# Structures and Low Dimensional Classifications of Hom-Poisson Superalgebras

Authors: Qingcheng Zhang; Chunyue Wang; Zhu Wei
Journal: Advances in Mathematical Physics
 (2015)

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## Abstract

Hom-Poisson superalgebras can be considered as a deformation of
Poisson superalgebras. We prove that Hom-Poisson superalgebras are
closed under tensor products. Moreover, we show that Hom-Poisson
superalgebras can be described using only the twisting map and one
binary operation. Finally, all algebra endomorphisms on 2-dimensional
complex Poisson superalgebras are computed, and their associated
Hom-Poisson superalgebras are described explicitly.


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## Body

## 1. Introduction

Poisson algebras are used in many fields in mathematics and physics. In mathematics, Poisson algebras play a fundamental role in Poisson geometry [1], quantum groups [2], and deformation of commutative associative algebras. In physics, Poisson algebras are a major part of deformation quantization, Hamiltonian mechanics [3], and topological field theories [4]. Poisson-like structures are also used in the study of vertex operator algebras [5]. Poisson superalgebras can be seen as a direct generalization of Poisson algebras. Remm show that a Poisson superalgebra can be described using only one binary operation in [6, 7].Recently, a twisted generalization of noncommutative Poisson algebras, called Hom-noncommutative Poisson algebras, are studied in [8]. In a noncommutative Hom-Poisson algebra, there exists a twisted map, a Hom-Lie bracket and a Hom-associative product. The associativity, the Jacobi identity, and the Leibniz identity are considered as their Hom-type analogues in a noncommutative Hom-Poisson algebra. The purpose of this paper is to introduce and study a twisted generalization of Poisson superalgebras, called Hom-Poisson superalgebras.This paper is organized as follows. In Section2, we recall the construction of the Hom-Lie superalgebras. In Section 3, we give the definition of Hom-Poisson superalgebras. We show that starting with a Poisson superalgebra and an even Poisson superalgebra endomorphism, a Hom-Poisson superalgebra can be constructed. Moreover, we prove that Hom-Poisson superalgebras are closed under tensor products. In Section 4, we show that a Hom-Poisson superalgebra can be described using only the twisting map and one binary operation. Section 5 is devoted to classifying all the algebra endomorphisms α on all the 2-dimensional complex Poisson superalgebras and 2-dimensional Hom-Poisson superalgebras.Throughout the paperC is the field of complex numbers. All algebras and vector spaces are considered over C.


## 2. Hom-Lie and Hom-Associative Superalgebras

In this section, we first recall the notion of Hom-Lie superalgebras and then give some construction of the Hom-Lie superalgebras.Definition 1 (see [9]).
A Hom-associative superalgebra is a triple(A,∘,α) consisting of a Z2-graded vector space A, an even bilinear map ∘:A×A→A, and an even homomorphism of algebras α:A→A satisfying(1)αx∘y∘z=x∘y∘αz for all homogeneous elements x,y,z∈A.Definition 2 (see [9]).
A Hom-Lie superalgebra is a triple(A,[·,],α) consisting of a Z2-graded vector space A, an even bilinear map [·,·]:A×A→A and an even homomorphism of algebras α:A→A satisfying(2)x,y=--1xyy,x,(3)-1xzαx,y,z+-1zyαz,x,y+-1xyαy,z,x=0for all homogeneous elements x,y,z∈A.Let(V,[·,·],α) and V′,·,·′,α′ be two Hom-Lie superalgebras. An even homomorphism f:V→V′ is said to be a morphism of Hom-Lie superalgebras if(4)fx,fy′=fx,y,∀x,y∈V,(5)f∘α=α′∘f.Morphisms of Hom-associative superalgebras are defined similarly.The following theorem provides a method to construct a Hom-Lie superalgebra by a Lie superalgebra and an even homomorphism of Lie superalgebras.Proposition 3 (see [9]).
Let(V,[·,·]) be a Lie superalgebra and let α:V→V be an even endomorphism of Lie superalgebras. Then (V,·,·α,α) is a Hom-Lie superalgebra, where x,yα=α(x,y).
Moreover, suppose that(V′,·,·′) is another Lie superalgebra and α′:V′→V′ is an even endomorphism of Lie superalgebras. If f:V→V′ is a morphism of Lie superalgebras satisfying f∘α=α′∘f, then(6)f:V,·,·α,α⟶V′,·,·α′′,α′ is a morphism of Hom-Lie superalgebras.Example 4 (see [9]).
From the orthosymplectic Lie superalgebraosp(1,2)=V0⊕V1, where V0 is spanned by (7)H=10000000-1,X=001000000,Y=000000100,and V1 is spanned by (8)F=000100010,X=01000-1000.The defining nonzero relations are (9)H,X=2X,H,Y=-2Y,X,Y=H,Y,G=F,X,F=G,H,F=-F,H,G=G,G,F=H,G,G=-2X,F,F=2Y. Let λ∈R∗ define a linear map αλ:osp(1,2)→osp(1,2) by (10)αλX=λ2X,αλY=1λ2Y,αλH=H,αλF=1λF,αλG=λG;then αλ is an even Lie superalgebra automorphism; by Proposition 3, we obtain a family of Hom-Lie superalgebras (osp(1,2),,λ,αλ). These Hom-Lie superalgebras are not Lie superalgebras for λ≠1.


