# Remarks on symmetric fusion categories of low rank in positive
characteristic

Agustina Czenky  Department of Mathematics, University of Oregon
<EMAIL_ADDRESS>
###### Abstract.

We give lower bounds for the rank of a symmetric fusion category in
characteristic $p\geq 5$ in terms of $p$. We prove that the second Adams
operation $\psi_{2}$ is not the identity for any non-trivial symmetric fusion
category, and that symmetric fusion categories satisfying
$\psi_{2}^{a}=\psi_{2}^{a-1}$ for some positive integer $a$ are super
Tannakian. As an application, we classify all symmetric fusion categories of
rank 3 and those of rank 4 with exactly two self dual simple objects.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Fusion categories
      1. 2.1.1 Frobenius-Perron dimension
    2. 2.2 Symmetric fusion categories
      1. 2.2.1 The second Adams operation
    3. 2.3 Non-degenerate fusion categories
    4. 2.4 Verlinde categories
      1. 2.4.1 The Verlinde fiber functor
  3. 3 Bounds for the ranks of symmetric fusion categories
  4. 4 Some properties of the Adams operation
    1. 4.1 Adams operation in $\operatorname{Ver}_{p}$
    2. 4.2 Powers of the Adams operation
    3. 4.3 Symmetric fusion categories with exactly two self-dual simple objects
  5. 5 Rank 3 symmetric fusion categories
  6. 6 Rank 4 symmetric fusion categories
    1. 6.1 Exactly two self-dual simple objects
    2. 6.2 All simple objects are self-dual

##  1\. Introduction

Fix an algebraically closed field k of characteristic $p\geq 0$. A _symmetric
fusion category_ $\mathcal{C}$ over k is a fusion category endowed with a
braiding $c_{X,Y}:X\otimes Y\to Y\otimes X$ such that
$c_{Y,X}c_{X,Y}=\operatorname{id}_{X\otimes Y}$ for all $X,Y\in\mathcal{C}$,
see [EGNO, Definition 8.1.12]. A well known theorem by Deligne [D] implies
that a symmetric fusion category in characteristic $0$ is _super Tannakian_ ,
that is, admits a symmetric tensor functor to the category
$\operatorname{sVec}$ of super vector spaces. As a consequence of this
theorem, a symmetric fusion category over a field of characteristic $0$ is
equivalent to the category $\operatorname{Rep}_{\textbf{k}}(G,z)$ of finite-
dimensional representations of a finite group $G$. Here $z\in G$ is a central
element of order $2$ that modifies the braiding, see [D, 8.19]. This results
gives a classification of symmetric fusion categories in characteristic zero
in terms of group data.

Examples of symmetric fusion categories in positive characteristic are the
Verlinde categories $\operatorname{Ver}_{p}$, defined as the
semisimplification of the category of finite dimensional k-representations of
the cyclic group $\mathbb{Z}_{p}$ for $p$ a positive prime, see [O2, Section
3.2]. For $p\geq 5$, these categories have no fiber functor to
$\operatorname{Vec}$ or $\operatorname{sVec}$, hence cannot be obtained as the
category of representations of a finite group, see [BEO].

An important result by Victor Ostrik in [O2] gives a new version of Deligne’s
Theorem for the case of symmetric fusion categories in positive
characteristic. He proved that any symmetric fusion category $\mathcal{C}$ in
characteristic $p>0$ admits a Verlinde fiber functor, that is, a symmetric
tensor functor

$\displaystyle F:\mathcal{C}\to\operatorname{Ver}_{p}.$

As a consequence, any k-linear symmetric fusion category is equivalent to the
category $\operatorname{Rep}_{\operatorname{Ver}_{p}}(G,\epsilon)$ of
representations of some finite group scheme $G$ in $\operatorname{Ver}_{p}$
[O2, Corollary 1.6]. However, this statement does not give an explicit
classification for $p\geq 5$, since the classification of finite group schemes
$G$ in $\operatorname{Ver}_{p}$ such that
$\operatorname{Rep}_{\operatorname{Ver}_{p}}(G)$ is semisimple is not known.

When $p>0$, Nagata [DG] and Masuoka [M] give a classification of finite group
schemes $G$ in $\operatorname{Vec}$ and $\operatorname{sVec}$, respectively,
such that $\operatorname{Rep}_{\textbf{k}}(G)$ is semisimple. This yields a
reasonable classification of symmetric fusion categories in the super
Tannakian case. Note that when $\operatorname{char}(\mathbf{k})=2$ or $3$,
symmetric fusion categories over k are Tannakian and super Tannakian,
respectively, so we know their classification.

In this paper we will focus on the non super Tannakian case. We will approach
the classification of symmetric fusion categories in positive characteristic
by rank, i.e., by the number of simple objects. Here is our first result.

######  Theorem 3.1.

Let $p\geq 5$. If $\mathcal{C}$ is a non super Tannakian symmetric fusion
category, then

$\operatorname{rank}(\mathcal{C})\geq\frac{p-1}{2}.$

We note that the statement does not hold for super Tannakian categories. For
example, for $p\geq 3$ the category $\operatorname{Rep}(\mathbb{Z}_{2})$ is
semisimple and has rank $2$ which is strictly less than $\frac{p-1}{2}$ for
$p>5$.

Note that equality in Theorem 3.1 is achieved by $\operatorname{Ver}_{p}^{+}$,
the fusion subcategory of $\operatorname{Ver}_{p}$ generated by simple objects
of odd index, see Section 2.4. In characteristic $5$, it is known that the
equality is only achieved by $\operatorname{Ver}_{5}^{+}$, see [EOV,
Proposition 4.6].

######  Question 1.1.

Let $p>5$ and $\mathcal{C}$ a symmetric fusion category of rank
$\frac{p-1}{2}$. If $\mathcal{C}$ is not super Tannakian, it true that
$\mathcal{C}\simeq\operatorname{Ver}_{p}^{+}$?

We give a positive answer for Question 1.1 for the case $p=7$ in Theorem 5.2.

We also know that there exist non super Tannakian symmetric fusion categories
of rank $\frac{p+3}{2}$. In fact, let $\delta\in\textbf{k}$ and consider
Karoubian envelope $\underline{\operatorname{Rep}}(O(\delta))$ of the _Brauer
category_ as defined in [D, 9.3]. Let
$\underline{\operatorname{Rep}}^{\text{s}s}(O(\delta))$ denote the
semisimplification of $\underline{\operatorname{Rep}}(O(\delta))$, i.e., the
quotient of $\underline{\operatorname{Rep}}(O(\delta))$ by the tensor ideal of
negligible morphisms, see e.g. [D, 6.1]. It turns out that when $\delta=-1$
this category contains a symmetric subcategory equivalent to
$\operatorname{Rep}(\mathbb{Z}_{2})$. The symmetric fusion category obtained
by de-equivariantization by $\mathbb{Z}_{2}$ of the neutral component of the
standard $\mathbb{Z}_{2}$-grading of
$\underline{\operatorname{Rep}}^{\text{ss}}(O(-1))$ has rank $\frac{p+3}{2}$,
see [O3] for details.

We thus have examples of non super Tannakian symmetric fusion categories in
ranks $\frac{p-1}{2}$ and $\frac{p+3}{2}$. A natural question follows.

######  Question 1.2.

Are there non super Tannakian symmetric fusion categories of rank
$\frac{p+1}{2}$?

For $p\geq 5$, the category $\operatorname{Ver}_{p}$ has precisely four fusion
subcategories: $\text{Vec},\text{sVec},\operatorname{Ver}_{p}^{+}$ and
$\operatorname{Ver}_{p}$, see [O2, Proposition 3.3]. Thus, if $\mathcal{C}$ is
not super Tannakian, its Verlinde functor
$F:\mathcal{C}\to\operatorname{Ver}_{p}$ is either surjective or its image is
$\operatorname{Ver}_{p}^{+}$. Our next result gives an improvement on the
bound for the former case.

######  Theorem 3.3.

Let $p\geq 5$ and let $\mathcal{C}$ be a symmetric fusion category with
Verlinde fiber functor $F\colon\mathcal{C}\to\operatorname{Ver}_{p}$. If $F$
is surjective then

$\operatorname{rank}(\mathcal{C})\geq p-1.$

The main tool in the proof of Theorems 3.1 and 3.3 is Galois theory. Another
useful tool for the classification of symmetric fusion categories in positive
characteristic is the _second Adams operation_. Let $p\neq 2$. For a symmetric
fusion category $\mathcal{C}$ with Grothendieck ring $\mathcal{K(C)}$, the
second Adams operation is the ring endomorphism
$\psi_{2}:\mathcal{K(C)}\to\mathcal{K(C)}$ given by

$\psi_{2}(X)=S^{2}(X)-\Lambda^{2}(X),$

for all $X$ in $\mathcal{C}$, see [EOV].

######  Theorem 4.6.

Let $p>2$ and let $\mathcal{C}$ be a symmetric fusion category. If
$\psi_{2}^{a}=\psi_{2}^{a-1}$ for some $a\geq 1$, then $\mathcal{C}$ is super
Tannakian.

The following comes as a consequence.

######  Theorem 4.8.

Let $p\neq 2$. If $\mathcal{C}$ is a non-trivial symmetric fusion category
then $\psi_{2}$ is not the identity.

We apply the second Adams operation to the problem of classification of
symmetric fusion categories of low rank in positive characteristic. In [BEO],
the second Adams operation was employed to give a complete clasification for
rank 2. We classify symmetric fusion categories of rank $3$, and symmetric
fusion categories of rank $4$ with exactly two self-dual simple objects, see
Theorems 5.2 and 6.7, respectively. We also note that by Theorem 3.1 non super
Tannakian symmetric fusion categories of rank $4$ are only possible in
characteristic $p=5$ or $7$.

Even though our results show that the second Adams operation is non-trivial
for non-trivial symmetric fusion categories, we note that it is useful for the
classification problem but definitely not sufficient on its own, see Remark
6.3.

This paper is organized as follows. A brief introduction to Verlinde
categories and the second Adams operation is given in Section 2. Proofs for
Theorems 3.1 and 3.3 are given in Section 3. In Section 4 we study properties
of the second Adams operation, give a formula for it in the Verlinde category,
and prove that it is not the identity for non trivial fusion categories.
Sections 5 and 6 describe the classification of fusion categories of rank 3
and those of rank 4 with exactly two self dual simple objects, respectively.
We also say some words about the classification of those of rank 4 where all
simple objects are self dual in Section 6.

### Acknowledgments

I am deeply grateful to my advisor Victor Ostrik for suggesting this project,
providing insightful advice in how to approach the proofs of the main results,
and for his numerous comments that made the exposition of this work much
clearer.

##  2\. Preliminaries

Throughout this paper k will denote an algebraically closed field of
characteristic $p\geq 0$.

For a ring $R$, we denote by $R_{\mathbb{Q}}$ the scalar extension
$R\otimes_{\mathbb{Z}}\mathbb{Q}$. If $z$ is a complex number, we denote by
$\mathbb{Q}(z)$ the field extension generated by $z$ over $\mathbb{Q}$, and by
$[\mathbb{Q}(z):\mathbb{Q}]$ the degree of said extension.

###  2.1. Fusion categories

We recall that a _symmetric tensor category_ $\mathcal{C}$ is a category
endowed with a functor $\otimes:\mathcal{C}\times\mathcal{C}\to\mathcal{C}$ of
tensor product, and with associativity and commutativity isomorphisms and unit
object satisfying suitable axioms, see e.g. [SR] or [EGNO].

A _fusion category_ $\mathcal{C}$ over k is a k-linear semisimple rigid tensor
category with a finite number of simple objects and finite dimensional Hom-
spaces, and such that
$\operatorname{End}_{\textbf{k}}(\textbf{1})\simeq\textbf{k}$.

A _fusion subcategory_ of a fusion category $\mathcal{C}$ is a full tensor
subcategory $\mathcal{C}^{\prime}\subset\mathcal{C}$ such that if
$X\in\mathcal{C}$ is isomorphic to a direct summand of an object of
$\mathcal{C}^{\prime}$, then $X\in\mathcal{C}^{\prime}$, see [DGNO, 2.1].

For a fusion category $\mathcal{C}$ we denote its unit object by 1 and by
$\mathcal{O(C)}$ the set of isomorphism classes of simple objects of
$\mathcal{C}$. We denote by Vec (respectively sVec) the symmetric fusion
category of finite dimensional vector spaces, (respectively super vector
spaces), over k.

For a tensor functor $F$ between fusion categories $\mathcal{C}$ and
$\mathcal{D}$ its image $F(\mathcal{C})$ is the fusion subcategory of
$\mathcal{D}$ generated by objects $F(X)$, $X\in\mathcal{C}$. The functor $F$
is called _injective_ if it is fully faithful, and _surjective_ if
$F(\mathcal{C})=\mathcal{D}$, see [ENO, 5.7]. Thus a tensor functor is an
equivalence if and only if it is both injective and surjective.

For two fusion categories $\mathcal{C}$ and $\mathcal{D}$, we can define their
_external tensor product_ , see [O2, Section 2.2], which we will denote by
$\mathcal{C}\boxtimes\mathcal{D}$.

####  2.1.1. Frobenius-Perron dimension

Let $\mathcal{C}$ be a fusion category. We will denote by $\mathcal{K(C)}$ its
Grothendieck ring, see e.g. [EGNO, 4.5]. For an object $X$ in $\mathcal{C}$ we
will use the same notation for its class $X$ in $\mathcal{K(C)}$. We recall
that there is a unique ring homomorphism
$\operatorname{FPdim}:\mathcal{K(C)}\to\mathbb{R}$ called _Frobenius-Perron
dimension_ such that $\operatorname{FPdim}(X)\geq 1$ for any object $X\neq 0$,
see [EGNO, Proposition 3.3.4]. The _Frobenius-Perron dimension_
$\operatorname{FPdim}(\mathcal{C})$ of $\mathcal{C}$ is defined as

$\displaystyle\operatorname{FPdim}(\mathcal{C})=\sum\limits_{X\in\mathcal{O(C)}}\operatorname{FPdim}(X)^{2}.$

We say that $\mathcal{C}$ is _weakly integral_ if
$\operatorname{FPdim}(\mathcal{C})$ is an integer, and _integral_ if
$\operatorname{FPdim}(X)$ is an integer for all simple $X$.

###  2.2. Symmetric fusion categories

Let $\mathcal{C}$ be a braided fusion category and denote by $c_{X,Y}$ the
braiding morphism $X\otimes Y\to Y\otimes X$. We say $\mathcal{C}$ is
_symmetric_ if

$\displaystyle c_{Y,X}c_{X,Y}=\operatorname{id}_{X\otimes Y}\ \ \text{for
all}\ X\in\ \mathcal{C},$

see [EGNO]. A _symmetric tensor functor_ between symmetric tensor categories
is a monoidal functor compatible with the commutativity isomorphism.

We say a symmetric fusion category $\mathcal{C}$ is _Tannakian_ (resp., _super
Tannakian_) if it admits a symmetric fiber functor, that is, a symmetric
tensor functor $\mathcal{C}\to\operatorname{Vec}$ (resp.,
$\mathcal{C}\to\operatorname{sVec}$), see [SR, DM, D].

For any object $X$ in a symmetric tensor category over a field of
characteristic $p>0$ , we can define its _symmetric powers_ $S^{i}(X)$ and
_exterior powers_ $\Lambda^{i}(X)$, see [EHO, 2.1].

####  2.2.1. The second Adams operation

Let $\mathcal{C}$ be a symmetric fusion category over a field of
characteristic $p\neq 2$. The _second Adams operation_
$\psi_{2}:\mathcal{K(C)}\to\mathcal{K(C)}$ is defined by

$\displaystyle\psi_{2}(X)=S^{2}(X)-\Lambda^{2}(X),$

for all $X\in\mathcal{K(C)}.$ This defines a ring endomorphism of
$\mathcal{K(C)}$, see [EOV, Lemma 4.4].

Since $X^{2}=S^{2}(X)+\Lambda^{2}(X)$ for all $X\in\mathcal{C}$, then

$\displaystyle X^{2}\equiv\psi_{2}(X)\mod 2\ \ \text{for all
}X\in\mathcal{K(C)}.$

We will use this fact repeatedly throughout this work. We also have that
$\psi_{2}$ commutes with duality, that is, $\psi_{2}(X)^{*}=\psi_{2}(X^{*})$
for all $X\in\mathcal{K(C)}$.

