Tubular groups and non positive curvature

Transcription

1 arxiv: v1 [math.gr] 1 Dec 2017 Tubular groups and non positive curvature J.O.Button Abstract We examine various different notions of non positive curvature for the case of tubular groups. We give a criterion for when a tubular group is CAT(0) which is then made into an algorithm. Along with various examples, we also show using results of Wise and Woodhouse that a tubular group is always virtually special if the underlying graph is a tree. We end by adapting Gardam and Woodhouse s argument on tubular groups that double cover 1-relator groups to show there exist 1-relator groups which are CAT(0) but not residually finite. 1 Introduction On attempting to assign suitable notions of curvature to a finitely presented group G, it is generally regarded that describing G as negatively curved means it is word hyperbolic. However coming up with a definition of G being non positively curved is more problematic as we do not have a version which is invariant under quasi-isometry. Probably the definition most in use is that G is CAT(0), meaning it acts geometrically(properly and cocompactly by isometries) on a CAT(0) space (though it is still open whether every word hyperbolic group is CAT(0), or in the above language whether every negatively curved group is non positively curved). CAT(0) groups obey strong conditions (see [5] Chapter III.Γ Theorem 1.1) but their closure properties are not especially nice, for instance the CAT(0) group F 2 F 2 F 2 contains finitely presented subgroups H whose finiteness properties rule them out from also being CAT(0) groups (see [5] Chapter III.Γ Section 5). A closely related notion, which is implied by being CAT(0), is that of acting properly and semisimply on a complete CAT(0) space and will be referred to here as weak CAT(0). A good feature of this property is that if G is a weak CAT(0) group then so are all subgroups and finite index supergroups 1

2 1 INTRODUCTION 2 L of G (the latter means that G has finite index in L and we will write this as G f L). But other variations present themselves, for instance we could ask for a geometric action not just on a CAT(0) space but on a CAT(0) cube complex, which might be thought of as strong geometric non positive curvature. Now by the big results of Wise and Agol ([16], [18], [1]) if G is also word hyperbolic then it is a virtually special group. Here we take this to mean that there is a subgroup H f G where H embeds into a right angled Artin group (a RAAG, which we always assume is finitely generated). This in turn implies very strong consequences for G such as being linear over Z. Hence we will think of being virtually special as strong group theoretic non positive curvature, though let us note here that if G is not word hyperbolic then this neither implies nor is implied by strong geometric non positive curvature. To see this, one way round follows from the finitely presented simple groups of BurgerandMozeswhereasthesubgroupH aboveoftheraagf 2 F 2 F 2 is virtually special but is not even CAT(0). Moreover note that being virtually special is another property preserved by arbitrary subgroups and finite index supergroups. In seeking further variations, we have already altered the cocompactness condition of a geometric action so we might now ask what happens if we remove it completely. Whilst acting properly on a CAT(0) space is just too general to be of use, we could regard weak geometric non positive curvature as acting properly on some CAT(0) cube complex (which is equivalent to acting freely if our group is torsion free). Whilst this is clearly implied by strong geometric non positive curvature, it also has our two closure properties and so is implied by strong group theoretic non positive curvature. Our interest here in this property is that Gersten in [12] gave an example of a freebycyclic groupwhichwasshownnottobecat(0), orevenweakcat(0) by [5] Chapter II Exercise However Wise in [19] demonstrated that this very group does have weak geometric non positive curvature. Moreover a group with weak geometric non positive curvature cannot have property (T), so examples of closed quaternionic hyperbolic 3-manifolds have fundamental groups which are CAT(0) but are not weakly geometrically non positively curved. This also ties in with Haglund s axis theorem in [15] which is used to show that if a group G has a proper action on a CAT(0) cube complex then G is balanced: that is, if we have an infinite order element g G where g m is conjugate in G to g n then m = n. Hence our final concept of non positive

3 1 INTRODUCTION 3 curvature is that of a group being balanced, which will be called weak group theoretic non positive curvature. By this result of Haglund, and also [5] Chapter 3.III Theorem 1.1 (iii) which shows that weak CAT(0) groups are also balanced, we see that weak group theoretic non positive curvature is implied by any of our other non positively curved properties that we have named. Thus we have considered six different notions of a group being non positively curved and have seen that there exist a few implications between them, but many cases where no containment holds either way. In this paper we consider tubular groups, which are defined to be the fundamental group of a finite graph of groups with all vertex groups a copy of Z 2 and all edge groups a copy of Z. These groups have occurred in various interesting contexts, for instance [3], [8], [19], [21]. It will be a consequence of this and the other papers that for these groups, CAT(0) and weak CAT(0) are the same, whereas all other properties are distinct but they form a total order, with strong geometric non positive curvature implying strong group theoretic non positive curvature which in turn implies being CAT(0). Then CAT(0), equivalently weak CAT(0), gives us weak geometric non positive curvature which finally results in weak group theoretic non positive curvature. We begin in Section 2 by looking at Gersten s argument that his free by cyclic group is not CAT(0), which involves (stable) translation lengths. This group also happens to be a tubular group and we show in Proposition 3 and Theorem 4 that this argument can be adapted and extended to give a straightforward criterion as to exactly when a tubular group is CAT(0), which is equivalent to being weak CAT(0) for this class of groups. The resulting criterion involves a finite number of equations and inequalities that must be satisfied in order that our tubular group is CAT(0). However we work over the reals in this case and so it need not follow that we have an algorithm in the strictly computational sense, where we require a finite input and finite calculations at each stage of the process. In Section 3 we show that in fact this criterion can be made into such an algorithm, with input the finite amount of data defining the graph of groups (G,Γ) and output yes or no according to whether or not the resulting fundamental group π 1 (G) is a CAT(0) group. This proceeds by utilising Tarski s results on quantifier elimination over real closed fields as well as finding an appropriate representation for real algebraic numbers which allows field and order computations, along with tests for equality and inequality. We remark here that there cannot be a general algorithm with input a finite presentation which always terminates

