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Steenrod squares do not all lift integrally

Statement refuted

It is false that every Steenrod square has a natural integral-valued lift. More precisely, assume AC. There is no degree-two cohomology operation

$Θ_{n} : H^{n} (-; F_{2}) ⟶ H^{n + 2} (-; Z)$

whose composite with coefficient reduction $ρ : H^{n + 2} (-; Z) \to H^{n + 2} (-; F_{2})$ is $S q^{2}$ for every space, degree, and class. In fact, the obstruction below rules out such a lift even before naturality is used.

This does not apply to $S q^{1}$ : the integral Bockstein is an integral-valued lift of $S q^{1}$ . The failed degree-one scaffold argument is therefore not used.

Facts & Assumptions

Given: A hypothetical family $Θ_{n}$ with the displayed lifting property.

[F1]

Steenrod squares on real projective space gives, under AC, $H^{*} ({R P}^{\infty}; F_{2}) = F_{2} [a]$ , $∣ a ∣ = 1$ , and $S q^{i} (a^{j}) = (\binom{j}{i}) a^{j + i}$ with coefficients reduced modulo two.

[F2]

On each standard finite skeleton from [F1], Real projective space cellular homology and the pinch map gives one cell in every dimension up to its top dimension and integral incidence maps $d_{j} = 2$ for positive even $j$ and $d_{j} = 0$ for odd $j$ . Axiomatic cellular boundaries are integral incidence matrices with coefficients identifies these incidence matrices with the cellular differential, and Cellular homology computes singular homology applies to the resulting infinite CW complex.

[F3]

Topological universal coefficient short exact sequence for cohomology gives, under AC, the exact sequence

$0 ⟶ {Ext}_{Z}^{1} (H_{n - 1} (X; Z), G) ⟶ H^{n} (X; G) ⟶ {Hom}_{Z} (H_{n} (X; Z), G) ⟶ 0.$

[F4]

Singular cohomology with coefficients makes the coefficient map $Z \to F_{2}$ induce the reduction homomorphism $ρ$ used in the statement.

[F5]

Bockstein connecting operation gives the integral and mod-two Bocksteins by canonical cyclic residue lifts.

[F6]

Assuming AC, Bocksteins are natural and stable makes these Bocksteins natural cohomology operations.

[F7]

Sq^1 is the mod-two Bockstein identifies the mod-two Bockstein with $S q^{1}$ .

[A1]

The Axiom of Choice is used exactly through [F1], [F3], and [F6]. The cellular homology and cyclic residue-lift calculations add no choice.

Counterexample

1.1

Choose the explicit mod-two input $u = a^{3}$ . [F1] It is nonzero in the polynomial ring and has degree three. Substitution of $i = 2, j = 3$ in [F1] gives

$S q^{2} (u) = (\binom{3}{2}) a^{5} = a^{5} \neq 0,$

since $(\binom{3}{2}) = 3 = 1$ in $F_{2}$ and every power of the polynomial generator is nonzero.

1.2

The required integral target is zero. [F1, F2, F3] The standard skeletal inclusions in [F1] preserve the cells in [F2], and a cellular differential in degree $j$ is already determined on the finite skeleton ${R P}^{j}$ . Hence their union gives the infinite integral cellular complex with the same alternating differentials. In particular, $d_{4} = 2$ , $d_{5} = 0$ , and $d_{6} = 2$ . Therefore

$H_{4} ({R P}^{\infty}; Z) = \ker (2 : Z \to Z) = 0$

and

$H_{5} ({R P}^{\infty}; Z) = Z / 2 Z .$

At $n = 5$ with $G = Z$ , [F3] therefore becomes

$0 ⟶ {Ext}_{Z}^{1} (0, Z) ⟶ H^{5} ({R P}^{\infty}; Z) ⟶ {Hom}_{Z} (Z / 2, Z) ⟶ 0.$

The left group is zero from the zero projective resolution. The right group is zero because the image in the torsion-free group $Z$ of an element killed by two must be zero. Exactness therefore gives $H^{5} ({R P}^{\infty}; Z) = 0$ .

1.3

The degree-one exception really has an integral natural lift. [F4, F5, F6, F7] For a mod-two cocycle $c$ , let $\hat{c}$ be its canonical valuewise zero/one integer lift and write $δ \hat{c} = 2 h$ . By [F5], the integral Bockstein is $[h]$ . The cochain $\hat{c} m o d 4$ is a lift through the mod-four coefficient sequence, and its coboundary is $2 h m o d 4$ . Pulling back along $2 : F_{2} \to Z / 4$ therefore gives $h m o d 2$ , so

$β_{2} ([c]) = ρ \tilde{β} ([c]) .$

By [F6] both sides are natural operations, and [F7] identifies the left side with $S q^{1}$ . Thus the integral Bockstein is precisely the promised integral-valued lift of $S q^{1}$ .

2.1

The lifting equation fails on $u$ . [F1, F4, step 1.1, step 1.2] Step 1.2 forces $Θ_{3} (u) = 0$ , so coefficient reduction gives $ρ Θ_{3} (u) = 0$ . Step 1.1 gives $S q^{2} (u) = a^{5} \neq 0$ . Hence $ρ Θ_{3} (u) \neq S q^{2} (u)$ , contradicting the defining property of $Θ$ . This single value rules out the lift without invoking naturality.

3.1

The boundary, qualification, and choice cases are explicit. [F1, F2, F3, F4, F5, F6, F7, A1, step 1.1, step 1.2, step 1.3, step 2.1] The witness space is nonempty and the input $a^{3}$ and failed output $a^{5}$ are nonzero; the zero class and unit are not witnesses. The indices $i = 2, j = 3$ lie inside the projective-space formula rather than an instability or truncation range, while the integral vanishing is calculated in the exact target degree five. Degenerate singular simplices remain in [F3]--[F5]. AC is inherited exactly from [F1], [F3], and [F6]. The item asserts nonexistence of one kind of lift and makes no biconditional claim; step 1.3 proves rather than merely asserts the integral Bockstein lift of $S q^{1}$ . ∎

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Sources

Hatcher, Algebraic Topology (standard reference, not scraped)
Miller, Algebraic Topology notes (standard reference, not scraped)