Proposition 10 — explicit exponent from local field data

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Statement

Let $S_{\mathbb Q}$ be a finite set of rational primes, let $k,e,f:S_{\mathbb Q}\to\mathbb Z_{>0}$, and let $\lambda>1$. Suppose there are Galois CM fields $K$ of arbitrarily large degree, with totally real subfields $F$ of degree $d=[F:\mathbb Q]$, such that

1. $\operatorname{rd}_{K/F}=\lambda$;
2. every prime of $F$ over each $p\in S_{\mathbb Q}$ splits in $K/F$;
3. $e(p)$ is the ramification index of $p$ in $F/\mathbb Q$; and
4. the inertia degree of $p$ in $F/\mathbb Q$ is at most $f(p)$.

For a finite planar set $U$, write $D_{\mathrm{ord}}(U)$ for the number of ordered pairs in $U^2$ at Euclidean distance one.

For $R>1$, define

$$\delta= \frac{ \log(1-1/R) +\frac12\log(2\pi/e) +\displaystyle\sum_{p\in S_{\mathbb Q}} \frac{\log(k(p)+1)}{2e(p)f(p)} -\frac14\log\lambda -\frac12\log\log\lambda }{ \log\left( 2R\displaystyle\prod_{p\in S_{\mathbb Q}} p^{k(p)/(2e(p))}+1 \right) }. \tag{1}$$

Then there are finite $U\subset\mathbb R^2$ of arbitrarily large cardinality for which

$$D_{\mathrm{ord}}(U) \geq\frac{|U|^{1+\delta}}{8\lambda^2}. \tag{2}$$

Proof

The later application has $\delta>0$, so first assume this. For one field in the family, use Lemma 8 and then Lemma 5. Since the actual inertia degree $f_p$ is at most $f(p)$ and $e_p=e(p)$, they give a set $U$ satisfying

$$|U|\leq A^{2d}, \qquad A=2R\prod_{p\in S_{\mathbb Q}}p^{k(p)/(2e(p))}+1, \tag{3}$$

and

$$\frac{D_{\mathrm{ord}}(U)}{|U|} \geq \left(1-\frac1R\right)^{2d} \frac{ \displaystyle\prod_{p\in S_{\mathbb Q}} (k(p)+1)^{d/(e(p)f(p))} }{2^d h^-(K)}. \tag{4}$$

Apply Lemma 9 to (4). With

$$B= \frac{ (1-1/R)^2 \displaystyle\prod_{p\in S_{\mathbb Q}} (k(p)+1)^{1/(e(p)f(p))} }{ 2\sqrt\lambda\log\lambda\,e/(4\pi) }, \tag{5}$$

the result is

$$\frac{D_{\mathrm{ord}}(U)}{|U|} \geq\frac{B^d}{8\lambda^2}. \tag{6}$$

Taking logarithms in (5) and dividing by two gives

$$\begin{aligned} \frac12\log B ={}&\log(1-1/R)+\frac12\log(2\pi/e)\\ &+\sum_{p\in S_{\mathbb Q}} \frac{\log(k(p)+1)}{2e(p)f(p)} -\frac14\log\lambda-\frac12\log\log\lambda. \end{aligned} \tag{7}$$

The term $\frac12\log(2\pi/e)$ in (7) equals $-\frac12\log\bigl(2\cdot e/(4\pi)\bigr)$, the contribution of the factors $2$ and $e/(4\pi)$ in the denominator of (5). Comparing (1), (3), and (7) yields

$$B=A^{2\delta}. \tag{8}$$

Since $\delta>0$, (3) and (8) imply

$$B^d=A^{2d\delta}\geq|U|^\delta.$$

Multiplying (6) by $|U|$ proves (2).

It also proves that the constructed cardinalities are unbounded. Indeed, $\delta>0$ makes $B>1$, so the lower bound in (6) tends to infinity as the available degrees $d$ tend to infinity. But $D_{\mathrm{ord}}(U)/|U|\leq|U|$, forcing $|U|\to\infty$ along a subfamily.

For completeness, if $\delta\leq0$, take arbitrarily long strings of equally spaced collinear points. Their ordered unit-pair count is $2(n-1)$, while $n^{1+\delta}/(8\lambda^2)\leq n/(8\lambda^2)$, so (2) is immediate.

Source scope

This is Proposition 10 and equation (8) on physical pp. 8--9 of the arXiv v1 manuscript. All factors from the point-count, norm-fiber, and class-number estimates are displayed. The field family required by the hypotheses is constructed in the next linked component.

Used by. Theorem 1.

Bears on. Problem 90.