The Geometry of Solvable Groups-Amin Saied

8/11/2019 The Geometry of Solvable Groups-Amin Saied

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Spring 2013

MATH7320:The Geometry of Solvable Groups

lecturer: Tim Rileyby: Amin Saied


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Contents

I Tims Lectures 1

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

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

3 Feb 05, 2013 53.1 Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

4 Feb 7, 2013 84.1 Idea behind Gromovs proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.1.1 Background: -algebras and Probability Measures . . . . . . . . . . . . . . . . . . . 94.1.2 Ultralters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.1.3 Construction of Y . (van den Dries-Wilke) . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Feb 12, 2013 115.1 Properties of Asymptotic Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

6 Feb 14, 2013 136.1 Filling Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

7 Feb 19, 2013 157.1 Combings of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

8 Feb 21, 2013 188.1 Quasi-Isometry Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

9 Feb 26, 2013 219.1 The Area Function of Nilpotent Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

10 Feb 29, 2013 2310.1 Lower Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

11 Mar 5, 2013 2511.1 Some Open Problems Relating to Dehn Functions . . . . . . . . . . . . . . . . . . . . . . . 2511.2 HNN-Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2611.3 Baumslag-Solitar Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

12 Mar 7, 2013 2812.1 The Geometry of Baumslag-Solitar Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

12.1.1 Normal Form for BS (1, n ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2812.1.2 Dehn Function of BS (1, m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3012.1.3 Cayley Graph of BS (1, 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3112.1.4 Treebolic Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

II Student Talks 35

13 Geometry of Sol (Chenxi Wu) 3513.1 Dehn Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3513.2 Quasi-Rigidity of Sol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3513.3 Area function becomes quadratic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35


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Tim Riley The Geometry of Solvable Groups 3

14 Geometry of the Lamplighter Group (Margarita Amchislavska) 3614.1 Dead-end Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3614.2 Horocyclic Product of Two Z-branching Trees . . . . . . . . . . . . . . . . . . . . . . . . . . 36

15 Generalisations of the Lamplighter Group (Margarita Amchislavska) 3815.1 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3815.2 Some Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3815.3 Horocyclic Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

16 Geometry of the Magnus Embedding (Andrew Sale) 4016.1 Magnus Embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4016.2 Applications of the Magnus Embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4016.3 Magnus Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4016.4 Fox Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4016.5 Wreath Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

16.5.1 Geometric Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4116.6 Geometric Denition of the Magnus Embedding . . . . . . . . . . . . . . . . . . . . . . . . . 42

17 Finitely Presented Metabelian Groups (Amin Saied) 4417.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4417.2 Baumslag/Margaritas Example: Embedding 1 = Z Z into 2 . . . . . . . . . . . . . . . . 4517.3 Proof of Baumslag-Remeslennikov Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

18 Sigma Invarients (Teddy Einstein) 5118.1 -invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5118.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

19 Sigma Invarients (Yash Lodha) 55

19.1 Finiteness Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5620 Automatic Groups (Scott Messick) 58

20.1 Closure Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6020.2 Biautomatic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6120.3 Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

21 Quasi-Isometric Rigidity (Iian Smythe) 62

22 Quasi Isometric Rigidity (Kristen Pueschel) 6522.1 Proof Form for QI Rigidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6522.2 X n and its Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

23 Random Walks on Solvable Groups (Johannes Cuno) 6723.1 Random Walks on Graphs and Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6723.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6723.3 Flows and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6823.4 Growth of Groups and Return Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6923.5 Advertisement for Next Talks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

24 Boundaries of Random Walks (Mathav Murugan) 7024.1 Martin Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7024.2 Poisson Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

25 Rate of Escape of Random Walks on Groups (Tianyi Zheng) 73


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Part I

Tims Lectures1 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 Dentions

Denition A group G is solvable if it has a nite abelian series - that is,

G = Gn Gn 1

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: metabeliane.g. G = G2 G1 1Get the short exact sequence

G1 G G/G 1where G1 and G/G 1 are abelian.

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 1K and H solvable. Now H = G/K . Now H is solvable, so take its abelain series. Lift this to one for G,and append it with the one for K . (also an exercise).

Denition The derived subgroup (or commutator subgroup ) of a group G is G = [G, G ]. Thederived series of G isG = 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 nite abelian series for G, and this G is solvable.


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Claim If we have a nite 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.

Denition A group G is nilpotent if it has a (nite) central series , that is

G = Gn Gn 1 G0 = 1such that the factor groups G i+1 /G i is contained in the center of G/G i (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

Denition The upper central series of a group G is

1 = Z 0(G) Z 1(G) Z 2(G) dened by Z n +1 (G)/Z n (G) = Z (G/Z n (G)). (In particular, since Z 0(G) = 1, we see that Z 1(G) = Z (G), isthe center) (Also note that this is clearly a central series, by denition)

The lower central series of a group G: 1(G) = G, then

n +1 (G) = [ n (G), G]

Notice that n (G)/ n +1 (G) Z (G/ n +1 (G)). (Exersize: check that this really is a central series) and n (G) G.Theses series are related in the following way:

1. i (G) Gn i+12. Gi Z i(G)

These can be proved by induction on i

Z n (G) Z n 1 Z 0(G) = 1 G = Gn Gn 1 G0 = 1

1(G) 2(G) n +1 (G)In this way we see that the lengths of these series agree if the central series was of minimal length.

Denition A group G is polycyclic if it has a nite normal series with cyclic factors.

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

1. Every subgroup is nitely generated (So F 2 doesnt satisfy the max condition)

2. Every ascending chain of subgroupsH 1 H 2 H 3

stabilises i.e. there exists N such that H N = H N +1 = (Noetherian condition)3. Every non-empty set of subgroups has a maximal element (a subgroup that is not a proper subgroup

of any other one of these subgroups)

We say G is max .


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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 nitely generated, and so the factors in the derivedseries are nitely generated abelian groups, and so the derived series can be rened to a cyclic series

2 Jan 31, 2013

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

Theorem (Hirsch) Finitely generated nilpotent groups are polycyclic.

