Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-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.

Tractable order parameter inequalities — Lemma 2.5

PDF p. 15: Lemma 2.5 and proof; PDF p. 16: Lemma 2.5 and proof

Statement

Assume countable choice. Let G=(V,E) be a finite connected undirected graph, possibly with loops, with n=∣V∣>0 and symmetric binary adjacency matrix A. Connected means that every two vertices are joined by a finite edge path. Let d∗≥0 and 0≤α≤1/5, and suppose G is an (n,d∗,α)-expander: ∥A−(d∗/n)J∥≤αd∗, where J is the all-ones matrix and the norm is the Euclidean operator norm. Let θ be a stable state, meaning a local minimum on the phase torus of the energy EG(θ)=12∑x,y∈VAxy(1−cos⁡(θx−θy)). For k≥1 put ρk(θ)=n−1∑x∈Veikθx, using real representatives of the phases, and choose a common rotation so that ρ1 is real and nonnegative. Then ρ12≥1−3α and ∣ρ2∣≥1−5α2.

Facts & Assumptions

[F1]

Under the stated hypotheses and normalization, X=ρ12>0.431 (Cubic estimates used for the order parameter bootstrap).

[F2]

For the same connected expander and stable state, when ρ1 is real and positive, ∣ρ2∣≥1−2α2/ρ12 (Lower bound for the second order parameter — Corollary 2.4).

[F3]

For the same graph and stable state, ρ12≥(1+∣ρ2∣2)/2−2α (First and second order parameter inequality — Lemma 2.1).

Proof

Given: The connected expander, stable state, normalized order parameters, and parameter range in the Statement, under countable choice.

Goal: Prove both stated order-parameter bounds.

1.1F1F2givenalgebra

By [F1], ρ1 is positive and 2/ρ12<2000/431<5. Therefore [F2] gives ∣ρ2∣≥1−5α2. Since 0≤α≤1/5, the right side is at least 4/5 and in particular is nonnegative.

2.1F3step 1.1givenalgebra∎

Because both sides of the bound in step 1.1 are nonnegative, squaring preserves it. Substituting into [F3] yields ρ12≥1−2α−5α2+(25/2)α4. The assumption 0≤α≤1/5 gives 5α2≤α, while the fourth-power term is nonnegative. Thus ρ12≥1−3α, completing both conclusions.

Source notes

The statement is attributed to Expander graphs are globally synchronizing, Lemma 2.5. The cubic supplier treats the endpoint α=0 separately; the argument above uses its proved estimate rather than importing the article's claim as a premise.

Depends on

Used by · 1

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.

Sources