Lattice points close to a helix

TL;DR

This paper establishes a lower bound on the maximum distance among three points in an affine lattice that are near a helix with small curvature and torsion, advancing understanding of lattice-helix interactions.

Contribution

It provides a new lower bound for the maximum distance between three lattice points close to a small-curvature helix, a novel geometric insight.

Findings

Lower bound for maximum distance among three lattice points near a helix

Insights into lattice configurations close to curved structures

Advancement in geometric lattice analysis

Abstract

We obtain lower bound for the maximum distance between any three distinct points in an affine lattice which are close to a helix with small curvature and torsion.

Equations

$$\mathcal{L}=\mathcal{L}(v_0,v_1,v_2,v_3)=\{v_0+m{v}_1+n{v}_2+p{v}_3:m,n,p\in \mathbb{Z}\}.$$

$$0\le \delta \le \min \left(\frac{{\mathcal{D}}_L}{4},\frac{{{\mathcal{D}}_L}^2}{11{\pi }^3}({\kappa }^2+{\tau }^2),\frac{2{\mathcal{A}}_L}{11\pi }({\kappa }^2+{\tau }^2{)}^{\frac{1}{2}}\right).$$

$$\min \left(1.2{{\mathcal{A}}_L}^{\frac{1}{3}}{\kappa }^{-\frac{1}{3}}-2\delta ,\frac{\pi \tau }{{\kappa }^2+{\tau }^2}-2\delta \right).$$

$$|S-S_1|\le \frac{\delta |AB|}{2}.$$

$$h_1\le h+|C{C}_1|\le h+\delta \text{mbox}andh\le h_1+|C{C}_1|\le h_1+\delta .$$

$$|A{A}_1|\le \delta ,|B{B}_1|\le \delta ,\text{mbox}and|C{C}_1|\le \delta .$$

$$|S-S_3|\le \frac{(|AB|+|AC|+|BC|)\delta }{2}+\frac{3{\delta }^2}{2}.$$

$$|S-S_3|$$

$$\le |S-S_1|+|S_1-S_2|+|S_2-S_3|.$$

$$|S-S_3|\le \frac{\delta (|AB|+|A{C}_1|+|B_1{C}_1|)}{2}.$$

$$\vec{r}(t)=⟨a\cos t,a\sin t,bt⟩.$$

$$A_0=(a\cos t_0,a\sin t_0,b{t}_0),A_1=(a\cos t_1,a\sin t_1,b{t}_1),\text{mbox}and{A}_2=(a\cos t_2,a\sin t_2,b{t}_2),$$

$$\overrightarrow{A_1{A}_0}$$

$$\overrightarrow{A_1{A}_2}$$

$${\left|\overrightarrow{A_1{A}_0}\right|}^2$$

$$=a^2(2-2\cos t_0\cos t_1-2\sin t_0\sin t_1)+b^2(t_0-t_1{)}^2$$

$$=a^2(2-2\cos (t_0-t_1))+b^2(t_0-t_1{)}^2$$

$$=a^2(2-2\cos h_1)+b^2{h}_1^2$$

$$=4a^2{\sin }^2\left(\frac{h_1}{2}\right)+b^2{h}_1^2.$$

$$\cos \theta$$

$$1-{\sin }^2\theta$$

$${\left|\overrightarrow{A_1{A}_0}\right|}^2{\left|\overrightarrow{A_1{A}_2}\right|}^2-{\left(\overrightarrow{A_1{A}_0}\cdot \overrightarrow{A_1{A}_2}\right)}^2$$

$$=4\text{Area}(△A_0{A}_1{A}_2{)}^2,$$

$$\overrightarrow{A_1{A}_0}\cdot \overrightarrow{A_1{A}_2}$$

$$+a^2(\sin t_0-\sin t_1)(\sin t_2-\sin t_1)+b^2(t_0-t_1)(t_2-t_1)$$

$$=a^2(\cos t_0\cos t_2-\cos t_1\cos t_2-\cos t_0\cos t_1+{\cos }^2{t}_1)$$

$$+a^2(\sin t_0\sin t_2-\sin t_1\sin t_2-\sin t_0\sin t_1+{\sin }^2{t}_1)+b^2(t_0-t_1)(t_2-t_1)$$

$$=a^2(\cos (h_2+h_1)-\cos h_2-\cos h_1+1)-b^2{h}_1{h}_2$$

$$=a^2((1-\cos h_2)(1-\cos h_1)-\sin h_2\sin h_1)-b^2{h}_1{h}_2$$

$$=4a^2\sin \frac{h_1}{2}\sin \frac{h_2}{2}\left(\sin \frac{h_1}{2}\sin \frac{h_2}{2}-\cos \frac{h_1}{2}\cos \frac{h_2}{2}\right)-b^2{h}_1{h}_2$$

$$=-4a^2\sin \frac{h_1}{2}\sin \frac{h_2}{2}\cos \left(\frac{h_1+h_2}{2}\right)-b^2{h}_1{h}_2.$$

$$4\text{Area}(△A_0{A}_1{A}_2{)}^2$$

$$-{\left(4a^2\sin \frac{h_1}{2}\sin \frac{h_2}{2}\cos \left(\frac{h_1+h_2}{2}\right)+b^2{h}_1{h}_2\right)}^2$$

$$=16a^4{\sin }^2\frac{h_1}{2}{\sin }^2\frac{h_2}{2}{\sin }^2\frac{h_1+h_2}{2}+4a^2{b}^2\left(h_1^2{\sin }^2\frac{h_2}{2}+h_2^2{\sin }^2\frac{h_1}{2}\right)$$

$$-8a^2{b}^2{h}_1{h}_2\sin \frac{h_1}{2}\sin \frac{h_2}{2}\cos \frac{h_1+h_2}{2}$$

$$=16a^4{\sin }^2\frac{h_1}{2}{\sin }^2\frac{h_2}{2}{\sin }^2\frac{h_1+h_2}{2}+4a^2{b}^2{\left(h_1\sin \frac{h_2}{2}-h_2\sin \frac{h_1}{2}\right)}^2$$

