THE HOMOLOGY OF INVARIANT GROUP CHAINS

KEVIN P. KNUDSON

Introduction

Let Q be a finite group acting on a group G as a group of automorphisms. This action induces an action of Q on the standard bar complex C_{•}(G) for computing the homology of G; denote the subcomplex of invariants by C_{•}(G)^{Q}. In this paper we study the homology of the complex C_{•}(G)^{Q}. The resulting homology groups are denoted H_{•}^{Q}(G).

Before summarizing the results, a brief history is in order. A quick literature search and consultation with experts in equivariant topology revealed nothing. Indeed, everyone I asked was more or less alarmed that I wanted to know about invariants of a chain complex (instead of a cochain complex). So I forged ahead with the cumbersome calculations using the bar complex and obtained a few results. For example, if A is an abelian group with trivial G- and Q-actions, then we can define homology groups with coefficients H_{•}^{Q}(G; A). The inclusion of complexes C_{•}(G; A)^{Q} → C_{•}(G; A) induces a homomorphism H_{•}^{Q}(G; A) → H_{•}(G; A) whose image clearly lies in the subgroup H_{•}(G; A)^{Q} of Q-invariant homology classes. The following is easily proved.

Proposition 3.3. Suppose that |Q| is invertible in A. Then the natural map

H_{•}^{Q}(G; A) ⟶ H_{•}(G; A)^{Q}

is an isomorphism.

In certain cases, we can construct transfer maps

H_{•}^{Q}(G; A) ⟶ H_{•}^{Q}(K; A)

for Q-stable subgroups K, satisfying the usual properties. One then obtains the following expected result.

Proposition 4.1. If G is finite, then H_{i}^{Q}(G; A) is annihilated by |G| for all i > 0.

The existence of transfers also allows the calculation of H_{•}^{ℤ/2}(ℤ/n) for certain n, where ℤ/2 acts on ℤ/n by x ↦ −x.

Date: September 15, 2003.

1991 Mathematics Subject Classification. Primary 20J06; Secondary 55N91.

Partially supported by NSF grant no. DMS-0242906.

Theorem 5.1. If n ≢ 0 mod 4, then

H_{•}^{ℤ/2}(ℤ/n; ℤ) ≅ H_{•}(ℤ/n; ℤ)^{ℤ/2}.

The case n ≡ 0 mod 4 presented difficulty, however. None of the techniques used to prove Theorem 5.1 work in this case. Moreover, it is easy to show that H_{1}^{ℤ/2}(ℤ/n) ≅ ℤ/2 ⊕ ℤ/2 in this case, and computer calculations of the low-dimensional groups for n = 4 indicated that something fairly complicated was going on. So I put this project on the back burner in hopes that a new idea would present itself.

The primary motivation for studying the H_{•}^{Q}(G) arose from joint work with Mark Walker [3]. There, we were confronted with calculating the groups H_{•}(BG(\overline{R})/Γ), where R is a nice affine domain over a field k, \overline{R} its absolute integral closure, G an algebraic group over k, and Γ the Galois group Gal(\overline{k(R)}/k(R)). The calculation is practically impossible in general, but we were able to handle R = ℝ, \overline{R} = ℂ, G = 𝔾_{m}, and Γ = Gal(ℂ/ℝ) = ℤ/2. For some time, I worked under the mistaken assumption that the complex C_{•}(G(\overline{R}))^{Γ} computed this homology (there is an isomorphism of the groups of chains in each degree, but it does not commute with the boundary map). However, the work in [3] did provide some insight into the problem considered here.

Consider the norm map N : C_{•}(BG/Q; A) ⟶ C_{•}(G; A)^{Q} defined by

σ ⊗ a ↦^{N} ∑_{q∈Q} qσ ⊗ a.

This provides a map

H_{•}(BG/Q; A) ⟶ H_{•}^{Q}(G; A)

whose composition with the obvious map

H_{•}^{Q}(G; A) ⟶ H_{•}(G; A)^{Q}

is the transfer map encountered in the study of the homology of quotient spaces. The homology of the quotient complex

D_{•} = C_{•}(G; A)^{Q}/C_{•}(BG/Q; A)

is computable in some cases. We have the following results.

Theorem 5.8. Let ℤ/2 act on ℤ via n ↦ −n. Then for i > 0, we have

H_{i}^{ℤ/2}(ℤ; ℤ) ≅ \begin{cases}ℤ/2 & i odd \\ 0 & i even. \end{cases}

One might have thought that H_{i}^{ℤ/2}(ℤ) would vanish for i ≥ 2 since Bℤ ≃ S^{1}. The problem arises, however, because S^{1} is not a ℤ/2-equivariant model for Bℤ.

If n ≡ 0 mod 4, the calculation of H_{•}^{ℤ/2}(ℤ/n) is possible using techniques in equivariant topology. One first reduces to n = 2^{s}, s ≥ 2.

Theorem 5.7. Assume s ≥ 2. Then for all ℓ > 0,

H_{ℓ}^{ℤ/2}(ℤ/2^{s}; ℤ) ≅ \begin{cases}(ℤ/2)^{k} & ℓ = 2k \\ (ℤ/2)^{2k} & ℓ = 4k − 3 \\ ℤ/2^{s} ⊕ (ℤ/2)^{2k} & ℓ = 4k − 1. \end{cases}

This paper is organized as follows. In Section 1 we develop the basic properties of H_{•}^{Q}(G; A) and the cohomological version H_{Q}^{•}(G; A). Section 2 relates our construction to the homology of quotient spaces. In Section 3 we make some basic calculations. Section 4 is devoted to the transfer maps and their application. Finally, in Section 5 we perform the above-mentioned calculations. The approach to calculations is to be as low-tech as possible for as long as possible; that is, we try to work with chains directly, resorting to more sophisticated techniques only when absolutely necessary.

Acknowledgements. I would like to thank Paul Goerss, John Greenlees, and Mark Walker for valuable discussions about this material. I also thank those who have listened to my talks about this for their suggestions; these include Eric Friedlander, Rick Jardine, and Stewart Priddy.

1. Basic Constructions

Denote by C_{•}(G) the bar complex for computing the homology of the group G:

C_{i}(G) = ℤ{[g_{1}| ⋯ |g_{i}] : g_{j} ∈ G}
d([g_{1}| ⋯ |g_{i}]) =

[g_{2}| ⋯ |g_{i}] + ∑_{k=1}^{i−1}(−1)^{k}[g_{1}| ⋯ |g_{k}g_{k+1}| ⋯ |g_{i}] + (−1)^{i}[g_{1}| ⋯ |g_{i−1}]

(that is, C_{•}(G) = B_{•} ⊗_{G} ℤ, where B_{•} is the bar resolution). If A is an abelian group, set C_{•}(G; A) = C_{•}(G) ⊗ A with boundary d ⊗ 1. Let Q be a finite group acting as a group of automorphisms of G. The action of Q on G induces an action on C_{•}(G; A) by

q([g_{1}| ⋯ |g_{i}] ⊗ a) = [q(g_{1})| ⋯ |q(g_{i})] ⊗ a.

