The Spectral Repulsion

Write two rows of four numbers. Each row adds to one.

Multiply down each column and add. The overlap is $0.20$.

Reverse the second row. Now the largest number meets the largest, the smallest meets the smallest. The overlap rises to $0.30$. We have added nothing to either row. We have changed who meets whom.

There are twenty-four arrangements of the second row. Their average overlap is exactly $0.25$. Our first arrangement falls twenty percent below it. The reversal takes us twenty percent above.

The collision spectrum and the primes supply two rows of their own. Their overlap is small in the calculations below. The question is whether it is small because of the weights themselves, or because of where they sit.

Keep the weights. Shuffle the positions.

Four pairs at twenty-five

Start with the centered collision table. Long division supplies a count of remainders that emit the same digit on consecutive steps. Subtract the bin scale, then the mean deviation within each last-digit family. The result is a signed weight $f(a)$ for each unit remainder modulo $b^2$.

A Dirichlet character is a multiplicative pattern on those remainders. Comparing the table with each character gives a coefficient. Square its magnitude and divide by the total squared magnitude. These are the weights in the first row.

For the second row, use the same characters to combine primes. Each prime contributes its character value divided by its square root.

$$P_X(\chi)=\sum_{b^2<p<X}\frac{\chi(p)}{\sqrt p}.$$

The character values can cancel. Some sums end large, others small. Square their magnitudes and normalize again. Both rows now add to one.

The coefficient-and-prime comparison measures magnitudes and their correlation. Here we use squared magnitudes as shares of a fixed total, then compare the actual assignment with every relabeling. The two statistics answer different questions.

There is a restriction before any shuffling begins. For an odd prime base, reflection removes the even characters from the collision spectrum. Family centering removes the characters inherited from modulus $b$. Both rows are normalized on the remaining primitive odd characters, the patterns that change sign under reflection and need the full modulus $b^2$.

Conjugate characters have equal weights. Keep each pair together. At base five, the modulus is twenty-five and eight eligible characters make four pairs. This time the four weights come from arithmetic, not from a convenient choice of decimals.

At a cutoff of two million, pair C carries about $65.2\%$ of the prime weight and $0.37\%$ of the collision weight. Most of the collision weight sits in A and B. Most of the prime weight sits in C.

The four pairs and their weights

Write $\chi_j(2)=e^{2\pi i j/20}$, since two generates the unit remainders modulo twenty-five. Pair A joins indices 1 and 19, B joins 3 and 17, C joins 7 and 13, and D joins 9 and 11. The weights below combine the two members of each pair. All $148{,}924$ primes strictly between $25$ and $2{,}000{,}000$ enter the prime sums.

Pair	Collision share	Prime share
A	38.7976%	5.7905%
B	57.0818%	22.4346%
C	0.3718%	65.2180%
D	3.7488%	6.5569%

The numbers are rounded for display. The figure and overlap use the unrounded values.

For $k$ pairs with weights $x_r$ and $y_r$, the average shuffled overlap is $1/k$. Each prime weight visits each position equally often, so the mean weight opposite any $x_r$ is $1/k$. Multiply by the first row’s total of one.

Scale the observed overlap by that reference value.

$$\rho=k\sum_{r=1}^{k}x_r y_r.$$

The shuffle mean is now one. At base five, $\rho\approx0.6216$. The actual pairing produces about sixty-two percent of the average shuffled overlap. It is not the smallest of the twenty-four. Eight arrangements give still less.

Nothing in this comparison makes the primes random. The weights stay fixed. The only random choice in the reference calculation is a permutation of their labels.

Give the primes more room

Increase the cutoff. The collision row stays exactly as it is. The prime row changes as new terms enter the character sums.

The figure keeps all six cutoffs and all ten nontrivial bases. At two million, base seven gives $\rho\approx0.4801$ and base nineteen gives $0.7690$. The mean across the ten bases is $0.5909$.

All sixty comparisons fall below one. Their sizes move. The mean rises slightly between the first two cutoffs, then falls to about $0.546$ at five million. Individual bases wander more than the mean does.

These windows share many of the same primes. They are not sixty independent trials, and their common sign does not prove an infinite law. Spectral repulsion names this measured deficit in overlap. Whether it persists as the cutoff grows is open.

Base three has only one eligible pair. There is nothing to rearrange. Its ratio is exactly one wherever the normalization is defined. A deficit computed by including excluded characters would mix two effects, how much prime weight reaches the eligible set and how that weight is placed inside it.

All sixty values and the exact shuffle variance

Each entry uses primes strictly between the square of its base and the column’s cutoff. The ratios are rounded to six decimal places.

