Comparability of random permutations in the strong Bruhat order

COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STR ONG BR UHA T ORDER NICHOLAS CHRISTO AND MARCUS MICHELEN Abstract. The (strong) Bruhat order for p ermutations provides a partial ordering defined as follows: tw o permutations are comparable if one can b e obtained from the other by a sequence of adjacen t transp ositions that each increase the n um ber of inversions by 1. Given t wo random permutations, what is the probabilit y that they are comparable in the Bruhat order? This problem w as first considered in a 2006 w ork of Hammett and Pittel, whic h show ed an exp onential low er bound and a p olynomial upp er b ound. The lo w er b ound was very recently improv ed to the sub exp onential bound of exp( − n 1 / 2+ o (1) ) by Boretsky , Cornejo, Hodges, Horn, Lesnevich, and McAllister. Hammett and Pittel predicted that the probabilit y should decrease polynomially . W e sho w that the probability decreases faster than any p olynomial and is on the order of exp( − Θ(log 2 n )). 1. Introduction The (str ong) Bruhat or der is a partial order in the symmetric group: tw o p ermutations are comparable if one can b e obtained from the other by a sequence of adjacen t transp ositions that increase the n um b er of in v ersions by 1. An easy-to-chec k criterion for comparability is the so-called (0 , 1)-matrix criterion: giv en p erm utations π and τ , let M π and M τ b e the corresponding p ermutation matrices. Then the Bruhat order ≤ is defined via [ 2 , Thm. 2.1.5] (1) π ≤ τ ⇐ ⇒ X i ≤ a, j ≤ b M τ ( i, j ) ≤ X i ≤ a, j ≤ b M π ( i, j ) for all a, b ≤ n . The Bruhat order is a central ob ject in algebraic combinatorics and plays a leading role in the study of Sc h ub ert v arieties. This line of w ork began with Ehresmann’s 1934 w ork [ 6 ] whic h included a c haracterization of the Bruhat order in terms of the increasing rearrangements of π and τ that is equiv alen t to ( 1 ). There are many further equiv alent descriptions of the Bruhat order [ 1 , 4 , 5 , 8 , 12 ] including criteria of a similar com binatorial flav or to ( 1 ) and of a more algebraic fla v or. In this note, we study the probabilit y that tw o random p ermutations are comparable. In particular, if we tak e π and τ to b e indep endent and uniformly chosen from the symmetric group S n , what is the probability that π ≤ τ ? This problem was introduced in a 2006 work of Hammett and Pittel [ 9 ] which show ed c ( . 708) n ≤ P ( π ≤ τ ) ≤ C n − 2 for constants C , c > 0. V ery recen tly 1 , Boretsky-Cornejo-Ho dges-Horn-Lesnevic h-McAllister [ 3 ] improv ed the low er b ound to the form exp( − c √ n log 3 / 2 n ). In their work, Hammett and Pittel stated that “Empirical estimates...suggest that P ( π ≤ τ ) is of order n − (2+ δ ) for δ close to 0 . 5.” Our main theorem shows that in fact the probability decreases faster than any p olynomial at the rate exp( − Θ(log 2 n )): Theorem 1. L et π , τ ∈ S n b e indep endently and uniformly chosen at r andom. Ther e ar e c onstants C , c > 0 so that for n sufficiently lar ge we have exp  − C log 2 n  ≤ P ( π ≤ τ ) ≤ exp  − c log 2 n  . 1 The work [ 3 ] app eared on arXiv while the present work was nearing completion. The main fo cus of the work [ 3 ] is showing that if one considers a different order on S n known as the we ak Bruhat or der ≤ W then P ( π ≤ W τ ) = exp( − (1 / 2 + o (1)) n log n ). 1 2 NICHOLAS CHRISTO AND MAR CUS MICHELEN The main idea is to define Z ( a, b ) = X i ≤ a, j ≤ b ( M π ( i, j ) − M τ ( i, j )) then ( 1 ) may b e rewritten as (2) Z ( a, b ) ≥ 0 for all a, b ≤ n . W e treat ( 2 ) as a t w o-dimensional p ersistence ev en t. As a useful point of comparison, a classical persistence ev en t is asking for the probability that a simple random w alk is non-negativ e for the first n steps. A simple random walk has only one time index, while ( 2 ) has tw o, and so there is not as simple of a combinatorial approac h as in the case of a simple random w alk. Ho wev er, t w o-dimensional persistence problems for Gaussian pro cesses—such as the Bro wnian sheet—hav e received a fair amoun t of attention. Our approaches to the upp er and lo w er b ounds are inspired b y these metho ds from probabilit y theory , and in fact our pro of of the upp er bound uses some of the machinery for analyzing Gaussian pro cesses to b ound our problem ( 2 ) in terms of an analogous Gaussian problem. W e pro ve the upper and lo w er bounds separately in Section 3 and Section 4 as Theorem 6 and Theorem 13 resp ectiv ely . W e b egin with a sketc h of our argumen ts along with a more detailed conjecture ab out the b eha vior of P ( π ≤ τ ) (see ( 6 )). 