A generating polynomial for the two-bridge knot with Conway's notation C(n,r)

TL;DR

This paper introduces a generating polynomial that encodes the Kauffman states of two-bridge knots specified by Conway's notation, providing a new algebraic tool for knot analysis.

Contribution

It constructs an explicit integer polynomial for two-bridge knots with Conway's notation, linking combinatorial knot states to algebraic expressions.

Findings

Polynomial coefficients enumerate Kauffman states

Provides a new algebraic representation for two-bridge knots

Facilitates computational analysis of knot invariants

Abstract

We construct an integer polynomial whose coefficients enumerate the Kauffman states of the two-bridge knot with Conway's notation C(n,r).

Equations

$$\displaystyle(\mathsf{A})\ \includegraphics[angle={90},origin=c,width=26.01613pt,valign={c}]{crossing}\Longrightarrow\includegraphics[angle={90},origin=c,width=26.01613pt,valign={c}]{A_split},\qquad(\mathsf{B})\ \includegraphics[width=26.01613pt,valign={c}]{crossing}\Longrightarrow\includegraphics[width=26.01613pt,valign={c}]{A_split}.$$

$$K(x):=\sum _S{x}^{|S|},$$

$$B_{n,0}(x)=x(x+1{)}^n$$

$$B_{n,\infty }(x)=(x+1{)}^n+x^2-1.$$

$$B_{n,0}(x)=\left(◯\bigsqcup B_{n-1,0}\right)(x)+B_{n-1,0}(x)$$

$$B_{n,\infty }(x)=B_{n-1,0}(x)+B_{n-1,\infty }(x),$$

$$B_{n,r}(x)=B_{n-1,r}(x)+(x+1{)}^{n-1}{B}_{\infty ,r}(x),$$

$$B_{n,r}(x)=\left(\frac{(x+1{)}^r+x^2-1}{x}\right)(x+1{)}^n+\left(x^2-1\right)\left(\frac{(x+1{)}^r-1}{x}\right).$$

$$B_{n,r}(x)$$

$$=B_{n-1,r}(x)+(x+1{)}^{n-1}{B}_{\infty ,r}(x),$$

$$B_{n,r}(x)=B_{0,r}(x)+B_{\infty ,r}(x)\left(\frac{(x+1{)}^n-1}{x}\right).$$

$$B_{n,0}(x)=x^2{\text{upalpha}}_n(x)+x$$

$$B_{n,\infty }(x)=x{\text{upalpha}}_n(x)+x^2,$$

$$B_{n,r}(x)=\left(x^2{\text{upalpha}}_n(x)+x\right)+\left(x^2{\text{upalpha}}_r(x)+x{\text{upalpha}}_n(x){\text{upalpha}}_r(x)\right).$$

$$b(n,r;k)=\left(\genfrac{}{}{0ex}{}{n+r}{k+1}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)+\left(\genfrac{}{}{0ex}{}{r}{k-1}\right)-\left(\genfrac{}{}{0ex}{}{n}{k+1}\right)-\left(\genfrac{}{}{0ex}{}{r}{k+1}\right)-{\delta }_{1,k},$$

$$\displaystyle\includegraphics[width=26.01613pt,valign={c}]{markersA1},\quad\includegraphics[angle={90},origin=c,width=26.01613pt,valign={c}]{markersA1},\quad\includegraphics[angle={180},origin=c,width=26.01613pt,valign={c}]{markersA1},\quad\includegraphics[angle={270},origin=c,width=26.01613pt,valign={c}]{markersA1},$$

$$\displaystyle\includegraphics[width=26.01613pt,valign={c}]{markersA1}\Longrightarrow\includegraphics[width=26.01613pt,valign={c}]{markersA2}\qquad\mbox{ and }\qquad\includegraphics[angle={90},origin=c,width=26.01613pt,valign={c}]{markersA1}\Longrightarrow\includegraphics[angle={90},origin=c,width=26.01613pt,valign={c}]{markersA2}.$$

$$\displaystyle B_{n,2}(x)=B_{n,0}(x)+B_{n,\infty}(x)+B_{n,\infty}(x)+\big{(}\bigcirc\sqcup B_{n,\infty}\big{)}(x).$$

Full text

Keywords: generating polynomial, shadow diagram, Kauffman state.

