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Almost Convex Groups and the Eight Geometries

arXiv:math/9306202v1 [math.GR] 16 Jun 1993Almost Convex Groups and the Eight GeometriesM. Shapiro and M. Stein*Abstract.

If M is a closed Nil geometry 3-manifold then π1(M) is almost convexwith respect to a fairly simple “geometric” generating set.If G is a centralextension or a Z extension of a word hyperbolic group, then G is also almostconvex with respect to some generating set. Combining these with previouslyknown results shows that if M is a closed 3-manifold with one of Thurston’s eightgeometries, π1(M) is almost convex with respect to some generating set if andonly if the geometry in question is not Sol.Introduction.In [C] , Cannon introduced the notion of a finitely generated group being almost convexwith respect to a given generating set.

This property is formulated in terms of the geometryof the Cayley graph, and gives a simple and efficient algorithm for constructing the Cayleygraph. One class of groups one would like to study in terms of this property is the class offundamental groups of closed 3-manifolds carrying one of Thurston’s eight geometries.

(Foran account of these, see [Sc] .) In most cases, the answer is already known.

If G = π1(M)where M is a Riemannian 3-manifold whose universal cover is S3, H3, or S2 ×R, then G isword hyperbolic and thus almost convex with respect to any finite generating set. If M iscovered by Euclidean 3-space, E3, then Cannon [C] shows that G is almost convex.

NowCannon et al. [CFGT] show that if M is a compact quotient of Sol, then G is not almostconvex with respect to any generating set.For the remaining three geometries, it is known that G is almost convex with respectto some generating set for special cases.

Specifically, if the universal cover is Nil and Mfibers over a torus, then G is almost convex with respect to a standard generating set[Sh1]. If the universal cover is ^PSL2R and M fibers over a closed orientable surface, thenG is almost convex with respect to a standard generating set [Sh2].

And if the universalcover is H2 × R and G = H × Z where H is a hyperbolic surface group, then G is almostconvex with respect to any split generating set. Note that in all the remaining cases, Gis a finite index supergroup of the Nil,^PSL2R and H2 × R cases we have just discussed.Now almost convexity is not a commensurability invariant.

Indeed, as Thiel has shown[T], it is not a group invariant, but does in fact depend on generating set. Still, it haslong seemed likely that the remaining Nil,^PSL2R and H2 × R groups would turn out tobe almost convex.

We will show this to be true, thus proving* Both authors thank the NSF for support.1

Theorem 1. Suppose G is the fundamental group of a closed 3-manifold M carrying oneof Thurston’s eight geometries.

Then G is almost convex with respect to some generatingset if and only if M is not a Sol geometry manifold.This paper is organized as follows. Section 1 contains background and definitions.In Section 2 we establish almost convexity for the remaining Nil groups.

In Section 3 weestablish almost convexity for the remaining^PSL2R and H2 × R cases. In the course ofthis, we prove the following general result.Theorem 2.

Suppose that H is word hyperbolic, that A is finitely generated abelian, andthat1 −→A −→Gπ−→H −→1is a central extension. Then there is a finite generating set G for G so that G is almostconvex with respect to G.1.

Background and definitions.Let G be a finitely generated group and G a finite set, and a 7→a a map of G to a monoidgenerating set G ⊂G. We use G∗to denote the free monoid on G. We refer to the elementsof G∗as words.

A word w = a1 . .

. an is said to have length n. This is denoted ℓ(w) = n.The map of G into G extends to a unique monoid homomorphism which we denote byw 7→w.

We will assume that G is supplied with an involution a 7→a−1 and that thisinvolution respects inverses in G, that is, a−1 = (a)−1. In all cases that we will consider,the map of G into G is an injection.

Thus we can make the following convention: any listof generators will be taken to include the inverses of those listed. Thus a definition suchas X = {x, y} will mean X = {x±1, y±1}.

If a generating set contains some element ρ sothat ρ has order two, we will take ρ = ρ−1.Given such a group G and G, we can form the Cayley graph of G with respect toG. This is a directed labelled graph Γ = ΓG(G).

The vertices of Γ are the elements ofG.There is a directed edge (g, a, g′) from g to g′ with label a exactly when g′ = gawith a ∈G. Since G generates G, Γ is connected.

One turns Γ into a metric space bydeclaring each edge isomorphic with the unit interval and taking the induced path metric.We denote this metric by d(·, ·) = dG(·, ·). This, in turn, gives each element x ∈Γ a length,ℓ(x) = ℓG = dG(1, x).

As usual, we take the ball of radius r, B(r) to be {x ∈Γ | ℓ(x) ≤1}.Each edge path in Γ is labelled by a unique element of G∗. We identify each wordwith the edge path it labels starting at 1 ∈G.

We take this path to be parameterized withunit speed, and extend each word w to a map of [0, ∞) by setting w(t) = w for t ≥ℓ(w).The translate of the path w by the group element g is denoted gw. This is the path basedat g bearing label w. We say that w is a geodesic if w|[0,ℓ(w)] is an isometry.

