Character varieties of some families of links

The set of representations of a finitely generated group $G$ into $SL_{2}(\mathbb{C})$ is an algebraic set defined over $\mathbb{C}$, on which $SL_{2}(\mathbb{C})$ 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 $X(G)$ (see [CS, LM]), has the structure of an algebraic set. There is a bijection between $X(G)$ and the set of all characters of representations of $G$ into $SL_{2}(\mathbb{C})$, hence $X(G)$ is usually called the character variety of $G$.

The character variety of a group $G$ is determined by the traces of some fixed elements $g_{1},\cdots,g_{k}$ in $G$. More precisely, one can find $g_{1},\cdots,g_{k}$ in $G$ such that for every element $g$ in $G$ there exists a polynomial $P_{g}$ in $k$ variables such that for any representation $\rho:G\to SL_{2}(\mathbb{C})$ one has $\operatorname{\mathrm{t}r}(\rho(g))=P_{g}(x_{1},\cdots,x_{k})$ where $x_{j}:=\operatorname{\mathrm{t}r}(\rho(g_{j}))$. It is known that the character variety of $G$ is equal to the zero set of the ideal of the polynomial ring $\mathbb{C}[x_{1},\cdots,x_{k}]$ generated by all expressions of the form $\operatorname{\mathrm{t}r}(\rho(u))-\operatorname{\mathrm{t}r}(\rho(v))$, where $u$ and $v$ are any two words in the letters $g_{1},\cdots,g_{k}$ which are equal in $G$.

In this paper we consider some families of links, including $(-2,2m+1,2n)$-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 $S_{k}(t)$. They are defined recursively by $S_{0}(t)=1$, $S_{1}(t)=t$ and $S_{k+1}(t)=tS_{k}(t)-S_{k-1}(t)$ for all integers $k$.

The character variety (the character ring, actually) of the $(-2,2m+1,2n)$-pretzel link has been calculated in [Tr]. Note that the $(-2,1,-2)$-pretzel link is the two-component unlink. Its link group is $\mathbb{Z}^{2}$ and hence its character variety is $\mathbb{C}^{3}$ 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.

Let $\mathfrak{b}(2p,m)$ be the two-bridge link associated to a pair of relatively prime integers $p>m>0$, where $m$ is odd (see [BZ]). It is known that $\mathfrak{b}(2p,1)$ is the $(2,2p)$-torus link, and $\mathfrak{b}(2p,m)$ is a hyperbolic link if $m\geq 3$. In the case $m=3$, we have the following.

For two-bridge knots $\mathfrak{b}(p,3)$, where $p>3$ 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 $\mathfrak{b}(6n+2,3)$ can be realized as $1/n$ Dehn filling on one cusp of the magic manifold, see e.g. [La2]. Note that $\mathfrak{b}(8,3)$ 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 $\mathbb{P}^{2}$ 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 $\mathfrak{b}(6n+2,3)$ is a rational surface isomorphic to $\mathbb{P}^{2}$ blown up at $9n+1$ points. In a forthcoming paper [PT], we will confirm Landes’ conjecture for all integers $n\geq 1$.

Similarly, Harada [Ha] identified the canonical component of the character variety of the two-bridge link $\mathfrak{b}(10,3)$ as a rational surface isomorphic to $\mathbb{P}^{2}$ blown up at 13 points. In [PT], we will also prove that the canonical component of the character variety of the two-bridge link $\mathfrak{b}(6n+4,3)$ is a rational surface isomorphic to $\mathbb{P}^{2}$ blown up at $9n+4$ points, for all integers $n\geq 1$.

For $k\geq 0$, the $k$-twisted Whitehead link $W_{k}$ is the two-component link depicted in Figure 2. Note that $W_{0}$ is the $(2,4)$-torus link, and $W_{1}$ is the Whitehead link. Moreover, $W_{k}$ is the two-bridge link $\mathfrak{b}(4k+4,2k+1)$ for all $k\geq 0$. These links are all hyperbolic except for $W_{0}$. Their character varieties are described as follows.

The twisted Whitehead link $W_{2n-1}$ can be realized as $1/n$ Dehn filling on one of the cusps of the Borromean rings. In [Ha], Harada identified the canonical component of the character variety of $W_{2}$, which is the two-bridge link $\mathfrak{b}(12,5)$, as a rational surface isomorphic to $\mathbb{P}^{2}$ blown up at 10 points. Hence it is an interesting problem to understand the canonical component of the character variety of $W_{k}$ for all integers $k\geq 0$.

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 $(-2,2m+1,2n)$-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.

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

Then $\alpha=xz-y$ and hence

If $n\geq 0$ then $Q=S_{n}(xz-y)=\prod_{j=1}^{n}(xz-y-2\cos\frac{j\pi}{n+1}),$ by Property 1.2. Similarly, if $n\leq-2$ then $Q=-S_{-(n+2)}(xz-y)=-\prod_{j=1}^{-(n+2)}(xz-y-2\cos\frac{j\pi}{-(n+1)}).$

In this case we have the following.

