# Bi-Frobenius algebra structure
on quantum complete intersections

Hai Jin, Pu Zhang∗

School of Mathematical Sciences
Shanghai Jiao Tong University, Shanghai 200240, China

###### Abstract.

This paper is to look for bi-Frobenius algebra structures on quantum complete
intersections. We find a class of comultiplications, such that if
$\sqrt{-1}\in k$, then a quantum complete intersection becomes a bi-Frobenius
algebra with comultiplication of this form if and only if all the parameters
$q_{ij}=\pm 1$. Also, it is proved that if $\sqrt{-1}\in k$ then a quantum
exterior algebra in two variables admits a bi-Frobenius algebra structure if
and only if the parameter $q=\pm 1$. While if $\sqrt{-1}\notin k$, then the
exterior algebra with two variables admits no bi-Frobenius algebra structures.
Since a quantum complete intersection over a field of characteristic zero
admits no bialgebra structures, this gives a class of examples of bi-Frobenius
algebras which are not bialgebras (and hence not Hopf algebras). On the other
hand, a quantum exterior algebra admits a bialgebra structure if and only if
${\rm char}\ k=2$. In commutative case, other two comultiplications on
complete intersection rings are given, such that they admit non-isomorphic bi-
Frobenius algebra structures.

Keywords: Bi-Frobenius algebras, coalgebras, bialgebras, Hopf algebras,
quantum complete intersections, quantum exterior algebras

∗ Corresponding author

jinhaifyh$\char 64\relax$sjtu.edu.cn pzhang$\char 64\relax$sjtu.edu.cn

2020 Mathematics Subject Classification. Primary 16T15, 16T30, 16S80, 81R50

Supported by National Natural Science Foundation of China, Grant No. 12131015,
11971304.

##  1\. Introduction

Frobenius algebras are of particular interest and have wide applications (see
e.g. [Y], [K]), their module categories are Frobenius categories, and the
corresponding stable categories are triangulated ([Hap]). A bi-Frobenius
algebra $A$, introduced by Y. Doi and M. Takeuchi [DT], is both a Frobenius
algebra and a Frobenius coalgebra, together with a linear map
$S:A\longrightarrow A$, called the antipode, which is an algebra anti-
homomorphism and also a coalgebra anti-homomorphism.

Finite-dimensional Hopf algebras are bi-Frobenius algebras, but the converse
is not true. A bi-Frobenius algebra is a Hopf algebra if and only if it is a
bialgebra ([Haim]). The theory of bi-Frobenius algebras is developed in [D1],
[D2], [Haim]. Some results in Hopf algebras have been generalized to bi-
Frobenius algebras, e.g. Radford’s $S^{4}$-formula ([Rad]), see [DT].

One of problems in this theory is to find bi-Frobenius algebras which are not
Hopf algebras. In [DT] a coalgebra structure on $k[X]/\langle X^{n}\rangle$ is
constructed such that it is a bi-Frobenius algebra but not a Hopf algebra.
Group-like algebras are introduced in [D2] such that they become cocommutative
bi-Frobenius algebras. This class of bi-Frobenius algebras are further studied
in [Haim], [D3], and [WLZ]. In particular, using group-like algebras, a class
of bi-Frobenius algebras which are not Hopf algebras are found in [Haim],
where the antipode $S$ is still the inverse of ${\rm Id}$ in the convolution
algebra; also the complexified stable Green algebra is a group-like algebra
([WLZ]). Using quiver approach, some bi-Frobenius algebras which are not Hopf
algebras are constructed in [WZ] and [WC].

Quantum complete intersections (or, quantum linear spaces) originate from Yu.
I. Manin’s quantum planes ([Man]), reach their present form by L. L. Avramov,
V. N. Gasharov and I. V. Peeva ([AGP]), and closely related to braided Hopf
algebras via quantum linear spaces by N. Andruskiewitsch and H.-J. Schneider
([AS1], [AS2]). They reveal many exotic homological and representation
properties of algebras (see e.g., [LS], [Rin], [BGMS], [BE1], [BO1], [BO2],
[BE2], [Mar], [YZ], [RZ]).

Quantum complete intersections are Frobenius algebras. Thus, it is natural to
look for bi-Frobenius algebra structures on them. In this paper, we construct
a class of comultiplications of an arbitrary quantum complete intersection
$A({\bf q},a_{1},\cdots,a_{n})$ over field $k$; and prove that if
$\sqrt{-1}\in k$, then it becomes a bi-Frobenius algebra, with any
comultiplication of this form, if and only if all the parameters $q_{ij}=\pm
1$. See Theorem 3.1. Also, it is proved that if $\sqrt{-1}\in k$, then the
quantum exterior algebra in two variables admits a bi-Frobenius algebra
structure (no restrictions on the form of comultiplication) if and only if
$q=\pm 1$ (see Theorem 5.1). While if $\sqrt{-1}\notin k$, then the exterior
algebra with two variables admits no bi-Frobenius algebra structures (see
Proposition 5.4).

We stress that the comultiplication we used in Theorem 3.1 is different from
the one in Example 1.9 of [AS2]. See Remark 3.4.

Using Kummer’s theorem in number theory, we know that if there is an $a_{i}$
such that it is not a power of char $k$, the characteristic of $k$, then the
quantum complete intersections admit no bialgebra structures (see Theorem
4.1). This gives a large class of examples of bi-Frobenius algebras which are
not bialgebras (and hence not Hopf algebras). On the other hand, a quantum
exterior algebra admits a bialgebra structure if and only if ${\rm char}\ k=2$
(see Corollary 4.3).

In commutative case, we find other two comultiplications on complete
intersection rings, such that they admit non-isomorphic bi-Frobenius algebra
structures. See Theorems 6.1 and 6.3.

##  2\. Preliminary

###  2.1. Frobenius algebras and Frobenius coalgebras

Let $A$ be a finite-dimensional algebra over field $k$. Then the $k$-dual
$A^{*}=\mathrm{Hom}_{k}(A,\ k)$ has a canonical $A$-bimodule structure. Denote
by $a\rightharpoonup f$ the left action of $a\in A$ on $f\in A^{*}$, i.e.,
$(a\rightharpoonup f)(x)=f(xa),\ \forall\ x\in A$. Similarly, denote by
$f\leftharpoonup a$ the right action of $a$ on $f$, i.e., $(f\leftharpoonup
a)(x)=f(ax)$.

Recall that $A$ is a Frobenius algebra provided that there is an isomorphism
$A\cong A^{*}$ of left $A$-modules, or equivalently, there is an isomorphism
$A\cong A^{*}$ of right $A$-modules; and that $A$ is a symmetric algebra,
provided that there is an isomorphism $A\cong A^{*}$ of $A$-bimodules.

By an elementary fact in linear algebra, a Frobenius algebra can also be
characterized as a pair $(A,\ \phi)$, where $\phi\in A^{*}$, such that
$A^{*}=A\rightharpoonup\phi$ (or equivalently, $A^{*}=\phi\leftharpoonup A$).
Such a $\phi$ is called a Frobenius homomorphism of $A$.

Let $(C,\ \Delta,\ \varepsilon)$ be a finite-dimensional coalgebra over $k$.
We use the Heyneman-Sweedler notation $\Delta(c)=\sum c_{1}\otimes c_{2}$.
Then $C^{*}:=\mathrm{Hom}_{k}(C,\ k)$ is an algebra and $C$ has a canonical
$C^{*}$-bimodule structure. Denote by $f\rightharpoonup c$ the left action of
$f\in C^{*}$ on $c\in C$, i.e., $f\rightharpoonup c=\sum c_{1}f(c_{2})$.
Similarly, $c\leftharpoonup f=\sum f(c_{1})c_{2}$.

A finite-dimensional coalgebra $C$ is a Frobenius coalgebra if there is an
element $t\in C$ such that $C=t\leftharpoonup C^{*}$, or equivalently,
$C=C^{*}\rightharpoonup t$. Usually we denote a Frobenius coalgebra by a pair
$(C,\ t)$.

It is clear that $(A,\ \phi)$ is a Frobenius algebra if and only if $(A^{*},\
\phi)$ is a Frobenius coalgebra.

For a coalgebra structure with comultiplication $\Delta$ on an algebra $A$, an
element $x\in A$ is primitive, if $\Delta(x)=1\otimes x+x\otimes 1$, where $1$
is the identity. All the primitive elements form a linear space, denoted by
$P(A)$.

###  2.2. Bi-Frobenius algebras

######  Definition 2.1.

${\rm([DT])}$ Let $A$ be a finite-dimensional algebra and coalgebra over a
field $k$ with $t\in A$ and $\phi\in A^{*}$. Let $S:A\longrightarrow A$ be the
$k$-linear map defined by $S(a)=\sum\phi(t_{1}a)t_{2}.$ The quadruple
$(A,\phi,t,S)$, or simply, $A$ is a bi-Frobenius algebra, if the following
conditions are satisfied$:$

(i) The counit $\varepsilon:A\longrightarrow k$ is an algebra homomorphism,
and the identity $1_{A}$ is a group-like element;

(ii) $(A,\ \phi)$ is a Frobenius algebra, and $(A,\ t)$ is a Frobenius
coalgebra;

(iii) $S$ is an anti-homomorphism of algebra, and $S$ is an anti-homomorphism
of coalgebra $A$ $($i.e., $\varepsilon\circ S=\varepsilon$, and
$\Delta(S(a))=\sum S(a_{2})\otimes S(a_{1}),\ \forall\ a\in A)$.

In this case, $S$ is called the antipode of bi-Frobenius algebra
$(A,\phi,t,S)$.

Let $(A,\ \phi,\ t,\ S)$ and $(A^{\prime},\ \phi^{\prime},\ t^{\prime},\
S^{\prime})$ be bi-Frobenius algebras. A linear map $F:A\longrightarrow
A^{\prime}$ is a homomorphism of bi-Frobenius algebra ([D1]), if $F$ is an
algebra homomorphism and a coalgebra homomorphism, such that $F\circ
S=S^{\prime}\circ F$.

An algebra $A$ is augmented, if there is an algebra homomorphism
$\varepsilon:A\longrightarrow k$. For an augmented algebra $(A,\varepsilon)$,
an element $t\in A$ is a right (respectively, left) integral of $A$ if
$tx=t\varepsilon(x)$ (respectively, $xt=\varepsilon(x)t$), for all $x\in A$.
If the left integral space coincides with the right integral space, then $A$
is called unimodular.

Dually, a coalgebra $C$ is coaugmented, if it has a group-like element $h\in
C$. For a coaugmented coalgebra $(C,\ h)$, an element $\phi\in C^{*}$ is a
right (respectively, left) cointegral of $C$ if $x\leftharpoonup\phi=\phi(x)h$
(respectively, $\phi\rightharpoonup x=h\phi(x)$), for all $x\in C$. If the
left cointegral space coincides with the right cointegral space, then $C$ is
called counimodular.

######  Lemma 2.2.

$($[DT], [Haim]$)$ $(1)$ If $A$ is an augmented Frobenius algebra, then the
space of right integrals is one-dimensional.

$(2)$ If $C$ is a Frobenius coalgebra with a group-like element, then the
space of right cointegrals is one-dimensional.

$(3)$ If $(A,\ \phi,\ t,\ S)$ is a bi-Frobenius algebra, then $t$ is a right
integral and $\phi$ is a right cointegral, and thus the space of right
integral is $kt$, the space of right cointegral is $k\phi$.

$(4)$ If $(A,\ \phi,\ t,S)$ is a bi-Frobenius algebra, then $(A,\ c\phi,\
\frac{1}{c}t,S)$ is also a bi-Frobenius algebra for any nonzero element $c\in
k$.

We need the following observation in [D1].

######  Lemma 2.3.

([D1]) Let A be a finite-dimensional algebra and coalgebra satisfying the
condition (i) in Definition 2.1. If there are $\phi\in A^{*}$ and $t\in A$
such that the $k$-linear map

$S:A\longrightarrow A,\ \ a\mapsto\sum\phi(t_{1}a)t_{2}$

is an algebra anti-homomorphism and a coalgebra anti-homomorphism, then $(A,\
\phi,\ t,\ S)$ is a bi-Frobenius algebra.

###  2.3. Quantum complete intersections

Let $k$ be a field, $a_{1},\cdots,a_{n}$ integers with all $a_{i}\geq 2$ and
$n\geq 2$, and ${\bf q}=(q_{ij})$ an $n\times n$ matrix over $k$ with
$q_{ii}=-1$ and $q_{ij}q_{ji}=1$ for $1\leq i,j\leq n$. A quantum complete
intersection $A({\bf q},a_{1},\cdots,a_{n})$ is the following $k$-algebra

$k\langle x_{1},\cdots,x_{n}\rangle/\langle x_{i}^{a_{i}},\
x_{i}x_{j}+q_{ij}x_{j}x_{i},\ 1\leq i,\ j\leq n\rangle.$

When all $a_{i}=2$, it is called a quantum exterior algebra. It is well-known
that a quantum complete intersection is a local Frobenius algebra and it is
symmetric if and only if $\prod\limits_{1\leq i\leq n}(-q_{ij})^{a_{i}-1}=1$
for all $j$ (see P. A. Bergh [B, Lemma 3.1]).

