Additive Boundaries and Conductor Energies of Periodic Remainder Orbits
Abstract
Long division has two finite geometries. A fixed digit condition selects a region in a finite carry table. A repeating tail closes into a multiplicative remainder orbit. Dirichlet characters read neither region through its total mass. They read the region’s additive signed boundary.
The squared boundary flux has a canonical decomposition into nonnegative energies of exact conductor. No response at a proper conductor can disappear through cancellation with another level. This gives an exact criterion for top-conductor concentration. For the base-15 terminal collision boundary modulo 225, the full odd energy is 1920. Its exact-conductor energies are 24, 240, 360, 400, and 896 at conductors 9, 25, 45, 75, and 225, respectively.
The same decomposition measures periodic remainder orbits. If multiplication by b gives the full unit orbit modulo an odd prime power p^e, every exact-conductor energy vanishes except the energy at conductor p. Increasing e thickens the same prime-level boundary without creating higher-conductor response. For an index-two orbit modulo an odd prime, the off-resonance energy has four exact forms determined by the prime modulo 8. Higher-index prime orbits admit an exact expansion in Jacobi-type sums. The orbit may grow in size, but its additive edge determines its arithmetic depth.
Remainder Orbits and Their Edges
Long division writes a digit and carries a remainder forward. When the denominator is prime to the base, multiplication by the base permutes the unit remainders. The repeating tail is therefore a finite multiplicative orbit.
A finite digit condition and a periodic tail select different kinds of finite regions. The first lies in a carry table. The second lies in a unit group. Their character theory begins at the same place. If O is a remainder orbit, its measured edge is the additive signed boundary \mu_O(x)=\mathbf 1_O(x-1)-\mathbf 1_O(x). The multiplication map creates the orbit. Addition by one exposes its edge. Dirichlet characters pair with that edge rather than with the size of the orbit.
Squaring the boundary flux separates a second structure. Every odd character comes from one primitive conductor, so the full energy splits into exact conductor levels. Each level is nonnegative. Proper-conductor response cannot be hidden by cancellation against the top level.
The base-15 terminal boundary makes this distinction concrete. Its signed top-conductor ledger gives the primitive moment 896 at modulus 225. The nonnegative decomposition finds the rest of the energy at conductors 9, 25, 45, and 75. Nothing has disappeared. The response occupies several arithmetic resolutions.
Periodic orbits sharpen the same point. A full unit orbit modulo p^e grows with e, but its additive edge is pulled back from modulo p. All of its energy remains at conductor p. An index-two orbit at prime modulus behaves differently. Its off-resonance energy has four closed forms according to the prime modulo 8.
The finite carry-boundary factorization and its signed conductor ledger are developed in [1]. Their short proofs are repeated so the exact-conductor decomposition and the periodic applications stand on their own.
Sawtooth transforms, generalized Bernoulli numbers, Gauss sums, Jacobi sums, and character orthogonality enter in their classical forms [2, 3, 4]. Long division supplies the finite boundary on which those tools act.
Consecutive power residues and additive patterns in multiplicative subgroups have their own established theory [7, 8]. The questions below are finite and exact. They ask how the additive edge of one orbit divides its energy among conductors.
Boundary Flux
A digit condition is not read through its total mass. Dirichlet characters read the signed places where membership changes. Carries are threshold crossings, so the threshold function supplies the sawtooth and its Bernoulli transform.
Fix m\ge2. For real t let \langle t\rangle=\{t\}-\tfrac12 for t\notin\mathbb{Z} and \langle t\rangle=0 for t\in\mathbb{Z}. For a weight w\colon\mathbb{Z}/m\mathbb{Z}\to\mathbb{C} and a residue a, the carry observable is F_w(a)=-\sum_{n=0}^{m-1}w(n)\Bigl(\Bigl\langle\tfrac{(n+1)a}{m}\Bigr\rangle -\Bigl\langle\tfrac{na}{m}\Bigr\rangle\Bigr). Dirichlet characters are extended by zero off (\mathbb{Z}/m\mathbb{Z})^{\times}, and \widehat F_w(\chi)=\sum_a F_w(a)\overline{\chi(a)}. The signed boundary chain is \mu_w(x)=w(x-1)-w(x), with flux S_w(\chi)=\sum_n w(n)(\chi(n+1)-\chi(n)) =\sum_x\mu_w(x)\chi(x). Let B_{1,\overline\chi}=\frac1m\sum_{a=1}^m a\,\overline{\chi(a)}.
Lemma 1 (Twisted sawtooth evaluation). For a primitive odd character \chi modulo m and every k\in\mathbb{Z}/m\mathbb{Z}, \sum_{a\bmod m}\Bigl\langle\tfrac{ka}{m}\Bigr\rangle\overline{\chi(a)} =\chi(k)\,B_{1,\overline\chi}. For every even Dirichlet character \chi modulo m, the left side vanishes for all k.
