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. Three different pairs give the same combined coverage of the digits.

The Cross-Alignment Matrix keeps every comparison between these rows. Its entry for two fractions is the proportion of digit positions where they agree. That article follows the comparisons into their eigenvalues. Here the question is why some eigenvalues must be zero, and what keeps others at the same level.

The binary table already contains two cancellations. We can find them before calculating an eigenvalue.

Three pairs with the same total

Give each position two slots, one for zero and one for one. Each row occupies the slot its digit uses. As in the matrix construction, give each occupied slot weight $1/\sqrt3$, making the row vector a unit vector. Call the vector for $k/7$ by the name $v_k$.

The pair 001 and 110 fills all six slots. So does 010 with 101. So does 011 with 100.

Subtract the second pair total from the first. Every slot cancels. Subtract the third from the first and every slot cancels again. These are two independent relations among six distinct rows.

The matrix is a Gram matrix, a table of inner products of the row vectors. It has exactly the same relations. The row weights that produce cancellation form its kernel. Its rank counts the independent directions that remain. With two independent cancellations already in hand, the rank can be no larger than four.

Two cancellations give an upper bound. We still have to show that there are no others.

Zero and one cannot vary independently

Follow the word 001. Its zero-indicator is $(1,1,0)$, and its one-indicator is $(0,0,1)$. At every position, exactly one indicator is on. Their sum is the constant sequence one.

Subtract each indicator’s average. The two remaining variations are exact opposites. Where the zero-indicator rises above its average, the one-indicator falls by the same amount.

A Fourier transform expresses the variations in frequencies around the repeating cycle. Frequency zero records the digit totals. The other frequencies record variation. Opposite variations have opposite Fourier coefficients, so those two digit channels supply only one independent direction at any nonzero frequency.

With more digits, the same restriction remains. If $\nu$ digits occur across the fractional field, their indicators sum to one in each row. Center the indicators and they sum to zero. Knowing $\nu-1$ variations determines the last.

This is the partition-of-unity constraint. The name is longer than the observation. Every position contains exactly one digit.

Counting the directions that fit

Take a prime $p$ not dividing the base $b$. Let $L$ be the repeating period and $C=(p-1)/L$ the number of multiplication cycles. The number of occupied digit bins is $\nu=\min(b,p-1)$.

The matrix article’s Fourier reduction gives a $C\times C$ problem at each of the $L$ frequencies. To construct it, record one digit-coefficient vector for each cycle. Each vector has $\nu$ coordinates, one for each occupied digit.

At frequency zero, at most $\min(C,\nu)$ of these vectors can be independent. At every other frequency, all their coordinates sum to zero. They lie in a space of dimension $\nu-1$, leaving room for at most $\min(C,\nu-1)$ independent vectors.

Add the available dimensions across the frequencies.

Seven in binary has two cycles of length three, represented by 001 and 011. At zero frequency their digit-count vectors are $(2,1)$ and $(1,2)$. Neither is a multiple of the other, so both directions survive.

At either nonzero frequency, the digit coefficients sum to zero. That leaves at most one direction. Each word has a digit occurring exactly once, whose Fourier coefficient cannot vanish, so there is exactly one.

The rank is therefore $2+1+1=4$. Six rows retain four independent directions, and the kernel has dimension two. The pair cancellations found at the start account for the entire kernel.

The exact rank and kernel identities

Let $Y_j$ have one column for each multiplication cycle and one row for each occupied digit. Its entries are the unnormalized Fourier coefficients of the digit indicators. The comparison matrix at frequency $j$ is

$$M_j=\frac1L Y_j^*Y_j.$$

A vector belongs to the kernel of $M_j$ exactly when it belongs to the kernel of $Y_j$. This follows from

$$\langle z,M_jz\rangle=\frac1L\|Y_jz\|^2.$$

The full Fourier decomposition therefore gives exact identities.

$$\operatorname{rank}\mathbf A_b(p)=\sum_{j=0}^{L-1}\operatorname{rank}Y_j,$$ $$\operatorname{nullity}\mathbf A_b(p)=\sum_{j=0}^{L-1}\bigl(C-\operatorname{rank}Y_j\bigr).$$

Only the subsequent dimension count is an upper bound. These Fourier ranks are over the complex numbers; the original real matrix has the same rank over either field. Conjugate frequencies combine to describe its real directions.

The frequency here follows the positions around a multiplication cycle. It is not the additive Fourier coordinate of the consecutive remainders.

A ceiling, not an exact count

The bound is attained at binary seven. It need not be attained at another prime. Fifths give a smaller table where it is not.

1/5 = 0.|0011|   in base two

All four rows belong to one cycle of length four. There are two occupied digits. The bound allows $1+3=4$ independent directions, which would be full rank.

