Frobenius-Perron dimensions of integral $\Bbb Z_+$-rings and applications

TL;DR

This paper introduces the Frobenius-Perron dimension for integral _+-rings and applies it to classify certain finite-dimensional quasi-Hopf and Hopf algebras, especially those with prime dimension or a single nontrivial simple module.

Contribution

It defines the Frobenius-Perron dimension for integral _+-rings and demonstrates its usefulness in classifying specific classes of quasi-Hopf and Hopf algebras.

Findings

Classification of finite-dimensional quasi-Hopf algebras with a unique nontrivial simple module.

Results on the structure of quasi-Hopf and Hopf algebras of prime dimension p.

Abstract

We introduce the notion of the Frobenius-Perron dimension of an integral $\Bbb Z_+$-ring and give some applications of this notion to classification of finite dimensional quasi-Hopf algebras with a unique nontrivial simple module, and of quasi-Hopf and Hopf algebras of prime dimension $p$.

Equations

$$d\ge N^{\frac{1}{r-1}}{\left(1-\frac{s+1}{r}\right)}^{1-\frac{s}{r-1}},$$

$$d\ge N^{\frac{1}{r-1}}{\left(\frac{\lceil \frac{r-1}{2}\rceil }{r}\right)}^{\frac{\lceil \frac{r-1}{2}\rceil }{r-1}}.$$

$$Q(d)=Q(d)p\cdot v=pQ(b)v=mp\cdot d=mN.$$

$$Q(d)=\prod _{i=1}^{r-2s-1}(d-{\lambda }_i)\prod _{j=1}^s|d-{\mu }_j{|}^2\ge N.$$

$$\prod _{i=1}^{r-2s-1}(1-{\alpha }_i)\prod _{j=1}^s|1-{\beta }_j{|}^2\ge \frac{N}{d^{r-1}}.$$

$$\prod _{i=1}^{r-2s-1}(1-{\alpha }_i)\prod _{j=1}^s(1-2\mathrm{Re}{\beta }_j+|{\beta }_j{|}^2)\ge \frac{N}{d^{r-1}}.$$

$$\prod _{i=1}^{r-2s-1}(1-{\alpha }_i)\prod _{j=1}^s(2-2\mathrm{Re}{\beta }_j)\ge \frac{N}{d^{r-1}}.$$

$$\sum _{i=1}^{r-2s-1}(1-{\alpha }_i)+\sum _{j=1}^s(2-2\mathrm{Re}{\beta }_j)\ge (r-s-1){\left(\frac{N}{d^{r-1}}\right)}^{\frac{1}{r-s-1}}.$$

$$r\ge (r-s-1){\left(\frac{N}{d^{r-1}}\right)}^{\frac{1}{r-s-1}}.$$

$$d\ge N^{\frac{1}{r-1}}{\left(1-\frac{s+1}{r}\right)}^{1-\frac{s}{r-1}},$$

$${\left(\frac{\lceil \frac{r-1}{2}\rceil }{r}\right)}^{\frac{\lceil \frac{r-1}{2}\rceil }{r-1}}\ge \frac{1}{2}.$$

$$(a-\lambda )hv=h(a-\lambda )v+(S^2-1)(h)av$$

$$(a-\lambda {)}^{N}hv=\sum _{i=0}^N(\genfrac{}{}{0ex}{}{N}{i})(S^2-1{)}^i(h)(a-\lambda {)}^{N-i}{a}^{i}v.$$

$$(a-\lambda )\sum _{i=0}^N\left(\genfrac{}{}{0ex}{}{N}{i}\right)(S^2-1{)}^i(h)(a-\lambda {)}^{N-i}{a}^{i}v=$$

$$=\sum _{i=0}^N\left(\genfrac{}{}{0ex}{}{N}{i}\right)(S^2-1{)}^i(h)(a-\lambda {)}^{N+1-i}{a}^{i}v+\sum _{i=0}^N\left(\genfrac{}{}{0ex}{}{N}{i}\right)(S^2-1{)}^{i+1}(h)(a-\lambda {)}^{N-i}{a}^{i+1}v,$$

$$V\otimes V^{\ast }\to V^{\ast \ast }\otimes V^{\ast }\to 1\to V\otimes V^{\ast }$$

$$P_X\otimes X^{\ast }=X\otimes X^{\ast }\oplus Y.$$

$$N-1\ge d\ge N/2,$$

$$|\det C|=|c_{00}{c}_{11}-c_{10}{c}_{01}|\le \max (c_{00}{c}_{11},c_{10}{c}_{01})\le$$

$$\le \frac{1}{4d^2}\max ((c_{00}+c_{11}{d}^2{)}^2,(c_{10}+c_{01}{)}^2{d}^2)\le$$

$$\le \frac{1}{4d^2}(c_{00}+(c_{01}+c_{10})d+c_{11}{d}^2{)}^2=\frac{\mathrm{FPdim}(\mathcal{C}{)}^2}{4d^2}.$$

$$\mathrm{End}(W\otimes X)=\mathrm{Hom}(W^{\ast }\otimes W,X\otimes X)=\mathrm{Hom}(P_1,X\otimes X)=\mathrm{Hom}(X\otimes P_1,X).$$

$$\mathrm{FPdim}(\mathcal{C})\ge 4+2\cdot 8+3\cdot 8+4s+16s+8(r-2s-3)=8r+20+4s\ge 8r+20,$$

$$p\ge 1+2d^2+m^2>3d^2\ge p,$$

$$p\ge 1+d^2+2m^2>3d^2\ge p,$$

$$d\ge 2^{-2/3}{p}^{\frac{1}{r-1}}.$$

$$f(r):={\left(\frac{\lceil \frac{r-1}{2}\rceil }{r}\right)}^{\frac{\lceil \frac{r-1}{2}\rceil }{r-1}}$$

$$p\ge (q+2)\left(d_i+\frac{1}{d_i}\sum _{k\ne i}\max (d_i{d}_k/d_{\ast },1)\right).$$

$$p\ge (q+2)\left(d_i+\frac{{\sum }_{k\ne i}{d}_k}{d_i}\right).$$

$$p\ge (q+2)\left(d_i+\frac{2r-3}{d_i}\right).$$

Full text

To Nicolás Andruskiewitsch on his 60th birthday with admiration

1. Introduction

The goal of this paper is to define and study the notion of the Frobenius-Perron dimension of an integral $\mathbb{Z}_{+}$-ring $A$ which is not necessarily a fusion ring (i.e., does not necessarily have a *-structure). Namely, if $N_{i}$ are the matrices of multiplication by the basis vectors of $A$ and $d_{i}$ are their largest positive eigenvalues, then we let $\mathbb{p}$ be the left eigenvector of $N_{i}$ with eigenvalues $d_{i}$, normalized to have positive integer entries with greatest common divisor $1$. Then we set ${\rm FPdim}(A):=\sum_{i}p_{i}d_{i}$. For fusion rings $p_{i}=d_{i}$ so ${\rm FPdim}(A)=\sum_{i}d_{i}^{2}$, which coincides with the standard definition in [EGNO]. Also, if $A$ is the Grothendieck ring of a finite tensor category $\mathcal{C}$ then ${\rm FPdim}(A)={\rm FPdim}(\mathcal{C})/D$, where $D$ is the greatest common divisor of the dimensions of the indecomposable projectives of $\mathcal{C}$.

We prove a number of properties of ${\rm FPdim}(A)$. In particular, we show that if $X\in A$ is a $\mathbb{Z}_{+}$-generator of $A$ and $r$ the rank of $A$ then $(2d)^{r-1}\geq{\rm FPdim}(A)$, where $d$ is the Frobenius-Perron dimension of a $\mathbb{Z}_{+}$-generator of $A$.

We then give two applications of this notion. The first application is to classification of quasi-Hopf and Hopf algebras with two simple modules. We give a number of examples of such quasi-Hopf and Hopf algebras and prove a number of restrictions on their structure. The second application is to quasi-Hopf and Hopf algebras $H$ of prime dimension $p$. It is conjectured that any such quasi-Hopf algebra is commutative and cocommutative, but this is not known even in characteristic zero, and we prove some restrictions on the structure of $H$ if it is not commutative and cocommutative; e.g., we show that $H$ has to have at least four simple modules. In particular, it follows that a quasi-Hopf algebra of prime dimension $\leq 31$ over any field is necessarily commutative and cocommutative. In the case of Hopf algebras, it is known that if $H$ is not commutative and cocommutative then $q>2$ and $\frac{p}{q}>4$ ([NW]). We improve this bound to $\frac{p}{q+2}>\frac{14}{3}$ (Corollary 8.2). Also, we show that $\frac{p}{q+2}>{\rm min}(9,\phi(p))$, where $\phi$ is a certain function such that $\phi(p)\sim 2\sqrt{2}\left(\frac{\log p}{\log\log p}\right)^{1/2}$ as $p\to\infty$ (Theorem 8.6). This means that if $p$ is large enough, we must have $\frac{p}{q+2}>9$. Moreover, we show that if $H$ has a nontrivial simple module $X$ with $X\cong X^{**}$ then $\frac{p}{q+2}>\phi(p)$.

The organization of the paper is as follows. In Section 2 we define the Frobenius-Perron dimension of an integral $\mathbb{Z}_{+}$-ring $A$ and prove basic properties of this notion. In Section 3 we prove the lower bound on the dimension of a $\mathbb{Z}_{+}$-generator of $A$. In Section 4 we prove some auxiliary lemmas. In Section 5 we give applications to quasi-Hopf algebras with two simple modules. In Section 6 we prove some bounds for dimensions of tensor categories with no nontrivial invertible objects. In Section 7 we give applications to classification of quasi-Hopf algebras of prime dimension. Finally, in Section 8 we give applications to classification of Hopf algebras of prime dimension, improving the bound on the characteristic from [NW].

