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Centered-adjacency pointwise phase bounds

PDF p. 14: Proof of Lemma 2.3

Statement

Assume countable choice. Let $G = (V, E)$ be a finite undirected graph with $n = ∣ V ∣ > 0$ , symmetric binary adjacency matrix $A$ , and possible self-loops. Let $d^{*} > 0$ and $Δ_{A} = A - (d^{*} / n) J$ , where $J$ is the all-ones matrix. Let $θ$ be a stable state, meaning a local minimum of $E_{G} (θ) = \frac{1}{2} \sum_{x, y} A_{x, y} (1 - \cos (θ_{x} - θ_{y}))$ on the phase torus. Choose representatives $θ_{x} \in (- π, π]$ after a common rotation for which $ρ_{1} = n^{- 1} \sum_{x} e^{i θ_{x}}$ is real and nonnegative. For $y \in V$ put $r_{y} = \sum_{x} A_{x, y} \cos (θ_{x} - θ_{y})$ and $S_{y} = \sum_{x} (Δ_{A})_{x, y} e^{i (θ_{x} - θ_{y})}$ .

Let $H (t) = \sin^{2} (∣ t ∣)$ for $∣ t ∣ \leq π / 2$ and $H (t) = 1$ for $π / 2 \leq ∣ t ∣ \leq π$ , on $(- π, π]$ . Then, for every $y \in V$ ,

$r_{y} = d^{*} ρ_{1} e^{- i θ_{y}} + S_{y} \geq 0,$

$\sum_{x} (Δ_{A})_{x, y} \sin (θ_{x} - θ_{y}) = d^{*} ρ_{1} \sin θ_{y}, \sum_{x} (Δ_{A})_{x, y} \cos (θ_{x} - θ_{y}) \geq - d^{*} ρ_{1} \cos θ_{y},$

and $∣ S_{y} ∣^{2} \geq (d^{*} ρ_{1})^{2} H (θ_{y})$ . The inequality includes the case $ρ_{1} = 0$ .

Facts & Assumptions

[F1]

The centered adjacency definition gives $A = Δ_{A} + (d^{*} / n) J$ with $J_{x, y} = 1$ (Centered graph matrices and reference degree).

[F2]

The first Daido parameter is $ρ_{1} = n^{- 1} \sum_{x} e^{i θ_{x}}$ ; a common rotation can make it real and nonnegative, and the rotated phases have representatives in $(- π, π]$ (Daido order parameters, Bounds, saturation, and rotation of order parameters).

[F3]

Under countable choice, a stable state is a local energy minimum. For the stated finite graph, stability gives $r_{y} \geq 0$ and $\sum_{x} A_{x, y} e^{i θ_{x}} = r_{y} e^{i θ_{y}}$ , including when there is a self-loop at $y$ (The Axiom of Countable Choice (AC_ω), Equilibria, stable states, synchronized states, and spurious minima, Second-order necessary conditions for stable states).

[F4]

For real angles $u, v$ , $e^{i u} = \cos u + i \sin u$ , $∣ e^{i u} ∣ = 1$ , and $e^{i (u - v)} = e^{i u} \overline{e^{i v}}$ ; these expressions are independent of representatives (Phasors, trigonometric identities and periodic derivatives).

[F5]

The displayed two branches define $H$ on $(- π, π]$ and agree at their overlap (Truncated sine-square function).

[F6]

Cosine decreases on $[0, π]$ , and $\cos (π / 2) = 0$ (Angular order and elementary inverse-trigonometric bounds, Quarter-turn values and shifts by pi/2 and pi).

Proof

Given: The graph, positive $d^{*}$ , stable state and normalized representatives in the statement, under countable choice.

Goal: Prove the displayed identity, real and imaginary part relations, and pointwise squared-modulus bound for each $y \in V$ .

1.1F1F2F3F4algebra

Fix $y$ . Multiply the equilibrium identity in [F3] by $e^{- i θ_{y}}$ . Substituting [F1] and using [F2] gives $r_{y} = (d^{*} / n) e^{- i θ_{y}} \sum_{x} e^{i θ_{x}} + S_{y} = d^{*} ρ_{1} e^{- i θ_{y}} + S_{y}$ . The left side is a nonnegative real number by [F3].

2.1F1F3F4step 1.1algebra

Write $S_{y} = C_{y} + i T_{y}$ , where $C_{y} = \sum_{x} (Δ_{A})_{x, y} \cos (θ_{x} - θ_{y})$ and $T_{y} = \sum_{x} (Δ_{A})_{x, y} \sin (θ_{x} - θ_{y})$ . The entries of $Δ_{A}$ are real. Taking imaginary and real parts in step 1.1 gives $T_{y} = d^{*} ρ_{1} \sin θ_{y}$ and $C_{y} = r_{y} - d^{*} ρ_{1} \cos θ_{y} \geq - d^{*} ρ_{1} \cos θ_{y}$ .

3.1F2F4F5step 2.1algebra

Put $a = d^{*} ρ_{1} \geq 0$ . From [F4], $\cos^{2} t + \sin^{2} t = 1$ and $\cos (- t) = \cos t$ , $\sin (- t) = - \sin t$ . If $∣ θ_{y} ∣ \leq π / 2$ , step 2.1 gives $∣ S_{y} ∣^{2} = C_{y}^{2} + a^{2} \sin^{2} θ_{y} \geq a^{2} H (θ_{y})$ by the first branch of [F5].

4.1F5F6step 2.1step 3.1algebra∎

If $π / 2 \leq ∣ θ_{y} ∣ \leq π$ , [F6] and evenness of cosine give $\cos θ_{y} \leq 0$ . Thus step 2.1 gives $C_{y} \geq - a \cos θ_{y} \geq 0$ , so $C_{y}^{2} \geq a^{2} \cos^{2} θ_{y}$ . Adding $T_{y}^{2} = a^{2} \sin^{2} θ_{y}$ yields $∣ S_{y} ∣^{2} \geq a^{2} = a^{2} H (θ_{y})$ by [F5]. The two cases cover the domain, including their common boundary and $a = 0$ , proving the result.

Source notes

The pointwise claim is attributed to Expander graphs are globally synchronizing, PDF p. 14, proof of Lemma 2.3. The proof above uses the stated mathematical suppliers.

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Sources

Expander graphs are globally synchronizing — Abdalla, Bandeira, Kassabov, Souza, Strogatz, Townsend (2026) · PDF p. 14, proof of Lemma 2.3