The Analytic Collision Transform
Abstract
Fix a prime base b, a lag \ell\ge 1, and m=b^{\ell+1}. Equality of the leading and trailing base-b digits selects a finite diagonal G_\ell modulo m. Its oriented character boundary is D_\ell(\chi) = \sum_{n\in G_\ell}\bigl[\chi(n+1)-\chi(n)\bigr]. Replacing the finite Bernoulli kernel by the periodic zeta kernel K(x;s)=F(x,1-s) gives a centered analytic transform with the exact factorization \widehat A^\circ(\chi;s) = \frac{2i\sin(\pi s/2)\Gamma(s)}{\varphi(m)} \left(\frac{m}{2\pi}\right)^{\!s} D_\ell(\chi)L(s,\overline\chi) for every primitive odd character modulo m and 0<\operatorname{Re}(s)<1. The finite diagonal factor is independent of s. At the left boundary, \widehat A^\circ(\chi;0) = -\frac{i\pi}{\varphi(m)} B_{1,\overline\chi}D_\ell(\chi). For an odd prime base at lag one, this is exactly i\pi times the centered finite collision coefficient. No Gauss-sum completion is needed. The opposite boundary has a separate Gauss-twisted value. Every active channel has exactly the same zeros, with multiplicity, as L(s,\overline\chi) in the open critical strip. The transform makes those zeros visible but does not locate them.
Introduction
Long division modulo a power of the base produces a finite diagonal. It consists of the residues whose leading base-b digit equals their trailing digit. The set is finite arithmetic data. It does not depend on an analytic parameter. A Dirichlet character measures its oriented boundary through the differences \chi(n+1)-\chi(n).
The periodic zeta function supplies an analytic kernel for every slice of that boundary. The kernel is placed on the finite diagonal before the character transform is taken. Primitive character cancellation then removes the nonunit fibers, and the surviving transform factors into two pieces. One is the finite diagonal sum. The other is a Dirichlet L-function with its standard gamma and sine factors.
The factorization has an exact anchor. At lag one, the value at s=0 is the centered finite collision coefficient multiplied by i\pi. The equality preserves the full complex coefficient with its character index. The finite coefficient is therefore a boundary value of the analytic family, not a numerical resemblance to it.
The same identity makes every zero of an active Dirichlet channel visible inside the strip. Visibility is not control. Nothing below places a zero on the critical line or compares different conductors. The result is an exact fixed-conductor bridge between a digit boundary and its analytic character channel.
The Finite Boundary
Fix an odd prime b and put m=b^2, \qquad U_m=(\mathbb Z/m\mathbb Z)^\times. Euler’s totient is denoted by \varphi. Dirichlet characters are extended by zero on nonunits. The lag-one digit diagonal is G_1 = \left\{0\le n<m: \left\lfloor\frac nb\right\rfloor=n\bmod b\right\} = \{r(b+1):0\le r\le b-1\}. For a character \chi modulo m, define its direct diagonal boundary D_1(\chi) = \sum_{n\in G_1} \bigl[\chi(n+1)-\chi(n)\bigr].
For a\in U_m and n\in G_1, put d_n(a) = \left\lfloor\frac{(n+1)a}{m}\right\rfloor - \left\lfloor\frac{na}{m}\right\rfloor. The finite collision function is S(a) = -1-\left\lfloor\frac ab\right\rfloor +\sum_{n\in G_1}d_n(a). For 1\le r\le b-1, let U_r=\{a\in U_m:a\equiv r\pmod b\}, \qquad \overline S_r=\frac1b\sum_{a\in U_r}S(a). Define S^\circ(a)=S(a)-\overline S_{a\bmod b} and \widehat S^\circ(\chi) = \frac1{\varphi(m)} \sum_{a\in U_m}S^\circ(a)\overline\chi(a). For any modulus q and any nonprincipal character \psi modulo q, the standard special-value formula [1] gives B_{1,\psi} = \frac1q\sum_{a=1}^{q}a\psi(a) = -L(0,\psi).
Theorem 1 (Finite collision coefficient). Let b be an odd prime, let m=b^2, and let \chi be primitive and odd modulo m. Then \widehat S^\circ(\chi) = -\frac{B_{1,\overline\chi}D_1(\chi)}{\varphi(m)}.
