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Criticality, Gaussian fields and exact block renormalization

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Infinite-volume nonanalytic-pressure criticality

Definition

Let I⊆R be a nonempty open interval of values of a scalar control parameter θ, expressed in one fixed unit if it is dimensional. A function g:I→R is real analytic if for every θ0∈I there are ϵ>0 and real coefficients (ak)k≥0 such that (θ0−ϵ,θ0+ϵ)⊆I, the series ∑k=0∞ak(θ−θ0)k converges there, and its sum equals g(θ) throughout that interval. Let (Nn)n≥1 be positive integers with Nn→∞, interpreted as finite system sizes such as site counts. A finite-volume partition-function sequence on I is a sequence of real-analytic functions Zn:I→(0,∞). For each n, define its dimensionless log partition density per site by pn(θ):=Nn−1log⁡Zn(θ), where log⁡ is the natural logarithm. If the pointwise limit p(θ):=lim⁡n→∞pn(θ) exists as a finite real number for every θ∈I, call p the infinite-volume pressure associated with this sequence. A point θc∈I is critical under the nonanalytic-pressure criterion exactly when there do not exist ϵ>0 with (θc−ϵ,θc+ϵ)⊆I and real coefficients (ak)k≥0 such that ∑k=0∞ak(θ−θc)k converges and equals p(θ) throughout that interval.

This definition specifies only the nonanalytic-pressure criterion and makes no claim that a particular model satisfies it.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite Gaussian probability with a positive precision matrix

Definition

Assume the Axiom of Countable Choice: every sequence (Xn)n∈N of nonempty sets has a function f with f(n)∈Xn for every n. For N≥1, a real symmetric positive-definite matrix K∈RN×N (meaning zTKz>0 for every nonzero z∈RN) defines the finite Gaussian probability γK on RN by

ZK:=∫RNe−zTKz/2 dλN(z),γK(dz):=ZK−1e−zTKz/2 dλN(z).

Under the stated Axiom of Countable Choice, ZK=(2π)N/2(det⁡K)−1/2. For the coordinate identity X(z)=z on RN, define its mean vector by μi:=∫RNzi γK(dz) and its covariance matrix by Σij:=∫RN(zi−μi)(zj−μj) γK(dz) for 1≤i,j≤N. These integrals are finite under the Gaussian theorem Gaussian covariance for positive quadratic forms and a boundary counterexample, which gives μ=0 and Σ=K−1. Thus γK is centered, meaning its mean vector is zero. Here K is called the precision matrix because it appears in the quadratic exponent; for this centered law its covariance matrix is its inverse. For N=0, define R0 to be the singleton containing the empty vector and define γ∅ to put mass one on that point; its partition factor and empty determinant are both 1.

The variables zi and exponent zTKz/2 are dimensionless in this mathematical definition. In a physical quadratic model, if each component of Φ has declared field unit Φ∗ and H(Φ)=12ΦTKphysΦ is energy in joules, then each entry of Kphys has units J/(Φ∗)2. With β=(kBT)−1, where in SI kB is Boltzmann's constant with units J K−1 and T>0 is absolute temperature in kelvins, β has units J−1. For zi=Φi/Φ∗, the dimensionless precision is K=βΦ∗2Kphys. This dimensional conversion does not assert an empirical model.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Fourier spectrum and positivity of the periodic lattice Laplacian

Statement

Let d,L≥1 be integers and let Λd,L:=(Z/LZ)d, where Z/LZ is the quotient set of integers modulo L. Use the quotient addition [a]L+[b]L=[a+b]L and additive inverse −[a]L=[−a]L, and define these operations coordinatewise on Λd,L. For x∈Λd,L, write x~=(x~1,…,x~d)∈{0,…,L−1}d for its unique coordinate representatives. For 1≤j≤d, let ej∈Λd,L have coordinate [1]L in position j and [0]L elsewhere. Count the two terms x+ej and x−ej separately even when they coincide for L=1 or L=2.

For a real-valued function f:Λd,L→R, define

(−Δf)(x):=∑j=1d(2f(x)−f(x+ej)−f(x−ej)).

For a complex-valued function, use the same formula for −Δ, which is its complex-linear extension. Write CΛd,L for the complex-valued functions on Λd,L with pointwise vector operations. For r=(r1,…,rd)∈{0,…,L−1}d, define the function χr:Λd,L→C by

χr(x):=exp⁡ ⁣(2πiL∑j=1drjx~j).

