Finite symmetric tensor categories with the Chevalley property in characteristic $2$

TL;DR

This paper classifies finite symmetric tensor categories with the Chevalley property in characteristic 2, showing they are equivalent to representation categories of certain finite group schemes and computing related cohomology groups.

Contribution

It extends Deligne's theorem to characteristic 2, establishing a classification of symmetric tensor categories with the Chevalley property and computing associated cohomology groups.

Findings

Categories admit symmetric fiber functors to a specific tensor category

Existence of a unique finite group scheme representing the category

Computed cohomology groups for group algebras and function algebras

Abstract

We prove an analog of Deligne's theorem for finite symmetric tensor categories $\mathcal{C}$ with the Chevalley property over an algebraically closed field $k$ of characteristic $2$. Namely, we prove that every such category $\mathcal{C}$ admits a symmetric fiber functor to the symmetric tensor category $\mathcal{D}$ of representations of the triangular Hopf algebra $(k[\dd]/(\dd^2),1\ot 1 + \dd\ot \dd)$. Equivalently, we prove that there exists a unique finite group scheme $G$ in $\mathcal{D}$ such that $\mathcal{C}$ is symmetric tensor equivalent to $\Rep_{\mathcal{D}}(G)$. Finally, we compute the group $H^2_{\rm inv}(A,K)$ of equivalence classes of twists for the group algebra $K[A]$ of a finite abelian $p$-group $A$ over an arbitrary field $K$ of characteristic $p>0$, and the Sweedler cohomology groups $H^i_{\rm{Sw}}(\mathcal{O}(A),K)$, $i\ge 1$, of the function algebra $\mathcal{O}(A)$ of $A$.

Equations

$$\Delta (h)=(1\otimes 1+\mathrm{d}\otimes \mathrm{d})(\Delta (h){)}_{21},h\in H.$$

$${\wedge }^{2}V:=\text{Im}(\text{id}+\tau )\subset {\Gamma }^{2}V:=\text{Ker}(\text{id}+\tau )\subset V^{\otimes 2},$$

$$S^{2}V:=V^{\otimes 2}/{\wedge }^{2}V,V^{(1)}:={\Gamma }^{2}V/{\wedge }^{2}V,$$

$${\Gamma }^{2}V\hookrightarrow V^{\otimes 2}\twoheadrightarrow S^{2}V.$$

$$\text{deg}({\mathrm{d}}_{n-1}^2\otimes {\mathrm{d}}_{n-1}\tilde{v})\ge n/2+n/2+p=n+p\ge n+1,$$

$$\varphi \circ (\mathrm{id}\otimes \mathrm{id}\otimes m)+\varphi \circ (m\otimes \mathrm{id}\otimes \mathrm{id})=\epsilon \otimes \varphi +\varphi \circ (\mathrm{id}\otimes m\otimes \mathrm{id})+\varphi \otimes \epsilon$$

$$\varphi \circ (\mathrm{id}\otimes \mathrm{id}\otimes 1)=\varphi \circ (1\otimes \mathrm{id}\otimes \mathrm{id})=\varphi \circ (\mathrm{id}\otimes 1\otimes \mathrm{id})=\epsilon \otimes \epsilon .$$

$${\beta }_j:=\sum _{l=1}^{2^{r_j}-1}{y}_j^{(l)}\otimes y_j^{(2^{r_j}-l)}.$$

$$\varphi =\sum _{1\le i<j<l\le n}{b}_{ijl}(y_i\otimes y_j\otimes y_l)+\sum _{i,j}{a}_{ij}{\beta }_i\otimes y_j+df,$$

$$\mathrm{Cyc}(df)$$

$$0=\mathrm{Alt}(\varphi )=\sum _{1\le i<j<l\le n}{b}_{ijl}\mathrm{Alt}(y_i\otimes y_j\otimes y_l).$$

$$\mathrm{Cyc}(df)$$

$$\mathrm{Cyc}(df)=\mathrm{Cyc}\left((\Delta \otimes \text{id})(f)+(\text{id}\otimes \Delta )(f)\right).$$

$$\Psi :=\pi \otimes \text{id}:{\Gamma }^{2}k[\mathcal{G}]\otimes k[\mathcal{G}]\to k[\mathcal{G}{]}^{(1)}\otimes k[\mathcal{G}],$$

$$\Psi (\mathrm{Cyc}(df))=0.$$

$$\Psi ({\beta }_i\otimes y_j)=y_i^{(2^{r_i-1})}\otimes y_j.$$

$$\sum _{i,j}{a}_{ij}{y}_i^{(2^{r_i-1})}\otimes y_j=0,$$

$$f+f_{21}=\sum _i{a}_i{\beta }_i+\sum _{i<j}{b}_{ij}(y_i\otimes y_j)+z\otimes 1+1\otimes z+\Delta (z)$$

$$f+f_{21}=z\otimes 1+1\otimes z+\Delta (z).$$

$$f_{12,3}+f_{1,3}+f_{2,3}+f_{3,12}+f_{3,1}+f_{3,2}=z_{123}+z_{12}+z_{23}+z_{13}+z_1+z_2+z_3.$$

$$\mathrm{Cyc}(df)=\mathrm{Cyc}(f_{12,3}+f_{1,3}+f_{2,3}+f_{3,12}+f_{3,1}+f_{3,2})=z_{123}+z_{12}+z_{23}+z_{13}+z_1+z_2+z_3.$$

$$z_{123}+z_{12}+z_{23}+z_{13}+z_1+z_2+z_3=0.$$

$$w_{12,3}+w_{1,3}+w_{2,3}=0.$$

$$\varphi =T\otimes \delta +df$$

$$\varphi =\sum _{1\le i<j<l\le n}{b}_{ijl}(y_i\otimes y_j\otimes y_l)+\sum _{i,j}{a}_{ij}{\beta }_i\otimes y_j+d{f}^{\prime },$$

$$0=\mathrm{Alt}(\varphi )=\sum _{1\le i<j<l\le n}{b}_{ijl}\mathrm{Alt}(y_i\otimes y_j\otimes y_l).$$

$$\mathrm{Cyc}(d{f}^{\prime })$$

$$\Psi (\mathrm{Cyc}(d{f}^{\prime }))=0.$$

$$\pi ({\beta }_i)=y_i^{(2^{r_i-1})}.$$

$$\sum _{i,j}{a}_{ij}{y}_i^{(2^{r_i-1})}\otimes y_j+\pi (T)\otimes \delta =0,$$

Full text

1. Introduction

The main objective of this paper is to classify finite symmetric tensor categories with the Chevalley property over an algebraically closed field $k$ of characteristic $2$. This completes the classification of finite integral symmetric tensor categories with the Chevalley property over an algebraically closed field of characteristic $p>0$, which for $p>2$ was established in [EG2], since by [O, Theorem 1.5], integrality follows from the rest of the conditions for $p=2,3$.

