Character varieties of odd classical pretzel knots

TL;DR

This paper computes the SL(2,C)-character varieties for odd classical pretzel knots and introduces a method to calculate their A-polynomials, advancing understanding of knot invariants.

Contribution

It provides explicit character varieties for odd classical pretzel knots and a novel method for computing their A-polynomials.

Findings

Explicit character varieties for all odd classical pretzel knots.

A new method for calculating A-polynomials of these knots.

Enhanced tools for studying knot invariants and their geometric properties.

Abstract

We determine the ${\rm SL}(2,\mathbb{C})$-character variety for each odd classical pretzel knot $P(2k_1+1,2k_2+1,2k_3+1)$, and present a method for computing its A-polynomial.

Equations

$$\mathcal{X}(K)=\{\chi (\rho ):\rho \in \mathcal{R}(K)\},$$

$${\mathcal{X}}^{\mathrm{irr}}(K)=\{\chi (\rho ):\rho \in {\mathcal{R}}^{\mathrm{irr}}(K)\}$$

$${\omega }_k(t)=\{\begin{array}{ll}(a^k-a^{-k})/(a-a^{-1}),& a\notin \{\pm 1\},\\ k{a}^{k-1},& a\in \{\pm 1\};\end{array}$$

$${\omega }_k(t)+{\omega }_{-k}(t)$$

$${\omega }_{k+1}(t)-t{\omega }_k(t)+{\omega }_{k-1}(t)$$

$${\omega }_k(t{)}^2-t{\omega }_k(t){\omega }_{k-1}(t)+{\omega }_{k-1}(t{)}^2$$

$$X^{-1}=tI-X,$$

$$X^k={\omega }_k(t)X-{\omega }_{k-1}(t)I.$$

$$XYX$$

$$XY+YX$$

$$X(t_{2}I-Y)+Y(t_{1}I-X)=X{Y}^{-1}+Y{X}^{-1}=\mathrm{tr}(X{Y}^{-1})\cdot I=(t_1{t}_2-t_{12})I.$$

$${\nu }_0$$

$${\nu }_1$$

$$t_{123}^2-{\nu }_1{t}_{123}+{\nu }_0=0.$$

$$\displaystyle U(a,b)=\left(\begin{array}[]{cc}a&b\\ 0&a^{-1}\end{array}\right),\qquad V(a,b)=\left(\begin{array}[]{cc}a&0\\ b&a^{-1}\end{array}\right).$$

$$z_{k+2}=z_{k+1}{z}_k{z}_{k+1}^{-1}=(z_{k+1}{z}_k)z_k(z_{k+1}{z}_k{)}^{-1}=(z_1{z}_0)z_k(z_1{z}_0{)}^{-1}.$$

$$z_{2h}=(z_1{z}_0{)}^h{z}_0(z_1{z}_0{)}^{-h},z_{2h+1}=(z_1{z}_0{)}^h{z}_1(z_1{z}_0{)}^{-h}.$$