Then $\alpha=y$ and hence

If $n=2$ then $Q=(z-1)(z+1)$. If $n=3$ then $Q=z(yz-x)$. Suppose now that $n\notin\{2,3\}$. Since $zS_{n-3}(y)\not\equiv 0$, $Q$ has degree 1 in $x$. This, together with the fact that $\gcd(zS_{n-3}(y),z^{2}S_{n-2}(y)+S_{n-4}(y))=1$, implies that $Q$ is irreducible in $\mathbb{C}[x,y,z].$

In this case we have the following.

Then $Q=S_{m}(\beta)-S_{m-1}(\beta)$, where $\beta=xyz+2-y^{2}-z^{2}.$

If $m\geq 0$ then $Q=\prod_{j=1}^{m}(xyz+2-y^{2}-z^{2}-2\cos\frac{(2j-1)\pi}{2m+1})$, by Property 1.2. Similarly, if $m\leq-1$ then $Q=S_{-m-1}(\beta)-S_{-m-2}(\beta)=\prod_{j=1}^{-(m+1)}(xyz+2-y^{2}-z^{2}-2\cos\frac{(2j-1)\pi}{-(2m+1)}).$

Note that, for any $\delta\in\mathbb{C}$, the polynomial $xyz-y^{2}-z^{2}+\delta$ is irreducible in $\mathbb{C}[x,y,z]$. Hence, in this case we have the following.

Then

where $\beta=xyz+2-y^{2}-z^{2}$. By Property 1.3, we have

It follows that $1+(S_{m}(\beta)-S_{m-1}(\beta))S_{m-2}(\beta)=S_{m-1}(\beta)(S_{m-1}(\beta)-S_{m-2}(\beta))$. Hence

In this case, we will prove the following.

This is equivalent to showing that

is a non-constant irreducible polynomial in $\mathbb{C}[x,y,z]$.

We first prove that $R$ is non-constant. Indeed, we have $R\mid_{z=0}\,=y(S_{m}(\gamma)-S_{m-2}(\gamma))$ where $\gamma=2-y^{2}$. It follows that $R\mid_{y=\pm 2,z=0}\,=\pm 4(-1)^{m}$. Hence $R$ is non-constant. Moreover we have $\gcd(z,R)=1$, since $R\mid_{z=0}\,\not\equiv 0.$

We now prove that $R$ is irreducible in $\mathbb{C}[x,y,z]$. Recall that $\beta=(xz-y)y+2-z^{2}$ and $R=y(S_{m}(\beta)-S_{m-1}(\beta))-(xz-y)(S_{m-1}(\beta)-S_{m-2}(\beta))$. Note that $x=\frac{(xz-y)+y}{z}$ (for $z\not=0$) and $\gcd(z,R)=1$.

Hence to prove the irreducibility of $R$ in $\mathbb{C}[x,y,z]$, we only need to prove that

where $\beta_{1}=x_{1}y+2-z^{2}$, is irreducible in $\mathbb{C}[x_{1},y,z]$.

Claim 1. $R_{1}$ is irreducible in $\mathbb{C}[x_{1},y,z^{2}]$.

Since $z^{2}=x_{1}y+2-\beta_{1}$, proving Claim 1 is equivalent to proving that

is irreducible in $\mathbb{C}[x_{1},y,\beta_{2}]$.

Since $S_{m}(\beta_{2})-S_{m-1}(\beta_{2})\not\equiv 0$, $R_{2}$ has degree 1 in $y$. This, together with the fact that $\gcd(S_{m}(\beta_{2})-S_{m-1}(\beta_{2}),S_{m-1}(\beta_{2})-S_{m-2}(\beta_{2}))=1$, implies that $R_{2}$ is irreducible in $\mathbb{C}[x_{1},y,\beta_{2}]$. Claim 1 follows.

Claim 2. $R_{1}$ is irreducible in $\mathbb{C}[x_{1},y,z]$.

Assume that $R_{1}$ is reducible in $\mathbb{C}[x_{1},y,z]$. Since $R_{1}$ is irreducible in $\mathbb{C}[x_{1},y,z^{2}]$, it must have the form $R_{1}=(fz+g)(-fz+g)$, for some $f,g\in\mathbb{C}[x_{1},y,z^{2}]\setminus\{0\}$ satisfying $(f,g)=1$. In particular, $R_{1}\mid_{z=0}$ is a perfect square in $\mathbb{C}[x_{1},y]$. This can not occur, since $R_{1}\mid_{x_{1}=0,z=0}\,=y$ is not a perfect square in $\mathbb{C}[y]$. Claim 2 follows.

This completes the proof of Proposition 2.4.

