Intersecting Families of Permutations
aa r X i v : . [ m a t h . C O ] J u l Intersecting Families of Permutations
David Ellis ∗ , Ehud Friedgut † and Haran Pilpel ‡ Abstract
A set of permutations I ⊂ S n is said to be k - intersecting if anytwo permutations in I agree on at least k points. We show that forany k ∈ N , if n is sufficiently large depending on k , then the largest k -intersecting subsets of S n are cosets of stabilizers of k points, proving aconjecture of Deza and Frankl. We also prove a similar result concern-ing k -cross-intersecting subsets. Our proofs are based on eigenvaluetechniques and the representation theory of the symmetric group. The classical Erd˝os-Ko-Rado theorem states that if r < n/
2, an intersect-ing family of r -subsets of { , , . . . , n } has size at most (cid:0) n − r − (cid:1) ; if equalityholds, the family must consist of all r -subsets containing a fixed element.The so-called ‘second Erd˝os-Ko-Rado theorem’ states that if n is sufficientlylarge depending on k and r , then any k -intersecting family of r -subsets of { , , . . . , n } has size at most (cid:0) n − kr − k (cid:1) ; if equality holds, the family must con-sist of all r -subsets containing k fixed elements. We deal with analogues ofthese results for permutations.As usual, [ n ] will denote the set { , , . . . , n } , and S n will denote thesymmetric group, the group of all permutations of [ n ]. Two permutations σ, τ ∈ S n are said to intersect if they agree at some point, i.e. if there exists i ∈ [ n ] such that σ ( i ) = τ ( i ). Similarly, they are said to k - intersect if they AMS 2000 subject classification: 05E10, 20C30, 05D99Key words and phrases: Intersecting families of permutations, Erd˝os-Ko-Rado, rep-resentation theory. ∗ DPMMS, University of Cambridge. † Hebrew University, Jerusalem, and University of Toronto. Research supported in partby the Israel Science Foundation, grant no. 0397684, and NSERC grant 341527. ‡ Hebrew University, Jerusalem. Research supported in part by the Giora Yoel Yashin-sky memorial grant. Current affiliation: Google, Inc. k points, i.e., if there exist i , i , . . . , i k ∈ [ n ] such that σ ( i t ) = τ ( i t ) for t = 1 , , . . . , k . A subset I ⊂ S n is said to be k - intersecting if any two permutations in I k -intersect. In this paper, we characterize thelargest k -intersecting subsets of S n for n sufficiently large depending on k ,proving a conjecture of Deza and Frankl. We also prove a similar resultconcerning k -cross-intersecting subsets, proving a conjecture of Leader.Our main tool in this paper is Fourier analysis on the symmetric group,which entails representation theory. Although Fourier analysis has becomea central tool in combinatorics and computer science in the last two decades(notably, since the landmark paper of [21]), it has not often been appliedto combinatorial problems in a non-Abelian setting. In retrospect, it seemsthat in this case it fits the task perfectly.As a bonus, we point out a nice aspect of Boolean (0/1 valued) func-tions on S n . One of the recurring themes in the applications of discreteFourier analysis to combinatorics over the last decade has been showing thatBoolean functions on { , } n are ‘juntas’ (i.e. depend essentially on few co-ordinates) precisely when their Fourier transform is concentrated mainly onsmall sets. Here, we study Boolean functions on S n whose Fourier trans-form is supported on the irreducible representations of low dimension, andconnect them to cosets of subgroups which are the pointwise stabilizers ofsmall subsets of { , . . . , n } . Along the way, we also prove an interestinggeneralization of Birkhoff’s theorem on bistochastic matrices. Let T i j = { σ ∈ S n , σ ( i ) = j } . Clearly, T i j is 1-intersecting subset of S n , with size ( n − T i j ,...,i k j k = k \ t =1 T i t j t = { σ ∈ S n : σ ( i t ) = j t (1 ≤ t ≤ k ) } . If i , . . . , i k are distinct and j , . . . , j k are distinct, then T i j ,...,i k j k is acoset of the stabilizer of k points; we will refer to it as a k - coset .Clearly, a k -coset is a k -intersecting family of size ( n − k )!. As observedby Deza and Frankl [11], it is easy to prove that a 1-intersecting subset of S n is no larger than a 1-coset: Theorem 1. [11] For any n ∈ N , if I ⊂ S n is -intersecting, then | I | ≤ ( n − . roof. Let H be the cyclic group generated by the n -cycle (123 . . . n ). Notwo permutations in H intersect, and the same is true for any left coset of H . The ( n − H partition S n . If I ⊂ S n is 1-intersecting,then I contains at most one permutation from each left coset of H , andtherefore | I | ≤ ( n − . Deza and Frankl conjectured in [11] that for any n ∈ N , the 1-cosets arethe only 1-intersecting subsets of S n with size ( n − k -intersecting families? For n small depending on k , the k -cosets need not be the largest k -intersecting subsets of S n . Indeed, { σ ∈ S n : σ has at least k + 1 fixed points in { , , . . . , k + 2 }} is a k -intersecting family with size( k + 2)( n − k − − ( k + 1)( n − k − , which is larger than ( n − k )! if k ≥ n ≤ k . However, Deza and Franklconjectured in [11] that if n is large enough depending on k , the k -cosets are the largest k -intersecting subsets of S n : Conjecture 1 (Deza-Frankl) . For any k ∈ N and any n sufficiently largedepending on k , if I ⊂ S n is k -intersecting, then | I | ≤ ( n − k )! . Equalityholds if and only if I is a k -coset of S n . This is our main result. Our proof uses eigenvalue techniques, togetherwith the representation theory of S n . In fact, it was proved by the firstauthor and the last two authors independently in 2008. Our two proofs ofthe upper bound are essentially equivalent, hence the joint paper. However,the latter two authors proved the equality statement directly, via their gen-eralization of Birkhoff’s Theorem (Theorem 29), whereas the first authordeduced it from the following ‘stability’ result, proved in [8]: Theorem 2 ([8]) . Let k ∈ N be fixed, let n be sufficiently large dependingon k , and let I ⊂ S n be a k -intersecting family which is not contained withina k coset. Then I is no larger than the family { σ ∈ S n : σ ( i ) = i ∀ i ≤ k, σ ( j ) = j for some j > k + 1 }∪ { (1 , k + 1) , (2 , k + 1) , . . . , ( k, k + 1) } , which has size (1 − /e + o (1))( n − k )! . r -sets; the k = 1 case was conjectured by Cameronand Ku in [6].When there exists a sharply k -transitive subset of S n , the ‘partitioning’argument in the above proof of Theorem 1 can be replaced by a Katona-typeaveraging argument, proving Conjecture 1 in this case.[Recall that a subset T ⊂ S n is said to be k - transitive if for any distinct i , . . . , i k ∈ [ n ] and any distinct j , . . . , j k ∈ [ n ], there exists σ ∈ T suchthat σ ( i t ) = j t for each t ∈ [ k ]; T it is said to be sharply k - transitive ifthere exists a unique such σ ∈ T . Note that a k -transitive subset T ⊂ S n issharply k -transitive if and only if it has size n ( n − . . . ( n − k + 1).]Indeed, suppose that S n has a sharply k -transitive subset T . Then anyleft translate σT of T is also sharply k -transitive, so any two distinct permu-tations in σT agree in at most k − I ⊂ S n be k -intersecting; then | I ∩ σT | ≤ σ ∈ S n . Averaging over all σ ∈ S n gives | I | ≤ ( n − k )!.For k = 2 and n = q a prime power, S n has a sharply 2-transitive subgroup : identify the ground set with the finite field F q of order q , and take H to be the group of all affine maps x ax + b ( a ∈ F q \ { } , b ∈ F q ). Anytwo distinct permutations in H agree in at most 1 point, and the same istrue for any left coset of H . So if I ⊂ S n is 2-intersecting, then I containsat most one permutation from each left coset of H . Since the ( n − H partition S n , this implies that | I | ≤ ( n − k = 3 and n = q + 1 (where q is a prime power), S n has asharply 3-transitive subgroup: identify the ground set with F q ∪ {∞} , andtake H to be the group of all M¨obius transformations x ax + bcx + d ( a, b, c, d ∈ F q , ad − bc = 0) . However, it is a classical result of C. Jordan [20] that the only sharply k -transitive permutation groups for k ≥ S k (for k ≥ A k − (for k ≥ M (for k = 4) and M (for k = 5), where M , M are the Matthieugroups. Moreover, sharply k -transitive subsets of S n have not been foundfor any other values of n and k . Thus, it seems unlikely that this approachcan work in general. Instead, we will use a different approach.Recall that if G is a group and X ⊂ G is inverse-closed, the Cayleygraph on G generated by X is the graph with vertex-set G and edge-set {{ g, h } : gh − ∈ X } . Let Γ be the Cayley graph on S n generated by theset of fixed-point-free permutations,FPF = { σ ∈ S n : σ ( i ) = i ∀ i ∈ [ n ] } . S n is precisely an independent set in Γ .It turns out that calculating the least eigenvalue of Γ (meaning the leasteigenvalue of its adjacency matrix) and applying Hoffman’s bound (Theorem11) yields an alternative proof of Theorem 1. Calculating the least eigenvalueof Γ is non-trivial, requiring use of the representation theory of S n . It wasfirst done by Renteln [26], using symmetric functions, and independently andslightly later by Friedgut and Pilpel [14], and also by Godsil and Meagher[15]. As observed in [14] and [15], this also leads to an alternative proof thatthe 1-cosets are the unique largest 1-intersecting subsets.The obvious generalization of this approach fails for k -intersecting sub-sets of S n . Let Γ k be the Cayley graph on S n generated by the setFPF k = { σ ∈ S n : σ has less than k fixed points } . A k -intersecting subset of S n is precisely an independent set in Γ k , so ourtask is to find the largest independent sets in Γ k . Unfortunately, for k fixedand n large, calculating the least eigenvalue of Γ k and applying Hoffman’sbound only gives an upper bound of Θ(( n − k -intersectingfamily.A key idea of our proof is to choose various subgraphs of Γ k , and to con-struct a ‘pseudo-adjacency matrix’ A for Γ k which is a suitable real linearcombination of the adjacency matrices of these subgraphs. We then apply aweighted version of Hoffman’s Theorem (Theorem 12) to this linear combi-nation, in order to prove the upper bound in Conjecture 1. The subgraphschosen will be Cayley graphs generated by various unions of conjugacy-classes of S n ; this will enable use to calculate their eigenvalues using therepresentation theory of S n . Most of the work of the proof is in showingthat an appropriate linear combination exists. A remark on terminology: we will often identify a subset A of S n , its char-acteristic function 1 A : S n → { , } , and its characteristic vector v A ∈ R | S n | ;so when we say that a set A is spanned by sets B , . . . , B t , we mean that v A ∈ Span { v B , . . . , v B t } .Our main results in this paper are as follows. Theorem 3.
