The mod 2 cohomology of the mapping class group for a surface of genus two

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

F. R. Cohen^{∗}

Department of Mathematics University of Kentucky Lexington, KY 40506 U.S.A.

1 Introduction

This is the third of a sequence of three papers aimed at calculation of the cohomology of the mapping class group Γ_{2,0}^{0} for a closed oriented surface of genus two. In the first two papers(Cohen [3], Benson [1]) it was shown that the cohomology is all 2, 3 and 5-torsion, the mod 3 and mod 5 cohomology were calculated, and partial information was given for the mod 2 cohomology. This paper completes the calculation of the mod 2 cohomology. The main theorem is as follows.

Theorem 1.1 The Poincaré series of the cohomology ring H^{∗}(Γ_{2,0}^{0}, F_{2}) is

(1 + t^{2} + 2t^{3} + t^{4} + t^{5})/(1 − t)(1 − t^{4}) = 1 + t + 2t^{2} + 4t^{3} + 6t^{4} + 7t^{5} + ⋯

The ring structure is the commutative ring generated by elements α, β, δ, ξ, η, ρ and ζ with

deg(δ) = 1, deg(α) = 2, deg(β) = deg(ξ) = 3, deg(η) = deg(ζ) = 4, deg(ρ) = 5

with one of the following two sets of relations.

(i)

α^{2} = δ^{4} ξα = ηδ ρ^{2} = 0
αβ = βδ^{2} ξδ^{3} = ηα ξ^{2} = ζδ^{2}
β^{2} = 0 ξβ = ρδ η^{2} = ζδ^{4}
ηβ = ρα ηξ = ζαδ
ρβ = 0 ρξ = ζβδ
ρα = ρδ^{2} ρη = ζβδ^{2}

∗Partially supported by a National Science Foundation grant

(ii)

α^{2} = δ^{4} ξα = ηδ  ρ^{2} = 0
αβ = βδ^{2} ξδ^{3} = ηα ξ^{2} = ζδ^{2} + ρδ
β^{2} = 0 ξβ = ρδ η^{2} = ζδ^{4} + ρδ^{3}
ηβ = ρα ηξ = ζαδ + ρα
ρβ = 0  ρξ = ζβδ
ρα = ρδ^{2} ρη = ζβδ^{2}

(we have not been able to work out which of these possibilities holds).

The method involved is to compute the cohomology of the mapping class group Γ_{0,0}^{6} of a 2-sphere with six punctures and then to analyse the spectral sequence of the group extension

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

The mod 2 cohomology of Γ_{0,0}^{6} was calculated in Benson [1], the ring structure was calculated, and the action of the Steenrod algebra was calculated but with one ambiguity. This ambiguity left two possibilities for the differentials in the spectral sequence of the above extension.

In this paper, we give an alternative method for computing the additive structure of the cohomology of Γ_{0,0}^{6}. Namely, there is a fibration

F (S^{2}, 6)/Σ_{6} → K(Γ_{0,0}^{6}, 1) →^{π} BSO(3)

given in Bödigheimer, Cohen and Peim [2], where it is also shown that the Serre spectral sequence of this fibration collapses. We find it comforting that the two answers for the additive structure, arrived at by these two very different methods, agree. Moreover, there is enough information, coming from the matching of these two answers, to resolve the ambiguity mentioned above, and hence to calculate the mod 2 cohomology of Γ_{2,0}^{0}. This involves identifying the characteristic class of the extension (∗) (and it is not given by lifting to BSpin(3) as one might guess), and then working out the spectral sequence for (∗).

One curious fact proven here is that the map π is not homotopic to B(g), for any group homomorphism g : Γ_{0,0}^{6} → SO(3).

Finally, it should be pointed out that in Section 2 we provide an explicit construction for the space K(Γ_{0,0}^{n}, 1) for n ≥ 3. It is a modification of a Grassmannian.

2 A construction for K(Γ_{0,0}^{n}, 1), characteristic classes

In Bödigheimer, Cohen and Peim [2] a direct construction was given for K(Γ_{0,0}^{n}, 1). We start this section by giving a more concrete description of this space. Recall that the Stiefel manifold V_{3}(R^{N}) of orthonormal 3-frames in R^{N} admits an action of SO(3) with quotient the Grassmannian manifold of oriented 3-planes in R^{N}, G_{3}(R^{N}) = V_{3}(R^{N})/SO(3). Each V_{3}(R^{N}) embeds via the first N coordinates in V_{3}(R^{N+1}), and the union ESO(3) = V_{3}(R^{∞}) (with the weak topology of the limit) is contractible so that BSO(3) = G_{3}(R^{∞}) is the classifying space of SO(3).