Denote by $\mathbb{N}_{0}$ the set of non-negative integers. Put

$V=\\{\mathbf{v}=(v_{1},\cdots,v_{n})\in\mathbb{N}_{0}^{n}\ |\ v_{i}\leq
a_{i}-1,\ 1\leq i\leq n\\}.$

For $\mathbf{v}\in V$, write $x_{\mathbf{v}}=x_{1}^{v_{1}}\cdots
x_{n}^{v_{n}}$. Then $\mathcal{B}=\\{x_{\mathbf{v}}\ |\ \mathbf{v}\in V\\}$ is
a basis of $A({\bf q},a_{1},\cdots,a_{n})$. The dual basis of $A^{*}$ is
$\\{x_{\bf v}^{*}\ |\ {\bf v}\in V\\}$, where $x_{\bf v}^{*}(x_{\bf
u})=\delta_{{\bf v},{\bf u}},\ \forall\ {\bf v},{\bf u}\in V$.

The degree of $x_{\mathbf{v}}\in\mathcal{B}$ is defined as
$|x_{\mathbf{v}}|=\sum\limits_{i=1}^{n}v_{i}$, and the degree of
$x_{\mathbf{u}}\otimes x_{\mathbf{v}}\in\mathcal{B}\otimes\mathcal{B}$ is
defined as $|x_{\mathbf{u}}|+|x_{\mathbf{v}}|$.

Put ${\bf a}=(a_{1},\cdots,a_{n}),\ \ {\bf 0}=(0,\cdots,0),\ \ {\bf
1}=(1,\cdots,1)$. For $1\leq i\leq n$, let ${\bf e_{i}}$ be the vector in
$\mathbb{N}_{0}^{n}$ with the $i$-th component $1$ and other components $0$.
Endow a partial order on $\mathbb{N}_{0}^{n}$ by $\mathbf{u}\leq\mathbf{v}$ if
and only if $u_{i}\leq v_{i}$ for all $1\leq i\leq n$. Then
$\mathbf{v}\leq\mathbf{a-1}$ for all $\mathbf{v}\in V.$

For $\mathbf{u,v}\in V$, put ${\bf
q}^{\langle\mathbf{u}|\mathbf{v}\rangle}=\prod\limits_{1\leq i<j\leq
n}(-\frac{1}{q_{ij}})^{u_{j}v_{i}}$. Then

$x_{\mathbf{u}}x_{\mathbf{v}}={\bf
q}^{\langle\mathbf{u}|\mathbf{v}\rangle}x_{\mathbf{u+v}}.$

See S. Oppermann [O, Section 2].

To look for bi-Frobenius algebra structures on $A=A({\bf q},\
a_{1},\cdots,a_{n})$, it is helpful to know all the Frobenius homomorphisms of
$A$. By the proof of Lemma 3.1 in [B], it is known that $x_{\mathbf{a-1}}^{*}$
is a Frobenius homomorphism. Thus, if $\phi\in A^{*}$ is a Frobenius
homomorphism, then $\phi=x_{\mathbf{a-1}}^{*}\leftharpoonup r$, where $r$ is
an invertible element of $A$. See M. Lorenz [L, 1.2.5]. Since $A$ is a finite-
dimensional local algebra,
$r=c^{\prime}_{\mathbf{a-1}}1+\sum\limits_{\mathbf{v}\neq\mathbf{0}}c^{\prime}_{\mathbf{a-1-v}}x_{\mathbf{v}}\in
A$ with each $c^{\prime}_{\mathbf{a-1-v}}\in k$ is invertible if and only if
$c^{\prime}_{\mathbf{a-1}}\neq 0$. It follows that a Frobenius homomorphism
$\phi$ is of the form

$\phi=x_{\mathbf{a-1}}^{*}\leftharpoonup
r=x_{\mathbf{a-1}}^{*}\leftharpoonup(c^{\prime}_{\mathbf{a-1}}1+\sum\limits_{\mathbf{v}\neq\mathbf{0}}c^{\prime}_{\mathbf{a-1-v}}x_{\mathbf{v}})=\sum\limits_{\mathbf{v}\in
V}c_{\mathbf{v}}\ x_{\mathbf{v}}^{*}$

with $c_{\mathbf{a-1}}=c^{\prime}_{\mathbf{a-1}}\neq 0$. This proves (1)
below.

######  Lemma 2.4.

(1) All the Frobenius homomorphisms of $A=A({\bf q},a_{1},\cdots,a_{n})$ are

$\phi=\sum_{\mathbf{v}\in V}c_{\mathbf{v}}x_{\mathbf{v}}^{*}\ \ \mbox{where}\
c_{\mathbf{v}}\in k\ \mbox{and}\ c_{\mathbf{a-1}}\neq 0.$

(2) Put $t=x_{\mathbf{a-1}}$ and $\phi=(x_{{\bf a-1}})^{*}$. Then $(A,\ \phi)$
is a unimodular, augmented Frobenius algebra, with the space of integrals
$kt.$

###### Proof.

(2) By (1), $\phi$ is a Frobenius homomorphism. Let
$\varepsilon:A\longrightarrow k$ be the algebra homomorphism given by
$\varepsilon(x_{\mathbf{v}})=\delta_{\mathbf{v,0}}$. Then $t\ x_{\bf v}=x_{\bf
a-1}\ x_{\bf v}=\delta_{\bf v,0}\ t=t\ \varepsilon(x_{\bf v})$ for ${\bf v}\in
V$. Thus $t$ is a right integral of augmented Frobenius algebra $(A,\ \phi)$.
Similarly, $t$ is a left integral. By Lemma 2.2(1), the space of right
(respectively, left) integrals is one-dimensional, thus $A$ is unimodular, and
the space of integrals $kt.$ ∎

##  3\. Bi-Frobenius algebra structures on quantum complete intersections

We will give a class of comultiplications on an arbitrary quantum complete
intersection

$A=A({\bf q},a_{1},\cdots,a_{n})=k\langle x_{1},\cdots,x_{n}\rangle/\langle
x_{i}^{a_{i}},\ x_{i}x_{j}+q_{ij}x_{j}x_{i},\ 1\leq i,\ j\leq n\rangle$

and then prove that if $\sqrt{-1}\in k$, then $A$ becomes a bi-Frobenius
algebra, with comultiplication of the given form, if and only if $q_{ij}=\pm
1$ for $1\leq i,j\leq n.$

### 3.1.

Put $g_{\mathbf{0,a-1}}=1=g_{\mathbf{a-1,0}},$ and choose arbitrary
$(\prod\limits_{1\leq i\leq n}a_{i})-2$ nonzero elements
$g_{\mathbf{v,a-1-v}}\in k$ for ${\bf v}\in V,\ {\bf v}\neq{\bf 0},\ {\bf
v}\neq{\bf a-1}$. Define $\Delta:A\longrightarrow A\otimes A$ and
$\varepsilon:A\longrightarrow k$ to be $k$-linear maps as

$\left\\{\begin{array}[]{l}\varepsilon(x_{\mathbf{v}})=\delta_{\mathbf{v,0}};\\\
\Delta(1)=1\otimes 1;\\\ \Delta(x_{\mathbf{v}})=1\otimes
x_{\mathbf{v}}+x_{\mathbf{v}}\otimes 1\ \ \text{for all}\ \ {\bf
0}\neq\mathbf{v}\in V\ \text{with}\ \ \mathbf{v}\neq\mathbf{a-1};\\\
\Delta(x_{\mathbf{a-1}})=\sum\limits_{{\bf v}\in V}\ g_{{\bf v},{\bf a-1-v}}\
x_{\mathbf{v}}\otimes x_{\mathbf{a-1-v}}\\\ \hskip 40.0pt=1\otimes
x_{\mathbf{a-1}}+x_{\mathbf{a-1}}\otimes
1+\sum\limits_{\begin{subarray}{c}{\bf 0}\neq{\bf v}\in V\\\ {\bf v}\neq{\bf
a-1}\end{subarray}}\ g_{{\bf v},{\bf a-1-v}}\ x_{\mathbf{v}}\otimes
x_{\mathbf{a-1-v}}.\end{array}\right.$ $None$

Put $t=x_{\mathbf{a-1}}$ and $\phi=(x_{{\bf a-1}})^{*}$. Let
$S:A\longrightarrow A$ be the $k$-linear map given by

$S(a)=\sum\phi(t_{1}a)t_{2}=\sum(x_{\bf a-1})^{*}(t_{1}a)t_{2}\ \ \text{for
all}\ \ a\in A.$

Using $x_{\bf a-1-v}\ x_{\bf v}={\bf q}^{\langle\mathbf{a-1-v|v}\rangle}\
x_{\bf a-1}$ one has

$S(x_{\mathbf{v}})=\sum(x_{\bf a-1})^{*}(t_{1}x_{\mathbf{v}})t_{2}=\
g_{\mathbf{a-1-v,v}}\ {\bf q}^{\langle\mathbf{a-1-v|v}\rangle}\
x_{\mathbf{v}}\ \ \text{for}\ \mathbf{v}\in V.$

Explicitly,

$\left\\{\begin{array}[]{l}S(1)=1,\ \ \
S(x_{\mathbf{a-1}})=x_{\mathbf{a-1}};\\\ \\\ S(x_{\mathbf{v}})=\
g_{\mathbf{a-1-v,v}}\ {\bf q}^{\langle\mathbf{a-1-v|v}\rangle}\
x_{\mathbf{v}},\ \ \text{for}\ {\bf 0}\neq\mathbf{v}\neq{\bf
a-1}.\end{array}\right.$ $None$

The main result of this section is as follows.

######  Theorem 3.1.

If $A=A({\bf q},\ a_{1},\ \cdots,\ a_{n})$ admits a bi-Frobenius algebra
structure, where the comultiplication $\Delta$ and the counit $\varepsilon$
are of the form $(3.1)$, then $q_{ij}^{2}=1$ for $1\leq i,j\leq n.$

Conversely, assume that $\sqrt{-1}\in k$ and $q_{ij}^{2}=1$ for $1\leq i,j\leq
n$. Then $(A,\ (x_{{\bf a-1}})^{*},\ x_{{\bf a-1}},\ S)$ is a bi-Frobenius
algebra, where the comultiplication $\Delta$ and the counit $\varepsilon$ are
of the form $(3.1)$ for some special nonzero elements $g_{\mathbf{v,a-1-v}}\in
k$ with ${\bf v}\in V$ and $g_{\mathbf{0,a-1}}=1=g_{\mathbf{a-1,0}},$ and $S$
is given by $(3.2)$.

######  Remark 3.2.

(1) From the proof of Theorem 3.1, we will see that the condition
$\sqrt{-1}\in k$ can be removed in some special cases. For example, in the
commutative case, i.e., $q_{ij}=-1$ for $1\leq i,j\leq n$.

On the other hand, if $\sqrt{-1}\notin k$, then indeed there is a quantum
complete intersection $A$, such that $A$ admits no bi-Frobenius algebra
structures, for any $\Delta$ and $\varepsilon$, even for $\Delta$ and
$\varepsilon$ given in $(3.3)$. See Proposition 5.4 below.

(2) All the bi-Frobenius algebras involved in Theorem 3.1 has the property of
$S^{4}={\rm Id}$.

### 3.2.

To prove Theorem 3.1 we need some preparations.

######  Lemma 3.3.

Put $t=x_{\mathbf{a-1}}$ and $\phi=(x_{{\bf a-1}})^{*}$. Then $(A,\ t)$ is a
counimodular Frobenius coalgebra with the space of cointegrals $k\phi$, where
the comultiplication $\Delta$ and the counit $\varepsilon$ are given by
$(3.1)$, and $g_{\mathbf{v,a-1-v}}\in k$ are arbitrary nonzero elements with
${\bf v}\in V$ and $g_{\mathbf{0,a-1}}=1=g_{\mathbf{a-1,0}}.$

Moreover, dim $P(A)=(\prod\limits_{1\leq i\leq n}a_{i})-2.$

###### Proof.

One directly verifies that $(A,\ \Delta,\ \varepsilon)$ is a coalgebra, i.e.,
$(\Delta\otimes{\rm Id})\Delta=({\rm Id}\otimes\Delta)\Delta$ and
$(\varepsilon\otimes{\rm Id})\Delta\cong{\rm Id}\cong({\rm
Id}\otimes\varepsilon)\Delta$. For example,

$\displaystyle(\Delta\otimes{\rm Id})\Delta(x_{\bf a-1})=$ $\displaystyle
1\otimes 1\otimes x_{\bf a-1}+1\otimes x_{\bf a-1}\otimes 1+x_{\bf a-1}\otimes
1\otimes 1$ $\displaystyle+\sum_{\begin{subarray}{c}{\bf 0}\neq{\bf v}\in V\\\
{\bf v}\neq{\bf a-1}\end{subarray}}g_{{\bf v},{\bf
a-1-v}}(x_{\mathbf{v}}\otimes x_{\mathbf{a-1-v}}\otimes 1+1\otimes
x_{\mathbf{v}}\otimes x_{\mathbf{a-1-v}}+x_{\mathbf{v}}\otimes 1\otimes
x_{\mathbf{a-1-v}})$ $\displaystyle=$ $\displaystyle({\rm
Id}\otimes\Delta)\Delta(x_{\bf a-1}).$

For $f\in A^{*}$, since $t\leftharpoonup f=\sum f(t_{1})t_{2}$, one has

$\displaystyle t\leftharpoonup 1^{*}=x_{\bf a-1};$ $\displaystyle
t\leftharpoonup x_{\bf a-1}^{*}=1;$ $\displaystyle t\leftharpoonup x_{\bf
v}^{*}=g_{\bf v,\ a-1-v}\ x_{\bf a-1-v}\ \ \text{with}\ {\bf
0}\neq\mathbf{v}\neq\mathbf{a-1}.$

Since $g_{\bf v,\ a-1-v}\neq 0$, $A=t\leftharpoonup A^{*}$, i.e., $(A,\ t)$ is
a Frobenius coalgebra.