For any k ∈ N , and any n sufficiently large depending on k ,if I ⊂ S n is k -intersecting, then | I | ≤ ( n − k )! . Equality holds if and only if I is a k -coset of S n . We also prove a cross-intersecting version of this theorem:5 efinition 1.
Two sets
I, J ⊂ S n are k -cross-intersecting if every permu-tation in I k -intersects every permutation in J . Theorem 4.
For any k ∈ N and any n sufficiently large depending on k , if I, J ⊂ S n are k -cross-intersecting, then | I || J | ≤ (( n − k )!) . Equality holdsif and only if I = J and I is a k -coset of S n . The k = 1 case of the above was a conjecture of Leader [23].Our argument proceeds in the following steps. (In order to not disruptthe flow of the paper, some of the representation-theoretic terms used willonly be defined later.)First, we bound the size of a k -intersecting family. Theorem 5.
For any k ∈ N and any n sufficiently large depending on k ,if I ⊂ S n is k -intersecting, then | I | ≤ ( n − k )! . Moreover, if I, J ⊂ S n are k -cross-intersecting, then | I || J | ≤ (( n − k )!) . Next, we describe the Fourier transform of the characteristic functions ofthe families which achieve this bound. Let V k be the linear subspace of real-valued functions on S n whose Fourier transform is supported on irreduciblerepresentations corresponding to partitions µ of n such that µ ≥ ( n − k, k ),where ≥ denotes the lexicographic order (see section 3). Theorem 6.
For k fixed and n sufficiently large depending on k , if I ⊂ S n is a k -intersecting family of size ( n − k )! , then I ∈ V k . Furthermore if I, J ⊂ S n are k -cross-intersecting and | I || J | = ( n − k )! , then I , J ∈ V k . We then prove the following:
Theorem 7. V k is spanned by the characteristic functions of the k -cosetsof S n . Finally, we complete the proof of our main theorem using the followingcombinatorial result:
Theorem 8.
For any k ∈ N , if f is a Boolean function on S n which isspanned by the k -cosets, then f is the characteristic function of a disjointunion of k -cosets. Clearly, the last four theorems immediately imply our main result, The-orem 3, and its cross-intersecting version, Theorem 4.6 .3 Structure of the paper
In section 2 we provide the background that we will use from general rep-resentation theory and graph theory. In section 3 we prove all necessaryresults and lemmas that pertain to representation theory of S n . Section4 ties together the results of the previous two sections in order to boundthe size and provide a Fourier characterization of the largest k -intersectingfamilies. Finally, in section 5 we show that this characterization holds onlyfor k -cosets. In this section, we recall the basic notions and results we need from generalrepresentation theory. For more background, the reader may consult [19].Let G be a finite group, and let F be a field. A representation of G over F is a pair ( ρ, V ), where V is a finite-dimensional vector space over F ,and ρ : G → GL ( V ) is a group homomorphism from G to the group of allinvertible linear endomorphisms of V . The vector space V , together withthe linear action of G defined by gv = ρ ( g )( v ), is sometimes called an F G - module . A homomorphism between two representations ( ρ, V ) and ( ρ ′ , V ′ ) isa linear map φ : V → V ′ such that such that φ ( ρ ( g )( v )) = ρ ′ ( g )( φ ( v )) for all g ∈ G and v ∈ V . If φ is a linear isomorphism, the two representations aresaid to be equivalent , and we write ( ρ, V ) ∼ = ( ρ ′ , V ′ ). If dim( V ) = n , we saythat ρ has dimension n . If V = F n , then we call ρ a matrix representation ;choosing an F -basis for a general V , one sees that any representation isequivalent to some matrix representation.The representation ( ρ, V ) is said to be irreducible if it has no proper sub-representation, i.e. there is no proper subspace of V which is ρ ( g )-invariantfor all g ∈ G .The group algebra F [ G ] denotes the F -vector space with basis G andmultiplication defined by extending the group multiplication linearly. Inother words, F [ G ] = X g ∈ G x g g : x g ∈ F ∀ g ∈ G , and X g ∈ G x g g X h ∈ G y h h ! = X g,h ∈ G x g y h ( gh ) . P g ∈ G x g g with the function g x g , we can view the vectorspace F [ G ] as the space of all F -valued functions on G . The representationdefined by ρ ( g )( x ) = gx ( g ∈ G, x ∈ F [ G ])is called the left regular representation of G ; the corresponding F G -moduleis called the group module . This will be useful when we describe irreduciblerepresentations of S n .When F = R or C , it turns out that there are only finitely many equiva-lence classes of irreducible representations of G , and any representation of G is isomorphic to a direct sum of irreducible representations of G . Hence, wemay choose a set of representatives R for the equivalence classes of complexirreducible representations of G . For the rest of section 2, R will be fixed,and will consist of matrix representations.If ( ρ, V ) is a representation of G , and α is a linear endomorphism of V ,we say that α commutes with ρ if α ◦ ( ρ ( g )) = ρ ( g ) ◦ α for every g ∈ G . (Soan isomorphism of ( ρ, V ) is simply an invertible linear endomorphism whichcommutes with ρ .) We will make use of the following: Lemma 1 (Schur’s Lemma) . If G is a finite group, and ( ρ, V ) is a complexirreducible representation of G , then the only linear endomorphisms of V which commute with ρ are scalar multiples of the identity. If F = R or C , we may define an inner product on F [ G ] as follows: h φ, ψ i = 1 | G | X g ∈ G φ ( g ) ψ ( g ) . If ( ρ, V ) is a complex representation of V , the character χ ρ of ρ is the mapdefined by χ ρ : G → C ; χ ρ ( g ) = Tr( ρ ( g )) , where Tr( α ) denotes the trace of the linear map α (i.e. the trace of anymatrix of α ). Note that χ ρ (Id) = dim( ρ ), and that χ ρ is a class function on G (meaning that it is constant on each conjugacy-class of G .)The usefulness of characters lies in the following Fact.
Two complex representations are isomorphic if and only if they havethe same character; the set of complex irreducible characters is an orthonor-mal basis for the space of class functions in C [ G ] . ρ is any complex representation of G , its character satisfies χ ρ ( g − ) = χ ρ ( g ) for every g ∈ G . In our case, G = S n , so g − is conjugate to g forevery g , and therefore χ ρ ( g ) = χ ρ ( g ), i.e. all the characters are real-valued.The irreducible characters of S n are therefore an orthonormal basis for thespace of class functions on R [ S n ].Given two representations ( ρ, V ) and ( ρ ′ , V ′ ) of G , we can form theirdirect sum, the representation ( ρ ⊕ ρ ′ , V ⊕ V ′ ), and their tensor product, therepresentation ( ρ ⊗ ρ ′ , V ⊗ V ′ ). We have χ ρ ⊕ ρ ′ = χ ρ + χ ρ ′ , and χ ρ ⊗ ρ ′ = χ ρ · χ ρ ′ . In this section, we recall the basic notions of Fourier analysis on finite non-Abelian groups. For more background, see for example [29]. Noting thatthe normalization chosen differs in various texts, we set out our conventionbelow.
Definition 2.
Let G be a finite group, and let f, g : G → R be two real-valued functions on G . Their convolution f ∗ g is the real-valued functiondefined by f ∗ g ( t ) = 1 | G | X s ∈ G f ( ts − ) g ( s ) ( t ∈ G ) . (1) Definition 3.
The
Fourier transform of a real-valued function f : G → R is a matrix-valued function on irreducible representations; its value at theirreducible representation ρ is the matrix b f ( ρ ) = 1 | G | X s ∈ G f ( s ) ρ ( s ) . (2)We now recall two related formulas we will need: the Fourier transformof a convolution, and the Fourier inversion formula. If f, g : G → R , and ρ is an irreducible representation of G , then [ f ∗ g ( ρ ) = b f ( ρ ) b g ( ρ ) . (3)The Fourier transform is invertible; we have: f ( s ) = X ρ ∈ R dim( ρ )Tr h b f ( ρ ) ρ ( s − ) i . (4)In other words, the Fourier transform contains all the information about afunction f . 9 .3 Cayley Graphs Recall that if G is a group, and X ⊂ G is inverse-closed, the Cayley graphon G generated by X is the graph with vertex-set G and edge-set {{ u, v } ∈ G (2) : uv − ∈ X } ; it is sometimes denoted by Cay( G, X ). In fact, we willonly be considering cases where G = S n and X is a union of conjugacyclasses (i.e., X is conjugation-invariant).The relevance of this notion to our problem stems from the followingobservation. Consider the Cayley graph Γ on S n generated by the set of fixed-point free permutations,FPF = { σ ∈ S n : σ ( i ) = i ∀ i ∈ [ n ] } . As observed in section 1.1, a 1-intersecting family of permutations is pre-cisely an independent set in this graph. More generally, a k -intersectingfamily of permutations is precisely an independent set in the Cayley graphΓ k on S n with generating setFPF k = { σ ∈ S n : σ has at most k fixed points } . For any real matrix A ∈ R [ G × G ], the left action of A on R [ G ] is definedas follows: ( Af )( σ ) = X τ ∈ G A σ,τ f ( τ ) . The main observation of this subsection is that the adjacency matrixof a Cayley graph operates on functions in R [ G ] by convolution with thecharacteristic function of the generating set. Theorem 9.
Let G be a finite group, let X ⊂ G be inverse-closed, letCay ( G, X ) be the Cayley graph on G generated by X , and let A be theadjacency matrix of Cay ( G, X ) . Then for any function f : G → R , Af = | G | (1 X ∗ f ) . (5) Proof.