When studying properties of the second Adams operation, we will often look at
its scalar extension

$(\psi_{2})_{\mathbb{Q}}:=\psi_{2}\otimes
1:\mathcal{K(C)}_{\mathbb{Q}}\to\mathcal{K(C)}_{\mathbb{Q}}.$

###  2.3. Non-degenerate fusion categories

Let $\mathcal{C}$ be a fusion category. Recall that a _pivotal structure_ on
$\mathcal{C}$ is a tensor isomorphism $X\simeq X^{**}$ for any
$X\in\mathcal{C}$, see [BW, EGNO]. Associated to a pivotal structure we can
define the left and right trace of a morphism $X\to X$. The pivotal structure
is called _spherical_ if for any such morphism its right trace equals its left
trace. A _spherical fusion category_ is a fusion category equipped with a
spherical structure. In the case when $\mathcal{C}$ is symmetric, there is a
canonical choice of spherical structure given by

${X}$${X\otimes X^{*}\otimes X^{**}}$${X^{*}\otimes X\otimes
X^{**}}$${X^{**},}$$\scriptstyle{\operatorname{Id}_{X}\otimes\operatorname{coev}_{X^{*}}}$$\scriptstyle{c_{X,X^{*}}\otimes\operatorname{id}_{X^{**}}}$$\scriptstyle{\operatorname{ev}_{X}\otimes\operatorname{Id}_{X^{**}}}$

see e.g. [EGNO, Secion 9.9].

Let $\mathcal{C}$ be a spherical fusion category. We recall the definition of
_dimension_ $\dim(X)\in\textbf{k}$ of an object $X$ as the trace of its
identity morphism. This determines a ring homomorphism
$\dim:\mathcal{K(C)}\to\textbf{k}$ sending $X$ to $\dim(X)$. By [EGNO,
Proposition 4.8.4] if $X\in\mathcal{O}(C)$ then $\dim(X)\neq 0$.

The _global dimension_ $\dim(\mathcal{C})\in\textbf{k}$ of a spherical fusion
category $\mathcal{C}$ is defined as

$\displaystyle\dim(\mathcal{C})=\sum\limits_{X\in\mathcal{O}(C)}\dim(X)^{2}.$

We say $\mathcal{C}$ is _non-degenerate_ if $\dim(\mathcal{C})\neq 0$, see
[ENO, Definition 9.1]. A crucial property of non-degenerate fusion categories
is that they can be lifted to characteristic zero, see [E] and [ENO, Section
9]. It is known that for $p=0$ any fusion category is non-degenerate, see
[ENO, Theorem 2.3].

###  2.4. Verlinde categories

Let $p>0$. Let $\mathbb{Z}_{p}$ be the cyclic group of $p$ elements with
generator $\sigma$. We have an equivalence of algebras
$\textbf{k}[\mathbb{Z}_{p}]=\textbf{k}[\sigma]/(\sigma^{p}-1)=\textbf{k}[\sigma]/(\sigma-1)^{p}$.
Thus isomorphism classes of indecomposable objects in the category
$\operatorname{Rep}_{\textbf{k}}(\mathbb{Z}_{p})$ are given by the
$\mathbb{Z}_{p}$-modules $\tilde{L}_{s}:=\textbf{k}[\sigma]/(1-\sigma)^{s}$,
for $s\in\mathbb{Z}$ satisfying $1\leq s\leq p$.

The Verlinde category $\operatorname{Ver}_{p}$ is the symmetric fusion
category over k obtained by quotienting
$\operatorname{Rep}_{\textbf{k}}(\mathbb{Z}_{p})$ by the tensor ideal of
negligible morphisms, see [O2] for details. Simple objects of this category
are precisely the images of the indecomposable objects $\tilde{L}_{s}$ for
$s=1,\dots,p-1$. We denote them by $\textbf{1}=L_{1},L_{2},\dots,L_{p-1}$. The
Verlinde fusion rules are given by

$\displaystyle L_{r}\otimes
L_{s}\simeq\sum\limits_{i=1}^{\min(r,s,p-r,p-s)}L_{|r-s|+2i-1}.$

We denote by $\operatorname{Ver}_{p}^{+}$ the abelian subcategory of
$\operatorname{Ver}_{p}$ generated by $L_{i}$ for $i$ odd. By the Verlinde
fusion rules, it turns out that $\operatorname{Ver}_{p}^{+}$ is a fusion
subcategory of $\operatorname{Ver}_{p}$. For $p>2$ the fusion subcategory
generated by $L_{1}$ and $L_{p-1}$ is tensor equivalent to sVec. We have an
equivalence of categories

(2.1)
$\displaystyle\operatorname{Ver}_{p}\simeq\operatorname{Ver}^{+}_{p}\boxtimes\
\text{sVec},$

see [O2].

####  2.4.1. The Verlinde fiber functor

Let $\mathcal{C}$ be a symmetric fusion category over k of characteristic
$p>0$. The main result of [O2] states:

######  Theorem 1.

[O2, Theorem 1.5] There exists a symmetric tensor functor
$F:\mathcal{C}\to\operatorname{Ver}_{p}$.

The functor $F$ is called the _Verlinde fiber functor_. It is shown in [EOV,
Theorem 2.6] that it is unique up to a non-unique isomorphism of tensor
functors.

##  3\. Bounds for the ranks of symmetric fusion categories

In this section we prove our two main results concerning the ranks of non
super Tannakian symmetric fusion categories. Throughout this section, we
assume $\operatorname{char}(\textbf{k})=p\geq 5$. Let $z=e^{2\pi i/p}$ be a
primitive $p$-th root of unity. Recall that we denote by $\mathbb{Q}(z)$ the
field extension generated by $z$ over $\mathbb{Q}$.

Let $\mathcal{C}$ be a symmetric fusion category and consider its Verlinde
fiber functor $F:\mathcal{C}\to\operatorname{Ver}_{p}$. Recall that by [O2,
Proposition 3.3] we have an equivalence of symmetric fusion categories

$\displaystyle\operatorname{Ver}_{p}\simeq\operatorname{Ver}_{p}^{+}\boxtimes\text{
sVec}.$

Consider the monoidal (non symmetric) forgetful functor
$\operatorname{Forget}:\operatorname{sVec}\to\operatorname{Vec}$. We have a
(possibly non symmetric) tensor functor

(3.1)
$\displaystyle\tilde{F}:=(\operatorname{id}\boxtimes\operatorname{Forget})\circ
F:\mathcal{C}\to\operatorname{Ver}_{p}^{+};$

we denote also by $\tilde{F}$ the induced ring homomorphism
$\mathcal{K(C)}\to\mathcal{K}(\operatorname{Ver}_{p}^{+})$. We are interested
in studying the image of this map. By [BEO, Theorem 4.5 (iv)], we have an
isomorphism

$\displaystyle\mathcal{K}(\operatorname{Ver}_{p}^{+})_{\mathbb{Q}}\simeq\mathbb{Q}(z+z^{-1}),$

and so $\tilde{F}(\mathcal{K(C)})_{\mathbb{Q}}$ is a subfield of
$\mathcal{K}(\operatorname{Ver}_{p}^{+})_{\mathbb{Q}}$. Hence to study the
image of $\mathcal{K(C)}$ through $\tilde{F}$ we start by looking at subfields
of $\mathbb{Q}(z+z^{-1})$.

We make the usual identification of the Galois group of $\mathbb{Q}(z)$ with
the multiplicative group $\mathbb{Z}_{p}^{\times}$. This is a cyclic group
with $p-1$ elements, where $j$ acts on $\mathbb{Q}(z)$ by $j\cdot z=z^{j}$ for
all $j\in\mathbb{Z}_{p}^{\times}$. We denote the Galois group of the maximal
real subextension $\mathbb{Q}(z+z^{-1})$ of $\mathbb{Q}(z)$ by $\mathcal{G}$,
which corresponds to the quotient of $\mathbb{Z}_{p}^{\times}$ by the subgroup
$\\{\pm 1\\}$. Thus $\mathcal{G}$ is a cyclic group of order $\frac{p-1}{2}$.

By Galois correspondence, subextensions of $\mathbb{Q}(z+z^{-1})$ are in
bijection with subgroups of $\mathcal{G}$. That is, for every positive integer
$k$ that divides $\frac{p-1}{2}$ there exists a unique subextension $A_{k}$ of
$\mathbb{Q}(z+z^{-1})$ such that $[A_{k}:\mathbb{Q}]=k$, and its Galois group
is exactly the quotient of $\mathcal{G}$ by the unique subgroup $H_{m}$ of
order $m$, where $mk=\frac{p-1}{2}$. Moreover, every subextension is of this
form, and $A_{k}$ is the set of elements fixed by every element in $H_{m}$.

Consider the basis $\left\\{z^{i}+z^{-i}\right\\}_{i=1}^{\frac{p-1}{2}}$ of
$\mathbb{Q}(z+z^{-1})$. Then the group $\mathcal{G}$ (and thus also all
subgroups $H_{m}$) acts on this set freely and transitively by permutation,

$a\cdot(z^{j}+z^{-j})=z^{aj}+z^{-aj},$

for all $a\in\mathcal{G}$. So the orbits of the action of $H_{m}$ on this set
have exactly $m$ elements. Let $\mathcal{O}_{1},\dots,\mathcal{O}_{k}$ denote
the orbits of the action of $H_{m}$, and define

(3.2) $\displaystyle
x_{i}:=\sum\limits_{z^{t}+z^{-t}\in\mathcal{O}_{i}}(z^{t}+z^{-t}),$

so that $\left\\{x_{1},\dots,x_{k}\right\\}$ is a basis of $A_{k}$. Without
loss of generality, we choose the labelling so that
$z+z^{-1}=z^{p-1}+z^{-(p-1)}\in\mathcal{O}_{k}$,

######  Theorem 3.1.

Let $p\geq 5$. If $\mathcal{C}$ is a non super Tannakian symmetric fusion
category, then

$\operatorname{rank}(\mathcal{C})\geq\frac{p-1}{2}.$

###### Proof.

Consider the tensor functor
$\tilde{F}:\mathcal{C}\to\operatorname{Ver}_{p}^{+}$ as defined in 3.1.
According to [BEO, Theorem 4.5 (iv)], under the isomorphism

$\displaystyle\mathcal{K}(\operatorname{Ver}_{p}^{+})_{\mathbb{Q}}\simeq\mathbb{Q}(z+z^{-1})$

we have identifications

(3.3) $\displaystyle L_{2j+1}=\sum\limits_{l=1}^{j}(z^{2l}+z^{-2l})+1,\ \
\text{for }j=0,\dots,(p-3)/2.$

Recall that $\tilde{F}(\mathcal{K(C)})_{\mathbb{Q}}$ is a subfield of
$\mathcal{K}(\operatorname{Ver}_{p}^{+})_{\mathbb{Q}}\simeq\mathbb{Q}(z+z^{-1})$.
Now,

$\displaystyle\operatorname{rank}(\mathcal{K(C)})\geq\operatorname{rank}(\tilde{F}(\mathcal{K(C)}))=\dim(\tilde{F}(\mathcal{K(C)})_{\mathbb{Q}}),$

where $\operatorname{rank}(\mathcal{K(C)})$ refers to the rank of
$\mathcal{K(C)}$ as a free abelian group. Thus we would like to show that
$\dim(\tilde{F}(\mathcal{K(C)})_{\mathbb{Q}})=\frac{p-1}{2}$, or in other
words, that $\tilde{F}(\mathcal{K(C)})_{\mathbb{Q}}=\mathbb{Q}(z+z^{-1})$.

By our discussion at the beginning of this section, we know that
$\tilde{F}(\mathcal{K(C)})_{\mathbb{Q}}$ is of the form $A_{k}$ for some $k$
that divides $\frac{p-1}{2}$, where $A_{k}$ is the unique subextension of
order $k$. Hence to prove the statement it is enough to show that
$k=\frac{p-1}{2}$. Note that $k>1$, since if $k=1$ the image of $\tilde{F}$
would be a multiple of the identity and we would have a symmetric tensor
functor from $\mathcal{C}$ to $\operatorname{sVec}$, which would mean that
$\mathcal{C}$ is super Tannakian.

For objects $X\in\mathcal{C}$, their images through $\tilde{F}$ are objects in
$\operatorname{Ver}_{p}$ and so can be written as $\mathbb{Z}_{\geq 0}$ linear
combinations of $L_{i}$’s with $i$ odd. Then $\tilde{F}(\mathcal{K(C)})$ has a
basis of elements of these form, and thus so does $A_{k}$. That is,

(3.4) $A_{k}$ has a basis given by $\mathbb{Z}_{\geq 0}$ linear combinations
of $L_{i}$’s for $i$ odd.

For the sake of contradiction, assume that $k<\frac{p-1}{2}$. We already know
that $k>1$.

Using formula (3.3) we compute

$\displaystyle z^{t}+z^{-t}\ =\ L_{t+1}-L_{t-1}\ \ \text{for }t\text{ even,}\
2\leq t<p-1.$

On the other hand,

$\displaystyle
L_{p-2}=\sum_{l=1}^{\frac{p-3}{2}}(z^{2l}+z^{-2l})+1=-(z^{p-1}+z^{-(p-1)}),$

since $\sum\limits_{\begin{subarray}{c}i=-(p-1)\\\ i\text{
even}\end{subarray}}^{p-1}z^{i}=0.$ Let
$\mathcal{O}_{1},\dots,\mathcal{O}_{k}$ and $x_{1},\dots,x_{k}$ be as in
(3.2). Since we can pick $t$ to be a positive even number for each summand
$z^{t}+z^{-t}$ of $x_{i}$ (if not, replace $t$ by $-t$ or $t-p$), then we can
identify each $x_{i}$ with a sum of $L_{s}$’s with multiplicity $\pm 1$, as
follows:

$\displaystyle x_{i}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}z^{t}+z^{-t}\in\mathcal{O}_{i}\\\
t\text{ even}\\\ 2\leq t<p-1\end{subarray}}(L_{t+1}-L_{t-1}),\text{ for }i\neq
k,$ and $\displaystyle
x_{k}=-L_{p-2}+\sum\limits_{\begin{subarray}{c}z^{t}+z^{-t}\in\mathcal{O}_{k}\\\
t\text{ even}\\\ 2\leq t<p-1\end{subarray}}(L_{t+1}-L_{t-1}).$

Let $s$ odd, $1<s\leq p-2$. Note that $L_{s}$ appears with nonzero
multiplicity in either two basis elements, with multiplicity $1$ and $-1$,
respectively, or in none (since it may cancel out with itself). On the other
hand, $L_{1}$ appears in only one basis element (explicitly, the basis element
$x_{j}$ such that $z^{2}+z^{-2}\in\mathcal{O}_{j}$), with multiplicity $-1$.
We will say $L_{s}$ is a ”positive” summand of $x_{i}$ if it has multiplicity
1 in $x_{i}$, and is a ”negative” summand if it has multiplicity $-1$.

We claim that every $x_{i}$ has at least one positive and one negative
summand. In fact, this is clear for $1\leq i<k$, since the number of positive
summands in $x_{i}$ is the same as the number of negative summands. Suppose
for contradiction that we have $x_{k}=-L_{s}$ for some even $s$, $2\leq s\leq
p-2$. Our assumption $k<\frac{p-1}{2}$ assures that every orbit has at least
two elements, so it is not possible to have $x_{k}=-L_{p-2}$. Since we are
assuming $x_{k}$ has only one negative summand, $L_{p-2}$ must cancel out with
a positive summand. This implies
$z^{3}+z^{-3}=z^{p-3}+z^{-(p-3)}\in\mathcal{O}_{k}$ as well, and so $H_{m}$
contains the class $\bar{3}$ of the number $3\in\mathbb{Z}_{p}^{\times}$. Now,
either $x_{k}=-L_{p-4}$, or $-L_{p-4}$ cancels out. In the latter case, we
have that $z^{5}+z^{-5}=z^{p-5}+z^{-(p-5)}\in\mathcal{O}_{k}$ and so
$\bar{5}\in H_{m}$. Recursively, we get that
$H_{m}=\\{\bar{1},\bar{3},\bar{5},\dots,\bar{j}\\}$ for some odd $3\leq j\leq
p-2$. We claim this contradicts that $H_{m}$ is a proper subgroup. In fact,
since $H_{m}$ is a subgroup, it must contain the classes of $3l$ for all $l$
odd, $1\leq l\leq j$. Let $l\in H_{m}$ such that $3l\leq j<3(l+2)$. Note that
$l+2$ is also in $H_{m}$ (if not, then $3l\leq j<l+2$, and so $l<1$, which is
not possible). So $3(l+2)$ must also be in $H_{m}$. But since $j<3(l+2)$, it
must be the case that $3(l+2)>p$ and $\overline{3(l+2)}=\bar{n}$ for some odd
$1\leq n<j$. So we have the inequalities

$\displaystyle 3l\leq j<p<3(l+2),$

which imply $p=3l+2$ or $p=3l+4$. If $p=3l+2$, since $H_{m}$ contains the
classes of all odd elements from $1$ to $3l=p-2$ we get that
$|H_{m}|=\frac{p-1}{2}$, a contradiction. If $p=3l+4$, then $H_{m}$ contains
all odd elements from $1$ to $3l=p-4$ (its missing at most one element) and
thus again $H_{m}$ must have all odd elements, a contradiction. Hence $x_{k}$
has at least one positive and one negative summand.