## 3. Hom-Poisson Superalgebras

Definition 5.
A Hom-Poisson superalgebra is a tuple(A,·,[·,·],α) consisting of a Z2-graded vector space V, two even bilinear maps ·,[·,·]:V×V→V, and an even homomorphism of algebras α:V→V, where α(x·y)=α(x)·α(y), α([x,y])=[α(x),α(y)], satisfying the following axioms.(1)
(A,·,α) is a supercommutative Hom-associative superalgebra.(2)
(A,[·,·],α) is a Hom-Lie superalgebra.(3)
The Hom-Leibniz superidentity[x·y,α(z)]=α(x)·[y,z]+-1yz[x,z]·α(y) holds for all homogeneous elements x,y,z∈A.Definition 6.
Let(A,·,[·,·],α) and (A′,·′,·,·′,α′) be two Hom-Poisson superalgebras. An even homomorphism f:A→A′ is said to be a morphism of Hom-Poisson superalgebras if(11)fx·′fy=fx·y,∀x,y∈A,fx,fy′=fx,y,∀x,y∈A,f·α=α′·f.Proposition 7.
Let(A,·,α) be a Hom-associative superalgebra and [·,·]:A×A→A be a binary operation on A defined by(12)x,y=x·y--1xyy·x,∀x,y∈A,then (A,·,[·,·],α) is a Hom-Poisson superalgebra with the same twisting map α.Proof.
It is straightforward.From Proposition7, there is the following construction of Hom-Poisson superalgebras by Poisson superalgebras and homomorphisms.Theorem 8.
Let(A,·,[·,·]) be a Poisson superalgebra and let α:A→A be an even endomorphism of Poisson superalgebras. Then A,·α,·,·α,α is a Hom-Poisson superalgebra, where x·αy=α(x·y) and [x,y]α=α([x,y]).
Moreover, suppose that(A′,·′,·,·′) is another Poisson superalgebra and α′:A′→A′ is an even endomorphism of Poisson superalgebras. If f:A→A′ is a morphism of Poisson superalgebras satisfying f·α=α′·f, then (13)f:A,·α,·,·α,α⟶A′,·α′,·,·α′′,α′is a morphism of Hom-Poisson superalgebras.Proof.
By Proposition3, (A,[·,·]α,α) is a Hom-Lie superalgebra.
We will show that(A,·α,α) satisfies the axioms (1) and (3) of Definition 5. In fact (14)αx·αy·αz=αx·ααy·z=ααx·y·z=αx·y·αz=αx·y·ααz=x·αy·ααz,x·αy,αzα=αx·αy,αz=ααx·y,αz=α2x·y,z=α2-1yzx,z·y+x·y,z=-1yzααx,z·αy+ααx·αy,z=-1yzαx,z·ααy+αx·ααy,z=-1yzx,zα·ααy+αx·αy,zα. Hence (A,·α,[·,·]α,α) is a Hom-Poisson superalgebra. Then second assertion follows from (15)fx,yα=fαx,y=f·αx,y=α′·fx,y=α′fx,fy=fx,fyα′;fx·αy=fαx·y=f·αx·y=α′·fx·y=α′fx·fy=fx·α′fy.This theorem provides a method to construct a Hom-Poisson superalgebra by a Poisson superalgebra and an even homomorphism of Poisson superalgebras.Example 9.