Use the inner product ⟨f,g⟩:=L−d∑x∈Λd,Lf(x)g(x)‾. Identify real functions with coordinate vectors using the lexicographic order on the representatives x~, and let IΛd,L be the identity operator on those functions. A real matrix A is symmetric when AT=A; a real symmetric matrix is positive semidefinite when vTAv≥0 for every real vector v, and positive definite when vTAv>0 for every nonzero real vector v. For all r,s∈{0,…,L−1}d, the inner products satisfy ⟨χr,χs⟩=1 if r=s and 0 otherwise; every complex-valued function on Λd,L is a unique linear combination of the functions χr.

For each r, the equation −Δχr=λrχr holds with λr:=4∑j=1dsin⁡2(πrj/L); this scalar is the eigenvalue associated with χr.

Consequently −Δ is real symmetric positive semidefinite, and for real m2>0 and κ>0 the matrix m2IΛd,L+κ(−Δ) is real symmetric positive definite with eigenvalues m2+4κ∑j=1dsin⁡2(πrj/L)>0.

Statement references

Facts & Assumptions

Given: Integers d,L≥1, the coordinatewise quotient operations, canonical representatives, and discrete Laplacian of the Statement.

[F2]

Quotient addition is [a]L+[b]L=[a+b]L and is independent of representatives; Z/LZ with this operation is an additive abelian group, with inverse −[a]L=[−a]L (Addition and multiplication on Z/n by [a]n+[b]n=[a+b]n and [a]n[b]n=[ab]n, For every natural n, (Z/n,+) is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[F3]

For N≥1 and integers k,ℓ, ∑u=0N−1e2πi(k−ℓ)u/N is N if k≡ℓ(modN) and is 0 otherwise (Orthogonality of the characters x↦e2πikx/N on Z/NZ).

[F6]

For real t, eit=cos⁡t+isin⁡t and cos⁡(2t)=cos⁡2t−sin⁡2t (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0, The addition formulas for sine and cosine).

[F7]

For real t, sin⁡2t+cos⁡2t=1, sin⁡(−t)=−sin⁡t, and cos⁡(−t)=cos⁡t (Pythagorean and parity identities for all six trigonometric functions on their natural domains).

[F8]

Complex conjugation fixes real numbers and sends i to −i (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive); together with [F6] and [F7], this gives eit‾=e−it for real t.

Proof

1.1givenF1F2F5algebra

Fix x∈Λd,L and put aj=x~j. By [F1]–[F2], the jth representative of x+ej is aj+1−Lϵj, where ϵj=1 if aj=L−1 and 0 otherwise; the jth representative of x−ej is aj−1+Lηj, where ηj=1 if aj=0 and 0 otherwise. All other representatives are unchanged. Substitution into the definition of χr, followed by the exponential addition law and exp⁡(2πin)=1 for integers n from [F5], gives χr(x+ej)=e2πirj/Lχr(x) and χr(x−ej)=e−2πirj/Lχr(x). This also holds for L=1 and L=2.

1.2F1F3F4F8algebra

For r,s∈{0,…,L−1}d, finite-sum factorization [F4] and the conjugation identity [F8] give ⟨χr,χs⟩=∏j=1dL−1∑u=0L−1e2πi(rj−sj)u/L. By [F1], distinct canonical representatives are incongruent modulo L, so [F3] makes each factor 1 when rj=sj and 0 otherwise. This is the inner-product identity stated above.

2.1F6F7step 1.1algebra

Substituting the shift identities of step 1.1 into −Δ gives −Δχr=∑j=1d(2−e2πirj/L−e−2πirj/L)χr. By [F6]–[F7], each coefficient equals 2−2cos⁡(2πrj/L)=4sin⁡2(πrj/L), proving the eigenvalue formula.

2.2F1F3F4F8step 1.2algebra

For any f:Λd,L→C, set f^(r)=L−d∑y∈Λd,Lf(y)χr(y)‾. By [F4] and [F8], the frequency sum factors as ∑rχr(x)χr(y)‾=∏j=1d∑rj=0L−1e2πirj(x~j−y~j)/L. The canonical-representative theorem [F1] and orthogonality [F3] make this product Ld if x=y in Λd,L and 0 otherwise. Therefore ∑rf^(r)χr(x)=L−d∑yf(y)∑rχr(x)χr(y)‾=f(x) for every x. If a linear combination ∑rcrχr vanishes, taking its inner product with χs and applying step 1.2 gives cs=0 for each s, so the expansion is unique.