Denote the subcomplex of invariant chains by C_{•}(G; A)^{Q}.

Definition 1.1.

H_{•}^{Q}(G; A) = h_{•}(C_{•}(G; A)^{Q})
H_{Q}^{•}(G; A) = h^{•}(Hom(C_{•}(G)^{Q}, A)).

First note the following basic fact.

Lemma 1.2. Denote by G^{Q} the subgroup of Q-invariant elements of G. Then there are natural maps

H_{•}(G^{Q}; A) ⟶ H_{•}^{Q}(G; A) ⟶ H_{•}(G; A)^{Q}.

Proof. Note that there are inclusions of complexes

C_{•}(G^{Q}; A) ⟶ C_{•}(G; A)^{Q} ⟶ C_{•}(G; A)

and these induce maps

H_{•}(G^{Q}; A) ⟶^{f_{∗}} H_{•}^{Q}(G; A) ⟶^{i_{∗}} H_{•}(G; A).

Note that the image of i_{∗} lies in the subgroup H_{•}(G; A)^{Q} of Q-invariants since a Q-invariant cycle z ∈ C_{•}(G; A)^{Q} gives rise to an invariant homology class [z] ∈ H_{•}(G; A)^{Q}. □

2. Homology of quotient spaces

Consider the CW-complex BG whose chain complex is C_{•}(G). The Qaction on G induces a cellular action on BG with the property that if an element of Q fixes a cell, then it fixes it pointwise. This gives BG the structure of a Q-CW-complex and the quotient space BG/Q is also a CWcomplex.

Observe that we have

C_{n}(BG/Q) = ℤ{[g_{1}| ⋯ |g_{n}] : g_{i} ∈ G}/⟨[q(g_{1})| ⋯ |q(g_{n})] = [g_{1}| ⋯ |g_{n}]⟩
= C_{n}(BG)_{Q}

with the boundary map being induced by the boundary map in C_{•}(G). Recall that for any Q-module M, there is a norm map M_{Q} → M^{Q} whose kernel and cokernel are annihilated by |Q|. This map is natural for maps M → M′ of Q-modules. In the situation at hand, we have

N : C_{n}(BG/Q) ⟶ C_{n}(G)^{Q}

of the form

[g_{1}| ⋯ |g_{n}] ↦^{N} ∑_{q∈Q}[q(g_{1})| ⋯ |q(g_{n})].

Since the domain is free abelian, N is injective, but it is not surjective. Denote the quotient complex coker(N) by D_{•}.

Proposition 2.1. Suppose Q = ℤ/p, where p is a prime. Then D_{0} = 0, and for n > 0, D_{n} = C_{n}(G^{Q}; ℤ/p). Consequently, there is an isomorphism

h_{n}(D_{•}) ≅ H̃_{n}(G^{Q}; ℤ/p).

Proof. The map N : C_{0}(BG/Q) → C_{0}(G)^{Q} is simply the identity map ℤ → ℤ, whence the assertion for D_{0}. If n > 0, then since p is prime, the orbit of any g ∈ G has length 1 or p, and hence the same is true of any [g_{1}| ⋯ |g_{n}]. The basis elements of C_{n}(G)^{Q} are the various

∑_{q∈Q}[q(g_{1})| ⋯ |q(g_{n})]

together with the elements [g_{1}| ⋯ |g_{n}] for g_{i} ∈ G^{Q}. If [g_{1}| ⋯ |g_{n}] is a generator of C_{n}(BG/Q), then

N : [g_{1}| ⋯ |g_{n}] ↦ ∑_{q∈Q}[q(g_{1})| ⋯ |q(g_{n})].

If the orbit length of [g_{1}| ⋯ |g_{n}] is p, then we obtain the corresponding basis element of C_{n}(G)^{Q}. If, however, each g_{i} ∈ G^{Q}, then

N : [g_{1}| ⋯ |g_{n}] ↦ p[g_{1}| ⋯ |g_{n}].

Thus,

D_{n} = \frac{ℤ\{[g_{1}| ⋯ |g_{n}] : g_{i} ∈ G^{Q}\}}{p[g_{1}| ⋯ |g_{n}] = 0} ≅ C_{n}(G^{Q}; ℤ/p).

The boundary map in D_{•} is clearly that of C_{•}(G^{Q}; ℤ/p); this proves the final assertion. □

Corollary 2.2. If Q = ℤ/p, then there is a long exact sequence

→ H_{n}(BG/Q; ℤ) → H_{n}^{Q}(G; ℤ) → H̃_{n}(G^{Q}; ℤ/p) → H_{n−1}(BG/Q; ℤ) → .

□

Corollary 2.3. Suppose that Q = ℤ/p and that G^{Q} = {e}. Then there is an isomorphism

H_{•}(BG/Q; ℤ) ≅ H_{•}^{Q}(G; ℤ).

Proof. In this case, we have H̃_{•}(G^{Q}; ℤ/p) = 0. □

For arbitrary Q, we also have the following.

Proposition 2.4. If |Q| is invertible in A, then the map

H_{•}(BG/Q; A) ⟶ H_{•}^{Q}(G; A)

is an isomorphism.

Proof. The kernel and cokernel of the map

N : C_{n}(BG/Q; A) ⟶ C_{n}(G; A)^{Q}

are both annihilated by |Q|. Since |Q| is invertible in A, we see that N is an isomorphism. □

3. Basic Calculations

We begin our low-tech approach to calculation with a few bare-hands examples.

Proposition 3.1. H_{0}^{Q}(G; A) = A.

Proof. We have C_{0}(G; A)^{Q} = A and the boundary map

C_{1}(G; A)^{Q} → C_{0}(G; A)^{Q}

is the zero map. □

The major difficulty in computing H_{•}^{Q}(G; A) is that the chain complex C_{•}(G; A) is so large (even if G is finite). Unfortunately, there is not much to be done about this. Here is an explicit example.

Let ℤ/2 act on ℤ via n ↦ −n. Then

C_{i}(ℤ)^{ℤ/2} = ℤ{[n_{1}| ⋯ |n_{i}] + [−n_{1}| ⋯ | − n_{i}], [0| ⋯ |0]}.

Proposition 3.2. H_{1}^{ℤ/2}(ℤ) ≅ ℤ/2.

Proof. We have

H_{1}^{ℤ/2}(ℤ) = \frac{ℤ\{[n] + [−n], [0] : n > 0\}}{im\{d : C_{2}(ℤ)^{ℤ/2} → C_{1}(ℤ)^{ℤ/2}\}}.

