On the \(S\)-equivalence for Genus One Knots
May 25, 2026
Abstract
In this paper, we construct a sequence of genus one knots that are both S-equivalent, yet can be distinguished by the Jones polynomial. This is related to the problem 1.6 in Kirby’s problem list (K3).
1 Introduction↩︎
Two knots are said to be S-equivalent [1] if their Seifert forms can be transformed into one another through a sequence of operations \(\varLambda_i^{\pm1}(i=1,2,3)\) defined as follows: \[\begin{align} \varLambda_1&: A\rightarrow TAT^T,withTintegral and unimodular,\\ \varLambda_2&: A\rightarrow \left(\begin{array}{ccc|cc} &A& &0&0\\ \hline 0&\cdots &0 &0&1\\ q_1&\cdots &q_n &0&0 \end{array}\right), q_ibeing integrals, \\ \varLambda_3&: A\rightarrow \left(\begin{array}{c|cc} &0&q_1\\ A&\vdots &\vdots \\ &0 & q_n\\ \hline 0&0&0\\ 0&1 &0 \end{array}\right) \end{align}\]
In S. Naik and T. Stanford’ work[2], they proved that S-equivalence is generated by the doubled-delta move on knot diagrams. And it can be seen that the doubled-delta move will change the knot type in general.
C. Livingston proposed the following problem about \(S\)-equivalence which is the problem 1.6 in Kirby’s problem list (K3)[3].
Problem 1.6: Suppose that \(M_1\) and \(M_2\) are \(S\)-equivalent Seifert forms. Does there exist a fixed knot bounding Seifert surfaces \(F_1\) and \(F_2\) for which that associated Seifert forms are \(M_1\) and \(M_2\)?
This problem is also refined by C. Livingston as following:
Given a Seifert form \(M\) of knot \(K\). Does there exist a fixed knot \(K\) such that the map \(\{\tilde{M}\mid \tilde{M} is a Seifert form of K\} \mapsto \{\tilde{M} is a Seifert form S-equivalent to M\}\) is surjective?
M. Aka, P. Feller, A. Miller, and A. Wieser[4] pose the genus-one version of this problem as Problem 7.7. The authors also claim an answer in [4] under additional hypothesis.
Another face of the problem 1.6 in K3[3] is also interesting. That is
How to find a knot such that the map \[\{\tilde{M}\mid \tilde{M}is a Seifert form ofK\} \mapsto \{\tilde{M} is a Seifert form S-equivalent toM\}\] is not surjective?
In this paper, we consider this problem for genus-one knot. Let \(K\) be a knot with genus one. Let \(S\) be the minimal genus Seifert surface of \(K\), and \(M=\begin{pmatrix} a_{11} &a_{12}\\ a_{21}& a_{22} \end{pmatrix}\) be the Seifert form of \(K\). Consider the operation we defined in Secion 2, we get resulting knot \(K_{(\ell,0)}\). By Theorem 4, we obtain the following.
Theorem 1. For a genus-one knot \(K\) with Jones polynomials \(V(K)\ne 1\) and Seifert form \(M=\begin{pmatrix} a_{11} &a_{12}\\ a_{21}& a_{22} \end{pmatrix}\) of \(K\) satisfying \(a_{22}=0\) \((or a_{11}=0)\), the map \[\{\tilde{M}\mid \tilde{M}is a Seifert form ofK\} \mapsto \{\tilde{M} is a Seifert form S-equivalent toM\}\] is not surjective.
Actually, our construction can also be applied for high genus knot. For example, do some speical connect sum of two genus-one knot, and we do the operation on one of the genus-one knots. Then we can get a pair of genus-two knots which is \(S\)-equivalent to each other, but is not equivalent to each other.
Our paper is organized as follows. In Section 2, we define an operation on knot \(K\), and get the resulting knot. Also we get a necessary and sufficient condition for that the resulting knot is \(S\)-equivalent to \(K\). And it is obtained a sufficient condition for that the resulting knot is not equivalent to \(K\). Here we calculate the Jones polynomial of the resulting knot and \(K\). In Section 3, some examples are given for our construction. We also give an example for high genus knot in Section 4.
