In this paper, we study the truncated polynomial algebra L in n variables , and discuss the following four problems in detail: 1 ) Homology complex and homology group of Poisson algebra L ; 2 ) Given a new Poisson bracket by calculation modular derivation of Frobenius Poisson algebra; 3 ) Calculate the twisted homology group of Poisson algebra L ; 4 ) Verify the theorem of twisted Poincaré duality between twisted Poisson homology and Poisson Cohomology.
Poisson structures appear in a large variety of different contexts, ranging from string theory, classical/quantum mechanics and differential geometry to abstract algebra, algebraic geometry and representation theory. They play an important role in Poisson geometry, in algebraic geometry and non-commutative. Poisson cohomologies are important invariants of Poisson structures. The set of Casimir elements of the Poisson structure is the 0-th cohomology; Poisson derivations modulo Hamiltonian derivations are the 1-st cohomology. Poisson cohomology appears as one considers deformations of Poisson algebras. Given a Poisson algebra, we can get some vital information about the Poisson algebra structure from calculate its Poisson (Co)homology.
C. Kassel started calculate the (Co)homology of linear Poisson structure (see [
The homology theorem of the singular algebra is few. Launois S and Richard L [
In this paper, we let k is a number field. We consider the truncated polynomial algebras in n variables Λ = k 〈 x 1 , x 2 , ⋯ , x n 〉 / 〈 x i x j − x j x i , x i 2 〉 with the Poisson bracket
{ x i , x j } = λ i j x i x j , ∀ 1 ≤ i < j ≤ n , λ i j ∈ k +
We get some mainly results. In part 4, we get the i-th homology group H P i ( Λ ) of algebra L
H P 0 ( Λ ) = k ⊕ k x 1 ⊕ k x 2 ⊕ ⋯ ⊕ k x n , H P i ( Λ ) = 0 ( i ≥ 1 ) .
In part 5, we calculate Poisson modular derivation after we get the Frobenius pairing
D ( x i ) = { 1 , x i } D = σ − 1 ( ( ( λ i i + 1 + λ i i + 2 + ⋯ + λ i n ) − ( λ 1 i + λ 2 i + ⋯ + λ i − 1 i ) ) x 1 * ⋯ x ^ i * ⋯ x n * ) = ( ( λ i i + 1 + λ i i + 2 + ⋯ + λ i n ) − ( λ 1 i + λ 2 i + ⋯ + λ i − 1 i ) ) x i
then we can get the new Poisson module structure
In part 6, we get the twisted Poisson homology group and we get the results
H P 0 ( Λ , Λ D ) = k , H P n ( Λ , Λ D ) = 0
the elements in H P m ( Λ , Λ D ) ( 1 ≤ m ≤ n − 1 ) with the length of 0 ~ m in ker ∂ m π .
In part 7, we check the twisted Poincaré duality between Poisson homology and Poisson Cohomology P . D . : H P i ( Λ , Λ ) ≅ ( H P i ( Λ , Λ D ) ) * by calculating the Twisted Poisson cohomology.
Definition 1 [
1) ( A , ⋅ ) is a commutative associative algebra over k, with unit 1;
2) ( A , { ⋅ , ⋅ } ) is a Lie algebra over F;
3) The two multiplications are compatible in the sense that
{ x ⋅ y , z } = x ⋅ { y , z } + y ⋅ { x , z } .
where x, y and z are arbitrary elements of A. The Lie bracket { ⋅ , ⋅ } is then called a Poisson bracket
Example 1. The commutative polynomial algebra in two variables S = k [ x , y ] is a Poisson algebra for the bracket defined on the generators by
{ x , y } = 1 ,
or equivalently for any P , Q ∈ S
{ P , Q } = ∂ P ∂ x ∂ Q ∂ y − ∂ Q ∂ x ∂ P ∂ y .
More generally, for any n ≥ 1 , S = k [ x 1 , ⋯ , x n , y 1 , ⋯ , y n ] is a Poisson algebra for the “symplectic” bracket defined on the generators by
{ x i , y j } = δ i , j , { x i , y j } = { y i , y j } = 0 , for all 1 ≤ i , j ≤ n
or equivalently for any P , Q ∈ S
{ P , Q } = ∑ i = 1 n ∂ P ∂ x i ∂ Q ∂ y i − ∂ Q ∂ x i ∂ P ∂ y i .
We refer to this Poisson algebra as the Poisson-Weyl algebra, denoted by S n ( k ) .
Example 2. For any λ ∈ k , the commutative polynomial algebra S = k [ x , y ] is a Poisson algebra for the “multiplicative” bracket defined on the generators by
{ x , y } = λ x y .
More generally, for any n ≥ 2 and for any n × n antisymmetric matrix λ = ( λ i j ) with entries in k, S = k [ x 1 , ⋯ , x n ] is a Poisson algebra for the bracket defined on the generators by
{ x i , x j } = λ i j x i x j for all 1 ≤ i < j ≤ n .
We refer to this Poisson algebras as the Poisson-quantum plane and Poisson-quantum space respectively, denoted by P 2 λ ( k ) and P n λ ( k ) .
Definition 2 [
1) ( M , { − , − } M ) is a right Lie-module over the Lie algebra ( R , { − , − } ) ;
2) { x ⋅ a , b } M = { x , b } M ⋅ a + x ⋅ { a , b } for any a , b ∈ R , x ∈ M ;
3) { x , a b } M = { x , a } M ⋅ b + { x , b } M ⋅ a for any a , b ∈ R , x ∈ M .
Left Poisson modules are defined similarly. In particular, any Poisson algebra R is naturally a right and left Poisson module over itself
For a Poisson algebra A, the space A has a natural right (also, left) Poisson module structure. Given two right Poisson modules ( M , ⋅ , { − , − } M ) and ( N , ⋅ , { − , − } N ) over A, a k-linear map f : M → N is called a morphism of Poisson modules if
f ( m ⋅ a ) = f ( m ) ⋅ a , f ( { m , a } M ) = { f ( m ) , a } N .
for each m ∈ M and a ∈ A . The following two properties on Poisson modules are straightforward.
Proposition 1 [
• a ⋅ α : M → k , ( a ⋅ α ) ( m ) : = α ( m ⋅ a ) .
• { a ⋅ α } M * : M → k , { a ⋅ α } M * ( m ) : = α ( { m , a } M ) .
for each a ∈ A , α ∈ M * , m ∈ M . Similarly, a left Poisson module M yields a right Poisson module M * .
Proposition 2 [
Definition 3 [
⋯ → M ⊗ R Ω p ( R ) → ∂ p M ⊗ R Ω p − 1 ( R ) → ∂ p − 1 ⋯ → ∂ 2 M ⊗ R Ω 1 ( R ) → ∂ 1 M ⊗ R R → ∂ 0 0 (1)
where for p ≥ 1 , ∂ p : M ⊗ R Ω p ( R ) → M ⊗ R Ω p − 1 ( R ) is defined as:
∂ p ( m ⊗ d a 1 ∧ ⋯ ∧ d a p ) = ∑ i = 1 p ( − 1 ) i − 1 { m , a i } M ⊗ d a 1 ∧ ⋯ d a i ^ ⋯ ∧ d a p + ∑ 1 ≤ i < j ≤ p ( − 1 ) i + j m ⊗ d { a i , a j } ∧ d a 1 ∧ ⋯ d a i ^ ⋯ d a j ^ ⋯ ∧ d a p .
It is easily check that ∂ p is well defined, that is ∂ p − 1 ∂ p = 0 .
The complex (1) is called the Poisson complex of R with values in M, and for p ≥ 0 its p-th homology is called the p-th Poisson homology of R with values in M, denoted by H P p ( R , M ) .
Definition 4 [
0 → M → δ 0 χ 1 ( M ) → δ 1 ⋯ → δ p − 1 χ p ( M ) → δ p χ p + 1 ( M ) → ⋯ (2)
where δ p : χ p ( M ) → χ p + 1 ( M ) is defined as F ↦ δ p ( F ) with
δ p ( F ) ( a 1 ∧ ⋯ ∧ a p + 1 ) = ( − 1 ) p + 1 ∑ i = 1 p + 1 ( − 1 ) i { F ( a 1 ∧ ⋯ a ^ i ⋯ ∧ a p + 1 ) , a i } M + ( − 1 ) p + 1 ∑ 1 ≤ i < j ≤ p + 1 ( − 1 ) i + j F ( { a i , a j } ∧ a 1 ∧ ⋯ a ^ i ⋯ a ^ j ⋯ ∧ a p + 1 )
It is easily check that ∂ p is well defined, that is δ p δ p − 1 = 0 .
The complex (2) is called the Poisson cochain complex of R with values in M, and for p ≥ 0 its p-th cohomology is called the p-th Poisson cohomology of R with values in M, denoted by H P p ( R , M ) .
The elements in ker δ 1 are called Poisson derivations, and the elements in Im δ 0 are called Hamiltonian derivations which are of the form { m , − } M , for m ∈ M .
Definition 5 [
Definition 6 [
Proposition 3 [
{ m , a } M D : = { m , a } M + m ⋅ D ( a ) .
Then the R-module M with { − , − } M D is a right Poisson R-module, which is called the twisted Poisson module of M twisted by the Poisson derivation D, denoted by M D .
Lemma 1 [
P . D . : H P i ( A , A ) ≅ ( H P i ( A , A D ) ) *
Before we calculate the homology, it is necessary to write the homology complex first. We give the basis in linear space in every point of the complex. Obviously, Ω 0 ( Λ ) = Λ . Ω 1 ( Λ ) = C 1 d x 1 ⊕ C 2 d x 2 ⊕ ⋯ ⊕ C n d x n , C j = x 1 i 1 x 2 i 2 ⋯ x n i n , i j = 0 and i k = 0 or 1, for i k = 0 . Ω 1 ( Λ ) = ⊕ C x 1 i 1 ⋯ x n i n d x j , i j = 0 , 0 ≤ i 1 , ⋯ , i n ≤ 1 , and C ∈ k .
d x j 2 = 0 , Ω 2 ( Λ ) = ⊕ C x 1 i 1 ⋯ x n i n d x j ∧ d x k , j < k , i j = i k = 0 , 0 ≤ i 1 , ⋯ , i n ≤ 1 , C ∈ k .
