The General Neutrality Theorem
Abstract
At fixed base and lag, the collision deviation is determined by a finite table on the units modulo a power of the base. Reflection gives that raw table mean negative one half. Centering over reduction fibers makes the table odd and gives every fiber sum zero.
An internal residue belongs to the canonical representative in the finite table. The true residue of an integer also depends on its quotient layer. Reflection fixes one internal class for every odd prime not dividing the base, while the true residue classes satisfy a stronger law. Every nonzero channel modulo every prime is independently centered-neutral. When the prime does not divide the base, each channel contains one complete copy of the finite table and has raw mean negative one half.
The channel law holds simultaneously for every squarefree modulus. Mertens’ theorem in arithmetic progressions then gives separate convergence at one in every true channel. An exact finite character decomposition isolates the remaining nonprincipal fluctuation. A base-ten lag-one calculation shows that its character contributions have mixed signs. Below one, the arithmetic assignment between the active collision coefficients and the finite prime-character sums remains open.
Every prime channel
The two nonzero residue classes modulo three are neutral on their own. Nothing in the joint-modulus argument asks for three. If a prime q does not divide the base, each nonzero q-channel contains one complete copy of the centered collision table. If q divides the base, each channel is a union of complete centered fibers. Either way its sum is zero.
This turns a residue-three law into a law at every prime. Several prime coordinates may be imposed at once without creating a principal term. The result matters because Mertens’ theorem in arithmetic progressions then acts inside each true channel separately. Convergence at one is built into the finite geometry before the remaining character fluctuations meet.
The finite table and its fiber centering arise in The Centered Collision Sum [2]. Its finite character expansion is developed in The Collision Transform and the Critical Strip [3]. The Neutrality Theorem [4] proves the independent residue-three law. The new step is the passage from that pair of channels to arbitrary prime channels and simultaneous squarefree coordinates. The definitions and proofs are included so the argument remains complete here.
The finite collision table
Fix a base b\geq2 and a lag \ell\geq1. Let p>b^{\ell+1} be coprime to b. For 1\leq r<p, put \delta_{p,b}(r)=\left\lfloor\frac{br}{p}\right\rfloor. If [x]_p denotes the representative of x\bmod p in \{1,\ldots,p-1\}, define C_{b,\ell}(p)= \#\left\{1\leq r<p\ \middle| \delta_{p,b}(r)=\delta_{p,b}([b^\ell r]_p)\right\} and S_{b,\ell}(p)=C_{b,\ell}(p) -\left\lfloor\frac{p-1}{b}\right\rfloor.
Set m=b^{\ell+1} and let G_{b,\ell}= \left\{0\leq n<m\ \middle| \left\lfloor\frac{n}{b^\ell}\right\rfloor=n\bmod b\right\}. The first and last base-b digits agree on G_{b,\ell}, while the intermediate digits are free. Hence |G_{b,\ell}|=b^\ell.
Theorem 1 (Finite determination). Let p=mt+a with 1\leq a<m and \gcd(a,m)=1. Then S_{b,\ell}(p)=T_{b,\ell}(a), where T_{b,\ell}(a)= -1-\left\lfloor\frac ab\right\rfloor +\sum_{n\in G_{b,\ell}} \left( \left\lfloor\frac{(n+1)a}{m}\right\rfloor -\left\lfloor\frac{na}{m}\right\rfloor \right).
Proof. Put n(r)=\lfloor mr/p\rfloor. The identity \left\lfloor\frac{\lfloor x\rfloor}{k}\right\rfloor =\left\lfloor\frac{x}{k}\right\rfloor for positive integral k gives \delta_{p,b}(r)=\left\lfloor\frac{n(r)}{b^\ell}\right\rfloor. Writing b^\ell r as a quotient and remainder modulo p gives \delta_{p,b}([b^\ell r]_p)=n(r)\bmod b. A collision therefore occurs exactly on the slices indexed by G_{b,\ell}.
