Forest Diagramsfor

Thompson’s Group

Jim Belk

Associative LawsConsider the following piecewise-linear

homeomorphism of :

Associative LawsThis homeomorphism is called the basic

associative law.

It corresponds to the operation

Associative LawsHere’s a different associative law.

It corresponds to .

Associative Laws

General Definition:

A dyadic subdivision of is obtained by repeatedly cutting intervals in half:

Associative LawsAn associative law is a PL-homeomorphism that

maps linearly between the intervals of two dyadic subdivisions.

Associative Laws

Associative LawsThompson’s Group is the group of all associative

laws.

Associative LawsThompson’s Group is the group of all associative

laws.

If , then:

• Every slope of is a power of 2.

• Every breakpoint of has dyadic rational coordinates.

The converse also holds.

2

½

1

(¼,½)

(½,¾)

Properties of

is an infinite, torsion-free group.

Properties of

is an infinite, torsion-free group.

is finitely generated:

Properties of

is an infinite, torsion-free group.

is finitely generated.

is finitely presented:

2 relations

Properties of

is an infinite, torsion-free group.

is finitely generated.

is finitely presented.

admits a complex with exactly two cells in each dimension.

Properties of is simple. Every proper quotient of

is abelian.

has exponential growth.

contains .

does not contain .

• Is amenable?

Tree Diagrams

Tree DiagramsWe can represent an associative law by a

pair of binary trees:

This is called a tree diagram.

Tree DiagramsUnfortunately, the tree diagram for an

element of is not unique.

Tree DiagramsUnfortunately, the tree diagram for an

element of is not unique.

Tree DiagramsUnfortunately, the tree diagram for an

element of is not unique.

Tree DiagramsUnfortunately, the tree diagram for an

element of is not unique.

We can always cancel opposing pairs of carets.

This is called a reduction of the tree diagram.

Multiplying Tree Diagrams

Multiplying Tree Diagrams

Multiplying Tree Diagrams

Multiplying Tree Diagrams

GeneratorsHere are the tree diagrams for the

generators:

The Action on

The Action on If we conjugate by the homeomorphism:

we get an action of on .

A dyadic subdivision of is obtained by repeatedly cutting intervals in half:

is the group of PL-homeomorphisms of that map linearly between the intervals of two dyadic subdivisions

The Action on

The Action on

What’s the point?

The generators become simpler.

Here’s the new picture for :

The Action on

What’s the point?

The generators become simpler.

And here’s :

Forest Diagrams

Forest DiagramsWe can represent an element of using a

pair of binary forests:

Forest Diagrams

This is called a forest diagram.

Forest Diagrams

This is called a forest diagram.

Here are the forest diagrams for the generators:

The Action of

Left-multiplication by moves the top pointer of a forest diagram:

The Action of

Left-multiplication by moves the top pointer of a forest diagram:

The Action of

Left-multiplication by moves the top pointer of a forest diagram:

The Action of

Left-multiplication by moves the top pointer of a forest diagram:

The Action of

Left-multiplication by moves the top pointer of a forest diagram:

The Action of

Left-multiplication by moves the top pointer of a forest diagram:

Here are the forest diagrams for the generators:

The Action of

Left-multiplication by adds a new caret on the top:

The Action of

Left-multiplication by adds a new caret on the top:

The Action of

Left-multiplication by adds a new caret on the top:

The Action of

Left-multiplication by adds a new caret on the top:

The Action of

Sometimes the new caret appears opposite a caret on the bottom:

The Action of

Sometimes the new caret appears opposite a caret on the bottom:

The Action of

Sometimes the new caret appears opposite a caret on the bottom:

The Action of

So can delete bottom carets (and can

create bottom carets.)

Lengths

Finding Lengths

Problem. Given an , find the -length of .

Example. Find the length of:

Finding Lengths

Finding Lengths

1

Finding Lengths

1

Finding Lengths

2

Finding Lengths

21

Finding Lengths

211

Finding Lengths

211

Finding Lengths

2111

Finding Lengths

2121

Finding Lengths

2221

Finding Lengths

2221

(1 + 2 + 2 + 2) + 2 = 9

This element has length 9.

Example 2

Example. Find the length of:

Example 2

Example 2

1

Example 2

1 1

Example 2

1 1

Example 2

1 1 1

Example 2

1 1 1 1

Example 2

1 1 1 1

Example 2

1 1 1 2

Example 2

1 1 2 2

Example 2

1 2 2 2

Example 2

2 2 2 2

Example 2

2 2 2 2

(2 + 2 + 2 + 2) + 2 = 10

This element has length 10.