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

Suppose G = x1, . . . , x m . As G/ c(G) is polycyclic, c 1/ c(G) is nitely generated (by max condition).Write c 1(G) = y1, . . . , yn c(G). Then

c(G) = [ x1, . . . , x m , y1, . . . , yn ] = [x i , y j ]|1 i m, 1 j nand so is nitely generated. But c(G) is abelian, and so is polycyclic. So as G/ c(G) and c(G) arepolycyclic, and hence G is polycyclic too.


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2.1 Examples of Solvable Groups

1. Abelian group: Zk the free abelian group

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

. . .0 1

These is nilpotent of class n 1. Indeed,

[eij , ekl ] = 1 j = keil j = k

eij = I n + E ij (The idea is that terms get pushed upwards and away from the diagonal)

In particular, U 3(Z) is the thee-dimensional integral Hiesenburg group , sometimes denoted

H3(Z) = a,b,c|[a, b] = c, [a, c] = [b, c] = 1Notice that this is metabelian, as the map which kills c takes H3(Z) ZZ. It is of course not abelian.Theorem. (P.Hall, 1969) Every nitely generated torsion free nilpotent group embeds in U n (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 havethe presentation,Z Z [

1m ]

where a generator for the Z-factors acts on Z[ 1m ] by multiplication by m. This is metabelian byconstruction.But this is not a polycyclic group: indeed Z[ 1m ] is not nitely generated, and is clearly (isomorphicto) a subgroup. This contradicts the max condition.

4. Sol = Z2 A Z, where A = 2 11 1 . This is metabelian and polycyclic by construction.

5. LamplightersBefore dening lamplighters we require the more general denition of a wreath product. For groupsA, B denote A (B ) :=

{nitely supported functions B

A

}= B A

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

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

where gb(x) = g(b 1x).

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

Z/m Z 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 nitely supported function from Z/ 2Z to Z. If we think


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of Z lamps arranged on a street, then f is a nite 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 denition 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 fg b

1 0 1 2 3 4 5Now we have (f, b )(g, c) = ( fg b, bc). fg b 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) = ( fg b, bc) = 1 0 1 2 3 4 5

To summarize this, we think of gb as a translating g by b, then fg b 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 1, r 0 for all r = 0 Now ( f, 0) and (0 , 1) generate Z/ 2Z wr Z.General result: A wr B is nitely generated if both A and B are.

For Z wr Z = Z Z is not polycyclic because the rst factor is not nitely generated.

6. F r a free group of rank r . Then F r /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

Denition. G a nitely 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 AA 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


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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 byyayxxax 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.

Denition. Growth function f is

f (n) : = #( AA 1{1})n

= # {g G : d(1, g) n}Examples

1. Z wrt 1 then f (n) = 2 n + 1

n

n

0

n

n

2n +12. Z2 wrt (1, 0), (0, 1) then f (n) n2

3. F 2 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:


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Proposition. For all groups f (n) (2|A|+ 1) n4. Z2wr Z

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 2 2n +1 group elements. Sof (6n) 22n+1

f (n) 2n5. BS (1, 2) = Z[12 ] 2 Z

We have 2 m b m abm01 + 121 + + k2k where i {0, 1} ki=0 (b i abi ) i and this has length k, so

f (n)

2n

6. Sol = Z2 A Z where A = 2 11 1

Exponential growth

2 11 1

k pq

0 0 1

7. The Heisenburg Group

H3 =

1 Z Z1 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 1

1 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 r1 1 q 0 0 1

where | p|, |q | k and |r | k2. Therefore f (n) n4Remark: If w is a word on {a,b,c} 1 then there exists a word

w representing the same group element

of the form a pbqcr where if l(w) = n then

| p

|,

|q

| n and

|r

| n2. This follows from the (nice) normal

form 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 as and bs at the expense of adding a c.Therefore we can bring our as to the front, and get some cs oating about, but because c is central,we can just send them all to the back.


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

G = 1(G) 2(G) n +1 (G) = 1the growth f (n)

nd where

d = k torsion-free rank( k(G)/ k+1 (G))

Example H3H3 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 g when there exists c > 1 such that f (n) cg(cn + c) + cn + c. Write f gwhen f g and g f . For example, n n for , 1 iff = , but n n for all , > 1.If A and B are nitely generated sets for a group G then f A f B . In fact the functions are Lipschitzequivalent i.e. there exists a c > 0 such that

f A(n/c ) f B (n) f A(cn)Lemma. If H is a nite index subgroup of a nitely generated group G then ( H is also nitely generated(!) and) their growth functions are equivalent.

Denition. A map : X Y between metric space is a (, )-quasi-isometry (with 1, 0)when1 d(a, b) d((a), (b)) d(a, b) +

and for all y Y there exists xX such thatd((x), y)

Examples

1. G = A = B with |A|, |B | . Then I d : (G, dA) (G, dB ) is a quasi-isometry2. H G is a quasi-isometry

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

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

f G1 f G2Note. 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 nitely generated quasi-isometric groups (i.e. there exists a quasi-isometrybetween them) then their respective growth functions are equivalent

f G f H


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Pf. : G H a (, )-quasi-isometry. Consider a ball of radius n about eG. The image of this ball inH under has radius n + (since is a quasi-isometry).There exists some c > 0 such that if x, y G have d(x, y) > c then (x) = (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

|. Dene

|B (c, e)

| = K , some constant. We have

|(M )

| =

|M

|, by

construction of M . But |M | f H (n + ) Putting this together gives,f G (n) = |B (n, e )| K |M | Kf H (n + )

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

Denition. A group is virtually nilpotent when it has a nite index subgroup which is nilpotent.

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

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

4.1 Idea behind Gromovs proof

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

Sketch of Proof of Gromovs Theorem.Suppose G is an innite nitely generated group of polynomial growth. Let d be a word metric on G withrespect to a nite generating set. Let

Y := limn (G, 1n d)(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, nite dimensional (topological dimension)

Gleason-Montgomery-Zippins Solution to Hilberts 5th Problem It can be shown that Isom (Y )is a Lie group. Gromov uses this to build a homomorphism from a nite 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 nite index. Then Milnor-Wolf nishes the proof.