LetA=A0¯⊕A1¯ be a 2-dimensional Z2-graded vector space, where A0¯ is generated by e1 and A1¯ is generated by e2 and nonzero products are given by (16)e1·e1=e1,e2·e2=e1,e1·e2=e2·e1=e2,e2,e2=2e1;then (A,·,[·,·]) is a Poisson superalgebra. For any a∈C, we consider the homomorphism α:A→A defined by α(e1)=ae1,α(e2)=ae2. By Theorem 8, for any a∈C, there is the corresponding Hom-Poisson superalgebra Aa=(A,·α,[·,·]α,α) with the nonzero products (17)e1·αe1=ae1,e2·αe2=ae1,e1·αeα=ae2,e2,e2α=2ae1. It is not a Poisson superalgebra when a≠0,1.Example 10.
LetA=A0¯⊕A1¯ be a 3-dimensional Z2-graded vector space, where A0¯ is generated by e1,e2 and A1¯ is generated by e3 and the nonzero products are given by (18)e1·e2=e1,e2·e2=e2,e3·e2=e3,e1,e2=ae1;then (A,·,[·,·]) is a Poisson superalgebra. For any a∈C, we consider the homomorphism α:A→A defined by (19)αe1=ae1,αe2=e1+e2.By Theorem 8, for any a∈C, there is the corresponding Hom-Poisson superalgebra Aα=(A,·α,[·,·]α,α) with the nonzero products (20)e1·αe2=ae1,e2·αe2=e1+e2,e1,e2α=ae1. It is not a Poisson superalgebra when a≠0,1.We know Hom-Poisson algebras are closed under tensor products ([8] Theorem 2.9). Here we aim to consider it in detail in the superalgebra case.Theorem 11.
Let(Ai,·i,[·,·]i,αi) be Hom-Poisson superalgebras for i=1,2, and let A=A1⊗A2. Define the operations α:A→A and ·,[·,·]:A⊗2→A by(21)α=α1⊗α2,x1⊗x2·y1⊗y2=-1x2y1x1·1y1⊗x2·2y2,x1⊗x2,y1⊗y2=-1x2y1x1,y11⊗x2·2y2+-1x2y1x1·1y1⊗x2,y22, for xi,yi∈Ai. Then (A,·,[·,·],α) is a Hom-Poisson superalgebra.Proof.
The(A,·,α) is a supercommutative Hom-associative superalgebra following from the supercommutativity and Hom-associativity of both ·i. Also, the supercommutativity of the ·i and the antisupersymmetry of the [·,·]i imply the antisupersymmetry of [·,·]. It remains to prove the Hom-Jacobi superidentity and the Hom-Leibniz superidentity in A.
To simplify the typography, we abbreviate·1,·2, and · using juxtaposition and drop the subscripts in [·,·]i and αi. Pick x=x1⊗x2,y=y1⊗y2,z=z1⊗z2∈A. Then (22)-1xzx,y,αz=-1s1x1,y1⊗x2y2,α1z1⊗α2z2+-1s1x1y1⊗x2,y2,α1z1⊗α2z2=-1s2x1,y1,α1z1⊗x2y2α2z2+-1s2x1,y1α1z1⊗x2y2,α2z2+-1s2x1y1,α1z1⊗x2,y2α2z2+-1s2x1y1α1z1⊗x2,y2,α2z2, where s1=x1z1+x1z2+x2z1+x2z2+x2y1, s2=x1z1+x1z2+x2z2 +x2y1+y2z1. Consider(23)-1xyy,z,αx=-1s3y1,z1,α1x1⊗y2z2α2x2+-1s3y1,z1α1x1⊗y2z2,α2x2+-1s3y1z1,α1x1⊗y2,z2α2x2+-1s3y1z1α1x1⊗y2,z2,α2x2, where s3=y1x1+y1x2+y2x2+y2z1+z2x1. Consider(24)-1yzz,x,αy=-1s4z1,x1,α1y1⊗z2x2α2y2+-1s4z1,x1α1y1⊗z2x2,α2y2+-1s4z1x1,α1y1⊗z2,x2α2y2+-1s4z1x1α1y1⊗z2,x2,α2y2,where s4=z1y1+z1y2+z2y2+z2x1+x2y1. Using the Hom-Jacobi superidentity in A1 and the supercommutativity and Hom-associativity in A2, we have(25)-1s2x1,y1,α1z1⊗x2y2α2z2+-1s3y1,z1,α1x1⊗y2z2α2x2+-1s4z1,x1,α1y1⊗z2x2α2y2=0.
