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The space of left orderings of a group with applications to topology Clay, Adam J. 2010

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The space of left orderings of a group With applications to topology by Adam J. Clay  B.Sc., The University of King’s College, 2003 M.Sc., The University of British Columbia, 2005  A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY in The Faculty of Graduate Studies (Mathematics)  THE UNIVERSITY OF BRITISH COLUMBIA (Vancouver) April 2010 c Adam J. Clay 2010  Abstract A group is left orderable if there exists a strict total ordering of its elements that is invariant under multiplication from the left. The set of all left orderings of a group comes equipped with a natural topological structure and group action, and is called the space of left orderings. This thesis investigates the topology of the space of left orderings for a given group, by analyzing those left orderings that correspond to isolated points and by characterizing the orbits of the natural group action using maps from a certain free object. Lastly, we find an application of the space of left orderings in the field of 3manifold topology, by using compactness to show that certain fundamental groups are not left orderable.  ii  Table of Contents Abstract  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  ii  Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . .  iii  List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  v  1 Introduction to orderable groups . . . . . . . . . . . . . . . .  1  1.1  Left orderability . . . . . . . . . . . . . . . . . . . . . . . . .  1  1.2  Conrad orderability  . . . . . . . . . . . . . . . . . . . . . . .  4  1.3  Bi-orderability . . . . . . . . . . . . . . . . . . . . . . . . . .  7  1.4  Criteria for orderability and examples . . . . . . . . . . . . .  9  1.4.1  Non-constructive criteria for orderability  . . . . . . .  9  1.4.2  Constructive criteria for orderability . . . . . . . . . .  11  1.4.3  Examples of left, Conrad, and bi-orderable groups . .  16  2 The space of left orderings . . . . . . . . . . . . . . . . . . . .  19  2.1  2.2  Definitions and elementary properties . . . . . . . . . . . . .  19  2.1.1  . . . . . . . . . . . .  25  Applications of the space of left orderings . . . . . . . . . . .  26  The infinite strand braid group  iii  Table of Contents 3 The topology of the space of left orderings . . . . . . . . . .  30  3.1  The case of Conradian orderings . . . . . . . . . . . . . . . .  32  3.2  Isolated points . . . . . . . . . . . . . . . . . . . . . . . . . .  33  3.3  Dense orderings and discrete orderings  . . . . . . . . . . . .  39  3.3.1  Abelian groups . . . . . . . . . . . . . . . . . . . . . .  41  3.3.2  Non-abelian groups  . . . . . . . . . . . . . . . . . . .  46  . . . . . . . . . . . . . . . . . . . . . . . .  50  3.4  The braid groups  4 Analysis of the natural action on the space of left orderings 56 4.1  Lattice-ordered groups  4.2  Free lattice ordered groups and the topology on the space of  4.3  . . . . . . . . . . . . . . . . . . . . .  56  left orderings . . . . . . . . . . . . . . . . . . . . . . . . . . .  62  Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  73  4.3.1  The free group . . . . . . . . . . . . . . . . . . . . . .  73  4.3.2  Left orderable groups with all left orderings Conradian 74  4.3.3  The braid groups  . . . . . . . . . . . . . . . . . . . .  5 3-manifolds and left orderability  77  . . . . . . . . . . . . . . . .  79  5.1  Examples of left orderable fundamental groups . . . . . . . .  79  5.2  Geometric applications of left orderability . . . . . . . . . . .  83  6 An application of the space of left orderings to 3-manifolds 86 6.1  6.2  The map ψG,H . . . . . . . . . . . . . . . . . . . . . . . . . .  86  6.1.1  95  Example: Torus knots . . . . . . . . . . . . . . . . . .  Analysis of the map ψG,H for bi-orderable groups  . . . . . .  98  Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 iv  List of Figures 5.1  The figure eight knot. . . . . . . . . . . . . . . . . . . . . . .  82  v  Chapter 1  Introduction to orderable groups In this chapter, we cover standard notions and equivalences pertaining to orderable groups. A good reference for this material is [5], or [30] and [24] for more modern results.  1.1  Left orderability  Definition 1.1.1. A group G is said to be left orderable if there exists a strict total ordering < of the its elements such that g < h implies f g < f h for all f, g, h in G. We can associate to every left ordering < of G its positive cone P , which is defined as P = {g ∈ G|g > 1}. The positive cone satisfies 1. P · P ⊂ P , and 2. P t P −1 t {1} = G, where A t B denotes the disjoint union of A and B. Conversely, given any subset P ⊂ G satisfying properties (1) and (2) 1  1.1. Left orderability above, we can define a strict, left invariant total ordering of G according to the rule g < h if and only if g −1 h ∈ P . It is natural to ask whether or not there is a notion of ‘right orderability’ of a group. The idea of totally ordering the elements of a group so that the ordering is invariant under multiplication from the right is equivalent to left orderability. Observe that if <R is a right ordering of G we may define a left ordering <L of G according to the rule g <L h if and only if g −1 <R h−1 . Therefore, whether we speak of left or right orderable groups is a matter of convention. As the group actions in this text are written from the left, our groups will be left orderable. It is an easy first exercise with left orderable groups to discover that they are all torsion free. For if G is left ordered and g is an element of G satisfying g > 1, then successive left multiplication yields g 2 > g, g 3 > g 2 , ... etc. Combining inequalities gives 1 < g < g 2 < g 3 < · · · so that no power of g can be equal to the identity. Definition 1.1.2. If G is a left ordered group with ordering <, a set S is said to be convex relative to the left ordering < if a, c ∈ S and a < b < c implies b ∈ S. The case when S is a subgroup is of particular importance. Proposition 1.1.3 ([30], Propositions 2.1.1 - 2.1.3). Suppose that C is a subgroup of the left ordered group G with ordering <. Then C is convex relative to the ordering < if and only if the prescription  g < h ⇒ gC ≺ hC 2  1.1. Left orderability provides a well-defined left invariant ordering ≺ of the set of left cosets G/C. This allows us to interpret left orderings with convex subgroups as ‘lexicographic’ orderings in the following sense: Proposition 1.1.4 ([30], Propositions 2.1.1 - 2.1.3). Suppose that C is a convex subgroup relative to the left ordering < of G, whose positive cone we denote by P ⊂ G. Set PC = P ∩ C, and let PG/C be the set of left cosets whose representatives are positive in the left ordering < of G. If we denote the quotient map onto the set of left cosets by q : G → G/C, then P = PC ∪ q −1 (PG/C ).  The previous proposition is closely related to the notion of orderability of extensions. Proposition 1.1.5 ([30], Proposition 1.2.6). Suppose that i  q  1→K→G→H→1  is a short exact sequence. If K and H are left orderable groups, then G is also left orderable. One of the most common applications of Propositions 1.1.3, 1.1.4 and 1.1.5 is to ’modify’ a given left ordering on a prescribed convex subgroup. Using the notation of Proposition 1.1.4, we can make this notion more precise: The positive cone P may be written as two disjoint parts, PC and q −1 (PG/C ). If we choose any other positive cone QC ⊂ C different from PC , 3  1.2. Conrad orderability then it is easy to check that QC ∪ q −1 (PG/H ) defines a new positive cone in the group G. The left ordering defined by this new positive cone differs from our original left ordering, as is easily seen by comparing the two left orderings when restricted to the subgroup C. We may also create new convex subgroups relative to a given left ordering, by taking the intersection or union of an arbitrary family of convex subgroups. Proposition 1.1.6 ([13]). Suppose that the subgroups {Ci }i∈I of G are T S convex relative to the left ordering <. Then i∈I Ci and i∈I Ci are also convex subgroups relative to the left ordering < of G. Finally, we introduce an important piece of terminology. Definition 1.1.7. A left ordering < of a group G is called Archimedian if for every pair of elements g, h ∈ G, with h different from the identity, there exists an integer n ∈ Z such that g < hn . Proposition 1.1.8 (Hölder, for a proof see [13]). If a group G admits an Archimedian left ordering <, then there exists an order-preserving isomorphism i : G → (R, +), so that G is isomorphic to a subgroup of the additive real numbers. Hence, G is abelian.  1.2  Conrad orderability  A stronger notion than that of left orderability is Conrad orderability. Definition 1.2.1 ([13]). A group G is said to be Conrad orderable if there exists a left ordering < of G such that the following implication holds: If 4  1.2. Conrad orderability g and h are positive, then there exists n ∈ N such that g < hg n . If this implication holds for some left ordering < of a group G, then < is called a Conradian ordering. It has since been observed that requiring n = 2 is equivalent to our definition above, so we may alternatively think of the Conradian orderings of a group G as those left orderings satisfying g < hg 2 for all pairs of positive elements g, h ∈ G [40]. Saying that a group G is Conrad orderable may also be interpreted as saying that G contains a positive cone P satisfying the third condition: 3C. g ∈ P implies g −1 P g 2 ⊂ P . The first investigation of Conradian orderings was motivated by their rich structure, in the sense that a Conradian ordering on a group G determines a family of convex subgroups of G. These convex subgroups can be index by elements of the group G, are ordered by inclusion, with successive subgroups in the ordering having abelian quotients. To be more specific, we first need a definition. Definition 1.2.2. Suppose that C ⊂ D are two convex subgroups in some left ordering < of a group G. Then D is said to cover C, or alternatively C → D is a convex jump, if there are no convex subgroups between C and D. Theorem 1.2.3 (Conrad, [13]). Let G be a group equipped with a Conradian ordering <. Then the following hold:  5  1.2. Conrad orderability 1. Every element g ∈ G determines a convex jump Cg → Dg with g ∈ Dg and g ∈ / Cg , by setting Dg =  \  C  C convex g∈C  and Cg =  [  C.  C convex g ∈C /  2. Every convex jump C → D in the ordering < arises from an element g ∈ G as above, that is, C = Cg and D = Dg for some g ∈ G. 3. For every convex jump C → D, C is normal in D, and the quotient D/C is abelian. The ordering < descends to a well-defined ordering on the quotient D/C (since C is convex), and this ordering is Archimedian. In particular this theorem allows one to define, for every convex jump C → D, an injective homomorphism τ : D/C → (R, +) by using Proposition 1.1.8. Often referred to as the ‘Conrad homomorphism’, these maps are essential to many of the modern dynamical approaches to left orderability [40, 41, 54]. With this theorem, we can also classify those Conradian orderings without convex subgroups. Proposition 1.2.4. Suppose < is a Conradian left ordering of G. Then the ordering < has no proper, nontrivial convex subgroups if and only if it is Archimedian. Proof. In one direction, it is easy to see that every Archimedian ordering is Conradian (c.f. Proposition 1.3.3). Conversely, given any non identity 6  1.3. Bi-orderability element g ∈ G, there exists a convex jump Cg → Dg , as defined in Theorem 1.2.3. In the case that G contains no proper, nontrivial convex subgroups, we must have Cg = {1} and Dg = G. The ordering on the quotient Dg /Cg = G is Archimedean by Theorem 1.2.3, and the proposition follows. In light of Proposition 1.1.8, we know that Conradian ordered groups that admit no convex subgroups are abelian groups. It should also be noted that modifying a left ordering on a convex subgroup, as explained following Proposition 1.1.5, can be done in a way that preserves the Conradian structure of a left ordering. If P defines a Conradian ordering of G relative to which C is a convex subgroup, then we may replace P ∩ C with any positive cone Q ⊂ C which defines a Conradian ordering of C. Then the set Q ∪ (P \ (C ∩ P )) defines a positive cone in G, whose associated left ordering is Conradian.  1.3  Bi-orderability  Definition 1.3.1. A group G is said to be bi-orderable if there exists a left ordering of G that is also invariant under multiplication from the right, so that g < h implies gf < hf for all f, g, h ∈ G. In the language of positive cones, this is equivalent to requiring that the positive cone P of a given left ordering satisfy the additional third condition 3. gP g −1 = P for all g in G. It is easy to see that bi-orderability is a strengthening of Conrad orderability. Indeed, a bi-ordering of a group G satisfies g < hg n for every 7  1.3. Bi-orderability positive pair g, h ∈ G by taking n = 1. In the case of left orderings, we found that we were able to left order a group G if it fit into a short exact sequence  1→K→G→H→1  with both K and H left orderable. In fact, we did not even require the set of left cosets of K to have a group structure. In the case of bi-orderings the technique of using short exact sequences to construct positive cones in groups is more restricted. Proposition 1.3.2. Suppose that i  q  1→K→G→H→1  is an exact sequence, and that K and H admit bi-orderings whose positive cones are PK and PH respectively. Then P = i(PK ) ∪ q −1 (PH ) defines the positive cone of a bi-ordering of G if and only if gPK g −1 = PK for all g ∈ G. Finally, there is an important connection between Archimedian, Conradian, and bi-orderings. Proposition 1.3.3. Let G be a left orderable group, and suppose that < is an Archimedian ordering. Then < is a bi-ordering. 8  1.4. Criteria for orderability and examples Proof. We apply Proposition 1.1.8 to conclude that G is abelian. Then, we observe that gf < hf if and only if f g < f h, left multiplying by f −1 yields g < h, establishing right invariance. This shows the underlying connection between Conrad orderability and bi-orderability: For every convex jump C → D in a Conradian ordered group G, the quotient D/C is bi-ordered by the inherited quotient ordering, because the ordering is Archimedean. These bi-orderings on successive quotients in G together do not determine a bi-ordering of G (instead merely a Conradian ordering), precisely because of the additional hypotheses required in Proposition 1.3.2 versus Proposition 1.1.4.  1.4  Criteria for orderability and examples  The primary purpose of this section is to provide a footing in the theory of orderable groups, by reviewing criteria that allow us to conclude that different groups are left, Conrad, or bi-orderable. In particular, while biorderable groups are Conrad orderable, and Conrad orderable groups are left orderable, we will give examples which show that neither implication is reversible.  1.4.1  Non-constructive criteria for orderability  Our first criteria is a theorem by Burns and Hale, the proof of which is non-constructive. Theorem 1.4.1 (Burns-Hale, [8]). A group is left orderable if and only  9  1.4. Criteria for orderability and examples if every nontrivial finitely generated subgroup maps onto a nontrivial left orderable group. Let T be a property of groups, and suppose that every finitely generated subgroup of a group G has property T . If this implies that G itself has property T , then T is called a local property. From the theorem of Burns and Hale, we conclude that left orderability is a local property (in fact, Conrad orderability and bi-orderability are also local properties). Theorem 1.4.2. Suppose that every finitely generated subgroup of G is left orderable (respectively Conrad orderable, bi-orderable). Then G is left orderable (respectively Conrad orderable, bi-orderable). This allows for us to easily bi-order torsion free abelian groups, by first dealing with the case of Zn geometrically. Let n be any positive integer, and let H be a hyperplane in Rn . We may choose H so that no element of Zn lies in H (c.f. the proof of Lemma 3.3.4). Therefore, we may create a positive cone P ⊂ Zn by declaring the elements of Zn on one side of the hyperplane H to be P . Hence, Zn is bi-orderable, and we can apply Theorem 1.4.2 to conclude: Corollary 1.4.3. Every torsion free abelian group is bi-orderable. The idea of Burns and Hale can be strengthened to guarantee the existence of a Conradian ordering. Definition 1.4.4. A group G is locally indicable if every finitely generated subgroup of G maps surjectively on to Z.  10  1.4. Criteria for orderability and examples Theorem 1.4.5 ([24, 40, 53]). A group is Conrad orderable if and only if it is locally indicable. It was unknown for some time whether or not left orderability is equivalent to local indicability, but it is now known that local indicability is a strictly stronger notion [4]. As local indicablity and Conrad orderability are very powerful restrictions on a group G and can provide great insight into its structure, it is natural to attempt to connect the weaker notion of left orderability with that of Conrad orderability. Efforts in this direction have focused on addressing the following question. Question 1.4.6. For what class of groups is left orderability equivalent to Conrad orderability (local indicability)? This question remains open at the time of writing. Modern approaches and results will be discussed in Section 2.2. A final non-constructive criterion is as follows. Theorem 1.4.7 ([26]). Every torsion free one relator group is Conrad orderable. This criterion is sometimes quite difficult to use, as the problem of determining when a one relator group is torsion free can be quite challenging. Perhaps most useful is the following corollary: Corollary 1.4.8. Every left orderable one relator group is Conrad orderable.  