# Diagramatics for cyclic pointed fusion categories

Agustina Czenky  Department of Mathematics, University of Oregon, Eugene, OR
97403, USA<EMAIL_ADDRESS>
###### Abstract.

We give a parametrization of cyclic pointed categories associated to the
cyclic group of order $n$ in terms of $n$-th roots of unity. We also provide a
diagramatic description of these categories by generators and relations, and
use it to characterize their $2$-group of automorphisms.

##  1\. Introduction

Let $\mathbf{k}$ be an algebraically closed field of characteristic zero. A
_fusion category_ over $\mathbf{k}$ is a semisimple rigid tensor category over
$\mathbf{k}$ with finitely many isomorphism classes of simple objects, finite
dimensional hom spaces, and such that the unit object is simple [ENO05]. They
are connected to many areas of mathematics and mathematical physics, which
makes the problem of classifying them both an important yet difficult task. At
the moment, a complete classification of fusion categories seems out of reach
as it would yield a classification of finite groups. Some progress has been
made towards the classification of fusion categories of certain classes; this
work concerns pointed fusion categories associated to finite cyclic groups.

An object $X$ in a fusion category $\operatorname{\mathcal{C}}$ is called
_invertible_ if its evaluation and coevaluation morphisms are isomorphisms. A
fusion category $\operatorname{\mathcal{C}}$ is called _pointed_ if every
simple object is invertible. Pointed fusion categories are well-understood in
terms of finite group data: any pointed fusion category is equivalent to the
category $\operatorname{Vec}_{G}^{\omega}$ of finite dimensional $G$-graded
vector spaces, where $G$ is a finite group and $\omega$ is a 3-cocycle on $G$
with coefficients in $\mathbf{k}$ codifying the associativity isomorphism.
Monoidal functors between these categories are also classified: a functor
$\operatorname{Vec}_{G_{1}}^{\omega_{1}}\to\operatorname{Vec}_{G_{2}}^{\omega_{2}}$
is determined by a group homomorphism $f:G_{1}\to G_{2}$ satisfying that
$\omega_{1}$ and $f^{*}\omega_{2}$ are cohomologous, along with the data of a
2-cocycle on $G_{1}$ [EGNO15, Section 2.6].

In this work, the hope is to provide a more efficient way of working with
pointed fusion categories associated to finite cyclic groups. To this end, we
show in Section 4 that for the cyclic group $\mathbb{Z}_{n}$ of order $n$,
pointed fusion categories associated to it have a classification which depends
on a unique parameter: an $n$-th root of unity $\zeta\in\mathbf{k}$. We refer
to the corresponding pointed category by
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$. We later use this
parametrization in Section 5 to give a diagramatic description of
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ by generators and relations.
With this description, we get the following universal property: tensor
functors $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\to\mathcal{C}$ are in
bijection with pairs $(X,\lambda)$, where $X\in\operatorname{\mathcal{C}}$ and
$\lambda:X^{\otimes n}\xrightarrow{\sim}\mathbf{1}$ satisfies
$\text{id}\otimes\lambda=\zeta\lambda\otimes\text{id}$, see Section 5.1. This
gives a way of measuring when two cyclic pointed fusion associated to
$\mathbb{Z}_{n}$ are equivalent, and we give an answer for the number of
equivalence classes for a fixed $n$ in Section 6. Lastly, we provide a
description of the automorphism 2-group of
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ in Section 6, summarized in the
following theorem.

For $j\in\mathbb{Z}_{n}$, denote by $\delta_{j}$ the simple object in
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ defined by $(\delta_{j})_{k}=0$
if $k\neq j$, and $(\delta_{j})_{k}=\mathbf{k}.$

######  Theorem 6.7.

The 2-group
$\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})$
has:

  1. (1)

Objects: monoidal automorphisms
$F_{X,\lambda}:\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\xrightarrow{\sim}\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$,
where

     * •

$X\simeq\delta_{j}$ for some $j\in\mathbb{Z}_{n}^{\times}$ such that
$\zeta^{j^{2}}=\zeta$,

     * •

$\lambda$ is an isomorphism $\delta_{j}^{\otimes
n}\xrightarrow{\sim}\mathbf{1},$

     * •

$F_{X,\lambda}$ is the tensor functor obtained from mapping the generator
$\delta_{1}\mapsto X$ and the canonical isomorphism $\delta_{1}^{\otimes
n}\xrightarrow{\sim}\mathbf{1}$ to $\lambda$.

  2. (2)

Morphisms: Given two automorphisms
$F_{X,\lambda},F_{Y,\lambda^{\prime}}:\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\xrightarrow{\sim}\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$,
where $X\simeq\delta_{j}$ and $Y\simeq\delta_{k}$ for some
$j,k\in\mathbb{Z}_{n}^{\times}$, we have that

$\displaystyle\operatorname{Hom}_{\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})}(F_{X,\lambda},F_{Y,\lambda^{\prime}})\simeq\begin{cases}\Theta&\text{if
}j=k,\\\ 0&\text{if }j\neq k,\end{cases}$

where $\Theta$ denotes the set of $n$-th roots of unity. Moreover,

$\displaystyle\operatorname{Aut}_{\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})}(F_{X,\lambda})\simeq\mathbb{Z}_{n},$

for all
$F_{X,\lambda}\in\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})$.

## Acknowledgements

I want to thank my advisor Victor Ostrik for presenting this problem to me and
for his guidance and advice.

##  2\. Preliminaries

We work over an algebraically closed field $\mathbf{k}$ of characteristic
zero. We denote the ring of non-negative integers by $\mathbb{N}$, and by
$\mathbb{Z}_{n}$ and $\mathbb{Z}_{n}^{\times}$ the rings of integers mod $n$
and its units, respectively. For a category $\operatorname{\mathcal{C}}$,
$\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ denotes the hom space
from $X$ to $Y$.

We recall some useful definitions regarding monoidal and tensor categories. We
refer the reader to [ENO05, EGNO15] for a more detailed study of these topics.

###  2.1. Monoidal categories

A _monoidal category_ is a collection
$(\mathcal{C},\otimes,a,\textbf{1},\emph{l},\emph{r})$ where $\mathcal{C}$ is
a category, $\otimes:\mathcal{C}\times\mathcal{C}\to\mathcal{C}$ is a
bifunctor, called _tensor product_ , $\textbf{1}\in\mathcal{C}$ is the _unit
object_ , and $a_{X,Y,Z}:(X\otimes Y)\otimes
Z\xrightarrow{\sim}X\otimes(Y\otimes Z),\
r_{X}:X\otimes\textbf{1}\xrightarrow{\sim}X,\ \text{and}\
l_{X}:\textbf{1}\otimes X\xrightarrow{\sim}X$ are natural isomorphisms, for
all $X,Y,Z\in\mathcal{C}$, called _associativity_ and _unit isomorphisms_ ,
respectively, such that the diagrams

${((X\otimes Y)\otimes Z)\otimes W}$${(X\otimes Y)\otimes(Z\otimes
W)}$${X\otimes(Y\otimes(Z\otimes W))}$${(X\otimes(Y\otimes Z))\otimes
W}$${X\otimes((Y\otimes Z)\otimes W),}$$\scriptstyle{a_{X\otimes
Y,Z,W}}$$\scriptstyle{a_{X,Y,Z}\otimes id_{W}}$$\scriptstyle{a_{X,Y,Z\otimes
W}}$$\scriptstyle{id_{X}\otimes a_{Y,Z,W}}$$\scriptstyle{a_{X,Y\otimes Z,W}}$

and

${(X\otimes\textbf{1})\otimes Y}$${X\otimes(\textbf{1}\otimes Y)}$${X\otimes
Y.}$$\scriptstyle{a_{X,\textbf{1},Y}}$$\scriptstyle{r_{X}\otimes
id_{Y}}$$\scriptstyle{id_{X}\otimes l_{Y}}$

commute for all $X,Y,Z,W\in\operatorname{\mathcal{C}}$.

