数理解析研究所講究録
第 1866 巻 2013 年 30-38

30

Left-orderable fundamental groups and Dehn surgery on
two-bridge knots
Masakazu Teragaito
Department of Mathematics Education, Hiroshima University

1

Introduction

In Heegaard Floer homology theory, -spaces introduced in [17] have an important role.
is called an -space if $HF(Y)$ is a free abelian group
A rational homology 3-sphere
whose rank is equal to the order of $H_{1}(Y)$ . Lens spaces are typical -spaces, and several
other families of -spaces are known so far. However, it is still an open problem to givea characterization of -spaces without involving Heegaard Floer homology.
In [4], Boyer, Gordon and Watson conjecture that an irreducible rational homology
3-sphere is an -space if and only if its fundamental group is not left-orderable. This
would be an algebraic characterization of -spaces. Here, a non-trivial group is said
which is invariant under leftto be left-orderable if it admits a strict total ordering$h\in G$ . As a convention,
multiplication. That is, if $g<h$ then $fg<fh$ for any
is leftthe trivial group is defined to be not left-orderable. It is easy to see thatorderable if and only if is right-orderable, which is defined similarly. The history of
research on orderable groups is long, and many groups which appear in topology are leftorderable. For example, free groups, free abelian groups, knot or link groups, braid groups
are left-orderable. Also, the fundamental groups of surfaces but the projective plane areleft-orderable. Since left-orderable groups are torsion-free, the fundamental groups of lensspaces, elliptic manifolds are not left-orderable. It is natural to ask which 3-manifolds haveleft-orderable fundamental groups. As a classical fact, the free products of left-orderable
groups are left-orderable. Hence we may restrict ourselves to prime 3-manifolds. Boyer,
Rolfsen and Wiest [5] prove that if a compact connected orientable prime 3-manifold has
non-zero first betti number, then its fundamental group is left-orderable. Thus irreducible
rational homology 3-spheres remain to be done.
Dehn surgery might be the easiest way to create rational homology 3-spheres. For a-surgery yields a rational homology sphere whenevergiven knot $K$ in the 3-sphere$r\neq 0$ .
By considering the cabling conjecture, the resulting rational homology sphere
$L$

$L$

$Y$

$L$

$L$

$L$

$L$

$G$

$L$

$<$

”

$f,$ $g,$

$G$

$G$

$S^{3},$$r$

31
would be irreducible if $K$ is not cabled. On the other hand, there are some strong
constraints for knots which admit Dehn surgery yielding -spaces. For example, suchknots are fibered ([16]), and their Alexander polynomials have a specffied form ([17]).
Thus the above conjecture by Boyer, Gordon and Watson suggests that any non-trivial
Dehn surgery on $K$ yields a 3-manifold with left-orderable fundamental group, unless $K$
passes such criteria.
Any knot group is left-orderable. The fundamental group of the resulting manifold byDehn surgery on a knot is a quotient of the knot group. Although any subgroup of a leftorderable group is left-orderable, a quotient may not be left-orderable. For torus knots,
the resulting manifold by Dehn surgery is either a Seifert fibered manifold or the connected sum of two lens spaces. Since Boyer, Gordon and Watson [4] solved the conjecture
affirmatively for Seifert fibered manifolds, the left-orderability of the fundamental groupsof the resulting manifolds by Dehn surgery is completely understandable for torus knots.
The simplest hyperbolic knot is the figure-eight knot. By [17], it does not admit Dehn
surgery yielding an -space. Hence we may expect that any non-trivial Dehn surgery
yields a 3-manifold whose fundamental group is left-orderable. Toward this direction,Boyer, Gordon and Watson [4] showed if the surgery slope lies in the interval $(-4,4)$ ,then -surgery yields a manifold with left-orderable fundamental group. Later, Clay,
Watson and Lidman [6] confirmed the same conclusion for $r=\pm 4$ . (We remark that as
$L$

