Transcript agg2004 4693

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
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.
Our Sequence of groups
Gn = Even symmetries of a (rooted) d-regular tree.
degree (#children) = d (fixed). Depth = n
1
1
2
3
4
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
1
1
2
3
4
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.
1
1
2
3
4
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)
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)
1
1
2
3
4
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)
1
1
2
3
4
Inductive definition of groups
G1 = Ad, the alternating group.
- Gn+1 = Gno Ad (wreath product)
- Gn+1 = {(,x1,x2,…,xd) | 2 Ad,xi 2 G }
Ad
Gn
Gn Gn Gn
Inductive definition of groups
G1 = Ad, the alternating group.
- Gn+1 = Gno Ad (wreath product)
- Gn+1 = {(,x1,x2,…,xd) | 2 Ad,xi 2 G }
Ad
Gn
Gn Gn Gn
Multiplication rule:
(,x1,x2,…,xd) ¢ (,y1,y2,…,yd) = (, x(1),y1, x (2)y2,…, x (d)yd)
Inductive definition of groups
G1 = Ad, the alternating group.
- Gn+1 = Gno Ad (wreath product)
- Gn+1 = {(,x1,x2,…,xd) | 2 Ad,xi 2 G }
Ad
Gn
Gn Gn Gn
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
U1 ½ Ad – “small” generating set
Un ½ Gn – “small” generating set
Goal: Un+1 ½ Gn+1 – “small” generating set
Generating sets
U1 ½ Ad – “small” generating set
Un ½ Gn – “small” generating set
Goal: Un+1 ½ Gn+1 – “small” generating set
Embed U1 ½ Gn : { (u,1,1,L,1) | u 2 U1}
Generates all {(,1,1,L,1) |  2 Ad}
U1 ½ Ad
1 2 Gn
1 2 Gn
1 2 Gn
1 2 Gn
Generating sets
Pick some x=(1,x1,x2,L,xd) 2 Gn+1
1
x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
Generating sets
Pick some x=(1,x1,x2,L,xd) 2 Gn+1
Conjugating x by {(,1,1,L,1) permutes x by 
We get all (even) permutations of x.
1
x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
Generating sets
Pick some x=(1,x1,x2,L,xd) 2 Gn+1
Conjugating x by {(,1,1,L,1) permutes x by 
We get all (even) permutations of x.
If the permutations of x generate (Gn)d
Then U1 [ {x} generates Gn+1
1
x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
Generating sets
Necessary condition: {x1,x2,L,xd} generates Gn
1
x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
Generating sets
Necessary condition: {x1,x2,L,xd} generates Gn
Un generates Gn. Un={u1,u2,L,up}. Suppose p divides d
Put x = (u1,u1,L,u1, u2,u2,L,u2,L, up,up,L,up)
1
x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
Generating sets
Necessary condition: {x1,x2,L,xd} generates Gn
Un generates Gn. Un={u1,u2,L,up}. Suppose p divides d
Put x = (u1,u1,L,u1, u2,u2,L,u2,L, up,up,L,up)
Define Un(d) = the orbit of x under Ad.
1
x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
Generating sets
Necessary condition: {x1,x2,L,xd} generates Gn
Un generates Gn. Un={u1,u2,L,up}. Suppose p divides d
Put x = (u1,u1,L,u1, u2,u2,L,u2,L, up,up,L,up)
Define Un(d) = the orbit of x under Ad.
Does Un(d) generate (Gn)d ?
Is it expanding ?
1
x1 2 Gn
x2 2 Gn
xd-1 2 Gn
xd 2 Gn
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
9 W½ Gn+1 (= Gn o Ad)
- (Gn+1, W) · 1/50 + 1/1000
Algebraic Zig-Zag theorem [ALW]
If
 ( (Gn)d, Un(d) ) · 1/50
 (Ad,U1) · 1/1000
Then
9 W½ Gn+1 (= Gn o Ad)
- (Gn+1, W) · 1/50 + 1/1000
- |W| = |U1|2 (Important! |W| is a func. of |U1|=const.)
Proof of main theorem
Induction assumption: 9 Un ½ Gn , (Gn,Un) · 1/1000
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)
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 Gn+1 (= Gn o Ad)
- (Gn+1, W) · 1/50 + 1/1000
- |W| = |U1|2
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 Gn+1 (= Gn o Ad)
- (Gn+1, W) · 1/50 + 1/1000
- |W| = |U1|2
Define Un+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
- 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| << 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) 2 U(d)
(u2-1, u1 –1 , u3-1 , L,ud-1) 2 U(d)
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 u1–1 , 1 , L , 1)
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
(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}
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
(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
Expansion on G £ G
X½G
Z½G£G
Z = {(x,x-1) | x 2 X} (completely correlated!)
Expansion on G £ G
X½G
Z½G£G
Z = {(x,x-1) | x 2 X} (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) | 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}
Expansion on G £ G
X½G
Z½G£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
Expansion on G £ G – decorrelating the gens.
BUT: If for every x 2 G, x = [ax,bx] = ax-1 bx-1 ax bx
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-1¢bx-1) | x 2 X} (+ inverses)
Z½G£G
Z = {(x,x-1) | x 2 Xc}
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-1¢bx-1) | x 2 X} (+ inverses)
Z½G£G
Z = {(x,x-1) | x 2 Xc}
(ax-1bx-1 , bxax) 2 Z
(ax
, ax-1) 2 Z
(bx
, bx-1) 2 Z
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-1¢bx-1) | x 2 X} (+ inverses)
Z½G£G
Z = {(x,x-1) | x 2 Xc}
(ax-1bx-1
(ax
(bx
+
(x
, bxax) 2 Z
, ax-1) 2 Z
, bx-1) 2 Z
, 1) is generated by Z
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-1¢bx-1) | x 2 X} (+ inverses)
Z½G£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-( /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 o 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 o Ad has (CP)
Proof: Reduce to the system of eqs
x1 yh(1) x(1)-1 y(1) = a1
x2 yh(2) x(2)-1 y(2) =a2
…
xd yh(d) x(d)-1 y(d) =ad
2d variables xi 2 G, yi 2 G, d constants ai 2 G.
3 arbitrary permutations h,, 2 Sd
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
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(log|G|) elements suffice [AR]
- log|G| =O(d¢logd) in our case)
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(log|G|) elements suffice [AR]
- log|G| =O(d¢logd) in our case)
- Abelian G needs (log |G|) generators
Expansion on alternating gp. Ad
Reasons to believe there are few expanding generators:
Expansion on alternating gp. Ad
Reasons to believe there are few expanding generators:
- 2 random elements generate Ad
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( log|Ad| ) [BL?]
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( log|Ad| ) [BKL]
- A conjecture (Aldous) implies O(d) transpositions
suffice for expansion (transpositions are weak!)
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( log|Ad| ) [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.
D &