No interior slice boundary is integral because p is coprime to m. The terminal slice counts the excluded endpoint p, so C_{b,\ell}(p)= -1+\sum_{n\in G_{b,\ell}} \left( \left\lfloor\frac{(n+1)p}{m}\right\rfloor -\left\lfloor\frac{np}{m}\right\rfloor \right). Substitution of p=mt+a contributes t on each of the b^\ell selected slices. Since a is a unit modulo b, \left\lfloor\frac{p-1}{b}\right\rfloor =b^\ell t+\left\lfloor\frac ab\right\rfloor. Subtracting proves (7). ◻
The exact formula defines T_{b,\ell} on U_m=(\mathbb Z/m\mathbb Z)^\times without choosing prime representatives.
Reflection and centering
Every unit will be written as its representative in \{1,\ldots,m-1\}.
Lemma 2 (Reflection). For every a\in U_m, T_{b,\ell}(a)+T_{b,\ell}(m-a)=-1. Consequently, \frac1{\varphi(m)}\sum_{a\in U_m}T_{b,\ell}(a)=-\frac12.
Proof. Write D_n(a)= \left\lfloor\frac{(n+1)a}{m}\right\rfloor -\left\lfloor\frac{na}{m}\right\rfloor. For 1\leq n\leq m-2, complementary floors give D_n(a)+D_n(m-a)=1. The corresponding sums at n=0 and n=m-1 are 0 and 2. Both endpoints belong to G_{b,\ell}, so \sum_{n\in G_{b,\ell}} \bigl(D_n(a)+D_n(m-a)\bigr)=b^\ell. Since a is a unit modulo b, \left\lfloor\frac ab\right\rfloor +\left\lfloor\frac{m-a}{b}\right\rfloor=b^\ell-1. Equation (8) follows from (7). Negation has no fixed point on U_m. Averaging the reflection identity proves (9). ◻
Let \rho:U_m\longrightarrow U_b be reduction modulo b. For u\in U_b, write A_u=\{a\in U_m\mid \rho(a)=u\}. Every fiber contains exactly b^\ell elements. Define \mu_{b,\ell}(u)= \frac1{b^\ell}\sum_{a\in A_u}T_{b,\ell}(a) and f_{b,\ell}(a)=T_{b,\ell}(a)-\mu_{b,\ell}(\rho(a)).
Proposition 3 (Centered reflection). For every u\in U_b and every a\in U_m, \begin{aligned} \mu_{b,\ell}(u)+\mu_{b,\ell}(-u)&=-1, \\ f_{b,\ell}(m-a)&=-f_{b,\ell}(a), \\ \sum_{a\in A_u}f_{b,\ell}(a)&=0. \end{aligned} In particular, \sum_{a\in U_m}f_{b,\ell}(a)=0.
Proof. Negation maps A_u bijectively onto A_{-u}. Averaging (8) over A_u proves (12). Subtracting that identity from the pointwise reflection law proves (13). Equation (14) is the definition of the fiber mean. Summing over the fibers proves (15). ◻
Internal residue classes
Let q be an odd prime not dividing b. For k\in\mathbb Z/q\mathbb Z, define the internal canonical slice I_q(k)= \left\{a\in\{1,\ldots,m-1\}\ \middle| \gcd(a,m)=1,\ a\equiv k\pmod q\right\}. This is a partition of the chosen representatives inside one finite table. It is not a partition of the integers carrying those representatives.
Theorem 4 (Internal reflection neutrality). Let k^* be the unique solution of 2k^*\equiv m\pmod q. Reflection preserves I_q(k^*) and exchanges every other slice with I_q(m-k). The fixed slice satisfies \begin{aligned} 2\sum_{a\in I_q(k^*)}T_{b,\ell}(a)&=-|I_q(k^*)|, \\ \sum_{a\in I_q(k^*)}f_{b,\ell}(a)&=0. \end{aligned} When I_q(k^*) is nonempty, its raw mean is -1/2.