$$+8a^2{b}^2{h}_1{h}_2\sin \frac{h_1}{2}\sin \frac{h_2}{2}\left(1-\cos \frac{h_1+h_2}{2}\right)$$

$$=16a^4{\sin }^2\frac{h_1}{2}{\sin }^2\frac{h_2}{2}{\sin }^2\frac{h_1+h_2}{2}+a^2{b}^2{h}_1^2{h}_2^2{\left(\frac{\sin \frac{h_2}{2}}{\frac{h_2}{2}}-\frac{\sin \frac{h_1}{2}}{\frac{h_1}{2}}\right)}^2$$

$$+16a^2{b}^2{h}_1{h}_2\sin \frac{h_1}{2}\sin \frac{h_2}{2}{\sin }^2\frac{h_1+h_2}{4}.$$

$${\left(\frac{\sin \frac{h_2}{2}}{\frac{h_2}{2}}-\frac{\sin \frac{h_1}{2}}{\frac{h_1}{2}}\right)}^2\le \frac{\max (h_1,h_2{)}^2}{144}(h_1-h_2{)}^2.$$

$${\left(\frac{\sin \frac{h_2}{2}}{\frac{h_2}{2}}-\frac{\sin \frac{h_1}{2}}{\frac{h_1}{2}}\right)}^2={\left(\frac{h_1}{2}-\frac{h_2}{2}\right)}^2{\left(\frac{\zeta \cos \zeta -\sin \zeta }{{\zeta }^2}\right)}^2.$$

$${\left(\frac{\sin \frac{h_2}{2}}{\frac{h_2}{2}}-\frac{\sin \frac{h_1}{2}}{\frac{h_1}{2}}\right)}^2<{\left(\frac{h_1}{2}-\frac{h_2}{2}\right)}^2{\left(\frac{\zeta }{3}\right)}^2\le \frac{\max (h_1,h_2{)}^2}{144}(h_1-h_2{)}^2.$$

Full text

1 Introduction

The problem of estimating the number of lattice points on or close to a curve has a rich history. In 1926, Jarník [7] proved that the number of integer points on a strictly convex closed curve of length $L>3$ does not exceed $3(2\pi)^{-1/3}L^{2/3}+O\left(L^{1/3}\right)$ and the exponent and the constant of the leading term are best possible. Assuming higher order smoothness conditions on the curve, a number of authors achieved sharper estimates, in particular Swinnerton-Dyer [9] and Bombieri and Pila [1].

Estimating the number of lattice points close to a curve is more recent topic. Of a survey of results on this topic and their applications one may see [3] and [4].

In 1972 Zygmund published a paper on spherical summability of Fourier series in two dimensions where an essential component was the following theorem.

Theorem 1**.**

(Schinzel) An arc of length $\sqrt[3]{2}R^{1/3}$ on a circle of radius $R$ contains no more than two lattice points.

The original proof is due to Schinzel but the following short proof was provided by Pelczynski.

Proof.

Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, and $C(x_{3},y_{3})$ be distinct lattice points on a circle of radius $R$.

Denote by $a$, $b$, and $c$ the lengths of the sides of $\triangle ABC$, and its area by $S$. Since the circle is strictly convex curve, we have S=\frac{1}{2}|\left|\begin{array}[]{ccc}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{array}\right||\ \geq{1\over 2}.

On the other hand, by a formula attributed to Heron of Alexandria, we have ${\displaystyle S={abc\over 4R}}$.

Thus, $abc\geq 2R$. ∎

In the paper [5] Howard and the second author generalized Schinzel’s result to the case of plane curve with bounded curvature and lattice points close to a plane affine lattice (where plane affine lattice is the set $\{{\bf v}_{0}+m{\bf v}_{1}+n{\bf v}_{2}:m\in{\mathbb{Z}},n\in{\mathbb{Z}},\}$ with ${\bf v}_{0},{\bf v}_{1},{\bf v}_{2}$, and plane vectors in such that ${\bf v}_{1}$ and ${\bf v}_{2}$ are linearly independent).

There are only a few papers on the topic of lattice points close to a three-dimensional curve.

Huang [6] obtains estimates for the number of lattice points close dilations of a curve with parametrization $(x,f_{1}(x),f_{2}(x))$, $x\in[a,b]$. Huang’s estimates depend on the dilation parameter $q$, the upper bound $\delta$ for the distance between the lattice points and the curve, and a constant depending on the functions $f_{1}$ and $f_{2}$ which is not explicitly computed.

In Chapter 4 of his PhD dissertation Letendre also obtains estimates for the number of lattice points close dilations of a curve with parametrization $(x,f_{1}(x),f_{2}(x))$, $x\in[a,b]$. His results only assume bounds on certain quantities involving derivatives of $f_{1}$ and $f_{2}$.

We will consider the special case when the curve is a helix and obtain lower bound for the maximal distance between any three distinct points in a general lattice close to a helix. We define a lattice in three-space as follows.

Definition 2**.**

Let $v_{0},v_{1},v_{2},v_{3}\in{\mathbb{R}}^{2}$ be vectors in ${\mathbb{R}}^{3}$ with $v_{1}$ , $v_{2}$, and $v_{3}$ linearly independent. Then, the affine lattice generated by $v_{1}$, $v_{2}$, and $v_{3}$ with origin $v_{0}$ is

[TABLE]

We will show in Section 2 that for each lattice ${\mathcal{L}}$ there exist positive constants ${\mathcal{D}_{L}}$ and ${\mathcal{A}_{L}}$ (depending on ${\mathcal{L}}$) such that if $A_{0}$ and $A_{1}$ are two distinct lattice points in ${\mathcal{L}}$ then $|A_{0}A_{1}|\geq{\mathcal{D}_{L}}$; furthermore, if $A_{0}$, $A_{1}$, and $A_{2}$ are three non-collinear lattice points in ${\mathcal{L}}$, then Area$(\triangle A_{0}A_{1}A_{2})\geq{\mathcal{A}_{L}}$.

