Cohomologies of finite-dimensional simple alternative superalgebras of characteristic 3

M. C. López-Díaz, Departamento de Matemáticas, Universidad de Oviedo, Oviedo, Spain

Abstract The second cohomology groups for certain simple finite-dimensional alternative superalgebras of characteristic 3 are calculated, with coefficients in regular superbimodules. Contrary to the case of algebras, the groups are not trivial since the Wedderburn splitting theorem is not always true for alternative superalgebras. The results complement the similar results by N.A.Pisarenko for the case of characteristic ≠ 2, 3.

1 Introduction

It is known that every separable alternative algebra A has trivial second cohomology group H^{2}(A, M) = 0, for every alternative A-bimodule M. This is just an equivalent formulation of Schafer’s result on validity of the Wedderburn splitting theorem for alternative algebras [6]. In [3], N.A.Pisarenko proved that this theorem does not hold for alternative superalgebras; hence simple alternative superalgebras may have nontrivial second cohomology groups. In [4], he described the indecomposable superbimodules for simple alternative superalgebras of characteristic ≠ 2, 3; and in [5] he calculated the second cohomology groups for these superalgebras, with coefficients in indecomposable superbimodules.

In our previous paper [2], we extended Pisarenko’s results on the Wedderburn splitting in alternative superalgebras to the case of characteristic 3. (Observe that nontrivial nonassociative simple alternative superalgebras appear only in this case.) We have found two new types of nonsplittable extensions of simple superalgebras; in one case it is an extension by a regular superbimodule, and in the other is an extension by a superbimodule opposed to the regular one.

In this paper, we calculate the second cohomology groups for these extensions.

If otherwise is not stating, all the algebras considered in the paper assumed to be finite dimensional, over a field F of characteristic ≠ 2.

2 Preliminaries

Recall that a superalgebra A = A_{0} + A_{1} is called an alternative superalgebra if it satisfies the superidentities

(x, y, z) = (−1)^{x̄ȳ+1}(y, x, z) = (−1)^{ȳz̄+1}(x, z, y),

where (x, y, z) = (xy)z − x(yz), x, y, z ∈ A_{0} ∩ A_{1}, and \overline{r} = i if r ∈ A_{i}.

An A-superbimodule M = M_{0} + M_{1} is called an alternative superbimodule if the corresponding split extension superalgebra E = A + M is alternative.

For an A-superbimodule M, the opposite superbimodule M^{op} = M_{0}^{op}+M_{1}^{op} is defined by the conditions M_{0}^{op} = M_{1}, M_{1}^{op} = M_{0}, and the following action of A: a·m = (−1)\overline{a}am, m· a = ma, for any a ∈ A_{0} ∪ A_{1}, m ∈ M^{op}. If M is an alternative A-superbimodule, then one can easily check that so is M^{op}. A regular superbimodule, Reg A, is defined on the vector superspace A with the action of A coinciding with the multiplication in A.

If A is a superalgebra and M, N are A-superbimodules, then a linear mapping α : M ⟶ N is called a homomorphism of superbimodules of degree j, if it is homogeneous of degree j, i.e., α(M_{i}) ⊆ N_{i+j}, and

α(am) = (−1)^{j \overline{a}}aα(m), and α(ma) = α(m)a,

for any m ∈ M, a ∈ A_{0} ∪ A_{1}.

Let A = A_{0} + A_{1} be a superalgebra and M be an A-supermodule. A superalgebra B is called an extension of A by M if there exists a short exact sequence of superalgebra homomorphisms

0 ⟶ M ⟶^{α} B ⟶^{β} A ⟶ 0.

Here M is considered as a superalgebra with zero multiplication.

Two extensions B and B′ of A by M are called equivalent, if there exists a homomorphism of superalgebras ϕ : B ⟶ B′ such that the diagram

0 ⟶ M ⟶^{α} B ⟶^{β} A ⟶ 0

↓ id ↓ ϕ ↓ id

0 ⟶ M ⟶^{α′} B′ ⟶^{β′} A ⟶ 0

is commutative.

Let A = A_{0} + A_{1} be an alternative superalgebra over a field F, M be an alternative A-supermodule, and B be an extension of A by M, with the corresponding short exact sequence 0 ⟶ M ⟶^{α} B ⟶^{β} A ⟶ 0. Let μ : A ⟶ B be a linear mapping such that β · μ = 1_{A}. Then B = α(M) + μ(A) is a vector space direct sum, and for any a, b ∈ A, h(a, b) = μ(a)μ(b) − μ(ab) ∈ M. So, if we identify A with μ(A) and M with α(M), then the product ∗ in B = A + M is defined by the action of A on M and by the conditions M^{2} = 0, a ∗ b = ab + h(a, b) for any a, b ∈ A. For a fixed A-superbimodule M, every extension of A by M is defined by a bilinear form h : A × A ⟶ M.

Consider F (a, b, c) = h(a, b)c + h(ab, c) − (−1)^{\overline{h} \overline{a}}ah(b, c) − h(a, bc), ∀a, b, c ∈ A. Then, since B is an alternative superalgebra, we have F (a, b, c) = (−1)^{\overline{a} \overline{b}+1}F (b, a, c) = (−1)^{\overline{b} \overline{c}+1}F (a, c, b) (see [5]).

Now, let μ′ : A ⟶ B be another linear mapping such that β · μ′ = 1_{A}. Consider ν = μ′ − μ : A ⟶ M, so μ′ = ν + μ. Then, h′(a, b) = μ′(a)μ′(b) − μ′(ab) = h(a, b) + μ(a)ν(b) + ν(a)μ(b) − ν(ab). Therefore, two extensions given by bilinear forms h and h′ are equivalent if and only if there exists a linear mapping μ : A ⟶ M such that h′(a, b) = h(a, b) + μ(a)b + μ(b)a − μ(ab).