Proof. Oddness of \langle\cdot\rangle forces the even case. For \chi odd primitive, the case k=1 is the standard generalized-Bernoulli identity [3] \sum_{a=1}^{m-1}\langle a/m\rangle\overline{\chi(a)}=B_{1,\overline\chi}, using \sum_a\overline{\chi(a)}=0. For (k,m)=1 substitute a\mapsto k^{-1}a.
Suppose (k,m)>1. Put e_m(t)=e^{2\pi i t/m} and write the finite Fourier expansion \Bigl\langle\frac{u}{m}\Bigr\rangle =\sum_{j\bmod m}c_j e_m(ju). Then \begin{aligned} \sum_{a\bmod m} \Bigl\langle\frac{ka}{m}\Bigr\rangle\overline{\chi(a)} &=\sum_{j\bmod m}c_j \sum_{a\bmod m}\overline{\chi(a)}e_m(jka)\\ &=\sum_{j\bmod m}c_j\,\tau(\overline\chi,jk). \end{aligned} For primitive \chi the additive Gauss sum obeys [2, 5] \tau(\overline\chi,c)=\chi(c)\tau(\overline\chi,1) for every integer c, with \chi(c)=0 when (c,m)>1. Since (k,m) divides (jk,m), every index jk is a nonunit. Every Gauss factor therefore vanishes. The result agrees with \chi(k)B_{1,\overline\chi}=0. ◻
Theorem 2 (Boundary flux factorization). For every weight w\colon\mathbb{Z}/m\mathbb{Z}\to\mathbb{C} and every primitive odd character \chi modulo m, \widehat F_w(\chi)=-\,B_{1,\overline\chi}\,S_w(\chi). For every even \chi, \widehat F_w(\chi)=0.
Proof. Exchange the finite sums in \widehat F_w(\chi) and apply Lemma 1 with k=n+1 and k=n. This gives \sum_a \left( \Bigl\langle\tfrac{(n+1)a}{m}\Bigr\rangle - \Bigl\langle\tfrac{na}{m}\Bigr\rangle \right) \overline{\chi(a)} = \bigl(\chi(n+1)-\chi(n)\bigr)B_{1,\overline\chi}. Summing against w(n) gives \widehat F_w(\chi)=-B_{1,\overline\chi}S_w(\chi). The even case follows from the even part of Lemma 1. No property of w beyond being a function on \mathbb{Z}/m\mathbb{Z} is used. ◻
Corollary 3 (Finite digit conditions). Let m=b^N and let G\subseteq\{0,\dots,m-1\} be selected by a fixed N-digit rule. For every primitive odd \chi, \widehat F_{\mathbf 1_G}(\chi) = -B_{1,\overline\chi}S_G(\chi). The collision diagonal is the case G=\{d_0=d_{N-1}\}.
Theorem 4 (Dirichlet-visible boundary). As \chi ranges over all Dirichlet characters modulo m, the family \{S_w(\chi)\}_\chi determines exactly the restriction of \mu_w to (\mathbb{Z}/m\mathbb{Z})^{\times}. Two weights have identical boundary-flux families if and only if their boundary chains agree on the unit group.
Proof. S_w(\chi)=\sum_x\mu_w(x)\chi(x) depends only on \mu_w|_{(\mathbb{Z}/m\mathbb{Z})^{\times}} since \chi vanishes off units, and the characters form an orthogonal basis of functions on (\mathbb{Z}/m\mathbb{Z})^{\times}. ◻
Energy by Exact Conductor
The boundary flux can now be grouped by the exact conductor of each character. Push the boundary down to every divisor and measure its primitive odd part. These pieces recover the full odd-character energy, and every piece is nonnegative.
Let \mu be a boundary chain on \mathbb{Z}/m\mathbb{Z}. For d\mid m and a unit r modulo d, the pushdown is M_d(r)=\sum_{x\equiv r\,(d),\ (x,m)=1}\mu(x). We use the trivial one-element unit group at d=1. A primitive character \chi modulo d is paired with \mu through the lift S_{\mu,d}(\chi)=\sum_{\substack{x\bmod m\\(x,m)=1}}\mu(x)\,\chi(x\bmod d), and P_d(\mu)=\sum_{\chi\ \mathrm{prim\,odd}\bmod d}|S_{\mu,d}(\chi)|^2 is the second moment of \mu over the primitive odd characters of conductor exactly d. At the top modulus we write S_\mu(\chi) = \sum_{x\bmod m}\mu(x)\chi(x) = \sum_{\substack{x\bmod m\\(x,m)=1}}\mu(x)\chi(x), where the second equality uses the convention that Dirichlet characters vanish off the unit group.
For any proposition P, let [P] denote 1 if P holds and 0 otherwise. Write \mu_{\mathrm{Mob}} for the ordinary Möbius function.
Lemma 5 (Odd character kernel). For units x,y modulo m, \sum_{\substack{\chi\bmod m\\ \chi(-1)=-1}} \chi(x)\overline{\chi(y)} = \frac{\varphi(m)}2 \bigl([x\equiv y\,(m)]-[x\equiv-y\,(m)]\bigr).