Equivalently,w is a geodesic if ℓ(w) = ℓ(w). We say that w is a (λ, ǫ) quasigeodesic if for every subwordu of w, ℓ(u) ≤λℓ(u) + ǫ.Following [C], we say that G is almost convex (m) with respect to G if there is aconstant K(m) with the following property: if ℓG(g) = ℓG(g′) = n and dG(g, g′) ≤mthen there is an edgepath p in Γ which runs from g to g′, lies inside B(n) and has lengthbounded by K(n).

We say that G is almost convex with respect to G if it is almost convex2

(m) with respect to G for all m. It is a result of [C] that if G is almost convex (2) withrespect to G, then G is almost convex with respect to G. We will say that G is almostconvex if there is some G so that G is almost convex with respect to G.In Section 3, we will need some standard results concerning word hyperbolic groups.For an account of these, see, for example, [Sho]. We say that G is word hyperbolic if thereis a generating set G and a constant δ so that if α, β and γ are geodesic edge paths forminga triangle in ΓG(G), then if p is any point on α, d(p, β ∪γ) ≤δ.

In fact, the existence ofsuch a δ is independent of choice of G. Given G and G, we say D = {r1, . .

., rk} ⊂G∗is aDehn’s algorithm if for each ri ∈D, we have ri = 1 and if for any w ∈G∗, if w = 1 thenw contains more than half of some ri ∈D as a subword. In fact, the existence of a Dehn’salgorithm can be taken as a definition of a word hyperbolic group.

That is, G is wordhyperbolic if and only if for any generating set G, there is a Dehn’s algorithm D ⊂G∗.Given a Dehn’s algorithm D, we say a word w is D-reduced if w does not contain morethan half of any word in D. By standard methods, one can check that given a Dehn’salgorithm D, there are λ and ǫ so that D reduced words are (λ, ǫ) quasigeodesics. It is astandard hyperbolic result that for any distance m, there is a constant k(m) so that if u, vare (λ, ǫ) quasigeodesics with d(u, v) ≤m then each point of u lies within distance k(m)of v and vice versa.2.

Nil manifold groups.It is shown in [Sh1] that if M in a closed 3-manifold with Nil geometry, and M fibersover a torus then π1(M) is almost convex. More specifically,Theorem 3.

LetN = ⟨x, y | [[x, y], x] = [[x, y], y] = 1⟩,N e = ⟨x, y, z | [x, z] = [y, z] = 1, [x, y] = ze⟩.We takeN = {x, y} ⊂N,N e = {x, y, z} ⊂N e.Then N is almost convex with respect to N . For e ≥1, N e is almost convex with respectto N e.Proof.

The assertions about N and N e for e > 1 are proven in [Sh1] . The assertion aboutN 1 uses the following lemma.Lemma 4.

Suppose G and H are generating sets for G such that G is almost convex withrespect to G, and there exists k such that for all g ∈G, |ℓG(g) −ℓH(g)| ≤k. Then G isalmost convex with respect to H.Proof.

Let λ = max({ℓG(h) | h ∈H} ∪{ℓH(g) | g ∈G}). Suppose ℓH(h) = ℓH(h′) = n anddH(h, h′) ≤2.

We must find a path of bounded length from h to h′ lying inside the H ballof radius n. We can assume that n ≥2k + λ + 1, for otherwise, h and h′ are connectedby a path of length at most 4k + 2λ + 2. Let a1 .

. .an and b1 .

. .bn be H geodesics for hand h′.

Then g = a1 . .

. an−2k−λ−1 and g′ = b1 .

. .bn−2k−λ−1 both lie inside the G ball of3

radius n −k −λ −1, and lie within G distance λ(4k + 2λ + 4) of each other. Since G isalmost convex with respect to G, they are connected by a G path p of length at most Kwhich lies entirely within the G ball of radius n −k −λ −1.

The path p is easily turnedinto an H path p′ of H length at most λK whose vertices all lie in the G ball of radiusn −k −1. In particular, p′ lies entirely within the H ball of radius n, and so does the pathfrom h to h′ labelled a−1n .

. .

a−1n−2k−λp′bn−2k−λ . .

. bn.Continuing with the proof of Theorem 3, we first note that N is isomorphic to N 1.We now observe that if g ∈N < N e thenℓN e(g) ≤ℓN(g) ≤ℓN e(g) + 20.The first inequality is obvious.

To see the second, notice that a geodesic cannot containboth z and z−1, for these could be commuted together and deleted. Nor can any geodesiccontain more than 25 z’s or z−1’s.

For these could be commuted to the end of the word andreplaced with [x5, y5]±1, thus shortening the word. Finally, check that for −25 ≤i ≤25,ℓN(zi) ≤20.

Now applying Lemma 4 completes the proof that N is almost convex withrespect to N 1.We have seen that if M is a Nil manifold that fibers over the torus, then π1(M) isalmost convex with respect to the above generating sets. We would like to extend thisresult to the general case.

