Ribbon category explained

In mathematics, a ribbon category, also called a tortile category, is a particular type of braided monoidal category.

Definition

lC

is, loosely speaking, a category equipped with a notion resembling the tensor product (of vector spaces, say). That is, for any two objects

C1,C2\inlC

, there is an object

C1C2\inlC

. The assignment

C1,C2\mapstoC1C2

is supposed to be functorial and needs to require a number of further properties such as a unit object 1 and an associativity isomorphism. Such a category is called braided if there are isomorphisms
c
C1,C2

:C1C2\stackrel\congC2C1.

A braided monoidal category is called a ribbon category if the category is left rigid and has a family of twists. The former means that for each object

C

there is another object (called the left dual),

C*

, with maps

1CC*,C*C1

such that the compositions

C*\congC*1C*(CC*)\cong(C*C)C*1C*\congC*

equals the identity of

C*

, and similarly with

C

. The twists are maps

C\inlC

,

\thetaC:CC

such that

\begin{align}

\theta
C1C2

&=

c
C2,C1
c
C1,C2
(\theta
C1

\theta
C2

)\\ \theta1&=id\\

\theta
C*

&=

*. \end{align}
(\theta
C)
To be a ribbon category, the duals have to be thus compatible with the braiding and the twists.

Concrete Example

Consider the category

FdVect(C)

of finite-dimensional vector spaces over

C

. Suppose that

C

is such a vector space, spanned by the basis vectors

\hat{e1},\hat{e2},,\hat{en}

. We assign to

C

the dual object

C\dagger

spanned by the basis vectors

\hat{e}1,\hat{e}2,,\hat{e}n

. Then let us define

\begin{align} :C\daggerC&\to1\\

i\hat{e
\hat{e}
j}

&\mapsto\begin{cases}1&i=j\ 0&ij\end{cases} \end{align}

and its dual

\begin{align} kIn:1&\toCC\dagger\\ k&\mapstok

n
\sum
i=1

\hat{ei}\hat{e}i\\ &=\begin{pmatrix}k&0&&0\ 0&k&&\vdots\&&\ddots&\\0&&&k\end{pmatrix} \end{align}

(which largely amounts to assigning a given

\hat{ei}

the dual

\hat{e}i

).

Then indeed we find that (for example)

\begin{align} \hat{e}i&\cong\hat{e}i1\\ &\underset{In}{\to}\hat{e}i

n
\sum
j=1

\hat{ej}\hat{e}j\\ &\cong

n
\sum
j=1

\left(\hat{e}i\hat{ej}\right)\hat{e}j\\ &\underset{}{\to}

n
\sum
j=1

\begin{cases}1\hat{e}j&i=j\ 0\hat{e}j&ij\end{cases}\\ &=1 ⊗ \hat{e}i\cong\hat{e}i \end{align}

and similarly for

\hat{ei}

. Since this proof applies to any finite-dimensional vector space, we have shown that our structure over

FdVect

defines a (left) rigid monoidal category.

Then, we must define braids and twists in such a way that they are compatible. In this case, this largely makes one determined given the other on the reals. For example, if we take the trivial braiding

\begin{align} c
C1,C2

:C1C2&\toC2C1\\

c
C1,C2

(a,b)&\mapsto(b,a) \end{align}

then

c
C1,C2
c
C2,C1
=id
C1 ⊗ C2
, so our twist must obey
\theta
C1C2

=

\theta
C1

\theta
C2
. In other words it must operate elementwise across tensor products. But any object

C\inFdVect

can be written in the form
n
C=otimes
i=1

1

for some

n

,

\thetaC=otimes

n
i=1

\theta1=

n
otimes
i=1

id=idC

, so our twists must also be trivial.

On the other hand, we can introduce any nonzero multiplicative factor into the above braiding rule without breaking isomorphism (at least in

C

). Let us for example take the braiding
\begin{align} c
C1,C2

:C1C2&\toC2C1\\

c
C1,C2

(a,b)&\mapstoi(b,a) \end{align}

Then

c
C1,C2
c
C2,C1
=-id
C1 ⊗ C2
. Since

\theta1=id

, then

\theta1=-id1 ⊗

; by induction, if

C

is

n

-dimensional, then

\thetaC=(-1)n+1idC

.

Other Examples

The name ribbon category is motivated by a graphical depiction of morphisms.

Variant

A strongly ribbon category is a ribbon category C equipped with a dagger structure such that the functor †: CopC coherently preserves the ribbon structure.

References