Likewise, using the supercommutativity and Hom-associativity inA1 and the Hom-Jacobi superidentity in A2, we obtain(26)-1s2x1y1α1z1⊗x2,y2,α2z2+-1s3y1z1α1x1⊗y2,z2,α2x2+-1s4z1x1α1y1⊗z2,x2,α2y2=0.
Using the Hom-Leibniz superidentity inAi, then we have (27)-1s2x1,y1α1z1⊗x2y2,α2z2+-1s2x1y1,α1z1⊗x2,y2α2z2+-1s3y1,z1α1x1⊗y2z2,α2x2+-1s3y1z1,α1x1⊗y2,z2α2x2+-1s4z1,x1α1y1⊗z2x2,α2y2+-1s4z1x1,α1y1⊗z2,x2α2y2=-1s2+y2z2x1,y1α1z1⊗x2,z2α2y2+-1s2x1,y1α1z1⊗α2x2y2,z2+-1s3+z2x2y1,z1α1x1⊗y2,x2α2z2+-1s3y1,z1α1x1⊗α2y2z2,x2+-1s3+z1x1y1,x1α1z1⊗y2,z2α2x2+-1s3α1y1z1,x1⊗y2,z2α2x2+-1s4+y2z2z1,x1α1y1⊗z2x2,α2y2+-1s4z1,x1α1y1⊗z2,y2α2x2+-1s4z1,x1α1y1⊗α2z2x2,y2+-1s4+x1y1z1,y1α1x1⊗z2,x2α2y2+-1s4α1z1x1,y1⊗z2,x2α2y2=-1s2+y2z2x1,y1α1z1⊗x2,z2α2y2+-1s4α1z1x1,y1⊗z2,x2α2y2+-1s2x1,y1α1z1⊗α2x2y2,z2+-1s3+z1x1y1,x1α1z1⊗y2,z2α2x2+-1s3y1,z1α1x1⊗α2y2z2,x2+-1s3+z1x1y1,x1α1z1⊗y2,z2α2x2+-1s3y1,z1α1x1⊗α2y2z2,x2+-1s4+x1y1z1,y1α1x1⊗z2,x2α2y2+-1s3α1y1z1,x1⊗y2,z2α2x2+-1s4+x2y2z1,x1α1y1⊗z2,y2α2x2+-1s2+y1z1x1,z1α1y1⊗x2,y2α2z2+-1s4z1,x1α1y1⊗α2z2x2,y2=0+0+0+0+0+0=0. This shows that (A,[·,·],α) satisfies the Hom-Jacobi superidentity: (28)-1xzx,y,αz+-1xyy,z,αx+-1yzz,x,αy=0.Finally, we check the Hom-Leibniz superidentity in A. Using the Hom-associativity and the Hom-Leibniz superidentity in the Ai, we have (29)xy,αz=x1⊗y1y1⊗y2,α1z1⊗α2z2=-1x2y1+x2z1+y2z1x1y1,α1z1⊗x2y2α2z2+-1x2y1+x2z1+y2z1x1y1α1z1⊗x2y2,α2z2=-1x2y1+x2z1+y2z1+y1z1x1,z1α1y1⊗x2y2α2z2+-1x2y1+x2z1+y2z1α1x1y1,z1⊗x2y2α2z2+-1x2y1+x2z1+y2z1+y2z2x1y1α1z1⊗x2,z2α2y2+-1x2y1+x2z1+y2z1x1y1α1z1⊗α2x2y2,z2,-1yzx,zαy+αxy,z=-1y1z1+y1z2+y2z1+y2z2+x2z1x1,z1⊗x2z2α1y1⊗α2y2+-1y1z1+y1z2+y2z1+y2z2+x2z1x1z1⊗x2,z2α1y1⊗α2y2+-1y2z1α1x1⊗α2x2y1,z1⊗y2z2+-1y2z1α1x1⊗α2x2y1z1⊗y2,z2=-1y1z1+y2z1+x2z1+x2y1x1,z1α1y1⊗x2z2α2y2+-1y2z1+y2z2+x2z1+x2y1x1z1α1y1⊗x2,z2α2y2+-1y2z1+x2y1+x2z1α1x1y1,z1⊗α2x2y2z2+-1y2z1+x2y1+x2z1α1x1y1z1⊗α2x2y2,z2.Therefore, we have (30)xy,αz=-1yzx,zαy+αxy,z.Settingαi=IdAi in Theorem 11, we obtain the result about Poisson superalgebras.Corollary 12.
Let(Ai,·i,[·,·]i,) be Poisson superalgebras for i=1,2, and let A=A1⊗A2. Define the operations ·,[·,·]:A⊗2→A by(31)x1⊗x2·y1⊗y2=-1x2y1x1·1y1⊗x2·2y2,x1⊗x2,y1⊗y2=-1x2y1x1,y11⊗x2·2y2+-1x2y1x1·1y1⊗x2,y22, for xi,yi∈Ai. Then (A,·,[·,·]) is a Poisson superalgebra.