A _monoidal subcategory_ of $\operatorname{\mathcal{C}}$ is a monoidal
category $(\mathcal{D},\otimes,a,\textbf{1},\emph{l},\emph{r})$, where
$\mathcal{D}\subset\mathcal{C}$ is a full subcategory such that
$\textbf{1}\in\mathcal{D}$, and $\mathcal{D}$ is closed by tensor product. For
any $X$ in a monoidal category $\operatorname{\mathcal{C}}$, we can consider
the _monoidal subcategory generated by $X$_, which we will denote by $\langle
X\rangle$, with objects $X^{\otimes n}$ for all $n\in\mathbb{N}$.

Given two monoidal categories
$(\mathcal{C},\otimes,a,\textbf{1},\emph{l},\emph{r})$ and
$(\mathcal{C^{\prime}},\otimes^{\prime},a^{\prime},\textbf{1}^{\prime},\emph{l}\
^{\prime},\emph{r}^{\prime})$, a _monoidal functor_
$\operatorname{\mathcal{C}}\to\operatorname{\mathcal{C}}^{\prime}$ is a triple
$(F,J,u),$ where $F:\mathcal{C}\to\mathcal{C^{\prime}}$ is a functor,
$u:\textbf{1}^{\prime}\xrightarrow{\sim}F(\textbf{1})$ is an isomorphism, and
$J:\otimes^{\prime}\circ(F\times F)\to F\circ\otimes$ is a natural isomorphism
satisfying certain commutative diagrams, see for example [EGNO15, Definition
2.4.1]. A monoidal functor $F:\operatorname{\mathcal{C}}\to\mathcal{D}$ is
said to be _full_ (respectively, _faithful_) if
$\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)\xrightarrow{F}\operatorname{Hom}_{\mathcal{D}}(F(X),F(Y))$
is surjective (respectively, injective) for all
$X,Y\in\operatorname{\mathcal{C}}.$ We say $F$ is _essentially surjective_ if
for all $Y\in\mathcal{D}$ there exists $X\in\operatorname{\mathcal{C}}$ such
that $F(X)\simeq Y$; $F$ is an _equivalence_ if it is both essentially
surjective and fully-faithful.

####  2.1.1. Duals

Let $\operatorname{\mathcal{C}}$ be a monoidal category and let
$X\in\operatorname{\mathcal{C}}$. A _left dual_ of $X$ (if it exists) is a
triple $(X^{*},\operatorname{ev}_{X},\operatorname{coev}_{X})$ where $X^{*}$
is an object in $\operatorname{\mathcal{C}}$ and
$\operatorname{ev}_{X}:X^{*}\otimes X\to\textbf{1}$ and
$\operatorname{coev}_{X}:\textbf{1}\to X\otimes X^{*}$ are morphisms in
$\operatorname{\mathcal{C}}$ satisfying that the following compositions yield
the identity of $X$ and $X^{*}$, respectively:

${X\simeq\textbf{1}\otimes X}$${(X\otimes X^{*})\otimes
X}$${X\otimes(X^{*}\otimes X)}$${X\otimes\textbf{1}\simeq X,}$${X^{*}\simeq
X^{*}\otimes\textbf{1}}$${X^{*}\otimes(X\otimes X^{*})}$${(X^{*}\otimes
X)\otimes X^{*}}$${\textbf{1}\otimes X^{*}\simeq
X^{*}.}$$\scriptstyle{\operatorname{coev}_{X}\otimes\text{id}_{X}}$$\scriptstyle{a_{X,X^{*},X}}$$\scriptstyle{\text{id}_{X}\otimes\operatorname{ev}_{X}}$$\scriptstyle{\text{id}_{X^{*}}\otimes\operatorname{coev}_{X}}$$\scriptstyle{a^{-1}_{X^{*},X,X^{*}}}$$\scriptstyle{\operatorname{ev}_{X^{*}}\otimes\text{id}_{X^{*}}}$

We call the morphisms $\operatorname{ev}_{X}$ and $\operatorname{coev}_{X}$
_evaluation_ and _co-evaluation_ , respectively. The definition of a _right
dual_ for $X$ is analogous. A monoidal category is said to be _rigid_ if every
object has a left and right dual.

An object $X$ in a rigid category $\operatorname{\mathcal{C}}$ is said to be
_invertible_ if its evaluation and coevaluation morphisms are isomorphisms. A
fusion category is said to be _pointed_ if all simple objects are invertible.

A _2-group_ is a rigid monoidal category in which every object is invertible
and all morphisms are isomorphisms. A $2$-group $\mathbf{G}$ is determined up
to monoidal equivalence by a triple
$(\pi_{0}(\mathbf{G}),\pi_{1}(\mathbf{G}),a),$ where $\pi_{0}(\mathbf{G})$ is
the group of isomorphism classes of objects in $\mathbf{G}$,
$\pi_{1}(\mathbf{G})$ is the group of automorphisms of the unit object, and
$a\in H^{3}(\pi_{0}(\mathbf{G}),\pi_{1}(\mathbf{G}))$ [BL04].

Given a monoidal category $\operatorname{\mathcal{C}}$, let
$\operatorname{Aut}_{\otimes}(\operatorname{\mathcal{C}})$ be the monoidal
category with objects the tensor autoequivalences of
$\operatorname{\mathcal{C}}$, morphisms monoidal natural isomorphisms, and
tensor product given by composition. This makes
$\operatorname{Aut}_{\otimes}(\operatorname{\mathcal{C}})$ a $2$-group, with
invariants $\pi_{0}(\operatorname{Aut}_{\otimes}(\operatorname{\mathcal{C}}))$
the group of isomorphism classes of automorphisms of
$\operatorname{\mathcal{C}}$ and
$\pi_{1}(\operatorname{Aut}_{\otimes}(\operatorname{\mathcal{C}}))$ the group
of monoidal natural isomorphisms of the identity functor.

####  2.1.2. Additive envelope

We call a monoidal category $\mathcal{C}$ _$\mathbf{k}$ -linear_ if for each
pair of objects $X,Y\in\mathcal{C}$ the set
$\operatorname{Hom}_{\mathcal{C}}(X,Y)$ has a k-module structure such that
composition of morphisms is k-bilinear. We say $\mathcal{C}$ is _additive_ if
it is k-linear and for every finite sequence of objects $X_{1},\dots,X_{n}$ in
$\mathcal{C}$, there exists their direct sum $X_{1}\oplus\dots\oplus X_{n}$ in
$\mathcal{C}$.

Given a monoidal category $\operatorname{\mathcal{C}}$, we can construct its
_additive envelope_ $\mathcal{A(C)}$ which has objects finite formal sums
$X_{1}\oplus\dots\oplus X_{m}$ of objects in $\mathcal{C}$, and morphisms
defined as follows. For every $X_{1}\oplus\dots X_{m},Y_{1}\oplus\dots Y_{n}$
in $\mathcal{A(C)}$,

$\operatorname{Hom}_{\mathcal{A(C)}}(X_{1}\oplus\dots
X_{m},Y_{1}\oplus\dots\oplus
Y_{n}):=\bigoplus\limits_{i,j}\operatorname{Hom}_{\mathcal{C}}(X_{i},Y_{j}).$

Composition of morphisms is given by matrix multiplication.

###  2.2. Tensor categories

A _tensor category_ $\mathcal{C}$ is a k-linear locally finite rigid abelian
monoidal category, such that
$\operatorname{End}_{\mathcal{C}}(\textbf{1})\simeq\textbf{k}$, see e.g.
[SR72] or [EGNO15]. For tensor categories $\mathcal{C}$ and $\mathcal{D}$, a
_tensor functor_ $F:\mathcal{C}\to\mathcal{D}$ is a k-linear exact and
faithful monoidal functor.

A _fusion category_ is a tensor category which is semisimple and has a finite
number of isomorphism classes of simple objects.

