1. Introduction
The automorphism group of an algebra has a long history and extensive study, and determining the fully automorphic group of an algebra is usually a very difficult problem. For instance, the automorphism group of the polynomial ring of one or two variables has been understood. Jung [
1] in 1942 first proved the case of two variables for characteristic zero and then van der Kulk [
2] in 1953 for an arbitrary characteristic. However, the automorphism group of the polynomial ring of three variables is not yet fully understood, and a remarkable result is given by Shestakov and Umirbaev [
3] in 2004, which shows that the Nagata automorphism is a wild automorphism. Many researchers have been successfully computing the automorphism groups of interesting infinite-dimensional noncommutative algebras, including certain quantum groups, generalized quantum Weyl algebras, skew polynomial rings, quantized algebras and many more—see [
4,
5,
6,
7,
8,
9,
10,
11,
12,
13,
14,
15], among others. The theory of algebra authomorphisms is very deep and related with mainstream, such as the Jacobian Conjecture [
16] and Holonomic
-modules [
17,
18].
In the theory of the classical Hopf algebras, Radford’s biproducts are very important Hopf algebras, which play an important role in the theory of classification of pointed Hopf algebra [
19,
20], and account for many examples of semisimple Hopf algebra. Therefore, the study of the automorphism group of Radford’s biproducts is very interesting, and Radford characterized Hopf algebra automorphisms in [
21]. There are many generalizations of Radford’s biproducts, such as [
22] for quasi-Hopf algebras, [
23] for multiplier Hopf algebras, Ref. [
24], for monoidal Hom–Hopf algebras, Ref. [
25], for Hom–Hopf algebras, and recently [
26] for
-twisted-Radford’s Hom-biproducts.
Let
be the Radford’s biproduct, where
A is both a left
H-module algebra and a left
H-comodule coalgebra. Define
,
, let
be the set of Hopf algebra automorphisms
F of
satisfying
. Radford [
21] characterized these Hopf algebra automorphisms, and factorized
F into two suitable maps. Inspired by Radford’s idea in [
21], the authors of [
27] discuss the automorphisms of twisted tensor biproducts, and reveal the relationships between results and Radford’s results, and the authors of [
28] study automorphisms of Radford’s Hom-biproducts introduced in [
24]. Motivated by these works, we want to characterize the Hom–Hopf algebra automorphisms of
-twisted-Radford’s Hom-biproduct introduced in [
26].
This paper is organized as follows. In
Section 2, we recall some definitions and basic results related to Hom-(co)algebras, Hom-bialgebras, Hom–Hopf algebras, Hom-(co)module, Hom-(co)module (co)algebras,
-twisted Hom-smash products,
-twisted Hom-smash coproducts and
-twisted-Radford’s Hom-biproduct. In
Section 3, we study the endomorphisms and automorphisms of
-twisted-Radford’s Hom-biproducts, and show that the automorphism has a factorization closely related to the factors
and
of the
-twisted-Radford’s Hom-biproduct
in [
26]. Finally, we characterize the automorphisms of a concrete example in
Section 4.
2. Preliminaries
We always assume that
is a base field. All algebras, linear spaces, etc. will be over
; the subscript of
is omitted for simplicity. Let
denote the identity map on a
-space
V. We now recall some useful definitions and terminology as follows; see [
25,
29,
30,
31,
32,
33,
34].
Definition 1 ([
29]).
A Hom-algebra is a quadruple (abbr. ), where A is a -linear space, and are linear maps with notation , , and α is an automorphism of A, satisfying the following conditions for all - (HA1)
,
- (HA2)
,
- (HA3)
,
- (HA4)
.
Let and be two Hom-algebras. Then, a morphism is called a Hom-algebra morphism if , and .
Definition 2 ([
30,
32,
34]).