Let $\alpha_{2}$ be the Frobenius kernel of the additive group $\mathbb{G}_{a}$. Then $k\alpha_{2}=k[{\rm d}]/({\rm d}^{2})$ with ${\rm d}$ primitive. Let $\mathcal{D}:={\rm Rep}(\alpha_{2},1\otimes 1+{\rm d}\otimes{\rm d})$ be the symmetric tensor category of finite dimensional representations of the triangular Hopf algebra $k[{\rm d}]/({\rm d}^{2})$ equipped with the $R$-matrix $1\otimes 1+{\rm d}\otimes{\rm d}$ 111$\mathcal{D}$ may be considered as a non-semisimple analog in characteristic $2$ of the category sVec of supervector spaces, see [V].. Recall [V] that an object in $\mathcal{D}$ is a finite dimensional $k$-vector space $V$ together with a linear map ${\rm d}:V\to V$ satisfying ${\rm d}^{2}=0$. In particular, $\mathcal{D}$ has two indecomposable objects, namely, the unit object (i.e., the vector space $k$ with ${\rm d}=0$), and the two dimensional vector space $k^{2}$ with ${\rm d}$ the strictly upper triangular matrix $E_{12}$.

Recall that a finite group scheme in $\mathcal{D}$ is, by definition, a finite dimensional cocommutative Hopf algebra $H$ in $\mathcal{D}$. In particular, this means that ${\rm d}:H\to H$ is a derivation of $H$ satisfying ${\rm d}^{2}=0$, and

[TABLE]

We can now state our main result (compare with [O, Conjecture 1.3]).

Theorem 1.1**.**

Let $\mathcal{C}$ be a finite symmetric tensor category with the Chevalley property over an algebraically closed field $k$ of characteristic $2$. Then $\mathcal{C}$ admits a symmetric fiber functor to $\mathcal{D}$. Thus, there exists a unique finite group scheme $G$ in $\mathcal{D}$ such that $\mathcal{C}$ is symmetric tensor equivalent to the category ${\rm Rep}_{\mathcal{D}}(G)$ of finite dimensional representations of $G$ which are compatible with the action of $\pi_{1}(\mathcal{D})$.

Remark 1.2**.**

Theorem 1.1 answers [BE, Question 1.2] for finite symmetric tensor categories with the Chevalley property over $k$, and we expect it to hold for every finite symmetric integral tensor category over $k$.

Finally, we note that the arguments used to prove [EG2, Theorem 1.1] and Theorem 1.1 in fact prove a stronger result (see Theorem 2.21).

The organization of the paper is as follows. Section 2 is devoted to the proof of Theorem 1.1. In Section 3 we compute the group $H^{2}_{\rm inv}(A,K)$ of equivalence classes of twists for the group algebra $K[A]$ of a finite abelian $p$-group $A$ over an arbitrary field $K$ of characteristic $p>0$ (see Theorem 3.5), and use it together with [EG2, Proposition 5.7] to compute the Sweedler cohomology groups $H^{i}_{\rm{Sw}}(\mathcal{O}(A),K)$ for every $i\geq 1$ (see Theorem 3.8).

Acknowledgments. P. E. was partially supported by NSF grant DMS 1502244. S. G. is grateful to the University of Michigan and MIT for their warm hospitality.

2. The proof of Theorem 1.1

All constructions in this section are done over an algebraically closed field $k$ of characteristic $2$ unless otherwise is explicitly stated. To lighten notation, we sometimes write $1$ for $1\otimes 1$ or $1\otimes 1\otimes 1$.

We refer the reader to [EGNO] for the general theory of finite tensor categories, to [Dr] for generalities on quasi-Hopf algebras (see also [EG2, 2.1]), and to [J, W] for the general theory of finite group schemes (see also [EG2, 2.4]).

By [O, Theorem 1.5], any finite symmetric tensor category with the Chevalley property in characteristic $2$ is integral (as ${\rm Ver}_{2}={\rm Vec}$). Thus by [EO, Theorem 2.6], $\mathcal{C}$ is symmetric tensor equivalent to ${\rm Rep}(H,R,\Phi)$ for some finite dimensional triangular quasi-Hopf algebra $(H,R,\Phi)$ with the Chevalley property over $k$. Thus, we have to prove the following theorem.

Theorem 2.1**.**

Let $(H,R,\Phi)$ be a finite dimensional triangular quasi-Hopf algebra with the Chevalley property over $k$. Then $(H,R,\Phi)$ is pseudotwist equivalent to a triangular Hopf algebra with $R$-matrix $1+{\rm d}\otimes{\rm d}$ for some ${\rm d}\in P(H)$ such that ${\rm d}^{2}=0$.

We will prove Theorem 2.1 in several steps.

2.1. ${\rm gr}(H)$

Let $(H,R,\Phi)$ be a finite dimensional triangular quasi-Hopf algebra with the Chevalley property over $k$. Let $I:={\rm Rad}(H)$ be the Jacobson radical of $H$. Since $I$ is a quasi-Hopf ideal of $H$, the associated graded algebra ${\rm gr}(H)=\bigoplus_{r\geq 0}H[r]$ has a natural structure of a graded triangular quasi-Hopf algebra with some $R$-matrix $R_{0}\in H[0]^{\otimes 2}$ and associator $\Phi_{0}\in H[0]^{\otimes 3}$ (see, e.g., [EG2, 2.2]).

Proposition 2.2**.**

[EG2, Proposition 3.2]** The following hold:

1. (1)

$H[0]$* is semisimple.* 2. (2)

$(H[0],R_{0},\Phi_{0})$* is a triangular quasi-Hopf subalgebra of $({\rm gr}(H),R_{0},\Phi_{0})$.* 3. (3)

${\rm Rep}(H[0],R_{0},\Phi_{0})$* is symmetric tensor equivalent to ${\rm Rep}(G)$ for some finite semisimple group scheme $G$ over $k$.* 4. (4)

$({\rm gr}(H),R_{0},\Phi_{0})$* is pseudotwist equivalent to a graded triangular Hopf algebra with $R$-matrix $1\otimes 1$, whose degree [math]-component is $(kG,1\otimes 1)$. ∎*

Corollary 2.3**.**

[EG2, Corollary 3.3]** Let $(H,R,\Phi)$ be a finite dimensional triangular quasi-Hopf algebra with the Chevalley property over $k$. Then ${\rm gr}(H)$ is pseudotwist equivalent to $k\mathcal{G}$ for some finite group scheme $\mathcal{G}$ over $k$ containing $G$ as a closed subgroup scheme. ∎

Remark 2.4**.**

By Nagata’s theorem (see, e.g, [A, p.223]), we have $G=\Gamma\ltimes P^{D}$, where $\Gamma$ is a finite group of odd order and $P$ is a finite abelian $2$-group. Hence, we have $kG=k\Gamma\ltimes\mathcal{O}(P)$.

Let $\Gamma:=\mathcal{G}/\mathcal{G}^{\circ}$. Then $\Gamma$ is a finite constant group of odd order, and we have $\mathcal{G}=\Gamma\ltimes\mathcal{G}^{\circ}$. Thus, we have $\mathcal{O}(\mathcal{G})=\mathcal{O}(\Gamma)\otimes\mathcal{O}(\mathcal{G}^{\circ})$ as algebras.