###  2.3. Pointed categories

For a group $G$, we can consider the category $\operatorname{\text{Vec}_{G}}$
of finite-dimensional $G$-graded vector spaces. That is, objects are vector
spaces $V$ with a decomposition $V=\bigoplus\limits_{g\in G}V_{g}$ and
morphisms are linear transformations that preserve the grading. Then
$\operatorname{\text{Vec}_{G}}$ is a monoidal category with tensor product
given by

$(V\otimes W)_{g}=\bigoplus_{x,y\in G\ :\ xy=g}V_{x}\otimes W_{y},$

unit object $\delta_{1}$ defined by $(\delta_{1})_{1}=\textbf{k}$ and
$(\delta_{1})_{g}=0$ for $g\neq 1$, and associativity given by the identity.

More generally, given a 3-cocycle $\omega:G\times G\times
G\to\textbf{k}^{\times}$, we have the monoidal category
$\operatorname{\text{Vec}_{G}^{\omega}}$ which modifies the associativity
isomorphism in $\operatorname{\text{Vec}_{G}}$ in the following way. For
$U,V,W\in\operatorname{\text{Vec}_{G}^{\omega}},$ define
$a_{U,V,W}^{\omega}:(U\otimes V)\otimes W\xrightarrow{\sim}U\otimes(V\otimes
W)$ as the linear extension of

$a_{U_{g},V_{h},W_{l}}^{\omega}:=\omega(g,h,l)\text{ id}_{U_{g}\otimes
V_{h}\otimes W_{l}}:(U_{g}\otimes V_{h})\otimes
W_{l}\xrightarrow{\sim}U_{g}\otimes(V_{h}\otimes W_{l}),$

for all $g,h,l\in G.$

When $G$ is a finite group, $\operatorname{\text{Vec}_{G}^{\omega}}$ is a
rigid fusion category. Here the simple objects are given by $\delta_{g}$
defined by $(\delta_{g})_{h}:=0$ if $h\neq g$, and
$(\delta_{g})_{g}:=\mathbf{k},$ for all $g\in G$. The dual of a simple object
$\delta_{g}$ is $(\delta_{g})^{*}=\delta_{g^{-1}}$, for all $g\in G$. In
particular, all the simple objects in $\operatorname{\text{Vec}_{G}^{\omega}}$
are invertible, and so $\operatorname{\text{Vec}_{G}^{\omega}}$ is pointed.

It is well known that any pointed fusion category is equivalent to
$\operatorname{\text{Vec}_{G}^{\omega}}$ for some finite group $G$ and
3-cocycle $\omega$ [ENO05]; cohomologically equivalent $\omega$’s give rise to
equivalent categories $\operatorname{\text{Vec}_{G}^{\omega}}$. Hence pointed
fusion categories are classified in terms of finite group data. We will assume
without loss of generality that the cocycle $\omega$ is normalized, i.e.,
$\omega(g,1,h)=1$, since any 3-cocycle $\omega$ is cohomologous to a
normalized one. This will help with computations, as it implies that
$l_{\delta_{g}}=r_{\delta_{g}}=\text{id}_{\delta_{g}}$, for all $g\in G$.

##  3\. Objects $X$ with an isomorphism $X^{\otimes n}\simeq\mathbf{1}$

Let $\operatorname{\mathcal{C}}$ be a tensor category and $X$ an object in
$\operatorname{\mathcal{C}},$ and suppose that there exists an isomorphism
$\lambda:X^{\otimes n}\xrightarrow{\sim}\textbf{1}$ for some $n\in\mathbb{N}.$
Since

$\displaystyle\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X^{\otimes(n+1)},X)\simeq\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,X)\simeq\mathbf{k},$

there exists a constant $\xi\in\mathbf{k}$ such that the diagram

(1) ${X^{\otimes(n+1)}}$${X\otimes\textbf{1}}$${\textbf{1}\otimes
X}$${X}$$\scriptstyle{\lambda\otimes\text{id}}$$\scriptstyle{\xi(\text{id}\otimes\lambda)}$$\scriptstyle{r_{X}}$$\scriptstyle{l_{X}}$

commutes. A priory, the constant $\xi$ depends on the choice of isomorphism
$X^{\otimes n}\xrightarrow{\sim}\textbf{1}.$ We show below this is not the
case.

######  Lemma 3.1.

The constant $\xi$ in Diagram (1) is independent of the choice of isomorphism
$X^{\otimes n}\xrightarrow{\sim}\textbf{1}$.

###### Proof.

Let $\lambda_{1}$ and $\lambda_{2}$ be isomorphisms $X^{\otimes
n}\xrightarrow{\sim}\textbf{1},$ and let $\xi_{1},\xi_{2}\in\mathbf{k}$ be the
respective constants making Diagram (1) commute. That is,

$\displaystyle
l_{X}(\lambda_{1}\otimes\text{id})=\xi_{1}r_{X}(\text{id}\otimes\lambda_{1})$
and $\displaystyle
l_{X}(\lambda_{2}\otimes\text{id})=\xi_{2}r_{X}(\text{id}\otimes\lambda_{2}).$

We show that $\xi_{1}=\xi_{2}$. By Schur’s Lemma, there exists
$\alpha\in\mathbf{k}$ such that $\lambda_{1}=\alpha\lambda_{2}$. Then

$\displaystyle l_{X}(\lambda_{1}\otimes\text{id})$
$\displaystyle=\xi_{1}r_{X}(\text{id}\otimes\lambda_{1})=\alpha\xi_{1}r_{X}(\text{id}\otimes\lambda_{2})$
$\displaystyle=\alpha\xi_{1}\xi_{2}^{-1}l_{X}(\lambda_{2}\otimes\text{id})=\xi_{1}\xi_{2}^{-1}l_{X}(\lambda_{1}\otimes\text{id}),$

where in the first and third equalities we are using Diagram (1) for
$\lambda_{1}$ and $\lambda_{2}$, respectively. It follows that
$\xi_{1}\xi_{2}^{-1}=1,$ or equivalentely, $\xi_{1}=\xi_{2}.$ ∎

Since $\xi$ does not depend on the isomorphism $X^{\otimes
n}\xrightarrow{\sim}\mathbf{1}$, we will sometimes refer to it as _the
constant associated to $X$_. We have the following property.

######  Lemma 3.2.

Let $\xi$ be as in Diagram $\eqref{main diagram}.$ Then $\xi$ is an
$n^{\text{th}}$ root of unity.

###### Proof.

Fix an isomorphism $\lambda:X^{\otimes n}\xrightarrow{\sim}\textbf{1}$. We
compute $\lambda\otimes\lambda:X^{\otimes n}\otimes X^{\otimes
n}\to\textbf{1}$ in two different ways, omitting the left and right unit
isomorphisms for simplicity.

On one hand, we will show by induction on $k$ that
$\lambda\otimes\lambda=\xi^{k}\lambda(\text{id}^{\otimes
k}\otimes\lambda\otimes\text{id}^{\otimes(n-k)})$ for all $0\leq k\leq n.$ By
functoriality of the tensor product,

$\lambda\otimes\lambda=\lambda\ (\lambda\otimes\text{id}^{\otimes n}),$

so the statement is true for $k=0$. Assume then that the claim holds for
$k<n$. By Diagram (1), we get that

$\displaystyle\text{id}^{\otimes
k}\otimes\lambda\otimes\text{id}^{\otimes(n-k)}=\xi\text{id}^{\otimes
k+1}\otimes\lambda\otimes\text{id}^{\otimes(n-k-1)}$

and thus using the inductive hypothesis we conclude

$\displaystyle\lambda\otimes\lambda=\xi^{k}\lambda(\text{id}^{\otimes
k}\otimes\lambda\otimes\text{id}^{\otimes(n-k)})=\xi^{k+1}\lambda(\text{id}^{\otimes
k+1}\otimes\lambda\otimes\text{id}^{\otimes(n-k-1)}),$

as desired. Hence, we have shown that

(2) $\displaystyle\lambda\otimes\lambda=\xi^{n}\lambda(\text{id}^{\otimes
n}\otimes\lambda),$

for all $n\in\mathbb{N}$.

