Equidistribution of primitive quartic torus packets for arbitrary orders

Proof. The additive group R/\OO has order q, so qR\subset\OO. For every p and every unit t_p, it follows that

At p\nmid q the middle lattice is I_p. At the remaining primes its image is a submodule of the finite module I_p/qI_p. The primary decomposition of I/qI identifies its p-primary part with I_p/qI_p. Thus a choice of these local submodules patches to a unique subgroup of I/qI, and its inverse image in I is the unique lattice with the prescribed completions.

The stabilizer of \OO_p under multiplication by R_p^\times is exactly \OO_p^\times. Indeed, t_p\OO_p=\OO_p implies that both t_p and t_p^{-1} belong to \OO_p, by applying the equality and its inverse to 1. The converse is immediate. The local choices therefore form the finite homogeneous space R_p^\times/\OO_p^\times. Its invariant probability is uniform and is the image of normalized unit Haar measure. Patching over the finitely many primes dividing q proves Equation (2.4) and the assertion about probabilities. The index of the lattice in Equation (2.3) is [R_p:\OO_p], and its R_p-span is I_p because 1\in\OO_p. Taking products of the indices and comparing all completions gives [I:L]=q and RL=I.

We record why the local condition describes exactly invertible ideals of the order. A full fractional \OO-ideal J is invertible if and only if

For the forward implication, complete an equality JJ^{-1}=\OO. The resulting equality over \OO_p expresses 1 as a finite sum \sum_i a_i b_i, with a_i\in J_p and b_i\in J_p^{-1}. At each maximal ideal \mathfrak m of the semilocal ring \OO_p, some a_i b_i is a unit after localization, and that a_i generates (J_p)_{\mathfrak m}. To obtain one generator for all the maximal ideals, choose such an a_{\mathfrak m} for each \mathfrak m, and use the Chinese remainder theorem [16, Theorem 1.14] to choose c_{\mathfrak m}\in\OO_p congruent to 1 at \mathfrak m and to 0 at every other maximal ideal. Then \eta_p=\sum_{\mathfrak m}c_{\mathfrak m}a_{\mathfrak m} generates (J_p)_{\mathfrak m} modulo \mathfrak m at each \mathfrak m; Nakayama’s lemma [16, Lemma 1.9] shows that it generates each localization. Hence it generates J_p. Since J_p spans K_p over \Q_p, \eta_p has no zero component and lies in K_p^\times.

Conversely, define

This definition commutes with completion: after choosing \Z-bases of J and \OO, the containment is a finite collection of integrality conditions on the coordinates of multiplication by a, and the same conditions over \Z_p describe the completed lattice. Under Equation (2.5) we therefore have (J^{-1})_p=\eta_p^{-1}\OO_p. Consequently (JJ^{-1})_p=\OO_p for every p, which implies JJ^{-1}=\OO. This proves the equivalence. It also proves properness: if aJ\subset J for an invertible J, multiplication by J^{-1} gives a\OO\subset\OO, and hence a\in\OO. Thus its multiplier order is \OO.

The lattices in Equation (2.3) are stable under \OO_p at every prime, so comparison of completions shows that \OO L\subset L. They are therefore fractional \OO-ideals satisfying Equation (2.5). In the other direction, let J be an invertible fractional \OO-ideal. Its R-span RJ is a fractional R-ideal. Multiplying J by an element of K^\times, we may arrange RJ=I for the chosen representative of this maximal-order class. If J_p=\eta_p\OO_p, then \eta_p R_p=I_p=\xi_pR_p, so \eta_p=\xi_p t_p for a unit t_p\in R_p^\times. Thus the resulting J belongs to \mathscr L_{\OO}(I), as required.□

Proof. The action in Equation (2.7) preserves the parameter set, since \epsilon I=I for \epsilon\in U, and preserves its image lattice. Lemma 2.1 shows that every ordinary invertible \OO-ideal class occurs. If two ideals differ by multiplication by \alpha\in K^\times, their normalized embedded lattices differ by the diagonal with coordinates \sigma_i(\alpha)/|\Nm_{K/\Q}(\alpha)|^{1/4}, an element of T_\infty. Finally T_\infty is the union of all sixteen sign translates of A_4. These facts prove surjectivity.

Suppose two displayed normalized lattices coincide. After absorbing the ratio of their positive scalars, this equality gives a diagonal map carrying \sigma(L) onto \sigma(L'). Choose 0\ne b\in L and let b'\in L' be its image. None of the four coordinates of \sigma(b) is zero. The diagonal entries must therefore be \sigma_i(b')/\sigma_i(b)=\sigma_i(b'/b). The map is multiplication by \alpha=b'/b, and L'=\alpha L. Taking R-spans gives I'=\alpha I. The chosen representatives then force I'=I and \alpha I=I, so \alpha\in U. The scalar normalizations are now equal, and the two parameters differ by exactly Equation (2.7). This proves injectivity, including all identifications involving signs. Equality of covolume-one row lattices is the same as equality in X_4: any two positively oriented bases of the same lattice differ by an element of \mathrm{SL}_4(\Z).

It remains to check that this identification gives the stipulated orbit weights. Write \OO^{\times,+}=\OO^\times\cap U^+. The U-stabilizer of a label L is \OO^\times, by its multiplier order. The stabilizer of L together with a sign pattern is therefore \OO^{\times,+}. This subgroup has finite index in U. For q=1 this follows directly from \OO=R. For q>1, the kernel of U\longrightarrow(R/qR)^\times lies in \OO^\times, since both a unit congruent to 1 modulo qR and its inverse belong to \OO, and imposing positive signs costs at most a further factor of sixteen. The same diagonal-identification argument shows that all periods of the corresponding A_4-orbit are precisely \ell(\OO^{\times,+}). They form a full lattice, so the orbit is compact.

Combining a label with a sign produces 16N_{\OO} finite choices above each I. Each U-orbit of these choices has [U:\OO^{\times,+}] elements and parametrizes one A_4-orbit. On that orbit the quotient of logarithmic Lebesgue measure is its Haar measure, with volume

The last equality uses \ker(\ell|_U)=\{1,-1\} and -1\notin\OO^{\times,+}. Consequently the number of distinct orbits times their common volume is

This proves both the assertion about the descended measure and Equation (2.8), by Equation (2.6).

One can also take \mathscr L_{\OO}(I)\times\mathscr F_U as a fundamental domain for the diagonal action in Equation (2.7): reducing v into \mathscr F_U transports L by the same unique unit, outside the usual null boundary set. Counting measure times dv is invariant under this action. These domains have total measure h_KN_{\OO}\vol(\mathscr F_U), equal to the total volume just computed. Dividing their integral by that quantity proves Equation (2.9).□

Proof. This is the unfolding of [18, Proposition 2.2] with the packet normalization of Proposition 2.2. Average the vector sum over L in Equation (2.9), and use Equation (3.3). The factor N_{\OO} cancels, and the left side becomes

For fixed I, the map b\mapsto(b)I^{-1} takes I\setminus\{0\} onto the integral ideals in the inverse class of I. Each fiber is one free U-orbit. Indeed, two generators of the same principal ideal differ by a unique unit. The coefficient P((b)I^{-1}) is constant on that orbit. Summing its unit translates and integrating over \mathscr F_U gives

The translates tile T_\infty, including its sign components. All changes in the order of integration and summation are justified by Tonelli’s theorem.

The absolute product of the coordinates of c_I\sigma(b) is

Translation of the first three logarithmic coordinates and relabeling of the signs therefore identify the last integral with \mathcal V_f(\Nm((b)I^{-1})/X), with Jacobian one. Each integral ideal belongs to the inverse class of exactly one chosen I. Equation (2.6) now gives the coefficient in Equation (3.5).□

Proof. Normalize additive Haar measure m_p on R_p by m_p(R_p)=1. In a component R_{\mathfrak p}, the set of elements of valuation l_{\mathfrak p} has measure (1-p^{-f_{\mathfrak p}})p^{-f_{\mathfrak p}l_{\mathfrak p}}. Consequently the shell \boldsymbol l_p in R_p has measure

Its conditional additive measure is invariant under the transitive unit action. It is therefore the image of unit Haar probability. By Equation (3.1), P_p(\boldsymbol l_p) is exactly the fraction of this shell lying in \OO_p. The elements with a zero component have measure zero, and m_p(\OO_p)=q_p^{-1}. Summing Equation (3.9) over the intersections with \OO_p gives

Multiplication over p\mid q proves Equation (3.6).

For the moment, choose x with normalized additive Haar probability on \OO_p, and let v_{\mathfrak p}=v_{\mathfrak p}(x_{\mathfrak p}). The projection H_{\mathfrak p} of \OO_p into R_{\mathfrak p} contains the embedded copy of \Z_p, because 1\in\OO_p. Its induced probability is normalized Haar on H_{\mathfrak p}. For an integer k\ge1, the image of this copy of \Z_p modulo the ideal of valuation k has p^{\lceil k/e_{\mathfrak p}\rceil} elements. Hence the image of H_{\mathfrak p} has at least that many elements, and

This uses only the size of the additive image of the projection; it does not require the projections to be independent.

Set

If j\ge a\ge1, then some v_{\mathfrak p} is at least e_{\mathfrak p}(a-1)+1. Equation (3.10) and a union bound over at most four components give \P_{\OO_p}\{j\ge a\}\le4p^{-a}. On the other hand,

These assertions hold outside the null sets where a component vanishes. The tail formula for the expectation of p^{j/2} now yields

Applying Equation (3.9) once more gives the exact factorization

Each factor is at most (1-p^{-1})^{-4}(1+4p^{-1/2}). For any fixed \eta>0, this is at most p^\eta at all sufficiently large primes. The product of the factors at the finitely many remaining primes is bounded by a constant depending only on \eta. Since \prod_{p\mid q}p\le q, this proves Equation (3.7). Finally, take \eta=1/128 there and use q\le X:

This proves Equation (3.8).□

Proof. We first treat large D, and put L=\log D. Stark’s zero theorem [25, Lemma 3] gives at most one nontrivial zero of \zeta_K in

any such zero is real and simple. His Lemma 8 [25, Lemma 8], valid also for nonnormal fields, states in degree four that a real zero \beta\ge1-(96L)^{-1} would force a quadratic subfield of K. Primitivity excludes that subfield. It follows that every nontrivial zero \rho satisfies

Indeed, a zero in the smaller open disk would lie in Stark’s rectangle, would be real, and would be in the forbidden real interval.

For real s>1, the logarithmic derivative of the completed zeta function gives

see [25, Equation (9)]. The real zero sum converges and has nonnegative terms, since \Re\rho\le1. At s_0=1+L^{-1}, the Euler product gives \zeta_K'/\zeta_K(s_0)\le0. The function G is bounded on [1,2], so for sufficiently large D,

This estimate persists for all 1<s\le1+4/L. To see this term by term, write \rho=\beta+i\gamma. Since \beta\le1,

Equation (3.15) consequently gives

Using \Re(s-\rho)^{-1}=(s-\beta)/|s-\rho|^2, we obtain

Thus S_K(s)\ll L on the entire interval.

Let \mathcal Z_K(s)=(s-1)\zeta_K(s). Rearranging Equation (3.16) shows that

The function \mathcal Z_K extends to s=1 with value \kappa>0. Integrating its logarithmic derivative across an interval of length at most 4/L proves Equation (3.12), with an absolute constant. At s_0, the inequality \zeta_K(s_0)\ge1 then gives \kappa\gg L^{-1}.

There are only finitely many fields with D<D_0, by Hermite’s finiteness theorem [16, Theorem 8.43]. For each of them, h\zeta_K(1+h) tends to the positive residue \kappa as h\downarrow0. Taking a common sufficiently small h_0 over this finite set proves the remaining comparison, and their positive residues have a common positive lower bound. Enlarging C proves Equation (3.13) for every field.

