Forest Diagrams for Thompsons Group Jim Belk Associative
Forest Diagrams for Thompson’s Group Jim Belk
Associative Laws Consider the following piecewise-linear homeomorphism of :
Associative Laws This homeomorphism is called the basic associative law. It corresponds to the operation
Associative Laws Here’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 Laws An associative law is a PL-homeomorphism that maps linearly between the intervals of two dyadic subdivisions.
Associative Laws
Associative Laws Thompson’s Group is the group of all associative laws.
Associative Laws Thompson’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. ½ 1 2 (½, ¾) (¼, ½)
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 Diagrams We can represent an associative law by a pair of binary trees: This is called a tree diagram.
Tree Diagrams Unfortunately, the tree diagram for an element of is not unique.
Tree Diagrams Unfortunately, the tree diagram for an element of is not unique.
Tree Diagrams Unfortunately, the tree diagram for an element of is not unique.
Tree Diagrams Unfortunately, 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
Generators Here 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 .
The Action 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 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 Diagrams We 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 1 2
Finding Lengths 1 1 2
Finding Lengths 1 1 2
Finding Lengths 1 1 1 2
Finding Lengths 1 2
Finding Lengths 1 2 2 2
Finding Lengths 1 2 2 2 (1 + 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
Example 2 1 1
Example 2 1 1 1 2
Example 2 1 1 2 2
Example 2 1 2 2 2
Example 2 2 2
Example 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 1 2 2 2
Example 3 1 2 2 2 (1 + 2 + 2) + 4 = 11 This element has length 11.
The Length Formula Labeling. Label each space as follows.
The Length Formula Labeling. Label each space as follows. L: Left of the pointer, exterior. L
The Length Formula Labeling. Label each space as follows. L: Left of the pointer, exterior. N: Next to a caret on the left. L N N N
The Length Formula 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. L N R N N
The Length Formula 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 I L N I R N N I
The Length Formula Theorem. The weight of a space is determined by its label pair. L N R L 2 1 1 N 1 2 2 R 1 2 2 I 1 2 0 0 L: N: R: I: Left exterior. Next to a caret. Right exterior. Interior
L N R L 2 1 1 N 1 2 2 R 1 2 2 I 1 2 0 0 L: N: R: I: Left exterior. Next to a caret. Right exterior. Interior
L L N R L 2 1 1 N 1 2 2 R 1 2 2 I 1 2 0 0 L: N: R: I: Left exterior. Next to a caret. Right exterior. Interior
N N L N R L 2 1 1 N 1 2 2 R 1 2 2 I 1 2 0 0 L: N: R: I: Left exterior. Next to a caret. Right exterior. Interior
R R R N N L N R R L N R L 2 1 1 N 1 2 2 R 1 2 2 I 1 2 0 0 L: N: R: I: Left exterior. Next to a caret. Right exterior. Interior
R R R N I L N I R R L N R L 2 1 1 N 1 2 2 R 1 2 2 I 1 2 0 0 L: N: R: I: Left exterior. Next to a caret. Right exterior. Interior
1 2 0 2 0 R R R N I L N I R R L N R L 2 1 1 N 1 2 2 R 1 2 2 I 1 2 0 0 L: N: R: I: Left exterior. Next to a caret. Right exterior. Interior
1 2 0 2 0 R R R N I L N I R R length weights + # of carets 7+4 11
Convexity
Convexity A 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
Convexity Theorem (Cleary and Taback). is not almost convex (using ).
Convexity Theorem (Cleary and Taback). is not almost convex (using ). 1 1 2 2 2 0 Length 16
Convexity Theorem (Cleary and Taback). is not almost convex (using ). 1 2 2 2 2 0 Length 17
Convexity Theorem (Cleary and Taback). is not almost convex (using ). 0 2 2 2 2 0 Length 16
Convexity Theorem (Cleary and Taback). is not almost convex (using ). 1 2 2 2 2 0 Length 17
Convexity Theorem (Cleary and Taback). is not almost convex (using ). 1 1 2 2 2 0 Length 16
Convexity Theorem (Cleary and Taback). is not almost convex (using ).
Convexity Theorem (Cleary and Taback). is not almost convex (using ). This gives us two elements of a distance two apart that have distance in .
Convexity Kai-Uwe and I have proven the following: Theorem (Belk and Bux). For even, there exist elements such that: • , and • The shortest path in from to has length .
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 The coefficients the Catalan numbers, and aresatisfies: current tree is trivial have growth rate . The polynomial above “converges” to the where is the number of binary trees with generating function for the Catalan numbers, leaves. which has a vertical asymptote at .
The End
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