On the other hand, again by functoriality of the tensor product, we know that

(3) $\lambda\otimes\lambda=\lambda(\text{id}^{\otimes n}\otimes\lambda).$

Combining Equations (2) and $\eqref{3},$ we obtain $\xi^{n}=1.$ ∎

For $j\in\mathbb{N}$, consider now the object $X^{\otimes j}\in\mathcal{C}$.
Given an isomorphism $\lambda:X^{\otimes n}\xrightarrow{\sim}\textbf{1},$ we
have an induced isomorphism $\lambda^{\otimes j}:X^{\otimes
nj}\xrightarrow{\sim}\textbf{1},$ so it makes sense to wonder about the
relationship between $\xi$ and the constant associated to $X^{\otimes j}$. We
arrive to the following result.

######  Lemma 3.3.

The constant associated to $X^{\otimes j}$ is $\xi^{j^{2}}.$

###### Proof.

We want to compute the constant $\zeta$ that makes the diagram

${X^{\otimes j(n+1)}}$${X^{\otimes j}\otimes\textbf{1}}$${\textbf{1}\otimes
X^{\otimes j}}$${X^{\otimes j}}$$\scriptstyle{\lambda^{\otimes
j}\otimes\text{id}}$$\scriptstyle{\zeta(\text{id}\otimes\lambda^{\otimes
j})}$$\scriptstyle{r_{X^{\otimes j}}}$$\scriptstyle{l_{X^{\otimes j}}}$

commute. We will omit the left and right unit isomorphisms for simplicity. By
Diagram $\eqref{main diagram},$

$\displaystyle\text{id}_{X}^{\otimes j}\otimes\lambda$
$\displaystyle=\xi\text{id}_{X}^{\otimes(j-1)}\otimes\lambda\otimes\text{id}_{X}.$

Repeating this step $j-1$ times, we obtain

$\displaystyle\text{id}_{X}^{\otimes j}\otimes\lambda$ $\displaystyle=\xi^{j}\
\lambda\otimes\text{id}_{X}^{\otimes j}.$

That is, switching the order of $\text{id}^{\otimes j}$ with $\lambda$ in the
tensor product results in multiplying by $\xi^{j}$, and so

$\displaystyle\text{id}_{X}^{\otimes j}\otimes\lambda^{\otimes j}$
$\displaystyle=\xi^{j^{2}}\lambda^{\otimes j}\otimes\text{id}_{X}^{j}$

since we do the switching $j$ times. Hence $\zeta=\xi^{j^{2}}$, as desired. ∎

##  4\. Application to cyclic pointed fusion categories

Let $n$ be a natural number and let $\zeta\in\mathbf{k}$ be an $n$-th root of
unity. Then $\zeta$ determines a $3$-cocyle
$\omega_{\zeta}:\mathbb{Z}_{n}\times\mathbb{Z}_{n}\times\mathbb{Z}_{n}\to\mathbf{k}^{\times}$
in the following way:

(4)
$\displaystyle\omega_{\zeta}(i,j,k)=\zeta^{\frac{i(j+k-\overline{j+k})}{n}},\
\ \text{for all }i,j,k\in\mathbb{Z}_{n},$

where for an integer $m$ we denote by $\bar{m}$ the remainder of the division
of $m$ by $n$. Moreover, all $3$-cocyles in $\mathbb{Z}_{n}$ (modulo
coboundaries) are of the form $\omega_{\zeta}$ for some $n$-th root of unity
$\zeta$, see for example [EGNO15, Example 2.6.4]. We denote by
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ the pointed fusion category
corresponding to the 3-cocycle $\omega_{\zeta}$ .

Let $X$ be a generator of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}.$ Then
there exists an isomorphism $X^{\otimes n}\xrightarrow{\sim}\textbf{1}$.
Recall that we showed in Section 3 that the constant associated to $X$ is an
$n$-th root of unity, and is independent of the choice of isomorphism
$X^{\otimes n}\xrightarrow{\sim}\textbf{1}$. So it makes sense to wonder if it
is an invariant in the category $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta},$
and what is its relation to $\zeta.$ In fact, we have the following result.

######  Lemma 4.1.

Consider the generator $\delta_{1}$ of
$\operatorname{Vec}^{\zeta}_{\mathbb{Z}_{n}}$. Then $\zeta$ is the unique
constant making the diagram

${\delta_{1}^{\otimes(n+1)}}$${\delta_{1}\otimes\mathbf{1}}$${\mathbf{1}\otimes\delta_{1}}$${\delta_{1}}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$$\scriptstyle{r_{\delta_{1}}}$$\scriptstyle{l_{\delta_{1}}}$

commute.

###### Proof.

Choose $\lambda$ to be the canonical isomorphism $\delta_{1}^{\otimes
n}\xrightarrow{\sim}\textbf{1}.$ Then the constant for which the diagram above
commutes is given by the associativity map from $\delta_{1}^{\otimes
n}\otimes\delta_{1}\to\delta_{1}\otimes\delta_{1}^{\otimes n}.$ That is, it is
determined by the following composition of associativity maps:

$\displaystyle(\dots((\delta_{1}\otimes\delta_{1})\otimes\delta_{1})\otimes\dots)\otimes\delta_{1}$
$\displaystyle\xrightarrow{a_{\delta_{1},\delta_{1},\delta_{1}}\otimes\text{id}^{\otimes(n-2)}}(\dots(\delta_{1}\otimes(\delta_{1}\otimes\delta_{1}))\otimes\dots)\otimes\delta_{1}$
$\displaystyle\xrightarrow{a_{\delta_{1},\delta_{1}^{\otimes
2},\delta_{1}}\otimes\text{id}^{\otimes(n-3)}}(\dots(\delta_{1}\otimes((\delta_{1}\otimes\delta_{1})\otimes\delta_{1})\otimes\dots)\otimes\delta_{1}$
$\displaystyle\ \ \ \ \vdots$
$\displaystyle\xrightarrow{a_{\delta_{1},\delta_{1}^{\otimes(n-1)},\delta_{1}}}\delta_{1}\otimes((\delta_{1}\otimes\delta_{1})\otimes\delta_{1})\otimes\dots)\otimes\delta_{1}).$

Recall that $a_{\delta_{1},\delta_{1}^{\otimes k},\delta_{1}}$ is determined
by multiplication by $\omega(1,k,1),$ see Equation (4). Hence the series of
compositions above is given by multiplication by the constant

$\displaystyle\prod_{k=1}^{n-1}\omega_{\zeta}(1,k,1)=\prod_{k=1}^{n-1}\zeta^{\frac{(k+1-\overline{k+1})}{n}}=\zeta,$

as desired. ∎

######  Remark 4.2.

Let $j\in\operatorname{\mathbb{Z}}_{n}$ such that $\gcd(j,n)=1$, so that
$\delta_{j}$ is a generator of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}.$
By Lemma 3.3, we know that the constant associated to $\delta_{j}$ is
$\zeta^{j^{2}}.$

##  5\. Diagramatics for $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$

In this section we find a description of the category
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ by generators and relations, and
give the corresponding diagramatics.

######  Definition 5.1.

For $\zeta\in\mathbf{k}$ and $n\in\mathbb{N},$ let $\mathcal{D}_{\zeta,n}$ be
the category defined as follows:

  * •

Objects are collections of $k$ side-by-side points:

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to18.59pt{\pgfpicture\makeatletter\hbox{\hskip-10.04283pt\lower-7.83301pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}} {}{{}}{}{{{}}
{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
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{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
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} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{
{}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
{{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} {}{{}}{}{{{}}
{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{1}\pgfsys@invoke{
}\definecolor[named]{pgffillcolor}{rgb}{0,0,1}\definecolor[named]{pgffillcolor}{rgb}{0,0,0}\pgfsys@color@gray@fill{0}\pgfsys@invoke{
}{}\pgfsys@moveto{45.52371pt}{0.0pt}\pgfsys@moveto{46.6618pt}{0.0pt}\pgfsys@curveto{46.6618pt}{0.62856pt}{46.15227pt}{1.13809pt}{45.52371pt}{1.13809pt}\pgfsys@curveto{44.89516pt}{1.13809pt}{44.38562pt}{0.62856pt}{44.38562pt}{0.0pt}\pgfsys@curveto{44.38562pt}{-0.62856pt}{44.89516pt}{-1.13809pt}{45.52371pt}{-1.13809pt}\pgfsys@curveto{46.15227pt}{-1.13809pt}{46.6618pt}{-0.62856pt}{46.6618pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{45.52371pt}{0.0pt}\pgfsys@fillstroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}\ \ ,$

for all $k\in\mathbb{N}$, with monoidal structure given by concatenation.

