Equidistribution of Prime-Degree Torus Packets with Arbitrary Local Type

Proof. Surjectivity follows from (2.3). A global homothety changes a normalized embedded lattice by an element of C. Since n is odd, each pair of opposite sign patterns has exactly one representative in W_n; a common negative sign does not change a lattice.

For injectivity, suppose two displayed normalized lattices coincide. Absorb the ratio of the normalizing scalars into the real diagonal map carrying \sigma(N) onto \sigma(N'). Applying this map to one nonzero \alpha\in N gives \sigma(\beta) for some \beta\in N'. None of the coordinates of \sigma(\alpha) vanishes, so the map is multiplication by \sigma(\beta/\alpha). Thus N'=bN for b\in K^\times. Their maximal-order spans represent the same ideal class and therefore equal the same chosen I. Hence bI=I, so b\in E, and the parameters differ by exactly the stated action.

Combine an N label with a sign component of C. Its stabilizer in E is the totally positive subgroup of \{\epsilon\in E:\epsilon N=N\}, and has finite index in E. It acts faithfully by logarithmic translations. By injectivity, these are all periods of the corresponding A_n-orbit. Each orbit of labels therefore contributes exactly one compact A_n-orbit, with precisely its prescribed Haar volume. This proves both finiteness and the measure assertion. One may equally use \mathcal N(I)\times F as a fundamental domain, since the E-action preserves counting measure times dt. The total measure of these fundamental domains is h_KT\vol(F). Dividing by this quantity gives precisely the uniform averages over ideal classes and finite labels, followed by integration against dt/\vol(F).□

Proof. Lemma 2.1 expresses the left side as

The map \alpha\mapsto(\alpha)I^{-1} runs over the integral ideals in the inverse class of I, each fiber being one free E-orbit. Summing that orbit and integrating over F unfolds to an integral over C, by Tonelli. The absolute coordinate product of the normalized vector is \N((\alpha)I^{-1})/Q. Translating the first n-1 logarithms and relabeling the signs identifies the C-integral with (2.5), with Jacobian one. Every integral ideal occurs in exactly one inverse class. Equation (2.4) supplies the coefficient in (2.6).□

Proof. Normalize additive Haar measure on R_\ell to have mass one. The valuation shell \mathbf k_\ell has mass Z_\ell^{-1}/d_\ell(\mathbf k_\ell). Its conditional law is invariant under the transitive unit action, so P_\ell is exactly the fraction of this shell lying in L_\ell. Since L_\ell has mass 1/q_\ell, summing over shells and then multiplying over S proves (3.1). Elements with a zero component have additive measure zero.

Choose an additive-Haar uniform element of L_\ell, and denote its component valuations by v_{\mathfrak p}. The condition R_\ell L_\ell=R_\ell implies that the projection to every component contains a unit. Modulo the kth power of its maximal ideal, the projection therefore contains all \Z_\ell multiples of that unit, at least \ell^{\lceil k/e_{\mathfrak p}\rceil} elements. Uniform measure on L_\ell induces uniform measure on this finite additive image. Consequently

This argument does not require the component projections to be independent, or even to equal their full integer rings.

Put j=\max_{\mathfrak p\mid\ell} \lceil v_{\mathfrak p}/e_{\mathfrak p}\rceil. Then d_\ell(\mathbf v_\ell)\le\ell^{nj}. If j\ge a\ge1, some v_{\mathfrak p} is at least e_{\mathfrak p}(a-1)+1. Equation (3.4) and a union bound give \Pr(j\ge a)\le n\ell^{-a}. Hence

Using the shell masses once more yields the exact factorization

Each factor is at most (1-\ell^{-1})^{-n}(1+C_n\ell^{-1/2}). For fixed \eps>0 this is at most \ell^\eps at all sufficiently large primes; the finitely many remaining primes contribute a constant depending only on n,\eps. As \prod_{\ell\in S}\ell\le q, this proves (3.2). Finally, take \eps=s/16 and use q\le Q to obtain

□

Proof. Put L=\log D. Stark’s Lemma 3 gives at most one zero in \operatorname{Re}s\ge1-(4L)^{-1}, |\operatorname{Im}s|\le(4L)^{-1}; such a zero is real and simple. His Lemma 8, which also applies to nonnormal fields, says that a real zero \beta\ge1-(4n!L)^{-1} forces a quadratic subfield of K [14, Lemmas 3 and 8]. Since n is odd, there is no such subfield. Consequently, for a constant c_n>0, every nontrivial zero \rho satisfies |1-\rho|\ge c_n/L.

The real logarithmic derivative of the completed zeta function gives

where zeros are counted with multiplicity and

This is the Hadamard identity in [14, Equation (9)]. The zero terms are nonnegative, G_n is bounded on [1,2], and the Euler product gives \zeta_K'/\zeta_K\le0 there. Thus, at \sigma_0=1+L^{-1}, S_K(\sigma_0)\ll_n L.