By the results of this subsection, we may assume without loss of generality in the proof of Theorem 2.1 that $R=1+$ terms of higher degree.

2.2. Trivializing $R$

Let $V$ be a $k$-vector space, and let $\tau:V^{\otimes 2}\to V^{\otimes 2}$ be the flip map. Recall that

[TABLE]

[TABLE]

and that $V^{(1)}$ is called the Frobenius twist of $V$ and $\Gamma^{2}V$ the divided second symmetric power of $V$. Note that $V^{(1)}$ is the image of the composition

[TABLE]

Let $\pi:\Gamma^{2}V\to V^{(1)}$ be the natural surjective map.

Let $(H,R,\Phi)$ be as in the end of Section 2.1.

Proposition 2.5**.**

The following hold:

1. (1)

Suppose $R=1+{\rm d}_{n-1}\otimes{\rm d}_{n-1}$ modulo terms of degree $\geq n\geq 1$ such that ${\rm d}_{n-1}\in{\rm Rad}(H)$. Then $(H,R,\Phi)$ can be twisted to a form such that $R=1+{\rm d}_{n}\otimes{\rm d}_{n}$ modulo terms of degree $\geq n+1$, where ${\rm d}_{n}\in{\rm Rad}(H)$ and ${\rm d}_{n}-{\rm d}_{n-1}$ has degree $\geq n/2$, by a pseudotwist $J_{n}$ such that $J_{n}-1$ has degree $\geq n$ if ${\rm d}_{n-1}=0$, and degree $\geq\frac{n}{2}+p$ if ${\rm deg}({\rm d}_{n-1})=p>0$. 2. (2)

If $R\neq 1$ then $(H,R,\Phi)$ can be twisted to the form $R=1+{\rm d}\otimes{\rm d}$, where ${\rm d}\in{\rm Rad}(H)$ is an element of positive degree. Moreover, if $R=1+{\rm d}^{\prime}\otimes{\rm d}^{\prime}$ modulo terms of degree $\geq n$, where ${\rm d}^{\prime}\in{\rm Rad}(H)$, then this can be achieved by a pseudotwist $J$ with $J-1$ of degree $\geq n$ if ${\rm d}^{\prime}=0$, and degree $\geq\frac{n}{2}+p$ if ${\rm d}^{\prime}\neq 0$ and has degree $p$, so that ${\rm d}-{\rm d}^{\prime}$ has degree $\geq n/2$. 3. (3)

If $R=1+{\rm d}\otimes{\rm d}$ then ${\rm d}^{2}=0$.

Proof.

(1) Let $R=1+{\rm d}_{n-1}\otimes{\rm d}_{n-1}$ modulo terms of degree $\geq n$, and consider $R$ modulo terms of degree $\geq n+1$. We have $R=1+{\rm d}_{n-1}\otimes{\rm d}_{n-1}+\widetilde{s}$ modulo terms of degree $\geq n+1$, where $\widetilde{s}\in H^{\otimes 2}$ has degree $\geq n$. Let $s\in{\rm gr}(H)^{\otimes 2}[n]$ be the leading part of $\widetilde{s}$. Then $s$ is symmetric because $R_{21}R=1\otimes 1$, so $s\in\Gamma^{2}{\rm gr}(H)[n]$. Moreover, if $t\in\wedge^{2}{\rm gr}(H)[n]$ then we can replace $s$ by $s+t$ by twisting.

Let $v:=\pi(s)$ be the image of $s$ in ${\rm gr}(H)^{(1)}[n]={\rm gr}(H)[n/2]^{(1)}$ (note that this space can be nonzero only if $n$ is even). Then we can twist $s$ into the form $v\otimes v$ by a pseudotwist $J$ with $J-1$ of degree $\geq n$. So we will get $R=1+{\rm d}_{n-1}\otimes{\rm d}_{n-1}+\widetilde{v}\otimes\widetilde{v}$ modulo terms of degree $\geq n+1$, where $\widetilde{v}$ is a lift of $v$ to $H$. If ${\rm d}_{n-1}=0$, this completes the proof (we can set ${\rm d}_{n}=\widetilde{v}$). Thus, it remains to consider the case when ${\rm d}_{n-1}\neq 0$ and has degree $p$; so we may assume that $n>2p$ (because for $n\leq 2p$, we can set ${\rm d}_{n}={\rm d}_{n-1}$ and $J=1$). In this case, let us twist by $J=1+{\rm d}_{n-1}\otimes\widetilde{v}$ (note that $\deg(J-1)\geq\frac{n}{2}+p$). Since $R_{21}R=1\otimes 1$, we have $\text{deg}({\rm d}_{n-1}^{2})\geq n/2$, hence

[TABLE]

so twisting by $J$ brings $R$ to the form $R=1+({\rm d}_{n-1}+\widetilde{v})\otimes({\rm d}_{n-1}+\widetilde{v})$ modulo terms of degree $\geq n+1$, i.e., we may take ${\rm d}_{n}={\rm d}_{n-1}+\widetilde{v}$, as desired.

(2) Follows immediately from (1). Namely, for the first statement we take ${\rm d}$ to be the stable limit of the ${\rm d}_{m}$’s and $J$ to be the product of the $J_{m}$’s, and for the second statement we take ${\rm d}_{n-1}={\rm d}^{\prime}$, ${\rm d}$ to be the stable limit of the ${\rm d}_{m}$’s, and $J$ to be the product of the $J_{m}$’s for $m\geq n$.

(3) Follows from the identity $R_{21}R=1\otimes 1$. ∎

Thus, from now on we may assume that $R=1+{\rm d}\otimes{\rm d}$ for some ${\rm d}\in{\rm Rad}(H)$ with ${\rm d}^{2}=0$ (but in general ${\rm d}$ is not a primitive element yet, as we have not made $\Phi=1$).

Remark 2.6**.**

Proposition 2.5 implies that the degree $p$ of ${\rm d}$ in Proposition 2.5(2) and its degree $p$ part $\delta\in{\rm gr}(H)[p]$ (when ${\rm d}\neq 0$) are uniquely determined. Indeed, if $(H,R,\Phi)$ is pseudotwist equivalent to $(H,R^{\prime},\Phi^{\prime})$ where $R=1+{\rm d}\otimes{\rm d}$ and $R^{\prime}=1+{\rm d}^{\prime}\otimes{\rm d}^{\prime}$ modulo terms of degree $\geq n$, and if ${\rm d}\neq 0$ and has degree $p<n/2$, then by Proposition 2.5(1) the pseudotwist $J$ can be chosen so that $J-1$ is of degree $\geq\frac{n}{2}+p>2p$, so ${\rm d}^{\prime}-{\rm d}$ has degree $\geq p+1$, as desired. In particular, if $R=J_{21}^{-1}J$ then whenever $R$ is twisted to $1+{\rm d}\otimes{\rm d}$, we must have ${\rm d}=0$. This is the case when ${\rm Rep}(H,R,\Phi)$ is Tannakian (as follows from Theorem 2.1). However, ${\rm d}$ itself is not unique (e.g., it can be conjugated by an invertible element $x$ of $1+{\rm Rad}(H)$, which results from applying the coboundary twist attached to $x$).