Similarly, we can get
Ω 3 ( Λ ) = ⊕ C x 1 i 1 ⋯ x n i n d x p ∧ d x q ∧ d x r , p < q < r , i p = i q = i r = 0 , 0 ≤ i 1 , ⋯ , i n ≤ 1 , C ∈ k ;
Ω 4 ( Λ ) = ⊕ C x 1 i 1 ⋯ x n i n d x p ∧ d x q ∧ d x r ∧ d x s , p < q < r < s , i p = i q = i r = i s = 0 , 0 ≤ i 1 , ⋯ , i n ≤ 1 , C ∈ k ;
⋯
Ω n ( Λ ) = C d x 1 ∧ ⋯ ∧ d x n , C ∈ k ;
Ω m ( Λ ) = 0 ( m > n ) .
So we can get the Poisson complex
→ Ω q + 1 ( Λ ) → ∂ q + 1 π Ω q ( Λ ) → ∂ q π Ω q − 1 ( Λ ) → ∂ q − 1 π ⋯ → Ω 2 ( Λ ) → ∂ 2 π Ω 1 ( Λ ) → ∂ 1 π Λ → ∂ 0 0 (3)
Then we calculate the Poisson homology group and get some results.
Proposition 4.1. H P 0 ( Λ ) = k ⊕ k x 1 ⊕ k x 2 ⊕ ⋯ ⊕ k x n
Proof:
∂ 1 π : Ω 1 ( Λ ) → Λ ∑ j = 1 n C x 1 i 1 ⋯ x j 0 ⋯ x n i n d x j ↦ ∑ j = 1 n C { x 1 i 1 x 2 i 2 ⋯ x j 0 ⋯ x n i n , x j }
With the definition of Poisson bracke, we can write { x i 1 ⋯ x i k | 1 ≤ i 1 < ⋯ < i k ≤ n } as the results of image { x 1 i 1 ⋯ x j 0 ⋯ x n i n , x j } . Next, we will prove that, when i k ≥ 2 , we can find preimage in Ω 1 ( Λ ) for all elements
1) The preimage of image with the length of 2
1 λ i j x i d x j ↦ 1 λ i j { x i , x j } = 1 λ i j λ i j x i x j = x i x j , ( i < j ) .
2) The preimage of image with the length of 3
1 λ i p + λ j p x i x j d x p ↦ 1 λ i p + λ j p { x i x j , x p } = 1 λ i p + λ j p ( x i { x j , x p } + x j { x i , x p } ) = x i x j x p , ( i < j < p ) .
3) The preimage of image with the length of n
1 λ 1 n + λ 2 n + ⋯ + λ n − 1 n x 1 x 2 ⋯ x n − 1 d x n ↦ 1 λ 1 n + λ 2 n + ⋯ + λ n − 1 n { x 1 x 2 ⋯ x n − 1 , x n } ↦ x 1 x 2 ⋯ x n .
4) Generally, The preimage of image with the length of m ( 1 < m < n )
1 λ i 1 i m + ⋯ + λ i m − 1 i m x i 1 ⋯ x i m − 1 d x i m ↦ 1 λ i 1 i m + ⋯ + λ i m − 1 i m { x i 1 ⋯ x i m − 1 , x i m } = 1 λ i 1 i m + ⋯ + λ i m − 1 i m ( x i 1 { x i 2 ⋯ x i m − 1 , x i m } + ⋯ + x i m − 1 { x i 1 ⋯ x i m − 2 , x i m } ) = x i 1 ⋯ x i m .
we can find preimage in Ω 1 ( Λ ) for all elements
Hence, we can find preimage of the image in Ω 1 ( Λ ) when image length i k satisfied i k ≥ 2 . However, we can not find the preimage of image with the length of 1 in Ω 1 ( Λ ) .
So we can get 0-th homology group H P 0 ( Λ ) = ker ∂ 0 π / Im ∂ 1 π = k ⊕ k x 1 ⊕ k x 2 ⊕ ⋯ ⊕ k x n .
The proof is completed.
Proposition 4.2. H P 1 ( Λ ) = 0
Proof:
Ω 2 ( Λ ) → ∂ 2 π Ω 1 ( Λ ) → ∂ 1 π Λ
Firstly, we calculate ker ∂ 1 π
∂ 1 π : Ω 1 ( Λ ) → Λ ∑ j = 1 n C x 1 i 1 ⋯ x j 0 ⋯ x n i n d x j ↦ ∑ j = 1 n C { x 1 i 1 x 2 i 2 ⋯ x j 0 ⋯ x n i n , x j }
the mapping between the elements is
x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 d x n + x 1 j 1 x 2 j 2 ⋯ x n − 2 j n − 2 x n j n d x n − 1 + ⋯ + x 1 p 1 x 3 p 3 ⋯ x n − 1 p n − 1 x n p n d x 2 + x 2 q 2 x 3 q 3 ⋯ x n − 1 q n − 1 x n q n d x 1 ↦ k i 1 , ⋯ , i n { x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 , x n } + k j 1 , ⋯ , j n { x 1 j 1 x 2 j 2 ⋯ x n − 2 j n − 2 x n j n , x n − 1 } + ⋯ + k p 1 , ⋯ , p n { x 1 p 1 ⋯ x n − 1 p n − 1 x n p n , x 2 } + k q 1 , ⋯ , q n { x 2 q 2 x 3 q 3 ⋯ x n − 1 q n − 1 x n q n , x 1 }
We have n preimage, each of them has the length of k ( 1 ≤ k ≤ n − 1 ) . The image has k + 1 in length, then we calculate the image in each item.
The first term:
The second item: k j 1 , ⋯ , j n { x 1 j 1 x 2 j 2 ⋯ x n − 2 j n − 2 x n j n , x n − 1 } = k k 1 , ⋯ , k n ( j 1 λ 1 n − 1 + j 2 λ 2 n − 1 + ⋯ + j n λ n n − 1 ) x 1 j 1 x 2 j 2 ⋯ x n − 1 x n j n
Similarly, we discuss all the situation about the second item in
......
The ( n − 1 ) -th item: k p 1 , ⋯ , p n { x 1 p 1 ⋯ x n − 1 p n − 1 x n p n , x 2 } = k p 1 , ⋯ , p n ( p 1 λ 12 + p 3 λ 32 + ⋯ + p n λ n 2 ) x 1 p 1 x 2 ⋯ x n − 1 p n − 1 x n p n
The n-th item: k q 1 , ⋯ , q n { x 2 q 2 ⋯ x n − 1 q n − 1 x n q n , x 1 } = k q 1 , ⋯ , q n ( q 2 λ 21 + q 3 λ 31 + ⋯ + q n λ n 1 ) x 1 x 2 q 2 ⋯ x n − 1 q n − 1 x n q n
Similarly, we can discuss all the situation about the ( n − 1 ) -th, n-th item.
Because the Poisson structure of L is homogeneous. We can discuss the image of ∂ 1 π by length.
1) The preimage of image with the length of 1
∑ C i j x i d x j ↦ ∑ C i j λ i j x i x j
Let ∑ C i j x i d x j ∈ ker ∂ 1 π , then ∑ C i j λ i j x i x j = 0 . we can get ( C i j − C j i ) λ i j = 0 for ∀ i < j , that is C i j = C j i . we have
| Image length | value | Denoted by | image |
|---|---|---|---|
| 2 | i 1 = 1 , i 2 = i 3 = ⋯ = i n − 1 = 0 | i 1 , 0 , 0 , ⋯ , 0 | λ 1 n x 1 x n |
| i 2 = 1 , i 1 = i 3 = ⋯ = i n − 1 = 0 | i 0 , 1 , 0 , ⋯ , 0 | λ 2 n x 2 x n | |
| ⋯ | ⋯ | ⋯ | |
| i n − 1 = 1 , i 1 = i 2 = ⋯ = i n − 2 = 0 | i 0 , 0 , 0 , ⋯ , 1 | λ n − 1 n x n − 1 x n | |
| 3 | i 1 = i 2 = 1, i 3 = ⋯ = i n − 1 = 0 | i 1 , 1 , 0 , ⋯ , 0 | ( λ 1 n + λ 2 n ) x 1 x 2 x n |
| i 1 = i 3 = 1, i 2 = ⋯ = i n − 1 = 0 | i 1 , 0 , 1 , ⋯ , 0 | ( λ 1 n + λ 3 n ) x 1 x 3 x n | |
| ⋯ | ⋯ | ⋯ | |
| i n − 2 = i n − 1 = 1 , i 1 = ⋯ = i n − 3 = 0 | i 0 , 0 , ⋯ , 1 , 1 | ( λ n − 2 n + λ n − 1 n ) x n − 2 x n − 1 x n | |
| ⋯ | |||
| n − 1 | i 1 = 0 , i 2 = i 3 = ⋯ = i n − 1 = 1 | i 0 , 1 , 1 , ⋯ , 1 | ( λ 2 n + ⋯ + λ n − 1 n ) x ^ 1 x 2 ⋯ x n − 1 x n |
| i 2 = 0 , i 1 = i 3 = ⋯ = i n − 1 = 1 | i 1 , 0 , 1 , ⋯ , 1 | ( λ 1 n + ⋯ + λ n − 1 n ) x 1 x ^ 2 ⋯ x n − 1 x n | |
| ⋯ | ⋯ | ⋯ | |
| i n − 1 = 0 , i 1 = i 2 = ⋯ = i n − 2 = 1 | i 1 , 1 , 1 , ⋯ , 0 | ( λ 1 n + ⋯ + λ n − 2 n ) x 1 ⋯ x n − 2 x ^ n − 1 x n | |
| n | i 1 = i 2 = ⋯ = i n − 2 = i n − 1 = 1 | i 1 , 1 , ⋯ , 1 , 1 | ( λ 1 n + ⋯ + λ n − 1 n ) x 1 ⋯ x n |
| Image length | value | Denoted by | image | |
|---|---|---|---|---|
| 2 | j 1 = 1 , j 2 = ⋯ = j n − 2 = j n = 0 | j 1 , 0 , 0 , ⋯ , 0 | λ 1 n − 1 x 1 x n − 1 | |
| j 2 = 1 , j 1 = ⋯ = j n − 2 = j n = 0 | j 0 , 1 , 0 , ⋯ , 0 | λ 2 n − 1 x 2 x n − 1 | ||
| ⋯ | ⋯ | ⋯ | ||
| j n = 1 , j 1 = ⋯ = j n - 2 = 0 | j 0 , 0 , ⋯ , 1 , 0 | λ n n − 1 x n − 1 x n | ||
| 3 | j 1 = j 2 = 1 , j 3 = ⋯ = j n = 0 | j 1 , 1 , 0 , ⋯ , 0 | ( λ 1 n − 1 + λ 2 n − 1 ) x 1 x 2 x n − 1 | |
| j 1 = j 2 = 1 , j 3 = ⋯ = j n = 0 | j 1 , 0 , 1 , ⋯ , 0 | ( λ 1 n − 1 + λ 3 n − 1 ) x 1 x 3 x n − 1 | ||
| ⋯ | ⋯ | ⋯ | ||
| j n − 2 = j n = 1 , j 1 = ⋯ = j n − 3 = 0 | j 0 , 0 , ⋯ , 1 , 1 | ( λ n − 2 n − 1 + λ n n − 1 ) x n − 2 x n − 1 x n | ||
| ⋯ | ||||
| n − 1 | j 1 = 0 , j 2 = j 3 = ⋯ = j n = 1 | j 0 , 1 , 1 , ⋯ , 1 | ( λ 2 n − 1 + ⋯ + λ n n − 1 ) x ^ 1 x 2 ⋯ x n − 1 x n | |
| j 2 = 0 , j 1 = j 3 = ⋯ = j n = 1 | j 1 , 0 , 1 , ⋯ , 1 | ( λ 1 n − 1 + ⋯ + λ n n − 1 ) x 1 x ^ 2 ⋯ x n − 1 x n | ||
| ⋯ | ⋯ | ⋯ | ||
| j n = 0 , j 1 = j 2 = ⋯ = j n − 2 = 1 | j 1 , 1 , 1 , ⋯ , 0 | ( λ 1 n − 1 + ⋯ + λ n − 2 n − 1 ) x 1 x 2 ⋯ x n − 1 x ^ n | ||
| n | j 1 = j 2 = ⋯ = j n − 2 = j n = 1 | j 1 , 1 , ⋯ , 1 , 1 | ( λ 1 n − 1 + ⋯ + λ n n − 1 ) x 1 ⋯ x n | |
x 1 d x 2 + x 2 d x 1 ∈ ker ∂ 1 π ⋮ x 1 d x n + x n d x 1 ∈ ker ∂ 1 π x 2 d x 3 + x 3 d x 2 ∈ ker ∂ 1 π ⋮ x n − 1 d x n + x n d x n − 1 ∈ ker ∂ 1 π
Obviously, for ∀ i < j , the preimage of x i d x j + x j d x i is 1 λ i j d x i ∧ d x j under the map ∂ 2 π .