$$K=P(2k_1+1,2k_2+1,2k_3+1),k_1,k_2,k_3\in \mathbb{Z}.$$

$$(X_3{X}_1^{-1}{)}^{k_2+1}{X}_1(X_3{X}_1^{-1}{)}^{-k_2-1}$$

$$(X_1{X}_2^{-1}{)}^{k_3+1}{X}_2(X_1{X}_2^{-1}{)}^{-k_3-1}$$

$$(X_2{X}_3^{-1}{)}^{k_1+1}{X}_3(X_2{X}_3^{-1}{)}^{-k_1-1}$$

$$Y_j$$

$$A_j$$

$$A_1=A_2=A_3.$$

$$A_j$$

$$=U(u^2,((k_j+1)u^{-1}-k_{j}u)a_{j+}+((k_j+1)u-k_j{u}^{-1})a_{j-}).$$

$$(k_1+k_3+1)(u^{-1}-u)(a_2-1)+((k_1+1)u-k_1{u}^{-1})(a_3-1)=0,$$

$$((k_1+1)u^{-1}-k_{1}u)(a_2-1)+(k_1+k_2+1)(u-u^{-1})(a_3-1)=0.$$

$$\mathrm{tr}(X_1{X}_2{X}_3)$$

$$\mathrm{tr}(Y_j)=s_j=v_j+v_j^{-1},j=1,2,3.$$

$$\mathrm{tr}(X_{j+}{X}_{j-})=\mathrm{tr}(X_{j+}(tI-X_{j-}^{-1}))=t^2-s_j.$$

$${\sigma }_1=s_1+s_2+s_3,{\sigma }_2$$

$$\tau$$

$$\delta$$

$$\kappa$$

$$\kappa =\delta .$$

$$\delta =-(s_3-v_1{v}_2-v_1^{-1}{v}_2^{-1})(s_3-v_1{v}_2^{-1}-v_1^{-1}{v}_2),$$

Full text

Keywords: ${\rm SL}(2,\mathbb{C})$-representation, character variety, odd classical pretzel knot, A-polynomial

MSC2010: 57M25, 57M27

1 Introduction

For a link $K\subset S^{3}$, let $E_{K}=S^{3}-N(K)$, with $N(K)$ a tubular neighborhood. The ${\rm SL}(2,\mathbb{C})$-representation variety of $K$ is the set $\mathcal{R}(K)$ of representations $\rho:\pi_{1}(E_{K})\to{\rm SL}(2,\mathbb{C})$, and the character variety of $K$ is

[TABLE]

where the character $\chi(\rho):\pi_{1}(E_{K})\to\mathbb{C}$ sends each $x\in\pi_{1}(E_{K})$ to ${\rm tr}(\rho(x))$.

It is well-known that (see [4]) both $\mathcal{R}(K)$ and $\mathcal{X}(K)$ can be defined by a finite set of polynomial equations, and this is why they are so named. Actually, $\mathcal{X}(K)$ is the geometric-invariant-theoretic quotient of $\mathcal{R}(K)$ under the action of ${\rm SL}(2,\mathbb{C})$ via conjugation. To be more plain, denoting the subset of $\mathcal{R}(K)$ consisting of irreducible representations by $\mathcal{R}^{\rm irr}(K)$, it is also well-known that up to conjugacy, each $\rho\in\mathcal{R}^{\rm irr}(K)$ is determined by $\chi(\rho)$. Call

[TABLE]

the irreducible character variety of $K$. Our main concern is to study $\mathcal{X}^{\rm irr}(K)$; the characters of reducible representations are relatively easy to understand.

Since the seminal paper [4], much attention has been attracted to representations of 3-manifold groups into ${\rm SL}(2,\mathbb{C})$. One reason is that, representation/character variety encodes much topological and geometric information on the underlying 3-manifolds.

Among the invariants extracted from character variety, the A-polynomial (the definition is recalled in Section 4) proposed in [3] is rather important, in that, it plays a key role not only in the problem of exceptional fillings, but also in the celebrated AJ Conjecture which involves the behavior of quantum invariants.

Till now, character varieties have been found for the following links: torus knots [11], double twist knots [12], double twist links [14], $(-2,2m+1,2n)$-pretzel links [15]. In this paper, we aim to determine $\mathcal{X}^{\rm irr}(K)$ when $K$ is an odd classical pretzel knot. This is the first time to deal with a 3-parameter family of knots. We are faced with the complications caused by three non-commuting variables. However, we can still find a route to bypass the difficulty, reducing equalities between matrices to those of traces.

As the result shows, $\mathcal{X}^{\rm irr}(K)$ consists of a finite set of isolated points, several conics and an algebraic curve of high genus. Based on this, we present a method for explicitly computing the A-polynomial.

Acknowledgement

The author is supported by NSFC (Grant No. 11401014). He is grateful to Prof. Xun Yu at Tianjin University for beneficial conversations.

2 Preliminary

Let $\mathcal{M}(2,\mathbb{C})$ denote the set of $2\times 2$ matrices with entries in $\mathbb{C}$; it is a 4-dimensional vector space over $\mathbb{C}$. Let $I$ denote the $2\times 2$ identity matrix.