4 1 INTRODUCTION 4 with yes or no according to whether the group defined by this presentation is CAT(0) or not. This is because being CAT(0) is a Markov property (as Baumslag - Solitar subgroups are poison subgroups), so that this decision problem is unsolvable amongst all finitely presented groups. We look at a range of examples in Section 4, thus demonstrating that our criterion can be exploited quickly to determine whether π 1 (G) is a CAT(0) group. Whilst this is always so if Γ is a tree, we need to look further at the edge inclusions whenever Γ contains loops and not just at the underlying graph. This is because having a Baumslag - Solitar subgroup a,t ta m t 1 = a n which is non Euclidean, that is when m n, means that π 1 (G) will fail every non positively curved property we have mentioned. Thus the best we can hope for is that either π 1 (G) is CAT(0) or it contains a non Euclidean Baumslag Solitar subgroup. We show in Proposition 8 that this alternative does hold if Γ has Euler characteristic zero, but it certainly does not if Γ is a single vertex with at least two loops, such as occurs for Gersten s group. Indeed Wise illustrates the start of [19] with examples of various tubular groups (including that one), all of which have Γ equal to this graph. We show how our criterion can detect immediately which of these examples are CAT(0); indeed this extends to the case when Γ is a single vertex with an arbitrary number of self loops. We finish the section with specific examples of tubular groups possessing each of our non positive curvature properties but without any of the stronger properties, thus these different classes are indeed distinct for tubular groups. To finish, we examine which tubular groups are virtually special. Although being CAT(0) might be taken as the standard definition of non positive curvature, for tubular groups being virtually special is strictly stronger and comes with a whole host of important group theoretic implications. For instance in [17] Wise gave an example of a CAT(0) tubular group (which incidentally is immediately seen to be CAT(0) by our criterion) that is not even Hopfian, so certainly it fails to be residually finite or linear. But a group known to virtually embed in a RAAG will inherit any property held by RAAGs and preserved under taking subgroups and finite index supergroups. In particular such a group will be linear even over Z (indeed there are various groups only known to be linear because they are virtually special), as well as having other good properties such as being virtually biorderable, virtually residually torsion free nilpotent and virtually residually finite rational solvable. The downside is that confirming directly that a group G is virtually spe-

5 2 CAT(0) TUBULAR GROUPS AND TRANSLATION LENGTHS 5 cial can be very involved, because one usually needs Wise s machinery to show not only that G has a finite index subgroup H which is the fundamental group of a non positively curved cube complex but also that the complex is special. However Woodhouse introduces in [21] a technique specifically for tubular groups which is used to show that a tubular group is virtually special if and only if it acts freely on some locally finite CAT(0) cube complex. This is achieved by adapting Wise s equitable sets condition in [19] (which is equivalent to the group having a free action on some CAT(0) cube complex) and imposing extra constraints on these equitable sets which imply that the group is virtually special. Therefore it seems appropriate to see for which tubular groups this result can be made to work. In the final section we show that Woodhouse s criterion is satisfied for any tubular group defined by (G,Γ) where Γ is a tree. Thus this gives a new class of virtually special groups and we can conclude that they all possess our strong group theoretic properties above. In particular these groups are all linear over Z, though they were not in general previously known to be linear at all. We end that section with a brief discussion of non positive curvature properties for free by cyclic groups. It is known by [6] exactly when a tubular group is free by cyclic and in particular this is true for our class of groups above where Γ is a tree. In section 6 we look at the construction in [11] of Gardam and Woodhouse, which shows that the particular class of tubular groups known as the snowflake groups are index 2 subgroups in 1-relator groups, thus allowing them to transfer various group theoretic and geometric properties from the tubular group to the 1-relator group. By varying their parameters, but where we still have this phenomenon of an index 2 tubular subgroup of a 1-relator group, we give the first examples of 1-relator groups which are CAT(0) but not residually finite. We do this by showing the tubular subgroup is non Hopfian using a very similar argument to Wise in [17], which itself is an adaptation of the original argument by Baumslag and Solitar. 2 CAT(0) tubular groups and translation length assignments A tubular group is defined to be the fundamental group π 1 (G) of a graph of groups (G,Γ) where every vertex group is a copy of Z 2 and every edge

6 2 CAT(0) TUBULAR GROUPS AND TRANSLATION LENGTHS 6 group is a copy of Z. In this paper we will always assume that this graph Γ is finite so that π 1 (G) is finitely generated. Indeed the standard procedure of deducing a presentation for the fundamental group of a graph of groups will result here in a finite presentation with 2n+b generators and 2n+b 1 relators, where n is the number of vertices and b is the first Betti number of Γ. When (G,Γ) is a graph of groups as above, we will sometimes refer to π 1 (G) as the tubular group defined by this graph of groups. We should bear in mind that it is perfectly possible for different graphs of groups to define isomorphic tubular groups, though this will not matter to us as it is the geometric and group theoretic properties of the resulting tubular group that will be of interest. Gersten s famous free by cyclic tubular group in [12] was shown there not to be a CAT(0) group by an argument involving translation lengths. As further noted in [5] Chapter II Exercise 7.18, it actually implies that the group is not weak CAT(0). In this section we will see that his argument can be generalised to all tubular groups, so we first briefly review the required properties of translation lengths and CAT(0) spaces. If a group G acts on a metric space (X,d) by isometries then the translationlengthofanelementg Gcansometimesmeanδ G (g) := inf x X d(x,g(x)), namelytheinfimum ofthedisplacement functionofg, orit might beτ G (g) := d(x,g lim n (x)) n n which exists and is independent of x; here called the stable translation length. The stable translation length is always a lower bound for the displacement function but it might not be the greatest lower bound, which in turn might not be attained by a point in X anyway. On regarding these as functions δ G,τ G : G [0, ), they are both invariant under conjugation. We also have δ G (g n ) n δ(g) and even τ G (g n ) = n τ(g) for any n Z. It is not true in general that f(gh) f(g)+f(h) when f = δ G or τ G, though this does hold for τ G if g and h commute (but even then it might not for δ G ). However we might hope for good behaviour from semisimple elements (those with a point attaining the infimum of the displacement function) and even better behaviour from semisimple elements where this infimum is equal to the stable translation length (so that both definitions of translation length defined above are equal). Now for an arbitrary isometry g of an arbitrary metric space X, these two conditions together are equivalent to the existence ofapoint x 0 inx withd(x 0,g n (x 0 )) = nd(x 0,g(x 0 ))foralln N. The power of the theory of CAT(0) spaces is that if g is a semisimple isometry and X is