Proof. The full character kernel on (\mathbb{Z}/m\mathbb{Z})^{\times} is \sum_{\chi\bmod m}\chi(x)\overline{\chi(y)} = \varphi(m)[x\equiv y\,(m)]. Applying the odd projector \tfrac12(\mathrm{id}-J) in the y variable gives the stated formula. ◻
Lemma 6 (Primitive-odd kernel). For units x,y modulo m, \begin{aligned} \sum_{\chi\ \mathrm{prim\,odd}}\chi(x)\overline{\chi(y)} &= \frac12 \sum_{d\mid m} \mu_{\mathrm{Mob}}\!\Bigl(\frac md\Bigr)\varphi(d) \\ &\quad\cdot \bigl([x\equiv y\,(d)]-[x\equiv-y\,(d)]\bigr). \end{aligned}
Proof. On units, character orthogonality gives \sum_{\chi\bmod d}\chi(x)\overline{\chi(y)} =\varphi(d)\,[x\equiv y\,(d)]. Since x and y are units modulo m, they are units modulo every divisor d\mid m. Every character modulo m is induced by a unique primitive character of some conductor d\mid m [3], and a primitive character modulo d evaluated at a unit x of m equals \chi(x\bmod d). Summing the full orthogonality kernel over d\mid m therefore groups characters by conductor. Möbius inversion over the divisor lattice isolates the primitive part. \sum_{\chi\ \mathrm{prim}\bmod m}\chi(x)\overline{\chi(y)} = \sum_{d\mid m} \mu_{\mathrm{Mob}}\!\Bigl(\tfrac md\Bigr)\varphi(d)\, [x\equiv y\,(d)]. The odd part is extracted by the projector \tfrac12(\mathrm{id}-J) applied in the variable y, where J is the antipodal map y\mapsto-y. Since \tfrac12(\chi(x)-\chi(-x))=\tfrac12(1-\chi(-1))\chi(x) equals \chi(x) for odd \chi and 0 for even \chi, the projector annihilates the even characters and fixes the odd ones. Applying it yields the bracket [x\equiv y\,(d)]-[x\equiv-y\,(d)] and the factor \tfrac12. The unprojected Möbius-graded identity is classical [6]. ◻
Theorem 7 (Conductor ledger). For every boundary chain \mu and every D\mid m, P_D(\mu) = \frac12 \sum_{e\mid D} \mu_{\mathrm{Mob}}\!\Bigl(\tfrac De\Bigr)\varphi(e) \bigl(\Delta_e-\mathcal A_e\bigr), with \Delta_e=\sum_{r}|M_e(r)|^2, \qquad \mathcal A_e=\sum_r M_e(r)\overline{M_e(-r)} .
Proof. Expanding the primitive odd moment at conductor D gives P_D(\mu) = \sum_{\chi\ \mathrm{prim\,odd}\bmod D} \left| \sum_{\substack{x\bmod m\\(x,m)=1}} \mu(x)\chi(x\bmod D) \right|^2. After expanding the square, Lemma 6, with D in place of m, gives \begin{aligned} P_D(\mu) &= \frac12 \sum_{e\mid D} \mu_{\mathrm{Mob}}\!\Bigl(\tfrac De\Bigr)\varphi(e) \\ &\quad\cdot \sum_{\substack{x,y\bmod m\\(x,m)=(y,m)=1}} \mu(x)\overline{\mu(y)} \bigl([x\equiv y\,(e)]-[x\equiv-y\,(e)]\bigr). \end{aligned} The congruence x\equiv y\pmod e groups x and y by the same residue r\in(\mathbb{Z}/e\mathbb{Z})^{\times} and gives \sum_r|M_e(r)|^2=\Delta_e. The congruence x\equiv-y\pmod e groups x by r and y by -r, giving \sum_rM_e(r)\overline{M_e(-r)}=\mathcal A_e. This proves the formula. ◻
Theorem 8 (Energy decomposition). For every boundary chain \mu, \sum_{\chi\ \mathrm{odd}}|S_\mu(\chi)|^2=\sum_{d\mid m}P_d(\mu), \qquad P_d(\mu)\ge0 .
Proof. Every odd character \chi modulo m is induced by a unique primitive odd character \chi^\ast of some conductor d\mid m. For unit x modulo m, \chi(x)=\chi^\ast(x\bmod d), and both characters vanish off the appropriate unit groups. Therefore S_\mu(\chi) = \sum_{\substack{x\bmod m\\(x,m)=1}}\mu(x)\chi^\ast(x\bmod d) = S_{\mu,d}(\chi^\ast). As \chi ranges over all odd characters modulo m, the inducing primitive character \chi^\ast ranges once over the primitive odd characters of each conductor d\mid m. Partitioning by conductor gives the identity. Each P_d(\mu) is a sum of squared magnitudes, so P_d(\mu)\ge0. ◻
Corollary 9 (Top-conductor concentration). The following are equivalent.
P_m(\mu) equals the full odd moment \sum_{\chi\ \mathrm{odd}}|S_\mu(\chi)|^2.
P_d(\mu)=0 for all proper d\mid m.
