Non left-orderable surgeries on L-space twisted torus knots

TL;DR

This paper proves that for certain L-space twisted torus knots, the fundamental group of the 3-manifold obtained by specific surgeries is not left-orderable when the surgery coefficient exceeds a bound related to the knot's genus.

Contribution

It establishes a new criterion linking surgery coefficients and left-orderability for a class of twisted torus knots, expanding understanding of their fundamental groups.

Findings

Fundamental groups are not left-orderable for surgeries with coefficient ≥ 2g(K)-1.

Applicable to L-space twisted torus knots with specified parameters.

Provides a bound connecting surgery slope and knot genus.

Abstract

We show that if $K$ is an L-space twisted torus knot $T^{l,m}_{p,pk \pm 1}$ with $p \ge 2$, $k \ge 1$, $m \ge 1$ and $1 \le l \le p-1$, then the fundamental group of the $3$-manifold obtained by $\frac{r}{s}$-surgery along $K$ is not left-orderable whenever $\frac{r}{s} \ge 2 g(K) -1$, where $g(K)$ is the genus of $K$.

Equations

$$a^{p-l+1}(z^{m}a{)}^{l-1}=b^{k(p-l+1)-1}(z^m{b}^k{)}^{l-1}$$

$$(yx{)}^{k-1}y((yx{)}^{k-1}y{)}^{p-l}(z^m(yx{)}^{k-1}y{)}^{l-1}=(yx{)}^k(yx{)}^{k(p-l)-1}(z^m(yx{)}^k{)}^{l-1}.$$

$$((yx{)}^{k-1}y{)}^{p-l}(z^m(yx{)}^{k-1}y{)}^{l-1}=x(yx{)}^{k(p-l)-1}(z^m(yx{)}^k{)}^{l-1}.$$

$$\rho (((yx{)}^{k-1}y{)}^{p-l})\alpha$$

$$\rho ((yx{)}^{k-1}{y}^n)\alpha$$

$$\rho (y^n(xy{)}^{k-1})\alpha$$

$$\rho (((yx{)}^{k-1}y{)}^{p-l})\alpha >\rho (x(yx{)}^{k(p-l)-1})\alpha$$

$$\rho (((yx{)}^{k-1}y{)}^{p-l})\alpha >\rho (x(yx{)}^{k(p-l)-1})\alpha >\rho (x(yx{)}^{k-1}((yx{)}^{k-1}y{)}^{p-l-1})\alpha .$$

$$\rho ((yx{)}^{k-1}{y}^n)\alpha$$

$$\rho (y^n(xy{)}^{k-1})\alpha$$

$$\frac{{\Delta }_{T_{p,pk-1}^{l,m}}(t)}{t-1}=\frac{\Phi (\frac{\partial r_1}{\partial x}-\frac{\partial r_2}{\partial x})}{\Phi (y)-1}$$

$$\frac{\partial uv}{\partial x}=\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial x},\frac{\partial u^{-1}}{\partial x}=-u^{-1}\frac{\partial u}{\partial x}\text{and}\frac{\partial u^n}{\partial x}=(1+u+\cdots +u^{n-1})\frac{\partial u}{\partial x}$$

$$\displaystyle\Phi\Big{(}\frac{\partial z}{\partial x}\Big{)}$$

$$\frac{\partial r_1}{\partial x}$$

$$\frac{\partial r_2}{\partial x}$$

$$\frac{\partial r_1}{\partial x}-\frac{\partial r_2}{\partial x}$$

$$a(z^{m}a{)}^{l-1}{a}^{p-l}=b^k(z^m{b}^k{)}^{l-1}{b}^{k(p-l)+1}$$

$$(xy{)}^{k}x(z^m(xy{)}^{k}x{)}^{l-1}((xy{)}^{k}x{)}^{p-l}=(xy{)}^k(z^m(xy{)}^k{)}^{l-1}(xy{)}^{k(p-l)+1}$$

$$(x{z}^m(xy{)}^k{)}^{l-1}x((xy{)}^{k}x{)}^{p-l}=(z^m(xy{)}^k{)}^{l-1}(xy{)}^{k(p-l)+1}.$$

$$\rho ((xy{)}^{k(p-l)+1})\alpha$$

$$\rho ((yx{)}^k{y}^n)\alpha$$

$$\rho (y^n(xy{)}^k)\alpha$$

$${\lambda }^{\prime }$$

$${\lambda }^{\prime }=((yx{)}^{k-1}y{)}^{p-2}[(yx{)}^{1-(p-2)k}((yx{)}^{k-1}y{)}^{p-1}{]}^2.$$

