations 
\begin{eqnarray*}
s_i^{} s_j^{}&=&s_j^{} s_i^{}\qquad\hbox{for }j<i-1\\
s_{i+1}^{}s_i^{}s_{i+1}^{}&=&s_i^{}s_{i+1}^{}s_i^{}
\end{eqnarray*}
Composition of braids is just multiplication in this group, 
represented diagramatically  by vertical
stacking of braids with the same number of strings.

\MissingPicture

\noindent
Tensor product of braids adds the number of strings
by placing one braid next to the other longitudinally.

\MissingPicture

\noindent
This makes\/ $\Braid$ a strict tensor category. 
A braiding\/ $\map\comm_\mkern3mu m\,n\from m+n\to{n+m}$ is given by
crossing the first\/ $m$ strings over the remaining\/ $n\,$.

\MissingPicture

\noindent
The axioms that show\/ $\Braid$ is braided are easily checked
diagramatically.
\end{example}

\begin{example} 
The category\/ $\Mod_R$ of modules over a commutative ring\/ $R$ 
is a symmetric tensor category with tensor product\/ $\xR\,$, 
with the canonical constraints,
and with symmetry\/ $\map\sigma\from A\xR B\to{B\xR A}$.
\end{example}

\begin{example} 
Let\/ $A$ be an\/ $R$-bialgebra. 
If\/ $M$ and\/ $N$ are\/ $A$-modules, we have an\/ $A$-module
structure on\/ $M\xR N$ given by 
\[
a\cdot(m\xO n) \;=\; \sum_{(a)}a_{(1)}^{}m\xO a_{(2)}^{}n
\]
as seen in the last section. 
So\/ $\Mod_R(A)$ becomes a tensor category 
with tensor product\/ $\xR\,$. 

If\/ $A$ is cocommutative, the switch morphism\/
$\map\sigma\from M\xR N\to{N\xR M}$ is a symmetry for\/ $\Mod_R(A)\,$. 
However, as in the rest of this book, we are more interested in
non-cocommutative $A\,$.

\medskip
\noindent
{\em We ask:} what are the possible braidings 
on the tensor category $\Mod_R(A)$? 