A sufficient condition for P_d(\mu)=0 at a given proper level d is that the pushdown M_d(r) vanish identically in r. This is not necessary, since P_d(\mu)=0 can hold with M_d\ne0 when the pushdown lies in the antipodally even subspace and has no primitive odd projection.
Proof. The difference of the two moments is \sum_{d<m}P_d(\mu), a sum of nonnegative terms by Theorem 8; it vanishes iff each P_d(\mu) does, which is where positivity is essential. No cancellation among levels can mask a nonzero proper energy. Vanishing of the pushdown M_d forces S_{\mu,d}(\chi)=0 for every character modulo d, hence P_d(\mu)=0. The converse need not hold. The equality P_d(\mu)=0 only says that the pushdown has zero projection onto the primitive odd character subspace at level d. The pushdown may still be nonzero, for instance if it lies entirely in an antipodally even subspace or in lower-conductor components. ◻
Remark 10. The top-conductor value is not, in general, twice a formal boundary count. That identification belongs to particular balanced boundary families. The general invariant is the level decomposition. A boundary whose energy leaks into a proper conductor, such as the base-15 collision diagonal, fails top-conductor concentration by exactly the leaked energy.
The Base-Fifteen Spectrum
The terminal base-15 collision boundary shows what the nonnegative decomposition adds. Its primitive top-conductor moment is only one part of the full odd energy.
The signed top-conductor ledger and the value P_{225}=896 are established in [1]. The calculation below determines every exact-conductor level and hence the full odd energy.
Proposition 11 (Base-fifteen conductor spectrum). Let \mu be the signed boundary of G=\{n<225\mid\text{the two padded base-$15$ digits agree}\}, restricted to the units and extended by zero off the units. Its exact-conductor energies are \begin{array}{c|rrrrrrrrr} d&1&3&5&9&15&25&45&75&225\\ P_d(\mu)&0&0&0&24&0&240&360&400&896. \end{array} Consequently, \sum_{\substack{\chi\bmod225\\ \chi(-1)=-1}}|S_\mu(\chi)|^2 =1920. The proper-conductor energy is 1024, and the remaining 896 lies at conductor 225.
Proof. The formal endpoints are L_a=16a, \qquad R_a=16a+1, \qquad 0\le a\le14, with signs -1 and +1. Put C=\{1,2,4,7,8,11,13,14\}=(\mathbb{Z}/15\mathbb{Z})^{\times}. The endpoint L_a is a unit modulo 225 exactly when a\in C. Since 16\equiv1\pmod{15}, the endpoint R_a is a unit exactly when a+1\in C. The unit-visible chain is therefore \mu = \sum_{c\in C} \bigl(\delta_{16c-15}-\delta_{16c}\bigr).
For every divisor e of 225, let M_e be the pushdown of this chain to (\mathbb{Z}/e\mathbb{Z})^{\times}, and put Q_e=\Delta_e-\mathcal A_e. The involution c\mapsto15-c preserves C, and 16(15-c)-15=225-16c. Thus M_e(-r)=-M_e(r) at every divisor level. The pushdowns are real, so \mathcal A_e=-\Delta_e, \qquad Q_e=2\Delta_e.
Modulo 15, the two endpoints belonging to the same c coincide because their difference is 15. The same cancellation holds modulo 1, 3, and 5. Hence \Delta_1=\Delta_3=\Delta_5=\Delta_{15}=0.
Modulo 9, exact reduction of the endpoint multisets gives \begin{aligned} \{16c:c\in C\} &=\{1,1,2,4,5,5,7,8\},\\ \{16c-15:c\in C\} &=\{1,2,4,4,5,7,8,8\}. \end{aligned} The pushdown is \begin{array}{c|rrrrrr} r&1&2&4&5&7&8\\ M_9(r)&-1&0&1&-1&0&1, \end{array} and \Delta_9=4.
For every nontrivial divisor e of 225, 16^{-1}\equiv-14\pmod e, because 16(-14)=1-225. At level 25, same-sign endpoints remain distinct. A cross-sign coincidence is equivalent to c-c'\equiv10\pmod{25}. The only possibilities in C are (c,c')=(11,1) and (14,4). After these two opposite-sign cancellations, twelve endpoints remain distinct. Therefore \Delta_{25}=12.
At levels 45, 75, and 225, same-sign endpoints remain distinct. A cross-sign coincidence would require c-c'\equiv-15\pmod e. This is impossible because |c-c'|\le13. All sixteen endpoints remain, and \Delta_{45}=\Delta_{75}=\Delta_{225}=16. The complete pushdown ledger is \begin{array}{c|rrrrrrrrr} e&1&3&5&9&15&25&45&75&225\\ \Delta_e&0&0&0&4&0&12&16&16&16\\ Q_e&0&0&0&8&0&24&32&32&32. \end{array}
Theorem 7 now gives \begin{aligned} P_9 &=\frac12(6\cdot8)=24,\\ P_{25} &=\frac12(20\cdot24)=240,\\ P_{45} &=\frac12(24\cdot32-8\cdot0-6\cdot8+2\cdot0)=360,\\ P_{75} &=\frac12(40\cdot32-20\cdot24-8\cdot0+4\cdot0)=400,\\ P_{225} &=\frac12(120\cdot32-40\cdot32-24\cdot32+8\cdot0)=896. \end{aligned} The levels 1, 3, 5, and 15 vanish because every divisor term in their ledger has Q_e=0. Summing the exact-conductor energies gives 24+240+360+400+896=1920. The top-level odd kernel gives the same total directly as \frac{\varphi(225)}2Q_{225} =\frac{120}{2}\cdot32 =1920. ◻
The signed primitive ledger has negative Möbius contributions of 384 from level 45 and 640 from level 75 while isolating conductor 225. The nonnegative spectrum opens those two corrections into 384=24+360, \qquad 640=240+400. The first pair belongs to conductors 9 and 45. The second belongs to conductors 25 and 75. The response missing from the top conductor is still present at lower arithmetic resolutions.
