Cyclic homology of Jordan superalgebras and related Lie superalgebras
Consuelo Martínez Efim Zelmanov Zezhou Zhang































label1]Consuelo Martínez
label2]Efim Zelmanov
label3]Zezhou Zhang

[label1]organization=Departamento de Matemáticas, Universidad de Oviedo,
            addressline=C/ Calvo Sotelo s/n, 
            city=Oviedo,
            postcode=33007,
            country=Spain
[label2]organization=SICM, Southern University of Science and Technology,
            city=Shenzhen,
            postcode=518055, 
            country=China
[label3]organization=Department of Mathematics, Beijing Normal University,
            city=Beijing,
            postcode=100875, 
            country=China

§ ABSTRACT 


We study the relationship between cyclic homology of Jordan superalgebras and second cohomologies of their Tits-Kantor-Koecher Lie superalgebras.

In particular, we focus on Jordan superalgebras that are Kantor doubles of bracket algebras.
The obtained results are applied to computation of second cohomologies and universal central extensions of Hamiltonian and contact type Lie superalgebras over arbitrary rings of coefficients.













Jordan algebra superalgebra superconformal algebra

[2020]  Primary 17B60, 17B66  Secondary 17A70  17C70, 17B68, 81R10









§ INTRODUCTION




Let L be a (super) algebra that is a Tits-Kantor-Koecher construction of J, a Jordan (super) algebra.
   S. Tan (in <cit.>) and, a bit later, B. Allison, G. Benkart, Y. Gao in <cit.> linked second cohomology of the (super) algebra L to cyclic homology of the (super) algebra  J, an analog of cyclic homology introduced by A. Connes (see <cit.>  and <cit.>).
  
For the classical case of superconformal algebras K(1:n), an explicit description of second cohomologies is due to V. G. Kac and J. van de Leur <cit.>.



We study, in this paper, the relationship between cyclic homology of Jordan superalgebras and second cohomologies of their Tits-Kantor-Koecher Lie superalgebras. We note that apart from this work and the above mentioned <cit.> and <cit.>, we are not aware of any other references concerning cyclic homology of Jordan (super) algebras.

In particular, we focus on Jordan superalgebras that are Kantor doubles of
bracket algebras. The main results related to cyclic homology of Kantor doubles appear in Section 4.  

The first 3 sections of the paper intend to provide the reader of all needed definitions and preliminary results concerning Lie superalgebras (Section 1), Jordan superalgebras (Section 2) and Brackets (Section 3).

Our interest in Kantor doubles stems from the fact that these Jordan superalgebras are related to superconformal algebras via the Tits-Kantor-Koecher construction.  

The obtained results are applied to the computation of second cohomologies and universal central extensions of Hamiltonian and contact type Lie superalgebras over arbitrary rings of coefficients, in sections 5 through 7. 

 Due to these applications to superconformal algebras – where the fact of zero characteristic is needed – we will assume that all considered vector spaces are over a field of zero characteristic, even though all results in Section 2 are valid over fields of characteristics not 2,3.



§ PRELIMINARIES (LIE SUPERALGEBRAS)


Let L=L_0̅+L_1̅ be a Lie superalgebra.
A bilinear mapping L× L→ F,  a× b↦ (a|b)∈ F is called a 2-cocycle if and only if it is super skew-symmetric and

    ([a, b]|c)+(-1)^|a|(|b|+|c|)([b, c]|a)+(-1)^|c|(|a|+|b|)([c, a]|b)=0

for arbitrary elements a,b,c∈ L.




For an arbitrary bilinear functional λ :L→ F,
the bilinear mapping (a|b)=λ ([a,b]) is a 2-cocycle.
Such cocycles are called 2-coboundaries.

Let C^2(L) be the vector space of all 2-cocycles of L, and
B^2(L) be the vector space of all 2-coboundaries of L.




A Lie superalgebra L is said to be perfect if L=[L,L].
Let L, L be perfect Lie superalgebras.
A surjective homomorphism Lφ⟶L
is called a central extension if the kernel of φ lies in the center
Z(L) of the superalgebra L.

Following ideas of I. Schur <cit.>, H. Garland <cit.> showed that
for an arbitrary perfect Lie (super)algebra L, there exists
a unique universal central extension Lu⟶L such that for an arbitrary central extension  Lφ⟶L,
there exists a homomorphism χ :L→L making the diagram

L[r,"u"] [d,"χ"]     L 
L[ur,"φ"]
commutative.




The vector space H^2(L) of second cohomologies can be identified with the dual space Z(L)^*, where Z(L) is the center of the universal central extension L.

The center Z(L) can be identified with the space (L⊗ L)1.25V, where V is the subspace spanned by all tensors 
    a⊗ b + (-1)^|a|·|b|b⊗ a,  [a,b]⊗ c + (-1)^|a|(|b|+|c|)[b, c]⊗ a + (-1)^|c|(|a|+|b|)[c, a]⊗ b
 with a,b,c∈ L.




For further references to universal central extensions see <cit.>.





Suppose that a Lie superalgebra L is ℤ-graded, L=∑ _i∈ℤL_i = L_-2+L_0+L_2, all homogeneous components L_i,  i≠ -2,0,2 are equal to 0.
Suppose also that L_0 = [L_-2,L_2].
Let (x  |  y) be a 2-cocycle on L.
Then for arbitrary elements a_-2,c_-2∈ L_-2; b_2,d_2∈ L_2, we have

    ([a_-2,[c_-2,d_2]]  |  b_2) + (-1)^|b|· (|c|+|d|)(a_-2  |  [b_2, [c_-2,d_2]]) 
       = (-1)^|b|· (|c|+|d|)([[a_-2,b_2],c_-2]  |  d_2) +
            (-1)^|a|· |c| + |b|· |d|(c_-2  |  [[a_-2,b_2],d_2]).


In fact, both sides are equal to (-1)^|b|· (|c|+|d|)([a_-2,b_2]  |  [c_-2,d_2]).



    A bilinear mapping L_-2× L_2→ F, a_-2× b_2↦ (a_-2  |  b_2) satisfying (<ref>) uniquely extends to a 2-cocycle on L.


    Without loss of generality, we assume that L is a Lie algebra.
    We define the 2-cocycle ψ(x|y) on L as follows:
    
    ψ (a_-2  |  b_2)=(a_-2  |  b_2),

    
    ψ ([a_-2,b_2]  |  [c_-2,d_2])= ([a_-2,[c_-2,d_2]]  |  b_2) + (-1)^|b|· (|c|+|d|)(a_-2  |  [b_2, [c_-2,d_2]]),

    
    ψ (L_i  |  L_j)=(0) for i+j≠ 0.

    The identity (<ref>) implies that
    
    ψ (∑ _i [a_-2^(i),b_2^(i)], ∑ _j [c_-2^(j),d_2^(j)])=∑ _i,jψ ([a_-2^(i),b_2^(i)],[c_-2^(j),d_2^(j)]))

    is well-defined.
    



    Let us check that
    
    ψ ([x,y]  |  z) + ψ ([y,z]  |  x) + ψ ([z,x]  |  y) = 0.

    We need to consider only two cases:
    
        
  * x=a_-2,y=b_2,z=[c_-2,d_2];
        
  * x,y,z∈ [L_-2,L_2].
    




    Case 1. We need to show that
    
    ψ ([a_-2,b_2]  |  [c_-2,d_2]) + ψ ([b_2, [c_-2,d_2]]  |  a_-2) + ψ ([[c_-2,d_2], a_-2]  |  b_2) = 0.





    This expression is equal to
    
    ([a_-2, [c_-2,d_2]]  |  b_2) + (a_-2  |  [b_2, [c_-2,d_2]]) 
       - (a_-2  |  [b_2, [c_-2,d_2]]) + ([[c_-2,d_2], a_-2]  |  b_2) = 0.

    



    Case 2. Let ρ_1,ρ_2,ρ_3∈ [L_-2,L_2],ρ_2=[a_-2,b_2].
    Consider the mapping 
    
    L_-2⊗ L_2ψ⟶F,   x_-2⊗ y_2ψ↦(x_-2  |  y_2).

    The tensor product L_-2⊗ L_2 is a module over the Lie algebra [L_-2,L_2].
    