In this case, we will prove the following.

This is equivalent to showing that $Q(x,y,z)$ is a non-constant irreducible polynomial in $\mathbb{C}[x,y,z]$.

We first prove that $Q$ is non-constant. Indeed, assume that $Q(x,y,z)$ is a constant polynomial. By [Tr, Prop. 2.5], we have $Q\mid_{z=0}\,=(-1)^{(m-1)(n-1)}S_{2mn-2m-n-2}(y)$. Hence $S_{2mn-2m-n-2}(y)$ is also a constant polynomial. It follows that $2mn-2m-n-2\in\{-2,-1,0\}$. Since $m\not\in\{0,1\}$ and $n\not\in\{-1,0\}$, we must have $(m,n)=(2,2)$. However, in the case $(m,n)=(2,2)$ we have $Q=-x^{2}z^{2}+xyz^{3}+xyz-y^{2}z^{2}-z^{4}+3z^{2}-1$. Hence $Q$ is non-constant. Moreover we have $\gcd(z,Q)=1$, since $Q\mid_{z=0}\not\equiv 0$.

We now prove that $Q$ is irreducible in $\mathbb{C}[x,y,z]$. Recall that $\alpha=yS_{m-1}(\beta)-(xz-y)S_{m-2}(\beta)$, $\beta=(xz-y)y+2-z^{2}$ and

Note that $x=\frac{(xz-y)+y}{z}$ (for $z\not=0$) and $\gcd(z,Q)=1$. Hence to prove the irreducibility of $Q$ in $\mathbb{C}[x,y,z]$, we only need to prove that

where $\alpha_{1}=yS_{m-1}(\beta_{1})-x_{1}S_{m-2}(\beta_{1})$ and $\beta_{1}=x_{1}y+2-z^{2}$, is irreducible in $\mathbb{C}[x_{1},y,z]$.

Claim 3. $Q_{1}$ is irreducible in $\mathbb{C}[x_{1},y,z^{2}]$.

Since $z^{2}=x_{1}y+2-\beta_{1}$, proving Claim 3 is equivalent to proving that

where $\alpha_{2}=yS_{m-1}(\beta_{2})-x_{1}S_{m-2}(\beta_{2})$, is irreducible in $\mathbb{C}[x_{1},y,\beta_{2}]$.

The proof of the irreducibility of $Q_{2}$ in $\mathbb{C}[x_{1},y,\beta_{2}]$ is divided into 2 steps.

Step 1. We first show that $\gcd(S_{m-2}(\beta_{2}),Q_{2})=1$. Indeed, since $m\not=1$ we have $S_{m-2}(\beta_{2})\not\equiv 0$. Suppose that $S_{m-2}(\beta_{2})=0$. By Property 1.3, we have $S^{2}_{m-1}(\beta_{2})+S^{2}_{m-2}(\beta_{2})-\beta_{2}S_{m-1}(\beta_{2})S_{m-2}(\beta_{2})=1$. It follows that $S_{m-1}(\beta_{2})=\varepsilon$ for some $\varepsilon\in\{\pm 1\}$. Hence $\alpha_{2}=\varepsilon y$ and $Q_{2}\mid_{S_{m-2}(\beta_{2})=0}\,=x_{1}S_{n-1}(\varepsilon y)-(\beta_{2}-1)\varepsilon S_{n-2}(\varepsilon y)$. Since $n\not=0$, we have $S_{n-1}(\varepsilon y)\not\equiv 0$. Therefore $Q_{2}\mid_{S_{m-2}(\beta_{2})=0}\,\not\equiv 0$, which implies that $\gcd(S_{m-2}(\beta_{2}),Q_{2})=1$.

Step 2. For $S_{m-2}(\beta_{2})\not=0$ we have $x_{1}=\frac{yS_{m-1}(\beta_{2})-\alpha_{2}}{S_{m-2}(\beta_{2})}.$ Since $\gcd(S_{m-2}(\beta_{2}),Q_{2})=1$, to prove the irreducibility of $Q_{2}$ in $\mathbb{C}[x_{1},y,\beta_{2}]$ we only need to prove that

is irreducible in $\mathbb{C}[x_{2},y,\beta_{2}]$.

Since $mn\not=0$ we have $S_{m-1}(\beta_{2})S_{n-1}(x_{2})\not\equiv 0$. It follows that $Q_{3}$ has degree 1 in $y$. We write $Q_{3}=yf-g$, where $f=S_{m-1}(\beta_{2})S_{n-1}(x_{2})$ and $g=x_{2}S_{n-1}(x_{2})+S_{m-2}(\beta_{2})(S_{m}(\beta_{2})-S_{m-1}(\beta_{2}))S_{n-2}(x_{2}).$ Then $Q_{3}$ is irreducible in $\mathbb{C}[x_{2},y,\beta_{2}]$ if $\gcd(f,g)=1$.