Our main result is

Theorem 3**.**

Let ${\mathcal{H}}$ be a helix of curvature $\kappa>0$ and torsion $\tau>0$, and let

[TABLE]

Let ${\mathcal{L}}$ be affine lattice, and let $A_{0}$, $A_{1}$, and $A_{2}$ be three distinct points in ${\mathcal{L}}$ which are within $\delta$ of the helix ${\mathcal{H}}$. Then, the maximal distance between $A_{0}$, $A_{1}$, and $A_{2}$ is at least

[TABLE]

The above theorem is a generalization of Schinzel’s result to the case of a helix.

Corollary 4**.**

Let ${\mathcal{H}}$ be a helix of curvature $\kappa>0$ and torsion $\tau>0$ with $\tau\kappa^{\frac{1}{3}}\geq 0.4(\kappa^{2}+\tau^{2})$. Let $A_{0}$, $A_{1}$, and $A_{2}$ be three distinct lattice points on ${\mathcal{H}}$. Then, the maximal distance between $A_{0}$, $A_{1}$, and $A_{2}$ is at least ${\displaystyle 1.1\kappa^{-\frac{1}{3}}}$.

The paper is organized as follows. In Section 2 we prove some auxiliary lemmas. In Section 3 we prove the main result.

We expect that an analogue of our main result will hold for curves such that $0<c_{1}K<\kappa(s)<c_{2}K$ and $0<c_{3}T<\tau(s)<c_{4}T$ where $c_{1},c_{2},c_{3}$, and $c_{4}$ are positive constants.

2 Some auxiliary results

In this section we prove several lemmas which will be needed in the proof of Theorem 3.

Lemma 5**.**

The area of a triangle in $3$-space with non-collinear lattice point vertices is at least $\frac{1}{2}$.

Proof.

Let $A,B$, and $C$ be non-collinear lattice points in $\mathbb{R}^{3}$. Then, the area of $\triangle ABC$ equals $\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|$. Since $A$, $B$, and $C$ are lattice points, the components of the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ are integers. Therefore, the components of the cross product $\overrightarrow{AB}\times\overrightarrow{AC}$ are integers, as well. Thus, Area$(\triangle ABC)=\frac{1}{2}\sqrt{n}$ for some integer $n$. Since the points $A$, $B$, and $C$ are non-collinear, Area$(\triangle ABC)\neq 0$, so $n\neq 0$. ∎

The next two lemmas where proved in [5] in the ${\mathbb{R}}^{2}$ case. The lemmas are easily extended to the $n$-dimensional case.

Lemma 6**.**

Let $A$, $B$, $C$, and $C_{1}$ be points in ${\mathbb{R}}^{n}$ where $n\geq 2$. Let $\delta\geq 0$ and $|CC_{1}|\leq\delta$. Denote by $S$ the area of $\triangle ABC$, and by $S_{1}$ the area of $\triangle ABC_{1}$. Then,

[TABLE]

Proof.

Let $h$ be the distance from the point $C$ to the line $AB$, and let $h_{1}$ be the distance from the point $C_{1}$ to the line $AB$. Then, ${\displaystyle S=\frac{|AB|h}{2}}$ and ${\displaystyle S_{1}=\frac{|AB|h_{1}}{2}}$.

By the triangle inequality,

[TABLE]

Therefore, $|h-h_{1}|\leq\delta$, so ${\displaystyle|S-S_{1}|=\frac{|AB||h-h_{1}|}{2}\leq\frac{\delta|AB|}{2}}$. ∎

Lemma 7**.**

Let $\triangle ABC$ and $\triangle A_{1}B_{1}C_{1}$ be triangles in ${\mathbb{R}}^{n}$ where $n\geq 2$, with areas $S$ and $S_{3}$, respectively. Let $\delta\geq 0$ and assume

[TABLE]

Then,

[TABLE]

Proof.

Let $S_{1}$ be the area of $\triangle ABC_{1}$, and let $S_{2}$ be the area of $\triangle AB_{1}C_{1}$. Then

[TABLE]

By Lemma 6, ${\displaystyle|S-S_{1}|\leq\frac{\delta|AB|}{2}}$. Applying Lemma 6 to the points $A$, $C_{1}$, $B$, and $B_{1}$, we get ${\displaystyle|S_{1}-S_{2}|\leq\frac{\delta|AC_{1}|}{2}}$. Finally, applying Lemma 6 to the points $B_{1}$, $C_{1}$, $A$, and $A_{1}$, we get ${\displaystyle|S_{2}-S_{3}|\leq\frac{\delta|B_{1}C_{1}|}{2}}$.

Therefore,

[TABLE]

Note that the the triangle inequality gives us $|AC_{1}|\leq|AC|+|CC_{1}|\leq|AC|+\delta$ and $|B_{1}C_{1}|\leq|B_{1}B|+|BC|+|CC_{1}|\leq|BC|+2\delta$. Using the upper bounds for $|AC_{1}|$ and $|B_{1}C_{1}|$ in (1) gives us the desired bound for $|S-S_{3}|$. ∎

We also need the following simple lemma.

Lemma 8**.**

For all $x$ in the interval $(0,\pi)$, we have $0<\sin x-x\cos x<\frac{x^{3}}{3}$. Also, for all $x$ in the interval $(0,\pi/2]$, we have ${\displaystyle\sin x\geq\frac{2}{\pi}x}$.

Proof.

Let $f(x)=\sin x-x\cos x$. Note $f(0)=0$. Consider $f^{\prime}(x)=x\sin x$. Clearly, $f^{\prime}(x)>0$ on $(0,\pi)$ and we have the first half of the inequality.