2.3. Trivializing $\Phi$

Let $(H,R,\Phi)$ be a finite dimensional triangular quasi-Hopf algebra with the Chevalley property over $k$, where $R=1+{\rm d}\otimes{\rm d}$ for some element ${\rm d}\in{\rm Rad}(H)$ with ${\rm d}^{2}=0$.

By Corollary 2.3, ${\rm gr}(H)=k[\mathcal{G}]=\bigoplus_{i\geq 0}k[\mathcal{G}][i]$, as graded Hopf algebras, for some finite group scheme $\mathcal{G}$ over $k$. We let $m$, $\varepsilon$ denote the multiplication and counit maps of $\mathcal{O}(\mathcal{G})$.

If $\Phi=1$ then ${\rm d}^{2}=0$ and $\Delta({\rm d})={\rm d}\otimes 1+1\otimes{\rm d}$, so we are done. Thus we may assume that $\Phi\neq 1$. Consider $\Phi-1$. If it has degree $\ell$ then let $\phi$ be its projection to ${\rm gr}(H)^{\otimes 3}[\ell]$.

For every permutation $(i_{1}i_{2}i_{3})$ of $(123)$, we will use $\phi_{i_{1}i_{2}i_{3}}$ to denote the $3$-tensor obtained by permuting the components of $\phi$ accordingly.

Lemma 2.7**.**

The following hold:

1. (1)

$\phi\in Z^{3}(\mathcal{O}(\mathcal{G}),k)$* is a normalized Hochschild $3$-cocycle of $\mathcal{O}(\mathcal{G})$ with coefficients in the trivial module $k$, i.e.,*

[TABLE]

and

[TABLE] 2. (2)

${\rm Alt}(\phi):=\phi_{312}+\phi_{132}+\phi_{123}+\phi_{231}+\phi_{213}+\phi_{321}=0$.

Proof.

(1) Follows from [EG2, (2.1)-(2.2)] in a straightforward manner.

(2) Follows from [EG2, (2.8)] in a straightforward manner. ∎

2.3.1. The case $R=1\otimes 1$.

In this subsection we will assume that $R=1\otimes 1$, i.e., ${\rm d}=0$.

Lemma 2.8**.**

The following hold:

1. (1)

$\phi_{312}+\phi_{132}+\phi_{123}=0=\phi_{231}+\phi_{213}+\phi_{123}$. 2. (2)

$\phi_{123}=\phi_{321}$. 3. (3)

${\rm Cyc}(\phi):=\phi_{312}+\phi_{231}+\phi_{123}=0$.

Proof.

(1) Follows from [EG2, (2.6)-(2.7)] in a straightforward manner.

(2) Using (1) and Lemma 2.7(2), we get

[TABLE]

as claimed.

(3) By (2), we have $\phi_{132}=\phi_{231}$. Thus the claim follows from (1). ∎

Following [EG2, 2.8]222In [EG2, 2.8], $y_{t}^{(l)}$ was denoted by $(x_{t}^{l})^{*}$., we set $y_{t}:=x_{t}^{*}$ and $y_{t}^{(l)}:=(x_{t}^{l})^{*}$, $1\leq t\leq n$, $1\leq l\leq r_{t}-1$ (so, $y_{t}^{(1)}=y_{t}$), and for every $1\leq i,j\leq n$, let

[TABLE]

Proposition 2.9**.**

The $3$-cocycle $\phi$ is a coboundary.

Proof.

By Lemma 2.7(1), $\phi\in Z^{3}(\mathcal{O}(\mathcal{G}),k)$ so we can express it in the following form:

[TABLE]

for some $b_{ijl},a_{ij}\in k$ and $f\in k[\mathcal{G}]^{\otimes 2}$.

Thus by Lemma 2.8(3), we have

[TABLE]

Also, since ${\rm Alt}(df)={\rm Alt}(\beta_{i}\otimes y_{j})=0$, it follows from Lemma 2.7(2) and the above that we have

[TABLE]

Therefore $b_{ijl}=0$ for every $i<j<l$, and we have

[TABLE]

It is also straightforward to verify that we have

[TABLE]

Consider the surjective homomorphism

[TABLE]

where $\pi:\Gamma^{2}k[\mathcal{G}]\to k[\mathcal{G}]^{(1)}$ is the natural surjective homomorphism. Observe that we have $\pi(\Delta(u))=0$ for every $u\in{\rm Rad}(H)$. Indeed this holds for $u=u_{1}\cdots u_{m}$, where $u_{1},\dots,u_{m}$ are primitive, and each element of ${\rm Rad}(H)$ is a linear combination of such with coefficients in $G$.

Now since $k[\mathcal{G}]$ is cocommutative, it follows from (2.2) that ${\rm Cyc}(df)$ is symmetric, hence we have

[TABLE]

We also have

[TABLE]

Thus

[TABLE]

which implies that $a_{ij}=0$ for all $i,j$. Thus $\phi=df$ is a coboundary, as claimed. ∎

Lemma 2.10**.**

In Proposition 2.9 we can choose $f\in\Gamma^{2}k[\mathcal{G}]$, i.e., we can choose $f$ to be symmetric.

Proof.

Since $\phi_{123}=\phi_{321}$ by Lemma 2.8(2), we have $df=d(f_{21})$. This implies that $f+f_{21}\in Z^{2}(\mathcal{O}(\mathcal{G}),k)$ is a $2$-cocycle, so it follows from [EG2, Proposition 2.4(2)] that we have

[TABLE]

for some $a_{i},b_{ij}\in k$ and $z\in k[\mathcal{G}]$. Since the left hand side is symmetric and $\Delta=\Delta^{cop}$, we must have $b_{ij}=0$ for every $i<j$. Applying the map $\Psi$ then yields $a_{i}=0$ for every $i$. Thus, we have

[TABLE]

Hence, applying the operator $y\mapsto y\otimes 1+1\otimes y+\Delta(y)$ to the first tensorand, we get

[TABLE]

Hence, the left hand side is symmetric, so

[TABLE]

Since ${\rm Cyc}(df)=0$, this implies that

[TABLE]

Let $w:=z_{1,2}+z_{1}+z_{2}$. Then Equation (2.5) implies

[TABLE]

This means that the tensorands of $w$ are primitive, hence $w=\sum_{i,j}c_{ij}p_{i}\otimes p_{j}$, where $p_{i}$ is a basis of primitive elements, with $c_{ij}=c_{ji}$. Moreover, $\pi(w)=0$, which implies that $c_{ii}=0$ for all $i$. Now replacing $f$ with $f+\sum_{i<j}c_{ij}p_{i}\otimes p_{j}$ (which is possible since this sum is a $2$-cocycle) we come to a situation where $f$ is symmetric, as desired. ∎

Choose $f\in(\mathcal{O}(\mathcal{G})^{*})^{\otimes 2}$ symmetric with the same degree $\ell$ as $\phi$ such that $\phi=df$, which is possible by Lemma 2.10. Let $\widetilde{f}$ be a symmetric lift of $f$ to $H$. Then the pseudotwist $F:=1+\widetilde{f}$ is symmetric, which implies that $(H,1,\Phi)^{F}=(H^{F},1,\Phi^{F})$, and the pseudotwisted associator $\Phi^{F}$ is equal to $1+$ terms of degree $\geq\ell+1$. By continuing this procedure, we will come to a situation where $(H,1,\Phi)^{F}=(H^{F},1,1)$ for some pseudotwist $F\in H^{\otimes 2}$, as desired. This concludes the proof of Theorem 2.1 in the case where $R=1$.

