We have equations for the boundary candidates. An equation can have many complex roots, and its endpoint equations can have many roots of unity. We still need to locate the relevant roots in the plane.
Karpelevich gets that information from angles in the invariant polygon. The integer congruences already controlling the vertex chains become determinant-one identities between rational angles. Those identities identify Farey neighbors.
This post connects the geometric angle estimates to the labels on the arcs. We will prove the Farey arithmetic, work through one angular estimate in full, and use the eight-state equation to see why the labels matter.
The unit circle supplies rational landmarks
Write a direction in turns:
From the polygon correspondence, the unit-circle points in $\mathcal K_n$ are exactly the roots of unity of order at most $n$. Thus their directions are the reduced fractions
In increasing order, these form the Farey sequence $F_n$. The fractions $0/1$ and $1/1$ represent the same point $1$ on the circle; keep both when discussing the interval $[0,1]$, and identify them when drawing the circle.
For example, the upper-half directions in $F_7$ are
Figure 1. Consecutive fractions $1/3$ and $2/5$ give directions $120^\circ$ and $144^\circ$. At order seven, no other unit-circle landmark lies between them. The picture marks directions and landmarks; it does not draw a boundary arc.
What makes two fractions neighbors?
Two reduced fractions $a/b<c/d$ are consecutive in $F_n$ exactly when
The determinant condition makes them a particularly close rational pair. The denominator condition says their gap contains no allowable new landmark at this order.
Here is the useful calculation. Suppose $bc-ad=1$ and a fraction $u/v$ lies strictly between them. The two integers
are positive. Moreover,
Every fraction between the pair therefore needs denominator at least $b+d$.
The bound is attained by the mediant:
It is reduced, because a common divisor of its numerator and denominator would divide $b(a+c)-a(b+d)=1$. It is also the unique fraction in the gap with the minimum denominator: equality in the boxed identity forces both positive integers to be one.
Why the determinant-one condition is necessary too
The preceding argument proves that determinant one together with $b+d>n$ is sufficient. To find the actual predecessor of a reduced $c/d\in F_n$ with $0<c/d<1$, use the Euclidean algorithm to solve
Choose the largest positive solution $b\leq n$, and put $a=(cb-1)/d$. Then $a/b$ is reduced, lies below $c/d$, and satisfies $bc-ad=1$. Maximality of $b$ gives $b+d>n$.
The denominator bound now proves that no fraction of $F_n$ lies between these two. Thus this construction finds the predecessor, and every neighboring pair has determinant one. The endpoint $1/1$ has predecessor $(n-1)/n$, which satisfies the same criterion.
The angular width of a neighboring gap is consequently exact:
A new order inserts a mediant
The pair $1/3<2/5$ is neighboring at order seven:
Its first new fraction is $3/8$. It appears at order eight, splitting the angular gap into two neighboring gaps:
Figure 2. The old gap has width $24^\circ$. Its new pieces have widths $15^\circ$ and $9^\circ$. The mediant is not generally the angular midpoint.
The neighbor relation depends on the matrix order. A pair stays neighboring through order $b+d-1$ and separates when order $b+d$ admits its mediant. This describes the changing angular labels. How the actual boundary curves change is part of the later induction.
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Filled points are current unit-circle landmarks. The dashed ray marks the future mediant direction; its hollow marker is inset for legibility and assigns no eigenvalue modulus. The mediant denominator exceeds the current order. Changing the order changes the neighbor list. The highlighted wedge describes directions.
The polygon’s angle bounds produce these exact neighbors
Return to a nonreal radial extremum new at order $n>3$, with $0<|\lambda|<1$. Choose the contact orientation from the earlier posts, and let $k$ be its cyclic shift. Initially the contact theorem gives the coarse sector
Karpelevich’s angular estimates, Theorems VIII–IX, refine this. Put $g=\gcd(k,n)$. Choose the unique integers $f,b$ satisfying
Then the chosen orientation lies in the open sector
The general angular refinement is a geometric input from §5 of the original proof. Its Farey meaning follows immediately. Reduce $k/n$ to $c/d$, where $c=k/g$ and $d=n/g$. Dividing the determinant identity by $g$ gives
and the bound on $b$ gives $b+d>n$. Thus $f/b$ and $c/d$ are neighbors in $F_n$.
| Order $n$ | Shift $k$ | Bézout identity | Selected angular sector |
|---|---|---|---|
| $8$ | $3$ | $3\cdot3-8\cdot1=1$ | $1/3<\theta<3/8$ |
| $7$ | $3$ | $3\cdot5-7\cdot2=1$ | $2/5<\theta<3/7$ |
| $8$ | $2$ | $2\cdot5-8\cdot1=2$ | $1/5<\theta<1/4$ |
The last row shows why reduction matters. The raw fraction $2/8$ gives determinant two; its reduced form $1/4$ gives determinant one with $1/5$.
