# Character varieties of some families of links

Anh T. Tran Department of Mathematics, The Ohio State University, Columbus, OH 43210, USA
###### Abstract.

In this paper we consider some families of links, including -pretzel links and twisted Whitehead links. We calculate the character varieties of these families, and determine the number of irreducible components of these character varieties.

2010 Mathematics Classification: 57M27.

## 0. Introduction

### 0.1. The character variety of a group

The set of representations of a finitely generated group into is an algebraic set defined over , on which acts by conjugation. The set-theoretic quotient of the representation space by that action does not have good topological properties, because two representations with the same character may belong to different orbits of that action. A better quotient, the algebro-geometric quotient denoted by (see [CS, LM]), has the structure of an algebraic set. There is a bijection between and the set of all characters of representations of into , hence is usually called the character variety of .

The character variety of a group is determined by the traces of some fixed elements in . More precisely, one can find in such that for every element in there exists a polynomial in variables such that for any representation one has where . It is known that the character variety of is equal to the zero set of the ideal of the polynomial ring generated by all expressions of the form , where and are any two words in the letters which are equal in .

### 0.2. Main results

In this paper we consider some families of links, including -pretzel links and twisted Whitehead links. We will calculate the character varieties of these families, and determine the number of irreducible components of these character varieties. To state our results, we first introduce the Chebyshev polynomials of the first kind . They are defined recursively by , and for all integers .

The character variety (the character ring, actually) of the -pretzel link has been calculated in [Tr]. Note that the -pretzel link is the two-component unlink. Its link group is and hence its character variety is by the Fricke-Klein-Vogt theorem, see [LM]. The new result of the following theorem is the determination of the number of irreducible components of the character variety.

###### Theorem 1.

(i) [Tr, Thm. 2] The character variety of the -pretzel link is the zero set of the polynomial

 (x2+y2+z2−xyz−4)[(xz−y)Sn−1(α)−(Sm(β)−Sm−1(β))Sn−2(α)],

where

 α=ySm−1(β)−(xz−y)Sm−2(β)andβ=xyz+2−y2−z2.

(ii) The number of irreducible components of the character variety of the -pretzel link is equal to

Let be the two-bridge link associated to a pair of relatively prime integers , where is odd (see [BZ]). It is known that is the -torus link, and is a hyperbolic link if . In the case , we have the following.

###### Theorem 2.

The character variety of the two-bridge link is the zero set of the polynomial , where

 Qp={(x2+y2)Sn(z)S2n−1(z)−xySn−1(z)(S2n(z)+S2n−1(z))+S3n(z)if p=3n+1(x2+y2)S2n(z)Sn−1(z)−xySn(z)(S2n(z)+S2n−1(z))+S3n+1(z)if p=3n+2.

It has exactly two irreducible components.

For two-bridge knots , where is an odd integer relatively prime with 3, a similar result about the number of irreducible components of their character varieties has been obtained in [Bu, MPL, NT].

The two-bridge link can be realized as Dehn filling on one cusp of the magic manifold, see e.g. [La2]. Note that is the Whitehead link. In [La1] Landes identified the canonical component of the character variety of the Whitehead link as a rational surface isomorphic to blown up at 10 points. Here the canonical component of the character variety of a hyperbolic link is the component that contains the character of a discrete faithful representation, see [Th]. In her thesis [La2], Landes conjectured that the canonical component of the character variety of the two-bridge link is a rational surface isomorphic to blown up at points. In a forthcoming paper [PT], we will confirm Landes’ conjecture for all integers .

Similarly, Harada [Ha] identified the canonical component of the character variety of the two-bridge link as a rational surface isomorphic to blown up at 13 points. In [PT], we will also prove that the canonical component of the character variety of the two-bridge link is a rational surface isomorphic to blown up at points, for all integers .

For , the -twisted Whitehead link is the two-component link depicted in Figure 2. Note that is the -torus link, and is the Whitehead link. Moreover, is the two-bridge link for all . These links are all hyperbolic except for . Their character varieties are described as follows.

###### Theorem 3.

Let .

(i) The character variety of is the zero set of the polynomial

 (γ−2)((xy−γz)Sn−1(γ)−(xy−2z)Sn−2(γ))n−1∏j=1(γ−2cosjπn).

It has exactly irreducible components.

(ii) The character variety of is the zero set of the polynomial

 (γ−2)(zSn(γ)−(xy−z)Sn−1(γ))n∏j=1(γ−2cos(2j−1)π2n+1).

It has exactly irreducible components.

The twisted Whitehead link can be realized as Dehn filling on one of the cusps of the Borromean rings. In [Ha], Harada identified the canonical component of the character variety of , which is the two-bridge link , as a rational surface isomorphic to blown up at 10 points. Hence it is an interesting problem to understand the canonical component of the character variety of for all integers .

