Magnitude Decorrelation in the Collision Spectrum

The Collision Spectrum ends with a question. Two magnitudes, both extracted from the digit function, keep rising and falling together. At base five the correspondence is exact. Their correlation is one. At base seven it is about $0.844$. By base seventy-one it is still about $0.675$.

Does that positive relation last as the base grows?

Its limit is zero.

There is a second result beside it. Square the magnitudes before comparing them, and a substantial relation persists. The average of their product approaches three times the product of their separate averages. The cubic law determines that three.

I want to show how both can be true.

The two numbers

A character assigns a complex weight to each residue class. Think of each weight as a little arrow. To add the weights, place the arrows head to tail. The magnitude of the sum is the distance from the starting point to the endpoint.

The short sum adds the character weights at the first $b-1$ integers. The bin-boundary calculation in The Collision Spectrum reduces the diagonal factor to this short sum, apart from a factor of two and a rotation. Those changes preserve its pattern of magnitudes across the characters.

The Bernoulli sum reads the whole set of residues modulo $b^2$. It multiplies each character weight by the position of its residue, adds, and divides by $b^2$. A classical identity connects its magnitude to the $L$-function value,

$$|B_{1,\overline\chi}|=\frac b\pi\,|L(1,\chi)|.$$

One sum uses a short initial interval. The other uses the full residue system. At base five, their magnitudes are proportional at every primitive odd character.

Here, primitive means the character pattern does not come from a smaller modulus. Odd means changing a residue to its negative reverses the character weight. These are the channels left by the collision table’s centering and reflection.

For each prime base, I compare all of them. Both members of every conjugate pair are included.

The finite picture

It helps to put the magnitudes on scales that stay comparable as the base changes. Divide the short-sum magnitude by $\sqrt b$ and call the result $X$. Write $Y=|L(1,\chi)|$. These rescalings do not change the correlation at a fixed base.

The average square of $X$ is exactly one. The average square of $Y$ approaches $\pi^2/6$.

The two plots use the same eighteen prime bases and the same 14,372 characters.

On the left is the ordinary correlation. It measures whether departures from one magnitude’s mean tend to accompany departures from the other’s mean.

On the right is a different calculation. Multiply $X^2$ by $Y^2$ for each character and average those products. Then divide by the product of the two separate square averages,

$$R_b=\frac{\operatorname{average}(X^2Y^2)} {\operatorname{average}(X^2)\operatorname{average}(Y^2)}.$$

A value of three means the joint square average is three times that separate-average baseline.

The finite plot is still a long way from zero on the left. Drawing a line through those points would not settle the question.

Squaring changes the balance

Here is a small example to keep in mind. It is an illustration of averaging, separate from the character data.

Take one hundred numbers. Make ninety-nine of them zero and the last one ten. Their average is $0.1$. Their average square is one.

Now take ten thousand numbers. Make all but one zero and the last one one hundred. Their average falls to $0.01$. Their average square is still one.

The large value becomes rarer and larger at the same time. Squaring gives it enough weight to keep the second average fixed.

Something of this kind happens to the normalized short sums. For any fixed positive threshold, the proportion of characters above it tends to zero. Yet those characters carry a share of the total square mass that tends to one.

The finite samples already let us inspect where that mass sits.

At base seventy-one, take the 245 characters with the largest short-sum magnitudes. They are one tenth of the family. Together they supply about $54.1$ percent of its square mass.

That finite imbalance is visible. The theorem goes further. As the prime base grows, a vanishing proportion carries essentially all of the square mass.

This is why an ordinary magnitude comparison and a squared comparison can behave so differently.

Getting from the table to the limit

The proof needs control of the rare large values. Showing that most values are small would leave open the possibility that a few enormous products keep the average high.

The useful estimate sits between the ordinary mean and the mean square. Average $X^{3/2}$. An argument of Adam Harper gives a way to control this fractional moment of short character sums. His stated theorem uses a prime modulus. I check the transfer to the prime-square modulus here, where every integer in the required comparisons remains a unit.

That fractional moment tends to zero.

The $L$-value side needs two controls of its own. Its third moment stays bounded, which limits the contribution from very large values. Its variance stays above a fixed positive number, so the correlation denominator does not collapse.

The variance bound comes from a short exact identity. Weight the square magnitude by the character’s value at two. The unweighted character values average to zero, but the weighted square average is exactly half the ordinary square average. If the magnitudes had almost no variation, that distinction could not persist. Cauchy–Schwarz turns it into a definite lower bound.

Hölder’s inequality then controls the average product of the two magnitudes. The resulting correlation bound is

$$|r_b|=O\!\left((\log\log b)^{-1/4}\right).$$

The double logarithm makes this a slow bound. A positive correlation at every base in the finite ledger is compatible with the theorem. The proof supplies the limit that the table alone cannot give.

The three comes from the energy

The squared calculation has an exact arithmetic expression. Let $E_b$ be the collision energy, obtained by centering the finite collision table and summing the squares of its entries. The spectrum factorization and Parseval give

$$R_b=\frac{3E_b}{(b-1)(b^2-1)}.$$

The Cubic Law gives $E_b/b^3\to1$. The denominator has the same cubic scale. The ratio therefore tends to three.

The Secondary Term supplies the leading correction. It describes how that square ratio approaches its limit. The proof that the ordinary magnitude correlation tends to zero uses the short-sum and $L$-value estimates above.

The exact proportionality at base five led to a positive relation throughout the finite ledger. Now its limiting behavior is determined. The magnitude correlation fades to zero, while the mixed square average retains a factor of three.

Long division produced both quantities. The answer is in how their mass is distributed across the characters. Averaging the magnitudes lets the relation fade. Squaring keeps enough weight on the rare large values for the collision energy to register it.