Existence of similar point configurations in thin subsets of ${\mathbb {R}}^d$

Abstract

We prove the existence of similar and multi-similar point configurations (or simplexes) in sets of fractional Hausdorff dimension in Euclidean space. Let $d \ge 2$ and $E\subset {{{\mathbb {R}}} }^d$ be a compact set. For $k\ge 1$, define

$$\begin{aligned} \Delta _k(E)=\left\{ \left( |x^1-x^2|, \dots , |x^i-x^j|,\dots , |x^k-x^{k+1}|\right) : \left\{ x^i\right\} _{i=1}^{k+1}\subset E\right\} \subset {{{\mathbb {R}}} }^{k(k+1)/2}, \end{aligned}$$

the $(k+1)$-point configuration set of E. For $k\le d$, this is (up to permutations) the set of congruences of $(k+1)$-point configurations in E; for $k>d$, it is the edge-length set of $(k+1)$-graphs whose vertices are in E. Previous works by a number of authors have found values $s_{k,d}<d$ so that if the Hausdorff dimension of E satisfies $\dim _{\mathcal H}(E)>s_{k,d}$, then $\Delta _k(E)$ has positive Lebesgue measure. In this paper we study more refined properties of $\Delta _k(E)$, namely the existence of similar or multi–similar configurations. For $r\in {\mathbb {R}},\, r>0$, let

$$\begin{aligned} \Delta _{k}^{r}(E):=\left\{ \mathbf {t}\,\in \Delta _k\left( E\right) : r\mathbf {t}\,\in \Delta _k\left( E\right) \right\} \subset \Delta _k\left( E\right) . \end{aligned}$$

We show that if $\dim _{\mathcal H}(E)>s_{k,d}$, for a natural measure $\nu _k$ on $\Delta _k(E)$, one has all $r\in {\mathbb {R}}_+$. Thus, in E there exist many pairs of $(k+1)$-point configurations which are similar by the scaling factor r. We extend this to show the existence of multi–similar configurations of any multiplicity. These results can be viewed as variants and extensions, for compact thin sets, of theorems of Furstenberg, Katznelson and Weiss [7], Bourgain [2] and Ziegler [11] for sets of positive density in ${\mathbb {R}}^d$.

Appendix: A measure-theoretic pigeon hole principle

Unable to find Lemma 1 in the literature, and believing that it should be useful for other problems, we prove it here. Without loss of generality the total measure $\sigma (X)$ can be normalized to be equal to 1, so for the proof we restate the result as

Lemma 2

Let $\mathcal X=(X,\mathcal M,\sigma )$ be a probability space. For $0<c<1$, let $\mathcal M_c=\{A\in \mathcal M: \sigma (A)\ge c\}$. Then, for every $n\in \mathbb {N}$, there exists an $N=N(\mathcal X,c,n)\in \mathbb {N}$ such that, for any collection $\{A_1,\dots ,A_N\}\subset \mathcal M_c$ of cardinality at least N, there is a subcollection $\{A_{i_1},\dots , A_{i_n}\}$ of cardinality n such that $\sigma (A_{i_1}\cap \cdots \cap A_{i_n})>0$ and hence $A_{i_1}\cap \cdots \cap A_{i_n}\ne \emptyset$.

To start the proof, first we establish the following claim, which is a quantitative strengthening of the statement for $n=2$:

Claim 1

Let $\mathcal X=(X,\mathcal M,\sigma )$ be a probability space. Then for any $0<c<1$ there exists $P_c\in \mathbb {N}$ such that for any $N>P_c$, if $\{A_1,\dots ,A_N\}\subset \mathcal {M}_c$, then there exist distinct $i,j\le N$ such that $\sigma (A_i\cap A_j)\ge c^3/3$.

