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The first two perfect Chern classes vanish on the third image piece in full scope

PDF p. 14: Lemma6.4; scheme conventions p3

Statement

Assume the Axiom of Choice and the fixed site and universe bounds of the supplied constructions. For every qcqs k-scheme X over any field, every α∈Fmot3K0(X) has rank zero and c1(α)=c2(α)=0. This uses actual stage maps and needs neither normality nor an image-stage isomorphism.

Facts & Assumptions

[A1]

Assume the Axiom of Choice as formulated in The Axiom of Choice and the stated size bounds. Choice and the fixed size bounds are inherited from the first-layer comparison in [F2] at step 1.1 and the actual second-fiber comparison in [F3] at step 2.1.

[F1]

Motivic filtration and image groups defines nested image groups. Stage projections, kernels and image groups gives the consecutive stage-to-layer fiber sequences.

[F2]

The globally normalized first-layer Picard map is determinant and First Chern class factors through determinant identify the actual first layer with determinant/first Chern on the rank-zero space for all qcqs field-schemes.

[F3]

Actual second-stage comparison on every qcqs field-scheme identifies the actual second layer with minus the actual second Chern operation in the same scope.

Proof

Given: X,α and a third-stage representative a∈π0Fil⁡3K(X), which exists by the image definition.

Goal: Prove lower-class vanishing without transporting projections through a finite-type image-stage identification.

1.1A1F1F2

Send a to the second, first and zeroth stages. Their common augmentation is α. Consecutive stage-to-layer composites are null by their fiber sequences. In particular its rank is zero, and the actual first-layer map of the first-stage representative is zero. By [F2] this gives c1(α)=0.

2.1A1F1F3step 1.1∎

Its second-stage representative is in the image of the third stage, so its actual second-layer value is zero. By [F3] that value is −c2(α). Thus c2(α)=0. No injectivity of any stage augmentation or representation of α by global bundles has been used.

Depends on

Used by · 3

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