Example. Quartic vertices on an alternating cycle [fdip-0061]
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  • November 1, 2026

On the alternating four-cycle with weights \((1,t,1,t)\), take \(t>0\), \(m>0\) and \(s=1+t+m^2\). Retain \(R=\{0,2\}\), eliminate \(U=\{1,3\}\), and choose the interaction \[\mathcal V(\phi )=\frac {\phi _1^4+\phi _3^4}{24}.\] In retained coordinates \(x=(x_0,x_2)\), the conditional rules give independent eliminated noises of variance \(1/s\), with means \[\mu _1=\frac {x_0+tx_2}{s},\qquad \mu _3=\frac {tx_0+x_2}{s}.\] The first coefficient \(a(x)\) of the fiber interaction is \[\begin {aligned} a(x)={}&\frac {(1+t^4)(x_0^4+x_2^4)}{24s^4}\\ &+\frac {4t(1+t^2)(x_0^3x_2+x_0x_2^3) +12t^2x_0^2x_2^2}{24s^4}\\ &+\frac {(1+t^2)(x_0^2+x_2^2)+4tx_0x_2}{4s^3} +\frac 1{4s^2}. \end {aligned}\] Thus eliminating a quartic supported on the removed vertices creates quartic and quadratic couplings on the retained vertices, as well as a constant.

Proof. 

  • November 1, 2026

The eliminated precision is \(D=sI_2\); its cross block gives the two means above. For a centered real Gaussian \(\eta \) of variance \(v\), Wick's identity gives \(\mathbb E\eta ^4=3v^2\). Expanding \((\mu +\eta )^4\) and discarding odd moments yields \[\mathbb E(\mu +\eta )^4=\mu ^4+6\mu ^2v+3v^2.\] Consequently \[a(x)=\frac {\mu _1^4+\mu _3^4}{24} +\frac {\mu _1^2+\mu _3^2}{4s}+\frac 1{4s^2}.\] The binomial expansions of \((x_0+tx_2)^4+(tx_0+x_2)^4\) give coefficients \(1+t^4\), \(4t(1+t^2)\) and \(12t^2\). Squaring the two numerators gives \((1+t^2)(x_0^2+x_2^2)+4tx_0x_2\). Substitution proves every coefficient in the displayed polynomial.

In these named vertex coordinates, the ratio of the \(x_0^3x_2\) coefficient to the \(x_0^4\) coefficient is \[r(t)=\frac {4t(1+t^2)}{1+t^4}.\] It is unchanged by exchanging the two retained vertices and obeys \[\begin {aligned} r(t)&=r(1/t),&r(2)=r(1/2)&=\frac {40}{17},\\ r'(t)&=\frac {4(1-t^2)(1+4t^2+t^4)}{(1+t^4)^2}. \end {aligned}\] The derivative is positive for \(0<t<1\). Therefore \(r\) separates the parameter for \(0<t\le 1\), and on all positive parameters it separates exactly modulo \(t\leftrightarrow1/t\). This is a vertex-coordinate shape ratio, not an invariant of arbitrary orthogonal changes of the retained basis. The complete coefficient tensors instead obey the naturality theorem.

The reciprocal equality resembles that of the trace ratio, but neither equality gives a strict fixed-mass background isomorphism at reciprocal parameters. The edge-weight products are \(t^2\) and \(t^{-2}\), so a weight-preserving edge bijection would force \(t=1\). Also \(s\) changes at fixed mass; equality of the shape ratio does not make all induced coefficients equal.

A finite illustration uses \(t=2\), \(m=1\) and \(x=(0.5,-0.3)\). The means are \((-1/40,7/40)\), and the exact coefficient is \[a(x)=\frac {541201}{30720000}.\] The polynomial identity is the calculation above. The same conditional moment is the one exercised by the project script revision-checks.py.