Our aim is to construct sequences of indexes, alternating between negative and
positive summands of different $x_{i}$’s. We have shown every basis element
has at least one positive and one negative summands. With this in mind, we
begin the construction of our sequences.

Fix $s_{0}\neq 1$ so that $L_{s_{0}}$ is a positive summand of some
$x_{j_{0}}$. Since $k>1$, then there exists $j_{1}\neq j_{0}$ such that
$L_{s_{0}}$ is a negative summand of $x_{j_{1}}$. By our preceding discussion,
there must exist a positive summand of $x_{j_{1}}$. So we can find $s_{1}\neq
1$ (since $L_{1}$ can only be a negative summand) such that $L_{s_{1}}$ is a
positive summand of $x_{j_{1}}$. Thus $L_{s_{1}}$ is a negative summand of
$x_{j_{2}}$ for some $j_{2}\neq j_{1}$. Again, there exists some $s_{2}\neq 1$
such that $L_{s_{2}}$ is a positive summand of $x_{j_{2}}$. Recursively, we
can construct sequences of indexes $\\{s_{t}\\}$ and $\\{j_{i}\\}$ such that
$s_{t}\neq 1$ for all $t$, $x_{i}\neq x_{i+1}$ for all $i$, and $L_{s_{t}}$ is
a positive summand of $x_{j_{t}}$ and a negative summand of $x_{j_{t+1}}$.
Since there are only finitely many indexes $j_{i}$, they must repeat at some
point. Without loss of generality, assume $j_{1}$ is the first one that
repeats, so our sequence is $\\{j_{1},j_{2},j_{3},\dots,j_{n},j_{1},\dots\\}$,
with $n\geq 2$.

Let $y:=a_{1}x_{1}+\dots+a_{k}x_{k}$ be an element in $A_{k}$ that can be
written as a positive linear combination of $L_{s}$’s. We show that $y$ is in
the subspace generated by $\\{x_{i}\\}_{i\neq j_{1},\dots j_{n}}$ and
$x_{j_{1}}+\dots+x_{j_{n}}$. Since $L_{s_{1}}$ has multiplicity
$a_{j_{1}}-a_{j_{2}}$ in $y$, it must happen that $a_{j_{1}}\geq a_{j_{2}}$.
Now, $L_{s_{2}}$ has multiplicity $a_{j_{2}}-a_{j_{3}}$ in $y$, which implies
$a_{j_{2}}\geq a_{j_{3}}$. Then we can obtain a sequence

$\displaystyle a_{j_{1}}\geq a_{j_{2}}\geq\dots\geq a_{j_{n}}\geq a_{j_{1}},$

which implies $a_{j_{1}}=a_{j_{2}}=\dots=a_{j_{n}}$, as desired.

Consequently, elements that can be written as a positive linear combination of
$L_{i}$’s are contained in a subset of dimension less than $k$, which
contradicts our statement (3.4). Hence $k=\frac{p-1}{2}$ and so

$\displaystyle\operatorname{rank}(\mathcal{K(C)})\geq\operatorname{rank}(\tilde{F}(\mathcal{K(C)}))=\dim(A_{k})=\frac{p-1}{2},$

which finishes the proof. ∎

Let $\mathcal{C}$ now be a symmetric fusion category with Verlinde fiber
functor $F:\mathcal{C}\to\operatorname{Ver}_{p}$, and suppose $F$ is
surjective. By [BEO, Theorem 4.5 (iv)] we have an isomorphism

(3.5)
$\displaystyle\mathcal{K}(\operatorname{Ver}_{p})_{\mathbb{Q}}\simeq\mathbb{Q}(z+z^{-1})\oplus\mathbb{Q}(z+z^{-1}).$

We denote also by $F$ the induced ring homomorphism
$\mathcal{K(C)}\to\mathcal{K}(\operatorname{Ver}_{p})$. Same as before,
$F(\mathcal{K(C)})_{\mathbb{Q}}$ is a subalgebra of
$\mathcal{K}(\operatorname{Ver}_{p})_{\mathbb{Q}}$, and we will show that it
is exactly $\mathcal{K}(\operatorname{Ver}_{p})_{\mathbb{Q}}$. We start by
looking at subalgebras $\mathbb{Q}(z+z^{-1})^{\oplus 2}$.

######  Lemma 3.2.

Subalgebras of $\mathbb{Q}(z+z^{-1})^{\oplus 2}$ of dimension greater than
$\frac{p-1}{2}$ are of the form $A_{k}\oplus A_{l}$, where $k,l$ are positive
integers dividing $\frac{p-1}{2}$.

###### Proof.

Let $\mathcal{A}$ be a subalgebra of $\mathbb{Q}(z+z^{-1})^{\oplus 2}$.
Suppose first that $\mathcal{A}$ has no nontrivial idempotents. Note that
$\mathbb{Q}(z+z^{-1})^{\oplus 2}$ has no nilpotent elements and thus neither
does $\mathcal{A}$. Hence $\mathcal{A}$ is semisimple and so by Artin-
Wedderburn’s Theorem it is isomorphic to a finite product of field extensions
over $\mathbb{Q}$. Since $\mathcal{A}$ has no idempotents, this implies that
$\mathcal{A}$ is isomorphic to a field extension over $\mathbb{Q}$.

Consider the projection map $p$ from $\mathbb{Q}(z+z^{-1})^{\oplus 2}$ to its
first summand, and let $q$ denote its restriction to $\mathcal{A}$. Then
$\ker(q)=0$ or $\mathcal{A}$. If $\ker(q)=\mathcal{A}$ then elements in
$\mathcal{A}$ are of the form $(0,a)$, which is only possible for $a=0$ since
$\mathcal{A}$ is a field. Hence if $\mathcal{A}\neq 0$ we must have
$\ker(q)=0$, that is, we have an injective map
$A\hookrightarrow\mathbb{Q}(z+z^{-1})$, and so
$\dim(\mathcal{A})\leq\frac{p-1}{2}$.

Suppose now that $\mathcal{A}$ contains a nontrivial idempotent $e$. Then $e$
is either $(1,0),(0,1)$, and

$\mathcal{A}\simeq e\cdot\mathcal{A}\oplus(1-e)\cdot\mathcal{A}.$

Hence $\mathcal{A}$ is a direct sum of subalgebras of $\mathbb{Q}(z+z^{-1})$,
as desired. ∎

The proof of the following Theorem follows the same steps as the proof of
Theorem 3.1.

######  Theorem 3.3.

Let $p\geq 5$ and let $\mathcal{C}$ be a symmetric fusion category with a
Verlinde fiber functor $F:\mathcal{C}\to\operatorname{Ver}_{p}$. If $F$ is
surjective, then

$\operatorname{rank}(\mathcal{C})\geq p-1.$

###### Proof.

Let $\mathcal{C}$ be as in the statement. Under the isomorphism

(3.6)
$\displaystyle\mathcal{K}(\operatorname{Ver}_{p})_{\mathbb{Q}}\simeq\mathbb{Q}(z+z^{-1})\oplus\mathbb{Q}(z+z^{-1}),$

we have identifications

(3.7) $\displaystyle L_{2j+1}$
$\displaystyle=\left(\sum\limits_{l=1}^{j}(z^{2l}+z^{-2l})+1,0\right)\ \
\text{for }j=0,\dots,(p-3)/2,\text{ and}$ (3.8) $\displaystyle L_{2j}$
$\displaystyle=\left(0,\sum\limits_{l=1}^{j}(z^{2l-1}+z^{-(2l-1)})\right)\ \
\text{for }j=1,\dots,(p-1)/2.$

Since $\mathcal{C}$ is not super Tannakian, by the proof of Theorem 3.1 we
know that the coposition

$\mathcal{C}\xrightarrow{F}\operatorname{Ver}_{p}\simeq\operatorname{Ver}_{p}^{+}\boxtimes\text{sVec}\xrightarrow{\operatorname{id}\boxtimes\operatorname{Forget}}\operatorname{Ver}_{p}^{+}$

is surjective. Moreover, since we are assuming $F$ is surjective,
$F(\mathcal{K(C)})$ cannot be equal to
$\mathcal{K}(\operatorname{Ver}_{p}^{+})$, and so

$\operatorname{rank}(F(\mathcal{K(C)})>\frac{p-1}{2}.$

This together with Lemma 3.2 implies that $F(\mathcal{K(C)})_{\mathbb{Q}}$ is
identified with a subalgebra of the form $\mathbb{Q}(z+z^{-1})\oplus A_{k}$,
for some $k$ that divides $\frac{p-1}{2}.$ Note that the rank of
$F(\mathcal{K(C)})$ is equal to the dimension of
$F(\mathcal{K(C)})_{\mathbb{Q}}$, and so we want to show that
$k=\frac{p-1}{2}$.

Analogous to the proof of Theorem 3.1, $A_{k}$ must have a basis such that its
elements can be written as linear combination of $L_{i}^{\prime}$, with $i$
even, with coefficients in $\mathbb{Z}_{\geq 0}$. For contradiction, assume
$k<\frac{p-1}{2}$. Note also that since $F$ is surjective we have that $k>1$.

Let $\mathcal{O}_{1},\dots,\mathcal{O}_{k}$ and $x_{1},\dots,x_{k}$ as in
(3.2), relabelling so that $z+z^{-1}\in\mathcal{O}_{1}$. Via the
identification (3.6) we have

$\displaystyle\left(0,z^{t}+z^{-t}\right)=L_{t+1}-L_{t-1}\ \ \text{for
}t\text{ odd,}\ 1<t<p-1,\text{ and}$
$\displaystyle\left(0,z+z^{-1}\right)=L_{2}.$

Since we can pick $t$ odd for each summand $z^{t}+z^{-t}$ of $x_{i}$ (if not
odd, replace $t$ by $t-p$), then we can identify each $x_{i}$ with a sum of
$L_{i}$’s with multiplicity $\pm 1$ or $0$, as follows

$\displaystyle\left(0,x_{1}\right)$
$\displaystyle=L_{2}+\sum\limits_{\begin{subarray}{c}z^{t}+z^{-t}\in\mathcal{O}_{1}\\\
t\text{ odd}\\\ 1<t<p-1\end{subarray}}(L_{t+1}-L_{t-1}),$ and
$\displaystyle\left(0,x_{i}\right)=\sum\limits_{\begin{subarray}{c}z^{t}+z^{-t}\in\mathcal{O}_{i}\\\
t\text{ odd}\\\ 1<t<p-1\end{subarray}}(L_{t+1}-L_{t-1})\text{ for }i\neq 1.$

Let $s$ odd, $2\leq s<p-1$. Note that $L_{s}$ appears with nonzero
multiplicity in either two basis elements, with multiplicity $1$ and $-1$,
respectively, or in none (since it may cancel out with itself). On the other
hand, $L_{p-1}$ appears in only one basis element (explicitly, the basis
element $x_{j}$ such that $z^{p-2}+z^{-(p-2)}\in\mathcal{O}_{j}$), with
multiplicity $1$.

Same as in the proof of Theorem 3.1, we will construct sequences of indexes,
alternating between negative and positive summands of different $x_{i}$’s. By
an analogous reasoning, the assumption $k<\frac{p-1}{2}$ implies that if
$L_{s}$, $s\neq p-1$, is a positive summand of some $x_{j}$, then we can also
find a negative summand $L_{t}$ of $x_{j}$.

Fix $s_{0}\neq p-1$ so that $L_{s_{0}}$ is a negative summand of some
$x_{j_{0}}$. Since $k>1$, then there exists $j_{1}\neq j_{0}$ such that
$L_{s_{0}}$ is a positive summand of $x_{j_{1}}$. By our preceding discussion,
there must exist a negative summand of $x_{j_{1}}$. So we can find $s_{1}\neq
p-1$ (since $L_{p-1}1$ can only be a positive summand) such that $L_{s_{1}}$
is a negative summand of $x_{j_{1}}$. Thus $L_{s_{1}}$ is a positive summand
of $x_{j_{2}}$ for some $j_{2}\neq j_{1}$. Recursively, we can construct
sequences of indexes $\\{s_{t}\\}$ and $\\{j_{i}\\}$ such that $s_{t}\neq p-1$
for all $t$, $x_{i}\neq x_{i+1}$ for all $i$, and $L_{s_{t}}$ is a negative
summand of $x_{j_{t}}$ and a positive summand of $x_{j_{t+1}}$. Since there
are only finitely many indexes $j_{i}$, they must repeat at some point.
Without loss of generality, assume $j_{1}$ is the first one that repeats, so
our sequence is $\\{j_{1},j_{2},j_{3},\dots,j_{n},j_{1},\dots\\}$, with $n\geq
2$.

Let $y:=a_{1}x_{1}+\dots+a_{k}x_{k}$ be an element in $A_{k}$ that can be
written as a positive linear combination of $L_{s}$’s. We show that $y$ is in
the subspace generated by $\\{x_{i}\\}_{i\neq j_{1},\dots j_{n}}$ and
$x_{j_{1}}+\dots+x_{j_{n}}$. Since $L_{s_{1}}$ has multiplicity
$a_{j_{2}}-a_{j_{1}}$ in $y$, it must happen that $a_{j_{2}}\geq a_{j_{1}}$.
Now, $L_{s_{2}}$ has multiplicity $a_{j_{3}}-a_{j_{2}}$ in $y$, which implies
$a_{j_{3}}\geq a_{j_{2}}$. Then we can obtain a sequence

$\displaystyle a_{j_{1}}\leq a_{j_{2}}\leq\dots\leq a_{j_{n}}\leq a_{j_{1}},$

which implies $a_{j_{1}}=a_{j_{2}}=\dots=a_{j_{n}}$, as desired. Thus elements
that can be written as a positive linear combination of $L_{i}$’s are
contained in a subset of dimension less than $k$, a contradiction, since we
should be able to obtain a basis for $A_{k}$ of that form. Hence
$k=\frac{p-1}{2}$ and so

$\operatorname{rank}(\mathcal{K(C)})\geq\operatorname{rank}(F(\mathcal{K(C)})=\dim(A_{k})=p-1,$

as desired. ∎

##  4\. Some properties of the Adams operation

Throughout this section, we assume $p>2$.

###  4.1. Adams operation in $\operatorname{Ver}_{p}$

Recall that the Adams operation of a symmetric fusion category $\mathcal{C}$
is defined as the ring endomorphism $\psi_{2}:\mathcal{K(C)}\to\mathcal{K(C)}$
given by

$\psi_{2}(X)=S^{2}(X)-\Lambda^{2}(X),$

for all $X\in\mathcal{K(C)}$, see section 2.2.1.

In this section we study some properties of the Adams operation in
$\operatorname{Ver}_{p}$. We first give an explicit formula for the second
Adams operation on simple objects $L_{t}$, $1\leq t\leq p-1$. We then use this
formula to show that if an object $X$ in $\operatorname{Ver}_{p}$ is fixed by
the Adams operation, then $X$ is in the abelian subcategory
$\operatorname{Vec}$ generated by $\mathbf{1}=L_{1}$.

######  Remark 4.1.

The image of
$\psi_{2}:\mathcal{K}(\operatorname{Ver}_{p})\to\mathcal{K}(\operatorname{Ver}_{p})$
is contained in $\mathcal{K}(\operatorname{Ver}_{p}^{+})$. In fact,

$\displaystyle
L_{t}^{2}=\sum\limits_{s=1}^{\min(t,p-t)}L_{2s-1}=S^{2}(L_{t})+\Lambda^{2}(L_{t}),$

for all $i=1,\dots,p-1$, and so the multiplicity of even simples is zero in
both $S^{2}(L_{i})$ and $\Lambda^{2}(L_{i}).$

Note that to compute $\psi_{2}$ in simple objects $L_{r}$ of
$\operatorname{Ver}_{p}$, it is enough to compute it for $r$ odd, since

$\psi_{2}(L_{r})=-\psi_{2}(L_{p-r}).$

This follows from

$\displaystyle\Lambda^{2}L_{r}=L_{p-1}^{2}\otimes S^{2}L_{p-r}=S^{2}L_{p-r},$

see [EOV, Proposition 2.4].

