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Steenrod squares commute with relative cohomology connectors

Statement

For every pair $(E, F)$ , the mod-two cohomology connector satisfies $δ S q^{a} (x) = S q^{a} δ (x)$ for every $x \in H^{m} (F; F_{2})$ and every nonnegative $a$ . Relative squares use the published relative cup- $i$ construction.

Facts & Assumptions

Given: a pair $(E, F)$ , an integer $m \geq 0$ , a class $x \in H^{m} (F; F_{2})$ represented by a cocycle $u$ , and a nonnegative integer $a$ .

[F1]

For $x \in H^{n}$ represented by a cocycle $a$ , the square is $S q^{k} (x) = [a ⌣_{n - k} a]$ for $0 \leq k \leq n$ , using the cup- $i$ products with their relative variants and the convention $⌣_{j} = 0$ for $j < 0$ (Steenrod squares from cup-i, Higher cup-i products).

[F2]

The cup- $i$ coboundary identity reads $δ (a ⌣_{i} b) = δ a ⌣_{i} b + a ⌣_{i} δ b + a ⌣_{i - 1} b + b ⌣_{i - 1} a$ , in both relative variants (Cup-i coboundary identity).

[F3]

The cohomology connector of a pair sends $[u]$ to $[δ \tilde{u}]$ for any extension $\tilde{u}$ of a cocycle representative to the ambient space, and fits in the exact pair sequence (Long exact sequence of a pair in singular cohomology); instability gives $S q^{k} = 0$ above the degree of the class (Steenrod normalization, instability, suspension, and top square).

Proof

technique · direct

1.1givenF1F3construct

Represent $x$ by a cocycle $u$ and extend $u$ by zero on singular simplices of $E$ not in $F$ , obtaining an absolute cochain $b$ . Then $c = d b$ restricts to zero on $F$ and represents $δ x$ . If $0 \leq a \leq m$ , set $j = m - a$ and $b^{'} = b ⌣_{j + 1} d b + b ⌣_{j} b .$

2.1step 1.1F1F2algebra

Its restriction to $F$ is $u ⌣_{j} u$ , which represents $S q^{a} x$ . The published cup- $i$ coboundary identity, $d^{2} = 0$ , and characteristic two give $d b^{'} = d b ⌣_{j + 1} d b + b ⌣_{j} d b + d b ⌣_{j} b + d b ⌣_{j} b + b ⌣_{j} d b + b ⌣_{j - 1} b + b ⌣_{j - 1} b = c ⌣_{j + 1} c .$

3.1step 2.1F2F3algebra∎

This is the relative representative of $S q^{a} [c]$ , since $∣ c ∣ = m + 1$ . Thus the two connector classes agree. For $a = m + 1$ , $S q^{m + 1} x = 0$ by instability, and $c ⌣_{0} c = d (b ⌣_{0} d b)$ ; the primitive restricts to zero on $F$ , proving that the other side also vanishes relatively. For $a > m + 1$ both sides vanish by instability. The relative-carrier property guarantees all relative cochains used above vanish on $F$ ; no representative-selection family or choice axiom is needed.

Depends on

Used by

Dependency tree · two levels

14 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

Allen Hatcher, Spectral Sequences in Algebraic Topology, Chapter 1 (standard reference, not scraped)