MATH 7320 Seminar in Algebra: The Geometry of Solvable Groups · Tim Riley The Geometry of Solvable Groups 1 Part I Tim’s Lectures 1 Jan 22, 2013 Two perspectives on studying solvable

MATH 7320Seminar in Algebra:The Geometry of Solvable Groups

lecturer: Tim Rileyby: Amin Saied

Spring 2013

Contents

I Tim’s Lectures 1

1 Jan 22, 2013 11.1 Basic Defintions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Jan 31, 2013 32.1 Examples of Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Feb 05, 2013 63.1 Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 Feb 7, 2013 94.1 Idea behind Gromov’s proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4.1.1 Background: σ-algebras and Probability Measures . . . . . . . . . . . . . . . . . . . 104.1.2 Ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.1.3 Construction of Y . (van den Dries-Wilke) . . . . . . . . . . . . . . . . . . . . . . . . 12

5 Feb 12, 2013 125.1 Properties of Asymptotic Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

6 Feb 14, 2013 146.1 Filling Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

7 Feb 19, 2013 167.1 Combings of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

8 Feb 21, 2013 198.1 Quasi-Isometry Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

9 Feb 26, 2013 229.1 The Area Function of Nilpotent Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

10 Feb 29, 2013 2410.1 Lower Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

11 Mar 5, 2013 2611.1 Some Open Problems Relating to Dehn Functions . . . . . . . . . . . . . . . . . . . . . . . 2611.2 HNN-Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2711.3 Baumslag-Solitar Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

12 Mar 7, 2013 2912.1 The Geometry of Baumslag-Solitar Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

12.1.1 Normal Form for BS(1, n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2912.1.2 Dehn Function of BS(1,m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3112.1.3 Cayley Graph of BS(1, 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3212.1.4 Treebolic Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

II Student Talks 36

13 Geometry of Sol (Chenxi Wu) 3613.1 Dehn Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3613.2 Quasi-Rigidity of Sol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3613.3 Area function becomes quadratic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Tim Riley The Geometry of Solvable Groups 3

14 Geometry of the Lamplighter Group (Margarita Amchislavska) 3714.1 Dead-end Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3714.2 Horocyclic Product of Two Z-branching Trees . . . . . . . . . . . . . . . . . . . . . . . . . . 37

15 Generalisations of the Lamplighter Group (Margarita Amchislavska) 3915.1 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3915.2 Some Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3915.3 Horocyclic Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

16 Geometry of the Magnus Embedding (Andrew Sale) 4116.1 Magnus Embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4116.2 Applications of the Magnus Embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4116.3 Magnus Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4116.4 Fox Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4116.5 Wreath Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

16.5.1 Geometric Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4216.6 Geometric Definition of the Magnus Embedding . . . . . . . . . . . . . . . . . . . . . . . . . 43

17 Finitely Presented Metabelian Groups (Amin Saied) 4517.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4517.2 Baumslag/Margarita’s Example: Embedding Γ1 = Z o Z into Γ2 . . . . . . . . . . . . . . . . 4617.3 Proof of Baumslag-Remeslennikov Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

18 Sigma Invarients (Teddy Einstein) 5218.1 Σ-invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5218.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

19 Sigma Invarients (Yash Lodha) 5619.1 Finiteness Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

20 Automatic Groups (Scott Messick) 5920.1 Closure Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6120.2 Biautomatic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6220.3 Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

21 Quasi-Isometric Rigidity (Iian Smythe) 63

22 Quasi Isometric Rigidity (Kristen Pueschel) 6622.1 Proof Form for QI Rigidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6622.2 Xn and its Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

23 Random Walks on Solvable Groups (Johannes Cuno) 6823.1 Random Walks on Graphs and Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6823.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6823.3 Flows and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6923.4 Growth of Groups and Return Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7023.5 Advertisement for Next Talks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

24 Boundaries of Random Walks (Mathav Murugan) 7124.1 Martin Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7124.2 Poisson Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

25 Rate of Escape of Random Walks on Groups (Tianyi Zheng) 74


Part I

Tim’s Lectures

1 Jan 22, 2013

Two perspectives on studying solvable groups

1. Classical perspective; of Phillip Hall and Gilbert Bammslang-Algebraic-Representations-Commutative Algebra-Cohomological methods-Bieri-Strebel Invarients

2. Geometric perspective;-Growth-Isoperimetry-Rigidity, quasi-isometry-Lattices-Random walks

The aim of this course is to try and create a bridge between these two perspectives.

1.1 Basic Defintions

Definition A group G is solvable if it has a finite abelian series - that is,

G = Gn D Gn−1 D · · · D G0 = 1 (1)

such that the factor groups are all abelian. The smallest such n is called the derived length of G.

1. Derived length 0: trivial groups

2. Derived length 1: abelain groups

3. Derived length ≥ 2: metabelian

e.g. G = G2 D G1 D 1Get the short exact sequence

G1 → G→ G/G1

where G1 and G/G1 are abelian.

Theorem. Let G be a solvable group. Then

1. Every subgroup H ≤ G is solvable.

2. Every quotient G/N is solvable.

In addition, if N CG and both N and G/N are solvable then G is solvable.

Proof.

1. Solvable series for G,G = G0 D G1 D · · · D Gn = 1

The second isomorphism theorem says that if A and B are subgroups of G with A C G then A ∩ Bis normal in B and B/(A ∩ B) ∼= AB/A. Applying this we see that, since Gi+1 E Gi (and thus alsoGi+1 E H ∩Gi), then Gi+1 E (H ∩Gi) ∩Gi+1


�

The class of solvable groups is closed under taking extensions, subgroups, homomorphic images. (Exersize:show this for subgroups and homomorphic images: for subgroups just take intersection of the whole series,for homomorphic images, just hit the whole series with the homomorphism f). For extensions consider,

1→ K → G→ H → 1

K and H solvable. Now H ∼= G/K. Now H is solvable, so take it’s abelain series. Lift this to one for G,and append it with the one for K. (also an exercise).

Definition The derived subgroup (or commutator subgroup) of a group G is G′ = [G,G]. Thederived series of G is

G = G(0) ≥ G(1) ≥ · · ·

where G(i+1) = (G(i))′. Clearly G(n)/G(n+1) is abelian (this is the abelianisation!). So if this terminatesthen we have a finite abelian series for G, and this G is solvable.

Claim If we have a finite abelian series (with the notation in 1) then G(i) ≤ Gn−iPf: By induction. For i = 0 it is immediate. And G(i+1) = (G(i))′ ≤ (Gn−i)

′ ≤ Gn−(i+1) where the last twoinequalities come from the induction hypothesis and because the last group has abelian factors (resp.)

Corollary The derived length of a group is the length of the derived series.

Definition A group G is nilpotent if it has a (finite) central series, that is

G = Gn D Gn−1 D · · · D G0 = 1

such that the factor groups Gi+1/Gi is contained in the center of G/Gi (for all i).Note// Since the center of a group is abelian (!) we see that nilpotent groups are examples of solvablegroups.We call the minimal length n of the central series the class of G

1. Class 0: trivial groups

2. Class 1: abelain groups

Definition The upper central series of a group G is

1 = Z0(G) ≤ Z1(G) ≤ Z2(G) ≤ · · ·

defined by Zn+1(G)/Zn(G) = Z(G/Zn(G)). (In particular, since Z0(G) = 1, we see that Z1(G) = Z(G), isthe center) (Also note that this is clearly a central series, by definition)

The lower central series of a group G: γ1(G) = G, then

γn+1(G) = [γn(G), G]

Notice that γn(G)/γn+1(G) ≤ Z(G/γn+1(G)). (Exersize: check that this really is a central series) andγn(G) E G.

Theses series are related in the following way:

1. γi(G) ≤ Gn−i+1

2. Gi ≤ Zi(G)

These can be proved by induction on i


Zn(G) ≥ Zn−1 ≥ · · · ≥ Z0(G) = 1‖ ∪ ‖

G = Gn D Gn−1 D · · · D G0 = 1‖ ∪ ‖

γ1(G) ≥ γ2(G) ≥ · · · ≥ γn+1(G)

In this way we see that the lengths of these series agree if the central series was of minimal length.

Definition A group G is polycyclic if it has a finite normal series with cyclic factors.

Definition The max condition on subgroups if a group G is any one of the following equivalent condi-tions:

1. Every subgroup is finitely generated (So F2 doesn’t satisfy the max condition)

2. Every ascending chain of subgroupsH1 ≤ H2 ≤ H3 ≤ ·

stabilises i.e. there exists N such that HN = HN+1 = · · · (Noetherian condition)

3. Every non-empty set of subgroups has a maximal element (a subgroup that is not a proper subgroupof any other one of these subgroups)

We say “G is max”.

Properties of Max Notice that cyclic groups are max. Max is inherited when taking extensions (if H,K are Max,and K → G→ H SES, then G is Max, exersize). So polycyclic groups are Max.

Theorem Polycyclic groups are precisely the solvable groups that are Max.

Pf: If G is solvable and Max, then its subgroups are finitely generated, and so the factors in the derivedseries are finitely generated abelian groups, and so the derived series can be refined to a cyclic series

2 Jan 31, 2013

The idea today is to fill in this map of different classes of groups with some examples.


Theorem (Hirsch) Finitely generated nilpotent groups are polycyclic.

Pf Recall γ1(G) = G, γn+1(G) = [G, γn(G)]. Let γc(G) be the last non-trivial term in the lower centralseries of a finitely generated nilpotent group G. By induction on c, we may assume that G/γc(G) is poly-cyclic (because by quotienting the lower central series for of G by γc(G) gives a central series for G/γc(G)with one fewer term, so by induction, polycyclic).

Suppose G = 〈x1, . . . , xm〉. As G/γc(G) is polycyclic, γc−1/γc(G) is finitely generated (by max condition).Write γc−1(G) = 〈y1, . . . , yn〉γc(G). Then

γc(G) = [〈x1, . . . , xm〉, 〈y1, . . . , yn〉] = 〈[xi, yj ]|1 ≤ i ≤ m, 1 ≤ j ≤ n〉

and so is finitely generated. But γc(G) is abelian, and so is polycyclic. So as G/γc(G) and γc(G) arepolycyclic, and hence G is polycyclic too. �

2.1 Examples of Solvable Groups

1. Abelian group: Zk the free abelian group

2. R a commutative ring with 1, Un(R) = (n × n) upper triangular matrices over R, with 1’s on thediagonal.

1 ∗1

. . .

0 1


These is nilpotent of class n− 1. Indeed,

[eij , ekl] =

{1 j 6= keil j = k

eij = In + Eij (The idea is that terms get pushed upwards and away from the diagonal)

In particular, U3(Z) is the thee-dimensional integral Hiesenburg group, sometimes denoted

H3(Z) = 〈a, b, c|[a, b] = c, [a, c] = [b, c] = 1〉

Notice that this is metabelian, as the map which kills c takesH3(Z)→ Z×Z. It is of course not abelian.

Theorem. (P.Hall, 1969) Every finitely generated torsion free nilpotent group embeds in Un(Z)for some n.

3. Baumslag-Solitar BS(1,m) = 〈a, b|b−1ab = am〉. b generates Z and a is the unit in Z[ 1m ]. We have

the presentation,Z n Z[ 1

m ]

where a generator for the Z-factors acts on Z[ 1m ] by multiplication by m. This is metabelian by

construction.But this is not a polycyclic group: indeed Z[ 1

m ] is not finitely generated, and is clearly (isomorphicto) a subgroup. This contradicts the max condition.

4. Sol = Z2 oA Z, where A =

(2 11 1

). This is metabelian and polycyclic by construction.

5. LamplightersBefore defining lamplighters we require the more general definition of a wreath product. For groupsA,B denote A(B) := {finitely supported functions B → A} =

⊕B A

Definition. The restricted wreath product A wr B = A oB = A(B) oB with the action

(f, b)(g, c) := (fgb, bc)

where gb(x) = g(b−1x).

Definition. A lamplighter is a restricted wreath product of the form

Z/mZ wr Zk

For example Z wr Z or Z/2Z wr Z. Lets look at Z/2Z wr Z more closely to see how the namelamplighter arises.

Example: A = Z/2Z wr ZAn element in A looks like (f, b) where f is a finitely supported function from Z/2Z to Z. If we thinkof Z lamps arranged on a street, then f is a finite set of these lamps which are turned on. We couldthen think of b as being someone at a given position on the street, say lighting the lamps. So

(f, b) = · · ·◦ ◦ • ◦ • • ◦

?−1 0 1 2 3 4 5

· · ·

means that f lights {1, 3, 4}, and b = 1 is the position of the lamplighter. So lets consider the groupproduct. Consider the element

(g, c) = · · ·◦ • • • ◦ ◦ ◦

?−1 0 1 2 3 4 5

· · ·


where g lights {0, 1, 2} and the lamplighter is at position c = 3. The definition of the group product ina wreath product uses gb(x) = g(b−1x). NB: that we are using multiplicative notation, but the groupoperation in Z is addition. So gb(0) = g(−1) = 0, gb(1) = g(0) = 1 etc., thus we get

gb ◦ ◦ • • • ◦ ◦f ◦ ◦ • ◦ • • ◦fgb ◦ ◦ ◦ • ◦ • ◦

−1 0 1 2 3 4 5

Now we have (f, b)(g, c) = (fgb, bc). fgb is just addition (mod 2) of f and gb, and bc is just regularaddition in Z of b and c, hence

(f, b) · (g, c) = (fgb, bc) = · · ·◦ ◦ ◦ • ◦ • ◦

?−1 0 1 2 3 4 5

· · ·

To summarize this, we think of gb as a translating g by b, then fgb is just addition mod 2 of theselights. The lamplighter just moves to the sum of other lamplighter positions.

Finite generation of Z/2Z wr Z:f : Z→ Z/2Z by 0 7→ 1, r 7→ 0 for all r 6= 0 Now (f, 0) and (0, 1) generate Z/2Z wr Z.

General result: A wr B is finitely generated if both A and B are.

For Z wr Z =⊕ZoZ is not polycyclic because the first factor is not finitely generated.

6. Fr a free group of rank r. Then Fr/F(d)r is a solvable group, where F

(d)r is the dth term in the derived

series(check?).

3 Feb 05, 2013

3.1 Growth

Definition. G a finitely generated group with G ≡ 〈A〉 where |A| <∞. The word metric on G wrt A isd(g, h) is the length of the shortest word on A∪A−1 representing g−1h. Minimal length words representinga given group element are called geodesic words.

Example The Lamplighter group Z2 wr Z2

Below is an example of a group element. Here o denotes the origin (0, 0), • denotes a lit light, and ◦ anon-lit light. ? is the position of the lamplighter.

◦ ◦ ◦ •◦ • ◦ ?◦ o ◦ ◦◦ ◦ ◦ ◦

A generating set here is A = {a, x, y} where x and y move you left and up, respectively, and a lights thelamp in that position. So the element above would be given by

yayxxax−1

The problem of identifying a geodesic representing a given group element here is NP-hard since it amountsto travelling salesman problem on a grip.

Definition. Growth function f is

f(n) : = #(A ∪A−1 ∪ {1})n

= #{g ∈ G : d(1, g) ≤ n}


Examples

1. Z wrt 〈1〉 then f(n) = 2n+ 1−n · · · · · · · · ·︸︷︷︸

n

0 · · · · · · · · ·n︸︷︷︸n︸︷︷︸

2n+1

2. Z2 wrt 〈(1, 0), (0, 1)〉 then f(n) ≈ n2

3. F2 wrt 〈a, b〉 notice f(0) = 1, f(1) = 5, f(2) = 17 in general f(n) = 2 · 3n − 1

The above example featured exponential growth. In fact this is the fastest that any group can grow:

Proposition. For all groups f(n) ≤ (2|A|+ 1)n

4. Z2wrZ◦ · · · ◦ ◦ ◦ · · · ◦−n · · · −1 0 1 · · · n︸︷︷︸

4n

Words of length ≤ 6n represent all group elements where the lamplighter is at the origin and theilluminated lights are in {−n, · · · , n}. That is 22n+1 group elements. So

f(6n) ≥ 22n+1

f(n) ≥ 2n


5. BS(1, 2) = Z[12 ]o×2 Z

We have 2m ↔ b−mabm

ε01 + ε121 + · · ·+ εk2

k where εi ∈ {0, 1} ↔∏ki=0(b

−iabi)εi and this has length ∼ k, so

f(n) ∼ 2n

6. Sol = Z2 oA Z where A =

(2 11 1

)Exponential growth

(2 11 1

)kpq

0 0 1

7. The Heisenburg Group H3 =

1 Z Z

1 1 Z0 0 1

with generators

a =

1 1 01 1 00 0 1

b =

1 0 01 1 10 0 1

We have am =

1 m 01 1 00 0 1

and bn =

1 0 01 1 n0 0 1

and [am, bn] =

1 0 mn1 1 00 0 1

= cmn where

c = [a, b] =

1 0 11 1 00 0 1

So if w is a word of length at most k. It corresponds to a matrix of the form 1 p r

1 1 q0 0 1

where |p|, |q| ≤ k and |r| ≤ k2. Therefore

f(n) ≈ n4

Remark: If w is a word on {a, b, c}±1 then there exists a word w representing the same group elementof the form apbqcr where if l(w) = n then |p|, |q| ≤ n and |r| ≤ n2. This follows from the (nice) normalform of elements in H3 coming form the fact that

[a, b] = c, [a, c] = 1, [b, c] = 1

This says that c is central, and that that we can swap a’s and b’s at the expense of adding a c.Therefore we can bring our a’s to the front, and get some c’s floating about, but because c is central,we can just send them all to the back.