3.1givenF2F4step 2.1step 2.2algebra∎

For every real f, the translations x↦x+ej are bijections with inverses x↦x−ej, so reindexing the finite sums [F4] gives ∑xf(x)(−Δf)(x)=∑j=1d∑x(f(x+ej)−f(x))2≥0. Thus −Δ is positive semidefinite; its matrix is symmetric because every shift term is paired with its inverse shift. For nonzero f, fT(m2I+κ(−Δ))f=m2∑xf(x)2+κ∑j=1d∑x(f(x+ej)−f(x))2>0, since m2>0 and κ>0. By steps 2.1 and 2.2, the spanning functions χr satisfy the eigenvalue equations in the Statement, so the combined operator has the displayed eigenvalues.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Chord and Lipschitz bounds for sine

Statement

For ∣u∣≤π/2, 2∣u∣/π≤∣sin⁡u∣≤∣u∣. For all real u,v, ∣sin⁡u−sin⁡v∣≤∣u−v∣.

Facts & Assumptions

Given: Real numbers u,v, with ∣u∣≤π/2 for the chord inequality.

[F1]

sin⁡′=cos⁡, cos⁡′=−sin⁡, sin⁡0=0, and sin⁡(π/2)=1 (The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi).

[F5]

sin⁡t≥0 on [0,π/2], ∣cos⁡t∣≤1 for all t, and sin⁡(−t)=−sin⁡t (Signs, monotonicity intervals, and ranges of sine and cosine, Pythagorean and parity identities for all six trigonometric functions on their natural domains).

Proof

1.1givenF1F2F3F4F5algebra

On [0,π/2] define h(t)=sin⁡t−tcos⁡t. By [F1]–[F3] and the product rule, h is differentiable and continuous, with h(0)=0 and h′(t)=tsin⁡t≥0 on (0,π/2). For 0<t≤π/2, the mean value theorem on [0,t] gives h(t)−h(0)=h′(c)t≥0 for some c∈(0,t); at t=0 equality holds.

1.2givenF1F3F4F5algebra

If u=0, then ∣sin⁡u∣=0=∣u∣ by [F1]. For 0<∣u∣≤π/2, sine is continuous on [0,∣u∣] and differentiable on its interior by [F1] and [F3], so the mean value theorem applies on this nondegenerate interval. Its derivative has modulus at most 1 by [F1] and [F5], so ∣sin⁡∣u∣∣≤∣u∣; parity gives ∣sin⁡u∣≤∣u∣.

2.1givenF1F2F4F5step 1.1algebra

For 0<t<π/2, put r(t)=sin⁡t/t. The quotient rule and step 1.1 give r′(t)=(tcos⁡t−sin⁡t)/t2=−h(t)/t2≤0. Applying the mean value theorem to r on [t,π/2] yields r(t)≥r(π/2)=2/π by [F1] and the quarter-turn value in [F1]; the endpoint t=π/2 is equality, and t=0 is immediate. Parity in [F5] extends the lower bound to negative u.

3.1givenF1F4F5algebra∎

If u≠v, apply the mean value theorem to sin⁡ on the closed interval with endpoints u,v. For some interior c, ∣sin⁡u−sin⁡v∣=∣cos⁡c∣ ∣u−v∣≤∣u−v∣ by [F1], [F4], and [F5]; if u=v, both sides are zero.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Positive Schur complements for finite Gaussian precisions

Statement

Let r≥0 and s≥1 be integers. For n≥1, a real matrix M∈Rn×n is symmetric positive definite if MT=M and vTMv>0 for every nonzero v∈Rn. Let A∈Rr×r, B∈Rr×s, and C∈Rs×s, and set

K:=(ABBTC)∈R(r+s)×(r+s).

Assume K is symmetric positive definite. Then C is symmetric positive definite, det⁡C>0, and C−1 exists. The Schur complement S:=A−BC−1BT is symmetric positive definite when r≥1; when r=0, S is the empty 0×0 matrix. For all x∈Rr and y∈Rs,

(xy)TK(xy)=xTSx+(y+C−1BTx)TC(y+C−1BTx).

For r=0, the first term and S are interpreted on the singleton space R0.

