Example. Edge derivatives and forests on an alternating cycle [fdip-005W]

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Take the four-cycle \(0\to 1\to 2\to 3\to 0\) with weights \((1,2,1,2)\) and \(m=1\). The orthonormal vectors \[\begin {aligned} u_0&=\tfrac 12(1,1,1,1),&u_2&=\tfrac 12(1,-1,-1,1),\\ u_4&=\tfrac 12(1,1,-1,-1),&u_6&=\tfrac 12(1,-1,1,-1) \end {aligned}\] have \(L_w\)-eigenvalues given by their subscripts, as direct multiplication verifies. Therefore \(Q\) has eigenvalues \(1,3,5,7\), and \[\det Q=105,\qquad \operatorname {tr}G =1+\frac 13+\frac 15+\frac 17=\frac {176}{105}.\] The edge vector \(d_{01}\) has squared projection \(1\) onto each of \(u_2,u_6\), and zero onto the other two. For \(d_{12}\) the nonzero squared projections are onto \(u_4,u_6\). Half-turn symmetry gives the other two edges. Thus \[\frac {\partial \mathcal W}{\partial w_e} =\begin {cases} \frac 12(\frac 13+\frac 17)=\frac 5{21}&w_e=1,\\ \frac 12(\frac 15+\frac 17)=\frac 6{35}&w_e=2. \end {cases}\] These are individual edge derivatives at the displayed weights; they are not derivatives of a simultaneous change of both edges of one weight. The weighted sum is \[2\cdot \frac 5{21}+4\cdot \frac 6{35} =\frac {122}{105} =\frac 12\left (4-\frac {176}{105}\right ).\] This is an instance of Proposition [fdip-005U].

The forest generating polynomial is \[\begin {aligned} \det (tI+L_w)&=t(t+2)(t+4)(t+6)\\ &=t^4+12t^3+44t^2+48t. \end {aligned}\] The 4 spanning trees have weights \(4,2,4,2\), so \(\tau _w=12\) and the coefficient of \(t\) is \(4\tau _w=48\). At \(t=1\), the forest law has \[\mathbb E[|R|]=\frac {176}{105},\qquad \mathbb E[|F|]=\frac {244}{105}.\] Its individual edge probabilities are \(10/21\) on weight-one edges and \(24/35\) on weight-two edges. They sum to \(244/105\), as the forest identity requires.

On the unit four-cycle, all 4 edges instead have the same effective resistance by rotation. One unit edge is in parallel with the other three in series, giving \(R_{\mathrm {eff}}=3/4\). Equivalently the nonconstant spectral calculation gives \(1/2+1/4=3/4\). Foster's sum is \(4(3/4)=3\), matching \(n-1\). This is an edge-observable limit, not a massless Gaussian law on all 4 vertex values.