A new family of expander Cayley graphs

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A new family of expander Cayley graphs (?)

• Eyal Rozenman, Hebrew University
• Aner Shalev, Hebrew University
• Avi Wigderson, IAS
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Definitions
Undirected, regular (multi)graphs. Definition.
The 2nd eigenvalue of a d-regular X ?(X) max
(AX /d) v v1 , v ? 1 ?(X) ?
0,1 Definition. Xn is an expander family if
?(Xn)? ?lt1 Equivalent RW on X converges in time
O(logX)
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Cayley graphs

• G a finite group
• U ½ G a (symmetric) set of generators.
• Definition Cayley graph C(G,U)
• Vertices elements of G
• Edges (g, gu) for all u?U.
• C(G,U) is regular with degree U.
• C(G,U) is connected , U generates G.

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Our Sequence of groups

• Gn Even symmetries of a (rooted) d-regular
  tree.
• degree (children) d (fixed). Depth n

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Our Sequence of groups

• Gn Even symmetries of a (rooted) d-regular
  tree.
• degree (children) d (fixed). Depth n
• Depth 1 tree symmetries alternating group Ad

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Our Sequence of groups

• Gn Even symmetries of a (rooted) d-regular
  tree.
• degree (children) d (fixed). Depth n
• Depth 1 tree symmetries alternating group Ad
• Theorem Under some assumption on Ad
• Every Gn has (const. ) expanding generators.

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Assumption on alternating gp. Ad

• Our construction is based on
• Assumption 9 U ½ Ad ( G1) such that
• U d1/30
• ?(Ad,U) 1/1000
• This is an Open problem (will discuss this more
  later)

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Main theorem

• Thm RSW 9 Un ½ Gn such that
• Un d1/7 (constant - independent of n)
• C(Gn,Un) is a good expander (?(GnUn) 1/1000)

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Main theorem

• Thm RSW 9 Un ½ Gn 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)

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Inductive definition of groups

• G1 Ad, the alternating group.
• Gn1 Gno Ad (wreath product)
• Gn1 (?,x1,x2,,xd) ?2 Ad,xi 2 G

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Inductive definition of groups

• G1 Ad, the alternating group.
• Gn1 Gno Ad (wreath product)
• Gn1 (?,x1,x2,,xd) ?2 Ad,xi 2 G
• Multiplication rule
• (?,x1,x2,,xd) (?,y1,y2,,yd) (??, x?(1),y1,
  x ?(2)y2,, x ?(d)yd)

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Inductive definition of groups

• G1 Ad, the alternating group.
• Gn1 Gno Ad (wreath product)
• Gn1 (?,x1,x2,,xd) ?2 Ad,xi 2 G
• 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))

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Generating sets

• U1 ½ Ad small generating set
• Un ½ Gn small generating set
• Goal Un1 ½ Gn1 small generating set

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Generating sets

• U1 ½ Ad small generating set
• Un ½ Gn small generating set
• Goal Un1 ½ Gn1 small generating set
• Embed U1 ½ Gn (u,1,1,L,1) u 2 U1
• Generates all (s,1,1,L,1) s 2 Ad

U1 ½ Ad
1 2 Gn
1 2 Gn
1 2 Gn
1 2 Gn
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Generating sets

• Pick some x(1,x1,x2,L,xd) 2 Gn1

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x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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Generating sets

• Pick some x(1,x1,x2,L,xd) 2 Gn1
• Conjugating x by (s,1,1,L,1) permutes x by s
• We get all (even) permutations of x.

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x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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Generating sets

• Pick some x(1,x1,x2,L,xd) 2 Gn1
• Conjugating x by (s,1,1,L,1) permutes x by s
• We get all (even) permutations of x.
• If the permutations of x generate (Gn)d
• Then U1 x generates Gn1

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x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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Generating sets

• Necessary condition x1,x2,L,xd generates Gn

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x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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Generating sets

• Necessary condition x1,x2,L,xd generates Gn
• Un generates Gn. Unu1,u2,?,up. Suppose p
  divides d
• Put x (u1,u1,?,u1, u2,u2,?,u2,?, up,up,?,up)

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x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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Generating sets

• Necessary condition x1,x2,L,xd generates Gn
• Un generates Gn. Unu1,u2,?,up. Suppose p
  divides d
• Put x (u1,u1,?,u1, u2,u2,?,u2,?, up,up,?,up)
• Define Un(d) the orbit of x under Ad.

