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

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Differential rank is lower semicontinuous

Statement

Let $f : U \subseteq R^{m} \to R^{n}$ be $C^{1}$ on an open set. For every natural $r$ , the locus $U_{\geq r} : = {x \in U : rank D f (x) \geq r}$ is open. Thus $x \mapsto rank D f (x)$ is lower semicontinuous. In particular the submersion locus, the immersion locus, and every locus on which the derivative has the largest possible rank are open.

Facts & Assumptions

Given: A $C^{1}$ map $f : U \subseteq R^{m} \to R^{n}$ and a natural number $r$ .

[L1]

For $1 \leq r \leq \min {m, n}$ , a matrix has rank at least $r$ exactly when it has a nonzero $r$ -rowed minor (A matrix has rank at least $r$ exactly when it has a nonzero $r$ -rowed minor, The rank of a derivative and constant-rank Euclidean maps, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L2]

Each fixed-size determinant is a polynomial in the matrix entries, while the first partial derivatives of a $C^{1}$ map are continuous; sums, products, and composites of continuous Euclidean maps are continuous, and the empty set and whole metric space are open (For every fixed finite size at least one, the determinant of a real square matrix is a polynomial in its matrix entries, $C^{k}$ Euclidean maps and diffeomorphisms, $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · direct

1.1

If $r = 0$ , then $U_{\geq r} = U$ ; if $r > \min {m, n}$ , then $U_{\geq r} = \emptyset$ . Both sets are open.

givenL2

1.2

Assume $1 \leq r \leq \min {m, n}$ and fix $a \in U_{\geq r}$ . By [L1], one $r$ -rowed minor $M (a)$ of $J f (a)$ is nonzero.

givenL1choose

2.1

By [L2], the same minor $M (x)$ is a continuous scalar function of $x$ . Its nonzero locus contains an open neighbourhood $V$ of $a$ , and [L1] gives $V \subseteq U_{\geq r}$ .

step 1.2L1L2

3.1

Every point of $U_{\geq r}$ therefore has an open neighbourhood inside it, and the two exceptional cases were settled in step 1.1. Hence $U_{\geq r}$ is open for every $r$ , proving all stated consequences.

step 1.1step 2.1∎

Depends on

Used by

A Euclidean immersion is locally the canonical inclusion and is locally an embedding Corollary
A Euclidean submersion is locally a coordinate projection Corollary
A regular level set is locally a Cᵏ graph of dimension m-n Corollary
FALSE: continuity of the derivative implies constant rank False statement
Lagrange multipliers for a regular vector-valued level-set constraint Theorem

Dependency tree · two levels

35 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, rank theorem discussion (standard reference, not scraped)
L. W. Tu, An Introduction to Manifolds, Section 11.1 (standard reference, not scraped)