Carry Boundaries and Bernoulli Spectra

Long division does not begin with a digit. It begins with a crossing.

Take a remainder, multiply it by the base, and divide by the denominator. The floor changes only when that motion crosses the next integer wall. The digit records the crossing. The new remainder carries the unfinished part forward.

I first met this structure by counting digit collisions. Two digits agreed or they did not. The centered counts formed a finite table, the table obeyed an exact reflection law, and its character transform reached special values of Dirichlet $L$-functions.

The collision table was real structure. It was not yet the source of the structure.

The source is the carry.

The carry under the digit

As a remainder moves through long division, its fractional part rises and resets. Rise, rise, cross an integer, reset. That is a sawtooth.

The centered sawtooth is odd. Negating its input negates its value. This simple fact determines which Dirichlet characters can hear it. Every even character vanishes. A primitive odd character returns a generalized Bernoulli number.

I call that Bernoulli value the floor potential. The name keeps the cause in view. The Bernoulli number is not inserted as an analytic decoration. It is what the floor becomes under the character transform.

The floor potential does not care which fixed digit pattern is being counted. A collision, a three-digit agreement, a transition, and a finite block pattern all receive the same response from the floor. What changes is the finite region selected by the digit rule.

Nine cells

The smallest useful example has nine cells.

Work in base 3 and look at the first two digits. There are nine possible two-digit words. The words whose digits agree are

As integers from 0 through 8, those cells are

Now use denominator 11. Among the ten fractions $r/11$, the first two base-3 digits agree exactly when

So the intrinsic digit count is 4. The carry reduction splits that four into three exact pieces.

The 3 is the bulk contribution from the three marked cells. The 2 is the carry count at the residue of 11 modulo 9. The final 1 removes the terminal endpoint.

That small identity is the finite-prefix theorem in plain view. Once the bulk and terminal endpoint are removed, the digit count depends only on a carry observable on the nine cells. The same argument works for any bounded weight on any fixed digit window when the denominator is coprime to the base.

The spectrum lives on the edge

Let $F_w$ be the centered carry count attached to a finite weight $w$. Let $S_w(\chi)$ be the signed character flux through its boundary. For every primitive odd Dirichlet character, the transform factors exactly.

That line is the center of the result.

The first factor is the floor potential. It is universal. The second factor is the boundary selected by the digit rule.

The interior cancels because consecutive character differences telescope. Step through a run of selected cells and every internal entrance is paired with the next exit. Only the places where the rule turns on or off remain.

This is a finite Stokes principle. The digit rule marks a region. The transform reads its edge.

The collision diagonal is therefore not an isolated construction. It is one boundary among many. Two-point collisions, three-point agreements, transitions, and arbitrary finite block patterns all enter through the same factorization. Their boundaries differ. Their floor potential does not.

The part characters cannot see

The boundary is more precise than the selected region, but even the boundary has a visible and an invisible part.

Dirichlet characters vanish on residues that are not units. The full family of characters modulo $m$ therefore determines exactly the signed boundary restricted to the unit group. Boundary mass supported entirely on nonunits disappears from every character channel.

For example, modulo 6 the set $G=\{0,1\}$ has boundary only at 0 and 2. Both residues are nonunits. The set is not constant, and its boundary is not empty, but every Dirichlet character sees zero flux.

Two digit regions can have completely different interiors and still have the same spectrum. What matters is whether their signed boundary derivatives agree on the units.

That is the exact blind spot. It is not an estimate and it is not missing information in a computation. It is the kernel of the character pairing.

The levels underneath

A character written modulo $m$ may really live at a smaller divisor of $m$. That smaller modulus is its conductor. It is the level where the character is genuinely primitive.

The boundary has the same layered behavior. Push a signed boundary down to a smaller conductor and ask what remains. The character at that level reads the pushed-down boundary through its own Bernoulli factor. The result is a conductor-stratified version of the same factorization.

The nine-cell example shows how a boundary can disappear at the wrong level. Lift $G=\{0,4,8\}$ from modulus 9 to modulus 27. The lifted cells are

Every surviving endpoint is divisible by 3. Primitive characters modulo 27 therefore see no boundary. The boundary has not vanished from the arithmetic. It remains visible at its native conductor 9.

This is the purpose of the conductor ledger. It records which boundary layers survive and the level at which each one can be heard.

Prime powers close

Take an odd prime $p$ and a depth of at least two. Among the $p^N$ padded base-$p$ words, select those whose first and last digits agree. Their signed boundary has three exact properties.

Its visible unit boundary has the right norm. Its mass cancels when pushed to the relevant lower conductor. Negation exchanges entrances and exits.

Together those three facts force an exact primitive-odd mean-square law.

The root-mean-square flux is therefore the square root of twice the formal boundary size. There is no error term. The identity controls the whole primitive odd family at once.

The nine-cell example makes the count visible. Its formal boundary has size 6. There are two primitive odd characters modulo 9, and each has squared flux magnitude 12. The mean is 12, exactly twice the formal boundary size.

The balance is not supplied by orthogonality alone. It comes from the geometry of this particular boundary at a prime power. The diagonal contribution and the surviving antipodal contribution match exactly.

Base 15 does not

Composite conductors can remember more than the top level.

For the terminal collision diagonal in base 15 at modulus 225, the formal boundary has size 30. If only the top conductor survived, the prime-power pattern would give a mean of 60.

It does not.

Boundary mass remains visible at the proper conductors 45 and 75. The exact conductor ledger records a total moment of 896 across 32 primitive odd characters.

The mean is 28, not 60. The lower-conductor terms remove exactly 32 from the top-level value.

This is not a failure of the boundary calculus. It is one of its sharpest tests. The same ledger proves the clean prime-power law and identifies the exact obstruction when the modulus has mixed prime structure.

Prime powers close because the required lower pushdown cancels. Base 15 does not close because two proper-conductor boundary layers survive.

The quadratic edge

The factorization is linear. Squaring it and summing over characters produces a conductor-stratified energy.

The geometric side is built from finite sawtooth correlations. At unit indices these are classical Dedekind and Rademacher objects. The analytic weight is the square of the Bernoulli factor. For primitive odd characters its magnitude is tied exactly to the value of the corresponding Dirichlet $L$-function at one.

The energy therefore has two layers. The finite digit rule supplies the boundary geometry. The floor supplies the arithmetic weight.

The linear theory is complete for fixed digit windows. Whole orbits, complete repetends, recurrences, and products of already-reduced carry observables still need finite boundary constructions of their own. The factorization is ready for them once those boundaries are found.

The source

The collision invariant did not disappear. It got located.

Digit equality is where the boundary first became visible. The carry is what created it. The floor leaves an odd sawtooth. The digit rule cuts out a finite region. Interior differences cancel. Dirichlet characters read the part of the edge that survives on the unit group.

That is why the even channels vanished. That is why generalized Bernoulli numbers appeared. It also shows where the collision table enters. Digit equality selects its boundary. The pieces were not separate coincidences. They were different views of one boundary mechanism.

Long division does more than produce digits. It produces boundaries, and those boundaries are what survive the transform.

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$ ./nfield finite-prefix-slice inspect 11 --base 3 --depth 2 --support 0,4,8

$ ./nfield boundary-bernoulli-factorization inspect 3 2 1 --support 0,4,8

$ ./nfield terminal-collision-conductor-ledger audit

$ ./nfield verify finite-prefix-slice
$ ./nfield verify boundary-bernoulli-factorization
$ ./nfield verify terminal-collision-conductor-ledger

The full research note is Carry Boundaries and Bernoulli Spectra in Long Division.

The code is in nfield.

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Alexander S. Petty .:.