Facts & Assumptions

Given: A symmetric positive-definite real block matrix K with block sizes r≥0 and s≥1 as in the Statement.

[F1]

Every real symmetric positive-definite s×s matrix has a factorization C=LLT with L lower triangular and positive diagonal; the determinant of a triangular matrix is the product of its diagonal entries, determinants are multiplicative, and a real square matrix with nonzero determinant is invertible (A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique, The determinant of a triangular matrix is the product of its diagonal entries, For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B), A finite square real matrix is invertible if and only if its determinant is nonzero).

Proof

1.1givenF1algebra

For nonzero z∈Rs, the vector (0,z) is nonzero, so zTCz=(0,z)TK(0,z)>0. Thus C is symmetric positive definite. By the Cholesky statement in [F1], C=LLT with positive diagonal; by the determinant laws in [F1], det⁡C=(det⁡L)2=(∏iLii)2>0, so the invertibility criterion in [F1] gives that C is invertible.

2.1step 1.1algebra

Since C is symmetric and invertible, C−1 is symmetric. Expanding the quadratic form on the right gives xT(A−BC−1BT)x+2xTBy+yTCy, which is the quadratic form on the left; this proves the displayed completion-of-the-square identity.

3.1givenstep 1.1step 2.1constructalgebra∎

If r≥1 and x≠0, choose y=−C−1BTx in step 2.1. The second term vanishes and (x,y)≠0, so positive definiteness of K gives xTSx>0. Symmetry of S follows from symmetry of A,C−1; hence S is symmetric positive definite. If r=0, the claimed empty-matrix convention holds by definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Unconstrained massless periodic Gaussian has infinite normalizer

Statement

Assume the Axiom of Countable Choice ACω. Let d,L≥1 be integers, let Λd,L=(Z/LZ)d have N=Ld sites, and let κ>0. Define the zero-mass precision K:=κ(−Δ) on RΛd,L, with the periodic Laplacian convention of Fourier spectrum and positivity of the periodic lattice Laplacian. Then the unrestricted Lebesgue normalizer diverges:

∫RΛd,Lexp⁡ ⁣(−12zTKz) dλN(z)=+∞.

Thus this quadratic weight does not define a probability density on the full coordinate space. A finite-volume massless Gaussian must first restrict the field, for example to {z:∑x∈Λd,Lzx=0}, before inverting the Laplacian.

Facts & Assumptions

Given: ACω, integers d,L≥1, the periodic site set Λd,L with N=Ld, and κ>0.

[F1]

The periodic operator −Δ is real symmetric and annihilates the constant vector 1; the r=0 function χ0(x)=1 for every x is that constant vector (Fourier spectrum and positivity of the periodic lattice Laplacian).

[F2]

Under ACω, a C1 diffeomorphism changes a nonnegative Lebesgue integral by its absolute Jacobian (The Axiom of Countable Choice (ACω), A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions).

[F3]

Euclidean Lebesgue measure is the product of its coordinate measures on Borel rectangles; the coordinate measures are sigma-finite, and Tonelli identifies the integral of a nonnegative measurable function with its iterated integral (Lebesgue measurable sets, the family L(Rn), and the restricted set function λn, The nonnegative Lebesgue integral, On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F4]

Boxes have their side-length product as Lebesgue measure, and increasing nonnegative functions pass to their limit under the integral (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included, Monotone convergence for the integral).

[F5]

The nonnegative Lebesgue integral is monotone (Monotonicity and nonnegative homogeneity of the nonnegative integral).

Proof

1.1givenF1algebra

By [F1], K1=0 and K is symmetric. Hence the quadratic form is unchanged by adding any constant multiple of 1: for every z and t∈R, (z+t1)TK(z+t1)=zTKz.

2.1givenF4F5step 1.1algebra

If N=1, then K=0 by step 1.1, and the integrand is identically 1 on R. Its integral is infinite because the intervals [−M,M] have measure 2M for every integer M≥1 by [F4], and [F5] bounds the full integral below by each of these unbounded values.

2.2givenF2step 1.1algebra

Suppose N≥2 and enumerate the lattice sites as x1,…,xN. Define the linear map T:RN−1×R→RN by T(w,t)=(w1+t,…,wN−1+t,t). Its matrix is triangular with all diagonal entries 1, so it is a C1 diffeomorphism with determinant 1. For w~=(w1,…,wN−1,0), one has T(w,t)=w~+t1. By step 1.1, T(w,t)TKT(w,t)=w~TKw~, which is independent of t. Applying [F2] changes the original integral to ∫RN−1×Rf(w) dλN(w,t),f(w):=exp⁡ ⁣(−12w~TKw~).

