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Unconstrained massless periodic Gaussian has infinite normalizer

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ . Let $d, L \geq 1$ be integers, let $Λ_{d, L} = (Z / L Z)^{d}$ have $N = L^{d}$ sites, and let $κ > 0$ . Define the zero-mass precision $K : = κ (- Δ)$ on $R^{Λ_{d, L}}$ , with the periodic Laplacian convention of Fourier spectrum and positivity of the periodic lattice Laplacian. Then the unrestricted Lebesgue normalizer diverges:

$\int_{R^{Λ_{d, L}}} \exp ⁣ (- \frac{1}{2} z^{T} K z) d λ_{N} (z) = + \infty .$

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 ${z : \sum_{x \in Λ_{d, L}} z_{x} = 0}$ , before inverting the Laplacian.

Statement references

Fourier spectrum and positivity of the periodic lattice Laplacian

Facts & Assumptions

Given: ${A C}_{ω}$ , integers $d, L \geq 1$ , the periodic site set $Λ_{d, L}$ with $N = L^{d}$ , and $κ > 0$ .

[F1]

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

[F2]

Under ${A C}_{ω}$ , a $C^{1}$ diffeomorphism changes a nonnegative Lebesgue integral by its absolute Jacobian (The Axiom of Countable Choice ( ${A C}_{ω}$ ), 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 $L (R^{n})$ , and the restricted set function $λ_{n}$ , 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 $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , 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], $K 1 = 0$ and $K$ is symmetric. Hence the quadratic form is unchanged by adding any constant multiple of $1$ : for every $z$ and $t \in R$ , $(z + t 1)^{T} K (z + t 1) = z^{T} K z$ .

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 $2 M$ for every integer $M \geq 1$ by [F4], and [F5] bounds the full integral below by each of these unbounded values.

2.2givenF2step 1.1algebra

Suppose $N \geq 2$ and enumerate the lattice sites as $x_{1}, \dots, x_{N}$ . Define the linear map $T : R^{N - 1} \times R \to R^{N}$ by $T (w, t) = (w_{1} + t, \dots, w_{N - 1} + t, t)$ . Its matrix is triangular with all diagonal entries $1$ , so it is a $C^{1}$ diffeomorphism with determinant $1$ . For $\tilde{w} = (w_{1}, \dots, w_{N - 1}, 0)$ , one has $T (w, t) = \tilde{w} + t 1$ . By step 1.1, $T (w, t)^{T} K T (w, t) = {\tilde{w}}^{T} K \tilde{w}$ , which is independent of $t$ . Applying [F2] changes the original integral to $\int_{R^{N - 1} \times R} f (w) d λ_{N} (w, t), f (w) : = \exp ⁣ (- \frac{1}{2} {\tilde{w}}^{T} K \tilde{w}) .$

3.1F3F4F5step 2.2algebra∎

The function $f$ is continuous and positive. On $[- 1, 1]^{N - 1}$ , $∣ {\tilde{w}}^{T} K \tilde{w} ∣ \leq M$ , where $M : = \sum_{i, j = 1}^{N - 1} ∣ K_{i j} ∣ < \infty$ ; hence $f (w) \geq e^{- M / 2}$ there. By [F4] and [F5], $\int_{R^{N - 1}} f (w) d λ_{N - 1} (w) \geq 2^{N - 1} e^{- M / 2} > 0$ . By [F3], Tonelli factors the nonnegative integral in step 2.2 as $\int f (w) d w \int_{R} 1 d t$ . The second factor is infinite because each interval $[- M^{'}, M^{'}]$ has measure $2 M^{'}$ for integer $M^{'} \geq 1$ by [F4], and [F5] bounds the full-line integral below by each such value. Thus the original normalizer is infinite, proving the claim.

Unconstrained massless periodic Gaussian has infinite normalizer

Statement

Statement references

Facts & Assumptions

Proof

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