Remarks on SU(2)-simple knots and SU(2)-cyclic 3-manifolds

TL;DR

This paper explores properties of SU(2)-simple knots and SU(2)-cyclic 3-manifolds, establishing a characterization of SU(2)-simple Montesinos knots and proposing a conjecture relating SU(2)-cyclic rational homology 3-spheres to L-spaces.

Contribution

It proves that a Montesinos knot is SU(2)-simple if and only if it is a 2-bridge knot, extending previous results, and conjectures a link between SU(2)-cyclic rational homology 3-spheres and L-spaces.

Findings

Montesinos knots are SU(2)-simple iff they are 2-bridge knots.

Conjecture: SU(2)-cyclic rational homology 3-spheres are L-spaces.

Extension of Zentner's result on pretzel knots.

Abstract

We give some remarks on two closely related issues as stated in the title. In particular we show that a Montesinos knot is SU(2)-simple if and only if it is a 2-bridge knot, extending a result of Zentner for 3-tangle summand pretzel knots. We conjecture with some evidence that an SU(2)-cyclic rational homology 3-sphere is an L-space.

Equations

$${\mathcal{L}}^{0}(M_{i})=\{\begin{array}[]{ll}\mbox{slopes in the open interval $(p_{i}q_{i}-p_{i}-q_{i},\infty)$},&\mbox{if $p_{i}>0$,}\\ \mbox{slopes in the open interval $(-\infty,p_{i}q_{i}-p_{i}+q_{i})$},&\mbox{if $p_{i}<0$.}\end{array}$$

$$h({\mu }_1^m{\lambda }_1^n)=h({\mu }_1^{m-p_1{q}_{1}n}({\mu }_1^{p_1{q}_2}{\lambda }_1{)}^n)=({\mu }_2^{p_2{q}_2}{\lambda }_2{)}^{m-p_1{q}_{1}n}{\mu }_2^n={\mu }_2^{p_2{q}_2(m-p_1{q}_{1}n)+n}{\lambda }_2^{m-p_1{q}_{1}n}.$$

$$\frac{m}{n}=p_1{q}_1+\frac{b}{a-p_2{q}_{2}b}=p_1{q}_1+\frac{1}{\frac{a}{b}-p_2{q}_2}.$$

$$\begin{array}[]{ccc}\bar{R}({\Gamma})&\stackrel{{\scriptstyle{\sigma}}}{{{\longrightarrow}}}&\bar{R}({\Gamma})\\ {\downarrow}t&&{\downarrow}t\\ \bar{X}({\Gamma})&\stackrel{{\scriptstyle{\sigma}}}{{{\longrightarrow}}}&\bar{X}({\Gamma}).\end{array}$$

$$\Delta (p_1,...,p_n)=<a_1,...,a_n;a_i^{p_i}=1,i=1,...,n,a_1{a}_2\cdots a_n=1>.$$

$${\pi }_1(\tilde{M})\to {\pi }_1(\tilde{M}(\tilde{\mu }))\to \Delta (p_1,p_2,...,p_n)\to \Delta (p_1,p_2,p_3).$$

$$\begin{array}[]{ccccccc}\pi_{1}(\tilde{M},\tilde{x})&{\longrightarrow}&\pi_{1}(\tilde{M}(\tilde{\mu}),\tilde{x})&{\longrightarrow}\pi_{1}(\tilde{M}(\tilde{\mu}),\tilde{x}_{0})&{\longrightarrow}&{\Delta}(p_{1},....,p_{n})\\ &&&&&&\\ {\downarrow}(\cdot)^{\mu}&&&{\downarrow}\bar{\tau}_{*}&&{\downarrow}\bar{\tau}_{*}\\ &&&&&&\\ \pi_{1}(\tilde{M},\tilde{x})&{\longrightarrow}&\pi_{1}(\tilde{M}(\tilde{\mu}),\tilde{x})&{\longrightarrow}\pi_{1}(\tilde{M}(\tilde{\mu}),\tilde{x}_{0})&{\longrightarrow}&{\Delta}(p_{1},....,p_{n})\end{array}$$

$$a_1=b_1,a_2=b_1^{-1}{b}_2,a_3=b_2^{-1}{b}_3,\cdots ,a_{n-1}=b_{n-2}^{-1}{b}_{n-1},a_n=b_{n-1}^{-1}$$

$$b_1=a_1,b_2=a_1{a}_2,b_3=a_1{a}_2{a}_3,\cdots ,b_{n-1}=a_1{a}_2\cdots a_{n-1},b_{n-1}=a_n^{-1}.$$

$$\begin{array}[]{l}{\Delta}(p_{1},\cdots,p_{n})=<a_{1},...,a_{n};a_{i}^{p_{i}}=1,i=1,...,n,a_{1}a_{2}\cdots a_{n}=1>\\ {\longrightarrow}{\Delta}(p_{1},p_{2},p_{3})=<\bar{a}_{1},\bar{a}_{2},\bar{a}_{3};\bar{a}_{i}^{p_{i}}=1,i=1,2,3,\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}=1>\end{array}$$