Proof. The kernel of reduction from U_m to (\mathbb Z/b\mathbb Z)^\times is H=\{1+jb:0\le j\le b-1\}. Primitivity makes \chi nontrivial on H. Every U_r is a multiplicative coset of H, so \sum_{a\in U_r}\overline\chi(a)=0. The fiber means and the constant term in S(a) therefore contribute zero to the character transform.
The fractional part of a/b is constant on every U_r. Fiber cancellation gives -\sum_{a\in U_m} \left\lfloor\frac ab\right\rfloor\overline\chi(a) = -\frac1b\sum_{a\in U_m}a\overline\chi(a) = -bB_{1,\overline\chi}.
If n is a unit modulo m, multiplication by n permutes U_m. Consequently, \sum_{a\in U_m} \overline\chi(a) \left\{\frac{na}{m}\right\} = \chi(n)B_{1,\overline\chi}, and hence \sum_{a\in U_m} \left\lfloor\frac{na}{m}\right\rfloor \overline\chi(a) = \bigl(n-\chi(n)\bigr)B_{1,\overline\chi}. Every interior element of G_1 has the form r(b+1) with 1\le r\le b-2. Both n and n+1 are units there, so its increment contributes \bigl[1+\chi(n)-\chi(n+1)\bigr]B_{1,\overline\chi}.
The endpoint increments d_0 and d_{m-1} contribute zero after the character sum. Their contributions to D_1(\chi) are both 1 because \chi(1)-\chi(0)=1 and \chi(m)-\chi(m-1)=0-\chi(-1)=1. It follows that the interior diagonal increments contribute B_{1,\overline\chi}\bigl[b-D_1(\chi)\bigr]. Combining this with the term -bB_{1,\overline\chi} and dividing by \varphi(m) proves the formula. ◻
The Periodic Zeta Lift
Now let b be any prime, let \ell\ge 1, and put m=b^{\ell+1}. The leading-to-trailing digit diagonal is G_\ell = \left\{n\in\{0,\ldots,m-1\}: \left\lfloor\frac{n}{b^\ell}\right\rfloor=n\bmod b\right\}. Both 0 and m-1 belong to G_\ell. Define D_\ell(\chi) = \sum_{n\in G_\ell} \bigl[\chi(n+1)-\chi(n)\bigr].
For every integer r\not\equiv0\pmod m, let \rho_m(r) be the least positive representative of r modulo m. For 0<x<1, define K(x;s)=F(x,1-s), \qquad F(x,z)=\sum_{n=1}^{\infty}\frac{e^{2\pi i n x}}{n^z}. The series defining F begins in its half-plane of convergence and continues analytically. Substitution in the periodic-zeta connection formula [1] gives, first for \operatorname{Re}(s)<0 and then by analytic continuation, K(x;s) = \frac{\Gamma(s)}{(2\pi)^s} \left[ e^{i\pi s/2}\zeta(s,x) + e^{-i\pi s/2}\zeta(s,1-x) \right]. For 0<x<1, Dirichlet convergence makes K(x;s) regular at s=0. At s=1, the two Hurwitz residues cancel. The removable value is K(x;1) = -\frac12+\frac{i}{2}\cot(\pi x).
The Hurwitz representation is singular at x=0. The two endpoint slices are paired before the transform is taken.
Definition 2. For a\in(\mathbb{Z}/m\mathbb{Z})^\times, define \begin{aligned} A(a;s) :={}& K\!\left(\frac{a}{m};s\right) - K\!\left(\frac{m-a}{m};s\right)\\ &+ \sum_{\substack{n\in G_\ell\\1\le n\le m-2}} \left[ K\!\left(\frac{\rho_m((n+1)a)}{m};s\right) - K\!\left(\frac{\rho_m(na)}{m};s\right) \right]. \end{aligned} For 1\le r\le b-1, put M_r(s) = \frac1{b^\ell} \sum_{\substack{a\in(\mathbb{Z}/m\mathbb{Z})^\times\\a\equiv r\,(\mathrm{mod}\,b)}}A(a;s). The centered observable and its character transform are A^\circ(a;s)=A(a;s)-M_{a\bmod b}(s) and \widehat A^\circ(\chi;s) = \frac1{\varphi(m)} \sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times}A^\circ(a;s)\overline\chi(a).
