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Stable triviality as the kernel of a right-invertible matrix

Statement

Assume AC. Let $X = Spec A$ , and let $E$ be a bundle of constant rank $r$ . Then $E$ is stably trivial if and only if, for some $N \geq 0$ , it is the bundle associated to $\ker q$ for an $A$ -linear map $q : A^{r + N} \to A^{N}$ admitting a right inverse $j : A^{N} \to A^{r + N}$ , $q j = id$ . Such a map is a right-invertible matrix after the standard free bases are fixed. Its kernel $P$ is finite projective of rank $r$ , with actual splitting $P \oplus A^{N} \tilde{\to} A^{r + N}$ given by $(v, w) \mapsto v + j (w)$ .

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]

Stable triviality means an actual isomorphism $E \oplus O_{X}^{N} ≅ O_{X}^{r + N}$ for some N. Stable triviality and stable bundle equivalence

[F2]

The affine equivalence preserves rank, direct sums and actual bundle/module isomorphisms. Affine bundles and finite projective modules

Proof

1.1F1F2construct

If $E$ is stably trivial, use [F2] to obtain an isomorphism $α : P \oplus A^{N} \to A^{r + N}$ , where $P = Γ (X, E)$ . Transport projection onto $A^{N}$ and its standard section through $α$ ; they give $q = {pr}_{2} α^{- 1}$ and $j (w) = α (0, w)$ . Then $q j = id$ and $α (P \oplus 0) = \ker q$ .

2.1F1F2step 1.1algebra∎

Conversely, if $q j = id$ , the displayed splitting has inverse $z \mapsto (z - j q (z), q (z))$ . The first coordinate is in $\ker q$ . Hence this kernel is a direct summand of a finite free module, thus finite projective. Localize the decomposition at each prime; its local free rank is $(r + N) - N = r$ . By [F2] its associated bundle, stabilized by the indicated trivial bundle, is trivial.

Depends on

Used by · 3

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