$L$

$r$

$r$

noted in [4], this is also true for any integral surgery by [9]. These two arguments arequite different. The former builds a non-trivial representation of the fundamental group of
the resulting manifold by -surgery into
, which is known to be left-orderable ([2]).
But the latter makes use of the torus decomposition of the resulting (graph) manifold into
two Seifert fibered pieces and some gluing technique of left-orderings ([3]). The argumentof [6] was generalized to all hyperbolic twist knots in [19]. We showed that 4-surgery ona hyperbolic twist knot yields a manifold with left-orderable fundamental group. (Here,the hook of a twist knot is assumed to be left-handed.) Furthermore, we extended theargument for any exceptional Dehn surgery on hyperbolic two-bridge knots in [7].
In this note, we report a generalization of the argument of [4] from the figure-eight knot
to hyperbolic genus one two-bridge knots. Details are found in [11]. Let $K=K(m, n)$ bea hyperbolic genus one two-bridge knot $S(4mn+1,2m)$ as shown in Figure 1. Here, thetwists in the vertical box is left-handed (resp. right-handed) if $m>0$ (resp. $m<0$ ), but
those in the horizontal box is right-handed (resp. left-handed) if $n>0$ (resp. $n<0$ ). By
symmetry, $K(m, n)$ is equivalent to $K(-n, -m)$ . Also, $K(-m, -n)$ is the mirror image
of $K(m, n)$ . Hence we may assume that $m>0$ . Thus $K(1,1)$ is the figure-eight knot,
and $K(1, -1)$ is the right-handed trefoil.
For a knot $K$ , a slope is said to be left-orierable if the resulting manifold $K(r)$ by
$)$

$r$

$r$

$SL_{2}(\mathbb{R})$32$\otimes 1$

$r$

:

$A$

genus one two-bridge knot $K(m, n)$-surgery has a left-orderable fundamental group.

Theorem 1.1 ([11]) Let $K(m, n)$ be a hyperbolic genus one two-bridge knot $S(4mn+$$1,2m)$ in the 3-sphere. Let I be the interval defined by$S^{3}$$I=\{\begin{array}{ll}(-4n, 4m) if n>0,{[}0, \max\{4m, -4n\}) if m>1 and n<-1,{[}0,4] otherwise.\end{array}$

Then any slope in I is left-ordemble. That is, the fundamental group of the resulting
manifold by -surgery on $K(m, n)$ is lefl-orderable if $r\in I.$$r$Among $K(m, n),$ $K(1, n)$ and $K(m, \pm 1)$ are twist knots. Moreover, $K(m, -1)$ is equivalent to $K(1, -m)$ , and $K(m, 1)$ is the mirror image of $K(1, m)$ .
Corollary 1.2 Let $K(1, n)$ be the -twist knot with $n\neq-1$ . If $n>0$ , then any slope
in the interval $(-4n, 4]$ is left-ordemble. If $n<-1$ , then then any slope in $[0,4]w$ leftordemble.
$n$

Our argument works for the figure-eight knot, and it is much simpler than one in [4],which involves character varieties. The fact that a knot has genus one is crucial in ourargument as well as that of [4]. In general, the longitude of a knot group is a product ofcommutators. If a knot has genus one, then the longitude is a single commutator. For, we need toa representation of a knot group into the universal covering groupcontrol the image of the longitude, by using Wood’s inequality [21]. See Lemma 2.7.
Anh Tran [20] obtained independently a similar result to Theorem 1.1.
$SL_{2}(\mathbb{R})$33
2OutlineLet $K=K(m, n)$ and let $G=\pi_{1}(S^{3}-K)$ be its knot group. We always assume that$m>0$ and $n\neq 0$ , unless specified otherwise.