Now let $g(x)=\sin x-x\cos x-\frac{x^{3}}{3}$. Note $g^{\prime}(x)=x(\sin x-x)$. If $x>0$, $\sin x-x<0$ and thus $g^{\prime}(x)<0$ on this interval also. Thus, because $g(0)=0$ and $g$ is decreasing on $(0,\pi)$, we have the second desired inequality. To prove the last inequality of the lemma, consider the function ${\displaystyle h(x)=\frac{\sin x}{x}}$. We have ${\displaystyle h^{\prime}(x)=\frac{x\cos x-\sin x}{x^{2}}}$. We proved above that $\sin x-x\cos x>0$ for $x\in(0,\pi)$. Thus $h(x)$ is decreasing in $(0,\pi/2]$ and its minimum is $h(\pi/2)=2/\pi$. ∎

Here, and throughout the end of the paper, when $A$ is point and $\vec{v}$ is a vector, the equation $A=\vec{v}$ will mean $\overrightarrow{OA}=\vec{v}$, where $O$ is the origin of the coordinate system.

A helix is a space curve with constant curvature and torsion. Let $A_{0}<A_{1},A_{2}$ be three points on a helix ${\mathcal{H}}$. In the next lemma we obtain formulas for the distances $|A_{0}A_{1}$, $|A_{1}A_{2}|$, $|A_{0}A_{2}|$, and for the area of triangle $\triangle A_{0}A_{1}A_{2}$.

Lemma 9**.**

Let ${\mathcal{H}}$ be a helix with curvature $\kappa>0$ and torsion $\tau>0$. Let $A_{0}$, $A_{1}$, and $A_{2}$ be three distinct points on the helix ${\mathcal{H}}$ whose arclengths from the origin of the helix are $s_{0}$, $s_{1}$, $s_{2}$ respectively, with $s_{0}<s_{1}<s_{2}$. Define $a=\frac{\kappa}{\kappa^{2}+\tau^{2}}$, $b=\frac{\tau}{\kappa^{2}+\tau^{2}}$, $t_{i}=\frac{s_{i}}{\sqrt{a^{2}+b^{2}}}$ for $i=0,1,2$, $h_{1}=t_{1}-t_{0}$, and $h_{2}=t_{2}-t_{1}$. Then,

(a) $|A_{0}A_{1}|^{2}=4a^{2}\sin^{2}\left(\frac{h_{1}}{2}\right)+b^{2}h_{1}^{2}$, $|A_{1}A_{2}|^{2}=4a^{2}\sin^{2}\left(\frac{h_{2}}{2}\right)+b^{2}h_{2}^{2}$, and $|A_{0}A_{2}|^{2}=4a^{2}\sin^{2}\left(\frac{h_{1}+h_{2}}{2}\right)+b^{2}(h_{1}+h_{2})^{2}$;

(b) $(\text{Area}(\triangle A_{0}A_{1}A_{2}))^{2}=(T_{1}+T_{2}+T_{3})/4$, where

$T_{1}:=16a^{4}\sin^{2}\frac{h_{1}}{2}\sin^{2}\frac{h_{2}}{2}\sin^{2}\frac{h_{1}+h_{2}}{2}$,

$T_{2}:=a^{2}b^{2}h_{1}^{2}h_{2}^{2}\left(\frac{\sin\frac{h_{2}}{2}}{\frac{h_{2}}{2}}-\frac{\sin\frac{h_{1}}{2}}{\frac{h_{1}}{2}}\right)^{2}$, and

$T_{3}:=16a^{2}b^{2}h_{1}h_{2}\sin\frac{h_{1}}{2}\sin\frac{h_{2}}{2}\sin^{2}\frac{h_{1}+h_{2}}{4}$.

Proof.

First, we consider a helix ${\mathcal{H}}(a,b)$ with parametrization

[TABLE]

It is easy to check that we have ${\displaystyle\kappa=\frac{a}{a^{2}+b^{2}}}$, ${\displaystyle\tau=\frac{b}{a^{2}+b^{2}}}$, and ${\displaystyle\kappa^{2}+\tau^{2}=\frac{1}{a^{2}+b^{2}}}$. From here, we get ${\displaystyle a=\frac{\kappa}{\kappa^{2}+\tau^{2}}}$, ${\displaystyle b=\frac{\tau}{\kappa^{2}+\tau^{2}}}$. Also, for the natural parameter $s$ we have $s=t\sqrt{a^{2}+b^{2}}$.

We have

[TABLE]

with $t_{0}<t_{1}<t_{2}$.

Therefore,

[TABLE]

Therefore,

[TABLE]

One can obtain the formulas for $|A_{1}A_{2}|$ and $|A_{0}A_{2}|$ similarly.

For the area of $\triangle A_{0}A_{1}A_{2}$ we use

[TABLE]

where $\theta$ is the angle between $\overrightarrow{A_{1}A_{0}}$ and $\overrightarrow{A_{1}A_{2}}$. Next we calculate the dot product:

[TABLE]

Thus, we have

[TABLE]

We proved the lemma in the case of a helix ${\mathcal{H}}(a,b)$ with parametrization 2.

Now, consider a helix ${\mathcal{H}^{\prime}}$ in general position, with curvature $\kappa>0$ and torsion $\tau>0$. Let $A_{0}$, $A_{1}$, and $A_{2}$ be three distinct points on the helix ${\mathcal{H}^{\prime}}$ whose arclengths from the origin of the helix are $s_{0}$, $s_{1}$, $s_{2}$ respectively, with $0<s_{0}<s_{1}<s_{2}$.

Here we use the uniqueness part of the Fundamental Theorem of the local theory of curves (for example, see do Carmo [2] p.19). It states that if $\alpha,\alpha^{\prime}:I\to{\mathbb{R}}$ are regular parametrized curves with natural parameter $s$, which have the same curvature $\kappa(s)$ and the same torsion $\tau(s)$ for all $s$, then $\alpha^{\prime}$ differs from $\alpha$ by a rigid motion, that is there exist an orthogonal linear map $\rho$ of ${\mathbb{R}}^{3}$, with positive determinant, and a vector $c$ such that $\alpha^{\prime}=\rho\circ\alpha+c$.

