Generalized derivations of Hom-Jordan algebras
Chenrui Yao, Yao Ma, Liangyun Chen
( School of Mathematics and Statistics, Northeast Normal University,
Changchun 130024, CHINA
)
Abstract
In this paper, we give some properties of generalized derivation algebras of Hom-Jordan algebras. In particular, we show that GDer(V)=QDer(V)+QC(V), the sum of the quasiderivation algebra and the quasicentroid. We also prove that QDer(V) can be embedded as derivations into a larger Hom-Jordan algebra. General results on centroids of Hom-Jordan algebras are also developed in this paper.
Keywords: Generalized derivations, quasiderivations, centroids, Hom-Jordan algebras.
2010 Mathematics Subject Classification: 17B40, 17C10, 17C99.
000 Corresponding author(L. Chen): [email protected].
000Supported by NNSF of China (Nos. 11771069 and 11801066), NSF of Jilin province (No. 20170101048JC), the project of Jilin province department of education (No. JJKH20180005K) and the Fundamental Research Funds for the Central Universities(No. 130014801).
1 Introduction
In 1932, the physicist P. Jordan proposed a program to discover a new algebraic setting for quantum mechanics, so Jordan algebras were created in this proceeding. Moreover, Jordan algebras were truned out to have illuminating connections with many areas of mathematics. A. A. Albert renamed them Jordan algebras and developed a successful structure theory for Jordan algebras over any field of characteristic zero in [1].
Hom-type algebras have been studied by many authors in [2], [3], [8], [5]. In general, Hom-type algebras are a kind of algebras which are obtained by twisting the identity defining the structure with one or several linear maps(called the twisted maps) on the basis of the original algebras. The concept of Hom-Jordan algebras was first introduced by A. Makhlouf in his paper [9]. He introduced a Hom-Jordan algebra and showed that it fits with the Hom-associative structure, that is a Hom-associative algebra leads to a Hom-Jordan algebra by considering a plus algebra.
As is well known, derivations and generalized derivations play an important role in the research of structure and property in algebraic system. The research on generalized derivations was started by Leger and Luks in [7]. They gave many properties about generalized derivations of Lie algebras. From then on, many authors generalized the results in [7] to other algebras. R. X. Zhang and Y. Z. Zhang generalized it to Lie superalgebras in [13]. And L. Y. Chen, Y. Ma and L. Ni generalized the results to Lie color algebras successfully in [4]. J. Zhou, Y. J. Niu and L. Y. chen generalized the above results to Hom-Lie algebras in [14]. As for Jordan algebras, A. I. Shestakov generalized the results into Jordan superalgebras in [12]. He showed that for semi-simple Jordan algebras over any field of characteristic [math] or simple Jordan algebras over any field of characteristic other than 2, the generalized derivations are the sum of the derivations and the centroids with some exceptions.
The purpose of this paper is to generalized some beautiful results in [14] and [11] to Hom-Jordan algebras. We proceed as follow. In Section 2, we recall some basic definitions and propositions which will be used in what follows. In Section 3, we’ll give some properties about generalized derivations of Hom-Jordan algebras and their Hom-subalgebras. In Section 4, we show that quasiderivations of a Hom-Jordan algebra can be embedded as derivations into a larger Hom-Jordan algebra and obtain a direct sum decomposition of Der(V˘) when the centralizer of V equals to zero. In Section 5, we give some propositions with respect to the centroids of Hom-Jordan algebras.
2 Preliminaries
Definition 2.1**.**
[10]**
An algebra J over a field F is a Jordan algebra satisfying for any x,y∈J,
- (1)
x∘y=y∘x;
2. (2)
(x2∘y)∘x=x2∘(y∘x).
Definition 2.2**.**
[9]**
A Hom-Jordan algebra over a field F is a triple (V,μ,α) consisting of a linear space V, a bilinear map μ:V×V→V which is commutative and a linear map α:V→V satisfying for any x,y∈V,
[TABLE]
where α2=α∘α.
Definition 2.3**.**
A Hom-Jordan algebra (V,μ,α) is called multiplicative if for any x,y∈V, α(μ(x,y))=μ(α(x),α(y)).
Definition 2.4**.**
[6]**
A subspace W⊆V is a Hom-subalgebra of (V,μ,α) if α(W)⊆W and
[TABLE]
Definition 2.5**.**
[6]**
A subspace W⊆V is a Hom-ideal of (V,μ,α) if α(W)⊆W and
[TABLE]
Definition 2.6**.**
[6]**
Let (V,μ,α) and (V′,μ′,β) be two Hom-Jordan algebras. A linear map ϕ:V→V′ is said to be a homomorphism of Hom-Jordan algebras if
- (1)
ϕ(μ(x,y))=μ′(ϕ(x),ϕ(y));
2. (2)
ϕ∘α=β∘ϕ.