2 The Main Construction↩︎
Let \(K\) be a genus-one knot and its Seifert form is denoted by \(M\). The minimal genus Seifert surface of \(K\) can be as a disk with two bands attached, which is drawed as following(see Figure 1). Here the dotted line means we do not know the situation about the crossing in that part.
Now, we will define the operation on a band of the Seifert surface. For convenient, the left band is called the first band, and another is called the second. Denote the closed curve through the first band (second band, respectively) by \(\alpha_1\) (\(\alpha_2\), respectively) when we calculate the Seifert form. Give the orientation of \(\alpha_1\) and \(\alpha_2\) by counterclockwise. See Figure 2.
Cut and move a no-twist part of the first band, and attached \(2\ell\) twists on the same band. Denote the result knot by \(K_{(\ell,0)}\). If the operation was on the second band, then denote the result knot by \(K_{(0,\ell)}\). Denote \(K_{(0,0)}=K\). It can be seen that the genus do not change after the operation.
Here \(\ell>0\) means the attached twist is positive, \(\ell<0\) means the attached twist is nagative after the knot was given the orientation. See Figure 3.
For example, the operation from \(K\) to \(K_{(-1,0)}\) is as the following Figure 4.
Remark 2. It can be seen that if do the operation on different bands and get \(K_{(\ell,0)}, K_{(0,\ell)}\), whether \(K_{(\ell,0)}\) is not equivalent to \(K_{(0,\ell)}\) depends on the “position” of no-twist part which was moved.
When saying \(S\)-equivalent of two knot, we need consider a sequence of operations of \(\varLambda_i^{\pm 1}\). We call two knot are first \(S\)-equivalent if two Seifert form of them can be transformed into one another by a sequence of operations of \(\varLambda_1^{\pm 1}\).
Lemma 1. \(K_{(\ell,0)}\) and \(K\) are first \(S\)-equivalent if and only if the Seifert form \(M=\begin{pmatrix} a_{11} &a_{12}\\ a_{21}& a_{22} \end{pmatrix}\) of \(K\) satisfying \(a_{22}=0\) and \(a_{12}+a_{21}\) is a factor of \(\ell\).
Proof. Given a genus-one knot \(K\), and the knot \(K_{(\ell,0)}\) after the operation, with their respective minimal genus Seifert surfaces being \(S, S'\) (one disk attaching two bands), and the Seifert matrix of \(K\) being \(M =\begin{pmatrix} a_{11} & a_{12} \\a_{21} &a_{22} \end{pmatrix}\) and that of \(K_{(\ell,0)}\) being \(M' =\begin{pmatrix} a_{11}-\ell & a_{12} \\a_{21} &a_{22} \end{pmatrix}\), \(\ell \neq0\) (By the definition of \(K_{(\ell,0)}\) and the choose of closed curve).
\(\Rightarrow\): First, if \(K_{(\ell,0)}\) and \(K\) are first \(S\)-equivalent, then there exists \(T=\begin{pmatrix} b_{11} & b_{12} \\b_{21} &b_{22} \end{pmatrix} \in GL_2(\mathbb{Z})\) such that \(TMT^T=M'\). Then \[\begin{align} \det(T)\det(M)\det(T^T) &= \det(M') \\ \det(M)&=\det(M')\\ a_{11}a_{22}-a_{12}a_{21} &= (a_{11}-\ell) a_{22} - a_{12}a_{21} \end{align}\] Since \(\ell \neq 0\), it follows that \(a_{22} = 0\).