This bound persists uniformly for 1<\sigma\le1+4/L. Indeed, for \rho=\beta+i\gamma, set X=L(1-\beta), Y=L\gamma, and U=L(\sigma-1). Then X\ge0, 0<U\le4, and |X+iY|\ge c_n. Hence

Summing gives 0\le S_K(\sigma)\ll_n L. For F_K(\sigma)=(\sigma-1)\zeta_K(\sigma), it follows that

Since F_K extends analytically to 1 with F_K(1)=\kappaK>0, integration over an interval of length at most 4/L proves the comparison in (4.1). At \sigma_0, the inequality \zeta_K(\sigma_0)\ge1 also gives \kappaK\gg_n L^{-1}.

For completeness, only finitely many fields of this fixed degree have discriminant below the cutoff. One can see this directly here: nonzero vectors of D^{-1/(2n)}\sigma(R) have absolute coordinate product at least D^{-1/2}, so Mahler’s criterion puts these lattices in a fixed compact set. Local basis lifts and a finite cover give bases of R with uniformly bounded conjugates. A basis element outside \Q generates K, since n is prime; its monic integral minimal polynomial therefore has bounded coefficients. There are only finitely many such polynomials. The simple pole of each of these finitely many zeta functions now gives the remaining claims uniformly for sufficiently small h>0.□

Proof. Set b(m)=a_K(m)\one_{(m,q)=1}. This function is nonnegative, multiplicative, and bounded by d_n. Indeed, at every prime \ell, including the ramified primes, the local generating series for a_K is \prod_{\mathfrak p\mid\ell}(1-T^{f_{\mathfrak p}})^{-1}; it is coefficientwise bounded by (1-T)^{-n}. Decomposing an ideal into its part over S and its coprime part, the sum in (4.3) becomes

First suppose d=d(\mathbf k)\le Q^{1/8}, and put u=x/d. Then u\ge Q^{5/8}\ge D^{1/4}. For h=1/\log u, Lemma 4.1 gives \zeta_K(1+h)\ll_n\kappaK\log u. There is no upper restriction on u: increasing u only decreases h, so the lemma applies even when D stays bounded and q grows. Positivity of the Euler logarithm yields

Since a_K(\ell)\le n and \sum_{\ell\le u}(\log\ell)/\ell\ll\log u, replacing \ell^{-1-h} by \ell^{-1} costs only O_n(1). The latter prime-sum bound follows, for example, by comparing \sum_{\ell\le u}\lfloor u/\ell\rfloor\log\ell with \log(\lfloor u\rfloor!). Therefore

Furthermore,

The first error is bounded by a constant times \sum_\ell\ell^{-2}. Removing the primes of S thus gives

For Q sufficiently large at fixed \eta, the length v=y/d\ge\eta u is at least u^{1/2}. Applying (4.2), then (4.5) and (3.1), bounds the contribution of these d by

For d>Q^{1/8}, terms with d>x vanish because the inner sum in (4.4) then contains no positive integer. For the remaining terms, x/d\ge1, and the elementary bound \sum_{m\le z}d_n(m)\le z(1+\log z)^{n-1} for z\ge1 follows by summing reciprocals in the first n-1 factors. Using (3.3), with s=1/(2n), the remaining contribution to (4.4) is at most

This is uniform in the stated range: Lemma 4.1 gives \kappaK^{-1}\ll_n\log(3D), and D\le Q^2. At fixed A_0,\eta, increase the threshold until this term is at most \kappaK\eta x/q\le\kappaK y/q. The implied constant has not changed with \eta, proving the proposition.□

Proof. For f=\one_{[-\theta,\theta]^n}, the first n-1 logarithms in (2.5) are at most \log\theta. The last coordinate bounds the sum of their gaps from \log\theta by \log(\theta^n/z). Thus

Put X=\theta^nQ and U=Q^{3/4}. Up to an n-dependent constant, Proposition 2.2 reduces the first moment to

If X<1, the sum is empty. Otherwise its logarithmic factor is at most (\log Q)^{n-1}, since \theta<1. The terms with \N\mathfrak a\le\min(U,X) contribute at most C_nQ^{-1/4}(\log Q)^{n-1}, by applying Proposition 4.2 at x=y=U. When X>U, define the truncated cumulative sum

Writing each logarithmic power as an integral and applying Tonelli gives

Here the same proposition at x=y=t bounds A_U(t), and the last inequality follows from \int_0^1(\log(1/v))^{n-2}\dd v<\infty. There is no lower boundary term, since A_U(U)=0, and the logarithmic weight vanishes at X. If X\le U, this part is absent. Dividing by \sqrt D\,\kappaK gives O_n(\theta^n). Since E_f\ge1 whenever the displayed cusp event occurs, this proves (5.1).