Conjugation reflects a sector and reverses its endpoints:
The determinant and denominator conditions are preserved, so the same labeling works in the opposite orientation.
One angular estimate, with the turn count kept visible
Here is a complete derivation for the forward branch from the rigidity post, in the standard notation: $\ell$ is the chain length, $m$ the number of repeated chains, and $\tau$ the tail. Its relations include
and $n=\ell m+\tau$ with $m\geq2$, $\tau\geq0$. In particular $\ell\leq n/2$. The arithmetic gives
Since $x_0,x_1$ are consecutive counterclockwise vertices and the origin is interior, their angular separation is strictly between zero and $\pi$. Dividing the contact equation by $x_0$ therefore gives
We must relate this principal argument to $\ell\theta$. The coarse sector and the integer identity imply
Because $\ell\leq n/2$ and $n>3$, this whole interval lies inside $(-1/2,1/2)$. There is no ambiguity about the principal turn:
Its positivity forces $\theta>f/\ell$. Also $\theta<k/n$: on the endpoint ray the root $e^{2\pi ik/n}$ is realizable, so a strict contraction there cannot be radially extremal. We obtain
This is precisely the Farey pair with $k\ell-fn=1$. For the eight-state example, $\ell=3$, $f=1$, and $k/n=3/8$.
Figure 3. At $\theta=17/48$, the original direction is $127.5^\circ$. Multiplication by three gives $3\theta=17/16$ turns. Removing the full turn leaves $1/16$ turn, or $22.5^\circ$. The integer $f=1$ is part of the argument.
The other branch uses the angle-packing estimate in Theorem VIII. We use the resulting general bound from section 4; the full packing argument is not reconstructed here.
The slider moves inside the coarse sector. Buttons set exact rational rays; their displayed slider positions may be rounded. This tests the angle condition and assigns no modulus or eigenvalue realization to a ray. The pale wedge is the coarse range; teal is the selected angular sector.
The angular sector fixes the exponents
The two families of Stop 6 each carry three integers: the chain length $\ell$, the repetition count $m$, and either a tail $\tau$ or a closing shift $\sigma$. Once the angular sector is known, all three are determined by the Farey pair that bounds it. They are exactly the integers in Ito’s equation for that pair.
Exponent dictionary
Proved hereLet $\lambda$ be a nonreal radial extremum, new at order $n>3$, with $0<|\lambda|<1$, in the chosen orientation with shift $k$ and $g=\gcd(n,k)$. Its sector $f/b<\theta<k/n$ is bounded by Farey neighbours. Write $\ell$ and $s$ for the smaller and larger of their denominators, and $m=\lfloor n/\ell\rfloor$.
- Forward branch ($\varphi>g$, so $g=1$). Here $s=n$, and $\ell=b$ is the inverse of $k$ modulo $n$. The relations of Stop 5 are the first family with tail $\tau=n-\ell m$, where $0\leq\tau<\ell$.
- gcd branch with $g\geq2$ ($\varphi=g$). Here $\ell=n/g$ is the reduced denominator of $k/n$, $m=g$, and $s=b$. The relations are the second family with closing shift $\sigma=n-b$.
In both cases the reduced equation is Ito’s equation for the pair.
Proof of the exponent dictionary
Forward branch. The window identity of Stop 5 reads $n=\ell m+\tau$. Here $\ell$ and $\tau$ are the record times $p$ and $q$ of Stop 4 at the stage $(L,U)=(1,m)$. Since $k\ell\equiv1\pmod n$, $\ell$ is the inverse of $k$ modulo $n$, and the previous section identified $f/\ell$ as the predecessor of $k/n$ in $F_n$. In particular $\ell<n=s$.
It remains to show $\tau<\ell$, which gives $m=\lfloor n/\ell\rfloor$. If $k=1$, the stage is the initial one, with $\ell=1$, $\tau=0$ and $m=n$. Otherwise $m=r_{2j-1}$ for some $j\geq2$. When $U$ first takes the value $r_{2j-1}$, the current value of $L$ is $r_{2j-2}>r_{2j-1}$. The algorithm therefore subtracts $U$ from $L$ at least once before $L$ reaches $r_{2j}=1$, and each such step replaces $p$ by $p+q$. Hence $\ell=p>q=\tau$. Finally $s=n\geq\ell m$, so Ito’s reduced form is the first family with $\tau=s-\ell m$.
gcd branch. The block has one target label in each residue class modulo $g$, so the $m=g$ chains all have length $\ell=n/g$. Label $0$ lies on the chain that starts at label $g$, at the step $\sigma$ with $g+\sigma k\equiv0\pmod n$; this is the closing relation $\lambda^\sigma x_g=x_0$ of Stop 5. The predecessor $f/b$ of $k/n$ satisfies $kb-nf=g$ with $0<n-b<\ell$. Hence $\sigma'=n-b$ satisfies $k\sigma'\equiv-g\pmod n$ and $0<\sigma'<\ell$. The congruence $k\sigma\equiv-g\pmod n$ determines $\sigma$ modulo $n/g=\ell$, so $\sigma=n-b$.