Proof. The map a\mapsto m-a sends the label k to m-k modulo q. The fixed label is the unique solution of (17) because 2 is invertible modulo q. Summing (8) over the fixed slice gives (18). Summing (13) gives (19). ◻
The word internal matters. The representative a and an integer p=mt+a have residues related by p\equiv a+mt\pmod q. The quotient layer t changes the true residue whenever q does not divide m. Thus the internal slice belongs to one finite table, while a true residue channel belongs to the joint modulus that carries both coordinates.
True residue channels
Let q now be any prime and put M_q=\operatorname{lcm}(m,q). For r\in U_q, define B_{q,r}=\{A\in U_{M_q}\mid A\equiv r\pmod q\} and lift the centered table by \widetilde f(A)=f_{b,\ell}(A\bmod m).
Theorem 5 (General neutrality). For every prime q and every r\in U_q, \sum_{A\in B_{q,r}}\widetilde f(A)=0. The channel size is |B_{q,r}|= \begin{cases} \varphi(m),&q\nmid b,\\ \varphi(m)/(q-1),&q\mid b. \end{cases} If q\nmid b, projection onto U_m is a bijection and \frac1{|B_{q,r}|} \sum_{A\in B_{q,r}}T_{b,\ell}(A\bmod m)=-\frac12.
Proof. Suppose first that q\nmid b. Then q\nmid m and M_q=mq. The Chinese remainder theorem gives U_{M_q}\cong U_m\times U_q. For fixed r, projection from B_{q,r} onto U_m is a bijection. Equations (24) and (27) follow from (15) and (9). The bijection also gives the first line of (25).
Suppose now that q\mid b. Then M_q=m. The set B_{q,r} is the union of the reduction fibers A_u satisfying u\equiv r\pmod q. Every one of those fibers has centered sum zero by (14). Their union therefore has centered sum zero. Reduction from U_m onto U_q is surjective with equal fibers, which gives the second line of (25). ◻
The theorem includes q=2. The failure of 2 to have an inverse modulo 2 affects the internal fixed-label statement, not the true-channel law. For primes not dividing the base, every true channel contains one complete copy of the finite table. No true channel is distinguished.
Simultaneous neutrality
The same argument applies to any finite set of prime coordinates at once.
Theorem 6 (Simultaneous neutrality). Let Q be squarefree, put M_Q=\operatorname{lcm}(m,Q), \qquad d=\gcd(Q,b), and let r\in U_Q. Define B_{Q,r}=\{A\in U_{M_Q}\mid A\equiv r\pmod Q\}. Then \sum_{A\in B_{Q,r}}f_{b,\ell}(A\bmod m)=0. Moreover, |B_{Q,r}|= \frac{\varphi(m)}{\varphi(d)} =\frac{\varphi(M_Q)}{\varphi(Q)}. If \gcd(Q,b)=1, the raw mean on B_{Q,r} is -1/2.
Proof. Since Q is squarefree, \gcd(Q,m)=\gcd(Q,b)=d. The generalized Chinese remainder theorem identifies B_{Q,r} with C_{d,r}=\{a\in U_m\mid a\equiv r\pmod d\}. Indeed, a unit a modulo m has a unique compatible lift modulo M_Q exactly when a\equiv r\pmod d. Because d divides b, the set C_{d,r} is a union of reduction fibers A_u. Equation (28) follows from (14). Reduction from U_m onto U_d has equal fibers, proving (29). When d=1, the set C_{d,r} is all of U_m, so the raw mean is -1/2 by (9). ◻
This is simultaneous neutrality of channel means. It is not a pointwise condition on the individual integers in a channel.
Separate convergence at one
Theorem 7 (Channel convergence). Let Q be squarefree and r\in U_Q. The limit \lim_{X\to\infty} \sum_{\substack{M_Q<p\leq X\\p\equiv r\pmod Q}} \frac{f_{b,\ell}(p\bmod m)}p exists.