## 4. Admissible Hom-Poisson Superalgebras

A Poisson algebra has two binary operations, the Lie bracket and the commutative associative product. It is shown in [10] that Poisson algebras can be described using only one binary operation via the polarization-depolarization process. Moreover, the result of Poisson algebras is extended to Hom-Poisson algebras in [8]. In other words, the paper shows that a Hom-Poisson algebra can be described using only the twisting map and one binary operation. The purpose of this section is to extend this alternative description of Poisson algebras to Hom-Poisson superalgebras.Definition 13.
An admissible Hom-Poisson superalgebraA is a Hom-superalgebra satisfying(32)3Ax,y,z+-1xyyxαz--1yzxzαy--1xy+xzyzαx+-1xz+yzzxαy=0,where A(x,y,z)=(xy)α(z)-α(x)(yz), for any homogeneous elements x,y,z∈A, the identity (32) is called the Hom-Remm identity.Remark 14.
In particular, takingα=IdA, we find the notion of admissible Poisson superalgebra presented in [8].Theorem 15.
Let(A,·,[·,·],α) be a double Hom-superalgebra. Then (A,·,[·,·],α) is a Hom-Poisson superalgebra if and only if there exists on A a nonassociative product xy such that (A,·,α) is an admissible Hom-Poisson superalgebra.Proof.
Assume that(A,·,[·,·],α) is a Hom-Poisson superalgebra. Consider the multiplication (33)xy=x·y+x,y.
We deduce that(34)x·y=12xy+-1xyyx.Thus the associativity condition can be denoted by (35)v1x,y,z=Ax,y,z--1xy+xz+yzAz,y,x+-1xyyxαz--1yzαxzy--1xy+xzyzαx+-1xz+yzαzxy=0, where A(x,y,z)=(xy)α(z)-α(x)(yz). Likewise, the Hom-Poisson bracket can be denoted by (36)x,y=12xy--1xyyx and the Hom-super Jacobi condition (37)v2x,y,z=-1xzAx,y,z--1xy+xzAy,x,z--1xy+yzAz,y,x--1xz+yzAx,z,y+-1xyAy,z,x+-1yzAz,x,y=0. The Hom-Leibniz superidentity can be denoted by (38)v3x,y,z=Ax,y,z--1xyAy,x,z+-1xy+xz+yzAz,y,x+-1yzAx,z,y+-1xy+xzAy,z,x--1xz+yzAz,x,y=0. Let us consider the vector (39)vx,y,z=13-1xyyxαz--1yzxzαy--1xy+xzyzαx+-1xz+yzzxαy+xyαz-αxyz. Then (40)vx,y,z=162v1x,y,z+-1xzv2x,y,z+v3x,y,z+2-1xz+yzv3z,x,y. We deduce that the product xy satisfies (41)vx,y,z=0for any homogeneous elements x,y,z∈A.