By $1\leftharpoonup\phi=0=\phi(1)1,$ $x_{\bf
v}\leftharpoonup\phi=0=\phi(x_{\bf v})1$ for ${\bf 0}\neq{\bf v}\in V$ with
${\bf v}\neq{\bf a-1}$, and $x_{\bf a-1}\leftharpoonup\phi=1=\phi(x_{\bf
a-1})1$, we see that $\phi$ is a right cointegral. Similarly, $\phi$ is a left
cointegral. By Lemma 2.2(2), the space of right (respectively, left)
cointegrals is one-dimensional, thus $A$ is counimodular, and the space of
cointegrals $k\phi.$

Moreover, by (3.1), all $x_{\bf v}$ with ${\bf v}\in V$ are primitive elements
of $A$, except ${\bf v}={\bf 0}$ or ${\bf v}={\bf a-1}$. Then, it is easy to
see dim $P(A)=(\prod\limits_{1\leq i\leq n}a_{i})-2.$ ∎

######  Remark 3.4.

The comultiplication $\Delta$ in (3.1) is different from the one given in
Example 1.9 by Andruskiewitsch and Schneider [AS2]: by Lemma 3.3, the
dimension of the space of primitive elements is $(\prod\limits_{1\leq i\leq
n}a_{i})-2;$ while the dimension of the space of primitive elements of the
algebra in Example 1.9 of [AS2] is just $n$.

### 3.3.

With $\Delta$ and $\varepsilon$ given in $(3.1)$ and $S$ given in $(3.2)$, we
look for a sufficient and necessary condition such that $S$ is an algebra
anti-homomorphism.

######  Lemma 3.5.

Let $\Delta$ and $\varepsilon$ be given by $(3.1)$, and $S:A\longrightarrow A$
the $k$-linear map given by $(3.2)$. Then $S$ is an algebra anti-homomorphism
of $A$ if and only if the following conditions are satisfied$:$

$q_{i,j}^{2}=1,\ \ \forall\ 1\leq i,j\leq n$ $None$
$g_{\mathbf{a-1-(u+v),u+v}}=g_{\mathbf{a-1-u,u}}\ g_{\mathbf{a-1-v,v}},\ \
\forall\ {\bf u},\ {\bf v}\in V,\ \ {\bf u}+{\bf v}\leq{\bf a-1}.$ $None$

###### Proof.

Since the actions of $S$ on elements in the basis $\mathcal{B}=\\{x_{\bf v}\
|\ {\bf v}\in V\\}$ are just scalar multiplications, to show $S$ is an anti-
homomorphism of algebra $A$, it is enough to show $S(x_{\bf u}x_{\bf
v})=S(x_{\bf v})S(x_{\bf u})$ for ${\bf u},{\bf v}\in V,\ {\bf u}+{\bf
v}\leq{\bf a-1}$. By (3.2) one has

$\displaystyle S(x_{\mathbf{u}}x_{\mathbf{v}})$ $\displaystyle=S({\bf
q}^{\langle\mathbf{u|v}\rangle}x_{\bf u+v})={\bf
q}^{\langle\mathbf{u|v}\rangle}S(x_{\bf u+v})$ $\displaystyle={\bf
q}^{\langle\mathbf{u|v}\rangle}{\bf
q}^{\langle\mathbf{a-1-(u+v)|u+v}\rangle}g_{\mathbf{a-1-(u+v),u+v}}\
x_{\mathbf{u+v}}$ $\displaystyle S(x_{\mathbf{u}})$ $\displaystyle={\bf
q}^{\langle\mathbf{a-1-u|u}\rangle}g_{\mathbf{a-1-u,u}}\ x_{\mathbf{u}}$
$\displaystyle S(x_{\mathbf{v}})$ $\displaystyle={\bf
q}^{\langle\mathbf{a-1-v|v}\rangle}g_{\mathbf{a-1-v,v}}\ x_{\mathbf{v}}$
$\displaystyle S(x_{\mathbf{v}})S(x_{\mathbf{u}})$ $\displaystyle={\bf
q}^{\langle\mathbf{a-1-u|u}\rangle}{\bf q}^{\langle\mathbf{a-1-v|v}\rangle}\
g_{\mathbf{a-1-u,u}}\ g_{\mathbf{a-1-v,v}}\ x_{\mathbf{v}}x_{\mathbf{u}}$
$\displaystyle={\bf q}^{\langle\mathbf{v|u}\rangle}{\bf
q}^{\langle\mathbf{a-1-u|u}\rangle}{\bf q}^{\langle\mathbf{a-1-v|v}\rangle}\
g_{\mathbf{a-1-u,u}}\ g_{\mathbf{a-1-v,v}}\ x_{\mathbf{u+v}}.$

Since

$\displaystyle{\bf q}^{\langle\mathbf{u|v}\rangle}{\bf
q}^{\langle\mathbf{a-1-(u+v)|u+v}\rangle}$ $\displaystyle=\prod_{1\leq i\leq
j\leq n}(-\frac{1}{q_{ij}})^{u_{j}v_{i}+(a_{j}-1-u_{j}-v_{j})(u_{i}+v_{i})}$
$\displaystyle=\prod_{1\leq i\leq j\leq
n}(-\frac{1}{q_{ij}})^{(a_{j}-1-u_{j})u_{i}+(a_{j}-1-v_{j})v_{i}-u_{i}v_{j}}$
$\displaystyle={\bf q}^{\langle\mathbf{a-1-u|u}\rangle}{\bf
q}^{\langle\mathbf{a-1-v|v}\rangle}{\bf q}^{-\langle\mathbf{v|u}\rangle}$

where ${\bf q}^{-\langle\mathbf{v|u}\rangle}=\frac{1}{{\bf
q}^{\langle\mathbf{v|u}\rangle}}$, it follows that $S(x_{\bf u}x_{\bf
v})=S(x_{\bf v})S(x_{\bf u})$ for $\forall\ {\bf u},\ {\bf v}\in V,\ \ {\bf
u}+{\bf v}\leq{\bf a-1}$ if and only if

$g_{\mathbf{a-1-(u+v),u+v}}=({\bf q}^{\langle\mathbf{v|u}\rangle})^{2}\
g_{\mathbf{a-1-u,u}}\ g_{\mathbf{a-1-v,v}},\ \ \forall\ {\bf u},\ {\bf v}\in
V,\ \ {\bf u}+{\bf v}\leq{\bf a-1}.$ $None$

Assume that (3.5) holds. Exchanging the position of ${\bf u}$ and ${\bf v}$,
one gets

$(q^{\langle\mathbf{u|v}\rangle})^{2}=(q^{\langle\mathbf{v|u}\rangle})^{2},\ \
\ \ \mbox{i.e.,}\ \ \ \ \prod_{1\leq i\leq j\leq
n}q_{ij}^{-2u_{j}v_{i}}=\prod_{1\leq i\leq j\leq n}q_{ij}^{-2v_{j}u_{i}}.$

Taking $\mathbf{v=e_{i},\ u=e_{j}}$ with $1\leq i<j\leq n$, one gets
$q_{i,j}^{2}=1$ for all $1\leq i,j\leq n$. Then $({\bf
q}^{\langle\mathbf{u|v}\rangle})^{2}=\prod\limits_{1\leq i\leq j\leq
n}q_{ij}^{-2u_{j}v_{i}}=1$, and by (3.6) one gets

$g_{\mathbf{a-1-(u+v),u+v}}=g_{\mathbf{a-1-u,u}}\ g_{\mathbf{a-1-v,v}},\ \
\forall\ {\bf u},\ {\bf v}\in V,\ \ {\bf u}+{\bf v}\leq{\bf a-1}.$

That is, the conditions (3.3) and (3.4) are satisfied.

Conversely, assume that the conditions (3.3) and (3.4) are satisfied. Since
$q_{i,j}^{2}=1$ for all $1\leq i,j\leq n$, $({\bf
q}^{\langle\mathbf{v|u}\rangle})^{2}=\prod\limits_{1\leq i\leq j\leq
n}q_{ij}^{-2u_{i}v_{j}}=1$, and then (3.5) holds, i.e., $S$ is an anti-
homomorphism of algebra.

In conclusion, $S$ is an anti-homomorphism of algebra if and only if the
conditions (3.3) and (3.4) are satisfied. ∎

### 3.4.

Next, we give a necessary and sufficient condition such that $S$ is an anti-
homomorphism of coalgebra.

######  Lemma 3.6.

Let $\Delta$ and $\varepsilon$ be given by $(3.1)$, and $S:A\longrightarrow A$
the $k$-linear map given by $(3.2)$. Then $S$ is an anti-homomorphism of
coalgebra $(A,\ \Delta,\ \varepsilon)$ if and only if

$g^{2}_{\mathbf{v,a-1-v}}={\bf q}^{-\langle\mathbf{v|a-1-v}\rangle}\ {\bf
q}^{-\langle\mathbf{a-1-v|v}\rangle},\ \ \forall\ {\bf v}\in V.$ $None$

###### Proof.

By (3.1) and (3.2) one has

$\displaystyle\varepsilon(S(x_{\mathbf{v}}))$
$\displaystyle=\delta_{\mathbf{v,0}}=\varepsilon(x_{\mathbf{v}}),\ \ \forall\
\mathbf{v}\in V;$ $\displaystyle\Delta(S(1))$ $\displaystyle=1\otimes
1=S\otimes S\circ\tau\circ(\Delta(1)),$

where $\tau$ is the twist map. For any ${\bf 0}\neq\mathbf{v\neq a-1}$, one
has

$\displaystyle\Delta(S(x_{\mathbf{v}}))$ $\displaystyle=g_{\mathbf{a-1-v,v}}\
{\bf q}^{\langle\mathbf{a-1-v|v}\rangle}\ \Delta(x_{\mathbf{v}})$
$\displaystyle=g_{\mathbf{a-1-v,v}}\
q^{\langle\mathbf{a-1-v|v}\rangle}(1\otimes
x_{\mathbf{v}}+x_{\mathbf{v}}\otimes 1)$ $\displaystyle=S\otimes
S\circ\tau\circ(\Delta(x_{\mathbf{v}})).$

For ${\bf v}=\mathbf{a-1}$, one has

$\Delta(S(x_{\mathbf{a-1}}))=\Delta(x_{\mathbf{a-1}})=1\otimes
x_{\mathbf{a-1}}+x_{\mathbf{a-1}}\otimes
1+\sum\limits_{\begin{subarray}{c}{\bf 0\neq v}\in V\\\ {\bf v}\neq{\bf
a-1}\end{subarray}}\ g_{{\bf v},{\bf a-1-v}}\ x_{\mathbf{v}}\otimes
x_{\mathbf{a-1-v}};$

while

$\displaystyle S\otimes S\circ\tau\circ(\Delta(x_{\mathbf{a-1}}))$
$\displaystyle=1\otimes x_{\mathbf{a-1}}+x_{\mathbf{a-1}}\otimes
1+\sum\limits_{\begin{subarray}{c}{\bf 0}\neq{\bf v}\in V\\\ {\bf v}\neq{\bf
a-1}\end{subarray}}\ g_{\mathbf{v,a-1-v}}\ S(x_{\mathbf{a-1-v}})\otimes
S(x_{\mathbf{v}})$ $\displaystyle=1\otimes
x_{\mathbf{a-1}}+x_{\mathbf{a-1}}\otimes 1$
$\displaystyle+\sum\limits_{\begin{subarray}{c}{\bf 0}\neq{\bf v}\in V\\\ {\bf
v}\neq{\bf a-1}\end{subarray}}\ g_{\mathbf{a-1-v,v}}\
g_{\mathbf{v,a-1-v}}^{2}\ {\bf q}^{\langle\mathbf{a-1-v|v}\rangle}\ {\bf
q}^{\langle\mathbf{v|a-1-v}\rangle}\ x_{\mathbf{a-1-v}}\otimes
x_{\mathbf{v}}.$

Comparing the coefficient of $x_{\mathbf{v}}\otimes x_{\mathbf{a-1-v}}$ with
${\bf 0}\neq\mathbf{v}\neq{\bf a-1}$, one gets

$g_{\mathbf{v,a-1-v}}=g_{\mathbf{v,a-1-v}}\ g^{2}_{\mathbf{a-1-v,v}}\ {\bf
q}^{\langle\mathbf{v|a-1-v}\rangle}\ {\bf
q}^{\langle\mathbf{a-1-v|v}\rangle},\ \ \forall\ \mathbf{v}\in V,\ {\bf
0}\neq\mathbf{v}\neq{\bf a-1}.$

That is

$g^{2}_{\mathbf{a-1-v,v}}={\bf q}^{-\langle\mathbf{a-1-v|v}\rangle}{\bf
q}^{-\langle\mathbf{v|a-1-v}\rangle},\ \ \forall\ \mathbf{v}\in V,\ {\bf
0}\neq\mathbf{v}\neq{\bf a-1}$

where $q^{-\langle\mathbf{u|v}\rangle}=\frac{1}{{\bf
q}^{\langle\mathbf{{u|v\rangle}}}}$, or equivalently,

$g^{2}_{\mathbf{v,a-1-v}}={\bf q}^{-\langle\mathbf{v|a-1-v}\rangle}\ {\bf
q}^{-\langle\mathbf{a-1-v|v}\rangle},\ \ \forall\ \mathbf{v}\in V,\ {\bf
0}\neq\mathbf{v}\neq{\bf a-1}.$

Since this equality also holds for $\mathbf{v}={\bf 0}$ and for
$\mathbf{v}={\bf a-1}$, $S$ is an anti-homomorphism of coalgebra
$(A,\Delta,\varepsilon)$ if and only if (3.6) holds. ∎

### 3.5.

Now, we prove the existence of $\prod\limits_{1\leq i\leq n}a_{i}$ nonzero
elements $g_{\mathbf{v,a-1-v}}\in k$ with ${\bf v}\in V$ and
$g_{\mathbf{0,a-1}}=1=g_{\mathbf{a-1,0}}$, satisfying $(3.4)$ and $(3.6)$,
under some conditions.

