The Orbit's Edge

The floor produces two things at every step. A digit and a remainder. The digit is the visible output. The remainder is the state that persists. It multiplies by the base, reduces mod the denominator, and drives the next step. It cycles. It orbits. And for two years I studied the digits without studying the orbit.

The digit side is settled. Carries create boundaries, boundaries carry character flux, the Bernoulli factor weighs them all the same way. That story is told.

This paper is about the other side. The remainder. The orbit. And what happens at its edge.

The part that keeps moving

The remainder cycles through a fixed set of values. Start at some remainder, multiply by the base, take the result mod the denominator, and you land at a new remainder. Do it again. Eventually you come back to where you started. The set of values you visited along the way is the orbit.

Sometimes the orbit visits every available remainder. Sometimes it visits only a fraction of them, and the rest form their own separate orbits. Either way, the orbit occupies some residues and not others. It is a region inside the integers mod $q$.

I had been looking at boundaries on the digit side for months. Collision diagonals have boundaries. Forbidden blocks have boundaries. Palindrome conditions have boundaries. Characters read all of those through the same Bernoulli factor.

Then I saw that the orbit has a boundary too. It is a region. Regions have edges.

The edge

Picture the integers mod $q$ arranged in a line: 0, 1, 2, 3, ..., $q-1$. Some of them belong to the orbit. Some do not. Now walk along that line, one step at a time. At certain steps you cross from outside the orbit to inside it. At other steps you cross from inside to outside. Those crossings are the boundary.

The signed boundary records each crossing with a sign. Step into the orbit: $+1$. Step out: $-1$. Stay inside or stay outside: $0$. In notation, that looks like

where $\mathbf{1}_H$ is 1 for residues in the orbit and 0 for residues outside it. The formula just computes the difference between consecutive membership values. If you were inside and now you are outside, the difference is $-1$. If you were outside and now you are inside, it is $+1$.

The characters do not watch the orbit move. They watch its edge. The places where the orbit's territory begins and ends, as seen by someone walking through the integers one step at a time.

Not the orbit itself. The orbit's edge.

The surprise

Here is the thing I did not expect, and it is the reason this paper exists.

The orbit is created by multiplication. Multiply by $b$, reduce mod $q$, repeat. That is multiplicative motion. But the boundary is measured by addition. It records where the indicator changes when you step from $x$ to $x+1$. Additive motion.

One operation builds the region. A different operation measures it.

That means the object cannot be only a multiplicative group, and it cannot be
only an additive line. If it were only multiplicative, the neighbor $x+1$ would
have no natural role. If it were only additive, the orbit generated by
multiplying by $b$ would disappear. The object has to live in a finite residue
ring, where both motions are present at once. Multiplication by $b$ creates the
region $H=\langle b\rangle$ inside the unit group, and addition by $1$ scans the
same residues for changes in membership. So the boundary is not an arbitrary
edge. It is the additive edge of a multiplicatively generated region. That is
the kind of object the math forces on us.

Make it concrete. Take the integers mod 7, laid out in order: 0, 1, 2, 3, 4, 5, 6. The quadratic residues, the perfect squares mod 7, are {1, 2, 4}. (Check: $1^2=1$, $2^2=4$, $3^2=2$, and then it repeats.) Mark those three. Now walk through the line one step at a time.

0 is out. Step to 1, you are in. Boundary crossing. 2 is in, still inside, no crossing. Step to 3, you are out. Crossing. Step to 4, you are in again. Crossing. Step to 5, you are out. Crossing. 6 is out, stay out. Four crossings total.

The region {1, 2, 4} was placed by multiplication. These are the squares, the index-two subgroup, a multiplicative object. But the four crossings were found by walking the number line in additive order. The squares do not sit in a single consecutive block. They are scattered: 1 and 2 are adjacent, then a gap at 3, then 4 alone, then two more gaps. Multiplication scattered the region. Addition found the edges.

If you could see this geometrically, it would look like a necklace with seven beads arranged in a circle: 0, 1, 2, 3, 4, 5, 6. Beads 1, 2, and 4 are lit. Beads 0, 3, 5, and 6 are dark. The lit beads are not clustered together. They are spread around the necklace with dark gaps between them. The boundary is every place where a lit bead sits next to a dark one. The pattern of lit and dark is determined by the multiplicative structure of the group. The boundary is determined by the additive structure of the circle. The two structures are different, and the places where they interact, the edges between lit and dark, are the object this paper measures.

