Lobachevskii Journal of Mathematics

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Vol. 14, 2004, 3–16 \begin{matrix}c \\ ◯ \end{matrix} A. Aljouiee

Abdulla Aljouiee

ON THE BRAUER MONOID OF S_{3}

(submitted by M. Arslanov)

Abstract. In [HLS], the authors showed that the Brauer monoid of a finite Galois group can be written as a disjoint union of smaller pieces (groups). Each group can be computed following Stimets by defining a chain complex and checking its exactness. However, this method is not so encouraging because of the difficulty of dealing with such computations even with small groups. Unfortunately, this is the only known method so far. This paper is to apply Stimets’ method to some idempotent weak 2-cocycles defined on S_{3}. In particular, the idempotent 2-cocycles whose associated graphs have two generators. Some nice results appear in the theory of noncommutative polynomials.

1. Preliminaries

Let K/F be a finite Galois extension of fields and let G be its Galois group. A weak 2–cocycle is a function f : G × G → K satisfying

(i) f^{σ}(τ, ν)f(σ, τν) = f(στ, ν)f(σ, τ)

(ii) f(σ, 1) = f(1, σ) = 1

for all σ, τ, ν ∈ G.

If we define an algebra A_{f} associated with f to be the algebra generated as a K-vector space by the indeterminates {x_{σ} : σ ∈ G} with the relations x_{σ}x_{τ} = f(σ, τ)x_{στ} and x_{σ}k = k^{σ}x_{σ} for k ∈ K, x_{1} = 1, then A_{f} is called a weak crossed product. Condition (i) above guarantees the associativity of A_{f}.

In the classical theory f does not take the value 0, and in that case A_{f} is a central simple F -algebra. The set H = {σ ∈ G : f(σ, σ^{−1}) ≠ 0} is a subgroup of G called the inertial subgroup of f. We define an order “≤” on G/H by σH ≤ τH if and only if f(σ, σ^{−1}τ) ≠ 0. This order is lower subtractive, that is if σH ≤ τH then σH ≤ αH ≤ τH if and only if σ^{−1}αH ≤ σ^{−1}τH. In this order, H is the unique minimal element (root) and has the property f(H × H) ⊆ K^{∗} [HLS]. The weak crossed product A_{f} can be written as

A_{f} = ⨁_{σ∈H} Kx_{σ} ⊕ ⨁_{σ∉H} Kx_{σ} = B ⊕ J

where J = ⨁_{σ∉H} Kx_{σ} is the radical of A_{f} and B = ⨁_{σ∈H} Kx_{σ} is a classical crossed product algebra for f|_{H×H}. In particular B is a central simple K^{H}–algebra.

Given a weak 2-cocycle f, define a function e : G × G → {0, 1} by e(σ, τ) = 0 if and only if f(σ, τ) = 0. Then e is a weak 2-cocycle called the idempotent weak 2-cocycle associated to f.

Two weak 2-cocycles f, g are called cohomologous (or equivalent) if there is a function α : G → K^{∗} such that

f(σ, τ) = \frac{α(σ)α^{σ}(τ)}{α(στ)} g(σ, τ) for all σ, τ ∈ G.

Any two cohomologous weak 2-cocycles have the same associated idempotent cocycle. Under the equivalence relation introduced above the set of classes of weak 2-cocycles from G × G to K forms a monoid denoted by M^{2}(G, K). The subgroup of invertible elements of this monoid is the usual cohomology group H^{2}(G, K^{∗}).

Let e be an idempotent weak 2-cocycle. If f is a weak 2-cocycle associated to e then we can define a function g : G × G → K by

g(σ, τ) = \begin{cases}(f(σ, τ))^{−1} & if f(σ, τ) ≠ 0 \\ 0 & otherwise. \end{cases}

Then g is a weak 2-cocycle associated to e. If [ ·] denotes the equivalence class in the relation above, let M_{e}^{2}(G, K) = {[f] ∈ M^{2}(G, K) | [f][e] = [f] and there is a weak 2-cocycle g such that [f][g] = [e]}. Then M_{e}^{2}(G, K) is a group with identity [e] and M^{2}(G, K) = ∪_{e} M_{e}^{2}(G, K) (disjoint) where the union is over all idempotent weak 2-cocycles, [H1]. In a similar way, we can define the group M_{e}^{i}(G, K) of i^{th} dimension.

The set of all idempotent cocycles on G × G with inertial subgroup H is in 1 − 1 correspondence with all lower subtractive orders (graphs) on G/H with unique root H. Whenever we refer to the graph for e, we mean the graph associated to the weak 2-cocycle e. If [f] ∈ M_{e}^{2}(G, K) then f and e have the same associated graph. The following results and definitions are from [S1] and [S2]. Each idempotent 2-cocycle e with trivial inertial subgroup H determines a ring R_{e} = ℤ{x_{σ} : σ ∈ G}/I_{e} where I_{e} is the ideal generated by {x_{σ}x_{τ} − x_{στ} | σ ≤ στ}. The ring R_{e} is called the derived ring of e. To define a graded R_{e}–module, it is more convenient to use the notation g_{i} for the elements of the group G and [g_{1}, g_{2}, …, g_{k}] for the free generators of the R_{e}–module, where g_{i} ≤ g_{i+1} according to the relation defined above.

