$c=1$ modular-bootstrap partition-function classification

Let $K$ be either the compact group $G$ in [F14] or a closed subgroup $H\le G$, and let ${\mathrm{\Gamma }}_K$ be the evaluation characters on $G$ or their restrictions to $H$. For a finite complex regular Borel measure $\mu$ on $K$, write $\widehat{\mu }(\gamma )=\int \bar{\gamma }d\mu$ and define modulation by $d(\gamma \mu )=\gamma d\mu$. Give $M(K)$ the total-variation norm and the weak-star topology as the dual of $C(K)$; Haar probability on each compact subgroup is normalized to mass one. Suppose $\mu \mathrm{\ne }0$ and all its ${\mathrm{\Gamma }}_K$-Fourier coefficients are integers. Let $A$ be any subset of the distinct measures $\{\gamma \mu :\gamma \in {\mathrm{\Gamma }}_K\}$, and let $\nu$ be an accumulation point of $A$ (every neighborhood of $\nu$ meets $A$ in infinitely many points). Put $M=\mathrm{\parallel }\mu \mathrm{\parallel }>0$. Every modulation and every weak-star limit has integer Fourier coefficients, since modulation shifts coefficients and coefficient evaluation is weak-star continuous. We prove $\mathrm{\parallel }\nu \mathrm{\parallel }\le M-1\mathrm{/}(8M)<M$. If $\nu =0$ this is immediate, since nonzero integer Fourier data and density in [F14] imply $M\ge 1$. Otherwise fix $0<\kappa <\mathrm{\parallel }\nu \mathrm{\parallel }\mathrm{/}M$. By [F4], the total-variation dual norm formula supplies $f\in C(K)$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{Re}\int fd\nu >\kappa M$. The resulting open half-space neighborhood contains two distinct members ${\gamma }_1\mu ,{\gamma }_2\mu$ of $A$. Since $\mu \ll \mathrm{\mid }\mu \mathrm{\mid }$, [F9] gives $d\mu =\theta d\mathrm{\mid }\mu \mathrm{\mid }$; [F10], applied to simple approximations of the measurable phase, gives $\mathrm{\mid }\mu \mathrm{\mid }=\mathrm{\mid }\theta \mathrm{\mid }\mathrm{\mid }\mu \mathrm{\mid }$, hence $\mathrm{\mid }\theta \mathrm{\mid }=1$ almost everywhere. For $j=1,2$, expanding the square gives $\int \mathrm{\mid }{\gamma }_j\theta -\bar{f}{\mathrm{\mid }}^{2}d\mathrm{\mid }\mu \mathrm{\mid }=M+\int \mathrm{\mid }f{\mathrm{\mid }}^{2}d\mathrm{\mid }\mu \mathrm{\mid }-2\mathrm{Re}\int f{\gamma }_{j}d\mu \le 2(1-\kappa )M\mathrm{.}$ The $L^2(\mathrm{\mid }\mu \mathrm{\mid })$ triangle inequality and Cauchy--Schwarz imply $\mathrm{\parallel }{\gamma }_1\mu -{\gamma }_2\mu \mathrm{\parallel }\le 2\sqrt{2(1-\kappa )}M$. Their difference is nonzero and still has integer Fourier coefficients. By [F14] its evaluation-character span is dense in $C(K)$; therefore Fourier uniqueness makes some coefficient a nonzero integer. That coefficient has modulus at most the total-variation norm, so the last norm is at least $1$. It follows that $\kappa \le 1-1\mathrm{/}(8M^2)$. Letting $\kappa \uparrow \mathrm{\parallel }\nu \mathrm{\parallel }\mathrm{/}M$ proves the asserted strict norm drop.

Write $x=P$, $y=bar$ P. Call a coordinate degenerate when it belongs to $(1\mathrm{/}2)Z_{\ge 0}$, and otherwise generic. Let $V_{ab}(X,Y)$ be the sum of $\mathrm{\mid }{d}_{xy}\mathrm{\mid }$ in $0\le x\le X$, $0\le y\le Y$ of type ab, a,b in {g,d}; $V_{ab}(R)$ means $V_{ab}(R,R)$. Constants depend on the fixed density and the fixed bump, not on growing parameters. Define the even signed density $\rho =(1\mathrm{/}4)\sum$ $d_{xy}({\delta }_{x,y}+{\delta }_{-x,y}+{\delta }_{x,-y}+{\delta }_{-x,-y})$; repeated axis atoms are counted with their multiplicities. This reproduces Z because the characters are separately even. Write $B_R:=\{x\in {\mathbb{R}}^2:\mathrm{\mid }x\mathrm{\mid }\le R\}$ for the Euclidean closed ball of radius $R$. Inverting the stipulated cumulative relation gives d_gg=D_gg; d_dg(x,y)=D(x,y)-D(x-1,y); d_gd(x,y)=D(x,y)-D(x,y-1); d_dd(x,y)=D(x,y)-D(x-1,y)-D(x,y-1)+D(x-1,y-1), where a negative coordinate means zero. Thus for real $b,c>0$, Theta(b,c)=sum D(x,y) A_x(b) A_y(c), A_x(b)=exp(-b x^2) (generic x), A_x(b)=exp(-b x^2)-exp(-b(x+1)^2) (degenerate x). Every term is nonnegative. Rearrangement is legitimate: cumulative sums defining D are bounded by cumulative |d| on the half-integer coordinates, and summation against any fixed Gaussian is bounded by summation of |d| against a slightly wider Gaussian times a convergent Gaussian tail. The assumed absolute convergence therefore gives absolute convergence of both displayed expressions, also locally uniformly for complex b,c with positive real parts. Here is the eta transformation used, rather than an extra bootstrap axiom. Define ${\theta }_3=\mathrm{\Theta }(0\mathrm{\mid }\tau )$, ${\theta }_4=\mathrm{\Theta }(1\mathrm{/}2\mathrm{\mid }\tau )$, ${\theta }_2=\exp (\pi i\tau \mathrm{/}4)\mathrm{\Theta }(\tau \mathrm{/}2\mathrm{\mid }\tau )$. Multiplying the products in [F2] and splitting even and odd indices gives $2\eta (\tau {)}^3={\theta }_2$ ${\theta }_3$ ${\theta }_4$. Periodize the explicit shifted Gaussian $\exp (-\pi t(x+a{)}^2)$, $t>0$, and integrate its Fourier coefficients by completing the square: the coefficient at n is $t^{-1\mathrm{/}2}\exp (-\pi n^2\mathrm{/}t)\exp (2\pi ina)$. Its Fourier series and all derivatives converge absolutely, so evaluation at the origin proves shifted Gaussian Poisson summation without a general summation theorem. Using $a=0$ or 1/2 and the half-frequency version gives ${\theta }_3(-1\mathrm{/}\tau )=\sqrt{-i\tau }{\theta }_3(\tau )$, with ${\theta }_2$ and ${\theta }_4$ exchanged, first for $\tau =it$ and then throughout the upper half-plane by holomorphy. Therefore $\eta (-1\mathrm{/}\tau )\mathrm{/}(\sqrt{-i\tau }\eta (\tau ))$ has cube1. The convergent eta product has no zeros; the ratio is holomorphic on the connected upper half-plane, so constant, and its value at $\tau =i$ is1. This proves $\eta (-1\mathrm{/}\tau )=\sqrt{-i\tau }\eta (\tau )$. For physical $\tau =\sigma +it$, the associated widths are $b=2\pi (t-i\sigma )$ and $c=2\pi (t+i\sigma )$; these form an open subset of a maximally totally real two-dimensional submanifold of $\{\mathrm{Re}b>0,\mathrm{Re}c>0\}$. Thus A1 and this transformation give, first on that locus and then by the local power-series identity principle, Theta(b,c)=2pi/sqrt(bc) Theta(4pi^2/b,4pi^2/c) for independent complex widths with positive real parts. In particular the identity holds for every pair of positive real widths. We also use the following uniqueness fact for the Laplace expansion: if a locally finite atomic array $a_{x,y}$ on ${\mathbb{R}}_{\ge 0}^2$ satisfies ${\sum }_{x,y}\mathrm{\mid }{a}_{x,y}\mathrm{\mid }{e}^{-bx-cy}<\mathrm{\infty }$ for every $b,c>0$ and $\sum a_{x,y}{e}^{-bx-cy}=0$ for all such $b,c$, then all $a_{x,y}=0$. Indeed, for fixed $b_0,c_0>0$, the finite complex measure $\sum a_{x,y}{e}^{-b_{0}x-c_{0}y}{\delta }_{(e^{-x},e^{-y})}$ on $[0,1{]}^2$ has every polynomial moment zero; polynomial density by [F11] and uniqueness in [F4] make it the zero measure, and injectivity of $(x,y)\mapsto (e^{-x},e^{-y})$ gives every coefficient zero. Absolute convergence of this weighted measure follows from the stipulated convergence at equal positive widths, using $\min (b,c)$. For the spin calculation take $x=P^2$ and $y={\bar{P}}^2$; local finiteness is preserved because squaring is proper on ${\mathbb{R}}_{\ge 0}$. Consequently $\mathrm{\Theta }(b,c)\le C\mathrm{/}\sqrt{bc}$ for $0<b,c\le 1$. On [X/2,X] the degenerate difference $A_x(X^{-2})$ is $\ge c_0\mathrm{/}X$, by the mean-value theorem; the bounded initial interval has the same estimate with $X=1$. Dyadic subdivision in each degenerate variable, followed by the preceding positive $\theta$ bound, gives V_gg(X,Y)≤C XY, sum_dg D(x,y), V_dg(X,Y)≤C X^2Y, sum_gd D(x,y), V_gd(X,Y)≤C XY^2, sum_dd D(x,y), V_dd(X,Y)≤C X^2Y^2. For example the dg dyadic shell x in $[2^{j-1},2^j]$ contributes at most C $2^j$ $\mathrm{\Theta }(2^{-2j},Y^{-2})\le C$ $2^{2j}$ Y; summing the geometric series gives the second line. Absolute values of d cost at most two (or four) sums of D. In particular $\mathrm{\mid }\rho \mathrm{\mid }(B_R)=O(R^4)$. Here $\mathcal{S}({\mathbb{R}}^2)$ is the space of smooth functions f for which every seminorm $su{p}_x$ $\mathrm{\mid }{x}^{\alpha }$ ${\mathrm{\partial }}^{\beta }$ f(x)| is finite, and S' is its continuous dual. Dyadic shells imply integral $(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-N}$ $d\mathrm{\mid }\rho \mathrm{\mid }(x)<\mathrm{\infty }$ for every $N>4$, so $\mathrm{\mid }<\rho ,f>\mathrm{\mid }$ $\le$ $C_N$ $su{p}_x$ $(1+\mathrm{\mid }x\mathrm{\mid }{)}^N$ |f(x)| and $\rho$ belongs to S'. Define S $f(x,y)=2$ integral $\exp (4\pi i(x{x}^{\mathrm{\prime }}-y{y}^{\mathrm{\prime }}))$ f(x',y') dx' dy' on S; [F16] gives continuity on S and S squared equal to reflection. For $T\in {\mathcal{S}}^{\mathrm{\prime }}$, define $ST$ by $⟨ST,f⟩:=⟨T,Sf⟩$ for every $f\in \mathcal{S}$; thus $S^{2}T$ is reflection as well. Gaussian transformation and A1 imply $<\rho ,f\ge <\rho ,Sf>$ for centered product Gaussians of every pair of positive widths, and also for their width derivatives. Here is a local separation proof. Put $T=\rho -S\rho$. At widths $b=c=1$, width derivatives give $⟨T,x^{2j}{y}^{2k}{e}^{-x^2-y^2}⟩=0$ for every $j,k\ge 0$; separate evenness gives the same equality for monomials with an odd exponent. The function $F(z,w)=⟨T,e^{-x^2-y^2+zx+wy}⟩$ is entire on ${\mathbb{C}}^2$: on every compact parameter set, the exponential Taylor series and each of its parameter derivatives converge in every Schwartz seminorm, bounded by a polynomial times $e^{-x^2-y^2+C(\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid })}$. All Taylor coefficients vanish, so $F=0$. For $f\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$, Fourier inversion on Schwartz functions (equivalently the identity $S^{2}f=f(-\cdot )$ above) writes $e^{-x^2-y^2}f$ as an integral of $e^{-x^2-y^2+i(tx+uy)}$ against the rapidly decreasing Fourier transform of $f$. The integral converges in Schwartz seminorms: the seminorm of each Gaussian exponential grows at most polynomially in $(t,u)$. Applying $T$ gives zero because $F(it,iu)=0$. For any $g\in C_c^{\mathrm{\infty }}$, take $f=e^{x^2+y^2}g$ to obtain $⟨T,g⟩=0$. Smooth cutoffs tending to one show $C_c^{\mathrm{\infty }}$ is dense in Schwartz space, since each weighted derivative of the discarded tail tends to zero. Thus $T=0$ and $\rho =S\rho$. Define the formal spin by $s(x,y):=x^2-y^2$. A2, after the analytic continuation above, gives a zero two-variable Laplace expansion with coefficients $d_{P,\bar{P}}(e^{2\pi i(P^2-{\bar{P}}^2)}-1)$. The uniqueness argument just proved forces every such coefficient to vanish; therefore $s(x,y):=x^2-y^2$ lies in $\mathbb{Z}$ at every occupied atom. Thus both Fourier self-duality and spin quantization have now been derived without presupposing the sharper growth bound.