Acknowledgements. The author is grateful to C. Negron for useful discussions and to V. Ostrik, S.-H. Ng and X. Wang for corrections and comments on the draft of this paper. The work of the author was partially supported by the NSF grant DMS-1502244.

2. The Frobenius-Perron dimension of an integral $\mathbb{Z}_{+}$-ring

2.1. Dimensions in integral $\mathbb{Z}_{+}$-rings.

Let $A$ be a transitive unital $\mathbb{Z}_{+}$-ring of finite rank ([EGNO], Definitions 3.1.1, 3.3.1). This means that $A$ is a free finitely generated abelian group with basis $b_{i},i\in I$ and a unital associative ring structure such that $b_{i}b_{j}=\sum_{k}N_{ij}^{k}b_{k}$ with $N_{ij}^{k}\in\mathbb{Z}_{+}$ and $b_{0}=1$ for some element $0\in I$, and for each $j,k$ there exists $i$ with $N_{ij}^{k}>0$. In this case we have a homomorphism ${\rm FPdim}:A\to\mathbb{R}$, determined uniquely by the condition that $d_{i}:={\rm FPdim}(b_{i})>0$, called the Frobenius-Perron dimension ([EGNO], Subsection 3.3).

Let $N_{i}$ be the matrix with entries $N_{ij}^{k}$. Let $\mathbb{d}=(d_{i})$ be the column vector with entries $d_{i}$. Then we have $N_{i}\mathbb{d}=d_{i}\mathbb{d}$, i.e., $\mathbb{d}$ is a common (right) eigenvector of the matrices $N_{i}$. The Frobenius-Perron theorem implies that we also have a left eigenvector $\mathbb{p}=(p_{i})$ with $p_{i}>0$ (defined uniquely up to a positive scalar) such that $\mathbb{p}N_{i}=d_{i}\mathbb{p}$.

Lemma 2.1**.**

For each $i,j,k$ one has $p_{k}\geq\frac{N_{ji}^{k}}{d_{j}}p_{i}$. In particular, one has $p_{k}\geq p_{0}/d_{k}$.

Proof.

One has $d_{j}p_{k}=\sum_{i}N_{ji}^{k}p_{i}$, which implies the statement. ∎

If $A$ is a fusion ring then $N_{i^{*}}=N_{i}^{T}$, so $\mathbb{p}$ is proportional to $\mathbb{d}$ and it is natural to normalize $\mathbb{p}$ by setting $\mathbb{p}=\mathbb{d}$; in this case the number ${\rm FPdim}(A):=\mathbb{p}\cdot\mathbb{d}=|\mathbb{d}|^{2}=\sum_{i}d_{i}^{2}$ is called the Frobenius-Perron dimension of $A$ ([EGNO], Section 3.3). However, in general $\mathbb{p}$ does not come with a natural normalization, which is why in [EGNO] the Frobenius-Perron dimension of a general $\mathbb{Z}_{+}$-ring of finite rank is not defined.

Note that when $A$ is the Grothendieck ring of a finite tensor category $\mathcal{C}$ then one may take $p_{i}$ to be the Frobenius-Perron dimensions of the indecomposable projectives, and $\mathbb{p}\cdot\mathbb{d}=\sum_{i}p_{i}d_{i}$ is then called the Frobenius-Perron dimension of $\mathcal{C}$. However, this normalization depends on the choice of $\mathcal{C}$: e.g., the categories $\mathcal{C}_{1}$ of representations of $\mathbb{Z}/2$ and $\mathcal{C}_{2}$ of representations of Sweedler’s 4-dimensional Hopf algebra categorify the same $\mathbb{Z}_{+}$-ring but give different normalizations of $\mathbb{p}$.

Now recall ([EGNO]) that $A$ is said to be integral if ${\rm FPdim}$ lands in $\mathbb{Z}$. In this case, the vector $\mathbb{p}$ can be uniquely normalized in such a way that $p_{i}$ are positive integers whose greatest common divisor is $1$. Let us call this normalization the canonical normalization.

Definition 2.2**.**

The Frobenius-Perron dimension of $A$ is ${\rm FPdim}(A):=\mathbb{p}\cdot\mathbb{d}=\sum_{i}p_{i}d_{i}$, where $p_{i}$ are normalized canonically.

In particular, we see that ${\rm FPdim}(A)\geq\sum_{i}d_{i}$.

Note that for fusion rings the canonical normalization just gives $\mathbb{p}=\mathbb{d}$ as above, so our definition of ${\rm FPdim}(A)$ agrees with the one in [EGNO]. However, in general $p_{0}\neq 1$, as shown by the following example.

Example 2.3**.**

Consider the ring $A$ with basis $1,X$ with $X^{2}=4$. Thus, $A$ is the Grothendieck ring of the representation category $\mathcal{C}=\mathop{\mathrm{Rep}}\nolimits(U)$ of the restricted enveloping algebra $U$ over a field $\mathbf{k}$ of characteristic $2$ generated by primitive elements $x,y,z$ with $z$ central, $[x,y]=z$ and $x^{2}=y^{2}=0,z^{2}=z$. Then $d_{0}=1,d_{1}=2$, $p_{0}=2,p_{1}=1$. This shows that the normalization with $p_{0}=1$ is not very natural: this would give $p_{1}=1/2$, which is not an algebraic integer.

2.2. Grothendieck rings of integral finite tensor categories

Now assume that $A$ is the Grothendieck ring of a finite tensor category $\mathcal{C}$. Then $\mathcal{C}$ is integral, i.e., of the form $\mathop{\mathrm{Rep}}\nolimits H$, where $H$ is a finite dimensional quasi-Hopf algebra ([EGNO]). Thus the Frobenius-Perron dimensions of objects of $\mathcal{C}$ are their usual vector space dimensions, and ${\rm FPdim}(\mathcal{C})=\dim H$. In this case $p_{i}$ divide the dimensions $\dim(P_{i})$ of the indecomposable projectives $P_{i}$ but aren’t necessarily equal to them. E.g., if $H$ is Sweedler’s 4-dimensional Hopf algebra, then $p_{0}=p_{1}=1$ but the indecomposable projectives $P_{0},P_{1}$ are 2-dimensional. It is easy to see that in general we have the following proposition.

Proposition 2.4**.**

Let $\mathcal{C}$ be an integral finite tensor category with Grothendieck ring $A$, and $D$ be the greatest common divisor of dimensions of the projective objects of $\mathcal{C}$. Then ${\rm FPdim}(\mathcal{C})=D\cdot{\rm FPdim}(A)$.

We also have

Proposition 2.5**.**

Let $X_{i}$ be the simple objects of $\mathcal{C}$. Then one has $p_{i}D\geq d_{i}$, and the equality holds for a given $i$ if and only if the object $X_{i}$ is projective. Otherwise, one has $p_{i}D\geq 2d_{i}$.

Proof.

The first statement follows from the fact that the projective cover $P_{i}$ of $X_{i}$ contains $X_{i}$ in its composition series. For the second statement, it is enough to note that by the Frobenius property of $\mathcal{C}$ we have $\dim P_{i}\geq 2\dim X_{i}$ if $X_{i}$ is not projective. ∎

Example 2.6**.**

A tensor category $\mathcal{C}$ is called unipotent if its only simple object is the unit object, i.e. $A=\mathbb{Z}$ or, equivalently, ${\rm FPdim}(A)=1$ ([EG1]). In other words, $\mathcal{C}={\rm Rep}H$, where $H$ is a local finite dimensional quasi-Hopf algebra. By [EGNO], Corollary 4.4.2, for $\dim H>1$ this can happen only in characteristic $p>0$. Moreover, in this case the associated graded ${\rm gr}(H)$ of $H$ with respect to the radical filtration is cocommutative 111This is a straightforward generalization of [W], Proposition 2.2(7) to the quasi-Hopf case, and I thank S. Gelaki for pointing this out., i.e. the group algebra of a finite unipotent group scheme over $k$, so $\dim H={\rm FPdim}(\mathcal{C})$ is a power of $p$ (and clearly any power can arise this way).

3. Lower bound on the dimension of a $\mathbb{Z}_{+}$-generator

Let $A$ be an integral $\mathbb{Z}_{+}$-ring of rank $r>1$ and $b=\sum_{i}m_{i}b_{i}\in A$ with $m_{i}\geq 0$. Let us say that $b$ is a $\mathbb{Z}_{+}$-generator of $A$ if for some $n$ all coefficients of $1+b+...+b^{n}$ are positive. Let $d={\rm FPdim}(b)$ and $N={\rm FPdim}(A)$.

Theorem 3.1**.**

(i) Let $\chi$ be the characteristic polynomial of $b$ on $A$, and $Q(z)=(z-d)^{-1}\chi(z)$. Then $Q(d)$ is a positive integer divisible by $N$. In particular, $Q(d)\geq N$.

(ii) Let $s$ be the number of roots of $Q$ with positive imaginary part (counted with multiplicities). Then

[TABLE]

(iii) We have

[TABLE]

In particular, $d\geq\frac{1}{2}N^{\frac{1}{r-1}}$.

Proof.

(i) First note that by the Frobenius-Perron theorem, $d$ is a simple eigenvalue of $b$, so $Q(d)$ is a positive integer. It remains to show that $Q(d)$ is divisible by $N$. Let $\mathbb{v}$ be an integral row vector such that $\mathbb{p}\cdot\mathbb{v}=1$ (it exists since $p_{i}$ are relatively prime). We have $Q(b)\mathbb{v}=m\mathbb{d}$, where $m=(Q(b)\mathbb{v})_{0}$ is an integer. Therefore

[TABLE]

This implies that $N$ divides $Q(d)$.