    Denote ρ_2=a_-2⊗ b_2.
    We need to show that
    
    ψ ([ρ_1,ρ_2]  | ρ_3) + ψ ([ρ_2,ρ_3]  | ρ_1) + ψ ([ρ_3,ρ_1]  | ρ_2) = 0.

    Taking into account that
    
    ψ ([ρ_1,ρ_2]  | ρ_3)=-ψ (ρ_3  |  [ρ_1,ρ_2]), ψ ([ρ_2,ρ_3]  | ρ_1)=ψ (ρ_1  |  [ρ_3,ρ_2]),

    the last equality follows from
    
    ψ(-[ρ_3, [ρ_1,ρ_2]]+[ρ_1, [ρ_3,ρ_2]]+[[ρ_3,ρ_1],ρ_2])=0,

    which is the Jacobi identity.
    This completes the proof of the lemma.





§ JORDAN SUPERALGEBRAS




 §.§ Preliminaries






We start with basic definitions.
A Jordan algebra is a vector space J with a binary bilinear equation (x,y)↦ xy satisfying the following identities:

    xy=yx,   (x^2y)x=x^2(yx).

See <cit.>.




Let V be a vector space with countable dimension and let G=G(V) denote the Grassmann (or exterior) algebra over V; that is, the quotient of the tensor algebra over the ideal generated by symmetric tensors.
Then G(V) is a ℤ/2ℤ-graded algebra,
G(V)=G(V)_0̅+G(V)_1̅.
Its even part G(V)_0̅ is the linear span of all tensors of even length, and the odd part G(V)_1̅ is the linear span of all tensors of odd length.




Let V be a variety of algebras defined by homogeneous identities (see <cit.>). A superalgebra A=A_0̅+A_1̅ is called a V-superalgebra if its Grassmann envelope G(A)=A_0̅⊗ G(V)_0̅ + A_1̅⊗ G(V)_1̅ lies in V.




Thus a Jordan superalgebra is a ℤ/2ℤ-graded algebra J=J_0̅+J_1̅ satisfying the graded identities

    xy=(-1)^|x|· |y|yx  (supercommutativity), and


    ((xy)z)t + (-1)^|y|· |z| + |y|· |t| + |z|· |t|((xt)z)y + (-1)^|x|· |y| + |x|· |z| + |x|· |t| + |z|· |t|((yt)z)x 
       = (xy)(zt) + (-1)^|y|· |z|(xz)(yt) + (-1)^|t|·(|y|+|z|)(xt)(yz).


For an element a∈ J_0̅∪ J_1̅, let R(a) denote the operator of right multiplication R(a):J∋ x↦ xa, where xR(a)R(b)=(xa)b. [In this paper operators are constantly applied to the right of vectors.]




For arbitrary elements a,b∈ J_0̅∪ J_1̅, the operator D(a,b)=R(a)R(b) - (-1)^|a|· |b|R(b)R(a) is a derivation of the superalgebra J. Such derivations are called inner derivations. The span of all inner derivations is a Lie superalgebra. We will denote it as Inder(J).




For more references on Jordan superalgebras see <cit.>.





J. Tits <cit.> made the following observation. Let L be a Lie (super) algebra, L_0̅ contains 𝔰𝔩(2) = Fe + Ff + Fh, [e,f]=h, [h,e]=2e, [h,f]=-2f (we call such triple e,f,h an 𝔰𝔩(2)-triple).
Suppose that the operator ad(h): L→ L is diagonalizable and has eigenvalues -2,0,2, so L=L_-2+L_0+L_2 is a direct sum of eigenspaces. Then J=(L_2, a· b=[[a,f],b]) is a Jordan (super)algebra.

Moreover, J. Tits <cit.>, I. Kantor <cit.> and M. Koecher <cit.> showed that every Jordan (super)algebra can be obtained in this way.
The corresponding Lie (super)algebra L is not unique. We will recall the constructions of two Lie superalgebras with these properties: the largest (universal) one and the smallest (reduced) one.




For elements x,y,z∈ J_0̅∪ J_1̅ of a Jordan superalgebra J, we consider their Jordan triple product 
    { x,y,z } = (xy)z + x(yz) - (-1)^|x|· |y|y(xz).

Fix elements y,z and consider the operator V(y,z): x↦{ x,y,z }. Then V(y,z)=D(y,z)+R(yz).




 §.§ The universal Tits-Kantor-Koecher construction








      We will introduce in the following, the TKK construction of a unital Jordan (super)algebra in the shortest way, using bases: even though it is possible to do it in  basis-free ways.


Let J be a Jordan (super)algebra with an identity 1. Consider a basis { e_i } _i∈ I of J. Let 
    { e_i,e_j,e_k } = ∑ _tγ _ijk^te_t, γ _ijk^t∈ F.

Define a Lie (super)algebra L by generators { x_i^-, x_j^+} _i,j and relations

    [[x_i^+, x_j^-],x_k^+]=2 ∑ _tγ _ijk^t x_t^+,


    [[x_i^-, x_j^+],x_k^-]=2 ∑ _tγ _ijk^t x_t^-,


    [x_i^-, x_j^-] = [x_i^+, x_j^+] = 0.


The resulting algebra L is ℤ-graded (let x_i^+=2,  x_i^-=-2).
Moreover, L is spanned by x_i^+, x_j^-, [x_i^+, x_j^-], which implies that L_i=(0) for i≠ -2,0,2.




Choose x_1=1. Then x_1^+, x_1^-, [x_1^+, x_1^-] is an 𝔰𝔩(2)-triple, J^+=span(x_i^+, i∈ I)=L_2,  J^-=span(x_i^-, i∈ I)=L_-2 are eigenspaces of L with respect to ad(h).




The (super)algebra L=TKK(J) is universal in the following sense. Let L' be a Lie (super)algebra, L'_0̅⊃𝔰𝔩(2)= Fe' + Ff' + Fh',  L'=L'_-2+L'_0+L'_2,  L'_0=[L'_-2, L'_2],  (L^-_2,∘ )≅ J.
Then there exists an epimorphism 
    φ :TKK(J)φ⟶L',  φ (x_1^+)=e', φ (x_1^-)=f',

the kerφ lies in the center of TKK(J).




It is easy to see that a Lie (super)algebra with this universal property is unique. In particular, the construction above does not depend on a choice of a basis in J.




 §.§ The reduced Tits-Kantor-Koecher construction






Again, let J be a Jordan (super)algebra with 1.
Consider two copies of the vector space J: J^-, J^+, and their direct sum J^-⊕ J^+.
For arbitrary elements a^-∈ J^-,b^+∈ J^+, consider the linear operator 
    δ (a^-,b^+):J^-⊕ J^+→ J^-⊕ J^+,   x^-↦ -(-1)^|a||b|{ x,b,a } ^-,  x^+↦{ x,a,b } ^+.





It follows from Jordan identities that the span δ (J^-,J^+) is a Lie (super)algebra.
The direct sum of vector spaces L=TKK(J)=J^-⊕δ (J^-,J^+)⊕ J^+ with the operations [J^-,J^-]=[J^+,J^+]=(0),  [a^-,b^+]=2δ (a^-,b^+) is a Lie superalgebra.
The elements 1^-,1^+,2δ (1^-,1^+) form an 𝔰𝔩(2)-triple. The Lie superalgebra TKK(J) is called the reduced Tits-Kantor-Koecher Lie (super)algebra of J. It is easy to see that Z(TKK(J))=(0).




 §.§ Cyclic homology of Jordan (super)algebras






Let J be a Jordan (super)algebra with the identity element e.
Let L=TKK(J)=J^++δ (J^+,J^-)+ J^-.
Let (x|y) be a 2-cocycle on the (super)algebra L.
The mapping J^+⊗ e^-→ [J^+, e^-],  a^+⊗ e^-↦ [a^+, e^-] is a bijection.
Indeed, if [a^+, e^-]=0, then [[a^+, e^-],e^+]=2a^+=0.
Hence we can define a linear functional λ : L→ F such that for an arbitrary element a∈ J we have (a^+  |  e^-)=λ ([a^+,e^-]).
Subtracting the coboundary corresponding to λ we can assume that

    (J^+  |  e^-)=0.





Let C_0^2(L) be the vector space of 2-cocycles satisfying (<ref>).
We have shown that C^2(L)=C_0^2(L) + B^2(L).





    Let (x|y) be a 2-cocycle from C_0^2(L). Then (J^-  |  e^+)=0.