1 Introduction

A state of a knot shadow diagram is a choice of splitting its crossings [2, Section 1]. There are two ways of splitting a crossing:

[TABLE]

By state of a crossing we understand either of the split of type $\displaystyle(\mathsf{A})$ or $\displaystyle(\mathsf{B})$. An example for the figure-eight knot is shown in Figure 1.

Let $\displaystyle K$ be a knot diagram. If $\displaystyle m$ denotes the initial number of crossings, then the final states form a collection of $\displaystyle 2^{m}$ diagrams of nonintersecting curves. We can enumerate those states with respect to the number of their components – called circles – by introducing the sum

[TABLE]

where $\displaystyle S$ browses the collection of states, and $\displaystyle|S|$ gives the number of circles in $\displaystyle S$. Here, $\displaystyle K(x)$ is an integer polynomial which we referred to as generating polynomial [6, 7] (in fact, it is a simplified formulation of what Kauffman calls “state polynomial” [2, Section 1–2] or “bracket polynomial” [3]). For instance, if $\displaystyle K$ is the figure-eight knot diagram, then we have $\displaystyle K(x)=5x+8x^{2}+3x^{3}$ (the states are illustrated in Figure 2).

In this note, we establish the generating polynomial for the two-bridge knot with Conway’s notation $\displaystyle\mathcal{C}(n,r)$ [4, 5]. We refer to the associated knot diagram as $\displaystyle B_{n,r}$, where $\displaystyle n$ and $\displaystyle r$ denote the number of half-twists. For example, the figure-eight knot has Conway’s notation $\displaystyle\mathcal{C}(2,2)$. Owing to the property of the shadow diagram which we draw on the sphere [1], we can continuously deform the diagram $\displaystyle B_{n,r}$ into $\displaystyle B_{r,n}$ without altering the crossings configuration. We let $\displaystyle B_{n,r}\rightleftharpoons B_{r,n}$ express such transformation (see Figure 3 3(a)). Besides, we let $\displaystyle B_{n,0}$ and $\displaystyle B_{n,\infty}$ denote the diagrams in Figure 3 3(b) and 3(c), respectively. Here, “$\displaystyle 0$” and “$\displaystyle\infty$” are symbolic notations – borrowed from tangle theory [2, p. 88] – that express the absence of half-twists. If $\displaystyle r=\infty$ and $\displaystyle n\geq 1$, then $\displaystyle B_{n,\infty}$ represents the diagram of a $\displaystyle(2,n)$-torus knot ($\displaystyle\rightleftharpoons B_{n-1,1}$). Correspondingly, we let $\displaystyle B_{0,r}$ and $\displaystyle B_{\infty,r}$ denote the diagrams pictured in Figure 3 3(d) and 3(e), respectively.

2 Generating polynomial

Let $\displaystyle K$, $\displaystyle K^{\prime}$ and $\displaystyle\bigcirc$ be knot diagrams, where $\displaystyle\bigcirc$ is the trivial knot, and let $\displaystyle\#$ and $\displaystyle\sqcup$ denote the connected sum and the disjoint union, respectively. The generating polynomial defined in (1) verifies the following basic properties:

1. (i)

$\displaystyle\bigcirc(x)=x$; 2. (ii)

\displaystyle\big{(}K\sqcup K^{\prime}\big{)}(x)=K(x)K^{\prime}(x); 3. (iii)

\displaystyle\big{(}K\#K^{\prime}\big{)}(x)=\dfrac{1}{x}K(x)K^{\prime}(x).

Furthermore, if $\displaystyle K\rightleftharpoons K^{\prime}$, then $\displaystyle K(x)=K^{\prime}(x)$ [6].