4.1.1 Background: -algebras and Probability Measures

Denition. Let X be a set, and P (X ) = 2 X its power set. Then a subset 2X is a -algebra if itsatises:1. is non-empty

2. is closed under compliment

3. is closed under countable unions

It thus follows by de Morganss laws that it is also closed under nite intersection.


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

Proof. By 1. is non-empty so there exists AX in . By 2. it is closed under taking compliments, soX \A. By 3. it is closed under countable unions, so AX \A = X . And nally it is closed undercompliment, so X \X = . Proposition. Given a nite collection of subsets of X , S = {X i}, there exists a unique smallest -algebracontaining S .Proof. Indeed, consider

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

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

Example. Let X = {1, 2, 3}. The -algebra generated by {1} is ({1}) = {, {1}, {2, 3}, {1, 2, 3}}Denition. Let X be a set and a -algebra over X . A function : R{}is called a measureif it satises:

1. Non-negativity: (E ) 0 for all E 2. Null empty set: () = 0

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

i

E i =i

((E i))

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

4.1.2 Ultralters

Denition. A non-principle ultralter : P (N) {0, 1} is a nitely additive probability measure onN taking values in {

0, 1

} and giving all singleton sets measure 0.

Remark. These exists by Zorns lemma: Consider the set of functions f : P (N) {0, 1} such thatf 1(1) is closed under taking nite intersections and supersets, and doesnt include the empty set, but doesinclude all conite sets. Example: f 1(1) = the conite sets. Now I want to take a maximal element usingZorns lemma, under the relation f f when f 1(1) f 1(1). This maximal element is an example of anon-principle ultralter.Denition. Let : P (N) {0, 1} be a non-principle ultralter. Then a R is an -ultralimit of asequence ( an )R when

> 0 ({i : |a a i | < }) = 1and say is an -ultralimit when

{i : a i > N }= 1 N > 0and say is an -ultralimit when

{i : a i < N }= 1 N > 0


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Lemma. Every sequence ( an ) in R has a unique -ultralimit in R{}. Call this lim an .Sketch. Assume (an ) is bounded, then within the bounded set of R (so an interval) there is a sequence of points. 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)

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

Cone (X, e, s) := (xn )X : limd(en , xn )

sn< /

where (xn )(yn ) iff limd(x n ,yn )

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

d((xn ), (yn )) := lim

d(xn , yn )

sn

(hence quotienting out by , as this identies sequences which would be distance 0 under the metric above).This is lim ia (X, ds i ), the limit of these metric spaces.

5 Feb 12, 2013

5.1 Properties of Asymptotic Cones

Proposition. Given ,e in X , s, then a quasi-isometry : X Y induces a bi-Lipshitz homeomorphismCone (X, e, s)

Cons (Y, (e), s)

where (e) = ( (ei )).

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

Proof of Prop. Suppose ( a i ) and ( bi) represents points in Cone(X, e, s). Then

1

d((a i), (bi)) d((a i ), (bi )) d((a i), (bi))(which is bi-Lipshitcz). Lets try and justify this:

d((a i ), (bi)) = limd((a i), (bi ))

s i

limd(a i , bi ) +

s i

= lim

d(a i , bi )s i

+ lim

s i

= d((a i ), (bi ))

since lim s i = 0 as is nite and s i . The other inequality is similar.Suppose ( yi ) representa a point in Cone(Y, (e), s). As is a quasi-isometry (and in particular we areincluding quadi-onto in that denition), for all i there exists x i X such that

d((x i ), yi ) We claim that


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1. (x i) represents a point in Cone (X, e, x)

2. ((x i )) = ( yi )

for (2.)

d(((x i )) , (yi )) = limd((x i ), yi )

s i

lims i

= 0

For (1.)

d((ei), (x i)) = limd(ei , x i )

s i

limd((x i ), (ei)) +

s i

lim

d((x i ), yi ) + d(yi , (ei )) +

s i

lim + d(yi , (ei )) +

s i= d((yi), ((ei ))) <

Proposition. For a nitely generated group , the asymptotic cone,

C := Cone (, e, s)

is:1. homogeneous

2. geodesic

3. complete

Proof.

1. Suppose ( gi ), (h i ) represents points in C. Then(h i g 1i ) :

C C(a i) (h ig 1i ai )is an isometry taking ( gi ) to ( h i ).

2. If = A with |A| < then Cay A() (its Cayley graph) is a (1 , 12 )-quasi-isometry (you map to the vertices of its Cayley graph, you miss the edges but, thats ne because we take edges to havelength 1). So C = Cone (, e, s)= Cone(Cay () , e, s). So I can just prove it for the latter space.

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

Indeed, suppose ( a i ), (bi) represent points in Cone(X, e, s). Then there exists a geodesic path i : [0, d(a i , bi)] Cone (X, e, s) from a i to bi . Dene

: [0, d(a , b )] Cone (X, e, s)by (r ) = ( i(s i r )). For r [0, d(a , b )], i(s i r ) is well dened for all i in some set of -measure 1.

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


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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 with a word metric,

C = Cone(, e, s)

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

Theorem. (Pansu) If is nitely 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 i g i where [g i , g j ] = g i+ j for all i, j ).

e.g. Real Heisenburg Group H3(R) =1 x y

1 z1

. Lie algebra is 3-dimensional x,y,z , where theonly non-trivial bracket is [ x, y] = z, so the grading is g 1g

2 where g 1 = x, y , g 2 = 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 inmal length of all at paths between the points, that is, paths that staytangent to some specied 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 a nitely generated nilpotent group, the topological dimension of Cone (, e, s) is the

cohomological dimension of /tor ()

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 of the 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 ll itin, and we are left with R3, which is contractable.