2.3.2. The case $R=1+{\rm d}\otimes{\rm d}$ with ${\rm d}\neq 0$.

In this subsection we will assume that $R=1+{\rm d}\otimes{\rm d}$ with ${\rm d}\neq 0$. Suppose ${\rm d}$ has degree $p$, and let $\delta$ be its projection to ${\rm gr}(H)[p]$.

The following lemma is the analogue of Lemma 2.8 in this case.

Lemma 2.11**.**

The following hold:

1. (1)

$\Delta(\delta)=\delta\otimes 1+1\otimes\delta$. 2. (2)

The degree of $\Delta({\rm d})-{\rm d}\otimes 1-1\otimes{\rm d}$ is $\geq\ell-p$. 3. (3)

Let $T\in{\rm gr}(H)^{\otimes 2}[\ell-p]$ be the part of $\Delta({\rm d})-{\rm d}\otimes 1-1\otimes{\rm d}$ of degree exactly $\ell-p$ (so $T=0$ if $\ell\leq 2p$). Then we have

1. (a)

$T\otimes\delta+\phi_{312}+\phi_{132}+\phi_{123}=0$. 2. (b)

$\delta\otimes T+\phi_{231}+\phi_{213}+\phi_{123}=0$. 3. (c)

$\phi_{123}+\phi_{321}=T\otimes\delta+\delta\otimes T$. 4. (d)

${\rm Cyc}(\phi)={\rm Cyc}(T\otimes\delta)$. 4. (4)

$T$* is a symmetric $2$-cocycle.*

Proof.

(1) is clear. (2) and (3) follow immediately from the hexagon relations [EG2, (2.6)-(2.7)] ((3)(d) is obtained by applying ${\rm Cyc}$ to (3)(a) and using that ${\rm Alt}(\phi)=0$). Also, let $Q:=T+T_{21}$. By (3)(c), we have $Q\otimes\delta=\delta\otimes Q$. Thus both left and right tensorands of $Q$ can only be multiples of $\delta$, i.e., $Q$ is a multiple of $\delta\otimes\delta$. But $\pi(Q)=0$, hence $Q=0$, proving (4). ∎

Proposition 2.12**.**

The $3$-cocycle $\phi$ has the form

[TABLE]

for some $f\in k[\mathcal{G}]^{\otimes 2}[\ell]$.

Proof.

By Lemma 2.7(1), $\phi\in Z^{3}(\mathcal{O}(\mathcal{G}),k)$ and we can express it in the following form:

[TABLE]

for some $b_{ijl},a_{ij}\in k$ and $f^{\prime}\in k[\mathcal{G}]^{\otimes 2}$.

Since ${\rm Alt}(df^{\prime})=0$, using Lemma 2.7(2) this implies that

[TABLE]

Therefore $b_{ijl}=0$ for every $i<j<l$. Thus by Lemma 2.11(2)(c), we have

[TABLE]

Now by (2.3), we have

[TABLE]

We also have

[TABLE]

Thus,

[TABLE]

which implies that

[TABLE]

for some $a_{i}\in k$. Hence,

[TABLE]

for some $h\in k[\mathcal{G}]$ (as the left hand side is a symmetric $2$-cocycle killed by $\pi$, hence a coboundary). So,

[TABLE]

(as $d\delta=0$). This implies that $\phi=T\otimes\delta+df$, where $f:=f^{\prime}+h\otimes\delta$, as desired. ∎

Lemma 2.13**.**

In Proposition 2.12 we can choose $f\in\Gamma^{2}k[\mathcal{G}]$, i.e., we can choose $f$ to be symmetric.

Proof.

Since $\phi_{123}=\phi_{321}+T\otimes\delta+\delta\otimes T$ by Lemma 2.11(3)(c), we have $df=d(f_{21})$. Thus, $f+f_{21}\in Z^{2}(\mathcal{O}(\mathcal{G}),k)$ is a $2$-cocycle, and we can proceed in exactly the same way as in the proof of Lemma 2.10 to get to a situation where $f$ is symmetric. ∎

Proposition 2.14**.**

The $4$-cocycle $T\otimes T$ is a coboundary.

Proof.

Let $f$ be a symmetric element provided by Lemma 2.13, and let $\widetilde{f}$ be a symmetric lift of $f$ to $H$. Then the pseudotwist $F:=1+\widetilde{f}$ is symmetric. Thus, $(H,R,\Phi)^{F}=(H^{F},R^{F},\Phi^{F})$, and $\Phi^{F}-1$ has degree $\geq\ell$ with degree $\ell$ part $T\otimes\delta$. Thus, we have

[TABLE]

where $U\in H^{\otimes 3}$ has degree $\geq\ell+1$.

The pentagon equation [EG2, (2.3)] for $\Phi^{F}$ yields that $dU$ has degree $\geq 2\ell-2p$, and its part of degree $2\ell-2p$ is $T\otimes T$. This means that $U$ has degree $s\leq 2\ell-2p$. Let $u$ be the leading part of $U$. If $s<2\ell-2p$ then the pentagon equation [EG2, (2.3)] yields that $du=0$, and arguing as above we see that $u=df$, where $f$ is symmetric. Thus, by a gauge transformation, we can make sure that $u=0$. Thus, we may assume that $s=2\ell-2p$. In this case [EG2, (2.3)] yields $du=T\otimes T$, i.e., $T\otimes T$ is a coboundary, as claimed. ∎

Proposition 2.15**.**

The $3$-cocycle $\phi$ is a coboundary.

Proof.

By [EG2, Proposition 2.4(2)] on the structure of cohomology, $\pi(T)=0$. Thus by (2.6), $a_{ij}=0$ for all $i,j$, so $\phi$ is a coboundary. ∎

We can now proceed as in the case $R=1$. Namely, by Proposition 2.15, we have $\phi=df$ for some $f\in(\mathcal{O}(\mathcal{G})^{*})^{\otimes 2}$ with the same degree $\ell$ as $\phi$, and by Lemma 2.13, we can choose $f$ to be symmetric. Then letting $\widetilde{f}$ be a symmetric lift of $f$ to $H$, we get the symmetric pseudotwist $F:=1+\widetilde{f}$, and by this pseudotwist we come to the situation where $\Phi-1$ has degree $\geq\ell+1$. Thus $\Delta({\rm d})-{\rm d}\otimes 1-1\otimes{\rm d}$ also has degree $\geq\ell+1$.

However, unlike in the case $R=1$, we are not done yet since the pseudotwist $F$ spoils the $R$-matrix. Namely, since $f$ is symmetric, $R$ has been brought to the form

[TABLE]

Thus, we need the following lemma.