Since $b>n-\ell=\ell(g-1)\geq\ell$, the smaller denominator is $\ell$ and $s=b$. Then $\ell m=n>s$, so Ito’s reduced form is the second family with $\sigma=\ell m-s=n-b$.
| $n,k$ | Branch | Sector | $\ell,\ s,\ m$ | Boundary equation |
|---|---|---|---|---|
| $8,3$ | Forward | $1/3<\theta<3/8$ | $3,\ 8,\ 2$ | $\lambda^2(\lambda^3-t)^2=(1-t)^2$ |
| $8,2$ | gcd, $g=2$ | $1/5<\theta<1/4$ | $4,\ 5,\ 2$ | $(\lambda^4-t)^2=(1-t)^2\lambda^3$ |
| $6,2$ | gcd, $g=2$ | $1/4<\theta<1/3$ | $3,\ 4,\ 2$ | $(\lambda^3-t)^2=(1-t)^2\lambda^2$ |
Route change: one case is settled by the modern statement
Ito 1997The dictionary leaves one configuration open: $g=1$ with a single edge contact ($\varphi=g=1$). Its relation from Stop 5 has $\ell=n$ and $m=1$, namely $\lambda^n-t=(1-t)\lambda^{n-b}$. Ito’s equation for the same pair, with smaller denominator $b$, is $\lambda^{\tau'}(\lambda^b-t')^{m'}=(1-t')^{m'}$ with $m'=\lfloor n/b\rfloor$.
If $b>n/2$, then $m'=1$ and the two equations coincide after $t'=1-t$. For example, $n=7$, $k=3$ gives the sector $2/5<\theta<3/7$ and $\lambda^2(\lambda^5-t)=1-t$. If $b<n/2$, they are different curves with the same endpoints, and only Ito’s is the boundary. Karpelevich’s final assembly decides between such candidates; we do not reconstruct that step.
From here on the series uses Karpelevich’s theorem in Ito’s form
The labels distinguish the endpoint roots
Our eight-state relations produced
The corresponding sector is $1/3<\theta<3/8$. At the parameter endpoints, the polynomial factors exactly:
| Parameter | Endpoint polynomial | Unit-circle root in the closed selected sector |
|---|---|---|
| $t=0$ | $\lambda^8-1$ | $e^{2\pi i(3/8)}$ |
| $t=1$ | $\lambda^2(\lambda^3-1)^2$ | $e^{2\pi i(1/3)}$ |
Figure 4. The gray points are other unit-circle roots of the endpoint polynomials. The selected sector distinguishes the eighth root at $135^\circ$ from the cube root at $120^\circ$. At $t=1$, zero is also a root, but it is not a unit-circle landmark.
An arc in this sector joining unit-circle landmarks must use those endpoint labels. The argument has identified their locations exactly. It has not certified every root of the equation, or every root path inside an angular sector, as an outer-boundary arc.
That final selection belongs to the induction on the order. The same distinction applies to the mediant picture: inserting a rational landmark organizes the sectors; establishing the new and inherited boundary pieces requires the remaining proof.
The algebra now has angular labels. The next stop follows the other conclusion of the original paper: Cyclic Polygons and the Whole Region.
With the algebraic boundary curves placed inside their exact Farey angular sectors, we turn to the other headline conclusion of the 1951 paper — Cyclic Polygons and the Whole Region →
Source and scope
[1] F. I. Karpelevich, On the characteristic roots of matrices with nonnegative elements, Izv. Akad. Nauk SSSR Ser. Mat. 15:4 (1951), 361–383. Original paper. The angular estimates are Theorems VIII–IX, pp. 377–379; their conversion to neighboring rational sectors begins the final assembly on p. 380.
This post uses the general angular refinement as a cited geometric input. It proves the Farey criterion, the mediant threshold, the forward-branch angular lift, the exponent dictionary, and the endpoint-root identification in the worked example. The selection between candidate equations in the single-contact case is taken from Karpelevich’s theorem in Ito’s form (H. Ito, A new statement about the theorem determining the region of eigenvalues of stochastic matrices, Linear Algebra Appl. 267 (1997), 241–246). All figures and explorers use exact rational or integer data, with trigonometric coordinates only for drawing. No numerical root tracking is used to establish a boundary claim.