bs theorie

Chapter 9

Bass-Serre theory

In this Chapter we introduce the notion of graphs of groups and their funda-mental group. Then we discuss the structure theorem of the Bass-Serre theory,i.e. the duality of graphs of groups and simplicial group actions on trees. For amore detailed account of Bass-Serre theory please refer to the article of H. Bass[?] or the book of J.P. Serre [Ser]. We further introduce the simple notions ofthe core and the free decomposition of a graph of groups.

9.1 Automorphisms of trees

In this section we briey study automorphisms of trees. Let T be a tree. Wewill say that an automorphism Aut (T ) inverts no edges if (e) = e1 forall e ET . From now on all automorphisms have this property.

Put := min{d(w,(w) |w V T}. If = 0 then (v) = v for somev V T . In this case we say that is an elliptic automorphism. If > 0 thenwe say that is a hyperbolic automorphsim. The following lemma describes thedynamics of a hyperbolic automorphism.

Lemma 9.1 Let T be a tree and be a hyperbolic automorphism of T . Put = . Then the following hold:

1. There exists a unique bi-innite -invariant line Y T and acts on Yby translation along .

2. d(v, z(v)) = |z| + 2d(v, Y ) for all v V T and z Z 0, in particulard(v, (v)) = if and only if v V Y .

Proof Choose v V T such that d(v, (v)) = . Put Y = zZ

[z(v), z+1(v)].

In order to show that Y is a bi-innte line it clearly suces to show that[z1(v), z(v)] [z(v), z+1(v)] = z(v) for all z Z. This is clearly equiva-lent to showing that [v, (v)] [(v), 2(v)] = (v).

83

84 CHAPTER 9. BASS-SERRE THEORY

Suppose that [v, (v)] [(v), 2(v)] = (v), i.e. that [v, (v)] [(v), 2(v)]contains a vertex w dierent form (v). This clearly implies that w and 1(w)lie in [v, (v)]. Thus d(w,(w)) = d(w,1(w)) < d(v, (v) = ( which contra-dicts the minimality of .

v

(v)

2(v)w

1(w)

It is now clear that acts as a translation by on Y . Let v be an arbitrary

vertex of T and w be its projection to Y , i.e. the unique vertex w V Y suchthat d(v, Y ) = d(v, w). It is now clear that the segment [v, (v)] equals theunion [v, w] [w,z(w)] [z(w), z(v)].

v z(v)

w z(w)Y

This implies that d(v, (v)) = d(v, w) + d(w,z(w)) + d(z(v), z(w)) =d(v, w)+ |z|+d(z(v), z(w)) = |z|+2d(v, w) = |z|+2d(v, Y ). The unique-ness of the set Y is obvious; thus the lemma is proven.

The proof of the above lemma gives the following simple characterization ofhyperbolic automorphisms:

Lemma 9.2 Let be an automorphism of a tree T . Then is hyperbolic if andonly if there exists a vertex v V T such that [v, (v)] [(v), 2(v)] = (v).Proof If is hyperbolic then any vertex of the -invariant line has this property.If a vertex wiht above properties exists then if follows from the proof of the abovelemma that acts on the line

zZ[z(v), z+1(v)] as a non-trivial translation.

In particular xes no vertex and is therefore hyperbolic.

The next lemma tells us that the product of two elliptic isometries is hyper-bolic unless it is elliptic for the obvious reason.

Lemma 9.3 Let T be a tree and , Aut T be elliptic automorphisms withoutcommon xed point. Then is hyperbolic.

Proof Choose a xed point x of and a xed point y of such that d(x, y) isminimal. This clearly implies that

[x, y] ([x, y]) = [x, y] [(x), (y)] = [x, y] [x, (y)] = x

9.2. GROUP ACTIONS ON TREES 85

and that

[x, y] ([x, y]) = [x, y] [(x), (y)] = [x, y] [(x), y] = yfor = 1, see Figure 9.1.

x y

(x) = (x)

1(x)

1(y)

11(y)

Figure 9.1: and are elliptic with xed points x and y, respectively.

It follows immediately that [11(y), x] [x, (x)] = x which impliesthat is hyperbolic by Lemma 9.2.