Periodic Remainder Orbits
The periodic regime begins after the terminating part of long division has been removed. What remains is not a finite prefix condition but a remainder orbit. The orbit is a region inside the unit group, so it has the same kind of signed boundary as a digit-defined set. The energy laws below measure that orbit boundary by the odd-character kernel.
Dividing n/q in base b splits by the Chinese Remainder Theorem into a terminating part on the b-part of q and a periodic part on the prime-to-b part q'. On q' the map T_b\colon R\mapsto bR is invertible. For a unit starting remainder R_0, the remainder sequence lies in the coset R_0\langle b\rangle and has period \mathop{\mathrm{ord}}_{q'}(b).
For the remainder of this section we assume (b,q)=1. For a general denominator, q should first be replaced by its prime-to-b part q'. We focus on the orbit of 1, O=\langle b\rangle\le(\mathbb{Z}/q\mathbb{Z})^{\times}. A unit numerator R_0 gives the coset R_0\langle b\rangle. The formulas use the orbit of 1. For the indicator \mathbf 1_O and chain \mu_O, set \mathcal E(b,q) = \sum_{\substack{\chi\bmod q\\ \chi(-1)=-1}}|S_{\mu_O}(\chi)|^2. This is the full odd-character energy at modulus q.
Theorem 12 (Prime-level support of a full orbit). Let p be an odd prime, let e\ge1, and put q=p^e. Assume that the remainder orbit is full, O=(\mathbb{Z}/q\mathbb{Z})^{\times}, as happens when b is a primitive root modulo q. Then \mu_O(x)=[x\equiv0\pmod p]-[x\equiv1\pmod p]. Its exact-conductor energies satisfy P_d(\mu_O)= \begin{cases} \dfrac{p-1}{2}\,p^{2e-2},&d=p,\\[4pt] 0,&d\mid p^e,\ d\ne p. \end{cases} Consequently, \mathcal E(b,p^e)=\frac{(p-1)p^{2e-2}}2. For e=1 the energy lies at the top conductor. For e\ge2 the top-conductor energy is zero, and all energy lies at the proper conductor p.
Proof. A residue modulo p^e is a unit exactly when its reduction modulo p is nonzero. Reduction commutes with the additive shift by one, so \mu_O(x)=\mathbf 1_O(x-1)-\mathbf 1_O(x) =[x\equiv0\pmod p]-[x\equiv1\pmod p]. The positive sheet is supported on nonunits. On the unit group, \mu_O(x)=-[x\equiv1\pmod p].
Fix 1\le f\le e and let \psi be a primitive odd character modulo p^f. Put K_f=\{r\in(\mathbb{Z}/p^f\mathbb{Z})^{\times}:r\equiv1\pmod p\}. Every r\in K_f has exactly p^{e-f} lifts modulo p^e, and every such lift is a unit. Therefore S_{\mu_O,p^f}(\psi) =-p^{e-f}\sum_{r\in K_f}\psi(r). For f=1, one has K_1=\{1\} and S_{\mu_O,p}(\psi)=-p^{e-1}. For f\ge2, the group K_f is the kernel of reduction from (\mathbb{Z}/p^f\mathbb{Z})^{\times} to (\mathbb{Z}/p\mathbb{Z})^{\times}. If a primitive character modulo p^f were trivial on K_f, it would factor through (\mathbb{Z}/p^f\mathbb{Z})^{\times}/K_f\cong(\mathbb{Z}/p\mathbb{Z})^{\times}. Its conductor would then divide p, contrary to primitivity at conductor p^f. Its restriction to K_f is therefore nontrivial. Subgroup orthogonality gives \sum_{r\in K_f}\psi(r)=0. Thus every primitive odd flux above conductor p vanishes.
There are (p-1)/2 odd characters modulo p. Each is primitive and each has squared flux p^{2e-2}. The conductor-one character is even. Every other exact-conductor energy vanishes, which proves the result. ◻
Greater prime-power depth thickens the prime-level flux without creating new conductor depth. Top-conductor concentration holds for the full orbit exactly when e=1.
