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Some properties of isoclinism in Lie superalgebras

Isoclinism of Lie superalgebras has been defined and studied currently. In this article it is shown that for finite dimensional Lie superalgebras of same dimension, the notation of isoclinism and isomorphism are equivalent. Furthermore we show that covers of finite dimensional Lie superalgebras are isomorphic using isoclinism concept.


Introduction
In 1940, P. Hall introduced an equivalence relation on the class of all groups called isoclinism, which is weaker than isomorphism and plays an important role in classification of finite p-groups [11]. In 1994, K. Moneyhun [3,4] gave a Lie algebra analogue of the concept of isoclinism. Further Saeedi and Veisi [6] have defined the same notation for n-Lie algebras. Similarly, isoclinism has been defined and studied for Lie superalgebras recently [7]. where µ((l,m)) := [l, m] for l, m ∈ L and similarly for ρ((r,s)) := [r, s] for r, s ∈ K. Or, equivalently ϕ and θ are defined in such a way that they are compatible, i.e., θ([l, m]) = [k, r], where l, k, m, r ∈ L in which k ∈ ϕ(l + Z(L)) and r ∈ ϕ(m + Z(L)). Then the pair (ϕ, θ) is called homoclinism and if they are both isomorphisms, then (ϕ, θ) is called isoclinism.
If (ϕ, θ) is an isoclinism between L and K, then L and K are said to be isoclinic, which is denoted by L ∼ K. Obviously isoclinism is an equivalence relation, and hence it produces a partition on the class of all Lie superalgebras into equivalence classes called isoclinism classes.
Here we listed some results on isoclinism of Lie superalgebras which are useful for us here and for details one can see [7].  A Lie superalgebra L is called a stem Lie superalgebra, whenever Z(L) ⊆ L ′ . The following properties of stem Lie superalgebras can be easily seen [7]. Lemma 1.5. Suppose C is an isoclinic family of Lie superalgebras. Then (1) C contains a stem Lie superalgebra.
(2) Each finite dimensional Lie superalgebra T ∈ C is stem if and only if T has minimal dimension in C.
The covers and multipliers for Lie algebras are defined and studied by Betten et al. It has been shown that unlike the case of groups the covers of a finite dimensional Lie algebras are isomorphic [1,2]. Likewise covers and multipliers for Lie superalgebras has recently been defined and studied [8].
An extension of a Lie superalgebra L is a short exact sequence Since e : M −→ e(M ) = ker(f ) is an isomorphism we identify M and e(M ). An extension of L is then same as an epimorphism f : K −→ L. If L is a Lie superalgebra generated by a Z 2 -graded set X = X0 ∪ X1 and φ : X → L is a degree zero map, then there exists a free Lie superalgebra F and ψ : F → L extending φ. Let R = Ker(ψ). The extension is called a free presentation of L and is denoted by (F, ψ). With this free presentation of L, we define multiplier of L as A homomorphism from an extension f : K −→ L to another extension f ′ : K ′ −→ L is a Lie superalgebra homomorphism g : K −→ K ′ satisfying f = f ′ • g; in other words, we have the following commutative diagram, The following is an important result which we use further.
The stem extension

equivalently dimension of K is maximal) if and only if it is a stem cover of L.
Finally below is another useful Lemma. Lemma 1.9. Let the pair (ϕ, θ) be isoclinism of Lie superalgebras L and M . Then Proof. We have the following commutative diagram, which proves (1). To prove the second part, consider for y ∈ L, we have z + Z(M ) = ϕ(y + Z(L)).  In Section 2 we recall some notations of Lie superalgebras. In Section 3 we show that isoclinism and isomorphism are equivalent for finite dimensional Lie superalgebras with the same dimension. Specifically we introduced factor set in Lie superalgebras to do the same. This result was first done for Lie algebras by Moneyhun [3] in 1994 and for n-Lie algebras by Eshrati et al. [5]. Moreover as an application to concept of isoclinism it is shown that covers for finite dimensional Lie superalgebras are isomorphic in Section 4. The same result hold for Lie algebras case see [3,4,1,2] and for n-Liealgebras see [5].

Preliminaries
Here we fix some terminology on Lie superalgebras (see [10,12]) and recall some notions. Set Z 2 = {0,1} is a field. A Z 2 -graded vector space V is simply a direct sum of vector spaces V0 and V1, i.e., V = V0 ⊕ V1. It is also referred as a superspace. We consider all vector superspaces and superalgebras are over F (characteristic of F = 2, 3). Elements in V0 (resp. V1) are called even (resp. odd) elements. Non-zero elements of V0 ∪ V1 are called homogeneous elements. For a homogeneous element v ∈ V σ , with σ ∈ Z 2 we set |v| = σ is the degree of v. A vector subspace U of V is called Z 2 -graded vector subspace(or superspace) if U = (V0 ∩ U ) ⊕ (V1 ∩ U ). We adopt the convention that whenever the degree function appeared in a formula, the corresponding elements are supposed to be homogeneous.
A Lie superalgebra is a superspace L = L0 ⊕ L1 with a bilinear mapping [., .] : L × L → L satisfying the following identities: For a Lie superalgebra L = L0 ⊕ L1, the even part L0 is a Lie algebra and L1 is a L0-module. If L1 = 0, then L is just Lie algebra. But in general a Lie superalgebra is not a Lie algebra. Lie superalgebra without even part, i.e., L0 = 0, is an abelian Lie superalgebra, as [x, y] = 0 for all x, y ∈ L. A sub superalgebra of L is a Z 2 -graded vector subspace which is closed under bracket is a graded ideal and it is called the center of L. Clearly, if I and J are graded ideals of L, then so is [I, J]. If I is an ideal of L, the quotient Lie superalgebra L/I inherits a canonical Lie superalgebra structure such that the natural projection map becomes a homomorphism.
By a homomorphism between superspaces f :