If we embed SO(2) in SO(3) as the stabiliser of a unit vector in R^{3}, then the quotient space BSO(2) = V_{3}(R^{N})/SO(2) is the set of pairs consisting of an oriented 3-plane V in R^{N} and a unit vector in V.

Denote by A_{n,3}(R^{N}) the set of pairs consisting of a 3-plane V in R^{N} and an ordered set of n distinct unit vectors in V, topologized as a subset of (V_{3}(R^{N})/SO(2))^{n}. Then Σ_{n} acts on A_{n,3}(R^{N}), each A_{n,3}(R^{N}) embeds via the first N coordinates in A_{n,3}(R^{N+1}), and we write A_{n,3}(R^{∞}) for the union (with the weak topology). Thus A_{n,3}(R^{∞})/Σ_{n} is the set of pairs consisting of an oriented 3-plane V in R^{∞} and an unordered set of n distinct unit vectors in V.

Theorem 2.1 Let F (S^{2}, n) denote the configuration space of ordered n-tuples of distinct points in S^{2}. Then for n ≥ 3, SO(3) acts freely on F (S^{2}, n), and the quotient is a classifying space BK_{n} = K(K_{n}, 1) for the kernel of the natural map Γ_{0,0}^{n} → → Σ_{n}.

For n ≥ 3, the space A_{n,3}(R^{∞})/Σ_{n} is a classifying space BΓ_{0,0}^{n} = K(Γ_{0,0}^{n}, 1).

Proof The second part of this theorem is essentially proved in Bödigheimer, Cohen and Peim [2], but we give here a more elementary proof which goes via the first part.

If n ≥ 3, then it is clear that SO(3) acts freely on F (S^{2}, n), since each non-identity element of SO(3) fixes exactly two points on S^{2}. The quotient F (S^{2}, n)/SO(3) is the same as F (R^{2}, n − 1)/SO(2). It is well known that F (R^{2}, n − 1) is a classifying space for the pure braid group P_{n−1} on n − 1 strings. We have a fibration

SO(2) → F (R^{2}, n − 1) → F (R^{2}, n − 1)/SO(2) = F (S^{2}, n)/SO(3).

The generator of π_{1}(SO(2)) ≅ Z goes to a full twist of all the strings, which is a generator for the centre of P_{n−1}. In particular the map π_{1}(SO(2)) → π_{1}F (R^{2}, n − 1) = P_{n−1} is injective, and so the long exact homotopy sequence shows that F (S^{2}, n)/SO(3) has no higher homotopy.

To show that π_{1}(F (S^{2}, n)/SO(3)) is the “pure” mapping class group K_{n}, we use the following argument. The inclusion SO(3) ↪ Diff^{+}(S^{2}) is a homotopy equivalence, and so the coset space Diff^{+}(S^{2})/SO(3) is contractible. We have a fibration

Diff^{+}(S^{2}, n) → Diff^{+}(S^{2})/SO(3) → F (S^{2}, n)/SO(3),

where Diff^{+}(S^{2}, n) denotes the group of orientation preserving self-diffeomorphisms of S^{2} fixing n points pointwise. Thus

π_{1}(F (S^{2}, n)/SO(3)) ≅ π_{0}Diff^{+}(S^{2}, n) = K_{n}.

This completes the proof of the first part of the theorem. Note that this also proves the following well known facts:

(i) K_{n} ≅ P_{n−1}/Z(P_{n−1})

(ii) The connected components of Diff^{+}(S^{2}, n) are contractible.

Turning to the second part of the theorem, we note that Σ_{n} acts freely on F (S^{2}, n) and commutes with the action of SO(3), but the induced action on the quotient is not free. We can free up this action by replacing F (S^{2}, n)/SO(3) by the homotopy equivalent space ESO(3) ×_{SO(3)} F (S^{2}, n). It follows that the quotient is a classifying space for Γ_{0,0}^{n}

ESO(3) ×_{SO(3)} F (S^{2}, n)/Σ_{n} = K(Γ_{0,0}^{n}, 1).