Lemma 2.16**.**

We can twist further to make sure that $R=1+{\rm d}\otimes{\rm d}$ and still $\Phi-1$ has degree $\geq\ell+1$.

Proof.

Let $v:=\pi(f)$. Then $f=v\otimes v+h+h_{21}$ for some $h\in k[\mathcal{G}]^{\otimes 2}$. Thus, by twisting by the pseudotwist $J:=1+[{\rm d}\otimes{\rm d},h]+{\rm d}v\otimes v{\rm d}$, we come to the situation where $\Phi-1$ still has degree $\geq\ell+1$, but

[TABLE]

Now, if $\ell<4p$ then $\ell/2+2p>\ell$, so twisting by $J:=1+{\rm d}\otimes[{\rm d},v]$, we get to a situation when $\Phi-1$ is of degree $\geq\ell+1$ and

[TABLE]

Now Proposition 2.5 implies that using twists $J$ with $J-1$ of degree $\geq\ell+1$ we can come to a situation where $\Phi=1$ modulo degree $\geq\ell+1$ and $R=1+{\rm d}\otimes{\rm d}$ on the nose, providing the desired induction step.

It remains to consider the situation $\ell\geq 4p$. By twisting by $J:=1+{\rm d}\otimes[{\rm d},v]$, we will get to a situation where $\Phi-1={\rm d}\otimes W$ + terms of degree $\geq\ell+1$ and $R=1+{\rm d}\otimes{\rm d}+{\rm terms\,\,of\,\,degree\,\,>2\ell-2p}$, where

[TABLE]

If $\deg(W)>\ell-p$ then we are done with the induction step, so it remains to consider the case $\deg(W)\leq\ell-p$. In this case the hexagon relations [EG2, (2.6)-(2.7)] yield $W=0$. Thus we come to a situation where $\Phi-1$ has degree $\geq\ell+1$ and $R-1-{\rm d}\otimes{\rm d}$ has degree $>2\ell-2p$. So by Proposition 2.5, by applying twists of degree $>\ell$, we can make sure that $R=1+{\rm d}\otimes{\rm d}$ and still $\Phi-1$ has degree $\geq\ell+1$, as desired. ∎

Thus it follows from the above that by continuing this procedure, we will come to a situation where

[TABLE]

for some pseudotwist $F\in H^{\otimes 2}$, as desired. This concludes the proof of Theorem 2.1 in the case where $R=1+{\rm d}\otimes{\rm d}$.

The proofs of Theorems 2.1 and 1.1 are complete. ∎

Remark 2.17**.**

Here is another short proof of the case when $R$ is twist equivalent to $1$, which uses the result of Coulembier. If $R=1$ then the symmetric square of a representation $V$ is the usual one, so for any injection $k\to V$ the induced map $k\to S^{2}V$ is injective. By [C, Theorem C], this implies that the category ${\rm Rep}(H,1,\Phi)$ is locally semisimple. Hence by [C, Proposition 6.2.2], the maximal Tannakian subcategory of ${\rm Rep}(H,1,\Phi)$ is a Serre subcategory. Since the subcategory of ${\rm Rep}(H,1,\Phi)$ generated by simple objects is Tannakian, we see that the whole category ${\rm Rep}(H,1,\Phi)$ is Tannakian, which implies the desired statement.

Remark 2.18**.**

The case when $R\neq 1$ is more subtle, as it is not captured by first order deformation theory. Indeed, the category $\mathcal{D}={\rm Rep}(k[{\rm d}]/({\rm d}^{2}),1+{\rm d}\otimes{\rm d})$ has a nontrivial first order deformation over $k[h]/(h^{2})$, with the same $R$-matrix $R$, but with $\Delta({\rm d})={\rm d}\otimes 1+1\otimes{\rm d}+h{\rm d}\otimes{\rm d}$ and associator $\Phi:=1+h{\rm d}\otimes{\rm d}\otimes{\rm d}$. This deformation is nontrivial because $\phi:={\rm d}\otimes{\rm d}\otimes{\rm d}$ is a nontrivial $3$-cocycle. However, it does not lift to $k[h]/(h^{3})$, as the difference between the left hand side and the right hand side of the pentagon equation [EG2, (2.3)] is $h^{2}{\rm d}^{\otimes 4}$.

The existence of such deformations is typical. For example, consider the category ${\rm Vec}(\mathbb{Z}/p\mathbb{Z})$ in characteristic $p>0$. Clearly, it has no nontrivial formal deformations, since $H^{3}(\mathbb{Z}/p\mathbb{Z},k^{\times})$ is trivial. However, it has a nontrivial first order deformation, since $H^{3}(\mathbb{Z}/p\mathbb{Z},k)=k$. This deformation in fact lifts modulo $h^{i}$ for any $i\leq p$, but does not lift modulo $h^{p+1}$. This is because $\mu_{p}$ and $\alpha_{p}$ are “the same” up to order $p-1$ inclusively, but differ in order $p$.

Corollary 2.19**.**

Let $(H,R)$ be a finite dimensional triangular Hopf algebra with the Chevalley property over $k$. Then $(H,R)$ is twist equivalent to a triangular Hopf algebra with $R$-matrix $1+{\rm d}\otimes{\rm d}$ for some ${\rm d}\in P(H)$ such that ${\rm d}^{2}=0$.

Proof.

Applying Theorem 2.1 to $(H,R,1)$ yields the existence of a pseudotwist $J$ for $H$ such that $(H,R,1)^{J}=(H^{J},1+{\rm d}\otimes{\rm d},1)$. In particular, we have $1^{J}=1$, which is equivalent to $J$ being a twist. ∎

Corollary 2.20**.**

Let $\mathcal{C}$ be a finite symmetric tensor category over $k$ such that $\text{\rm FPdim}(\mathcal{C})=2$. Then $\mathcal{C}$ is symmetric tensor equivalent to either ${\rm Vec}(\mathbb{Z}/2\mathbb{Z})$, ${\rm Rep}(\mathbb{Z}/2\mathbb{Z})$, ${\rm Rep}(\alpha_{2})$ or $\mathcal{D}$.

Proof.

Follows immediately from Theorem 1.1. ∎

2.4. Strengthening of [EG2, Theorem 1.1] and Theorem 1.1

The arguments used in this section and [EG2, Section 3] in fact prove a stronger result. Namely, we have the following theorem.

Theorem 2.21**.**

Let $\mathcal{E}\subset\mathcal{C}$ be finite symmetric tensor categories over an algebraically closed field $k$ with characteristic $p>0$, such that $\mathcal{E}$ contains all the simples of $\mathcal{C}$. The following hold:

1. (1)

Suppose $p>2$. If $\mathcal{E}$ has a fiber functor to ${\rm sVec}$, then so does $\mathcal{C}$. 2. (2)

Suppose $p=2$. If $\mathcal{E}$ has a fiber functor to ${\rm Vec}$, then $\mathcal{C}$ has a fiber functor to $\mathcal{D}$.