$${\lambda }^{\prime }$$

$${\lambda }^{\prime }=[(xy{)}^{(p-2)k+1}((xy{)}^{k}x{)}^{-(p-3)}{]}^2((xy{)}^{k}x{)}^{p-2}.$$

$${\lambda }^{\prime }$$

$$\alpha ={\lambda }^{\prime }\alpha =xux{u}^{-1}(yx{)}^{k}ux\alpha$$

Full text

2000 Mathematics Subject Classification. Primary 57M27, Secondary 57M25.

Key words and phrases. Dehn surgery, left-orderable, L-space, twisted torus knot.

1. Introduction

Heegaard Floer homology is a package of 3-manifold invariants introduced by Ozsvath and Szabo [OS]. Manifolds with minimal Heegaard Floer homology are called L-spaces. More precisely, a rational homology 3-sphere $Y$ is an L-space if the hat version of its Heegaard Floer homology, denoted by $\widehat{HF}(Y)$, has rank equal to the order of $H_{1}(Y;\mathbb{Z})$. Since computing $\widehat{HF}$ is difficult, it is desirable to know if there are characterizations of L-spaces which do not refer to Heegaard Floer homology. The L-space conjecture of Boyer, Gordon and Watson [BGW] proposes such a characterization. It states that an irreducible rational homology 3-sphere is an L-space if and only if its fundamental group is not left-orderable. Here a nontrivial group $G$ is left-orderable if it admits a total ordering $<$ such that $g<h$ implies $fg<fh$ for all elements $f,g,h$ in $G$.

Many hyperbolic L-spaces can be obtained via Dehn surgery. A knot $K$ in $S^{3}$ is called an L-space knot if it admits a positive Dehn surgery yielding an L-space. For an L-space knot $K$, it was shown in [OS] that the $\frac{r}{s}$-surgery of $K$ is an L-space if and only if $\frac{r}{s}\geq 2g(K)-1$, where $g(K)$ is the genus of $K$. In view of the L-space conjecture, one would expect that the fundamental group of the $\frac{r}{s}$-surgery of an L-space knot $K$ is not left-orderable if and only if $\frac{r}{s}\geq 2g(K)-1$. When an L-space knot $K$ is the $(-2,3,2n+1)$-pretzel knot (with $n\geq 3$) or the $(n-2)$-twisted $(3,3m+2)$-torus knot (with $n,m\geq 1$), it was shown in [Ni, Tr] that the fundamental group of the $\frac{r}{s}$-surgery of $K$ is not left-orderable if $\frac{r}{s}\geq 2g(K)-1$.

We denote by $T^{l,m}_{p,q}$ the twisted torus knot obtained from the $(p,q)$-torus knot by twisting $l$ strands $m$ full times. Determining all L-space knots among twisted torus knots is an open problem. However, when $q\equiv\pm 1\pmod{p}$, this problem was already solved. For $p\geq 2$, $k\geq 1$, $m\geq 1$ and $1\leq l\leq p-1$, it was shown in [Va] that the twisted torus knot $T^{l,m}_{p,pk\pm 1}$ is an L-space knot if and only if $l=p-1$ or $l\in\{p-2,2\}$ and $m=1.$ Some results on the non left-orderability for surgeries of L-space twisted torus knots $T^{p-1,m}_{p,pk\pm 1}$ and $T^{p-2,1}_{p,pk\pm 1}$ were obtained in [Ju, Na, CW, IT1, IT2, CGHV]. In this paper, we generalize the above results. More precisely, we will show the following.

Theorem 1**.**

Let $K$ be an L-space twisted torus knot $T^{l,m}_{p,pk\pm 1}$ with $p\geq 2$, $k\geq 1$, $m\geq 1$ and $1\leq l\leq p-1$. Then the fundamental group of the $3$-manifold obtained by $\frac{r}{s}$-surgery along $K$ is not left-orderable if $\frac{r}{s}\geq 2g(K)-1.$

The paper is organized as follows. In Section 2 we prove some preliminary lemmas on the knot groups of twisted torus knots $T^{l,m}_{p,pk\pm 1}$ and we compute the genera of the L-space knots among these knots. In Section 3 we use the results of the previous section to prove Theorem 1.

