Lecture 4, Geometric and asymptotic group theorymath.hunter.cuny.edu/olgak/Slides_lecture4.pdf · Preliminaries Presentations of groups Finitely generated groups viewed as metric

PreliminariesPresentations of groups

Finitely generated groups viewed as metric spacesQuasi-isometry

Limit groupsFree actions

Lecture 4, Geometric and asymptotic grouptheory

Olga Kharlampovich

NYC, Sep 16

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Homomorphisms of groups

The universal property of free groups allows one to describearbitrary groups in terms of generators and relators. Let G be agroup with a generating set S . By the universal property of freegroups there exists a homomorphism ϕ : F (S)→ G such thatϕ(s) = s for s ∈ S . It follows that ϕ is onto, so by the firstisomorphism theorem

G ' F (S)/ker(ϕ).

In this event ker(ϕ) is viewed as the set of relators of G , and agroup word w ∈ ker(ϕ) is called a relator of G in generators S . If asubset R ⊂ ker(ϕ) generates ker(ϕ) as a normal subgroup of F (S)then it is termed a set of defining relations of G relative to S .

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The pair 〈S | R〉 is called a presentation of G , it determines Guniquely up to isomorphism. The presentation 〈S | R〉 is finite ifboth sets S and R are finite. A group is finitely presented if it hasat least one finite presentation. Presentations provide a universalmethod to describe groups. Example of finite presentations

1 G = 〈s1, . . . , sn | [si , sj ], ∀1 ≤ i < j ≤ n〉 is the free abeliangroup of rank n.

2 Cn = 〈s | sn = 1〉 is the cyclic group of order n.

3 Both presentations 〈a, b | ba2b−1a−3〉 and〈a, b | ba2b−1a−3, [bab−1, a3]〉 define the Baumslag-Solitargroup BS(2, 3) (HWP7 Prove that these presentations defineisomorphic groups).

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If a group G is defined by a presentation, then one can try to findhomomorphisms from G into other groups.

Lemma

Let G = 〈S | R〉 be a group defined by a (finite) presentation with

the set of relators R = {rj = y(j)i1. . . y

(j)ij| y

(j)i ∈ S±1, 1 ≤ j ≤ m},

and let H be an arbitrary group. A map ψ : S±1 → H extends to ahomomorphism ψ : G → H, if and only if

ψ(rj) = ψ(y(j)i1

) . . . ψ(y(j)ij

) = 1 in H for all rj ∈ R.

Proof Define the map ψ : G → H by

ψ(yn1 . . . ynt ) = ψ(yn1) . . . ψ(ynt ),

whenever yni ∈ S±1. If ψ is a homomorphism, then obviouslyψ(rj) = 1 for all rj ∈ R.

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The converse follows from

Lemma (Mapping property of quotient groups)

Let N be a normal subgroup of G , let G = G/N, and letπ : G → G be the canonical map, π(g) = g = gN. Letφ : G → G ′ be a homomorphism such that N ≤ Ker(φ). Thenthere is a unique homomorphism φ : G → G ′ such that φ ◦ π = φ.This map is defined by the rule φ(g) = φ(g).

π

G → G

φ↘ ↓ φG ′

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Let G be a group, by the commutant (or derived subgroup) G ′ ofG we mean the subgroup generated by all the commutators[g , b] = gbg−1b−1 in G . Since a[g , b]a−1 = [aga−1, aba−1], thecommutant is a normal subgroup of G . The quotient G/G ′ iscalled the abelianization of G . G/G ′ is an abelian group.For example, the abelianization of a free group Fn is the freeabelian group of rank n. In general, if

G = 〈s1, . . . , sn | r1, . . . , rm〉, then

G/G ′ = 〈s1, . . . , sn | r1, . . . , rm, [si , sj ](1 ≤ i < j ≤ n)〉

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As the following corollary shows, the abelianization G/G ′ is thelargest abelian quotient of G , in a sense.