The bilinear forms h are called cocycles and they form the space of cocycles Z^{2}(A, M) = Z_{0}^{2}(A, M) + Z_{1}^{2}(A, M), where Z_{k}^{2}(A, M) = { h ∈ Z^{2}(A, M) | h(A_{i}, A_{j}) ⊆ M_{i+j+k} }, k, i, j ∈ {0, 1}. Now, denote by B^{2}(A, M) = B_{0}^{2}(A, M) + B_{1}^{2}(A, M) the set of cocycles which are equivalent to the zero bilinear form. They are called coboundaries, and B^{2}(A, M) is the space of coboundaries. If h ∈ B^{2}(A, M), then h(a, b) = −μ(ab) + (−1)^{\overline{a} \overline{μ}}aμ(b) + μ(a)b for some linear mapping μ : A ⟶ M. Note that every coboundary defines an extension of A by M isomorphic to the split extension E = A+M. Moreover, elements h, h′ ∈ Z^{2}(A, M) define equivalent extensions if and only if h − h′ ∈ B^{2}(A, M).

Definition 1 Let A = A_{0} + A_{1} be an alternative superalgebra over a field F and M be an alternative A-supermodule. Then, the second group of cohomology of A with coefficients in M is defined as a quotient group H^{2}(A, M) = Z^{2}(A, M)/B^{2}(A, M).

The next proposition shows that a calculation of H^{2}(A, M) provides H^{2}(A, M^{op}) as well.

Proposition 1 Let A be an alternative superalgebra over a field F. Let M be an alternative A-supermodule. Then

H_{i}^{2}(A, M) = H_{1−i}^{2}(A, M^{op}), ∀ i ∈ {0, 1}.

Proof. Consider Z^{2}(A, M). We have Z_{k}^{2}(A, M) = { h ∈ Z^{2}(A, M) | h(A_{i}, A_{j}) ⊆ M_{i+j+k} } = { h ∈ Z^{2}(A, M^{op}) | h(A_{i}, A_{j}) ⊆ M_{i+j+k+1}^{op} } = Z_{k+1}^{2}(A, M^{op}), ∀ k, i, j ∈ {0, 1}. The Proposition is now evident. •

3 The second cohomologies of semisimple alternative superalgebras

According to [7] and [8], simple finite dimensional alternative superalgebras B = B_{0} + B_{1} over an algebraically closed field F, charF ≠ 2, are of one of the following types:

1) B = B_{0} = C, a Cayley-Dickson algebra over F (see [6, 9]).

2) B = M_{n}(F)[u], where, for an algebra A, a superalgebra A[u] = A+Au is defined as an extension of A by an odd central element u such that u^{2} = 1. (In [3] this superalgebra was denoted by (M_{n}(F), M_{n}(F))).

3) B = M(p|q), B_{0} = \{( \begin{matrix}X & 0 \\ 0 & Y \end{matrix} ) ∈ M_{n}(F), | X ∈ M_{p}(F), Y ∈ M_{q}(F)\}, B_{1} =

\{( \begin{matrix}0 & A \\ B & 0 \end{matrix} ) ∈ M_{n}(F) | A ∈ M_{p×q}(F), B ∈ M_{q×p}(F)\}, where p + q = n and p ≤[\frac{n}{2}].

4) B = B(1, 2), where char F = 3, B_{0} = F · 1, B_{1} = F · x + F · y, with 1 being the unit of B and xy = −yx = 1, x^{2} = y^{2} = 0.

5) B = B(4, 2), where char F = 3, B_{0} = M_{2}(F), B_{1} = F ·m_{1}+F ·m_{2} is a 2-dimensional irreducible Cayley bimodule over B_{0}; that is, B_{0} acts on B_{1} by

e_{ij} · m_{k} = δ_{ik}m_{j}, i, j, k ∈ {1, 2},
m · a = \overline{a} · m,

where a ∈ B_{0}, m ∈ B_{1}, a → \overline{a} is the symplectic involution in B_{0} = M_{2}(F). The odd multiplication on B_{1} is defined by

m_{1}^{2} = −e_{21}, m_{2}^{2} = e_{12}, m_{1}m_{2} = e_{11}, m_{2}m_{1} = −e_{22}.

6) The twisted superalgebra of vector type B = B(E, D, γ). Let E = F [t_{1}, t_{2}, ..., t_{n} | t_{i}^{3} = 0] be the algebra of truncated polynomials over F (char F = 3), D be a nonzero derivation of E such that E is D-simple, and γ ∈ E. Observe that a derivation D for which E is D-simple always exists: one may take, for instance, D = ∂/∂t_{1} + t_{1}^{2}∂/∂t_{2} + ... + (t_{1}...t_{n−1})^{2}∂/∂t_{n}.

Denote by \overline{E} an isomorphic copy of the vector space E, with an isomorphism mapping a → \overline{a}. Consider the vector space direct sum B(E, D, γ) = E+\overline{E} and define multiplication on it by the rules

a · b = ab, a · \overline{b} = \overline{a} · b = \overline{ab}, \overline{a} · \overline{b} = γab + 2D(a)b + aD(b),

where a, b ∈ E and ab is the product in E. Define a Z_{2}-grading on B = B(E, D, γ) such that B_{0} = E and B_{1} = \overline{E}. Then B is a simple alternative superalgebra of dimension 2 · 3^{n} over F.

We will say that a simple alternative superalgebra A is of type X, where X is one of the superalgebras 1) - 6), if some scalar extension of A is isomorphic to X.

The results of [3, 2] imply the following

Theorem 1 Let A be a separable alternative superalgebra over a field F of characteristic ≠ 2, isomorphic to a direct sum of simple superalgebras. If no superalgebras of types F [u], M(1|1), B(1, 2), and B(E, D, γ) are found among simple direct summands of A, then H^{2}(A, M) = 0 for every alternative A-superbimodule M.