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

Proof. is quasi-isometric to Zn

is of polynomial growth

is virtually nilpotent (by Gromov).

Let 0 be a nite-index subgroup of which is nilpotent. The topological dimension of the cones of a groupquasi-isomteric to Zn is n. So 0/tor (0) has cd = n . But the growth of 0 is f (r ) = r n , a polynomial of degree n. So 0/tor (0) is Zn .

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


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

6.1 Filling Functions

Given a nite presentation x1, . . . , x m |r 1, . . . , r n of a group .Denition. 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

Denition. The Cayley 2-complex is the universal cover of K . 1(K ) = by Seifert-van Kampen.Can identify with K

(0) , and identify K (1) with the Cayley graph of with respect to {x1, . . . , x m }.

Example. Z2 =

a, b

|[a, b] , then K is the rose with 2 petals, R

2 with a square with edges labelled

a,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 1(Torus) = Z2 acts on R2 by translations(freely).

Denition. A van Kampen diagram of = x i |R is a nite contractable planar 2-complex withall the edges directed and labelled by some x i in such a way that around each 2-cell one reads the deningrelator.

Equivalently, a van Kampen diagram is a nite 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) .Denition. Let X be a topological space with basepoint e, let : (S 1,) (X, e ), be rectiable (nitelength). The lling length of is

F L ( ) := inf {L 0 : there is a nullhomotopy H t (s) of with l(H t ) L}Below are two contrasting examples: in the plane, a nullhomotopy can always be performed via paths of length 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:


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3. The lling length of ( ,), where is a choice of basepoint on , F L( ,), is the minimal L suchthat there is a combinatorial nullhomotopy ( i) of = 0 to = r , with max i 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 of the 2-complex at each stage in this process.

Denition. For M = Area, Diam or F L, dene M : N N byM (n) = max {M ( ) : an edge loop of length n in K }where

M ( ) = min {M () : is a van Kampen diagram such that = }Example Z

2

Area (n) = n2, Diam (n) = n, F L(n) = n.

7 Feb 19, 2013

7.1 Combings of Groups

Denition. G a group with nite generating set X . A combing is a section : G (X X 1) of thenatural surjection ( X X 1) G. : G (X X 1)

g gcan be considered as a unit speed path in the Cayley graph C ayX (G) from e to g.

1. is synchronously k-fellow-travelling when g , hG : dX (g, h) = 1 t[0, )d(g(t), h (t)) k

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


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3. is asynchronously k-fellow travelling when g , hG : dX (g, h) = 1 reparametrisations , such that tN d(g((t)) , h ( (t)))

4. The length functionL(n) = max {l(g) : 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 F n . The geodesics form a synchronously 1-fellow travelling combing with L(n) = n.

3. Zn = a1, . . . , a n : [a i , a j ] = 1 i, j . Theng = ar 11 a

r 22 a r nn

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

{br ua s : r, s Z, u {ab 1, b 1} and the rst letter of u is not b 1}denes an asynchronously fellow-travelling combing. The re-write rules:

(a) ab ba2(b) a 1b ba 2(c) a2b 1 b 1a(d) a 1b 1 ab 1a 1(e) aa 1 (f) a 1a (g) bb

1

(h) b 1b Using these rules one can convert any given word into a combing word. e.g.

abn ba2bn 1 b(ab)a2bn 2 bn a (2n )

Here L(n) 2n .


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Denition. Synchronously automatic groups are nitely generated groups with synchronous k-fellow-travelling combings for which the language {g : g G} is regular (low complexity).For example: Zn with the rst 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 nite generating set X has a combing that is asynchronously k-fellow-travelling and has length function L(n), then G is nitely presentable and

Area (n) CnL (n)Diam (n) F L (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 really just 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 2 k + 2.Let R be the set of words in ( X X

1)which represent 1 in G and which have length 2k+2. Note |R| < .

Claim 1: is a van Kampen diagram for w wrt X |RClaim 2: X |R = GIndeed 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 2 L(n) 2-cells. So in totalArea (n) 2nL (n).


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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 F L With that point 2 becomes redundant.

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

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

8 Feb 21, 2013

We can view a nite presentation = X |R as a short exact sequence,1 R F (X ) 1

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

w =N

i=1

u 1i ri

i ui ()

with r i R, i = 1, u i 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 if and only if w admits a van Kampen

diagram over X |R . Moreover, Area (w) is the minimum N amongst all expressions ().Proof. Suppose is a van Kampen diagram for w


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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 if and only if it can be reduced to the empty word using the moves

1. free reductionx(a 1a)y xy

2. free expansionxy x(aa 1)y

3. applying relationsxay xby where a 1bR

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

Example. Zn = a1, . . . , a n |[a i , a j ] i, jThe Dehn functions Area (n) n2 and F L(n) n.Lemma. w = 1

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

2. F L(w) is the minimum of max i l(wi) such that there is a null-sequence

w = w0 w1 w2 wn = 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 lling length will tell us how many moves it will take. Hence time andspace.


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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 F L(n) are all quasi-isometry invariants (up to ) among nitelygenerated groups.Corollary 1. Among nitely generated groups being nitely presentable is a quasi-isometry invariant

Corollary 2. If P 1 and P 2 are nite presentations for the same group, then

1. Area P 1 Area P 22. Diam P 1 Diam P 23. F LP 1 F L P 2

Proofs. Suppose that 1 = X 1|R1 nite presentation, and let 2 be a group with nite generating setX 2. Suppose, : 2 1

is a (, )-quasi-isometry. Suppose w a word on X 2X 12 and represents 1 2. Aim to show that thereexists a van Kampen diagram for w in 2. The idea is to

1. Represent the word w in C ayX 2 (2)

2. Use the quasi-isometry to push the loop from C ayX 2 (2) to C ayX 1 (1)

3. 1 has a nite presentation, so construct a van Kampen diagram in C ayX 1 (1).