$$\begin{array}[]{ccccccccc}\pi_{1}(\tilde{M})&{\longrightarrow}&\pi_{1}(\tilde{M}(\tilde{\mu}))&{\longrightarrow}&{\Delta}(p_{1},....,p_{n})&{\longrightarrow}&{\Delta}(p_{1},p_{2},p_{3})&\stackrel{{\scriptstyle\phi}}{{{\longrightarrow}}}PSL_{2}({\mathbb{C}})\\ &&&&&&&&\\ {\downarrow}(\cdot)^{\mu}&&{\downarrow}{\tau}_{*}&&{\downarrow}\bar{\tau}_{*}&&{\downarrow}\bar{\tau}_{\#}&&\\ &&&&&&&&\\ \pi_{1}(\tilde{M})&{\longrightarrow}&\pi_{1}(\tilde{M}(\tilde{\mu}))&{\longrightarrow}&{\Delta}(p_{1},....,p_{n})&{\longrightarrow}&{\Delta}(p_{1},p_{2},p_{3})&\stackrel{{\scriptstyle\phi}}{{{\longrightarrow}}}PSL_{2}({\mathbb{C}})\end{array}$$

Full text

Dedicated to Steve Boyer on the occasion of his 65th birthday

For a knot $K$ in $S^{3}$, $M_{{\mbox{\tiny K}}}$ will be its exterior and ${\mu}$ a meridian slope of $K$. Up to a choice of an orientation for ${\mu}$ and a choice of the base point for $\pi_{1}(M_{{\mbox{\tiny K}}})$, we may also consider ${\mu}$ as an element of $\pi_{1}(M_{{\mbox{\tiny K}}})$. A representation ${\rho}:\pi_{1}(M_{{\mbox{\tiny K}}}){\rightarrow}SU(2)$ is called trace free if the trace of ${\rho}({\mu})$ is zero (which is obviously well defined). An $SU(2)$-representation of $\pi_{1}(M_{{\mbox{\tiny K}}})$ is called binary dihedral if its image is isomorphic to a binary dihedral group. Note that every binary dihedral representation of $\pi_{1}(M_{{\mbox{\tiny K}}})$ is trace free [K, Proof of Theorem 10]. A knot $K$ is called $SU(2)$-simple if every irreducible trace free $SU(2)$-representation of $\pi_{1}(M_{{\mbox{\tiny K}}})$ is binary dihedral.

A Montesinos link is usually denoted by $K(e;q_{1}/p_{1},q_{2}/p_{2},...,q_{n}/p_{n})$ where $q_{i}/p_{i}$ represents a rational tangle, $|p_{i}|>1$ and $(q_{i},p_{i})=1$ for all $i$ (see Figure 1). By combining the $e$ twists in the figure with one of the tangles, we may assume that $e=0$, and we will simply write a Montesinos link as $K(q_{1}/p_{1},\cdots,q_{n}/p_{n})$ and sometimes we refer it as a cyclic tangle sum of $n$ rational tangles. When $q_{i}=1,i=1,...,n$, we get a pretzel link. In [Z1] it was shown that every pretzel knot $K(1/p,1/q,1/r)$, $p,q,r$ pairwise coprime, is not $SU(2)$-simple. In this paper we extend this result to all Montesinos knots of at least three rational tangle summands.

For any $SU(2)$-representation ${\rho}:\pi_{1}(M_{{\mbox{\tiny K}}}){\rightarrow}SU(2)$, let $\bar{\rho}:\pi_{1}(M_{{\mbox{\tiny K}}}){\rightarrow}PSU(2)$ be the induced $PSU(2)$-representation. If ${\rho}:\pi_{1}(M_{{\mbox{\tiny K}}}){\rightarrow}SU(2)$ is a trace free representation, then ${\rho}({\mu})$ is an order $4$ matrix and ${\rho}({\mu}^{2})=-I$, where $I$ is the identity matrix of $SU(2)$. So $\bar{\rho}:\pi_{1}(M_{{\mbox{\tiny K}}}){\rightarrow}PSU(2)$ factors through the quotient group $\pi_{1}(M_{{\mbox{\tiny K}}})/<{\mu}^{2}>$, where $<{\mu}^{2}>$ denote the normal subgroup of $\pi_{1}(M_{{\mbox{\tiny K}}})$ generated by ${\mu}^{2}$. Let $\Sigma_{2}(K)$ denote the double branched cover of $(S^{3},K)$, then $\pi_{1}(\Sigma_{2}(K))$ is an index two subgroup of $\pi_{1}(M_{{\mbox{\tiny K}}})/<{\mu}^{2}>$. It is known that a trace free irreducible representation ${\rho}:\pi_{1}(M_{{\mbox{\tiny K}}}){\rightarrow}SU(2)$ is binary dihedral if and only if the restriction of $\bar{\rho}$ on $\pi_{1}(\Sigma_{2}(K))$ has nontrivial cyclic image [K, Section I.E]. For any $2$-bridge knot $K$, $\Sigma_{2}(K)$ is a lens space and so every irreducible trace free SU(2)-representation of $\pi_{1}(M_{{\mbox{\tiny K}}})$ is binary dihedral, that is, every $2$-bridge knot is $SU(2)$-simple. As a Montesinos knot is a $2$-bridge knot if and only if it has lass than three rational tangle summands, we have