Given $t\in\mathbb{C}$ and $k\in\mathbb{Z}$, take $a$ with $a+a^{-1}=t$ and put

[TABLE]

note that the right-hand-side is unchanged when $a$ is replaced by $a^{-1}$. It is easy to verify that for all $k\in\mathbb{Z}$,

[TABLE]

If $X\in{\rm SL}(2,\mathbb{C})$ with ${\rm tr}(X)=t$, then by Cayley-Hamilton Theorem,

[TABLE]

and inductively we can obtain that for all $k\in\mathbb{Z}$,

[TABLE]

Lemma 2.1**.**

For any $X,Y\in{\rm SL}(2,\mathbb{C})$ with ${\rm tr}(X)=t_{1}$, ${\rm tr}(Y)=t_{2}$ and ${\rm tr}(XY)=t_{12}$, one has

[TABLE]

Proof.

The first identity follows from $XYXY=t_{12}XY-I$, and the second one can be deduced from

[TABLE]

∎

Lemma 2.2**.**

Given $t,t_{12},t_{23},t_{13},t_{123}\in\mathbb{C}$, let

[TABLE]

(i)* There exist $X_{1},X_{2},X_{3}\in{\rm SL}(2,\mathbb{C})$ with ${\rm tr}(X_{1})={\rm tr}(X_{2})={\rm tr}(X_{3})=t$, ${\rm tr}(X_{i}X_{j})=t_{ij}$, $1\leq i<j\leq 3$, and ${\rm tr}(X_{1}X_{2}X_{3})=t_{123}$ if and only if*

[TABLE]

(ii)* If (9) holds and $X_{1},X_{2},X_{3}$ are required to have no common eigenvector, then the ordered triple $(X_{1},X_{2},X_{3})$ is unique up to simultaneous conjugacy.*

Here (i) is a special case of the statement next to Lemma 2.3 in [9], and (ii) follows from Lemma 2.4 in [9] (referred to Lemma 1.5.2 in [4]), restating that up to conjugacy, an irreducible representation of the free group of rank 3 is determined by the tuple $(t,t_{12},t_{23},t_{13},t_{123})$ satisfying (9). Also see [7].

For $a,b\in\mathbb{C}$ with $a\neq 0$, put

[TABLE]

3 The Character variety

3.1 Set up

For the Wirtinger presentation of a link $L$, refer to Theorem 3.4 in [1]. Suppose a part of $L$ is an integral tangle as shown in Figure 1. Let $z_{0},z_{1},\ldots,z_{n}$ denote the elements of $\pi_{1}(E_{L})$ corresponding to the labeled directed arcs. Since $z_{k+1}z_{k}=z_{k}z_{k-1}=\cdots=z_{1}z_{0}$, we have

[TABLE]

Hence for all $h\in\mathbb{Z}$,

[TABLE]

In this paper we consider the odd classical pretzel knot

[TABLE]

A diagram for $K$ is shown in Figure 2 when $k_{1}=1,k_{2}=k_{2}=2$.

Let $\rho:\pi_{1}(E_{K})\to{\rm SL}(2,\mathbb{C})$ be a representation, and let $X_{j}=\rho(x_{j}),j=1,2,3$. The following can be deduced using (15):

[TABLE]

Conversely, any $X_{1},X_{2},X_{3}\in{\rm SL}(2,\mathbb{C})$ satisfying these relations give rise to a representation.

Remark 3.1**.**

Since the $X_{j}$’s are conjugate to each other, none of them can be $\pm I$, unless $\rho$ is trivial. Let us assume $\rho$ to be nontrivial.

Notation 3.2**.**

To simplify the writing, by $X_{j+}$ we mean $X_{j+1}$ for $j\in\{1,2\}$ and $X_{1}$ for $j=3$; by $X_{j-}$ we mean $X_{j-1}$ for $j\in\{2,3\}$ and $X_{3}$ for $j=1$. Similarly for other situations.

