Example. The positive action of the alternating cycle [fdip-002P]
Example. The positive action of the alternating cycle [fdip-002P]
Use the four-cycle oriented \(0\to 1\to 2\to 3\to 0\), with edge weights \((1,t,1,t)\), \(t>0\). Let \(D:X_V\to X_E\) evaluate endpoint differences and \(W\) multiply by the supplied edge weights. With counting adjoint \(D^\dagger \), put \[L=D^\dagger WD,\qquad Q=L+m^2\operatorname {id},\qquad m>0.\] For every \(x,y\), the energy interface gives \[h_V(x,Ly)=\sum _e w(e)(Dx)(e)(Dy)(e).\] It is symmetric in \(x,y\), so \(Q\) is self-adjoint. On the diagonal all edge terms are nonnegative and \(m^2h_V(x,x)>0\) off zero. Thus \(Q\) is a positive precision parameter. At zero mass constant fields instead expose the Laplacian kernel.
In the specified vertex order \((0,1,2,3)\), write \(x=(a,b,c,d)\). The 4 differences are \(b-a,c-b,d-c,a-d\), so the action is exactly \[\begin {aligned} 2S_{t,m}(a,b,c,d) &=(b-a)^2+t(c-b)^2\\ &\quad +(d-c)^2+t(a-d)^2\\ &\quad +m^2(a^2+b^2+c^2+d^2). \end {aligned}\] Each edge occurs once, with one half of its weighted energy in \(S\). Every summand of \(2S\) is nonnegative, with strict positivity off zero from the mass term.
Set \(s=1+t+m^2\). The 4 edge outer products, equivalently expansion of the 4 squares, give diagonal coefficient \(s\), cross terms \(-2ab\), \(-2tbc\), \(-2cd\), \(-2tad\), and no \(ac\) or \(bd\) term. The counting array is therefore \[Q_{t,m}=\begin {pmatrix} s&-1&0&-t\\ -1&s&-t&0\\ 0&-t&s&-1\\ -t&0&-1&s \end {pmatrix}.\] Multiplication recovers \(x^TQ_{t,m}x=2S_{t,m}(x)\). For \(x=(0,1,0,0)\), the edge energy is \(1+t\) and the mass energy is \(m^2\); hence \(S_{t,m}(0,1,0,0)=s/2\), explicitly \(3/2\) when \(t=m=1\). No change of basis or reference measure occurs.
This is a direct calculation in the supplied cycle data, with the edge-once convention of [spielman2009effective, Section 2.1, pp. 4-5] and the positive kinetic sign and half-action of [wingate2025lattice, Section 4, Equation (4.5), p. 3]. All coefficients here are dimensionless. No Gaussian normalizer or covariance is inferred from the energy calculation.