######  Lemma 3.7.

Assume that $\sqrt{-1}\in k$ and $q_{ij}^{2}=1$ for $1\leq i,j\leq n$. Then
there are indeed $\prod\limits_{1\leq i\leq n}a_{i}$ nonzero elements $g_{{\bf
v},{\bf a-1-v}}$ in $k$ with ${\bf v}\in V$ and
$g_{\mathbf{0,a-1}}=1=g_{\mathbf{a-1,0}}$, such that $(3.4)$ and $(3.6)$ are
satisfied, namely

$g_{\mathbf{a-1-(u+v),u+v}}=g_{\mathbf{a-1-u,u}}\ g_{\mathbf{a-1-v,v}},\ \
\forall\ {\bf u},\ {\bf v}\in V,\ {\bf u}+{\bf v}\leq{\bf a-1},$ $None$

$g^{2}_{\mathbf{v,a-1-v}}={\bf q}^{-\langle\mathbf{a-1-v|v}\rangle}{\bf
q}^{-\langle\mathbf{v|a-1-v}\rangle},\ \ \forall\ {\bf v}\in V.$ $None$

In particular, $g^{2}_{\mathbf{v,a-1-v}}=\pm 1,\ \forall\ {\bf v}\in V.$

###### Proof.

Since $\sqrt{-1}\in k$ and $\prod\limits_{1\leq j\leq
n}(-q_{ij})^{a_{j}-1}=\pm 1$, thus one can choose

$g_{\bf a-1-e_{i},e_{i}}=h_{i}\prod_{1\leq j\leq
n}(-q_{ij})^{\frac{a_{j}-1}{2}}\in k,\ \ \text{where}\ h_{i}\in\\{1,-1\\},\ \
1\leq i\leq n.$

(If ${\bf q=-1}$, i.e., $q_{ij}=-1$ for $1\leq i,j\leq n$, then
$\prod\limits_{1\leq j\leq n}(-q_{ij})^{a_{j}-1}=1$ and $g_{\bf
a-1-e_{i},e_{i}}=h_{i}\prod\limits_{1\leq j\leq
n}(-q_{ij})^{\frac{a_{j}-1}{2}}$ is always in $k$. Thus, in this case the
assumption $\sqrt{-1}\in k$ can be removed.)

Then, one takes $g_{\mathbf{a-1-v,v}}$ with ${\bf 0\neq v}\in V,\ {\bf v\neq
a-1}$ as

$g_{\mathbf{a-1-v,v}}=\prod_{1\leq i\leq
n}(g_{\mathbf{a-1-e_{i},e_{i}}})^{v_{i}}.$

Since $g_{\mathbf{a-1,0}}=1$, it is clear that this also holds for ${\bf v=0}$
(note that in this case all $v_{i}=0$).

Claim: One can choose $h_{i}\in\\{1,-1\\}$ for $1\leq i\leq n$, such that
$g_{\mathbf{a-1-v,v}}=\prod\limits_{1\leq i\leq
n}(g_{\mathbf{a-1-e_{i},e_{i}}})^{v_{i}}$ is also true for ${\bf v=a-1}$,
i.e.,

$\prod\limits_{1\leq i\leq n}(g_{\mathbf{a-1-e_{i},e_{i}}})^{a_{i}-1}=1.$

In fact, notice that $q_{ij}^{2}=1$ and $q_{ij}q_{ji}=1$, hence
$q_{ij}=q_{ji}.$ Thus one has

$g_{\bf a-1-e_{i},e_{i}}^{a_{i}-1}=h_{i}^{a_{i}-1}\prod_{1\leq j\leq
n}(-q_{ij})^{\frac{(a_{i}-1)(a_{j}-1)}{2}},\ \ 1\leq i\leq n.$

Also note that $q_{ii}=-1$ for $1\leq i\leq n$. One has

$\displaystyle\prod_{1\leq i\leq n}g_{\bf a-1-e_{i},e_{i}}^{a_{i}-1}$
$\displaystyle=\prod_{1\leq i\leq n}(h_{i}^{a_{i}-1}\prod_{1\leq j\leq
n}(-q_{ij})^{\frac{(a_{i}-1)(a_{j}-1)}{2}})$ $\displaystyle=(\prod_{1\leq
i\leq n}h_{i}^{a_{i}-1})(\prod_{1\leq i\leq n}\prod_{1\leq j\leq
n}(-q_{ij})^{\frac{(a_{i}-1)(a_{j}-1)}{2}})$ $\displaystyle=(\prod_{1\leq
i\leq n}h_{i}^{a_{i}-1})(\prod_{1\leq i,j\leq
n}(-q_{ij})^{\frac{(a_{i}-1)(a_{j}-1)}{2}})$ $\displaystyle=(\prod_{1\leq
i\leq n}h_{i}^{a_{i}-1})(\prod_{1\leq i<j\leq
n}(-q_{ij})^{\frac{(a_{i}-1)(a_{j}-1)}{2}})(\prod_{1\leq j<i\leq
n}(-q_{ij})^{\frac{(a_{i}-1)(a_{j}-1)}{2}})$ $\displaystyle=(\prod_{1\leq
i\leq n}h_{i}^{a_{i}-1})(\prod_{1\leq i<j\leq
n}(-q_{ij})^{\frac{(a_{i}-1)(a_{j}-1)}{2}})(\prod_{1\leq i<j\leq
n}(-q_{ji})^{\frac{(a_{i}-1)(a_{j}-1)}{2}})$ $\displaystyle=(\prod_{1\leq
i\leq n}h_{i}^{a_{i}-1})(\prod_{1\leq i<j\leq
n}(-q_{ij})^{(a_{i}-1)(a_{j}-1)}).$

Since $q_{ij}=\pm 1$, it follows that $\prod\limits_{1\leq i<j\leq
n}(-q_{ij})^{(a_{i}-1)(a_{j}-1)}=\pm 1$. Choose $h_{i}\in\\{1,-1\\}$ for
$1\leq i\leq n$ as follows.

If $\prod\limits_{1\leq i<j\leq n}(-q_{ij})^{(a_{i}-1)(a_{j}-1)}=1$, then we
choose $h_{i}=1$ for all $1\leq i\leq n$.

If $\prod\limits_{1\leq i<j\leq n}(-q_{ij})^{(a_{i}-1)(a_{j}-1)}=-1$, then
there is $i_{0}$ with $1\leq i_{0}\leq n$ such that $a_{i_{0}}$ is even, then
we can choose $h_{i_{0}}=-1$ and $h_{i}=1$ for all $i\neq i_{0}$.

In the both cases, one has $\prod\limits_{1\leq i\leq
n}(g_{\mathbf{a-1-e_{i},e_{i}}})^{a_{i}-1}=1.$ This proves Claim.

Up to now, we have already fixed $\prod\limits_{1\leq i\leq n}a_{i}$ nonzero
elements $g_{\mathbf{v,a-1-v}}\in k$ in the way

$g_{\mathbf{a-1-v,v}}=\prod_{1\leq i\leq
n}(g_{\mathbf{a-1-e_{i},e_{i}}})^{v_{i}},\ \ \forall\ {\bf v}\in V.$ $None$

It remains to show that these nonzero elements $g_{\mathbf{v,a-1-v}}\in k$
with ${\bf v}\in V$ satisfy $(3.4)$ and $(3.6)$. Indeed, for any $\ {\bf u},\
{\bf v}\in V,\ {\bf u}+{\bf v}\leq{\bf a-1},$ by (3.7) one has

$\displaystyle g_{\mathbf{a-1-(u+v),u+v}}$ $\displaystyle=\prod_{1\leq i\leq
n}g_{\mathbf{a-1-e_{i},e_{i}}}^{u_{i}+v_{i}}$ $\displaystyle=\prod_{1\leq
i\leq n}g_{\mathbf{a-1-e_{i},e_{i}}}^{u_{i}}\prod_{1\leq i\leq
n}g_{\mathbf{a-1-e_{i},e_{i}}}^{v_{i}}$ $\displaystyle=g_{\mathbf{a-1-u,u}}\
g_{\mathbf{a-1-v,v}}$

i.e., (3.4) is satisfied. Also, for ${\bf v}\in V$ by (3.7) one has

$\displaystyle g_{\mathbf{v,a-1-v}}^{2}$ $\displaystyle=\prod_{1\leq i\leq
n}(g_{\mathbf{a-1-e_{i},e_{i}}}^{2})^{a_{i}-1-v_{i}}$
$\displaystyle=\prod_{1\leq i\leq n}(\prod_{1\leq j\leq
n}(-q_{ij})^{a_{j}-1})^{a_{i}-1-v_{i}}$ $\displaystyle=\prod_{1\leq i,j\leq
n}(-q_{ij})^{(a_{j}-1)(a_{i}-1-v_{i})}.$

On the other hand, $q_{ij}=q_{ji}$ for $1\leq i,j\leq n$. By $q_{ii}=-1$ for
$1\leq i\leq n$, one has

$\displaystyle{\bf q}^{-\langle\mathbf{a-1-v|v}\rangle}{\bf
q}^{-\langle\mathbf{v|a-1-v}\rangle}$ $\displaystyle=\prod_{1\leq i<j\leq
n}(-q_{ij})^{(a_{j}-1-v_{j})v_{i}}\prod_{1\leq i<j\leq
n}(-q_{ij})^{(a_{i}-1-v_{i})v_{j}}$ $\displaystyle=\prod_{1\leq i<j\leq
n}(-q_{ij})^{(a_{j}-1-v_{j})v_{i}}\prod_{1\leq j<i\leq
n}(-q_{ji})^{(a_{j}-1-v_{j})v_{i}}$ $\displaystyle=\prod_{1\leq i<j\leq
n}(-q_{ij})^{(a_{j}-1-v_{j})v_{i}}\prod_{1\leq j<i\leq
n}(-q_{ij})^{(a_{j}-1-v_{j})v_{i}}$ $\displaystyle=\prod_{1\leq i,j\leq
n}(-q_{ij})^{(a_{j}-1-v_{j})v_{i}}.$

To compare

$g_{\mathbf{v,a-1-v}}^{2}=\prod_{1\leq i,j\leq
n}(-q_{ij})^{(a_{j}-1)(a_{i}-1-v_{i})}$

with

${\bf q}^{-\langle\mathbf{a-1-v|v}\rangle}{\bf
q}^{-\langle\mathbf{v|a-1-v}\rangle}=\prod_{1\leq i,j\leq
n}(-q_{ij})^{(a_{j}-1-v_{j})v_{i}}$

one has

$\displaystyle\prod_{1\leq i,j\leq
n}(-q_{ij})^{(a_{j}-1)(a_{i}-1-v_{i})-(a_{j}-1-v_{j})v_{i}}$ $\displaystyle=$
$\displaystyle\prod_{1\leq i,j\leq
n}(-q_{ij})^{v_{i}v_{j}+(a_{i}-1)(a_{j}-1)-2v_{i}(a_{j}-1)}$ $\displaystyle=$
$\displaystyle\prod_{1\leq i,j\leq
n}(-q_{ij})^{v_{i}v_{j}+(a_{i}-1)(a_{j}-1)}$ $\displaystyle=$
$\displaystyle\prod_{1\leq i<j\leq
n}(-q_{ij})^{v_{i}v_{j}+(a_{i}-1)(a_{j}-1)}+\prod_{1\leq j<i\leq
n}(-q_{ij})^{v_{i}v_{j}+(a_{i}-1)(a_{j}-1)}$ $\displaystyle=$
$\displaystyle\prod_{1\leq i<j\leq
n}(-q_{ij})^{v_{i}v_{j}+(a_{i}-1)(a_{j}-1)}+\prod_{1\leq i<j\leq
n}(-q_{ji})^{v_{i}v_{j}+(a_{i}-1)(a_{j}-1)}$ $\displaystyle=$
$\displaystyle\prod_{1\leq i<j\leq
n}(-q_{ij})^{2v_{i}v_{j}+2(a_{i}-1)(a_{j}-1)}=1.$

This proves $g^{2}_{\mathbf{v,a-1-v}}={\bf
q}^{-\langle\mathbf{a-1-v|v}\rangle}{\bf
q}^{-\langle\mathbf{v|a-1-v}\rangle},\ \ \forall\ {\bf v}\in V,$ i.e., (3.6)
is satisfied. This completes the proof. ∎

### 3.6.

We are now in position to prove the main result.