By [KM, Corollary 7.17], every nontrivial knot in $S^{3}$ has an irreducible trace free $SU(2)$-representation. It follows that if the double branched cover of a nontrivial knot $K$ is a homology $3$-sphere, i.e. if the knot determinant $|{\Delta}_{{\mbox{\tiny K}}}(-1)|=1$ where ${\Delta}_{{\mbox{\tiny K}}}(t)$ is the Alexander polynomial of $K$, then $K$ is not $SU(2)$-simple.

A $3$-manifold $Y$ is called $SU(2)$-cyclic (resp. $PSU(2)$-cyclic) if every $SU(2)$-representation (resp. $PSU(2)$-representation) of $\pi_{1}(Y)$ has cyclic image. In general $PSU(2)$-cyclic is a stronger condition than $SU(2)$-cyclic, that is, $PSU(2)$-cyclic implies $SU(2)$-cyclic but not the other way around. Since $\Sigma_{2}(K)$ is an ${\mathbb{Z}}_{2}$-homology $3$-sphere [R, Cor 3 of 8D], every $PSU(2)$-representation of $\pi_{1}(\Sigma_{2}(K))$ lifts to an $SU(2)$-representation [BZ, Page 752] and thus $\Sigma_{2}(K)$ is $SU(2)$-cyclic if and only if it is $PSU(2)$-cyclic. So if $\Sigma_{2}(K)$ is $SU(2)$-cyclic, then $K$ is an $SU(2)$-simple knot. The following question concerns the converse.

One may consider an $SU(2)$-cyclic $3$-manifold as an $SU(2)$-representation L-space. The following conjecture suggests that for a rational homology $3$-sphere being $SU(2)$-cyclic is more restrictive than being a usual $L$-space in the Heegaard Floer homology sense.

Here are some evidences for the conjecture. Let $K_{1}=T(p_{1},q_{1})$ and $K_{2}=T(p_{2},q_{2})$ be two torus knots in $S^{3}$, and let $M_{1}$ and $M_{2}$ be their exteriors. Let $Y(T(p_{1},q_{1}),T(p_{2},q_{2}))$ be the graph manifold obtained by gluing $M_{1}$ and $M_{2}$ along their boundary tori by an orientation reversing homeomorphism $h:{\partial}M_{1}{\rightarrow}{\partial}M_{2}$ which identifies the meridian slope in ${\partial}M_{1}$ to the Seifert fiber slope in ${\partial}M_{2}$ and identifies the Seifert fiber slope in ${\partial}M_{1}$ with the meridian slope in ${\partial}M_{2}$. By [Mot, Proposition 5] $Y(T(p_{1},q_{2}),T(P_{2},q_{2}))$ has only cyclic $PSL_{2}({\mathbb{C}})$-representations. (Although it was assumed in [Mot] that all $p_{1},q_{1},p_{2},q_{2}$ are positive, the same argument with obvious modification works without this assumption). Hence $Y(T(p_{1},q_{1}),T(p_{2},q_{2}))$ is $SU(2)$-cyclic.

**Proof.  **We prove this assertion by applying [HW, Theorem 1.6]. By that theorem, we just need to verify that $h(\mathcal{L}^{\circ}_{M_{1}})\cup\mathcal{L}^{\circ}_{M_{2}}\cong\mathbb{Q}\cup\{1/0\}$, where $\mathcal{L}^{0}_{M_{i}}$ is the interior of the set of $L$-space filling slopes of $M_{i}$, $i=1,2$. Note that a general torus knot can be expressed as $T(p,q)$ with $(p,q)=1$ and $|p|,q\geq 2$. By [OS, Corollary 1.4]

[TABLE]

Let ${\mu}_{i},{\lambda}_{i}$ be the meridian and longitude of $K_{i}$. Note that $p_{i}q_{i}$ is the Seifert fiber slope in ${\partial}M_{i}$. We have $h({\mu}_{1})={\mu}_{2}^{p_{2}q_{2}}{\lambda}_{2}$ and $h({\mu}_{1}^{p_{1}q_{1}}{\lambda}_{1})={\mu}_{2}$. Hence for a general slope $m/n$ in ${\partial}M_{1}$, where $m,n$ are relative prime,

[TABLE]

Now suppose $a/b$ is a slope in ${\partial}M_{2}$, where $a,b$ are relatively prime. Choose $n=a-p_{2}q_{2}b$ and $m=p_{1}q_{1}(a-p_{2}q_{2}b)+b$, then $m,n$ are relatively prime, $h(m/n)=a/b$, and

[TABLE]

Case 1. $p_{1}>0$ and $p_{2}>0$.