Define a graded R_{e}–module M by M = ⨁_{n∈ℤ} M_{n} where

M_{n} = \begin{cases}⨁_{g_{1}<···<g_{n}} R_{e}[g_{1}, …, g_{n}] & n ≥ 1, g_{i} ≠ 1 \\ R_{e} & n = 0 \\ ℤ & n = −1 \\ 0 & n ≤ −2 \end{cases}

with differentials d_{n}[g_{1}, …, g_{n}] = x_{g_{1}}[g_{1}^{−1}g_{2}, …, g_{1}^{−1}g_{n}] + ∑(−1)^{i}[g_{1}, …, ĝ_{i}, …, g_{n}]. We call the pair (M, d) the chain complex of e. This definition can be given in a similar manner if H ≠ {1}. Suppose f is a function from G^{n} to a field K which satisfies

(i) in case

f(g_{1}, g_{2}, …, g_{i−1}, 1, g_{i+1}, …, g_{n}) ≠ 0,

we have f(g_{1}, g_{2}, …, g_{i−1}, 1, g_{i+1}, …, g_{n}) = 1 for all g_{1}, g_{2}, …, g_{n} in G and all 1 ≤ i ≤ n,

(ii) f(1, 1, …, 1, g, 1, …, 1) = 1 for all g ∈ G, and (iii) for each g_{1}, g_{2}, …, g_{n+1} in G,

f^{g_{1}}(g_{2}, …, g_{n+1}) ∏_{i even} f(g_{1}, …, g_{i}g_{i+1}, …, g_{n+1})

= f(g_{1}, …, g_{n}) ∏_{i odd} f(g_{1}, …, g_{i}g_{i+1}, …, g_{n+1})

if n is even, and

f^{g_{1}}(g_{2}, …, g_{n+1})f(g_{1}, …, g_{n}) ∏_{i even} f(g_{1}, …, g_{i}g_{i+1}, …, g_{n+1})

= ∏_{i odd} f(g_{1}, …, g_{i}g_{i+1}, …, g_{n+1})

if n is odd. We then call f a weak n-cocycle. The first condition is analogous to the standard degeneracy conditions used in homological algebra, but still allows for the possibility that certain cochains may take on non-invertible values.

The second condition ensures that the cochains (cocycles) have a sufficient amount of invertibility.

If f is a weak n-cocycle and there is another weak n-cocycle h and an invertible cochain β : G^{n−1} → K^{∗} such that f = ∂β · h, where

∂β(g_{1}, …, g_{n})
= β^{g_{1}}(g_{2}, …, g_{n})β^{(−1)^{n}}(g_{1}, …, g_{n−1}) ∏ β^{(−1)^{i}}(g_{1}, …, g_{i}g_{i+1}, …, g_{n}),

then we say f is cohomologous to h and write f ∼ h. Let M^{n}(G, K) be the monoid of weak n-cocycles modulo the equivalence relation ∼. Then M^{n}(G, K) is called the weak Galois cohomology monoid. The class of cocycles equivalent (cohomologous) to the identity are known as weak coboundaries.

Since the groups M_{e}^{i}(G, K) are components of the required monoid, we are interested in computing these groups. Stimets ([S1],[S2]) has shown that under a sharp condition (exactness of (M, d)) we have the following isomorphism M_{e}^{i}(G, K) ≃ Ext_{R_{e}}^{i}(ℤ, K^{∗}) for all i, where the groups Ext_{R_{e}}^{i}(ℤ, K^{∗}) are relatively easier to deal with because they are well known and R_{e} acts on K^{∗} in the obvious way. (See [S1],[S2] for details). In this paper, we investigate exactness in some special cases. Exactness is always guaranteed at M_{0} ([S1]). It is quite difficult to check exactness in general but using the “contraction process” makes the situation somewhat easier. If g_{1} < ⋯ < g_{n}, call [g_{1}, ⋯, g_{n}] (n)–cell in M_{n}.

Proposition 1.1 (Contraction process) (Stimets). Let C_{n} ∈ M_{n}, C_{n−1} ∈ M_{n−1} be two cells such that C_{n} does not appear in the boundary of M_{n+1} and in the boundary of an n–cell not equal to C_{n}, the cell C_{n−1} does not appear. But if C_{n−1} appears in the boundary of C_{n}, then the chain complex (M′, d) obtained by removing C_{n} and C_{n−1} is exact at M′_{n} if and only if (M, d) is exact at M_{n}. Moreover, (M, d) and (M′, d) have the same homology groups.