Lemma 2.7**.**
[6]**
Let (V,μ,α) be a Hom-Jordan algebra over a field F, we define a subspace W of End(V) where W={w∈End(V)∣w∘α=α∘w}, σ:W→W is a linear map satisfying σ(w)=α∘w.
- (1)
A triple (W,ν,σ), where the multiplication ν:W×W→W is defined for w1,w2∈W by
[TABLE]
is a Hom-Jordan algebra over F.
2. (2)
A triple (W,ν′,σ), where the multiplication ν′:W×W→W is defined for w1,w2∈W by
[TABLE]
is a Hom-Lie algebra over F.
Definition 2.8**.**
[6]**
For any nonnegative integer k, a linear map D:V→V is called an αk-derivation of the Hom-Jordan algebra (V,μ,α), if
- (1)
D∘α=α∘D;
2. (2)
D(μ(x,y))=μ(D(x),αk(y))+μ(αk(x),D(y)),∀x,y∈V.
Lemma 2.9**.**
[6]**
For any D∈Derαk(V) and D′∈Derαs(V), define their commutator ν′(D,D′) as usual:
[TABLE]
Then ν′(D,D′)∈Derαk+s(V).
Let (V,μ,α) be a multiplicative Hom-Jordan algebra. We denote the set of all αk-derivations by Derαk(V), then Der(V)=∔k≥0Derαk(V) provided with the commutator and the following map
[TABLE]
is a Hom-subalgebra of (W,ν′,σ) according to Lemma 2.9 and is called the derivation algebra of (V,μ,α).
Definition 2.10**.**
For any nonnegative integer k, a linear map D∈End(V) is said to be a generalized αk-derivation of (V,μ,α), if there exist two linear maps D′,D′′∈End(V) such that
- (1)
D∘α=α∘D,D′∘α=α∘D′,D′′∘α=α∘D′′;
2. (2)
μ(D(x),αk(y))+μ(αk(x),D′(y))=D′′(μ(x,y)),∀x,y∈V.
Definition 2.11**.**
For any nonnegative integer k, a linear map D∈End(V) is said to be an αk-quasiderivation of (V,μ,α), if there exist a linear map D′∈End(V) such that
- (1)
D∘α=α∘D,D′∘α=α∘D′;
2. (2)
μ(D(x),αk(y))+μ(αk(x),D(y))=D′(μ(x,y)),∀x,y∈V.
Let GDerαk(V) and QDerαk(V) be the sets of generalized αk-derivations and of αk-quasiderivations, respectively. That is
[TABLE]
It’s easy to verify that both GDer(V) and QDer(V) are Hom-subalgebras of (W,ν′,σ)(See Proposition 3.1).
Definition 2.12**.**
For any nonnegative integer k, a linear map D∈End(V) is said to be an αk-centroid of (V,μ,α), if
- (1)
D∘α=α∘D;
2. (2)
μ(D(x),αk(y))=μ(αk(x),D(y))=D(μ(x,y)),∀x,y∈V.
Definition 2.13**.**
For any nonnegative integer k, a linear map D∈End(V) is said to be an αk-quasicentroid of (V,μ,α), if
- (1)
D∘α=α∘D;
2. (2)
μ(D(x),αk(y))=μ(αk(x),D(y)),∀x,y∈V.
Let Cαk(V) and QCαk(V) be the sets of αk-centroids and of αk-quasicentroids, respectively. That is
[TABLE]
Definition 2.14**.**
For any nonnegative integer k, a linear map D∈End(V) is said to be an αk-central derivation of (V,μ,α), if
- (1)
D∘α=α∘D;
2. (2)
μ(D(x),αk(y))=D(μ(x,y))=0,∀x,y∈V.
We denote the set of all αk-central derivations by ZDerαk(V), then ZDer(V)=∔k≥0ZDerαk(V).
According to the definitions, it’s easy to show that ZDer(V)⊆Der(V)⊆QDer(V)⊆GDer(V)⊆End(V).
Definition 2.15**.**
Let (V,μ,α) be a Hom-Jordan algebra. If Z(V)={x∈V∣μ(x,y)=0,∀y∈V}, then Z(V) is called the centralizer of (V,μ,α).
3 Generalized derivation algebras and their subalgebras
At first, we give some basic properties of center derivation algebras, quasiderivation algebras and generalized derivation algebras of a Hom-Jordan algebra.
Proposition 3.1**.**
Suppose that (V,μ,α) is a multiplicative Hom-Jordan algebra. Then the following statements hold:
- (1)
GDer(V), QDer(V) and C(V) are Hom-subalgebras of (W,ν′,σ);
2. (2)
ZDer(V)* is a Hom-ideal of Der(V).*
Proof. .
(1). Suppose that D1∈GDerαk(V), D2∈GDerαs(V). Then for any x,y∈V,
[TABLE]
Since σ(D1′′),σ(D1′)∈End(V), we have σ(D1)∈GDerαk+1(V).