7 2 CAT(0) TUBULAR GROUPS AND TRANSLATION LENGTHS 7 a CAT(0) geodesic metric space then ([5] Chapter II Theorem 6.8) either g has a fixed point or there is an isometric embedding in X of the real line such that this image is invariant under g, which restricts to a translation on this line. Thus any point x 0 on this line will satisfy the equation above, meaning that the two given definitions of translation length are equal. Consequently if a group G acts by semisimple isometries on a CAT(0) space X then our function τ G has the property that τ G (g) = 0 if and only if g has a fixed point. Thus if our action further satisfies a rather weak properness assumption, namely every point in X is fixed by only finitely many g G, then τ G (g) = 0 if and only if g has finite order. Moreover in the CAT(0) case we have even stronger properties holding for translation length functions when we restrict to abelian groups. This is a consequence of Bridson s Flat Torus Theorem which we state here as: Proposition 1 ([5] Chapter II, 7.1 and 7.17) Suppose that G is a group acting properly by semisimple isometries on a complete CAT(0) space X and H is a subgroup of G which is isomorphic to Z n. Then there is an isometrically embedded copy of E n in X invariant under H and on which H acts faithfully as a discrete group of translations. Consequently there is an injective homomorphism ι of H = Z n to E n with discrete image such that the restriction to H of our translation function τ G is merely ι(g) for ι(g) E n. Here E n will always mean R n equipped with the standard Euclidean norm. Now Gersten s well known argument for the group G = t,a,b,c tat 1 = a,tbt 1 = ba,tct 1 = ca 2 is to consider τ G : G [0, ) restricted to H = a,t = Z 2. We have that τ G is constant on t,at,a 2 t as these elements are all conjugate in G, thus contradicting Proposition 1. (Here it is crucial that our CAT(0) space provides us with a flat torus: on regarding a = (1,0) and t = (0,1) in Z 2, the non Euclidean norm (x,y) = max( x /2, y ) on R 2 would not lead to a contradiction.) We now present our generalisation to all tubular groups. As in Gersten s argument, we use the existence of the translation length function having the properties discussed above as an obstruction to our tubular group being CAT(0) or even weakly CAT(0). This motivates the next definition.

8 2 CAT(0) TUBULAR GROUPS AND TRANSLATION LENGTHS 8 Definition 2 If (G,Γ) is a graph of groups defining a tubular group then a translation length assignment for (G, Γ) consists of the following data: A maximal tree T of Γ For each vertex v V(Γ), an injective group homomorphism θ v from the vertex group G v = Z 2 to E 2 with discrete image For each edge e E(Γ) with edge group z e = Z, endpoints v ± and inclusions j ± from z e to G v±, we have: (i) If e T then θ v (j (z e )) = θ v+ (j + (z e )) E 2 \{(0,0)}. (ii) If e / T then θ v (j (z e )) = θ v+ (j + (z e )) (0, ). Proposition 3 If (G,Γ) defines a tubular group which is a CAT(0) group, or is even weakly CAT(0), then there exists a translation length assignment on (G,Γ). Proof. If so then as above we obtain the corresponding translation length function τ : π 1 (G) [0, ) from the action of π 1 (G) on our CAT(0) space X say. Now take any maximal tree T of Γ with some root vertex v 0 T. By Bridson s Flat Torus Theorem above, we have for the vertex group G v0 = Z 2 anisometricallyembeddedcopyofe 2 inx whichisinvariantundertheaction of G v0 and such that G v0 acts as translations on E 2. Take any isometric identification of this invariant plane with E 2 and define θ v0 (g) to be the translation component of g G v0 on this plane. Note that θ v0 (g) is the translation length of g and that θ v0 (G v0 ) cannot lie in a line as π 1 (G) acts properly on X, thus G v0 does so on E 2. On moving to a neighbouring vertex v 1, we do the same to obtain a map θ v1 from G v1 to E 2. Now if e is the edge joining v 0 and v 1 with edge inclusions j 0,j 1 of the edge group z e into G v0,g v1 respectively then we certainly have θ v0 (j 0 (z e )) = θ v1 (j 1 (z e )) because these are just the translation lengths of j 0 (z e ) and j 1 (z e ) which are the same group element in π 1 (G). Thus we can rotate our copy of E 2 based at v 1 so that these translation components θ v0 (j 0 (z e )) and θ v1 (j 1 (z e )) are actually equal. We now continue down the tree T, defining a new map θ v for every new vertex v. We define this inductively for every vertex of Γ until we have our translation length assignment which satisfies (i). It remains to show (ii): on

