Proof. The additive group \(R/\mathcal O\) has order \(q\), so \(qR\subset\mathcal O\). For every \(p\) and every unit \(t_p\), it follows that \[qI_p=\xi_p t_p qR_p\subset \xi_p t_p\mathcal O_p\subset I_p.\] 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 \(\mathcal O_p\) under multiplication by \(R_p^\times\) is exactly \(\mathcal O_p^\times\). Indeed, \(t_p\mathcal O_p=\mathcal O_p\) implies that both \(t_p\) and \(t_p^{-1}\) belong to \(\mathcal O_p\), by applying the equality and its inverse to \(1\). The converse is immediate. The local choices therefore form the finite homogeneous space \(R_p^\times/\mathcal O_p^\times\). Its invariant probability is uniform and is the image of normalized unit Haar measure. Patching over the finitely many primes dividing \(q\) proves Equation (6) and the assertion about probabilities. The index of the lattice in Equation (5) is \([R_p:\mathcal O_p]\), and its \(R_p\)-span is \(I_p\) because \(1\in\mathcal O_p\). 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 \(\mathcal O\)-ideal \(J\) is invertible if and only if \[ J_p=\eta_p\mathcal O_p\quad\text{for some }\eta_p\in K_p^\times \quad\text{at every }p. \tag{7}\] For the forward implication, complete an equality \(JJ^{-1}=\mathcal O\). The resulting equality over \(\mathcal O_p\) 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 \(\mathcal O_p\), 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 (Milne 2020a, Theorem 1.14) to choose \(c_{\mathfrak m}\in\mathcal O_p\) 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 (Milne 2020a, Lemma 1.9) shows that it generates each localization. Hence it generates \(J_p\). Since \(J_p\) spans \(K_p\) over \(\mathbb Q_p\), \(\eta_p\) has no zero component and lies in \(K_p^\times\).

Conversely, define \[J^{-1}=\{a\in K:aJ\subset\mathcal O\}.\] This definition commutes with completion: after choosing \(\mathbb Z\)-bases of \(J\) and \(\mathcal O\), the containment is a finite collection of integrality conditions on the coordinates of multiplication by \(a\), and the same conditions over \(\mathbb Z_p\) describe the completed lattice. Under Equation (7) we therefore have \((J^{-1})_p=\eta_p^{-1}\mathcal O_p\). Consequently \((JJ^{-1})_p=\mathcal O_p\) for every \(p\), which implies \(JJ^{-1}=\mathcal O\). This proves the equivalence. It also proves properness: if \(aJ\subset J\) for an invertible \(J\), multiplication by \(J^{-1}\) gives \(a\mathcal O\subset\mathcal O\), and hence \(a\in\mathcal O\). Thus its multiplier order is \(\mathcal O\).

The lattices in Equation (5) are stable under \(\mathcal O_p\) at every prime, so comparison of completions shows that \(\mathcal OL\subset L\). They are therefore fractional \(\mathcal O\)-ideals satisfying Equation (7). In the other direction, let \(J\) be an invertible fractional \(\mathcal O\)-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 \(J_p=\eta_p\mathcal O_p\), 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 \(\mathscr L_{\mathcal O}(I)\), as required. ◻

Proof. The action in Equation (9) preserves the parameter set, since \(\epsilon I=I\) for \(\epsilon\in U\), and preserves its image lattice. Lemma 2 shows that every ordinary invertible \(\mathcal O\)-ideal class occurs. If two ideals differ by multiplication by \(\alpha\in K^\times\), their normalized embedded lattices differ by the diagonal with coordinates \(\sigma_i(\alpha)/|\mathop{\mathrm{N}}_{K/\mathbb Q}(\alpha)|^{1/4}\), 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 (9). 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 \(\mathrm{SL}_4(\mathbb Z)\).

It remains to check that this identification gives the stipulated orbit weights. Write \(\mathcal O^{\times,+}=\mathcal O^\times\cap U^+\). The \(U\)-stabilizer of a label \(L\) is \(\mathcal O^\times\), by its multiplier order. The stabilizer of \(L\) together with a sign pattern is therefore \(\mathcal O^{\times,+}\). This subgroup has finite index in \(U\). For \(q=1\) this follows directly from \(\mathcal O=R\). For \(q>1\), the kernel of \(U\longrightarrow(R/qR)^\times\) lies in \(\mathcal O^\times\), since both a unit congruent to \(1\) modulo \(qR\) and its inverse belong to \(\mathcal O\), 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 \(\ell(\mathcal O^{\times,+})\). They form a full lattice, so the orbit is compact.

Combining a label with a sign produces \(16N_{\mathcal O}\) finite choices above each \(I\). Each \(U\)-orbit of these choices has \([U:\mathcal O^{\times,+}]\) elements and parametrizes one \(A_4\)-orbit. On that orbit the quotient of logarithmic Lebesgue measure is its Haar measure, with volume \[\mathop{\mathrm{covol}}\bigl(\ell(\mathcal O^{\times,+})\bigr) =[\ell(U):\ell(\mathcal O^{\times,+})]R_K =\frac{[U:\mathcal O^{\times,+}]}{2}R_K .\] The last equality uses \(\ker(\ell|_U)=\{1,-1\}\) and \(-1\notin\mathcal O^{\times,+}\). Consequently the number of distinct orbits times their common volume is \[\frac{16h_KN_{\mathcal O}}{[U:\mathcal O^{\times,+}]} \cdot \frac{[U:\mathcal O^{\times,+}]}2 R_K =8h_KN_{\mathcal O}R_K.\] This proves both the assertion about the descended measure and Equation (10), by Equation (8).

One can also take \(\mathscr L_{\mathcal O}(I)\times\mathscr F_U\) as a fundamental domain for the diagonal action in Equation (9): 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 \(h_KN_{\mathcal O}\mathop{\mathrm{vol}}(\mathscr F_U)\), equal to the total volume just computed. Dividing their integral by that quantity proves Equation (11). ◻

Proof. This is the unfolding of (OpenAI 2026, Proposition 2.2) with the packet normalization of Proposition 3. Average the vector sum over \(L\) in Equation (11), and use Equation (14). The factor \(N_{\mathcal O}\) cancels, and the left side becomes \[\frac1{h_K\mathop{\mathrm{vol}}(\mathscr F_U)} \sum_I\int_{\mathscr F_U} \sum_{0\ne b\in I}P((b)I^{-1})f(c_I\sigma(b)v)\,dv .\] 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 \[\int_{\mathscr F_U}\sum_{\epsilon\in U} f(c_I\sigma(\epsilon b)v)\,dv =\int_{T_\infty}f(c_I\sigma(b)v)\,dv .\] 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 \[c_I^4|\mathop{\mathrm{N}}_{K/\mathbb Q}(b)| =\frac{\mathop{\mathrm{N}}((b)I^{-1})}{X}.\] Translation of the first three logarithmic coordinates and relabeling of the signs therefore identify the last integral with \(\mathcal V_f(\mathop{\mathrm{N}}((b)I^{-1})/X)\), with Jacobian one. Each integral ideal belongs to the inverse class of exactly one chosen \(I\). Equation (8) now gives the coefficient in Equation (16). ◻

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 \[ m_p(\text{shell }\boldsymbol l_p) =\frac1{Z_p\,N_p(\boldsymbol l_p)}. \tag{20}\] Its conditional additive measure is invariant under the transitive unit action. It is therefore the image of unit Haar probability. By Equation (12), \(P_p(\boldsymbol l_p)\) is exactly the fraction of this shell lying in \(\mathcal O_p\). The elements with a zero component have measure zero, and \(m_p(\mathcal O_p)=q_p^{-1}\). Summing Equation (20) over the intersections with \(\mathcal O_p\) gives \[\sum_{\boldsymbol l_p} \frac{P_p(\boldsymbol l_p)}{N_p(\boldsymbol l_p)} =\frac{Z_p}{q_p}.\] Multiplication over \(p\mid q\) proves Equation (17).

For the moment, choose \(x\) with normalized additive Haar probability on \(\mathcal O_p\), and let \(v_{\mathfrak p}=v_{\mathfrak p}(x_{\mathfrak p})\). The projection \(H_{\mathfrak p}\) of \(\mathcal O_p\) into \(R_{\mathfrak p}\) contains the embedded copy of \(\mathbb Z_p\), because \(1\in\mathcal O_p\). Its induced probability is normalized Haar on \(H_{\mathfrak p}\). For an integer \(k\ge1\), the image of this copy of \(\mathbb Z_p\) 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 \[ \mathbb P_{\mathcal O_p}\{v_{\mathfrak p}\ge k\} \le p^{-\lceil k/e_{\mathfrak p}\rceil}. \tag{21}\] This uses only the size of the additive image of the projection; it does not require the projections to be independent.