Define f : C_{1}(ℤ)^{ℤ/2} → ℤ/2 by f([n] + [−n]) = n mod 2 and f([0]) = 0. Then f is surjective. Moreover, if [n_{1}|n_{2}] + [−n_{1}| − n_{2}] ∈ C_{2}(ℤ)^{ℤ/2}, then

f(d([n_{1}|n_{2}] + [−n_{1}| − n_{2}])) = f(([n_{1}] + [−n_{1}]) + ([n_{2}] + [−n_{2}]) −
([n_{1} + n_{2}] + [−(n_{1} + n_{2})]))
= n_{1} − (n_{1} + n_{2}) + n_{2} mod 2
= 0 mod 2

and f(d([0|0])) = f([0] − [0] + [0]) = 0. Thus, f induces a surjection \overline{f} : H_{1}^{ℤ/2}(ℤ) → ℤ/2. It remains to show that \overline{f} is injective. For this it suffices to show

(1) [2k] + [−2k] = [0] in H_{1}^{ℤ/2}(ℤ), and

(2) [2k + 1] + [−(2k + 1)] = [1] + [−1] in H_{1}^{ℤ/2}(ℤ).

Observe that [0] is the identity element in H_{1}^{ℤ/2}(ℤ) as [0] = d([0|0]). Consider the relations

d([m|m] + [−m| − m]) = 2([m] + [−m]) − ([2m] + [−2m])
d([m| − m] + [−m|m]) = 2([m] + [−m]) − 2[0].

Note that the second equation implies that H_{1}^{ℤ/2}(ℤ) is 2-torsion. Subtracting these shows that [2m]+[−2m] = 2[0] and hence [2m]+[−2m] = 0 in H_{1}^{ℤ/2}(ℤ).

We also have the following equations

d([m|m + 3] + [−m| − (m + 3)]) =

([m] + [−m]) + ([m + 3] + [−(m + 3)]) − ([2m + 3] + [−(2m + 3)]) d([m + 1|m + 2] + [−(m + 1)| − (m + 2)]) =

([m + 1] + [−(m + 1)]) + ([m + 2] + [−(m + 2)])

− ([2m + 3] + [−(2m + 3)])

so that in H_{1}^{ℤ/2}(ℤ) we have (by subtracting these)

([m] + [−m]) + ([m + 3] + [−(m + 3)]) =
([m + 1] + [−(m + 1)]) + ([m + 2] + [−(m + 2)]).

Setting m = 0, 2, 4, … we obtain (since [m]+[−m] = 0 = [m+2]+[−(m+2)])

[1] + [−1] = [3] + [−3] = ⋯ = [2m + 1] + [−(2m + 1)] = · · · .

This completes the proof. □

Now consider the resolution F_{•} →^{ε} ℤ obtained from S^{1} ≃ Bℤ:

F_{•} : 0 ⟶ ℤ[t, t^{−1}] ⟶^{t−1} ℤ[t, t^{−1}] ⟶^{ε} ℤ ⟶ 0.

To compute the complex (F_{•} ⊗_{G} ℤ)^{ℤ/2} of ℤ/2-invariants, we need an augmentation preserving chain map α_{∗} : F_{•} → F_{•} commuting with the action. The map α_{0} : t^{n} ↦ t^{−n} and α_{1} : t^{n} ↦ −t^{−(n+1)} does the job and one finds that the complex (F_{•} ⊗_{G} ℤ)^{ℤ/2} is

0 ⟶ 0 ⟶ ℤ.

The first homology of this complex vanishes, of course, and so we cannot replace C_{•}(ℤ) with F_{•} ⊗_{G} ℤ to calculate H_{•}^{ℤ/2}(ℤ).

A moment’s reflection reveals why this happens: the invariants functor is left exact, but not right exact, so there is no reason to believe that we can switch complexes at will. From the point of view of equivariant topology, there is a clearer answer: the homotopy equivalence Bℤ → S^{1} cannot be a ℤ/2-equivariant equivalence. Indeed, a map f : X → Y of Q-spaces is a Q-homotopy equivalence if the induced map f^{H} : X^{H} → Y^{H} is a homotopy equivalence for each H ≤ Q. In the case of Bℤ → S^{1}, Q = ℤ/2, the ℤ/2-fixed points are (Bℤ)^{ℤ/2} ≃ ∗ while (S^{1})^{ℤ/2} = S^{0}.

Let us proceed with some general calculations.

Proposition 3.3. Let G be a group and let A be an abelian group with trivial G- and Q-actions. Assume that |Q| is invertible in A. Then the natural map

i_{∗} : H_{•}^{Q}(G; A) ⟶ H_{•}(G; A)^{Q}

is an isomorphism.

Proof. This is a standard fact in equivariant topology, proved using the transfer map. We work directly with chains, however. Let [z] ∈ H_{ℓ}^{Q}(G; A) and suppose i_{∗}[z] = 0. Then z = dτ for some τ ∈ C_{ℓ+1}(G; A). Since z ∈ C_{ℓ}(G; A)^{Q}, we have qz = z for any q ∈ Q. Thus, if the Q-orbit of τ has length n, we have

d(∑_{q∈Q} qτ) = ∑_{q∈Q} q(dτ) = ∑_{q∈Q} qz = nz.

Write σ = ∑_{q∈Q} qτ. Then qσ = σ for all q ∈ Q. Since n is invertible in A by assumption (note n | |Q|), we have \frac{1}{n}σ ∈ C_{ℓ+1}(G; A)^{Q} and d(\frac{1}{n}σ) = z. So [z] = 0 and i_{∗} is injective.

To show that i_{∗} is surjective, let [z] ∈ H_{ℓ}(G; A)^{Q}. Then for each q ∈ Q, qz = z + dτ_{q} for some τ_{q} ∈ C_{ℓ+1}(G; A). Write w = ∑_{q∈Q} qz. Then w ∈ C_{ℓ}(G; A)^{Q} and dw = 0. We have

i_{∗}[w] = ∑_{q∈Q}[qz]
= ∑_{q∈Q}[z]
= nz

for some n invertible in A. So i_{∗}(\frac{1}{n}[w]) = [z] and i_{∗} is surjective. □

Note that as a corollary of the proof we have the following.

Proposition 3.4. The subgroup ker(i_{∗}) is annihilated by |Q|. □

In particular, if H_{ℓ}(G; A)^{Q} = 0 for some ℓ, we see that H_{ℓ}^{Q}(G; A) is annihilated by the order of Q. Thus, we have the following result.

Proposition 3.5. For each i ≥ 1, the group H_{i}^{ℤ/2}(ℤ) is 2-torsion. □

Calculating H_{i}^{ℤ/2}(ℤ) by hand from the bar resolution is essentially impossible, however. We will complete the computation in Section 5.

Proposition 3.6. If G is a divisible abelian group, then

H_{1}^{Q}(G; ℤ) = H_{1}(G; ℤ)^{Q} = G^{Q}.

Proof. We have the natural map

f_{∗} : H_{1}(G^{Q}) ⟶ H_{1}^{Q}(G).