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x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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Generating sets

• Necessary condition x1,x2,L,xd generates Gn
• Un generates Gn. Unu1,u2,?,up. Suppose p
  divides d
• Put x (u1,u1,?,u1, u2,u2,?,u2,?, up,up,?,up)
• Define Un(d) the orbit of x under Ad.
• Does Un(d) generate (Gn)d ?
• Is it expanding ?

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x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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Algebraic Zig-Zag theorem ALW

• If
• ?( (Gn)d, Un(d) ) 1/50
• ?(Ad,U1) 1/1000

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Algebraic Zig-Zag theorem ALW

• If
• ?( (Gn)d, Un(d) ) 1/50
• ?(Ad,U1) 1/1000
• Then
• 9 W½ Gn1 ( Gn o Ad)
• ?(Gn1, W) 1/50 1/1000

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Algebraic Zig-Zag theorem ALW

• If
• ?( (Gn)d, Un(d) ) 1/50
• ?(Ad,U1) 1/1000
• Then
• 9 W½ Gn1 ( Gn o Ad)
• ?(Gn1, W) 1/50 1/1000
• W U12 (Important! W is a func. of
  U1const.)

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Proof of main theorem

• Induction assumption 9 Un ½ Gn , ?(Gn,Un)
  1/1000

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Proof of main theorem

• Induction assumption 9 Un ½ Gn , ?(Gn,Un)
  1/1000
• Main Lemma ?( (Gn)d, Un(d) ) 1/50 (for good
  G,U)

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Proof of main theorem

• Induction assumption 9 Un ½ Gn , ?(Gn,Un)
  1/1000
• Main Lemma ?( (Gn)d, Un(d) ) 1/50 (for good
  G,U)
• Zigzag thm RVW,ALW
• 9 W½ Gn1 ( Gn o Ad)
• ?(Gn1, W) 1/50 1/1000
• W U12

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Proof of main theorem

• Induction assumption 9 Un ½ Gn , ?(Gn,Un)
  1/1000
• Main Lemma ?( (Gn)d, Un(d) ) 1/50 (for good
  G,U)
• Zigzag thm RVW,ALW
• 9 W½ Gn1 ( Gn o Ad)
• ?(Gn1, W) 1/50 1/1000
• W U12
• Define Un1 W2 all words of length 2 in W
• Un1 U14 d1/7
• ?(Gn1, Un1) (1/50 1/1000)2 1/50

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Main Lemma

• Main Lemma Given
• U ½ G
• ?(G, U) 1/1000
• then
• - ?( (G)d, U(d)) 1/50 (for good G,U)

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Main Lemma

• Main Lemma Given
• U ½ G
• ?(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

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Main Lemma

• Main Lemma Given
• U ½ G 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 ltlt d
• A random element of Ud is more or less in U(d).
• So U(d) approximates Ud well.

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Main Lemma-reduction to G G

• When/Why is C( Gd, U(d)) connected ?

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Main Lemma-reduction to G G

• When/Why is C( Gd, U(d)) connected ?
• (u1 , u2 , u3 , L,ud) 2 U(d)
• (u2-1, u1 1 , u3-1 , L,ud-1) 2 U(d)

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Main Lemma-reduction to G G

• When/Why is C( Gd, U(d)) connected ?
• (u1 , u2 , u3 , L,ud) 2 U(d)
• (u2-1, u1 1 , u3-1 , L,ud-1) 2 U(d)
• multiply to get
• (u1, u2-1 , u2 u11 , 1 , L , 1)

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Main Lemma-reduction to G G

• When/Why is C( Gd, U(d)) connected ?
• (u1 , u2 , u3 , L,ud) 2 U(d)
• (u2-1, u1 1 , u3-1 , L,ud-1) 2 U(d)
• multiply to get
• (u1, u2-1 , u2 u11 , 1 , L , 1)
• Setting u2 1 (assume 1 2 U) we generate all
  elements
• (g , g-1 , 1 , 1, , 1) g 2 U

U1
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Main Lemma-reduction to G G

• When/Why is C( Gd, U(d)) connected ?
• (u1 , u2 , u3 , L,ud) 2 U(d)
• (u2-1, u1 1 , u3-1 , L,ud-1) 2 U(d)
• multiply to get
• (u1, u2-1 , u2 u11 , 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
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Expansion on G G

• X ½ G
• Z ½ G G
• Z (x,x-1) x 2 X (completely correlated!)

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Expansion on G G

• X ½ G
• Z ½ G G
• Z (x,x-1) x 2 X (completely correlated!)
• Suppose ?(G,X) 1-?
• Is ?(G G, Z) 1-f(?) for some f?