Set \[j=\max_{\mathfrak p\mid p} \left\lceil\frac{v_{\mathfrak p}}{e_{\mathfrak p}}\right\rceil.\] If \(j\ge a\ge1\), then some \(v_{\mathfrak p}\) is at least \(e_{\mathfrak p}(a-1)+1\). Equation (21) and a union bound over at most four components give \(\mathbb P_{\mathcal O_p}\{j\ge a\}\le4p^{-a}\). On the other hand, \[N_p(x)=p^{\sum_{\mathfrak p\mid p}f_{\mathfrak p}v_{\mathfrak p}} \le p^{j\sum_{\mathfrak p\mid p}e_{\mathfrak p}f_{\mathfrak p}} =p^{4j}.\] These assertions hold outside the null sets where a component vanishes. The tail formula for the expectation of \(p^{j/2}\) now yields \[\begin{align*} \mathbb E_{\mathcal O_p}N_p(x)^{1/8} &\le \mathbb E_{\mathcal O_p}p^{j/2} \\ &=1+\sum_{a\ge1} (p^{a/2}-p^{(a-1)/2})\mathbb P_{\mathcal O_p}\{j\ge a\} \le 1+4p^{-1/2}. \tag{22}\end{align*}\]

Applying Equation (20) once more gives the exact factorization \[\sum_{\boldsymbol l} P(\boldsymbol l)d(\boldsymbol l)^{-1+1/8} =\frac1q\prod_{p\mid q} \left(Z_p\,\mathbb E_{\mathcal O_p}N_p(x)^{1/8}\right).\] 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 (18). Finally, take \(\eta=1/128\) there and use \(q\le X\): \[\sum_{d(\boldsymbol l)>X^{1/8}}\frac{P(\boldsymbol l)}{d(\boldsymbol l)} \le X^{-1/64} \sum_{\boldsymbol l}P(\boldsymbol l)d(\boldsymbol l)^{-1+1/8} \le \frac Cq X^{-1/128}.\] This proves Equation (19). ◻

Proof. We first treat large \(D\), and put \(L=\log D\). Stark’s zero theorem (Stark 1974, Lemma 3) gives at most one nontrivial zero of \(\zeta_K\) in \[\Re s\ge1-\frac1{4L},\qquad |\Im s|\le\frac1{4L};\] any such zero is real and simple. His Lemma 8 (Stark 1974, 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 \[ |1-\rho|\ge \frac{c_0}{L},\qquad c_0=\frac1{192}. \tag{26}\] 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 \[ \begin{aligned} S_K(s):=\sum_\rho\Re\frac1{s-\rho} &=\frac1s+\frac1{s-1}+\frac L2+ \frac{\zeta_K'}{\zeta_K}(s)+G(s),\\ G(s)&=2\left(\frac{\Gamma'(s/2)}{\Gamma(s/2)}-\log\pi\right); \end{aligned} \tag{27}\] see (Stark 1974, 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 \(D\), \[0\le S_K(s_0)\ll L.\] This estimate persists for all \(1<s\le1+4/L\). To see this term by term, write \(\rho=\beta+i\gamma\). Since \(\beta\le1\), \[s-\beta\le4(s_0-\beta),\qquad |s-\rho|\ge|1-\rho|.\] Equation (26) consequently gives \[|s_0-\rho|\le |s-\rho|+\frac3L \le (1+3/c_0)|s-\rho|.\] Using \(\Re(s-\rho)^{-1}=(s-\beta)/|s-\rho|^2\), we obtain \[\Re\frac1{s-\rho} \le4(1+3/c_0)^2\Re\frac1{s_0-\rho}.\] Thus \(S_K(s)\ll L\) on the entire interval.

Let \(\mathcal Z_K(s)=(s-1)\zeta_K(s)\). Rearranging Equation (27) shows that \[\frac{\mathcal Z_K'}{\mathcal Z_K}(s) =S_K(s)-\frac1s-\frac L2-G(s)=O(L) \qquad(1<s\le1+4/L).\] 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 (23), 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 (Milne 2020a, 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 (24) 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 \[\sum_{p\le Y}a_K(p)p^{-1-h}\le\log\zeta_K(1+h).\] Moreover, \(a_K(p)\le4\), \(1-p^{-h}\le h\log p\), and the elementary Chebyshev bound followed by partial summation imply \[\sum_{p\le Y}a_K(p)(p^{-1}-p^{-1-h}) \le4h\sum_{p\le Y}\frac{\log p}{p} \ll h\log Y=1.\] It follows that \[\sum_{p\le Y}\frac{a_K(p)}p \le \log\zeta_K(1+h)+O(1) \le \log(\kappa\log Y)+O(1),\] which proves Equation (25). 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 (Pollack 2020, Theorem 1.1) with no excluded residue classes and with its parameter \(\beta=1/4\). Its two growth conditions hold with bounds independent of \(b\): \[b(p^j)\le d_4(p^j)=\binom{j+3}{3}\le4^j,\qquad b(n)\le d_4(n)\ll_\epsilon n^\epsilon \quad(\epsilon>0).\] 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 (28), 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 (OpenAI 2026, Proposition 4.2). Put \[b(n)=a_K(n)\mathbf 1_{(n,q)=1}.\] 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 (29) is therefore \[ \sum_{\boldsymbol l}P(\boldsymbol l) \sum_{(x-y)/d(\boldsymbol l)<n\le x/d(\boldsymbol l)}b(n). \tag{30}\]

First take \(d=d(\boldsymbol l)\le X^{1/8}\), and put \(u=x/d\), \(v=y/d\). Then \[u\ge X^{5/8}\ge D^{1/4},\qquad v\ge cu.\] 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 (25), once \(X\) is above an absolute threshold. To remove the primes dividing \(q\) from that Euler bound, observe that \[ \log Z_q=\sum_{p\mid q}\frac{a_K(p)}p+O(1),\qquad \sum_{\substack{p\mid q\\p>u}}\frac{a_K(p)}p \ll\frac{\log(2q)}u\ll1 . \tag{31}\] 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 (25) and (31) give \[\exp\left(\sum_{p\le u}\frac{b(p)}p\right) \ll \frac{\kappa\log u}{Z_q}.\] Lemma 7 and Equation (17) now bound the contribution of these tuples to Equation (30) by \[C\frac{\kappa y}{Z_q} \sum_{d(\boldsymbol l)\le X^{1/8}} \frac{P(\boldsymbol l)}{d(\boldsymbol l)} \le C\frac{\kappa y}{q}.\]

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 \[\sum_{n\le z}d_4(n) =\sum_{a_1a_2a_3a_4\le z}1 \le z\left(\sum_{a\le z}\frac1a\right)^3 \le z(1+\log z)^3 .\] Use this with \(z=x/d\), and then Equation (19). The remaining contribution is at most \[Cx(1+\log x)^3 \sum_{d(\boldsymbol l)>X^{1/8}} \frac{P(\boldsymbol l)}{d(\boldsymbol l)} \le C\frac{x}{q}X^{-1/128}(1+\log x)^3 .\] For fixed \(A\), once \(X\ge A\) the condition \(x\le AX\) gives \(\log x\le2\log X\). By Equation (24) and \(D\le X^2\), the ratio of this last bound to \(\kappa x/q\) is at most \[C X^{-1/128}(\log(3X))^4=o(1).\] 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 (33). Put \(T=r^4X\) and \(T_0=X^{3/4}\). By Equations (16) and (32), it suffices to bound \[ \frac1{\sqrt D\,\kappa} \sum_{\mathop{\mathrm{N}}\mathfrak a\le T} P(\mathfrak a) \bigl(\log(T/\mathop{\mathrm{N}}\mathfrak a)\bigr)^3 . \tag{34}\] The sum is empty when \(T<1\). Otherwise its logarithmic factor is at most \((\log X)^3\), since \(\mathop{\mathrm{N}}\mathfrak a\ge1\) and \(r<1\). Apply Proposition 8 at \(x=y=T_0\), with \(A=c=1\). The part with \(\mathop{\mathrm{N}}\mathfrak a\le\min\{T,T_0\}\) is at most \[C\frac{\kappa T_0}{q}(\log X)^3.\] 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 \[A_{T_0}(t)=\sum_{T_0<\mathop{\mathrm{N}}\mathfrak a\le t}P(\mathfrak a) \qquad(T_0\le t\le T).\] 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 \[\begin{align*} \sum_{T_0<\mathop{\mathrm{N}}\mathfrak a\le T} P(\mathfrak a)\bigl(\log(T/\mathop{\mathrm{N}}\mathfrak a)\bigr)^3 &=3\int_{T_0}^T A_{T_0}(t) \bigl(\log(T/t)\bigr)^2\frac{dt}{t}\\ &\le C\frac{\kappa}{q} \int_0^T\bigl(\log(T/t)\bigr)^2\,dt \le C\frac{\kappa T}{q}. \end{align*}\] The last integral is finite after the substitution \(t=Ts\). This part of Equation (34) 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 (33). This is the compactness argument of (OpenAI 2026, Proposition 5.1), now using the quartic weighted interval bound.

Mahler’s compactness criterion (Mahler 1946, sec. 3, Theorem 2) says that \[\mathcal K_r=\{\Lambda\in X_4:\|v\|_\infty\ge r \text{ for all }0\ne v\in\Lambda\}\] is compact. Choose \(r\) so that \(Cr^4<\epsilon/2\). Equation (33) 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 3. 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 \[\int f(xa)\,d\mu_i(x)=\int f(x)\,d\mu_i(x)\] 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 \[\log\frac{|v_i|+br}{|v_i|-br}\ll_{V_0,b}r .\] 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 (15) that, for one \(C_1\ge1\), \[ \mathcal V_{\mathbf 1_{v+[-br,br]^4}}(z) \le C_1r^3\,\mathbf 1_{[t_v-C_1r,t_v+C_1r]}(z) \qquad(v\in V_0,\ 0<r<r_0). \tag{36}\] 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 \[x=(t_v+2C_1r)X,\qquad y=4C_1rX.\] The possible ideal norms in Equation (36) 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 8 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 (16) and (36), and using \(\sqrt D=X/q\), proves Equation (35). 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 (OpenAI 2026, Proposition 5.2). First, every full lattice has a vector with no zero coordinate. If \(\Lambda=\mathbb Z^4g\), 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 \(\mathrm{SL}_4(\mathbb R)\), 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 \(V_0\subset(\mathbb R^\times)^4\). 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 \[|(v_xh-v_x)_j| \le \sum_i |(v_x)_i|\,|h_{ij}-\delta_{ij}|<br.\] Thus every lattice in \(xB(r)\) has a nonzero vector in \(v_x+[-br,br]^4\). Lemma 10 gives, for each fixed small \(r>0\) and all sufficiently large \(i\), \[\mu_i(xB(r)) \le \int E_{\mathbf 1_{v_x+[-br,br]^4}}\,d\mu_i \le C_\Omega r^4 \qquad(x\in\Omega).\] 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 \[\mu(xB(r))\le\liminf_{i\to\infty}\mu_i(xB(r)) \le C_\Omega r^4.\] 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 (37). Fix \(0<b<1/4\) and put \[T_t=a(t)B(b)a(-t)\cap a(-t)B(b)a(t), \qquad r=e^{-\lambda_a t}\quad(t\ge0).\] For \(h\in T_t\), the two conjugations give \[ |h_{ii}-1|<b,\qquad |h_{ij}|<b e^{-|\lambda_i-\lambda_j|t}\le br \quad(i\ne j). \tag{39}\] 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 (39) implies \(\prod_i h_{ii}=1+O(r^2)\). The diagonal matrix \[d(h)=\mathop{\mathrm{diag}}\bigl(h_{11},h_{22},h_{33}, (h_{11}h_{22}h_{33})^{-1}\bigr)\] 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 \(\mathbb R^3\). A mesh of spacing comparable to \(r\), followed by a fixed enlargement to account for matrix multiplication, gives \[ T_t\subset\bigcup_{j=1}^{N_t}d_jB(Cr), \qquad d_j\in\mathcal D_\infty,\qquad N_t\le Cr^{-3}. \tag{40}\] For a compact \(\Omega\subset X_4\), the centers \(xd_j\) lie in the compact set \(\Omega\mathcal D_\infty\). Domination and Equation (38) now imply \[ \xi(xT_t)\le C_{\Omega,c}r^{-3}r^4 =C_{\Omega,c}e^{-\lambda_a t} \qquad(x\in\Omega) \tag{41}\] for all sufficiently large \(t\). Apply Theorem 12 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 \(A_4\simeq\mathbb R^3\) is locally compact and second countable, and acts continuously on the standard Borel space \(X_4\). The ergodic decomposition theorem (Greschonig and Schmidt 2000, Theorem 5.2), specialized to invariant probabilities, gives a measurable decomposition \[\mu=\int_Z\nu_z\,d\tau(z)\] 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 \[h_\nu(T)=\sup_k\inf_{N\ge1}\frac1N H_\nu\left(\bigvee_{j=0}^{N-1}T^{-j}\mathcal P_k\right),\] 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 (Rokhlin 1967, Theorem 9.8) gives \[ h_\mu(T)=\int_Z h_{\nu_z}(T)\,d\tau(z), \tag{42}\] 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 \[\xi=\frac1w\int_{Z_0}\nu_z\,d\tau(z)\] would be an \(A_4\)-invariant probability with \(\xi\le w^{-1}\mu\). The restricted form of Equation (42) would give \(h_\xi(T)=0\), contrary to Proposition 13. ◻