Since the helix ${\mathcal{H}^{\prime}}$ and the helix ${\mathcal{H}(a,b)}$ have the same curvature $\kappa$ and the same torsion $\tau$, there exists a rigid motion which maps the helix ${\mathcal{H}^{\prime}}$ onto the helix ${\mathcal{H}(a,b)}$; and it maps the points $A_{0},A_{1},A_{2}$ to some points $A_{0}^{\prime},A_{1}^{\prime},A_{2}^{\prime}$ respectively, which are on ${\mathcal{H}(a,b)}$ and have corresponding values of the natural parameter $s_{0},s_{1},s_{2}$.

Note that the rigid motions do not change the distance between two points, and preserve the area of a triangle. Therefore, $|A_{0}A_{1}|=|A_{0}^{\prime}A_{1}^{\prime}|$, $|A_{1}A_{2}|=|A_{1}^{\prime}A_{2}^{\prime}|$, $|A_{0}A_{2}|=|A_{0}^{\prime}A_{2}^{\prime}|$, Area$(\triangle A_{0}A_{1}A_{2})=$ Area $(\triangle A_{0}^{\prime}A_{1}^{\prime}A_{2}^{\prime})$. Moreover, $a$,$b$, $h_{1}$, and $h_{2}$ are uniquely determined by $s_{0}$, $s_{1}$, $s_{2}$, $\kappa$ and $\tau$. Therefore, the lemma holds in the case of a helix ${\mathcal{H}^{\prime}}$ in general position.

∎

It is easy to estimate the terms $T_{1}$ and $T_{3}$. To estimate $T_{2}$ we use the following lemma.

Lemma 10**.**

Let $h_{1}$ and $h_{2}$ be real numbers, both in the interval $(0,2\pi)$. Then,

[TABLE]

Proof.

Applying the mean value theorem to the function ${\displaystyle\frac{\sin x}{x}}$, we have that there exists $\zeta$ between $\frac{h_{1}}{2}$ and $\frac{h_{2}}{2}$ such that

[TABLE]

By Lemma 8, we get that ${\displaystyle 0<\frac{\sin x-x\cos x}{x^{2}}<\frac{x}{3}}$ for every $x\in(0,\pi)$. Therefore,

[TABLE]

∎

In the next two lemmas we prove the existence of the positive constants ${\mathcal{D}_{L}}$ and ${\mathcal{A}_{L}}$ mentioned in the introduction.

Lemma 11**.**

Let $v_{1}$, $v_{2}$, and $v_{3}$ be linearly independent vectors in ${\mathbb{R}}^{3}$. Then, there exists a constant $c>0$ (depending on $v_{1}$, $v_{2}$, and $v_{3}$) such that $||m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}||\geq c$ for any integers $m_{1}$, $m_{2}$, and $m_{3}$ with $(m_{1},m_{2},m_{3})\neq(0,0,0)$.

Proof.

We have $||m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}||^{2}=(m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3})\cdot(m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3})={\displaystyle\sum_{i=1}^{3}\sum_{j=1}^{3}(v_{i}\cdot v_{j})m_{i}m_{j}:=Q(m_{1},m_{2},m_{3})}$.

The quadratic form $Q(m_{1},m_{2},m_{3})$ is positive definite, since $v_{1}$, $v_{2}$, and $v_{3}$ are linearly independent. Denote by $M(Q)$ the matrix of the quadratic form $Q(m_{1},m_{2},m_{3})$. Since $M(Q)$ is symmetric and $Q(m_{1},m_{2},m_{3})$ is positive definite, the eigenvalues of $Q(M)$ are real positive numbers, say $0<\lambda_{1}\leq\lambda_{2}\leq\lambda_{3}$. Moreover, by the Spectral Theorem for real symmetric matrices, there exists an orthonormal basis of vectors $w_{1},w_{2},w_{3}$ for ${\mathbb{R}}^{3}$, where $w_{1}$, $w_{2}$, and $w_{3}$ are eigenvectors corresponding to the eigenvalues $\lambda_{1}$, $\lambda_{2}$, and $\lambda_{3}$. This implies that the minimum of $Q(m_{1},m_{2},m_{3})$ on the unit sphere is $\lambda_{1}$. Therefore, the lemma holds with $c=\sqrt{\lambda_{1}}$.

∎

Lemma 12**.**

Consider a lattice $\mathcal{L}=\mathcal{L}(v_{0},v_{1},v_{2})=\{v_{0}+mv_{1}+nv_{2}+pv_{3}:m,n,p\in{\mathbb{Z}}\}$. There exist positive constants ${\mathcal{D}_{L}}$ and ${\mathcal{A}_{L}}$ (depending on ${\mathcal{L}}$) such that

(i) if $A_{0}$ and $A_{1}$ are any two distinct lattice points in ${\mathcal{L}}$, then $|A_{0}A_{1}|\geq{\mathcal{D}_{L}}$;

and

(ii) if $A_{0}$, $A_{1}$, and $A_{2}$ are any three non-collinear lattice points in ${\mathcal{L}}$, then Area$(\triangle A_{0}A_{1}A_{2})\geq{\mathcal{A}_{L}}$.

Proof.

If $A_{0}$ and $A_{1}$ are two distinct lattice points in ${\mathcal{L}}$, then $\overrightarrow{A_{0}A_{1}}=m^{\prime}v_{1}+n^{\prime}v_{2}+p^{\prime}v_{3}$ with $m^{\prime}$, $n^{\prime}$, $p^{\prime}$ integers such that $(m^{\prime},n^{\prime},p^{\prime})\neq(0,0,0)$. Now, (i) follows from Lemma 11.

Next, let $A_{0}$, $A_{1}$, and $A_{2}$ be three non-collinear lattice points in ${\mathcal{L}}$. We have that Area$(\triangle A_{0}A_{1}A_{2})=||\overrightarrow{A_{0}A_{1}}\times\overrightarrow{A_{0}A_{2}}||/2$.

