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.

A tensor category is a category $\scrV$ together with functor $\map\XO\from{\!\scrV\times\!\scrV}\to{\!\scrV}$ called tensor product, an object I of $\scrV$ called the unit object, and natural families of isomorphisms

$$\begin{eqnarray*} &\relax\map a_{\!A,B,C}^{\phj}\from(A\XO B)\XO C\to{A\XO(B\XO ... ...hj}\from A\XO I\to A\quad,\quad \map l_{\!A}^{}\from I\XO A\to A \end{eqnarray*}$$

called respectively the associativity constraint, the right unit constraint and the left unit constraint, subject to the two conditions:

$$\spreaddiagramcolumns {-2pc}\spreaddiagramrows {.5pc}\diagra... ...((B\XO C)\XO D)$}\urto_-{1\xo a_{\!B,C,D}^{\phj}} \enddiagram$$

$$\spreaddiagramcolumns{-.5pc}\spreaddiagramrows{.5pc}\diagram... ...\XO(I\XO C)\dlto^-{\qquad1\xo l_{C}^{}}\\ & A\XO C \enddiagram$$

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)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 strict. The 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 $\scrV$.

A 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

$$\spreaddiagramcolumns{-1pc}\spreaddiagramrows{.5pc}\diagram ... ...& B\XO\rlap{$(A\XO C)$}\urto_(.6){1\xo\comm_A\,C} \enddiagram$$

$$\spreaddiagramcolumns{-1pc}\spreaddiagramrows{.5pc}\diagram ... ... & (A\XO\rlap{$C)\XO B$}\urto_(.6){\comm_A\,C\xo1}\enddiagram$$

A braided tensor categoryis a tensor category with a chosen braiding.

A symmetry for a tensor category is a braiding which satisfies the following extra condition:

$$\spreaddiagramcolumns{-1pc}\spreaddiagramrows{1pc}\diagram ... ...-{1\;} & & A\xO B\\ & B\xO A \urto_-{\comm_B\,A}& \enddiagram$$

symmetric tensor categoryis a tensor category with a chosen symmetry.

Example 12.1  

The 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 $\alpha$

$$\mbox{Picture missing here.}$$

on n strings can be regarded as an element of the Artin braid group $\Braid_n$ with generators $s_{\!1}^{},\,\dots\,, s_{n-1}^{}$ subject to the relations

$$\begin{eqnarray*} s_i^{} s_j^{}&=&s_j^{} s_i^{}\qquad\hbox{for }j<i-1\\ s_{i+1}^{}s_i^{}s_{i+1}^{}&=&s_i^{}s_{i+1}^{}s_i^{} \end{eqnarray*}$$

where s_i is the braid depicted as follows.

$$\mbox{Picture missing here.}$$

Composition of braids is just multiplication in this group, represented diagramatically by vertical stacking of braids with the same number of strings.

$$\mbox{Picture missing here.}$$

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

$$\mbox{Picture missing here.}$$

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 .

$$\mbox{Picture missing here.}$$

The axioms that show $\Braid$ is braided are easily checked diagramatically.

Example 12.2  

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 $\map\sigma\from A\xR B\to{B\xR A}$.

Example 12.3  

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 .

We ask: what are the possible braidings on the tensor category $\Mod_R(A)$ ?

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 $\hat m(1)=m\,$.

$$\spreaddiagramcolumns{1pc}\spreaddiagramrows{.5pc}\diagram R... ...\xo\hat m}\\ & M\xR N \rto^-{\comm_M\,N\;}& N\xR M \enddiagram$$

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

$$\begin{eqnarray*} c\bigl(a\cdot(m\xO n)\bigr)&=& c\Bigl(\sum_{(a)}(a_{(1)}^{}m)... ...{(2)}^{}n)\Bigr)\\ &=& \sum_{(a)}\sum_i(u_i^{}n) \xO (v_i^{}m) \end{eqnarray*}$$

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

$$\hbox to\hsize{\rlap{\hbox{\rm (B0)}}\hss \spreaddiagramcolu... ...^{}\;} & A^{\xo4}\rto^-{\mu\xo\mu\;}& A^{\xo2}\enddiagram\hss}$$

For a braiding, we require two more conditions:

$$\begin{eqnarray*} && \comm_M\,{N\xO L}(m\xO n\xO l)\;\;=\;\; (1_{\!N}\xO \comm_... ...\comm_M\,L\xO 1_{\!N}^{})(1_{\!M}^{}\xO \comm_N\,L)(m\xO n\xO l) \end{eqnarray*}$$

that is,

$$\begin{eqnarray*} && \sum_{i,(u_i)}u_{i(1)}n\,\xO\,u_{i(2)}l\,\xO\,v_im\;\;=\;\;... ... \;\;=\;\; \sum_{i,j}u_j^{}u_i^{}l\,\xO\,v_j^{}m\,\xO\,v_i^{}n. \end{eqnarray*}$$