######  Example 4.2.

In $\operatorname{Ver}_{5}$,

$\psi_{2}(L_{3})=L_{1}-L_{3}=-\psi_{2}(L_{2}).$

In fact, we know that $\psi_{2}=S^{2}(L_{3})-\Lambda^{2}(L_{3})$ and
$L_{3}^{2}=L_{1}+L_{3}=S^{2}(L_{3})+\Lambda^{2}(L_{3})$. Hence there must
exist $\epsilon_{1},\epsilon_{3}\in\\{\pm 1\\}$ such that
$\psi_{2}(L_{3})=\epsilon_{1}L_{1}+\epsilon_{3}L_{3}$. Now,

$\displaystyle
L_{1}+2\epsilon_{1}\epsilon_{3}L_{3}+L_{3}^{2}=\psi_{2}(L_{3})^{2}=\psi_{2}(L_{3}^{2})=(1+\epsilon_{1})L_{1}+\epsilon_{3}L_{3},$

and so $2=1+\epsilon_{1}$, which implies $\epsilon_{1}=1$. It follows that
$\epsilon_{3}=-1$.

######  Proposition 4.3.

The second Adams operation
$\psi_{2}:\mathcal{K}(\operatorname{Ver}_{p})\to\mathcal{K}(\operatorname{Ver}_{p})$
is given by

$\displaystyle\psi_{2}(L_{t})=\sum\limits_{s=1}^{\min(t,p-t)}(-1)^{s+1}L_{2s-1}\text{
for $t$ odd, }1\leq t\leq p-1,\text{ and}$
$\displaystyle\psi_{2}(L_{t})=\sum\limits_{s=1}^{\min(t,p-t)}(-1)^{s}L_{2s-1}\text{
for $t$ even, }1\leq t\leq p-1.$

###### Proof.

Note that the second formula follows from the first by the equality
$\psi_{2}(L_{r})=-\psi_{2}(L_{p-r})$. We will use the isomorphism

$\mathcal{K}(\operatorname{Ver}_{p}^{+})_{\mathbb{Q}}\simeq\mathbb{Q}(z+z^{-1}),$

for $z$ a primitive $p$-th root of unity, see [BEO, Theorem 4.5 (iv)] and
Section 3. Consider the basis
$\left\\{z^{2i}+z^{-2i}\right\\}_{i=1,\dots,\frac{p-1}{2}}$ of
$\mathbb{Q}(z+z^{-1})$. Via this isomorphism, we have identifications

$\displaystyle L_{2j+1}=\sum\limits_{l=1}^{j}(z^{2l}+z^{-2l})+1,\ \ \text{for
}j=0,\dots,(p-3)/2,$

from which we compute

(4.1) $\displaystyle\begin{aligned} &z^{t}+z^{-t}\ =\ L_{t+1}-L_{t-1}\ \
\text{for }t\text{ even,}\ 2\leq t<p-1,\text{ and}\\\
&z^{p-1}+z^{-(p-1)}=-L_{p-2},\end{aligned}$

see Section 3 and the proof of Theorem 3.1 for details.

Note that
$(\psi_{2})_{\mathbb{Q}}:\mathbb{Q}(z+z^{-1})\to\mathbb{Q}(z+z^{-1})$ maps
$z+z^{-1}=z^{p-1}+z^{-(p-1)}$ to $z^{2}+z^{-2}$. In fact, since
$\dim(S^{2}(L_{p-2}))=\frac{(p-2)(p-1)}{2}=1\mod p$, it must be the case that
$S^{2}(L_{p-2})=L_{1}$. On the other hand,
$\dim(\Lambda^{2}(L_{p-2}))=\frac{(p-2)(p-3)}{2}=3\mod p$ and $L_{1}$ can only
be a summand of either $S^{2}(L_{p-2})$ or $\Lambda^{2}(L_{p-2})$ but not
both, then $\Lambda^{2}(L_{p-2})=L_{3}$. Thus using identification (4.1) we
get

$\displaystyle\psi_{2}(z^{p-1}+z^{-(p-1)})=-\psi_{2}(L_{p-2})=-S^{2}(L_{p-2})+\Lambda^{2}(L_{p-2})=L_{3}-L_{1}=z^{2}+z^{-2},$

as desired.

We prove now our formulas for $\psi_{2}(L_{t})$, $t$ odd, by induction. We do
the case $1\leq t\leq\frac{p-1}{2}$ first. We know $\psi_{2}(L_{1})=L_{1}$, so
the formula works for $t=1$. Fix $1<t\leq\frac{p-1}{2}$ odd, and suppose the
formula is true for all odd $1\leq r<t$. Then

$\displaystyle\psi_{2}(z^{t-1}+z^{-(t-1)})=z^{2(t-1)}+z^{-2(t-1)}=L_{2(t-1)+1}-L_{2(t-1)-1}=L_{2t-1}-L_{2t-3},$

where in the third equality we are using the identification (4.1). Since
$z^{t-1}+z^{-(t-1)}=L_{t}-L_{t-2}$, we compute using induction

$\displaystyle\psi_{2}(L_{t})$
$\displaystyle=\psi_{2}(L_{t-2})+L_{2t-1}-L_{2t-3}$
$\displaystyle=\sum\limits_{s=1}^{t-2}(-1)^{s+1}L_{2s-1}+L_{2t-1}-L_{2t-3}$
$\displaystyle=\sum\limits_{s=1}^{t}(-1)^{s+1}L_{2s-1}=\sum\limits_{s=1}^{\min(t,p-t)}(-1)^{s+1}L_{2s-1},$

and so the formula holds for $t$.

It remains to show that the formula holds for the case of odd
$\frac{p-1}{2}<t<p-1$ . We already computed $\psi_{2}(L_{p-2})=L_{1}-L_{3}$,
so it works for $t=p-2$. Fix odd $\frac{p-1}{2}<t=p-l<p-1$ and assume the
formula holds for all odd $t<r<p-1$. Since
$z^{p-l+1}+z^{-(p-l+1)}=L_{p-(l-2)}-L_{p-l}$, we compute same as before

$\displaystyle\psi_{2}(L_{p-l})$
$\displaystyle=\psi_{2}(L_{p-(l-2)})-(L_{2l-1}-L_{2l-3})$
$\displaystyle=\sum\limits_{s=1}^{l-2}(-1)^{s+1}L_{2s-1}-L_{2l-3}+L_{2l-1}$
$\displaystyle=\sum\limits_{s=1}^{l}(-1)^{s+1}L_{2s-1}=\sum\limits_{s=1}^{\min(t,p-t)}(-1)^{s+1}L_{2s-1},$

as desired. ∎

######  Remark 4.4.

The statement of Proposition 4.3 only works for actual objects in
$\operatorname{Ver}_{p}$, that is, objects that can be written as
$\mathbb{Z}_{\geq 0}$ linear combinations of $L_{1},\dots,L_{p-1}$. There can
exist objects in $\mathcal{K}(\operatorname{Ver}_{p})$ that are fixed by the
second Adams operation but are not multiples of $L_{1}$. For example, consider
$p=17$ and $L_{5}-L_{7}+L_{9}-L_{15}\in\mathcal{K}(\operatorname{Ver}_{p})$.
Then

$\displaystyle\psi_{2}(L_{5}-L_{7}+L_{9}-L_{15})$
$\displaystyle=\psi_{2}(L_{5})-\psi_{2}(L_{7})+\psi_{2}(L_{9})-\psi_{2}(L_{15})$
$\displaystyle=(L_{1}-L_{3}+L_{5}-L_{7}+L_{9})-(L_{1}+L_{3}-L_{5}+L_{7}-L_{9}+L_{11}-L_{13})+$
$\displaystyle+(L_{1}+L_{3}-L_{5}+L_{7}-L_{9}+L_{11}-L_{13}-L_{15})-(L_{1}-L_{3})$
$\displaystyle=L_{5}-L_{7}+L_{9}-L_{15}$

and so $L_{5}-L_{7}+L_{9}-L_{15}$ is fixed by $\psi_{2}$ but is not in
$\operatorname{Vec}$.

We now study objects in $\operatorname{Ver}_{p}$ that are fixed by the second
Adams operation. For a simple object $X$ in a symmetric fusion category
$\mathcal{C}$, we denote by $[Y:X]$ the multiplicity of $X$ in $Y$ for all
$Y\in\mathcal{C}$.

######  Corollary 4.5.

An object $X\in\operatorname{Ver}_{p}$ is fixed by $\psi_{2}$ if and only if
$X\in\operatorname{Vec}$.

###### Proof.

Let $X\in\operatorname{Ver}_{p}$ such that $\psi_{2}(X)=X$ and let
$a_{1},\dots,a_{p-1}$ be non-negative integers such that
$X=\sum\limits_{j=1}^{p-1}a_{j}L_{j}$. Since the image of $\psi_{2}$ is
contained in $\mathcal{K}(\operatorname{Ver}_{p}^{+})$ (see Remark 4.1) then
$a_{2}=a_{4}=\dots=a_{p-1}=0$.

Using the formulas from Proposition 4.3 we compute

$\displaystyle
a_{1}=[X:L_{1}]=[\psi_{2}(X):L_{1}]=\sum\limits_{j=1}^{\frac{p-1}{2}}a_{2j-1}.$

Since $a_{i}$ is non-negative for all $i$ this implies
$a_{3}=a_{5}=\dots=a_{p-2}=0$, as desired. ∎

###  4.2. Powers of the Adams operation

Recall that in this section we assume $p>2$. Here we classify symmetric fusion
categories $\mathcal{C}$ such that $\psi_{2}^{a}=\psi_{2}^{a-1}$ for some
$a\geq 1$. Namely, we show that such categories are super Tannakian and thus
classified by group data, see [DM, D, DG]. Moreover, we show that the case
$a=1$ is only possible for the trivial category. That is, we prove that if
$\psi_{2}=\text{Id}$ in $\mathcal{K(C)}$ then
$\mathcal{C}=\operatorname{Vec}$.

######  Theorem 4.6.

Let $p\neq 2$ and let $\mathcal{C}$ be a symmetric fusion category. If
$\psi_{2}^{a}=\psi_{2}^{a-1}$ for some $a\in\mathbb{Z}_{\geq 1}$, then
$\mathcal{C}$ is super Tannakian.

###### Proof.

We denote also by $F$ the induced ring homomorphism
$\mathcal{K(C)}\to\mathcal{K}(\operatorname{Ver}_{p})$. Suppose first that
$a=1$. Let $X\in\mathcal{C}$. Since $F$ preserves the symmetric structure, we
have that

$\displaystyle\psi_{2}(FX)=F(\psi_{2}X)=F(X),$

and so $\psi_{2}$ fixes $FX\in\operatorname{Ver}_{p}$. Thus by Lemma 4.5 we
have that $FX\in\text{Vec}$ for all $X\in\mathcal{C}$, and so $\mathcal{C}$ is
Tannakian, as desired.

Suppose now that $a>1$. Similarly, we have that

$\displaystyle\psi_{2}^{a}(FX)=F(\psi_{2}^{a}X)=F(\psi_{2}^{a-1}X)=\psi_{2}^{a-1}(FX),\text{
for all }X\in\mathcal{K(C)}.$

That is, $\psi_{2}^{a}=\psi_{2}^{a-1}$ in the image of $\mathcal{K(C)}$
through $F$. Suppose $\mathcal{C}$ is not super Tannakian (and thus, in
particular, $p>3$). The proofs of Theorems 3.1 and 3.3 show that
$F(\mathcal{K(C)})_{\mathbb{Q}}$ is either $\mathbb{Q}(z+z^{-1})$ or
$\mathbb{Q}(z+z^{-1})^{\oplus 2}$. Recall that
$\psi_{2}(z+z^{-1})=z^{2}+z^{-2}$. Then
$(\psi_{2}^{a})_{\mathbb{Q}}=(\psi_{2}^{a-1})_{\mathbb{Q}}$ in
$F(\mathcal{K(C)})$ would imply that
$z^{2^{a}}+z^{-2^{a}}=z^{2^{a-1}}+z^{-2^{a-1}}$, and so $2^{a}\equiv\pm
2^{a-1}\mod p$, which implies $2\equiv\pm 1\mod p$. This is a contradiction
since $p>3$, and thus $\mathcal{C}$ must be super Tannakian. ∎

######  Remark 4.7.

The same proof as in Theorem 4.6 shows that if $\mathcal{C}$ is a non-
Tannakian symmetric fusion category, then $\psi_{2}^{a}=\psi_{2}^{b}$ is only
possible when $2^{a}\equiv\pm 2^{b}\mod p$.

Note that for the case $a=1$, i.e., $\psi_{2}=\text{Id}$, we showed that
$\mathcal{C}$ should be Tannakian, and thus equivalent to
$\operatorname{Rep}_{\textbf{k}}(G)$ for a finite group scheme $G$. A
classification of finite group schemes $G$ such that
$\operatorname{Rep}_{\textbf{k}}(G)$ is semisimple is given by Nagata’s
theorem [DG, IV, 3.6]; thus Theorem 4.6 yields a classification of symmetric
fusion categories such that $\psi_{2}=\text{Id}$. Namely, any such category is
an equivariantization (see [DGNO, Section 4]) of a pointed category associated
with a $p$-group (see e.g. [EGNO, 8.4]) by the action of a group $H$ of order
relatively prime to $p$.

######  Theorem 4.8.

Let $p>2$. If $\mathcal{C}$ is a non-trivial symmetric fusion category then
$\psi_{2}$ is not the identity.

###### Proof.

Suppose $\psi_{2}$ is the identity in $\mathcal{C}$. By the discussion above
we know that $\mathcal{C}$ is an equivariantization of a pointed category
associated with a $p$-group by a group $H$. Suppose $H$ is non-trivial and
consider the subcategory $\operatorname{Rep}_{\textbf{k}}(H)$ of
$\mathcal{C}$.

The Adams operation acts on $\operatorname{Rep}_{\textbf{k}}(H)$ by mapping a
character $\chi(g)$ to $\chi(g^{2})$ for all $g\in H$. Thus if
$\psi_{2}=\text{Id}$ then

$\chi(g^{2})=\chi(g)\text{ for all }g\in H.$

Hence for all $g$ in $H$, $g$ is conjugate to $g^{2}$. So if $|H|$ is even,
there is an element $h\in H$ of order 2, which is conjugated to $h^{2}=1$, a
contradiction.

Suppose then $|H|$ is odd. We have that for all $g$ in $H$ there exists some
$h\in H$ such that $g^{2}=hgh^{-1}$ and so

$g=hgh^{-1}g^{-1}.$

Thus $g$ is in the commutator subgroup of $H$, and so $H\subseteq[H,H]$. This
contradicts the Feit-Thompson Theorem, which states that every finite group of
odd order is solvable. So $H$ must be trivial.

We thus have that $\mathcal{C}$ is a pointed category associated with an
abelian $p$-group $P$. Hence $\psi_{2}$ maps $g\mapsto g^{2}$ for all $g\in P$
and so $g=g^{2}$ for all $g\in P$. Then $P$ is trivial and $\mathcal{C}$ is
equivalent to $\operatorname{Vec}$. ∎

Note that the previous Theorem also holds in characteristic $0$, since the
Adams operation acts on $\operatorname{Rep}_{\textbf{k}}(G)$ by mapping a
character $\chi(g)$ to $\chi(g^{2})$ for all $g\in G$.

###  4.3. Symmetric fusion categories with exactly two self-dual simple
objects

In this subsection we prove some general properties of the Adams operation
$\psi_{2}:\mathcal{K(C)}\to\mathcal{K(C)}$ for the case when $\mathcal{C}$ is
a symmetric fusion category with exactly two self-dual simple objects. These
results will be useful for the classification of symmetric fusion categories
of ranks 3 and 4 in Sections 5 and 6. In particular, we show that if
$\psi_{2}$ is an automorphism then it has even order, see Theorem 4.11.

Recall that throughout this section we assume $p>2$.