9.2 Group actions on trees

Let G be a group and T a simplicial tree. We say that G acts on T or that Tis a G-tree if G acts on T from the left by tree automorphisms that invert noedges. Mory precisely we have a map (G) Aut T such that for any g Gand v V T and e ET we have

gv = (g)(v) and ge = (g)(e) = e1.Let T be a G-tree. We say that g G is elliptic if (g) is elliptic and we say

that g is hyperbolic if (g) is hyperbolic.

For any g G we put |g| = min{d(v, gv) | v V T}. We call |g| the transla-tion length of g. Clearly g is hyperbolic i |g| > 0. For any hyperbolic elementg we further denote the unique g invariant bi-innte line by Ag, also called theaxis of g. The following observation is a simple consequence of Lemma 9.1.

Lemma 9.4 Let T be a G-tree and g G be a hyperbolic element. Then thefollowing hold:

1. g is of innite order.

2. d(v, gv) = |g| if and only if v Ag.3. |gz| = |z||g| for all z Z.We next observe that most interesting group actions on trees always have

hyperbolic elements.


Lemma 9.5 Let G be a nitely generated group and T be a G-tree such thatevery g G is elliptic. Then G acts with a global xed point, i.e. Gx = x forsome x T .

Proof Let {g1, . . . , gn} be a generating set of G. By assumption each gi iselliptic. Thus Ti := {x T | gix = x} T is a non-empty subtree of T for1 i n. It follows from Lemma 9.3 that Ti Tj = for all 1 i, j nas otherwise gigj is hyerbolic contradicting the hypothesis that all elements areelliptic. It is a simple exercise to verify that nitely many subtree of a giventree with pairwise non-trivial intersection have non-trivial intersection. Thusi=1,...,n

Ti = , i.e. G = g1, . . . , gn has a xed point.

If we drop the assumption that G is nitely generated then the conclusion ofLemma 9.5 remains no longer true, see Lemma 9.9 below. The following lemmatells us that groups that act on trees often contain free subgroups, the proofuses a variation of the famous ping pong argument.

Lemma 9.6 Let T be a G-tree and g, h G be hyperbolic such that the length ofAgAh is strictly less than max(|g|, |h|). Then g and h generate a free subgroupof rank 2.

Proof Let : F (g, h) G be the extension of the map g g and h h. Thisalso gives an action of F (g, h) on T via wx = (w)x for every w F (g, h). Tosee that g and h generate a free subgroup of G we have to show that the kernelof is trivial. To do this it clearly suces to show that every w F (g, h) actsnon-trivially on T .

Let w F (g, h). If w is a power of g or h then the assertion is trivial. Thuswe can assume that after conjugation w = gz1hz

1 . . . gzkhzk with zi, zi = 0 for

1 i k. Let p be a vertex of the segment Ag Ah. We show by induction onk 0 that p = wp if k 1 and that

[p, wp] Ah Ag Ah.

which clearly implies the claim.

Ag

Ah

q q wpwp g

z1 q

wphz1q hz1wp

Figure 9.2: Two hyperbolic elements g and h wit Ag Ah bounded


For k = 0 there is nothing to show as w = 1 implies that [p, wp] = [p, p] AgAh. Suppose that k 1, i.e. that w = gz1hz1w with w = gz2hz2 . . .gzkhzk .By induction we have [p, wp] Ah Ag Ah, in particular the projection q ofwp to Ah lies in Ag Ah.

As |h| is assumed to be greater than the length of Ag Ah this implies thatthe projection q of hz

1wp to Ag lies in Ag Ah. It follows that the projection

gz1 q of gz1hz1wp = wp to Ag does not lie in in Ag Ah. This proves the

claim. .

As Ag = Agn for any n 1 and |gn| = n|g| for all n we get the followingimmediate application.

Corollary 9.7 Let T be a G-tree and g, h G such that Ag Ah is bounded.Then there exist numbers n,m N such that gn and hm generate a free subgroupof rank 2.

The following lemma tells us that there are ve types of actions; the mostinteresting one being the hyperbolic actions.