2) The preimage of image with the length of 2
∑ C i j k x i x j d x k ↦ ∑ C i j k ( λ i k + λ j k ) x i x j x k
Let ∑ C i j k x i x j d x k ∈ ker ∂ 1 π , then ∑ C i j k ( λ i k + λ j k ) x i x j x k = 0 . we can get C i j k ( λ i k + λ j k ) + C j k i ( λ j i + λ k i ) = 0 for ∀ i < j < k , that is C i j k = λ i k − λ j i λ i k + λ j k C j k i .
we have:
λ 1 n − λ 21 λ 1 n + λ 2 n x 1 x 2 d x n + x 2 x n d x 1 ∈ ker ∂ 1 π ⋮ λ n − 1 n − λ 1 n − 1 λ 1 n + λ n − 1 n x 1 x n − 1 d x n + x 1 x n d x n − 1 ∈ ker ∂ 1 π
Obviously, for ∀ i < j < k , the preimage of λ i k − λ j i λ i k + λ j k x i x j d x k + x j x k d x i is − x j d x i ∧ d x k under the map ∂ 2 π .
3) Generally, the preimage of image with the length of m ( 1 < m < n − 1 )
∑ C i 1 ⋯ i m − 1 i x i 1 ⋯ x i m − 1 x i d x j ↦ ∑ C i 1 ⋯ i m − 1 i ( λ i 1 j + ⋯ + λ i j ) x i 1 ⋯ x i m − 1 x i x j
Let ∑ C i 1 ⋯ i m − 1 i x i 1 ⋯ x i m − 1 x i d x j ∈ ker ∂ 1 π , then ∑ C i 1 ⋯ i m − 1 i ( λ i 1 j + ⋯ + λ i j ) x i 1 ⋯ x i m − 1 x i x j = 0 .
we have:
− ( λ i 1 i + λ i 2 i + ⋯ + λ i m − 1 i + λ i j ) x i 1 x i 2 ⋯ x i m − 1 x i d x j + ( λ i 1 j + λ i 2 j + ⋯ + λ i m − 1 j + λ i j ) x i 1 x i 2 ⋯ x i m − 1 x j d x i ∈ ker ∂ 1 π
Obviously, under the map ∂ 2 π , for ∀ i < j , the preimage of
− ( λ i 1 i + λ i 2 i + ⋯ + λ i m − 1 i + λ j i ) x i 1 x i 2 ⋯ x i m − 1 x i d x j + ( λ i 1 j + λ i 2 j + ⋯ + λ i m − 1 j + λ i j ) x i 1 x i 2 ⋯ x i m − 1 x j d x i is − x i 1 x i 2 ⋯ x i m − 1 d x i ∧ d x j .
(4)The preimage of image with the length of n − 1
∑ C 1 ⋯ k ⌢ ⋯ n x 1 ⋯ x ⌢ k ⋯ x n d x k ↦ ∑ C 1 ⋯ k ⌢ ⋯ n ( λ 1 k + ⋯ + λ n k ) x 1 ⋯ x k ⋯ x n
Let ∑ C 1 ⋯ k ⌢ ⋯ n x 1 ⋯ x ⌢ k ⋯ x n d x k ∈ ker ∂ 1 π , then ∑ C 1 ⋯ k ⌢ ⋯ n ( λ 1 k + ⋯ + λ n k ) x 1 ⋯ x k ⋯ x n = 0 .
we have:
λ 1 n + λ 2 n + ⋯ + λ n − 1 n λ 12 + λ 13 + ⋯ + λ 1 n x 2 x 3 ⋯ x n d x 1 + x 1 x 2 ⋯ x n − 1 d x n ∈ ker ∂ 1 π ⋮ λ 1 n + ⋯ λ j n + ⋯ + λ k n + ⋯ + λ n − 1 n λ 12 + ⋯ λ 1 j + ⋯ + λ 1 k + ⋯ + λ 1 n x 1 ⋯ x ⌢ k ⋯ x n d x k + x 1 ⋯ x ⌢ j ⋯ x n d x j ∈ ker ∂ 1 π
Obviously, under the map ∂ 2 π , for ∀ 1 ≤ j < k ≤ n , the preimage of ( λ 1 n + ⋯ λ j n + ⋯ + λ k n + ⋯ + λ n − 1 n ) x 1 ⋯ x ⌢ k ⋯ x n d x k + ( λ 12 + ⋯ λ 1 j + ⋯ + λ 1 k + ⋯ + λ 1 n ) x 1 ⋯ x ⌢ j ⋯ x n d x j is − x 1 x 2 ⋯ x ⌢ j ⋯ x ⌢ k ⋯ x n d x j ∧ d x k .
In conclusion, ∂ 2 π is surjection. So we have the 1-th homology group
H P 1 ( Λ ) = ker ∂ 1 π Im ∂ 2 π = 0 .
The proof is completed.
Now, we write the general condition in Proposition 4.3, the process of proof is similar with Proposition 4.2.
Proposition 4.3. H P p ( Λ ) = 0 , p ( 1 ≤ p ≤ n ) .
Proposition 4.4. H P n ( Λ ) = 0
Proof: lastly, we calculate the n-th homology group H P n ( Λ ) = ker ∂ n π Im ∂ n + 1 π
Ω n + 1 ( Λ ) → ∂ n + 1 π Ω n ( Λ ) → ∂ n π Ω n − 1 ( Λ )
Firstly, we calculate ker ∂ n π ,
∂ n π : Ω n ( Λ ) → Ω n − 1 ( Λ ) d x 1 ∧ ⋯ ∧ d x n ↦ ∑ 1 ≤ i < j < ⋯ < k ≤ n ( − 1 ) i + j d { F i , F j } ∧ d F 1 ∧ ⋯ ∧ d F ^ i ∧ ⋯ ∧ d F ^ j ∧ ⋯ ∧ d F ^ k ∧ ⋯ ∧ d F n = ( − 1 ) 1 + 2 d { x 1 , x 2 } ∧ d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + ( − 1 ) 1 + 3 d { x 1 , x 3 } ∧ d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ + ( − 1 ) ( n − 1 ) + n d { x n − 1 , x n } ∧ d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2
the image is:
( − 1 ) 1 + 2 d { x 1 , x 2 } ∧ d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + ( − 1 ) 1 + 3 d { x 1 , x 3 } ∧ d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ + ( − 1 ) ( n − 1 ) + n d { x n − 1 , x n } ∧ d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2 = − d { x 1 , x 2 } ∧ d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + d { x 1 , x 3 } ∧ d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ − d { x n − 1 , x n } ∧ d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2
= − λ 12 ( x 1 d x 2 + x 2 d x 1 ) ∧ d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + λ 13 ( x 1 d x 3 + x 3 d x 1 ) ∧ d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ − λ n − 1 n ( x n − 1 d x n + x n d x n − 1 ) ∧ d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2
We can easily see that this element can not be 0, that is ker ∂ n π = 0 , so
H P n ( Λ ) = ker ∂ n π Im ∂ n + 1 π = 0
The proof is completed.