This procedure allows us to cancel cells gradually until we reach a point after which we cannot proceed any further. Then we can investigate exactness at fewer modules.

2. Contracting to a Simpler Graph

In some cases, we can simplify the given graph to a smaller one. This section is devoted to demonstrating a class of graphs that can be contracted to a specific form of Z_{4} and showing that such graphs possess exact chain complex.

Lemma 2.1. In the ring R = ℤ{x, y}/(x^{2} = y^{2}), if α(x − 1) = β(y − 1), then α = h(x + 1), β = h(y + 1) for some h ∈ R.

Proof. We claim that the set V = {y^{ε}xyxy … xyx^{i} | ε ∈ {0, 1}, i ≥ 0} forms a ℤ-basis for R. Clearly V generates R. We show the independence. Let M be a free module on T = {Y^{ε}XY ⋯ Y X^{i} | ε ∈ {0, 1}, i ≥ 0}. The module M can be viewed as a right R-module by defining the action:

(Y^{ε}XY_{1}X ⋯ XY_{j}X^{i}) · x = Y^{ε}XY_{1}X ⋯ XY_{j}X^{i+1}
(Y^{ε}XY_{1}X ⋯ XY_{j}X^{i}) · y = \begin{cases}Y^{ε}XY_{1}X ⋯ XY_{j+1}X^{i−1} & if i is odd \\ Y^{ε}XY_{1}X ⋯ XY_{j−1}X^{i+3} & if i is even. \end{cases}

This action is well-defined and notice that

(Y^{ε}XY_{1}X ⋯ XY_{j}X^{i}) · y^{2} = (Y^{ε}XY_{1}X ⋯ XY_{j}X^{i}y) · y
= \begin{cases}Y^{ε}XY_{1}X ⋯ XY_{j+1}X^{i−1} · y & if i is odd \\ Y^{ε}XY_{1}X ⋯ XY_{j−1}X^{i+3} · y & if i is even \end{cases}
= \{ \begin{matrix}(Y^{ε}XY_{1}X ⋯ XY_{j}XY_{j+1} · y)X^{i−1} \\ Y^{ε}XY_{1}X ⋯ XY_{j}X^{i+2} \end{matrix}
= Y^{ε}XY_{1}X ⋯ XY_{j}X^{i+2}
= (Y^{ε}XY_{1}X ⋯ XY_{j}X^{i}) · x^{2}.

So we can define a homomorphism φ : R → End_{ℤ}(M) by

φ(x) = right multiplication by x.
φ(y) = right multiplication by y.

Now, for z_{i} ∈ V, if ∑ a_{i}z_{i} = 0, then φ(∑ a_{i}z_{i})(1) = 0 = ∑ a_{i}Z_{i} ⟹ a_{i} = 0 for all i, and we showed the claim. Now, we define a total order on the basis elements as follows: To each y^{ε}xy_{1}x ⋯ xy_{j}x^{i} ∈ V, assign a degree

(ε, \underbrace{2, 2, …, 2}_{j}, \underbrace{1, 1, …, 1}_{i}) and let deg m = (\frac{1}{2}) for all m ∈ ℤ^{∗}, deg 0 = 0.

If u, v ∈ V, we define u ≤ v if the length of the corresponding tuple of v is greater than the length of the corresponding tuple of u. If they have the same length, then compare the first numbers on the right, if they are the same go to the next numbers and so on. Notice that

(ε, \underbrace{2, 2, …, 2}_{j}, \underbrace{1, 1, …, 1}_{i}) · x = (ε, \underbrace{2, 2, …, 2}_{j}, \underbrace{1, 1, …, 1}_{i+1})

and

(ε, \underbrace{2, 2, …, 2}_{j}, \underbrace{1, 1, …, 1}_{i}) · y = \begin{cases}(ε, \underbrace{2, 2, …, 2}_{j−1}, \underbrace{1, 1, …, 1}_{i+3}) & i is even \\ (ε, \underbrace{2, 2, …, 2}_{j+1}, \underbrace{1, 1, …, 1}_{i−1}) & i is odd. \end{cases}

This shows that multiplying an element of V by x or y increases the degree. Since any monomial in V ends from the right with either y or x^{i}, i > 0, we can write any element in V as a sum f + gy where f and g are sums of monomials of V ending with x^{i}, i > 0. We say f is the component of 1 and g is the component of y. Remember that in this form any term in the sum g must end with x^{1}. Let α = f_{1} + g_{1}y, β = h_{1} + k_{1}y where f_{1}, g_{1}h_{1} and k_{1} are sums of elements in V which end with x^{i}, i > 0. Let f, g, h and k be of the largest degrees in f_{1}, g_{1}, h_{1} and k_{1} respectively. The equation α(x − 1) = β(y − 1) gives that

fx + gyx − gy + lower terms in both components

= kx^{2} − h + hy − ky + lower terms in both components. (2.1)

Let us denote by (g) for the degree of g, (g, 2, 1) for the degree of gyx and (k, 1, 1) for the degree of kx^{2}. From (2.1), if we assume that g ≠ 0, and h ends with x^{1}, we get

Case 1. If (h) < (k) then (g) = (k) and (g, 2, 1) ≤ (k, 1, 1) (impossible).

