Two-tone colorings and surjective dihedral representations for links

TL;DR

This paper introduces a two-tone coloring method for links and establishes conditions under which link groups admit surjective dihedral representations, showing that links with three or more components can map onto dihedral groups of any degree.

Contribution

It presents a novel two-tone coloring technique and proves that links with at least three components can have surjective homomorphisms to dihedral groups of arbitrary degree.

Findings

Links with three or more components admit surjective dihedral representations of any degree.

The two-tone coloring provides a new criterion for such surjective representations.

The method extends the understanding of link group representations beyond knots.

Abstract

It is well-known that a knot is Fox $n$-colorable for a prime $n$ if and only if the knot group admits a surjective homomorphism to the dihedral group of degree $n$. However, this is not the case for links with two or more components. In this paper, we introduce a two-tone coloring on a link diagram, and give a condition for links so that the link groups admit surjective representations to the dihedral groups. In particular, it is shown that the link group of any link with at least 3 components admits a surjective homomorphism to the dihedral group of arbitrary degree.

Equations

$$D_n=⟨a,b\mid a^2=b^n=(ab{)}^2=e⟩.$$

$$\Gamma (z)$$

$$=a{b}^{i+j-n}a{b}^i=b^{n-i-j+i}=b^{n-j}=b_{n-j}$$

$$H_2(M)\to H_1(N-\tilde{K})\to H_1(N)\oplus H_1(M-\tilde{K})\to H_1(M)\to H_0(N-\tilde{K})$$

$$0\to H_1(N-\tilde{K})\to H_1(N)\oplus H_1(M-\tilde{K})\to H_1(M)\to 0.$$

$$0\to H_1(\partial \tilde{D})\to H_1(M-\tilde{K})\to H_1(M)\to 0$$

$$0\to H_1(\partial \tilde{D})/(x+y)\to H_1(M-\tilde{K})/(x+y)\to H_1(M)\to 0.$$

$${\pi }_1(X-\tilde{K})\to {\pi }_1(M-\tilde{K})\to H_1(M-\tilde{K})\to H_1(M-\tilde{K})/(x+y)\to \mathbb{Z}.$$

$$\tilde{f}(g)=\left\{\begin{array}[]{ll}\bar{f}(g)&(g\in\pi_{1}(X-\tilde{K})),\\ a\bar{f}(m^{-1}g)&(g\not\in\pi_{1}(X-\tilde{K})).\end{array}\right.$$

$$\Gamma (\{\text{arcs of ℓ1}\})\subset \{a_1,a_3\}\text{and}\Gamma (\{\text{arcs of ℓ2}\})\subset \{a_0,a_2\}.$$

Full text

This work was supported by JSPS KAKENHI Grant Numbers 22K03301, 20K14309, 20K03596.

1. Introduction

One of the most well-known invariants of knots in 3-space would be the Fox’s 3-colorablity. In general, for a prime $n\geq 3$, it is well-known that a knot is Fox $n$-colorable if and only if the knot group admits a surjective homomorphism to the dihedral group $D_{n}$ of degree $n$. For instance, it is stated in [2, Chap. VI, Exercises, 6, pp.92–93]. However, this is not the case for links with two or more components. Some examples are given in [3] for $D_{3}$-coloring, which is the coloring by the symmetric group of degree three.

In this paper, we introduce a two-tone coloring on a link diagram, and give a condition for links that the link groups admit surjective homomorphisms to the dihedral groups. In particular, we show that the link group of any link with at least 3 components admits a surjective homomorphism to the dihedral group of arbitrary degree.

Remark 1*.*

It is well-known that, for $n\geq 3$, a link is Fox $n$-colorable, i.e., a diagram of the link admits a non-trivial Fox $n$-coloring (a coloring with at least two colors), if and only if $\det(L)=0$ or $(n,\det(L))\neq 1$, where $\det(L)$ denotes the determinant of the link. See [4, Proposition 2.1] for example. Also a condition for knot groups to admit a surjective homomorphism to the dihedral groups in terms of the homology of the double branched covering is also known. See [1, 14.8] for example.