4. Use quasi-inverse to pull this van Kampen diagram back to C ayX 2 (2)Given a van Kampen diagram for w we have a nite set of relations that reduce w to 1 2. Thus wehave shown that 2 is nitely 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 CayX 2 (2),and compare it to that in CayX 1 (1).

Use to push the word w from CayX 2 (2) to CayX 1 (1). Here we have a nite presentation, and sohave a van Kampen diagram representing the word:


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Now is a quasi-isometry, so it has a quasi inverse such that d(x, (x)) C, d(x, (x)) C and isa quasi-isometry. Push this van Kampen diagram back to C ayX 2 (2) with :

Since 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, nite) amount. Indeed - consider a loop in the vanKampen diagram in CayX 1 (1). This has a perimeter length bounded above by the largest length of adening relation in R2 (which is nite, so this perimeter is bounded). Hence when we push diagram backto C ayX 2 (2), 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 CayX 2 (2). But thefact that is a quasi-isometry means that this added bit are bounded, hence we are done.

9 Feb 26, 20139.1 The Area Function of Nilpotent Groups

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

Area (n) n c+1 and F L(n) 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 n3dening relations and with max i l(w) n (hence simultaneously).Dene 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 + Bn 2) = u(A)u(n2)B for all integers A, B with 0 A n2 1 and B > 0.

Note: l(u(A + Bn 2))

K 0n for some K 0 depending on B .

Lemma. Fix K 1 > 0. Then there exists K 2 > 0 such that for all 0 s K 1n2, there is a concatenationof sequences converting zs to u(s):zs zs u(0) zs 1u(1) zs 2u(2) zu (s 1) u(s)

using K 2n3 relations.


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Proof of Lemma Key idea:

zn [xn , yk] = zn x n y kxn yk

= zn x n y k(y 1y)xn yk (insert 1 = y 1y)

= zn x n y (k+1) (xz 1)n yk + 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 nitely 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/ c(G) which is a nitely presentable nilpotent group of class c 1. Let X bea nite generating set for G and X its image in G/ c(G). Let

X

|R be a nite presentation. Let X be a

nite set of length c commutators (entries in X ) in G that generate c(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 centralis a nite 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/ c(G),so there is a null-sequence,

w = w0 wk = with l(wi) n and the number of relations used n c (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/ 2(H3) = H3/ z = x, y |[x, y] = 1 .First

w (z) 1w0zby carrying all the zs to the right end, and all the z 1s 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 zs to the right, and all z 1s to the left. Compress the powers of z ateach end as the z 1s arrive. Eventually we get

u(s) 1 wk

=1u(t)

for some s, t 0. In H3 this is z s z t = 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 zs to the endsKn 3 - compression at each end

Total: n2 + n2 n + Kn 3


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10 Feb 29, 2013

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

Remark. This would imply the nc+1 Dehn function and n lling 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 nc+1 : Suppose G is a nitely generated nilpotent group of class c,

Cone (X, e, s )

are graded (homogeneous) nilpotent Lie groups. In this case loops can be lled with discs of area n.The problem is how do we pull this result back to the group. There is a partial result of Papasoglu

Area (n) nc+1+ , which is close.

10.1 Lower BoundsThese 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.Gerstens 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.


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Mad Proof that Area ([an , bn ]) n2Suppose W = N i=1 u 1i [a, b] i u i , (where i = 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 n 2. We have W = N i=1 c i inH3 = a,b,c|[a, b] = c, [a, c] = 1, [b, c] = 1

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

, and c has inniteorder. Now W = [an , bn ], by hypothesis, and so equating gives,

W = [an , bn ] = cn2

=N

i=1

c i = W

whence comparing the powers of c gives N n2. (Recall, i = 1, so we get the lower bound n2 preciselywhen these are all +1, but it will be bigger then n 2 is some of them are 1s)

Centralised Isoperimetric FunctionsP = A|R a nite presentation of , R= R in F (A).

= F (A)/ RSuppose w = 1 in .

Denition. The centralised area Area cent (w) = min {N |w( N i=1 u 1i r ii ui )[R, F ]}

This is just, the number of times, up to sign, that each relator occurs in a product of conjugates freely equal

to w, ignoring conjugation involved. This is a calculation in

R/ [R, F ]which is nitely generated abelian group, as it is generated by R, say {r 1, . . . , r k}. Write w = r s11 r skk inR/ [R, F ]. Then the centralised area, Area cent (w) = min { ki=1 |s i|}Lemma. Suppose B = {y1, . . . , ym } generates the abelian group,

R/ [R, F ]For a word w(A) = 1 in , let lB (w) be the minimal length of a word on B representing w[

R, F ]. Let

K = max {lB (r ) : rR}, then1. lB (W ) K Area cent (w) K Area (w)2. c > 0 such that if w[R, F ] = ym [R, F ], where y[R, F ] has innite order in R/ [R, F ] then

m c Area cent (w)Proof.

1.

2. R/ [R, F ] = T Zk

for some nite T . Let B = T { a basis for Z}. Then m lB (ym

) (by projectingto the Zk factor). Then m lB (ym ) = lB (w) K Area cent (w).

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


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Theorem. F/F c+1 has Dehn function n c+1 (so nc+1 )Proof.Let u = [

c

[[a, b], a], , a]

Lemma uF c\{1}, and ukc

= [ [[ak , bk], a k], . . . , a k ] mod F c+1This is just a more elaborate version of the same notion we showed in the example with H3 above. Notgoing to prove it.

Let R= F c+1 . Then [R, F ] = F c+ k and R/ [R, F ] is F c+1 /F c+2 is generated by simple commutators. Nowwrite ukc = [ c[[ak , bk ], a k ], . . . , a k ] mod F c+1 = wc,k . Now wc+1 ,k F c+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 /F c+1 .

Area (wc+1 ,k ) Area cent (wc+1 ,k ) constant l(wk+1 ,c)c+1In the Heisenburg group example this is

Area ([ak , bk ]) Area cent ([ak , bk ]) constant (Lk)c+1

11 Mar 5, 2013

11.1 Some Open Problems Relating to Dehn Functions

Recall, nitely generated nilpotent class c groups have Dehn function (i.e. area) n c+1 and F L(n) n.