[TABLE]
Since ν′(D1′′,D2′′),ν′(D1′,D2′)∈End(V), we have ν′(D1,D2)∈GDerαk+s(V).
Therefore, GDer(V) is a Hom-subalgebra of (W,ν′,σ).
Similarly, we have QDer(V) is a Hom-subalgebra of (W,ν′,σ).
Suppose that D1∈Cαk(V), D2∈Cαs(V). Then for any x,y∈V,
[TABLE]
Similarly, we have σ(D1)(μ(x,y))=μ(αk+1(x),σ(D1)(y)). Hence, we have σ(D1)∈Cαk+1(V).
[TABLE]
Similarly, we have ν′(D1,D2)(μ(x,y))=μ(αk+s(x),ν′(D1,D2)(y)).
Hence, we have ν′(D1,D2)∈Cαk+s(V).
Therefore, we have C(V) is a Hom-subalgebras of (W,ν′,σ).
(2). Suppose that D1∈ZDerαk(V), D2∈Derαs(V). Then for any x,y∈V,
[TABLE]
[TABLE]
Hence, σ(D1)∈ZDerαk+1(V).
[TABLE]
[TABLE]
Hence, we have ν′(D1,D2)∈ZDerαk+s(V).
Therefore, ZDer(V) is a Hom-ideal of Der(V).
∎
Proposition 3.2**.**
Let (V,μ,α) be a multiplicative Hom-Jordan algebra, then
- (1)
ν′(Der(V),C(V))⊆C(V);
2. (2)
ν′(QDer(V),QC(V))⊆QC(V);
3. (3)
ν′(QC(V),QC(V))⊆QDer(V);
4. (4)
C(V)⊆QDer(V);
5. (5)
QDer(V)+QC(V)⊆GDer(V);
6. (6)
C(V)∘Der(V)⊆Der(V).
Proof. .
(1). Suppose that D1∈Derαk(V), D2∈Cαs(V). Then for any x,y∈V, we have
[TABLE]
Similarly, we have ν′(D1,D2)(μ(x,y))=μ(αk+s(x),ν′(D1,D2)(y)). Hence, ν′(D1,D2)∈Cαk+s(V). Therefore ν′(Der(V),C(V))⊆C(V).
(2). Suppose that D1∈QDerαk(V), D2∈QCαs(V). Then for any x,y∈V, we have
[TABLE]
Hence, we have ν′(D1,D2)∈QCαk+s(V). So ν′(QDer(V),QC(V))⊆QC(V).
(3). Suppose that D1∈QCαk(V), D2∈QCαs(V). Then for any x,y∈V, we have
[TABLE]
i.e., μ(ν′(D1,D2)(x),αk+s(y))+μ(αk+s(x),ν′(D1,D2)(y))=0.
Hence, we have ν′(D1,D2)∈QDerαk+s(V), which implies that ν′(QC(V),QC(V))⊆QDer(V).
(4). Suppose that D∈Cαk(V). Then for any x,y∈V, we have
[TABLE]
Hence, we have
[TABLE]
which implies that D∈QDerαk(V). So C(V)⊆QDer(V).
(5). Suppose that D1∈QDerαk(V), D2∈QCαk(V). Then for any x,y∈V, we have
[TABLE]
Since D1′, D1−D2∈End(V), we have D1+D2∈GDerαk(V). So QDer(V)+QC(V)⊆GDer(V).
(6). Suppose that D1∈Cαk(V), D2∈Derαs(V). Then for any x,y∈V, we have
[TABLE]
which implies that D1∘D2∈Derαk+s(V). So C(V)∘Der(V)⊆Der(V).
∎
Theorem 3.3**.**
Suppose that (V,μ,α) is a multiplicative Hom-Jordan algebra. Then GDer(V)=QDer(V)+QC(V).
Proof. .
According to Proposition 3.2 (5), we need only to show that GDer(V)⊆QDer(V)+QC(V).
Suppose that D∈GDerαk(V). Then there exist D′,D′′∈End(V) such that
[TABLE]
Since μ is commutative, we have
[TABLE]
which implies that D′∈GDerαk(V).
Moreover, we have
[TABLE]
which implies 2D+D′∈QDerαk(V).
[TABLE]
which implies that 2D−D′∈QCαk(V).
Hence, D=2D+D′+2D−D′∈QDer(V)+QC(V), i.e., GDer(V)⊆QDer(V)+QC(V).
Therefore, GDer(V)=QDer(V)+QC(V).
∎
Proposition 3.4**.**
Suppose that (V,μ,α) is a multiplicative Hom-Jordan algebra where V can be decomposed into the direct sum of two Hom-ideals, i.e., V=V1⊕V2 and α is surjective. Then we have
- (1)
Z(V)=Z(V1)⊕Z(V2);
2. (2)
If Z(V)={0}, then we have
- (a)
Der(V)=Der(V1)⊕Der(V2);
2. (b)
GDer(V)=GDer(V1)⊕GDer(V2);
3. (c)
QDer(V)=QDer(V1)⊕QDer(V2);
4. (d)
C(V)=C(V1)⊕C(V2).