To state our results, we prepare some notations. Let $D_{n}$ be the dihedral group of degree $n$. It is well-known that $D_{n}$ has the following presentation with $e$ the identity element:

[TABLE]

Note that any element in $D_{n}$ is represented as $a^{x}b^{y}$ ($x=0,1$, $0\leq y\leq n-1$). Thus, by setting $a_{i}:=ab^{i}$ ($0\leq i\leq n-1$) and $b_{j}:=b^{j}$ ($1\leq j\leq n-1$), we see that $D_{n}=\{e,a_{0},a_{1},\dots,a_{n-1},b_{1},\dots,b_{n-1}\}$ as a set. In geometric view point, the $a_{i}$’s represent reflections and $b_{j}$’s represent rotations as the symmetries of a regular polygon ($n$-gon).

In the following, let $L$ be an oriented link in the 3-sphere $S^{3}$ with a link diagram $D$. We call a map \Gamma:\{\mbox{arcs on D}\}\rightarrow D_{n} a $D_{n}$-coloring on $D$ if it satisfies $\Gamma(x)\Gamma(z)=\Gamma(y)\Gamma(x)$ (respectively, $\Gamma(z)\Gamma(x)=\Gamma(x)\Gamma(y)$) in $D_{n}$ at each positive (resp. negative) crossing on $D$, where $x$ denotes the over arc, $y$ and $z$ the under arcs at the crossing supposing $y$ is the under arc before passing through the crossing and $z$ is the other. (See Figure 1.)

Remark 2*.*

The $D_{n}$-colorings and the Fox $n$-colorings are related in terms of representations of link groups to $D_{n}$ as follows. For a link diagram $D$ with $c$ crossings of a link $L$, set $g_{1},\dots,g_{c}$ the Wirtinger generators of the link group $G_{L}$, i.e., $G_{L}=\pi_{1}(S^{3}-L)$. Then a $D_{n}$-coloring on $D$ corresponds to a map $\{g_{1},\dots,g_{c}\}\to D_{n}$ which extends to a homomorphism of $G_{L}$ to $D_{n}$. When a $D_{n}$-coloring sends $g_{k}$’s to $a_{i}$’s (reflections, $0\leq i\leq n-1$) in $D_{n}$, it induces a map $\{\text{arcs of }D\}\to\{0,1,\dots,n-1\}$, which gives a Fox $n$-coloring.

Let $\Gamma$ be a $D_{n}$-coloring on a link diagram $D$ of an oriented link $L$. We say that $\Gamma$ is two-tone if $\mathrm{Im}(\Gamma)\cap\{a_{0},\dots,a_{n-1}\}\neq\emptyset$ and $\mathrm{Im}(\Gamma)\cap\{b_{1},\dots,b_{n-1}\}\neq\emptyset$, that is, the coloring uses colors from both $\{a_{i}\}$ and $\{b_{j}\}$. We say that a link is two-tone $D_{n}$-colorable if, with some orientation, it has a diagram $D$ admitting a two-tone $D_{n}$-coloring.

Now the following are our main results. Here $D_{\infty}$ denote the group presented by $\left<a,b\mid a^{2}=(ab)^{2}=e\right>$, and two-tone $D_{\infty}$-colorable for links are defined in the same way as above.

Theorem 1.1**.**

For a 2-component link $L=\ell_{1}\cup\ell_{2}$, the following are equivalent.

• (i)

$lk(\ell_{1},\ell_{2})$* is even.*

• (ii)

$L$* is two-tone $D_{n}$-colorable for some odd $n\geq 3$.*

• (iii)

$L$* is two-tone $D_{\infty}$-colorable.*

• (iv)

The link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$.

• (v)

The link group $G_{L}$ admits a surjective homomorphism to $D_{\infty}$.

We remark that (ii) in Theorem 1.1 can be also equivalent to that $L$ is two-tone $D_{n}$-colorable for some even $n\geq 3$ except for the case that $L$ admits a $D_{n}$-coloring with only the color $b_{n/2}$. Also remark that (iii) in Theorem 1.1 does not imply that $L$ is two-tone $D_{n}$-colorable for every odd $n\geq 3$. See Remark 3. For simplicity, we omit the details.