Case 2. If (h) ≥ (k) then (g) = (h) and (g, 2, 1) ≤ max{(h), (k, 1, 1,)} (impossible).

Now assume g ≠ 0 and h ends with x^{i}, i > 1. This implies that g = k and fx = hy. Let ∑_{ℓ=0}^{t} h_{ℓ} be the component of 1 in β where h_{ℓ} ends with x^{i}, i > 1, for all ℓ and h_{t} = h. Equation(2.1) implies that there exists h_{j} for some j such that gyx = h_{j}y or gyx = h_{j}. The latter is clearly impossible since h_{j} ends with x^{i}, i > 1. The first is also impossible since

h_{j}y = h′_{j}x^{i}y = \begin{cases}h′_{j}yx^{i} & i is even \\ h′_{j}xyx^{i−1} & i is odd ≥ 3, \end{cases}

and in both cases h_{j}y ends with x^{r}, r > 1. So, gyx ≠ h_{j}y for all j.

Thus g must be 0 and either k = 0 or k = h. In the first case, ∑_{ℓ=0}^{t} h_{ℓ} = β where h_{ℓ} ends with x^{i}, i > 1 for all ℓ. Equation (2.1) implies that fx = hy. But hy always ends with x^{2i}. so f ends with x^{2i−1}

and f = uyx^{2i−1} for some u. So h = \begin{cases}ux^{2i} \\ u′yx^{2i+1} \end{cases} .

Figure 1

Cancel fx with hy in (2.1) we get the term of the highest degree in the left hand side is f, but f = h_{ℓ}y for any ℓ. Therefore f = h which is impossible. This forces h = k.

Hence the original equation takes the form f_{1}(x − 1) = h′(1 + y)(y − 1) where h′ is the sum of either components. This implies that f_{1}(x − 1) = h′(y^{2} − 1) = h′(x^{2} − 1) = h′(x + 1)(x − 1) ⟹ α = h′(x + 1) and β = h′(y + 1). □

Theorem 2.2. The chain complex of the idempotent weak 2-cocycle e whose graph is given by: over ℤ_{4} is exact.

Proof. We have M_{0} = ℤ{x, y}/(x^{2} = y^{2}) = R, M_{1} = R[1] ⊕ R[2] ⊕ R[3]. M_{2} = R[1, 2] ⊕ R[3, 2], where x = x_{1}, y = x_{3}. We need to check the exactness of the following chain complex:

0 ⟶ M_{2} ⟶^{d_{2}} M_{1}⟶^{d_{1}} M_{0} ⟶ 0.

As stated earlier, we always have exactness at M_{0}. Now d_{2}(α[1, 2] + β[3, 2]) = 0 ⟹ α(x_{1}[1]−[2]+[1])+β(x_{3}[3]−[2]+[3]) = 0 ⟹ αx_{1}+α = 0, −α − β = 0 and βx_{3} + β = 0 ⟹ α = β = 0 and d_{2} is injective. To check exactness at M_{1}, let z = a[1] + b[2] + c[3] ∈ ker d_{1}, then z ∈ imd_{2} if and only if z′ = z + d_{2}(b[1, 2]) = (a + bx_{1} + b)[1] + c[3] ∈ imd_{2}. Let −s = a + bx_{1} + b. Now, z′ ∈ ker d_{1} ⟹ s(x_{1} − 1) = c(x_{3} − 1). By the lemma above, the unique solution for such an equation is s = h(x_{1} + 1) and c = h(x_{3} + 1) for some h ∈ R. So z′ = −h(x_{1} + 1)[1] + h(x_{3} + 1)[3] = d_{2}(−h[1, 2] + h[3, 2]). Hence z ∈ imd_{2}, so imd_{2} = ker d_{1}. □

Remark 2.3. We joint each weak 2–cocycle with a chain complex and a unique graph, so excising cells in the complex is equivalent to cancelling edges in the corresponding graph, and we are free to talk about one of these two contractions instead of the other.

Figure 2

Figure 3

Corollary 2.4. Let e be an idempotent weak 2-cocycle and let us denote by Γ_{e} its lower subtractive graph on a finite group G with a derived ring ℤ{x, y}/(x^{2} = y^{2}). If Γ_{e} can be contracted to a graph of the form: with σ^{2} = τ^{2} = ν, then the chain complex of e is exact.