For any $a/b\notin{\mathcal{L}}^{0}(M_{2})$, i.e. either $a/b=1/0$ or $a/b$ is finite and $a/b\leq p_{2}q_{2}-p_{2}-q_{2}$ by (1), choose correspondingly in ${\partial}M_{1}$ the slope $m/n=p_{1}q_{1}$ or as in (2) which yields $m/n\geq p_{1}q_{1}+\frac{1}{-p_{2}-q_{2}}>p_{1}q_{1}-1$. So in either case $m/n\in{\mathcal{L}}^{0}(M_{1})$ by (1) and $h(m/n)=a/b$, which means $h(\mathcal{L}^{\circ}_{M_{1}})\cup\mathcal{L}^{\circ}_{M_{2}}\cong\mathbb{Q}\cup\{1/0\}$ in this case.

Case 2. $p_{1}>0$ and $p_{2}<0$.

For any $a/b\notin{\mathcal{L}}^{0}(M_{2})$, we may assume that $a/b$ is finite and so $a/b\geq p_{2}q_{2}-p_{2}+q_{2}$ by (1). So $\frac{a}{b}-p_{2}q_{2}$ is positive. Choose the slope $m/n$ in ${\partial}M_{1}$ as in (2) which yields $m/n>p_{1}q_{1}$ in this case. So $m/n\in{\mathcal{L}}^{0}(M_{1})$ by (1) and $h(m/n)=a/b$. Thus $h(\mathcal{L}^{\circ}_{M_{1}})\cup\mathcal{L}^{\circ}_{M_{2}}\cong\mathbb{Q}\cup\{1/0\}$ holds in this case.

Case 3. $p_{1}<0$ and $p_{2}>0$.

This case is really Case 2 if we switch $K_{1}$ and $K_{2}$.

Case 4. $p_{1}<0$ and $p_{2}<0$.

For any $a/b\notin{\mathcal{L}}^{0}(M_{2})$, again we may assume $a/b$ is finite and so $a/b\geq p_{2}q_{2}-p_{2}+q_{2}$ by (1). Choose the slope $m/n$ in ${\partial}M_{1}$ as in (2) which yields $m/n\leq p_{1}q_{1}+\frac{1}{-p_{2}+q_{2}}<p_{1}q_{1}+1$. So $m/n\in{\mathcal{L}}^{0}(M_{1})$, $h(m/n)=a/b$ and we have $h(\mathcal{L}^{\circ}_{M_{1}})\cup\mathcal{L}^{\circ}_{M_{2}}\cong\mathbb{Q}\cup\{1/0\}$.

The proof of Proposition 5 is now completed.  $\diamondsuit$

It was shown in [Z2] that if $p_{1}q_{1}p_{2}q_{2}-1$ is odd, then $Y(T(p_{1},q_{1}),T(p_{2},q_{2}))$ is the double branched cover of an alternating knot in $S^{3}$, so $Y(T(p_{1},q_{1}),T(p_{2},q_{2}))$ is an $L$-space and the knot in $S^{3}$ is an $SU(2)$-simple knot (and is an arborescent knot) but is not a $2$-bridge knot.

There are also examples of hyperbolic rational homology $3$-spheres which are $SU(2)$-cyclic [C]. These examples are also double branched covers of alternating knots in $S^{3}$ and thus are $L$-spaces. These alternating knots are thus $SU(2)$-simple but are not arborescent.

Proof of Theorem 1. Let $K=K(q_{1}/p_{1},\cdots,q_{n}/p_{n})$ be a Montesinos knot with $n\geq 3$. We need to show that $\pi_{1}(M_{{\mbox{\tiny K}}})$ has an irreducible trace free $SU(2)$-representation which is not binary dihedral. Here is an outline of how the proof goes. We show that the double branched cover $\Sigma_{2}(K)$ has an irreducible $PSU(2)$-representation $\bar{\rho}_{0}$ which can be extended to an $PSU(2)$-representation $\bar{\rho}$ of $\pi_{1}(M_{{\mbox{\tiny K}}})$ up to conjugation. This $PSU(2)$-representation $\bar{\rho}$ lifts to an $SU(2)$-representation ${\rho}$ of $\pi_{1}(M_{{\mbox{\tiny K}}})$ which is automatically trace free. Since $\bar{\rho}_{0}$ is an irreducible representation, ${\rho}$ is not binary dihedral. The existence of $\bar{\rho}_{0}$ is provided by [B]. We first apply some ideas from [Mat] to show that $\bar{\rho}_{0}$ extends to a unique $PSL_{2}({\mathbb{C}})$-representation $\bar{\rho}$ of $\pi_{1}(M_{{\mbox{\tiny K}}})$. Then we further show that this $\bar{\rho}$ is conjugate to an $PSU(2)$-representation by applying some results from [HP][CD].