Proof of Theorem 3.1. Assume that $A$ admits a bi-Frobenius algebra structure
$(A,\ \phi,\ t,\ S)$, where the comultiplication $\Delta$ and the counit
$\varepsilon$ are of the form $(3.1)$. By Lemma 2.2(3), $t$ is a right
integral of $A$. On the other hand, by Lemma 2.4(2), the space of right
integrals of $A$ is $k\ x_{{\bf a-1}}$. It follows that $t=c\ x_{{\bf a-1}}$
for some $0\neq c\in k$.

By Lemma 2.2(3), $\phi$ is a right cointegral of $A$. On the other hand, by
Lemma 3.3, the space of right cointegrals of $A$ is $k\ (x_{{\bf a-1}})^{*}$.
It follows that $\phi=d\ (x_{{\bf a-1}})^{*}$ for some $0\neq d\in k$.

By

$\Delta(t)=\Delta(cx_{{\bf a-1}})=c(1_{A}\otimes x_{{\bf a-1}}+x_{{\bf
a-1}}\otimes 1_{A}+\sum\limits_{\begin{subarray}{c}{\bf 0}\neq{\bf v}\in V\\\
{\bf v}\neq{\bf a-1}\end{subarray}}\ g_{{\bf v},{\bf a-1-v}}\
x_{\mathbf{v}}\otimes x_{\mathbf{a-1-v}})$

and

$1_{A}=S(1_{A})=\sum\phi(t_{1})t_{2}=\sum dx_{{\bf
a-1}}^{*}(t_{1})t_{2}=cd1_{A}$

one gets $d=\frac{1}{c}$. Thus, $(A,\ \frac{1}{c}(x_{{\bf a-1}})^{*},\
cx_{{\bf a-1}},\ S)$ is a bi-Frobenius algebra. By Lemma 2.2(4), $(A,\
(x_{{\bf a-1}})^{*},\ x_{{\bf a-1}},\ S)$ is also a bi-Frobenius algebra. In
particular, $S$ is an algebra anti-homomorphism, and hence by Lemma 3.5,
$q_{ij}^{2}=1$ for $1\leq i,j\leq n$.

Conversely, assume that $\sqrt{-1}\in k$ and $q_{ij}^{2}=1$ for $1\leq i,j\leq
n$. By Lemma 2.4(2), $(A,\ (x_{{\bf a-1}})^{*})$ is a Frobenius algebra.

Choose nonzero elements $g_{\mathbf{v,a-1-v}}\in k$ with ${\bf v}\in V$ and
$g_{\mathbf{0,a-1}}=1=g_{\mathbf{a-1,0}}$, such that $(3.4)$ and $(3.6)$ are
satisfied. Lemma 3.7 guarantees the existence of these
$g_{\mathbf{v,a-1-v}}$’s.

Using these chosen $g_{\mathbf{v,a-1-v}}$’s, consider the $k$-linear maps
$\Delta$ and $\varepsilon$, as defined in $(3.1)$. Then $\varepsilon$ is an
algebra homomorphism. By Lemma 3.3, $(A,\ x_{{\bf a-1}})$ is a Frobenius
coalgebra. Clearly, the identity $1$ is a group-like element.

By Lemmas 3.5 and 3.6, the $k$-linear map $S:A\longrightarrow A$ given by
$(3.2)$ is an algebra anti-homomorphism of $A$ and a coalgebra anti-
homomorphism of $(A,\ \Delta,\ \varepsilon)$. By definition, $(A,\ x_{{\bf
a-1}}^{*}$, $x_{{\bf a-1}},\ S)$ is a bi-Frobenius algebra.

This completes the proof. $\square$

By ${\bf q=-1}$ we mean $q_{ij}=-1$ for $1\leq i,j\leq n$. From the proof of
Lemma 3.7 we see

######  Corollary 3.8.

Complete intersection rings $A=A({\bf-1},\ a_{1},\ \cdots,\ a_{n})$ admits a
bi-Frobenius algebra structure $(A,\ (x_{{\bf a-1}})^{*},\ x_{{\bf a-1}},\
S)$, where the comultiplication $\Delta$ and the counit $\varepsilon$ are of
the form $(3.1)$.

##  4\. Quantum complete intersections and bialgebras

### 4.1.

The following result shows that in many cases quantum complete intersections
admits no bialgebra structures.

######  Theorem 4.1.

If there is an $a_{i}$ such that it is not a power of char $k$, then quantum
complete intersection $A({\bf q},a_{1},\cdots,a_{n})$ admits no bialgebra
structures.

In particular, if ${\rm char}\ k=0$, then any quantum complete intersection
admits no bialgebra structures.

To prove Theorem 4.1, we need a consequence of Kummer’s theorem.

Let $p$ be a prime, and $\nu_{p}(x)$ the $p$-adic valuation of a nonzero
integer $x$, i.e., the largest non-negative integer $s$ such that $p^{s}$
divides $x$. For $n\in\mathbb{N}_{0}$, write $n$ as the expansion in base $p$:

$n=n_{0}+n_{1}p+\cdots+n_{r}p^{r}$

where $0\leq n_{0},\cdots,n_{r}\leq p-1.$ For $n,m\in\mathbb{N}_{0}$ with
$n\geq m$, write $n,\ m$ and $t=n-m$ as the expansion in base $p$, with the
digits $n_{j}$, $m_{j}$ and $t_{j}$ as above, respectively. Let
$\epsilon_{j}=1$ if there is a carry in the $j$-th digit when adding $m$ and
$t$ in base $p$, and let $\epsilon_{j}=0$ otherwise. Then Kummer’s theorem
claims that the $p$-adic valuation of binomial coefficient $\binom{n}{m}$ is

$\nu_{p}(\binom{n}{m})=\sum_{j\geq 0}\epsilon_{j}.$

See e.g. [Mol, Theorem 2.6.7].

Assume that $p\mid n$ with $\nu_{p}(n)=r$. Consider $\binom{n}{p^{r}}$. For
convenience write $m=p^{r}$. Then the expansion of $m$ in base $p$ is just
$m=p^{r}$, i.e.,

$m_{j}=\delta_{j,r}=\begin{cases}0,&j\neq r;\\\ 1,&j=r.\end{cases}$

To know $\nu_{p}(\binom{n}{p^{r}})$, let $t=n-m=p^{r}(\frac{n}{m}-1)$. Then
the expansion of $t$ in base $p$ looks like

$t=t_{r}p^{r}+t_{r+1}p^{r+1}+\cdots+n_{r+s}p^{r+s}.$

It is clear that $t_{r}\neq p-1$. Otherwise, $n=m+t$ is divided by $p^{r+1}$,
contradicts $\nu_{p}(n)=r$.

Thus, when adding $m$ and $t$ in base $p$, there is no carry in $r$-th digit,
and hence there are no carries in any digit. By Kummer’s theorem,
$\nu_{p}(\binom{n}{p^{r}})=0.$ This proves

######  Fact 4.2.

(E. Kummer) Let $n$ be a positive integer and $p$ a prime. Assume that $p\mid
n$ with $\nu_{p}(n)=r$. Then $p\nmid\binom{n}{p^{r}}.$

Proof of Theorem 4.1. Without loss of generality, one may assume that $a_{1}$
is not a power of char $k$. Assume otherwise that $A=A({\bf
q},a_{1},\cdots,a_{n})$ has a bialgebra structure, say, with multiplication
$\Delta$ and counit $\varepsilon$. Thus $\Delta$ is an algebra homomorphism.
Write

$\Delta(x_{1})=\sum_{\mathbf{u,v}\in V}{a_{\mathbf{u,v}}}x_{\mathbf{u}}\otimes
x_{\mathbf{v}}\ \ \mbox{with}\ \ a_{\mathbf{u,v}}\in k.$

Since $A$ is a finite-dimensional local algebra, the kernel of the algebra
homomorphism $\varepsilon:A\longrightarrow k$ is the unique maximal ideal of
$A$. Thus $\varepsilon(x_{\mathbf{u}})=\delta_{\mathbf{u},\mathbf{0}}$, where
$\delta$ is the Kronecker symbol. By $(\varepsilon\otimes{\rm
Id})\Delta\cong{\rm Id}\cong({\rm Id}\otimes\varepsilon)\Delta$, one has

$\sum_{\mathbf{v}\in
V}a_{\mathbf{0,v}}x_{\mathbf{v}}=x_{1}=\sum_{\mathbf{u}\in
V}a_{\mathbf{u,0}}x_{\mathbf{u}}.$

Thus $a_{\mathbf{0,v}}=\delta_{\mathbf{v,e_{1}}}$ and
$a_{\mathbf{u,0}}=\delta_{\mathbf{u,e_{1}}}$, and hence

$\Delta(x_{1})=1\otimes x_{1}+x_{1}\otimes 1+\sum_{\mathbf{u,v}\in
V/\\{\mathbf{0}\\}}{a_{\mathbf{u,v}}}x_{\mathbf{u}}\otimes x_{\mathbf{v}}.$

Write

$\displaystyle\Delta(x_{1})^{a_{1}}$ $\displaystyle=(1\otimes
x_{1}+x_{1}\otimes 1+\sum\limits_{\mathbf{u,v}\in
V/\\{\mathbf{0}\\}}{a_{\mathbf{u,v}}}x_{\mathbf{u}}\otimes
x_{\mathbf{v}})^{a_{1}}$ $\displaystyle=(1\otimes x_{1}+x_{1}\otimes
1)^{a_{1}}+\Sigma_{1}$ $\displaystyle=\sum_{1\leq i\leq
a_{1}-1}\binom{a_{1}}{i}x_{1}^{i}\otimes x_{1}^{a_{1}-i}+\Sigma_{1}$

and write $\Sigma_{1}$ as a $k$-combination of elements in
$\mathcal{B}\otimes\mathcal{B}$. Notice that the degree of each term in
$x_{1}^{i}\otimes x_{1}^{a_{1}-i}\ (1\leq i\leq a_{1}-1)$ is $a_{1}$; and the
degree of each term in $\Sigma_{1}$ is bigger than or equal to

$a_{1}-i+2i=a_{1}+i$

with $1\leq i\leq a_{1}$. Thus the degree of each term in $\Sigma_{1}$ is
bigger than $a_{1}$. Since $\Delta$ is an algebra homomorphism, $(\Delta
x_{1})^{a_{1}}=\Delta(x_{1}^{a_{1}})=0$. It follows that $\binom{a_{1}}{i}=0,\
1\leq i\leq a_{1}-1$. This is absurd if char $k=0$.

So, suppose that ${\rm char}\ k=p>0$ and $a_{1}$ is not a power of $p$. If
$p\nmid a_{1}$, then $\binom{a_{1}}{1}=a_{1}\neq 0$ in $k$, a contradiction.
If the largest positive integer $t$ such that $p^{t}\mid a_{1}$ is $s$, then
by Fact 4.2, $p\nmid\binom{a_{1}}{p^{s}}$, and hence $\binom{a_{1}}{p^{s}}\neq
0$ in $k$, again a contradiction. This completes the proof. $\square$

By Theorem 4.1, if ${\rm char}\ k\neq 2$, then any quantum exterior algebra
admits no bialgebra structures. On the other hand, it is well-known that any
quantum exterior algebra over field $k$ with ${\rm char}\ k=2$ has a Hopf
algebra structure, with each $x_{i}$ a primitive element and $S={\rm Id}$.
That is, one has

######  Corollary 4.3.

A quantum exterior algebra admits a bialgebra structure if and only if ${\rm
char}\ k=2$. In this case, it is a Hopf algebra.

### 4.2.

Combined Theorem 3.1 with Theorem 4.1, one gets a large class of examples of
bi-Frobenius algebras which are not bialgebras (and hence not Hopf algebras).