\smallskip\noindent
A braiding\/ $\map{\comm_M\,N}\from M\xR N\to{N\xR M}$ 
gives, for each\/ $A\,$, a morphism\/
$\map{\comm_A\,A}\from A\xR A\to{A\xR A}$ 
which gives an element\/ 
$\gamma=\comm_A\,A(1\xO 1)\in A\XO A\,$. 
Conversely, each element\/
$\gamma=\sum_i u_i^{}\xO v_i^{}\in A\XO A$
determines a natural morphism\/
$\map{\comm_M\,N}\from M\xR N\to{N\xR M}$ via the formula 
\[
 \comm_M\,N (m\xO n)=\sum_i(u_i^{}n)\xO(v_i^{}m)\,.
\]
This is a bijection, as can be seen from the following
diagram in which\/ $\map{\hat m}\from A\to M$ is the
unique module morphism with\/ $\hat m(1)=m\,$.
$$
\spreaddiagramcolumns{1pc}
\spreaddiagramrows{.5pc}%
\diagram
R \xR R\rto^{\eta\xo\eta\;}& A\xR A\dto_-{\hat m\xo\hat n}
	\rto^-{\comm_A\,A\;}& A\xR A\dto^-{\hat n\xo\hat m}\\
& M\xR N \rto^-{\comm_M\,N\;}& N\xR M \\
\enddiagram
$$
%\unitlength=0.08pt
%\displaydiagram yy
%{\begin{picture}(2000,618)(0,0)
%\hmap{0}{618}{R\otimes _RR}
%     {1000}{A\otimes_RA}
%     {\eta\otimes \eta}u
%\hmap{1000}{618}{\phantom{A\otimes_RA}}
%     {2000}{A\otimes_RA}
%     {c_{A,A}}u
%\hmap{1000}{0}{M\otimes_RN}
%     {2000}{N\otimes_RM}
%     {c_{M,N}}d
%\vvector{1000}{618}d{618}{\hat m\otimes \hat n}11l
%\vvector{2000}{618}d{618}{\hat n\otimes \hat m}11r
%\end{picture}}
In order for each\/ $\comm_M\,N$ to be an isomorphism it is
necessary for\/ $\gamma\in A\xR A$ to be invertible. 
In order for each\/ $\comm_M\,N$ to be a module morphism we need 
\begin{eqnarray*}
c(a\cdot(m\xO n))&=&
 c\left(\sum_{(a)}(a_{(1)}^{}m)\xO (a_{(2)}^{}n)\right)\\
 &=& \sum_{(a)}\sum_i(u_i^{}n)	\xO (v_i^{}m)
\end{eqnarray*}
to be equal to 
\begin{eqnarray*}
 a\cdot c(m\xO n) &=& a\cdot \sum_i(u_i^{}n)\xO(v_i^{}m)\\
 &=&\sum_{(a)}\sum_i(a_{(1)}^{}u_i^{}n)\xO(a_{(2)}^{}v_i^{}m)\,.
\end{eqnarray*}
This is equivalent to the requirement
\[
\sum_{i,(a)}(u_i^{}a_{(2)}^{})\xO(v_i^{}a_{(1)}^{})
=\sum_{i,(a)}(a_{(1)}^{}u_i^{})\xO(a_{(2)}^{}v_i^{})\,.
\]
Regarding\/ $\gamma\in A\xR A$ as a morphism\/ 
$\map\gamma\from R\to{A\xR A}$ 
whose value at\/ $1\in R$ is the given\/ $\gamma\,$.
We can express this condition diagramatically as
%
\[
\hbox to\hsize{\rlap{\hbox{\rm(B0)}}\hss
\spreaddiagramcolumns{1pc}
\spreaddiagramrows{.5pc}
\diagram
A\rto^-{\gamma\xo\delta\;}& A^{\xo4}
\rto<.5ex>^-{\sigma_{1432}^{}\;}
\rto<-.5ex>_-{\sigma_{3142}^{}\;}
& A^{\xo4}\rto^-{\mu\xo\mu\;}& A^{\xo2} 
\enddiagram\hss}
\]
%
%\unitlength=0.08pt
%\displaydiagram yy
%{\rlap{\begin{picture}(0,100)(0,-50)
%\put(0,0){\cent{\rm \phantom{(B0)}(B0)}}
%\end{picture}}
%\hskip 0 pt plus
%1000 fill 
%\begin{picture}(2400,100)(0,-50) 
%\hmap{0}{0}{R}
%     {800}{A^{\otimes4}}
%     {\gamma\otimes \gamma}u
%\hmap{800}{50}{\phantom{A^{\otimes4}}}
%     {1600}{\phantom{A^{\otimes4}}}
%     {\sigma_{1432}}u
%\hmap{800}{-50}{\phantom{A^{\otimes4}}}
%     {1600}{\phantom{A^{\otimes4}}}
%     {\sigma_{3142}}d
%\hmap{1600}{0}{A^{\otimes4}}
%     {2400}{A^{\otimes2}}
%     {\mu\otimes \mu}u
%\end{picture}\hskip 0 pt plus 1000 fill}
