On the Regularity Radius of Delone Sets in \({\mathbb {R}}^3\)

We complete the proof of the upper bound \({\hat{\rho }}_3\le 10R\) for the regularity radius of Delone sets in three-dimensional Euclidean space. Namely, summing up the results obtained earlier, and adding the missing cases, we show that if all \(10R\)-clusters of a Delone set X in \({\mathbb {R}}^3\) with parameters (r, R) are equivalent, then X is regular (has a transitive symmetry group).

1. The tower height of a group is the number of subgroups of the longest chain of strictly nested subgroups of G.

2. This was recently noticed by Garber.

3. Sometimes clusters of Delone sets are also called patches or centered patches, see e.g. [1].

4. Historically, however, Theorem 3.5 preceded Theorem 3.4 and inspired it.

5. Available at https://faculty.utrgv.edu/alexey.garber/papers/10R/.

6. https://faculty.utrgv.edu/alexey.garber/papers/10R/

We would like to thank the American Institute for Mathematics (AIM) for hosting a weeklong workshop in 2016 on “Soft Packings, Nested Clusters, and Condensed Matter”, as well as the authors’ ongoing SQuaRe research project on “Delaunay Sets: Local Rules in Crystalline Structures” that grew out of it. The present paper resulted from the discussions at the first SQuaRE meeting at AIM in November 2018. We greatly appreciated the opportunity to meet at AIM and are grateful to AIM for its hospitality. The work of N.D. was supported by the Russian Science Foundation under grant 20-11-20141 and performed in the Steklov Mathematical Institute of Russian Academy of Sciences. The work of E.S. was partially supported by Simons Foundation award no. 420718.

Authors and Affiliations

1. Steklov Mathematical Institute of Russian Academy of Sciences, 8 Gubkina St., Moscow, Russia, 119991

   Nikolay Dolbilin

2. School of Mathematical & Statistical Sciences, The University of Texas Rio Grande Valley, 1 West University Blvd, Brownsville, TX, 78520, USA

   Alexey Garber

3. Department of Mathematics, Northeastern University, Boston, MA, 02115, USA

   Undine Leopold & Egon Schulte

4. Department of Mathematics, Smith College, Northampton, MA, 01063, USA

   Marjorie Senechal

Corresponding author

Correspondence to Alexey Garber.

Editor in Charge: Kenneth Clarkson

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Appendix A: Computational Details

Lemma A.1

In the notation of Case 1 of the proof of Theorem 5.1, there exist two different vertices of the antiprisms \(P_{{\mathbf {x}}}\) and \(P_{{\mathbf {y}}}\) that are at distance less than \(r_1\).

Proof

Recall that \({{\mathbf {x}}}=(0,0,0)\) and \(r_1=1\). Without loss of generality we can assume that \({{\mathbf {y}}}=(a,0,b)\) with \(a^2+b^2=1\) and \(a\ne 0\), \(b\ne 0\). Then the antiprism \(P_{{\mathbf {x}}}\) can be written as the convex hull of its vertices,

obtained using the action of \(\langle \sigma \rangle \) on \(\mathbf{y}\).

Next we exploit the fact that \(r_1=1\) is the smallest distance between two points of X. Particularly, the two vertices (a, 0, b) and (0, a, b) in the same base of \(P_{{\mathbf {x}}}\) must be at least 1 apart, and thus

or equivalently, \(a^2\ge b^2\). Similarly, the distance between the two vertices (a, 0, b) and \(({a}/{\sqrt{2}}, {a}/{\sqrt{2}}, -b)\) in different bases of \(P_{\mathbf {x}}\) must be at least 1, giving

Next we will find coordinates for the vertices of \(P_{\mathbf {y}}\). We know that \({{\mathbf {x}}}=(0,0,0)\) is a vertex of \(P_{\mathbf {y}}\); let \({\mathbf {z}}\) be the vertex of \(P_{\mathbf {y}}\) opposite to \({\mathbf {x}}\) in the same base as \({\mathbf {x}}\). Suppose \({\mathbf {z}}=(a+x,y,b+z)\) for some numbers x, y, z, so \(\mathbf {yz}=(x,y,z)\). The distance between \({\mathbf {y}}\) and \({\mathbf {z}}\) must be \(r_1=1\), so

Also the scalar product of the vectors \(\mathbf {yx}\) and \(\mathbf {yz}\) must be equal to the scalar product of the vectors (a, 0, b) and \((-a,0,b)\). This turns into the equality

which completes our list of conditions for the parameters a, b, x, y, z describing all possible positions of \(P_{{\mathbf {x}}}\) and \(P_{\mathbf {y}}\).

The remaining vertices of \(P_{\mathbf {y}}\) can be computed as follows. The midpoint

of the segment \([{\mathbf {x}},{\mathbf {z}}]\) is the center of the base of \(P_{\mathbf {y}}\) that contains \({\mathbf {x}}\). The two remaining vertices of this base can be obtained as the endpoints of the vectors

placed at \({\mathbf {t}}\), where here \(\times \) indicates the cross product of vectors. The length of the numerator of this fraction is twice the area of the triangle \(\mathbf {xyz}\), and the denominator is twice the height of the same triangle from the vertex \(\mathbf{y}\). Thus, the overall vectors will represent the vectors \(\pm \mathbf {tx}\) rotated by \(90^\circ \) about the line \(\overline{\mathbf {ty}}\). We also note that the denominator is actually the distance from the center of the antiprism \(P_{\mathbf {y}}\) to its base, which is equal to |b|. Thus, the two remaining vertices of \(P_{\mathbf {y}}\) in this base can be written as

Finally, if we set

then the four vertices of the other base of \(P_{\mathbf {y}}\) can be written as

Now that we know all vertices of the two antiprisms, we can set up the following optimization problem.