Now, every Nil manifold fibers* over a 2-dimensional Euclideanorbifold. In fact, the list of orbifolds E which occur in this role is quite restrictive.

To seethis, first recall from [Sc] that there are no orientation reversing isometries of Nil. Thisimplies that E can not have any reflector curves, for each reflector curve must lift to anorientation reversing element of π1(M).

Thus the underlying surface of E must be closedand its singularities (if any) are all cone points. Now E must have 0 as its orbifold Eulercharacteristic.

Thus, if E is orientable, then E is either the torus or one of S(2, 2, 2, 2),S(2, 4, 4), S(3, 3, 3), S(2, 3, 6). (These denote the sphere with cone points of the orderslisted.) If E is not orientable, then E is either the Klein bottle or P(2, 2), the projecitveplane with two cone points of order 2.In each of these cases, if G = π1(M), we have the diagram11↓↓1→Z→N e→Z2→1↓↓1→Z→G→πorb1(E)→1↓↓Q∼=Q↓↓11If E is orientable, each of the Z kernels is central.

In the case where E is not orientable,each element of πorb1(E) acts trivially or nontrivially on the Z kernel depending on whetherit is orientation preserving or reversing. * More properly, we should say “Seifert fibers,” but we will not keep up this distinction.4

Our strategy for all of these groups is to use the extension1 →N e →G →Q →1to lift the almost convexity of N e up to G. We first establish the lemmas we will need.Lemma 5. Suppose1 →Hi→Gp→Q →1with Q finite.

Suppose H and G are generating sets for H and G respectively so that H ⊂G,i is an isometry, and the elements of G \ H permute the elements of H when acting byconjugation. Then there is a finite set T ⊂(G \ H)∗so that every element of G has Ggeodesic lying in H∗T.Proof.

We will take T = {t ∈(G \ H)∗: ℓ(t) ≤#Q}.Let g ∈G and suppose w is a geodesic with w = g. Since the elements of G\H permutethe elements of H, it is easy to see that we can find w′ so that w = w′, ℓ(w) = ℓ(w′) andw′ = h1 . .

. hmg1 .

. .

gn where hi ∈H for 1 ≤i ≤m and gi ∈G \ H for 1 ≤i ≤n.If n ≤#Q, we are done. If not, then by the pigeon hole principle, there are j and k,1 ≤j < k ≤n so that p(g1 .

. .gj) = p(g1 .

. .

gk). Then gj+1 .

. .

gk ∈H. Since i is anisometry, we can replace gj+1 .

. .

gk by an expression of equal length in H∗. Continuingwith these two processes produces a geodesic of the desired form.Corollary 6.

Suppose1 →Hi→Gp→Q →1with Q finite. Suppose H and G are generating sets for H and G so that H ⊂G, theelements of G \ H permute the elements of H when acting by conjugation, and i is anisometry.

If H is almost convex with respect to H, then G is almost convex with respect toG.Proof. Suppose that g, g′ ∈G, ℓ(g) = ℓ(g′) and d(g, g′) ≤2.

We invoke the previous lemmaand perhaps interchange g and g′ to find geodesics g = ut, g′ = u′v′t′, where u, u′v′ ∈H∗,t, t′ ∈T, and ℓ(v′t′) = ℓ(t) ≤#Q. Consequently, u, u′ ∈H, with d(u, u′) ≤2#Q + 2.Using the fact that H is almost convex, there is a path of length at most K = K(2#Q+2)connecting u to u′ inside the ball of radius ℓ(u) in H. This path lies inside the ball ofradius ℓ(u) in G. But this gives us a path of length at most K + 2#Q connecting g to g′inside the ball of radius ℓ(g).Now we will use Corollary 6 to show that when E is the Klein bottle, S(2, 2, 2, 2),S(2, 2, 4) or P(2, 2), G is almost convex.

First we note that in these cases, the action ofQ on Z2 preserves the standard generating set. Now in each of these cases, we can finda generating set for G by lifting the orbifold generating set for πorb1(E) and appending z.Conjugation by these generators of G preserves a generating set of N e of the formN es = {x, xzi1, .

. ., xzia, y, yzj1, .

. ., yzjb, zk1, .

. ., zkc}.We shall call such a generating set saturation of N e. In order to use Corollary 6, we needto establish5

Theorem 7. N e is almost convex with respect to any saturation N es of N .This follows from Lemma 4 together with the followingLemma 8.

Let N es be a saturation of N e. Then there is a constant K so that for allg ∈N e, |ℓN es (g) −ℓN e(g)| ≤K.In order to do this, we need to recall some facts about geodesics in N in the generatingset N , established in [Sh1]. We review this since we will need to use these methods laterfor the (3, 3, 3) and (3, 6, 6) groups.