1.1. Pro of outline. T o first see heuristically that exp( − Θ(log 2 n )) is the correct probabilit y for the even t in ( 2 ), note that Z ( a, b ) is a mean-zero random v ariable with v ariance ≈ ab/n . F urther, while Z ( a, b ) is not strictly a sum of indep endent random v ariables, one ma y show that Z ( a, b ) obeys a cen tral limit theorem and so is close to a Gaussian random v ariable of v ariance ≈ ab/n . As a consequence, P ( Z ( a, b ) ≥ 0) ≈ 1 / 2 for each fixed ( a, b ) (once ab/n ≫ 1). When ( a 1 , b 1 ) and ( a 2 , b 2 ) are close together, the random v ariables Z ( a 1 , b 1 ) and Z ( a 2 , b 2 ) are quite correlated. The idea is to find a set of ( a, b ) so that all pairwise correlations are uniformly b ounded a w a y from 1. In particular, if w e consider S 0 = { ρ i : i ∈ { 0 , 1 , 2 , . . . }} ∩ [ n ] for some fixed integer ρ ≥ 2, then for all pairs ( a 1 , b 1 ) , ( a 2 , b 2 ) ∈ S 2 0 with ( a 1 , b 1 )  = ( a 2 , b 2 ) we hav e (3) | E [ Z ( a 1 , b 1 ) Z ( a 2 , b 2 )] | p V ar( Z ( a 1 , b 1 )) V ar( Z ( a 2 , b 2 )) ≤ 1 − δ for some δ > 0 depending on ρ . Here one migh t imagine that across all ( a, b ) ∈ S 2 0 , the random v ariables ( Z ( a, b )) are sufficiently indep endent so that (4) P ( Z ( a, b ) > 0 for all ( a, b ) ∈ S 2 0 ) ≈ Y ( a,b ) ∈S 2 0 P ( Z ( a, b ) > 0) = exp( − Θ( |S 2 0 | )) = exp( − Θ(log 2 n )) . While such a strong approximate indep endent statemen t do es not precisely hold, this heuristic provides the correct shap e of the pro of of b oth the upp er b ound and low er b ound, whose techni cal details are quite differen t. F or the upp er b ound, it turns out that if ( Z ( a, b )) ( a,b ) ∈S 2 0 is a Gaussian vector, then an assumption similar to ( 3 ) do es imply an upp er b ound similar to ( 4 ), with a loss that is exp onen tial in |S 0 | 2 . This is due to a theorem of Li and Shao [ 13 ] whic h we restate in Theorem 12 . Ho w ev er, it is a highly nontrivial task to approximate ( Z ( a, b )) ( a,b ) ∈S 2 0 with Gaussians simultaneously , esp ecially since Z ( a, b ) are not sums of indep enden t and identically distributed (i.i.d.) random v ariables. Our first step tow ards a Gaussian comparison is to show that if we only consider submatrices of M π ( i, j ) up to N = ⌊ n 7 / 12 ⌋ , then the distribution of M π is essentially the same as an N × N random matrix whose en tries are i.i.d. Bernoulli(1 /n ) v ariables. In order to understand our choice of N , note that the exp ected n um b er of 1’s in the first N × N submatrix of M π is N 2 /n . If w e hop e to show that ( 2 ) is rare, we need to hav e that the num b er of 1’s is diverging, and so we require N ≫ n 1 / 2 . Note ho wev er that in an N × N matrix with Bernoulli(1 /n ) entries, the n um b er of rows or columns with at least t w o 1’s is of order ≈ N · ( N/n ) 2 = N 3 /n 2 COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STRONG BRUHA T ORDER 3 and so w e need N ≪ n 2 / 3 . Balancing the requiremen ts of N ≫ n 1 / 2 and N ≪ n 2 / 3 yields the c hoice of N = ⌊ n 7 / 12 ⌋ . W e sho w that we may replace an N × N submatrix of M π and M τ with i.i.d. Bernoulli(1 /n ) en tries, which thus allo ws us to view Z ( a, b ) as a sum indep endent random v ariables. This is p erformed in Theorem 7 . Finally , we lean on a strong appro ximation theorem of Rio [ 15 ] (see Theorem 8 ) whic h allows us to replace the underlying random v ariables with standard Gaussian random v ariables, once we group enough random v ariables together to attain v ariance of order 1. The work of Rio [ 15 ] may b e viewed as a tw o-dimensional analogue of the Koml´ os-Ma jor-T usnady (KMT) coupling, which shows that a random walk and Bro wnian motion ma y b e coupled together so the entire tra jectories remain close. Altogether, this completes the upp er b ound. F or the low er b ound, we again will consider dyadic scales. Ho w ever, we hav e an additional to ol that is helpful for sho wing a version of ( 4 ): each ev ent Z ( a, b ) > 0 is an incr e asing ev ent (in the Bruhat order) in the p erm utation π and de cr e asing in τ . A correlation inequalit y of Johnson-Leader-Long [ 10 ] (see Theorem 16 ) states that suc h even ts are p ositively c orr elate d in the sense that P ( A ∩ B ) ≥ P ( A ) P ( B ) . This may b e understo od as a v ersion of the F ortuin-Kasteleyn-Ginibre (FKG) inequality for the Bruhat order. Our first use of this correlation inequality is to sho w P ( Z ( a, b ) ≥ 0 for all a, b ≤ n ) ≥ P ( Z ( a, b ) ≥ 0 for all a, b ≤ ⌈ n/ 2 ⌉ ) 4 (see Theorem 18 ). F or a low er b ound on the resulting probability , we will give ourselves Ω(log n ) of wiggle ro om at a cost of exp( − Θ(log 2 n )) by conditioning on the first and last Θ(log n ) entries of the p erm uta- tions underlying Z . After breaking in to dy adic rectangles and iterativ ely using the correlation inequality Theorem 16 , we see it is enough to show P  min ( a,b ) ∈ [ x, 5 x/ 4] × [ y , 5 y/ 4] Z ( a, b ) ≥ − C log n  ≥ c . uniformly in x, y ≤ n/ 2 . If for a set S ⊂ [ n ] × [ n ] w e write Z ( S ) = P ( i,j ) ∈ S ( M π ( i, j ) − M τ ( i, j )) then for ( a, b ) ∈ [ x, 5 x/ 4] × [ y , 5 y / 4] w e can break up Z ( a, b ) = Z ( x, y ) + Z ([ x ] × [ y , b ]) + Z ([ x, a ] × [ y ]) + Z ([ x, a ] × [ y , b ]) . The idea is that t ypically the maximum of the last three terms is order p xy /n + log n . Conditioning on a typical even t for these last three terms, w e then may use a cen tral limit theorem to show that Z ( x, y ) is larger than any fixed multiple of p xy /n with probabilit y b ounded b elow. The cen tral challenge is showing that the maximum of the last three terms is not to o large. The middle tw o terms are easily handled by F reedman’s inequality for martingales (see Theorem 19 ). The last requires a more delicate approach. T o show E max ( a,b ) ∈ [ x, (5 / 4) x ] × [ y , (5 / 4) y ] | Z ([ x, a ] × [ y , b ]) | ≤ C  p xy /n + log n  w e use a chaining approac h, a technique used to b ound the exp ected maximum of a sto chastic pro cess. W e no w provide a quick description of how chaining works in our context; for a more extensive background on c haining, se e [ 16 , Chapter 8] and the references therein. F or simplicity of notation, supp ose we are interested in b ounding E max ( a,b ) ∈ [ x ] × [ y ] | Z ( a, b ) | . The strategy is to break up the set of ( a, b ) ∈ [ x ] × [ y ] into dyadic scales, e.g. the k th scale consists of all ( i · x/ 2 k , j · y / 2 k ) for i ≤ 2 k , j ≤ 2 k . The idea is to approximate each p oint ( a, b ) at eac h dy adic scale and note that the difference b etw een the approximations at scales k and k + 1 is a random v ariable of v ariance O ( xy / ( n 2 k )). F urther, the num ber of p oin ts at scale k is increasing exponentially . The exp onential decrease in v ariance leads to a strong enough tail b ound (due to F reedman’s inequality) that one can union b ound o v er all p oin ts at scale k and k + 1 to uniformly con trol this error. W e note t wo subtle p oin ts: one reason why c haining works in this case is implicitly b ecause the set of rectangles has b ounde d V C (V apnik–Chervonenkis) dimension , which is an assumption on the complexit y 4 NICHOLAS CHRISTO AND MAR CUS MICHELEN on the set of indices that allows one to construct efficien t nets; in this case, our efficien t nets are simply the dyadic rectangles. Chaining in general works w ell on classes of sets of b ounded VC dimension and w e refer the reader to [ 16 , Chapter 8.3] for more context. The other subtle p oint is that since w e are dealing with sums of sparse random v ariables, the tail b ehavior of Z ( a, b ) is Poisson-lik e rather than sub-gaussian: in some regimes the tail is gaussian-lik e and in others it is only exp onen tial. This complicates the c haining picture slightly . Essentially , w e only c hain down to the scale at whic h the exponential tail ov ertak es the sub-gaussian tail; at this p oin t, we use the exponential tail to obtain a logarithmic error, whic h we are able to tolerate due to our Ω(log n ) wiggle ro om. W e handle the c haining argumen t in Section 4.3 . As a final note, we sp eculate on the asymptotic b ehavior of P ( π ≤ τ ) . F or the Brownian sheet B ( s, t ) it w as pro v en by Molchan [ 14 ] that there is some ψ > 0 so that (5) P  min s,t ≤ T B ( s, t ) ≥ − 1  = exp  − ( ψ + o (1)) log 2 T  . W e conjecture that in fact (6) P ( π ≤ τ ) = exp ( − ( ψ + o (1)) log 2 n ) for the same ψ as in ( 5 ). As a justification for ( 6 ), we first suggest heuristically that P ( π ≤ τ ) = exp( − o (log 2 n )) P ( Z ( a, b ) ≥ 0 for all a, b ∈ [ n 1 − o (1) ]) 4 . Since Z ( n 1 − o (1) , n 1 − o (1) ) has v ariance on the order of n 1+ o (1) , one exp ects that (7) ( Z ( a, b )) ( a,b ) ∈ [ n 1+ o (1) ] × [ n 1+ o (1) ] ≈ ( B ( s, t )) ( s,t ) ∈ [ n 1 / 2+ o (1) ] × [ n 1 / 2+ o (1) ] . In fact, our pro of of the upp er b ound of Theorem 1 can b e understo od as showing that the approximate distributional identit y in ( 7 ) holds up to the smaller scale of ( a, b ) ∈ [ n 7 / 12 ] × [ n 7 / 12 ]. Since log 2 ( n 1 / 2+ o (1) ) = (1 / 4 + o (1)) log 2 n , this provides a heuristic justification for our conjecture ( 6 ). 2. Preliminaries Let π , τ ∈ S n b e indep endently and uniformly chosen at random. W e will often write P n in order to denote the dep endence on n . Let M π and M τ denote the p ermutation matrices corresp onding to π and τ . F or a set A ⊂ [ n ] × [ n ] let X ( A ) = P ( i,j ) ∈ A M π ( i, j ), Y ( A ) = P ( i,j ) ∈ A M τ ( i, j ) and Z ( A ) = X ( A ) − Y ( A ). W e will also write X ( i, j ) = X ([ i ] × [ j ]) and define Y ( i, j ) and Z ( i, j ) similarly . Then Theorem 1 is equiv alent to showing P n ( Z ( i, j ) ≥ 0 for all i, j ∈ [ n ]) = exp  − Θ(log 2 n )  . F or a set A ⊂ [ n ] × [ n ] we hav e E X ( A ) = | A | /n and so we define e X ( A ) = X ( A ) − | A | /n and define e Y ( A ) analogously . As a result, for each fixed A w e hav e E e X ( A ) = 0. F or eac h rectangle A , we will see that X ( A ) is a hyp er ge ometric random v ariable. F or parameters N and A, B ≤ N , a hypergeometric random v ariable ξ ∼ Hyp erGeom( N , B , A ) may be defined com binatorially as follo ws: given N total ob jects with A red ob jects and N − A blue ob jects, pick B ob jects without replacemen t and let ξ be the num b er of red ob jects pick ed. F or ξ ∼ Hyp erGeom( N , B , A ) we recall (8) E ξ = AB N , V ar( ξ ) = AB ( N − A )( N − B ) N 2 ( N − 1) . W e first claim that for any b o x B , X ( B ) is a hypergeometric random v ariable. Throughout, for a p erm utation matrix P and a set S ⊂ [ n ] × [ n ] we write P S to b e the p ortion of the matrix indexed by S and | P S | to b e the n um b er of ones in P S . Lemma 2. L et P b e a r andom p ermutation matrix. F or any b ox B = ( a 1 , a 2 ] × ( b 1 , b 2 ] we have that | P B | ∼ HyperGeom( n, b 2 − b 1 , a 2 − a 1 ) . COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STRONG BRUHA T ORDER 5 Pr o of. Consider the n p ossible columns given by n co ordinate v ectors. The set of columns giving B consists of b 2 − b 1 columns sampled uniformly without replacement from these n total. A column contains a 1 if it is from the a 2 − a 1 columns con taining a 1 in a ro w in ( a 1 , a 2 ]. This precisely matches the definition of the h yp ergeometric distribution. □ W e will require tail bounds for h ypergeometric random v ariables as well as a cen tral limit theorem. W e will ultimately deduce a Bernstein-like tail b ound from F reedman’s inequality . W e recall a version of F reedman’s inequalit y that not only b ounds a martingale but also bounds the running maxim um of a martingale [ 7 , (1.6)]: Theorem 3 (F reedman’s inequalit y) . L et ( X k , F k ) k ≥ 0 b e a martingale with incr ements ξ k = X k − X k − 1 . Supp