A Kind of Infinite-Dimensional Novikov Algebras and Its Realizations

Definition 1 (see [17].)Let (𝒜, ∘) be an algebra over F such that

and then 𝒜 is called a Novikov algebra over F.

Remark 2. An algebra 𝒜 is called a left symmetric algebra if it only satisfies (4). It is clear that left symmetric algebras contain Novikov algebras.

Remark 3. (1) If (𝒜, ∘) is a left symmetric algebra satisfying

then (𝒜, [, ]) is a Lie algebra. Usually, it is called an adjoining Lie algebra.

(2) Let (𝒜, ·) be a commutative algebra, and then (𝒜, d₀, ∘) is a Novikov algebra if d₀ is a derivation of 𝒜 with a bilinear operator ∘ such that

Lemma 4. Let {b₀, a₁, b₁, a₂, b₂, …  a_n, b_n, …} be a basis of the linear space 𝒜 over a field F of characteristic p ≠ 2 satisfying

where b_−m = b_m, a_−m = −a_m. Then 𝒜 is a commutative and associative algebra.

Proof. It is clear that 𝒜 is a commutative algebra over F:

Similarly, we have that (b_k, b_n, b_m) = (a_k, a_n, b_m) = (a_k, b_n, a_m) = (b_k, a_n, a_m) = (b_k, b_n, a_m) = (b_k, a_n, b_m) = (a_k, b_n, b_m) = 0. Then (a, b, c) = 0, ∀a, b, c ∈ 𝒜. The result follows.

Corollary 5. b₀ of Lemma 4 is a unity of 𝒜.

Lemma 6. Let 𝒜 be a commutative and associative algebra satisfying Lemma 4. Then the following statements hold:

• (1)

  If D₀ is a linear transformation of 𝒜 such that

  $$ ()

•  

  then D₀ is a derivation of 𝒜.

• (2)

  If aD₀ is a linear transformation of 𝒜 such that

  $$ ()

•  

  then aD₀ is a derivation of 𝒜.

• (3)

  𝒟₁ = {aD₀∣a ∈ 𝒜} is a subalgebra of Lie algebra Der𝒜.

Proof. (1) We have

So D₀ is a derivation of 𝒜.

(2) For ∀a, b, c ∈ 𝒜, we have

so aD₀ is a derivation of 𝒜.

(3) For ∀a, b, c ∈ 𝒜, we have

Then [aD₀, bD₀] = (aD₀(b) − bD₀(a))D₀ ∈ 𝒟₁, and so (3) holds.

Theorem 7. Let 𝒜 be a commutative and associative algebra satisfying Lemma 4, and let a be an element of 𝒜. If D₀ satisfies Lemma 6 and ∘ satisfies

then the following statements hold:

• (1)

  (𝒜, aD₀, ∘) is a Novikov algebra.

• (2)

  (𝒜, aD₀, [, ]) is an adjoining Lie algebra of (𝒜, aD₀, ∘) and [, ] such that

  $$ ()

Proof. (1) By Lemma 6, aD₀ is a derivation of the commutative algebra 𝒜. So (𝒜, aD₀, ∘) is a Novikov algebra by Remark 3(2).

(2) (𝒜, aD₀, [, ]) is an adjoining Lie algebra of (𝒜, aD₀, ∘) by Remark 3(1). For ∀b, c ∈ 𝒜, ∃a ∈ 𝒜, we have

since 𝒜 is commutative. Hence we obtain the desired result.

Corollary 8. Let 𝒜 be a commutative and associative algebra satisfying Lemma 4. Then the following statements hold:

Lemma 9. 𝒯 satisfying the above product is a commutative associative algebra.

Proof. Since the above product is commutative and associative, we only need 𝒯 to be closed for the product. In fact,

So 𝒯 is a commutative and associative algebra.

Lemma 10. Let 𝒯 be a linear space generated by {sinh  mx, cosh  nx∣m, n ∈ N} over F, and then {1, sinh  mx, cosh  nx∣m, n ∈ N₀} is a basis of 𝒯.

Proof. For ∀n ∈ N₀, suppose that there are c₀, a_i, b_j ∈ F, i, j ∈ N₀ such that

We take derivative for (20) such that its derivative order is 2k − 1  (k ∈ N₀), and put x = 0. Then we have Let k = 1,2, …, n, and then we obtain the following system of n linear equations: If a₁, …, a_n are seen to be unknown, then the coefficient matrix of (22) is the Vandermonde matrix whose determinant is not 0, so a_i = 0, i = 1, …, n.

We take derivative for (20) such that its derivative order is 2k (k ∈ N₀), and put x = 0. Then we have

Let k = 1,2, …, n, and then we obtain the following system of n linear equations: If b₁, …, b_n are seen to be unknown, then the coefficient matrix of (24) is the Vandermonde matrix whose determinant is not 0, so b_i = 0, i = 1, …, n. Since, for any i ∈ N₀, a_i = 0 and b_i = 0 satisfy (20), we have c₀ = 0. Hence {1, sinh  x, cosh x,…, sinh  nx, cosh  nx} are linearly independent for any n ∈ N₀, and then {1, sinh  nx, cosh  mx∣n, m ∈ N₀} are linearly independent and so they form a basis of 𝒯 as desired.

Theorem 11. Let 𝒜₁, 𝒜₂ be commutative and associative algebras over F. If φ: 𝒜₁ → 𝒜₂ is an isomorphism and D₁ ∈ Der𝒜₁, then the following statements hold:

• (1)

  D₂ : = φD₁φ⁻¹ ∈ Der𝒜₂,

• (2)

  φ: (𝒜₁, D₁, ∘)→(𝒜₂, D₂, ∘) is also an isomorphism of Novikov algebras.

Proof. (1) For any a, b ∈ 𝒜₁, we have

So (1) holds.

(2) For any a, b ∈ 𝒜₁, we have

So (2) holds.

Theorem 12. Let 𝒜 be a commutative and associative algebra over F satisfying Lemma 4, let D₀ be its derivation satisfying (10), and let 𝒯 be a commutative and associative algebra over F satisfying Lemmas 9 and 10. If φ : 𝒜 → 𝒯 satisfies

then the following statements hold:

• (1)

  φ is an isomorphism of commutative and associative algebras,

• (2)

  φD₀φ⁻¹ = d/dx,

• (3)

  φ : (𝒜, aD₀, ∘)→(𝒯, φ(a)(d/dx), ∘) is an isomorphism of Novikov algebras.

Proof. It is clear by Lemma 10, (8), and (19).

(2) By Lemma 6, we have

So (2) holds.

(3) It is clear that φ(aD₀)φ⁻¹ = φ(a)d/dx. By (27) and (10), we have

Similarly, we have φ(aD₀)φ⁻¹(cosh nx) = φ(a)d(cosh nx)/dx. So φ(aD₀)φ⁻¹ = φ(a)d/dx.

By Theorems 7 and 11 and Remark 3(2), we have

So φ : (𝒜₀, aD₀, ∘)→(𝒯, φ(a)(d/dx), ∘) is an isomorphism of Novikov algebras.