For any f ∈ R [ G ], and any σ ∈ G , we have( Af )( σ ) = X τ ∈ G A σ,τ f ( τ ) = X τ ∈ G X ( στ − ) f ( τ ) = | G | (1 X ∗ f )( σ ) , as required. 10aking the Fourier transform of both sides of (5), we obtain: c Af = | G | · c X b f . (6)We will see shortly that if X is conjugation-invariant, as in our case,then c X is a scalar matrix, so (6) essentially reveals all eigenfunctions of theoperator A : it is well known that for any finite group G , the entries of thematrices of a complete set of complex irreducible representations of G forman orthogonal basis for the space of all complex-valued functions on G . Soin our case, these are also a complete set of eigenfunctions of A . Theorem 10 (Schur; Babai; Diaconis, Shahshahani; Roichman; others [3,7, 27]) . Let G be a finite group, and let R be a complete set of complexirreducible matrix representations of G , as above. Let X ⊂ G be inverse-closed and conjugation-invariant, and let Cay ( G, X ) be the Cayley graph on G with generating set X . Let A be the adjacency matrix of Cay ( G, X ) . Forany ρ ∈ R , and any i, j ≤ dim( ρ ) , consider the function ρ ij ( σ ) = ρ ( σ ) ij .Then { ρ ij } ρ,i,j is a complete set of eigenvectors of A . Furthermore, theeigenvalue of ρ ij is λ ρ = 1dim( ρ ) X τ ∈ X χ ρ ( τ ) = | G |h χ ρ , X i dim( ρ ) , (7) which depends only on ρ .Proof. Note that, due to (6), the claim regarding eigenvectors will followimmediately once we have shown that c X ( ρ ) is a scalar matrix for everyirreducible representation ρ . To do this, we will show that c X ( ρ ) commuteswith every ρ ( σ ), which will imply the result (by Schur’s Lemma.) Indeed,for every σ ∈ G , ρ ( σ ) − c X ( ρ ) ρ ( σ ) = ρ ( σ − ) c X ( ρ ) ρ ( σ )= 1 | G | X τ ∈ G ρ ( σ − ) ρ ( τ )1 X ( τ ) ρ ( σ )= 1 | G | X τ ∈ X ρ ( σ − τ σ )= 1 | G | X τ ∈ X ρ ( τ )= c X ( ρ ) , X is conjugation-invariant for the fourthequality. Hence, by Schur’s Lemma, c X ( ρ ) is indeed a scalar matrix; write c X ( ρ ) = c ρ I . To calculate c ρ , note that for any i ≤ dim( ρ ), c ρ = 1 | G | X τ ∈ X X ( τ ) ρ i,i ( τ ) . Summing over i , we obtain: c ρ dim( ρ ) = 1 | G | X τ ∈ X Tr[ ρ ( τ )] = 1 | G | X τ ∈ X χ ρ ( τ ) . From (6), λ ρ = | G | c ρ , completing the proof. First, a word regarding normalization, as this is always a potential source ofconfusion when doing Fourier analysis. Given a graph G , we use the uniformprobability measure on the vertex-set V of G , not the counting measure. Theuniform measure induces the following inner product on R [ V ]: h f, g i = 1 | V | X v ∈ V f ( v ) g ( v );this induces the Euclean norm || f || = p h f, f i . If G = ( V, E ) is a graph, the adjacency matrix A of G is defined by A v,w = 1 { vw ∈ E ( G ) } ( v, w ∈ V ( G )) . This is a real, symmetric, n × n matrix, so there exists an orthonormalsystem of n eigenvectors of A , which forms a basis for R [ V ]. (Note that theeigenvalues of A are often referred to as the eigenvalues of G .)Hoffman [18] observed the following useful bound on the measure of anindependent set in a regular graph, in terms of the eigenvalues of the graph: Theorem 11. [18] Let G = ( V, E ) be a d -regular, n -vertex graph. Let A bethe adjacency matrix of G . Let { v , v , . . . , v n } be an orthonormal system ofeigenvectors of A , with corresponding eigenvalues d = λ ≥ λ ≥ . . . ≥ λ n =12 min (so that v is the all-1’s vector). If I ⊂ V is an independent set in G ,then | I || V | ≤ − λ min λ − λ min . (8) If equality holds, then I ∈ Span ( { v } ∪ { v i : λ i = λ min } ) . Proof.
Let f = 1 I , and let α = | I || V | . Observe that f t Af = X v,w ∈ I A v,w = 2 e ( G [ I ]) = 0 , since I is independent. Write f as a linear combination of the eigenvectors: f = n X i =1 a i v i . Then α = h f, v i = a . Moreover, by Parseval’s identity, we have P i a i = || f || = α . Now,0 = f t Af = n X i =1 a i λ i ≥ a λ + n X i =2 a i λ min = α λ + ( α − α ) λ min . Rearranging gives (8). If equality holds, then a i = 0 implies that i = 1 or λ i = λ min , completing the proof.A variant of Hoffman’s theorem, which will be crucial for us, comes fromweighting the edges of the graph G with real (possibly negative) weights. Theorem 12.
Let G = ( V, E ) be an n -vertex graph. Let G , . . . , G t beregular, spanning subgraphs of G , all having { v , v . . . , v n } as an orthonor-mal system of eigenvectors (where v is the all-1’s vector). Let λ ( j ) i be theeigenvalue of v i in G j . Let β , . . . , β t ∈ R , and let λ i = P j β j λ ( j ) i , and let λ min = min i λ i . If I ⊂ V is an independent set in G , then | I || V | ≤ − λ min λ − λ min . (9) If equality holds, then I ∈ Span ( { v } ∪ { v i : λ i = λ min } ) . roof. The proof is a simple generalization of that of Theorem 11. For each j , let A j be the adjacency matrix of G j , and let A = P j β j A j . We have0 = f t Af = t X j =1 β j f t A j f = t X j =1 β j n X i =1 a i λ ( j ) i = n X i =1 a i λ i ≥ λ a + X i a i λ min = α λ + ( α − α ) λ min . Rearranging gives (9). If equality holds, then a i = 0 implies that i = 1or λ i = λ min , completing the proof.We may think of the λ i ’s above as the eigenvalues of the linear combi-nation of graphs Y = t X j =1 β j G j . The corresponding linear combination of adjacency matrices A = t X j =1 β j A j is a so-called pseudo-adjacency matrix for G , meaning a symmetric matrixsuch that A v,w = 0 whenever vw / ∈ E ( G ); the λ i ’s are the eigenvalues of A .Finally, we will need the following cross-independent version of Hoffman’sTheorem. Variants of this theorem can be found in several sources, e.g. [2]. Theorem 13.
Let G = ( V, E ) be a d -regular, n -vertex graph, and let { v , v , . . . , v n } be an orthonormal system of eigenvectors of G , with cor-responding eigenvalues d = λ , λ , . . . , λ n ordered by descending absolutevalue (so that v is again the all-1’s vector). Let I, J ⊂ V be (not necessar-ily disjoint) sets of vertices in G such that there are no edges of G between I and J . Then s | I || V | · | J || V | ≤ | λ | λ + | λ | . (10) If equality holds, then I , J ∈ Span ( { v } ∪ { v i : | λ i | = | λ |} ) . roof. Let f and g be the characteristic functions of I and J respectively.As in the proof of Hoffman’s Theorem, write f = n X i =1 a i v i , g = n X i =1 b i v i . Let α = | I || V | , β = | J || V | . We have0 = 2 e ( I, J ) = f t Ag = n X i =1 a i b i λ i = αβλ + n X i =2 a i b i λ i . (11)Hence, by the Cauchy-Schwarz inequality, αβλ = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X i =2 a i b i λ i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n X i =2 | a i b i λ i | ≤ | λ | p α − α p β − β . Rearranging gives s αβ (1 − α )(1 − β ) ≤ | λ | λ . Recall the AM-GM inequality for two real variables (see for example [16]):( α + β ) / ≥ p αβ. This implies that(1 − α )(1 − β ) = 1 − α − β + αβ ≤ − p αβ + αβ = (1 − p αβ ) , and therefore p (1 − α )(1 − β ) ≤ − p αβ. Hence, √ αβ − √ αβ ≤ | λ | λ . Rearranging gives p αβ ≤ | λ | λ + | λ | , as required. The case of equality is dealt with as in the original Hoffmantheorem.Note that the generalization of Hoffman’s theorem that we mentionabove holds, with the obvious modifications, in the cross-independent case.15 Representation Theory of S n In this section we gather all the necessary background and results regardingthe representation theory of S n . Readers familiar with the basics of thistheory are invited to skip the following subsection. Our treatment follows Sagan [28].A partition of n is a non-increasing sequence of integers summing to n ,i.e. a sequence λ = ( λ , . . . , λ k ) with λ ≥ λ ≥ . . . ≥ λ k and P ki =1 λ i = n ;we write λ ⊢ n . For example, (3 , , ⊢
7; we sometimes use the shorthand(3 , ,
2) = (3 , ). The following two orders on partitions of n will be useful. Definition 4. (Dominance order) Let λ = ( λ , . . . , λ r ) and µ = ( µ , . . . , µ s ) be partitions of n . We say that λ D µ ( λ dominates µ ) if P ij =1 λ i ≥ P ij =1 µ i for all i (where we define λ i = 0 for all i > r , and µ j = 0 for all j > s ). It is easy to see that this is a partial order.
Definition 5. (Lexicographic order) Let λ = ( λ , . . . , λ r ) and µ = ( µ , . . . , µ s ) be partitions of n . We say that λ > µ if λ j > µ j , where j = min { i ∈ [ n ] : λ i = µ i } . It is easy to see that this is a total order which extends the dominanceorder.The cycle-type of a permutation σ ∈ S n is the partition of n obtainedby expressing σ as a product of disjoint cycles and listing its cycle-lengthsin non-increasing order. The conjugacy-classes of S n are precisely { σ ∈ S n : cycle-type( σ ) = λ } λ ⊢ n . Moreover, there is an explicit one-to-one correspondence between irreduciblerepresentations of S n (up to isomorphism) and partitions of n , which we nowdescribe.Let λ = ( λ , . . . , λ k ) be a partiton of n . The Young diagram of λ is anarray of n boxes, or cells , having k left-justified rows, where row i contains λ i cells. For example, the Young diagram of the partition (3 , ) is:16f the array contains the numbers { , , . . . , n } inside the cells, we call ita λ - tableau , or a tableau of shape λ ; for example:6 1 75 43 2is a (3 , )-tableau. Two λ -tableaux are said to be row-equivalent if theyhave the same numbers in each row; if a λ -tableau t has rows R , . . . , R l ⊂ [ n ]and columns C , . . . , C l ⊂ [ n ], then we let R t = S R × S R × . . . × S R l be therow-stablizer of t and C t = S C × S C × . . . × S C k be the column-stabilizer.A λ - tabloid is a λ -tableau with unordered row entries (or formally, arow-equivalence class of λ -tableaux); given a tableau t , we write [ t ] for thetabloid it produces. For example, the (3 , )-tableau above produces thefollowing (3 , )-tabloid: { }{ }{ } Consider the natural permutation action of S n on the set X λ of all λ -tabloids; let M λ = R [ X λ ] be the corresponding permutation module, i.e.the real vector space with basis X λ and S n action given by extending thepermutation action linearly. In general, M λ is reducible. However, we candescribe a complete set of real irreducible representations, as follows.If t is a tableau, let κ t = P π ∈ C t sgn( π ) π ; this is an element of the groupmodule R [ S n ]. Let e t = κ t { t } . This is a ( ± M λ ; we call the e t ’s polytabloids . Definition 6.