Theorem 1.2**.**

Let $L=\ell_{1}\cup\ell_{2}$ be a 2-component link with $lk(\ell_{1},\ell_{2})$ odd. Then the following hold.

• (i)

The link $L$ admits no two-tone $D_{n}$-colorings for any odd $n\geq 3$.

• (ii)

If the link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for $n\geq 3$, then the homomorphism is induced from a Fox $n$-coloring on $\ell_{1}$, $\ell_{2}$ or $L$, i.e., the homomorphism sends a meridional element in $G_{L}$ to the trivial element or a reflection in $D_{n}$.

Theorem 1.3**.**

Let $L$ be a link with at least 3 components. Then the link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$.

We remark that even if the link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$, the link $L$ may not be two-tone $D_{n}$-colorable for every $n\geq 3$. For example, pretzel links of type $(2m,2m,2m)$ with odd $m$ admit no two-tone $D_{m}$-colorings.

As a corollary of the theorems, we have the following.

Corollary 1.4**.**

If a link $L$ is two-tone $D_{m}$-colorable for some odd $m$, then $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$. If $G_{L}$ admits a surjective homomorphism to $D_{n}$ for some $n$, then $L$ is two-tone $D_{n}$-colorable or contains a Fox $n$-colorable sub-link.

Note that even if a link admits a nontrivial Fox $n$-coloring, it may not induce a surjective homomorphism from $G_{L}$ to $D_{n}$. See the example given in Section 4.

On the other hand, even if a link $L$ is known to be two-tone $D_{n}$-colorable for some $n$, finding a two-tone $D_{n}$-coloring on a given diagram of $L$, or, finding a surjective homomorphism of $G_{L}$ to $D_{n}$, is a tedious task in general. The next proposition and its proof give a simple way to find a two-tone $D_{n}$-coloring on a give link diagram for any $n\geq 3$.

Proposition 1.5**.**

Suppose that there exists a trivial component $\ell_{0}$ of a link $L$ and, for every component $\ell\subset L-\ell_{0}$, $lk(\ell_{0},\ell)$ is even. Then any diagram of $L$ admits a two-tone $D_{n}$-coloring for every odd $n\geq 3$ which assigns the arcs on $\ell_{0}$ to $a_{i}$’s and the other arcs to $b_{j}$’s.

2. Properties of $D_{n}$-coloring

In this section, we study some properties of $D_{n}$-colorings, and give lemmas which will be used in the next sections. In the following, we set $A_{n}:=\{a_{i}\}$ and $B_{n}:=\{b_{j}\}$ for $D_{n}$.

Lemma 2.1**.**

Let $\Gamma$ be a $D_{n}$-coloring on a diagram $D$ of an oriented link $L$ in $S^{3}$. At a crossing on $D$, $x$ denotes the over arc, $y$ and $z$ the under arcs at the crossing supposing $y$ is the under arc before passing through the crossing and $z$ is the other. Then the following hold.

1. (1)

If $\Gamma(x)\in B_{n}$ and $\Gamma(y)\in B_{n}$, then $\Gamma(z)=\Gamma(y)$. 2. (2)

If $\Gamma(x)=a_{i}$ and $\Gamma(y)=a_{i^{\prime}}$, then $\Gamma(z)=a_{k}$ and $k\equiv 2i^{\prime}-i\pmod{n}$. 3. (3)

If $\Gamma(x)=a_{i}$ and $\Gamma(y)=b_{j}$, then $\Gamma(z)=b_{k}$ and $k\equiv n-j\pmod{n}$. 4. (4)

If $\Gamma(x)=b_{j}$ and $\Gamma(y)=a_{j}$, then $\Gamma(z)=a_{k}$ and $k\equiv i+2j$ (resp. $k\equiv i-2j$ ) $\pmod{n}$ if the crossing is a positive (resp. negative) crossing.

Proof.

We give a proof of the case (3) and the crossing is a positive crossing. The others are proved in the same way. Suppose that $\Gamma(x)=a_{i}$ and $\Gamma(y)=b_{j}$. By the definition of a $D_{n}$-coloring, we have the following.