2. Preliminary lemmas

For a knot $K\subset S^{3}$, the knot group of $K$ is defined to be the fundamental group of the complement of $K$ in $S^{3}$. We denote by $G^{l,m}_{p,q}$ the knot group of $T^{l,m}_{p,q}$. From now on we consider the twisted torus knots $T^{l,m}_{p,pk\pm 1}$ with $p\geq 2$, $k\geq 1$, $m\geq 1$ and $1\leq l\leq p-1$.

2.1. The case $T^{l,m}_{p,pk-1}$

By [CGHV] $G^{l,m}_{p,pk-1}$ has a presentation with two generators $a,b$ and one relation

[TABLE]

where $z=b^{1-k(p-l)}a^{p-l}.$ Moreover, $\mu=a^{-1}b^{k}$ is a meridian and the preferred longitude corresponding to $\mu$ is $\lambda=\mu^{-p(pk-1)-l^{2}m}\lambda^{\prime}$ where $\lambda^{\prime}=a^{p-l}(z^{m}a)^{l}.$

Let $x=\mu=a^{-1}b^{k}$ and $y=b^{1-k}a$. Then $b=yx$, $a=b^{k-1}y=(yx)^{k-1}y$ and $z=(yx)^{1-k(p-l)}((yx)^{k-1}y)^{p-l}$. Moreover, $G^{l,m}_{p,pk-1}$ has a presentation with two generators $x,y$ and one relation

[TABLE]

This relation is equivalent to

[TABLE]

By abelianizing $G^{l,m}_{p,pk-1}$, the homology class of $y$ is $p-1$ times that of $x$.

Let $\text{Homeo}^{+}(\mathbb{R})$ be the group of order-preserving homeomorphisms of $\mathbb{R}.$

Lemma 2.1**.**

Suppose $\rho:G^{l,m}_{p,pk-1}\to\text{Homeo}^{+}(\mathbb{R})$ is a homomorphism such that $\rho(x)\alpha>\alpha$ for all $\alpha\in\mathbb{R}$. Then $\rho(y)\alpha>\alpha$ and

[TABLE]

for all $\alpha\in\mathbb{R}$ and $n\geq 1$.

Proof.

Since $\rho(x)\alpha>\alpha$ for all $\alpha\in\mathbb{R}$ we have $\rho(z^{m}(yx)^{k})\alpha>\rho(z^{m}(yx)^{k-1}y)\alpha.$ The relation (2.2) then implies that

[TABLE]

for all $\alpha\in\mathbb{R}$. Since $\rho((yx)^{k})\alpha>\rho((yx)^{k-1}y)\alpha$ we have

[TABLE]

Hence $\rho((yx)^{k-1}y)\alpha>\rho(x(yx)^{k-1})\alpha$ for all $\alpha\in\mathbb{R}$. By [Tr, Lemma 2.4], this implies that

[TABLE]

for all $\alpha\in\mathbb{R}$ and $n\geq 1$.

Finally, since $\rho((yx)^{k-1}y)\alpha>\rho(x(yx)^{k-1})\alpha>\rho((yx)^{k-1})\alpha$ we have $\rho(y)\alpha>\alpha.$ ∎

Lemma 2.2**.**

If $T^{l,m}_{p,pk-1}$ is an L-space knot, then its genus is equal to $\frac{1}{2}l(l-1)m+\frac{1}{2}p(p-1)k-(p-1).$

Proof.

For a Laurent polynomial $f(t)\in\mathbb{C}[t^{\pm 1}]$, the degree of $f$ is defined to be the difference of the highest degree and the lowest degree of $f$.

By [OS], the degree of the Alexander polynomial of an L-space knot is twice the knot genus. Hence, to prove Lemma 2.2 it suffices to show that the degree of the Alexander polynomial of $T^{l,m}_{p,pk-1}$ is equal to $l(l-1)m+p(p-1)k-(2p-2).$ We will compute the Alexander polynomial via Fox’s free calculus.

Since $z=b^{1-k(p-l)}a^{p-l}$, the relation (2.1) is equivalent to $za(z^{m}a)^{l-1}=b^{k}(z^{m}b^{k})^{l-1}.$ Let $r_{1}=za(z^{m}a)^{l-1}$ and $r_{2}=ax(z^{m}ax)^{l-1}.$ Since $b^{k}=ax$ we have $G^{l,m}_{p,pk-1}=\langle x,y\mid r_{1}=r_{2}\rangle$. The Alexander polynomial of $T^{l,m}_{p,pk-1}$ can be computed via the formula

[TABLE]

where $\Phi:G^{l,m}_{p,pk-1}\to\mathbb{Z}\cong\langle t\rangle$ is the abelianization. See, for example, [Tr, Section 3].