Proof. This is a direct consequence of Theorem 2.2 and Proposition 1.1. □

Example 2.1. Consider the idempotent e which is given by its graph over the quaternion group Q = {±1, ±i, ±j, ±k}:

It is easy to check that R_{e} = ℤ{x, y}/(x^{2} = y^{2}) where x = x_{i}, y = x_{j}. d_{3}[i, k, −i] = x_{i}[j, −1] − [k, −i] + [i, −i] − [i, k] d_{3}[j, −k, −j] = x_{j}[i, −1] − [−k, −j] + [j, −j] − [j, −k] The cells [i, k, −i] and [j, −k, −j] can be excised with [i, k] and [−k, −j]

respectively. We now only have the relations:

d_{2}[i, −1] = x_{i}[i] − [−1] + [i] d_{2}[i, −i] = x_{i}[−1] − [−i] + [i] d_{2}[j, −k] = x_{j}[i] − [−k] + [j] d_{2}[j, −1] = x_{j}[j] − [−1] + [j] d_{2}[j, −j] = x_{j}[−1] − [−j] + [j] d_{2}[k, −i] = x_{k}[j] − [−i] + [k]

Figure 4

Notice that the cells [k, −i], [j, −j], [j, −k] and [i, −i] can be excised with [k], [−j], [−k] and [−i] respectively. And then we are left with a graph of the form: where i^{2} = j^{2} = −1. So, the chain complex of e is exact.

Corollary 2.5. In the previous example M_{e}^{i}(Q, K) ≃ Ext_{R_{e}}^{i}(ℤ, K^{∗}) where Q = {±1, ±i, ±j, ±k} is the Galois group of the extension K/F for some base field F.

3. S_{3} with Two Generators

We will need some lemmas before we state and prove the main result in this work.

Lemma 3.1. There is no non-trivial solution for the equation α(x−1) = β(y − 1) over the ring R = ℤ{x, y}/(yx = xy^{2}).

Proof. First, we show that the set {x^{i}y^{j}|i, j ≥ 0} forms a basis for R over ℤ. Let M be the free ℤ–module generated by {X^{i}Y^{j}|i, j ≥ 0}. Define the following action to make M a right R–module. X^{i}Y^{j}x = X^{i+1}Y^{2j}, X^{i}Y^{j}y = X^{i}Y^{j+1} for all i, j. This action is well-defined as we have seen in Lemma 2.1 and we can define a homomorphism φ : R → End_{ℤ}(M) by x ↦ multiplication from the right by x, y ↦ multiplication from the right by y. So, for any combination t = ∑ a_{ij}x^{i}y^{j} = 0, we have φ(t)(1) = 0 or ∑ a_{ij}X^{i}Y^{j} = 0 and hence a_{ij} = 0 for all i, j. Let x^{s}y^{t} be denoted by (s, t).

We define a degree function on the basis elements in R by (s, t) < (s′, t′) if s < s′. If s = s′, then (s, t) < (s′, t′) if t < t′. Notice that multiplying by x, y from the right gives (s, t)x ↦ (s + 1, 2t), (s, t)y ↦ (s, t + 1). Let α = (s_{1}, t_{1}) + (s_{2}, t_{2})+ lower degrees, where (s_{1}, t_{1}) > (s_{2}, t_{2}), β = (s′, t′)+ lower degrees. α(x − 1) = β(y − 1) gives that (s_{1} + 1, 2t_{1}) − (s_{1}, t_{1}) + (s_{2} + 1, 2t_{2})+ lower degrees = (s′, t′ + 1) − (s′, t′)+ lower degrees ⟹ (s_{1} + 1, 2t_{1}) = (s′, t′ + 1), that is, s′ = s_{1} + 1 and t′ = 2t_{1} − 1. We claim that (s′, t′) > max{(s_{1}, t_{1}), (s_{2} + 1, 2t_{2})}. Clearly s′ > s_{1} since s′ = s_{1} + 1. s_{2} ≤ s_{1} ⟹ s_{2} + 1 ≤ s′, if s_{2} + 1 < s′, we are done. So, let s_{2} + 1 = s′, thus, s_{1} = s_{2}. This implies t_{1} > t_{2} … (*). Assume that t′ ≤ 2t_{2}, so 2t_{1} − 1 ≤ 2t_{2} or t_{2} ≥ t_{1} which contradicts (*). □

Lemma 3.2. In the ring R = ℤ{x, y}/(xyx = y^{2}), α(x − 1) = β(y − 1) implies that α = k_{0}(x^{n−1} + x^{n−2} + ⋯ + 1)(1 + xy) and β = k_{0}(1 − x^{n} + (x^{n−1} + x^{n−2} + ⋯ + 1)y), for any k_{0} ∈ R and n ∈ ℕ.