A Hom-coalgebra is a quadruple (abbr. ), where C is a -linear space, and are linear maps with notation and α is an automorphism of C, satisfying the following conditions for all - (HC1)
,
- (HC2)
,
- (HC3)
,
- (HC4)
.
Let and be two Hom-coalgebras. Then, a morphism is called a Hom-coalgebra morphism if , and .
Definition 3 ([
31,
32]).
A Hom-bialgebra is a sextuple (abbr. ), where is a Hom-algebra and is a Hom-coalgebra, such that Δ and ε are Hom-algebra morphisms, i.e.,- (HB1)
,
- (HB2)
,
- (HB3)
,
- (HB4)
.
Furthermore, if there is a linear map such that
- (HS1)
,
- (HS2)
.
Then, we call (abbr. ) a Hom–Hopf algebra.
The antipode S is both a Hom-anti-algebra morphism and a Hom-anti-coalgebra morphism. A morphism between two Hom-bialgebras is called a Hom-bialgebra morphism if it is both a Hom-algebra and a Hom-coalgebra morphism. A morphism between two Hom–Hopf algebras is called a Hom–Hopf algebra morphism if it is a Hom-bialgebra morphism and compatible with the antipodes. A useful remark is that Hom-bialgebra morphisms of Hom–Hopf algebras are Hom–Hopf algebra morphisms.
Definition 4 ([
32]).
Let be a Hom-algebra. A left -Hom-module is a triple , where X is a -linear space, (write , ) is a linear map, and α is an automorphism of X, such that- (HM1)
,
- (HM2)
,
- (HM3)
are satisfied for and .
Let and be two left -Hom-modules. Then, a morphism is called a -Hom-module morphism if and .
Definition 5 ([
32]).
Let be a Hom-coalgebra. A left -Hom-comodule is a triple , where X is a -linear space, (write , ) is a linear map, and α is an automorphism of X, such that- (HCM1)
,
- (HCM2)
,
- (HCM3)
are satisfied for .
Let and be two left -Hom-comodules. Then, a morphism is called a -Hom-comodule morphism if and .
Definition 6 ([
33]).
Let be a Hom-bialgebra and be a Hom-algebra. If is a left -Hom-module, and for all and ,- (HMA1)
,
- (HMA2)
,
then, is called a left -Hom-module Hom-algebra.
Definition 7 ([
25,
34]).
Let be a Hom-bialgebra and a Hom-algebra. If is a left -Hom-comodule and for all ,- (HCMA1)
,
- (HCMA2)
,
then, is called a left -Hom-comodule Hom-algebra.
Definition 8 ([
25]).
Let be a Hom-bialgebra and be a Hom-coalgebra. If is a left -Hom-module and for all and ,- (HMC1)
,
- (HMC2)
,
then is called a left -Hom-module Hom-coalgebra.
Definition 9 ([
25]).
Let be a Hom-bialgebra and be a Hom-coalgebra. If is a left -Hom-comodule and for all ,- (HCMC1)
,
- (HCMC2)
,
then is called a left -Hom-comodule Hom-coalgebra.
Now, we recall the definition of
-twisted-Radford’s Hom-biproduct from [
26]. First, if
is a Hom-bialgebra, we denote the group of all Hom-bialgebra automorphisms of
H by
.
Definition 10 ([
26]).
Let be a Hom-bialgebra, and be a left -Hom-module Hom-algebra. A θ-twisted smash Hom-product of and is defined as follows. For all :- (i)
as a vector space, ,
- (ii)
the multiplication is given by
The θ-twisted smash Hom-product is a Hom-algebra with the unit .
Definition 11 ([
26]).
Let be a Hom-bialgebra, and be a left -Hom-comodule Hom-coalgebra. A ω-twisted smash Hom-coproduct of and is defined as follows. For all :- (i)
as a vector space, ,
- (ii)
the comultiplication is given by
The ω-twisted smash Hom-coproduct is a Hom-coalgebra with the counit .
Theorem 1 ([
26]).
