On the Expansion of Group based Lifts

Transcription

1 On the Expansion of Group base Lifts Naman Agarwal, Karthekeyan Chanrasekaran, Alexanra Kolla, Vivek Maan July 7, 015 Abstract A k-lift of an n-vertex base graph G is a graph H on n k vertices, where each vertex v of G is replace by k vertices v 1,, v k an each ege u, v in G is replace by a matching representing a bijection π uv so that the eges of H are of the form u i, v πuvi. Lifts have been stuie as a means to efficiently construct expaners. In this work, we stuy lifts obtaine from groups an group actions. We erive the spectrum of such lifts via the representation theory principles of the unerlying group. Our main results are: 1. There is a constant c 1 such that for every k c1n, there oes not exist an Abelian k-lift H of any n-vertex -regular base graph such that H is almost Ramanujan nontrivial eigenvalues of the ajacency matrix at most O in magnitue. This can be viewe as an ana of the well-known no-expansion result for Abelian Cayley graphs.. A uniform ranom lift in a cyclic group of orer k of any n-vertex -regular base graph G, with the nontrivial eigenvalues of the ajacency matrix of G boune by λ in magnitue, has the new nontrivial eigenvalues also boune by λ + O in magnitue with probability 1 ke Ωn/. In particular, there is a constant c such that there exists a lift of a Ramanujan graph in a cyclic group of orer k which is almost Ramanujan, for every k cn/. We use this fact to esign a quasi-polynomial time algorithm to construct almost Ramanujan expaners eterministically. The existence of expaning lifts in cyclic groups of orer k = On/ can be viewe as a lower boun on the orer k 0 of the largest cyclic group that prouces expaning lifts. Our two results show that the lower boun closely matches the upper boun for k 0 upto a factor of 3 in the exponent, thus suggesting a threshol phenomenon. We believe that our results coul prove crucial in constructing families of almost Ramanujan expaners of all egrees in polynomial time. naman@cs.princeton.eu, Princeton University karthe@illinois.eu, University of Illinois Urbana-Champaign akolla@illinois.eu, University of Illinois Urbana-Champaign vmaan@illinois.eu, University of Illinois Urbana-Champaign

2 1 Introuction Expaner graphs have spawne research in pure an applie mathematics uring the last several years, with several applications to multiple fiels incluing complexity theory, the esign of robust computer networks, the esign of error-correcting coes, e-ranomization of ranomize algorithms, compresse sensing an the stuy of metric embeings. For a comprehensive survey of expaner graphs see [Sar06, HLW06]. Informally, an expaner is a graph where every small subset of the vertices has a relatively large ege bounary. Most applications are concerne with sparse -regular graphs G, where the largest eigenvalue of the ajacency matrix A G is. In case of a bipartite graph, the largest an smallest eigenvalues of A G are an, which are referre to as trivial eigenvalues. The expansion of the graph is relate to the ifference between an λ, the first largest in absolute value non-trivial eigenvalue of A G. Roughly, the smaller λ is, the better the graph expansion. The Alon-Boppana boun [Nil91] states that λ 1 o1, thus graphs with λ 1 are optimal expaners an are calle Ramanujan. A simple probabilistic argument can show the existence of infinite families of expaner graphs [Pin73]. However, constructing such infinite families explicitly has proven to be a challenging an important task. It is easy to construct Ramanujan graphs with a small number of vertices: -regular complete graphs an complete bipartite graphs are Ramanujan. The challenge is to construct an infinite family of -regular graphs that are all Ramanujan, which was first achieve by Lubotzky, Phillips an Sarnak [LPS88] an Margulis [Mar88]. They built Ramanujan graphs from Cayley graphs. All of their graphs are regular, have egrees p + 1 where p is a prime, an their proofs rely on eep number theoretic facts. In two recent breakthrough papers, Marcus, Spielman, an Srivastava showe the existence of bipartite Ramanujan graphs of all egrees [MSS13, MSS15]. However their results o not provie an efficient algorithm to construct those graphs. A striking result of Frieman [Fri08] an a slightly weaker but more general result of Puer [Pu13], shows that almost every -regular graph on n vertices is very close to being Ramanujan i.e. for every ɛ > 0, asymptotically almost surely, λ < 1 + ɛ. It is still unknown whether the event that a ranom - regular graph is exactly Ramanujan happens with constant probability. Despite the large boy of work on the topic, all attempts to efficiently construct large Ramanujan expaners of any given egree have faile, an exhibiting such constructions remains an intriguing open problem. A combinatorial approach to constructing expaners, initiate by Frieman [Fri03], is to prove that one may obtain new larger Ramanujan graphs from smaller ones. In this approach, one starts with a base graph G which one lifts to obtain a larger graph H. More concretely, a k-lift of an n-vertex base-graph G is a graph H on k n vertices, where each vertex u of G is replace by k vertices u 1,, u k an each ege u, v in G is replace by a matching between u 1,, u k an v 1,, v k. In other wors, for each ege u, v of G there is a permutation π uv so that the corresponing k eges of H are of the form u i, v πuvi. The graph H is a uniformly ranom lift of G if for every ege u, v the bijection π uv is chosen uniformly an inepenently at ranom from the set of permutations of k elements, S k. Since we are focusing on Ramanujan graphs, we will restrict our attention to lifts of -regular graphs. It is easy to see that any lift H of a -regular base-graph G is itself -regular an inherits all the eigenvalues of G which, hereafter we refer to as ol eigenvalues, whereas the rest of the eigenvalues are referre to as new eigenvalues. In orer to use lifts for builing expaners, it is necessary that the lift woul also inherit the expansion properties of its base graph. One hopes that a ranom lift of a Ramanujan graph will also be almost Ramanujan with high probability or even that there exists a k-lift which is almost Ramanujan for some boune k. Karthik: last part of the statement is unclear to me.?? Frieman [Fri03] first stuie the eigenvalues of ranom k-lifts of regular graphs an prove that every new eigenvalue of H is O 3/4 with high probability. He conjecture a boun of 1+o1, which woul be tight see, e.g. [Gre95]. Linial an Puer [LP10] improve Frieman s boun to O /3. Lubetzky, Suakov an Vu [LSV11] showe that the absolute value of every nontrivial eigenvalue of the lift is Oλ, where λ is the secon largest in absolute value eigenvalue of the base graph, improving on the previous results when G is significantly expaning. Aarrio-Berry an Griffiths [ABG10] further improve the bouns above by showing that every new eigenvalue of H is O, an very recently, Puer [Pu13] prove the nearly-optimal boun of All those results hol with probability tening to 1 as k, thus the orer k of the lift in question nees to be large. Nearly no results were known in the regime where k is boune with respect to the number of noes n of the graph. Bilu an Linial [BL06] were the first to stuy k-lifts of graphs with boune k, an suggeste constructing 1

