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Braids, orderings and minimal volume cusped hyperbolic 3-manifolds

It is well-known that there is a faithful representation of braid groups on automorphism groups of free groups, and it is also well-known that free groups are bi-orderable. We investigate which n-strand braids give rise to automorphisms which preserve some bi-ordering of the free group rank n. As a consequence of our work we find that of the two minimal volume hyperbolic 2-cusped orientable 3-manifolds, one has bi-orderable fundamental group whereas the other does not. We prove a similar result for the 1-cusped case, and have further results for more cusps. In addition, we study pseudo-Anosov braids and find that typically those with minimal dilatation are not order-preserving.


Introduction
If < is a strict total ordering of the elements of a group G such that g < h implies f g < f h for all f, g, h ∈ G, we call (G, <) a left-ordered group. If the leftordering < is also invariant under right-multiplication, we call (G, <) a bi-ordered group (sometimes known in the literature simply as "ordered" group). If a group admits such an ordering it is said to be left-or bi-orderable. It is easy to see that a group is left-orderable if and only if it is right-orderable. If (G, <) is a bi-ordered group, then < is invariant under conjugation: g < h if and only if f gf −1 < f hf −1 for all f, g, h ∈ G. Nontrivial examples of bi-orderable groups are the free groups F n of rank n, as discussed in Appendix A.
An automorphism φ of a group G is said to preserve an ordering < of G if for every f, g ∈ G we have f < g =⇒ φ(f ) < φ(g); we also say < is φ-invariant. If φ : G → G preserves an ordering < of G, then the nth power φ n preserves the ordering < of G for each integer n. We note that if φ n preserves an ordering it does not necessarily follow that φ does. E. Artin [2,3] observed that each n-strand braid corresponds to an automorphism of F n . This paper concerns the question of which braids give rise to automorphisms which preserve some bi-ordering of F n . In turn this is related to the orderability of the fundamental group of the complement of certain links in S 3 , namely the braid closure together with its axis, which we call a braided link. We pay special attention to pseudo-Anosov mapping classes and their stretch factors (dilatations), and cusped hyperbolic 3-manifolds of small volume. This paper is organized as follows: Section 2 reviews the relation among braids, mapping class groups, free group automorphisms and certain links in the 3-sphere S 3 . In Section 3 we recall basic properties of orderable groups, with explicit bi-ordering of free groups further described in Appendix A. The study of order-preserving braids and their relation to bi-ordering the group of the corresponding braided links is initiated in Section 4. Applications to cusped hyperbolic 3-manifolds of minimal volume are considered in Section 5. In Section 6 we give many examples of non-orderpreserving braids, including pseudo-Anosov braids with minimal dilation as well as large dilatations. We also find a family of pretzel links whose fundamental groups can not be bi-orderable. Appendix B is devoted to a proof that the fundamental group of the Whitehead link complement is bi-orderable.
which is not necessarily commutative. However, we leave it to the reader to check the following Proposition 2.2. The above diagram commutes up to conjugation. Specifically, h • φ * p equals φ * q • h followed by a conjugation in π 1 (D n , q), the conjugating element being the class of the loop ℓ(φ • ℓ) −1 in π 1 (D n , q). Corollary 2.3. The map φ * p preserves a bi-ordering of π 1 (D n , p) if and only if φ * q preserves a bi-ordering of π 1 (D n , q).
Then one checks that f < q g =⇒ φ * q (f ) < q φ * q (g) using conjugation invariance of bi-orderings. The converse is proved similarly.
If one passes from Mod(D n , ∂D) to Mod(D n ), there is a further ambiguity regarding the action of a braid on π 1 (D n ). However, this ambiguity corresponds to conjugation by a power of ∆ 2 n , so again it is irrelevant to the question of preserving a bi-ordering. More concretely, given an n-strand braid β, let φ : D n → D n be any representative of the mapping class β ∈ Mod(D n ). We take any basepoint q of D n possibly φ(q) = q. By choosing a path ℓ in D n from q to φ(q), we have the induced map (2.2) φ * q : π 1 (D n , q) → π 1 (D n , q) which sends the class of a loop α in D n based at q to the class of the loop ℓαℓ −1 based at the same point. By using Corollary 2.3, one sees that β is order-preserving if and only if φ * q : π 1 (D n , q) → π(D n , q) preserves a bi-ordering of π 1 (D n , q). By abuse of notation again, we denote the induced map φ * q by β, when q is specified.
These remarks show that if one allows different choices of basepoint, say a basepoint q, the action of B n on F n should really be regarded as a representation B n → Out(π 1 (D n , q)) ∼ = Out(F n ), the group of outer automorphisms. Recall that Out(G) = Aut(G)/Inn(G), where Inn(G) is the (normal) subgroup of inner automorphisms of a group G.
The closure β of a braid β is a knot or link in the 3-sphere S 3 and the braided link, denote by br(β), is the closure β, together with the braid axis A, which is an unknotted curve that β runs around in a monotone manner: br(β) = β ∪ A. See Figure 3(1) (2). Whereas all links can be realized as β, this is not true for br(β), as each component of β has nonzero linking number with the braid axis A. As an example the Whitehead link considered in Appendix B is not a braided link.  (1) (3) i i+1 1 n (2) 1 2 3 Figure 1.
We see that T β is homeomorphic with the complement of the braid closure β in the solid torus D 2 × S 1 . The interior Int(T β ) can be identified with the complement of β ∪ A in S 3 , so we have the following.
Of course the fundamental group of Int(T β ) is isomorphic with that of T β , which in turn is the semidirect product There is a (split) exact sequence One may consider F n ⋊ β Z as the set of ordered pairs with the multiplication given by

Orderable groups
In this section we outline a few well-known facts about orderable groups. For more details, see [11,6]. Further details regarding bi-ordering of F n may be found in Appendix A.