Let µ be a partition of n . The Specht module S µ is thesubmodule of M µ spanned by the µ -polytabloids: S µ = Span { e t : t is a µ -tabloid } . Theorem 14.
The Specht modules are a complete set of pairwise inequiva-lent, irreducible representations of S n . Hence, any irreducible representation ρ of S n is isomorphic to some S λ ;in this case, we say that ρ has Young diagram λ . For example, S ( n ) = M ( n ) is the trivial representation; M (1 n ) is the left-regular representation, and S (1 n ) is the sign representation sgn . 17rom now on we will write [ λ ] for the equivalence class of the irreduciblerepresentation S λ , χ λ for the character χ S λ , and ξ λ for the character of M λ .Notice that the set of λ -tabloids forms a basis for M λ , and therefore ξ λ ( σ ),the trace of the corresponding permutation representation, is precisely thenumber of λ -tableaux fixed by σ .We now explain how the permutation modules M µ decompose into irre-ducibles. Definition 7.
Let λ, µ be partitions of n . A λ -tableau is produced by placinga number between 1 and n in each cell of the Young diagram of λ ; if it has µ i i ’s (1 ≤ i ≤ n ) it is said to have content µ . A generalized λ -tableau issaid to be semistandard if the numbers are non-decreasing along each rowand strictly increasing down each column. Definition 8.
Let λ, µ be partitions of n . The Kostka number K λ,µ is thenumber of semistandard generalized λ -tableaux with content µ . Theorem 15. (Young’s rule) Let µ be a partition of n . Then the permuta-tion module M µ decomposes as M µ ∼ = ⊕ λ ⊢ n K λ,µ S λ . Hence, ξ µ = X λ ⊢ n K λ,µ χ λ . For example, M ( n − , , which corresponds to the natural permutationaction of S n on [ n ], decomposes as M ( n − , ∼ = S ( n − , ⊕ S ( n ) , and therefore ξ ( n − , = χ ( n − , + 1.The restriction of an irreducible representation of S n to the subgroup { σ ∈ S n : σ ( i ) = i ∀ i > n − k } = S n − k can be decomposed into irreduciblerepresentations of S n − k as follows: Theorem 16. (The branching rule.) Let α be a partition of n − k , and λ apartition of n . We write α ⊂ k λ if the Young diagram of α can be producedfrom that of λ by sequentially removing k corners (so that after removingthe i th corner, we have the Young diagram of a partition of n − i .) Let a α,λ be the number of ways of doing this; then we have [ λ ] ↓ S n − k = X α ⊢ n − k : α ⊂ k λ a α,λ [ α ] , nd therefore χ λ ↓ S n − k = X α ⊢ n − k : α ⊂ k λ a α,λ χ α . Definition 9.
Let λ = ( λ , . . . , λ k ) be a partition of n ; if its Young diagramhas columns of lengths λ ′ ≥ λ ′ ≥ . . . ≥ λ ′ l ≥ , then the partition λ t =( λ ′ . . . , λ ′ l ) is called the transpose of λ , as its Young diagram is the transposeof that of λ . Theorem 17.
Let λ be a partition of [ n ] ; then [ λ ] ⊗ [1 n ] = [ λ t ] . Hence, χ λ t = χ λ · sgn, and dim[ λ ] = dim[ λ t ] . Definition 10.
The hook of a cell ( i, j ) in the Young diagram of a partition µ is H i,j = { ( i, j ′ ) : j ′ ≥ j } ∪ { ( i ′ , j ) : i ′ ≥ i } . The hook length of ( i, j ) is h i,j = | H i,j | . Theorem 18 (Frame, Robinson, Thrall [10]) . If λ is a partition of n withhook lengths ( h i,j ) , then dim[ λ ] = n ! Q i,j h i,j . (12) In this subsection we state and prove several lemmas regarding representa-tions of S n and their characters; these will be instrumental in proving ourmain theorem. Let k ∈ N . Then there exists E k > depending on k alone suchthat for any irreducible representation [ λ ] of S n with all rows and columnsof length greater than n − k , dim[ λ ] ≥ E k n k +1 . To prove this lemma, we need two simple claims dealing with irreduciblerepresentations with a relatively long row or column, and a separate resultdealing with the rest.
Claim 1.
Let [ λ ] be an irreducible representation whose first row or columnis of length n − t . Then dim[ λ ] ≥ (cid:18) nt (cid:19) e − t . (13)19 roof. Note that if λ has first column of length n − t , then λ t has first rowof length n − t . Since dim[ λ ] = dim[ λ t ], we may assume that λ has first rowof length n − t .By the hook formula (12), it suffices to prove that Y i,j h i,j ≤ t !( n − t )! e t . Delete the first row R of the Young diagram of λ ; the resulting Youngdiagram D corresponds to a partition of t , and therefore a representation of S t , which has dimension t ! Q ( i,j ) ∈ D h i,j ≥ . Hence, Y ( i,j ) ∈ D h i,j ≤ t ! . We now bound the product of the hook lengths of the cells in the first row;this is of the form Y ( i,j ) ∈ R h i,j = n − t Y j =1 ( j + c j ) , where P n − tj =1 c j = t . Using the AM/GM inequality, we obtain: n − t Y j =1 j + c j j = n − t Y j =1 (cid:18) c j j (cid:19) ≤ n − t X j =1 c j j n − t n − t ≤ n − t + P n − tj =1 c j n − t ! n − t = (cid:18) tn − t (cid:19) n − t < e t . Hence, Y i,j h i,j ≤ t !( n − t )! e t , as desired. 20ote that, if t is suffiently small depending on n , (cid:0) nt (cid:1) e − t is an increasingfunction of t : Claim 2.
Let L ( n, t ) = (cid:0) nt (cid:1) e − t . Then L ( n, t ) ≤ L ( n, t + 1) for all t ≤ ( n − e ) / ( e + 1) .Proof. Observe that L ( n, t ) L ( n, t + 1) = e ( t + 1) n − t . Solving for when this expression is at most 1 proves the claim.For the representations not covered by Claim 1, we use the following.
Theorem 19 ([25]) . If α, ǫ > , then there exists N ( α, ǫ ) ∈ N such that forall n > N ( α, ǫ ) , any irreducible representation [ λ ] of S n which has all rowsand columns of length at most n/α has dim[ λ ] ≥ ( α − ǫ ) n . The proof of Lemma 2 is now immediate.
Proof of Lemma 2:
If the partition λ contains a row or column of lengthbetween n − ( n − e ) / ( e + 1) and n − k −
1, then by Claims 1 and 2,dim[ λ ] ≥ e − ( k +1) (cid:18) nk + 1 (cid:19) ≥ E k n k +1 , provided we choose E k > α = ( e + 1) /e − ǫ ′ for some small ǫ ′ >
0, and therefore,by Theorem 19, dim[ λ ] ≥ n k +1 , provided n is sufficiently large depending on k , completing the proof. We will be working with certain minors of the character table of S n . Thefollowing lemmas imply that certain related matrices are upper-triangularwith 1’s all along the diagonal. Lemma 3. If λ , µ are partitions of n , let K λ,µ denote the Kostka number,the number of semistandard λ -tableaux of content µ . If K λ,µ ≥ , then λ ≥ µ . Moreover, K λ,λ = 1 for all λ . roof. Observe that if there exists a semistandard generalized λ -tableau ofcontent µ , then all µ i i ’s must appear in the first i rows, so P ij =0 µ j ≤ P ij =0 λ j for each i . Hence, λ D µ . Since the lexicographic order extendsthe dominance order, it follows that λ ≥ µ . Observe that the generalized λ -tableau with λ i i ’s in the i th row is the unique semistandard generalized λ -tableau of content λ , so K λ,λ = 1 for every partition λ . Lemma 4.
Let λ be a partition of n , and let ξ λ be the character of thepermutation module M λ . Let σ ∈ S n . If ξ λ ( σ ) = 0 , then cycle-type ( σ ) E λ .Moreover, if cycle-type ( σ ) = λ , then ξ λ ( σ ) = 1 .Proof. The set of λ -tabloids is a basis for the permutation module M λ .Thus, ξ λ ( σ ), which is the trace of the corresponding representation on thepermutation σ , is simply the number of λ -tabloids fixed by σ . If ξ λ ( σ ) = 0,then σ fixes some λ -tabloid [ t ]. Hence, every row of length l in [ t ] is a unionof the sets of numbers in a collection of disjoint cycles of total length l in σ .Thus, the cycle-type of σ is a refinement of λ , and therefore cycle-type( σ ) E λ ,as required. If σ has cycle-type λ , then it fixes just one λ -tabloid, the onewhose rows correspond to the cycles of σ , so ξ λ ( σ ) = 1. Theorem 20.