######  Corollary 4.4.

Let $\sqrt{-1}\in k$ and $q_{ij}^{2}=1$ for $1\leq i,j\leq n$. Then quantum
complete intersection $A=A({\bf q},\ a_{1},\ \cdots,\ a_{n})$ admits a bi-
Frobenius algebra structure; moreover, if there is an $a_{i}$ such that it is
not a power of ${\rm char}\ k$ $($in particular, if ${\rm char}\ k=0)$, then
it admits no bialgebra structures, hence no Hopf algebra structures.

##  5\. Bi-Frobenius algebra structures on quantum exterior algebras in two
variables

In the special case of quantum exterior algebras in two variables, Theorem 3.1
has stronger form. Namely, let $A_{q}=A(q,2,2)=k\langle
x_{1},x_{2}\rangle/\langle x_{1}^{2},\ x_{2}^{2},\
x_{1}x_{2}+qx_{2}x_{1}\rangle$. Then one has

######  Theorem 5.1.

If $A_{q}$ admits a bi-Frobenius algebra structure, then $q=\pm 1.$

Conversely, if $\sqrt{-1}\in k$ and $q=\pm 1,$ then $A_{q}$ admits a bi-
Frobenius algebra structure.

Moreover, $A_{q}$ admits a Hopf algebra structure if and only if ${\rm char}\
k=2$.

### 5.1.

Denote a basis of $A_{q}$ by

$b_{0}=1,\ \ b_{1}=x_{1},\ \ b_{2}=x_{2},\ \ b_{3}=x_{1}x_{2}.$

Consider the dual basis of $A_{q}^{*}$: $b_{i}^{*},\ i=0,1,2,3$, where
$b_{i}^{*}(b_{j})=\delta_{ij},\ \forall\ 0\leq i,j\leq 3$. By Lemma 2.4 one
has

######  Fact 5.2.

All the Frobenius homomorphisms of $A_{q}$ are
$\phi=c_{0}b_{0}^{*}+c_{1}b_{1}^{*}+c_{2}b_{2}^{*}+c_{3}b_{3}^{*}$, where
$c_{i}\in k\ \text{and}\ c_{3}\neq 0$.

The following lemma is not a consequence of Theorem 3.1, since the coalgebra
structure in Theorem 3.1 is special, as given in (3.1); but in the following
lemma there are no restrictions on coalgebra structures.

######  Lemma 5.3.

If $A_{q}$ admits a bi-Frobenius algebra structure, then $q=\pm 1.$

###### Proof.

Suppose that $(A_{q},\ \phi,\ t,\ S)$ is a bi-Frobenius algebra, with
comultiplication $\Delta$ and counit $\varepsilon$. By Fact 5.2,
$\phi=\sum\limits_{0\leq i\leq 3}c_{i}b_{i}^{*}$ with $c_{3}\neq 0$. By Lemma
2.2(3), $t$ is a right integral, i.e., $ta=t\ \varepsilon(a)$ for any $a\in
A_{q}$. Then $tx_{1}=tx_{2}=0$, and hence $t=cx_{1}x_{2}$ with $0\neq c\in k$.

By definition, the identity $1$ is a group-like element. Write

$\Delta(b_{i})=1\otimes b_{i}+b_{i}\otimes 1+\Sigma_{i},\ \ 1\leq i\leq 3,$

where

$\Sigma_{1}=\sum_{1\leq i,j\leq 3}a_{ij}b_{i}\otimes b_{j},\ \ \
\Sigma_{2}=\sum_{1\leq i,j\leq 3}b_{ij}b_{i}\otimes b_{j},\ \ \
\Sigma_{3}=\sum_{1\leq i,j\leq 3}c_{ij}b_{i}\otimes b_{j}$

with all $a_{ij},\ b_{ij},\ c_{ij}\in k$. By Lemma 2.2(3), $\phi$ is a right
cointegral, i.e., $\phi(a)1=\sum\phi(a_{1})a_{2}$ for any $a\in A_{q}$. Thus
$\phi(t)1=\sum\phi(t_{1})t_{2}=S(1)=1$. On the other hand,

$\phi(t)=(\sum\limits_{0\leq i\leq 3}c_{i}b_{i}^{*})(cx_{1}x_{2})=cc_{3},$

so $c=1/c_{3}$. By Lemma 2.2(4), without lose of generality, one can assume
$c_{3}=1=c$, i.e., $A_{q}$ has a bi-Frobenius algebra structure $(A_{q},\
\phi=c_{0}b_{0}^{*}+c_{1}b_{1}^{*}+c_{2}b_{2}^{*}+b_{3}^{*},\ t=x_{1}x_{2},\
S)$.

By $a\leftharpoonup f=\sum f(a_{1})a_{2},\ \forall\ a\in A_{q},\ f\in
A_{q}^{*}$, and

$\Delta(t)=\Delta(x_{1}x_{2})=1\otimes x_{1}x_{2}+x_{1}x_{2}\otimes
1+\sum_{1\leq i,j\leq 3}c_{ij}b_{i}\otimes b_{j}=\sum t_{1}\otimes t_{2}$

one has

$\displaystyle x_{1}x_{2}\leftharpoonup b_{0}^{*}=x_{1}x_{2}$ $\displaystyle
x_{1}x_{2}\leftharpoonup b_{1}^{*}=c_{11}x_{1}+c_{12}x_{2}+c_{13}x_{1}x_{2}$
$\displaystyle x_{1}x_{2}\leftharpoonup
b_{2}^{*}=c_{21}x_{1}+c_{22}x_{2}+c_{23}x_{1}x_{2}$ $\displaystyle
x_{1}x_{2}\leftharpoonup
b_{3}^{*}=c_{31}x_{1}+c_{32}x_{2}+c_{33}x_{1}x_{2}+1.$

Since $(A_{q},\ t)$ is a Frobenius coalgebra, $A_{q}=t\leftharpoonup
A_{q}^{*}=x_{1}x_{2}\leftharpoonup A_{q}^{*}$. It follows that
$\left(\begin{smallmatrix}c_{11}&c_{12}\\\
c_{21}&c_{22}\end{smallmatrix}\right)$ is an invertible matrix, i.e.,
$c_{11}c_{22}\neq c_{12}c_{21}$.

By definition $S:A\longrightarrow A$ is the $k$-linear map given by
$S(a)=\sum\phi(t_{1}a)t_{2}.$ Thus

$\displaystyle S(x_{1})$
$\displaystyle=\sum(c_{0}b_{0}^{*}+c_{1}b_{1}^{*}+c_{2}b_{2}^{*}+b_{3}^{*})(t_{1}x_{1})t_{2}$
$\displaystyle=c_{1}x_{1}x_{2}+(c_{0}b_{0}^{*}+c_{1}b_{1}^{*}+c_{2}b_{2}^{*}+b_{3}^{*})(x_{2}x_{1})\sum_{1\leq
j\leq 3}c_{2j}b_{j}$
$\displaystyle=c_{1}x_{1}x_{2}+b_{3}^{*}(x_{2}x_{1})\sum_{1\leq j\leq
3}c_{2j}b_{j}$
$\displaystyle=-\frac{1}{q}[c_{21}x_{1}+c_{22}x_{2}+(c_{23}-qc_{1})x_{1}x_{2}];$
$\displaystyle S(x_{2})$
$\displaystyle=\sum(c_{0}b_{0}^{*}+c_{1}b_{1}^{*}+c_{2}b_{2}^{*}+b_{3}^{*})(t_{1}x_{2})t_{2}$
$\displaystyle=c_{2}x_{1}x_{2}+(c_{0}b_{0}^{*}+c_{1}b_{1}^{*}+c_{2}b_{2}^{*}+b_{3}^{*})(x_{1}x_{2})\sum_{1\leq
j\leq 3}c_{1j}b_{j}$ $\displaystyle=c_{2}x_{1}x_{2}+\sum_{1\leq j\leq
3}c_{1j}b_{j}$
$\displaystyle=c_{11}x_{1}+c_{12}x_{2}+(c_{13}+c_{2})x_{1}x_{2};$
$S(x_{1}x_{2})=\sum(c_{0}b_{0}^{*}+c_{1}b_{1}^{*}+c_{2}b_{2}^{*}+b_{3}^{*})(t_{1}x_{1}x_{2})t_{2}=b_{3}^{*}(x_{1}x_{2})x_{1}x_{2}=x_{1}x_{2}.$

So one has

$\left\\{\begin{array}[]{l}S(x_{1})=-\frac{1}{q}[c_{21}x_{1}+c_{22}x_{2}+(c_{23}-qc_{1})x_{1}x_{2}];\\\
S(x_{2})=c_{11}x_{1}+c_{12}x_{2}+(c_{2}+c_{13})x_{1}x_{2};\\\
S(x_{1}x_{2})=x_{1}x_{2}.\end{array}\right.$ $None$

Since $S$ is an anti-homomorphism of coalgebra, i.e., $\Delta(S(a))=\sum
S(a_{2})\otimes S(a_{1})$ for any $a\in A_{q}$, one has

$\Delta(S(x_{1}x_{2}))=\Delta(x_{1}x_{2})=1\otimes
x_{1}x_{2}+x_{1}x_{2}\otimes 1+\Sigma_{3},\\\ $

and

$\displaystyle\sum S(t_{2})\otimes S(t_{1})$ $\displaystyle=S(1)\otimes
S(x_{1}x_{2})+S(x_{1}x_{2})\otimes S(1)+\sum_{1\leq i,j\leq
3}c_{ij}S(b_{j})\otimes S(b_{i})$ $\displaystyle=1\otimes
x_{1}x_{2}+x_{1}x_{2}\otimes 1+\sum_{1\leq i,j\leq 3}c_{ij}S(b_{j})\otimes
S(b_{i}).$

Thus $\Sigma_{3}=\sum\limits_{1\leq i,j\leq 3}c_{ij}S(b_{j})\otimes S(b_{i})$,
that is

$\sum\limits_{1\leq i,j\leq 3}c_{ij}b_{i}\otimes b_{j}=\sum\limits_{1\leq
i,j\leq 3}c_{ij}S(b_{j})\otimes S(b_{i}).$

Using (5.1) and comparing the coefficients of $x_{1}\otimes x_{1},\ \
x_{1}\otimes x_{2}$ and $x_{2}\otimes x_{1}$, one gets

$c_{11}=\frac{1}{q^{2}}c_{11}c_{21}^{2}-\frac{1}{q}c_{11}c_{12}c_{21}-\frac{1}{q}c_{11}c_{21}^{2}+c_{22}c_{11}^{2},$
$None$
$c_{12}=\frac{1}{q^{2}}c_{11}c_{21}c_{22}-\frac{1}{q}c_{11}c_{12}c_{22}-\frac{1}{q}c_{12}c_{21}^{2}+c_{22}c_{11}c_{12},$
$None$
$c_{21}=\frac{1}{q^{2}}c_{11}c_{21}c_{22}-\frac{1}{q}c_{21}c_{12}^{2}-\frac{1}{q}c_{11}c_{21}c_{22}+c_{11}c_{12}c_{22}.$
$None$

By $S(x_{1})^{2}=0$ one has $(c_{21}c_{22})(1-q)=0$. By $S(x_{2})^{2}=0$ one
has $c_{11}c_{12}(1-q)=0$. By $S(x_{1}x_{2})=S(x_{2})S(x_{1})$ one has
$c_{12}c_{21}-qc_{11}c_{22}=q^{2}.$ All together one has the following:

(i) $c_{11}c_{22}\neq c_{12}c_{21};$

(ii) $(c_{21}c_{22})(1-q)=0;$

(iii) $c_{11}c_{12}(1-q)=0;$

(iv) $c_{12}c_{21}-qc_{11}c_{22}=q^{2}.$

Now, assume that $q\neq 1$. By (ii) and (iii), $c_{21}c_{22}=0=c_{11}c_{12}.$
By (5.2) and (5.3) one has

$c_{11}=\frac{1}{q^{2}}c_{11}c_{21}^{2}-\frac{1}{q}c_{11}c_{21}^{2}+c_{22}c_{11}^{2}$
$None$ $c_{12}=-\frac{1}{q}c_{12}c_{21}^{2}$ $None$

By $c_{11}c_{22}\neq c_{12}c_{21}$ and $c_{21}c_{22}=0=c_{11}c_{12}$, there
are two situations:

  1. (1)

$c_{11}=0=c_{22}$ and $c_{12}\neq 0\neq c_{21};$

  2. (2)

$c_{12}=0=c_{21}$ and $c_{11}\neq 0\neq c_{22}.$

If (1) happens, then by (iv) one has $c_{12}c_{21}=q^{2}$. By $(5.5)$ and
(5.6) one gets $c_{21}^{2}=-q=c_{12}^{2}$. Thus $q^{4}=q^{2}$, and hence
$q=-1$.

If (2) happens, then by (iv) one has $c_{11}c_{22}=-q$. By $(5.5)$ one gets
$c_{11}c_{22}=1$. Thus $q=-1$.

In conclusion, if $A_{q}$ has a bi-Frobenius algebra structure, then either
$q=1$, or $q=-1$. That is, if $q^{2}\neq 1$, then $A_{q}$ can not be a bi-
Frobenius algebra.∎

###  5.2. Proof of Theorem 5.1.

In Lemma 5.3 we already know that if $A_{q}$ admits a bi-Frobenius algebra
structure, then $q=\pm 1.$ Conversely, suppose that $k$ contains $\sqrt{-1}$
and that $q=\pm 1.$ Then by Theorem 3.1, $A_{q}$ has a bi-Frobenius algebra
structure, with $\Delta$ and $\varepsilon$ given by (3.1).

