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## Science of Spherical Arrangements

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### Science of Spherical Arrangements

- Well-distributed points on the sphere
- Motivation from Chemistry, Biology, Physics
- Survey of results in the literature
- New results for logarithmic points

Peter Dragnev

Mathematical Sciences, IPFW

Well-distributed points on the sphere

Let N ={X1 , … , XN} Sd-1 - the unit sphere in

d-dimensional space Rd : { X = (x1 , x2 , … , xd ) }.

Sd-1 = { X : x12 + x22 + … + xd2=1 }

How do we distribute “well” the points?

- For d=2, this problem is simple.The solution is up to rotations the roots of unity.

Reason - direction and order

For d3 - no direction or order exists. Other methods and criteria are needed.

To well-distribute means to minimize some energy.

We distinguish

- Best packing points;

- Fekete points;

- Logarithmic points, etc.

Minimum Energy Problem on the Sphere

GivenanN-point configuration N ={X1 , … , XN}on the sphere S2 we define its generalized energy as

E(N )=Σij |Xi - Xj|.

- MaximizeE(N ) when >0;
- Minimize it when <0;
- When =0minimize the logarithmic energy
- E0( N )=Σij log(1/|Xi - Xj|)
- or maximize the product P( N )=exp(-E0 ( N )).

Denote the extremal energy with E(N,d) .

= 1Except for small N a long standing open problem in discrete geometry (L. Fejes Toth - 1956)

-Tammes problem; maximize the minimum distance between any pair of points. Known for N= 1-12 and 24.(Best packing points)

= -1Thompson problem; recent discovery of fullerenes attracted the attention of researchers in chemistry, physics, crystallography. Answer known for N=1-4, 6, 12. (Fekete Points)

= 0 The problem was posed by L. L. Whyte in 1952. Until recently the answer was known only for N=1-4. (Logarithmic Points)

C70

Motivation from Chemistry

- Large carbon molecules discovered in 1985 by Richard Smalley et. el.

- Fullerenes (Buckyballs)

Nanowire -- “a giant single fullerene molecule”, “a truly metallic electrical conductor only a few nanometers in diameter, but hundreds of microns (and ultimately meters) in length”, “expected to have an electrical conductivity similar to copper's, a thermal conductivity about as high as diamond, and a tensile strength about 100 times higher than steel” (R. Smalley).

- Problem of Tammes: ( -)
- Questions, raised by the Dutch botanist Tammes in 1930 in connection with the distribution of pores on pollen grains.
- What is the largest diameter of n equal circles that can be packed on the surface of a unit sphere without overlap?
- How to arrange the circles to achieve this maximum, and when is the arrangement essentially unique?
- This is the same as to ask to maximize the minimum distance between the points in the arrangement.

Starshine 3 satellite was launched in 2001 to study Earth’s upper atmosphere. The satellite was covered by 1500 small mirrors, which reflected the sun light during its free fall, allowing a large group of students nationwide to track the satellite.

Image credit: Michael A.Savell and Gayle R. Fullerton.

Electrons in Equilibrium

- N electrons orbit the nucleus

- Electrons repel

- Equilibrium will occur at minimum energy

Problem:

If Coulomb’s Law is assumed, then we minimize the sum of the reciprocals of the mutual distances, i. e.

Σij|Xi - Xj|-1

over all possible configurations of N points X1 , … , XN on the unit sphere. This corresponds to the case = -1.

Distribution of Dirichlet Cells (School Districts)

Dj:={Xє S2: |X-Xj|=mink |X-Xk|} j=1,…,N

The D-cells of 32 electrons at equilibrium are the tiles of the Soccer Ball. Soccer Ball designs occur in Nature frequently. The vertices of the Soccer Ball form C60.

There has to be exactly 12 pentagons in a soccer ball design.

Q. How are D-cells distributed for large N?

Tammes problem (best packing) = -

The solution is known for N=1,2, …, 12, 24.

N=4 - regular tetrahedron

N=5,6 - south, north and the rest on the equator

N=8 - skewed cube

N=12 - regular icosahedron (12 pentagons)

Survey of results for small NThompson problem = -1 (known for N=4,6,12)

N=4 - regular tetrahedron

N= 6 - regular octahedron

N=12 - regular icosahedron

Whyte’s problem = 0 (Logarithmic points)

N=4 - regular tetrahedron - {1,3}

N=12 - regular icosahedron - {1,5,5,1} [A] ‘96

N=6 - regular octahedron - {1,4,1} [KY] ‘97

N=5 - D3h {1,3,1} - [D+Legg+Townsend] ‘01

Remark: The method of proof is different from the other two results - a mixture of analytical and geometrical methods.

Definition:A collection of points which minimizes the logarithmic energy E0( N )=Σij log(1/|Xi - Xj|)is calledoptimal configuration. The points are referred to as logarithmic points.

Remark:Rakhmanov, Saff, Zhou were first to show the separation condition with constant 3/5. Dubickas obtained a constant 7/4. It is not known whether exists.

Logarithmic points - new results

Let N ={X1 , … , XN } be an optimal configurationon the sphere S2. Define dN :=minij |Xi - Xj|.

Theorem 1(Dragnev, Legg, Townsend - ‘01)

{1,3,1} is the only optimal 5-point configuration.

O is center of mass for {X1, X2,… , XN }.

(ii)

Proposition (Properties of optimal configurations)

Let N={X1, X2 … , XN } be an optimal configuration. Then

(iii)

d+2 configuration on Sd-1

Let N={X1, X2 … , XN } Sd-1be optimal, i.e.

E0 ( N )= E0(N,d). Derivative conditions on the energy functional (when N-1 points are fixed) yield:

Corollary:Theregular d-simplex is the only optimal d+1-configuration on Sd-1 for any 0.

Definitions: Configuration is called

- critical, if it satisfies (i) (recall (i) implies (ii) and (iii)).
- degenerate, if it does not span Rd.
- A vertex Xi is mirror related toXj (Xi ~ Xj) if Xi Xk= Xj Xkfor all k i,j. Then N\ {Xi , Xj } lie in the orthogonal bisector hyperspace of XiXj.

Note: Mirror relation is equivalence relation.

Theorem 3 (Dragnev ‘02)For fixed N and 0 the extremal energy E(N,d) is strictly decreasing for d<N, and for dN, E(N,d) = E(N,N-1).

CorollaryOptimal configurations are non-degenerate.

Theorem 4 (Dragnev ‘02)If N=d+2, then any critical configurations satisfies at least one of the following:

(a) it is degenerate;

(b) a vertex with all edges stemming out equal;

(c) every vertex has a mirror related partner.

Example:Let N=5, d=3. By Theorem 3, optimal configurations will satisfy at least one of the following:

(a) degenerate {5}

(b) EA=EB=EC=ED {1,4}

(c) Every vertex has a mirror related partner. In this case we arrive at A~B~C and D~E {1,3,1}

Comparing {5}, {1,4}, and {1,3,1} proves Theorem 2.

- N=4, d=2; diagonals of a square, form orthogonal simplexes.
- N=5, d=3; equilateral triangle on Equator and diameter.
- N=6, d=4; (cos k/3, sin k/3,0,0); (0,0,cos k/3, sin k/3), k=0,1,2;

In progress: If N=d+2 the optimal configurations is unique up to rotations and consists of two mutually orthogonal regular [d/2]- and [(d+1)/2]-simplexes. So,

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