    An arbitrary element from J^- can be represented as [[a^+, e^-],e^-],  a^+∈ J^+.
    Applying the identity ([x,y]  |  z) = (x  |  [y,z])- (-1)^|x|· |y|(y  |  [x,z]) twice, we get
    
    ([[a^+, e^-],e^-]  |  e^+)    = ([a^+, e^-]  |  [e^-,e^+]) - (e^-  |  [[a^+, e^-],e^+]) 
       = (a^+  |  [e^-,[ e^-,e^+]]) - (e^-  |  [a^+, [e^-,e^+]]) - (e^-  |  [[a^+, e^-],e^+]) 
       = -2(a^+  |  e^-) - 4(e^-  |  a^+) = 2(a^+  |  e^-) = 0.



From now on we consider a 2-cocycle (x  |  y) such that

    (J^-  |  e^+)=(J^+  |  e^-)=(0).


Define the bilinear mapping J× J→ F as (a  |  b)=(a^+  |  b^-).



    For arbitrary elements a,b∈ J_0̅∪ J_1̅, we have (a  |  b) + (-1)^|a|· |b| (b  |  a)=0.


    
    (a^+  |  b^-)    = -1/2 ([[a^-,e^+],e^+]  |  b^-)
       = -1/2 ( ([a^-,e^+]  |  [e^+,b^-]) - (e^+  |  [[a^-,e^+],b^-]) ) 
       = -1/2 ([a^-,e^+]  |  [e^+,b^-]) 
       = -1/2 ( (a^-  |  [e^+,[e^+,b^-]]) - (e^+  |  [a^-,[e^+,b^-]]) ) 
       = (a^-  |  b^+),

    since [e^+,[e^+,b^-]]=-2b^+. Now, (a^-  |  b^+)=-(-1)^|a|· |b|(b^+  |  a^-), which completes the proof of the lemma.




    For arbitrary elements a,b,c∈ J, we have
    
    (ab  |  c) + (-1)^|a|(|b|+|c|)(bc  |  a) + (-1)^|b|· |c|(ac  |  b) = 0.



    Let a,b,c,d∈ J_0̅∪ J_1̅. The equality (<ref>) applied to ([a^+,b^-]  |  [c^+,d^-]) yields:
     
    ([[a^+,b^-],c^+]  |  d^-) - (-1)^|c|· |b| + |c|· |a| + |a|· |b| (c^+  |  [[b^-,a^+],d^-])
       = - (-1)^|b|· |c| + |b|· |d| + |c|· |d| ([a^+,[d^-,c^+]]  |  b^-) + (a^+  |  [b^+,[c^+,d^-]]).

    In the language of Jordan triple products, it looks as:
     
    ({ a,b,c }  |  d) - (-1)^|c|· |b| + |c|· |a| + |a|· |b| (c  | { b,a,d } )
       = - (-1)^|b|· |c| + |b|· |d| + |c|· |d| ({ a,d,c }  |  b) + (a  | { b,c,d }).

    Equivalently,
    
    ({ a,b,c }  |  d) + (-1)^|a|· |b| + |d|· |c| ({ b,a,d }  |  c)
       + (-1)^|b|· |c| + |b|· |d| + |c|· |d| ({ a,d,c }  |  b)
       + (-1)^|a|(|b|+|c|+|d|)({ b,c,d }  |  a) = 0.

    Let d=e. Then we get the assertion of the lemma.




Let J be a Jordan (super) algebra. We call a (super) skew-symmetric bilinear mapping J× J→ F a cyclic cocycle if the identity (<ref>) holds.



    The identity (<ref>) immediately implies that (J  |  e)=(0).


Let C(J) be the vector space of all cyclic cocycles on J.
We have described the linear mapping C_0^2(L)μ⟶C(J).
Since a 2-cocycle on L is uniquely determined by its values on J^+⊗ J^-, it follows that the mapping μ is injective.



    The mapping μ is bijective.


    We need to show that (<ref>) implies (<ref>).
    Again without loss of generality, we will consider the case of algebras, not superalgebras.
    Choose arbitrary elements a,b,c,d∈ J.
    The left hand side of (<ref>) is:
    
    ((ab)c^(1) + a(bc)^(2) - b(ac)^(3)  |  d) + ((ab)d^(1) + b(ad)^(4) - (bd)a^(5)  |  c) 
       + ((ad)c^(4) + a(dc)^(6) - (ac)d^(3)  |  b) + ((bc)d^(2) + b(cd)^(6) - (bd)c^(5)  |  a).

    The upper script is the number of the “grouping".
    For example, in the group (1), the identity (<ref>) is applied to three elements ab,c,d.
    We get 
    - (dc  |  ab)^(1) - (ad  |  bc)^(2) + (bd  |  ac)^(3)
       - (bc  |  ad)^(4) + (ac  |  bd)^(5) - (ab  |  dc)^(6).

    This expression is equal to 0 since the cocycle (x  |  y) is skew-symmetric.


Recall that Inder(J) is the span of all inner derivations of a Jordan superalgebra J, and Inder(J) is a Lie superalgebra.
For arbitrary elements a,b,c∈ J_0̅∪ J_1̅, we have

    D(ab,c) + (-1)^|a|(|b|+|c|)D(bc,a) + (-1)^|b|· |c|D(ac,b) = 0

(see <cit.>).




Let λ :Inder(J)→ F be a linear functional.
From (<ref>) it follows that (a  |  b)=λ (D(a,b)) is a cyclic cocycle of the Jordan super-algebra J.
We call such cocycles cyclic coboundaries.




Let B(J) be the vector space of all cyclic coboundaries of J.
Following S. Tan <cit.> and B. Allison, G. Benkart, Y. Gao <cit.>[In <cit.>, HC(J) is called the full skew-dihedral homology group.], we call HC(J)=C(J)/B(J) the cyclic homology space of the Jordan (super) algebra J.




In the important case of J coming from associative algebra, HC(J) is the first cyclic homology group of A. Connes.



    μ (B^2(L)∩ C_0^2(L)) = B(J).


    Let a 2-cocycle φ∈ C_0^2(L) be a coboundary. It means that there exists a linear functional λ :L→ F such that φ (a  |  b)=λ ([a,b]). In particular, λ ([J^+,e^-])=λ ([J^-,e^+])=(0).




    Let ψ = μ (φ). Then for arbitrary elements a,b∈ J, we have ψ (a  |  b)=λ ([a^+,b^-]).
    By Lemma 3, λ ([a^+,b^-]) = - (-1)^|a|· |b|λ ([b^+,a^-])=1/2(λ([a^+,b^-])-(-1)^|a|· |b|λ ([b^+,a^-])).




    Define the linear functional δ : Inder(J)→ F,  δ (D(a,b))=λ ([a^+,b^-]).
    



    We need to show that ∑ _i D(a_i,b_i)=0 implies λ (∑ _i [a_i^+,b_i^-])=0. Without loss of generality, we assume that |a_i|+|b_i| does not depend on i. We will show that ∑ _i D(a_i,b_i)=0 implies ∑ _i ([a_i^+,b_i^-] - (-1)^|a_i|· |b_i|[b_i^+,a_i^-]) ∈ Z(L). Therefore ∑ _i ([a_i^+,b_i^-] - (-1)^|a_i|· |b_i|[b_i^+,a_i^-]) = 0 as L is the reduced Tits-Kantor-Koecher Lie superalgebra of J.




    For an arbitrary element c∈ J, we have
    
    [∑ _i ([a_i^+,b_i^-] - (-1)^|a_i|· |b_i|[b_i^+,a_i^-]),c^+] 
       =2 ∑ _i ({ a_i,b_i,c } ^+ - (-1)^|a_i|· |b_i|{ b_i,a_i,c } ^+ ) 
       = -4(-1)^|c|(|a_i|+|b_i|)(c∑ _i D(a_i,b_i))^+ = 0.


    Similarly,
    
    [∑ _i ([a_i^+,b_i^-] - (-1)^|a_i|· |b_i|[b_i^+,a_i^-])),J^-]=(0).


    We showed that
    
    μ (B^2(L)∩ C_0^2(L)) ⊆ B(J).


    Now let δ : Inder(J)→ F be a linear functional. We will show that for arbitrary elements a_i,b_i∈ J, ∑ _i [a_i^+,b_i^-]=0 implies ∑ _i D(a_i,b_i)=0.