Let C be the character table of S n , with rows and columnsindexed by partitions / conjugacy classes in decreasing lexicographic order(so C λ,µ = χ λ ( σ µ ) where σ µ is a permutation with cycle-type µ , and the top-left corner of C is χ ( n ) ( σ ( n ) ) .) Then the contiguous square minor ˜ C of C with rows and columns ψ : ψ > ( n − k, k ) is invertible and does not dependon n , provided n > k .Proof. Let K be the Kostka matrix, and let D be the matrix of permutationcharacters, D λ,µ = ξ λ ( σ µ ) , (14)where σ µ denotes a permutation with cycle-type µ . Let ˜ K and ˜ D denotethe top-left minor of K and D respectively (i.e. the minor with rows andcolumns ψ : ψ > ( n − k, k )).Recall that by Young’s rule (Theorem 15), we have M µ ∼ = ⊕ λ K λ,µ S λ , (15)and therefore ξ µ = X λ K λ,µ χ λ . (16)22ence,( K t C ) λ,µ = X τ K τ,λ C τ,µ = X τ K τ,λ χ τ ( σ µ ) = ξ λ ( σ µ ) = D λ,µ , (17)and therefore K t C = D. (18)Since the rows and columns of K are sorted in decreasing lexicographicorder, K is upper-triangular with 1’s all along the diagonal, by Lemma 3.Therefore, K t is lower-triangular with 1’s all along the diagonal.Since K t is lower-triangular, in addition to (18), we also have˜ K t ˜ C = ˜ D. (19)Since ˜ K t is lower-triangular with 1’s all along the diagonal, it is invertible,and therefore ˜ C = ( ˜ K t ) − ˜ D. By Lemma 4, ˜ D is upper-triangular with 1’s all along the diagonal, and istherefore invertible. It follows that ˜ C is invertible also.We will now show that ˜ K and ˜ D are independent of n , provided n > k ;this will prove that ˜ C is also independent of n .Let λ > ( n − k, k ) be a partition. Then λ ≥ n − k . Write λ ′ =( λ − ( n − k ) , λ , . . . ). (Note that this may not be a bona fide partition, asit may not be in non-increasing order.) Now the mapping λ λ ′ has thesame image over { λ : λ > ( n − k, k ) } for all n ≥ k : namely, ‘partitions’ of k where the first row is not necessarily the longest.We first consider K . Recall once again that K λ,µ is the number ofsemistandard λ -tableaux of content µ . Let t be a semistandard λ -tableau ofcontent µ ; we now count the number of choices for t . Since the numbers ina semistandard tableau are strictly increasing down each column and non-increasing along each row, and µ ≥ n − k , we must always place 1’s in thefirst n − k cells of the first row of t . We must now fill the rest of the cellswith content µ ′ . Provided n ≥ k , µ ′ is independent of n , and the remainingcells in the first row have no cells below them, so the number of ways ofdoing this is independent of n . Hence, the entire minor ˜ K is independent of n . Now consider D . Recall that D λ,µ = ξ λ ( σ µ ) is simply the number of λ -tabloids fixed by σ µ . To count these, first note that the numbers in thelong cycle of σ (which has length at least n − k ) must all be in the first rowof the λ -tabloid (otherwise the long cycle of σ must intersect two or more23ows, as n − k > k .) This leaves us with a ( λ − µ , λ , . . . , λ r )-‘tableau’,which we need to fill with the remaining n − µ elements in such a way that σ fixes it. It is easy to see that, again, the number of ways of doing this isindependent of n .In particular, if n ≥ k , the number of partitions λ of n such that λ ≥ ( n − k, k ) is independent of n ; we denote it by q k . Note that q k = k X t =0 p t , where p t denotes the number of partitions of t .We need a slightly more general result, which allows us to split some ofthe partitions. Definition 11.
Assume that n > k + 1 , and let µ = ( µ , . . . , µ r ) be apartition of n with µ ≥ n − k . We defineSplit ( µ ) = ( µ − k − , k + 1 , µ , . . . , µ r ) . It is easy to see that Split( µ ) is indeed a partition (i.e. it is in descendingorder). Further, exactly one of µ and Split( µ ) is even. Theorem 21.
Let C be as above, and let n > k + 1 . Let φ , . . . , φ q k − bethe partitions > ( n − k, k ) . Let µ , . . . , µ q k − be partitions such that, foreach j , either µ j = φ j or else µ j = Split ( φ j ) . Then the square minor ˘ C of C with i th row φ i and j th column µ j is independent of the choices of the µ j ’s,so is always equal to the top-left minor ˜ C .Proof. It is easy to see that if λ, µ > ( n − k, k ), σ has cycle-type µ and σ ′ has cycle-type Split( µ ), then in fact, ξ λ ( σ ) = ξ λ ( σ ′ ). (All rows of a λ -tabloidbelow the first have length at most k . Hence, if a permutation σ ′ with cycle-type Split( µ ) fixes a λ -tabloid [ t ], the numbers in ( µ − k − k +1)-cycle must all lie in the first row of [ t ]. It follows that a permutation σ produced by merging these two cycles of σ ′ fixes exactly the same λ -tabloidsas σ ′ does.) Since the Kostka matrix K is upper-triangular with 1’s all downthe diagonal, { ξ λ : λ > ( n − k, k ) } and { χ λ : λ > ( n − k, k ) } are basesfor the same linear space, and therefore χ λ ( σ ) = χ λ ( σ ′ )for each λ > ( n − k, k ). Thus, C λ, Split( µ ) = C λ,µ ∀ λ, µ > ( n − k, k ) , as required. 24 .2.3 Functions with Fourier transform concentrated on the ‘fat’irreducible representations One of the recurring themes in applications of discrete Fourier analysis tocombinatorics is proving that certain functions depend on few coordinates,by showing that their Fourier transform is concentrated on the ‘low frequen-cies’, the characters indexed by small sets. Examples of this can be found forexample in [5], [12] and [13]. In this paper, we need a non-Abelian analogue.We show that functions on S n whose Fourier transform is supported on ir-reducible representations which are large with respect to the lexicographicorder, are spanned by the cosets of the pointwise stabilizers of small sets. Definition 12.
Let V k be the linear space of functions whose Fourier trans-form is supported only on representations ≥ ( n − k, k ) . We are now ready to prove
Theorem 7. V k is the span of the k -cosets.Proof. First, we show that the characteristic function of any k -coset is indeedin V k . Let T = T a b ,...,a k b k . It is easy to check that if f : S n → R , τ, π ∈ S n , and g : S n → R σ f ( πστ ) , i.e. the function g is a ‘double-translate’ of f , then ˆ f ( ρ ) = 0 ⇒ ˆ g ( ρ ) = 0.Hence, if f ∈ V k , so is g . (This is saying that V k is a ‘two-sided ideal’ of thegroup algebra R [ S n ].)Hence, by double-translation, without loss of generality, we may assumethat a i = b i = i for each i ∈ [ k ], i.e. T = T ,...,k k , so T ∼ = S n − k . Weuse the branching rule (Theorem 16.) If ρ < ( n − k, k ), then ρ has at least k + 1 cells outside the first row. In that case, every µ ⊂ k ρ is a nontrivialirreducible representation of S n − k . We claim that this implies X σ ∈ S n − k µ ( σ ) = 0 . To see this, observe that the linear map P σ ∈ S n − k µ ( σ ) commutes with µ ,and therefore by Schur’s Lemma, it must be a scalar multiple of the identitymap. Its trace is zero, since χ µ is orthogonal to the trivial character, so itmust be the zero map. Hence, c T ( ρ ) = 0 for each ρ < ( n − k, k ).25Note also that if ρ = ( n − k, k ), then the only µ ⊂ k ρ with nonzerosum on T is that corresponding to the partition ( n − k ), i.e. the trivialrepresentation, which has sum ( n − k )! on T . Therefore, c T (( n − k, k )) = 0.This will be useful in the proof of Theorem 22.)In the other direction, let f ∈ V k . Using the Fourier inversion formula,we have f ( σ ) = X ρ ≥ ( n − k, k ) dim[ ρ ]Tr h ˆ f ( ρ ) ρ ( σ − ) i . (20)Consider the permutation module M ( n − k, k ) . By Young’s Rule (Theorem15), M ( n − k, k ) must contain at least one copy of every Schur module S ρ with ρ ≥ ( n − k, k ), and no others. We can therefore rewrite the previousformula as: f ( σ ) = Tr (cid:2) Aψ ( σ − ) (cid:3) , (21)where ψ is a matrix representation corresponding to M ( n − k, k ) , and A is ablock-diagonal matrix whose blocks correspond to the irreducible modules S ρ , i.e. it has K ρ, ( n − k, k ) blocks corresponding to ρ , all equal todim[ ρ ] K ρ, ( n − k, k ) ˆ f ( ρ ) , where the Kostka number K ρ, ( n − k, k ) is the multiplicity of S ρ in M ( n − k, k ) .Next, recall that the permutation module M ( n − k, k ) corresponds to thepermutation action of S n on ( n − k, k )-tabloids, which can be identifiedwith ordered k -tuples of distinct numbers between 1 and n . Since trace isconjugation-invariant, we can perform a change of basis, replacing A and ψ by similar matrices B and φ , so that f ( σ ) = Tr (cid:2) Bφ ( σ − ) (cid:3) , (22)where φ α,β ( τ ) is 1 if τ takes the ordered k -tuple α to the ordered k -tuple β ,and 0 otherwise. But then equation (22) means precisely that f = X α,β B α,β T α β , completing the proof. 26 Proof of main theorems
In this section we finally proceed to eat the pudding. In view of Theorem 7,we would like to prove that whenever I is a maximum-sized k -intersectingfamily, the Fourier transform of its characteristic function is supported onthe irreducible representations corresponding to partitions ρ ≥ ( n − k, k ),i.e. those whose Young diagram has first row of length at least n − k . Letus call these representations the ‘fat’ representations, and their transposes(those whose Young diagram has first column of height at least n − k ) the‘tall’ representations. The rest will be called ‘medium’ representations.Given a k -intersecting family I ⊂ S n , we will first consider it as anindependent set in the Cayley graph Γ k on S n generated by FPF k , the setof permutations with less than k fixed points. Since FPF k is a union ofconjugacy classes, by Theorems 10 and 14, the eigenvalues of Γ k are givenby λ ( k ) ρ = 1dim[ ρ ] X σ ∈ FPF k χ ρ ( σ ) ( ρ ⊢ n ) . (23)Fixed-point-free permutations are also called derangements , and Γ isalso called the derangement graph . Let d n = | FPF ( n ) | ; it is well-known(and easy to see, using the inclusion-exclusion formula), that d n = n X i =0 ( − i (cid:18) ni (cid:19) ( n − i )! = n X i =0 ( − i n ! i ! = (1 /e + o (1)) n ! . For n ≥
5, the eigenvalues of Γ satisfy: λ (1)( n ) = d n λ (1)( n − , = − d n / ( n − | λ (1) ρ | < cd n /n < d n / ( n −
1) for all other ρ ⊢ n (where c is an absolute constant). Hence, the matrix Γ has λ min /λ = − / ( n − . As observed in [15] and [26], this implies via Hoffman’s bound (Theorem11) that any 1-intersecting family I ⊂ S n satisfies | I | ≤ ( n − I ∈ V , from which we may conclude (e.g. from the k = 1case of Theorems 7 and 8) that I is a 1-coset. Also, we may conclude via27heorem 13 that any 1-cross-intersecting pair of families I, J ⊂ S n safisfy | I || J | ≤ (( n − . If equality holds, then 1 I , J ∈ V , which enables us toconclude that I = J is a 1-coset of S n . This proves Leader’s conjecture on1-cross-intersecting families in S n for n ≥ n = 4).However, for k fixed and n large, calculating the least eigenvalue of Γ k and applying Hoffman’s bound only gives an upper bound of Θ(( n − k -intersecting subset of S n . Indeed, λ ( k )( n − , = 1dim[ n − , X σ ∈ FPF k χ ( n − , ( σ )= 1 n − X σ ∈ FPF k ( ξ ( n − , ( σ ) − n − k − X i =0 (cid:18) ni (cid:19) d n − i ( i − − n − /e + o (1)) n ! − k − X i =1 i ( i + 1)! ! . Note that ∞ X i =1 i ( i + 1)! = ddx e x − x (cid:12)(cid:12)(cid:12)(cid:12) x =1 = 1 , so for any k ∈ N , 1 − k − X i =1 i ( i + 1)! > , and therefore λ ( k )( n − , = − Θ k (( n − | λ ( k ) ρ | ≤ a k n ! /n for all partitions ρ = ( n ) , ( n − , a k > k alone. Hence,if k is fixed and n is large, Γ k has λ min = λ ( k )( n − , = − Θ k (( n − . Hence, applying Hoffman’s bound only gives | I | ≤ Θ k (( n − , for a k -intersecting I ⊂ S n .Instead, we will construct a linear combination Y of subgraphs of Γ k (each a Cayley graph generated by a conjugacy-class within FPF k ) which28as the correct eigenvalues for us to prove Theorems 5 and 6 from Theorem12. By Theorem 10, if X , . . . , X t are conjugacy-classes within FPF k , and β , . . . , β t ∈ R , then the eigenvalues of the linear combination Y = t X j =1 d j Cay( S n , X j )are λ ρ = 1dim[ ρ ] t X j =1 d j X σ ∈ X j χ ρ ( σ ) = 1dim[ ρ ] t X j =1 d j | X j | χ ρ ( τ j ) , ( ρ ⊢ n ) (24)where τ j denotes any permutation in X j . So the eigenvalues of Y stillcorrespond to partitions of n , and are therefore relatively easy to handle.The value of λ min /λ required in Theorem 12 to produce the upper boundin Theorem 5 is as follows: Corollary 1.