Proof. .
(1). Obviously, Z(V1)∩Z(V2)={0}. And it’s easy to show that Z(Vi)(i=1,2) are Hom-ideals of Z(V). For all zi∈Z(Vi)(i=1,2), take x=x1+x2 where x1∈J1, x2∈J2. We have
[TABLE]
since zi∈Z(Vi)(i=1,2), we have
[TABLE]
Hence,
[TABLE]
which implies that z1+z2∈Z(V), i.e., Z(V1)⊕Z(V2)⊆Z(V).
On the other hand, for all a∈Z(V), suppose that a=a1+a2 where ai∈Vi(i=1,2). Then for all x1∈V1,
[TABLE]
since a∈Z(V), we have
[TABLE]
Hence,
[TABLE]
which implies that a1∈Z(V1). Similarly, we have a2∈Z(V2). Hence, Z(V)⊆Z(V1)⊕Z(V2).
Therefore, we have Z(V)=Z(V1)⊕Z(V2).
(2). Step1. We’ll show that ∀i=1,2, D(Vi)⊆Vi,∀D∈Derαk(V)(k≥0).
Suppose that xi∈Vi, then
[TABLE]
since V1, V2 are Hom-ideals of (V,μ,α).
Suppose that D(x1)=u1+u2 where u1∈V1, u2∈V2. Then
[TABLE]
which implies that u2∈Z(V2) since α is surjective. Note that Z(V)={0}, we have Z(Vi)={0}(i=1,2). Hence, u2=0. That is to say D(x1)∈V1. Similarly, we have D(x2)∈V2.
Step2. We’ll show that Derαk(V1)∔Derαk(V2)⊆Derαk(V)(k≥0).
For any D∈Derαk(V1), we extend it to a linear map on V as follow
[TABLE]
Then for any x,y∈V, suppose that x=x1+x2, y=y1+y2∈V, where x1,y1∈V1, x2,y2∈V2, we have
[TABLE]
[TABLE]
since D∈Derαk(V1),
[TABLE]
we have
[TABLE]
which implies that D∈Derαk(V), i.e., Derαk(V1)⊆Derαk(V). Moreover, D∈Derαk(V1) if and only if D(x2)=0,∀x2∈V2.
Similarly, we have Derαk(V2)⊆Derαk(V) and D∈Derαk(V2) if and only if D(x1)=0,∀x1∈V1.
Then we have Derαk(V1)+Derαk(V2)⊆Derαk(V) and Derαk(V1)∩Derαk(V2)={0}. Hence, Derαk(V1)∔Derαk(V2)⊆Derαk(V).
Step3. We’ll prove that Derαk(V1)∔Derαk(V2)=Derαk(V).
Suppose that D∈Derαk(V). Set x=x1+x2,xi∈Vi. Define D1,D2 as follows
[TABLE]
Obviously, D=D1+D2.
For any u1,v1∈V1,
[TABLE]
Hence, D1∈Derαk(V1). Similarly, D2∈Derαk(V2).
Therefore, Derαk(V1)∔Derαk(V2)=Derαk(V) as a vector space.
Hence, we have
[TABLE]
Step4. We’ll show that Der(Vi)(i=1,2) are Hom-ideals of Der(V).
Suppose that D1∈Derαk(V1), D2∈Derαs(V). Then for any x1,y1∈V1, we have
[TABLE]
which implies that σ(D1)∈Derαk+1(V1).
[TABLE]
which implies that ν′(D1,D2)∈Derαk+s(V1). So Der(V1) is a Hom-ideal of Der(V).
Similarly, we have Der(V2) is a Hom-ideal of Der(V).
Therefore, Der(V)=Der(V1)⊕Der(V2).
(3), (4), (5) similar to the proof of (2).
∎
Theorem 3.5**.**
Let (V,μ,α) be a multiplicative Hom-Jordan algebra, α a surjection and Z(V) the centralizer of (V,μ,α). Then ν′(C(V),QC(V))⊆End(V,Z(V)). Moreover, if Z(V)={0}, then ν′(C(V),QC(V))={0}.
Proof. .
Suppose that D1∈Cαk(V), D2∈QCαs(V) and x∈V. Since α is surjective, there exists y′∈V such that y=αk+s(y′) for any y∈V.
[TABLE]
which implies that ν′(D1,D2)(x)∈Z(V), i.e., ν′(C(V),QC(V))⊆End(V,Z(V)).
Furthermore, if Z(V)={0}, it’s clearly that ν′(C(V),QC(V))={0}.
∎
Theorem 3.6**.**
Let (V,μ,α) be a multiplicative Hom-Jordan algebra. Then ZDer(V)=C(V)∩Der(V).
Proof. .