[TABLE]

Thus $\Gamma(z)=b_{k}$ and $k\equiv n-j\pmod{n}$ holds. ∎

Lemma 2.2**.**

Let $\Gamma$ be a $D_{n}$-coloring on a diagram $D$ of an oriented link $L$ in $S^{3}$. Let $x,y,z,w$ be the arcs depicted in Figure 2. If $\Gamma(x)=a_{i}$ and $\Gamma(y)=b_{j}$, then $\Gamma(z)=a_{k}$ with $k\equiv i-2j\pmod{n}$ and $\Gamma(w)=b_{l}$ with $l\equiv n-j\pmod{n}$

Proof.

We only give a proof for the case that both crossings in the figure are positive crossings. A proof for the other case is similar. In that case, by Lemma 2.1(3), $\Gamma(u)=b_{l}$ with $l\equiv n-j\pmod{n}$ and $\Gamma(v)=\Gamma(x)=a_{i}$. Then, by Lemma 2.1(4), $\Gamma(z)=a_{k}$ and $k\equiv i+2(n-j)\equiv i-2j\pmod{n}$ and $\Gamma(w)=\Gamma(u)=b_{l}$ with $l\equiv n-j\pmod{n}$. ∎

3. Two-tone colorings and surjective homomorphisms to $D_{\infty}$

In this section, we study some conditions for two-tone $D_{n}$-colorings and surjective homomorphisms to $D_{\infty}$.

Let $lk(L,L^{\prime})$ denote the (total) linking number of oriented links $L,L^{\prime}$, i.e., $lk(L,L^{\prime})=\sum_{\ell\subset L,\ell^{\prime}\subset L^{\prime}}lk(\ell,\ell^{\prime})$. In the following, the linking number is calculated for the link with arbitrarily chosen orientations. Note that the parity of such a linking number is independent of the choice of orientations.

Lemma 3.1**.**

If $L$ is two-tone $D_{n}$-colorable for some odd $n\geq 3$, then there exists a sub-link $L_{b}$ of $L$ such that, for every component $\ell\subset L_{b}$, $lk(\ell,L-L_{b})$ is even.

Proof.

Take a two-tone $D_{n}$-coloring $\gamma$ on a diagram of $L$ for some $n\geq 3$. Let $L_{b}$ be a maximal sub-link of $L$ such that each arc in a diagram of $L_{b}$ is colored by $b_{j}$’s by $\Gamma$. By the definition of the $D_{n}$-coloring, $L_{b}$ is well-defined for $\Gamma$, independent of the choice of a diagram.

Now we consider the diagram of $L$ depicted in Figure 3, where $D_{b}$ is a sub-diagram corresponding to $L_{b}$, $D_{a}$ is the remaining sub-diagram, and each box between $D_{a}$ and $D_{b}$ contains a full-twist. For the sub-diagram $D_{\ell}$ corresponding to a component $\ell\subset L_{b}$, we consider the arcs $\beta$ and $\beta^{\prime}$ as in Figure 3.

Since $\beta$ and $\beta^{\prime}$ are connected in $D_{b}$, we see $\Gamma(\beta)=\Gamma(\beta^{\prime})$ by Lemma 2.1(1). On the other hand, if $\Gamma(\beta)=\Gamma(\beta^{\prime})$, then the number $N$ of the boxes (full-twists) $\ell_{i}$ runs through have to be even by Lemma 2.2 together with $n$ is odd. This number $N$ is equal to $lk(\ell,L-L_{b})$, and so the lemma holds. ∎

The next is a key proposition to prove the theorems.

Proposition 3.2**.**

Suppose that a link $L$ contains a component $\ell_{0}$ and a sub-link $L^{\prime}$ with $lk(\ell_{0},L^{\prime})$ is even and $\det L^{\prime}\neq 0$. Then $G_{L}$ admits a surjective homomorphism to $D_{\infty}$ and $L$ is two-tone $D_{\infty}$-colorable.

Proof.