Maximize: minimal distance between vertices of \(P_{\mathbf {x}}\) and \(P_{\mathbf {y}}\) that are at least 0.01 apart.

Under conditions:

• \(a^2+b^2=1\);

• \(a^2\ge b^2\);

• \(a^2(1-\sqrt{2})+3b^2\ge 0\);

• \(x^2+y^2+z^2=1\);

• \(ax+bz=a^2-b^2\).

The (numerical) computations using Wolfram Mathematica (v. 11.2.0.0)^{Footnote 5} show that the maximal value of the corresponding function is 0.598, which means that there is a vertex of \(P_{\mathbf {x}}\) and a vertex of \(P_\mathbf{y}\) that are at distance at least 0.01 but closer than \(r_1=1\). \(\square \)

Lemma A.2

In the notation of Case 3 of the proof of Theorem 5.1, if the point \({\mathbf {z}}\) does not lie in the perpendicular bisector of the segment \([{\mathbf {x}},{\mathbf {y}}]\), then there exist two vertices of the antiprism \(P'_{{\mathbf {y}}}\) in the same base that are closer to \({\mathbf {z}}\) than \(r_2\).

Proof

Recall that \(\mathbf {{\mathbf {x}}}=(0,0,0)\) and \(r_1=1\). Without loss of generality we can assume \({\mathbf {y}}=(0,0,1)\). We can also assume that \({\mathbf {z}}=(a,0,b)\) with \(r_2^2=a^2+b^2>1\) and \(a\ge 0\), and that \(b>0\), because the base of \(P'_{{\mathbf {x}}}\) that contains \({\mathbf {z}}\) is closer to \({\mathbf {y}}\) than the other base of \(P'_{{\mathbf {x}}}\). The point \({\mathbf {z}}\) is not on the perpendicular bisector of \([{\mathbf {x}},{\mathbf {y}}]\), so \(b<1/2\); otherwise \(r_2\) is not the shortest distance from \({\mathbf {x}}\) among all points of \(C_{{\mathbf {x}}}(2R)\) off \(\ell \).

The distance between the two vertices \({\mathbf {z}}=(a,0,b)\) and (0, a, b) of the same base of \(P'_\mathbf {{\mathbf {x}}}\) should be at least \(r_2\), as otherwise the sides of the base meeting at \(\mathbf{z}\) would give a non-collinear triple of points in \(C_{\mathbf{z}}(2R)\) in which two points are at distance less than \(r_2\) from \({\mathbf {z}}\), thus contradicting our case assumption. For similar reasons, the two vertices \({\mathbf {z}}=(a,0,b)\) and \(({a}/{\sqrt{2}},{a}/{\sqrt{2}},-b)\) of \(P'_{{\mathbf {x}}}\) in different bases must also be at least \(r_2\) apart. These two conditions give the inequalities

Next we look at the 2R-cluster of \({\mathbf {y}}=(0,0,1)\). The point \({\mathbf {x}}\) is at distance \(r_1=1\) from \({\mathbf {y}}\), so the line corresponding to \(\ell \) in \(C_{{\mathbf {y}}}(2R)\) must coincide with the line \(\overline{\mathbf {yx}}=\ell \). Therefore, the line \(\ell \) is also the axis of rotation of the antiprism \(P'_{\mathbf {y}}\) and the two bases of this antiprism lie in the planes \(z=1-b\) and \(z=1+b\). Without loss of generality we can assume that one of the vertices of \(P'_{\mathbf {y}}\) is a point \({\mathbf {u}}_1:=(x,y,1-b)\) with \(x,y\ge 0\). Then \({\mathbf {u}}_2:=(y,-x,1-b)\) is a vertex of \(P'_{\mathbf {y}}\) as well. Note that the distance of \({\mathbf {u}}_1\) from \(\ell \) is equal to the distance of \({\mathbf {z}}\) from \(\ell \), so \(x^2+y^2=a^2\). We claim that

In order to establish this inequality we set up the following optimization problem.

Maximize: \(|\mathbf {zu}_1|+|\mathbf {zu}_2|-1-\sqrt{a^2+b^2}\)

Under conditions:

• \(a^2+b^2>1\);

• \(a\ge 0\);

• \(b>0\);

• \(b<1/2\);

• \(a^2\ge b^2\);

• \(a^2(1-\sqrt{2})+3b^2\ge 0\);

• \(x^2+y^2=a^2\);

• \(x\ge 0\);

• \(y\ge 0\).

The (numerical) computations using Wolfram Mathematica (v. 11.2.0.0)^{Footnote 6} show that the maximal value of the corresponding function (the difference of the two sides of the inequality) is \(-0.3367\), which means that for all possible situations,

Since the distance between any two points of X is at least \(r_1\), both distances \(|\mathbf {zu}_1|\) and \(|\mathbf {zu}_2|\) are less than \(r_2\). Thus \({\mathbf {u}}_1\) and \({\mathbf {u}}_2\) are vertices of the antiprism \(P'_{{\mathbf {y}}}\) in the same base that are closer to \({\mathbf {z}}\) than \(r_2\). \(\square \)

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