QUASIRANDOMNESS AND GOWERS THEOREM QIAN ZHANG August 16, 2007 Abstract. Quasiradomess will be described ad the Quasiradomess Theorem will be used to prove Gowers Theorem. This article assumes some familiarity with liear algebra ad elemetary probability theory. Cotets 1. Lidsey s Lemma: A Illustratio of quasiradomess 1 1.1. How Lidsey s Lemma is a Quasiradomess result 2 2. The Quasiradomess Theorem 4 2.1. How the Quasiradomess Theorem is a quasiradomess result. 8 3. Gowers theorem 9 3.1. Traslatig Gowers Theorem: Provig m Proves Gowers Theorem 9 3.2. Provig m 11 Refereces 15 1. Lidsey s Lemma: A Illustratio of quasiradomess Defiitio 1.1. A is a Hadamard matrix of size if it is a matrix with each etry (a ij ) either +1 or -1. Moreover, its rows are orthogoal, i.e. ay two row vectors have ier product = 0. Remark 1.2. If A is a matrix with orthogoal rows, the AA T = I, so ( 1 A)( 1 A) T = I ( 1 A) T = ( 1 A) 1 A T A = I, so the colums of A are also orthogoal. Notatio 1.3. 1 deotes the colum vector that has 1 as every compoet ad 1 T deotes the row vector with 1 as every compoet. Remark 1.4. For ay matrix A, 1 T A 1 is the sum of the etries of A. 1 Defiitio 1.5. Give a matrix A ad a submatrix T of A, let X be the set of rows of T ad let Y be the set of colums of T. Let x i be ay compoet of x ad let y i be ay compoet of y. x is a icidece vector of X whe x i = 1 if the ith row vector of A is a row vector of T ad x i = 0 otherwise. y is a icidece 1 1 selects colums of A ad sums their correspodig compoets. 1 T selects rows of A, selectig ad addig together certai compoet sums. Replacig the ith compoet of 1 with a 0 would deselect the ith colum of A ad replacig the ith compoet of 1 T with 0 would deselect the ith row of A. 1
2 QIAN ZHANG vector of Y whe y i = 1 if the ith colum vector of A is a colum vector of T ad y i = 0 otherwise. Lemma 1.6. (Lidsey s Lemma) If A = (a ij ) is a Hadamard matrix ad T is a k l-submatrix, the (i,j) T a ij kl Proof. Let X be the set of rows of T ad let Y be the set of colums of T. Note that X = k ad Y = l. Let x be the icidece vector of X ad let y be the icidece vector of Y. x T A y is (i,j)ɛt a ij, which is the sum of all etries of T, so x T A y = (i,j)ɛt a ij. By the Cauchy-Schwarz iequality, x T A y x A y. 1 A y = y T ( 1 A) T ( 1 A) y = y T y = y A y = ( 1 A y) = 1 A y = y Substitutig for A y i the Cauchy-Schwarz iequality ad otig that x = k ad y = l, x T A y x ( y ) = kl. 1.1. How Lidsey s Lemma is a Quasiradomess result. The followig corollary illustrates how Lidsey s Lemma is a quasiradomess result. It says that if T is a sufficietly large submatrix, the the umber of +1 s ad the umber of -1 s i T are about equal. Corollary 1.7. Let T be a k l submatrix of a Hadamard matrix A. If kl 100, the the umber of +1 s ad the umber of -1 s each occupy at least 45% ad at most 55% of the cells of T. Proof. Let x be the umber of +1 s i T ad let y be the umber of -1 s i T. Suppose kl 100. We wat to show that (0.45)kl x (0.55)kl ad (0.45)kl y (0.55)kl. By Lidsey s Lemma, (i,j)ɛt a ij kl. Note that x y = (i,j)ɛt a ij, so (i,j)ɛt a ij = x y kl. We kow that k > 0 ad l > 0, so kl > 0. x y kl kl (kl) 2 = kl 100 = 1 10 where the last iequality holds because kl 100. Sice all etries of T are either +1 or -1, the sum of the umber of +1 s ad the umber of -1 s is the umber of etries i T, so x + y = kl, hece y = kl x. Substitutig i for y,
QUASIRANDOMNESS AND GOWERS THEOREM 3 x (kl x) 1 kl 10 2x kl kl 10 kl 10 9kl 20 9kl 20 9kl 20 2x kl kl 10 x 11kl 20 kl x 11kl 20 y 11kl 20 Defiitio 1.8. A radom matrix is a matrix whose etries are radomly assiged values. Etries assigmets are idepedet of each other. To see how Corollary 1.7 shows Hadamard matrix A to be like a radom matrix but ot a radom matrix, cosider a radom matrix B whose etries are assiged either +1 or -1 with probability p ad 1 p respectively. Cosider U, a k l submatrix of B. U has kl etries, ad w, the umber of etries of the kl etries that are +1, would be a radom variable. Cosiderig U s etries assigmets idepedet trials that result i either success or failure ad callig the occurrece of +1 a success, P (w = s) is the probability of s successes i kl idepedet trials, which is the product of the probability of a particular sequece of s successes, p s (1 p) kl s, ad the umber of such se- ( ) ( ) queces, kl s, so P (w = s) = kl s p s (1 p) kl s. I other words, w has a biomial probability distributio. Hece, w has expected value klp. If each etry has equal probability of beig assiged +1 or -1, p = 1 2 so E(w) = kl( 1 2 ). Note that w ca take values far from E(w), sice P (w = s) shows w has ozero probability of beig ay iteger s where 0 s kl. Now cosider Hadamard matrix A, its k l submatrix T, ad x, the umber of +1 s i T. Corollary 1.7 shows that x must take values close to E(w). More precisely, if kl > 100, x must be withi 5% of E(w). x is like radom w i that we ca expect x to take values close to the expected value of w. However, x is ot radom because it must be withi 5% of E(w), while radom w ca take values farther from E(w), ay value ragig from 0 to kl. 2 The above argumet is symmetrical: It ca be used to compare y, the umber of -1 s i T, ad z, the umber of -1 s i U. I derivig P (w = s), we called the occurrece of +1 a success. We could have arbitrarily called the occurrece of -1 a success. The P (z = s) = ( kl s ) p s (1 p), E(z) = kl( 1 2 ) if p = 1 2, ad y would be like radom z, but ot radom, i the same way that x would be like radom w, but ot radom. 2 If kl 100, x must be withi 5% of E(w). 100 was used i the hypothesis of 1.7 for the sake of cocreteess. Ay arbitrary costat c could have replaced 100, so that kl c. So log as c > 1, x is more limited tha w i the values it ca take.