In conclusion, we have the homology group of L H P i ( Λ )
H P i ( Λ ) = { k ⊕ k x 1 ⊕ ⋯ ⊕ k x n , ( i = 0 ) 0 , ( i ≥ 1 ) (4)
In this part, we should calculate the Frobenius modular derivation and get the twisted Poisson structure. Firstly, we need get the Frobenius isomorphism and Frobenius pairing.
The Frobenius algebra L has dimension 2n, with basis { 1 } ∪ { x i 1 ⋯ x i k | 1 ≤ i 1 < ⋯ < i k < n } . The dual space Λ * has the basis { 1 } ∪ { x i 1 * ⋯ x i k * | 1 ≤ i 1 < ⋯ < i k < n } , satisfied
x i * : Λ → k x i ↦ 1 x j ↦ 0 ( j ≠ i )
Frobenius isomorphism σ : Λ → Λ * is given by
σ ( 1 ) : ∑ c i 1 , ⋯ , i n x i 1 ⋯ x i n ↦ c i 1 , ⋯ , i n .
Choosing a basis { x 1 i 1 x 2 i 2 ⋯ x n i n | 0 ≤ i 1 , i 2 , ⋯ , i n ≤ 1 } .
σ : Λ → Λ * 1 ↦ σ ( 1 ) .
We can easily get the Frobenius paring is
〈 x 1 q 1 ⋯ x n q n , x 1 p 1 ⋯ x n p n 〉 = { 1 , q i + p i = 1 , ∀ i 0 , else
Now we calculate the Modular derivation D ( x i )
D ( x i ) = { 1 , x i } D = σ − 1 ( { σ ( 1 ) , x i } A * ) = σ − 1 ( ∑ 1 + k i x i * + k i j x i * x j * + ⋯ + k 1 ⋯ n x 1 * ⋯ x n * ) .
Compute it by each item
{ σ ( 1 ) , x i } ( 1 ) = σ ( 1 ) { x i , 1 } = 0 (5)
{ σ ( 1 ) , x i } ( x j ) = σ ( 1 ) { x i , x j } = σ ( 1 ) ( λ i j x i x j ) = 0 (6)
⋯
{ σ ( 1 ) , x i } ( x 1 ⋯ x ^ j ⋯ x n − 1 ) = σ ( 1 ) { x i , x 1 ⋯ x ^ j ⋯ x n − 1 } = 0 (7)
{ σ ( 1 ) , x i } ( x 1 ⋯ x ^ j ⋯ x n ) = σ ( 1 ) { x i , x 1 ⋯ x ^ j ⋯ x n } (8)
By the definition of σ ( 1 ) , Equations (5)-(7) equal to 0. Then we discuss (8).
when i = j ,
{ x i , x 1 ⋯ x ^ i ⋯ x n } = x 2 x 3 ⋯ x ^ i ⋯ x n { x i , x 1 } + x 1 x 3 ⋯ x ^ i ⋯ x n { x i , x 2 } + ⋯ + x 1 x 2 ⋯ x i − 2 x ^ i ⋯ x n { x i , x i − 1 } + x 1 x 2 ⋯ x ^ i x i + 2 ⋯ x n { x i , x i + 1 } + x 1 x 2 ⋯ x ^ i x i + 1 x i + 3 ⋯ x n { x i , x i + 2 } + ⋯ + x 1 x 2 ⋯ x ^ i ⋯ x n − 1 { x i , x n } = ( ( λ i i + 1 + λ i i + 2 + ⋯ + λ i n ) − ( λ 1 i + λ 2 i + ⋯ + λ i − 1 i ) ) x 1 ⋯ x i ⋯ x n .
We have { σ ( 1 ) , x i } ( x 1 ⋯ x ^ i ⋯ x n ) = ( λ i i + 1 + λ i i + 2 + ⋯ + λ i n ) − ( λ 1 i + λ 2 i + ⋯ + λ i − 1 i )
⇒ { σ ( 1 ) , x i } A * = ( ( λ i i + 1 + λ i i + 2 + ⋯ + λ i n ) − ( λ 1 i + λ 2 i + ⋯ + λ i − 1 i ) ) x 1 * ⋯ x ⌢ i * ⋯ x n * .
We can get
D ( x i ) = { 1 , x i } D = σ − 1 { ( ( λ i i + 1 + λ i i + 2 + ⋯ + λ i n ) − ( λ 1 i + λ 2 i + ⋯ + λ i − 1 i ) ) x 1 * ⋯ x ^ i * ⋯ x n * } = ( ( λ i i + 1 + λ i i + 2 + ⋯ + λ i n ) − ( λ 1 i + λ 2 i + ⋯ + λ i − 1 i ) ) x i . (9)
Then we give the twisted Poisson module of L
{ x i , x j } Λ D = { ( ( λ j j + 1 + ⋯ + λ j n ) − ( λ 1 j + λ 2 j + ⋯ + λ i − 1 j + λ i + 1 j + ⋯ + λ j − 1 j ) ) x i x j , ( i < j ) ( ( λ j j + 1 + ⋯ + λ j i − 1 + λ j i + 1 + ⋯ + λ j n ) − ( λ 1 j + λ 2 j + ⋯ + λ j − 1 j ) ) x i x j , ( i > j ) 0 , ( i = j ) (10)
In this part, we calculate the twisted Poisson homology after we get the new poisson module structure { x i , x j } Λ D . give a right Poisson module Λ D , we have the new canonical chain complex
⋯ → Λ D ⊗ Λ Ω p ( Λ ) → ∂ p π Λ D ⊗ Λ Ω p − 1 ( Λ ) → ∂ p − 1 π ⋯ → ∂ 2 π Λ D ⊗ Λ Ω 1 ( Λ ) → ∂ 1 π Λ D ⊗ Λ Λ → ∂ 0 π 0 . (11)
where p ≥ 1 , ∂ p π : Λ D ⊗ Λ Ω p ( Λ ) → Λ D ⊗ Λ Ω p − 1 ( Λ ) is defined by:
∂ p π ( m ⊗ d a 1 ∧ ⋯ ∧ d a p ) = ∑ i = 1 p ( − 1 ) i − 1 { m , a i } Λ D ⊗ d a 1 ∧ ⋯ d a i ^ ⋯ ∧ d a p + ∑ 1 ≤ i < j ≤ p ( − 1 ) i + j m ⊗ d { a i , a j } ∧ d a 1 ∧ ⋯ d a i ^ ⋯ d a j ^ ⋯ ∧ d a p .
Then we calculate the twisted Poisson homology group and get some results.
∂ 1 π : Ω 1 ( Λ ) → Λ ∑ x 1 i 1 ⋯ x j 0 ⋯ x n i n d x j ↦ { x 1 i 1 ⋯ x j 0 ⋯ x n i n , x j } Λ D .
we will prove that we can find preimage in Ω 1 ( Λ ) for all elements when i k ≥ 1
1) The preimage of image with the length of 1
1 ( ( λ i i + 1 + ⋯ + λ 1 n ) − ( λ 1 i + λ 2 i ⋯ + λ i − 1 i ) ) d x i ↦ x i , ( 1 ≤ i ≤ n ) .
2) The preimage of image with the length of 2
1 ( λ j j + 1 + ⋯ + λ j n ) − ( λ 1 j + ⋯ + λ i − 1 j + λ i + 1 j + ⋯ + λ j − 1 j ) x i d x j ↦ x i x j , ( i < j ) .
3) The preimage of image with the length of 3
1 2 ( λ p p + 1 + ⋯ + λ p n ) − ( λ 1 p + ⋯ + λ j − 1 p + λ j + 1 p + ⋯ + λ p − 1 p ) − ( λ 1 p + ⋯ + λ i − 1 p + λ i + 1 p + ⋯ + λ p − 1 p ) x i x j d x p ↦ x i x j x p , ( ∀ 1 ≤ i < j < p ≤ n )
......
4) The preimage of image with the length of n
x 1 x 2 ⋯ x n − 1 − ( λ 2 n + ⋯ + λ n − 1 n ) − ⋯ − ( λ 1 n + ⋯ + λ n − 2 n ) d x n ↦ x 1 x 2 ⋯ x n − 1 x n .
Hence, we can find preimage of the image x 1 ⋯ x m ( 1 ≤ m ≤ n ) in Ω 1 ( Λ ) , the preimage is
x 1 x 2 ⋯ x m − 1 ( m − 1 ) ( λ m m + 1 + ⋯ + λ m n ) − ( λ 2 m + ⋯ + λ m − 1 m ) − ⋯ − ( λ 1 m + ⋯ + λ m − 2 m ) d x m ↦ x 1 ⋯ x m .
But for the constant term k (with the length of 0), we can not find the preimage. So we can get 0-th twisted homology group
H P 0 ( Λ , Λ D ) = ker ∂ 0 π Im ∂ 1 π = k .
∂ 1 π : Ω 1 ( Λ ) → Λ ∑ j = 1 n C x 1 i 1 ⋯ x j 0 ⋯ x n i n d x j ↦ { x 1 i 1 x 2 i 2 ⋯ x j 0 ⋯ x n i n , x j } Λ D .
the mapping between the elements is
C x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 d x n + x 1 j 1 x 2 j 2 ⋯ x n − 2 j n − 2 x n j n d x n − 1 + ⋯ + x 2 k 2 x 3 k 3 ⋯ x n − 1 k n − 1 x n k n d x 1 ↦ k i 1 , ⋯ , i n { x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 , x n } Λ D + k j 1 , ⋯ , j n { x 1 j 1 x 2 j 2 ⋯ x n − 2 j n − 2 x n j n , x n − 1 } Λ D + ⋯ + k p 1 , ⋯ , p n { x 1 p 1 x 3 p 3 ⋯ x n − 1 p n − 1 x n p n , x 2 } Λ D k p 1 , ⋯ , p n + { x 2 p 2 x 3 p 3 ⋯ x n − 1 p n − 1 x n p n , x 1 } Λ D .
We have n preimage, and each of them has the length of k ( 1 ≤ k ≤ n − 1 ) , and the image has k + 1 in length. Similarly, we can calculate the image in each item like in part 3, we do not repeat.
Because the Poisson structure of L is homogeneous. We can discuss the image of ∂ 1 π by length.