For a braiding, we require two more conditions:
\begin{eqnarray*}
&&\comm_M\,{N\xO L}(m\xO n\xO l)\;\;=\;\; 
(1_{\!N}\xO \comm_M\,L)(\comm_M\,N \xO 1_{\!L}^{})(m\xO n\xO l)\\
&&\comm_M\xO N\,L (m\xO n\xO l)\;\;=\;\;
(\comm_M\,L\xO 1_{\!N}^{})(1_{\!M}^{}\xO \comm_N\,L)(m\xO n\xO l)
\end{eqnarray*}
that is, 
\begin{eqnarray*}
&&\sum_{i,(u_i)}u_{i(1)}n\,\xO\,u_{i(2)}l\,\xO\,v_im\;\;=\;\; 
\sum_{i,j}u_i^{}n\,\xO\,u_j^{}l\,\xO\,v_j^{}v_i^{}m\\
&&\sum_{i,(v_i)}u_{i}^{}l\,\xO\,v_{i(1)}^{}m\,\xO\,v_{i(2)}^{}n
\;\;=\;\; 
\sum_{i,j}u_j^{}u_i^{}l\,\xO\,v_j^{}m\,\xO\,v_i^{}n.
\end{eqnarray*}
These are equivalent to the two conditions
\begin{eqnarray*}
&&\sum_{i,(u_i)}u_{i(1)}^{}\xO u_{i(2)}^{}\xO v_i^{}\;\;=\;\;
\sum_{i,j}u_i^{}\xO u_j^{}\xO v_j^{}v_i^{}\\
&&\sum_{i,(v_i)}u_{i}^{}\xO v_{i(1)}^{}\xO v_{i(2)}^{}\;\;=\;\; 
\sum_{i,j}u_j^{}u_i^{}\xO v_j^{}\xO v_i^{}\,.
\end{eqnarray*}
Diagramatically, these conditions become:
%
\[
\hbox to\hsize{\rlap{\hbox{\rm(B1)}}\hss
\spreaddiagramcolumns{1pc}%
\spreaddiagramrows{.5pc}%
\diagram
R \dto_\gamma \rto^-{\gamma\xo\gamma\;}
& A^{\xo4}\rto^-{\sigma_{1432}^{}\;}&
A^{\xo4}\dto^-{1\xo1\xo\mu}\\
A^{\xo4}\rrto^-{\delta\xo1\;}& & A^{\xo3}\\
\enddiagram\hss}
\]
\[
\hbox to\hsize{\rlap{\hbox{\rm(B2)}}\hss
\spreaddiagramcolumns{1pc}%
\spreaddiagramrows{.5pc}%
\diagram
R \dto_\gamma \rto^{\gamma\xo\gamma\;}
 & A^{\xo4}\rto^{\sigma_{3142}^{}\;}&
 A^{\xo4}\dto^{\mu\xo1\xo1}\\
A^{\xo4}\rrto^{1\xo\delta\;}& & A^{\xo3}\\
\enddiagram\hss}
\]
%\unitlength=0.067pt
%\displaydiagram yy
%{\rlap{\begin{picture}(0,618)(0,0)
%\put(0,309){\cent{\rm \phantom{(B1)}(B1)}}
%\end{picture}}
%\hskip 0 pt plus
%1000 fill 
%\begin{picture}(2000,618)(0,0)
%\hmap{0}{618}{R}
%     {1000}{A^{\otimes4}}
%     {\gamma\otimes \gamma}u
%\hmap{1000}{618}{\phantom{A^{\otimes4}}}
%     {2000}{A^{\otimes4}}
%     {\sigma_{1342}}u
%\hmap{0}{0}{A^{\otimes2}}
%     {2000}{A^{\otimes3}}
%     {\delta\otimes 1}d
%\vvector{0}{618}d{618}{\gamma}11l
%\vvector{2000}{618}d{618}{1\otimes 1\otimes \mu}11r
%\end{picture}\hskip 0 pt plus
%1000 fill 
%}
%\unitlength=0.067pt
%\displaydiagram yy
%{\rlap{\begin{picture}(0,618)(0,0)
%\put(0,309){\cent{\rm \phantom{(B2)}(B2)}}
%\end{picture}}
%\hskip 0 pt plus
%1000 fill 
%\begin{picture}(2000,618)(0,0)
%\hmap{0}{618}{R}
%     {1000}{A^{\otimes4}}
%     {\gamma\otimes \gamma}u
%\hmap{1000}{618}{\phantom{A^{\otimes4}}}
%     {2000}{A^{\otimes4}}
%     {\sigma_{3142}}u
%\hmap{0}{0}{A^{\otimes2}}
%     {2000}{A^{\otimes3}}
%     {1\otimes \delta}d
%\vvector{0}{618}d{618}{\gamma}11l
%\vvector{2000}{618}d{618}{\mu\otimes 1\otimes 1}11r
%\end{picture}\hskip 0 pt plus
%1000 fill 
%}
Hence, we define a {\em braiding element\/}%
\index{braiding \subitem element}
for a bialgebra $A$ to be an invertible element\/ $\gamma\in A\xR A$ 
which satisfies\/ {\rm(B0), (B1), (B2)}.
We have proved above that {\em braiding elements for $A$ 
are in bijection with braidings\/} on the tensor category $\Mod_R(A)\,$.