This map has a splitting τ : H_{1}^{Q}(G) → H_{1}(G^{Q}) defined by τ([z_{1}]+⋯ [z_{n}]) = [z_{1} ⋯ z_{n}], where z_{1}, …, z_{n} is a Q-orbit. This is well-defined since

q(z_{1} ⋯ z_{n}) = q(z_{1}) ⋯ q(z_{n}) = z_{1} ⋯ z_{n}

so that z_{1} ⋯ z_{n} ∈ G^{Q}.

We now need the following relation in H_{1}^{Q}(G):

(1)  [z_{1}^{n}] + ⋯ + [z_{n}^{n}] = n[z_{1} ⋯ z_{n}].

Assuming this, we show that f_{∗} is surjective by noting that if [w_{1}]+⋯+[w_{m}] is an orbit, write w_{1} = z_{1}^{m}, …, w_{m} = z_{m}^{m} and then

[w_{1}] + ⋯ [w_{m}] = [z_{1}^{m}] + ⋯ + [z_{m}^{m}]
= m[z_{1} ⋯ z_{m}]
= f_{∗}(m[z_{1} ⋯ z_{m}]).

But f_{∗} is injective since τf_{∗}([z]) = τ[z] = [z].

It remains to show (1). If [z_{1}] + ⋯ + [z_{n}] ∈ G_{1}(G)^{Q}, set z_{i} = q_{i}(z_{1}), i = 2, …, n. Consider the two relations

∑_{j=1}^{m−1}([z_{1}^{j}, z_{i}] + ⋯ [z_{m}^{j}, z_{m}]) ↦^{d} m([z_{1}] + ⋯ + [z_{m}]) − ([z_{1}^{m}] + ⋯ [z_{m}^{m}])

and

∑_{j=1}^{m−1}([z_{1} ⋯ z_{j}, z_{j+1}] + [q_{2}(z_{1} ⋯ z_{j}), q_{2}(z_{j+1})] + ⋯ + [q_{m}(z_{1} ⋯ z_{j}), q_{m}(z_{j+1})])

↦^{d} m([z_{1}] + ⋯ + [z_{m}]) − m[z_{1} ⋯ z_{m}].

Subtracting these yields (1). □

For example, if G = 𝔾_{m}(ℂ) with Q = Gal(ℂ/ℝ), then H_{1}^{Q}(𝔾_{m}(ℂ)) = 𝔾_{m}(ℝ).

4. Transfer

Before we can make more calculations, we need to discuss transfer maps. Recall that if K is a subgroup of finite index in G, then there is a transfer map

tr : H_{•}(G; A) ⟶ H_{•}(K; A).

If j_{∗} : H_{•}(K; A) → H_{•}(G; A) is the map induced by the inclusion j : K → G, then the composite j_{∗} ∘ tr is (G : K)id. There is a double coset formula for the composite tr ∘ j_{∗}; we shall need only a special case of it in Section 5. It will be recalled there.

We want to know that we have transfer maps in our setting. To this end, assume that K is a Q-stable subgroup of finite index in G (that is, qK = K for each q ∈ Q, but it is not necessarily the case that K ⊆ G^{Q}). For arbitrary K and G, the transfer map is induced by a chain map τ : C_{•}(G) → C_{•}(K) defined by choosing a set E of right coset representatives of K and setting

τ[g_{1}| ⋯ |g_{n}] =

∑_{x_{j}∈E}[x_{j}g_{1}\overline{x_{j}g_{1}}^{−1}|\overline{x_{j}g_{1}}g_{2}\overline{x_{j}g_{1}g_{2}}^{−1}| ⋯ |\overline{x_{j}g_{1} ⋯ g_{n−1}}g_{n}\overline{x_{j}g_{1} ⋯ g_{n}}^{−1}],

where \overline{a} is the representative of Ka in E.

In order for this to commute with the Q-action, we need a set of representatives E such that q(\overline{g}) = \overline{q(g)} for every q ∈ Q. If we can do this, then since the homotopy from j_{∗} ∘ tr to (G : K)id maps orbits to orbits, the formula j_{∗} ∘ tr = (G : K)id will hold in this setting. In the case K = {e}, we can do this since the coset representative of g ∈ G is just g itself and hence the transfer map exists on the chain level C_{•}(G; A)^{Q} → C_{•}({e}; A)^{Q}. We therefore obtain the following.

Proposition 4.1. If G is finite, then H_{i}^{Q}(G; A) is annihilated by |G| for all i > 0.

Proof. Let j : {e} → G be the inclusion. Then j_{∗} ∘ tr = |G|id, and H_{i}^{Q}({e}; A) = 0 for all i > 0. □

Another example is the following. Let Q = ℤ/2 act on ℤ/2^{s}k, k odd, by n ↦ −n. Let K be the subgroup ⟨k⟩. The set

E = \{0, 1, 2k − 1, 2, 2k − 2, …, \frac{k − 1}{2}, \frac{k + 1}{2}\}

is a set of coset representatives of K in ℤ/2^{s}k and if g denotes the nonidentity element of ℤ/2, then for 1 ≤ i ≤ \frac{k−1}{2}, g(\overline{i}) = g(i) = 2k − i and \overline{g(i)} = \overline{2k − i} = 2k − i. Moreover, g(\overline{2k − i}) = g(2k − i) = i and \overline{g(2k − i)} = \overline{i} = i. So we get a transfer map on the level of chains in this case as well:

tr : H_{•}^{Q}(ℤ/2^{s}k; A) ⟶ H_{•}^{Q}(K; A)

and this satisfies j_{∗} ∘ tr = kid, where j : K → ℤ/2^{s}k is the inclusion.

5. Computations

In this section we carry out some calculations. We begin with the case Q = ℤ/2 acting on G = ℤ/n by x ↦ −x. In keeping with our desire to remain as low-tech as possible for as long as possible, we break this into cases.

5.1. ℤ/2 acting on ℤ/n, n odd. We claim the following:

H_{ℓ}^{ℤ/2}(ℤ/n) = \begin{cases}ℤ & ℓ = 0 \\ ℤ/n & ℓ = 4k − 1, k ≥ 1 \\ 0 & otherwise. \end{cases}

The case ℓ = 0 is clear. If ℓ is even, then H_{ℓ}(ℤ/n) = 0. By Proposition 3.4, we see that H_{ℓ}^{ℤ/2}(ℤ/n) is 2-torsion. On the other hand, H_{ℓ}^{ℤ/2}(ℤ/n) is annihilated by n (Proposition 4.1) and since n is odd, H_{ℓ}^{ℤ/2}(ℤ/n) must vanish for ℓ even.

Now suppose that ℓ is odd and consider the map i_{∗} : H_{ℓ}^{ℤ/2}(ℤ/n) → H_{ℓ}(ℤ/n)^{ℤ/2}. The kernel of this is 2-torsion, but it is also n-torsion and hence i_{∗} is injective.