4 QIAN ZHANG I short, Hadamard matrix A is quasiradom because it is like a radom matrix B, but ot itself a radom matrix. Characteristics (x ad y) of k l T, a sufficietly large 3 submatrix of A, are similar to characteristics (w ad z) of k l U, a submatrix of B. A is like, but ot, a radom matrix B because submatrices of A have properties similar to, but ot the same as, submatrices of B. 2. The Quasiradomess Theorem Defiitio 2.1. A graph G = (V, E) is a pair of sets. Elemets of V are called vertices ad elemets of E are called edges. E cosists of uordered pairs of vertices such that o vertex forms a edge with itself: v V, E V V \{v, v}. v 1, v 2 V are adjacet whe {v 1, v 2 } E, deoted v 1 v 2. The degree of a vertex is the umber of vertices with which it forms a edge. Notatio 2.2. If x is a vertex, deg(x) deotes its degree. Remark 2.3. Vertices ca be visualized as poits ad a edge ca be visualized as a lie segmet coectig two poits. Defiitio 2.4. Cosider a graph G = (V, E) ad let deote G s maximum umber of possible edges, i.e. the umber of edges there ( would ) be if every vertex V were coected with every other vertex, so that =. E is the umber 2 of edges i the graph. The desity p of G is E. Defiitio 2.5. A bipartite graph Γ(L, R, E) is a graph cosistig of two sets of vertices L ad R such that a edge ca oly exist betwee a vertex i L ad a vertex i R. Call L the left set ad R the right set. Notatio 2.6. Give two sets of vertices V 1 ad V 2, E(V 1, V 2 ) deotes the set of edges betwee vertices i V 1 ad vertices i V 2. E(V 1, V 2 ) deotes the umber of elemets i E(V 1, V 2 ). Defiitio 2.7. A bipartite adjacecy matrix of a bipartite graph that has k vertices i the left set ad l vertices i the right set is a k l matrix such that { 1 if i j, where i L ad j R a ij = 0 otherwise Remark 2.8. Let A be a k x l bipartite adjacecy matrix. (A T A) T = A T (A T ) T = A T A. Sice A T A is symmetric, it has l real eigevalues, deoted λ 1,..., λ l i decreasig order. A T A is positive semidefiite because x R l, x T A T Ax = Ax 2 0. Sice A T A is positive semidefiite, its eigevalues are oegative. Defiitio 2.9. A biregular bipartite graph Γ(L, R, E) is a bipartite graph where every vertex i L has the same degree s r ad every vertex i R has the same degree s c. Remark 2.10. E = L s r = R s c. Fact 2.11. (Rayleigh Priciple) Let symmetric matrix A have eigevalues λ 1,..., λ i decreasig order. Defie the Rayleigh quotiet R A (x) = xt A x x T x. The λ 1 = R A (x). max x R, x 0 3 kl >
QUASIRANDOMNESS AND GOWERS THEOREM 5 Notatio 2.12. Subscripts of the form m o matrices ad vectors give their dimesios: m rows ad colums. (x 1,.., x ) 1, deotes a 1 row vector where the x i are compoets of x. 1 deotes a vector with 1 for every compoet. Lemma 2.13. Let Γ(L, R, E) be a biregular bipartite graph with L = k ad R = l. Let each vertex i L have degree s r ad let each vertex i R have degree s c. Let A be the k l adjacecy matrix of Γ, ad let λ 1 be the largest eigevalue of A T A. The λ 1 = s r s c. Proof. Let r 1,..., r k be the row vectors of A. Recall that A has oly 1 or 0 for etries ad that each r i cotais s r 1 s, so dottig r i with some vector adds together s r compoets of that vector. A 1 l 1 2 2 = 1 l 1 ( r 1 1 l 1,..., r k 1 l 1 ) l 2 = (s r,..., s r ) 1 k 2 l = ks2 r l = ( ks r )s r = s c s r l where the last equality follows from s r k = s c l 2.10. We have that A x 2 = s x 2 c s r whe x = 1 l l. If we could show that x R l, A x 2 s x 2 c s r, the we would have that A x 2 reaches its upper boud s x 2 c s r, so its max must be s c s r, ad by 2.11, λ 1 = x T A T A x A x 2 max x R l, x 0 x T = max x x R l, x 0 x 2 = s c s r It remais to show that x R l, A x 2 x 2 s c s r. Let x 1,...x l deote the compoets of x. A x = ( r 1 x,..., r k x) T, so (2.14) A x 2 = k ( r i x) 2 