Using the usual Stiefel variety V_{3}(R^{∞}) as our model for ESO(3), it is clear that the above space is the same as the space A_{n,3}(R^{∞}) described above. □

Theorem 2.2 For n ≥ 6, the fibration π : BΓ_{0,0}^{n} → BSO(3) given above is not homotopic to B(g) for any group homomorphism g : Γ_{0,0}^{n} → SO(3).

Proof Recall that Γ_{0,0}^{n} is generated by elements σ_{1}, …, σ_{n−1} satisfying braid relations σ_{i}σ_{i+1}σ_{i} = σ_{i+1}σ_{i}σ_{i+1} and σ_{i}σ_{j} = σ_{j}σ_{i} if |i − j| ≥ 2 (together with further relations). In particular the σ_{i} are all conjugate since (σ_{i}σ_{i+1})σ_{i}(σ_{i}σ_{i+1})^{−1} = σ_{i+1}. If g : Γ_{0,0}^{n} → SO(3) is a homomorphism with B(g) = π then g does not factor through the abelianisation of Γ_{0,0}^{n}, since the exact sequence 1 → Z/2 → π_{1}F (S^{2}, n)/Σ_{n} → Γ_{0,0}^{n} → 1, when restricted to the commutator subgroup of Γ_{0,0}^{n}, is not a product. Indeed, the preimage of the dihedral group of order 2n if n is odd and n if n is even, acting on n equatorial points with one or two orbits, is not a product. Note that this preimage is the same as the preimage in SU(2). Thus g(σ_{i}) ≠ 1, g(σ_{i}) is conjugate to g(σ_{j}), and g(σ_{i}) commutes with g(σ_{j}) if |i − j| ≥ 2.

Since the g(σ_{i}) are conjugate elements of SO(3), they are rotations of the same angle θ about axes x_{i}. If θ ≠ 180^{o} then the centraliser in SO(3) of g(σ_{i}) is the group of all rotations about the axis x_{i}. Thus x_{3} = x_{1} = x_{4} and so g(σ_{3}) commutes with g(σ_{4}). But this forces g(σ_{3}) = g(σ_{4}), and hence all the g(σ_{i}) are equal (for example g(σ_{2})g(σ_{4})g(σ_{2}) = g(σ_{4})g(σ_{2})g(σ_{4}) so that g(σ_{2}) = g(σ_{4})) contradicting the fact that g does not factor through the abelianisation of Γ_{0,0}^{n}. Thus θ = 180^{o} and the centraliser of g(σ_{1}) is O(2). Since g(σ_{3}), g(σ_{4}) and g(σ_{5}) are all in this centraliser and satisfy braid relations, x_{3}, x_{4} and x_{5} are all perpendicular to x_{1} and are at 60^{o} to each other. But then g(σ_{3}) does not commute with g(σ_{5}). This contradiction completes the proof. □

Next, recall from Cohen [3] that there is a non-split central extension

1 → Z/2 → Δ_{g} → Γ_{0,0}^{2g+2} → 1

with Δ_{g} ⊆ Γ_{g,0}^{0} and Δ_{2} = Γ_{2,0}^{0}. At the level of classifying spaces this gives a non-trivial fibration

RP^{∞} → BΔ_{g} → BΓ_{0,0}^{2g+2}.

We now work out the characteristic class χ_{g} ∈ H^{2}(Γ_{0,0}^{2g+2}, F_{2}) of this extension. Recall the following theorem from Bödigheimer, Cohen and Peim [2].

Theorem 2.3 The Serre spectral sequence

H^{∗}(BSO(3), H^{∗}(F (S^{2}, n)/Σ_{n}, F_{2})) ⇒ H^{∗}(BΓ_{0,0}^{n}, F_{2})

collapses at E_{2} when n is even. Thus there is an isomorphism of H^{∗}(BSO(3), F_{2})-modules

H^{∗}(BΓ_{0,0}^{2q}, F_{2}) ≅ H^{∗}(BSO(3), F_{2}) ⊗ H^{∗}(F (S^{2}, 2q)/Σ_{2q}, F_{2})

and H^{∗}(F (S^{2}, 2q)/Σ_{2q}, F_{2}) ≅ H^{∗}(F (R^{2}, 2q)/Σ_{2q} ∨ Σ^{2}(F (R^{2}, 2q − 1)/Σ_{2q−1})_{+}, F_{2}). □

Thus H^{1}(Γ_{0,0}^{2q}, F_{2}) is one dimensional, and we write x_{1} for its unique non-zero element. It is well known that

H^{∗}(BSO(3), F_{2}) = F_{2}[w_{2}, w_{3}]

where w_{2}, w_{3} are the Stiefel-Whitney classes of degrees 2 and 3 respectively and Sq^{1}(w_{2}) = w_{3}, Sq^{2}(w_{3}) = w_{2}w_{3} (Wu relations).