Since $A_{0}$, $A_{1}$, and $A_{2}$ are non-collinear, then Area$(\triangle A_{0}A_{1}A_{2})\neq 0$. Therefore, $\overrightarrow{A_{0}A_{1}}\times\overrightarrow{A_{0}A_{2}}\neq\overrightarrow{0}$.

Furthermore, since $\overrightarrow{A_{0}A_{1}}=m^{\prime}v_{1}+n^{\prime}v_{2}+p^{\prime}v_{3}$ and $\overrightarrow{A_{0}A_{2}}=m^{\prime\prime}v_{1}+n^{\prime\prime}v_{2}+p^{\prime\prime}v_{3}$ for some integers $m^{\prime},m^{\prime\prime},n^{\prime},n^{\prime\prime},p^{\prime}$, and $p^{\prime\prime}$, then

[TABLE]

with $q=m^{\prime}n^{\prime\prime}-m^{\prime}n^{\prime\prime}$, $r=n^{\prime}p^{\prime\prime}-n^{\prime\prime}p^{\prime}$, and $s=m^{\prime}p^{\prime\prime}-m^{\prime\prime}p^{\prime}$.

Since, $\overrightarrow{A_{0}A_{1}}\times\overrightarrow{A_{0}A_{2}}\neq\overrightarrow{0}$ we have $(q,r,s)\neq(0,0,0)$.

Moreover, the vectors $v_{1}\times v_{2}$, $v_{2}\times v_{3}$, and $v_{1}\times v_{3}$ are linearly independent.

Indeed, if $c_{1}(v_{1}\times v_{2})+c_{2}(v_{2}\times v_{3})+c_{3}(v_{1}\times v_{3})=0$, then after taking the dot product of the last equation with $v_{3}$, we get $c_{1}((v_{1}\times v_{2})\cdot v_{3})=0$. However, the triple product $(v_{1}\times v_{2})\cdot v_{3}\neq 0$, since the vectors $v_{1}$, $v_{2}$, and $v_{3}$ are linearly independent. We show similarly that $c_{2}=0$ and $c_{3}=0$.

Thus, the vectors $v_{1}\times v_{2}$, $v_{2}\times v_{3}$, and $v_{1}\times v_{3}$ are linearly independent.

Now (ii) follows from Lemma 11. ∎

3 Proof of the main result

First, we consider the case when the lattice points are on the helix.

Theorem 13**.**

Let ${\mathcal{H}}$ be a helix with curvature $\kappa>0$ and torsion $\tau>0$. Let $\mathcal{L}=\mathcal{L}(v_{0},v_{1},v_{2},v_{3})$ be an affine lattice in ${\mathbb{R}}^{3}$, and let $A_{0}$, $A_{1}$, and $A_{2}$ be three distinct points in ${\mathcal{L}}$ which are also on the helix ${\mathcal{H}}$ and whose arclengths from the origin of the helix are $s_{0}$, $s_{1}$, $s_{2}$ respectively, with $s_{0}<s_{1}<s_{2}$. Then ${\displaystyle|A_{0}A_{2}|\geq\min\left(\frac{\pi\tau}{\kappa^{2}+\tau^{2}},1.5{\mathcal{A}_{L}}^{\frac{1}{3}}\kappa^{-\frac{1}{3}}\right)}$ and the length of the arc \stackrel{{\scriptstyle\mbox{\large\frown}}}{{A_{0}A_{2}}} is at least ${\displaystyle\min\left(\frac{\pi}{\sqrt{\kappa^{2}+\tau^{2}}},2.4{\mathcal{A}_{L}}^{\frac{1}{3}}\kappa^{-\frac{1}{3}}\right)}$.

Proof.

Define ${\displaystyle a=\frac{\kappa}{\kappa^{2}+\tau^{2}}}$ and ${\displaystyle b=\frac{\tau}{\kappa^{2}+\tau^{2}}}$. Recall that $s=\sqrt{a^{2}+b^{2}}t$. Let ${\displaystyle t_{i}=\frac{s_{i}}{\sqrt{a^{2}+b^{2}}}}$, for $i=0,1,2$. Then $0<t_{0}<t_{1}<t_{2}$ and $A_{i}=\vec{r}(t_{i})$ for $i=0,1,2$. Let $h_{1}=t_{1}-t_{0}$ and $h_{2}=t_{2}-t_{1}$.

We consider two cases.

Case I: $h_{1}+h_{2}\geq\pi$.

Then the arclength of the arc \stackrel{{\scriptstyle\mbox{\large\frown}}}{{A_{0}A_{2}}} is

[TABLE]

Also, by Lemma 9 (i)

[TABLE]

Note that one cannot expect to get a nontrivial lower bound on $|A_{0}A_{2}|$ depending only on $\kappa$. For example, consider the case when $a$ is a large integer, $b=\frac{1}{2\pi}$, $t_{0}=0$, $t_{1}=2\pi$, and $t_{2}=4\pi$.

Case II: $h_{1}+h_{2}<\pi$.

Denote by $S$ the area of $\triangle A_{0}A_{1}A_{2}$. From Lemma 9 (ii) we have $4S^{2}=T_{1}+T_{2}+T_{3}$, where

$T_{1}=16a^{4}\sin^{2}\frac{h_{1}}{2}\sin^{2}\frac{h_{2}}{2}\sin^{2}\frac{h_{1}+h_{2}}{2}$,

$T_{2}=a^{2}b^{2}h_{1}^{2}h_{2}^{2}\left(\frac{\sin\frac{h_{2}}{2}}{\frac{h_{2}}{2}}-\frac{\sin\frac{h_{1}}{2}}{\frac{h_{1}}{2}}\right)^{2}$, and

$T_{3}=16a^{2}b^{2}h_{1}h_{2}\sin\frac{h_{1}}{2}\sin\frac{h_{2}}{2}\sin^{2}\frac{h_{1}+h_{2}}{4}$.