r i x is the sum of s r compoets of x. Let x i1,..., x isr be the s r compoets of x that r i selects to sum. The r i x = sr x ij. r i x = s r x ij = (x i1,..., x isr ) 1 sr 1 1 sr 1 (x i1,..., x isr ) = s r s r (x ij ) 2 where the iequality follows from the Cauchy-Schwarz Iequality, so we have that ( r i x) 2 s r s r (x ij ) 2. Substitutig ito 2.14, (2.15) A x 2 s r k s r (x ij ) 2 Observe that the first summatio cycles through all the row vectors ad, for each row vector r i, the secod summatio cycles through the compoets of x chose by r i. Recall that A has s c 1 s i every colum, so i multiplyig A ad x, every compoet of x is selected by exactly s c row vectors. Hece,
6 QIAN ZHANG k s r l (x ij ) 2 = s c (x i ) 2 = s c x 2 Substitutig ito 2.15, A x 2 s r s c x 2, so x R l, A x 2 x 2 s c s r. Lemma 2.16. Uder the assumptios of 2.13, 1 l l is a eigevector of A T A correspodig to eigevalue λ 1. Proof. Each etry of A 1 l 1 is the sum of a row of A, which is s r, so A 1 l 1 = s r 1 k 1. Similarly, A T 1 k 1 = s c 1 l 1. Hece, A T A 1 l 1 = A T (s r 1 k 1 ) = s r (A T 1 k 1 ) = s r s c 1 l 1 = λ 1 1 l, where the last equality follows by 2.13. We have that A T A 1 l 1 = λ 1 1 l 1, so 1 l 1 is a eigevector of A T A correspodig to eigevalue λ 1. Notatio 2.17. J deotes a matrix with 1 for every etry. Theore.18. (Quasiradomess Theorem) Suppose Γ(L, R, E) is a biregular bipartite graph with L = k ad R = l. Let the degree of every vertex i L be s r ad the degree of every vertex i R be s c. Let X L ad Z R, let p be the desity of Γ, let A be the k l adjacecy matrix of Γ, ad let λ i be the i th eigevalue of A T A i decreasig order. The E(X, Z) p X Z λ 2 X Z Proof. Let x be the icidece vector of X ad let z be the icidece vector of Z. E(X, Z) = x T A z. Cosider the subgraph Γ(X, Z, E(X, Z)). If all vertices i X were coected with all vertices i Z, the umber of edges i the subgraph would be X Z = x T J k l z. E(X, Z) p X Z = x T A z p( x T J k l z) = x T (A pj k l ) z x T (A pjk l ) z = X (A pj k l ) z where the iequality follows by the Cauchy-Schwarz iequality. It remais to show that (A pj k l ) z λ 2 Z i.e. (A pj k l ) z 2 λ 2 Z = λ 2 z 2. (A pj k l ) z 2 = z T (A pj k l ) T (A pj k l ) z = z T (A T pj T k l)(a pj k l ) z = z T (A T A pa T J k l pj T k la + p 2 J T k lj k l ) z We will simplify A T A pa T J k l pjk l T A + p2 Jk l T J k l term-by-term. (Simplifyig Jk l T A) Γ is biregular: Every vertex i R is coected to s c vertices i L, so s c = E l, ad every vertex i L is coected to s r vertices i R, so s r = E k. Put aother way, the etries of each colum of A sum to s c ad the etries of each row of A sum to s r. p = E kl, so: s c = E l s r = E k E kl = (kl) = pkl = pk l l E = kl (kl) = pkl k k = pl
QUASIRANDOMNESS AND GOWERS THEOREM 7 Notice that each etry of J T k l A is s c, which is pk, so J T k l A = pkj l l. (Simplifyig A T J k l ) A T J k l = (J T k l A)T = (pkj l l ) T = pkj l l, where the last equality holds because J l l is symmetric. (Simplifyig J T k l J k l) Each etry of J T k l J k l is the sum of a colum of J k l, which is k, so J T k l J k l = kj l l. Substitutig i for J T k l A, AT J k l, ad J T k l J k l: A T A pa T J k l pj T k la + p 2 J T k lj k l = A T A p(pkj l l ) p(pkj l l ) + p 2 (kj l l ) = A T A p 2 kj l l M By 2.16, 1 is a eigevector of A T A to eigevalue λ 1 = s r s c = (pk)(pl) = p 2 kl. Sice J l l 1 = l 1, (p 2 kj l l ) 1 = p 2 k(j l l 1) = p 2 k(l 1) = (p 2 kl) 1 = λ 1 1. Now cosider M = A T A p 2 kj l l. M 1 = A T A 1 p 2 kj l l 1 = λ 1 1 λ 1 1 = 0 = 0 1 so 1 is a eigevector of M correspodig to eigevalue 0. Also, M = A T A p 2 kj l l = (A T A) T (p 2 kj l l ) T = (A T A p 2 kj l l ) T = M T. Sice M is a symmetric matrix, by the Spectral Theorem, there exists a orthogoal eigebasis to M. Let e i be a vector i this