Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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.

An irrational line as a dense immersed Lie subgroup of a torus

Example

Assume ACω and fix αRQ. Then

i:RT2,i(t)=(e2πit,e2πiαt)

identifies R with a one-dimensional immersed Lie subgroup whose image is dense, proper, nonclosed, and nonembedded in T2.

Facts & Assumptions

Given: ACω, an irrational real number α, and the displayed winding homomorphism i.

[A1]

The winding map is an injective immersion and homomorphism, and its image is dense. The irrational torus flow is free with dense orbits.

[A2]

The homomorphism-image theorem equips its image with the unique intrinsic immersed-subgroup structure for which the corestriction is a submersion. The Axiom of Countable Choice (ACω), Images are immersed Lie subgroups.

[F1]

Embeddedness means that this intrinsic topology agrees with the ambient subspace topology. Immersed, embedded, and closed Lie subgroups.

Verification

Proof technique: calculate the image and compare its intrinsic and ambient topologies.

1.1

By [A1], i is an injective immersed homomorphism with dense image. Since its kernel is trivial, the canonical image structure in [A2] is transported from the one-dimensional source R.

A1A2
1.2

The image is proper. The point (1,eπiα) is not in it: equality of the first coordinate would force t=nZ, while equality of the second would make α(n12) an integer, impossible because a nonzero rational multiple of irrational α is irrational. A proper dense subset is not closed.

A1algebra
2.1

For each j1, let qj be the least positive integer satisfying qjα<1/j, whose existence is the finite-pigeonhole calculation in [A1]. Irrationality makes every fixed qα positive, so qj. Nevertheless i(qj)=(1,e2πiαqj)(1,1)=i(0) in the ambient subspace. Hence the inverse of i on its image is not continuous, so [F1] shows that the subgroup is not embedded. Leastness makes the sequence choice-free; ACω is inherited only through the general image supplier [A2]. The source dimension is exactly one, its tangent (1,α) is nonzero, and no endpoint is present.

A1A2F1step 1.1algebra

Depends on

Used by

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Dependency tree · two levels

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Sources