(ii) Let $\lambda_{i}$, $i=1,...,r-2s-1$ be the real roots of $Q$, and $\mu_{1},...\mu_{s}$ the roots with positive imaginary part. We have

[TABLE]

Let $\alpha_{i}=\lambda_{i}/d$ and $\beta_{j}=\mu_{j}/d$. Then

[TABLE]

This can be written as

[TABLE]

By the Frobenius-Perron theorem, $|\beta_{j}|\leq 1$, so we get

[TABLE]

Thus by the arithmetic and geometric mean inequality,

[TABLE]

But the left hand side is the trace of $1-b/d$, so it is $\leq r$, as ${\rm Tr}(b)\geq 0$. Thus we get

[TABLE]

This implies that

[TABLE]

as desired.

(iii) Analyzing the function of $s$ in (ii) and using that $2s\leq r-1$, we find easily that the worst case scenario is the maximal possible value of $s$, i.e., $s=\lfloor\frac{r-1}{2}\rfloor$. This gives the result. The last statement follows from the fact that

[TABLE]

∎

4. Auxiliary lemmas

Lemma 4.1**.**

Let $H$ be a local finite dimensional quasi-Hopf algebra, $M$ a finite dimensional $H$-module, and $a:M\to M^{**}$ a homomorphism. Then generalized eigenspaces of $a$ (regarded as a linear map $M\to M$) are $H$-submodules of $M$. In particular, if $M$ is indecomposable, then all eigenvalues of $a$ are the same.

Proof.

The associated graded ${\rm gr}(H)$ of $H$ with respect to the radical filtration is a Hopf quotient of an enveloping algebra, so it has $S^{2}=1$. Thus the operator $S^{2}-1$ strictly preserves the radical filtration. We have $ah=S^{2}(h)a$, so

[TABLE]

We claim that

[TABLE]

The proof is by induction in $N$. Namely, the case $N=0$ is clear, and for the induction step note that

[TABLE]

[TABLE]

so the induction step follows from the identity $\binom{N}{i}+\binom{N}{i-1}=\binom{N+1}{i}$.

Therefore, since $(S^{2}-1)^{i}(h)$ is zero for sufficiently large $i$, the generalized eigenspaces of $a$ are $H$-submodules of $M$. This implies the statement. ∎

Remark 4.2**.**

Note that in Lemma 4.1, the ordinary eigenspaces of $a$ (as opposed to generalized eigenspaces) don’t have to be submodules of $M$, e.g. $M=H$ and $a=S^{2}$ when $S^{2}\neq 1$. An example of a local Hopf algebra with $S^{2}\neq 1$ is given in [NWW], Theorem 1.3, case B3.

Lemma 4.3**.**

Let $V$ be a simple object of a tensor category $\mathcal{C}$, and $a:V\to V^{**}$ be an isomorphism. Then the morphism $a\otimes(a^{*})^{-1}$ commutes with the composition of $a\otimes 1$, evaluation and coevaluation

[TABLE]

Proof.

It is easy to see that the morphism $a\otimes(a^{*})^{-1}$ preserves the evaluation and the coevaluation moprhism, so it commutes with their composition. It also obviously commutes with $a\otimes 1$. ∎

Now let $H$ be a finite dimensional Hopf algebra over an algebraically closed field $\mathbf{k}$ with antipode $S$ such that $S^{2}\neq 1$.

Lemma 4.4**.**

Let $M$ be a finite dimensional $H$-module which tensor generates $\mathop{\mathrm{Rep}}\nolimits(H)$ and $a:M\to M^{**}$ be an isomorphism. Then $a$ is not a scalar (when regarded as a linear map $M\to M$).

Proof.

Assume the contrary. Then for any $h\in H$ we have $h|_{M}=S^{2}(h)|_{M}$. Hence the same holds for any tensor product of copies of $M$ and $M^{*}$. But since $M$ tensor generates $\mathop{\mathrm{Rep}}\nolimits(H)$, some such product contains the regular representation of $H$, which implies that $S^{2}=1$ on $H$, a contradiction. ∎

Recall that a Hopf algebra $H$ is called simple if it has no nontrivial Hopf quotients.

Lemma 4.5**.**

Let $H$ be a simple finite dimensional Hopf algebra over an algebraically closed field $\mathbf{k}$ of characteristic $\neq 2$ with $S^{2}\neq 1$. Let $V$ be an irreducible 2-dimensional $H$-module such that $V\cong V^{**}$. Then $S^{4}\neq 1$. In particular, the distinguished grouplike element of $H$ is nontrivial.

Proof.

Since $H$ is simple and has a 2-dimensional simple module, $H$ cannot have nontrivial 1-dimensional modules: otherwise $H$ would have a nontrivial commutative Hopf quotient. For the same reason, ${\rm Ext}^{1}(\mathbf{k},\mathbf{k})=0$, since otherwise $H$ would have a nontrivial local Hopf quotient. Also it is clear that $V$ tensor generates $\mathop{\mathrm{Rep}}\nolimits(H)$.

Assume the contrary, i.e. that $S^{4}=1$. Pick an isomorphism $a:V\to V^{**}$ such that $a^{2}=1$ (this can be done since $S^{4}=1$). Then $a$ is semisimple with eigenvalues $\pm 1$ (as ${\rm char}(\mathbf{k})\neq 2$) and not a scalar by Lemma 4.4, so ${\rm Tr}(a)=0$. Since $H$ has no nontrivial 1-dimensional modules and ${\rm Ext}_{H}^{1}(\mathbf{k},\mathbf{k})=0$, this implies that the Loewy series of $V\otimes V^{*}$ must be $\mathbb{1},W,\mathbb{1}$, where $W$ is an irreducible 2-dimensional module. By Lemma 4.3, the morphism $a\otimes(a^{*})^{-1}$ commutes with the composition of $a\otimes 1$, evaluation and coevaluation $V\otimes V^{*}\to V\otimes V^{*}$. Hence the trace of the block of the morphism $a\otimes(a^{*})^{-1}$ on the constituent $W$ is $-2$, which implies that $a\otimes(a^{*})^{-1}$ must act by $=-1$ on $W$. But since $H$ is simple, $W$ must tensor generate $H$, which gives a contradiction with Lemma 4.4. ∎

Example 4.6**.**

Let $n\geq 3$, $q$ be a primitive $n$-th root of unity, and $H=\mathfrak{u}_{q}(\mathfrak{sl}_{2})$ be the corresponding small quantum group over $\mathbb{C}$, generated by a grouplike element $K$ and skew-primitive elements $E$, $F$ with $K^{n}=1$, $E^{n}=0$, $F^{n}=0$ and the usual commutation relations $KE=q^{2}EK$, $KF=q^{-2}EK$, $[E,F]=\frac{K-K^{-1}}{q-q^{-1}}$. This Hopf algebra has a 2-dimensional tautological representation $V$, which is simple, self-dual and tensor generates $\mathop{\mathrm{Rep}}\nolimits(H)$, Also $S^{2}(a)=KaK^{-1}$, so the order of $S^{2}$ is the order of $q^{2}$, which is $n$ for odd $n$ and $n/2$ for even $n$. Thus $S^{4}=1$ for $n=4$. However, $H$ is simple if and only if $n$ is odd: otherwise it has a 1-dimensional representation $\psi$ such that $\psi(E),\psi(F)=0$, $\psi(K)=-1$. So the only reason the proof of Lemma 4.5 fails for $n=4$ is that the representation $W$ is not irreducible but rather isomorphic to $\psi\oplus\psi$ and therefore does not tensor generate $\mathop{\mathrm{Rep}}\nolimits(H)$.

Lemma 4.7**.**

Let $X$ be a simple object of a finite tensor category $\mathcal{C}$, such that $X\otimes X^{*}$ is projective. Then $X$ is projective.

Proof.

Assume the contrary. Then we have

[TABLE]

Hence $\mathrm{Hom}(X,P_{X})=\mathrm{Hom}(\mathbb{1},P_{X}\otimes X^{*})\neq 0$. Thus $P_{X}$ is the injective hull of $X$, i.e. the socle of $P_{X}$ is $X$. Hence $P_{X}\otimes X^{*}=2X\otimes X^{*}\oplus Z$, so $\dim\mathrm{Hom}(X,P_{X})=\dim\mathrm{Hom}(\mathbb{1},P_{X},\otimes X^{*})\geq 2$, a contradiction. ∎

5. Finite dimensional quasi-Hopf algebras with two simple modules

In this section we consider integral finite tensor categories of rank $r=2$ unless otherwise specified.

5.1. The lower bound for the Frobenius-Perron dimension of a $\mathbb{Z}_{+}$-generator

For $r=2$ the $\mathbb{Z}_{+}$-ring $A$ is $\mathbb{Z}_{+}$-generated by the nontrivial basis element $X$, so we may take $d={\rm FPdim}(X)$. Theorem 3.1 then gives the inequality $d\geq N/2$. More explicitly, let $b_{0}=1,b_{1}=X$. We have $X^{2}=aX+b$ for some $a\geq 0$, $b\geq 1$. Hence $d^{2}=ad+b$. It is easy to see that $p_{0}=d-a$ and $p_{1}=1$, so ${\rm FPdim}(A)=p_{0}+p_{1}d=2d-a$. Thus we get

Lemma 5.1**.**

Let $A$ be a transitive $\mathbb{Z}_{+}$-ring of rank $2$, and let the dimension of the nontrivial basis element $X$ be $d$. Then

[TABLE]

where $N={\rm FPdim}(A)$.

5.2. The case of large characteristic

Now let $\mathcal{C}=\mathop{\mathrm{Rep}}\nolimits H$ be the representation category of a quasi-Hopf algebra $H$ over $\mathbf{k}$ with two simple modules. If $\mathcal{C}$ is semisimple, it is easy to see that $\mathcal{C}={\rm Vec}(\mathbb{Z}/2,\omega)$ is the category of $\mathbb{Z}/2$-graded vector spaces twisted by a $3$-cocycle $\omega$. So from now till the end of the section we assume that $\mathcal{C}$ is non-semisimple.

Let $C$ be the Cartan matrix of $\mathcal{C}$.