The viewpoint followed in [Sh1] is to use the shortexact sequence1 →Z →Np→Z2 →1to view words in N ∗as lifts of paths in the Cayley graph of Z2 based at the identity.Two such paths α and β represent the same element of N if and only if they end at thesame point in Z2, and the concatenation αβ−1 encloses zero signed area.For a giveng = (a, b) ∈Z2, we let Bg(n) = Bn−ℓ(g)g= B(n) ∩p−1(g). (Recall that B(n) is the ball ofradius n in Γ.

Since p−1(g) consists only of group elements, so does Bg(n) = Bn−ℓ(g)g.)For each g = (a, b) ∈Z2, we identify p−1(g) with Z by carrying xayb[x, y]t to t. Weobserve(1) Geodesics for elements in p−1(0, 0) project to closed paths; in fact, to simple closedcurves.If you can enclose n squares with a loop of length l, you can enclose n −1 squareswith a loop of length at most l. This shows that Bn(0,0) is an interval. In fact,(2) For any g ∈Z2, Bng is an interval.We will call the elements of Bng at the extremes of the interval extremals.

A geodesicor a projection of a geodesic for an extremal will also be called extremal. We also observe(3) The number of squares enclosed by a closed extremal is strictly monotone in the lengthof the extremal.¿From this we deduce that(4) Closed extremals are rectangles.

Indeed, their sides differ in length by at most one.We call such a rectangle an almost square.For suppose r is any simple closed curve. Let R be the unique minimal rectanglecontaining r. R encloses at least as many squares as little r, but R is no longer than r.This is easy to see: first notice that minimality of R ensures that r meets all sides of R.But now given two points on successive sides of R, then the subpath of R that connectsthem is geodesic, so it is no longer than the corresponding subpath of r. Now, since thelargest rectangles for a fixed perimeter are almost squares, (4) follows.We are now prepared to describe a sublanguage of the geodesics in N which containsat least one geodesic for each element of N and suffices for our purposes.First notice that for each g ∈Z2, all words representing elements of p−1(g) differ inlength from g by an even amount.

This follows from the fact that all relators are of evenlength. (G1) For each g ∈Z2, the elements of B0g are all represented by geodesics which project togeodesics in Z2.

(Indeed, every geodesic in Z2 occurs in this role. )(G2) Any geodesic for an extremal element projects to a subpath of an almost square.6

(G3) For every element of B2kg , k ≥1, there is a geodesic whose projection first follows thealmost square given by an extremal of B2k−2g, then crosses via a single edge to followthe almost square given by an extremal of B2kg .We are now ready to prove Lemma 8.Proof of Lemma 8. Recall our generating sets N e = {x, y, z}, andN es = {x, xzi1, .

. ., xzia, y, yzj1, .

. ., yzjb, zk1, .

. ., zkc}.Choose g ∈N e. Let w be a geodesic word in N es with w = g. Now let w′ be the wordin N e∗obtained by deleting all generators of type za in w and replacing each x±1zb andy±1zc generator by x±1 and y±1 respectively.

Let g′ = w′ ∈N < N e. Now, g = g′z±twhere 0 ≤t ≤kℓ(w) withk = max{|i1|, . .

., |ia|, |j1|, . .

., |jb|, |k1|, . .

.|kc|}.Notice that [xm, ym] = [x, y]m2 = zem2. Hence the N e-length of z±t is at most 4lpt/em+e ≤4pt/e + e + 4.

So now if v ∈{x, y}∗is an N geodesic for g′ ∈N, thenℓ(w) ≤ℓ(v) + 4pt/e + e + 4 ≤ℓ(v) + 4rkℓ(w)e+ e + 4.So if ℓ(w) = ℓG(g) is sufficiently large, we will have1/2ℓ(w) ≤ℓ(v) ≤ℓ(w).Now g = vz±t where t ≤kℓ(w) ≤2kℓ(v). But we can assume that the projection of vlooks like one of the projections described in (G1) — (G3), and hence has a straight pieceof length ℓ(v)/4.

So by inserting either x and x−1 or y and y−1, around the piece of vcorresponding to this side, we increase the word length by 2, but increase enclosed areaby ℓ(v)/4 squares. Continuing in this way, and introducing at most ⌈8k/e⌉such pairs ofgenerators and at most e z’s, we produce a word for g. This has increased length by atmost ⌈16k/e⌉+ e. Thus we haveℓN es (g) ≤ℓN e(g) ≤ℓ(v) + ⌈16k/e⌉+ e ≤ℓN es (g) + ⌈16k/e⌉+ e.This proves Lemma 8 with K = ⌈16k/e⌉+ e.We are now ready to establish the almost convexity of the Nil groups G = π1(M)such that M fibers over E where E is the Klein bottle, S(2, 2, 2, 2), S(2, 4, 4) or P(2, 2).We will use a generating set of the form G = N es ∪S where S is a lift to G of a standardgenerating set for πorb1(E), and N es is a saturated generating set preserved by S.Weobserved that there are saturated generating sets preserved by S, and that N e is almostconvex with respect to any saturated generating set.