Proposition 2.4 The Stiefel-Whitney class w_{2} is not in the image of the inflation map

H^{∗}(Σ_{2g+2}, F_{2}) → H^{∗}(Γ_{0,0}^{2g+2}, F_{2}).

If g = 2 then χ_{g} = w_{2} + x_{1}^{2}.

Proof Recall from Cohen [3] that there are maps

B_{2g+2} → → B̄_{2g+2} → → Δ_{g} → → Γ_{0,0}^{2g+2} → → Σ_{2g+2}

where B_{2g+2} is the braid group on 2g +2 strings, B̄_{2g+2} = B_{2g+2}/(centre) and the composite map is the natural epimorphism. It is also shown in Cohen-Lada-May [4, p. 226–227] that inflation H^{∗}(Σ_{2g+2}, F_{2}) → H^{∗}(B_{2g+2}, F_{2}) is an epimorphism. But

H_{1}(B̄_{2g+2}, Z) = Z/(2g + 2)(2g + 1)

and thus if g is even, H_{1}(B̄_{2g+2}, Z) ≅ Z/2 ⊕ (odd torsion). Thus as observed in Cohen [3], H^{∗}(Σ_{2g+2}, F_{2}) → H^{∗}(B̄_{2g+2}, F_{2}) is an epimorphism for g even.

Next, notice that if BΓ_{0,0}^{2g+2} → K(Z/2, 2) is the map representing the characteristic class χ_{g}, then the composite map

B(B_{2g+2}) → BΓ_{0,0}^{2g+2} → K(Z/2, 2)

is null (because B_{2g+2} → Γ_{0,0}^{2g+2} factors through Δ_{g}) and so we have a lifting

F (R^{2}, 2g + 2)/Σ_{2g+2} ⊆ F (S^{2}, 2g + 2)/Σ_{2g+2}
↓  (†)
BΔ_{g}  →  BΓ_{0,0}^{2g+2}  → K(Z/2, 2)

We now claim that H^{2}(Σ_{2g+2}, F_{2}) → H^{2}(F (S^{2}, 2g + 2)/Σ_{2g+2}, F_{2}) is an isomorphism. Both groups are two dimensional over F_{2}. We have maps

B(B_{2g+2}) = F (R^{2}, 2g+2)/Σ_{2g+2} → F (S^{2}, 2g+2)/Σ_{2g+2} → BΣ_{2g+2} = F (S^{∞}, 2g+2)/Σ_{2g+2}.

There is a class in H^{2}(Σ_{2g+2}, F_{2}) which restricts to a non-zero element in H^{2}(B_{2g+2}, F_{2}). Moreover, Sq^{1} acts trivially on the group H^{1}(B_{2g+2}, F_{2}) and non-trivially on the group H^{1}(F (S^{2}, 2g +2)/Σ_{2g+2}, F_{2}), and the map H^{1}(Σ_{2g+2}, F_{2}) → H^{1}(F (S^{2}, 2g +2)/Σ_{2g+2}, F_{2}) is an isomorphism. This proves the claim. Thus the first part of the proposition follows; w_{2} and the image of inflation H^{2}(Σ_{2g+2}, F_{2}) → H^{2}(Γ_{0,0}^{2g+2}, F_{2}) generate H^{2}(Γ_{0,0}^{2g+2}, F_{2}), which is three dimensional over F_{2}.