1) The preimage of image with the length of 1
∑ C i j x i d x j ↦ ∑ C i j ( ( λ j j + 1 + ⋯ + λ j j + n ) − ( λ 1 j + ⋯ + λ i − 1 j + λ i + 1 j + λ j − 1 j ) ) x i x j
Let ∑ C i j x i d x j ∈ ker ∂ 1 π , then ∑ C i j ( ( λ j j + 1 + ⋯ + λ j j + n ) − ( λ 1 j + ⋯ + λ i − 1 j + λ i + 1 j + λ j − 1 j ) ) x i x j = 0 .
for ∀ i < j , C i j λ j j + 1 + ⋯ + λ j i − 1 + ⋯ + λ j i + 1 + ⋯ + λ j n − ( λ 1 j + ⋯ + λ j − 1 j ) λ j j + 1 + ⋯ + λ j n − ( λ 1 j + ⋯ + λ i − 1 j + ⋯ + λ i + 1 j + ⋯ + λ j − 1 j ) = − C j i .
We have
λ 12 + ⋯ + λ 1 n − 1 λ 2 n + ⋯ + λ n − 1 n x 1 d x n + x n d x 1 ∈ ker ∂ 1 π ⋮ ( λ j j + 1 + ⋯ + λ j i − 1 + ⋯ + λ j i + 1 + ⋯ + λ j n ) − ( λ 1 j + ⋯ + λ j − 1 j ) λ j j + 1 + ⋯ + λ j n − ( λ 1 j + ⋯ + λ i − 1 j + ⋯ + λ i + 1 j + ⋯ + λ j − 1 j ) x i d x j + x j d x i ∈ ker ∂ 1 π ⋮ ( λ 1 i + ⋯ + λ i − 1 i ) − λ i i + 1 + ⋯ + λ i n − 1 λ 1 n + ⋯ + λ i − 1 n + ⋯ + λ i + 1 n + λ n − 1 n x i d x n + x n d x i ∈ ker ∂ 1 π
We can see that the image x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 d x n and x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n i n d x n − 1 in Ω 2 ( Λ ) has the same coefficient, it is conflict with our results. So the image with length 1 in ker ∂ 1 π can not find the preimage in Ω 2 ( Λ ) . In the same time, the constant term k (with the length of 0) also can not find the preimage. So we can get 1-th twisted homology group.
2) The preimage of image with the length of 2
∑ C i j k x i x j d x k ↦ ∑ C i j k ( 2 ( λ k k + 1 + ⋯ + λ k k + n ) − 2 ( λ 1 k + ⋯ + λ ⌢ i k + λ ⌢ j k + ⋯ + λ k − 1 j ) − λ i k − λ j k ) x i x j x k
Let ∑ C i j k x i x j d x k ∈ ker ∂ 1 π , then
for ∀ i < j < k , that is
C j i k = − ( 2 ( λ k k + 1 + ⋯ + λ k n ) − 2 ( λ 1 k + ⋯ + λ ⌢ i k + λ ⌢ j k + ⋯ + λ k − 1 k ) − λ i k − λ j k ) ( 2 ( λ k k + 1 + ⋯ + λ k n ) − 2 ( λ 1 k + ⋯ + λ ⌢ j k + λ ⌢ i k + ⋯ + λ k − 1 k ) − λ j k − λ i k ) C i j k .
We have
λ 1 2 + λ 13 ⋯ + λ 1 n − 1 λ 1 n + ⋯ + λ n − 1 n x 1 x 2 d x n + x 2 x n d x 1 ∈ ker ∂ 1 π ⋮ − λ 2 n − 1 + ⋯ + λ n − 2 n − 1 λ 2 n + ⋯ + λ n − 1 n x 1 x n − 1 d x n + x 1 x n d x n − 1 ∈ ker ∂ 1 π
Obviously, for ∀ i < j < k , the preimage of this image is − x j d x i ∧ d x k under the map ∂ 2 π .
3) Generally, the preimage of image with the length of m ( 1 < m < n − 1 )
∑ C i 1 ⋯ i m − 1 i x i 1 x i 2 ⋯ x i m − 1 x i d x j ↦ ∑ C i 1 ⋯ i m − 1 i ( 2 ( λ j j + 1 + ⋯ + λ j j + n ) − 2 ( λ 1 j + ⋯ + λ ⌢ i 1 j + λ ⌢ i 2 j + ⋯ + λ j − 1 i m + λ j − 1 i ) − λ i 1 j − ⋯ − λ i j ) x i 1 ⋯ x i m − 1 x i x j
Let ∑ C i 1 ⋯ i m − 1 i x i 1 x i 2 ⋯ x i m − 1 x i d x j ∈ ker ∂ 1 π , then
∑ C i 1 ⋯ i m − 1 i ( 2 ( λ j j + 1 + ⋯ + λ j j + n + λ 1 j + ⋯ + λ ⌢ i 1 j + λ ⌢ i 2 j + ⋯ + λ j − 1 i m + λ j − 1 i ) − λ i 1 j − ⋯ − λ i j ) x i 1 ⋯ x i m − 1 x i x j = 0
We have,
x i 1 ⋯ x i m − 1 x i d x j + 2 ( λ i i + 1 + ⋯ + λ i i + n λ 1 i + ⋯ + λ i − 1 i m + λ i − 1 j ) − λ i 1 i − ⋯ − λ j i 2 ( λ j j + 1 + ⋯ + λ j j + n λ 1 j + ⋯ + λ j − 1 i m + λ j − 1 i ) − λ i 1 j − ⋯ − λ i j x i 1 ⋯ x i m − 1 x j d x i ∈ ker ∂ 1 π
for ∀ i 1 < ⋯ < i m − 1 < i < j , the preimage of this element is − x i 1 x i 2 ⋯ x i m − 1 d x i ∧ d x j .
4) The preimage of image with the length of n − 1
Let ∑ C 1 x 1 ⋯ x n − 1 d x n ∈ ker ∂ 1 π ,
then
C 1 ( − ( n − 2 ) ( λ 1 n + ⋯ + λ n − 1 1 ) − λ 1 n − ⋯ − λ n − 1 n ) x 1 ⋯ x n + C n ( − ( n − 2 ) ( λ n 1 + ⋯ + λ 1 n − 1 ) − λ n 1 − ⋯ − λ n − 1 n ) x 1 ⋯ x n = 0 .
We have
− λ 1 n + λ 2 n + ⋯ + λ n − 1 n λ 12 + λ 13 + ⋯ + λ 1 n x 2 x 3 ⋯ x n d x 1 + x 1 x 2 ⋯ x n − 1 d x n ∈ ker ∂ 1 π ⋮ − λ 1 n + ⋯ λ j n + ⋯ + λ k n + ⋯ + λ n − 1 n λ 12 + ⋯ λ 1 j + ⋯ + λ 1 k + ⋯ + λ 1 n x 1 ⋯ x ⌢ k ⋯ x n x j d x k + x 1 ⋯ x ⌢ j ⋯ x n x k d x j ∈ ker ∂ 1 π
for ∀ 1 ≤ j < k ≤ n , the preimage of this element is
x 2 x 3 ⋯ x n − 1 d x n ∧ d x 1 − λ 1 n + λ 2 n + ⋯ + λ n − 1 n λ 12 + λ 13 + ⋯ + λ 1 n x 2 x 3 ⋯ x n − 1 d x 1 ∧ d x n ⋮ − x 1 ⋯ x ⌢ j ⋯ x ⌢ k ⋯ x n d x j ∧ d x k − λ 1 k + λ 2 k + ⋯ + λ n − 1 k λ j 2 + λ j 3 + ⋯ + λ j n − x 1 ⋯ x ⌢ j ⋯ x ⌢ k ⋯ x n d x j ∧ d x k
In conclusion, ∂ 2 π is not surjection. So we have the 1-th homology group
H P 1 ( Λ , Λ D ) = ker ∂ 1 π Im ∂ 2 π = { k λ 12 + ⋯ + λ 1 n − 1 λ 2 n + ⋯ + λ n − 1 n x 1 d x n + x n d x 1 λ 21 + ⋯ + λ 2 n − 1 λ 1 n + λ 3 n ⋯ + λ n − 1 n x 2 d x n + x n d x 2 ⋮ ( λ j j + 1 + ⋯ + λ ⌢ j i + ⋯ + λ j n ) − ( λ 1 j + ⋯ + λ j − 1 j ) λ j j + 1 + ⋯ + λ j n − ( λ 1 j + ⋯ + λ ⌢ i j + ⋯ + λ j − 1 j ) x i d x j + x j d x i ⋮ ( λ 1 i + ⋯ + λ i − 1 i ) − λ i i + 1 + ⋯ + λ i n − 1 λ 1 n + ⋯ + λ i − 1 n + ⋯ + λ i + 1 n + λ n − 1 n x i d x n + x n d x i
∂ 2 π : Ω 2 ( Λ ) → Ω 1 ( Λ ) ∑ 1 ≤ j < k ≤ n C x 1 i 1 x 2 i 2 ⋯ x n i n d x j ∧ d x k ↦ { x 1 i 1 ⋯ x n i n , x j } Λ D d x k − { x 1 i 1 ⋯ x n i n , x k } Λ D d x j − x 1 i 1 ⋯ x n i n d { x j , x k }
That is
∑ 1 ≤ j < k ≤ n C x 1 i 1 x 2 i 2 ⋯ x n i n d x j ∧ d x k = C 1 x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 d x n − 1 ∧ d x n + C 2 x 1 i 1 x 2 i 2 ⋯ x n − 3 i n − 3 x n − 1 i n − 1 d x n − 2 ∧ d x n + ⋯ + C c n 2 x 3 i 3 x 4 i 4 ⋯ x n − 1 i n − 1 x n i n d x 1 ∧ d x 2
we calculate the image in each item
The first term
x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 d x n − 1 ∧ d x n ↦ { x 1 i 1 ⋯ x n − 2 i n − 2 , x n − 1 } Λ D d x n − { x 1 i 1 ⋯ x n − 2 i n − 2 , x n } Λ D d x n − 1 − x 1 i 1 ⋯ x n − 2 i n − 2 d { x n − 1 , x n } = x 2 i 2 x 3 i 3 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 { x 1 i 1 , x n − 1 } Λ D + x 1 i 1 x 3 i 3 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 { x 2 i 2 , x n − 1 } Λ D + ⋯ + x 1 i 1 x 2 i 2 ⋯ x n − 3 i n − 3 { x n − 2 i n − 2 , x n − 1 } Λ D − x 2 i 2 x 3 i 3 ⋯ x n − 2 i n − 2 { x 1 i 1 , x n } Λ D
+ x 1 i 1 x 3 i 3 ⋯ x n − 2 i n − 2 { x 2 i 2 , x n } Λ D + ⋯ + x 1 i 1 x 2 i 2 ⋯ x n − 3 i n − 3 x n − 1 i n − 1 { x n − 2 i n − 2 , x n } Λ D − x 1 i 1 ⋯ x n − 2 i n − 2 d { x n − 1 , x n } = − ( λ n n − 1 + λ n n − 2 + λ n n 1 ) ( x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 d x n + x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n i n d x n − 1 )
......
