The Analytic Collision Transform

Write out the two-digit numbers in base five, including a leading zero where needed. The first row is 00, 01, 02, 03, 04. The second is 10, 11, 12, 13, 14. Continue to 44.

Now mark the entries whose two digits agree.

You have marked a diagonal. There are five entries on it, and their values in ordinary decimal notation are 0, 6, 12, 18 and 24.

This small diagonal already contributes to the collision spectrum. Its character sum is one factor in the exact formula for a collision coefficient. I want to follow that factor while an analytic parameter changes. Can the finite count become part of a family of functions, with a precise way to recover the number we started from?

It can. The construction uses the periodic zeta function, and the place where it returns the finite coefficient is $s=0$.

Five positions

A character assigns complex weights to the residues, with multiplication of residues becoming multiplication of weights. At each marked position $n$, take the change in weight from $n$ to $n+1$. Add the five changes.

Call the result $D(\chi)$,

$$D(\chi)=\sum_{n\in\{0,6,12,18,24\}}[\chi(n+1)-\chi(n)].$$

The arithmetic wraps around at 25. Character values at multiples of five are zero.

For a concrete choice of character, start at $\chi(1)=1$. Each time a residue is multiplied by 2 modulo 25, turn its weight through 18 degrees on the unit circle. Twenty such steps return to the start and visit every residue coprime to 25. With these weights, the sum is approximately

$$D(\chi)\approx5.520147+0.206396i.$$

That is a single fixed complex number. The two coordinates record its horizontal and vertical components. There is no $s$ in the calculation.

At greater lag, more digits lie between the first and last. In base five, for example, the three-digit strings with matching ends include 000, 010, 020, 030, 040 and 101. There are twenty-five of them altogether. The same construction works with their twenty-five boundary differences. In general, lag $\ell$ uses modulus $m=b^{\ell+1}$ and a diagonal of $b^\ell$ positions. I will keep the five-position example in view.

A parameter in the weights

The periodic zeta function starts with a familiar kind of sum. Put a turning arrow at each positive integer. The arrow at integer $j$ has made $jx$ full turns, and its length is $1/j^{1-s}$ when $s$ is real. Add the arrows,

$$K(x;s)=\sum_{j=1}^{\infty}\frac{e^{2\pi i jx}}{j^{1-s}}.$$

At $s=0$, the lengths are $1,1/2,1/3,\ldots$. At $s=1/2$, they are $1,1/\sqrt2,1/\sqrt3,\ldots$. Increasing $s$ gives the more distant terms greater weight. Their directions provide cancellation.

For $0<x<1$, this series converges when the real part of $s$ is less than one. Its continuation supplies the values beyond that region. This is a classical periodic zeta function, written with parameter $1-s$ so that its boundary agrees with the finite calculation.

There is a concrete reason to use it. At $s=0$, the imaginary part of the kernel is

$$\operatorname{Im}K(x;0)=\pi\left(\frac12-x\right).$$

After dividing by $\pi$, that is a straight line falling from $1/2$ to $-1/2$. It is the negative of the first Bernoulli polynomial. The Bernoulli factor in the collision spectrum can therefore be recovered from the boundary of this moving kernel.

To build the transform, use the kernel at the two ends of each marked step. For every residue $a$ coprime to 25, compare its value at $(n+1)a$ with its value at $na$, wrapping the arguments modulo 25 and dividing by 25. Add those differences over the diagonal. The two endpoint steps are paired first, so their potentially singular terms at zero cancel before evaluation.

As with the finite collision table, subtract the mean within each class having the same final base-five digit. Then average with the conjugate character weights. Write this resulting function as $\mathcal A_\chi(s)$. It is the centered analytic collision transform.

The exact product

The critical strip consists of the complex numbers $s$ whose real part lies between zero and one. Its horizontal coordinate can vary between those two edges while its imaginary coordinate extends in either direction.

For a prime base and a primitive odd character, the transform has an exact factorization. Here primitive means that the character needs the full modulus, and odd means that reflection changes the sign of its value.

The identity is

$$\mathcal A_\chi(s)=C_m(s)\,D_\ell(\chi)\,L(s,\overline\chi),$$

where the common analytic factor is

$$C_m(s)=\frac{2i\sin(\pi s/2)\Gamma(s)}{\varphi(m)} \left(\frac{m}{2\pi}\right)^s.$$

The notation $\varphi(m)$ counts the residues coprime to $m$, so it is 20 in our example. The gamma and sine functions belong to the classical analytic factor. The bar over $\chi$ conjugates its complex weights. The character-weighted number-theoretic series is the Dirichlet $L$-function.

There are three factors to follow. One depends on $s$ and the modulus. One depends on the digit diagonal and the character. One is the $L$-function itself.

The proof separates them. Multiplication by a residue coprime to the modulus rearranges the character-weighted kernel sum. If that residue is not coprime, primitivity makes the corresponding weights cancel. Oddness combines the two reflected terms in the kernel into the sine factor. After summing over the marked steps, the only digit-dependent term left outside the $L$-function is $D_\ell(\chi)$.

This constructs an analytic family from the diagonal. An identification with a prime-weighted collision sum would be a separate statement.

The part that does not move

Return to the character with the 18-degree turn. Evaluate its transform at different real values of $s$ between zero and one. The values change. Remove the analytic and $L$-function factors at points where they are nonzero, and the quotient is always the same number we computed from five positions.

The figure uses nineteen evaluations from nfield. These are numerical checks of the identity, with the kernel sums evaluated directly on the finite diagonal. The identity itself is proved for complex $s$ throughout the open strip.

The distinction is useful. We can change the analytic parameter without recomputing the digit geometry. For a fixed base, lag and character, its entire contribution is already in $D_\ell(\chi)$.

The two edges

At $s=0$, the common factor has the limiting value $i\pi/\varphi(m)$. The classical identity $L(0,\overline\chi)=-B_{1,\overline\chi}$ then gives

$$\mathcal A_\chi(0)=-\frac{i\pi}{\varphi(m)} B_{1,\overline\chi}D_\ell(\chi).$$

At lag one in an odd prime base, the product on the right is exactly $i\pi$ times the finite collision coefficient. Thus

$$\frac{\mathcal A_\chi(0)}{i\pi}=\widehat S^\circ(\chi).$$

This recovers the full complex coefficient, including its phase. For our base-five character, it is approximately $0.441164-0.520524i$.

The opposite edge has a different expression,

$$\mathcal A_\chi(1)=-\frac{\tau(\overline\chi)}{\varphi(m)} B_{1,\chi}D_\ell(\chi).$$

Here $\tau$ is a Gauss sum, a finite sum combining character weights with equally spaced points on a circle. Notice the changed Bernoulli index as well. The Dirichlet functional equation relates the two edges. It does not make their values identical.

Zeros of the product

The factor $C_m(s)$ never vanishes inside the open strip. If the fixed diagonal factor is also nonzero, the product can vanish exactly where its $L$-function vanishes. Even the multiplicities agree. If the diagonal factor is zero, the whole character component is identically zero there.

This tells us exactly which zeros an active component carries. Their location remains the $L$-function problem. The factorization supplies no new zero-free region and no comparison uniform in the growing modulus.

I can now follow the contribution of the digits through the whole calculation. It begins at five marked positions, enters an identity in a complex variable, and returns the finite collision coefficient at the left edge. The character sum never needs to be adjusted to make that recovery work. It is the one calculated from the diagonal at the start.