    For an arbitrary element x∈ J, we have { x,a,b } = (xa)b + x(ab) - (-1)^|a|· |b|(xb)a = x(D(a,b)+R(ab)). Hence 
    [x^-,∑ _i ([a_i^+,b_i^-]]=2(x(∑ _i D(a_i,b_i) + R(∑ _i a_ib_i)))^-,
 which implies 
    ∑ _i D(a_i,b_i) + R(∑ _i a_ib_i) = 0.


    Applying the left hand side operator to x=e, we will get ∑ _i a_ib_i = 0, and therefore, ∑ _i D(a_i,b_i) = 0.




    Define the linear functional λ : L→ F as follows: 
    λ (J^+) = λ (J^-) = (0),  λ ([a_i^+,b_i^-])=δ (D(a_i^+,b_i^-)).

    The cocycle corresponding to λ lies in C_0^2(L) and μ (λ )=δ.


We proved the following theorem.


    H^2(L)≅ HC(J).


    Let J be a finite dimensional simple Jordan algebra with the trace tr: J→ F (see <cit.>). Let F[t,t^-1] be the algebra of Laurent polynomials, let Resf(t) denote the coefficient of the polynomial f(t) at t^-1.
    Then,
		
    (J⊗ F[t,t^-1])× (J⊗ F[t,t^-1]) → F,  (af(t)  |  bg(t))=tr(ab)Res(f'(t)g(t)),
 
		a,b∈ J, is a cyclic cocycle on the Jordan algebra J⊗ F[t,t^-1] and 
		
    _F HC(J⊗ F[t,t^-1]) = 1.






§ BRACKETS



Let A=A_0̅+ A_1̅ be an associative commutative superalgebra. A binary bilinear operation [  ,  ]:A× A→ A is called a Poisson bracket if

    
  * (A,[  ,  ]) is a Lie superalgebra,
    
  * [ab,c]=a[b,c]+(-1)^|b|· |c|[a,c]b.


    Let A=F[p_1,… ,p_n,q_1,… ,q_n] be a polynomial algebra in 2n variables. Then 
    [f,g]=∑ _i=1^n(∂ f/∂ p_i∂ g/∂ q_i -∂ f/∂ q_i∂ g/∂ p_i )
 is a Poisson bracket in A.


    Let G(n) be the Grassmann algebra on an n-dimensional vector space V. Let ξ _1,… ,ξ _n be a basis of V. The bracket [ξ _i,ξ _j]=δ _ij,  1≤ i,j≤ n, uniquely extends to a Poisson bracket on the superalgebra G(n)=G(n)_0̅+ G(n)_1̅.


An associative commutative superalgebra with a Poisson bracket is called a Poisson superalgebra.




The Poisson superalgebra of Example 2 is denoted as H_n.




Given two Poisson superalgebras A,B, their tensor product is again a Poisson superalgebra: 
    [a_1⊗ b_1, a_2⊗ b_2]=(-1)^|b_1|· |a_2|([a_1,a_2]⊗ b_1b_2 + a_1a_2⊗ [b_1,b_2]).


I. Kantor <cit.> noticed that if A is a Poisson superalgebra then the vector space J=A+Av with the operation that extends the multiplication in A and a(bv)=abv,  (bv)a=(-1)^|a|bav,  (av)(bv)=(-1)^|b|[a,b];  a,b∈ A_0̅∪ A_1̅, is a Jordan superalgebra.
We call it the Kantor double of the bracket [  ,  ] and denote it as K(A, [  ,  ]) or simply K(A).




There exists, however, non-Poisson brackets whose Kantor doubles are Jordan superalgebras. We call such brackets Jordan brackets.
D. King and K. McCrimmon <cit.> characterized Jordan brackets in terms of identities.



    Let A be an associative commutative superalgebra with an even derivation D. Then the bracket [a,b]=D(a)b-aD(b) is Jordan, though not Poisson. We say that [a,b] is a bracket of vector type.

We notice that if A is an associative commutative superalgebra with the identity element 1 and [  ,  ] is a Jordan bracket on A, then a'=[a,1] is an even derivation on A. For arbitrary elements a,b,c∈ A_0̅∪ A_1̅, we have 
    [ab,c]=a[b,c]+(-1)^|b|· |c|[a,c]b+abc' .


    Let A=A_0̅+ A_1̅ be an associative commutative superalgebra with a Jordan bracket. Then there exists a unique Jordan bracket on A⊗ G(n) that
    
        
  * extends the Jordan bracket on A,
        
  * extends the Poisson bracket (see Example 3) on G(n),
        
  * [ξ _1⋯ξ _k,a]=(k-1)ξ _1⋯ξ _k a' for an arbitrary element a∈ A.
    


    Straightforward computation.




    A binary bilinear product [  ,  ] : A× A→ A is called a contact bracket if
    
        
  * (A, [  ,  ]) is a Lie superalgebra,
        
  * the linear transformation D:a↦ [a,1], a∈ A, is an even derivation of A,
        
  * [ab,c] = a[b,c] + (-1)^|b|· |c|[a,c]b + ab D(c) for arbitrary elements a,b,c∈ A_0̅∪ A_1̅.
    


N. Cantarini and V. Kac <cit.> noticed that Jordan brackets are in 1-1 correspondence with contact brackets. More precisely, if [a,b] is a contact bracket with derivation D(a)=[a,1], then ⟨ a,b ⟩ = [a,b]-1/2(D(a)b-aD(b)) is a Jordan bracket. Even derivations corresponding to the brackets [  ,  ],⟨  , ⟩ are different: ⟨ a,1 ⟩ = 1/2[a,1].

An associative commutative superalgebra A with a contact bracket is called a contact algebra.

Now we are ready to formulate the analog of Lemma 7 for contact algebras.

    Let A be a contact algebra. Then there exists a unique contact bracket on A⊗ G(n) that
    
        
  * extends the contact bracket on A,
        
  * extends the Poisson bracket on G(n),
        
  * [ξ _1⋯ξ _k,a]=k-2/2ξ _1⋯ξ _k a' for an arbitrary element a∈ A.
    


    The lemma above is a partial answer to the Question 1 from <cit.>.



Let A be an associative commutative superalgebra with a contact bracket [a,b]. Following N. Cantarini and V. Kac <cit.>, we define the Jordan bracket ⟨ a,b ⟩ = [a,b]-1/2(D(a)b-aD(b)). Extend the Jordan bracket ⟨ a,b ⟩ to a Jordan bracket on A⊗ G(n) as in Lemma 7. Extend the contact bracket [a,b] to a contact bracket on A⊗ G(n+3) as in Lemma 8.
Let L=(A⊗ G(n+3),[  ,  ]).




Then (refn. <cit.>) 
    TKK(K(A⊗ G(n),⟨  , ⟩ ))≅ [L,L] .





§ CYCLIC HOMOLOGY OF KANTOR DOUBLES


Let A be an associative commutative superalgebra with a Jordan bracket [  ,  ]. Let J=K(A, [  ,  ]) be the Kantor double, J=A+Av,  J_0̅=A_0̅+ A_1v,  J_1̅=A_1̅+ A_0v.



 §.§ Poisson centers






Let A be an associative commutative superalgebra with a Jordan bracket [x,y]. We define the Poisson center of A as Z_p={ u∈ A  |  u'=0, (cu)'+[c,u]=0, ∀ c∈ A }.
If [x,y] is a Poisson bracket then Z_p is the well known Poisson center.




Let J = A + Av the Kantor double Jordan algebra and λ : A→ F be a linear functional. Define a bilinear mapping J× J→ F,   x× y↦ (x  |  y)_λ via (A  |  A)_λ=(0),(Av  |  A)_λ=(0), (av  |  bv)_λ=(-1)^|b|λ (ab);  a,b∈ A.



(1) For an arbitrary linear functional λ∈ A^*, (x  |  y)_λ is a cyclic cocycle of J;


(2) (x  |  y)_λ is a coboundary if and only if λ (Z_p)=(0).


    The assertion (1) is straightforward.
    Let us check the assertion (2).
    