So all that remains is to choose asaturated generating set N es preserved by S so that the inclusion N e ֒→G is geodesic withrespect to the generating sets N es and G = N es ∪S.7

We begin by noting that the inclusion Z2 ֒→πorb1(E) is geodesic with respect to thegenerating sets X = {x, y} and E = {ρ, x, y}, {a, b, c, d, x, y}, {p, q, r, x, y}, {a, b, ρ, x, y}depending on whether E is the Klein bottle, S(2, 2, 2, 2) or S(2, 4, 4) or P(2, 2) (In eachcase, ρ is orientation reversing, a, b, c, d, p are rotations of order 2 and q and r are rotationsof order 4.) We spell out the argument for E = S(2, 2, 2, 2).

We need to show that nogeodesic in {x, y}∗may be shortened by rewriting it in the {a, b, c, d, x, y} generating set.So suppose w′ is an {x, y} geodesic which can be shortened to a {a, b, c, d, x, y} geodesic w.We can push the {a, b, c, d} letters all to the right without changing the number of {x, y}letters. Thus we assume can assume w = uv, where u is composed of {x, y} letters and vis composed of {a, b, c, d} letters.

But now, since Q = Z2, any substring v of length 3 ormore includes a substring which evaluates into Z2. One checks that words in {a, b, c, d}∗of length less than 3 which evaluate into Z2 are not shorter than the corresponding {x, y}words for the elements they represent.By this process we can eliminate all {a, b, c, d}letters from w without increasing length.

Since w′ was assumed to be an X geodesic, thiscontradicts the assumption that w is shorter than w′.The argument for the other cases proceeds similarly. When E is the Klein bottle, wecheck {ρ} words of length 2.

Since ρ2 = x the result is immediate. For E = S(2, 4, 4), onechecks all {p, q, r} words of length at most 4.

Similarly for the case E = P(2, 2), we mustcheck {a, b, ρ} words of lengths at most 4.We now wish to lift this to N e < G. Choose a saturated generating N ′s for N e whichis preserved by conjugation by S. Suppose the inclusion N e ֒→G is not geodesic withrespect to N ′s and G′ = N ′s ∪S. As above any failure of this inclusion to be geodesiccan be observed in some short word v ∈S∗with ℓ(v) ≤#Q.

Consider the projectionof v into πorb1(E). Since the inclusion down below is geodesic, there is an {x, y} word v′which evaluates to this projection and ℓ(v′) ≤ℓ(v).

We now consider v′ as a word in thegenerators for N e. Note that v = v′zk for some k. Further, only finitely many such koccur since there are only finitely many short v’s. Let K be the maximum of the absolutevalues of those k which arise.

We takeN es = {azt | a ∈N ′s, |t| ≤K}.Clearly the inclusion of N e into G is geodesic with respect to N es and G = N es ∪S, andN es is preserved by the action of S.Thus we have shown that if G = π1(M) is a Nil group so that M fibers over the Kleinbottle, S(2, 2, 2, 2), S(2, 4, 4) or P(2, 2), there is a generating set G so that G is almostconvex with respect to G.It is now easy to see that G is almost convex with respect to the generating set N e ∪Sor N ∪S if this latter generates G. This is because every element of G has a G geodesic inwhich there are at most #Q S letters and these occur at the end. But now Lemma 8 tellsus that replacing the N es part by the corresponding N e or N geodesic increases length byat most a bounded amount.

Thus by Lemma 4, G is almost convex with respect to thisreduced generating set.We are now prepared to pursue the case where M fibers over S(3, 3, 3) or S(2, 3, 6).We wish to pursue the same program. However, the action of our finite quotient no longerpreserves the x and y directions in the plane, and hence does not preserve the generating8

set {x, y}. Rather, it preserves hexagonal symmetry, and hence preserves generating setsof the form {x, y, xy}.

(We may think of our x and y directions as labelling nonadjacentrays in this hexagonal symmetry.) Thus we will need to work with generating sets of theformX = {x, y, t} ⊂Z2,N = {x, y, t} ⊂N,N e = {x, y, t, z} ⊂N e.(In each case, we take t = xy.) To carry out our program, we must first proveTheorem 9.

N is almost convex with respect to N .Theorem 10. N e is almost convex with respect to N e.Any generating set of the formN es = {x, xzi1, .

. .xzia, y, yzj1, .

. .yzjb, t, tzk1, .

. .tzkc, z, zl1, .

. .

, zld}is called a saturation of N e.Corollary 11. N e is almost convex with respect to any saturation N es .Once again, we study N by studying the projection of paths in N into Z2.

This time,however, we have taken the generating set N = {x, y, t} and its projection X in Z2. TheCayley graph of Z2 with respect to X is the 1-skeleton of the tessellation of the plane byequilateral triangles.

Now πorb1S(3, 3, 3) acts as the orientation preserving subgroup of thesymmetries of this tessellation. There are two orbits of triangles under this action, thosewith boundary label xyt−1 (which we color white) and those with boundary label tx−1y−1(which we color black).