9 2 CAT(0) TUBULAR GROUPS AND TRANSLATION LENGTHS 9 taking the remaining edges e with inclusions j ± into G v±, we have that j + (z e ) and j (z e ) are elements of π 1 (G) which are conjugate in π 1 (G) via a stable letter, so they have the same translation length, giving us θ v (j (z e )) = θ v+ (j + (z e )). We can now argue in the other direction. Theorem 4 If (G,Γ) is a graph of groups defining a tubular group which has a translation length assignment then π 1 (G) is a CAT(0) group. Proof. We will show that in fact G = π 1 (C) where C is a non positively curved geodesic metric space which is also compact. This then implies that G is a CAT(0) group by realising it as the group of deck transformations on the universal cover C equipped with the induced length metric. This results in an action by isometries which will be proper, free and cocompact. Wewill beusing thegluing theorems asin [5] Chapter II.11. Notethat we will be gluing with a tube because our edge subgroups might be generated by proper powers in the vertex groups, so we use [5] Corollary This states that if X 0,X 1 and A are non positively curved geodesic spaces with A compact and we have two local isometries φ i : A X i (i = 0,1) then the quotient of X 0 (A [0,1]) X 1 by the equivalence relation generated by (a,0) φ 0 (a) and (a,1) φ 1 (a) is a non positively curved geodesic metric space C. In all applications, A will be a circle of length l and φ i will map A to a closed local geodesic in X i, also necessarily of length l as we are mapping by a local isometry (although it could for instance wrap many times around a simple closed geodesic). This means that, as X 0 and X 1 will always be compact in our applications, so will C. We first take a copy S v of E 2 for each v V(Γ) along with the quotient X v = S v /Z 2 where the subgroup Z 2 of E 2 is given by the translation length assignment θ v on S v. Of course X v is a torus which is compact and non positively curved, with π 1 (X v ) = Z 2. Moreover, for any non identity element g v G v = Z 2, we have a geodesic in S v from the origin to g v of length

10 2 CAT(0) TUBULAR GROUPS AND TRANSLATION LENGTHS 10 θ v (g v ), which will map under the projection π v : S v X v to a closed local geodesic in X v of the same length, as π v is a local isometry. We now take our maximal tree T of Γ and choose a root vertex v 0 and neighbouring vertex v 1, joined by the edge e. Thus we have non positively curved spaces X v0,x v1 and for A above we have the interval [0,l] where l = θ v0 (j 0 (z e )) = θ v1 (j 1 (z e ) by Property (i). Now θ v0 (j 0 (z e )) defines a closed local geodesic of length l in X v0 with the obvious map from [0,l] to X v0 a local isometry, and the same is true for X v1. Thus the conditions of the gluing theorem are satisfied, giving us the non positively curved space Y which is compact and has fundamental group equal to the amalgamation of π 1 (X v0 ) and π 1 (X v1 ) over z e = Z by Seifert - Van Kampen. ItmightbethecasethattheinclusionofX v0 (say)iny isnotanisometric embedding (for instance if A {0} is attached to a long simple closed geodesic in X v0 but A {1} wraps many times round a short simple closed geodesic in X v1 ). However we can replace the interval [0,1] parametrising the length of the tube by [0,r] in the gluing construction and for suitably large r we will find that X v0 and X v1 isometrically embed in Y. In particular, given any group element g v0 of π 1 (X v0 ) = Z 2 (say), we have that the closed local geodesic defined above in X v0 is also a closed local geodesic in Y. Consequently we can now continue this process by successive gluing over the edges in the maximal tree T, whereupon at each stage we glue the space Y constructed so far to the space X v where v is a new vertex which is the other endpoint of the next edge we are taking in T. These two spaces are again glued along closed local geodesics which will have matching lengths, by our translation length assignment on (G, Γ) and the fact that these lengths do not change under any of the earlier embeddings. This results in a compact non positively curved space with fundamental group the corresponding amalgamated free product. Having finished with the tree T, the process for the remaining edges is similar. Here we use [5] Proposition 11.13, which is the same as the corollary quoted above except that X 0 and X 1 are the same metric space X, with the resulting non positively curved complex C obtained from the quotient of X (A [0,1]) by the appropriate equivalence relation. This means that π 1 (C) is obtained from π 1 (X) by the resulting HNN extension over the two infinite cyclic subgroups generated by the group elements used to define the closed local subgroups. Again we need to take care with the length of the gluing tube used

11 3 ALGORITHMIC TRANSLATION LENGTH ASSIGNMENTS 11 (which corresponds to the stable letter of the HNN extension), but for long enough tubes we will have that X embeds isometrically in C which is again compact. Once we have done this for the remaining edges lying outside T, the space we end up with has fundamental group equal to π 1 (G). We note that these two results immediately combine to give the following, which was previously shown in [19] Lemma 4.4: Corollary 5 If G is a tubular group acting properly and semisimply by isometries on a CAT(0) space then G is a CAT(0) group; indeed it is the fundamental group of a compact non positively curved space. 3 Algorithmic translation length assignments In the last section we saw that we have a criterion which tells us when a tubular group is CAT(0) but we now look at how we can decide this algorithmically. First of all, if we merely examine the conditions that define this criterion then we can think about it in the following way. Given (G,Γ) with Γ having n vertices v 1,...,v n, our translation length assignment gives rise to an ordered pair of vectors (x 2i 1,x 2i ) R 2 R 2 for each vertex v i where 1 i n. This is obtained by setting x 2i 1 = θ vi (1,0) and x 2i = θ vi (0,1) where (1,0),(0,1) is the standard basis for the vertex group Z 2. Moreover this completely determines the assignment because Z 2 is free abelian, thus knowing x 2i 1 and x 2i tells us θ vi. But of course setting x 1,...,x 2n to be arbitrary vectors in R 2 does not necessarily give us a translation length assignment as we have various conditions that must be satisfied. At this point we can write x i = (x i,y i ) R 2 to give us 4n real numbers x 1,...,x 2n,y 1,...,y 2n that we will regard as our variables. We then have to satisfy the conditions (i) and (ii) for a translation length assignment in Definition 2. Each of these two cases gives rise to finitely many equations: if we take an edge running between the vertices v i and v j then the equation in (i) will be of the form ax 2i 1 +bx 2i = cx 2j 1 +dx 2j where a,b,c,d are all integers, given by the two inclusions of the generator of theedgesubgroupwhich wetaketobe(a,b) Z 2 atthevertex v i and(c,d)