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 \(\mathfrak{sl}_4(\mathbb R)\) are the distinct one-dimensional spaces \(\mathbb RE_{ij}\), \(i\ne j\), 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 \(\mathbb R_{>0}\), 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 \(\mathop{\mathrm{diag}}(u_1,\ldots,u_4)\), 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 \(W_\mathbb Q\). Let \[\mathcal C_\mathbb Q =\{M\in\mathop{\mathrm{End}}_\mathbb Q(W_\mathbb Q):M\gamma=\gamma M \text{ for every }\gamma\in\Gamma_\Lambda\}.\] 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 \(\mathcal C_\mathbb Q\otimes_\mathbb Q\mathbb R\) 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 \[\Gamma_\Lambda\backslash J\longrightarrow\mathop{\mathrm{End}}_\mathbb R(W_\mathbb R), \qquad \Gamma_\Lambda j\longmapsto j^{-1}Mj\] 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 \(\mathbb R\) invariant under such a dilation is supported at zero: the annuli \(\{x:r^k\le |x|<r^{k+1}\}\), \(k\in\mathbb Z\), for a fixed \(r>1\), are disjoint dilates covering \(\mathbb R\setminus\{0\}\). 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 \(\mathcal C_\mathbb Q\otimes_\mathbb Q\mathbb R\simeq\mathbb R^s\). In particular, \(\mathcal C_\mathbb Q\) is commutative and has no nonzero nilpotents. The structure theorem for finite reduced commutative algebras over \(\mathbb Q\) 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 \(e\in\mathcal C_\mathbb Q\) were an idempotent other than \(0\) and \(1\), then \(W=e(W_\mathbb R)\) 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 \[\chi:J\longrightarrow\mathbb R_{>0},\qquad \chi(j)=\lvert\det(j|_W)\rvert\] 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 \(\mathbb R_{>0}\) invariant under all multiplicative translations. The disjoint-interval argument above rules this out. Thus \(\mathcal C_\mathbb Q\) is a field, which we denote by \(E\).

Since \(E\otimes_\mathbb Q\mathbb R\simeq\mathbb R^s\), its degree is \(s\). The space \(W_\mathbb Q\) is a vector space over \(E\), and \[W_\mathbb Q\otimes_\mathbb Q\mathbb R \simeq\bigoplus_{\tau:E\hookrightarrow\mathbb R} W_\mathbb Q\otimes_{E,\tau}\mathbb R.\] Each summand has dimension \(\dim_E W_\mathbb Q=4/s\). These summands are the eigenspaces of the block-scalar algebra, namely the coordinate blocks. The blocks therefore have equal size. ◻

Proof. Theorem 15, 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 16 and the positive-entropy assumption, either \(J=\mathrm{SL}_4(\mathbb R)\), 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 16 is quadratic. Choose a nonzero \(\alpha\in E\) with \(\mathop{\mathrm{Tr}}_{E/\mathbb Q}(\alpha)=0\). Its two scalar values on the blocks are \(c\) and \(-c\), with \(c\ne0\). The rational endomorphism \(\alpha\) of \(W_\mathbb Q=\mathbb Q\Lambda\) induces a rational endomorphism \[D_\alpha:\bigwedge\nolimits^2 W_\mathbb Q\longrightarrow \bigwedge\nolimits^2 W_\mathbb Q,\qquad D_\alpha(v\wedge w)=\alpha(v)\wedge w+v\wedge\alpha(w).\] 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 \[\bigwedge\nolimits^2\mathbb R^{\{i,j\}} \oplus \bigwedge\nolimits^2\mathbb R^{\{k,l\}}=V_b.\] Because \(D_\alpha\) is rational, \(V_b\) is a rational subspace relative to \(\bigwedge^2 W_\mathbb Q\).

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 (45) 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 \(A_{\mathcal O}\), so it multiplies \(\delta_{\mathcal O}\) by the sign of that permutation. Permuting the four indices in the three signed products in Equation (50) permutes the partitions and introduces the same common sign. Division by \(\delta_{\mathcal O}\) gives precisely the equivariance in Equation (48). 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 \[Q\bigl((\bigwedge\nolimits^2 A_{\mathcal O})t\bigr) =\det(A_{\mathcal O})Q(t), \qquad Q(t)=t_{12}t_{34}-t_{13}t_{24}+t_{14}t_{23}.\] It follows either by expanding the six minors or by evaluating \(\frac12 t\wedge t\) in the top exterior power. This proves Equation (51). 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 \(\prod_i a_i=\mathop{\mathrm{N}}(a)\), and the scalar \(c\) contributes \(c^2\). This proves Equation (52) after splitting and hence over \(k\). ◻

Proof. Over the splitting field set \[c_0=\frac{r_{12}r_{13}}{r_{23}},\qquad (a_1,a_2,a_3,a_4)= \left(1,\frac{r_{23}}{r_{13}},\frac{r_{23}}{r_{12}}, \frac{r_{14}r_{23}}{r_{12}r_{13}}\right).\] 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 \(c^2\mathop{\mathrm{N}}(a)=s\). ◻

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 \(\mathbb Q_p\). 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 \(L_{V,p}\times\mathbb Z_p^\times\) 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 \(\mathbb Q_p\) 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 \(\mathbb Q_p\) 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 \(\delta_{\mathcal O}^{\,3}\). Consequently its measure form in the six embedding coordinates is \[|\delta_{\mathcal O}|_p^{-3}\left|\bigwedge_{i<j}dx_{ij}\right|.\] 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 (62).

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 \(|\delta_{\mathcal O}/(u x_{12})|_p\), \(|\delta_{\mathcal O}/(u x_{13})|_p\), and \(|\delta_{\mathcal O}/(u x_{14})|_p\). They cancel \(|\delta_{\mathcal O}|_p^{-3}\), since \(u\) is a unit. Haar probability on \(\mathbb Z_p^\times\) is \((1-p^{-1})^{-1}|du/u|\). The resulting fiber form is \[ \frac{1}{1-p^{-1}} \left|\frac{dx_{12}}{x_{12}}\wedge \frac{dx_{13}}{x_{13}}\wedge \frac{dx_{14}}{x_{14}}\wedge\frac{du}{u}\right|. \tag{65}\] The fiber action preserves this measure. Indeed its determinant on \(V_p\) is \(c^6\mathop{\mathrm{N}}(a)^3=(c^2\mathop{\mathrm{N}}(a))^3\), 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 (65). Put \(\ell_i=da_i/a_i\) and \(s=dc/c\) over the splitting field. The pullbacks of its four logarithmic differentials under Equation (61) are \[\ell_1+\ell_2+s,\quad \ell_1+\ell_3+s,\quad \ell_1+\ell_4+s,\quad -\ell_1-\ell_2-\ell_3-\ell_4-2s.\] 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 \[\begin{pmatrix} 1&0&0&1\\ 0&1&0&1\\ 0&0&1&1\\ -1&-1&-1&-2 \end{pmatrix}, \qquad \det=1.\] 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 \(\mathbb Q_p\) tangent lifts of a quotient basis gives the same absolute quotient form for nonsplit \(K_p\). On a patch where a \(\mathbb Q_p\)-linear coefficient \(\lambda(a)\) is nonzero, normalizing it to its value at a reference point gives a genuine \(\mathbb Q_p\) 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 \(\mathop{\mathrm{N}}(a)\). On \(R_p^\times\) the latter is a unit. The mass of this unit set is consequently \[|D|_p^{1/2} \prod_{\mathfrak p\mid p,\ \mathfrak p\subset R} (1-\mathop{\mathrm{N}}(\mathfrak p)^{-1}).\] 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 (65), the mass of each \(\mathscr U_p\)-coset is \[ \frac{|D|_p^{1/2}}{1-p^{-1}} \prod_{\mathfrak p\mid p,\ \mathfrak p\subset R} (1-\mathop{\mathrm{N}}(\mathfrak p)^{-1}). \tag{66}\] 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 \(L_{V,p}\times\mathbb Z_p^\times\). 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 (66) by \(\sum_{\mathfrak c}\alpha_p(\mathfrak c;z)\). The result is the density \(f_p^*(z)\). Solving for the sum proves Equation (64). ◻