If ℓ = 4k − 3, then the ℤ/2-action on H_{ℓ}(ℤ/n) = ℤ/n is nontrivial: x ↦ −x (this is an easy exercise in group cohomology). Since n is odd, we have H_{4k−3}(ℤ/n)^{ℤ/2} = 0 and since i_{∗} is injective, H_{4k−3}^{ℤ/2}(ℤ/n) = 0.

It remains to treat the case ℓ = 4k−1. Here the ℤ/2-action on H_{ℓ}(ℤ/n) = ℤ/n is trivial so that H_{ℓ}(ℤ/n)^{ℤ/2} = ℤ/n. We must show that i_{∗} is surjective. Let z be a generating cycle of H_{ℓ}(ℤ/n). Denote by \overline{z} the ℤ/2-translate of z. Since ℤ/2 acts trivially on H_{ℓ}(ℤ/n), we have \overline{z} = z+dσ for some (ℓ+1)-chain σ. Now, z + \overline{z} is an element of C_{ℓ}(ℤ/n)^{ℤ/2} and

i_{∗}([z + \overline{z}]) = [z] + [\overline{z}] = 2[z].

Since n is odd, 2[z] also generates H_{ℓ}(ℤ/n), and thus i_{∗} is an isomorphism. To summarize, then, we have proved that if n is odd,

i_{∗} : H_{•}^{ℤ/2}(ℤ/n; ℤ) ⟶^{≅} H_{•}(ℤ/n; ℤ)^{ℤ/2}.

5.2. ℤ/2 acting on ℤ/2k, k odd. Write k = p_{1}^{i_{1}} ⋯ p_{m}^{i_{m}}. By Proposition 3.3, the map

i_{∗} : H_{•}^{ℤ/2}(ℤ/2k; ℤ/p^{i}) ⟶ H_{•}(ℤ/2k; ℤ/p^{i})^{ℤ/2}

is an isomorphism for each p = p_{j}, i = i_{j}. The groups H_{ℓ}(ℤ/2k; ℤ/p^{i})^{ℤ/2} are easily calculated: they vanish if p ≠ p_{j} and

H_{ℓ}(ℤ/2k; ℤ/p_{j}^{i_{j}})^{ℤ/2} = \begin{cases}ℤ/p_{j}^{i_{j}} & ℓ = 4k − 1, 4k \\ 0 & otherwise. \end{cases}

It remains to calculate H_{•}^{ℤ/2}(ℤ/2k; ℤ/2). Let K = ⟨k⟩ be the subgroup of ℤ/2-invariants. Note that (ℤ/2k : K) = k. Let j : K → ℤ/2k be the inclusion. We claim that

j_{∗} : H_{•}^{ℤ/2}(K; ℤ/2) ⟶ H_{•}^{ℤ/2}(ℤ/2k; ℤ/2)

is an isomorphism. Indeed, since ℤ/2 acts trivially on K, we have

H_{•}^{ℤ/2}(K; ℤ/2) = H_{•}(K; ℤ/2).

Consider the transfer map

tr : H_{•}^{ℤ/2}(ℤ/2k; ℤ/2) ⟶ H_{•}(K; ℤ/2).

We have j_{∗} ∘ tr = kid = id. Thus, j_{∗} is surjective. To see that j_{∗} is injective, consider the composite

H_{•}(K; ℤ/2) ⟶^{j_{∗}} H_{•}^{ℤ/2}(ℤ/2k; ℤ/2) ⟶^{i_{∗}} H_{•}(ℤ/2k; ℤ/2)^{ℤ/2}

(note that the last is simply H_{•}(ℤ/2k; ℤ/2) since ℤ/2 acts trivially on the mod 2 homology). We claim that i_{∗} ∘ j_{∗} is an isomorphism. Indeed, i_{∗} ∘ j_{∗} is the map induced by the inclusion K → ℤ/2k. Consider the commutative diagram

\xymatrix{H_{2m+2}(ℤ/2k; ℤ/2) \ar[r] & H_{2m+1}(ℤ/2k; ℤ) \ar[r]^{×2} & H_{2m+1}(ℤ/2k; ℤ) \ar[r] & H_{2m+1}(ℤ/2k; ℤ/2) \\ H_{2m+2}(K; ℤ/2) \ar[u] \ar[r] & H_{2m+1}(K; ℤ) \ar[u] \ar[r]^{×2} & H_{2m+1}(K; ℤ) \ar[u] \ar[r] & H_{2m+1}(K; ℤ/2) \ar[u]}

where the rows are the long exact sequences associated to the short exact coefficient sequence

0 ⟶ ℤ ⟶^{×2} ℤ ⟶ ℤ/2 ⟶ 0.

This diagram takes the form

\xymatrix{ℤ/2 \ar[r] & ℤ/2k \ar[r]^{×2} & ℤ/2k \ar[r]^{k↦1} & ℤ/2 \\ ℤ/2 \ar[u]^{i_{∗}∘j_{∗}} \ar[r]^{≅} & ℤ/2 \ar[u]^{1↦k} \ar[r]^{0} & ℤ/2 \ar[u]^{1↦k} \ar[r]^{≅} & ℤ/2 \ar[u]^{i_{∗}∘j_{∗}}}

An easy diagram chase shows that i_{∗} ∘ j_{∗} is an isomorphism. Thus, j_{∗} is injective and hence an isomorphism.

We can now piece this together to get a calculation of H_{•}^{ℤ/2}(ℤ/2k; ℤ). Consider the diagram of universal coefficient sequences

\xymatrix{H_{ℓ}(K) ⊗ ℤ/2 \ar[d] \ar[r] & H_{ℓ}(K; ℤ/2) \ar[d]^{≅} \ar[r] & Tor(H_{ℓ−1}(K), ℤ/2) \ar[d] \\ H_{ℓ}^{ℤ/2}(ℤ/2k) ⊗ ℤ/2 \ar[r] & H_{ℓ}^{ℤ/2}(ℤ/2k; ℤ/2) \ar[r] & Tor(H_{ℓ−1}^{ℤ/2}(ℤ/2k), ℤ/2)}

Note that j_{∗} : H_{ℓ}(K) → H_{ℓ}^{ℤ/2}(ℤ/2k) satisfies

j_{∗} ∘ tr = kid_{H_{ℓ}^{ℤ/2}(ℤ/2k)}.

