LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-14

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Free cyclic resolution, group cohomology, and cochain transfer

Statement

Let $p$ be prime, $C_{p} = ⟨ T ⟩$ , $R = F_{p} [C_{p}]$ , and $N = 1 + T + \dots + T^{p - 1}$ . The augmented complex of free left $R$ -modules

$\dots ⟶ R e_{2} \overset{N}{\to} R e_{1} \overset{T - 1}{\to} R e_{0} \overset{ε}{\to} F_{p} ⟶ 0,$

with $d (e_{2 r + 1}) = (T - 1) e_{2 r}$ and $d (e_{2 r}) = N e_{2 r - 1}$ for $r \geq 1$ , is exact. If $F_{p}$ has the trivial $R$ -action, then the cohomology of ${Hom}_{R} (W, F_{p})$ , with the cup product induced by the standard equivariant diagonal, is

$H^{*} (C_{p}; F_{p}) = {\begin{cases} F_{2} [t], & p = 2, ∣ t ∣ = 1, \\ F_{p} [u] \otimes Λ (v), & p odd, ∣ u ∣ = 2, ∣ v ∣ = 1. \end{cases}$

With the positive connecting convention, one may take $v = [w_{1}]$ and $u = [w_{2}] = β v$ when $p$ is odd; for $p = 2$ , $t = [w_{1}]$ and $β t = t^{2}$ .

On quotient cellular chains, the standard equivariant diagonal has the exact form

$\overline{D} (e_{2 r}) = \sum_{a = 0}^{r} e_{2 a} \otimes e_{2 r - 2 a} + \frac{p (p - 1)}{2} \sum_{a = 0}^{r - 1} e_{2 a + 1} \otimes e_{2 r - 2 a - 1},$

$\overline{D} (e_{2 r + 1}) = \sum_{a = 0}^{2 r + 1} e_{a} \otimes e_{2 r + 1 - a} .$

In particular, at $p = 2$ every split of the total resolution degree occurs with coefficient one.

More generally, let $H \leq G$ with $G$ finite, let $C_{*}$ be a chain complex of left $F_{p} [G]$ -modules, and let $A$ be a left $F_{p} [G]$ -module. There is a cochain map

${Tr}_{H}^{G} : {Hom}_{H} (C_{*}, A) ⟶ {Hom}_{G} (C_{*}, A)$

such that ${Tr}_{H}^{G} {Res}_{H}^{G} = [G : H]$ on $G$ -equivariant cochains. Consequently this composite is zero over $F_{p}$ whenever $p$ divides $[G : H]$ ; the transfer of an arbitrary $H$ -equivariant class need not itself be zero.

Facts & Assumptions

Given: A prime $p$ , the displayed cyclic resolution, and, for the transfer clause, $H \leq G$ , $C_{*}$ , and $A$ as in the statement.

[F1]

Cohomology is the quotient of cocycles by coboundaries (Singular cohomology with coefficients).

[F2]

For the mod- $p$ Bockstein, least nonnegative residue lifts are canonical and require no AC (Bockstein connecting operation).

[F3]

The cochain external product evaluates a tensor functional on tensor chains, with no extra sign in that evaluation (Additive singular cohomology cross product). The cyclic chain diagonal used to define the internal product is the explicit Steenrod--Epstein construction quoted in Step 4.1, not a claim of the cross-product definition.

Proof

Proof technique: compute kernels in the truncated polynomial group ring, then evaluate the explicit cyclic diagonal and define transfer directly on the finite quotient set.

1.1

Identify the group ring and the two differentials. [given] Put $s = T - 1$ . In characteristic $p$ , $(1 + s)^{p} = 1 + s^{p}$ , so the basis $1, T, \dots, T^{p - 1}$ gives

$R ≅ F_{p} [s] / (s^{p}) .$

Expanding $(T - 1)^{p - 1}$ and using $(\binom{p - 1}{j}) \equiv (- 1)^{j} (m o d p)$ gives $N = s^{p - 1}$ . Hence $s N = N s = s^{p} = 0$ , which proves $d^{2} = 0$ .

1.2

Define transfer without choosing coset representatives. [given] For a left coset $g H \in G / H$ and $f \in {Hom}_{H} (C_{n}, A)$ , define

$Φ_{g H} (f) (c) = g f (g^{- 1} c) .$

This depends only on the coset: replacing $g$ by $g h$ gives

$g h f (h^{- 1} g^{- 1} c) = g h h^{- 1} f (g^{- 1} c) = g f (g^{- 1} c)$

by $H$ -equivariance. Hence ${Tr}_{H}^{G} f = \sum_{g H \in G / H} Φ_{g H} (f)$ is a specified finite sum over the quotient set, not a sum requiring a chosen transversal. For $k \in G$ , substitution $g = k r$ permutes $G / H$ and gives $(Tr f) (k c) = k (Tr f) (c)$ , so the result is $G$ -equivariant.