Mahler’s compactness criterion [9, Section 3, Theorem 2] says that the complement of this cusp event is compact. Given a desired mass error, choose \theta first and then Q sufficiently large. The finitely many earlier packet supports are compact by Lemma 2.1, so the whole sequence is tight. The lattice space is locally compact and second countable; weak compactness for tight probabilities gives probability subsequential limits. Invariance passes to them by testing against continuous compactly supported functions.□

Proof. First fix a compact V\subset(\R^\times)^n and b_0\ge1. For w\in V and small r, take f=\one_{w+[-b_0r,b_0r]^n}. Each coordinate has a fixed sign, is uniformly separated from zero, and has logarithmic width O_{V,b_0}(r). Its absolute product lies within Cr of z_w=\prod_j|w_j|, so

for some C\ge1 depending only on V,b_0,n. Choose r_0 small enough that z_w>4Cr_0 on V. Use Proposition 4.2 with

This slightly enlarged interval contains every contributing ideal norm. Uniformly in w, for sufficiently large Q it has Q^{3/4}\le x\le A_0Q with a fixed A_0\ge1, and \eta x\le y\le x with \eta=4Cr/A_0\le1, after decreasing r_0 if necessary. For every fixed r\in(0,r_0), Proposition 2.2 therefore gives, eventually,

Here Q=q\sqrt D. The implicit constant is independent of r. The threshold for Q may depend on r, which is all that will be needed.

We next choose a suitable lattice vector at each x\in\Omega. Every full lattice has a vector with no zero coordinate: for any basis, its combination with coefficients (1,N,\ldots,N^{n-1}) has coordinates that are nonzero polynomials in N, so an integer N can avoid their finitely many roots. Near each center choose a continuous local lift to basis matrices, and continue the chosen vector using its fixed integer basis coefficients. On a sufficiently small neighborhood with compact closure in that chart, its coordinates remain bounded and bounded away from zero. A finite cover of \Omega by such neighborhoods therefore gives choices w_x lying in one compact V\subset(\R^\times)^n. This pointwise selection need not be continuous or measurable.

For h in B(r), each coordinate of w_xh-w_x has absolute value at most r\sum_i|(w_x)_i|. Thus one fixed b_0 makes every lattice in xB(r) contain a nonzero vector in w_x+[-b_0r,b_0r]^n. Equation (5.3) bounds the packet mass of xB(r) by C_\Omega r^n. This set is open, so Portmanteau gives

The argument holds for every fixed x and r with the same constant on \Omega. It proves (5.2) for all 0<r<r_\Omega, without any exchange of shrinking-radius and packet limits.□

Proof. Fix 0<b<1/4 and consider the two-sided neighborhoods

If h\in T_t, then

Set r=e^{-\lambda t}. Every term in the determinant expansion other than the product of the diagonal entries contains at least two off-diagonal entries. Since \det h=1, it follows that \prod_i h_{ii}=1+O_{n,b}(r^2). Thus

belongs to a fixed compact subset \mathcal D\subset A_n, and d(h)^{-1}h\in B(Cr) for a constant depending only on n,b.

In logarithmic coordinates, \mathcal D is a bounded subset of \R^{n-1}. Covering it by a mesh of size comparable to r, and enlarging C to allow for matrix multiplication, gives

For x in a compact set \Omega, all the centers xd_j lie in the fixed compact set \Omega\mathcal D. The assumed ball estimate and domination therefore imply, for all sufficiently large t,

Since \xi is a probability, the constant can be enlarged to cover the finitely many integer times before this estimate holds.

We use the entropy criterion of Einsiedler–Lindenstrauss–Michel–Venkatesh [3, Corollary 3.3]. In the notation

the criterion states the following: if flow-invariant probabilities \xi_i converge weakly to a probability \xi, and t_i\to\infty, then h_\xi(a(1))\ge\eta provided that, for every compact set \Omega, there are an identity neighborhood B and a constant C_\Omega such that

This criterion applies on the noncompact lattice space; its requirement that the limit be a probability is essential. Here take the constant sequence \xi_i=\xi, the neighborhood B(b), t_i=i, and \eta=\lambda/3. Since B(b)^{(-t,t)}=T_t, Equation (6.3) proves the required estimate.□

Proof. The group A_n\simeq\R^{n-1} is locally compact and second countable, and its action on the standard Borel space X_n is continuous. The ergodic decomposition theorem [5, Theorem 5.2], specialized to invariant probabilities, therefore supplies a measurable decomposition

into A_n-invariant, A_n-ergodic probabilities over a factor fixed by A_n. Set T(x)=xa(1). The function z\mapsto h_{\nu_z}(T) is measurable. Indeed, choose increasing finite Borel partitions \mathcal P_k generating the Borel \sigma-algebra of X_n. For every T-invariant probability \nu,

where H_\nu denotes the Shannon entropy of a finite partition. The right side is a countable combination of Borel functions of \nu.