1.4.2  Constructive criteria for orderability  In this section we will investigate those groups which are known to be left, Conrad, or bi-orderable by virtue of the construction of a left, Conrad, 11  1.4. Criteria for orderability and examples or bi-ordering of their elements. As a first class of examples, and as a useful theorem in many applications to come, we investigate groups of orderpreserving automorphisms of sets. Definition 1.4.9. Suppose that a group G acts on a set S. The action of G on S is said to be effective if for every g ∈ G different from the identity, there exists x ∈ S such that gx 6= x. Theorem 1.4.10. A group G is left orderable if and only if G acts effectively on some linearly ordered set (Ω, <) by order preserving bijections. Proof. One direction is trivial: If G is left orderable, then G acts on itself in an effective, order preserving manner by left multiplication. Conversely, suppose that G acts effectively on the linearly ordered set (Ω, <). Choose some well-ordering ≺ of Ω, and for each g ∈ G, let xg be the smallest element in Ω (with respect to our chosen well ordering) such that g(x) 6= x. Declare g > 1 in G if g(xg ) > xg . We must check that this rule defines a set P satisfying P · P ⊂ P , and G = P t P −1 t {1}. To this end, suppose that g and h are both positive, and so satisfy g(xg ) > xg and h(xh ) > xh . If xg ≺ xh , then xg = xgh and h(xgh ) = xgh since xgh ≺ xh , so we compute  gh(xgh ) = g(xgh ) = g(xg ) > xg = xgh ,  so that gh > 1. On the other hand, if xh ≺ xg then xh = xgh and we compute gh(xgh ) = gh(xh ) > g(xh ) = g(xgh ) = xgh , 12  1.4. Criteria for orderability and examples so that gh > 1. This shows that the positive cone we have defined satisfies property (1). Property (2) is clear, we need only observe that since the action on Ω is effective the identity is the only element of G which acts trivially. In the case that G is a countable group, this result can be surprisingly strengthened by taking Ω = R. For a proof of the following theorem, see [34]. Theorem 1.4.11. Suppose that G is a countable group. Then the following are equivalent: 1. G is left orderable. 2. There exists an injective homomorphism G → Homeo+ (R). 3. There exists an injective homomorphism G → Homeo+ (Q). These criteria demonstrate that left orderability of a countable group can be interpreted geometrically, and allow for the interpretation of many properties of countable left orderable groups in the context of dynamical systems [40]. We now have our first class of examples of left orderable groups. Corollary 1.4.12. Every group of order preserving bijections of a linearly ordered set is left orderable. In order to review one of the simplest ways of constructing bi-orderings, recall the definition of the lower central series {Gk }k>0 in a group G. Set  13  1.4. Criteria for orderability and examples G1 = G and Gk+1 = [G, Gk ], the subgroup of G generated by all commutators ghg −1 h−1 where g ∈ G and h ∈ Gk . The quotients Gk /Gk+1 are abelian groups, and for every k the short exact sequence  1 → Gk /Gk+1 → G/Gk+1 → G/Gk → 1  is a central extension. Theorem 1.4.13 ([44]). Let G be a group. Suppose that ∞ \  Gi = {1},  i=1  and that every quotient Gk /Gk+1 is torsion free. Then G is bi-orderable. Proof. Define a bi-ordering of G as follows. For each k, let qk : Gk → Gk /Gk+1 denote the quotient map. As each quotient is a torsion free abelian group, Gk /Gk+1 is bi-orderable by Corollary 1.4.3. Thus we may fix a positive cone Pk ⊂ Gk /Gk+1 for every k > 0. Define a positive cone P ⊂ G according to the formula:  P =  ∞ [  qi−1 (Pi ).  i=1  We must check that P satisfies the required properties. For ease of exposition, denote by n(g) the unique integer n for which g ∈ Gn \ Gn+1 (such T an integer exists for all g since we are assuming that ∞ i=1 Gi = {1}). Thus g ∈ G lies in P if and only if qn(g) (g) lies in Pn(g) . 1. P · P ⊂ P . Given g, h ∈ P , set n = min{n(g), n(h)}. Then qn (g) and 14  1.4. Criteria for orderability and examples qn (h) are both non-negative elements in the ordered group Gn /Gn+1 , thus their product is non-negative, i.e. qn (gh) ∈ Pn . Hence gh ∈ qn−1 (Pn ), and gh ∈ P . 2. G = P t P −1 t {1}. Given g ∈ G different from the identity, observe that g ∈ P and g −1 ∈ P if and only if qn(g) (g) ∈ Pn(g) and qn(g) (g −1 ) ∈ Pn(g) (here, we use the fact that n(g) = n(g −1 )). This is not possible since Pn(g) is a positive cone. Similarly, we find that g ∈ / P and g −1 ∈ / P is not possible, because this would imply that qn(g) (g) ∈ / Pn(g) and qn(g) (g)−1 ∈ / Pn(g) . 3. gP g −1 = P . Let h ∈ P be given. There are three cases to consider. (a) n(g) = n(h). In this case, qn(g) (ghg −1 ) = qn(g) (h) ∈ Pn(g) , since the image of qn(g) is abelian. Hence ghg −1 ∈ P . (b) n(g) > n(h). Then qn(h) (g) = 1, and so qn(h) (ghg −1 ) = qn(h) (h) ∈ Pn(h) , and we conclude that ghg −1 ∈ P . (c) n(g) < n(h). In this case we may use the central extension  1 → Gn(h) /Gn(h)+1 → G/Gn(h)+1 → G/Gn(h) → 1 to write ghg −1 = hc for some c ∈ Gn(h)+1 . Then c satisfies qn(h) (c) = 1, and we calculate qn(h) (ghg −1 ) = qn(h) (hc) = qn(h) (h) ∈ Pn(h) , and hence ghg −1 ∈ P .  Bi-orderings of a group G that are constructed as in the theorem above 15  1.4. Criteria for orderability and examples are called standard bi-orderings. It is well known that free groups satisfy the hypotheses of Theorem 1.4.13, namely ∞ \  Gi = {1} and Gk /Gk+1 is torsion free.  i=1  Thus we arrive at the useful corollary: Corollary 1.4.14. Free groups are bi-orderable.  1.4.3  Examples of left, Conrad, and bi-orderable groups  With some preliminary theorems behind us, we turn to the question of finding a left orderable group that is not locally indicable. The most classical method of finding a group that is left orderable, but not locally indicable, is by finding a group that is left orderable and contains a finitely generated perfect subgroup. If a group G contains a finitely generated perfect subgroup H, then G cannot be locally indicable since there is no surjective map H → ^ Z. This is the method applied in [4] to show that SL 2 (R) is not locally indicable, and in [53] to show that the braid groups are not locally indicable for n > 4 (see Section 3.4 for an explicit construction of a left ordering ^ of the braid groups). In the case of SL 2 (R), the finitely generated perfect subgroup is isomorphic to  x, y, z|x2 = y 3 = z 7 = xyz ,  in the case of the braid groups the commutator subgroup [Bn , Bn ] is finitely generated and perfect for n > 4 [39]. 16  1.4. Criteria for orderability and examples This is not the only method for showing that a group is left orderable but not locally indicable. The subgroup of Aut(Q) of functions f satisfying 1. f (x) + 1 = f (x + 1), and 2. f is piecewise linear on [0, 1] is also left orderable, but not locally indicable. The proof of this fact is delicate and does not rely upon the construction of a finitely generated perfect subgroup, see [61]. While it is generally difficult to find a left orderable group that is not locally indicable, it is quite easy to find examples of locally indicable groups that are not bi-orderable. As an easy example, consider the group G = x, y|xyx−1 = y −1 .  This group is easily seen to be left orderable, since it fits into the short exact sequence 1 → Z → G → Z → 1. As G is a one relator group, it is locally indicable by Corollary 1.4.8, and so it is Conrad orderable. However, G cannot be bi-ordered, because the element y ∈ G is conjugate to its own inverse. Supposing G were bi-ordered so that y > 1, we could left multiply by x and right multiply by x−1 to yield 1 < xyx−1 = y −1 , a contradiction. As a final example, recall that the braid group B3 is defined by the presentation hσ1 , σ2 |σ1 σ2 σ1 = σ2 σ1 σ2 i . 17  1.4. Criteria for orderability and examples The commutator subgroup of B3 is isomorphic to the free group on two generators, with σ2 σ1−1 , σ1 σ2 σ1−2 as a generating set [39]. Thus, we may construct a Conradian ordering of B3 by using the short exact sequence  1 → F2 → B3 → Z → 1,  and a bi-ordering of the free group F2 . However, B3 is easily seen to be not bi-orderable, as it contains an element that is conjugate to its own inverse. It is easy to check that (σ1−1 σ2−1 σ1−1 )(σ1 σ2−1 )(σ1 σ2 σ1 ) = σ2 σ1−1 .  18  Chapter 2  The space of left orderings The idea of topologizing the set of all left orderings of a given group first appears in [58], although the results found there primarily deal with the case of countable groups. When G is a countable group, we will see that the topology on the set of all left orderings arises from a metric, and the space of left orderings has many properties in common with the Cantor set. The introduction to the space of left orderings that we provide here does not rely on countability of the underlying group, and closely follows [21]. While the topology from this perspective can be less intuitive, elementary properties such as compactness are much easier to prove.  2.1  Definitions and elementary properties  Definition 2.1.1. Let G be a group. The space of left orderings is denoted by  LO(G) = {P ⊂ G|P is the positive cone of a left ordering of G }.  We also define the space of Conradian orderings and bi-orderings, denoted CO(G) and BO(G), to be the set of all positive cones of Conradian orderings  19  2.1. Definitions and elementary properties and bi-orderings respectively. Since all bi-orderings are Conradian, and all Conradian orderings are left orderings, we have inclusions BO(G) ⊂ CO(G) ⊂ LO(G). Defined as the set of all positive cones, instead of as the set of all left orderings, there is a natural inclusion of LO(G) into the power set 2G . The power set 2G comes equipped with a topology, a subbasis for which is given by sets of the form Ug = {S ∈ 2G |g ∈ S} and Ug0 = {S ∈ 2G |g ∈ / S} where g ranges over all elements of G. Basic open sets in 2G are finite intersections of the subbasic sets, and therefore are of the form  {S ∈ 2G |g1 , · · · , gn ∈ S and h1 , · · · , hm ∈ / S}  where {g1 , · · · , gn } and {h1 , · · · , hn } are arbitrary finite sets of elements in G. The set LO(G) becomes a topological space when equipped with the subspace topology inherited from 2G . Subbasic open sets in LO(G) are therefore of the form  Ug = {P ∈ LO(G)|g ∈ P }  20  2.1. Definitions and elementary properties and / P }, Ug0 = {P ∈ LO(G)|g ∈ where g ranges over all elements of G. Since P is a positive cone, the decomposition G = P t P −1 t {1} allows us to rewrite the sets Ug0 as Ug0 = {P ∈ LO(G)|g ∈ / P } = {P ∈ LO(G)|g −1 ∈ P } = Ug−1 . Thus, as a subbasis for the topology on LO(G) the family of sets {Ug }g∈G T suffices. A basic open set in LO(G) therefore has the form ni=1 Ugi , for T some finite set of elements {g1 , · · · , gn } in G. The basic open set ni=1 Ugi contains all those positive cones whose corresponding left orderings of G satisfy 1 < gi for i = 1, · · · , n. Proposition 2.1.2 ([15, 21, 58]). With the topology described above, the space LO(G) is compact. Proof. Those sets which are not positive cones must not satisfy the two criteria: (1) P · P ⊂ P , and (2) G = P t P −1 t {1}. For every set S which violates (1), there must exist elements g, h ∈ G such that g and h lie in S, while gh does not. In other words, the set of all subsets violating (1) is given by the union U=  [  {S ∈ 2G |g, h ∈ S and gh ∈ / S}.  g,h∈G  Similarly, if a set S is to violate condition (2) there must exist some non identity element g ∈ G such that g, g −1 ∈ S or g, g −1 ∈ / S, or we could also have 1 ∈ S. The set of all sets satisfying one of these three conditions can  21  2.1. Definitions and elementary properties be written as [  V =  {S ∈ 2G |g, g −1 ∈ S} ∪  g∈G  [  {S ∈ 2G |g, g −1 ∈ / S}.  g∈G\{1}  Both U and V are a union of basic open sets, and so U ∪ V is open, and thus its complement LO(G) is closed. Recall that by Tychonoff’s theorem, 2G is compact, and hence the closed subset LO(G) must also be compact. We can show that the spaces CO(G) and BO(G) are compact, by checking that the conditions gP g −1 = P and g ∈ P implies g −1 P g 2 ⊂ P both define closed subsets of 2G as well. By using basic open sets, we can also see that LO(G), CO(G) and BO(G) are Hausdorff (in fact, totally disconnected–the only connected components are singletons). Given two distinct positive cones P1 and P2 in a set U , suppose that they disagree on an element g ∈ G, that is, g ∈ P1 and g ∈ / P2 . Then P1 ∈ Ug and P2 ∈ Ug−1 , while Ug ∪ Ug−1 is the entire space, so that Ug and Ug−1 disconnect U . Thus LO(G), CO(G) and BO(G) are totally disconnected. If the group G is countable, we have already indicated that there is a metric on the space LO(G) that induces the same topology as the subbasis provided above. Proposition 2.1.3 ([58]). If G is countable, then LO(G), CO(G) and BO(G) are metrizable. Proof. Let g0 , g1 , · · · be an enumeration of the elements of G. Given two positive cones P1 , P2 ∈ LO(G) (or in CO(G), or BO(G)), suppose that the 22  2.1. Definitions and elementary properties orderings determined by P1 and P2 agree on the elements g0 , · · · , gn−1 (that is, gi ∈ P1 if and only if gi ∈ P2 for all i < n), and disagree on gn . In this case, define d(P1 , P2 ) =  1 2n .  It is easy to check that this defines a metric on  the space of left, Conrad or bi-orderings. In fact, we may check that the above definition yields a metric satisfying d(P1 , P2 ) ≤ max{d(P1 , P3 ), d(P2 , P3 )}, so that it is a ultrametric. It is a mechanical check to verify that this ultrametric defines a topology agreeing with the subspace topology inherited from 2G , for full details see [58]. We recall a proposition from elementary topology. Proposition 2.1.4 (E.g., [25] Corollary 2.98). Any nonempty, compact, totally disconnected metric space without isolated points is homeomorphic to the Cantor set. The spaces of left, Conradian and bi-orderings therefore have many properties in common with the Cantor set when the underlying group is countable. Proposition 2.1.5 ([58]). Let G be a countable group, and suppose that LO(G) (resp. CO(G), BO(G)) contains no isolated points. Then LO(G) (resp. CO(G), BO(G)) is homeomorphic to the Cantor set. As a first attempt to better understand these spaces for a given countable group, we can therefore attempt to determine their homeomorphism types by determining whether or not the spaces LO(G), CO(G) or BO(G) contain one-point open sets.  23  2.1. Definitions and elementary properties A one-point open set corresponds to a positive P cone satisfying  {P } =  n \  Ugi  i=1  for some finite set of elements {g1 , · · · , gn } in G. Such a positive cone in the space LO(G) (or CO(G), or BO(G)) corresponds to the unique left (or Conradian, or bi-) ordering of the group G in which the elements g1 , · · · , gn are positive. Determining the necessary conditions for the existence of such a left ordering will be the main focus of Chapter 3. There is also a natural group action by homeomorphisms on the space of left orderings. For each element g in a group G, the action on a positive cone P is by conjugation, P 7→ gP g −1 . In terms of left orderings of the group G, this action may be interpreted as follows: if < is the left ordering of G arising from the positive cone P , and <g is the left ordering with positive cone gP g −1 , then ag < bg if and only if a <g b for all a, b ∈ G. The condition that distinguishes the positive cones of bi-orderings from those of left orderings is invariance under conjugation, recall that P is the positive cone of a bi-ordering if gP g −1 = P for all g ∈ G. Therefore, the bi-orderings of a group G are exactly the fixed points of the natural action of G on LO(G). The role played by the Conradian orderings of a group with respect to this action is not yet clear, and is an active area of research. The positive cone of a Conradian ordering is required to satisfy g −1 P g 2 ⊂ P for all g ∈ P , which can almost certainly be given a clean topological or dynamical interpretation in terms of the G-action on LO(G). As yet, however, no 24  2.1. Definitions and elementary properties satisfactory topological or dynamical characterization of the subspace of Conradian orderings exists (see [43] for the latest efforts in this direction). The task of better understanding the G-action on LO(G) will be the focus of Chapter 4.  