A {\em braided bialgebra\/}\index{braided bialgebra}%
\index{bialgebra \subitem braided}
(also called ``quasitriangular bialgebra'')%
\index{quasitriangular bialgebra}\index{bialgebra \subitem quasitriangular}
is a bialgebra equipped with a braiding element $\gamma\in A\xR A$. 
A braiding element $\gamma$ is
called a symmetry element when $\gamma^2=1\in A\xR A$; 
these are in bijection with symmetries on $\Mod_R(A)$. 
A {\em symmetric bialgebra\/}\index{bialgebra \subitem symmetric}
(also sometimes called ``triangular algebra'')%
\index{bialgebra \subitem triangular}
is a bialgebra equipped with a symmetry element.

Before leaving this example, we point out that
conditions\/ {\rm(B1), (B2)} can be put in a more
familiar form in the case where\/ $A$ is cauchy as an\/ $R$-module. 
For in this case, elements $\gamma=\sum_i u_i^{}\xO v_i^{}\in A\xR A$ 
are in bijection with\/ $R$-module
morphisms $\map g\from A^*\to A$ via the formula
$$
\spreaddiagramcolumns{.5pc}
\gamma\quad=\quad \Bigl(\diagram
R \rto^-{d\;}& A^*\xR A\rto^-{g\xo1_{\!A}^{}\,}& A\xR A 
\enddiagram\Bigr)
$$
Condition {\rm(B1)} precisely says that\/ $g$ preserves comultiplication,  
while condition {\rm(B2)} says that\/ $g$ reverses multiplication. 
In fact, if $\gamma$ is a braiding element, 
$\map g \from A^* \to {A'^\op}$
is a bialgebra morphism; preservation of unit and
counit follows from $\comm_M\,I =\comm_I\,M =\Id_{\!M}^{}$ 
(see Assignment 4).