Since $\Phi(x)=t$ and $\Phi(y)=t^{p-1}$, we have $\Phi(a)=t^{kp-1}$, $\Phi(b)=t^{p}$ and $\Phi(z)=\Phi(b^{1-k(p-l)}a^{p-l})=t^{l}.$ We will use the following formulas

[TABLE]

for $u,v\in G^{l,m}_{p,pk-1}$ and $n\geq 1$. Note that $\frac{\partial x}{\partial x}=1$ and $\frac{\partial y}{\partial x}=0$.

For $n\geq 0$ and $u\in G^{l,m}_{p,pk-1}$ we let $\delta_{n}(u)=\sum_{i=0}^{n}u^{i}.$ By direct calculations we have $\Phi(\frac{\partial a}{\partial x})=\Phi(\delta_{k-2}(yx)y)=\frac{t^{p(k-1)}-1}{t^{p}-1}t^{p-1}$, $\Phi(\frac{\partial b}{\partial x})=\Phi(y)=t^{p-1}$ and

[TABLE]

Since

[TABLE]

we have

[TABLE]

By comparing degrees, the highest degree of the Laurent polynomial \Phi\big{(}\frac{\partial r_{1}}{\partial x}-\frac{\partial r_{2}}{\partial x}\big{)} is $ml(l-1)+kpl-1$, which is the highest degree of $\Phi(ax\delta_{l-2}(z^{m}ax)z^{m}a)$. Moreover, the lowest degree is $-kp(p-l-1)+p-1$, which is the lowest degree of \Phi\big{(}\frac{\partial z}{\partial x}\big{)}. Hence, the degree of \Phi\big{(}\frac{\partial r_{1}}{\partial x}-\frac{\partial r_{2}}{\partial x}\big{)} is $l(l-1)m+p(p-1)k-p$. Equation (2.3) then implies that the degree of the Alexander polynomial of $T^{l,m}_{p,pk-1}$ is equal to $l(l-1)m+p(p-1)k-(2p-2).$ ∎

2.2. The case $T^{l,m}_{p,pk+1}$

By [CGHV] $G^{l,m}_{p,pk+1}$ has a presentation with two generators $a,b$ and one relation

[TABLE]

where $z=b^{k(p-l)+1}a^{l-p}.$ Moreover, $\mu=b^{-k}a$ is a meridian and the preferred longitude corresponding to $\mu$ is $\lambda=\mu^{-p(pk+1)-l^{2}m}\lambda^{\prime}$ where $\lambda^{\prime}=(z^{m}a)^{l}a^{p-l}.$

Let $x=\mu=b^{-k}a$ and $y=a^{-1}b^{k+1}$. Then $b=xy$ and $a=b^{k}x=(xy)^{k}x.$ Moreover, $G^{l,m}_{p,pk+1}$ has a presentation with two generators $x,y$ and one relation

[TABLE]

where $z=(xy)^{k(p-l)+1}((xy)^{k}x)^{l-p}.$ The group relation is equivalent to

[TABLE]

By abelianizing $G^{l,m}_{p,pk+1}$, the homology class of $y$ is $p-1$ times that of $x$. Similar to the previous case, we have the following lemmas.

Lemma 2.3**.**

Suppose $\rho:G^{l,m}_{p,pk+1}\to\text{Homeo}^{+}(\mathbb{R})$ is a homomorphism such that $\rho(x)\alpha>\alpha$ for all $\alpha\in\mathbb{R}$. Then $\rho(y)\alpha>\alpha$ and

[TABLE]

for all $\alpha\in\mathbb{R}$ and $n\geq 1$.

Lemma 2.4**.**

If $T^{l,m}_{p,pk+1}$ is an L-space knot, then its genus is equal to $\frac{1}{2}l(l-1)m+\frac{1}{2}p(p-1)k.$

2.3. Longtitude of L-space $T^{l,m}_{p,pk\pm 1}$

From now on we let $K$ denote an L-space twisted torus knot $T^{l,m}_{p,pk\pm 1}.$ Since $K$ is an L-space, by [Va], one of the following conditions holds

(1) $l=p-1$,

(2) $l=p-2$ and $m=1$,

(3) $l=2$ and $m=1$.