orthogoal eigebasis, so M e i = u i e i, where u i R is a eigevalue of M. Let e 1 1 l, so u 1 = 0. Sice the e i are orthogoal, 1 is orthogoal to e i, i 2. Notice that for i 2, each etry of J l l e i is 1 e i = 0, so J l l e i = 0. Hece, for i 2, M e i = (A T A p 2 kj l l ) e i = A T A e i p 2 k(j l l e i ) = A T A e i. For i 2, u i e i = M e i = A T A e i = λ i e i so u i = λ i for i 2. This implies that the largest eigevalue of M is λ 2, NOT λ 1 : Sice λ i s are ordered by size ad o u i = λ 1 for i 2 ad u 1 = 0, which is ot geerally equal to λ 1 = s r s c 0, o u i ever is λ 1. The ext largest value that a u i ca be is λ 2. (I particular, the largest eigevalue of M is u 2 = λ 2.) z By 2.11, the largest eigevalue of M is max T M z z z T z. zt M z z max T M z z T z z = λ z T z 2 z T M z λ 2 z T z, ad z T z = z z = z 2, so z T M z λ 2 z 2. Recall, (A pj) z 2 = z T (A pj) T (A pj) z = z T (A T A pa T J k l pj T k la + p 2 J T k lj k l ) z = z T M z λ 2 z 2 which is what we eeded to fiish the proof. The smaller λ 2 is, the closer E(X, Z) is to p X Z, so the closer E(X,Z) is to X Z p X Z X Z = p. Notice that E(X,Z) X Z is the desity of the bipartite subgraph formed by X ad Z, Γ(X L, Z R, E(X, Z)). Hece, the Quasiradomess Theorem says that the desity of Γ(X, Z, E(X, Z)) is approximately the desity of the larger graph Γ(L, R, E). Corollary 2.19. Uder the same hypotheses as Theore.18, if p 2 X Z > λ 2, the E(X, Z) > 0.
8 QIAN ZHANG Proof. By 2.18, p 2 X Z > λ 2 p 2 ( X Z ) 2 > λ 2 X Z p X Z > λ 2 X Z p X Z λ 2 X Z > 0 E(X, Z) p X Z λ 2 X Z λ 2 X Z E(X, Z) p X Z p X Z λ 2 X Z E(X, Z) Combiig the above results, 0 < p X Z λ 2 X Z E(X, Z) 0 < E(X, Z) 2.1. How the Quasiradomess Theorem is a quasiradomess result. Defiitio 2.20. A radom graph is a graph whose every pair of vertices is radomly assiged a edge. Pairs assigmets are idepedet of each other. Remark 2.21. A radom bipartite graph is a radom graph such that ay two vertices i the same set have 0 probability of formig a edge. Cosider a radom situatio. Let G(L, R, E ) be a radom bipartite graph, ad let each pair {l, r}, l L ad r R, have probability p of beig a edge. Let X L ad let Z R. Cosider the subgraph g(x, Z, E(X, Z )). The umber of pairs of vertices of g that ca form edges is X Z. Cosiderig the desigatio of edge a success, E(X, Z ), the umber of successes i X Z idepedet trials, would follow a biomial distributio: P ( E(X, Z ) = s) = ( X Z s ) p s (1 p) X Z s. E(X, Z ) would have expected value p X Z, so the desity of g, E(X,Z ) X Z, would have expected value ( ) p X Z X Z = p. By the same argumet, P ( E = s) = L R s p s (1 p) L R s, the expected value of E would be p L R, so the desity of G, E(L,R ) L R, would have expected value p. The desity of G ad the desity of g have the same expected value, but there is o guaratee that the desities be withi some rage of each other. The probability that the desities are wildly differet, say a desity of 0 ad a desity of 1, is ozero. Now cosider biregular bipartite graph Γ(L, R, E) described i the hypotheses of 2.18. The Quasiradomess Theorem says that the desity of subgraph Γ(X L, Z R, E(X, Z)) must be withi some rage 4 of the desity of Γ(L, R, E), so i this sese oe ca expect the desity of Γ(X, Z, E(X, Z)) to be approximately the desity of Γ(L, R, E). Similarly, oe ca expect the desity of G ad the desity of g to be close to each other (i the sese that their expected values are the same), but ulike the desity of Γ(L, R, E) ad the desity of Γ(X, Z, E(X, Z)), the desity of G ad the desity of g are ot ecessarily withi some rage (other tha 1) of each other. 4 The rage is cotrolled by λ2 ad the sizes of X ad Z, ad could be less tha 1. The larger X ad Z are ad the smaller λ 2 is, the closer the desity of the subgraph is to the desity of the larger graph.