The component partition is fixed by T, since a(1)\in A_n. The entropy integral formula for a fixed invariant partition [12, Theorem 9.8] consequently gives

The same formula holds after restricting to any measurable collection of components. It does not require the \nu_z to be T-ergodic: the induced transformation on the component space is the identity and contributes zero entropy.

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

would be an A_n-invariant probability with \xi\le w^{-1}\mu. Equation (6.4), applied to this restricted decomposition, would give h_\xi(T)=0, contradicting Proposition 6.1.□

Proof of Theorem 1.1. Let \mu_i be the packet probabilities for a sequence in the theorem. Equation (2.2) gives Q_i\to\infty. By Proposition 5.1, this sequence is tight and every weak subsequential limit \mu is an A_n-invariant probability. Proposition 5.2 supplies the hypotheses of Proposition 6.1. Lemma 6.2 shows that almost every A_n-ergodic component of \mu has positive entropy for a(1). As n is prime, the classification just stated makes every such component equal to m_n. Integrating the components gives \mu=m_n.

Thus every convergent subsequence has the same probability limit. Together with tightness, this proves \mu_i\to m_n and excludes loss of mass. All preceding estimates were for each prescribed local homothety packet separately, so no averaging over different local module types is used.□

Proof. Write M=M(g_f) and \widehat M=\widehat M(g_f). First identify the ambient quotient with X_n. Given a finite matrix lift g_f, choose a basis matrix c\in\operatorname{GL}_n(\Q) for \Q^n\cap\widehat\Z^{\,n}g_f^{-1}. Then cg_f\in\operatorname{GL}_n(\widehat\Z), and the corresponding real lattice is \Z^ncg_\infty, normalized to covolume one. Changing this basis or either matrix lift has no effect on that lattice. This constructs the usual quotient identification directly. For odd n, the signed real nth root gives [h]\mapsto(\det h)^{-1/n}h, identifying G(\R) with \SL_n(\R) and G(\Z) with \SL_n(\Z).

Multiplication matrices modulo scalars give

Here the pointwise lifts can also be seen by elementary linear algebra: over any ground field F_0 containing \Q, a matrix whose projective class lies in \mathbf T_K becomes a scalar multiple of a multiplication operator over an algebraic closure. It therefore commutes exactly with the regular representation already over F_0. The centralizer of that representation is the algebra of multiplication operators, so the matrix belongs to (K\otimes_\Q F_0)^\times, with scalar ambiguity F_0^\times. Finite lifts may be chosen integral at almost all primes.

A finite idele t_f acts on the completed lattice by \widehat M\mapsto t_f^{-1}\widehat M. Hence its translates are exactly the prescribed local homothety type. The stabilizer of \widehat M among finite ideles is

equality holds precisely when both z and z^{-1} preserve \widehat M. In the projective torus the stabilizer is the image of this group, since a representative that preserves the lattice up to a scalar idele can be rescaled to preserve it exactly. The unit subgroup is compact open in the finite ideles and has finite index in \widehat R^\times; at each prime not dividing q it equals R_\ell^\times. Its projective image is compact open as well. No invertibility of M over \cO(M) is used.

To describe the quotient and its measure, first divide (A.2) by the image of \widehat R^\times. Fractional-ideal factorization and \mathbb A_\Q^\times =\Q^\times\R^\times\widehat\Z^\times give one real fiber E\backslash C for each ordinary ideal class. These fibers are compact by the unit theorem, and the omitted finite-unit group is compact. Thus the torus quotient is compact. At the stabilizer of \widehat M, the finite fibers instead retain the translates \mathcal N(I). The same decomposition becomes

with exactly the action in Lemma 2.1. Haar measure pushes to uniform counting measure on each finite unit orbit and to the stated logarithmic Haar measure on C. After normalization it is therefore precisely the probability in that lemma. For g_\infty=[B_\sigma], (A.1) identifies the resulting real lattices with its displayed map. This proves both the packet identification and the equality of measures, including the orbit volume weights.

Finally, every full M is obtained by choosing a basis matrix for its completed lattice and taking its inverse as g_f. Any other allowed real diagonalizer has the form [B_\sigma dP], where d is diagonal and P is a permutation matrix, because it must permute the joint eigenspaces. The diagonal factor translates the real torus and permutes its sign components, preserving its Haar probability; the permutation reorders the embeddings. If a rational change of basis has matrix c, then B'_\sigma=cB_\sigma and \rho'=c\rho c^{-1}, while the same packet data have representatives g'_\infty=cg_\infty and g'_f=cg_f. Thus the torus is conjugated and the translated packet is left-multiplied by c, which disappears in the ambient rational quotient.□