Proof. For two such pairs, Lemma 20 over \(\mathbb Q\) gives \(w'=c(\bigwedge^2a)w\) with \(a\in K^\times\), \(c\in\mathbb Q^\times\), since the common opposite ratio is \(\mathop{\mathrm{N}}(I')/\mathop{\mathrm{N}}(I)\). 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 \(d_p\in\mathbb Q_p^\times\) satisfies \[d_pa\xi_p/\xi'_p\in R_p^\times,\qquad d_p^{-2}c\in\mathbb Z_p^\times.\] In particular \(v_p(c)=2v_p(d_p)\) is even at every finite prime. Therefore \(c=\epsilon s^2\) with \(s\in\mathbb Q^\times\) 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 \(\delta_{\mathcal O}\) are units, \[k_{ij}+k_{kl}=j_{\mathfrak q} \quad\text{for the partition }ij|kl\text{ over }\mathfrak q .\] 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 \[v_p(c)+v_p(a_i)+v_p(a_j)=0\qquad(i<j).\] Subtraction makes all four \(v_p(a_i)\) equal to a common \(r\), and \(v_p(c)=-2r\). Because \(K_p\) is unramified, \(r\in\mathbb Z\). 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 (68). 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 \(\mathbb Z_p\) 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 \[\begin{cases} t^{\sum_{\mathfrak q}f_{\mathfrak q}r_{\mathfrak q}}, &\min_{\mathfrak q}r_{\mathfrak q}=0,\\ 0,&\min_{\mathfrak q}r_{\mathfrak q}>0. \end{cases}\] 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 \[ \mathbb E_p b_p= \frac{1-(\prod_{\mathfrak q}\epsilon_{\mathfrak q})t^3} {\prod_{\mathfrak q}(1-\epsilon_{\mathfrak q}t^{f_{\mathfrak q}})}. \tag{71}\] The five Frobenius types give the following data. A pair \((f,\epsilon)\) in the second column is attached to one prime of \(F\). \[\begin{array}{c|c|c|c} \text{cycle type on four labels} & (f_{\mathfrak q},\epsilon_{\mathfrak q}) & \prod_{\mathfrak p\mid p}(1-t^{f_{\mathfrak p}}) & a_K(p)-1\\ \hline 1^4 &(1,+),(1,+),(1,+)&(1-t)^4&3\\ 2\,1^2 &(1,+),(2,+)&(1-t)^2(1-t^2)&1\\ 2^2 &(1,+),(1,-),(1,-)&(1-t^2)^2&-1\\ 3\,1 &(3,+)&(1-t)(1-t^3)&0\\ 4 &(1,-),(2,-)&1-t^4&-1 \end{array}\] In every row \(\prod_{\mathfrak q}\epsilon_{\mathfrak q}=1\) and \[\prod_{\mathfrak q}(1-\epsilon_{\mathfrak q}t^{f_{\mathfrak q}}) =\frac{\prod_{\mathfrak p\mid p}(1-t^{f_{\mathfrak p}})}{1-t}.\] Substitution in Equation (71) 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 \(\mathbb P_p(v_{\mathfrak q}(z)\ge k)=p^{-f_{\mathfrak q}k}\). Taking the union over at most three factors proves Equation (70). 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 \(\mathop{\mathrm{Tr}}z=m\ne0\) imply \(z\in F^\times\). At every prime the set of accepted cosets is finite by Lemma 21. At \(p\notin S\), Lemma 23 proves the integrality requirement and bounds the local sum by \(b_p(z)\). The set \[E_z=S\cup\{p:z\notin R_{F,p}^\times\}\] 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 (59) identifies the weight of a global pair at each prime with its coset fraction \(\alpha_p\). By Lemma 22, 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 (64) at \(p\in S\) and Equation (68) outside \(S\). This proves Equation (72) 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 (51) 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 (73). 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 \(E=\bigwedge^2 A_{\mathcal O}\) be the exterior embedding matrix. Its entries are integral in every local splitting ring, and \(\det E=\delta_{\mathcal O}^{\,3}\). Since \(d=\delta_{\mathcal O}^{\,2}\), \[d^2 E^{-1}=\delta_{\mathcal O}\operatorname{adj}(E)\] also has integral entries. For \(\alpha\in R_{F,p}\), its plane-scalar matrix over the splitting ring is integral. Consequently \(d^2\rho(\alpha)=d^2E^{-1}\mathop{\mathrm{diag}}(\alpha_b,\alpha_b)E\) is integral over the splitting ring. Its entries are in \(\mathbb Q_p\), so they lie in \(\mathbb Z_p\). 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 \(\mathbb Z_p\); 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 (Milne 2020a, Proposition 7.64 and Remark 7.65). It follows that \(D_F\ll D^2\). The discriminant-index formula now gives \[\Delta_F=D_F[R_F:C_F]^2\ll D^2d^{12}\le X^{28}.\] 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 \(\mathop{\mathrm{Disc}}(\mathcal O)\). Write \(\operatorname{tr}\) for matrix trace, put \(\mathscr E=\mathop{\mathrm{End}}_\mathbb Z(\mathcal O)\), and let \(\mathscr W_K\subset\mathop{\mathrm{End}}_\mathbb Q(K)\) be the rational subspace of multiplication matrices. Its intersection \(\mathscr L=\mathscr W_K\cap\mathscr E\) is saturated. It is multiplication by \(\mathcal O\): if \(\alpha\mathcal O\subset\mathcal O\), then \(\alpha=\alpha\cdot1\in\mathcal O\), and the converse follows because \(\mathcal O\) 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 \[ \mathcal E_K(\mathscr E)=\mathscr L^\vee. \tag{78}\] 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 \[\mathscr L^\vee\subset N^{-1}\mathscr L,\qquad X^2=\mathop{\mathrm{Disc}}(\mathcal O)=[\mathscr L^\vee:\mathscr L]\le N^4.\] 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 \[C_F^\vee=\{\beta\in F:\mathop{\mathrm{Tr}}_{F/\mathbb Q}(\beta C_F)\subset\mathbb Z\}.\] The adjugate of the trace Gram matrix gives \(\Delta_F b_i\in C_F\). Define on \(\mathop{\mathrm{End}}_\mathbb Q(V)\) \[ \mathcal E_F(T)=\sum_{i=1}^3\rho(a_i)T\rho(b_i). \tag{79}\] For two embeddings \(\sigma,\tau\) of \(F\), the dual-basis identity is \[\sum_i\sigma(a_i)\tau(b_i)= \begin{cases}1,&\sigma=\tau,\\0,&\sigma\ne\tau.\end{cases}\] Thus, after splitting, \(\mathcal E_F\) extracts exactly the three diagonal blocks \(\mathop{\mathrm{End}}(V_b)\). It is the orthogonal projection for the matrix trace pairing, and \[ \Delta_F\mathcal E_F(\mathop{\mathrm{End}}_\mathbb Z(L_V))\subset\mathop{\mathrm{End}}_\mathbb Z(L_V). \tag{80}\] 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 \[j(T)(v\wedge w)=Tv\wedge w+v\wedge Tw \qquad(T\in\mathop{\mathrm{End}}_\mathbb Q(K)).\] There is a natural contraction \(\mathfrak c:\mathop{\mathrm{End}}_\mathbb Q(\bigwedge^2K)\to\mathop{\mathrm{End}}_\mathbb Q(K)\). For clarity, in a basis \(e_1,\ldots,e_4\) write \[S(e_j\wedge e_l)=\sum_{i<k}S^{ik}_{jl}e_i\wedge e_k\qquad(j<l),\] and extend these coefficients antisymmetrically in \(i,k\) and in \(j,l\), with zero coefficients when either pair repeats. Then \[\mathfrak c(S)^i_j=\sum_{k=1}^4 S^{ik}_{jk}.\] 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 \(\mathcal O\) and exterior \(\mathcal O\)-bases. Expanding their coefficients gives \[ \mathfrak c(j(T))=2T+\operatorname{tr}(T)\mathop{\mathrm{id}}_K. \tag{81}\] 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 \(\mathcal E_F(j(T))=j(\mathop{\mathrm{diag}}T)\) over the splitting field. Equation (81) now proves the formula \[ \mathcal E_K(T)= \tfrac12\bigl(\mathfrak c(\mathcal E_F(j(T)))-\operatorname{tr}(T)\mathop{\mathrm{id}}_K\bigr) \tag{82}\] for the projection defined above: its right side is \(\mathop{\mathrm{diag}}T\) 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 (80) and integrality of \(j\) and \(\mathfrak c\) give \[2\Delta_F\mathcal E_K(\mathop{\mathrm{End}}_\mathbb Z(\mathcal O))\subset\mathop{\mathrm{End}}_\mathbb Z(\mathcal O).\] Taking \(N=2\Delta_F\) in the preceding lattice bound gives \[X^2\le(2\Delta_F)^4.\] This is the lower inequality in Equation (77). ◻