Choose Y_0 sufficiently large that 1/\log Y_0\le h_0, and let Y\ge\max\{Y_0,D^{1/4}\}. For h=1/\log Y, the comparison just proved applies: in the large D case h\le4/\log D, and in the bounded D case h\le h_0. Positivity in the Euler logarithm gives

Moreover, a_K(p)\le4, 1-p^{-h}\le h\log p, and the elementary Chebyshev bound followed by partial summation imply

It follows that

which proves Equation (3.14). There is no upper restriction on Y. This observation is what permits D to stay bounded while the order index, and hence X, grows.□

Proof. Apply [19, Theorem 1.1] with no excluded residue classes and with its parameter \beta=1/4. Its two growth conditions hold with bounds independent of b:

The theorem’s constant and threshold depend only on these growth bounds and on the fixed parameter. Its strict length condition u^\beta<v is satisfied even at v=u^{1/2} once u>1. This gives Equation (3.17), including both endpoints.□

Proof. The range is empty for c>1, so suppose 0<c\le1. We adapt the separation of the local norm from the remaining ideal in [18, Proposition 4.2]. Put

This is nonnegative and multiplicative, and it is at most d_4. Every integral ideal splits uniquely into its part above the primes dividing q, described by one tuple \boldsymbol l, and a part coprime to qR. The left side of Equation (3.18) is therefore

First take d=d(\boldsymbol l)\le X^{1/8}, and put u=x/d, v=y/d. Then

The second inequality implies v\ge u^{1/2} when X is large in terms of c. The first inequality puts u in the range of Equation (3.14), once X is above an absolute threshold. To remove the primes dividing q from that Euler bound, observe that

For the first identity, expand \log Z_p=\sum_{\mathfrak p\mid p}\sum_{j\ge1}p^{-jf_{\mathfrak p}}/j. The terms of order p^{-1} count exactly the primes of residue degree one, and the remaining terms are O(p^{-2}), uniformly in the splitting type. Their sum over rational primes is bounded. For the second estimate, use a_K(p)\le4, at most \log(2q)/\log2 primes dividing q, and u\ge X^{5/8}, q\le X. Consequently Equations (3.14) and (3.20) give

Lemma 3.4 and Equation (3.6) now bound the contribution of these tuples to Equation (3.19) by

It remains to control tuples with d>X^{1/8}. Those with d>x contribute nothing. For z\ge1, summing the first three factors of the divisor convolution gives

Use this with z=x/d, and then Equation (3.8). The remaining contribution is at most

For fixed A, once X\ge A the condition x\le AX gives \log x\le2\log X. By Equation (3.13) and D\le X^2, the ratio of this last bound to \kappa x/q is at most

This bound is independent of the field and the order. Increase the threshold in terms of A,c until the ratio is at most c. The tail is then at most \kappa cx/q\le\kappa y/q. The constant in the resulting bound has not changed with A or c, proving the proposition.□

Proof. The count E_{f_r} dominates the indicator in Equation (3.22). Put T=r^4X and T_0=X^{3/4}. By Equations (3.5) and (3.21), it suffices to bound

The sum is empty when T<1. Otherwise its logarithmic factor is at most (\log X)^3, since \Nm\mathfrak a\ge1 and r<1. Apply Proposition 3.5 at x=y=T_0, with A=c=1. The part with \Nm\mathfrak a\le\min\{T,T_0\} is at most

After division by \sqrt D\,\kappa=X\kappa/q, this contributes O(X^{-1/4}(\log X)^3)=o(1).

If T>T_0, put

The same interval estimate at x=y=t gives A_{T_0}(t)\le C\kappa t/q, since T\le X. Writing each logarithmic cube as an integral and using Tonelli yields

The last integral is finite after the substitution t=Ts. This part of Equation (3.23) is therefore O(r^4). If T\le T_0, it is absent. The constants used at A=c=1 are absolute, so the resulting estimate proves Equation (3.22). This is the compactness argument of [18, Proposition 5.1], now using the quartic weighted interval bound.

Mahler’s compactness criterion [15, Section 3, Theorem 2] says that

is compact. Choose r so that Cr^4<\epsilon/2. Equation (3.22) then gives \mu_i(\mathcal K_r)\ge1-\epsilon for all sufficiently large i. Each of the finitely many preceding packet measures is supported on a finite union of compact orbits, by Proposition 2.2. Enlarging the compact set to include those supports proves tightness of the entire sequence. The space X_4 is locally compact, second countable, and metrizable; the weak compactness theorem for tight probabilities therefore gives probability subsequential limits. For a\in A_4 and f\in C_c(X_4), the equality

passes to a weak limit, since both test functions are continuous with compact support. This proves A_4-invariance.□

Proof. The assertion is immediate if V_0 is empty. Otherwise choose 0<\epsilon\le M such that \epsilon\le|v_i|\le M for every v\in V_0 and i. For r_0<\epsilon/(2b), each box in the statement lies in one coordinate sign component. Its first three logarithmic widths satisfy

The product map x\mapsto\prod_i|x_i| has uniformly bounded first derivatives on these boxes. If t_v=\prod_i|v_i|, it follows from Equation (3.4) that, for one C_1\ge1,

Let t_->0 and t_+<\infty be the minimum and maximum of t_v on V_0, and shrink r_0 so that 6C_1r_0<t_-. For a fixed r in this range set

The possible ideal norms in Equation (3.25) lie strictly inside (x-y,x). The enlargement avoids any issue with integer endpoints. There is a fixed A\ge1, depending only on V_0,b, such that x\le AX for all centers and radii under consideration. For sufficiently large X, uniformly in the centers, x\ge X^{3/4}. Also y\le x by the choice of r_0, and y\ge cx with c=4C_1r/A>0, after decreasing r_0 if necessary so that c\le1.

Proposition 3.5 applies for every fixed r, with this c and the fixed A. Its constant is independent of c, so the sum of the relevant ideal weights is at most C\kappa rX/q, with a constant independent of r. Combining this with Equations (3.5) and (3.25), and using \sqrt D=X/q, proves Equation (3.24). Only the threshold for X depends on the fixed radius.□

Proof. The assertion is immediate if \Omega is empty, so assume otherwise. We use the passage from vector boxes to ordinary neighborhoods in [18, Proposition 5.2]. First, every full lattice has a vector with no zero coordinate. If \Lambda=\Z^4g, each coordinate of (1,N,N^2,N^3)g is a nonzero polynomial in N: its coefficients are a column of the invertible matrix g. An integer N outside their finitely many roots gives the required vector.

For each x_0\in\Omega, choose a continuous local lift g(x) of nearby lattices to \mathrm{SL}_4(\R), and an integer row z for which zg(x_0) has no zero coordinate. On a sufficiently small neighborhood with compact closure in this chart, zg(x) stays bounded and its coordinates stay bounded away from zero. A finite collection of these neighborhoods covers \Omega. Consequently we may select, for every x\in\Omega, a vector v_x\in\Lambda_x so that all selected vectors lie in one compact set V_0\subset(\R^\times)^4. The selection is used pointwise and need not be measurable.

Choose M with |(v_x)_i|\le M for all selected vectors, and fix b>4M. If y=xh with h\in B(r), then v_xh\in\Lambda_y, and

Thus every lattice in xB(r) has a nonzero vector in v_x+[-br,br]^4. Lemma 3.7 gives, for each fixed small r>0 and all sufficiently large i,

The constants are common to all x and r in the stated ranges; the index after which this estimate holds may depend on r. The set xB(r) is open in X_4. For each fixed x,r, Portmanteau’s inequality therefore yields

This proves the bound for every r with the same constant. It uses no interchange between a shrinking radius and the packet limit.□

Proof. Use the entrywise neighborhoods B(r) defined in Equation (3.26). Fix 0<b<1/4 and put

For h\in T_t, the two conjugations give

In the determinant expansion, every term except the product of the diagonal entries contains at least two off-diagonal entries. Since \det h=1, Equation (4.1) implies \prod_i h_{ii}=1+O(r^2). The diagonal matrix

therefore belongs to a fixed compact subset \mathcal D_\infty\subset A_4, and d(h)^{-1}h\in B(Cr). The constant C here and below may depend on b and the fixed flow. Indeed, the first three diagonal entries of d(h)^{-1}h are one, the fourth differs from one by O(r^2), and all its off-diagonal entries are O(r).

The logarithmic coordinates on A_4 identify \mathcal D_\infty with a bounded subset of \R^3. A mesh of spacing comparable to r, followed by a fixed enlargement to account for matrix multiplication, gives

For a compact \Omega\subset X_4, the centers xd_j lie in the compact set \Omega\mathcal D_\infty. Domination and Equation (3.27) now imply

for all sufficiently large t. Apply Theorem 4.1 to the constant sequence \xi_i=\xi, with t_i=i, B=B(b), and \eta=\lambda_a/3. The identity B(b)^{(-t,t)}=T_t and 2\lambda_a/3<\lambda_a verify its estimate for large i. For the finitely many earlier i, enlarge C_{\Omega,c}, using that \xi is a probability. The criterion gives the assertion.□

Proof. The group A_4\simeq\R^3 is locally compact and second countable, and acts continuously on the standard Borel space X_4. The ergodic decomposition theorem [7, Theorem 5.2], specialized to invariant probabilities, gives a measurable decomposition

into A_4-invariant, A_4-ergodic probabilities over a factor fixed by A_4. Put T(x)=xa(1).

We record why a restriction to the zero-entropy components is measurable. Choose increasing finite Borel partitions \mathcal P_k generating the Borel \sigma-algebra of X_4. For any T-invariant probability \nu, the Kolmogorov–Sinai approximation theorem and subadditivity give

where H_\nu is the Shannon entropy of a finite partition. Each finite-partition entropy is a Borel function of \nu, so z\mapsto h_{\nu_z}(T) is measurable. For each probability under consideration, complete the standard Borel probability space; it is then a normalized Lebesgue space. The entropy integral formula for an invariant component factor [21, Theorem 9.8] gives

and the same formula after restricting and normalizing \tau on any measurable subset of positive measure. This formula applies because the component factor is fixed by T. It does not require the individual \nu_z to be T-ergodic.

If Z_0=\{z:h_{\nu_z}(T)=0\} had measure w>0, then

would be an A_4-invariant probability with \xi\le w^{-1}\mu. The restricted form of Equation (4.4) would give h_\xi(T)=0, contrary to Proposition 4.2.□

Proof. Let \mathfrak j be the Lie algebra of J. It contains the trace-zero diagonal algebra and is invariant under diagonal conjugation. The other weight spaces in \mathfrak{sl}_4(\R) are the distinct one-dimensional spaces \R E_{ij}, i\ne j, where E_{ij} is a matrix unit. An invariant subspace is a sum of these weight spaces, so \mathfrak j consists of the diagonal algebra and a selection of the root spaces.

The group J is unimodular. One way to see this standard fact is to use the quotient criterion for Haar measures: existence of an invariant quotient measure and triviality of the modular character of the discrete group \Gamma_\Lambda imply that the modular character of J is trivial on \Gamma_\Lambda. It therefore descends to \Gamma_\Lambda\backslash J. Its pushforward of invariant probability would be invariant under translations by its image. A nontrivial positive dilation admits no invariant probability on \R_{>0}, as its disjoint dilates of a fundamental interval all have the same mass. Hence the modular character is trivial on J.

Draw a directed edge i\to j when E_{ij}\in\mathfrak j. The trace of the adjoint action of \diag(u_1,\ldots,u_4), with \sum_i u_i=0, on \mathfrak j is the sum of the selected roots u_i-u_j. Unimodularity makes this sum zero. Its coefficient at u_i is the outgoing degree of i minus its incoming degree. These coefficients sum to zero, and a linear form vanishing on \sum_i u_i=0 has all coefficients equal. Thus every vertex has equal outgoing and incoming degrees.

Every edge now lies within a strongly connected component. Otherwise the finite directed acyclic graph of components would have a source with an outgoing edge; summing degree balance over that source would give a positive outgoing degree and zero incoming degree. For distinct indices, the bracket identity [E_{ij},E_{jk}]=E_{ik} makes the edge relation transitive. Each strongly connected component is consequently a complete directed graph. These components form a partition \mathcal I, and \mathfrak j is the Lie algebra of J_{\mathcal I}. Connected Lie subgroups with the same Lie algebra agree, giving J=J_{\mathcal I}.

We use \Gamma_\Lambda to recover a rational structure on the block scalars. Its elements preserve \Lambda, hence act rationally on W_\Q. Let

In a lattice basis the commutation equations have integer coefficients. A finite subcollection of these linear equations defines the same solution space, by finite dimensionality. It follows that \mathcal C_\Q\otimes_\Q\R is the real commutant of \Gamma_\Lambda.