QUASIRANDOMNESS AND GOWERS THEOREM 9 Γ(L, R, E) is a quasiradom graph because it is like a radom graph G(L, R, E ). Oe ca expect sufficietly large subgraphs of Γ(L, R, E) to have characteristics (amely desities) similar to characteristics of subgraphs of a radom graph. 3. Gowers theorem Theorem 3.1. (Gowers Theorem - GT) Let G be a group of order G ad let m be the miimum degree of otrivial represetatios of G over the reals. If X, Y, Z G ad X Y Z G 3 m, the x X, y Y, z Z s.t. xy = z. Corollary 3.2. 3.1 would still be true if its coclusio were replaced by XY Z = G Proof. Take X, Y, Z G such that X Y Z G 3 m. XY Z = G meas x X, y Y, z Z, g G s.t. xyz = g ad g G, x X, y Y, z Z, s.t. xyz = g. The first statemet holds by closure of G, so it remais to show the secod statemet. Take g G. Let Z = gz 1. By closure of G, Z G. Sice Z = Z, X Y Z G 3 m. By 3.1, x X, y Y, z Z s.t. xy = z xy(z 1 ) = z (z 1 ) = 1 xy(z 1 g) = g xyz = g. 3.1. Traslatig Gowers Theorem: Provig m Proves Gowers Theorem. Variables i this subsectio refer to those defied i the cotext of Γ(G 2, G 2, E): To prove 3.1, we take a graph theoretic view of it. Let G be a group. Let Γ(G 1, G 2, E) be a bipartite graph with two sets of vertices G 1 ad G 2, which are copies of G. Let there be a edge betwee g 1 G 1 ad g 2 G 2 oly if y Y G s.t. g 1 y = g 2, let A be the G G adjacecy matrix of Γ, let λ 2 be the secod largest eigevalue of A T A, let p be the desity of Γ, let X G 1, ad let Z G 2. 3.1 says that, for sufficietly large X ad Z, there is at least oe edge betwee a member of X ad a member of Z, i.e. E(X, Z) > 0. Curiously, which particular vertices are chose to costitute X ad Z is irrelevat to guarateeig a edge betwee them. Rather, the sizes of X ad Z are all that matter. I this graph theoretic view of Gowers Theorem, the hypotheses of the Quasiradomess Thrm hold. If, i additio, p 2 X Z > λ 2 were to hold, the by 2.19, E(X, Z) > 0, provig Gowers Theorem. To traslate provig GT ito provig some other statemet, we use the followig results: Notatio 3.3. g 1 deotes ay vertex i G 1 ad g 2 deotes ay vertex i G 2. Lemma 3.4. The degree of every vertex of Γ(G 1, G 2, E) is Y Proof. We will show that every vertex i G 1 has degree Y ad every vertex i G 2 has degree Y, so every vertex of Γ has degree Y. Claim: Every g 1 G 1 has degree Y. Sice G is a group, g, y G, gy G so g 1 G 1 = G ad y Y G, g 1 y G = G 2 so g 1 y = g 2 G 2. Every g 1 ca be multiplied by every elemet i Y to get a g 2. g 1, multiplyig g 1 by differet y leads to distict products. Take distict y 1, y 2 Y ad suppose, for a cotradictio, that g 1 y 1 = h ad g 1 y 2 = h. The
10 QIAN ZHANG y 1 = g1 1 h ad y 2 = g1 1 h, so y 1 = y 2, cotradictig the assumptio that y 1 ad y 2 are distict, so g 1 y 1 g 1 y 2. Hece, for each g 1, multiplyig by every y yields Y distict products i G 2. Sice {g 1, g 2 } E iff y Y s.t. g 1 y = g 2, g 1 ca form o other edges, so the degree of every g 1 is Y. Claim: Every g 2 has degree Y. Every g 2 has Y preimages i G 1 : y Y, uique g 1 G 1 s.t. g 1 y = g 2. Take y Y G so y G. Sice G is a group, y 1 G. Take g 2 G 2 = G. By closure, g 2 y 1 G = G 1 so g 1 = g 2 y 1. To cout the umber of g 1 s that form a edge with a g 2, it suffices to cout the umber of y s, which is Y. Corollary 3.5. E = G Y Proof. Every g 1 G 1 forms Y edges, ad there are G g 1 s, so E = G Y Fact 3.6. If A is a real matrix with eigevalues λ 1,..., λ, the T r(a) = λ i Notatio 3.7. λ i deotes oe of the G eigevalues of A T A: { λ 1,..., λ G }, listed i decreasig order. m i deotes the multiplicity of λ i. Corollary 3.8. λ 2 < T r(at