Proof. In fact, in this case the Wedderburn decomposition holds in any extension of A by a superbimodule, hence every extension is equivalent to the trivial one. •

If A is isomorphic to one of the superalgebras F [u], M(1|1), B(1, 2), or B(E, D, γ), then the examples of nonsplittable extensions in [3] and [2] show that H^{2}(A, M) may by nonzero for certain M.

The cases F [u] and M(1|1) were investigated by Pisarenko [4, 5]. He described the indecomposable superbimodules over these superalgebras and then calculated the second cohomology groups with coefficients in these superbimodules. We repeat some of his results for the convenience of the reader, since the publication [5] is not quite available.

Theorem 2 [4, 5]. Let A = M(1|1)(F). Then,

(i) Every unital alternative A-superbimodule is associative and completely reducible,

(ii) Any irreducible A-superbimodule is isomorphic to Reg A or to (Reg A)^{op},

(iii) H^{2}(A, Reg A) = 0 ⊕ F^{2}.

Theorem 3 [4, 5]. Let A = F [u], where F is an algebraically closed field of characteristic ≠ 2. Then,

(i) The set of irreducible alternative A-superbimodules is parametrized by the points (α, δ) of the ellipse α^{2} + αδ + δ^{2} = 1. Every such a superbimodule is of the form M_{(α,δ)} = F n + F m, where n is even and m is odd, and an action of A is defined as follows.

– M_{(α,δ)} (δ ≠ 0) : nu = \frac{1+α}{δ} m, mu = δn, un = m, um = (1 − α)n;

– M_{(1,0)} : nu = m, mu = 2n, un = 0, um = n;

– M_{(−1,0)} : nu = m, mu = 0, un = −2m, um = −n.

(ii) If char F ≠ 2, 3, then

– H^{2}(A, M_{(α,δ)}) = 0 if δ ≠ 0,

– H^{2}(A, M_{(1,0)}) = 0 ⊕ F,

– H^{2}(A, M_{(−1,0)}) = F ⊕ 0.

In fact, Pisarenko has obtained more general results: he described all indecomposable alternative superbimodules for F [u] and calculated the second cohomologies for them. Every indecomposable bimodule M has a socle isomorphic to one of the bimodules M_{(α,δ)}, and in this case H^{2}(A, M) = H^{2}(A, M_{(α,δ)}) (see [5]).

4 H^{2}(A, Reg A) for A = B(1, 2)

In this section we calculate H^{2}(A, Reg A) for the superalgebra A = B(1, 2) over any field of characteristic 3.

Theorem 4 Let A = B(1, 2), then H^{2}(A, Reg A) = 0 ⊕ F^{4}.

Proof. Let A = B(1, 2) = F · 1 + F · x + F · y which is defined in Section 3. Let M = Reg(B(1, 2)^{op}) = F · a_{1} + F · a_{2} + F · e where M_{0} = F · a_{1} + F · a_{2} and M_{1} = F · e. Let us construct H^{2}(A, (RegA)^{op}).

Consider h ∈ Z^{2}(A, (RegA)^{op})_{0}. We have h(A_{i}, A_{j}) ⊆ M_{i+j(mod 2)}. So, h(1, 1) = α_{1}a_{1} + α_{2}a_{2}, h(1, x) = αe, h(1, y) = βe, h(x, 1) = δe, h(x, x) = β_{1}a_{1} + β_{2}a_{2}, h(x, y) = δ_{1}a_{1} + δ_{2}a_{2},h(y, 1) = κe, h(y, x) = κ_{1}a_{1} + κ_{2}a_{2}, h(y, y) = λ_{1}a_{1} + λ_{2}a_{2}.

Now, let F (a, b, c) = h(a, b)c + h(ab, c) − (−1)^{\overline{h} \overline{a}}ah(b, c) − h(a, bc) be the function defined in Section 2. Applying the properties of this function 0 = F (1, 1, x) = h(1, 1)x+h(1, x)−h(1, x)−h(1, x) = (α_{1}a_{1}+α_{2}a_{2})x−αe = (−α−α_{2})e ⇒ α_{2} = −α.

In a similar way, if we consider F (a, b, c) for all a, b, c ∈ A, we have δ = α, α_{1} = β, κ = β, κ_{2} = β_{1} − δ_{2}, λ_{2} = κ_{1} + δ_{1}.

In conditions above, ∀ a, b, c ∈ A, F (a, b, c) satisfies all the properties of the F function.

Now, if h ∈ Z^{2}(A, (RegA)^{op})_{0}, then h(1, 1) = βa_{1} − αa_{2}, h(1, x) = αe, h(1, y) = βe, h(x, 1) = αe, h(x, x) = δa_{1} + γa_{2}, h(x, y) = λa_{1} + πa_{2}, h(y, 1) = βe, h(y, x) = σa_{1} + (δ − π)a_{2}, h(y, y) = ρa_{1} + (σ + λ)a_{2} where α, β, δ, γ, λ, π, σ, ρ ∈ F.

Consider h_{i} ∈ Z^{2}(A, (RegA)^{op})_{0}, i ∈ {1, ..., 8} such that the nonzero images are h_{1}(1, 1) = −a_{2}, h_{1}(1, x) = e = h_{1}(x, 1), h_{2}(1, 1) = a_{1}, h_{2}(1, y) = e = h_{2}(y, 1), h_{3}(x, x) = a_{1}, h_{3}(y, x) = a_{2}, h_{4}(x, x) = a_{2}, h_{5}(x, y) = a_{1}, h_{5}(y, y) = a_{2}, h_{6}(x, y) = a_{2} = −h_{6}(y, x), h_{7}(y, x) = a_{1}, h_{7}(y, y) = a_{2}, h_{8}(y, y) = a_{1}.