ose that | ξ k | ≤ M almost sur ely. Define the quadr atic variation V n = n X k =1 E [ ξ 2 k | F k − 1 ] . Then for al l t, s > 0 we have P ( ∃ k ≤ n : | X k − X 0 | ≥ t and V k ≤ s 2 ) ≤ 2 exp  − t 2 2( s 2 + M t/ 3)  . F rom F reedman’s inequality , a b ound on hypergeometric random v ariables follows quickly: Corollary 4. L et a ≤ b ≤ 3 n/ 4 and set X ∼ Hyp erGeom( n, b, a ) . Then P ( | X − E X | ≥ t ) ≤ 2 exp  − 1 16 min  t 2 ab/n , t  . Pr o of. Recall that we may sample X by drawing b ob jects total from a collection of n ob jects consisting of a red ob jects and n − a blue ob jects. Set F j to b e the σ -field generated b y the first j dra ws and define X k = E [ X | F k ]. Note that X k − X k − 1 is given by a centered Bernoulli v ariable, whose success parameter p k is at most a/ ( n − b ) ≤ 4 a n . This sho ws | X k − X k − 1 | ≤ 1 and that V b ≤ 4 ab n almost surely . Applying Theorem 3 sho ws P ( | X − E X | ≥ t ) = P ( | X b − X 0 | ≥ t ) ≤ 2 exp  − t 2 2(4 ab/n + t/ 3)  ≤ 2 exp  − 1 16 min  t 2 ab/n , t  . □ W e require a central limit theorem for h yp ergeometric random v ariables. W e isolate the following state- men t, whic h follows from e.g. [ 11 , Thm. 2.2] Theorem 5. F or e ach M ≥ 0 ther e ar e c onstants c, C > 0 so that the fol lowing holds. L et W ∼ Hyp erGeom( n, a, b ) and set σ 2 = V ar( W ) . If V ar( W ) ≥ C then P ( W − E W ≥ M σ ) ≥ c . 3. Upper Bound W e prov e the upp er b ound of Theorem 1 . Prop osition 6. Ther e is a universal c onstant c > 0 so that for n ≥ 2 the fol lowing holds. L et π , τ ∈ S n b e indep endently and uniformly chosen at r andom. Then P n ( π ≤ τ ) ≤ exp( − c (log n ) 2 ) . W e first compare the permutation matrices underlying Z with indep endent matrices with i.i.d. Bernoulli(1 /n ) en tries. W e show that on an appropriately chosen submatrix, the likelihoo d ratio is essentially 1. 6 NICHOLAS CHRISTO AND MAR CUS MICHELEN Lemma 7. Set N = ⌊ n 7 / 12 ⌋ and let M b e an N × N matrix with entries in { 0 , 1 } . L et P b e a r andom n × n p ermutation matrix and Q a r andom N × N matrix with i.i.d. Bernoulli(1 /n ) entries. Then uniformly among al l M with P i,j M ( i, j ) ≤ n 1 / 5 so that ther e is at most a single 1 in e ach r ow and c olumn we have P ( P [ N ] × [ N ] = M ) = (1 + o (1)) P ( Q = M ) . W e prov e Theorem 7 in Section 3.1 . Our next step is to use a theorem of Rio in order to couple sums o v er rectangles in i.i.d. matrices with sub exponential entries with a corresponding sum with Gauss ian entries. The following theorem is a consequence of [ 15 , Theorem 2.4]: 2 Theorem 8. L et ξ b e a r andom variable with E ξ = 0 , E ξ 2 = 1 and E e tξ ≤ 2 for al l | t | ≤ c 0 . L et ξ i,j b e an arr ay of i.i.d. c opies of ξ . Then ther e is a c onstant C > 0 dep ending only on c 0 so that we may c ouple ξ i,j with an arr ay of i.i.d. standar d Gaussian r andom variables g i,j so that P   max a,b ≤ n       X i ≤ a,j ≤ b ( ξ i,j − g i,j )       ≥ C log 2 n   ≤ 2 e − log 2 n . Finally , we will require an upp er bound on the Gaussian p ersistence problem: Lemma 9. F or e ach C > 0 ther e is a c onstant c > 0 so that P   min a,b ≤ n X i ≤ a,j ≤ b g i,j ≥ − C log 2 n   ≤ 2 exp  − c log 2 n  . W e show that Theorem 9 follows from a general theorem by Li and Shao [ 13 ] for Gaussian pro cesses in Section 3.2 . W e no w deduce the upp er b ound Theorem 6 . Pr o of of The or em 6 . W e first perform our Bernoulli replacemen t. Set N = ⌊ n 7 / 12 ⌋ and define T = { X ( N , N ) ≤ n 1 / 5 , Y ( N , N ) ≤ n 1 / 5 } . Theorem 4 shows P ( T c ) ≤ exp( − n Ω(1) ) ≤ exp( − log 2 n ) . W e then b ound: (9) P n ( π ≤ τ ) = P n  min a ≤ n,b ≤ n Z ( a, b ) ≥ 0  ≤ P n  min a ≤ N ,b ≤ N Z ( a, b ) ≥ 0 ∩ T  + e − log 2 n . Consider the random v ariables ζ i,j with common distribution ζ given by P ( ζ = 1) = P ( ζ = − 1) = 1 n  1 − 1 n  and P ( ζ = 0) = 1 − 2 n  1 − 1 n  and note that ζ is the difference of t wo indep endent copies of Bernoulli(1 /n ) random v ariables. By Theorem 7 , for n sufficiently large we hav e P n  min a ≤ N ,b ≤ N Z ( a, b ) ≥ 0 ∩ T  ≤ 2 P   min a ≤ N ,b ≤ N X i ∈ [ a ] ,j ∈ [ b ] ζ i,j ≥ 0   . (10) W e no w group the random v ariables in to blo cks with v ariance 1 + o (1). F or this, define R = ⌊ √ n ⌋ and M = ⌊ N /R ⌋ = Θ( n 1 / 12 ). F or i, j ∈ { 0 , 1 , . . . , M − 1 } , define the random v ariables ξ i,j and ξ i,j via ξ i,j = X k ∈ [ iR, ( i +1) R )) ,ℓ ∈ [ j R, ( j +1) R ) ζ k,ℓ , ξ i,j = ξ i,j p V ar( ξ i,j ) 2 Rio’s work is stated in terms of a b ound on the num b er of moments and the distribution function. Under our assumption of an exponential moment, one may uniformly bound the distribution function with Markov’s inequality . COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STRONG BRUHA T ORDER 7 and b ound (11) P   min a ≤ N ,b ≤ N X i ∈ [ a ] ,j ∈ [ b ] ζ i,j ≥ 0   ≤ P   min a ≤ M ,b ≤ M X i ∈ [ a ] ,j ∈ [ b ] ξ i,j ≥ 0   . Seeking to apply Theorem 8 , note that E ξ i,j = 0 and E ξ 2 i,j = 1. Set α = 1 /n (1 − 1 /n ) and b ound E exp( tξ i,j ) = (2 α cosh( t ) + 1 − 2 α ) R 2 ≤ exp  2 αR 2 (cosh( t ) − 1)  = exp((2 + o (1))(cosh( t ) − 1)) whic h can b e made ≤ 2 by requiring, sa y , | t | ≤ 1 / 2. Theorem 8 and Theorem 9 sho w (12) P   min a ≤ M ,b ≤ M X i ∈ [ a ] ,j ∈ [ b ] ξ i,j ≥ 0   ≤ e − Ω(log 2 n ) + P   min a ≤ M ,b ≤ M X i ∈ [ a ] ,j ∈ [ b ] g i,j ≥ − C log 2 n   ≤ e − Ω(log 2 n ) . Com bining equations ( 9 ), ( 10 ), ( 11 ) and ( 12 ) completes the proof. □ 3.1. Pro of of Theorem 7 . The main step is to sho