Theorem (Dixmier-Wolf-Bass-Guivarch)For a finitely generated nilpotent group with lower central series

G = γ1(G) ≥ γ2(G) ≥ · · · ≥ γn+1(G) = 1

the growth f(n) ≈ nd where

d =∑

k torsion-free rank(γk(G)/γk+1(G))


Example H3

H3 ≥ Z ≥ 1

So d = 1 · 2 + 2 · 1 = 4, in agreement with example 7 above.

Some Facts About GrowthFor f, g : N → N write f 4 g when there exists c > 1 such that f(n) ≤ cg(cn + c) + cn + c. Write f ' gwhen f 4 g and g 4 f . For example, nα ' nβ for α, β ≥ 1 iff α = β, but αn ' βn for all α, β > 1.

If A and B are finitely generated sets for a group G then fA ' fB. In fact the functions are Lipschitzequivalent i.e. there exists a c > 0 such that

fA(n/c) ≤ fB(n) ≤ fA(cn)

Lemma. If H is a finite index subgroup of a finitely generated group G then (H is also finitely generated(!) and) their growth functions are equivalent.

Definition. A map φ : X → Y between metric space is a (λ, µ)-quasi-isometry (with λ ≥ 1, µ ≥ 0)when

1

λd(a, b)− µ ≤ d(φ(a), φ(b)) ≤ λd(a, b) + µ

and for all y ∈ Y there exists x ∈ X such that

d(φ(x), y) ≤ µ

Examples

1. G = 〈A〉 = 〈B〉 with |A|, |B| ≤ ∞. Then Id : (G, dA)→ (G, dB) is a quasi-isometry

2. H ↪→ G is a quasi-isometry

Lemma. Quasi-isometry is an equivalence relation. (Exersize)

Lemma. If G1 and G2 are finitely generated quasi-isometric groups (i.e. there exists a quasi-isometrybetween them) then their respective growth functions are equivalent

fG1 ' fG2

Note. Proving this lemma in combination with example 2 above is sufficient to prove the lemma above.

4 Feb 7, 2013

Recall,

Lemma. If G and H are finitely generated quasi-isometric groups (i.e. there exists a quasi-isometrybetween them) then their respective growth functions are equivalent

fG ' fH

Pf. φ : G→ H a (λ, µ)-quasi-isometry. Consider a ball of radius n about e ∈ G. The image of this ball inH under φ has radius λn+ µ (since φ is a quasi-isometry).

There exists some c > 0 such that if x, y ∈ G have d(x, y) > c then φ(x) 6= φ(y). Now take a maxi-mal subset of G such that no two points are ≤ c apart. Call this subset a mesh, M . Now |M | ∼ |B(n, e)|.But |M | ≤ |B(n, e)| ≤ |B(c, e)||M |. Define |B(c, e)| = K, some constant. We have |φ(M)| = |M |, byconstruction of M . But |M | ≤ fH(λn+ µ) Putting this together gives,

fG(n) = |B(n, e)| ≤ K|M | ≤ KfH(λn+ µ)

Now since quasi-isometry is an equivalence relation there is an inverse quasi-isometry. Doing the same asabove will give the other direction. �.


Definition. A group is virtually nilpotent when it has a finite index subgroup which is nilpotent.

Theorem. (Gromov) A finitely generated group G has polynomial growth iff G is virtually nilpotent.

Corollary. Being virtually nilpotent is a geometric property among finitely generated groups (that is,if G is quasi-isometric to a virtually nilpotent group H, then G is virtually nilpotent).

4.1 Idea behind Gromov’s proof

Theorem. (Milnor-Wolf) A finitely generated solvable group is of polynomial growth iff it is virtuallynilpotent. It is of exponential growth otherwise.

Sketch of Proof of Gromov’s Theorem.Suppose G is an infinite finitely generated group of polynomial growth. Let d be a word metric on G withrespect to a finite generating set. Let

Y := limn→∞

(G, 1nd)

(This is a clever idea: Think of Zn lattice. The idea of Y is that as we move further away from this it startsto look like Rn)Properties of Y :

1. Homogeneous

2. Connected, locally connected

3. Complete

4. Locally compact, finite dimensional (topological dimension)

Gleason-Montgomery-Zippin’s Solution to Hilbert’s 5th Problem It can be shown that Isom(Y )is a Lie group. Gromov uses this to build a homomorphism from a finite index subgroup of G onto Z. (Thisis hard)

The kernel of this homomorphism is of polynomial growth of lower degree. Induction will allow us todeduce that G has a solvable subgroup of finite index. Then Milnor-Wolf finishes the proof. �

4.1.1 Background: σ-algebras and Probability Measures

Definition. Let X be a set, and P(X) = 2X its power set. Then a subset Σ ⊆ 2X is a σ-algebra if itsatisfies:

1. Σ is non-empty

2. Σ is closed under compliment

3. Σ is closed under countable unions

It thus follows by de Morgans’s laws that it is also closed under finite intersection.

Proposition. X and ∅ are both contained in Σ for any σ-algebra (of X).

Proof. By 1. Σ is non-empty so there exists A ⊂ X in Σ. By 2. it is closed under taking compliments, soX\A ∈ Σ. By 3. it is closed under countable unions, so A ∪X\A = X ∈ Σ. And finally it is closed undercompliment, so X\X = ∅ ∈ Σ. �

Proposition. Given a finite collection of subsets of X, S = {Xi}, there exists a unique smallest σ-algebracontaining S.


Proof. Indeed, consider

Φ := {E ⊆ 2X : E is a σ-algebra of X containing S}

First note that Φ is non-empty, as 2X ∈ Φ. Now take the intersection of every element in Φ. Since eachelement contains S, so too will the intersection. Moreover, since Φ consists of σ-algebras, and since they areclosed under intersections, then this intersection will itself be a σ-algebra, and thus is the minimal elementin Φ. �

Definition. Denote this intersection σ(S) and call it the σ-algebra generated by S.

Example. Let X = {1, 2, 3}. The σ-algebra generated by {1} is σ({1}) = {∅, {1}, {2, 3}, {1, 2, 3}}

Definition. Let X be a set and Σ a σ-algebra over X. A function µ : Σ→ R∪{±∞} is called a measureif it satisfies:

1. Non-negativity: µ(E) ≥ 0 for all E ∈ Σ

2. Null empty set: µ(∅) = 0

3. Countably additive: for all countable collections {Ei} of pairwise disjoint sets in Σ

µ

(⋃i

Ei

)=∑i

(µ(Ei))

Definition. A probability measure is a measure with total measure 1, that is, µ(X) = 1.

4.1.2 Ultrafilters

Definition. A non-principle ultrafilter ω : P(N)→ {0, 1} is a finitely additive probability measure onN taking values in {0, 1} and giving all singleton sets measure 0.

Remark. These exists by Zorn’s lemma: Consider the set of functions f : P(N) → {0, 1} such thatf−1(1) is closed under taking finite intersections and supersets, and doesn’t include the empty set, but doesinclude all cofinite sets. Example: f−1(1) = the cofinite sets. Now I want to take a maximal element usingZorn’s lemma, under the relation f ≤ f when f−1(1) ⊆ f−1(1). This maximal element is an example of anon-principle ultrafilter.

Definition. Let ω : P(N) → {0, 1} be a non-principle ultrafilter. Then a ∈ R is an ω-ultralimit of asequence (an) ∈ R when

∀ ε > 0 ω ({i : |a− ai| < ε}) = 1

and say ∞ is an ω-ultralimit when

ω{i : ai > N} = 1 ∀ N > 0

and say −∞ is an ω-ultralimit when

ω{i : ai < −N} = 1 ∀ N > 0

Lemma. Every sequence (an) in R has a unique ω-ultralimit in R ∪ {±∞}. Call this limω an.

Sketch. Assume (an) is bounded, then within the bounded set of R (so an interval) there is a sequence ofpoints. Cut this interval in half. One interval will have measure 1, the other measure 0. On the half withmeasure 1 repeat this process.


4.1.3 Construction of Y . (van den Dries-Wilke)

Definition. (X, d) a metric space, ω a non-principle ultrafilter, e = (en) a sequence of basepoints in X,s = (sn) a sequnce in R>0 with sn →∞ as n→∞. Then an asymptotic cone is

Coneω(X, e, s) :=

{(xn) ∈ X : lim

ω

d(en, xn)

sn<∞

}/ ∼

where (xn) ∼ (yn) iff limωd(xn,yn)

sn= 0. The metric on Coneω(X, e, s) is

d((xn), (yn)) := limω

d(xn, yn)

sn

(hence quotienting out by ∼, as this identifies sequences which would be distance 0 under the metric above).This is limi→a(X,

dsi

), the ‘limit’ of these metric spaces.

5 Feb 12, 2013

5.1 Properties of Asymptotic Cones

Proposition. Given ω, e in X, s, then a quasi-isometry φ : X → Y induces a bi-Lipshitz homeomorphism

Coneω(X, e, s)→ Consω(Y, φ(e), s)

where φ(e) = (φ(ei)).

Corollary. Topological invarients of cones give quasi-isometry invarients of spaces e.g. finitely generatedgroups. (So for instance, cones being contractable, say, or simply connected, is a quasi-isometry invariant)

Proof of Prop. Suppose (ai) and (bi) represents points in Coneω(X, e, s). Then

1

λd((ai), (bi)) ≤ d(φ(ai), φ(bi)) ≤ λd((ai), (bi))

(which is bi-Lipshitcz). Lets try and justify this:

d(φ(ai), φ(bi)) = limω

d(φ(ai), φ(bi))

si

≤ limω

λd(ai, bi) + µ

si

= λ limω

d(ai, bi)

si+ lim

ω

µ

si= λd((ai), (bi))

since limωµsi

= 0 as µ is finite and si →∞. The other inequality is similar.

Suppose (yi) representa a point in Coneω(Y, φ(e), s). As φ is a quasi-isometry (and in particular we areincluding quadi-onto in that definition), for all i there exists xi ∈ X such that

d(φ(xi), yi) ≤ µ

We claim that

1. (xi) represents a point in Coneω(X, e, x)

2. (φ(xi)) = (yi)


for (2.)

d((φ(xi)), (yi)) = limω

d(φ(xi), yi)

si

≤ limω

µ

si= 0

For (1.)

d((ei), (xi)) = limω

d(ei, xi)

si

≤ limω

λd(φ(xi), φ(ei)) + λµ

si

≤ limω

λd(φ(xi), yi) + λd(yi, φ(ei)) + λµ

si

≤ limω

λµ+ λd(yi, φ(ei)) + λµ

si= λd((yi), (φ(ei))) <∞

�

Proposition. For a finitely generated group Γ, the asymptotic cone,

C := Coneω(Γ, e, s)

is:

1. homogeneous

2. geodesic

3. complete

Proof.

1. Suppose (gi), (hi) represents points in C. Then

(hig−1i ) : C → C

(ai) 7→ (hig−1i ai)

is an isometry taking (gi) to (hi).

2. If Γ = 〈A〉 with |A| <∞ then Γ→ CayA(Γ) (its Cayley graph) is a (1, 12)-quasi-isometry (you map Γto the vertices of its Cayley graph, you miss the edges but, that’s fine because we take edges to havelength 1). So C = Coneω(Γ, e, s) ∼= Coneω(Cay(Γ), e, s). So I can just prove it for the latter space.

Claim. If X is a geodesic space, then Coneω(X, e, s) is a geodesic space.

Indeed, suppose (ai), (bi) represent points in Coneω(X, e, s). Then there exists a geodesic pathγi : [0, d(ai, bi)]→ Coneω(X, e, s) from ai to bi. Define

γ : [0, d(a,b)]→ Coneω(X, e, s)

by γ(r) = (γi(sir)). For r ∈ [0, d(a,b)], γi(sir) is well defined for all i in some set of ω-measure 1.

3. So Cauchy sequences converge. Use an appropriate diagonal argument.

Remark. C geodesic implies that it is both connected and locally connected. However, there are exampleswhere they are not locally simply connected, so that can look very ugly locally...


Theorem. (Drutu-Gromov) For a group Γ with a word metric,

C = Coneω(Γ, e, s)

is proper (closed balls are compact) for all ω and s iff Γ has polynomial growth.

Theorem. (Pansu) If Γ is finitely generated and virtually nilpotent, then all its asymptotic cones areisometric to the same graded nilpotent Lie group with Carnot-Cartheodory metric (i.e. its Lie algebra hasa decomposition ⊕igi where [gi, gj ] = gi+j for all i, j).

e.g. Real Heisenburg Group H3(R) =

1 x y1 z

1

. Lie algebra is 3-dimensional 〈x, y, z〉, where the

only non-trivial bracket is [x, y] = z, so the grading is g1 ⊕ g2 where g1 = 〈x, y〉, g2 = 〈z〉.

A Carnot-Cartheodory metric (here on a C-C metric) on a manifold is a metric in which the distancebetween two points is the infimal length of all ‘flat’ paths between the points, that is, paths that staytangent to some specified chosen sub-bundle of the tangent bundle.

e.g. Thinking still of the Heisenburg group H3 the paths can only travel in the x and y directions (sothat is the sub-bundle).

6 Feb 14, 2013

Theorem. For Γ a finitely generated nilpotent group, the topological dimension of Coneω(Γ, e, s) is thecohomological dimension of Γ/tor(Γ)

Remark. Torsion ruins cohomological dimension, so kill that (in a nilpotent group torsion forms a normalsubgroup).

Here is how to think of cohomological dimension: think of the Cayley graph, and add cells in dimen-sion 1, say, then 2 etc. until the space is contractable. The cohomological dimension is the dimension ofthe cells at which this stops. For example, think of Zn, add 2 cells, 3 cells, etc. up to n-cells, when we getRn which is contractable. Thus cohomological dimension of Zn is n.e.g. n=3,

Here we think of adding 2-cells to the lattice Z3, but this is still not contractable, so add 3-cells to ‘fill itin’, and we are left with R3, which is contractable.

Corollary. If a group Γ is quasi-isometric to Zn then it is virtually Zn.

Proof. Γ is quasi-isometric to Zn ⇒ Γ is of polynomial growth ⇒ Γ is virtually nilpotent (by Gromov).Let Γ0 be a finite-index subgroup of Γ which is nilpotent. The topological dimension of the cones of a groupquasi-isomteric to Zn is n. So Γ0/tor(Γ0) has cd = n. But the growth of Γ0 is f(r) = rn, a polynomial ofdegree n. So Γ0/tor(Γ0) is Zn.

(To see this last step recall the formula Σ k torsionfreerank(γk(G)/γk+1(G)). We see that k has tobe one as we also have n = Σ torsionfreerank(γk(G)/γk+1(G)).)

Definition. We say that Zn is quasi-isometry-rigid.


6.1 Filling Functions

Given a finite presentation 〈x1, . . . , xm|r1, . . . , rn〉 of a group Γ.

Definition. The presentation 2-complex, K has:

1. 0-cells: 1

2. 1-cells: a directed edge for each generator

3. 2-cells: glue in corresponding to relations

Definition. The Cayley 2-complex is the universal cover of K. π1(K) = Γ by Seifert-van Kampen.Can identify Γ with K(0), and identify K(1) with the Cayley graph of Γ with respect to {x1, . . . , xm}.

Example. Z2 = 〈a, b|[a, b]〉, then K is the rose with 2 petals, R2 with a square with edges labelleda, b, a−1, b−1 glued to R2 in such a way that the edges and their orientations match up. The result of thisis a torus. The universal cover of of the torus is R2. We see that π1(Torus) = Z2 acts on R2 by translations(freely).

Definition. A van Kampen diagram ∆ of Γ = 〈xi|R〉 is a finite contractable planar 2-complex withall the edges directed and labelled by some xi in such a way that around each 2-cell one reads the definingrelator.

Equivalently, a van Kampen diagram is a finite contractable planar 2-complex ∆ together with a combina-torial map f : ∆→ K (meaning that it sends i-cells homeomorphically to i-cells of the Cayley 2-complex).

1. The area of ∆ is the number of 2-cells of ∆, Area(∆).

2. The diameter of ∆ is Diam(∆) := max{d(p, q) : p, q in∆(0)}, where distance here is the combinatorialdistance in ∆(1).

Definition. Let X be a topological space with basepoint e, let γ : (S1, ∗) → (X, e), be rectifiable (finitelength). The filling length of γ is

FL(γ) := inf{L ≥ 0 : there is a nullhomotopy Ht(s) of γ with l(Ht) ≤ L}

Below are two contrasting examples: in the plane, a nullhomotopy can always be performed via paths oflength strictly less then that of the original loop. However, on a sphere with a puncture the loops can haveto get much larger then the original loop during the homotopy e.g. see the red loop below:


3. The filling length of (∆, ∗), where ∗ is a choice of basepoint on ∂∆, FL(∆, ∗), is the minimal L suchthat there is a combinatorial nullhomotopy (∆i) of ∆ = ∆0 to ∗ = ∆r, with maxi l(∆i ≤ L).

A combinatorial nullhomotopy is thought of as reducing our planar 2-complex by ‘obvious’ moves,e.g. can cast of a free edge (think of this as contracting a free edge) or removing a 2-cell whichshares a boundary component with the boundary. Then we think of the length of the boundary ofthe 2-complex at each stage in this process.

Definition. For M = Area,Diam or FL, define M : N→ N by

M(n) = max{M(γ) : γ an edge loop of length ≤ n in K}

whereM(γ) = min{M(∆) : ∆ is a van Kampen diagram such that γ = ∂∆}

Example Z2

Area(n) = n2, Diam(n) = n, FL(n) = n.