These are the dynamics that govern the appearance of the fractional field in every base. Change the base and you change the multiplication. Change the prime and you change the group. But the interplay between the multiplicative orbit and its additive boundary is always there, and it is always what produces the digit expansion you see. The digits themselves are a symbolic interface we are accustomed to. The structure underneath, the orbit and its edge, is the thing being studied.

Jacobi sums had appeared before in the collision program. This paper shows why.

A Jacobi sum is a finite sum that pairs two different kinds of characters. One character is multiplicative. It assigns a value to each residue based on its multiplicative structure, which orbit it belongs to, which coset it sits in. The other character is additive. It cares about where you are on the number line, not which group you belong to.

The Jacobi sum $J(\rho, \chi) = \sum_y \rho(y)\chi(y+1)$ runs through every residue $y$, evaluates the multiplicative character $\rho$ at $y$ (which tells you about the orbit), and the additive character $\chi$ at $y+1$ (which takes one step forward and crosses whatever boundary is there). Then it adds everything up.

The multiplicative character sees the orbit. The additive shift crosses the boundary. The Jacobi sum measures the interaction between the two. It is the natural finite object that stands at the intersection of multiplication and addition, measuring one through the other.

Once I saw that, the rest of the paper followed. The orbit is multiplicative. The boundary is additive. The Jacobi sum is the bridge.

The ledger

Once you have a boundary, you can ask a natural question. How much of the total character energy sits at each level of the conductor?

The conductor is the depth at which a character lives. Some characters are primitive at the full modulus $m$. Others are lifted from a smaller modulus $d$ that divides $m$. They look like characters mod $m$, but they really live at level $d$. The conductor is the true home address.

The energy of a boundary decomposes by conductor. Each level $d$ carries a piece $P_d(\mu)$ of the total, and every piece is nonnegative. That nonnegativity matters. It means no cancellation between levels can hide a contribution. If there is energy at a lower conductor, the ledger shows it. There is no way to balance one level against another.

In words: the total energy (left side) equals the sum of the energies at each conductor level (right side), and each of those pieces is zero or positive. If all the energy sits at the top level, the boundary is spectrally clean. If energy leaks to a lower conductor, the boundary has leakage, and the ledger records exactly how much.

The criterion for cleanliness has a story behind it. I first conjectured that the boundary's pushdown to a lower level had to vanish for the energy there to be zero. That seemed natural. If nothing pushes down, nothing leaks.

Then I found a counterexample. The "first digit less than second digit" language at modulus 25 has nonvanishing pushdowns but still obeys the clean mean law. The energy at the lower conductor is zero even though the pushdown is not. The reason: the pushdown is antipodally even. It assigns the same value to $r$ and to $-r$. Odd characters cannot see an even function, so the odd energy at that level is zero even though the pushdown itself is nonzero.

The corrected criterion survived every test after that. Two hundred and forty random digit languages across six moduli, zero mismatches. The counterexample did not break the theory. It sharpened it into the right statement.

Thickness

When the orbit fills the entire unit group at a prime power $p^e$, something simple happens to the boundary. It lives at just two residue classes mod $p$. Zero, where you step out of the units (because multiples of $p$ are not units). And one, where you step back in.

Those two boundary points do not move as the prime power grows. Going from mod $p$ to mod $p^2$ to mod $p^3$, the boundary stays at the same two residue classes. But behind each boundary point, the number of units stacked up grows. At mod $p$ there is one unit per class. At mod $p^2$ there are $p$ units per class. At mod $p^e$ there are $p^{e-1}$.

The boundary is fixed. The mass behind it thickens.

The energy formula captures this cleanly. The symbol $\varphi(p^e)$ is Euler's totient, the count of units mod $p^e$, which equals $(p-1)p^{e-1}$.

Half the units, times the thickness. For $p = 3$ and $e = 2$, there are $\varphi(9) = 6$ units mod 9, and the thickness is $p^{e-1} = 3$. The energy is $\frac{6}{2} \cdot 3 = 9$. Three units per residue class, stacked behind each of two boundary points.