Proof. Since \(C_F\) is an order, \(C_F\cap\mathbb Q=\mathbb Z\), 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 (83). 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 \(\mathbb Z\). Lemma 26, 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 (63) is supported on \((C_{F,p}^\sharp)^\vee\). For a contributing global pair, use its local representation \(z=u\Psi(x)\) and Equation (61) 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 \(\mathop{\mathrm{Tr}}z=m\ne0\), so \(z\ne0\); since \(F\) is a field, all its embeddings are nonzero. The norm \(\mathop{\mathrm{N}}(\Delta_F^\sharp z)\) is consequently a nonzero integer. The boundedness of \(\Omega\) and \(\Delta_F^\sharp\ll X^{28}\) give \[1\le|\mathop{\mathrm{N}}(\Delta_F^\sharp z)| =(\Delta_F^\sharp)^3\prod_{b\in\mathcal B}|z_b| \ll_\Omega X^{84}.\] At \(p\notin S\), \(z\) is integral and the local algebra is unramified. Hence \[n_p(z)\le\sum_{\mathfrak q\mid p} f_{\mathfrak q}v_{\mathfrak q}(z) \le v_p\bigl(\mathop{\mathrm{N}}(\Delta_F^\sharp z)\bigr).\] The product of these inequalities gives the depth bound because \(\mathop{\mathrm{N}}(\Delta_F^\sharp z)\) is an integer. Finally, if the coordinates of \(\Omega\) have absolute value at most \(B\), with \(B\ge1\), then for each \(b\) \[|z_b|=\frac{|\mathop{\mathrm{N}}(\Delta_F^\sharp z)|} {(\Delta_F^\sharp)^3\prod_{b'\ne b}|z_{b'}|} \ge\frac{1}{(\Delta_F^\sharp)^3 B^2}\gg_\Omega X^{-84}.\] This proves Equation (85). ◻

Proof. Define an \(E\)-valued bilinear form \(b_E\) by \[\mathop{\mathrm{Tr}}_{E/\mathbb Q_p}(a b_E(x,y))=B(\rho(a)x,y)\qquad(a\in E).\] Self-adjointness of the action makes \(b_E\) symmetric and \(E\)-bilinear. It is nondegenerate because \(B\) is nondegenerate. Equation (87) 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 \(\mathbb Q_p\)-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 \(\mathbb Q_p\)-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 \(L\times\mathbb Z_p^\times\) 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 \(r\in\mathbb Z_p\), \[\mathop{\mathrm{Tr}}_{E/\mathbb Q_p}((r+2c)\Psi_E(x)) =rQ_1(x)+B(\rho(c)x,x)\in\mathbb Z_p\qquad(x\in L).\] Multiplication by \(u\in\mathbb Z_p^\times\) 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 \(u\in\mathbb Z_p^\times\) and \(v_p(a)=-k<0\), integration in the second coordinate gives \[\int_{\mathbb Z_p^2}\psi_p(ua xy)\,dx\,dy =\int_{\mathbb Z_p}\mathbf1_{\{ua x\in\mathbb Z_p\}}\,dx=p^{-k}.\] For \(a\in\mathbb Z_p\) the integral is \(1\). The sign of \(a\) has no effect, so the three independent pairs in Equation (86) give Equation (91), even before averaging \(u\). This calculation uses the integral split expression and does not divide by \(2\) in \(\mathbb Z_p\).

The scalar coefficient is the Fourier coefficient of the trace law, since \(\mathop{\mathrm{Tr}}(a z)=a\mathop{\mathrm{Tr}}(z)\). That law is supported on \(\mathbb Z_p\) by Lemma 28. The character group of \(\mathbb Z_p\) is \(\mathbb Q_p/\mathbb Z_p\), and it has \(p^k-p^{k-1}\) characters of exact denominator \(p^k\). Hence its nonconstant Fourier coefficients have total absolute sum \[\sum_{k\ge1}(p^k-p^{k-1})p^{-3k} =\frac{1-p^{-1}}{p^2-1}=\frac1{p(p+1)}.\] Fourier inversion on the compact group \(\mathbb Z_p\) is therefore uniformly convergent and gives the continuous density and Equation (92). 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 (94). 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 \[ p^{g_p}\mathfrak h_p\subset\mathcal C_L. \tag{96}\] 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 (Weil 1964, I, §14, Theorem 2, p. 161). We record its normalization for the transform \[\mathcal F_{B_0}h(y)=\int h(x)\psi_p(B_0(x,y))\,dx.\] Completing the square gives \[q_P(x)+B_0(x,y)=q_P(x+P^{-1}y)-q_{P^{-1}}(y).\] 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 \[|c_N|^2=\mathop{\mathrm{vol}}(B_N) \int_{B_N\cap P^{-1}B_N^\perp}\psi_p(q_P(h))\,dh.\] For sufficiently large \(N\), the intersection is \(P^{-1}p^NL_0\) and \(q_P\) is integral on it. The right side then equals \(\mathop{\mathrm{vol}}(p^{-N}L_0)\mathop{\mathrm{vol}}(P^{-1}p^NL_0)=|\det P|_p^{-1}\). 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 (98).

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 (Weil 1964, II, §25, p. 173). For \(P=u\rho(\alpha)\) its inverse is \(u^{-1}\rho(\beta)\). Rank two over \(E\) gives \(\det\rho(\alpha)=\mathop{\mathrm{N}}_{E/\mathbb Q_p}(\alpha)^2\), and \(|u|_p=1\). This proves Equation (99). 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 \(\mathbb Z_p\). ◻

Proof. For \(c\in\mathcal C_L^\sharp\), Equation (87) shows that \(Q_c(L)\subset\mathbb Z_p\). 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 (100) 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 \[|I_\beta(u)|^2 =\int_{\substack{h\in L\\\rho(\beta)h\in L}} \psi_p(-u^{-1}Q_\beta(h))\,dh.\] 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 \[j_p=\max\{0,-\min_i a_i\},\qquad [L:L\cap\rho(\beta)^{-1}L] =p^{\sum_i\max(0,-a_i)}\ge p^{j_p}.\] 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 (97) and (99), then use Equation (94) and average over \(u\). This proves the first bound in Equation (101). The shell is contained in \(\mathfrak h_p^{-1}\), which has volume \(H_p\mathop{\mathrm{vol}}(R_E)\). Integration of the first bound, with \(p^{-j_p/2}\le1\), proves the second.

We next prove Equation (102). 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 30, 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 (94) gives \[\zeta\in p^{N-\iota_p}R_E,\qquad \alpha+t=\alpha(1+\zeta),\qquad (\alpha+t)^{-1}-\beta =-\frac{t\beta^2}{1+\zeta} \in p^{N-2\iota_p}\mathfrak h_p.\] 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 (96) therefore places the inverse difference in \(2\mathcal C_L\). Its \(Q\)-value on \(L\) is integral by Equation (87). In Equation (99), 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 (102) 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 (90) 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 (98) 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 31 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 \[c_{\boldsymbol\varepsilon}H_p^{-1} \prod_{h_\nu=0}\mathbf1_{R_\nu}(A_\nu) \prod_{h_\nu>0} \mathbf1_{\pi_\nu^{-h_\nu}R_\nu^\times}(A_\nu) \chi_\nu(\overline{\pi_\nu^{h_\nu}A_\nu})^{\varepsilon_\nu}, \qquad |c_{\boldsymbol\varepsilon}|\le1,\] 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 \(\psi_\nu=\psi_p\circ\mathop{\mathrm{Tr}}_{E_\nu/\mathbb Q_p}\) and use the component self-dual measure. For \(h_\nu=0\), the integral over \(R_\nu\) is \(\mathop{\mathrm{vol}}(R_\nu)\mathbf1_{\mathfrak d_\nu^{-1}}(z_\nu)\). For \(h=h_\nu>0\), the normalized inverse of the trivial unit character is \[q_\nu^{-h}\int_{\pi_\nu^{-h}R_\nu^\times} \psi_\nu(-Az)\,dA =\mathop{\mathrm{vol}}(R_\nu)\left( \mathbf1_{\mathfrak q_\nu^h\mathfrak d_\nu^{-1}}(z) -q_\nu^{-1}\mathbf1_{\mathfrak q_\nu^{h-1}\mathfrak d_\nu^{-1}}(z) \right).\] 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 \(\mathop{\mathrm{vol}}(R_\nu)q_\nu^{-1}\) 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 \[\left|\sum_{a\in k^\times}\chi(a)\eta(a)\right|^2 =\sum_{t\in k^\times}\chi(t) \sum_{b\in k^\times}\eta((t-1)b) =(q-1)-\sum_{\substack{t\in k^\times\\t\ne1}}\chi(t)=q.\] Thus the quadratic factor has magnitude \(\mathop{\mathrm{vol}}(R_\nu)q_\nu^{-1/2}\) at the boundary. The trivial factor there has the smaller magnitude \(\mathop{\mathrm{vol}}(R_\nu)q_\nu^{-1}\). Multiplying these component estimates, using \(H_p=\prod_{h_\nu>0}q_\nu^{h_\nu}\) and \(\prod_\nu\mathop{\mathrm{vol}}(R_\nu)=\mathop{\mathrm{vol}}(R_E)\), and summing at most eight terms proves Equations (103) and (104). The annihilator computation retains the different in every ramified factor. ◻

Proof. Let \(B_n=p^{-n}R_E\). Fubini and additive orthogonality give \[\int_{B_n}\gamma_p(A)\psi_p(-\mathop{\mathrm{Tr}}_{E/\mathbb Q_p}(Az))\,dA =\mathop{\mathrm{vol}}(B_n)\,\mu_p(z+p^n\mathfrak d_E^{-1}).\] The self-dual normalizations give \[B_n^\perp=p^n\mathfrak d_E^{-1},\qquad \mathop{\mathrm{vol}}(B_n)\mathop{\mathrm{vol}}(B_n^\perp)=1.\] Thus the right side is the average density on that neighborhood. By Lemma 28 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 (102), a nonzero value requires \[h_\nu\le v_\nu(z_\nu)+d_\nu+4e_\nu(g_p+1).\] In Equation (103), 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 (105). ◻

Proof. Let \(M_p=\mathfrak h_p\cap C_{F,p}\). The additive lattice identity \[[R_{F,p}:C_{F,p}] =[R_{F,p}:C_{F,p}+\mathfrak h_p]\,[\mathfrak h_p:M_p]\] 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 26; multiplying gives Equation (108).