Now, h_{i} ∈ Z^{2}(A, (RegA)^{op})_{0}, i ∈ {1, ..., 8} generate Z^{2}(A, (RegA)^{op})_{0}.

Consider h ∈ B^{2}(A, (RegA)^{op})_{0}. Then, h ∈ Z^{2}(A, (RegA)^{op})_{0} such that for all a, b, c ∈ A, h(a, b) = −μ(ab)+(−1)^{\overline{a} \overline{μ}}aμ(b)+μ(a)b, where μ : A ⟶ M is an even linear mapping. So, μ(1) = x_{1}a_{1} + x_{2}a_{2}, μ(x) = ze and μ(y) = te. Now, βa_{1} − αa_{2} = h(1, 1) = −μ(1) + μ(1) + μ(1) = μ(1) ⇒ x_{1} = β and x_{2} = α.

In a similar way, if we consider h(a, b) for all a, b ∈ A, we obtain t = −β −λ, z = π −α and δ = γ = σ + λ = ρ = 0.

We have that if h ∈ Z^{2}(A, (RegA)^{op})_{0} such that δ = γ = ρ = 0 and σ = −λ then h ∈ B^{2}(A, (RegA)^{op})_{0}. So, if h ∈ B^{2}(A, (RegA)^{op})_{0}, then h(1, 1) = βa_{1} − αa_{2}, h(1, x) = αe, h(1, y) = βe, h(x, 1) = αe, h(x, x) = 0, h(x, y) = λa_{1} + πa_{2}, h(y, 1) = βe, h(y, x) = −λa_{1}−πa_{2}, h(y, y) = 0 where α, β, λ, π, ∈ F. Now, {h_{1}, h_{2}, h_{5}−h_{7}, h_{6}} generates B^{2}(A, (RegA)^{op})_{0}.

So, H^{2}(A, (RegA)^{op})_{0} = Z^{2}(A, (RegA)^{op})_{0}/B^{2}(A, (RegA)^{op})_{0} is isomorphic to F^{4}. Let us check Z^{2}(A, (RegA)^{op})_{1}. Consider h ∈ Z^{2}(A, (RegA)^{op})_{1}. Then h(A_{i}, A_{j}) ⊆

M_{i+j+1(mod 2)}, and h(1, 1) = αe, h(1, x) = α_{1}a_{1} + α_{2}a_{2}, h(1, y) = β_{1}a_{1} + β_{2}a_{2}, h(x, 1) = δ_{1}a_{1} + δ_{2}a_{2}, h(x, x) = βe, h(x, y) = δe, h(y, 1) = λ_{1}a_{1} + λ_{2}a_{2}, h(y, x) = λe, h(y, y) = κe.

In a similar way to case Z^{2}(A, (RegA)^{op})_{0}, using the properties of the F function, we have α_{1} = α, α_{2} = 0, δ_{1} = α, δ_{2} = 0, β_{1} = 0, β_{2} = α, λ_{1} = 0, λ_{2} = α, λ = −δ, β = 0, κ = 0.

In conditions above, for all a, b, c ∈ A, F (a, b, c) satisfies all the properties of the F function.

Now, if h ∈ Z^{2}(A, (RegA)^{op})_{1}, then h(1, 1) = αe, h(1, x) = αa_{1}, h(1, y) = αa_{2}, h(x, 1) = αa_{1}, h(x, x) = 0, h(x, y) = δe, h(y, 1) = αa_{2}, h(y, x) = −δe, h(y, y) = 0 where α, δ ∈ F.

Consider h_{i} ∈ Z^{2}(A, (RegA)^{op})_{1}, i ∈ {1, 2} such that the nonzero images are h_{1}(1, 1) = e, h_{1}(1, x) = a_{1} = h_{1}(x, 1), h_{1}(1, y) = a_{2} = h_{1}(y, 1), h_{2}(x, y) = e = −h_{2}(y, x).

We have that h_{1} and h_{2} generate Z^{2}(A, (RegA)^{op})_{1}.

Consider h ∈ B^{2}(A, (RegA)^{op})_{1}. Then, h ∈ Z^{2}(A, (RegA)^{op})_{1} such that for all a, b, c ∈ A, h(a, b) = −μ(ab) + (−1)^{\overline{a} \overline{μ}}aμ(b) + μ(a)b, where μ : A ⟶ M is an odd linear mapping. So, μ(1) = σe, μ(x) = x_{1}a_{1} + x_{2}a_{2} and μ(y) = y_{1}a_{1} + y_{2}a_{2}. Now, in a similar way to B^{2}(A, (RegA)^{op})_{0}, we obtain σ = α, y_{2} = δ + α − x_{1}, and we have that if h ∈ Z^{2}(A, (RegA)^{op})_{1} then h ∈ B^{2}(A, (RegA)^{op})_{1}. Consequently, Z^{2}(A, (RegA)^{op})_{1} = B^{2}(A, (RegA)^{op})_{1} and

H^{2}(A, (RegA)^{op})_{1} = Z^{2}(A, (RegA)^{op})_{1}/B^{2}(A, (RegA)^{op})_{1} = 0.

Finally, H^{2}(A, (RegA)^{op}) = F^{4} ⊕ 0 and by Proposition 1 H^{2}(A, (RegA)) = 0 ⊕ F^{4}. •

5 H^{2}(A, Reg A) for A = B(E, D, 0)

Here we calculate H^{2}(A, Reg A) for the algebra A = B(E, D, 0).

Theorem 5 Let A = B(E, D, 0), where F is a field of characteristic 3, E = F [x | x^{3} = 0], and D = ∂/∂x. Then, H^{2}(A, RegA) = F^{4} ⊕ F^{6}.

Proof. Let A = B(E, D, 0) and M = RegA = F ·e+F ·a_{1} +F ·a_{2} +F ·\overline{e}+F ·\overline{a}_{1} +F ·\overline{a}_{2} where M_{0} = F · e + F · a_{1} + F · a_{2} and M_{1} = F · \overline{e} + F · \overline{a}_{1} + F · \overline{a}_{2}.

