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Lower bound for the second order parameter — Corollary 2.4

PDF p. 15: Corollary 2.4, equation (2.10)

Statement

Assume countable choice. Let G=(V,E) be a connected finite undirected graph with n=∣V∣>0, symmetric binary adjacency matrix A and possible self-loops. Connected means that every two vertices can be joined by a finite edge path. Let d∗,α∈R, 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 EG(θ)=12∑x,y∈VAxy(1−cos⁡(θx−θy)). For k≥1, define its Daido order parameter by ρk(θ)=n−1∑x∈Veikθx, using real representatives of the phases. If ρ1(θ) is real and positive, then ∣ρ2(θ)∣≥1−2α2/ρ1(θ)2.

Facts & Assumptions

[F1]

Under the statement's hypotheses, after choosing representatives θx∈(−π,π], Lemma 2.3 gives nα2≥ρ1(θ)2∑xH(θx). It assumes countable choice and applies to connected graphs, including zero reference degree (Order parameter control of total H mass — Lemma 2.3, The Axiom of Countable Choice (AC_ω)).

[F2]

On (−π,π], H(t)=sin⁡2(∣t∣) for ∣t∣≤π/2 and H(t)=1 for π/2≤∣t∣≤π; in particular H(t)≥sin⁡2t (Truncated sine-square function, Elementary properties of the function H).

[F3]

The Daido parameter is the stated average, independent of the representatives. For real t, eit=cos⁡t+isin⁡t, ∣eit∣=1, and ei(u+v)=eiueiv; the real part of a finite sum is the sum of its real parts (Daido order parameters, Phasors, trigonometric identities and periodic derivatives).

[F4]

For z=a+ib with a,b∈R, Re⁡z=a and ∣z∣=a2+b2, where the square root is nonnegative (Real and imaginary parts, complex conjugation, and modulus).

Proof

Given: The connected expander and stable state in the statement, with ρ1(θ)>0 real, under countable choice.

Goal: Prove ∣ρ2(θ)∣≥1−2α2/ρ1(θ)2.

1.1F2F3F4givenalgebra

Choose each phase representative in (−π,π]. By [F3] and [F4], cos⁡2t+sin⁡2t=∣eit∣2=1; squaring eit=cos⁡t+isin⁡t and using the addition law gives cos⁡(2t)=cos⁡2t−sin⁡2t=1−2sin⁡2t. For any z=a+ib, put m=∣z∣. By [F4], m≥0 and m2=a2+b2. If m<a, then a>0 and 0<(a−m)(a+m)=−b2≤0, a contradiction. Thus ∣z∣≥Re⁡z. Hence ∣ρ2∣≥Re⁡ρ2=1−(2/n)∑xsin⁡2θx≥1−(2/n)∑xH(θx).

2.1F1step 1.1givenalgebra∎

Because n>0 and ρ1>0, [F1] gives (1/n)∑xH(θx)≤α2/ρ12. Substitution into step 1.1 proves the claimed bound.

Source notes

The corollary and its standing connectedness assumption are attributed to Expander graphs are globally synchronizing, PDF pp. 11 and 15. The proof above derives the conclusion from the stated suppliers.

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