LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Matrix completion detects the actual nontrivial bundle

Statement

Assume AC. For $q : A^{r + N} \to A^{N}$ with right inverse as above, say that $q$ is completable if it is the last $N$ rows of an invertible $(r + N) \times (r + N)$ matrix $B$ , equivalently $q = {pr}_{2} B$ under $A^{r + N} = A^{r} \oplus A^{N}$ . The associated rank- $r$ kernel bundle is trivial if and only if $q$ is completable. Thus failure of completion is a precise obstruction to trivializing this bundle, even though its displayed free stabilization 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 kernel is finite projective and the actual free stabilization is explicit. Stable triviality as the kernel of a right-invertible matrix

[F2]

The affine equivalence detects isomorphisms with finite free modules. Affine bundles and finite projective modules

Proof

1.1F1F2given

If $q = {pr}_{2} B$ and $B$ is invertible, restriction of $B$ identifies $\ker q$ with $\ker {pr}_{2} = A^{r} \oplus 0$ . Thus the kernel is free and its associated bundle is trivial by [F2].

2.1F1F2step 1.1construct∎

Conversely suppose $\ker q$ is free, and choose an isomorphism $u : A^{r} \to \ker q$ . The map $T : A^{r} \oplus A^{N} \to A^{r + N}$ , $T (v, w) = u (v) + j (w)$ , is an isomorphism by [F1] and satisfies $q T = {pr}_{2}$ . Hence $B = T^{- 1}$ is invertible and $q = {pr}_{2} B$ . This supplies the completion and proves the equivalence with actual triviality.

Depends on

Used by · 2

Dependency tree · 3 levels

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