The C n 2 -th term
x 3 i 3 x 4 i 4 ⋯ x n i n d x 1 ∧ d x 2 ↦ { x 3 i 3 ⋯ x n i n , x 1 } Λ D d x 2 − { x 3 i 3 ⋯ x n i n , x 2 } Λ D d x 1 − x 3 i 3 ⋯ x n i n d { x 1 , x 2 } = x 4 i 4 x 5 i 5 ⋯ x n − 1 i n − 1 x n i n { x 3 i 3 , x 1 } Λ D + x 3 i 3 x 5 i 5 ⋯ x n − 1 i n − 1 x n i n { x 4 i 4 , x 1 } Λ D + ⋯ + x 3 i 3 x 4 i 4 ⋯ x n − 1 i n − 1 { x n i n , x 1 } Λ D − x 4 i 4 x 5 i 5 ⋯ x n − 1 i n − 1 x n i n { x 3 i 3 , x 2 } Λ D
+ x 3 i 3 x 5 i 5 ⋯ x n − 1 i n − 1 x n i n { x 4 i 4 , x 2 } Λ D + ⋯ + x 3 i 3 x 4 i 4 ⋯ x n − 1 i n − 1 { x n i n , x 2 } Λ D − x 3 i 3 ⋯ x n i n d { x 1 , x 2 } = − ( λ 1 n + λ 2 n + λ n − 1 n ) ( x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n − 1 i n − 1 d x n + x 1 i 1 x 2 i 2 ⋯ x n − 2 i n − 2 x n i n d x n − 1 )
Next, For every element in Ω 2 ( Λ ) , we try to find the preimage in Ω 3 ( Λ ) .
∂ 3 π : Ω 3 ( Λ ) → Ω 2 ( Λ ) ∑ 1 ≤ p < q < r ≤ n x 1 i 1 x 2 i 2 ⋯ x n i n d x p ∧ d x q ∧ d x r ↦ { x 1 i 1 ⋯ x n i n , x p } Λ D d x q ∧ d x r − { x 1 i 1 ⋯ x n i n , x q } Λ D d x p ∧ d x r + { x 1 i 1 ⋯ x n i n , x r } Λ D d x p ∧ d x q − x 1 i 1 ⋯ x n i n d { x p , x q } ∧ d x r + x 1 i 1 ⋯ x n i n d { x p , x r } ∧ d x q − x 1 i 1 ⋯ x n i n d { x q , x r } ∧ d x p
We just take one item for example
∑ 1 ≤ p < q < r ≤ n x 1 i 1 x 2 i 2 ⋯ x n i n d x p ∧ d x q ∧ d x r ↦ { x 1 i 1 ⋯ x n − 3 i n − 3 , x n − 2 } Λ D d x n − 1 ∧ d x n − { x 1 i 1 ⋯ x n − 3 i n − 3 , x n − 1 } Λ D d x n − 2 ∧ d x n + { x 1 i 1 ⋯ x n − 3 i n − 3 , x n } Λ D d x n − 1 ∧ d x n − 2 − x 1 i 1 ⋯ x n − 3 i n − 3 d { x n − 2 , x n − 1 } ∧ d x n + x 1 i 1 ⋯ x n − 3 i n − 3 d { x n − 2 , x n } ∧ d x n − 1 − x 1 i 1 ⋯ x n i n d { x n − 1 , x n } ∧ d x n − 2 = ( i 1 λ 1 n − 2 + i 2 λ 2 n − 2 + ⋯ + i n − 3 λ n − 3 n − 2 ) x 1 i 1 ⋯ x n − 3 i n − 3 d x n − 1 ∧ d x n
− ( i 1 λ 1 n − 1 + i 2 λ 2 n − 1 + ⋯ + i n − 3 λ n − 3 n − 1 ) x 1 i 1 ⋯ x n − 3 i n − 3 d x n − 2 ∧ d x n + ( i 1 λ 1 n + ⋯ + i n − 3 λ n − 3 n ) x 1 i 1 ⋯ x n − 3 i n − 3 d x n − 1 ∧ d x n − 2 − λ n − 2 n − 1 x 1 i 1 ⋯ x n − 3 i n − 3 ( x n − 2 d x n − 1 ∧ d x n + x n d x n − 2 ∧ d x n − 1 ) + λ n − 2 n x 1 i 1 ⋯ x n − 3 i n − 3 ( x n − 2 d x n ∧ d x n − 1 + x n d x n − 2 ∧ d x n − 1 ) − λ n − 1 n x 1 i 1 ⋯ x n i n ( x n − 1 d x n ∧ d x n − 2 + x n d x n − 1 ∧ d x n − 2 )
Similarly, because the Poisson structure of L is homogeneous. We can discuss the image of ∂ 2 π by length.
1) The preimage of image with the length of 1
∑ C p q x i d x p ∧ d x q ↦ ∑ C p q ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ ⌢ i p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ p q ) x i x p d x q − ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ p q ) x i x q d x p )
Let ∑ C p q x i d x p ∧ d x q ∈ ker ∂ 2 π , then
∑ C p q ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ ⌢ i p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ p q ) x i x p d x q − ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ p q ) x i x q d x p ) = 0
Then for ∀ i < p < q ,
C p q ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ ⌢ i p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ p q ) x i x p d x q − ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ p q ) x i x q d x p ) C i q ( ( 2 ( λ i i + 1 + ⋯ + λ i n ) − 2 ( λ 1 i + ⋯ + λ ⌢ p i + ⋯ + λ i − 1 i ) − λ ⌢ p i − λ ⌢ i q ) x p x i d x q
− ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ i q ) x p x q d x i ) C i p ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ ⌢ i p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ p i ) x q x p d x i − ( 2 ( λ i i + 1 + ⋯ + λ i n ) − 2 ( λ 1 i + ⋯ + ⋯ + λ i − 1 i ) − λ ⌢ q i − λ ⌢ p i ) x q x i d x p ) = 0
That is
C i j λ j j + 1 + ⋯ + λ j i − 1 + ⋯ + λ j i + 1 + ⋯ + λ j n − ( λ 1 j + ⋯ + λ j − 1 j ) λ j j + 1 + ⋯ + λ j n − ( λ 1 j + ⋯ + λ i − 1 j + ⋯ + λ i + 1 j + ⋯ + λ j − 1 j ) = − C j i .
we have
λ 1 n + ⋯ + λ n − 1 n λ n 1 + ⋯ + λ 21 x 1 d x 2 ∧ d x n − x n d x 1 ∧ d x 2 + x 2 d x 1 ∧ d x n ∈ ker ∂ 2 π ⋮ λ i q + ⋯ + λ p q λ q i + ⋯ + λ p i x i d x p ∧ d x q − x q d x i ∧ d x p + x p d x i ∧ d x q ∈ ker ∂ 2 π ⋮ λ n − 11 + ⋯ + λ n − 1 n λ n 1 + ⋯ + λ n − 11 x 1 d x n − 1 ∧ d x n − x n d x 1 ∧ d x n − 1 + x n − 1 d x 1 ∧ d x n ∈ ker ∂ 2 π
2) The preimage of image with the length of 2
∑ C p q x i x j d x p ∧ d x q ↦ ∑ C p q ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ j p − λ ⌢ p q ) x i x j x p d x q − ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ p q − λ ⌢ j p ) x i x j x q d x p )
Let ∑ C p q x i x j d x p ∧ d x q ∈ ker ∂ 2 π , then
∑ C p q ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ j p − λ ⌢ p q ) x i x j x p d x q − ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ p q − λ ⌢ j p ) x i x j x q d x p ) = 0
for ∀ i < j < p < q ,
C i j k ( 2 ( λ k k + 1 + ⋯ + λ k n ) − 2 ( λ 1 k + ⋯ + λ ⌢ i k + λ ⌢ j k + ⋯ + λ k − 1 k ) − λ i k − λ j k ) C j i k ( 2 ( λ k k + 1 + ⋯ + λ k n ) − 2 ( λ 1 k + ⋯ + λ ⌢ j k + λ ⌢ i k + ⋯ + λ k − 1 k ) − λ j k − λ i k ) = 0
that is
C j i k = − ( 2 ( λ k k + 1 + ⋯ + λ k n ) − 2 ( λ 1 k + ⋯ + λ ⌢ i k + λ ⌢ j k + ⋯ + λ k − 1 k ) − λ i k − λ j k ) ( 2 ( λ k k + 1 + ⋯ + λ k n ) − 2 ( λ 1 k + ⋯ + λ ⌢ j k + λ ⌢ i k + ⋯ + λ k − 1 k ) − λ j k − λ i k ) C i j k
We have
λ 31 + ⋯ + λ n 1 − λ 42 λ 34 − λ 41 + ⋯ + λ 4 n x 3 x 4 d x 1 ∧ d x 2 − x 3 x 1 d x 4 ∧ d x 2 + x 3 x 2 d x 4 ∧ d x 1 ∈ ker ∂ 2 π ⋮ λ i q + ⋯ + λ n q − λ p r λ i p − λ p q + ⋯ + λ p n x i x p d x q ∧ d x r − x i x q d x p ∧ d x r + x i x r d x p ∧ d x q ∈ ker ∂ 2 π
We can see that the image x 1 i 1 x 2 i 2 ⋯ x n − 3 i n − 3 x n − 2 i n − 2 d x n − 1 ∧ d x n and must have the same coefficient, it is conflict with our results. So the image with length 1 and 2 in ker ∂ 2 π can not find the preimage in Ω 3 ( Λ ) . In the same time, the constant term k (with the length of 0) also can not find the preimage. Now we discuss the elements in ker ∂ 2 π when the length ≥ 3, we will prove that we can get the preimage in Ω 3 ( Λ )
3) The preimage of image with the length of 3
∑ C p q r x i x j x r d x p ∧ d x q ↦ ∑ C p q ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ j p − λ ⌢ p q − λ ⌢ r q ) x i x j x p d x q − ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ p q − λ ⌢ j p − λ ⌢ r p ) x i x j x q d x p )
Let ∑ C p q r x i x j d x p ∧ d x q ∧ d x r ∈ ker ∂ 2 π , then
∑ C p q ( ( 2 ( λ p p + 1 + ⋯ + λ p n ) − 2 ( λ 1 p + ⋯ + λ p − 1 p ) − λ ⌢ i p − λ ⌢ j p − λ ⌢ p q − λ ⌢ r q ) x i x j x p d x q − ( 2 ( λ q q + 1 + ⋯ + λ q n ) − 2 ( λ 1 q + ⋯ + λ q − 1 q ) − λ ⌢ i q − λ ⌢ p q − λ ⌢ j p − λ ⌢ r p ) x i x j x q d x p ) = 0
We have
λ 31 + ⋯ + λ n 1 − λ 42 λ 34 − λ 41 + ⋯ + λ n 2 x 3 x 4 x 5 d x 1 ∧ d x 2 − x 3 x 5 x 1 d x 4 ∧ d x 2 + x 3 x 5 x 2 d x 4 ∧ d x 1 ∈ ker ∂ 2 π ⋮ λ i q + ⋯ + λ n q − λ p r λ i p − λ p q + ⋯ + λ n r x i x p x j d x q ∧ d x r − x i x q x j d x p ∧ d x r + x i x j x r d x p ∧ d x q ∈ ker ∂ 2 π
Obviously, for ∀ i < j < k , the preimage of this image is
− x 5 d x 1 ∧ d x 2 ∧ d x 4 ⋮ − x j d x q ∧ d x r ∧ d x p
under the ∂ 2 π .