We shall just look at the translation of {\rm(B2)} to $g$.
Begin with the defining diagram
$$
\spreaddiagramcolumns{1.5pc}
\spreaddiagramrows{.5pc}
\diagram
R \dto_d \rto^-{d\;}& A^*\xR A \rto^-{1\xo d\xo1\;}&
A^*\xR A^*\xR A\rlap{$\xR A$}\dto^{\delta^*\xo1\xo1}\\
 A^*\xR A \rrto^-{1\xo\delta\;}&& A^*\xR A\xR A\\
\enddiagram
$$
%\unitlength=0.09pt
%\displaydiagram yy
%{\begin{picture}(2000,618)(0,0)
%\hmap{0}{618}{R}
%     {600}{A^*\otimes _RA}
%     {d}u
%\hmap{600}{618}{\phantom{A^*\otimes _RA}}
%     {2000}{A^*\otimes _RA^*\otimes _RA\otimes _RA}
%     {1\otimes d\otimes 1}u
%\hmap{0}{0}{A^*\otimes _RA}
%     {2000}{A^*\otimes _RA\otimes _RA}
%     {1\otimes \delta}d
%\vvector{0}{618}d{618}{d}11l
%\vvector{2000}{618}d{618}{\delta^*\otimes 1\otimes 1}11r
%\end{picture}}
for\/ $\delta^*$,  which is the multiplication for\/ $A^*$.
To prove $g$ reverses multiplication is to prove
$$
\spreaddiagramcolumns{.5pc}
\spreaddiagramrows{.5pc}
\diagram
A^*\xR A^* \dto_{d^*} \rto^-{g\xo g\;}& A\xR A \rto^-{\sigma\;}
& A\xR A \dto^{\mu}\\
 A^* \rrto^-{g\;}& & A \\
\enddiagram
$$
%\unitlength=0.08pt
%\displaydiagram yy
%{\begin{picture}(2000,618)(0,0)
%\hmap{0}{618}{A^*\otimes _RA^*}
%     {1000}{A\otimes _RA}
%     {g\otimes g}u
%\hmap{1000}{618}{\phantom{A\otimes _RA}}
%     {2000}{A\otimes _RA}
%     {\sigma}u
%\hmap{0}{0}{A^*}
%     {2000}{A}
%     {g}d
%\vvector{0}{618}d{618}{\delta^*}11l
%\vvector{2000}{618}d{618}{\mu}11r
%\end{picture}}
This is equivalent to proving the legs are equal after
applying $\slot\xR A\xR A$
and composing with
$$
\diagram
\spreaddiagramcolumns{1pc}
R \rto^-d & A^*\xR A^* \rto^-{1\xo d\xo1\;} & A^*\xR A^*\xR A\xR A 
\enddiagram
$$
%\unitlength=0.08pt
%\displaydiagram yy
%{\begin{picture}(3000,0)(0,0) 
%\hmap{0}{0}{R}
%     {800}{A^*\otimes _RA}
%     {d}u
%\hmap{800}{0}{\phantom{A^*\otimes _RA}}
%     {2400}{A^*\otimes _RA^*\otimes_R A\otimes _RA
%.}
%     {1\otimes d\otimes 1}u
%\end{picture}}
From the defining diagram for $\delta^*$, this amounts to
$$
\spreaddiagramcolumns{-.5pc}
\spreaddiagramrows{.5pc}
\diagram
R \rto^-d & A^*\xR A^* \dto_{1\xo\delta}\rto^-{1\xo d\xo1\;}
 & A^*\xR A^*\xR A\xR A 
 \rto^-{g\xo g\xo1\xo1\;}& A^{\xo4}\rto^-{\sigma\,}
 & A\rlap{$^{\xo4}$}\dto_{\mu\xo1\xo1}\\
& A^*\xR A\xR A \arrow[rrr]^-{g\xo1\xo1\;}&\hskip10pc&& A^{\xo3} \\
\enddiagram
$$
%\unitlength=0.08pt
%\displaydiagram ny
%{\begin{picture}(3700,1300)(300,0)
%\put(400,1300){\cent{R}}
%\vvector{400}{1300}d{500}{d}11l
%\hmap{400}{800}{A^*\shotimes _RA}
%     {1700}{A^*\shotimes _RA^*\shotimes _RA\shotimes _RA}
%     {1\shotimes d\shotimes 1}u
%\hmap{1700}{800}{\phantom{A^*\shotimes _RA^*\shotimes
%_RA\shotimes _RA}}
%     {3150}{A^{\shotimes4}}
%     {g\shotimes g\shotimes 1\shotimes 1}u
%\hmap{3150}{800}{\phantom{A^{\shotimes4}}}
%     {4000}{A^{\shotimes4}}
%     {\sigma\shotimes 1\shotimes 1}u
%\hmap{550}{0}{A^*\shotimes _RA\shotimes _RA}
%     {4000}{A^{\shotimes 3}}
%     {g\shotimes 1\shotimes 1}d
%\vvector{400}{800}d{800}{1\shotimes \delta}11r
%\vvector{4000}{800}d{800}{\mu\shotimes 1\shotimes 1}11l
%\end{picture}}
Using $\gamma=(g\xO 1_{\!A})\circ d\,$, 
we easily see that this is equivalent to {\rm(B2)}.
\end{example}

Although a braiding is as useful as a symmetry for
most purposes, there is sometimes further structure
on a braiding which makes it even more like a
symmetry without actually forcing it to be one.