Choose elements a,b ∈ A_0̅∪ A_1̅ and consider the inner derivation D(av,bv)=R(av)R(bv)-(-1)^(|a|+1)(|b|+1)R(bv)R(av) of the Jordan superalgebra J=Kan(A). For an arbitrary element c∈ A we have
    
    cD(av,bv)    = cav· bv + (-1)^|a||b|+|a|+|b|cbv· av 
       = (-1)^|b|[ca,b]+(-1)^|a||b|+|b|[cb,a] 
       = (-1)^|b|+|a||b|[c,b]a+(-1)^|b|c[a,b]+(-1)^|b|cab'
        .4cm +(-1)^|b|[c,a]b+(-1)^|b|+|a||b|c[b,a] +(-1)^|a||b|+|b|cba' 
       =(-1)^|b|((-1)^|a||b|[c,b]a+[c,a]b)+(-1)^|b|c(ab)' .

    On the other hand, 
    [c,ab]=[c,a]b+(-1)^|a||b|[c,b]a-c'ab.

    Hence 
    cD(av,bv)=(-1)^|b|([c,ab]+(cab)') .

    Furthermore, 
    vD(av,bv)=(-1)^|a|[1,a]bv+(-1)^|a||b|+|a|[1,b]av=(-1)^|a|[1,ab]v=-(-1)^|b|v(ab)'.

    This implies that ∑ _i D(a_iv,b_iv)=0 if and only if ∑ _i (-1)^|b_i|a_ib_i∈ Z_p.




   Suppose that a cyclic cocycle (x  |  y)_λ is a coboundary. Then there exists a linear functional μ : Inder(J)→ F, such that (x  |  y)_λ = μ (D(x,y)). Let u∈ Z_p. Then D(uv,v)=0, which implies (uv  |  v)_λ=λ (u)=μ (0) = 0. We proved that λ (Z_p)=(0).




    Conversely, suppose that λ (Z_p)=(0). Define μ : D(Av,Av) → F via μ (D(av,bv))=(-1)^|b|λ (ab). To show that this mapping is well-defined, we need to verify that ∑ _i D(a_iv,b_iv)=0 implies ∑ _i(-1)^|b_i|λ (a_ib_i)=0. But we know that ∑ _i D(a_iv,b_iv)=0 if and only if  ∑ _i (-1)^|b_i| a_ib_i ∈ Z_p.  And  ∑ _i (-1)^|b_i|  a_ib_i ∈ Z_p implies that ∑ _i(-1)^|b_i|λ (a_ib_i)=0 by our assumption.




    Extend μ to a mapping Inder(J)→ F via μ (D(A,A))=μ (D(Av,A)) = (0). Hence the cocycle  (x|y)_λ is a coboundary. This completes the proof of the lemma.





 §.§ Bracket cyclic cocycles






Let (x  |  y) be a cyclic cocycle on J=Kan(A). Consider the linear functional λ : A→ F defined via λ (a)=(av  |  v). Now,

    (av  |  bv) = (a |  v· bv) - (-1)^|a|(v  |  a· bv) = (-1)^|b|+1(a  |  b') + (-1)^|b|+1λ (ab).


It follows that

    (a'  |  b) = (-1)^|a|· |b|(b'  |  a).


Let us explore the cocycle condition for av,bv,c;  a,b,c∈ A_0̅∪ A_1̅:

    (av· bv  |  c) + (-1)^(|a|+1)(|b|+|c|+1)(bv· c  |  av) + (-1)^|c|(|a|+|b|)(c· av  |  bv) = 0.


In view of (<ref>), it is equivalent to

    ([a,b]  |  c) = (a'  |  bc) - (-1)^|a|· |b|(b'  |  ac).



    We call a cyclic cocycle (x  |  y) a bracket cyclic cocycle if (<ref>) and (<ref>) hold.



    (f  |  g) = Resf'g is a bracket cyclic cocycle on F[t,t^-1] relative to the bracket [f,g]=f'g-fg'.


We showed that if (x  |  y) is a cyclic cocycle on Kan(A), then 
    
  * the restriction of (x  |  y) to A× A is a bracket cyclic cocycle,
    
  * (av  |  bv) = (-1)^|b|+1(a  |  b') -(a  |  b)_λ.




    Let (x  |  y) be a bracket cyclic cocycle on A; let λ∈ A^*, J=Kan(A)=A+Av. Then the super skew-symmetric mapping J× J→ F that
    
        
  * extends (x  |  y) on A× A,
        
  * (av  |  bv) = (-1)^|b|+1(a  |  b') -(a  |  b)_λ,
        
  * (Av  |  A)=(0),
    
    is a cyclic cocycle on J.


    Choose three elements for the cocycle identity. If all three elements lie in A or two elements of them lie in Av, then the identity follows from the computation above. If one or three elements lie in Av, then the identity follows from (Av  |  A)=(0).



Let C_br(A) be the vector space of all bracket cyclic cocycles A× A→ F.




Each bracket cyclic cocycle on A can be extended to a cyclic cocycle on J via (av  |  bv) = (-1)^|b|(a  |  b'), (Av  |  A)=(0). Hence C_br(A)⊆ C(J).




For arbitrary elements a,b∈ A, the inner derivation D(a,b) of J is zero. Hence C_br(A)∩ B(J)=(0) and therefore HC_br(A)=C_br(A)⊆ HC(J).



 §.§ Mixed cocycles







    We call a cyclic cocycle (x  |  y) on J mixed if (A  |  A)=(Av  |  Av)=(0).




    Let ⟨  | ⟩ :A× A→ F be a bilinear mapping. The mapping (  |  ) : J× J→ F, (a  |  bv) = ⟨ a  |  b⟩ , (bv  |  a) = -(-1)^|a|(|b|+1)⟨ a  |  b⟩ , (A  |  A)=(Av  |  Av)=(0) is a cyclic cocycle if and only if




    (i) ⟨ [a,b]  |  c⟩ + (-1)^|a|(|b|+|c|)⟨ [b,c]  |  a⟩ + (-1)^|c|(|a|+|b|)⟨ [c,a]  |  b⟩ = 0,

    (ii) ⟨ a  |  bc⟩ + (-1)^|a|(|b|+|c|)⟨ b  |  ca⟩ + (-1)^|c|(|a|+|b|)⟨ c  |  ab⟩ = ⟨ abc  |  1⟩,
    



for arbitrary elements a,b,c∈ A_0̅∪ A_1̅.


    The bilinear mapping (x  |  y) is super skew-symmetric by definition. We need to examine the cyclic cocycle identity for the following two triples:
    
        
  * av,bv,cv;
        
  * av,b,c,
    
    where a,b,c∈ A_0̅∪ A_1̅.

    Again to simplify computation, without loss of generality, we will assume that A=A_0̅.
    Then 
		
		 
    (av· b   |  c) + (bc   |  av) + (c· av   |  b) 
       = -(c   |  abv) + (bc   |  av) - (b   |  cav) 
       = - ⟨ c  |  ab⟩ + ⟨ bc  |  a⟩ - ⟨ b  |  ca⟩ .

		
		Furthermore,
		
    (av· bv   |  cv) + (bv· cv   |  av) + (cv· av   |  bv) 
       = ([a,b]   |  cv) + ([b,c]   |  av) + ([c,a]   |  bv) 
       = ⟨ [a,b]  |  c⟩ + ⟨ [b,c]  |  a⟩ + ⟨ [c,a]  |  b⟩ .

    So, the cocycle identity is satisfied for elements av,   bv,   cv, a,b,c ∈ A if and only if the identity 
		(i) is satisfied.
		


    Suppose that (x  |  y) is a cyclic cocycle. Then
		
		
    (av· b   |  c) + (bc   |  av) + (c· av   |  b) = 0.

		
		Doing b = c = 1 in (<ref>) we get
		
    
    ⟨ 1 |  a ⟩ = (1  |  av) = 0, a∈ A.

    
    Applying the cocycle identity to elements a,b,v, we get (a  |  bv) + (b  |  va) + (v  |  ab) = 0, which implies
    
    ⟨ a |  b ⟩ + ⟨ b |  a ⟩ = ⟨ ab |  1 ⟩ .


    Applying (<ref>) to the right hand side of (<ref>) we get
		
    ⟨ ab |  c ⟩ - ⟨ abc |  1 ⟩ + ⟨ bc |  a ⟩ + ⟨ ca |  b ⟩ - ⟨ abc |  1 ⟩ =0.

		
		That is,
		
    ⟨ ab |  c ⟩ + ⟨ bc |  a ⟩ + ⟨ ca |  b ⟩ = 2 ⟨ abc |  1 ⟩ .