Lemma 1**.**

The generating polynomial for the knots $\displaystyle B_{n,0}$ and $\displaystyle B_{n,\infty}$ are given by

[TABLE]

and

[TABLE]

The key ingredient for establishing (2) and (3) consists of the states of specific crossings which produce the recurrences

[TABLE]

and

[TABLE]

respectively, with initial values $\displaystyle B_{0,0}(x)=x$ and $\displaystyle B_{0,\infty}(x)=x^{2}$ [6]. Note that the lemma still holds if we replace index $\displaystyle n$ by $\displaystyle r$.

Proposition 2**.**

The generating polynomial for the two-bridge knot $\displaystyle B_{n,r}$ is given by the recurrence

[TABLE]

and has the following closed form:

[TABLE]

Proof.

By Figure 4 we have

[TABLE]

where the last relation follows from property (iii). Solving the recurrence for $\displaystyle n$ yields

[TABLE]

We conclude by the closed forms in Lemma 2. ∎

Remark 3*.*

We can write

[TABLE]

and

[TABLE]

where $\displaystyle\upalpha_{n}(x):=\dfrac{(x+1)^{n}-1}{x}$, so that identity (5) becomes

[TABLE]

Since the coefficients of $\displaystyle\upalpha_{n}(x)$ are all nonnegative, it is clear, by (6), that the polynomial $\displaystyle x^{2}\upalpha_{n}(x)$ counts the states of $\displaystyle B_{n,0}$ that have at least $\displaystyle 2$ circles. This is illustrated in Figure 5 5(a). Likewise, we have an interpretation of (7) in Figure 5 5(b). In Figure 5 and 6, the dashed diagrams represent all possible disjoint union of $\displaystyle\ell-1$ circles ($\displaystyle\ell=n$ or $\displaystyle r$, depending on the context), counted by $\displaystyle\upalpha_{\ell}(x)$ and eventually empty.

Therefore, for $\displaystyle n,r\notin\left\{0,\infty\right\}$, identity (8) means that we can classify the states into $\displaystyle 4$ subset as shown in Figure 6. In these illustrations, there are $\displaystyle 2^{n}-1$ and $\displaystyle 2^{r}-1$ states of 6 and 6 kind, respectively, and $\displaystyle\binom{n}{1}\times\binom{r}{1}+1$ one-component states of 6 and 6 kind. The remaining states are of 6 kind, bringing the total number of states to $\displaystyle 2^{n+r}$.

3 Particular values

Let $\displaystyle\sum_{k\geq 0}b{(n,r;k)}x^{k}:=B_{n,r}(x)$, or $\displaystyle b(n,r;k):=\left[x^{k}\right]B_{n,r}(x)$. Then

[TABLE]

where $\displaystyle\delta_{1,k}$ is the Kronecker symbol. By (1), we recognize $\displaystyle b(n,r;k)$ as the cardinal of the set $\displaystyle\{|S|=k:S\textit{ is a state of }B_{n,r}\}$, i.e., the number of states having $\displaystyle k$ circles. In this section, the coefficients $\displaystyle b(n,r;k)$ are tabulated for some values of $\displaystyle n$, $\displaystyle r$ and $\displaystyle k$. We give as well the corresponding A-numbers in the On-Line Encyclopedia of Integer Sequences [8].

• •

$\displaystyle b(n,0;k)=\left[x^{k}\right]x(x+1)^{n}$, essentially giving entries in Pascal’s triangle A007318 (see Table 1).

• •

$\displaystyle b(n,1;k)=\left[x^{k}\right]\left((x+1)^{n+1}+x^{2}-1\right)$, generating a subtriangle in A300453 (see Table 2).

• •

$\displaystyle b(n,2;k)=\left[x^{k}\right]\left((2x+2)(x+1)^{n}+\left(x^{2}-1\right)(x+2)\right)$, giving triangle in A300454 (see Table 3).