Conversely, assume that the produce of the nonassociative productA satisfies v(x,y,z)=0 for any homogeneous elements x,y,z∈A. Let v1(x,y,z), v2(x,y,z), v3(x,y,z) be the elements of A defined in the first part, respectively, in relation to the Hom-associativity, Hom-super Jacobi, and Hom-super Leibniz relations. We have (42)v1x,y,z=vx,y,z--1xy+xz+yzvz,y,x+-1yzvx,z,y--1xz+yzvz,x,y,v2x,y,z=-1xzvx,y,z--1xy+xzvy,x,z--1xy+yzvz,y,x--1xz+yzvx,z,y+-1xyvy,z,x+-1yzvz,x,y,v3x,y,z=vx,y,z--1xyvy,x,z+-1xy+xz+yzvz,y,x+-1yzvx,z,y+-1xy+xzvy,z,x--1xz+yzvz,x,y.Takingα=IdA in Theorem 15, we obtain the following result, which is Theorem 1 in [7].Corollary 16.
Let(A,·,[·,·]) be a double superalgebra. Then (A,·,[·,·]) is a Poisson superalgebra if and only if there exists on A a nonassociative product xy satisfying (43)3Ax,y,z+-1xyyxz--1yzxzy--1xy+xzyzx+-1xz+yzzxy=0, where A(x,y,z)=(xy)z-x(yz), for any homogeneous elements x,y,z∈A.Definition 17.
A Hom-nonassociative superalgebra(A,·,α) is called Hom-superflexive if the multiplication xy satisfies(44)Ax,y,z+-1xy+xz+yzAz,y,x=0 for any homogeneous elements x,y,z∈A, where Aα(x,y,z)=(xy)α(z)-α(x)(yz) is called a Hom-associator of the multiplication.Proposition 18.
Let(A,·,[·,·],α) be a Hom-Poisson superalgebra. Then the Hom-Remm product defining the Hom-Poisson superalgebra structure is Hom-superflexive.Proof.
Let(45)Bx,y,z=3Ax,y,z+-1xy+xz+yzAz,y,x.We have (46)Bx,y,z=--1xyyxαz+-1yzxzαy+-1xy+xzyzαx--1xz+yzzxαy+-1xy+xz+yz--1zyyzαx+-1yxzxαy+-1zy+zxyxαy--1zx+yzxzαy=--1xy+-1xyyxαz+-1yz--1yzxzαy+-1xy+xz--1xy+xzyzαx+--1xz+yz+-1xz+yzzxαy=0.Takingα=IdA in Proposition 18, we obtain the following result, which is Proposition 3 in [7].Corollary 19.
Let(A,·,[·,·]) be a Poisson superalgebra. Then the Remm product defining Poisson superalgebra structure is superflexive.Remark 20.
The deformation cohomology of Hom-Poisson superalgebras can be computed with the Hom-Remm identity, which is similar to the method in [11]. This content is not primary in the paper, we do not have a detailed discussion here.