A circuit around a white triangle lifts to the identity in N. Acircuit around a black triangle lifts to [x, y] in N. Consequently, a closed curve in theplane lifts to [x, y]n, where n is the signed number of black triangles enclosed. As before,we can use this technology to treat N as equivalence classes of based edgepaths in theCayley graph of Z2.

Then after replacing the words “almost square” with “almost regularhexagon” (1) — (4) and (G1) — (G3) hold. (A hexagon is an almost regular hexagon ifno two of its sides differ in length by more than 2.

)¿From this description of geodesics in N we can deduceLemma 12. Suppose that g, g′ ∈N and that ℓ(g) = ℓ(g′) = n and that g′ = gr withℓ(r) ≤2.

Then one of the following occurs1) Both g and g′ are small, i.e., each has length less than 50.2) There are standard geodesics for g and g′ whose projections in Z2 lie in an 25 neigh-borhood of each other and 50-fellow travel.3) The projection of gr crosses an axis. If it crosses exactly one axis, then there is g′′ sothat ℓ(g′′) = n, d(g, g′′) ≤2, the projection of g′′ lies on the axis, d(g′, g′′) ≤2, and2) above holds for each of the pairs (g, g′′) and (g′′, g′).The above estimates are not sharp.9

Sketch of the proof of Lemma 12. We suppose that g, g′, and r are given as above, andsuppose that α′ and β′ are geodesics for g and g′.

If g ∈B0, then the projection of α′ liesentirely in the parallelogram determined by the tessellation of the plane with corners at(0, 0) and p(g). In this case we can further demand that the projection of α′ have a specificform: we demand that it stay along one side of the parallelogram as long as possible.

Inthis case we take α to be the path in the plane given by the projection of α′. If g /∈B0,then α′ can be taken to lie (in projection) close to the boundary of an almost regularhexagon.

We take α to be the path along the boundary of this hexagon determined by α′.We choose β similarly.One can then perform a systematic enumeration of the possibilities using the followingfacts:(E1) Each of α and β is either a geodesic of the special form we have given or consists ofbetween one and six sides of an almost regular hexagon. (E2) Each of α and β starts along one of the 6 axial directions.

(E3) If α or β lies along an almost regular hexagon, it turns either clockwise or counter-clockwise.Without loss of generality, one picks an axial direction for α and quickly eliminatesthe cases in which α and β turn in opposite directions.One has the following data:(D1) The lengths of α and β differ by at most 4. (D2) The end points of α and β are separated by a path s of length at most 4.

(D3) The path given by α′rβ′−1 in projection encloses 0 signed area, where area is measuredby the number of black triangles enclosed. (D4) The projection of α′ and α are either identical or lie separated by a straight strip ofwidth 1.

In the case where α is more than 2 sides of an almost regular hexagon, wecan take α to have the same endpoint as the projection of α′, and similarly for β.Thus, when α and β are more than 2 sides of an almost regular hexagon, s is simplythe projection of gr.One then proceeds to examine each of these cases using elementary Euclidean geom-etry. In each of the cases where at least one of α or β does not consist of 6 sides of analmost regular hexagon, we find that we are in case 1) or 2) of the Lemma.

That is, eitherg and g′ are both short, or there are geodesic paths for g and g′ which fellow travel inprojection.The case where both α and β are 6 sides of an almost regular hexagon is more in-teresting. If α and β both end in the interior of a common sextant (i.e., in the regionbetween two axial rays), then α and β must start in the same axial direction and haveapproximately the same size.

Thus, they fellow travel. On the other hand, if, say, α endsin the interior of a sextant and β ends on an axis which defines that sextant, α and βneed not fellow travel in projection.

For here, α may start out in (say) the t direction,while β can start out in the y direction, so that we have, say, β = yitjxjy−jt−jx−j, withi < j. However, the fact that β starts in the y direction and ends on the y axis allows us to“slide” β in the following manner: for any δ ≤i, the path yi−δtjxjy−jt−jx−jyδ evaluatesto the same element of N as β.

By taking δ = j we replace β with an equivalent geodesicwhich starts in the same direction as α. Their almost regular hexagons are approximately10

the same size, and thus once again, α and β fellow travel.Finally, if the path gs from the end of α to the end of β crosses exactly one axis, weneed only note that there is an element g′′ which is within distance 2 of each of g and g′,has the same length as these and projects to a point lying on this axis.Proof of Theorem 9 from Lemma 12.Let ℓ(g) = ℓ(g′) = n and d(g, g′) ≤2, say g′ = gr, with ℓ(r) ≤2. Thus, we are in thesituation of Lemma 12.

We must exhibit a path of bounded length which connects g to g′and lies in B(n).If situation 1) of the Lemma holds, then we can connect g to g′ by going through theidentity, and this path has length at most 100.Suppose situation 2) of the Lemma holds. We shall also assume that g and g′ havelength at least 900 for otherwise, we proceed as above.