We claim that this is also the real commutant of J. Let M be a real endomorphism commuting with \Gamma_\Lambda. The continuous map

is well-defined. Its pushforward of invariant probability is invariant under conjugation by A_4. Each off-diagonal coordinate is multiplied by a nontrivial positive dilation under some element of A_4. A probability on \R invariant under such a dilation is supported at zero: the annuli \{x:r^k\le |x|<r^{k+1}\}, k\in\Z, for a fixed r>1, are disjoint dilates covering \R\setminus\{0\}. Thus every off-diagonal coordinate of j^{-1}Mj is zero almost everywhere. Invariant probability on the transitive space \Gamma_\Lambda\backslash J has full support; continuity makes j^{-1}Mj diagonal for every j\in J. In particular M is diagonal. Conjugating by 1+tE_{ij} for i,j in the same block forces its i-th and j-th diagonal entries to agree. Therefore M is scalar on each block, which is exactly the condition that it commute with J. The reverse inclusion is immediate, proving the claim.

It follows that \mathcal C_\Q\otimes_\Q\R\simeq\R^s. In particular, \mathcal C_\Q is commutative and has no nonzero nilpotents. The structure theorem for finite reduced commutative algebras over \Q expresses it as a product of number fields, all totally real because its real scalar extension has no complex factor.

This product has only one factor. If e\in\mathcal C_\Q were an idempotent other than 0 and 1, then W=e(W_\R) would be a nonempty proper union of coordinate blocks, rational relative to \Lambda. Every \gamma\in\Gamma_\Lambda preserves the full lattice \Lambda\cap W in W, so \lvert\det(\gamma|_W)\rvert=1. The continuous character

is therefore trivial on \Gamma_\Lambda. It is surjective: if k=\dim W, a diagonal matrix may scale the coordinates in W by e^{(4-k)t} and those in its complement by e^{-kt}. Its total determinant is one and its determinant on W is e^{k(4-k)t}. The character \chi would consequently push the invariant probability on \Gamma_\Lambda\backslash J to a probability on \R_{>0} invariant under all multiplicative translations. The disjoint-interval argument above rules this out. Thus \mathcal C_\Q is a field, which we denote by E.

Since E\otimes_\Q\R\simeq\R^s, its degree is s. The space W_\Q is a vector space over E, and

Each summand has dimension \dim_E W_\Q=4/s. These summands are the eigenspaces of the block-scalar algebra, namely the coordinate blocks. The blocks therefore have equal size.□

Proof. Theorem 4.4, in the right-quotient form explained above, represents \nu as invariant probability on \Lambda J for a connected closed subgroup J containing A_4. By Lemma 4.5 and the positive-entropy assumption, either J=\mathrm{SL}_4(\R), which gives \nu=m_4, or J has two coordinate blocks of size two.

In the second case write the blocks as \{i,j\} and \{k,l\}, and put b=ij|kl. The commutant field E in Lemma 4.5 is quadratic. Choose a nonzero \alpha\in E with \Tr_{E/\Q}(\alpha)=0. Its two scalar values on the blocks are c and -c, with c\ne0. The rational endomorphism \alpha of W_\Q=\Q\Lambda induces a rational endomorphism

On the exterior square of the first block it acts by 2c, and on that of the second by -2c. It vanishes on the cross-block summand. Its real image is therefore exactly

Because D_\alpha is rational, V_b is a rational subspace relative to \bigwedge^2 W_\Q.

It follows that V_b\cap\bigwedge^2\Lambda is a rank-two lattice in V_b. It contains a vector off both coordinate axes: take two independent lattice vectors; if neither is already off the axes, their sum is. For this vector Q(w)\ne0, and Equation (4.7) makes its value a nonzero integer. Every element of J preserves each of the two exterior lines, multiplying its coefficient by the nonzero determinant of the corresponding block. The translated vector therefore has both coordinates nonzero at every lattice of \Lambda J. This proves \nu(\mathcal E_b)=1.□

Proof. An element g\in G permutes the rows of A_{\OO}, so it multiplies \delta_{\OO} by the sign of that permutation. Permuting the four indices in the three signed products in Equation (5.3) permutes the partitions and introduces the same common sign. Division by \delta_{\OO} gives precisely the equivariance in Equation (5.1). This is an identity of quadratic polynomials in the rational coordinates of w, and therefore defines a rational quadratic map to F.

The Pfaffian identity for the exterior square of a four-dimensional linear map is

It follows either by expanding the six minors or by evaluating \frac12 t\wedge t in the top exterior power. This proves Equation (5.4). If a coordinate of a rational w vanishes, Galois transitivity on the unordered pairs makes all six coordinates vanish. The exterior embedding map is injective, so w=0. Finally, multiplication by a multiplies w_{ij} by a_i a_j after splitting, where a_i are its four embeddings. Each opposite product is multiplied by \prod_i a_i=\Nm(a), and the scalar c contributes c^2. This proves Equation (5.5) after splitting and hence over k.□

Proof. Over the splitting field set

The three equal opposite products give r_{ij}=c_0a_i a_j for every pair. If two such factorizations are divided, all products of two of the four coordinatewise quotients are equal. Comparing pairs that share one index shows that the four quotients are equal. Hence the projective tuple [a_1:a_2:a_3:a_4] is uniquely determined.

The ratios r_{ij} are Galois-equivariant without alternating signs. Uniqueness makes the projective tuple equivariant for the permutation action on the four embeddings. Express it in a k-basis of K_k. The resulting projective coefficient vector is fixed by the usual Galois action. Dividing by one nonzero coefficient makes every coefficient fixed, hence in k. It gives a\in K_k^\times, since all four embedding components are nonzero. The remaining coefficient c in r_{ij}=c a_i a_j is Galois-fixed as well. This proves existence over k, including when K_k is a product of nonsplit fields. The projective uniqueness proves the stated ambiguity. Multiplying two opposite equalities gives c^2\Nm(a)=s.□

Proof. Let \Phi_p(x,u)=u\Psi(x). If \Phi_p(x,u)\in F_p^\times, all six split coordinates of x are nonzero. After splitting, the derivative in x is the derivative of three products on three disjoint two-dimensional planes. It has rank three, and therefore has rank three over \Q_p. The map is a submersion there. The remaining input set is a finite union of proper algebraic zero loci and has additive measure zero. Local change of variables gives an absolutely continuous pushforward on the submersion locus, hence a density for the full law. To obtain a locally constant version at a fixed z\in F_p^\times, cover its inverse image in the compact set L_{V,p}\times\Z_p^\times by finitely many submersion charts, refined to disjoint compact open pieces. The compact complement has image avoiding a sufficiently small neighborhood of z. On that neighborhood each chart contributes a locally constant density. This also treats an empty inverse image.

Give V_p its additive measure for which L_{V,p} has mass one, before restricting to that lattice. We first compute the fiber form for the product input measure after disintegration against dz, and then the mass of \mathscr U_p for that form. In a \Q_p submersion chart, the relative absolute four-form is obtained by evaluating the input top form on four fiber basis vectors and three lifts of a target basis, then dividing by the dz-form on that target basis. We may compute this identity after extension to a splitting field, still evaluating on \Q_p tangent vectors and using the fixed extension of |p|_p=p^{-1}. We use the same interpretation for the scalar quotient below.

Over a splitting field, the exterior embedding matrix has determinant \delta_{\OO}^{\,3}. Consequently its measure form in the six embedding coordinates is

Likewise dz is the absolute embedding-coordinate form \left|\bigwedge_{b\in\mathcal B}dz_b\right|. To verify the latter normalization, evaluate this form on an R_{F,p}-basis. The square of its embedding determinant is the discriminant, so its absolute value is |D_F|_p^{1/2}, as in Equation (5.15).

After splitting, compute the relative form by retaining x_{12},x_{13},x_{14},u and substituting the three product equations for x_{34},x_{24},x_{23}. The three change-of-variable factors are |\delta_{\OO}/(u x_{12})|_p, |\delta_{\OO}/(u x_{13})|_p, and |\delta_{\OO}/(u x_{14})|_p. They cancel |\delta_{\OO}|_p^{-3}, since u is a unit. Haar probability on \Z_p^\times is (1-p^{-1})^{-1}|du/u|. The resulting fiber form is

The fiber action preserves this measure. Indeed its determinant on V_p is c^6\Nm(a)^3=(c^2\Nm(a))^3, of absolute value one, and it multiplies u by a unit while fixing z.

It remains to compute the mass of \mathscr U_p in Equation (5.18). Put \ell_i=da_i/a_i and s=dc/c over the splitting field. The pullbacks of its four logarithmic differentials under Equation (5.14) are

The vertical scalar direction d has vector (1,1,1,1,-2). Taking the first \ell_i as its coordinate and using the slice \ell_1=0, the matrix on the four remaining differentials is

The vertical coordinate has coefficient 1. Equivalently, the wedge of the four displayed pullbacks is the contraction of (\bigwedge_i\ell_i)\wedge s by the vertical vector, up to sign. Evaluating this identity on four \Q_p tangent lifts of a quotient basis gives the same absolute quotient form for nonsplit K_p. On a patch where a \Q_p-linear coefficient \lambda(a) is nonzero, normalizing it to its value at a reference point gives a genuine \Q_p section; after shrinking a unit-group patch, the normalizing scalar is a unit. Since \lambda(da)=d\lambda(a), the vertical logarithmic derivative is 1. Thus the contraction is the quotient of the parameter form by |dd/d|, with quotient Jacobian 1, including at p=2.

For an R_p-basis, the multiplicative embedding form on K_p^\times is the additive basis form multiplied by its embedding determinant and divided by \Nm(a). On R_p^\times the latter is a unit. The mass of this unit set is consequently

The c-unit mass and the scalar-kernel d-unit mass are both 1-p^{-1}, and cancel when taking the quotient. Haar integration on the compact unit groups assembles the local sections. Including the factor (1-p^{-1})^{-1} in Equation (5.18), the mass of each \mathscr U_p-coset is

This calculation uses the actual image of the unit group, so it requires no surjectivity assertion about the norm on units.

The accepted part of the fiber is the inverse image of z in the compact set L_{V,p}\times\Z_p^\times. Since \mathscr U_p is open, this compact set meets only finitely many of its cosets. Integrating its indicator in the fiber multiplies the mass in Equation (5.19) by \sum_{\mathfrak c}\alpha_p(\mathfrak c;z). The result is the density f_p^*(z). Solving for the sum proves Equation (5.17).□

Proof. For two such pairs, Lemma 5.3 over \Q gives w'=c(\bigwedge^2a)w with a\in K^\times, c\in\Q^\times, since the common opposite ratio is \Nm(I')/\Nm(I). Relative to the local generators \xi_p,\xi'_p, their local comparison is represented by (a\xi_p/\xi'_p,c). Equality of the local cosets means that some d_p\in\Q_p^\times satisfies

In particular v_p(c)=2v_p(d_p) is even at every finite prime. Therefore c=\epsilon s^2 with s\in\Q^\times and \epsilon\in\{1,-1\}. Absorbing s into a makes a\xi_p/\xi'_p a unit for every p. Hence I'=aI as fractional maximal-order ideals. The chosen representatives of the ordinary ideal classes then give I'=I and a\in R^\times=U.

There are at most [U:U^+]\le16 possibilities for the unit action modulo U^+, distinguished by the four real signs, and two possibilities for \epsilon. This proves the bound 32.□

Proof. An accepted coset has a representative x\in L_{V,p}=\bigwedge^2 R_p. Use the extended valuation with v_p(p)=1 on a splitting field. Its six valuations k_{ij}=v_p(x_{ij}) are nonnegative integers, invariant under Frobenius. Since u and \delta_{\OO} are units,

This proves integrality of z. If the pair labels over \mathfrak q have two Frobenius orbits, one of their two valuations can be chosen from 0,\ldots,j_{\mathfrak q}. If there is one orbit, the two valuations are equal, which requires even j_{\mathfrak q}. The number of records is therefore at most b_p(z).

Two representatives with the same six valuations have comparison (a,c) satisfying

Subtraction makes all four v_p(a_i) equal to a common r, and v_p(c)=-2r. Because K_p is unramified, r\in\Z. Replacing a by p^{-r}a and c by p^{2r}c puts both in the unit subgroup. Thus a record specifies at most one coset, also for nonsplit unramified factors. The subgroup \mathscr U_p preserves \bigwedge^2R_p, so its coset fraction is either 0 or 1. This proves Equation (5.21). The bounds at depth zero and at general depth follow immediately from the definition and the at most three factors.

