The Double Transversality

At the prime seventeen, the centered collision weight in base two is $+1/2$. In base three it is $-2/3$. Add them and the result is small.

$$\frac12-\frac23=-\frac16.$$

Both counts use lag one. Each nonzero remainder is compared with the remainder one long-division step later, and we count how often the two emit the same digit. The centering removes the baseline count and the average deviation for the relevant remainder class. What is left is a signed weight.

The two weights at seventeen partly cancel. But that one total tells us very little about the patterns they came from. To find out whether combining bases loses information, we need to keep the whole table.

A table for two bases

In base two, the lag-one weight depends only on the remainder modulo four. There are two allowed remainders. At one the weight is $+1/2$, and at three it is $-1/2$.

Base three needs the remainder modulo nine. Its six allowed remainders are $1,2,4,5,7,8$.

Put the binary choices down the side of a table and the ternary choices across the top. There are twelve cells. Each cell represents one possible pair of remainders, and each contains the sum of the two weights.

Seventeen belongs in the upper-right cell. It leaves remainder one modulo four and remainder eight modulo nine. The entry is our $-1/6$.

Take the average across the top row. The binary contribution stays at $+1/2$ all the way across. The six ternary contributions average to zero. So the row average is exactly $+1/2$.

The bottom row gives $-1/2$. We have recovered the binary pattern.

Now average down a column. The two binary contributions cancel, leaving the ternary weight for that column. We recover all six ternary values in the same way.

Nothing about this calculation depends on a supply of primes. The Chinese remainder theorem says that every pair of allowed remainders occurs exactly once among the allowed classes modulo thirty-six. It supplies the complete rectangle on which the averaging works.

The combined table contains both patterns, and we can separate them by hand.

The size of the combined pattern

A zero average does not mean a table is empty. The binary values $+1/2$ and $-1/2$ average to zero, but each has a definite size.

Square the entries before averaging and that size is retained. Call this average squared weight the energy. The binary table has energy $1/4$. The ternary table has energy $8/9$.

For the twelve-cell combined table, the energy is

$$\frac14+\frac89=\frac{41}{36}.$$

There is a short reason. Squaring a sum gives the two squares and a cross term. Across the complete rectangle, that cross term averages to the product of the separate means. Both means are zero. The cross term disappears, leaving the two energies added together.

This is orthogonality. It is the same algebra as the Pythagorean theorem, applied to tables of numbers.

The result holds for any finite collection of pairwise coprime bases, at independently chosen lags. Coprime means that no two bases share a prime factor. We can give the tables different weights before adding them. Their energies still add with the corresponding squared weights, and averaging over the other coordinates recovers each weighted component.

When bases share a factor, some remainder choices are linked. An exact formula reduces the cross term to averages over that shared remainder information. It can be nonzero. The independence in the coprime case has an arithmetic reason.

These statements concern complete finite tables. A prime sum visits selected entries, in the order the primes supply them. A small value of that sum is a different observation from a zero cross term across the whole rectangle.

Two numbers for each character

Within a single base, there is another comparison to make.

A Dirichlet character assigns consistent signs or complex phases to the remainder classes. Use it to read the centered collision table and it gives a Fourier coefficient. Use the same character to read a finite list of primes and it gives a prime-character sum.

Each character now has two numbers attached to it. One measures the strength of its component in the collision table. The other is the size of its prime sum. Are the large values usually paired with large values, or with small ones?

I compared the two lists of magnitudes using Pearson correlation. A negative value means that larger collision coefficients tend to be paired with smaller prime sums in that finite sample. It does not require every pair to follow the trend.

The calculation uses primes between the table modulus and two million, each weighted by $1/\sqrt p$. The bases are three, five and seven. Increasing the lag compares digits farther apart and enlarges the table from a modulus $b^2$ to $b^{\ell+1}$.

There are fourteen tested base-and-lag combinations. Thirteen have a defined correlation, and twelve of those are negative. The exception is base three at lag two, where the value is approximately $+0.416$.

At base three and lag one, the two active characters have equal coefficient magnitudes. There is no variation in that list, so Pearson correlation is undefined. Reporting zero there would give the statistic a value it does not have.

At the deepest tested ternary lag, there are $2{,}186$ active characters. One further odd character has an exactly zero collision coefficient and is excluded. Conjugate characters count separately throughout. The corresponding correlation is approximately $-0.202$.

The negative association is visible in much larger tables than the first examples. The positive row is part of the result too. There is no all-lag sign theorem here.

Changing the cutoff

Keep the deepest tested lag in each base fixed, and increase the prime cutoff from a quarter of a million to five million.

All three correlations become more negative through these five cutoffs. The ternary value moves from about $-0.052$ to $-0.247$.

The signed total behaves differently. In the same ternary calculation it moves from about $+7.17$ to $-13.77$, then $-1.32$, $+10.36$ and $+4.62$.

The two measurements answer different questions. Correlation compares the magnitudes attached to matching characters. The total adds their signed contributions. A steady trend in the first does not make the second steady.

Every row was checked against the direct finite prime sum in nfield. The character expansion and the direct count agree. The cutoff is still part of the answer. Extending it changes the object being measured.

Nor is prime-sum magnitude a measurement of distance to an $L$-function zero. The calculation contains no zero locations. Describing the observed correlation as perpendicularity to the zeros would give it a geometric conclusion that has not been established.

Two different separations

The name double transversality brings these two comparisons together.

Across coprime bases, the separation is exact. One table varies along the rows, another along the columns. Their cross term vanishes, their energies add, and their combined pattern can be separated again.

Within one base, the coefficient assignment shows a predominantly negative correlation in the stated windows. Its persistence under changing lag, cutoff and normalization is an open question. The complete-table theorem does not settle that question about primes.

At seventeen, the combined weight is still $-1/6$. Across the twelve-cell table, the half and the two thirds are recoverable. A small total at one entry has not erased either pattern.

That gives me a precise distinction to hold onto. I can recover the separate base patterns exactly. I am still trying to understand the pairing inside each one.