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A new family of expander Cayley graphs (?). Eyal Rozenman, Hebrew University Aner Shalev, Hebrew University Avi Wigderson, IAS You need the TexPoint PowerPoint add-in to view this presentation properly Download it from http://raw.cs.berkeley.edu/texpoint/. Definitions.
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A new family of expander Cayley graphs (?) Eyal Rozenman, Hebrew University Aner Shalev, Hebrew University Avi Wigderson, IAS You need the TexPoint PowerPoint add-in to view this presentation properly Download it from http://raw.cs.berkeley.edu/texpoint/
Definitions Undirected, regular (multi)graphs. Definition. The 2nd eigenvalue of a d-regular X (X) = max { || (AX /d) v || : ||v||=1 , v 1 } (X) [0,1] Definition. {Xn}is an expander family if (Xn) <1 Equivalent: RW on X converges in time O(log|X|)
Cayley graphs • Ga finite group • U½ Ga (symmetric) set of generators. • Definition: Cayley graph C(G,U): • Vertices: elements of G • Edges :(g, gu)for all uU. • C(G,U)is regular with degree |U|. • C(G,U) is connected ,U generates G.
1 2 3 4 1 Our Sequence of groups Gn = Even symmetries of a (rooted) d-regular tree. degree (#children) = d (fixed). Depth = n
1 2 3 4 1 Our Sequence of groups Gn = Even symmetries of a (rooted) d-regular tree. degree (#children) = d (fixed). Depth = n Depth 1: tree symmetries = alternating groupAd
1 2 3 4 1 Our Sequence of groups Gn = Even symmetries of a (rooted) d-regular tree. degree (#children) = d (fixed). Depth = n Depth 1: tree symmetries = alternating groupAd Theorem: Under some assumption on Ad: Every Gn has (const. #) expanding generators.
Assumption on alternating gp. Ad Our construction is based on Assumption:9U½Ad (= G1) such that • |U| ·d1/30 • (Ad,U) ·1/1000 This is an Open problem (will discuss this more later)
1 2 3 4 1 Main theorem Thm [RSW]: 9Un½Gn such that • |Un| ·d1/7 (constant -independent of n) • C(Gn,Un) is a good expander ((GnUn) · 1/1000)
1 2 3 4 1 Main theorem Thm [RSW]: 9Un½Gn such that • |Un| ·d1/7 (constant -independent of n) • C(Gn,Un) is a good expander ((GnUn) · 1/1000) Proof is by induction on n (base of induction is the assumption on Ad)
Ad Gn Gn Gn Gn Inductive definition of groups G1 = Ad, the alternating group. • Gn+1 = GnoAd (wreath product) • Gn+1= {(,x1,x2,…,xd) | 2Ad,xi2G }
Ad Gn Gn Gn Gn Inductive definition of groups G1 = Ad, the alternating group. • Gn+1 = GnoAd (wreath product) • Gn+1= {(,x1,x2,…,xd) | 2Ad,xi2G } Multiplication rule: (,x1,x2,…,xd) ¢ (,y1,y2,…,yd) = (, x(1),y1, x (2)y2,…, x (d)yd)
Ad Gn Gn Gn Gn Inductive definition of groups G1 = Ad, the alternating group. • Gn+1 = GnoAd (wreath product) • Gn+1= {(,x1,x2,…,xd) | 2Ad,xi2G } Multiplication rule: (,x1,x2,…,xd) ¢ (,y1,y2,…,yd) = (, x(1),y1, x (2)y2,…, x (d)yd) Example: conjugation g=(,1,…,1) , x = (1,x1,…,xd) g-1 x g=(1,x(1),…,x(d))
Generating sets U1½Ad – “small” generating set Un½Gn – “small” generating set Goal: Un+1½Gn+1 – “small” generating set