Primitive Character Cancellation
Lemma 3 (Base-fiber cancellation). Let \chi be primitive modulo m=b^{\ell+1}. For every residue r coprime to b, \sum_{\substack{a\in(\mathbb{Z}/m\mathbb{Z})^\times\\a\equiv r\,(\mathrm{mod}\,b)}} \overline\chi(a)=0. Consequently, \widehat A^\circ(\chi;s) = \frac1{\varphi(m)} \sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times}A(a;s)\overline\chi(a).
Proof. Let H=\{u\in(\mathbb{Z}/m\mathbb{Z})^\times:u\equiv1\pmod b\}. If \chi were trivial on H, it would factor through reduction modulo b, contrary to primitivity modulo b^{\ell+1}. Every coprime residue class modulo b is a multiplicative coset of H. Character orthogonality gives the first identity. The fiber mean is constant on that coset, which gives the second. ◻
Lemma 4 (Congruence-fiber cancellation). Let \chi be primitive modulo m=b^{\ell+1}. For every 1\le j\le\ell and every residue t coprime to b, \sum_{\substack{a\in(\mathbb{Z}/m\mathbb{Z})^\times\\a\equiv t\,(\mathrm{mod}\,b^j)}} \overline\chi(a)=0.
Proof. Put H_j=\{u\in(\mathbb{Z}/m\mathbb{Z})^\times:u\equiv1\pmod{b^j}\}. Triviality of \chi on H_j would make its conductor divide b^j. This contradicts primitivity modulo b^{\ell+1}. Every coprime congruence class modulo b^j is a multiplicative coset of H_j, so character orthogonality gives the result. ◻
Lemma 5 (Periodic-zeta model sum). Let \chi be primitive and odd modulo m=b^{\ell+1}. For every 1\le n\le m-1 and every 0<\operatorname{Re}(s)<1, \sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times} \overline\chi(a) K\!\left(\frac{\rho_m(na)}{m};s\right) = \chi(n)\,2i\sin(\pi s/2)\Gamma(s) \left(\frac{m}{2\pi}\right)^{\!s} L(s,\overline\chi). Here \chi(n)=0 when n is not a unit modulo m.
Proof. Write the left side as \Sigma_n(s). The Hurwitz representation gives \Sigma_n(s) = \frac{\Gamma(s)}{(2\pi)^s} \left[e^{i\pi s/2}U_n(s)+e^{-i\pi s/2}V_n(s)\right], where U_n(s) = \sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times} \overline\chi(a) \zeta\!\left(s,\frac{\rho_m(na)}{m}\right) and V_n(s) = \sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times} \overline\chi(a) \zeta\!\left(s,1-\frac{\rho_m(na)}{m}\right).
Suppose first that n is a unit. Substitution by u\equiv na\pmod m gives U_n(s) = \chi(n) \sum_{u\in(\mathbb{Z}/m\mathbb{Z})^\times}\overline\chi(u) \zeta\!\left(s,\frac um\right) = \chi(n)m^sL(s,\overline\chi) by the Hurwitz decomposition of a Dirichlet L-function [1].
If n is not a unit, write \gcd(n,m)=b^j, \qquad q=b^{\ell+1-j}. The residue \rho_m(na) depends only on a modulo q. Every fiber of reduction modulo q has total \overline\chi-weight zero by Lemma 4. Hence U_n(s)=0. Both cases give U_n(s)=\chi(n)m^sL(s,\overline\chi).
Since 1-\frac{\rho_m(na)}m=\frac{\rho_m(-na)}m, the same argument with -n gives V_n(s) = \chi(-n)m^sL(s,\overline\chi) = -\chi(n)m^sL(s,\overline\chi). Substitution completes the proof. ◻
The Analytic Transform
Theorem 6 (Analytic collision transform). Let b be prime, let m=b^{\ell+1} with \ell\ge1, and let \chi be primitive and odd modulo m. For every 0<\operatorname{Re}(s)<1, \widehat A^\circ(\chi;s) = \frac{2i\sin(\pi s/2)\Gamma(s)}{\varphi(m)} \left(\frac{m}{2\pi}\right)^{\!s} D_\ell(\chi)L(s,\overline\chi).
Proof. Lemma 3 removes the fiber means. Both endpoints 0 and m-1 lie in G_\ell. Their formal contributions are K\!\left(\frac am;s\right)-K(0;s) and K(0;s)-K\!\left(\frac{m-a}{m};s\right). The singular terms cancel before evaluation. The remaining terms are the interior diagonal differences already present in A(a;s).