Proof. Using an idea similar to what is in Lemma 3.1, we can show that S = {y^{ε}x^{i_{1}}yx^{i_{2}}y ⋯ yx^{i_{ℓ}}, ε ∈ {0, 1}, i_{j}, ℓ ∈ ℤ^{+} for j ≠ ℓ, i_{ℓ} ∈ {0, 1, … }} forms a ℤ-basis for R. Assign to each element y^{ε}x^{i_{1}}yx^{i_{2}}y … yx^{i_{ℓ}} a degree (ε, i_{1}, i_{2}, …, i_{ℓ}) if i_{ℓ} > 0 and (ε, i_{1}, i_{2}, …, i_{ℓ−1}) for y^{ε}x^{i_{2}}y … yx^{i_{ℓ−1}}y. If f ends from the right with x^{i}, i > 0, denote by (f) for degree of f, (f + k) for fx^{k}, (f, 1) for fyx. Notice also that we can express any element in R as f_{0} + g_{0}y where f_{0} and g_{0} are sums of monomials from S ending with x^{i}, i > 0. Call f_{0} the component of 1 and g_{0} the component of y.

Let f, g, h and k be of the largest degree in α, β in both components. So

(f + gy + ⋯)(x − 1) = (h + ky + ⋯)(y − 1)  (3.1)

⟹ fx + gyx − gy+ lower terms = kxyx − h + hy − ky+ lower terms ⟹

(i) (g) = (h − k)

(ii) max((f + 1), (g, 1)) = max((k + 1, 1), (h)).

We discuss the following cases:

Case 1. If g = 0 then h = k and (f + 1) = (k + 1, 1) (impossible).

Case 2. labelcs2.6 If g ≠ 0 and (h) < (k) then (g) = (k) and from (ii) (f + 1) = (k + 1, 1) (impossible).

Case 3. If g ≠ 0 and (h) > (k) then (g, 1) = (k + 1, 1) ⟹ g = mkx and h = rkx for some m, r ∈ ℤ^{∗}. Substituting in the equation (3.1) yields m = 1 = −r and so, g = kx = −h. Let g′, k′, h′ be of the next largest degree (g′) < (g), (k′) < (k), (h′) < (h), then we get

fx + kxyx + g′yx − kxy − g′y + lower terms
= −kxy + kx + h′y + kxyx − ky + k′xyx + lower terms ⟹

(i)′ (g′) = (h′ − k)

(ii)′ max((f + 1), (g′, 1)) = max((k + 1), (k′ + 1, 1)).

Again if g′ = 0 then since h′ = k = tfx, t ∈ ℤ^{∗} and equation (3.1) gives fx+ lower terms = tfx^{2}+ lower terms ⟹ f = 0 and hence h′ = k = f = g′ = g = h = 0.

If g′ ≠ 0, then (h′) ≤ (k) ⟹ (g′) = (k) and hence (g′) = (k′ + 1) ⟹ g′ = sk = tk′x, s, t ∈ ℤ^{∗}. As above we find s = t = 1. Repeat this step to get g_{i} = k_{i}x, k_{i+1} = k_{i}x where (k_{i+1}) > (k_{i}) and equation (3.1) becomes

(F +k_{0}(x^{n}+x^{n−1}+⋯+x)y)(x−1) = (−k_{0}x^{n}+H′+k_{0}(x^{n−1}+⋯+1)y)(y−1)

(3.2)

where F is the sum of components of 1 in α, H′ the sum of components of 1 in β that have degrees less than h. Let A := (x^{n−1} + ⋯ + 1), so equation (3.2) reads

(F + k_{0}Axy)(x − 1) = (−k_{0}x^{n} + H′ + k_{0}Ay)(y − 1) ⟹

F x − F + k_{0}Axyx − k_{0}Axy

= H′y − H′ + k_{0}Axyx − k_{0}x^{n}y − k_{0}Ay + k_{0}x^{n} ⟹
F x − F = H′y − H′ − k_{0}y + k_{0}x^{n}
(since − x^{n} − A = −Ax − 1) ⟹
k_{0} = H′ ⟹
F (x − 1) = k_{0}(x^{n} − 1) ⟹
F (x − 1) = k_{0}(x^{n−1} + ⋯ + 1)(x − 1) = k_{0}A(x − 1) ⟹
F = k_{0}A.

Therefore

α(x − 1) = (k_{0}A + k_{0}Axy)(x − 1) = k_{0}A(1 + xy)(x − 1)  (3.3)

and

β(y − 1) = (−k_{0}x^{n} + k_{0} + k_{0}Ay)(y − 1) = k_{0}(−x^{n} + 1 + Ay)(y − 1)
= k_{0}(−x^{n}y + x^{n} + y − 1 + Axyx − Ay)
= k_{0}(−Axy − y + x^{n} + y − 1 + Axyx)
(−x^{n}y − Ay = −Axy − y)
= k_{0}((x^{n} − 1) + Axy(x − 1))
= k_{0}(A(x − 1) + Axy(x − 1))
= k_{0}A(1 + xy)(x − 1) = α(x − 1).
□

Lemma 3.3. In the ring R = ℤ{x, y}/(xyx = yxy), the solutions of α(x − 1) = β(y − 1) are α = h(xy − y + 1) and β = h(yx − x + 1) for some h ∈ R.

