Hostname: page-component-cb9f654ff-plnhv Total loading time: 0.001 Render date: 2025-08-27T12:35:30.809Z Has data issue: false hasContentIssue false

$\mathbb {Z}$-disks in ${\mathbb {C}} P^2$

Published online by Cambridge University Press:  27 August 2025

Anthony Conway
Affiliation:
The University of Texas at Austin, Austin, TX 78712, USA anthony.conway@austin.utexas.edu
Irving Dai
Affiliation:
The University of Texas at Austin, Austin, TX 78712, USA irving.dai@math.utexas.edu
Maggie Miller
Affiliation:
The University of Texas at Austin, Austin, TX 78712, USA maggie.miller.math@gmail.com
Rights & Permissions [Opens in a new window]

Abstract

We study locally flat disks in $(\mathbb {C}P^2)^\circ :=({\mathbb {C}} P^2)\setminus \mathring {B}^4$ with boundary a fixed knot $K$ and whose complement has fundamental group $\mathbb {Z}$. We show that, up to topological isotopy relative to the (rel.) boundary, such disks necessarily arise by performing a positive crossing change on $K$ to an Alexander polynomial one knot and capping off with a $\mathbb {Z}$-disk in $D^4.$ Such a crossing change determines a loop in $S^3 \setminus K$ and we prove that the homology class of its lift to the infinite cyclic cover leads to a complete invariant of the disk. We prove that this determines a bijection between the set of rel. boundary topological isotopy classes of $\mathbb {Z}$-disks with boundary $K$ and a quotient of the set of unitary units of the ring $\mathbb {Z}[t^{\pm 1}]/(\Delta _K)$. Number-theoretic considerations allow us to deduce that a knot $K \subset S^3$ with quadratic Alexander polynomial bounds $0,1,2,4$, or infinitely many $\mathbb {Z}$-disks in $(\mathbb {C}P^2)^\circ$. This leads to the first examples of knots bounding infinitely many topologically distinct disks whose exteriors have the same fundamental group and equivariant intersection form. Finally, we give several examples where these disks are realized smoothly.

MSC classification

Information

Type
Research Article
Creative Commons
Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
Copyright
© The Author(s), 2025.

1. Introduction

Freedman proved that a knot $K \subset S^3$ bounds a locally flat disk $D \subset D^4$ with $\pi _1(D^4 \setminus D) \cong \mathbb {Z}$ if and only if $K$ has Alexander polynomial one [Reference FreedmanFre82]. It is additionally known that if $K$ bounds such a disk, then it is unique up to isotopy relative to the (rel.) boundary [Reference Conway and PowellCP21]. In other words, the set $\mathcal {D}_{\mathbb {Z}}(K,D^4)$ of rel. boundary isotopy classes of $\mathbb {Z}$ -disks in $D^4$ with boundary $K$ is non-empty if and only if $\Delta _K \doteq 1$ , in which case it contains a single element.

This paper describes the classification of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ :={\mathbb {C}} P^2 \setminus \mathring {B}^4$ with boundary a knot $K$ : we list necessary and sufficient conditions for the existence of such disks (Theorem1.1) and construct an explicit bijection from $\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ )$ to a subset of the Alexander module (Theorem1.2). The outcome is quite different that in $D^4$ : when $\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ )$ is non-empty, it rarely consists of a single element and we use some number theory to show that if $K$ has trivial or quadratic Alexander polynomial (e.g. if $K$ has genus one), then $\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ )$ must have cardinality $0,1,2,4$ , or be infinite (Theorem1.3).

In what follows, a $4$ -manifold is understood to mean a compact, connected, oriented, topological $4$ -manifold and embeddings are understood to be locally flat. Knots are assumed to be oriented. Given a closed simply-connected $4$ -manifold $X$ , we write $X^\circ :=X \setminus \mathring {B}^4$ , and a $\mathbb {Z}$ -disk in $X^\circ$ refers to a properly embedded disk $D \subset X^\circ$ whose complement has fundamental group $\mathbb {Z}$ . The knot $K=\partial D$ is then called $\mathbb {Z}$ -slice in $X^\circ$ .

1.1 Existence of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$

Necessary and sufficient criteria for deciding whether a knot $K$ is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ can be deduced from the work of Fogel [Reference FogelFog94]. To state this result, use $E_K$ to denote the exterior of $K$ , and recall that the Blanchfield form of $K$ is a nonsingular, sesquilinear, Hermitian form

\begin{equation*} {\textrm {Bl}}_K \colon H_1(E_K^\infty ) \times H_1(E_K^\infty ) \to \mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}],\end{equation*}

which, roughly speaking, measures equivariant linking in the infinite cyclic cover $E_K^\infty$ of $E_K$ . We say that ${\textrm {Bl}}_K$ is represented by a size $n$ Hermitian matrix $A$ with $\det (A) \neq 0$ and coefficients in $\mathbb {Z}[t^{\pm 1}]$ if ${\textrm {Bl}}_K$ is isometric to the following linking form:

(1.1) \begin{align}\begin{array}{rl} \mathbb {Z}[t^{\pm 1}]^n/A^T\mathbb {Z}[t^{\pm 1}]^n \times \mathbb {Z}[t^{\pm 1}]^n/A^T\mathbb {Z}[t^{\pm 1}]^n &\to \mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}]\\[4pt] ([x],[y]) &\mapsto -x^TA^{-1}\overline {y}. \end{array} \end{align}

If such a matrix has size $1$ , then it is a symmetric polynomial $p(t)$ and, in this case, we say that ${\textrm {Bl}}_K$ is represented by $p(t)$ . Here, and in what follows, we write $\Delta _K$ for the unique symmetric representative of the Alexander polynomial of $K$ that evaluates to $-1$ at $t=1$ ; the reason for this convention will be elucidated in § 1.3. Recalling that we are working in the topological category, $\mathbb {Z}$ -sliceness in $(\mathbb {C}P^2)^\circ$ can be characterized as follows; see § 3 for the proof.