Consider h ∈ Z^{2}(A, RegA)_{0}. We have h(A_{i}, A_{j}) ⊆ M_{i+j(mod 2)}.

In a similar way to Theorem 4, applying the properties of the F function given in Section 2, we have h(1, 1) = α_{1}e+α_{2}a_{1} +α_{3}a_{2}, h(1, x) = α_{1}a_{1} +α_{2}a_{2} = h(x, 1), h(1, x^{2}) = α_{1}a_{2} = h(x^{2}, 1), h(1, \overline{1}) = α_{1}\overline{e} + α_{2}\overline{a}_{1} + α_{3}\overline{a}_{2} = h(\overline{1}, 1), h(1, \overline{x}) = α_{1}\overline{a}_{1} + α_{2}\overline{a}_{2} = h(\overline{x}, 1), h(1, \overline{x^{2}}) = α_{1}\overline{a}_{2} = h(\overline{x^{2}}, 1), h(x, x) = 0, h(x, x^{2}) = π_{1}e = h(x^{2}, x), h(x, \overline{1}) = K_{1}\overline{e}+K_{2}\overline{a}_{1} = h(\overline{1}, x), h(x, \overline{x}) = E_{1}\overline{e} + E_{2}\overline{a}_{1} + E_{3}\overline{a}_{2} = h(\overline{x}, x), h(x, \overline{x^{2}}) = π_{1}\overline{e} − E_{1}\overline{a}_{1} + (−E_{2} − K_{1})\overline{a}_{2} = h(\overline{x^{2}}, x), h(x^{2}, x^{2}) = π_{1}a_{1}, h(x^{2}, \overline{1}) = E_{1}\overline{e} + (E_{2} + K_{1})\overline{a}_{1} + (E_{3} + K_{2})\overline{a}_{2} = h(\overline{1}, x^{2}), h(x^{2}, \overline{x}) = π_{1}\overline{e} − K_{1}\overline{a}_{2} = h(\overline{x}, x^{2}), h(x^{2}, \overline{x^{2}}) = π_{1}\overline{a}_{1} − E_{1}\overline{a}_{2} = h(\overline{x^{2}}, x^{2}), h(\overline{1}, \overline{1}) = ζ_{1}e + ζ_{2}a_{1} +ζ_{3}a_{2}, h(\overline{1}, \overline{x}) = ω_{1}e−α_{2}a_{1} +ω_{3}a_{2}, h(\overline{1}, \overline{x^{2}}) = (−E_{2} −K_{1})e+(E_{3} −ω_{1} −α_{1})a_{1} −ζ_{1}a_{2}, h(\overline{x}, \overline{1}) = −ω_{1}e+(α_{2} −ζ_{1})a_{1} +(−ω_{3} −ζ_{2})a_{2}, h(\overline{x}, \overline{x}) = ζ_{1}a_{2}, h(\overline{x}, \overline{x^{2}}) = E_{1}e+K_{1}a_{1} +(ω_{1} + α_{1} − K_{2} − E_{3})a_{2} = −h(\overline{x^{2}}, \overline{x}), h(\overline{x^{2}}, \overline{1}) = (K_{1} + E_{2})e + (−E_{3} + ω_{1} + α_{1})a_{1}, h(\overline{x^{2}}, \overline{x^{2}}) = 0.

Now, consider h ∈ B^{2}(A, RegA)_{0}. Then we have h ∈ Z^{2}(A, RegA)_{0} and h(a, b) = −μ(ab) + (−1)^{\overline{a} \overline{μ}}aμ(b) + μ(a)b where μ : A ⟶ M is an even linear function. Now, studying all the possibilities for h(a, b) with a, b ∈ A in a similar way to Theorem 4, we obtain that h ∈ B^{2}(A, RegA)_{0} if h ∈ Z^{2}(A, RegA)_{0} with π_{1} = ζ_{1} = ζ_{2} = ζ_{3} = 0.

Now, consider the elements h_{i} ∈ Z^{2}(A, RegA)_{0} where i ∈ {1, ..., 14} such that the nonzero images are: h_{1}(1, 1) = e, h_{1}(1, x) = a_{1} = h_{1}(x, 1), h_{1}(1, x^{2}) = a_{2} = h_{1}(x^{2}, 1), h_{1}(1, \overline{1}) = \overline{e} = h_{1}(\overline{1}, 1), h_{1}(1, \overline{x}) = \overline{a}_{1} = h_{1}(\overline{x}, 1), h_{1}(1, \overline{x^{2}}) = \overline{a}_{2} = h_{1}(\overline{x^{2}}, 1), h_{1}(\overline{1}, \overline{x^{2}}) = −a_{1} = −h_{1}(\overline{x^{2}}, \overline{1}), h_{1}(\overline{x}, \overline{x^{2}}) = a_{2} = −h_{1}(\overline{x^{2}}, \overline{x}),

h_{2}(1, 1) = a_{1}, h_{2}(1, x) = a_{2} = h_{2}(x, 1), h_{2}(1, \overline{1}) = \overline{a}_{1} = h_{2}(\overline{1}, 1), h_{2}(1, \overline{x}) = \overline{a}_{2} = h_{2}(\overline{x}, 1), h_{2}(\overline{1}, \overline{x}) = −a_{1} = −h_{2}(\overline{x}, \overline{1}),

h_{3}(1, 1) = a_{2}, h_{3}(1, \overline{1}) = \overline{a}_{2} = h_{3}(\overline{1}, 1)

h_{4}(x, x^{2}) = e = h_{4}(x^{2}, x), h_{4}(x, \overline{x^{2}}) = \overline{e} = h_{4}(\overline{x^{2}}, x), h_{4}(x^{2}, x^{2}) = a_{1}, h_{4}(x^{2}, \overline{x}) = \overline{e} = h_{4}(\overline{x}, x^{2}), h_{4}(x^{2}, \overline{x^{2}}) = \overline{a}_{1} = h_{4}(\overline{x^{2}}, x^{2}),