(We remind the reader that ourestimates are in fact very crude!). Let w = uv and w′ = u′v′ be geodesics for g and g′with ℓ(v) = ℓ(v′) = 900.

Let q be the lift to N of a path connecting p(u) to p(u′). We canassume ℓ(q) ≤50.

Then the path given by v−1qv′r−1 in projection encloses an area A with|A| ≤(50)(900). We let γ be a geodesic for z−A.

This has length at most 4√A + 4 < 900.We then let P be the path based at g bearing the label v−1qγv′. It is easy to see that thisconnects g to g′ and lies inside B(n).We now suppose that we are in situation 3) of the Lemma.

If the projection of grcrosses only one axis, then we can perform the previous process twice, once connecting gto g′′, and once connecting g′′ to g′. But in fact, r has length at most 2, so the projectionof gr can cross at most 2 axes, and this happens only when g and g′ project to pointswithin distance 1 of (0, 0).

In this case we repeat the previous method twice and are done.Proof of Theorem 10. This follows from Theorem 9 by observing that as usual an N egeodesic in N e can have very few z’s.

Then if g, g′ ∈N e with ℓ(g) = ℓ(g′) and d(g, g′) ≤2,we write geodesics w = uzm and w′ = u′zm′ for g and g′, and |m| and |m′| are bounded,and u and u′ are free of z’s. Without loss of generality we can assume |m| ≤|m′|.

We letr be the terminal segment of u of length |m′| −|m|, so that u = u′′r with ℓ(u′′) = ℓ(u′).Now the almost convexity of N gives us a path Q of bounded length connecting u′′ to u′inside the ball of radius ℓ(g) −|m′|. The path we seek is the one which starts at g and islabelled z−mr−1Qzm′.The Corollary now follows by the methods of Lemma 8.We now have all the tools in place that we used for the proof in the square case.The proof in the present case follows along exactly the same lines.One checks thatthe embedding of Z2 into the appropriate orbifold groups is geodesic with respect to thegenerating set X = {x, y, t} for Z2 and the appropriate orbifold generating sets S ∪X .The action of S preserves X .

Once again, this lifts to the inclusion N e ֒→G giving almostconvexity with respect to a generating set of the form N es ∪S, where N es is a saturatedgenerating set. As before, this gives almost convexity with respect to the generating setN e ∪S or N ∪S if this latter generates.3.

Central extensions of word hyperbolic groups.11

We will now prove Theorem 2.Proof of Theorem 2. We take ρ : H × H →A to be the cocycle defining G. Thus we canidentify G with the set A × H endowed with the multiplication (a, h)(a′, h′) = (a + a′ +ρ(h, h′), hh′).

Let G′ be a generating set for G. Then H = πG′ is a generating set for H.We will extend G′ to a generating set G so that G is almost convex with respect to G.Lemma 13.Suppose G′ is a generating set for G with πG′ = H.Suppose that D is a Dehn’salgorithm for H with respect to H. Then there are sets G′′ ⊂π−1H and A′ ⊂A so that ifA′ ⊂A and G = G′ ∪G′′ ∪A, then for any g ∈G there is a geodesic w ∈G∗so that w = gand πw is D-reduced.Proof. We start by takingG′′ = {(1, h) | h ∈H}.We enlarge D if necessary to make sure it is closed under inversion and cyclic per-mutation.

We wish to lift the words in D to A. For each d = h1 .

. .hk ∈D, we take˜d = (1, h1) .

. .

(1, hk) ∈A. For each d = h1 .

. .hk ∈D, letRd = {(a1, h1) .

. .

(ak, hk) : (ai, hi) ∈G′ ∪G′′ for 1 ≤i ≤k}.For each r = (a1, h1) . .

. (ak, hk) ∈Rd and each i ≤k, let ai(r) = a1 + .

. .

+ ai. We takeA′ = {ai(r) + ˜d : r ∈Rd, d ∈D}.We take an arbitrary geodesic w′′ = (a1, h1) .

. .

(ap, hp) for g ∈G. Since A is centralin G, we may replace w′′ with w′ of the formw′ = (a1, 1) .

. .

(ai, 1)(ai+1, hi+1) . .

. (ap, hp)where hj ̸= 1 for j > i.

We then have π(w′) = hi+1 . .

.hp. If this is D-reduced we are done.If not, there are m and n with i < m < n ≤p so that hm .

. .

hn is more than half a relatorin D. Hence hm . .

. hn = h′m .

. .

h′n′ with n′ < n and with d = hm . .

. hn(h′m .

. .

h′n′)−1 ∈D.We then have(am, hm) . .

. (an, hn) = (am + .

. .

+ an, 1)(1, hm) . .

. (1, hn)= (am + .

. .

+ an + ˜d, 1)(1, h′m) . .

. (1, h′n′).By construction, (am + .

. .

+ an + ˜d, 1) ∈G, so this last expression lies in G∗, and sincen′ < n, it is no longer than the first expression. Thus, we may use it to replace the firstexpression in w′.

