The boundary equations use one contact weight. Our polygon begins with a different weight at every edge contact. The missing step is rigidity: unequal neighboring weights would let us move the polygon while keeping it invariant, with one image now strictly inside.
That would contradict what extremality forces. The polygon can change, but its multiplier stays fixed throughout the argument.
Karpelevich uses this strategy twice. A projective variation first eliminates certain return patterns. A simpler slide then forces the surviving weights to agree. We will explain the projective mechanism and prove the weight calculation explicitly, including the zero-weight case that finishes the reduction.
Move a generator and its whole chain
Continue with a nonreal radial extremum $\lambda$ of least realizing order $n>3$, with $0<|\lambda|<1$. The origin is interior to every invariant polygon under consideration. Every image vertex lies on its boundary.
The return arithmetic organized the vertices into chains
To move a generating vertex $x_i$ to $x_i'$, move every vertex it generates to $\lambda^j x_i'$. All power relations within the chosen chains remain exact. The remaining checks concern the terminal images at the chain ends.
Figure 1. An exact chain illustration with $\lambda=(1+i)/2$, $z=1$, and $z_s=1+i/5$. Each displacement is the preceding displacement multiplied by $\lambda$. This illustrates propagation along a chain; no polygon-invariance claim is needed for this picture.
A sufficiently small motion keeps the vertices extreme and in the same cyclic order: a finite set of strict convexity inequalities remains strict. To contradict extremality, we must then verify two things:
- Every new vertex image still belongs to the new polygon.
- At least one belongs to its interior.
An interior image alone is insufficient if some other image has escaped. This is why the directions of motion and the supporting edges matter.
Projective geometry eliminates a return pattern
Write $\varphi$ for the contact-block size and $g=\gcd(n,k)$. In the proper-block case $g<\varphi<n$, Stop 4 found an even-position Euclidean remainder $r$ with
Karpelevich’s geometric reduction, Theorem V. Under the extremal hypotheses above, $r=1$. Every Euclidean remainder is a multiple of $g$, so this forces $g=1$. The same part of the paper shows that a full block ($\varphi=n$) has shift $k=1$, which puts it in the forward contact form described below
This removes some integer patterns that the arithmetic allowed:
| $n,k,\varphi$ | Selected $r$ | Remainder test |
|---|---|---|
| $8,3,2$ | $1$ | Passes this test |
| $13,5,3$ | $2$ | Eliminated as an extremal configuration |
| $8,2,2$ | Separate case $\varphi=g$ | Uses the other branch |
The proof moves selected generators along lines, then uses successive projections through fixed contact points to determine the neighboring motions. At the end, the question is whether the last image lies on the inward side of a closing edge.
With suitable line coordinates, the starting configuration is $s=0$ and the returned coordinate is a fractional linear function
The desired inward displacement is exactly $s-u(s)>0$. Central projection between lines is fractional linear; a composition of such projections remains fractional linear.
Figure 2. Project $(s,0)$ from $C=(0,1)$ onto the line $y=x$. Its intersection is $(u,u)$ with $u=s/(1+s)$. At $s=1/2$, the returned coordinate is $1/3$. This exact local model displays the projective mechanism.
The return-map calculation has two cases. Exactly,
If $a\ne1$, choose a sufficiently small $s$ with the sign of $1-a$. The numerator and denominator then give $s-u(s)>0$.
If $a=1$, the first-order difference vanishes. Karpelevich’s geometric ordering gives $0<u(1)<1$ in the relevant nondegenerate configuration. Since $u(1)=1/(1+c)$, this forces $c>0$. Therefore
for every sufficiently small nonzero $s$.
Figure 3. For $u(s)=as/(1+s)$, the gray line is the starting coordinate $s$. An inward gap occurs on one side when $a\ne1$ and on both sides near zero when $a=1$. The critical case still has a strictly positive quadratic gap.
The global placement of the projection lines, the control of all other contacts, and the strict inequality $0<u(1)<1$ are the geometric content of Karpelevich’s Theorem V
Why the low-order exception is real
Let $\zeta=e^{2\pi i/3}$ and take the regular triangle $Q=\operatorname{conv}\{1,\zeta,\zeta^2\}$. The multiplier $\lambda=-1/2-i/2$ lies strictly inside the edge $[\zeta,\zeta^2]$. Rotational symmetry puts every image vertex inside its opposite edge.
Thus $n=3$, $k=2$, and $\varphi=3$: the full block has selected remainder $r=k=2$. This multiplier is radially extremal by the three-state trace bound $1+2\operatorname{Re}\lambda\geq0$, and two-state stochastic spectra are real. The conclusion that the selected remainder is one cannot be extended to this order.
This models the return-map inequality. The global construction still requires Karpelevich’s geometric ordering and verification of every other contact. Here all three maps have 0 < u(1) < 1.
Put the surviving relations in a usable form
After $r=1$, let $m$ be the preceding odd Euclidean remainder. The contact block has $\varphi\leq m$. Apply the return arithmetic to the larger marked window $1,\ldots,m$, and write $\ell$ and $\tau$ for that window’s two return times (the $p$ and $q$ of Stop 4). Stop 7 shows that $\tau<\ell$, so $m=\lfloor n/\ell\rfloor$.
Some added marks may be vertex contacts. Encode those by $\alpha_j=0$, $\beta_j=1$. The resulting relations are
where $\alpha_j\geq0$, $\beta_j>0$, and $\alpha_j+\beta_j=1$. The full-block case has $\ell=1$, $\tau=0$, $m=n$, with $x_m=x_0$.