h_{5}(x, \overline{1}) = \overline{e} = h_{5}(\overline{1}, x), h_{5}(x, \overline{x^{2}}) = −\overline{a}_{2} = h_{5}(\overline{x^{2}}, x), h_{5}(x^{2}, \overline{1}) = \overline{a}_{1} = h_{5}(\overline{1}, x^{2}), h_{5}(x^{2}, \overline{x}) = −\overline{a}_{2} = h_{5}(\overline{x}, x^{2}), h_{5}(\overline{1}, \overline{x^{2}}) = −e = −h_{5}(\overline{x^{2}}, \overline{1}), h_{5}(\overline{x}, \overline{x^{2}}) = a_{1} = −h_{5}(\overline{x^{2}}, \overline{x}),

h_{6}(x, \overline{1}) = \overline{a}_{1} = h_{6}(\overline{1}, x), h_{6}(x^{2}, \overline{1}) = \overline{a}_{2} = h_{6}(\overline{1}, x^{2}), h_{6}(\overline{x}, \overline{x^{2}}) = −a_{2} = −h_{6}(\overline{x^{2}}, \overline{x}),

h_{7}(x, \overline{x}) = \overline{e} = h_{7}(\overline{x}, x), h_{7}(x, \overline{x^{2}}) = −\overline{a}_{1} = h_{7}(\overline{x^{2}}, x), h_{7}(x^{2}, \overline{1}) = \overline{e} = h_{7}(\overline{1}, x^{2}), h_{7}(x^{2}, \overline{x^{2}}) = −\overline{a}_{2} = h_{7}(\overline{x^{2}}, x^{2}), h_{7}(\overline{x}, \overline{x^{2}}) = e = −h_{7}(\overline{x^{2}}, \overline{x}),

h_{8}(x, \overline{x}) = \overline{a}_{1} = h_{8}(\overline{x}, x), h_{8}(x, \overline{x^{2}}) = −\overline{a}_{2} = h_{8}(\overline{x^{2}}, x), h_{8}(x^{2}, \overline{1}) = \overline{a}_{1} = h_{8}(\overline{1}, x^{2}), h_{8}(\overline{1}, \overline{x^{2}}) = −e = −h_{8}(\overline{x^{2}}, \overline{1}),

h_{9}(x, \overline{x}) = \overline{a}_{2} = h_{9}(\overline{x}, x), h_{9}(x^{2}, \overline{1}) = \overline{a}_{2} = h_{9}(\overline{1}, x^{2}), h_{9}(\overline{1}, \overline{x^{2}}) = a_{1} = −h_{9}(\overline{x^{2}}, \overline{1}), h_{9}(\overline{x}, \overline{x^{2}}) = −a_{2} = −h_{9}(\overline{x^{2}}, \overline{x}),

h_{10}(\overline{1}, \overline{1}) = e, h_{10}(\overline{1}, \overline{x^{2}}) = −a_{2}, h_{10}(\overline{x}, \overline{1}) = −a_{1}, h_{10}(\overline{x}, \overline{x}) = a_{2},

h_{11}(\overline{1}, \overline{1}) = a_{1}, h_{11}(\overline{x}, \overline{1}) = −a_{2},

h_{12}(\overline{1}, \overline{1}) = a_{2},

h_{13}(\overline{1}, \overline{x}) = e = −h_{13}(\overline{x}, \overline{1}), h_{13}(\overline{1}, \overline{x^{2}}) = −a_{1} = −h_{13}(\overline{x^{2}}, \overline{1}), h_{13}(\overline{x}, \overline{x^{2}}) = a_{2} = −h_{13}(\overline{x^{2}}, \overline{x}),

h_{14}(\overline{1}, \overline{x}) = a_{2} = −h_{14}(\overline{x}, \overline{1}).

It is clear that these elements generate Z^{2}(A, RegA)_{0}. Moreover, we have that h_{i} ∈ B^{2}(A, RegA)_{0} if i ∈ {1, 2, 3, 5, 6, 7, 8, 9, 13, 14}, and they generate B^{2}(A, RegA)_{0}. So, H^{2}(A, RegA)_{0} is isomorphic to F^{4}.