12 3 ALGORITHMIC TRANSLATION LENGTH ASSIGNMENTS 12 Z 2 at v j. On taking x and y components, this results in two linear equations with integer coefficients involving x 2i 1,x 2i,x 2j 1,x 2j and y 2i 1,y 2i,y 2j 1,y 2j respectively. However the equations from (ii), being of the form ax 2i 1 +bx 2i = cx 2j 1 +dx 2j with notation as in (i), will not be linear. But here they will be quadratic, since if we take vectors x,y in a vector space R N then x = y is equivalent to x.x = y.y provided we areusing the Euclidean norm. Thus each condition as above in case (ii) is equivalent to the equation a 2 (x 2 2i 1 +y2 2i 1 )+2ab(x 2i 1x 2i +y 2i 1 y 2i )+b 2 (x 2 2i +y2 2i ) = c 2 (x 2 2j 1 +y2 2j 1 )+2cd(x 2j 1x 2j +y 2j 1 y 2j )+d 2 (x 2 2j +y2 2j ) in our variables and this again has integer coefficients. Finally we do have finitely many inequalities, because at each vertex v i Γ we require θ vi (Z 2 ) to be discretely embedded in R 2. But this is equivalent to requiring that θ vi (1,0) and θ vi (0,1) are linearly independent vectors, which is just the quadratic inequality (again with integer coefficients in our variables) x 2i 1 y 2i x 2i y 2i 1 0. This allows us to use Tarski s quantifier elimination over real closed fields. Theorem 6 Given a graph of groups (G,Γ) defining a tubular group, there exists an algorithm to determine whether or not π 1 (G) is a CAT(0) group. Proof. We first work over the field R and imagine that we can perform exact arithmetic in this field. For details and proofs of results quoted, we refer to [2], specifically Section 2.3. Suppose that R[X 1,...,X k ] is the multivariate polynomial ring over R and that P R[X 1,...,X k ] is a finite set of polynomials. This gives rise to the algebraic sets of R k, namely the common zeros {x R k P PP(x) = 0}. More generally we have the constructible sets, which certainly include {x R k P P P(x) = 0 Q QQ(x) 0}

13 3 ALGORITHMIC TRANSLATION LENGTH ASSIGNMENTS 13 wherep,q R[X 1,...,X k ]arefinite(thesearethebasicconstructiblesets). Consequently if we take k = 4n for n = V(Γ) then our condition on the existence of a translation length assignment becomes X 1 X 2... X 4n P PP(X 1,...,X 4n ) = 0 Q(X 1,...,X 4n ) 0 Q Q where P and Q are the polynomials defined by the equations given above. This is a sentence (that is, a formula with no free variables in the language of R) with coefficients in Z. Whether π 1 (G) is a CAT(0) group or not is the same as asking whether there exists x = (x 1,...,x 2n,y 1,...,y 2n ) R 4n such that this formula is satisfied when X 1 = x 1,...,X 2n = x 2n,X 2n+1 = y 1,...,X 4n = y 2n. But R, considered as an ordered field, admits quantifier elimination with [2] Theorem 2.77 telling us that if Φ is a sentence in the language of ordered fields with coefficients in Z then there is a quantifier free sentence Ψ with coefficients in Z such that Φ is true if and only if Ψ is true. Moreover Ψ can be ordained algorithmically from Φ. As Ψ contains no free variables, this is just a finite list of statements of the form r = 0 or r 0 for r R, combined using the Boolean operations and, which can therefore be checked to see if it holds or not (provided we can determine equality and inequality in R). Thus we can decide whether or not π 1 (G) is a CAT(0) group. However this assumes that our decision process can perform arithmetic operations and tests for equality and inequality using real numbers. To get round this, we note that in fact Tarski s elimination of quantifiers works for any real closed field. This is defined in [2] Section 2.1 but for our purposes we note that R A, which consists of the real algebraic numbers, is such a field. An important point is the Tarski-Seidenberg principle ([2] Theorem 2.80): If R R for R,R real closed fields and Φ is a sentence in the language of ordered fields with coefficients in R then Φ is true in R if and only if it is true in R. On taking R = R A and R = R we see that the question of whether π 1 (G) is CAT(0) is equivalent to whether there exists a solution in R A to our equations. Thus we have our decision problem provided we know that the operations of R A, regarded as an ordered field, can be performed algorithmically. This is the case, for instance see [9] which shows how to perform exact addition, subtraction, multiplication and division in R A, as well as an exact test for equality/inequality and also an exact test for <,>,,.

14 4 EXAMPLES 14 4 Examples Although we now have an algorithm for when (G,Γ) defines a CAT(0) group, it could be somewhat complicated to implement in practice. We would like some results that ensure the answer is yes without going through the whole process. The next proposition records the case of a tree, whereupon the argument is immediate. Proposition 7 If (G,Γ) defines a tubular group and Γ is a tree then π 1 (Γ) is CAT(0). However there can be no other definitive results which merely look at the graph Γ without considering the various edge inclusions. To see this, first note that for any connected graph Γ with n vertices and m edges, if at both ends of every edge we always choose the inclusion of the edge group to equal (1,0) Z 2 in the vertex group then the following presentation results for π 1 (G): x i,y i,t k [x i,y i ],x j+ = x j,t k x k+ t 1 k = x k where 1 i n, j ± are the endpoints of the jth edge in the maximal tree T (so 1 j n 1) and x k± are the endpoints of the kth edge outside T (so 1 k m n+1). As Γ is connected we conclude that all x i are equal, giving us the equivalent presentation x,y i,t k [x,y i ],[x,t k ] which is just the RAAG group F m+1 Z and so is non positively curved for all of our definitions of non positive curvature. But whenever Γ contains a simple loop v 0...v l (as written in terms of vertices with v 0 = v l ), we can form another tubular group (G,Γ) but with the same graph Γ where the edge group joining v 0 and v 1 injects in the vertex group G v0 = Z 2 as say (r,0) and into G v1 arbitrarily, but now the next edge group from v 1 to v 2 injects in G v1 as the same subgroup (but arbitrarily in G v2 ), and so on with the same edge groups as we proceed around this loop until we reach the last edge. Here we inject the edge group in G v0 as say (s,0). Consequently, on forming the fundamental group π 1 (G), we see that the infinite order element