We have not assumed that the marked window consists entirely of edge contacts. Showing that its weights agree will force every $\alpha_j$ to be positive, because at least one genuine edge contact is present. It will also show $\varphi=m$.
The other branch, $\varphi=g$, has $m=g$ equal-length chains from the outset:
Here all contact weights are positive, and $\sigma$ is the closing shift. Notice the different edge: the first branch uses the generator and its successor; the second uses its predecessor and the generator.
Unequal weights create an interior image
Consider neighboring coefficients $a=\alpha_i$ and $b=\alpha_{i+1}$ in the forward branch. For this local calculation write
The two contacts are
Move the middle generator along the preceding edge line:
Move its whole generated chain with it; keep the other chains fixed. The first contact remains on the same line. Its exact new expression is
For sufficiently small allowed $s$, both coefficients are nonnegative. So this contact remains inside the new edge.
The moving chain’s terminal image is
Express it in the triangle with vertices $A,B_s,C$:
where
The coefficients sum to one. If $a\ne b$, choose a sufficiently small $s$ with the sign of $b-a$. Then all three coefficients are strictly positive. Thus $\eta_s$ lies strictly inside that triangle, and hence strictly inside the new polygon.
Only these two contacts need new checks. Every other terminal contact is unchanged; every image within a chain follows the propagated vertices. Smallness preserves the polygon’s convexity. We have therefore retained invariance and created an interior image of a vertex—a contradiction.
It follows that $\alpha_i=\alpha_{i+1}$. Apply this for $i=1,\ldots,m-1$:
At equal weights, this attempted slide has $w_A=-(1-a)s^2/(1+s)<0$ for nonzero small $s$. The same motion now points outside the supporting edge.
The coefficient calculation, including a possible zero weight
Substitute
into
Collect the coefficients of $A,B_s,C$. This gives the boxed formulas and their sum one.
Since $a,b<1$, the coefficient $w_C$ is positive for small $s$. The chosen sign makes $w_A>0$. If $b>0$, then $w_B>0$ by continuity. If $b=0$, then $a>0$ and the chosen $s$ is negative, so $w_B=-s(1-a)/(1+s)>0$ as well.
The fixed contact’s coefficient $(a+s)/(1+s)$ also stays positive: use smallness if $a>0$; if $a=0$, then $b>0$ and the chosen sign is positive. Thus the proof includes unequal pairs in which one alpha coefficient is zero.
This rules out zero coefficients in the enlarged window once any alpha is positive. Consequently every mark is a genuine edge contact, and $\varphi=m$.
The exact local model uses A=1, B=(T−a)/(1−a), C=(T−b)B/(1−b), with T=1/4+i/2. It verifies the two local contact identities. In a complete eigenpolygon the motion is propagated along the selected chain and chosen small enough to preserve every other required condition.
The gcd branch uses the opposite slide
For $\varphi=g$, put $T=\lambda^\ell$ and take three neighboring generators $A,B,C$. The contact equations now read
Slide toward the opposite edge line:
The fixed image $TC$ remains on the new edge $[B_s,C]$, since
The moved image has coefficients
Again they sum to one. All contact weights in this branch are positive. If $a\ne b$, a sufficiently small $s$ with the sign of $b-a$ makes all three coefficients positive. This gives the same contradiction, with the same chain and convexity checks.
Compare consecutive pairs for $i=1,\ldots,g-1$, leaving the last generating chain and its closing relation fixed. All $g$ weights agree. If $g=1$, there is only one weight and nothing to compare.
Figure 4. Exact local models with $a=1/4$, $b=3/4$, and $s=1/2$. In the forward picture the moved image has coefficients $(1/8,1/2,3/8)$ at $A,B_s,C$; in the backward picture they are $(3/8,1/2,1/8)$. Both images are interior. For a global extremal polygon, the proof uses a sufficiently small motion so every other required condition persists.
What is rigid now?
The geometric step has supplied two conclusions that arithmetic alone could not give:
| Branch | Surviving structure | Common weights |
|---|---|---|
| $\varphi>g$ | $g=1$, selected remainder $1$, $\varphi=m$ chains of length $\ell$ | $\alpha_j=\alpha$, $\beta_j=\beta$ |
| $\varphi=g$ | $m=g$ chains of length $\ell=n/g$ | $\alpha_j=\alpha$, $\beta_j=\beta$ |
In both cases $\alpha,\beta>0$ and $\alpha+\beta=1$. The terminal equations now have one shared parameter. Eliminating the generating coordinates gives
in the forward case, and
in the gcd case. In both branches $t$ is the weight on the chain’s own generator, the local return. That is $\alpha$ in the forward branch and $\beta$ in the gcd branch, because the two branches use opposite edges. This is the convention for $t$ throughout the series.
The boundary-equations post carries out those eliminations. The work here explains why the repeated factors have the same coefficient: every unequal neighboring pair would admit an invariant variation that extremality forbids.
With $r=1$ and one common local-return weight $t$, eliminate the polygon coordinates altogether and construct the realizing stochastic matrices: Where the Boundary Equations Come From →
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, §4, Theorems V–VII, pp. 372–377. Original paper.
This post explains Theorem V’s projective contradiction, using the theorem for the global ordering and configuration checks. It proves the fractional-linear inequality and spells out the neighboring-weight variations in both branches. The local models are exact illustrations of those calculations; they do not certify complete extremal eigenpolygons. The earlier contact and return theorems remain inputs. No numerical optimization or graphical experiment replaces a proof.