w that the random matrices P [ N ] × [ N ] and Q hav e asymptotically the same distribution for the n umber of 1’s. W e recall that by Theorem 2 the n um b er of 1’s in P [ N ] × [ N ] is a hypergeometric random v ariable. Lemma 10. Set N = ⌊ n 7 / 12 ⌋ an d let X ∼ HyperGeom( n, N , N ) and Y ∼ Binomial( N 2 , 1 /n ) . Then uniformly for k ≤ n 1 / 5 we have P ( X = k ) P ( Y = k ) = 1 + o (1) for n sufficiently lar ge. Pr o of. First compute P ( Y = k ) =  N 2 k  n − k (1 − n − 1 ) N 2 − k = (1 + O ( k 2 / N 2 )) ( N 2 /n ) k k ! (1 + O ( N 2 /n 2 )) e − N 2 /n = (1 + o (1)) ( N 2 /n ) k k ! e − N 2 /n . W e then compute P ( X = k ) =  N k  n − N N − k   n N  = ( N ) 2 k k ! · ( n − N ) N − k ( n ) N . First note that ( N ) 2 k = N 2 k (1 + O ( k 2 / N )) = N 2 k (1 + o (1)) . Second, note ( n − N ) N − k ( n ) N = (1 + o (1)) n − k · ( n − N ) N ( n ) N = (1 + o (1)) n − k · N − 1 Y j =0 n − N − j n − j = (1 + o (1)) n − k (1 − N/n ) N = (1 + o (1)) n − k e − N 2 /n where in the last asymptotic equality we used that N 3 /n 2 = o (1). □ W e now need to see that typically we do not hav e any rows or columns in Q with tw o or more 1’s. Set Q = {∃ i : P j Q ( i, j ) ≥ 2 or P j Q ( j, i ) ≥ 2 } . Lemma 11. L et N = ⌊ n 7 / 12 ⌋ , and Q a r andom N × N matrix with i.i.d. Bernoulli(1 /n ) entries. L et | Q | = P i,j Q ( i, j ) b e the total numb er of 1 ’s. Then uniformly in k ≤ n 1 / 5 for n sufficiently lar ge we have P ( Q | | Q | = k ) ≥ 1 − n − 1 / 6 . Pr o of. By exchangeabilit y of the N 2 en tries, note that conditioned on | Q | = k , Q is uniform on all 0-1 matrices with exactly k many 1’s. W e may sample such a matrix by placing k many 1’s one-at-a-time 8 NICHOLAS CHRISTO AND MAR CUS MICHELEN without ever placing tw o in the same entry . F or each pair of 1’s, the probability w e place them in the same ro w or column is O (1 / N ) . By union b ounding ov er all pairs, w e see that P ( Q c | | Q | = k ) ≤ O ( k 2 / N ) = O ( n − 11 / 60 ) ≤ n − 1 / 6 . □ Pr o of of The or em 7 . Set k = P i,j M ( i, j ). By exchangeabilit y we hav e that P ( P N × N = M ) = P ( | P N × N | = k ) P ( P N × N = M | | P N × N | = k ) = P ( | P N × N | = k ) P ( Q = M | | Q | = k , Q ) (13) where the last equality uses the fact that ( Q | | Q | = k , Q ) and ( P N × N | | P N × N | = k ) hav e the same distribu- tion. W rite (14) P ( Q = M | | Q | = k , Q ) = P ( Q = M ) P ( | Q | = k , Q ) = P ( Q = M ) P ( | Q | = k ) P ( Q | | Q | = k ) = (1 + o (1)) P ( Q = M ) P ( | P N × N | = k ) where in the last equality w e used Theorem 10 and Theorem 11 . Combining ( 13 ) and ( 14 ) completes the pro of. □ 3.2. Pro of of Theorem 9 . W e use the following theorem of Li-Shao [ 13 , Theorem 2.2]: Theorem 12. L et ( X j ) j ∈ [ M ] b e a me an-zer o Gaussian ve ctor so that V ar( X j ) ≥ x 2 for al l j . Supp ose also that for e ach i we have X j ∈ [ M ] | E [ X j X i ] | p V ar( X j ) V ar( X i ) ≤ 5 4 . Then P ( min j ∈ [ M ] X j ≥ − x ) ≤ e − M / 10 . Pr o of of The or em 9 . Define G ( a, b ) = P i ≤ a,j ≤ b g i,j and note that ( G ( a, b )) a,b ≤ n is a mean-zero Gaussian v ector. Compute E [ G ( a 1 , b 1 ) G ( a 2 , b 2 )] = E X i 1 ≤ a 1 ,j 1 ≤ b 1 i 2 ≤ a 2 ,j 2 ≤ b 2 g i 1 ,j 1 g i 2 ,j 2 = X i 1 ≤ a 1 ,j 1 ≤ b 1 i 2 ≤ a 2 ,j 2 ≤ b 2 1 { i 1 = i 2 , j 1 = j 2 } = min { a 1 , a 2 } min { b 1 , b 2 } . F or ρ = 400 and each fixed ( i, j ), b ound X ( a,b ) E [ G ( ρ i , ρ j ) G ( ρ a , ρ b )] p V ar( G ( ρ i , ρ j )) V ar( G ( ρ a , ρ b )) = X ( a,b ) ρ − i/ 2 − j / 2 − a/ 2 − b/ 2 ρ min { i,a } min { j,b } = X ( a,b ) ρ −| i − a | / 2 −| j − b | / 2 ≤ X k ∈ Z ρ −| k | / 2 ! 2 =  1 + ρ − 1 / 2 1 − ρ − 1 / 2  2 ≤ 5 4 (15) where the last inequality is true for ρ = 400 . W e now define the set S 0 = { i : ρ i ∈ [log 3 n, n ] } and note that |S 0 | = Θ(log n ) . F or all ( i, j ) ∈ S 0 w e hav e that V ar( G ( ρ i , ρ j )) ≥ log 6 n ≥ C 2 log 4 n for n sufficien tly large. By Theorem 12 we hav e P ( min a,b ≤ n G ( a, b ) ≥ − C log 2 n ) ≤ P ( min i,j ∈S 2 0 G ( ρ i , ρ j ) ≥ − C log 2 n ) ≤ e −|S 0 | 2 / 10 = e − Ω(log 2 n ) where we used ( 15 ) to v erify the hypotheses of Theorem 12 . □ 4. Lower bound The goal is to pro ve the low er b ound of Theorem 1 . Prop osition 13. Ther e is a universal c onstant C > 0 so that for n ≥ 2 the fol lowing holds. L et π , τ ∈ S n b e indep endently and uniformly chosen at r andom. Then P n ( π ≤ τ ) ≥ exp  − C (log n ) 2  . COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STRONG BRUHA T ORDER 9 W e prov e Theorem 13 in tw o main steps. First, w e will use a correlation inequality to show to reduce to the pro duct of probabilities o v er v arious boxes. Define x i = (5 / 4) i and y j = (5 / 4) j . Define k 0 = max { i : x i ≤ n/ 2 } . W e will first show the following low er b ound using a correlation inequality together with conditioning on the first and last O (log n ) elements of the p ermutations. Lemma 14. F or e ach C > 0 ther e is a C ′ > 0 so that for n sufficiently lar ge we have P n ( π ≤ τ ) ≥ e − C ′ (log n ) 2 Y i,j ≤ k 0 P n − 2 C log n  min ( a,b ) ∈ [ x i , 5 x i / 4] × [ y j , 5 y j / 4] Z ( a, b ) ≥ − C log n  4 . W e will pro v e Theorem 14 in Section 4.1 using a correlation inequalit y of Johnson-Leader-Long for random p erm utations in the strong Bruhat order. W e then uniformly b ound the probabilities in Theorem 14 . Lemma 15. Ther e ar e c onstants C, c > 0 so that for al l x, y ≤ n/ 2 we have P  min ( a,b ) ∈ [ x, 5 x/ 4] × [ y , 5 y/ 4] Z ( a, b ) ≥ − C