  * •

Morphisms: consider the set of diagrams $\mathcal{S}$ constructed by
composition (vertical stacking) and tensor product (concatenation) out of the
generating diagrams below:

(5)

For $k,l\in\mathbb{N}$, let $\mathcal{S}_{k,l}$ be the subset of $\mathcal{S}$
of diagrams going from $k$ points to $l$ points (read from top to bottom).
Define $\operatorname{Hom}_{\mathcal{D}_{\zeta,n}}(k,l)$ to be the
$\mathbf{k}$-linear span of $\mathcal{S}_{k,l}$, modulo the relations below:

  * •

Relations:

(6) (7)

and

(8)

######  Remark 5.2.

Since the non-identity generating maps are in
$\operatorname{Hom}_{\mathcal{D}_{\xi}}(0,n)$ and
$\operatorname{Hom}_{\mathcal{D}_{\xi}}(n,0)$, respectively, it follows that
$\operatorname{Hom}_{\mathcal{D}_{\xi}}(k,l)=0$ if $k$ and $l$ are not
congruent modulo $n$. Also, for all $k\in\mathbb{N}$ we have an isomorphism
$k\xrightarrow{\sim}k+n$ given by

######  Example 5.3.

For $n=2$, we have two choices $\zeta=\pm 1$. Hence the generating morphisms
for $\mathcal{D}_{\pm 1,n}$ are

with relations

Tensor product in $\mathcal{D}_{\zeta,n}$ is given by concatenation of points
and diagrams, respectively. Hence $k\otimes l=k+l$ for all $k,l\in\mathbb{N}$.
Thus the objects in the category $\mathcal{D}_{\zeta,n}$ are tensor-generated
by one point. We show next that $\mathcal{D}_{\zeta,n}$ is a rigid monoidal
category, with the following choices.

  1. (1)

_Associativity:_ $\alpha_{k,l,j}:(k\otimes l)\otimes j\to k\otimes(l\otimes
j)$ is the identity map $\text{id}_{k+l+j}$ for all $k,l,j\in\mathbb{N}$,

  2. (2)

_Unit:_ given by $0\in\mathbb{N},$ and represented by “no points”. The left
and right unit isomorphisms $0\otimes k\to k$ and $k\otimes 0\to k$ are the
identity $\text{id}_{k}$ for all $k\in\mathbb{N}$.

  3. (3)

_Duals_ : recall that for all $k\in\mathbb{N},$ we have an isomorphism
$k\xrightarrow{\sim}k+n$. Hence it is enough to define duals for $k$ such that
$0\leq k<n$. In that case, the dual of $k$ is $n-k$ with evaluation and
coevaluation maps given by:

and

respectively.

The triangle and pentagon axioms are trivially satisfied, as all morphisms
involved are identities. We show below that duals are well defined for all
$k\in\mathbb{N}$. In fact, using Relations (6) and (8), we obtain

$=\zeta^{-k+k}$$n$$\dots$

$k$$\dots$=

$k$$\dots$,

and

$n$=$\dots$

$=\zeta^{-k-(n-k)}$$n-k$$\dots$

$n-k$$\dots$,

as desired.

We show now that $\mathcal{D}_{\zeta,n}$ is a diagramatic description of
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$. That is, we show that
$\mathcal{D}_{\zeta,n}$ is monoidally equivalent to the monoidal subcategory
$\langle\delta_{1}\rangle$ of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$
generated by $\delta_{1}$, whose additive envelope is equivalent to
$\operatorname{Vec}_{\operatorname{\mathbb{Z}}_{n}}^{\zeta}$.

In Section 4 we proved that $\zeta$ is the constant associated to the
generator $\delta_{1}$. We also know by Lemma 3.1 that this constant is
independent of the choice of isomorphism $\delta_{1}^{\otimes
n}\xrightarrow{\sim}\mathbf{1}$. Fix then the canonical isomorphism
$\lambda:\delta_{1}^{\otimes n}\xrightarrow{\sim}\mathbf{1}$, and let
$\mu:\mathbf{1}\xrightarrow{\sim}\delta_{1}^{\otimes n}$ be its inverse.
Define a functor $F:\mathcal{D}_{\zeta,n}\to\langle\delta_{1}\rangle$ on
generating objects and morphisms by

(9) $\displaystyle\begin{aligned}
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}&~{}\longmapsto~{}\left(\mu:\mathbf{1}\to\delta_{1}^{\otimes
n}\right),\end{aligned}$

and extend it $\mathbf{k}$-linearly and monoidally to all morphisms. We want
$F$ to be $\mathbf{k}$-linear monoidal functor that is both essentially
surjective and fully-faithful. We prove this in what follows by a series of
Lemmas.

######  Lemma 5.4.

The functor $F$ as above is well-defined.

###### Proof.

It is enough to check that $F$ vanishes Relations (6) and (8). In fact,

$\displaystyle\mu\circ\lambda=\text{id}_{\mathbf{1}}$ and
$\displaystyle\lambda\circ\mu=\text{id}_{\delta_{1}^{\otimes n}}$

since $\lambda$ and $\mu$ are inverses, and

$\displaystyle\text{id}_{\delta_{1}}\otimes\lambda^{n}=\zeta\lambda^{n}\otimes\text{id}_{\delta_{1}}$

by Lemma 4.1. ∎

######  Lemma 5.5.

$F$ is surjective on objects.

###### Proof.

This is immediate from the definition of $F$. ∎

To show $F$ is fully-faithful, we must check that the map

$\displaystyle\operatorname{Hom}_{\mathcal{D}_{\zeta,n}}(k,l)\xrightarrow{F}\operatorname{Hom}_{\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}}(\delta_{1}^{\otimes
k},\delta_{1}^{\otimes l})$

induced by $F$ is bijective for all $k,l\in\mathbb{N}$. By duality, it is
enough to check this for

(10)
$\displaystyle\operatorname{Hom}_{\mathcal{D}_{\zeta,n}}(k,0)\xrightarrow{F}\operatorname{Hom}_{\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}}(\delta_{1}^{\otimes
k},\mathbf{1})$

for all $k\in\mathbb{N}$.

######  Lemma 5.6.

The map in Equation $\eqref{eq: surj of F}$ is surjective for all
$k\in\mathbb{N}$.

###### Proof.

When $n$ does not divide $k$, both Hom spaces in Equation (10) are zero, so
there is nothing to show. Suppose then that $n$ divides $k$, that is, that
there exists $l\in\mathbb{N}$ so that $k=ln$. Then $\delta_{1}^{\otimes
k}=\mathbf{1}$, so

$\displaystyle\operatorname{Hom}_{\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}}(\delta_{1}^{\otimes
k},\mathbf{1})\simeq\operatorname{Hom}_{\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}}(\mathbf{1},\mathbf{1})\simeq\mathbf{k}$

and it is enough to check the map in Equation (10) is non-zero. In fact, the
morphism in $\operatorname{Hom}_{\mathcal{D}_{\zeta,n}}(k,0)$ given by

is mapped via $F_{1}$ to the isomorphism $\lambda^{\otimes
l}:(\delta_{1})^{\otimes nl}\xrightarrow{\sim}\mathbf{1}.$ ∎

It remains to check that (10) is injective. For this, we inspect
$\operatorname{Hom}_{\mathcal{D}_{\zeta,n}}(k,0)$ for all $k\in\mathbb{N}$.

######  Lemma 5.7.

Let $k\in\mathbb{N}$ be divisible by $n$. Then
$\operatorname{Hom}_{\mathcal{D}_{\zeta,n}}(k,0)$ is the $\mathbf{k}-$span of

where $l\in\mathbb{N}$ is such that $k=ln$.