Example 13. Take b=2 and q=9. Multiplication by 2 gives the six-state cycle 1\longmapsto2\longmapsto4\longmapsto8\longmapsto7 \longmapsto5\longmapsto1. The same set read in additive order is O=\{1,2,4,5,7,8\}. Its signed boundary is \mu_O =\delta_0+\delta_3+\delta_6 -\delta_1-\delta_4-\delta_7. Dirichlet characters modulo 9 vanish at the three positive boundary points 0, 3, and 6. The visible boundary is therefore \mu_O\big|_{(\mathbb{Z}/9\mathbb{Z})^{\times}} =-\delta_1-\delta_4-\delta_7. All three visible points reduce to 1 modulo 3. The odd character induced from modulo 3 sees the same value at each point, so its flux is -3 and its energy is 9. A primitive character modulo 9 is nontrivial on \{1,4,7\}, the kernel of reduction to (\mathbb{Z}/3\mathbb{Z})^{\times}. Subgroup orthogonality makes its flux zero. The orbit has six states, but its entire odd energy lives at conductor 3.
For H=\langle b\rangle, the annihilator H^\perp marks the resonant characters. The odd energy splits as \mathcal E(b,q) = \sum_{\chi\in H^\perp,\ \chi\ \mathrm{odd}} |S_{\mu_H}(\chi)|^2 + \sum_{\chi\notin H^\perp,\ \chi\ \mathrm{odd}} |S_{\mu_H}(\chi)|^2, and the resonant odd family is \{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}/H}:\chi(-1H)=-1\}. If -1\in H there are none. A character is trivial on H if and only if it factors through the quotient (\mathbb{Z}/q\mathbb{Z})^{\times}/H. Such a quotient character is odd precisely when its value on the coset -1H is -1. If -1\in H, then -1H=H, so every quotient character takes value 1 on -1H, and no resonant odd character exists.
Theorem 14 (Higher-index Jacobi expansion). Let q be an odd prime, let H\le(\mathbb{Z}/q\mathbb{Z})^{\times} have index r, and let X=\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}/H} be the group of characters trivial on H, viewed as Dirichlet characters modulo q. For any odd character \chi\bmod q, S_{\mu_H}(\chi) = \frac1r \sum_{\rho\in X} \left( J_+(\rho,\chi) -(q-1)\mathbf 1_{\rho\chi=1} \right), where J_+(\rho,\chi)=\sum_{y\bmod q}\rho(y)\chi(y+1). Here \mathbf 1_{\rho\chi=1} denotes the indicator that \rho\chi is the principal Dirichlet character modulo q. Consequently every subgroup-orbit boundary flux modulo a prime is an explicit finite combination of Jacobi-type sums over the quotient character group.
Proof. For x\in\mathbb{Z}/q\mathbb{Z}, \mathbf 1_H(x)=\frac1r\sum_{\rho\in X}\rho(x), with Dirichlet characters extended by zero at 0. Since 0\notin H, this identity already includes the endpoint convention. Therefore \mu_H(x) = \frac1r \sum_{\rho\in X}\bigl(\rho(x-1)-\rho(x)\bigr). Pairing with \chi and changing variables y=x-1 in the first term gives S_{\mu_H}(\chi) = \frac1r \sum_{\rho\in X} \left( \sum_{y\bmod q}\rho(y)\chi(y+1) - \sum_{x\bmod q}\rho(x)\chi(x) \right). The second sum is q-1 when \rho\chi is the principal character and 0 otherwise. This gives the displayed formula. ◻
Index-Two Orbits
Lemma 15 (Index-two endpoint reduction). Let q be an odd prime, let H=\ker\psi be the quadratic-residue subgroup, and let \chi_0 be the principal Dirichlet character modulo q. If \chi is odd and \chi\notin\{\chi_0,\psi\}, then S_{\mu_H}(\chi) = \frac12\bigl(J(\chi)-1\bigr), \qquad J(\chi)=\sum_y\psi(y)\chi(y+1).