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 \[\mathfrak b_{\mathfrak h,z} =\mathfrak h\mathfrak d_F^{-1} \prod_{p\in\mathscr L_{\mathfrak h}} (pR_F)^{-4(g_p+1)} \prod_{\substack{p\in S\setminus\mathscr L_{\mathfrak h}\\ \mathfrak q\in\partial_{\mathfrak h_p}(z)}} \mathfrak q^{-1}.\] 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 \[|\mathop{\mathrm{N}}_{F/\mathbb Q}z| =\mathop{\mathrm{N}}(\mathfrak b_{\mathfrak h,z}) [\mathfrak b_{\mathfrak h,z}:(z)] \ge \frac{H} {D_FT_\partial\prod_{p\in\mathscr L_{\mathfrak h}}p^{12(g_p+1)}}.\] 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 \[\prod_{p\in\mathscr L_{\mathfrak h}}p^{12(g_p+1)} \le30^{12}G_0^{24}.\] This proves the first inequality of Equation (109). 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 (110) is the product of the boundary bounds of Lemma 32; 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 \[\mathfrak a_{\mathfrak h} =\mathfrak d_F^{-1} \prod_{p\in\mathscr L_{\mathfrak h}} (pR_F)^{-4(g_p+1)} ,\] which gives Equation (111).

Lemma 26 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 \(\eqref{eq:add-profile-indices}\)–\(\eqref{eq:add-profile-norm}\) 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 \(F\otimes\mathbb Q_p\). Lemma 33 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 (112). Finally, for any fixed \(u>0\), Rankin’s inequality bounds the number of profiles with \(H\le X^{C_3}\) by \[X^{C_3u}\prod_{p\in S}(1-p^{-u})^{-3}.\] 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 \(|\mathop{\mathrm{N}}_{F/\mathbb Q}A|\ge \mathop{\mathrm{N}}\mathfrak a=P^{-1}\). The product inequality in three real coordinates gives \(|\mathop{\mathrm{N}}A|\le(\|A\|/\sqrt3)^3\), 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 (Mahler 1946, sec. 2, Theorem 1). For such a basis, order its lengths increasingly and write its defect as \(\omega=\prod_i\|b_i\|/\mathop{\mathrm{covol}}(\mathfrak a)\ll1\). 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 \[\#(\mathfrak a\cap B_Y)\ll \prod_{i=1}^3(1+Y/\lambda_i).\] Minkowski’s second theorem for the Euclidean ball, in the formulation of (Widmer 2010, Theorem 4.3), gives \(\lambda_1\lambda_2\lambda_3\asymp\mathop{\mathrm{covol}}(\mathfrak a)\). Expanding the last product, using \(\lambda_2\ge\lambda_1\), and inserting the first-minimum and covolume bounds proves Equation (116). The same coordinate functional bound with \(\|b_i\|\ge\lambda_1\) proves the final assertion. ◻

Proof. We use \(\lambda=1/3\). Lemma 26, which gives \(\Delta_F\ge X^{1/2}/2\), implies \[ \Delta_F\ge X^\lambda \tag{118}\] 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, \[ |\widehat\phi(A)|\ll_r X^{r\tau}(1+\|A\|)^{-r}. \tag{119}\] Indeed each derivative of order \(r\) costs at most a constant times \(X^{r\tau}\), and all supports are in one compact set. Lemma 32 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 (119), for \(r>3\), yields \[ \sum_{\substack{A\in F\\\|A\|>Y}} |\widehat\phi(A)\widehat\eta_{\mathfrak h,l}(A)| \ll_r X^{r\tau}l^3Y^{3-r} \le X^{r\tau+3\theta+w(3-r)}\,O_r(1). \tag{120}\] This also proves absolute convergence in Equation (114). 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 \(\mathbb Z[lHA]\) 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 \[D_F\ll(lHY)^6,\qquad H\gg X^{d/6-s}.\] Apply Equation (116) to \(\Gamma_{l,\mathfrak h}\) and multiply it by \(H^{-1}\). The four resulting terms are bounded by constants times \[ X^{-d/6+s} +X^{-d/9+5s/3} +X^{-d/18+7s/3} +X^{-d/2+3s}. \tag{121}\] 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 (118) gives \[[R_F:C_F]=(\Delta_F/D_F)^{1/2}\ge X^{\lambda/3}.\] Profiles with \(G_0\le X^e\) are best estimated in their original form \(S_{\mathfrak h}\). Equation (108) 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 (111). The first-minimum bound of Lemma 36, applied in the fixed real region, gives \(O(D_FG_0^{24})\) possible points. At each such point, Equations \(\eqref{eq:add-profile-norm}\)–\(\eqref{eq:add-profile-inverse}\) give \[\left|\prod_{p\in S}K_{p,\mathfrak h_p}(z)\right| \ll X^{o(1)} \left(\frac{D_FG_0^{24}}{H}\right)^{1/2}.\] 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 \[ |S_{\mathfrak h}| \ll X^{-\lambda/6+(3/2)d+(73/2)e+o(1)}. \tag{122}\] 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 \[ O\bigl(H^{-1}+YlH^{-2/3}\bigr). \tag{123}\] By \(H\ge X^{\lambda/3-e}\) this also saves a power for the choices below. Equation (114) expresses the full dual sum as \(S_{\mathfrak h}\). Subtracting its scalar frequencies in the ball and using Equation (120) for the scalar tail therefore bounds the full nonscalar sum for these profiles by Equations (122)–(123) 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 30, and write \[\alpha_p=A+s_p,\qquad \beta_p=\alpha_p^{-1},\qquad J=\prod_{p\in S}p^{j_p},\] 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 \[|\widehat\eta_{\mathfrak h,l}(A)| \ll H^{-1}J^{-1/2}.\] For frequencies with \(J\ge X^{e_2}\), the rough lattice count immediately gives \[ \sum_{\substack{\|A\|\le Y\\J\ge X^{e_2}}} |\widehat\phi(A)\widehat\eta_{\mathfrak h,l}(A)| \ll \frac{(1+YP^{1/3})^3}{H}\,X^{-e_2/2} \ll X^{3s-e_2/2}. \tag{124}\] 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 \(\operatorname{disc}(A)=\mathop{\mathrm{Disc}}(1,A,A^2)\in\mathbb Q^\times\). The equality \(F=\mathbb Q[A]\) remains \(F\otimes\mathbb Q_p=\mathbb Q_p[A]\) after tensoring. Shifting by a scalar and inverting an element with nonzero components preserve the generated algebra, so \(p^{j_p}\beta_p\) generates \(F\otimes\mathbb Q_p\). The full-rank order \(\mathbb Z_p[p^{j_p}\beta_p]\) is contained in \(C_{F,p}\). The identities \[\operatorname{disc}(\beta_p) =\frac{\operatorname{disc}(A)} {\mathop{\mathrm{N}}_{F_p/\mathbb Q_p}(\alpha_p)^4}, \qquad \operatorname{disc}(p^{j_p}\beta_p) =p^{6j_p}\operatorname{disc}(\beta_p)\] follow by taking the three differences of conjugates and by scaling. Discriminant divisibility therefore gives \[v_p(\operatorname{disc}(A)) -4v_p(\mathop{\mathrm{N}}\alpha_p)+6j_p\ge v_p(\Delta_F).\] The shift lemma says \(v_p(\mathop{\mathrm{N}}\alpha_p)=-\log_pH_p\) 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 \[ Y^6\gg |\operatorname{disc}(A)| \gg \Delta_F\,J^{-6}H^{-4}l^{-6}. \tag{125}\] The first inequality uses \(\|A\|\le Y\). Thus for \(J<X^{e_2}\), Equations \(\eqref{eq:add-order-lower}\) and \(\eqref{eq:add-frequency-discriminant}\) imply \[ H\gg X^{\lambda/4-(3/2)(s+e_2)}. \tag{126}\] 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 36 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 (126), gives \[ L\asymp YP^{1/3},\qquad L\gg X^{\lambda/24},\qquad L^3/H\ll X^{3s}. \tag{127}\]

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 \(\mathbb Z_p\)-basis of \(R_{F,p}\), because \(p\nmid l\). Accordingly the map from the three integer coordinates to \(x\) is an affine \(\operatorname{GL}_3(\mathbb Z_p)\) 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 \(\ell_p:R_{F,p}\to\mathbb Z_p\), meaning that its image is \(\mathbb Z_p\), and an integer \(g_p'\ge g_p/3\) such that \[\ell_p\bigl(i_p^{-1}(\mathfrak h_p\cap C_{F,p})\bigr) \subset p^{g_p'}\mathbb Z_p.\] 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 \[ \ell_p(x^{-1})\equiv0\pmod {p^k} \quad\text{for every }1\le k\le(g_p'-j_p)_+. \tag{128}\] 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 (128) 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 (128) holds for \(y\) in place of \(x^{-1}\). On the permitted locus the inverse identities \[y(x)-y(x')=p x^{-1}(x'-x)(x')^{-1},\qquad x-x'=p^{-1}x\bigl(y(x')-y(x)\bigr)x'\] 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 \[ \prod_{p\in S}p^{a_p} \ge G_0^{1/3}/J > X^{e/3-e_2}. \tag{129}\] 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 \(\eqref{eq:add-available-congruences}\) ensures this is possible, and \[X^{e_1/2}\le M_1\le X^{3e_1/2}<X^{2e_1}.\] 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 (127). Counting each admissible residue class in the cube then gives \[ O(L^3/M_1)=O(L^3X^{-e_1/2}) \tag{130}\] 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 \(\mathcal N_B\ne0\), \[\ell_p(x^{-1})=\frac{P_\ell(x)}{\mathcal N_B(x)},\qquad P_\ell(x)=\ell_p\bigl(\operatorname{adj}(M_x)1_B\bigr),\] 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 \[ O\bigl(L^3/p+L^2\bigr) =O\bigl(L^3(X^{-e_1}+X^{-\lambda/24})\bigr). \tag{131}\] This bounds the subset for which \(x\) is a unit as well. Multiplying Equations \(\eqref{eq:add-composite-congruence-count}\)–\(\eqref{eq:add-prime-congruence-count}\) by \(H^{-1}\), and using Equation (127), gives respectively the bounds \[O(X^{3s-e_1/2}),\qquad O(X^{3s-e_1}+X^{3s-\lambda/24})\] 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 \[ \begin{aligned} d=e&=\frac{\lambda_0}{1000},& e_1&=\frac e{100},& e_2&=\frac{e_1}{100},\\ w&=\frac{e_2}{100},& \theta=\tau&=\frac w{100},& \epsilon_F&=\frac{e_2}{4}. \end{aligned} \tag{132}\] Then \(s=101e_2/10000\), \[s<d/100,\quad d<\lambda/3,\quad e/3-e_2>2e_1,\quad 2e_1<\lambda/24,\quad s+e_2<\lambda/12.\] The saving in Equation (122) is at least \[\lambda/6-(3/2)d-(73/2)e \ge (1/6-0.038)\lambda_0,\] and Equation (123) has a still larger margin than \(\epsilon_F\), since \(e\le\lambda/1000\). All four exponents in Equation (121) also have margins at least \(29d/900>\epsilon_F\), by \(s<d/100\). Among the remaining ball estimates the smallest margin is in Equation (124), namely \[e_2/2-3s=\frac{4697}{10000}e_2>\epsilon_F;\] the congruence margins \(e_1/2-3s\), \(e_1-3s\), and \(\lambda/24-3s\) are larger. With \(r=100\), the exponent in Equation (120) is \[r\tau+3\theta+w(3-r)=-\frac{9597}{10000}e_2.\] 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 (117). Finally \(2\tau+3\theta=e_2/2000<\epsilon_F\), as asserted. ◻