Consider h ∈ Z^{2}(A, Reg(A))_{1}. We have h(A_{i}, A_{j}) ⊆ M_{i+j+1(mod 2)}. Applying the properties of the F function, we have h(1, 1) = α_{1}\overline{e} + α_{2}\overline{a}_{1} + α_{3}\overline{a}_{2}, h(1, x) = α_{1}\overline{a}_{1} + α_{2}\overline{a}_{2} = h(x, 1), h(1, x^{2}) = α_{1}\overline{a}_{2} = h(x^{2}, 1), h(1, \overline{1}) = −α_{2}e+α_{3}a_{1} = h(\overline{1}, 1), h(1, \overline{x}) = α_{1}e−α_{3}a_{2} = h(\overline{x}, 1), h(1, \overline{x^{2}}) = −α_{1}a_{1}+α_{2}a_{2} = h(\overline{x^{2}}, 1), h(x, x) = −π_{2}\overline{e}−π_{3}\overline{a}_{1}+κ_{3}\overline{a}_{2}, h(x, x^{2}) = π_{2}\overline{a}_{1}+ π_{3}\overline{a}_{2} = h(x^{2}, x), h(x, \overline{1}) = D_{1}e + (K_{2} + ζ_{1})a_{1} + (K_{3} − ζ_{2})a_{2}, h(x, \overline{x}) = E_{1}e + E_{2}a_{1} + E_{3}a_{2}, h(x, \overline{x^{2}}) = −E_{1}a_{1} + (−E_{2} − D_{1})a_{2}, h(x^{2}, x^{2}) = −π_{2}\overline{a}_{2}, h(x^{2}, \overline{1}) = (E_{1} − π_{3})e + (E_{2} − κ_{3} + D_{1})a_{1} + (E_{3} + K_{2} + ζ_{1})a_{2}, h(x^{2}, \overline{x}) = π_{2}e + (κ_{3} − D_{1})a_{2}, h(x^{2}, \overline{x^{2}}) = −π_{2}a_{1} + (π_{3} − E_{1})a_{2} = h(\overline{x^{2}}, x^{2}), h(\overline{1}, x) = K_{2}a_{1}+K_{3}a_{2}, h(\overline{1}, x^{2}) = (E_{1}−π_{3})e+(E_{2}−κ_{3}−D_{1})a_{1}+(E_{3}+K_{2}−ζ_{1})a_{2}, h(\overline{1}, \overline{1}) = ζ_{1}\overline{e} + ζ_{2}\overline{a}_{1}, h(\overline{1}, \overline{x}) = ω_{1}\overline{e} + ω_{2}\overline{a}_{1} + ω_{3}\overline{a}_{2}, h(\overline{1}, \overline{x^{2}}) = E_{1}\overline{e} + (−D_{1} + E_{2} − ω_{1} − α_{1})\overline{a}_{1} + (−K_{2} + ζ_{1} + E_{3} − ω_{2} − α_{2})\overline{a}_{2}, h(\overline{x}, x) = E_{1}e + (E_{2} − D_{1})a_{1} + (E_{3} − ζ_{1})a_{2}, h(\overline{x}, x^{2}) = π_{2}e + κ_{3}a_{2}, h(\overline{x}, \overline{1}) = (−D_{1} − ω_{1})\overline{e} + (ζ_{1} − ω_{2})\overline{a}_{1} − ω_{3}\overline{a}_{2}, h(\overline{x}, \overline{x}) = −D_{1}\overline{a}_{1}, h(\overline{x}, \overline{x^{2}}) = −π_{2}\overline{e} + (E_{1} − π_{3})\overline{a}_{1} + (ω_{1} + α_{1} + κ_{3} + D_{1} + E_{2})\overline{a}_{2} = −h(\overline{x^{2}}, \overline{x}), h(\overline{x^{2}}, x) = −E_{1}a_{1}+(−E_{2}+D_{1})a_{2}, h(\overline{x^{2}}, \overline{1}) = −E_{1}\overline{e}+(−D_{1}−E_{2}+ω_{1}+α_{1})\overline{a}_{1}+(K_{2}−ζ_{1}−E_{3}+ω_{2}+α_{2})\overline{a}_{2}, h(\overline{x^{2}}, \overline{x^{2}}) = 0.

Now consider h ∈ B^{2}(A, RegA)_{1}. Then we have h ∈ Z^{2}(A, RegA)_{1} and h(a, b) = −μ(ab) + (−1)^{\overline{a} \overline{μ}}aμ(b) + μ(a)b where μ : A ⟶ M is an odd linear function. Studying all the possibilities for h(a, b) with a, b ∈ A, we obtain that h ∈ B^{2}(A, RegA)_{1} if h ∈ Z^{2}(A, RegA)_{1} with 0 = α_{2} = α_{3} = D_{1} = ζ_{1} = ζ_{2} = ω_{3} − K_{3}.

Now, consider the elements h_{i} ∈ Z^{2}(A, Reg(A))_{1} where i ∈ {1, ..., 17} such that the nonzero images are: h_{1}(1, 1) = \overline{e}, h_{1}(1, x) = \overline{a}_{1} = h_{1}(x, 1), h_{1}(1, x^{2}) = \overline{a}_{2} = h_{1}(x^{2}, 1), h_{1}(1, \overline{x}) = e = h_{1}(\overline{x}, 1), h_{1}(1, \overline{x^{2}}) = −a_{1} = h_{1}(\overline{x^{2}}, 1), h_{1}(\overline{1}, \overline{x^{2}}) = −\overline{a}_{1} = −h_{1}(\overline{x^{2}}, \overline{1}), h_{1}(\overline{x}, \overline{x^{2}}) = \overline{a}_{2} = −h_{1}(\overline{x^{2}}, \overline{x}),

h_{2}(1, 1) = \overline{a}_{1}, h_{2}(1, x) = \overline{a}_{2} = h_{2}(x, 1), h_{2}(1, \overline{1}) = −e = h_{2}(\overline{1}, 1), h_{2}(1, \overline{x^{2}}) = a_{2} = h_{2}(\overline{x^{2}}, 1), h_{2}(\overline{1}, \overline{x^{2}}) = −\overline{a}_{2} = −h_{2}(\overline{x^{2}}, \overline{1}),

h_{3}(1, 1) = \overline{a}_{2}, h_{3}(1, \overline{1}) = a_{1} = h_{3}(\overline{1}, 1), h_{3}(1, \overline{x}) = −a_{2} = h_{3}(\overline{x}, 1)

h_{4}(x, x) = −\overline{e}, h_{4}(x, x^{2}) = \overline{a}_{1} = h_{4}(x^{2}, x), h_{4}(x^{2}, x^{2}) = −\overline{a}_{2}, h_{4}(x^{2}, \overline{x}) = e = h_{4}(\overline{x}, x^{2}), h_{4}(x^{2}, \overline{x^{2}}) = −a_{1} = h_{4}(\overline{x^{2}}, x^{2}), h_{4}(\overline{x}, \overline{x^{2}}) = −\overline{e} = −h_{4}(\overline{x^{2}}, \overline{x}),