7 Feb 19, 2013

7.1 Combings of Groups

Definition. G a group with finite generating set X. A combing is a section σ : G → (X ∪X−1) of thenatural surjection (X ∪X−1)→ G.

σ : G→ (X ∪X−1)g 7→ σg

can be considered as a unit speed path in the Cayley graph CayX(G) from e to g.

1. σ is synchronously k-fellow-travelling when ∀ g, h ∈ G : dX(g, h) = 1⇒ ∀ t ∈ [0,∞)

d(σg(t), σh(t)) ≤ k

2. A reparametrisation ρ is an unbounded function N→ N with ρ(0) = 0 and ρ(n+1) ∈ {ρ(n), ρ(n)+1}for all n

3. σ is asynchronously k-fellow travelling when ∀ g, h ∈ G : dX(g, h) = 1 ∃ reparametrisations ρ, ρ′

such that∀ t ∈ N d(σg(ρ(t)), σh(ρ′(t)))

4. The length functionL(n) = max{l(σg) : g ∈ G, d(e, g) ≤ n}


Examples. (Combable Groups)

1. Finite groups all have synchronously k-fellow-travelling combings with L(n) ≤ constant. (Just take kto be the diameter).

2. The free group Fn. The geodesics form a synchronously 1-fellow travelling combing with L(n) = n.

3. Zn = 〈a1, . . . , an : [ai, aj ] = 1 ∀ i, j〉. Then

σg = ar11 ar22 · · · a

rnn

4. Baumslag-Solitar group, BS(1, 2) = 〈a, b : b−1ab = a2〉

{bruas : r, s ∈ Z, u ∈ {ab−1, b−1}∗ and the first letter of u is not b−1}

defines an asynchronously fellow-travelling combing. The re-write rules:

(a) ab 7→ ba2

(b) a−1b 7→ ba−2

(c) a2b−1 7→ b−1a

(d) a−1b−1 7→ ab−1a−1

(e) aa−1 7→ ε

(f) a−1a 7→ ε

(g) bb−1 7→ ε

(h) b−1b 7→ ε

Using these rules one can convert any given word into a combing word. e.g.

abn 7→ ba2bn−1 7→ b(ab)a2bn−2 7→ · · · 7→ bna(2n)

Here L(n) ≈ 2n.

Definition. Synchronously automatic groups are finitely generated groups with synchronous k-fellow-travelling combings for which the language {σg : g ∈ G} is regular (low complexity).

For example: Zn with the first combing is automatic (language can be realised by an automaton), butif one uses the straight line combing then it is not automatic. Other examples of synchronously automaticgroups are:

1. Hyperbolic groups

2. Braid groups

3. Many 3-manifold groups

4. Almost no nilopotent groups are automatic: Exceptions: the only torsion free nilpotent groups whichare automatic are Zn.


Remark. The length functions of synchronously automatic groups satisfy L(n) ≤ Cn for a constant C.

Remark. CAT (0) groups admit synchronously k-fellow travelling combings with L(n) ≤ Cn for a constantC.

Theorem. If a group G with finite generating set X has a combing σ that is asynchronously k-fellow-travelling and has length function L(n), then G is finitely presentable and

Area(n) ≤ CnL(n)

Diam(n) ≤ FL(n) ≤ Cn

Proof. Suppose w ∈ (X ∪ X−1)∗ represents 1 in G. Then w represents a loop in the Cayley graph.Moreover there is a combing path between e and every group element in w:

Consider this as a planar graph and send this into the Cayley graph in a combinatorial way (which nowmight not be planar). (Note: there is a subtle point here that the combing paths drawn above are reallyjust representations of the combing paths in the Cayley graph, and we draw them in such a way that theyare disjoint. When we map this into CayX(G) they may intersect). Faces have perimeter at most 2k + 2.LetR be the set of words in (X∪X−1)∗ which represent 1 inG and which have length≤ 2k+2. Note |R| <∞.

Claim 1: ∆ is a van Kampen diagram for w wrt 〈X|R〉Claim 2: 〈X|R〉 = G

Indeed claim 1 is true by construction. Will come back to Claim 2.

1. Area claim: There are l(w) icicles (i.e. combing paths) each with at most 2L(n) 2-cells. So in totalArea(n) ≤ 2nL(n).

2. Diameter claim: Will show that Diam(n) ≤ Cn. Take a general vertex. Travel orthogonal to theicicles. There are at most k vertices between you and the icicle adjacent to you, and there are at mostn paths, hence the diameter is ≤ kn.

3. It is a general fact that Diam ≤ FL With that point 2 becomes redundant.

4. Filling length claim: null-homotope by contracting ∂∆ orthogonal to the icicles


Open Question. Do finitely generated class c nilpotent groups all admit (a)synchronous k-fellow-travellingcombings with L(n) ≤ Cn for a constant C. (And if it’s asynchronous can you do it with a regular language).

8 Feb 21, 2013

We can view a finite presentation Γ = 〈X|R〉 as a short exact sequence,

1→ 〈〈R〉〉 → F (X)→ Γ→ 1

so a word w on X ∪X−1 represents 1 in Γ if and only if w can be written as a product,

w =N∏i=1

u−1i rεii ui (∗)

with ri ∈ R, εi = ±1, ui words on X ∪X−1. (Note: (∗) reduces to 1 in the quotient F (X)/〈〈R〉〉)

Lemma. (van Kampen) A word w on X ∪X−1 represents 1 in Γ if and only if w admits a van Kampendiagram over 〈X|R〉. Moreover, Area(w) is the minimum N amongst all expressions (∗).

Proof. Suppose ∆ is a van Kampen diagram for w

For the reverse implication, start with the lollipop diagram associated to the expression in (∗). Morally weshould perform the above operation in reverse.


We have, however, to be careful about maintaining planarity.

That said, if we are careful, then it works.

�

Lemma. If w = 1 ∈ Γ if and only if it can be reduced to the empty word ε using the moves

1. free reductionx(a−1a)y 7→ xy

2. free expansionxy 7→ x(aa−1)y

3. applying relationsxay 7→ xby where a−1b ∈ R

Such a sequence of moves is called a null-sequence.

Example. Zn = 〈a1, . . . , an|[ai, aj ] ∀ i, j〉The Dehn functions Area(n) ≤ n2 and FL(n) ≤ n.

Lemma. w = 1 ∈ Γ

1. Area(w) is the minimal N such that there is a null-seuqnce for w using at most N applications ofrelations.

2. FL(w) is the minimum of maxi l(wi) such that there is a null-sequence

w = w0 7→ w1 7→ w2 7→ · · · 7→ wn = ε

for w.

Remark. This is a computer science type of notion. Re-label (1.)=Time, (2.)=Space. That is, thinkingof performing the moves 1, 2, 3 to a word w in a non-deterministic way (so no prescribed rule for when toapply a given word), the idea above is that counting 3. tells us how long the algorithm will take to get thethe empty word, and counting the filling length will tell us how many moves it will take. Hence time andspace.

Proposition. The following are equivalent:

1. Area(n) is recursive

2. There is an algorithm solving the word problem

3. ∃ a recursive function f(n) such that Area(n) ≤ f(n) for all n


8.1 Quasi-Isometry Invariants

Proposition. Area(n), Diam(n) and FL(n) are all quasi-isometry invariants (up to ') among finitelygenerated groups.

Corollary 1. Among finitely generated groups being finitely presentable is a quasi-isometry invariant

Corollary 2. If P1 and P2 are finite presentations for the same group, then

1. AreaP1 ' AreaP2

2. DiamP1 ' DiamP2

3. FLP1 ' FLP2

Proofs. Suppose that Γ1 = 〈X1|R1〉 finite presentation, and let Γ2 be a group with finite generating setX2. Suppose,

φ : Γ2 → Γ1

is a (λ, µ)-quasi-isometry. Suppose w a word on X2 ∪X−12 and represents 1 ∈ Γ2. Aim to show that thereexists a van Kampen diagram for w in Γ2. The idea is to

1. Represent the word w in CayX2(Γ2)

2. Use the quasi-isometry to push the loop from CayX2(Γ2) to CayX1(Γ1)

3. Γ1 has a finite presentation, so construct a van Kampen diagram in CayX1(Γ1).

4. Use quasi-inverse φ to pull this van Kampen diagram back to CayX2(Γ2)

Given a van Kampen diagram for w we have a finite set of relations that reduce w to 1 ∈ Γ2. Thus wehave shown that Γ2 is finitely presented. This will prove Corollary 1. In fact it is just the same idea toprove the Proposition, we just need to analyse the van Kampen diagram obtained for w in CayX2(Γ2),and compare it to that in CayX1(Γ1).

Use φ to push the word w from CayX2(Γ2) to CayX1(Γ1). Here we have a finite presentation, and sohave a van Kampen diagram representing the word:

Now φ is a quasi-isometry, so it has a quasi inverse φ such that d(x, φφ(x)) ≤ C, d(x, φφ(x)) ≤ C and φ isa quasi-isometry. Push this van Kampen diagram back to CayX2(Γ2) with φ:


Since φ is only a quasi-inverse, the van Kampen diagram might get shifted about some. But the idea isthat it only gets shifted by a bounded (in particular, finite) amount. Indeed - consider a loop in the vanKampen diagram in CayX1(Γ1). This has a perimeter length bounded above by the largest length of adefining relation in R2 (which is finite, so this perimeter is bounded). Hence when we push diagram backto CayX2(Γ2), the perimeter length is bounded. The only remaining problem is complete the diagram (thered bit) by adding cells (the light blue bits) to get a van Kampen diagram for w in CayX2(Γ2). But the

fact that φ is a quasi-isometry means that this added bit are bounded, hence we are done.

�

9 Feb 26, 2013

9.1 The Area Function of Nilpotent Groups

Theorem. (Gromov; Gersten-Holt-Riley) If G is a finitely generated nilpotent group of class c then

Area(n) 4 nc+1 and FL(n) 4 n

Moreover, these bounds are realisable simultaneously.

Proof. (Will prove this for the group H3 as this captures all the ideas of the general proof)

H3 = 〈x, y, z|[x, y] = z, [x, z] = 1, [y, z] = 1〉

Suppose w = 1 ∈ H3. Let n = l(w). We will show that there is a null-sequence {wi} for w using 4 n3

defining relations and with maxi l(w) 4 n (hence simultaneously).

Define compression words u(s) representing zs for s ≥ 0:

• for 0 ≤ s ≤ n2 − 1, u(s) = zs0 [xn, ys1 ] where s = s0 + s1n and s0, s1 ∈ {0, 1, . . . , n− 1}

• u(n2) = [xn, yn]

• u(A+Bn2) = u(A)u(n2)B for all integers A,B with 0 ≤ A ≤ n2 − 1 and B > 0.

Note: l(u(A+Bn2)) ≤ K0n for some K0 depending on B.

Lemma. Fix K1 > 0. Then there exists K2 > 0 such that for all 0 ≤ s ≤ K1n2, there is a concatenation

of sequences converting zs to u(s):

zs → zsu(0)→ zs−1u(1)→ zs−2u(2)→ · · · → zu(s− 1)→ u(s)

using ≤ K2n3 relations.


Proof of Lemma Key idea:

zn[xn, yk] = znx−ny−kxnyk

= znx−ny−k(y−1y)xnyk (insert 1 = y−1y)

= znx−ny−(k+1)(xz−1)nyk + 1 ([x, y] = z n times)

= [xn+1, yk+1] (n(2n+ k + 1) relations)

So the total cost is ≤ n2.

�

Remark. Finitely generated nilpotent groups are all finitely presentable, a fact which we will prove later.

Induction on class c : Base case c = 1 i.e. abelian group. (easy)Induction step: Consider G/γc(G) which is a finitely presentable nilpotent group of class c − 1. Let X bea finite generating set for G and X its image in G/γc(G). Let 〈X|R〉 be a finite presentation. Let X ′ be afinite set of length c commutators (entries in X) in G that generate γc(G) (a fact that itself can be provedby induction on c). Let R be a set of |R| relations (words on X) each expressing elements of R as equal toa word on X ′. So,

〈X ∪X ′|R,X ′central〉

is a finite presentation for G.

Suppose w is a word on X and w = 1 in G. Let w be its image as a word on X. w = 1 in G/γc(G),so there is a null-sequence,

w = w0 → · · · → wk = ε

with l(wi) 4 n and the number of relations used 4 nc (by induction hypothesis). We seek to lift this to anull-sequence for w. Prove this for H3, but the idea is the same in general.

Let G = H3 we have G = H3/γ2(H3) = H3/〈z〉 = 〈x, y|[x, y] = 1〉.

First

w 7→ (z∗)−1w0z∗

by carrying all the z’s to the right end, and all the z−1’s to the left.

Each move wi → wi+1 that applies the relation [x, y] = 1 lifts to a move which applies [x, y] = z, sointroduces a z±1. Then shuffle all z’s to the right, and all z−1’s to the left. Compress the powers of z ateach end as the z±1’s arrive. Eventually we get

u(s)−1 wk︸︷︷︸=1

u(t)

for some s, t ≥ 0. In H3 this isz−szt = zt−s = 1 (2)

so s = t. The total cost of this procedure is:

n2 - the initial step

n2 · n - lifting the wi null-sequence and carrying the z’s to the ends

Kn3 - compression at each end

Total: n2 + n2 · n+Kn3


�

10 Feb 29, 2013

Question. (Open) Do class c, finitely generated, nilpotent groups have (a)synchronous fellow-travellingcombings with L(n) 4 nc

Remark. This would imply the 4 nc+1 Dehn function and 4 n filling length result we proved last lecture.

Question. (Open) Does H3 admit a synchronously fellow-travelling combing?

Remark. There is an alternative proof to the result from last lecture using asymptotic cones.

Asymptotic cones approach to 4 nc+1: Suppose G is a finitely generated nilpotent group of class c,

Coneω(X, e, s)

are graded (homogeneous) nilpotent Lie groups. In this case loops can be filled with discs of area 4 n.The problem is how do we pull this result back to the group. There is a partial result of Papasoglu⇒ Area(n) 4 nc+1+ε, which is close.

10.1 Lower Bounds

These can be tricky. Here is a warm up problem: Prove that for Z× Z ∼= 〈a, b : [a, b]〉 the Area(n) ≥ n2. Aguess might be that Area([an, bn]) ≥ n2.

Gersten’s Lemma If the Cayley 2-complex is contractable then a van Kampen diagram which is 1-1 onthe compliment of its 1-skeleton is of minimal area.


Mad Proof that Area([an, bn]) ≥ n2Suppose W =

∏Ni=1 u

−1i [a, b]εiui, (where εi = ±1, ui are words on a, b, a−1, b−1) equals [an, bn] in F (a, b).

We want to show that N is at least n2. We have W =∏Ni=1 c

εi in

H3 = 〈a, b, c|[a, b] = c, [a, c] = 1, [b, c] = 1〉

(since [a, b] = c and c is central, so the ui’s cancel). But in H3 we have [an, bn] = cn2, and c has infinite

order. Now W = [an, bn], by hypothesis, and so equating gives,

W = [an, bn] = cn2

=N∏i=1

cεi = W

whence comparing the powers of c gives N ≥ n2. (Recall, εi = ±1, so we get the lower bound n2 preciselywhen these are all +1, but it will be bigger then n2 is some of them are −1’s)

�

Centralised Isoperimetric FunctionsP = 〈A|R〉 a finite presentation of Γ, R = 〈〈R〉〉 in F (A).

Γ = F (A)/R

Suppose w = 1 in Γ.

Definition. The centralised area Areacent(w) = min{N |w ∈ (∏Ni=1 u

−1i rεii ui)[R, F ]}

This is just, the number of times, up to sign, that each relator occurs in a product of conjugates freely equalto w, ignoring conjugation involved. This is a calculation in

R/[R, F ]

which is finitely generated abelian group, as it is generated by R, say {r1, . . . , rk}. Write w = rs11 · · · rskk in

R/[R, F ]. Then the centralised area, Areacent(w) = min{∑k

i=1 |si|}

Lemma. Suppose B = {y1, . . . , ym} generates the abelian group,

R/[R, F ]

For a word w(A) = 1 in Γ, let lB(w) be the minimal length of a word on B representing w[R, F ]. LetK = max{lB(r) : r ∈ R}, then

1. lB(W ) ≤ K ·Areacent(w) ≤ K ·Area(w)

2. ∃c > 0 such that if w[R, F ] = ym[R, F ], where y[R, F ] has infinite order in R/[R, F ] then

m ≤ c ·Areacent(w)

Proof.

1.

2. R/[R, F ] = T ⊕Zk for some finite T . Let B = T ∪ { a basis for Z}. Then m ≤ lB(ym) (by projectingto the Zk factor). Then m ≤ lB(ym) = lB(w) ≤ K ·Areacent(w).

�

Definition. Let F = F (a, b). Write F1 = F, Fi+1 = [Fi, F ]. Then F/Fc+1 is the free nilpotent groupof class c on two generators .


Theorem. F/Fc+1 has Dehn function < nc+1 (so ' nc+1)

Proof.Let u = [ · · ·︸︷︷︸

c

[[a, b], a], · · · , a]

Lemma u ∈ Fc\{1}, and ukc

= [· · · [[ak, bk], ak], . . . , ak] mod Fc+1

This is just a more elaborate version of the same notion we showed in the example with H3 above. Notgoing to prove it.

Let R = Fc+1. Then [R, F ] = Fc+k and R/[R, F ] is Fc+1/Fc+2 is generated by ‘simple commutators’. Nowwrite uk

c= [ · · ·︸︷︷︸

c

[[ak, bk], ak], . . . , ak] mod Fc+1 = wc,k. Now wc+1,k ∈ Fc+1 and is a kc+1 power of a simple

commutator (e.g. wc+1,1). So wc+1,k is a word representing 1 in F/Fc+1.