If ℓ ≥ 2 is even, then H_{ℓ}^{ℤ/2}(ℤ/2k; ℤ) is 2-torsion since H_{ℓ}(ℤ/2k; ℤ) = 0. But then k is invertible in H_{ℓ}^{ℤ/2}(ℤ/2k; ℤ) so that j_{∗} ∘ tr is an isomorphism. But H_{ℓ}(K; ℤ) = 0 and hence j_{∗} ∘tr is the zero map. This forces H_{ℓ}^{ℤ/2}(ℤ/2k; ℤ) = 0. The above diagram then shows that the map

H_{ℓ}^{ℤ/2}(ℤ/2k; ℤ/2) ⟶ Tor(H_{ℓ−1}^{ℤ/2}(ℤ/2k), ℤ/2)

is an isomorphism and so if r is odd, H_{r}^{ℤ/2}(ℤ/2k; ℤ) contains a single copy of ℤ/2. If r = 4s − 3, this is all H_{r}^{ℤ/2}(ℤ/2k; ℤ) contains (see the calculation above). If r = 4s−1, then H_{r}^{ℤ/2}(ℤ/2k; ℤ) also contains a copy of ℤ/p^{i}. Thus

H_{4s−3}^{ℤ/2}(ℤ/2k; ℤ) = ℤ/2

and

H_{4s−1}^{ℤ/2}(ℤ/2k; ℤ) = ℤ/2 ⊕ ℤ/p_{i}^{i_{1}} ⊕ ⋯ ⊕ ℤ/p_{m}^{i_{m}} = ℤ/2k.

Note that ℤ/2 acts trivially on H_{4s−1}(ℤ/2k) and nontrivially on H_{4s−3}(ℤ/2k); the invariant subgroup of the latter is precisely H_{4s−3}(K) = ℤ/2. In summary, then, in the last two subsections we have proved the following.

Theorem 5.1. If n ≢ 0 mod 4, then

i_{∗} : H_{•}^{ℤ/2}(ℤ/n; ℤ) ⟶ H_{•}(ℤ/n; ℤ)^{ℤ/2}

is an isomorphism. □

5.3. ℤ/2 acting on ℤ/n, n ≡ 0 mod 4. The technique used in the previous section breaks down when n ≡ 0 mod 4 because the subgroup of invariants has even index. This then tells us that certain maps are zero, but not that they are isomorphisms. Again, away from p = 2, we have

i_{∗} : H_{•}^{ℤ/2}(ℤ/n; ℤ/p^{i}) ⟶^{≅} H_{•}(ℤ/n; ℤ/p^{i})^{ℤ/2}

(p^{i} | n) and this is easily calculated. Therefore we shall confine our attention to the calculation of H_{•}^{ℤ/2}(ℤ/n; ℤ/2).

Write n = 2^{s}m, where s ≥ 2 and m is odd. Denote by M the subgroup ⟨m⟩. Then M ≅ ℤ/2^{s} and (ℤ/n : M) = m. Let α : M → ℤ/n be the inclusion. Then α_{∗} ∘ tr is mid. Taking ℤ/2 coefficients, we see that

α_{∗} : H_{•}^{ℤ/2}(M; ℤ/2) ⟶ H_{•}^{ℤ/2}(ℤ/n; ℤ/2)

is surjective. We claim that α_{∗} is injective as well. For this we need a special case of the transfer double coset formula: If H is a normal subgroup of G, and φ : H → G is the inclusion, then

(tr ∘ α_{∗})(z) = ∑_{g∈G/H} gz,

where gz is obtained from z via the conjugation action of G on H. In our case, everything in sight is abelian and so conjugation is trivial. It follows that

(tr ∘ α_{∗})(z) = ∑_{g∈(ℤ/n)/M} gz = mz.

Since m is odd, we have tr ∘ α_{∗} = id. Thus, α_{∗} is injective as well.

Thus, it remains to calculate H_{•}^{ℤ/2}(ℤ/2^{s}; ℤ), and for this we need heavier machinery. To this end, let Γ = ℤ/2, X = B(ℤ/2^{s}), Y = X^{Γ} = B(⟨2^{s−1}⟩). We first calculate the homology of X/Γ. If p ≠ 2, we have H_{•}(X/Γ; ℤ/p) = H_{•}(X; ℤ/p)^{ℤ/2} = 0. We therefore devote our attention to the case of ℤ/2 coefficients, and since homology and cohomology are dual over a field, we calculate cohomology instead. In what follows, unless otherwise specified, cohomology is with ℤ/2 coefficients.

Denote by π the quotient map X → X/Γ. Let β : Y → X be the inclusion and note that π∘β maps Y homeomorphically into X/Γ as a closed subspace. We first note the following fact: β^{∗} : H^{r}(X) → H^{r}(Y) is the zero map for r odd and the identity for r even. This is easily seen via a diagram chase using the long exact coefficient sequences

\xymatrix{H^{2k−1}(X; ℤ/2) \ar[r] \ar[d]^{β^{∗}} & H^{2k}(X; ℤ) \ar[r]^{×2} \ar[d]^{β^{∗}} & H^{2k}(X; ℤ) \ar[r] \ar[d]^{β^{∗}} & H^{2k}(X; ℤ/2) \ar[d]^{β^{∗}} \\ H^{2k−1}(Y; ℤ/2) \ar[r]^{≅} & H^{2k}(Y; ℤ) \ar[r]^{0} & H^{2k}(Y; ℤ) \ar[r]^{≅} & H^{2k}(Y; ℤ/2)}

Denote by γ the inclusion π ∘ β : Y → X/Γ. We need the following result.

Lemma 5.2. γ^{∗} : H^{q}(X/Γ) ⟶ H^{q}(Y) is the zero map for all q > 0.

Proof. Consider the commutative diagram obtained from the long exact sequences of the pairs (X/Γ, Y) and (X, Y):

\xymatrix{H^{2k+1}(X/Γ) \ar[d]^{π^{∗}} \ar[r]^{γ^{∗}} & H^{2k+1}(Y) \ar[d]^{≅} \ar[r] & H^{2k+2}(X/Γ, Y) \ar[d] \\ H^{2k+1}(X) \ar[r]^{0} & H^{2k+1}(Y) \ar[r]^{≅} & H^{2k+2}(X, Y)}

A diagram chase shows that γ^{∗} : H^{2k+1}(X/Γ) → H^{2k+1}(Y) is the zero map.