## 5. A Classification of 2-Dimensional Hom-Poisson Superalgebras

In this section, we only consider thatA1¯ is nontrivial. A2i denotes one of the 2-dimensional admissible Poisson superalgebra types. α2i denotes one of the homomorphism types corresponding to A2i. A~2i denotes one of the 2-dimensional admissible Hom-Poisson superalgebra types corresponding to α2i. In the following, the products equal to zero are omitted.Lemma 21.
Let(A,()) be an admissible Poisson superalgebra and let α:A→A be an even Poisson superalgebra endomorphism. Then (A,()α,α) is an admissible Hom-Poisson superalgebra, where (xy)α=α(xy).Proof.
It is straightforward by Definition13.Lemma 22 (see [7]).
Let(A,()) be a 2-dimensional admissible Poisson superalgebra with a basis {e0,e1}, where e0∈A0¯,e1∈A1¯. Then A is one of the following types:A21:e0e0=ae0, e0e1=ae1, e1e0=ae1, e1e1=de0, d≠0.A22:e0e0=ae0.A23:e0e0=ae0, e0e1=ae1, e1e0=ae1, a≠0.A24:e0e1=be1, e1e0=-be1, b≠0.Proof.
Let(47)e0e0=ae0,e0e1=be1,e1e0=ce1,e1e1=de0.By Corollary 16, we have(48)3a-bb+ab-2bc+c2=0,db-a=0,3a-cc+ab-2bc+c2=0,d3c-b-2a=0,db-c=0.Now we consider the cases as follows.
Case  1. If d≠0, then b=a=c; hence we have A21.
Case  2. If d=0, then (b-c)(a-b-c)=0.
Subcase  2.1. If b=c, then b(a-b)=0.
Ifb=0, then we have A22.
Ifb≠0, then a=b=c≠0; hence we have A23.
Subcase  2.2. If b≠c, then b=-c≠0, a=0; hence we have A24.Remark 23.
Lemma5.2 in [6] has been obtained; however, there is one minor inaccuracy on the product operation in that proof.Lemma 24.
Let(A,()) be a 2-dimensional admissible Poisson superalgebra with dim A0¯=1 and dim A1¯=1. Then an even homomorphism α of type A21 is as follows: (49)α211:αe0=k2e0αe1=ke1;α212:αe0=0αe1=0;α213:αe0=e0αe1=±e1.Proof.
Let(50)αe0=a10e0,αe1=a11e1. From α is an even homomorphism, we obtain(51)αe0e0=αe0αe0,αe0e1=αe0αe1,αe1e0=αe1αe0,αe1e1=αe1αe1. By Lemma 22 and (50) and (51), we obtain (52)aa10a10-1=0,aa11a10-1=0,a10=a112.Case  1. If a=0, then we have α21(1).
Case  2. If a≠0, then we consider two cases as follows.
Ifa10=0, then a11=0; hence we have α21(2).
Ifa10≠0, then a10=1, a11=±1; hence we have α21(3).Corollary 25.
Let(A~,()α,α) be a 2-dimensional admissible Hom-Poisson superalgebra. Then A~ with respect to α21 is as follows:A~2(1)1:e1e1α=dk2e0, d≠0.A~2(2)1:AAα=0.A~2(3)1:(e0e0)α=ae0, (e0e1)α=±ae1, (e1e0)α=±ae1, (e1e1)α=de0, a≠0, d≠0.Proof.
Apply Lemmas21, 22, and 24.Lemma 26.
Let(A,()) be a 2-dimensional admissible Poisson superalgebra with dim⁡A0¯=1 and dim⁡A1¯=1. Then an even homomorphism α of type A22 is as follows: (53)α221:αe0=k0e0αe1=ke1;α222:αe0=0αe1=ke1;α223:αe0=e0αe1=ke1.Proof.
By Lemma22 and (20) and (21), we obtain aa10(a10-1)=0.
Case  1. Suppose that a=0, we have α22(1).
Case  2. Suppose that a≠0, we consider two cases as follows.
Ifa10=0, then we have α22(2).
Ifa10≠0, then a10=1; hence we have α22(3).Corollary 27.
Let(A~,()α,α) be a 2-dimensional admissible Hom-Poisson superalgebra. Then A~ with respect to α22 is as follows:A~2(1)2:AAα=0.A~2(2)2:AAα=0.A~2(3)2:e0e0α=ae0, a≠0.Proof.
Apply Lemmas21, 22, and 26.Lemma 28.
