Specht Modules and the Cohomology of Mapping Class Groups

D. J. Benson Mathematical Institute 24–29 St. Giles Oxford OX1 3LB England

1 Introduction

Denote by Γ_{g,k}^{n} the mapping class group of orientation preserving self-diffeomorphisms of an orientable surface M_{g,k}^{n} of genus g with k boundary components and n punctures, where the boundary is required to be fixed pointwise, but the punctures are allowed to be permuted. Our goal is to calculate the cohomology of Γ_{2,0}^{0} with coefficients in an arbitrary field k. Our methods depend very heavily on work of F. Cohen [3], where in particular it is shown that H^{∗}(Γ_{2,0}^{0}, k) = 0 for char k ≠ 2, 3 or 5, and H^{∗}(Γ_{2,0}^{0}, F_{5}) is also computed. Our answer in characteristic 3 is as follows.

Theorem 1.1 (= Theorem 8.6)

The cohomology ring of the mapping class group of a closed oriented surface of genus two

H^{∗}(Γ_{2,0}^{0}, F_{3}) ≅ H^{∗}(Γ_{0,0}^{6}, F_{3})

is the graded commutative ring generated by elements β, γ, w and x with

deg(β) = 3, deg(γ) = deg(w) = 4, deg(x) = 5

subject to the relations

β^{2} = 0, βw = 0, w^{2} = 0, γw = βx, wx = 0, x^{2} = 0.

The Poincaré series of this ring is

∑_{n=0}^{∞} t^{n} dim H^{n}(Γ_{2,0}^{0}, F_{3}) = \frac{1 + t^{3} + t^{4} + t^{5}}{1 − t^{4}}
= 1 + t^{3} + 2t^{4} + t^{5} + t^{7} + 2t^{8} + t^{9} + t^{11} + · · · .

To compute this, we examine the central extension

1 → Z/2 → Γ_{2,0}^{0} → Γ_{0,0}^{6} → 1.

In characteristics other than 2, the Lyndon-Hochschild-Serre spectral sequence of this central extension collapses and we have H^{∗}(Γ_{2,0}^{0}, k) ≅ H^{∗}(Γ_{0,0}^{6}, k). In characteristic 2, we have managed to get as far as calculating H^{∗}(Γ_{0,0}^{6}, k), but an ambiguity in the action of the Steenrod algebra prevents us from completing^{1} the calculation of the Lyndon-Hochs-child-Serre spectral sequence

H^{∗}(Γ_{0,0}^{6}, H^{∗}(Z/2, F_{2})) ⇒ H^{∗}(Γ_{2,0}^{0}, F_{2}).

Theorem 1.2 ( = Theorem 7.3)

The cohomology H^{∗}(Γ_{0,0}^{6}, F_{2}) of the mapping class group of a sphere with six punctures is generated by elements α, β, γ, δ and u with deg(α) = deg(u) = 2, deg(β) = deg(γ) = 3 and deg(δ) = 1, subject to the following relations.

\begin{matrix}γδ = 0 & βγ = 0 & (δ^{3} + αδ + β)^{2} = 0 \\ γu = 0 & u^{2} = δ^{4} & (de^{3} + αδ + β)(u + δ^{2}) = 0. \end{matrix}

The action of the Steenrod algebra is given as follows.

Sq^{1}(α) = β + γ Sq^{1}(β) = Sq^{1}(γ) = 0
Sq^{1}(δ) = δ^{2}  Sq^{1}(u) = either δu + αδ + β or  δu + δ^{3}

Sq^{2}(β) = αβ  Sq^{2}(γ) = αγ.

The element of H^{2}(Γ_{0,0}^{6}, F_{2}) corresponding to the central extension Γ_{2,0}^{0} is u + α.

We shall indicate in Section 7 what needs to be done to resolve this ambiguity in the action of the Steenrod algebra. In the first case (Sq^{1}(u) = δu + αδ + β) we obtain a Poincaré series

∑_{n=0}^{∞} t^{n} dim H^{n}(Γ_{2,0}^{0}, F_{2}) = \frac{1 + t^{2} + 2t^{3} + t^{4} + t^{5}}{(1 − t)(1 − t^{4})}
= 1 + t + 2t^{2} + 4t^{3} + 6t^{4} + 7t^{5} + 8t^{6} + 10t^{7} + ⋯

while in the second case (Sq^{1}(u) = δu + δ^{3}) we obtain

∑_{n=0}^{∞} t^{n} dim H^{n}(Γ_{2,0}^{0}, F_{2}) = \frac{1 + t^{2} + t^{3} + t^{5} − t^{6}}{(1 − t)(1 − t^{4})}
= 1 + t + 2t^{2} + 3t^{3} + 4t^{4} + 5t^{5} + 5t^{6} + 6t^{7} + ⋯

error of my ways. Sq^{1}(u) = δu + αδ + β) is correct.

1In a circulated preprint of this paper there was a mistake in Theorem 7.3, which enabled us to give a complete, but wrong answer for H^{∗}(Γ_{2,0}^{0}, F_{2}). I am grateful to Fred Cohen for pointing out to me the

In the sequel to this paper, contained in the same volume, the ambiguity in the Steenrod algebra action is resolved using configuration space methods, and the Lyndon-Hochschild-Serre spectral sequence calculation for H^{∗}(Γ_{2,0}^{0}, F_{2}) is completed. It turns out that the first of the given possibilities (namely

In order to calculate H^{∗}(Γ_{0,0}^{6}, k), we examine more generally the Lyndon-Hochschild-Serre spectral sequence of the extension

1 → K_{n} → Γ_{0,0}^{n} →^{λ} Σ_{n} → 1

where λ is the natural map to the symmetric group on the set of punctures, and K_{n} is its kernel.

Cohen [3] found generators and relators for H^{∗}(K_{n}, Z) together with the action of Σ_{n}. It turns out in particular that H^{j}(K_{n}, Z) is a finitely generated free Z-module which vanishes for j ≥ n − 2.

Since we are mostly interested in the case n = 6, it suffices to identify the modules H^{j}(K_{n}, Z) for j ≤ 3 as representations of the symmetric group Σ_{n}. We do this in Section 2, in terms of Gordon James’ theory of Specht modules S^{λ} (see James [5]; for a brief summary of the theory of Specht modules, see the end of this introduction). Our result (Theorems 2.1, 2.2 and 2.3) may be summarised as follows.

Theorem 1.3 1. H^{1}(K_{n}, Z) ≅ S^{(n−2,2)}

2. There is a short exact sequence of Σ_{n}-lattices

0 → S^{(n−3,1^{3})} → Λ^{2}H^{1}(K_{n}, Z) → H^{2}(K_{n}, Z) → 0

3. There is an exact sequence of Σ_{n}-lattices

0 → ((S^{(2,2)} ⊗ S^{(1^{4})}) ˆ⊗S^{(n−4)})_{Σ_{4}×Σ_{n−4}} ↑^{Σ_{n}}
→ S^{(n−2,2)} ⊗ S^{(n−3,1^{3})} → Λ^{3}H^{1}(K_{n}, Z) → H^{3}(K_{n}, Z) → 0

In Section 3, we give preliminary information on modules for Σ_{6} and A_{6} in characteristic two. In Section 4, we use this information together with the above theorem to determine the modules H^{j}(K_{6}, F_{2}) in terms of the diagrammatic notation of Benson and Carlson [2].

In Section 5, we calculate the E_{2} term of Lyndon-Hochschild-Serre spectral sequence

H^{∗}(Σ_{6}, H^{∗}(K_{6}, F_{2})) ⇒ H^{∗}(Γ_{0,0}^{6}, F_{2})

We calculate using injective resolutions according to the methods of Benson and Carlson [2]. It turns out to be easier in our case to use injective rather than projective resolutions. Thus we shall be using the isomorphism

H^{i}(G, M) ≅ Ext_{kG}^{i}(k, M)
≅ Hom_{kG}(k, Ω^{−i}(M))

for M a kG-module. Here, Ω^{−i}(M) denotes the ith cokernel in a minimal injective resolution of M as a kG-module (we also denote by Ω^{i}(M) the ith kernel in a minimal projective resolution of M). Cohomology elements will therefore appear as copies of k at the bottom of the diagram for Ω^{−i}(M). If we had been using projective resolutions of k instead, we would have had to quotient out homomorphisms factoring through a projective module. The way we have set it up, every cochain is a cocycle, and every coboundary is zero.

In order to calculate the differentials in the spectral sequence, we use a theorem of Quillen [7, Theorem 14.1], which tells us that an element of H^{∗}(Γ_{0,0}^{6}, F_{2}) is nilpotent if and only if its restriction to every finite elementary abelian subgroup is nilpotent. Thus in Section 6, we investigate the finite subgroups of Γ_{0,0}^{6}, and their lift to Γ_{2,0}^{0}. In Section 7, we use this information to calculate the differentials in the spectral sequence, and hence complete the proof of Theorem 1.2.

Finally in Section 8, we sketch the analogous calculation in characteristic three. The computations are more complex, but mostly routine.

Since the bulk of this paper consists of applications of the diagrammatic methods of Benson and Carlson [2] for computing projective resolutions and cohomology of modules for finite groups, we consider it worthwhile^{2} to include here a brief introduction to these methods.

A kG-module diagram X consists of a finite directed graph with no directed cycles or multiple edges, often drawn on the page in such a way that the direction of the arrows is towards the bottom of the page, so that the arrow may be represented by a simple line without an arrowhead. To each vertex x ∈ X of the graph is associated a simple kG-module S_{x}, drawn from a standard list of the simples (i.e., a list containing one representative of each isomorphism class of simple modules). To each edge x → y is associated a non-zero extension class in Ext_{kG}^{1}(S_{x}, S_{y}). The following conditions should be satisfied.

(a) If there are edges x_{1} → x_{2} → ⋯ → x_{n} with n ≥ 3 then there is no edge x_{1} → x_{n}.

(b) If there are edges x → y_{1}, …, x → y_{n} with S_{y_{1}} = ⋯ = S_{y_{n}}, then the corresponding elements of Ext_{kG}^{1}(S_{x}, S_{y_{i}}) are linearly independent. Dually, if there are edges y_{1} → x, …, y_{n} → x with S_{y_{1}} = ⋯ = S_{y_{n}}, then the corresponding elements of Ext_{kG}^{1}(S_{y_{i}}, S_{x}) are linearly independent.

The vertices of the diagram X are given a topology in which the open sets are the subsets of the set of vertices U ⊆ X with the property that if x ∈ U and x → y is an edge then y ∈ U. Thus an open set determines a subgraph of U.

A representation of a module diagram consists of a kG-module M and a function U ↦ M_{U} from the open sets to the submodules of M, in such a way that the following conditions are satisfied.

(i) M_{X} = M, M_{∅} = {0}, and M_{U} ⊆ M_{V} whenever U ⊆ V.

(ii) M_{U∩V} = M_{U} ∩ M_{V} and M_{U∪V} = M_{U} + M_{V}.

(iii) If V = U ∪ {x} with x ∉ U then there is an exact sequence

0 → M_{U} → M_{V} → S_{x} → 0

2I would like to thank the referee for suggesting the inclusion of this discussion.

in such a way that if U ⊆ U′, V ⊆ V′ and V′ = U′ ∪ {x} with x ∉ U′ then the following diagram commutes.

0 → M_{U} → M_{V} → S_{x} → 0
↓  ↓  ∥
0 → M_{U′} → M_{V′} → S_{x} → 0

(iv) If V = U ∪ {x} with x ∉ U and W = V ∪ {y} with y ∉ V, and there is an edge y → x, then in the diagram

0  0
↓  ↓
M_{U} =  M_{U}
↓  ↓
0 → M_{V} → M_{W}  → S_{y} → 0
↓  ↓  ∥
0 → S_{x} → M_{W} /M_{U} → S_{y} → 0
↓  ↓
0  0

the bottom row represents the corresponding extension class in Ext_{kG}^{1}(S_{y}, S_{x}).

Thus in a diagram for a module in the above sense, the vertices correspond to the composition factors of the module and the edges tell us how the composition factors are “glued together.” The reader should be warned that the diagram does not give complete information about module structures. In particular, there can be several modules, or indeed no modules at all, which are representations of a given diagram. Also, there are modules which are not representations of any diagram. However, quite a lot of information may be recovered from the diagram. For example, the socle (maximal semisimple submodule) of a module corresponds to the open set consisting of points x for which there is no arrow x → y. Dually, the radical of the module corresponds to the open set consisting of points x for which there is some arrow y → x.

A uniserial module is one with a unique composition series. These are always representations of module diagrams of the form • → • → ⋯ → •. More generally, a string diagram is one in which two vertices have total (in + out) valency of one, and the remaining vertices have total valency two. A string diagram is said to be rigid if for each edge x → y we have dim_{k} Ext_{kG}^{1}(S_{x}, S_{y}) = 1. It is shown in Sections 6, 7 and 8 of Benson and Carlson [2] that a string diagram has a unique representation if every uniserial subdiagram has a unique representation, and that every representation of a rigid string diagram is indecomposable.

A diagrammatical homomorphism or D-homomorphism from a representation of a module diagram to another is, roughly speaking, a homomorphism of modules which corresponds to a map of diagrams. For the exact definition, see Section 2 of Benson and Carlson [2]. It is shown in Section 7 of that paper that if uniserial diagrams have unique representations then every homomorphism between representations of rigid string diagrams is a linear combination of D-homomorphisms.

Projective and injective resolutions may be formed by a process of “cutting and pasting.” This is best described by an example, and the reader is referred to Benson and Carlson [2] for the technical details. Note that for the group algebra of a finite group, projective and injective modules coincide. The projective cover of a module M is the same as the projective cover of M/Rad(M), and the injective hull of M is the same as the injective hull of Soc(M).

Example Let G = GL_{3}(2) and k = F_{2}. There are four simple kG-modules k, A, B, C of dimensions 1, 3, 3, 8 with projective covers P_{k}, P_{A}, P_{B}, P_{C} which are representations of the following diagrams.

\begin{matrix}k \\ / \ \\ A B \\ \ / \\ C \end{matrix}  \begin{matrix}A \\ / \ \\ k \begin{matrix}B \\ | \\ A \\ | \\ B \end{matrix} \\ \ / \\ A \end{matrix}  \begin{matrix}B \\ / \ \\ k \begin{matrix}A \\ | \\ B \\ | \\ A \end{matrix} \\ \ / \\ B \end{matrix}  C.

This means P_{k} has a unique top and bottom composition factor each isomorphic to k, and Rad(P_{k})/Soc(P_{k}) ≅ A ⊕ B. Similarly, P_{A} has a unique top and bottom composition factor each isomorphic to A, and Rad(P_{A})/Soc(P_{A}) is a direct sum of k and a uniserial module with composition factors B, A, B.

To obtain a projective resolution of k, we begin with P_{k}, so that the first kernel is

Ω(k) = \begin{matrix}A B \\ \ / \\ k \end{matrix} . From this diagram we can see that Ω(k)/RadΩ(k) ≅ A ⊕ B, so that the

projective cover of Ω(k) is P_{A} ⊕ P_{B}. The next kernel is obtained by cutting and pasting the diagrams for P_{A} and P_{B} as follows.

0 → \begin{matrix}B  A \\ |  | \\ A  B \\ |  | \\ B  k  A \\ \ / \ / \\ A B \end{matrix} → \begin{matrix}A \\ / \ \\ \begin{matrix}B \\ | \\ A \\ | \\ B \end{matrix} k \\ \ / \\ A \end{matrix} ⊕ \begin{matrix}B \\ / \ \\ k \begin{matrix}A \\ | \\ B \\ | \\ A \end{matrix} \\ \ / \\ B \end{matrix} → \begin{matrix}A B \\ \ / \\ k \end{matrix} → 0.

The reader is advised to continue this projective resolution at least a couple of stages further in order to get the hang of the diagrammatic method. Injective resolutions are calculated dually in exactly the same way.

The reader may get the impression that to calculate an entire projective or injective resolution this way would take infinitely much work. However, after a while the patterns start “repeating.” This is formalised, at least for groups of p-rank at most two, in Section 9 of Benson and Carlson [2].

