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Loop removal in the weighted Kuramoto calculus

Statement

Let $A$ be a network on a nonempty finite set $V$ , with $W = A - diag (A_{x x})$ , row degrees $d_{A} (x)$ and $d_{W} (x)$ , and Laplacians $L_{A}$ and $L_{W}$ , as in Finite symmetric networks and loop removal. For phases $θ \in R^{V}$ define $E_{A} (θ) = \frac{1}{2} \sum_{x, y} A_{x y} (1 - \cos (θ_{x} - θ_{y}))$ and $F_{A} (θ)_{x} = - \sum_{y} A_{x y} \sin (θ_{x} - θ_{y})$ . Then $E_{A} = E_{W}$ , $F_{A} = F_{W} = - \nabla E_{A}$ , and $L_{A} = L_{W}$ . Moreover $f^{T} \nabla^{2} E_{A} (θ) f = \frac{1}{2} \sum_{x, y} A_{x y} \cos (θ_{x} - θ_{y}) (f_{x} - f_{y})^{2}$ for every real $f$ . Thus arbitrary diagonal weights affect row degrees but none of these energy or dynamical expressions.

Facts & Assumptions

[F1]

The network degree and Laplacian are row sum and degree matrix minus adjacency; removing the diagonal leaves the Laplacian unchanged (Finite symmetric networks and loop removal).

[F2]

The gradient and Hessian of a symmetric periodic pair potential are given by the finite pair-potential formulas (Differentiation of symmetric periodic pair potentials).

[F3]

For real angles, $z (t) = \cos t + i \sin t$ has modulus one and satisfies $z (u - v) = z (u) \overline{z (v)}$ . Also $\sin^{'} (t) = \cos t$ and $\cos^{'} (t) = - \sin t$ ; these identities and derivatives are $2 π$ -periodic (Phasors, trigonometric identities and periodic derivatives).

Proof

Given: The data in the Statement.

Goal: Establish the stated conclusions.

1.1F1F3algebra

Taking $u = v$ in [F3] gives $z (0) = ∣ z (u) ∣^{2} = 1$ , hence $\cos 0 = 1$ and $\sin 0 = 0$ . For $x = y$ , both $1 - \cos (θ_{x} - θ_{y})$ and $\sin (θ_{x} - θ_{y})$ vanish. Deleting exactly those summands therefore leaves $E_{A}$ and $F_{A}$ unchanged. Also $d_{A} (x) = d_{W} (x) + A_{x x}$ , so the identical diagonal correction in the degree matrix and in $A$ cancels in the Laplacian, giving $L_{A} = L_{W}$ .

2.1F2F3step 1.1algebra∎

Apply [F2] with $v (t) = 1 - \cos t$ , whose derivatives are $v' (t) = \sin t$ and $v'' (t) = \cos t$ . By [F3] and step 1.1, $z (- t) = z (0) \overline{z (t)} = \overline{z (t)}$ , so equality of real parts gives $\cos (- t) = \cos t$ . Thus $v$ is even. The derivative identities in [F3] make $v$ twice continuously differentiable, and its periodicity follows from [F3]. The resulting gradient is $- F_{A}$ , and its Hessian quadratic form is the displayed formula. Every diagonal Hessian term vanishes because $f_{x} - f_{x} = 0$ . This proves all assertions.

Depends on

Used by · 2

Dependency tree · 2 levels

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