Corollary

Let H be an abelian quotient of G, and let ν : G → G/G ′ andψ : G → H be the natural homomorphisms. Then there is ahomomorphism ϕ : G/G ′ → H so that the following diagramcommutes:

G → G/G ′

ψ ↘ ↓ ϕH

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Proof.

Let G be generated by S = {s1, . . . , sn}, then G/G ′ is generatedby ν(S) = {ν(s1), . . . , ν(sn)}. We still denote ν(si ) by si , since wewant to fix the alphabet S±1 for both G and G/G ′. Hence, G/G ′

has the presentation above. Define a map ϕ′ : ν(S)→ H byϕ′(si ) = ψ(si ) for all i . Observe that ϕ′(rj) = ψ(rj) = 1 in H, sinceψ is a homomorphism and rj = 1 in G . Also,ϕ′([si , sj ]) = ψ([si , sj ]) = [ψ(si ), ψ(sj)] = 1, since H is abelian. Itfollows now from the previous lemma that the map ϕ′ extends to ahomomorphism from G/G ′ to H.

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More groups given by generators and relations

The free Burnside group of exponent n with two generators isgiven by the presentation

〈a, b | un = 1〉

for all words u in the alphabet a, b.The fundamental group of the orientable surface of genus n isgiven by the presentation

〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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〈a, b | un = 1〉


〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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〈a, b | un = 1〉


〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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〈a, b | un = 1〉


〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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〈a, b | un = 1〉


〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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〈a, b | un = 1〉


〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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〈a, b | un = 1〉


〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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〈a, b | un = 1〉


〈a1, b1, ..., an, bn | [a1, b1]...[an, bn] = 1〉.

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Group presentations

The object of study in Geometric Group Theory- finitely generatedgroups given by presentations 〈a1, ..., an | r1, r2, ...〉, where ri is aword in a1, ..., an. That is groups generated by a1, ..., an withrelations r1 = 1, r2 = 1, ... imposed.

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Group presentations


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Group presentations


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Group presentations


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Some classical results

Theorem. (Boone-Novikov’s solution of Dehn’s problem) Thereexists a finitely presented group with undecidable word problem.

Theorem. (Higman) A group has recursively enumerable wordproblem iff it is a subgroup of a finitely presented group.

Theorem. (Adian-Novikov’s solution of Burnside problem) Thefree Burnside group of exponent n with at least two generators isinfinite for large enough odd n.It is still unknown if such a group is infinite for n = 5, 8 etc

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The growth rate of a group is a well-defined notion fromasymptotic analysis. To say that a finitely generated group haspolynomial growth means the number of elements of length(relative to a symmetric generating set) at most n is boundedabove by a polynomial function p(n). The order of growth is thenthe least degree of any such polynomial function p.A nilpotent group G is a group with a lower central seriesterminating in the identity subgroup.Theorem. (Gromov’s solution of Milnor’s problem) Any group ofpolynomial growth has a nilpotent subgroup of finite index.

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Action of a group on a set

A group G acts on a set X if for each g ∈ G there is a bijectionx → gx defined on X such that

ex = x , (g1(g2(x)) = (g1g2)(x).

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Finitely generated groups viewed as metric spaces

Let G be a group given as a quotient π : F (S)→ G of the freegroup on a set S . Therefore G = 〈S |R〉. The word length |g | of anelement g ∈ G is the smallest integer n for which there exists asequence s1, . . . , sn of elements in S ∪ S−1 such thatg = π(s1 . . . sn). The word metric dS(g1, g2) is defined on G by

dS(g1, g2) = |g−11 g2|.

G acts on itself from the left by isometries.

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Cayley graph

Note, that if S and S are two finite generating sets of G then dS

and dS are bi-Lipschitz equivalent, namely ∃C∀g1, g2 ∈ G( 1C dS(g1, g2) ≤ dS(g1, g2) ≤ CdS(g1, g2).)