Proof. Since xyx = yxy in R, any element in R can be written in a unique way as a combination of elements of the form x^{i_{1}}y^{j_{1}}x^{i_{2}}y^{j_{2}} … x^{i_{ℓ}}y^{j_{ℓ}} where i_{1}, j_{ℓ} ≥ 0, i_{2}, …, i_{ℓ−1} ≥ 2, j_{1}, … j_{ℓ−1} ≥ 1, i_{ℓ} ≥ 2 if j_{ℓ} > 0 and i_{ℓ} ≥ 1 if j_{ℓ} = 0. The form yxy is always replaced by xyx. This set of monomials forms a ℤ-basis for R by applying the same trick in Lemma 3.1. Assign to each such element x^{i_{1}}y^{j_{1}}x^{i_{2}}y^{j_{2}} … x^{i_{ℓ}}y^{j_{ℓ}} a degree (i_{1}, j_{2}, i_{2}, j_{2}, …, i_{ℓ}, j_{ℓ}) and notice that

(i_{1}, j_{1}, i_{2}, j_{2}, …, i_{ℓ}, j_{ℓ})x = \begin{cases}(i_{1}, j_{1}, …, i_{ℓ} + 1, 0) & if j_{ℓ} = 0 \\ (i_{1}, j_{1}, …, i_{ℓ}, j_{ℓ}, 1, 0) & if j_{ℓ} ≠ 0 \end{cases}

(n_{1}, m_{1}, …, m_{t−1}, n_{t}, m_{t})y =

= \begin{cases}(n_{1}, m_{1}, …, n_{t}, m_{t} + 1) & if m_{t} ≠ 0 \\ (n_{1}, m_{1}, …, m_{t−1}, n_{t}, 1) & if m_{t} = 0 and n_{t} > 1 \\ (n_{1}, m_{1}, …, n_{t−1} + 1, 1, m_{t−1}, 0) & if m_{t} = 0 and n_{t} = 1 \end{cases}

Obviously, these two terms can not be equal except if m_{t} = 0 and n_{t} = 1. If j_{ℓ} = 0 then the equality implies (h′x^{i_{ℓ−1}}yx^{i_{ℓ}})x = (h′x^{i_{ℓ−1}−1}y^{i_{ℓ}+1}x)y for some h′ ∈ R. If j_{ℓ} ≠ 0 then the equality gives j_{ℓ} = m_{t−1} = 1 and i_{ℓ} = n_{t−1} + 1. So, (h′′x^{i_{1}}y)x = (h′′x^{i_{ℓ−1}}yx)y for some h′′ ∈ R. But in the first case we note that h′x^{i_{ℓ−1}}yx^{i_{ℓ}} = h′x^{i_{ℓ−1}−1}y^{i_{ℓ}}xy, so in all cases if (i_{1}, j_{1}, i_{2}, j_{2}, …, i_{ℓ}, j_{ℓ})x = (n_{1}, m_{1}, …, m_{t−1}, n_{t}, m_{t})y then there is h ∈ R such that hxy = (i_{1}, j_{1}, i_{2}, j_{2}, …, i_{ℓ}, j_{ℓ}), hyx = (n_{1}, m_{1}, …, m_{t−1}, n_{t}, m_{t}). Now, let α = ∑_{i=1}^{r} α_{i}, β = ∑_{i=1}^{s} β_{i}, so by a suitable rearrangement, α(x−1) = β(y−1) implies that α_{1}x = β_{1}y ⟹ α_{1} = h_{1}xy and β_{1} = h_{1}yx and hence h_{1}xyx − h_{1}xy + α_{2}x − α_{2} + α_{3}x − α_{3} + ⋯ = h_{1}yxy − h_{1}yx + β_{2}y − β_{2} + β_{3}y − β_{3} + … ⟹ α_{2}x = −h_{1}yx or α_{2} = −h_{1}y. Similarly, we find β_{2} = −h_{1}x, α_{3} = h_{1}, β_{3} = h_{1}, α_{4} = h_{2}xy, β_{4} = h_{2}yx, … ⟹ α = ∑_{i=1}^{q} h_{i}(xy −y +1) and β = ∑_{i=1}^{q} h_{i}(yx−x+1) for some integer q. Take h = ∑_{i=1}^{q} h_{i}. □

Definition 3.1. Let e be a weak 2–cocycle and let Γ_{e} be its graph. Then, we call the elements of G of level 1 generators of Γ_{e}. That is the elements lie right above the root of Γ_{e} are called generators.

Note that generators of the graph certainly generate the group itself.

Figure 5

Theorem 3.4. Let e be an idempotent 2-cocycle whose graph has two generators over S_{3} = {σ, τ|σ^{2}τ = τσ, σ^{3} = τ^{2} = 1}. Then e admits an exact chain complex.

Proof. Using a computer program, it can be shown that there are nine distinct lower subtractive graphs with two generators over S_{3}. Five of them are trees which admit exact graded modules (see [S2]). The remaining four are:

By excising cells, we end up with M_{j}^{i} = 0 for all j > 2, i = 1, 2, 3, 4.