Assume that D∈C(V)αk∩Derαk(V). Then for any x,y∈V, we have
[TABLE]
since D∈Derαk(V).
[TABLE]
since D∈Cαk(V).
Hence we have D(μ(x,y))=μ(D(x),αk(y))=0, which implies that D∈ZDerαk(V). Therefore, C(V)∩Der(V)⊆ZDer(V).
On the other hand, assume that D∈ZDerαk(V). Then for any x,y∈V, we have
[TABLE]
hence we have
[TABLE]
Therefore
[TABLE]
which implies that D∈Derαk(V).
And
[TABLE]
which implies that D∈Cαk(V).
Therefore D∈C(V)αk∩Derαk(V), i.e., ZDer(V)⊆C(V)∩Der(V).
Hence, we have ZDer(V)=C(V)∩Der(V).
∎
Theorem 3.7**.**
Let (V,μ,α) be a multiplicative Hom-Jordan algebra. Then (QC(V),ν,σ) is a Hom-Jordan algebra.
Proof. .
According to Lemma 2.7 (1), we need only to show that QC(V) is a Hom-subalgebra of (W,ν,σ).
Suppose that D1∈QCαk(V), D2∈QCαs(V). Then for any x,y∈V, we have
[TABLE]
which implies that σ(D1)∈QCαk+1(V).
[TABLE]
which implies that ν(D1,D2)∈QCαk+s(V). Therefore, QC(V) is a Hom-subalgebra of (W,ν,σ), i.e., (QC(V),ν,σ) is a Hom-Jordan algebra.
∎
Theorem 3.8**.**
Suppose that (V,μ,α) be a multiplicative Hom-Jordan algebra over a field F of characteristic other than 2.
- (1)
(QC(V),ν′,σ)* is a Hom-Lie algebra if and only if (QC(V),ι,σ) is a Hom-associative algebra where ι(D1,D2)=D1∘D2,∀D1,D2∈QC(V).*
2. (2)
If α is a surjection and Z(V)={0}, then (QC(V),ν′,σ) is a Hom-Lie algebra if and only if ν′(QC(V),QC(V))={0}.
Proof. .
(1). (⇐). For all D1,D2∈QC(V), we have D1∘D2∈QC(V), D2∘D1∈QC(V). So ν′(D1,D2)∈QC(V). Moreover, σ(D1)∈QC(V). Therefore, (QC(V),ν′,σ) is a Hom-Lie algebra.
(⇒). For all D1,D2∈QC(V), we have ι(D1,D2)=21(ν′(D1,D2)+ν(D1,D2)). According to Theorem 3.7, we have ν(D1,D2)∈QC(V). Note that ν′(D1,D2)∈QC(V), we have ι(D1,D2)∈QC(V). Hence, (QC(V),ι,σ) is a Hom-associative algebra.
(2). (⇒). Suppose that D1∈QCαk(V), D2∈QCαs(V) and x∈V. Since α is a surjection, there exists y′∈V such that y=αk+s(y′) for any y∈V. Note that (QC(V),ν′,σ) is a Hom-Lie algebra, then ν′(D1,D2)∈QCαk+s(V). Then
[TABLE]
Note that
[TABLE]
Hence, we have μ(ν′(D1,D2)(x),y)=μ(ν′(D1,D2)(x),αk+s(y′))=0, which implies that ν′(D1,D2)(x)∈Z(V). Note that Z(V)={0}, we have ν′(D1,D2)(x)=0, i.e., ν′(D1,D2)=0. Therefore, ν′(QC(V),QC(V))={0}.
(⇐). Obviously.
∎
4 The quasiderivations of Hom-Jordan algebras
In this section, we will prove that the quasiderivations of (V,μ,α) can be embedded as derivations in a larger Hom-Jordan agebra and obtain a direct sum decomposition of Der(V˘) when the centralizer of (V,μ,α) is equal to zero.
Proposition 4.1**.**
Suppose that (V,μ,α) is a Hom-Jordan algebra over F and t is an indeterminate. We define V˘:={∑(x⊗t+y⊗t2)∣x,y∈V}, α˘(V˘):={∑(α(x)⊗t+α(y)⊗t2)∣x,y∈V} and μ˘(x⊗ti,y⊗tj):=μ(x,y)⊗ti+j where x,y∈V,i,j∈{1,2}. Then (V˘,μ˘,α˘) is a Hom-Jordan algebra.
Proof. .
It’s obvious that μ˘ is a bilinear map and commutative since μ is a bilinear map and commutative.
For any x⊗ti,y⊗tj∈V˘, we have
[TABLE]
Therefore, (V˘,μ˘,α˘) is a Hom-Jordan algebra.
∎
For convenience, we write xt(xt2) in place of x⊗t(x⊗t2).
If U is a subspace of V such that V=U∔μ(V,V), then
[TABLE]
Now we define a map φ:QDer(V)→End(V˘) satisfying
[TABLE]
where D∈QDerαk(V), and D′ is in Definition 2.11 (2), a∈V,u∈U,b∈μ(V,V).