Second, \(TMT^T=M'\), then we obtain \[\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & 0 \end{pmatrix} \begin{pmatrix} b_{11} & b_{21} \\ b_{12} & b_{22} \end{pmatrix} = \begin{pmatrix} a_{11}-\ell & a_{12} \\ a_{21} & 0 \end{pmatrix}\] Then we get that \[\begin{cases} b_{11}^2a_{11}+b_{11}b_{12}a_{21} + b_{11}b_{12}a_{12} = a_{11}-\ell \\[4pt] b_{11}b_{21}a_{11}+b_{12}b_{21}a_{21} + b_{11}b_{22}a_{12} = a_{12} \\[4pt] b_{21}b_{11}a_{11}+b_{11}b_{22}a_{21} + b_{12}b_{21}a_{12} = a_{21} \\[4pt] b_{21}^2a_{11}+b_{21}b_{22}a_{21} + b_{21}b_{22}a_{12} = 0 \end{cases}\] Thus \[\begin{cases} (b_{11}^2-1)a_{11}+b_{11}b_{12}a_{21} + b_{11}b_{12}a_{12} = -\ell \\[4pt] b_{11}b_{21}a_{11}+b_{12}b_{21}a_{21} + (b_{11}b_{22}-1)a_{12} = 0 \\[4pt] (b_{12}b_{21}-b_{11}b_{22}+1)(a_{21}-a_{12})= 0\\[4pt] b_{21}^2a_{11}+b_{21}b_{22}a_{21} + b_{21}b_{22}a_{12} = 0 \end{cases}\] Since \(|a_{12} - a_{21}|=1\) by the property of seifert form [5], \[\begin{cases} (b_{11}^2-1)a_{11}+b_{11}b_{12}(a_{21} + a_{12}) =- \ell \quad (1)\\[4pt] b_{11}b_{21}a_{11}+b_{12}b_{21}a_{21} + (b_{11}b_{22}-1)a_{12} = 0 \quad (2)\\[4pt] b_{12}b_{21}-b_{11}b_{22}+1= 0\quad (3)\\[4pt] b_{21}\left( b_{21} a_{11}+b_{22}(a_{21} +a_{12})\right) =0 \quad (4) \end{cases}\]
From (3): \[b_{11}b_{22}=b_{12}b_{21}+1\] then \[\begin{cases} (b_{11}^2-1)a_{11}+b_{11}b_{12}(a_{21} + a_{12}) =- \ell \\[4pt] b_{11}b_{21}a_{11}+b_{12}b_{21}(a_{21} +a_{12}) = 0 \\[4pt] b_{12}b_{21}-b_{11}b_{22}+1= 0\\[4pt] b_{21}\left( b_{21} a_{11}+b_{22}(a_{21} +a_{12})\right) =0 \end{cases}\]
Let \(s=a_{21}+a_{12}\), then \[\begin{cases} (b_{11}^2-1)a_{11}+b_{11}b_{12}s = -\ell \\[4pt] b_{21}(b_{11}a_{11}+b_{12}s) = 0 \\[4pt] b_{12}b_{21}-b_{11}b_{22}+1= 0\\[4pt] b_{21}\left( b_{21} a_{11}+b_{22}s\right) =0 \end{cases}\]
Because we suppose that \(T\) exists, the equation must have a solution, and there will be two possible results. One is \(b_{21}=0\), another one is \(b_{21}\ne0\).
If \(b_{21}=0\), then \(b_{11}b_{22}=1\). So \(b_{11}^2=1\). Thus \[\begin{cases} b_{11}b_{12}s = -\ell \\[4pt] 0\times (b_{11}a_{11}+b_{12}s)= 0 \\[4pt] -b_{11}b_{22}+1= 0\\[4pt] 0 \times \left( b_{21} a_{11}+b_{22}s\right) =0 \end{cases}\] Thus we get \(s=a_{12}+a_{21}\) shall be a factor of \(\ell\).
If \(b_{21}\ne 0\), then \(b_{21}a_{11}+b_{22}s\) and \(b_{11}a_{11}+b_{12}s\) shall be \(0\). Then \[\frac{b_{11}}{b_{12}}=\frac{b_{21}}{b_{22}}=\frac{-s}{a_{11}}\] which makes \(b_{12}b_{21}-b_{11}b_{22}=0\). Contradiction.