log n  ≥ c . The pro of of Theorem 13 now follows quickly: Pr o of of The or em 13 . W e note that k 0 = Θ(log n ), and so com bining Theorem 14 with Theorem 15 completes the pro of. □ 4.1. Pro of of Theorem 14 . W e note that S n under the strong Bruhat order is not a lattice and so do es not satisfy the usual assumptions of the F ortuin–Kasteleyn–Ginibre (FK G) inequalit y . How ever, a theorem of Johnson, Leader and Long [ 10 ] prov es that the corresp onding FKG inequality holds for the strong Bruhat order. Recall that A ⊂ S n is incr e asing if whenev er π 1 ∈ A and π 2 ≥ π 1 then π 2 ∈ A . Theorem 16 (Johnson-Leader-Long, [ 10 ]) . L et A, B ⊂ S n b e incr e asing events in the str ong Bruhat or der and let P denote the uniform pr ob ability on S n . Then P ( A ∩ B ) ≥ P ( A ) P ( B ) . W e quickly note that for even ts A, B w e hav e P ( A c ∩ B c ) − P ( A c ) P ( B c ) = 1 − ( P ( A ) + P ( B ) − P ( A ∩ B )) − (1 − P ( A ))(1 − P ( B )) = P ( A ∩ B ) − P ( A ) P ( B ) and so Theorem 16 holds for decreasing even ts as well. Corollary 17. L et A, B ⊂ S n × S n b e incr e asing in the first c o or dinate and de cr e asing in the se c ond c o or dinate, and let P denote the uniform pr ob ability on S n × S n . Then P ( A ∩ B ) ≥ P ( A ) P ( B ) . Pr o of. Let ( π , τ ) ∈ S n × S n b e chosen uniformly at random. Set A π = { τ : ( π , τ ) ∈ A } and define B π similarly . Note that A π , B π are decreasing even ts by assumption. W e can then write P ( A ∩ B ) = E π [ P τ ( A π ∩ B π )] ≥ E π [ P τ ( A π ) P τ ( B π )] where in the last inequalit y w e used Theorem 16 for the decreasing even ts A π , B π . W e now may write E π [ P τ ( A π ) P τ ( B π )] = Z ∞ 0 Z ∞ 0 P π ( P τ ( A π ) > s, P τ ( B π ) > t ) ds dt ≥ Z ∞ 0 Z ∞ 0 P π ( P τ ( A π ) > s ) P π ( P τ ( B π ) > t ) ds dt = E π [ P τ ( A π )] · E π [ P τ ( B π )] = P ( A ) P ( B ) where in the inequality we used that A and B are increasing in π and applied Theorem 16 . □ 10 NICHOLAS CHRISTO AND MAR CUS MICHELEN F or N = ⌈ n/ 2 ⌉ define the ev en ts E 1 =  min a,b ∈ [ N ] × [ N ] Z ( a, b ) ≥ 0 }  , E 2 =  min a,b ∈ [ n − N ,n ] × [ N ] Z ( a, b ) ≥ 0 }  E 3 =  min a,b ∈ [ N ] × [ n − N ,n ] Z ( a, b ) ≥ 0 }  , E 4 =  min a,b ∈ [ n − N ,n ] × [ n − N ,n ] Z ( a, b ) ≥ 0 }  . Since { π ≤ τ } = E 1 ∩ E 2 ∩ E 3 ∩ E 4 w e ha v e (16) P n ( π ≤ τ ) = P ( E 1 ∩ E 2 ∩ E 3 ∩ E 4 ) ≥ 4 Y j =1 P ( E j ) where in the last inequalit y w e used Theorem 17 . W e now sho w that the even ts E j ha v e the same probabilit y . Lemma 18. We have P ( E 1 ) = P ( E 2 ) = P ( E 3 ) = P ( E 4 ) . Pr o of. F or all a, b we hav e that X ([ a ] × [ b ]) = b − X (( a, n ] × [ b ]). This means that min a,b ∈ [ n − N ,n ] × [ N ] Z ( a, b ) = min a,b ∈ [ n − N ,n ] × [ N ] ( − Z (( n − a, n ] × [ b ])) . Ho w ever, b y replacing the first a rows with the last a rows, we hav e that min a,b ∈ [ n − N ,n ] × [ N ] ( − Z (( n − a, n ] × [ b ])) d = min a ′ × b ∈ [ N ] × [ N ] ( − Z ( a ′ , b )) d = min a × b ∈ [ N ] × [ N ] Z ( a ′ , b ) where in the second equality we used that Z and − Z hav e the same distribution. This shows P ( E 1 ) = P ( E 2 ). By taking the transp ose we see P ( E 2 ) = P ( E 3 ). Replacing the first N rows with the last N rows and similarly for the columns shows P ( E 1 ) = P ( E 4 ) . □ W e are no w ready to prov e Theorem 14 : Pr o of of The or em 14 . By ( 16 ) and Theorem 18 it is sufficient to show that P ( E 1 ) ≥ e − C ′ (log n ) 2 Y i,j ≤ k 0 P n − 2 C log n  min ( a,b ) ∈ [ x i , 5 x i / 4] × [ y j , 5 y j / 4] Z ( a, b ) ≥ − C log n  . Recall that Z ( a, b ) = X ( a, b ) − Y ( a, b ). Let π denote the p ermutation corresponding to X and τ the p erm utation corresp onding to Y . Define the even ts G ′ := { π ( j ) = j ∀ j ∈ [ C log n ] ∪ [ n − C log n, n ] } ∩ { τ ( j ) = n − j + 1 ∀ j ∈ [ C log n ] ∪ [ n − C log n, n ] } and compute P ( G ′ ) = n − (4+ o (1)) C log n = exp( − (4 C + o (1)) log 2 n ). If we condition on G ′ ,the remaining p ortion of the p erm utations π and τ are uniform at random p ermutations on n − 2 C log n many elemen ts. Let π ′ and τ ′ denote these p ermutations. F or i, j ≤ k 0 define G i,j =  min ( a,b ) ∈ [ x i , 5 x i / 4] × [ y j , 5 y j / 4] Z π ′ ,τ ′ ( a, b ) ≥ − C log n  and note that Z π ′ ,τ ′ ( a, b ) = Z π ,τ ([ C log n, C log n + a ] × [ C log n, C log n + b ]) = Z π ,τ ( C log n + a, C log n + b ) − C log n . Th us we hav e P ( E 1 ) ≥ P ( G ′ ∩ \ i,j ≤ k 0 G i,j ) = e − (4+ o (1) C ) log 2 n P ( \ i,j ≤ k 0 G i,j | G ′ ) ≥ e − C ′ (log n ) 2 Y i,j ≤ k 0 P n − 2 C log n  min ( a,b ) ∈ [ x i , 5 x i / 4] × [ y j , 5 y j / 4] Z ( a, b ) ≥ − C log n  where in the last inequalit y w e used Theorem 17 . □ COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STRONG BRUHA T ORDER 11 4.2. Pro of of Theorem 15 . I n order to pro v e Theorem 15 , w e will break the box [ a ] × [ b ] in to four sections, and write Z ( a, b ) = Z ( x, y ) + Z ([ x ] × [ y , b ]) + Z ([ x, a ] × [ y ]) + Z ([ x, a ] × [ y , b ]) ≥ Z ( x, y ) − max b ∈ [ y , (5 / 4) y ] | Z ([ x ] × [ y , b ]) | − max a ∈ [ x, (5 / 4) x ] | Z ([ x, a ] × [ y ]) | − max ( a,b ) ∈ [ x, (5 / 4) x ] × [ y , (5 / 4) y ] | Z ([ x, a ] × [ y , b ]) | . (17) W e will first prov e that in exp ectation, the last three terms are of order p xy /n + log n . W e then will sho w that with constant probability Z ( x, y ) is p ositive enough to make up for it. W e handle the middle t w o terms in ( 17 ) with F reedman’s inequality: Lemma 19. Ther e is a universal