Lemma 5.2**.**

$|\det C|\leq\frac{{\rm FPdim}(\mathcal{C})^{2}}{4d^{2}}$.

Proof.

Using the arithmetic and geometric mean inequality, we have

[TABLE]

[TABLE]

[TABLE]

∎

Let $A$ be the Grothendieck ring of $\mathcal{C}$.

Proposition 5.3**.**

Suppose that the characteristic of $\mathbf{k}$ is [math] or $p>\frac{{\rm FPdim}(\mathcal{C})^{2}}{4d^{2}}$, $\mathcal{C}$ is not pointed, and ${\rm Id}\cong**$ as an additive functor. Then ${\rm FPdim}(\mathcal{C})=m{\rm FPdim}(A)^{2}$, where $m>1$ is an integer. In particular, ${\rm FPdim}(\mathcal{C})$ cannot be square free.

Proof.

By [EGNO], Theorem 6.6.1 and Lemma 5.2, $P_{0}$ and $P_{1}$ must be proportional in the Grothendieck ring of $\mathcal{C}$, so $P_{0}=(d-a)P_{1}$. This implies that $P_{1}=m((d-a)\mathbb{1}+X)$, where $m$ is a positive integer. Hence ${\rm FPdim}(\mathcal{C})=\dim P_{0}+d\dim P_{1}=m(2d-a)^{2}$.

Assume that $m=1$. Since $\mathcal{C}$ is not pointed, it is unimodular. Also we have $[P_{1}:X]=1$. By the Frobenius property of $\mathcal{C}$ this implies that $P_{1}=X$, so $d-a=0$, a contradiction. Thus, $m>1$, as desired. ∎

Remark 5.4**.**

1. Note that the statement that $m>1$ in Proposition 5.3 is false in the pointed case, e.g. for the representation category of Sweedler’s 4-dimensional Hopf algebra.

2. Pointed finite tensor categories with two simple objects in characteristic zero are classified in [EG2]. They are either $\mathop{\mathrm{Rep}}\nolimits H$ where $H$ is a Nichols Hopf algebra of dimension $2^{n}$ ([N]) or exceptional quasi-Hopf algebras of dimension $8$ or $32$.

5.3. Arbitrary characteristic

Corollary 5.5**.**

In any characteristic one has ${\rm FPdim}(\mathcal{C})\geq d\cdot{\rm FPdim}(A)$, i.e., $D\geq d$, with the equality if and only if $X$ is projective. In particular, ${\rm FPdim}(\mathcal{C})\geq\frac{1}{2}{\rm FPdim}(A)^{2}$. Moreover, if $X$ is not projective then $D\geq 2d$ and ${\rm FPdim}(\mathcal{C})\geq{\rm FPdim}(A)^{2}$.

Proof.

This follows from Proposition 2.5, since $p_{1}=1$. ∎

5.4. Minimal categories

Let us say that $\mathcal{C}$ is minimal if $X$ is projective.

Proposition 5.6**.**

Suppose $\mathcal{C}$ is minimal and $d>1$. Then $d=p^{r}$ where $p={\rm char}(\mathbf{k})$, and $a=p^{r}-p^{s}$ for some $s\leq r$, so ${\rm FPdim}(A)=p^{s}(p^{r-s}+1)$, ${\rm FPdim}(\mathcal{C})=p^{r+s}(p^{r-s}+1)$, $D=p^{r}$ and $\dim P_{0}=b=p^{r+s}$.

Proof.

We have $X^{2}=aX\oplus P_{0}$, and $P_{0}$ is an iterated extension of $\mathbb{1}$, i.e. it coincides with the projective cover of $\mathbb{1}$ in the unipotent subcategory of $\mathcal{C}$. This implies the statement, since the Frobenius-Perron dimension of any unipotent category is the power of the characteristic (Example 2.6). ∎

Corollary 5.7**.**

Let $\mathcal{C}$ be a minimal category of Frobenius-Perron dimension $pn>2$ where $p$ is a prime and $p>n$. Then $p=n+1$ and $n$ is a power of $2$, i.e., $p=2^{2^{m}}+1$ is a Fermat prime. Moreover, ${\rm char}(\mathbf{k})=2$, $d=p-1$ and $X^{2}=(p-2)X+p-1$.

Proof.

By Proposition 5.6 we have $pn=q^{r+s}(q^{r-s}+1)$, where $q$ is a prime. Since $p>n$, we must have $s=0$, $n=q^{r}$ and $p=q^{r}+1$. Thus $q$ is a power of $2$ and $p$ is a Fermat prime, as desired. ∎

5.5. Examples

Example 5.8**.**

Minimal categories as in Corollary 5.7 exist for every Fermat prime $p$. Namely, let $G={\rm Aff}(p)$ be the group of affine linear transformations of the field $\mathbb{F}_{p}$, then we can take $\mathcal{C}={\rm Rep}_{\mathbf{k}}(G)$. This shows that Proposition 5.3 fails in small characteristic.

Example 5.9**.**

Consider the case $a=0$, i.e., $X^{2}=d^{2}$ for some positive integer $d$. We claim that for $d>1$ such a ring $A$ admits a categorification by a finite tensor category in characteristic $p$ if and only if $d$ is a power of $p$, and in particular it never admits such categorification in characteristic [math].

Indeed, let $\mathcal{C}$ be a finite tensor category with Grothendieck ring $A$. We may assume that $\mathcal{C}$ is tensor generated by $X$. Then $\mathcal{C}=\mathcal{C}_{0}\oplus\mathcal{C}_{1}$ is $\mathbb{Z}/2$-graded and ${\rm Ext}^{1}(\mathbb{1},X)={\rm Ext}^{1}(X,\mathbb{1})=0$. Thus $P_{\mathbb{1}}$ is an iterated extension of $\mathbb{1}$ and $\dim P_{\mathbb{1}}={\rm FPdim}(\mathcal{C}_{0})$ is a power of $p$, since $\mathcal{C}_{0}$ is unipotent (Example 2.6). Also, $P_{X}$ is an iterated extension of $X$, and $X\otimes P_{X}=P_{\mathbb{1}}$. This implies that $d^{2}[P_{X}:X]=p^{m}$ for some $m$, which implies that $d=p^{s}$ for some $s$.

On the other hand, such categories exist for every $s$. Indeed, it is enough to construct such a category $\mathcal{C}$ for $s=1$, then for any $s$ we can take the subcategory $\mathcal{C}_{s}$ in $\mathcal{C}^{\boxtimes s}$ generated by $X^{\boxtimes s}$. For $p=2$, an example of $\mathcal{C}$ is given in Example 2.3. If $p>2$, let $H$ be the Hopf algebra $\mathbb{Z}/2\ltimes\mathbf{k}[x,y]/(x^{p},y^{p})$, where $x,y$ are primitive and the generator $g\in\mathbb{Z}/2$ acts by $gxg^{-1}=-x,gyg^{-1}=-y$. Let $J=\sum_{j=0}^{p-1}\frac{x^{j}\otimes y^{j}}{j!}$ be a twist for $H$, and $H^{J}$ be the corresponding twisted triangular Hopf algebra. Then we can take $\mathcal{C}$ to be the category of comodules over $H^{J}$.

5.6. Categorifications of $X^{2}=X+d(d-1)$

Suppose $a=1$, i.e., $X^{2}=X+d(d-1)$. Let $\mathcal{C}$ be a finite tensor category with Grothendieck ring $A$. Since $X$ is self-dual, this implies that any morphism $a:X\to X^{**}$ has zero trace. Indeed, otherwise $X\otimes X=\mathbb{1}\oplus Y$ where $Y$ has only one copy of $X$ in the composition series. Thus $\mathrm{Hom}(Y,\mathbb{1})\neq 0$ (as $Y$ is self-dual), so $\dim\mathrm{Hom}(X\otimes X,\mathbb{1})\geq 2$, a contradiction.

Proposition 5.10**.**

The characteristic of $\mathbf{k}$ is a prime $p$ dividing $d(d-1)$. Moreover, if $H$ is a Hopf algebra then $p$ divides $d$.

Proof.

Consider the Loewy series of $X\otimes X$. Exactly one of its terms contains $X$. So we have a 3-step filtration of $X\otimes X$ with composition factors $M,X,N$ invariant under any automorphism of $X\otimes X$, where $M,N$ are indecomposable iterated self-extensions of $\mathbb{1}$ (in fact, $M$ and $N^{*}$ are cyclic). The $H$-modules $M$ and $N$ factor through a local quasi-Hopf algebra $H^{\prime}$. Therefore, by Lemma 4.1, for any isomorphism $a:X\to X^{**}$ the morphism $a\otimes a$ on $X\otimes X$ has only one eigenvalue on $M$ and only one on $N$. Moreover, these two coincide by Lemma 4.3; we will normalize $a$ in such a way that this eigenvalue is $1$. Then, computing the trace of $a\otimes a$, we get $d(d-1)=0$ in $\mathbf{k}$, as claimed.

Now assume that $H$ is a Hopf algebra and let $a_{1},...,a_{d}$ be the eigenvalues of $a$. We may assume that $d>2$. Then the eigenvalues of $a\otimes a$ are $a_{i}a_{j}$. At least $d(d-1)$ of these numbers are $1$, so at most $d$ of them are $\neq 1$. We claim that $a_{i}=1$ for all $i$ or $a_{i}=-1$ for all $i$. Indeed, if none of $a_{i}$ are $1$ or $-1$ then $a_{i}^{2}\neq 1$ for all $i$, so $a_{i}a_{j}=1$ for $i\neq j$, hence $a_{i}=a_{j}^{-1}$, which is impossible since $d>2$. So at least one $a_{i}$ is $1$ or $-1$, and we can assume it is $1$ by multiplying $a$ by $-1$ if needed. Let $n>0$ be the number of $i$ such that $a_{i}=1$. If $a_{i}=1,a_{j}\neq 1$ then $a_{i}a_{j},a_{j}a_{i}\neq 1$, so $2n(d-n)\leq d$. Thus for $d>2$ we get $n=d$, i.e. $a_{i}=1$ for all $i$. Thus, $d={\rm Tr}(a)=0$ in $\mathbf{k}$, as desired. ∎

5.7. Categorifications of $X^{2}=X+2$

Now consider the case $d=2$, i.e. the fusion rule $X^{2}=X+2$. Let $\mathcal{C}_{X}$ be the tensor subcategory of $\mathcal{C}$ generated by $X$.

