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 , 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

n n(n−1)(n−2)(n−3) and so on with degree four, with the last one being xixjxkx`, of which there are 4 = 24 . This already tells us that the number of these vectors is on the order of n4. Exercise 12.4. Of course, we don’t want something on the order of n4; it turns out we don’t need all the polynomials we’ve been calculating. Find a quadratic number of polynomials that span all the fj. If that (n+1)(n+4) quadratic number is 2 , you’re done.

Exercise 12.5. Two vectors a and b span a parallelogram of area | det(a, b)|. P Exercise 12.6. In three dimensions, we have a parallelepiped spanned by a1, a2, a3, which is the set { αiai | 0 ≤ αi ≤ 1}. The volume Vol(a1, a2, a3) of such a parallelepiped is equal to | det(a1, a2, a3)|. Exercise 12.7. Now n dimensions: First, make the following assumptions about volume: 1. additive

2. translation invariant 3. Vol(rectangle) = Q side lengths.

2 Based on these assumptions on volume, prove:

Vol(a1, . . . , an) = | det(a1, . . . , an)|.

The area of a parallelogram with vertices on the integer lattice is itself an integer, since the of an integer is an integer. In three dimensions, the volume of a parallelogram spanned by three integer vectors is an integer for the same reason, and this holds into n dimensions. Suppose we have two integer vectors in three dimensions. In R2,3, then a · b = |a| · |b| · cos θ. In R3, then a × b is the vector z where |z| = |a| · |b| · | sin θ| and z is pairwise orthogonal to a and b, according to the | {z } Area right-hand rule. In matrix form, if we take the matrix   a1 b1 a2 b2 a3 b3 encoding a and b, we can take the of minors of this matrix to get the coordinates of z. Hence  the vector z = z1 z2 ze is an , but its norm is not necessarily an integer. That is, the area of a parallelogram spanned by two integer vectors in three dimensions is the square root of an integer, but is not necessarily an integer itself.

n 2 Exercise 12.8. In R , we can take k integral vectors and know that Vol(a1, . . . , ak) ∈ Z. 2 Exercise 12.9. Vol(a1, . . . , ak) = det(G(a1, . . . , ak)) is equal to the Gram determinant, where

< a , a > G(a1, . . . , ak) = i j k×k . To prove this, use only our assumptions on volume.

Given v1, . . . , vk, perform Gram-Schmidt orthogonalization to obtain b1, . . . , bk. Exercise 12.10. Y Vol(v1, . . . , vk) = Vol(b1, . . . , bk) = ||bi||.  Exercise 12.11. Hadamard’s inequality: Given an n × n matrix A = a1 ··· an , then

n Y | det A| ≤ ||ai||, i=1 where equality holds iff they are orthogonal or the RHS is equal to zero.

Definition: A Hadamard matrix is a matrix each of whose entries is ±1 and whose columns are orthogonal. 1 1  For example, is a Hadamard matrix. 1 −1 Exercise 12.12. A(±1) matrix is Hadamard iff the rows are orthogonal. Is there a 3 × 3 Hadamard matrix? No, since the determinant would be of the form ±1 ± 1 ± 1 ≡ 1 + 1 + 1 (mod 2) ≡ 1 (mod 2). Exercise 12.13. If there exists an n × n Hadamard matrix, then n = 2 or 4 | n.

Conjecture: If 4 | n, then there exists an n × n Hadamard matrix. 1 Let H = {n | ∃n × n Hadamard matrix}. The conjecture is that the density of this set is 4 in Z, but we don’t even know if this density is positive.

3 Exercise 12.14. If n = 2k, then n ∈ H . Exercise 12.15. If p is a prime satisfying p ≡ −1 (mod 4), then p + 1 ∈ H . (Hint: Think Paley.)

Definition: The discrepancy of a ±1 matrix is:

X max aij ,

(i,j)∈T combinatorial rectangle where a combinatorial rectangle T satisfies R ⊆ [k], S ⊆ [`], and T = R × S. Notice that the discrepancy of a n × n such matrix is at most n2.

Lindsay’s Lemma: If T is an r × s combinatorial rectangle in an n × n Hadamard matrix H, then

X √ hij ≤ rsn.

(i,j)∈T

3 Hence the discrepancy of H is at most n 2 . This doesn’t beat random (that would be n log n), but it’s an explicit de-randomization tool.

Gale-Shepley game: This game is played on a (±1) matrix. Player 1 creates the matrix. Player 2 is allowed to flip the sign of any number of rows and columns. Then the sum of the entries in the output is the win for player 2. First, we know that player 2 always wins something, since if the sum of the matrix is negative then player 2 can flip the rows with negative sums. Since player 2 always wins, player 1 should start off with some money. What is the fair value of this game? i.e., if both players play optimally (first player makes a good matrix, second player makes the best of it), then what original sum to player 1 would make the expected outcome fair for both people? The maximum possible gain for player 2 would be n2. What would be a good strategy for player 1? The checkerboard, which we might think would work, is in fact the worst possible play for player 1, since player 2 can convert it to the matrix J. Next, let’s see what happens if player 1 produces an Hadamard matrix. Since player 2’s moves turn a Hadamard matrix into a Hadamard matrix, then Lindsay’s Lemma keep the 3 3 payoff under n 2 . Is that optimal? i.e., can player 2 realize n 2 ?

3 Exercise 12.16. (1) The value of the game is O(n 2 ) for every n (not just for those n for which there exists | {z } 3 ≤c·n 2 3 3 a Hadamard matrix). In fact, (∀ > 0)(∃n0)(∀n > n0)(≤ (1+)n 2 ). (2) The value of the game is Ω(n 2 ) . | {z } 3 ≥c0·n 2 ,c0>0

n t n Definition: For A ∈ Mn(R) and x ∈ R , define the function qA(x) = x Ax ∈ R, where qA(x): R 7→ R. This P 2 is said to be a “quadratic form,” as qA(x) = aijxixj has n terms. A matrix A is called positive-definite if Pn 2 (∀x 6= 0)(qA(x) > 0). Note that qI (x) = i=1 xi , so I is positive definite. The matrix D = diag(λ1, . . . , λn) Pn 2 gives rise to qD(x) = i=1 λixi , so the necessary and sufficient condition for D to be positive-definite is that (∀i)(λi > 0). Exercise 12.17. Prove that A is positive definite iff all eigenvalues are strictly positive iff all corner deter- minants (i.e., determinants of square minors flush with the upper-left-hand corner) are strictly positive.

For the last equivalence, note that if λ1 ≥ · · · ≥ λn are the eigenvalues of the full matrix and µ1 ≥ · · · ≥ µn−1 are the eigenvalues of the n − 1 × n − 1 corner , then the values interlace. Hence =⇒ is easy in the last equivalence.

4 Exercise 12.18. A matrix is called positive semi-definite if (∀x)(qA(x) ≥ 0). For example, a with nonnegative eigenvalues, at least one of which is zero, is positive semi-definite but not positive definite. The exercise is: A matrix is positive semi-definite iff all eigenvalues are nonnegative. Exercise 12.19. If A is positive definite and B positive semi-definite, then A + B is positive definite.

Exercise 12.20. If A is positive definite, then A is non-singular. (This is clear, since we determined that the eigenvalues are all nonnegative). Use this to prove Fisher’s.

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