Proof. For x\in\mathbb{Z}/q\mathbb{Z}, \mathbf 1_H(x) = \frac12\bigl(\chi_0(x)+\psi(x)\bigr) = \frac12\bigl(1+\psi(x)\bigr)-\frac12\delta_0(x), where the 1 in the final expression is the constant function and \delta_a denotes the indicator of the residue a. Therefore \mu_H(x) = \frac12\bigl(\psi(x-1)-\psi(x)\bigr) - \frac12\bigl(\delta_1(x)-\delta_0(x)\bigr). Pairing with \chi gives an endpoint contribution -\frac12\chi(1)+\frac12\chi(0)=-\frac12, because \chi(1)=1 and \chi(0)=0. The term -\tfrac12\sum_x\psi(x)\chi(x) vanishes by orthogonality, since \psi\chi\ne\chi_0. The remaining term is \frac12\sum_x\psi(x-1)\chi(x) = \frac12\sum_y\psi(y)\chi(y+1) = \frac12J(\chi). Adding the contributions proves the lemma. ◻
Theorem 16 (Index-two off-resonance law). Let q be an odd prime, H=\ker\psi the quadratic-residue subgroup, \psi the quadratic character, and \chi_0 the principal character. The off-resonance energy is \mathcal E_{\mathrm{off}}(q) =\sum_{\substack{\chi\ \mathrm{odd}\\ \chi\notin\{\chi_0,\psi\}}} |S_{\mu_H}(\chi)|^2. It is given by one of four polynomials in q. The branch is determined by the residue class of q modulo 8. \mathcal E_{\mathrm{off}}(q)= \begin{cases} (q-1)(q+3)/8, & q\equiv1\ (8),\\ (q-3)(q+3)/8, & q\equiv3\ (8),\\ (q-1)^2/8, & q\equiv5\ (8),\\ (q-5)(q+1)/8, & q\equiv7\ (8). \end{cases}
Proof. By Lemma 15, every off-resonance odd character satisfies S_{\mu_H}(\chi)=\tfrac12(J(\chi)-1), with J(\chi)=\sum_y\psi(y)\chi(y+1). With J_{\mathrm{std}}(\alpha,\beta) =\sum_t\alpha(t)\beta(1-t), the substitution t=-y gives J(\chi)=\psi(-1)J_{\mathrm{std}}(\psi,\chi). For an off-resonance odd character, \psi, \chi, and \psi\chi are nonprincipal. Therefore |J(\chi)|^2=q [4, 5]. Hence \begin{aligned} \mathcal E_{\mathrm{off}}(q) &=\tfrac14\sum_{\chi}\bigl|J(\chi)-1\bigr|^2\\ &=\tfrac14\Bigl(\sum_\chi|J(\chi)|^2 -2\,\mathrm{Re}\sum_\chi J(\chi)+N\Bigr)\\ &=\tfrac14\bigl((q+1)N-2\,\mathrm{Re}\,\Sigma\bigr), \end{aligned} where N is the number of off-resonance odd characters and \Sigma=\sum_\chi J(\chi). There are (q-1)/2 odd characters. The resonant odd ones are those trivial on H, namely the odd members of \{\chi_0,\psi\}. Since \psi is odd exactly when q\equiv3\ (4), N= \begin{cases} (q-1)/2, & q\equiv1\ (4),\\ (q-3)/2, & q\equiv3\ (4). \end{cases} Interchanging the sums gives \Sigma=\sum_y\psi(y)\sum_{\chi\ \mathrm{off}}\chi(y+1). The off-resonance family is the odd characters with the resonant ones removed, so the inner sum is the odd kernel less the resonant term, \sum_{\chi\ \mathrm{off}}\chi(z) =\tfrac{q-1}{2}\bigl([z\equiv1]-[z\equiv-1]\bigr) -[q\equiv3\ (4)]\,\psi(z), the subtraction of \psi occurring exactly when \psi is odd, i.e. q\equiv3\ (4). Substituting and using \psi(0)=0 in the kernel term, \begin{aligned} \Sigma &= -\frac{q-1}{2}\psi(-2) - [q\equiv3\ (4)]\sum_y\psi(y)\psi(y+1) \\ &= -\frac{q-1}{2}\psi(-2)+[q\equiv3\ (4)], \end{aligned} where the last step uses the standard evaluation \sum_y\psi(y(y+1))=-1 [4]. The minus sign on the Jacobi term, inherited from subtracting \psi, turns -(-1) into +1. In particular, \Sigma is real. By the supplementary laws [4], \psi(-1)=(-1)^{(q-1)/2} \qquad\text{and}\qquad \psi(2)=(-1)^{(q^2-1)/8}. Thus the product \psi(-2) and the parity of q\bmod4 are fixed by q\bmod8. Substituting the four cases of (N,\Sigma) into \mathcal E_{\mathrm{off}}=\tfrac14((q+1)N-2\Sigma) yields the stated values. ◻
Example 17. Take q=7\equiv7\ (8). The quadratic-residue subgroup is H=\{1,2,4\}, and \psi is odd since 7\equiv3\ (4), so among \{\chi_0,\psi\} the resonant odd character is \psi alone and N=(7-3)/2=2 off-resonance odd characters remain. Theorem 16 gives \mathcal E_{\mathrm{off}}(7)=(7-5)(7+1)/8=2. Direct computation of \sum|S_{\mu_H}(\chi)|^2 over the two off-resonance characters gives the value 2, and the reduction S_{\mu_H}(\chi)=\tfrac12(J(\chi)-1) with |J(\chi)|^2=7 recovers it through \tfrac14((q+1)N-2\Sigma) with \Sigma=4.
The Modulus-Eight Split
The modulus 8 is intrinsic. The boundary shift produces the quadratic value \psi(-2), and \psi(-2)=\psi(-1)\psi(2). The factor \psi(-1) depends on q\bmod4, while the supplementary law for \psi(2) depends on q\bmod8. The four cases record the quadratic data of -1 and 2 seen by the shifted boundary. The flux is a Jacobi sum, and the odd kernel closes its second moment over the off-resonance family.
Finite-State Orbit Boundaries
A digit statistic with finite memory also closes on a finite cycle. This produces a dynamical boundary on the lifted state space. It does not, by itself, produce an additive boundary on \mathbb{Z}/q\mathbb{Z}.