Proposition 5.11**.**

One has $X\otimes X=P\oplus X$, where $P$ is a nontrivial extension of $\mathbb{1}$ by $\mathbb{1}$. Moreover, $P$ and $X$ are the indecomposable projectives of $\mathcal{C}_{X}$, so ${\rm FPdim}(\mathcal{C}_{X})=6$.

Proof.

Let us first show that $X\otimes X=P\oplus X$. Assume the contrary. Since $X\otimes X$ is self-dual and $\dim\mathrm{Hom}(\mathbb{1},X\otimes X)=1$, we get that $X\otimes X$ must be indecomposable with Loewy series $\mathbb{1},X,\mathbb{1}$. Then $\dim\mathrm{Hom}(X,X\otimes X\otimes X)=\dim{\rm End}(X\otimes X)=2$. But $X\otimes X\otimes X$ is self-dual, has composition series $X,X\otimes X,X$, and $\mathrm{Hom}(X,X\otimes X)=0$. This means that $X\otimes X\otimes X=2X\oplus X\otimes X$. Thus for each $n$, $X^{\otimes n}$ is a direct sum of copies of $X$ and $X\otimes X$. But for some $n$, the object $X^{\otimes n}$ has a direct summand which is projective in $\mathcal{C}_{X}$. Hence $X\otimes X$ is projective. Therefore, so is $X$ (as it is a direct summand in $X\otimes X\otimes X$). But then $X$ must be a direct summand in $X\otimes X$, a contradiction.

Note that $\dim\mathrm{Hom}(X,P\otimes X)=\dim\mathrm{Hom}(X\otimes X,P)=\dim\mathrm{Hom}(P,P)=2$. Thus, $P\otimes X=2X$. Thus, for any $n$, $X^{\otimes n}$ is a direct sum of copies of $P$ and $X$. This implies that $P$ and $X$ are the indecomposable projectives of $\mathcal{C}_{X}$. ∎

We note that such a category of dimension $6$ does exist, e.g. ${\rm Rep}_{\mathbf{k}}(S_{3})$, with ${\rm char}(\mathbf{k})=2$. It can also be realized as the $\mathbb{Z}/2$-equivariantization ${\rm Vec}_{\mathbf{k}}(\mathbb{Z}/3)^{\mathbb{Z}/2}$ of ${\rm Vec}_{\mathbf{k}}(\mathbb{Z}/3)$. A generalization is ${\rm Vec}_{\mathbf{k}}(\mathbb{Z}/3,\omega)^{\mathbb{Z}/2}$, where $\omega\in H^{3}(\mathbb{Z}/3,\mathbf{k}^{\times})$.

5.8. Categories of dimension $sn$, where $s$ is square free and coprime to $n$

Proposition 5.12**.**

Let ${\rm char}(\mathbf{k})$ be zero or $p>\frac{{\rm FPdim}(\mathcal{C})^{2}}{4d^{2}}$, and ${\rm FPdim}(\mathcal{C})=sn$, where $s$ is a square free number coprime to $n$. Assume that $\mathcal{C}$ is not pointed, and ${\rm Id}\cong**$ as additive functors. Then $n=m\ell^{2}$, where $\ell\geq 4$ and $m\geq 2$, so $n\geq 32$.

Proof.

It follows from Proposition 5.3 that $n=m\ell^{2}$ where $\ell={\rm FPdim}(A)$ and $m\geq 2$. So it remains to show that $\ell\geq 4$. To do so, note that $\ell\geq 2$, and if $\ell=2$ then $\mathcal{C}$ is pointed. Also, if $\ell=3$ then $X^{2}=X+2$, so ${\rm char}(\mathbf{k})=2$ (Proposition 5.10). ∎

Example 5.13**.**

(see [Ne], Subsection 9.4) The bound $n\geq 32$ in Proposition 5.12 is sharp. Namely, let ${\rm char}(\mathbf{k})\neq 2$ and let $B$ be the small quantum group $\mathfrak{u}_{q}(\mathfrak{sl}_{2})$ at a primitive 8-th root of unity $q$. It is a Hopf algebra of dimension $2^{7}=128$ generated by $E,F,K$ with relations $KE=q^{2}EK$, $KF=q^{-2}EK$, $[E,F]=\frac{K-K^{-1}}{q-q^{-1}}$, $E^{4}=F^{4}=K^{8}-1=0$. With its usual Hopf structure, $B$ does not admit an $R$-matrix, but it admits a quasi-Hopf structure $\Phi$ with a factorizable R-matrix ([CGR]). Let $\mathcal{D}=\mathop{\mathrm{Rep}}\nolimits(B^{\Phi})$ be the corresponding nondegenerate braided category. This category contains a unique 1-dimensional nontrivial representation $\psi$ such that $\psi(E)=\psi(F)=0$, $\psi(K)=-1$ (it has highest weight $4$). It is easy to show using formula (4.10) of [CGR] that the subcategory $\mathcal{E}$ generated by $\mathbb{1}$ and $\psi$ has trivial braiding, i.e., it is equivalent to $\mathop{\mathrm{Rep}}\nolimits(\mathbb{Z}/2)$. Let $\mathcal{C}:=\mathcal{E}^{\perp}/\mathcal{E}$ be the de-equivariantization of the centralizer $\mathcal{E}^{\perp}$ of $\mathcal{E}$ by $\mathbb{Z}/2$. The centralizer consists of representations with even highest weights ($0,2,4,6$), so modding out by $\psi$ gives a category $\mathcal{C}$ with just two simple objects, $\mathbb{1}$ and $X$ (the quantum adjoint representation).

The category $\mathcal{C}$ has Frobenius-Perron dimension $32$ and is nondegenerate braided. Its Grothendieck ring is generated by a basis element $X$ with $X^{2}=2X+3$. The indecomposable projectives $P_{\mathbb{1}},P_{X}$ have the Loewy series $\mathbb{1},X\oplus X,\mathbb{1}$ and $X,\mathbb{1}\oplus\mathbb{1},X$, and we have $X\otimes X=\mathbb{1}\oplus P_{X}$.

Also $\mathcal{C}$ does not admit a pivotal structure: indeed, if $g$ is a pivotal structure and $u:{\rm Id}\to**$ the Drinfeld isomorphism then $g^{-1}u$ is an automorphism of the identity functor, hence acts by $1$ on $X$ (as $X$ is linked to $1$). But then the squared braiding $c^{2}:X\otimes X\to X\otimes X$ is unipotent, which is a contradiction with formula (4.10) of [CGR], as this formula implies that $-1$ is an eigenvalue of $c^{2}$ on the tensor product of highest weight vectors.

Moreover, any pivotal finite nondegenerate braided tensor category $\mathcal{C}$ over $\mathbb{C}$ with two simple objects is equivalent to ${\rm Vec}(\mathbb{Z}/2)$ with braiding defined by the quadratic form $Q$ on $\mathbb{Z}/2$ such that $Q(1)=\pm i$. Indeed, by the result of [GR], a pivotal finite nondegenerate braided tensor category over $\mathbb{C}$ must contain a simple projective object, so $X$ is projective, which implies that $\mathcal{C}$ is semisimple. Thus, the above example shows that the pivotality assumption in [GR] is essential.

Remark 5.14**.**

I don’t know any other finite dimensional quasi-Hopf algebras over $\mathbb{C}$ with a unique nontrivial simple module $X$, such that $\dim X>1$. In particular, I don’t know if there exist Hopf algebras with this property.

5.9. Non-minimal categories

Corollary 5.15**.**

(i) Suppose $\mathcal{C}$ is non-minimal and ${\rm FPdim}(\mathcal{C})=pn$, where $p$ is a prime and $p>n$. Then ${\rm FPdim}(A)$ divides $n$ and the dimension of every projective object in $\mathcal{C}$ is divisible by $p$.

Proof.

By Corollary 5.5 we have ${\rm FPdim}(A)\leq\sqrt{{\rm FPdim}(\mathcal{C})}<p$, so since ${\rm FPdim}(A)$ divides $pn$, it must divide $n$. Hence $D$ is divisible by $p$, as desired. ∎

Corollary 5.15 has strong implications for the structure of $\mathcal{C}$ if $n$ is small. First of all note that if $\mathcal{C}$ is pointed then its Frobenius-Perron dimension has to be even.

Proposition 5.16**.**

Let $\mathcal{C}$ be an integral category with two simple objects of Frobenius-Perron dimension $pn$ where $p$ is a prime. If $n\leq 3$ then $n=2$ and $\mathcal{C}$ is pointed or ${\rm FPdim}(\mathcal{C})=6$ and $\mathcal{C}$ is a minimal category which categorifies the fusion rule $X^{2}=X+2$ in characteristic $2$.

Proof.

By Corollary 5.7, we may assume that $\mathcal{C}$ is non-minimal.

Suppose $n=1$. If $p>2$, Corollary 5.15 implies that ${\rm FPdim}(A)=1$, which is impossible, and if $p=2$, the category is semisimple, hence pointed.

Suppose $n=2$. If $p>2$, Corollary 5.15 implies that ${\rm FPdim}(A)=2$, i.e., $\mathcal{C}$ is pointed. If $p=2$ then $d<2$, so $d=1$ and again $\mathcal{C}$ is pointed.