Area(wc+1,k) ≥ Areacent(wc+1,k) ≥ constant · l(wk+1,c)c+1

In the Heisenburg group example this is

Area([ak, bk]) ≥ Areacent([ak, bk]) ≥ constant · (Lk)c+1

11 Mar 5, 2013

11.1 Some Open Problems Relating to Dehn Functions

Recall, finitely generated nilpotent class c groups have Dehn function (i.e. area) 4 nc+1 and FL(n) 4 n.Free nilpotent class c groups on ≥ 2 generators have Dehn function < nc+1.

Theorem. (Gromov, Allcock, Olshanskii-Sapir) The (2k + 1)-dimensional Heisenberg group fork ≥ 2 has Dehn function 4 n2.

H5 =

1 Z Z Z

1 0 Z1 Z

1

H7 =

1 Z Z Z Z

1 0 0 Z1 0 Z

1 Z1

of class 2.

Theorem. (R. Young) For all c there exists nilpotent class c group with Dehn function' n2, n3, . . . , nc+1.

Theorem. (S. Wenger) There exists a nilpotent group with Dehn function � nα for any α.

Sapir’s Example 〈a1, b1, . . . , as, bs| class 2, [a1, b1] · · · [as, bs] = 1〉. n2ρ(n) ≤ Area(n) ≤ n2 log n whereρ(n)→∞ as n→∞.

Open Question.

1. Classify finitely generated nilpotent groups up to quasi-isometry

2. Classify them by Dehn function

3. Do there exist uniform upper bounds on the Dehn function and filling length functions on other classesof groups, like:

(a) polycyclic groups (NB always finitely presented)


(b) finitely presentable metabelian groups

(c) finitely presentable linear groups

There is a problem:

Theorem. (Kharlampovich) There exist finitely presented 3-step solvable groups with undeciable wordproblem.

11.2 HNN-Extensions

Definition. G a group, A,B subgroups and φ : A→ B an isomorphism. Then

Gφ := 〈G, t|t−1at = φ(a) ∀a ∈ A〉

Britton’s Lemma Suppose w = g0te1g1t

e2 · · · tengn where ei = ±1, n ≥ 1, gi ∈ G, and w = 1 ∈ G. Thenit contains a subword of the form

1. t−1gt for some g ∈ A, or

2. tgt−1 for some g ∈ B

This subword is called a pinch.

Proof. Draw a van Kampen diagram for w. Consider the cell attached to te1 . This cell represents arelation in Gφ, and so must be of the form t−1atφ(a)−1 reading either clockwise or anticlockwise dependingon e1.

This leaves an open t-edge on the cell (possibly), and by the same argument, this must have a cell of thesame form attached. If we repeat this process we see that we form a ‘corridor’ across the van Kampendiagram ending in another t edge on the boundary, with elements in A along one side, and elements of Balong the other.

All t’s along the boundary must pair up in this way. Consider now the outermost t-corridor. Suppose thatthe t is oriented as shown in the diagram:


then we see that along the right side of the corridor we get w word in a. Now the green cell gives a relationg2 = a1a2a3a4. More generally we get that g2 is a word in either A or B, depending on the orientation of t(i.e. ei = ±1), and this gives us the desired subword of one form or the other.

�

Proposition. (Normal Form) Every g ∈ Gφ can be represented uniquely as

w = g0te1g1t

e2 · · · tengnwhere ei = ±1, n ≥ 1, gi ∈ G, and such that w contains no pinches.

Proof. Existence: easy - just represent the word in that form and remove pinches.Uniqueness: Suppose two such w1, w2 both representing g. Then by Britton’s lemma w1w

−12 contains a

pinch as it represents the identity in G. If one of wi does not contain a t then nor does the other. If bothcontain no t then the result is immediate. So suppose that at lease one of the wi’s contains a t. Now w1w

−12

contains a pinch and both contain a t.

They must pair the last t±1 with w1 with the last t±1 in w2. By induction on the max of the number of t±1

in w1 and w2, u and v are the same. It follows that w1 and w2 are the same.

11.3 Baumslag-Solitar Groups

These were introduced in 1963, as

BS(m,n) = 〈a, t|t−1amt = an〉 = Zφ

This is an HNN-extension where φ : mZ→ nZ by am 7→ an, where a is a generator. Trivial examples thatwere known prior to Baumsalg and Solitar’s paper are:

1. BS(±1,±1) = Z× Z

2. BS(±1,∓1) = Z o Z


Definition. A group G is Hopfian if it is not isomorphic to any one of its proper quotients i.e. everyepimorphsim G→ G is an isomorphism.

For example: Z∞, consider infinite sequences of integers. f : (a1, a2, a3, . . .) 7→ (a2, a3, a4 . . . , ) is a surjectivehomomorphism with infinite kernel. Baumslag-Solitar groups were constructed to give an easy example ofnon-Hopfian groups.

Theorem. BS(2, 3) is non-Hopfian.

Proof. We haveBS(2, 3) = 〈a, t|t−1a2t = a3〉

Now a 7→ a2, t 7→ t defines an epimorphism G→ G with non-trivial kernel.

Well defined: t−1a2t 7→ t−1a4t = (t−1a2t)2 = (a3)2 = a6 so a3 7→ a6.

Surjective: a−1t−1at 7→ a−2t−1a2t = a−1a3 = a

Kernel: a−1t−1ata−1t−1ata−1 7→ a−1t−1a2ta−2t−1a2ta−2 = a−2a3a−2a3a−2 = 1

�

Definition. G is residually finite when for all g ∈ G\{1} there exists a finite group H and a homomor-phism φ : G→ H with φ(g) 6= 1.

Remark. Linear ⇒ Residually Finite

Proposition. BS(m,n) is residually finite if and only if |m| = |n| or |m| = 1 or |n| = 1 (if and only iflinear)

Proof. BS(m,n)→ Z[ 1mn ]o Z by a 7→ (1, 1), t 7→ (m/n, 0)

And Z[ 1mn ]o Z

∼=−→(

(m/n)Z Z[ 1mn ]

0 1

)⊆ GL2(Q). Not injective when m 6= 1, n 6= 1.

Proposition. BS(m,n) is

1. Hopfian if and only if it is residually finite, or m and n have the same prime divisors.

2. Solvable if and only if it is metabelian if and only if |m| = 1 or |n| = 1.

12 Mar 7, 2013

12.1 The Geometry of Baumslag-Solitar Groups

12.1.1 Normal Form for BS(1, n)

BS(1, n) = 〈a, t|t−1at = an〉.

1. HNN-normal form: Use the HNN-normal form to give a normal form for BS(1, n),

{tiakt−j : i, j ≥ 0, n - k}

This is not an (a)synchronous k-fellow travelling combing for n ≥ 2: For example in BS(1, 2),

t−ja = t−jatjt−j = a2jt−j

Now t−j and a2jt−j are both in normal form, and differ by one generator. Consider a van Kampen

diagram: here each cell represents the relation t−1at = a2, and we get


Now consider the paths γ for t−j (in purple) and µ for t−ja = a2jt−j (in green) defined by this

HNN-normal form:

Claim: d(p, γ) ≥ j. This is only a claim because I have not got drawn the Cayley graph, so thisdiagram only gives an idea of whats going on. We will see the Cayley graph later this lecture.


2. Asynchronous k-fellow travelling normal form: Use the rewrite rules

a2t−1 7→ t−1a

a−1t−1 7→ at−1a−1

to convert it to an asynchronous fellow travelling combings. For example consider the words tiakt−j

and tiakt−ja written in the new normal forms. Well tiakt−ja = tiak+2j t−j by the first rewrite rule.Now compare their normal forms with the rewrite rules above (not these are in HNN normal form,not our new normal form) (exersize) to see asynchronous combing of length 2n.

12.1.2 Dehn Function of BS(1,m)

Proposition. BS(1,m) has Dehn function ' 2n

Proof. Prove this for BS(1, 2), but the idea is the same for m ≥ 2.Upper bound: Area(n) 4 nL(n) for any asynchronous combing, and we have just given an example whereL(n) ' 2n. (Alternatively, think of Britton’s lemma: identifying a pinch in a word equal to the identity.)Lower bound: Let wn = [a, t−natn]. Now l(wn) = 4n + 4. wn = 1 ∈ BS(1, 2) because t−natn = a2

n.

Consider the van Kampen diagram for wn:

Consider the t-corridor starting as shown from the ‘rightmost’ t (corridor starting in green). We know thismust end at a t on the boundary. Simple orientation considerations prevent the grey paths from beingvalid corridors, leaving only the purple and the red as potential corridors. I claim it cannot be the redcorridor. Indeed if it were the red corridor shown, then the only option for any t-corridor contained withinthis corridor (e.g. the blue t-corridor in the diagram) can only start and end in the red highlighted area.But this is impossible as the orientation would agree. So the only possible choice for the red t-corridor isto end up in the ‘leftmost’ t, as shown below:


The blue cell here gives a defining relation in the a, namely that a = a2k for some k. This is not a validrelation, showing that this is not a valid t-corridor. This leaves only the purple t-corridor as an option:

I claim that this t-corridor must end up in the ‘rightmost’ position (as shown): indeed, all n copies of t wehave forming the top edge and the bottom edge must line up one-to-one, for if they didn’t we would getthat one t-corridor had to loop back on itself (contradicting orientation considerations), or would have topass through another t-corridor (illegal). And so the black word highlighed is freely equal to ar for some r.But reading around the diagram we get

ar = t−natn

so r = 2n and so Area(wn) < 2n−1.

�

Corollary. BS(1,m) admits no combing with length function L(n) ' n, and in particular is not automatic.

Proof. Since automatic implies at most quadratic Dehn function.

�

12.1.3 Cayley Graph of BS(1, 2)

The Cayley graph is the 1-skeleton of the Cayley 2-complex, which is the universal cover of

the space obtained from gluing the cell representing the relation t−1at = a2 to the rose R2. The universalcover of the resulting space is


Indeed, consider the bottom most row of a’s. We glue in the relation a2 = t−1at everywhere we can, westart with:

Notice that if we start one further along the bottom row we can glue in another relation, and thus getanother ‘strip’:

We can now repeat this process along each line of a’s, at each point we get two folds emanating, and theresult is:

the Cayley diagram for BS(1, 2). Notice:

1. Let T be the infinite binary tree. Then we get that the Cayley graph is T × R - a Bass-Serre tree(vertices correspond to cosets of 〈a〉):


2. This is the universal cover of the aforementioned space: indeed it is simply connected as is takes theform (Tree)×R

12.1.4 Treebolic Space

Fix q > 0. Height function h : H→ R by z 7→ logq(Im z) (this gives our heights in red):

We have a tree Tp = infinite rooted tree with braching number p, e.g. the infinite binary tree has p = 2.We can embed Tp in this picture according to the height function. Now the height function restricts toh : Tp → R (note this function depends on q):

Definition. Treebolic space is the ‘horocyclic product’ HT (q, p) := {(u, v) ∈ H× Tp : h(u) + h(v) = 0}(to be clear - this depends on q as it is used to define the function h).

Example: Let v be the root of the tree, having height 1. Then we have that pairs {(u, v) ∈ H×Tp : Im(u) =p} ∈ H(p, p), so we get that point v ∈ Tp crossed with the line {u ∈ H : h(u) = −1}.


Similarly for all points v ∈ Tp we get a line homeomorphic to R in HT (p, p)

Theorem. BS(1, p) acts properly discontinuously and cocompactly on HT (p, p).


Part II

Student Talks

13 Geometry of Sol (Chenxi Wu)

Sol = R2 oφ R where φ : R→ GL2R by t 7→(et 00 e−t

)Discrete subgroup in Sol: Z2 oφ Z where φ : 1 7→ A =

(λ 00 λ−1

)(Note: the mapping torus of T 2 glued with an Anasov map will also have a Sol structure)

13.1 Dehn Function

Definition. Area(l) = suplength(γ)=l

infD,γ=∂D

Area(D)

We get that Z2 oφ Z is quasi-isometric to Sol. Hence we get

AreaZ2oZ ' AreaSol

Proposition. AreaSol ' exp

13.2 Quasi-Rigidity of Sol

Γ is quasi-isometric to Sol then there exists K C Γ, |K| < ∞ such that Z2 oA Z ≤f.i Γ/K where hereA : Z→ Aut(Z2) by 1 7→ A.

13.3 Area function becomes quadratic

Sol3 =

et xe−t y

1

= R2 oR = 〈x, y〉o 〈t〉

Sol5 =

et1 x

et2 yet3 z

1

= R3 oR2 = 〈x, y, z〉o 〈t1 − t2, t1 − t3〉


14 Geometry of the Lamplighter Group (Margarita Amchislavska)

Definition. The Lamplighter has numerous equivalent definitions:

1. Γ1 = Z[x, x−1]o Z, generator of Z factor acts on Z[x, x−1] by multiplication by x

2. Γ1 = ⊕i∈ZZ o Z = Z wr Z

3. Γ1 =

{(xk f0 1

): k ∈ Z, f ∈ Z[x, x−1

}4. Γ1 = 〈a, t|[a, atk ] = 1, k ∈ Z〉 finitely generated, not finitely presented.

5. Picture: Lamps at integer locations, each lamp has an integer worth of possible ‘brightness’ or values.Lamplighter moves around adjusting the brightness (as we saw in the lectures)

14.1 Dead-end Depth

Let G be a group with generating set X. g ∈ G is a dead-end element if no geodesic ray from the identityto g in the Cayley graph Cay(G,X) can be extended past g. That is, ∀ x ∈ X, d(1, gx) ≤ d(1, g). Thedepth of a dead-end element g with d(1, g) = n is the length of the shortest path from g to an elementin B

cn (in other words, a dead-end element g means that the geodesic from 1 to g if extended by 1 (i.e.

multiplication by any other element x ∈ X results in moving back inside the ball of radius d(1, g). It stillcould leave this ball if you extend far enough, and the minimum distance for which this happens is calledthe dead-end depth).

Examples

1. G = Z, X = {2, 3}. 1 has dead end depth 2.

2. Γ1,2 = 〈a, t|a2 = 1, [a, atk] = 1, k ∈ Z〉

Theorem (Clearly, Taback 2003) Γ1,2 has elements of unbounded dead-end depth.

d(1, gn) = 4n + 2n + 1 = 6n + 1, but d(gn, h) ≥ n + 1 ∀ h ∈ Bc6n+1 since the lamplighter will

have to go to at least one lamp at n+ 1 to get out of the ball.

14.2 Horocyclic Product of Two Z-branching Trees

There is a bijection between ‘vertex set of infinite Z-branching trees’ and ‘cosets of Laurent polynomials insome variable’.

Remark

1. Each level partitions the polynomials.

2. Given a polynomial and its height there is a unique vertex corresponding to it.

3. Coefficients of the polynomials give sequences of edge labels when following successive downwardedges.

Consider two such trees: one using Z[x] cosets (as exemplified above), the other using Z[x−1] cosets.

Define the height functions

1. H(f + xkZ[x]) = k

2. H(f + xkZ[x−1]) = −(k + 1)


The horocyclic product of these two trees T1, T2 has vertex set {(v1, v2) ∈ T1×T2 : H(v1) +H(v2) = 0} andedges (v1, v2) ∼ (w1, w2) are connected if and only if v1 ∼ w1 and v2 ∼ w2. Let H1 denote this horocyclicproduct of the two trees with Z[x] and Z[x−1]. Note: (g+ bxn−1 + xnZ[x], g+ dxn + xn−1Z[x−1]) ∈ V (H1).We will show that H1 ' Cay(Γ1, Y ).

Lets look at Γ1 again: g ∈ Γ1 is given by k ∈ Z and f =∑fix

i in the matrix description say. OR:

1. k ∈ Z and a pair of sequences

(a) {fk−1, fk−2, . . .}(b) {fk, fk+1, . . .}

OR:

2. A point (f + xkZ[x], f + xk+1Z[x−1]}

Another presentation of Γ1 is

Γ1 = 〈λi, i ∈ Z|λmi λ−mj = λm−jλm−i, i, j,m ∈ Z〉

(check these are equivalent via λi = ait). In trees λi corresponds to “go down in the second tree T2 (withoutchoice by fk), and “go up in the first tree T1 (by fk + i). This gives a graph isomorphism between theCayley graph Cay(Γ1, Y ) and H1 (where Y is the λi’s).


15 Generalisations of the Lamplighter Group (Margarita Amchislavska)

Γ2 = Z[x, x−1, (1 + x)−1]o Z2 =

{(xk(1 + x)l f

0 1

): k, l ∈ Z, f ∈ Z[x, x−1, (1 + x)−1]

}= 〈a, s, t|[a, at] = 1, [s, t] = 1, as = aat〉

Γ2 was constructed by Baumslag in ’72 as the first example of a finitely presented group with an abeliannormal subgroup of infinite rank.

15.1 Model

2 dimensional rhombic grid. Lamplighter travels on the lattice points. There are lamps on the integerpositions of the t-axis, and on the negative integer positions of the s-axis. The instructions are then: a flipsa switch, t says move one unit to the right, and s says move one unit up.

Elements in Γ2 correspond to integers on the grid points (only finitely many non-zero) and the positionof the lamplighter. Call a finite collection of integers on the lattice points a configuration. The identitycorresponds to zeros everywhere and the lamplighter at (0, 0). g ∈ Γ2 is given by (k, l, f) → (k, l) is finalposition of lamplighter and f =

∑fijx

i(1 + x)j where fij is the integer at position (i, j).