Suppose $\scrV$ is a braided tensor category. 
A {\em twist\/} for $\scrV$ is a natural family of isomorphisms  
\[
\map{\theta_{\!A}^{}}\from A\to A
\]
such that $\theta_{\!I}^{}=1_{\!I}^{}$ and
$$\spreaddiagramcolumns{1pc}\spreaddiagramrows{.5pc}
\diagram
A\XO B \dto_{\theta_{\!A\xo B}\quad} \rto^{\comm_A\,B\;}& B\XO A 
\dto^{\theta_{\!B}^{}\xo\theta_{\!A}^{}\;}\\
 A\XO B & B\XO A \lto^{\;\comm_B\,A} \\
\enddiagram
$$
%\unitlength=0.08pt
%\displaydiagram yy
%{\begin{picture}(1000,618)(0,0)
%\hmap{0}{618}{A\otimes B}
%     {1000}{B\otimes A}
%     {c_{A,B}}u
%\hmap{1000}{0}{B\otimes A\rlap{$\quad .$}}
%     {0}{A\otimes B}
%     {c_{B,A}}d
%\vvector{0}{618}d{618}{\theta_{A\otimes B}}11l
%\vvector{1000}{618}d{618}{\theta_B\otimes \theta_A}11r
%\end{picture}}
A {\em balanced tensor category\/}%
\index{tensor category \subitem balanced}
is a braided tensor
category with a chosen twist. 
(A braiding is a symmetry if and only if 
the identity arrows provide a twist.)

\begin{example}
The braid category\/ $\Braid$ is canonically balanced. 
The twist\/ $\map\theta_n\from n\to n$ is obtained by
taking\/ $n$ vertical parallel strings with ends tied to
two horizontal parallel rods, and rotating the bottom
rod through a full $2\pi$ twist in the right-hand
screw direction with thumb vertical. 
Then $\theta_0^{}$,
$\theta_1^{}$ are identities, while\/ $\theta_2^{}$ is

\MissingPicture

\end{example}

\begin{example}
There is a tensor category $\tilde\Braid$ which is defined 
similarly to $\Braid$, except that the arrows are braids on
ribbons (instead of on strings) and it is permissible
to twist the ribbons through full\/ $2\pi$ turns (see diagram). 

\MissingPicture

The homsets\/ $\tilde\Braid(n,n)=\tilde\Braid_n$ are groups
under composition. A presentation of this group\/
$\tilde\Braid_n$ is given by generators\/ $s_1^{},\,\dots\,,s_n^{}$
where\/ $s_1^{},\,\dots\,,s_{n-1}^{}$ satisfy the relations as for\/
$\Braid_n$, while here there is the extra relation
\[
s_{n-1}^{}s_n^{}s_{n-1}^{}s_n^{}=s_n^{}s_{n-1}^{}s_n^{}s_{n-1}^{}\,.
\]
Composition in\/ $\tilde\Braid$ is vertical stacking of
digrams, and tensor product for\/ $\tilde\Braid$ is
horizontal placement of diagrams, much as for\/ $\Braid$.
The braiding\/ $\map\comm_\mkern3mu m\,n\from m+n\to{n+m}$ for\/
$\tilde\Braid$ is obtaining the first\/ $m$ ribbons over the
remaining\/ $n$ without introducing any twists. 
The twist\/ $\map\theta_n^{}\from n\to n$ for $\tilde\Braid$ 
is obtained by regarding the two boundary edges of the
ribbons as extra strings and taking
$\map\theta_{2n}^{}\from 2n\to{2n}$ in\/ $\Braid$. 
Then in $\tilde\Braid$ we have