    Again applying (<ref>) to each summand on the identity above, we get 
		
    - ⟨ c |  ab ⟩ - ⟨ a |  bc ⟩ - ⟨ b |  ca ⟩ + 3 ⟨ abc |  1 ⟩ = 2 ⟨ abc |  1 ⟩ ,

    which implies (ii).




    Now suppose that ⟨ x |  y ⟩ satisfies the identities (i) and (ii). As mentioned above, the identity (i) implies that the cocycle identity is satisfied for elements av,   bv,   cv, a,b,c ∈ A. Substituting b=c=1 in (ii), we get (<ref>). Substituting c=1 in (ii), we get (<ref>).




    Now the identity (ii) together with (<ref>), (<ref>) imply that the last line of (<ref>) is equal to 0, hence the cyclic cocycle identity holds for elements av,b,c. This completes the proof that (x  |  y) is a cyclic cocycle.



Question. Which mixed cocycles are coboundaries?




Consider the linear mapping 
    μ : A⊗_F A ↦ A,   (a ⊗ b)↦ [a,b]+(-1)^|a|· |b|b'a.

Denote W=Kerμ.

    
        
  * ∑ _i D(a_i,b_iv)=0 if and only if ∑ _i a_i⊗ b_i∈ W;
        
  * let ⟨  |  ⟩ : A× A→ F be a bilinear mapping satisfying i) and ii) in Lemma 11. We can identify it with ⟨  |  ⟩ : A⊗ A→ F, the unique linear map that defines.  The cyclic cohomology (a  |  bv) = ⟨ a |  b ⟩ ,  (A  |  A)=(Av  |  Av)=(0); a,b∈ A is a coboundary if and only if ⟨  , ⟩ vanishes on W.
    


    We have AD(A,Av)=(0). Hence ∑ _i D(a_i,b_iv)=0 if and only if v∑ _i D(a_i,b_iv)=0.




    Let a,b∈ A_0̅∪ A_1̅. Then 
    vD(a,bv)    = (va)(bv) - (-1)^|a|(|b|+1)(v(bv))a  
       = (-1)^|a| + |b| [a,b] + (-1)^|a|· |b| + |a| + |b| b'a
       = (-1)^|a| + |b| ([a,b] + (-1)^|a|· |b| b'a) .

    It implies the assertion 1 of the lemma. The assertion 2 directly follows from the assertion 1, arguing as in the Proof of Lemma 6.



Let us summarize the obtained results.



 §.§ Section Summary






Let Z_p be the Poisson center of an associative commutative superalgebra with a Jordan bracket and J = Kan(A). For an arbitrary linear functional λ : Z_p→ F, consider an extension λ : A→ F. By Lemma 9, the cohomology class (x  |  y)_λ + B(J) does not depend on a choice of  an extension λ.




The vector space HC_Z(J)={ (x  |  y)_λ + B(J)  | λ∈ Z_p^*} can be identified with the dual space Z_p^*.




Recall that C_br(A) denotes the vector space of bracket cyclic cocycle on the algebra A. An arbitrary bracket cyclic cocycle (x  |  y) extends to a cyclic cocycle on J via (av  |  bv) = (-1)^|a|· |b|(b'  |  a), (Av  |  A)=(0). This defines an embedding of C_br(A) into the vector space C(J) and, furthermore, into the cyclic homology space HC(J).




Let C_M(J) denote the vector space of mixed cyclic cocycles on J, 
    HC_M(J)=(C_M(J)+B(J))/B(J).


    HC(J)=HC_Z(J)⊕ C_br(A)⊕ HC_M(J).





§ POISSON BRACKETS AND HAMILTONIAN SUPERALGEBRAS


In this section we compute cyclic homology of Kantor doubles K(A), where A=(A,[  ,  ]) is a Poisson superalgebra.




Recall that H_n=(F[p_1,… ,p_n,q_1,… ,q_n],[p_i,q_j]=δ _ij,[p_i,p_j]=[q_i,q_j]=0),n≥ 1 is a family of Poisson algebras.




Let B be an arbitrary Poisson superalgebra and let A=H_1⊗ B.

    HC(A)= Z_p(B)^*, where Z_p(B) denotes the Poisson center of B.



Consider the Grassmann superalgebra G(3) with the standard Poisson bracket and the tensor product A⊗ G(3) of Poison superalgebras.
Abusing notation we denote the Poisson bracket on A⊗ G(3) as [  ,  ].
Consider the Lie superalgebra L=(A⊗ G(3), [  ,  ]).
The superalgebra [L,L] is isomorphic to the (universal) Tits-Kantor-Koecher construction of the Jordan superalgebra K(A).
Theorems 1, 2, 3 yield description of the vector space H^2([L̅,L̅])=H^2(TKK(K(A))), where L is the quotient of the superalgebra L modulo the center.
In particular we compute second homologies of Hamiltonian superalgebras.




Let A=H_n⊗ G(m),n≥ 1,m≥ 3. Let H(n,m) denote the Lie superalgebra (A,[  ,  ]).

    _F H^2(H(n,m)) = 1.



Throughout this section we assume that A=H_1⊗ _F B, where B is a Poisson superalgebra and H_1=(F[p,q],[p,q]=1).

    C_br(A)=(0).


    If (a  |  b) is a bracket cyclic cocycle on A, then the identity (<ref>) implies that ([A,A]  |  A)=(0).
    It remains to show that [A,A]=A.
    Indeed, the equality p^iq^j=1/j+1[p,p^iq^j+1] implies that [H_1,H_1]=H_1. Consider an arbitrary tensor h⊗ b, h∈ H_1, b∈ B. Suppose that h=∑ _i [h_i',h_i”]. Then h⊗ b=∑ _i [h_i'⊗ b,h_i”⊗ 1].




    Z_p(H_1⊗ B)=Z_p(B).


    The inclusion 1⊗ Z_p(B) ⊆ Z_p(H_1⊗ B) is straightforward.




    Suppose that an element c=∑ _i h_i⊗ b_i lies in Z_p(H_1⊗ B);h_i∈ H_1, b_i∈ B, the elements { b_i } _i are linearly independent. For an arbitrary element h∈ H_1 we have 
    [h⊗ 1,∑ _i h_i⊗ b_i]=∑ _i [h,h_i]⊗ b_i,
 which implies that all elements [h,h_i] are equal to zero, h_i∈ Z_p(H_1)=F· 1. Hence c=1⊗ b,b∈ B. Again for an arbitrary element b'∈ B we have [1⊗ b',1⊗ b]=1⊗ [b',b]=0. Hence b∈ Z_p(B).



Now our aim is

    HC_M(K(A))=(0).


Proof of this proposition requires several lemmas.


    A mixed cyclic cocycle (  | ) on K(A) is a coboundary if and only if ∑ _i [a_i,b_i]=0; a_i,b_i∈ A_0̅∪ A_1̅ implies ∑ _i ⟨ a_i  |  b_i ⟩ = 0. Here ⟨  | ⟩ is the bilinear map defining (  | ).

This lemma immediately follows from Lemma 12 (2).

    Let (x  |  y) be a cyclic cocycle on K(A). Then (A  |  v)=(0).


    We showed in the proof of Lemma 13 that A=[A,A]=(Av)(Av). By the cyclic cocycle identity (av· bv  |  v) + (-1)^|b|(|a|+1)(bv· v  |  av) + (-1)^|a|+|b|(v· av  |  bv)=0. Since [  ,  ] is a Poisson bracket, it follows that bv· v=[b,1]=-b'=0,v· av=(-1)^|a|[1,a]=0.



Let (x  |  y) be a mixed cocycle on K(A). As above let ⟨ a  |  b ⟩ = (a  |  bv); a,b∈ A. By Lemma 16 we have ⟨ A  |  1 ⟩ = (0).




Now the identity (<ref>) and Lemma 11 imply that

    ⟨ a  |  b ⟩ = - (-1)^|a|· |b|⟨ b | a ⟩ ,


    ⟨ a  |  bc ⟩ + (-1)^|a|(|b|+|c|)⟨ b  |  ca ⟩ + (-1)^|c|(|a|+|b|)⟨ c  |  ab ⟩ = 0,


    ⟨ [a,b]  |  c ⟩ + (-1)^|a|(|b|+|c|)⟨ [b,c]  |  a ⟩ + (-1)^|c|(|a|+|b|)⟨ [c,a]  |  b ⟩ = 0.