Now we give the details of the proof. For a finitely generated group ${\Gamma}$, $\bar{R}({\Gamma})=Hom({\Gamma},PSL_{2}({\mathbb{C}}))$ denotes the $PSL_{2}({\mathbb{C}})$ representation variety of ${\Gamma}$ and $\bar{X}({\Gamma})$ the $PSL_{2}({\mathbb{C}})$ character variety of ${\Gamma}$. Let $t:\bar{R}({\Gamma}){\rightarrow}\bar{X}({\Gamma})$ be the map which sends a representation $\bar{\rho}$ to its character $\chi_{\bar{\rho}}$. We shall write an element in $PSL_{2}({\mathbb{C}})$ as $\bar{A}$ which is the image of an element $A$ in $SL_{2}({\mathbb{C}})$ under the quotient map $SL_{2}({\mathbb{C}}){\rightarrow}PSL_{2}({\mathbb{C}})$ and for convenience we sometimes call elements in $PSL_{2}({\mathbb{C}})$ as matrices. For any $\bar{A}\in PSL_{2}({\mathbb{C}})$ define $tr^{2}(\bar{A})=(trace(A))^{2}$ which is obviously well defined. Recall that the character $\chi_{\bar{\rho}}$ of an $PSL_{2}({\mathbb{C}})$-representation $\bar{\rho}$ is the function $\chi_{\bar{\rho}}:{\Gamma}{\rightarrow}{\mathbb{C}}$ defined by $\chi_{\bar{\rho}}({\gamma})=tr^{2}(\bar{\rho}({\gamma}))$.

A character $\chi_{\bar{\rho}}$ is real if $\chi_{\bar{\rho}}({\gamma})\in\mathbb{R}$ for all ${\gamma}\in{\Gamma}$. If we consider $\bar{X}({\Gamma})$ as an algebraic subset in ${\mathbb{C}}^{n}$ (for some $n$), then real characters of $\bar{X}({\Gamma})$ correspond to real points of $\bar{X}({\Gamma})$, i.e. points of $\bar{X}({\Gamma})\cap\mathbb{R}^{n}$. If ${\sigma}:{\mathbb{C}}^{n}{\rightarrow}{\mathbb{C}}^{n}$ (for each $n\geq 1$) denotes the operation of coordinatewise taking complex conjugation, then any complex affine algebraic set $Y$ in ${\mathbb{C}}^{n}$ defined over $\mathbb{Q}$ is invariant under ${\sigma}$ and the set of real points of $Y$ is precisely the fixed point set of ${\sigma}$ in $Y$. Note that $\bar{R}({\Gamma})$ and $\bar{X}({\Gamma})$ are both algebraic sets defined over $\mathbb{Q}$ and that the map $t:\bar{R}({\Gamma}){\rightarrow}\bar{X}({\Gamma})$ is an algebraic map defined over $\mathbb{Q}$, we thus have the following commutative diagram of maps:

[TABLE]

It follows that ${\sigma}(\chi_{\bar{\rho}})=\chi_{{\sigma}(\bar{\rho})}$.

A representation $\bar{\rho}\in\bar{R}({\Gamma})$ is irreducible if the image of $\bar{\rho}$ cannot be conjugated into the set {$\bar{A}$; $A$ upper triangular}. Two irreducible representations in $\bar{R}({\Gamma})$ are conjugate if and only if they have the same character.

If $W$ is a compact manifold, $\bar{R}(W)$ and $\bar{X}(W)$ denote $\bar{R}(\pi_{1}W)$ and $\bar{X}(\pi_{1}W)$ respectively.

