Example. Eliminating alternate vertices of a ring-valued cycle [fdip-001Z]
Example. Eliminating alternate vertices of a ring-valued cycle [fdip-001Z]
Use the four-cycle endpoints \(s(e_i)=i\), \(h(e_i)=i+1\) on \(\mathbb Z/4\mathbb Z\), but now let \(R\) be any commutative unital ring and take \(t,\mu \in R\). Give successive edges weights \((1,t,1,t)\). On the finite function modules with their counting pairings, set \[Q=d^\dagger Wd+\mu I,\qquad s=1+t+\mu .\] Here \(d\) is incidence and \(W\) multiplies each edge field by its weight. This extends the positive-real weight choice in Example [fdip-0003]; no weight is assumed positive or invertible.
First compute from endpoints. The difference is \((df)(e_i)=f(i+1)-f(i)\). Grouping the incoming and outgoing edge terms gives \[(d^\dagger Wdf)(i) =w(e_{i-1})(f(i)-f(i-1))+w(e_i)(f(i)-f(i+1)).\] Adding \(\mu f(i)\) yields \[\begin {aligned} (Qf)(0)&=sf(0)-f(1)-tf(3),\\ (Qf)(1)&=sf(1)-f(0)-tf(2),\\ (Qf)(2)&=sf(2)-tf(1)-f(3),\\ (Qf)(3)&=sf(3)-tf(0)-f(2). \end {aligned}\]
Choose the additional partition \(V_0=\{0,2\}\), \(V_1=\{1,3\}\), and put \(X=\mathcal F_R(V_0)\), \(Y=\mathcal F_R(V_1)\). Restriction and extension by zero give the chosen splitting. The resulting component maps are \[\begin {aligned} A&=sI_X,&C&=sI_Y,\\ (Ex)(1)&=-x(0)-tx(2),&(Ex)(3)&=-tx(0)-x(2),\\ (By)(0)&=-y(1)-ty(3),&(By)(2)&=-ty(1)-y(3). \end {aligned}\] The pivot \(C\) is invertible exactly when \(s\) is a unit. A unit gives \(C^{-1}=s^{-1}I_Y\). Conversely, if \(CJ=I_Y\), evaluation of \(CJ\delta _1=\delta _1\) at vertex \(1\) gives \(s(J\delta _1)(1)=1\); commutativity makes this a two-sided inverse of \(s\). In the zero ring \(0=1\) is a unit, consistently with this argument.
Assume this unit condition. Applying \(B\) to \(Ex\) gives \[\begin {aligned} (BEx)(0)&=(1+t^2)x(0)+2tx(2),\\ (BEx)(2)&=2tx(0)+(1+t^2)x(2). \end {aligned}\] Consequently the split Schur map is \[\begin {aligned} (Sx)(0)&=[s-s^{-1}(1+t^2)]x(0)-2ts^{-1}x(2),\\ (Sx)(2)&=[s-s^{-1}(1+t^2)]x(2)-2ts^{-1}x(0). \end {aligned}\] Here \(2=1+1\); neither \(2\), \(t\), \(\mu \) nor \(S\) is asserted to be invertible. In coordinates ordered \((0,2)\), the same map is \[[S]=\begin {pmatrix} s-s^{-1}(1+t^2)&-2ts^{-1}\\ -2ts^{-1}&s-s^{-1}(1+t^2) \end {pmatrix}.\] This array presents the map just computed from endpoints.
The half-turn exchanges \(0\leftrightarrow 2\) and \(1\leftrightarrow 3\). Let \(r_X,r_Y\) exchange the corresponding field values. The component formulas give \(Ar_X=r_XA\), \(Cr_Y=r_YC\), \(Er_X=r_YE\) and \(Br_Y=r_XB\); hence \(Sr_X=r_XS\). A one-step shift exchanges the two parts and fails to preserve the alternating weights when \(t\ne 1\). No retained partition is implicit in general incidence data.
Now specialize to \(R=\mathbb R\), \(t>0\), \(\mu =m^2>0\). The weighted energy identity gives, for \(f\ne 0\), \[b(f,Qf)=\sum _e w(e)(df(e))^2+m^2\sum _v f(v)^2>0.\] Thus \(Q\) is positive definite and invertible, and the positive-real criterion gives \(S>0\). At \(t=m=1\), \(s=3\) and \[[S]=\frac 13\begin {pmatrix}7&-2\\-2&7\end {pmatrix}.\] To solve \(Sx=\delta _2\), solve \(7x(0)-2x(2)=0\) and \(-2x(0)+7x(2)=3\). These give \(x(2)=7/15\) and \(x(0)=2/15\). Therefore \[(\pi _XQ^{-1}\iota _X\delta _2)(0) =(S^{-1}\delta _2)(0)=\frac {2}{15}.\] By contrast, fixing the eliminated field to zero leaves the retained map \(A=3I_X\), and \((A^{-1}\delta _2)(0)=0\). These are different map operations.
At real \(\mu =0,t>0\), the pivot \(C=(1+t)I_Y\) is still invertible. But \(s-s^{-1}(1+t^2)=2t/(1+t)\), so the nonzero constant retained field satisfies \(S1=0\). The kernel correspondence then recovers the constant full field in \(\ker Q\), since \(-C^{-1}E1=1\). An invertible pivot does not imply an invertible full precision.
[spielman2009effective, Section 2.1, pp. 4-5] gives the incidence sign convention for positive real weights. The arbitrary-ring identities use the direct counting construction of Definition [fdip-0007].
[boyd2004convex, Appendix C.4.1, pp. 672-673] gives the real elimination convention. All coefficients and inverse entries here were computed from the specified endpoints and weights. They recover the algebraic values in real alternate-vertex elimination; the covariance and marginal-law interpretation there additionally uses the Gaussian measure. No massless Gaussian formula follows by substituting \(\mu =0\) into a massive normalization.