REU 2007 · Apprentice Class · Lecture 12 Instructor: L´aszl´oBabai Scribes: Courtney Morris and Gabriel Kerr July 11, 2007. Revised by instructor. Last updated July 11, 9:30 p.m. A12.1 Rational roots of polynomials Consider the polynomial f(x) = x4 + ax3 + bx2 + cx − 15, a, b, c ∈ Z. What can we say about the roots? Claim A12.1.1. If f(m) = 0, m ∈ Z, then m ∈ {±1, ±3, ±5, ±15}. Proof. We have 15 = −m(m3 + am2 + bm + c), thus m divides 15. m Claim A12.1.2. If f( n ) = 0, m, n ∈ Z, n 6= 0, gcd(m, n) = 1, then n = 1. m m4 m3 4 Proof. We have 0 = f( n ) = n4 + a n3 + ··· + 15. Multiply through by n to get 0 = m4 + am3n + bm2n2 + cmn3 + 15n4. Moving m4 to the other side and factoring out an n we get m4 = −n(am3 + bm2n + cmn3 + 15n3). Thus n divides m4. Since we assumed gcd(m, n) = 1, n = ±1. Without any loss we can assume n = 1 (move the minus sign to the numerator. k Exercise A12.1.3. Assume f(x) = a0 + a1x + ··· + akx ∈ Z[x] with a0 6= 0 and ak 6= 0. m Show that if f( n ) = 0 and gcd(m, n) = 1 then m | a0 and n | ak. A12.2 Minimal graphs of girth 5 In this section we describe a mathematical gem due to Alan J. Hoffman and R. R. Singleton. Definition A12.2.1. The girth of a graph G, girth(G), is the length of the shortest cycle. If G is a tree (or a forest) then girth(G) is smallest member (the inf) of the empty set, i. e. +∞. With this convention, we have: Observation A12.2.2. A graph has girth ≥ k exactly if it has no cycles of length < k. Definition A12.2.3. A graph is r-regular if every vertex has degree r. 1 Lemma A12.2.4. If an r-regular graph has girth ≥ 5 then n ≥ r2 + 1, where n is the number of vertices. Remark A12.2.5. The Petersen graph shows that there exists a graph with r = 3 and n = 32 + 1 = 10. Proof of Lemma A12.2.4. We illustrate the proof on the case r = 3. To show that you cannot have fewer than 10 vertices when r = 3, start with one of your vertices, v0. It has three neighbors. Let S be the set consisting of v0 and its thre neighbors. Each of the 3 neighbors has 2 more neighbors. None of these additional neighbors belong to S (because then we would have a 3-cycle), and they are all distinct, because otherwise we would have a 4-cycle. Thus we have a minimum of 10 vertices. To show the inequality n ≥ r2 + 1 for general r, make the same argument, except replace 3 by r. You will see that it is necessary to have at least 1 + r + r(r − 1) = r2 + 1 vertices. Question A12.2.6. For which values of r is equality possible? If r = 1 then the graph with two vertices and one edge between them satisfies n = 12 + 1. If r = 2 then the pentagon satisfies n = 22 + 1. If n = 3 then the Petersen graph satisfies n = 32 + 1. For r = 4, 5, or 6, there does not exist a graph satisfying n = r2 + 1, however, for r = 7, there does exist one with n = 50. The graph is called the Hoffman-Singleton graph and just as the Petersen graph is a bunch of pentagons glued together in a clever way, the Hoffman-Singleton graph is a bunch of Petersen graphs glued together in a very clever way. (You can see pictures of this graph and explanations on Wolfram MathWorld.) The answer to the question is a theorem due to Hoffman-Singleton: Theorem A12.2.7 (Hoffman-Singleton Theorem). If there exists a graph of girth ≥ 5, degree r, and n = r2 + 1, then r ∈ {1, 2, 3, 7, 57}. Remark A12.2.8. It is unknown whether or not there exists a graph for r = 57. Theorem A12.2.9. (M. Aschbacher) If there exists a graph with r = 57 then it cannot be vertex-symmetrical. A graph is vertex-symmetrical if for each pair x, y of vertices there exists an automorphism that maps x to y. Definition A12.2.10. Let G be a graph with n vertices, numbered 1 to n. The adjacency matrix of G is an n × n (0, 1)-matrix, A = (aij), with aij = 1 if i ∼ j and 0 if i j, where i ∼ j if i is adjacent to j. Observation A12.2.11. A has zeros along the diagonal and is symmetric (AT = A). The number of nonzero entries is equal to twice the number of edges. 