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Polynomial growth

A ball of radius n in Cay(G , S) is

Bn = {g ∈ G ||g | ≤ n}.

G has polynomial growth iff the number of elements in Bn isbounded by a polynomial p(n).

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Hyperbolic groups

A geodesic metric space is called δ-hyperbolic if for every geodesictriangle, each edge is contained in the δ neighborhood of the unionof the other two edges. If δ = 0 the space is called a real tree orR-tree.A group G is hyperbolic Cay(G ,X ) is hyperbolic.

A finitely generated group is called hyperbolic if its Cayley graph ishyperbolic.

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Quasi-isometry

Definition Let (X , dX ) and (Y , dY ) be metric spaces. Given realnumbers k ≥ 1 and C ≥ 0, a map f : X → Y is called a(k ,C )-quasi-isometry if

1 1k dX (x1, x2)− C ≤ dY (f (x1), f (x2)) ≤ kdX (x1, x2) + C for allx1, x2 ∈ X ,

2 the C neighborhood of f (X ) is all of Y .

Examples of quasi-isometries1. (Z; d) and (R; d) are quasi-isometric.The natural embedding of Z in R is isometry. It is not surjective,but each point of R is at most 1/2 away from Z.

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All regular trees of valence at least 3 are quasi-isometric. Wedenote by Tk the regular tree of valence k and we show that T3 isquasi-isometric to Tk for every k ≥ 4. We define the mapq : T3 → Tk , sending all edges drawn in thin lines isometricallyonto edges and all paths of length k − 3 drawn in thick lines ontoone vertex. The map q thus defined is surjective and it satisfiesthe inequality

1

k − 2dist(x , y)− 1 ≤ dist(q(x), q(y)) ≤ dist(x , y).

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3. Free groups of finite rank are quasi-isometric

All non-Abelian free groups of finite rank are quasi-isometric toeach other. The Cayley graph of the free group of rank n withrespect to a set of n generators and their inverses is the regularsimplicial tree of valence 2n.

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4. Let G be a group with a finite generating set S , and letCay(G , S) be the corresponding Cayley graph. We can makeCay(G ,X ) into a metric space by identifying each edge with a unitinterval [0, 1] in R and defining d(x , y) to be the length of theshortest path joining x to y . This coincides with the path-lengthmetric when x and y are vertices. Since every point of Cay(G ,X )is in the 1/2 -neighbourhood of some vertex, we see that G andCay(G , S) are quasi-isometric for this choice of d .

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Quasi-isometry

5. Every bounded metric space is quasi-isometric to a point.6. If S and T are finite generating sets for a group G , then (G , dS)and (G , dT ) are quasi-isometric.7. The main example, which partly justifies the interest inquasi-isometries, is the following. Given M a compact Riemannianmanifold, let M be its universal covering and let π1(M) be itsfundamental group. The group π1(M) is finitely generated, in facteven finitely presented.The metric space M with the Riemannian metric is quasi-isometricto π1(M) with some word metric.8. If G1 is a finite index subgroup of G , then G and G1 arequasi-isometrically equivalent (HW8).

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Quasi-isometric rigidity

Classes of groups K complete with respect toquasi-isometries (every group quasi-isometric to a groupfrom K has a finite index subgroup in K)

Finitely presented groups,

Nilpotent groups,

Abelian groups,

Hyperbolic groups,

nonabelian free groups of finite rank (follows from the factthat their Cayley graphs are trees).

Amenable groups (see below)

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Quasi-isometric rigidity

Eskin, Fisher, Whyte obtained first results on quasi-isometricrigidity of non-nilpotent polycyclic groups.Theorem Any group quasi-isometric to the three dimensionalsolvable Lie group Sol is virtually a lattice in Sol.That completed the classification of three-dimensional manifoldsup to quasi-isometry started by Thurston, Schwartz and others.Conjecture Let G be a solvable Lie group, and Γ a lattice in G .Any finitely generated group Γ′ quasi-isometric to Γ is virtually alattice in a solvable Lie group G ′.Equivalently, any f.g. group quasi-isometric to a polycyclic group isvirtually polycyclic.