To show γ^{∗} : H^{2k}(X/Γ) → H^{2k}(Y) is zero, we work directly with chains. Since homology and cohomology are dual with ℤ/2 coefficients, it suffices to show that γ_{∗} : H_{2k}(Y) → H_{2k}(X/Γ) is zero. By constructing a chain homotopy from the usual periodic resolution of ℤ over ℤ/2 to the bar resolution, it is easy to see that

z = ∑_{i_{k}=0}^{1} ∑_{i_{k−1}=0}^{1} ⋯ ∑_{i_{1}=0}^{1}[i_{k} · 2^{s−1}|2^{s−1}|i_{k−1} · 2^{s−1}| ⋯ |i_{1} · 2^{s−1}|2^{s−1}]

is a generator for H_{2k}(Y; ℤ/2). Let σ_{0} = 0|2^{s−1} and σ_{1} = 2^{s−1}|2^{s−1} and consider the (2k + 1)-chain

w = ∑_{i_{k−1}=0}^{1} ⋯ ∑_{i_{1}=0}^{1}([2^{s−2}|2^{s−2}|2^{s−1}|σ_{i_{k−1}}| ⋯ |σ_{i_{1}}]+[2^{s−2}|−2^{s−2}|2^{s−1}|σ_{i_{k−1}}| ⋯ |σ_{i_{1}}])

in C_{•}(X/Γ; ℤ/2). A straightforward (if tedious) verification shows that z = dw in C_{•}(X/Γ; ℤ/2) (one must keep in mind that, for example, [2^{s−2}|2^{s−1}] = [−2^{s−2}|2^{s−1}] in C_{•}(X/Γ; ℤ/2)). Thus, γ_{∗}([z]) = 0. □

Now, the lemma implies that the long exact sequence of the pair (X/Γ, Y) breaks into short exact sequences

(2)  0 ⟶ H^{q−1}(Y) ⟶ H^{q}(X/Γ, Y) ⟶ H^{q}(X/Γ) ⟶ 0.

According to [4], p. 34, there is a long exact sequence (with ℤ/2 coefficients) (3) H̃^{n}((X/Y)/Γ) → H^{n}(X) → H̃^{n}((X/Y)/Γ) ⊕ H^{n}(Y) → H̃^{n+1}((X/Y)/Γ).

Note, however, that

(X/Y)/Γ = (X/Γ)/Y

and

H̃^{•}((X/Γ)/Y) ≅ H^{•}(X/Γ, Y).

Thus, this sequence becomes

H^{n}(X/Γ, Y) → H^{n}(X) → H^{n}(X/Γ, Y) ⊕ H^{n}(Y) → H^{n+1}(X/Γ, Y).

In this sequence, the map H^{n}(X) → H^{n}(Y) is easily seen to be the usual restriction map.

Since the map H^{2k}(X) → H^{2k}(Y) is an isomorphism, the map

H^{2k}(X) ⟶ H^{2k}(X/Γ, Y) ⊕ H^{2k}(Y)

is injective. We claim that the map H^{2k+1}(X/Γ, Y) → H^{2k+1}(X) is surjective. Indeed, consider the long exact sequences of the pairs (X/Γ, Y) and (X, Y):

\xymatrix{H^{2k}(Y) \ar[r]^{0} \ar[d]^{≅} & H^{2k+1}(X/Γ, Y) \ar[r] \ar[d]^{π^{∗}} & H^{2k+1}(X/Γ) \ar[r] \ar[d]^{π^{∗}} & H^{2k+1}(Y) \ar[r]^{0} \ar[d]^{≅} & H^{2k+2}(X/Γ, Y) \ar[d]^{π^{∗}} \\ H^{2k}(Y) \ar[r]^{0} & H^{2k+1}(X, Y) \ar[r]^{≅} & H^{2k+1}(X) \ar[r]^{0} & H^{2k+1}(Y) \ar[r]^{≅} & H^{2k+2}(X, Y)}

The map H^{2k+1}(X/Γ, Y) → H^{2k+1}(X) is the composite

H^{2k+1}(X/Γ, Y) ⟶^{π^{∗}} H^{2k+1}(X, Y) ⟶^{≅} H^{2k+1}(X).

The generator of H^{2k+1}(X) is the function f ∈ Hom(C_{2k+1}(X), ℤ/2) given by f(z) = 1, f(w) = 0, w ≠ z, where z is the generating cycle

z = ∑_{i_{k}=0}^{2^{s}−1} ∑_{i_{k−1}=0}^{2^{s}−1} ⋯ ∑_{i_{1}=0}^{2^{s}−1}[1|i_{k}|1| ⋯ |i_{1}|1]

of H_{2k+1}(X; ℤ/2). Define h : C_{2k+1}(X/Γ) → ℤ/2 by h({z}) = 1 and h({w}) = 0 for {w} ≠ {z}, where {z} is the Γ-orbit of z. Then π^{∗}(h) = f and so π^{∗} is surjective.

From this, we conclude via (3) that we have the following exact sequences for all k ≥ 0

0 → H^{2k}(X) → H^{2k}(X/Γ, Y) ⊕ H^{2k}(Y) → H^{2k+1}(X/Γ, Y) → H^{2k+1}(X) → 0;

and

0 → H^{2k+1}(X/Γ, Y) ⊕ H^{2k+1}(Y) → H^{2k+1}(X/Γ, Y) → 0.

Thus, H^{0}(X/Γ, Y) = 0, H^{1}(X/Γ, Y) ≅ ℤ/2, H^{2}(X/Γ, Y) ≅ (ℤ/2)^{2},…, H^{q}(X/Γ, Y) ≅ (ℤ/2)^{q}. Substituting this into (2) yields the following result.

Theorem 5.3. If s ≥ 2, then H_{1}(X/Γ; ℤ/2) ≅ ℤ/2 and for all i ≥ 2

H_{i}(X/Γ; ℤ/2) ≅ (ℤ/2)^{i−1}.
□

Now, we need to compute the integral homology H_{•}(X/Γ; ℤ). We first note the following fact.

Proposition 5.4. If ℓ ≠ 4k − 1, then H_{ℓ}(X/Γ; ℤ) is 2-torsion.

Proof. Note that we have a transfer map

μ : H_{•}(X/Γ; ℤ) ⟶ H_{•}(X; ℤ)

satisfying the following

π_{∗} ∘ μ(α) = 2α μ ∘ π_{∗}(β) = β + \overline{β}

where \overline{β} is the image of β under the action of ℤ/2 (see [1], p. 37). If ℓ is even, then since H_{ℓ}(X; ℤ) = 0, we see that 2 · H_{ℓ}(X/Γ; ℤ) = 0.

If ℓ = 4k − 3, then μ([z]) = [z] + [\overline{z}] = 0 (since \overline{β} = −β in H_{ℓ}(X; ℤ)), but π_{∗}(μ([z])) = 2[z] so that 2[z] = 0, unless z is actually a chain over Y = X^{Γ}. In that case, μ([z]) = [z], but 2[z] = 0 anyway since [z] lies in the image of H_{ℓ}(Y; ℤ) = ℤ/2. □

Proposition 5.5. H_{4k−1}(X/Γ; ℤ) contains a single copy of ℤ/2^{s}.

Proof. Let [z] be any generator of H_{4k−1}(X; ℤ) ≅ ℤ/2^{s}. Let [w] be a generator of H_{4k−1}(Y; ℤ) ≅ ℤ/2. Then 2^{s−1}[z] = [w] and

μ ∘ π_{∗}(2^{s−1}[z]) = μ ∘ π_{∗}([w])
= μ([w])
= [w]
≠ 0.