###### Proof.

Let $f_{n}$ and $g_{n}$ denote the generating morphisms $n\to 0$ and $0\to n$
in Equation (5), respectively. We call a map in $\mathcal{D}_{\zeta,n}$ a
_building block_ if it is a tensor product of either $f_{n}$’s and identity
maps, or $g_{n}$’s and identity maps. Then any morphism in
$\mathcal{D}_{\zeta,n}$ is given by vertical stacking of building blocks.

Let $b_{1}:k_{1}\to k_{2}$ and $b_{2}:k_{2}\to k_{3}$ be two building blocks,
constructed out of tensor products of $f_{n}$’s and $g_{n}$’s, respectively.
Then we can turn the composition $b_{2}b_{1}$ into a scalar product of a
unique block. In fact, using Relation (8) we can move any component $f_{n}$ in
$b_{1}$ left or right to cancel out with a component $g_{n}$ of $b_{2}$ using
Relation (6), until we run out of either $f_{n}$’s or $g_{n}$’s. Hence we
reduce $b_{2}b_{1}$ to a scalar multiple of a unique building block.
Analogously, if $b_{1}$ and $b_{2}$, are constructed out of tensor products of
$g_{n}$’s and $f_{n}$’s, respectively, we can do the same using Relations
$\eqref{relations 2}$ and $\eqref{relations 3}$.

Hence, any map in $\mathcal{D}_{\zeta,n}$ can be reduced to a map built with
only $f_{n}$’s or $g_{n}$’s (and identity maps). So if we want a map $k\to 0$,
it must be the case that it is built with only $f_{n}$’s and identities. Let
$f:k\to 0$, and let $b_{1},\dots,b_{l}$ be building blocks such that
$f=b_{l}\dots b_{1}$. Using Relation (8), we can write $b_{i}$ as a scalar
multiple of $f_{n}^{\otimes j_{i}}\otimes\text{id}^{\otimes n(l-(j_{1}+\dots
j_{i}))}$ for some $j_{i}\in\mathbb{N}$ such that $\sum_{i=1}^{l}j_{i}=l$, for
all $i=1,\dots,l$. Hence $f$ is a scalar multiple of the composition

$\displaystyle(f_{n}^{\otimes j_{l}})\dots(f_{n}^{\otimes
j_{2}}\otimes\text{id}^{\otimes n(l-(j_{1}+j_{2}))})(f_{n}^{\otimes
j_{1}}\otimes\text{id}^{\otimes n(l-j_{1})})$ $\displaystyle=f_{n}^{\otimes
j_{1}}\otimes f_{n}^{\otimes j_{1}}\otimes\dots\otimes f_{n}^{\otimes j_{l}}$
$\displaystyle=f_{n}^{\otimes l},$

as desired. ∎

######  Lemma 5.8.

The map in Equation (10) is injective for all $k\in\mathbb{N}$.

###### Proof.

We already know by Lemma 5.6 that
$\operatorname{Hom}_{\mathcal{D}_{\zeta,n}}(k,0)\xrightarrow{F}\operatorname{Hom}_{\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}}(\delta_{1}^{\otimes
k},\mathbf{1})$ is surjective for all $k\in\mathbb{N}$. By Lemma 5.7 both Hom-
spaces are one-dimensional, so it is also injective. ∎

######  Corollary 5.9.

The functor $F:\mathcal{D}_{\zeta,n}\to\langle\delta_{1}\rangle$ defined in
Equation (9) is an equivalence.

###### Proof.

This follows from Lemmas 5.6, 5.6 and 5.8. ∎

###  5.1. Universal property of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$

Let $\operatorname{\mathcal{C}}$ be a tensor category. Now that we have a
description of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ by generators and
relations (see Definition 5.1), to define a tensor functor
$F:\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\to\operatorname{\mathcal{C}}$
it is enough to make a choice of:

  1. (1)

an object $X\in\operatorname{\mathcal{C}}$, and

  2. (2)

an isomorphism $\lambda:X^{\otimes n}\xrightarrow{\sim}\mathbf{1}$ such that
$\text{id}\otimes\lambda=\zeta\lambda\otimes\text{id}$.

We refer to the corresponding functor by $F_{X,\lambda}$. That is,
$F_{X,\lambda}:\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\to\operatorname{\mathcal{C}}$
is the tensor functor obtained from mapping $\delta_{1}\mapsto X$ and the
canonical isomorphism $\delta_{1}^{\otimes n}\xrightarrow{\sim}\mathbf{1}$ to
$\lambda$. We then have a bijection:

$\displaystyle\begin{Bmatrix}\text{Tensor functors}\\\
\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\to\mathcal{C}\end{Bmatrix}\longleftrightarrow\begin{Bmatrix}\text{pairs
}(X,\lambda),\text{ where }X\in\operatorname{\mathcal{C}}\text{ and
}\lambda:X^{\otimes n}\xrightarrow{\sim}\mathbf{1}\\\ \ \text{is such that
}\text{id}\otimes\lambda=\zeta\lambda\otimes\text{id}\end{Bmatrix}.$
$\displaystyle\hskip 71.13188ptF_{X,\lambda}\longleftrightarrow(X,\lambda)$

Formally, we can consider the category
$\operatorname{Fun}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta},\operatorname{\mathcal{C}})$
with objects tensor functors
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\to\operatorname{\mathcal{C}}$ and
morphisms monoidal natural transformations between functors. Then the
bijection above can be promoted to an equivalence of categories in the
following way. Let $\mathcal{C}_{\zeta}$ denote the category with:

  * •

Objects are pairs $(X,\lambda),$ where $X$ is an object in
$\operatorname{\mathcal{C}}$ and $\lambda:X^{\otimes
n}\xrightarrow{\sim}\mathbf{1}$ satisfies
$\text{id}\otimes\lambda=\zeta\lambda\otimes\text{id}$.

  * •

Maps $(X,\lambda)\to(X^{\prime},\lambda^{\prime})$ are morphisms $f:X\to
X^{\prime}$ in $\operatorname{\mathcal{C}}$ such that
$\lambda^{\prime}f^{\otimes n}=\lambda$.

Then we have an equivalence of categories

$\displaystyle\Phi:\operatorname{Fun}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta},\operatorname{\mathcal{C}})$
$\displaystyle\longrightarrow\mathcal{C}_{\zeta}$ $\displaystyle
F_{X,\lambda}$ $\displaystyle\mapsto(X,\lambda)$
$\displaystyle(\tau:F_{X,\lambda}\to F_{X^{\prime},\lambda^{\prime}})$
$\displaystyle\mapsto(\tau_{\delta_{1}}:X\to X^{\prime}).$

######  Proposition 5.10.

The map defined above is in fact an equivalence of categories.

###### Proof.

To show that $\Phi$ is well defined, we need to prove that any monoidal
natural transformation $\tau:F_{X,\lambda}\to F_{X^{\prime},\lambda^{\prime}}$
satisfies $\lambda^{\prime}\tau_{\delta_{1}}^{\otimes n}=\lambda$. In fact,
this follows from naturality of $\tau$, which gives the following commutative
diagram

${X^{\otimes n}}$${X^{\prime\otimes
n}}$${\mathbf{1}}$${\mathbf{1}.}$$\scriptstyle{\lambda}$$\scriptstyle{\tau_{\delta_{1}}^{\otimes
n}}$$\scriptstyle{\lambda^{\prime}}$$\scriptstyle{\tau_{\mathbf{1}}=\text{id}}$

It is clear that $\Phi$ is surjective on objects. It remains to show that
$\Phi$ is fully-faithful. For this, we prove that

(11)
$\displaystyle\operatorname{Hom}_{\operatorname{Fun}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta},\operatorname{\mathcal{C}})}(F_{X,\lambda},F_{X^{\prime},\lambda^{\prime}})\xrightarrow{\Phi}\operatorname{Hom}_{\operatorname{\mathcal{C}}_{\zeta}}((X,\lambda),(X^{\prime},\lambda^{\prime}))$

is bijective for every suitable pair $(X,\lambda)$. Let
$f:(X,\lambda)\to(X^{\prime},\lambda^{\prime})$ be a morphism in
$\operatorname{\mathcal{C}}_{\zeta}$, and define a monoidal natural
transformation $\tau:F_{X,\lambda}\to F_{X^{\prime},\lambda^{\prime}}$ by
$\tau_{\delta_{1}}=f$ and extend monoidally everywhere else. It is a quick
check that this gives a well defined inverse
$\operatorname{Hom}_{\operatorname{\mathcal{C}}_{\zeta}}((X,\lambda),(X^{\prime},\lambda^{\prime}))\to\operatorname{Hom}_{\operatorname{Fun}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta},\operatorname{\mathcal{C}})}(F_{X,\lambda},F_{X^{\prime},\lambda^{\prime}})$
for $\Phi$, and so the statement follows. ∎

##  6\. Automorphism group of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$

In this section, we give a description of the 2-group
$\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})$ of
tensor automorphisms of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$. We
start by describing when two such categories are equivalent.