Proposition 3.1. A group G is left-orderable if and only if there exists a subsemigroup P ⊂ G such that for every g ∈ G exactly one of g = 1, g ∈ P or g −1 ∈ P holds.
Indeed such a P defines a left-ordering < = < P by the rule g < h ⇐⇒ g −1 h ∈ P. Conversely, a left-ordering < defines a positive cone P = P < := {g ∈ G | 1 < g} satisfying the conditions of Proposition 3.1.
Proposition 3.2. A group G is bi-orderable if and only if it posesses P ⊂ G as in Proposition 3.1, and in addition g −1 Pg = P for all g ∈ G.
We note that a left-or bi-ordering of G is preserved by an automorphism φ : G → G if and only if φ(P) = P, where P is the positive cone of the ordering.
is an exact sequence of groups. If K and H are left-ordered with positive cones P K and P H respectively, then G is left-orderable using the positive cone P G := i(P K ) ∪ p −1 (P H ). This is sometimes called the lexicographic order of G, considered as an extension.
Proposition 3.4. In Proposition 3.3, if K and H are bi-ordered, then the formula P G = i(P K ) ∪ p −1 (P H ) defines a bi-ordering of G if and only if the bi-ordering of K is respected under conjugation by elements of G; equivalently g −1 P K g = P K for all g ∈ G.
A subset S of a left-ordered group (G, <) is said to be convex if for all s, s ′ ∈ S and g ∈ G satisfying s < g < s ′ , we have g ∈ S.
is an exact sequence of groups with (G, <) a left-ordered group and K a convex subgroup of (G, <), then there is a leftordering ≺ of H in which 1 ≺ h if and only if some (and hence every) element g ∈ p −1 (h) satisfies 1 < g. If (G, <) is a bi-ordered group, so is (H, ≺).
Being bi-orderable is a much stronger property than being left-orderable. An intermediate property of groups is to be locally-indicable, which means that every finitely-generated nontrivial subgroup has the infinite cyclic group Z as a quotient.
Proposition 3.7 ( [7]). If L is a knot or link in S 3 , then its group π 1 (S 3 \ L) is locally indicable and therefore left-orderable.
Certain knot groups and link groups are bi-orderable, and this is a particular focus of this paper. For example, torus knot groups are not bi-orderable, but the figure-eight knot 4 1 has bi-orderable group [33], and we shall see that the same is true of the Whitehead link and many links constructed from braids as above. One of the reasons bi-orderability of knot groups is of interest has to do with surgery and L-spaces, which were introduced by Ozsváth and Szabó [32] and include all 3-manifolds with finite fundamental group.
Theorem 3.8 ( [10]). If K is a knot in S 3 for which π 1 (S 3 \ K) is bi-orderable, then surgery on K cannot produce an L-space.
We note that this is not true in general for links. As we shall see in Theorem B.1, the Whitehead link has bi-orderable fundamental group, but one can construct lens spaces by certain surgeries on the link.
Consider a group K, an automorphism φ : K → K and the semidirect product Proposition 3.9. Suppose K is bi-orderable and φ : K → K is an automorphism. Then G = K ⋊ φ Z is bi-orderable if and only if there exists a bi-ordering of K which is preserved by φ.
Proof. Suppose that G is bi-ordered by <. Since bi-orderings are invariant under conjugation, the equation φ(k) = tkt −1 implies that <, restricted to K, is φ-invariant. For the converse, suppose φ preserves a bi-ordering ≺ of K. Then we use the lexicographic ordering defined by (k 1 , t p ) < (k 2 , t q ) if and only if p < q as integers or p = q and k 1 ≺ k 2 , to order G. It is easily checked that this is a bi-ordering using the identity t p kt −p = φ p (k) and the assumption that φ preserves the ordering ≺.
If φ(x) = x for x ∈ G, then we say that the orbit of x (under φ) is trivial.
Proposition 3.11. Let G be a left-orderable group. If an automorphism φ : G → G preserves a left-ordering < of G, then φ cannot have any nontrivial finite orbits.
Proof. If the orbit of x ∈ G is nontrivial, we may assume x < φ(x). Then φ(x) < φ 2 (x), and by transitivity and induction x < φ m (x) for all m ≥ 1 and therefore the orbit of x, {φ n (x) | n ∈ Z}, is infinite.
Proposition 3.11 says that if φ : G → G has a nontrivial finite orbit, then φ does not preserve any left-ordering of G.
Consider an automorphism φ : F n → F n of a free group. The following two criteria for φ being order-preserving (or not) will be useful; they involve the abelianization φ ab : Z n → Z n and its eigenvalues, which are, a priori, n complex numbers, possibly with multiplicity.
Theorem 3.12 ([33]). Let φ : F n → F n be an automorphism. If every eigenvalue of φ ab : Z n → Z n is real and positive, then there is a bi-ordering of F n which is φ-invariant. This is useful in showing that certain fibred 3-manifolds have fundamental groups which are not bi-orderable. However, we note that in the case of braids it does not apply in that way, for if φ is an automorphism of F n induced by a braid β ∈ B n , or more generally a symmetric automorphism, then φ ab is simply a permutation of the generators of Z n and therefore has at least one eigenvalue equal to one.
We note that Theorem 3.12 cannot have a full converse. It has been observed in [28] that for n ≥ 3 there exist automorphisms of F n which preserve a bi-ordering of F n , but whose eigenvalues are precisely the nth roots of unity in C. Such examples appear in the discussion in Section 4.6.
In Appendix A a certain class of bi-orderings of the free group F n , called standard orderings is defined, using the lower central series.