h_{5}(x, x) = −\overline{a}_{1}, h_{5}(x, x^{2}) = \overline{a}_{2} = h_{5}(x^{2}, x), h_{5}(x^{2}, \overline{1}) = −e = h_{5}(\overline{1}, x^{2}), h_{5}(x^{2}, \overline{x^{2}}) = a_{2} = h_{5}(\overline{x^{2}}, x^{2}), h_{5}(\overline{x}, \overline{x^{2}}) = −\overline{a}_{1} = −h_{5}(\overline{x^{2}}, \overline{x}),

h_{6}(x, x) = \overline{a}_{2}, h_{6}(x^{2}, \overline{1}) = −a_{1} = h_{6}(\overline{1}, x^{2}), h_{6}(x^{2}, \overline{x}) = a_{2} = h_{6}(\overline{x}, x^{2}), h_{6}(\overline{x}, \overline{x^{2}}) = \overline{a}_{2} = −h_{6}(\overline{x^{2}}, \overline{x}),

h_{7}(x, \overline{1}) = e, h_{7}(x, \overline{x^{2}}) = −a_{2} = −h_{7}(\overline{x^{2}}, x), h_{7}(x^{2}, \overline{1}) = a_{1} = −h_{7}(\overline{1}, x^{2}), h_{7}(x^{2}, \overline{x}) = −a_{2}, h_{7}(\overline{1}, \overline{x^{2}}) = −\overline{a}_{1} = h_{7}(\overline{x^{2}}, \overline{1}), h_{7}(\overline{x}, x) = −a_{1}, h_{7}(\overline{x}, \overline{1}) = −\overline{e}, h_{7}(\overline{x}, \overline{x}) = −\overline{a}_{1}, h_{7}(\overline{x}, \overline{x^{2}}) = \overline{a}_{2} = −h_{7}(\overline{x^{2}}, \overline{x}),

h_{8}(x, \overline{1}) = a_{1} = h_{8}(\overline{1}, x), h_{8}(x^{2}, \overline{1}) = a_{2} = h_{8}(\overline{1}, x^{2}), h_{8}(\overline{1}, \overline{x^{2}}) = −\overline{a}_{2} = −h_{8}(\overline{x^{2}}, \overline{1}),

h_{9}(x, \overline{1}) = a_{2} = h_{9}(\overline{1}, x),

h_{10}(x, \overline{1}) = a_{1}, h_{10}(x^{2}, \overline{1}) = a_{2} = −h_{10}(\overline{1}, x^{2}), h_{10}(\overline{1}, \overline{1}) = \overline{e}, h_{10}(\overline{1}, \overline{x^{2}}) = \overline{a}_{2} = −h_{10}(\overline{x^{2}}, \overline{1}), h_{10}(\overline{x}, x) = −a_{2}, h_{10}(\overline{x}, \overline{1}) = \overline{a}_{1},

h_{11}(x, \overline{1}) = −a_{2}, h_{11}(\overline{1}, \overline{1}) = \overline{a}_{1}, h_{12}(x, \overline{x}) = e = h_{12}(\overline{x}, x), h_{12}(x, \overline{x^{2}}) = −a_{1} = h_{12}(\overline{x^{2}}, x), h_{12}(x^{2}, \overline{1}) = e = h_{12}(\overline{1}, x^{2}), h_{12}(x^{2}, \overline{x^{2}}) = −a_{2} = h_{12}(\overline{x^{2}}, x^{2}), h_{12}(\overline{1}, \overline{x^{2}}) = \overline{e} = −h_{12}(\overline{x^{2}}, \overline{1}), h_{12}(\overline{x}, \overline{x^{2}}) = \overline{a}_{1} = −h_{12}(\overline{x^{2}}, \overline{x}),

h_{13}(x, \overline{x}) = a_{1} = h_{13}(\overline{x}, x), h_{13}(x, \overline{x^{2}}) = −a_{2} = h_{13}(\overline{x^{2}}, x), h_{13}(x^{2}, \overline{1}) = a_{1} = h_{13}(\overline{1}, x^{2}), h_{13}(\overline{1}, \overline{x^{2}}) = \overline{a}_{1} = −h_{13}(\overline{x^{2}}, \overline{1}), h_{13}(\overline{x}, \overline{x^{2}}) = \overline{a}_{2} = −h_{13}(\overline{x^{2}}, \overline{x}),

h_{14}(x, \overline{x}) = a_{2} = h_{14}(\overline{x}, x), h_{14}(x^{2}, \overline{1}) = a_{2} = h_{14}(\overline{1}, x^{2}), h_{14}(\overline{1}, \overline{x^{2}}) = \overline{a}_{2} = −h_{14}(\overline{x^{2}}, \overline{1}),

h_{15}(\overline{1}, \overline{x}) = \overline{e} = −h_{15}(\overline{x}, \overline{1}), h_{15}(\overline{1}, \overline{x^{2}}) = −\overline{a}_{1} = −h_{15}(\overline{x^{2}}, \overline{1}), h_{15}(\overline{x}, \overline{x^{2}}) = \overline{a}_{2} = −h_{15}(\overline{x^{2}}, \overline{x}),

h_{16}(\overline{1}, \overline{x}) = \overline{a}_{1} = −h_{16}(\overline{x}, \overline{1}), h_{16}(\overline{1}, \overline{x^{2}}) = −\overline{a}_{2} = −h_{16}(\overline{x^{2}}, \overline{1}),

h_{17}(\overline{1}, \overline{x}) = \overline{a}_{2} = −h_{17}(\overline{x}, \overline{1}).

It is clear that these elements generate Z^{2}(A, Reg(A))_{1}. Moreover, we have that h_{i} ∈ B^{2}(A, Reg(A))_{1} with i ∈ {1, 4, 5, 6, 8, 9, 12, 13, 14, 15, 16} generate B^{2}(A, Reg(A))_{1}. So, H^{2}(A, RegA)_{1} = F^{6}.

Finally, H^{2}(A, RegA) = F^{4} ⊕ F^{6}. •

References

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