######  Lemma 4.9.

Let $\mathcal{C}$ be a symmetric fusion category with exactly two self-dual
objects $\mathbf{1}$ and $Y$. Then

$\displaystyle[\psi_{2}^{2k+1}(Y):\mathbf{1}]\equiv 1\mod 2$

for all $k\geq 0$.

###### Proof.

We proceed by induction on $k$. Let
$\mathbf{1},Y,X_{1},X_{1}^{*},\dots,X_{n},X_{n}^{*}$ denote the simple objects
in $\mathcal{C}$. Since $\psi_{2}(Y)\equiv Y^{2}\mod 2$ and
$[Y^{2}:\mathbf{1}]=1$ then $[\psi_{2}(Y):\mathbf{1}]\equiv 1\mod 2$, which
proves the base case.

Fix $k>1$ and suppose that

$\displaystyle[\psi_{2}^{2l+1}(Y):\mathbf{1}]\equiv 1\mod 2,\text{ for all
}l<k.$

We want to show it also holds for $l=k$. Since $Y$ is self-dual then
$[Y^{2}:X_{i}]=[Y^{2}:X_{i}^{*}]$ for all $i=1,\dots,n$, and so

$\displaystyle\psi_{2}(Y)\equiv
Y^{2}\equiv\mathbf{1}+\sum\limits_{i=1}^{n}[Y^{2}:X_{i}]\
(X_{i}+X_{i}^{*})+[Y^{2}:Y]\ Y\mod 2,$

for all $i=1,\dots,n$. Applying $\psi_{2}^{2k}$ on both sides of the previous
equation we get

(4.2) $\displaystyle\psi_{2}^{2k+1}(Y)$
$\displaystyle\equiv\mathbf{1}+\sum\limits_{i=1}^{n}[Y^{2}:X_{i}]\left(\psi_{2}^{2k}(X_{i})+\psi_{2}^{2k}(X_{i}^{*})\right)+[Y^{2}:Y]\psi_{2}^{2k}(Y)\mod
2.$

Recall that $\psi_{2}$ commutes with duality. That is,
$\psi_{2}(X_{i})^{*}=\psi_{2}(X_{i}^{*})$ and so

(4.3)
$\displaystyle[\psi_{2}^{l}(X_{i}):\mathbf{1}]=[\psi_{2}^{l}(X_{i})^{*}:\mathbf{1}]\equiv[\psi^{l}_{2}(X_{i}^{*}):\mathbf{1}]\mod
2,$

for all $l\geq 1$. From equations (4.2) and (4.3) we get

$\displaystyle[\psi_{2}^{2k+1}(Y):\mathbf{1}]$ $\displaystyle\equiv
1+2\sum\limits_{i=1}^{n}[Y^{2}:X_{i}]\
[\psi_{2}^{2k}(X_{i}):\mathbf{1}]+[Y^{2}:Y]\ [\psi_{2}^{2k}(Y):\mathbf{1}]\mod
2$ $\displaystyle\equiv 1+[Y^{2}:Y]\ [\psi_{2}^{2k}(Y):\mathbf{1}]\mod 2.$

Analogously,

$\displaystyle[\psi_{2}^{2k}(Y):\mathbf{1}]$ $\displaystyle\equiv
1+[Y^{2}:Y][\psi_{2}^{2k-1}(Y):\mathbf{1}]\mod 2$ $\displaystyle\equiv
1+[Y^{2}:Y]\mod 2,$

since we assumed $[\psi^{2k-1}_{2}(Y):\mathbf{1}]\equiv 1\mod 2$. Hence

$\displaystyle[\psi_{2}^{2k+1}(Y):\mathbf{1}]$ $\displaystyle\equiv
1+[Y^{2}:Y]\left(1+[Y^{2}:Y]\right)\mod 2$ $\displaystyle\equiv
1+[Y^{2}:Y]+[Y^{2}:Y]^{2}\mod 2$ $\displaystyle\equiv 1\mod 2,$

as desired. ∎

######  Remark 4.10.

In the proof of Lemma 4.9 we showed that

$[\psi_{2}^{l}(Y):Y]\equiv 1+[Y^{2}:Y][\psi_{2}^{l-1}(Y):Y]\mod 2,$

for all $l\geq 1$. This implies also that

$[\psi_{2}^{2k}(Y):\mathbf{1}]\equiv 0\mod 2,$

for all $k\geq 0$.

The following is a direct application of Lemma 4.9.

######  Theorem 4.11.

Let $\mathcal{C}$ be a symmetric fusion category with exactly two self-dual
objects. If $\psi_{2}$ is an automorphism of $\mathcal{K(C)}$ then it has even
order.

###### Proof.

Let $k\geq 0$ such that $\psi_{2}^{k}=\text{Id}$, and let $\textbf{1},Y$
denote the self-dual objects. By Theorem 4.9 the multiplicity of $\mathbf{1}$
in $\psi_{2}^{k}(Y)=Y$ is positive whenever $k$ is odd, and thus $k$ must be
even. ∎

In the following proposition we restrict to the case when
$\psi_{2}:\mathcal{K(C)}\to\mathcal{K(C)}$ has trivial image.

######  Proposition 4.12.

Let $\mathcal{C}$ be a symmetric fusion category with exactly two self-dual
simple objects $\mathbf{1}$ and $Y$. If
$\text{Im}({\psi_{2}})\simeq\mathbb{Z}$ then $Y^{2}=\mathbf{1}$ and
$\psi_{2}(Y)=-\mathbf{1}$. Moreover, $[X\otimes X^{*}:Y]=1$ and $[XY:Y]=0$ for
all non-self-dual simple $X$.

###### Proof.

Let $X$ be a non-self-dual simple object. Since $\mathbf{1}$ is not a summand
of $X^{2}$ then $\mathbf{1}$ has multiplicity zero in $\psi_{2}(X)$. On the
other hand, the multiplicity of $\mathbf{1}$ in $Y^{2}$ is $1$ and so its
coefficient in $\psi_{2}(Y)$ is $\pm 1$. Thus if ${\psi_{2}}$ has trivial
image we get that $\psi_{2}(X)=0$ for all non-self-dual simple $X$ and
$\psi_{2}(Y)=\epsilon$, where $\epsilon=\pm 1$. Hence

$\displaystyle\mathbf{1}=\psi_{2}(Y)^{2}=\psi_{2}(Y^{2})=\sum\limits_{\text{Simple
}Z}[Y^{2}:Z]\
\psi_{2}(Z)=\left(1+\epsilon\cdot[Y^{2}:Y]\right)\cdot\mathbf{1},$

which implies

(4.4) $\displaystyle[Y^{2}:Y]=0.$

Similarly,

(4.5) $\displaystyle 0=\psi_{2}(XY)=\sum\limits_{\text{ simple
}Z}[XY:Z]\psi_{2}(Z)=\epsilon\cdot[XY:Y]\cdot\mathbf{1},$

and thus

$\displaystyle[XY:Y]=0,\text{ for all non-self-dual simple }X.$

From the fusion rule $[XY:Y]=[Y^{2}:X]$ and equations (4.4) and (4.5) we
conclude $Y^{2}=\mathbf{1}$. Lastly, note that

$\displaystyle 0=\psi_{2}(XX^{*})=\mathbf{1}+\epsilon\cdot[XX^{*}:Y],$

and thus we must have $\epsilon=-1$ and $[XX^{*}:Y]=1$. ∎

##  5\. Rank 3 symmetric fusion categories

In this section we classify symmetric fusion categories $\mathcal{C}$ of rank
3, making use of the properties of the Adams operation. Namely, we show that
$\mathcal{C}$ is equivalent to one of the following:

  * •

If $p=2$, $\mathcal{C}\simeq\operatorname{Rep}_{\mathbf{k}}(\mathbb{Z}_{3})$.

  * •

If $p=3$,
$\mathcal{C}\simeq\operatorname{Vec}_{\mathbb{Z}_{3}}^{\mathbb{Z}_{2}}$.

  * •

If $p=7$, $\mathcal{C}\simeq\operatorname{Rep}_{\mathbf{k}}(S_{3})$,
$\mathcal{C}\simeq\operatorname{Rep}_{\textbf{k}}(\mathbb{Z}_{3})$ or
$\mathcal{C}\simeq\operatorname{Ver}_{7}^{+}$.

  * •

If $p=5$ or $p>7$, $\mathcal{C}\simeq\operatorname{Rep}_{\mathbf{k}}(S_{3})$
or $\mathcal{C}\simeq\operatorname{Rep}_{\textbf{k}}(\mathbb{Z}_{3})$.

We note that if $\mathcal{C}$ is non super Tannakian, then by Theorem 3.1 we
know that $p\leq 7$ . Hence the only possibilities are $p=5$ or $7$, since in
the cases $p=2$ or $p=3$ the category would be super Tannakian.

We make use of the parametrization of self dual based rings of rank 3 as given
in [O1]. Let $k,l,m,n$ be non negative integers satisfying

(5.1) $\displaystyle k^{2}+l^{2}=kn+lm+1,$

and consider the ring $K(k,l,m,n)$ with basis $1,X,Y$ and multiplication rules

(5.2) $\displaystyle X^{2}=1+mX+kY,$ $\displaystyle Y^{2}=1+lX+nY,$
$\displaystyle XY=YX=kX+lY.$

Note that we have a based ring isomorphism $K(k,l,m,n)\simeq K(l,k,n,m)$
obtained by the interchange $X\leftrightarrow Y$. Hence we will assume $l\geq
k$. By [O1, Proposition 3.1] any unital based ring of rank 3 is isomorphic to
either $K(k,l,m,n)$ or $K(\mathbb{Z}_{3})$.

######  Theorem 5.1.

There is an integral symmetric fusion category $\mathcal{C}$ with Grothendieck
ring $K(k,l,m,n)$ if and only if $(k,l,m,n)=(0,1,0,1)$ and $p\geq 3$.
Moreover, in such case

  1. (1)

$\mathcal{C}\simeq\operatorname{Rep}(S_{3})$ if $p>3$, or

  2. (2)

$\mathcal{C}\simeq\operatorname{Vec}_{\mathbb{Z}_{3}}^{\mathbb{Z}_{2}}$ if
$p=3$.

###### Proof.

Suppose $\mathcal{C}$ is an integral fusion category with Grothendick ring
given by $K(k,l,m,n)$ with $l\geq k$. Taking Frobenius-Perron dimension on the
multiplication rules for $X^{2}$ and $XY$ (equation (5.2)), we get that

$\displaystyle\operatorname{FPdim}(X)^{2}=1+m\operatorname{FPdim}(X)+k\operatorname{FPdim}(Y)$
and $\displaystyle
1=\frac{k}{\operatorname{FPdim}(Y)}+\frac{l}{\operatorname{FPdim}(X)},$

respectively. From the first equality we deduce that $\operatorname{FPdim}(X)$
and $\operatorname{FPdim}(Y)$ are coprime. Hence the second equality is only
possible if $k=0$ and $l=\operatorname{FPdim}(X)$. So the multiplication rules
are

$\displaystyle X^{2}=1+mX,$ $\displaystyle
Y^{2}=1+\operatorname{FPdim}(X)X+nY,$ $\displaystyle
XY=YX=\operatorname{FPdim}(X)Y.$

Taking Frobenius-Perron dimension on both sides of the equation for $X^{2}$ we
get that $\operatorname{FPdim}(X)^{2}-m\operatorname{FPdim}(X)-1=0$, which
implies

$\displaystyle\operatorname{FPdim}(X)=\frac{m+\sqrt{m^{2}+4}}{2}.$

Since $\operatorname{FPdim}(X)$ is an integer, we need for $m^{2}+4$ to be a
square. It is easy to check this is only possible for $m=0$, and so the
multiplication rules are

$\displaystyle X^{2}=1,$ $\displaystyle Y^{2}=1+X+nY,$ $\displaystyle
XY=YX=Y.$

Taking Frobenius-Perron dimension on both sides of the equation for $Y^{2}$ we
get that

$\displaystyle\operatorname{FPdim}(Y)=\frac{n+\sqrt{n^{2}+8}}{2}.$

We thus need $n^{2}+8$ to be a square, which is only possible for $n=1$. Hence

(5.3) $\displaystyle X^{2}=1,$ $\displaystyle Y^{2}=1+X+Y,$ $\displaystyle
XY=YX=Y.$

So far we have showed that if $\mathcal{C}$ is integral, then it must have
fusion rules as above. We have not used the assumption that $\mathcal{C}$ is
symmetric yet. We will use it in what follows to complete the proof.

When $p>3$ we get that $\dim(\mathcal{C})=6\neq 0$, so $\mathcal{C}$ is non-
degenerate. Hence we can lift $\mathcal{C}$ to a symmetric fusion category
$\widetilde{\mathcal{C}}$ over a field f in characteristic zero, which has the
same Grothendick ring as $\mathcal{C}$, see [EGNO, Subsection 9.16] and [E].
Thus $\widetilde{\mathcal{C}}$ is equivalent to
$\operatorname{Rep}_{\textbf{f}}(S_{3})$ (see [D]), and so by uniqueness of
the lifting we get that $\mathcal{C}$ is equivalent to
$\operatorname{Rep}_{\textbf{k}}(S_{3})$.

For the case $p=3$ we have that $\mathcal{C}$ contains a copy of
$\operatorname{Rep}(\mathbb{Z}_{2})$. Doing de-equivariantization by
$\mathbb{Z}_{2}$ we obtain a symmetric fusion category
$\mathcal{C}^{\mathbb{Z}_{2}}$ of dimension $3$ [EGNO, Section 2.7]. Hence
$\mathcal{C}^{\mathbb{Z}_{2}}$ is a symmetric pointed category and must be
equivalent to $\operatorname{Vec}_{\mathbb{Z}_{3}}$. There’s only one action
of $\mathbb{Z}_{2}$ on $\mathbb{Z}_{3}$, and so doing equivariantization by
$\mathbb{Z}_{2}$ gives us back $\mathcal{C}$. Hence
$\mathcal{C}\simeq\text{Vec}_{\mathbb{Z}_{3}}^{\mathbb{Z}_{2}}$.

On the other hand, for $p=2$ there is no category realizing these fusion
rules. In fact, in such case we would have that the category is Tannakian, and
so we have a fiber functor $F:\mathcal{C}\to\operatorname{Vec}$. From equation
$\eqref{dim eq}$ we know that $X$ is invertible and $Y$ is not, and thus the
same is true for $F(X)$ and $F(Y)$, respectively. Thus $d:=\dim(F(Y))$ is not
1 and must satisfy $d^{2}=2+d=d$. The only other possible solution is thus
$d=0$, and since $F$ preserves dimensions we obtain $\dim(Y)=0$, which is not
possible. ∎

######  Theorem 5.2.

If $\mathcal{C}$ is a non-integral symmetric fusion category with Grothendieck
ring $K(k,l,m,n)$ then $p=7$ and $\mathcal{C}=\operatorname{Ver}_{7}^{+}$.

###### Proof.

Note that $p\neq 2,3$ since in that case $\mathcal{C}$ is super Tannakian and
thus integral. Moreover, we also know that $p\leq 7$ by Theorem 3.1. However,
our proof does not require use of this fact.

Consider the Adams operation $\psi_{2}:K(k,l,m,n)\to K(k,l,m,n)$. Since $X$ is
self-dual, either $S^{2}(X)$ or $\Lambda^{2}(X)$ contains a copy of the unit
object (but not both). The same is true for $Y$. Hence the multiplicity of
$\mathbf{1}$ in both $\psi_{2}(X)$ and $\psi_{2}(Y)$ is 1 or -1, and so

$\displaystyle\psi_{2}(X)=\epsilon_{1}\textbf{1}+\alpha X+\beta Y,$
$\displaystyle\psi_{2}(Y)=\epsilon_{2}\textbf{1}+\gamma X+\delta Y,$

for some $\epsilon_{1},\epsilon_{2}\in\\{1,-1\\}$ and
$\alpha,\beta,\gamma,\delta\in\mathbb{Z}$.

Since $\mathcal{K(C)}$ is a based ring of rank $3$ then
$\mathcal{K(C)}_{\mathbb{Q}}:=\mathcal{K(C)}\otimes\mathbb{Q}$ is a semisimple
algebra of dimension 3. Hence we have three distinct possibilities for
$\mathcal{K(C)}_{\mathbb{Q}}$, and we proceed by looking at them separetly.

$\blacktriangledown$ Case 1:
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}$.
In this case the homomorphism
$\operatorname{FPdim}_{\mathbb{Q}}:\mathcal{K(C)}_{\mathbb{Q}}\to\mathbb{R}$
can only have rational image and thus $\mathcal{C}$ is integral.