Theorem 1.1. Given a knot $K$ , the following assertions are equivalent:

  1. (1) $K$ is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ ;

  2. (2) the Blanchfield form ${\textrm {Bl}}_K$ is presented by $-\Delta _K(t)$ ;

  3. (3) $K$ can be converted into an Alexander polynomial one knot by switching a single positive crossing to a negative crossing; and

  4. (4) $K$ can be converted into an Alexander polynomial one knot by a single positive generalized crossing change.

Here, given a curve $\gamma \subset S^3 \setminus K$ with $\ell k(K,\gamma )=0$ that is unknotted in $S^3$ , we say that a knot $K'$ in $S^3$ is obtained from $K$ by a positive generalized crossing change along $\gamma$ if $(S^3_{-1}(\gamma ),K)\cong (S^3,K')$ .As illustrated in Figure 1, when only two strands of $K$ are involved, we omit the word ‘generalized’. We note that we are also considering $K \subset S^3$ as a knot in $S^3\setminus \nu (\gamma )$ and hence in $S^3_{-1}(\gamma )$ . We emphasize that the last three items of Theorem1.1 were already known to be equivalent by work of Fogel [Reference FogelFog94]; see also [Reference Borodzik and FriedlBF14] for the case of multiple crossing changes.

Figure 1 Top: a positive crossing change from $K$ to $K'$ realized as a (generalized) crossing change about a curve $\gamma$ (bold). Bottom: a generalized positive crossing change about $\gamma$ transforms $K$ into $K'$ . In this example, $\gamma$ links $K$ geometrically more than two times (although as always links $K$ zero times algebraically).

In summary, Theorem1.1 describes conditions for obstructing a knot from bounding a $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ as well as ways to construct $\mathbb {Z}$ -slice knots in $(\mathbb {C}P^2)^\circ$ . In § 1.6 we list related work on the existence question for $\mathbb {Z}$ -surfaces in $4$ -manifolds with boundary $S^3$ such as [Reference Feller and LewarkFL18, Reference Feller and LewarkFL24, Reference Kjuchukova, Miller, Ray and SakalliKMRS24, Reference Conway, Piccirillo and PowellCPP23] but first we discuss the classification of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ .

1.2 Classification of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$

Assume now that a generalized crossing change along the surgery curve $\gamma$ results in an Alexander polynomial one knot. As we recall in more detail in § 4, one can associate to $K$ and $\gamma$ a $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ , which we refer to as a generalized crossing change $\mathbb {Z}$ -disk.

From now on, we fix a basepoint $z \in \partial E_K$ and a lift $\widetilde {z} \in \partial E_K^\infty$ of $z$ to the infinite cyclic cover. When we refer to loops and lifts, it is with respect to these basepoints. We note in Proposition 4.10 below that the homology class of the lift to $E_K^\infty$ of the surgery curve $\gamma \subset E_K$ belongs to

(1.2) \begin{equation} \mathcal {G}_K:= \left \lbrace x \in H_1(E_K^\infty ) \ \Big | \ x \text { is a generator and } {\textrm {Bl}}_K(x,x) =\frac {1}{\Delta _K} \right \rbrace . \end{equation}

Observe that the set $\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ acts on $ \mathcal {G}_K$ by multiplication. The following theorem (which, recall, takes place in the topological category) gives a complete characterization of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ as well as a complete invariant; see § 5 for the proof.

Theorem 1.2. Let $K$ be a knot.

  1. (1) Every $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ with boundary $K$ is isotopic rel. boundary to a crossing change $\mathbb {Z}$ -disk.

  2. (2) Two generalized crossing change $\mathbb {Z}$ -disks with boundary $K$ are isotopic rel. boundary if and only if the homology classes of the lifts of their surgery curves agree in the Alexander module $H_1(E_K^\infty )$ up to multiplication by $t^k$ for some $k \in \mathbb {Z}$ .

  3. (3) Mapping a generalized crossing change $\mathbb {Z}$ -disk with boundary $K$ to the homology class of the lift of its surgery curve defines a bijection

    \begin{equation*}\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ ) \xrightarrow {\cong } \mathcal {G}_K/\lbrace t^k \rbrace _{k \in \mathbb {Z}}.\end{equation*}

Note that $\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ )$ is non-empty if and only if $K$ is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ , whereas $\mathcal {G}_K$ can be non-empty as soon as $K$ has cyclic Alexander module; the bijection only holds if $K$ is $\mathbb {Z}$ -slice.

In summary, Theorem1.2 shows that all $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ arise as crossing change $\mathbb {Z}$ -disks and that the lift of the surgery curve leads to a complete invariant. In § 1.6 we compare Theorem1.2 to the classification of $\mathbb {Z}$ -surfaces obtained in [Reference Conway, Piccirillo and PowellCPP23] but first we describe how the bijection with $\mathcal {G}_K/\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ allows us to explicitly enumerate $\mathbb {Z}$ -disks when $K$ has trivial or quadratic Alexander polynomial.

Figure 2. Left: the twist knot $K_n$ where the box denotes $n$ full twists. Middle: the figure eight knot $K_1$ . Right: the right-handed trefoil $K_{-1}$ .