But the alternating frequency gives successive positions the weights $+1,-1,+1,-1$. In 0011, the zero-indicator occupies the first two positions. Its weighted sum is $1-1=0$. The one-indicator occupies the last two positions and cancels the same way. Both digit channels vanish at that frequency.

The four frequency ranks are $1,1,0,1$, so the matrix has rank three. Its eigenvalues are $2,1,0,1$. In the original rows, the missing direction is the relation

The general count allowed four directions. The placement of the digits supplied only three. At seven the partition constraint accounts for all the rank loss; at five an additional cancellation comes from the word itself.

Six binary tables, from four rows to 126

These matrices use the same pair-score scale and ordinary numerator order. Their ranks are computed by exact rational elimination, not by rounding small numerical eigenvalues to zero. The labels compare the actual ranks with the partition-of-unity ceilings.

At seventeen the ceiling is nine, while the exact rank is five. At 127 it is attained. The 126 rows retain eight independent directions, leaving 118 dimensions in the kernel. The size of the table and the number of independent directions in it can be very different.

Two cycles do not guarantee two levels

Counting the surviving directions is one question. Their eigenvalue levels pose another.

Recall thirteen in decimal from the matrix article. Its two cycle words have no repeated digit within either word, and all their cross-cycle matches occur at one shift. This gives two flat levels, $2/3$ and $4/3$.

Now divide by seven in base four. There are again two cycles, represented by

1/7 = 0.|021|   in base four
3/7 = 0.|123|   in base four

Compare 021 with 123. The middle digits agree. Shift the second word to 231 and the last digits agree. Shift once more to 312 and none do. The cross-cycle counts are $(1,1,0)$.

Both examples have two matches in their cross-cycle count, though their periods differ. At thirteen, both matches occupy one shift. In base four they occupy two. That placement determines whether the Fourier magnitude can stay constant.

One arrow, or two

In a Fourier sum, each occupied shift contributes an arrow. Its length is the match count there; its angle depends on the shift and the frequency.

At thirteen there is one arrow of length two. Changing frequency turns it, without changing its length. Dividing by the period six gives the constant split $1\pm1/3$.

In the base-four example, there are two unit arrows. At frequency zero they point the same way, giving a resultant of length two. At each of the other frequencies they make an angle of 120 degrees, giving a resultant of length one.

Divide those lengths by the period three. The eigenvalues are $1\pm2/3$ at frequency zero and $1\pm1/3$ at the other two frequencies. Across the whole spectrum, there are four levels,

There is an exact criterion here. With two cycles, and no repeated digit within either cycle word, there are exactly two global eigenvalue levels if and only if the cross-cycle count is nonzero and concentrated at a single shift.

One occupied shift gives one arrow of fixed length. For the converse, look at frequency zero, where all arrows point together and the resultant is as long as it can be. At the next frequency, distinct shifts give distinct angles. Two occupied shifts would make that resultant shorter. Constant length at every frequency therefore requires a single occupied shift.

Two cycles give two eigenvalues at each frequency. They do not require those eigenvalues to stay put as the frequency changes.

The two-level formula and its assumptions

Let $Q(\ell)$ count matches between the first word and the second word shifted by $\ell$ positions. Define

$$\widehat Q(j)=\sum_{\ell=0}^{L-1}Q(\ell)e^{-2\pi ij\ell/L}.$$

When both words have distinct digits, the two eigenvalues at frequency $j$ are

$$\lambda_j^\pm=1\pm\frac{|\widehat Q(j)|}{L}.$$

Exactly two global levels require the same nonzero magnitude at every frequency. A count of $m>0$ at a single shift gives levels $1-m/L$ and $1+m/L$, each repeated $L$ times. If the words share no digits at any shift, the matrix is the identity and there is only one level. If either word repeats a digit, its own autocorrelation also enters the calculation and this simplified criterion no longer applies.

The companion paper gives the general two-cycle formula, the exact kernel decomposition, and the rank bound. The classical correlation and Fourier sources are collected there and in The Cross-Alignment Matrix. The restrictions derived here come from applying that machinery to the digit partition across the complete table of fractions.

Pair cancellations and orbit-frequency profiles

The pair diagram and two-cycle comparison can also be inspected in these wider plates. The decimal profile plate keeps the eigenvalues from the preceding article available for comparison, without recalculating them in the main argument.

Return to 001 and 110. Exchange them for 010 and 101. Every digit slot is still filled exactly once. The rows change; the pair total does not.

Now lay out the 126 binary rows at 127. Every row is different. Choose eight independent row vectors. Every one of the other 118 is a weighted sum of those eight.

Those 118 rows fill the page. They do not enlarge the space.