Let be a Hom-bialgebra, be a left -Hom-module Hom-algebra with Hom-module structure ▹, and a left -Hom-comodule Hom-coalgebra with Hom-comodule structure ρ, and , be a θ-twisted smash Hom-product, and be a ω-twisted smash Hom-coproduct. Then, the following two conditions are equivalent:- (i)
is a Hom-bialgebra.
- (ii)
The following conditions hold:
- (T1)
is a Hom-algebra map and ,
- (T2)
is a left -Hom-module Hom-coalgebra,
- (T3)
is a left -Hom-comodule Hom-algebra,
- (T4)
,
- (T5)
,
for all and .
Theorem 2 ([
26]).
Suppose is a -twisted-Radford’s Hom-biproduct. If H is a Hom–Hopf algebra with antipode and is a linear map, such that is a convolution inverse of and , then is a Hom–Hopf algebra with antipode S described by The -twisted-Radford’s Hom-biproduct is a Hom–Hopf algebra and is denoted by , for short, and write elements .
3. Factorization of Certain Twisted Hom-Biproduct Endomorphisms
Let be a -twisted-Radford’s Hom-biproduct. We define by for all and by for all , are Hom–Hopf algebra maps which satisfy . Let be the set of all Hom–Hopf algebra endomorphisms F of such that and let be its set of units, which is the group of Hom–Hopf algebra automorphisms F of such that under composition. The purpose of this section is to show that F has a factorization closely related to the factors and of .
We define
and
by
, for all
and
, for all
, respectively. There is a fundamental relationship between these four maps given by:
The factorization of
F is given in terms of
and
defined by:
First, we shall reveal the relationships among F, and in the following lemma.
Lemma 1. Let . Then:for all and . Proof. According to the definition of these four maps and
F, we have
and
Next, we need to calculate
. For
and
, we use (
4) to compute
It follows that
for
and
. Equations (
7) and (
8) follow from the above equation via setting
and
respectively. As for (
9), we calculate
as desired. □
By (
7) and (
8) of the above Lemma:
Since
by Equation (
7). By (
9) of Lemma 1:
for all
. Now, we can calculate the factors of a composite.
Corollary 1. Let . Then:
- (i)
,
- (ii)
.
Proof. (i) For any
, by (
7) of Lemma 1, we have
(ii) For all
. Using Equation (
11), the fact
F is multiplicative and Equation (
7), we obtain that
i.e.,
Applying to both sides of the above equation, and replacing h with , we can obtain part (ii). □
According to Lemma 1, in order to characterize F, we must characterize and . First, we shall characterize in the following lemma.
Lemma 2. Let . Then:
- (i)
is a Hom-algebra endomorphism.
- (ii)
.
- (iii)
- (iv)
- (v)
For all and ,
Proof. (i) We need to check three aspects. From the above discussion, we have known that
and
have checked in Lemma 1. Finally, we shall check that
preserves the multiplication. In fact, for
, we have
It is easy to check part (ii), since , and F is a Hom-coalgebra morphism.
Next, we will check that parts (iii) and (iv) hold. As a matter of fact, we compute the coproduct of
in two ways. Firstly, we have
and secondly, since
F is a Hom-coalgebra map, we have
Applying
to both sides of the above equation yields (
12). It follows easily that
from (
11). Applying
to both sides of the above equation again, we can gain
and applying
to both sides, we obtain (
13).
(v) Finally, it is left to us to check (
14). Indeed, for
and
, we have
On the other hand, Since
F preserves the multiplication, we compute:
Applying
to the above equation and replacing
a with
, we obtain (
14). □
As the reader might suspect, whether or not is a Hom-coalgebra morphism is explained in terms of .
Corollary 2. Let . Then, is a Hom-coalgebra morphism if, and only if, , for any .