1.3 Knots with quadratic Alexander polynomial: an explicit enumeration

Alexander polynomial one knots bound a unique $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary; this is a consequence of Theorem1.2 (and should be compared to there being a unique $\mathbb {Z}$ -sphere in ${\mathbb {C}} P^2$ [Reference Conway and OrsonCO25, Proposition A.1]) but also follows from the combination of [Reference Conway and PowellCP23, Theorem1.2] and [Reference Orson and PowellOP22, Theorem C]. When the polynomial is nontrivial, some number theory is needed to understand the cardinality of $\mathcal {G}_K/\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ , as we now explain. If the Alexander polynomial of a knot is trivial or quadratic (e.g. if $K$ is a genus one knot), then up to multiplication by $\pm t^k$ , it is of the form

\begin{equation*}\Delta _n:=nt-(2n+1)+nt^{-1},\end{equation*}

for some $n \in \mathbb {Z}$ . Note that $\Delta _0=1, \Delta _{-1}=-t+1-t^{-1}$ is the Alexander polynomial of the trefoil, $\Delta _1$ is the Alexander polynomial of the figure eight knot, and more generally, $\Delta _n$ is the Alexander polynomial of the twist knot $K_n$ with $n$ full twists illustrated in Figure 2. Twists knots are genus one knots that are $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ because they can be unknotted by switching a single positive crossing change; recall Theorem1.1. We find it more convenient to index the $\Delta _n$ using their leading coefficient which is why we work with a symmetric representative of the Alexander polynomial that satisfies $\Delta _K(1)=-1$ : this way $\Delta _{K_n}=\Delta _n$ .

Our next result enumerates $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ for knots with trivial or quadratic Alexander polynomial; its proof follows by combining Theorem1.2 with the number-theoretic Theorem1.7.

Theorem 1.3. If a knot $K$ with trivial or quadratic Alexander polynomial is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ , then it bounds $1,2,4$ or infinitely many $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary. More precisely, up to isotopy rel. boundary,

\begin{equation*} K \text{ bounds } \begin{cases} 1 \text { disk} & \text {if $\Delta _K=\Delta _0,\Delta _{-1}$}, \\ 2 \text { disks} & \text {if $\Delta _K=\Delta _1,\Delta _2$, or $\Delta _K=\Delta _n$ for $n=-p^k$ with $p$ prime and $k$ odd}, \\ 4 \text { disks} & \text {if $\Delta _K=\Delta _n$ for $n=-p^k$ with $p$ prime and $k$ even}, \\ \infty \text { disks} & \text {otherwise}.\\ \end{cases} \end{equation*}

We emphasize the difference with $\mathbb {Z}$ -disks in $D^4$ where $|\mathcal {D}_{\mathbb {Z}}(K,D^4)| \in \lbrace 0,1\rbrace$ ; in contrast, when $\Delta _K$ is quadratic, Theorem1.3 shows that, generically, $\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ )$ is infinite.

Remark 1.4. Theorems 1.2 and 1.3 admit analogues where the word ‘isotopy’ is replaced by ‘equivalence’. To form the analogue of Theorem 1.2, one substitutes multiplication by $t^k$ for multiplication by $\pm t^k$ , whereas for Theorem 1.3, the outcome is that if $K$ is a knot with trivial or quadratic Alexander polynomial that is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ , then, up to equivalence rel. boundary,

\begin{equation*} K \text { bounds } \begin{cases} 1 \text { disk} & \text {if $\Delta _K=\Delta _0,\Delta _{-1},\Delta _1,\Delta _2$}, \\ & \text {or $\Delta _K=\Delta _n$ for $n=-p^k$ with $p$ prime and $k$ odd}, \\ 2 \text { disks} & \text {if $\Delta _K=\Delta _n$ for $n=-p^k$ with $p$ prime and $k$ even}, \\ \infty \text { disks} & \text {otherwise}.\\ \end{cases} \end{equation*}

We refer to Remarks 1.8 and 5.5 for further details.

We describe some further context surrounding Theorem1.3 in § 1.6 but presently we discuss examples of Theorem1.3 with an emphasis on realizing the disks smoothly.

As we explain in § 4, the construction of the $\mathbb {Z}$ -disk $D_\gamma$ associated to a generalized crossing change curve $\gamma$ depends on a choice of orientation for $\gamma$ . Reversing the orientation on $\gamma$ results in a disk $D_{-\gamma }$ which is equivalent, but not necessarily isotopic, to the original rel. boundary. Each generalized crossing change curve thus gives two $\mathbb {Z}$ -disks which are equivalent rel. boundary, but usually distinct in the sense of Theorem1.3. For knots with trivial or quadratic Alexander polynomial, we note in Remark 8.3 that $D_{\pm \gamma }$ are isotopic rel. boundary if and only if $\Delta _K = \Delta _0$ or $\Delta _{-1}$ .

Example 1.5. We illustrate Theorem 1.3 using twist knots and include several examples of explicit smooth disks properly embedded in $(\mathbb {C}P^2)^\circ$ . The point here is that if $K$ can be unknotted by a single positive generalized crossing change, then the associated $\mathbb {Z}$ -disk is smoothly embedded.

  1. The right-handed trefoil $K_{-1}$ bounds a unique $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary. As we discuss in Example 8.2, this disk can be smoothly realized as a crossing change $\mathbb {Z}$ -disk $D_\gamma$ . We explicitly show that, in this case, the disks $D_{\pm \gamma }$ are smoothly isotopic rel. boundary.

  2. The figure eight knot $K_1$ and the stevedore knot $K_2$ each bound two $\mathbb {Z}$ -disks up to isotopy rel. boundary and a single disk up to equivalence rel. boundary. Example 8.4 focuses on the figure eight knot and shows that the disks are smoothly realized by crossing change $\mathbb {Z}$ -disks $D_{\pm \gamma }$ .

  3. The twist knot $K_{-4}$ bounds four $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary and two $\mathbb {Z}$ -disks up to equivalence rel. boundary. As we discuss in Example 8.5, these disks are smoothly realized by generalized crossing change $\mathbb {Z}$ -disks $D_{\pm \gamma _1},D_{\pm \gamma _2}$ .