3.1F3F4F5step 2.2algebra∎

The function f is continuous and positive. On [−1,1]N−1, ∣w~TKw~∣≤M, where M:=∑i,j=1N−1∣Kij∣<∞; hence f(w)≥e−M/2 there. By [F4] and [F5], ∫RN−1f(w) dλN−1(w)≥2N−1e−M/2>0. By [F3], Tonelli factors the nonnegative integral in step 2.2 as ∫f(w) dw∫R1 dt. The second factor is infinite because each interval [−M′,M′] has measure 2M′ for integer M′≥1 by [F4], and [F5] bounds the full-line integral below by each such value. Thus the original normalizer is infinite, proving the claim.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite periodic Gaussian lattice field with positive mass

Definition

Assume the Axiom of Countable Choice from The Axiom of Countable Choice (ACω). Let integers d,L≥1 specify the periodic lattice Λd,L=(Z/LZ)d and let m2>0 and κ>0 be dimensionless real parameters in lattice units. On the real coordinate space RΛd,L, let

Kd,L,m,κ:=m2I+κ(−Δ),

where −Δ is the periodic Laplacian of Fourier spectrum and positivity of the periodic lattice Laplacian. Its eigenvalues are

m2+4κ∑j=1dsin⁡2(πrj/L),r∈{0,…,L−1}d,

so it is symmetric positive definite. The finite periodic Gaussian lattice field is the centered probability γKd,L,m,κ of Finite Gaussian probability with a positive precision matrix, with density proportional to exp⁡(−zTKd,L,m,κz/2) against Lebesgue measure on RLd and covariance Kd,L,m,κ−1 by Gaussian covariance for positive quadratic forms and a boundary counterexample at β=1.

The lattice coordinates zx and the quadratic exponent are dimensionless. Under the physical quadratic convention of Finite Gaussian probability with a positive precision matrix, where H(Φ)=12ΦTKphysΦ and each field component has unit Φ∗, choosing zx=Φx/Φ∗ gives dimensionless precision K=βΦ∗2Kphys when the energy is divided by kBT. Here β=(kBT)−1 with the SI units for kB and T specified in Finite Gaussian probability with a positive precision matrix; equivalently the physical precision for this lattice model is Kphys=(kBT/Φ∗2)Kd,L,m,κ. This conversion makes no empirical claim. At m2=0, the constant field is a zero mode and the unrestricted normalizer diverges by Unconstrained massless periodic Gaussian has infinite normalizer, so the full-space positive-precision definition does not apply. A finite-volume massless model must first specify a restriction such as the subspace ∑x∈Λd,Lzx=0 before inverting the Laplacian.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Dimension threshold for the infinite-lattice massless Gaussian Green integral

Statement

Assume the Axiom of Countable Choice ACω, meaning every sequence (Xn)n∈N of nonempty sets has a function f with f(n)∈Xn for every n. Let integer d≥1, parameter κ>0, and k=(k1,…,kd)∈[−π,π]d, and put

Dd(k):=4∑j=1dsin⁡2(kj/2).

Here λd is d-dimensional Lebesgue measure restricted to the cube. For a measurable function h:[−π,π]d→R, write h+=max⁡{h,0} and h−=max⁡{−h,0}; its signed Lebesgue integral is ∫h+ dλd−∫h− dλd when at most one of these two integrals is infinite, and h is integrable when ∫∣h∣ dλd<+∞. For x=(x1,…,xd)∈Zd, define k⋅x:=∑j=1dkjxj and ∣k∣22:=∑j=1dkj2. Define the massless infinite-lattice Green integral when it is finite by

Gd,κ(x):=(2π)−d∫[−π,π]dcos⁡(k⋅x)κDd(k) dλd(k),

where the integrand is assigned value 0 at k=0, a choice that does not affect the integral because a singleton has Lebesgue measure zero. If d>2, this integral is absolutely finite for every fixed x∈Zd. If 1≤d≤2, the nonnegative diagonal integral with x=0 is +∞. Thus the dimension threshold concerns this free Gaussian Fourier integral; it does not assert an interacting critical exponent or a continuum scaling limit.