Proposition 4.2**.**
V,V˘,φ* are defined as above. Then*
- (1)
φ* is injective and φ(D) does not depend on the choice of D′.*
2. (2)
φ(QDer(V))⊆Der(V˘).
Proof. .
(1). If φ(D1)=φ(D2), then for all a∈V,u∈U,b∈μ(V,V), we have
[TABLE]
which implies that
[TABLE]
Hence we have
[TABLE]
which implies that D1=D2. Therefore, φ is injective.
Suppose that there exists D′′ such that φ(D)(at+ut2+bt2)=D(a)t+D′′(b)t2 and μ(D(x),αk(y))+μ(αk(x),D(y))=D′′(μ(x,y)), then we have
[TABLE]
which implies that D′′(b)=D′(b).
Hence, we have
[TABLE]
That is to say φ(D) does not depend on the choice of D′.
(2). ∀i+j≥3, we have μ˘(x⊗ti,y⊗tj)=μ(x,y)⊗ti+j=0. Hence, we need only to show that
[TABLE]
For all x,y∈V, we have
[TABLE]
Hence, φ(D)∈Derα˘k(V˘). Therefore, φ(QDer(V))⊆Der(V˘).
∎
Proposition 4.3**.**
Let (V,μ,α) be a Hom-Jordan algebra. Z(V)={0} and V˘,φ are defined as above. Then Der(V˘)=φ(QDer(V))∔ZDer(V˘).
Proof. .
Assume that xt+yt2∈Z(V˘), then for any x′t+y′t2∈V˘, we have
[TABLE]
which implies that μ(x,x′)=0. Note that Z(V)={0}, we have x=0. Hence, Z(V˘)⊆Vt2. Obviously, Vt2⊆Z(V˘). Therefore, Z(V˘)=Vt2.
Suppose that g∈Derα˘k(V˘), at2∈Z(V˘). Since α is surjective, α˘ is also surjective. For any xt+yt2∈V˘, there exists x′t+y′t2∈V˘ such that xt+yt2=α˘k(x′t+y′t2).
[TABLE]
which implies that g(at2)∈Z(V˘). Therefore, g(Z(V˘))⊆Z(V˘).
Hence, we have g(Ut2)⊆g(Z(V˘))⊆Z(V˘)=Vt2.
Now we define a map f:Vt+Ut2+μ(V,V)t2→Vt2 by
[TABLE]
It’s clear that f is linear.
Note that
[TABLE]
[TABLE]
we have f∈ZDerα˘k(V˘).
Since
[TABLE]
[TABLE]
[TABLE]
hence there exist D,D′∈End(V) such that ∀a∈V,b∈μ(V,V),
[TABLE]
Since g−f∈Derα˘k(V˘), we have
[TABLE]
which implies that
[TABLE]
Hence, we have D∈QDerαk(V)⊆QDer(V).
Therefore, g−f=φ(D)∈φ(QDer(V)), which implies that Der(V˘)⊆φ(QDer(V))+ZDer(V˘). According to Proposition 4.2 (2), we have Der(V˘)=φ(QDer(V))+ZDer(V˘).
∀f∈φ(QDer(V))∩ZDer(V˘), there exists D∈QDer(V) such that f=φ(D). For all a∈V,b∈μ(V,V),u∈U,
[TABLE]
On the other hand, for any xt+yt2∈V˘, there exists x′t+y′t2∈V˘ such that xt+yt2=α˘k(x′t+y′t2) since α˘ is surjective. Then we have
[TABLE]
since f∈ZDer(V˘). Hence, f(at+bt2+ut2)∈Z(V˘)=Vt2.
Therefore, D(a)=0, i.e., D=0. Hence, f=0.
Therefore, Der(V˘)=φ(QDer(V))∔ZDer(V˘).
∎
5 The centroids of Hom-Jordan algebras
Proposition 5.1**.**
Suppose that (V,μ,α) is a simple multiplicative Hom-Jordan algebra over an algebraically closed field F of characteristic [math]. If C(V)=Fid, then α=id.
Proof. .
For all k∈N+, ∀0=ψ∈Cαk(V), we have ψ=pid,p∈F,p=0. So ∀x,y∈V,
[TABLE]
which implies that μ(x,y)=μ(x,αk(y)) for all k∈N+.
Since F is algebraically closed, α has an eigenvalue λ. We denote the corresponding eigenspace by Eλ(α). So Eλ(α)=0. Let k=1. For any x∈Eλ(α), y∈V, we have
[TABLE]
which implies that μ(x,y)∈Eλ(α).
Moreover, for any x∈Eλ(α), we have
[TABLE]
which implies that α(x)∈Eλ(α), i.e., α(Eλ(α))⊆Eλ(α).