Proof

Suppose not. Let $S\subset (0,1)$ be the set of all $c\in (0,1)$ such that the statement of the claim is false, and suppose $c\in S$. Then for every $N\in \mathbb {N}$ there exists a subset $\{A_1,\dots ,A_{2N}\}\subset \mathcal {M}_c$ such that $\sigma (A_i)\ge c$ for all i but $\sigma (A_i\cap A_j)<c^3/3$ for all $i\ne j$. Consider the sets $A_{2i-1}\cup A_{2i}$, $i=1,\dots ,N$. We have

$$\begin{aligned} \sigma (A_{2i-1}\cup A_{2i})=\sigma (A_{2i-1})+\sigma (A_{2i})-\sigma (A_{2i-1}\cap A_{2i}) > c+c-\frac{c^3}{3}=2c-\frac{c^3}{3}. \end{aligned}$$

Since $\sigma (X)=1\ge \sigma (A_{2i-1}\cup A_{2i})$, this implies $1> 2c-\frac{c^3}{3}$. In particular, since $c<1$, we have $c\lesssim 0.52<3/5$; hence $[3/5,1)\cap S=\emptyset$. Moreover,

$$\begin{aligned} \sigma \big ((A_{2i-1}&\cup A_{2i})\cap (A_{2j-1}\cup A_{2j})\big )\\&=\sigma \big ((A_{2i-1}\cap A_{2j}) \cup (A_{2i-1}\cap A_{2j-1}) \cup (A_{2i}\cap A_{2j}) \cup (A_{2i}\cap A_{2j-1})\big )\\&\le \sigma (A_{2i-1}\cap A_{2j}) +\sigma (A_{2i-1}\cap A_{2j-1}) +\sigma (A_{2i}\cap A_{2j}) +\sigma (A_{2i}\cap A_{2j-1})\\&< 4\frac{c^3}{3}\le \frac{(2c-c^3/3)^3}{3}\hbox { since } 0<c<1. \end{aligned}$$

Thus, there exist N sets, namely $A_1\cup A_2,\dots ,A_{2N-1}\cup A_{2N}$, such that each has measure at least $f(c):=2c-\frac{c^3}{3}$ but all pairwise intersections have measure less than $\frac{f(c)^3}{3}$.

Thus, we have shown that if $c\in S$ then $f(c)\in S$ as well. However if $0<c<1$, then there exists $k\in \mathbb {N}$ such that $f^k(c)>3/5$ and is thus $\notin S$ (where $f^k$ denotes f composed with itself k times). It follows that S must be empty. $\square$

We use Claim 1 as a building block for the proof of Lemma  2, which is by induction on n. If $n=1$, then we can take $N=1$, since any $A_{i_1}\in \mathcal {M}_c$ satisfies the statement. If $n=2$ then any $N\ge \lceil { 1/c}\rceil$ suffices, since there cannot be more than 1/c pairwise disjoint sets of measure $\ge c$ each; alternatively, one may simply invoke Claim  1.

Now suppose that the conclusion of Lemma 2 holds for some n, $n\ge 2$. Set $N=2N(\mathcal {X},c^3/3,n)+P_c$, and suppose $\{A_1,\dots ,A_N\}\subset \mathcal {M}_c$ is a collection of cardinality N. Since $N>P_c$, by Claim 1 there exist distinct $i,j\le N$ such that $\sigma (A_i\cap A_j)>\frac{c^3}{3}$. Let $B_1=A_i\cap A_j$. Removing $A_i$ and $A_j$ from the collection we still have $N-2>P_c$ sets, so can find another pair whose intersection has measure at least $\frac{c^3}{3}$. Repeating this procedure $N(\mathcal {X},c^3/3,n)$ times, one finds sets $B_1,\dots ,B_{N(\mathcal {X},c^3/3,n)}\in \mathcal {M}_{c^3/3}$. By the induction hypothesis there exist $0<i_1<i_2<\dots <i_n\le N(\mathcal {X},c^3/3,n)$ such that $\sigma (B_{i_1}\cap \dots \cap B_{i_n})>0$. Since $B_{i_1}\cap \dots \cap B_{i_n}$ is the intersection of 2n distinct sets from the collection $\{A_1,\dots ,A_N\}$, the intersection of any $n+1$ of those 2n will have positive measure, completing the induction step. $\square$