Let $\lambda =\mathrm{\mid }\mu \mathrm{\mid }$ and let a net of modulations ${\gamma }_i\mu$ converge weak-star to $\nu$. For each $f\in C(G)$, $\mid \int fd\nu \mid ={\lim }_i\mid \int f{\gamma }_{i}d\mu \mid \le \int \mathrm{\mid }f\mathrm{\mid }d\lambda \mathrm{.}$ If $U\subseteq G$ is open, every $f\in C_c(U)$ extends by zero to a continuous function on $G$. The same inequality and the Riesz--Markov norm formula [F4] on the locally compact space $U$ give $\mathrm{\mid }\nu \mathrm{\mid }(U)\le \lambda (U)$. Outer regularity then gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\le \lambda (E)$ for every Borel $E\subseteq G$. Thus the limit is dominated in variation by the common modulation bound. For any closed subgroup $H\le G$ and $a\in C(H)$, the bounded Borel function $a{\mathbf{1}}_H$ on $G$ is in $L^1(\lambda )$. By [F7] approximate it in $L^1(\lambda )$ by continuous functions on $G$. Their integral errors are uniform for all ${\gamma }_i\mu$ and $\nu$, since $\mathrm{\mid }{\gamma }_i\mu \mathrm{\mid }=\lambda$ and $\mathrm{\mid }\nu \mathrm{\mid }\le \lambda$. Weak-star convergence therefore implies ${\gamma }_i(\mu {\mathrm{\mid }}_H)\to \nu {\mathrm{\mid }}_H$ weak-star on $H$. This proves the needed continuity of restriction along this dominated net. For a finite complex measure $\sigma$ on $H$, [F8] applied to real and imaginary parts and [F9] give $\sigma =f{m}_H+{\sigma }_s$, where $m_H$ is normalized Haar measure, $f\in L^1(m_H)$, and ${\sigma }_s$ is singular to $m_H$. Suppose distinct restricted evaluation characters ${\eta }_i\in \mathrm{\Gamma }{\mathrm{\mid }}_H$ escape every finite subset. For each $a\in C(H)$, approximate $af\in L^1(m_H)$ first by a continuous function using [F7], then uniformly by a finite linear combination of restricted evaluation characters using [F11] and [F14]. Haar orthogonality [F6], [F14] makes ${\int }_{H}a{\eta }_{i}fd{m}_H$ tend to zero, so ${\eta }_{i}f{m}_H\to 0$ weak-star. Every weak-star limit of ${\eta }_i{\sigma }_s$ is dominated by $\mathrm{\mid }{\sigma }_s\mathrm{\mid }$ by the preceding argument and is therefore still singular to $m_H$. Hence every limit along an escaping net of restricted evaluation characters is singular. Consequently, for any modulation-orbit limit $\nu$ and closed subgroup $H$, either $\nu {\mathrm{\mid }}_H=\eta (\mu {\mathrm{\mid }}_H)$ for some restricted evaluation character $\eta$, or $\nu {\mathrm{\mid }}_H$ is singular to $m_H$: if it is not an orbit measure, direct the neighborhoods of the limit together with finite subsets of $\mathrm{\Gamma }{\mathrm{\mid }}_H$, and choose approximating orbit measures outside each finite subset.

Fix an even smooth bump $\psi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ supported on $[-1,1]$, with $0\le \psi \le 1$ and $\psi (0)=1$, and use the Fourier convention $\widehat{\psi }(\xi ):={\int }_{\mathbb{R}}{e}^{2\pi i\xi t}\psi (t)dt$. For a doubly degenerate point (u,v) with $u,v\ge 1$, use F_uv(x,y)=sum_{σ,τ=±1} psi(b(σx-u))psi(b(τy-v)), b=4. It selects only (u,v) among all dd atoms in the quadrant. Its Fourier transform is SF_uv=8 b^-2 cos(4pi xu)cos(4pi yv) psi-hat(2x/b)psi-hat(2y/b). Since $\rho$ is tempered, the sum of the absolute values of its coefficients against the last Schwartz product is a finite constant independent of (u,v). Fourier self-duality therefore bounds $\mathrm{\mid }{d}_{uv}\mathrm{\mid }$ by that constant plus the absolute non-dd mass in the bump square. Summing over $1\le u,v\le R$, every non-dd atom lies in at most one such square (their centers have spacing 1/2, their radii are 1/4, and the bump vanishes at its boundary). The dd strips $u<1$ or $v<1$ cost $O(R^2)$ by step 1.2. Hence V_dd(R)≤C R^2+C[V_gg(R+1)+V_dg(R+1)+V_gd(R+1)]. It suffices to prove $V_{dg}(R),V_{gd}(R)=O(R^2)$.