Finally, we include here a brief introduction to Specht modules for symmetric groups. For further details see James [5] or James and Kerber [6]. The Specht module S^{λ} corresponding to a partition λ of n is a certain ZΣ_{n}-submodule of a permutation module, constructed as follows. The Young subgroup Σ_{λ} of Σ_{n} corresponding to a partition λ = (λ_{1}, …, λ_{s}) of n (i.e., λ_{1} ≥ ⋯ ≥ λ_{s} > 0 and ∑_{i=1}^{s} λ_{i} = n) is the subgroup Σ_{λ_{1}} × Σ_{λ_{2}} × ⋯ × Σ_{λ_{s}}, embedded in the obvious way (up to conjugation) in Σ_{n}. We denote by M^{λ} the permutation module on the cosets of Σ_{λ}. In other words, M^{λ} = Z_{Σ_{λ}} ↑^{Σ_{n}}= ZΣ_{n} ⊗_{ZΣ_{λ}} Z.

Corresponding to a partition λ we draw a diagram with λ_{1} boxes in the first row, λ_{2} in the second, and so on. So for example the diagram for (4, 2, 1) is

A λ-tableau is defined to be a way of filling the boxes in the diagram corresponding to λ with the numbers from 1 to n. The group Σ_{n} acts on the λ-tableaux by permuting the entries, and the permutations which keep the numbers in the same row form a conjugate of Σ_{λ}. A λ-tabloid is defined to be an equivalence class of λ-tableaux under these row permutations. Thus the λ-tabloids form a Z-basis for M^{λ}.

The λ-polytabloid corresponding to a λ-tableau is the element of M^{λ} given by the alternating sum of the images in M^{λ} of the λ-tableaux obtained by applying column permutations. The sign is given by the sign of the corresponding column permutation.

Thus for example the polytabloid corresponding to the (2, 2)-tableau 4 2 5 1 3 is

4 2 5 1 3 − 1 2 5 4 3 − 4 3 5 1 2 + 1 3 5 4 2

The Specht module S^{λ} is defined to be the submodule of M^{λ} given by the Z-span of the polytabloids. It is a cyclic module generated by any one polytabloid, and has a basis consisting of the polytabloids for tableaux in which the numbers are increasing down each column and along each row. The number of such polytabloids is given by the hook formula (James [5, Section 20]).

If K is a field of characteristic zero, then the modules KS^{λ} = K ⊗_{Z} S^{λ} form a complete set of irreducible KΣ_{n} modules without repetition. For a field k of characteristic p, the module kS^{λ} = k ⊗_{Z} S^{λ} is usually not irreducible. However, if λ is a p-regular partition (i.e., no p consecutive λ_{i} are equal) then kS^{λ} has a unique maximal submodule, so that D^{λ} = kS^{λ}/Rad(kS^{λ}) is irreducible. The modules D^{λ} for λ p-regular form a complete set of irreducible kΣ_{n}-modules, without repetition. One of the major unresolved problems in the subject is to find the multiplicities of the composition factors of the modules kS^{λ} (or even the dimensions of the modules D^{λ}). However, for small n, the answer is known, and tables of decomposition matrices may be found in James [5].

2 The modules H^{j}(K_{n}, Z), j ≤ 3

Theorem 2.1 As a module for Σ_{n} = Γ_{0,0}^{n}/K_{n}, H^{1}(K_{n}, Z) is isomorphic to the Specht module S^{(n−2,2)}.

Proof Denote by P_{n} the pure braid group on n strings. By Cohen [3, Section 3] we have a monomorphism 0 → H^{1}(K_{n}, Z) →^{α^{∗}} H^{1}(P_{n}, Z), where the module H^{1}(P_{n}, Z) is freely generated as a Z-module by symbols A_{i,j} = A_{j,i} (i ≠ j) with Σ_{n}-action given by

σA_{i,j} = A_{σ(i),σ(j)} σ ∈ Σ_{n}.

It is therefore the permutation module on cosets of Σ_{n−2} × Σ_{2}. The submodule H^{1}(K_{n}, Z) is spanned by symbols B_{i,j} with n ≥ i > j ≥ 2 and

α^{∗}(B_{i,j}) = \{ \begin{matrix}A_{i,j} − A_{i,j−1} + A_{j−1,1} − A_{j,1} & if j ≥ 3 \\ A_{i,2} − A_{i,1} + A_{3,1} − A_{3,2} & if j = 2, i ≥ 4. \end{matrix}

In James’ terminology (see the summary at the end of Section 1) these are the polytabloids corresponding to the tableaux

\begin{matrix}j − i & 1 & ⋯ \\ j & i \end{matrix} if j ≥ 3, and \begin{matrix}1 & 3 & ⋯ \\ 2 & i \end{matrix} if j = 2, i ≥ 4.

Since the Z-span of the polytabloids forms a cyclic ZΣ_{n}-lattice generated by any one polytabloid, it follows that H^{1}(K_{n}, Z) is simply the Specht module S^{(n−2,2)}. □

Theorem 2.2 There is a Z-split short exact sequence of ZΣ_{n}-lattices

0 → S^{(n−3,1^{3})} →^{ϕ} Λ^{2}H^{1}(K_{n}, Z) → H^{2}(K_{n}, Z) → 0.

Proof It is shown in Cohen [3] that H^{∗}(K_{n}, Z) is generated as an algebra by H^{1}(K_{n}, Z). Thus H^{j}(K_{n}, Z) is a quotient of Λ^{j}H^{1}(K_{n}, Z). It merely remains to identify the kernel.

We define a map ϕ : S^{(n−3,1^{3})} → Λ^{2}H^{1}(K_{n}, Z) by stipulating that the polytabloid corresponding to the tableau

\begin{matrix}a & ⋯ \\ b \\ c \\ d \end{matrix}

is taken to the wedge of polytabloids

\begin{matrix}a & b & ⋯ \\ c & d \end{matrix} ∧ \begin{matrix}a & c & ⋯ \\ b & d \end{matrix}

Note that the span of all the polytabloids with a, b, c and d in these four places has Z-rank two, and so any other recipe consisting of a wedge of two polytabloids with a, b, c and d in these four places is either zero or equal to plus or minus this recipe.

To check that this map ϕ is well defined, we must check the column alternation relations and the Garnir relations (see the remark in James and Kerber [6, bottom of p. 302]).

A typical example of an alternation relation is that the image of the polytabloid with a and b interchanged

\begin{matrix}b & ⋯ \\ a \\ c \\ d \end{matrix}

must be the negative of the original expression. This is checked as follows.

\begin{matrix}b & a & ⋯ \\ c & d \end{matrix} ∧ \begin{matrix}b & c & ⋯ \\ a & d \end{matrix} = ( \begin{matrix}a & b & ⋯ \\ c & d \end{matrix} + \begin{matrix}b & c & ⋯ \\ a & d \end{matrix} ) ∧ (− \begin{matrix}a & c & ⋯ \\ b & d \end{matrix} )

= − ( \begin{matrix}a & b & ⋯ \\ c & d \end{matrix} ∧ \begin{matrix}a & c & ⋯ \\ b & d \end{matrix} )

It is necessary to make three such checks, for the three pairs {a, b}, {b, c} and {c, d}. The remaining two are left to the reader.

Finally we must check the Garnir relation. The image of the original polytabloid should equal the image of

\begin{matrix}e & a & ⋯ \\ b \\ c \\ d \end{matrix} + \begin{matrix}a & b & ⋯ \\ e \\ c \\ d \end{matrix} + \begin{matrix}a & c & ⋯ \\ b \\ e \\ d \end{matrix} + \begin{matrix}a & d & ⋯ \\ b \\ c \\ e \end{matrix}

which is

( \begin{matrix}e & b & a & ⋯ \\ c & d \end{matrix} ∧ \begin{matrix}e & c & a & ⋯ \\ b & d \end{matrix} ) + ( \begin{matrix}a & e & b & ⋯ \\ c & d \end{matrix} ∧ \begin{matrix}a & c & b & ⋯ \\ e & d \end{matrix} )

+ ( \begin{matrix}a & b & c & ⋯ \\ e & d \end{matrix} ∧ \begin{matrix}a & e & c & ⋯ \\ b & d \end{matrix} ) + ( \begin{matrix}a & b & d & ⋯ \\ c & e \end{matrix} ∧ \begin{matrix}a & c & d & ⋯ \\ b & e \end{matrix} )

The first term in this sum is

( \begin{matrix}a & b & e \\ c & d \end{matrix} + \begin{matrix}e & b & c \\ a & d \end{matrix} ) ∧ ( \begin{matrix}a & c & e \\ b & d \end{matrix} + \begin{matrix}e & c & b \\ a & d \end{matrix} ) = ( \begin{matrix}a & b & e \\ c & d \end{matrix} ∧ \begin{matrix}a & c & e \\ b & d \end{matrix} )

+ \begin{matrix}e & b & c \\ a & d \end{matrix} ∧ ( \begin{matrix}a & e & c \\ b & d \end{matrix} + \begin{matrix}a & c & d \\ b & e \end{matrix} ) + ( \begin{matrix}a & e & b \\ c & d \end{matrix} + \begin{matrix}a & b & d \\ c & e \end{matrix} ) ∧ \begin{matrix}e & c & b \\ a & d \end{matrix}

so that after adding the other three terms we obtain

( \begin{matrix}a & b & e \\ c & d \end{matrix} ∧ \begin{matrix}e & b & c \\ a & d \end{matrix} ) + ( \begin{matrix}e & b & c \\ a & d \end{matrix} ∧ \begin{matrix}a & c & d \\ b & e \end{matrix} ) + ( \begin{matrix}a & b & d \\ c & e \end{matrix} ∧ \begin{matrix}e & c & b \\ a & d \end{matrix} )

+ ( \begin{matrix}e & b & c \\ a & d \end{matrix} ∧ \begin{matrix}e & c & b \\ a & d \end{matrix} ) + ( \begin{matrix}a & b & d \\ c & e \end{matrix} ∧ \begin{matrix}a & c & d \\ b & e \end{matrix} )

= ( \begin{matrix}a & b & e \\ c & d \end{matrix} ∧ \begin{matrix}a & c & e \\ b & d \end{matrix} ) + ( \begin{matrix}e & b & c \\ a & d \end{matrix} + \begin{matrix}a & b & d \\ c & e \end{matrix} ) ∧ ( \begin{matrix}a & c & d \\ b & e \end{matrix} + \begin{matrix}e & c & b \\ a & d \end{matrix} )

= ( \begin{matrix}a & b & e \\ c & d \end{matrix} ∧ \begin{matrix}a & c & e \\ b & d \end{matrix} ) + ( \begin{matrix}e & c & b \\ a & d \end{matrix} + \begin{matrix}e & b & d \\ a & c \end{matrix} + \begin{matrix}b & a & d \\ c & e \end{matrix}

+ \begin{matrix}a & c & d \\ b & e \end{matrix} ) ∧ ( \begin{matrix}a & c & d \\ b & e \end{matrix} + \begin{matrix}e & c & b \\ a & d \end{matrix} )

= \begin{matrix}a & b & e \\ c & d \end{matrix} ∧ \begin{matrix}a & c & e \\ b & d \end{matrix}

as required.

The injectivity of ϕ follows from the irreducibility of S^{(n−3,1^{3})} as a ZΣ_{n}-lattice.

Since B_{4,2}B_{4,3} = 0 (Cohen [3, Lemma 3.8]), the image of the polytabloid

\begin{matrix}1 & ⋯ \\ 2 \\ 3 \\ 4 \end{matrix}

under ϕ is in the kernel of Λ^{2}H^{1}(K_{n}, Z) → H^{2}(K_{n}, Z). Since this polytabloid generates S^{(n−3,1^{3})} as a ZΣ_{n}-lattice it follows that the image of ϕ is in the kernel of this map. By Cohen [3, Lemma 3.8], all relations follow from B_{4,2}B_{4,3} = 0 together with equivariance, and so the image is equal to the kernel. □

Theorem 2.3 There is an exact sequence

0 → ((S^{(2,2)} ⊗ S^{(1^{4})}) ˆ⊗S^{(n−4)})_{Σ_{4}×Σ_{n−4}} ↑^{Σ_{n}}

→^{ξ} S^{(n−2,2)} ⊗ S^{(n−3,1^{3})} →^{ψ} Λ^{3}H^{1}(K_{n}, Z) → H^{3}(K_{n}, Z) → 0

(Here, ˆ⊗ denotes the outer tensor product of modules, as a module for the direct product of the groups)

Proof It follows from Cohen [3, Section 3], that the generators for H^{∗}(K_{n}, Z) are all in degree one, while the relators are all in degree two. This gives us the sequence

S^{(n−2,2)} ⊗ S^{(n−3,1^{3})} →^{ψ} Λ^{3}H^{1}(K_{n}, Z) → H^{3}(K_{n}, Z) → 0

and it remains to identify the kernel of the map ψ.

We define the map ξ by

(( \begin{matrix}a & b \\ c & d \end{matrix} ⊗ \begin{matrix}a \\ b \\ c \\ d \end{matrix} ) ˆ⊗ ⋯) ⊗ 1_{G} ↦ \begin{matrix}a & b & ⋯ \\ c & d \end{matrix} ⊗ \begin{matrix}a & ⋯ \\ b \\ c \\ d \end{matrix}

This clearly gives a well defined map. It is also clear that Ker(ψ) is generated by elements of the form

\begin{matrix}a & b & ⋯ \\ c & d \end{matrix} ⊗ \begin{matrix}a & ⋯ \\ b \\ c \\ d \end{matrix} ↦^{ψ} \begin{matrix}a & b & ⋯ \\ c & d \end{matrix} ∧ \begin{matrix}a & b & ⋯ \\ c & d \end{matrix} ∧ \begin{matrix}a & b & ⋯ \\ c & d \end{matrix} = 0

It remains to show that ξ is injective. We do this by comparing Z-ranks.

Z − rank H^{3}(K_{n}, Z) = 2.3.4 + 2.3.5 + ⋯ + (n − 4)(n − 3)(n − 2)

(Cohen [3, Section 3])

= n(n − 3)(n − 4)(n − 5)(n^{2} − n + 2)/48

Z − rank Λ^{3}H^{1}(K_{n}, Z) = \frac{n(n − 3)}{2} (\frac{n(n − 3)}{2} − 1) (\frac{n(n − 3)}{2} − 2) /6
= (n + 1)n(n − 3)(n − 4)(n^{2} − 3n − 2)/48

Z − rank(S^{(n−2,2)} ⊗ S^{(n−3,1^{3})}) = \frac{n(n − 3)}{2} (\frac{(n − 1)(n − 2)(n − 3)}{6})
= n(n − 1)(n − 2)(n − 3)^{2}/12

Z − rank((S^{(2,2)} ⊗ S^{(1^{4})}) ˆ⊗S^{(n−4)})_{Σ_{4}×Σ_{n−4}} ↑^{Σ_{n}}= 2.1.1.\binom{n}{4}

= n(n − 1)(n − 2)(n − 3)/12.

It is easily checked that n(n − 1)(n − 2)(n − 3)/12 + (n + 1)n(n − 3)(n − 4)(n^{2} − 3n − 2)/48 = n(n − 1)(n − 2)(n − 3)^{2}/12 + n(n − 3)(n − 4)(n − 5)(n^{2} − n + 2)/48. □

We are particularly interested in the case n = 6. In this case, H^{j}(K_{6}, Z) = 0 for j > 3, and so we have enough information to study all H^{j} in this case. The first step is to calculate the composition factors of H^{j}(K_{6}, C) as CΣ_{6}-modules. By Brauer character theory, this gives us information about the composition factors over any field of any characteristic, since we know the decomposition matrices for Σ_{6} for all primes (only 2, 3 and 5 are relevant, of course).

Theorem 2.4 The composition factors of H^{j}(K_{6}, C) as CΣ_{n}-modules are as follows. (i) H^{1}(K_{6}, C) = CS^{(4,2)} (ii) H^{2}(K_{6}, C) = CS^{(4,1^{2})} ⊕ CS^{(3,2,1)} (iii) H^{3}(K_{6}, C) = CS^{(4,1^{2})} ⊕ CS^{(2^{2},1^{2})}

Proof (i) follows immediately from Theorem 2.1.

(ii) follows from Theorem 2.2, since

Λ^{2}H^{1}(K_{6}, C) = Λ^{2}(CS^{(4,2)}) = CS^{(4,1^{2})} ⊕ CS^{(3,2,1)} ⊕ CS^{(3,1^{3})}

by ordinary character theory.