Lemma 2.5**.**

If $K=T^{p-1,m}_{p,pk-1}$ then $\lambda^{\prime}=(yx)^{k-1}y^{m+1}x(yx)^{k-1}(y^{m}(yx)^{k})^{p-2}.$

If $K=T^{p-1,m}_{p,pk+1}$ then $\lambda^{\prime}=((xy)^{k}xy^{m})^{p-1}(xy)^{k}x.$

If $K=T^{p-2,1}_{p,pk-1}$ then $\lambda^{\prime}=x(yx)^{k-1}(y(yx)^{k-1}y)^{p-2}(yx)^{k-1}y.$

If $K=T^{p-2,1}_{p,pk+1}$ then $\lambda^{\prime}=(xy)^{k}x(y(xy)^{k})^{p-2}(xy)^{k}x.$

If $K=T^{2,1}_{p,pk-1}$ then

[TABLE]

If $K=T^{2,1}_{p,pk+1}$ then

[TABLE]

Proof.

If $K=T^{p-1,m}_{p,pk-1}$ then, by Section 2.1, we have $z=y$ and $\lambda^{\prime}=(yx)^{k-1}y(y^{m}(yx)^{k-1}y)^{p-1}.$ The relation (2.2) is $(yx)^{k-1}y(y^{m}(yx)^{k-1}y)^{p-2}=x(yx)^{k-1}(y^{m}(yx)^{k})^{p-2}.$ Hence

[TABLE]

If $K=T^{p-1,m}_{p,pk+1}$ then, by Section 2.2, we have $z=((xy)^{k}x)y((xy)^{k}x)^{-1}$ and $\lambda^{\prime}=((xy)^{k}xy^{m})^{p-1}(xy)^{k}x.$

If $K=T^{p-2,1}_{p,pk-1}$ then, by Section 2.1, we have $z=(yx)^{1-2k}((yx)^{k-1}y)^{2}=(yx)^{-k}y(yx)^{k-1}y$ and $\lambda^{\prime}=((yx)^{k-1}y)^{2}(z(yx)^{k-1}y)^{p-2}.$ The relation (2.2) is $((yx)^{k-1}y)^{2}(z(yx)^{k-1}y)^{p-3}=x(yx)^{2k-1}(z(yx)^{k})^{p-3}.$ Hence

[TABLE]

If $K=T^{p-2,1}_{p,pk+1}$ then, by Section 2.2, we have $z=(xy)^{2k+1}((xy)^{k}x)^{-2}$ and

[TABLE]

If $K=T^{2,1}_{p,pk-1}$ then, by Section 2.1, we have $z=(yx)^{1-(p-2)k}((yx)^{k-1}y)^{p-2}$ and

[TABLE]

The relation (2.2) is $((yx)^{k-1}y)^{p-2}(yx)^{1-(p-2)k}((yx)^{k-1}y)^{p-1}=x((yx)^{k-1}y)^{p-1}x.$ Hence

[TABLE]

If $K=T^{2,1}_{p,pk+1}$ then, by Section 2.2, we have $z=(xy)^{(p-2)k+1}((xy)^{k}x)^{-(p-2)}$ and

[TABLE]

The relation (2.5) is $x(xy)^{(p-1)k+1}x=(xy)^{(p-2)k+1}((xy)^{k}x)^{-(p-2)}(xy)^{(p-1)k+1}.$ Hence

[TABLE]

This completes the proof of Lemma 2.5. ∎

3. Non left-orderable surgeries

In this section we prove Theorem 1. Recall that $K$ is an L-space twisted torus knot $T^{l,m}_{p,pk\pm 1}.$ Let $M_{\frac{r}{s}}$ be the 3-manifold obtained by $\frac{r}{s}$-surgery along $K$. Assume $\pi_{1}(M_{\frac{r}{s}})$ is left-orderable for some $\frac{r}{s}\geq 2g(K)-1$, where $s>0$. Then there exists a homomorphism $\rho:\pi_{1}(M_{\frac{r}{s}})\to\text{Homeo}^{+}(\mathbb{R})$ such that there is no $\alpha\in\mathbb{R}$ satisfying $\rho(h)(\alpha)=\alpha$ for all $h\in\pi_{1}(M)$, see e.g. [CR, Problem 2.25]. From now on we write $h\alpha$ for $\rho(h)(\alpha)$.

