aa r X i v : . [ m a t h . C O ] D ec Loose Laplacian spectra of random hypergraphs
Linyuan Lu ∗ Xing Peng † June 29, 2018
Abstract
Let H = ( V, E ) be an r -uniform hypergraph with the vertex set V and the edge set E .For 1 ≤ s ≤ r/
2, we define a weighted graph G ( s ) on the vertex set (cid:0) Vs (cid:1) as follows. Everypair of s -sets I and J is associated with a weight w ( I, J ), which is the number of edgesin H passing through I and J if I ∩ J = ∅ , and 0 if I ∩ J = ∅ . The s -th Laplacian L ( s ) of H is defined to be the normalized Laplacian of G ( s ) . The eigenvalues of L ( s ) are listedas λ ( s )0 , λ ( s )1 , . . . , λ ( s ) ( ns ) − in non-decreasing order. Let ¯ λ ( s ) ( H ) = max i =0 {| − λ ( s ) i |} . Theparameters ¯ λ ( s ) ( H ) and λ ( s )1 ( H ), which were introduced in our previous paper, have anumber of connections to the mixing rate of high-ordered random walks, the generalizeddistances/diameters, and the edge expansions.For 0 < p <
1, let H r ( n, p ) be a random r -uniform hypergraph over [ n ] := { , , . . . , n } ,where each r -set of [ n ] has probability p to be an edge independently. For 1 ≤ s ≤ r/ p (1 − p ) ≫ log nn r − s , and 1 − p ≫ log nn , we prove that almost surely¯ λ ( s ) ( H r ( n, p )) ≤ sn − s + (3 + o (1)) s − p (cid:0) n − sr − s (cid:1) p . We also prove that the empirical distribution of the eigenvalues of L ( s ) for H r ( n, p )follows the Semicircle Law if p (1 − p ) ≫ log / nn r − s and 1 − p ≫ log nn r − s . The spectrum of the adjacency matrix (and/or the Laplacian matrix) of a random graphwas well-studied in the literature [1, 10, 11, 13, 14, 15, 17, 18, 21]. Given a graph G , let µ ( G ) , . . . , µ n ( G ) be the eigenvalues of the adjacency matrix of G in the non-decreasingorder, and λ ( G ) , . . . , λ n − ( G ) be the eigenvalues of (normalized) Laplacian matrix of G respectively. Let G ( n, p ) be the Ed˝os-R´enyi random graph model. F¨uredi and Koml´os [21]showed that if np (1 − p ) ≫ log n then almost surely µ n = (1+ o (1)) np and max {− µ , µ n − } ≤ (2 + o (1)) p np (1 − p ). The results are extended to sparse random graphs [17, 25] and generalrandom matrices [15, 21]. Alon, Krivelevich, and Vu [1] proved the concentration of the s -th largest eigenvalue of a random symmetric matrix with independent random entries ofabsolute value at most 1. Friedman (in a series of papers [18, 19, 20]) proved that thesecond largest eigenvalue of random d -regular graphs is almost surely (2 + o (1)) √ d − d ≥
4. Chung, Lu, and Vu [11] studied the Laplacian eigenvalues of random graphs with ∗ University of South Carolina, Columbia, SC 29208, ( [email protected] ). This author was supported in partby NSF grant DMS 1000475. † University of South Carolina, Columbia, SC 29208, ( [email protected] ).This author was supportedin part by NSF grant DMS 1000475. H = ( V, E ) be an r -uniform hypergraph on n vertices. We can associate r − L ( s ) (1 ≤ s ≤ r −
1) to H ; roughly speaking, L ( s ) captures the incidencerelations between s -sets and edges in H . Our definition of the Laplacian at the spacial case s = 1 is the same as the Laplacian considered by Rodr´ıguez [28, 29]. The s -th Laplacian is loose if 1 ≤ s ≤ r/
2, and is tight if r/ < s ≤ r −
1. Here we consider only the spectra ofloose Laplacians.For 1 ≤ s ≤ r/
2, we consider an auxiliary weighted graph G ( s ) defined as follows: thevertex set of G ( s ) is (cid:0) Vs (cid:1) while the weight function W : (cid:0) Vs (cid:1) × (cid:0) Vs (cid:1) → Z is defined as W ( S, T ) = (cid:26) |{ F ∈ E ( H ) : S ∪ T ⊂ F }| if S ∩ T = ∅ ;0 otherwise. (1)The s -th Laplacian of H , denoted by L ( s ) , is the normalized Laplacian of G ( s ) . For any s -set S , let d S be the number of edges in H passing through S ; the degree of S in G ( s ) is (cid:0) r − ss (cid:1) d S . Let D be the diagonal matrix of the degrees { d S } and W be the weight matrix { w ( S, T ) } . Note that T := (cid:0) r − ss (cid:1) D is the diagonal matrix of degrees in G ( s ) . We have L ( s ) = I − T − / W T − / . (2)The eigenvalues of L ( s ) are listed as λ ( s )0 , λ ( s )1 , . . . , λ ( s ) ( ns ) − in non-decreasing order. Wehave 0 = λ ( s )0 ≤ λ ( s )1 ≤ · · · ≤ λ ( s ) ( ns ) − ≤ . (3)The first non-trivial eigenvalue λ ( s )1 > G ( s ) is connected. When this occurs,we say H is s -connected . The diameter of G ( s ) is called the s -th diameter of H . The largesteigenvalue λ ( s ) ( ns ) − is also denoted by λ ( s ) max . The (Laplacian) spectral radius, denoted by ¯ λ ( s ) ,is the maximum of 1 − λ ( s )1 and λ ( s ) max − G ( s ) ′ ) is the set of all distinct s -tuples instead. Note that G ( s ) ′ is theblow-up of G ( s ) . Their Laplacian spectra differ only by the multiplicity of 1’s. Therefore, twodifferent definitions give the same values of λ ( s )1 , λ ( s ) max , and ¯ λ ( s ) .For different s , the following inequalities were proved in [26]. λ (1)1 ≥ λ (2)1 ≥ . . . ≥ λ ( ⌊ r/ ⌋ )1 ; (4) λ (1)max ≤ λ (2)max ≤ . . . ≤ λ ( ⌊ r/ ⌋ )max . (5)The s -th Laplacian has a number of connections to the mixing rate of high-ordered randomwalks, the generalized distances/diameters, and the edge expansions. Here we list someapplications, which are similar to results in [26], and results for graphs [4, 6, 7, 8, 9, 12]. Random s -Walks: The mixing rate of the random s -walk on H is at most ¯ λ ( s ) .2 he s -Diameter: The s -diameter of H is at most log | E ( H ) | ( rs ) δ log λ ( s )max + λ ( s )1 λ ( s )max − λ ( s )1 . Here δ = min S ∈ ( Vs ) d S is the minimum degree among all s -sets. Edge expansion:
For 1 ≤ t ≤ s ≤ r , S ⊂ (cid:0) Vt (cid:1) , and T ⊂ (cid:0) Vt (cid:1) , define E ( S , T ) = { F ∈ E ( H ) : ∃ S ∈ S , ∃ T ∈ T such that S ∩ T = ∅ , and S ∪ T ⊂ F } ,e ( S , T ) = | E ( S , T ) | (cid:12)(cid:12)(cid:12) E ( (cid:0) Vs (cid:1) , (cid:0) Vt (cid:1) ) (cid:12)(cid:12)(cid:12) ,e ( S ) = P S ∈S d S P S ∈ ( Vs ) d S ,e ( T ) = P T ∈T d T P T ∈ ( Vt ) d T . Then we have | e ( S , T ) − e ( S ) e ( T ) | ≤ ¯ λ ( s ) q e ( S ) e ( T ) e ( ¯ S ) e ( ¯ T ) . The proofs of these claims are very similar to those in [26] and are omitted here.Our first result is the eigenvalues of the s -th Laplacian of the complete r -uniform hyper-graph K rn . Theorem 1