Factor set in Lie Superalgebras
The notation for factor set in Lie algebras was defined by Moneyhun in [3]. Here we start with defining factor set in Lie superalgebras and study some of its properties.
Proof. Here we have Z(L) is a graded ideal and L/Z(L) is a quotient Lie superalgebra. Hence there is a natural Z 2 -grading on set R. Any (x, a) ∈ R, is an even element when x ∈ Z(L)0, a ∈ (L/Z(L)) 0 and it is an odd element when x ∈ Z(L)1, a ∈ (L/Z(L)) 1 . Now it is easy to check that R is a vector superspace with comonent wise addition and furthermore with the multiplication defined in (1) it is a superalgebra. Now we intend to show that R is a Lie superalgebra.
(1) To check graded skew-symmetric property consider (3.1) (2) To check the graded Jacobi identity, consider Hence, R is a Lie superalgebra with the given multiplication. Also proof of part(ii) is obvious.
Below is the Lemma which states that factor sets exist for any Lie superalgebras. Proof. Let K be the vector superspace and let it be the complement of Z(L) in L, i.e., L = K ⊕ Z(L). Let us define T : L/Z(L) → L by T (a) = T (a + Z(L)) = T (k + l + Z(L)) = k, for all a ∈ L/Z(L) with a = k + l where k ∈ K and l ∈ Z(L). Here, T is a well-defined map and is also homogeneous linear map of even degree. Clearly we have T (a) = a. Now for We show that r is a factor set.
Since T is a homogeneous linea map of even degree and bracket respects gradation, so r(a, b) ∈ (L/Z(L)) α+β povided a ∈ L/Z(L)) α and b ∈ L/Z(L)) β . To check skew-symmetric property of r, for all a, b ∈ L/Z(L). To check Jacobi identity, take Hence, θ is well-defined map and it is injective. Evidently θ is surjective. Also θ is a homogeneous linear map of even degree and as so it is a Lie superalgebra homomorphism, hence θ is the required isomorphism.
Following is the Lemma which gives a connection between two stem Lie superalgebras.
So, φ is a homomorphism and similarly ψ is also a homomorphism. Since, λ is an isomorphism, µ is a homogeneous linear map of even degree. Now, to check the map µ is well-defined and injective, let   Hence, we get γ(a+b) = γ(a)+γ(b) implies γ is a linear map. Further as λ and µ are Lie superalgebra isomorphisms, that immediately implies γ is a homogeneous linear map of even degree. We have for x ∈ Z(L), λ(x, 0) = (ν(x), 0).
On one hand, and ∀ x ∈ L τ ∩ Z(L) with τ ∈ Z 2 . Here λ is a well defined map and also a bijection. Further as ν, δ, µ are all homogeneous linear maps of even degree hence so is λ. Now,  where the equality holds if (0, µ(a)) ∈ S satisfying λ(Z R ) = Z S which proves our second assertion.
The following Theorem plays an important role in proving the main result of this section. This result was first proved by Moneyhun for Lie algebras [3]. As Z R ⊆ Z(R), we get Z R = Z(R). Let us consider the following commutative diagram; To apply Lemma 2.6, we may extend δ to L/Z(L) by defining 0 on the complement of L ′ /Z(L) in L/Z(L). Then, we get R ∼ = S.
Hence T is a stem Lie superalgebra. As M is an ideal of L and M ∩ L ′ = 0, we get a Z 2 -graded vector subspace K of L containing L ′ complementary to M . Now, It is easy to check K is a stem Lie superalgebra. Finally T ∼ L and L ∼ K implies T ∼ K. Using Below is an example which shows that, two isoclinic Lie superalgebras of different dimension may not be isomorphic.
If (K, λ) ∈ C(L) is universal element, then for each (M, τ ) ∈ C(L), there exists homomorphism ρ ∈ Hom(K, M ) such that τ • ρ = λ. By Lemma 4.3, we have ρ is onto, hence dim M ≤ dim K. This implies K is a cover ( using Lemma 1.8). So, to show all covers for a finite dimensional Lie superalgebra L are isomorphic it is enough to show universal elements for L are isomorphic.