Apply Lemma 5 to every surviving kernel value. The common analytic factor can be taken outside the finite sum. What remains is \sum_{n\in G_\ell} \bigl[\chi(n+1)-\chi(n)\bigr] = D_\ell(\chi). Division by \varphi(m) proves the identity. ◻
Recovery at the Left Boundary
Proposition 7 (General left-boundary value). Under the hypotheses of Theorem 6, \widehat A^\circ(\chi;0) = -\frac{i\pi}{\varphi(m)} B_{1,\overline\chi}D_\ell(\chi).
Proof. Every kernel argument in A(a;s) lies strictly between 0 and 1, so the finite transform is regular at s=0. The analytic factor has the removable limit \lim_{s\to0} 2i\sin(\pi s/2)\Gamma(s) \left(\frac{m}{2\pi}\right)^{\!s} = i\pi. Also, L(0,\overline\chi)=-B_{1,\overline\chi}. Taking the limit in Theorem 6 gives the result. ◻
Theorem 8 (Exact collision recovery). Let b be an odd prime, let \ell=1, and let \chi be primitive and odd modulo b^2. Then \widehat A^\circ(\chi;0) = i\pi\,\widehat S^\circ(\chi). Equivalently, \widehat S^\circ(\chi) = \frac{\widehat A^\circ(\chi;0)}{i\pi}.
Proof. At lag one, Proposition 7 and Theorem 1 have the same Bernoulli factor, the same direct diagonal boundary, and the same normalization. ◻
The recovery requires no character-dependent normalization. The analytic family reaches the finite collision coefficient directly at the left boundary.
The Opposite Boundary
For a primitive character modulo m, define the Gauss sum \tau(\chi) = \sum_{r\,(\mathrm{mod}\,m)} \chi(r)e^{2\pi i r/m}.
Proposition 9 (Right-boundary value). Let \chi be primitive and odd modulo m=b^{\ell+1}. Then \widehat A^\circ(\chi;1) = -\frac{\tau(\overline\chi)}{\varphi(m)} B_{1,\chi}D_\ell(\chi).
Proof. The finite transform is regular at s=1. Theorem 6 gives \widehat A^\circ(\chi;1) = \frac{im}{\pi\varphi(m)} D_\ell(\chi)L(1,\overline\chi). The odd functional equation [1] gives B_{1,\chi} = \frac{i}{\pi}\tau(\chi)L(1,\overline\chi) . For a primitive character, |\tau(\chi)|^2=m [1]. Conjugating the defining sum and using oddness gives \tau(\chi)\tau(\overline\chi)=-m. Substitution proves the formula. ◻
The right-boundary value is exact, but it is not the finite collision coefficient. The Gauss sum and the opposite Bernoulli index record the functional equation at the other edge of the strip.
Active Channels
Theorem 10 (Zero visibility). Let b be prime, let m=b^{\ell+1}, and let \chi be primitive and odd modulo m.
If D_\ell(\chi)=0, then \widehat A^\circ(\chi;s)=0 throughout the open critical strip. If D_\ell(\chi)\ne0, then \widehat A^\circ(\chi;s) and L(s,\overline\chi) have exactly the same zeros there, with the same multiplicities.
Proof. Theorem 6 separates the transform into D_\ell(\chi)L(s,\overline\chi) and the factor \frac{2i\sin(\pi s/2)\Gamma(s)}{\varphi(m)} \left(\frac{m}{2\pi}\right)^{\!s}. The gamma function has no zeros. The sine factor has no zeros in 0<\operatorname{Re}(s)<1, and the remaining factors never vanish. The stated alternatives follow. ◻
The theorem identifies the active analytic channels exactly. It does not control the location of their zeros. The conductor is fixed, and no positivity statement or uniform estimate across conductors is proved here.
The Fixed Diagonal
The periodic-zeta lift gives the finite collision coefficient an exact analytic carrier. At lag one, its value at zero is not merely proportional in magnitude. It is the full complex coefficient multiplied by i\pi.
Inside the strip, the gamma factor, sine factor, and L-function move. The digit diagonal does not. An active collision channel therefore records every zero of its associated Dirichlet L-function with the correct multiplicity. The remaining problem is control of those zeros, not their visibility.
References
[1]NIST Digital Library of Mathematical Functions, \S25.13, \S25.15, and \S27.10, https://dlmf.nist.gov/.