Let $K=K(q_{1}/p_{1},\cdots,q_{n}/p_{n})$ and $M=M_{{\mbox{\tiny K}}}$. We may assume that all $p_{i}$ are positive (by changing the sign of $q_{i}$ if necessary). Let $p:\tilde{M}{\rightarrow}M$ be the $2$-fold cyclic covering and let $\tilde{\mu}=p^{-1}({\mu})$ which is a connected simple closed essential curve in ${\partial}\tilde{M}$ which double covers ${\mu}$. Then $p_{*}:\pi_{1}(\tilde{M}){\rightarrow}\pi_{1}(M)$ is an injection and we may consider $\pi_{1}(\tilde{M})$ as an index two normal subgroup of $\pi_{1}(M)$, in which $\tilde{\mu}={\mu}^{2}$. Dehn filling $\tilde{M}(\tilde{\mu})$ of $\tilde{M}$ with the slope $\tilde{\mu}$ is the double branched cover $\Sigma_{2}(K)$ of $(S^{3},K)$. The covering involution ${\tau}$ on $\tilde{M}$ extends to one on $\tilde{M}(\tilde{\mu})$ which we still denote by ${\tau}$. Montesinos proved in [Mont1][Mont2] that $\tilde{M}(\tilde{\mu})$ admits a Seifert fibering invariant under the covering involution ${\tau}$, the base orbifold of the Seifert fibred space is $S^{2}(p_{1},...,p_{n})$ which is the $2$-sphere with $n$ cone points of orders $p_{1},...,p_{n}$, and ${\tau}$ descends down to an involution $\bar{\tau}$ on $S^{2}(p_{1},...,p_{n})$ which is a reflection in a circle passing through all the cone points (see Figure 2).

We denote the orbifold fundamental group of $S^{2}(p_{1},...,p_{n})$ by ${\Delta}(p_{1},...,p_{n})$ which has the following presentation:

[TABLE]

It was shown in [Mat, Section 3.3] that when $n=3$ any irreducible $PSL_{2}({\mathbb{C}})$-representation of $\pi_{1}(\tilde{M})$ which factors through $\pi_{1}(\tilde{M}(\tilde{\mu}))$ has a unique extension to $\pi_{1}(M)$. Note that this extended representation can be lifted to a trace free $SL_{2}({\mathbb{C}})$-representation of $\pi_{1}(M)$. We shall slightly extend this result to the following

[TABLE]

Let $\phi:{\Delta}(p_{1},p_{2},p_{3}){\rightarrow}PSL_{2}({\mathbb{C}})$ be any irreducible representation. Then $\bar{\rho}_{0}=\phi\circ{\delta}$ has a unique extension to $\pi_{1}(M_{{\mbox{\tiny K}}})$.

**Proof.  **The proof for uniqueness is verbatim as that given in [Mat] on page 38-39. We need to note that as $K(q_{1}/p_{1},\cdots,q_{n}/p_{n})$ is a knot at most one of $p_{i}$’s is even. So [Mat, Lemma 2.4.9] still applies to our current case.

Again this claim can be proved verbatim as that of [Mat, Claim 3.3.2].

So to finish the proof of Proposition 6, we just need to find an $\bar{A}\in PSL_{2}({\mathbb{C}})$ with the properties stated in Claim 7, which is what we are going to do in the rest of the proof of Proposition 6. Recall that $\tilde{M}(\tilde{\mu})$ is the Dehn filling of $\tilde{M}$ with a solid torus $N$ whose meridian slope is identified with the slope $\tilde{\mu}$. The core circle of $N$ is the fixed point set of ${\tau}$ in $\tilde{M}(\tilde{\mu})$. Let $D$ be a meridian disk of $N$ such that the fixed point of ${\tau}$ in $D$ (the center point of $D$) is disjoint from the singular fibers of the Seifert fibred space $\tilde{M}(\tilde{\mu})$. Choose a point $\tilde{x}$ in ${\partial}D$ and let $\tilde{x}_{0}$ be the center point of $D$. Then arguing as on [Mat, Page 40] we have the following commutative diagram:

[TABLE]

where $(\cdot)^{\mu}:\pi_{1}(\tilde{M},\tilde{x}){\rightarrow}\pi_{1}(\tilde{M},\tilde{x})$ corresponds to the conjugation action by ${\mu}$, i.e. $(\beta)^{\mu}={\mu}\beta{\mu}^{-1}$ and ${\Delta}(p_{1},...,p_{n})$ is the orbifold fundamental group of $S^{2}(p_{1},...,p_{n})$ whose base point is the image $x_{0}$ of the point $\tilde{x}_{0}$ under the quotient map $\tilde{M}(\tilde{\mu}){\rightarrow}S^{2}(p_{1},...,p_{n})$.

Figure 2 shows the generating set $b_{1},...,b_{n-1}$ of the orbifold fundamental group ${\Delta}(p_{1},...,p_{n})$ of $S^{2}(p_{1},...,p_{n})$. In fact we have

[TABLE]

and conversely

[TABLE]