Proof. For every A\in U_{M_Q}, Mertens’ theorem in arithmetic progressions gives [1] \sum_{\substack{p\leq X\\p\equiv A\pmod{M_Q}}}\frac1p =\frac1{\varphi(M_Q)}\log\log X+C_{M_Q}(A)+o(1). Multiply by f_{b,\ell}(A\bmod m) and sum over A\in B_{Q,r}. The coefficient of \log\log X vanishes by (28). The remaining expression is a finite sum of constants and terms tending to zero. Omitting the primes at or below M_Q does not affect convergence. ◻
Neutrality removes the principal logarithmic growth. It supplies no estimate for a real exponent below one. That problem belongs to the nonprincipal character fluctuations.
The finite character decomposition
For a Dirichlet character \chi\bmod m, define \widehat f(\chi)= \frac1{\varphi(m)}\sum_{a\in U_m} f_{b,\ell}(a)\overline{\chi(a)}.
Proposition 8 (The active channels). The coefficient \widehat f(\chi) vanishes when \chi is even. It also vanishes when \chi factors through reduction from U_m to U_b.
Proof. If \chi(-1)=1, the substitution a\mapsto m-a and (13) give \widehat f(\chi)=-\widehat f(\chi), so the coefficient is zero. If \chi=\psi\circ\rho, then \varphi(m)\widehat f(\chi) =\sum_{u\in U_b}\overline{\psi(u)} \sum_{a\in A_u}f_{b,\ell}(a)=0 by (14). ◻
Call a character active when \widehat f(\chi)\ne0. Every active character is odd and does not descend to the base.
Base ten has one further exact silence. Let \chi_4 be the primitive character modulo four, lifted to modulus one hundred. It does not descend to U_{10} because 1 and 11 agree modulo ten but have opposite \chi_4-values.
Proposition 9 (The conductor-four silence). For b=10 and \ell=1, \widehat f(\chi_4)=0.
Proof. Write \mathfrak s(x)= \begin{cases} \{x\}-1/2,&x\notin\mathbb Z,\\ 0,&x\in\mathbb Z. \end{cases} For a positive odd integer L and an integer k, put J_L(k)= \sum_{\substack{a\bmod 4L\\a\ \mathrm{odd}}} \chi_4(a)\mathfrak s\left(\frac{ka}{4L}\right). Let d=\gcd(k,L), k=dk', and L=dL'. Writing a=4t+1 or a=4t+3 and using \sum_{t=0}^{n-1}\mathfrak s\left(x+\frac tn\right) =\mathfrak s(nx) gives J_L(k)=d\left( \mathfrak s\left(\frac{k'}4\right) -\mathfrak s\left(\frac{3k'}4\right) \right) =-\frac d2\chi_4(k').
Now put H(k)=\sum_{a\in U_{100}} \chi_4(a)\mathfrak s\left(\frac{ka}{100}\right). The units modulo one hundred are the odd residues with the odd multiples of five removed. Hence H(k)=J_{25}(k)-J_5(k). Equation (34) shows that H(k)= \begin{cases} -10\chi_4(k/25),&25\mid k,\\ 0,&25\nmid k. \end{cases}
Let F(k)=\sum_{a\in U_{100}}\chi_4(a) \left\lfloor\frac{ka}{100}\right\rfloor. Since \sum_{a\in U_{100}}\chi_4(a)=0, the floor and sawtooth forms give F(k)=\frac{k}{100} \sum_{a\in U_{100}}\chi_4(a)a-H(k). For lag one, G_{10,1}=\{0,11,22,33,44,55,66,77,88,99\}. The linear terms cancel in -F(10)+\sum_{n\in G_{10,1}}\bigl(F(n+1)-F(n)\bigr) because |G_{10,1}|=10. The remaining terms are values of H at 10, at members of G_{10,1}, and at their successors. None is 25 or 75 modulo one hundred, so every term vanishes by (35). This proves that the raw table is orthogonal to \chi_4.
