Proposition. Formal interaction under Gaussian elimination [fdip-0046]
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October 30, 2026

Eliminating a field with a quartic interaction generally creates quadratic and sextic terms. The coefficients through second order can be found by conditional Gaussian moments, while preserving a supplied symmetry and composition of eliminations.

Use the finite orthogonal sum \(E=X\mathbin \perp Y\), positive precision \(Q\), components \(A,B,C\), Schur precision \(S\) and scalar \(c_C\) of Gaussian elimination. Let \(T:E^4\to \mathbb R\) be a symmetric four-linear form and put \[V(u)=\frac {T(u,u,u,u)}{4!}.\] Write \(\mathcal P(X)\) for the polynomial function algebra. For \(x\in X\), let \(\eta \) have centered Gaussian law with covariance \(C^{-1}\) on \(Y\), and set \[v_x=(x,-C^{-1}B^\dagger x+\eta ).\] This is the conditional law of the eliminated field given the retained value. It is the conditional kernel with its two named summands exchanged; its covariance is \(C^{-1}\), whereas that proposition's retained-given-eliminated kernel has covariance \(A^{-1}\).

For a formal variable \(\lambda \), use ordinary power-series coefficients in \(\lambda \) and define the fiber effective interaction by \[\begin {aligned} F_\lambda (x)&=-\log \mathbb E_\eta e^{-\lambda V(v_x)},\\ \mathbb E_\eta e^{-\lambda V(v_x)} &=\sum _{n\geq 0}\frac {(-\lambda )^n}{n!} \mathbb E_\eta V(v_x)^n. \end {aligned}\] These are formal series in \(\lambda \) with coefficients in \(\mathcal P(X)\); the logarithm is the formal logarithm of a series with constant coefficient one. Every coefficient is finite by polynomial integrability. The notation \(O(\lambda ^3)\) means a series divisible by \(\lambda ^3\), not an analytic error estimate. With \[\begin {aligned} a(x)&=\mathbb E[V\mid x],\\ b(x)&=\mathbb E[V^2\mid x]-a(x)^2, \end {aligned}\] the first two coefficients are \[F_\lambda =\lambda a-\frac {\lambda ^2}{2}b+O(\lambda ^3).\] The degree-zero terms are retained: coefficientwise integration gives \[\begin {aligned} &\int _Y e^{-\frac 12\langle (x,y),Q(x,y)\rangle } e^{-\lambda V(x,y)}\,dy\\ &\qquad =c_Ce^{-\frac 12\langle x,Sx\rangle }e^{-F_\lambda (x)}. \end {aligned}\] Here \(dy\) is the specified Euclidean volume. This fiber logarithm integrates eliminated variables; it does not take the Legendre transform of a source-generating function used for a one-particle irreducible (1PI) effective action in [etingof2024mathematical, v2, Proposition 3.23, p. 50].

For a specified orthogonal flag \(E=X\mathbin \perp Y\mathbin \perp Z\), first eliminate \(Z\) and then \(Y\). All conditional expectations below use the original Gaussian law, with the \(Y\)-given-\(X\) kernel obtained after eliminating \(Z\). Put \[\begin {aligned} a_1(x,y)&=\mathbb E[V\mid x,y],\\ b_1(x,y)&=\mathbb E[V^2\mid x,y]-a_1(x,y)^2. \end {aligned}\] Both stages together give \[\begin {aligned} F_\lambda (x) ={}&\lambda \mathbb E[a_1\mid x]\\ &-\frac {\lambda ^2}{2}\bigl ( \mathbb E[b_1\mid x]+\operatorname {Var}(a_1\mid x)\bigr )\\ &+O(\lambda ^3), \end {aligned}\] which agrees with direct elimination of \(Y\oplus Z\). The first coefficient has degrees among \(4,2,0\); the second has degrees among \(6,4,2,0\). Particular components may vanish.

Proof.

October 30, 2026

Expand the expectation as \(1-\lambda a+\lambda ^2\mathbb E[V^2\mid x]/2+O(\lambda ^3)\). The identity \(\log (1+u)=u-u^2/2+O(u^3)\) gives the displayed coefficient \(-b/2\). Square completion in Theorem [fdip-000D] identifies the normalized conditional Gaussian and the raw scalar \(c_C\); applying it to each polynomial coefficient proves the fiber identity. No convergence or interchange of an infinite series and an integral is needed.

The first-stage interaction is \(F_1=\lambda a_1-\lambda ^2b_1/2+O(\lambda ^3)\). Hence \[e^{-F_1}=1-\lambda a_1+ \frac {\lambda ^2}{2}(a_1^2+b_1)+O(\lambda ^3).\] Take the \(Y\)-given-\(X\) expectation and the negative logarithm. This yields the stated two-stage formula. Polynomial integrability permits signed Fubini for every coefficient, so the tower identities for these Gaussian kernels give \[\begin {aligned} &\mathbb E[a_1\mid x]=\mathbb E[V\mid x],\\ &\mathbb E[b_1\mid x]+\operatorname {Var}(a_1\mid x)\\ &\qquad =\mathbb E[V^2\mid x]-\mathbb E[V\mid x]^2. \end {aligned}\] These identities also follow directly by expanding \(b_1\) and cancelling \(\mathbb E[a_1^2\mid x]\). They prove composition through the stated order without discarding any generated term.

The mean of \(v_x\) is linear in \(x\), and its centered noise is independent of \(x\). Expanding the four-linear tensor and pairing centered Gaussian factors by Wick's theorem leaves degrees \(4,2,0\) in \(a\). In \(\mathbb E[V^2\mid x]-\mathbb E[V\mid x]^2\), pairings with no contraction between the two copies of \(V\) cancel. Every remaining pairing uses at least two noise factors. The 8 original factors therefore leave at most 6 mean factors. An even number of noise factors is paired, giving only degrees \(6,4,2,0\). The homogeneous components are symmetric tensors on \(X\) by polarization; they need not be diagonal in a chosen coordinate basis.

Suppose a second such model is related by a linear isometric isomorphism \(L=L_X\oplus L_Y:E\to E'\) satisfying \(Q'L=LQ\) and \(V'\circ L=V\). Then \[F'_\lambda (L_Xx)=F_\lambda (x)\pmod {\lambda ^3}.\] Indeed the component identities give \(C'L_Y=L_YC\) and \((B')^\dagger L_X=L_YB^\dagger \). Thus \(L_Y\) transports the conditional mean and covariance, hence its Gaussian law and both moments \(a,b\). For two-stage elimination also require preservation of the supplied flag. In particular, a supplied finite group of these isometries preserves each induced coefficient. The result transports a given action; it does not select that action.

Wick contractions are [etingof2024mathematical, v2, Theorem 3.1 and Example 3.2, p. 29]. The perturbation convention in [etingof2024mathematical, v2, Section 3.3, p. 33] is \(B(u,u)/2-\sum _i g_iB_i(u,\ldots ,u)/i!\); here \(g_4=-\lambda \) and \(B_4=T\), at \(\hbar =1\). The conditional formulas and composition above are direct deductions using these contractions and the specified Gaussian kernels. The three-coordinate example exhibits every indicated degree.