For $j=1,2,3$, put

[TABLE]

The relations (17)–(19) are equivalent to

[TABLE]

Remark 3.3**.**

If $\rho$ is reducible, then taking conjugation if necessary, we may assume that the elements of the image of $\rho$ are all upper-triangular, and furthermore, $X_{1}=U(u,1)$ for some $u\neq 0$. Since $X_{1},X_{2},X_{3}$ are conjugate to each other via upper-triangular elements, we have $X_{j}=U(u,a_{j}),j=2,3$ for some $a_{j}$; let $a_{1}=1$. Then $Y_{j}=U(1,u(a_{j+}-a_{j-}))$ so that ${\rm tr}(Y_{j})=2$, and

[TABLE]

The condition (22) is equivalent to

[TABLE]

All reducible representations can be found this way. Note that $\rho$ is abelian if and only if $a_{2}=a_{3}=1$.

3.2 Irreducible representations

In this subsection, $\rho$ is assumed to be irreducible; equivalently, $X_{1},X_{2},X_{3}$ have no common eigenvector.

Suppose

[TABLE]

Clearly,

[TABLE]

Let

[TABLE]

Then (9) can be rewritten as

[TABLE]

Remark 3.4**.**

Note that

[TABLE]

so $\delta=0$ if and only if $v_{3}=v_{1}^{\epsilon_{1}}v_{2}^{\epsilon_{2}}$ with $\epsilon_{1},\epsilon_{2}\in\{\pm 1\}$.

For each $j$, denote

[TABLE]

Using (6) and (7), we compute

[TABLE]

so that

[TABLE]

where in the last line, (8) is used.

Lemma 3.5**.**

$s_{j}\neq 2$* for at least one $j$.*

Proof.

Assume on the contrary that $s_{j}=2$ for all $j$.

If $u\notin\{\pm 1\}$, then taking conjugation if necessary, we may assume $X_{1}=U(u,0)$. It follows from ${\rm tr}(X_{1}^{-1}X_{2})=2$ that $X_{2}=U(u,a)$ or $X_{2}=V(u,a)$ for some $a$. Similar situation occurs for $X_{3}$. Since $\rho$ is irreducible, we have $\{X_{2},X_{3}\}=\{U(u,a),V(u,b)\}$ for some $a,b\neq 0$. But then $s_{1}={\rm tr}(X_{2}X_{3}^{-1})=2-ab\neq 2$, a contradiction.

If $u\in\{\pm 1\}$, then taking conjugation if necessary, we may assume $X_{1}=U(u,u)$. It would follow from ${\rm tr}(X_{1}^{-1}X_{2})={\rm tr}(X_{1}^{-1}X_{3})=u^{2}+u^{-2}=2$ that $X_{2},X_{3}$ are both upper-triangular, contradicting the irreducibility of $\rho$. ∎

Remark 3.6**.**

Referred to Remark 3.3, the condition that $s_{j}\neq 2$ for at least one $j$ is also sufficient for $\rho$ to be irreducible.

Representations of $\pi_{1}(E_{K})$ with $t=0$ were determined by the author in [2]; recall the result of Section 3 there, stated in a different form:

Proposition 3.7**.**

Suppose $t=0$ and (30) is satisfied. Then (22) holds if and only if one of the following cases occurs:

• •

$\gamma_{j}=\beta_{j}$, $j=1,2,3$;

• •

$\gamma_{j}=-\beta_{j}$, $j=1,2,3$;

• •

$\delta=0$, and there exist $\theta_{j}\in\mathbb{R},j=1,2,3$, such that $s_{j}=2\cos\theta_{j}$ and $\cos(k_{1}+1)\theta_{1}=\cos(k_{2}+1)\theta_{2}=\cos(k_{3}+1)\theta_{3}\neq\pm 1$.

Lemma 3.8**.**

For each $j$, $A_{j}=I$ if and only if $\gamma_{j}=\beta_{j}$ or $X_{j+}X_{j-}=I$.

Proof.

By (32), $A_{j}=I$ if and only if

[TABLE]

which, by (5), is equivalent to

[TABLE]

i.e., $(\gamma_{j}-\beta_{j})(X_{j+}+X_{j-}-tI)=0$. Hence $A_{j}=I$ if and only if either $\gamma_{j}=\beta_{j}$ or $X_{j+}=tI-X_{j-}=X_{j-}^{-1}$. ∎

Proposition 3.9**.**

If $\gamma_{\ell}\neq\beta_{\ell}$ and $\gamma_{\ell^{\prime}}=\beta_{\ell^{\prime}}$ for some $\ell,\ell^{\prime}$, then (22) holds if and only if

[TABLE]

Proof.