The trace from an unramified integer ring to \Z_p is onto, as is seen after reduction from the nonzero finite-field trace. If F_p has more than one factor, the trace-m coset therefore projects uniformly onto each chosen factor: the other factors can supply any required trace. More generally, for integers r_{\mathfrak q}\ge0, the probability that v_{\mathfrak q}(z)\ge r_{\mathfrak q} for all \mathfrak q is

The first line follows by projecting onto the constrained factors while a factor with r_{\mathfrak q}=0 supplies the trace. The second uses p\nmid m. It also covers the case that F_p is a field.

Expanding b_{\mathfrak q}(j)=\sum_{r=0}^j\epsilon_{\mathfrak q}^r and summing the absolutely convergent geometric series gives

The five Frobenius types give the following data. A pair (f,\epsilon) in the second column is attached to one prime of F.

In every row \prod_{\mathfrak q}\epsilon_{\mathfrak q}=1 and

Substitution in Equation (5.24) proves the exact mean. Its linear term is the number of degree-one primes of K minus one, which is a_K(p)-1; the five fixed rows give the uniform error term.

If F_p is a field, n_p\ge1 would make its trace divisible by p, and is impossible. Otherwise the uniform projection above gives \P_p(v_{\mathfrak q}(z)\ge k)=p^{-f_{\mathfrak q}k}. Taking the union over at most three factors proves Equation (5.23). Finally, the trace map is onto, so the affine trace lattice has two integral coordinates. Projection to each unramified factor modulo p^{k+1} is determined by these coordinates modulo p^{k+1}. The valuations truncated at k+1, and hence the asserted threshold and finite-depth functions, have the stated congruence dependence. This remains true on the component-vanishing set with the convention above.□

Proof. The field F and the condition \Tr z=m\ne0 imply z\in F^\times. At every prime the set of accepted cosets is finite by Lemma 5.4. At p\notin S, Lemma 5.6 proves the integrality requirement and bounds the local sum by b_p(z). The set

is finite, since a nonzero global z is a unit at all but finitely many primes. At p\notin E_z, b_p=1, and each coset fraction is 0 or 1, so there is at most one accepted coset, of fraction 1. If some prime has no accepted coset, then M(z)=0 and the claimed inequality is immediate. Otherwise the accepted coset is unique with fraction 1 outside E_z, and each prime inside the finite set E_z has finitely many accepted cosets. Thus there are only finitely many assignments of accepted cosets, and all local sums outside E_z equal 1.

Choose local references. Equation (5.12) identifies the weight of a global pair at each prime with its coset fraction \alpha_p. By Lemma 5.5, at most 32 global pairs occupy any assignment of accepted cosets. Summing over the finite set of assignments, including ones not realized globally, bounds M(z) by 32 times the product of the local sums of \alpha_p. Use Equation (5.17) at p\in S and Equation (5.21) outside S. This proves Equation (5.25) and finiteness.□

Proof. Galois permutes the planes V_b, including possible alternating signs on their coordinate axes, and permutes the scalars \alpha_b in the same way. The plane-scalar action therefore commutes with Galois and descends. After splitting there are three planes of dimension two, proving the rank assertion. Equation (5.4) shows that the matrix of B_V pairs the three opposite pairs by unit off-diagonal entries, with the displayed signs, so it is unimodular. On each plane, multiplying both coordinates by \alpha_b is self-adjoint for its split pairing. The contribution of that plane to \frac12B_V(\rho(\alpha)x,x) is \alpha_b\Psi_b(x), which proves Equation (5.26). Polarization gives the next identity. The determinant and the last quadratic identity follow by multiplying, respectively, the two equal eigenvalues and the two coordinate scalars on each plane. All these are rational identities, hence persist under base change.□

Proof. Let E=\bigwedge^2 A_{\OO} be the exterior embedding matrix. Its entries are integral in every local splitting ring, and \det E=\delta_{\OO}^{\,3}. Since d=\delta_{\OO}^{\,2},

also has integral entries. For \alpha\in R_{F,p}, its plane-scalar matrix over the splitting ring is integral. Consequently d^2\rho(\alpha)=d^2E^{-1}\diag(\alpha_b,\alpha_b)E is integral over the splitting ring. Its entries are in \Q_p, so they lie in \Z_p. This proves d^2R_F\subset C_F and [R_F:C_F]\le d^6.

If p\nmid D, the etale algebra R_p splits over an unramified extension of \Z_p; its splitting field and cubic resolvent are unramified. Hence every prime dividing D_F divides D. For p>3, a cubic algebra is tamely ramified and its discriminant exponent is at most 2. At 2 and 3, the finitely many local extensions of degree at most three have bounded discriminant exponents [16, Proposition 7.64 and Remark 7.65]. It follows that D_F\ll D^2. The discriminant-index formula now gives

The containment d^2R_F\subset C_F also proves equality of the local orders away from Dq.

For the reverse inequality, we first explain why a projection denominator controls \Disc(\OO). Write \operatorname{tr} for matrix trace, put \mathscr E=\End_\Z(\OO), and let \mathscr W_K\subset\End_\Q(K) be the rational subspace of multiplication matrices. Its intersection \mathscr L=\mathscr W_K\cap\mathscr E is saturated. It is multiplication by \OO: if \alpha\OO\subset\OO, then \alpha=\alpha\cdot1\in\OO, and the converse follows because \OO is a ring. The pairing (T,S)\mapsto\operatorname{tr}(TS) is unimodular on \mathscr E, since its matrix units are dual. Its restriction to \mathscr W_K is the nondegenerate field trace pairing. Let \mathcal E_K be the orthogonal projection onto \mathscr W_K.

For the restricted pairing one has

One inclusion follows by pairing a projected integral matrix with \mathscr L. Conversely, an integer functional on the primitive sublattice \mathscr L extends to one on \mathscr E. Unimodularity represents the extension by an element of \mathscr E; its orthogonal projection represents the original functional on \mathscr L. Nondegeneracy identifies that projection with the prescribed element of \mathscr L^\vee. Consequently, if a positive integer N satisfies N\mathcal E_K(\mathscr E)\subset\mathscr E, then

It remains to produce the denominator N=2\Delta_F.

Choose trace-dual bases a_i of C_F and b_i of C_F^\vee, where

The adjugate of the trace Gram matrix gives \Delta_F b_i\in C_F. Define on \End_\Q(V)

For two embeddings \sigma,\tau of F, the dual-basis identity is

Thus, after splitting, \mathcal E_F extracts exactly the three diagonal blocks \End(V_b). It is the orthogonal projection for the matrix trace pairing, and

This denominator conclusion uses the trace-dual lattice only; it does not require that lattice to be an invertible C_F-ideal.

We transfer this projection to matrices on K explicitly. Let

There is a natural contraction \mathfrak c:\End_\Q(\bigwedge^2K)\to\End_\Q(K). For clarity, in a basis e_1,\ldots,e_4 write

and extend these coefficients antisymmetrically in i,k and in j,l, with zero coefficients when either pair repeats. Then

This is the contraction of \bigwedge^2K\otimes\bigwedge^2K^\vee in one covariant and one contravariant index, so it is independent of the basis. Both j and \mathfrak c are integral in the \OO and exterior \OO-bases. Expanding their coefficients gives

Indeed an off-diagonal coefficient of T occurs for the two remaining indices k; a diagonal coefficient gives \sum_{k\ne i}(T^i_i+T^k_k)=2T^i_i+\operatorname{tr}(T).

In embedding coordinates, an off-diagonal entry of j(T) changes one index of a two-element subset. Such subsets belong to distinct opposite-pair planes. It follows that \mathcal E_F(j(T))=j(\diag T) over the splitting field. Equation (5.34) now proves the formula

for the projection defined above: its right side is \diag T over the splitting field. A Galois-equivariant diagonal tuple is the tuple of embeddings of an element of K, so these diagonal matrices are precisely \mathscr W_K. Diagonal extraction is orthogonal for matrix trace. Finally, Equation (5.33) and integrality of j and \mathfrak c give

Taking N=2\Delta_F in the preceding lattice bound gives

This is the lower inequality in Equation (5.30).□

Proof. Since C_F is an order, C_F\cap\Q=\Z, so 1 is primitive in its lattice. A basis 1,a_2,a_3 of C_F gives the basis 1,2a_2,2a_3 of C_F^\sharp. This proves the index and discriminant assertions in Equation (5.36). The functional taking 1 to 1 and the other two basis vectors to 0 is represented in the trace dual. Hence trace from that dual is onto \Z. Lemma 5.9, and 2\in S, give C_{F,p}^\sharp=R_{F,p} outside S.

The two integrality identities preceding the lemma show that the law in Equation (5.16) is supported on (C_{F,p}^\sharp)^\vee. For a contributing global pair, use its local representation z=u\Psi(x) and Equation (5.14) to translate x into L_{V,p}, adjusting u by the inverse norm of that unit. This proves local trace-dual membership at every prime. Intersecting the local lattices proves the stated global support. For the second support assertion, the density factors give this membership at p\in S. At p\notin S, the order is R_{F,p}, whose trace pairing is unimodular because F_p is unramified; thus R_{F,p}^\vee=R_{F,p}, and the assumed integrality gives membership there as well.

The adjugate of the trace Gram matrix of C_F^\sharp gives \Delta_F^\sharp(C_F^\sharp)^\vee\subset C_F^\sharp\subset R_F. If z\in\Omega, then \Tr z=m\ne0, so z\ne0; since F is a field, all its embeddings are nonzero. The norm \Nm(\Delta_F^\sharp z) is consequently a nonzero integer. The boundedness of \Omega and \Delta_F^\sharp\ll X^{28} give

At p\notin S, z is integral and the local algebra is unramified. Hence

The product of these inequalities gives the depth bound because \Nm(\Delta_F^\sharp z) is an integer. Finally, if the coordinates of \Omega have absolute value at most B, with B\ge1, then for each b

This proves Equation (5.38).□

Proof. Define an E-valued bilinear form b_E by

Self-adjointness of the action makes b_E symmetric and E-bilinear. It is nondegenerate because B is nondegenerate. Equation (6.2) then gives \Psi_E(x)=b_E(x,x)/2 and d\Psi_{E,x}(h)=b_E(x,h). On the \nu-th field factor, a nonzero component of \Psi_E(x) implies that the component x_\nu of x is nonzero. The nondegenerate form on the rank-two E_\nu-space makes h_\nu\mapsto b_E(x_\nu,h_\nu) a nonzero, hence onto, E_\nu-linear functional. It follows that \Psi_E is a \Q_p-submersion at every preimage of an output with nonzero components. Multiplication by a unit u is invertible on E, so the map (x,u)\mapsto u\Psi_E(x) is a submersion at the same points.

For each \nu the polynomial \Psi_{E,\nu} is nonzero: a nondegenerate symmetric form in characteristic zero has a nonzero quadratic form. Choose a nonzero \Q_p-coordinate polynomial of this component. Its zero set in the open lattice L has Haar measure zero, by the usual one-variable root bound followed by induction and Fubini. Off the union of these null sets, the local submersion theorem gives absolute continuity of the pushforward. For a fixed output with nonzero components, its preimage in the compact set L\times\Z_p^\times is compact. Cover this preimage by finitely many compact-open submersion charts, refined to disjoint compact-open pieces within those charts. The image of the remaining compact set avoids a sufficiently small neighborhood of the output. After shrinking that neighborhood, integration in the fiber coordinates of the pieces gives a finite constant density on it. This proves the asserted pointwise version, including value zero when the neighborhood misses the image.

Finally, for c\in\mathcal C_L and r\in\Z_p,

Multiplication by u\in\Z_p^\times preserves this inclusion. The image of the compact domain is therefore contained in the displayed trace dual, which proves the support and trace assertions.□

Proof. For fixed u\in\Z_p^\times and v_p(a)=-k<0, integration in the second coordinate gives

For a\in\Z_p the integral is 1. The sign of a has no effect, so the three independent pairs in Equation (6.1) give Equation (6.6), even before averaging u. This calculation uses the integral split expression and does not divide by 2 in \Z_p.