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Expansion on G G

• X ½ G
• Z ½ G G
• Z (x,x-1) x 2 X (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) g 2 G

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Expansion on G G

• X ½ G
• Z ½ G G
• Z (x,x-1) x 2 X (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) g 2 G
• If Y (x,x) x 2 X then C(G G, Y) is
  disconnected

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Expansion on G G decorrelating the gens.

• BUT If for every x 2 G, x ax,bx ax-1 bx-1
  ax bx

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Expansion on G G decorrelating the gens.

• BUT If for every x 2 G, x ax,bx ax-1 bx-1
  ax bx
• Xc ax, bx, (ax-1bx-1) x 2 X (
  inverses)
• Z ½ G G
• Z (x,x-1) x 2 Xc

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Expansion on G G decorrelating the gens.

• BUT If for every x 2 G, x ax,bx ax-1 bx-1
  ax bx
• Xc ax, bx, (ax-1bx-1) x 2 X (
  inverses)
• Z ½ G G
• Z (x,x-1) x 2 Xc
• (ax-1bx-1 , bxax) 2 Z
• (ax , ax-1) 2 Z
• (bx , bx-1) 2 Z

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Expansion on G G decorrelating the gens.

• BUT If for every x 2 G, x ax,bx ax-1 bx-1
  ax bx
• Xc ax, bx, (ax-1bx-1) x 2 X (
  inverses)
• Z ½ G G
• Z (x,x-1) x 2 Xc
• (ax-1bx-1 , bxax) 2 Z
• (ax , ax-1) 2 Z
• (bx , bx-1) 2 Z
• (x , 1) is generated by Z

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Expansion on G G decorrelating the gens.

• BUT If for every x 2 G, x ax,bx ax-1 bx-1
  ax bx
• Xc ax, bx, (ax-1bx-1) x 2 X (
  inverses)
• Z ½ G G
• Z (x,x-1) x 2 Xc
• (ax-1bx-1 , bxax) 2 Z
• (ax , ax-1) 2 Z
• (bx , bx-1) 2 Z
• (x , 1) is generated by Z
• If ?(G,X) 1-? then ?(G G, Z) 1-(? /500X2)

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Commutator representation in Gn

• Def a group G has the commutator property (CP) if
  every element in G is a commutator.

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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)

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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 o Ad has
  (CP)

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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 o Ad 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 xi 2 G, yi 2 G, d constants ai 2
  G.
• 3 arbitrary permutations h,s,t 2 Sd

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Expansion on alternating gp. Ad

• Our construction was based on
• Assumption 9 U ½ Ad ( G1) such that
• U d1/30
• ?(Ad,U) 1/1000

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Expansion on alternating gp. Ad

• Our construction was based on
• Assumption 9 U ½ Ad ( G1) such that
• U d1/30
• ?(Ad,U) 1/1000
• For any group G, O(logG) elements suffice AR
  
• logG O(dlogd) in our case)

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Expansion on alternating gp. Ad

• Our construction was based on
• Assumption 9 U ½ Ad ( G1) such that
• U d1/30
• ?(Ad,U) 1/1000
• For any group G, O(logG) elements suffice AR
  
• logG O(dlogd) in our case)
• Abelian G needs ?(log G) generators

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Expansion on alternating gp. Ad

• Reasons to believe there are few expanding
  generators

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Expansion on alternating gp. Ad

• Reasons to believe there are few expanding
  generators
• 2 random elements generate Ad

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Expansion on alternating gp. Ad

• Reasons to believe there are few expanding
  generators
• 2 random elements generate Ad D
• 9 7 elements in Ad such that Cayley graph has
  diameter O( logAd ) BL?

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Expansion on alternating gp. Ad

• Reasons to believe there are few expanding
  generators
• 2 random elements generate Ad D
• 9 7 elements in Ad such that Cayley graph has
  diameter O( logAd ) BKL
• A conjecture (Aldous) implies O(d) transpositions
  suffice for expansion (transpositions are weak!)

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Expansion on alternating gp. Ad

• Reasons to believe there are few expanding
  generators
• 2 random elements generate Ad D
• 9 7 elements in Ad such that Cayley graph has
  diameter O( logAd ) BL?
• A conjecture (Aldous) implies O(d) transpositions
  suffice for expansion (transpositions are weak!)
• A conjecture (Wigderson) implies O(d1/2)
  permutations suffice for expansion.

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D