Suppose $n=3$. If $p>3$, Corollary 5.15 implies that ${\rm FPdim}(A)=2d-a=3>d$, which implies that $d=2$, $a=1$, so $b=2$ and $X^{2}=X+2$. In the latter case, ${\rm char}(\mathbf{k})=2$ (Subsection 5.7). Consider an isomorphism $g:P_{\mathbb{1}}\to P_{\mathbb{1}}^{**}$, and let $w\in H$ be the element realizing the canonical isomorphism $V\to V^{****}$ such that $\varepsilon(w)=1$. Then $g^{-2}w:P_{\mathbb{1}}\to P_{\mathbb{1}}$ is an automorphism, and we may normalize $g$ so that this automorphism is unipotent. Then the eigenvalues of $g$ on the invariant part of the graded pieces of the Loewy filtration of $P_{\mathbb{1}}$ are all equal to $1$. Since the trace of any morphism $X\to X^{**}$ is zero, we see that ${\rm Tr}(g)$ cannot be zero since $\dim P_{\mathbb{1}}=p$ and hence $[P_{\mathbb{1}}:\mathbb{1}]$ is odd. This is a contradiction, since this implies that $\mathbb{1}$ is projective so $\mathcal{C}$ is semisimple. ∎

5.10. Representations of finite groups

Recall that ${\rm Aff}(q)$ denotes the group of affine transformation $x\to\alpha x+\beta$ over the finite field $\mathbb{F}_{q}$.

Theorem 5.17**.**

Let $G$ be a finite group. Then:

(i) If $G$ has exactly two irreducible representations over $\mathbf{k}$ then either $G=\mathbb{Z}/2$ or ${\rm char}(\mathbf{k})=p>0$.

(ii) Let ${\rm char}(\mathbf{k})=p>0$ and let $N$ be the largest normal $p$-subgroup of $G$. Then $G$ has exactly two irreducible representations over $\mathbf{k}$ if and only if $G/N$ is one of the following:

(1) $G={\rm Aff}(q)$ where $q$ is a Fermat prime, and $p=2$;

(2) $G={\rm Aff}(p+1)$ where $p$ is a Mersenne prime.

(3) $G={\rm Aff}(9)$ or $G=\mathbb{Z}/2\ltimes{\rm Aff}(9)$ (where the generator of $\mathbb{Z}/2$ acts by the Galois automorphism of $\mathbb{F}_{9}$) and $p=2$.

(4) $G=\mathbb{Z}/2$, $p\neq 2$.

Proof.

(i) Let ${\rm char}(\mathbf{k})=0$ and $X$ be the nontrivial irreducible representation of $G$. Then $X\otimes X$ has to contain $\dim X$ copies of the trivial representation. Thus $\dim X=1$ and $G=\mathbb{Z}/2$.

(ii) This follows by specializing Theorem A in [DN] to the case when the only linear representation of $G$ is trivial and using the easy fact that consecutive prime powers are $q-1,q$ for a Fermat prime $q$, or $p,p+1$ for a Mersenne prime $p$, or $8,9$. See e.g. [CL] for a short proof of this statement. ∎

6. Lower bounds for integral finite tensor categories with no nontrivial invertible objects

Proposition 6.1**.**

Let $\mathcal{C}=\mathop{\mathrm{Rep}}\nolimits(H)$ be a non-semisimple integral finite tensor category of rank $r$ such that ${\rm Ext}^{1}(\mathbb{1},\mathbb{1})=0$ (e.g., ${\rm char}(\mathbf{k})=0$) and $\mathcal{C}$ has no nontrivial invertible objects. Then:

(i) $\mathop{\mathrm{FPdim}}\nolimits\mathcal{C}\geq 8r+3$;

(ii) If $r=2$ then $\mathop{\mathrm{FPdim}}\nolimits(\mathcal{C})\geq 32$.

In particular, these bounds hold if $\mathcal{C}$ is not pointed and simple (has no nontrivial tensor subcategories).

Proof.

(i) It is clear that $\dim P_{\mathbb{1}}\geq 4$, since it has head and socle $\mathbb{1}$.

Assume first that $\dim P_{\mathbb{1}}=4$. Then $P_{\mathbb{1}}$ is uniserial with Loewy series $\mathbb{1},V,\mathbb{1}$ for some 2-dimensional simple module $V$, so $V^{*}\cong V$ and $\dim\mathop{\mathrm{Ext}}\nolimits^{1}(\mathbb{1},V)=1$, while ${\rm Ext}^{1}(\mathbb{1},Y)=0$ for any simple $Y\neq V$. So if $V\otimes V$ has Loewy series $\mathbb{1},Y,\mathbb{1}$, then $Y$ must be simple, and we must have $Y\cong V$ (as otherwise ${\rm Ext}^{1}(\mathbb{1},Y)=0$). Then $V\otimes V=P_{\mathbb{1}}$, so $V$ is projective by Lemma 4.7, a contradiction. 222Another way to get a contradiction is to note that $V^{2}=V+2$ in the Grothendieck ring and apply Proposition 5.11.

Thus, $V\otimes V=\mathbb{1}\oplus X$, where $X$ is a self-dual simple 3-dimensional module. Consider the tensor product $P_{\mathbb{1}}\otimes V=P_{V}\oplus T$, which has composition factors $V,\mathbb{1},X,V$. Note that $P_{V}$ necessarily includes composition factors $V,\mathbb{1},V$. If $P_{V}$ contains no other composition factors, i.e. it is 5-dimensional, then $T=X$, so $X=P_{X}$ is projective. But then $X\otimes V=P_{V}\oplus Q$ (as $\mathrm{Hom}(X\otimes V,V)\neq 0$), so $Q$ is projective and $\dim Q=1$, which is impossible. Thus, $P_{V}$ is indecomposable 8-dimensional with Loewy series $V,\mathbb{1}\oplus X,V$.

Let $W$ be a 2-dimensional projective module. Then $W^{*}\otimes W=P_{\mathbb{1}}$, so $\mathrm{End}(W\otimes V)=\mathrm{Hom}(W^{*}\otimes W,V\otimes V)=\mathrm{Hom}(P_{\mathbb{1}},\mathbb{1}\oplus X)=\mathbf{k}$. But $\mathcal{C}$ is unimodular, so the head and socle of any indecomposable projective coincide, and hence if $P$ is a non-simple indecomposable projective then $\dim\mathrm{End}(P)\geq 2$. Thus, $W\otimes V=Q$ is a 4-dimensional simple projective module. Moreover, $Q\otimes V=W\otimes V\otimes V=W\oplus W\otimes X$, and

[TABLE]

But ${\rm Ext}^{1}(V,X)\neq 0$, so $P_{X}$ includes composition factors $X,V,X$, hence $\dim P_{X}\geq 8$. Thus $X\otimes P_{\mathbb{1}}$, which has dimension $12$, cannot contain more than one copy of $P_{X}$. Hence ${\rm End}(W\otimes X)=\mathrm{Hom}(X\otimes P_{\mathbb{1}},X)=\mathbf{k}.$ Thus $W\otimes X$ is a 6-dimensional simple projective module. So $W$ can be recovered from $Q$ as the unique 2-dimensional summand in $Q\otimes V$, i.e., the assignment $W\mapsto Q$ is injective.

Now suppose we have $s$ projective 2-dimensional objects. Then they contribute $4s$ to ${\rm FPdim}(\mathcal{C})$, and the corresponding modules $Q$ contribute $16s$. The rest of the simple modules except $\mathbb{1},V,X$ contribute at least $8$ each, so in total $8(r-2s-3)$. So altogether we have (using that $\dim P_{\mathbb{1}}=4$, $\dim P_{V}=8$, $\dim P_{X}\geq 8$):

[TABLE]

as claimed.

Now assume that $\dim P_{\mathbb{1}}\geq 5$. Then by Lemma 2.1, for each 2-dimensional simple module $W$ we have $\dim P_{W}\geq 5/2$, so $W$ is not projective, and $\dim P_{W}\geq 4$.

If $\dim P_{\mathbb{1}}\geq 5$ then let $X$ be a simple submodule of $P_{\mathbb{1}}/\mathbb{1}$. If $\dim X\geq 3$ then $\dim P_{X}\geq 7$ (as it has composition factors $X,\mathbb{1},X$). Hence ${\rm FPdim}(\mathcal{C})\geq 5+3\cdot 7+8(r-2)=8r+10$, as claimed.

It remains to consider the case $\dim X=2$, in which case we must have $\dim P_{\mathbb{1}}\geq 6$. Then $\dim P_{X}\geq 5$. If $\dim P_{X}=5$ then the Loewy series of $P_{X}$ is $X,\mathbb{1},X$, so $X^{*}\otimes P_{X}$ has composition series $X^{*}\otimes X,X^{*},X^{*}\otimes X$. But $X^{*}\otimes P_{X}=P_{\mathbb{1}}\oplus Q$, and $P_{\mathbb{1}}$ has composition series $\mathbb{1},X,X^{\prime},\mathbb{1}$ where $\dim X^{\prime}=2$, and $Q$ is a 4-dimensional projective, so cannot involve $\mathbb{1}$. Thus $X^{*}\otimes X=\mathbb{1}\oplus Y$, where $Y$ is a 3-dimensional simple, i.e. we have only one 2-dimensional simple constituent in $X^{*}\otimes P_{X}$, a contradiction. Thus, $\dim P_{X}\geq 6$. If $\dim P_{X}>6$ or $\dim P_{\mathbb{1}}>6$ then they contribute at least $19$ into ${\rm FPdim}(\mathcal{C})$, so ${\rm FPdim}(\mathcal{C})\geq 19+8(r-2)=8r+3$, as claimed.

