Lens spaces which are realizable as closures of homology cobordisms over planar surfaces. (English) Zbl 1452.57015
In [Ill. J. Math. 62, No. 1–4, 99–111 (2018; Zbl 1409.57017)] Y. Nozaki showed by solving algebraic equations that every lens space contains a genus-one homologically fibered knot. Like him, the author gives us a relation between an algebraic equation and the existence of a homologically fibered link of a given number of components whose Seifert surface is a planar surface in a lens space. Denote by \(L(p, q)\) a lens space and by \(\Sigma_{g, n}\) a surface of genus \(g\) having \(n\) boundary components; we first consider the following
Proposition 1. \(L(p, q)\) has a realization as a closure of a homology cobordism over \(\Sigma_{0, n+1}\) for \(n \in \mathbb N\) if and only if there are integers \(a_h\), \(l_{i, j}\) and \(t_k\), where \(1 \leq h, i, j, k \leq n\) with \(i \leq j\), satisfying \[ \det\begin{bmatrix} p &-qa_1&-qa_2& \dotsb &-qa_n\\ a_1&t_1&l_{1, 2}& \dotsb& l_{1,n}&\\a_2&l_{1, 2}&t_2& \dotsb &l_{2, n}\\ \vdots&\vdots & \vdots &\ddots &\vdots\\ a_n& l_{1, n}& l_{2, n}& \dotsb &t_n\end{bmatrix}=\pm1. \]
Since any manifold that has a realization as a closure of a homology cobordism over \(\Sigma_{0, 1}\) is a homology \(3\)-sphere, none of the lens spaces are realizable as such. Further, if the determinant of the \(n\times n\) matrix above has a solution, then we may let \(a_{n+1} = l_{i, n+1}=0\) for all \(1 \leq i \leq n\) and \(t_{n+1}=1\) to obtain a solution for a matrix having a larger dimension than the one in the \(n\times n\) case. In other words, a lens space having a realization as a closure of a homology cobordism over \(\Sigma_{0, n+1}\) has one over \(\Sigma_{0, n+2}\). In this sense, the best way to think about the value \(n\) is to find its minimal number in the above matrix which still has a solution. In this article, the author answers this question as stated below.
Theorem 1. \(L(p, q)\) has a realization as a closure of a homology cobordism over \(\Sigma_{0, 2}\) if and only if \(q\) or \(-q\) is a quadratic residue modulo \(p\).
Proof. The proof of this theorem completely relies on the first proposition which is interestingly a straightforward idea. \(L(p, q)\) is realizable as a closure of a homology cobordism over \(\Sigma_{0, 2}\) is equivalent to the existence of integers \(a\) and \(t\) satisfying \(\det\begin{bmatrix} p &-qa\\ a& t\end{bmatrix}=\pm1\).
Hence, \(tp +qa^2 = \pm1\). Thus, \(p\) and \(a^2\) are relatively prime meaning \(p\) and \(a\) are coprime as well. Further, \(a\) satisfies \(qa^2 \equiv \pm 1 \pmod p\). As \(gcd\{p, a\}=1\), we obtain \(q \equiv \pm (a^{-1})^2 \pmod p\). Consequently, the condition for \(L(p, q)\) being realizable as a closure of a homology cobordism over \(\Sigma_{0, 2}\) is equivalent to \(q\) or \(-q\) becoming a quadratic residue modulo \(p\).
Theorem 2. Every lens space has a realization as a closure of a homology cobordism over \(\Sigma_{0, 3}\).
Finally, I think that [H. Goda and T. Sakasai, Tokyo J. Math. 36, No. 1, 85–111 (2013; Zbl 1287.57022)] and [T. Sakasai, Winter Braids Lect. Notes 3, Exp. No. 4, 25 p. (2016; Zbl 1422.57051)] are excellent references to understand related problems discussed in this paper.
Proposition 1. \(L(p, q)\) has a realization as a closure of a homology cobordism over \(\Sigma_{0, n+1}\) for \(n \in \mathbb N\) if and only if there are integers \(a_h\), \(l_{i, j}\) and \(t_k\), where \(1 \leq h, i, j, k \leq n\) with \(i \leq j\), satisfying \[ \det\begin{bmatrix} p &-qa_1&-qa_2& \dotsb &-qa_n\\ a_1&t_1&l_{1, 2}& \dotsb& l_{1,n}&\\a_2&l_{1, 2}&t_2& \dotsb &l_{2, n}\\ \vdots&\vdots & \vdots &\ddots &\vdots\\ a_n& l_{1, n}& l_{2, n}& \dotsb &t_n\end{bmatrix}=\pm1. \]
Since any manifold that has a realization as a closure of a homology cobordism over \(\Sigma_{0, 1}\) is a homology \(3\)-sphere, none of the lens spaces are realizable as such. Further, if the determinant of the \(n\times n\) matrix above has a solution, then we may let \(a_{n+1} = l_{i, n+1}=0\) for all \(1 \leq i \leq n\) and \(t_{n+1}=1\) to obtain a solution for a matrix having a larger dimension than the one in the \(n\times n\) case. In other words, a lens space having a realization as a closure of a homology cobordism over \(\Sigma_{0, n+1}\) has one over \(\Sigma_{0, n+2}\). In this sense, the best way to think about the value \(n\) is to find its minimal number in the above matrix which still has a solution. In this article, the author answers this question as stated below.
Theorem 1. \(L(p, q)\) has a realization as a closure of a homology cobordism over \(\Sigma_{0, 2}\) if and only if \(q\) or \(-q\) is a quadratic residue modulo \(p\).
Proof. The proof of this theorem completely relies on the first proposition which is interestingly a straightforward idea. \(L(p, q)\) is realizable as a closure of a homology cobordism over \(\Sigma_{0, 2}\) is equivalent to the existence of integers \(a\) and \(t\) satisfying \(\det\begin{bmatrix} p &-qa\\ a& t\end{bmatrix}=\pm1\).
Hence, \(tp +qa^2 = \pm1\). Thus, \(p\) and \(a^2\) are relatively prime meaning \(p\) and \(a\) are coprime as well. Further, \(a\) satisfies \(qa^2 \equiv \pm 1 \pmod p\). As \(gcd\{p, a\}=1\), we obtain \(q \equiv \pm (a^{-1})^2 \pmod p\). Consequently, the condition for \(L(p, q)\) being realizable as a closure of a homology cobordism over \(\Sigma_{0, 2}\) is equivalent to \(q\) or \(-q\) becoming a quadratic residue modulo \(p\).
Theorem 2. Every lens space has a realization as a closure of a homology cobordism over \(\Sigma_{0, 3}\).
Finally, I think that [H. Goda and T. Sakasai, Tokyo J. Math. 36, No. 1, 85–111 (2013; Zbl 1287.57022)] and [T. Sakasai, Winter Braids Lect. Notes 3, Exp. No. 4, 25 p. (2016; Zbl 1422.57051)] are excellent references to understand related problems discussed in this paper.
Reviewer: Ryo Ohashi (Wilkes-Barre)
MSC:
| 57K30 | General topology of 3-manifolds |
| 57M50 | General geometric structures on low-dimensional manifolds |
Keywords:
lens spacesReferences:
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