Finally, we show that χ_{g} = w_{2} + x_{1}^{2} in case g = 2. Notice that χ_{g} is a linear combination of w_{2}, x_{1}^{2} and x_{2}, where x_{2} is any element which restricts to a generator of H^{2}(F (R^{2}, 2g + 2)/Σ_{2g+2}, F_{2}). But by the above, w_{2} and x_{1}^{2} restrict to zero in H^{2}(F (R^{2}, 2g + 2)/Σ_{2g+2}, F_{2}) and so the coefficient of x_{2} is zero by the diagram (†). Since the extension is non-split, we have χ_{g} = Aw_{2} + Bx_{1}^{2} ≠ 0. So if we assume that A = 0, then B = 1. But then this contradicts Theorem 1.4 of Cohen [3], as the dimensions of H^{2}(Σ_{2g+2}, F_{2}) and H^{2}(B̄_{2g+2}, F_{2}) are both two. Hence we have A = 1. If B = 0, then there is a diagram of fibrations

F (S^{2}, 2g + 2)/Σ_{2g+2} → F (S^{2}, 2g + 2)/Σ_{2g+2} →  ∗
↓  ↓  ↓
BΔ_{g}  →  BΓ_{0,0}^{2g+2}  → K(Z/2, 2)
↓  ↓  ↓
BS^{3}  →  BSO(3)  → K(Z/2, 2)

Now H^{∗}(BS^{3}, F_{2}) = H^{∗}(BSU(2), F_{2}) = F_{2}[c_{2}] with c_{2} the second Chern class, of degree four. Thus using the Serre spectral sequence

H^{∗}(BS^{3}, H^{∗}(F (S^{2}, 2g + 2)/Σ_{2g+2}, F_{2})) ⇒ H^{∗}(BΔ_{g}, F_{2})

and Theorem 2.3 we have

dim H^{5}(Δ_{g}, F_{2}) ≤ dim H^{5}(F (S^{2}, 2g + 2)/Σ_{2g+2}, F_{2})
+ dim H^{1}(F (S^{2}, 2g + 2)/Σ_{2g+2}, F_{2})
= dim H^{5}(F (R^{2}, 2g + 2)/Σ_{2g+2}, F_{2})
+ dim H^{3}(F (R^{2}, 2g + 1)/Σ_{2g+1}, F_{2})
+ dim H^{1}(F (R^{2}, 2g + 2)/Σ_{2g+2}, F_{2})
= 2 if g = 2

since the Poincaré series for the mod 2 cohomology of F (S^{2}, 6)/Σ_{6} is (1 + t + 2t^{2} + 3t^{3} + 2t^{4} + t^{5}). However,

dim H^{5}(B̄_{6}, F_{2}) = 3,

contradicting the surjectivity of the inflation H^{∗}(Δ_{2}, F_{2}) → H^{∗}(B̄_{6}, F_{2}). Thus B = 1 and χ_{g} = w_{2} + x_{1}^{2}. □

3 Steenrod operations on H^{∗}(F (S^{2}, 6)/Σ_{6}, F_{2})

In this section we work out the Steenrod operations in the mod 2 cohomology of F (S^{2}, 6)/Σ_{6}. Our method is to construct a map

h : F (S^{2}, 6)/Σ_{6} → Ω^{∞}Σ^{∞}RP^{2}

which is onto in mod 2 cohomology. As the Steenrod operations in the mod 2 cohomology of Ω^{∞}Σ^{∞}RP^{2} are well understood, the required results follow.

The main calculation needed here is the following theorem.

Theorem 3.1 If x_{1} is the generator of H^{1}(F (S^{2}, 6)/Σ_{6}, F_{2}) ≅ F_{2}, then x_{1}^{4} = 0.

It is this theorem which enables us to resolve the ambiguity in the Steenrod operations on H^{∗}(Γ_{0,0}^{6}, F_{2}) given in Benson [1] and hence determine the differentials in the spectral sequence for (∗).

First we list the elements in the mod 2 homology of F (S^{2}, 6)/Σ_{6} and Ω^{∞}Σ^{∞}RP^{2}.

Degree  1 2  3  4  5
Basis for  e_{1} e_{1}^{2} e_{1}^{3} e_{1}.Q_{1}e_{1}
H_{∗}(F (S^{2}, 6)/Σ_{6}, F_{2})  Q_{1}e_{1}
z_{2} z_{3}  z_{4}  z_{5}
Basis for  f_{1} f_{1}^{2} f_{1}^{3}  f_{1}^{4}  f_{1}^{5}
H_{∗}(Ω^{∞}Σ^{∞}RP^{2}, F_{2})  f_{2} f_{1}f_{2} f_{1}^{2}f_{2}  f_{1}^{3}f_{2}
in degree ≤ 5  Q_{1}f_{1} f_{1}.Q_{1}f_{1} f_{1}^{2}.Q_{1}f_{1}
f_{2}^{2}  f_{1}.f_{2}^{2}
Q_{2}f_{1} f_{1}.Q_{2}f_{1}
Q_{3}f_{1}
Q_{1}f_{2}

In this chart, the following notation is used. The Q_{i} denote Dyer-Lashof operations. The monomials in e_{1} provide a basis for the image of the monomorphism

H_{∗}(F (R^{2}, 6)/Σ_{6}, F_{2}) → H_{∗}(F (S^{2}, 6)/Σ_{6}, F_{2}).