Fiber centering does not change the coefficient. In each fixed residue class modulo ten, \chi_4 takes each sign five times. Every fiber-constant subtraction is therefore orthogonal to \chi_4. Equation (33) follows. ◻
For a finite cutoff X>m and a real exponent \sigma>0, put \begin{aligned} P_X(\sigma,\chi)&= \sum_{m<p\leq X}\frac{\chi(p)}{p^\sigma}, \\ F_X(\sigma)&= \sum_{m<p\leq X}\frac{f_{b,\ell}(p\bmod m)}{p^\sigma}. \end{aligned}
Proposition 10 (Exact finite decomposition). For every finite X and every \sigma>0, F_X(\sigma)= \sum_{\chi\ \mathrm{active}} \widehat f(\chi)P_X(\sigma,\chi).
Proof. Fourier inversion on U_m gives f_{b,\ell}(a)=\sum_{\chi\bmod m}\widehat f(\chi)\chi(a). Substitution into the finite prime sum proves the identity. Proposition 8 removes every inactive term. ◻
The cutoff is essential at \sigma=1/2. No infinite value of P(1/2,\chi) is assumed.
A finite lag-one assignment
The finite calculations were performed with nfield [5]. The collision table was evaluated directly from (7). For the numerical tables, a character was retained when its coefficient magnitude exceeded 10^{-12} and was otherwise excluded from correlations. The smallest retained base-ten magnitude was 0.1226. The largest numerical residual among its three proved zeros was less than 5.1\mathbin{\cdot}10^{-19}.
An exact lag-one audit over bases from two through twenty used the internal primes 3, 5, 7, 11, 13, 17, and 19. It checked 1{,}744 reflection values and 209 fiber identities. Among the 117 internal cases covered by Theorem 4, 101 were nonempty and 16 were empty. Another 16 cases had the prime dividing the base and lay outside that theorem. All had an empty formal fixed slice. The true-channel audit used 2, 3, 5, 7, 11, 13, 17, and 19. It checked 1{,}311 prime channels and 2{,}014 simultaneous channels with squarefree moduli 6, 30, 105, and 210. These finite checks support the implementation. The proofs are symbolic and do not depend on them.
Table 1 uses base ten, lag one, and all 348{,}488 primes in the interval 100<p\leq5{,}000{,}000.
| \sigma | positive | negative | exact zero | net | correlation |
|---|---|---|---|---|---|
| 1.0 | 10 | 7 | 3 | +0.077446 | -0.4081 |
| 0.8 | 12 | 5 | 3 | +0.185838 | -0.5169 |
| 0.6 | 10 | 7 | 3 | +0.444176 | -0.4925 |
| 0.5 | 10 | 7 | 3 | +0.725464 | -0.4192 |
The mixed signs rule out any argument that treats the displayed finite decomposition as termwise positive. They do not show that cancellation continues as the cutoff grows. The changing signs and magnitudes leave an exact finite assignment whose behavior with lag and base is not yet known.
The surviving assignment
Every true prime channel is centered-neutral, and the law survives any finite squarefree collection of prime coordinates. Mertens’ theorem in arithmetic progressions therefore gives convergence at one separately in each channel. The principal logarithmic drift is gone before the character terms are assembled.
Below one, the exact finite decomposition leaves finitely many active odd collision coefficients paired with finite prime-character sums. The base-ten lag-one data show that these terms already point in both directions. Their finite signs do not determine their course as the cutoff grows.
The open object is the assignment itself. How do the active collision coefficients meet the finite prime-character sums as the lag changes and across coprime bases? Neutrality removes the principal direction. The surviving assignment carries the next arithmetic question.
References
[1]H. L. Montgomery and R. C. Vaughan, Multiplicative Number Theory I, Cambridge University Press, 2007.
[2]A. S. Petty, The Centered Collision Sum, research note, October 2023 (revised August 2026). doi:10.5281/zenodo.21852556
[3]A. S. Petty, The Collision Transform and the Critical Strip, research note, January 2024 (revised August 2026). doi:10.5281/zenodo.21853435
[4]A. S. Petty, The Neutrality Theorem, research note, April 2024 (revised August 2026). doi:10.5281/zenodo.21853667
[5]A. S. Petty, nfield, software repository. https://github.com/alexspetty/nfield