Generating sets U1½Ad – “small” generating set Un½Gn – “small” generating set Goal: Un+1½Gn+1 – “small” generating set Embed U1½Gn : {(u,1,1,L,1) |u2U1} Generates all {(s,1,1,L,1) |s2Ad} U1½Ad 12Gn 12Gn 12Gn 12Gn
Generating sets Pick some x=(1,x1,x2,L,xd) 2Gn+1 1 x12Gn x22Gn xd-12Gn xd2Gn
Generating sets Pick some x=(1,x1,x2,L,xd) 2Gn+1 Conjugating x by {(s,1,1,L,1) permutes xby s We get all (even)permutations of x. 1 x12Gn x22Gn xd-12Gn xd2Gn
Generating sets Pick some x=(1,x1,x2,L,xd) 2Gn+1 Conjugating x by {(s,1,1,L,1) permutes xby s We get all (even)permutations of x. If the permutations of xgenerate (Gn)d Then U1[ {x} generates Gn+1 1 x12Gn x22Gn xd-12Gn xd2Gn
Generating sets Necessary condition: {x1,x2,L,xd} generates Gn 1 x12Gn x22Gn xd-12Gn xd2Gn
Generating sets Necessary condition: {x1,x2,L,xd} generates Gn Un generatesGn. Un={u1,u2,,up}. Suppose p dividesd Put x = (u1,u1,,u1, u2,u2,,u2,, up,up,,up) 1 x12Gn x22Gn xd-12Gn xd2Gn
Generating sets Necessary condition: {x1,x2,L,xd} generates Gn Un generatesGn. Un={u1,u2,,up}. Suppose p dividesd Put x = (u1,u1,,u1, u2,u2,,u2,, up,up,,up) DefineUn(d) = the orbit of x under Ad. 1 x12Gn x22Gn xd-12Gn xd2Gn
Generating sets Necessary condition: {x1,x2,L,xd} generates Gn Un generatesGn. Un={u1,u2,,up}. Suppose p dividesd Put x = (u1,u1,,u1, u2,u2,,u2,, up,up,,up) DefineUn(d) = the orbit of x under Ad. Does Un(d) generate (Gn)d ? Is it expanding ? 1 x12Gn x22Gn xd-12Gn xd2Gn
Algebraic Zig-Zag theorem [ALW] If • ( (Gn)d, Un(d) ) · 1/50 • (Ad,U1) · 1/1000
Algebraic Zig-Zag theorem [ALW] If • ( (Gn)d, Un(d) ) · 1/50 • (Ad,U1) · 1/1000 Then 9W½Gn+1(= GnoAd) • (Gn+1, W) · 1/50 + 1/1000
Algebraic Zig-Zag theorem [ALW] If • ( (Gn)d, Un(d) ) · 1/50 • (Ad,U1) · 1/1000 Then 9W½Gn+1(= GnoAd) • (Gn+1, W) · 1/50 + 1/1000 • |W| = |U1|2 (Important! |W| is a func. of |U1|=const.)
Proof of main theorem Induction assumption: 9Un½Gn , (Gn,Un) · 1/1000
Proof of main theorem Induction assumption: 9Un½Gn , (Gn,Un) · 1/1000 Main Lemma:( (Gn)d, Un(d) ) · 1/50 (for good G,U)
Proof of main theorem Induction assumption: 9Un½Gn , (Gn,Un) · 1/1000 Main Lemma:( (Gn)d, Un(d) ) · 1/50 (for good G,U) Zigzag thm [RVW,ALW]: 9W½Gn+1(= GnoAd) • (Gn+1, W) · 1/50 + 1/1000 • |W| = |U1|2
Proof of main theorem Induction assumption: 9Un½Gn , (Gn,Un) · 1/1000 Main Lemma:( (Gn)d, Un(d) ) · 1/50 (for good G,U) Zigzag thm [RVW,ALW]: 9W½Gn+1(= GnoAd) • (Gn+1, W) · 1/50 + 1/1000 • |W| = |U1|2 DefineUn+1 = W*2 := all words of length 2 in W • |Un+1| = |U1|4·d1/7 • (Gn+1, Un+1) · (1/50 + 1/1000)2· 1/50
Main Lemma Main Lemma: Given • U½G • (G, U)· 1/1000 then - ( (G)d, U(d)) · 1/50 (for good G,U)
Main Lemma Main Lemma: Given • U½G • (G, U)· 1/1000 then - ( (G)d, U(d)) · 1/50 (for good G,U) Example: ( (G)d, Ud ) · 1/1000 But: Ud is NOT one Ad orbit
Main Lemma Main Lemma: Given • U½G small enough • (G, U)· 1/1000 then - ( (G)d, U(d)) · 1/50 (for good G,U) Example: ( (G)d, Ud ) · 1/1000 But: Ud is NOT one Ad orbit The idea – when |U|<< d • A random element of Ud is “more or less” in U(d). • So U(d) ”approximates” Ud well.
Main Lemma-reduction to G £ G When/Why is C( Gd, U(d)) connected ?