Example 3

Example. Find the length of:

Example 3

Example 3

1

Example 3

1 1

Example 3

1 1

Example 3

1 1 1

Example 3

1 1 1

Example 3

1 1 2

Example 3

1 1 2

Example 3

1 2 2

Example 3

1 2 2

Example 3

2 2 2

Example 3

2 2 21

Example 3

2 2 21

(1 + 2 + 2 + 2) + 4 = 11

This element has length 11.

Labeling. Label each space as follows.

The Length Formula

Labeling. Label each space as follows.

L: Left of the pointer, exterior.

The Length Formula

L

Labeling. Label each space as follows.

L: Left of the pointer, exterior.

N: Next to a caret on the left.

The Length Formula

L NN N

Labeling. Label each space as follows.

L: Left of the pointer, exterior.

N: Next to a caret on the left.

R: Right of the pointer, exterior.

The Length Formula

L N NNR

Labeling. Label each space as follows.

L: Left of the pointer, exterior.

N: Next to a caret on the left.

R: Right of the pointer, exterior.

I: Interior

The Length Formula

L N R NNI I I

The Length Formula

Theorem. The weight of a space is determined by its label pair.

L N R I

L 2 1 1 1

N 1 2 2 2

R 1 2 2 0

I 1 2 0 0

L: Left exterior.

N: Next to a caret.

R: Right exterior.

I: Interior

L N R I

L 2 1 1 1

N 1 2 2 2

R 1 2 2 0

I 1 2 0 0

L: Left exterior.

N: Next to a caret.

R: Right exterior.

I: Interior

L N R I

L 2 1 1 1

N 1 2 2 2

R 1 2 2 0

I 1 2 0 0

L: Left exterior.

N: Next to a caret.

R: Right exterior.

I: Interior

L

L N R I

L 2 1 1 1

N 1 2 2 2

R 1 2 2 0

I 1 2 0 0

L: Left exterior.

N: Next to a caret.

R: Right exterior.

I: Interior

NL N

N

L N R I

L 2 1 1 1

N 1 2 2 2

R 1 2 2 0

I 1 2 0 0

L: Left exterior.

N: Next to a caret.

R: Right exterior.

I: Interior

LR

NN NR RR R R R

L N R I

L 2 1 1 1

N 1 2 2 2

R 1 2 2 0

I 1 2 0 0

L: Left exterior.

N: Next to a caret.

R: Right exterior.

I: Interior

L NN N II

IR R R

R R R R

L N R I

L 2 1 1 1

N 1 2 2 2

R 1 2 2 0

I 1 2 0 0

L: Left exterior.

N: Next to a caret.

R: Right exterior.

I: Interior

L NN N

1

R R RR R R RI

I I

2 0 2 0 2 0

L NN NR R RR R R RI

I I

1 2 0 2 0 2 0

length weights + # of carets 7 + 4

11

Convexity

ConvexityA group is convex if is convex for each

.

id

Convexity is almost convex if any two elements in

a distance two apart are connected by a path of length .

id

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

1 1 2 2 2 2 2 2 0

Length 16

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

1 2 2 2 2 2 2 2 0

Length 17

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

0 2 2 2 2 2 2 2 0

Length 16

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

1 2 2 2 2 2 2 2 0

Length 17

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

1 1 2 2 2 2 2 2 0

Length 16

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

ConvexityTheorem (Cleary and Taback). is not

almost convex (using ).

This gives us two elements of a distance two apart that have distance in .

Convexity

Theorem (Belk and Bux). For even, there exist elements such that:

, and

• The shortest path in from to has length .

Kai-Uwe and I have proven the following:

Amenability

The Isoperimetric Constant

Let be the Cayley graph of a group .

If is a finite subset of ,

its boundary consists of

all edges between and

.

The Isoperimetric Constant

Let be the Cayley graph of a group .

The isoperimetric constant is:

is amenable if .

Theorem (Belk and Brown). .Proof: Let be all elements of the form:

We claim that:

Given a random , we must compute:

exits the current tree of is trivial.

Claim. Given a random element of :

current tree is trivial

as .

Theorem. As , the probability that the current

tree is trivial satisfies:

where is the number of binary trees with

leaves.

Theorem. As , the probability that the current

tree is trivial satisfies:

where is the number of binary trees with

leaves.

The coefficients are the Catalan numbers, and have growth rate .

The polynomial above “converges” to the generating function for the Catalan numbers, which has a vertical asymptote at .

The End