######  Theorem 6.1.

Let $\zeta$ and $\xi$ be $n$-th roots of unity. Then

$\displaystyle\operatorname{Vec}^{\zeta}_{\mathbb{Z}_{n}}\simeq\operatorname{Vec}^{\xi}_{\mathbb{Z}_{n}}$

if and only if there exists $j\in\mathbb{Z}_{n}^{\times}$ such that
$\xi^{j^{2}}=\zeta$.

###### Proof.

Recall from Proposition 5.10 that to define a functor
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\to\operatorname{Vec}_{\mathbb{Z}_{n}}^{\xi}$
it is enough to make a choice of an object
$X\in\operatorname{Vec}_{\mathbb{Z}_{n}}^{\xi}$ and an isomorphism
$\lambda:X^{\otimes n}\xrightarrow{\sim}\mathbf{1}$ such that
$\text{id}\otimes\lambda=\zeta\lambda\otimes\text{id}$. In particular, this
implies that $X$ is invertible, hence $X\simeq\delta_{j}$ for some $0\leq
j<n.$ Moreover, if we want to have an equivalence, the functor should be
surjective and thus $X\simeq\delta_{j}$ should be a generator. The statement
then follows from Remark 4.2. ∎

Now that we have a way of measuring when two cyclic pointed fusion associated
to $\mathbb{Z}_{n}$ are equivalent, we give an answer for the number of
equivalence classes for a fixed $n$.

######  Theorem 6.2.

Let $n\in\mathbb{N}$ and let $p_{1},\cdots,p_{l}$ be distinct odd primes such
that $n=2^{k_{0}}p_{1}^{k_{1}}\cdots p_{l}^{k_{l}}$ for some
$k_{0},\cdots,k_{l}\in\mathbb{N}$. There are $c(n)$ categories of the form
$\operatorname{Vec}^{\zeta}_{\mathbb{Z}_{n}}$ up to equivalence, where

$\displaystyle c(n)=\begin{cases}\prod\limits_{i=1}^{k}(2k_{i}+1)&\text{if}\
k_{0}=0,\\\ 2\prod\limits_{i=1}^{k}(2k_{i}+1)&\text{if}\ k_{0}=1,\\\
4\prod\limits_{i=1}^{k}(2k_{i}+1)&\text{if}\ k_{0}=2,\\\
4(k_{0}-1)\prod\limits_{i=1}^{k}(2k_{i}+1)&\text{if}\ k_{0}\geq 3.\end{cases}$

###### Proof.

Let $\theta$ be a primitive $n$-th root of unity. Then any pointed fusion
category associated to $\mathbb{Z}_{n}$ is of the form
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\theta^{a}}$ for some $a\in\mathbb{N}$.
By Corollary 6.1, we know that any two pointed categories
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\theta^{a}}$ and
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\theta^{b}}$ are equivalent if and only
if there exists $j\in\mathbb{Z}_{n}^{\times}$ such that
$\theta^{aj^{2}}=\theta^{b}$, or equivalentely, such that $j^{2}a\equiv b\mod
n$. Consider the action of $\mathbb{Z}_{n}^{\times}$ in $\mathbb{Z}_{n}$ given
by

$\displaystyle\mathbb{Z}_{n}^{\times}$
$\displaystyle\to\operatorname{End}(\mathbb{Z}_{n})$ $\displaystyle j$
$\displaystyle\mapsto(a\mapsto j^{2}a).$

That is, the action of left multiplication by squares. Then counting
equivalence classes of pointed fusion categories associated to
$\mathbb{Z}_{n}$ reduces to computing the number of orbits of this action. The
rest of the proof is dedicated to this computation.

Assume for now that $n=p^{k}$ for some prime $p$ and $k\in\mathbb{N}$, and let
$x,y\in\mathbb{Z}_{p^{k}}$. Then one can get from $x$ to $y$ multiplying by a
unit if and only if $x$ and $y$ are divisible by the same power of $p$. So if
we are looking at the action of multiplication by squares of units, it is
enough to compute the orbits of the action restricted to
$\mathcal{H}_{i}:=\\{x\in\mathbb{Z}_{p^{k}}/x\text{ is divisible by
}p^{i}\text{ but not }p^{i+1}\\}$ for all $i=0,\dots,k-1$.

Let $x,y\in\mathcal{H}_{i}$. Then there exist
$x_{1},y_{1}\in\mathbb{Z}_{p^{k-i}}^{\times}$ such that $x=p^{i}x_{1}$ and
$y=p^{i}y_{1}$. So $x,y$ are in the same orbit if and only if there exists
$j\in\mathbb{Z}_{p^{k}}^{\times}$ such that

(12) $\displaystyle p^{i}x_{1}\equiv j^{2}p^{i}y_{1}\mod{p^{k}}.$

That is,

$\displaystyle x_{1}\equiv j^{2}y_{1}\mod{p^{k-i}}$

which is equivalent to

$\displaystyle y_{1}y_{2}^{-1}\equiv j^{2}\mod{p^{k-i}}.$

So if $\mathcal{R}_{k-i}$ denotes the subgroup of quadratic residues of
$\mathbb{Z}_{p^{k-i}}$, the number of orbits of the action restricted to
$\mathcal{H}_{i}$ is exactly

$\displaystyle\Big{|}\mathbb{Z}_{p^{k-i}}/\mathcal{R}_{k-i}\Big{|}$

for all $0\leq i<k.$ When $p$ is odd,
$\Big{|}\mathbb{Z}_{p^{k-i}}/\mathcal{R}_{k-i}\Big{|}=2$ for all $0\leq i<k$,
and this together with the fact that $\mathcal{H}_{k}=\\{0\\}$ gives a total
number of $2k+1$ orbits. If $p=2,$ then

$\displaystyle\Big{|}\mathbb{Z}_{p^{k-i}}/\mathcal{R}_{k-i}\Big{|}=\begin{cases}1&\text{if}\
i=k-1\\\ 2&\text{if}\ i=k-2\\\ 4&\text{if}\ 0\leq i\leq k-3.\end{cases}$

Hence if $k=1$ the action has exactly two orbits, if $k=2$ the action has four
orbits and if $k\geq 3$ the action has $4(k-1)$ orbits.