This reduces the length of πw′, so continuing in this way produces ageodesic w whose projection is D-reduced.12

Notice that this means that A′ is a generating set for A. For suppose w is as guaranteedby Lemma 13.

If w ∈A, then πw = 1 ∈H, and a D-reduced path for the identity is theempty path. In particular A is geodesic in G, that is, given a generating set G of the formG′ ∪G′′ ∪A as above, the inclusion of A into G is an isometry.We continue with our proof that G is almost convex.

We must choose our generatingset. Recall that D-reduced words are (λ, ǫ) quasigeodesics in H, and that there is a k so thatif two (λ, ǫ) quasigeodesics end at most distance 2 apart, then they lie in k neighborhoodsof each other.

We will takeA = A′ ∪{a ∈A : a = r with r ∈(G′ ∪G′′)∗and ℓ(r) ≤λ(2k + 3) + ǫ + 2k + 3}.We take G = G′ ∪G′′ ∪A.We now suppose that g, g′ ∈G with ℓ(g) = ℓ(g′) = n and g′ = gq with ℓ(q) ≤2. Wewrite g = w where w = uv so that each letter of u projects to 1 ∈H and no letter of vdoes.

Similarly, we write g′ = w′ with w′ = u′v′. Notice that d(πv, πv′) ≤2.

Since πv andπv′ are both D-reduced, they are (λ, ǫ) quasigeodesics lying in k neighborhoods of eachother.Suppose first that ℓ(v) ≥k+1, say v = xy with ℓ(y) = k+1. Then ℓ(ux) = n−k−1 andd(πx, πw′) ≤k.

Choose z so that ℓ(z) ≤k and πz labels a path from πx to a point on πv′.We will suppose this point to be πx′ where v′ = x′y′. Notice that the path uxz stays insidethe ball of radius n−1 Since πv′ is a (λ, ǫ)-quasigeodesic, it follows that ℓ(y′) ≤λ(2k+3)+ǫ.Hence the path y′qy−1z has length at most ℓ(y′) ≤λ(2k + 3) + ǫ + 2k + 3.

It projects to aclosed path, so, in particular y′qy−1z = (a, 1) ∈A. We now have gy−1z(a, 1) = u′x′.

Butthe path labelled y−1x(a, 1) based at g stays inside the ball of radius n. Clearly the pathlabelled y′ based at u′x′ also stays inside this ball. Thus, the path labelled y−1z(a, 1)−1y′runs from g to g′ staying inside the ball of radius n. Its length is clearly bounded.We must now check that we can produce such a path when ℓ(v) ≤k.

Suppose ℓ(v) ≥1.Then “backing up along πv” takes us to 1 ∈H, and we can perform the same argumentas above taking y = v and z to be trivial.We are now left with the case where ℓ(v) = 0 and by symmetry, we may assumeℓ(v′) = 0. In that case g and g′ lie in the abelian group A.

We have seen that A is geodesicin G, so we are done.This gives the followingScholium 14. Suppose that H is word hyperbolic and that1 −→Z −→G −→H −→1.Then there is a generating set G so that G is almost convex with respect to G.Proof.

There are only two actions on Z, namely the trivial action and the action whichinverts elements of Z. Thus at the possible cost of inverting elements of Z, we can move eachof these to the beginning of a word.

Now this process cannot increase length. Consequently,each element of G has a geodesic in which every Z generator appears the beginning.

Nowone can proceed as above.13

Corollary 15. Let M be a closed 3-manifold with ^PSL2 R or H2×R geometry.

Then thereis a generating set G so that π1(M) is almost convex with respect to G.Proof. In this case1 −→Z −→π1(M) −→H −→1,where H is the orbifold fundamental group of a hyperbolic surface orbifold.

Since this isnecessarily word hyperbolic, the result follows.This completes the proof of Theorem 1.References[C]J. Cannon, Almost convex groups, Geometrae Dedicata 22, 197—210 (1987). [CFGT] J. Cannon, W. Floyd, M. Grayson, W. Thurston, Solvgroups are not almost con-vex, Geometrae Dedicata 31, 291—300 (1989).[Sc]P.

Scott, The Geometries of three-manifolds, Bull. London Math.

Soc. 15, 401—487(1983).[Sh1]M.

Shapiro, A Geometric approach to the almost convexity and growth of somenilpotent groups, Mathematische Annalen, 285, 601—624 (1989).[Sh2]M. Shapiro, Growth of a^PSL2 R manifold group, to appear in MathematischeNachrichten.[Sho]H.

Short, ed., Notes on word hyperbolic groups, in Group Theory from a GeometricViewpoint, E. Ghys, A. Haefliger, A. Verjovsky eds., World Scientific.[T]C. Thiel, Zur Fast-Konvexit¨at einiger nilpotenter Gruppen, Bonner MathematischeSchriften, 1992.City CollegeNew York, NY 10031Ohio State UniversityColumbus, OH 4321014

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