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Top squares do not determine lower squares

Statement refuted

The identity $S q^{∣ z ∣} (z) = z^{2}$ does not determine the lower Steenrod squares, even when the degree and the value of the top square are fixed.

More explicitly, assume AC, base ${R P}^{2}$ at its zero-cell, and let $a$ be its nonzero class in $H^{1} ({R P}^{2}; F_{2})$ . If

$x = σ a \in {\tilde{H}}^{2} (Σ {R P}^{2}; F_{2})$

under the standard reduced cohomology-suspension isomorphism, and if $y$ is the nonzero class in $H^{2} (S^{2}; F_{2})$ , then

$x^{2} = 0 = y^{2}, S q^{1} (x) \neq 0, S q^{1} (y) = 0.$

Facts & Assumptions

Given: The based projective plane, its class $a$ , and the classes $x, y$ specified above.

[F1]

Bockstein detects integral two-torsion in real projective space states that, under AC, $S q^{1} (a) = a^{2}$ and that this class is nonzero.

[F2]

Under AC, Mod-two cohomology ring of infinite real projective space gives the infinite polynomial generator and says restriction to ${R P}^{2}$ is an isomorphism through degree two. The one-cell-per-degree construction and the integral incidence coefficients zero or two come from Real projective space cellular homology and the pinch map. By Axiomatic cellular boundaries are integral incidence matrices with coefficients, these coefficients act on $F_{2}$ , so they all vanish; then Cellular homology computes singular homology and field duality [F5] give $H^{q} ({R P}^{2}; F_{2}) = 0$ for $q > 2$ . In particular $a^{2}$ is nonzero and $a^{3} = 0$ .

[F3]

Steenrod normalization, instability, suspension, and top square gives $S q^{2} (z) = z^{2}$ for every degree-two class, makes $S q^{1}$ commute with the standard reduced cohomology suspension.

[F4]

Homology of spheres computes $H_{k} (S^{2}; F_{2})$ as $F_{2}$ for $k = 0, 2$ and zero otherwise.

[F5]

Cohomology over a field is dual to homology over that field turns [F4] into the corresponding mod-two cohomology calculation, under AC.

[A1]

The Axiom of Choice is used exactly through [F1], the infinite-ring and field-duality clauses in [F2], and [F5]. The cone-pair suspension and the Steenrod calculation in [F3] add no use of choice.

Counterexample

1.1

By definition, the standard reduced cohomology suspension $σ : {\tilde{H}}^{q} (Z; F_{2}) \to {\tilde{H}}^{q + 1} (Σ Z; F_{2})$ is an isomorphism; [F3] fixes this same standard suspension in its stability formula. In particular $x = σ a$ is nonzero.

2.1

The first square distinguishes the two classes. [F1, F2, F3, F4, F5, step 1.1] Stability and [F1] give

$S q^{1} (x) = S q^{1} (σ a) = σ S q^{1} (a) = σ (a^{2}) .$

The class $a^{2}$ is nonzero by [F1] (and explicitly by the ring [F2]); the degree-two instance of the isomorphism in step 1.1 therefore makes $S q^{1} (x)$ nonzero. On the other hand [F4]--[F5] give $H^{3} (S^{2}; F_{2}) = 0$ , so $S q^{1} (y) = 0$ .

2.2

Both top squares vanish. [F2, F3, F4, F5, step 1.1] The degree-three projective group is zero by [F2], so the degree-three instance of the suspension isomorphism gives ${\tilde{H}}^{4} (Σ {R P}^{2}; F_{2}) = 0$ . Thus [F3] gives

$x^{2} = S q^{2} (x) = 0.$

Likewise [F4]--[F5] give $H^{4} (S^{2}; F_{2}) = 0$ , whence $y^{2} = S q^{2} (y) = 0$ .

3.1

These computations refute determination by the top-square formula. [F3, step 2.1, step 2.2] The two nonzero degree-two classes have the same top-square value, namely zero, but different $S q^{1}$ values. Therefore knowing only $S q^{∣ z ∣} (z) = z^{2}$ cannot recover all lower squares.

4.1

The boundary and choice cases do not hide an exception. [F1, F2, F3, F4, F5, A1, step 1.1, step 2.1, step 2.2, step 3.1] Both spaces and both displayed input classes are nonempty and nonzero; the unit and zero classes are not the witnesses. The degree endpoint is exactly $∣ x ∣ = ∣ y ∣ = 2$ , so $S q^{1}$ is genuinely lower and $S q^{2}$ is genuinely top. The vanishing statements come from zero target groups, not from omitting degenerate singular simplices. AC is inherited exactly from the projective and field-duality computations [F1], [F2], and [F5]; suspension and all remaining calculations are choice-free. This is an explicit pair of witnesses, not either direction of a biconditional. ∎

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Sources

May, A Concise Course in Algebraic Topology (standard reference, not scraped)