15 4 EXAMPLES 15 x = (1,0) G v0 has the property that x r is conjugate to x ±s in π 1 (G). Thus if r s then this is not a balanced group and so is not non positively curved in any sense. Thus in order to obtain any type of non positive curvature, we require that π 1 (G) is a balanced group, which for tubular groups is equivalent to not containing a Baumslag - Solitar group BS(m,n) = x,y yx m y 1 = x n for m,n Z\{0} that is non Euclidean, meaning that m n (see [7] Corollary 9.6). Consequently we now might hope for results saying that under appropriate conditions for (G,Γ), either π 1 (G) is CAT(0) or it contains a non Euclidean Baumslag Solitar subgroup. On moving to the simplest graphs which are not trees, let us first examine the case of a single vertex with self loops. Here in order to have a translation length assignment, we need to find (x,y) R 2 R 2 with x,y linearly independent for our single vertex group. As we can scale and rotateany translation length assignment, which corresponds to rescaling a given CAT(0) metric, we can assume without loss of generality that x = 1 = (1,0). Let us first consider the case of a single edge so that (1,y) is required to satisfy m1 + ny = p1 + qy for (m,n),(p,q) Z 2 \ {(0,0)} which are given by the edge group inclusions. If (m,n) and (p,q) are linearly dependent, which is exactly when the two edge subgroups lie in the same cyclic subgroup of the vertex group, then we have no solutions (as π 1 (G) will contain a non Euclidean Baumslag - Solitar subgroup) unless (m,n) = ±(p,q) in which case any y off the x-axis is a solution. At this point it is better to move to the notation of complex numbers, so we write 1,z C in place of 1,y R 2. We then have the equation nz +m qz +p with m,n,p,q Z where now np mq 0 and we are looking for solutions z which are not real. But of course this is merely f 1 (S 1 )\R where S 1 C is the unit circle and f(z) = (nz + m)/(qz + p) is a Möbius map with all entries integers. Thus our solution set is either a circle with centre on the real axis, minus its two real points, or a vertical line minus its intersection with the x-axis. So for a vertex with one self loop, our alternative does hold in that either π 1 (G) contains a non Euclidean Baumslag - Solitar subgroup or it is CAT(0). = 1

16 4 EXAMPLES 16 We now look at when Γ has one vertex but an arbitrary number of self loops. This gives us several equations of the form n i z +m i q i z +p i = 1 where (m i,n i ),(p i,q i ) Z 2 \{(0,0)} with n i p i m i q i which need to be satisfied simultaneously for a non real z C in order for π 1 (G) to be CAT(0). (We can ignore the linearly dependent equations as either this gives no restriction or we immediately have the empty set and thus no solutions.) Thus we have Möbius maps f i defined by f(z) = (n i z + m i )/(q i z + p i ) and our solution set to each equation is the circle f 1 i (S 1 ) perpendicular to the real axis, where here we regard a line as a circle through infinity, with the real axisremoved. Thus C 1 = f1 1 (S1 ) provides us withacircle andc 2 = f2 1 (S 1 ) then defines a circle that is either the same (in which case we can ignore this equation), or which does not meet C 1 or is tangent to C 1 (which will be at a real point), in which case we have no simultaneous solutions whereupon π 1 (G) is not CAT(0), or C 1 C 2 consists of two complex conjugate points. In this last case we have that π 1 (G) is CAT(0) if and only if all remaining circles pass through these two points. We can now see how this works for the examples in [19], where Γ is always a single point with vertex group a,b = Z 2 joined by two self loops. Example 1: Gersten s free by cyclic group can be expressed as a tubular group where one edge provides inclusions ab and b, and the other a 1 b and b. Thus this results in the complex equations 1+z = z and 1+z = z which is two disjoint parallel lines, so we see yet again that this group is not CAT(0). Moreover it does not contain a non Euclidean Baumslag - Solitar subgroup (for instance by [7] Proposition 2.5) so we also see that our alternative now fails as soon as we allow more than one self loop. Example 2: Wise s non Hopfian CAT(0) group in [17] is given by the two pairs of inclusions a and a 2 b 2, b and a 2 b 2. Thus we have 2+2z = 1 = z which has two solutions, confirming it is CAT(0). Indeed this works more generallyfortheinclusions aanda n b n, banda n b n whenever n 1asitisasking for a unit complex number such that 1+z = 1/n. As { 1+z : z = 1}

17 4 EXAMPLES 17 varies from 2 to 0, we are done by the Intermediate Value Theorem. Example 3: Wise s next family of examples has inclusions a n and ab, b n and ab, whereupon it is pointed out that the resulting tubular group acts freely on a CAT(0) cube complex when n = 2 but not (hence is not a CAT(0) group by [19] Theorem 1.2) when n > 2. This gives us the equations n = 1+z = n z, resulting in no solutions for n 3 by the triangle inequality, nor any valid solutions for n = 2. Thus none of these groups are CAT(0). Example 4: The last family has inclusions a n b and ab, ab n and ab, which were shown in [19] to have no free actions on any CAT(0) complex when n 2. This gives us n+z = 1+z = 1+nz, which is easily seen to have no solutions and thus confirms these groups are not CAT(0). Consequently the best we can now hope for is to have results giving conditions under which either π 1 (G) is CAT(0) or it contains a non Euclidean Baumslag Solitar subgroup. However as this does not hold even for Γ a single vertex with two self loops as seen in Example 1 above, showing this is true whenγisatreeunionanextraedgeisasfaraswecangoinfindingconditions which only depend on the graph Γ. Proposition 8 If (G,Γ) defines a tubular group where Γ is a connected graph with an equal number of vertices and edges then either π 1 (Γ) is CAT(0) or it contains a non Euclidean Baumslag Solitar subgroup. Proof. First suppose that Γ is a simple loop of k vertices v 0,v 1,...,v k 1 labelled clockwise, so that v i 1 is joined to v i (modulo k) and the edges are given their clockwise orientation. At each vertex group we have two inclusions from the neighbouring edge groups. For our starting case we will assume that at every vertex v the images of these two edge groups lie in the same maximal cyclic subgroup g v of G v = Z 2. In particular we will have non zero integers i,j such that the generator of the edge group coming in to v is sent to gv, i with gv j the image of the edge group generator leaving v. Thus on picking the maximal tree which includes all edges other than the one joining v k 1 and v 0 and performing the group amalgamations, we obtain the relations g j 0 v 0 = g i 1 v1,g j 1 v 1 = g i 2 v2,...,g j k 2 v k 2 = g i k 1 v k 1. Hence if we try to define a translation length assignment for (G,Γ) such that (without loss of generality) g v0 is sent to (1,0) R 2 then we see that the