Free nilpotent class c groups on 2 generators have Dehn function nc+1

.

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

H5 =1 Z Z Z

1 0 Z1 Z

1 H7 =

1 Z Z Z Z1 0 0 Z

1 0 Z1 Z

1

of class 2.

Theorem. (R. Young) For all c there exists nilpotent class c group with Dehn function n2, n 3, . . . , n c+1 .

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

Sapirs Example a1, b1, . . . , a s , bs | class 2, [a1, b1] [a s , bs ] = 1 . n2(n) Area (n) n2 log n where(n) as n .Open Question.

1. Classify nitely 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 lling length functions on other classesof groups, like:

(a) polycyclic groups (NB always nitely presented)


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(b) nitely presentable metabelian groups(c) nitely presentable linear groups

There is a problem:

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

11.2 HNN-Extensions

Denition. G a group, A, B subgroups and : A B an isomorphism. ThenG := G, t |t 1at = (a) a A

Brittons Lemma Suppose w = g0te1 g1te2 ten gn where ei = 1, n 1, gi G, and w = 1G. Thenit contains a subword of the form1. t

1gt for some g A, or

2. tgt 1 for some g BThis subword is called a pinch .

Proof. Draw a van Kampen diagram for w. Consider the cell attached to te1 . This cell represents arelation in G , and so must be of the form t 1at (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 the

same 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:


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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 G can be represented uniquely asw = g0te1 g1te2 ten gn

where 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 Brittons lemma w1w 1

2 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 12contains 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 1am t = an = ZThis is an HNN-extension where : mZ nZ by am an , where a is a generator. Trivial examples thatwere known prior to Baumsalg and Solitars paper are:

1. BS (1, 1) = Z Z2. BS (1,1) = Z Z


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

For example: Z , consider innite sequences of integers. f : (a1, a 2, a 3, . . . ) (a2, a 3, a 4 . . . , ) is a surjectivehomomorphism with innite kernel. Baumslag-Solitar groups were constructed to give an easy example of non-Hopan groups.

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

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

Now a a2, t t denes an epimorphism G G with non-trivial kernel.Well dened: t 1a2t t 1a4t = ( t 1a2t)2 = ( a3)2 = a6 so a3 a6.

Surjective: a 1t 1at a 2t 1a2t = a 1a3 = aKernel: a 1t 1ata 1t 1ata 1 a 1t 1a2ta 2t 1a2ta 2 = a 2a3a 2a3a 2 = 1

Denition. G is residually nite when for all g G\{1} there exists a nite group H and a homomor-phism : G H with (g) = 1.Remark. Linear

Residually Finite

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

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

=

(m/n )Z Z[ 1mn ]0 1 GL 2(Q). Not injective when m = 1 , n = 1.

Proposition. BS(m,n) is

1. Hopan if and only if it is residually nite, 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 ),

{t iak t 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 j a = t j at j t 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


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Now consider the paths for t j (in purple) and for t j a = a2jt j (in green) dened by this

HNN-normal form:

Claim : d( p, ) 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.


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

a2t 1 t 1aa 1t 1 at 1a 1

to convert it to an asynchronous fellow travelling combings. For example consider the words ti ak t j

and tiak t j a written in the new normal forms. Well tiak t j a = ti ak+2 j t j by the rst 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 2 n .

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) nL (n) for any asynchronous combing, and we have just given an example whereL(n) 2n . (Alternatively, think of Brittons lemma: identifying a pinch in a word equal to the identity.)Lower bound: Let wn = [a, t n at n ]. Now l(wn ) = 4 n + 4. wn = 1 BS (1, 2) because t nat n = a2n .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:


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The blue cell here gives a dening 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 didnt 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 a r for some r .But reading around the diagram we get

a r = t n at n

so r = 2 n 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


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Indeed, consider the bottom most row of as. 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 as, at each point we get two folds emanating, and theresult is:

the Cayley diagram for BS (1, 2). Notice:1. Let T be the innite binary tree. Then we get that the Cayley graph is T R - a Bass-Serre tree(vertices correspond to cosets of a ):


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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 logq(Im z ) (this gives our heights in red):

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

Denition. Treebolic space is the horocyclic product H T (q, p) := {(u, v)H T p : h(u) + h(v) = 0}(to be clear - this depends on q as it is used to dene the function h).Example: Let v be the root of the tree, having height 1. Then we have that pairs {(u, v)HT p : Im (u) = p}H ( p, p), so we get that point v T p crossed with the line {uH : h(u) = 1}.


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Similarly for all points v T p we get a line homeomorphic to R in H T ( p, p)

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


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Part II

Student Talks13 Geometry of Sol (Chenxi Wu)

Sol = R2 R where : R GL 2R by t et 00 e t

Discrete subgroup in Sol: Z2 Z where : 1 A = 00 1

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

13.1 Dehn Function

Denition. Area (l) = suplength ( )= l

inf D, = D

Area (D )

We get that Z2

Z is quasi-isometric to Sol. Hence we getArea Z2 Z Area Sol

Proposition. Area Sol exp13.2 Quasi-Rigidity of Sol

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

13.3 Area function becomes quadratic

Sol 3 =et x

e t y1

= R2 R = x, y t

Sol 5 =

et1 xet 2 y

et3 z1

= R3 R 2 = x,y,z t1 t2, t 1 t3


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14 Geometry of the Lamplighter Group (Margarita Amchislavska)

Denition. The Lamplighter has numerous equivalent denitions:

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

2. 1 = iZZ Z = Z wr Z

3. 1 = xk f

0 1 : k Z, f Z[x, x 1

4. 1 = a, t |[a, a tk] = 1, k Z nitely generated, not nitely 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 identity

to 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 Bcn (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. 1,2 = a, t |a2 = 1 , [a, a t

k

] = 1, k ZTheorem (Clearly, Taback 2003) 1,2 has elements of unbounded dead-end depth.

d(1, gn ) = 4 n + 2n + 1 = 6 n + 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 innite 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 exemplied above), the other using Z[x 1] cosets.

