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.

In long division, I multiply the current remainder by the base. The quotient gives the next digit; what is left becomes the next remainder. When the denominator shares no factor with the base, the remainders return to their starting point. The repeating digits are the visible trace of that cycle.

The carry-boundary calculation made me look again at the set of remainders in a cycle. A digit rule selects some cells and leaves others empty. A remainder orbit also selects a region. It has an edge that can be measured by the same finite boundary calculus.

Multiplication and neighbors

Take base 2 and denominator 7. Starting at remainder 1, multiplication by 2 gives the cycle $1\to2\to4\to1$. The orbit is $H=\{1,2,4\}$, the three nonzero squares modulo 7.

Now arrange all seven residues around a circle in their ordinary order. Light beads 1, 2 and 4. Leave 0, 3, 5 and 6 dark. Walk around the circle by adding one at a time. There are four changes between light and dark, although multiplication visits only three lit beads.

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

The signed boundary records those changes. I use predecessor minus current membership,

$$\mu_H(x)=\mathbf 1_H(x-1)-\mathbf 1_H(x).$$

With this convention an entrance has sign $-1$ and an exit has sign $+1$. The entries at 1 and 4 are negative; those at 3 and 5 are positive. All other entries vanish. The signs tell us which way each crossing goes.

To measure the boundary, I pair it with a Dirichlet character. A character assigns compatible complex phases to the invertible residues: multiplying residues multiplies their phases. It gives zero to the other residues. Each character therefore measures a particular arithmetic component of the signed edge. The squared magnitude of that measurement is its boundary-flux energy.

The shift inside a Jacobi sum

The orbit comes from multiplication, but its boundary compares neighbors. That combination leads to a classical Jacobi sum,

$$J_+(\rho,\chi)=\sum_{y\bmod q}\rho(y)\chi(y+1).$$

Both $\rho$ and $\chi$ are multiplicative characters. Addition enters through the argument $y+1$. The sum compares one multiplication-compatible labeling at a residue with another at its neighbor.

For the square orbit modulo an odd prime, let $\psi$ assign $+1$ to nonzero squares and $-1$ to nonsquares. Apart from the characters that are constant on the orbit, the odd boundary flux has the exact form

$$S_{\mu_H}(\chi)=\frac{J_+(\psi,\chi)-1}{2}.$$

The subtraction of one accounts for an endpoint at the edge of the unit set. Keeping that small term is essential when the flux is squared.

These ingredients have a substantial history. Girstmair’s work on digit variance and Dedekind sums studies statistics along remainder orbits. Lehmer, Lehmer and Mills study consecutive power residues. Here I am measuring the signed orbit boundary and asking how its energy is distributed across arithmetic resolutions.

A ledger of resolutions

A character displayed modulo 225 may already be determined by residues modulo 9, or 25, or another divisor. Its conductor is the smallest modulus that determines it. Grouping characters by conductor tells us how much resolution each part of the boundary actually needs.

I use odd characters, which assign opposite values to a residue and its negative. For each conductor $d$, let $P_d$ be the sum of their squared boundary fluxes at exactly that conductor. Every entry is nonnegative. Their sum is the full odd-character energy.

The equal-digit rule in base 15 gives a concrete comparison with the carry boundary. Its selected cells are $0,16,32,\ldots,224$ modulo 225. The conductor ledger is

Conductor	Boundary-flux energy
9	24
25	240
45	360
75	400
225	896
Total	1920

The other divisor levels carry zero energy. Of the total, 1024 lies below the full conductor. A calculation that kept only primitive characters modulo 225 would retain 896 and omit the rest.

This ledger differs from the signed inclusion-exclusion calculation used to isolate the top conductor. Its entries are actual sums of squares. An occupied lower level cannot be canceled by a negative entry elsewhere in the ledger.

There is a useful subtlety in deciding when a lower level is silent. A pushdown adds all the unit-boundary weights that reduce to the same smaller residue. A zero pushdown is certainly silent. An even pushdown is also invisible to odd characters, because equal weights at $r$ and $-r$ cancel in every odd pairing.

For example, select the two-digit base-5 cells whose first digit is smaller than the second. At conductor 5 the unit pushdown is $(-1,-1,-1,-1)$ on residues $1,2,3,4$. It is nonzero and even. Its odd energy is zero, while conductor 25 carries all 40 units of odd energy. At any single conductor, the exact test is vanishing of its primitive odd component. For concentration entirely at the top, this must hold at every proper conductor. It is an energy criterion, not a count of boundary points.

A larger orbit at the same conductor

Multiplication by 2 modulo 9 visits every unit,

$$1\to2\to4\to8\to7\to5\to1.$$

Read those six residues in additive order instead. The boundary is positive at $0,3,6$ and negative at $1,4,7$. Dirichlet characters vanish at the multiples of 3, so only the three negative entries contribute to the flux.

All three reduce to 1 modulo 3. The odd character from conductor 3 reads them with the same phase. Its flux is $-3$ and its energy is 9. The primitive characters modulo 9 sum to zero across those three entries.

The same mechanism holds for a full unit orbit modulo an odd prime power $p^e$. Its edge occupies the two residue classes 0 and 1 modulo $p$. Each class has $p^{e-1}$ lifts. The nonunit class is invisible to characters; the visible class gives flux $-p^{e-1}$ to each of the $(p-1)/2$ odd characters at conductor $p$.

The full odd energy is therefore

$$\mathcal E(b,p^e)=\frac{p-1}{2}\,p^{2e-2}.$$

Every higher-conductor contribution is zero. The orbit grows, and the flux grows, while the conductor carrying that energy stays at $p$. Size and arithmetic depth separate in an exact, testable way.

The quadratic value at minus two

For the square orbit modulo a prime, I can also remove the characters that are constant on the orbit and calculate the remaining energy exactly. The result has four branches, selected by the prime modulo 8. For instance, the off-resonance energies at primes $7,17,23,31$ are $2,40,54,104$.

The modulus 8 comes from a specific step of the proof. The odd-character sum singles out the shifted argument $y+1=-1$, hence $y=-2$. That forces the quadratic value $\psi(-2)=\psi(-1)\psi(2)$. Gauss’s supplementary laws determine those two signs from the prime modulo 8.

I would not have predicted that from the digit side of the program. A walk between neighboring residues has made the energy depend on the quadratic behavior of minus one and two.

The paper gives the four formulas, the complete conductor ledger and the expansion for higher-index orbits. Its boundary energy is the squared flux before the Bernoulli factor is applied; the carry transform then adds that classical weight. Keeping those two measurements distinct makes the comparison between digit regions and remainder orbits precise.

The remainder cycle tells me which states long division visits. Its additive boundary tells me something else: how finely arithmetic must resolve that set. Six states modulo 9 can be measured entirely at conductor 3. That is the distinction I wanted to understand.