These are equivalent to the two conditions

$$\begin{eqnarray*} &&\sum_{i,(u_i)}u_{i(1)}^{}\xO u_{i(2)}^{}\xO v_i^{}\;\;=\;\; ... ...i(2)}^{}\;\;=\;\; \sum_{i,j}u_j^{}u_i^{}\xO v_j^{}\xO v_i^{}\,. \end{eqnarray*}$$

Diagramatically, these conditions become:

$$\hbox to\hsize{\rlap{\hbox{\rm (B1)}}\hss \spreaddiagramcolu... ...}\\ A^{\xo4}\rrto^-{\delta\xo1\;}& & A^{\xo3}\enddiagram\hss}$$

$$\hbox to\hsize{\rlap{\hbox{\rm (B2)}}\hss \spreaddiagramcolu... ...1}\\ A^{\xo4}\rrto^{1\xo\delta\;}& & A^{\xo3}\enddiagram\hss}$$

Hence, we define a braiding element for a bialgebra A to be an invertible element $\gamma\in A\xR A$ which satisfies (B0), (B1), (B2). We have proved above that braiding elements for A are in bijection with braidings on the tensor category $\Mod_R(A)\,$.

A braided bialgebra (also called ``quasitriangular bialgebra'') 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 symmetric bialgebra (also sometimes called ``triangular algebra'') is a bialgebra equipped with a symmetry element.

Before leaving this example, we point out that conditions (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

$$\spreaddiagramcolumns{.5pc}\gamma \;\;=\;\; \Bigl(\raise 1ex... ... A^*\xR A\rto^-{g\xo1_{\!A}^{}\,}& A\xR A \enddiagram }\Bigr)$$

Condition (B1) precisely says that g preserves comultiplication, while condition (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 one of the assignments).

We shall just look at the translation of (B2) to g. Begin with the defining diagram

$$\spreaddiagramcolumns{1.5pc}\spreaddiagramrows{.5pc}\diagram... ...\\ A^*\xR A \rrto^-{1\xo\delta\;}&& A^*\xR A\xR A \enddiagram$$

for $\delta^*$, which is the multiplication for A^*. To prove g reverses multiplication is to prove

$$\spreaddiagramcolumns{.5pc}\spreaddiagramrows{.5pc}\diagram ... ...ma\;} & A\xR A \dto^{\mu}\\ A^* \rrto^-{g\;}& & A \enddiagram$$

This is equivalent to proving the legs are equal after applying $\slot\xR A\xR A$ and composing with

$$\spreaddiagramcolumns{1pc}\diagram R \rto^-d & A^*\xR A^* \rto^-{1\xo d\xo1\;} & A^*\xR A^*\xR A\xR A \enddiagram$$

From the defining diagram for $\delta^*$, this amounts to

$$\spreaddiagramcolumns{-.5pc}\spreaddiagramrows{.5pc}\diagram... ...row[rrr]^-{g\xo1\xo1\;}&\relax\hskip10pc&& A^{\xo3} \enddiagram$$

Using $\gamma=(g\xO 1_{\!A})\circ d\,$, we easily see that this is equivalent to (B2).

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 twist for $\scrV$ is a natural family of isomorphisms

$$\map{\theta_{\!A}^{}}\from A\to A$$

such that $\theta_{\!I}^{}=1_{\!I}^{}$ and

$$\spreaddiagramcolumns{1pc}\spreaddiagramrows{.5pc}\diagram A... ...{\!A}^{}\;}\\ A\XO B & B\XO A \lto^{\;\comm_B\,A} \enddiagram$$

A balanced tensor category is a braided tensor category with a chosen twist. (A braiding is a symmetry if and only if the identity arrows provide a twist.)

Example 12.4  

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

$$\mbox{Picture missing here.}$$

Example 12.5  

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 (as in the following diagram).

$$\mbox{Picture missing here.}$$

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 $\Braid_n$. These are depicted by thickened versions of the diagrams in the first example, along with the extra relation

s_n-1s_ns_n-1s_n=s_ns_n-1s_ns_n-1

where s_n is depicted as follows

$$\mbox{Picture missing here.}$$

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

$$\mbox{Picture missing here.}$$

Example 12.6  

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 $\thetaX = \map f(x,x)\from x\to x$.

Example 12.7  

Let A be a braided R-bialgebra with braiding element $\gamma=\sum_i u_i^{}\xO v_i^{}\in A\xR A\,$. A 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:

$$\spreaddiagramcolumns{2pc}%\spreaddiagramrows{1pc} \diagram ... ...o^-{\mu\xo\mu}\\ A \rrto^-{\delta\;} & & A^{\xo2} \enddiagram$$

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.

A balanced bialgebra is a braided bialgebra with a twist.