Dene the height functions1. H (f + xkZ[x]) = k

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


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The horocyclic product of these two trees T 1, T 2 has vertex set {(v1, v2)T 1 T 2 : 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 + xn Z[x], g + dxn + xn 1Z[x 1])V (H1).We will show that H1 Cay (1, Y ).

Lets look at 1 again: g 1 is given by k Z and f = f i x i in the matrix description say. OR:1. k Z and a pair of sequences

(a) {f k 1, f k 2, . . .}(b) {f k , f k+1 , . . .}

OR:

2. A point ( f + xkZ[x], f + xk+1 Z[x 1]}Another presentation of 1 is

1 = i , iZ|mi m j = m j m i , i , j ,m Z(check these are equivalent via i = a i t). In trees i corresponds to go down in the second tree T 2 (withoutchoice by f k ), and go up in the rst tree T 1 (by f k + i). This gives a graph isomorphism between theCayley graph C ay(1, Y ) and H1 (where Y is the is).


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15 Generalisations of the Lamplighter Group (Margarita Amchislavska)

2 = Z[x, x 1, (1 + x) 1] Z 2 = xk (1 + x)l f

0 1 : k, lZ, f Z[x, x 1, (1 + x) 1]= a,s,t |[a, a t ] = 1, [s, t ] = 1, a s = aa t

2 was constructed by Baumslag in 72 as the rst example of a nitely presented group with an abeliannormal subgroup of innite 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 ipsa switch, t says move one unit to the right, and s says move one unit up.

Elements in 2 correspond to integers on the grid points (only nitely many non-zero) and the positionof the lamplighter. Call a nite collection of integers on the lattice points a conguration . The identity

corresponds to zeros everywhere and the lamplighter at (0 , 0). g 2 is given by (k, l , f ) (k, l) is nalposition of lamplighter and f = f ij x i(1 + x) j where f ij is the integer at position ( i, j ).Two congurations 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 conguration there exists a unique conguration equivalent to it supported only on thelampstand.

15.2 Some Results

Get a basis for Z[x, x 1, (1 + x) 1] taking {x i , (1 + x) j : iZ, j Z< 0}. One can use this to give a normalform for 2.Let 2,r be the group 2 with the additional relation that a r = 1.

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

This was the rst example of a nitely presented group with this property.

Theorem (Grigorchuk, Linnell, Schick, Zuk) Gave 2,2 as a counter example to Atiyahs conjecture(by constructing a 7-dimensional manifold with fundamental group 2,2 and the third L2-Betti number 13not an integer).

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

Theorem (Kassabov, Riley) 2 has exponential Dehn function

Remark

1. Filling length function unknown.

2. 2 = ,,c,d |[, ] = 1, 1c2 = c, 1d2 = d


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15.3 Horocyclic Product

Claim Cay(2, Y ) is to the 1-skeleton of the horocyclic product of three innite Z-branching trees.Use another presentation of 2:

2 = i , i , i (iZ)| i = imu i , i+ j = i j (i, j Z)with i = a i t, i = a i s, i = a i ts 1a i

Recall

1. Vertices of an innite 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 successive

downward edges.

Combinatorially Use the model. Given g 2 want to see how to get a vertex in H2. Well, g 2 (k, l) and a conguration of integers on the grid.1. Use the equivalence relation on triangles to propagate the conguration 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.


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16 Geometry of the Magnus Embedding (Andrew Sale)

What is a free solvable group? F r = x1, . . . , x r is the free group in the variety of r -generated groups.Therefore any r-generated group is a quotient of F r . 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 . S r,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 S r,d . So,

S r,d = F r /F (d)r

where F r = F 1r = [F r , F r ] and F (i+1)r = [F

(i)r , F

(i)r ].

16.1 Magnus Embedding

N F r . Let : F r F r /N,N = [N, N ]. The aim is to embed, : F /N Zr F r /N

If N = F (d)r then S r,d +1 Zr S r,d16.2 Applications of the Magnus Embedding

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

2. (Remeslennikov-Sokolov, 70) S r,d is conjugacy separable

3. Baumslags embedding theorem (that every nitely generated metabelian group embeds in a nitelypresented metabelian group) (see Amins talk)

4. Random walks on S r,d (see Tianyis talk)

Classical Denition We will realise Z F/N as a matrix group. Let Rbe the free abelian Z[F/N ]-modulegenerated by t1, . . . , t r .M (F/N ) = g p0 1 : g F/N,P = R

16.3 Magnus Representation

: F M (F/N ) by w (w)

wx i t i

0 1 which is involving something called the Fox derivative .

(Magnus 30s) ker = N the Magnus embedding.

16.4 Fox Calculus

A Fox derivative is a derivation

D: Z[F ] Z[F ]where (recall) a derivation must satisfy, for a, bZ[F ]

1. D(a + b) = D(a) + D(b)2.

D(ab) =

D(a) (b) + a

D(b)

where : Z[F ] Z by sending g F 1.F = x1, . . . , x r . Dene

x i

(x j ) = ij


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

x 1

(x1x2x 11 + x 11 ) =

x 1

(x1x2x 11 ) + x 1

(x 11 )

= x 1

(x1x2) (x 11 ) + x1x2

x 1(x 11 ) x 11

Lemma. (Fox) Let D: Z[F ] Z[F ] by any derivation. Then K i Z[F ] such that D= ri=1 K i x iZ[F ]

D

D Z[F ]

Z[F/N ]

16.5 Wreath Products

A = Y , B = X ,

|X

|,

|Y

| 1. Indeed, suppose [ a, a s

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

follows from [a, a s ] = 1. Then

1 = [a, a si]t = [a t , (a s

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

i] = [aa s , (aa s )s

i] = [aa s , a s

ia s

i +1] = [a, a s

i +1]