Two configurations are equivalent if they differ in the following way: if there is a triangle of integers a, b, cwith a sitting above b and c, then it can be replaced with a− 1, b+ 1, c+ 1. If one follows this all the waydown to the t-axis we get a row from pascals triangle sitting on the lampstand (the t-axis together with thenegative 2-axis).

Fact For each configuration there exists a unique configuration equivalent to it supported only on thelampstand.

15.2 Some Results

Get a basis for Z[x, x−1, (1 + x)−1] taking {xi, (1 + x)j : i ∈ Z, j ∈ Z<0}. One can use this to give a normalform for Γ2.

Let Γ2,r be the group Γ2 with the additional relation that ar = 1.

Theorem (Cleary, Riley) Γ2,2 has elements of unbounded dead-end depth.

This was the first example of a finitely presented group with this property.

Theorem (Grigorchuk, Linnell, Schick, Zuk) Gave Γ2,2 as a counter example to Atiyah’s conjecture(by constructing a 7-dimensional manifold with fundamental group Γ2,2 and the third L2-Betti number 1

3not an integer).

Theorem (de Cornulier, Tessene) Γ2,2 has quadratic Dehn function.

Theorem (Kassabov, Riley) Γ2 has exponential Dehn function

Remark

1. Filling length function unknown.

2. Γ2 = 〈µ, ν, c, d|[ν, µ] = 1, µ−1c2ν = c, ν−1d2µ = d〉


15.3 Horocyclic Product

Claim Cay(Γ2, Y ) is ' to the 1-skeleton of the horocyclic product of three infinite Z-branching trees.

Use another presentation of Γ2:

Γ2 = 〈λi, µi, νi(i ∈ Z)|λi = νimui, λi+j = µiνj(i, j ∈ Z)〉

with λi = ait, µi = ais, νi = aits−1a−i

Recall

1. Vertices of an infinite Z-branching trees correspond to cosets of Laurent polynomials in some variableZ[x],Z[1 + x],Z[x−1].

2. Given a Laurent polynomial and height, there exists a unique vertex corresponding to it.

3. Coefficients of polynomials in the given variables give sequences of edge labels when following successivedownward edges.

Combinatorially Use the model. Given g ∈ Γ2 want to see how to get a vertex in H2. Well, g ∈ Γ2 →(k, l) and a configuration of integers on the grid.

1. Use the equivalence relation on triangles to propagate the configuration to the lampstand based at(k, l).

2. Project portions to the axis: Split the lampstand at (k, l) into three portions, split horizontal portioninto two, left of (k, l) (not inclusively) and to the right of (k, l) (inclusively), and then the thirdpiece is the vertical portion. Now for the left portion project this down the the t-axis using thetriangle relation (rows of pascal triangle). Only consider the image on the t-axis to the left of k (notinclusively). Similarly for the right portion project down and consider only piece to the right of k + l(inclusively). Finally project the vertical portion to the s-axis and consider only the image up to l(not inclusively).

3. Read of three heights and three sequences from these three projections: In the left projection h1 =k, {a1}, in the right most projection h3 = −(k + l), {a3} in the vertical protection h2 = l, {a2}

Claim this gives an element in the horocyclic product of three trees. This is actually a bijection.


16 Geometry of the Magnus Embedding (Andrew Sale)

What is a free solvable group? Fr = 〈x1, . . . , xr〉 is the free group in the variety of r-generated groups.Therefore any r-generated group is a quotient of Fr. Similarly, Zr is the free group in the variety of r-generated free abelian groups, so any r-generated free abelian group is a quotient of Zr. Sr,d is going to bethe free group in the variety of r-generated solvable groups of derived length d, and thereforeany r-generated solvable group (of derived length d) is a quotient of Sr,d. So,

Sr,d = Fr/F(d)r

where F ′r = F 1r = [Fr, Fr] and F

(i+1)r = [F

(i)r , F

(i)r ].

16.1 Magnus Embedding

N C Fr. Let α : Fr → Fr/N,N′ = [N,N ]. The aim is to embed,

ψ : F/N ′ → Zr o Fr/N

If N = F(d)r then Sr,d+1 ≤ Zr o Sr,d

16.2 Applications of the Magnus Embedding

1. Various algorithm results for F/N e.g. (Kargapolov-Remeslennikov, ’66) Conjugacy Problem is solv-able in Sr,d

2. (Remeslennikov-Sokolov, ’70) Sr,d is conjugacy separable

3. Baumslag’s embedding theorem (that every finitely generated metabelian group embeds in a finitelypresented metabelian group) (see Amin’s talk)

4. Random walks on Sr,d (see Tianyi’s talk)

Classical Definition We will realise ZoF/N as a matrix group. Let R be the free abelian Z[F/N ]-modulegenerated by t1, . . . , tr.

M(F/N) =

{(g p0 1

): g ∈ F/N,P = R

}16.3 Magnus Representation

φ : F →M(F/N) by w 7→(α(w) ∂∗w

∂xiti

0 1

)which is involving something called the Fox derivative.

(Magnus ’30s) kerφ = N ′ ⇒ the Magnus embedding.

16.4 Fox Calculus

A Fox derivative is a derivationD : Z[F ]→ Z[F ]

where (recall) a derivation must satisfy, for a, b ∈ Z[F ]

1. D(a+ b) = D(a) +D(b)

2. D(ab) = D(a)ε(b) + aD(b)

where ε : Z[F ]→ Z by sending g ∈ F 7→ 1.

F = 〈x1, . . . , xr〉. Define∂

∂xi(xj) = δij


extend to F by (2), extend to Z[F ] by (1). Here’s a concrete example:

∂

∂x1(x1x2x

−11 + x−11 ) =

∂

∂x1(x1x2x

−11 ) +

∂

∂x1(x−11 )

=∂

∂x1(x1x2)ε(x

−11 ) + x1x2

∂

∂x1(x−11 )− x−11

Lemma. (Fox) Let D : Z[F ]→ Z[F ] by any derivation. Then ∃ Ki ∈ Z[F ] such that D =∑r

i=1Ki∂∂xi

Z[F ]

D∗ $$

D // Z[F ]

α

��

Z[F/N ]

16.5 Wreath Products

A = 〈Y 〉, B = 〈X〉, |X|, |Y | <∞. Let Γ = A oB = ⊕BAoB.

An element γ ∈ Γ is a pair (f, b) where f : B → A with finite support, and b ∈ B.

Multiplication: (f, b)(g, c) = (fg(b), bc) where g(b)(x) = g(b−1x)

16.5.1 Geometric Interpretation

Generating set for A oB, S = {(1, x), (fy, eb) : x ∈ X, y ∈ Y } where fy(g) =

{y g ∈ ebeA else

Let w be a word on S ∪S−1 representing (f, b) ∈ A oB. We can group together the letters of w which comefrom X ∪X−1 and those from Y ∪ Y −1.

w = (f0, eB)(1, u1)(f1, eB) · · · (fm, eB). Cay(B,X):

Supp(fi) = {eB} (except possibly for for i = 0,m where it may be empty).

Notice (f, b) = w = (f0f(u1)1 f

(u1u2)2 · · · , u1u2 · · · )⇒ Supp(f) ⊆ {eB, u1, u1u2, . . .}

So a word w for (f, b) describes a path in Cay(B,X) from eB to b via Supp(f). Notice, this is just ageneralisation of the lamplighter model. Let K(Supp(f), b) be the length of the shortest such path, then:

Lemma. |(f, b)|S = K(Supp(f, )b) +∑

x∈B |f(x)|A


Lemma. M(F/N) ∼= Zr o F/N

Proof. Take (b p0 1

)7→ (fP , b)

where fP : F/N → Zr, P =∑r

i=1(∑

g∈F/N a(i)g )ti with a

(i)g ∈ Z, then fP (g) = (a

(1)g , . . . , a

(r)g )

�

16.6 Geometric Definition of the Magnus Embedding

Define φgeo : F → Zr o F/N by w 7→ (fw, α(w)) (where α : F → F/N is the quotient map). What is fw?The word w will describe a path ρw in Cay(F/N,α(X)):

Definte a function πw on the edges by setting πw(g, gxi) to count the net number of times ρw crosses (g, gxi)i.e. count +1 when ρw goes from g to gxi and count −1 when it goes from gxi to g:

fw : F/N → Zr

g 7→ (πw(g, gx1), . . . , πw(g, gxr))

The function πw is a flow on Cay(F/N,α(X)) i.e. it determines a family of paths from e to α(w).

(Drons-Lewin-Servatis ’95) πw = πu if and only if uN ′ = wN ′ This result implies that kerφgeo = N ′.So φgeo : F/N → Zr o F/N is an embedding.

Proposition. φgeo = φ, that is, the two embeddings are equivalent.

Proof. NTS fw = fP . P =∑r

i=1∂∗w∂xi

ti. That is, NTS

∂∗w

∂xi=

∑g∈F/N

πw(g, gxi)g

Induction on l(w). If w = xi


therefore πw(g, gxj) =

{1 g = e, i = j0 else

and ∂∗w∂xj

= δij .

The inductive step considers w = w′xi:

Hence πw(g, gxj) =

{πw′(g, gxi) + 1 g = α(w), i = jπw′(g, gxj) else

, and

∂∗

∂xj(w′xi) =

∑g

πw′(g, gxj)g + w′∂ij

�


17 Finitely Presented Metabelian Groups (Amin Saied)

Here is an outline of the talk:

1. Introduction: Higman’s Embedding Theorem, Metabelian groups, Baumslag-Remeslennikov’s Theo-rem, Baumslag’s sketch of proof

2. Baumslag/Margarita’s Example: embedding Γ1 into Γ2 using an HNN-extension

3. Sketch of Proof:

(a) Magnus Embedding Theorem: reduce proof to case where G = H nA where A,H are abelian.

(b) Special Polynomial Lemma: construct injective endomorphisms of A - will be used to constructHNN-extensions

(c) Construct ascending HNN-extensions of G = H nA where H = 〈h1〉 × · · · × 〈hn〉 culminating inG = QnA

(d) Show G is metabelian

(e) Show G is finitely presented

17.1 Introduction

Motivating Question There are uncountably many finitely generated groups, but there are only count-ably many finitely presented ones. Which countably many finitely generated groups then appear as sub-groups of finitely presented ones?

Higman’s Embedding Theorem (’60s) A finitely generated group is recursively presentable if andonly if it is isomorphic to a subgroup of a finitely presented group.

Recursively Presentable: G = 〈a1, . . . , an︸︷︷︸finite

| w1, w2, w3, w4, w5 . . .︸︷︷︸recursively enumerable

〉

where a set is recursively enumerable if there exists an algorithm (a Turing machine) that can producea complete list of its elements.

Remark.

1. Higman’s embedding theorem provides a complete answer to the motivating question. Its proof how-ever would take us too far afield from the theory of solvable groups.

2. There is, somewhat astonishingly, a related theorem for solvable groups, or more specifically, formetabelian groups.

Definition. A group G is metabelian if it has derived length ≤ 2, that is it has an abelian series,

1CH CG

The following theorem was proved independently by Baumslag and Remeslennikov,

Baumslag-Remeslennikov’s Thorem (’73) Every finitely generated metabelian group embeds into afinitely presented metabelian group.


Remark. In a paper titled ‘Finitely Presented Metabelian Groups’ by Baumslag he sketches a proof ofthis theorem in three steps:

(Andrew) Reduce the proof with the Magnus embedding: embed G into W/N where W = A o H where A,Habelian

(Margarita) Show that W is embedded into a finitely presented metabelian group in the case

W = Z o Z = Γ1

(Amin) Complete the proof...

I am going to use Margarita’s step as an example both of the theorem and of how the proof works. ThenI will use Andrew’s step to reduce the problem, demonstrate Margarita’s step in full generality (not justwhen A ∼= H ∼= Z, as Baumslag did) and sketch out the remainder of the proof.

17.2 Baumslag/Margarita’s Example: Embedding Γ1 = Z o Z into Γ2

In Margarita’s first talk we saw the following presentation for the lamplighter group,

Γ1 = 〈a, s|[a, ask ] = 1, k = 1, 2, . . .〉

This is clearly an infinitely presented group. I claim it is metabelian:

Claim. Γ1 is metabelian

Proof. Surjective homomorphism

φ : Γ1 → Za 7→ 1

s 7→ s

Then ker(φ) ∼= Z∞ is abelian. Thus we have the following abelian series for Γ1,

1C 〈kerφ〉C Γ1

�

So Γ1 is an infinitely presented metabelian group. Baumslag-Remeslennikov’s theorem says that we shouldbe able to embed this into a finitely presented metabelian group.

Construct G

1. Define an injective endomorphism of Γ1, σ : Γ1 → Γ1 by σ(a) = aas, σ(s) = s.

2. Form HNN extensions of Γ1:G = 〈t,Γ1|gt = σ(g), g ∈ Γ1〉

Then G has the presentation,

G = 〈a, s, t|at = aas, st = s, [a, ask] = 1, k = 1, 2, . . .〉

Proposition. G is metabelian


Proof. Claim: G′ = 〈asi : i ∈ Z〉

(⊇) aas = at ⇒ as = a−1t−1at = [a, t] ∈ G′. Simple induction to complete (e.g. s−1ass = s−1(a−1t−1at)s =(as)−1t−1(as)t since [s, t] = 1)

(⊆) Basic fact: G/H abelain ⇒ G′ ⊆ H, since G/G′ is the largest abelain quotient of G.

Consider G modulo 〈asi : i ∈ Z〉. Get

G/〈asi : i ∈ Z〉 = 〈s, t|[s, t] = 1〉 = Z2

Proving the claim.

Now G/G′ is abelian (by proceeding proof), and G′ is free abelian of infinite rank. Therefore

1CG′ CG

is an abelian series for G of derived length 2, that is, G is metabelian.

�

Proposition. G is finitely presented. In fact, G = Γ2 = 〈a, s, t|at = aas, [a, as] = 1 = [s, t]〉

Proof. The relations [a, asi] = 1 are redundant for i > 1. Indeed, suppose [a, as

j] = 1 for j = 2, . . . , i

follows from [a, as] = 1. Then

1 = [a, asi]t = [at, (as

i)t] = [at, (at)s

i] = [aas, (aas)s

i] = [aas, as

iasi+1

] = [a, asi+1

]

Where the equalities follow (respectively) from:

1. By hypothesis that [a, asi] = 1

2. Conjugation is a homomorphism, so [x, y]t = [xt, yt]

3. [s, t] = 1 commute

4. at = aas

5. Again conjugation is a homomorphism

6. By the inductive hypothesis. (Exersize)

[aas, asiasi+1

] = aasasiasi+1

(as)−1a−1(asi)−1(as

i+1)−1

The underlined terms give [a, asi+1

]. Use the inductive hypothesis to show that the other stuff com-mutes and cancels.

Remark. So we have embedded a finitely generated, infinitely presented metabelian group Γ1 into afinitely presented metabelian group Γ2 by constructing an HNN-extension and showing that it was finitelypresented. This is the underlying idea behind the proof Baumslag-Remeslennikov theorem. Perhaps knowingnow, as we do, that Γ1 is the horocyclic product of two trees, and Γ2 is the horocyclic product of threetrees, this embedding is not so surprising. It is then perhaps more surprising that this works in general.

17.3 Proof of Baumslag-Remeslennikov Theorem

Magnus Embedding Theorem F a free group on {xi| ∈ I}, RC F . Given an isomorphism F/R→ Hby xiR 7→ hi. Let A be the free abelian group on {ai|i ∈ I}. Then the assignment xiR

′ 7→ hiai determinesan embedding of F/R′ into the wreath product A oH.

Lemma 1. G a finitely generated metabelian group. Then G can be embedded into Gab n A where A isa finitely generated ZGab-module.


Proof. G = 〈g1, . . . , gn〉, F = F 〈x1, . . . , xn〉, θ : F → G,K := ker θ

G metabelian ⇒ F ′′ ≤ K (for f ′′ ∈ F ′′, θ(f ′′) ∈ G′′ = {1})R := θ−1(G′)⇒ R = F ′KR′ ≤ K (indeed, f ∈ F ′, k ∈ K, θ[f, f ] = θ[f, k] = θ[k, k] = 1)

Let A0 be the free abelian group on {a1, . . . , an}. We can apply the Magnus embedding with H = Gab sinceF/R ∼= G/G′ = Gab,

ψ : F/R′ → A0 oGab = A(Gab)0 oGab = W B := A

(Gab)0 base group

xiR′ 7→

(xiR ∗

0 1

)where ∗ ∈ B, the base group.

Claim 1 W is finitely generated metabelian groupB and Gab both abelian.

Define N = ψ(K/R′)

Claim 2 N ≤ B

ψ(wR′) 7→(wR ∗0 1

). But R = F ′K contains K. So if wR′ ∈ N then w ∈ K ⊂ R, thus upper left entry

is trivial: (1R ∗0 1

)∼=−→ ∗ ≤ B

Claim 3 N C Im(ψ)Third isomorphism theorem: K/R′ C F/R′. Hence N = ψ(K/R′)C ψ(F/R′) = Im(ψ).

Claim 4 N CWIm(ψ) contains all xiR. W is generated by B (abelian) and by (xiR).

Finally, by the third isomorphism theorem again,

G ∼= (F/R′)/(K/R′)

so ψ induces an embedding of G into W/N = Gabn (B/N) where B/N is a finitely generated ZGab-module.

�

Remark The key point here is that we embedded G into a finitely generated metabelian group of theform H nA where H,A were abelian. So if we can prove Baumslag-Remeslennikov theorem in that case weare done.