c onstant C > 0 so that for al l x, y ≤ n/ 2 we have E max b ∈ [ y , (5 / 4) y ] | e X ([ x ] × [ y , b ]) | ≤ C  r xy n + 1  . Pr o of. Define M j = e X ([ x ] × [ y , y + j ]). Note that M j is a martingale with b ounded increments. F urther, as in the pro of of Theorem 4 , we hav e that the v ariance of the increments is b ounded by C x/n . In particular, Theorem 3 shows P  max b ∈ [ y , (5 / 4) y ] | e X ([ x ] × [ y , b ]) | ≥ t r xy n  ≤ exp  − c min { t 2 , t p xy /n }  . In tegrating gives E max b ∈ [ y , (5 / 4) y ] | e X ([ x ] × [ y , b ]) | ≤ r xy n + r xy n Z ∞ 1 exp( − c { t 2 , t p xy /n } ) dt ≤ C  r xy n + 1  . □ Handling the last term in ( 17 ) is significan tly trickier. W e prov e an upp er b ound on the exp ectation via a chaining approach: Lemma 20. Ther e is a universal c onstant C > 0 so that for al l x, y ≤ n/ 2 we have E max ( a,b ) ∈ [ x, (5 / 4) x ] × [ y , (5 / 4) y ] | e X ([ x, a ] × [ y , b ]) | ≤ C  p xy /n + log n  . W e prov e Theorem 20 in Section 4.3 . F or a universal constant C 1 > 0 define the ev ents G 1 :=  max b ∈ [ y , (5 / 4) y ] | Z ([ x ] × [ y , b ]) | ≤ C 1  p xy /n + log n   G 2 :=  max a ∈ [ x, (5 / 4) x ] | Z ([ x, a ] × [ y ]) | ≤ C 1  p xy /n + log n   G 3 =  max ( a,b ) ∈ [ x, (5 / 4) x ] × [ y , (5 / 4) y ] | Z ([ x, a ] × [ y , b ]) | ≤ C 1  p xy /n + log n   By Marko v’s inequality together with Theorem 19 and Theorem 20 , w e ma y choose C 1 large enough so that (18) P ( G 1 ∩ G 2 ∩ G 3 ) ≥ 1 / 2 . In order to sho w that Z ( x, y ) can b e large even conditioned on G 1 ∩ G 2 ∩ G 3 , w e identify the conditional distribution for the num ber of 1’s in a b ox of a random p ermutation conditioned on a p ortion out of the b ox. Lemma 21. L et P b e a r andom p ermutation matrix. F or x 1 ≤ x 2 and y 1 ≤ y 2 , we have  | P [ x 1 ] × [ y 1 ] | | P [ x 2 ] × [ y 2 ] \ [ x 1 ] × [ y 1 ]  ∼ HyperGeom ( n − ( x 2 − x 1 ) − ( y 2 − y 1 ) + m 3 , y 1 − m 2 , x 1 − m 1 ) wher e m 1 = | P [ x 1 ] × ( y 1 ,y 2 ] | , m 2 = | P ( x 1 ,x 2 ] × [ y 1 ] | , m 3 = | P ( x 1 ,x 2 ] × ( y 1 ,y 2 ] | . 12 NICHOLAS CHRISTO AND MAR CUS MICHELEN Pr o of. Let π ∈ S n b e the permutation corresp onding to P , meaning row j has its 1 in column π ( j ). W e view π as a map from the set of ro ws to the set of columns. Define S 1 = [ x 1 ] , S 2 = ( x 1 , x 2 ] , T 1 = [ y 1 ] , T 2 = ( y 1 , y 2 ] . Condition on the pattern of P on [ x 2 ] × [ y 2 ] \ ([ x 1 ] × [ y 1 ]). This reveals: which rows of S 1 map into T 2 (there are m 1 of them); whic h columns of T 1 are already used by rows in S 2 (there are m 2 of them); and which ro ws of S 2 map into T 2 (there are m 3 of them). Let S 0 ⊂ S 1 b e the set of rows not mapp ed in to T 2 and note that | S 0 | = x 1 − m 1 . W e then see that all ro ws in S 0 m ust map into the set of columns C := [ n ] \ ( π ( S 2 ) ∪ T 2 ) . Since | π ( S 2 ) | = | S 2 | = x 2 − x 1 and | π ( S 2 ) ∩ T 2 | = m 3 , the principle of inclusion-exclusion gives | C | = n − ( x 2 − x 1 ) − ( y 2 − y 1 ) + m 3 . Among the av ailable columns C , the ones lying in T 1 are exactly C T := T 1 \ π ( S 2 ) and so | C T | = y 1 − m 2 . By exc hangeabilit y , the image π ( S 0 ) is a uniformly random ( x 1 − m 1 )-subset of C , and | P [ x 1 ] × [ y ] | = | π ( S 0 ) ∩ C T | . This sho ws  | P [ x 1 ] × [ y 1 ] | | P [ x 2 ] × [ y 2 ] \ [ x 1 ] × [ y 1 ]  ∼ HyperGeom( | C | , | C T | , | S 0 | ) whic h completes the pro of. □ Set σ 2 = ab/n and for a constan t C > 0 define G X 4 via G X 4 :=      X ([ a ] × ( b, 5 b/ 4]) − ab 4 n     ≤ C σ  ∩      | X (( a, 5 a/ 4] × [ b ]) − ab 4 n     ≤ C σ  ∩      X (( a, 5 a/ 4] × ( b, 5 b/ 4]) − ab 16 n     ≤ C σ  . Define G Y 4 analogously for Y , and set G 4 = G X 4 ∩ G Y 4 . By Theorem 4 and Theorem 2 we may increase C so that P ( G c 4 ) ≤ 1 4 and so (19) P ( G 1 ∩ G 2 ∩ G 3 ∩ G 4 ) ≥ 1 / 4 . Lemma 22. F or e ach M , C > 0 ther e is a c > 0 so that P ( Z ( a, b ) ≥ M σ − C | G 1 ∩ G 2 ∩ G 3 ∩ G 4 ) ≥ c . Pr o of. First note that if σ is smaller than a constant C 0 then we hav e P ( | Z ( a, b ) | ≤ 2 C 0 ) ≥ 1 / 2 by Marko v’s inequalit y . W e may thus assume that σ > C 0 for some C 0 dep ending on M to b e chosen later. F or P = M π , conditioned on P [5 a/ 4] × [5 b/ 4] \ [ a ] × [ b ] , Theorem 21 tells us that | P [ a ] × [ b ] | is a hypergeometric random v ariable. By Theorem 5 , it is sufficient to sho w that for all possible patterns P [(5 / 4) a ] × [(5 / 4) b ] \ [ a ] × [ b ] in G 4 w e ha v e that the conditioned mean of | P [ a ] × [ b ] | is at most O ( σ ) from the unconditional mean of P [ a ] × [ b ] . Using the notation of Theorem 21 , define m 1 = | P [ a ] × ( b, 5 b/ 4] | , m 2 = | P ( a, 5 a/ 4] × [ b ] | , m 3 = | P ( a, 5 a/ 4] × ( b, 5 b/ 4] | . W rite e m 1 = m 1 − ab 4 n , e m 2 = m 2 − ab 4 n , e m 3 = m 3 − ab 16 n and note that on G 4 w e hav e that | e m j | ≤ C σ for all j ∈ { 1 , 2 , 3 } . Set A = a − ab 4 n , B = b − ab 4 n and N = n − a 4 − b 4 + ab 16 n and note first that AB N = ab n . By Theorem 21 , the distance from the conditional mean COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STRONG BRUHA T ORDER 13 of | P [ a 1 ] × [ b 1 ] | to the unconditional mean is     ( A − e m 1 )( B − e m 2 ) N + e m 3 − AB N     ≤     AB N + e m 3 − AB N     +     A e m 2 N + e m 3     +     B e m 1 N + e m 3     +     e m 1 e m 2 N + e m 3     ≤ 4 C σ . Seeking to apply Theorem 5 , recall that we ha v e assumed σ ≥ C 0 ; taking C 0 large enough so that we ma y apply Theorem 5 shows