### 0.3. Plan of the paper

In Section 1 we review some properties of the Chebyshev polynomials of the first kind. In Section 2 we recall the calculation of the character variety of the -pretzel link from [Tr], and prove the part of Theorem 1 on the determination of the number of irreducible components of its character variety. In Section 3 we review character varieties of two-bridge links and prove Theorems 2 and 3.

### 0.4. Acknowledgment

We would like to thank K. Petersen for helpful discussions.

## 1. Properties of Chebyshev polynomials

Recall from the Introduction that the Chebyshev polynomials are recursively defined by , and for all integers .

In this section we list some properties of which will be repeatedly used in the rest of the paper.

Property 1.1. One has and for all integers . Moreover if , where , then .

Property 1.2. One has for all integers . For ,

 Sk(t)=k∏j=1(t−2cosjπk+1)andSk(t)−Sk−1(t)=k∏j=1(t−2cos(2j−1)π2k+1).

Property 1.3. One has for all integers . As a consequence, in .

Property 1.4. One has for all integers .

The proof of Property 1.1 is elementary and hence is omitted. Properties 1.2–1.4 are proved by applying Property 1.1. We will only prove Property 1.4, and leave the proofs of Properties 1.2–1.3 for the reader.

It is easy to see that we only need to check Property 1.4 for . Write for some . Then, by Property 1.1, we have

 S3k(t)−3Sk(t)S2k−1(t)+tS3k−1(t) = (qk+1−q−k−1q−q−1)3−3(qk+1−q−k−1q−q−1)(qk−q−kq−q−1)2+(q+q−1)(qk−q−kq−q−1)3 = q3k+1−q−3k−1q−q−1=S3k(t).

This completes the proof of Property 1.4.

In this section, we let denote the -pretzel link in Figure 1. From [Tr, Thm. 2], we have that the character ring of is the quotient of the polynomial ring by the principal ideal generated by the polynomial

 (x2+y2+z2−xyz−4)[(xz−y)Sn−1(α)−(Sm(β)−Sm−1(β))Sn−2(α)],

where and

Let

 Q(x,y,z)=(xz−y)Sn−1(α)−(Sm(β)−Sm−1(β))Sn−2(α).

Then the character variety of is the zero set of .

To determine the number of components of the character variety of , we need to study the factorization of in .

### 2.1. The case m=0

Then and hence

 Q=(xz−y)Sn−1(xz−y)−Sn−2(xz−y)=Sn(xz−y).

If then by Property 1.2. Similarly, if then

In this case we have the following.

###### Proposition 2.1.

The number of components of the non-abelian character variety of the -pretzel link, where , is equal to

### 2.2. The case m=1

Then and hence

 Q = (xz−y)Sn−1(y)−(xyz+1−y2−z2)Sn−2(y) = −xzSn−3(y)+z2Sn−2(y)+Sn−4(y).

If then . If then . Suppose now that . Since , has degree 1 in . This, together with the fact that , implies that is irreducible in

In this case we have the following.

###### Proposition 2.2.

The number of irreducible components of the non-abelian character variety of the -pretzel knot is equal to

### 2.3. The case n=0

Then , where

If then , by Property 1.2. Similarly, if then

Note that, for any , the polynomial is irreducible in . Hence, in this case we have the following.

###### Proposition 2.3.

The number of irreducible components of the non-abelian character variety of the -pretzel link is equal to

### 2.4. The case n=−1

Then

 Q = −(xz−y)+(Sm(β)−Sm−1(β))(ySm−1(β)−(xz−y)Sm−2(β)) = (y−xz)[1+(Sm(β)−Sm−1(β))Sm−2(β)]+y(Sm(β)−Sm−1(β))Sm−1(β),

where . By Property 1.3, we have

 1+Sm(β)Sm−2(β)=1+Sm(β)(tSm−1(β)−Sm(β))=S2m−1(β).

It follows that . Hence

 Q=Sm−1(β)[y(Sm(β)−Sm−1(β))−(xz−y)(Sm−1(β)−Sm−2(β))].

In this case, we will prove the following.

###### Proposition 2.4.

The number of irreducible components of the non-abelian character variety of the -pretzel link is equal to if .

This is equivalent to showing that

 R(x,y,z):=y(Sm(β)−Sm−1(β))−(xz−y)(Sm−1(β)−Sm−2(β))

is a non-constant irreducible polynomial in .

We first prove that is non-constant. Indeed, we have where . It follows that . Hence is non-constant. Moreover we have , since

We now prove that is irreducible in . Recall that and . Note that (for ) and .

Hence to prove the irreducibility of in , we only need to prove that

 R1(x1,y,z):=y(Sm(β1)−Sm−1(β1))−x1(Sm−1(β1)−Sm−2(β1)),

where , is irreducible in .

Claim 1. is irreducible in .

Since , proving Claim 1 is equivalent to proving that

 R2(x1,y,β2):=y(Sm(β2)−Sm−1(β2))−x1(Sm−1(β2)−Sm−2(β2)),

is irreducible in .