Proposition 2.1 The knot group$G$admits a presentation$G=\langle x, y|w^{n}x=yw^{n}\rangle,$

where and are meridians and $w=(xy^{-1})^{m}(x^{-1}y)^{m}$ . Further.more, the longitudeis given as, where $w_{*}=(yx^{-1})^{m}(y^{-1}x)^{m}$ is obtained fromby reversing theorder of letters.
$x$

$y$

$\mathcal{L}$$\mathcal{L}=w_{*}^{n}w^{n}$

$w$

$2$ :

$A$

surgery diagram of $K(m, n)$This is slightly different from that in [13, Proposition 1], but both are isomorphic. It isderived from a surgery diagram of $K$ as illustrated in Figure 2, where $1/m$ -surgery and$-1/n$-surgery are performed along the second and third components, respectively.
Let and be real numbers such that $s>0$ and $t>1$ . Let :be arepresentation of defined by
$t$

$s$

$\rho$$Garrow SL_{2}(\mathbb{R})$$G$$\rho(x)=(_{0}^{\sqrt{t}} 1/\sqrt{t}1/\sqrt{t}) , \rho(y)=(_{-s\sqrt{t}}\sqrt{t}1/\sqrt{t}0)$

By [18], gives a non-abelian representation if and are a pair of solutions of the Riley
polynomial. Let $W=\rho(w)$ andbe the $(i, j)$ -entry of. Then the Riley polynomial
of $K$ is given by $\phi_{K}(s, t)=z_{1,1}+(1-t)z_{1,2}$ . (See also [8].) Since and are limited
to be positive real numbers in our setting, it is not obvious that there exist solutions for
Riley’s equation $\phi_{K}(s, t)=0$ . However, this will be verified in Proposition 2.3 under somecondition.
To describe the Riley polynomial of $K$ explicitly, we need two sequences of polynomials
with a single variable$s$$\rho$$z_{i,j}$$t$$W^{n}$

$s$

$s.$

$t$

34
For non-negative integer $m$ , let$f_{m}\in \mathbb{Z}[s]$be defined by the recursion(2.1)$f_{m+2}-(s+2)f_{m+1}+f_{m}=0$with initial conditions $f_{0}=1$ and $f_{1}=s+1$ . Also, letrecursion$g_{m}\in \mathbb{Z}[s]$be defined by the same
(2.2)

$g_{m+2}-(s+2)g_{m+1}+g_{m}=0$with slightly different initial conditions $g_{0}=1$ and $g_{1}=s+2$ . We remark thatequivalent to the Chebyshev polynomial of the second kind.
are
The closed formulae forand$f_{m}$$g_{m}$

is$g_{m}$$f_{m}= \sum_{i=0}^{m}(\begin{array}{ll}m +im -i\end{array})s^{i}, g_{m}= \sum_{i=0}^{m}(\begin{array}{ll}m +l+i -im\end{array})s^{i}.$In particular, all coefficients ofare positive integers, and the degree ofandis $m$ . Also,and
are monic.
Let
be the eigenvalues of $W=\rho(w)$ . For any integer , set.
$f_{m}$$f_{m}$$g_{m}$$g_{m}$$f_{m}$

and$g_{m}$$k$$\lambda_{\pm}\in \mathbb{C}$$\tau_{k}=(\lambda_{+}^{k}-\lambda^{\underline{k}})/(\lambda_{+}-$$\lambda_{-})$

Proposition 2.2 The Riley polynomial of $K$ is$\phi_{K}(s, t)=(\tau_{n+1}-\tau_{n})+(s+2-t-1/t)f_{m-1}g_{m-1^{\mathcal{T}}n}.$For convenience, we introduce a variable $T=t+1/t$ . Then the Riley polynomial of $K$
is $\phi_{K}(s, T)=(\tau_{n+1}-\tau_{n})+(s+2-T)f_{m-1}g_{m-1^{\mathcal{T}}n}.$For example, if $n=1$ then$\phi_{K}(s, T) = (\tau_{2}-\tau_{1})+(s+2-T)f_{m-1}g_{m-1}\tau_{1}$$= (trW-1)+(s+2-T)f_{m-1}g_{m-1}$$= s(s+2-T)g_{m-1}^{2}+1+(s+2-T)f_{m-1}g_{m-1}$$= (s+2-T)g_{m-1}(sg_{m-1}+f_{m-1})+1$$= (s+2-T)g_{m-1}f_{m}+1.$

has the unique solution $T=s+2+1/(f_{m}g_{m-1})$ for
Thus Riley’s equation
any $s>0$ . Then $T>s+2>2$ , because $f_{m}>0$ and $g_{m-1}>0$ . Hence we have a real
solution $t=(T+\sqrt{T^{2}-4})/2>1$ . In fact, we have $s+2<T<s+2+4/(sg_{m-1}^{2})$ .
$\phi_{K}(\mathcal{S}, T)=0$