Define ω = ω n,k = − ( n − k )! n ! − ( n − k )! = − n ( n − . . . ( n − k + 1) − . Assume the conditions of Theorem 12. If λ min /λ = ω , and I is an inde-pendent set in G , then | I | ≤ ( n − k )! .Proof. Immediate from Theorem 12.Since rescaling the linear combination of graphs makes no differenceto the above, our aim will be to construct a linear combination Y with λ = 1 and λ min = ω . Since equality has to hold in Theorem 12 when theindependent set is a k -coset, we know that for any k -coset T , we must have1 T ∈ Span ( { v } ∪ { v i : λ i = ω } ) . By Theorem 7, it follows that we must have λ ρ = ω for each fat partition ρ = ( n ).We will in fact construct two linear combinations, Y even and Y odd , fromCayley graphs generated by respectively even/odd conjugacy-classes withinFPF k . We design these linear combinations so that λ ( n ) = 1 and λ ρ = ω forall fat ρ = ( n ). Recalling for any partition ρ , we have χ ρ t = χ ρ · sgn , where sgn is the sign character, we see from (24) that for any partition ρ , we have: • λ ρ t = λ ρ in Y even , whereas 29 n ) fat, = ( n ) (1 n ) tall, = (1 n ) medium Y even ω ω o ( | ω | ) Y odd ω − − ω o ( | ω | ) Y ω o ( | ω | )Table 1: Eigenvalues • λ ρ t = − λ ρ in Y odd .Hence, simply taking Y = ( Y even + Y odd )will ensure that the eigenvalues of Y corresponding to the tall representationsare all zero.Moreover, we will show that in both Y even and Y odd (and consequentlyin Y ), the eigenvalues corresponding to the medium representations all haveabsolute value | λ | ≤ c k | ω | / √ n = o ( | ω | ), where c k > k alone.So provided n is sufficiently large, ω is both the minimal eigenvalue of Y andthe second-largest in absolute value, and is attained only on the non-trivialfat irreducible representations. The situation is summarized in Table 4.1;note that the o ( | ω | ) function is always the same function.Applying Theorem 12 to Y will not only prove that | I | ≤ ( n − k )!, butalso that if equality holds, then the Fourier transform of the characteristicfunction of I is totally supported on the fat representations, yielding theproof of our main theorem (pending the results of section 5). Since ω isalso the eigenvalue of second-largest absolute value, we will also be able todeduce Theorem 4, concerning k -cross-intersecting families.In order to carry out our plan, we will identify appropriate conjugacy-classes X j to use as the generating sets for our Cayley graphs, and appro-priate coefficients d j . The set of linear equations the d j ’s must satisfy willalways correspond to a specific square minor of the character table of S n ,which turns out to be non-singular and independent of n (for n sufficientlylarge). The latter statement is precisely the content of Theorems 20 and 21.As for the medium representations, we will show that their eigenvalueshave small absolute value using the lower bound we proved on their dimen-sions in Lemma 2. The Cayley graphs we will use to construct our linear combination will begenerated by conjugacy-classes of permutations with specific cycle-types.30hese cycle-types will fall into two categories: those corresponding to par-titions > ( n − k, k ), and those corresponding to partitions π , where( n − k − , k + 1) ≥ π > ( n − k − , k + 1 , k ) . Note that the second category is exactly the split of the first category(see definition 11). We need to use the split partitions in order to ensurethat our conjugacy-classes have the correct sign.We will need the following bounds on the sizes of the above conjugacy-classes:
Lemma 5.
Let X be a conjugacy class of S n with a cycle of length n − t ,where t < n/ . Then n !( n − t ) t t ≤ | X | ≤ n − . Proof.
Suppose the cycle-type of X is ( n − t, c , . . . , c l ), where l ≤ t and P li =1 c i = t . Define a mapping F : S n → X by taking a permutation σ ∈ S n , writing it in sequence-notation σ (1) , σ (2) , . . . , σ ( n ) , and then placing parentheses at the appropriate points, producing a per-mutation in X (written in disjoint cycle notation). This mapping is clearlysurjective, and each element of X has the same number of preimages, N say. Note that the preimages of a permutation in X are obtained by rotat-ing each cycle and then permuting the cycles of length j for each j . Let a j = |{ i : c i = j }| for each j ; then N = ( n − t ) Y j a j ! l Y i =1 c i . Observe that n/ < n − t ≤ N ≤ ( n − t ) l ! l Y i =1 c i ≤ ( n − t ) l !( t/l ) l ≤ ( n − t ) t l ≤ ( n − t ) t t , using the AM/GM inequality. Hence, n !( n − t ) t t ≤ | X | ≤ n − , as required. 31e now proceed to the key step of the proof. Theorem 22. If ρ is a partition of n , let X ρ denote the conjugacy classof permutations with cycle-type ρ . Assume that n > k + 1 , and let q k denote the number of partitions ≥ ( n − k, k ) . Let φ , . . . , φ q k − be thepartitions of n which are > ( n − k, k ) (in decreasing lexicographic order).Let µ , . . . , µ q k − be such that for all j , either µ j = φ j or else µ j = Split ( φ j ) .Let G j = Cay ( S n , X µ j ) , and let ( λ ( j ) i ) i be the eigenvalues of G j , in decreasinglexicographic order of their representations (so that λ ( j )1 = | X µ j | ).Then there exist d , . . . , d q k − ∈ R such that: X j d j λ ( j ) i = (cid:26) , for i = 1 ω, for < i ≤ q k (25) Moreover, there exists B k > depending only on k such that max j | d j | ≤ B k ( n − . Theorem 22 will enable us to get the correct eigenvalues on the fat rep-resentations, but we will also need the following, which ensures that theeigenvalues for the medium representations are smaller in absolute value:
Theorem 23.
Under the assumptions and notation of Theorem 22, let ρ bea medium partition, and let λ ρ = P q k − j =1 d j λ ( j ) ρ . Then | λ ρ | ≤ c k | ω | / √ n = o ( | ω | ) , where c k > depends only on k . We defer the proof of this until later, and turn first to the proof ofTheorem 22.
Proof of Theorem 22:
There are three steps in the proof. Step 1 is showingthe existence of coefficients d j such that equation (25) holds for i < q k . Step2 is showing that with these coefficients, the equation also holds for i = q k .Step 3 is obtaining the bound on max j | d j | . Step 1 ( i < q k ): Consider the system of linear equations
M d = b , where d = ( d , . . . , d q k − ), b = (1 , ω, . . . , ω ), and M ji = λ ( j ) i . By Theorem 10, λ ( j ) i = 1dim[ φ i ] X τ ∈ X j χ φ i ( τ ) = | X j | dim[ φ i ] χ φ i ( τ j ) , (26)32here τ j is a representative for the conjugacy-class X j (recall that charactersare class-functions). Equivalently, M = N ˘ C t N , where ˘ C denotes the minor of the character table of S n with rows φ , . . . , φ q k − and columns µ , . . . , µ q k − , and N and N are the diagonal row- and column-normalization matrices respectively; explicitly,( N ) i,j = δ i,j dim[ φ i ] , ( N ) i,j = δ i,j | X j | . (27)Recall from Theorem 21 that ˘ C is always equal to the top-left minor ˜ C ,and that ˜ C (and therefore ˜ C t ) is invertible. Therefore, there is a (unique)solution for d , so we can find appropriate values for d , . . . , d q k − . Step 2 ( i = q k ): Write λ = X j d j λ ( j ) q k . We must show that λ = ω. This will follow from analyzing the proof of thegeneralized Hoffman bound, Theorem 12, when G = Γ k and the independentset I is a k -coset.Let T be a k -coset of S n , and let f = 1 T be its characteristic function.Choose the orthonormal basis of eigenvectors { v i } in Theorem 12 to bethe normalization of the orthogonal basis formed by the entries of (matrix-equivalents of) the irreducible representations [ ρ ] (see section 2.3). Observethat α = E f = ( n − k )! n ! . By the argument in the proof of Theorem 7, we see that b f ( ρ ) = 0 ⇒ ρ ≥ ( n − k, k ) , (28)and moreover, b f (( n − k, k )) = 0.Let W ( ρ ) = T r [ ˆ f ( ρ )( ˆ f ( ρ )) t ]. This is simply the L -weight of the coeffi-cients of eigenvectors v i which correspond to entries of (the matrix-equivalentof) [ ρ ], i.e. if we write as usual f = P a i v i , then W ( ρ ) = P a i , where thesum is over all i such that v i is (the normalization of) one of the (dim ρ ) eigenvectors which are entries of [ ρ ]. Hence, W ( ρ ) = 0 ⇒ ρ ≥ ( n − k, k ) , W (( n − k, k )) = 0.Using the coefficients d , . . . , d q k − in Theorem 12, we have λ = 1 and λ t = ω for 1 < t < q k . Since T is an independent set in the graph G = Γ k ,we have0 = X ρ ≥ ( n − k, k ) λ ρ W ( ρ )= α + ω ( α − α − W (( n − k, k ))) + λW (( n − k, k )) . By definition of ω , we have ω ( α − α ) = − α ; combining this with the aboveyields λ = ω , as required. Step 3 (bounding the coefficients d j ) : Recall that M = N ˜ C t N ,and by Theorem 21, ˜ C is invertible and independent of n . Hence, the entriesof ( ˜ C t ) − are uniformly bounded by some function of k alone. Since M d = b ,we have d = M − b = ( N ) − ( ˜ C t ) − ( N − b ) . (29)We now proceed to bound uniformly the entries of the vector N − b . Wehave ( N − ) ij = δ ij dim[ φ i ] , and therefore ( N − b ) i = dim[ φ i ] b i . For i = 1, we have ( N − b ) = dim[ φ ] b = 1 · . For i >
1, we havedim[ φ i ] + 1 ≤ dim( M ( n − k, k ) ) = n ( n − . . . ( n − k + 1) , since both φ i and the trivial representation are constituents of M ( n − k, k ) .Hence, | ( N − b ) i | = | dim[ φ i ] || ω | ≤ . We now bound uniformly the entries of the matrix ( N ) − . Using Lemma5, we have( N − ) ij = δ ij | X j | ≤ ( n − k − k − k − n ! ≤ b k ( n − , where b k > k . 34ombining these bounds with (29), we see that there exists B k > k such that | d j | ≤ B k ( n − ∀ j < q k , completing the proof.Next, we need two more lemmas to assist in the proof of Theorem 23. Lemma 6.