that w e hav e P ( e X ( a, b ) ≥ 2 M σ | G 1 ∩ G 2 ∩ G 3 ∩ G 4 ) ≥ c . Arguing similarly for e Y completes the pro of. □ 4.3. Pro of of Theorem 20 . F or simplicit y , write e X ( a, b ) = e X ([ a ] × [ b ]) . W e hav e the distributional equalit y  e X ([ x, a ] × [ y , b ])  ( a,b ) ∈ [ x, (5 / 4) x ] × [ y , (5 / 4) y ] d =  e X ( a, b )  ( a,b ) ∈ [ x/ 4] × [ y / 4] . By p oten tially increasing the constant C in the statement, it is sufficient to show that for x ≤ y ≤ n/ 4 eac h giv en by a p o wers of 2 that we ha ve (20) E max a ≤ x,b ≤ y | e X ( a, b ) | ≤ C  p xy /n + log n  . W rite x = 2 N x , y = 2 N y . F or each k ≤ N y define the set of p oin ts (21) S k = n i · x 2 k ∧ N x , j · y 2 k o i ≤ 2 k ∧ N x ,j ≤ 2 k . Define K = N y ∧ max { k : xy n 2 k ≥ 1 } . F or ( a, b ) ∈ [ x ] × [ y ] and k ≤ K we ma y choose ( a k , b k ) ∈ S k so that (22) | a k − a | ≤ x 2 k ∧ N x , | b k − b | ≤ y 2 k . W rite | e X ( a, b ) | ≤ | e X ( a K , b K ) | + | e X ( a, b ) − e X ( a K , b K ) | whic h implies (23) E max a ≤ x,b ≤ y | e X ( a, b ) | ≤ E max ( a K ,b K ) ∈S K | e X ( a K , b K ) | + E max a ≤ x,b ≤ y min ( a K ,b K ) ∈S K | e X ( a, b ) − e X ( a K , b K ) | . W e b egin with the second term. Lemma 23. In the notation ab ove, ther e is a universal c onstant C > 0 so that for n ≥ 2 we have E max a ≤ x,b ≤ y min ( a K ,b K ) ∈S K | e X ( a, b ) − e X ( a K , b K ) | ≤ C log n . Pr o of. T o b egin with, note that for each a ≤ x, b ≤ y there is some ( a K , b K ) ∈ S K so that (24) | a K − a | ≤ x 2 K ∧ N x and | b K − b | ≤ y 2 K . As suc h, if ( a, b ) and ( a K , b K ) satisfy ( 24 ), then we may write the symmetric difference of the rectangles [ a ] × [ b ] and [ a K , b K ] as the union of at most tw o rectangles of area at most xy 2 K . Note that xy 2 K n ≤ 2 b y assumption. By Theorem 4 , for all t ≥ 2 we may th us bound P  | e X ( a, b ) − e X ( a K , b K ) | ≥ t  ≤ 4 exp  − t 32  . Since there are at most n 4 c hoices for ( a, b ) , ( a K , b K ) we obtain P  max a ≤ x,b ≤ y min ( a K ,b K ) ∈S K | e X ( a, b ) − e X ( a K , b K ) | ≥ t  ≤ 4 n 4 exp  − t 32  . 14 NICHOLAS CHRISTO AND MAR CUS MICHELEN Upp er b ounding the exp ectation by integrating the tail probability we see E max a ≤ x,b ≤ y min ( a K ,b K ) ∈S K | e X ( a, b ) − e X ( a K , b K ) | ≤ 128 log n + Z ∞ 128 log n 4 n 4 exp  − t 32  dt = O (log n ) . □ W e now handle the first term in ( 23 ). Lemma 24. Ther e is a universal c onstant C > 0 so that for n ≥ 2 we have E max ( a K ,b K ) ∈S K | e X ( a K , b K ) | ≤ C  r xy n + log n  . Pr o of. Note first that if xy n ≤ 2 then the argument in Theorem 23 shows an upp er b ound o f log n . W e may th us assume without loss of generality that xy n ≥ 2 . Letting ( a k , b k ) denoting the choice from ( 22 ) we ma y write the telescoping sum: e X ( a K , b K ) = e X ( a 0 , b 0 ) + K X k =1 ( e X ( a k , b k ) − e X ( a k − 1 , b k − 1 )) . and so (25) E max ( a K ,b K ) ∈S K | e X ( a K , b K ) | ≤ E | e X ( a 0 , b 0 ) | + K X k =1 E max ( a k ,b k ) , ( a k − 1 ,b k − 1 ) ∈ ( 22 ) | e X ( a k , b k ) − e X ( a k − 1 , b k − 1 ) | where in the maximum w e write ( a k , b k ) , ( a k − 1 , b k − 1 ) ∈ ( 22 ) to denote that ( a k , b k ) ∈ S k , ( a k − 1 , b k − 1 ) ∈ S k − 1 and b oth satisfy ( 22 ) for the sam e ( a, b ). W e will sho w that there is an absolute constant C > 0 so that (26) E max ( a k ,b k ) , ( a k − 1 ,b k − 1 ) ∈ ( 22 ) | e X ( a k , b k ) − e X ( a k − 1 , b k − 1 ) | ≤ C 2 − k/ 8 r xy n for all k ∈ [0 , K ] (where we interpret S − 1 = ∅ ). F or pairs ( a k , b k ) , ( a k − 1 , b k − 1 ) ∈ ( 22 ) we hav e that | a k − a k − 1 | ≤ 4 x 2 k ∧ N x and | b k − b k − 1 | ≤ 4 y 2 k . In particular, the symmetric difference of the rectangles [ a k , b k ] and [ a k − 1 , b k − 1 ] may b e written as the union of at most t w o rectangles of area at most 4 xy 2 k . Thus for a univ ersal c onstan t c > 0 if we set σ 2 = xy /n P  | e X ( a k , b k ) − e X ( a k − 1 , b k − 1 ) | ≥ θσ 2 − k/ 8  ≤ exp  − c min  θ 2 σ 2 / 2 k/ 4 σ 2 / 2 k , θ σ 2 − k/ 8  ≤ exp  − c min n θ 2 2 3 k/ 4 , θ 2 3 k/ 8 o where in the second line we used that σ 2 2 − k ≥ 1 by the definition of K . There are at most, say , 16 k suc h c hoices for ( a k , b k ) , ( a k − 1 , b k − 1 ) and so we hav e P  max ( a k ,b k ) , ( a k − 1 ,b k − 1 ) ∈ ( 22 ) | e X ( a k , b k ) − e X ( a k − 1 , b k − 1 ) | ≥ θσ 2 − k/ 8  ≤ 16 k exp  − c min n θ 2 2 3 k/ 4 , θ 2 3 k/ 8 o . W e may then b ound E max ( a k ,b k ) , ( a k − 1 ,b k − 1 ) ∈ ( 22 ) | e X ( a k , b k ) − e X ( a k − 1 , b k − 1 ) | ≤ σ 2 − k/ 8 + σ 2 − k/ 8 Z ∞ 1 16 k exp  − c min n θ 2 2 3 k/ 4 , θ 2 3 k/ 8 o dθ ≤ C σ 2 − k/ 8 for some absolute constant C > 0 . This establishes ( 26 ). Com bining with ( 25 ) completes the pro of. □ Pr o of of The or em 20 . The lemma is equiv alen t to ( 20 ), which follows from combining Theorem 23 and Theorem 24 with ( 23 ). □ COMP ARABILITY OF RANDOM PERMUT A TIONS IN THE STRONG BRUHA T ORDER 15 A cknowledgments M.M. is supp orted in part by NSF CAREER grant DMS-2336788. Both authors are supp orted in part by NSF grants DMS-2137623 and DMS-2246624. The authors thank Igor Pak for bringing this problem to their atten tion. References [1] A. Bj¨ orner. Orderings of Coxeter groups. In Combinatorics and algebr a (Boulder, Colo., 1983) , v olume 34 of Contemp. Math. , pages 175–195. Amer. Math. Soc., Providence, RI, 1984. 1 [2] A. Bj¨ orner and F. Brenti. 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