Lemma 2 — lattice averaging for unit pairs

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Statement

Let $d\geq1$, let $\Lambda$ be a full-rank lattice in $\mathbb R^{2d}$, and let $\|\cdot\|$ be any norm on that space. Suppose $\pi:\Lambda\to\mathbb R^2$ is an injective group homomorphism, and write $|\cdot|$ for the Euclidean norm on $\mathbb R^2$. Set

$$\rho=\min\{\|v\|:v\in\Lambda\setminus\{0\}\}$$

and

$$M=\#\{v\in\Lambda:\|v\|\leq1,\ |\pi(v)|=1\}.$$

For every real $R>1$ there is a finite nonempty $U\subset\mathbb R^2$ such that

$$|U|\leq\left(\frac{2R}{\rho}+1\right)^{2d} \tag{1}$$

and, for the ordered unit-pair count

$$D_{\mathrm{ord}}(U) =\#\{(u_1,u_2)\in U^2:|u_1-u_2|=1\},$$

one has

$$\frac{D_{\mathrm{ord}}(U)}{|U|} \geq\left(1-\frac1R\right)^{2d}M. \tag{2}$$

Proof

For $w\in\mathbb R^{2d}$ and $r>0$, let

$$B(r,w)=\{x\in\mathbb R^{2d}:\|x-w\|\leq r\}.$$

We will take

$$U=\pi(B(R,w)\cap\Lambda)$$

for a suitable center $w$. Injectivity of $\pi$ makes its cardinality equal to $\#(B(R,w)\cap\Lambda)$.

The open norm-balls of radius $\rho/2$ about the lattice points in $B(R,w)$ are pairwise disjoint. All lie in $B(R+\rho/2,w)$. Since volume in $2d$ dimensions scales by the $2d$-th power of the radius,

$$\#(B(R,w)\cap\Lambda) \leq\left(\frac{R+\rho/2}{\rho/2}\right)^{2d} =\left(\frac{2R}{\rho}+1\right)^{2d},$$

which proves (1).

Now fix $v_1\in B(R-1,w)\cap\Lambda$. For every vector $v$ counted by $M$, the point $v_2=v_1+v$ lies in $B(R,w)\cap\Lambda$ by the triangle inequality, and

$$|\pi(v_2)-\pi(v_1)|=|\pi(v)|=1.$$

Distinct choices give distinct ordered pairs after projection. Therefore

$$\frac{D_{\mathrm{ord}}(U)}{|U|} \geq \frac{\#(B(R-1,w)\cap\Lambda)} {\#(B(R,w)\cap\Lambda)}M. \tag{3}$$

It remains to choose $w$. Choose it uniformly in a fundamental domain of $\Lambda$. Unfolding the translates of that domain shows that, for every $r>0$,

$$\mathbb E_w\#(B(r,w)\cap\Lambda) =\frac{\operatorname{vol}(B(r,0))} {\operatorname{covol}(\Lambda)} =\frac{r^{2d}\operatorname{vol}(B(1,0))} {\operatorname{covol}(\Lambda)}. \tag{4}$$

Consequently the expectation of

$$\#(B(R-1,w)\cap\Lambda) -\left(1-\frac1R\right)^{2d} \#(B(R,w)\cap\Lambda)$$

is zero. Put $N(w)=\#(B(R,w)\cap\Lambda)$. Equation (4) gives $\mathbb E N(w)>0$, so $N(w)>0$ on a set of positive measure. When $N(w)=0$, the inner count and the displayed difference are also zero. If that difference were negative at every center with $N(w)>0$, its expectation would be strictly negative, a contradiction. Thus some center has both $N(w)>0$ and nonnegative difference. For that center $U$ is nonempty, and (3) gives (2), completing the proof.

Source scope

This is Lemma 2 on physical pp. 3--4 of the arXiv v1 manuscript. The notation $D_{\mathrm{ord}}$ makes explicit that the source construction counts oriented pairs.

Used by. Lemma 5.

Bears on. Problem 90.