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Limit groups (fully residually free groups)

A marked group (G , S) is a group G with a prescribed family ofgenerators S = (s1, . . . , sn).Two marked groups (G , (s1, . . . , sn)) and (G ′, (s ′1, . . . , s

′n)) are

isomorphic as marked groups if the bijection si ←→ s ′i extends toan isomorphism. For example, (〈a〉, (1, a)) and (〈a〉, (a, 1)) are notisomorphic as marked groups. Denote by Gn the set of groupsmarked by n elements up to isomorphism of marked groups.One can define a metric on Gn by setting the distance between twomarked groups (G ,S) and (G ′,S ′) to be e−N if they have exactlythe same relations of length at most N (under the bijectionS ←→ S ′) (Grigorchuk, Gromov’s metric)Finally, a limit group is a limit (with respect to the metric above)of marked free groups in Gn.

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limits of free groups

Example: A free abelian group of rank 2 is a limit of a sequence ofcyclic groups with marking

(〈a〉, (a, an)), n→∞.

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limits of free groups

Theorem Let G be a finitely generated group. Then the followingconditions are equivalent:

1) G is fully residually free (that is for finitely many non-trivialelements g1, . . . , gn ∈ G there exists a homomorphism φ fromG to a free group such that φ(gi ) 6= 1 for i = 1, . . . , n);

2) [Champetier and Guirardel] G is a limit of free groups inGromov-Grigorchuk metric.

3) [Remeslennikov] G is universally equivalent to F (in thelanguage without constants);

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Free actions on metric spaces

Theorem. A group G is free if and only if it acts freely byisometries on a tree.

Free action = no inversion of edges and stabilizers of vertices aretrivial.

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R-trees

An R-tree is a metric space (X , p) (where p : X × X → R) whichsatisfies the following properties:

1) (X , p) is geodesic,

2) if two segments of (X , p) intersect in a single point, which isan endpoint of both, then their union is a segment,

3) the intersection of two segments with a common endpoint isalso a segment.

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R-trees

An R-tree is a metric space (X , p) (where p : X × X → R) whichsatisfies the following properties:

1) (X , p) is geodesic,

2) if two segments of (X , p) intersect in a single point, which isan endpoint of both, then their union is a segment,

3) the intersection of two segments with a common endpoint isalso a segment.

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Examples

X = R with usual metric.

A geometric realization of a simplicial tree.

X = R2 with metric d defined by

d((x1, y1), (x2, y2)) =

{|y1|+ |y2|+ |x1 − x2| if x1 6= x2

|y1 − y2| if x1 = x2

x

(x1,y1)

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Examples




d((x1, y1), (x2, y2)) =

{|y1|+ |y2|+ |x1 − x2| if x1 6= x2

|y1 − y2| if x1 = x2

x

(x1,y1)

(x2,y2)30 / 32




Examples




d((x1, y1), (x2, y2)) =

{|y1|+ |y2|+ |x1 − x2| if x1 6= x2

|y1 − y2| if x1 = x2

x

(x1,y1)

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X = R2 with SNCF metric (French Railway System)

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Finitely generated R-free groups

Rips’ Theorem [Rips, 1991 - not published]

A f.g. group is R-free (acts freely on an R-tree by isometries) ifand only if it is a free product of surface groups (except for thenon-orientable surfaces of genus 1,2, 3) and free abelian groups offinite rank.

Gaboriau, Levitt, Paulin (1994) gave a complete proof of Rips’Theorem.

Bestvina, Feighn (1995) gave another proof of Rips’ Theoremproving a more general result for stable actions on R-trees.

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Lecture 4, Geometric and asymptotic group theorymath.hunter.cuny.edu/olgak/Slides_lecture4.pdf · Preliminaries Presentations of groups Finitely generated groups viewed as metric