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The irrational torus flow is free with dense orbits
Statement
Let . The formula
defines a smooth free left action of on , and every orbit is dense. Its orbit map through ,
is an injective immersion and a Lie-group homomorphism.
Facts & Assumptions
Given: An irrational real number , the additive Lie group , and the usual torus Lie group .
A smooth left action is jointly smooth and satisfies the identity and associativity laws. Smooth left actions of Lie groups.
Among points placed in sets, two points lie in the same set. The pigeonhole principle on .
Proof
The displayed map is jointly smooth. Its phase factors satisfy and likewise with , so acts as the identity and ; hence it is a smooth left action by [F1]. If fixes any , then , so and . Irrationality forces , proving freeness.
The integer rotation orbit is dense in . Indeed, given , choose an integer and partition into half-open intervals of length . Applying [F2] to the fractional parts of gives integers with . Replacing by if necessary makes satisfy . For any , the integer satisfies and , so the orbit point is within of .
Fix a source point and a target . Parameters , , put the first coordinate exactly at , while their second-coordinate phases are . Step 1.2 makes the latter dense in , so some such parameters put the action point arbitrarily close to the target. Hence every orbit is dense.
The orbit map is a homomorphism by the phase calculation in step 1.1. It is injective because implies by the freeness calculation. Its differential sends to the nonzero tangent vector ; translating the homomorphism identity shows its differential is injective everywhere. Thus is an injective immersion. All choices above are finite or single choices, so no choice principle is used.
Depends on
Used by
- A free irrational torus action that is not proper Counterexample
- An irrational line as a dense immersed Lie subgroup of a torus Example
- A free action need not have a manifold quotient False statement
- A homomorphism image need not be embedded False statement
- A Lie subalgebra need not integrate to a closed subgroup False statement
- Not every Lie subgroup is embedded and closed False statement
Dependency tree · two levels
13 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, 2nd ed. (standard reference, not scraped)
- Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I (standard reference, not scraped)