2.1.1  The infinite strand braid group  One of the primary uses of the natural action on LO(G) has been to disprove the existence of isolated points in LO(G) by showing that a given positive cone P is an accumulation point of OrbG (P ). As an example of this approach, we will investigate the infinite strand braid group B∞ , which is defined by the presentation  hσ1 , σ2 , σ3 , · · · |σi σj σi = σj σi σj if |i − j| = 1, σi σj = σj σi if |i − j| > 1i .  To show that B∞ is left orderable is quite difficult, an explicit left ordering will be described in Section 3.4, for full details see [21]. The group B∞ is an example of a left orderable group with the property that every point in LO(B∞ ) is an accumulation point of its orbit under the B∞ -action. Proposition 2.1.6 ([21]). The space of left orderings of B∞ is homeomorphic to the Cantor set, in fact every positive cone P ∈ LO(B∞ ) is an accumulation point of its conjugates. Proof. Let P ∈ LO(B∞ ), and suppose that P lies in the basic open set T U = ni=1 Uβi where {β1 , · · · , βn } is some finite set of elements in B∞ . We will show that U contains a conjugate of P that is different from P itself.  25  2.2. Applications of the space of left orderings As the braids β1 , · · · , βn are represented by finite words in the generators σ1 , σ2 , · · · , there exists k > 0 such that every generator σi for i ≥ k commutes with every braid βj , for j = 1, · · · , n. We claim that there exists σl with l ≥ k such that σl P σl−1 6= P . For contradiction, suppose not, and consider the embedding shk : B∞ → B∞ given by σi 7→ σi+k , i > 0. The intersection Q = shk (B∞ ) ∩ P is a positive cone that defines a left ordering of the image shk (B∞ ). Moreover, as we are assuming that σl P σl−1 = P for all l ≥ k, the positive cone Q is invariant under conjugation by elements of shk (B∞ ). Thus, Q defines a bi-ordering of shk (B∞ ), which is a group isomorphic to B∞ . This is not possible, as we have already observed that B3 is not bi-orderable, and B3 is a subgroup of B∞ . Therefore we may choose σl , with l ≥ k (so that σl commutes with the braids βi , for i = 1, · · · , n) such that σl P σl−1 6= P . Then we observe that βi ∈ P ⇒ σl βi σl−1 ∈ σl P σl−1 ⇒ βi ∈ σl P σl−1 , so that σl P σl−1 ∈ U . Since we have chosen l so that σl P σl−1 6= P , we conclude that P is not isolated and LO(B∞ ) is homeomorphic to the Cantor set by Proposition 2.1.5.  2.2  Applications of the space of left orderings  Two open problems have recently been solved using the space of left orderings and the natural group action. We will outline their solutions and the  26  2.2. Applications of the space of left orderings role played by the space of left orderings, ending with an indication of where we may find future applications of LO(G). The first question we will address is concerned with the cardinality of LO(G). There exist groups for which LO(G), CO(G) and BO(G) are finite (see [30, 54], and [5] respectively), and it is known that BO(G) may be countably infinite [9]. Moreover, there are many easy cases where LO(G), CO(G) and BO(G) are uncountable, this happens whenever the there is a subset of the space that is homeomorphic to the Cantor set (which is common, as we will see). We find the following question in [35]: Question 2.2.1. Does there exist a group admitting countably infinitely many left orderings? Theorem 2.2.2 (Conjectured by Tararin [30], resolved by Linnell [33]). No group has a countable infinity of left orderings. Sketch of proof. If X is a Hausdorff topological space, define X 0 to be the set of limit points of X. Then for each ordinal α, define X (α) by transfinite induction as follows: • X (0) = X, • X (α+1) = (X (α) )0 • X (α) = ∩λ<α X (λ) if α is a limit ordinal. For contradiction, we suppose that LO(G) is countable. Using compactness and a theorem of Poincaré and Bendixson, it follows that if LO(G) is countable then LO(G)(α) is finite for some ordinal α, say LO(G)(α) = {P1 , · · · Pn }. 27  2.2. Applications of the space of left orderings Then the action of G on LO(G) permutes this finite set {P1 , · · · Pn }, since each set X (α) is invariant under the action by homeomorphisms. Thus StabG (Pi ) is finite index in G, and bi-orderable by Lemma 3.2.3. It is known that groups admitting a finite index bi-orderable subgroup do not have countably many left orderings, [63]. The next theorem outlines the most recent advancement towards answering Question 1.4.6. Theorem 2.2.3 (Conjectured by Linnell [34], resolved by Morris [37]). Suppose that G is an amenable group. Then G is left orderable if and only if G is locally indicable (Conrad orderable). Sketch of proof. If we are to check local indicability of a given amenable, left orderable group, it is sufficient to check on a finitely generated (hence countable) subgroup. Thus, we assume that G is countable. Recall that a discrete group G is amenable if and only if for every continuous action of G on a compact, Hausdorff space X, there is a G-invariant probability measure on X. Thus, the definition of amenablility supplies us with a probability measure on LO(G) that is invariant under the G-action. With the proper setup, an application of the Poincaré recurrence theorem [60] yields a set Ag ⊂ LO(G) for every g ∈ G that consists of positive cones that satisfy g −1 P g 2 ⊂ P (in fact, they satisfy a stronger condition that Morris calls ‘recurrent for the cyclic subgroup hgi’). The conditions of the Poincaré recurrence theorem also guarantee that the complement LO(G)\Ag is measurable, and has measure zero.  28  2.2. Applications of the space of left orderings We conclude that the union  S  g∈G (LO(G)\Ag )  has measure zero, since it T is a countable union of sets of measure zero, and thus g∈G Ag is nonempty. T An element in the intersection g∈G Ag satisfies g −1 P g 2 ⊂ P for all g ∈ G, and thus the corresponding ordering is Conradian. Denote by N F S the family of all groups that contain no non-abelian free subgroups. Amenable groups do not contain non-abelian free subgroups, so every amenable group is in N F S. Peter Linnell has made the following stronger conjecture. Conjecture 2.2.4. Suppose that G is in N F S. Then G is left orderable if and only if G is Conrad orderable. It is thought that a proof of this conjecture may employ the space of left orderings in one of two ways. The condition of being left orderable, but not Conrad orderable places a restriction on the natural action of G on LO(G), and this restriction may allow one to create a free subgroup of G. For example, a deeper understanding of the G-action and its relationship with Conradian orderings may allow for an application of the ping-pong lemma [18], or the construction of a group action on a tree so that we may apply results from Bass-Serre theory [57].  29  Chapter 3  The topology of the space of left orderings As in the case of B∞ , it is possible that a positive cone P in LO(G) may be approximable by its orbit under the G-action. In this chapter, we develop conditions on a positive cone P that guarantee it is an accumulation point of its conjugates. By imposing these conditions on the positive cones of dense left orderings of G, we arrive at a dense Gδ set within a Cantor set in LO(G). Following [40], we define the Conradian soul C< (G) in a left ordered group G with ordering < to be the largest convex subgroup C ⊂ G such that the restriction of < to C is a Conradian ordering. Similarly, we use the notation B< (G) to denote the largest convex subgroup C ⊂ G such that the restriction of < to C is a bi-ordering. Note that we always have B< (G) ⊂ C< (G), since all bi-orderings are also Conradian orderings. Using this notation, one of the strongest results of [40], which we will extend here to the case of uncountable groups, can be stated as follows. Theorem 3.0.1. Let G be a nontrivial group, and let P ∈ LO(G) be an isolated point with associated ordering < of G. Then B< (G) is abelian of rank 30  Chapter 3. The topology of the space of left orderings one, and C< (G) is non-trivial and admits only finitely many left orderings. While Theorem 3.0.1 is proven for the case of countable groups in [40], the dynamical approach used therein is entirely different than our approach, and does not seem to generalize to the case of uncountable groups. Finally, recall that a left ordering of a group G is dense if whenever g < h, then there exists f ∈ G such that g < f < h. If a left ordering < of G is not dense, then it is discrete, meaning that in the ordering < of G there is a least positive element  > 1. We explore the structure of LO(G) by considering the cases of dense and discrete left orderings separately, and we will find the following. Recall that the rank of an abelian group A is the dimension of Q ⊗ A considered as a vector space over Q. Theorem 3.0.2. Let DLO(G) ⊂ LO(G) denote the set of all dense left orderings of a countable group G, and suppose that all rank one abelian subgroups of G are isomorphic to Z. Then if DLO(G) is non-empty, its closure DLO(G) is homeomorphic to the Cantor set, and the set DLO(G) is a Gδ set in DLO(G). Note that our requirement that all rank one abelian subgroups be isomorphic to Z is necessary, as the conclusion of the theorem clearly fails to hold for the abelian group G = Q. In the case of abelian groups, our result will be slightly stronger than Theorem 3.0.2. Specfically, in the case that G is countable and abelian, we will show that DLO(G) = LO(G).  31  3.1. The case of Conradian orderings  3.1  The case of Conradian orderings  We first review known results concerning Conradian orderings, and consider also the case of bi-orderings. Note that the results of this section concerning C< (G) appear in [40], and rely on the following difficult lemma ([40] Lemma 4.4), the bulk of which appeared first in [32], and partially in [30]. Lemma 3.1.1. Suppose that P is the positive cone of a Conradian ordering of the group G, and that there is exactly one proper, nontrivial convex subgroup C ⊂ G. Further suppose that both C and G/C are rank one abelian groups. If P is isolated in LO(G), then G is not biorderable. The next two theorems require the following work of Tararin ([30], Theorem 5.2.1). Recall that a group G admits a finite rational series if 1 = G 0  G 1  · · ·  Gn = G  is a finite normal series with all quotients Gi+1 /Gi rank one abelian. Theorem 3.1.2. Let G be a left-ordered group. 1. If LO(G) is finite, then G has a finite rational series. 2. Suppose that G has a finite rational series. Then LO(G) is finite if and only if Gi  G for all i, and none of the quotients Gi+2 /Gi are bi-orderable. Furthermore, in this case the rational series is unique, and for every left ordering of G, the convex subgroups are precisely G0 , G 1 , · · · , G n .  32  3.2. Isolated points Theorem 3.1.3 ([40] Proposition 4.1). Suppose that P is the positive cone of a Conradian ordering of G. Then P is not an isolated point in the space LO(G), unless LO(G) is finite. Theorem 3.1.4. Suppose that P is the positive cone of a bi-ordering of G. Then P is not isolated in LO(G) unless G is rank 1 abelian. Proof. In the case that G is bi-ordered by the ordering < associated to P , we have C< (G) = G. From Theorem 3.1.3, it follows that G itself must have only finitely many left orderings if the bi-ordering < is to have a positive cone that is isolated in LO(G). However, by Theorem 3.1.2, we see that no group G admitting only finitely many left orders is bi-orderable, except in the case that G is rank one abelian.  3.2  Isolated points  We have already explicitly seen one method for approximating a given positive cone, in Section 2.1.1 the conjugation action is used to approximate a given positive cone in LO(B∞ ). We have also seen, following Proposition 1.1.5, that a left ordering of a group G may be modified on a convex subgroup C in G to create a new left ordering whose positive cone differs from the original only on C. This method of creating new left orderings can also be used as a tool for approximating a given positive cone, as outlined in the following lemmas. Lemma 3.2.1. Suppose P ∈ LO(G) and that C is a convex subgroup of G. Then if PC = P ∩ C is not an isolated point in LO(C), P is not an isolated point in LO(G). 33  3.2. Isolated points Proof. Suppose that P ∈  m \  Ugi ,  i=1  and suppose also that we have numbered the elements gi so that gi ∈ C for i ≤ k (possibly k = 0, in the case that no gi lies in C). Now in LO(C), we have that PC ∈  k \  Ugi ,  i=1  and since PC is not an isolated point, we can choose PC0 ∈  Tk  i=1 Ugi ,  with  PC0 6= PC . We can now construct a positive cone P 0 6= P on G as follows: Given g ∈ G, g ∈ P 0 if g ∈ C and g ∈ PC0 , or if g ∈ / C and g ∈ P . The positive cone P 0 is different from P , since P and P 0 disagree on C, T and by construction, P 0 ∈ m i=1 Ugi . It follows that P is not isolated. Lemma 3.2.2. Suppose P ⊂ G and that C is a normal, convex subgroup of G. Let P 0 denote the positive cone of the ordering inherited by the quotient G/C. If P 0 is not an isolated point in LO(G/C), P is not an isolated point in LO(G). The proof is routine. Lemma 3.2.3. Let G be a left ordered group with ordering <, whose positive cone we denote as P . Then the subgroup StabG (P ) = {g ∈ G : gP g −1 = P }  is bi-ordered by the restriction of < to H = StabG (P ). 34  3.2. Isolated points Proof. To see that the restriction of < is a bi-ordering, consider its positive cone PH = P ∩ H. If g ∈ PH and h ∈ H, then • hgh−1 ∈ H since H is a subgroup, and • hgh−1 ∈ P since, by definition, every element of H fixes the positive cone P under conjugation. Therefore H is bi-ordered. The main difficulty in characterizing the Conradian soul of an isolated point in LO(G) is in showing that the Conradian soul is necessarily nontrivial. If P is an isolated point in LO(G) with associated ordering < of G, then P is certainly not an accumulation point of its conjugates in LO(G). It turns out that knowing P is not an accumulation point of its conjugates gP g −1 ∈ LO(G) is enough to deduce that B< (G) (and hence C< (G)) is non-trivial. Observe that for any group G, if 1 < h < g in the ordering corresponding to P , then left multiplication yields 1 < h−1 g, and then using the fact that h is positive, we conclude that 1 < h−1 gh. Translating this observation into a topological language, we have observed that if P ∈ Ug , then hP h−1 ∈ Ug for any h with 1 < h < g. Supposing that  {P } =  m \  Ugi ,  i=1  is an isolated point, applying the above trick to the set of elements {g1 , · · · , gm } allows us to conclude that for any h with 1 < h < gi for all i ∈ {1, · · · , m},  35  3.2. Isolated points we must have hP h  −1  ∈  m \  Ugi .  i=1  However, since P is isolated, this means that hP h−1 = P , so that (in a sense soon to be made more precise) “small elements in G are bi-ordered,” as they fix the positive cone P under conjugation. Lemma 3.2.4. Suppose that  P ∈  m \  Ugi ,  i=1  where {g1 , · · · , gm } is some finite set of elements of G, yet no conjugates of P (different from P itself ) are in this open set. Then there exists gi ∈ {g1 , · · · , gm } such that the set  Ci = {g ∈ G : gi−k ≤ g ≤ gik for some k} contains only elements of G that fix the positive cone P under conjugation, that is, g ∈ Ci ⇒ gP g −1 = P . Proof. First, we show that there exists gi such that all elements in the set Ci+ = {g ∈ G : 1 < g ≤ gik for some k}  fix P under conjugation. To this end, suppose not. Then for each gi there exists hi with 1 < hi ≤ giki for some ki , and hi P h−1 i 6= P . Choose h = min{h1 , · · · , hm }. Then for  36  3.2. Isolated points each i, we have h ≤ giki ⇒ 1 ≤ h−1 giki ⇒ 1 < h−1 giki h, and therefore giki ∈ hP h−1 . Now since the element giki is positive in the order determined by the positive cone hP h−1 , its ki -th root gi is also positive. This shows that hP h−1 ∈  m \  Ugi ,  i=1  and by our choice of h, hP h−1 6= P , a contradiction. Therefore there exists i such that all elements in Ci+ fix P under conjugation. To prove that all elements g ∈ Ci fix the positive cone P , suppose that g ∈ G satisfies gi−k ≤ g < 1 for some k. Then 1 ≤ gik g < gik , so that either g = gi−k or gik g ∈ Ci+ . 1. In the case g = gi−k , then g −1 ∈ Ci+ and so fixes g −1 P g = P , which implies gP g −1 = P . 2. If gik g ∈ Ci+ , then gik gP g −1 gi−k = P, so that we multiply by powers of gi from both sides and find gP g −1 = gi−k P gik = P.  Note that case (1) has been used to yield the final equality. Therefore we have found gi such that all elements in Ci fix P as claimed.  