\(\Leftarrow\): Let \(a_{22}=0\) and \(s=a_{12}+a_{21}\) is an factor of \(\ell\). Then we need to check whether the following equation has a solution. \[\begin{cases} b_{11}^2a_{11}+b_{11}b_{12}a_{21} + b_{11}b_{12}a_{12} = a_{11}-\ell \\[4pt] b_{11}b_{21}a_{11}+b_{12}b_{21}a_{21} + b_{11}b_{22}a_{12} = a_{12} \\[4pt] b_{21}b_{11}a_{11}+b_{11}b_{22}a_{21} + b_{12}b_{21}a_{12} = a_{21} \\[4pt] b_{21}^2a_{11}+b_{21}b_{22}a_{21} + b_{21}b_{22}a_{12} = 0 \end{cases}\] Do the same as before, and we get \[\begin{cases} (b_{11}^2-1)a_{11}+b_{11}b_{12}s =- \ell \\[4pt] b_{21}(b_{11}a_{11}+b_{12}s) = 0 \\[4pt] b_{12}b_{21}-b_{11}b_{22}+1= 0\\[4pt] b_{21}\left( b_{21} a_{11}+b_{22}s\right) =0 \end{cases}\] And we can see that there must have a solution \(b_{21}=0, b_{11}=1, b_{22}=1, b_{12}=\frac{-\ell}{s}\). ◻
It is obvious that if two knots are first \(S\)-equivalent, they are also \(S\)-equivalent. Thus we have
Lemma 2. If the Seifert form \(M=\begin{pmatrix} a_{11} &a_{12}\\ a_{21}& a_{22} \end{pmatrix}\) of \(K\) satisfying \(a_{22}=0\) and \(a_{12}+a_{21}\) is a factor of \(\ell\), then \(K_{(\ell,0)}\) and \(K\) are \(S\)-equivalent.
Example 1. Give a knot \(K\) with the Seifert form \(M=\begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix}\), and the knot \(K_{(\ell,0)}\) after the operation on \(K\). Then the Seifert form \(M'\) of \(K_{(\ell,0)}\) is equal to \(M=\begin{pmatrix} -\ell & 1 \\ 2 & 0 \end{pmatrix}\). Because \(a_{22}=0, a_{11}+a_{12}=1+(2)=3\), let \(\ell =3k\), then by Lemma 2, \(K\) and \(K_{\ell,0}\) must be first \(S\)-equivalent, furthermore, \(K\) and \(K_{\ell,0}\) are \(S\)-equivalent . By some easy calculation, we can get that \[\begin{pmatrix} 1 & -k \\ 0 & 1 \end{pmatrix} M \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix} =M'\]
Lemma 3. If the Jones polynomial \(V(K)\)of \(K\) is not equal to \(1\), then \(K_{(\ell,0)}\) is not equivalent to \(K\).
Proof. Let \(\ell>0\), considering the skein relation of Jones polynomial, we have, \[t^{-1}V(K_{(\ell,0)}) - tV(K_{(\ell-1,0)}) = (t^{\frac{1}{2}} - t^{-\frac{1}{2}})\left(-(t^{-\frac{1}{2}} + t^{\frac{1}{2}})\right)=t^{-1}-t,\] then \[V(K_{(\ell,0)})=t^2V(K_{(\ell-1,0)}) - t^2 + 1\] Then \[\begin{align} V(K_{(\ell,0)})&=t^{2\times \ell} V(K_{(\ell-1-(\ell-1),0)})+(1-t^2) (t^{2\times (\ell-1)}+\cdots +t^2+1)\\ &=t^{2\times \ell} V(K_{(0,0)})+(1-t^2) (t^{2\times (\ell-1)}+\cdots +t^2+1)\\ &=t^{2\times \ell} V(K)+(1-t^2) \frac{1-(t^2)^{\ell}}{1-t^2}\\ &=t^{2\times \ell} V(K)+1-t^{2\times \ell}\\ \end{align}\] If \(V(K_{(\ell,0)})=V(K)\),then \(V(K_{(\ell,0)})=V(K)=1\).