Since $0<h_{1}+h_{2}<\pi$, we have $T_{1}>0$, $T_{2}\geq 0$, and $T_{3}>0$. Therefore, $S>0$, so the points $A_{0}$, $A_{1}$, and $A_{2}$ are non-collinear. Now, by Lemma 12 we have $S\geq{\mathcal{A}_{L}}>0$. We obtain

[TABLE]

Next we get bounds for $T_{1}$, $T_{2}$, and $T_{3}$.

Since $0<h_{1}+h_{2}$ and $\sin x<x$ when $x>0$ we obtain

[TABLE]

(Here and below, we use the inequality $ab\leq(a+b)^{2}/4$.)

Next, by Lemma 10 we have

[TABLE]

Since, $h_{1}+h_{2}<\pi$, we obtain,

[TABLE]

Again, using $six<x$ for $x>0$ again, we obtain

[TABLE]

Combining equation (3) with the bounds (4), (5), and (6), we obtain

[TABLE]

Now $\frac{1}{64}+\frac{\pi^{2}}{2304}<0.02$. Therefore,

[TABLE]

Also, $200^{1/6}>2.4$, so

[TABLE]

in Case 2.

Recall that $s_{2}-s_{0}=(h_{1}+h_{2})\sqrt{a^{2}+b^{2}}$, ${\displaystyle a=\frac{\kappa}{\kappa^{2}+\tau^{2}}}$, and ${\displaystyle a^{2}+b^{2}=\frac{1}{\kappa^{2}+\tau^{2}}}$.

We get,

[TABLE]

completing the stated lower bound for the arclength of \stackrel{{\scriptstyle\mbox{\large\frown}}}{{A_{0}A_{2}}}.

Next, we obtain a lower bound for $|A_{0}A_{2}|$.

In this case, using Lemma 8 we obtain $\sin\left(\frac{h_{1}+h_{2}}{2}\right)\geq\frac{1}{\pi}(h_{1}+h_{2})$. By Lemma 9 (i) $|A_{0}A_{2}|^{2}=4a^{2}\sin^{2}\left(\frac{h_{1}+h_{2}}{2}\right)+b^{2}(h_{1}+h_{2})^{2}$. Therefore,

[TABLE]

in Case 2. Using (9) we get ${\displaystyle|A_{0}A_{2}|\geq\frac{4.8}{\pi}{\mathcal{A}_{L}}^{\frac{1}{3}}\kappa^{-\frac{1}{3}}}$.

Since $\frac{4.8}{\pi}>1.5$, we obtain the stated bound for $|A_{0}A_{2}|$.

We have established the theorem in the case of a helix ${\mathcal{H}}(a,b)$ with parametrization $\vec{r}(t)=\langle a\cos t,a\sin t,bt\rangle$.

∎

Next we prove Corollary 4.

Proof.

First, by Lemma 5 in the case of the standard lattice in ${\mathbb{R}}^{3}$, ${\mathcal{A}_{L}}=\frac{1}{2}$. Therefore, the condition $\tau\kappa^{\frac{1}{3}}\geq 0.4(\kappa^{2}+\tau^{2})$ implies ${\displaystyle\frac{\pi\tau}{\kappa^{2}+\tau^{2}}>1.5{\mathcal{A}_{L}}^{\frac{1}{3}}\kappa^{-\frac{1}{3}}}$ in the case of standard lattice ${\mathcal{L}}$. The corollary follows by noting that $\frac{1.5}{\sqrt[3]{2}}>1.1$. ∎

.

Finally, we prove Theorem 3.

Proof.

Let

[TABLE]

Let $A_{0}^{\prime}$, $A_{1}^{\prime}$, and $A_{2}^{\prime}$ be three distinct lattice points within $\delta$ of the helix ${\mathcal{H}}$. Therefore, there exist points $A_{0}$, $A_{1}$, and $A_{2}$ on the helix, such that

[TABLE]

Since $A_{0}^{\prime}$, $A_{1}^{\prime}$, and $A_{2}^{\prime}$ are in ${\mathcal{L}}$, by Lemma 12 we have $|A_{0}^{\prime}A_{1}^{\prime}|\geq{\mathcal{D}_{L}}$, $|A_{0}^{\prime}A_{2}^{\prime}|\geq{\mathcal{D}_{L}}$, and $|A_{1}^{\prime}A_{2}^{\prime}|\geq{\mathcal{D}_{L}}$. Since $\delta\leq{\mathcal{D}_{L}}/4$, by the triangle inequality, we get

[TABLE]

Similarly, $|A_{0}A_{1}|\geq{\mathcal{D}_{L}}/2$ and $|A_{1}A_{2}|\geq{\mathcal{D}_{L}}/2$ Thus, the points $A_{0}$, $A_{1}$, and $A_{2}$ are distinct. Let the values of the natural parameter corresponding to $A_{0}$, $A_{1}$, and $A_{2}$ be $s_{0}$, $s_{1}$, and $s_{2}$, respectively. Without loss of generality we can assume $s_{0}<s_{1}<s_{2}$ (otherwise we relabel $A_{0}^{\prime}$, $A_{1}^{\prime}$, and $A_{2}^{\prime}$). As before, let $t_{i}=s_{i}(a^{2}+b^{2})^{-\frac{1}{2}}$ for $i=0,1,2$, $h_{1}=t_{1}-t_{0}>0$ and $h_{2}=t_{2}-t_{1}>0$.

We consider two cases.

Case I: $h_{1}+h_{2}\geq\pi$.

In this case, in exactly the same way as in the proof of Theorem 13, we obtain

[TABLE]

Now, by the triangle inequality, we get

[TABLE]

so

[TABLE]

Case II. $h_{1}+h_{2}\leq\pi$.

Here, first we show that $A_{0}^{\prime}$, $A_{1}^{\prime}$ and $A_{2}^{\prime}$ are not on a straight line.

