LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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.

In source rank coordinates, the remaining components depend only on the rank coordinates

Statement

Assume the hypotheses of A nonzero rank minor supplies the source coordinates for the constant-rank theorem and that $f$ has constant rank $r$ near $a$ . In the source coordinates $y = Φ (x)$ , shrink to a rectangular neighbourhood $P \times Q \subseteq R^{r} \times R^{m - r}$ and write $g = f \circ Φ^{- 1}, g (u, v) = (u, h (u, v)) .$ Then $h (u, v)$ is independent of $v$ . When $r = 0$ , $g$ is locally constant; when $r = m$ , the $v$ block is empty; and when $r = n$ , the $h$ block is empty.

Facts & Assumptions

Given: A $C^{k}$ map of constant rank $r$ near $a$ and the local coordinates $Φ$ from the source-coordinate lemma.

[L1]

The source-coordinate map is a local $C^{k}$ diffeomorphism, the chain rule computes the derivative of $g = f \circ Φ^{- 1}$ , and a nonzero $(r + 1)$ -rowed minor forces rank at least $r + 1$ (A nonzero rank minor supplies the source coordinates for the constant-rank theorem, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, A matrix has rank at least $r$ exactly when it has a nonzero $r$ -rowed minor).

[L2]

A differentiable map with zero derivative on a nonempty connected open Euclidean set is constant (A differentiable map on a connected open Euclidean set has zero derivative exactly when it is constant).

Proof

technique · direct

1.1

Shrink the coordinate image to a product of open rectangles $P \times Q$ around $Φ (a)$ . By [L1], $g$ is $C^{k}$ , has constant rank $r$ , and its first $r$ components are the coordinates $u$ .

givenL1

2.1

If $r < n$ and $r < m$ , any nonzero partial derivative $\partial h_{i} / \partial v_{j}$ would join the identity $r \times r$ block of $D g$ to form a nonzero $(r + 1)$ -rowed minor, contradicting constant rank $r$ by [L1]. Hence all derivatives of $v \mapsto h (u, v)$ vanish.

step 1.1L1algebra

3.1

For each fixed $u$ , the rectangle $Q$ is connected and [L2] makes $v \mapsto h (u, v)$ constant. If $r = 0$ , the same argument applies to all components on the connected rectangle; if $r = m$ or $r = n$ , the relevant block is empty and the conclusion is immediate.

step 2.1L2

4.1

Thus, after shrinking, $g (u, v) = (u, h (u))$ in every rank regime.

step 3.1∎

Depends on

Used by

The Euclidean constant-rank normal form Theorem

Dependency tree · two levels

30 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

J. M. Lee, Introduction to Smooth Manifolds, proof of Theorem 7.13 (standard reference, not scraped)
L. W. Tu, An Introduction to Manifolds, proof of Theorem 11.1 (standard reference, not scraped)