3 Ramanujan graphs through a sequence of -lifts of a base graph: start with a small -regular Ramanujan graph on some finite number of noes e.g. K +1. Every time the -lift operation is performe, the size of the graph oubles. If there is a way to preserve expansion after lifting, then repeating this operation will give large goo expaners of the same boune egree. The authors in [BL06] showe that if the starting graph G is significantly expaning so that λg = O, then there exists a ranom -lift of G that has all its new eigenvalues upper-boune in absolute value by O 3. In the recent breakthrough work of Marcus, Spielman an Srivastava [MSS13], the authors showe that for every bipartite graph G, there exists a -lift of G, such that the new eigenvalues achieve the Ramanujan boun of 1, but their result still oes not provie any efficient algorithm to fin such lifts. 1.1 Our Results In this work, we stuy lifts as a means to efficiently construct almost Ramanujan expaners of all egrees. We erive these lifts from groups. This is a natural generalization of Cayley graphs. Definition 1 Γ-lift. Let Γ be a group of orer k with enoting the group operation. A Γ-lift of an n-vertex base graph GV, E is a graph H = V Γ, E obtaine as follows: it has k n vertices, where each vertex u of G is replace by k vertices {u} Γ. For each ege u, v of G, we choose an element g u,v Γ an replace that ege by a perfect matching between {u} Γ an {v} Γ that is given by the eges u i, v j for which g u,v i = j. We enote Γ = k to be the orer of the lift. We refer to Γ-lifts obtaine using Γ = Z/kZ, the aitive group of integers moulo k, as shift k-lifts. Since every cyclic group of orer k is isomorphic to Z/kZ, we have that Γ-lifts are shift k-lifts whenever Γ is a cyclic group. A tight connection between the spectrum of Γ-lifts an the representation theory of the unerlying group Γ is known [MS95, FKL04]. This connection tells us that the lift graph incurs the eigenvalues of the base graph, while its new eigenvalues are the union of eigenvalues of a collection of matrices arising from the irreucible representations of the group an the group elements assigne to the eges. This connection has been recently use in [HPS15] in the context of expansion of lifts, aiming to generalize the results in [MSS15]. In orer to unerstan the expansion properties of the lifts, we focus on the new eigenvalues of the lifte graph. We aress the expansion of Γ-lifts obtaine from cyclic groups an abelian groups. We present a high probability boun on the expansion of ranom shift k-lifts for boune k. Our results for -lifts an more generally, for shift k-lifts are as follows: Theorem 1. Let G be a -regular graph with non-trivial eigenvalues at most λ in absolute value, an H be a uniformly ranom -lift of G. Let λ new be the largest new eigenvalue of H in magnitue. Then λ new = Oλ with probability 1 e Ωn/. Moreover, if G is moerately expaning such that λ with probability 1 e Ωn/. λ new λ = O, then Theorem. Let G be a -regular graph with non-trivial eigenvalues at most λ in absolute value, an H be a ranom shift k-lift of G. Let λ new be the largest new eigenvalue of H in magnitue. Then λ new = Oλ with probability 1 k e Ωn/. Moreover, if G is moerately expaning such that λ with probability 1 k e Ωn/. λ new λ = O, then

4 In particular, if we start with G being a Ramanujan expaner, then w.h.p. a ranom shift k-lift will be almost Ramanujan, having all its new eigenvalues boune by O. Remark 1. In contrast to the case of lifts of orer k, the epenency on λ is necessary for boune k. This has previously been observe by the authors in [BL06] who gave the following example: Let G be a isconnecte graph on n vertices that consists of n/ + 1 copies of K +1, an let H be a ranom -lift of G. Then the largest non-trivial eigenvalue of G is λ = an it can be shown that with high probability, λ new = λ =. Therefore, our eigenvalue bouns are nearly tight. Remark. Our result for -lifts improves upon the factor present in the result of Bilu-Linial [BL06]. This factor arises in their analysis ue to the use of the converse of the Expaner Mixing Lemma EML along with an ɛ-net style argument. The converse EML is provably tight, so straightforwar use of the converse EML will inee incur the factor. We are able to improve the eigenvalue boun by performing a eeper analysis of the ɛ-net argument, avoiing irect use of the converse EML. Karthik: Bilu-Linial also o an ɛ-net type argument; change the remark. Please verify.?? Lifts base on groups immeiately suggest an algorithm towars builing -regular n-vertex Ramanujan expaners. In orer to escribe this algorithm, we first recall the brute-force algorithm that follows from the existential result of [MSS13]. The iea is to start with the complete bipartite graph K, an lift the graph n/ times. At each stage, brute force searching over the space of all possible -lifts an picking the best most expaning one. However, since a graph V, E has E possible -lifts, it follows that the final lift will be chosen from among n/4 possible -lifts, which means that the brute force algorithm will run in time exponential in n. Next, suppose that for every k, we are guarantee the existence of a group Γ of orer k such that for every base graph there exists a Γ-lift that has all its new eigenvalues at most 1 in absolute value e.g., suppose for every k an for every base graph, there exists a shift k-lift that has all new eigenvalues with absolute value at most 1; then a brute force algorithm similar to the one above, woul perform only one lift operation of the base graph K, to create a Γ-lift with n = k vertices. This algorithm woul only have to choose the best among k possibilities k ifferent choices of group element per ege of the base graph, which is polynomial in n, the size of the constructe graph. Here we have assume that is a constant. This motivates the following question: what is the largest possible group Γ that might prouce expaning Γ-lifts? Our next result rules out the existence of large abelian groups that might lea to even slightly expaning lifts. Theorem 3. For every n-vertex -regular graph G, ɛ 0, 1, an abelian group Γ of size at least n 1 k = exp ɛ + n 1 eɛ all Γ-lifts of G have secon largest eigenvalue at least ɛ. In particular, when k = Ωn, there is no Γ-lift H of any n-vertex -regular graph G with λh = O whenever Γ is an abelian group of orer k. Remark 3. The first an only known efficient constructions of Ramanujan expaners are Cayley graphs of certain groups [LPS88]. We observe that a Cayley graph for a group Γ with generator set S can be obtaine as a Γ-lift of the bouquet graph a graph that consists of one vertex with multiple self loops [Mak15]. Our no-expansion result for abelian groups complements the known result on no-expansion of abelian Cayley graphs [FMT06]. Remark 4. Our Theorems 3 an can be viewe as lower an upper bouns on the largest orer k 0 of a group Γ such that for every n-vertex graph, there exists a Γ-lift for which all the new eigenvalues are small. On the one han, Theorem shows that, for k = On/, most of the shift k-lifts of a Ramanujan graph have their new eigenvalues upper-boune by O. On the other han, Theorem 3 shows that for k Ωn, there is no shift k-lift that achieves such expansion guarantees. This suggests a threshol behaviour for k 0. Moreover, Theorem leas to a eterministic quasi-polynomial time algorithm for constructing almost Ramanujan with λ = O families of graphs., 3