Moreover, by Corollary 4.3, $A_{q}$ admits a Hopf algebra structure if and
only if ${\rm char}\ k=2$. $\square$

### 5.3.

We point out that, if $k$ does not contain $\sqrt{-1}$, then the exterior
algebra with two variables $A_{1}=k\langle x_{1},x_{2}\rangle/\langle
x_{1}^{2},\ x_{2}^{2},\ x_{1}x_{2}+x_{2}x_{1}\rangle$ admits no bi-Frobenius
algebra structures.

######  Proposition 5.4.

If $\sqrt{-1}\notin k$, then the exterior algebra with two variables $A_{1}$
admits no bi-Frobenius algebra structures.

###### Proof.

Let $k$ be a field such that $\sqrt{-1}\notin k$. Then ${\rm char}\ k\neq 2$.
Assume otherwise that $A_{1}$ has a bi-Frobenius algebra structure $(A_{1},\
\phi,\ t,\ S)$, with comultiplication $\Delta$ and counit $\varepsilon$. As in
the proof of Lemma 5.3 one gets (5.2), (5.3), and (5.4), using $q=1$ and
rewriting them one has

$c_{11}=-c_{11}c_{12}c_{21}+c_{22}c_{11}^{2}=c_{11}(c_{11}c_{22}-c_{12}c_{21})$
$None$
$c_{12}=c_{11}c_{21}c_{22}-c_{12}c_{21}^{2}=c_{21}(c_{11}c_{22}-c_{12}c_{21})$
$None$
$c_{21}=-c_{21}c_{12}^{2}+c_{11}c_{12}c_{22}=c_{12}(c_{11}c_{22}-c_{12}c_{21}),$
$None$

and the condition (iv) becomes

$c_{12}c_{21}-c_{11}c_{22}=1.$ $None$

By substituting $({\rm iv}^{\prime})$ into $(5.2^{\prime})$, $(5.3^{\prime})$
and $(5.4^{\prime})$, one gets $c_{11}=c_{22}=0$, and then $c_{12}^{2}=-1$.
But $c_{12}\in k$, this contradicts the assumption $c_{12}=\sqrt{-1}\notin k$.
This completes the proof. ∎

##  6\. Non-isomorphic bi-Frobenius algebra structures on an algebra

We give two coalgebra structures on a complete intersection ring, such that it
admits non-isomorphic bi-Frobenius algebra structures.

### 6.1.

Taking ${\bf q=-1}$ (i.e., $q_{ij}=-1$ for $1\leq i,j\leq n$) in quantum
complete intersections, one gets complete intersection ring

$A=A({\bf-1},a_{1},\cdots,a_{n})=k\langle x_{1},\cdots,x_{n}\rangle/\langle
x_{i}^{a_{i}},\ x_{i}x_{j}-x_{j}x_{i},\ \ 1\leq i,\ j\leq n\rangle.$

Thus $\mathcal{B}=\\{x_{\mathbf{v}}=x_{1}^{v_{1}}\cdots x_{n}^{v_{n}}\ |\
\mathbf{v}\in V\\}$ is a basis of $A$, where
$V=\\{\mathbf{v}=(v_{1},\cdots,v_{n})\in\mathbb{N}_{0}^{n}\ |\ v_{i}\leq
a_{i}-1,\ 1\leq i\leq n\\}.$ Inspired by the construction of path coalgebra of
quivers (see D. Simson [S]; also C. Cibils and M. Rosso [CR]), we consider the
$k$-linear maps $\varepsilon:A\longrightarrow k$ and $\Delta:A\longrightarrow
A\otimes A$ given by

$\left\\{\begin{array}[]{l}\varepsilon(x_{\mathbf{v}})=\delta_{\mathbf{v,0}},\
\ \forall\ \mathbf{v}\in V;\\\ \\\
\Delta(x_{\mathbf{v}})=\sum\limits_{\begin{subarray}{c}\mathbf{v_{1}+v_{2}=v},\\\
\mathbf{v_{1},v_{2}}\in V\end{subarray}}x_{\mathbf{v_{1}}}\otimes
x_{\mathbf{v_{2}}},\ \ \forall\ \mathbf{v}\in V.\end{array}\right.$ $None$

######  Theorem 6.1.

With counit $\varepsilon$ and comultiplication $\Delta$ as in $(6.1)$,
complete intersection ring $A=A({\bf-1},a_{1},\cdots,a_{n})$ admits a bi-
Frobenius algebra $(A,\ (x_{\mathbf{a-1}})^{*},\ x_{\mathbf{a-1}},\ {\rm
Id})$.

Moreover, in general this bi-Frobenius algebra is not isomorphic $($as a bi-
Frobenius algebra$)$ to the one in Corollary 3.8, with coalgebra structure
given by $(3.1)$.

###### Proof.

It is straightforward to verify that $(A,\ \Delta,\ \varepsilon)$ is a
coalgebra, the identity $1$ is a group-like element, and that $\varepsilon$ is
an algebra map. Put $t=x_{\bf a-1}$. Since $A$ is commutative and
cocommutative, ${\rm Id}$ is an algebra anti-homomorphism and a coalgebra
anti-homomorphism. To prove $(A,\ (x_{\mathbf{a-1}})^{*},\ x_{\mathbf{a-1}},\
{\rm Id})$ is a bi-Frobenius algebra, by Lemma 2.3, it remains to check

$\sum(x_{\bf a-1})^{*}(t_{1}x_{\bf v})t_{2}=x_{\bf v},\ \ \forall\ {\bf v}\in
V.$

In fact, by $\Delta(t)=\Delta(x_{\bf
a-1})=\sum\limits_{\begin{subarray}{c}\mathbf{v_{1}+v_{2}=a-1},\\\
\mathbf{v_{1},v_{2}}\in V\end{subarray}}x_{\mathbf{v_{1}}}\otimes
x_{\mathbf{v_{2}}}$, one has

$\displaystyle\sum(x_{\bf a-1})^{*}(t_{1}x_{\bf v})t_{2}=(x_{\bf
a-1})^{*}(x_{\bf a-1-v}\ x_{\bf v})\ x_{\bf v}=x_{\bf v},\ \forall\ {\bf v}\in
V.$

This proves $(A,\ (x_{\mathbf{a-1}})^{*},\ x_{\mathbf{a-1}},\ {\rm Id})$ is a
bi-Frobenius algebra. For convenience, denote this bi-Frobenius algebra by
$B_{1}$.

When $\sqrt{-1}\in k$, to show that $B_{1}$ is not isomorphic to the bi-
Frobenius algebra $B_{2}=(A,\ (x_{{\bf a-1}})^{*},\ x_{{\bf a-1}},\ S)$ in
Theorem 3.1, we consider the spaces $P(B_{1})$ and $P(B_{2})$, of the
primitive elements of $B_{1}$ and $B_{2}$, respectively.

By Lemma 3.3, dim $P(B_{2})=(\prod\limits_{1\leq i\leq n}a_{i})-2.$

On the other hand, each $x_{i}$ is a primitive element of $B_{1}$. We claim
that ${\rm dim}\ P(B_{1})=n$. Otherwise, there exists a primitive element
$x=\sum\limits_{{\bf v}\in W}c_{\bf v}\ x_{\bf v}$, where $W$ is a non empty
subset of $V$ such that $c_{\bf v}\neq 0$ and ${\bf v\neq e_{i}}$ for $1\leq
i\leq n$, for all ${\bf v}\in W$. From the proof below, one can see that,
without loss of generality, one may assume that ${\bf 0}\notin W$. From the
right hand side of the equality

$\sum\limits_{{\bf v}\in W}c_{\bf v}\ \Delta(x_{\bf v})=\Delta(x)=1\otimes
x+x\otimes 1=\sum\limits_{{\bf v}\in W}c_{\bf v}\ (1\otimes x_{\bf
v})+\sum\limits_{{\bf v}\in W}c_{\bf v}\ (x_{\bf v}\otimes 1)$

and since ${\bf 0}\notin W$, one sees that $\Delta(x)$ is a $k$-linear
combination of $2|W|$ elements in the basis $\mathcal{B}\otimes\mathcal{B}$ of
$A\otimes A$, with each coefficient nonzero, where $|W|$ is the number of
elements in $W$. While by (6.1), each $\Delta(x_{\bf v})$ in the left hand
side is a $k$-linear combination of $t$ elements in the basis
$\mathcal{B}\otimes\mathcal{B}$, where $t>2$. For ${\bf u\neq v}\in V$, there
are no the same summands in $\Delta(x_{\bf u})$ and $\Delta(x_{\bf v})$, since
$x_{\bf u_{1}}\otimes x_{\bf u_{2}}=x_{\bf v_{1}}\otimes x_{\bf v_{2}}$
implies ${\bf u=u_{1}+u_{2}=v_{1}+v_{2}=v}$. In this way we see that
$\Delta(x)$ is a $k$-linear combination of $s$ elements in the basis
$\mathcal{B}\otimes\mathcal{B}$, with each coefficient nonzero, where
$s>2|W|$. A contradiction! This proves dim $P(B_{1})=n$.

Thus, one has ${\rm dim}\ P(B_{2})=(\prod\limits_{1\leq i\leq
n}a_{i})-2>n={\rm dim}\ P(B_{1})$, except the case $n=2=a_{1}=a_{2}$. So
$B_{1}\ncong B_{2}$ as coalgebras, except $n=2=a_{1}=a_{2}$. This completes
the proof. ∎

######  Remark 6.2.

If ${\bf q}\neq{\bf-1}$ in Theorem 6.1 $($i.e., the algebra $A$ considered is
not complete intersection ring$)$, then the corresponding result is no longer
true. For example, when $\sqrt{-1}\notin k$, the exterior algebra with two
variables $A_{1}=k\langle x_{1},x_{2}\rangle/\langle x_{1}^{2},\ x_{2}^{2},\
x_{1}x_{2}+x_{2}x_{1}\rangle$ admits no bi-Frobenius algebra structures. See
Proposition 5.4.

### 6.2.

In Subsection 6.1, taking $(a_{1},\ \cdots,\ a_{n})=(2,\ \cdots,\ 2)$, namely

$A=A({\bf-1},\ 2,\ \cdots,\ 2)=k\langle x_{1},\cdots,x_{n}\rangle/\langle
x_{i}^{2},\ x_{i}x_{j}-x_{j}x_{i},\ 1\leq i,\ j\leq n\rangle.$

Consider the $k$-linear maps $\varepsilon:A\longrightarrow k$ and
$\Delta:A\longrightarrow A\otimes A$ given by

$\left\\{\begin{array}[]{l}\varepsilon(x_{\mathbf{v}})=\delta_{\mathbf{0,v}},\
\ \forall\ \mathbf{v}\in V;\\\ \\\
\Delta(x_{\mathbf{v}})=\sum\limits_{\begin{subarray}{c}\mathbf{v_{1},v_{2}\leq
v}\\\ \mathbf{v\leq
v_{1}+v_{2}}\end{subarray}}(-1)^{|\mathbf{v_{1}+v_{2}-v}|}\
x_{\mathbf{v_{1}}}\otimes x_{\mathbf{v_{2}}},\ \ \forall\ \mathbf{v}\in
V.\end{array}\right.$ $None$

######  Theorem 6.3.

With counit $\varepsilon$ and comultiplication $\Delta$ given by $(6.2)$,
$(A,\ \phi,\ t=x_{1}\cdots x_{n},\ {\rm Id})$ is a bi-Frobenius algebra, where
$\phi=\sum\limits_{\mathbf{v}\in V}(x_{\mathbf{v}})^{*}$.

Moreover, this bi-Frobenius algebra has no primitive elements, and hence it
can not be isomorphic to the bi-Frobenius algebras given in Corollary 3.8 and
in Theorem 6.1.