Base / cutoff	250,000	500,000	1,000,000	2,000,000	3,000,000	5,000,000
5	0.358604	0.623585	0.569230	0.621637	0.623558	0.698186
7	0.517232	0.466571	0.531456	0.480109	0.481618	0.487967
11	0.497731	0.599124	0.527714	0.554413	0.538091	0.461928
13	0.565944	0.502696	0.484069	0.571188	0.588751	0.513094
17	0.568194	0.553257	0.551279	0.535310	0.531447	0.431643
19	0.666823	0.749630	0.715464	0.769019	0.608098	0.578852
23	0.646340	0.587253	0.595957	0.580464	0.598449	0.539563
29	0.847195	0.763904	0.697677	0.609580	0.581497	0.663482
31	0.766327	0.761552	0.738612	0.613236	0.651921	0.516871
37	0.802902	0.659609	0.560271	0.573913	0.544948	0.571037

For a uniformly chosen permutation $\pi$, put $R_\pi=k\sum_r x_r y_{\pi(r)}$. When $k>1$, its mean and variance are

$$\mathbb E[R_\pi]=1,$$

$$\begin{aligned} \operatorname{Var}(R_\pi)&=\frac{k^2}{k-1}A_xA_y,\\ A_x&=\sum_r(x_r-1/k)^2,\\ A_y&=\sum_r(y_r-1/k)^2. \end{aligned}$$

The variance is exact for these two lists. One shuffled entry has the list’s mean-square deviation. Two distinct shuffled entries have negative covariance because they cannot occupy the same position. Expanding the square of the weighted sum gives the formula.

At two million, base five sits $0.71$ shuffle standard deviations below one. Base thirty-seven sits $2.93$ below. These distances describe the relabeling distribution. They are not, by themselves, probabilities that the arithmetic is accidental. A single pair has zero variance.

Where the weight goes

There is another difference between the rows. The collision weight is more concentrated.

At base thirty-seven, order the 324 pairs by decreasing collision weight. Give the prime row exactly the same order. The first sixteen pairs hold about seventy percent of the collision weight, but less than two and a half percent of the prime weight.

We can also measure spread without keeping track of the positions. The effective support is the number of equal weights that would have the same entropy. Four equal weights give four. One occupied position gives one. Unequal weights can give a value between integers.

These figures count equivalent individual characters, not pairs. At base twenty-nine, the collision weights have the entropy of roughly sixty-four equal weights. The prime weights have the entropy of roughly two hundred and ninety. Neither number counts the nonzero entries.

A concentrated row can still put its largest weight opposite the largest weight in a broad row. Concentration alone cannot explain the deficit. That is why we shuffle. Both lists keep their shapes while the partners change.

A partner with the opposite sign

Return to the collision table and ask a different question. Instead of matching characters, match remainders whose sum is a chosen target.

The base-three table fits in nine places. Nonunit remainders receive zero weight.

For target zero, one meets eight, two meets seven, and four meets five. The weights in each pair are opposites. Each product is a negative square. Count every starting remainder, so both orders appear, and the total is $-16/3$.

For target one, remainder $a$ meets $1-a$ modulo nine. Continue through all nine targets. The resulting sums form the additive convolution.

$$K(t)=\sum_{a\bmod b^2}f(a)f(t-a).$$

The whole grid adds to zero. Its first column does not. Reflection forces that column to be nonpositive cell by cell, because $f(-a)=-f(a)$.

$$K(0)=-\sum_{a\bmod b^2}f(a)^2.$$

The value at zero is the negative of the table’s energy. All the other targets together supply the opposite amount. Some of those targets are negative too. Their total, not each individual value, compensates for the first column.

From nine remainders to 1,369

The same construction below uses bases three, five, seven, ten, nineteen and thirty-seven. Every cell is a product $f(a)f(t-a)$, with $a$ down the page and target $t$ across. The largest grid contains $1{,}874{,}161$ products. The first column remains nonpositive at every size.

Select the image, choose 100%, and drag to inspect the individual cells. Color retains the sign and uses $\log(1+|f(a)f(t-a)|)$ for intensity, rescaled in each panel. The colors make small products visible; the column sums use the original weights.

Why the additive Fourier coefficients are nonpositive

These identities hold for every base $b\ge2$, including composite bases. With $m=b^2$, the zero extension of $f$ is real and odd. Its additive Fourier transform

$$\mathcal F(j)=\sum_{a\bmod m}f(a)e^{-2\pi ija/m}$$

is purely imaginary. The cosine contributions cancel between $a$ and $-a$. Convolution becomes multiplication, so

$$\widehat K(j)=\mathcal F(j)^2=-|\mathcal F(j)|^2\le0.$$

This is the square, not the squared magnitude. The sign follows from the purely imaginary transform. Also $\sum_t K(t)=(\sum_a f(a))^2=0$, which proves the compensation without requiring the individual convolution values to vanish.

The sign reaches prime pairs directly. Seventeen and nineteen add to thirty-six, a multiple of nine. Their table weights are $-2/3$ and $2/3$. Their product is $-4/9$.

For any even target divisible by $b^2$, the same reflection gives a nonpositive product for every prime pair that represents it. A nonunit prime contributes zero. This proves a sign, not the existence of a prime representation.

Back to the four pairs

In the additive grid, we can point to the partner that forces each negative product. That argument does not explain the multiplicative overlap. Adding remainders and matching character weights are different operations on the same table.

The multiplicative question is still there at base five. At two million, more than sixty-five percent of the prime weight sits opposite less than four tenths of one percent of the collision weight. We can enumerate all twenty-four shuffles and locate the actual assignment. We do not yet have the arithmetic explanation for the repeated deficit, or a theorem that carries it beyond the tested cutoffs.

The largest prime weight meets the smallest collision weight. Nobody has reversed a row.

Three companion views

The bar comparison keeps the opening weights side by side. The cutoff plot compresses the ten bases to a range and a mean. The final plot gives all nine additive sums without the individual products.