4) The preimage of image with the length of n − 2
∑ C n − 1 n x 1 ⋯ x n - 2 d x n − 1 ∧ d x n ↦ ∑ C 1 ( ( − ( n − 2 ) ( λ 1 n + ⋯ + λ n − 1 1 ) − λ 1 n − ⋯ − λ n − 1 n ) x 1 ⋯ x n − 1 d x n + ( − ( n − 2 ) ( λ 1 - 1 n + ⋯ + λ 1 1 ) − λ 1 n - 1 − ⋯ − λ 1 n - 1 ) x 1 ⋯ x n d x n − 1 )
Let ∑ C n − 1 n x 1 ⋯ x n - 2 d x n - 1 ∧ d x n ∈ ker ∂ 1 π , then
∑ C 1 ( ( − ( n − 2 ) ( λ 1 n + ⋯ + λ n − 1 1 ) − λ 1 n − ⋯ − λ n − 1 n ) x 1 ⋯ x n − 1 d x n + ( − ( n − 2 ) ( λ 1 - 1 n + ⋯ + λ 1 1 ) − λ 1 n - 1 − ⋯ − λ 1 n - 1 ) x 1 ⋯ x n d x n − 1 ) = 0
We have
λ 12 + λ 32 + ⋯ + λ n 2 λ 21 + λ 31 + ⋯ + λ n 1 x 1 x ⌢ 2 x 3 ⋯ x n − 1 d x 2 ∧ d x n − x 2 x 3 ⋯ x n − 1 d x 1 ∧ d x n + x 3 x 4 ⋯ x n d x 1 ∧ d x 2 λ 13 + λ 23 + ⋯ + λ n 3 λ 21 + λ 31 + ⋯ + λ n 1 x 1 x 2 x ⌢ 3 ⋯ x n − 1 d x 3 ∧ d x n − x 2 x 3 ⋯ x n − 1 d x 1 ∧ d x n + x 2 x 4 ⋯ x n d x 1 ∧ d x 3 ⋮ λ 1 n − 1 + λ 2 n − 1 + ⋯ + λ n n − 1 λ 21 + λ 31 + ⋯ + λ n 1 x 1 x 2 x 3 ⋯ x n − 2 d x n − 1 ∧ d x n − x 2 x 3 ⋯ x n − 1 d x 1 ∧ d x n + x 2 x 4 ⋯ x n d x 1 ∧ d x n − 1
Obviously, the preimage of this image is
x 3 x 4 ⋯ x n − 1 d x 1 ∧ d x 2 ∧ d x n x 2 x 4 ⋯ x n − 1 d x 1 ∧ d x 3 ∧ d x n ⋮ x 2 x 3 ⋯ x n − 2 d x 1 ∧ d x n − 1 ∧ d x n
under the ∂ 2 π .
In conclusion, ∂ 2 π is not surjection. So we have the 2-th twisted homology group
H P 2 ( Λ , Λ D ) = ker ∂ 2 π Im ∂ 3 π = { k λ 1 n + ⋯ + λ n − 1 n λ n 1 + ⋯ + λ 21 x 1 d x 2 ∧ d x n − x n d x 1 ∧ d x 2 + x 2 d x 1 ∧ d x n ⋮ λ i q + ⋯ + λ p q λ q i + ⋯ + λ p i x i d x p ∧ d x q − x q d x i ∧ d x p + x p d x i ∧ d x q ⋮ λ n − 11 + ⋯ + λ n − 1 n λ n 1 + ⋯ + λ n − 11 x 1 d x n − 1 ∧ d x n − x n d x 1 ∧ d x n − 1 + x n − 1 d x 1 ∧ d x n } length of 1 λ 31 + ⋯ + λ n 1 − λ 42 λ 34 − λ 41 + ⋯ + λ 4 n x 3 x 4 d x 1 ∧ d x 2 − x 3 x 1 d x 4 ∧ d x 2 + x 3 x 2 d x 4 ∧ d x 1 ⋮ λ i q + ⋯ + λ n q − λ p r λ i p − λ p q + ⋯ + λ p n x i x p d x q ∧ d x r − x i x q d x p ∧ d x r + x i x r d x p ∧ d x q } length of2
In general situation, for 1 ≤ m ≤ n − 1 , the m-th twisted homology H P m ( Λ , Λ D ) has the elements in ker ∂ m π with length of 0 ~ m .
Ω n + 1 ( Λ ) → ∂ n + 1 π Ω n ( Λ ) → ∂ n π Ω n − 1 ( Λ )
∂ n π : Ω n ( Λ ) → Ω n − 1 ( Λ ) C d x 1 ∧ ⋯ ∧ d x n ↦ ∑ 1 ≤ i < j < ⋯ < k ≤ n ( − 1 ) i + j d { F i , F j } ∧ d F 1 ∧ ⋯ ∧ d F ^ i ∧ ⋯ ∧ d F ^ j ∧ ⋯ ∧ d F ^ k ∧ ⋯ ∧ d F n = ( − 1 ) 1 + 2 d { x 1 , x 2 } ∧ d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + ( − 1 ) 1 + 3 d { x 1 , x 3 } ∧ d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ + ( − 1 ) ( n − 1 ) + n d { x n − 1 , x n } ∧ d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2
We calculate the Poisson bracket
( − 1 ) 1 + 2 d { x 1 , x 2 } ∧ d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + ( − 1 ) 1 + 3 d { x 1 , x 3 } ∧ d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ + ( − 1 ) ( n − 1 ) + n d { x n − 1 , x n } ∧ d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2 = − d { x 1 , x 2 } ∧ d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + d { x 1 , x 3 } ∧ d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ − d { x n − 1 , x n } ∧ d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2
= − λ 12 ( x 1 d x 2 + x 2 d x 1 ) d x 3 ∧ d x 4 ∧ ⋯ ∧ d x n + λ 13 ( x 1 d x 3 + x 3 d x 1 ) d x 2 ∧ d x 4 ∧ ⋯ ∧ d x n + ⋯ − λ n − 1 n ( x n − 1 d x n + x n d x n − 1 ) d x 1 ∧ d x 2 ∧ ⋯ ∧ d x n − 2
We can easily see that this element can not be 0, that is ker ∂ n π = 0 , so
H P n ( Λ , Λ D ) = ker ∂ n π Im ∂ n + 1 π = 0
In this part, we will check the Twisted Poincaré duality between Poisson homology and cohomology, that is for i ∈ N , we have
P . D . : H P i ( A , A ) ≅ ( H P i ( A , A D ) ) *
Next, we need to calculate the cohomology group.
obviously, for all i , j , the equation established
{ x 1 i 1 ⋯ x k i k ⋯ x n i n , x k } = ( λ 1 k i 1 + ⋯ + λ k − 1 k i k − 1 + λ k + 1 k i k + 1 + ⋯ + λ n k i n ) x 1 i 1 ⋯ x k + 1 i k + 1 ⋯ x n i n . (12)
For every p ∈ N , p-fold polyderivations from L to L, denoted χ p ( Λ ) , χ i ( Λ ) = H o m ( Ω i , Λ ) . recall the canonical cochain complex
0 → Λ → δ 0 χ 1 ( Λ ) → δ 1 ⋯ → δ k − 1 χ k ( Λ ) → δ k χ k + 1 ( Λ ) → ⋯
For every P ∈ χ k and f 0 , ⋯ , f k ∈ Λ . δ k : χ k → χ k + 1 is defined by
δ p ( P ) ( f 0 , ⋯ , f k ) = ∑ i = 0 k ( − 1 ) i { f i , P ( f 0 , ⋯ , f ⌢ i , ⋯ , f k ) } + ∑ 0 ≤ i < j ≤ k ( − 1 ) i + j P ( { f i , f j } , f 0 , ⋯ , f ⌢ i , ⋯ , f ⌢ j ⋯ , f k )
obviously, δ k ( P ) ∈ χ k + 1 and δ k + 1 ∘ δ k = 0 .