Proof of Proposition 34. Use the constants in Equation (132). Equations (113) and \(\eqref{eq:add-poisson}\), together with Lemma 37, leave the scalar frequencies with absolute error \(O(X^{-\epsilon_F})\). We first remove the profile cutoff from those frequencies.

For \(a\in\mathbb Q\), 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 29 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 \((tl)^{-1}\mathbb Z\). The change of real variables \((z_i,z_j,z_k)\mapsto(z_i,z_j,\mathop{\mathrm{Tr}}z)\) has absolute Jacobian one. Writing \(\mathbf 1=(1,1,1)\), it gives the exact identity \[\widehat\phi(a\mathbf 1)=A_\phi\,\widehat\chi(a).\] For any fixed \(r>1\), rapid decay of the fixed function \(\chi\) implies \(\sum_{n\in\mathbb Z}|\widehat\chi(n/(tl))|\ll_r tl\). The scalars omitted by \(H\le X^{C_3}\) therefore contribute at most \[ \sum_{t>X^{C_3/3}}t^{-3} \sum_{n\in\mathbb Z}|\widehat\phi((n/(tl))\mathbf 1)| \ll A_\phi\,l\,X^{-C_3/3}. \tag{133}\] 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 \(d_p(t)\mathbf 1_{\mathbb Z_p}(t)\). 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 \[p^{-2v_p(l)}\mathbf 1_{m+p^{v_p(l)}\mathbb Z_p}(t)\] with respect to Haar measure of mass one on \(\mathbb Z_p\). At every other finite prime the marginal is \(\mathbf 1_{\mathbb Z_p}\). Their Fourier transforms at a rational \(a\) are exactly the finite factors in Equation (114) evaluated at the scalar \(a\), because \(\mathop{\mathrm{Tr}}(az)=a\mathop{\mathrm{Tr}}z\).

For completeness, this one-dimensional application of Poisson does not require a smoothness assumption on \(d_p\). Its Fourier coefficients on \(\mathbb Q_p/\mathbb Z_p\) 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 \(\mathbb Z_p\), to which Poisson applies. The densities converge uniformly by Lemma 29, 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 \[A_\phi\,l^{-2}\prod_{p\in S}d_p(m).\]

Lemma 29 bounds the product of \(d_p(m)\) above and below by positive absolute constants. Also \(A_\phi\gg X^{-2\tau}\) by Equation (106), 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 (107) with its stated uniformity. ◻

Proof. For each residue class \(a\), Equation (107) gives \[\nu_\phi(a)=V_\phi l^{-2}(1+\epsilon_a), \qquad |\epsilon_a|\le C_{\mathrm{add}}X^{-\theta},\] 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 \(C_{\mathrm{add}}V_\phi X^{-\theta}l^{-2}\sum_a W(a) =C_{\mathrm{add}}V_\phi X^{-\theta}\mathbb E_l W\). This also covers \(\mathbb E_l W=0\). In particular the error retains the local mean even when that mean is very small. ◻

Proof. We use the Selberg upper-bound square (Selberg 1947); see also (Koukoulopoulos 2019, chap. 21). Its calculation is included because the relative error in Lemma 38 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 \[\vartheta(d)=\prod_{p\mid d}\vartheta_p,\qquad h(d)=\prod_{p\mid d}\frac{\vartheta_p}{1-\vartheta_p}\] 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 \[\left(\sum_{d\in\mathcal D}\lambda_d \prod_{p\mid d}\mathbf 1_{\{n_p\ge1\}}\right)^2\] majorizes the indicator that every prime of \(\mathcal I\) is inactive. Its expectation in the independent product of the local probability spaces is \[ \sum_{d,e\in\mathcal D}\lambda_d\lambda_e\vartheta([d,e]) =\sum_{l\in\mathcal D}\frac1{h(l)} \left(\sum_{\substack{d\in\mathcal D\\l\mid d}} \lambda_d\vartheta(d)\right)^2. \tag{146}\] 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 \[\sum_{\substack{d\in\mathcal D\\l\mid d}}\lambda_d\vartheta(d) =\frac{\mu_{\mathrm{Mob}}(l)h(l)}{\mathcal H}, \qquad l\in\mathcal D,\] where \(\mu_{\mathrm{Mob}}\) is the Möbius function. Explicitly, \[\lambda_d= \frac{\mu_{\mathrm{Mob}}(d)h(d)} {\vartheta(d)\mathcal H} \sum_{\substack{r\le T_0/d\\(r,d)=1\\dr\in\mathcal D}}h(r).\] Consequently \(\lambda_1=1\), the value of Equation (146) is \(\mathcal H^{-1}\), and \[ |\lambda_d| \le\frac{h(d)}{\vartheta(d)} =\prod_{p\mid d}(1-\vartheta_p)^{-1} \le2^{\omega(d)}\le T_0. \tag{147}\]

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 \[\mathbb E\log d=\sum_{p\in\mathcal I}\vartheta_p\log p \le C_*\log Y<\tfrac14\log T_0.\] Markov’s inequality shows that at least half the mass lies at \(d\le T_0\). It follows that \[ \mathcal H\ge\tfrac12\prod_{p\in\mathcal I}(1+h(p)), \qquad \mathcal H^{-1}\le2\prod_{p\in\mathcal I}(1-\vartheta_p). \tag{148}\]

It remains to average this square against \(\nu_\phi\) together with the assigned weights. Write \[A_{\mathcal A}= \prod_{p\in\mathcal A}\mathbb E_p \bigl(b_p\mathbf 1_{\{n_p=j_p\}}\bigr).\] The assigned weight is defined modulo \[\prod_{p\in\mathcal A}p^{C(j_p+1)} \le\left(\prod_{p\in\mathcal A}p^{j_p}\right)^{2C} \le T_0^{4C}.\] 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 \[ T_0^{8C}\le X^\theta \tag{149}\] 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 38 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 (147), each coefficient product has absolute value at most \(T_0^2\). Summing the errors absolutely costs at most \[ C V_\phi A_{\mathcal A}X^{-\theta}T_0^4. \tag{150}\] 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 (146). Finally, \(X^{-\theta}T_0^4=o((\log X)^{-C_1})\) by Equation (143). Equations (144) and (148) absorb Equation (150) and prove the lemma. ◻

Proof of Proposition 39. Put \(P_Y=\prod_{p\in\mathcal P_Y}\mathbb E_p b_p\). Equations (139) and (144) imply \[ P_Y\gg(\log X)^{-C_1}. \tag{151}\] First observe that primes exceeding \(Y\) cost only a fixed factor. Equations (140) and (141), together with \(b_p\le A_0^{n_p}\), give \[ \prod_{\substack{p>Y\\p\notin S}}b_p(z) \le A_0^{\sum_{p>Y,\ p\notin S}n_p(z)} \le \exp(B\log A_0/\delta). \tag{152}\]

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 40. 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 \[V_\phi\prod_{p\in\mathcal P_Y} \left(1-\vartheta_p+ \sum_{j\ge1}\mathbb E_p(b_p\mathbf 1_{\{n_p=j\}})\right) =V_\phi P_Y.\] Together with Equation (152), 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 \[p^{C(J_p+1)}\le T_0^C Y^{2C}\le X^\theta.\] Lemma 38 and Equation (138) 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 (141), 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 (151).

