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Regular critical PSD phases: kernel and half-circle

Statement

For a regular critical PSD phase configuration as in the preceding lemma, put $K (u, v) = \sin (∣ u ∣ - \min (∣ v ∣, π / 2))$ . Then $\sum_{x} A_{x y} K (θ_{x}, θ_{y}) \geq 0$ for every $y$ . If the off-diagonal graph is connected and every $∣ θ_{x} ∣ < π / 2$ , all phases equal zero under the normalization $ρ \geq 0$ . Thus a normalized nonsynchronized critical PSD state on a connected graph has $∣ {x : ∣ θ_{x} ∣ \geq π / 2} ∣ > 0$ .

Facts & Assumptions

[F1]

Use the graph, phase, algebraic and analytic interfaces of Regular critical PSD phases: a direct mass bootstrap at 0.0816, Angular order and elementary inverse-trigonometric bounds. Every configuration assumed here is a critical point with positive-semidefinite Hessian; local minimality is never inferred from that condition.

[F2]

The off-diagonal graph is the support graph with vertex set $V$ and edges ${x, y}$ where $x \neq y$ and $A_{x y} > 0$ (Finite symmetric networks and loop removal). It is connected when every two vertices can be joined by a vertex path (Connected graphs and connected components defined by the existence of vertex paths).

[F3]

A synchronized state means that all its phases are equal modulo $2 π$ ; nonsynchronized means that this condition fails. The phase quotient, $e^{i u} = \cos u + i \sin u$ , the phasor difference identities, and real and imaginary parts of finite sums are supplied by Phasors, trigonometric identities and periodic derivatives.

[F4]

Use the real sine and cosine addition formulas (The addition formulas for sine and cosine), oddness of sine and evenness of cosine (Parity and the Pythagorean identity for sine and cosine), quarter-turn values and shift formulas (Quarter-turn values and shifts by pi/2 and pi), and the sine and cosine zero sets (The zero sets of sine and cosine and the least positive common period 2 pi). Together with the nonnegativity of sine and monotonicity of cosine in [F1], these give $\sin t > 0$ for $0 < t < π$ , $\sin t < 0$ for $- π < t < 0$ , and the signs of cosine on either side of $π / 2$ for principal absolute phases.

Proof

Given: The hypotheses of the Statement.

Goal: Establish its conclusions directly for critical PSD phases.

1.1F1F3F4algebra

The phasor identity $(A e^{i θ})_{y} = r_{y} e^{i θ_{y}}$ with $r_{y} \geq 0$ implies $\sum_{x} A_{x y} \sin (θ_{x} - θ_{y}) = 0$ and $\sum_{x} A_{x y} \cos θ_{x} = r_{y} \cos θ_{y}$ by symmetry of $A$ and taking imaginary and real parts. If $0 \leq θ_{y} \leq π / 2$ , expansion of $\sin (∣ θ_{x} ∣ - θ_{y})$ and $\sin ∣ θ_{x} ∣ \geq \sin θ_{x}$ gives $K (θ_{x}, θ_{y}) \geq \sin (θ_{x} - θ_{y})$ , because $\cos θ_{y} \geq 0$ . For $- π / 2 \leq θ_{y} \leq 0$ , the corresponding inequality is $K (θ_{x}, θ_{y}) \geq - \sin (θ_{x} - θ_{y})$ , using $\sin ∣ θ_{x} ∣ \geq - \sin θ_{x}$ . Summing against $A_{x y} \geq 0$ gives a nonnegative sum. For $∣ θ_{y} ∣ \geq π / 2$ , the quarter-turn formula and cosine parity give $K (θ_{x}, θ_{y}) = - \cos θ_{x}$ , so the sum is $- r_{y} \cos θ_{y} \geq 0$ .

2.1F1F2F3F4step 1.1algebra∎

If every absolute phase is below $π / 2$ , choose a vertex with maximal absolute phase $c$ . At any vertex attaining $c$ , step 1.1 gives $\sum_{x} A_{x y} \sin (∣ θ_{x} ∣ - c) \geq 0$ . Every term is nonpositive and is negative for an adjacent smaller absolute phase. Thus every neighbor also attains $c$ , and propagation along paths gives $∣ θ_{x} ∣ = c$ at every vertex. If both signs $c, - c$ occur with $c > 0$ , put $S_{\pm} = {x : θ_{x} = \pm c}$ . A path between the two nonempty classes has an edge across them, so $\sum_{x \in S_{+}, y \in S_{-}} A_{x y} > 0$ . Summing the sine identities from step 1.1 over $y \in S_{-}$ gives $0 = \sin (2 c) \sum_{x \in S_{+}, y \in S_{-}} A_{x y}$ , impossible since $0 < 2 c < π$ . If only one sign occurs, all phases have a common value $t$ with $∣ t ∣ < π / 2$ ; real nonnegative $ρ = e^{i t}$ gives $\sin t = 0$ , and the sine zero set forces $t = 0$ . This also covers $c = 0$ . Thus all phases are equal, in the phase quotient of [F3], and the stated contrapositive follows.

Depends on

Used by · 3

Dependency tree · 2 levels

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