Set $A=\{\gamma \mu :\int \gamma d\mu \mathrm{\ne }0\}$. If $\mu \mathrm{\ne }0$, density of the evaluation characters in [F14] shows $A\mathrm{\ne }\mathrm{\varnothing }$. Its weak-star closure $C$ is compact: it lies in the norm-$M$ dual ball, compact by [F5], and is closed. It does not contain $0$, since evaluation at the constant function $1$ takes values in the closed finite set of nonzero integers of modulus at most $M$. The norm is weak-star lower semicontinuous, so it attains a minimum on $C$ at some $\nu \mathrm{\ne }0$. Whenever $\int \eta d\nu \mathrm{\ne }0$, the modulation $\eta \nu$ lies in $C$: if ${\gamma }_i\mu \to \nu$, then $\int \eta {\gamma }_{i}d\mu$ converges to the nonzero integer $\int \eta d\nu$, and hence eventually equals it, so $\eta {\gamma }_i\mu \in A$. The set $B=\{\eta \nu :\int \eta d\nu \mathrm{\ne }0\}$ is therefore contained in $C$. If $B$ had an accumulation point, [1.1] applied to $\nu$ would give a point of $C$ with norm strictly below the minimum. Thus $B$ is finite, since every infinite subset of a compact Hausdorff space has an accumulation point. Let $H$ be the closed subgroup of $G$ generated by $\mathrm{supp}\nu$. Two modulations of $\nu$ agree exactly when their characters restrict equally to $H$. Therefore only finitely many restricted evaluation characters $\eta \in \mathrm{\Gamma }{\mathrm{\mid }}_H$ have ${\widehat{\nu }}_H(\eta )\mathrm{\ne }0$. Each such Fourier coefficient is an integer, because $\eta$ is the restriction of some $\gamma \in \mathrm{\Gamma }$ and $\nu$ is supported on $H$. If these distinct characters are ${\eta }_1,\dots ,{\eta }_s$, put $n_j={\widehat{\nu }}_H({\eta }_j)\in \mathbb{Z}$ and ${\nu }_0={\sum }_{j=1}^s{n}_j{\eta }_j{m}_H$. Haar orthogonality shows that ${\nu }_0$ and $\nu$ have identical Fourier coefficients on every character in $\mathrm{\Gamma }{\mathrm{\mid }}_H$; their difference integrates to zero against the uniformly dense span from [F14]. By [F4], representing measures are unique, so $\nu ={\nu }_0$. Thus $\nu$ is a nonzero finite integer combination of evaluation characters times Haar measure on $H$, and is absolutely continuous with respect to $m_H$. For this proof, a canonical measure on $H$ means a finite sum ${\sum }_j{n}_j{\eta }_j{m}_H$ with distinct restricted evaluation characters ${\eta }_j$ and integers $n_j$. This argument uses only the canonical evaluation characters; it does not assume that every abstract character of $H$ extends to $G$.

Assume R is an integer, sufficiently large; bounded R and nonintegral R follow by enlargement. Fix an occupied dg point (u,v) with $2\le u,v\le R$. Put $b=R^3$ and, for $1\le n\le R$ and a freely variable $h>0$, set G_uv(x,y)=cos(4pi h(y-v))psi(b(x-u))psi(b(y-v)), F_uv(x,y)=sum_{σ,τ=±1} sum_{s=0,1;t=-1,0,1} (-1)^s c_t G_uv(σx-sn,τy-t), c_0=1, c_1=c_-1=-1/2. Direct finite Fourier expansion yields SF_uv(x,y)=16b^-2 sin(2pi nx)sin(2pi(2u+n)x)psi-hat(2x/b) cos(4pi yv)sin^2(2pi y) [psi-hat(2(y-h)/b)+psi-hat(2(y+h)/b)]. The transform vanishes whenever either coordinate is degenerate. On the primal side, only positive reflection signs can meet the spectrum. The $t=\pm 1$ squares contain no integer-spin point: write $v=\sqrt{m}\mathrm{/}2$ with m a nonsquare integer. Then $dist(2v,Z)\ge 1\mathrm{/}(2\sqrt{m}+1)\ge c\mathrm{/}R$, whereas varying a point within a square of radius $R^{-3}$ changes its spin by $O(R^{-2})$. At the $t=0$ square, write an atom as $(u+sn+\xi ,v+\eta )$ with $\mathrm{\mid }\xi \mathrm{\mid },\mathrm{\mid }\eta \mathrm{\mid }\le R^{-3}$. The center has integer spin because $u^2-v^2\in \mathbb{Z}$ and $2un+n^2\in \mathbb{Z}$; every occupied atom has integer spin by A2. Thus the correction 2(u+sn)ξ-2vη+ξ^2-η^2 is an integer. Since $u,v,n\le R$ and $\mathrm{\mid }\xi \mathrm{\mid },\mathrm{\mid }\eta \mathrm{\mid }\le R^{-3}$, its absolute value is at most $6R^{-2}+2R^{-6}<1$ for sufficiently large $R$; it is therefore zero. If either perturbed coordinate is degenerate, its displacement is zero (the square is much narrower than half-integer spacing), and the equality then forces both displacements to vanish. Every other occupied atom in that square is gg. Thus exactly d_uv-d_(u+n,v) =-sum_gg d_xy sum_s(-1)^s G_uv(x-sn,y)+sum_gg d_xy SF_uv(x,y). To choose n, multiply the difference by $\mathrm{sign}(d_{uv})$, average $1\le n\le R$, and telescope $d_{u+n,v}=D_{u+n,v}-D_{u+n-1,v}$. Summing over occupied dg points in the indicated box gives (1/R)sum_n sum_occupied |d_uv-d_(u+n,v)| ≥V_dg(R)-C R^2. Indeed the error is bounded by $(1\mathrm{/}R){\sum }_{0\le u\le 2R,0\le v\le R,dg}{D}_{uv}=O(R^2)$, using step 1.2. The omitted strips also cost $O(R^2)$. Choose one n attaining this average, and put $e_{uv}=\mathrm{sign}(d_{uv}-d_{u+n,v})$. The primal gg error, summed over (u,v), is $O(R^2)$: different dg centers are separated in u by 1/2 or in v by c/R, hence each gg atom meets at most one square for each s; all these atoms lie in $[0,2R+1{]}^2$. Its coefficients have absolute value at most one, uniformly in h. Since psi-hat is Schwartz, the Fourier term consequently gives V_dg(R)≤C R^2+C b^-2 sum_gg |d_xy| (1+x/b)^-5 (1+|y-h|/b)^-5 |Psi(x,y)|, Psi(x,y)=sum_occupied e_uv sin(2pi(2u+n)x)cos(4pi vy). Here and below constants are uniform in h. The elementary bound $\mathrm{\mid }\mathrm{\Psi }\mathrm{\mid }\le V_{dg}(R)$ uses integer degeneracies: every occupied atom has $\mathrm{\mid }{d}_{uv}\mathrm{\mid }\ge 1$. Psi is 1-periodic in x.