Proof. Suppose
, for any
. Then, we have
Conversely, suppose that
is a Hom-coalgebra morphism. For all
, we compute
as desired. □
From Lemma 2, we have characterized the conditions that satisfies. Pay special attention to its part (v) describing a commutation relation between and . It is left to us to characterize as follows.
Lemma 3. Let . Then:
- (i)
.
- (ii)
- (iii)
is a Hom-coalgebra morphism.
- (iv)
Proof. (i) By (
11), we have
for
. Since
F is a Hom-algebra morphism and
is an automorphism,
, which implies
.
(ii) For
, we have
and on the other hand,
Applying
to both expressions for
, it follows that (
15) holds.
To prove parts (iii) and (iv), we calculate
, for all
, in two ways as follows.
Applying to the expressions for gives . since and have been discussed before, and hence, part (iii) holds.
Applying
to the expressions for
again yields
Therefore,
it follows that
Applying to both sides of the equation gives part (iv). □
Corollary 3. Let . Then, is a left -Hom-module morphism if and only if the condition holds for all and .
Proof. The necessary condition can be followed easily from (
14) of Lemma 2. Now, we shall prove a sufficient part. Suppose that the condition holds. Note that
is a Hom-coalgebra morphism by Lemma 3(iii). Using this fact and (
14), for all
and
, we have
and since
, we show that
is a left
-Hom-module morphism. □
Lemma 4. Let . Then, is a Hom-algebra morphism if, and only if, for all .
Proof. Suppose that
is a Hom-algebra morphism. Using (ii) and (iii) of Lemma 3, for
, we have
i.e.,
. Replacing
h by
yields the condition.
Conversely, if
holds, by using Equation (
15), we have
The proof is completed. □
Corollary 4. Let . Then, has a convolution inverse defined by for .
Proof. Let
. Then, by parts (i) and (ii) of Lemma 3, we have
and using the fact that
is a left
-Hom-module algebra, we have
The proof is completed. □
Using the above lemmas and corollaries that we have, we can gain the main result.
Theorem 3. Let be the -twisted-Radford’s Hom-biproduct, let be the projection from onto H, and let be the set of pairs , where , are morphisms which satisfy the conclusions of Lemmas 2 and 3 for and , respectively. Then,
- (i)
The function , described by , where , for all and , is a bijection. Furthermore, and .
- (ii)
Suppose , then, if, and only if, L is a bijection.
Proof. (i) Assume that the function is well-defined. We define
by
. It is easily proven that
and
are mutually inverse; see Equations (
5) and (
9). According to the previous results, to complete the proof of part (a), we only need to prove that elements of
give rise to elements of
, as indicated. As the reader might suspect, the proof that
is somewhat tedious. We will use Lemmas 1–3, which do not need special citations initially.
It is easy to see that
. Obviously,
. Note that
and
for
and
, which means that
.
Let
and
. Then,
Therefore, F is a Hom-algebra morphism.
Next, we shall check that
. Indeed, for all
,
,
We have shown that F preserves comultiplication. Therefore, F is a Hom-coalgebra morphism, and hence, is a Hom-bialgebra morphism. Since Hom-bialgebra morphisms of Hom–Hopf algebras are Hom-Hopf algebra morphisms, this completes the proof of part (i).
(ii) Suppose
. Then,
and
are inverse by (
10) and Corollary 1(i). Thus,
is bijective and
.
Conversely, suppose that
and
are bijective. Set
. From Corollary 4,
has a convolution inverse
. Set
and define
for all
and
. Since
is a Hom-algebra morphism, we compute
and
Thus, we have shown that . □
Now, let denote the set of , such that L is bijective. Then, the correspondence of the above theorem induces a bijection .
When , the structure of F is particularly simple.
Proposition 1. Let . Then, the following conditions are equivalent:
- (i)
.
- (ii)
for all .
- (iii)
for all .
- (iv)
.
- (v)
for all and .
- (vi)
for all .