Facts & Assumptions

Given: ACω, an integer d≥1, κ>0, x∈Zd, the cube [−π,π]d with Euclidean Lebesgue measure, and Dd as in the Statement.

[F1]

For ∣u∣≤π/2, 2∣u∣/π≤∣sin⁡u∣≤∣u∣ (Chord and Lipschitz bounds for sine).

[F3]

The nonnegative Lebesgue integral is monotone and additive for finite sums; increasing nonnegative measurable functions pass to their limit under the integral (Monotonicity and nonnegative homogeneity of the nonnegative integral, Additivity of the nonnegative Lebesgue integral, Monotone convergence for the integral).

[F4]

A continuous real function is Borel measurable, and a geometric series with ratio in (0,1) has finite sum (A continuous map has Borel preimages of Borel sets, For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges).

[F5]

∣cos⁡t∣≤1 for every real t (Signs, monotonicity intervals, and ranges of sine and cosine).

[F6]

The real signed Lebesgue integral and the criterion ∫∣h∣ dλd<+∞ for finite integrability are as stated in the real-integral definition linked above; the nonnegative integrals used in the comparison are those of The nonnegative Lebesgue integral.

Proof

1.1givenF1F4algebra

For k≠0, apply [F1] to u=kj/2 in each coordinate and square the bounds. Summing yields 4π2∣k∣22≤Dd(k)≤∣k∣22, where ∣k∣22=∑jkj2. The denominator is positive off the origin, so its reciprocal is continuous on the cube with the origin removed. Extending it by the stated value at the closed singleton {0} makes it Borel measurable: each open-set preimage is the union of a Borel subset of that complement and, when applicable, the singleton. The cosine numerator is continuous, so all nonnegative comparison integrands below are measurable by [F4].

2.1givenF2F3F4step 1.1algebra

For j≥0, let Qj=[−2−j,2−j]d∖[−2−(j+1),2−(j+1)]d. The sets are pairwise disjoint, are contained in [−1,1]d, and by [F2] have measure (2d−1)2−jd. On Qj, ∣k∣2≥∥k∥∞>2−(j+1), so step 1.1 gives (κDd(k))−1≤(π2/(4κ))22j+2. Hence ∫Qj(κDd(k))−1 dλd(k)≤π2κ(2d−1)2(2−d)j. For d>2, these upper bounds sum to a finite geometric series by [F4]. On the remaining part of [−π,π]d outside [−1,1]d, ∣k∣2≥1, so step 1.1 bounds (κDd)−1 by π2/(4κ); that part has finite measure by [F2]. Countable additivity and monotone convergence in [F3] therefore show that (κDd)−1 is integrable on the whole cube.

2.2givenF3step 1.1algebra

For 1≤d≤2, step 1.1 gives (κDd(k))−1≥(κ∣k∣22)−1 on every Qj. There ∣k∣2≤d 2−j, and hence ∫Qj(κDd(k))−1 dλd(k)≥1κd(2d−1)2(2−d)j. The right side is bounded below by the positive constant (2d−1)/(κd) for all j≥0 when d≤2. Since the Qj are disjoint, finite additivity in [F3] shows that the integral over their first M+1 layers is at least (M+1)(2d−1)/(κd); letting M→∞ and using monotonicity gives ∫(κDd)−1=+∞.

3.1givenF5F6step 2.1step 2.2algebra∎

If d>2, then ∣cos⁡(k⋅x)∣(κDd(k))−1≤(κDd(k))−1 away from the origin by [F5], and step 2.1 proves ∫∣hx∣ dλd<+∞; hence the signed integral in the Statement is finite by [F6], and multiplication by the finite constant (2π)−d preserves it. If x=0 and d≤2, the cosine is 1, so step 2.2 gives the stated infinite nonnegative diagonal integral.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Exact finite Gaussian block elimination

Definition

The exact Gaussian block-elimination algorithm has input:

  • an integer N≥1 and a real matrix K0∈RN×N satisfying K0T=K0 and vTK0v>0 for every nonzero v∈RN. In the finite Gaussian convention, K0 is the precision because it appears in the density proportional to exp⁡(−zTK0z/2);
  • an integer q≥0 and nested retained coordinate sets R0={1,…,N}⊇R1⊇⋯⊇Rq, each given in increasing coordinate order.