A) Proof. By 3.6, T r(a T A) = G λ i = m 1 λ 1 + λ 2 +... > λ 2, where the last iequality follows from A T A havig oegative eigevalues (by 2.8). Lemma 3.9. T r(a T A) = E(X, Z) Proof. Let c 1,..., c G be the colum vectors of A. G T r(a T A) = c j c j = ( c ij ) This double summatio adds all the etries of A, hece couts the umber of edges of Γ(G 1, G 2, E). A alterative view: The secod summatio gives the degree of a particular g 2. The first summatio cycles through all vertices i G 2. Hece, the double summatio couts all the edges that vertices i G 2 are members of, so it couts all the edges of Γ. Corollary 3.10. λ 2 < G Y Proof. λ 2 < T r(at A). The first iequality holds by 3.8, the secod equality holds by 3.9, ad the third equality holds by 3.5. Remark 3.11. p = G Y = E(X,Z) = G Y G G = Y G G G, where the first equality follows from 3.5 ad 2.4. Propositio 3.12. To prove Gowers Theorem, it remais to show that m. Proof. From 3.10, we have that λ 2 < G Y. If we could show that G Y p 2 X Z, the λ 2 < p 2 X Z, fulfillig the hypothesis of 2.19 ad reachig the coclusio of Gowers Theorem. I other words, to prove GT, it remais to prove p 2 X Z. G Y
QUASIRANDOMNESS AND GOWERS THEOREM 11 G Y p 2 X Z G Y ( ) 2 Y G X Z G 3 X Y Z, where the first iff follows from 3.11. To prove GT it remais to prove G 3 X Y Z. Give GT s hypothesis X Y Z G 3 m, if we could show m, the X Y Z G 3. Hece, all we eed to prove GT is m. 3.2. Provig m. Recall that is the multiplicity of λ 2 ad m is the miimum dimesio of otrivial represetatios of G over R i.e. the smallest dimesio of a real vector space i which G has otrivial represetatio. To show that m, we will eed some prelimiary defiitios ad results. Defiitio 3.13. For a group G ad a iteger d 1, a d-dimesioal represetatio of G is a homomorphic map ϕ : G GL(V ), where V is a d- dimesioal vector space, so V = F d, where F is a field. GL(V ) = GL d (F ), which is the geeral liear group, the set of d x d ivertible matrices whose etries are elemets of F ; the set forms a group uder matrix multiplicatio. Sice GL(V ) = GL d (F ), ϕ is a mappig G GL d (F ), so we say ϕ is a represetatio of G over F. d is the dimesio of ϕ. Remark 3.14. A represetatio of G over R is a represetatio of G, ϕ : G GL d (R). To clarify, such a ϕ maps elemets of G to d d ivertible matrices with etries from R. Such matrices correspod to ivertible mappigs from R d to R d. Defiitio 3.15. Let V be a d-dimesioal vector space. U V is ivariat uder ϕ : G GL(V ) if g G, U is ivariat uder ϕ(g), i.e. u U, g G, ϕ(g)u U. I other words, every mappig that ϕ associates with a elemet of G maps U to U. The trivial ivariat subspaces are the zero subspace (whose oly elemet is 0 R d ) ad V. Defiitio 3.16. ϕ : G GL d (R) is a trivial represetatio if it maps every elemet of G to the idetity trasformatio. Defiitio 3.17. If λ F ad A is a x matrix over F, the the eigespace to eigevalue λ is U λ = { x F s.t. A x = λ x}. A member of the eigespace is called a eigevector correspodig to λ. Lemma 3.18. If AB = BA, the every eigespace of A is ivariat uder B. Proof. Let U λ be a eigespace of A. We wat to show that x U λ, B x U λ. Sice x U λ, A x = λ x, so AB x = BA x = B(λ x) = λb x. Defiitio 3.19. A eigebasis of a matrix A is a set of eigevectors of A that forms a basis for the domai of the liear trasformatio correspodig to A. Theorem 3.20. (Spectral Theorem) Every real symmetric matrix has a orthogoal eigebasis. Notatio 3.21. Give mappig f : A B ad C A, f C deotes the mappig that is the same as f, except with domai restricted to C. Hom(A,B) deotes the set of homomorphisms from A to B.