The elements z_{i} run over any choice of elements giving a basis for the reduced homology of the cofibre. A basis for the reduced homology of RP^{2} is given by f_{1} and f_{2}. The rest of H_{∗}(Ω^{∞}Σ^{∞}RP^{2}, F_{2}) is given in Cohen-Lada-May [4]. In particular, we have the following information about the action of the Steenrod operations. We have

Sq^{1}_{∗}e_{1} = Sq^{1}_{∗}e_{1}^{2} = Sq^{1}e_{1}^{3} = 0
Sq^{1}_{∗}Q_{1}e_{1} = e_{1}^{2}, Sq^{1}_{∗}e_{1}.Q_{1}e_{1} = e_{1}^{3},

and Sq^{2}_{∗} annihilates H_{∗}(F (R^{2}, 6)/Σ_{6}, F_{2}). We have

Sq^{1}_{∗}f_{1}^{2} = 0, Sq^{1}_{∗}f_{2} = f_{1}, Sq^{2}_{∗}(f_{2})^{2} = f_{1}^{2}

and Sq^{2}_{∗} annihilates all other basis elements in H_{4}(Ω^{∞}Σ^{∞}RP^{2}, F_{2}).

Thus in the dual basis for H^{∗}(Ω^{∞}Σ^{∞}RP^{2}, F_{2}) we have (f_{1}^{∗})^{2} = Sq^{1}f_{1}^{∗} = f_{2}^{∗} ≠ 0, while (f_{1}^{∗})^{4} = Sq^{2}f_{2}^{∗} = 0. Finally, Sq^{1}_{∗}z_{2} = e_{1} in H_{1}(F (S^{2}, 6)/Σ_{6}, F_{2}) as H_{1}(F (S^{2}, 6)/Σ_{6}, Z) ≅ Z/10 by abelianising π_{1}F (S^{2}, 6)/Σ_{6}.

Now Theorem 3.1 follows from the following lemma.

Lemma 3.2 There is a map h : F (S^{2}, 6)/Σ_{6} → Ω^{∞}Σ^{∞}RP^{2} which induces an isomorphism on degree one cohomology with F_{2}-coefficients.

Proof Consider the pretransfer to get a commutative diagram

F (R^{2}, 6)/Σ_{6} →^{i} EΣ_{15} ×_{Σ_{15}} (F (R^{2}, 2)/Σ_{2})^{15} → Ω^{∞}Σ^{∞}F (R^{2}, 2)/Σ_{2}
↓  ↓  ↓
F (S^{2}, 6)/Σ_{6} →^{i} EΣ_{15} ×_{Σ_{15}} (F (S^{2}, 2)/Σ_{2})^{15} → Ω^{∞}Σ^{∞}F (S^{2}, 2)/Σ_{2}
↓  ↓  ↓
BΣ_{6}  →^{i}  EΣ_{15} ×_{Σ_{15}} (RP^{∞})^{15}  →  Ω^{∞}Σ^{∞}RP^{∞}

where i is given as follows. There are \binom{k}{2} projection maps X^{k} → X^{2}/Σ_{2}. Thus we get

F (X, k)/Σ_{k} → EΣ\binom{k}{2} ×_{Σ_{\binom{k}{2}}} (F (X, 2)/Σ_{2})^{\binom{k}{2}}.

Next notice that F (S^{2}, 2)/Σ_{2} is homotopy equivalent to RP^{2}. Finally, Kahn and Priddy [5] have shown that the composite map

BΣ_{k} → EΣ\binom{k}{2} ×_{Σ_{\binom{k}{2}}} (RP^{∞})^{\binom{k}{2}} → Ω^{∞}Σ^{∞}RP^{∞}

is a monomorphism in homology. This suffices to prove the lemma. □

4 Steenrod operations on H^{∗}(Γ_{0,0}^{6}, F_{2})

In this section, we resolve the ambiguity in the action of the Steenrod operations on H^{∗}(Γ_{0,0}^{6}, F_{2}) given in Benson [1], by comparing the calculations made there with the calculations made in Sections 2 and 3 of this paper.