Since $x^{r}\lambda^{s}=1$ in $\pi_{1}(M)$ and $x\lambda=\lambda x$, there exists an element $v\in\pi_{1}(M)$ such that $x=v^{s}$ and $\lambda=v^{-r}$, see e.g. [Na, Lemma 3.1].

Proposition 3.1**.**

We have $v\alpha\not=\alpha$ for all $\alpha\in\mathbb{R}$.

Proof.

Assume $v\alpha=\alpha$ for some $\alpha\in\mathbb{R}$. Then $x\alpha=v^{s}\alpha=\alpha$ and $\lambda\alpha=v^{-r}\alpha=\alpha.$ Since $\lambda^{\prime}=x^{p(pk\pm 1)+l^{2}m}\lambda$, we have $\lambda^{\prime}\alpha=\alpha.$ If $y\alpha=\alpha$ then $h\alpha=\alpha$ for all $h\in\pi_{1}(M)$, a contradiction. Without loss of generality, we assume that $y\alpha>\alpha$.

If $K=T^{p-1,m}_{p,pk\pm 1}$ or $K=T^{p-2,1}_{p,pk\pm 1}$ then, by Lemma 2.5, all the exponents of $y$ in $\lambda^{\prime}$ are positive. Since $x\alpha=\alpha$ and $y\alpha>\alpha$, we conclude that $\lambda^{\prime}\alpha>\alpha.$ This contradicts $\lambda^{\prime}\alpha=\alpha.$

If $K=T^{2,1}_{p,pk-1}$ then, by Lemma 2.5, we have

[TABLE]

Let $u=((yx)^{k-1}y)^{p-1}$. Since $x\alpha=\alpha$ and $y\alpha>\alpha$ we have $u\alpha>\alpha$. Hence

[TABLE]

This implies that $(yx)^{1-(p-2)k}((yx)^{k-1}y)^{p-1}\alpha=(yx)^{1-(p-2)k}u\alpha>u\alpha>\alpha.$ Hence

[TABLE]

This contradicts $\lambda^{\prime}\alpha=\alpha.$

If $K=T^{2,1}_{p,pk+1}$ then, by Lemma 2.5, we have

[TABLE]

Let $u=(xy)^{(p-1)k+1}$. Since $x\alpha=\alpha$ and $y\alpha>\alpha$ we have $u\alpha>\alpha$. Hence

[TABLE]

This implies that $((xy)^{k}x)^{-(p-3)}(xy)^{(p-1)k+1}\alpha=((xy)^{k}x)^{-(p-3)}u\alpha>u\alpha>\alpha.$ Hence

[TABLE]

This contradicts $\lambda^{\prime}\alpha=\alpha.$ ∎

Since $v\alpha\not=\alpha$ for all $\alpha\in\mathbb{R}$ and $v\alpha$ is a continuous function of $\alpha$, without loss of generality, we may assume $v\alpha>\alpha$ for all $\alpha\in\mathbb{R}$. Then $x\alpha=v^{s}\alpha>\alpha.$

Proposition 3.2**.**

With $\frac{r}{s}\geq 2g(K)-1$ we have $\alpha<y\alpha<x\alpha$ for all $\alpha\in\mathbb{R}$.

Proof.

With $\frac{r}{s}\geq 2g(K)-1$ and $s>0$, we have $-r+(2g(K)-1)s\leq 0$. Since $x=v^{s}$, $\lambda=v^{-r}$ and $v\alpha>\alpha$ for all $\alpha\in\mathbb{R}$, we have

[TABLE]

Case 1a: $K=T^{p-1,m}_{p,pk-1}.$ By Lemmas 2.2 and 2.5 we have

[TABLE]

By Lemma 2.1 we have $\rho(y)\alpha>\alpha$, $(yx)^{k-1}y^{n}\alpha>x^{n}(yx)^{k-1}\alpha$, $y^{n}(xy)^{k-1}\alpha>(xy)^{k-1}x^{n}\alpha$ for all $\alpha\in\mathbb{R}$ and $n\geq 1$. Hence

[TABLE]

Case 1b: $K=T^{p-1,m}_{p,pk+1}.$ By Lemmas 2.4 and 2.5 we have

[TABLE]