Recall that $|A_{0}A_{1}|\geq{\mathcal{D}_{L}}/2$. Therefore, $s_{1}-s_{0}\geq|A_{0}A_{1}|\geq{\mathcal{D}_{L}}/2$. Since $s_{1}-s_{0}=h_{1}(a^{2}+b^{2})^{\frac{1}{2}}$, we get $h_{1}\geq\frac{{\mathcal{D}_{L}}}{2}(a^{2}+b^{2})^{-\frac{1}{2}}$. Similarly, $s_{2}-s_{1}\geq{\mathcal{D}_{L}}/2$ and $h_{2}\geq\frac{\mathcal{D}_{L}}{2}(a^{2}+b^{2})^{-\frac{1}{2}}$.

Denote the area of $\triangle A_{0}A_{1}A_{2}$ by $S$.

By Lemma 9 (ii), we have $4S^{2}=T_{1}+T_{2}+T_{3}$, where $T_{1}=16a^{4}\sin^{2}\frac{h_{1}}{2}\sin^{2}\frac{h_{2}}{2}\sin^{2}\frac{h_{1}+h_{2}}{2}$, $T_{2}\geq 0$, and $T_{3}>0$.

Since $h_{1}+h_{2}\leq\pi$, by Lemma 8

[TABLE]

Therefore, ${\displaystyle 4S^{2}\geq\frac{16}{\pi^{6}}h_{1}^{2}h_{2}^{2}(h_{1}+h_{2})^{2}}$. We get,

[TABLE]

Denote the area of $\triangle A_{0}^{\prime}A_{1}^{\prime}A_{2}^{\prime}$ by $S_{3}$.

Assume that the points $A_{0}^{\prime}$, $A_{1}^{\prime}$ and $A_{2}^{\prime}$ are on a straight line. Then, $S_{3}=0$.

Applying Lemma 7 to $\triangle A_{0}A_{1}A_{2}$ and $\triangle A_{0}^{\prime}A_{1}^{\prime}A_{2}^{\prime}$ we obtain

[TABLE]

Now, $|A_{0}A_{1}|<s_{1}-s_{0}$, $|A_{1}A_{2}|<s_{2}-s_{1}$, and $|A_{0}A_{2}|<s_{2}-s_{0}$, so we obtain

[TABLE]

Since, $s_{2}-s_{0}\geq|A_{0}A_{1}|+|A_{1}A_{2}|\geq{\mathcal{D}_{L}}$ and $\delta\leq{\mathcal{D}_{L}}/4$, we have $\delta\leq(s_{2}-s_{0})/4$, so

[TABLE]

Combining (12) and (10) we obtain

[TABLE]

Therefore,

[TABLE]

Now, $h_{1}=t_{1}-t_{0}=(s_{1}-s_{0})(a^{2}+b^{2})^{-\frac{1}{2}}>|A_{0}A_{1}|(a^{2}+b^{2})^{-\frac{1}{2}}\geq\frac{\mathcal{D}_{L}}{2}(a^{2}+b^{2})^{-\frac{1}{2}}$. Similarly, $h_{2}>\frac{\mathcal{D}_{L}}{2}(a^{2}+b^{2})^{-\frac{1}{2}}$. Therefore,

[TABLE]

which contradicts the upper bound for $\delta$, so our assumption is false. Hence, the points $A_{0}^{\prime}$, $A_{1}^{\prime}$ and $A_{2}^{\prime}$ are not on a straight line. By Lemma 12 $S_{3}$, the area of $\triangle A_{0}^{\prime}A_{1}^{\prime}A_{2}^{\prime}$ is at least ${\mathcal{A}_{L}}$.

Next, we get a lower bound for the area of $\triangle A_{0}A_{1}A_{2}$. Using (11) we obtain

[TABLE]

Now, $|A_{0}A_{1}|+|A_{1}A_{2}|+|A_{0}A_{2}|<2(s_{2}-s_{0})$ and we showed above that $\delta\leq(s_{2}-s_{0})/4$. Therefore,

[TABLE]

We have $s_{2}-s_{0}=2(h_{1}+h_{2})(a^{2}+b^{2})^{\frac{1}{2}}\leq 2\pi(a^{2}+b^{2})^{\frac{1}{2}}$.

We obtain,

[TABLE]

Now, by the upper bound on $\delta$, we have

[TABLE]

Therefore,

[TABLE]

Now, we get lower bound for $h_{1}+h_{2}$ the same way we did in the case of lattice points on the helix, in Case 2 of the proof of Theorem 13. The upper bounds of $T_{1}$, $T_{2}$, and $T_{3}$ are the same as before, and the estimates (4), (5), and (6) still hold.

The only difference is that the lower bound for the area of $\triangle A_{0}A_{1}A_{2}$ now is ${\mathcal{A}_{L}}/2$ rather than ${\mathcal{A}_{L}}$.

The analogue equation (7) is

[TABLE]

(with the difference that now the left-hand-side of (14) is $1/4$ of the left-hand-side of (7)).

Also, $50^{1/6}>2.4$, so

[TABLE]

Using Lemma 8 we obtain $\sin\left(\frac{h_{1}+h_{2}}{2}\right)\geq\frac{1}{\pi}(h_{1}+h_{2})$. By Lemma 9 (i) $|A_{0}A_{2}|^{2}=4a^{2}\sin^{2}\left(\frac{h_{1}+h_{2}}{2}\right)+b^{2}(h_{1}+h_{2})^{2}$. Therefore,

[TABLE]

Using (15) and ${\displaystyle\kappa=\frac{a}{a^{2}+b^{2}}}$ we get

[TABLE]

Since $\frac{3.8}{\pi}>1.2$, and $|A_{0}^{\prime}A_{2}^{\prime}|\geq|A_{0}A_{2}|-2\delta$ we obtain the stated bound for $|A_{0}^{\prime}A_{2}^{\prime}|$.

∎