Lemma and Denition 9.8 Let G be a group and T a G-tree. Then one ofthe following holds:

1. G acts trivially on T , i.e. Gv = v for some vertex v V T . We say thatthe action is elliptic.

2. There are two hyperbolic elements g and h such that Ag Ah is bounded.In this case some powers of gn and hm of g and h generate a free group.We say that the action is hyperbolic.

3. T contains an invariant bi-innite G-invariant line Y and G acts on Yby translations. In this case we say that the action is cyclic.

4. T contains an invariant bi-innite G-invariant line Y and G acts on Y bytranslations and reections. In this case we say that the action is dihedral.

5. The action is not elliptic, cyclic or dihedral and there exists an inniteray R in T such that gRR is again an innite ray for all g G. In thiscase we say that the action is parabolic.

Proof Case A: Suppose rst that all elements of G are elliptic. If G is nitelygenerated then we are in case 1 by Lemma 9.5. Thus we can assume that G isnot nitely generated. We show that the action is either elliptic or parabolic.

Suppose that the action is not elliptic. We construct a ray R such thatRgR is an innite ray for all g G. Choose an element g0 and a point x0 Tsuch that gx0 = x0. Then we show inductively that there exist elements gi andpoints xi for all i 1 such that the following hold:

1. For each i 1 the element gi is such that gixi1 = xi1.


2. For each i 1 the point xi is such that gjxi = xi for 0 j i and thatd(xi1, xi) is minimal.

Note that these choices are always possible as each nitely generated sub-group has a xed point by Lemma 9.5 and we assume that G has no xed point.It is clear that the innite union

R := [x0, x1] [x1, x2] [x2, x3] . . .denes a ray in T . To see that the action of G on T is parabolic it suces toshow that any g G xes all but nitely many vertices of R. Suppose that this

x0 x1 xi xi+1 = gi+1xi+1y = gy

py

Figure 9.3: The xed point sets of g and gi+1 are disjoint as gxi = xi = gi+1xi

does not hold for some g G. Choose a vertex y such that gy = y let py bethe vertex of R that is in minimal distance from y. Choose a vertex xi of Rsuch that gxi = xi and that d(x0, xi) > d(x0, py). Such a vertex must exist aswe assume that innitely many vertices of R are not xed by g. It follows thatneither gi+1 nor g x xi and their xed point sets lie in dierent components ofT {xi}. Thus ggi+1 is hyperbolic by Lemma 9.5, a contradiction.

Case B: Suppose that G contains hyperbolic elements. If there exists aG-invariant line then we are clearly in situation 3 or 4. Suppose now that theaction is neither elliptic nor cyclic, dihedral or parabolic. We rst show thatthere exist elements g, h G such that Ag Ah is bounded.

Suppose that Ag Ah is unbounded for all hyperbolic g, h G. We rstshow that there exist g, h such that R = Ag Ah is a ray. If no such elementsexist then all hyperbolic element have the same axis A which must then also bexed by the elliptic elements as Ahgh1 = hAg and we are in situation 3 or 4.

Ag Ahgh1

Ah

Figure 9.4: The axes of the elements g, h and f = hgh1 intersect in the ray R

Thus there are g, h G such that R = Ag Ah is a ray. Put f = hgh1. Itis clear that Ag Af and Ah Af are also infnitite rays that intersect R in an


innite ray, i.e. R = AgAhAf is an innite ray. As we assume that the actionis not parabolic it follows that gR R is not an innite ray for some g G. Itfollows that gR R is bounded. Note that gR = Aggg1 Aghg1 Agf g1 .It is further clear that one the axis Aggg1 , Aghg1 or Agf g1 , say Aggg1 , hasbounded intersection with R. It follows that Aggg1 has bounded intersectionwith Ag, Ah or Af , thus the rst assertion of (2) follows. The remaining claimnow follows from Lemma 9.6.

Note that an action of a non-nitely generated group G on a tree T canbe parabolic even if all elements are elliptic. In this situation the proof ofLemma 9.8 implies that G =

nNStab vn if R = [v0, v1] [v1, v2] . . . is a ray

such that gR R is a ray for all g G. There is partial converse:

Lemma 9.9 Any non-nitely generated countable group G admits a parabolicaction on a tree such that any g G is elliptic.