Proof. Firstly, we prove that part (i) implies part (ii) implies part (iii). Suppose part (i) holds. Then, we have
by Equation (
11), and hence,
for all
, because
. Secondly, it is easy to see that parts (iii) and (iv) are equivalent. Thirdly, part (iv) implies part (v) by Equation (
9), and part (v) implies part (vi) as
. At last, part (vi) trivially implies part (i). □
Let
be a Hom-algebra and
be a Hom-coalgebra over
. Let
be the set of
-linear maps
satisfying
. The group
acts on the (non-associative) convolution algebra
by
for all
and
. Obviously, this action satisfies:
for all
and
. Then,
is a Hom-algebra with the unit
under the convolution *. Indeed, for
and
,
and
Thus, it follows that and . Moreover, can be checked similarly.
It is easy to check that
is an ordinary associative algebra. Let
be the group of units of the algebra
. Then
; thus, there is a group homomorphism
given by
for all
and
. The resulting group
has product given by
Note that the action of on by group homomorphisms is also one on . For a group G, the group is the group whose underlying set is G and the product is given by for all . Therefore, we have the following result of Theorem 3 and Corollary 1.
Theorem 4. Let be the -twisted-Radford’s Hom-biproduct, let be the projection from onto H. Then, there is an injective group homomorphism , which is given by for all .
Proof. For
, it is easy to see that
and
are bijective by Theorem 3(ii) and Equation (
11). If
, then, using (
9), we have
for all
and
. Thus, it follows that
, i.e.,
is injective.
For
, since
it follows that
preserves the multiplication. Thus,
is an injective group homomorphism. □
In conclusion, by Theorem 3, we obtain the characterization of the automorphism group of a -twisted-Radford’s Hom-biproduct via the automorphisms of the factors and . Furthermore, by Theorem 4, the automorphism group can be regarded as a subgroup of some semidirect product , when is a projection. These are important for the study of automorphism groups of twisted-Radford’s Hom-biproducts.
4. Example
In this section, we consider an example in [
26]. Let
and the automorphism
via
and
. Then,
is both a Hom-algebra with product:
,
,
, and a Hom-coalgebra with coproduct and counit given by
We define by , , which is the convolution inverse of .
Let
be the group Hopf algebra with
,
,
and
. Define an automorphism
by
,
. Then, we can define a new product
and a new coproduct
on
H. So, we obtain a Hom–Hopf algebra
. Then,
is a left
-Hom-module Hom-algebra and Hom-module Hom-coalgebra with the action
given by
Furthermore,
is a left
-Hom-comodule Hom-algebra and Hom-comodule Hom-coalgebra with the coaction
given by
Then, we can obtain many twisted Radford’s Hom-biproducts, such as -twisted Radford’s Hom-biproduct, -twisted Radford’s Hom-biproduct, -twisted Radford’s Hom-biproduct, -twisted Radford’s Hom-biproduct, and so on.
Now, we clearly write the structures of the
-twisted Radford’s Hom-biproduct
and characterize the automorphisms of it. Its product is defined as follows:
Its coproduct and counit are defined as follows:
Its antipodes are as follows:
Now, we compute the morphisms
satisfying the conclusions of Lemma 2. This is taking a base of
,
, given respectively by
Let , where . If L satisfies Lemma 2(ii), it follows that and . By part (iv) of Lemma 2, we can obtain . Therefore, for some . So, there is a bijection between the set of the morphisms satisfying the conclusions of Lemma 2 and the set .
Now, we will discuss the morphisms of
which satisfy Lemma 3 in a similar way as above. Taking a base of
,
, given respectively by
Let
, where
,
. Since
R satisfies part (i) of Lemma 3, we have
and
. By Equation (
16), it follows that
and
. Thus,
. Next, we shall check part (ii) of Lemma 3, and we obtain
and
. Therefore,
, i.e.,
,
,
. Hence,
. We can obtain the concrete characterization of
. Let
. By Theorem 3, we have
where
.