By Lemma 3.8, $A_{\ell^{\prime}}=I$.

If (22) holds, then $A_{j}=I$ for all $j$. Since $\gamma_{\ell}\neq\beta_{\ell}$, by Lemma 3.8 we have $X_{\ell+}X_{\ell-}=I$, implying $s_{\ell}=t^{2}-2$. Let $\{\ell_{0}\}=\{1,2,3\}-\{\ell,\ell^{\prime}\}$. Then $X_{\ell_{0}+}X_{\ell_{0}-}\neq I$, as otherwise $X_{1},X_{2},X_{3}$ would have a common eigenvector, so $\gamma_{\ell_{0}}=\beta_{\ell_{0}}$. Thus $\gamma_{\ell\pm}=\beta_{\ell\pm}$, which is equivalent to $v_{\ell\pm}^{2k_{\ell\pm}+1}=-1$; in particular, $|v_{\ell\pm}|=1$, so that $s_{\ell\pm}=2{\rm Re}(v_{\ell\pm})<2$. Moreover,

[TABLE]

If $t=0$, then $v_{\ell+}=-v_{\ell-}$ or $v_{\ell+}=-v_{\ell-}^{-1}$, either implying $v_{\ell+}^{(2k_{\ell+}+1)(2k_{\ell-}+1)}=-v_{\ell-}^{(2k_{\ell+}+1)(2k_{\ell-}+1)}$, which contradicts $v_{\ell\pm}^{2k_{\ell\pm}+1}=-1$ as just obtained.

If (34) is satisfied, then by Lemma 3.8, $A_{\ell\pm}=I$. Taking conjugation if necessary, we may assume $X_{\ell+}=U(u,0)$. Assume

[TABLE]

From ${\rm tr}(X_{\ell-}X_{\ell+}^{-1})=s_{\ell}=t^{2}-2$ we obtain $u^{-1}a+u(t-a)=t^{2}-2$, implying $a=u^{-1}$, hence $bc=0$. From ${\rm tr}(X_{\ell}X_{\ell+}^{-1})=s_{\ell-}\neq 2$ we obtain $b^{\prime}c^{\prime}\neq 0$. It follows from ${\rm tr}(X_{\ell}^{-1}(X_{\ell+}+X_{\ell-}))=s_{\ell+}+s_{\ell-}=t^{2}$ that $b^{\prime}c+bc^{\prime}=0$. Thus $b=c=0$, and $X_{\ell-}=U(u^{-1},0)=X_{\ell+}^{-1}$, establishing (22). ∎

Proposition 3.10**.**

Suppose $t\neq 0$, $\gamma_{j}\neq\beta_{j}$ for all $j$, and (30) is satisfied. Let $\lambda=\tau/t$. Then (22) holds if and only if

[TABLE]

Proof.

Using (33) and $\gamma_{j}^{2}-s_{j}\gamma_{j}\beta_{j}+\beta_{j}^{2}=1$ which is a special case of (4), the following can be computed directly:

[TABLE]

If (22) holds, then

[TABLE]

due to $t\neq 0$ and $\gamma_{j}\neq\beta_{j}$, from (37) and (38) we see that (40) is equivalent to (35).

Now suppose (35) holds (so that (40) is satisfied). Then for $j=1,2,3$,

[TABLE]

hence

[TABLE]

where in the last line we use $s_{j}^{3}-\sigma_{1}s_{j}^{2}+\sigma_{2}s_{j}-\sigma_{3}=0$. By (30),

[TABLE]

If $\sigma_{1}+2-2\lambda=0$, then (41) and (42) imply $s_{1}=s_{2}=s_{3}=2$, contradicting Lemma 3.5. Hence $\sigma_{1}+2-2\lambda\neq 0$, so that $\kappa\neq 0$ and

[TABLE]

Consequently,

[TABLE]

which is independent of $j$. Thus (39) is established.

