# Ribbon category

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

## Definition

A monoidal category ${\displaystyle {\mathcal {C}}}$ is, loosely speaking, a category equipped with a notion resembling the tensor product (of vector spaces, say). That is, for any two objects ${\displaystyle C_{1},C_{2}\in {\mathcal {C}}}$, there is an object ${\displaystyle C_{1}\otimes C_{2}\in {\mathcal {C}}}$. The assignment ${\displaystyle C_{1},C_{2}\mapsto C_{1}\otimes C_{2}}$ 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

${\displaystyle c_{C_{1},C_{2}}:C_{1}\otimes C_{2}{\stackrel {\cong }{\rightarrow }}C_{2}\otimes C_{1}.}$

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 ${\displaystyle C}$ there is another object (called the left dual), ${\displaystyle C^{*}}$, with maps

${\displaystyle 1\rightarrow C\otimes C^{*},C^{*}\otimes C\rightarrow 1}$

such that the compositions

${\displaystyle C^{*}\cong C^{*}\otimes 1\rightarrow C^{*}\otimes (C\otimes C^{*})\cong (C^{*}\otimes C)\otimes C^{*}\rightarrow 1\otimes C^{*}\cong C^{*}}$

equals the identity of ${\displaystyle C^{*}}$, and similarly with ${\displaystyle C}$. The twists are maps

${\displaystyle C\in {\mathcal {C}}}$, ${\displaystyle \theta _{C}:C\rightarrow C}$

such that

{\displaystyle {\begin{aligned}\theta _{C_{1}\otimes C_{2}}&=c_{C_{2},C_{1}}c_{C_{1},C_{2}}(\theta _{C_{1}}\otimes \theta _{C_{2}})\\\theta _{1}&=\mathrm {id} \\\theta _{C^{*}}&=(\theta _{C})^{*}.\end{aligned}}}

To be a ribbon category, the duals have to be thus compatible with the braiding and the twists.

### Concrete Example

Consider the category ${\displaystyle \mathbf {FdVect} (\mathbb {C} )}$ of finite-dimensional vector spaces over ${\displaystyle \mathbb {C} }$. Suppose that ${\displaystyle C}$ is such a vector space, spanned by the basis vectors ${\displaystyle {\hat {e_{1}}},{\hat {e_{2}}},\cdots ,{\hat {e_{n}}}}$. We assign to ${\displaystyle C}$ the dual object ${\displaystyle C^{\dagger }}$ spanned by the basis vectors ${\displaystyle {\hat {e}}^{1},{\hat {e}}^{2},\cdots ,{\hat {e}}^{n}}$. Then let us define

{\displaystyle {\begin{aligned}\cdot :\ C^{\dagger }\otimes C&\to 1\\{\hat {e}}^{i}\cdot {\hat {e_{j}}}&\mapsto {\begin{cases}1&i=j\\0&i\neq j\end{cases}}\end{aligned}}}

and its dual

{\displaystyle {\begin{aligned}kI_{n}:1&\to C\otimes C^{\dagger }\\k&\mapsto k\sum _{i=1}^{n}{\hat {e_{i}}}\otimes {\hat {e}}^{i}\\&={\begin{pmatrix}k&0&\cdots &0\\0&k&&\vdots \\&&\ddots &\\0&\cdots &&k\end{pmatrix}}\end{aligned}}}

(which largely amounts to assigning a given ${\displaystyle {\hat {e_{i}}}}$ the dual ${\displaystyle {\hat {e}}^{i}}$).

Then indeed we find that (for example)

{\displaystyle {\begin{aligned}{\hat {e}}^{i}&\cong {\hat {e}}^{i}\otimes 1\\&{\underset {I_{n}}{\to }}{\hat {e}}^{i}\otimes \sum _{j=1}^{n}{\hat {e_{j}}}\otimes {\hat {e}}^{j}\\&\cong \sum _{j=1}^{n}\left({\hat {e}}^{i}\otimes {\hat {e_{j}}}\right)\otimes {\hat {e}}^{j}\\&{\underset {\cdot }{\to }}\sum _{j=1}^{n}{\begin{cases}1\otimes {\hat {e}}^{j}&i=j\\0\otimes {\hat {e}}^{j}&i\neq j\end{cases}}\\&=1\otimes {\hat {e}}^{i}\cong {\hat {e}}^{i}\end{aligned}}}

and similarly for ${\displaystyle {\hat {e_{i}}}}$. Since this proof applies to any finite-dimensional vector space, we have shown that our structure over ${\displaystyle \mathbf {FdVect} }$ 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

{\displaystyle {\begin{aligned}c_{C_{1},C_{2}}:C_{1}\otimes C_{2}&\to C_{2}\otimes C_{1}\\c_{C_{1},C_{2}}(a,b)&\mapsto (b,a)\end{aligned}}}

then ${\displaystyle c_{C_{1},C_{2}}c_{C_{2},C_{1}}=\mathrm {id} _{C_{1}\otimes C_{2}}}$, so our twist must obey ${\displaystyle \theta _{C_{1}\otimes C_{2}}=\theta _{C_{1}}\otimes \theta _{C_{2}}}$. In other words it must operate elementwise across tensor products. But any object ${\displaystyle C\in \mathbf {FdVect} }$ can be written in the form ${\displaystyle C=\bigotimes _{i=1}^{n}1}$ for some ${\displaystyle n}$, ${\displaystyle \theta _{C}=\bigotimes _{i=1}^{n}\theta _{1}=\bigotimes _{i=1}^{n}\mathrm {id} =\mathrm {id} _{C}}$, 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 ${\displaystyle \mathbb {C} }$). Let us for example take the braiding

{\displaystyle {\begin{aligned}c_{C_{1},C_{2}}:C_{1}\otimes C_{2}&\to C_{2}\otimes C_{1}\\c_{C_{1},C_{2}}(a,b)&\mapsto i(b,a)\end{aligned}}}

Then ${\displaystyle c_{C_{1},C_{2}}c_{C_{2},C_{1}}=-\mathrm {id} _{C_{1}\otimes C_{2}}}$. Since ${\displaystyle \theta _{1}=\mathrm {id} }$, then ${\displaystyle \theta _{1\otimes 1}=-\mathrm {id} _{1\otimes 1}}$; by induction, if ${\displaystyle C}$ is ${\displaystyle n}$-dimensional, then ${\displaystyle \theta _{C}=(-1)^{n+1}\mathrm {id} _{C}}$.

### Other Examples

The name ribbon category is motivated by a graphical depiction of morphisms.[2]

## 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

1. ^ Turaev 2020, XI. An algebraic construction of modular categories
2. ^ Turaev 2020, p. 25
• Turaev, V.G. (2020) [1994]. Quantum Invariants of Knots and 3-Manifolds. de Gruyter. ISBN 978-3-11-088327-5.
• Yetter, David N. (2001). Functorial Knot Theory. World Scientific. ISBN 978-981-281-046-5. OCLC 1149402321.
• Ribbon category at the nLab