By Lemma 2.3 we have $\rho(y)\alpha>\alpha$, $(yx)^{k}y^{n}\alpha>x^{n}(yx)^{k}\alpha$, $y^{n}(xy)^{k}\alpha>(xy)^{k}x^{n}\alpha$ for all $\alpha\in\mathbb{R}$ and $n\geq 1$. Hence

[TABLE]

Case 2a: $K=T^{p-2,1}_{p,pk-1}.$ By Lemmas 2.2 and 2.5 we have

[TABLE]

By Lemma 2.1 we have $\rho(y)\alpha>\alpha$, $(yx)^{k-1}y^{n}\alpha>x^{n}(yx)^{k-1}\alpha$, $y^{n}(xy)^{k-1}\alpha>(xy)^{k-1}x^{n}\alpha$, $((yx)^{k-1}y)^{2}\alpha>x(yx)^{2k-1}\alpha$ for all $\alpha\in\mathbb{R}$ and $n\geq 1$. The latter implies that $y(yx)^{k-1}y\alpha>(yx)^{-(k-1)}x(yx)^{k}(yx)^{k-1}\alpha.$ Hence

[TABLE]

Case 2b: $K=T^{p-2,1}_{p,pk+1}.$ By Lemmas 2.4 and 2.5 we have

[TABLE]

By Lemma 2.3 we have $\rho(y)\alpha>\alpha$, $(yx)^{k}y^{n}\alpha>x^{n}(yx)^{k}\alpha$, $y^{n}(xy)^{k}\alpha>(xy)^{k}x^{n}\alpha$, $(xy)^{2k+1}\alpha>x((xy)^{k}x)^{2}\alpha$ for all $\alpha\in\mathbb{R}$ and $n\geq 1.$ The latter is equivalent to $y(xy)^{2k}\alpha>((xy)^{k}x)^{2}\alpha$, which implies that $y(xy)^{k}\alpha>((xy)^{k}x)^{2}(xy)^{-k}\alpha.$ Hence

[TABLE]

Case 3a: $K=T^{2,1}_{p,pk-1}.$ By Lemmas 2.2 and 2.5 we have

[TABLE]

By Lemma 2.1 we have $\rho(y)\alpha>\alpha$, $(yx)^{k-1}y^{n}\alpha>x^{n}(yx)^{k-1}\alpha$, $y^{n}(xy)^{k-1}\alpha>(xy)^{k-1}x^{n}\alpha$, $((yx)^{k-1}y)^{p-2}\alpha>x(yx)^{k(p-2)-1}\alpha$ for all $\alpha\in\mathbb{R}$ and $n\geq 1.$ The latter implies that $((yx)^{k-1}y)^{p-1}x\alpha>x(yx)^{k(p-2)-1}(yx)^{k}\alpha=x(yx)^{k-1}(yx)^{k(p-2)}\alpha.$ Hence

[TABLE]

Here we use $(yx)^{k}\alpha=(yx)^{k-1}yx\alpha>x(yx)^{k-1}x\alpha>xyx^{k}\alpha$ for all $\alpha\in\mathbb{R}$.

Case 3b: $K=T^{2,1}_{p,pk+1}.$ By Lemmas 2.4 and 2.5 we have

[TABLE]

By Lemma 2.3 we have $\rho(y)\alpha>\alpha$, $(yx)^{k}y^{n}\alpha>x^{n}(yx)^{k}\alpha$, $y^{n}(xy)^{k}\alpha>(xy)^{k}x^{n}\alpha$, $(xy)^{k(p-2)+1}\alpha>x((xy)^{k}x)^{p-2}\alpha$ for all $\alpha\in\mathbb{R}$ and $n\geq 1$. The latter implies that $(xy)^{k(p-1)+1}\alpha>((xy)^{k}x)^{p-1}\alpha.$ Hence

[TABLE]

This completes the proof of Proposition 3.2. ∎

By Lemmas 2.1 and 2.3 we have $(xy)^{j}x\alpha<y(xy)^{j}\alpha$ for all $\alpha\in\mathbb{R}$, where $j=k-1$ if $K=T^{l,m}_{p,pk-1}$ and $j=k$ if $K=T^{l,m}_{p,pk+1}.$ With $\frac{r}{s}\geq 2g(K)-1$, by Proposition 3.2 we have $x\alpha>y\alpha$ for all $\alpha\in\mathbb{R}$. Hence

[TABLE]

a contradiction. This completes the proof of Theorem 1.

Acknowledgements

The author has been partially supported by a grant from the Simons Foundation (#354595 to AT).