dehnfunctionsslamy/teaching/m2... · 2020. 11. 28. · a2 = 1canchangea mintoa...

Puzzle model for a finitely presented groupDehn function

Isoperimetry in the Cayley complex

DEHN FUNCTIONS

DEHN FUNCTIONS 1 / 19



I. Puzzle model for a finitely presented group

Let G ' 〈S |R〉 be a finite groupe presentation. Then G ' FS/HR where HR isthe normal subgroup generated by R. For all ω ∈ FS , denote [ω] its classmodulo HR .

Definition

ω ∈ FS is reducible if [ω] is the class of the empty word i.e. ω represents 1G .

Claim : If ω is reducible we can shrink it to the empty word using relations.

Question : Given ω ∈ FS reducible of length n, how many steps are needed toshrink it ?

We will introduce a nice geometric model to address the question in the formof a puzzle kit consisting of edges and pieces.




For each s ∈ S , consider an oriented edge labelled by s. We can represent wordin FS with closed loops.

Let’s adopt the convention to read words counter-clockwise from the bottowleft vertex to avoid ambiguity.




For each r ∈ R, attach a face inside the loop representing r ∈ FS . We obtain apuzzle piece.

We now have a set of edges and a set of puzzle pieces (puzzle kit).




Cancelling pairs ss−1 are free puzzle pieces which come naturally and don’tdepend on R.

This is why we won’t consider cancelling a pair ss−1 to be a step in theshrinking of a reducible word.




Proposition

Let ω ∈ FS . ω is reducible iff the loop representing ω can be filled with puzzlepieces.

We say that the loop representing ω is a solvable puzzle.




Let ω be a reducible word. How can we shrink it to the empty word ?

• Cancelling a pair ss−1 ↔ removing a free piece (doesn’t cost anything)• Removing a relation r ∈ R ↔ removing a piece (costs 1)

The fastest reduction possible is the solution to the puzzle with the least pieces.




II. Dehn function

G ' 〈S |R〉. Pick n ∈ N.

Ln = {ω ∈ FS reducible | `(ω) ≤ n}

Definition

The Dehn function maps n to the smallest integer k such that all ω ∈ Ln canbe reduced in k steps or less.

d : n ∈ N 7→ min {k ∈ N | ∀ω ∈ Ln, ω can be reduced in k stepsor less}

Observation : We can define the Dehn function of a puzzle as well.

Ln = {solvable loops ω | `(ω) ≤ n}

d : n ∈ N 7→ min {k ∈ N | ∀ω ∈ Ln, ω can be filled using k piecesor less}

d is a difficulty measure for the puzzle. d depends on the choice of S and R.




Example : Z2 '〈a, b|aba−1b−1

〉.

This is the abelianization of F2. Relations are used to shuffle letters (changethe order). For instance :

ab −→︸︷︷︸free pieces

ab(a−1b−1ba) = (aba−1b−1)ba −→︸︷︷︸1 relation

ba

Let ω ∈ F2 reducible such that `(ω) ≤ n. Then

#(a) = #(a−1) and #(b) = #(b−1)

To reduce ω we just need to shuffle each letter at most n times before freelycancelling the pairs. At most n/2 letters need to be shuffled so ω can be reducedin less than n2/2 steps.

First estimation (brutal) : d(n) ≤ n2/2

Question : What does the puzzle tell us about d ?




Let ω be a solvable loop of length ≤ n. it can be filled with a collection ofsquares so we can delimit it in the smallest rectangle possible.

2(L + h) ≤ `(ω) ≤ n and the number of pieces is smaller than Lh. Afterstudying the function f : x 7→ x(n/2− x) on [0, n/2] we can see that Lh ≤ n2/16.So d(n) ≤ n2/16.

The word akbka−kb−k needs k2 steps to be reduced. So the upper bound n2/16is attained when n = 4k. Therefore the inequality d(n) ≤ n2/16 is optimal.




Another example : Z/2Z '〈a|a2

〉Let ω = am. `(ω) = m. ω is reducible iff m is even. Applying the relationa2 = 1 can change am into am±2. So the fastest way to shrink am to the emptyword needs |m|2 steps. One can check that d(n) = b

n2c.