37  3.2. Isolated points Lemma 3.2.5. For any group G, if  P ∈  m \  Ugi ,  i=1  and no conjugates of P distinct from P lie in this open set, then there exists gi such that the set Ci = {g ∈ G : gi−k ≤ g ≤ gik for some k}  is a convex, bi-ordered subgroup of G. Proof. Convexity of Ci is clear from the definition. By Lemma 3.2.4, Ci is a subset of the bi-ordered group StabG (P ), it follows that Ci is bi-ordered by the restriction ordering as well. Being bi-ordered, we can then conclude that Ci is a subgroup of G: If 1 < g ≤ gik for some k, then gi−k ≤ g −1 < 1, and similarly the implication a < b and c < d ⇒ ac < bd (these implications do not hold in general for left ordered groups) shows closure under multiplication. Corollary 3.2.6. Suppose that the left ordering < of G has positive cone P which is not an accumulation point of its conjugates in LO(G). Then both B< (G) and C< (G) are non-trivial. In particular, we have proven that if < corresponds to an isolated point in LO(G), then both B< (G) and C< (G) are non-trivial. We are now ready to complete the proof of Theorem 3.0.1. Proof of Theorem 3.0.1. Let P be the positive cone of a left ordering < of a 38  3.3. Dense orderings and discrete orderings group G, and suppose that P is an isolated point in LO(G). We know that B< (G) and C< (G) are non-trivial by Corollary 3.2.6, so it remains to show that B< (G) is rank one abelian, and that C< (G) admits only finitely many left orderings. Since we have assumed that P is isolated in LO(G), it follows from Lemma 3.2.1 that the restriction of P to B< (G) must define a bi-ordering that is isolated in LO(B< (G)). However, by Theorem 3.1.4, this is only possible in the case when B< (G) is rank one abelian. Similarly, It follows that the restriction of P to C< (G) must define a Conradian ordering that is isolated in LO(C< (G)), which by Theorem 3.1.3 is only possible in the case that LO(C< (G)) is finite.  3.3  Dense orderings and discrete orderings  In this section, we consider dense orderings of a given group. Note that the word “dense” is being used in this section in two ways: The word “dense” may refer to a property of a subset of LO(G), or it may refer to a property of an ordering of a group G. Similarly, the word “discrete” is being used in this section to describe both a property of an ordering of a group, and a property of a subset of LO(G). The reader should be careful of the context when these words are used. In recent work ([12, 51]), it has proven fruitful to consider discrete and dense group orderings separately, as they reflect different structures of the underlying group. In considering the structure of LO(G), dense orderings of a given group G (with minor restrictions on the group G) are in some sense  39  3.3. Dense orderings and discrete orderings “generic” in LO(G), in that dense orderings of G constitute a dense Gδ set inside of a Cantor set within LO(G). Recall that a set U in a topological space X is a Gδ set if U can be written as a countable intersection of open sets. Lemma 3.3.1. Let DLO(G) ⊂ LO(G) denote the set of dense left orderings of G. If G is countable, then DLO(G) is a Gδ set. Proof. Observe that if  > 1 is the least positive element in a left ordering < of G with positive cone P , then for all g ∈ G (with g 6= 1 different from  and −1 ) either g < −1 or  < g. In other words, either P ∈ Ug−1 −1 or P ∈ U−1 g for all 1 6= g ∈ G different from . That is to say, if V denotes the set of all discrete left orderings of G with least element , then we have observed that V =  \  (Ug−1 −1 ∪ U−1 g ) ∩ U .  g∈G g6=,1  Note that V is closed, as it is an intersection of closed sets, and consists of those positive cones that define an ordering of G with  as least positive element. Therefore, the set of dense orderings is given by  DLO(G) =  \  (LO(G) \ V ),  16=∈G  a countable intersection of open sets. The remaining difficulty is to show that any dense ordering is an accumulation point of other dense orderings. We first consider the case of abelian groups. 40  3.3. Dense orderings and discrete orderings  3.3.1  Abelian groups  From [2], we have the following fact: Proposition 3.3.2. If A is a torsion-free abelian group with rank(A) > 1, then the space LO(A) has no isolated points. For a given torsion-free abelian group A, we can deduce much more about the structure of LO(A) by examining the set of all dense orderings of A. Proposition 3.3.3. Let P be any positive cone in LO(A), where A is a torsion-free abelian group with rank(A) > 1. Then P is an accumulation point of positive cones whose associated orderings are dense orderings. We begin by proving a special case. Lemma 3.3.4. Let P be any ordering in LO(Zk ), where k > 1. Then P is an accumulation point of dense orderings. Proof. We follow the ideas of Sikora in [58], making modifications where necessary. For contradiction, let k > 1 be the smallest k for which the claim fails. Suppose that P ∈  n \  Ugi ,  i=1  with no dense orderings in this open set. Note that we may assume that none of the gi ’s are integer multiples of one another. Extend the ordering < defined by P to an ordering of Qk by declaring v1 < v2 for v1 , v2 ∈ Qk if nv1 < nv2 whenever nv1 , nv2 ∈ Zk . Let H ⊂ Qk ⊗ R = Rk be the subset 41  3.3. Dense orderings and discrete orderings of elements x ∈ Rk such that every Euclidean neighbourhood of x contains both positive and negative elements. Then H is a hyperplane, and Rk \ H is divided into two open components H− and H+ having the property that H+ contains only positive elements, and H− contains only negative elements. Therefore the elements gi lie either in H+ or H itself. Suppose that two or more of the elements {g1 , · · · , gn } lie inside H. In this case, H ∩ Zk = Zm for some m > 1 with m < k, and in this case the positive cone P ∩ Zm ⊂ Zm cannot be an accumulation point of dense orderings in LO(Zm ), for then we could change the positive cone P using Lemma 3.2.1. This change will produce a new positive cone in LO(Zk ) whose associated ordering of Zk is dense, because it is a dense ordering when restricted to the convex subgroup Zm . This contradicts the minimality of k. The remaining possibilities are that exactly one (or none) of the elements {g1 , · · · , gn } lie inside H. In this case, by slight perturbations of the hyperplane H, we can produce a new hyperplane H 0 containing none of the lattice points Zk ⊂ Rk , and with all points gi lying on one side of the hyperplane H 0. Specifically, suppose that  n = (x1 , x2 , · · · , xk )  is the normal vector defining H. Choose real numbers y1 , y2 , · · · , yk that are linearly independent over Q, this is possible for every k > 0 because R is an abelian group of infinite rank. For each i = 1, · · · , k, we may choose a rational number qi such that qi yi is as near to xi as we please, the real 42  3.3. Dense orderings and discrete orderings numbers {qi yi }ki=1 are again linearly independent over Q. Therefore we have shown that we can choose a vector n0 = (q1 y1 , q2 y2 , · · · , qk yk ) arbitrarily close to n such that the hyperplane H 0 with normal vector n0 satisfies H 0 ∩ Zk = {0}. If none of the elements {g1 , · · · , gn } lie inside H, then we may choose n0 to be sufficiently close to n so that {g1 , · · · , gn } lie on one side of H 0 . This new hyperplane H 0 defines a new ordering P 0 on Zk by declaring P 0 = 0 ∩ Zk , where H 0 is the component of Rk \ H 0 containing all g . H+ i +  If one of the elements {g1 , · · · , gn } lies inside H, say gj , then we must take more care in choosing our normal vector n0 . In this case, we choose n0 to be close to n, and inside of the open set  {x ∈ Rk |the angle between x and gj is less than  π 2 }.  Again we see that the new hyperplane H 0 with normal vector n0 defines a positive cone P 0 that contains all of the elements {g1 , · · · , gn }. To see that this ordering is dense, suppose that  ∈ P 0 is the smallest element in P 0 . Recall that the normal vector to H 0 is n0 , and therefore the distance between a vector v and the hyperplane H 0 is given by |projn0 (v)|, the length of the projection of v onto n0 . Since  is the smallest positive element in our ordering, the vector  must 0 for all satisfy  < v for all v ∈ P 0 , which happens if and only if v −  ∈ H+  43  3.3. Dense orderings and discrete orderings v in P 0 . Hence, for all v ∈ P 0 we have |projn0 (v)| > |projn0 ()|, and so  0 < ε = |projn0 ()| = infv∈P {|projn0 (v)|}. If we restrict our attention to the two-dimensional subspace of Rk that is spanned by  and some other vector u ∈ P 0 , this means that all lattice points in the span are at least a distance ε from the line H 0 ∩ span{, u}. This is not possible, as any line in R2 of irrational slope has on either side an infinite number of lattice points lying closer than any positive distance ε > 0 [Theorem 1.1, [45]].  Proof of Proposition 3.3.3. To prove the statement for an arbitrary torsionfree abelian group A with rank(A) > 1, we let g1 , · · · , gm ∈ A be any finite family of elements in a given positive cone P . We will show that there exist infinitely many positive cones with associated dense orderings on A in which all gi are positive. Let N be the subgroup of A generated by the elements g1 , · · · , gm . Then N∼ = Zk for k ≥ 1. Assume that k > 1, for if it is the case that N ∼ = Z, add an additional generator gm+1 none of whose powers lie in N –we may do this since rank(A) > 1. By Lemma 3.3.4, N admits infinitely many dense orderings in which all of g1 , · · · , gm are positive, each constructed by perturbations of the hyperplane associated to the restriction ordering whose positive cone is PN = N ∩P . Fix a positive cone PN0 with a dense associated ordering of N , with PN0 6= PN .  44  3.3. Dense orderings and discrete orderings We may extend PN0 to a distinct ordering Q on the isolator of N I(N ) = {g ∈ A : g l ∈ N for some l}  by declaring g ∈ Q iff g l ∈ PN0 for some l. Observe that the ordering of I(N ) with positive cone Q is dense, for suppose not, say Q had least element . Then  ∈ P 0 is not possible since P 0 is a dense ordering, so let l > 1 be the least positive integer such that l ∈ P 0 . By density of P 0 , we may then choose g ∈ P 0 ⊂ Q with 1 < g < l . Since the only positive elements less than l are , 2 , · · · , l−1 , we have that g = i for i < l. This contradicts our choice of l. Now I(N ) is normal, and the quotient A/I(N ) is torsion-free abelian, so we may order the quotient. Using any ordering on the quotient, we can extend the dense ordering of I(N ) with positive cone Q to give a dense ordering of A with the required properties. Therefore, when A is an abelian group with rank(A) > 1, we know that the closure of the set of dense orderings in LO(A) is the entire space LO(A). Thus, Proposition 3.3.3 and Lemma 3.3.1 together give us the following theorem. Theorem 3.3.5. Suppose that A is a torsion free countable abelian group with rank(A) > 1. Then LO(A) is a Cantor set, and the set DLO(A) of all dense left orderings of A is a dense Gδ set within LO(A). Note that the case of discrete orderings must necessarily be different than this, for there exist abelian groups admitting no discrete orderings: divisible 45  3.3. Dense orderings and discrete orderings torsion free abelian groups are such an example. Question 3.3.6. Let A be a torsion free abelian group with rank(A) > 1. What is the closure of the set of the discrete orderings in LO(A)?  3.3.2  Non-abelian groups  Our results concerning dense orderings of abelian groups generalize to some non-abelian groups. Proposition 3.3.7. Let G be any group in which all rank one abelian subgroups are isomorphic to Z. If P ∈ LO(G) corresponds to a dense left ordering < of G, then P is an accumulation point of positive cones whose associated left orderings are dense orderings. Proof. Let U =  Tm  i=1 Ugi  be an open set in LO(G) containing P , the positive  cone of a dense left ordering < of G. If U contains any conjugates of P (different from P itself), then we are done, so suppose that no conjugate orderings lie in U . Then by proposition 3.2.5, G contains a convex, bi-ordered subgroup C of the form C = Ci = {g ∈ G : gi−k ≤ g ≤ gik  for some k},  where gi ∈ {g1 , · · · , gm }. Denote by C 0 the intersection of all non-trivial convex subgroups of C. There are now two cases to consider. 1. C 0 6= {1}. In this case, since C 0 is bi-ordered and contains no nontrivial convex subgroups, we can use a theorem of Conrad (Theorem 1.2.3) which tells us the order must be Archimedian, and so C 0 must be 46  3.3. Dense orderings and discrete orderings abelian. From our assumption on G, if rank(C 0 ) = 1, we have C 0 ∼ = Z, meaning our ordering is discrete. Therefore rank(C 0 ) > 1. Now the restriction ordering on C 0 with positive cone P ∩ C 0 is a dense ordering, and we know from Theorem 3.3.3 that every dense ordering in LO(C 0 ) is an accumulation point of other dense orderings. Therefore we may change the positive cone P as in the proof of Lemma 3.2.1, creating a new positive cone P 0 containing all gi , and corresponding to a dense ordering of G. 2. C 0 = {1}. In this case, C must have infinitely many convex subgroups whose intersection is trivial. Therefore, we may choose a convex subgroup K, that is non-trivial and contains no gi . Define the −1 positive cone of the “flipped ordering” of K to be (P −1 ∩ K) = PK .  Then we define a new positive cone P 0 ⊂ G, with P 0 ∈ U , by setting −1 P 0 = PK ∪ (P ∩ (G \ K)). Again, the new ordering <0 of K with  positive cone P 0 is dense, and so the ordering we have defined on G is dense.  In the case of an abelian group A, the closure of the set of dense orderings was the entire space LO(A), which is known to be homeomorphic to the Cantor set when A is countable. In the non-abelian case, Theorem 3.0.2 gives us a similar result. Proof of Theorem 3.0.2. Let G be any countable group with all rank one abelian subgroups isomorphic to Z. Then since G is countable, LO(G) is 47  3.3. Dense orderings and discrete orderings metrizable, as is the space DLO(G) ⊂ LO(G). Proposition 3.3.7 shows that the set DLO(G) contains no isolated points, and since it is closed, it is compact. Therefore DLO(G) is a compact, metrizable, totally disconnected perfect space, and so is homeomorphic to the Cantor set [25]. By Lemma 3.3.1, the set DLO(G) is also a dense Gδ set within DLO(G). With the restriction that all rank one abelian subgroups of G be isomorphic to Z, it also follows readily that any isolated point in LO(G) must correspond to a discrete left ordering of G. This can be seen by appealing to either Theorem 3.0.2 (which is stronger than what we need), or by appealing to Theorem 3.0.1, and remarking that the smallest convex subgroup in the Conradian soul of an isolated left ordering must be a rank one abelian group. Next we turn our attention to discrete orderings, and observe conditions under which a discrete ordering of G is not an isolated point in LO(G). We no longer need the restriction that all rank one abelian subgroups be isomorphic to Z. Lemma 3.3.8. Let G be an arbitrary nontrivial group. Suppose that P ⊂ G is the positive cone of a discrete left ordering < with least positive element . Then if gg −1 > 1 for all g ∈ G and  P ∈  m \  Ugi  i=1  contains no conjugates of P , there exists gi which is not a power of  such  48  3.3. Dense orderings and discrete orderings that Ci = {g ∈ G : gi−k ≤ g ≤ gik for some k} is a convex, bi-ordered subgroup which properly contains the convex subgroup hi. Proof. Suppose that U =  Tm  i=1 Ugi  contains P , but no conjugates of P . If  no gi is equal to a power of , then we are done, as we may apply proposition 3.2.5. On the other hand, suppose that some gi is a power of , say g1 = l . Then the condition gg −1 > 1 for all g ∈ G guarantees that the open set U contains every conjugate of P . Therefore, if m \  ! Ugi  ∩ Ul  i=2  contains no conjugates of P , neither does the open set  Tm  i=2 Ugi .  Continuing  to eliminate powers of  in this way, we can eventually find an open set Tm i=r Ugi containing no conjugates of P , and with no gi equal to a power of . From here we may apply Proposition 3.2.5. Theorem 3.3.9. Let G be a group, and P the positive cone of a discrete left ordering < with least positive element . If gg −1 ∈ P for all g ∈ G, then P is not isolated in LO(G). Proof. We proceed very similarly to the proof of Theorem 3.0.2. Let U = Tm i=1 Ugi be an open set in LO(G) containing P . If U contains any conjugates of P distinct from P , then we are done, so by Lemma 3.3.8, we may suppose  49  3.4. The braid groups that there exists convex subgroup C properly containing hi, which is biordered by the restriction of P . Note that the convex subgroup C is not rank one abelian: Suppose that rank(C) = 1. As the containment hi ⊂ C is proper, we can choose c ∈ C with C 6= 1, that is not a power of . If we then assume that C is rank one abelian, we arrive at k = cl for some integers k, l, we may take k and l to be relatively prime. Choose integers s, t such that sk + tl = 1, and observe that  = sk+tl = (k )s · tl = (cl )s · tl = (cs t )l . Therefore  may be written as a power of some (necessarily positive) element, contradicting the fact that  is the least positive element. Thus, by Theorem 3.1.4, we know that the restriction of P to the subgroup C is not isolated in LO(C), and it follows from Lemma 3.2.1 that P is not isolated in LO(G).  3.4  The braid groups  As a sample application of these results, we turn our focus to the braid groups. It is known that the space of left orders LO(Bn ) is not homeomorphic to the Cantor set for n ≥ 2. We begin by defining the Dehornoy left ordering of the braid groups (also known as the ‘standard’ ordering), whose positive cone we shall denote PD [20, 21]. Recall that for each integer n ≥ 2, the Artin braid group Bn is the group generated by σ1 , σ2 , . . . , σn−1 , subject  50  3.4. The braid groups to the relations  σi σj = σj σi if |i − j| > 1,  σi σj σi = σj σi σj if |i − j| = 1.  Definition 3.4.1. Let w be a word in the generators σi , · · · , σn−1 . Then w is said to be: i-positive if the generator σi occurs in W with only positive exponents, i-negative if σi occurs with only negative exponents, and i-neutral if σi does not occur in w. It is shown in [20] that every nontrivial braid β ∈ Bn admits a representative word that is i-positive or i-negative for some i = 1, · · · , n − 1. If a braid admits a i-positive (resp. i-negative) representative word, then it is called an i-positive braid (resp. i-negative braid). We may then define the positive cone of the Dehornoy ordering as Definition 3.4.2. The positive cone PD ⊂ Bn of the Dehornoy ordering is the set PD = {β ∈ Bn : β is i-positive for some i ≤ n − 1}. There is also a second positive cone of interest, discovered by the authors of [23], which we shall denote by PDD . Denote by Pi ⊂ Bn the set of all i-positive braids. Note that the set of all i-negative braids is simply Pi−1 . Definition 3.4.3. The positive cone PDD ⊂ Bn is the set (−1)n  PDD = P1 ∪ P2−1 ∪ · · · ∪ Pn−1 .  That either of these notions defines a positive cone in Bn is difficult to 51  3.4. The braid groups show, as it is not clear that no braid is both i-positive and i-negative. This was the main idea introduced to braid theorists in Dehornoy’s seminal paper [20]. The positive cone PDD was originally defined in light of the following property: Proposition 3.4.4 (Dubrovina, Dubrovin [23]). The positive cone PDD is generated as a semigroup by the braids (−1)n  y1 = σ1 · · · σn−1 , y2 = (σ2 · · · σn−1 )−1 , y3 = σ3 · · · σn−1 , · · · , yn−1 = σn−1 . Note that for two positive cones P and Q, if P ⊂ Q then necessarily P = Q. Therefore Corollary 3.4.5. The positive cone PDD is an isolated point in LO(Bn ), in particular, {PDD } =  n−1 \  Uyi .  i=1  Knowing that LO(Bn ) has isolated points for n ≥ 2, it makes sense to ask the question: Is the standard ordering PD an isolated point in LO(Bn )? This question is answered in [21], using a very explicit calculation. That PD is not isolated, however, was originally proven in [40], though the techniques are different than those used here. First, we begin with a proposition which establishes a very important property of the ordering PD . Recall the Garside monoid Bn+ ⊂ Bn is the monoid generated by the elements σ1 , · · · , σn−1 . Proposition 3.4.6. Let β ∈ Bn and α ∈ Bn+ be given. If α 6= 1, then 52  3.4. The braid groups βαβ −1 ∈ PD . This property of the Dehornoy ordering is referred to as the subword property, or property S [21] . Next, we must know that the Dehornoy ordering is discrete [12]. Proposition 3.4.7. The Dehornoy ordering of Bn is discrete, with smallest positive element σn−1 . k Proof. Suppose that β > 1. There are two possibilities: First, if β = σn−1 k is a power of σn−1 , then k ≥ 1 and so σn−1 ≤ σn−1 = β. On the other hand,  if β is not a power of σn−1 , then β must be i-positive for some i < n − 1. −1 −1 β β must also be i-positive, and hence 1 < σn−1 In this case, the braid σn−1  and so σn−1 < β as claimed. These two propositions together show us that PD satisfies the hypotheses of Theorem 3.3.9. If we can additionally show that PD has no biorderable convex subgroups properly containing hσn−1 i, then we can conclude that PD is an accumulation point of its orbit under the Bn -action on LO(Bn ). Recall the natural inclusions Bm ⊂ Bn whenever m ≤ n which takes σi ∈ Bm to σi ∈ Bn . A useful operation is the shift homomorphism sh : Bm → Bn , m < n defined by sh(σi ) = σi+1 . This is clearly injective and orderpreserving. The shift may be iterated, and we note that shr (Bn−r ) is just the subgroup hσr+1 , . . . , σn−1 i of Bn , or in other words, the subgroup of all elements which are i-neutral for all i ≤ r. Lemma 3.4.8. The subgroups shr (Bn−r ), r > 0, are the only convex subgroups of Bn under the left ordering defined by PD . 53  3.4. The braid groups Proof. Set Hr = shr (Bn−r ), and let C be a convex subgroup in the Dehornoy ordering. Choose i to be the smallest integer such that C contains an ipositive braid. Then clearly C ⊂ Hi−1 , our aim is to show the opposite inclusion, which establishes the lemma. Let β ∈ C be an i-positive braid. The braid σj−1 β is i-positive for j > i, so that 1 < σj < β ⇒ σj ∈ C, and so Hi ⊂ C. Considering the generator σi , we write β = w1 σi w2 , where w1 is an empty or i-neutral word, and w2 is an empty, i-neutral, or i-positive word. We claim that this implies σi ∈ C. First, we note that the the braid represented by the word σi w2 lies in C, as w1 contains only σi+1 , · · · , σn−1 , all of which are in C. If w2 is empty, the claim is proven, if w2 is i-neutral, then we may right multiply by appropriate σj or σj−1 for j > i to arrive at σi ∈ C, and again the claim is proven. Lastly, if w2 is i-positive, then we get:  1 < w2 ⇒ 1 < σi < σi w2 ∈ C,  and the claim follows from convexity of C. Since all convex subgroups are isomorphic to a shifted copy of the braid groups, we conclude that Corollary 3.4.9. No subgroup that is convex under the ordering PD is biorderable, except for the subgroup hσn−1 i. Of course, this relies upon the fact that the braid groups are not biorderable for n > 2, as we have already seen. Theorem 3.4.10. For every n > 2, the positive cone PD in Bn is an 54  3.4. The braid groups accumulation point of its conjugates in LO(Bn ). Proof. Apply Corollary 3.4.9 and Lemma 3.3.8.  55  Chapter 4  Analysis of the natural action on the space of left orderings In this chapter we will use the category of lattice ordered groups to investigate the topological properties of the orbits in LO(G) under the G-action. In particular, we are able to characterize those points lying in the closure of a given orbit by using a free object in the category of lattice ordered groups. Appealing to known results in the field of lattice ordered groups, this characterization yields a proof that the space of left orderings of the non abelian free group contains a dense orbit.  4.1  Lattice-ordered groups  A group G is said to be lattice ordered, referred to as an l-group, if there exists a partial ordering < of the elements of G satisfying: 1. g < h implies f g < f h and gf < hf for all f, g, h in G, and 2. the ordering < admits a lattice structure, that is, every finite set has a least upper bound and a greatest lower bound.  56  4.1. Lattice-ordered groups As is standard, we denote the greatest lower bound and the least upper W V bound of g1 , · · · , gn by ni=1 gi and ni=1 gi respectively. Recall that a lattice L is distributive if the following identities hold for all x, y, z in L:  x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z),  and x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). We will show that all lattice ordered groups must necessarily be distributive lattices, by beginning with the following lemma. Lemma 4.1.1. [17] Let L be a lattice-ordered group, and let x, y ∈ L be given. Then: 1. (x ∨ y)−1 = x−1 ∧ y −1 , and 2. (x ∧ y)−1 = x−1 ∨ y −1 . Proof. We prove (1), the proof of (2) is dual. Observe that x ∨ y ≥ x and x ∨ y ≥ y, and so (x ∨ y)−1 ≤ x−1 and (x ∨ y)−1 ≤ y −1 . Hence (x ∨ y)−1 ≤ x−1 ∧ y −1 . On the other hand, we have x−1 ∧y −1 ≤ x−1 and x−1 ∧y −1 ≤ y −1 , so that (x−1 ∧ y −1 )−1 ≥ x and (x−1 ∧ y −1 )−1 ≥ y. Therefore (x−1 ∧ y −1 )−1 ≥ x ∨ y, and so x−1 ∧ y −1 ≤ (x ∨ y)−1 . It follows that x−1 ∧ y −1 = (x ∨ y)−1 . Proposition 4.1.2. [17] Let L be a lattice ordered group. Then L is a distributive lattice. 57  4.1. Lattice-ordered groups Proof. We use Lemma 4.1.1 without reference. Let x, y, z in L be given, and write g = x ∨ y. Then 1 ≤ gx−1 , and so z ≤ gx−1 z, and hence z ∧ g ≤ gx−1 z ∧ g = (gx−1 )(z ∧ x).  Similarly, we find that z ∧ g ≤ gy −1 z ∧ g = (gy −1 )(z ∧ y).  Right multiplying the inequality (z ∧ x) ≤ z ∧ g ≤ (gx−1 )(z ∧ x)  by (z ∧ x)−1 , we find 1 ≤ (z ∧ g)(z ∧ x)−1 ≤ gx−1 . We also find 1 ≤ (z ∧ g)(z ∧ y)−1 ≤ gy −1 . However, we may compute 1 = g(x ∨ y)−1 = g(x−1 ∧ y −1 ) = gx−1 ∧ gy −1 ,  and thus 1 = (z ∧ g)(z ∧ y)−1 ∧ (z ∧ g)(z ∧ x)−1 = (z ∧ g)((z ∧ y)−1 ∧ (z ∧ x)−1 ) = (z ∧ g)((z ∧ y) ∨ (z ∧ x))−1 .  The dual is proved similarly.  58  4.1. Lattice-ordered groups A homomorphism h of lattice ordered groups from L1 to L2 (often called an l-homomorphism) is a map h : L1 → L2 that is simultaneously a group homomorphism and a morphism of lattices, so that h respects the partial ordering <, as well as distributing over all finite meets and joins. The following two examples are standard constructions which are of great importance in what follows. First, given a totally ordered set Ω with ordering <, the group of all order preserving automorphisms of Ω forms a lattice ordered group, which we denote by Aut(Ω, <). The lattice ordering ≺ of Aut(Ω, <) is defined pointwise: for any f, g in Aut(Ω, <), we declare f ≺ g if f (x) < g(x) for all x ∈ Ω. As a second example, let {Li }i∈I , be an arbitrary collection of lattice ordered groups, with Li having the lattice ordering ≺i . We can form a new lattice ordered group L by setting  L=  Y  Li  i∈I  and for any x, y in L we declare x ≺ y if πi (x) ≺i πi (y) for all i ∈ I. Here, πi : L → Li is projection onto the i-th factor in the product. We are now ready to introduce the main object of concern in this chapter. A lattice ordered group F (G) is said to be the free lattice ordered group over G if F (G) satisfies: 1. There exists an injective homomorphism i : G → F (G) such that i(G) generates F (G) as an l-group, so that no proper l-subgroup of F (G) contains the set i(G). 59  4.1. Lattice-ordered groups 2. For any lattice ordered group L and any homomorphism of groups φ : G → L there exists a unique l-homomorphism φ̄ : F (G) → L such that φ̄ ◦ i = φ. Obviously any such group is unique up to l-isomorphism. We present a construction of the free lattice ordered group F (G) due to Conrad [14], in the case that G is a left orderable group. For each positive cone P ∈ LO(G), denote the natural lattice ordering of Aut(G, <P ) by ≺P , defined as in the previous discussion. We may embed the group G into Aut(G, <P ) by sending every g ∈ G to the orderpreserving automorphism of the totally ordered set (G, <P ) defined by left multiplication by g. Denote this order preserving automorphism by gP , so that gP ∈ Aut(G, <P ) has action gP (h) = gh. Define a map Y  i:G→  Aut(G, <P ),  P ∈LO(G)  according to the rule πP (i(g)) = gP , where  πP :  Y  Aut(G, <Q ) → Aut(G, <P )  Q∈LO(G)  is projection. Thus, on each factor in the product, g ∈ G acts by left multiplication. Denote by F (G) the smallest lattice ordered subgroup of Y  Aut(G, <P )  P ∈LO(G)  60  4.1. Lattice-ordered groups containing the set i(G). Then F (G), together with the map i defined above, is the free lattice ordered group over the left orderable group G. The lattice ordering of F (G) will be denoted by ≺, and is the restriction of the canonical product ordering to the lattice ordered subgroup F (G). Essential in showing that this construction produces a group satisfying the required universal property is the following proposition, due to Conrad (though if we assume the construction of F (G) above yields a group with the desired universal property, the proposition is obvious) [14]. Proposition 4.1.3. Let G be a left orderable group, and suppose that x is any non-identity element of F (G). Then there exists P ∈ LO(G) such that πP (x) 6= 1. From this point forward we will simplify our notation by writing g ∈ F (G) in place of i(g) ∈ F (G), for every element g of G. Thus, every element W V of F (G) can be (non-uniquely) written in the form i∈I j∈J gij for suitable gij ∈ G. The map πP : F (G) → Aut(G, <P ) sends such an element to the order-preserving map whose action on an element h ∈ G is defined according to the rule: πP (  _^  gij )(h) = maxi∈I minj∈J {gij h}.  i∈I j∈J  Here, the max and min are taken relative to the total ordering <P of G. Corollary 4.1.4. Let x ∈ F (G) be any element satisfying πP (x)(1) = 1 for all P in LO(G). Then x = 1. Proof. Suppose that x is a non-identity element of F (G), from Conrad’s proposition there exists P such that πP (x)(h) 6= h for some h in G. Set 61  4.2. Free lattice ordered groups and the topology on the space of left orderings Q = hP h−1 , we will show that πQ (x)(1) 6= 1. W V We may write x = i∈I j∈J gij for some elements gij ∈ G, so that the condition πP (x)(h) 6= h becomes  πP (  _^  gij )(h) = maxPi∈I minPj∈J {gij h} = 6 h,  i∈I j∈J  and thus maxPi∈I minPj∈J {h−1 gij h} 6= 1. Here we have added a superscript P to indicate that the maxima and minima are relative to the left ordering of G with positive cone P . Recall that in the left ordering <h of G defined by the positive cone Q = hP h−1 , we have g <h f if and only if h−1 gh < h−1 f h. Thus we may reQ write the inequality maxPi∈I minPj∈J {h−1 gij h} 6= 1 as maxQ i∈I minj∈J {gij } 6=  1, where the maxima and minima are now taken relative to the left ordering defined by the positive cone Q. It follows that πQ (x)(1) 6= 1.  4.2  Free lattice ordered groups and the topology on the space of left orderings  In this section, we establish several connections between the topology of LO(G) and the structure of the group F (G). We begin with a generalization of a known result, which was originally proven in the case of F (Fn ), the free lattice ordered group over the free group on n generators [1]. We observe, however, that the same result holds for any left orderable group G. Recall that an element x in a lattice ordered group L is said to be a basic element if the set {y ∈ L|1 ≤ y ≤ x} is totally ordered by the restriction of 62  4.2. Free lattice ordered groups and the topology on the space of left orderings the lattice ordering. Lemma 4.2.1. [1] Suppose that 1 ≺ x is an element of F (G). Then x is a basic element if and only if there exists a unique left ordering <P of G such that πP (x)(1) >P 1. Proof. Suppose that <P is the unique left ordering of G for which πP (x)(1) >P 1, and suppose that y1 , y2 are two distinct elements of F (G) that satisfy 1 ≺ yi ≺ x. Without loss of generality, we may assume that πP (y1 )(1) ≤P πP (y2 )(1), and hence 1 ≤P πP (y1−1 y2 )(1). Therefore, considering the element y1−1 y2 ∧ 1 ∈ F (G), we compute that πP (y1−1 y2 ∧ 1)(1) = min{πP (y1−1 y2 )(1), 1} = 1. Now in any left ordering <Q with Q 6= P , we have 1 ≤Q πQ (yi )(1) ≤Q πQ (x)(1) ≤Q 1, where the final inequality follows from our assumption that <P is the unique left ordering with πP (x)(1) >P 1. Therefore πQ (y1−1 y2 ∧ 1)(1) = 1 is true for all those positive cones Q different from P and in the case Q = P . It follows that y1−1 y2 ∧ 1 = 1 by Corollary 4.1.4, and hence 1 ≺ y1−1 y2 , and y1 ≺ y2 as desired. On the other hand, suppose that x is a (positive) basic element, and suppose that P and Q are distinct positive cones such that πP (x)(1) >P 1 and πQ (x)(1) >Q 1. Choose an element h of G such that h >P 1 and h−1 >Q 1. Then the elements y1 = (x ∧ h) ∨ 1 and y2 = (x ∧ h−1 ) ∨ 1 satisfy 1 ≺ yi ≺ x, yet are not comparable in the partial ordering ≺ of F (G). This  63  4.2. Free lattice ordered groups and the topology on the space of left orderings follows from computing  πP (y1 )(1) = max{min{πP (x)(1), h}, 1} >P 1,  and πP (y2 )(1) = max{min{πP (x)(1), h−1 }, 1} = 1, while πQ (y1 )(1) = 1 and πQ (y2 )(1) >Q 1.  Theorem 4.2.2. [1] Let G be a left orderable group. Then F (G) contains a basic element if and only if LO(G) contains an isolated point. Proof. Suppose that {P } = Ug1 ∩ · · · ∩ Ugn is an isolated point in LO(G). Then P is the unique left ordering for which 1 <P gj for all j, and is therefore the unique ordering for which πP (gj )(1) >P 1 for all j. Therefore, <P is the unique left ordering for which V πP ( nj=1 gj )(1) >P 1. By definition, a basic element must be positive, and V so we consider the element ( nj=1 gj ) ∨ 1. The left ordering <P is the unique V left ordering for which πP (( nj=1 gj ) ∨ 1)(1) >P 1, and it now follows that V ( nj=1 gj ) ∨ 1 is a basic element of F (G), by Lemma 4.2.1. W Vn Conversely, if m i=1 j=1 gij is a basic element in F (G), then there is a W Vn unique left ordering <P of G such that πP ( m i=1 j=1 gij )(1) >P 1. ThereV fore, for some index i, πP ( nj=1 gij )(1) = minnj=1 {gij } >P 1, and <P is the unique left ordering of G for which this inequality holds, for our chosen index  64  4.2. Free lattice ordered groups and the topology on the space of left orderings i. In other words, <P is the unique left ordering in which all the elements gij are positive for our chosen index i, so that  {P } =  \  Ugij  j∈J  is an isolated point in LO(G). Recall that for any left-orderable group G, the space LO(G) comes equipped with a G-action by conjugation, which is an action by homeomorphisms. Set OrbG (P ) = {gP g −1 |g ∈ G}, and we denote the closure of each such orbit by OrbG (P ). The group action on the space of left orderings has been used to great success in investigating the structure of the space of left orderings, see [11, 40, 42, 43]. However, in most applications, one often asks if a positive cone P is an accumulation point of its own conjugates in order to show that P is not an isolated point. More useful would be an answer to the question: When is Q an accumulation point of the conjugates of some different positive cone P , that is, when is Q ∈ OrbG (P )? We find that this question is equivalent to an algebraic question about the free lattice-ordered group F (G). Theorem 4.2.3. Let G be a left orderable group, and let P, Q ∈ LO(G) be given. Then Q ∈ OrbG (P ) if and only if ker(πP ) ⊂ ker(πQ ). Proof. Suppose that ker(πP ) ⊂ ker(πQ ), and that Q lies in the basic open T set nj=1 Ugj . We must show that some conjugate of P lies in this open set as well. 65  4.2. Free lattice ordered groups and the topology on the space of left orderings T V Consider the element ( nj=1 gj ) ∨ 1 in F (G). As Q ∈ nj=1 Ugj , we know that gj >Q 1 for all j, and hence we find that  πQ ((  n ^  gj ) ∨ 1)(1) = max{min{gj }, 1} = min{gj } >Q 1.  j=1  V Therefore, ( nj=1 gj ) ∨ 1 is not in the kernel of the map πQ , and so from our assumption it is not in the kernel of the map πP . Thus, there exists h ∈ G V such that πP (( nj=1 gj ) ∨ 1)(h) 6= h, and we compute  πP ((  n ^  gj ) ∨ 1)(h) = max{min{gj h}, h} >P h,  j=1  so that gj h >P h for j = 1, · · · , n. Therefore, h−1 gj h >P 1, and hence h−1 gj h ∈ P for all j. This is equivalent to gj ∈ hP h−1 for all j, or hP h−1 ∈ Tn j=1 Ugj , so that Q is an accumulation point of the orbit of P . W V On the other hand, suppose that Q ∈ OrbG (P ), and let i∈I j∈J gij be W V any element of F (G) such that πQ ( i∈I j∈J gij ) 6= 1. There are two cases W V / ker(πP ). to consider, in order to show that i∈I j∈J gij ∈ V Case 1. There exists i such that πQ ( j∈J gij )(h) >Q h for some h ∈ G. Then minj gij h >Q h, and therefore h−1 gij h ∈ Q for all j, hence T Q ∈ j∈J Uh−1 gij h . By assumption, we can choose f ∈ G such that f P f −1 ∈ T −1 −1 for all j. In other words, f −1 h−1 g hf > ij P j∈J Uh−1 gij h , so that h gij h ∈ f P f 1 for all j, so that gij hf >P hf for all j. Now, we may compute  πP (  _^  i∈I j∈J  gij )(hf ) >P πP (  ^  j∈J  gij )(hf ) = min{gij hf } >P hf. j  66  4.2. Free lattice ordered groups and the topology on the space of left orderings W  We conclude that  i∈I  V  j∈J  gij ∈ / ker(πP ).  V Case 2. For all i, πQ ( j∈J gij )(h) ≤Q h for all h ∈ G. Then in particV ular, we may choose h ∈ G such that πQ ( j∈J gij )(h) <Q h for every i ∈ I W V (strict inequality), since the image of i∈I j∈J gij must act nontrivially on (G, <Q ). Now with h as above, we observe that for every index i there exists an index j such that gij h <Q h. Thus, for every i, we may choose hi from the set of elements {gij }j∈J , so that hi satisfies hi h <Q h. Then each hi satisfies T h−1 hi h <Q 1, so that Q ∈ i∈I Uh−1 h−1 h . We may therefore choose f ∈ G i T so that f P f −1 ∈ i∈I Uh−1 h−1 h , in other words, f −1 h−1 h−1 i hf >P 1 for all i  i ∈ I. It follows that  f −1 h−1 h  i hf  <P 1 and so hi hf <P hf for all i ∈ I.  Thus, we find that for every i ∈ I,  πP (  ^  gij )(hf ) = min{gij hf } ≤P hi hf <P hf,  j∈J  j  where the inequality minj {gij hf } ≤P hi hf follows from the fact that hi lies in the set {gij }j∈J . Thus, when we take a (finite) maximum over all i, we compute that W V πP ( i∈I j∈J gij )(hf ) = minj {gij hf } for some i, and hence  πP (  _^  i∈I j∈J  gij )(hf ) = min{gij hf } ≤P hi hf <P hf. j  W V It follows that πP ( i∈I j∈J gij ) is nontrivial, and the claim is proven. Corollary 4.2.4. For a given positive cone P in a left orderable group G, OrbG (P ) = LO(G) if and only if πP : F (G) → Aut(G, <P ) is injective. 67  4.2. Free lattice ordered groups and the topology on the space of left orderings Proof. It is clear that if πP : F (G) → Aut(G, <P ) has trivial kernel, then OrbG (P ) = LO(G) by Theorem 4.2.3. For the other direction, suppose that OrbG (P ) = LO(G). From TheoT rem 4.2.3 we deduce the containment ker(πP ) ⊂ Q∈LO(G) ker(πQ ). HowT ever, from Proposition 4.1.3 we find that Q∈LO(G) ker(πQ ) = {1}, so that ker(πP ) = {1}. Thus, for a given left orderable group G, injectivity of the map πP : F (G) → Aut(G, <P ) for some P ∈ LO(G) tells us a great deal about the structure of LO(G). Proposition 4.2.5. Let G be a left orderable group, and suppose that there exists P ∈ LO(G) such that πP is injective. Then LO(G) contains no isolated points. Proof. If the map πP is injective, then we know that we may write LO(G) = OrbG (P ), and so only those points in OrbG (P ) itself are possibly isolated in LO(G), which can only happen if P itself is isolated. Supposing that P is an isolated point, it follows that P −1 is also an isolated point in LO(G), and hence P −1 ∈ OrbG (P ); so we may write P −1 = gP g −1 for some g ∈ G, with g different from the identity. This is impossible, for supposing g ∈ P yields (upon conjugation by g) g ∈ gP g −1 = P −1 . Similarly, g ∈ P −1 is impossible, so that P is not isolated. From Theorem 4.2.3, we now have a bijection between certain normal subgroups of F (G) and certain closed subsets of LO(G). Specifically, if K is the kernel of the map πP , we can associate to K the closed set OrbG (P ). 68  4.2. Free lattice ordered groups and the topology on the space of left orderings Note that if πQ is some other map with kernel K, then OrbG (Q) = OrbG (P ), so that the closed set associated to K is well-defined. Inclusion of kernels ker(πP ) ⊂ ker(πQ ) yields a reverse inclusion of associated subsets, OrbG (Q) ⊂ OrbG (P ). It is well known that the space LO(G) is compact (Theorem 2.1.2), and compactness has thus far proven to be one of the most useful properties of LO(G). In [32], compactness is the key ingredient in showing that no group has countably infinitely many left orderings (which has been proven again recently in [43]), and in [37], compactness is used to show that an amenable group is left orderable if and only if it is locally indicable. In our present setting, compactness of the space of left orderings yields the following: If G is a left orderable group, let SG denote the set of all normal subgroups of F (G) that occur as the kernel of some map πP : F (G) → Aut(G, <P ), where P ranges over all positive cones in LO(G). The set SG is partially ordered by inclusion. Proposition 4.2.6. Every chain in SG has an upper bound. In particular, SG has a maximal element. Proof. Let T be a subset of LO(G) such that {ker(πP )}P ∈T is a totally ordered subset of SG . Observe that OrbG (P ) ⊂ OrbG (Q) if and only if ker(πQ ) ⊂ ker(πP ), and thus {OrbG (P )}P ∈T is a nested collection of closed subsets of LO(G). In particular, this nested collection of sets has the finite intersection property. For if P1 , · · · , Pn is some finite subset of T , upon renumbering if necessary, we may assume that OrbG (P1 ) ⊂ · · · ⊂ OrbG (Pn ), from which it is obvious that P1 is contained in their intersection. 69  4.2. Free lattice ordered groups and the topology on the space of left orderings Thus, the intersection \  OrbG (P )  P ∈T  is nonempty, as LO(G) is compact. Choosing any positive cone R from this intersection yields a closed set OrbG (R) that lies in OrbG (P ) for every P in T , and hence ker(πP ) ⊂ ker(πR ) for every P in T . It now follows from Zorn’s lemma that SG contains a maximal element. The following is a standard definition in the theory of dynamical systems. Definition 4.2.7. A nonempty set U in LO(G) is said to be a minimal invariant set if U is closed and G-invariant, and for every closed G-invariant set V in LO(G), U ∩ V 6= ∅ implies U ⊂ V . The equivalence of (1) and (2) in the following proposition is a standard result from the theory of dynamical systems ([19], pp. 69-70). Proposition 4.2.8. For any nonempty closed subset U of LO(G), the following are equivalent: 1. U is a minimal invariant set 2. for every P ∈ U , U = OrbG (P ) 3. U = OrbG (P ) for some P ∈ LO(G) whose kernel is maximal in SG . Proof. (1) if and only if (2). Suppose that U is a minimal invariant set, and let P ∈ U be given. Then OrbG (P ) ⊂ U , since U is closed and Ginvariant. Since U is small, this implies U ⊂ OrbG (P ), and so U = OrbG (P ). Conversely, suppose that (2) is satisfied and let V be some other closed, Ginvariant set such that U ∩ V is nonempty. Choose Q ∈ U ∩ V , and observe 70  4.2. Free lattice ordered groups and the topology on the space of left orderings that U = OrbG (Q) ⊂ V , since V is closed and G-invariant. Therefore U is a minimal invariant set. (2) if and only if (3). Suppose property (2) holds, and let P ∈ U be given, and suppose that ker(πP ) ⊂ ker(πQ ) for some Q ∈ LO(G). Then by Theorem 4.2.3, Q ∈ OrbG (P ) = U , and hence, by condition (2), OrbG (Q) = U = OrbG (P ). It follows that ker(πP ) = ker(πQ ) is maximal. Conversely, suppose (3) and let P ∈ U be given. Then for any other Q in U = OrbG (P ), we have ker(πP ) ⊂ ker(πQ ) by Theorem 4.2.3. Since ker(πP ) is maximal, this gives ker(πQ ) = ker(πP ) and (2) follows. We can now see that Proposition 4.2.6 mirrors a standard proof of the existence of minimal invariant sets, see for example [19] Theorem 3.12. It is also clear from the above characterization that ker(πP ) is maximal if P is the positive cone of a bi-ordering (so OrbG (P ) = {P }) , or if <P is a bi-ordering when restricted to some finite index subgroup H ⊂ G (for then OrbG (P ) is finite). We may apply the notion of minimal invariant sets to provide yet another proof that no left orderable group has countably infinitely many left orderings. Proposition 4.2.9. Let G be a left orderable group, and let U be a minimal invariant subset of LO(G). Then U is finite, or uncountable. Proof. Suppose that U is not finite. If U is infinite and contains no isolated points, then U is uncountable, as U is a compact Hausdorff space [Theorem 2-80, [25]]. Thus, suppose that U contains an isolated point, and we will  71  4.2. Free lattice ordered groups and the topology on the space of left orderings arrive at a contradiction. Note that in this context, “isolated point” means isolated in U , not isolated in LO(G). Choose P in U . By (2) of Proposition 4.2.8, U = OrbG (P ), and it follows that P itself must be an isolated point (and hence every point in OrbG (P ) is isolated). Since U is compact, the set of conjugates OrbG (P ) must accumulate on some Q ∈ U , moreover, Q does not lie in OrbG (P ), since Q is not isolated. We again apply (2) of Proposition 4.2.8 to find that U = OrbG (Q), and it follows that P cannot be an isolated point, a contradiction. Corollary 4.2.10. For any group G, LO(G) is either finite or uncountable. Proof. Given any left orderable group G, by Proposition 4.2.6, there exists P ∈ LO(G) for which ker(πP ) is maximal in SG . Correspondingly, the set OrbG (P ) is a minimal invariant set, and hence it is either finite, or uncountable. Assuming OrbG (P ) is finite, we must have that OrbG (P ) is finite, and hence the stabilizer StabG (P ) is a finite index subgroup of G that is biordered by the restriction of the left ordering <P . It follows that G is locally indicable [53], and hence has uncountably many left orderings, by [63]. It should be noted that this proof is very similar to the proof of Peter Linnell, given in [32]. The crucial difference in our proof is that some difficult topological arguments have been replaced by an application of Zorn’s lemma. It does not appear that a clean topological statement will characterize precisely those closed sets that occur as OrbG (P ) for some ordering whose 72  4.3. Examples associated kernel is minimal in SG . We do, however, have the following observation. Proposition 4.2.11. Let G be a left orderable group, and let P in LO(G) be an isolated point. Then ker(πP ) is minimal in SG . Proof. Suppose that ker(πQ ) ⊂ ker(πP ) for some Q ∈ LO(G). Then P ∈ OrbG (Q) by Theorem 4.2.3, but P is isolated, so P ∈ OrbG (Q). Therefore Q is conjugate to P , and so Q ∈ OrbG (P ), and ker(πP ) ⊂ ker(πQ ) by Theorem 4.2.3. Thus ker(πQ ) is equal to ker(πP ), so that ker(πP ) is minimal. Not every minimal kernel in SG corresponds to an isolated point in LO(G). In the next section, we will see that with G = Fn , the free group on n generators, SFn contains a minimal kernel ([28, 29]) that does not correspond to an isolated point.  4.3 4.3.1  Examples The free group  It now follows easily from work of Kopytov that LO(Fn ) contains a dense orbit under the conjugation action by Fn . This case appears to be the first known example of a left ordering (of any group) whose orbit is dense in the space of left orderings. Corollary 4.3.1. Let Fn denote the free group on n > 1 generators. There exists P such that OrbFn (P ) = LO(Fn ).  73  4.3. Examples Proof. There exists a left ordering of Fn with positive cone P such that πP : F (G) → Aut(G, <P ) is injective ([28, 29]), and we apply Corollary 4.2.4. We may apply Proposition 4.2.5 and Theorem 4.2.2 to arrive at a new proof of the following corollary (see [36, 40] for alternate proofs). Corollary 4.3.2. The space LO(Fn ) has no isolated points for n > 1. Equivalently, F (Fn ) has no basic elements for n > 1. Recall that the Conradian soul of a left ordering is the largest convex subgroup on which the restriction ordering is Conradian [40]. It was shown in [36] and [22] that the construction of Kopytov can be improved so that the map πP is injective, and the left ordering <P of Fn has no convex subgroups. Consequently, the Conradian soul of the ordering <P must be trivial, and so P is an accumulation point of its own conjugates in LO(Fn ) [11, 40]. It is also worth noting that McCleary’s construction in [36] of a faithful o−2 transitive action of F (Fn ) on some linearly ordered set is much stronger than is needed to conclude that F (Fn ) has no basic elements; yet [36] appears to contain the first proof of this fact appearing in the literature.  4.3.2  Left orderable groups with all left orderings Conradian  In the case that all left orderings of a given left orderable group are Conradian, we may highlight two cases of interest. First we will observe that no finitely generated group G, all of whose left orderings are Conradian, can have a dense orbit in LO(G). Second, we will show that if we allow the 74  4.3. Examples group to be infinitely generated, then it may be the case that every orbit in LO(G) is dense. Recall that, for example, every left ordering of a torsion free locally nilpotent group is Conradian [3, 5]. An element g in a left ordered group G with ordering < is cofinal if for every h in G, there exists n ∈ Z such that g −n < h < g n . Proposition 4.3.3. Suppose that G is a left orderable group, and suppose that g is cofinal in <P , for some P ∈ LO(G). Then P −1 is not in OrbG (P ). Proof. Suppose that g >P 1 is cofinal, and let 1 <P h ∈ G be given. Choose an integer n so that h <P g n , and observe that 1 <P h−1 g n , and hence 1 <P h−1 g n h, since h is positive. Therefore 1 <P h−1 gh, and so h <P gh. On the other hand, if h is negative, we similarly conclude that h <P gh, and thus πP (g ∧ 1)(h) = h for all h ∈ G, so that g ∧ 1 ∈ ker(πP ). On the other hand, in the reverse ordering with positive cone P −1 , we find h >P −1 gh for all h ∈ G, so that πP −1 (g ∧ 1)(h) <P −1 h for all h ∈ G, so that g ∧ 1 6= 1 in F (G). Thus the map πP is not injective, and in particular, ker(πP ) is not contained in ker(πP −1 ). By Theorem 4.2.3, P −1 is not in OrbG (P ). Corollary 4.3.4. Let G be a finitely generated group, all of whose left orderings are Conradian. Then LO(G) does not contain a dense orbit. Proof. Let P be any positive cone in a finitely generated group G. Since the associated left ordering <P is Conradian, there exists a convex subgroup C ⊂ G such that G/C is abelian, and the induced ordering on G/C is Archimedean. Every element in G \ C is then cofinal in the left ordering <P , and the claim follows. 75  4.3. Examples Next, we consider the group T∞ = hxi , i ∈ N : xi+1 xi xi+1 = x−1 i , xi xj = xj xi for |i − j| > 1i. Note that each subgroup Tn = hx1 , · · · , xn i has only 2n left orderings, as it is one of the Tararin groups, as described in [30], Theorem 5.2.1. Moreover, the convex subgroups of any left ordering of Tn are precisely Ti ⊂ Tn for i ≤ n, and so the convex subgroups of T∞ are exactly Ti for i ∈ N, for any left ordering of T∞ . The orderings of T∞ are all Conradian, with convex jumps Ti+1 /Ti ∼ = Z. Given any positive cone P ∈ LO(T∞ ), it is therefore determined by the signs of the generators xi , which we record in a sequence  ε = (±1, ±1, · · · )  writing +1 in the i-th position if xi is in P , and −1 otherwise. We then write P = Pε . With this notation, we observe that xi+1 Pε x−1 i+1 = Pε0 , where ε0 differs from ε only in the sign of the i-th entry (this follows from the defining relations of the group T∞ , and corresponds to the idea of “flipping” the ordering on the i-th convex jump). Proposition 4.3.5. Let P ∈ LO(T∞ ) be any positive cone. Then LO(T∞ ) = OrbT∞ (P ). Proof. Let Pε1 and Pε2 be two positive cones in LO(T∞ ). It is enough to show that for every n ∈ N, there exists g ∈ T∞ such that gPε1 g −1 ∩ Tn = Pε2 ∩ Tn , so that the associated left orderings agree upon restriction to the subgroup Tn . 76  4.3. Examples The proof is a simple induction. First, note that there exists a conjugate of Pε1 that agrees with Pε2 upon restriction to the subgroup T1 = hx1 i: if ε1 and ε2 agree in the first entry, then Pε1 and Pε2 agree on T1 , whereas if ε1 and ε2 disagree in the first entry, then x2 Pε1 x−1 2 and Pε2 agree on T1 . For induction, suppose that Q is a conjugate of Pε1 with associated sequence ε01 first differing from ε2 in the n-th entry, so that Q ∩ Tn−1 = 00 Pε2 ∩ Tn−1 . Then xn+1 Qx−1 n+1 will have associated sequence ε1 that agrees  with ε2 in the sign of the n-th entry. Thus the conjugate xn+1 Qx−1 n+1 of Pε1 agrees with Pε2 upon restriction to the subgroup Tn , and the result follows by induction.  4.3.3  The braid groups  The braid groups Bn , n > 2 also provide an interesting class of examples, as their spaces of left orderings are known to contain isolated points [11, 23, 40]. Proposition 4.3.6. For n > 2, the space LO(Bn ) contains no dense orbit. Proof. This follows immediately from Proposition 4.2.5 and Corollary 4.2.4, in light of the fact that LO(Bn ) contains isolated points. As an alternative way of proving Proposition 4.3.6, recall that the center of the braid group Bn , for n > 2, is infinite cyclic with generator ∆2n . Here, ∆k := (σk−1 σk−2 · · · σ1 )(σk−1 σk−2 · · · σ2 ) · · · (σk−1 σk−2 )(σk−1 ) is the Garside half-twist. It is well known that ∆2n is cofinal in any left ordering of Bn [21], from which is follows by Proposition 4.3.3 that LO(Bn ) 77  4.3. Examples contains no dense orbit.  78  Chapter 5  3-manifolds and left orderability As left orderability of a countable group is equivalent to an action on the real line by order-preserving homeomorphisms, it is not surprising that groups arising in a geometric context behave exceptionally with respect to left orderability. The class of groups that occur as the fundamental groups of 3-manifolds exhibit such exceptional behaviour with respect to left orderability. In turn, left (or bi-) orderability of the fundamental group of a manifold M can be connected to the underlying geometry of M .  5.1  Examples of 3-manifolds with left orderable fundamental group  An example of such behaviour is the following variation of Theorem 1.4.1. Recall that a 3-manifold M is P 2 -irreducible if M is irreducible and does not contain any two-sided, properly embedded projective planes. Theorem 5.1.1 (See [7], attributed to Howie and Short [27]). Suppose that M is a compact, connected, P 2 -irreducible 3-manifold and that π1 (M ) 79  5.1. Examples of left orderable fundamental groups is nontrivial. Then π1 (M ) is left orderable if and only if there exists a surjective homomorphism h : π1 (M ) → L onto a nontrivial left orderable group L. Sketch of proof. We will show that π1 (M ) is left orderable by showing that every finitely generated subgroup surjects onto a nontrivial left orderable group. This will allow us to apply the Burns-Hale Theorem (Theorem 1.4.1). Thus, suppose that h : π1 (M ) → L is a homomorphism onto a nontrivial left orderable group, and let H be a finitely generated subgroup of π1 (M ). If H is finite index, then the image h(H) is finite index in L, and so h(H) is a nontrivial left orderable group. Hence, the restriction of h to the subgroup H provides us with a homomorphism onto a nontrivial left orderable group in this case. f of In the case that H is infinite index, we consider the covering space M M corresponding to the subgroup H. As H is infinite index in π1 (M ), the f is noncompact. Here, we apply the compact core theorem of Scott cover M f (necessarily with boundary) [56] to produce a compact submanifold C ⊂ M f is such that π1 (C) = H. Since M is P 2 -irreducible, the covering space M P 2 -irreducible. We can use P 2 -irreducibility to show that we may assume f contains no spheres or projective planes. The the boundary of C in M analysis concludes by using an Euler characteristic argument to find that H1 (C) is infinite, and thus surjects onto the integers. We now have a map ab H∼ = π1 (C) → H1 (C) → Z onto the left orderable group Z. This allows us to  apply the Burns-Hale theorem to conclude that π1 (M ) is left orderable. An important corollary follows from this proof. Observe that the finitely 80  5.1. Examples of left orderable fundamental groups generated subgroups of infinite index in π1 (M ) must all surject onto the integers. Thus if we impose a condition that guarantees that the finitely generated subgroups of finite index map onto the integers, we will be able to conclude that π1 (M ) is locally indicable, and hence Conrad orderable. Thus, we require that H1 (M ) is infinite. Corollary 5.1.2. Suppose that M is a compact, connected, P 2 -irreducible 3-manifold and that H1 (M ) is infinite. Then π1 (M ) is Conrad orderable. Calegari and Dunfield have also succeeded in left ordering the fundamental groups of a large class of 3-manifolds whose homology is finite. See [10] for background on taut foliations. Theorem 5.1.3 ([10]). Let M be an irreducible, atoroidal Q-homology 3sphere which admits a transversely orientable taut foliation. Then the commutator subgroup [π1 (M ), π1 (M )] is left orderable, and hence the covering f of M corresponding to [π1 (M ), π1 (M )] has left orderable fundaspace M f is also compact since the mental group. Moreover, if M is compact, then M cover is finite-sheeted. Finally, it has been observed very recently that left orderability of fundamental groups behaves in an interesting way with respect to Dehn surgery. We recall the basics of Dehn surgery, see [55] for full details. Let M be a 3-manifold with torus boundary, and let m, l be a choice of meridian and longitude basis for π1 (∂M ). Let T ∼ = S 1 × D2 be a torus, and h : S 1 × D2 → ∂M be the homeomorphism that sends ∂D2 to a curve representing the class pm + ql ∈ π1 (∂M ). Denote by Mp/q the manifold obtained by gluing M and T along their boundaries, using the homeomorphism h. 81  5.1. Examples of left orderable fundamental groups  Figure 5.1: The figure eight knot.  The manifold Mp/q is said to be the manifold obtained from Dehn surgery on M with surgery coefficient p/q. In the case that M is a knot or link complement in S 3 , the boundary ∂M admits a preferred meridian and longitude basis for π1 (∂M ), and so the manifold Mp/q is determined by the pair of (relatively prime) numbers p, q. Recall that the figure eight knot complement is the complement in S 3 of the embedding of the torus depicted in Figure 5.1. Theorem 5.1.4 ([7]). Let M be the figure eight knot complement, and let p/q ∈ (−4, 4) be given. Then there exists a finite index subgroup of π1 (Mp/q ) which is left orderable. It has also recently been announced that this theorem may be strengthened, to show that π1 (Mp/q ) is left orderable whenever p/q ∈ (−4, 4) [6]. We will further investigate the connection between Dehn surgery and left orderability of the fundamental group π1 (Mp/q ) in Chapter 6.  82  5.2. Geometric applications of left orderability  5.2  Geometric applications of left orderability  At present, there are several known geometric consequences of left orderability of the fundamental group of a given 3-manifold. We recall the definition of an R-covered foliation. Let F be a codimension one foliation of the 3f → M be the universal covering space. Denote manifold M , and let p : M by Fe the pullback foliation to the universal cover. If we identify every leaf f/F, e and is a of Fe to a point, the resulting topological space is denoted by M f/Fe (not necessarily Hausdorff) 1-dimensional manifold. In the case that M is homeomorphic to R, the foliation F of M is said to be R-covered. There f by deck transformations, which descends is a natural action of π1 (M ) on M f/F. e It is from this action, and Theorem to an action on the leaf space M 1.4.11 that we arrive at the following. Theorem 5.2.1 ([7]). Let M be a compact, connected P 2 -irreducible 3manifold. If π1 (M ) is not left orderable, then M does not admit a transversely oriented R-covered foliation. We may also use orderability as an obstruction to the existence of a map of nonzero degree between closed, oriented manifolds. Theorem 5.2.2 ([7]). Let M and N be closed, oriented 3-manifolds and suppose that M is prime. Suppose that π1 (M ) is not left orderable, and that π1 (N ) is left orderable and is nontrivial. Then every continuous map h : M → N must be of degree zero. There has also been a very recent discovery that connects non-left orderability of π1 (M ) for a given 3-manifold M to its Heegaard-Floer homology 83  5.2. Geometric applications of left orderability groups (see [46, 47] for definitions). Roughly speaking, a manifold M is called an L-space if it is a rational homology 3-sphere whose Heegaard-Floer homology groups are as simple as possible [49]. Theorem 5.2.3 ([52], [6], [62]). Suppose that M is a closed, connected, orientable Seifert fibered 3-manifold. Then M is an L-space if and only if π1 (M ) is not left orderable. Remarkable is that the evidence accumulated thus far seems to point towards a stronger version of this theorem which may cover the case of hyperbolic 3-manifolds. For example, from [16] we have the following infinite families of 3-manifolds, all of whose fundamental groups are shown to be not left orderable. Theorem 5.2.4 ([16]). Denote by M n (L) the n-fold branched cyclic cover of S 3 , branched along the link L ⊂ S 3 , with n > 1. The fundamental group π1 (M n (L)) is not left orderable for the following (oriented) links L: 1. L = T(20 ,2k) is the torus link of type (2, 2k) where the strings are given anti-parallel orientation, and where n is arbitrary. 2. L = P (n1 , n2 , · · · , nk ) is the pretzel link of type (n1 , n2 , · · · , nk ), k > 2, where either n1 , n2 , · · · , nk > 0 or n1 = n2 = · · · = nk−1 = 2, nk = −1 and k > 3, with n = 2. 3. L = L[2k,2m] is the 2-bridge knot of type p/q = 2m +  1 2k  = [2k, 2m]  where k, m > 0 and n is arbitrary. 4. L = L[n1 ,1,n3 ] is the 2-bridge knot of type p/q = n3 +  1 1+ n1  , where  1  84  5.2. Geometric applications of left orderability n1 and n3 are odd, positive integers. The degree of the covering is n = 2, 3. Many of these manifolds are hyperbolic, see [16] for details. The main theorem of [52] is the following: Theorem 5.2.5. All of those spaces listed above are L-spaces. Boyer, Watson and Cameron have also recently announced that the 2fold branched cover of S 3 , branched along an alternating link, also has nonleft orderable fundamental group [6]. As above, these manifolds are known to be L-spaces [48].  85  Chapter 6  An application of the space of left orderings to 3-manifolds Our goal in this chapter will be to use the topology of LO(G) in order to establish non-left orderability of the fundamental groups of manifolds obtained from Dehn surgery.  6.1  The map ψG,H  Let H be a subgroup of G. We may define a map r : LO(G) → LO(H) by restricting a given left ordering of G to the subgroup H. In terms of positive cones, the map r acts according to the formula r(P ) = P ∩ H. Note that if H is a subgroup of G, then for every h ∈ H the notation Uh can ambiguously refer to an open set in LO(G) and an open set in LO(H). We therefore introduce the notation UhH and UhG to distinguish between these two sets. Proposition 6.1.1. The map r is continuous.  86  6.1. The map ψG,H Proof. We will show that the inverse image of any basic open set is a basic open set. Let UhH ⊂ LO(H) be given, where h is a nonidentity element in H. Then r−1 (UhH ) = {P ∈ LO(G)|h ∈ P ∩ H} = {P ∈ LO(G)|h ∈ P } = UhG .  It follows that r is continuous. We now wish to consider the case when H = Z × Z, for which we need a structure theorem, due to Sikora [58]. Denote by R(] the real line with the topology whose basis consists of sets of the form (a, b], and define R[) analagously. These topologies on R descend to two different topologies on S 1 ⊂ R2 , via the map z 7→ e2πiz , yielding quotient topologies via R(] /Z → S(]1 and R[) /Z → S[)1 . A point (p, q) ∈ S 1 is said to be rational if p/q ∈ Q. Let X denote the union of S[)1 and S(]1 with corresponding irrational points identified. Define a map φ : X → LO(Z2 ) as follows. Given any point x ∈ S[) , φ will map x to the positive cone in Z2 (which we have identified with the integer points in R2 ) consisting of all those pairs (p, q) such that the counter-clockwise oriented angle between the vector x and (p, q) lies in the interval (0, π]. Similarly given x ∈ S(] , φ will map x to the positive cone of elements (p, q) whose oriented angle with x lies in the interval [0, π). Note that this assignment respects the equivalence relation used in defining X: If x is irrational, then our definition of φ(x) is well-defined regardless of whether we consider x to be a point in S[)1 or S(]1 , as no points of Z × Z lie on the line determined by x. 87  6.1. The map ψG,H Theorem 6.1.2. [58] The map φ is a homeomorphism. We use the notation  [ (p, q) =  p q , ||(p, q)|| ||(p, q)||    to denote rational points in any of S 1 , S[)1 or S(]1 , for relatively prime integers p, q ∈ Z. In the case that x ∈ S 1 is a rational point we will also consider x ∈ S[)1 (resp. S(]1 ) as lying in the space X, as the rational points in x ∈ S[)1 (resp. S(]1 ) unambiguously correspond to unique points in X via the quotient map S[)1 t S(]1 → X. In the following theorem, we write the group operation in Z2 multiplicatively. Lemma 6.1.3. Let P ∈ LO(Z2 ) be a positive cone, and let φ : X → LO(Z2 ) be defined as above. Suppose that Z2 has generators m, l, and that p and q are relatively prime integers. The positive cone P defines an ordering of Z2 relative to which 1. the subgroup hmp lq i is convex, and 2. mp lq is positive if and only if [ φ−1 (P ) = (p, q) considered as a point in S(]1 , or [ φ−1 (P ) = −(p, q) 88  6.1. The map ψG,H considered as a point in S[)1 . Proof. Let x denote the point  [ (p, q) ∈ S(]1 .  We will show that φ(x) defines a bi-ordering of Z×Z for which the subgroup hmp lq i is convex, and the element mp lq is positive. By definition, the positive cone φ(x) corresponds to all those elements ma lb such that the oriented angle between the vector (a, b) and x lies in the interval [0, π). Observe that (p, q) has oriented angle zero with the vector x, therefore mp lq lies in φ(x), and property (2) above is satisfied. Next, let h ∈ Z2 be an element in the positive cone φ(x) not of the form (mp lq )n , and suppose that h = mc ld . Since p and q are relatively prime, mp lq is not a power of h, so the oriented angle between (c, d) and x lies in the open interval (0, π). As the vectors x and (p, q) are parallel, any point of the form (c, d) + n(p, q) = (c + np, d + nq) also has an oriented angle with x lying in the interval (0, π). Therefore any element in Z2 of the form mc+np ld+nq lies in the positive cone φ(x), and hence the bi-ordering defined by φ(x) satisfies 1 < mc+np ld+nq = (mp lq )n mc ld ⇒ (mp lq )−n < mc ld .  Thus, every element h ∈ φ(x) that is not a power of mp lq must be greater than all powers of mp lq . This shows that property (1) holds for the ordering 89  6.1. The map ψG,H of Z2 defined by the positive cone φ(x). Next, consider the case  [ y = −(p, q) ∈ S[)1 . Here, the positive cone φ(y) is defined as all those elements ma lb such that the oriented angle between the vector (a, b) and y lies in the interval (0, π]. Observe that the oriented angle between y and (p, q) is π, so that mp lq ∈ φ(y), and φ(y) satisfies property (1). The proof of property (2) in this case is similar to the case of φ(x). To prove the converse, we must show that any positive cone that defines an ordering of Z × Z satisfying properties (1) and (2) must be mapped by φ−1 to one of the two points of the required form. Observe that there are exactly two positive cones in LO(Z2 ) which define an ordering satisfying properties (1) and (2), namely those positive cones arising from the short exact sequence 1 → hmp lq i → Z × Z → Z → 1, where hmp lq i ∼ = Z is ordered so that mp lq is positive. Call these two positive cones P1 and P2 . We have already proven that the positive cones φ(x) and φ(y) above correspond to orderings satisfying properties (1) and (2), and it is easy to see that φ(x) and φ(y) are distinct (for example, m−p lq lies in φ(x) but not in φ(y)). Therefore the sets {P1 , P2 } and {φ(x), φ(y)} must be equal, so that P1 and P2 are mapped by φ−1 to the points y ∈ S[)1 and x ∈ S(]1 , which are of the required form. 90  6.1. The map ψG,H  Let f : X → S 1 denote the map obtained by identifying the rational points of S(]1 with the rational points of S[)1 . The map f is continuous, since the topologies on S(]1 and S[)1 are both refinements of the standard topology on S 1 . Let G be a left orderable group with subgroup H isomorphic to Z × Z. We have seen that there exists a continuous map ψG,H : LO(G) → S 1 , which is the composition of the maps r  φ−1  f  LO(G) → LO(H) → X → S 1 . Proposition 6.1.4. Let G be a left orderable group with subgroup H ∼ = Z × Z having basis m and l, and let (p, q) be any relatively prime pair of integers. With the map ψG,H : LO(G) → S 1 ⊂ R2 defined as above, the [ points ±(p, q) ∈ S 1 lie in the image of ψG,H if and only if there exists a left ordering < of G satisfying: 1. 1 < mp lq , and 2. (mp lq )n < h for all positive integers n and all positive h ∈ H not of the form (mp lq )k . [ Proof. Suppose that ±(p, q) ∈ S 1 both lie in the image of ψG,H , and let P −1 [ be any positive cone in ψG,H ((p, q)), we will show that either P or P −1 has  the desired properties. Considering φ−1 (r(P )), we see that φ−1 (r(P )) lies in the set f −1 (x) and  91  6.1. The map ψG,H is therefore equal to either  [ [ q) ∈ S(]1 in X, (p, q) ∈ S[)1 or (p,  we will deal with these two cases separately. [ If φ−1 (r(P )) = (p, q) ∈ S(]1 , by Lemma 6.1.3 the positive cone P defines a left ordering satisfying properties (1) and (2) when restricted to H. This is sufficient to guarantee that P defines a left ordering of G satisfying properties (1) and (2), so we are done. [ In the case that φ−1 (r(P )) = (p, q) ∈ S[)1 , it follows that [ φ−1 (r(P −1 )) = −(p, q) ∈ S[)1 ,  and by an application of Lemma 6.1.3 we conclude that the positive cone P −1 defines a left ordering of G satisfying properties (1) and (2). Conversely, suppose that < is a left ordering of G satisfying properties (1) and (2), with associated positive cone P ∈ LO(G). In this case, r(P ) defines an ordering of H in which mp lq is positive, and relative to which the subgroup hmp lq i is convex (in H). Thus we may apply Lemma 6.1.3 to [ conclude that the points ±(p, q) lie in the image of ψG,H .  Proposition 6.1.5. Let M 3 be a P 2 -irreducible 3-manifold with torus boundary, set G = π1 (M 3 ) and H = π1 (∂M 3 ), and let {m, l} be a meridian and 3 denote the 3-manifold obtained by p/q longitude basis of π1 (∂M 3 ). Let Mp/q  Dehn surgery on M 3 , where p and q are relatively prime. If  92  6.1. The map ψG,H 3 ) is left orderable, and 1. π1 (Mp/q 3 ), 2. π1 (∂M 3 ) is not killed in the quotient π1 (Mp/q  [ then the points ±(p, q) lie in the image of ψG,H . 3 ) is left orderable. Denote by hhmp lq ii the Proof. Suppose that π1 (Mp/q  normal closure of mp lq in G. From the short exact sequence  3 1 → hhmp lq ii → G → π1 (Mp/q )→1  we may left order G so that the subgroup hhmp lq ii is convex, and mp lq is positive. Here, we are applying Theorem 5.1.1 to conclude that the kernel hhmp lq ii may be left ordered. Observe that in any ordering of H ∼ = Z × Z, if C is convex then either 1. C = {0}, 2. C ∼ = Z, or 3. C = H. In our situation, the subgroup hm, li ∩ hhmp lq ii is nontrivial and convex in the restriction ordering of H, and so there are two possiblities: 1. hm, li ∩ hhmp lq ii = H, or 2. hm, li ∩ hhmp lq ii is infinite cyclic, generated by mp lq . Case (1) is not possible, as we have assumed that π1 (∂M 3 ) is not killed in 3 ). Therefore, we are in case (2), and so the restriction the quotient π1 (Mp/q  93  6.1. The map ψG,H ordering to H satisfies (1) mp lq > 1, and (2) hmp lq i is convex. We now [ apply Proposition 6.1.4 to conclude that the points ±(p, q) lie in the image of ψG,H .  Proposition 6.1.6. Let M 3 be an irreducible 3-manifold with torus boundary, and let {m, l} be a meridian and longitude basis of π1 (∂M 3 ). Suppose 3 ), further that π1 (∂M 3 ) is not killed by the quotient map π1 (M 3 ) → π1 (Mp/q  for every surgery slope p/q ∈ Q. If there is no left ordering of π1 (M 3 ) such 3 ) is not left that 1 < m  l, then there exists N ∈ R such that π1 (Mp/q  orderable whenever p/q > N . Proof. With G = π1 (M 3 ), and H = π1 (∂M ), we consider the map ψG,H . If there exists no left ordering of π1 (M 3 ) such that 1 < m  l, then by Proposition 6.1.4, the points ±(1, 0) do not lie in the image of ψG,H . Since LO(π1 (M 3 )) is compact, its image under the map ψG,H is compact in S 1 and hence closed. As ±(1, 0) are not contained in the closed set ψG,H ((LO(π1 (M 3 ))), there must be open sets U1 about (1, 0) and U2 about (−1, 0) in S 1 such that no point in U1 ∪ U2 lies in the image of the map [ ψG,H . This proves the claim: Points of the form (p, q) in S 1 that are in the neighbourhood U1 ∪ U2 of {(1, 0), (−1, 0)} ⊂ S 1 cannot lie in the image 3 ) cannot be left orderable by Proposition of the map ψG,H , and so π1 (Mp/q  6.1.5. The condition on M 3 above, namely that π1 (∂M 3 ) is not killed by the 3 ), is satisfied by all knot complements. For quotient map π1 (M 3 ) → π1 (Mp/q 3 ) for if π1 (∂M 3 ) is sent to {1} under the quotient map π1 (M 3 ) → π1 (Mp/q  94  6.1. The map ψG,H some knot complement M 3 , then 3 )∼ π1 (Mp/q = π1 (M 3 )/ hhm, lii ∼ = {1}.  This cannot happen for nontrivial Dehn surgeries on a nontrivial knot, since all nontrivial knots satisfy property P : 3 is Theorem 6.1.7 ([31]). If K ⊂ S 3 is a nontrivial knot in S 3 , and Mp/q  the 3-manifold obtained by p/q Dehn surgery on M 3 = S 3 \ K with q 6= 0, 3 ) must be nontrivial. then π1 (Mp/q  In particular, this means that we may apply Proposition 6.1.6 to knot complements. If we start with the complement of a hyperbolic knot (a knot whose complement admits a complete Riemannian metric of constant negative curvature), then Thurston’s hyperbolic Dehn surgery theorem guarantees that only finitely many Dehn surgeries yield non-hyperbolic 3-manifolds. Thus Proposition 6.1.6 yields infinitely many hyperbolic 3-manifolds with non-left orderable fundamental group, if we can find one knot complement that satisfies the necessary hypotheses.  6.1.1  Example: Torus knots  Recall that for an (r, s) torus knot, the fundamental group of the complement has presentation ha, b|ar = bs i, where r and s are relatively prime. With this presentation, the meridian corresponds to the element m = ai bj , where i and j are relatively prime integers such that is + jr = 1, and the longitude is given by l = m−sr ar . An element g in a left ordered group G with ordering  95  6.1. The map ψG,H < is called cofinal if G = {h ∈ G|g −n < h < g n for some n ∈ Z}. Lemma 6.1.8. In every left ordering of the group G = ha, b|ar = bs i, the central element ar is cofinal. Proof. Suppose that < is a left ordering of G, and let C = {x ∈ G|there exists n ∈ Z such that a−rn < x < arn }.  We check that C is a subgroup of G. Given x, y ∈ C, choose integers n, m such that a−rn < x < arn and a−rm < y < arm . Then x < arn ⇒ yx < yarn = arn y, while y < arm ⇒ arn y < ar(n+m) ; combining these two inequalities gives xy < ar(n+m) . Similarly, we may compute that a−r(n+m) is a lower bound for xy, so that C is closed under multiplication. Next, we observe that C is closed under taking inverses, for if a−rn < x < arn , then x < arn ⇒ 1 < x−1 arn ⇒ a−rn < a−rn x−1 arn = x−1 ,  so that a−rn is a lower bound for x−1 . Similarly we show that arn is an upper bound for x−1 , so that C is closed under taking inverses. It is easy to see from the definition that C is convex in the given left ordering, so that C is a convex subgroup of G. Finally, we may observe that both a and b lie in C, since a and b are 96  6.1. The map ψG,H clearly both bounded by powers of a in any left ordering of G. This shows that the generators of G are contained in C, and hence G is contained in C. It follows that G = C and the element ar is confinal in the left ordering <. Proposition 6.1.9. If M 3 is the complement of an (r, s) torus knot, then there exists an open neighbourhood U ⊂ R of rs ∈ R such that for every 3 ) is not left orderable. surgery slope p/q in U , the group π1 (Mp/q  Proof. Let G = π1 (M 3 ) = ha, b|ar = bs i, and let H = π1 (∂M 3 ). From Lemma 6.1.8, the powers of mrs l = ar are not bounded above by any element of the group G in any left ordering, so by Proposition 6.1.4 the points \ ±(rs, 1) are not in the image of ψG,H . Since the image of ψG,H is compact, and hence closed, it follows that there is an open set U containing \ ±(rs, 1) ∈ S 1 such that no points in the image of ψG,H lie inside of U . By 3 ) is not left orderable Proposition 6.1.5, this means that the group π1 (Mp/q  whenever p/q ∈ U . Thus, Proposition 6.1.9 yields infinitely many examples of 3-manifolds with non-left orderable fundamental group, all of which are known to be Seifert fibered or the connect sum of two lens spaces. Theorem 6.1.10 ([38]). Suppose that M is the complement of a torus knot in S 3 . For every surgery coefficient p/q ∈ Q, the manifold Mp/q is either a Seifert fibered space, or a connect sum of two lens spaces.  97  6.2. Analysis of the map ψG,H for bi-orderable groups  6.2  Analysis of the map ψG,H for bi-orderable groups  The goal of this section is to prove the following theorem, and discuss its consequences when applied to Dehn surgery on knots. Recall that a subgroup H of a group G is isolated if for every g ∈ G the following implication holds: If there exists a nonzero integer k ∈ Z, such that g k ∈ H, then g ∈ H. Theorem 6.2.1. Suppose that G is a bi-orderable group, and that H ∼ = Z × Z is an isolated subgroup of G. Then the map ψG,H : LO(G) → S 1 is surjective. The following theorem is essentially an application of Theorem 1.4.10. Below, we are considering a bi-ordered group G, together with the action of G on itself by conjugation. This action is necessarily order preserving if G is equipped with a bi-ordering. While the proof below is essentially a special case of the proof of Theorem 1.4.10, we fully cover the details of this special case. Theorem 6.2.2. Let G be a bi-orderable group with bi-ordering <, and let A be any nontrivial subgroup of G. Denote by CG (A) the centralizer of A in G. Then there exists a left ordering of G in which the centralizer CG (A) is convex. Proof. Let ≺ be any well-ordering of A, and for any g ∈ / CG (A), set ag = min≺ {a ∈ A : g −1 ag 6= a}.  98  6.2. Analysis of the map ψG,H for bi-orderable groups Define a positive cone P ⊂ G according to the rule: g ∈ P if g ∈ / CG (A) and g −1 ag g > ag , or if g ∈ CG (A) and g > 1. We check that P satisfies the required properties. 1. P · P ⊂ P . Let g, h ∈ P . There are three cases to consider. (a) Suppose g, h ∈ CG (A). Then gh ∈ CG (A), and gh > 1, so gh ∈ P . (b) Suppose that g ∈ CG (A) and h ∈ / Cg (A). Consider the product hg. Observe that ahg = ah , for suppose not, say ahg ≺ ah . Then both h and g commute with ahg , while hg does not, a contradiction. Similarly, if ah ≺ ahg , then hg commutes with ah , and g commutes with ah , while h does not. This is again a contradiction, as we compute ah = (hg)−1 ah hg = g −1 h−1 ah hg > g −1 ah g = ah , where we have used that h−1 ah h > ah , and > is a bi-ordering. Thus ahg = ah implies (hg)−1 ahg hg > ahg , so that hg ∈ P . Next, consider the product gh. Then gh ∈ / CG (A), and agh = ah , for suppose that agh ≺ ah . Then h commutes with agh , and g commutes with agh . Therefore gh commutes with agh , a contradiction. We arrive at a similar contradiction if ah ≺ agh : then gh commutes with ah , as does g, while h does not. We have already dealt with a similar case above, showing a contradiction,  99  6.2. Analysis of the map ψG,H for bi-orderable groups and so we conclude that ah = agh . Then we compute that (gh)−1 agh gh = h−1 g −1 ah gh = h−1 ah h > ah = agh ,  so that gh ∈ P . (c) Suppose that g, h ∈ / CG (A), and consider the product gh. Then agh = min≺ {ag , ah }, for suppose not, say min≺ {ag , ah } = ah ≺ agh . Then gh commutes with agh , and so does g, while h does not. We have already dealt with such a situation above, showing a contradiction. Similarly, agh ≺ ah reduces to a case previously considered, and we arrive at a contradiction, concluding that agh = min≺ {ag , ah } = ah . Then using the fact that ah ≺ ag , so g commutes with ah , we compute (gh)−1 agh gh = h−1 g −1 ah gh = h−1 ah h > ah = agh ,  so that gh ∈ P . The case of gh ∈ P when min≺ {ag , ah } = ag is nearly identical. Lastly, we must consider the case when ag = ah . In this case, we have agh = ag = ah , and it follows that (gh)−1 agh (gh) = (gh)−1 ag (gh) > h−1 ag h = h−1 ah h > ah = agh .  2. We must show that P t P −1 t {1} = G. This follows immediately from the fact that ag = ag−1 for any g ∈ / CG (A), and the observation that g −1 ag g > ag ⇒ gag−1 g −1 < ag−1 , since < is a bi-ordering of G. 100  6.2. Analysis of the map ψG,H for bi-orderable groups Lastly, we must show that CG (A) is convex, so suppose that g ∈ CG (A), and that h ∈ / CG (A) is positive, and consider the element g −1 h. Observe that g −1 h ∈ / CG (A), and that ag−1 h = ah , by the arguments used in case 1. (b). Therefore we compute (g −1 h)−1 ag−1 h g −1 h = h−1 gah g −1 h > ah = ag−1 h , so that g −1 h ∈ P and CG (A) is convex. Theorem 6.2.3. Suppose that A is a nontrivial isolated abelian subgroup in G, and that G is a bi-orderable group. Then there exists a left ordering of G in which the subgroup A is convex. Proof. We will construct an ordering of CG (A) in which A is convex. The constructed ordering can be extended to an ordering of the whole group G, by using Theorem 6.2.2 to construct a left ordering of G in which CG (A) is convex. By [[5], Theorem 2.2.4], if G is bi-orderable, then so is G/Z(G), where Z(G) denotes the center of the group G. Therefore we can construct a bi-ordering of CG (A) in which Z(CG (A)) is convex. In the abelian group Z(CG (A)) the subgroup A is isolated, and so the quotient by A is a torsion free abelian group, and hence bi-orderable. Therefore we can bi-order the convex subgroup Z(CG (A)) lexicographically using the short exact sequence  1 → A → Z(CG (A)) → Z(CG (A))/A → 1,  thus producing a bi-ordering of CG (A) with A as a convex subgroup. 101  6.2. Analysis of the map ψG,H for bi-orderable groups Proof of Theorem 6.2.1. Recall that ψG,H is the composition r  φ−1  f  LO(G) → LO(H) → X → S 1 ,  where φ−1 and f are both surjective. Thus, to show surjectivity of ψG,H it is enough to show surjectivity of the restriction map r. To this end, let G be a bi-orderable group, H an isolated subgroup isomorphic to Z × Z, and let P ∈ LO(H) be any positive cone. By Theorem 6.2.3 there exists a left ordering of G relative to which H is a convex subgroup, so the set of left cosets G/H can be ordered in a way that is preserved by multiplication from the left. Following Proposition 1.1.4, let q : G → G/H denote the quotient map onto the set of left cosets, and let PG/H denote those cosets which are greater than the identity coset in the ordering of G/H. Then Q = P ∪ q −1 (PG/H ) defines a positive cone in LO(G) satisfying r(Q) = Q ∩ H = P , so that the restriction map is surjective. Hence, ψG,H is a composition of surjective maps, and is itself surjective. In order to apply this theorem to Dehn surgery on knots, we recall the following. Theorem 6.2.4 ([50]). If K is a fibered knot in S 3 and all the roots of its Alexander polynomial are real and positive, then π1 (S 3 \ K) is bi-orderable. In particular, this means that the complement of the figure eight knot has bi-orderable fundamental group, as the Alexander polynomial of the figure eight knot is ∆(t) = t2 − 3t + 1, which has real, positive roots  √ 3± 5 2 .  Finally, we need the following result concerning peripheral elements of knot 102  6.2. Analysis of the map ψG,H for bi-orderable groups groups. Theorem 6.2.5 ([59]). If K is a hyperbolic knot, the group π1 (∂(S 3 \ K)) is isolated in π1 (S 3 \ K). It is well known that the figure eight knot is hyperbolic, so the fundamental group of its boundary is isolated. Therefore, we are in a position to apply Theorem 6.2.1 to the complement of the figure eight knot, and we conclude the following. Theorem 6.2.6. Suppose that K is the figure eight knot. If G = π1 (S 3 \ K) and H = π1 (∂(S 3 \ K)), then the map ψG,H is surjective. 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