(iii) follows from Theorem 2.3 in the same way. □

3 Modules for A_{6} and Σ_{6} in characteristic two

Throughout this section and Sections 4 and 5, the field k is F_{2}. In this section, we collect some facts about the representation theory of A_{6} and Σ_{6} in characteristic two. This is in preparation for an investigation of the structure of the H^{j}(K_{6}, k) as kA_{6}-modules in Section 4, and the construction of injective resolutions in Section 5. The 2-modular decomposition matrix for Σ_{6} may be found in James [5, p. 137]. For the convenience of the reader, we reproduce it here.

dim  1  4  4  16
D^{(6)} D^{(5,1)} D^{(4,2)} D^{(3,2,1)}
1 S^{(6)}  1  0  0  0
5 S^{(5,1)}  1  1  0  0
9 S^{(4,2)}  1  1  1  0
16 S^{(3,2,1)} 0  0  0  1
10 S^{(4,1^{2})}  2  1  1  0
5 S^{(3^{2})}  1  0  1  0
10 S^{(3,1^{3})}  2  1  1  0
5 S^{(2^{3})}  1  0  1  0
9 S^{(2^{2},1^{2})} 1  1  1  0
5 S^{(2,1^{4})}  1  0  0  0
1 S^{(1^{6})}  1  0  0  0

The module D^{(3,2,1)} is in a block of defect zero, which means that it is both projective and irreducible. Therefore it makes no contribution to cohomological calculations. On restriction to A_{6}, it remains irreducible, but not absolutely irreducible.

The remaining three 2-modular irreducibles remain (absolutely) irreducible on restriction to A_{6}. We shall use the notation k (i.e., the field k = F_{2} with trivial group action) for D^{(6)}, M_{1} for D^{(5,1)}, M_{2} for D^{(4,2)} and D for D^{(3,2,1)}. In the notation of Benson and Carlson [2] (see also the discussion at the end of Section 1), the projective covers P_{k}, P_{M_{1}} and P_{M_{2}} of k, M_{1} and M_{2} as kA_{6}-modules are representations of the following diagrams.

\begin{matrix}k & M_{1} & M_{2} \\ / \ & | & | \\ M_{1} M_{2} & k & k \\ | | & | & | \\ k k & M_{2} & M_{1} \\ | | & | & | \\ M_{2} M_{1} & k & k \\ | | & | & | \\ k k & M_{1} & M_{2} \\ | | & | & | \\ M_{1} M_{2} & k & k \\ | | & | & | \\ k k & M_{2} & M_{1} \\ | | & | & | \\ M_{2} M_{1} & k & k \\ \ / & | & | \\ k & M_{1} & M_{2} \\ P_{k} & P_{M_{1}} & P_{M_{2}} \end{matrix}

Remarks :

1. There is an outer automorphism of A_{6} (which extends to an outer automorphism of Σ_{6}), interchanging the modules M_{1} and M_{2}.

2. It is not hard to show that uniserial kA_{6}-module diagrams which have representations have unique representations. If the diagram has M_{1} or M_{2} as its socle or head, this follows from the uniseriality of P_{M_{1}} and P_{M_{2}}. Otherwise it follows from the existence of sufficiently many endomorphisms of quotients of P_{k}. For example, to show that

k
|
M_{1}
|
k
|
M_{2}
|
k

has a unique representation, it suffices to show that the two quotients of

k
/ \
M_{1} M_{2}
| |
k k
|
M_{2}
|
k

with this diagram are isomorphic. They are, because there is an endomorphism taking one to the other.

Since all irreducible modules in the principal block are absolutely indecomposable, it follows by Benson and Carlson [2, Proposition 6.1] that string diagrams which have representations have unique representations.

Lemma 3.1 As kA_{6}-modules, we have

(i) Λ^{2}(M_{1}) = \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \end{matrix} (ii) Λ^{2}(M_{2}) = \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \end{matrix}

(iii) Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) = \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} (iv) Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} ) = \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix}

Proof (i) By character theory, Λ^{2}(M_{1}) has composition factors k, k and M_{2}. It is self-dual since M_{1} is, and has at most a one-dimensional fixed space since an irreducible module cannot support two linearly independent symplectic forms, by Schur’s lemma. Since Ext_{kA_{6}}^{1}(k, k) = 0, the given diagram is the only possible structure.

(ii) This is the same argument as in (i).

(iii) Since Λ^{2}(k) = 0, there is a short exact sequence

0 → Λ^{2}(M_{1}) → Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) → M_{1} → 0.

If we can show that there is no non-zero homomorphism

Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) → k ,

the result will follow. But such a homomorphism corresponds to an invariant symplectic

form on \begin{matrix}k \\ | \\ M_{1} \end{matrix} . Such a form cannot be non-singular since the module is not self-dual. Thus

at least M_{1} is in the radical of the form. But the k on top certainly doesn’t support a non-singular symplectic form.

(iv) This follows from an argument dual to (iii). □

Lemma 3.2 As a kA_{6}-module, the natural permutation module on six points is isomorphic to Λ^{2}(M_{2}).

Proof This permutation module has composition factors k, k and M_{1}, and unique invariant subspaces of dimension one and codimension one, one contained in the other, and hence has the same diagram as that given in Lemma 3.1 for Λ^{2}(M_{2}). By the remarks above, uniserial diagrams have unique representations, and so the lemma is proved. □

Lemma 3.3 As kA_{6}-modules, we have

(i) S^{2}(M_{1}) = \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} (ii) S^{2}(M_{2}) = \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} (iii) M_{1} ⊗ M_{2} ≅ D

(iv) M_{1} ⊗ M_{1} = \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} (v) M_{2} ⊗ M_{2} = \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} (vi) \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{1} = \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} ≅ Ω^{−1}(M_{1})

(vii) Λ^{2}(M_{2}) ⊗ M_{1} ≅ P_{M_{1}} (viii) Λ^{2}(M_{1}) ⊗ M_{2} ≅ P_{M_{2}}

(ix) \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \end{matrix} ≅ k ⊕ P_{M_{1}} (x) \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ \begin{matrix}M_{2} \\ | \\ k \end{matrix} ≅ k ⊕ P_{M_{2}}

(xi) \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \end{matrix} ≅ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ D

Proof (i) Since A_{6} does not fix an orthogonal form on M_{1} (the order of A_{6} is greater than the order of O_{4}^{+}(2) or O_{4}^{−}(2)!), the “symmetric square” S^{2}(M_{1}) = ⊗^{2}(M_{1})/Λ^{2}(M_{1}) does not have an invariant one-dimensional subspace. Using the short exact sequence

0 → F^{2}(M_{1}) → S^{2}(M_{1}) → Λ^{2}(M_{1}) → 0

(where F^{2}(M_{1}) denotes the Frobenius square of M_{1}), we see that (i) holds.

(ii) This is the same argument as (i) with M_{1} and M_{2} interchanged.

(iii) This follows just by computing characters.

(iv) For any module A in characteristic two we always have a filtration

⊗^{2}(A) = \begin{matrix}Λ^{2}(A) \\ F^{2}(A) \\ Λ^{2}(A) \end{matrix}

The result now follows from the fact that M_{1} and M_{2} are self-dual.

(v) This is the same argument as in (iv), with M_{1} and M_{2} interchanged.

(vi) There is a short exact sequence

0 → M_{1} ⊗ M_{1} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{1} → M_{1} → 0.

Thus the result follows from the fact that

dim_{k} Hom_{kA_{6}}( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{1}, k) = dim_{k} Hom_{kA_{6}}( \begin{matrix}k \\ | \\ M_{1} \end{matrix} , M_{1}) = 0.

The isomorphism with Ω^{−1}(M_{1}) follows from the diagram for P_{M_{1}}.

(vii) There is a short exact sequence

0 → M_{1} → Λ^{2}(M_{2}) ⊗ M_{1} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{1} → 0.

Thus the result follows from the fact that

dim_{k} Hom_{kA_{6}}(k, Λ^{2}(M_{2}) ⊗ M_{1}) = dim_{k} Hom_{kA_{6}}(M_{1}, Λ^{2}(M_{2})) = 0.

(viii) This is the same argument as in (vi) and (vii), with M_{1} and M_{2} interchanged. (ix) There is a short exact sequence

0 → \begin{matrix}k \\ | \\ M_{1} \end{matrix} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \end{matrix} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{1} → 0.

Now by (vi),

dim_{k} Ext_{kA_{6}}^{1}( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{1}, \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) = dim_{k} Ext_{kA_{6}}^{1}(Ω^{−1}(M_{1}), \begin{matrix}k \\ | \\ M_{1} \end{matrix} )

= dim_{k} Hom_{kA_{6}}(M_{1}, \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) = 1

and so the only non-split extension is

0 → \begin{matrix}k \\ | \\ M_{1} \end{matrix} → k ⊕ P_{M_{1}} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{1} → 0

(the existence of such a non-split sequence is easily seen by examining the diagrams for these modules). The fact that the extension does not split follows from the fact that k is a direct summand of the tensor product, by Benson and Carlson [1, Theorem 2.1].

(x) Again, this is the same argument with M_{1} and M_{2} interchanged.

(xi) Since D is projective, the sequence

0 → M_{2} ⊗ M_{1} → \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ M_{1} → M_{1} → 0

splits, so that

\begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ M_{1} = M_{1} ⊕ D.

We now examine the short exact sequence

0 → \begin{matrix}k \\ | \\ M_{2} \end{matrix} → \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \end{matrix} → M_{1} ⊕ D → 0.

This sequence does not split, since

dim_{k} Hom_{kA_{6}}(M_{1}, \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \end{matrix} ) = dim_{k} Hom_{kA_{6}}( \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ M_{1}, \begin{matrix}M_{1} \\ | \\ k \end{matrix} )

= dim_{k} Hom_{kA_{6}}(M_{1} ⊕ D, \begin{matrix}M_{1} \\ | \\ k \end{matrix} ) = 0.

Thus the only possibility is the one given. □

We now examine extendability of modules to Σ_{6}. The projective indecomposable kΣ_{6}- modules are representations of the following diagrams.

k  M_{1}  M_{2}
/ | \  |\  /|
M_{1} k M_{2}  k M_{1}  M_{2} k
|\ / \ /|  |\|  |/|
k M_{1} M_{2} k  M_{2} k  k M_{1}
|\| |/|  |\|  |/|
M_{2} k k M_{1}  k M_{2}  M_{1} k
|\| |/|  |\|  |/|
k M_{2} M_{1} k  M_{1} k  k M_{2}
|\| |/|  |\|  |/|  D
M_{1} k k M_{2}  k M_{1}  M_{2} k
|\| |/|  |\|  |/|
k M_{1} M_{2} k  M_{2} k  k M_{1}
|\| |/|  |\|  |/|
M_{2} k k M_{1}  k M_{2}  M_{1} k
\| |/  |\|  |/|
M_{2}kM_{1}  M_{1} k  k M_{2}
\|/  \|  |/
k  M_{1}  M_{2}

It follows that the projective indecomposable modules P_{k}, P_{M_{1}} and P_{M_{2}} for kA_{6} extend to kΣ_{6}-modules. It also follows that representations of uniserial diagrams as kA_{6}-modules extend to representations as kΣ_{6}-modules. The same is also true of string diagrams and of D-homomorphisms between representations of them.

Restriction to A_{5} Since k = F_{2} is not a splitting field for the principal block of A_{5}, we set k′ = F_{4}, and we use the notations M′_{1} = M_{1} ⊗_{k} k′, M′_{2} = M_{2} ⊗_{k} k′. The restrictions of these k′A_{6}-modules to A_{5} are as follows. We have M′_{1} ↓_{A_{5}}= P, the projective irreducible of dimension four, and M′_{2} ↓_{A_{5}}= N_{1} ⊕ N_{2}, where N_{1} and N_{2} are the two-dimensional irreducibles for A_{5}.

The projective indecomposable k′A_{5}-modules are

\begin{matrix}k′ \\ / \ \\ N_{1} N_{2} \\ | | \\ k′ k′ \\ | | \\ N_{2} N_{1} \\ \ / \\ k′ \end{matrix}  \begin{matrix}N_{1} \\ | \\ k′ \\ | \\ N_{2} \\ | \\ k′ \\ | \\ N_{1} \end{matrix}  \begin{matrix}N_{2} \\ | \\ k′ \\ | \\ N_{1} \\ | \\ k′ \\ | \\ N_{2} \end{matrix}  P

Lemma 3.4 (i) ( \begin{matrix}k′ \\ | \\ M′_{2} \end{matrix} ) ↓_{A_{5}}= \begin{matrix}k′ \\ / \ \\ N_{1} N_{2} \end{matrix}  ( \begin{matrix}M′_{2} \\ | \\ k′ \end{matrix} ) ↓_{A_{5}}= \begin{matrix}N_{1} N_{2} \\ \ / \\ k′ \end{matrix}

(ii) ( \begin{matrix}k′ \\ | \\ M′_{2} \\ | \\ k′ \end{matrix} ) ↓_{A_{5}}= \begin{matrix}k′ \\ / \ \\ N_{1} N_{2} \\ \ / \\ k′ \end{matrix} (iii) ( \begin{matrix}M′_{1} \\ | \\ k′ \\ | \\ M′_{2} \\ | \\ k′ \end{matrix} ) ↓_{A_{5}}= \begin{matrix}k′ \\ / \ \\ N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ⊕ P

(iv) ( \begin{matrix}k′ \\ | \\ M′_{1} \\ | \\ k′ \\ | \\ M′_{2} \end{matrix} ) ↓_{A_{5}}= \begin{matrix}k′ \\ | \\ N_{1} \end{matrix} ⊕ \begin{matrix}k′ \\ | \\ N_{2} \end{matrix} ⊕ P

(v) Λ^{2} ( \begin{matrix}N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ) = \begin{matrix}k′ \\ | \\ N_{1} \end{matrix} ⊕ \begin{matrix}k′ \\ | \\ N_{2} \end{matrix} ⊕ P

Proof This is an easy exercise in the use of the Frobenius reciprocity law

Hom_{kG}(A↑^{G}, B) = Hom_{kH}(A, B ↓_{H}). □

4 Diagrams for H^{j}(K_{6}, F_{2}) as F_{2}A_{6}-modules

In this section, we use Theorem 2.1 and the exact sequences described in Section 3, to describe the modules H^{j}(K_{6}, k) for j = 1, 2, 3 (the field k is taken to be F_{2} throughout this setion). Recall also the H^{j}(K_{6}, k) = 0 for j > 3.

By Theorem 2.1, the module H^{1}(K_{6}, k) (k = F_{2} throughout this section) is the Specht module kS^{(4,2)}. By the decomposition matrix for Σ_{6} (Section 3), this has composition factors k, M_{1} and M_{2}. By the general theory of Specht modules, it has M_{2} = D^{(4,2)} as its unique top composition factor, as a kΣ_{6}-module. Thus by Frobenius reciprocity

dim_{k} Hom_{kA_{6}}(kS^{(4,2)}, M_{1}) = dim_{k} Hom_{kΣ_{6}}(kS^{(4,2)}, \begin{matrix}M_{1} \\ | \\ M_{1} \end{matrix} ) = 0

and similarly

dim_{k} Hom_{kA_{6}}(kS^{(4,2)}, k) = dim_{k} Hom_{kΣ_{6}}(kS^{(4,2)}, \begin{matrix}k \\ | \\ k \end{matrix} ) = 0

so that as a kA_{6}-module it still has M_{2} as unique top composition factor. It is therefore a quotient of P_{M_{2}}, and so it consists of just the top three composition factors of P_{M_{2}}. So we have proved the following.

Theorem 4.1 As kA_{6}-modules, we have

H^{1}(K_{6}, k) ≅ kS^{(4,2)} = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} . □

Next, we examine the module H^{2}(K_{6}, k), by means of Theorem 2.2. So we first need to examine Λ^{2}H^{1}(K_{6}, k).

Lemma 4.2 As kA_{6}-modules, we have

Λ^{2}H^{1}(K_{6}, k) = D ⊕ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} ⊕ \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} .

Proof In general, if we have a module of the form \begin{matrix}A \\ | \\ B \\ | \\ C \end{matrix} then the exterior square has a Mapping Class Groups 51 filtration

Λ^{2}(A)
A ⊗ B
A ⊗ C ⊕ Λ^{2}(B)
B ⊗ C
Λ^{2}(C)

In our case, M_{1} ⊗ M_{2} ≅ D is projective (= injective), and Λ^{2}(k) = 0, and so

Λ^{2}H^{1}(K_{6}, k) = D ⊕ \frac{Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} )}{Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} )}

Now the structures of Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} ) and Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) are given in Lemma 3.1, and we have

dim_{k} Ext_{kA_{6}}^{1}(Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} ) , Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ))

= dim_{k} \underline{Hom}_{kA_{6}}(Ω (Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} )) , Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ))

= dim_{k} \underline{Hom}_{kA_{6}}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k k \\ | | \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \\ \ / \\ k \end{matrix} , \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} ) = 1.

