Free group automorphisms, invariant orderingsand topological applications
Abstract
We are concerned with orderable groups and particularly those with orderings invariant not only under multiplication, but also under a given automorphism or family of automorphisms. Several applications to topology are given: we prove that the fundamental groups of hyperbolic nonorientable surfaces, and the groups of certain fibred knots are biorderable. Moreover, we show that the pure braid groups associated with hyperbolic nonorientable surfaces are leftorderable.
6F15 \secondaryclass57M05 \agt ATG Volume 1 (2001) 311–319\nlPublished: 24 May 2001
Abstract\stdspace\theabstract
AMS Classification\stdspace\theprimaryclass; \thesecondaryclass
Keywords\stdspace\thekeywords
1 Introduction
If is a group, and a strict total ordering of its elements, we say that is a leftordered group if for all , and biordered if the ordering is also rightinvariant: . Surprisingly many groups are leftorderable or even biorderable; for example free groups are biorderable [3], although this is by no means obvious (one method is described in Section 3).
If is an automorphism, then we say that an ordering of is invariant under (or respected by ) if for all . As we will see, it may or may not be possible to find such an invariant ordering, depending on the nature of the . In fact all our results on such invariant orderings involve free groups.
The goal of this paper is to establish new orderability results for several families of groups which arise in topology. We use a sort of bootstrap process involving extensions and orderings of free subgroups which also invariant under certain automorphisms. We show the following results:
(1)\quaThe fundamental groups of all closed surfaces, orientable or not, are biorderable, with the exceptions of the projective plane and the Klein bottle (cf. [4]).
(2)\quaThe pure braid groups associated with nonorientable surfaces are leftorderable.
(3)\quaThe fundamental groups of certain puncturedtorus bundles over the circle are biorderable; examples include the figureofeight knot group.
Some of our technical arguments are directly adapted from [5] (which in turn built on ideas from [6]). In these papers it was proved that the pure braid groups, and more generally all pure surface braid groups (in any orientable compact surface) are biorderable. Moreover, [5] shows that is not biorderable for nonorientable surfaces .
More general surveys on orderable groups and their rôle in lowdimensional topology, which place our results in their proper context, can be found in the introductory sections of [4] and [12].
We thank Steven Boyer for many helpful discussions; some of the present results arose directly from our work on the orderability of 3manifold groups. Thanks also to Juan GonzalesMeneses for inspiring discussions. Dale Rolfsen was partially supported by research grant of the Canadian Natural Sciences and Engineering Research Council, and Bert Wiest by a PIMS postdoctoral fellowship.
2 Orderable groups and extensions
Groups which are leftorderable are easily seen to be torsionfree. However, leftorderable groups enjoy certain advantages over merely torsionfree ones; for instance, they are known to satisfy the (still open) zerodivisor conjecture, which states that the group ring of a torsion free group should have no zerodivisors. The group rings of biorderable groups are known to have an even stronger property, due to Malcev and Neumann (and conjectured to be true for leftorderable groups as well): if is a biorderable group, then embeds in a division algebra. We refer the reader to [3] for proofs and more general statements.
It is easy to verify that a group is leftorderable if and only if there is a subset which does not contain the identity element, is closed under multiplication and such that for every , exactly one of and belongs to . Given such a , define . On the other hand, given a leftordering , define to be the positive cone . Note that if one instead used the criterion , a rightinvariant ordering would result: a group is rightorderable if and only if it is leftorderable.
A leftordering is biinvariant if and only if its positive cone is normal: . Moreover, the ordering is invariant under an automorphism if and only . Following is one of the reasons to be interested in orderings invariant under automorphisms. Its straightforward proof is left to the reader.
Lemma 1.
Suppose we have a short exact sequence of groups
If and are leftorderable, then is leftorderable, with positive cone . If and are biordered, then the same formula defines the positive cone for a biordering of if and only if for all , that is, if and only if the ordering of is invariant under conjugation by elements of .
This gives us a strategy to prove that a group is biorderable. One finds a convenient normal subgroup such that is biorderable and in addition can be given a biordering invariant under conjugation by elements of .
3 Surface groups
It has been known for a long time [1, 8] that the fundamental groups of orientable surfaces are biorderable (and in particular leftorderable – an interesting nonconstructive argument in [2] also shows this). Our aim is to generalize these results to nonorientable surfaces, using very different techniques.
We shall denote the connected sum of projective planes by . We recall that is homeomorphic to the connected sum of a torus and projective planes. Consider , the nonorientable surface with Euler characteristic ; this surface will be the key to our analysis. Note that has a hyperbolic structure for , whereas is spherical and the Klein bottle is Euclidean.
Proposition 2.
The group is biorderable.
Theorem 3.
If is any connected surface other than the projective plane or Klein bottle , then is biorderable. For Klein bottle, is leftorderable.
Proof of theorem 3.
Let us first see how theorem 3 follows from proposition 2. If is noncompact, or if is nonempty, then is a free group, and therefore biorderable. Thus we are reduced to considering closed surfaces. According to the standard classification, such surfaces are either a connected sum of tori, or of projective planes in the nonorientable case.
We shall first consider nonorientable surfaces. Of course is certainly not leftorderable. For , the Klein bottle,
is a wellknown example of a group which is leftorderable (being an extension of by ), but not biorderable, as the defining relation would lead to a contradiction.
By proposition 2, the surface has biorderable fundamental group. We shall picture it as a torus with a small disk removed, and replaced by sewing in a Möbius band. Consider an fold cover of the torus by itself, and modify the covering by replacing one disk downstairs, and disks upstairs, by Möbius bands. This gives a covering of by the connected sum of a torus with copies of . Thus the fundamental group of injects in that of , and is therefore biordered.
We now turn to orientable closed surfaces. The cases of genus zero or one being easy, we consider a closed surface of genus . This surface is the oriented double cover of . Therefore its fundamental group is a subgroup of a biorderable group. This completes the proof of theorem 3, assuming proposition 2. ∎
Corollary 4.
The pure braid group on strands in a compact surface is leftorderable.
If is nonorientable, this is the strongest possible result, because by [5] the groups are definitely not biorderable. If is orientable, the result is redundant, since in this case is known to be even biorderable [5]. We also remark that it is not known which nonpure surface braid groups are leftorderable.
Proof of corollary 4.
We shall proceed by induction on . For , we have that is leftorderable by theorem 3. Moreover, we have a short exact sequence
where is induced by forgetting one of the strands. Now is free and hence leftorderable, and is leftorderable by induction; an application of lemma 1 completes the induction step. ∎
Proof of proposition 2.
Let . Our strategy for constructing a biinvariant order on is to apply lemma 1, where the normal subgroup of will be chosen so that .
To define the subgroup , we note that has presentation
where and represent free generators of the punctured torus in and the generator corresponding to the central curve of the Möbius band in . We define , the normal subgroup generated by . Note that a word in belongs to if and only its exponent sums in and are both zero.
The covering of with is very easy to imagine: consider the universal covering , and modify by taking a family of small disks (say ) centered at the integral points . Remove each of these and replace by a Möbius band . This defines a covering . The group of covering translations of is just , with acting by translation on the part of , and taking each to . Therefore we have (as required) an exact sequence
We now turn to the task of proving that the orderability hypotheses of lemma 1 are satisfied. The group can be biordered, say lexicographically. (In fact, there are uncountably many different biorders on ; e.g. there are already two for each line of irrational slope in through .)
All that remains to be proven is that has a biordering which is invariant under conjugation by elements in . We note that is an infinitelygenerated free group. There is a free basis for consisting of the generators represented by a loop that goes around the central curve of the Möbius band , connected by a tail to the basepoint in some (noncanonical) way; for definiteness we may take
as a free generating set for . Now acts upon by conjugation, which may be described in terms of the generators as follows.
Lemma 5.
Suppose has exponent sums and in and , respectively. Then
where .
Proof of the lemma.
We just take ; by calculating the exponent sums of and in we can verify that indeed . ∎
For the following, denotes the abelianization of , which is an infinitely generated free abelian group, with generators, say ; the abelianization map is just . Any automorphism of induces a unique automorphism of . For example, in the above lemma, under abelianization the conjugation map is just the shift . Now lemma 6 completes the proof of proposition 2.∎
Lemma 6.
There is a biordering of the free group which is invariant under every automorphism which induces, on , a uniform shift automorphism .
Proof.
We use the Magnus expansion [9], sending into the ring of formal power series in the infinitely many noncommuting variables . Since there are infinitely many generators, some care must be taken in defining the ring , which we take to be the ring consisting of formal power series in the , but we consider only such series which involve just finitely many different variables. The Magnus map is given by
Clearly the image of lies in the group of units of the form inside , and it is an embedding of groups, by the same proof as in [9]. Elements of may be written in standard form, arranged in ascending degree, and within a degree terms are arranged lexicographically by their subscripts (which in turn are ordered lexicographically). Then two series are compared according to the coefficient of the first term at which they differ (here is where the finiteness assumption is necessary). The proof that this defines a (multiplicative) biinvariant ordering of , is routine  cf. [6]. Via the injection , we may regard as a subgroup, and hence it is also biordered.
Finally, we argue that this ordering has the desired invariance property; equivalently, that preserves the positive cone of . Consider an automorphism such that is a shift This means that , where is in the commutator subgroup Since maps into under the Magnus embedding, the effect of is reflected in by the substitution Therefore, if the Magnus expansion of is , where is the sum of all degree terms, then terms of higher degree. Therefore, the lowest degree nonzero terms of the Magnus expansions of and are identical, except that the subscripts are shifted. Thus the “first” nonconstant terms of both and have the same coefficient, and we conclude that preserves the positive cone of in the ordering we described. ∎
4 Puncturedtorus bundles over the circle
Suppose that a 3manifold is a fibre bundle over the circle, with fibre some surface , and monodromy map . A necessary and sufficient condition for to be biordered is that there exists some biorder on which is invariant under . This can be seen by considering the short exact sequence of the fibration
and applying lemma 1. It is an important and apparently hard problem in general to decide which automorphisms of free groups or surface groups leave some biordering invariant.
In the present paper we restrict our attention to the case where is a oncepunctured torus  for a different line of attack see [11]. We shall apply a slightly more sophisticated version of lemma 6 in order to prove
Theorem 7.
Suppose the 3manifold is a fibre bundle over the circle, with fibre a oncepunctured torus . Suppose that the monodromy map is orientation preserving and induces in homology a homomorphism which preserves some biordering on . Then is biorderable.
It is known which automorphisms of leave some biorder invariant. This is due to Levitt [7]: suppose is a linear transformation that restricts to an automorphism (i.e. it is represented by some matrix with integer entries, and determinant ). Consider a basis of such that the matrix of with respect to is in Jordan normal form. We can split uniquely as a direct sum , where we define to be spanned by the vectors of belonging to Jordan blocks with positive real eigenvalue, and to be spanned by the vectors of with negative or complex eigenvalues. Now Levitt’s criterion is: the homomorphims leaves some biorder on invariant if and only if does not intersect the integer lattice except in . The proof is not difficult, and left to the reader  the basic observation is that for any biordering of there exists a hyperplane in such that all integer lattice points to one side of the plane are in the positive, and those on the other side in the negative cone.
We remark that if the monodromy map is orientation reversing, then is definitely not biorderable, because it contains a Klein bottle group. Also, if the monodromy map is periodic, then cannot be biorderable, because a periodic automorphism of cannot leave a biordering of invariant. This proves, for example, that the group of the trefoil knot, which is a fibred knot with a monodromy map of period , is not biorderable. This fact, which was pointed out in [10], seems to have dissuaded mathematicians from studying biorderability of knot groups for the last 25 years.
Examples for theorem 7\quaAn orientation preserving (i.e. determinant 1) automorphism of preserves a biorder if and only if its eigenvalues are both positive (but not necessarily distinct). For if has eigenvalues , where is the eigenvector for and is the other basis vector in a Jordan normal basis, then we can define the positive cone of a biordering of by
We observe that this order is invariant under .
For instance, theorem 7 implies that the complement of the figureofeight knot has biorderable fundamental group, because it is a fibred knot with punctured torus fibre , and the matrix of the monodromy action on is is the only classical knot (in ) which is covered by theorem 7, because the only classical fibred knots of genus are the knots and . A few more classical knots are biorderable by the main theorem of [11]. , which has two positive eigenvalues. The knot
If, on the other hand, has two negative, or two complex conjugate eigenvalues, then it cannot respect any decomposition of into a “positive” and a “negative” halfspace; thus cannot preserve any ordering on .
Proof of theorem 7\quaConsider the short exact sequence
where is the abelianisation homomorphism (which geometrically corresponds to “patching the puncture in ”), and is the covering space of whose fundamental group is the commutator subgroup . One can picture as the plane with all integer lattice points removed.
In particular, is an infinitely generated free group, with generating set , where the loop has winding number one around the puncture at the point , and winding number zero around the punctures at all the other integer lattice points (many different choices are possible here). Note that the abelianisation is an infinitely generated free abelian group with generators .
We have to prove that leaves some biorder on invariant. We already have a invariant biorder on , and in fact we shall take it to be the order constructed in the example for theorem 7. Thus it suffices by lemma 1 to find a biorder on which is

invariant under conjugation by elements in , and

invariant under (the restriction of) .
Let’s study the effect of conjugation by elements in and of . Conjugation by an element with sends a generator to , where depends on , as in lemma 5. Thus, on the abelianisation , conjugation by induces a uniform shift automorphism . (Note that the mapping , preserves our ordering of .)
Moreover, the restriction of to is given geometrically by the action of the lift which fixes the point . This sends the generator to a conjugate of the generator , and induces on the abelianisation an automorphism determinded by another simple permutation of the generators: . The proof of theorem 7 is now completed by the following
Lemma 8.
There is a biordering of the free group which is invariant under every automorphism which induces, on the abelianisation , an automorphism that acts simply by permuting the variables: , where the permutation preserves the biordering of .
Proof.
The proof is virtually the same as for lemma 6. We order the variables by defining to “come before” if in the invariant biorder of . Then in order to define which of two words in is the larger we compare their images under the Magnus map . Finally the invariance property is proved precisely as in lemma 6.∎
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Received:\qua8 February 2001 Revised:\qua16 May 2001