An Explicit Construction of an Expander Family (Margulis-Gaber-Galil)

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1 An Explct Constructon of an Expander Famly Marguls-Gaber-Gall) Orr Paradse July 4th 08 Prepared for a Theorst's Toolkt, a course taught by Irt Dnur at the Wezmann Insttute of Scence. The purpose of these notes s to help the students ncludng, most mportantly, me) durng the presentaton, so nformaltes and naccuraces are to be expected. Introducton We present an smple constructon of an expander famly due to M75] and GG8]. As s often the case, a smple constructon requres complcated analyss, but we present the analyss of T4, 6.], whch s composed of three straghtforward steps and s beautful n ts own rght. As you wll see, t also ncorporates and generalzes many of the concepts we saw n the course.. Overvew Our alleged expander wll be {G n } n N, where G n s a graph on vertces Z n Z n, where a, b) s connected to S a, b), S a, b), T a, b), T a, b), a ±, b), a, b ± ) where S a, b) := a, a + b), T a, b) := a + b, b), S a, b) = a, b a), T a, b) = a b, b) wth addton and subtracton modulus n. To show that {G n } n N s an expander famly, we must show that λ G n ) = mn f,f 0 ε f) Var f) = E u v f u) f v)) ] mn f,f 0 = Ω ) ] E u π f u) and we do ths by constructng an nnte verson of ths graph whch would be easer to analyze n terms of expanson, and showng that the nnte verson has smlar expanson propertes to G n. In fact, we wll have an addtonal nnte) ntermedate graph. For a more detaled explanaton, we must ntroduce two addtonal actors. The graph famly {R n } n N. Each R n s a graph on vertces 0, n) 0, n), where each x, y) s connected to S x, y), S x, y), T x, y), T x, y). Graph Z on vertces Z Z\{0, 0}, Each vertex a, b) s connected to S x, y), S x, y), T x, y), T x, y), only now we dene S and T usng regular addton not addton modulus n). Informally, we wll then prove each of the followng nequaltes, orr.paradse@wezmann.ac.l As we have not yet dened λ R n) and λ Z). Θ λ G n )) λ R n ) λ Z) = Ω )

2 The countable case Let's start at the end. The noton of conductence of a graph can be generalzed to countably nnte graphs by lettng ) E A, A Φ G) := We then reason about Z's conductence. Proposton.. Φ Z) /3. nf A nte and nonempty Proof. Let A be a nte nonempty subset of Z \{0, 0}. Let A be the ntersecton of A wth the -th quadrant of Z, and let A 0 be the ntersecton of A wth each axs. E A \ A 0, A ) A \ A 0 = A A 0 Clam.. Proof. Consder A. The followng three propertes hold: S A ) = T A ) = A S A ) T A ) =. Otherwse we'd have a, a + b) = a + b, b ) whch mples that b = a, whch the fact that a, b) and a, b ) resde n the rst quadrant. S A ) and T A ) are contaned n the rst quadrant. S A ) T A ) s twce A 's sze and s contaned n the rst quadrant, so at least half of ts elemnts aren't n A. Snce the set of edges wth one pont n A contans S A ) T A ), ths means that there are at least A edges leavng A and landng n A. Repeatng the above clam for A, A 3 and A 4 whle consderng S, T ), S, T ) and S, T ) respectvely) gves that for all 4], A of edges leavng A land n A. In other words, E A \ A0, A ) = E A, A ) = A E A, A ) A = A \ A 0 Clam.3. E A 0, A ) 5 A 0 3 A Proof. Each vertex n A 0 has two outgong edges that land outsde A 0 e.g. a, 0) s connected to a, ±a)). Some of these edges land n A \ A 0 and some land n A. A \ A 0 can account for at most 4 A \ A 0 edges, therefore the number of edges landng n A s at least A 0 4 A \ A 0. But usng the prevous clam we see that A \ A 0 can account for at most 3 A \ A 0 edges, as at least A \ A 0 edges come from A and speccally do not come from A 0 ), gvng us that E A 0, A ) A 0 3 A \ A 0 = 5 A 0 3 A. To conclude the proof, add 5/6 of the rst nequalty to /6 of the second. Formally, observe that ) E A, A = E A \ A0, A ) + E A0, A ) 5 E A \ A0, A ) + E A0, A ) 5 A 5 A A 0 3 A = A Smlarly to the nte case followng the Courant-Fscher theorem), for a countably nnte graph G we dene λ G) to be the nmum of the Raylegh quotent taken over all zero-meaned functons that are nce enough) 3. u,v) EG) f u) f v)) λ G) := nf u V G) f u) The other two edges are self-loops. 3 We could have dened a Laplacan for countably nnte graphs and proven that ts second egenvalue s equal to sad nmum, but ths s not needed for our proof.

