Angular interpolation identities and failure of shell-sign arguments

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This was a different approach from convex layers, valuations, and scalar radial positivity. Its exact identities are valid, but neither of the proposed sign arguments solves Problem 132.

Identities retaining the edge directions

Identify $A$ with distinct complex numbers $z_1,\ldots,z_n$, and put

$$f(z)=\prod_i(z-z_i),\qquad w_i=1/f'(z_i).$$

Lagrange interpolation, followed by taking the coefficient of $z^{n-1}$, gives

$$M_\ell:=\sum_i w_i z_i^\ell=0\quad(0\le\ell\le n-2), \qquad M_{n-1}=1.$$

For an occurring distance $r$ and integer $k\ge1$, define

$$B_r(k)=\sum_{\substack{i<j\\|z_i-z_j|=r}} w_iw_j(z_i-z_j)^{2k}.$$

The even exponent makes this independent of how unordered pairs are oriented. Expanding the power yields

$$2\sum_r B_r(k)= \sum_{a=0}^{2k}(-1)^a\binom{2k}{a}M_{2k-a}M_a.$$

For $1\le k\le n-2$, at least one exponent in every product is at most $n-2$, so the sum is zero. For $k=n-1$, only $a=n-1$ survives. Consequently

$$\sum_rB_r(k)=0\quad(1\le k\le n-2),$$ $$\sum_rB_r(n-1)= \frac{(-1)^{n-1}}2\binom{2n-2}{n-1}. \tag{1}$$

Under a direct similarity $z\mapsto az+b$, the weights acquire the factor $a^{-(n-1)}$. Hence each terminal charge $B_r(n-1)$ is unchanged for its corresponding transformed class. Reflection conjugates it. These charges retain angular information discarded by scalar radial averaging.

A real Hermitian alternative is

$$H_r=\operatorname{Re}\sum_{\substack{i<j\\|z_i-z_j|=r}} w_i\overline{w_j}.$$

For $n\ge3$, expanding the full ordered sums and using $\sum_iw_i=\sum_iw_iz_i=0$ gives

$$\sum_rH_r=-\frac12\sum_i|w_i|^2,$$ $$\sum_r r^2H_r=-\frac12 \left|\sum_i w_i\overline{z_i}\right|^2. \tag{2}$$

A rich class can have either sign

Let $\omega=e^{\pi i/3}$ and

$$A_p=\{0,1,2,\omega,1+\omega,p\}.$$

Among the five fixed points there are exactly seven unit pairs: $01,0\omega,12,1\omega,1(1+\omega),2(1+\omega), \omega(1+\omega)$. Along any fixed nonzero ray $p=\varepsilon v$ with $\varepsilon\to0^+$, no additional unit pair occurs for all sufficiently small $\varepsilon>0$. For the fixed points of modulus one this follows directly from

$$|\varepsilon v-a|^2=1 \iff \varepsilon^2|v|^2-2\varepsilon \operatorname{Re}(v\bar a)=0;$$

the other fixed points follow by continuity. Thus the unit class has $7>6$ pairs throughout these small configurations.

Put

$$F(z)=z(z-1)(z-2)(z-\omega)(z-1-\omega),\qquad u_a=1/F'(a).$$

Using $\omega^2=\omega-1$ gives

$$u_0=\frac1{2(2\omega-1)},\qquad u_1=1, \qquad u_\omega=-\frac{1+\omega}{3}.$$

The new weight at a fixed point $a$ is $u_a/(a-p)$. Only the unit pairs $01$ and $0\omega$ have weights singular as $p\to0$. Their contributions give

$$B_1(5)=\frac{C}{p}+O(1), \qquad C=u_0(u_\omega-1)=\frac\omega2-\frac13,$$ $$H_1=\operatorname{Re}\frac{\omega}{6p}+O(1). \tag{3}$$

For the second identity the coefficient is $-u_0(1+\overline{u_\omega/\omega})=\omega/6$. All other unit-pair terms stay bounded.

Taking $p=\pm\varepsilon C$ makes $\operatorname{Re}B_1(5)\to\pm\infty$; taking $p=i\varepsilon C$ makes its imaginary part unbounded. Taking $p=\pm\varepsilon\omega$ similarly makes $H_1\to\pm\infty$.

Thus richness alone gives no fixed sign, even for the real part, in either proposed charge. The failure is established by exact algebra, not by numerical examples.

Exact conclusion

The example does not have bounded total rarity and is not an Erdős counterexample. It rules out the proposed inference that each rich class contributes a charge of a prescribed sign to (1) or (2). It does not rule out a future argument controlling several classes jointly under a bounded-rarity hypothesis. No such joint control was proved here; proposing it is not a reduction of the original problem to an established lemma.

The author checked the interpolation expansions, similarity normalization, seven-edge count, and both coefficients in (3); the argument is author-recorded.