For the same reason, if \(\ell <0\), then \[V(K_{(\ell,0)})=V(K) \Rightarrow V(K_{(\ell,0)})=V(K)=1\]
Thus we complete the proof of this lemma. ◻
Remark 3. A well-known property of the Jones polynomial is its invariance under Conway mutation. The fact that our knots exhibit different Jones polynomials implies: The knots \(K\) and \(K_{(\ell,0)}\) are not mutants of each other.
By Lemma 2 and Lemma 3, it is obtained the following theorem.
Theorem 4. For the knot \(K\) with Jones polynomials \(V(K)\ne 1\), and Seifert form satisfying the condition of Lemma 2, \(K_{(\ell,0)}\) is \(S\)-equivalent to \(K\), and \(K_{(\ell,0)}\) is not equivalent to \(K\) .
Remark 5. Lemma 1, Lemma 2, Lemma 3 and Theorem 4 are proved for \(K_{(\ell,0)}\). The same result can be proved for \(K_{(0,\ell)}\) (the condition \(a_{22}=0\) shall be changed to \(a_{11}=0\)).
3 sec:example↩︎
In this section, we will give some examples by our construction. The Seifert surface \(S\) of a genus-one knot can be as a disk attaching two bands. Denote the closed curve through the band by \(\alpha_1, \alpha_2\) (the generator of \(H_1(S)\)) when we calculate the Seifert form.
Definition 1. Define the types of double crossing between two bands as the following figure where the dashed line is the band, and the solid line with arrow is the closed curve \(\alpha_1\) or \(\alpha_2\). See Figure 5.
For convenient, as in Section 2, the left band is called the first band, and another is called the second. Also we denote the closed curve through the first band (second band, respectively) by \(\alpha_1\) (\(\alpha_2\), respectively) when we calculate the Seifert form. Give the orientation of \(\alpha_1\) and \(\alpha_2\) by counterclockwise.
Definition 2. Define a type of knot \(\lambda(n,m,p)\) as follows.
Let \(n\) be the number of twists of the first band of \(S\), with \(\lvert n \lvert\) being an even number (because the Seifert surface is orientable).
Let \(m\) be the number of twists of the second band, with \(\lvert m \lvert\) also being an even number.
Let \(p\) be the number of double crossings between two bands, \(\lvert p \lvert\) is an odd number and \(\lvert p \lvert \geq 3\).
Additionally, there are no other types of crossing of the two bands.
Example 2. The following Figure 6 are some examles for \(\lambda(n,m,p)\).
It can be seen that \(\lambda(2\ell,0,3)\) is exactly the resulting knot \(\lambda(0,0,3)_{(\ell,0)}\) after our operation on \(\lambda(0,0,3)\), and \(\lambda(0,2\ell,3)\) is exactly the resulting knot \(\lambda(0,0,3)_{(0,\ell)}\) after our operation on \(\lambda(0,0,3)\) .
3.1 \(S\)-equivalence↩︎
We calculate the Seifert forms \(M(\lambda(m,n,p))\) of the knot \(\lambda(m,n,p)\) as follows. \[\begin{align} &M(\lambda(0,0,3)) = \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix},\\ &M(\lambda(6,0,3))= \begin{pmatrix} -3 & 2 \\ 1 & 0 \end{pmatrix}, M(\lambda(-6,0,3)) = \begin{pmatrix} 3 & 2 \\ 1 & 0 \end{pmatrix}, \\ &M(\lambda(0,6,3)) = \begin{pmatrix} 0 & 2 \\1 & -3 \end{pmatrix}, M(\lambda(0,-6,3)) = \begin{pmatrix} 0 & 2 \\ 1 & -3 \end{pmatrix}. \end{align}\] And it is obtained that \[\begin{pmatrix} 1 &-1 \\ 0 & 1 \end{pmatrix} M(\lambda(0,0,3)) \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} = M(\lambda(6,0,3))\] \[\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} M(\lambda(0,0,3)) \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = M(\lambda(-6,0,3))\] \[\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} M(\lambda(0,0,3)) \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = M(\lambda(0,6,3))\] \[\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} M(\lambda(0,0,3)) \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = M(\lambda(0,-6,3))\] Thus we can get \(\lambda(0,0,3)\),\(\lambda(0,0,3)_{(3,0)}\), \(\lambda(0,0,3)_{(0,3)}\), \(\lambda(0,0,3)_{(-3,0)}\), and \(\lambda(0,0,3)_{(0,-3)}\) is \(S\)-equivalent to each other.