$\blacktriangledown$ Case 2:
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}\times\mathbb{Q}(\sqrt{m})$ for
some $m\in\mathbb{Z}$. Since $(\psi_{2})_{\mathbb{Q}}$ is an endomorphism of
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}\times\mathbb{Q}(\sqrt{m})$
mapping $(1,1)\mapsto(1,1)$, then it is either an automorphism of order $1$ or
$2$, or has image the diagonal copy of $\mathbb{Q}$ inside
$\mathbb{Q}\times\mathbb{Q}(\sqrt{m})$. Hence we have three possibilities for
$\psi_{2}:\mathcal{K(C)}\to\mathcal{K(C)}$: it satisfies $\psi_{2}=\text{Id}$,
$\psi_{2}^{2}=\text{Id}$ or $\text{Im}(\psi_{2})=\mathbb{Z}$. We show none of
these are possible.

We assume first that $\text{Im}(\psi_{2})=\mathbb{Z}$. So we have
$\psi_{2}(X)=\epsilon_{1}\textbf{1}$ and $\psi_{2}(Y)=\epsilon_{2}\textbf{1}$.
The equalities

$\displaystyle 1=[\psi_{2}(X)^{2}:\mathbf{1}]=[\psi_{2}(X^{2}):\mathbf{1}]$
$\displaystyle=[\mathbf{1}+m\psi_{2}(X)+k\psi_{2}(Y):\mathbf{1}]=1+m\epsilon_{1}+k\epsilon_{2},$

imply that $\epsilon_{1}$ and $\epsilon_{2}$ must have opposite signs and that
$m=k$. The analogous computation with $Y$ shows that $l=n$. Moreover, since

$\displaystyle\epsilon_{1}\epsilon_{2}=[\psi_{2}(X)\psi_{2}(Y):\mathbf{1}]=[\psi_{2}(XY):\mathbf{1}]=[k\psi_{2}(X)+l\psi_{2}(Y):\textbf{1}]=k\epsilon_{1}+l\epsilon_{2},$

we have that $l=k\pm 1$. Recall that we are assuming $l\geq k$ and thus
$l=k+1$. Then

$X^{2}=1+kX+(k+1)Y,$

and so

$\displaystyle X^{2}\equiv 1+X\mod 2$ or $\displaystyle X^{2}\equiv 1+Y\mod
2,$

which contradicts $X^{2}\equiv\psi_{2}(X)\equiv 1\mod 2$. So the case
$\text{Im}(\psi_{2})=\mathbb{Z}$ is not possible.

On the other hand, $\psi_{2}=\text{Id}$ is not possible by Theorem 4.8. Thus
it remains to show that we cannot have $\psi_{2}^{2}=\text{Id}$ either.
Suppose $\psi_{2}^{2}=\text{Id}$. Then

$\displaystyle
0=[\psi_{2}^{2}(X):\mathbf{1}]=\epsilon_{1}+\alpha\epsilon_{1}+\beta\epsilon_{2}$
and $\displaystyle
0=[\psi_{2}^{2}(Y):\mathbf{1}]=\epsilon_{2}+\gamma\epsilon_{1}+\delta\epsilon_{2},$

which imply

(5.4) $\displaystyle\beta=-(\alpha+1)\epsilon_{1}\epsilon_{2}$ and
$\displaystyle\gamma=-(\delta+1)\epsilon_{1}\epsilon_{2}.$

Also

$\displaystyle 0=[\psi^{2}_{2}(X):Y]=\beta(\alpha+\delta)$ and $\displaystyle
0=[\psi_{2}^{2}(Y):X]=\gamma(\alpha+\delta).$

If $\alpha+\delta\neq 0$ then $\beta=\gamma=0$, which implies
$\alpha=\delta=-1$. That is,

$\displaystyle\psi_{2}(X)=\epsilon_{1}\textbf{1}-X$ and
$\displaystyle\psi_{2}(Y)=\epsilon_{2}\textbf{1}-Y.$

We compute

$\displaystyle\psi_{2}(X)\psi_{2}(Y)=\epsilon_{1}\epsilon_{2}\textbf{1}+(k-\epsilon_{2})X+(l-\epsilon_{1})Y,$
$\displaystyle\psi_{2}(XY)=k\psi_{2}(X)+l\psi_{2}(Y)=(k\epsilon_{1}+l\epsilon_{2})\textbf{1}-kX-
lY.$

Since the previous equations must be equal, we have that $k-\epsilon_{2}=-k$
and so $2k=\pm 1$, which is not possible.

Hence we should have $\alpha+\delta=0$, and using this together with (5.4) we
get

(5.5) $\displaystyle\alpha\equiv\delta\mod 2,$ $\displaystyle\beta=\gamma\mod
2$ and $\displaystyle\alpha\equiv\beta+1\mod 2.$

On the other hand, from the equation

$\displaystyle
m\alpha+k\gamma=[\psi_{2}(X^{2}):X]=[\psi_{2}(X)^{2}:X]=2\epsilon_{1}\alpha+\alpha^{2}m+2\alpha\beta
k+\beta^{2}l,$

we obtain

$\displaystyle m\alpha+k\gamma\equiv m\alpha+l\beta\mod 2.$

This together with congruences (5.5) gives $k\beta\equiv l\beta\mod 2$.
Lastly,

$\displaystyle[\psi_{2}(XY):X]=k\alpha+l\gamma,$
$\displaystyle[\psi_{2}(X)\psi_{2}(Y):X]=\epsilon_{1}\gamma+\epsilon_{2}\alpha+m\gamma\alpha+k\alpha\delta++k\gamma\beta+l\beta\delta,$

and so

$\displaystyle\gamma+\alpha+m\gamma\alpha+k\alpha\delta+k\gamma\beta+l\beta\delta\equiv
k\alpha+l\gamma\mod 2.$

Note that $\alpha\gamma\equiv 0\equiv\beta\delta\mod 2$, see (5.5). Hence the
equation above is

$\displaystyle\gamma+\alpha+k\alpha\delta+k\gamma\beta\equiv
k\alpha+l\gamma\mod 2.$

Now, using the congruences in (5.5) and $k\beta\equiv l\beta\mod 2$ in the
equation above, we obtain $1\equiv 0\mod 2$. Thus $(\psi_{2})_{\mathbb{Q}}$
cannot be an automorphism of order 2 and this case is not possible.

$\blacktriangledown$ Case 3: $\mathcal{K(C)}$ is a field extension of degree 3
over $\mathbb{Q}$. In this case $(\psi_{2})_{\mathbb{Q}}$ must be an
automorphism of order 3, since it cannot be the identity by Theorem (4.8).
Hence, $X,\psi_{2}(X)$ and $\psi_{2}^{2}(X)$ are distinct roots of the minimal
polynomial of $X$, given by

$m_{X}(t)=t^{3}-(m+l)t^{2}-(1+k^{2}-ml)t+l.$

By the Vieta formulas we have that

(5.6) $\displaystyle
m+l=[X+\psi_{2}(X)+\psi_{2}^{2}(X):\mathbf{1}]=2\epsilon_{1}+\alpha\epsilon_{1}+\beta\epsilon_{2}.$

Repeating this for $Y$ we get

(5.7) $\displaystyle
k+n=[Y+\psi_{2}(Y)+\psi_{2}^{2}(Y):\mathbf{1}]=2\epsilon_{2}+\gamma\epsilon_{1}+\delta\epsilon_{2}.$

On the other hand,

$\displaystyle
1+m\epsilon_{1}+k\epsilon_{2}=[\psi_{2}(X^{2}):\mathbf{1}]=[\psi_{2}(X)^{2}:\mathbf{1}]=1+\alpha^{2}+\beta^{2},\
\ \text{and}$ $\displaystyle
1+l\epsilon_{1}+n\epsilon_{2}=[\psi_{2}(Y^{2}):\mathbf{1}]=[\psi_{2}(Y)^{2}:\mathbf{1}]=1+\gamma^{2}+\delta^{2},$

and thus

$\displaystyle\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}=(m+l)\epsilon_{1}+(k+n)\epsilon_{2}.$

Combining this together with equations (5.6) and (5.7) we get

$\displaystyle\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$
$\displaystyle=(2\epsilon_{1}+\alpha\epsilon_{1}+\beta\epsilon_{2})\epsilon_{1}+(2\epsilon_{2}+\gamma\epsilon_{1}+\delta\epsilon_{2})\epsilon_{2}$
$\displaystyle=2+\alpha+\beta\epsilon_{1}\epsilon_{2}+2+\delta+\gamma\epsilon_{1}\epsilon_{2},$

and so

(5.8)
$\displaystyle\alpha^{2}-\alpha+\beta^{2}-\beta\epsilon_{1}\epsilon_{2}+\gamma^{2}-\gamma\epsilon_{1}\epsilon_{2}+\delta^{2}-\delta=4.$

Hence $|\alpha|,|\beta|,|\gamma|$ and $|\delta|$ are at most 2, which implies
by (5.6) and $\eqref{vieta 2}$ that $m+l,k+n\leq 6$.

From each of the equalities $\psi_{2}(X^{2})=\psi_{2}(X)^{2}$,
$\psi_{2}(Y^{2})=\psi_{2}(Y)^{2}$ and $\psi_{2}(XY)=\psi_{2}(X)\psi_{2}(Y)$ we
get three equations on the parameters
$\alpha,\beta,\gamma,\delta,k,l,m,n,\epsilon_{1},\epsilon_{2}$. The bound
$k,n\leq 6$ allows us to verify in Sage that the only solutions to said
equations fulfilling (5.1) and (5.8) are

$\displaystyle k=1,l=1,m=1,n=0,\alpha=-1,\beta=1,\gamma=-1,\delta=0,\
\text{and}$ $\displaystyle
k=1,l=1,m=0,n=1,\alpha=0,\beta=-1,\gamma=1,\delta=-1.$

By simmetry, it is enough to consider the second case. We have multiplication
rules

$\displaystyle X^{2}=\mathbf{1}+Y$ $\displaystyle Y^{2}=\mathbf{1}+X+Y,$
$\displaystyle XY=YX=X+Y.$

Note that these are the same fusion rules as $\operatorname{Ver}_{7}^{+}$.
Taking dimension on the equalities above we arrive at the equation

(5.9) $\displaystyle\dim(Y)^{3}-2\dim(Y)^{2}-\dim(Y)+1=0.$

On the other hand, note that $\mathcal{C}$ must be degenerate. In fact, if it
was non-degenerate (see Section 2.3) it would lift to a symmetric category
over a field of characteristic zero [E] and thus would have integer Frobenius-
Perron dimensions [EGNO, Theorem 9.9.26]. We compute

$\displaystyle 0=\dim(\mathcal{C})$ $\displaystyle=1+\dim(X)^{2}+\dim(Y)^{2}$
$\displaystyle=1+1+\dim(Y)+\dim(Y)^{2},$

and so

$\displaystyle\dim(Y)^{2}=-2-\dim(Y).$

Replacing this into (5.9) we get

$\displaystyle 0$ $\displaystyle=\dim(Y)^{3}-2\dim(Y)^{2}-\dim(Y)+1$
$\displaystyle=\dim(Y)(-2-\dim(Y))-2(-2-\dim(Y))-\dim(Y)+1$
$\displaystyle=-2\dim(Y)+2+\dim(Y)+4+2\dim(Y)-\dim(Y)+1$ $\displaystyle=7.$

Thus we must have $p=7$. Since
$\mathcal{K(C)}\simeq\mathcal{K}(\operatorname{Ver}_{7}^{+})$ we see that the
fiber functor $F:\mathcal{C}\to\operatorname{Ver}_{7}$ gives an equivalence
onto $\operatorname{Ver}_{7}^{+}$. ∎

######  Remark 5.3.

Theorem 5.2 gives a positive answer for Question 1.1 in the case $p=7$.

##  6\. Rank 4 symmetric fusion categories

###  6.1. Exactly two self-dual simple objects

In this section we classify symmetric fusion categories $\mathcal{C}$ of rank
$4$ with exactly two self-dual simple objects. Namely, we show that
$\mathcal{C}$ is equivalent to one of the following:

  * •

If $p=2$,
$\mathcal{C}\simeq\operatorname{Vec}_{\mathbb{Z}_{4}}^{\mathbb{Z}_{3}}$.

  * •

If $p=3$, $\mathcal{C}\simeq\mathcal{C}\left(\mathbb{Z}_{4},q\right)$, where
$q:\mathbb{Z}_{4}\to\textbf{k}^{\times}$ is a group map satisfying
$q(g)^{2}=1$ for all $g\in\mathbb{Z}_{4}$, see [EGNO, Section 8.4].

  * •

If $p>3$, $\mathcal{C}\simeq\operatorname{Rep}(A_{4})$ or
$\mathcal{C}\simeq\mathcal{C}\left(\mathbb{Z}_{4},q\right)$.

We use the parametrization of based rings of rank 4 with exactly two self-dual
basis elements as given in [L]. Let $c,e,k,l,p,q$ be non-negative integers
satisfying the following equations:

(6.1) $\displaystyle kl+lc=lp+kq,$ (6.2) $\displaystyle kp+le+kc=2lq+k^{2},$
(6.3) $\displaystyle l^{2}+c^{2}=1+q^{2}+p^{2},$ (6.4) $\displaystyle
l^{2}+k^{2}+q^{2}=1+2pk+qe.$

We denote by $K(c,e,k,l,p,q)$ the based ring with basis $1,X,Y,Z$ and
multiplication given by

(6.5) $\displaystyle\begin{aligned} &X^{2}=pX+lY+cZ&&XY=YX=qX+kY+lZ\\\
&Y^{2}=\mathbf{1}+kX+eY+kZ&&YZ=ZY=lX+kY+qZ\\\
&Z^{2}=cX+lY+pZ&&XZ=ZX=\mathbf{1}+pX+qY+pZ.\end{aligned}$

Any based ring of rank 4 with exactly two self-dual basis elements is of the
form $K(c,e,k,l,p,q)$ for some non-negative integers $c,e,k,l,p,q$ satisfying
(6.1)-(6.4), see [L].

When $p>2$, recall that for the second Adams operation
$\psi_{2}:K(c,e,k,l,p,q)\to K(c,e,k,l,p,q)$ we have that $\psi_{2}(W)\equiv
W^{2}\mod 2$ for any $W\in K(c,e,k,l,p,q)$. Hence

(6.6) $\displaystyle\psi_{2}(X)=\alpha_{1}X+\alpha_{2}Y+\alpha_{3}Z,$
$\displaystyle\psi_{2}(Y)=\epsilon\textbf{1}+\beta_{1}X+\beta_{2}Y+\beta_{3}Z,$
$\displaystyle\psi_{2}(Z)=\gamma_{1}X+\gamma_{2}Y+\gamma_{3}Z,$

for some $\epsilon=\pm 1$ and $\alpha_{i},\beta_{i},\gamma_{i}\in\mathbb{Z}$,
$i=1,2,3.$ Recall that $\psi_{2}(W)\equiv W^{2}\mod 2$ and thus

$\displaystyle\alpha_{1}\equiv p\equiv\gamma_{3}\mod 2$
$\displaystyle\alpha_{2}\equiv l\equiv\gamma_{2}\mod 2$
$\displaystyle\alpha_{3}\equiv c\equiv\gamma_{1}\mod 2$
$\displaystyle\beta_{1}\equiv k\equiv\beta_{3}\mod 2$
$\displaystyle\beta_{2}\equiv e\mod 2.$

We will use the previous congruences repeatedly throughout this section.

We start by discarding some possibilities for the Adams operation in the
Lemmas below.

######  Lemma 6.1.

Let $\mathcal{C}$ be a symmetric fusion category of rank $4$ with exactly two
self-dual simple objects. Then $\text{Im}({\psi_{2}})\neq\mathbb{Z}$ .

###### Proof.

Suppose that $\text{Im}({\psi_{2}})=\mathbb{Z}$. Then by Proposition 4.12 we
have that $Y^{2}=\textbf{1}$ and $[X\otimes Z:Y]=1$, which imply $k=e=0$ and
$q=1$. But then from equation (6.4) we get that $l^{2}+1=1$, thus $l=0$. Thus
equation (6.3) is $c^{2}=2+p^{2}$, which has no integer solutions. ∎

######  Lemma 6.2.

Let $\mathcal{C}$ be a symmetric fusion category of rank $4$ with exactly two
self-dual simple objects. Then $\psi_{2}$ does not have order 2.

