Example. Separator reconstruction on the four-cycle [fdip-006B]

Apply separator gluing to the unit four-cycle with mass \(m=1\).

The doubled vertices form the separator \(S=\{0,2\}\). The cover sets are \(V_1=\{0,1,2\}\) and \(V_2=\{0,2,3\}\); their private vertices \(1,3\) have no connecting edge. The graph has only the 4 drawn edges, with no diagonal edge. Its precision is \[Q=\begin {pmatrix} 3&-1&0&-1\\-1&3&-1&0\\0&-1&3&-1\\-1&0&-1&3 \end {pmatrix}.\] Use \(V_1=(0,1,2)\), \(V_2=(0,2,3)\) and \(S=(0,2)\), with the listed coordinate orders. Each omitted vertex has precision \(3\), so the Schur formula gives \[\begin {aligned} \Lambda _1&=\begin {pmatrix} 8/3&-1&-1/3\\-1&3&-1\\-1/3&-1&8/3 \end {pmatrix},\\ \Lambda _2&=\begin {pmatrix} 8/3&-1/3&-1\\-1/3&8/3&-1\\-1&-1&3 \end {pmatrix},\\ \Lambda _S&=\begin {pmatrix}7/3&-2/3\\-2/3&7/3\end {pmatrix}. \end {aligned}\] Both triple laws have this same overlap marginal. Padding and subtracting gives \(Q\) entry by entry: each overlap diagonal is \(8/3+8/3-7/3=3\), and the induced overlap entry is \(-1/3-1/3+2/3=0\). The remaining edges retain weight \(-1\). The determinants are \(15,15,5\), respectively, so the glued normalization gives \(\det Q=15\cdot15/5=45\), agreeing with its eigenvalues \(1,3,3,5\). This uses one separator of a non-chordal cycle; it does not assert that arbitrary edge laws on that cycle glue.