5 Theorem 4. There exists an algorithm to construct a -regular n-vertex graph G such that λg = O in O4 n time. Algorithm 1 Quasi-polynomial time algorithm to construct expaners of arbitrary size n 1: Pick an r such that cr/ r = n, for a constant c given by Theorem. Do an exhaustive search to fin a -regular graph G on r vertices with λ = O. : For k = cr/, o an exhaustive search to fin a shift k-lift G of G with minimum λg. Proof of Theorem 4. We use Algorithm 1. We note that the choice of r in the first step ensures that r = O n. By Theorem, there exists a lift G of G such that λg = O. Thus, the exhaustive search in the secon step gives a graph G with λg = O. For the running time, we note that the first step can be implemente to run in time Or = O4 n. To boun the running time of the secon step, we observe that for each ege in G, there are k possible choices. Therefore the total search space is at most k r/ = cr / = O3 n an for each k-lift, it takes polyn time to compute λg. Thus, the overall running time of the algorithm is O4 n. Organization. We give some preliminary efinitions, notations, facts an lemmas in Section. In particular, we recall the tight connection between the spectrum of Γ-lifts an the representation of Γ in Section.4. We prove Theorem 3 in Section 3. For the purpose of intuition, we present an prove a slightly weaker version of Theorem 1 see Theorem 11 in Section 4. We prove the concentration inequality Lemma 3 neee for the weaker version in Section 5. We use a stronger version of the concentration inequality an prove Theorems 1 an in Section 6. Preliminaries.1 Notations Let G = V, E be a graph with vertex set V, V = n an ege set E. Let A be the ajacency matrix of the graph an let λ 1 λ... λ n be its n eigenvalues. Let λg = max λ i. Note that since A is a real, i:[,n] symmetric matrix its eigenvalues are also real. Moreover if G is regular with egree it is well-known that λ 1 = an that λg. If G is bipartite, then λ n = an we efine λg = max λ i. Throughout i:[,n 1] the paper, G will be a -regular graph an we will be concerne with eigenvalues of ajacency matrices. For any two subsets S, T V let ES, T be the number of eges between S an T. For a matrix M, we enote by M its spectral raius. For a vector x the set Sx enotes its support, i.e. the set of coorinates of x with a non-zero value. We efine to be the function with base. We represent e x by expx. Given a vector x {0, ±1/, ±1/4...} we efine the iaic ecomposition of x as the set { i u i } where each u i efine as We use the following combinatorial ientities. 1, if x j = i [u i ] j = 1, if x j = i 0, otherwise Lemma 1 Discretization Lemma. For any x R n, x 1/ an M such that the iagonal entries of M are 0, there exists y {±1/, ±1/4,...} n such that x T Mx y T My an y 4 x. Moreover, each entry of x between ± i an ± i 1 is roune to either ± i or ± i 1. Similarly, for any x 1, x R n, x 1, x 1/, there exists y 1, y {±1/, ±1/4,... } n such that x T 1 Mx y T 1 My, y 1 4 x 1, y 4 x an each entry of x 1, x between i an i 1 is roune to either i or i 1. 4

6 Proof of Lemma 1. To obtain such a vector y we take a vector x an roun its coorinates inepenently with the following probabilistic rule. Let x i = ±1 + δ i i be the i th coorinate of x. We roun x i to signx i i+1 with probability δ i an signx i i with probability 1 δ i. Let the roune vector be x. Note that E[x i ] = x i. Now since each coorinate is roune inepenently an the iagonal entries of M are 0, we get that E[x T Mx ] = x T Mx. This implies there exists a y {±1/, ±1/4,...} n that can be generate by this rouning such that x T Mx y T My. Also it is easy to see that y 4 x an by efinition every coorinate in y with value between ± i an ± i 1 is roune to either ± i or ± i 1. The proof of the secon part of the lemma is the same as the first part. Here we obtain x 1 an x by the same proceure an follow the same argument to get y 1 an y. Lemma. Assuming that r t z/, r, x > 1/, we have the following inequality: i=t r i z/r i x crr t z/r t x i=0 where cr is a constant epening only on r an c < 9. Proof of Lemma. For all i efine a i = r i z/r i x. Let us consier the ratio of consecutive terms a i+1 /a i for i [0, t 1]. a i+1 a i = = r i+1 z/r i+1 r r i z/r i 1 r 1 x r z i r r 1 + t i r x x r t z/ x If i t, we get that a i+1 /a i r x 1+r 1+ r = αr. It is easy to see that αr > > 1 for r. 3 Also for i = t 1, we get that a i+1 /a i r/1 + r x 1. Now consier the sum S 1 efine as S 1 = a 0 + a a t 1 αrs 1 = αra 0 + a a t 1 αr 1S 1 = a 0 + αra 0 a 1 + αra 1 a... + a t 1 αr αr 1S 1 a t 1 αr a i+1 αra i αr S 1 a t 1 αr 1 Therefore Setting cr = c < 9. a i S 1 + a t i [t] 1 + αr αr 1 αr 1 + a t. αr 1 we get the require result. α is greater than 3 which implies that Fact 1. For every c 1 0, there exists c s.t. x 1 x c 1 + c x where 0 x 1. We will make use of the well-known Hoeffing inequality: Theorem 5. Let X 1,..., X n be inepenent ranom variables such that X i is strictly boune within the interval [a i, b i ], then n n P X i E[X i ] t e t ni=1 b i a i. i=1 i=1 5