The proof is completed once $I,X_{1}^{-1},X_{2}^{-1},X_{3}^{-1}$ are shown to form a basis of $\mathcal{M}(2,\mathbb{C})$; then (39), (40) will imply that ${\rm tr}(A_{1}Z)={\rm tr}(A_{2}Z)={\rm tr}(A_{3}Z)$ for all $Z\in\mathcal{M}(2,\mathbb{C})$, forcing $A_{1}=A_{2}=A_{3}$.

Assume on the contrary that $I,X_{1}^{-1},X_{2}^{-1},X_{3}^{-1}$ are linearly dependent, so are $I,X_{1},X_{2},X_{3}$. Without loss of generality, assume $X_{3}=aI+bX_{1}+cX_{2}$. Then

[TABLE]

The first equation implies

[TABLE]

The following can be computed using (8):

[TABLE]

Then (33) can be re-written as

[TABLE]

where the $(\cdots)$’s stand for coefficients that are irrelevant.

By (39), (40), ${\rm tr}(A_{1}\overline{X}_{\ell})={\rm tr}(A_{2}\overline{X}_{\ell})={\rm tr}(A_{3}\overline{X}_{\ell})$, $\ell=1,2$. Consequently,

[TABLE]

Comparing ${\rm tr}(A_{j}\overline{X}_{1})$ and ${\rm tr}(A_{j}\overline{X}_{2})$ for $j=1,2,3$, we respectively obtain

[TABLE]

which lead to

[TABLE]

These force $p=2d$, so that $s_{3}=2$. With (45), one can deduce from (46)–(49) that $s_{1}=s_{2}=2$, contradicting Lemma 3.5. ∎

Remark 3.11**.**

It is worth highlighting some points contained in the proof. With (35) satisfied, we have the following implications:

1. (i)

If $\sigma_{1}+2-2\lambda\neq 0$ and (30) holds, then $\kappa\neq 0$, $\delta\neq 0$ and $\gamma_{j}\neq\beta_{j}$, $j=1,2,3$, as seen from (44); also, $s_{j}\neq 2$ for some $j$, as seen from (41) and (42). 2. (ii)

(30) implies (39) which reads

[TABLE]

conversely, if $\sigma_{1}+2-2\lambda\neq 0$ and $s_{\ell}\neq s_{\ell^{\prime}}$ for some $\ell,\ell^{\prime}$, then (30) can follow from (53), since by (43),

[TABLE]

3.3 The character variety

The results of Proposition 3.7, Lemma 3.8, Proposition 3.9 and Proposition 3.10 are summarized as:

Theorem 3.12**.**

The irreducible character variety of $K$ can be embedded in $\{(t,s_{1},s_{2},s_{3},\tau)\in\mathbb{C}^{5}\}$ and is the disjoint union of four parts:

[TABLE]

where

• •

$\mathcal{X}_{0}=\mathcal{X}_{0,1}\sqcup\mathcal{X}_{0,2}$, where $\mathcal{X}_{0,1}$ consists of $(0,s_{1},s_{2},s_{3},\tau)$ with

[TABLE]

and $\mathcal{X}_{0,2}$ consists of $(0,2\cos\theta_{1},2\cos\theta_{2},2\cos\theta_{3},0)$ with

[TABLE]

• •

$\mathcal{X}_{1}=\mathcal{X}_{1,1}\sqcup\mathcal{X}_{1,2}\sqcup\mathcal{X}_{1,3}$, where $\mathcal{X}_{1,\ell}$ consists of $(t,s_{1},s_{2},s_{3},\tau)$ with

[TABLE]

• •

$\mathcal{X}_{2}$* consists of $(t,s_{1},s_{2},s_{3},\tau)$ with*

[TABLE]

so $\mathcal{X}_{2}$ has $(k_{1}+1)(k_{2}+1)(k_{3}+1)$ components, each being a conic;

• •

$\mathcal{X}_{3}$* consists of $(t,s_{1},s_{2},s_{3},t\lambda)$ with*

[TABLE]

The dimensions are: $\dim\mathcal{X}_{0}=\dim\mathcal{X}_{1}=0$, $\dim\mathcal{X}_{2}=\dim\mathcal{X}_{3}=1$.