The scalar coefficient is the Fourier coefficient of the trace law, since \Tr(a z)=a\Tr(z). That law is supported on \Z_p by Lemma 6.1. The character group of \Z_p is \Q_p/\Z_p, and it has p^k-p^{k-1} characters of exact denominator p^k. Hence its nonconstant Fourier coefficients have total absolute sum

Fourier inversion on the compact group \Z_p is therefore uniformly convergent and gives the continuous density and Equation (6.7). The lower bound is positive at every prime, and \sum_p1/(p(p+1))<\infty. Taking logarithms of the upper and lower bounds proves the last assertion.□

Proof. An integral scalar shift leaves every pole valuation -h_\nu<0 unchanged. For an integral component and p>5, the condition that A_\nu+s fail to be a unit forbids at most one residue s\in\mathbf F_p: it forbids none if the residue of -A_\nu does not belong to the scalar subfield. There are at most three components, and 0,1,2,3 are distinct modulo p. Some choice therefore makes all integral components units.

For p\le5, normalize the extension of v_p by v_p(p)=1, so that on E_\nu it is v_\nu/e_\nu. Distinct elements s,s' of the list satisfy v_p(s-s')\le1. The ultrametric inequality shows that for a fixed integral component at most one candidate can satisfy v_p(A_\nu+s)>1. Again there are at most three components, so a candidate remains for which all their valuations are at most 1. Thus A_\nu+s is nonzero in every integral component.

The bounds on \beta follow componentwise. In the norm, the possible loss from the integral components is at most p^{-\iota_p\sum_\nu e_\nu f_\nu}=p^{-3\iota_p}, proving Equation (6.9). The group \mathfrak h_p/(\mathfrak h_p\cap\mathcal C_L) has order p^{g_p} and is killed by p^{g_p}. Thus

Multiplying the containment for \beta by p^{g_p+\iota_p} proves the claim about j_p. Finally v_\nu(i_\nu)=h_\nu. The components of i\alpha consequently have valuation zero on the pole factors and valuation between zero and e_\nu\iota_p on the others. This proves the assertions about x and x^{-1}.□

Proof. We use Weil’s Fourier formula for a quadratic character [27, Chapter I, §14, Theorem 2, p. 161]. We record its normalization for the transform

Completing the square gives

Let B_N=p^{-N}L_0 and c_N=\int_{B_N}\psi_p(q_P(x))\,dx. On any fixed compact set of y, translation by P^{-1}y preserves B_N for all sufficiently large N. The transform of \mathbf1_{B_N}\psi_p(q_P) on that set is therefore c_N\psi_p(-q_{P^{-1}}). If this identity is paired with a Schwartz–Bruhat function, its left side stabilizes with N, since the Fourier transform of the test function has compact support. Choosing a test function whose integral against \psi_p(-q_{P^{-1}}) is nonzero shows that c_N stabilizes as well. This proves the distributional shape of the formula.

Differencing the absolute square of c_N and integrating in one variable gives

For sufficiently large N, the intersection is P^{-1}p^NL_0 and q_P is integral on it. The right side then equals \vol(p^{-N}L_0)\vol(P^{-1}p^NL_0)=|\det P|_p^{-1}. Hence the stabilized constant is \omega(P)|\det P|_p^{-1/2} with unit phase. The transform of \mathbf1_{L_0} is \mathbf1_{L_0}, so pairing the distribution identity with this function gives Equation (6.13).

Under q_{T^*PT}=q_P\circ T, changing variables multiplies the Gaussian constant by |\det T|_p^{-1} and its normalizing determinant square root by |\det T|_p. Their product, the phase, is unchanged; this is also the isometry invariance in Weil [27, Chapter II, §25, p. 173]. For P=u\rho(\alpha) its inverse is u^{-1}\rho(\beta). Rank two over E gives \det\rho(\alpha)=\Nm_{E/\Q_p}(\alpha)^2, and |u|_p=1. This proves Equation (6.14). Taking T=\rho(r) proves the last phase identity. The polar operator of q_P is P, because its polarization is B_0(Px,y). In particular this proof never assumes that 2 is a unit of \Z_p.□

Proof. For c\in\mathcal C_L^\sharp, Equation (6.2) shows that Q_c(L)\subset\Z_p. Consequently \gamma_p(A+c)=\gamma_p(A). Each denominator shell is invariant under translation by R_E: adding an integral element preserves every negative component valuation and preserves the set of integral components. Since \mathcal C_L^\sharp\subset R_E, the zero-extended shell restriction retains this period. Its inverse is therefore supported in the annihilator (\mathcal C_L^\sharp)^\vee, whose elements have integral trace because 1\in\mathcal C_L^\sharp. The order \mathcal C_L^\sharp is an open lattice, and the shell is compact open. Thus the zero extension in Equation (6.15) is Schwartz–Bruhat, as is its Fourier inverse.

For a fixed unit u put I_\beta(u)=\int_L\psi_p(-u^{-1}Q_\beta(x))\,dx. Polarization and additive orthogonality on the self-dual lattice give

Indeed, in the difference Q_\beta(y+h)-Q_\beta(y) the term linear in y is B(\rho(\beta)h,y), whose integral is the indicator of \rho(\beta)h\in L. The absolute value of the last display is at most the relative mass of L\cap\rho(\beta)^{-1}L. If p^{a_1},\ldots,p^{a_6} are the Smith elementary divisors of \rho(\beta) on L, then

The equality for j_p is exactly the definition of \mathcal C_L. It follows that |I_\beta(u)|\le p^{-j_p/2}. Apply Equations (6.12) and (6.14), then use Equation (6.9) and average over u. This proves the first bound in Equation (6.16). The shell is contained in \mathfrak h_p^{-1}, which has volume H_p\vol(R_E). Integration of the first bound, with p^{-j_p/2}\le1, proves the second.

We next prove Equation (6.17). Put N=4(g_p+1) and take t\in p^N\mathfrak h_p^{-1}. For A in the shell choose s as in Lemma 6.3, and write \alpha=A+s, \beta=\alpha^{-1}. The component valuations of t are at least -h_\nu+e_\nu N. Translation by t therefore preserves the shell, and the same s still has the valuation properties of that lemma. Set \zeta=t\beta. Equation (6.9) gives

Here 1+\zeta is a unit. It is a square unit in each factor when p is odd, by Hensel’s lemma on principal units. When p=2, one has \zeta\in8R_E. In the \nu-th field the Hensel inequality for Y^2-(1+\zeta_\nu) at Y=1 is v_\nu(\zeta_\nu)\ge3e_\nu>2e_\nu=2v_\nu(2), so it is a square unit there as well, including for ramified dyadic factors.

Moreover N-2\iota_p\ge g_p+v_p(2). Equation (6.11) therefore places the inverse difference in 2\mathcal C_L. Its Q-value on L is integral by Equation (6.2). In Equation (6.14), the inverse integral is unchanged by this difference. The norm is unchanged by the unit 1+\zeta, and the phase is unchanged because that unit is a square. The original integral, for each fixed u, is consequently unchanged under A\mapsto A+t. So is its unit average. Translation by p^N\mathfrak h_p^{-1} preserves the shell as a set, hence preserves its zero-extended function. The inverse transform is supported on the annihilator of this group, namely p^{-N}\mathfrak h_p\mathfrak d_E^{-1}. This proves Equation (6.17) with the absolute exponent 4.

It remains to prove the sharper assertions for g_p=0 and p>5. The equality g_p=0 says \mathfrak h_p\subset\mathcal C_L. For each factor with h_\nu=0, its component idempotent 1_\nu and its whole component ring 1_\nu R_\nu lie in \mathfrak h_p. The self-adjoint projections \rho(1_\nu) preserve L and split off their lattices orthogonally. The complementary projection 1-\sum_{h_\nu=0}\rho(1_\nu) also preserves L. Each resulting summand is self-dual, because the splitting is orthogonal and L is self-dual.

On an h_\nu=0 summand, an integral A_\nu preserves its lattice, so Q_{A_\nu} is integral there since 2 is a unit. That factor of the integral in Equation (6.5) is 1. On the combined summand of all h_\nu>0, the frequency A_+ is invertible. Its inverse, extended by zero on the other factors, lies in \mathfrak h_p\subset\mathcal C_L. Its inverse quadratic integral in Equation (6.13) is therefore also 1. On that summand the square root of the polar determinant has absolute value H_p, because each positive factor still has rank two. Applying the formula therefore shows that the remaining frequency function is H_p^{-1} times a unit-average of Gaussian phases. Its absolute value is at most H_p^{-1}. If there are no positive h_\nu, this conclusion has H_p=1 and the empty summand contributes 1.

Let \chi_\nu be the quadratic character of (R_\nu/\mathfrak q_\nu)^\times for a factor with h_\nu>0. At odd residue characteristic every principal unit is a square, so unit square classes are exactly the two residue square classes. Phase invariance in Lemma 6.4 shows that the bounded unit-average just obtained depends only on the tuple of these classes. This use of phase invariance concerns the ambient quadratic form; a unit need not preserve the lattice. Fourier expansion on the product of the two-element square-class groups expresses the shell function as a sum of at most 2^{|\{\nu:h_\nu>0\}|}\le8 terms of the form

where \varepsilon_\nu\in\{0,1\}. The bound for the coefficients follows from the normalized finite Fourier expansion of a function of absolute value at most 1. It also covers any coupling of components caused by the common average over u.

We compute the inverse of each factor. Put \psi_\nu=\psi_p\circ\Tr_{E_\nu/\Q_p} and use the component self-dual measure. For h_\nu=0, the integral over R_\nu is \vol(R_\nu)\mathbf1_{\mathfrak d_\nu^{-1}}(z_\nu). For h=h_\nu>0, the normalized inverse of the trivial unit character is

This follows by subtracting the integral over \pi_\nu^{-h+1}R_\nu from that over \pi_\nu^{-h}R_\nu and using their annihilators. For the quadratic unit character, its function of a=\pi_\nu^hA is invariant under addition by \mathfrak q_\nu. This forces the normalized inverse to vanish outside \mathfrak q_\nu^{h-1}\mathfrak d_\nu^{-1}. It also vanishes on \mathfrak q_\nu^h\mathfrak d_\nu^{-1}, where the additive character on R_\nu is trivial and the sum of \chi_\nu is zero. At the intervening boundary the additive character descends to a nontrivial character \eta of the residue field k_\nu, and the normalized integral is \vol(R_\nu)q_\nu^{-1} times its quadratic Gauss sum.

For completeness, if \chi is the quadratic character of a finite field k of odd order q and \eta is a nontrivial additive character, then

Thus the quadratic factor has magnitude \vol(R_\nu)q_\nu^{-1/2} at the boundary. The trivial factor there has the smaller magnitude \vol(R_\nu)q_\nu^{-1}. Multiplying these component estimates, using H_p=\prod_{h_\nu>0}q_\nu^{h_\nu} and \prod_\nu\vol(R_\nu)=\vol(R_E), and summing at most eight terms proves Equations (6.18) and (6.19). The annihilator computation retains the different in every ramified factor.□

Proof. Let B_n=p^{-n}R_E. Fubini and additive orthogonality give

The self-dual normalizations give

Thus the right side is the average density on that neighborhood. By Lemma 6.1 it equals f_p^*(z) for sufficiently large n. The ball B_n is the disjoint union of the finitely many shells with h_\nu\le ne_\nu.

There can in fact be only finitely many nonzero shell inverses at this z. The injection \mathfrak h_p/(\mathfrak h_p\cap\mathcal C_L)\hookrightarrow R_E/\mathcal C_L gives g_p\le v_p([R_E:\mathcal C_L]). In Equation (6.17), a nonzero value requires

In Equation (6.18), it requires h_\nu\le v_\nu(z_\nu)+d_\nu+1 whenever h_\nu>0. All right sides are bounded for this z, because its component valuations are finite. The shell sum therefore stabilizes, and the preceding ball inversion proves Equation (6.20).□

Proof. Let M_p=\mathfrak h_p\cap C_{F,p}. The additive lattice identity

has first factor between 1 and [R_{F,p}:\mathfrak h_p]=H_p. The index of C_F is supported on S by Lemma 5.9; multiplying gives Equation (7.3).