###### Proof.

One directly verify that $(A,\ \Delta,\ \varepsilon)$ is a coalgebra:

$\displaystyle(\Delta\otimes{\rm Id})\Delta(x_{\mathbf{v}})$
$\displaystyle=\sum_{\begin{subarray}{c}\mathbf{v_{1},v_{2}\leq v}\\\
\mathbf{v\leq v_{1}+v_{2}}\end{subarray}}(-1)^{|\mathbf{v_{1}+v_{2}-v}|}\
\Delta(x_{\mathbf{v_{1}}})\otimes x_{\mathbf{v_{2}}}$
$\displaystyle=\sum_{\begin{subarray}{c}\mathbf{v_{1},v_{2},v_{3}\leq v}\\\
\mathbf{v\leq
v_{1}+v_{2}+v_{3}}\end{subarray}}(-1)^{|\mathbf{v_{1}+v_{2}+v_{3}-v}|}\
x_{\mathbf{v_{1}}}\otimes x_{\mathbf{v_{2}}}\otimes x_{\mathbf{v_{3}}}$
$\displaystyle=({\rm Id}\otimes\Delta)\Delta(x_{\mathbf{v}}),\ \ \forall\
\mathbf{v}\in V$

and

$\displaystyle(\varepsilon\otimes{\rm Id})\Delta(x_{\mathbf{v}})$
$\displaystyle=\sum_{\mathbf{v\leq v_{2}\leq v}}(-1)^{|\mathbf{v_{2}-v}|}\
1\otimes x_{\mathbf{v_{2}}}=1\otimes x_{\mathbf{v}}=x_{\mathbf{v}}$
$\displaystyle=x_{\mathbf{v}}\otimes 1=\sum_{\mathbf{v\leq v_{1}\leq
v}}(-1)^{|\mathbf{v_{1}-v}|}\ x_{\mathbf{v_{1}}}\otimes 1=({\rm
Id}\otimes\varepsilon)\Delta(x_{\mathbf{v}}),\ \ \forall\ \mathbf{v}\in V.$

Put $t=x_{1}\cdots x_{n}$. Since $A$ is commutative and cocommutative, ${\rm
Id}$ is an algebra anti-homomorphism and a coalgebra anti-homomorphism. By
Lemma 2.3, it remains to prove

$\sum\phi(t_{1}x_{\bf u})t_{2}=x_{\bf u},\ \ \forall\ {\bf u}\in V.$

In fact, by

$\Delta(t)=\Delta(x_{1}\cdots
x_{n})=\sum_{\begin{subarray}{c}\mathbf{v_{1},v_{2}}\leq(1,\cdots,1)\\\
(1,\cdots,1)\leq\mathbf{v_{1}+v_{2}}\end{subarray}}(-1)^{|\mathbf{v_{1}+v_{2}-1}|}\
x_{\mathbf{v_{1}}}\otimes x_{\mathbf{v_{2}}}$

one has

$\displaystyle\sum\phi(t_{1}x_{\bf u})t_{2}$
$\displaystyle=\sum_{\begin{subarray}{c}\mathbf{v_{1},v_{2}}\leq(1,\cdots,1)\\\
(1,\cdots,1)\leq{\bf v_{1}+v_{2}}\end{subarray}}(-1)^{\bf|{v_{1}}+{v_{2}}-1|}\
\phi(x_{\mathbf{v_{1}}}\ x_{\mathbf{u}})\ x_{\mathbf{v_{2}}},\ \forall\
\mathbf{u}\in V.$

If $v_{1i}>1-u_{i}$ for some $i$ (here $v_{1i}$ means the $i$th component of
${\bf v_{1}}$), then $v_{1i}+u_{i}\geq 2$, and then $x_{\mathbf{v_{1}}}\
x_{\mathbf{u}}=x_{\mathbf{v_{1}+u}}=0$. Thus, we only need to consider those
$\mathbf{v_{1}}$ with $\mathbf{v_{1}}\leq\mathbf{1-u}$, and hence ${\bf
v_{2}\geq 1-v_{1}\geq u}$. Fixing ${\bf v_{2}}$, then the coefficient of
$x_{\mathbf{v_{2}}}$ in the sum above is (note that
$\phi=\sum\limits_{\mathbf{v}\in V}(x_{\mathbf{v}})^{*}$)

$\displaystyle\sum_{\mathbf{1-v_{2}\leq v_{1}\leq 1-u}}(-1)^{|{\bf
v_{1}+v_{2}-1}|}\ \phi(x_{\mathbf{v_{1}+u}})$
$\displaystyle=\sum_{\mathbf{1-v_{2}\leq v_{1}\leq
1-u}}(-1)^{\bf|{v_{1}}+{v_{2}}-1|}$ $\displaystyle=\sum_{\mathbf{0\leq
v_{1}+v_{2}-1\leq v_{2}-u}}(-1)^{\bf|{v_{1}}+{v_{2}}-1|}.$

If $\mathbf{v_{2}=u}$, then the vector ${\bf v_{1}}$ in the sum is unique, so
the coefficient of $x_{\mathbf{v_{2}}}$ in the sum is

$\sum\limits_{\mathbf{0\leq v_{1}+v_{2}-1\leq v_{2}-u}}(-1)^{|{\bf
v_{1}+v_{2}}-1|}=(-1)^{|{\bf 0}|}=1.$

Assume that ${\bf v_{2}}\neq\mathbf{u}$. Set $m=|{\bf v_{2}-u}|$. Note that
${\bf v_{2}-u}$ is a vector with $m$ components being $1$ and $n-m$ components
being $0$. We count the number of vectors ${\bf v_{1}}$ in the sum via $|{\bf
v_{1}+v_{2}-1}|$ as follows.

When $|{\bf v_{1}+v_{2}-1}|=0$, then the choice of ${\bf v_{1}}$ is unique,
i.e., ${\bf v_{1}=1-v_{2}}$. That is, the number of ${\bf v_{1}}$ is
$\binom{m}{0}=1$.

When $|{\bf v_{1}+v_{2}-1}|=1$, since $\mathbf{v_{1}+v_{2}-1\leq v_{2}-u}$ and
${\bf v_{2}-u}$ has $m$ components being $1$, it follows that the number of
the choices of ${\bf v_{1}}$ such that $|{\bf v_{1}+v_{2}-1}|=1$ is just
$\binom{m}{1}$;

In general, when $|{\bf v_{1}+v_{2}-1}|=i$ with $1\leq i\leq n$, then the
number of the choices of ${\bf v_{1}}$ such that $|{\bf v_{1}+v_{2}-1}|=i$ is
just $\binom{m}{i}$.

All together, if ${\bf v_{2}}\neq\mathbf{u}$, then we see that the coefficient
of $x_{\mathbf{v_{2}}}$ in the sum above is

$\sum_{\mathbf{0\leq v_{1}+v_{2}-1\leq
v_{2}-u}}(-1)^{\bf|{v_{1}}+{v_{2}}-1|}=\sum_{i=0}^{m}\binom{m}{i}(-1)^{i}=0.$

In this way we see that

$\displaystyle\sum\phi(t_{1}x_{\bf
u})t_{2}=\sum_{\begin{subarray}{c}\mathbf{v_{1},v_{2}}\leq(1,\cdots,1)\\\
(1,\cdots,1)\leq{\bf v_{1}+v_{2}}\end{subarray}}(-1)^{\bf|{v_{1}}+{v_{2}}-1|}\
\phi(x_{\mathbf{v_{1}}}\ x_{\mathbf{u}})\ x_{\mathbf{v_{2}}}=x_{\bf u},\ \
\forall\ {\bf u}\in V.$

This proves that $(A,\ \phi,\ x_{\mathbf{1}}=x_{1}\cdots x_{n},\ {\rm Id})$ is
a bi-Frobenius algebra.

Using the similar idea as in the proof of Theorem 6.1, one claim that this bi-
Frobenius algebra has no primitive elements. For completeness, we include a
proof.

In fact, assume otherwise that $x=\sum\limits_{{\bf v}\in W}c_{\bf v}\ x_{\bf
v}$ is a primitive element, where $W$ is a non empty subset of $V$ such that
$c_{\bf v}\neq 0$ for all ${\bf v}\in W$.

First, assume that ${\bf 0}\notin W$. From the right hand side of the equality

$\Delta(x)=1\otimes x+x\otimes 1=\sum\limits_{{\bf v}\in W}c_{\bf v}\
(1\otimes x_{\bf v})+\sum\limits_{{\bf v}\in W}c_{\bf v}\ (x_{\bf v}\otimes
1)$

and since ${\bf 0}\notin W$, one sees that $\Delta(x)$ is a $k$-linear
combination of $2|W|$ elements in the basis $\mathcal{B}\otimes\mathcal{B}$ of
$A\otimes A$, with each coefficient nonzero. On the other hand,

$\Delta(x)=\sum\limits_{{\bf v}\in W}c_{\bf v}\ \Delta(x_{\bf v})$

and by (6.2) one has

$\Delta(x_{\mathbf{v}})=\sum\limits_{\begin{subarray}{c}\mathbf{v_{1},v_{2}\leq
v}\\\ \mathbf{v\leq
v_{1}+v_{2}}\end{subarray}}(-1)^{|\mathbf{v_{1}+v_{2}-v}|}\
x_{\mathbf{v_{1}}}\otimes x_{\mathbf{v_{2}}}.$

Since ${\bf 0}\notin W$ and $\Delta(x_{\bf v})$ at least includes the
summands:

$x_{\bf v}\otimes x_{\bf v},\ \ x_{\bf v-e_{i}}\otimes x_{\bf v},\ \ x_{\bf
v}\otimes x_{\bf v-e_{i}}$

for some ${\bf e_{i}}$, it follows that $\Delta(x_{\bf v})$ is a $k$-linear
combination of $t$ elements in the basis $\mathcal{B}\otimes\mathcal{B}$,
where $t\geq 3$.

For ${\bf u\neq v}\in V$, there are no the same summands in $\Delta(x_{\bf
u})$ and $\Delta(x_{\bf v})$. Actually, if $x_{\bf u_{1}}\otimes x_{\bf
u_{2}}=x_{\bf v_{1}}\otimes x_{\bf v_{2}}$, then

${\bf u_{1}}={\bf v_{1}},\ \ {\bf u_{2}}={\bf v_{2}},\ \ {\bf u_{1},u_{2}\leq
u\leq u_{1}+u_{2}},\ \ {\bf v_{1},v_{2}\leq v\leq v_{1}+v_{2}}.$

Since $a_{i}=2$ for $1\leq i\leq n$, one has

$u_{i}=\begin{cases}0&\text{if}\ \ {u_{1}}_{i}={u_{2}}_{i}=0;\\\ 1&\text{if \
\ otherwise}\end{cases}$

and

$v_{i}=\begin{cases}0&\text{if}\ \ {v_{1}}_{i}={v_{2}}_{i}=0;\\\ 1&\text{if \
\ otherwise}.\end{cases}$

It follows that ${\bf u=v}.$ In this way we see that $\Delta(x)$ is a
$k$-linear combination of $s$ elements in the basis
$\mathcal{B}\otimes\mathcal{B}$, with each coefficient nonzero, where $s\geq
3|W|$. A contradiction!

Next, assume that ${\bf 0}\in W$. Since $1$ is not a primitive element,
$|W|\geq 2$. By the similar argument as above, one sees that

$\sum\limits_{{\bf v}\in W}c_{\bf v}\ \Delta(x_{\bf
v})=\Delta(x)=\sum\limits_{{\bf v}\in W}c_{\bf v}\ (1\otimes x_{\bf
v})+\sum\limits_{{\bf v}\in W}c_{\bf v}\ (x_{\bf v}\otimes 1).$

The left hand side is a $k$-linear combination of $s$ elements in the basis
$\mathcal{B}\otimes\mathcal{B}$, with each coefficient nonzero, where $s\geq
3(|W|-1)+1=3|W|-2$; while the right hand side is a $k$-linear combination of
$2|W|-1$ elements in the basis $\mathcal{B}\otimes\mathcal{B}$, with each
coefficient nonzero. A contradiction!

This completes the proof. ∎

######  Remark 6.4.

If $(a_{1},\cdots,a_{n})\neq(2,\cdots,2)$ in Theorem 6.3, then the
corresponding result is no longer true.

For example, let $A=k\langle x_{1},x_{2}\rangle/\langle x_{1}^{2},\
x_{2}^{3},\ x_{1}x_{2}-x_{2}x_{1}\rangle$. Then $A$ becomes a coalgebra with
comultiplication $\Delta$ and counit $\varepsilon$ as given by $(6.2)$, but
$(A,\ \phi=\sum\limits_{\mathbf{v}\in V}(x_{\mathbf{v}})^{*},\
t=x_{1}x_{2}^{2},\ S)$ is not a bi-Frobenius algebra, where
$S:A\longrightarrow A$ is the $k$-linear map given by
$S(a)=\sum\phi(t_{1}a)t_{2}.$

In fact, in this case $V=\\{(0,0)$, $(0,1)$, $(0,2)$, $(1,0)$, $(1,1)$,
$(1,2)\\}$,

$\displaystyle\Delta(t)$
$\displaystyle=\Delta(x_{1}x_{2}^{2})=\sum_{{\begin{subarray}{c}{\bf
v_{1},v_{2}}\leq(1,2),\\\ (1,2)\leq{\bf
v_{1}+v_{2}}\end{subarray}}}(-1)^{|{\bf v_{1}+v_{2}}-(1,2)|}\ x_{\bf
v_{1}}\otimes x_{\bf v_{2}},$

and

$\displaystyle S(1)$ $\displaystyle=\sum_{{\begin{subarray}{c}{\bf
v_{1},v_{2}}\leq(1,2),\\\ (1,2)\leq{\bf
v_{1}+v_{2}}\end{subarray}}}(-1)^{|{\bf v_{1}+v_{2}}-(1,2)|}\phi(x_{\bf
v_{1}})x_{\bf v_{2}}=1+x_{2}^{2}.$

So $S$ is not an algebra homomorphism, and hence $A$ is not a bi-Frobenius
algebra.

Acknowledgements. We thank Libin Li and Yanhua Wang for valuable suggestions
concerning the presentation of the results.

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