7 . Spectral Graph Theory Basics Expaner graphs are often seen as graphs which are close to ranom graphs. This iea is quantifie by the following well-known fact known as the Expaner Mixing Lemma which bouns the eviation between the number of eges between two subsets an the expecte number in a ranom graph. Theorem 6 Expaner-Mixing Lemma [LW03]. For non-bipartite graph with maximum non-trivial eigenvalue λ, S, T V ES, T S T λ S T n We can also get an anaue for bipartite graphs from the proof of the Expaner Mixing Lemma. The following theorem states the general boun. Theorem 7. For a graph with maximum non-trivial eigenvalue λ, S, T V ES, T S T n + λ S T. We nee the following theorem showing that expaners have small iameter in orer to show no-expansion of large abelian lifts. Theorem 8. [Chu89] The iameter of a -regular graph G with n vertices is upper boune by n/ /λ..3 Lifts In this section we formally efine lifts of graphs an state some of their properties. Definition Γ, S, -lift. Let Γ be a group, S be a set of size k an be a faithful group action of Γ on S. A Γ, S, -lift of an n-vertex base graph GV, E is a graph H = V S, E obtaine as follows: it has k n vertices, where each vertex u of G is replace by k vertices {u} S. For each ege u, v of G, we choose an element g u,v Γ an replace that ege by a perfect matching between {u} S an {v} S that is given by the eges u i, v j for which g u,v i = j. We enote S = k to be the orer of the lift. We note that if S = Γ an the group action is the left group operation itself, then Γ, S, -lifts are just Γ-lifts. Remark 5 Group Elements as Permutations. A faithful action of a group Γ on a set S inuces an embeing from Γ to SymS, where SymS is the symmetric group of S group of all permutations of S. Thus, we can ientify group elements with permutations of S = k objects. Using this language, the set of eges of the lift H can be rewritten as E = {u i, v j u, v E, π u,v i = j}, where π u,v is the permutation corresponing to the group element we have chosen for ege u, v. Besies Γ-lifts another interesting case of Γ, S, -lifts is when Γ = Sym[k] the symmetric group on k elements, S = [k] an the group action : Γ S S is efine by σ t = σt, i.e., the action of the permutation on the corresponing element. Such lifts are known as general lifts or simply k-lifts. Recall that shift k-lifts are Γ-lifts where the group Γ is a cyclic group. We will use the term abelian lifts to refer to Γ-lifts where the group Γ is an abelian group. Some initial easy observations can be mae about the structure of any lift: i the lifte graph is also regular with the same egree as the base graph an ii the eigenvalues of A are also eigenvalues of A H. Therefore we call the n eigenvalues of A the ol eigenvalues an nk 1 other eigenvalues of A H the new eigenvalues. We will enote by λ new the largest new eigenvalue of H in magnitue, which we also refer to as the first new eigenvalue for simplicity. Definition 3 Generalize Signing. Given a base graph GV, E, a group Γ, a set S an an action of Γ on S as in the above efinition, we efine a generalize signing of GV, E as a function s : EG Γ. We use the convention that su, v = g then sv, u = g 1. There is a bijection between signings an Γ, S, -lifts. 6

8 .4 Spectrum of Lifts via Representation Theory In this section, we characterize the spectrum of Γ-lifts as a union of the spectrum of certain matrices. We begin with some elementary facts on the representation theory of finite groups see also [Art98, Ser97]. Definition 4 Representation. A representation of a finite group Γ on a finite-imensional vector space V is a homomorphism ρ : Γ GLV, where GLV is the general linear group of V. If the imension of V is, then we efine the imension of ρ to be. A trivial representation is one where V = C an ρg = 1 for all g Γ. A permutation representation is one where the matrices ρg correspon to permutation matrices. We next consier an interesting special case of permutation representations. Definition 5 Regular Representation. For a group element g Γ, let e g be the Γ -imensional inicator vector of g an let C Γ enote the vector space efine by the basis vectors {e g } g Γ. Let P g enote the permutation matrix associate with the left action of g on Γ. Then ρg = P g is a representation of Γ on V = C Γ. This is known as the left regular representation of Γ on C Γ. Definition 6 Irreucible Representation. For a representation ρ : Γ GLV, a subspace W V is invariant uner ρ if ρgw W for all g Γ. The representation ρ is irreucible hereafter calle irrep if it has no proper invariant subspace. A well-known theorem of Maschke shows that every permutation representation can be ecompose into a finite number of irreps. Our next theorem is a consequence of this result as applie to the regular representation. Theorem 9 Decomposition into irreps for Regular Representation [Ser97]. Let ρ be the regular representation of Γ on C Γ. Then there exists a unitary matrix U C Γ Γ, an orthogonal ecomposition C Γ = V i an irreps ρ i : Γ GLV i such that UρgU 1 = i ρ i g for every g Γ. Moreover, the invariant subspaces V i are unique an the trivial representation is always one of the irreps. We next state a few properties of the irreps arising in Theorem 9 for abelian groups an cyclic groups. Fact. For abelian groups, the irreps in Theorem 9 are one-imensional. In particular, for a cyclic group Γ = {c, c,..., c k }, the irreps are given by ρ 1,..., ρ k : Γ GLC, where ρ i c j = ω j i, where ω i is a primitive k-th root of unity. We note that when k =, the two roots of unity are ω 1 = 1 an ω = 1, an the only non-trivial irrep is ρ, where ρ 0 = 1, ρ 1 = 1. We now characterize the eigenvalues of Γ-lifts. We observe that the ajacency matrix of a Γ-lift is a nk nk symmetric matrix, which has n n blocks B u,v, each of size k k; the block B u,v is the zero k k matrix if u, v is not an ege in G; for every ege u, v of G, we have B u,v = P u,v, which is the permutation representation of the element g = su, v Γ. The following theorem characterizes the spectrum of the lift in terms of the spectrum of certain smaller matrices. We note that even though G is an unirecte graph, for the purposes of the theorem, we view it as a irecte graph where if u, v E then v, u E. Recall that when su, v = g, then sv, u = g 1. Theorem 10. [MS95] For g Γ, let G g be the inuce subgraph of G consisting of irecte eges u, v E such that su, v = g, an let A g be its ajacency matrix. The ajacency matrix of the lifte graph H is equal to A H = g Γ A g P g = U i g Γ A g ρ i g U 1, for some unitary matrix U. Here ρ i are the irreps of the regular left representation of Γ given in Theorem 9. The above theorem shows that there is some basis given by the columns of the matrix U such that A H is block-iagonal in that basis, with blocks D i = g Γ A g ρ i g. In particular, the spectrum of H is equal to the spectrum of the set of matrices D i. We note that since for any group ρ 1 is the trivial, one-imensional representation, it follows that D 1 = A G, the ajacency matrix of the original graph. This is consistent with the observation in Section.3 that all the ol eigenvalues of G are also eigenvalues of H. We now specialize Theorem 10 to the case of cyclic groups to characterize the spectrum of shift k- lifts. For a shift k-lift of a graph G = V, E with ajacency matrix A, which is given by the signing 7