Thus it remains to consider the case $\dim P_{\mathbb{1}}=\dim P_{X}=6$. Then the Loewy series of $P_{X}$ must have the form $X,\mathbb{1}\oplus\mathbb{1},X$ (so ${\rm Ext}^{1}(\mathbb{1},X)=\mathbf{k}^{2}$) and the Loewy series of $P_{\mathbb{1}}$ therefore looks like $\mathbb{1},X\oplus X,\mathbb{1}$, where $X$ is self-dual. Now consider the module $X\otimes X$. If it has Loewy series $\mathbb{1},X^{\prime},\mathbb{1}$ where $X^{\prime}$ is simple 2-dimensional then $\mathrm{Hom}(P_{\mathbb{1}},X\otimes X)=\mathbf{k}$ (as then $X\otimes X$ is not a quotient of $P_{\mathbb{1}}$), which is a contradiction since we have $[X\otimes X,\mathbb{1}]=2$. Thus, $X\otimes X=\mathbb{1}\oplus Y$, where $Y$ is a 3-dimensional simple. So $Y$ contributes at least $9$ into ${\rm FPdim}(\mathcal{C})$. Thus we get ${\rm FPdim}(\mathcal{C})\geq 6+2\cdot 6+9+8(r-3)=8r+3$, as claimed. This proves (i).

To prove (ii), note that $\dim P_{\mathbb{1}}$ is divisible by ${\rm dim}P_{X}$. Let $X$ be the nontrivial simple object. If $\dim X\geq 4$ then $\dim P_{X}\geq 9$, so $\dim P_{\mathbb{1}}\geq 9$ and $\mathop{\mathrm{FPdim}}\nolimits(\mathcal{C})\geq 9+4\cdot 9=45$, as claimed. If $\dim X=3$ then $\dim P_{X}\geq 7$, so $\dim P_{\mathbb{1}}\geq 7$, which means that $[P_{\mathbb{1}}:X]\geq 2$, so $\dim P_{\mathbb{1}}\geq 8$, $\dim P_{X}\geq 8$ and ${\rm FPdim}(\mathcal{C})\geq 32$, as claimed.

Finally, the case $\dim X=2$ is impossible by Proposition 5.11, as ${\rm Ext}^{1}(\mathbb{1},\mathbb{1})\neq 0$. ∎

Example 6.2**.**

1. The bound ${\rm FPdim}(\mathcal{C})\geq 27$ for $r=3$ in Proposition 6.1 is sharp in any characteristic. Namely, it is attained for the category $\mathcal{C}$ of representations of the small quantum group $H=\mathfrak{u}_{q}(\mathfrak{sl}_{2})$ where $q$ is a primitive cubic root of $1$ if ${\rm char}(\mathbf{k})\neq 3$ and of the restricted enveloping algebra $H=\mathfrak{u}(\mathfrak{sl}_{2})$ in characteristic $3$. In this case there are three simple modules: $\mathbb{1}$, the 2-dimensional tautological module $V$ and the 3-dimensional Steinberg module $X$, which is projective. Moreover, the projective covers of $\mathbb{1}$ and $V$ have dimension $6$ and Loewy series $\mathbb{1},V\oplus V,\mathbb{1}$ and $V,\mathbb{1}\oplus\mathbb{1},V$.

2. The bound ${\rm FPdim}(\mathcal{C})\geq 32$ for $r=2$ in Proposition 6.1 is also sharp, as shown by Example 5.13.

3. There exists a category as in Proposition 6.1 with $\dim P_{\mathbb{1}}=4$. Namely, let $\mathbf{k}=\mathbb{C}$, $G$ be the binary icosahedral group (of order $120$), and $V$ its 2-dimensional tautological representation. Let $z$ be the nontrivial central element of $G$. Consider the supergroup $G\ltimes V$ and the category $\mathcal{C}={\rm Rep}(G\ltimes V,z)$ of representations of $G$ on superspaces such that $z$ acts by parity. Then $P_{\mathbb{1}}=\wedge V={\rm Ind}_{G}^{G\ltimes V}\mathbb{1}$. The simple objects of this category are just simple $G$-modules, which are labeled by vertices of the affine Dynkin diagram $\widetilde{E}_{8}$ and have dimensions $1,2,3,4,5,6,3,4,2$, and ${\rm FPdim}(\mathcal{C})=480$.

4. Examples of categories of small dimension ${\rm FPdim}(\mathcal{C})=60$ satisfying the assumptions of Proposition 6.1 are given by the representation categories $\mathop{\mathrm{Rep}}\nolimits_{p}(A_{5})$ of the alternating group $A_{5}$ in characteristics $p=2,3,5$. Namely:

$\mathop{\mathrm{Rep}}\nolimits_{2}(A_{5})$ has simple objects of dimension $1,2,2,4$ with projective covers of dimension $12,8,8,4$;

$\mathop{\mathrm{Rep}}\nolimits_{3}(A_{5})$ has simple objects of dimension $1,3,3,4$ with projective covers of dimension $6,3,3,9$; The Loewy series of $P_{\mathbb{1}}$ is $\mathbb{1},X,\mathbb{1}$, where $X$ is the 4-dimensional simple module.

$\mathop{\mathrm{Rep}}\nolimits_{5}(A_{5})$ has simple objects of dimension $1,3,5$ with projective covers of dimension $5,10,5$. The Loewy series of $P_{\mathbb{1}}$ is $\mathbb{1},X,\mathbb{1}$, where $X$ is the 3-dimensional simple module.

In particular, these examples show that $P_{\mathbb{1}}$ can have Loewy series $\mathbb{1},V,\mathbb{1}$ with $V$ of dimensions $2,3,4$.

7. Tensor categories of dimension $p$

Now let $\mathcal{C}$ be a finite tensor category over an algebraically closed field $\mathbf{k}$ of any characteristic $q\geq 0$ of Frobenius-Perron dimension $p>2$ (a prime). It is known that such a category is automatically integral ([EGNO]). If $\mathcal{C}$ is pointed then either it is semisimple (hence $\mathcal{C}=\mathrm{Vec}(\mathbb{Z}/p,\omega)$) or $\mathcal{C}$ is unipotent and ${\rm char}(\mathbf{k})=p$.

It is conjectured that $\mathcal{C}$ is always pointed. The goal of this section is to prove it under some assumptions.

Suppose $\mathcal{C}$ is not pointed, and let $A$ be the Grothendieck ring of $\mathcal{C}$. Let $r$ be its rank, and $d$ the smallest dimension of a nontrivial simple object (call it $X$). Clearly $d>1$. Let $d_{i}$ be the dimensions of simple objects and $p_{i}$ the dimensions of the corresponding indecomposable projectives. So $\sum_{i}p_{i}d_{i}=p$. Also, since ${\rm FPdim}(A)>1$ and divides $p$, we must have ${\rm FPdim}(A)=p$ and $D=1$. Also we have ${\rm Ext}^{1}(\mathbb{1},\mathbb{1})=0$, since otherwise the tensor subcategory of $\mathcal{C}$ formed by iterated extensions of $\mathbb{1}$ is nontrivial, and its dimension must divide $p$ by [EGNO], Theorem 7.17.6.

Proposition 7.1**.**

(i) $\mathcal{C}$ has at least $4$ simple objects.

(ii) $p\geq 37$.

Proof.

(i) Suppose $\mathcal{C}$ has two simple objects. Then by Theorem 3.1, $d\geq p/2$, so $p\geq 1+d^{2}\geq 1+p^{2}/4$, hence $p\leq 2$, a contradiction.

Now assume that $\mathcal{C}$ has three simple objects, $\mathbb{1}$ and $X,Y$ of Frobenius-Perron dimensions $d,m$. By Theorem 3.1, we have $d\geq(p/3)^{1/2}$, so $3d^{2}\geq p$. If $P_{X}\neq X$ then $\dim P_{X}\geq 2d$ so

[TABLE]

a contradiction. Similarly, if $P_{Y}\neq Y$ then $\dim P_{Y}\geq 2m$ so

[TABLE]

again a contradiction. Thus $X$ and $Y$ are projective. This means that the block of $\mathbb{1}$ is a tensor subcategory of $\mathcal{C}$. The Frobenius-Perron dimension of this subcategory must divide $p$, so this subcategory must be trivial. Hence $\mathcal{C}$ is semisimple. Then $X^{2}=1+aX+bY$ with $b\neq 0$, so $d^{2}=1+ad+bm$, so $(d,m)=1$. Now $X\otimes Y=sX+tY$, so $dm=sd+tm$, hence $sd=(d-t)m$. But $s,t>0$ since $X$ occurs in $Y\otimes Y$ and $Y$ occurs in $X\otimes X$. This gives a contradiction.

Thus $\mathcal{C}$ has at least $4$ simple objects.

(ii) It follows from (i) and Proposition 6.1 that $p\geq 8\cdot 4+3=35$. Hence $p\geq 37$. ∎

Corollary 7.2**.**

One has

[TABLE]

Proof.

This follows from Proposition 7.1, Theorem 3.1(iii) and the fact that the function

[TABLE]

satisfies $f(r)\geq f(4)=2^{-2/3}$ if $r\geq 4$. ∎

Proposition 7.3**.**

Let $H$ be a quasi-Hopf algebra of prime dimension $p$ with $r$ simple modules, where $r>1$ is not of the form $4k+1$ with $k\in\mathbb{Z}_{+},k\geq 2$ (e.g., $r\leq 8$). Then there exists a simple $H$-module $X\neq\mathbb{1}$ such that $X\cong X^{**}$.

Proof.

Assume the contrary. Then the distinguished invertible object $\chi$ of $H$ must be $\mathbb{1}$ (as $\chi^{**}\cong\chi$), so $X^{****}\cong X$ for all $X$ by categorical Radford’s formula ([EGNO], Subsection 7.19). Thus the group $\mathbb{Z}/4$ acts freely on nontrivial simple $H$-modules by taking the dual. This implies that $r=4k+1$.