  4. More generally, for $n = -k^2$ , $k$ a prime power, the twist knot $K_n$ bounds four $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary and two $\mathbb {Z}$ -disks up to equivalence rel. boundary. In Example 8.5, we realize two isotopy classes smoothly as crossing change $\mathbb {Z}$ -disks $D_{\pm \gamma _1}$ with $\gamma _1$ an unknotting curve, and describe the other two disks as $D_{\pm \gamma _2}$ with $\gamma _2$ a generalized unknotting curve to an Alexander polynomial knot. We do not know if all four isotopy classes can be realized smoothly. The construction also holds for $k$ not a prime power, but in this case $K_n$ has infinitely many $\mathbb {Z}$ -disks, out of which we have only attempted to describe four.

  5. Finally, for $n = -k^2$ , $k$ a prime power, the $n$ -twisted negative Whitehead double of the torus knot $T(-k,k-1)$ bounds four $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary and two $\mathbb {Z}$ -disks up to equivalence rel. boundary. In Example 8.6, we realize all of these smoothly via a pair of distinct generalized unknotting curves (together with their orientation reversals). The construction also holds for $k$ not a prime power, but in this case the knot in question has infinitely many $\mathbb {Z}$ -disks, out of which we have only attempted to describe four.

This shows that each case of Theorem 1.3 where $K$ bounds a finite number of $\mathbb {Z}$ -disks can be illustrated by an example in which all isotopy classes can be realized smoothly.

We further illustrate Theorems1.1 and 1.3 by calculating the number $d_K$ of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ bounded by a knot $K$ . Thanks to Theorem1.1, we know that if a knot or its mirror is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ , then its Blanchfield form is presented by a size one nondegenerate Hermitian matrix. Work of Fogel [Reference FogelFog94, Theorem] therefore implies that if a knot $K$ or its mirror is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ , then $u_a(K)=1$ , where $u_a(K)$ denotes the algebraic unknotting number of $K$ . The algebraic unknotting number is listed on KnotInfo [Reference Livingston and MooreLM24].

Example 1.6. We list the number $d_K$ of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ up to isotopy for knots $K$ up to five crossings. As mentioned in Example 1.5, the unknot and the trefoil satisfy $d_K=1$ , whereas the figure eight has $d_K=2$ . Since the cinquefoil $K=5_1$ satisfies $u_a(K)=2$ , we have $d_K=0$ . Finally, the knot $K=5_2$ is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ with $\Delta _{K}=-2t+3-2t^{-1}=\Delta _{-2}$ so $d_K=2$ .

1.4 The underlying number theory

Next we state the number-theoretic result that underlies Theorem1.3. First we recast $\mathcal {G}_K$ in a more algebraic way. Assume that $K$ is a knot that is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ so that by Theorem1.1 its Alexander module is cyclic (i.e. $H_1(E_K^\infty )\cong \mathbb {Z}[t^{\pm 1}]/(\Delta _K)$ ) and ${\textrm {Bl}}_K(x,y)={x\overline {y}}/({\Delta _K})$ for every $x,y \in H_1(E_K^\infty )$ . Here $x \mapsto \overline {x}$ denotes the involution $p(t) \mapsto p(t^{-1}).$ This implies that polynomial multiplication induces a free and transitive action of

\begin{equation*} U\left (\frac {\mathbb {Z}[t^{\pm 1}]}{(\Delta _K)}\right ):= \left \lbrace u \in \frac {\mathbb {Z}[t^{\pm 1}]}{(\Delta _K)} \ \Big | \ u\overline {u}=1 \right \rbrace, \end{equation*}

on $\mathcal {G}_K$ . In particular, there is a noncanonical bijection

\begin{equation*} \mathcal {G}_K \xrightarrow {\approx } U\left (\frac {\mathbb {Z}[t^{\pm 1}]}{(\Delta _K)}\right ).\end{equation*}

The general setting is as follows. Given a ring $R$ with involution $x \mapsto \overline {x}$ , the group of unitary units $U(R)$ refers to those $x \in R$ such that $x \overline {x}=1$ . For example, when $R=\mathbb {Z}[t^{\pm 1}]$ with the involution $p(t) \mapsto p(t^{-1})$ , all units are unitary and are of the form $\pm t^{k}$ with $k \in \mathbb {Z}$ . More generally, if $R=\mathbb {Z}[t^{\pm 1}]/p(t)$ for some $p(t) \in \mathbb {Z}[t^{\pm 1}]$ , then the group $U(R)$ always contains the image of the set $\lbrace \pm t^k \rbrace _{k \in \mathbb {Z}} \subset \mathbb {Z}[t^{\pm 1}]$ under the projection map $\mathbb {Z}[t^{\pm 1}] \to \mathbb {Z}[t^{\pm 1}]/p(t).$

Theorem1.3 therefore follows from Theorem1.2 and the study of $U(\mathbb {Z}[t^{\pm 1}]/p(t))/\lbrace t^k\rbrace _{k \in \mathbb {Z}}$ for quadratic Laurent polynomials $p(t)$ that arise as Alexander polynomials, i.e. for the polynomials $\Delta _n=nt-(2n+1)+nt^{-1}$ .

We write

\begin{equation*}U_n:=U\left (\frac {\mathbb {Z}[t^{\pm 1}]}{(\Delta _n)}\right ),\end{equation*}

as a shorthand for the set of unitary units of $\mathbb {Z}[t^{\pm 1}]/(\Delta _n)$ . In order to prove Theorem1.3, it suffices to determine for what values of $n$ the set $U_n/\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ is infinite and to determine its cardinality when it is finite. The next theorem (which is proved in § 7) achieves slightly more as it also determines the rank of $U_n/\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ in the infinite case in terms of the number $\Omega (n)$ of positive primes dividing $n$ .