9 si, j = g i,j i,j E, efine the following family of Hermitian matrices A s ω parameterize by ω where ω is a k-th primitive root of unity. { 0, if A ij = 0 [A s ω] ij = ω gi,j, if A ij = 1 The following corollary regaring A s t follows from Theorem 10 an Fact. Corollary 1. Let G = V, E be a graph an H be a shift k-lift of G with the corresponing signing of the eges si, j = g i,j i,j E, where g i,j C k. Then the set of eigenvalues of H are given by ω: ω is a k-th primitive root of unity eigenvalues A s ω. The above simplifies significantly for -lifts as note in the next corollary. Corollary. When k =, the set of eigenvalues of a -lift H is given by the eigenvalues of A an the eigenvalues of A s, where A s is the signe ajacency matrix corresponing to the signing s, with entries from {0, 1, 1}. 3 No-expansion of Abelian Lifts In this section we show that it is impossible to fin even slightly expaning graphs which are lifts by large abelian groups Γ. By Theorem 8, we know that if a graph is an expaner, then it has small iameter. We show that if the size of the abelian group Γ is large, then all Γ-lifts of any base graph have large iameter, an hence they cannot be expaners. We prove Theorem 3. Theorem 3. For every n-vertex -regular graph G, ɛ 0, 1, an abelian group Γ of size at least n 1 k = exp ɛ + n 1 eɛ all Γ-lifts of G have secon largest eigenvalue at least ɛ. In particular, when k = Ωn, there is no Γ-lift H of any n-vertex -regular graph G with λh = O whenever Γ is an abelian group of orer k. Proof. Let Γ be an abelian group of orer k an G = V, E be a base graph on n-vertices that is -regular. Let e 1,..., e n/ be an arbitrarily chosen orering of the eges E. Let H be a lift graph obtaine using a Γ-lift. Recall that the signing of the eges of the base graph correspon to group elements, which in turn correspon to permutations of k elements. Let these signing of the eges be σ e e EG. For notational convenience, let us efine a layer L i of H to be the set of vertices {v i : v V }. We note that H has k layers. Let us fix an arbitrary vertex v in G. Let enote the iameter of H. This implies that for every j =,..., k there exists a path of length at most in H from v 1 to a vertex in L j. A layer j is reachable within istance in H iff there exists a walk e 1, e,..., e t from v of length t in G such that σ et σ et 1... σ e σ e1 1 = j. Thus the set of layers reachable within istance in H is containe in the set S = {σ et... σ e1 1 : e 1,..., e t is a walk from v in G of length t }. Since the group Γ is abelian, S {σe a1 1 σe a... σ a n/ e n/ 1 n/ i=1 a i } =: T. Since H has k layers, the carinality of S is at least k. The number of integral a i s satisfying n/ i=1 a i is at most n/+ n/ n/. Therefore, k T n + n n, e 1 + n e n e. n If λ G > ɛ, then it follows that λ H > ɛ for every Γ an the result follows. So, we may assume that λ G ɛ. Since H has nk vertices, using Theorem 8, we have nk/ /λ. Thus, if 8

10 λ new ɛ, then nk/ 1/ɛ an consequently, Rearranging the terms, we obtain that k e n ɛ exp 1 ɛ k e n e nk 1 ɛ. n ɛ n 1 exp ɛ + n 1 eɛ. 4 Expansion of Ranom -lifts 4.1 Overview In this section, we sketch a proof of Theorem 11 that is a slightly weaker version of Theorem 1 weaker by a multiplicative factor of four. The proof of this weaker result captures the main ieas involve in the proof of Theorems 1 an. Theorem 1 follows from Theorem as a special case. We present the full proof of Theorem in Section 6. Theorem 11. Let G be a -regular n-vertex graph with non-trivial eigenvalues at most λ in absolute value where λ, n 3 ln n, an H be a uniformly ranom -lift of G. Let λ new be the largest new eigenvalue of H in magnitue. Then, there exists a constant c such that λ new 4λ + c max λ, with probability at least 1 e n/. We note that G is moerately expaning such that λ, then we get λ new = 4λ + O. To prove this theorem, we require the following concentration inequality. Karthik: We are assuming that λ. We nee to remove this from the theorem statement an lemma statements an a a justification.?? Lemma 3. Let G be a -regular graph with non-trivial eigenvalues at most λ in absolute value where λ,. Let H be a uniformly ranom -lift of G, with corresponing signe ajacency n 3 ln n matrix A s. The following statements hol with probability at least 1 e n/ over the choice of the ranom signing: 1. For all u 1,..., u r {0, ±1} n, an v 1,..., v l {0, ±1} n satisfying I Su i Su j = for every i, j [r] an Sv i Sv j = for every i, j [l], an II Either Su i > n/ for every i [r] with non-zero u i, or Sv i > n/ for every i [l] with non-zero v i, we have i u T i A s j v j 377 max λ, ij r i=1 l λ Su i i Sv j j.. For all u 1,..., u r {0, ±1} n, an v 1,..., v l {0, ±1} n satisfying I, II an III Su i > Sv j for every i [r], j [l] with non-zero u i, j=1 9

11 we have i u T i A s j v j ij = 31 max λ, We will now prove Theorem 11 using the lemma above. r Su i i + i=1 l Sv j i. Proof of Theorem 11. The first new eigenvalue of the lift is λ new = max x R n x T A s x/x T x. To prove an upper boun on λ new, we will boun x T A s x/x T x for all x with high probabliity. In particular, assuming that the concentration inequalities given by Lemma 3 hols, we will show that x T A s x 4 λ x. By re-scaling we may assume that the maximum entry of x is less than 1/ in absolute value. Next, we use Lemma 1 to fin a vector y {±1/, ±1/4,...} n such that x T A s x y T A s y an y 4 x. We will prove a boun on y T A s y, which in turn will imply the esire boun on x T A s x. Let us consier the iaic ecomposition of y = i=1 i u i obtaine as follows: a coorinate of u i is 1 if the corresponing coorinate of y is i, it is 1 if the corresponing coorinate in y is i, an is zero otherwise. We note that Su i Su j = for every pair i, j N. Next, we partition the set of vectors u i s base on their support sizes. Let M := {i N : Su i n/ } an L := {i N : Su i > n/ } M an L for mini an large supports respectively. Corresponingly, efine y M := i M i u i an y L = i L i u i. We note that y = y M + y L, y = y M + y L = i N Su i i, an y T A s y y T M A s y M + y T M A s y L + y T LA s y L. We next boun each term in the following three claims. Claim 1. y T M A s y M λ + 4 y M. Proof. Let y M be a vector obtaine from y M by taking the absolute values of the coorinates. Then y M = y M an ym T A sy M y M T Ay M. Let J be the n n matrix with all entries being 1. We have y M T Ay M = y M A T n J y M + y M T n J y M λ y M + y M T n J y M. Above, we have use the fact that A nj has the same set of eigenvalues as A except for the first eigenvalue which was for the matrix A an is now zero. It remains to boun y M T n J y M. Consier the iaic ecomposition of y M = i M i u i, where the coorinates of u i are the absolute values of the coorinates of u i. y M T n J y M n i Su i j Su j i M j M:j i 1 i Su i i j since Su j n/ j M i M 4 y M. j M:j i j=1 Claim. y T LA s y L λ max λ, y L. 10