Here, \underline{Hom} denotes homomorphisms modulo those which factor through a projective module. Note that a homomorphism from Ω (Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} )) to Λ^{2} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) factors through

the projective cover P_{k} of Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} ). Thus the D-homomorphism with image \begin{matrix}M_{2} \\ | \\ k \end{matrix} factors through a projective, while the D-homomorphism with image k does not. It follows

that the only two possible extensions are the split one and \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} .

We can distinguish between the two possibilities by restriction to A_{5} (after tensoring with k′ = F_{4}).

Λ^{2} ( \begin{matrix}M′_{2} \\ | \\ k′ \\ | \\ M′_{1} \end{matrix} ) ↓_{A_{5}} = Λ^{2} ( \begin{matrix}M′_{2} \\ | \\ k′ \\ | \\ M′_{1} \end{matrix} ↓_{A_{5}} )
= Λ^{2} ( \begin{matrix}N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ⊕ P )
= \begin{matrix}k′ \\ | \\ N_{1} \end{matrix} ⊕ \begin{matrix}k′ \\ | \\ N_{2} \end{matrix} ⊕ \begin{matrix}k′ \\ / \ \\ N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ⊕ (projective)

so that ( \begin{matrix}M′_{1} \\ | \\ k′ \\ | \\ M′_{2} \end{matrix} ) ↓_{A_{5}}= \begin{matrix}k′ \\ / \ \\ N_{1} N_{2} \end{matrix} ⊕ P is not a summand. In fact the observed summands are correctly accounted for by

( \begin{matrix}M′_{1} \\ | \\ k′ \\ | \\ M′_{2} \\ | \\ k′ \end{matrix} ) ↓_{A_{5}}= \begin{matrix}k′ \\ / \ \\ N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ⊕ P

and

( \begin{matrix}k′ \\ | \\ M′_{1} \\ | \\ k′ \\ | \\ M′_{2} \end{matrix} ) ↓_{A_{5}}= \begin{matrix}k′ \\ | \\ N_{1} \end{matrix} ⊕ \begin{matrix}k′ \\ | \\ N_{2} \end{matrix} ⊕ P. □

Theorem 4.3 As kA_{6}-modules, we have

(i) kS^{(3,1^{3})} = \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} (ii) H^{2}(K_{6}, k) = \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ D

Proof By Theorem 2.2 with n = 6, we have a short exact sequence

0 → kS^{(3,1^{3})} → Λ^{2}H^{1}(K_{6}, k) → H^{2}(K_{6}, k) → 0.

Now from the decomposition matrix of Σ_{6}, we can see that kS^{(3,1^{3})} has composition factors k, k, M_{1} and M_{2}. By Lemma 4.2, the only submodules of Λ^{2}H^{1}(K_{6}, k) with these composition factors are isomorphic to

k ⊕ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} and \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} .

The first of these is impossible since this would imply the existence of complementary invariant subspaces of dimension one and codimension one under the action of A_{6}, and hence under the action of Σ_{6}. In particular kS^{(3,1^{3})} would be decomposable as a kΣ_{6}- module, contradicting James [5, Corollary 13.18]. Thus

kS^{(3,1^{3})} = \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} .

There are two submodules of Λ^{2}H^{1}(K_{6}, k) isomorphic to kS^{(3,1^{3})}, but the quotients are also isomorphic, and are as in the statement of the theorem. □

Having completed the calculation of H^{2}(K_{6}, k), we now turn to the structure of the module H^{3}(K_{6}, k). According to Theorem 2.3, we should begin by examining the modules ((kS^{(2,2)} ⊗ kS^{(1^{4})}) ˆ⊗kS^{(2)})_{Σ_{4}×Σ_{2}} ↑^{Σ_{6}} and kS^{(4,2)} ⊗ kS^{(3,1^{3})}.

Lemma 4.4 As kA_{6}-modules, we have

((kS^{(2,2)} ⊗ kS^{(1^{4})}) ˆ⊗kS^{(2)})_{Σ_{4}×Σ_{2}} ↑^{Σ_{6}}↓_{A_{6}}= \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ D

Proof By the Mackey decomposition theorem, this module is the two dimensional irreducible module R, say, for (Σ_{4} × Σ_{2}) ∩ A_{6} (≅ Σ_{4}) induced up to A_{6}. By character theory, the composition factors are as given. By Frobenius reciprocity, we have

dim_{k} Hom_{kA_{6}}(k, R ↑^{A_{6}}) = dim_{k} Hom_{k((Σ_{4}×Σ_{2})∩A_{6})}(k, R) = 0

and similarly dim_{k} Hom_{kA_{6}}(R ↑^{A_{6}}, k) = 0. The module is also self-dual, and by the diagrams for P_{M_{1}} and P_{M_{2}}, we see that the only possibility is the one given. □

Proposition 4.5 As kA_{6}-modules, we have

kS^{(4,2)} ⊗ kS^{(3,1^{3})} = D ⊕ D ⊕ P_{M_{1}} ⊕ P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix}

Proof By Theorem 4.1 and Theorem 4.3 (i) we have

kS^{(4,2)} ⊗ kS^{(3,1^{3})} = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} .

We shall creep up on this module in stages. Stage 1 : Since M_{1} ⊗ M_{2} = D is injective ( = projective) by Lemma 3.3 (iii), the sequence

0 → M_{1} ⊗ M_{2} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{2} → M_{2} → 0

splits, and we have \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{2} = M_{2} ⊕ D.

Stage 2 : Using the short exact sequence

0 → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ M_{2} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \end{matrix} → 0

and Lemma 3.3 (ix), we see that we have a short exact sequence

0 → M_{2} ⊕ D → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} → k ⊕ P_{M_{1}} → 0.

Since

dim_{k} Hom_{kA_{6}}(k, \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = dim_{k} Hom_{kA_{6}}( \begin{matrix}M_{1} \\ | \\ k \end{matrix} , \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = 0

this sequence is non-split. It is easy to see that there is only one non-split extension 56 D. J. Benson & F. R. Cohen possible, and so we have

\begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} = \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊕ D ⊕ P_{M_{1}}.

Stage 3 : By the result of Stage 2, the short exact sequence

0 → \begin{matrix}k \\ | \\ M_{1} \end{matrix} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} → 0

can be written

0 → \begin{matrix}k \\ | \\ M_{1} \end{matrix} → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} → \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊕ D ⊕ P_{M_{1}} → 0.

Since by Lemma 3.3 (xi)

dim_{k} Hom_{kA_{6}}( \begin{matrix}k \\ | \\ M_{2} \end{matrix} , \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} ) = dim_{k} Hom_{kA_{6}}( \begin{matrix}k \\ | \\ M_{2} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \end{matrix} , \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} )

= dim_{k} Hom_{kA_{6}}( \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ D, \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} ) = 0

the sequence does not split. So the only possibility is

\begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} = \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ D ⊕ P_{M_{1}} .

Stage 4 : Since M_{1} ⊗ M_{2} = D is projective by Lemma 3.3 (iii), the sequence

0 → M_{2} ⊗ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \end{matrix} → M_{2} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} → M_{2} ⊗ M_{1} → 0

splits. By Lemma 3.1 (i) and Lemma 3.3 (viii), it follows that

M_{2} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} = P_{M_{2}} ⊕ D .

Stage 5 : We now put together Stages 3 and 4. Since M_{2} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} is projective, the

sequence

0 → \begin{matrix}k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} → \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} → M_{2} ⊗ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \end{matrix} → 0

splits, and the proposition follows. □

Proposition 4.6 As kA_{6}-modules, we have

Λ^{3}H^{1}(K_{6}, k) = D ⊕ P_{M_{1}} ⊕ P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix}

Proof By Theorem 4.1, kS^{(4,2)} = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} . In general, if we have a module of the form \begin{matrix}A \\ | \\ B \\ | \\ C \end{matrix} then the exterior cube has a filtration

Λ^{3}(A)
Λ^{2}(A) ⊗ B
A ⊗ Λ^{2}(B) ⊕ Λ^{2}(A) ⊗ C
A ⊗ B ⊗ C ⊕ Λ^{3}(B)
Λ^{2}(B) ⊗ C ⊕ A ⊗ Λ^{2}(C)
B ⊗ Λ^{2}(C)
Λ^{3}(C)

In our case some of these terms are zero and this reduces to

\begin{matrix}Λ^{3} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} ) \\ Λ^{2}(M_{2}) ⊗ M_{1} \\ M_{2} ⊗ k ⊗ M_{1} \\ M_{2} ⊗ Λ^{2}(M_{1}) \\ Λ^{3} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) \end{matrix} .

Since dim ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} ) = 5 we have

Λ^{3} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} ) = Λ^{2} ( \begin{matrix}M_{2} \\ | \\ k \end{matrix} )^{∗} = Λ^{2} ( \begin{matrix}k \\ | \\ M_{2} \end{matrix} ) = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix}

by Lemma 3.1 (iii). Similarly by Lemma 3.1 (iv),

Λ^{3} ( \begin{matrix}k \\ | \\ M_{1} \end{matrix} ) = \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} .

By Lemma 3.3 (iii), (vi) and (vii), the middle three layers are projective = injective, and so

Λ^{3} ( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = D ⊕ P_{M_{1}} ⊕ P_{M_{2}} ⊕ X

where X is some extension

0 → \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} → X → \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} → 0.

Now

dim_{k} Ext_{kA_{6}}^{1}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} , \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = dim_{k} \underline{Hom}_{kA_{6}}(Ω ( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} ) , \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} )

= dim_{k} \underline{Hom}_{kA_{6}}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} , \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = 1

and so the only non-split extension possible is

0 → \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} → k ⊕ \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} → \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} → 0.

We can again distinguish between these two possibilities by restriction to A_{5}.

Λ^{3} ( \begin{matrix}M′_{2} \\ | \\ k′ \\ | \\ M′_{1} \end{matrix} ) ↓_{A_{5}} = Λ^{3} ( \begin{matrix}M′_{2} \\ | \\ k′ \\ | \\ M′_{1} \end{matrix} ↓_{A_{5}} )

= Λ^{3} ( \begin{matrix}N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ⊕ P )

= Λ^{3} ( \begin{matrix}N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ) ⊕ Λ^{2} ( \begin{matrix}N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ) ⊗ P

⊕ \begin{matrix}N_{1} N_{2} \\ \ / \\ k′ \end{matrix} ⊗ Λ^{2}(P) ⊕ Λ^{3}(P)

= \begin{matrix}N_{1} \\ | \\ k′ \end{matrix} ⊕ \begin{matrix}N_{2} \\ | \\ k′ \end{matrix} ⊕ X′

(by Lemma 3.4 (v)) where, by Benson and Carlson [1, Proposition 2.2], X′ is a sum of indecomposable modules of even dimension.

Thus the restriction to A_{5} does not have k as a direct summand, and hence the extension splits, proving the proposition. □

Theorem 4.7 As kA_{6}-modules, we have H^{3}(K_{6}, k) ≅ P_{M_{2}}.

Proof By Theorem 2.3 with n = 6, Lemma 4.4, Proposition 4.5 and Proposition 4.6, we have an exact sequence as follows.

0 → \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ D →^{ξ} D ⊕ D ⊕ P_{M_{1}} ⊕ P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} →^{ψ} D ⊕ P_{M_{1}} ⊕ P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix}

→ H^{3}(K_{6}, k) → 0

Since D is projective and injective, we may remove all copies of it from this sequence to obtain

0 → \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} →^{ξ′} P_{M_{1}} ⊕ P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} →^{ψ′} P_{M_{1}} ⊕ P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} → H^{3}(K_{6}, k) → 0 .

Now write

ξ′ = ( \begin{matrix}a \\ b \\ c \end{matrix} ) ,

where a, b and c are maps into the first, second and third summand of the second module respectively. Since ξ′ is injective, b is injective. Let α be the homomorphism from P_{M_{2}} to P_{M_{1}} with a = αb. Then we may change our basis for the second module by conjugating by the automorphism

( \begin{matrix}1 & α & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) ,

and ξ′ will have matrix

( \begin{matrix}1 & α & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}a \\ b \\ c \end{matrix} ) = ( \begin{matrix}0 \\ b \\ c \end{matrix} )

with respect to the new basis. Thus without loss of generality ψ′ is injective on the summand P_{M_{1}}, and so we may remove this module from the second and third terms to obtain

0 → \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} →^{ξ′′} P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} →^{ψ′′} P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} → H^{3}(K_{6}, k) → 0 .

Next, we claim that the second component of the map ξ′′ is zero. Certainly the first component is injective as before, so that if the second component were the unique non-zero map, then we would have

coker ξ′′ = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ k

which is impossible since there is no embedding of this into the third module in the sequence.

So the given sequence is the Yoneda splice of the following two sequences.

0 → \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} →^{ξ′′} P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} → \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} → 0

0 → \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} →^{i} P_{M_{2}} ⊕ \begin{matrix}k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ⊕ \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} → H^{3}(K_{6}, k) → 0

We now observe that the latter of these sequences splits. Write

i = ( \begin{matrix}a \\ b \end{matrix} )

where a and b are maps to the first and the sum of the second and third components respectively. Since i is injective, b is an isomorphism, and so we may change our basis for the second module by conjugating by the automorphism

( \begin{matrix}1 & ab^{−1} \\ 0 & b^{−1} \end{matrix} ) .

With respect to the new basis, i will have matrix

( \begin{matrix}1 & ab^{−1} \\ 0 & b^{−1} \end{matrix} ) ( \begin{matrix}a \\ b \end{matrix} ) = ( \begin{matrix}0 \\ 1 \end{matrix} )

and so the sequence splits. This completes the proof of the theorem. □

5 Calculation of H^{∗}(Σ_{6}, H^{∗}(K_{6}, F_{2})) ⇒ H^{∗}(Γ_{0,0}^{6}, F_{2})

In this section, we calculate the E_{2} term of the Lyndon-Hochschild-Serre spectral sequence

H^{∗}(Σ_{6}, H^{∗}(K_{6}, k)) ⇒ H^{∗}(Γ_{0,0}^{6}, k).

Throughout this section, k denotes the field F_{2}. We proceed in two stages. The first stage is to use the diagrammatic methods of Benson and Carlson [2] (see also the discussion at the end of Section 1) to calculate the minimal injective resolutions of the kA_{6}-modules H^{j}(K_{6}, k) for 0 ≤ j ≤ 3, and hence calculate H^{∗}(A_{6}, H^{∗}(K_{6}, k)). The second stage then consists of going up from A_{6} to Σ_{6} using an appropriate double complex.

Recall from Section 4 that we have

H^{0}(K_{6}, k) ≅ k, H^{1}(K_{6}, k) = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} , H^{2}(K_{6}, k) = \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ D, H^{3}(K_{6}, k) ≅ P_{M_{2}}

and H^{j}(K_{6}, k) = 0 for j > 3.

From the diagrams given in Section 3 for the projective = injective indecomposable kA_{6}-modules, it is easy to compute injective resolutions for these modules, and hence compute the cohomology H^{∗}(A_{6}, H^{j}(K_{6}, k)) as a module over H^{∗}(A_{6}, H^{0}(K_{6}, k)) = H^{∗}(A_{6}, k) for j ≤ 3. Note that because we are using minimal injective resolutions, every cochain is a cocycle, and every coboundary is zero, so that

H^{i}(A_{6}, H^{j}(K_{6}, k)) ≅ Hom_{kA_{6}}(k, Ω^{−i}H^{j}(K_{6}, k))

so that cohomology elements appear as copies of k at the bottom of the diagram for Ω^{−i}H^{j}(K_{6}, k). We label such a copy of k by a greek letter in a box, so that this greek letter is the name of a cohomology element, well defined up to non-zero scalar multiplication. Since we are working with F_{2} coefficients, there is no ambiguity in this method of naming elements of cohomology.

We first form an injective resolution of H^{0}(K_{6}, k) ≅ k as a kA_{6}-module. The results are as follows.

