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Stable triviality fixes the K-zero class and determinant

Statement

Assume AC. If a rank- $r$ bundle $E$ on a noetherian affine scheme $X$ has a stable trivialization $E \oplus O_{X}^{N} ≅ O_{X}^{r + N}$ , then $[E] = r [O_{X}]$ in the article's perfect spectral $K_{0} (X)$ , and $\det E ≅ O_{X}$ . Here $\det E = Λ^{r} E$ is the determinant line. These conclusions use the actual stable isomorphism; they do not assert that the bundle itself is trivial.

Facts & Assumptions

Given: The data and hypotheses in the statement, AC and the article's fixed universe bounds where its K-theory is used.

Goal: Establish the stated conclusion.

[F1]

The affine comparison takes vector-bundle exact-sequence classes to the article perfect spectral object classes. Isomorphism classes of vector bundles

[F2]

Determinants are the top exterior powers with their local wedge bases. Exterior powers and determinant of an algebraic bundle

[F3]

The stable trivialization is the actual isomorphism in Stable triviality and stable bundle equivalence.

Proof

1.1F1F3algebra

The split exact-sequence relation gives $[E] + N [O_{X}] = (r + N) [O_{X}]$ . Subtract the common class in the abelian group to obtain $[E] = r [O_{X}]$ , and use [F1] to read it in the spectral group.

2.1F2F3step 1.1construct∎

Take top exterior powers of the stable isomorphism. Wedge the local basis of $E$ before the standard basis of the trivial summand; this gives $\det (E \oplus O_{X}^{N}) ≅ \det E$ . The construction glues because changes of frame act by their determinant. The determinant of $O_{X}^{r + N}$ is canonically trivial, so $\det E$ is trivial.

Depends on

Used by · 1

A nontrivial rank-two bundle with trivial Chern classes on a smooth complex affine fourfold

Dependency tree · 3 levels

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