Indeed, in both cases it follows that $\mathcal{C}$ is integral, so we have $\mathcal{C}={\rm Rep}(H,R,\Phi)$ for some finite dimensional triangular quasi-Hopf algebra over $k$. Now the arguments are exactly the same, except the radical of $H$ should be replaced by the annihilator of $\mathcal{E}$ inside $\mathcal{C}$, which is a nilpotent quasi-Hopf ideal of $H$ since $\mathcal{E}$ contains all the simples of $\mathcal{C}$.

3. Twists and Sweedler cohomology for finite abelian $p$-groups

In this section we let $K$ be an arbitrary field of characteristic $p>0$, and $\mathbb{F}_{q}$ be a finite field of characteristic $p>0$.

3.1. Truncated Witt vectors

Let $W_{n}(K)$ be the ring of truncated Witt vectors of length $n$ with coefficients in $K$. Recall that $W_{n}(K)=K^{n}$ as a set, with nontrivial addition and multiplication given, e.g., in [L, VI, p.330-332].

Example 3.1**.**

We have the following:

1. (1)

$W_{1}(K)=K$ as rings. 2. (2)

The addition and multiplication in $W_{2}(K)$ are given as follows

[TABLE]

and

[TABLE] 3. (3)

$W_{n}(\mathbb{F}_{p})=\mathbb{Z}/p^{n}\mathbb{Z}$ for every $n\geq 1$.

For $x:=(x_{0},\dots,x_{n-1})\in W_{n}(K)$, let $F(x)=(x_{0}^{p},\dots,x_{n-1}^{p})$. (Note that if $n>1$ then $F(x)\neq x^{p}$.) Recall that $F:W_{n}(K)\to W_{n}(K)$ is a ring homomorphism, and we have an additive homomorphism

[TABLE]

The kernel of $\mathscr{P}$ is the cyclic group $W_{n}(\mathbb{F}_{p})=\mathbb{Z}/p^{n}\mathbb{Z}$.

Lemma 3.2**.**

The following hold:

1. (1)

If $K$ is perfect then $W_{n}(K)/\mathscr{P}(W_{n}(K))$ is a free $\mathbb{Z}/p^{n}\mathbb{Z}$-module. 2. (2)

$W_{n}(\mathbb{F}_{q})/\mathscr{P}(W_{n}(\mathbb{F}_{q}))\cong\mathbb{Z}/p^{n}\mathbb{Z}$.

Proof.

(1) First note that since $K$ is perfect, we have $W_{n}(K)/p^{s}W_{n}(K)\cong W_{s}(K)$ for every $0\leq s\leq n$.

Secondly, let $a\in W_{n}(K)$ be an element such that its image $a_{0}$ in $K$ is not in $\mathscr{P}(K)$. We claim that the order of $a$ in $W_{n}(K)/\mathscr{P}(W_{n}(K))$ is $p^{n}$. Indeed, suppose $s<n$ is such that $p^{s}a=0$ in $W_{n}(K)/\mathscr{P}(W_{n}(K))$, i.e., $p^{s}a=\mathscr{P}(y)$ for some $y\in W_{n}(K)$. Then $\mathscr{P}(y)=0$ in $W_{n}(K)/p^{s}W_{n}(K)=W_{s}(K)$. Thus $y=k\in\mathbb{Z}/p^{s}\mathbb{Z}\subseteq W_{n}(K)/p^{s}W_{n}(K)$ (as ${\rm ker}(\mathscr{P})=\mathbb{Z}/p^{n}\mathbb{Z}$), so $y=k+p^{s}z$ for some integer $k$ and $z\in W_{n}(K)$. But then $p^{s}a=\mathscr{P}(y)=\mathscr{P}(p^{s}z)$, so if $z_{0}$ is the image of $z$ in $K$ then $a_{0}=\mathscr{P}(z_{0})$, which is a contradiction.

Finally, take $a\in W_{n}(K)$ such that $p^{n-1}a=0$ in $W_{n}(K)/\mathscr{P}(W_{n}(K))$, and consider its image $a_{0}$ in $K$. We have shown that $a_{0}$ must be in $\mathscr{P}(K)$, i.e., $a_{0}=x_{0}^{p}-x_{0}$ for some $x_{0}$ in $K$. Let $x:=(x_{0},0,\dots,0)\in W_{n}(K)$. We have $a-\mathscr{P}(x)=py$ for some $y\in W_{n}(K)$ (again using that $K$ is perfect). Thus $a=py$ in $W_{n}(K)/\mathscr{P}(W_{n}(K))$, proving freeness.

(2) Since the kernel of $\mathscr{P}:W_{n}(\mathbb{F}_{q})\to W_{n}(\mathbb{F}_{q})$ is $\mathbb{Z}/p^{n}\mathbb{Z}$, it follows that the cokernel of $\mathscr{P}$ has order $p^{n}$. Thus $W_{n}(\mathbb{F}_{q})/\mathscr{P}(W_{n}(\mathbb{F}_{q}))$ is abelian of order $p^{n}$, so the claim follows from Part (1). ∎

Remark 3.3**.**

If $K$ is not perfect then for instance $W_{2}(K)$ is not a free $\mathbb{Z}/p^{2}\mathbb{Z}$-module. Indeed, take an element $(0,a)$ in $W_{2}(K)$, where $a\in K$ is not a $p$th power. Then $p(0,a)=(0,1)(0,a)=0$, but $(0,a)\neq p(x,y)$ for any $x,y$, since $p(x,y)=(0,x^{p})$.

3.2. Twists for abelian groups and torsors

Recall that an interesting invariant of a tensor category $\mathcal{C}$ over $K$ is the group of tensor structures on the identity functor of $\mathcal{C}$ (i.e., the group of isomorphism classes of tensor autoequivalences of $\mathcal{C}$ which act trivially on the underlying abelian category) up to an isomorphism [Da, BC]. This group is called the second invariant (or lazy) cohomology group of $\mathcal{C}$ and denoted by $H^{2}_{\rm inv}(\mathcal{C},K)$.

In particular, if $\mathcal{C}:={\rm Rep}_{K}(A)$ is the representation category of a finite abelian group $A$ then $H^{2}_{\rm inv}(A,K):=H^{2}_{\rm inv}(\mathcal{C},K)$ is the group of gauge equivalence classes of twists for the Hopf algebra $K[A]$ [EG1].

Lemma 3.4**.**

Let $A$ be a finite abelian group. We have a canonical group isomorphism $H^{2}_{\rm inv}(A,K)\cong{\rm Hom}(G,A)$, where $G:={\rm Aut}(\overline{K}/K)={\rm Gal}(K^{s}/K)$. 333When considering Hom from a profinite group, as usual it means continuous homomorphisms. 444$K^{s}$ is the separable closure of $K$.

Proof.

Let $J$ be a twist for $K[A]$, and consider the twisted $K$-algebra $(K[A]_{J})^{*}$. Observe that (up to $K$-algebra isomorphism) this algebra depends only on $[J]$. Since by [AEGN, Theorem 6.5] every twist for $\overline{K}[A]$ is trivial, it follows that $(K[A]_{J})^{*}\otimes_{K}\overline{K}$ and ${\rm Fun}(A,\overline{K})$ are isomorphic as $\overline{K}$-algebras. Thus, $(K[A]_{J})^{*}$ is a semisimple commutative $K$-algebra. Furthermore, $(K[A]_{J})^{*}$ is an $A$-algebra, which is isomorphic to the regular representation of $A$ as an $A$-module. Thus $(K[A]_{J})^{*}$ is an $A$-torsor.