We assume $L=\ell_{0}\cup L^{\prime}$; if $L$ has other components, we can obtain a desired representation on $G_{L}$ from one on $G_{\ell_{0}\cup L^{\prime}}$ by sending their meridians to the identity.

Let $p:X\to S^{3}-L^{\prime}$ be the cyclic double covering on the total linking number with $L^{\prime}$, and $\bar{p}:M\to S^{3}$ the cyclic double branched covering. Let $\tilde{K}=K_{1}\cup K_{2}$ denote the inverse image $p^{-1}(\ell_{0})\subset X$; because $lk(\ell_{0},L^{\prime})$ is even, $\tilde{K}$ is a $2$-component link in $X$ (or in $M$). We shall construct a surjective group homomorphism $\pi_{1}(M-\tilde{K})\to\mathbb{Z}$ and extend the composition $\pi_{1}(X-\tilde{K})\to\pi_{1}(M-\tilde{K})\to\mathbb{Z}$ to obtain a $D_{\infty}$-coloring.

Taking a regular neighborhood $N$ of $\tilde{K}$, we consider the Mayer-Vietoris exact sequence for $M=N\cup(M-\tilde{K})$:

[TABLE]

is exact. The rightmost map is zero as usual and the leftmost one is also zero because $\det L^{\prime}\neq 0$ (hence $|H_{1}(M)|=|\det(L^{\prime})|<\infty$); by the Poincaré duality $H_{2}(M)\cong H^{1}(M;\mathbb{Z})=0$. Thus, we obtain a short exact sequence

[TABLE]

Take a meridional disc $D_{1}\subset N$ of $K_{1}$ and let $D_{2}$ denote $\varphi(D_{1})$, where $\varphi:M\to M$ is the nontrivial covering transformation. We denote $D_{1}\cup D_{2}$ by $\tilde{D}$. Because the kernel of the surjective homomorphism $H_{1}(N-\tilde{K})\to H_{1}(N)$ is the image of the injective map $H_{1}(\partial\tilde{D})\to H_{1}(N-\tilde{K})$, the short exact sequence above shows that

[TABLE]

is also exact. We should remark that (1) is compatible with $\varphi_{*}$.

Let $x\in H_{1}(\partial D_{1})$ be a generator and set $y=\varphi_{*}(x)\in H_{1}(\partial D_{2})$. We use the same symbols $x,y$ for their images in $H_{1}(\partial{D})$ or $H_{1}(M-\tilde{K})$. By taking the quotient of (1) by the $\varphi_{*}$-invariant part of $H_{1}(\partial\tilde{D})$, we obtain an exact sequence

[TABLE]

Since $H_{1}(\partial\tilde{D})/(x+y)\cong\mathbb{Z}$ and $|H_{1}(M)|<\infty$, the rank of $H_{1}(M-\tilde{K})/(x+y)$ equals $1$ and there exists a surjective homomorphism $f:H_{1}(M-\tilde{K})/(x+y)\to\mathbb{Z}$, which satisfies $f\circ\varphi_{*}=-f$. Let $\bar{f}:\pi_{1}(X-\tilde{K})\to\mathbb{Z}$ denote the composition

[TABLE]

Let $m\in G_{L}$ be a meridian of a component of $L^{\prime}$. Identifying $\langle b\rangle\subset D_{\infty}$ with $\mathbb{Z}$, we define $\tilde{f}:G_{L}\to D_{\infty}$ by

[TABLE]

Since $a^{2}=e$, $\tilde{f}$ is well defined as a map. Furthermore, we have $\bar{f}(mgm^{-1})=f\circ\varphi_{*}(g)=f(g)^{-1}=\bar{f}(g)^{-1}\in D_{\infty}$ for $g\in\pi_{1}(X-\tilde{K})$. By this equation, we can easily check that $\tilde{f}$ is a group homomorphism. Because $\bar{f}$ is surjective and $\tilde{f}(m)=a$, the homomorphism $\tilde{f}:G_{L}\to D_{\infty}$ is surjective. ∎

The next is an immediate corollary of the proposition above, since any knot has odd determinant.

