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.
The fibrewise quotient of a vector bundle by arbitrary varying subspaces is a vector bundle
Statement
The fibrewise quotient of a vector bundle by arbitrary varying subspaces is always a smooth vector bundle.
Facts & Assumptions
Given: The displayed claim.
A quotient bundle theorem requires a smooth vector subbundle, in particular constant fibre dimension and smooth local frames (A vector bundle quotient by a subbundle is a smooth vector bundle, Vector subbundles).
Refutation
In the trivial line bundle , let for and let . Then the quotient fibre is one-dimensional for and zero-dimensional at .
A smooth vector bundle has locally constant fibre dimension, so this family of quotients cannot be a vector bundle. The missing hypothesis is exactly that the subspaces form a smooth subbundle as in [L1].
Depends on
Used by
- A rank-jumping kernel is not a vector subbundle Counterexample
Dependency tree · two levels
7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- John M. Lee, Introduction to Smooth Manifolds (standard reference, not scraped)
- Will J. Merry, Differential Geometry (standard reference, not scraped)