Let 0 → M_{2}^{i} →^{d_{2}^{i}} M_{1}^{i} →^{d_{1}^{i}} M_{0}^{i} → ℤ → 0 be the chain complex of e_{i}, i = 1, 2, 3, 4, which we get after excision. We always have exactness at M_{0}^{i}. At M_{2}^{i}, it is easy to show that d_{2}^{i} is injective for all i. To show exactness at M_{1}^{3}, let z = a[σ^{2}τ] + b[σ^{2}] + c[σ] + d[στ] ∈ ker d_{1}^{3}, so z ∈ imd_{2}^{3} if and only if z′ = z + d_{2}^{3}(c[σ^{2}, σ] + d[σ^{2}, στ]) = α[σ^{2}τ] + β[σ^{2}] ∈ imd_{2}^{3}, for some α, β ∈ R_{3}. But z ∈ ker d_{1}^{3} ⟹ z′ ∈ ker d_{1}^{3} ⟹ α(x − 1) = β(y − 1) ⟹ by Lemma 3.1 α = β = 0. Thus, z ∈ imd_{2}^{3} if and only if 0 ∈ imd_{2}^{3} which is always true. A similar idea can be applied to show exactness at M_{1}^{2}. For M_{1}^{4}, let u = p[σ^{2}τ] + q[σ^{2}] + r[τ] + s[στ] ∈ ker d_{1}^{4}, so u ∈ imd_{2}^{4} if and only if u′ = γ[σ^{2}τ] − δ[τ] ∈ imd_{2}^{4} for some γ, δ ∈ R_{e_{4}}. Since u′ ∈ ker d_{1}^{4} ⟹ by Lemma 3.3 γ = h(xy − y + 1), δ = h(yx − x + 1) for some h ∈ R_{e_{4}} ⟹ u ∈ imd_{2}^{4} if and only if u′ = h(xy − y + 1)[σ^{2}τ] − h(yx − x + 1)[τ] ∈ imd_{2}^{4}, but clearly d_{2}^{4}{h(1 − y)[σ^{2}τ, σ^{2}] + h[σ^{2}, στ] − h[τ, στ]} = u′. For M_{1}^{1}, let z = a[σ^{2}τ] + b[σ^{2}] + c[σ] + d[στ] ∈ ker d_{1}^{1} ⇒ z ∈ imd_{2}^{1} if and only if z′ = z + d_{2}^{1}(c[σ^{2}, σ] + d[σ^{2}, στ]) = α(σ^{2}τ) − β[σ^{2}] ∈ imd_{2}^{1} for some α, β ∈ R_{e_{1}}. Since z ∈ ker d_{1}^{1} ⇒ z′ ∈ ker d_{1}^{1} and α(x − 1) = β(x − 1) in the ring ℤ{x, y}/(xyx = y^{2}) = R_{e_{1}}. This implies that α = k_{0}(x^{n−1} + ⋯ + 1)(1+xy), β = k_{0}(1−x^{n}+(x^{n−1}+⋯+1)y) (by Lemma 3.2). So z ∈ imd_{2}^{1} 16 A. ALJOUIEE

if and only if z′ = k_{0}A(1 + xy)[σ^{2}τ] − k_{0}(1 − x^{n} + Ay)[σ^{2}] ∈ imd_{2}^{1}. Notice that 1−x^{n} = A−Ax. So z′ ∈ imd_{2}^{1} if and only if z′′ = k_{0}A((1+xy)[σ^{2}τ]− (1 − x + y)[σ^{2}]) ∈ imd_{2}^{1}. But z′′ = d_{2}^{1}(x[σ^{2}, στ] + [σ^{2}τ, σ] − [σ^{2}, σ]). □

Corollary 3.5. For any idempotent weak 2-cocycle e over S_{3} with two generators, M_{e}^{i}(S_{3}, K) ≈ Ext_{R_{e}}^{i}(ℤ, K^{∗}).

References

[A] A. Aljouiee, On Weak Crossed Products, Frobenius Algebras and Weak Bruhat Ordering, J. Algebra, to appear.

[H1] D. Haile, On Crossed Product Algebras Arising from Weak Cocycles, J. Algebra, 74, 1982, 270-279.

[H2] D. Haile, The Brauer Monoid of a Field, J. Algebra, 81, 1983, 521-539.

[HLS] D. Haile, R. Larson, M. Sweedler, Almost Invertible Cohomology Theory and the Classification of Idempotent Cohomology Classes and Algebras by Partially Ordered Sets with a Galois Group Action, Amer. J. Math., 105, 1983, 689-814.

[S1] R. Stimets, Weak Galois Cohomology and Group Extension, Communications in Algebra, 2000.

[S2] R. Stimets, Relative Weak Cohomology and Extension, J. Algebra, to appear.

Teachers College, P.O. Box 32644, Riyadh 11438, Saudi Arabia E-mail address: joa111@lycos.com

Received December 12, 2003