Corollary 3.3**.**

Let $L=\ell_{1}\cup\ell_{2}$ be a $2$-component link. If $lk(\ell_{1},\ell_{2})$ is even, then $G_{L}$ admits a surjective homomorphism to $D_{\infty}$ and $L$ is two-tone $D_{\infty}$-colorable. ∎

4. Proof of theorems

To prove the theorems, we first prepare the next lemma.

Lemma 4.1**.**

Let $L=\ell_{1}\cup\ell_{2}$ be a $2$-component link. If $\det(L)=0$, then $lk(\ell_{1},\ell_{2})$ is even.

Proof.

Let $D$ be a diagram of $L$. Since $\det(L)=0$, there exists a Fox $4$-coloring $\Gamma$ on $D$ which induces a surjective group homomorphism. By the definition of Fox colorings, if $\Gamma(x)$ equals $a_{0}$ or $a_{2}$ (resp. $a_{1}$ or $a_{3}$) for an arc $x$ belonging to $\ell_{i}\;(i=1,2)$, it holds for any arc $x$ of $\ell_{i}$. Then, we may assume

[TABLE]

For a crossing point of $D$, let $x$ be the over arc and $y,z$ the under arcs. Again by the definition of Fox colorings, we find that $\Gamma(y)=\Gamma(z)$ holds if and only if $x$ and $y$ belong to the same component. In particular, the crossing switches the color of the under arc if $x$ belongs to $\ell_{1}$ and $y$ to $\ell_{2}$. This implies that $D$ has an even number of such crossings, and hence the linking number $lk(\ell_{1},\ell_{2})$ is even. ∎

Proof of Theorem 1.1.

Let $L=\ell_{1}\cup\ell_{2}$ a 2-component link. We show that all the following are equivalent.

• (i)

$lk(\ell_{1},\ell_{2})$ is even.

• (ii)

$L$ is two-tone $D_{n}$-colorable for some odd $n\geq 3$.

• (iii)

$L$ is two-tone $D_{\infty}$-colorable.

• (iv)

The link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$.

• (v)

The link group $G_{L}$ admits a surjective homomorphism to $D_{\infty}$.

We see that (i)$\Rightarrow$(iii) follows from Corollary 3.3 and (ii)$\Rightarrow$(i) follows from Lemma 3.1.

(iii)$\Rightarrow$(ii): Suppose that $L$ is two-tone $D_{\infty}$-colorable, that is, a diagram of $L$ admits a two-tone $D_{\infty}$-coloring. Since there is a surjection from $D_{\infty}$ to $D_{n}$ for every $n\geq 3$ defined by $a\in D_{\infty}\mapsto a\in D_{n}$ and $b\in D_{\infty}\mapsto b\in D_{n}$, this implies that the diagram of $L$ admits a $D_{n}$-coloring for every $n$. By taking odd $n$ sufficiently large, the $D_{n}$-coloring uses at least two colors from $a_{i}$’s. Furthermore, if necessary, retaking $n$ to satisfy $(n,\det(L))=1,(n,\det(\ell_{1}))=1,(n,\det(\ell_{2}))=1$, the coloring cannot come from Fox $n$-colorings on $L$, $\ell_{1}$, $\ell_{2}$. Thus the coloring have to be two-tone, and so, $L$ is two-tone $D_{n}$-colorable for some odd $n\geq 3$.

We also see that (i)$\Rightarrow$(v) follows from Corollary 3.3.

(v)$\Rightarrow$(iv): By a surjection from $D_{\infty}$ to $D_{n}$ for every $n\geq 3$ defined as above, if the link group $G_{L}$ admits a surjective homomorphism to $D_{\infty}$, then the link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$.