It remains to show that $r\neq 5$. Indeed, for $r=5$ the nontrivial simple objects are $X_{1}=X$, $X_{2}=X^{*}$, $X_{3}=X^{**}$, $X_{4}=X^{***}$, so $d_{j}=d$ and $p_{j}=p_{1}$ for $1\leq j\leq 4$. Consider the product $P_{0}\otimes X^{*}$, of dimension $dp_{0}$. It decomposes in a direct sum of $P_{j}$, $1\leq j\leq 4$. Thus we have $dp_{0}=np_{1}$, where $n$ is the number of summands. But $p_{0}$ and $p_{1}$ are coprime (since $p_{0}+4dp_{1}=p$), so $d$ is divisible by $p_{1}$. Hence $d=p_{1}$ and $X_{j}$ are projective for $1\leq j\leq 4$. Thus ${\rm Ext}^{1}(\mathbb{1},\mathbb{1})$ cannot be zero, a contradiction. ∎

8. Hopf algebras of dimension $p$

Now let $H$ be a Hopf algebra of dimension $p$ (a prime) over a field $\mathbf{k}$ of characteristic $q$. It was recently proved by R. Ng and X. Wang [NW] that if $4q\geq p$ or $q=2$ then $H={\rm Fun}(\mathbb{Z}/p)$ or $q=p$ and $H=\mathbf{k}[x]/x^{p}$ with $\Delta(x)=x\otimes 1+1\otimes x$ or $\Delta(x)=x\otimes 1+1\otimes x+x\otimes x$. This is conjectured to hold for any $q$, and our goal is to improve the bound $4q\geq p$.

Assume that $q>2$ and $H$ is not of this form. If $H$ is semisimple and cosemisimple then by [EG3], Theorem 3.4, $H$ is commutative and cocommutative, so this is impossible. Thus, by replacing $H$ with $H^{*}$ if needed (as done in [NW]), we may (and will) assume that $H$ is not semisimple. Also, by a standard argument, $H$ and $H^{*}$ have no nontrivial $1$-dimensional modules (as their order must divide $p$). In particular, the distinguished group-like elements of $H$ and $H^{*}$ must be trivial , so $S^{4}=1$. Thus $S^{2}\neq 1$ (as $\dim H=0$ in $\mathbf{k}$) and defines an involution on the simple $H$-modules $X_{i}$, sending $X_{j}$ to $X_{j}^{**}$. Also, $H$ is simple, so ${\rm Rep}(H)$ is tensor generated by any nontrivial simple module and ${\rm Ext}^{1}(\mathbb{1},\mathbb{1})=0$.

Let $r$ be the number of simple $H$-modules, $d_{j}$ be the dimension of $X_{j}$ and $p_{j}$ be the dimension of the projective cover $P_{j}$ of $X_{j}$.

Proposition 8.1**.**

(i) (cf. [EG3], Lemma 2.10) There exists $i$ such that $X_{i}^{**}=X_{i}$ and $p_{i}$ and $d_{i}$ are both odd. Moreover, if there exists $j\neq 0$ with $X_{j}\cong X_{j}^{**}$ then there exists $i\neq 0$ with $p_{i}$ odd.

(ii) ([NW], Lemma 2.3) Suppose $p_{i}$ is odd and $X_{i}^{**}=X_{i}$. Then $p_{i}\geq q+2$.

(iii) Let $d_{*}$ be the maximum of all $d_{j}$, and let $i$ be such that $X_{i}\cong X_{i}^{**}$ and $p_{i}$ is odd, with largest possible $d_{i}$. Then we have

[TABLE]

(iv) If $d_{i}=d_{*}$ then

[TABLE]

In particular,

[TABLE]

If $d_{i}<d_{*}$ then

[TABLE]

(v) One has $p\geq 2(q+2)\sqrt{r-1}$.

(vi) If $d_{i}>1$ then $d_{i}\geq 3$.

Proof.

(i) Let $J$ be the set of $j$ such that $X_{j}^{**}\cong X_{j}$. Since double dual is an involution, we have that $\sum_{j\notin J}p_{j}d_{j}$ is even. Hence $\sum_{j\in J}p_{j}d_{j}$ is odd (as they add up to $p$). So there exists $i$ such that $p_{i}d_{i}$ is odd, which proves the first statement.

To prove the second statement, we may assume that $d_{j}$ or $p_{j}$ is even, otherwise we can take $i=j$. We may also assume that $p_{0}$ is odd; otherwise $i\neq 0$ automatically. Now consider the even-dimensional object $X_{j}^{*}\otimes P_{j}=P_{0}\oplus\oplus_{k\neq 0}m_{k}P_{k}$. We see that $\sum_{k\neq 0}m_{k}p_{k}$ is odd. So there exists $k$ such that $p_{k}$ is odd, and moreover, there is an odd number of such $k$. Hence the involution $**$ has a fixed point on the set of such $k$, which gives the statement.

(ii) Let $a:P_{i}\to P_{i}^{**}$ be an isomorphism. We may choose $a$ so that $a^{2}=1$. Then ${\rm Tr}(a)=0$, otherwise $\mathbb{1}$ is a direct summand in $P_{i}$, hence is projective and $H$ is semisimple, a contradiction. Let $m_{+},m_{-}$ be the multiplicities of the eigenvalues $1$ and $-1$ for $a$. Then $m_{+}+m_{-}=p_{i}$, $m_{+}-m_{-}=sq$ for some integer $s$. It is clear that $s$ has to be odd, so $|m_{+}-m_{-}|\geq q$, and $m_{-},m_{+}>0$ by Lemma 4.4. Thus $p_{i}\geq q+2$.

(iii) By Lemma 2.1, $p_{k}\geq N_{ji}^{k}p_{i}/d_{j}\geq N_{ji}^{k}(q+2)/d_{j}$ for any $j,k$. It is clear that for each $k$ there exists $j$ such that $X_{k}$ is a composition factor of $X_{j}\otimes X_{i}$ (so that $N_{ji}^{k}\geq 1$) and $d_{j}\leq d_{i}d_{k}$. Thus

[TABLE]

Hence

[TABLE]

as desired.

(iv) The first inequality follows immediately from (iii). The second inequality follows from the first one, as $d_{k}\geq 2$ for $k\neq 0$. To prove the third inequality, take $j$ such that $d_{j}=d_{*}$, and separate the corresponding term in the sum in (iii), while replacing all the other terms by $1$. This gives the statement.

(v) It follows from (iv) that

[TABLE]

so the statement follows by the arithmetic-geometric mean inequality.

(vi) This follows from Lemma 4.5. ∎

The following corollary improves the bound $p>4q$ from [NW].

Corollary 8.2**.**

One has $p>\frac{14}{3}(q+2)$.

Proof.

By Proposition 7.1, $r\geq 4$. Assume first that $r=4$. It is clear that $X^{**}=X$ for any simple object $X$ (otherwise we would have objects $X,X^{*},X^{**},X^{***}$ which are pairwise nonisomorphic, which would give $r\geq 5$). Thus by Proposition 8.1(i), we have $d_{i}\geq 2$, hence by Proposition 8.1(vi) $d_{i}\geq 3$. Thus, by Proposition 8.1(iv) we get $p>14(q+2)/3$.

It remains to consider the case $r\geq 5$. In this case, if $d_{i}\geq 2$ then by Proposition 8.1(vi) $d_{i}\geq 3$, so if $d_{i}=d_{*}$ then by Proposition 8.1(iv) we have $p>16(q+2)/3$, and if $d_{i}\neq d_{*}$ then Proposition 8.1(iv) yields $p>5(q+2)$. On the other hand, if $d_{i}=1$ then Proposition 8.1(iv) yields $p>r(q+2)\geq 5(q+2)$.

So in all cases we have $p>14(q+2)/3$, which proves the statement. ∎

Proposition 8.3**.**

Suppose that for some $i$ we have $X_{i}\cong X_{i}^{**}$, $p_{i}$ is odd and $d_{i}>1$. Then

[TABLE]

Proof.

By Corollary 7.2, we have

[TABLE]

Thus by Proposition 8.1(ii) we have

[TABLE]

which implies the statement. ∎

Corollary 8.4**.**

(i) Let $H$ be a non-semisimple Hopf algebra of dimension $p$ in characteristic $q$ for which there exists $i\neq 0$ such that $X_{i}^{**}\cong X_{i}$ (for instance, $**$ is isomorphic to ${\rm Id}$ as an additive functor on $\mathop{\mathrm{Rep}}\nolimits(H)$, i.e., $S^{2}$ is an inner automorphism). Then if

[TABLE]

then $H$ is commutative and cocommutative.

(ii) Let $W$ be the Lambert $W$-function (the inverse to the function $f(z)=ze^{z}$), and

[TABLE]

If

[TABLE]

then $H$ is commutative and cocommutative.

Proof.

(i) By Proposition 8.3, it suffices to show that there exists $i$ such that $X_{i}^{**}\cong X_{i}$, $d_{i}>1$ and $p_{i}$ is odd. But this follows from Proposition 8.1(i).

(ii) Fix $p$ and note that the function $2\sqrt{x}$ is increasing and $2^{-2/3}p^{1/x}$ is decreasing in $x$. Thus if $s$ is the solution of the equation

[TABLE]

then by Proposition 8.1(v) and part (i),

[TABLE]

Let $y=2^{5/3}\sqrt{s}$. We have

[TABLE]

Thus

[TABLE]

So $\log(y^{2})=W(2^{13/3}\log p)$. Thus we get

[TABLE]

so

[TABLE]

This implies (ii). ∎

Remark 8.5**.**

It is easy to show that

[TABLE]

as $x\to\infty$.

Theorem 8.6**.**

Let $H$ be a Hopf algebra of prime dimension $p$ over a field $\mathbf{k}$ characteristic $q>2$, which is not commutative and cocommutative. Then

[TABLE]

Proof.

If there exists simple $X\neq\mathbb{1}$ with $X\cong X^{**}$ then Corollary 8.4 applies. Otherwise, Proposition 7.3 and Proposition 8.1(v) imply that $p>r(q+2)\geq 9(q+2)$. This together with Corollary 8.2 implies the theorem. ∎