Conversely, suppose $B$ is an $A$-torsor, i.e., a commutative semisimple $K$-algebra with an $A$-action such that $B\otimes_{K}\overline{K}\cong{\rm Fun}(A,\overline{K})$. By Wedderburn theorem, $B$ decomposes uniquely into a direct sum of field extensions $L_{i}$ of $K$: $B=\bigoplus_{i}L_{i}$. Since the space of $A$-invariants in $B$ is $1$-dimensional, $A$ acts transitively on the set of fields $L_{i}$. Let $H\subseteq A$ be the stabilizer of $L:=L_{1}$. Clearly $L$ is a cyclic extension of $K$ with Galois group $H$. Then it is well known that $L\cong(K[H]^{*})_{J}$ for a unique (up to gauge equivalence) Hopf $2$-cocycle $J$ for $K[H]^{*}$. Viewing $J$ as a twist for $K[H]$ (hence for $K[A]$), it is easy to see that the class $[J]$ is uniquely determined by the isomorphism class of the $A$-torsor $B$.

Finally we note that $A$-torsors form an abelian group under the product rule $(B_{1},B_{2})\mapsto(B_{1}\otimes B_{2})^{A}$, where $a\in A$ acts on $B_{1}$ by $a$ and on $B_{2}$ by $a^{-1}$, and that $(K[A]_{IJ})^{*}\cong((K[A]_{I})^{*}\otimes(K[A]_{J})^{*})^{A}$ (see, e.g., [AEGN, Remark 3.12]).

It now follows from the above that the group $H^{2}_{\rm inv}(A,K)$ is canonically isomorphic to the group of $A$-torsors over $K$. Since the latter is canonically isomorphic to the Galois cohomology group $H^{1}(G,A)=\text{Hom}(G,A)$, the claim follows. ∎

3.3. Invariant cohomology of abelian groups

Let $A$ be a finite abelian group of exponent dividing $p^{n}$. Let $G$ be as in Section 3.2, and let $G_{n}$ be its maximal abelian quotient of exponent dividing $p^{n}$. Then $\text{Hom}(G,A)=\text{Hom}(G_{n},A)$. Thus by Lemma 3.4, we have a canonical group isomorphism

[TABLE]

Theorem 3.5**.**

Let $A$ be a finite abelian group of exponent dividing $p^{n}$. Then the following hold:

1. (1)

We have a canonical group isomorphism

[TABLE]

where $A^{\vee}:={\rm Hom}(A,\mathbb{Z}/p^{n}\mathbb{Z})$. 2. (2)

If moreover $K$ is perfect then we have a canonical group isomorphism

[TABLE]

Proof.

(1) Recall that Artin-Schreier-Witt theory provides a canonical group isomorphism

[TABLE]

(see, e.g., [L, VI, p.330–332]). Thus we get from (3.1) a canonical group isomorphism

[TABLE]

The claim follows now from the fact that ${\rm Hom}(B^{\vee},A)={\rm Hom}(A^{\vee},B)$ for every $B$.

(2) By Lemma 3.2(1), $W_{n}(K)/\mathscr{P}(W_{n}(K))$ is a free $\mathbb{Z}/p^{n}\mathbb{Z}$-module. Therefore the group

[TABLE]

is the same as the group $A\otimes_{\mathbb{Z}/p^{n}\mathbb{Z}}(W_{n}(K)/\mathscr{P}(W_{n}(K)))$, as desired. ∎

Corollary 3.6**.**

We have a group isomorphism

[TABLE]

In particular, we have a group isomorphism

[TABLE]

Proof.

By Theorem 3.5(1), $H^{2}_{\rm{inv}}(\mathbb{Z}/p^{n}\mathbb{Z},\mathbb{F}_{q})\cong W_{n}(\mathbb{F}_{q})/\mathscr{P}(W_{n}(\mathbb{F}_{q}))$, so the second claim follows from Lemma 3.2(2). ∎

Remark 3.7**.**

(1) Theorem 3.5(1) implies that if $K$ is algebraically closed then $H^{2}_{\rm{inv}}(A,K)=0$, which agrees with [EG2, Proposition 5.7] for $i=2$.

(2) Theorem 3.5(1) was obtained by Guillot [G] for $p=2$ and $n=1$.

3.4. Sweedler cohomology of algebras of functions on abelian groups

Let $A$ be a finite abelian group, and let $\mathcal{O}(A)$ be the Hopf algebra of functions on $A$ with values in $K$. Recall that $H^{2}_{\rm{inv}}(A,K)$ coincides with the second Sweedler cohomology group $H^{2}_{\rm{Sw}}(\mathcal{O}(A),K)$ with coefficients in $K$.

Theorem 3.8**.**

Let $A$ be a finite abelian group of exponent dividing $p^{n}$. Then the Sweedler cohomology of $\mathcal{O}(A)$ with coefficients in $K$ is as follows:

1. (1)

$H^{1}_{\rm Sw}(\mathcal{O}(A),K)=A$. 2. (2)

$H^{2}_{\rm Sw}(\mathcal{O}(A),K)={\rm Hom}(A^{\vee},W_{n}(K)/\mathscr{P}(W_{n}(K)))$. 3. (3)

$H^{i}_{\rm Sw}(\mathcal{O}(A),K)=0$* for every $i\geq 3$.*

Proof.

(1) is clear and (2) is Theorem 3.5(1). To prove (3) consider the normalized complex computing $H^{i}_{\rm Sw}(\mathcal{O}(A),K)$:

[TABLE]

where $C^{i}$ is the algebraic group such that for any field $L$, $C^{i}(L)=(L[A]^{\otimes i})^{\times}_{1}$ is the group of invertible elements $a$ in $L[A]^{\otimes i}$ with $\varepsilon(a)=1$. Then $C^{i}$ is a connected commutative unipotent algebraic group over $K$ (i.e., an iterated extension of $\mathbb{G}_{a}$).

Now fix $n\geq 2$. Since by [EG2, Proposition 5.7],

[TABLE]

we have a short exact sequence

[TABLE]

where $D^{i}\subseteq C^{i}$ is the kernel of the differential map $d:C^{i}\to C^{i+1}$. Thus we have an exact sequence

[TABLE]

where

[TABLE]

is the Galois cohomology group. But since $C^{n-1}/D^{n-1}$ is an iterated extension of $\mathbb{G}_{a}$, and $H^{1}(K,\mathbb{G}_{a})=0$, the Galois cohomology group $H^{1}(K,C^{n-1}/D^{n-1})$ vanishes. Thus we have a short exact sequence

[TABLE]

which implies that $H^{n+1}_{\rm Sw}(\mathcal{O}(A),K)=D^{n+1}(K)/d(C^{n}(K))=0$, as claimed. ∎