At p\in\mathscr L_{\mathfrak h}, the local support is contained in p^{-4(g_p+1)}\mathfrak h_p\mathfrak d_p^{-1}. At every other p\in S, the support loses at most one power of each boundary prime \mathfrak q, and loses no power in the other components. It follows that a supporting z belongs to

The real trace condition excludes z=0. Since F is a field, its norm is nonzero, and containment of the principal ideal (z) in \mathfrak b_{\mathfrak h,z} gives

The left side is bounded above on \Omega_\infty. For g_p>0 we have g_p+1\le2g_p; the only additional factors, for g_p=0, are at 2,3,5. Hence

This proves the first inequality of Equation (7.4). Each boundary prime occurs once, and \sum_{\mathfrak q\mid p}f_{\mathfrak q}\le \sum_{\mathfrak q\mid p}e_{\mathfrak q}f_{\mathfrak q}=3; this proves the bound for T_\partial. Equation (7.5) is the product of the boundary bounds of Lemma 6.5; its absolute implied constant accounts for the three small primes.

Dropping \mathfrak h_p in the first support condition enlarges that support. In the second, the exponent v_{\mathfrak q}(\mathfrak h_p)-1 at a boundary is nonnegative. Thus every supporting point belongs to the ideal

which gives Equation (7.6).

Lemma 5.9 gives \Delta_F\ll X^{28}, D_F\le\Delta_F, and [R_F:C_F]\le\Delta_F^{1/2}. The definition of S gives \prod_{p\in S}p\ll_m X^2. Equations \text{(7.3)}–\text{(7.4)} therefore give H\ll X^{28+24\cdot14+6}=X^{370}. The fixed implied constant is absorbed by X^{371} for sufficiently large X.

A nonzero field element has nonzero image in every component of F\otimes\Q_p. Lemma 6.6 therefore identifies the density f_p^*(z) with its pointwise finite sum of inverse shells at all the present z. The norm bound just proved excludes every product with H>X^{371}, uniformly in z, and proves Equation (7.7). Finally, for any fixed u>0, Rankin’s inequality bounds the number of profiles with H\le X^{C_3} by

There are at most three prime-ideal components over each p. The radical bound on S implies |S|=O(\log X/\log\log X), since the product of k distinct primes is at least k!. The displayed Euler product is then X^{o(1)} for fixed u. Taking u arbitrarily small proves the asserted bound for every \eta>0.□

Proof. The covolume formula follows from the discriminant. For 0\ne A\in \mathfrak a, containment (A)\subset\mathfrak a implies |\Nm_{F/\Q}A|\ge \Nm\mathfrak a=P^{-1}. The product inequality in three real coordinates gives |\Nm A|\le(\|A\|/\sqrt3)^3, proving the first-minimum bound.

Let \lambda_1\le\lambda_2\le\lambda_3 be the Euclidean successive minima. A lattice in fixed dimension has a basis of bounded orthogonality defect, by Mahler’s basis theorem [15, Section 2, Theorem 1]. For such a basis, order its lengths increasingly and write its defect as \omega=\prod_i\|b_i\|/\covol(\mathfrak a)\ll1. Cramer’s rule and Hadamard’s inequality bound the norm of the i-th coordinate functional by \omega/\|b_i\|. Since \|b_i\|\ge\lambda_i, counting its possible integer coordinates gives

Minkowski’s second theorem for the Euclidean ball, in the formulation of [28, Theorem 4.3], gives \lambda_1\lambda_2\lambda_3\asymp\covol(\mathfrak a). Expanding the last product, using \lambda_2\ge\lambda_1, and inserting the first-minimum and covolume bounds proves Equation (7.11). The same coordinate functional bound with \|b_i\|\ge\lambda_1 proves the final assertion.□

Proof. We use \lambda=1/3. Lemma 5.9, which gives \Delta_F\ge X^{1/2}/2, implies

for all sufficiently large X.

After bounding the real Fourier tail, we split according to the field discriminant. If D_F is large, comparing it with the discriminant of the order generated by a retained nonscalar frequency after clearing its denominators forces H large. Otherwise the lower bound for \Delta_F makes the index [R_F:C_F] large. For profiles with small G_0, this forces H large; inverse support and boundary decay then bound the possibly signed original sum S_{\mathfrak h}. Poisson expresses its full nonscalar Fourier sum as S_{\mathfrak h} minus the separately estimated scalar series. For profiles with large G_0, the inverse depths attached to each frequency and its chosen shifts provide either sufficient pointwise decay or congruences; in the latter range a further discriminant comparison supplies the required denominator size.

The positive exponents d,e,e_1,e_2,w,\theta,\tau below will be fixed at the end of the proof; put Y=X^w and s=w+\theta.

First the frequencies outside \|A\|\le Y can be discarded. For every fixed integer r\ge0, integration by parts in the fixed real coordinates gives, uniformly over the box family,

Indeed each derivative of order r costs at most a constant times X^{r\tau}, and all supports are in one compact set. Lemma 6.5 and the bounds for p\notin S give |\widehat\eta_{\mathfrak h,l}(A)|\ll H^{-1}. The first-minimum bound also gives the rough count \#(\Gamma_{l,\mathfrak h}\cap B_R)\ll(1+RP^{1/3})^3. For R\ge Y\ge1 this is O(R^3P), since P=l^3H\ge1. Dyadic summation of Equation (7.14), for r>3, yields

This also proves absolute convergence in Equation (7.9). The profile count costs only X^{o(1)}. When bounding Fourier terms directly, we may therefore restrict to the ball and use the uniform bound |\widehat\phi|\ll1 there.

Suppose first that D_F\ge X^d. A nonscalar A generates F, since the field has degree three. Moreover HR_F\subset\mathfrak h: the integer H=|R_F/\mathfrak h| annihilates this finite group. Thus lHA\in R_F for A\in\Gamma_{l,\mathfrak h}. The discriminant of the order \Z[lHA] is a positive integer multiple of D_F, whereas the product of the squared differences of its three real conjugates is O((lHY)^6). Existence of a nonscalar frequency in the ball therefore implies

Apply Equation (7.11) to \Gamma_{l,\mathfrak h} and multiply it by H^{-1}. The four resulting terms are bounded by constants times

All have a fixed power saving once s<d/100. If the ball contains no nonscalar frequency there is nothing to count for that profile.

We now assume D_F<X^d. When d\le\lambda/3, Equation (7.13) gives

Profiles with G_0\le X^e are best estimated in their original form S_{\mathfrak h}. Equation (7.3) implies H\ge X^{\lambda/3-e}. All supporting points lie in \mathfrak a_{\mathfrak h}, whose reciprocal norm is O(D_FG_0^{24}) by Equation (7.6). The first-minimum bound of Lemma 7.3, applied in the fixed real region, gives O(D_FG_0^{24}) possible points. At each such point, Equations \text{(7.4)}–\text{(7.5)} give

Here and below X^{o(1)} denotes O_\eta(X^\eta) for every fixed \eta>0, uniformly in the arithmetic data; it includes factors 8^{|S|}. It follows that

The use of the original sum has gained the boundary factor from the inverse transform. It remains to check that its scalar portion is small as well, so that this bounds the nonscalar portion.

If a rational scalar has denominator p^k at p, its denominator ideal in R_{F,p} is p^kR_{F,p}; its component exponents are e_{\mathfrak q}k, so its norm is p^{3k}, including at ramified primes. Thus a scalar in this profile has its S-part denominator t=H^{1/3}, or the profile contains no scalar. Its denominator away from S divides l, so there are O(1+YlH^{1/3}) such rationals in the ball. Their absolute contribution is at most

By H\ge X^{\lambda/3-e} this also saves a power for the choices below. Equation (7.9) expresses the full dual sum as S_{\mathfrak h}. Subtracting its scalar frequencies in the ball and using Equation (7.15) for the scalar tail therefore bounds the full nonscalar sum for these profiles by Equations (7.17)–(7.18) and the tail. No positivity of an individual inverse shell is needed.

It remains to treat D_F<X^d and G_0>X^e. For each frequency in the denominator shell of the profile choose at each p\in S one of the shifts s_p of Lemma 6.3, and write

where j_p is the least nonnegative integer with p^{j_p}\beta_p\in C_{F,p}. All sums involving J below are over frequencies in this profile. The pointwise local estimate now reads

For frequencies with J\ge X^{e_2}, the rough lattice count immediately gives

We used H,l,Y\ge1. This leaves J<X^{e_2}, where the extra factor in the Fourier bound may be too small. The order discriminant then forces a large denominator ideal, and the definition of j_p supplies congruences among its possible frequencies.

For a nonscalar A let \operatorname{disc}(A)=\Disc(1,A,A^2)\in\Q^\times. The equality F=\Q[A] remains F\otimes\Q_p=\Q_p[A] after tensoring. Shifting by a scalar and inverting an element with nonzero components preserve the generated algebra, so p^{j_p}\beta_p generates F\otimes\Q_p. The full-rank order \Z_p[p^{j_p}\beta_p] is contained in C_{F,p}. The identities

follow by taking the three differences of conjugates and by scaling. Discriminant divisibility therefore gives

The shift lemma says v_p(\Nm\alpha_p)=-\log_pH_p for p>5, and differs from this by at most three at 2,3,5. For p\notin S, lA is integral and C_{F,p}=R_{F,p} is unramified, so v_p(\operatorname{disc}(A))\ge-6v_p(l). Combining these valuation bounds for the nonzero rational \operatorname{disc}(A), with only an absolute factor from 2,3,5, gives

The first inequality uses \|A\|\le Y. Thus for J<X^{e_2}, Equations \text{(7.13)} and \text{(7.20)} imply

In particular, once s+e_2<\lambda/12 with a margin, every nonscalar frequency under consideration has H\ge X^{\lambda/8} for large X.

For the remaining count, group frequencies according to their j_p’s and the chosen s_p’s. The number of exponent lists with J<X^{e_2} is X^{o(1)}, by the same Rankin argument as for profiles; the bounded lists of shifts cost 4^{|S|}=X^{o(1)}. Groups containing no nonscalar frequency contribute zero and are omitted. Fix one of the other groups. Apply the basis part of Lemma 7.3 to \Gamma_{l,\mathfrak h}. All frequencies in the ball have integer coordinates in a cube [-L,L]^3, where we choose L comparable to the upper bound YP^{1/3}. This choice, together with Equation (7.21), gives

At p\in S choose a generator i_p of the maximal-order ideal \mathfrak h_p, and set x=i_p(A+s_p). Localizing the global basis of \Gamma_{l,\mathfrak h} at p, then multiplying it by i_p, gives a \Z_p-basis of R_{F,p}, because p\nmid l. Accordingly the map from the three integer coordinates to x is an affine \operatorname{GL}_3(\Z_p) change of coordinates. This statement holds also for a product of ramified local fields.

Smith normal form for \mathfrak h_p\cap C_{F,p} inside \mathfrak h_p gives a primitive linear form \ell_p:R_{F,p}\to\Z_p, meaning that its image is \Z_p, and an integer g_p'\ge g_p/3 such that

Indeed the three elementary-divisor exponents sum to g_p, and we use the coordinate with largest exponent. If p>5, the shift lemma makes x a unit and \beta_p=i_px^{-1}\in\mathfrak h_p. The condition p^{j_p}\beta_p\in C_{F,p} implies

Inversion permutes the units in the finite algebra R_{F,p}/p^kR_{F,p}. A primitive linear form has p^{2k} zeros among the p^{3k} additive residues. Thus the unit condition and Equation (7.23) retain at most the fraction p^{-k} of all residues modulo p^k. Nilpotents in the finite algebra do not affect this argument.

For p=2,3,5, the shift lemma gives x\in R_{F,p} and x^{-1}\in p^{-1}R_{F,p}. Put y=p x^{-1}. Then p^{j_p}i_py=p(p^{j_p}\beta_p) still lies in \mathfrak h_p\cap C_{F,p}, and the congruence in Equation (7.23) holds for y in place of x^{-1}. On the permitted locus the inverse identities

show respectively that x\bmod p^{k+1} determines y\bmod p^k, and that a fixed residue of y\bmod p^k confines x to one residue modulo p^{k-1}. The permitted locus has v_{\mathfrak q}(x)=0 on the pole components and 0\le v_{\mathfrak q}(x)\le e_{\mathfrak q} on the other components. These conditions are determined modulo p^2, since a change in p^2R_{F,p} has component valuation at least 2e_{\mathfrak q}. At modulus p^{k+1}, each of the at most p^{2k} residues in the linear kernel therefore allows at most p^6 residues of x. The retained fraction is at most p^{3-k}, hence O(p^{-k}) with an absolute constant at these three primes. This proves the required congruence fraction with only one extra power of a small prime in its modulus.