Let G be a finite group, let X ⊂ G be inverse-closed andconjugation-invariant, and let Cay ( G, X ) be the Cayley graph on G withgenerating set X . Let ρ be an irreducible representation of G with dimension d , and let λ ρ be the corresponding eigenvalue of Cay ( G, X ) , as in Theorem10. Then | λ ρ | ≤ p | G || X | d . Proof.
Since the irreducible characters of G are orthonormal, we have h χ ρ , χ ρ i = 1 | G | X g ∈ G | χ ρ ( g ) | = 1 . By the Cauchy-Schwarz inequality, |h χ ρ , X i| ≤ q h χ ρ , χ ρ ih X , X i = p | X | / | G | . Substituting into Equation 7, we obtain | λ ρ | = | G ||h χ ρ , X i| d ≤ p | G || X | d , as required.Combining this with Lemma 5 yields: Lemma 7.
Let X be a conjugacy-class of S n , with cycle-type having a cycleof length at least n − k − . Let Cay ( S n , X ) be the Cayley graph on S n generated by X . For any C k > , there exists D k > , depending only on k ,such that if ρ is an irreducible representation of dimension at least C k n k +1 ,then the corresponding eigenvalue λ ρ of Cay ( S n , X ) satisfies | λ ρ | < D k ( n − | ω n,k |√ n . roof. Note that | ω n,k | = Θ(1 /n k ). By choosing D k large enough, we mayassume that n > k + 2, so n − k − > n/
2, and therefore the hypothesesof Lemma 5 hold. Assume that dim ρ ≥ C k n k +1 . Then | λ ρ | ≤ p | G || X | C k n k +1 ≤ p n !( n − C k n k +1 = √ n − C k n k +1 / ≤ D k ( n − | ω n,k |√ n , as required.We can now prove Theorem 23. Proof of Theorem 23:
Assume the hypotheses of Theorem 23. By Lemma2, we have dim[ ρ ] ≥ E k n k +1 . Hence, we see that | λ ρ | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q k − X j =1 d j λ ( j ) ρ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( q k −
1) max j | d j | max j | λ ( j ) ρ |≤ ( q k − B k ( n − D k ( n − | ω n,k |√ n = c k | ω | / √ n, as required. Finally, we can now construct the linear combination of Cayley graphs wewill use to prove our main theorem.
Theorem 24.
There exists a linear combination Y even of Cayley graphsgenerated by conjugacy-classes of even permutations within FPF k , such thatits eigenvalues are as described in the first line of Table 1, i.e., • λ ( n ) = 1 , λ ρ = ω for each fat ρ = ( n ) , • λ (1 n ) = 1 , • λ ρ = ω for each tall ρ = (1 n ) , and • | λ ρ | ≤ c k | ω | / √ n ≤ o ( | ω | ) for each medium ρ , where c k > dependsonly on k .Proof. Recall that for each partition φ j > ( n − k, k ), exactly one of φ j andSplit( φ j ) is even. For each partition φ j > ( n − k, k ), define µ j = (cid:26) φ j , if X φ j is an even conjugacy class;Split( φ j ) , if X φ j is an odd conjugacy class . Then each X µ j consists of even permutations. Take Y even = q k − X j =1 d j Cay( S n , X µ j ) , where the d j ’s are as defined in Theorem 22. Then we have λ ( n ) = 1, andfor each fat ρ = ( n ), we have λ ρ = ω . By Theorem 17, for any partition ρ ,we have χ ρ t = χ ρ · sgn. So by equation (24), for any partition ρ , we have λ ρ t = λ ρ , since our Cayley graphs are all generated by conjugacy-classes ofeven permutations. Therefore, λ (1 n ) = 1, and λ ρ = ω for each tall ρ = (1 n ).By Theorem 23, | λ ρ | ≤ c k | ω | / √ n ≤ o ( | ω | ) for each medium ρ , completingthe proof. Theorem 25.
There exists a linear combination Y odd of Cayley graphs gen-erated by conjugacy-classes of odd permutations within FPF k , such that itseigenvalues satisfy • λ ( n ) = 1 , • λ ρ = ω for all fat ρ = ( n ) , • λ (1 n ) = − , • λ ρ = − ω for each tall ρ = (1 n ) , and • | λ ρ | ≤ c k | ω | / √ n ≤ o ( | ω | ) for each medium ρ , where c k > dependsonly on k . roof. For each partition φ j > ( n − k, k ), define µ j = (cid:26) φ j , if X φ j is an odd conjugacy class;Split( φ j ) , if X φ j is an even conjugacy class . Then each X µ j consists of odd permutations. Take Y odd = q k − X j =1 d j Cay( S n , X µ j ) , where the d j ’s are as defined in Theorem 22. Then we have λ ( n ) = 1, andfor each fat ρ = ( n ), we have λ ρ = ω . This time, our Cayley graphs are allgenerated by odd permutations, so we have λ ρ t = − λ ρ for every partition ρ .Hence, λ (1 n ) = −
1, and λ ρ = − ω for each tall ρ = (1 n ). Again, by Theorem23, | λ ρ | ≤ c k | ω | / √ n ≤ o ( | ω | ) for each medium ρ , completing the proof. Theorem 26.
There exists a linear combination Y of Cayley graphs gener-ated by conjugacy-classes within FPF k , such that its eigenvalues satisfy: • λ ( n ) = 1 , • λ ρ = ω for each fat ρ = ( n ) , • λ ρ = 0 for each tall ρ , and • | λ ρ | ≤ c k | ω | / √ n ≤ o ( | ω | ) for each medium ρ , where c k > dependsonly on k .Proof. Set Y = ( Y even + Y odd ). Proof of Theorems 5 and 6: If n is sufficiently large depending on k , then c k | ω | / √ n < | ω | , so ω is both the minimum eigenvalue of Y and the second-largest in absolute value. By our choice of ω , applying the generalizedHoffman Theorem 12 to Y , we see that if I ⊂ S n is k -intersecting, then | I | ≤ ( n − k )!, and that if equality holds, then the characteristic function1 I ∈ V k . Similarly, applying Theorem 13 to Y , we see that if I, J ⊂ S n are k -cross-intersecting, then | I || J | ≤ (( n − k )!) , and that if equality holds,then 1 I , J ∈ V k . 38 Boolean functions
Recall that V k is the linear space of real-valued functions on S n whose Fouriertransform is supported only on irreducible representations ≥ ( n − k, k ).Theorem 7 states that V k is spanned by the k -cosets of S n . We now wishto characterize the Boolean functions in V k , proving (a strengthened versionof) Theorem 8.We will show that every non-negative function in V k can be written asa linear combination of the characteristic functions of k -cosets with non-negative coefficients , and every 0 / V k can be writtenwith 0 / A k be the set of ordered k -tuples of distinct numbers between 1 and n , and let ( n ) k = |A k | = n !( n − k )! = n ( n − . . . ( n − k + 1). We will prove thefollowing Theorem 27.