Theorem 1.7. For every $n \in \mathbb {Z}$ , the group $U_n/\{t^k\}_{k \in \mathbb {Z}}$ can be described as follows.

  1. (1) The group $U_n/\{t^k\}_{k \in \mathbb {Z}}$ is finite precisely when $n = 2, 1, 0, -1$ , or $-p^k$ for a prime $p$ , and

    \begin{equation*} U_n/\{t^k\}_{k \in \mathbb {Z}} \cong \begin{cases} \lbrace 1 \rbrace & \text {if $n=-1,0$}, \\ \mathbb {Z}/2\mathbb {Z} & \text {if $n=1,2$ or $n=-p^k$ with $k$ odd}, \\ \mathbb {Z}/4\mathbb {Z} & \text {if $n=-p^k$ with $k$ even}. \\ \end{cases} \end{equation*}
  2. (2) In all other cases, $U_n/\{t^k\}_{k \in \mathbb {Z}}$ is infinite and

    \begin{equation*} {\textrm {rk}} ( U_n/\{t^k\}_{k \in \mathbb {Z}}) =\begin{cases} \Omega (n) & \text {if $\Delta _n$ is irreducible and $n\gt 0$}, \\ \Omega (n)-1 & \text {if $\Delta _n$ is irreducible and $n\lt 0$ or if $\Delta _n$ is reducible.} \end{cases} \end{equation*}

We use the convention that strictly negative rank should be interpreted as zero rank; we write this for simplicity of the theorem statement. The cases in which $(2)$ gives a rank less than or equal to zero correspond precisely to the cases in $(1)$ ; see Remark 7.1.

Remark 1.8. As explained in Remark 7.3, Theorem 1.7 can be modified to obtain the following description of $U_n/\{ \pm t^k\}_{k \in \mathbb {Z}}$ .

  1. (1) The group $U_n/\{\pm t^k\}_{k \in \mathbb {Z}}$ is finite precisely when $n = 2, 1, 0, -1$ , or $-p^k$ for a prime $p$ , and

    \begin{equation*} U_n/\{\pm t^k\}_{k \in \mathbb {Z}} \cong \begin{cases} \lbrace 1 \rbrace & \textrm {if $n=-1,0,1,2$ or $n=-p^k$ with $k$ odd}, \\ \mathbb {Z}/2\mathbb {Z} & \textrm{if $n=-p^k$ with $k$ even}. \\ \end{cases} \end{equation*}
  2. (2) In all other cases, $U_n/\{ \pm t^k\}_{k \in \mathbb {Z}}$ is infinite and

    \begin{equation*} {\textrm {rk}} ( U_n/\{ \pm t^k\}_{k \in \mathbb {Z}}) =\begin{cases} \Omega (n) & \textrm {if $\Delta _n$ is irreducible and $n\gt 0$}, \\ \Omega (n)-1 & \textrm {if $\Delta _n$ is irreducible and $n\lt 0$ or if $\Delta _n$ is reducible.} \end{cases} \end{equation*}

As noted in Remark 1.4, for $K$ a knot with trivial or quadratic Alexander polynomial, this result leads to the classification of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ with boundary $K$ up to equivalence rel. boundary (instead of isotopy rel. boundary).

1.5 Challenges in the smooth category

We now briefly discuss the results of Theorem1.3 in the context of the smooth category. The most immediate question in this regard is to determine which of the isotopy classes in Theorem1.3 are realized by smooth disks. As we have seen in Example 1.5, there are knots illustrating the first through third cases of Theorem1.3 for which all isotopy classes are realized smoothly.

Question 1.9. Is there a genus one knot with an infinite number of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ such that each class is realized smoothly?

Question 1.10. Is there a genus one knot such that only a strict, non-empty subset of the isotopy classes from Theorem 1.3 have a smooth representative?

Recall that two smoothly embedded disks in a $4$ -manifold with boundary $S^3$ are exotic if they are topologically but not smoothly isotopic rel. boundary. Exotic disks are known to exist in $D^4$ [Reference AkbulutAkb91, Reference HaydenHay20] (see also [Reference Hayden and SundbergHS24, Reference Dai, Mallick and StoffregenDMS23]) as well as in many $2$ -handlebodies with $S^3$ boundary, including $(\mathbb {C}P^2)^\circ$ [Reference Conway, Piccirillo and PowellCPP23, Theorem 1.13]. On the other hand, given a knot $K$ that is smoothly slice in a $4$ -manifold $N$ , it remains challenging to decide whether or not $K$ bounds any exotic slice disks in $N$ . Theorem1.2 implies that two smooth crossing change disks $D_{\gamma _1}$ and $D_{\gamma _2}$ with boundary a knot $K$ are exotic if they are smoothly distinct and the classes $[\widetilde {\gamma }_1]$ and $[\widetilde {\gamma }_2]$ agree in $\mathcal {G}_K/\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ .

One broad approach towards generating pairs of exotic disks would thus be to choose a knot $K$ falling into the first three cases of Theorem1.3 and find several distinct unknotting curves that turn $K$ into a smoothly $\mathbb {Z}$ -slice knot. If sufficiently many curves are found, then certainly $[\widetilde {\gamma }_1]=[\widetilde {\gamma }_2] \in \mathcal {G}_K/\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ for some pair $(\gamma _1, \gamma _2)$ . One could then attempt to leverage smooth invariants such as knot Floer homology to distinguish the corresponding pair of disks (see for example [Reference Dai, Mallick and StoffregenDMS23]).

For instance, the trefoil bounds a unique crossing change $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ up to topological isotopy rel. boundary. Thus, any pair of smooth $\mathbb {Z}$ -disks for the trefoil which can be distinguished in the smooth category would constitute an exotic pair. Unfortunately, the authors have not been able to find any such candidates for exotic pairs of disks, although a systematic investigation of this strategy is beyond the scope of this paper.

