Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Expander mixing lemma

Statement

For any subsets S,T of a finite d-regular adjacency-slot graph on n≥1 vertices, let e(S,T)=∑u∈S,v∈TAuv count ordered slots. Then ∣e(S,T)−d∣S∣∣T∣n∣≤αd∣S∣(1−∣S∣/n) ∣T∣(1−∣T∣/n). Overlap and loop slots are allowed.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

For the regular multigraph and spectral conventions in the stated convention, put α=∥M∣1⊥∥. For n≥2 order the eigenvalues 1=μ1≥μ2≥⋯≥μn, counting multiplicity, and put γ=1−μ2. Thus α=max⁡j≥2∣μj∣, which also controls negative eigenvalues. Write cut⁡(S)=∑u∈S,v∉SAuv and ∂VS={v∉S:Auv>0 for some u∈S}. Normalized edge expansion and external vertex expansion are h=min⁡0<∣S∣≤n/2cut⁡(S)d∣S∣,hV=min⁡0<∣S∣≤n/2∣∂VS∣∣S∣. For n=1, put α=0 and leave μ2,γ,h,hV undefined; cut-expansion assertions are vacuous. A bounded-degree family is an expander family when its normalized edge expansion has a positive uniform lower bound for n≥2. Polynomial-time constructibility means a uniform algorithm outputs the adjacency list in time polynomial in n; neighbor computation in time polynomial in log⁡n is a stronger requirement. (Spectral edge and vertex expansion).

[F2]

For vectors u,v in a real or complex inner product space, ∣⟨u,v⟩∣≤∥u∥∥v∥. Equality holds if and only if u and v are linearly dependent, including the case in which either vector is zero. (Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

Proof

1.1F1

Set s=∣S∣/n, t=∣T∣/n, f=1S−s1 and g=1T−t1. Both are mean zero and have normalized squared norms s(1−s) and t(1−t). Since M preserves constants and their orthogonal complement, e(S,T)/(nd)−st=⟨f,Mg⟩. This counts loops and overlap exactly as specified.

2.1F2step 1.1∎

Cauchy–Schwarz and the defining operator bound give ∣⟨f,Mg⟩∣≤αs(1−s)t(1−t). Multiply by nd. Empty or full sets give zero centered vectors and equality; at n=1 all sets are of that form. No division by a set size or by α is made.

Depends on

Used by

Dependency tree · two levels

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Sources