12 QIAN ZHANG Propositio 3.22. Let A = A T be a real d d matrix, ad G a group. Let m = mi{s : φ otrivial Hom(G, GL s (R))}, i.e. m is the miimum dimesio of otrivial represetatios of G over the reals. Let ϕ Hom(G, GL d (R)) be otrivial. Suppose that A commutes with all matrices i GL d (R). The there is a eigevalue of A with multiplicity at least m. Proof. By 3.20, we ca choose a particular eigebasis of A. Call this basis B A = { e 1,..., e d }. Pick g 0 G, such that ϕ(g 0 ) is ot the idetity matrix. Let ψ : R d R d be the uique liear map whose trasformatio matrix with respect to B A is ϕ(g 0 ). ϕ(g 0 ) is ot the idetity matrix, so ψ is ot the idetity map o R d. Sice A commutes with every elemet of GL d (R), i particular it commutes with ϕ(g 0 ), so by 3.18, ψ seds each eigespace of A to itself. ψ caot act as the idetity o every U λ, because if it did, the v R d, v = d α i e i where α i R, ad ψ( v) = ψ( d α i e i ) = d α i ψ( e i ) = d α i e i = v so ψ would act as the idetity o R d, which is cotrary to the choice of ψ. We ve show by cotradictio that there must be a eigespace U λ such that ψ : U λ U λ is ot the idetity map. Because ψ Uλ is ot the idetity map, ϕ(g 0 ) Uλ is ot the idetity matrix, so ϕ : g ϕ(g) Uλ is a otrivial represetatio of G. Note that ϕ : g ϕ(g) Uλ meas ϕ : G GL(U λ ) = GL dim(uλ )R so the dimesio of ϕ is the dimesio of U λ. By defiitio, m is the miimum dimesio of otrivial represetatios of G, so the dimesio of ϕ (which is the dimesio of U λ ) is at least m. Sice A is symmetric, the dimesio of U λ is the multipliticy of λ, so the multiplicity of λ is at least m as desired. Defiitio 3.23. σ : V V is a permutatio o set V if it is a bijectio from V to V. Defiitio 3.24. Cosider a graph G = (V, E). A graph automorphism is a mappig σ : V V that preserves adjacecy, i.e. i, j V, i j σ(i) σ(j) Remark 3.25. A graph automorphism of a bipartite graph Γ(V 1, V 2, E) cosists of permutatios σ 1 : V 1 V 1 ad σ 2 : V 2 V 2 s.t. v 1 V 1 ad v 2 V 2, v 1 v 2 σ 1 (v 1 ) σ 2 (v 2 ). Defiitio 3.26. P (σ) is a permutatio matrix of permutatio σ if { 1 if σ(i) = j P (σ) ij = 0 otherwise. Lemma 3.27. Let Γ(V 1, V 2, E) be a biregular bipartite graph, let A be its adjacecy matrix, let σ 1 be a permutatio of V 1, ad let σ 2 be a permutatio of V 2. The σ 1 ad σ 2 costitute a bipartite graph automorphism iff P (σ 1 )A = AP (σ 2 ) Proof. The claim is that iɛv 1, jɛv 2, i j σ 1 (i) σ 2 (j) P (σ 1 )A = AP (σ 2 ) We will traslate the right-had side ito some other statemet. By defiitio, P (σ 1 )A = AP (σ 2 ) i, j, [P (σ 1 )A] ij = [AP (σ 2 )] ij.
QUASIRANDOMNESS AND GOWERS THEOREM 13 For all i, j, [AP (σ 2 )] ij = L l=1 A ilp (σ 2 ) lj. Notice that cells of A ad cells of P oly take values 1 or 0, so terms of the sum are either 1 or 0. The summatio is equivalet to summig oly the terms that are 1. For a term to be 1, A il ad P (σ 2 ) lj must both be 1. By defiitio, A il = 1 iff i l, ad P (σ 2 ) lj = 1 iff σ 2 (l) = j. Hece, A il P (σ 2 ) lj = 1 iff i l ad σ 2 (l) = j, so L A il P (σ 2 ) lj = A ilp (σ 2 ) lj. l=1 l s.t. i l=σ 1 2 (j) Multiple l s ca be adjacet to i, but sice σ 2 is oe-to-oe, oly oe l ca equal σ 1 2 (j), so [AP (σ 2 )] ij = For all i,j,[p (σ 1 )A] ij = l s.t. i l=σ 1 2 (j) A ilp (σ 2 ) lj = K k=1 { 1 if i σ 1 2 (j) 0 otherwise P (σ 1 ) ik A kj. The terms of this sum are either 1 or 0, so the sum is equivalet to summig oly the terms that are 1. For a term to be 1, P (σ 1 ) ik = 1 iff σ 1 (i) = k, ad A kj = 1 iff k j. Hece, P (σ 1 ) ik A kj = 1 iff σ 1 (i) = k ad k j, so K P (σ 1 ) ik A kj = k=1 k s.t. σ 1(i)=k j P (σ 1 ) ik A kj. Multiple k could be adjacet to j, but sice σ 1 is oe-to-oe, oly oe k = σ 1 (i). Hece, the summatio ca have oly oe term that is 1, so { 1 if σ1 (i) j [P (σ 1 )A] ij = P (σ 1 ) ik A kj = 0 otherwise k s.t.σ 1(i)=k j For all i,j [P (σ 1 )A] ij = [AP (σ 2 )] ij iff the cells are both 1 or both 0 iff ( σ 1 (i) j ad i σ2 1 (j)) or ( σ 1 (i) j ad i σ2 1 (j)) Hece, σ 1 (i) j is equivalet to i σ2 1 (j). To summarize, P (σ 1 )A = AP (σ 2 ) meas i, j, σ 1 (i) j iff i σ2 1 (j), so the lemma says: i V 1, j V 2, i j σ 1 (i) σ 2 (j) i V 1, j V 2, σ 1 (i) j i σ 1 2 (j) ( )Suppose (3.28) i V 1, j V 2, i j σ 1 (i) σ 2 (j). We wat to show σ 1 (i) j i σ 1 2 (j). (3.29) σ 1 (i) j σ 1 (i) σ 2 (σ2 1 (j)) i σ 1 2 (j) where the last equivalece comes from the directio of 3.28