Let(A,()) be a 2-dimensional admissible Poisson superalgebra with dim A0¯=1 and dim A1¯=1. Then an even homomorphism α of type A23 is as follows: (54)α231:αe0=0αe1=0;α232:αe0=e0αe1=ke1;α233:αe0=0αe1=0;α234:αe0=e0αe1=0;α235:αe0=e0αe1=ke1,k≠0.Proof.
By Lemma22 and (20) and (21), we obtain (55)a10a10-1=0,a11a10-1=0.Case  1. If a10=0, then a11=0; hence we have α23(1).
Case  2. If a10≠0, then a10=1; hence we have α23(2).
Case  3. If a11=0, then we consider two cases as follows.
Ifa10=0, we have α23(3).
Ifa10≠0, then a10=1; hence we have α23(4).
Case  4. If a11≠0, then a10=1; hence we have α23(5).Corollary 29.
Let(A~,()α,α) be a 2-dimensional admissible Hom-Poisson superalgebra. Then A~ with respect to α23 is as follows:A~2(1)3:(AA)α=0.A~2(2)3:(e0e0)α=ae0, (e0e1)α=kae1, (e1e0)α=kae1, a≠0.A~2(3)3:(AA)α=0.A~2(4)3:(e0e0)α=ae0, a≠0.A~2(5)3:(e0e0)α=ae0, (e0e1)α=kae1, (e1e0)α=kae1, a≠0, k≠0.Proof.
Apply Lemmas21, 22, and 28.Lemma 30.
Let(A,()) be a 2-dimensional admissible Poisson superalgebra with dim⁡A0¯=1 and dim⁡A1¯=1. Then an even homomorphism α of type A24 is as follows: (56)α241:αe0=ke0αe1=0;α242:αe0=e0αe1=ke1,k≠0.Proof.
By Lemma22 and (20) and (21), we obtain a11(a10-1)=0.
Case  1. If a10=0, then we have α24(1).
Case  2. If a10≠0, then a10=1; hence we have α24(2).Corollary 31.
Let(A~,()α,α) be a 2-dimensional admissible Hom-Poisson superalgebra. Then A~ with respect to α24 is as follows:A~2(1)4:(AA)α=0.A~2(2)4:(e0e1)α=kbe1, (e1e0)α=-kbe1, b≠0, k≠0.Proof.
Apply Lemmas21, 22, and 30.Theorem 32.
Let(A,()) be a 2-dimensional admissible Poisson superalgebra with dim⁡A0¯=1 and dim⁡A1¯=1. Then an even homomorphism α of A is as follows:(57)α211:αe0=k2e0αe1=ke1;α212:αe0=0αe1=0;α213:αe0=e0αe1=±e1;α221:αe0=k0e0αe1=ke1;α222:αe0=0αe1=ke1;α223:αe0=e0αe1=ke1;α231:αe0=0αe1=0;α232:αe0=e0αe1=ke1;α233:αe0=0αe1=0;α234:αe0=e0αe1=0;α235:αe0=e0αe1=ke1,k≠0;α241:αe0=ke0αe1=0;σ242:αe0=e0αe1=ke1,k≠0.Proof.
Apply Lemmas24–30.Corollary 33.
Let(A~,()α,α) be a 2-dimensional admissible Hom-Poisson superalgebra. Then A~ with respect to α2i is as follows:A~2(1)1:(e1e1)α=dk2e0, d≠0.A~2(2)1:(AA)α=0.A~2(3)1:(e0e0)α=ae0, (e0e1)α=±ae1, (e1e0)α=±ae1, (e1e1)α=de0, a≠0, d≠0.A~2(1)2:(AA)α=0.A~2(2)2:(AA)α=0.A~2(3)2:(e0e0)α=ae0, a≠0.A~2(1)3:(AA)α=0.A~2(2)3:(e0e0)α=ae0, (e0e1)α=kae1, (e1e0)α=kae1, a≠0.A~2(3)3:(AA)α=0.A~2(4)3:(e0e0)α=ae0, a≠0.A~2(5)3:(e0e0)α=ae0, (e0e1)α=kae1, (e1e0)α=kae1, a≠0, k≠0.A~2(1)4:(AA)α=0.A~2(2)4:(e0e1)σ=kbe1, (e1e0)α=-kbe1, b≠0, k≠0.Proof.
Apply Corollaries25–31.Remark 34.
(
1
) Some nonisomorphic 2-dimensional admissible Poisson superalgebras have isomorphic admissible Hom-Poisson deformations. For example, the admissible Poisson superalgebras A22 (with k=0 in its algebra homomorphism α2(3)2) and A23 can be deformed into isomorphic admissible Hom-Poisson superalgebras A~2(3)2 and A~2(4)3. There are several other such pairs in the cases above.
(
2
) We give a classification of 2-dimensional admissible Hom-Poisson superalgebras in Corollary 33. Using Theorem 15, a classification of 2-dimensional Hom-Poisson superalgebras will be obviously obtained.



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