Example. Breaking the weights breaks the half-turn [fdip-003K]

✍️sourceAGENTDRAFTED

In the alternating four-cycle, take \(t,m>0\) and the half-turn. It fails to preserve the action after a specific positive change of weights. Replace \((1,t,1,t)\) by \((1,t,2,t)\) while keeping the proposed half-turn. Then \(L_{00}=1+t\) and \(L_{22}=2+t\), so \(PLP^T\ne L\) and \(\mathcal A(\delta _2)-\mathcal A(\delta _0)=1/2\). If its Gaussian law were invariant, second moments would give \(PGP^T=G\); inversion would give \(PQP^T=Q\), contradicting those diagonal entries. At \(t=m=1\), retain the opposite pair \(R=(0,2)\) and eliminate \(D=(1,3)\). The blocks in that order are \[\begin {aligned} A=C&=\operatorname {diag}(3,4),\\ H&=-\begin {pmatrix}1&1\\1&2\end {pmatrix},\\ S_{\mathrm {broken}} &=\begin {pmatrix}29/12&-5/6\\-5/6&8/3\end {pmatrix}. \end {aligned}\] Indeed \[HC^{-1}H^T=\begin {pmatrix}7/12&5/6\\5/6&4/3\end {pmatrix},\] and subtracting from \(A\) gives the displayed precision. Its diagonal entries differ by \(1/4\), so its marginal also fails swap invariance. Its positive Gaussian still exists.

[spielman2009effective, Section 2.1, pp. 4-5] gives the edge-once convention.

[boyd2004convex, Appendix A.5.5, pp. 650-651] gives the positive Gaussian elimination convention. The action difference and matrix products are direct calculations with these conventions. Weight preservation is sufficient in the general transport theorem; it need not be necessary for every presentation with redundant loops or parallel edges. This example establishes failure for the stated weights, not a universal necessity.