The Spectral Structure of Fractional Fields

Write the fractions of seven in base two. Their repeating blocks contain only zeros and ones.

Fraction	Repeating block
$1/7$	001
$2/7$	010
$3/7$	011
$4/7$	100
$5/7$	101
$6/7$	110

Pair the first row with the last, the second with the fifth, and the third with the fourth.

In every pair, every position contains one zero and one one. The individual rows differ. Their combined coverage of the digits is identical.

This gives the comparison matrix two exact redundancies. Six different fractions do not supply six independent patterns of agreement.

Three pairs with the same total

In The Cross-Alignment Matrix, we gave each row a geometric representation. Every digit position has separate slots for the possible digits. In binary, there is a slot for zero and a slot for one. A row marks the slot it uses.

Take the pair 001 and 110. Together they mark both slots at all three positions. So do 010 and 101. So do 011 and 100.

If $v_k$ records the marks for $k/7$, the observation is

$$v_1+v_6=v_2+v_5=v_3+v_4.$$

The first pair total minus the second is zero. The second minus the third is zero. These are two independent ways of combining different rows and getting complete cancellation.

The vectors in the comparison matrix retain exactly four independent directions. This is its rank. The two directions of cancellation form its kernel, which has dimension two.

$$\operatorname{rank}\mathbf A_2(7)=4, \qquad \operatorname{nullity}\mathbf A_2(7)=2.$$

The rows have not become identical. Their relations have become visible.

Zero and one cannot vary independently

The binary example suggests a more general question. How many independent patterns can the digits support?

Follow one repeating block. Make one indicator that turns on wherever a zero occurs and another that turns on wherever a one occurs. At every position, exactly one is on. Their sum is the constant sequence one.

Subtract each indicator’s average. Their remaining variations are exact opposites. If the zero-indicator rises above its average at a position, the one-indicator falls by the same amount.

A Fourier transform expresses those variations in repeating patterns of different frequencies. Frequency zero keeps the totals. Every other frequency describes variation around them. Since the two variations are negatives, their Fourier coefficients are negatives too.

There is only one independent changing digit pattern, although there are two digits.

The same argument works with more digits. If $\nu$ different digits occur, their indicators sum to one. After removing their averages, they sum to zero. Knowing $\nu-1$ determines the last.

This is the partition-of-unity constraint. The name records the simple fact we started with. Every position contains exactly one digit.

Counting the directions that fit

Now put the remainder cycles back together.

Let $L$ be their common length and $C$ their number. Each cycle contributes a digit pattern at each frequency. There are $\nu$ occupied digit bins available to record those patterns.

At frequency zero, at most $\min(C,\nu)$ cycle patterns can be independent. At each other frequency, at most $\min(C,\nu-1)$ can be independent. Add across the $L$ frequencies,

$$\begin{aligned} \operatorname{rank}\mathbf A_b(p) \le{}&\min(C,\nu)\\ &+(L-1)\min(C,\nu-1). \end{aligned}$$

Here $p$ is prime and does not divide the base. The quantities are all finite counts. The period is $L$, there are $C=(p-1)/L$ cycles, and $\nu=\min(b,p-1)$ digits occur.

The formula gives a ceiling. Particular digit arrangements can introduce more relations and lower the rank further. When there are fewer cycles than occupied digits, the ceiling can also allow full rank.

For seven in binary, there are two cycles of length three, using two digits. The frequency counts are especially small.

Frequency	Independent directions
Zero	2
First nonzero frequency	1
Second nonzero frequency	1
Total	4

At zero, the two cycles have different totals of zeros and ones, so both directions survive. At either other frequency, the zero and one variations must oppose each other. Only one direction remains.

The two missing directions are the same cancellations we found by pairing the rows.

Seven, thirteen, seventeen

An eigenvalue records how much the comparison matrix stretches an independent direction. It lets us describe the directions that survive as well as count them.

Return to decimal notation. The block 142857 has six distinct digits. None of its five nontrivial rotations agrees with it anywhere. Its equality counts are

$$(6,0,0,0,0,0).$$

Dividing by six gives the first row of the comparison matrix. The Fourier transform gives six eigenvalues, all equal to one. Every direction survives unchanged.

Thirteen has two distinct levels. Seventeen has nine.

The flat spectrum at seven resembles the spectral shape associated with white noise. Here it comes from exact equality counts in a deterministic repeating block. A full-length repetend need not have this property. Seventeen also has a full-length decimal repetend, but several digits repeat.

The number of cycles tells us how many eigenvalues each frequency contributes. Their values depend on where the digits agree.

Two cycles do not guarantee two levels

At thirteen, the two cycles can be represented by

$$076923,\qquad153846.$$

Compare the first block with each rotation of the second. They agree only after three shifts, when the second block becomes 846153. Two digits then line up.

All the cross-cycle agreement is concentrated at that one shift. The resulting eigenvalue levels are $2/3$ and $4/3$.

Now divide by seven in base four. There are again two cycles. Their blocks are

$$021,\qquad123.$$

There is one match without shifting. After shifting the second block to 231, there is another. The third rotation gives none.

Example	Matches at successive shifts
Thirteen in base ten	$0,0,0,2,0,0$
Seven in base four	$1,1,0$

The total is two in both cases. Its placement differs.

In the Fourier calculation, each occupied shift contributes an arrow. At zero frequency they all point the same way. At other frequencies their directions depend on their shifts.

With only one occupied shift, the arrow turns but its length stays fixed. With two occupied shifts, the arrows can reinforce or partially cancel. Their sum changes length. That is why the base-four example has four levels, $1/3$, $2/3$, $4/3$, and $5/3$.

The exact criterion applies when both cycle words have distinct digits. Under that condition, there are exactly two global eigenvalue levels if and only if the cross-cycle match count is nonzero and concentrated at one shift.

Two cycles determine the size of the small comparison problems. The positions of the matching digits determine their answers.

The constraint was already there

The tools are classical. Lempel and Greenberger study cyclic equality counts, and Kak and Chatterjee apply such comparisons to reciprocal digit sequences. Repetend structure is discussed by Armstrong and Armstrong. The Fourier methods are treated by Davis and Terras.

Here those methods account for two restrictions in the complete table of fractions. The digit indicators limit how many independent directions can survive. The placement of cross-cycle matches determines whether the two-cycle spectrum stays at two levels.

I find the binary table the clearest way to see the first restriction. Pair 001 with 110. Then pair 010 with 101. The same digit slots are covered in both cases. There is already an exact relation among the rows before we calculate an eigenvalue.

The Fourier calculation tells us how that relation fits into the whole matrix. Six distinct rows retain four independent directions, and the missing two have a visible cause.