Claim : For Z/dZ '〈a|ad

〉we have d(n) = b n

dc.




III. Isoperimetry in the Cayley complex

Let Γ(G ,S) be the Cayley graph of G ' 〈S |R〉. Then a word ω in FScorresponds bijectively to a path in Γ(G , S) from 1G to [ω].

ω is reducible ⇐⇒ [ω] = 1G ⇐⇒ the path is a loop

So there is a one-to-one correspondence between reducible words and loopsstarting at 1G in Γ(G , S).




Definition

The Cayley 2−complex of G ' 〈S |R〉 is obtained from Γ(G , S) after gluing aface inside each loop representing a relation. It is denoted Γ̃(G , S).

Let ω be an edge loop in Γ(G , S) starting at 1G . The area of ω is the smallestnumber of faces needed to solve the loop.




Isoperimetry problem : Pick n ∈ N. Ln is the set of all the loops ω in Γ(G , S)such that `(ω) ≤ n. Set a(n) = maxω∈Ln area(ω).

Proposition

d(n) = a(n)

∼∼∼

Example : S3 '〈τ, σ|τ2, σ2, (τσ)3

〉with τ = (12) and σ = (23).

• Step 1 : The Cayley graph Γ(S3, {τ, σ})




• Step 2 : The Cayley 2−complex

There are no loops of odd length. So if n is odd we have d(n) = d(n − 1).• n = 2 the biggest area is 1 (ττ is a loop of area 1)• n = 4 the biggest area is 2 (τσστ is a loop of area 2)• n = 6 the biggest area is 4 (τστσ−1τ−1σ−1 is a loop of area 4)

So d(2) = 1, d(4) = 2 and d(6) = 4.




What happens if we change the presentation ? Let us considerS3 '

〈τ, σ|τ2, σ3, τστσ

〉with τ = (12) and σ = (123).

• Step 1 : The Cayley graph




• Step 2 : The cayley 2−complex

• n = 2 the biggest area is 1 (ττ is a loop of area 1)• n = 4 the biggest area is 3 (στ−1στ−1 is a loop of area 3)• n = 5 the biggest area is 4 (στσ−1τσ is a loop of area 4)

We can already tell that the Dehn function is different : d(5) = 4 now whereasd(5) was equal to 2 in the previous example.




Proposition

Let G be a finite group. Given any finite presentation G ' 〈S |R〉 and theassociated Dehn function d, there is a positive constant λ such that d(n) ≤ λn.

Proof : Γ̃(G , S) has a finite number of faces, edges and vertices. Let us denoteλ the number of faces. Let ω ∈ FS ; it is represented by a loop in Γ̃(G , S). Ifthe loop is simple (ie. injective) then its area is at most λ.

Now assume the loop isn’t simple. Denote n its length. After freely cancellingthe pairs ss−1 the loop can be written as the concatenation of at most n simpleloops. Each simple loop has an area which is smaller than λ so the area of ω isless than λn.




Definition

Let us define the order relation � on the set of all functions N→ N by

f � g ⇐⇒ ∃C > 0 such that f (n) ≤ Cg(Cn + C) + Cn + C ∀n ∈ N

And the equivalence relation ∼ by

f ∼ g ⇐⇒ f � g and g � f

Theorem

The Dehn functions of two finite presentations for the same group areequivalent for the relation ∼.

Consider the two presentations Z ' 〈a|∅〉 and Z ' 〈a, b|b〉. Their respectiveDehn functions are n 7→ 0 and n 7→ n. So 0 ∼ n.

Little landscape : The maps n 7→ αn are all equivalent for α > 1. For instance,the Dehn function of the group presentation BS(1, 2) =

〈a, b|ab = ba2

〉is

equivalent to n 7→ 2n.


Puzzle model for a finitely presented groupDehn functionIsoperimetry in the Cayley complex

dehnfunctionsslamy/teaching/m2... · 2020. 11. 28. · a2 = 1canchangea mintoa...

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