Recall from Benson [1, Theorem 7.3] that 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.

Now by Theorem 2.3 we have an injective map

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

and we shall write w_{2} and w_{3} for the images in H^{∗}(Γ_{0,0}^{6}, F_{2}) of the Stiefel-Whitney classes with the same names in H^{∗}(BSO(3), F_{2}). Lemma 4.1 w_{2} = u + α + δ^{2} ∈ H^{2}(Γ_{0,0}^{6}, F_{2}). Proof Since H^{1}(Γ_{0,0}^{6}, F_{2}) is one dimensional we have δ = x_{1}. By Proposition 2.4 we have χ_{g} = w_{2} +x_{1}^{2} and by the definition of u ∈ H^{2}(Γ_{0,0}^{6}, F_{2}) in the paragraph before Lemma 7.2 of Benson [1] we have χ_{g} = u + α. Hence we have w_{2} = u + α + δ^{2} as required. □

We now examine the two possibilities for Sq^{1}(u) in detail.

Case 1 Sq^{1}(u) = δu + αδ + β. In this case we have

w_{3} = Sq^{1}(w_{2}) = Sq^{1}(u + α + δ^{2}) = δu + αδ + γ,

and so by Theorem 2.3

H^{∗}(F (S^{2}, 6)/Σ_{6}, F_{2}) = H^{∗}(Γ_{0,0}^{6}, F_{2}) ⊗_{F_{2}[w_{2},w_{3}]} F_{2}

has a basis as follows

degree  1 2 3 4 5
basis element δ δ^{2} δ^{3} αδ^{2} βδ^{2}
α β βδ
αδ

and in particular δ^{4} = δ^{2}w_{2} + δw_{3} ≡ 0 in H^{∗}(F (S^{2}, 6)/Σ_{6}, F_{2}), thereby satisfying Theorem 3.1.

Case 2 Sq^{1}(u) = δu + δ^{3}. In this case we have

w_{3} = Sq^{1}(w_{2}) = δu + δ^{3} + β + γ,

and so by Theorem 2.3

H^{∗}(F (S^{2}, 6)/Σ_{6}, F_{2}) = H^{∗}(Γ_{0,0}^{6}, F_{2}) ⊗_{F_{2}[w_{2},w_{3}]} F_{2}

has a basis as follows

degree  1 2 3 4 5
basis element δ δ^{2} δ^{3} δ^{4} αδ^{3}
α β βδ
γ

and in particular δ^{4} ≠ 0 in H^{∗}(F (S^{2}, 6)/Σ_{6}, F_{2}), contradicting Theorem 3.1.

Thus we have proved the following. Proposition 4.2 Sq^{1}(u) = δu + αδ + β in H^{3}(Γ_{0,0}^{6}, F_{2}). □

5 The spectral sequence for Γ_{2,0}^{0}

Now that we know the action of the Steenrod algebra on H^{∗}(Γ_{0,0}^{6}, F_{2}), we can work out the differentials in the Lyndon-Hochschild-Serre spectral sequence for Γ_{2,0}^{0}.

Proposition 5.1 The differentials in the spectral sequence

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

are given by

d_{2}(z) = u + α
d_{3}(z^{2}) = γ
d_{n} = 0 (n ≥ 4)

where z is the generator in degree one of H^{∗}(Z/2, F_{2}).

Proof Since u was chosen so that the extension 1 → Z/2 → Γ_{2,0}^{0} → Γ_{0,0}^{6} → 1 corresponds to u + α ∈ H^{2}(Γ_{0,0}^{6}, F_{2}), we have d_{2}(z) = u + α. Hence by Proposition 4.2 we have

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

which is equal to γ in E_{3}. Finally

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

which is zero in E_{5} and so E_{4} = E_{∞}. □

Here is a picture of E_{4} = E_{∞} in this spectral sequence.