Ω^{−1}(k) = \begin{matrix}k \\ / \ \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \\ | | \\ k k \\ | | \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \end{matrix} Ω^{−2}(k) = \begin{matrix}M_{2} M_{1} \\ \ / \\ k \end{matrix} ; α

Ω^{−3}(k) = \begin{matrix}k \\ / \ \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \\ | | \\ k k \\ | | \\ M_{1} M_{2} \\ | | \\ k k \\ β & γ \end{matrix}  Ω^{−4}(k) = \begin{matrix}k  k \\ / \  / \ \\ M_{1} M_{2} M_{1} M_{2} \\ | \ / | \\ k  k  k \\ | α^{2}  | \\ M_{2}  M_{1} \\ |  | \\ k  k \\ |  | \\ M_{1}  M_{2} \\ |  | \\ k  k \\ |  | \\ M_{2}  M_{1} \end{matrix}

Ω^{−5}(k) = \begin{matrix}k \\ / \ \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \\ | | \\ k k \\ | | \\ M_{2} M_{1} M_{2} M_{1} \\ \ /  \ / \\ k  k \\ αβ & αγ \end{matrix} Ω^{−6}(k) = \begin{matrix}k  k \\ / \  / \ \\ M_{1} M_{2} M_{1} M_{2} \\ | \ / | \\ k  k  k \\ | α^{3}  | \\ M_{2}  M_{1} \\ |  | \\ k  k \\ |  | \\ M_{1}  M_{2} \\ |  | \\ k  k \\ β^{2} & γ^{2} \end{matrix}  βγ = 0

It is worth making a few remarks at this point about calculation of products of cohomology elements. For the purpose of these calculations, the Yoneda definition of products is the most convenient (and agrees with cup product when both are defined). If f : M → M′ is a map of kG-modules, and P and P′ are the projective covers of M and M′, then Ω(f) is defined by completing the following diagram.

0 → Ω(M) → P → M → 0
↓Ω(f)  ↓  ↓f
0 → Ω(M′) → P′ → M′ → 0

The map obtained this way is well defined modulo maps factoring through a projective, i.e., as an element of \underline{Hom}_{kG}(Ω(M), Ω(M′)). Dually, if we use injective hulls (remember projective and injective modules are the same thing) we obtain a well defined element Ω^{−1}(f) of \underline{Hom}_{kG}(Ω^{−1}(M), Ω^{−1}(M′)). The maps Ω and Ω^{−1} establish an isomorphism

\underline{Hom}_{kG}(M, M′) ≅ \underline{Hom}_{kG}(Ω(M), Ω(M′)).

Now if f : Ω^{−i}(M′) represents a class [f] in Ext_{kG}^{i}(M, M′) ≅ \underline{Hom}_{kG}(M, Ω^{−i}(M′)), and g : M′ → Ω^{−j}(M′′) represents [g] in Ext_{kG}^{j}(M′, M′′) ≅ \underline{Hom}_{kG}(M′, Ω^{−j}(M′′)), then (Ω^{−i}g) ∘ f : M → Ω^{−i−j}(M′′) represents the Yoneda composite [g][f] in Ext_{kG}^{i+j}(M, M′′) ≅ \underline{Hom}_{kG}(M, Ω^{−i−j}(M′′)).

This method of composition is easy to perform diagrammatically (see also Benson and Carlson [2, Section 10]), and we have used this to give the composite elements (for example α^{2}, αβ, αγ) in the above diagrams. Similarly we compute this way for example that βγ is the zero map from k to Ω^{−6}(k), and so we write the relation βγ = 0 in a box beside Ω^{−6}(k).

Clearly the outer automorphism of A_{6} which swaps M_{1} and M_{2}, fixes α and swaps β and γ.

Proposition 5.1 H^{∗}(A_{6}, k) = k[α, β, γ]/(βγ) with deg(α) = 2, deg(β) = deg(γ) = 3.

Proof We use the dual of the method described in Benson and Carlson [2, Sections 9 and 10]. The module L_{α(β+γ)}^{∗} = coker(α(β + γ) : k → Ω^{−5}(k)) decomposes as a direct sum of modules isomorphic to L_{α}^{∗} (≅ M_{1} ⊕ M_{2}) and L_{β+γ}^{∗}. It follows that

V_{A_{6}}(α) ∩ V_{A_{6}}(β + γ) = {0},

and so α generates the periodicity of L_{α+β}^{∗}, which is hence periodic with period two (this is also easily verified directly). It is now easy to check that the conditions of Benson and Carlson [2, Theorem 9.3] hold, so that for all j ≥ 0 we have a short exact sequence

0 → Ω^{−j}(k) → Ω^{−j−3}(k) → Ω^{−j}(L_{β+γ}^{∗}) → 0.

Using Benson and Carlson [2, Theorem 9.5], we obtain the Poincaré series

∑_{j=0}^{∞} dim_{k} H^{j}(A_{6}, k) = \frac{1}{1 − t^{3}} (1 + t^{2} + \frac{t^{3}}{1 − t})

= \frac{(1 − t^{6})}{(1 − t^{2})(1 − t^{3})^{2}} .

and we see that multiplication by β + γ is injective.

It follows that in order to show that the homomorphism θ : k[x, y, z]/(yz) → H^{∗}(A_{6}, k) given by x ↦ α, y ↦ β, z ↦ γ is an isomorphism, we only need to show that it is injective since the two rings have the same Poincaré series. By induction, the fact that multiplication by β + γ is injective, and the fact that dim_{k} H^{j}(A_{6}, k) = dim_{k} H^{j−3}(A_{6}, k) + 1, it suffices to find a non-trivial element of the image of θ in H^{j}(A_{6}, k)/((β + γ).H^{j−3}(A_{6}, k)).

(i) If j = 2t is even then α^{t} is such an element since V_{A_{6}}(α^{t}) ⊉ V_{A_{6}}(β + γ).

(ii) If j = 2t + 1 is odd then α^{t−1}β is such an element. For if α^{t−1}β = λ(β + γ) is in the image of multiplication by β + γ, then by applying the outer automorphism σ which swaps β and γ, we have α^{t−1}γ = λ^{σ}(β + γ). So since multiplication by β + γ is injective, α^{t−1} = λ + λ^{σ}. This equation is impossible to satisfy since by the inductive hypothesis for any value of λ, λ + λ^{σ} must be in the image of multiplication by (β + γ) by (i). □

We next apply the same method to calculate the minimal resolution and cohomology

of the kA_{6}-module H^{1}(K_{6}, k) = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} . Again, we use the Yoneda method to calculate the

products of these elements with elements of H^{∗}(A_{6}, k).

Ω^{−1}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = M_{1} ; | ; k ; | ; M_{2} ; | ; k ; | ; M_{1} ; | ; k ; u Ω^{−2}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = \begin{matrix}k \\ / \ \\ M_{1} M_{2} \\ | | \\ k k \\ | v \\ M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix}

Ω^{−3}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = \begin{matrix}k \\ / \ \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \\ | | \\ k k \\ | | \\ M_{2} M_{1} M_{2} \\ \ / | \\ k  k \\ αu | \\ M_{1} \end{matrix}  Ω^{−4}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = k ; / \ ; M_{1} M_{2} M_{1} ; | \ / ; k  k ; | αv ; M_{2} ; | ; k ; | ; M_{1} ; | ; k ; βu γu = 0

Ω^{−5}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = \begin{matrix}k \\ / \ \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \\ | | \\ k  k k \\ / \ | | \\ M_{1} M_{2} M_{1} M_{2} \\ | \ / | \\ k  k  k \\ \begin{matrix}| \\ M_{2} \\ | \end{matrix} & \begin{matrix}βv = \\ α^{2}u \end{matrix} & γv \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} Ω^{−6}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = \begin{matrix}k  k \\ / \  / \ \\ M_{1} M_{2} M_{1} M_{2} \\ | \ / | \\ k  k  k \\ | α^{2}v  | \\ M_{2}  M_{1} \\ |  | \\ k  k \\ |  | \\ M_{2} M_{1}  M_{2} \\ \ /  | \\ k  k \\ αβu  | \\ M_{1} \end{matrix}

Proposition 5.2 As a module over H^{∗}(A_{6}, k) we have

H^{∗}(A_{6}, H^{1}(K_{6}, k)) = ⟨u, v⟩/(γu, α^{2}u + βv)

Proof Consider the element

ζ ∈ Ext_{kA_{6}}^{3}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} , \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = \underline{Hom}_{kA_{6}}( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} , Ω^{−3} ( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ))

represented by the diagonal of the two D-homomorphisms corresponding to the left and right hand ends of the diagram for

Ω^{−3} ( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) .

The cokernel of this homomorphism is isomorphic to L_{β+γ}^{∗}, and as noted in the proof of Theorem 5.1, α generates the periodicity of this module. It is now easy to see that the conditions of Benson and Carlson [2, Theorem 9.3] hold, so that for all j ≥ 0 we have a short exact sequence

0 → Ω^{−j} ( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) → Ω^{−j−3} ( \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) → Ω^{−j}(L_{β+γ}^{∗}) → 0.

Using Benson and Carlson [2, Theorem 9.5], we obtain the Poincaré series

∑_{j=0}^{∞} t^{j} dim_{k} H^{j}(A_{6}, \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \end{matrix} ) = \frac{1}{(1 − t^{3})} (t + t^{2} + \frac{t^{3}}{1 − t})

= \frac{t}{(1 − t)(1 − t^{3})}

and we see that multiplication by ζ is injective.

It follows that in order to show that the homomorphism θ : ⟨x, y⟩/(γx, α^{2}x + βy) → H^{∗}(A_{6}, H^{1}(K_{6}, k)) given by x ↦ u, y ↦ v is an isomorphism, we only need to show that it is surjective since the two modules have the same Poincaré series.

We calculate that uζ = βu = (β + γ)u and vζ = (β + γ)v, so that on the submodule generated by u and v, multiplication by (β + γ) is injective. Thus we have a short exact sequence of H^{∗}(A_{6}, k)-modules

0 → H^{∗}(A_{6}, H^{1}(K_{6}, k)) ⟶^{(β+γ)∪−} H^{∗+3}(A_{6}, H^{1}(K_{6}, k)) → H^{∗}(A_{6}, L_{β+γ}^{∗}) → 0.

70 D. J. Benson & F. R. Cohen

Since H^{∗}(A_{6}, L_{β+γ}^{∗}) is generated as an H^{∗}(A_{6}, k)-module by elements of degree zero and one, it follows that H^{∗}(A_{6}, H^{1}(K_{6}, k)) is generated by elements of degree at most four, and hence by u and v. □

Now we turn our attention to the kA_{6}-module H^{2}(K_{6}, k) = \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ⊕ D, in order

to calculate H^{∗}(A_{6}, H^{2}(K_{6}, k)). Since D is projective, we have H^{∗}(A_{6}, H^{2}(K_{6}, k)) =

H^{∗}(A_{6}, \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ).

Ω^{−1}( \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} Ω^{−2}( \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = M_{2} ; | ; k ; | ; M_{1} ; | ; k ; w

Ω^{−3}( \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = \begin{matrix}k \\ / \ \\ M_{1} M_{2} \\ | | \\ k k \\ | | \\ M_{2} M_{1} \\ | | \\ k k \\ | x \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix}  Ω^{−4}( \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = \begin{matrix}k \\ / \ \\ M_{2} M_{1} \\ | | \\ k k \\ | | \\ M_{2} M_{1} M_{2} \\ \ / | \\ k  k \\ αw  | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix}

Ω^{−5}( \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = k ; / \ ; M_{1} M_{2} ; | | ; k k ; | | ; M_{2} M_{1} M_{2} ; | \ / ; k  k ; | αx ; M_{1} ; | ; k ; βw  Ω^{−6}( \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} ) = \begin{matrix}k \\ / \ \\ M_{2} M_{1} \\ | | \\ k  k k \\ / \ | | \\ M_{1} M_{2} M_{1} M_{2} \\ | \ / | \\ k  k  k \\ | α^{2}w  | \\ M_{2}  M_{1} \\ |  | \\ k  k \\ |  βx \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix}  γx = 0

γw = 0

Proposition 5.3 As a module over H^{∗}(A_{6}, k) we have

H^{∗}(A_{6}, H^{2}(K_{6}, k)) = ⟨w, x⟩/(γw, γx)

Proof The proof is essentially the same as the proof of Proposition 5.2. For the element ζ we take the homomorphism

\begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} → Ω^{−3} ( \begin{matrix}k \\ | \\ M_{1} \\ | \\ k \\ | \\ M_{2} \end{matrix} )

which is again the diagonal of two D-homomorphisms, namely the injective one and the one with kernel M_{2}. The cokernel of ζ is the periodic module

k
/ \
M_{1} M_{2}
| |
k k
| |
M_{2} M_{1}
\ /
k

whose periodicity is again generated by α. Again, multiplication by ζ has the same effect as multiplication by (β + γ), the conditions of Benson and Carlson [2, Theorem 9.3] are satisfied, and the rest of the proof carries through identically. □

Lemma 5.4 u^{2} = 0 as an element of H^{2}(A_{6}, H^{2}(K_{6}, k))

Proof In general if H ⊴ G, and we have elements

x ∈ H^{i}(G/H, H^{r}(H, k)) ≅ Hom_{G/H}(k, Ω^{−i}H^{r}(H, k))

and

y ∈ H^{j}(G/H, H^{s}(H, k)) ≅ Hom_{G/H}(k, Ω^{−j}H^{s}(H, k))

then the cup product

xy ∈ H^{i+j}(G/H, H^{r+s}(H, k)) ≅ Hom_{G/H}(k, Ω^{−i+j}H^{r+s}(H, k))

is given by the composite map

k →^{x} Ω^{−i}H^{r}(H, k) ⟶^{H^{r}(H,k)⊗Ω^{−i}(y)} Ω^{−i−j}(H^{r}(H, k) ⊗ H^{s}(H, k))

⟶^{Ω^{−i−j}(_{∪})} Ω^{−i−j}H^{r+s}(H, k)

(note that H^{r}(H, k) ⊗ Ω^{−i}Ω^{−j}H^{s}(H, k) ≅ Ω^{−i−j}(H^{r}(H, k) ⊗ H^{s}(H, k)) ⊕ (projective) ). In our case this works out as follows.

k → Ω^{−1}H^{1}(K_{6}, k) = \begin{matrix}M_{1} \\ | \\ k \\ | \\ M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} → Ω^{−2}H^{2}(K_{6}, k) = \begin{matrix}M_{2} \\ | \\ k \\ | \\ M_{1} \\ | \\ k \end{matrix} .

Any composite of maps between these modules is zero. □

Lemma 5.5 (i) uv = 0 as an element of H^{3}(A_{6}, H^{2}(K_{6}, k))

(ii) v^{2} = 0 as an element of H^{4}(A_{6}, H^{2}(K_{6}, k))

Proof (i) Since the only non-zero element of the same degree as uv is x, the other possibility is uv = x. If this were the case then since βv = α^{2}u we would have βv^{2} = α^{2}uv = α^{2}x, contradicting the fact that α^{2}x is not in the image of multiplication by β, by Proposition 5.3.

(ii) Since uv = 0, we have βv^{2} = α^{2}uv = 0 and since by Proposition 5.3 multiplication by β is injective, it follows that v^{2} = 0. □

Our next task is to go from A_{6} to Σ_{6}.

Lemma 5.6

H^{∗}(Σ_{6}, k) ≅ H^{∗}(A_{6}, k)⊗H^{∗}(Σ_{2}, k) = k[α, β, γ, δ]/(βγ = 0) with deg(α) = 2, deg(β) = deg(γ) = 3 and deg(δ) = 1.

In particular in the Lyndon-Hochschild-Serre spectral sequence H^{∗}(Σ_{2}, H^{∗}(A_{6}, k)) ⇒ H^{∗}(Σ_{6}, k) we have E_{2}^{∗∗} = E_{∞}^{∗∗} ≅ H^{∗}(Σ_{6}, k).

Proof This follows from the fact that at the level of Sylow 2-subgroups the extension is a direct product.

\begin{aligned}1 → & A_{6} & → & Σ_{6} & → & Σ_{2} & → 1 \\ & ∪ & & ∪ & & ∥ \\ 1 → & D_{8} & → & D_{8} × Σ_{2} & → & Σ_{2} & → 1 \end{aligned} . □

Proposition 5.7 As modules over H^{∗}(Σ_{6}, k) we have (i) H^{∗}(Σ_{6}, H^{1}(K_{6}, k)) = ⟨u, v⟩/(γu, α^{2}u + βv) (ii) H^{∗}(Σ_{6}, H^{2}(K_{6}, k)) = ⟨w, x⟩/(γw, γx) (iii) H^{∗}(Σ_{6}, H^{3}(K_{6}, k)) = 0.

