Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generated
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.

Density mixing preserves admissibility but need not preserve idempotence

Statement

If D∈Dr and the gapped map Φ(D) of The occupied-projector response map on its strict-gap domain is defined, then for every α∈[0,1], D+=(1−α)D+αΦ(D) belongs to Dr. The conclusion does not assert D+∈Pr.

Facts & Assumptions

Given: D∈Dr, a defined occupied projector Φ(D) of rank r, and 0≤α≤1.

[F1]

Dr is the set of Hermitian positive contractions of trace r (The fixed-particle admissible Hermitian density domain).

[F2]

The gapped response construction gives a unique orthogonal projector Φ(D)∈Pr (The occupied-projector response map on its strict-gap domain, A strict occupied gap defines a unique rank-r spectral projector).

[F4]

Every rank-r orthogonal projector is an admissible density (Every rank-r orthogonal projector is an admissible density).

Proof

1.1F1F2F4givenalgebra

Since D and Φ(D) are Hermitian, D+ is Hermitian. For every x, x∗D+x=(1−α)x∗Dx+αx∗Φ(D)x≥0. Likewise x∗(I−D+)x=(1−α)x∗(I−D)x+αx∗(I−Φ(D))x≥0, since D∈Dr and Φ(D)∈Dr by [F1], [F2], [F4].

2.1step 1.1F1F2algebra

The trace is linear, so tr⁡D+=(1−α)r+αr=r. The two positive-semidefinite inequalities and this trace identity give D+∈Dr.

3.1step 2.1F2givenalgebra∎

To verify that idempotence need not be preserved, take n=2, r=1, the constant Hermitian response F(X)=diag⁡(0,1) on U=H2, D=diag⁡(0,1), and α=1/2. Directly, D is Hermitian, lies between 0 and I2, and has trace 1, so D∈D1. The response matrix has gap 1, so [F2] gives Φ(D)=diag⁡(1,0). Then D+=12I2 and (D+)2=14I2≠D+. Thus the update belongs to D1 by step 2.1 but is not an orthogonal projector.

Depends on

Used by

Dependency tree · two levels

10 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