Let λ = ∑ C i 1 ⋯ i n x 1 i 1 ⋯ x n i n ∈ Λ , and assume it satisfies
{ λ , x k } = { ∑ C i 1 ⋯ i n x 1 i 1 ⋯ x n i n , x k } = 0 ( k = 1 , ⋯ , n ) .
Then Equation (12) lead successively to
For i k ≠ 1 and i 1 = i 2 = ⋯ = i k − 1 = i k + 1 = ⋯ = i n ≠ 0 , we have C i 1 ⋯ i n = 0 .
Hence, we can get the Proposition 7.1.
Proposition 7.1. H P 0 ( Λ ) = C .
Proof: for every d ∈ χ 1 , d is uniquely determined by the values of
d ( x 1 ) = ∑ C i 1 ⋯ i n 1 x 1 i 1 x 2 i 2 ⋯ x n − 1 i n − 1 x n i n ,
d ( x 2 ) = ∑ C i 1 ⋯ i n 2 x 1 i 1 x 2 i 2 ⋯ x n − 1 i n − 1 x n i n ,
⋯
d ( x n − 1 ) = ∑ C i 1 ⋯ i n n − 1 x 1 i 1 x 2 i 2 ⋯ x n − 1 i n − 1 x n i n ,
d ( x n ) = ∑ C i 1 ⋯ i n n x 1 i 1 x 2 i 2 ⋯ x n − 1 i n − 1 x n i n .
Moreover, d must satisfy the relations d ( x 1 2 ) = d ( x 2 2 ) = ⋯ = d ( x n 2 ) = 0 , since d is a derivation, so we have x 1 d ( x 1 ) = x 2 d ( x 2 ) = ⋯ = x n d ( x n ) = 0 .
So we have the that for ∀ i 1 , ⋯ , i j , C 0 i 2 ⋯ i n 1 = C i 1 0 ⋯ i n 2 = ⋯ = C i 1 ⋯ i n − 1 0 n = 0 . Hence the space χ 1 = D e r ( Λ , Λ ) has basis { d i 1 ⋯ i n 1 } ∪ { d i 1 ⋯ i n 2 } ∪ ⋯ ∪ { d i 1 ⋯ i n n } , where
1) for 1 ≤ i 1 < 2 , 0 ≤ i 2 , i 3 , ⋯ , i n < 2 , the derivation d i 1 ⋯ i n 1 is defined by
d i 1 ⋯ i n ( 1 ) ( x 1 ) = x 1 i 1 x 2 i 2 ⋯ x n i n and d i 1 ⋯ i n ( 1 ) ( x k ) = 0 ( when k ≠ 1 ) ;
2) for 1 ≤ i 2 < 2 , 0 ≤ i 1 , i 3 , ⋯ , i n < 2 , the derivation d i 1 ⋯ i n 2 is defined by
d i 1 ⋯ i n ( 2 ) ( x 2 ) = x 1 i 1 x 2 i 2 ⋯ x n i n and d i 1 ⋯ i n ( 2 ) ( x k ) = 0 ( when k ≠ 2 ) ;
......
(n) for 1 ≤ i n < 2 , 0 ≤ i 1 , i 2 , ⋯ , i n − 1 < 2 , the derivation d i 1 ⋯ i n n is defined by
d i 1 ⋯ i n ( n ) ( x n ) = x 1 i 1 x 2 i 2 ⋯ x n i n and d i 1 ⋯ i n ( n ) ( x k ) = 0 ( when k ≠ n ) ;
In particular, dim ( χ 1 ) = n ⋅ 2 n . Let
d = ∑ i 1 ≠ 0 α i 1 ⋯ i n ( 1 ) d i 1 ⋯ i n ( 1 ) + ∑ i 2 ≠ 0 α i 1 ⋯ i n ( 2 ) d i 1 ⋯ i n ( 2 ) + ⋯ + ∑ i n ≠ 0 α i 1 ⋯ i n ( n ) d i 1 ⋯ i n ( n ) ∈ χ 1
d ∈ ker δ 1 if and only if d satisfied:
d ( { x k , x 1 ⋯ x ^ k ⋯ x n } ) = { d ( x k ) , x 1 ⋯ x ^ k ⋯ x n } + { x k , d ( x 1 ⋯ x ^ k ⋯ x n ) } . (13)
In Equation (13), we have:
R H S = { d ( x k ) , x 1 ⋯ x ^ k ⋯ x n } + { x k , d ( x 1 ⋯ x ^ k ⋯ x n ) } = { x 1 i 1 ⋯ x n i n , x 1 ⋯ x ^ k ⋯ x n } + { x k , d ( x 1 ⋯ x ^ k ⋯ x n ) } = ( λ 12 + ⋯ + λ 1 n + λ 23 + ⋯ + λ 2 n + ⋯ + λ n − 1 n ) x 1 i 1 + 1 ⋯ x k i k ⋯ x n i n + 1 + { x k , x 2 ⋯ x n d ( x 1 ) + ⋯ + x 1 ⋯ x n − 1 d ( x n ) }
= ( λ 12 + ⋯ + λ 1 n + λ 23 + ⋯ + λ 2 n + ⋯ + λ n − 1 n ) x 1 i 1 + 1 ⋯ x k i k ⋯ x n i n + 1 + { x k , α i 1 ⋯ i n ( 1 ) x 1 i 1 x 2 i 2 + 1 ⋯ x n i n + 1 + ⋯ + α i 1 ⋯ i n ( n ) x 1 i 1 + 1 x 2 i 2 + 1 ⋯ x n i n }
From the coefficient of x 1 i 1 x 2 i 2 ⋯ x n i n we can get for ∀ 0 ≤ i 1 , ⋯ , i n ≤ 1 , we have
d ∈ ker δ 1 ⇔ ( n − 2 − ( λ 12 + λ 13 + ⋯ + λ 1 n ) ) α i 1 ⋯ i n ( 1 ) + ( n − 2 − ( λ 21 + λ 23 + ⋯ + λ 2 n ) ) α i 1 ⋯ i n ( 2 ) + ⋯ + ( n − 2 − ( λ n − 11 + λ n − 12 + ⋯ + λ n − 1 n ) ) α i 1 ⋯ i n ( n − 1 ) + ( n − 2 − ( λ n 1 + λ n 2 + ⋯ + λ n n − 1 ) ) α i 1 ⋯ i n ( n ) = 0
The proof is completed.
Proposition 7.2. H P 1 ( Λ ) = C ⊕ C d 1 , 1 , ⋯ , 0 ( 1 ) ⊕ C d 0 , 1 , 1 , ⋯ , 0 ( 2 ) ⊕ ⋯ ⊕ C d 0 , 0 , ⋯ , 1 , 1 ( n ) .
Proof: let d ( { x k , x 1 ⋯ x ^ k ⋯ x n } ) = { d ( x k ) , x 1 ⋯ x ^ k ⋯ x n } + { x k , d ( x 1 ⋯ x ^ k ⋯ x n ) } ∈ ker δ 1 .
Set λ : = ∑ i 2 , i 3 , ⋯ , i n ≠ 0 α i 1 ⋯ i n ( 1 ) x 1 i 1 ⋯ x n i n ∈ Λ , from Equation (12) we can get d 1 = d + { λ , − } is a Poisson derivation and satisfied d 1 ( x 1 ) = ∑ i 1 ≥ 0 α i 1 , 0 ⋯ 0 ( 1 ) x 1 i 1 . From Equation (13) we can get that for all i 1 ≠ 0 , we have α i 1 , 0 ⋯ 0 ( 1 ) . That is d 1 ( x 1 ) = α i 1 , 0 ⋯ 0 ( 1 ) x 1 . Similarly,
Let d 2 ( x 2 ) = ∑ i 1 = 0 1 α i 1 , 0 ⋯ 0 ( 1 ) x 1 i 1 + ⋯ + ∑ i n = 0 1 α i 1 , 0 ⋯ 0 ( n ) x n i n .
Let a new set
u : = ∑ i 1 = 0 1 α i 1 ⋯ i n ( 1 ) x 1 i 1 ⋯ x n i n + ∑ i 2 = 0 1 α i 1 ⋯ i n ( 1 ) x 1 i 1 ⋯ x n i n + ⋯ ∈ Λ .
d 2 ( x 2 ) = d 1 ( x 1 ) + ⋯ + d n ( x n ) − { u , − } , so we have
d 2 ( x 2 ) = α i 1 ⋯ i n ( 1 ) d 1 , 0 , ⋯ , 0 + ⋯ + α i 1 ⋯ i n ( 2 ) d 0 , 0 , ⋯ , 1 .
That is the image of C d 1 , 1 , ⋯ , 0 ( 1 ) ⊕ C d 0 , 1 , 1 , ⋯ , 0 ( 2 ) ⊕ ⋯ ⊕ C d 0 , 0 , ⋯ , 1 , 1 ( n ) in H P 1 ( Λ ) .
The proof is completed.
Similarly, we can calculate the m-th cohomology H P m ( Λ ) .
From the propostion 7.1 and propostion 7.2 we can easily get that the demension of m-th cohomology group H P m ( Λ ) equal to the demension of m-th Twisted Poisson homology, H P m ( Λ , Λ D ) , we check the twisted Poincaréduality between them.
In this paper, we successfully calculate the i-th homology group of the algebra Λ = k 〈 x 1 , x 2 , ⋯ , x n 〉 / 〈 x i x j − x j x i , x i 2 〉 in part 4. In part 5, after getting Frobenius pairing, we calculate the modular derivation and then have the twisted Poisson module structure. Furthermore, we calculate the twisted Poisson homology group in part 6. Lastly, we verified the twisted Poincaré duality between Poisson homology and Poisson Cohomology through the cohomology group of L in part 7. In future studies, we will discuss whether all these conclusions hold up for general algebra.
Wang, M.Y. (2018) Poisson (Co)homology of a Class of Frobenius Poisson Algbras. Journal of Applied Mathematics and Physics, 6, 530-553. https://doi.org/10.4236/jamp.2018.63048