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A smooth map of locally maximal rank has locally constant rank
Statement
Let be smooth and let . Suppose there is a neighbourhood of such that for every . Then has constant rank on some neighbourhood of .
Facts & Assumptions
Given: A smooth map , a point , and a neighbourhood on which the rank never exceeds .
is the rank of the differential at (The rank of a smooth map at a point).
For Euclidean smooth maps, the set where the differential has rank at least a fixed value is open (Differential rank is lower semicontinuous).
Charts identify neighbourhoods in manifolds with open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).
Proof
Choose charts around and as in [L2], and let be the coordinate representative of . By [F1], has rank , and the hypothesis says nearby ranks are at most .
By [L1], the locus where has rank at least is open. Since lies in that locus and nearby ranks are never above , there is a smaller Euclidean neighbourhood on which the rank is both at least and at most , hence exactly .
Pulling that neighbourhood back through the source chart, has constant rank on a neighbourhood of .
Depends on
Used by
- Rank at one point need not determine nearby rank False statement
Dependency tree · two levels
17 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, Maps of Constant Rank (standard reference, not scraped)