Let K rn be the complete r -uniform hypergraph on n vertices. For ≤ s ≤ r/ ,the eigenvalues of s -th Laplacian of K rn are given by − ( − i (cid:0) n − s − i ) s − i (cid:1)(cid:0) n − ss (cid:1) with multiplicity (cid:18) ni (cid:19) − (cid:18) ni − (cid:19) for ≤ i ≤ s. Here we point out an application of this theorem to the celebrated Erd˝os-Ko-Rado Theorem,which states “if the n ≥ s , then the size of the maximum intersecting family of s -sets in [ n ]is at most (cid:0) n − s − (cid:1) .” (The theorem was originally proved by Erd˝os-Ko-Rado [16] for sufficientlylarge n ; the simplest proof was due to Katona [24].) Here we present a proof adapted fromCalderbank-Frankl [2], where they use the eigenvalues of Kneser graph instead. (The relationbetween L ( s ) ( K rn ) and the Laplacian of the Kneser graph is explained in section 2.)It suffices to show for any intersecting family U of s -sets, | U | ≤ (cid:0) n − s − (cid:1) . Note that U is anindependent set of G ( s ) ( K rn ). Restricting to U , L ( s ) ( K rn ) becomes an identity matrix; whoseeigenvalues are all equal to 1. By Cauchy’s interlace theorem, we have λ ( s ) k ≤ ≤ λ ( s ) ( ns ) −| U | + k (6)for 0 ≤ k ≤ | U | −
1. Let N + (or N − ) be the number of eigenvalues of L ( s ) ( K rn ) which is ≥ ≤
1) respectively. Inequality (6) implies that | U | ≤ N + and | U | ≤ N − . By Theorem 1, N + = P ⌊ ( s − / ⌋ i =0 (cid:16)(cid:0) n i +1 (cid:1) − (cid:0) n i (cid:1)(cid:17) and N − = P ⌊ s/ ⌋ i =0 (cid:16)(cid:0) n i (cid:1) − (cid:0) n i − (cid:1)(cid:17) . We have | U | ≤ min { N + , N − } = s − X i =0 ( − s − − i (cid:18) ni (cid:19) = (cid:18) n − s − (cid:19) . < p <
1, let H r ( n, p ) be a random r -uniform hypergraph over [ n ] = { , , . . . , n } ,where each r -set of [ n ] has probability p to be an edge independently. We can estimate theLaplacian spectrum of H r ( n, p ) using the Laplacian spectrum of K rn as follows. Theorem 2
Let H r ( n, p ) be a random r -uniform hypergraph. For ≤ s ≤ r/ , if p (1 − p ) ≫ log nn r − s and − p ≫ log nn , then almost surely the s -th spectral radius ¯ λ ( s ) ( H r ( n, p )) satisfies ¯ λ ( s ) ( H r ( n, p )) ≤ sn − s + (3 + o (1)) s − p (cid:0) n − sr − s (cid:1) p . (7) Moreover, for ≤ k ≤ (cid:0) ns (cid:1) − , almost surely we have | λ ( s ) k ( H r ( n, p )) − λ ( s ) k ( K rn ) | ≤ (3 + o (1)) s − p (cid:0) n − sr − s (cid:1) p . (8)Note that G ( n, p ) is a special case of H r ( n, p ) with r = 2. By choosing s = 1, Theorem 2implies that ¯ λ ( G ( n, p )) ≤ (3 + o (1)) s − p ( n − p for p (1 − p ) ≫ log nn . (9)Chung, Lu, and Vu’s result[11], when restricted to G ( n, p ), implies¯ λ ( G ( n, p )) ≤ (4 + o (1)) 1 √ np for 1 − ǫ ≥ p ≫ log nn . (10)Inequality (9) has a smaller constant and works for a larger range of p than inequality(10).F¨uredi and Koml´os [21] proved the empirical distribution of the eigenvalues of G ( n, p )follows the Semicircle Law. Chung, Lu, and Vu [11] proved a similar result for the randomgraphs with given expected degrees. Here we prove a similar result for random hypergraphs. Theorem 3
For ≤ s ≤ r/ , if p (1 − p ) ≫ log / nn r − s and − p ≫ log nn r − s , then almostsurely the empirical distribution of eigenvalues of the s -th Laplacian of H r ( n, p ) follows theSemicircle Law centered at and with radius (2 + o (1)) q − p ( r − ss )( n − sr − s ) p . Remark 1
The proof of Theorem 3 actually implies the eigenvalues of L ( s ) ( H r ( n, p )) −L ( s ) ( K rn ) follows the Semicircle Law centered at and with radius (2 + o (1)) q − p ( r − ss )( n − sr − s ) p .Thus we have max ≤ k ≤ ( ns ) − | λ ( s ) k ( H r ( n, p )) − λ ( s ) k ( K rn ) | ≥ q(cid:0) r − ss (cid:1) + o (1) s − p (cid:0) n − sr − s (cid:1) p . (11) This shows that the upper bound of | λ ( s ) k ( H r ( n, p )) − λ ( s ) k ( K rn ) | in Theorem 2 is best up to aconstant multiplicative factor. The rest of the paper is organized as follows. In section 2, we introduce the notation andprove some basic lemmas. We will prove Theorem 1 in section 3 and Theorem 2 in section 4.4
Notation and Lemmas
Let H = ( V, E ) be an r -uniform hypergraph. For any subset S ( | S | < r ), the degree of S ,denoted by d S , is the number of edges passing through S . For each 1 ≤ s ≤ r/
2, we associatea weighted graph G ( s ) on the vertex set (cid:0) Vs (cid:1) to H as follows. Every pair of s -sets S and T isassociated with a weight w ( S, T ), which is given by w ( S, T ) = (cid:26) d S ∪ T if S ∩ T = ∅ , . The s -th Laplacian L ( s ) of H is defined to be the normalized Laplacian of G ( s ) . The degreeof S in G ( s ) is P T w ( S, T ) = (cid:0) r − ss (cid:1) d S .We assume that the s -sets in (cid:0) Vs (cid:1) are ordered alphabetically. Let N := (cid:0) ns (cid:1) ; all squarematrices considered in the paper have the dimension N × N and all vectors have dimension N . Let W := ( W ( S, T )) be the weight matrix, D be the diagonal matrix with diagonalentries D ( S, S ) = d S , d be the column vector with entries d S at position S ∈ (cid:0) VS (cid:1) , J be thesquare matrix of all 1’s, and be the column vector of all 1’s. Let T := (cid:0) r − ss (cid:1) D ; here T isthe diagonal matrix of degrees in G ( s ) . Then, we have L ( s ) = I − T − / W T − / . We list the eigenvalues of L ( s ) as0 = λ ( s )0 ≤ λ ( s )1 , . . . , λ ( s ) ( ns ) − ≤ . We aim to compute the spectral radius ¯ λ ( s ) ( H ) = max i =0 | − λ ( s ) i | . Let vol ( s ) ( H ) := P S ∈ ( Vs ) d s and φ := √ vol ( s ) ( H ) D / . Note that φ is the unit eigenvector corresponding tothe trivial eigenvalue 0 of L ( s ) .We are ready to prove theorem 1. Proof of Theorem 1:
We can express L ( s ) ( K rn ) using the following notation. TheKneser graph K ( n, s ) is a graph over the vertex set (cid:0) [ n ] s (cid:1) ; two s -sets S and T form an edge of K ( n, s ) if and only if S ∩ T = 0. Let K be the adjacency matrix of K ( n, s ); the eigenvalues of K are ( − i (cid:0) n − s − i ) s − i (cid:1) with multiplicity (cid:0) ni (cid:1) − (cid:0) ni − (cid:1) for 0 ≤ i ≤ s (see [22]). Note that K ( n, s )is a regular graph; so the Laplacian eigenvalues can be determined from the eigenvalues ofits adjacency matrix. We observe that the associated weighted graph G ( s ) for the complete r -uniform hypergraph K rn is essentially the Kneser graph with each edge associated with aweight (cid:0) n − sr − s (cid:1) . Note that the multiplicative factor (cid:0) n − sr − s (cid:1) is canceled after normalization.The L ( s ) (for K rn ) is exactly the Laplacian of Kneser graph. Hence, L ( s ) ( K rn ) = I − (cid:0) n − ss (cid:1) K. Thus, the eigenvalues of s -th Laplacian of K rn are given by1 − ( − i (cid:0) n − s − i ) s − i (cid:1)(cid:0) n − ss (cid:1) with multiplicity (cid:18) ni (cid:19) − (cid:18) ni − (cid:19) for 0 ≤ i ≤ s. (cid:3) emark 2 For ≤ s ≤ r/ , we have λ ( s )1 ( K rn ) = 1 − s ( s − n − s )( n − s − , (12) λ ( s ) max ( K rn ) = 1 + sn − s , (13)¯ λ ( s ) ( K rn ) = sn − s . (14) Let H r ( n, p ) be a random r -uniform hypergraph over the vertex set V = [ n ] and each r -sethas probability p to be an edge independently. We would like to bound the spectral radiusof the s -th Laplacian of H r ( n, p ) for 1 ≤ s ≤ r/ F ∈ (cid:0) Vr (cid:1) , let X F be the random indicator variable for F being an edge in H r ( n, p );all X F ’s are independent to each other. For any S, T ∈ (cid:0) Vs (cid:1) , we have W ( S, T ) = ( P F ∈ ( nr ) S ∪ T ⊂ F X F if S ∩ T = ∅ ;0 otherwise.Thus, E( W ( S, T )) = (cid:26) (cid:0) n − sr − s (cid:1) p if S ∩ T = ∅ ;0 otherwise. (15)The degree d S = P S ⊂ F ∈ ( Vr ) X F ; we have E( d S ) = (cid:0) n − sr − s (cid:1) p . For simplicity, let d := (cid:0) n − sr − s (cid:1) p .We use the following Lemma to compare the eigenvalues of two matrices. Lemma 1
Given any two ( N × N ) -Hermitian matrices A and B , for ≤ k ≤ N , let µ k ( A ) (or µ k ( B ) ) be the k -th eigenvalues of A (or B ) in the increasing order. We have | µ k ( A ) − µ k ( B ) | ≤ k A − B k . Proof:
By the Min-Max Theorem (see [27]), we have µ k ( A ) = min S k max x ∈ S k , k x k =1 x ′ Ax,µ k ( B ) = min S k max x ∈ S k , k x k =1 x ′ Bx. where the minimum is taken over all k -th dimensional subspace S k ⊂ R N . We have µ k ( A ) = min S k max x ∈ S k , k x k =1 x ′ Ax = min S k max x ∈ S k , k x k =1 ( x ′ Bx + x ′ ( A − B ) x ) ≤ min S k max x ∈ S k , k x k =1 ( x ′ Bx + k A − B k )= µ k ( B ) + k A − B k . Similarly, we can show µ k ( A ) ≥ µ k ( B ) − k A − B k . The proof of the Lemma is finished. (cid:3) Our idea is to bound the spectral norm of the difference of L ( s ) ( H r ( n, p )) and L ( s ) ( K rn ).Let M := L ( s ) ( K rn ) − L ( s ) ( H r ( n, p )) = T − / W T − / − ( n − ss ) K . We write M = M + M +6 + M , where M = 1 (cid:0) r − ss (cid:1) (cid:16) D − / ( W − E( W )) D − / − d − ( W − E( W )) (cid:17) ,M = 1 (cid:0) r − ss (cid:1) d ( W − E( W )) ,M = 1 (cid:0) r − ss (cid:1) D − / E( W ) D − / − d (cid:0) ns (cid:1) D − / JD − / − (cid:0) n − ss (cid:1) K + 1 (cid:0) ns (cid:1) J,M = 1 (cid:0) ns (cid:1) ( dD − / JD − / − J ) . By the triangular inequality of matrix norms, we have k M k ≤ k M k + k M k + k M k + k M k . Through this paper, the norm of any square matrix is the spectral norm. We would like tobound k M i k for i = 1 , , ,
4. We use the following Chernoff inequality.
Theorem 4 [3] Let X , . . . , X n be independent random variables with Pr( X i = 1) = p, Pr( X i = 0) = 1 − p. We consider the sum X = P ni =1 X i , with expectation E( X ) = np . Then we have(Lower tail) Pr( X ≤ E( X ) − λ ) ≤ e − λ / X ) , (Upper tail) Pr( X ≥ E( X ) + λ ) ≤ e − λ X )+ λ/ . Lemma 2
Suppose d ≥ log N . With probability at least − N , we have d S ∈ ( d − √ d log N , d + 3 √ d log N ) for all S ∈ (cid:0) Vs (cid:1) . Proof:
Note d s = P F : S ⊂ F X F and E( d S ) = d . Applying the lower tail of Chernoff’sinequality with λ = 3 p E( X ) log N , we havePr ( X − E( X ) ≤ − λ ) ≤ e − λ / X ) ≤ N / . Applying the upper tail of Chernoff’s inequality with λ = 3 p E( X ) log N , we havePr ( X − E( X ) ≥ λ ) ≤ e − λ X )+ λ/ ≤ N / . The probability that d S ( d − √ d log N , d +3 √ d log N ) is at most N . Thus, with probabilityat least 1 − N , we have d S ∈ ( d − √ d log N, d + 3 √ d log N ) for all S ∈ (cid:0) Vs (cid:1) . (cid:3) For convenience, let d min := d − √ d log N , d max := d + 3 √ d log N ; almost surely we have d min ≤ d S ≤ d max for all S . Lemma 3 If d ≥ log N , then almost surely k M k = O (cid:16) √ log Nn √ d (cid:17) . Proof:
Note E( W ) = (cid:0) n − sr − s (cid:1) pK , where K is the adjacency matrix of the Kneser graph K ( n, s ). Let M := ( n − ss ) K − ( ns ) J . We can rewrite M as M = dD − / M D − / − M . k M k = ¯ λ ( s ) ( K rn ) = sn − s . We have k M k = k dD − / M D − / − M k≤ k ( dD − / − d / I ) M D − / k + k M ( d / D − / − I ) k≤ k ( d / I − dD − / ) kk M kk D − / k + k M kk ( d / D − / − I ) k≤ (cid:12)(cid:12)(cid:12) d / − dd − / min (cid:12)(cid:12)(cid:12) sn − s d − / min + sn − s (cid:12)(cid:12)(cid:12) d / d − / min − (cid:12)(cid:12)(cid:12) = O (cid:18) √ log Nn √ d (cid:19) . (cid:3) Lemma 4 If p (1 − p ) ≫ log nn r − s , then almost surely X S ∈ ( Vs )( d S − d ) = (1 + o (1)) (cid:18) ns (cid:19) d (1 − p ) . Proof:
For S ∈ (cid:0) Vs (cid:1) , let X S = ( d S − d ) . We haveE( X S ) = E(( d S − d ) ) = Var( d S ) = (cid:18) n − sr − s (cid:19) p (1 − p ) = d (1 − p ) . We use the second moment method to prove that P S X s concentrates around its expectation (cid:0) ns (cid:1) d (1 − p ). For any S, T ∈ (cid:0) Vs (cid:1) , the covariance can be calculated as follows.Cov( X S , X T ) = E( X S X T ) − E( X S )E( X T )= E(( d S − d ) ( d T − d ) ) − d (1 − p ) . For F ∈ (cid:0) Vr (cid:1) , let Y F = X F − E( X F ). Then we have d S − d = P S ⊂ F Y F .E(( d S − d ) ( d T − d ) ) = X F ,F : S ⊂ F ∩ F F ,F : T ⊂ F ∩ F E( Y F Y F Y F Y F ) . Since E( Y F i ) = 0, the non-zero terms occur only if1. F = F = F = F . In this case, we haveE( Y F Y F Y F Y F ) = E( Y F ) = (1 − p ) p + ( − p ) (1 − p ) = p (1 − p )(1 − p + 3 p ) . The number of choices is (cid:0) n −| S ∪ T | r −| S ∪ T | (cid:1) .2. F = F = F = F . In this case, we haveE( Y F Y F Y F Y F ) = E( Y F )E( Y F ) = p (1 − p ) . The number of choices is (cid:0) n − sr −| S | (cid:1)(cid:0) n − sr −| T | (cid:1) − (cid:0) n −| S ∪ T | r −| S ∪ T | (cid:1) .3. F = F = F = F . In this case, we haveE( Y F Y F Y F Y F ) = E( Y F )E( Y F ) = p (1 − p ) . The number of choices is (cid:0) n −| S ∪ T | r −| S ∪ T | (cid:1) − (cid:0) n −| S ∪ T | r −| S ∪ T | (cid:1) .8. F = F = F = F . This is the same as item 3.Thus, we haveE( X S X T ) = (cid:18) n − | S ∪ T | r − | S ∪ T | (cid:19) p (1 − p )(1 − p + 3 p )+ (cid:18) n − sr − s (cid:19) + 2 (cid:18) n − | S ∪ T | r − | S ∪ T | (cid:19) − (cid:18) n − | S ∪ T | r − | S ∪ T | (cid:19)! p (1 − p ) = (cid:18) n − | S ∪ T | r − | S ∪ T | (cid:19) p (1 − p )(1 − p + 6 p ) + (cid:18) n − sr − s (cid:19) + 2 (cid:18) n − | S ∪ T | r − | S ∪ T | (cid:19) ! p (1 − p ) . This expression on the right depends only on the size of S ∪ T . Putting together, we getVar X S ∈ ( Vs ) X S = X S,T ∈ ( Vs ) Cov( X S , X T )= X S,T ∈ ( Vs )(E( X S X T ) − d (1 − p ) )= X S,T ∈ ( Vs ) E( X S X T ) − (cid:18) n − sr − s (cid:19) p (1 − p ) ! = s X i = s X | S ∪ T | = i (cid:18) n − ir − i (cid:19) p (1 − p )(1 − p + 6 p ) + 2 (cid:18) n − ir − i (cid:19) p (1 − p ) ! ≤ s X i = s X | S ∪ T | = i (cid:18) n − ir − i (cid:19) p (1 − p ) (cid:18) − p + 6 p + 2 (cid:18) n − sr − s (cid:19) p (1 − p ) (cid:19) ≤ s X i = s X | S ∪ T | = i (cid:18) n − ir − i (cid:19) dp (1 − p ) = (cid:18) nr (cid:19) dp (1 − p ) s X i = s r !( i − s )! (2 s − i )!( r − i )! < · r (cid:18) nr (cid:19) dp (1 − p ) = O (cid:18)(cid:18) ns (cid:19) d (1 − p ) (cid:19) . Let X = P S X S . We have E[ X ] = (cid:0) ns (cid:1) d (1 − p ) and Var( X ) = O (cid:0)(cid:0) ns (cid:1) d (1 − p ) (cid:1) . ApplyingChebyshev’s inequality to X = P S ∈ ( Vs ), we havePr (cid:16) | X − E( X ) | ≥ log n p Var( X ) (cid:17) ≤ n . Thus, almost surely X = E( X ) + O (log n p Var( X )) = (1 + o (1)) (cid:0) ns (cid:1) d (1 − p ). (cid:3) Lemma 5 If p (1 − p ) ≫ log nn r − s , then almost surely k M k ≤ (1 + o (1)) q − pd . roof: We can rewrite M as M = 1 (cid:0) ns (cid:1) ( dD − / JD − / − J )= 1 (cid:0) ns (cid:1) (cid:16)(cid:16) d / D − / − I (cid:17) JD − / d / + J (cid:16) d / D − / − I (cid:17)(cid:17) = 1 (cid:0) ns (cid:1) (cid:16) α ′ D − / d / + α ′ (cid:17) . Here α := d / D − / − . Note that the spectral norm of a vector is the same as the L -norm. We have k α k = k d / D − / − k = vuuut X S ∈ ( Vs ) √ d √ d S − ! = vuut X S ∈ ( Vs ) ( d S − d ) d S ( √ d + √ d S ) ≤ qP S ∈ ( Vs )( d S − d ) √ d min ( √ d + √ d min )= ( 12 + o (1)) s (1 − p ) (cid:0) ns (cid:1) d . In the last step, we applied Lemma 4. Therefore, we have k M k = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) (cid:0) ns (cid:1) (cid:16) α ′ D − / d / + α ′ (cid:17)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = 1 (cid:0) ns (cid:1) (cid:16)(cid:13)(cid:13)(cid:13) α ′ D − / d / (cid:13)(cid:13)(cid:13) + k α ′ k (cid:17) ≤ (cid:0) ns (cid:1) k α k (cid:16) k ′ D − / d / k + k k (cid:17) = 1 (cid:0) ns (cid:1) k α k vuut X S ∈ ( ns ) dd S + s(cid:18) ns (cid:19) ≤ (cid:0) ns (cid:1) (cid:18)
12 + o (1) (cid:19) s (1 − p ) (cid:0) ns (cid:1) d (2 + o (1)) s(cid:18) ns (cid:19)! = (1 + o (1)) r − pd . To estimate the spectral norm of M and M , we need consider the matrix C := W − E( W ).We estimate the expectation of the trace of C t as follows.10 emma 6 For any k satisfying k ≪ p n r − s p (1 − p ) , we have E (cid:0) Trace( C k ) (cid:1) ≤ (1 + o (1)) n s + k ( r − s ) (cid:0) r − ss (cid:1) k ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k , (16)E (cid:0) Trace( C k +1 ) (cid:1) = O k n s + k ( r − s ) (cid:0) r − ss (cid:1) k +1 ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k ! . (17) If further k = o (log( n r − s p (1 − p ))) , then we have E (cid:0) Trace( C k ) (cid:1) = (1 + o (1)) n s + k ( r − s ) ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k . (18)The proof of this technical Lemma is quite long. We will delay its proof until the end of thissection. Lemma 7 If p (1 − p ) ≫ log nn r − s , then we have k C k ≤ (cid:0) (cid:0) r − ss (cid:1) + o (1) (cid:1) p d (1 − p ) almost surely. Proof:
By Lemma 6, we have E(Trace( C k )) ≤ (1 + o (1)) n s + k ( r − s ) ( r − ss ) k ( k +1)( s !) k +1 (( r − s )!) k (cid:0) kk (cid:1) p k (1 − p ) k .As E( k C k k ) ≤ E(Trace( C k )), we haveE( k C k k ) ≤ (1 + o (1)) n s + k ( r − s ) (cid:0) r − ss (cid:1) k ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k . Let U := n s + k ( r − s ) ( r − ss ) k ( k +1)( s !) k +1 (( r − s )!) k (cid:0) kk (cid:1) p k (1 − p ) k . By Markov’s inequality,Pr (cid:16) k C k ≥ (1 + ǫ ) k √ U (cid:17) = Pr (cid:0) k C k k ≥ (1 + ǫ ) k U (cid:1) ≤ E( k C k k )(1 + ǫ ) k U ≤ (1 + o (1)) U (1 + ǫ ) k U = 1 + o (1)(1 + ǫ ) k . Let g ( n ) be a slowly growing function such that g ( n ) → ∞ as n approaches the infinityand g ( n ) ≪ ( n r − s p (1 − p )) / s log n . This is possible because n r − s p (1 − p ) ≫ log n . Choose k = sg ( n ) log n and ǫ = 1 /g ( n ). We have k ≪ ( n r − s p (1 − p )) / and ǫ →
0. Then we have(1 + o (1)) / (1 + ǫ ) k = O ( n − s ), which implies that almost surely |k C k ≤ (1 + o (1)) k √ U = (1 + o (1)) n s + k ( r − s ) (cid:0) r − ss (cid:1) k ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k ! k < n s k s n r − s (cid:0) r − ss (cid:1) p (1 − p ) s !( r − s )!= (cid:18) (cid:18) r − ss (cid:19) + o (1) (cid:19) p d (1 − p ) . (cid:3) Recall M = ( r − ss ) d C . We have 11 emma 8 If p (1 − p ) ≫ log nn r − s , then we have k M k ≤ (2 + o (1)) q − pd almost surely. Lemma 9 If p (1 − p ) ≫ log nn r − s , then we have k M k = O (cid:18) √ (1 − p ) log Nd (cid:19) almost surely. Proof:
We have M = 1 (cid:0) r − ss (cid:1) (cid:16) D − / CD − / − d − C (cid:17) = 1 (cid:0) r − ss (cid:1) (cid:16) ( D − / − d − / I ) CD − / + d − / C ( D − / − d − / I ) (cid:17) . Note k D − / − d − / I k ≤ | d − / min − d − / | = O ( √ log Nd ), k D − / k ≤ d − / min = (1+ o (1)) d − / ,and k C k = (cid:0) (cid:0) r − ss (cid:1) + o (1) (cid:1) p d (1 − p ). We have k M k = 1 (cid:0) r − ss (cid:1) (cid:13)(cid:13)(cid:13) ( D − / − d − / I ) CD − / + d − / C ( D − / − d − / I ) (cid:13)(cid:13)(cid:13) = O p (1 − p ) log Nd ! . (cid:3) Proof of Theorem 2:
Combining Lemmas 3, 5, 8, and 9, we have k M k = k M + M + M + M k≤ k M k + k M k + k M k + k M k≤ O p (1 − p ) log Nd ! + (2 + o (1)) √ − p √ d + O (cid:18) √ log Nn √ d (cid:19) + (1 + o (1)) r − pd = (3 + o (1)) r − pd . In the last step, we use the fact √ log Nn √ d = o (cid:18)q − pd (cid:19) since 1 − p ≫ log nn .By Lemma 1, for 1 ≤ k ≤ (cid:0) ns (cid:1) −
1, we have | λ ( s ) k ( H r ( n, p )) − λ ( s ) k ( K rn ) | ≤ k M k ≤ (3 + o (1)) r − pd . (cid:3) Reall that X F is the random indicator variable for F being an edge in H r ( n, p ). For anyfixed positive integer t , the terms in Trace( C t ) are of the form c S S c S S . . . c S t S S . Here c ST = W ( S, T ) − E( W ( S, T )) = P F ∈ ( Vr ) S ∪ T ⊂ F ( X F − E( X F )) if S ∩ T = ∅ ; c ST = 0 otherwise.Note c S i S j = 0 if S i ∩ S j = ∅ . Thus we need only to consider the sequence S S . . . S t S such that S i ∩ S i +1 = ∅ for each 1 ≤ i ≤ t , here t + 1 = 1 . F ∈ (cid:0) Vr (cid:1) and S, T ∈ (cid:0) Vs (cid:1) , we define a random variable c FST as follows. c FST = (cid:26) X F − E( X F ) if S ∩ T = ∅ and S ∪ T ⊆ F ;0 otherwise.The sequence w := S F S F S . . . S t F t S is called a closed s -walk of length t if1. S , . . . , S t ∈ (cid:0) Vs (cid:1) ,2. F , . . . , F t ∈ (cid:0) Vr (cid:1) ,3. S i ∩ S i +1 = ∅ , for i = 1 , , . . . , t ,4. S i ∪ S i +1 ⊂ F i , for i = 1 , , . . . , t .Here we use the convention S t +1 = S . Those r -sets F i ’s are referred as edges while those s -sets S i ’s are referred as stops. For 1 ≤ i ≤ t , we say w walks from S i to S i +1 at step i viathe edge F i .Using the notation above, we rewrite the trace asTrace( C t ) = X closed s -walks c F S S c F S S . . . c F t S t S , where the summation is over all possible closed s -walks of length t .Taking the expectation on both sides, we getE(Trace( C t )) = X closed s -walks E( c F S S c F S S . . . c F t S t S ) . The terms in the product above can be regrouped according to the values of F i ’s; thoseterms with distinct F ’s are independent to each other. Since E( c FS,T ) = 0, the contributionof a closed walk is 0 if some F appears only once. Thus we need only to consider the setof closed walks where each edge appears at least twice or do not occur; we call these closedwalks as good closed walks. A good closed walk can contain at most ⌊ t ⌋ distinct edges.Let G i be the set of good closed walks of length t with i distinct edges. For 1 ≤ i ≤ ⌊ t ⌋ ,let G ji be the set of good closed walks with exactly i distinct edges and j distinct vertices; wehave G i := ∪ j G ji .We consider a good closed walk in G i . When a new edge comes in the walk, it can bring inat most ( r − s ) new vertices. Thus such a good closed walk covers at most m i := s + i ( r − s )vertices. Any walk contains at least one edge. Hence, the number of vertices in a walk from G i is in the interval [ r, m i ].We have E(Trace( C t )) = ⌊ t ⌋ X i =1 X S F S ...S t S ∈G i E( c F S S c F S S . . . c F t S t S ) . (19)Assume that an edge F occurs q times in a good closed walk and T := { i : 1 ≤ i ≤ t and F i = F } . We have Pr (cid:16) Π i ∈ T c FS i S i +1 = (1 − p ) q (cid:17) = p and Pr (cid:16) Π i ∈ T c FS i S i +1 = ( − p ) q (cid:17) =1 − p . Thus, for each positive integer l ≥
2, we haveE (cid:16) Π i ∈ T c FS i S i +1 (cid:17) = (1 − p ) q p + ( − p ) q (1 − p ) ≤ p (1 − p ) . The equality holds for q = 2. 13ick a good closed walk w := S F S F S . . . S t F t S in G i . Let F , . . . , F i be the list ofdistinct edges in the order as they appear in w .For each 1 ≤ l ≤ i , let T l := { ≤ j ≤ t : F j = F l } ; then P il =1 | T l | = t. We haveE( c F S S c F S S . . . c F t S t S ) = Π il =1 Π j ∈ T l E( c F l S j S j +1 ) ≤ Π il =1 p (1 − p ) = p i (1 − p ) i . This implies X S F S ...S t S ∈G i E( c F S S c F S S . . . c F t S t S ) ≤ |G i | p i (1 − p ) i (20)for all 1 ≤ i ≤ ⌊ t ⌋ . In particular, the equality holds when t = 2 i . Combining equation (19)and inequality (20), we get E(Trace( C t )) ≤ ⌊ t ⌋ X i =1 |G i | p i (1 − p ) i . (21)Now we estimate the value of |G ji | , the number of good closed walks of length t on i edgesand j vertices. Let w be a good closed walk in G ji . For 2 ≤ k ≤ i , let · · · SF k S ′ · · · be a pieceof sequence in w where the edge F k occurs first time; S is called the in-stop of F k and S ′ iscalled the out-stop of F k .The following lemma will state the hypergraph structure of these i edges; it is independentof the walk w . We will use the following notation. Let S = ∪ il =1 (cid:0) F l s (cid:1) . For any s -set S ∈ S ,the degree of S , denoted by d S , is the number of edges in { F , F , . . . , F i } containing S .Define d ′ S = (cid:26) d S − k such that S = F k ∩ ( ∪ k − l =1 F l ) ,d S otherwise . Lemma 10
Assume that F , . . . , F i is the list of distinct edges in the order as they appearin w ∈ G ji . Then we have X S ∈S ( d ′ S − ≤ (cid:18) s (cid:18) rs − (cid:19)(cid:19) ( m i − j ) . Proof:
For 2 ≤ k ≤ i , let x k = | F k \ ( ∪ k − l =1 F l ) | ; we have0 ≤ x k ≤ r − s. Thus, j = r + x + x + · · · + x i ≤ r + ( i − r − s ) = m i . Since a new edge F k can contribute at most (cid:0) r − x k s (cid:1) to P S ∈S ( d S − X S ∈S ( d S − ≤ i X k =2 (cid:18) r − x k s (cid:19) . Let K := { k : x k = r − s, ≤ k ≤ i } and K = { , . . . , i }\ K . The edges in the set { F k : k ∈ K } are called forward edges while the edges in the set { F k : k ∈ K } are called backward edges.Note each backward edge contribute at least one to m i − j ; thus m i − j ≥ | K | . k ∈ K , (cid:0) r − x k s (cid:1) = 1. We have X S ∈S ( d S − ≤ i X k =2 (cid:18) r − x k s (cid:19) = | K | + X k ∈ K (cid:18) r − x k s (cid:19) = | K | + X k ∈ K r − x k − s + 1 s (cid:18) r − x k s − (cid:19) ≤ | K | + X k ∈ K s (cid:18) rs − (cid:19) ( r − x k − s )= | K | + 2 s (cid:18) rs − (cid:19) ( m i − j ) . For any k ∈ K , let S ( F k ) := F k ∩ ( ∪ k − l =1 F l ) be the starting stop of F k when F k firstoccurs in w . List the elements in K as k , k , . . . , k | K | in an increasing order. Consider thesequence of stops S ( F k ) , S ( F k ) , . . . , S ( F k | K | ) (not necessarily distinct). Let z be the numberof distinct stops in the sequence. If S ( F k l ) does not appear the first time in the sequenceabove, then we consider the partial walk S k l − F k l − . . . S k l F k l . Since F k l − is a forward edge,there exists at least one backward edge F l ′ for some l ′ ∈ ( k l − , k l ). Thus, | K | ≤ z + | K | ≤ z + m i − j. Hence, X S ∈S ( d ′ S −
1) = X S ∈S ( d S − − z ≤ | K | + 2 s (cid:18) rs − (cid:19) ( m i − j ) − z ≤ s (cid:18) rs − (cid:19) ( m i − j ) + m i − j = (cid:18) s (cid:18) rs − (cid:19)(cid:19) ( m i − j ) . The proof of this Lemma is finished. (cid:3)
Lemma 11
For ≤ i ≤ ⌊ t ⌋ and r ≤ j ≤ m i , we have |G ji | ≤ (cid:18) t − t − i (cid:19) i t − i i + 1 (cid:18) ii (cid:19)(cid:18) r − ss (cid:19) t − i n m i ( s !) i +1 (( r − s )!) i (cid:18) C i C n (cid:19) m i − j . Here C and C depend only on r and s , independent of i , j , and n . Corollary 1
For ≤ i ≤ ⌊ t ⌋ and n ≫ i C , we have |G i | ≤ (1 + o (1)) (cid:18) t − t − i (cid:19) i t − i i + 1 (cid:18) ii (cid:19)(cid:18) r − ss (cid:19) t − i n m i ( s !) i +1 (( r − s )!) i . (22)15 roof: We can associate a walk w ∈ G ji with a code of length t consisting of three symbols:‘(’, ‘)’, and ‘ ∗ ’. We scan the edges of the walk w from left to right; if an edge appears firsttime, then we assign the code ‘(’; if an edge appears second time, then we assign the code ‘)’;otherwise, we assign the code ‘ ∗ ’.For example, consider the following good walk with i = 3, j = 8, and t = 8: w = S F S F S F S F S F S F S F S F S Here edges are: F = (1 , , , , F = (4 , , , , F = (7 , , , , S =(1 , S = (4 , S = (7 , S = (3 , , S = (2 , ∗ )) ∗ .Since w has i distinct edges, there are i ‘(’s, i ‘)’s, and ( t − i ) ‘ ∗ ’s. Note that the numberof ‘(’ is always greater than or equal to the number of ‘)’ at any point when the sequence isread from left to right; each ‘(’ has a matched ‘)’ in the sequence. The symbol ’ ∗ ’ starts atposition three and up. There are (cid:0) t − t − i (cid:1) ways to choose the ’ ∗ ’-positions and i +1 (cid:0) ii (cid:1) ways tochoose i matched parentheses (the Catalan number). The number of such codes is (cid:18) t − t − i (cid:19) i + 1 (cid:18) ii (cid:19) . To construct a walk from a given code, we scan the symbols from left to right. The firstsymbol is always ‘(’. There are (cid:0) ns (cid:1) ways to choose the first stop S and (cid:0) n − sr − s (cid:1) ways to choosethe rest of vertices in the first edge F . Suppose that we already build a partial walk andneed to decide the next stop and the next edge. There are at most (cid:0) r − ss (cid:1) ways to choose thenext stop S . The choices of selecting the next edge depends on the next available symbol inthe code sequence. Let b ( , b ) , and b ∗ be the product of the number of ways to choose thenext edge at the ‘(’, ‘)’, and ‘ ∗ ’ positions respectively. We have |G ji | ≤ (cid:18) t − t − i (cid:19) i + 1 (cid:18) ii (cid:19)(cid:18) ns (cid:19)(cid:18) r − ss (cid:19) t b ( · b ∗ · b ) . (23)First we estimate b ( , the number of ways to choose new edges F , . . . , F i given the firststop S . Besides the s vertices selected at the first stop, there are (cid:0) n − sj − s (cid:1) ways to chooseremaining j − s vertices. Recall that F , . . . , F i is the list of distinct edges in the order asthey appear in w . For 2 ≤ l ≤ i , let ˜ F l := F l \ ( ∪ l − l ′ =1 F l ′ ), x l := | ˜ F l | , and y l := r − s − x l . Wealso define ˜ F := F \ S ; x := | ˜ F | = r − s , and y = 0. Note that ∪ il =1 ˜ F l forms a partitionof the remaining ( j − s ) selected vertices. The number of ways to choose such a partition is( j − s )! x ! x ! · · · x i ! . To choose F l , we need select x l new vertices and y l old vertices; each old vertex has at most j choices. We have b ( ≤ X x ,...,x i (cid:18) n − sj − s (cid:19) ( j − s )!( r − s )! x ! · · · x i ! j P il =2 y l . Observe that P il =2 y l = m i − j and( j − s )!( r − s )! x ! · · · x i ! ≤ (cid:18) j − sr − s (cid:19) ( m i − r )!(( r − s )!) i − . x , . . . , x i is the same as the number of ways to choose y , . . . , y i ,which is (cid:0) m i − j + i − m i − j (cid:1) ≤ ( m i − j + i − m i − j . Therefore, b ( ≤ X x ,...,x i (cid:18) n − sj − s (cid:19) ( j − s )!( r − s )! x ! · · · x i ! j P il =2 y l ≤ (cid:18) n − sj − s (cid:19)(cid:18) j − sr − s (cid:19) ( m i − r )!(( r − s )!) i − ( m i − j + i − m i − j j m i − j ≤ (cid:18) n − sj − s (cid:19)(cid:18) j − sr − s (cid:19) ( m i − r )!(( r − s )!) i − (cid:18) m i + i − (cid:19) m i − j ) . (24)There is at most i choices of edges at each ‘ ∗ ’ position. Thus b ∗ ≤ i t − i . (25)It remains to bound b ) . We first present an easy bound for b ) . Edge F can be chosen atmost one ‘)’-position. For any possible stop S ∈ S , S can appear at the )-positions at most d S times; each occurrence of S involves different edges since we are considering the secondoccurrence of edges. Thus, b ) ≤ Y S ∈S d S ! ≤ Y S ∈S d d S − S ≤ i P S ∈S ( d S − . We need a better upper bound for b ) . Consider a stop S which is first chosen at a ‘)’-position. Let F be the edge on the walk right before the ‘)’-position; i.e., the walk w enters S through F . If this F occurred before, then the choices of edges at ‘)’-positions starting with S is at most ( d S − ≤ i d S − . If this F occurs first time and F is an forward edge, then there is only one choice for the nextedge leaving S ; namely F itself. In this case, the choices of edges at ‘)’-positions startingwith S is at most ( d S − ≤ i d S − . In the remaining case, F must be a backward edge. The number of backward edges is at most m i − j . Since F contains at most (cid:0) rs (cid:1) stops, the number of such S is at most (cid:0) rs (cid:1) ( m i − j ). Aadditional factor i ( rs ) ( m i − j ) is enough. We have b ) ≤ i ( rs ) ( m i − j ) Y S : d S ≥ i d S − ≤ i ( rs ) ( m i − j ) Y S i d ′ S − ≤ i ( rs ) ( m i − j ) i (1+ s ( rs − ) )( m i − j ) = i (( rs ) +1+ s ( rs − )) ( m i − j ) . (26)17ombining equations (23), (24), (25), and (26), we get |G ji | ≤ (cid:18) t − t − i (cid:19) i + 1 (cid:18) ii (cid:19)(cid:18) ns (cid:19)(cid:18) r − ss (cid:19) t (cid:18) n − sj − s (cid:19)(cid:18) j − sr − s (cid:19) ( m i − r )!