Proposition 2.3 Suppose $n\neq\pm 1$ . For any $s>0$ , Riley’s equation $\phi_{K}(s, T)=0$ hasa solution satisfying $s+2+c/(sg_{m-1}^{2})<T<s+2+d/(sg_{m-1}^{2})$ , where and are
constants in $(0,4)$ depending only on . In particular, $\phi_{K}(s, t)=0$ has a solution $t>1$
for any $s>0.$
$T$

$c$

$n$

$d$

35
Now, we introduce a continuous family of representations ofbe the representation defined by the correspondence$G$. For $s>0$ , let$\rho_{s}:Garrow$$SL_{2}(\mathbb{R})$(2.3)$\rho_{S}(x)=(\begin{array}{ll}\sqrt{t} 00 \frac{1}{\sqrt{t}}\end{array}), \rho_{s}(y)=(^{\frac{t-s-1}{\sqrt{t}-\frac{1}{r_{t}}}}-\mathcal{S} \frac{}{(\sqrt{t}-\frac{S-\underline{1}S\frac{1}{+1\gamma_{t}})^{2}}{\sqrt{t}-\frac{1}{r_{t}}}}-1)$

Since is conjugate with , if and satisfy Riley’s equation $\phi_{K}(s, t)=0$ thena non-abelian representation of as well as (see [8, 14]).
$\rho_{s}$$\rho$

$t$

$s$

$G$

$\rho_{s}$gives$\rho$

Proposition 2.4 For the longitude ofentry ofis a positive real number.
$\mathcal{L}$$G$, the matrix$\rho_{S}(\mathcal{L})$is diagonal, and the $(1, 1)-$$\rho_{s}(\mathcal{L})$The first conclusion is easy, but the second is important. To show it, the character variety theory was used in [4, Lemma 7], but we can establish it through a direct calculation.
Letbe the (1, 1)-entry of the matrix.
$B_{s}$$\rho_{s}(\mathcal{L})$Proposition 2.5$B_{s}= \frac{-f_{m}+tf_{m-1}}{-f_{m-1}+tf_{m}}.$This conclusion is interesting, because the parameter disappears.
Let $r=p/q$ be a rational number, and let $K(r)$ denote the resulting manifold by rsurgery on $K$ . In other words, $K(r)$ is obtained by attaching a solid torus to the knot
exterior $E(K)$ along their boundaries so that the loop
bounds a meridian disk of $V,$
where and are a meridian and longitude of $K.$
Our representationinduces a homomorphism $\pi_{1}(K(r))arrow SL_{2}(\mathbb{R})$ if
and only if. Since both ofandare diagonal (see (2.3) and
Proposition 2.4), this is equivalent to the single equation
$n$

$V$

$x^{p}\mathcal{L}^{q}$$x$$\mathcal{L}$$\rho_{s}:Garrow SL_{2}(\mathbb{R})$$\rho_{s}(x)^{p}\rho_{S}(\mathcal{L})^{q}=I$$\rho_{s}(x)$$A_{s}^{p}B_{s}^{q}=1$$\rho_{s}(\mathcal{L})$

,

(2.4)

are the (1, 1)-entries of, respectively. We remark that
and$A_{s}=\sqrt{t}(>1)$ is a positive real number, so is
by Proposition 2.4. Hence the equation(2.4) is furthermore equivalent to the equation
where$A_{s}$

and$B_{S}$$\rho_{s}(x)$$\rho_{s}(\mathcal{L})$$B_{s}$$- \frac{\log B_{S}}{\log A_{s}}=\frac{p}{q}$Let$g:(0, \infty)arrow \mathbb{R}$.

be a function defined by$g(s)=- \frac{\log B_{s}}{\log A_{s}}.$By calculating limits, we obtain the following.