Since , has degree 1 in . This, together with the fact that , implies that is irreducible in . Claim 1 follows.

Claim 2. is irreducible in .

Assume that is reducible in . Since is irreducible in , it must have the form , for some satisfying . In particular, is a perfect square in . This can not occur, since is not a perfect square in . Claim 2 follows.

This completes the proof of Proposition 2.4.

### 2.5. The case m∉{0,1} and n∉{−1,0}

In this case, we will prove the following.

###### Proposition 2.5.

The non-abelian character variety of the -pretzel link, where and , is irreducible.

This is equivalent to showing that is a non-constant irreducible polynomial in .

We first prove that is non-constant. Indeed, assume that is a constant polynomial. By [Tr, Prop. 2.5], we have . Hence is also a constant polynomial. It follows that . Since and , we must have . However, in the case we have . Hence is non-constant. Moreover we have , since .

We now prove that is irreducible in . Recall that , and

 Q=(xz−y)Sn−1(α)−(Sm(β)−Sm−1(β))Sn−2(α).

Note that (for ) and . Hence to prove the irreducibility of in , we only need to prove that

 Q1(x1,y,z):=x1Sn−1(α1)−(Sm(β1)−Sm−1(β1))Sn−2(α1),

where and , is irreducible in .

Claim 3. is irreducible in .

Since , proving Claim 3 is equivalent to proving that

 Q2(x1,y,β2):=x1Sn−1(α2)−(Sm(β2)−Sm−1(β2))Sn−2(α2),

where , is irreducible in .

The proof of the irreducibility of in is divided into 2 steps.

Step 1. We first show that . Indeed, since we have . Suppose that . By Property 1.3, we have . It follows that for some . Hence and . Since , we have . Therefore , which implies that .

Step 2. For we have Since , to prove the irreducibility of in we only need to prove that

 Q3(x2,y,β2):=(ySm−1(β2)−x2)Sn−1(x2)−Sm−2(β2)(Sm(β1)−Sm−1(β1))Sn−2(x2)

is irreducible in .

Since we have . It follows that has degree 1 in . We write , where and Then is irreducible in if .

Since , we have . This, together with the fact that , implies that

By Property 1.3, we have

 Sm(β2)Sm−2(β2)=Sm(β2)(tSm−1(β2)−Sm(β2))=S2m−1(β2)−1.

It follows that

 gcd(Sm−1(β2),g)=gcd(Sm−1(β2),x2Sn−1(x2)−Sn−2(x2))=gcd(Sm−1(β2),Sn(x2)).

Since , we have and hence . Therefore , which implies the irreducibility of in . Claim 3 follows.

Claim 4. is irreducible in .

Assume that is reducible in . Since is irreducible in , it must have the form , for some satisfying . In particular, is a perfect square in .

Since and , is not a perfect square in unless . In the case we have is not a perfect square in , since . This shows that can not be a perfect square in . Claim 4 follows.

This completes the proof of Proposition 2.5.

### 2.6. Proof of Theorem 1(ii)

If then Theorem 1(ii) follows from Propositions 2.12.5. If then is the -pretzel link, which is the two-component unlink. Its link group is and hence its character variety is by the Fricke-Klein-Vogt theorem, see [LM]. This completes the proof of Theorem 1(ii).

We first review character varieties of two-bridge links. Let be the two-bridge link associated to a pair of relatively prime integers , where is odd (see [BZ]). The link group of is , where and Here and are 2 meridians of .

Let be the free group in 2 letters and . The character variety of is isomorphic to by the Fricke-Klein-Vogt theorem. For every word in there is a unique polynomial in 3 variables such that for any representation one has where and . By [CS, Prop. 1.4.1], the polynomial can be calculated inductively using the following identity for traces of matrices :

 (3.1) tr(AB)+tr(AB−1)=tr(A)tr(B).

Suppose be a group generated by 2 elements and . For every representation , we consider and as functions of . By abuse of notation, we will identify with its image .

By [LT], the character ring of is the quotient of the polynomial ring by the principal ideal generated by the polynomial . It follows that the character variety of is the zero set of .

The following lemma will be frequently used in the proofs of Theorems 2 and 3.

###### Lemma 3.1.

Suppose . Then, for all integers ,

 (3.2) Mk=Sk(trM)I−Sk−1(trM)M−1.
###### Proof.

For , the Caley-Hamilton theorem implies that , where denotes the identity matrix in . By induction on , we can show that Eq. (3.2) holds true for all and .

For , by applying Eq. (3.2) for and we have

 (M−1)−k = S−k(trM−1)I−S−k−1(trM−1)M = −Sk−2(trM)I+Sk−1(trM)((trM)I−M−1) = Sk(trM)I−Sk−1(trM)M−1.

The lemma follows. ∎

We will also need the following.

One has .

###### Proof.

By applying Eq. (3.1), we have