Proof (i) We have a Lyndon-Hochschild-Serre spectral sequence

H^{∗}(Σ_{2}, H^{∗}(A_{6}, H^{1}(K_{6}, k))) ⇒ H^{∗}(Σ_{6}, H^{1}(K_{6}, k))

Since H^{1}(A_{6}, H^{1}(K_{6}, k)) and H^{2}(A_{6}, H^{1}(K_{6}, k)) are one-dimensional, generated by u and v respectively, and generate H^{∗}(A_{6}, H^{1}(K_{6}, k)) as a module, it follows that Σ_{2} acts trivially on H^{∗}(A_{6}, H^{1}(K_{6}, k)). So the E_{2} page of the above spectral sequence is isomorphic to H^{∗}(Σ_{2}, k) ⊗ H^{∗}(A_{6}, H^{1}(K_{6}, k)).

Next, we claim that E_{2} = E_{∞}. Using the structure of this spectral sequence as a module over the spectral sequence

H^{∗}(Σ_{2}, H^{∗}(A_{6}, k)) ⇒ H^{∗}(Σ_{6}, k)

we see that it suffices to prove that d_{2}(v) = 0. The only other possibility is that d_{2}(v) = δ^{2}u, which is absurd since it would imply that

0 ≠ βδ^{2}u = d_{2}(βv) = d_{2}(α^{2}u) = 0.

The reader will observe that the best information we can get from the spectral sequence is that γu = 0 and α^{2}u = βv + λ_{1}αδ^{2}u + λ_{2}δ^{4}u for some values of λ_{1}, λ_{2} ∈ {0, 1}. To get any further, we must construct a double complex.

It follows from the remarks on extendability at the end of Section 3 that the minimal injective resolution of H^{1}(K_{6}, k) as a kA_{6}-module extends to a long exact sequence of kΣ_{6}-modules,

J = 0 → H^{1}(K_{6}, k) → J_{0} → J_{1} → ⋯

with the property that J_{i} is injective when restricted to kA_{6}. Now let

Q = 0 → k → Q_{0} → Q_{1} → ⋯

be the minimal kΣ_{2}-resolution of k, so that each Q_{i} has k-dimension two, and regard this as a sequence of kΣ_{6}-modules with A_{6} in the kernel.

Each J_{i} ⊗ Q_{j} is injective as a kΣ_{6}-module, and so J ⊗ Q is a double complex whose total complex is an injective resolution of H^{1}(K_{6}, k). Moreover

dim_{k} Hom_{kΣ_{6}}(k, (Tot(J ⊗ Q))_{j}) = dim_{k} H^{j}(Σ_{6}, H^{1}(K_{6}, k))

by the above spectral sequence calculation (where “Tot” denotes the total complex of a double complex), and so every cochain is a cocycle and every coboundary is zero. Thus H^{j}(Σ_{6}, H^{1}(K_{6}, k)) is bigraded, with bideg(u) = (1, 0), bideg(v) = (2, 0) and multiplication by δ has bidegree (0, 1).

Keeping track of bidegrees, it is now easy to see that γu = 0 and α^{2}u = βv, completing the proof of (i).

(ii) We can use the same proof as in (i), provided we can show that in the spectral sequence

H^{∗}(Σ_{2}, H^{∗}(A_{6}, H^{2}(K_{6}, k))) ⇒ H^{∗}(Σ_{6}, H^{2}(K_{6}, k))

we have d_{2}(x) = 0. The only other possibility is d_{2}(x) = δ^{2}w. Elimination of this possibility is slightly more subtle than in (i). Suppose that d_{2}(x) = δ^{2}w. Then δ^{2} would annihilate the whole of H^{∗}(Σ_{6}, H^{2}(K_{6}, k)), and so the variety V_{Σ_{6}}(H^{2}(K_{6}, k)) would be contained in the image of the map res^{∗} : V_{A_{6}} → V_{Σ_{6}}. It would follow that the restriction of H^{2}(K_{6}, k) to an involution in Σ_{6} not lying in A_{6} would have to be projective. However, by Lemma 3.3 (ii) we have H^{2}(K_{6}, k) ≅ S^{2}(M_{2}), so that whatever the Jordan blocks of this involution in its action on M_{2}, there must be a Jordan block of size one in the action on S^{2}(M_{2}), which is therefore not projective for this involution.

(iii) Tensoring the short exact sequence

0 → k_{Σ_{6}} → k_{A_{6}} ↑^{Σ_{6}}→ k_{Σ_{6}} → 0

of kΣ_{6}-modules with H^{3}(K_{6}, k), we obtain a short exact sequence of kΣ_{6}-modules

0 → H^{3}(K_{6}, k) → (P_{M_{2}})_{A_{6}} ↑^{Σ_{6}}→ H^{3}(K_{6}, k) → 0

(since H^{3}(K_{6}, k) ⊗ k_{A_{6}} ↑^{Σ_{6}} ≅ (H^{3}(K_{6}, k)↓_{A_{6}} ⊗k)↑^{Σ_{6}} ≅ (P_{M_{2}})_{A_{6}} ↑^{Σ_{6}}).

The second module in this sequence is injective = projective and so as kΣ_{6}-modules we have Ω^{−j}H^{3}(K_{6}, k) ≅ H^{3}(K_{6}, k). In particular

H^{j}(Σ_{6}, H^{3}(K_{6}, k)) ≅ Hom_{kΣ_{6}}(k, Ω^{−j}H^{3}(K_{6}, k)) = 0. □

This completes the calculation of the E_{2} term of the spectral sequence

H^{∗}(Σ_{6}, H^{∗}(K_{6}, k)) ⇒ H^{∗}(Γ_{0,0}^{6}, k).

Here is a picture of the answer.

↑ 0
•  •  •  •  •
w  x  αw  βw, αx α^{2}w, βx
•  •  •  •  •  •
u  v  αu βu, αv α^{2}u = βv, αβu, α^{2}v
γv
• •  •  •  •  •  •
δ  α  β, γ  α^{2}  αβ, αγ α^{3}, β^{2}, γ^{2}
βγ = 0, u^{2} = 0, uv = 0, v^{2} = 0,

γu = 0, α^{2}u = βv, γw = 0, γx = 0.

Multiplication by δ is injective, and so products with delta are not indicated. Relations are those indicated together with the relations which follow from E_{2}^{pq} = 0 for q ≥ 3.

6 Finite subgroups of Γ_{0,0}^{6} and Γ_{2,0}^{0}

In order to calculate the differentials in the spectral sequence

H^{∗}(Σ_{6}, H^{∗}(K_{6}, k)) ⇒ H^{∗}(Γ_{0,0}^{6}, k)

for k = F_{2} or F_{3}, we use a theorem of Quillen [7, Theorem 14.1]. This says that for a group of finite virtual cohomological dimension (i.e., a group with a subgroup of finite index and finite cohomological dimension; Γ_{0,0}^{n} is of this type since K_{n} has finite cohomological dimension), an element of cohomology is nilpotent if and only if its restriction to every finite elementary abelian subgroup is nilpotent. For this purpose, we need to understand the elementary abelian subgroups of Γ_{0,0}^{6}. In fact we shall list all the finite subgroups.

Lemma 6.1 Given a finite subgroup G of Γ_{0,0}^{n}, we may embed the sphere with n punctures as the unit sphere in R^{3} in such a way that G acts as orientation preserving orthogonal transformations.

Proof We have a map Diffeo^{+}(M_{0,0}^{n}) → Γ_{0,0}^{n} ≅ π_{0}Diffeo^{+}(M_{0,0}^{n}), which is epimorphic with contractible kernel, and every finite subgroup G of Γ_{0,0}^{n} lifts to Diffeo^{+}(M_{0,0}^{n}). Choose a Riemannian metric on the sphere compatible with the smooth structure, and average its images under G to obtain a G-invariant Riemannian metric. Then G also preserves the underlying complex structure on the sphere, and so it acts as Möbius transformations, i.e., as a subgroup of P SL_{2}(C), and hence as a subgroup of P SU(2) = SO(3). □

Now every finite subgroup of SO(3) is contained in either the icosahedral group A_{5}, the octahedral group Σ_{4} or the dihedral group O(2). The icosahedral group does not have an orbit of size as small as six, and so every finite subgroup of Γ_{0,0}^{6} is contained in either the octahedral or dihedral group.

Theorem 6.2 The subgroups of Σ_{6} which lift to subgroups of Γ_{0,0}^{6} are either

(a) contained in the octahedral group Σ_{4} —in this case the six punctures are the vertices of a regular octahedron, or

(b) contained in the dihedral groups D_{10} or D_{12} (of orders ten and twelve respectively). □

In particular, every finite two-subgroup and three-subgroup of Γ_{0,0}^{6} is contained in the octahedral group Σ_{4}. By Quillen [7, Theorem 14.1], we therefore have the following.

Corollary 6.3 An element of H^{∗}(Γ_{0,0}^{6}, k), k = F_{2} or F_{3}, is nilpotent if and only if its restriction to the octahedral group Σ_{4} ⊆ Γ_{0,0}^{6} is nilpotent. □

We have H^{∗}(Σ_{4}, F_{2}) = F_{2}[X, Y, Z]/(XZ) as follows.

degree Sq^{1}  Sq^{2}
X  1  X^{2}  0
Y  2  Z + XY Y^{2}
Z  3  0  Y Z

It is easily checked that the restriction map

H^{∗}(Σ_{6}, F_{2}) → H^{∗}(Σ_{4}, F_{2})

(where Σ_{4} is embedded as the octahedral group in Σ_{6}) is given by

α ↦ Y + X^{2}, β ↦ XY, γ ↦ Z, δ ↦ X.

Applying Corollary 6.3, we obtain the following.

Proposition 6.4 The images of (δ^{3} + αδ + β) and γδ under inflation from H^{∗}(Σ_{6}, F_{2}) to H^{∗}(Γ_{0,0}^{6}, F_{2}) are nilpotent. □

We now need to examine the lift of the octahedral group to Γ_{2,0}^{0}. This defines for us a double cover

1 → Z/2 → Γ_{2,0}^{0} → Γ_{0,0}^{6} → 1
∥ ∪ ∪
1 → Z/2 → Σ̂_{4} → Σ_{4} → 1

In order to decide which double cover this is, it suffices to examine the orders in Σ̂_{4} of preimages of elements of Σ_{4} of orders two and four not lying in A_{4}.

Lemma 6.5 The preimages in Σ̂_{4} of elements of Σ_{4} \ A_{4} of orders two and four have orders two and eight respectively.

Proof We use the presentations of Γ_{0,0}^{6} and Γ_{2,0}^{0} as quotients of the Braid group B_{6} on six strings, as given in Cohen [3, Section 2]. A representative of an element of order two in Σ_{4} \ A_{4} is given by the braid

(σ_{1}σ_{2}σ_{3}σ_{4}σ_{5})(σ_{1}σ_{2}σ_{3}σ_{4})(σ_{1}σ_{2}σ_{3})(σ_{1}σ_{2})σ_{1}

(a half-twist of the entire set of strings) whose square is (σ_{1}σ_{2}σ_{3}σ_{4}σ_{5})^{6} and is hence equal to one both in Γ_{0,0}^{6} and in Γ_{2,0}^{0}.

A representative of an element of order four in Σ_{4} \ A_{4} is given by the braid σ_{5}^{2}σ_{4}σ_{3}σ_{2}. It is easy to check that

(σ_{5}^{2}σ_{4}σ_{3}σ_{2})^{4}(σ_{1}σ_{2}σ_{3}σ_{4}σ_{5})(σ_{5}σ_{4}σ_{3}σ_{2}σ_{1}) = (σ_{5}σ_{4}σ_{3}σ_{2}σ_{1})^{6}

in the following sequence of steps.

(σ_{5}^{2}σ_{4}σ_{3}σ_{2})^{4}(σ_{1}σ_{2}σ_{3}σ_{4}σ_{5})(σ_{5}σ_{4}σ_{3}σ_{2}σ_{1}) = (σ_{5}^{2}σ_{4}σ_{3}σ_{2})^{3}(σ_{5}σ_{1}σ_{2}σ_{3}σ_{4})(σ_{5}σ_{4}σ_{3}σ_{2}σ_{1})^{2}
= (σ_{5}^{2}σ_{4}σ_{3}σ_{2})^{2}(σ_{5}σ_{4}σ_{1}σ_{2}σ_{3})(σ_{5}σ_{4}σ_{3}σ_{2}σ_{1})^{3}
= (σ_{5}^{2}σ_{4}σ_{3}σ_{2})(σ_{5}σ_{4}σ_{3}σ_{1}σ_{2})(σ_{5}σ_{4}σ_{3}σ_{2}σ_{1})^{4}
= (σ_{5}σ_{4}σ_{3}σ_{2}σ_{1})^{6}.

Since (σ_{1}σ_{2}σ_{3}σ_{4}σ_{5})(σ_{5}σ_{4}σ_{3}σ_{2}σ_{1}) is the identity element in Γ_{0,0}^{6} and has order two in Γ_{2,0}^{0} it follows that σ_{5}^{2}σ_{4}σ_{3}σ_{2} has order eight in Γ_{2,0}^{0}. □

Thus Σ̂_{4} is the double cover which restricts to a proper double cover of A_{4}, and which has semidihedral rather than generalised quaternion Sylow two-subgroups. In particular, note that it is not the binary octahedral group. The corresponding element of H^{2}(Σ_{4}, F_{2}) is the element Y.

Again using Quillen [7, Theorem 14.1], we obtain the following.

Proposition 6.6 An element of H^{∗}(Γ_{2,0}^{0}, k), k = F_{2} or F_{3}, is nilpotent if and only if its restriction to Σ̂_{4} is nilpotent. In particular, the images of the elements β, γ and α + δ^{2} under inflation from H^{∗}(Σ_{6}, F_{2}) to H^{∗}(Γ_{2,0}^{0}, F_{2}) are nilpotent. □

7 Calculation of the spectral sequences for H^{∗}(Γ_{0,0}^{6}, F_{2}) and H^{∗}(Γ_{2,0}^{0}, F_{2})

At the end of Section 5, we completed the calculation of the E_{2} term of the spectral sequence H^{∗}(Σ_{6}, H^{∗}(K_{6}, F_{2})) ⇒ H^{∗}(Γ_{0,0}^{6}, F_{2}). In this section we use Proposition 6.4 to complete the computation of this spectral sequence, and also use Proposition 6.6 to provide a partial computation for the spectral sequence H^{∗}(Γ_{0,0}^{6}, H^{∗}(Z/2, F_{2})) ⇒ H^{∗}(Γ_{2,0}^{0}, F_{2}).

Proposition 7.1 In the spectral sequence H^{∗}(Σ_{6}, H^{∗}(K_{6}, F_{2})) ⇒ H^{∗}(Γ_{0,0}^{6}, F_{2}) the differentials are given by

d_{2}(u) = 0  d_{3}(x) = (δ^{3} + αδ + β)^{2}
d_{2}(v) = γδ d_{n} = 0 n ≥ 4.
d_{2}(w) = (δ^{3} + αδ + β)u
d_{2}(x) = 0

Proof Since uv = 0, we have ud_{2}(v) = vd_{2}(u) and so d_{2}(u) is 0 or β. However, uw = 0, and so wd_{2}(u) = 0 which forces d_{2}(u) = 0.

Again using ud_{2}(v) = vd_{2}(u) this tells us that d_{2}(v) is either 0 or γδ. If d_{2}(v) = 0 then it is not hard to check that there is no way of killing any power of γδ in the spectral sequence, as Proposition 6.4 tells us we must. We therefore have d_{2}(v) = γδ.

Similarly the only way of killing some power of (δ^{3}+αδ+β), in accord with Proposition 6.4, is to have d_{2}(w) = (δ^{3} + αδ + β)u, d_{2}(x) = 0 and d_{3}(x) = (δ^{3} + αδ + β)^{2}. □

This completes the calculation of the E_{∞} term of the first spectral sequence. A diagram of the answer is as follows.