Question 1.11. Does the right-handed trefoil bound an exotic pair of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ ?

1.6 Further context and related work

Let $N$ be a simply-connected $4$ -manifold with boundary $S^3$ , and let $K \subset S^3$ be a knot. $\mathbb {Z}$ -surfaces with boundary $K$ considered up to equivalence rel. boundary were classified in [Reference Conway and PowellCP23, Reference Conway, Piccirillo and PowellCPP23]. Complete invariants are given by the equivariant intersection form and automorphism invariant of their exterior. Prior work on the existence question includes [Reference Feller and LewarkFL18, Reference Feller and LewarkFL24] (for $\mathbb {Z}$ -sliceness in $N=D^4 \#_m S^2 \times S^2 \#_n {\mathbb {C}} P^2 \# \overline {{\mathbb {C}} P}^2)$ and [Reference Kjuchukova, Miller, Ray and SakalliKMRS24] (for $\mathbb {Z}$ -sliceness in $N=(\#_\ell {\mathbb {C}} P^2)^\circ$ ). Concerning this latter result, to the best of our knowledge, [Reference Kjuchukova, Miller, Ray and SakalliKMRS24, Theorem1.3] does not recover Theorem1.1. We emphasize again that Theorem1.1 follows fairly promptly from [Reference FogelFog94]. We also note that since $\mathbb {Z}$ -disks are nullhomologous (see for example [Reference Conway and PowellCP23, Lemma 5.1]), Theorem1.1 also yields criteria for a knot to be $H$ -slice, a topic that has attracted some attention recently.

We now focus on $\mathbb {Z}$ -disks. In what follows, the exterior of a disk $D \subset N$ is denoted by $N_D$ . Given a Hermitian form $\lambda$ over $\mathbb {Z}[t^{\pm 1}]$ , we write $\mathcal {D}_\lambda (K,N)^{\textrm {equiv}}$ for the set of rel. boundary equivalence classes of $\mathbb {Z}$ -disks in $N$ with boundary $K$ and equivariant intersection form $\lambda _{N_D}\cong \lambda$ ; we refer to § 2.3 for a brief review of equivariant intersection forms. The automorphism invariant from [Reference Conway, Piccirillo and PowellCPP23] gives rise to a (noncanonical) bijection

\begin{equation*}b \colon \mathcal {D}_\lambda (K,N)^{\textrm {equiv}} \xrightarrow {\approx } {\textrm {Aut}}({\textrm {Bl}}_K)/{\textrm {Aut}}(\lambda ). \end{equation*}

The target of $b$ is the orbit set of the left action of the automorphism group ${\textrm {Aut}}(\lambda )$ on the automorphism group ${\textrm {Aut}}({\textrm {Bl}}_K)$ by $F \cdot h=h \circ \partial F^{-1}$ ; we refer to the introductions of [Reference Conway and PowellCP23, Reference Conway, Piccirillo and PowellCPP23] for more details.

A statement involving isotopy classes of disks instead of equivalence classes of disks has not yet appeared in the literature but can be obtained by tracing through the proofs of [Reference Conway, Piccirillo and PowellCPP23] and applying [Reference Orson and PowellOP22, Theorem C] to upgrade equivalences to isotopies.

We now restrict to $N=(\mathbb {C}P^2)^\circ$ . We observe in Proposition 3.2 that the exterior of a $\mathbb {Z}$ -disk $D \subset (\mathbb {C}P^2)^\circ$ with boundary $K$ necessarily has equivariant intersection form $\lambda _{N_D}(x,y)=x\Delta _K\overline {y}$ . This implies that ${\textrm {Aut}}({\textrm {Bl}}_K)/{\textrm {Aut}}(\lambda ) \approx U(\mathbb {Z}[t^{\pm 1}]/(\Delta _K))/\lbrace \pm t^k \rbrace _{k \in \mathbb {Z}}$ and so $b$ induces a bijection

\begin{equation*}b \colon \mathcal {D}(K,(\mathbb {C}P^2)^\circ )^{\textrm {equiv}} \xrightarrow {\approx } {\textrm {Aut}}({\textrm {Bl}}_K)/{\textrm {Aut}}(\lambda ) \xrightarrow {\approx } U(\mathbb {Z}[t^{\pm 1}]/(\Delta _K))/\lbrace \pm t^k \rbrace _{k \in \mathbb {Z}}. \end{equation*}

Although this has not appeared in print, if one traces through [Reference Conway, Piccirillo and PowellCPP23] and applies [Reference Orson and PowellOP22, Theorem C] when necessary, one can verify that $b$ induces a bijection

(1.3) \begin{equation} b \colon \mathcal {D}(K,(\mathbb {C}P^2)^\circ ) \xrightarrow {\approx } U(\mathbb {Z}[t^{\pm 1}]/(\Delta _K))/\lbrace t^k \rbrace _{k \in \mathbb {Z}}. \end{equation}

In practice, the nature of the automorphism invariant $b_D$ can make it difficult to distinguish concrete $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ with boundary $K$ .

Thus the main novelty of Theorem1.2 resides less in the existence of a bijection as in (1.3) than in the explicit nature of the invariant as well as in the fact that every $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ is isotopic to a crossing change $\mathbb {Z}$ -disk. Another feature of Theorem1.2 is that while the bijection in (1.3) can be derived by modifying the lengthy argument of [Reference Conway, Piccirillo and PowellCPP23, § 5], our argument is more self-contained and instead relies on [Reference FogelFog94]. In summary, our main insight is that, for generalized crossing change $\mathbb {Z}$ -disks, the automorphism invariant boils down to the homology class of the lift of the surgery curve in the Alexander module.