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 38 therefore bounds the measure of the union by \(O_{\mathcal Q}(V_\phi T_0^{-1/N})\). After multiplication by Equation (141) 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 \[ \eta M_0>2B\log A_0+1,\qquad M_0/\log y_0<\tfrac14. \tag{153}\] 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: \[ \sqrt y<p\le y,\qquad y=Y^{1/2^i},\quad i=0,1,2,\ldots. \tag{154}\] 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 \[ \exp\left(\frac{2B\log A_0}{\log y}\log X\right). \tag{155}\] For each assignment in the group, require inactivity at its unassigned primes in \(\mathcal P_{\sqrt y}\) and apply Lemma 40. We use Rankin’s power majorant (Rankin 1938): since the assigned product exceeds \(T_0\), its upper bound can be multiplied by \[T_0^{-s}\prod_{\text{assigned }p}p^{s j_p}\ge1, \qquad s=M_0/\log y.\] 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 \(\sum_{j\ge1}\mathbb E_p(b_p\mathbf 1_{\{n_p=j\}})p^{sj}\). Their sum is \(\mathbb E_p(b_p p^{s n_p})\). 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 \(\mathbb E_p(b_p p^{s n_p})+\vartheta_p\). Thus the group’s contribution is at most \[ C V_\phi T_0^{-s} \exp\left(\frac{2B\log A_0}{\log y}\log X\right) \prod_{p\in\mathcal P_{\sqrt y}}\mathbb E_p(b_p p^{s n_p}) \prod_{\substack{\sqrt y<p\le y\\p\notin S}} \bigl(\mathbb E_p(b_p p^{s n_p})+\vartheta_p\bigr). \tag{156}\]

We estimate these local moments uniformly for \(y>y_0\). By Equation (153), \(s<1/4\). The total contribution of depths \(j\ge2\) to \(\mathbb E_p(b_p p^{s n_p})\) is at most \[C^2\sum_{j\ge2}(1+j)^C p^{-(1-s)j}\ll_C p^{-3/2}.\] At depth one, the increase over \(\mathbb E_p b_p\) is bounded by \[\frac{C'}p(p^s-1) \ll_{M_0}\frac{\log p}{p\log y} \qquad(p\le y),\] 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 \(\mathbb E_p b_p\ge1/2\), comparison of the products in Equation (156) with their base means gives \[\prod_{p\in\mathcal P_{\sqrt y}}\mathbb E_p(b_p p^{s n_p}) \prod_{\substack{\sqrt y<p\le y\\p\notin S}} \bigl(\mathbb E_p(b_p p^{s n_p})+\vartheta_p\bigr) \ll_{M_0}\prod_{p\in\mathcal P_y}\mathbb E_p b_p.\] For a fixed \(C_2\), a final use of Equation (139) and Mertens’ estimate yields \[\prod_{p\in\mathcal P_y}\mathbb E_p b_p \le P_Y\prod_{\substack{y<p\le Y\\p\notin S}} (1-\vartheta_p)^{-1} \ll P_Y\left(\frac{\log Y}{\log y}\right)^{C_2}.\] By Equation (153), the bound for the group in Equation (156) is consequently at most \[C_{M_0}V_\phi P_Y \exp\left(-\frac{\log X}{\log y}\right) \left(\frac{\log Y}{\log y}\right)^{C_2}.\] For the values of \(y\) in Equation (154), its sum is bounded by \[C_{M_0}V_\phi P_Y \sum_{i\ge0}\exp(-2^i/\delta)\,2^{iC_2} \ll V_\phi P_Y.\] Together with the small-product part and the two discarded sets, this proves Equation (142). ◻

Proof of Theorem 19. 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 \[A_\phi\asymp\operatorname{area}(\mathcal R).\] The constants are uniform over the positions, side lengths, and aspect ratios of the rectangles. Proposition 24 and Proposition 39 give \[ \sum_{\substack{\mathop{\mathrm{Tr}}z=m\\z\in\mathcal R}}M(z) \ll \sqrt D\, A_\phi \prod_{p\in S}(1-p^{-1})Z_p \prod_{p\in\mathcal P_Y}\mathbb E_p b_p, \qquad Z_p=\prod_{\substack{\mathfrak p\mid p\\\mathfrak p\subset R}} (1-\mathop{\mathrm{N}}(\mathfrak p)^{-1})^{-1}. \tag{157}\] Here \(\prod_{p\in S}|D|_p^{-1/2}=\sqrt D\), because \(S\) contains every prime divisor of \(D\), and Equation (135) 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 23 is \[\mathbb E_p b_p=(1-p^{-1})(1-p^{-3})Z_p\qquad(p\notin S).\] Since \(1-p^{-3}\le1\), the two Euler products on the right of Equation (157) 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, \[\log Z_p=\frac{a_K(p)}p+O(p^{-2}).\] Thus \[ \prod_{p\in S}(1-p^{-1})Z_p \prod_{p\in\mathcal P_Y}\mathbb E_p b_p \ll \exp\left(\sum_{p\in S\ \mathrm{or}\ p\le Y} \frac{a_K(p)-1}{p}\right). \tag{158}\]

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 (158) is at most \[\sum_{p\le Z}\frac{a_K(p)}p-\sum_{p\le Y}\frac1p.\] We have \(Z\ge D^{1/4}\) because \(X=q\sqrt D\ge\sqrt D\). Equation (25) and Mertens’ estimate now give \[\exp\left(\sum_{p\in S\ \mathrm{or}\ p\le Y} \frac{a_K(p)-1}{p}\right) \ll \frac{\kappa\log Z}{\log Y} =\frac c\delta\,\kappa.\] This also applies when \(D\) remains bounded and \(q\) tends to infinity, by the stated uniform range of Lemma 6. Substitution in Equation (157) proves \[\sum_{\substack{\mathop{\mathrm{Tr}}z=m\\z\in\mathcal R}}M(z) \ll\kappa\sqrt D\,\operatorname{area}(\mathcal R).\] 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 19. ◻

Proof. We use the packet integral in Equation (11). For its maximal-order ideal representative \(I\) and local label \(L'\in\mathscr L_{\mathcal O}(I)\), the normalized lattice at the sign vector \(s=(s_1,\ldots,s_4)\in\{\pm1\}^4\) and the logarithmic parameter \((u_1,\ldots,u_4)\in\mathbb R^4\) with \(u_1+\cdots+u_4=0\) is \[(X\mathop{\mathrm{N}}I)^{-1/4}\sigma(L') \mathop{\mathrm{diag}}(s_1e^{u_1},\ldots,s_4e^{u_4}).\] For a raw vector \(w\in\bigwedge^2 I\), its normalized exterior coordinates are \[y_{ab}=(X\mathop{\mathrm{N}}I)^{-1/2}s_as_b e^{u_a+u_b}w_{ab}.\] If \(z=\Psi(w)/\mathop{\mathrm{N}}I\) and \(\epsilon(s)=\prod_a s_a\), then Equation (53) gives \[ (y_{12}y_{34},-y_{13}y_{24},y_{14}y_{23}) =\epsilon(s)z,\qquad Q(y)=\epsilon(s)\mathop{\mathrm{Tr}}z. \tag{162}\] 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 (11) 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 \[\frac1{N_{\mathcal O}}\sum_{L'\in\mathscr L_{\mathcal O}(I)} \mathbf 1_{\{w\in\bigwedge^2L'\}} =\prod_p P_p(I,w),\] the weight in Equation (55). This weight is constant on its \(U^+\)-orbit. That action is free: every embedding coordinate of the nonzero raw vector \(w\in\bigwedge^2_{\mathbb Z}I\subset\bigwedge^2_{\mathbb Q}K\) is nonzero by Lemma 18. 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 \(h_K\mathop{\mathrm{vol}}(\mathscr F_U)=\sqrt D\,\kappa\) by Equation (8). 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 \[u_1+u_2,\qquad u_1+u_3,\qquad -u_2-u_3.\] 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 (55). Equation (162) 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 (161). 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 \[\Omega_t=\{z\in\mathcal H_t:|z_{b_2}|,|z_{b_3}|\le1\}.\] Now let \(0<r<r_0\). In the trace-\(t\) sum in Equation (161), 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 \(t\), \[|z_{b_1}|\ge |m|-2r^2\ge |m|/2.\] Thus the first logarithm in Equation (160) 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 27 gives \(|z_b|\gg_{\Omega_t}X^{-84}\) for each \(b\). Hence the last two logarithms in Equation (160) are \(O(\log X)\), uniformly for \(0<r<1\).

We first remove the points in the two strips \[ |z_{b_2}|\le4X^{-\tau} \quad\text{or}\quad |z_{b_3}|\le4X^{-\tau}. \tag{164}\] 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 19 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 (161) is \[O_{m,L}\bigl(X^{-\tau}(\log X)^2\bigr)=o(1).\] 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 \[r^2 2^{-a-1}<|z_{b_2}|\le r^2 2^{-a},\qquad r^2 2^{-b-1}<|z_{b_3}|\le r^2 2^{-b}, \qquad a,b\ge0.\] 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 19; any boundary rectangle may be included in full. Its area is \(O(r^4 2^{-a-b})\), while the last two factors of Equation (160) are at most \((a+1)\log2\) and \((b+1)\log2\). Theorem 19 gives \[\frac1{\sqrt D\,\kappa} \sum_{z\text{ in that rectangle}} M(z) \ll_{m,L} r^4 2^{-a-b}.\] The sum over the four choices of signs and over all \(a,b\) is therefore at most \[C_{m,L}r^4 \sum_{a,b\ge0}(a+1)(b+1)2^{-a-b} \ll_{m,L}r^4.\] 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 (161). These observations prove Equation (163) for both traces and all three partitions. ◻

Proof of Theorem 1. For the sequence in that theorem, \(X_i=\sqrt{|\mathop{\mathrm{Disc}}(\mathcal O_i)|}\) tends to infinity. By Proposition 9, the packet probabilities are tight, and every subsequence has a further weakly convergent subsequence with an \(A_4\)-invariant probability limit \(\mu\). Proposition 11 gives the local ball bound required by Corollary 14. Hence almost every component in the \(A_4\)-ergodic decomposition of \(\mu\) has positive entropy for a diagonal flow. Proposition 17 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 \[E_{m,L,r}=\{\Lambda\in X_4:N_{m,L,r}(\Lambda)\ge1\}\] is open by the observation following Equation (159).

The open-set inequality for weak convergence and Proposition 42 now give, along the convergent subsequence, \[\mu(E_{m,L,r}) \le\liminf_i\mu_{\mathcal O_i,\sigma_i}(E_{m,L,r}) \le\limsup_i\int N_{m,L,r}\,d\mu_{\mathcal O_i,\sigma_i} \le C_{m,L}r^4\] for every sufficiently small fixed \(r>0\). In particular, \[\mu\left(\bigcap_{n\ge2}E_{m,L,1/n}\right) \le \inf_{\substack{n\ge2\\1/n<r_0}} C_{m,L}n^{-4}=0.\] Every vector in one of the obstruction planes with \(Q\ne0\) has a nonzero integer \(Q\)-value by Equation (45), 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 (47) satisfies \[\mathcal E_{b_1} \subset \bigcup_{m\in\mathbb Z\setminus\{0\}}\ \bigcup_{\substack{L\in\mathbb Z\\L\ge2}} \bigcap_{n\ge2}E_{m,L,1/n}.\] 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 17 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. ◻