Proof As G is countable there exists an innite strictly ascending sequence ofsubgroup (Un) such that G =

nNUn. G now acts on a graph with vertex set

consisting of cosets, namely {gUn | g G,n N}. Furthermore two cosets g1Unand g2Um are joined by an edge i |n m| = 1 and g1Un g2Um or g2Um g1Un. G acts on this graph from the left in the natural way by g gUn = (gg)Un.As any g G is containd in some Un it follows that all elements are elliptic. Itis further easily veried that the graph is a tree.

Some groups cannot act non-trivially on a tree; we say that such a grouphas the xed point property or simply say that it has property FA. Lemma 9.5clearly implies the following.

Lemma 9.10 Finite groups have property FA.

We further say that a group G has property AR if any G-tree is eitherelliptic, cyclic or dihedral. Recall further that a group G is noetherian if thereis no innite strictly ascending sequence of subgroups.

Lemma 9.11 Let G be a group. Then the following are equivalent:

1. G is noetherian.

2. Every subgroup of G is nitely generated.

3. Every subgroup of G has property AR.

Proof (1)(2) It is clear that all subgroups of Noetherian groups are nitelygenerated as any non-nitely generated group contains an innite strictly as-cending sequence of subgroups.

(2)(3) Suppose that some subgroup U does not have property AR. Thusthere exists a hyperbolic or a parabolic U -tree T . If the action is hyperbolic thenU contains a free group and therefore also a non-nitely generated free subgroup


contradicting our assumption. Assume now that there exists a parabolic U -treeT . Choose a ray R = [x0, x1] [x1, x2] . . . such that R uR is a ray for allu U . By assumption the group U :=

nNStab xn is nitely generated, thus

there exists some N N such that U = Stab xn for n N . Let further u Ube a hyperbolic element of minimal translation length. Note that R Au isan innite ray. It follows that any element of U can be written as uzu withz Z and u U . Thus Au is U -invariant, i.e. the action is not parabolic, acontradiction.

(3)(1) Suppose that U0, U1, . . . is an innite strictly ascending sequence ofsubgroups. We can assume that all Ui are countable. It follows that the groupU =

nNUn is countable but not nitely generated as any nite set of elements

lies in some Un. As every countable non-nitely generated subgroup admits aparabolic action by Lemma 9.9 this implies that U does not have property ARcontradicting our assumption.

9.3 The universal covering tree

In this section we show how to construct the universal covering tree of a graph ofgroups and describe the action of the fundamental group of the graph of groupson this tree. We discuss two dierent ways of doing this, namely vie graphs ofspaces and their universal covering space and combinatorialy using equvialenceclasses of A-graphs. The rst one is more intuitive, the second one is howevertechnically easier to handle.

Construction B Let A be a graph of groups with base vertex v0 V A. Wedene an equivalence relation on the set of A-paths originating at v0 by sayingthat p p if

1. p and p are both A-path from v0 to v for some v V A.2. p pa for some a Av.

For an A-path p from v0 to v, we shall denote the -equivalence class ofp by pAv. As in the proof of the reduced form theorem we see that we eachequivalence class contains a unique representative of type

a0, e1, a1, . . . , ak1, ek, 1

with ai Rei+1 where Re is a set of left coset representatives of e(Ae) in A(e),i.e. A(e) =

rRer e(Ae).

We now dene the graph (A, v0) as follows.

1. The vertices of (A, v0) are -equivalence classes of A-paths originating atv0. Thus each vertex of (A, v0) has the form pAv, where p is an A-pathfrom v0 to a vertex v V A. We put x0 = 1Av0 .

9.3. THE UNIVERSAL COVERING TREE 91

2. Two vertices x, x of (A, v0) are connected by an edge f = (x, e, x) with(f) = x, (f) = x and e EA if we can express x, x as x = pAv, x =paeAv , where p is an A-path from v0 to v, a Av, (e) = v and (e) = v.