Lastly, note that for the general case $n=2^{k_{0}}p_{1}^{k_{1}}\cdots
p_{l}^{k_{l}}$ we have identifications

$\displaystyle\mathbb{Z}_{m}^{\times}\simeq\mathbb{Z}_{2^{k_{0}}}^{\times}\oplus\mathbb{Z}_{p_{1}^{k_{1}}}^{\times}\oplus\cdots\oplus\mathbb{Z}_{p_{l}^{k_{l}}}^{\times}$

and

$\displaystyle\operatorname{End}(\mathbb{Z}_{m})\simeq\operatorname{End}(\mathbb{Z}_{2^{k_{0}}})\oplus\operatorname{End}(\mathbb{Z}_{p_{1}^{k_{1}}})\oplus\cdots\oplus\operatorname{End}(\mathbb{Z}_{p_{l}^{k_{l}}}).$

Thus if we consider the action of multiplication by a square, the orbit of an
element $x=(x_{0},\dots,x_{l})$ in $\mathbb{Z}_{m}$ is identified with the
product of the orbits of each $x_{i}\in\mathbb{Z}_{p_{i}^{n_{i}}}$. Hence the
number of orbits of the action on $\mathbb{Z}_{m}$ is the product of the
number of orbits restricted to each of the terms $\mathbb{Z}_{p_{i}^{n_{i}}}$
which we computed previously, and the result follows. ∎

######  Remark 6.3.

The first 10 terms of the sequence $c(n)$ are $1,2,3,4,3,6,3,8,5,6$. See
sequence A092089 at OEIS for more information.

We now wish to give a description of the 2-group
$\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})$,
with objects monoidal automorphisms of
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ and morphisms their monoidal
natural isomorphisms.

It follows from the previous results in this section that monoidal
automorphisms of $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$ are of the form
$F_{X,\lambda}$, where $X\simeq\delta_{j}$ for some
$j\in\mathbb{Z}_{n}^{\times}$ such that $\zeta^{j^{2}}=\zeta$, $\lambda$ is
any isomorphism $\delta_{j}^{\otimes n}\xrightarrow{\sim}\mathbf{1}$, and
$F_{X,\lambda}$ is the tensor functor obtained from mapping $\delta_{1}\mapsto
X$ and the canonical isomorphism $\delta_{1}^{\otimes
n}\xrightarrow{\sim}\mathbf{1}$ to $\lambda$, see Section 5.1.

For the rest of this section, let $X\simeq\delta_{j}$ and $Y\simeq\delta_{k}$
in $\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta},$ for
$j,k\in\mathbb{Z}_{n}^{\times}$ satisfying
$\zeta^{j^{2}}=\zeta=\zeta^{k^{2}}$, and let
$F_{X,\lambda},F_{Y,\lambda^{\prime}}$ be two automorphisms
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\to\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$.

######  Proposition 6.4.

If $j\neq k$, then
$\operatorname{Hom}_{\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})}(F_{X,\lambda},F_{Y,\lambda^{\prime}})=0.$

###### Proof.

To define a natural transformation $F_{j,\lambda}\to F_{k,\lambda^{\prime}}$
we need to choose a map $\tau_{\delta_{1}}:X\to Y$ in
$\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$. If $j\not\equiv k\mod n,$ the
only possible choice is zero. ∎

######  Proposition 6.5.

If $j=k$, then
$\operatorname{Hom}_{\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})}(F_{X,\lambda},F_{Y,\lambda^{\prime}})\simeq\Theta$
as sets, where $\Theta$ denotes the set of $n$-th roots of unity.

###### Proof.

Since $j=k$, we have that $X\simeq Y$. Fix an isomorphism
$\Gamma:X\xrightarrow{\sim}Y$. By Proposition 5.10, defining a monoidal
transformation $\tau:F_{X,\lambda}\xrightarrow{\sim}F_{Y,\lambda^{\prime}}$ is
equivalent to choosing a map $\tau_{\delta_{1}}:X\to Y$ such that
$\lambda^{\prime}\tau_{\delta_{1}}^{\otimes n}=\lambda$. Now, since
$\lambda,\lambda^{\prime}\Gamma^{\otimes n}\in\operatorname{Hom}(X^{\otimes
n},\mathbf{1})\simeq\mathbf{k},$ there exists $\alpha\in\mathbf{k}^{\times}$
such that $\lambda=\alpha\lambda^{\prime}\Gamma^{\otimes n}$. Hence if
$\lambda^{\prime}\tau_{\delta_{1}}^{\otimes n}=\lambda$, we get that

$\displaystyle\lambda^{\prime}\tau_{\delta_{1}}^{\otimes
n}=\alpha\lambda^{\prime}\Gamma^{\otimes n},$

and so

(13) $\displaystyle\tau_{\delta_{1}}^{\otimes n}=\alpha\Gamma^{\otimes n}.$

Since
$\Gamma^{-1}\tau_{\delta_{1}}\in\operatorname{Hom}(X,X)\simeq\mathbf{k}$, then
$\Gamma^{-1}\tau_{\delta_{1}}$ is a multiple of the identity, and so it
follows from Equation (13) that

$\tau_{\delta_{1}}=\alpha_{n}\Gamma,$

where $\alpha_{n}$ is some $n$-th root of $\alpha$.

Note that this does not depend on the choice of isomorphism $\Gamma:X\to Y$.
In fact, if $\Gamma^{\prime}:X\to Y$ is another isomorphism, then since
$\operatorname{Hom}(X,Y)\simeq\mathbf{k}$ there exists
$a\in\mathbf{k}^{\prime}$ such that $\Gamma=a\Gamma^{\prime}$. Following the
same computations for $\Gamma^{\prime}$, we get that
$\tau_{\delta_{1}}=\beta\Gamma^{\prime}$ for $\beta$ and $n$-th root of
$\alpha a^{n}$, and so again $\tau_{\delta_{1}}=\alpha_{n}\Gamma$ for
$\alpha_{n}$ an $n$-th root of $\alpha$.

Since we now know that our computations do not depend on the choice of
$\Gamma$, we may choose $\Gamma$ so that
$\lambda=\lambda^{\prime}\Gamma^{\otimes n}.$ Hence

$\displaystyle\tau_{\delta_{1}}=\theta_{n}^{k}\Gamma,\text{ for
}k=0,1\dots,n-1,$

are all the possible choices for $\tau_{\delta_{1}}$, where $\theta$ is a
primitive $n$-th root of unity. ∎

######  Remark 6.6.

It follows from the proof that
$\operatorname{Aut}_{\otimes}(F_{X,\lambda})\simeq\mathbb{Z}_{n}$ as groups.

We summarize the results of this section in the following theorem.

######  Theorem 6.7.

The 2-group
$\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})$
has:

  1. (1)

Objects: monoidal automorphisms
$F_{X,\lambda}:\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\xrightarrow{\sim}\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$,
where

     * •

$X\simeq\delta_{j}$ for some $j\in\mathbb{Z}_{n}^{\times}$ such that
$\zeta^{j^{2}}=\zeta$,

     * •

$\lambda$ is an isomorphism $\delta_{j}^{\otimes
n}\xrightarrow{\sim}\mathbf{1},$

     * •

$F_{X,\lambda}$ is the tensor functor obtained from mapping the generator
$\delta_{1}\mapsto X$ and the canonical isomorphism $\delta_{1}^{\otimes
n}\xrightarrow{\sim}\mathbf{1}$ to $\lambda$.

  2. (2)

Morphisms: Given two automorphisms
$F_{X,\lambda},F_{Y,\lambda^{\prime}}:\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}\xrightarrow{\sim}\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}$,
where $X\simeq\delta_{j}$ and $Y\simeq\delta_{k}$ for some
$j,k\in\mathbb{Z}_{n}^{\times}$, we have that

$\displaystyle\operatorname{Hom}_{\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})}(F_{X,\lambda},F_{Y,\lambda^{\prime}})\simeq\begin{cases}\Theta&\text{if
}j=k,\\\ 0&\text{if }j\neq k,\end{cases}$

where $\Theta$ denotes the set of $n$-th roots of unity. Moreover,

$\displaystyle\operatorname{Aut}_{\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})}(F_{X,\lambda})\simeq\mathbb{Z}_{n},$

for all
$F_{X,\lambda}\in\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})$.

######  Corollary 6.8.

The $2$-group
$\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta})$
has the following invariants:

  1. (1)

$\pi_{0}(\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}))=\\{j\in\mathbb{Z}_{n}/\zeta^{j^{2}}=\zeta\\}<\mathbb{Z}_{n}$,
and

  2. (2)

$\pi_{1}(\operatorname{Aut}_{\otimes}(\operatorname{Vec}_{\mathbb{Z}_{n}}^{\zeta}))\simeq\mathbb{Z}_{n}$.

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