Write a_p=(g_p'-j_p)_+. The available congruence powers satisfy

We will choose e/3-e_2>2e_1. If every prime with a_p>0 is at most X^{e_1}, multiply available factors p, one at a time, until their product M_1 first reaches X^{e_1/2}. Equation \text{(7.24)} ensures this is possible, and

Impose just the corresponding congruences. The Chinese remainder theorem combines their retained fractions into O(M_1^{-1}). Their joint modulus is M_1 times at most 2\cdot3\cdot5, because only the small-prime tests need one extra power. The affine integral coordinate changes above preserve these fractions in the global coordinates. If 2e_1<\lambda/24, this joint modulus is o(L) by Equation (7.22). Counting each admissible residue class in the cube then gives

possible frequencies.

Otherwise some available prime satisfies p>X^{e_1}; for large X it exceeds 5. Use its congruence modulo p. Set B=R_{F,p}/pR_{F,p}, a three-dimensional commutative \mathbf F_p-algebra, possibly with nilpotents, and reduce \ell_p modulo p. Choose an \mathbf F_p-basis of B and let x be the universal element of B\otimes_{\mathbf F_p}\mathbf F_p[X_1,X_2,X_3] in this basis. Let M_x be its multiplication matrix and put \mathcal N_B(x)=\det M_x. This is a nonzero polynomial of degree three, since \mathcal N_B(1)=1. On the unit open set \mathcal N_B\ne0,

where P_\ell has degree at most two. This numerator is nonzero. Indeed the affine space parametrizing elements of the vector space B has coordinate ring \mathbf F_p[X_1,X_2,X_3], which is a domain even when multiplication in B has nilpotents. The unit open set is dense, and inversion is an automorphism of that open set. The nonzero linear polynomial \ell_p stays nonzero on the open set and after composition with inversion. The displayed formula, or injectivity of localization at \mathcal N_B, then forces P_\ell\ne0. The affine \operatorname{GL}_3(\mathbf F_p) change from global lattice coordinates preserves nonvanishing.

Let N_L=2\lfloor L\rfloor+1\asymp L be the number of possible integers in one coordinate of the cube. A nonzero polynomial of bounded degree has O(N_L^2) zeros on a product of three residue intervals of length N_L<p; this follows by induction on the number of variables, viewing it as a polynomial in the last variable and bounding the slices on which all its coefficients vanish. If N_L\ge p, partition the cube into boxes of side at most p and use the same bound there. The number of global integer coordinates satisfying P_\ell=0\bmod p is consequently

This bounds the subset for which x is a unit as well. Multiplying Equations \text{(7.25)}–\text{(7.26)} by H^{-1}, and using Equation (7.22), gives respectively the bounds

for the fixed group of j_p’s and shifts. Their X^{o(1)} number is harmless once these exponents are negative with a margin. This completes the count of nonscalar frequencies in the ball.

We finish by making the parameter choices and all margins explicit. Put \lambda_0=\min(\lambda,1) and take

Then s=101e_2/10000,

The saving in Equation (7.17) is at least

and Equation (7.18) has a still larger margin than \epsilon_F, since e\le\lambda/1000. All four exponents in Equation (7.16) also have margins at least 29d/900>\epsilon_F, by s<d/100. Among the remaining ball estimates the smallest margin is in Equation (7.19), namely

the congruence margins e_1/2-3s, e_1-3s, and \lambda/24-3s are larger. With r=100, the exponent in Equation (7.15) is

All X^{o(1)} costs from profiles, shifts, and exponent lists can thus be absorbed while retaining \epsilon_F=e_2/4. This proves Equation (7.12). Finally 2\tau+3\theta=e_2/2000<\epsilon_F, as asserted.□

Proof of Proposition 7.1. Use the constants in Equation (7.27). Equations (7.8) and \text{(7.9)}, together with Lemma 7.4, leave the scalar frequencies with absolute error O(X^{-\epsilon_F}). We first remove the profile cutoff from those frequencies.

For a\in\Q, let t=\prod_{p\in S}p^{\max(0,-v_p(a))} be its S-part denominator, with t=1 for a=0. As noted above, its unique profile has H=t^3. Lemma 6.2 gives the exact product \prod_{p\in S}\gamma_p(a)=t^{-3}, including at 2,3,5. The other denominator divides l, so all such scalars with a given t lie in (tl)^{-1}\Z. The change of real variables (z_i,z_j,z_k)\mapsto(z_i,z_j,\Tr z) has absolute Jacobian one. Writing \mathbf 1=(1,1,1), it gives the exact identity

For any fixed r>1, rapid decay of the fixed function \chi implies \sum_{n\in\Z}|\widehat\chi(n/(tl))|\ll_r tl. The scalars omitted by H\le X^{C_3} therefore contribute at most

We have enlarged the set of t’s to all positive integers, which is an upper bound. With C_3=371, this is O(X^{-\epsilon_F}), since A_\phi\ll1, l\le X^\theta, and 371/3-\theta>\epsilon_F. The same calculation without the cutoff proves absolute convergence of the unrestricted scalar series.

That series is one-dimensional Poisson summation for the trace marginals. At infinity the marginal is A_\phi\chi(t). At p\in S it is d_p(t)\mathbf 1_{\Z_p}(t). At p\mid l, the onto integral trace map and two kernel coordinates show that the indicator z_p^0+p^{v_p(l)}R_{F,p} has trace density

with respect to Haar measure of mass one on \Z_p. At every other finite prime the marginal is \mathbf 1_{\Z_p}. Their Fourier transforms at a rational a are exactly the finite factors in Equation (7.9) evaluated at the scalar a, because \Tr(az)=a\Tr z.

For completeness, this one-dimensional application of Poisson does not require a smoothness assumption on d_p. Its Fourier coefficients on \Q_p/\Z_p of denominator p^k are p^{-3k}, and there are p^k-p^{k-1} such characters. Their absolute sum is finite. Truncating each expansion gives a compactly supported locally constant density on \Z_p, to which Poisson applies. The densities converge uniformly by Lemma 6.2, and the preceding absolute scalar bound permits passage to the limit on the Fourier side. On the original side, the finite marginals force the rational trace to be an integer. Only m meets the real cutoff. Hence the unrestricted scalar contribution is precisely

Lemma 6.2 bounds the product of d_p(m) above and below by positive absolute constants. Also A_\phi\gg X^{-2\tau} by Equation (7.1), and l^{-2}\ge X^{-2\theta}. Dividing the combined absolute error O(X^{-\epsilon_F}) by the main term therefore costs at most O(X^{2\tau+2\theta}). Since \epsilon_F>2\tau+3\theta, the relative error is O(X^{-\theta}). Every estimate used fixed real derivative and support bounds and constants independent of the fields, orders, residues, and box locations and scales. This proves Equation (7.2) with its stated uniformity.□

Proof. For each residue class a, Equation (7.2) gives

with the same C_{\mathrm{add}} for every a and l in the stated range. Multiply by W(a) and sum. Since W(a)\ge0, the absolute value of the summed error is at most C_{\mathrm{add}}V_\phi X^{-\theta}l^{-2}\sum_a W(a) =C_{\mathrm{add}}V_\phi X^{-\theta}\E_l W. This also covers \E_l W=0. In particular the error retains the local mean even when that mean is very small.□

Proof. We use the Selberg upper-bound square [22]; see also [11, Chapter 21]. Its calculation is included because the relative error in Lemma 8.1 must preserve the assigned means. If \vartheta_p=0, the activity event at p is a union of residue classes of Haar measure zero, hence is empty on global points. Such a prime can be removed from \mathcal I. For the remaining primes put

for squarefree d supported on \mathcal I. Both functions take value one at d=1. Let \mathcal D be the set of those integers with d\le T_0, and put \mathcal H=\sum_{d\in\mathcal D}h(d).

For coefficients \lambda_d supported on \mathcal D with \lambda_1=1, the square

majorizes the indicator that every prime of \mathcal I is inactive. Its expectation in the independent product of the local probability spaces is

Here [d,e] denotes the least common multiple. The identity follows by expanding the right side and using \vartheta([d,e]) =\vartheta(d)\vartheta(e)\sum_{l\mid(d,e)}h(l)^{-1}.

The set \mathcal D is closed under divisors. Finite Möbius inversion therefore permits the choice

where \mu_{\mathrm{Mob}} is the Möbius function. Explicitly,

Consequently \lambda_1=1, the value of Equation (8.13) is \mathcal H^{-1}, and

To estimate \mathcal H, give every squarefree integer supported on \mathcal I probability proportional to h(d), without the bound d\le T_0. In this finite product distribution, prime p is selected with probability \vartheta_p. Thus

Markov’s inequality shows that at least half the mass lies at d\le T_0. It follows that

It remains to average this square against \nu_\phi together with the assigned weights. Write

The assigned weight is defined modulo

For a pair d,e\in\mathcal D, its activity indicators require only the additional modulus [d,e]^{2C}\le T_0^{4C}. The two prime sets are disjoint, so the total modulus is at most

and is prime to every prime of S. The function before multiplication by \lambda_d\lambda_e is nonnegative, and its residue mean is A_{\mathcal A}\vartheta([d,e]) by the Chinese remainder theorem. Lemma 8.1 therefore gives the corresponding main term V_\phi A_{\mathcal A}\vartheta([d,e]) and an error of absolute value at most C V_\phi X^{-\theta}A_{\mathcal A}\vartheta([d,e]). There are at most T_0^2 pairs and, by Equation (8.14), each coefficient product has absolute value at most T_0^2. Summing the errors absolutely costs at most

This estimate keeps A_{\mathcal A} even if it is very small. The main term is V_\phi A_{\mathcal A}\mathcal H^{-1} by Equation (8.13). Finally, X^{-\theta}T_0^4=o((\log X)^{-C_1}) by Equation (8.10). Equations (8.11) and (8.15) absorb Equation (8.17) and prove the lemma.□

Proof of Proposition 8.2. Put P_Y=\prod_{p\in\mathcal P_Y}\E_p b_p. Equations (8.6) and (8.11) imply

First observe that primes exceeding Y cost only a fixed factor. Equations (8.7) and (8.8), together with b_p\le A_0^{n_p}, give

Consider the points for which \prod_{p\in\mathcal P_Y}p^{n_p(z)}\le T_0. Assign all their active primes in \mathcal P_Y and require inactivity at the other primes in \mathcal P_Y. These assignments satisfy Lemma 8.3. After applying it, we may enlarge the sum of its nonnegative upper bounds by allowing every subset and every positive depth. The resulting product is

Together with Equation (8.19), this bounds the contribution from these points.

For a remaining point, the product through \mathcal P_Y exceeds T_0. Order its active primes in \mathcal P_Y increasingly and stop at the first prefix whose product exceeds T_0. If every individual factor p^{n_p(z)} for p\in\mathcal P_Y is at most T_0, the prefix has product in (T_0,T_0^2], since its preceding product and its last factor are both at most T_0. If also \prod_{p\in\mathcal P_{y_0}}p^{n_p(z)}\le T_0, for a fixed cutoff y_0>4 to be chosen below independently of X, its last prime exceeds y_0. The next two estimates allow these restrictions at negligible weighted cost.

First, if p^{n_p(z)}>T_0 for p\in\mathcal P_Y, then n_p(z)\ge J_p, where J_p=\lfloor\log T_0/\log p\rfloor+1. The threshold is determined modulo

Lemma 8.1 and Equation (8.5) bound its \nu_\phi-measure by O(V_\phi p^{-J_p})=O(V_\phi/T_0). Summing over p\in\mathcal P_Y gives O(V_\phi Y/T_0). By Equation (8.8), the contribution after multiplication by \mathcal W is O_\epsilon(V_\phi X^{\epsilon+\delta-\eta}). Choose \epsilon<\eta-\delta. This is smaller than V_\phi P_Y by a power of X, in view of Equation (8.18).