3. The involution on the set of edges is dened the obvious way by putting(v, e, v)1 = (v, e1, v).

If a0, e1, a1, . . . , ak1, ek, ak and a0, e1, a

1, . . . , a

k1, ek , a

k are reduced rep-

resentatives for vertices x and x and k k then it follows from Proposition 6.11that x and x are joined by an edge if and only if k = k + 1 and

a0, e1, a1, . . . , ak1, ek , ak a0, e1, a1, . . . , ak1, ek , ak .

The edge joining x and x is then (x, ek, x).

Theorem 9.12 The graph (A, v0) is a tree. It is called the universal coveringtree or the Bass-Serre tree of (A, v0).

Proof Let q = f1, . . . , fk be a non-trivial closed path in (A, v0); it clearly sucesto show that q is not cyclically reduced, i.e. that either fi+1 = f

1i for some

1 i k 1 or that f1 = f1k . Put yi1 = (fi) for 1 i k and representyi by a reduced path pi for 0 i k 1. Choose i such that the length of pi ismaximal. After a cyclic permutation of the path q we may assume that i = 0.

As yi is represented by the reduced path pi = a0, e1, a1, . . . , ak1, ek, ak andthe length of pi is maximal it follows from the remark before the Theorem thatboth yi1 and yi+1 are represented by the path a0, e1, a1, . . . , ak2, ek1, ak1,i.e. that yi1 = yi+1. It further follows that yi1 is joined to yi by the edgefi = (yi1, ek, yi) and yi+1 it joined to yi by the edge f1i+1 = (yi+1, ek, yi) =(yi1, ek, yi) = fi which proves the assertion.

We give two examples of universal covering trees, namely the free productZ2 Z3 and the Baumslag Solitar group BS(1, 2) which is a HNN-extension.Example 1: The graph of groups A corresponding to the free product Z2 Z3consist of two vertices v0 and v1 with vertex groups Av0 = Z2 and Av1 = Z3and a single edge pair {e, e1} with (e) = (e1) = v0, (e) = (e1) = v1and Ae = 1.

Example 2: The graph of groups A corresponding to the group BS(1, 2)consists of a single vertex v0 with vertex group Av0 = Z and a single edge pair{e, e1} with Ae = Z and boundary monomorphisms e : Z Z, z z ande : Z Z, z 2z.

We next show that the fundamental group G = 1(A, v0) acts in a natu-ral way on T = (A, v0). In the theory of covering spaces this corresponds tothe action of the fundamental group of a space on the universal covering viaDecktransformations.


Figure 9.5: The Bass-Serre tree of Z2 Z3

If g = [q] G (where q is an A-path from v0 to v0) and u = pAv (where p isan A-path from v0 to v V A), then we dene

g u = [q] pAv := qpAv.

This action is cleary well-dened on the set of vertices of (A, v0) and it pre-serves the adjacency relation. Thus G has a canonical simplicial action without

inversions on (A, v0). The following is immediate:

Lemma 9.13 Consider the action of 1(A, v0) of (A, v0).

1. The stabilizer of the vertex x = pAv is pAvp1.

2. The stabilizer of the edge (pAv, e, paeAv) is p(ae(Ae)a1)p1.

9.4 The structure theorem

In the above section we have constructed a tree on which the fundamental groupof a given graph of groups acts. In this section we show that for any group Gand G-tree T we can associate a graph of groups. The structure theorem of theBass-Serre theory says that these two constructions are mutual inverses.

Suppose we have a G-tree T . Recall that we always assume that G acts onT by tree-automorphisms that do not invert edges. In particular we have that

9.4. THE STRUCTURE THEOREM 93

Figure 9.6: The Bass-Serre tree of BS(1, 2)

g(e1) = (ge)1, g(e) = (ge), g((e) = (ge) and ge = e1 for all g G ande ET .

We show how to construct a graph of groups A with underlying graph A =G\T . Let Y be a maximal subtree of A. It is clear that Y can be lifted to T , i.e.that there exists an injective graph morphism i : Y T such that i = idYwhere : T G\T is the canonical projection. We can clearly extend i tothe set EA such that for every edge e EA EY either i((e)) = (i(e)) ori((e)) = (i(e)).

