Rich distance classes invisible to planar topology

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This approach does not resolve either assertion of Problem 132. What it establishes is an exact limitation of charging rich distance classes to changes in planar topology. No computation or external theorem about distance multiplicities is used.

Let

$$\omega=e^{i\pi/3},\qquad A_N=\{a+b\omega:a,b\in\mathbb Z_{\ge0},\ a+b\le N\}, \qquad n=\binom{N+2}{2}.$$

For every prescribed integer $K\ge1$, sufficiently large $N$ has arbitrarily many initial rich distance classes, but all their occurring thresholds have the same planar Rips shadow and trivial Rips fundamental group. All but the first also occur after the disk union is contractible.

Planar shadow and fundamental group

Write $C_N=\operatorname{conv}(A_N)$. For $r\ge0$, let $\operatorname{VR}_r(A_N)$ denote the abstract complex whose simplices are the subsets of diameter at most $r$. Its planar shadow is the union of the geometric convex hulls of those simplices.

The standard triangulation of $C_N$ consists of equilateral unit triangles. These are simplices of $\operatorname{VR}_r(A_N)$ whenever $r\ge1$. Every simplex shadow is contained in $C_N$. Therefore the planar shadow is exactly the same closed triangle $C_N$ for all $r\ge1$.

In fact,

$$\pi_1(\operatorname{VR}_r(A_N))=0\qquad(r\ge1). \tag{1}$$

For an edge $pq$, write $q-p=a+b\omega$. There is a unit-edge lattice path from $p$ to $q$ in $A_N$ using two adjacent lattice directions, with the distances from $p$ increasing along the path. Here are the details ensuring it stays in the finite triangle.

If $a,b$ have the same sign, use the corresponding directions $1,\omega$ or their negatives. Both coordinates and their sum stay between their endpoint values. If $a\ge0$ and $b=-c\le0$, first take $\min(a,c)$ steps in direction $1-\omega$, then the remaining steps in direction $1$ if $a\ge c$, or $-\omega$ if $c\ge a$. Both coordinates stay between their endpoint values; the first part preserves their sum, and the second changes it monotonically to its final value. The other mixed-sign case is symmetric. Thus all path vertices satisfy the three defining inequalities of $A_N$.

The two directions used in each case make angle $60^\circ$. For nonnegative integer coefficients, adding either direction strictly increases the squared norm of the partial displacement. Consequently, writing the path as $p=p_0,p_1,\ldots,p_s=q$,

$$|p-p_i|\le |p-q|\le r.$$

Every nondegenerate fan triangle $(p,p_i,p_{i+1})$ is therefore a Rips simplex: its two radial edges have length at most $r$, and its third edge has length one. This fan homotopes $pq$ to the unit-edge path, fixing its endpoints. Every loop in the Rips complex can first be represented in its one-skeleton, then homotoped to unit edges by these fans. The resulting loop contracts in the standard triangular triangulation of $C_N$, already a subcomplex. This proves (1), and hence $H_1(\operatorname{VR}_r(A_N);\mathbb Z)=0$.

This argument does not show that the abstract Rips complex is contractible. Its higher homology and full Euler characteristic have not been controlled.

Disk unions stabilize before the second distance

Put $U_\rho=\bigcup_{a\in A_N}\overline B(a,\rho)$. Every point of a unit equilateral triangle is within $1/\sqrt3$ of a vertex, so

$$C_N\subseteq U_\rho\qquad(\rho\ge1/\sqrt3).$$

For these radii, nearest-point projection $\pi$ onto the closed convex set $C_N$ defines a strong deformation retraction of $U_\rho$ onto $C_N$. Indeed, if $x\in\overline B(a,\rho)$ with $a\in A_N$, the convex-projection inequality gives

$$|\pi(x)-a|\le|x-a|\le\rho.$$

The segment joining $x$ to $\pi(x)$ stays in that same convex disk. Thus $H(x,t)=(1-t)x+t\pi(x)$ is the required homotopy. In particular, $U_\rho$ is contractible for every $\rho\ge1/\sqrt3$.

For $N\ge2$ the first two distances are $1,\sqrt3$. Squared lattice lengths are $a^2+ab+b^2$; they cannot equal $2$, since modulo $3$ this form equals $(a-b)^2$, while $1$ and $3$ both occur. The pair-contact radius of the second distance satisfies

$$\sqrt3/2>1/\sqrt3.$$

Hence every later distance-class contact occurs while disk-union homology is already trivial. The same obstruction applies to a complex model whenever it is known to be homotopy equivalent to that union; no additional statement about such a model is needed here.

Arbitrarily many initial classes are rich

For a nonzero lattice displacement $v=a+b\omega$, put

$$S(v)=\max\{|a|,|b|,|a+b|\}.$$

The number of pairs with the chosen oriented displacement $v$ is exactly $\binom{N-S(v)+2}{2}$ when $N\ge S(v)$. To check this, a starting point $x+y\omega$ must satisfy

$$x\ge\max(0,-a),\quad y\ge\max(0,-b),\quad x+y\le N-\max(0,a+b).$$

Translate the two lower bounds to zero and use

$$\max(0,-a)+\max(0,-b)+\max(0,a+b)=S(v).$$

There remain two nonnegative integer coordinates with sum at most $N-S(v)$, giving the formula.

Rotation through $60^\circ$ sends $(a,b)$ to $(-b,a+b)$ and preserves $S(v)$. The three resulting unoriented directions are distinct, and their unordered pair sets are disjoint. Thus

$$m_{A_N}(|v|)\ge3\binom{N-S(v)+2}{2}. \tag{2}$$

If $S(v)\le N/3$, the right side is strictly greater than $\binom{N+2}{2}=n$. For example, after multiplying by two, the lower bound from $N-S(v)\ge2N/3$ exceeds $(N+1)(N+2)$ by $N^2/3+3N+4>0$.

Also $S(v)\le2|v|/\sqrt3$, by completing the square in $a^2+ab+b^2$ for each of $a,b,a+b$. Therefore

$$N\ge3\left\lceil\frac{2K}{\sqrt3}\right\rceil$$

ensures that every occurring distance at most $K$ is rich. There are at least $K$ such distances, since $0,1,\ldots,N$ belong to $A_N$. This proves the claimed richness of the first $K$ classes.

Exact scope of the obstruction

There can be arbitrarily many consecutive rich class additions with constant planar Rips shadow, zero Rips $H_1$, and constant disk-union homotopy type. Thus none of these invariants can require each rich class to spend a nonzero amount of topology. This does not rule out more elaborate arguments using additional geometry or higher abstract Rips homology. No such successful argument was established.

These configurations are not counterexamples to Erdős 132: nothing here bounds their total number of rare classes.