Thus, π_{∗}(2^{s−1}[z]) ≠ 0 and hence 2^{s−1}[z] ≠ 0 in H_{4k−1}(X/Γ; ℤ). On the other hand,

2^{s}[z] = π_{∗} ∘ μ(2^{s−1}[z])
= π_{∗}(2^{s−1}[z] + 2^{s−1}[\overline{z}])
= π_{∗}(2^{s}[z])(since [\overline{z}] = [z])
= π_{∗}(0)
= 0.

Thus, every generator of H_{4k−1}(X; ℤ) maps to an element of order 2^{s} in H_{4k−1}(X/Γ; ℤ) and since these generators are all multiples of each other in H_{4k−1}(X; ℤ), the same is true in H_{4k−1}(X/Γ; ℤ). □

Using the previous two propositions and the universal coefficient theorem, we then obtain the following result.

Theorem 5.6. For all ℓ > 0,

H_{ℓ}(X/Γ; ℤ) ≅ \begin{cases}(ℤ/2)^{k−1} & ℓ = 2k \\ (ℤ/2)^{2k−1} & ℓ = 4k − 3 \\ ℤ/2^{s} ⊕ (ℤ/2)^{2k−1} & ℓ = 4k − 1. \end{cases}

□

Now to complete the calculation of H_{•}^{ℤ/2}(ℤ/2^{s}; ℤ), we use Corollary 2.2. There is a long exact sequence

→ H_{n}(X/Γ; ℤ) → H_{n}^{ℤ/2}(ℤ/2^{s}; ℤ) →^{p_{∗}} H̃_{n}(ℤ/2; ℤ/2) →^{∂} H_{n−1}(X/Γ; ℤ) → .

We claim that the map p_{∗} is surjective. Indeed, let z be a generating cycle for H_{n}(ℤ/2; ℤ/2). If n is odd, then z is an integral cycle and z ∈ Z_{n}(ℤ/2^{s}; ℤ)^{ℤ/2} so that p_{∗}[z] = [z]. If n is even, then dz = 2w where w generates H_{n−1}(ℤ/2; ℤ). But 2[w] = 0 in H_{n−1}(X/Γ; ℤ) (since 2[w] = 0 in H_{n−1}(X; ℤ)) and so ∂[z] = [2w] = 0. It follows that p_{∗} is surjective in this case as well.

Thus, for each q ≥ 1, we have a short exact sequence

0 → H_{q}(X/Γ; ℤ) → H_{q}^{ℤ/2}(ℤ/2^{s}; ℤ) → H_{q}(ℤ/2; ℤ/2) → 0.

This clearly splits for q ≠ 4k−1 since each term is 2-torsion. When q = 4k−1 this still splits since H_{4k−1}(X/Γ; ℤ) contains a copy of ℤ/2^{s} together with some copies of ℤ/2, while Proposition 4.1 implies that H_{4k−1}^{ℤ/2}(ℤ/2^{s}; ℤ) is annihilated by 2^{s} and hence there are no copies of ℤ/2^{s+1} to provide a nonsplit quotient. We therefore obtain the following calculation.

Theorem 5.7. For all ℓ > 0,

H_{ℓ}^{ℤ/2}(ℤ/2^{s}; ℤ) ≅ \begin{cases}(ℤ/2)^{k} & ℓ = 2k \\ (ℤ/2)^{2k} & ℓ = 4k − 3 \\ ℤ/2^{s} ⊕ (ℤ/2)^{2k} & ℓ = 4k − 1. \end{cases}

□

The interested reader may now use this to calculate H_{•}^{ℤ/2}(ℤ/n; ℤ) for any n ≡ 0 mod 4.

5.4. ℤ/2 acting on ℤ via n ↦ −n. We are now able to calculate H_{•}^{ℤ/2}(ℤ). We know this is all 2-torsion in positive degrees (Proposition 3.5). Set Γ = ℤ/2, X = Bℤ, Y = X^{Γ} ≃ ∗. The long exact sequence of the pair (X/Γ, Y) (with ℤ/2 coefficients)

H^{q}(X/Γ, Y) ⟶ H^{q}(X/Γ) ⟶ H^{q}(Y) ⟶ H^{q+1}(X/Γ, Y)

shows that H^{q}(X/Γ, Y) ⟶^{≅} H^{q}(X/Γ) for all q > 0. Again, using [4], p. 34, we obtain a long exact sequence

H^{q}(X/Γ, Y) → H^{q}(X) → H^{q}(X/Γ, Y) ⊕ H^{q}(Y) → H^{q+1}(X/Γ, Y).

Since H^{q}(X) = 0 for q ≥ 2 and H^{q}(Y) = 0 for q ≥ 1, we see that for q ≥ 2

H^{q}(X/Γ, Y) ⟶^{≅} H^{q+1}(X/Γ, Y).

Also, H^{0}(X/Γ, Y) = 0 and we have an exact sequence

0 → H^{1}(X/Γ, Y) → H^{1}(X) → H^{1}(X/Γ, Y) → H^{2}(X/Γ, Y) → 0.

If H^{1}(X/Γ, Y) = 0, then we would have the exact sequence

0 ⟶ ℤ/2 ⟶ 0,

which is absurd. So H^{1}(X/Γ, Y) ≅ ℤ/2 and the map

H^{1}(X/Γ, Y) → H^{2}(X/Γ, Y)

is an isomorphism. Thus, H^{q}(X/Γ, Y) ≅ ℤ/2 for all q ≥ 1 and so

H^{q}(X/Γ; ℤ/2) ≅ ℤ/2

for all q ≥ 1.

Theorem 5.8.

H̃_{i}^{ℤ/2}(ℤ; ℤ) ≅ \begin{cases}ℤ/2 & i odd \\ 0 & i even. \end{cases}

Proof. We know that H_{i}^{ℤ/2}(ℤ; ℤ) is 2-torsion for i ≥ 1. According to Corollary 2.3, there is an isomorphism

H_{i}(X/Γ; ℤ) ≅ H_{i}^{ℤ/2}(ℤ; ℤ).

The universal coefficient theorem applied to the homology of X/Γ then yields the desired result. □

References

[1] A. Borel, et. al., Seminar on transformation groups, Annals of Mathematics Studies 46, Princeton University Press, 1960.

[2] K. S. Brown, Cohomology of groups, Springer–Verlag Graduate Texts in Mathematics 87, 1982.

[3] K. Knudson, M. Walker, Homology of linear groups via cycles in BG × X, preprint, 2003.

[4] J. P. May, Equivariant homotopy and cohomology theory (with contributions by M. Cole, G. Comenzaña, S. Costenoble, A. D. Elmendorf, J. P. C. Greenlees, L. G. Lewis, R. J. Piacenza, G. Triantafillou, and S. Waner), CBMS Regional Conference Series in Mathematics 91, American Mathematical Society, Providence, 1996.

Department of Mathematics and Statistics, Mississippi State University, P. O. Drawer MA, Mississippi State, MS 39762

E-mail address: knudson@math.msstate.edu