That number 9 is not a coincidence. It is the kernel size (3, from the odd-character count) times the thickness (3, from the prime-power depth). Boundary times depth. That is the pattern at every prime power.

The prime remembers modulo 8

This is the result that surprised me most.

When the orbit fills half the group (the quadratic residues, the squares mod $q$), the boundary energy splits into a resonant part and an off-resonance part. The resonant part comes from characters that are trivial on the orbit. The off-resonance part comes from all the others.

The off-resonance energy depends on $q$ modulo 8. Not modulo 4. Not modulo 3 or 12. Eight.

I found this numerically first, by computing the energy at small primes and watching the pattern. At $q = 7$ the off-resonance energy is 2. At $q = 17$ it is 40. At $q = 23$ it is 65. At $q = 31$ it is 128. I tried to fit one formula and could not. There were four cases.

The proof, when it came, was one of the more satisfying moments in this program. The boundary shift, stepping from $x$ to $x+1$, forces the quadratic character $\psi$ to evaluate at the specific argument $-2$. Not at $-1$. Not at $2$. At their product, $-2$. And the value $\psi(-2)$ factors as $\psi(-1) \cdot \psi(2)$.

Those two factors are classical. Gauss proved them in the early 1800s. The value of $\psi(-1)$ depends on $q$ mod 4 (the first supplementary law of quadratic reciprocity). The value of $\psi(2)$ depends on $q$ mod 8 (the second supplementary law). Together they produce four cases, and those four cases are exactly the four formulas I had found numerically.

Two facts from the 1800s, appearing because the additive boundary of a multiplicative orbit forced the quadratic character to look at $-2$. The prime remembers its residue class mod 8 through the edge of its own orbit. I would not have predicted that from the digit side of the program. It came from the orbit side, from following the remainder instead of the quotient.

Both sides of the floor

The floor function takes a real number and rounds it down. The integer part is the digit. The fractional part becomes the next remainder. One input, two outputs, at every step.

The digit side of this program produced a carry-boundary calculus. Digit patterns create regions. Regions have signed boundaries. Characters read boundaries through the Bernoulli factor. The Bernoulli factor is universal.

The orbit side, this paper, produces the same structure from the other output. The remainder's orbit is a region. The region has a signed boundary. Characters read it through the same kernel. The conductor ledger decomposes the energy the same way.

The two sides meet at the floor. The digit is what the floor keeps. The remainder is what the floor carries forward. Both leave edges. Both are measured by the same finite machinery.

I started this program by counting digit collisions. Twenty-four posts later, I am measuring the edges of multiplicative orbits with Jacobi sums. The object changed. The mechanism did not.

None of the ingredients here are new to number theory. Multiplicative orbits, additive boundaries, Jacobi sums, quadratic reciprocity, conductor decompositions. All classical. All well studied.

The new thing is the assembly. The signed boundary chain of a digit condition or a remainder orbit, read through a universal Bernoulli factor, decomposed by conductor into a finite ledger. That specific route, from digit pattern to boundary to flux to conductor energy, with the Bernoulli weight connecting them, did not exist before this program. The four-case energy law mod 8 for index-two orbits is a new closed form, even though the Jacobi sums and supplementary laws inside it are two centuries old. The conductor ledger with its corrected antipodal criterion is new. The collision invariant itself is recent, from the earlier papers in this program. The recognition that it was one boundary instance among a structured class came from this work.

The ingredients belong to Gauss, Jacobi, Dedekind, and their successors. The framework that connects them through the signed boundary of a long-division state is what these papers contribute.

Try it yourself

Watch the remainder orbit. The division table shows each step of the recurrence: the remainder, the product, the emitted digit. The remainder column is the orbit.

$ ./nfield div 7

Six remainders: 1, 3, 2, 6, 4, 5. Every unit mod 7 visited. One orbit, one cycle, six edges.

Now try a prime where the orbit does not fill the whole group.

$ ./nfield div 13

Twelve units but only six in the orbit of 1. The other six form a separate orbit. Two lanes on the same track.

See the repeating decimals themselves. Each fraction $k/7$ produces the same six digits in a different rotation.

$ ./nfield field 7

The digits are the visible trace. The remainder orbit underneath is the motion that produces them.

Code: github.com/alexspetty/nfield

Alexander S. Petty
May 2026
.:.