2.1

Prove exactness in every degree. [step 1.1] Every element of $R$ has a unique form $a_{0} + a_{1} s + \dots + a_{p - 1} s^{p - 1}$ . Multiplication by $s$ has kernel $F_{p} s^{p - 1} = N R$ and image $s R$ ; multiplication by $N = s^{p - 1}$ has kernel $s R$ and image $F_{p} s^{p - 1}$ . Finally $\ker ε = s R$ , the image of the first map $T - 1 = s$ . These equalities prove exactness at $F_{p}$ , at $R e_{0}$ , and alternately at every positive degree. They also cover $p = 2$ , where the two displayed multipliers coincide.

2.2

Verify the cochain and restriction identities. [step 1.2] Because the $G$ -action commutes with the differential of $C_{*}$ ,

$δ Φ_{g H} (f) (c) = g f (g^{- 1} d c) = Φ_{g H} (δ f) (c) .$

The finite sum therefore commutes with $δ$ . If $f$ is already $G$ -equivariant, every summand satisfies $g f (g^{- 1} c) = f (c)$ , whence

${Tr}_{H}^{G} {Res}_{H}^{G} (f) = [G : H] f .$

When $p ∣ [G : H]$ , that scalar is zero in $F_{p}$ . This proves only the stated composite identity, not vanishing of transfer on an arbitrary class.

3.1

Compute the equivariant cochain groups. [F1, step 2.1] Let $w_{j} \in {Hom}_{R} (R e_{j}, F_{p})$ have $w_{j} (e_{j}) = 1$ . Each cochain group is the one-dimensional span of $w_{j}$ . Since the trivial action sends $s$ to $0$ and $N$ to $p = 0$ , precomposition with every differential is zero. Thus every $w_{j}$ is a cocycle, there are no nonzero coboundaries, and [F1] gives one basis class $[w_{j}]$ in each degree.

4.1

Evaluate the cyclic diagonal. [F3, step 3.1] Steenrod--Epstein's cyclic-diagonal lemma in Chapter V, section 5, on printed page 67 constructs the equivariant cellular diagonal. After passing to the quotient and writing the single cell in degree $j$ again as $e_{j}$ , its formulas on printed page 68 are

$\overline{D} (e_{2 r}) = \sum_{a = 0}^{r} e_{2 a} \otimes e_{2 r - 2 a} + \frac{p (p - 1)}{2} \sum_{a = 0}^{r - 1} e_{2 a + 1} \otimes e_{2 r - 2 a - 1},$

$\overline{D} (e_{2 r + 1}) = \sum_{a = 0}^{2 r + 1} e_{a} \otimes e_{2 r + 1 - a} .$

The source verifies before quotienting that this is a chain map and a diagonal approximation; reduction modulo $p$ therefore defines the cup product. Evaluating by the tensor functional of [F3] gives

$w_{1}^{2} = \frac{p (p - 1)}{2} w_{2}, w_{2}^{r} = w_{2 r}, w_{2}^{r} w_{1} = w_{2 r + 1} .$

For $p = 2$ the first coefficient is $1$ , so induction gives $w_{j} = w_{1}^{j}$ for every $j$ . For odd $p$ the coefficient is divisible by $p$ , so $w_{1}^{2} = 0$ , while the other two equations show that the displayed $u = [w_{2}]$ and $v = [w_{1}]$ produce the unique basis class in every degree. There can be no further relation, because a nonzero polynomial monomial $u^{r}$ or $u^{r} v$ is exactly the nonzero basis vector in its degree. This proves the two asserted graded-algebra descriptions.

5.1

Fix the Bockstein sign. [F2, step 4.1] For the quotient cellular model over $Z / p^{2}$ , the relevant boundary is $\partial e_{2} = p e_{1}$ . Lift $w_{1}$ by the canonical residues of [F2]. Its coboundary takes $e_{2}$ to $p$ ; division by the coefficient inclusion $a \mapsto p a$ therefore gives $w_{2}$ . Thus the positive convention here is $β [w_{1}] = [w_{2}]$ . This says $β t = t^{2}$ at $p = 2$ and permits $u = β v$ at odd $p$ . This sign differs from sources that build a minus sign into their cochain connector.

6.1

If $H = G$ transfer is the identity, while zero coefficients make it zero. [step 2.1, step 2.2, step 5.1] The prime endpoint $p = 2$ was treated separately, and every odd prime uses the same divisible odd-odd coefficient. If $H = G$ , the quotient has one element and transfer is the identity; if $H = {1}$ , the same formula is the full finite group sum. Zero chain groups, zero cochains, and zero modules make all maps zero. There are no negative resolution degrees; $e_{0}$ and the augmentation were checked in step 2.1. The source's geometric model retains all cells, while the algebraic computation depends only on the displayed free modules and so has no separate degenerate-simplex exception. The only coefficient lift in step 5.1 is the canonical residue lift singled out in [F2], and step 1.2 sums representative-independent functions over a finite set. Thus the proof makes no arbitrary choice and uses no AC. ∎

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Sources

N. E. Steenrod and D. B. A. Epstein, Cohomology Operations (standard reference, not scraped)