Theorem 18 (Finite-state lift). Let b\ge2 and q\ge2 satisfy (q,b)=1, and represent each R\in(\mathbb{Z}/q\mathbb{Z})^{\times} by its least positive residue. Let d(R)=\left\lfloor\frac{bR}{q}\right\rfloor be the emitted base-b digit. Let \mathcal A=(Q,s_0,\delta) be a deterministic finite automaton over the digit alphabet, and define T_{\mathcal A}(R,s) =\bigl(bR\bmod q,\delta(s,d(R))\bigr) \qquad (R,s)\in(\mathbb{Z}/q\mathbb{Z})^{\times}\times Q. Every orbit of T_{\mathcal A} is eventually periodic. On its eventual cycle C, a Boolean finite-state digit predicate is the indicator of a subset G\subseteq C with dynamical boundary \mu^{\mathrm{dyn}}_G(z) =\mathbf 1_G(T_C^{-1}z)-\mathbf 1_G(z). Every complex-valued finite-state statistic is likewise a weight w on C with boundary \mu^{\mathrm{dyn}}_w(z)=w(T_C^{-1}z)-w(z).
Proof. The remainder recurrence and the automaton recurrence together iterate T_{\mathcal A} on the finite set (\mathbb{Z}/q\mathbb{Z})^{\times}\times Q. Every orbit in a finite set is eventually periodic. On its eventual cycle, T_{\mathcal A} is a cyclic permutation. Taking predecessor value minus current value gives the displayed boundaries. ◻
The dynamical boundary uses the orbit predecessor T_C^{-1}. The additive boundary used in the character-energy theorems is instead \mu_{\widetilde w}(x)=\widetilde w(x-1)-\widetilde w(x) for a specified weight \widetilde w on all of \mathbb{Z}/q\mathbb{Z}. If the automaton state is determined by the current remainder, its values on the projected cycle define a weight there. An extension away from that cycle must still be specified before the additive boundary is formed. Different extensions can change its flux. In general \mu_w^{\mathrm{dyn}} is not the restriction or pushforward of \mu_{\widetilde w}. No Dirichlet-character identity involving \mu_w^{\mathrm{dyn}} is asserted.
Arithmetic Depth of the Edge
A finite digit region and a periodic remainder orbit enter the character system through the same object. Each leaves a signed additive boundary. Squared boundary flux then separates into nonnegative energies of exact conductor.
The base-15 spectrum shows why that separation matters. The signed top-conductor ledger isolates 896 units of energy at conductor 225. The full spectrum recovers another 1024 units at four proper conductors. Those lower levels are not an error term or a failed symmetry. They are part of the exact arithmetic structure of the boundary.
The full prime-power orbit gives the opposite lesson. It fills every unit class modulo p^e, but its additive edge comes from modulo p. Higher depth increases the size of the prime-level flux while every higher-conductor channel remains zero. Orbit size and conductor depth are different quantities.
Finite memory widens the state space without changing that distinction. The lifted cycle always has a dynamical predecessor boundary. Conductor energy begins only after an additive remainder weight has been specified on the whole residue space.
At prime modulus, an index-two orbit brings the additive shift into contact with the quadratic data of -1 and 2. The result is the four-case split modulo 8. For higher-index subgroups, the boundary flux already has an exact quotient-character expansion in Jacobi-type sums. The distribution of Jacobi-sum families is studied in [9]. The remaining problem is to decide when these particular finite boundary moments admit comparably simple closed forms.
The orbit’s apparent complexity does not set its arithmetic depth. Its edge does.
References
[1]A. S. Petty, Carry Boundaries and Bernoulli Spectra in Long Division, Zenodo, 2026. doi:10.5281/zenodo.20451178.
[2]H. Davenport, Multiplicative Number Theory, 3rd ed., Graduate Texts in Mathematics 74, Springer, 2000.
[3]L. C. Washington, Introduction to Cyclotomic Fields, 2nd ed., Graduate Texts in Mathematics 83, Springer, 1997.
[4]K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, 2nd ed., Graduate Texts in Mathematics 84, Springer, 1990.
[5]B. C. Berndt, R. J. Evans, and K. S. Williams, Gauss and Jacobi Sums, Wiley, 1998.
[6]H. Iwaniec and E. Kowalski, Analytic Number Theory, American Mathematical Society Colloquium Publications 53, American Mathematical Society, 2004.
[7]D. H. Lehmer, E. Lehmer, and W. H. Mills, ``Pairs of consecutive power residues,'' Canadian Journal of Mathematics 15 (1963), 172–177. doi:10.4153/CJM-1963-020-4.
[8]N. Alon and J. Bourgain, ``Additive patterns in multiplicative subgroups,'' Geometric and Functional Analysis 24 (2014), no. 3, 721–739. doi:10.1007/s00039-014-0270-y.
[9]Q. Lu, W. Zheng, and Z. Zheng, ``On the distribution of Jacobi sums,'' Journal für die reine und angewandte Mathematik 741 (2018), 67–86. doi:10.1515/crelle-2015-0087.