If $\nu \in A$, then $\nu =\gamma \mu$ for some evaluation character $\gamma$; modulation does not change support, so $\mu$ is supported on the same $H$. The finite Haar expansion in [2.3] then shows that $\mu =\bar{\gamma }\nu$ is canonical with integer Fourier coefficients, so this stage terminates without another extraction. Otherwise $\nu$ is an accumulation point of $A$. Apply the dichotomy [2.1] on $H$. Since $\nu {\mathrm{\mid }}_H=\nu$ is a nonzero Haar-absolutely-continuous measure by [2.3], the singular alternative is impossible; hence $\nu =\eta (\mu {\mathrm{\mid }}_H)$ for some restricted evaluation character $\eta$. Therefore ${\mu }_1:=\mu {\mathrm{\mid }}_H=\bar{\eta }\nu$ is a nonzero finite Haar expansion with integer coefficients. Its Fourier coefficients on $G$ are integers: each restricted evaluation character in the expansion pairs orthogonally with any other such restriction. The remainder $\mu -{\mu }_1=\mu {\mathrm{\mid }}_{G\setminus H}$ is mutually singular with ${\mu }_1$, so $\mathrm{\parallel }\mu -{\mu }_1\mathrm{\parallel }=\mathrm{\parallel }\mu \mathrm{\parallel }-\mathrm{\parallel }{\mu }_1\mathrm{\parallel }\le \mathrm{\parallel }\mu \mathrm{\parallel }-1,$ because a nonzero integer Fourier coefficient of ${\mu }_1$ has modulus at least $1$. The remainder retains integer Fourier coefficients. Repeat the same construction; every nonzero remainder has norm at least $1$, so after at most $\lfloor \mathrm{\parallel }\mu \mathrm{\parallel }\rfloor$ extractions the remainder is zero. Thus $\mu$ is a finite sum of canonical components. The Fourier transform on the evaluation characters of $\eta m_H$ is exactly the coset indicator ${\mathbf{1}}_{v_0+H^{\perp }}$, where $v_0$ represents $\eta$ and $H^{\perp }=\{v\in E:{\chi }_v{\mathrm{\mid }}_H=1\}$. This proves the needed integer-Fourier-measure result for the Bohr group [F14] and its closed subgroups, without invoking a general idempotent-measure theorem.

Take $h=B$ l b for $l=1,\mathrm{.}\mathrm{.}\mathrm{.},L$, $B\ge 1$. The rectangle estimate $V_{gg}(X,Y)\le CXY$ gives sum_{l≤L}sum_gg |d_xy|(1+x/b)^-3(1+|y/b-Bl|)^-3 ≤C LB b^2. For completeness, summing the second factor in l is bounded by $C(1+max(y\mathrm{/}b-BL,0){)}^{-2}$. This follows by comparison of the decreasing tail with its integral, while on $y\mathrm{/}b\le BL$ the separated centers give a bounded sum. Subdivide x at $2^i$ b and y at $2^j$ LBb; rectangle growth bounds each shell, and the series ${\sum }_i{2}^{-2i}$ ${\sum }_j{2}^{-j}$ converges. This proves the stated estimate with constants independent of B,L. At least half the l therefore satisfy the same inner sum $\le C$ B $b^2$. Call these good. In step 3.1's Fourier sum discard the region $\mathrm{\mid }y-Blb\mathrm{\mid }>Bb\mathrm{/}4$: two unused decay powers give its bound C $B^{-1}$ $V_{dg}(R)$. Fix B large enough to absorb this into the left hand side. On the remaining region, periodicity in x and the good-height bound give constants $c,\delta >0$, independent of R,l, such that 2delta V_dg(R)≤cR^2+|Psi(x_l,y_l)|, 0≤x_l≤1, |y_l-Blb|≤Bb/4, for at least half the positive integers l. More precisely the finite-L pigeonhole estimate yields at least L/2 good indices among the first L, sufficient for every limiting lower-density assertion that follows. The derivatives satisfy $\mathrm{\mid }{\mathrm{\Psi }}_x\mathrm{\mid }\le 6\pi$ R $V_{dg}(R)$, $\mathrm{\mid }{\mathrm{\Psi }}_y\mathrm{\mid }\le 4\pi$ R $V_{dg}(R)$. On a square of half-side delta/(100R) about each $(x_l,y_l)$, interpreted modulo 1 horizontally, therefore $\mathrm{\mid }\mathrm{\Psi }\mathrm{\mid }\ge \delta$ $V_{dg}(R)-c{R}^2$. The squares are disjoint because distinct selected heights differ by at least Bb/2. Their lower area proportion in [0,1]×[0,T], as T tends to infinity, is at least $c_{1R}^{-5}$. This is a genuine lower bound, with fixed positive $c_1$, despite the possibly very large value of R.