Some supplements are in order:

• •

$\mathcal{X}_{0}\subset\partial\mathcal{X}_{3}$; actually, we could have defined $\mathcal{X}_{0}\cup\mathcal{X}_{3}$ by replacing (56) and (57) respectively with (55) and

[TABLE]

and adding (53) which, by Remark 3.11 (ii), is redundant for $\mathcal{X}_{3}$.

• •

Each of the points in $\mathcal{X}_{1}$ is isolated.

• •

$\mathcal{X}_{2}\cap\partial\mathcal{X}_{3}$ consists of $(t,s_{1},s_{2},s_{3},\tau)$ with

[TABLE]

We investigate $\mathcal{X}_{3}$ in more detail. Recall Remark 3.11 (i) that $\delta\neq 0$.

Let $\ell\in\{1,2,3\}$. By (57) for $j=\ell\pm$,

[TABLE]

Assertion 3.13**.**

On $\mathcal{X}_{3}$, $\gamma_{\ell-}\beta_{\ell+}=\gamma_{\ell+}\beta_{\ell-}$ if and only if $s_{\ell+}=s_{\ell-}$ and $v_{\ell+}^{2(k_{\ell+}-k_{\ell-})}=1$.

Proof.

The “if” part is trivial. Suppose $\gamma_{\ell-}\beta_{\ell+}=\gamma_{\ell+}\beta_{\ell-}$. Then $(s_{\ell-}-s_{\ell+})(\gamma_{\ell+}-\beta_{\ell+})\beta_{\ell-}=0$. If $\beta_{\ell-}\neq 0$, then since $\gamma_{\ell+}\neq\beta_{\ell+}$, we have $s_{\ell+}=s_{\ell-}$. If $\beta_{\ell-}=0$, then $\gamma_{\ell-}\beta_{\ell+}=0$, which, due to $\gamma_{\ell-}\neq\beta_{\ell-}$, implies $\beta_{\ell+}=0$; by (57) for $j=\ell\pm$, $\lambda=2+s_{\ell+}=2+s_{\ell-}$, hence also $s_{\ell+}=s_{\ell-}$. Thus $v_{\ell+}=v_{\ell-}$ or $v_{\ell+}=v_{\ell-}^{-1}$. Then $\gamma_{\ell-}\beta_{\ell+}=\gamma_{\ell+}\beta_{\ell-}$ becomes $v_{\ell+}^{2(k_{\ell+}-k_{\ell-})}=1$. ∎

Decompose $\mathcal{X}_{3}$ as the “singular” part and the “regular” part: let

[TABLE]

Obviously, $\mathcal{X}_{3}^{\rm sin}$ consists of finitely many points, unless $k_{1}=k_{2}=k_{3}$, in which case $\mathcal{X}_{3}^{\rm sin}$ is a curve with two components, each being parameterized by $v+v^{-1}$.

Let $(t,s_{1},s_{2},s_{3},t\lambda)\in\mathcal{X}_{3}^{\rm reg}$, then $\gamma_{\ell-}\beta_{\ell+}\neq\gamma_{\ell+}\beta_{\ell-}$ for some $\ell$. From (58), (59) we obtain

[TABLE]

With $\lambda$ and $s_{\ell}$ substituted, (57) for $j=\ell$ can be re-written as one polynomial equation in $s_{\ell+},s_{\ell-}$, displaying $\mathcal{X}_{3}^{\rm reg}$ as an affine algebraic curve. We may regard $t^{2}$ as a rational function on $\mathcal{X}_{3}^{\rm reg}$ through (56); moreover, when $s_{\ell+}\neq s_{\ell-}$, by Remark 3.11 (ii), the condition (56) is equivalent to

[TABLE]

4 The A-polynomial

Choose a meridian-longitude pair $(\mathfrak{m},\mathfrak{l})$ of $K$, and denote the corresponding elements of $\pi_{1}(E_{K})$ by the same notations. Following [8] (see Page 303), let $\mathcal{R}_{U}(K)$ denote the subset of $\mathcal{R}(K)$ consisting of representations which send $\mathfrak{m}$ and $\mathfrak{l}$ to upper-triangular matrices. Define

[TABLE]

by setting $\xi_{1}(\rho)$ (resp. $\xi_{2}(\rho)$) to be the upper-left entry of $\rho(\mathfrak{m})$ (resp. $\rho(\mathfrak{l})$). Taking the Zariski closure of the image of $\xi$, we obtain an affine algebraic curve $\mathcal{V}$, which is known to have the property that each component has dimension zero or one. The A-polynomial $A_{K}$ is defined to be the defining polynomial of the one-dimensional part of $\mathcal{V}$.