• •

\displaystyle b(n,n;k)=\left[x^{k}\right]\left(\dfrac{(x+1)^{2n}+\big{(}x^{2}-1\big{)}\big{(}2(x+1)^{n}-1)\big{)}}{x}\right), giving triangle in A321127 (see Table 4).

• •

$\displaystyle b(n,r;1)=nr+1$, giving A077028, and displayed as square array in Table 5.

In Kauffman’s language, $\displaystyle b(n,r;1)$ is, for a fixed choice of star region, the number of ways of placing state markers at the crossings of the diagram $\displaystyle B_{n,r}$, i.e., of the forms

[TABLE]

so that the resulting states are “Jordan trails” [2, Section 1–2]. Note that a state marker is interpreted as an instruction to split a crossing as shown below:

[TABLE]

The process is illustrated in Figure 7 for the figure-eight knot.

• •

$\displaystyle b(n,r;2)=n\left(\binom{r}{2}+1\right)+r\left(\binom{n}{2}+1\right)$, giving square array in A300401 (see Table 6).

We paid a special attention to the case $\displaystyle k=2$ because, surprisingly, columns \displaystyle\big{(}b(n,1;2)\big{)}_{n} and \displaystyle\big{(}b(n,2;2)\big{)}_{n} match sequences A000124 and A014206, respectively [6]. The former gives the maximal number of regions into which the plane is divided by $\displaystyle n$ lines, and the latter the maximal number of regions into which the plane is divided by $\displaystyle(n+1)$ circles.

• •

\displaystyle b\big{(}n,r;d(n,r)\big{)}=\textit{leading coefficient of }B_{n,r}(x), giving square array in A321125 (see Table 7). Here, $\displaystyle d(n,r)=\max(n+1,r+1,n+r-1)$ denotes the degree of $\displaystyle B_{n,r}(x)$, and gives entries in A321126. We have Table 8 giving the numbers $\displaystyle d(n,r)$ for $\displaystyle 0\leq n\leq 7$ and $\displaystyle 0\leq r\leq 7$.

We have the following properties:

• –

$\displaystyle d(n,r)=d(r,n)$;

• –

if $\displaystyle r=0$, then $\displaystyle d(n,r)=n+1$;

• –

if $\displaystyle r=\infty$, then sequence \displaystyle\big{(}d(n,r)\big{)}_{n} begins: $\displaystyle 2,2,2,3,4,5,6,7,8,\ldots$ (A233583 with offset 1).

Diagramatically, we give the corresponding illustration for some values of $\displaystyle n$ and $\displaystyle r$ in Figure 8.

Correspondingly, we have

• –

\displaystyle b\big{(}n,r;d(n,r)\big{)}=b\big{(}r,n;d(r,n)\big{)};

• –

if $\displaystyle r=0$, then \displaystyle b\big{(}n,r;d(n,r)\big{)}=1;

• –

if $\displaystyle r=\infty$, then sequence \displaystyle\Big{(}b\big{(}n,r;d(n,r)\big{)}\Big{)}_{n} begins: $\displaystyle 1,1,2,1,1,1,1,1,\ldots$ (A294619 with initial term equals to $\displaystyle 0$).

Remarquable values in Table 7 correspond to knots $\displaystyle B_{1,1}$ (“Hopf link”, see Figure 9), $\displaystyle B_{2,2}$ (figure-eight knot, see Figure 1, 2) and $\displaystyle B_{n,2}$ (“twist knot” [6]) for $\displaystyle n\geq 3$. The latter case can be observed from identity (8) for which the leading coefficient is larger than $\displaystyle 1$ when $\displaystyle n+1=n+r-1$ is satisfied. Also, considere the identity below:

[TABLE]

We notice that the leading coefficient is deduced from the contribution of the summands $\displaystyle B_{n,0}(x)$ and \displaystyle\big{(}\bigcirc\sqcup B_{n,\infty}\big{)}(x) [6].