The resulting map will however not respect the graph structure of A unlessY = A, i.e. unless A is a tree. It follows that we can assume that i((e)) =(i(e)) for some e EA EY if A is not a tree (Figure 9.7).

ie

(e) i((e))

i(e) (i(e))

Figure 9.7: The lift a maximal subtree Y of A to T

We then dene Av := StabGi(v) for v V Y = V A and Ae := StabGi(e) for


e EA. The boundary monomorphisms e are dened as follows, recall that itsuces to dene the e as e = e1 for all e EA.

1. e is the inclusion map if (i(e)) = i((e)).

2. If (i(e)) = i((e)) then we choose an element ge G such that (i(e)) =gei((e)) and dene e : Ae A(e) by e(g) = g1e gge.

For those e EA EY with (i(e)) = i((e)) we put ge = g1e1 . As anumber of arbitrary choices were made during the construction of A the resultis not unique. It is however not dicult to verify that any two graphs of groupsobtained this way are equivalent where we say two graphs of groups A and Aare equivalent and write A A if there exists a graph isomorphism f : A A,isomorphisms e : Ae Af(e) for all e EA, isomorphisms v : Av Af(v) forall v V A and elements xe A(e) such that (e)(e(g)) = x1e f(e)(e(g))xefor any e EA and g Ae.

Let A, Y be as above and v0 V A. We will show that there exists an isomor-phism : 1(A, v0) G and -equivariant graph isomorphism f : (A, v0) T .

For any v A let pv = ev1, . . . , evmv be the unique reduced path in Y Ajoining v0 and v and pv = 1, ev1, 1, . . . , 1, evmv , 1 the associated A-path. Recallthat 1(A, v0) is generated by the [p1v Avpv] for v V A together with theelements he := [p(e), e, p

1(e)] for e EA EY . We dene a homomorphism

: 1(A, v0) G

by [p1v avpv] av for all av Av and he ge for e EAEY . It is immediatethat this maps extends to a homomorphism. We further dene

f : (A, v0) T

by f(pvAv) = i(v) and extending this map equivariantly to set of all verticesby putting f(g pvAv) = (g)f(pvAv) for all g 1(A, v0) and v V A. Thismap is clearly well-dened and extends to the set of edges as adjacent verticesget mapped to adjacent vertices. Note further that it is immediate that isbijective on edge and vertex stabilizers.

Theorem 9.14 (The structure theorem) The pair (, f) is an isomorphismof group actions on tree. Thus is an isomorphism, f is a tree isomorphismand

f(gx) = (g)f(x)

for any g 1(A, v0) and x (A, v0)

Proof The equivariance follows from the denition of and f . Thus it sucesto show that f and are bijective.

9.5. MINIMAL ACTIONS AND THE CORE OF A GRAPH OF GROUPS95

To see that is surjective we have to verify that

G = H := vV Y

Av {ge | e EA EY }.

Note that H contains Stab v for all v V Y . To see that G = H it thereforesuces to show thatHV Y = V T . Suppose thatHV Y = V T . As T is connectedand both HV Y and V T are H-invariant it follows that there exist adjacentvertices v V Y and w V T HV Y , thus there exists an edge e1 with(e1) = v and (e1) = w. By construction of Y and H there exists an edgee2 with (e2) = v and w

= (e2) HV Y that is Stab v-equivalent to e1,i.e. there exists a g Stab v such that ge1 = e2. This however implies thatgw = w HV Y as Stab v H, a contradiciton.

The surjectivity of together with the denition of f clearly implies thesurjectivity of f . To see that f is injective it suces to verify that f is locallyinjective. If f is not locally injective then f identies two edges e1 and e2 withv := (e1) = (e2), in particular ge1 = e2 for some g Stab v Stab e1. Note(g) Stab f(e1) = Stab f(e2) contradicting the fact that is bijective onedge stabilizers. Thus f is a bijection.

It remains to verify that is injective. Let g 1(A, v0) 1. If g acts witha xed point that (g) = 1 as is bijective on stabilizers. If g is hyperbolicthen gv = v for some vertex v and therefore (g)f(v) = f(gv) = f(v) as f isinjective. Thus (g) acts non-trivially on T which implies that (g) = 1.