Finally, we describe some additional context surrounding Theorem1.3. In [Reference BoyerBoy86, Reference StongSto93, Reference BoyerBoy93, Reference Crowley and SixtCS11, Reference Conway, Crowley, Powell and SixtCCPS21, Reference Conway, Piccirillo and PowellCPP23, Reference Conway, Crowley and PowellCCP23], isometries of linking forms have been used to study surfaces in $4$ -manifolds as well as the (stable) classification of $4k$ -manifolds. A subset of the applications in these papers rely on being able to find examples of Hermitian forms $(H,\lambda )$ such that the automorphism set ${\textrm {Aut}}(\partial \lambda )/{\textrm {Aut}}(\lambda )$ is large; here $\partial \lambda$ denotes the boundary linking form of $\lambda$ , see for example [Reference Conway and PowellCP23, § 2].

In [Reference Conway, Piccirillo and PowellCPP23], examples were given of $3$ -manifolds $Y$ that have arbitrarily large ${\textrm {Aut}}({\textrm {Bl}}_Y)/{\textrm {Aut}}(\lambda )$ , and this was later improved to examples with infinite ${\textrm {Aut}}({\textrm {Bl}}_Y)/{\textrm {Aut}}(\lambda )$ [Reference Conway, Crowley and PowellCCP23]; we refer to [Reference Conway, Crowley, Powell and SixtCCPS21] for applications of this sort of algebra to the stable classification of $4k$ -manifolds with $k\gt 1$ . Prior to this article, however, few examples of knots $K$ had been produced for which ${\textrm {Aut}}({\textrm {Bl}}_K)/{\textrm {Aut}}(\lambda )$ is nontrivial (i.e. for which the equivariant intersection form does not determine the isotopy type of the disk) and no example for which it is infinite. Surprisingly, Theorem1.3 suggests that the orbit set ${\textrm {Aut}}({\textrm {Bl}}_K)/{\textrm {Aut}}(\lambda )$ might in fact be, generically, infinite.

1.6.1 Organization

Section 2 collects some background homological material. In § 3 we prove Theorem1.1 which concerns criteria for $\mathbb {Z}$ -sliceness in $(\mathbb {C}P^2)^\circ$ . In § 4 we study the properties of generalized crossing change $\mathbb {Z}$ -disks. In § 5 we prove the classification stated in Theorem1.2. In § 6, we review number-theoretic background. In § 7, we prove Theorem1.7 which concerns the unitary units of $\mathbb {Z}[t^{\pm 1}]/(\Delta _n)$ . Section 8 is concerned with examples and the smooth realization of $\mathbb {Z}$ -disks.

1.6.2 Conventions

We work in the topological category with locally flat embeddings unless otherwise stated. From now on, all manifolds are assumed to be compact, connected, based, and oriented; if a manifold has a non-empty, connected boundary, then the basepoint is assumed to be in the boundary. The boundary of a manifold is oriented according to the ‘outwards normal vector first’ rule. We write $\Delta _K$ for the unique symmetric representative of the Alexander polynomial of a knot $K$ that evaluates to $-1$ at $t=1$ .

2. Background material

We fix some notation concerning the (twisted) homology of infinite cyclic covers (§ 2.1), and review the definition of the Blanchfield form (§ 2.2) and of the equivariant intersection form (§ 2.3). In the interest of expediency, we give the homological definitions of these pairings, referring to [Reference Conway, Piccirillo and PowellCPP23, §§ 2.1 and 2.2] and the references within for further details and geometric intuition. We also refer to [Reference Friedl and PowellFP17, § 2] for a treatment of twisted homology and the Blanchfield form that is similar to ours but contains further details.

2.1 Twisted homology

In what follows, spaces are assumed to have the homotopy type of a finite CW complex. Given a space $X$ together with an epimorphism $\varphi \colon \pi _1(X) \twoheadrightarrow \mathbb {Z}$ , we write $p\colon X^\infty \to X$ for the infinite cyclic cover corresponding to $\ker (\varphi )$ . If $A \subset X$ is a subspace, then we set $A^\infty :=p^{-1}(A)$ and often write $H_*(X,A;\mathbb {Z}[t^{\pm 1}])$ instead of $H_*(X^\infty ,A^\infty )$ . Note that these are finitely generated $\mathbb {Z}[t^{\pm 1}]$ -modules because $X$ and $A$ have the homotopy type of finite CW-complexes and $\mathbb {Z}[t^{\pm 1}]$ is Noetherian; see for example [Reference Friedl, Nagel, Orson and PowellFNOP19, Proposition A.9].

Remark 2.1. The Alexander polynomial of $X$ is the order of the Alexander module $H_1(X;\mathbb {Z}[t^{\pm 1}])$ . Note that $\Delta _X$ is a Laurent polynomial that is well defined up to multiplication by $\pm t^k$ with $k \in \mathbb {Z}$ . If $K$ is a knot with exterior $E_K$ and $0$ -framed surgery $M_K$ , and we take $\varphi$ to be the abelianization homomorphism, then a Mayer–Vietoris calculation shows that the inclusion $E_K \subset M_K$ induces a $\mathbb {Z}[t^{\pm 1}]$ -isomorphism $H_1(E_K;\mathbb {Z}[t^{\pm 1}]) \to H_1(M_K;\mathbb {Z}[t^{\pm 1}])$ . The Alexander polynomial of $K$ , denoted $\Delta _K$ , can therefore equivalently be defined as the order of $E_K$ or as the order of $M_K$ .

We write $H^*(X,A;\mathbb {Z}[t^{\pm 1}])$ for the homology of ${\textrm {Hom}}_{\mathbb {Z}[t^{\pm 1}]}(\overline {C_*(X^\infty ,A^\infty )},\mathbb {Z}[t^{\pm 1}])$ and remind the reader that $H^*(X,A;\mathbb {Z}[t^{\pm 1}])$ is isomorphic to the cohomology with compact support of the pair $(X^\infty ,A^\infty )$ , not to $H^*(X^\infty ,A^\infty ).$ Here, given a $\mathbb {Z}[t^{\pm 1}]$ -module $V$ , we write $\overline {V}$ for the $\mathbb {Z}[t^{\pm 1}]$ -module whose underlying group agrees with that of $V$ but with the $\mathbb {Z}[t^{\pm 1}]$ -module structure induced by $t \cdot v=t^{-1}v$ for $v \in V$ .