Where the equalities follow (respectively) from:

1. By hypothesis that [ a, as i

] = 12. Conjugation is a homomorphism, so [ x, y ]t = [x t , yt ]

3. [s, t ] = 1 commute

4. at = aa s

5. Again conjugation is a homomorphism

6. By the inductive hypothesis. (Exersize)

[aa s , a sias

i +1] = aa s as

ias

i +1(a s ) 1a 1(as

i) 1(as

i +1) 1

The underlined terms give [ a, a si +1

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

Remark. So we have embedded a nitely generated, innitely presented metabelian group 1 into anitely presented metabelian group 2 by constructing an HNN-extension and showing that it was nitelypresented. This is the underlying idea behind the proof Baumslag-Remeslennikov theorem. Perhaps knowingnow, as we do, that 1 is the horocyclic product of two trees, and 2 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 {x i |I }, R F . Given an isomorphism F/R H by x i R h i . Let A be the free abelian group on {a i |iI }. Then the assignment x iR h i a i determinesan embedding of F /R into the wreath product A H .Lemma 1. G a nitely generated metabelian group. Then G can be embedded into Gab A where A isa nitely generated ZGab -module.


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Proof. G = g1, . . . , gn , F = F x1, . . . , x n , : F G, K := ker G metabelian F K (for f F , (f )G = {1})R := 1(G )R = F K R K (indeed, f F , k K, [f, f ] = [f, k ] = [k, k] = 1)Let A0 be the free abelian group on {a1, . . . , a n}. We can apply the Magnus embedding with H = Gab sinceF/R = G/G = Gab ,

: F /R A0 Gab = A(G ab )0 Gab = W B := A

(G ab )0 base group

x iR xi R

0 1

where B , the base group.Claim 1 W is nitely generated metabelian group

B and Gab both abelian.Dene N = (K/R )

Claim 2 N B(wR )

wR 0 1 . But R = F K contains K . So if wR N then w K R, thus upper left entryis trivial:

1R 0 1 = BClaim 3 N Im ()Third isomorphism theorem: K/R F/R . Hence N = (K/R ) (F/R ) = Im ().

Claim 4 N W Im () contains all x i R. W is generated by B (abelian) and by ( x i R).

Finally, by the third isomorphism theorem again,

G= (F/R )/ (K/R )

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

Remark The key point here is that we embedded G into a nitely generated metabelian group of theform H A 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 nitely presented metabelian group. We need a way to construct these HNN-extensions, namely, weneed an endomorphism:

Lemma 2. H a nitely generated abelian group, A a nitely generated ZH -module. For each h

H

there exists a polynomial p = 1 + c1x + + cr 1x r 1 + xr Z[x]

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


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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. Dene

A0 = {aA : 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 nitely generated, then A0 is nitely generated. Say it is generated by

b1, . . . , bs

Then there exist special polynomials pi such that bi pi (h) = 0. Dene

p = xp1 ps + 1This is clearly a special polynomial. Suppose ap(h) = 0 some aA. Then aA0. Therefore

a = b1f 1 + + bs f ssome f i ZH . Notice that

bi p(h) = bi(hp1(h) ps (h) + 1) = hp1(h) bi pi(h) =0 ps (h) + bi = bi

Therefore0 = ap(h) = ( b1f 1 + + bs f s ) p(h) = b1f 1 + + bs f s = a

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

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

H = h1 hr hnwhere h1, . . . , h r have innite order, and hqii = 0 , i = r +1 , . . . , n . By lemma 2 there exist special polynomials,

p1, . . . , p r

such that a ap i(h i ) determines an injective ZH -endomorphism of A, say i .Constructing the HNN-extensions

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

G1 = t, G 0|gt 10 = g 10 , g0 G02. Extend 2 to G1 by requiring it to act as identity on the abelian subgroup H, t 1 . Dene,

G2 = t2, G1|gt 21 = g

21 , g1 G1

3. Repeat this r times resulting in G r .


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Claim 1 Gr = G = Q A, where Q = H t1 t r and A = A t1 ,...,t r is the normal closure of Ain A, t 1, . . . , t r .Pf: Starting with G0 = H A. Adding t is and forcing them to commute with H and with a dened 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 t i . For example, if g = ( h, a )

G1

t1gt 11 = 11 (g)

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

Claim 2 G embeds in G and G is metabelian.

Pf: G is generated by the elements

h1, . . . , h n t1, . . . , t r a1, . . . , a m

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

Constructing G

What relations do we have?

1. [h i , h j ] = [t i , t j ] = [h i , t j ] = [a i , a j ] = 1

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

3. at ji = a i p j (h j ) for i = 1 , . . . , m , j = 1 , . . . , r

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

4. ar i 11 ar i 22 a r imm = 1 for i = 1 , . . . , k with r ij ZH

Finally need relations to ensure the normal closure of a1, . . . , a m in G is abelian,

5. [a i , a

j ] = 1 for , of the form hu11 h

u nn where 0 u i di for di the degree of pi , when 1 i r ,and 0 u i < q i for r + 1 i n

DeneG= h1, . . . , h n , t 1, . . . , t r , a 1, . . . , a m |1, 2, 3, 4, 5

The Conclusion There is a surjective homomorphism

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

1. Prove Gmetabelian

2. Then G satises max-n

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

satises max-n N is nitely generated. Therefore G is too.


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Lemma. G is metabelian.

Sketch of Proof:

1. If we show that A= a1, . . . , a m G

is abelian, then

G= h1, . . . , h n , t 1, . . . , t r , a 1, . . . , a m |1, 2, 3, 4, 5G/A = h1, . . . , h n , t 1, . . . , t r |[h i , h j ] = [h i , t j ] = [t i , t j ] = 1 is abelainhence 1 A G is an abelian series of derived length 2 and therefore G is metabelian.

2. To prove A is abelian one uses the special polynomials

The Geometry of Solvable Groups-Amin Saied

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Transcript of The Geometry of Solvable Groups-Amin Saied