The idea of the proof is to embed G into ascending HNN-extensions and show that eventually we are leftwith a finitely presented metabelian group. We need a way to construct these HNN-extensions, namely, weneed an endomorphism:

Lemma 2. H a finitely generated abelian group, A a finitely generated ZH-module. For each h ∈ Hthere exists a polynomial

p = 1 + c1x+ · · ·+ cr−1xr−1 + xr ∈ Z[x]

such that a 7→ ap(h) is an injective ZH-endomorphism of A.


Proof. A is a ZH-module, so multiplication by p(h) is clearly an endomorphism. Need to show it isinjective. Call polynomials of the form above ‘special polynomials’. Define

A0 = {a ∈ A : ap(h) = 0 some special p}

Then A0 is a ZH-submodule (indeed multiplication of two special polynomials is again special).

Fact: G virtually polycyclic then R = ZG is a Noetherian R-module.Since H is abelian, and since A is finitely generated, then A0 is finitely generated. Say it is generated by

b1, . . . , bs

Then there exist special polynomials pi such that bipi(h) = 0. Define

p = xp1 · · · ps + 1

This is clearly a special polynomial. Suppose ap(h) = 0 some a ∈ A. Then a ∈ A0. Therefore

a = b1f1 + · · ·+ bsfs

some fi ∈ ZH. Notice that

bip(h) = bi(hp1(h) · · · ps(h) + 1) = hp1(h) · · · bipi(h)︸︷︷︸=0

· · · ps(h) + bi = bi

Therefore0 = ap(h) = (b1f1 + · · ·+ bsfs)p(h) = b1f1 + · · ·+ bsfs = a

Thus a = 0 and we see that the map is injective.

�

Proof of Baumslag-Remeslennikov TheoremAssume G = H n A where A,H abelian, is a finitely generated metabelian group. If H were finite then Gwould be polycyclic and therefore finitely presented. So assume H infinite,

H = 〈h1〉 × · · · × 〈hr〉 × · · · × 〈hn〉

where h1, . . . , hr have infinite order, and hqii = 0, i = r+1, . . . , n. By lemma 2 there exist special polynomials,

p1, . . . , pr

such that a 7→ api(hi) determines an injective ZH-endomorphism of A, say τi.

Constructing the HNN-extensions

1. G0 = G = H nAExtend τ1 to an injective endomorphism of G0 acting as the identity on H (H abelian). Define,

G1 = 〈t, G0|gt10 = gτ10 , g0 ∈ G0〉

2. Extend τ2 to G1 by requiring it to act as identity on the abelian subgroup 〈H, t1〉. Define,

G2 = 〈t2, G1|gt21 = gτ21 , g1 ∈ G1〉

3. Repeat this r times resulting in Gr.


Claim 1 Gr = G = QnA, where Q = H × 〈t1〉 × · · · × 〈tr〉 and A = A〈t1,...,tr〉 is the normal closure of Ain 〈A, t1, . . . , tr〉.

Pf: Starting with G0 = HnA. Adding ti’s and forcing them to commute with H and with a defined actionof ti on A. So we expect a semi-direct product of this form. The slight question is, why A. To see thisconsider conjugation by a negative power of ti. For example, if g = (h, a) ∈ G1

t1gt−11 = τ−11 (g)

τi are not necessarily surjective. Hence we take the normal closure. This is easier to visualise with thesimpler example BS(1, 2) = 〈a, t|at = a2〉.

�

Claim 2 G embeds in G and G is metabelian.

Pf: G is generated by the elements

h1, . . . , hn t1, . . . , tr a1, . . . , am︸︷︷︸generators for A

Similarly to before, 1CQCG is an abelian normal series for G because both Q and A are abelian.

�

Constructing G∗

What relations do we have?

1. [hi, hj ] = [ti, tj ] = [hi, tj ] = [ai, aj ] = 1

2. hqii = 1 for i = r + 1, . . . , n

3. atji = aipj(hj) for i = 1, . . . ,m, j = 1, . . . , r

Fact: ZH Noetherian⇒ A (a ZH-module) is finitely presented as finitely generated is equivalent to finitelypresented for modules over Noetherian rings. Put these relations in,

4. ari11 ari22 · · · arimm = 1 for i = 1, . . . , k with rij ∈ ZH

Finally need relations to ensure the normal closure of 〈a1, . . . , am〉 in G is abelian,

5. [aνi , aµj ] = 1 for ν, µ of the form hu11 · · ·hunn where 0 ≤ ui ≤ di for di the degree of pi, when 1 ≤ i ≤ r,

and 0 ≤ ui < qi for r + 1 ≤ i ≤ n

DefineG∗ = 〈h1, . . . , hn, t1, . . . , tr, a1, . . . , am|1, 2, 3, 4, 5〉

The Conclusion There is a surjective homomorphism

G∗ → G

Hall (’54) Finitely generated metabelian groups satisfy max-n, the maximum condition on normal sub-groups. That is, every normal subgroup is finitely generated.

1. Prove G∗ metabelian

2. Then G∗ satisfies max-n

3. Hence G is finitely presented: Indeed, G = G∗/N some normal subgroup N C G∗. Now since G∗

satisfies max-n N is finitely generated. Therefore G is too.


Lemma. G∗ is metabelian.

Sketch of Proof:

1. If we show that A∗ = 〈a1, . . . , am〉G∗

is abelian, then

G∗ = 〈h1, . . . , hn, t1, . . . , tr, a1, . . . , am|1, 2, 3, 4, 5〉G∗/A∗ = 〈h1, . . . , hn, t1, . . . , tr|[hi, hj ] = [hi, tj ] = [ti, tj ] = 1〉 is abelain

hence 1CA∗ CG∗ is an abelian series of derived length 2 and therefore G∗ is metabelian.

2. To prove A∗ is abelian one uses the special polynomials pi. It involves bashing out even largercommutators then in the example of embedding Γ1 into Γ2. The idea is similar enough.

�


18 Sigma Invarients (Teddy Einstein)

18.1 Σ-invariants

Definition. A character of a group G is a homomorphism χ : G→ R. Observe

G

��

χ// R

Gab

==

where Gab = Zk ⊕ torsion. Hom(G,R) ∼= Rk

Definition. (Character Sphere) χ, ψ characters on G. χ ∼ ψ ⇔ χ = cψ some c > 0. [χ] is theequivalence class of χ. The character sphere is,

S(G) = {[χ] : χ 6= 0}

Remark. Thinking of this equivalence relation as identifying rays in the space of characters providesmotivation for the name ‘character sphere’.

Definition. For each χ : G→ R define

Gχ = {g ∈ G : χ(g) ≥ 0}

Clearly if χ ∼ ψ then Gχ = Gψ.

Definition. (Σ0-invariant) Let G be a finitely generated group, and let A be finitely generated overZG. Define

Σ0(G;A) = {[χ] : A f.g. as a ZGχ-module}

Let M = G′/G′′. We consider this as a module over ZG where scalar multiplication is by conjugation:g ∈ G,m = [g1, g2]G

′′ ∈M then

g ·m = mg = ([g1, g2]G′′)g = [g1, g2]

gG′′ = [gg1 , gg2 ]G′′ ∈M

Theorem. M = G′/G′′ a module over ZG by conjugation. If G is finitely presented and solvable then

Σ0(G/G′;M) ∪ −Σ0(G/G′;M) = S(G)

(where we are taking the union with the antipode on the sphere).

Remark. The converse holds if G is metabelian.

We can generalise this,

ΣG′ = {[χ] ∈ S(G) : G′ f.g over a f.g. submonoid of Gχ}

Say G is finitely generated by X. Γ(G,X) the Caley graph wrt X.

Definition. (Σ1-invariant) Let Γχ be the subgraph of Γ(G,X) with vertices g such that χ(g) ≥ 0, andwith edges: if g, h ∈ Γχ connect an edge if there was one in Γ(G,X).

Σ1(G) = {[χ] : Γχconnected}


Examples.

1. Z = 〈t〉. The character χ is completely determined by the choice χ(t). If we choose χ(t) = 1 then [χ]represents all characters ψ where ψ(t) > 0. Then −[χ] represents all characters ψ where ψ(t) < 0.

S(Z) = {±[χ]} where χ(t) = 1.

2. Z2 − 〈a, b|[a, b] = 1〉. Character χ is determined by χ(a), χ(b). Eg χ(a) = 1 = χ(b). So χ vanishes ony = −x.

Γχ is connected. Here,S(Z2) = Σ1(Z2)

3. G = BS(1, 2) = 〈a, u|au = a2〉 = Z[12 ] o Z where ((x, 2m)(y, 2m′) = (x + 2my, 2m+m′), u = (0, 2), a =

(1, 20))

Consider the Cayley graph drawn on a plane. We build it inductively: We need to represent therelation u−1au = a2. Fill out the plane with this relation starting at a chosen point, say, the origin.

Now, if we chose a different we still need to see this relation, so add it in. Think of this as projectingdifferent leaves into the same plane. At ‘height’ h we project 2h leaves into the plane.


The character is determined by χ(a) and χ(u). But,

χ(a) = χ(u−1au) = χ(a2) = 2χ(a)

⇒ χ(a) = 0

So we get that the character is determined by χ(u), and we have the same as in example 1,

S(G) = {±[χ0]} where χ0(u) = 1

Claim. [χ0] /∈ S(G) but −[χ0] is.

(a) Γχ0 not connected: (1

2+ z, 2m

)m ≥ 0

check that (1

2+ z, 2m

)(0, 2±1

)=

(1

2+ z, 2m±1

)check (

1

2+ z, 2m

)(±1, 20

)=

(1

2+ z ± 2m, 2m

)m ≥ 0 ⇒ z ± 2m ∈ Z. First coordinate is an integer. No path from (12 , 2

m) to the identity inΓχ0 . Γχ0 is not connected.

(b) −Γχ0 is connected in Z[12 ]: Arbitrary point (t, 2−m) where m ≥ 0. There exists n ∈ N, z ∈ Zsuch that 2−nz = t. Point (2−nz, 2−m). Walk to (2−nz, 2−n) by a ‘−z’ times,

(2−nz, z−n)(±1, 20) = (2−n(z ± 1), 2−n)

Get to (0, 2−n). Walk by u ‘n’-times to get to the identity. Hence Γχ0 is connected, therefore−[χ0] ∈ Σ′(G).

Theorem. Σ1 is invariant under choice of generators.

Proof. X1, X2 finite generating sets. WLOG X2 = X1 ∪ {z}. Write Γ = Γ(G,X1) and Γ′ = Γ(G,X2). IfΓχ is connected then Γχ is a subgraph of Γ′χ, hence Γ′χ is connected.

Suppose Γ′χ is connected.

Definition. p a path. Let vχ(p) = min{χ(g) : g vertex on p}

So vχ ≥ 0⇔ p lies on Γχ. Let z = w1 · · ·wn be a word over X±1 . There exists t ∈ X±1 with χ(t) > 0. Wantk such that χ(tk) = k. χ(t) ≥ |vχ(w1, . . . , wn)|. Take g, h ∈ Γχ. Then there exists a path p′ from t−kgtk tot−khtk in Γ′χ.

Take

p′ = x1x2 · · ·xm xi ∈ X2

toq = y1y2 · · · yl

by replacing all z±1 by (w1w2 · · ·wn)±1. Then vχ(g) ≥ −|vχ(w1w2 · · ·wn)|. Shift a by tk. Path tkq from gtk

to htk gives vχ(tkq) = vχ(q) + kχ(t) ≥ 0.

�


18.2 Results

Theorem. G a f.g. group with no non-abelian free subgroups. G finitely presented then

Σ′(G) ∪ −Σ′(G) = S(G)

Example. −[χ0] ∈ BS(1, 2) ∈ S(BS(1, 2))

Corollary. If G is finitely presented as above, and G/G′ has rank ≥ 2 over Z, then there exists a normalsubgroup N CG such that G/N is cyclic and N = kerχ for some character χ.

Theorem. χ is rank 1, t ∈ G with χ(t) = 1. Then [χ] ∈ Σ1(G) if and only if ∃ H ≤ N = kerχ such that

t−1Ht ⊆ H

andN =

⋃l∈N

tlHt

(Here rank of χ is the rank of χ(G) in R over Z)


19 Sigma Invarients (Yash Lodha)

There are two kinds of sigma invarients:

1. Homotopical: input the group G (usually finitely generated)

2. Homological: input (G,A) for a group G and a ZG-module A

Consider the vector space Hom(G,R) consisting of characters χ : G → R. This is a real vector spaceisomorphic to Rk where k is the torsion free rank of G. If X is a path connected topological space we haveπ1(X) = G, Hom(π1(X),R) ∼= Hom(H1(X),R) ∼= H1(X,R).

Recall from last time,Gχ = {g ∈ G : χ(g) ≥ 0}

and recall that Gχ = Gcχ for all c > 0. So this set of mono ids is parametrised by the set of rays inHom(G,R). Identify this with the unit sphere, call it S(G).

Now consider the pair (G,A) where G is a finitely generated group and A is a ZG-module. Then

ΣA = {[v] ∈ S(G) : A finitely generated as a ZG−module}

Lets try and clarify this somewhat opaque definition. Let C(A) = {y ∈ ZG : ∀ a ∈ A, g · a = a} be thecentraliser (suppose action of g is by conjugation). Let I(A) = C(A)− 1 = {g − 1 : g ∈ C(A)}. Now for allg ∈ I(A) and for all a ∈ A, g · a = 0.

Proposition. [v] ∈ ΣA if and only if there is a d ∈ C(A) such that v(d) > 0.

Proof. Assume there exists d ∈ C(A) such that v(d) > 0. For all g ∈ G there exists m > 0 such thatv(g) < v(dm) = mv(d). Hence gdm · a = g · a and v(gdm) > 0⇒ gdm ∈ ZGv.

If {a1, . . . , ak} is a finite generating set for A as a ZG-module.

Σigiai = Σigidmiai

for gidmi ∈ Gv.

Conversely, assume [v] ∈ ΣA. Let {a1, · · · , ak} be a generating set for A as a ZGv-module. Take d ∈ ZGvsuch that v(d) > 0. Say

d−1ai = Σiµi,jai

Hence

Σj(δij − dµi,j)aj = 0

Therefore det(δij − dµij) ∈ I(A). But δij − dµi,j = 1 − dµ for µ ∈ ZGv. Therefore dµ ∈ C(A). v(dµ) =v(d) + v(µ) > 0.

�

Definition. ΣA = ∪λ∈C(A){[v] ∈ S(G) : v(d) > 0}

Proposition.

1. ΣA is an open subset of S(G)

2. ΣA = ΣZG/I where I = I(A).

3. A′ → A→ A′′ of ZG-modules then ΣA = ΣA′ ∩ ΣA′′


Proof.

1. d : S(G)→ R by d(v) = v(d). So 1 follows since the sets in the definition are open sets.

2. ZG/I as a ZG-module has annihilation I = I(A). Therefore C(A) = C(ZG/I), and ΣA = ΣZG/I .

3. ΣA ⊆ ΣA′′ follows since if A is finitely generated as a ZGv-module then so is A′. Similarly, ΣA′′∩ΣA′ ⊆ΣA, since if A′′, A′ finitely generated wrt ZGv then so is A. We need ΣA ⊆ ΣA′ . I(A) ⊆ I(A′) thereforeC(A) ⊆ C(A′). So this follows by definition.

19.1 Finiteness Properties

If A is a ZG-module, a projective resolution for A is an exact sequence

· · · → Pi → Pi−1 → · · · → P0 → A→ 0

where each Pi is a projective ZG-module.

We are interested in the case A = Z. Examples:

1. One always has the free resolution.

2. Take X a K(G, 1), that is, a connected CW complex such that π1(X) = G and πi(X) = 0 for i > 1.This is unique up to homotopy type and X is contractable.An example of a free resolution of Z overZG is the chain complex of X.

Definition. (FPn) A group G is of type FPn if Z has a projective resolution

· · · → Pi → · · · → P0 → Z→ 0

Remark.

1. FP1 ⇔ finitely generated

2. FP2 ⇐ finitely presented. There is an example (by Bestvina-Feign) of an FP2 which is not finitelypresentable.

Definition. (Fn) A group G is of type Fn if it admits a K(G, 1) with a finite n-skeleton.

Remark.

1. F1 ⇔ FP1

2. F2 ⇔ finitely presentable i.e ⇒ FP2.

3. In general Fn ⇒ FPn

Theorem. If G is an abelian group and A is a ZG-module then ΣA = S(G) if and only if A is finitelygeneratable as a group.

Recall that a group G is metabelian if and only if there is an exact sequence

1→ A→ Gφ−→ Q→ 1

where A and Q are abelian. A can be viewed as a ZQ-module as follows: qa = l−1al where l ∈ φ−1(Q), andwhere q ∈ Q, a ∈ A.


Theorem. (Bieri-Strebel) G is finitely presentable if and only if ΣA ∪ −ΣA = S(Q)

A is 2-tame as a ZQ-module if and only if ΣA ∪ −ΣA = S(Q). In general, for A a ZQ-module, then A ism-tame if any m point subset of ΣC

A is contained in an open hemisphere.

Conjecture. (Bieri-Groves) Given the short exact sequence

1→ A→ G→ Q→ 1

G is of type FPm if and only if A is m-tame as a ZQ-module.

Remark. This conjecture is open. There are some special cases in which it has been proved:

1. If G is of type FPm then A⊗Z k is m-tame as a KQ-module for any field k

2. The full conjecture holds for groups of finite profer rank

3. The only if part holds true if the additive group of A is torsion or if the extension is split.

Will end with a criterion for FPn:

Definition. Let G be a group. Then X is a G-Space if X is a connected CW-complex with G acting onX by cell preserving homeomorphisms.