So Eλ(α) is a Hom-ideal of (V,μ,α). Since (V,μ,α) is simple, we have Eλ(α)=V, i.e., α=λid.
Then for any x,y∈V,k=1, we have
[TABLE]
which implies that λ=1, i.e., α=id.
∎
Proposition 5.2**.**
Let (V,μ,α) be a multiplicative Hom-Jordan algebra.
- (1)
If α is a surjection, then V is indecomposable if and only if C(V) does not contain idempotents except [math] and id.
2. (2)
If (V,μ,α) is perfect(i.e., V=μ(V,V)), then every ψ∈Cαk(V) is αk-symmetric with respect to any invariant form on V.
Proof. .
(1). (⇒). If there exists ψ∈Cαk(V) is an idempotent and satisfies that ψ=0,id, then ψ2(x)=ψ(x),x∈V.
For any x∈ker(ψ),y∈V, we have
[TABLE]
which implies that μ(x,y)∈ker(ψ).
Moreover, ψ(α(x))=α(ψ(x))=α(0)=0, so α(x)∈ker(ψ), i.e., α(ker(ψ))⊆ker(ψ).
Hence, ker(ψ) is a Hom-ideal of (V,μ,α).
For any y∈Im(ψ), there exists y′∈V such that y=ψ(y′). And for any z∈V, there exists z′∈V such that z=αk(z′) since α is surjective. Then
[TABLE]
which implies that μ(y,z)∈Im(ψ).
Moreover, α(y)=α(ψ(y′))=ψ(α(y′)), so α(y)∈Im(ψ), i.e., α(Im(ψ))⊆Im(ψ).
Hence, Im(ψ) is a Hom-ideal of (V,μ,α).
∀x∈ker(ψ)∩Im(ψ), ∃x′∈V such that x=ψ(x′). So 0=ψ(x)=ψ2(y)=ψ(y), which implies that x=0. Hence, ker(ψ)∩Im(ψ)={0}.
We have a decomposition x=(x−ψ(x))+ψ(x), ∀x∈V. So we have V=ker(ψ)⊕Im(ψ). Contradiction.
(⇐). Suppose that V=V1⊕V2 where Vi are Hom-ideals of (V,μ,α). Then for any x∈V, ∃xi∈Vi such that x=x1+x2.
Define ψ:V→V by ψ(x)=ψ(x1+x2)=x1−x2. It’s obvious that ψ2(x)=ψ(x).
For any x,y∈V, suppose that x=x1+x2,y=y1+y2,
[TABLE]
[TABLE]
we have
[TABLE]
Similarly, we have ψ(μ(x,y))=μ(x,ψ(y)). So ψ∈Cα0(V)⊆C(V). Contradiction.
(2). Let f be an invariant F-bilinear form on V. Then we have f(μ(x,y),z)=f(x,μ(y,z)) for any x,y,z∈V.
Since (V,μ,α) is perfect,let ψ∈Cαk(V), then for any a,b,c∈V we have
[TABLE]
∎
Proposition 5.3**.**
Let (V,μ,α) be a Hom-Jordan algebra and I an α-invariant subspace of V where α∣I is surjective. Then ZV(I)={x∈V∣μ(x,y)=0,∀y∈I} is invariant under C(V). So is any perfect Hom-ideal of (V,μ,α).
Proof. .
For any ψ∈Cαk(V) and x∈ZV(I), ∀y∈I, ∃y′∈I such that y=αk(y′) since α∣I is surjective, we have
[TABLE]
which implies that ψ(x)∈ZV(I). So ZV(I) is invariant under C(V).
Suppose that J is a perfect Hom-ideal of (V,μ,α). Then J=μ(J,J). For any y∈J, there exist a,b∈J such that y=μ(a,b), then we have
[TABLE]
So J is invariant under C(V).
∎
Theorem 5.4**.**
Suppose that (V1,μ1,α1) and (V2,μ2,α2) are two Hom-Jordan algebras over field F with α1 is surjective. Let π:V1→V2 be an epimorphism of Hom-Jordan algebras. Then for any f∈EndF(V1,ker(π)):={g∈W1∣g(ker(π))⊆ker(π)}, there exists a unique fˉ∈W2 satisfying π∘f=fˉ∘π where Wi(i=1,2) are defined as Lemma 2.7. Moreover, the following results hold:
- (1)
The map πEnd:EndF(V1,ker(π))→W2, f↦fˉ is a Hom-algebra homomorphism with the following proporties, where (EndF(V1,ker(π)),∘,σ1) and (W2,∘,σ2) are two Hom-associative algebras and σ1:EndF(V1,ker(π))→EndF(V1,ker(π)),f↦α1∘f and σ2:W2→W2,g↦α2∘g.
- (a)
πEnd(Mult(V1))=Mult(V2), πEnd(C(V1)∩EndF(V1,ker(π)))⊆C(V2).