(2.5)

36
Proposition 2.6 The image of$g$contains an open interval $(0,4m)$ .

The next is the key in [4], which is originally claimed in [14], for the figure-eight knot.
Our proof most follows that ofcan be described as
The universal covering group$[4\underline{].}$$SL_{2}(\mathbb{R})$$S\overline{L_{2}(\mathbb{R}})=\{(\gamma,\omega)||\gamma|<1, -\infty<\omega<\infty\}.$See [1, 14]. Let$\chi$:$S\overline{L_{2}(\mathbb{R}}$)be the covering projection. Then $ker\chi=\{(0,2j\pi)|$$arrow SL_{2}(\mathbb{R})$$j\in \mathbb{Z}\}.$

Lemma 2.7 Let :for some
the subgroup $(-1,1)\cross\{0\}$ of$\tilde{\rho}$$\tilde{\rho}’=h\cdot\tilde{\rho}$) be a lift of . Then replacing, we can suppose that

$Garrow S\overline{L_{2}\underline{(\mathbb{R}}}$$\rho_{s}$by a representationis contained in$\tilde{\rho}(\pi_{1}(\partial E(K)))$$h:Garrow S\underline{L_{2}(\mathbb{R})}$$SL_{2}(\mathbb{R})$$\tilde{\rho}$.

Proof of Theorem 1.1 Suppose $n\neq-1$ . Let $r=p/q\in(0,4m)$ . By Proposition 2.6, we
can find so that $g(s)=r$ . Choose a lift of so thatlies in. This means that, so(Lemma 2.7). Then
induce a homomorphism. ThenHence$\pi_{1}(K(r))arrow SL_{2}(\mathbb{R})$ with non-abelian image.
is left-orderable ([2])
Recall thatand any (non-trivial) subgroup of a left-orderable group is left-orderable. Since $K(r)$is irreducible [12], $\pi_{1}(K(r))$ is left-orderable by [5, Theorem 1.1]. For $r=0,$ $K(O)$ isirreducible ([10]) and has positive betti number. Hence $\pi_{1}(K(O))$ is left-orderable by[5, Corollary 3.4]. Thus we have shown that any slope in $[0,4m)$ is left-orderable for$K=K(m, n)$ .
Suppose $n>0$ . If we apply the above argument for $K(n, m)$ , then any slope in $[0,4n)$ isshown to be left-orderable. Since $K(n, m)$ is equivalent to the mirror image of $K(m, n)$ ,any slope in $(-4n, 0] is left-$ orderable $for K(m, n)$ . Thus we can conclude that $(-4n, 4m)$
consists of left-orderable slopes for $K=K(m, n)$ with $n>0.$
Suppose $m>1$ and $n<-1$ . Since $K(m, n)$ is equivalent to $K(-n, -m)$ , the argumentin the first paragraph shows that any slope in $[0, -4n)$ is left-orderable. In this case, we
obtain $[0, \max\{4m, -4n\})$ consisting of left-orderable slopes.
Finally, consider the remaining cases. They are $K(1, n)$ with $n<-1$ and $K(m, -1)$
with $m>1$ . Since $K(m, -1)$ is isotopic to $K(1, -m)$ , two cases coincide. We obtain $[0,4)$
consisting of left-orderable slopes by the argument in the first paragraph. Furthermore,since these knots are twist knots, the slope 4 is also left-orderable by [19].
$s$$\tilde{\rho}_{s}$$\rho_{s}(x^{p}\mathcal{L}^{q})=I$$ker\chi=\{(0,2j\underline{\pi)|j}\in \mathbb{Z}\}.$$\rho_{s}$$\tilde{\rho}_{s}(\pi_{1}(\partial E(K)))\subset(-1,1)\cross\{0\}$$\chi(\tilde{\rho}_{s}(x^{p}\mathcal{L}^{q}))=I$$\tilde{\rho}_{s}(x^{p}\mathcal{L}^{q})=(0,0)$$\tilde{\rho}_{s}(x^{p}\mathcal{L}^{q})$$\tilde{\rho}_{s}\underline{can}$$SL_{2}(\mathbb{R})$$\square$37
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Department of Mathematics Education
Hiroshima University
Higashi-hiroshima 739-8524

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