↑ 0
•
u
•  •  •  •
δ  α  β, γ

\begin{matrix}βγ = 0 & γδ = 0 & (δ^{3} + αδ + β)^{2} = 0 \\ γu = 0 & u^{2} = 0 & (δ^{3} + αδ + β)u = 0 \end{matrix}

Now it is easy to check that the central extension

1 → Z/2 → Γ_{2,0}^{0} → Γ_{0,0}^{6} → 1

is not a pullback of the form

Γ_{2,0}^{0} → Γ_{0,0}^{6}
↓  ↓
Σ̂_{6} → Σ_{6}

and so the corresponding element of H^{2}(Γ_{0,0}^{6}, F_{2}) is not in the image of inflation from Σ_{6}. We may therefore choose a representative of u in H^{2}(Γ_{0,0}^{6}, F_{2}), which we shall also denote by u, in such a way that this central extension corresponds to u+α. We choose this rather than u in order to simplify the multiplication on H^{∗}(Γ_{0,0}^{6}, F_{2}), as we shall see. In order to calculate this multiplicative structure, we first calculate the action of the Steenrod algebra on H^{∗}(Σ_{6}, F_{2}).

Lemma 7.2 The action of the Steenrod algebra on H^{∗}(Σ_{6}, F_{2}) is given by

Sq^{1}(α) = β + γ Sq^{1}(β) = Sq^{1}(γ) = 0 Sq^{1}(δ) = δ^{2}
Sq^{2}(β) = αβ  Sq^{2}(γ) = αγ

Proof Since the double cover of A_{6} does not lift to a 4-fold cover, the element α must be killed at the E_{2} term of the Bockstein spectral sequence for A_{6}. By symmetry under the outer automorphism swapping β and γ, we have Sq^{1}(α) = β + γ. Hence Sq^{1}(β) = Sq^{1}(γ) is either 0 or α^{2}. If it is α^{2} then 0 = Sq^{1}(βγ) = α^{2}β + α^{2}γ ≠ 0, so we have Sq^{1}(β) = Sq^{1}(γ) = 0. Finally since Sq^{3} = Sq^{1}Sq^{2}, Sq^{2}(β) is some element whose Sq^{1} is β^{2}, and so Sq^{2}(β) = αβ. Similarly Sq^{2}(γ) = αγ. □

Theorem 7.3 H^{∗}(Γ_{0,0}^{6}, F_{2}) is generated by elements α, β, γ, δ and u with deg(α) = deg(u) = 2, deg(β) = deg(γ) = 3 and deg(δ) = 1, subject to the following relations.

\begin{matrix}γδ = 0 & βγ = 0 & (δ^{3} + αδ + β)^{2} = 0 \\ γu = 0 & u^{2} = δ^{4} & (δ^{3} + αδ + β)(u + δ^{2}) = 0 \end{matrix}

The action of the Steenrod algebra is given as follows.

Sq^{1}(α) = β + γ Sq^{1}(β) = Sq^{1}(γ) = 0
Sq^{1}(δ) = δ^{2}  Sq^{1}(u) = either δu + αδ + β or  δu + δ^{3}
Sq^{2}(β) = αβ  Sq^{2}(γ) = αγ.

Proof According to the calculation at the end of Section 6, the restriction of u + α to the octahedral group is Y. Thus the restriction of u + δ^{2} is zero, and so by Corollary 6.3, u + δ^{2} is nilpotent. We know that u satisfies some quadratic equation over the image of inflation from Σ_{6}, and so this quadratic expression must be u^{2} = δ^{4}.

Now Sq^{1}(u) restricts to Sq^{1}(X^{2}) = 0 on the octahedral group, and so

Sq^{1}(u) ∈ {0, δ^{3} + αδ + β, δu + αδ + β, δ^{3} + δu}.

Since Sq^{1}(δ^{3} + αδ + β) ≠ 0, Sq^{1}(u) ≠ δ^{3} + αδ + β.

We next show that Sq^{1}(u) ≠ 0. This is the same as saying that the 2-fold cover corresponding to u does not lift to a 4-fold cover. This 2-fold cover is the central extension given by

σ_{j}σ_{j+1}σ_{j} = σ_{j+1}σ_{j}σ_{j+1}
σ_{j}σ_{k} = σ_{k}σ_{j} if |j − k| ≥ 2
(σ_{1}σ_{2}σ_{3}σ_{4}σ_{5})(σ_{5}σ_{4}σ_{3}σ_{2}σ_{1}) = −1
and (σ_{1}σ_{2})^{3} = −(σ_{4}σ_{5})^{3}.

If it lifts to a 4-fold cover then in this 4-fold cover we have

σ_{j}σ_{k} = i.σ_{k}σ_{j} for k − j ≥ 2

where i denotes the square root of minus one, and

(σ_{1}σ_{2})^{3} = θ.(σ_{4}σ_{5})^{3} = θ.(σ_{5}σ_{4})^{3}

with θ ∈ {i, −i}. Then

σ_{4}(σ_{1}σ_{2})^{3} = −(σ_{1}σ_{2})^{3}σ_{4} = −θ.(σ_{4}σ_{5})^{3}σ_{4}
= −θ.σ_{4}(σ_{5}σ_{4})^{3} = −σ_{4}(σ_{1}σ_{2})^{3}.

This contradiction shows that Sq^{1}(u) ≠ 0, and hence it has one of the two given values.

In either case we have Sq^{1}(γu) = 0, and γu restricts to zero on the octahedral group, and so γu is either zero or δ^{5} + βδ^{2} + α^{2}δ + αβ. The latter expression is not killed by multiplication by β, and so we have γu = 0.

Now Sq^{1}(δ^{3} + αδ + β)u is (δ^{3} + αδ + β)(αδ + β) = (δ^{3} + αδ + β)δ^{3} in both cases, and (δ^{3} + αδ + β)u restricts to zero on the octahedral group, and so (δ^{3} + αδ + β)u is either zero or (δ^{3} +αδ +β)δ^{2}. But (δ^{3} +αδ +β)u^{2} = (δ^{3} +αδ +β)δ^{4}, and so the latter possibility holds. □

Remark : Since the two possibilities for Sq^{1}(u) restrict to β and 0 respectively on the subgroup of index two, in order to distinguish between the possibilities it suffices to ascertain whether the central extension given in the above proof lifts to a 4-fold cover on the subgroup of index two. We have not been able to get anywhere with this question^{3}.

We now discuss briefly the spectral sequence

H^{∗}(Γ_{0,0}^{6}, H^{∗}(Z/2, F_{2})) ⇒ H^{∗}(Γ_{2,0}^{0}, F_{2}).

Let H^{∗}(Z/2, F_{2}) = F_{2}[z] with deg(z) = 1. Since the central extension corresponds to the element u + α, we have d_{2}(z) = u + α. Using the fact that Steenrod operations commute with transgressions, we have

d_{3}(z^{2}) = d_{3}Sq^{1}(z) = Sq^{1}d_{2}(z) = Sq^{1}(u + α)

is either δu + αδ + γ or δu + δ^{3} + β + γ. In either case the element

d_{5}(z^{4}) = d_{5}Sq^{2}(z^{2}) = Sq^{2}d_{3}(z^{2})

has already been hit earlier in the spectral sequence, and so the spectral sequence stops here. The Poincaré series given in the introduction are easy to compute from this information.

8 Calculations in characteristic three

In this section we present an outline only of the calculation of H^{∗}(Γ_{2,0}^{0}, F_{3}), but with enough details so that the particularly assiduous reader could supply the rest. The cohomology calculations are considerable more complicated than in characteristic two, so that complete details would fill too much space. In a sense the calculations are more straightforward than in characteristic two, because Σ_{6}/A_{6} and Ker(Γ_{2,0}^{0} → Γ_{0,0}^{6}) both have order two, which is coprime to the characteristic.

Thus rather than just looking at the spectral sequence

H^{∗}(Σ_{6}, H^{∗}(K_{6}, F_{3})) ⇒ H^{∗}(Γ_{0,0}^{6}, F_{3}) ≅ H^{∗}(Γ_{2,0}^{0}, F_{3}),

we look at H^{∗}(A_{6}, H^{∗}(K_{6}, F_{3})), but keeping track of the Σ_{2}-action, so that at the end of the calculation we can take Σ_{2}-invariants.

There are four irreducible modules in the principal block of A_{6} in characteristic three, which we shall denote by the symbols ◇, ▷, ◁ and ◯. They have dimensions 1, 3, 3 and 4 respectively. Since

dim_{F_{3}} Ext_{F_{3}A_{6}}^{1}(◇, ◯) = 2,

3Note that this ambiguity is settled in the final paper in this series, later in this volume.

we must be careful about labelling the edges. We have chosen a basis for this Ext group, and represented the basis elements by a single edge and a double edge. Since ◇^{∗} ≅ ◇ and ◯^{∗} ≅ ◯, we have a natural isomorphism

Ext_{F_{3}A_{6}}^{1}(◇, ◯) ≅ Ext_{F_{3}A_{6}}^{1}(◯, ◇),

and we use the same symbols for the basis elements here.

As stated in Benson and Carlson [2], the projective indecomposable modules in the principal block are representations of the following diagrams.

[figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; \▷; ◇; ◇; /; ◁; ^{\ ▷}; ^{/◁}; ◇; ◇; \; ▷; / ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; \; ▷; /; ◁]

By dualising the projective resolution of the trivial module ◇ given in Benson and Carlson [2], we obtain the cokernels in the minimal injective resolution of the trivial module as representations of the following diagrams. Note that this time, when we regard a copy of ◇ at the bottom of a diagram as a cohomology element, since F_{3}^{×} ≅ {±1}, we always have a choice of sign to make. Our policy is to keep the generality of choice until we find a relation. It turns out that in every case, there are two possibilities. Either we can lose some generality to choose the sign in the relation, or associativity (and graded commutativity) forces the sign in the relation.

Ω^{−1}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁] + +

Ω^{−2}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; \▷; /; ◁; ◇; / ◁; \; ▷; −; −; α]

Ω^{−3}(◇) = [figure: ◇; ◇; ◇; ^{\ ▷}; ^{/◁}; ◇; −; β_{1}; β_{2}]

Ω^{−4}(◇) =  [figure: ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; +; γ]  α^{2} = 0

Ω^{−5}(◇) =  [figure: ◇; \▷; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; +; +; /; ◁; \; ▷; ◇; / ◁]  \begin{matrix}αβ_{1} = 0 \\ αβ_{2} = 0 \end{matrix}

Ω^{−6}(◇) =  [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ^{\ ▷}; /; ◁; ◇; ^{/◁}; \; ▷; −; −; \begin{matrix}β_{1}β_{2} = −β_{2}β_{1} \\ = αγ \end{matrix}]  \begin{matrix}β_{1}^{2} = 0 \\ β_{2}^{2} = 0 \end{matrix}

Ω^{−7}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; δ_{1}; β_{1}γ; β_{2}γ; δ_{2}; −]

Ω^{−8}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; +; ε_{1}; γ^{2}; ε_{2}]

Ω^{−9}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; /; ◁; \; ▷; /; ◁; \; ▷; +; +] \begin{matrix}αδ_{1} = 0 \\ αδ_{2} = 0 \end{matrix}

Ω^{−10}(◇) =[figure: ◇; \▷; /; ◁; ◇; / ◁; \; ▷; −; −; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; \; ▷; /; ◁; ◇; \; ▷; /; ◁; αε_{1} = β_{1}δ_{1}; αγ^{2} = β_{1}β_{2}γ; αε_{2} = β_{2}δ_{2}] \begin{matrix}β_{1}δ_{2} = 0 \\ β_{2}δ_{1} = 0 \end{matrix}

Ω^{−11}(◇) = [figure: ◇; ^{\ ▷}; ^{/◁}; ◇; −; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; β_{1}ε_{1}; \begin{matrix}γδ_{1} \\ = β_{2}ε_{1} \end{matrix}; β_{1}γ^{2}; β_{2}γ^{2}; \begin{matrix}γδ_{2} \\ = β_{1}ε_{2} \end{matrix}; β_{2}ε_{2}]

Ω^{−12}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; +; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; γε_{1}; γ^{3}; γε_{2}]

Ω^{−13}(◇) = [figure: ◇; \▷; /; ◁; \; ▷; /; ◁; \; ▷; /; ◁; \; ▷; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; +; +; ◇; / ◁]

Ω^{−14}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; \; ▷; /; ◁; ◇; \; ▷; /; ◁; ◇; ^{\ ▷}; /; ◁; ◇; ^{/◁}; \; ▷; αγε_{1} = β_{1}γδ_{1}; \begin{matrix}δ_{1}δ_{2} = αγ^{3} \\ = β_{1}β_{2}γ^{2} \end{matrix}; αγε_{2} = β_{2}γδ_{2}; −; −]

Ω^{−15}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; −; δ_{1}ε_{1}; β_{1}γε_{1}; \begin{matrix}β_{2}γε_{1} \\ = γ^{2}δ_{1} \end{matrix}; \begin{matrix}β_{1}γ^{3} \\ = −δ_{2}ε_{1} \end{matrix}; \begin{matrix}β_{2}γ^{3} \\ = δ_{1}ε_{2} \end{matrix}; \begin{matrix}β_{1}γε_{2} \\ = γ^{2}δ_{2} \end{matrix}; β_{2}γε_{2}; δ_{2}ε_{2}]

Ω^{−16}(◇) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ε_{1}^{2}; γ^{2}ε_{1}; ε_{1}ε_{2} = γ^{4}; γ^{2}ε_{2}; ε_{2}^{2}; +]

In these diagrams, we have indicated the action of the automorphism of order two in Σ_{6}/A_{6} as follows. The action is given by reflection about a vertical axis of symmetry. In the case of vertices of the diagram lying on this axis of symmetry, we need to indicate which extension of the action of A_{6} to Σ_{6} we have. Since in all cases, the composition factor involved is the trivial module for A_{6}, the two extensions to Σ_{6} can be labelled “+” and “−” without ambiguity of meaning.

In this way, it is easy to keep track of the action of Σ_{6}/A_{6} on H^{∗}(A_{6}, F_{3}). We have chosen the notation in such a way that when elements are interchanged by this action, they are given the same names, but with subscripts 1 and 2.

Proposition 8.1 H^{∗}(A_{6}, F_{3}) is the graded commutative ring with generators α, β_{1}, β_{2}, γ, δ_{1}, δ_{2}, ε_{1} and ε_{2} subject to the following relations

α^{2} = 0 β_{1}β_{2} = −β_{2}β_{1} = αγ β_{1}δ_{2} = 0 δ_{1}ε_{2} = β_{2}γ^{3}
αβ_{1} = 0  αδ_{1} = 0  β_{2}δ_{1} = 0 δ_{2}ε_{1} = −β_{1}γ^{3}
αβ_{2} = 0  αδ_{2} = 0  γδ_{1} = β_{2}ε_{1} ε_{1}ε_{2} = γ^{4}
β_{1}^{2} = 0  αε_{1} = β_{1}δ_{1}  γδ_{2} = β_{1}ε_{2}
β_{2}^{2} = 0  αε_{2} = β_{2}δ_{2}  δ_{1}δ_{2} = αγ^{3}
where deg(α) = 2, deg(β_{1}) = deg(β_{2}) = 3, deg(γ) = 4,
deg(δ_{1}) = deg(δ_{2}) = 7, deg(ε_{1}) = deg(ε_{2}) = 8.

The action of the automorphism of order two in Σ_{6}/A_{6} is given by

\begin{matrix}α ↦ −α & γ ↦ γ & ε_{1} ↦ ε_{2} \\ β_{1} ↦ β_{2} & δ_{1} ↦ δ_{2} & ε_{2} ↦ ε_{1} \\ β_{2} ↦ β_{1} & δ_{2} ↦ δ_{1} & . \end{matrix} □

Note that as in Benson and Carlson [2], we may give H^{∗}(A_{6}, F_{3}) a double grading with

bideg(α) = (1, 1) bideg(β_{1}) = (1, 2) bideg(β_{2}) = (2, 1)
bideg(γ) = (2, 2) bideg(δ_{1}) = (2, 5) bideg(δ_{2}) = (5, 2)
bideg(ε_{1}) = (2, 6) bideg(ε_{2}) = (6, 2)  .

Next, we examine the structure of the F_{3}Σ_{6}-modules H^{j}(K_{6}, F_{3}), 1 ≤ j ≤ 3.

Proposition 8.2 The following are diagrams for the modules H^{j}(K_{6}, F_{3}), 1 ≤ j ≤ 3, as F_{3}A_{6}-modules.