Write $u=k\mathrm{/}2$, $v=j$ $\sqrt{s}\mathrm{/}2$ with $s>1$ squarefree and $j>0$. These are forced by integer spin. Then Psi(x,y)=sum_s Phi_s(x,sqrt(s)y), Phi_s(x,z)=sum_(k,j) e_(k/2,j sqrt(s)/2) sin(2pi(k+n)x)cos(2pi jz). For fixed x, equip the finite product torus indexed by the occurring squarefree s with product Haar probability and let $X_s$ be its coordinate variables; they are independent and individually uniform on R/Z. The terms ${\mathrm{\Phi }}_s(x,X_s)$ are independent, mean zero, and satisfy |Phi_s|≤C R^2, sum_s E Phi_s^2≤C' R V_dg(R). The first inequality counts at most $O(R^2)$ pairs (k,j). For the second, orthogonality of cosines gives $(1\mathrm{/}2){\sum }_{s,j}({\sum }_{k}esin{)}^2$; Cauchy-Schwarz bounds each inner square by O(R) times its number of occupied k. The total number of occupied pairs is at most $V_{dg}(R)$. Distinct squarefree square roots are rationally independent. One elementary proof places them in the multiquadratic field generated by the finitely many occurring primes; independent sign changes of prime square roots give distinct characters of its elementary abelian Galois group, and character orthogonality isolates every squarefree monomial coefficient. Here is the needed degree induction: after adjoining the first r prime square roots, the rational numbers that have a square root in the extension have square classes generated by those r primes. For the next prime p, unique factorization therefore shows $\sqrt{p}$ is not already present. If $(a+b\sqrt{p}{)}^2$ is rational, comparison of the two basis components forces $ab=0$. Thus its rational square class either already belonged to the previous list or is p times one that did. This proves the assertion and the quadratic degree increase by induction. Each independent prime sign change is now an automorphism of the resulting field, proving the character-orthogonality argument above. Continuous flow y $mapsto(\sqrt{s}ymod1{)}_s$ consequently equidistributes: integrate each nonconstant torus character explicitly, obtaining limit zero, and approximate continuous functions uniformly by finite trigonometric polynomials. After integration in x, this applies to any continuous function of Psi. Here is the exponential inequality used, including its proof. If independent mean-zero real $Y_i$ satisfy $\mathrm{\mid }{Y}_i\mathrm{\mid }\le a$ and sum E $Y_i^2\le {\sigma }^2$, then for $t>0$ P(|sum Y_i|≥t)≤2 exp[-sigma^2/a^2 h(at/sigma^2)], h(u)=(1+u)log(1+u)-u. For $\lambda >0$, the Taylor series and $\mathrm{\mid }{Y}_i{\mathrm{\mid }}^k\le a^{k-2}{Y}_i^2$ bound E $\exp (\lambda Y_i)$ by exp[(E $Y_i^2\mathrm{/}{a}^2)(\exp (\lambda a)-1-\lambda$ a)]. Multiply expectations, apply Markov, and minimize at $\exp (\lambda a)=1+at\mathrm{/}{\sigma }^2$. Apply the same bound to $-Y_i$. The exponent is decreasing as ${\sigma }^2$ increases, so an upper variance bound is legitimate. Set $V=V_{dg}(R)$, $a=C{R}^2$, ${\sigma }^2=C^{\mathrm{\prime }}RV$, $t=\delta$ $V\mathrm{/}2-c{R}^2$. If $V\le (4c\mathrm{/}\delta )R^2$ there is nothing to prove. Otherwise $t\ge \delta$ V/4. The formula and $h(u)\ge (u\mathrm{/}2)\log$ u imply, for large R, P(|sum_s Phi_s(x,X_s)|≥delta V/2-cR^2) ≤C_2 exp[-c_2(delta V/R^2-2c)log R], uniformly in x. To pass from equidistribution to the level-set area, use a continuous even majorant equal to 1 for $\mathrm{\mid }z\mathrm{\mid }\ge \delta$ V-$c{R}^2$ and zero for $\mathrm{\mid }z\mathrm{\mid }<\delta$ $V\mathrm{/}2-c{R}^2$. The limiting proportion where $\mathrm{\mid }\mathrm{\Psi }\mathrm{\mid }\ge \delta$ V-$c{R}^2$ is bounded by the last probability. Combining this upper bound with step 4.1's lower bound $c_{1R}^{-5}$ and taking logarithms gives $V\le C_{3R}^2$. The gd argument is identical with coordinates exchanged; step 2.2 now gives $V_{dd}(R)=O(R^2)$. Therefore |rho|(B_R)≤C R^2. This closes the sharp-density prerequisite from precisely the target hypotheses, without positivity of the quadrant multiplicity coefficients and without treating the whole spectral measure as positive.

Define the Bohr compactification concretely as the compact group of all homomorphisms from the additive group $R^2$ with discrete topology to U(1), a closed subgroup of the product $U(1{)}^{R^2}$. Its evaluation characters ${\chi }_v$, indexed by v in $R^2$, separate points and are closed under products and conjugation, so their span is uniformly dense in its continuous functions. They are all its characters: a different character would be Haar-orthogonal to this dense span. The canonical continuous map b sends x to the homomorphism v mapsto $\exp (4\pi i(v_1{x}_1-v_2{x}_2))$. Apply the construction to $4\rho$, whose atoms have integer coefficients. Choose an even compactly supported smooth phi that equals 1 on a neighborhood of 0, and let n increase to infinity; define ${\rho }_n=(xmapsto\varphi (nx))\ast 4\rho$, where * denotes convolution on $R^2$. For ${\varphi }_n(x)=\varphi (nx)$, changing variables gives $S{\varphi }_n(x)=n^{-2}(S\varphi )(x\mathrm{/}n)$. For $T\in {\mathcal{S}}^{\mathrm{\prime }}$ and ${\varphi }_n\in C_c^{\mathrm{\infty }}$, testing against a Schwartz function and applying Fubini gives $S({\varphi }_n\ast T)=\frac{1}{2}(S{\varphi }_n)(ST)$ under the normalization of $S$ in [1.2]. Hence $S{\rho }_n=(1\mathrm{/}(2n^2))(S\varphi )(\cdot \mathrm{/}n)S(4\rho )$, where the Schwartz factor multiplies the measure. Since $S(4\rho )=4\rho$, its total variation is bounded by $(2\mathrm{/}{n}^2)$ integral |Sphi(x/n)| $d\mathrm{\mid }\rho \mathrm{\mid }(x)$. The quadratic bound $\mathrm{\mid }\rho \mathrm{\mid }(B_R)=O(R^2)$ from [5.1] and dyadic shells show that this integral is $O(n^2)$: the ball $\mathrm{\mid }x\mathrm{\mid }\le n$ contributes $O(n^2)$, and on $2^j$ $n\le \mathrm{\mid }x\mathrm{\mid }\le 2^{j+1}n$ Schwartz decay contributes $O(n^2{2}^{-2j})$ after choosing four decay powers. Thus ||S ${\rho }_n\mathrm{\mid }{\mathrm{\mid }}_{TV}\le C$ uniformly for $n\ge 1$. The functional $L_n(F)=2$ integral (F composed with b) $d(S{\rho }_n)$ on the Bohr compactification therefore has uniformly bounded norm. On the character ${\chi }_{-}v$, $L_n$ equals ${\rho }_n(v)=(S^2{\rho }_n)(-v)=\int$ phi(n(v-x)) $d(4\rho )(x)$. Since $4\rho$ is locally finite and phi has compact support, for each fixed v and all sufficiently large n only $x=v$ contributes; $\varphi (0)=1$, so $L_n({\chi }_{-}v)=4\rho (\{v\})$ eventually. Finite linear combinations of evaluation characters are uniformly dense by [F14]. For each such polynomial the values $L_n$ stabilize, since it uses only finitely many fixed frequencies; their limit defines a linear functional L on the polynomial algebra with $\mathrm{\mid }L(P)\mathrm{\mid }\le C\mathrm{\mid }\mathrm{\mid }P\mathrm{\mid }{\mathrm{\mid }}_{\mathrm{\infty }}$. Density extends L uniquely to a bounded functional on C(G), and [F4] represents it by a finite regular Borel measure mu. Its Fourier coefficients are exactly the integers $4\rho (\{v\})$. The locally proved integer-Fourier-measure theorem says these coefficients are a finite sum of integer multiples of coset indicators in $R_{disc}^2$. They take only finitely many values since $\mathrm{\mid }\mu -hat\mathrm{\mid }\le \mathrm{\mid }\mathrm{\mid }\mu \mathrm{\mid }\mathrm{\mid }$. For each nonzero value k, its level set is in the coset ring (finite intersections/complements of the finitely many coset indicators); it is locally finite by the assumed spectrum. The existing proved lemma lem-cft-r5-discrete-elements-of-coset-ring replaces it by a finite Boolean combination of discrete cosets. Inclusion-exclusion converts its indicator to a finite integer linear combination of discrete coset indicators. Summing over the finitely many k gives a finite lattice-comb expansion. Here a generalized comb means a finite complex linear combination of measures ${\sum }_{x\in a+L}{\delta }_x$, where $L\le {\mathbb{R}}^2$ is a discrete subgroup and $a\in {\mathbb{R}}^2$; its rank is the rank of $L$, which is $0$, $1$, or $2$. A rank-two lattice is a discrete subgroup spanning ${\mathbb{R}}^2$. No converse or unproved multiplicity assertion is being inserted at this interface.