18 4 EXAMPLES 18 other generator of the vertex group G v0 can be sent to an arbitrary element of R 2 off the x-axis but that we must send g v1 to (j 0 /i 1,0) R 2 and so on around the vertex groups G v1,...,g vk 1. However when we put back the last edge, we have a final relation involving a stable letter t say which says that tg j k 1 v k 1 t 1 = g i 0 0. As g vk 1 maps to ( ) j0 j 1...j k 2,0, i 1 i 2...i k 1 weconcludethatπ 1 (G)isCAT(0)ifandonlyif i = j,wherei = i 0 i 1...i k 1 and j = j 0 j 1...j k 1. But now in π 1 (G) the infinite order element g 0 satisfies tg i 0 t 1 = g j 0. Thus if i j then π 1 (G) is not balanced and hence contains a non Euclidean Baumslag - Solitar subgroup. In the other case we now suppose that (by relabelling) the root vertex v 0 has the property that the two edge inclusions are not parallel in G v0 = Z 2, so that without loss of generality we can take a free basis g v0,h v0 for G v0 where the two edge groups include as g l v 0 going in to v 0 and h m v k 1 coming out. We also choose a free basis g vi,h vi Z 2 for every other vertex group G vi such that the edge group coming into v i lies in g vi. Then we can form a translation length assignment in the same way as above, starting by sending h v0 to (1,0) R 2 and h v0 to any x R 2 \R, as yet not specified. The images in G v0 and G v1 of the first edge group then define the translation length assignment for g v1 in terms of that for h v0 and we can choose it arbitrarily for h v1. We can then proceed around the loop in this fashion until we reach the final edge, which gives rise to the relation tg m v k 1 h n v k 1 t 1 = g l v 0 for l 0 and m,n not both zero. This imposes the condition mx k 1 +ny k 1 = lx where x k 1,y k 1 are the translation length assignments already chosen for g vk 1,h vk 1 and x is that for g v0 which is yet to be determined. However if there is such an x then we will have satisfied all conditions for our translation length assignment and hence we will have a CAT(0) group. Now this equation in x merely defines a circle with centre 0 and the radius is non zero because y k 1 was already picked to be linearly independent from x k 1 and we know (m,n) (0,0). Thus there exist solutions for x which are not on the real line. For an arbitrary connected graph Γ with an equal number of vertices and edges, there will be an embedded simple loop L. We can apply the above

19 4 EXAMPLES 19 proof replacing Γ by L, which tells us that either there is a non Euclidean Baumslag - Solitar subgroup in the fundamental group of the graph of groups given by L, but this is a subgroup of π 1 (G), or we obtain a translation length assignment on the vertex subgroups of L. But in the latter case, if we now remove the edges of L from Γ we are left with a disjoint union of trees, each with a root vertex where the vertex group has been assigned a translation length. We can now work outwards from the root over all vertices in each tree, ensuring condition (i) in Definition 2 is always satisfied, with no further conditions of the form (ii) left over. Thus we obtain a translation length assignment for (G,Γ) and we have shown that π 1 (G) is CAT(0). We finish this section by illustrating how our six different notions of non positive curvature work in the case of tubular groups, where they form an order. First recall from the introduction the implications between our different definitions which hold in general. Now let us restrict to tubular groups where we can use the results of [19] to obtain additional implications. First Corollary5.9theretellsusthatactingproperlyandcocompactlyonaCAT(0)cube complex implies being virtually special. This in turn implies weak CAT(0) which we have already seen is equivalent to being CAT(0) for tubular groups. Then [19] Theorem 1.2 says that in this case weak CAT(0) implies the existence of a proper action on some CAT(0) cube complex. Finally we know that this property implies in general that our group is balanced. We now give examples to demonstrate that, apart from CAT(0) and weak CAT(0), these properties can be distinguished amongst tubular groups. Acting properly and compactly on a CAT(0) cube complex: We can take G to be a RAAG itself, for instance F 2 Z is obtained by taking two Z 2 vertex groups a,b and c,d and one edge group which identifies b with c. In fact we have a complete classification of these tubular groups in [19] Corollary 5.10: they have exactly one or two parallelism classes of edge groups in each vertex group and do not contain any non Euclidean Baumslag - Solitar subgroup. Virtually special but does not act properly and compactly on any CAT(0) cube complex: For this we require a method that can deduce a group is virtually special without requiring cocompactness of such an action. We have exactly this