Proposition 3.14. If φ : F n → F n is a non-pure symmetric automorphism, then φ cannot preserve any standard bi-ordering of F n .
Proof. Assume φ : F n → F n is not pure, but preserves a standard bi-ordering < of F n . Then φ ab is a nontrivial permutation of the generators of the abelianization Z n . By Proposition A.1, the commutator subgroup [F n , F n ] is convex relative to <, and therefore F n /[F n , F n ] ∼ = Z n inherits a bi-ordering < 1 according to Proposition 3.5. Since φ preserves <, one easily checks that φ ab preserves the order < 1 of Z n . But, φ being non-pure symmetric implies that φ ab is a nontrivial permutation of the generators of Z n which are the images of the x i . By Proposition 3.11, φ ab cannot preserve any ordering of Z n , a contradiction.
Proposition 3.15. If φ : F n → F n is a pure symmetric automorphism, then φ is order-preserving. In fact, it preserves every standard ordering of F n . Proof. A pure symmetric automorphism φ : F n → F n induces the identity map φ ab = id : Z n → Z n , so Proposition A.3 applies.
For any symmetric automorphism φ : F n → F n , there exists k ≥ 1 so that φ k is pure symmetric. By Proposition 3.15, we have the following.
Corollary 3. 16. If φ : F n → F n is a symmetric automorphism, then there exists k ≥ 1 such that φ k is order-preserving. Proposition 4.1. A braid β ∈ B n is order-preserving if and only if π 1 (S 3 \ br(β)) is bi-orderable.
The full twist ∆ 2 is written by ∆ 2 = δ n n = (δ n σ 1 ) n−1 , which means that δ n and δ n σ 1 are roots of ∆ 2 . We note that ∆ 2 commutes with all n-strand braids and in fact generates the centre of B n when n ≥ 3, see [24,Theorem 1.24].

Corollary 4.2.
A braid β ∈ B n is order-preserving if and only if β∆ 2k is orderpreserving for some (hence all) k ∈ Z. Moreover, they preserve exactly the same bi-orderings of F n .
Proof. This follows since S 3 \ br(β) and S 3 \ br(β∆ 2k ) are homeomorphic: By using the disk twist as we shall define in Section 4.4, we see that kth power of the disk twist t k about the disk bounded by the braid axis A of β sends the exterior E(br(β)) of the link br(β) to the exterior E(br(β∆ 2k )). An alternative argument is that ∆ 2 acts by conjugation of F n , so it preserves every bi-order of F n .
As noted by Garside [16] every n-strand braid has an expression β∆ 2k where β is a Garside positive braid word, meaning that β can be written as a word in the σ i generators without negative exponents. Thus a question of a braid being order-preserving can always be reduced to the case of positive braids. Notice that changing a braid β by conjugation does not change the link br(β) up to isotopy, so we have:  Proof. Suppose that σ i preserves a bi-order < of F n . We may assume . But conjugation invariance of the ordering then yields x i+1 < x i , which is a contradiction.
An alternative argument is as follows. We take a basepoint p 1 in the interior of D n . There exists a homeomorphism φ : D n → D n which represents σ i ∈ Mod(D n ) such that the induced map φ * : π 1 (D n , p 1 ) → π 1 (D n , p 1 ) gives rise to the following automorphism on F n = x 1 , · · · , x n : Figure 4(2). (In the figures, we denote by x ′ , the image of x ∈ F n under the automorphism of F n , and denote by x, the inverse preserves a bi-order < of F n . We may assume since σ i preserves <. This implies that x i+1 < x i by conjugation invariance of the bi-ordering. This is a contradiction.

Pure braids.
As is well-known, there is a homomorphism B n → S n of the n-strand braid group onto the permutation group of n letters, and the kernel is the group of pure braids P n .
Proposition 4.6. Every pure braid β ∈ P n is order-preserving, and in fact preserves every standard bi-order of F n .
Proof. For a pure braid β, the image of the Artin representatation is pure symmetric. This completes the proof by Proposition 3.15. Figure 6. Links whose groups are bi-orderable.
The 2nd power σ 2 i of each generator σ i is a pure braid. Thus we have the following.
By Corollary 3.16, we immediately obtain the following. Corollary 4.9. For every braid β some power β k is order-preserving. The fundamental group π 1 (T β k ) is bi-orderable and may be regarded as a normal subgroup of index k in π 1 (T β ).
(1) If β is conjugate to (δ n σ 1 ) k for some k, then β is order-preserving and π 1 (T β ) is bi-orderable. (2) If β is conjugate to δ k n for some k with k ≡ 0 (mod n), then β is not order-preserving and π 1 (T β ) can not be bi-orderable.
Note that δ k n is order-preserving when k ≡ 0 (mod n) since δ n n = ∆ 2 . Theorem 4.10 is a consequence of [7, Theorem 1.5]: If β is a periodic braid, then T β is a Seifert fibered 3-manifold. If β is of type 1, then T β has no exceptional fibres. If β is of type 2 and if k is not a multiple of n, then T β has an exceptional fibre.
We will give an alternative, more explicit proof of Theorem 4.10 in Section 4.5.

Disk twists.
We review a method to construct links in S 3 whose complements are homeomorphic to each other. Let L be a link in S 3 . We denote a tubular neighborhood of L by N (L), and the exterior of L, that is S 3 \ int(N (L)) by E(L). Suppose that L contains an unknot K as a sublink. Then E(K) (resp. ∂E(K)) is homeomorphic to a solid torus (resp. torus). We denote the link L \ K by L K . Taking a disk D bounded by the longitude of N (K), we define two homeomorphisms ) as follows. We cut E(K) along D. We have resulting two sides obtained from D. Then we reglue the two sides by rotating either of the sides 360 degrees so that the mapping class of the restriction T D | ∂E(K) : ∂E(K) → ∂E(K) defines the right-handed Dehn twist about ∂D, see Figure 7(1). Such an operation defines the homeomorphism T D : In the case m = 2, see Figure 7(2). For example, if L is equivalent to br(β) for some β ∈ B n and if K is taken to be the braid axis A of β, then T D (L K ) is equivalent to the closure of β∆ 2 n . Notice that T D : E(K) → E(K) determines the latter homeomorphism We call t D the (right-handed) disk twist about D.
For any integer m = 0, we have a homeomorphism of the mth power T m D :

Alternative proof of Theorem 4.10.
Proof of Theorem 4.10. We prove the claim (1). Consider the pure 2-strand braid σ 2 1 . Then br(σ 2 1 ) is a link consisting of three unknotted components, including the axis. Performing a disk twist n times on one of the components of the closed braid σ 2 1 converts the braided link of the 2-strand braid σ 2 1 to the braided link of the (n + 2)strand braid β ′ , which is conjugate to the type 1 braid σ 1 σ 2 · · · σ n+1 σ 1 , see Figure 8. But the disk twist being a homeomorphism of the complement of the link, we see that T σ 2 1 ≃ T β ′ . But since σ 2 1 is pure, T σ 2 1 has bi-orderable fundamental group, in fact isomorphic with F 2 × Z. Hence the fundamental group of T β ′ is bi-orderable and σ 1 σ 2 · · · σ n+1 σ 1 is order-preserving. Thus the kth power (σ 1 σ 2 · · · σ n+1 σ 1 ) k is also order-preserving. 1 to that of σ 1 σ 2 · · · σ n+1 σ 1 . (n = 2 in this case.) We turn to the claim (2). We use the basepoint p 1 in the interior of D n . There exists a homeomorphism φ : D n → D n which represents δ n ∈ Mod(D n ) such that the induced map φ * : π 1 (D n , p 1 ) → π 1 (D n , p 1 ) gives rise to the following automorphism 1 on F n = x 1 , · · · , x n : see Figure 4(2). The orbit of x 1 is nontrivial (since x 1 = x n ) and finite. By Proposition 3.11, δ n is not order-preserving. Then k ≡ 0 (mod n) if and only if δ k n has a nontrivial finite orbit of x 1 . Thus δ k n is not order-preserving. Proposition 4.11. The half twist ∆ n ∈ B n is order-preserving if and only if n is odd.
As a special case of Theorem 4.10(2), we see that the periodic 3-strand braid σ 1 σ 2 is not order-preserving. Another way to see this is to observe that the 3-strand braid σ 1 σ 2 gives the following automorphism on F 3 = x, y, z by using the Artin One can show this doesn't preserve a bi-order of F 3 as follows. Assume some biorder < were preserved by the map, then supposing without loss of generality that x < y, we would have xyzy −1 x −1 < x (applying σ 1 σ 2 ), hence yzy −1 < x (conjugation invariance), so xyx −1 < xyzy −1 x −1 (applying σ 1 σ 2 ) and then xyx −1 < x (transitivity) and finally the contradiction y < x (conjugation again).
On the other hand σ 1 σ −1 2 ∈ B 3 is pseudo-Anosov, and in fact it is the simplest pseudo-Anosov 3-strand braid; see Section 6.1. We thank George Bergman for pointing out the following argument. It will also follow from more general result in Section 6.2; see Corollary 6.4.
Proof. Let x, y, z be the free generators of F 3 . Using the Artin representation B 3 → Aut(F 3 ), one sees that the action of Consider the orbit of the element w = y −1 x under this action. Assuming there is a bi-order < of F 3 invariant under this action, we may assume without loss of generality that 1 < w, and therefore all elements of the orbit of w are positive. Moreover 1 < w implies y < x and since w → zx −1 we have y < x < z.
Now the calculation and the facts that z −1 y and x(xz −1 )x −1 are negative show that the action is decreasing on the orbit of zx −1 , which is also the orbit of w.
Calculating the preimages of the generators, we have the action of and therefore y and since the expressions in the parentheses are < 1 (i.e, yz −1 < 1 and y −1 (x −1 y)y < 1), we conclude that the action is increasing on the orbit of w. This contradiction shows that an invariant bi-order of F 3 cannot exist.
4.6. Explicit orderings preserved by periodic braids of type 1. We have seen that for n ≥ 3 the root δ n σ 1 of the full twist ∆ 2 ∈ B n preserves an ordering of F n . In this section we will explicitly construct uncountably many such orderings.
Recall that δ n ∈ B n induces the following automorphism on F n , using the basepoint in the interior of D n δ n : . By (4.1) σ 1 ∈ B n induces the following automorphism on F n , using the same basepoint: Thus the automorphism δ n σ 1 on F n is given by (For the case n = 3, just take the first three terms above.) Here is another way to realize this automorphism of F n , elaborating on Example 3.6 in [28]. Fix n ≥ 3 and consider the free group F 2 = u, v of rank 2 and the homomorphism of F 2 onto the cyclic group G = t | t n−1 = 1 given by u → t and v → 1. The kernel K = K n of this map is a normal subgroup of F 2 , of index n − 1. If we realize F 2 as the fundamental group of a bouquet of two circles labelled u and v, the covering space corresponding to K is a finite planar graph as pictured in Figure 9 whose fundamental group is free of rank n. Using the basepoint in the (3) β Figure 10.
Now consider the automorphism of the normal subgroup K of F 2 given by φ(x) = uxu −1 . This is not an inner automorphism of K but it is an inner automorphism of the larger group F 2 . Therefore, if we take any bi-ordering of F 2 , then its restriction to K will be preserved by φ. By inspection, the action of φ on K is given by Under the isomorphism K ∼ = F n given by z i → x i for each i, we see that φ corresponds to δ n σ 1 . One can use this isomorphism to see that any bi-ordering of F 2 restricted to K provides an ordering of F n invariant under δ n σ 1 . Finally note that any ordering of F n respected by δ n σ 1 will also be respected by (δ n σ 1 ) k for any integer k.
Note that we can use, for example, any standard ordering of F 2 . However, this ordering restricted to K cannot be a standard ordering (defined using the lower central series of K) by Proposition 3.14.

4.7.
Tensor product of braids. Given braids α ∈ B m and β ∈ B n one can form the (m + n)-strand braid α ⊗ β ∈ B m+n with α on the first m strings and β on the last n strings, but no crossings between any of the first m strings with any of the last n strings ( Figure 10); see for example [24, page 69]. The action of α ⊗ β on F m+n ∼ = F m ⋆ F n is just the free product α ⋆ β : F m ⋆ F n → F m ⋆ F n .
The following Lemma and Proposition are proved in [34,Corollary 4].  There is a natural inclusion B m ⊂ B m+n (n ≥ 1) given by β → β ⊗ 1 n , where 1 n is the identity braid of B n . 4.8. Do order-preserving braids form a subgroup? Let OP n ⊂ B n denote the set of n-strand braids which are order-preserving. It is clear that β ∈ OP n if and only if β −1 ∈ OP n and that the identity braid belongs to OP n . Moreover, P n ⊂ OP n by Proposition 4.6. It is natural to ask whether OP n forms a subgroup of B n , in other words whether OP n is closed under multiplication. For n = 2 the answer is affirmative. Noting that ∆ 2 2 = σ 2 1 we conclude from Propositions 4.4 and 4.6 and Corollary 4.2 that OP 2 consists of exactly the 2-strand braids σ k 1 with k even. Therefore OP 2 is exactly the subgroup P 2 . Proof. Consider the n-strand braids α = σ 1 σ 2 σ 1 and β = σ −2 1 . Then α is orderpreserving, being an extension in B n of type 1 periodic braid σ 1 σ 2 σ 1 ∈ B 3 (see Corollary 4.15), and β is order-preserving, being a pure braid. But αβ = σ 1 σ 2 σ −1 1 is not order-preserving, as it is conjugate to σ 2 which is not order-preserving by Proposition 4.4.
Although not a subgroup for n > 2, OP n is a large subset of B n : Proposition 4.17. For n > 2 the set OP n of order-preserving n-braids generates B n .
Proof. We saw above that σ 1 σ 2 σ −1 1 is a product of braids α, β ∈ OP n ; therefore σ 2 is also a product of appropriate conjugates of α and β; these conjugates are also in OP n . But all the generators σ i of B n are conjugate to each other, and therefore are also products of elements of OP n . It follows that all braids are products of elements of OP n .

Small volume cusped hyperbolic 3-manifolds
It is known by Gabai-Meyerhoff-Milley [15] that the Weeks manifold is the unique closed orientable hyperbolic 3-manifold of smallest volume. Its fundamental group is not left-orderable; see Calegari-Dunfield [8]. In this section we will see that certain minimum volume orientable n-cusped 3-manifolds can be distinguished by orderability properties of their fundamental groups. We also prove that some orientable hyperbolic n-cusped 3-manifolds with the smallest known volumes have bi-orderable fundamental groups.
Let C 3 and C 4 be the chain links with 3 and 4 components as in Figure 11(1) and (2). For n ≥ 5, let C n be the minimally twisted n-chain link; see [23, Section 1] for the definition of such a link. Figure 11(3) and (4) show C 5 and C 6 . Let W n be the n-fold cyclic cover over one component of the Whitehead link complement W. (See Figure 22(2) for W 3 .) It seems that S 3 \ C n and W n−1 play an important role for the study of the minimal volume hyperbolic 3-manifolds with n cusps, where n ≥ 3, as we shall see below.
5.1. One cusp. Cao-Meyerhoff [9] proved that a minimal volume orientable hyperbolic 3-manifold with 1 cusp is homeomorphic to either the complement of figureeight knot in S 3 or its sibling manifold (which can be described as 5/1 Dehn surgery on one component of the Whitehead link). We note that the sibling manifold cannot be described as a knot complement, as its first homology group is Z ⊕ Z/5Z.
Like the figure-eight knot complement, it can be described as a punctured torus bundle over S 1 .

Theorem 5.1. The figure-eight knot complement has bi-orderable fundamental group. The fundamental group of its sibling is not bi-orderable.
Proof. The first assertion is proved in [33], using the monodromy 2 1 1 1 associated with the fibration, which has two positive eigenvalues, and Theorem 3.12. The sibling has the monodromy −2 −1 −1 −1 , see [22, Proposition 3 and Note]. This has the two negative eigenvalues (−3 ± √ 5)/2. By Theorem 3.13, the sibling has non-bi-orderable fundamental group.
Theorem 5.2. The fundamental group of W is bi-orderable. The fundamental group of W ′ is not bi-orderable.
Theorem 5.5. The complement S 3 \ C 5 has bi-orderable fundamental group.
5.6. n cusps for n ≥ 6. It is a conjecture by Agol [1] that for n ≤ 10, the minimally twisted n chain link complement S 3 \ C n has the minimal volume among orientable hyperbolic 3-manifold with n cusps. The following theorem follows from Lemmas 5.8 and 5.9 below.
Theorem 5.7. The complement S 3 \ C 6 has bi-orderable fundamental group.
Let L 1 be the 6-circle link as shown in Figure 15(2).
Proof. We consider the two trivial components K, K ′ ⊂ br(β) bounded by the closure of the first string and the closure of the last string. Let D and D ′ be the disk bounded by K and K ′ . We do the disk twists t D and t D ′ . Then the resulting link is equivalent to L 1 . Thus S 3 \ br(β) is homeomorphic to S 3 \ L 1 . Since β is a pure braid, this implies that π 1 (S 3 \ L 1 ) is bi-orderable.
We thank Ken'ichi Yoshida who conveys the following argument to the authors. Lemma 5.9. S 3 \ L 1 is homeomorphic to S 3 \ C 6 .
Proof. We take four 3-punctured spheres embedded in the exterior E(L 1 ) as shown in Figure 16(2). The four 3-punctured spheres are also embedded in E(C 6 ) as shown in Figure 18(2). Let U (resp. U ′ ) be a subset of E(L 1 ) (resp. a subset of E(C 6 )) bounded by these 3-punctured spheres. Note that E(L 1 ) (resp. E(C 6 )) are the double of U (resp. U ′ ) with respect to the four 3-punctured spheres, see Figure 16 and Figure 18(3). We now see that U is homeomorphic to U ′ . This is enough to prove the lemma since the double of a manifold is uniquely determined.
We push the bottom shaded 3-punctured sphere in U as shown in Figure 16(3), and deform it into the shaded 3-punctured sphere as in Figure 16 (6). (The middle red colored annulus in (3) is modified into the bottom red colored annulus in (6).) As a result, we get U ′ .

A result by M. Baker and a question.
We learned the following theorem from Hidetoshi Masai.
Theorem 5.11 (Baker [4]). The 6-circle link L 1 is an arithmetic link. Every link L in S 3 occurs as a sublink of a link J such that S 3 \ J is a covering space of S 3 \ L 1 .
In particular L is a sublink of an arithmetic link in S 3 .
Theorem 5.11 and Lemma 5.8 immediately give us the following.
Theorem 5.12. Let L be a link in S 3 . Then L is a sublink of a link whose complement has bi-orderable fundamental group.
The link with 4 components shown in Figure 22(2) is obtained from C 3 adding a trivial knot. Its complement which is homeomorphic to W 3 has bi-orderable fundamental group (Proposition 5.10), although we have not decided yet N = S 3 \C 3 has bi-orderable fundamental group. We ask the following.
Question 5.13. Let M be a 3-manifold. Does there exist a knot K in M such that π 1 (M \ K) is bi-orderable?

Non-order-preserving braids
In this section, we give some sequences of non-order-preserving braids. Our sequences include examples of pseudo-Anosov braids with small dilatations. We also provide examples of non-order-preserving pseudo-Anosov braids with arbitrary large dilatations.
6.1. Pseudo-Anosov braids with smallest dilatations. Let β ∈ B n be a pseudo-Anosov braid. Let λ(β) > 1 be the dilatation (i.e, stretch factor) of the corresponding pseudo-Anosov mapping class β ∈ Mod(D n ). (See [12] for the definition of dilatations, for example.) It is known that there exists a minimum, denoted by δ(D n ) among dilatations for all pseudo-Anosov n-strand braids. The explicit values of minimal dilatations δ(D n )'s are determined for 3 ≤ n ≤ 8 by Ko-Los-Song, Ham-Song and Lanneau-Thiffeault. The following n-strand braids realize δ(D n ).
If n ≥ 5, then δ n σ 1 σ 2 ∈ B n is pseudo-Anosov, see [21,Theorem 3.11]. This claim can be proved by using the criterion by Bestvina-Handel [5]. (See also [21,Section 2.4].) Since δ n σ 2 1 ∼ δ n σ 1 σ 2 , the braid δ n σ 2 1 is pseudo-Anosov. By using the same criterion, it is not hard to see that σ 2k−2 1 δ n σ 1 σ 2 ∼ δ n σ 2k 1 is pseudo-Anosov for n ≥ 5 and k ≥ 1. Viewing the transition matrix associated to the pseudo-Anosov braid δ n σ 2k 1 , we see that the largest eigenvalue of the transition matrix, which is equal to λ(δ n σ 2k 1 ), goes to ∞ as n goes to ∞. Proof of Theorem 6.1. We fix a basepoint of π 1 (D n ) in the interior of D n . By (4.2) we have the automorphism We turn to braid σ 2k 1 ∈ B n . By (4.1) we have the following automorphism induced by σ 1 ∈ B n using the same basepoint of π 1 (D n ): 2 , x j → x j for j = 1, 2. Computing the automorphism σ 2 1 on F n , we get σ 2 1 : In the same manner, the automorphism σ 2k 1 for k ≥ 1 is given by σ 2k 1 : Thus δ n σ 2k 1 induces the following automorphism on F n . φ := δ n σ 2k 1 : We assume that φ preserves some bi-ordering < on F n . Without loss of generality, we may assume that x 1 < x 2 . Then we have by conjugation invariance of the bi-ordering. This implies that , since φ −1 also preserves the same bi-ordering <. Hence we have x 1 < x 2 < x 3 . Notice that x 3 is contained in the orbit O(x 1 ) under φ, i.e, 2 ) k → · · · Since x 1 < x 3 , this implies that the ordering < is increasing on O(x 1 ), i.e, This together with x 1 < x n tells us that (6.1) We multiply the both sides of (6.1) by (x 2 x 1 ) k−1 x 2 on the right.
Multiply the both sides of (6.2) by (x 2 x 1 ) k on the right.
We turn to the last sequence of braids which plays an important role for the study of pseudo-Anosov minimal dilatations, see [21,26]. Theorem 6.7. For each n ≥ 5, The braid δ 2 n σ −1 n−1 σ −1 n−2 is reducible if n is even. (In fact it is easy to find an essential simple closed curve on D n containing the punctures labelled 2, 4, · · · , n − 2 which is invariant under the corresponding mapping class of Mod(D n ).) If n is odd, the braid δ 2 n σ −1 n−1 σ −1 n−2 is pseudo-Anosov, see [21,Theorem 3.11, Figure 18(c)]. Proof of Theorem 6.7. We fix a basepoint of π 1 (D n ) in the interior of D n . From the automorphism σ i of F n for each i = 1, · · · , n − 1 (see (4.1)), we calculate the automorphism σ −1 i on F n : 2) together with automorphisms σ −1 i for i = n − 1, n − 2, we see that the following automorphism is arisen from the braid δ n σ −1 Figure 20. Thus the automorphism ψ := δ n (δ n σ −1 n−1 σ −1 n−2 ) : F n → F n is given by x n → x −1 n−2 x n−1 x n−2 . Suppose that n is even. Then the orbit of x 2 under ψ is nontrivial and finite: Thus ψ does not preserve any left-ordering on F n by Proposition 3.11.
Suppose that n is odd. Then the above automorphism ψ : F n → F n is described as follows.
We suppose that ψ preserves a bi-ordering < of F n . Without of loss of generality, we may assume that x n−1 < x n . By the conjugation invariance of the bi-ordering, x −1 n−2 x n−1 x n−2 < x −1 n−2 x n x n−2 holds. Since ψ −1 preserves the same ordering <, this implies that x n < x 1 , see the above definition of ψ. Hence we have x n−1 < x n < x 1 . This tells us that the ordering < is increasing on the orbit O(x n−1 ), i.e, x n−1 < x n−3 < · · · < x 2 < x n−2 < x n−4 < x 1 < x −1 n−2 x n x n−2 < · · · In particular x n−2 < x −1 n−2 x n x n−2 which implies that x n−2 < x n by the conjugation invariance of the bi-ordering. This together with the ψ-invariance of < gives us that x n−4 < x −1 n−2 x n−1 x n−2 . On the other hand, by x n−1 < x n−2 , we have x −1 n−2 x n−1 x n−2 < x n−2 by the conjugation invariance again. However we have x n−2 < x n−4 < x −1 n−2 x n−1 x n−2 , in particular x n−2 < x −1 n−2 x n−1 x n−2 (< x n−2 ), which is a contradiction. This completes the proof.
Finally we claim that the following pseudo-Anosov braid with the minimal dilatation δ(D 8 ) is not order-preserving.
If (m, n) = (0, 0), braids γ m,n 's are non-pure and order-preserving. By Proposition 3.14, the braid γ m,n preserves some bi-ordering which is not standard ordering of the free group. Notice that for m, n ≥ 0, the permutation of S m+n+3 associated to γ m,n has more than 1 cycle, i.e, the closure of the braid γ m,n is a link with more than 1 component. We ask the following. Question 6.10. Does there exist an order-preserving braid β ∈ B n (n ≥ 3) whose permutation is cyclic (i.e, the closure β is a knot)?
Appendix A. Ordering free groups For any group G the lower central series G = γ 1 G ⊃ γ 2 G ⊃ · · · is defined inductively by the formula γ k+1 G := [G, γ k G]. It is well known (see for example [30]) that if F is a free group (or hyperbolic surface group) then the lower central quotients γ k F/γ k+1 F are free abelian of finite rank and also that ∩ ∞ k=1 γ k F = {1}. Following [31] one can bi-order F as we will now describe. Choose an arbitrary bi-order < k of the (free abelian) group γ k F/γ k+1 F and define the positive cone for an ordering < of F as follows. For 1 = g ∈ F , declare g to be positive if 1 < k [g] in γ k F/γ k+1 F , where k is the unique integer for which g ∈ γ k F \ γ k+1 F . It is routine to check that this defines a bi-ordering < of F . We shall say that an ordering of F defined in this way is a standard ordering of F . If the rank of F is greater than one, there are uncountably many standard orderings. However there are uncountably many non-standard orderings of F as well, for example as in the end of Section 4.6.
Proposition A.1. For any standard ordering of the free group F , all the lower central subgroups γ k F are convex.
Proof. Using invariance under multiplication it is sufficient to suppose 1 < f < g and g ∈ γ k F , and show that f ∈ γ k F . Now f ∈ γ j F \ γ j+1 F for some unique positive integer j. Assume for contradiction that j < k. By the definition of < we have that 1 < j [f ] in γ j F/γ j+1 F . But since j < k we also have g ∈ γ j F and we see that , which implies the contradiction that 1 < j [f −1 ]. Therefore we conclude that j ≥ k and therefore f ∈ γ j F ⊂ γ k F .
Note that for any automorphism of a group φ : G → G we have φ(γ k G) = γ k G and so there are induced automorphisms φ k : γ k G/γ k+1 G → γ k G/γ k+1 G. In particular, φ 1 is the same as the abelianization φ ab : G/[G, G] → G/[G, G]. The following is a well-known group theoretic result, but we include a proof, suggested by Thomas Koberda, for the reader's convenience.
Lemma A.2. Suppose φ : G → G is an automorphism of a group G such that the abelianization φ ab : G/[G, G] → G/[G, G] is the identity map. Then for each k ≥ 1, the homomorphism φ k : γ k G/γ k+1 G → γ k G/γ k+1 G is also the identity map.
Proof. We prove φ k = id by induction on k, the hypothesis being the base case. Note that the hypothesis implies that, for every g ∈ G we have φ(g) = gc where c ∈ γ 2 G may depend upon g. Now, suppose that φ k : γ k G/γ k+1 G → γ k G/γ k+1 G is the identity. This implies that for h ∈ γ k G we have that φ(h) = hd, where d ∈ γ k+1 G, also dependent upon h.
To show that φ k+1 = id, consider any nonidentity element of γ k+1 G. Such an element is a product of commutators of the form [g, h] with g ∈ G, h ∈ γ k G, so it suffices to show that the latter equivalence following because d ∈ γ k+1 G commutes with every element of G, modulo γ k+2 G. But note that Observe that the expression in parentheses is a commutator of h −1 ∈ γ k G and c ∈ γ 2 G. It is well-known (see for example [30, page 293]) that [γ m G, γ n G] ⊂ γ m+n G. Therefore, h −1 chc −1 ∈ γ k+2 G. Being a conjugate, x also belongs to γ k+2 G, and therefore φ( Proposition A.3. Suppose φ : F n → F n is such that φ ab : Z n → Z n is the identity map φ ab = id : Z n → Z n . Then φ preserves every standard bi-ordering of F n .
Proof. By Lemma A.2, with G = F n we see that φ k is the identity automorphism of γ k F n /γ k+1 F n and therefore preserves every ordering of γ k F n /γ k+1 F n . It follows that φ preserves the positive cone of every standard ordering of F n . (1) (2) Let W be the Whitehead link in S 3 , see Figure 22(1). Our goal is to prove: Theorem B.1. The fundamental group of the Whitehead link complement W = S 3 \ W is bi-orderable.
We first recall the Murasugi sum of surfaces. See [25, Section 4.2] for more details. Let R 1 , R 2 and R be compact oriented surface embedded in S 3 . We say that R is a (2n-)Murasugi sum of R 1 and R 2 if we have the following.
(1) R = R 1 ∪ R 2 , and R 1 ∩ R 2 is a disk D which satisfies the following.
(1.1) ∂D is a 2n-gon with edges a 1 , b 1 , a 2 , b 2 , · · · , a n , b n enumerated in this order. (1.2) a i ⊂ ∂R 1 , and a i is a proper arc in R 2 for all i. Theorem B.2 (Theorem 1.3 and Corollary 1.4 in Gabai [13]). Suppose that R is a Murasugi sum of R 1 and R 2 . Then L is a fibered link with a fiber R if and only if L i is a fibered link with a fiber R i for i = 1, 2.
Let R i , R and L i be as in Theorem B.2, and let f i : R i → R i denote the monodromy. We may assume that f i | ∂R i equals the identity map id. Let R be a Murasugi sum of R 1 and R 2 . By Theorem B.2, L = ∂R is a fibered link with a fiber R if L i is a fibered link with a fiber R i for i = 1, 2. The following theorem tells us what the monodromy f : R → R looks like.  (1) (3) Convention B.4. The product f ′ 2 f ′ 1 means that we first apply f ′ 2 , then apply f ′ 1 . A Hopf band is an unknotted annulus in S 3 . Two kinds of hopf bands S + and S − as in Figure 23(1) and (2) are called positive and negative respectively. The links L + = ∂S + and L − = ∂S − are called the Hopf links. It is known that L ± is a fibered link with a fiber S ± . The monodromy f + : S + → S + (resp. f − : S − → S − ) is the right handed Dehn twist (resp. left handed Dehn twist) about the core circle of the annulus, see [14, Figure 1].
Proof of Theorem B.1. We will find a fibered surface S of the Whitehead link W and its monodromy f : S → S. LetŜ be the 2-Murasugi sum of hopf bands S + and S − , see Figure 24(1). Figure 24(3) is a surface which is isotopic toŜ. By Theorem B.2, ∂Ŝ is a fibered link with a fiberŜ. Theorem B.3 tells us that f ′ + f ′ − :Ŝ →Ŝ serves the monodromyf :Ŝ →Ŝ. Figure 25(1)(2) shows the images of two proper arcs (solid and broken arcs) underf :Ŝ →Ŝ. This is the so-called point-pushing map, see [12,Section 4.2].
A 4-Murasugi sum ofŜ and S + gives rise to a fibered surface S of W , which is a torus with 2 boundary components, see Figure 26. By Theorem B.3, the product (f ) ′ f ′ + : S → S serves the monodromy f : S → S. Note that (f ) ′ : S → S is a pushing map along the arc m, and f ′ + : S → S is the right handed Dehn twist about a simple closed curve ℓ, see Figure 27(1).
Shrinking one of the boundary components to a puncture, one can take a simple model as a representative of S, which is a torus with one boundary component and with a puncture as in Figure 27(2). Abusing the notation, we denote such a simple model by the same notation S. We also denote the corresponding loop based at the puncture and the corresponding closed loop in the simple model by the same notations m and ℓ, see Figure 27(2).
We choose a basepoint p on ∂S, and take oriented loops c 0 , c 1 and c 2 based at this point so that c 0 is a loop surrounding the puncture, and c 1 , c 2 are the meridian and the longitude of the torus, see Figure 28(1). Abusing the notations, we denote the equivalence class of c i in π 1 (S, p) = F 3 by the same notation c i . (Then {c 0 , c 1 , c 2 } is a generating set of π 1 (S, p).) The images of c 1 , c 2 and c 3 under   By calculation one sees that the characteristic polynomial of (f * ) ab equals (t − 1) 3 , and all eigenvalues of (f * ) ab are 1. By Theorem 3. 12, it follows that f * is orderpreserving. Note that π 1 (W) = F 3 ⋊ f * Z. By Proposition 3.10, we conclude that π 1 (W) is bi-orderable.