Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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A nonzero d-three in the K-AHSS for RP-two times RP-four

Example

Assume AC. In the complex K-theory Atiyah–Hirzebruch spectral sequence for RP2×RP4, let u and v be the degree-one mod-two generators and let z=βZ(uv)H3(;Z). Then d3(z)=βZ(uv4)0; equivalently d3 agrees with βSq2ρ2 on this class. No complex K-theory Künneth theorem or group-order argument is used.

Facts & Assumptions

[A1]

Assume AC. In the complex K-AHSS, d2=0 and d3=βZSq2ρ2 on all even coefficient rows (The first possible complex K-theory AHSS differential is integral Sq-three).

[A2]

Assume AC. For u,v the degree-one mod-two generators one has βZSq2ρ2(βZ(uv))=βZ(uv4)0 (A Bockstein class on RP-two times RP-four has nonzero integral Sq-three).

Verification

technique · direct

Given: Assume AC, the space RP2×RP4, its K-AHSS, and z=βZ(uv)H3(;Z).

1.1

The class z lies in E33,2: the coefficient row 2 is even and H3 is the integral cohomology degree of the target, and by [A1] the differential d3 on this row is the operation βZSq2ρ2.

A1given
2.1

By [A2] the value of that operation on z=βZ(uv) is βZ(uv4)0.

A2step 1.1
3.1

Therefore d3(z)=βZ(uv4) is nonzero, which exhibits a nonzero d3 and shows that the K-AHSS does not collapse for this space.

A1step 2.1
4.1

Steps 1.1 and 3.1 verify the displayed nonzero value of d3 without using a K-theory Künneth theorem.

step 3.1

Source notes

Compare Ji, §3.2.4, Figure 2 and Proposition 3.12, printed pp. 11–12, for the nonzero d3 on RP2×RP4.

Depends on

Used by

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Dependency tree · two levels

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Sources