\MissingPicture

\end{example}

\begin{example} 
Let $A$ and $B$ be abelian groups and
$\map f\from A\times A\to B$ be a bilinear function.
There is a balanced strict tensor category\/ $\scrC_f^{}$ 
constructed as follows. 
The objects are the elements of $A$. 
The homset\/ $\scrC_f^{}(x,y)$ is empty unless\/ $x=y\,$, 
in which case\/ $\scrC_f^{}(x,x)=B\,$. The tensor
product is given by
\[
(\mapnamed\alpha\from x\to x)\xo(\mapnamed\beta\from y\to y) 
\;\;=\;\;
(\longmapnamed<2.5pc>\alpha+\beta\from x+y\to{x+y})
\] 
The braiding is
$\comm_x\,y = \map f(x,y)\from x+y\to{y+x}$ 
and the twist is
$\thetaX = \map f(x,x)\from x\to x$.

\end{example}

\begin{example}
Let $A$ be a braided $R$-bialgebra with braiding
element $\gamma=\sum_i u_i^{}\xO v_i^{}\in A\xR A\,$. 
A {\em twist element\/}\index{twist element}
for $A$ is an invertible central element\/ $\tau\in A$ 
such that\/ $\varepsilon(\tau)=1$ and
\[
\delta(\tau)=\sum_{i,j} (u_i^{}\tau v_j^{})\xO(v_i^{}\tau u_j^{})\,.
\]
Diagrammatically the last equation becomes:
$$
\spreaddiagramcolumns{2pc}%\spreaddiagramrows{1pc}
\diagram
R \ddto_-\tau \rto^-{\gamma\xo\tau\xo\tau\xo\gamma\;} 
& A^{\xo6} \rto^-{\sigma_{136245}^{}\;} 
&  A^{\xo6}\dto^-{\mu\xo1\mu\xo1}\\
& & A^{\xo4}\dto^-{\mu\xo\mu}\\
A \rrto^-{\delta\;} & & A^{\xo2} \\
\enddiagram
$$
%\unitlength=0.09pt
%\displaydiagram yy
%{\begin{picture}(2000,1000)(0,0)
%\hmap{0}{1000}{R}
%     {1000}{A^{\otimes 6}}
%     {\gamma\otimes \tau\otimes \tau\otimes \gamma}u
%\hmap{1000}{1000}{\phantom{A^{\otimes 6}}}
%     {1750}{A^{\otimes 6}}
%     {\sigma_{136245}}u
%\hmap{0}{0}{A}
%     {1750}{A^{\otimes 2}\rlap{\quad .}}
%     {\delta}d
%\vvector{0}{1000}d{1000}{\tau}11l
%\vvector{1750}{1000}d{500}{\mu\otimes 1\otimes \mu\otimes 1}11r
%\put(1750,500){\cent{A^{\otimes 4}}}
%\vvector{1750}{500}d{500}{\mu\otimes \mu}11r
%\end{picture}}
{\em Twist elements\/ $\tau$ for\/ $A$ are in bijection with
twists\/ $\theta$} for the braided tensor category\/ $\Mod_R(A)\,$. 
Naturality of
$\map{\theta_{\!M}^{}}\from M\to M$ means it has the form\/
$\theta_{\!M}^{}(m)=\tau m$ for some $\tau\in A\,$; 
for\/ $\theta_{\!M}^{}$ to be an\/ $A$-module morphism, 
$\tau$ needs to be central (meaning\/ $\tau\cdot a=a\cdot\tau$
for all\/ $a\in A$); 
for\/ $\theta_{\!M}^{}$ to be an isomorphism, 
$\tau$ needs to be invertible; 
for\/ $\theta_{\!M}^{}=1_{\!R}^{}$, 
the condition\/ $\varepsilon(\tau)=1$ is needed; 
and of course the remaining twist conditions correspond.
\end{example}

\medskip
\noindent
A {\em balanced bialgebra\/}\index{balanced bialgebra}%
\index{bialgebra \subitem balanced}
is a braided bialgebra with a twist.


\clearpage
\bogusendpage