Definition. A G-space X is m-good if

1. Hi(X) = 0 for i < r

2. Stabilisers of each cell in G is of type Fn−p for 1 ≤ p ≤ n− 1

Let D be a collection of G-invariant subcomplexes such that D = {Xα}A where A is a poset and Xα/G hasa finite n-skeleton.

Theorem. (Brown) G is of type FPn if and only if H1(Xα) is essentially trivial.

D here is a directed system of subcomplexes: that is α ≤ β ∈ A if Xα → Xβ. For all α, β ∃ γ such thatXα → Xγ and Xβ → Xγ .

Hi(Xα) is essentially trivial if for all α there exists β such that Hi(Xα)→ Hi(Xβ) is trivial.


20 Automatic Groups (Scott Messick)

Here is an example of a finite automata: this is a word acceptor for Z:

Any double circle is an accepting state. The idea is input your word on the left (hence the arrow comingin), follow the arrows depending on your word and if you are a reduced word in Z you will end up in anaccepting state. e.g. aaaa is a reduced word, aaa−1a−1aa−1 is not a reduced word.

Given words σ ∈ Σ∗, τ ∈ ∆∗ we have⊗ (σ, τ) ∈ (Σ′ ×∆′)∗

where Σ′ = Σ ∪ {�} where � represents the the empty word. Consider σ = abs, τ = cccdc. Then,

⊗ (σ, τ) =

(a b a � �c c c d c

)Thus we can say a relation on strings is FA-recognisable.

(a a a �a a a a

)


Definition. (CGA) A Cayley graph automatic (CGA or graph automatic) structure for a group G withsemigroup generators X consists of

1. A regular (FA-recognisable) language R over some alphabet Σ

2. A surjection ν : R → G

3. An FA-recognising equality i.e. {(σ, τ) : ν(σ) = ν(τ)}

4. For each x ∈ X an FA-recognising right multiplication by x i.e. {(σ, τ) : ν(σ)x = ν(τ)}We get an automatic structure if X = Σ and ν interprets strings as group words.

Lemma. In reverse binary representation, integer addition is FA-recognisable.

Proof. Here is an automaton for non-negative integers: first consider an example of reverse binary addi-tion:

Starting from the left, we add 1 and 0 to get 1. This is represented by

101

. Then we add 1 and 1, we get

a 0 in this column, but we ‘carry the 1’, this is represented by

110

, and so on... We get the automaton:

Proposition. H3(Z) is CGA

Proof. Represent

1 a c0 1 b0 0 1

as ⊗(a, b, c) Multiplication is given by,

⊗(a, b, c)A−→ ⊗(1 + a, b, c)

⊗(a, b, c)B−→ ⊗(a, 1 + b, c)

⊗(a, b, c)C−→ ⊗(a, a+ b, 1 + c)

�


Remark. Any regular subgroup of the full monoid of integer matrices, Mn(Z) is CGA.

Lemma. Suppose f : Dn → D is FA-recognisable. Then the value f(x1, . . . , xn) can be computed inlinear time.

Proof. Fix an FA M recognising f . Calculate the set of states

S = {s : ∃ y such that |y| ≤ maxi|xi| and M running ⊗(x1, . . . , xn, y) finishes in state s}

To do this recursively calculate

Sj = {s : ∃ y |y| = j and M runs ⊗(x|j , . . . , xn|j , y) finishes in s}

Getting Sj+1 from Sj required O(1) time. Actually, also track witnesses (at least one for each s ∈ Si).

If S has an accepting state, we’re done. If not, continue finding Sj for j > maxi |xi| (pad xi’s with �’s). Forsome j < maxi |xi|+ (number of states), Sj will have an accepting states.

�

Theorem. If G is a CGA then the word problem for G is solvable in O(n2) time.

Remark. The proof follows from the above lemma and the fact that finite automata can be done in lineartime.

20.1 Closure Properties

The class of graph automatic groups is closed under:

1. Finite extensions

2. Free products

3. Amalgamated free produce over finite groups

4. Direct products

5. Semi-direct products Goφ H where for all g ∈ G, φ(g) ∈ Aut(H) is automatic.

6. Regular finitely generated subgroups.

Remark. If G is finite, G o Z is CGA.

Theorem. For n ≥ 1, BS(1, n) is CGA

Proof. Recall BS(1, 2) ∼= Z n Z[12 ] where (n, p) · (m, q) = (n+m, 2mp+ q).

Write the elements as (n, m2k

) where m is odd. Represent this as ⊗(n,m, 2k), which we will represent in

reverse binary again (note that this works nicely for 2k).(n,m

2k

)· (0, 1) =

(n,m+ 2k

2k

)(n,m

2k

)· (1, 0) =

(n+ 1,

m

2k−1

)�

Theorem. If G is nilpotent of class ≤ 2 then G is CGA.


20.2 Biautomatic Groups

Definition. A (graph) biautomatic structure for G is a (graph) automatic structure for left and rightmultiplication over the same language.

Theorem. Graph biautomatic groups have solvable conjugacy problem.

Proof. (Sketch) Regular languages are closed under existential quantifiers. So for p ∈ G

{(σ, τ) : ν(σ) · p = ν(τ)}

is regular. Because it’s biautomatic we get,

{(σ, τ) : ν(σ) = p · ν(τ)}

is regular. Fix g, h ∈ G.{(σ, τ) : ν(σ) = ν(τ) and g · ν(σ) = ν(τ) · h}

then that is regular.

�

20.3 Open Problems

1. How many (solvable) groups are graph automatic?

2. Is there are geometric relationship, i.e. a necessary and/or sufficient condition?

3. Is every (graph) automatic group (graph) biautomatic?

4. Is the automatic group isomorphism problem decidable?

5. Is Thompson’s group F (graph) automatic?


21 Quasi-Isometric Rigidity (Iian Smythe)

Assume all groups finitely generated throughout.

Basic Quasi-Isometry Facts

1. If S, S′ are finite generating sets for G then Cay(G,S) is quasi-isometric to Cay(G,S′)

2. Quasi-isometry is an equivalence relation on the class of metric spaces

3. IF G ∼= H then G is quasi-isometric to H

4. G is quasi-isometric to a point if and only if G is finite.

Fundamental Lemma Suppose G is countable, discrete and acts geometrically (i.e. properly discon-tinuously, cocompactly by isometries) on a proper, geodesic metroc space (X, d), then G is finitely generatedand quasi-isometric to (X, d). In fact, for any fixed x0 ∈ X the map g 7→ g · x0 is a quasi-isometry.

(?) Suppose G and H are such that G′ ≤ G,H ′ ≤ H are of finite index and F CG′,K CH ′ are finite withG′/F ∼= H ′/K

Claim. G is quasi-isometric to H

Proof. Let G′ act on Cay(G,S), this is geometric. Let G′ act on Cay(G′/F, T ), this is geometric. Similarlyfor H,H ′, H ′/K

�

Definition. If G and H are as in (?), we say G and H are virtually isomorphic.

The claim says that virtually isomorphic groups are quasi-isometric.

Fact. Virtually isomorphic is an equivalence relation on the class of all groups (transitivity takes a bit ofwork).

We should note that quasi-isometric does not in general imply virtually isomorphic. Here’s an example ofthis.

Example. Let Hyp(2,Z) be all matrices A ∈ SL2Z such that A is diagnosable over R and A2 6= I. So ifλ, λ′ are eigenvalues of A, neither has absolute value 1.

Let GA = Z2 oA Z where 〈a〉 = Z, and a acts on Z2 by multiplication by A. By diagonalising A, we get anembedding

GA → Sol = R2 oD R

where

D(t) =

(et 00 e−t

)t ∈ R

It follows that GA is torsion-free (since Sol is). This embedding makes GA a lattice in Sol, and its actionon Sol is geometric. So GA is quasi-isometric to Sol.

Pick A,B ∈ Hyp(2,Z) such that Am is not conjugate to Bm for any n,m ∈ Z\{0}

A =

(2 11 1

)B =

(3 21 1

)1

2(3±

√5) 2±

√3


Let HA be a finite index subgroup of GA. HA will intersect the maximal rank 2 abelian subgroup Z2 of GAin a rank 2 abelian subgroup, LA. The image of HA under the quotient map GA → GA/Z2 ∼= Z must beinfinite cyclic, say generated by n. It follows that

HA∼= Z2 oAn Z

and similarlyHB∼= Z2 oBm Z

But any isomorphism HA → HB will carry LA → LB and induces a conjugation of An to Bm. Thereforewe have shown that these quasi-isometric groups are note virtually isomorphic.

Quasi-isometric rigidity is when quasi-isometry does imply virtual isomorphism.

Definition.

1. A group G is quasi-isometrically rigid if whenever H is quasi-isometric to G, H is virtually iso-morphic to G

2. A class of groups C is quasi-isometrically rigid if whenever H is quasi-isometric to some C ∈ Cthen there exists C ′ ∈ C such that H is virtually isomorphic to G′.

Remark We get increasingly coarse partitions by the equivalence relations

∼= ⇒ virtually isomorphic ⇒ quasi-isometric

and we want to understand when the last implication goes both ways.

In general, if a class of groups is defined by a quasi-isometric invariant, then the class is quasi-isometricallyrigid.

Examples.

1. The class of finite groups

2. Any class defined by its growth rate. For example, nilpotent groups (by Gromov’s theorem - i.e.polynomial growth gives you nilpotent)

3. Any class defined by its Dehn function/ isoperimetric inequality. For example solvable word problem(linear), hyperbolic groups (linear isperimetric inequality).

4. Amenable groups (G with finite generating set A is amenable if and inly if for every ε > 0 ∃ F ⊆ Gfinite such that |∂SF | ≤ ε|F |)

5. Finitely presented groups. Sketch: Let S be a finite generating set for G, let L(k) denote the set ofall loops in Cay(G,S) of the form φ · ψ · ψ−1 where φ is a path e→ x and ψ is a loop at x of length≤ k. Fact: G is finitely presented if and only if L(k) generates π1(Cay(G,S) for some k︸︷︷︸

q.i invariant

.

Theorem. (Gromov) Finitely generated abelian groups is a quasi-isometrically rigid class, in fact, Znis quasi-isometrically rigid.

Theorem. (Stallings) The class of non-abelian groups is quasi-isometrically rigid.

Theorem. (Farb-Mosher) Each solvable Baumslag-Solitar groups BS(1, q) is quasi-isometrically rigid.

Theorem. (Eskin-Fisher-Whyte)

1. Being a lattice in a Sol is quasi-isometrically rigid

2. F o Z and F ′ o Z are quasi-isometric if and only if there exists d, r, s such that |F | = ds and |F ′| = dr


Non-Examples

1. Theorem. (Burger-Moses) The class of simple groups is not quasi-isometrically rigid

2. Theorem. (Dyubina) The class of finitely generated solvable groups is not quasi-isometrically rigid.

Pf: We start with a lemma,

Lemma. Let A,B be bi-Lipschitz equivalent. Then for any group C, C oA and C oB are bi-Lipschitzequivalent. (where C oA means C is acting on A)

Let D be a finite non-solvable group (e.g. A5). Let G = Z o Z, H = Z o (Z ⊕ D). So G and H arebi-Lipschitz. G is solvable, but H contains ⊕ZD, so any finite index subgroup of H will contain acopy of D (it will actually contain ⊕CD where |C| = S\S0). So H is not virtually isomorphic to anysolvable group.

�

Remark. By taking A and B to be a solvable and a non-solvable finite group (resp.) of the sameorder, one can get Z o (Z⊕A) and Z o (Z⊕B) to be isometric, but the latter not virtually isomorphicto a solvable group. (So solvability is not a geometric property)

Open Questions.

1. Is the class of polycyclic groups quasi-isometrically rigid?

2. Is the class of finitely presented solvable groups quasi-isometrically rigid?


22 Quasi Isometric Rigidity (Kristen Pueschel)

Positive results for QI rigidity

Mostow Rigidity Finite volume hyperbolic manifolds of dimension at least 3 are determined up toisometry by their fundamental groups

Eskin, Fischer, Whyte Let Λ be quasi-isometric to the discrete cocompact lattice Γ ⊆ Sol. Then Λ isvirtually a discrete cocompact subgroup of Sol.

Farb-Mosher

1. BS(1, n) ' QIBS(1,m) if and only if n = rα,m = rβ if and only if commensurable

2. BS(1, 2) is QI rigid

22.1 Proof Form for QI Rigidity

Step 1 Work out QI(X) = {f : X → X : f quasi-isometry}/f ∼ g if they differ within a constant

Example (EFW) Bulk of the paper spent doing this.

Any (K,C)-quasi-isometry φ of Sol, is within bounded distance of a height-respecting quasi-isometryφ.

Step 2 If f : Γ → X is a quasi-isometry, then there is an induced homomorphism Γ → QI(X) by g 7→f · Lg · f−1. Try to show this map has finite kernel.

Step 3 Show QI(X) acts on the boundary of X. Study the dynamics of this action.

22.2 Xn and its Boundary

BS(1, n) has a Cayley 2-complex, Xn.

Facts

1. Xn has embedded hyperbolic planes - describe these as lines in the tree Tn (the cross section)

2. There are trees in Xn isometric to Tn - describe these as sections of Xn to H2 (think of collapsingeverything up to a given height). These correspond to vertical lines in H2. So there is a tree for eachelement of R. Code these as elements of Qn - the n-adic rationals,

∑i∈Z ξi − ni for i ∈ {0, . . . , n− 1}.

(Note: Eventually zero in the negative direction)

Definition

1. The upper boundary ∂u(Xn) = {set of hyperbolic planes}

2. The lower boundary ∂l(Xn) = {set of trees} ' R

BS(1, n) acts on these boundaries: BS(1, n) = 〈a, t : tat−1 = an〉, action is defined as:

∂l(Xn) : a · x = x+ 1

t · x = nx

∂u(Xn) : a · x = x+ 1

t · x = nx


Proposition. If f : Xn → Xm is a quasi-isometry then for each hyperbolic plane Q ∈ Xn (reps. tree T )there is a unique hyperbolic plane Q′ ∈ Xm (resp. T ′) such that

dH(f(Q), Q′) ≤ C

This induces a map between boundaries fu : ∂l(Xn)→ ∂l(Xm). fu and fl are bi-Lipschitz maps.

Step 1 Identify QI(BS(1, n))

Theorem. QI(BS(1, n)) ' Bilip(R)×Bilip(Qn)

Step 2 QI rigidity condition: We have an induced homomorphism g 7→ f · Lg · f−1 This action is properlydiscontinuous and cocompact (note: this is actually a quasi-action, and as such we have things closeto properly discontinuous and close to cocompact, but in QI(Xn) this is fine).

Definition. Suppose X,Y are two topological spaces. T (X,Y, Y ) = {(x, ν, ξ) : x ∈ X, ν, ξ ∈ Y, ν 6= ξ}.Similarly T (X,X, Y ) = {(x, y, ξ) : x, y ∈ X, ξ ∈ Y, x 6= y}. Then G acts uniformly biconvergently on(X,Y ) if the diagonal actions T (X,X, Y ) and T (X,Y, Y ) are properly discontinuous and cocompact.

Proposition. BS(1, n) acts uniformly biconvergently on (∂l(Xn), ∂u(Xn)).

Pf (Sketch). We can send T (X,Y, Y )π−→ Xn continuous, proper isometry.

d(π · (fl(x), fl(y), fl(ξ))) f(π(x, y, ξ)) ≤ c

�

Theorem. (Contraction Property) Let G be a finitely generated group quasi-isometric to Xn. Let(ρl, ρ

u) : G → (Bilip(R), Bilip(Qn)). For every compact K ∈ Qn, and every open subset U ⊆ Qn thereexists g ∈ G such that g ·K ⊆ U .

ρl : G→ Bilip(R) projection. Result: there is a φ ∈ Bilip(R) such that for all g ∈ G φ(g) = φ ·ρl(g) ·φ−1 ∈Aff(R), φ(g)x = mgx+ bx

G

θ��

1 // A

��

// Γ

��

// Stretch(Γ)

��

// 1

1 // Isom(R) // Aff(R) // SL(1,R) // 1

Construction principle:

1. θ has finite kernel

2. Stretch(Γ) ∼= Z. Γ = A〉Z ...some work... Γ = 〈B, t : t−1Bt = B1 where B1 ⊆ B ⊆ A.


23 Random Walks on Solvable Groups (Johannes Cuno)

23.1 Random Walks on Graphs and Groups

On a Graph Require

1. locally finite

2. connected

3. a probability function that determines the probability of moving to a given adjacent vertex:

p : V × V → [0, 1]

∀ v ∈ V :∑

w∈V p(v, w) = 1.

4. concentrated on neighbours. That is, supp(p(v, ·)) = {neighbours of v}

k-step transition function:

p(k)(v, w) =∑w∈V

p(k−1))(v, w′) · p(w′, w)

Definition. The random walk is irreducible if and only if ∀ v, w ∈ V ∃ k ∈ N0 : p(k)(v, w) > 0

A random walk X = (X0, X1, . . .) takes values in V∞ = {(v0, v1, . . . , ) : vi ∈ V }. Define a σ-algebra: acylinder set

Pv = {(v1, v2, . . . , vn, ∗, ∗, · · · )}= δv,v0p(v0, v1) · p(v1, v2) · · · p(vn−1, vn)

where we have the first, say, n values determined, and afterwards anything can happen. Define

Σ = σ(cylinder sets) = smallest σ-algebra containing the cylinder sets

This makes (V∞,Σ,Pv) a measure space.