(iv)$\Rightarrow$(i) or (ii): Suppose that the link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$. Such a surjective homomorphism induces a $D_{n}$-coloring on a diagram of $L$ for $n\geq 3$ by considering the Wirtinger generators for the diagram. If $\det(L)=0$, then $lk(\ell_{1},\ell_{2})$ is even by Lemma 3.1, and so (i) holds. If $\det(L)\neq 0$, then for some odd $n$ which is coprime to $\det(L),\det(\ell_{1}),\det(\ell_{2})$, the $D_{n}$-coloring does not come from a Fox $n$-coloring, and so, it have to be two-tone. This implies (ii). ∎

Remark 3*.*

We remark that (iii) in Theorem 1.1 does not imply that $L$ is two-tone $D_{n}$-colorable for every odd $n\geq 3$. Actually even if there is a two-tone $D_{\infty}$-coloring on a diagram of a link $L$, the coloring may not give a two-tone $D_{n}$-coloring, but a Fox $n$-coloring on a sub-diagram of $L$. For example, pretzel links of type $(m,2,m,2)$ with odd $m$ admit a two-tone $D_{\infty}$-coloring on a diagram, but no two-tone $D_{m}$-colorings. Moreover, for any $n\geq 3$, there exists a two-tone $D_{\infty}$-colorable link $L$ that admits no two-tone $D_{n}$-colorings.

Proof of Theorem 1.2.

Let $L=\ell_{1}\cup\ell_{2}$ be a 2-component link with $lk(\ell_{1},\ell_{2})$ is odd.

(i) Then $L$ admits no two-tone $D_{n}$-colorings for any $n\geq 3$ by Theorem 1.1 (by the contraposition of (ii)$\Rightarrow$(i)).

(ii) By (i), if the link group $G_{L}$ admits a surjective homomorphism to $D_{n}$ for $n\geq 3$, then it is not induced from two-tone $D_{n}$-colorings. That is, the homomorphism must send Wirtinger generators to either the trivial element and reflections in $D_{n}$ or the trivial element and rotations in $D_{n}$. However the latter cannot be surjective, and so, it is impossible. Therefore the homomorphism sends Wirtinger generators to either the trivial element and reflections in $D_{n}$. Such a homomorphism is induced from a Fox $n$-coloring on $\ell_{1}$, $\ell_{2}$ or $L$. ∎

Proof of Theorem 1.3.

Let $L$ be a link with at least 3 components. We show that $G_{L}$ admits a surjective homomorphism to $D_{n}$ and $L$ is two-tone $D_{n}$-colorable for every $n\geq 3$.

Consider sub-links of 2 components in $L$. If some of them, say $L^{\prime}=\ell^{\prime}_{1}\cup\ell^{\prime}_{2}$, satisfies $lk(\ell^{\prime}_{1},\ell^{\prime}_{2})$ is even, then by Theorem 1.2, $G_{L^{\prime}}$ admits a surjective homomorphism to $D_{n}$ and $L^{\prime}$ is two-tone $D_{n}$-colorable for $n$, and also does $G_{L}$ via a surjection $G_{L}\to G_{L^{\prime}}$ and also is $L$.

Suppose that for all the 2 component sub-links of $L$, the linking numbers of the two components are odd. Then, by Lemma 4.1, no such links have the determinant 0. Since $L$ has at least 3 components, we can consider a sub-link of $L$ with 3 components, say $L^{\prime}=\ell_{1}\cup\ell_{2}\cup\ell_{3}$. For this link, $lk(\ell_{1},\ell_{2}\cup\ell_{3})$ is even and $\det(\ell_{2}\cup\ell_{3})\neq 0$ holds. Then, by Proposition 3.2, $G_{L^{\prime}}$ admits a surjective homomorphism to $D_{\infty}$ and so a surjective homomorphism to $D_{n}$ for every $n\geq 3$. This implies that $G_{L}$ admits a surjective homomorphism to $D_{n}$ for every $n\geq 3$. ∎

Proof of Corollary 1.4.

Suppose that $L$ is two-tone $D_{m}$-colorable for some odd $m\geq 3$. If $L$ is a link with 2 components, then $G_{L}$ admits a homomorphism to $D_{n}$ for every $n\geq 3$ by Theorem 1.2 ((ii)$\Rightarrow$(iv)). If $L$ has at least 3 components, then $G_{L}$ admits a homomorphism to $D_{n}$ for every $n\geq 3$ by Theorem 1.3.