(i) H^{1}(K_{6}, F_{3}) ≅ D^{(4,2)} is a 9-dimensional projective irreducible modules in a nonprincipal block. In particular H^{∗}(A_{6}(H^{1}(K_{6}, F_{3})) = 0 and H^{∗}(Σ_{6}, H^{1}(K_{6}, F_{3})) = 0.

(ii)

H^{2}(K_{6}, F_{3}) = [figure: ◇; \; ▷; /; ◁; ◇; ◇; ^{/◁}; ^{\ ▷}]

(iii)

H^{3}(K_{6}, F_{3}) = D^{(2^{2},1^{2})}⊕

where D^{(2^{2},1^{2})} is again a 9-dimensional projective irreducible in a non-principal block. For A_{6}, it is isomorphic to D^{(4,2)}, but for Σ_{6} they are related by tensoring with the sign representation.

Again, the action of Σ_{6}/A_{6} is given by reflection in a vertical axis. For the F_{3}A_{6}- module ◯, the extensions ◯^{+} and ◯^{−} to Σ_{6} are D^{(5,1)} and D^{(3,2,1)} respectively.

The triple lines in (iii) indicate the sum of the extension classes corresponding to single and double lines; namely the element invariant under Σ_{6}/A_{6}.

Sketch of proof : (i) This is clear from Theorem 2.1, since the Specht module S^{(4,2)} lies in a 3-block of defect zero.

(ii) From the exact sequence of Theorem 2.2, and the fact that Λ^{2} of a projective module is projective in odd characteristic (it is a summand of the tensor square), it follows that H^{2}(K_{6}, F_{3}) is the cokernel of the injective hull of the Specht module F_{3}S^{(3,1^{3})}. It is not hard to check that as an F_{3}A_{6}-module,

F_{3}S^{(3,1^{3})} = − / ◁\▷

from which the result follows by the usual method.

(iii) The modules in the sequence of Theorem 2.3, taken mod 3, are periodic, except possibly for the last term, and hence the last term is periodic too. It is not hard to check that the given diagram is the only possibility for a periodic module with the given composition factors. □

We can now compute H^{∗}(A_{6}, H^{j}(K_{6}, F_{3})) using injective resolutions in the usual way, for j = 2 and 3, and keeping track of the action of Σ_{6}/A_{6}. The injective resolutions are as follows.

H^{2}(K_{6}, F_{3}) = [figure: ◇; \; ▷; /; ◁; ◇; ◇; ^{/◁}; ^{\ ▷}]

Ω^{−1}H^{2}(K_{6}, F_{3}) = [figure: ◇; ◇; ◇]

Ω^{−2}H^{2}(K_{6}, F_{3}) = [figure: /; ◁; \; ▷; ◇; ◇; \▷; / ◁]

Ω^{−3}H^{2}(K_{6}, F_{3}) = /◁ \ ▷

Ω^{−4}H^{2}(K_{6}, F_{3}) =  [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷] u_{1}  u_{2}

Ω^{−5}H^{2}(K_{6}, F_{3}) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; v_{1}; v_{2}]

Ω^{−6}H^{2}(K_{6}, F_{3}) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; /; ◁; \; ▷] \begin{matrix}αu_{1} = 0 \\ αu_{2} = 0 \end{matrix}

Ω^{−7}H^{2}(K_{6}, F_{3}) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; \▷; /; ◁; ◇; / ◁; \; ▷; ◇; \; ▷; /; ◁; β_{1}u_{1} = αv_{1}; β_{2}u_{2} = αv_{2}]  \begin{matrix}β_{1}u_{2} = 0 \\ β_{2}u_{1} = 0 \end{matrix}

Ω^{−8}H^{2}(K_{6}, F_{3}) = [figure: ◇; ^{\ ▷}; ^{/◁}; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; β_{1}v_{1}; \begin{matrix}γu_{1} \\ = −β_{2}v_{1} \end{matrix}; \begin{matrix}γu_{2} \\ = β_{1}v_{2} \end{matrix}; β_{2}v_{2}]

Ω^{−9}H^{2}(K_{6}, F_{3}) = [figure: ◇; ◇; ◇; ◇; ◇; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; γv_{1}; γv_{2}]

Ω^{−10}H^{2}(K_{6}, F_{3}) = [figure: ◇; \▷; /; ◁; \; ▷; /; ◁; \; ▷; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; / ◁]

Ω^{−11}H^{2}(K_{6}, F_{3}) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; \; ▷; /; ◁; ◇; ^{\ ▷}; /; ◁; ◇; ^{/◁}; \; ▷; \begin{matrix}αγv_{1} = β_{1}β_{2}v_{1} \\ = −β_{1}γu_{1} \end{matrix}; \begin{matrix}αγv_{2} = β_{1}β_{2}v_{2} \\ = −β_{2}γu_{2} \end{matrix}] \begin{matrix}δ_{1}u_{1} = 0 \\ δ_{1}u_{2} = 0 \\ δ_{2}u_{1} = 0 \\ δ_{2}u_{2} = 0 \end{matrix}

Ω^{−12}H^{2}(K_{6}, F_{3}) =[figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; /; ◁; \; ▷; ◇; ◇; ◇; ◇; \begin{matrix}δ_{1}v_{1} \\ = ε_{1}u_{1} \end{matrix}; \begin{matrix}β_{1}γv_{1} \\ = −ε_{1}u_{2} \end{matrix}; \begin{matrix}β_{2}γv_{1} \\ = δ_{1}v_{2} \\ = −γ^{2}u_{1} \end{matrix}; \begin{matrix}β_{1}γv_{2} \\ = −δ_{2}v_{1} \\ = γ^{2}u_{2} \end{matrix}; \begin{matrix}β_{2}γv_{2} \\ = −ε_{2}u_{1} \end{matrix}; \begin{matrix}δ_{2}v_{2} \\ = ε_{2}u_{2} \end{matrix}]

Ω^{−13}H^{2}(K_{6}, F_{3}) = [figure: ◇; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ◇; ◇; ◇; ◇; \; ▷; /; ◁; ◇; ◇; ◇; ε_{1}v_{1}; γ^{2}v_{1} = ε_{1}v_{2}; γ^{2}v_{2} = ε_{2}v_{1}; ε_{2}v_{2}]

H^{3}(K_{6}, F_{3}) = D^{(2^{2},1^{2})} ⊕

Ω^{−1}H^{3}(K_{6}, F_{3}) =

Ω^{−2}H^{3}(K_{6}, F_{3}) =

Ω^{−3}H^{3}(K_{6}, F_{3}) =

Ω^{−4}H^{3}(K_{6}, F_{3}) =

Ω^{−5}H^{3}(K_{6}, F_{3}) =  ≅ Ω^{−1}H^{3}(K_{6}, F_{3})

Ω^{−6}H^{3}(K_{6}, F_{3}) =  ≅ Ω^{−2}H^{3}(K_{6}, F_{3})

etc.

Proposition 8.3 As a module over H^{∗}(A_{6}, F_{3}), we have (i) H^{∗}(A_{6}, H^{1}(K_{6}, F_{3})) = 0 (ii) H^{∗}(A_{6}, H^{2}(K_{6}, F_{3})) is generated by elements u_{1} and u_{2} in degree 4 and v_{1} and v_{2}

in degree 5, subject to the following relations

αu_{1} = 0  δ_{1}u_{1} = 0  δ_{2}v_{1} = −γ^{2}u_{2}
αu_{2} = 0  δ_{1}u_{2} = 0  δ_{1}v_{2} = −γ^{2}u_{1}
β_{2}u_{1} = 0  δ_{2}u_{1} = 0  ε_{2}v_{1} = γ^{2}v_{2}
β_{1}u_{2} = 0  δ_{2}u_{2} = 0  ε_{1}v_{2} = γ^{2}v_{1}
β_{1}u_{1} = αv_{1} ε_{1}u_{2} = −β_{1}γv_{1}
β_{2}u_{2} = αv_{2} ε_{2}u_{1} = −β_{2}γv_{2}
γu_{1} = −β_{2}v_{1} δ_{1}v_{1} = ε_{1}u_{1}
γu_{2} = β_{1}v_{2}  δ_{2}v_{2} = ε_{2}u_{2}

(iii) H^{∗}(A_{6}, H^{3}(K_{6}, F_{3})) is generated by elements w in degree 1 and x in degree 2 subject to the following relations

\begin{matrix}αw = 0 & αx = 0 & β_{1}x = β_{2}x = −γw \\ β_{1}w = 0 & β_{2}w = 0 & ε_{1}w = ε_{2}w = −γ^{2}w = −δ_{1}x = δ_{2}x \\ δ_{1}w = 0 & δ_{2}w = 0 & ε_{1}x = ε_{2}x = −γ^{2}x \end{matrix}

The action of Σ_{6}/A_{6} is given by

\begin{matrix}u_{1} ↦ u_{2} & v_{1} ↦ v_{2} & w ↦ w \\ u_{2} ↦ u_{1} & v_{2} ↦ v_{1} & x ↦ x \end{matrix} . □

Next we compute the differentials.

Proposition 8.4 In the spectral sequence

′E_{2}^{∗∗} = H^{∗}(A_{6}, H^{∗}(K_{6}, F_{3})) ⇒ H^{∗}(′Γ_{0,0}^{6}, F_{3})

(where ′Γ_{0,0}^{6} is the subgroup of index two in Γ_{0,0}^{6} corresponding to the even permutations) we have

(a) d_{2} = 0

(b) \begin{matrix}d_{3}(u_{1}) = β_{2}γ + δ_{1} & d_{3}(u_{2}) = β_{1}γ + δ_{2} \\ d_{3}(v_{1}) = γ^{2} + ε_{1} & d_{3}(v_{2}) = γ^{2} + ε_{2} \\ d_{3}(w) = 0 & d_{3}(x) = 0 \end{matrix}

(c) d_{4} = 0

Proof The fact that d_{2} = 0 and d_{4} = 0 is easy to see because there is no other possibility. However, by the results of Section 6, there has got to be a non-zero differential, because the elementary abelian subgroup of order 3^{2} in A_{6} does not lift to ′Γ_{0,0}^{6}. It is now not hard to check that there is, up to multiplication by ±1, only one possibility for a non-zero d_{3}. By rechoosing the signs of the generators for ′E_{2}^{∗,2} if necessary, without loss of generality d_{3} is as given. □

By taking invariants under Σ_{6}/A_{6}, we obtain the following.

Proposition 8.5 (i) H^{∗}(Σ_{6}, H^{0}(K_{6}, F_{3})) = H^{∗}(Σ_{6}, F_{3}) is generated by β = β_{1} + β_{2}, γ, δ = δ_{1} + δ_{2} and ε = ε_{1} + ε_{2} subject only to the relations β^{2} = 0 and δ^{2} = 0.

(ii) H^{∗}(Σ_{6}, H^{1}(K_{6}, F_{3})) = 0.

(iii) H^{∗}(Σ_{6}, H^{2}(K_{6}, F_{3})) is generated as a module for H^{∗}(Σ_{6}, F_{3}) by w and x subject to the relations

\begin{matrix}βw = 0 & βx = γw & δw = 0 \\ εw = −δx & (ε − γ^{2})w = 0 & (ε − γ^{2})x = 0 \end{matrix}

(iv) \begin{matrix}d_{2} = 0 & d_{3}(u) = βγ + δ & d_{4} = 0 \\ & d_{3}(v) = ε − γ^{2} \\ & d_{3}(w) = 0 \\ & d_{3}(x) = 0. & □ \end{matrix}

It remains to calculate the ring structure.

Theorem 8.6 The ring H^{∗}(Γ_{0,0}^{6}, F_{3}) = H^{∗}(Γ_{2,0}^{0}, F_{3}) is generated by elements β, γ, w and x with deg(β) = 3, deg(γ) = deg(w) = 4, deg(x) = 5 subject to the relations

\begin{matrix}β^{2} = 0 & βw = 0 & w^{2} = 0 \\ γw = βx & wx = 0 & x^{2} = 0. \end{matrix}

In particular, the Poincaré series is

∑_{n=0}^{∞} t^{n} dim_{F_{3}} H^{n}(Γ_{2,0}^{0}, F_{3}) = \frac{1 + t^{3} + t^{4} + t^{5}}{1 − t^{4}}
= 1 + t^{3} + 2t^{4} + t^{5} + t^{7} + 2t^{8} + t^{9} + t^{11} + ⋯

Proof The generators β, γ, w and x are representatives of the elements of E_{∞} of the same names. The Poincaré series is easy to calculate from E_{∞}, so it remains to verify the exact form of the relations.

The relations β^{2} = 0 and x^{2} = 0 come from graded commutativity. The relation βw = 0 comes from replacing w by w + γ if necessary. The relation γw = βx is true modulo γ^{2} since it is true at E_{∞}. But if γw = βx + γ^{2} then multiplication by β gives βγ^{2} = 0, which is not true in E_{∞}. Hence γw = βx. Thus H^{8} has γw = βx and γ^{2} as a basis, since neither is zero and only one of them is annihilated by multiplication by β. This implies that multiplication by γ^{2} is an isomorphism from H^{8} to H^{16}. But γ^{2}w^{2} = β^{2}x^{2} = 0, so w^{2} = 0. Similarly H^{9} is spanned by γx, and so multiplication by γ is an isomorphism from H^{9} to H^{13}. Since γwx = βx^{2} = 0, this implies that wx = 0. □

References

[1] D. J. Benson and J. F. Carlson. Nilpotent elements in the Green ring. J. Algebra 104, 329–350 (1986).

[2] D. J. Benson and J. F. Carlson. Diagrammatic methods for modular representation theory and cohomology. Comm. in Algebra 15 (1&2) (1987), 53–121.

[3] F. R. Cohen. Artin’s braid group and the homology of certain subgroups of the mapping class group — I. This volume.

[4] L. Evens. On the Chern classes of representations of finite groups. Trans. A. M. S. 115, 180–193 (1965).

[5] G. D. James. The representation theory of the symmetric groups. Springer lecture notes in Mathematics 682, Springer-Verlag, Berlin/New York 1978.

[6] G. D. James and A. Kerber. The representation theory of the symmetric group. Encyclopædia of Mathematics Vol. 16, Addison-Wesley 1981.

[7] D. Quillen. The spectrum of an equivariant cohomology ring, II. Ann. of Math. 94, 573–602, 1971.

Index of Notation

Symbol  Meaning
Σ_{n}  symmetric group of degree n
A_{n}  alternating group of degree n
B_{n}  Artin’s braid group on n strings
P_{n}  the pure braid group; namely the kernel of B_{n} → Σ_{n}
A_{i,j}  basis elements of H^{1}(P_{n}, Z) as a free Z-module

M_{g,k}^{n} surface of genus g with k boundary components and n punctures (only the case k = 0 is considered here, so no confusion should arise between this and the field k)

Γ_{g,k}^{n} mapping class group π_{0}Diff^{+}(M_{g,k}^{n} of orientation preserving self diffeomorphisms fixing the boundary pointwise, but allowing permutations of the punctures

K_{n}  kernel of the map Γ_{0,0}^{n} → Σ_{n}

S^{λ} Specht module for Σ_{n} corresponding to a partition λ

of n

D^{λ} the simple F_{p}Σ_{n}-module corresponding to a p-regular

partition λ

P_{M}  projective cover of a module M
I_{M}  injective hull of a module M
Ω(M)  kernel of the projective cover P_{M} → M
Ω^{−1}(M)  cokernel of the injective hull M → I_{M}

⊗ inner tensor product (over the coefficient ring) of modules with diagonal group action g(m_{1} ⊗ m_{2}) = gm_{1} ⊗ gm_{2}

ˆ⊗ outer tensor product (over the coefficient ring) of modules as a module for the direct product of groups, with action (g_{1}, g_{2})(m_{1} ⊗ m_{2}) = g_{1}m_{1} ⊗ g_{2}m_{2}

k field of coefficients (equal to F_{2} throughout Sections

3, 4 and 5)

Hom_{kG}  kG-module homomorphisms
\underline{Hom}_{kG}

homomorphisms modulo those factoring through a projective module

k, M_{1}, M_{2}, D the simple kA_{6}-modules with k = F_{2}, of dimensions

1, 4, 4, 16 respectively

k′, M′_{1}, M′_{2}, D′ the simple k′A_{6}-modules with k′ = F_{4}, obtained from

the above by extension of scalars

k′, N_{1}, N_{2}, P the simple k′A_{5}-modules with k′ = F_{4}, of dimensions 1, 2, 2, 4 respectively