REU Apprentice Fifth Monday
Benjamin McKenna
12 Twelfth Class: Mon. 7/21/14 12.1 First session: Erd˝os-deBruijn, two-distance spaces, parallelogram span- ning, Hadamard matrices, Gale-Shepley game
Erd˝os-deBruijn: Let A1,...,Am ⊆ [n] satisfy (∀i 6= j)(|Ai ∩ Aj| = 1) and ∀|Ai| ≥ 2. Then m ≤ n. The case m = n is possible. Alternately, Fisher’s Inequality, generalized by RC Bose, would read: For t ≥ 2 given and |Ai ∩ Aj| = t such that ∀|Ai| ≥ t + 1. Prove that under the Erd˝os-de Bruijn conditions, the incidence vectors are linearly independent. How many points in the plane can have pairwise equal distance? Three. In space, you can have four. In Rn, you can have n+1, namely forming an n+1-simplex. The coordinates to describe these are complicated in Rn; they become easier in Rn+1, however. For example, the regular triangle can be easily described by the standard basis in R3, and the interior of the triangle described by points (x, y, z) satisfying x+y +z = 1. Now suppose we permit two distances: Call a set with two such distances a 2-distance set. In the plane, squares and pentagons are two-distance sets. Exercise 12.1. In the plane, a two-distance set has ≤ 5 points.
How about in space? Well, you can take a pentagon and put a point above it the length of a diagonal from each of the five other points. The octahedron also forms a two-distance set with six points.1 What is the maximum number of points in a two-distance set in Rn? Well, we can at least construct n+2 by taking the one-distance set of n + 1 points and adding one point in the middle. Generally, we do better with tables of numbers than with imagining n-dimensional space. So let’s go back√ to our n + 1-dimensional simplex. If ei and ej are distinct basis points in this simplex, then ||ei − ej|| = 2. n Find a two-distance set√ in R with 2n vectors,√ and then with a lot more. For the first, take ±e1,..., ±en (the distances are 2 and 2, since ||±ei ± ej|| = 2 and ||±ei ∓ ei|| = 2). To get a lot more, take the set n of vectors ei + ej, of which there are 2 many. Often, in (ei + ej) − (ek + e`), we will have two ones and two negative ones, with norm two. Sometimes, we’ll have a single one and a single negative one, with norm √ P 2. These vectors satisfy the equation xi = 2, so they lie in an n − 1-dimensional hyperplane. Hence, in n+1 n dimensions, we get 2 vectors in a two-distance set. Is this maximum?
n (n+1)(n+4) Theorem: The number of points in R with two distances is ≤ 2 . (That is, we gave an almost tight bound.) x1 y1 . . Proof. Suppose the two distances are α and β. We have two vectors x = . and y = . . How can xn yn we express that the distance between x and y is either α or β? Well, we have ||x − y|| ∈ {α, β}, but without
1Easy way to construct the octahedron: Take a cube, put a point in the center of each face, and connect those.
1 using brackets we can write
n ! n ! X 2 2 X 2 2 F (x, y) := (xi − yi) − α (xi − yi) − β = 0. i=1 i=1 This is a polynomial of degree four. n Suppose a1, . . . , am ∈ R are a two-distance set with distances α and β. Now consider
fj(x) = F (aj, x) ∈ R[x1, . . . , xn]. If think of this univariate polynomial space as a vector space, then we could do the following:
Exercise 12.2. f1, . . . , fm are linearly independent.
Now prove that all of the fj are spanned by a few polynomials. We know that each of the fj has ≤4 degree four. We also know that the space V = R [x1, . . . , xn] of polynomials of degree at most four is a finite-dimensional vector space. Exercise 12.3. Find dim V , and the dimension of the similar set of polynomials of degree at most k (this is a very simple formula). A basis would have the following:
n + 1 → 1, x1, . . . , xn, 2 2 n → x1, . . . , xn, n → x x 2 i j 3 3 n → x1, . . . , xn, 2 2 2 n(n − 1) → x1x2, x1x3, . . . , xi xj, n → x x x , 3 i j k