Suppose that $G_{L}$ admits a surjective homomorphism to $D_{n}$ for $n\geq 3$. Then there is a $D_{n}$-coloring on a diagram of $L$. See Remark 2. If the coloring is two-tone, then $L$ is two-tone $D_{n}$-colorable. Otherwise, since the homomorphism is surjective, the coloring comes from a nontrivial Fox $n$-coloring on a diagram of a sub-link of $L$ as in the proof of Theorem 1.2. ∎

Note that even if a link admits a nontrivial Fox $n$-coloring, it may not induce a surjective homomorphism from $G_{L}$ to $D_{n}$. See the example illustrated in Figure 4. In this case, the image of the Wirtinger generators by the homomorphism induced by the Fox $4$-coloring is the set $\{a_{0},a_{2}\}\subset D_{4}$, but the elements $a_{0}$ and $a_{2}$ do not generate $D_{4}$. Thus the induced homomorphism is not surjective.

5. Finding two-tone colorings

Proof of Proposition 1.5.

Suppose that there exists a trivial component $\ell_{0}$ of a link $L$ and, for every component $\ell\subset L-\ell_{0}$, $lk(\ell_{0},\ell)$ is even. If a diagram of a link $L$ admits a two-tone $D_{n}$-coloring for every odd $n\geq 3$ which assigns the arcs on $\ell_{0}$ to $a_{i}$’s and the other arcs to $b_{j}$’s, then also does any diagram of $L$. Thus, to prove the proposition, it suffices to show that a particular diagram of $L$ admits such a $D_{n}$-coloring.

Now we take a diagram $D$ of $L$ depicted in Figure 5. In the figure, $D_{0}$ is a sub-diagram corresponding to $\ell_{0}$, which is a trivial knot diagram, and each box between $D_{0}$ and the remaining sub-diagram $D_{b}$ contains a full-twist.

Consider the arc $\alpha$ in the figure, take an arc $\beta_{i}$ from each component of $L-\ell_{0}$, and assign $a_{0}$ to $\alpha$ and $b_{1}$ to $\beta_{i}$’s. Let us show that this assignment induces a two-tone $D_{n}$-coloring.

For the arc $\beta_{i}$, let $\ell$ be the component of $L-\ell_{0}$ containing $\beta_{i}$. Since $lk(\ell_{0},\ell)$ is even for every component $\ell\subset L-\ell_{0}$, due to Lemma 2.2, the assignment for $\beta_{i}$ to be $b_{1}$ induces a $D_{n}$-coloring on $\ell$. In the same way, we can find a $D_{n}$-coloring on $L-\ell_{0}$.

Note that, on the sub-diagram corresponding to each component of $L-\ell_{0}$, an arc in the lower right of a box in center is colored by $b_{1}$ or $b_{n-1}$. Beside, when the arc in the lower right is colored by $b_{1}$, then the arc in the upper right is colored in $b_{n-1}$, and vice versa.

Thus, by Lemma 2.2, for each component of $L-\ell_{0}$, the number of the boxes in center with the arc in the lower right colored by $b_{1}$ is equal to the number of those with the arc colored in $b_{n-1}$.

Let $m$ be the half of the linking number $lk(\ell_{0},L-\ell_{0})$. (Note that $lk(\ell_{0},L-\ell_{0})$ must be even, since $lk(\ell_{0},\ell)$ is even for each component $\ell$ of $L-\ell_{0}$.) Then the number of the boxes in center with the arc in the lower right colored by $b_{1}$ is $m$ and the number of those with the arc colored in $b_{n-1}$ is also $m$.

Again by Lemma 2.2, the assignment for $\alpha$ to be $a_{0}$ induces the assignment for the arc in the upper left of the top box in center to be $a_{0-2(m\cdot 1+m\cdot(-1))}=a_{0}$. This implies that the assignment induces a $D_{n}$-coloring on whole the diagram. By construction, the $D_{n}$-coloring is obviously two-tone.

Thus any diagram of $L$ admits a two-tone $D_{n}$-coloring for every odd $n\geq 3$ which assigns the arcs on $\ell_{0}$ to $a_{i}$’s and the other arcs to $b_{j}$’s. ∎