3.2 Jones Polynomial Computation↩︎
It is easy to know that \(\lambda(0,0,3)\) is the knot \(9_{46}\) in the knot table of Rolfsen[6]. And we have \[V(\lambda(0,0,3))=-1/t + 1/t^2 - 2/t^3 + 1/t^4 - 1/t^5 + 1/t^6 + 2.\].Then by Theorem 3, it is obtained that \[\begin{align} V(\lambda(-6,0,3))&=1/t^6 - 1/t^7 + 1/t^8 - 2/t^9 + 1/t^{10} - 1/t^{11} + 1/t^{12} + 1\\ V(\lambda(0,-6,3))&=1/t^6 - 1/t^7 + 1/t^8 - 2/t^9 + 1/t^{10} - 1/t^{11} + 1/t^{12} + 1\\ V(\lambda(6,0,3))&= t^6 - t^5 + t^4 - 2t^3 + t^2 - t + 2\\ V(\lambda(0,6,3))&= t^6 - t^5 + t^4 - 2t^3 + t^2 - t + 2\\ \end{align}\] Clearly, \[\begin{align} &V(\lambda(0,0,3)) \neq V(\lambda(6,0,3))=V(\lambda(0,6,3))\\ &V(\lambda(0,0,3)) \neq V(\lambda(-6,0,3)) = V(\lambda(0,-6,3)) \end{align}\] Thus we can see that \(\lambda(0,0,3)\) is not equivalent to \(\lambda(0,0,3)_{(3,0)}\), \(\lambda(0,0,3)_{(0,3)}\), \(\lambda(0,0,3)_{(-3,0)}\), and \(\lambda(0,0,3)_{(0,-3)}\).
4 Example for high genus knot↩︎
Actually, our construction can be applied to high genus knot. We can also get a pair of knots which is \(S\)-equivalent to each other and is not equivalent to each other. We will give an example here.
Example 3. Consider \(K_1=\lambda(0,0,3)\sharp \lambda(0,0,3)\) and \(K_2=\lambda(0,0,3)_{(-3,0)}\sharp \lambda(0,0,3)\). Actually, \(\lambda(0,0,3)_{(-3,0)}\) is exactly \(\lambda(-6,0,3)\). See Figure 7.
Because \(\lambda(0,0,3)\) is \(S\)-equivalent to \(\lambda(0,0,3)_{(-3,0)}\), \(K_1\) is \(S\)-equivalent to \(K_2\). By easy calculation, we get \(T=\begin{pmatrix} 1 & 1 & & \\ 0 & 1 &&\\ &&1&0\\ &&0&1 \end{pmatrix}\), and \(T\begin{pmatrix} M(\lambda(0,0,3)) & & \\ && M(\lambda(0,0,3)) \end{pmatrix} T^T= \begin{pmatrix} M(\lambda(-6,0,3)) & & \\ && M(\lambda(0,0,3)) \end{pmatrix}.\)
For the Jones polynomial of \(K_1\) and \(K_2\), because the Jones Polynomial is multiplicative over connect sums [7], we have \(V(K_1)=V(\lambda(0,0,3))\times V(\lambda(0,0,3))\), and \(V(K_1)=V(\lambda(-6,0,3))\times V(\lambda(0,0,3))\). By the calculation of Jones polynomial of \(\lambda(0,0,3)\) and \(\lambda(-6,0,3)\) in Subsection 3.2, it is obtained that \(V(K_1)\ne V(K_2)\). Thus \(K_1\) is \(S\)-equivalent to \(K_2\), but they are different knot.
Acknowledges↩︎
The project was funded by Science and Technology Project of Hebei Education Department (Grant No. QN2023030), Science Foundation of Hebei Normal University (Grant Nos. L2022B02, L2025ZD04).