14 QIAN ZHANG ( )Suppose (3.30) i V 1, j V 2, σ 1 (i) j i σ 1 2 (j) We wat to show i j σ 1 (i) σ 2 (j). (3.31) i j i σ 1 2 (σ 2(j)) σ 1 (i) σ 2 (j) where the last equivalece comes from the of 3.30 Claim 3.32. Let σ be a permutatio ad let P be its permutatio matrix. P T = P 1. Proof. The claim is that P P T = P T P = I. Recall that P (σ) ij is 1 if σ(i) = j ad is 0 otherwise. Sice σ is a fuctio, every row vector of P has oly oe etry that is 1. Sice σ is bijective, every colum vector of P has oly oe etry that is 1. No two row vectors ca have same the same compoet be 1, because if there were two such row vectors, there would be a colum vector with more tha oe 1-etry, cotradictig that every colum vector has oly oe 1-etry. Similarly, o two colum vectors ca have the same compoet be 1. Hece, every pair of distict row vectors of P is orthogoal ad every pair of distict colum vectors of P is orthogoal. Let the rows of P be r 1,..., r. (P P T ) ii = r i r i = r ij = 1, sice every row vector has oly oe etry that is 1. For i j, (P P T ) ij = r i r j = 0 sice row vectors are orthogoal. Hece, P P T = I. Let the colums of P be c 1,..., c. (P T P ) ii = c i c i = c ij = 1, sice every colum vector has oly oe etry that is 1. For i j, (P P T ) ij = c i c j = 0, sice colum vectors are orthogoal. Hece, P T P = I. Lemma 3.33. Let σ 1, σ 2 costitute a graph automorphism. The P (σ 2 ) commutes with A T A. Proof. P (σ 2 ) 1 A T AP (σ 2 ) = P (σ 2 ) T A T (I k k )AP (σ 2 ) = P (σ 2 ) T A T (P (σ 1 )P (σ 1 ) 1 )AP (σ 2 ) = P (σ 2 ) T A T (P (σ 1 )P (σ 1 ) T )AP (σ 2 ) = (P (σ 2 ) T A T P (σ 1 ))(P (σ 1 ) T AP (σ 2 )) = (P (σ 2 ) 1 A T P (σ 1 ))(P (σ 1 ) 1 AP (σ 2 )) = A T A We have P (σ 2 ) 1 A T AP (σ 2 ) = A T A, so A T AP (σ 2 ) = P (σ 2 )A T A. Now cosider the particular bipartite graph Γ(G 2, G 2, E) ivolved i the proof of Gowers Theorem. A is its bipartite adjacecy matrix, which is a G x G real matrix, so A T A is a G x G real matrix. Let λ 2 deote the secod largest eigevalue of A T A. Choose σ 1, σ 2 that costitute a graph automorphism. Let ϕ : g P (σ 2 ) be a otrivial represetatio of G, i.e. let ϕ map some g to a P (σ 2 ) that is ot the idetity matrix. Let ψ be the liear trasformatio correspodig to this P (σ 2 ). Usig a argumet similar to that i the proof of Propositio 3.27, we will show that m.
QUASIRANDOMNESS AND GOWERS THEOREM 15 Remark 3.34. Recall defiitio 3.23. U λ2 (A T A) { x R G s.t. A T A x = λ 2 x } Propositio 3.35. m Proof. P (σ 2 ) is ot the idetity matrix, so ψ does ot act as the idetity o R G. Suppose we could show that ψ does ot act as the idetity o U λ2 (A T A). The P (σ 2 ) Uλ2 (AT A) would ot be the idetity matrix. Hece, ϕ Uλ2 (AT A) : G P (σ 2 ) Uλ2 (AT A) would be a otrivial represetatio of G, so its dimesio would be at least the miimum dimesio of a otrivial represetatio of G i.e. m. By 3.33, P (σ 2 ) commutes with A T A, so by 3.18, U λ2 (A T A) is ivariat uder P (σ 2 ), so P (σ 2 ) Uλ2 (AT A) = GL(U λ2 (AT A)), so ϕ Uλ2 (AT A) : G GL(U λ2 (AT A)), so dim(u λ2 (AT A)) = the dimesio of ϕ Uλ2 (AT A), which we already showed is at least m. Sice A T A is symmetric, the = dim(u λ2 (AT A)), which is at least m, so m. It remais to show that ψ does ot act as the idetity o U λ2 (A T A). The oly way a ψ that is ot the idetity trasformatio ca act as the idetity o U λ2 (A T A) is if each vector i U λ2 (A T A) has idetical compoets, i.e. is a multiple of 1. (To see this, ote that ψ does ot act as the idetity o U λ2 (A T A) iff P (σ 2 ) Uλ2 (A T A) is ot the idetity matrix.) By 2.16, A T A 1 = λ 1 1 λ 2 1, so for c R, A T A(c 1) λ 2 (c 1) so o multiple of 1 is i U λ2 (A T A), so ψ caot act as the idetity o U λ2 (A T A). 3.35 fiishes the proof of Gowers Theorem. Ackowledgemets Thaks to my metors Irie Peg ad Marius Beceau for their help. Refereces [1] L. Babai. Discrete Math Lecture Notes. http://people.cs.uchicago.edu/ laci/reu07/. [2] J. A. Rice. Mathematical Statistics ad Data Aalysis. Duxbury Press. 2006.