Let f ∈ V k be non-negative. Then there exist non-negativecoefficients ( b α,β ) α,β ∈A k such that f = P b α,β T α β . Furthermore, if f isBoolean, then f is the characteristic function of a disjoint union of k -cosets. For didactic reasons, we begin by dealing with the case k = 1. Theorem 28. If f ∈ V is nonnegative, then there exist b i,j ≥ such that f = P i,j b i,j T i j .Proof. We say a real n × n matrix A = ( a i,j ) i,j ∈ [ n ] represents a function f ∈ V if f = X i,j a i,j T i j , or equivalently, f ( σ ) = n X i =1 a i,σ ( i ) ∀ σ ∈ S n . Let A be a matrix representing f . Our task is to find a non-negative matrix B which also represents f . Let x , . . . , x n and y , . . . , y n be real numberssuch that n X i =1 x i + n X j =1 y j = 0 . Let B = ( b i,j ) i,j ∈ [ n ] be the matrix produced from A by adding x i to row i for each i , and then adding y j to column j for each j , i.e. b i,j = a i,j + x i + y j ( i, j ∈ [ n ]) . B also represents f . We wish to show that there exists a choiceof x i ’s and y j ’s such that the matrix B has all its entries non-negative, i.e.we wish to solve the system of inequalities x i + y j ≥ − a i,j (1 ≤ i, j ≤ n ) subject to X i x i + X j y j = 0 . (30)By the strong duality theorem of linear programming (see [24]), this is un-solvable if and only if there exist c i,j ≥ X j c i,j = 1 (1 ≤ i ≤ n ) , X i c i,j = 1 (1 ≤ j ≤ n ) , X i,j c i,j a i,j < . Suppose for a contradiction that this holds. The matrix C = ( c i,j ) i,j ∈ [ n ] isbistochastic, and therefore by Birkhoff’s theorem (see [4]), it can be writtenas a convex combination of permutation matrices, C = r X t =1 s t P σ t , where s t ≥ ≤ t ≤ r ), P rt =1 s t = 1, σ , . . . , σ t ∈ S n , and P σ denotes thepermutation matrix of σ , i.e. ( P σ ) i,j = 1 { σ ( i )= j } . But then P i,j c i,j a i,j is aconvex combination of values of f : X i,j c i,j a i,j = r X t =1 s t X i,j ( P σ t ) i,j a i,j = r X t =1 s t n X i =1 a i,σ t ( i ) = r X t =1 s t f ( σ t ) (31)and is therefore non-negative, a contradiction. Hence, the system (30) issolvable, proving the theorem.The following corollary is immediate. Corollary 2. If f ∈ V is Boolean, then it is the characteristic function ofa disjoint union of 1-cosets.Proof. By induction on the number of non-zero values of f . Let f ∈ V be a Boolean function, and suppose the statement is true for all Booleanfunctions with fewer non-zero values. By Theorem 28, there exist b i,j ≥ f = X i,j b i,j T i j . Choose i, j such that b i,j >
0. Then f > T i j , so f = 1 on T i,j . Hence, f − T i j is also Boolean, with fewer non-zero values than f , and so by theinduction hypothesis, it is the characteristic function of a disjoint union of1-cosets. Hence, the same is true of f , as required.40e now extend this proof to the case k >
1, proving Theorem 27. How-ever, some preliminaries are necessary.Let A be an ( n ) k × ( n ) k matrix with rows and columns indexed by A k .Choose an ordering of the k coordinates of the k -tuples, or equivalently apermutation π ∈ S k , and consider the natural lexicographic ordering on A k induced by this ordering, i.e. α < β iff α π ( m ) < β π ( m ) , where m = min { l : α π ( l ) = β π ( l ) } . This lexicographic ordering on A k recursively partitions A into blocks: first it partitions A into n ( n − k − × ( n − k − blocks B i,j according to the π (1)-coordinate of each k -tuple; then it partitions eachblock B i,j into ( n − ( n − k − × ( n − k − sub-blocks according to the π (2)-coordinate of each k -tuple, and so on.The following two recursive definitions concerning such matrices will becrucial. Definition 13.
We define a k -line in an ( n ) k × ( n ) k matrix A as follows.If k = 1 , i.e. A is an n × n matrix, a 1-line in A is just a row or columnof A . If k > , and A is an ( n ) k × ( n ) k matrix, a k -line in A is given bychoosing an ordering π of the k -coordinates, partitioning A into n blocksaccording to the π (1) -coordinate of each k -tuple as above, choosing a row orcolumn of ( n − k − × ( n − k − blocks B i,j , and then taking a union of ( k − -lines, one from each block. Definition 14.
We say that an ( n ) k × ( n ) k matrix A is k -bistochastic ifthe following holds. If k = 1 , an n × n matrix A is if itis bistochastic in the usual sense, i.e. all its entries are non-negative, andall its row and column sums are 1. If k > , an ( n ) k × ( n ) k matrix A is k -bistochastic if, for any partition into n blocks B i,j of size ( n − k − × ( n − k − according to a lexicographic order on A k induced by any of the k ! orderings of the coordinates, there exists a bistochastic n × n matrix R =( r i,j ) and n ( k − -bistochastic matrices M i,j of order ( n − , such that B i,j = r i,j M i,j . The following two self-evident observations indicate the relevance of thesetwo notions.
Observation 1. An ( n ) k × ( n ) k matrix with non-negative entries is k -bistochastic if and only if the sum of the entries of every k -line in it (comingfrom any of the k ! recursive partitions into blocks) is 1. Observation 2.
Given a k -line ℓ , the corresponding k -cosets { T α β : ( α, β ) ∈ ℓ } partition S n . σ ∈ S n induces a permutation of A k ; we write P ( k ) σ forthe corresponding ( n ) k × ( n ) k permutation matrix, i.e.( P ( k ) σ ) α,β = 1 { σ ( α )= β } . These n ! permutations are only a small fraction of all (( n ) k )! permutations of A k . It is not hard to show that any P ( k ) σ is k -bistochastic. It is a bit harderto show that any other permutation matrix is not k -bistochastic (for thisit is important to note that we demand taking into account lexicographicorderings induced by all possible orderings of the coordinates.)The authors thank Siavosh Benabbas for suggesting the correct formu-lation of the following theorem and for help in proving it. Theorem 29. (Generalized Birkhoff theorem)[Benabbas, Friedgut, Pilpel]An ( n ) k × ( n ) k matrix M is k -bistochastic if and only if it is a convexcombination of ( n ) k × ( n ) k matrices of permutations of A k which are inducedby permutations of [ n ] .Proof. By induction on k .For k = 1, this is Birkhoff’s theorem.For the induction step, let M = r i,j · M i,j be a block decomposition of M as in the definition of k -bistochasticity, according to the natural orderingof the coordinates, i.e. π = id. By Birkhoff’s theorem, R = ( r i,j ) is aconvex combination of permutation matrices, and therefore it is either apermutation matrix, or else it can be written as a convex combination R = sP + (1 − s ) T , with P a permutation matrix, and T a bistochastic matrixwith more zero entries than R . Treating P separately, and proceeding in thismanner by induction, we may assume that R is a permutation matrix, andwithout loss of generality, we may assume that R = I , the identity matrix.So in M , every non-zero entry is indexed by a pair of k -tuples of the form(( i, ∗ , ∗ , . . . , ∗ ) , ( i, ∗ , ∗ , . . . , ∗ )) for some i .Now, we reorder the rows and columns of M with a lexicographic orderon A k determined by the transposition π = (1 2). Consider now an off-diagonal block in this ordering of the form (( ∗ , i, ∗ , ∗ , . . . , ∗ ) , ( ∗ , j, ∗ , ∗ , . . . , ∗ ))with i = j . It breaks into ( n − sub-blocks of size ( n − k − × ( n − k − ,but only n − l, i, ∗ , ∗ , . . . , ∗ ) , ( l, j, ∗ , ∗ , . . . , ∗ )) for some l = i, j . Hence, the blockhas a row of zero sub-blocks, namely the sub-blocks(( j, i, ∗ , ∗ , . . . , ∗ ) , ( m, j, ∗ , ∗ , . . . , ∗ ))42or m = j , and a column of zero sub-blocks, namely the sub-blocks(( m, i, ∗ , ∗ , . . . , ∗ ) , ( i, j, ∗ , ∗ , . . . , ∗ ))for m = i . Since the block is ( k − M in the new ordering of A k , blocks of theform (( ∗ , i, ∗ , ∗ , . . . , ∗ ) , ( ∗ , i, ∗ , ∗ , . . . , ∗ )), they can only have nonzero sub-blocks on the diagonal. Since they are ( k − P ( k − σ ’swhere each σ is a permutation of [ n ] \{ i } . But if any of these σ ’s has σ ( l ) = l for some l , then the off-diagonal sub-block(( l, i, ∗ , ∗ , . . . , ∗ ) , ( σ ( l ) , i, ∗ , ∗ , . . . , ∗ ))is non-zero, a contradiction. Hence, each permutation σ must be the identity,and therefore all the diagonal blocks are copies of the ( n − k − × ( n − k − identity matrix. Hence, M is the identity matrix, and we are done.We now can prove Theorem 27. Proof of Theorem 27:
We follow the lines of the proof of Theorem 28. Let f ∈ V k be non-negative, and let M = ( M α,β ) α,β ∈A k be an ( n ) k × ( n ) k matrix(with rows and columns indexed by A k ) which represents f , meaning that f = X α,β ∈A k M α,β T α β , i.e. f ( σ ) = X α ∈A k M α,σ ( α ) ∀ σ ∈ S n . Let L k be the set of all k -lines. Recall from Observation 2 that every k -linecorresponds to a partition of S n into k -cosets. Therefore, adding a constant x ∈ R to a single k -line in M corresponds to increasing f by x . We thereforeassociate with every line ℓ ∈ L k a variable x ℓ . We wish to solve the followingsystem of linear inequalities X ℓ :( α,β ) ∈ ℓ x ℓ ≥ − M α,β ( ℓ ∈ L k ) subject to X ℓ ∈L k x ℓ = 0 . Again, by the strong duality theorem of linear programming, this ispossible unless there exists an ( n ) k × ( n ) k matrix C = ( c α,β ) α,β ∈A k withnon-negative entries, such that X ( α,β ) ∈ ℓ c α,β = 1 ∀ ℓ ∈ L k , X α,β ∈A k c α,β M α,β < . By Observation 1, C is k -bistochastic. By the Generalized Birkhoff Theo-rem, C is in the convex hull of the permutation matrices induced by permu-tations of [ n ]. This means, as before, that X α,β ∈A k c α,β M α,β is a convex linear combination of values of f , and is therefore non-negative, acontradiction. Therefore, the system of linear inequalities above is solvable,proving the first part of the theorem. The second part follows as in the k = 1 case, by induction on the number of non-zero values of f .This completes the proof of Theorems 3 and 4. In their landmark paper [1], Ahlswede and Khachatrian characterized thelargest k -intersecting subsets of [ n ] ( r ) for every value of k , r and n . In ouropinion, the most natural open problem in the area is to characterize thelargest k -intersecting subsets of S n for every value of n and k . We make thefollowing Conjecture 2. A k -intersecting family in S n with maximum size must bea ‘ double translate ’ of one of the families M i = { σ ∈ S n : σ has ≥ k + i fixed points in [ k + 2 i ] } (0 ≤ i ≤ ( n − k ) / , i.e. of the form σM i τ where σ, τ ∈ S n . This would imply that the maximum size is ( n − k )! for n > k . It isnatural to ask how large n must be for our method to give this bound, i.e.when there exists a weighted graph Y which is a real linear combination ofCayley graphs on S n generated by conjugacy-classes in FPF k , such that Y has maximum eigenvalue 1 and minimum eigenvalue ω n,k = − n ( n − . . . ( n − k + 1) − . It turns out that this need not be the case when n = 2 k + 2; indeed, it failsfor k = 2 and n = 6. We believe that new techniques will be required toprove Conjecture 2. 44 cknowledgements The authors would like to thank Noga Alon, Siavosh Benabbas, Imre Leader,Timothy Gowers, Yuval Roichman, and John Stembridge for various usefulremarks.
Erratum, added 7th July 2017
Recently, Yuval Filmus [9] has pointed out that Theorem 27 is false for each k ≥
2; the above proof of Theorem 29 contains a hole, and in fact Theorem29 is false for each k ≥
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