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Cartan formula for Steenrod squares

Statement

For $x \in H^{p} (X; F_{2})$ , $y \in H^{q} (X; F_{2})$ , and every integer $k$ , the Steenrod squares satisfy the Cartan formula

S q^{k} (x ⌣ y) = \sum_{i + j = k} S q^{i} (x) ⌣ S q^{j} (y) .

Only finitely many terms are nonzero. More generally, for classes on two spaces the external formula is

S q^{k} (x \times y) = \sum_{i + j = k} S q^{i} (x) \times S q^{j} (y) .

Facts & Assumptions

Given: Mod-two cocycles $a, b$ representing classes of degrees $p, q$ .

[F1]

Squares vanish above the degree of their input (Steenrod normalization, instability, suspension, and top square).

[F2]

The before-Alexander--Whitney and convolution families have the coherent homotopy with its exact $(1 + Q) H_{i - 1}$ correction (Cartan coherence for higher diagonals).

[F3]

Squares are independent of the coherent higher-diagonal system (Steenrod squares are well-defined and natural).

[F4]

A square of a degree- $n$ class is represented by $a ⌣_{n - k} a$ in its defining range (Steenrod squares from cup-i).

[F5]

The Alexander--Whitney external cochain represents the cohomology cross product, and its pullback along the diagonal is the cup product (Singular cup product on cochains).

[F6]

Alexander--Whitney and shuffle are chain-homotopy inverses (Alexander--Whitney and shuffle are natural chain-homotopy inverses).

[F7]

Squares are natural for maps of spaces (Steenrod squares are well-defined and natural).

Proof

technique · evaluate Cartan coherence and pull back the external formula along the diagonal

1.1F1F3F4F5F6

Reduce to the normalized face-formula system. Because of [F3], compute all squares with the explicit system in Medina--Mardones, Definition 7 and Theorem 10 (printed pages 8--9). It has $D_{r} (s) = 0$ when $r$ exceeds the dimension of the simplex $s$ . By [F5], the external class $[a] \times [b]$ is represented by the Alexander--Whitney external cochain. Since shuffle induces the inverse cohomology isomorphism by [F6], it suffices to compare the two sides after precomposition with shuffle. If $k < 0$ , every square in the formula is zero by definition. If $k > p + q$ , at least one of $i > p$ or $j > q$ holds in each summand, so [F1] makes both sides zero. Hence assume $0 \leq k \leq p + q$ and put $ℓ = p + q - k$ .

2.1F2step 1.1

Evaluate the coherent comparison. Pair the equation for $L_{ℓ} - R_{ℓ}$ in [F2] with $λ = a \otimes b \otimes a \otimes b$ . Since $a, b$ are cocycles, $λ d = 0$ . Also $λ Q = λ$ , so the two evaluations of $Q H_{ℓ - 1}$ and $H_{ℓ - 1}$ cancel in characteristic two. Therefore

$λ L_{ℓ} - λ R_{ℓ} = δ (λ H_{ℓ}) .$

The left term is the shuffled cochain representing $S q^{k} ([a] \times [b])$ ; the right term is cohomologous to it.

3.1F4step 2.1

Identify every convolution term. For $r + s = ℓ$ , regrouping the four factors gives

$λ τ (D_{r} \otimes T^{r} D_{s}) = (a ⌣_{r} a) \otimes (b ⌣_{s} b),$

because $(b \otimes b) T^{r} = b \otimes b$ . The normalized face formula makes the first factor zero when $r > p$ and the second zero when $s > q$ : on the only chain degree where it could be evaluated, respectively $2 p - r$ or $2 q - s$ , the higher-diagonal index exceeds the simplex dimension. For the remaining terms put $i = p - r$ and $j = q - s$ . Then $i, j \geq 0$ , $i + j = k$ , and [F4] identifies their classes as $S q^{i} [a]$ and $S q^{j} [b]$ . Step 2.1 and the shuffle isomorphism prove the external Cartan formula.

4.1F5F7step 3.1

Pull back along the diagonal. For two classes $x, y$ on $X$ , [F5] gives $x ⌣ y = Δ^{*} (x \times y)$ . Naturality [F7] and step 3.1 give

$S q^{k} (x ⌣ y) = Δ^{*} S q^{k} (x \times y) = \sum_{i + j = k} Δ^{*} (S q^{i} x \times S q^{j} y) = \sum_{i + j = k} S q^{i} x ⌣ S q^{j} y .$

Instability [F1] leaves at most $(p + 1) (q + 1)$ possible pairs, so the sum is finite. Empty spaces, zero classes, degree-zero factors, one-point spaces, and degenerate singular simplices were retained throughout. Every formula is a specified finite sum, and no AC is used. ∎

Cartan formula for Steenrod squares

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources