NON-LEFT-ORDERABLE SURGERIES ON 1-BRIDGE BRAIDS

arXiv:1711.11389v1 [math.GT] 30 Nov 2017

SHIYU LIANG

Abstract. Boyer, Gordon, and Watson [BGW13] have conjectured that an
irreducible rational homology 3-sphere is an L-space if and only if its fundamental group is not left-orderable. Since Dehn surgeries on knots in S 3
can produce large families of L-spaces, it is natural to examine the conjecture on these 3-manifolds. Greene, Lewallen, and Vafaee [GLV16] have proved
that all 1-bridge braids are L-space knots. In this paper, we consider three
families of 1-bridge braids. First we calculate the knot groups and peripheral subgroups. We then verify the conjecture on the three cases by applying
the criterion [CGHV16] developed by Christianson, Goluboff, Hamann, and
Varadaraj, when they verified the same conjecture for certain twisted torus
knots and generalized the criteria in [CW13] and [IT15].

1. Introduction
c ) the “hat” version of
Let Y be a rational homology sphere, and denote by HF(Y
Heegaard Floer homology, as defined in [OS04b]. The following result is shown by
d (Y ) ≥ |H1 (Y ; Z)|. As a space with minimal
Ozsváth and Szabó in [OS04a]: rk HF
Heegaard Floer homology, an L-space is defined as follows:
Definition 1. A closed, connected, orientable 3-manifold Y is an L-space if it is a
rational homology sphere with the property
d (Y ) = |H1 (Y ; Z)|.
rk HF
It is interesting that L-spaces might be characterized by properties of their fundamental groups, which seem to be unrelated to Heegaard Floer homology. Recall
the following definition:
Definition 2. A non-trivial group G is called left-orderable if there exists a strict
total ordering < on G which is left-invariant, i.e.,
g < h ⇒ f g < f h, ∀f, g, h ∈ G.
The identity element is always denoted by symbol 1 in this paper, and the
symbols >, ≤, and ≥ have the usual meaning.
In [BGW13], Boyer, Gordon, and Watson make the following conjecture that
indicates a connection between L-spaces and left-orderability of their fundamental
groups.
Conjecture 3 ([BGW13, Conjecture 3]). An irreducible rational homology 3-sphere
is an L-space if and only if its fundamental group is not left-orderable.
2010 Mathematics Subject Classification. 57M25 (20F60 57M50).
Key words and phrases. left-orderability, 1-bridge braid, Dehn surgery.
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Figure 1. The (4, 2, 1)-1-bridge braid, denoted by B(4, 2, 1).
Since Dehn surgeries on knots in S 3 provide large families of 3-manifolds, it is
natural to consider if the conjecture can be verified on them. The concept of an
L-space knot is needed to simplify our discussion:
Definition 4. An knot K in S 3 is an L-space knot if it admits some non-trivial
Dehn surgery yielding an L-space.
Conjecture 3 has been verified for certain families of Dehn surgeries. For instance,
it has been verified for sufficiently large surgery on some twisted torus knots in
[CGHV16, Theorem 14]. Our goal of this paper is to verify the conjecture on
another similar family of knots. The specific family of knots that will be worked on
is 1-bridge braids, which are first studied by Berge and Gabai in [Ber] and [Gab90],
and are a natural subset of a broad family of (1, 1) knots that are L-space knots
([GLV16]). They are defined as follows:
Definition 5 ([GLV16, Definition 1.3]). A knot in the solid torus D2 × S 1 is a
1-bridge braid if it is isotopic to a union of two arcs ρ ∪ τ such that
• ρ ⊂ ∂(D2 × S 1 ) is braided, i.e., transverse to each meridian ∂D2 × pt, and
• τ is a bridge, i.e., properly embedded in some meridional disk D2 × pt.
It is positive if ρ is a positive braid in the usual sense. A knot in S 3 is a 1-bridge
braid if it is isotopic to a 1-bridge braid supported in a solid torus coming from a
genus-1 Heegaard splitting of S 3 .
To present a 1-bridge braid, let
Bω = hσ1 , σ2 , · · · , σω−1 |σi σi+1 σi = σi+1 σi σi+1 , 1 ≤ i ≤ ω − 2,
σj σk = σk σj , 1 ≤ j, k ≤ ω − 1, |j − k| ≥ 2i
denote the braid group with ω strands, where σi gives strands i, i + 1 a right-hand
half twist, and the braids are composed from right to left. It follows from Gabai’s
classification theorem of 1-bridge braids in the solid torus ([Gab90, Proposition 2.3])
that every 1-bridge braid in S 3 is of the form B(ω, t + mω, b) = B(ω, t + mω, b),
i.e., the braid closure of
B(ω, t + mω, b) = (σ1 σ2 · · · σb )(σ1 σ2 · · · σω−1 )t+mω ∈ Bω ,
1 ≤ b ≤ ω − 2, 1 ≤ t ≤ ω − 2, m ∈ Z.
We will be interested in the case that B(ω, t + mω, b) is a knot rather than a link.
J. and S. Rasmussen conjectured that every 1-bridge braid is an L-space knot at
the end of [RR15], and then Greene, Lewallen, and Vafaee proved a more general
result, of which the following is a special case:

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Theorem 6 ([GLV16, Theorem 1.4]). All 1-bridge braids in S 3 are L-space knots.
In particular, positive 1-bridge braids admit positive L-space surgeries. We will
verify Conjecture 3 on three families of 1-bridge braids. We first claim that they
are all knots, and the proof will be given in Section 2.
Proposition 7. B(ω, t + mω, b) is a knot if
t = 1, b = 2k, 1 ≤ k ≤ [(ω − 2)/2].
Proposition 8. B(ω, t + mω, b) is a knot if
ω = 2n + 1, t = 2n − 1, b = 2k, 1 ≤ k ≤ n − 1.
Proposition 9. B(ω, t + mω, b) is a knot if
ω = 2n, t = 2n − 2, b = 2k − 1, 1 ≤ k ≤ n − 1.
The following result due to Ozsváth and Szabó completely characterizes which
surgeries on an L-space knot yield L-spaces.
Theorem 10 ([OS11]). Let K ⊂ S 3 , and suppose that there exists
3
Sp/q
(K)

that
2g(K) − 1.

is an L-space. Then

Sp30 /q0 (K)

p
q

∈ Q≥0 such

is an L-space if and only if

p0
q0

≥

3
In the theorem, Sp/q
(K) denotes p/q-surgery on knot K in S 3 . In order to
provide a sufficient condition on the knot group of K in S 3 to imply that r-surgery
on K yields a manifold with non-left-orderable fundamental group, the following
equivalent condition for left-orderability is required.

Theorem 11 ([Ghy01, Theorem 6.8]). Let G be a countable group. Then the
following are equivalent:
• G acts faithfully on the real line by order-preserving homeomorphisms.
• G is left-orderable.
With this theorem, Christianson, Goluboff, Hamann, and Varadaraj generalized
Ichihara and Temma’s result in [IT15], and got the following criterion for non-leftorderability:
Theorem 12 ([CGHV16, Theorem 10]). Let K be a non-trivial knot in S 3 . Let
G denote the knot group of K, and let G(p/q) be the quotient of G resulting from
p/q-surgery. Let µ be a meridian of K and s be a v-framed longitude with v > 0.
Suppose that G has two generators, x and y, such that x = µ and s is a word which
excludes x−1 and y −1 and contains at least one x. Suppose further that every
homomorphism Φ : G(p/q) → Homeo+ (R) satisfies Φ(x)t > t for all t ∈ R implies
Φ(y)t ≥ t for all t ∈ R. If p, q > 0, then, for p/q ≥ v, G(p/q) is not left-orderable.
We will use similar notations to [CGHV16] and work with the usual order on
R. Denote by M (ω, t + mω, b, r) the 3-manifold produced by r-surgery on B(ω, t +
mω, b), and let G(ω, t + mω, b) = π1 (S 3 \ B(ω, t + mω, b)), G(ω, t + mω, b, r) =
π1 (M (ω, t + mω, b, r)). Throughout, we will assume ω ≥ 3, 1 ≤ t ≤ ω − 2, 1 ≤ b ≤
ω − 2, m ≥ 0, r ∈ Q≥0 .
In this paper, we apply Theorem 12 to prove the following result.
Theorem 13. For sufficiently large r, the L-spaces produced by r-surgery on B(ω, t+
mω, b), where ω, t, and b satisfy the conditions in Propositions 7, 8, 9, and m ≥ 0,

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Figure 2. The braid B(5, 1, 2), which induces the permutation (1, 3, 4, 5, 2).
all have non-left-orderable fundamental groups, i.e., G(ω, 1 + mω, 2k, r), G(2n +
1, 2n−1+m(2n+1), 2k, r), and G(2n, 2n−2+2mn, 2k−1, r) are non-left-orderable if
r ≥ ω+2k−1, r ≥ 2n[2n−1+m(2n+1)]+2k, and r ≥ (2n−1)(2n−2+2mn)+2k−1
respectively.
Acknowledgements. We thank Jennifer Hom and Sudipta Kolay for providing
the resources and background material necessary for our work and for their guidance
throughout. We also thank the School of Mathematics of Georgia Institute of
Technology for their support and giving us the opportunity to pursue this research
project. This REU program was partially funded by NSF grant DMS-1552285.
2. Proving All the Three Families of 1-bridge Braids are Knots
First, we define a surjective group homomorphism h : Bω → Sω by mapping
every braid σi to si in symmetric group
Sω = hs1 , s2 , · · · , sω−1 |si si+1 si = si+1 si si+1 , 1 ≤ i ≤ ω − 2,
sj sk = sk sj , s2j = 1, 1 ≤ j, k ≤ ω − 1, |j − k| ≥ 2i.
For any braid σ ∈ Bω , h(σ) is called the permutation induced by σ, so σ acts on the
ω endpoints of the braid by permutation h(σ). The permutations are composed
from right to left. Denote h(B(ω, t + mω, b))(i) briefly by B(ω, t + mω, b)(i).
In this section, we prove Propositions 7, 8, and 9.
Proof of Proposition 7. We claim that B(ω, 1 + mω, 2k) induces the action on the
endpoints by the permutation
p = (1, 3, 5, · · · , 2k − 1, 2k + 1, 2k + 2, 2k + 3, · · · , ω, 2, 4, 6, · · · , 2k),
by forgetting how the strands twist and cross, i.e., if strand i is sent to be strand j
by B(ω, 1 + mω, 2k), then p(i) = j (see Figure 2).
In fact, B(ω, 1 + mω, 2k) has the expression:
B(ω, 1 + mω, 2k) = (σ1 σ2 · · · σ2k )(σ1 σ2 · · · σω−1 )1+mω .
If 1 ≤ i ≤ 2k − 1 is odd, (σ1 σ2 · · · σω−1 )1+mω (i) = i + 1 ≤ 2k, and (σ1 σ2 · · · σ2k )(i +
1) = i + 2, so B(ω, 1 + mω, 2k)(i) = i + 2 = p(i).
Similarly, if 2k + 1 ≤ i ≤ ω − 1, (σ1 σ2 · · · σω−1 )1+mω (i) = i + 1 ≥ 2k + 2 > 2k + 1,
so (σ1 σ2 · · · σ2k )(i + 1) = i + 1. Thus, B(ω, 1 + mω, 2k)(i) = i + 1 = p(i).
For strand ω, (σ1 σ2 · · · σω−1 )1+mω (ω) = 1, and then (σ1 σ2 · · · σ2k )(1) = 2, so
B(ω, 1 + mω, 2k)(ω) = 2 = p(ω).

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Figure 3. The braid B(5, 3, 2).
If 2 ≤ i ≤ 2k − 2 is even, (σ1 σ2 · · · σω−1 )1+mω (i) = i + 1 ≤ 2k − 1, and
(σ1 σ2 · · · σ2k )(i + 1) = i + 2, so B(ω, 1 + mω, 2k)(i) = i + 2 = p(i).
Finally, for strand 2k, (σ1 σ2 · · · σω−1 )1+mω (2k) = 2k + 1, and (σ1 σ2 · · · σ2k )(2k +
1) = 1, so B(ω, 1 + mω, 2k)(2k) = 1 = p(2k).
Therefore, B(ω, 1 + mω, 2k) is a knot.

Proof of Proposition 8. We claim that B(2n + 1, 2n − 1 + m(2n + 1), 2k) induces
the action on the endpoints by the permutation
p = (1, 2n, 2n−2, 2n−4, · · · , 2k+2, 2k+1, 2k, · · · , 2, 2n+1, 2n−1, 2n−3, · · · , 2k+3),
as shown in Figure 3.
In fact, B(2n + 1, 2n − 1 + m(2n + 1), 2k) has the expression:
B(2n + 1, 2n − 1 + m(2n + 1), 2k) = (σ1 σ2 · · · σ2k )(σ1 σ2 · · · σ2n )2n−1+m(2n+1) .
(σ1 σ2 · · · σ2n )2n−1+m(2n+1) (1) = 2n. Then, since n ≥ k + 1, 2n ≥ 2k + 2 > 2k + 1,
(σ1 σ2 · · · σ2k )(2n) = 2n. Thus, B(2n + 1, 2n − 1 + m(2n + 1), 2k)(1) = 2n = p(1).
For strand 2k + 4 ≤ i ≤ 2n an even number, (σ1 σ2 · · · σ2n )2n−1+m(2n+1) (i) ≡
i + 2n − 1(mod 2n + 1). Since 1 < 2k + 2 ≤ i + 2n − 1 − ω = i − 2 ≤ 2n − 2 < ω,
(σ1 σ2 · · · σ2n )2n−1+m(2n+1) (i) = i − 2. Then, since i − 2 ≥ 2k + 2 > 2k + 1,
(σ1 σ2 · · · σ2k )(i − 2) = i − 2. Therefore, B(2n + 1, 2n − 1 + m(2n + 1), 2k)(i) =
i − 2 = p(i).
For strand 3 ≤ i ≤ 2k+2, (σ1 σ2 · · · σ2n )2n−1+m(2n+1) (i) ≡ i+2n−1(mod 2n+1).
Since 1 ≤ i + 2n − 1 − ω = i − 2 ≤ 2k < ω, (σ1 σ2 · · · σ2n )2n−1+m(2n+1) (i) = i − 2.
Then, since 1 ≤ i − 2 ≤ b, (σ1 σ2 · · · σ2k )(i − 2) = i − 1. Therefore, B(2n + 1, 2n −
1 + m(2n + 1), 2k)(i) = i − 1 = p(i).
For strand 2, (σ1 σ2 · · · σ2n )2n−1+m(2n+1) (2) = 2n+1, and (σ1 σ2 · · · σ2k )(2n+1) =
2n + 1, so B(2n + 1, 2n − 1 + m(2n + 1), 2k)(2) = 2n + 1 = p(2).
For strand 2k + 5 ≤ i ≤ 2n + 1 an odd number, (σ1 σ2 · · · σ2n )2n−1+m(2n+1) (i) ≡
i + 2n − 1(mod 2n + 1). Since 1 ≤ 2k + 3 ≤ i + 2n − 1 − ω = i − 2 ≤ 2n − 1 < ω,
(σ1 σ2 · · · σ2n )2n−1+m(2n+1) (i) = i − 2. Then, since i − 2 ≥ 2k + 3 > 2k + 1,
(σ1 σ2 · · · σ2k )(i − 2) = i − 2. Therefore, B(2n + 1, 2n − 1 + m(2n + 1), 2k)(i) =
i − 2 = p(i).
Finally, for strand 2k + 3, we have (σ1 σ2 · · · σ2n )2n−1+m(2n+1) (2k + 3) ≡ 2k +
2n + 2(mod 2n + 1). Then, since 1 ≤ 2k + 2n + 2 − ω = 2k + 1 ≤ 2n − 1 < ω,
(σ1 σ2 · · · σ2n )2n−1+m(2n+1) (2k + 3) = 2k + 1. Since (σ1 σ2 · · · σ2k )(2k + 1) = 1,
B(2n + 1, 2n − 1 + m(2n + 1), 2k)(2k + 3) = 1 = p(2k + 3).
Therefore, B(2n + 1, 2n − 1 + m(2n + 1), 2k) is a knot.


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Figure 4. The braid B(6, 4, 3).
Proof of Proposition 9. We claim that B(2n, 2n−2+2mn, 2k−1) induces the action
on the endpoints by the permutation
p = (1, 2n−1, 2n−3, 2n−5, · · · , 2k+1, 2k, 2k−1, · · · , 2, 2n, 2n−2, 2n−4, · · · , 2k+2),
as shown in Figure 4.
In fact, since B(2n, 2n − 2 + 2mn, 2k − 1) has the expression:
B(2n, 2n − 2 + 2mn, 2k − 1) = (σ1 σ2 · · · σ2k−1 )(σ1 σ2 · · · σ2n−1 )2n−2+2mn ,
(σ1 σ2 · · · σ2n−1 )2n−2+2mn (1) = 2n + 1. Then, since 2n − 1 ≥ 2k + 1 > 2k,
(σ1 σ2 · · · σ2k−1 )(2n + 1) = 2n + 1. Thus, B(2n, 2n − 2 + 2mn, 2k − 1)(1) = 2n + 1 =
p(1).
For strand 2k + 3 ≤ i ≤ 2n − 1 an odd number, (σ1 σ2 · · · σ2n−1 )2n−2+2mn (i) ≡
i + 2n − 2(mod 2n). Since 1 < 2k + 1 ≤ i + 2n − 2 − ω = i − 2 ≤ 2n − 3 < ω,
(σ1 σ2 · · · σ2n−1 )2n−2 (i) = i−2. Then, since i−2 ≥ 2k +1 > 2k, (σ1 σ2 · · · σ2k−1 )(i−
2) = i − 2. Therefore, B(2n, 2n − 2 + 2mn, 2k − 1)(i) = i − 2 = p(i).
The remainder of the proof of the claim follows analogously to the proof of
Proposition 8.
Therefore, B(2n, 2n − 2 + 2mn, 2k − 1) is a knot.

3. Computing Knot Groups and Peripheral Subgroups
It is a fact that (see [Lic97]), for a non-trivial knot, the fundamental group of the
boundary of the knot complement injects into the knot group, and its image up to
conjugation is the peripheral subgroup, so all peripheral subgroups are abelian. We
first derive the knot groups of B(ω, 1 + mω, 2k), B(2n + 1, 2n − 1 + m(2n + 1), 2k),
and B(2n, 2n − 2 + 2mn, 2k − 1). We will achieve this for all three cases by the same
way both Clay and Watson in [CW13] and Christianson et al. in [CGHV16] did: by
finding a genus-2 Heegaard splitting of S 3 with the knot embedded on the Heegaard
surface, we can apply Seifert-van Kampen Theorem on the two handlebodies of
genus 2 (see Figures 6, 10, and 14).
Proposition 14. For B(ω, 1 + mω, 2k),
(a) The knot group is
G(ω, 1 + mω, 2k) = hα, β|(βα)k β = {(βα)k β ω−2k [(βα)k βα−1 (αβ)−k ]−ω+k }m (αβ)k αi.

(b) The peripheral subgroup is generated by the meridian
µ=α

NON-LEFT-ORDERABLE SURGERIES ON 1-BRIDGE BRAIDS

Figure 5. Basepoint x0 and generators α, β, δ, γ for the fundamental group of the Heegaard surface Σ.

Figure 6. The red graph is homotopy equivalent to the complement of B(ω, 1 + mω, 2k) on the Heegaard surface Σ. Here, we
show B(5, 1, 2) as an example. The knot is shown in gray.

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Figure 7. The first generator for the fundamental group of Σ \
ν(B(ω, 1 + mω, 2k)).
and the surface framing
s = µ(ω−1)(1+mω)+2k λ = α(βα)k β ω−2k (αβ)k .
Proof. Let S 3 = U ∪Σ V be the genus-2 Heegaard splitting of S 3 with the knot
B(ω, 1 + mω, 2k) embedded on the Heegaard surface, as shown in Figure 6. Then
π1 (U ) is the free group on the generators α and β, and π1 (V ) is the free group
on the generators δ and γ (see Figure 5). Using Seifert-van Kampen Theorem, we
can then express G(ω, 1 + mω, 2k) as a free product with amalgamation of π1 (U )
and π1 (V ). In order to figure out the amalgamation, we need the images of the
generators of π1 (Σ \ ν(B(ω, 1 + mω, 2k))) under the map induced by inclusion into
π1 (U ) and π1 (V ).
Now, Σ\ν(B(ω, 1+mω, 2k)) is homotopy equivalent to a twice-punctured genus-1
surface whose fundamental group is generated by the green, purple, and olive loops
in Figures 7, 8, and 9. The green loop has image α(βα)k β ω−2k α−1 in π1 (U ) and
image γ m(ω−k)+1 in π1 (V ), so we get the relation
α(βα)k β ω−2k α−1 = γ m(ω−k)+1 .
Likewise, the purple loop gives
α(βα)k βα−1 = γ m(k+1) δ,
and the olive loop gives
α(αβ)k = γ mk δ.
The second and the third relation give us the expression for γ m in terms of α and
β:
γ m = α(βα)k βα−1 (αβ)−k α−1 .

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Figure 8. The second generator for the fundamental group of
Σ \ ν(B(ω, 1 + mω, 2k)).
Then by the first relation, replacing γ m(ω−k) of the right side, we can get a expression for γ in terms of α and β:
γ = α(βα)k β ω−2k α−1 γ −m(ω−k) = α(βα)k β ω−2k α−1 [α(βα)k βα−1 (αβ)−k α−1 ]−ω+k
= α(βα)k β ω−2k [(βα)k βα−1 (αβ)−k ]−ω+k α−1 .
Also, from the third relation, we can write δ in terms of α and β:
δ = γ −mk α(αβ)k = [α(βα)k βα−1 (αβ)−k α−1 ]−k α(αβ)k
= α[(βα)k βα−1 (αβ)−k ]−k (αβ)k
We can substitute these three relations for the original ones. The last two relations
provide us with expressions for γ and δ in terms of α and β. By getting rid of γ of
the expression for γ m , we are left with only one relation:
(βα)k β = {(βα)k β ω−2k [(βα)k βα−1 (αβ)−k ]−ω+k }m (αβ)k α.
Thus, we have
G(ω, 1 + mω, 2k) = hα, β|(βα)k β = {(βα)k β ω−2k [(βα)k βα−1 (αβ)−k ]−ω+k }m (αβ)k αi.

For the peripheral subgroup, we will compute the meridian µ and the surface
framing s which is a push-off of the knot B(ω, 1 + mω, 2k) in Σ. It is clear that

s = γ 1+mω δ
= α{(βα)k β ω−2k [(βα)k βα−1 (αβ)−k ]−ω+k }1+mω [(βα)k βα−1 (αβ)−k ]−k (αβ)k .
Using the relation in the expression of G(ω, 1 + mω, 2k), s can be simplified as
follows
s = γ 1+mω δ = α(βα)k β ω−2k (αβ)k .

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Figure 9. The third generator for the fundamental group of Σ \
ν(B(ω, 1 + mω, 2k)).
In order to compute µ, consider the meridian based at x0 . It can be isotoped
to be sitting in the handlebody U , and then it goes around the right genus of the
handlebody once, so is isotopic to the generator α. Thus,
µ = α.
Finally, we note that the linking number between B(ω, 1 + mω, 2k) and a pushoff in Σ, by the construction of the 1-bridge braid, is (ω − 1)(1 + mω) + 2k, which
gives us
s = µ(ω−1)(1+mω)+2k λ.

Proposition 15. For B(2n + 1, 2n − 1 + m(2n + 1), 2k),
(a) The knot group is
G(2n + 1, 2n − 1 + m(2n + 1), 2k)
= hα, β|[(αβ n−k+1 αβ −n+k )n+k (αβ)−2k+1 β −n+k−1 ]m+1 = αβ n−k+1 αβ −n+k i.
(b) The peripheral subgroup is generated by the meridian
µ=α
and the surface framing
s = µ2n[2n−1+m(2n+1)]+2k λ = αβ n−k+1 (αβ)2k−1 αβ n−k+1 .
Proof. Let S 3 = U ∪Σ V again be the genus-2 Heegaard splitting of S 3 specified in
Figure 10. Then π1 (U ) is the free group on the generators α and β, and π1 (V ) is
the free group on the generators δ and γ (see Figure 5). Using Seifert-van Kampen
Theorem, we can then write G(2n + 1, 2n − 1 + m(2n + 1), 2k) as a free product

NON-LEFT-ORDERABLE SURGERIES ON 1-BRIDGE BRAIDS

11

Figure 10. The red graph is homotopy equivalent to the complement of B(2n + 1, 2n − 1 + m(2n + 1), 2k) on the Heegaard surface
Σ. Here, we show B(5, 3, 2) as an example. The knot is shown in
gray.

with amalgamation of π1 (U ) and π1 (V ). To work out the amalgamation, we need
the images of the generators of π1 (Σ \ ν(B(2n + 1, 2n − 1 + m(2n + 1), 2k))) under
the maps induced by inclusion into π1 (U ) and π1 (V ).
Now, Σ \ ν(B(2n + 1, 2n − 1 + m(2n + 1), 2k)) is homotopy equivalent to a twicepunctured genus-1 surface whose fundamental group is generated by the green,
purple, and olive loops in Figures 11, 12, and 13. The image of the green loop
is αβ n−k+1 (αβ)2k−1 α−1 under the map induced by inclusion to π1 (U ), and is
γ n+k−1+m(n+k) under the map induced by inclusion to π1 (V ), so we get the relation
αβ n−k+1 (αβ)2k−1 α−1 = γ (m+1)(n+k)−1 .
Likewise, from the purple loop we get
αβ n−k α−1 = γ (m+1)(n−k)−1 δ,
and from the olive loop we get
α2 β n−k+1 = γ (m+1)(n−k+1)−1 δ.
Using the second and the third relation, we have
α2 β n−k+1 = γ (m+1)(n−k+1)−1 δ = γ m+1 γ (m+1)(n−k)−1 δ = γ m+1 αβ n−k α−1 .
⇔ γ m+1 = α2 β n−k+1 αβ −n+k α−1 .

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Figure 11. The first generator for the fundamental group of Σ \
ν(B(2n + 1, 2n − 1 + m(2n + 1), 2k)).

Figure 12. The second generator for the fundamental group of
Σ \ ν(B(2n + 1, 2n − 1 + m(2n + 1), 2k)).

NON-LEFT-ORDERABLE SURGERIES ON 1-BRIDGE BRAIDS

13

Figure 13. The third generator for the fundamental group of Σ \
ν(B(2n + 1, 2n − 1 + m(2n + 1), 2k)).
This relation can replace the second relation. Moreover, it provides us with an
expression for γ m+1 in terms of α and β, so with the first relation, we have
γ = γ (m+1)(n+k) α(αβ)−2k+1 β −n+k−1 α−1
= (α2 β n−k+1 αβ −n+k α−1 )n+k α(αβ)−2k+1 β −n+k−1 α−1
= α(αβ n−k+1 αβ −n+k )n+k (αβ)−2k+1 β −n+k−1 α−1 ,
which can replace the first relation. We can also write δ in terms of α and β by
replacing γ of the second relation with its expression in terms of α and β:
δ = γ −(m+1)(n−k)+1 αβ n−k α−1
= [α(αβ n−k+1 αβ −n+k )n+k (αβ)−2k+1 β −n+k−1 α−1 ]−(m+1)(n−k)+1 αβ n−k α−1
= α[(αβ n−k+1 αβ −n+k )n+k (αβ)−2k+1 β −n+k−1 ]−(m+1)(n−k)+1 β n−k α−1 .
By substitution, we are left with only one relation:
[(αβ n−k+1 αβ −n+k )n+k (αβ)−2k+1 β −n+k−1 ]m+1 = αβ n−k+1 αβ −n+k .
Thus, we have
G(2n + 1, 2n − 1 + m(2n + 1), 2k)
= hα, β|[(αβ

n−k+1

αβ −n+k )n+k (αβ)−2k+1 β −n+k−1 ]m+1 = αβ n−k+1 αβ −n+k i.

For the peripheral subgroup, the surface framing s is a push-off of the knot
B(2n + 1, 2n − 1 + m(2n + 1), 2k) along Σ, so
s = γ 2n−1+m(2n+1) δ.
We note that this is actually the product of the right sides of the first and the

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Figure 14. The red graph is homotopy equivalent to the complement of B(2n, 2n − 2 + 2mn, 2k − 1) on the Heegaard surface Σ.
Here, we show B(6, 4, 3) as an example. The knot is shown in gray.
third relations, so we get
s = αβ n−k+1 (αβ)2k−1 αβ n−k+1 .
In order to compute µ, for the same reason as in the proof of Proposition 14,
since the meridian based at x0 can be isotoped to be going around the right genus
of the handlebody U once as in Figure 10, it is isotopic to the generator α. Thus,
µ = α.
Finally, we note that the linking number between B(2n+1, 2n−1+m(2n+1), 2k)
and a push-off along Σ, by the construction of the 1-bridge braid, is 2n[2n − 1 +
m(2n + 1)] + 2k, which gives us
s = µ2n[2n−1+m(2n+1)]+2k λ.

Proposition 16. For B(2n, 2n − 2 + 2mn, 2k − 1),
(a) The knot group is
G(2n, 2n − 2 + 2mn, 2k − 1)
= hα, β|[(αβ n−k+1 αβ −n+k )n+k−1 (αβ)−2k+2 β −n+k−1 ]m+1 = αβ n−k+1 αβ −n+k i.
(b) The peripheral subgroup is generated by the meridian
µ=α
and the surface framing
s = µ(2n−1)(2n−2+2mn)+2k−1 λ = β n−k+1 (αβ)2k−2 αβ n−k+1 α.

NON-LEFT-ORDERABLE SURGERIES ON 1-BRIDGE BRAIDS

15

Figure 15. The first generator for the fundamental group of Σ \
ν(B(2n, 2n − 2 + 2mn, 2k − 1)).
Proof. We use the same method as in the proof of the Propositions 14 and 15. Let
S 3 = U ∪Σ V be the genus-2 Heegaard splitting. Then π1 (U ) and π1 (V ) are the
free group on the generators α and β and the free group on the generators δ and
γ respectively(see Figure 5). Using Seifert-van Kampen Theorem, G(2n, 2n − 2 +
2mn, 2k − 1) is a free product with amalgamation of π1 (U ) and π1 (V ). Now we will
give the images of the generators of π1 (Σ \ ν(B(2n, 2n − 2 + 2mn, 2k − 1))) under
the maps induced by inclusion into π1 (U ) and π1 (V ).
Σ \ ν(B(2n, 2n − 2 + 2mn, 2k − 1)) is homotopy equivalent to a twice-punctured
genus-1 surface whose fundamental group is generated by the green, purple, and
olive loops in Figures 15, 16, and 17. The image of the green loop is αβ n−k+1 (αβ)2k−2
α−1 under the map induced by inclusion to π1 (U ), and is γ (m+1)(n+k−1)−1 under
the map induced by inclusion to π1 (V ), so we get the relation
αβ n−k+1 (αβ)2k−2 α−1 = γ (m+1)(n+k−1)−1 .
As for the purple loop, we have
αβ n−k α−1 = γ (m+1)(n−k)−1 δ,
and for the olive loop we have
α2 β n−k+1 = γ (m+1)(n−k+1)−1 δ.
Using the second and the third relation, we have
α2 β n−k+1 = γ (m+1)(n−k+1)−1 δ = γ m+1 γ (m+1)(n−k)−1 δ = γ m+1 αβ n−k α−1 .
⇔ γ m+1 = α2 β n−k+1 αβ −n+k α−1 .

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Figure 16. The second generator for the fundamental group of
Σ \ ν(B(2n, 2n − 2 + 2mn, 2k − 1)).
This relation can replace the second relation. Moreover, it provides us with an
expression for γ m+1 in terms of α and β, so with the first relation, we have
γ = γ (m+1)(n+k−1) α(αβ)−2k+2 β −n+k−1 α−1
= (α2 β n−k+1 αβ −n+k α−1 )n+k−1 α(αβ)−2k+2 β −n+k−1 α−1
= α(αβ n−k+1 αβ −n+k )n+k−1 (αβ)−2k+2 β −n+k−1 α−1 ,
which can replace the first relation. We can also write δ in terms of α and β by
replacing γ of the second relation with its expression in terms of α and β:
δ = γ −(m+1)(n−k)+1 αβ n−k α−1
= [α(αβ n−k+1 αβ −n+k )n+k−1 (αβ)−2k+2 β −n+k−1 α−1 ]−(m+1)(n−k)+1 αβ n−k α−1
= α[(αβ n−k+1 αβ −n+k )n+k−1 (αβ)−2k+2 β −n+k−1 ]−(m+1)(n−k)+1 β n−k α−1 .
By substitution, we are left with only one relation:
[(αβ n−k+1 αβ −n+k )n+k−1 (αβ)−2k+2 β −n+k−1 ]m+1 = αβ n−k+1 αβ −n+k .
Thus, we have
G(2n, 2n − 2 + 2mn, 2k − 1)
n−k+1

−n+k n+k−1

= hα, β|[(αβ
αβ
)
For the peripheral subgroup,

(αβ)−2k+2 β −n+k−1 ]m+1 = αβ n−k+1 αβ −n+k i.

s = γ 2n+2mn−2 δ,
which is actually the product of the right sides of the first and the third relations,
so we get
s = αβ n−k+1 (αβ)2k−2 αβ n−k+1 .

NON-LEFT-ORDERABLE SURGERIES ON 1-BRIDGE BRAIDS

17

Figure 17. The third generator for the fundamental group of Σ \
ν(B(2n, 2n − 2 + 2mn, 2k − 1)).
µ = α.
Finally, we note that the linking number between B(2n, 2n − 2 + 2mn, 2k − 1)
and a push-off along Σ is (2n − 1)(2n − 2 + 2mn) + 2k − 1, i.e.,
s = µ(2n−1)(2n−2+2mn)+2k−1 λ.


4. Non-Left-Orderability of Certain 1-bridge Braids
In this section, we prove Theorem 13 by applying Theorem 12 on the three
families of 1-bridge braids specified in Propositions 14, 15, and 16.
First, from Propositions 14, 15, and 16, we note that in each of the three cases,
the knot group G(ω, t + mω, b) can be generated by α, β, satisfying µ = α, s is
a word that excludes α−1 and β −1 and contains at least one α. Furthermore, the
framing of the longitude s is always (ω − 1)(t + mω) + b > 0. Therefore, it is
sufficient to prove that for any homomorphism Φ : G(ω, t + mω, b) → Homeo+ (R),
αt > t for all t ∈ R implies βt ≥ t for all t ∈ R, where wt denotes a shorthand for
Φ(w)t for any word w.
It is a fact that if a and b are real numbers, a < b, and w ∈ Homeo+ (R), then
wa < wb. We give the following two lemmas about groups acting on the real line
that are needed in our proof of Theorem 13.
Lemma 17. Let Φ : G → Homeo+ (R) and α ∈ G. If αt > t for all t ∈ R, then
w1 w2 t > w1 α−1 w2 t for all t ∈ R and any w1 , w2 ∈ G.

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Proof. For any words w1 , w2 ∈ G, w1 t > α−1 w1 t, since αt0 > t0 for t0 = α−1 w1 t.
Then since Φ(w2 ) ∈ Homeo+ (R), applying Φ(w2 ) on both sides, we have w2 w1 t >
w2 α−1 w1 t.

Lemma 18. Let Φ : G → Homeo+ (R) and wi ∈ G, i = 1, 2, 3, 4. If w1 t > w2 t for
all t ∈ R and w3 t > w4 t for all t ∈ R, then w1 w3 t > w2 w4 t for all t ∈ R.
Proof. First we have w1 w3 t > w2 w3 t, since w1 t0 > w2 t0 for t0 = w3 t. Since Φ(w2 ) ∈
Homeo+ (R), we get w2 w3 t > w2 w4 t. Thus w1 w3 t > w2 w3 t > w2 w4 t for all t ∈
R.

Lemma 19. For any homomorphism Φ : G(ω, 1 + mω, 2k) → Homeo+ (R), 1 ≤
k ≤ [(ω − 2)/2], m ≥ 0, αt > t for all t ∈ R implies βt ≥ t for all t ∈ R.
Proof. Recall the expression for the knot group:
G(ω, 1 + mω, 2k) = hα, β|(βα)k β = {(βα)k β ω−2k [(βα)k βα−1 (αβ)−k ]−ω+k }m (αβ)k αi.

The relation can be rewritten slightly as:
1 = {(βα)k β ω−2k [(αβ)k α(αβ)−k β −1 ]ω−k }m (αβ)k α(αβ)−k β −1 .
Applying both sides of the relation on t, we have
1t = [(βα)k β ω−2k C0ω−k ]m C0 t,
where C0 = (αβ)k α(αβ)−k β −1 . Assume αt > t for all t ∈ R. By Lemmas 17 and 18,
we can remove all α’s from (βα)k and the α in the middle from C0 and get β k and
β −1 at the expense of getting an appropriate strict inequality on any t, so we have
1t > (β k β ω−2k β −ω+k )m β −1 t = β −1 t.
Since β is order-preserving, applying it on both sides, we get
βt > t
for all t ∈ R.



Remark. If m = 0, by applying Markov’s Theorem and the Type II Markov Move,
B(ω, 1, 2k) is actually the torus knot T (2, 2k + 1). The fact the conjecture is true
for torus knots follows from a result of Moser [Mos71], which shows that surgery
along a torus knot is always either a lens space, a connected sum of lens spaces,
or Seifert fibered. A result of Boyer, Gordon, and Watson [BGW13, Theorem 4]
shows that the conjecture is true for Seifert fibered spaces.
Lemma 20. For any homomorphism Φ : G(2n + 1, 2n − 1 + m(2n + 1), 2k, r) →
Homeo+ (R), 1 ≤ k ≤ n − 1, m ≥ 0, αt > t for all t ∈ R implies βt ≥ t for all
t ∈ R.
For any homomorphism Φ : G(2n, 2n − 2 + 2mn, 2k − 1, r) → Homeo+ (R),
1 ≤ k ≤ n − 1, m ≥ 0, αt > t for all t ∈ R implies βt ≥ t for all t ∈ R.
Proof. We note that the two knot groups can be written in a unified form:
G(ω, t + mω, b) = hα, β|[(C1 C2 )n−k+b (αβ)−b+1 β −n+k−1 ]m+1 = C1 C2 i,

NON-LEFT-ORDERABLE SURGERIES ON 1-BRIDGE BRAIDS

19

where C1 = αβ n−k+1 and C2 = αβ −n+k , and in the first case, ω = 2n+1, t = 2n−1,
b = 2k, while ω = 2n, t = 2n − 2, b = 2k − 1 in the second. We can rewrite the left
side in a slightly different way:
(C1 C2 )n−k+b [(αβ)−b+1 β −n+k−1 (C1 C2 )n−k+b ]m (αβ)−b+1 β −n+k−1 = C1 C2 .
⇒ (C1 C2 )n−k+b−1 [(αβ)−b+1 β −n+k−1 (C1 C2 )n−k+b ]m = β n−k+1 (αβ)b−1 .
For any t0 ∈ R, applying both sides of the relation on t0 , we have
β n−k+1 (αβ)b−1 t0 = (C1 C2 )n−k+b−1 [(αβ)−b+1 β −n+k−1 (C1 C2 )n−k+b ]m t0 .
Assume αt > t for all t ∈ R. By Lemmas 17 and 18, we can add α−1 on one side
of the equation to get a strict inequality on any t0 . Thus,
(C1 C2 )n−k+b−1 t0 = (C1 C2 )n−k (C1 C2 )b−1 t0
> (α−1 C1 α−1 C2 )n−k (C1 α−1 C2 )b−1 t0 = β n−k (αβ)b−1 t0
for all t0 ∈ R. Similarly,
(C1 C2 )n−k+b t0 > β n−k+1 (αβ)b−1 t0
for all t0 ∈ R. Thus,
(C1 C2 )n−k+b−1 [(αβ)−b+1 β −n+k−1 (C1 C2 )n−k+b ]m t0
> β n−k (αβ)b−1 [(αβ)−b+1 β −n+k−1 β n−k+1 (αβ)b−1 ]t0 = β n−k (αβ)b−1 t0 .
⇒ β n−k+1 (αβ)b−1 t0 > β n−k (αβ)b−1 t0 ,
for all t0 ∈ R. Given any t ∈ R, let t0 = (αβ)−b+1 β −n+k t. We get
βt > t.

Therefore, we have proved Theorem 13.
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Department of Mathematics, Jilin University, Changchun, Jilin 130012, China
E-mail address: liangshiyu23@outlook.com