Use the Lorentzian pairing $⟨x,y{⟩}_L:=2(x_1{y}_1-x_2{y}_2)$ and norm $\mathrm{\parallel }x{\mathrm{\parallel }}_L^2:=⟨x,x{⟩}_L=2(x_1^2-x_2^2)$, so the spin condition from step 1.2 is exactly $\mathrm{\parallel }x{\mathrm{\parallel }}_L^2\in 2\mathbb{Z}$. For a lattice $L$, define its Lorentzian dual by $L^{\vee }:=\{y:⟨x,y{⟩}_L\in \mathbb{Z}\text{ for all }x\in L\}$. Two rank-two lattices are commensurate when their intersection has finite index in each. A lattice is even when $\mathrm{\parallel }x{\mathrm{\parallel }}_L^2\in 2\mathbb{Z}$ for every lattice vector and unimodular when its Gram determinant has absolute value one. Split the finite generalized-comb expansion as $\rho ={\rho }_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}+{\rho }_{\mathrm{l}\mathrm{o}\mathrm{w}}$, where the first sum has rank-two lattice cosets and the second has ranks zero or one. By [F15] after an invertible linear change of coordinates, the Fourier transform of a full-rank lattice comb is an atomic dual-lattice comb. For a rank-$r<2$ comb, apply the same Poisson formula in a basis of its rank-$r$ span; its transform has a discrete comb in the transformed span and Lebesgue measure in a positive-dimensional transverse direction, with a phase from the coset translation, so it has no point masses (rank zero gives a smooth density). The original $\rho$ and ${\rho }_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$ are atomic, hence ${\rho }_{\mathrm{l}\mathrm{o}\mathrm{w}}$ is atomic. Since $S\rho =\rho$, also $S{\rho }_{\mathrm{l}\mathrm{o}\mathrm{w}}=\rho -S{\rho }_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$ is atomic, while the rank calculation makes it atomless. Therefore $S{\rho }_{\mathrm{l}\mathrm{o}\mathrm{w}}=0$; injectivity of $S$ (its square is reflection) gives ${\rho }_{\mathrm{l}\mathrm{o}\mathrm{w}}=0$. Group full-rank lattices by commensurability and replace each class by its finite-index intersection lattice. Thus $\rho ={\sum }_c$ ${\rho }_c$, each ${\rho }_c$ a finite sum of cosets of one rank-two lattice ${\mathrm{\Lambda }}_c$. Different classes intersect only on finitely many rank-zero or rank-one cosets. By combining identical cosets within a class, every coset of nonzero coefficient belongs to the actual support except for such lower-rank intersections. For v,w in ${\mathrm{\Lambda }}_c$ choose x in an occupied coset such that x,x+v,x+w,x+v+w all avoid those intersections. Existence follows because a rank-two lattice is not covered by finitely many cosets of rank at most one: in lattice coordinates, the box $[-n,n{]}^2$ contains $O(n^2)$ points, while each such coset contains O(n). Spin quantization gives 2\langle v,w\rangle_L=||x+v+w||_L^2-||x+v||_L^2-||x+w||_L^2+||x||_L^2 in 2Z. Hence ${\mathrm{\Lambda }}_c$ is integral and commensurable with its dual. Symmetrize each class under joint inversion, which preserves its class and the total density. ${\mathrm{\Delta }}_c=S{\rho }_c-{\rho }_c$ is supported both on finitely many cosets in that class and, since sum ${\mathrm{\Delta }}_c=0$, on finitely many lower-rank intersections. Restrict the finite generalized-comb expression to those intersections. Restriction of a lattice exponential to a sublattice is another lattice exponential, and finite inclusion-exclusion handles intersections; thus ${\mathrm{\Delta }}_c$ is a finite sum of lower-rank generalized combs. But $S{\mathrm{\Delta }}_c=-{\mathrm{\Delta }}_c$, since ${\rho }_c$ is inversion-even. The left side has no atoms and the right side is atomic, so ${\mathrm{\Delta }}_c=0$. Let $K:=\{k\in {\mathbb{R}}^2:{\tau }_k{\rho }_c={\rho }_c\}$ be its full period group, where ${\tau }_k$ denotes translation by $k$, and let $M$ be the subgroup generated by its support. Choose $x\in \mathrm{supp}{\rho }_c$. Then $x+K\subseteq \mathrm{supp}{\rho }_c$, which lies in a finite union of cosets of ${\mathrm{\Lambda }}_c$; hence $K$ is a subgroup of a discrete set and is discrete. It contains ${\mathrm{\Lambda }}_c$, so it is a rank-two lattice. Because ${\rho }_c$ is $K$-periodic, its Lorentzian Fourier transform is supported on $K^{\vee }$; self-duality $S{\rho }_c={\rho }_c$ gives $\mathrm{supp}{\rho }_c\subseteq K^{\vee }$, hence $M\subseteq K^{\vee }$. Also $K\subseteq M$, because both $x$ and $x+k$ belong to the support for every period $k$. Conversely, for $k\in M^{\vee }$, the Fourier multiplier $x\mapsto e^{2\pi i⟨x,k{⟩}_L}$ is $1$ on $\mathrm{supp}{\rho }_c\subseteq M$, so $S{\rho }_c$ is unchanged by translation by $k$. Fourier injectivity and $S{\rho }_c={\rho }_c$ show $M^{\vee }\subseteq K$. Dualizing $M\subseteq K^{\vee }$ gives $K\subseteq M^{\vee }$; hence $K=M^{\vee }$ and $M=K^{\vee }$. For any $k\in K$, choose $x$ in a support coset of ${\rho }_c$ avoiding the finite union of rank-at-most-one intersections both at $x$ and at $x+k$; a rank-two $K$-coset is not covered by that union, so $x,x+k\in \mathrm{supp}\rho$. Since $x\in M=K^{\vee }$ and $k\in K$, $⟨x,k{⟩}_L$ is integral; spin quantization gives $\mathrm{\parallel }k{\mathrm{\parallel }}_L^2\in 2\mathbb{Z}$. For any $y\in \mathrm{supp}{\rho }_c$, the same rank-two-coset argument chooses a period $k$ with $y+k\in \mathrm{supp}\rho$; since $y\in M$ and $k\in K$, the same identity gives $\mathrm{\parallel }y{\mathrm{\parallel }}_L^2\in 2\mathbb{Z}$. Thus K is even. The quotient $G=M\mathrm{/}K$ is finite. Define $q(x+K):=\mathrm{\parallel }x{\mathrm{\parallel }}_L^2\mathrm{/}2(\mathrm{m}\mathrm{o}\mathrm{d}\mathbb{Z})$: if $k\in K$, then $q(x+k)-q(x)=⟨x,k{⟩}_L+\mathrm{\parallel }k{\mathrm{\parallel }}_L^2\mathrm{/}2\in \mathbb{Z}$, since $M=K^{\vee }$ and $K$ is even. The induced bilinear pairing on $G$ is nondegenerate because $K=M^{\vee }$, so $(G,q)$ is a finite quadratic module. A subset is isotropic if $q$ vanishes on it. The support of ${\rho }_c$ modulo $K$ is isotropic and generates $G$, which is generated by at most two elements. Write ${\rho }_c={\sum }_{x\in G}f(x){\delta }_{x+K}$. Its self-duality is finite Fourier invariance f(y)=|G|^-1/2 sum_x f(x)exp(2pi i(x,y)_L), and T-invariance says f is supported on $q(x)=\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }{\mathrm{\mid }}_L^2\mathrm{/}2=0$ in Q/Z. A generating set of isotropic elements can be replaced by two isotropic generators. To see this, decompose into p-primary groups. Orthogonality of distinct primary components follows by coprimality of their orders. Isotropy splits into its primary components because $q(x_p)$ has p-power order. On $G_p\mathrm{/}p{G}_p$, of dimension at most two, choose two images of isotropic generators forming a basis, and lift them to isotropic elements. They generate $G_p$: if their subgroup H has $G_p=H+p{G}_p$, iterate this equality until $p^r{G}_p=0$. Sum the chosen first and second generators over primes. Their sums are isotropic and generate G by the Chinese remainder projections. Call them $e_1,e_2$ and $\lambda =⟨$ $e_1,e_2{⟩}_L$. Nondegeneracy shows $\mathrm{ord}(e_1)=\mathrm{ord}(e_2)=\mathrm{ord}(\lambda )=N$: $\mathrm{ord}(\lambda )e_i$ pairs trivially with both generators; conversely $\mathrm{ord}(e_i)\lambda =0$. A relation $a{e}_1+b{e}_2=0$ pairs with $e_1,e_2$ to force N|a,N|b. Rescaling $e_1$ by a unit gives G=(Z/N)^2, q(a,b)=ab/N, ((a,b),(c,d))=(ad+bc)/N. (The case $N=1$ is included.) Define $F(a,b)={\sum }_w$ $f(a,w)\exp (2\pi ibw\mathrm{/}N)$. Direct finite summation turns S and T into $F(b,-a)=F(a,b)$ and $F(a,a+b)=F(a,b)$. Their orbits are classified by gcd(a,b,N). For the converse orbit assertion, divide by that gcd and choose integral representatives a',b' coprime: one can keep a' fixed and change b' by a multiple of N' to avoid every prime divisor of a' not dividing N', by the Chinese remainder theorem; prime divisors of N' are already avoided by $\gcd (a^{\mathrm{\prime }},b^{\mathrm{\prime }},N^{\mathrm{\prime }})=1$. If $a^{\mathrm{\prime }}=0$, first replace it by N'. Complete the coprime pair to an SL(2,Z) matrix by Bezout. Thus the orbit contains (g,0). The orbit-indicator basis can equivalently be replaced by the triangular basis $1_{g\mathrm{\mid }a,g\mathrm{\mid }b}$, g|N. Inverting the b-transform maps this basis to $(1\mathrm{/}g)1_{g\mathrm{\mid }a,(N\mathrm{/}g)\mathrm{\mid }w}$. Therefore f is a linear combination of indicators of the isotropic order-N subgroups H_g=g(Z/N) x (N/g)(Z/N). The inverse images Lambda(g) of $H_g$ in M are even self-dual lattices: isotropy gives evenness and $H_g=H_g^{\perp }$ by nondegeneracy and equal orders. Every such lattice is a Narain lattice. Indeed an even unimodular rank-two Gram matrix [[2a,b],[b,2c]] of signature (1,1) has $b^2-4ac=1$. Its quadratic polynomial has a rational zero (if $a=0$ there is already an evident zero), so the lattice has a primitive null vector e. Unimodularity gives f with $(e,f)=1$. Replace f by f-(f,f)e/2; this is integral, null, and still pairs to one with e. In the given ambient Lorentzian plane, the two null generators have form (R/2)(1,1) and (1/(2R))(1,-1), after signs and interchange. Thus the original density, after harmless separate sign symmetrization and R↔1/R, is a FINITE sum rho=sum_{R≥1} c_R rho_R.