12 Proof. By triangle inequality, yla T s y L = i u T i A s j u j i,j L i u i A s j u j + i,j L:ij i,j L:i>j i u i A s j u j. We boun each term using the first part of Lemma 3. For both terms, our choice is r min{i L}, l = r, u i u i if i L an u i 0 if i L, v i = u i for every i [r], where 0 is the all-zeroes vector. We note that the conitions i an ii of the lemma are satisfie by this choice since every pair Su i, Su j is mutually isjoint an Su i > n for all i L. Consequently, λ yla T s y L 754 max λ, Su i i Su j j i L j L λ max λ, y L. Claim 3. ym T A s y L 408 max λ, y M + Proof. By triangle inequality, ym T A s y L = i u T i A s j u j i u i A s j u j + i M,j L i M,j L:ij λ max λ, y L. i M,j L:i>j i u i A s j u j. We boun the first an secon terms by the first an secon parts of Lemma 3 respectively. Let 0 be the all-zeroes vector. For the first term, our choice is r min{i M}, l min{i L}, u i u i if i M an u i 0 if i M, an v i u i if i L an v i 0 if i L. For the secon term, our choice is r min{i L}, l min{i M}, u i u i if i L an u i 0 if i L, an v i u i if i M an v i 0 if i M. The conitions i, ii an iii of the lemma are satisfie for the respective choices since every pair Su i, Su j is mutually isjoint, Su i > n for all i L an Su i > n/ Su j for every i L, j M. Consequently, ym T A s y L 377 max λ, i M λ Su i i j L + 31 max λ, Su j j + Su j j j L j M 408 max λ, y M + Su j j λ max λ, y L. 11

13 From the above three claims, we have 4λ y T A s y λ max λ, y M max λ, y L λ max λ, y. Therefore, we have x T A s x y T A s y λ max λ, y 4 λ max λ, x. We note that in the above proof, the multiplicative factor of 4 is a by-prouct of the iscretization of x. This can be avoie if we o not iscretize x straightaway, but instea push the iscretization a little eeper into the proof. Inee, we can see that the proof of Claim 1 where we boun y T M A n Jy M by λ y M oes not require y M to be a iscretize vector. This is how we are able to prevent the multiplicative factor loss to obtain Theorem 1. 5 Concentration Inequality In orer to prove Lemma 3 we nee to upper boun the sum ij i j u T i A sv j for all sets of vectors {u 1,..., u r }, {v 1,..., v l } satisfying the assumptions of the lemma. To begin with, one coul try to use the triangle inequality an upper boun each term u T i A sv j separately for each i, j. We note that u T i A sv j is a sum of ESu i, Sv j ii ranom variables with mean zero one for each ege between Su i an Sv j. By the expaner mixing lemma Theorem 7, we may upper boun the size of ESu i, Sv j by Su i Sv j /n + λ Su i Sv j. Depening on which of these two terms in the RHS ominates, we have two cases. For each case, we use a ifferent concentration boun Lemma 4 an Corollary 3. We begin with the neee concentration bouns. 5.1 Concentration Bouns Lemma 4. Let G be a -regular, n-vertex graph with non-trivial eigenvalues at most λ in absolute value where n 3 ln n an λ. Let H be a uniformly ranom -lift of G, with corresponing signe ajacency matrix A s. The following property hols with probability atleast 1 e n / over the ranom choice of signings: For every r {0, 1,..., 1/ }, every a, b 0, b 1,..., b r {0, ±1} n satisfying i Sb i Sb j = i, j [r], i j, ii Sa i Sb i i [r], an iii λ Sbi Sa n i [r] with non-zero b i, we have r r at A s i b i 14 n n Sa Sb i i. Sa i=0 i=0 Proof. For notational convenience, let b = r i=0 i b i. Fix a, b 1, b,..., b r {0, ±1} n. Then a T A s b is a sum of inepenent ranom variables with mean 0 one for each ege between Sa an Sb i. This is because the intersection between the support of any two vectors b i an b j is empty. The sum of squares of the ifference between the maximum an the minimum values of these variables is at most 1

14 r i=1 4ESb i, Sa i. For vectors a, b 1,..., b r satisfying i an ii, by the Expaner Mixing Lemma, we have ESb i, Sa 3 Sbi Sa n. We note that this inequality hols even if b i is a zero vector. By Theorem 5, P r a T A s b > 14 n n Sa Sb i i exp 98 Sa n Sa 3 Sa i Now fixing the values of the support sizes α = Sa, β i = Sb i, the number of possible choices for a is at most n α α exp 3α n α. Similarly the number of possible choices for each b i is atmost r exp 3β i n β i. Therefore the total number of choices for b is at most exp i=1 3β i n β i. Since each α, β i n, we can replace each β i by its upper boun α i. Hence, using Lemma, r n exp 3β i exp 3 i=1 β i r α i n α i i=1 exp 7α n. α Therefore, the total number of choices of a, b 1,..., b r of sizes α, β 1,... β r respectively is at most n exp 30α. α By taking a union boun over the choices of vectors with the fixe support sizes, the probability of the existence of a set of vectors a, b 1,..., b r with sizes α, β 1,..., β r respectively an satisfying i an ii is boune by exp 8α n 3 exp 8n α 3. Above, we have use that α n/ which follows since α = Sa nλ/ n/ by ii an iii. Next, let us boun the number of choices for the support sizes of the vectors a, b 1,..., b r. The number of choices for the support sizes is at most n +1/. Therefore taking the union boun over the choice of the support sizes, we get that the total probability is at most exp + 1/ ln n exp 8n 3 exp n. Lemma 5. Let G be a -regular, n-vertex graph with non-trivial eigenvalues at most λ in absolute value, where n, an H be a uniformly ranom -lift of G, with corresponing signe ajacency matrix 3 ln n A s. The following property hols with probability at least 1 e 3n/ over the ranom choice of signings: For every a, b {0, ±1} n, q, w {1,..., n} satisfying i Sa q, Sb w, Sb N G Sa, ii q w q, iii w > n, an iv λ qw < n, we have a T A s b 10 λ qw 3/ q. 1 w Here, N G Sa enotes the set of neighbors of Sa formally efine as {v u Sa with u, v E}. 13