9.5 Minimal actions and the core of a graph ofgroups

We will often restrict ourselves to situations where the action of a group G ona tree T is minimal, i.e. where T contains no proper subtree that is invariantunder the action of G. If G acts without xed point of T this is simply achievedby replacing T with the minimal G-invariant subtree of T . If the action is ellipticwe replace T with an arbitrary vertex of T that is xed by G. We say that agraph of groups is minimal if the corresponding action of the fundamental groupon the Bass-Serre tree is minimal.

Given a minimal graph of groups A, the structure theorem makes it easyto determine of what type the action of the fundamental group on the univer-sal covering tree is. Proposition 9.15 below implies in particular that in theabove examples the action of Z2 Z3 is hyperbolic and the action of BS(1, 2) isparabolic.

To simplify the statement we restrict attention to reduced graphs of groups.We say that a graph of groups A is reduced if for every vertex v V A of valence2 and non-loop edge e with (e) = v we have that e is not surjective. If agraph of groups is not reduced and v and e are as before then we say that thepair (v, e) is inessential.


Note that any nite graph of groups can be easily modied to be reducedby repeatedly collapsing inessential pairs: Suppose that (v, e) is an inessentialpair. We then replace the graph of groups A by the graph of groups A obtainedfrom A by omitting the edge e and vertex v, putting (f) = (f1) = (e) forall e with (e) = v and putting e = e1 = e 1e f . It is easily seen thatthis procedure does not change the fundamental group of the graph of groups.

Proposition 9.15 Let A be a minimal reduced graph of groups. Let G =1(A, v0) and T = (A, v0). The action of G on T is

1. elliptic i the graph A underlying A consists of a single vertex.

2. cyclic i A consists of a single loop-edge and both boundary monomor-phisms are surjective. Thus there is a short exact sequence 1 Av G Z 1.

3. dihedral i A consists of a single non-loop edge and the image of bothboundary monomorphisms is of index 2 in the vertex group. Thus G =A C B wih |A : C| = |B : C| = 2.

4. parabolic i A consists of a single loop edge and one boundary monomor-phism is surjective.

5. hyperbolic otherwise.

Proof Part (1) is trivial. Suppose now the action is cyclic. The minimalityimplies that T is a line and that G acts on T by translation, it follows that G\Tis a circle. It is further clear that any element that xes a vertex already xesall of T thus all boundary monomorphisms are surjective. As we assume thatA is reduced this implies that the loop G\T consistes of a single loop-edge.

Suppose next that the action is dihedral. Thus T is a line and G acts bytranslations and reection. In particular G\T is a segment, i.e. has verticesv0, . . . , vk and edges e1, . . . , ek such that (ei) = vi1 and (ei) = vi for 1 i k. Note that any element that stabilizes an edge stabilizes all of T , that thestabilizer of T is of index two in the stabilizer of those vertices that along whichreections occurs and coincides with the stabilizer of the remaining vertices.As the vertices v1, . . . , vk1 correspond to vertices without reections it followsthat the reducedness assumption implies that they cannot exist, i.e. that k = 1.This proves the claim.

We conclude by investigating the parabolic case with ray R. If all elementsare elliptic then R gets embedded into G\T under the quotient map and itis easily veried that the graph of groups A cannot be reduced. Thus we canassume that G contain a hyperbolic element. Choose a hyperbolic element g Gof minimal translation length. Pick vertices v, w R such that d(v, w) = |g|. Itis clear that the segment [v, w] projects onto G\T and that the map is injectiveexcept that v is being identied with w. This is true as G must be generatedby g and Stab v. The assertion now follows easily.

9.5. MINIMAL ACTIONS AND THE CORE OF A GRAPH OF GROUPS97

An immediate consequence is the following characterisation of groups thathave the xed point property FA.

Lemma 9.16 Let G be group. Then G has property FA if and only if thefollowing hold:

1. G is nitely generated.

2. G does not split as a proper amalgamated product.

3. G does not split as an HNN-extension.

Proof Note rst that if G has property FA then it must be nitely generatedby Lemma 9.9. It further cannot split as an amalgamated product or an HNN-extension as G would then admit a non-elliptic action on the correspondingBass-Serre tree.