2 Exercise A12.2.12. If G is any graph and A is its adjacency matrix, then A = (bij), where bii = degi and bij = the number of common neighbors of i and j. 2 Exercise A12.2.13. If G is regular of degree r, has girth ≥ 5, and n = r2 + 1 vertices the number of common neighbors of a pair x, y of vertices is zero if x, y are adjacent and 1 if x, y are not adjacent. Hint: let x = v0 in the proof of Lemma A12.2.4. Note that in particular, diam(G) = 2. 2 proof of Theorem A12.2.7. Let A = B = (bij) where A is the adjacency matrix. Combining Exx A12.2.12 and A12.2.13, we obtain that bii = r; and bij = 0 if i ∼ j and bij = 1 if i j. ¯ ¯ Let A = (¯aij) be the adjacency matrix of G. Thena ¯ii = 0 anda ¯ij = 1 − aij if i 6= j. We see that A + A¯ + I = J and that A2 = B = rI + A¯ = rI + J − A − I = (r − 1)I − A + J or A2 + A − (r − 1)I = J. (1) In general, P row i = degi. Thus deg1 . A · 1 = . , degn where 1 is the vector of all 1’s. In our case, A · 1 = r · 1, and we see that 1 is an eigenvector with eigenvalue r. From the Spectral Theorem (adjacency matrices are always real symmetric matrices) we know that A has an orthogonal eigenbasis, e1 = 1, e2,..., en. What can we say about the ei, i ≥ 2? Well, ei⊥1. It follows that for i ≥ 2 we have Jei = 0. Multiplying equation (1) by ei to the right (i ≥ 2), we see that 2 (A + A − (r − 1)I)ei = Jei = 0. 2 2 2 Now Aei = λiei and A ei = λi ei, thus λi ei + λiei − (r − 1)ei = 0 or 2 λi + λi − (r − 1) = 0, for i ≥ 2. Since this is a quadratic equation it has at most 2 roots. It follows that the matrix has 3 eigenvalues, call them r, µ1, and µ2 with multiplicities 1, m1 and m2, respectively. 2 µ1 and µ2 are roots of the equation t + t − (r − 1) = 0. Thus √ −1 ± p1 + 4(r − 1) −1 ± 4r − 3 µ = = . (2) 1,2 2 2 √ Letting s = 4r − 1 we have that s2 = 4r − 3, i. e., s2 + 3 r = . (3) 4 Note that s is the square root of an integer. 3 We can make the following table for the eigenvalues. eigenvalue multiplicity r 1 µ1 m1 µ2 m2 Our unknowns are m1 and m2; we need two equations connecting them. Counting multi- plicites we have 1 + m1 + m2 = n. (4) Pn Consider the trace. We know that i=1 λi = Tr(A) = 0. Thus 1 · r + m1µ1 + m2µ2 = 0. (5) Rewrite the first equation as 2 m1 + m2 = r (6) and the second as 2r + m1(−1 + s) + m2(−1 − s) = 0 or 2r − (m1 + m2) + s(m1 − m2) = 0. (7) From this equation we ask whether s is rational or not. If it is irrational then clearly 2 m1 − m2 = 0 and then (after substituting r for m1 + m2) our equation reads 2r − r2 = 0 or r = 2 (since r 6= 0). This number occurs in the theorem. Henceforth we may assume that s is rational, and therefore (being the square root of an integer), s is an integer. 2 Again, substituting r for m1 + m2, our equation reads 2 r − 2r − (m1 − m2)s = 0. s2 + 3 Putting r = into this equation and multiplying by 16 yields 4 2 2 2 (s + 3) − 8(s + 3) − 16(m1 − m2)s = 0. Expanding and simplifying gives 4 2 s − 2s − 16(m1 − m2)s − 15 = 0, s2 + 3 so that s must divide 15, i. e., s = ±1, ±3, ±5, ±15. Again using r = we obtain 4 r = 1, 3, 7, 57, as required. 4 A12.3 Helly’s Theorem We state Helly’s theorem: n Theorem A12.3.1. If C1,...,Cm are convex sets in R and every n + 1 of them intersect then all do, i. e., there is a point contained in all the Ci. T We outline a proof of this for m = n + 2. First take a pi ∈ j6=i Cj. Then if we replace Ci with conv({pj : j 6= i}) the result will still hold (since conv({pj : j 6= i}) ⊆ Ci (where conv(S) denotes the convex hull of S ⊆ Rn). n Lemma A12.3.2. (Radon’s Lemma) If p1, . , pn+2 are distinct points in R then it is possible to divide them into two disjoint subsets A and B such that A ∪ B = {p1, .