3 Where the nmum s taken over f : Z R such that c Z f c) <, c Z f c) = 0, f 0. The dllgent reader can try to generalze the total varaton and varance over nte graphs to countably nnte ones, and nd that ther rato.e. the Raylegh quotent) s exactly as above. Pluggng Z nto our denton, we get λ Z) := nf f u) f S u))) + f u) f T u))) ] f u) We gnore the factor of / and keep t n mnd for the nal tally f we wsh to make t). Followng these denton, t s natural to prove a Cheeger nequalty for them, and ndeed such a relaton holds. Fact.4. For any countably nnte d-regular graph G, Φ G) dλ G) The proof s smlar to the proof for the nte case we saw n class use Fedler's algorthm). From ths and the above we get that λ G) Φ G) /6 /44. 3 Countable to contnuous Ths s where thngs get nterestng. Agan, we dene an analogue to the Raylegh quotent, ths tme for contnuous uncountable) graphs. To avod many techncaltes, not only do we skp the denton of a Laplacan operator and proof of the egenvalue and Raylegh qutent correspondence), but also plug R n straght nto the denton from the get-go. 4 Let L denote. Let 0,n) λ R n ) := nf f u) f S u))) + f u) f T u))) du f L 0,n) f u) du Where the nmum s taken over the space of functons f : 0, n) R such that 0,n) f z) dz s welldened.e. f s Lebesgue ntegrable), nte and nonzero, and that f z) dz = 0. 0,n) Proposton 3.. λ R n ) λ Z) Proof. Each f L has a Fourer decomposton gven by f z) = f c Z c) χ c z), where χ a,b x, y) = n eπax+by) f c) = f, χc = f z) χ c z) dz 0,n) Our goal s to show a correspondance between the expresson mnmzed n the LHS and the expresson mnmzed n the RHS. Speccally, we wll show that the Raylegh quotent of f : 0, n) R w.r.t R n ) s equal to the Raylegh quotent of f : Z R w.r.t Z), whch gves the requres result. For the denomnators, we use the Parseval equalty for ths contnuous product space) to get f c) = f c) = f z) dz c Z 0,n) c Z \{0,0)} where the leftmost equalty s from the assumpton that f 0, 0) = f z) dz = 0. 0,n) For the numerators, let s z) := f z) f S z)) and smlarly t z) := f z) f T Z)), then usng Parseval's equalty and lnearty of Fourer coecents.e. lnearty of an nner product), s z) + t z) dz = ) ŝ c) + t c) = f c) f S c) + f c) f T c)) 0,n) c Z c Z \{0,0)} 4 What's mportant s that we're be able to reason about ths expresson both now and n the followng secton. 3

4 To conclude, we observe that f S a, b) = f S a, b)) e πax+by) d x, y) = f x, x + y) e πax+by) d x, y) n 0,n) n 0,n) y =x+y = f x, y ) e πa b)x+by ) = f a b, b) = f T a, b) n 0,n) and smlarly f T a, b) = f S a, b). As such, ) f c) f S c) + f c) f T c)) 0,n) s z) + t z) dz = c Z = c Z f c) f T c)) + f c) f S c) = c Z f T c) f c) ) + f S c) f c) ) ) so the numerators are equal as well. 4 λ G n ) Ω λ R n )) We have all the tools we need, so we can jump straght to the proof. Proof. Let f be the mnmzer of G n 's Raylegh quotent. That s, f : Z n R such that f 0, n f u) = 0, and 5 λ G) = 4 f u) f Su)) + f u) f T u))) + f u) f u + e )) + f u) f u + e )) n f u) n Agan, we wsh to show a correspondence between functons f : Z n R and functons f : 0, n) R. Ths tme, we extend f : Z n R nonzero, zero meaned) to all of 0, n) by breakng 0, n) nto a grd of squares, and lettng the value of the f on each square be equal to the value of the f on the square's bottom-left vertex. That s, we let f x, y) := f x, y ). Snce the area of each square n the grd s exactly, then f z) dz = 0,n) f u), and so the n denomnators are equal up to factor 4). We show that the numerators are equal. Let's focus on the left addend of the RHS, 0,n) f z) f S z))). Integratng over each square of the grd) seperately, we have ) ) f z) f S z)) dz = f z) f S z)) dz 0,n) a,b) Z n a,a+) b,b+) Now, n a gven square, the top-left to bottom-rght dagonal splts the square n half, and n the bottom trangle t holds that a + b = a + b, whereas n the top trangle t holds that a + b = a + b +. 5 In very few detals, ths follows from E u v f u) f v)) ] = 8 φ {S ±,T ±, )±e, )±e } E u π f u) f φ u))) = 8 and snce π G n's statonary dstrbuton) s unform as G n s d-regular. φ {S,T, )+e, )+e } E u π f u) f φ u))) ] 4

5 As such, a,a+) b,b+) ) f z) f S z)) dz = f a, b) f a, a + b)) + f a, b) f a, a + b + )) ) c=a,b) = f c) f S c))) + f c) f S c) + e )) ) f c) f S c))) + f c) f c) + e )) ) + f c + e ) f S c) + e )) α γ) α β) +β γ) ) = f c) f S c))) + f c) f c) + e )) + f c + e ) f S c) + e )) now summng over all squares, a,b) Z n a,a+) b,b+) f z) f S z)) ) dz = Changng order of summaton = 3 a,b) Z n a,b) Z n f c) f S c))) + f c) f c) + e )) + f c + e ) f S c) + e )) 3 f c) f S c))) + f c) f c) + e )) f c) f S c))) + f c) f c) + e )) ) Makng the analgous observatons for T, we have ) 3 f z) f T z)) dz 0,n) a,b) Z n a,b) Z n f c) f T c))) + f c) f c) + e )) Summng both nequaltes, we get that the numerator of G n 's Raylegh quotent s at least two-thrds the numerator of R n 's Raylegh quotent. Puttng everythng together, we have that λ G n ) 3 4 λ R n ) = 6 λ R n ) References M75] G.A. Marguls. Explct Constructon of Concentrators. Prob. Per. Infor., Vol. 9 4), pages 780, 973 n Russan). Englsh translaton n Problems of Infor. Trans., pages 3533, 975. GG8] O. Gabber and Z. Gall. Explct Constructons of Lnear Sze Superconcentrators. Journal of Computer and System Scence, Vol., pages 40740, 98. T4] L. Trevsan. Lecture Notes on Expanson, Sparsest Cut, and Spectral Graph Theory