Suppose now that G fullls conditions (1)-(3) and that T is a G-tree. Wehave to show that the action of G is elliptic; because of Lemma 9.5 it sucesto show that every element g G is elliptic. If some element is hyperbolicthen the action is not elliptic and G splits as an amalgamated product or anHNN-extension by Proposition 9.15.

In the following we establish the notion of the core of a graph of groups, asubgraph of groups that captures the essential part of a graph of groups. Let Abe a graph of groups. We then dene

core(A) = A()

where the subgraph A is chosen as follows.

1. If there exists a vertex v V A such that the inclusion of Av in 1(A) issurjective then is the subtree of A that is spanned by all vertices thathave this property. Notice that in this case A must be a tree thus is awell-dened subtree. In this case we say that A has a simple core.

2. Otherwise we choose to be the minimal (with respect to inclusion)subgraph of A such that the inclusion of 1(A(), v0) into 1(A, v0) issurjective for one/any v0 V. The uniqueness of is easily veried.

We associate to any G-tree T the subtree TG T in the following way: IfG acts without xed point we dene TG to be the minimal G-invariant subtreeof T . If G acts with xed point then one could choose TG to be any point thatis xed under the action of G. In order to have a well-dened object we will inthe latter case dene TG T to be the subtree that consists of all points thatare xed under the action of G. It is easily veried that the tree TG has thefollowing relationship with the core of a graph of groups.

Lemma 9.17 Let A be a graph of groups, G = 1(A, v0) and T = (A, v0).Choose TG as above. Then core(A) is equivalent to A(G\TG).


The following lemma describes the relationship between a graph of groupsand its core in somewhat more detail, it essentially means that a graph of groupscan be obtained from its core by attaching inessential trees.

Lemma 9.18 Let A be a graph of groups and A() its core. Suppose that thecore is not simple.

Then for every v V AV there exists a unique edge path e1, . . . , ek suchthat the following hold.

1. ei EA E for 1 i k.

2. (e1) V A and (ei) = (ei+1) V AV for 1 i < k and (ek) = v.

3. ei(Aei) = A(ei) for 1 i k.

We will occasionally depict a graph of groups A by drawing A such that theedges of , i.e. the edges corresponding to its core are thicker. Note that inFigure 9.8 we have Av1 , Av2 Av3 and Av7 , Av8 , Av9 Av6

v1

v2

v3 v4

v5

v6 v7

v8

v9

Figure 9.8: A graph of groups and its core

9.6 The free decomposition of a graph of groups

If the core of a graph of group has a trivial edge group then its fundamentalgroup decomposes as a free product. This simple observation gives rise to thenotion a the free decomposition of a graph of groups.

Let A be a graph of groups, E EA the set of edges with trivial edge groupand A = AE. Let Ai A with i I for some index set I be the componentsof A. We call the subgraphs of groups Ai = A(Ai) the free factors of A. We willsay that a free factor is cyclic if its fundamental group is cyclic.

Let further n be the number of edges outside a maximal tree of the graphA obtained from A by contracting the subgraphs Ai to a point. We call n thefree rank of A. It is easy to see that

1(A) = ri=1

1(Ai) Fn.

9.6. THE FREE DECOMPOSITION OF A GRAPH OF GROUPS 99

Note however that this decomposition does not necessarily coincide with thecanonical decomposition of 1(A) into free factors and a free group as we do notexclude that 1(Ai) is innite cyclic or a proper free product for some i I.We call the pair

cf (A) := (r, n)

the free complexity of A.We will depict the free decomposition of A by drawing all edges with non-

trivial edge groups as solid lines and edges with trivial edge groups as dottedlines. We furthermore draw vertices that have non-trivial vertex groups but areonly incident with edges with trivial edge groups as fat dots. Clearly every fatvertex corresponds to a free factor all by itself.

Figure 9.9: A graph of groups A with free complexity cf (A) = (3, 2)

Occasionally we will combine this type of gure with the gures introducedin Section 9.5, i.e. we will draw the subgraph corresponding to a free factorwith thick and thin lines, where the thick lines correspond to the core of thefree factor.

Figure 9.10: A graph of groups with three free factors and their cores

bs theorie

Documents