###### Proof.

Suppose for the sake of contradiction that $\psi_{2}^{2}=\text{Id}$. Then from
$\psi_{2}^{2}(X)=X$ and (6.6) we get the equations

$\displaystyle 0=\alpha_{2}\epsilon,$ $\displaystyle
1=\alpha_{1}^{2}+\alpha_{2}\beta_{1}+\alpha_{3}\gamma_{1},$ $\displaystyle
0=\alpha_{1}\alpha_{3}+\alpha_{2}\beta_{3}+\alpha_{3}\gamma_{3}.$

Thus $\alpha_{2}=0$ and

(6.7) $\displaystyle 1=\alpha_{1}^{2}+\alpha_{3}\gamma_{1},$ $\displaystyle
0=\alpha_{3}(\alpha_{1}+\gamma_{3}).$

Similarly, from $\psi_{2}^{2}(Z)=Z$ we get that $\gamma_{2}=0$ and

(6.8) $\displaystyle 1=\gamma_{1}\alpha_{3}+\gamma_{3}^{2},$ $\displaystyle
0=\gamma_{1}(\alpha_{1}+\gamma_{3}).$

We divide the rest of the proof in two cases.

$\blacktriangledown$ Case 1: $\alpha_{1}\neq-\gamma_{3}$. Then by equations
(6.7) and (6.8) we must have $\alpha_{3}=0=\gamma_{1}$ and
$\alpha_{1}^{2}=1=\gamma_{3}^{2}$, so $\gamma_{3}=\alpha_{1}=\pm 1$. Then

$\displaystyle\epsilon
l=[\psi_{2}(X^{2}):\mathbf{1}]=[\psi_{2}(X)^{2}:\mathbf{1}]=\alpha_{2}^{2}+2\alpha_{1}\alpha_{3}=0,$

which implies $l=0$. On the other hand,

$\displaystyle
1+q\epsilon=[\psi_{2}(XZ):\mathbf{1}]=[\psi_{2}(X)\psi_{2}(Z):\mathbf{1}]=\alpha_{1}\gamma_{3}=1,$

which implies $q=0$. But then by equation (6.3) we have $p=0$ which
contradicts $p\equiv\alpha_{1}\mod 2$.

$\blacktriangledown$ Case 2: $\alpha_{1}=-\gamma_{3}$. Then

$\displaystyle\epsilon
l=[\psi_{2}(X^{2}):\mathbf{1}]=[\psi_{2}(X)^{2}:\mathbf{1}]=2\alpha_{1}\alpha_{3}=-2\gamma_{3}\alpha_{3},$
$\displaystyle\epsilon
l=[\psi_{2}(Z^{2}):\mathbf{1}]=[\psi_{2}(Z)^{2}:\mathbf{1}]=2\gamma_{1}\gamma_{3},$

and so $\gamma_{1}\gamma_{3}=-\gamma_{3}\alpha_{3}$. Suppose first that
$\gamma_{3}\neq 0$. Then $\gamma_{1}=-\alpha_{3}$ and so by equation (6.7)

$\displaystyle 1=\alpha_{1}^{2}-\alpha_{3}^{2},$

which implies $\alpha_{1}=\pm 1$ and $\alpha_{3}=0=\gamma_{1}$. But then

$\displaystyle\epsilon
l=[\psi_{2}(X^{2}):\mathbf{1}]=[\psi_{2}(X)^{2}:\mathbf{1}]=0,$

so $l=0$ and

$\displaystyle p\alpha_{1}=[\psi_{2}(X^{2}):X]=[\psi_{2}(X)^{2}:X]=p.$

Since $p\equiv\alpha_{1}\equiv 1\mod 2$ this implies $\alpha_{1}=1$, and thus
$\gamma_{3}=-1$. But then

$\displaystyle[\psi_{2}(Z^{2}):Z]=[\psi_{2}(cX+pZ):Z]=c\alpha_{3}+p\gamma_{3}=-p,$
$\displaystyle[\psi_{2}(Z)^{2}:Z]=[(\gamma_{3}Z)^{2}:Z]=\gamma_{3}^{2}p=p,$

since $l,\alpha_{3},\gamma_{1}$ and $\gamma_{2}$ are all 0 and
$\gamma_{3}=-1$. Since the two equations above should be equal, then $p=0$
which contradicts $p\equiv 1\mod 2$.

The contradiction came from assuming $\gamma_{3}\neq 0$, and thus we should
have $\gamma_{3}=0=\alpha_{1}$. Hence from the equations

$\displaystyle[\psi_{2}(X^{2}):\mathbf{1}]=[\psi_{2}(pX+lY+cZ):\mathbf{1}]=\epsilon
l,$
$\displaystyle[\psi_{2}(X)^{2}:\mathbf{1}]=[(\alpha_{3}Z)^{2}:\mathbf{1}]=0$

we get that $l=0$. On the other hand, by equation (6.7) we have that
$\alpha_{3}\gamma_{1}=1$, and so

$\displaystyle[\psi_{2}(XZ):\mathbf{1}]=[\psi_{2}(\textbf{1}+pX+qY+pZ):\mathbf{1}]=1+q\epsilon,$
$\displaystyle[\psi_{2}(X)\psi_{2}(Z):\mathbf{1}]=[(\alpha_{3}Z)(\gamma_{1}X):\mathbf{1}]=\alpha_{3}\gamma_{1}=1,$

hence $q=0$. Then from equations (6.3) and (6.4) we conclude $k=1=c$ and
$p=0$. Moreover, $\psi_{2}^{2}(Y)=Y$ implies $0=\epsilon(\beta_{2}+1)$ and so
$\beta_{2}=-1$. Hence

$\displaystyle\epsilon$
$\displaystyle=[\psi_{2}(XY):\mathbf{1}]=[\psi_{2}(X)\psi_{2}(Y):\mathbf{1}]=\beta_{1},$
$\displaystyle\beta_{1}$
$\displaystyle=[\psi_{2}(XY):X]=[\psi_{2}(X)\psi_{2}(Y):X]=\alpha_{3}\beta_{3},$
$\displaystyle-1$
$\displaystyle=[\psi_{2}(XY):Y]=[\psi_{2}(X)\psi_{2}(Y):Y]=-\alpha_{3}$
$\displaystyle\beta_{3}$
$\displaystyle=[\psi_{2}(XY):Z]=[\psi_{2}(X)\psi_{2}(Y):Z]=\epsilon\alpha_{3},$

which imply $1=\alpha_{3}=\gamma_{1}$ and $\epsilon=\beta_{1}=\beta_{3}$.
Lastly

$\displaystyle e\epsilon$
$\displaystyle=[\psi_{2}(Y^{2}):1]=[\psi_{2}(Y)^{2}:1]=\beta_{2}^{2}+2\beta_{1}\beta_{3}=3.$

Thus the fusion rules for $\mathcal{C}$ would be

(6.9) $\displaystyle\begin{aligned} &X^{2}=Z&&XY=Y,\\\
&Y^{2}=1+X+3Y+Z&&ZY=Y,\\\ &Z^{2}=X&&ZX=1.\end{aligned}$

These are the fusion rules of the Izumi-Xu category (see [CMS]). The Frobenius
Perron dimension of $\mathcal{C}$ is $\frac{21+2\sqrt{21}}{2}$, which is not
possible in positive characteristic. In fact, dimensions in
$\operatorname{Ver}_{p}$ are in the field $\mathbb{Q}(z+z^{-1})$, see [BEO,
Theorem 4.5 (iv)]. Since we have a symmetric fiber functor
$F:\mathcal{C}\to\operatorname{Ver}_{p}$, the same should be true for
$\mathcal{C}$, which makes the obtained dimension not possible. ∎

######  Remark 6.3.

The Adams operation is not enough on its own to classify symmetric fusion
categories in positive characteristic. In fact, in the proof of the previous
Lemma we found a possible based ring (6.9) and a suitable Adams operation,
given by

$\displaystyle\psi_{2}(X)=Z,$ $\displaystyle\psi_{2}(Y)=\mathbf{1}+X-Y+Z,$
$\displaystyle\psi_{2}(Z)=X.$

However, as stated before this ring is not categorificable.

######  Lemma 6.4.

Let $\mathcal{C}$ be a symmetric fusion category of rank $4$ with exactly two
self-dual simple objects. Then $\psi_{2}^{2}\neq\psi_{2}$.

###### Proof.

Suppose that $\psi_{2}^{2}=\psi_{2}$. From the equality
$\psi_{2}^{2}(X)=\psi_{2}(X)$ we get the equations

$\displaystyle 0=\alpha_{2}\epsilon,$
$\displaystyle\alpha_{1}=\alpha_{1}^{2}+\alpha_{2}\beta_{1}+\alpha_{3}\gamma_{1},$
$\displaystyle\alpha_{3}=\alpha_{1}\alpha_{3}+\alpha_{2}\beta_{3}+\alpha_{3}\gamma_{3}.$

Thus $\alpha_{2}=0$ and $\alpha_{3}=\alpha_{3}(\alpha_{1}+\gamma_{3})$. If
$\alpha_{3}\neq 0$ then $\alpha_{1}+\gamma_{3}=1$ and so $p+p\equiv 1\mod 2$
which is a contradiction. Thus we have $\alpha_{3}=0$ and $\alpha_{1}=1$ or
$0$. Analogously, from $\psi_{2}^{2}(Z)=\psi_{2}(Z)$ we get
$\gamma_{2}=0=\gamma_{1}$ and $\gamma_{3}=1$ or $0$. Lastly, from
$\psi_{2}^{2}(Y)=\psi_{2}(Y)$ we have that $\beta_{2}=0$.

On the other hand,

$\displaystyle\epsilon
l=[\psi_{2}(X^{2}):\mathbf{1}]=[\psi_{2}(X)^{2}:\mathbf{1}]=\alpha_{2}^{2}+2\alpha_{1}\alpha_{3}=0,$

which implies $l=0$, and so

$\displaystyle
c\gamma_{3}=[\psi_{2}(X^{2}):Z]=[\psi_{2}(X)^{2}:Z]=c\alpha_{1}^{2}.$

If $c=0$ then $0=1+q^{2}+p^{2}$ by equation (6.3), which is not possible. Thus
$\gamma_{3}=\alpha_{1}^{2}=\alpha_{1}$. We divide the rest of the proof in two
cases:

$\blacktriangledown$ Case 1: $\alpha_{1}=\gamma_{3}=0$. Then

$\displaystyle\epsilon
k=[\psi_{2}(XY):1]=[\psi_{2}(X)\psi_{2}(Y):1]=\alpha_{1}\beta_{3}+\alpha_{3}\beta_{1}+\alpha_{2}\beta_{2}=0,$

and so $k=0$. Then by equation (6.4) we have $q^{2}=1+qe$, thus $q=1$ and so
$c^{2}=2+p^{2}$ by equation (6.3), which has no integer solutions.

$\blacktriangledown$ Case 2: $\alpha_{1}=\gamma_{3}=1$. Then

$\displaystyle 0=[\psi_{2}(XY):Y]=[\psi_{2}(X)\psi_{2}(Y):Y]=q\beta_{3},$

so $q=0$ or $0=\beta_{3}=\beta_{1}$. If $q=0$ then $c^{2}=1+p^{2}$ by Equation
(6.3) and so $c=1$ and $p=0$. Then $1\equiv c\equiv\alpha_{3}\equiv 0\mod 2$.
Thus we must have $0=\beta_{3}$ and so

$\displaystyle\epsilon
k=[\psi_{2}(XY):\mathbf{1}]=[\psi_{2}(X)\psi_{2}(Y):\mathbf{1}]=\beta_{3}=0,$

which implies $k=0$. Since $q^{2}=1+qe$ by Equation (6.4) we get $q=1$. But
then $c^{2}=2+p^{2}$ by equation (6.3), which has no integer solutions. ∎

We will need the following auxiliary Lemma.

######  Lemma 6.5.

If $\mathcal{C}$ is a fusion category with commutative $\mathcal{K(C)}$ and a
non-self-dual object, then there exists an algebra homomorphism
$\mathcal{K(C)}\to\mathbb{C}$ whose image is not contained in $\mathbb{R}$.

###### Proof.

Let $X\in\mathcal{C}$ be a non-self-dual object, and consider the map of
multiplication by $X-X^{*}$ in $\mathcal{K(C)}_{\mathbb{C}}$. In the basis
given by simple objects, we can represent this map by a non-trivial skew
symmetric matrix. Thus its eigenvalues are zero or non-real. Since
$X\not\simeq X^{*}$ there must exist at least one non-real eigenvalue
$\lambda\neq 0$. Hence $\mathcal{K(C)}_{\mathbb{C}}$ has a $1$ dimensional
representation where $X$ acts as multiplication by $\lambda$. ∎

######  Theorem 6.6.

Let $\mathcal{C}$ be a symmetric fusion category of rank 4 with exactly 2
self-dual simple objects. Then $\mathcal{C}$ is integral.

###### Proof.

We proceed by looking at the different possibilities for
$\mathcal{K(C)}_{\mathbb{Q}}$. Since $\mathcal{K(C)}_{\mathbb{Q}}$ is a
semisimple commutative algebra of dimension $4$, we have five cases:

$\blacktriangledown$ Case 1:
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}$.
Note that field maps $\mathcal{K(C)}_{\mathbb{Q}}\to\mathbb{C}$ are
projections to $\mathbb{Q}$ and so
$\operatorname{FPdim}_{\mathbb{Q}}:\mathcal{K(C)}_{\mathbb{Q}}\to\mathbb{C}$
has rational image. Thus $\mathcal{C}$ is integral.

$\blacktriangledown$ Case 2:
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}(\sqrt{n})\times\mathbb{Q}\times\mathbb{Q}$.
By Remark 6.5 we must have $n<0$. But then
$\operatorname{FPdim}_{\mathbb{Q}}:\mathcal{K(C)}_{\mathbb{Q}}\to\mathbb{R}$
can only have rational image and so $\mathcal{C}$ is integral.

$\blacktriangledown$ Case 3:
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}(\sqrt{n})\times\mathbb{Q}(\sqrt{m})$
with $\mathbb{Q}(\sqrt{n})\not\simeq\mathbb{Q}(\sqrt{m})$. Endomorpshisms of
$\mathbb{Q}(\sqrt{n})\times\mathbb{Q}(\sqrt{m})$ are given by

$\displaystyle(1,0)\mapsto(1,0)$ $\displaystyle(0,1)\mapsto(0,1)$
$\displaystyle(\sqrt{n},0)\mapsto(\pm\sqrt{n},0)$
$\displaystyle(0,\sqrt{m})\mapsto(0,\pm\sqrt{m}).$

These are all automorphisms of order 1 or 2, which is not possible for
$(\psi_{2})_{\mathbb{Q}}$ by Lemma 6.2. Hence this case is discarded.

$\blacktriangledown$ Case 3:
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}(\sqrt{n})\times\mathbb{Q}(\sqrt{n})$.
By Lemma 6.5 we have $n<0$. But morphisms from
$\mathbb{Q}(\sqrt{n})\times\mathbb{Q}(\sqrt{n})$ to $\mathbb{C}$ are
embeddings onto $\mathbb{Q}(\sqrt{n})$. This contradicts the fact that
$\operatorname{FPdim}:\mathbb{Q}(\sqrt{n})\times\mathbb{Q}(\sqrt{n})\to\mathbb{C}$
should have real image, and so this case is discarded.

$\blacktriangledown$ Case 4:
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\textbf{F}\times\mathbb{Q}$, where F is a
field extension of degree 3 over $\mathbb{Q}$. The only endomorphism of
$\textbf{F}\times\mathbb{Q}$ with non-trivial kernel is given by

$\displaystyle(1,0)\mapsto(1,1)$ $\displaystyle(0,1)\mapsto(0,0)$
$\displaystyle(0,a)\mapsto(0,0).$

By Lemma 6.1 we know that $(\psi_{2})_{\mathbb{Q}}$ is not of this form.

On the other hand, non trivial automorphisms of $\textbf{F}\times\mathbb{Q}$
have order 3 which is not possible for $(\psi_{2})_{\mathbb{Q}}$ by Theorem
4.11. Hence this case is also discarded.