20 5 VIRTUALLY SPECIAL TUBULAR GROUPS FROM TREES 20 in [21] where Woodhouse gives a criterion for tubular groups to be virtually special. Wewill applythis inthenext section toshow that any tubular group π 1 (G) obtained from (G,Γ) with Γ a tree is virtually special. Thus on taking (say)γtobeastarwiththreeedgesandsuchthateachinclusionintothecentral vertex group is a different primitive element, we obtain such an example. CAT(0) but not virtually special: Here we can use Wise s non Hopfian CAT(0) group from [17], which is Example 2 above. If this group G were virtually special then the resulting finite index special group would be residually finite as a subgroup of a RAAG, hence so would G. However it is well known that for finitely generated groups, being residually finite implies being Hopfian. Acting properly on a CAT(0) cube complex but not CAT(0): Example 1 above, or Example 3 in the case when n = 2. Balanced but does not act properly on any CAT(0) cube complex: Example 3 above for n > 2, or Example 4 (these will all be balanced by [7] Theorem 8.3). Not balanced: On taking one vertex with vertex group a,b, we can add in a self loop so that the stable letter conjugates a m to a n but m n. 5 Virtually special tubular groups from trees The key definition in[19] which is used forresults ontubular groupsisthat of an equitable set for (G,Γ). This is defined to be a choice at each vertex v in Γ of a finite number of closed curves c (v) 1,...,c(v) m(v) on a torus S1 S 1 placed at v, which therefore can be regarded as elements of π 1 (S 1 S 1 ) = G v = Z 2, so that on taking any edge e with endpoints e ± and edge group z e embedding ing e+ asz e+ Z 2 \{(0,0)}and ing e as z e, thenthe sum of the(algebraic, unsigned) intersection numbers at e + is equal to the sum #[c (e +) 1,z e+ ]+...+#[c (e +) m(e + ),z e + ] #[c (e ) 1,z e ]+...+#[c (e ) m(e ),z e ]

21 5 VIRTUALLY SPECIAL TUBULAR GROUPS FROM TREES 21 of intersection numbers at the other end. The importance of this definition of Wise is his result in [19] stating that the tubular group defined by (G,Γ) acts freely on a CAT(0) cube complex (which might well be infinite dimensional and/or not locally finite) if and only if we can find an equitable set for (G,Γ). Now we can regard z e+ as a closed curve lying in the torus at e + and similarly z e at e, with these two tori joined by a cylinder corresponding to the edge e in order to obtain a graph of spaces. Then this condition on intersection numbers is needed to ensure a bijection between the intersection points of z e+ with the equitable set at v + and those of z e with the equitable set at v. (It is also required as part of the definition of an equitable set that the elements at each vertex of this set are not all parallel.) Here we want to show that there are a range of tubular groups which are virtually special, in addition to those which act properly and cocompactly on acat(0) cubecomplex. Nowthereseemtobefewcriteriainexistence which show a group is virtually special without requiring this geometric action. However we have a result of Woodhouse in [21] which does exactly this for tubular groups. As we know from the discussion above that a tubular group which is virtually special also acts freely on a CAT(0) complex, it is not surprising that Woodhouse s result begins by taking an equitable set for a tubular group but then requires further conditions on this set to hold, which we now review. If a closed curve c is not primitive, so that it is of the form n(a,b) Z 2 \ {(0,0)} with n 2 and a,b coprime, then in the Wise result we can either regard c as a primitive curve traversed n times, or as n parallel disjoint primitive curves. In the Woodhouse work the second approach is taken and we will do that here too. For Woodhouse s virtually special criterion, we first require an equitable set consisting only of primitive elements (which can be ensured by the comment above) but which is also fortified, meaning that every time we inject an edge group z e into a vertex group G e+ (or G e ), the image z e+ of z e is parallel to something in the equitable set at e +, or alternatively there is at least one element c in this equitable set such that the intersection number #[c,z e+ ] = 0. A primitive, fortified equitable set is then used to create (primitive, fortified immersed) walls. This can be thought of as using the equality between intersection points on either side of an edge to create a graph Ω where each vertex of Ω corresponds to an element of the equitable set (and so can be thought of as lying over a vertex v of the graph Γ). The edges of Ω are

22 5 VIRTUALLY SPECIAL TUBULAR GROUPS FROM TREES 22 given by taking an edge e of Γ and choosing a bijection b between the intersection points on either side of e, then adding an edge in Ω between the two elements of the equitable set in which the pair of intersection points p ± lie, for each pair where b satisfies b(p ) = p +. The resulting graph Ω, which certainly need not be connected, is the space of immersed walls (actually this space is where each vertex is not just a point, rather a copy of S 1 but for the result that follows in [21] each of these copies of S 1 can be replaced with a single point and we will use this construction). Note that our graph Ω can be thought of as having the graph Γ as a quotient, in that we have a projection respecting edges which sends an element of our equitable set to the vertex of Γ in which it lies, and the same with the edges of Ω and of Γ. Let the connected components of Ω be Ω 1,...,Ω k. The criterion that we will now use is [21] Proposition 4.8, which states that if Ω 1,...,Ω k are primitive, fortified, undilated immersed walls then the tubular group G = π 1 (G) is virtually special, with only the condition of being undilated left to explain. We describe this as follows: suppose we have a directed edge path e 1,...,e n in the graph Ω. The dilation function from such edge paths to Q is calculated as follows: we start with value 1 at the vertex (e 1 ), which is an element of our equitable set, and then as we traverse the edge e 1 to arrive at the vertex (e 1 ) +, we multiply our value by the ratio of intersection numbers #[(e 1 ),z (e1 ) ]/#[(e 1 ) +,z (e1 ) + ]. Here e 1 is the edge in Γ lying below the edge e 1 Ω and z (e1 ) ± are the respective inclusions at each end of the generator z (e1 ) of the edge group for e 1 in Γ. We then multiply our value by #[(e 2 ),z (e2 ) ]/#[(e 2 ) +,z (e2 ) + ] as we cross the edge e 2 in Ω and so on, thus obtaining the value of our dilation function for this edge path when we arrive at the final vertex (e n ) +. We then say that Ω has undilated immersed walls if the dilation function of every closed path in Ω is 1. Now although our original graph Γ is a tree, the components of Ω need not be. Even if they are, this does not ensure that Ω has undilated immersed walls, as demonstrated in [20] Example where the tubular group a,b,c,d [a,b],[b,c],[c,d] is taken, along with an equitable set and pairing of intersection points to create a graph Ω which is in fact connected but where there exists a dilated immersed wall. As this group is clearly a RAAG, we see that we have to take care both in providing an appropriate equitable set and in choosing suitable bijections between intersection points in order to conclude a group is virtually special using this criterion. The following is the crucial point that we will use in order to obtain undilated immersed walls.