It starts at stage 0 with the ordered coordinates R0 and matrix K0. For each j=1,…,q, let Ej=Rj−1∖Rj be the coordinates to eliminate at that stage.

  1. If Ej=∅, set Kj=Kj−1 and ℓj=0.
  2. If Ej≠∅, reorder the current coordinates as the increasing list of Rj followed by the increasing list of Ej, and write the correspondingly permuted matrix Kj−1=(AjBjBjTCj). Solve the nonsingular system CjXj=BjT, set Kj:=Aj−BjXj,ℓj:=∣Ej∣2ln⁡(2π)−12ln⁡(det⁡Cj). Interpret Kj in the increasing coordinate order on Rj.

The output is (Kq,ℓ1,…,ℓq), with the accumulated log normalization ∑j=1qℓj. If q=0, this output means K0 and an empty list of increments. When Rj=∅, Kj is the unique empty 0×0 matrix and its determinant is defined to be 1. This is an exact-arithmetic procedure on finite matrices.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Correctness of finite Gaussian block elimination

Statement

Assume the Axiom of Countable Choice ACω and the hypotheses of Exact finite Gaussian block elimination with input K0 and retained sets R0,…,Rq. Let Kj,Cj,ℓj be the algorithm outputs. At each stage write Ej=Rj−1∖Rj, and use the empty-set conventions of the algorithm. Then the algorithm terminates after q stages, every nonempty retained precision Kj is positive definite, and for every fixed retained vector x∈RRq,

∫R⋃j=1qEjexp⁡ ⁣(−12z~(x,y)TK0z~(x,y)) dλ(y)=exp⁡ ⁣(∑j=1qℓj)exp⁡ ⁣(−12xTKqx),

where y lists the eliminated coordinates in increasing order and z~(x,y)∈RR0 is the unique full coordinate vector agreeing with x on Rq and with y on R0∖Rq. The measure λ is Euclidean Lebesgue measure when the eliminated-coordinate space has positive dimension; if ⋃jEj=∅, identify R⋃jEj with the singleton R0={()} and let λ assign mass one to it. If Rq≠∅, the normalized marginal law of the retained coordinates under γK0 is exactly γKq; if Rq=∅, the final retained space is the singleton R0 and the identity gives ZK0=exp⁡(∑jℓj). In all cases,

ln⁡ZK0=∑j=1qℓj+ln⁡ZKq,

with ZKq=1 under the empty-space convention. These are exact finite-dimensional Gaussian identities and make no approximation or empirical claim.

Facts & Assumptions

Given: ACω, a finite symmetric positive-definite K0, and a finite nested retained-coordinate list as in the algorithm.

[F1]

For a positive-definite precision matrix K on Rn with n≥1, the density proportional to e−zTKz/2 has normalizer (2π)n/2(det⁡K)−1/2; the empty-space normalizer is 1 by definition (Gaussian covariance for positive quadratic forms and a boundary counterexample, Finite Gaussian probability with a positive precision matrix).

[F2]

Completing the square gives the Schur-complement identity, and for a nonempty retained block its Schur complement is positive definite (Positive Schur complements for finite Gaussian precisions).

Proof

1.1givenF2algebra

Initially K0 is positive definite. At any later stage with nonempty Ej, the current Kj−1 is positive definite by induction, so its principal block Cj is positive definite. The Schur identity in [F2] then gives exactly the algorithm's Kj=Aj−BjCj−1BjT and proves it positive definite when Rj is nonempty. If Rj is empty, the empty-matrix convention applies. Stages with Ej=∅ leave Kj=Kj−1. The algorithm is consequently well-defined at every one of its finitely many stages.

2.1F1F2F3step 1.1algebra

Fix one stage with Ej≠∅ and fix its retained coordinates x. By step 1.1 the current matrix is well-defined and positive definite. Reordering the coordinates is a linear permutation with Jacobian determinant of absolute value 1, so the change-of-variables formula in [F3] preserves the integral. With eliminated coordinates denoted by y, [F2] gives (xy)TKj−1(xy)=xTKjx+(y+Cj−1BjTx)TCj(y+Cj−1BjTx). For fixed x, the translation y↦y+Cj−1BjTx is a C1 diffeomorphism of Euclidean space with Jacobian 1. The change-of-variables formula [F3] and the Gaussian normalizer [F1] therefore give ∫REje−12(x,y)TKj−1(x,y) dy=(2π)∣Ej∣/2det⁡Cje−12xTKjx=eℓje−12xTKjx. The last equality follows from the definition of ℓj and positivity of det⁡Cj. If Ej=∅, the same identity holds with factor 1=eℓj by definition.