15 Proof. For a pair of vectors a, b {0, ±1} n an q, w {1,..., n}, let Baa, b, q, w enote the event that inequality 1 is violate. We nee to upper boun the probabillity that there exists a, b, q, w satisfying i, ii, iii an iv such that Baa, b, q, w happens. We note that the sum a T A s b over ranom choices of A s is a sum of inepenent ranom variables chosen from {±, ±1}, all of which have mean 0. The number of such ranom variables being summe is at most ESa, Sb, i.e. the number of eges between Sa an Sb. Therefore for a fixe a, b, q, w by applying the Hoeffing inequality Theorem 5, we get that q w 50λ qw 3/ P Baa, b, q, w exp ESa, Sb Now using iv an the expaner mixing lemma Theorem 7, we have ESa, Sb Sa Sb /n + λ Sa Sb qw/n + λ qw 3λ qw. Substituting this in the previous expression, we obtain q P Baa, b, q, w exp 50/3w. w We will use the union boun now. For this purpose, we will first fix q, w an the size of the support of a an b. We take a union boun over all possible choices of a, b of that fixe size, an then take a union boun over all choices of the support sizes. For fixe support sizes α = Sa, β = Sb, we observe that the total number of choices for the support sets for a are n α. Now, since Sb is a subset of NG Sa, the number of choices of Sb is boune by α β. Also, since each entry in a, b is 0 or ±1 the total number of choices for a an b is at most n α α α β β exp 3α n exp 3β α. α We will first show upper bouns on each of these terms. Since w n, by ii, we have q n 3. Also, α = Sa q, β = Sb w. Therefore, β n exp 3α α n exp 3q q exp 9q q w = exp 9 q w exp 9w q w w. q w q w The last line follows from the fact that x / x is boune by 1 for x [1/, 1] an that [1/, 1]. Further, α exp 3β β exp 3β q exp 3w β q. w The last inequality follows by the fact that x c x is an increasing function if x < c. Therefore, by union boun we get that the probability of a ba event for fixe q, w an support sizes α = Sa, β = Sb is at most exp 14/3w 4q exp 14n 4q exp 14n w 3 w 3. 14

16 Now the number of choices of the supports is at most n, number of choices for q, w is at most n an therefore, P a, b, q, w satisfying i, ii, iii, an iv:baa, b, q, w n 4 exp 14n 3 exp 3n. Corollary 3. Let G be a -regular, n-vertex graph with non-trivial eigenvalues at most λ in absolute value, where n, an H be a uniformly ranom -lift of G, with corresponing signe ajacency 3 ln n matrix A s. The following property hols with probability at least 1 e 3n/ over the ranom choice of signings: For every a, b {0, ±1} n satisfying i Sa Sb Sa, ii Sb > n, an iii λ Sa Sb < n, we have a T A s b 10 λ Sa Sb Sb Sa. Sb Proof. For every a, b, we apply the boun from Lemma 5 on a T A s b with q = Sa, w = Sb where b is the same as b restricte to the coorinates in Sb N G Sa. We observe that a T A s b = a T A s b an hence the corollary. 5. Proof of Lemma 3 Next, we use Corollary 3 an Lemma 4 to prove Lemma 3. We restate the Lemma for the sake of presentation. Lemma 3. Let G be a -regular graph with non-trivial eigenvalues at most λ in absolute value where λ,. Let H be a uniformly ranom -lift of G, with corresponing signe ajacency n 3 ln n matrix A s. The following statements hol with probability at least 1 e n/ over the choice of the ranom signing: 1. For all u 1,..., u r {0, ±1} n, an v 1,..., v l {0, ±1} n satisfying I Su i Su j = for every i, j [r] an Sv i Sv j = for every i, j [l], an II Either Su i > n/ for every i [r] with non-zero u i, or Sv i > n/ for every i [l] with non-zero v i, we have i u T i A s j v j 377 max λ, ij r i=1 l λ Su i i Sv j j.. For all u 1,..., u r {0, ±1} n, an v 1,..., v l {0, ±1} n satisfying I, II an III Su i > Sv j for every i [r], j [l] with non-zero u i, we have i u T i A s j v j ij = 31 max λ, r Su i i + i=1 j=1 l Sv j i. j=1 15

17 Proof. For notational convenience, we will replace Su i by s i an Sv j by t j. We split the sum i u T i A s j v j ij into several subcases epening on i, j an the sizes of Su i an Sv j an use the triangle inequality. Figure 5. summarizes the splitting of i, j into various terms epening on the various values of i, j, s i an t j. Next, we boun each of the terms separately. By Lemma 4 an Corollary 3, we know that A s satisfies the property mentione in both of them with probability atleast 1 e 3n/. We boun the terms assuming that A s satisfies the property mentione in Lemma 4 an Corollary 3. i, j [r] [l] C 1 i j < i + 1 maxs i, t j < mins i, t j j i + 1 maxs i, t j mins i, t j C s i t j s i < t j λ si t j < n λ si t j n C 3 C 4 C 5 C 6 s i i < λ t j j s i i λ t j j s i i < t j j s i i t j j Claim 4. i u T i A s j v j i,j C 1 3 s i i + t j j. i [r] j [l] Proof. The sum is conitione over the set of tuples i, j in C 1, where { C 1 = i [r], j [l] j i + 1 } or maxs i, t j mins i, t j. By triangle inequality, i j u T i A s v j i,j C 1 i j u T i A s v j i,j [r] [l]:j i+ 1 + i,j [r] [l]:ij<i+ 1, maxs i,t j mins i,t j i j u T i A s v j We note that the number of eges out of any set S is boune by S. So, u T i A sv j mins i, t j for 16