A-polynomial is notoriously difficult to compute. Till now, in the literary, formulas for A-polynomials have been obtained for few families of knots; see [5, 6, 10, 13] and the references therein.

Here we present a method for deriving a formula for the A-polynomial of an odd classical pretzel knot, leaving practical computations to possible future work.

For the pretzel knot $K$, take $\mathfrak{m}=x_{1}$. Let $\rho$ be a representation of $\pi_{1}(E_{K})$ as in Section 3, but this time we require $\rho\in\mathcal{R}_{U}(K)$.

For $j=1,2,3$, let

[TABLE]

Observe that

[TABLE]

Let $L_{j}$ denote the image under $\rho$ of the longitude associated to $x_{j}$ (so $L_{1}=\rho(\mathfrak{l})$). Using (15), we find

[TABLE]

The 1-dimensional part of $\mathcal{X}^{\rm irr}(K)$ consists of $\mathcal{X}_{3}$ and $\mathcal{X}_{2}$, the latter having $n=(k_{1}+1)(k_{2}+1)(k_{3}+1)$ components, so $A_{K}$ can be factorized as

[TABLE]

Note that $s_{1},s_{2},s_{3}$ are all known constants on each component of $\mathcal{X}_{2}$, thus the corresponding factor $A_{K}^{(h)}$ is easy to understand, as explained in Remark 4.2.

The main achievement of this section is concerned with the “hard” factor $\overline{A}_{K}$ contributed by $\mathcal{X}_{3}$.

Suppose the upper-left entries of $X_{1},L_{1}$ are $u,w$, respectively, so that ${\rm tr}(L_{j})=w+w^{-1}$, $j=1,2,3$.

Proposition 4.1**.**

If $\chi(\rho)\in\mathcal{X}_{3}$, then

[TABLE]

Proof.

It suffices to prove (70) when $\chi(\rho)$ belongs to an open subset of $\mathcal{X}_{3}$. Let us assume $u\notin\{\pm 1\}$ and $\lambda\neq 2+s_{j},j=1,2$.

It is easy to see that

[TABLE]

Let $q={\rm tr}(B_{3})$. By (67), ${\rm tr}(B_{2}B_{1})={\rm tr}(B_{3}^{-1})=q$. By (68), $B_{2}B_{1}=L_{1}B_{3}^{-1}=B_{3}^{-1}L_{2}$, hence

[TABLE]

As a consequence,

[TABLE]

By (66),

[TABLE]

hence

[TABLE]

where

[TABLE]

By (66) again,

[TABLE]

hence

[TABLE]

Thus (73) implies

[TABLE]

Combining (72), (75) and (78), we obtain (70). ∎

Then $\overline{A}_{K}$ can be obtained by computing the multi-variable resultant of the following (remembering $t=u+u^{-1}$):

[TABLE]

Remark 4.2**.**

If $\chi(\rho)$ lies in the $h$-th component of $\mathcal{X}_{2}$, then (72), (74) and (80) are still valid (whenever $t\notin\{0,\pm 2\}$). Eliminating $\lambda$ from these, we may obtain a polynomial in $u,w$, which is the very $A^{(h)}_{K}$.

Example 4.3**.**

As an illustration, consider the case $k_{1}=k_{2}=1$. Then $\beta_{1}=\beta_{2}=1,\gamma_{1}=s_{1},\gamma_{2}=s_{2}$. From (79) for $j=1,2$ we obtain

[TABLE]

then use them to re-write (80), (81) as, respectively,

[TABLE]

Use (83), we can re-write (79) for $j=3$ as

[TABLE]

Then $\overline{A}_{K}$ (for $K=P(3,3,2k_{3}+1)$) may be obtained from (82)–(84) by eliminating $s_{1}+s_{2}$ and $s_{3}$.