(( r − s )!) i − (cid:18) m i + i − (cid:19) m i − j ) i t − i i (( rs ) +1+ s ( rs − )) ( m i − j ) ≤ (cid:18) t − t − i (cid:19) i t − i i + 1 (cid:18) ii (cid:19)(cid:18) r − ss (cid:19) t − i n j ( s !) i +1 (( r − s )!) i ( m i − r )!( j − r )! (cid:18) m i + i − (cid:19) m i − j ) i (( rs ) +1+ s ( rs − )) ( m i − j ) ≤ (cid:18) t − t − i (cid:19) i t − i i + 1 (cid:18) ii (cid:19)(cid:18) r − ss (cid:19) t − i n m i ( s !) i +1 (( r − s )!) i (cid:18) C i C n (cid:19) m i − j Here we set C = 4( r − s ) and C = (cid:0) rs (cid:1) + 4 + s (cid:0) rs − (cid:1) . (cid:3) Lemma 12 If t = 2 k is even, then we have |G m k k | = (cid:18) nm k (cid:19) m k !( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) . (27) Proof:
We will construct a bijection from G m k k to a triple ( U, P , C ), where U is a set of m k vertices, P is a partition of U into ( k + 1) s -sets and k ( r − s )-sets, and C is a code consistingof k pairs valid parentheses.For any good walk w ∈ G m k k , let U be the set of vertices covered by w . Note each edgeappears exactly twice. We define a graph T , whose vertices are the stops in w . Two stopsare connected if they belong to one edge. Observe that T is acyclic and connected; T mustbe a tree. Since T has exactly k edges, T must have k + 1 vertices. Hence w has exactly k + 1stops; we list them as S , S , . . . , S k . For 1 ≤ i ≤ k , let E i be the set of ( r − s ) vertices in F i but not in any stops. We get a partition: U = (cid:0) ∪ ij =0 S j (cid:1) ∪ (cid:0) ∪ ij =1 E j (cid:1) . A code consists of k ‘(’and k ‘)’ is generated as follows. When we scan the walk from left to right, if an edge appearsthe first time, we append the code by a ‘(’; otherwise, we append the code by a ‘)’. The codeis a valid sequence of k pairs of parentheses. (In this case, the number of ‘ ∗ ’s is zero.) Itsuffices to recover a walk from a partition of [ m k ] and a sequence of valid parentheses.Given a partition of U (cid:0) ∪ ij =0 S j (cid:1) ∪ (cid:0) ∪ ij =1 E j (cid:1) and a sequence of k pairs valid parentheses, we first build a rooted tree T as follows. At eachtime, we maintain a tree T , a current stop S , a set of unused stops S . Initially T containsnothing but the root stop S , S := S , and S = { S , S , . . . , S k } . At each time, read a symbolfrom the sequence. If the symbol is an open parenthesis, then find an S i in S with index i assmall as possible, delete S i from S , attach S i to T as a child stop of S , and let S := S i ; if thesymbol is “)”, then let S point to the the parent stop of the current S . Repeat this processuntil all symbols from the sequence are processed.Since every closed parenthesis has a matching open parenthesis, this process never getstuck. When the process ends, a rooted tree T on the vertex set { S , . . . , S k } is created. For1 ≤ i ≤ k , let F i be the union of E i and two ends of i -th edge, which created in the process.For example, for k = 3, if the sequence is (())(), then the corresponding good closed walk is S F S F S F S F S F S F S F = S ∪ S ∪ E , F = S ∪ S ∪ E , and F = S ∪ S ∪ E .Thus, this is a bijection from G m k k to all triples { U, P , C } . The number of ways to choose m k vertices is (cid:0) nm k (cid:1) . The number of ways to choose these sets S , S , . . . , S k , E , . . . , E k as apartition of U is (cid:18) m k s, . . . , s, r − s, . . . , r − s (cid:19) = m k !( s !) k +1 (( r − s )!) k . The number of sequences of k pairs valid parentheses is the Catalan number k +1 (cid:0) kk (cid:1) . Bytaking product of these three numbers, we get equation (27). (cid:3) Proof of Lemma 6:
By equations (21) and (22), we haveE(Trace( C t )) ≤ ⌊ t ⌋ X i =1 |G i | p i (1 − p ) i ≤ (1 + o (1)) ⌊ t ⌋ X i =1 a i . Here a i := (cid:0) t − t − i (cid:1) i t − i i +1 (cid:0) ii (cid:1)(cid:0) r − ss (cid:1) t − i n mi p i (1 − p ) i ( s !) i +1 (( r − s )!) i . We get a i a i +1 = (cid:0) t − t − i (cid:1) i t − i i +1 (cid:0) ii (cid:1)(cid:0) r − ss (cid:1) t − i n mi p i (1 − p ) i ( s !) i +1 (( r − s )!) i (cid:0) t − t − i − (cid:1) i t − i − i +2 (cid:0) i +2 i +1 (cid:1)(cid:0) r − ss (cid:1) t − i − n mi +1 p i +1 (1 − p ) i +1 ( s !) i +2 (( r − s )!) i +1 = i (2 i − i + 2)(2 i + 1)( t − i )( t − i −
1) ( r − s )! n r − s p (1 − p ) < i ( r − s )! n r − s p (1 − p ) . When n r − s p (1 − p ) ≫ t , we have a i = o ( a i +1 ). Thus,E(Trace( C t )) ≤ (1 + o (1)) a ⌊ t ⌋ . When t = 2 k , we getE (cid:0) Trace( C k ) (cid:1) ≤ (1 + o (1)) n s + k ( r − s ) (cid:0) r − ss (cid:1) k ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k . For t = 2 k + 1, we haveE(Trace( C k +1 )) ≤ (1 + o (1)) a k ≤ (1 + o (1)) 2 k n s + k ( r − s ) (cid:0) r − ss (cid:1) k +1 ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k . Now we assume k = o (log( n r − s p (1 − p ))). For t = 2 k , let b k := |G m k k | p k (1 − p ) k = (cid:18) nm k (cid:19) m k !( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k . It is clear that E(Trace( C k )) ≥ b k . We also haveE(Trace( C k )) − b k ≤ k − X i =1 |G i | p i (1 − p ) i + m k − X j = r |G jk | p k (1 − p ) k = (1 + o (1)) k − X i =1 a i + a k m k − X j = r (cid:18) C k C n (cid:19) m k − j . a k = (1 + o (1)) (cid:0) r − ss (cid:1) k b k and a i = O (cid:18) a k (cid:16) k n r − s p (1 − p ) (cid:17) k − i (cid:19) . We concludeE(Trace( C k )) − b k = O b k (cid:18) r − ss (cid:19) k (cid:18) k n r − s p (1 − p ) + C k C n (cid:19)! = o ( b k ) . Here we use the fact (cid:0) r − ss (cid:1) k (cid:16) k n r − s p (1 − p ) + C k C n (cid:17) = o (1) since k = o (log( n r − s p (1 − p ))). (cid:3) Let us review the definition of the Semicircle Law. Let F ( x ) be the continuous distributionfunction with density f ( x ) such that f ( x ) = (2 /π ) √ − x when | x | ≤ f ( x ) = 0 when | x | > . Let A be a Hermitian matrix of dimension N × N . The empirical distribution of theeigenvalues of A is F ( A, x ) := 1 N |{ eigenvalues of A less than x }| . We say, the empirical distribution of the eigenvalues of A asymptotically follows theSemicircle Law centered at c with radius R if F ( R ( A − cI ) , x ) tends to F ( x ) in probabilityas N goes to infinity. (In this case, we write F ( R ( A − cI ) , x ) p → F ( x ).) If c is the center ofthe Semicircle Law, then any c ′ = c + o ( R ) is also the center of the Semicircle Law. Theorem 5 If n r − s p (1 − p ) → ∞ , then the empirical distribution of the eigenvalues of W − E( W ) follows the semicircle law centered at with radius q(cid:0) r − ss (cid:1)(cid:0) n − sr − s (cid:1) p (1 − p ) . Proof:
Let R := 2 q(cid:0) r − ss (cid:1)(cid:0) n − sr − s (cid:1) p (1 − p ), C := W − E( W ), and C nor := R C .To prove the theorem, we need to show that for any fixed t , the t -th moment of F ( C nor , x )(with n goes to infinity) is asymptotically equal to the t -th moment of F ( x ). We know the t -th moment of F ( C nor , x ) equals (cid:0) ns (cid:1) − E(Trace( C tnor )). For even t = 2 k , the t -th moment of F ( x ) is (2 k )! / k k !( k + 1)!. For odd t , the t -th moment of F ( x ) is 0.In order to prove the theorem, we need to show for any fixed k ,1 (cid:0) ns (cid:1) E(Trace( C knor )) = (1 + o (1)) (2 k )!2 k k !( k + 1)!and 1 (cid:0) ns (cid:1) E(Trace( C k +1 nor )) = o (1) . We know E(Trace( C tnor )) = 1 R t E(Trace( C t ))for any t . By Lemma 6, we haveE(Trace( C k )) = (1 + o (1)) n s + k ( r − s ) ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19) p k (1 − p ) k . Then 1 (cid:0) ns (cid:1) E(Trace( C knor )) = (1 + o (1)) (2 k )!2 k k !( k + 1)!20s desired.By Lemma 6 again, we haveE(Trace( C k +1 )) = O k n s + k ( r − s ) p k (1 − p ) k (cid:0) r − ss (cid:1) k +1 ( k + 1)( s !) k +1 (( r − s )!) k (cid:18) kk (cid:19)! . Thus 1 (cid:0) ns (cid:1) E(Trace( C k +1 nor )) = O k (cid:0) kk (cid:1)(cid:0) r − ss (cid:1) k +1 k ( k + 1) R ! = o (1) . Here k is any constant but R → ∞ . The theorem is proved. (cid:3) The following Lemma is useful to derive the Semicircle Law from one matrix to the other.
Lemma 13
Let A and B be two ( N × N ) -Hermitian matrices. Suppose that the empiricaldistribution of the eigenvalues of A follows the Semicircle Law centered at c with radius R . Ifeither k B k = o ( R ) or the rank of B is o ( N ) , then the empirical distribution of the eigenvaluesof A + B also follows the Semicircle Law centered at c with radius R . Proof:
It suffices to show F ( R ( A + B − cI ) , x ) p → F ( x ). First we assume k B k = o ( R ). ByLemma 1, for 1 ≤ k ≤ N , we have (cid:12)(cid:12)(cid:12)(cid:12) µ k (cid:18) R ( A + B − cI ) (cid:19) − µ k (cid:18) R ( A − cI ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ k B k R = o (1) . Hence F (cid:18) R ( A − cI ) , x − k B k R (cid:19) ≤ F (cid:18) R ( A + B − cI ) , x (cid:19) ≤ F (cid:18) R ( A − cI ) , x + k B k R (cid:19) . Since k B k = o ( R ), we have F (cid:16) R ( A − cI ) , x − k B k R (cid:17) p → F ( x ) and F (cid:16) R ( A − cI ) , x + k B k R (cid:17) p → F ( x ). By the Squeeze theorem, we have F ( R ( A + B − cI ) , x ) p → F ( x ).Now we assume rank( B ) = o ( N ). Let U be the kernel of B (i.e. B | U = 0); U hasco-dimension rank(B). Let Z := R ( A − cI ) | U = R ( A + B − cI ) | U . By Cauchy’s interlacetheorem [23], for 1 ≤ j ≤ N − rank( B ), we have µ j (cid:18) R ( A − cI ) (cid:19) ≤ µ j ( Z ) ≤ µ j +rank( B ) (cid:18) R ( A − cI ) (cid:19) ,µ j (cid:18) R ( A + B − cI ) (cid:19) ≤ µ j ( Z ) ≤ µ j +rank( B ) (cid:18) R ( A + B − cI ) (cid:19) . Thus, for rank( B ) + 1 ≤ j ≤ N − rank( B ), we have µ j − rank( B ) (cid:18) R ( A − cI ) (cid:19) ≤ µ j (cid:18) R ( A + B − cI ) (cid:19) ≤ µ j +rank( B ) (cid:18) R ( A − cI ) (cid:19) . It implies F (cid:18) R ( A − cI ) , x (cid:19) − rank( B ) N ≤ F (cid:18) R ( A + B − cI ) , x (cid:19) ≤ F (cid:18) R ( A − cI ) , x (cid:19) + rank( B ) N .
Since rank( B ) = o ( N ), we have F (cid:0) R ( A − cI ) , x (cid:1) ± rank( B ) N p → F ( x ). By the Squeezetheorem, we have F ( R ( A + B − cI ) , x ) p → F ( x ). (cid:3) roof of Theorem 3 : Recall L ( s ) ( K rn ) − L ( s ) ( H r ( n, p )) = M + M + M + M . We can write L ( s ) ( H r ( n, p )) as − M + (cid:18) − ( − s ( ns ) (cid:19) I + B − M − M − M , where B = L ( s ) ( K rn ) − (cid:18) − ( − s ( ns ) (cid:19) I .By Theorem 5, the empirical distribution of the spectrum of W − E( W ) follows theSemicircle Law centered at 0 with radius (2 + o (1)) q(cid:0) r − ss (cid:1)(cid:0) n − sr − s (cid:1) p (1 − p ). Since M = ( r − ss ) d ( W − E( W )), (cid:18) − ( − s ( ns ) (cid:19) I − M follows the Semicircle Law centered at c := 1 − ( − s ( ns )with radius R := (2 + o (1)) q − p ( r − ss )( n − sr − s ) p . Note ( − s ( ns ) = o ( R ). We can change the center to 1.By Theorem 1, L ( s ) ( K rn ) has an eigenvalue 1 − ( − s ( n − ss )( ns ) with multiplicity (cid:0) ns (cid:1) − (cid:0) ns − (cid:1) .Thus B has rank (cid:0) ns − (cid:1) = o (cid:0)(cid:0) ns (cid:1)(cid:1) . We also observe that M has rank at most 2, k M k = O (cid:18) √ (1 − p ) log Nd (cid:19) = o ( R ), and k M k = O (cid:16) √ log Nn √ d (cid:17) = o ( R ). Here we notice d ≫ log / n and1 − p ≫ log nn d .By Lemma 13, the matrices B , M , M , and M will not affect the Semicircle Law. Theproof of this Lemma is finished. (cid:3) References [1] N. Alon, M. Krivelevich, and V. H. Vu, Concentration of eigenvalues of random matrices
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