On a Groups G a finitely generated group, S ⊆ G a finite generating set, µ a measure on G with supportS ∪ S−1 (tells you the probability of going to an adjacent edge). Here,

Pe = {g0, g1, . . . , gn, ∗, ∗, . . .)}= δe,g0µ(g−10 g1)µ(g−11 g2) · · ·µ(g−1n−1gn)

23.2 Examples

Riffle ShuffleG = S52 µ according to Gilbert-Shannon-Reeds.After one Rips shuffle we can’t get too far away from the starting arrangement. Indeed, the relative orderof each of the cut decks remains the same. After repeatedly shuffling we would like to know how long ittakes to get ‘close to a uniform distribution’. Need a notion of distance between probability distributions:

dTV (p(n)(e, ·), unifS52(·)) =1

2

∑g∈G|p(n)(e, g)− unif(g)|

Here are some values of this function:

n 1 2 3 4 5 6 7 8 9 10

dTV 1.000 1.000 1.000 1.000 0.924 0.624 0.312 0.161 0.083 0.041

So we get exponential decay after a certain cut-off time. That is, we see that after 7 or 8 riffle shuffles weget very close to the uniform distribution.


Simple Random Walk on ZdG = Zd, S = standard generating set.

µ(g) =

{12d g ∈ S ∪ S−10 else

Definition. w ∈ V is recurrent if one of the following equivalent conditions holds:

1. Pv{∃ ∞ many n ∈ N0 : Xn = w} = 1

2. Pv{∃ n ∈ N0 : Xn = w} = 1

3. Ev[|{n ∈ N0 : Xn = w}|] =∑∞

n=0 Pv{Xn = w} =∞

Remark Intuitively recurrence says that the random walker will return to the given point infinitely manytimes. Recurrence does not depend on w ∈ V . If not recurrent then say you are transient. We seetransience then as saying that after a point the walker never returns to a given point.

Theorem. (Polya) The simple random walk X = (X0, X1, X2, . . .) on Zd is recurrent for d ∈ {1, 2} andtransient for d ≥ 3.

23.3 Flows and Energy

Definition. A flow is a function from the oriented edges

f : E → R

We think of the flow as follows: suppose we ‘pump water’ through our vertex, and imagine the orientededges connected to the vertex as pipes - then the flow tells us how mush water flows through each pipe.

Definiiton. The energy is∑

e∈E f(e)2.

Theorem (Lyons) X reversible random walk (e.g. simple random walk on a graph). Then transient ifand only if ∃ f with finite energy.

We can use this to prove Polya’s theorem.

Proof of Polya’s Theorem

• d = 1: Consider an arbitrary flow f : E → R. There must a constant non-zero flow in at least onedirection (left or right), indeed there is just one choice at the beginning (how much flows left, howmuch flows right) and then everything is fixed. Therefore the energy must be infinite and thus therandom walk is recurrent.

• d = 2: Consider the spheres of size n and n+ 1. One observes that there are 8n+ 4 edges connectedthe n-sphere to the (n+ 1)-sphere. Say |En| = 8n+ 4|.

∑e∈E

f(e)2 =

∞∑n=0

∑e∈E

f(e)2

≥∞∑n=0

1

|En|

(∑e∈E

f(e)2

)︸︷︷︸

=1

=∞∑n=0

1

8n+ 4=∞

�


Remark. Using the flow criterion from Lyon’s theorem one sees that recurrence/transience is independentof choice of finite generating set S.

23.4 Growth of Groups and Return Probability

Another question one can ask is, what are the return probabilities. That is, define the return probability

φ(n) = Pe{X2n = e}

Motivating the forcing of parity above (need n even) is the example of simple random walks on Zd. Indeed,it is impossible to return to the origin in Zd after an odd number of steps. In this example one obtains theresult that,

φ(n) � n−d/2

Where a(n) � b(n) means : a(n) � b(n) (that is ∃ c > 0 : a(n) ≤ cb(n)), and vice versa.

How does φ(n) behave?

1. V (n) = |Be(n)| � nd ⇔ φ(n) � n−d/2

2. V (n) � exp(nα) for some α ∈ [0, 1]⇒ φ(n) � exp(−nαα+2 )

In the case of exponential growthφ(n) � exp(−n1/3)

23.5 Advertisement for Next Talks

Define a functionh : V → R

Think of this as saying how much money you get at a given vertex. Want to understand how the expectedpayoff varies as you random walk. Define the function

Ph(g) =∑g′∈G

p(g, g′)h(g′)

which tells you the expected earning of the next step. We think of defining a discrete analogue of theLaplacian as Ph− h, and then a harmonic function is one where this is 0, that is the set

H = {h : Ph = h}

and also superharmonic functions

S = {h : h ≥ Ph}

Remark. Notice first that any constant function h is harmonic (and therefore superharmonic). Whatabout non-constant harmonic functions?

Theorem A random walk on (G,Σ,Pv) is recurrent if and only if S≥0 = {constants}.

Questions.

1. Martin boundary: ∃ non-constant h ∈ H≥0

2. Poisson boundary ∃ non-constant bounded h ∈ H

Tail σ-algebraΣ = σ(cylinder sets). (V∞,Σ,Pe). Consider

Σ∞n = {elements of Σ being determined by (gn, gn+1, · · · )}

Define a new σ-algebraΣ∞ = ∩n∈NΣ∞n

and the triple(G∞,Σ∞,Pe)


24 Boundaries of Random Walks (Mathav Murugan)

Recall from last time, (G∞,Σ∞,P)

1. Martin Boundary:

- Topological space

- Potential Theory

- Integral Representation for non-negative harmonics

2. Poisson Boundary:

- Measure Space

- Ergodic/Measure theory

- Representation for bounded harmonic function

Definition. p1(x, y) = p(x, y) the probability measure, p(x, y) = µ(x−1y) We say µ is non-degenerateif supp(µ) generates G as a semi-group.n ≥ 2 then

pn(x, y) =∑z∈G

pn−1(x, z)p(z, y)

Definition. A function f : G→ R is said to be harmonic on (G,µ) if

f(x) =∑g∈G

µ(g)f(xg)

24.1 Martin Boundary

Definition. Green’s function G(x, y) =∑∞

n=0 pn(x, y)

If G(x, y) <∞ the random walk is said to be transient. Fix o ∈ G.

K0(x, y) =G(x, y)

G(o, y)∀ x, y

G(x, y) = F (x, y))G(y, y) where F (x, y) = probability that a random walk starting at x ever visits y.

1 ≥F (x, y) > 0

F (x, y) ≥ F (x, z)F (z, y)

F (o, x) ≤ K0(x, y) =F (x, y)

F (o, y)≤ 1

F (o, x)

{K0(·, y) : y ∈ G} is relatively compact in topology of point wise convergence.

Definition. Martin compactification is the pointwise closure of {K0(·, y) : y ∈ G}

Definition. The Martin boundary, ∂MG is the topological boundary of {K0(·, y) : y ∈ G} Denoteγ ∈ ∂MG as function K0(·, γ).

Definition. A positive harmonic function is said to be minimal if for all harmonic functions f ′ such that0 ≤ f ′ ≤ f then f ′ = tf for t ∈ R.


Facts

1. All minimal harmonic functions belong to the Martin boundary (with normalisation)

2. Minimal harmonic functions form a Borel subset of ∂MG called the minimal Martin boundary,∂mG.

3. Every harmonic function can be uniquely represented as an integral of minimal harmonic functions

f(x) =

∫γ∈∂mG

K0(x, y)dνo,f (γ)

The measure corresponding to the constant function νo = ν0,1 is the harmonic measure.

4. (∂m, G, ν0) is isomorphic to the Poisson boundary (to be defined).

Examples

Theorem. (Choquet-Deny) All minimal harmonic functions in an abelian group are R+-valued multi-plicative characters

Proof. Let f be a minimal harmonic function.

T xf(g) = f(x, g) = f(g, x)

Suppose µ(x) > 0.f(g) ≥ µ(x)f(gx)

NowT xf(g) = f(xg) =

∑µ(y)f(xgy) =

∑µ(y)T xf(gy)

using that f is a minimal harmonic function.

f(gx) = λf(g) ∀ g ∈ G ∀ x : µ(x) > 0

hence f is an R+-valued multiplicative character.

�

1 is a minimal harmonic for abelian groups. It is true for nilpotent groups a result of (Dynkin-Maljutov).

Example. Free group generated by a1, . . . , an write a−1i = a−i

f(ai1 · · · ain) =

{ ∑n−1k=0

1(2n−1)k i1 > 0

−∑n−1

k=01

(2n−1)k i1 < 0

Martin boundary is space of ends is infinite irreducible words. Let γ = ai1ai2 · · · . Call the first n terms inthis expression γn.

limn→∞

K0(x, γn) = K0(x, γ) = K0(·, γ)

24.2 Poisson Boundary

T : G∞ → G∞ by (xn) 7→ (xn+1). x, x′ ∈ G∞ are equivalent (write x ∼ x′) if

Tn(x) = Tn′(x′)

for some n, n′ ∈ N. Let S be the measurable union of the ∼-equivalence classes. S forms a σ-algebra (Γ, ν).There exists a map b : G∞ → Γ such that § coincides with pre images of measurable sets in Γ.


Definition. Entropy of the probability measure p = (pi) is defined as

H(p) =∑−pi log pi

If n is the size of the support then H(p) ≤ log n if∑n

i=1 pi = 1.

Define measure

µn(g) =

{µ(g) n = 1∑

z∈G µn−1(z)µ(z−1g) n > 1

Define

h(G,µ) = limn→∞

1

nH(µn)

H(µ ∗ γ) ≤ H(µ) +H(γ)

Theorem. (Kaimanovich-Vershik) (r, µ) is trivial if and only if h(G,µ) = 0

Corollary. For any group of sub exponential growth and any finitely supported measure (r, µ) is trivial.

Proof.1

nH(µn) ≤ 1

nlog |supp(µn)|

�

Theorem. (Furstenbery) Let (G,µ) be a non-degenerate markov chain. Let G0 be a subgroup of Gwhich is a recurrent set. Define a measure µ0 on G0 as the distribution of the point where the walk startingfrom the identity hits G0 for the first time. Then r(G,µ) ∼= r(G0, µ0)

Proof. L∞( Γ︸︷︷︸f

, ν) ∼= H∞( G︸︷︷︸f

, µ). f(g) = 〈f , gν〉. ∆f = 0 in U , f = g in ∂U .

Alternatively, f(g) = Egf(X∞). How does Γ(G,µ) ∼= Γ(G0, µ0) work?

f 7→ f0 by restriction

f0 7→ f f(g) = Eg(f(xi))

�

Theorem. G polycyclic, µ finitely supported symmetric measure. Then the Poisson boundary is trivial.

Theorem. (Kaimanovich-Vershik) IfG is non-amenable then (Γ, ν) is non-trivial every non-degenerate(G,µ)

If G is amenable then there exists a symmetric non-degenerate probability measure such that (Γ, µ) is trivial.


25 Rate of Escape of Random Walks on Groups (Tianyi Zheng)

2 diffusive lower bound

Eischler, James and Peres

E|Sn| ≥ C√

n

2|S|

where C is the universal constant and G is amenable. (G finitely generated amenable then there exists anon-constant equivalent harmonic map Ψ : G→ H where H is some hilbert space.

Question. (Vershik-Peres) For what β ∈ [12 , 1] does there exist G,µ such that

E|Sn| � nβ+o(1)

where f(n) � g(n) if there exists c, C > 0 such that

cg(cn) ≤ f(n) ≤ C(GCn)

Erschler G1 = (Z/2Z) o Z, G2 = G1 o Z, Gi+1 = Gi o Z. Denote by LG,µ(n) = E|Sn|

Theorem. LGi,switch-walk-switch(n) � n1−1

2i

The range of the simple random walk on Z at time n � n1

2+o(1) . On Z2 the range is like nlogn .

Now repeat this construction with:H1 = (Z/2Z) o Z2, Hi+1 = Hi o Z2

Theorem. LHi,switch-walk-switch(n) � nlogi n

where logi n = log · · · log︸︷︷︸i

n

Amir-Virag Given any function f such that

a34 f(n) ≤ f(an) ≤ af(n)

for example f(n) = nγ with γ ∈ [34 , 1]. Then there exists G = Z o∂ImMm such that

E|Xn| � f(n)

(where Mm is the piecewise-linear mother group)

Questions.

1. Are there examples with β ∈ (12 ,34)?

2. What are the possible behaviours of rate of escape on finitely presented groups? (Polycyclic groupshave diffuse behaviour)

3. What are the possible behaviours in solvable groups? (Free solvable groups, β = 1, e.g. free metabeliangroup F2/F

′′2 )


Proof of Diffuse Lower Bound E|Sn|2. π : G→ Affine Isometries of H

Ψ : G→ H

with Ψ harmonic equivalriant.

Ψ(g) =∑s∈S±1

Ψ(gs)µ(s)

Ψ(gh) = π(g)Ψ(h)

Ψ(Sn) is a martingale E[Ψ(Sn+1)|S1 . . . Sn] = Ψ|Sn

E||Ψ(Sn)||2H = E(||Ψ(Sn)−Ψ(Sn−1 + · · ·+ Ψ(1)−Ψ(e)||2H)

= E||Ψ(Sn)−Ψ(Sn−1)||2H + · · ·+ E||Ψ(S1)−Ψ(e)||2H= nE||Ψ(X1)−Ψ(e)||2H

normalise Ψ such that E||Ψ(X1)−Ψ(e)||2H = 1. Ψ is 1√2|S|

Lipshitz.

E|Sn|2 ≥1

2|S|E||Ψ(Sn)||2 =

1

2|S|· n

E|Sn| ≥ c√

n

2|S|

�

In the upper bound direction E|Sn| � nβ. One can use equivalent embedding into a Banach space (Naor-Peres)

∂#Lp ≤1

p · β(G)

Tessera establishes a connection between L2-isoperimetry profile and Hilbert compression.

Corollary. “Class L” containing polycyclic groups F o Z, BS(1, n) diffuse speed exponent n12+o(1).

Question. Characterise groups with diffuse behaviour.

Revelle Zd o Z. Switch-walk-switch ν ∗ µ ∗ νX1, . . . , Xn, · · · on the lamps νZ1, . . . , Zn, . . . on the base µ.Wn = (fn, Vn) ∼ (ν ∗ µ ∗ ν)∗n

Writing Sn = X1 + · · ·+Xn

Vn = Z1 · · ·Zn on base. fn(i) ∼∑

(Xj +Xj+1){1Zj=i} ∼ S2θ(i,n) where θ(1, n) =∑n

i=1 1Zj=i|Wn|

∑i∈Z |Fn(i)|+ size of range{V1, . . . , Vn}

Method of Revelle: Exponential tightness of random walk on the lamps gives tightness normal of Wn

P (|Sn| ≥ xn2) ≤ c1 · exp(−c2xβ) valid for all x.

Example. Simple random walk on Zd, P (|Sn| ≥ xn1/2 ≤ c1 exp(−c2x2) satisfies a Gaussian tail.∑i∈Z|fn(i)| =

∑i∈Z|fn(i)|θ−αi,n · θ

αi,n

≤

[∑i∈Z

(|fn(i)|θ−αi,n )1

1−α

]1−α︸︷︷︸

†

(∑i∈Z

θαi,n1

α

)α︸︷︷︸

=nα

Lemma. X1, . . . , Xn conditionally independent, each of them P (|Xi| > x) ≤ c1 exp(−xβ) then P (∑n

i=1Xi >λn) ≤ c1 exp(−c2λβ). In the end † gives nβ/(1−α)


Example. Z2 oA Z where A =

(2 11 1

). Character polynomial λ2 − 3λ+ 1 = 0

Z o Z ν ∗ µ ∗ νWn =

∑i∈Z

AiS(i)2θ(n,i)


Index

Σ0-invariant, 52Σ1-invariant, 52ω-ultralimit, 10σ-algebra, 9σ-algebra generated by S, 10

Asymptotic Cone, 11

Baumslag-Solitar Groups, 4Biautomatic Structure, 62

Cayley 2-Complex, 14Cayley Graph Automatic Structure, 60Centralised Area, 24Centralised Isoperimetric Functions, 24Character, 52Character Sphere, 52Commutator Subgroup, 1Compression Words, 21

Dead-end Element, 37Depth, 37Derivation, 41Derived Length, 2Derived Subgroup, 1

Energy, 69

Filling Length, 14Flow, 69Fox Derivative, 41Free Nilpotent Group, 24

G-Space, 58Geodesic Words, 5Growth Function, 6

Harmonic Functions, 70

Irreducible Random Walk, 68

K(G, 1), 57

Lamplighter, 4Lower Central Series, 2

m-Tame, 58Max Condition, 2

Nilpotent Group, 2Non-principle Ultrafilter, 10Null-sequence, 19

Poisson Boundary, 70Polycyclic Group, 2Presentation 2-Complex, 14Probability Measure, 10

Projective Resolution, 57

Quasi-isometry, 8Quasi-Isometry Rigid, 14

Recurrent Random Walk, 69

Solvable Group, 1Superharmonic Functions, 70

Transient, 69Treebolic Space, 33Type Fn, 57Type FPn, 57

Upper Central Series, 2

van Kampen Diagram, 14Area, 14Diameter, 14Filling Length, 15

van Kampen’s Lemma, 18Virtually Nilpotent, 9

Word Metric, 5Wreath Product, 4

78

Documents

MATH 7320 Seminar in Algebra: The Geometry of Solvable Groups · Tim Riley The Geometry of Solvable Groups 1 Part I Tim’s Lectures 1 Jan 22, 2013 Two perspectives on studying solvable