$\blacktriangledown$ Case 5: $\mathcal{K(C)}_{\mathbb{Q}}$ is a fied extension
of degree $4$ over $\mathbb{Q}$. Since $(\psi_{2})_{\mathbb{Q}}\neq\text{Id}$
by Theorem 4.8, then it must be an automorphisms of order 2 or 4. Using Lemma
6.2 we can discard the former possibility. If $(\psi_{2})_{\mathbb{Q}}$ has
order $4$ then $Y,\psi_{2}(Y),\psi_{2}^{2}(Y)$ and $\psi_{2}^{3}(Y)$ are
distinct roots of the minimal polynomial of $Y$. Since the minimal polynomial
of $Y$ is given by

$m_{Y}(t)=t^{4}+(-2q-e)t^{3}+(2qe+q^{2}+-k^{2}-l^{2}-1)t^{3}+(-q^{2}e+qk^{2}-lk^{2}+l^{2}e+2q)t+l^{2}-q^{2},$

then by the Vieta formulas the sum of the roots equals $2q+e$. Hence

$\displaystyle[Y+\psi_{2}(Y)+\psi_{2}^{2}(Y)+\psi_{2}^{3}(Y):\mathbf{1}]=2q+e.$

We compute

$\displaystyle[\psi_{2}(Y):\mathbf{1}]=\epsilon,$
$\displaystyle[\psi_{2}^{2}(Y):\mathbf{1}]=\epsilon+\beta_{2}\epsilon,$
$\displaystyle[\psi_{2}^{3}(Y):\mathbf{1}]=\epsilon+\beta_{2}\epsilon+\epsilon(\beta_{1}\alpha_{2}+\beta_{2}^{2}+\beta_{3}\gamma_{2}),$

and thus

$\displaystyle\epsilon+\epsilon+\beta_{2}\epsilon+\epsilon+\beta_{2}\epsilon+\epsilon(\beta_{1}\alpha_{2}+\beta_{2}^{2}+\beta_{3}\gamma_{2})=2q+e.$

Taking congruence mod 2 on both sides we get

$\displaystyle e\equiv 1+1+e+1+e+kl+e+kl\equiv 1+e\mod 2,$

which is not possible, so this case is also discarded. ∎

######  Theorem 6.7.

Let $\mathcal{C}$ be an integral symmetric fusion category of rank 4 with
exactly 2 self-dual simple objects. Then either

  * •

$\mathcal{C}\simeq\mathcal{C}\simeq\mathcal{C}\left(\mathbb{Z}_{4},q\right)$
and $p>2$, or

  * •

$\mathcal{C}\simeq\operatorname{Rep}(A_{4})$ and $p>3$, or

  * •

$\mathcal{C}\simeq\operatorname{Vec}_{\mathbb{Z}_{4}}^{\mathbb{Z}_{3}}$ and
$p=2$.

###### Proof.

Let $\mathcal{C}$ be as in the statement with Grothendieck ring
$K(c,e,k,l,p,q)$, see equation 6.5. Since $X^{*}=Z$ we have that
$\operatorname{FPdim}(X)=\operatorname{FPdim}(Z)$. Thus taking Frobenius-
Perron dimensions on both sides of the fusion rule $Y^{2}=\mathbf{1}+kX+eY+kZ$
we get

$\displaystyle\operatorname{FPdim}(Y)(\operatorname{FPdim}(Y)-e)=1+2k\operatorname{FPdim}(X),$

and so $\gcd(\operatorname{FPdim}(X),\operatorname{FPdim}(Y))=1$. From the
fusion rule $XY=qX+kY+lZ$ we get
$1=\frac{q+l}{\operatorname{FPdim}(Y)}+\frac{k}{\operatorname{FPdim}(X)}$ and
so since the denominators are coprime either $\operatorname{FPdim}(Y)=q+l$ and
$k=0$ or $q+l=0$ and $\operatorname{FPdim}(X)=k$. We split the rest of the
proof in two cases.

$\blacktriangledown$ Case 1: $\operatorname{FPdim}(Y)=q+l$ and $k=0$. Since
$Y^{2}=\mathbf{1}+eY$ then

$\operatorname{FPdim}(Y)=\frac{e+\sqrt{e^{2}+4}}{2}.$

But $\operatorname{FPdim}(Y)$ is an integer and then $e^{2}+4$ must be a
square, so $e=0$. Hence $Y^{2}=\mathbf{1}$ which implies
$q+l=\operatorname{FPdim}(Y)=1$. Recall that $q$ and $l$ are non-negative
integers, and so $q=1$ and $l=0$ or $q=0$ and $l=1$. The former is not
possible, since in that case $c^{2}=2+p^{2}$ by equation (6.3), which has no
integer solutions. Hence we have that $q=0$ and $l=1$, and so
$\operatorname{FPdim}(Y)=1$. Moreover, the fusion rule $X^{2}=pX+Y+pZ$ implies
that

$\operatorname{FPdim}(X)^{2}=2p\operatorname{FPdim}(X)+1.$

This has integer solution only for $p=0$, in which case
$\operatorname{FPdim}(X)=1$. Lastly by equation (6.3) we have $c=0$. Thus the
fusion rules are

$\displaystyle X^{2}=Y$ $\displaystyle XY=YX=Z$ $\displaystyle
Y^{2}=\mathbf{1}$ $\displaystyle YZ=ZY=X$ $\displaystyle Z^{2}=Y$
$\displaystyle XZ=ZX=\mathbf{1}.$

Hence the category is pointed and
$\mathcal{C}\simeq\mathcal{C}\left(\mathbb{Z}_{4},q\right)$, where
$q:\mathbb{Z}_{4}\to\textbf{k}^{\times}$ is a group map satisfying
$q(g)^{2}=1$ for all $g\in\mathbb{Z}_{4}$, see [EGNO, Section 8.4].

$\blacktriangledown$ Case 2: $\operatorname{FPdim}(X)=k$ and $q+l=0$. Since
$q$ and $l$ are non-negative integers this implies $q=0=l$. The fusion rule
$XZ=\mathbf{1}+pX+pZ$ implies that
$\operatorname{FPdim}(X)^{2}=1+2p\operatorname{FPdim}(X)$ and so
$\operatorname{FPdim}(X)=1$ and $p=0$. On the other hand, from equations (6.3)
and (6.4) we get that $k=1=c$. Lastly, since $Y^{2}=\mathbf{1}+X+eY+Z$ we have
that

$\operatorname{FPdim}(Y)=\frac{e+\sqrt{e^{2}+12}}{2}.$

Thus for $\operatorname{FPdim}(Y)$ to be an integer we need $e=2$.
Consequently. the fusion rules are

$\displaystyle X^{2}=Z$ $\displaystyle XY=YX=Y$ $\displaystyle
Y^{2}=\mathbf{1}+X+2Y+Z$ $\displaystyle YZ=ZY=Y$ $\displaystyle Z^{2}=X$
$\displaystyle XZ=ZX=\mathbf{1}.$

When $p>3$ we have that $\dim(\mathcal{C})=12\neq 0$ and thus we can lift
$\mathcal{C}$ to a symmetric fusion category $\widetilde{\mathcal{C}}$ over a
field f in characteristic zero [E], which has the same Grothendick ring as
$\mathcal{C}$. Thus $\widetilde{\mathcal{C}}$ is equivalent to
$\operatorname{Rep}_{\textbf{f}}(A_{4})$ [D], and so by uniqueness of the
lifting we get that $\mathcal{C}$ is equivalent to
$\operatorname{Rep}_{\textbf{k}}(A_{4})$.

For the case $p=2$ we have that $\mathcal{C}$ contains a copy of
$\operatorname{Rep}(\mathbb{Z}_{3})$. Doing de-equivariantization by
$\mathbb{Z}_{3}$ we obtain a symmetric fusion category
$\mathcal{C}^{\mathbb{Z}_{3}}$ of dimension $4$ [EGNO, Section 2.7]. Hence
$\mathcal{C}^{\mathbb{Z}_{3}}$ is pointed and thus equivalent to
$\operatorname{Vec}_{\mathbb{Z}_{4}}$. There’s only one action of
$\mathbb{Z}_{3}$ on $\mathbb{Z}_{4}$, and so doing equivariantization by
$\mathbb{Z}_{3}$ gives us back $\mathcal{C}$. Hence
$\mathcal{C}\simeq\text{Vec}_{\mathbb{Z}_{4}}^{\mathbb{Z}_{3}}$.

Lastly, for $p=3$ there is no category realizing these fusion rules. In fact,
we know all symmetric fusion categories in characteristic $3$. Any such
category is an equivariantization of a pointed category associated with a
$3$-group by the action of a group of order relatively prime to $3$, see [EOV,
Section 8]. Thus the category should contain a Tannakian subcategory of order
prime to $3$, which is not possible with the fusion rules above.

∎

###  6.2. All simple objects are self-dual

We are not able to provide a classification of symmetric fusion categories of
rank $4$, but here are some comments for the remaining case, in which all
simple objects are self dual.It follows from Theorem 3.1 that we can have
examples of such categories that are non super Tannakian only in
characteristics $p=5$ or $7$.

We take a look first at the case $p=5$.

######  Proposition 6.8.

Let $\mathcal{C}$ be a non super Tannakian symmetric fusion category of rank
$4$ in characteristic $p=5$. Then
$\mathcal{K}(\mathcal{C})_{\mathbb{Q}}\simeq\mathbb{Q}(\sqrt{5})\oplus\mathbb{Q}^{\oplus
2}$ or
$\mathcal{K}(\mathcal{C})_{\mathbb{Q}}\simeq\mathbb{Q}(\sqrt{5})\oplus\mathbb{Q}(\sqrt{m})$
for some $m\in\mathbb{Z}$. In any case, the Adams operation
$\psi_{2}:\mathcal{K}(\mathcal{C})\to\mathcal{K(C)}$ satisfies
$\psi_{2}^{2}=\text{Id}$.

###### Proof.

Since $\mathcal{K(C)}_{\mathbb{Q}}$ is a semisimple commutative algebra of
dimension $4$, it can either be $\mathbb{Q}^{\oplus 4}$, $\mathbb{Q}^{\oplus
2}\oplus\mathbb{Q}(\sqrt{n})$,
$\mathbb{Q}(\sqrt{n})\oplus\mathbb{Q}(\sqrt{m})$,
$\mathbb{Q}(a)\oplus\mathbb{Q}$ or $\mathbb{Q}(b)$, for $n,m\in\mathbb{Z}$,
and $a,b\in\mathbb{Q}$ such that $[\mathbb{Q}(a):\mathbb{Q}]=3$ and
$[\mathbb{Q}(b):\mathbb{Q}]=4$.

Consider the Verlinde fiber functor $F:\mathcal{C}\to\operatorname{Ver}_{5}$.
We denote also by $F$ the induced algebra homomorphism

$\displaystyle
F:\mathcal{K(C)}_{\mathbb{Q}}\to\mathcal{K}(\operatorname{Ver}_{p})_{\mathbb{Q}}.$

By the proof of Theorem 3.1, since $\mathcal{C}$ is not super Tannakian then

$F(\mathcal{K(C)}_{\mathbb{Q}})=\mathcal{K}(\operatorname{Ver}_{p}^{+})_{\mathbb{Q}}\simeq\mathbb{Q}(\xi_{5}+\xi_{5}^{-1})=\mathbb{Q}(\sqrt{5}).$

That is, the image of $\mathcal{K(C)}_{\mathbb{Q}}$ through $F$ is
$\mathbb{Q}(\sqrt{5})$. Then the only remaining possibilities for
$\mathcal{K(C)}_{\mathbb{Q}}$ are $\mathbb{Q}^{\oplus
2}\oplus\mathbb{Q}(\sqrt{5})$ or
$\mathbb{Q}(\sqrt{5})\oplus\mathbb{Q}(\sqrt{m})$ for some $m\in\mathbb{Z}$.

In any case, if $f\in\operatorname{End}(\mathcal{K(C)}_{\mathbb{Q}})$, then
$f=1$, $f^{2}=f$ or $f^{2}=1$. The first two are not possible for
$(\psi_{2})_{\mathbb{Q}}\in\operatorname{End}(\mathcal{K(C)}_{\mathbb{Q}})$ by
Theorems 4.8 and 4.6. Hence we must have that $(\psi_{2})_{\mathbb{Q}}$ is an
automorphism of order $2$. ∎

For $p=7$, we have the following result.

######  Proposition 6.9.

Let $\mathcal{C}$ be a non-Tannakian symmetric fusion category of rank $4$ in
characteristic $p=7$. Then
$\mathcal{K(C)}_{\mathbb{Q}}\simeq\mathbb{Q}\times\mathbb{Q}(a)$ for some $a$
such that $[\mathbb{Q}(a):\mathbb{Q}]=3$. Moreover, $\psi_{2}^{3}=\text{Id}$.

###### Proof.

Since $\mathcal{K(C)}_{\mathbb{Q}}$ is a semisimple commutative algebra of
dimension $4$, it can either be
$\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}$,
$\mathbb{Q}(\sqrt{n})\times\mathbb{Q}\times\mathbb{Q}$,
$\mathbb{Q}(\sqrt{n})\times\mathbb{Q}(\sqrt{m})$,
$\mathbb{Q}(a)\times\mathbb{Q}$ or $\mathbb{Q}(b)$, for $a,b\in\mathbb{C}$
such that $[\mathbb{Q}(a):\mathbb{Q}]=3$ and $[\mathbb{Q}(b):\mathbb{Q}]=4.$

Thus we have that if $f\in\operatorname{End}(\mathcal{K(C)}_{\mathbb{Q}})$
then either $f^{n}=\operatorname{Id}$ for $n=1,2,3,4$, $f^{k}=f$ for $k=2,3$
or $f^{3}=f^{2}$. By Remark 4.7, the only possibility for
$(\psi_{2})_{\mathbb{Q}}\in\operatorname{End}(\mathcal{K(C)}_{\mathbb{Q}})$ is
that $(\psi_{2})_{\mathbb{Q}}^{3}=\text{Id}$, which can only happen if
$\mathcal{K(C)}\simeq\mathbb{Q}\times\mathbb{Q}(a)$, as desired. ∎

## References

  * [BEO] D. Benson, P. Etingof, V. Ostrik. _New incompressible symmetric tensor categories in positive characteristic,_ Duke Math. J. (to appear).
  * [BW] J. Barrett, B. Westbury. _Spherical categories_ , Adv. Math. 143, 357-375 (1999).
  * [CMS] F. Calegari, S. Morrison, and N. Snyder. _Cyclotomic integers, fusion categories, and subfactors._ Comm. Math. Phys., 303(3):845–896, (2011).
  * [D] P. Deligne. _Catégories tannakiennes_ , The Grothendieck Festschrift, Vol. II, 111-195, Progr. Math., 87, Birkhäuser Boston (1990).
  * [DG] M. Demazure, P. Gabriel. _Groupes algébriques, tome I_ , Paris / Amsterdam (1970).
  * [DGNO] V. Drinfeld, S. Gelaki, D. Nikshych, V. Ostrik. _On braided fusion categories I_ , Selecta Mathematica (N. S.) 16, 1-119 (2010).
  * [DM] P. Deligne, J. Milne _Tannakian categories_ , Lecture Notes in Mathematics 900 (1982).
  * [E] Etingof, P. _On faithfulness of the lifting for Hopf algebras and fusion categories._ Algebra Number Theory 12 (3), 551–569 (2018)
  * [EGNO] P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik. _Tensor categories_ , AMS, Providence (2015).
  * [EHO] P. Etingof, N. Harman, V. Ostrik. _p-adic dimensions in symmetric tensor categories in characteristic p_ , Quantum Topol. 9, no. 1, pp. 119–140 (2018).
  * [ENO] P. Etingof, D. Nikshych, V. Ostrik. _On fusion categories_ , Annals of Mathematics 162, 581-642 (2005).
  * [EOV] P. Etingof, V. Ostrik, S. Venkatesh. _Computations on symmetric fusion categories in characteristic $p$_, IMRN 2:468–489 (2017).
  * [L] H. K. Larson. _Pseudo-unitary non-self-dual fusion categories of rank 4_ , Journal of Algebra, Vol. 415 (2014), 184-213.
  * [M] A. Masuoka. _Harish-Chandra pairs for algebraic affine supergroup schemes over an arbitrary field_ Transformation Groups 17, 1085–1121 (2012).
  * [O1] V. Ostrik. _Pivotal fusion categories of rank 3_ , Mosc. Math. J., 15:2 (2015), 373–396.
  * [O2] V. Ostrik. _On symmetric fusion categories in positive characteristic_ , Selecta Mathematica (2020).
  * [O3] V. Ostrik. Private communication (2022).
  * [SR] N. Saavedra Rivano. _Catégories Tannakiennes_ , Lecture Notes in Math 265, Springer-Verlag, Berlin-New York (1972).