The identity (<ref>) repeats the first identity of Lemma 11 for convenience of a reader.





Consider the tensor square A⊗ A and the subspace S⊆ A⊗ A spanned by  elements a⊗ b + (-1)^|a|· |b|b⊗ a,ab⊗ c + (-1)^|b|· |c|ac⊗ b + (-1)^|a|(|b|+|c|)bc⊗ a, [a,b]⊗ c + (-1)^|b|· |c|[a,c]⊗ b  + (-1)^|a|(|b|+|c|)[b,c]⊗ a; a,b,c∈  A_0̅∪ A_1̅.





    A⊗ A = p⊗ A + S.


    We will prove the lemma in several steps.
    
        
  * We notice that A=[p,A]. Indeed, p^iq^jb=1/j+1[p,p^iq^j+1b] for an arbitrary element b∈ B. Similarly, A=[q,A].
        
  * For an arbitrary element a∈ A, 1⊗ a∈ S.
        
  * For an arbitrary element a∈ A we have [p,q]⊗ a=p⊗ [q,a]-q⊗ [p,a] S, which implies that
        
    p⊗ [q,a] - q⊗ [p,a] ∈ S.

        In view of Step 1 we conclude that p⊗ A= q⊗ A S.
        
  * Consider the linear operator P: A→ A, a↦ a+q[p,a].
        This operator is a bijection. Indeed, if a=p^iq^jb,b∈ B, then P(a)=(1+j)p^iq^jb.
        
  * Let a∈ A,b∈ B, then 
    b⊗ a = [p,qb]⊗ a =  p⊗ [qb,a] - (-1)^|a|· |b|qb⊗ [p,a] S,

    
    qb⊗ [p,a] = q⊗ b[p,a] + b⊗ q[p,a] S,

    
    b[p,a]=[p,ba],

    
    q⊗ b[p,a] = q⊗ [p,ba] = p⊗ [q,ba] S.

    Hence, 
    b⊗ a    = p⊗ [qb,a] - (p⊗ [q,ba] + b⊗ q[p,a])  S 
       = p⊗ ([q,ba] - [q,ba]) - b⊗ q[p,a]  S .

    Now, 
    
    b⊗ (a+q[p,a]) = p⊗ ([qb,a]-q[b,a])  S,

    
    [qb,a]-q[b,a]=(-1)^|a|· |b|[q,a]b=b[q,a],

    
    b⊗ P(a)=p⊗ b[q,a] S.

    Finally, b⊗ a=p⊗ b[q,P^-1(a)] S.
    
  * Obviously, A⊗ A= p⊗ A + q⊗ A + B⊗ A  S = p⊗ A  S by Step 3 and Step 5.

    This completes the proof of the lemma.
    





    Let (x  |  y) be a mixed cocycle on K(A). Let us check the condition of Lemma 15, that is, ∑ _i [a_i,b_i]=0; a_i,b_i∈ A_0̅∪ A_1̅ implies ∑ _i ⟨ a_i  |  b_i ⟩ = 0.




    By Lemma 17, there exists an element c∈ A such that ∑ _i a_i⊗ b_i = p⊗ c  S. If ∑ _i x_i⊗ y_i ∈ S, then ∑ _i [x_i, y_i] = 0. By (<ref>), (<ref>) and (<ref>), we also have ∑ _i ⟨ x_i  |  y_i ⟩ = 0. Hence we need to show that [p,c]=0 implies ⟨ p  |  c ⟩ = 0.




     The equality [p,c]=0 implies c=∑ _i≥ 0 p^ib_i,b_i∈ B.
     Let us show that ⟨ p  |  p^ib ⟩ = 0. We have p=1/2[p^2,q]. Hence, ⟨ p  |  p^ib ⟩ = 1/2⟨ [p^2,q]  |  p^ib ⟩ = 1/2 ( ⟨ p^2  |  [q,p^ib] ⟩ - ⟨ q  |  [p^2,p^ib] ⟩ ).




     If i≥ 1, then [q,p^ib] = -ip^i-1b. If i=0, then [q,b]=0. We also have [p^2,p^ib]=0. Assuming i≥ 1, we have ⟨ p  |  p^ib ⟩ = -i/2⟨ p^2  |  p^i-1b ⟩ = -i ⟨ p  |  p^ib ⟩ and (i+1)⟨ p  |  p^ib ⟩ = 0. Hence ⟨ p  |  p^ib ⟩ = 0.





§ BRACKETS OF VECTOR TYPE


Let A be an associative commutative superalgebra with an even derivation ' : A→ A.
Consider a Jordan bracket [a,b]=a'b-ab'.
We make the additional assumption that 1∈ A_0̅'A_0̅.
We will determine bracket cyclic cocycles and Poisson center of the Kantor double K(A,[  ,  ]).




Let (a  |  b) be a bracket cyclic cocycle on A. By (<ref>) we have 
    (a'b-ab'  |  c) = (a'  |  bc) - (-1)^|a|· |b|(b'  |  ac).

On the other hand, the cocycle identity implies

    (a'b  |  c) = (a'  |  bc) + (-1)^|a|· |b|(b  |  a'c),


    (ab'  |  c) = (a  |  b'c) + (-1)^|a|· |b|(b'  |  ac).

Substituting these equalities to the left hand side of (<ref>) we get (a'  |  bc) + (-1)^|a|· |b|(b  |  a'c) - (a  |  b'c) - (-1)^|a|· |b|(b'  |  ac) = (a'  |  bc) - (-1)^|a|· |b|(b'  |  ac), which implies 
    (a  |  b'c) = (-1)^|a|· |b|(b  |  a'c).





We claim that there exists a linear functional λ :A→ F such that (a  |  b) = λ (a'b) for arbitrary elements a,b∈ A.




To prove the claim we need to show that ∑ _i a_i'b_i=0;a_i,b_i∈ A implies ∑ _i (a_i  |  b_i) = 0. 

By (<ref>) for an arbitrary element c∈ A, we have 
    (c  |  a_i'b_i) = (-1)^|a_i|· |c|(a_i  |  c'b_i).
  
Hence ∑ _i (-1)^|a_i|· |c|(a_i  |  c'b_i) = 0.



If ∑ _i a_i'b_i=0, then for an arbitrary element x∈ A, we have ∑ _i a_i'b_ix=0. Hence ∑ _i (-1)^|a_i|· |c|(a_i  |  c'b_ix) = ∑ _i (-1)^|a_i|· |c| + |b_i|· |x|(a_i  |  c'xb_i) = 0.



Since 1∈ A_0̅'A_0̅, there exist elements c_j,x_j∈ A_0̅ such that ∑ _j c_j'x_j=1. Now ∑ _i,j (a_i  |  c_j'x_jb_i) = ∑ _i (a_i  |  b_i) = 0.



We have proved that (a  |  b) = λ (a'b); a,b∈ A for some linear functional λ∈ A^*.




For an arbitrary element a∈ A, (a  |  1) = λ (a') = 0. Hence λ (A') = (0).
It is easy to see that if λ is a linear functional on A/A', then (a  |  b) = λ (a'b) is a cyclic cocycle on A. Hence C_br(A) can be identified with the dual space (A/A')^*.




If a lies in the Poisson center of A, then a'=0 and for an arbitrary element b∈ A, we have [b,a]+(ba)'=2b'a=0. Again from 1∈ A_0̅'A_0̅, it follows that a=0, so Z_p(A)=(0).




Remark.  We don't have a description of the space of mixed cyclic cocycles HC_M(K(A)).




§ SUPERALGEBRAS A⊗ G(N),N≥ 1


Let A be an associative commutative superalgebra with a Jordan bracket. Let G(n)= ⟨ 1,ξ _1,… ,ξ _n   | ξ _i ξ _j + ξ _j ξ _i = 0 ⟩ be the Grassmann algebra with a standard Poisson bracket. We have already mentioned above that both brackets on A and G(n) extend to a Jordan bracket on A⊗ G(n) via 
    [A, ξ _i]=(0); ξ _i'=0,  1≤ i≤ n.





In this section we assume that the bracket on A is of vector type [a,b]=a'b-ab', and 1∈ A_0̅'A_0̅.




Let us start with n=1. 