2. (b)
By restriction, there is a Hom-algebra homomorphism
[TABLE]
3. (c)
If ker(π)=Z(V1), then every φ∈C(V1) leaves ker(π) invariant.
2. (2)
Suppose that V1 is perfect and ker(π)⊆Z(V1). Then πC:C(V1)∩EndF(V1,ker(π))→C(V2),f↦fˉ is injective.
3. (3)
If V1 is perfect, Z(V2)={0} and ker(π)⊆Z(V1), then πC:C(V1)→C(V2) is a Hom-algebra homomorphism.
Proof. .
For any y∈V2, there exists x∈V1 such that y=π(x) since π is an epimorphism. Then for any f∈EndF(V1,ker(π)), define fˉ:V2→V2 by fˉ(y)=π(f(x)) where x satisfies that y=π(x). It’s obvious that π∘f=fˉ∘π.
[TABLE]
Since π is surjective, we have α2∘fˉ=fˉ∘α2, i.e., fˉ∈W2.
If there exist fˉ and fˉ′ such that π∘f=fˉ∘π and π∘f=fˉ′∘π, then we have fˉ∘π=fˉ′∘π. For any y∈V2, there exists x∈V1 such that y=π(x). Hence, fˉ(y)=fˉ(π(x))=fˉ′(π(x))=fˉ′(y). So fˉ=fˉ′.
(1). For all f,g∈EndF(V1,ker(π)), we have
[TABLE]
which implies that πEnd(f∘g)=fˉ∘gˉ=πEnd(f)∘πEnd(g).
[TABLE]
which implies that πEnd(σ1(f))=α2∘fˉ=σ2(πEnd(f)), i.e., πEnd∘σ1=σ2∘πEnd. Therefore, πEnd is a Hom-algebra homomorphism.
(a). For all x∈V1, z∈ker(π), we have
[TABLE]
which implies that Lx(z)∈ker(π), i.e., Lx∈EndF(V1,ker(π)).
Moreover, we have π∘Lx=Lπ(x)∘π. So πEnd(Lx)=Lπ(x)∈Mult(V2). Hence, πEnd(Mult(V1))⊆Mult(V2).
On the other hand, for any Ly∈Mult(V2), ∃x∈V1 such that y=π(x). So
[TABLE]
which implies that Mult(V2)⊆πEnd(Mult(V1)). Therefore, πEnd(Mult(V1))=Mult(V2).
For any φ∈Cα1k(V1)∩EndF(V1,ker(π)), ∀x′,y′∈V2, ∃x,y∈V1 such that x′=π(x),y′=π(y), then we have
[TABLE]
Similarly, we have φˉ(μ2(x′,y′))=μ2(α2k(x′),φˉ(y′)). Hence, φˉ∈Cα2k(V2). Therefore, πEnd(C(V1)∩EndF(V1,ker(π)))⊆C(V2).
(b). ∀f,g∈C(V1)∩EndF(V1,ker(π)), we have
[TABLE]
which implies that πC(f∘g)=fˉ∘gˉ=πC(f)∘πC(g).
[TABLE]
which implies that πC(σ1(f))=α2∘fˉ=σ2(πC(f)), i.e., πC∘σ1=σ2∘πC. Therefore, πC is a Hom-algebra homomorphism.
(c). If ker(π)=Z(V1), ∀x∈ker(π),φ∈Cα1k(V1), for any y∈V1 there exists y′∈V1 such that y=α1k(y′) since α1 is surjective. We have
[TABLE]
which implies that φ(x)∈Z(V1)=ker(π). Hence, every φ∈C(V1) leaves ker(π) invariant.
(2). If φˉ=0 for φ∈Cα1k(V1)∩EndF(V1,ker(π)), then π(φ(V1))=φˉ(π(V1))=0. Hence, φ(V1)⊆ker(π)⊆Z(V1).
Hence, φ(μ1(x,y))=μ1(φ(x),α1k(y))=0. Note that V1 is perfect, we have φ=0. Therefore, πC:C(V1)∩EndF(V1,ker(π))→C(V2),f↦fˉ is injective.
(3). ∀y∈V2,∃y′∈V1 such that y=π(y′). For all x∈Z(V1),
[TABLE]
which implies that π(x)∈Z(V2). So π(Z(V1))⊆Z(V2)={0}. Therefore, Z(V1)⊆ker(π). Note that ker(π)⊆Z(V1), we have Z(V1)=ker(π).
For all φ∈Cα1k(V1),x∈Z(V1),y∈V1, ∃y′∈V1 such that y=α1k(y′),
[TABLE]
which implies that φ(x)∈Z(V1)=ker(π). So φ(ker(π))⊆ker(π), i.e., φ∈EndF(V1,ker(π)). So C(V1)⊆EndF(V1,ker(π)). Therefore, C(V1)∩EndF(V1,ker(π))=C(V1). According to (1) (b), we have πC:C(V1)→C(V2) is a Hom-algebra homomorphism.
∎