We also need the analogous removal for a fixed nonempty finite set \mathcal Q of primes, even though some of them may belong to the varying set S. Put N=|\mathcal Q|. If \prod_{p\in\mathcal Q\setminus S}p^{n_p(z)}>T_0, some prime in \mathcal Q\setminus S has p^{n_p(z)}>T_0^{1/N}. Put J_{p,\mathcal Q}=\lfloor\log T_0/(N\log p)\rfloor+1. The event n_p\ge J_{p,\mathcal Q} is defined modulo at most T_0^{C/N}p^{2C}\le X^\theta for large X, and its local probability is at most Cp^{-J_{p,\mathcal Q}}\le C T_0^{-1/N}. Lemma 8.1 therefore bounds the measure of the union by O_{\mathcal Q}(V_\phi T_0^{-1/N}). After multiplication by Equation (8.8) with \epsilon<\eta/N, its weighted contribution again saves a power of X relative to V_\phi. If \mathcal Q\setminus S is empty the event is empty. Thus this removal is uniform in the possible intersections of \mathcal Q with S.

Choose a fixed M_0>0 and then a fixed y_0>4 such that

Apply the finite-set removal with \mathcal Q=\{p:p\le y_0\}. These choices precede the choice of the small \epsilon used in that removal, and all are independent of X.

The retained first-crossing prefixes therefore have product in (T_0,T_0^2] and last prime greater than y_0. Group these assignments by the interval containing their last prime:

Only groups with y>y_0 occur. Every unassigned active prime is greater than the last assigned prime and hence greater than \sqrt y. The total weight of those primes, including primes beyond Y, is therefore at most

For each assignment in the group, require inactivity at its unassigned primes in \mathcal P_{\sqrt y} and apply Lemma 8.3. We use Rankin’s power majorant [20]: since the assigned product exceeds T_0, its upper bound can be multiplied by

Now enlarge the sum of these nonnegative upper bounds to all subsets and all positive depths through y. This enlarges only a numerical sum after the lemma has been applied to the genuine prefixes; it invokes no residue estimate for the newly admitted products above T_0^2.

For p\in\mathcal P_{\sqrt y}, the option of no assignment has weight 1-\vartheta_p and the assigned options have total weight \sum_{j\ge1}\E_p(b_p\mathbf 1_{\{n_p=j\}})p^{sj}. Their sum is \E_p(b_p p^{s n_p}). For \sqrt y<p\le y outside S, inactivity was not imposed, so the option of no assignment has weight one. The total there is \E_p(b_p p^{s n_p})+\vartheta_p. Thus the group’s contribution is at most

We estimate these local moments uniformly for y>y_0. By Equation (8.20), s<1/4. The total contribution of depths j\ge2 to \E_p(b_p p^{s n_p}) is at most

At depth one, the increase over \E_p b_p is bounded by

because 0\le s\log p\le M_0. Chebyshev’s estimate shows that these increases have bounded sum over p\le y. Mertens’ estimate also gives \sum_{\substack{\sqrt y<p\le y\\p\notin S}}\vartheta_p\ll1. Since \E_p b_p\ge1/2, comparison of the products in Equation (8.23) with their base means gives

For a fixed C_2, a final use of Equation (8.6) and Mertens’ estimate yields

By Equation (8.20), the bound for the group in Equation (8.23) is consequently at most

For the values of y in Equation (8.21), its sum is bounded by

Together with the small-product part and the two discarded sets, this proves Equation (8.9).□

Proof of Theorem 5.2. Fix the nonzero trace and bounded region in that theorem. Let \mathcal R be any of its rectangles, in the specified two partition coordinates. Choose an allowed smooth function \phi that majorizes \mathbf 1_{\mathcal R}, with its two scales equal to the side lengths of \mathcal R. Its support lies in a fixed enlargement of the bounded region and

The constants are uniform over the positions, side lengths, and aspect ratios of the rectangles. Proposition 5.7 and Proposition 8.2 give

Here \prod_{p\in S}|D|_p^{-1/2}=\sqrt D, because S contains every prime divisor of D, and Equation (8.2) absorbed the trace densities. The restriction to integral points outside S in the multiplicity proposition agrees with the support of \nu_\phi.

The exact unramified mean in Lemma 5.6 is

Since 1-p^{-3}\le1, the two Euler products on the right of Equation (8.24) are at most the product of (1-p^{-1})Z_p over p\in S\cup\mathcal P_Y. Ideal factorization in the degree-four maximal order gives, uniformly also at ramified primes,

Thus

For fixed m, the radical of S is polynomial in X. Choose a fixed c\ge1/2 large enough that, for all sufficiently large X, every prime in S and every prime at most Y is at most Z=X^c. Since a_K(p)\ge0, the exponent in Equation (8.25) is at most

We have Z\ge D^{1/4} because X=q\sqrt D\ge\sqrt D. Equation (3.14) and Mertens’ estimate now give

This also applies when D remains bounded and q tends to infinity, by the stated uniform range of Lemma 3.3. Substitution in Equation (8.24) proves

The only box-dependent quantity in the proof is A_\phi, which is comparable to the area with a fixed constant. All other estimates are uniform over the box translations and the two scales between X^{-\tau} and the fixed upper bounds. The local multiplicity majorant has the same constant for every chosen ideal representative system, and its right side is independent of that choice. Thus the bound is uniform in the representative system as well. This proves the precise uniformity asserted in Theorem 5.2.□

Proof. We use the packet integral in Equation (2.9). For its maximal-order ideal representative I and local label L'\in\mathscr L_{\OO}(I), the normalized lattice at the sign vector s=(s_1,\ldots,s_4)\in\{\pm1\}^4 and the logarithmic parameter (u_1,\ldots,u_4)\in\R^4 with u_1+\cdots+u_4=0 is

For a raw vector w\in\bigwedge^2 I, its normalized exterior coordinates are

If z=\Psi(w)/\Nm I and \epsilon(s)=\prod_a s_a, then Equation (5.6) gives

There are eight sign vectors for each value of \epsilon(s).

The kernel of the signature map on U is U^+, so d_U=|\operatorname{sgn}U|\le16; the signature of -1 gives d_U\ge2. After averaging over L', the vector-count integrand in Equation (2.9) is invariant under v\mapsto\sigma(e)v for e\in U, since L'\mapsto eL' permutes the labels and transports their exterior lattices. Replacing the fundamental domain for U in T_\infty by one for U^+ therefore multiplies its integral by d_U, which we compensate by division by d_U. The latter domain consists of all sign components, with a fundamental domain for the positive unit logarithms in each component.

For a given raw w, its average membership over the local labels is

the weight in Equation (5.8). This weight is constant on its U^+-orbit. That action is free: every embedding coordinate of the nonzero raw vector w\in\bigwedge^2_{\Z}I\subset\bigwedge^2_{\Q}K is nonzero by Lemma 5.1. A stabilizing positive unit e would therefore satisfy \sigma_a(e)\sigma_b(e)=1 for every pair ab, forcing e=1. Summing the translates of each raw vector thus unfolds the positive-unit fundamental domain to all of the three-dimensional space u_4=-u_1-u_2-u_3, with multiplicity one. The packet normalization remaining after the uniform label average is h_K\vol(\mathscr F_U)=\sqrt D\,\kappa by Equation (2.6). Hence this unfolding has coefficient 1/(d_U\sqrt D\,\kappa) before the sign sum and the integral in u. All the sums and integrals are nonnegative, so Tonelli’s theorem applies.

Choose one coordinate from each opposite pair. The logarithms of the absolute values of these coordinates are affine functions of (u_1,u_2,u_3). For the choices 12,13,14, their linear parts are

Their matrix has determinant of absolute value two. Choosing the opposite member of a pair negates its row, and ordering the partitions only permutes rows, so the same absolute determinant holds for every choice. If the two entries in an opposite pair have product of absolute value |z_b| and both must be less than h>0 in absolute value, the logarithm of the chosen entry lies in an interval of length \log_+(h^2/|z_b|). Therefore, for a raw vector with the required trace after the sign change, its integral over (u_1,u_2,u_3) is \frac12\mathcal L_{L,r}(z).

The U^+-orbit sum of the weights, including the sum over I, is exactly M(z) by Equation (5.8). Equation (9.4) allows raw trace m for the eight positive determinant signs and raw trace -m for the other eight. Multiplying eight by the Jacobian factor 1/2 gives the factor 4 in Equation (9.3). The unit -1, although it acts trivially on exterior vectors, has already been accounted for in the index d_U.□

Proof. Take r_0<1 so small that 2r_0^2\le |m|/2. For each t\in\{m,-m\}, fix the coordinate rectangle

Now let 0<r<r_0. In the trace-t sum in Equation (9.3), only points with |z_{b_2}|,|z_{b_3}|<r^2 contribute, and these points lie in \Omega_t. Since their three coordinates have sum t,

Thus the first logarithm in Equation (9.2) is bounded in terms of m,L on the contributing points, independently of r. For every z\in\Omega_t with M(z)>0, Lemma 5.10 gives |z_b|\gg_{\Omega_t}X^{-84} for each b. Hence the last two logarithms in Equation (9.2) are O(\log X), uniformly for 0<r<1.

We first remove the points in the two strips

For large X, each strip within \Omega_t is covered by a rectangle of width 8X^{-\tau} and length 2 in the coordinates z_{b_2},z_{b_3}, and these rectangles lie in \Omega_t. Both sides meet the minimum length in Theorem 5.2 for large X. That theorem bounds the sum of M(z) over their union, divided by \sqrt D\,\kappa, by O(X^{-\tau}). The logarithmic bounds just proved show that their contribution to Equation (9.3) is

The use of the rectangle estimate here controls the weighted mass directly; it does not require a pointwise bound for M(z).

For the remaining points, separate the signs of z_{b_2},z_{b_3} and use the dyadic intervals

Each signed dyadic rectangle lies in \Omega_t. Each interval has width half its upper endpoint. If it meets |z_{b_j}|>4X^{-\tau}, that width exceeds 2X^{-\tau}. Hence every dyadic rectangle meeting the remaining region satisfies the minimum side lengths of Theorem 5.2; any boundary rectangle may be included in full. Its area is O(r^4 2^{-a-b}), while the last two factors of Equation (9.2) are at most (a+1)\log2 and (b+1)\log2. Theorem 5.2 gives

The sum over the four choices of signs and over all a,b is therefore at most

All constants are independent of r: the rectangles \Omega_t and the uniform rectangle constant were fixed before choosing r. The two strips tend to zero for each fixed r, and 4/d_U\le2 in Equation (9.3). These observations prove Equation (9.5) for both traces and all three partitions.□

Proof of Theorem 1.1. For the sequence in that theorem, X_i=\sqrt{|\Disc(\OO_i)|} tends to infinity. By Proposition 3.6, the packet probabilities are tight, and every subsequence has a further weakly convergent subsequence with an A_4-invariant probability limit \mu. Proposition 3.8 gives the local ball bound required by Corollary 4.3. Hence almost every component in the A_4-ergodic decomposition of \mu has positive entropy for a diagonal flow. Proposition 4.6 reduces the identification of these components to the exclusion of its three exterior-plane events.

Fix a partition b_1, a nonzero integer m, and L>1. The event

is open by the observation following Equation (9.1).

The open-set inequality for weak convergence and Proposition 9.2 now give, along the convergent subsequence,

for every sufficiently small fixed r>0. In particular,

Every vector in one of the obstruction planes with Q\ne0 has a nonzero integer Q-value by Equation (4.7), and its two entries are less than some integer L\ge2 in absolute value. An exact witness belongs to E_{m,L,1/n} for every n\ge2. Thus the Borel event in Equation (4.9) satisfies

The right side is a countable union of null sets, so \mu(\mathcal E_{b_1})=0. The same argument applies to all three partitions. Integrating these null events over the ergodic decomposition shows that the proper alternative in Proposition 4.6 has zero total component weight. Almost every component is therefore m_4, and hence \mu=m_4.

Every subsequential probability limit is now identified as m_4. Tightness implies that the entire sequence converges weakly to m_4, which gives the asserted convergence against every f\in C_c(X_4). The same tightness statement supplies, for each \epsilon>0, one compact set carrying at least 1-\epsilon of every sufficiently late packet probability. This is the required no-escape assertion.□