z^{4} • •  •  ⋯
z^{4}δ z^{4}α
z^{4}δ^{2}
•  •  •  •  ⋯
z^{2}δ z^{2}α z^{2}β z^{2}βδ
z^{2}δ^{2} z^{2}αδ z^{2}αδ^{2}
z^{2}δ^{3} z^{2}δ^{4}
• •  •  •  •  •  •
δ  α  β  βδ  αβ = βδ^{2} βδ^{3}
δ^{2} αδ  αδ^{2}  αδ^{3}  αδ^{4}
δ^{3} α^{2} = δ^{4}  δ^{5}  δ^{6}
β^{2} = 0

We now attack the problem of determining the product structure in H^{∗}(Γ_{2,0}^{0}, F_{2}). We shall see that there are two possible answers, and we have not been able to determine which is correct.

Proof of Theorem 1.1 We first observe that multiplication by δ is injective on E_{∞} and hence on H^{∗}(Γ_{2,0}^{0}, F_{2}).

Choose a representative ξ ∈ H^{3}(Γ_{2,0}^{0}, F_{2}) of the element z^{2}δ ∈ E_{∞}^{1,2}. Then there is a unique element η ∈ H^{4}(Γ_{2,0}^{0}, F_{2}) representing z^{2}α in E_{∞}^{2,2} such that ξα + ηδ = 0. Since (ξδ^{3} + ηα)δ = ξα^{2} + ηαδ = 0 we have ξδ^{3} = ηα.

There is now a unique element ρ ∈ H^{5}(Γ_{2,0}^{0}, F_{2}) representing z^{2}β ∈ E_{∞}^{3,2} and satisfying ξβ = ρδ. We have

(ηβ + ρα)δ = (ηδ)β + (ρδ)α = (ξα)β + (ξβ)α = 0

and hence ηβ = ρα. Next,

(ρβ)δ^{2} = (ρδ)βδ = ξβ^{2}δ = 0

and hence ρβ = 0. We have

ρ(α + δ^{2})δ = ξβ(α + δ^{2}) = 0

and hence ρα = ρδ^{2}. This completes the calculation of products between representatives of E_{∞}^{∗,2} and E_{∞}^{∗,0}, and we now calculate the products among the elements ξ, η and ρ.

We have ρ^{2}δ^{2} = ξ^{2}β^{2} = 0 and hence ρ^{2} = 0. Next, we turn to the choice of a representative ζ ∈ H^{4}(Γ_{2,0}^{0}, F_{2}) for the element z^{4} ∈ E_{∞}^{4,0}. For any such choice, the element ξ^{2} + ζδ^{2} is in the linear span of the elements ρδ, ηδ^{2}, ξδ^{3}, βδ^{3}, αδ^{4}, δ^{6}. Thus by altering our choice of ζ by a suitable linear combination of η, ξδ, βδ, αδ^{2}, δ^{4}, we can ensure that ξ^{2} + ζδ^{2} is either zero or equal to ρδ. This is the origin of the ambiguity in Theorem 1.1, and so we now separate into two cases.

Case 1 ξ^{2} = ζδ^{2}.

In this case we have

η^{2}δ^{2} = ξ^{2}α^{2} = (ζδ^{2})δ^{4}

and so η^{2} = ζδ^{4}. Next,

ηξδ = ξ^{2}α = ζαδ^{2}

and so ηξ = ζαδ. We have

ρξδ = ξ^{2}β = ζδ^{2}β

and so ρξ = ζβδ. Finally we have

ρηδ^{2} = (ρδ)(ηδ) = (ξβ)(ξα) = ζδ^{2}αβ = ζβδ^{4}

and so ρη = ζβδ^{2}.

Case 2 ξ^{2} = ζδ^{2} + ρδ.

In this case we have

η^{2}δ^{2} = ξ^{2}α^{2} = (ζδ^{2} + ρδ)δ^{4}

and so η^{2} = ζδ^{4} + ρδ^{3}. Next,

ηξδ = ξ^{2}α = (ζδ^{2} + ρδ)α

and so ηξ = ζαδ + ρα. We have

ρξδ = ξ^{2}β = (ζδ^{2} + ρδ)β = ζδ^{2}β

and so ρξ = ζβδ. Finally we have

ρηδ^{2} = (ρδ)(ηδ) = (ξβ)(ξα) = (ζδ^{2} + ρδ)αβ = ζβδ^{4}

and so ρη = ζβδ^{2}.

This completes the proof of Theorem 1.1. □

References

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

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