18 any u i, v j { 1, 0, +1} n. We now boun the two terms above. For the first term, i j u T i A s v j l i j u T i A s v j i,j [r] [l]:j i+ 1 i [r] j=i+ 1 l i j mins i, t j i [r] j=i+ 1 l i j s i i [r] j=i+ 1 i s i. i [r] For the secon term, i,j [r] [l]:ij<i+ 1, maxs i,t j mins i,t j i j u T i A s v j i [r],j [l]:ij<i+ 1, maxs i,t j mins i,t j i [r],j [l]:ij<i+ 1, maxs i,t j mins i,t j i [r],j [l]:ij<i+ 1, maxs i,t j mins i,t j i [r],j [l]:ij<i+ 1, maxs i,t j mins i,t j i j u T i A s v j i j mins i, t j i j maxs i, t j i j s i + t j = i [r] i s i i+ 1 j=i j + j [l] i [r] s i i + j [l] i=j j t j t j j. i=j 1 i Claim 5. i u T i A s j v j i,j C 8 max, λ i [r] s i i. Proof. The sum is conitione over the set of tuples i, j in C, where C = {i, j [r] [l] i j < i + 1 an t j s i < t j }. By triangle inequality the require sum is at most i,j C i j u T i A sv j. We note that u i, v j 0 since t j s i < t j. Consier the term u T i A sv j where i, j is in C. We have two cases: Case 1: If /λ s i t j n, then we use Lemma 4 for the choice a u i, b 0 v j. This choice satisfies the 17

19 conitions of Lemma 4. Hence, u T i A s v j 14 s i tj n n 14 s i. Here, the last inequality follows by using x x 1 for x < 1. Case : If /λ s i t j < n, then we use Corollary 3 for the choice a v j, b u i. This choice satisfies the conitions of Corollary 3 since t j s i < t j, conition I of the Lemma implies s i > n/, an /λ s i t j < n. Hence, u T i A s v j 14 λ tj t j s i s i 14 λ s i. The last inequality follows since t j s i. Thus, for i, j C, we have u T i A sv j 14 max, λ s i. Therefore, i u T i A s j v j i,j C i j u T i A s v j i,j C 14 i j max, λ s i i [r] j=i 8 max, λ s i i. s i t j i [r] Claim 6. i u T i A s j v j i,j C 3 = λ j [l] t j j. Proof. The sum is conitione over the set of tuples i, j in C 3, where { C 3 = i, j i j i + 1 s i t j < s i si t j < n λ By triangle inequality, i u T i A s j v j i,j C 3 i j u T i A s v j. i,j C 3 s i i < λ } t j j. We note that u i, v j 0 since s i t j < s i. We use Corollary 3 to boun each term u T i A sv j. We use Corollary 3 with the choice a u i an b v j. This choice satisfies the conitions of Corollary 3 since 18

20 s i t j s i, conition I of the Lemma implies t j > n/, an /λ s i t j < n. Hence, i,j C 3 i j u T i A s v j 10 i,j C 3 i j < 10 λ3/4 1/8 i,j C 3 10 λ3/4 1/8 = 90 λ3/4 1/8 t j j j [l] λ si s i t j t j t j t j i j j i i=j i=j 1 +1 λ j i j i s i i < λ t j j λ j i λ j [l] t j j by Lemma an λ λ = 90λ λ t j j. j [l] By Fact 1, we can chose an appropriate constant c 1 such that the above quantity is boune by λ t j j. j [l] Claim 7. i u T i A s j v j i,j C 4 = 136 s i i. i [r] Proof. The sum is conitione over the set of tuples i, j in C 4, where { C 4 = i, j i j < i + 1 s i t j < s i si t j < n λ s i i λ } t j j. By triangle inequality, i u T i A s j v j i,j C 4 i j u T i A s v j. i,j C 4 We note that u i, v j 0 since s i t j < s i. We use Corollary 3 to boun each term u T i A sv j. We use Corollary 3 with the choice a u i an b v j. This choice satisfies the conitions of Corollary 3 since 19

21 s i t j s i, conition I of the Lemma implies t j > n/, an /λ s i t j < n. Hence, i u T i A s j v j i,j C 4 i j u T i A s v j i,j C 4 10 λ si s i t j t j i,j C 4 i j = i,j C 4 i j i j 3/8 i,j C 4 λsi tj s i 3 t j si λ s 1/4 i 3j 3i t j λ j i Above we use the fact that x 3 c x is an increasoing function if x c an s i j λ t j j. Therefore, i u T i A s j v j i,j C /8 j=i+ 1 1 λ s i i j i 1/4 i [r] j=i = 90 3/8 λ λ s i i 1/4 i [r] = 90 1 λ si i λ i [r] = 136 i [r] s i i. λ j i by Lemma That last equality is because, λ for every -regular graph an hence λ λ Claim 8. i u T i A s j v j i,j C 5 56 t j j + s i i. j [l] i [r] Proof. The sum is conitione over the set of tuples i, j in C 5, where { C 5 = i, j i j < i + 1 s i t j < s i si t j n λ By triangle inequality, i u T i A s j v j i,j C 5 j [l]: i [r] with i,j C 5 j s i i < t j j}. i+j u T i A s v j i:i,j C 5. We note that u i, v j 0 since s i t j < s i for every i, j C 5. Let us fix j such that there exists i, j C 5. We boun i {j 1/,...,j}: i,j C 5 i+j u T i A s v j 0

22 using Lemma 4. We will use Lemma 4 for the choice a v j an for every k = 0, 1,..., 1/, we take b k u j k if j k, j C 5 an b k 0 if j k, j C 5. This choice satisfies the conitions of Lemma 4 since i conition I of the Lemma implies Sb k are mutually non-intersecting, ii Sv j = t j j i s i = j i Su i for every i, j C 5 implies Sa k Sb k for every k = 0, 1,..., 1/, an iii b k is non-zero if an only if j k, j C 5 implies /λ Sb k Sa n for every non-zero b k. Hence, by Lemma 4, we have j i+j u T i A s v j j [l] i:i,j C 5 14 j t i=j j n s i n i+j t j j [l] i=j 1 = 14 j t j n n i=j s i t i. j j [l] i=j 1 Next, we group v j accoring to their support sizes an then sum them together. For c = 0, 1,,..., n, let J c be the set of inices j [l] s.t. n/ c t j < n/ c an for non-empty sets J c, efine j c := minj J c. With this notation, the above sum is 14 n 4n j c c c=0 j J c i=j i=j 1/ n 1 4n j jc c + j+jc c c c=0 j J c = 8 n c=0 8 n c=0 j J c j J c n j jc c + 7 n n c j jc + 7 i [r] i=j s i i c=0 j J c i=j 1/ +1 n s i i c=0 i=j i=j 1/ +1 c c s i i j+jc c c s i i j J c j+jc. n c t j < n c G.M. A.M. We observe that Moreover, n c=0 j J c j J c j+jc n n c j jc an n c=0 c=0 t j j jc j J c n c=0 t j j = t j j. j J c j [l]. c c 4. Substituting these we have the claim. Claim 9. i u T i A s j v j i,j C 6 = 154 s i i. i [r] Proof. The sum is conitione over the set of tuples i, j in C 6, where { C 6 = i, j i j i + 1 s i t j < s i λ si t j n s i i t j j}. 1