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A split trivial line annihilates the exterior and top Chow classes

Statement

Assume AC. Let $r \geq 1$ and let $E ≅ O_{X} \oplus F$ be a rank- $r$ bundle on a noetherian affine scheme, with $F$ of rank $r - 1$ . Then $ε (E) = 0$ . If $X$ is smooth and equidimensional over a field, then also $c_{r} (E) = 0$ in the ordinary integral Chow group ${C H}^{r} (X)$ .

Facts & Assumptions

Given: The hypotheses and data in the statement, with AC and the article's fixed universes wherever perfect K-theory is used.

Goal: Establish the asserted conclusion.

[F1]

The actual exterior-dual class is defined in The actual exterior-dual class in affine perfect K zero.

[F2]

Exterior powers have their increasing-wedge bases. Exterior powers and determinant of an algebraic bundle

[F3]

Whitney gives $c_{t} (E) = c_{t} (O_{X}) c_{t} (F) = c_{t} (F)$ , and Chern classes vanish above the rank. Operational Chern classes and the Whitney formula, Chern classes of a vector bundle on a smooth scheme

Proof

1.1F1F2algebra

Dualize the splitting. For each $i$ , exterior multiplication identifies $Λ^{i} (O_{X} \oplus F^{\lor})$ with $Λ^{i} F^{\lor} \oplus Λ^{i - 1} F^{\lor}$ , where nonexistent exterior degrees are zero: wedges either contain the chosen generator 1 of the trivial line or do not. The basis formula proves the identification locally and its naturality in $F$ glues it. In the finite alternating sum, each $[Λ^{i} F^{\lor}]$ occurs once with each opposite sign, so the total is zero.

2.1F3step 1.1∎

By [F3], $c_{t} (E) = c_{t} (F)$ and the rank- $(r - 1)$ bundle $F$ has zero coefficient in degree $r$ . Hence $c_{r} (E) = 0$ . This is the necessary direction for an actual splitting.

Depends on

Used by · 2

Dependency tree · 3 levels

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