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The mod-two Bockstein is a derivation

Statement

Let $m \geq 1$ and let $β$ be the Bockstein associated to $0 \to Z / m \overset{m}{\to} Z / m^{2} \to Z / m \to 0$ . If $x \in H^{p} (X; Z / m)$ and $y \in H^{q} (X; Z / m)$ , then

β (x ⌣ y) = β (x) ⌣ y + (- 1)^{p} x ⌣ β (y) .

In particular, for $m = 2$ the sign disappears.

Facts & Assumptions

Given: Cocycle representatives $φ$ of $x$ and $ψ$ of $y$ .

[F1]

For the mod- $m$ coefficient sequence, least nonnegative residue representatives give canonical cochain lifts without AC (Bockstein connecting operation).

[F2]

The positive-coboundary convention satisfies $δ (φ ⌣ ψ) = δ φ ⌣ ψ + (- 1)^{p} φ ⌣ δ ψ$ (Cup product Leibniz identity).

Proof

technique · residue-lift calculation

1.1

Choose the canonical lifts $\tilde{φ}$ and $\tilde{ψ}$ with values in $Z / m^{2}$ . [given, F1] Because $φ$ and $ψ$ are cocycles, there are uniquely determined $Z / m$ -cochains $η$ and $μ$ such that $δ \tilde{φ} = m η$ and $δ \tilde{ψ} = m μ$ in $Z / m^{2}$ .

2.1

These cochains represent the two Bocksteins. [F1, step 1.1] By the defining lift-and-coboundary construction, $[η] = β (x)$ and $[μ] = β (y)$ .

3.1

Compute the Bockstein of the product. [F2, step 1.1, step 2.1] The cochain $\tilde{φ} ⌣ \tilde{ψ}$ lifts $φ ⌣ ψ$ , and [F2] gives

δ (\tilde{φ} ⌣ \tilde{ψ}) = m (η ⌣ ψ + (- 1)^{p} φ ⌣ μ) .

Here multiplication by $m$ makes the expression depend only on the reductions of the displayed lifts modulo $m$ . This product lift need not be the canonical residue lift used in [F1], so compare them explicitly. Their difference takes values in the kernel of reduction $Z / m^{2} \to Z / m$ and hence is uniquely $m h$ for a $Z / m$ -cochain $h$ . Their coboundaries differ by $m δ h$ , so division by the injective copy of $Z / m$ changes the resulting cocycle by the coboundary $δ h$ . Thus this noncanonical lift computes the same Bockstein class as the canonical lift.

4.1

Divide by the injective copy of $Z / m$ and pass to cohomology. [F1, step 2.1, step 3.1] This yields the stated derivation identity. When $m = 2$ , $- 1 = 1$ in the coefficient ring, so the parity sign is invisible. ∎

Depends on

Used by

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Hatcher, Algebraic Topology (standard reference, not scraped)