LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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PTFA localized first homology of Betti-one manifolds vanishes

Statement

Let $N$ be a compact connected Hausdorff topological manifold, with boundary allowed, and let $k$ be a commutative field with $\dim_{k} H_{1} (N; k) = 1$ . Let $ϕ : π_{1} (N) \to G$ be nontrivial with $G$ PTFA, and let $K$ be the Ore fraction division ring of $k [G]$ . Then $H_{0} (N; K) = H_{1} (N; K) = 0$ . No choice principle or finite CW model for $N$ is assumed.

Facts & Assumptions

Given: The compact connected manifold, coefficient field, nontrivial homomorphism and PTFA fraction division ring.

[F1]

Actual singular right-coefficient chains of a compact manifold have explicit finite-free domination, and Wall's finite algebraic construction gives a nonnegative bounded finite-projective chain model. Finite geometric chains and compact manifold chain domination without choice, Choice-free Wall chain finiteness from an explicit finite domination

[F2]

For such a projective model the PTFA basechange inequality is $b_{1}^{K} \leq b_{1}^{k} - 1$ if $b_{0}^{k} = 1$ and $b_{0}^{K} = 0$ . PTFA finite projective base change and Euler characteristic

Proof

1.1givenconstructalgebra

Degree-zero local homology identifies all point fibers with the base fiber by paths. A based loop with image $g \neq 1$ imposes $(g - 1) K = K$ , since $g - 1$ is a nonzero element of the embedded group ring and is invertible in $K$ . Hence $H_{0} (N; K) = 0$ , while ordinary connectedness gives $H_{0} (N; k) = k$ . Only a finite set of paths is needed for any given zero-cycle.

2.1F1F2step 1.1constructalgebra∎

Apply [F1] to the actual $k [G]$ coefficient complex. Tensoring its specified chain equivalence and homotopies with $k$ or $K$ preserves their identities; augmentation gives the ordinary complex because every group transport augments to $1$ . Thus its two models have the actual $k$ and $K$ homology of $N$ . By [F2] and step 1.1, $\dim_{K} H_{1} (N; K) \leq 1 - 1 = 0$ , so it is zero. The models and every subsequent elimination are bounded finite projective; no arbitrary infinite vector-space splitting or triangulation theorem is used.

Depends on

Used by

Dependency tree · two levels

13 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

Cochran, Orr and Teichner, Knot concordance, Whitney towers and L2-signatures (standard reference, not scraped)