3.1F3F4step 2.1algebra

Apply step 2.1 successively for the finite stages. At each stage the integrand is nonnegative, so Tonelli [F3] permits the iterated eliminations in the stated order. Multiplying the one-stage factors yields, for each fixed final retained vector x, ∫R⋃jEje−12z~(x,y)TK0z~(x,y) dy=(∏j=1qeℓj)e−12xTKqx=e∑jℓje−12xTKqx, using exponential addition in [F4]. If ⋃jEj=∅, including q=0, the integral is with respect to the unit mass on R0 and the equality is the identity with an empty product.

4.1F1F3F4step 1.1step 3.1algebra∎

If no coordinates are eliminated, then every Ej=∅, Kq=K0, and ℓj=0, so the normalizer identity is immediate. If Rq=∅, then ⋃jEj=R0 has N≥1 coordinates; evaluating step 3.1 at the unique empty retained vector gives ZK0=e∑jℓj, while ZKq=1. In the remaining case both Rq and ⋃jEj are nonempty, so Tonelli [F3] and the Euclidean product-measure identity in [F3] allow integration of step 3.1 over the retained vector and give ZK0=e∑jℓjZKq. The nonempty Kq is positive definite by step 1.1, so [F1] makes ZKq finite and positive. In each case, dividing step 3.1 by ZK0 gives the retained density ZKq−1e−xTKqx/2 when Rq≠∅, exactly γKq; for empty Rq the marginal is the unit mass γ∅. Taking natural logarithms of the positive normalizer identity and using [F4] proves the logarithmic formula, including the empty-space case.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Engineering scaling of the Gaussian gradient and quartic actions

Statement

Assume the Axiom of Countable Choice ACω, meaning every sequence (Xn)n∈N of nonempty sets has a function f with f(n)∈Xn for every n. Let integer d≥1, dimensionless scale factor b>0, and real function φ′∈Cc∞(Rd) be given, where Cc∞(Rd) denotes the functions ψ:Rd→R whose partial derivatives of every order exist and are continuous and whose support supp⁡ψ:={x:ψ(x)≠0}‾ is compact. Define

φb(x):=b−(d−2)/2φ′(x/b),x∈Rd.

For any ψ∈Cc∞(Rd), define its Gaussian gradient action and quartic integral by

S2[ψ]:=12∫Rd∣∇ψ(x)∣2 dλd(x),Q4[ψ]:=14!∫Rdψ(x)4 dλd(x),

where ∇ is the vector of first partial derivatives. Then

S2[φb]=S2[φ′],Q4[φb]=b4−dQ4[φ′].

Consequently a coefficient g multiplying Q4 transforms to g′=b4−dg when the action is re-expressed in the rescaled coordinate x′=x/b. This is engineering power counting on smooth test fields; it neither constructs an interacting probability measure nor proves an interacting critical exponent.

Facts & Assumptions

Given: ACω, integer d≥1, b>0, and a real smooth compactly supported φ′ on Rd.

[F1]

Under ACω, a C1 diffeomorphism changes a nonnegative Lebesgue integral by its absolute Jacobian (The Axiom of Countable Choice (ACω), A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions).

Proof

1.1givenF1F2algebra

The map T(x′)=bx′ is a C1 diffeomorphism of Rd with Jacobian determinant bd. By [F2], each partial derivative satisfies ∂xjφb(x)=b−(d−2)/2b−1(∂xj′φ′)(x/b)=b−d/2(∂xj′φ′)(x/b). Thus ∣∇φb(x)∣2=b−d∣∇φ′(x/b)∣2. Applying [F1] with T to this nonnegative integrand gives S2[φb]=12∫Rdb−d∣∇φ′(x′)∣2bd dλd(x′)=S2[φ′]. The compact support and smoothness assumptions make these integrands measurable and finite.

2.1givenF1algebra∎

The same change of variables gives Q4[φb]=14!∫Rdb−2(d−2)φ′(x/b)4 dλd(x)=bd−2(d−2)Q4[φ′]=b4−dQ4[φ′]. Therefore gQ4[φb]=g′Q4[φ′] with g′=b4−dg, as claimed.

5 · Examples, experiments and counterexamples

None yet.

Sources