Obviously from Figure 2 the induced isomorphism $\bar{\tau}_{*}:{\Delta}(p_{1},...,p_{n}){\rightarrow}{\Delta}(p_{1},...,p_{n})$ sends $b_{i}$ to $b_{i}^{-1}$, $i=1,...,n-1$. So we have $\bar{\tau}_{*}(a_{1})=a_{1}^{-1}$, $\bar{\tau}_{*}(a_{2})=b_{1}b_{2}^{-1}=a_{1}a_{2}^{-1}a_{1}^{-1}$, $\bar{\tau}_{*}(a_{3})=b_{2}b_{3}^{-1}=a_{1}a_{2}a_{3}^{-1}a_{2}^{-1}a_{1}^{-1}$, $\cdots$, $\bar{\tau}_{*}(a_{n-1})=b_{n-2}b_{n-1}^{-1}=a_{1}a_{2}\cdots a_{n-2}a_{n-1}^{-1}a_{n-2}^{-1}\cdots a_{2}^{-1}a_{1}^{-1}$, $\bar{\tau}_{*}(a_{n})=\bar{\tau}_{*}(b_{n-1}^{-1})=b_{n-1}=a_{n}^{-1}$. Since the quotient homomorphism

[TABLE]

sends $a_{i}$ to $\bar{a}_{i}$, $i=1,2,3$, and send $a_{i}$ to $1$, $i=4,...,n$. we see that $\bar{\tau}_{*}$ descents to an isomorphism $\bar{\tau}_{\#}:{\Delta}(p_{1},p_{2},p_{3}){\rightarrow}{\Delta}(p_{1},p_{2},p_{3})$ such that $\bar{\tau}_{\#}(\bar{a}_{1})=\bar{a}_{1}^{-1}$, $\bar{\tau}_{\#}(\bar{a}_{2})=\bar{a}_{1}\bar{a}_{2}^{-1}\bar{a}_{1}^{-1}$, $\bar{\tau}_{\#}(\bar{a}_{3})=\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}^{-1}\bar{a}_{2}^{-1}\bar{a}_{1}^{-1}$ and we have the following commutative diagram:

[TABLE]

So $\bar{\tau}_{\#}(\bar{a}_{1}\bar{a}_{2})=\bar{a}_{2}^{-1}\bar{a}_{1}^{-1}=(\bar{a}_{1}\bar{a}_{2})^{-1}$. Since ${\Delta}(p_{1},p_{2},p_{3})$ is generated by $\bar{a}_{1},\bar{\alpha}_{2}$, we see by applying [BZ, Lemma 3.1] that $\phi$ and $\phi\circ\bar{\tau}_{\#}$ have the same $PSL_{2}({\mathbb{C}})$ character. (In fact if $\phi(\bar{a}_{1})=\bar{A}_{1}$ and $\phi(\bar{a}_{2})=\bar{A}_{2}$, then $\phi(\bar{a}_{1}\bar{a}_{2})=\bar{A}_{1}\bar{A}_{2}=\overline{A_{1}A_{2}}$, $(\phi\circ\bar{\tau}_{\#})(\bar{a}_{1})=(\bar{A}_{1})^{-1}=\overline{A_{1}^{-1}}$, $(\phi\circ\bar{\tau}_{\#})(\bar{a}_{2})=\bar{A}_{1}(\bar{A}_{2})^{-1}(\bar{A}_{1})^{-1}=\overline{A_{1}A_{2}^{-1}A_{1}^{-1}}$ and $(\phi\circ\bar{\tau}_{\#})(\bar{a}_{1}\bar{a}_{2})=(\bar{A}_{2})^{-1}(\bar{A}_{1})^{-1}=\overline{(A_{1}A_{2})^{-1}}$. Now let $F_{2}$ be the free group on two generators $\xi_{1}$ and $\xi_{2}$. Let ${\rho}_{1}$ and ${\rho}_{2}$ be the $SL_{2}({\mathbb{C}})$ representations of $F_{2}$ defined by ${\rho}_{1}(\xi_{i})=A_{i}$, $i=1,2$, and ${\rho}_{2}(\xi_{1})=A_{1}^{-1}$, ${\rho}_{2}(\xi_{2})=A_{1}A_{2}^{-1}A_{1}^{-1}$. Then one can easily verify that $tr({\rho}_{1}(\xi_{1}))=tr({\rho}_{2}({\xi}_{1}))$, $tr({\rho}_{1}(\xi_{2}))=tr({\rho}_{2}({\xi}_{2}))$ and $tr({\rho}_{1}(\xi_{1}\xi_{2}))=tr({\rho}_{2}({\xi}_{1}\xi_{2}))$. So [BZ, Lemma 3.1] applies.) So $\phi$ and $\phi\circ\bar{\tau}_{\#}$ are conjugate $PSL_{2}({\mathbb{C}})$ representations, that is, there is $\bar{A}\in PSL_{2}({\mathbb{C}})$ with $\bar{A}\phi\bar{A}^{-1}=\phi\circ\bar{\tau}_{\#}$. Combining this with the definition of $\bar{\rho}_{0}$ and the last commutative diagram, we see that $\bar{A}\bar{\rho}_{0}({\beta})\bar{A}^{-1}=\bar{\rho}_{0}({\mu}{\beta}{\mu}^{-1})$ for each ${\beta}\in\pi_{1}(\tilde{M})$. The proof of Proposition 6 is now finished.  $\diamondsuit$

Now by [B], every triangle group ${\Delta}(p_{1},p_{2},p_{3})$ has an irreducible $SO(3)\cong PSU(2)$-representation. Therefore there is an irreducible representation $\bar{\rho}_{0}$ as given in Proposition 6 with its image contained in $PSU(2)$. So the character $\chi_{\bar{\rho}_{0}}$ of $\bar{\rho}_{0}$ is real valued. Let $\bar{\rho}$ be the unique extension of $\bar{\rho}_{0}$ to $\pi_{1}(M)$ as guaranteed by Proposition 6.