Let (x  |  y) be a bracket cyclic cocycle on A⊗ G(1), G(1)=F· 1+Fξ, ξ ^2=0, [ξ ,ξ ]=1.

As  shown in the previous section, there exists a linear functional λ∈ (A/A')^* such that (a  |  b) = λ (a'b);  a,b∈ A. Choose a∈ A, x∈ A∪ Aξ. Then a=[aξ , ξ ]. Hence 
    (a  |  x) = ([aξ , ξ ]  |  x) = (a'ξ  | ξ x) = (-1)^|a|(ξ  |  a'ξ x).
 This implies that (A  |  Aξ )=(0). 

Now let x=b∈ A. Then (a  |  b) = (-1)^|a|(ξ  |  a'ξb ). Since A'A=A, it follows that 
    (ξ  |  aξ )=(-1)^|a|λ (a), (ξ a   | ξ b)=(-1)^|a|(ξ  | ξ ab)=(-1)^|a|λ (ab),
 for arbitrary elements a,b∈ A.




Again we conclude that C_br(A[ξ ])=(A/A')^*.




Let us show that Z_p(A[ξ ])=Z_p (A). Indeed, the inclusions Z_p (A)⊆ Z_p(A[ξ ]) and Z_p(A[ξ ])∩ A⊆ Z_p (A) are obvious. Let a∈ A such that aξ∈ Z_p(A[ξ ]). Then a=[aξ ,ξ ]=0. This proves the claim.




If [a,b]=a'b-ab' is a vector type bracket and A'A=A, then Z_p (A)=(0).




We have nothing to say about mixed cyclic cocycles of A.





Now let n≥ 2.

    For an arbitrary associative commutative superalgebra A with a Jordan bracket, we have C_br(A⊗ G(n))=(0) for n≥ 2.


     Let (x  |  y) be a bracket cyclic cocycle on A⊗ G(n). We have 
    (ξ _1   |  x )=([ξ _1ξ _2 ,ξ _2 ]   |  x )=((ξ _1ξ _2)'   | ξ _2 x )-(ξ _2'   | ξ _1ξ _2 x )=0.

     We have shown that (ξ _i   |  A⊗ G(n) )= 0, 1≤ i≤ n.




     For an arbitrary element a∈ A, we have a=[aξ _1 ,ξ _1 ]. Hence, 
    (a  |  x) = ([aξ _1 ,ξ _1 ]   |  x ) = (a'ξ _1  | ξ _1 x) + (ξ _1'   |  aξ _1 x) = (a'ξ _1  | ξ _1 x).


     Furthermore, 
    (a'ξ _1  | ξ _1 x) = (a'  | ξ _1·ξ _1 x) + (ξ _1   |  a'ξ _1 x) = 0+0=0.

     We have shown that (A   |  A⊗ G(n) )= (0).



As in the case n=1, we get Z_p(A⊗ G(n))=Z_p(A).






    Let L be a perfect Lie superalgebra. Suppose that there is h∈ L_0̅ such that L=⊕ _i∈ℤL_i,  [h,a_i]=ia_i, ∀ a_i∈ L_i. Let (x  |  y) be a 2-cocycle on L. Then (L_i  |  L_j)=(0) whenever i+j≠ 0.




    Let L be a universal central extension of L. It is known that L=L+Z, where
    Z={∑ _i a_i⊗ b_i  | ∑ _i [a_i,b_i]=0 } /I, and I is the linear span of the set { a⊗ b + (-1)^|a|· |b|b⊗ a, [a,b]⊗ c + (-1)^|a|(|b|+|c|)[b,c]⊗ a + (-1)^|c|(|a|+|b|)[c,a]⊗ b  |  a,b,c∈ L }.
    The ℤ-grading L=⊕ _i∈ℤL_i extends to Z and to the superalgebra L.




    Let us show that Z_n=(0) for n≠ 0. Indeed, Let (x  |  y) be a 2-cocycle of L and let z= ∑ _i a_i⊗ b_i, ∑ _i [a_i,b_i]=0,  (a_i)+ (b_i)=n. 
 
    Since L is perfect, it follows that h=∑ _j [e_j,f_j];e_j,f_j∈ L. Now, 
    ∑ _i ([h,a_i ]  |  b_i) + ∑ _i (a_i  |  [h,b_i ]) =

		
    ∑ _j ([e_j,∑ _i [a_i,b_i]]  |  f_j) + ∑ _j (e_j  |  [f_j,∑ _i [a_i,b_i]]]) = 0.

    
    On the other hand, ∑ _i ([h,a_i ]  |  b_i) + ∑ _i (a_i  |  [h,b_i ]) = n∑ _i (a_i  |  b_i). 
		
		This completes the proof of the lemma.




The following result shed some light on mixed cyclic cocycles of A⊗ G(n).

Mixed cyclic cocycles on J=K(A⊗ G(n)) correspond to 2-cocycles (x  |  y) on Lie superalgebras (A⊗ G(n+3),[  ,  ]) such that (aξ _i_1⋯ξ _i_p  |  bξ _j_1⋯ξ _j_q)≠ 0,   { i_1,… , i_p}≠{ j_1,… , j_q}. 

For a subset π⊆{ 1,… ,n },π ={ i_1<… <i_p }, denote ξ _π = ξ _i_1⋯ξ _i_p.




Let (x  |  y) be a 2-cocycle on a Lie superalgebra (A⊗ G(n),[  ,  ]), where A is a contact superalgebra, [  ,  ] is a contact bracket on A⊗ G(n) that extends the contact bracket on A and the standard Poisson bracket on G(n).


    Let (aξ _π  |  bξ _τ)≠ 0;a,b∈ A;π , τ⊆{ 1,… ,n } ,π≠τ. Then π∪̇τ = { 1,… ,n }.




    Suppose that k∈τ∖π.
    
        
  * Suppose at first that |π∩τ |≥ 2. Then π = π ' ∪̇π” is a disjoint union of two subsets, such that π '∩τ≠∅ and π”∩τ≠∅. 
				
				We have [aξ _π 'ξ _k , ξ _π”ξ _k]=± aξ _π. Hence 
				
    (aξ _π  |  bξ _τ) = ± ([aξ _π 'ξ _k , ξ _π”ξ _k]  |  bξ _τ) =
 
    ± (aξ _π 'ξ _k  |  [ξ _π”ξ _k , bξ _τ]) ± (ξ _π 'ξ _k  |  [aξ _π 'ξ _k , bξ _τ]) = 0.

        



        
  * Now suppose that π∩τ = { i } , π '=π∖{ i } , τ '=τ∖{ i,k }, ξ _π=±ξ _iξ _π ', ξ _τ = ±ξ _i ξ _k ξ _τ '. Let 
    ξ = 1/2(ξ _i + √(-1)ξ _k),η = 1/2(ξ _i - √(-1)ξ _k),[ξ ,ξ ]=[η ,η ]=0,[ξ ,η ]=1/2.

        Then ξ _i = ξ + η , ξ _k = -√(-1)(ξ - η ), h= 2ξη ,[h,ξ ]=ξ ,[h,η ]=-η.
        
				The element aξ _π is a linear combination of elements aξ _π 'ξ and aξ _π 'η. We also have bξ _τ = ± bξ _τ 'ξ _i ξ _k=± 2bξ _τ 'ξη. Now, 
    [h,aξ _π 'ξ ]=± aξ _π 'ξ,   [h,aξ _π 'η ] = ± aξ _π 'η,  [h,bξ _τ 'ξη ]=0.
 Hence (aξ _π  |  bξ _τ)=0 by Lemma 19.
        



        
  * Finally, suppose that π∩τ = ∅ but π∪̇τ≠{ 1,… ,n }. Let 
 i∈{ 1,… ,n }∖ (π∪̇τ ). Consider again the elements ξ = 1/2(ξ _i + √(-1)ξ _k),η = 1/2(ξ _i - √(-1)ξ _k),   h= 2ξη. Then the element bξ _τ is a linear combination of elements bξ _τ 'ξ,   bξ _τ 'η,  [h,bξ _τ 'ξ ]=± bξ _τ 'ξ,  [h,bξ _τ 'η ]=± bξ _τ 'η,  [h,aξ _π] = 0. Again by Lemma 19, (aξ _π  |  bξ _τ)=0.
    

    This completes the proof of the lemma.

















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