The Collision Transform

The collision invariant assigns a signed integer to every coprime class modulo $b^2$. One number per class. Positive here, negative there, locked in complement pairs that sum to $-1$. A pattern on a finite group.

But what kind of pattern? Is it smooth or jagged? Does it favor certain frequencies or spread its energy evenly? Is there a dominant tone, or is it noise?

To answer that, you decompose it. The same way you decompose a sound into its constituent frequencies, you decompose the collision invariant into Dirichlet characters. Each character is a frequency. Each frequency carries some amplitude. The collection of amplitudes is the collision transform.

When I computed it, two things happened that I did not expect.

The silence

Half the frequencies are dead.

The collision invariant satisfies $S^\circ(a) + S^\circ(b2 - a) = 0$. The complement map flips the sign. This acts like clamping a vibrating string at both ends: only certain vibration patterns are compatible. A string clamped at both ends cannot vibrate in even harmonics. The nodes fall in the wrong places. Only the odd harmonics survive.

Dirichlet characters come in two parities. Even characters satisfy $\chi(-1) = +1$. Odd characters satisfy $\chi(-1) = -1$. The complement antisymmetry kills every even character's contribution to the collision transform. Completely. Not approximately, not in the limit. The coefficient is exactly zero.

In base 10, there are 40 coprime classes modulo 100. The character group has 40 frequencies. Twenty are even. Twenty are odd. The collision invariant is inaudible at twenty frequencies. It speaks only in the odd half of the spectrum.

I did not build this in. The digit function floor(br/p) has no obvious parity. The complement involution creates it. The antisymmetry is a consequence of how remainders pair under $r \mapsto p - r$, and it reaches up through the collision count to silence half the Fourier spectrum.

The cancellation

Now sum the collision deviations over primes. Each prime $p$ carries a centered value $S^\circ(p)$: how much its collision count deviates from the class mean. Weight by $1/p^s$ and add them all up:

At $s = 2$: easy convergence. The weights $1/p^2$ shrink fast. Nothing to see.

At $s = 1$: each prime gets weight $1/p$. The sum of $1/p$ over primes diverges. Euler proved this in 1737. So the raw collision counts, summed with weight $1/p$, would diverge too. But the centered values are not all positive. They oscillate. Positive at some primes, negative at others, driven by the antisymmetry and the class structure.

The oscillation produces cancellation. The positive terms and the negative terms wash each other out, and the sum converges. At $s = 1$. At the edge of the critical strip. At the exact boundary where the prime number theorem lives.

The boundary $s = 1$ is where the prime number theorem lives. The classical PNT is equivalent to $\sum \mu(n)/n$ converging: the Mobius function's cancellation at this same boundary. The collision invariant cancels here too, with a different function, for a different reason. The digit function, which knows nothing about primes, which is a floor function applied to remainders, produces deviations that cancel across primes at the PNT boundary.

I did not expect this. The collision invariant was defined by counting bin coincidences. There is no reason, from the definition, to expect cancellation at $s = 1$. But the decomposition into odd characters connects the collision sum to prime character sums $\sum \chi(p)/p^s$, and the classical theory of those sums gives convergence at $s = 1$ for free. The cancellation is inherited. The digit function, through its Fourier decomposition, taps into the same cancellation mechanism that drives the prime number theorem.

Below the boundary

At $s < 1$: unconditional convergence is no longer guaranteed. The cancellation at $s = 1$ is the tightest the collision invariant can manage without further hypotheses.

But under a hypothesis, the convergence extends deeper. If the L-functions attached to the odd characters have no zeros near the real axis, the collision sum continues to cancel below $s = 1$, into the critical strip. The further the zeros stay away, the deeper the cancellation reaches.

Each zero of an L-function acts like a resonance point. At that height in the complex plane, the prime character sum amplifies instead of cancelling. A zero near $s = \sigma$ disrupts the convergence of $F^\circ(\sigma)$. A zero far away leaves it undisturbed.

The collision fluctuation, defined by the digit function, has its analytic behavior controlled by the zeros of L-functions. The digit function does not see the zeros. But its Fourier spectrum is tuned to the same frequencies, and the convergence of its prime sum depends on whether those frequencies resonate or cancel.

The mod-3 structure

One more fact falls out of the algebra. In base 10, the coprime classes modulo 100 partition into three families by their residue modulo 3. The neutrality theorem says: one family (the neutral class, $a \equiv 2 \pmod{3}$) has mean collision deviation exactly $-1/2$. The other two families are swapped by the complement map, their means summing to $-1$. The prime 3 is special in the digit function's arithmetic. It organizes the class means of the collision invariant into a three-part structure with a neutral center.

The collision transform takes the signed pattern on the finite group and reveals its inner architecture: half the spectrum silent, the surviving half cancelling across primes at the PNT boundary, the class means organized by the prime 3, the convergence controlled by L-function zeros deep in the strip. The digit function is a floor function. Its Fourier decomposition is a window into the analytic theory of primes. The distance between those two facts is the content of this paper.

Code: github.com/alexspetty/nfield
Paper: The Collision Transform

Alexander S. Petty
March 2026
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