Suppose otherwise. Recall that ${\sigma}:\bar{X}(M){\rightarrow}\bar{X}(M)$ is the operation of taking complex conjugation and a character is real valued if and only if it is a fixed point of ${\sigma}$. So $\chi_{\bar{\rho}}\neq{\sigma}(\chi_{\bar{\rho}})=\chi_{{\sigma}(\bar{\rho})}$ are two different characters of irreducible representations and thus $\bar{\rho}$ and ${\sigma}(\bar{\rho})$ are non-conjugate representations. But $\chi_{\bar{\rho}_{0}}={\sigma}(\chi_{\bar{\rho}_{0}})=\chi_{{\sigma}(\bar{\rho}_{0})}$ and $\bar{\rho}_{0}$ is irreducible. Hence $\bar{\rho}_{0}$ and ${\sigma}(\bar{\rho}_{0})$ are conjugate representations, that is, there is $\bar{B}\in PSL_{2}({\mathbb{C}})$ such that $\bar{\rho}_{0}=\bar{B}{\sigma}(\bar{\rho}_{0})\bar{B}^{-1}$. Hence $\bar{\rho}$ and $\bar{B}{\sigma}(\bar{\rho})\bar{B}^{-1}$ are non-conjugate representations which have the same restriction on $\pi_{1}(\tilde{M})$. We get a contradiction with Proposition 6.

By [HP, Lemma 10.1] an $PSL_{2}({\mathbb{C}})$-character $\chi_{\bar{\rho}}$ of a finitely generated group is real valued if and only if the image of $\bar{\rho}$ can be conjugated into $PSU(2)$ or $PGL_{2}(\mathbb{R})$. So our current representation $\bar{\rho}$ can be conjugated into $PSU(2)$ or $PGL_{2}(\mathbb{R})$. If it can be conjugated into $PSU(2)$, then we are done because this conjugated representation lifts to a trace free $SU(2)$-representation which is not binary dihedral. So suppose that $\bar{\rho}$ is conjugate to an $PGL_{2}(\mathbb{R})$-representation $\bar{\rho}^{\prime}$. As noted in [HP] right after Lemma 10.1, $PGL_{2}(\mathbb{R})\subset PGL_{2}({\mathbb{C}})\cong PSL_{2}({\mathbb{C}})$ has two components, the identity component is $PSL_{2}(\mathbb{R})$ and the other component consists of matrices of determinant $-1$ (which in $PSL_{2}({\mathbb{C}})$ are represented by matrices with entries in ${\mathbb{C}}$ with zero real part). Considering the action of $PSL_{2}({\mathbb{C}})$ on hyperbolic space $\mathbb{H}^{3}$ by orientation preserving isometries, the group $PGL_{2}(\mathbb{R})$ is the stabilizer of a total geodesic plane in $\mathbb{H}^{3}$, and an element of $PGL_{2}(\mathbb{R})$ is contained in $PSL_{2}(\mathbb{R})$ if and only if it preserves the orientation of the plane. Since $\pi_{1}(\tilde{M})$ is the unique index two normal subgroup of $\pi_{1}(M)$, $\pi_{1}(\tilde{M})$ is generated by elements ${\gamma}^{2}$, ${\gamma}\in\pi_{1}(M)$. Therefore the image of the restriction $\bar{\rho}_{0}^{\prime}$ of $\bar{\rho}^{\prime}$ on $\pi_{1}(\tilde{M})$ consists of elements preserving the orientation of the total geodesic plane mentioned above, i.e. the image of $\bar{\rho}_{0}^{\prime}$ is contained in $PSL_{2}(\mathbb{R})$. So the image of $\bar{\rho}_{0}$ can be conjugated into $PSL_{2}(\mathbb{R})$. But the image of $\bar{\rho}_{0}$ is contained in $PSU(2)$. [CD, Lemma 2.10] says that if an $PSU(2)$-presentation can be conjugated into an $PSL_{2}(\mathbb{R})$-representation, then it is a reducible representation. So our $\bar{\rho}_{0}$ is a reducible representation. We arrive at a contradiction, which completes the proof of Theorem 1.