Main Lemma-reduction to G £ G When/Why is C( Gd, U(d)) connected ? (u1 , u2 , u3 , L,ud) 2U(d) (u2-1, u1 –1 , u3-1 , L,ud-1) 2U(d)
Main Lemma-reduction to G £ G When/Why is C( Gd, U(d)) connected ? (u1 , u2 , u3 , L,ud) 2U(d) (u2-1, u1 –1 , u3-1 , L,ud-1) 2U(d) multiply to get (u1, u2-1 ,u2 u1–1 , 1 , L , 1)
Main Lemma-reduction to G £ G When/Why is C( Gd, U(d)) connected ? (u1 , u2 , u3 , L,ud) 2U(d) (u2-1, u1 –1 , u3-1 , L,ud-1) 2U(d) multiply to get (u1, u2-1 ,u2 u1–1 , 1 , L , 1) Setting u2 =1 (assume 1 2 U) we generate all elements { (g , g-1 , 1 , 1, … , 1) | g 2 U} U1
Main Lemma-reduction to G £ G When/Why is C( Gd, U(d)) connected ? (u1 , u2 , u3 , L,ud) 2U(d) (u2-1, u1 –1 , u3-1 , L,ud-1) 2U(d) multiply to get (u1, u2-1 ,u2 u1–1 , 1 , L , 1) Setting u2 =1 (assume 1 2 U) we generate all elements { (g , g-1 , 1 , 1, … , 1) | g 2 U} • If first two coordinates generate G£G we are done • Expansion also follows U1
Expansion on G £ G X½G Z½G£G Z = {(x,x-1) | x2X} (completely correlated!)
Expansion on G £ G X½G Z½G£G Z = {(x,x-1) | x2X} (completely correlated!) Suppose (G,X) · 1- Is (G£G, Z) · 1-f() for some f?
Expansion on G £ G X½G Z½G£G Z = {(x,x-1) | x2X} (completely correlated!) Suppose (G,X) · 1- Is (G£G, Z) · 1-f() for some f? G abelian – NO. • C (G£G, Z) is disconnected. • Connected component of (1,1) is {(g,g-1) | g2G}
Expansion on G £ G X½G Z½G£G Z = {(x,x-1) | x2X} (completely correlated!) Suppose (G,X) · 1- Is (G£G, Z) · 1-f() for some f? G abelian – NO. • C (G£G, Z) is disconnected. • Connected component of (1,1) is {(g,g-1) | g2G} If Y = {(x,x) | x2X} then C(G£G, Y) is disconnected
Expansion on G £ G – decorrelating the gens. BUT: If for every x2G, x = [ax,bx] = ax-1 bx-1 ax bx
Expansion on G £ G – decorrelating the gens. BUT: If for every x2G, x = [ax,bx] = ax-1 bx-1 ax bx Xc = [ {ax,bx,(ax-1¢bx-1) | x2X} (+ inverses) Z½G£G Z = {(x,x-1) | x2Xc}
Expansion on G £ G – decorrelating the gens. BUT: If for every x2G, x = [ax,bx] = ax-1 bx-1 ax bx Xc = [ {ax,bx,(ax-1¢bx-1) | x2X} (+ inverses) Z½G£G Z = {(x,x-1) | x2Xc} (ax-1bx-1 , bxax) 2Z (ax , ax-1) 2Z (bx , bx-1) 2Z
Expansion on G £ G – decorrelating the gens. BUT: If for every x2G, x = [ax,bx] = ax-1 bx-1 ax bx Xc = [ {ax,bx,(ax-1¢bx-1) | x2X} (+ inverses) Z½G£G Z = {(x,x-1) | x2Xc} (ax-1bx-1 , bxax) 2Z (ax , ax-1) 2Z (bx , bx-1) 2Z + (x , 1) is generated by Z
Expansion on G £ G – decorrelating the gens. BUT: If for every x2G, x = [ax,bx] = ax-1 bx-1 ax bx Xc = [ {ax,bx,(ax-1¢bx-1) | x2X} (+ inverses) Z½G£G Z = {(x,x-1) | x2Xc} (ax-1bx-1 , bxax) 2Z (ax , ax-1) 2Z (bx , bx-1) 2Z + (x , 1) is generated by Z If (G,X) · 1- then (G£G, Z) · 1-( /500|X|2)
Commutator representation in Gn Def a group G has the commutator property (CP) if every element in G is a commutator.
Commutator representation in Gn Def a group G has the commutator property (CP) if every element in G is a commutator. Theorem [ORR, 50s] Ad has (CP)
Commutator representation in Gn Def a group G has the commutator property (CP) if every element in G is a commutator. Theorem [ORR, 50s] Ad has (CP) Theorem [N 03’] If G has (CP) then G oAd has (CP)
Commutator representation in Gn Def a group G has the commutator property (CP) if every element in G is a commutator. Theorem [ORR, 50s] Ad has (CP) Theorem [N 03’] If G has (CP) then G oAd has (CP) Proof: Reduce to the system of eqs x1 yh(1) xs(1)-1 yt(1) = a1 x2 yh(2) xs(2)-1 yt(2) =a2 … xd yh(d) xs(d)-1 yt(d) =ad 2d variables xi2G, yi2G, d constants ai2G. 3 arbitrary permutations h,s,t2Sd
Expansion on alternating gp. Ad Our construction was based on Assumption:9U½Ad (= G1) such that • |U| ·d1/30 • (Ad,U) ·1/1000