More generally, if $M$ is a $\mathbb {Z}[t^{\pm 1}]$ -module, then we write $H_*(X,A;M)$ and $H^*(X,A;M)$ for the homology of the $\mathbb {Z}[t^{\pm 1}]$ -chain complexes $M \otimes _{\mathbb {Z}[t^{\pm 1}]} C_*(X^\infty ,A^\infty )$ and ${\textrm {Hom}}_{\mathbb {Z}[t^{\pm 1}]}(\overline {C_*(X^\infty ,A^\infty )},M)$ , respectively. Apart from the case $M=\mathbb {Z}[t^{\pm 1 }]$ which we have already considered, we will occasionally consider the case where $M=\mathbb {Q}(t)$ is the field of fractions of $\mathbb {Z}[t^{\pm 1}]$ as well as $M=\mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}].$

For the $\mathbb {Z}[t^{\pm 1}]$ -modules $M=\mathbb {Z}[t^{\pm 1}],\mathbb {Q}(t)$ and $\mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}]$ , there is an evaluation homomorphism

\begin{equation*}\textrm {ev} \colon H^*(X,A;M) \to \overline {{\textrm {Hom}}_{\mathbb {Z}[t^{\pm 1}]}(H_*(X,A;\mathbb {Z}[t^{\pm 1}]),M)}.\end{equation*}

A more thorough discussion of this evaluation map can be found for example in [Reference Friedl and PowellFP17, § 2.3].

When $X$ is an $n$ -manifold, there are Poincaré duality isomorphisms

\begin{equation*}{\textrm {PD}} \colon H^{n-*}(X;M) \xrightarrow {\cong } H_*(X,\partial X;M) \ \ \ \text { and} \ \ \ {\textrm {PD}} \colon H^{n-*}(X,\partial X;M) \xrightarrow {\cong } H_*(X;M).\end{equation*}

2.2 The Blanchfield form

We very briefly review the homological definition of the Blanchfield form. Given a $3$ -manifold $Y$ and an epimorphism $\varphi \colon \pi _1(Y) \twoheadrightarrow \mathbb {Z}$ such that the Alexander module $H_1(Y;\mathbb {Z}[t^{\pm 1}])$ is torsion, the Blanchfield form is the pairing

\begin{align*} {\textrm {Bl}}_Y \colon H_1(Y;\mathbb {Z}[t^{\pm 1}]) \times H_1(Y;\mathbb {Z}[t^{\pm 1}]) &\to \mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}] \\ (x,y) &\mapsto \textrm {ev}( \textrm {BS}^{-1} \circ {\textrm {PD}}^{-1}(i_*(y)))(x). \end{align*}

Here $i_* \colon H_1(Y;\mathbb {Z}[t^{\pm 1}]) \to H_1(Y,\partial Y;\mathbb {Z}[t^{\pm 1}])$ is the homomorphism induced by the inclusion, ${\textrm {PD}}^{-1} \colon H_1(Y,\partial Y;\mathbb {Z}[t^{\pm 1}]) \to H^2(Y;\mathbb {Z}[t^{\pm 1}])$ denotes the inverse of the Poincaré duality isomorphism, ${\textrm {BS}}^{-1}$ denotes the inverse of the Bockstein map ${\textrm {BS}} \colon H^2(Y;\mathbb {Z}[t^{\pm 1}]) \to H^1(Y;\mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}])$ associated to the short exact sequence $0 \to \mathbb {Z}[t^{\pm 1}] \to \mathbb {Q}(t) \to \mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}] \to 0$ of coefficients (here the homomorphism $\textrm {BS}$ is an isomorphism because we assumed that $H_1(Y;\mathbb {Z}[t^{\pm 1}])$ is torsion), and $\textrm {ev} \colon H^1(Y;\mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}]) \to \overline {{\textrm {Hom}}(H_1(Y;\mathbb {Z}[t^{\pm 1}]),\mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}])}$ is the evaluation homomorphism. The Blanchfield form is sesquilinear and Hermitian; see for example [Reference PowellPow16]. If $Y$ is closed, then ${\textrm {Bl}}_Y$ is nonsingular.

Remark 2.2. Given a knot $K$ , the inclusion $E_K \subset M_K$ induces an isometry

\begin{equation*}(H_1(E_K^\infty ),{\textrm {Bl}}_{E_K}) \xrightarrow {\cong } (H_1(M_K^\infty ),{\textrm {Bl}}_{M_K}).\end{equation*}

In what follows, we will refer to either pairing as the Blanchfield form of $K$ and write ${\textrm {Bl}}_K$ .

2.3 The equivariant intersection form

We briefly review the homological definition of the equivariant intersection form. Given a $4$ -manifold $X$ with (possibly empty) boundary, and an epimorphism $\varphi \colon \pi _1(X) \twoheadrightarrow \mathbb {Z}$ , the equivariant intersection form is the pairing

\begin{align*} \lambda _X \colon H_2(X;\mathbb {Z}[t^{\pm 1}]) \times H_2(X;\mathbb {Z}[t^{\pm 1}]) &\to \mathbb {Z}[t^{\pm 1}] \\ (x,y) &\mapsto \textrm {ev}({\textrm {PD}}^{-1} \circ i_*(y))(x), \end{align*}

where $i_*\colon H_2(X;\mathbb {Z}[t^{\pm 1}]) \to H_2(X,\partial X;\mathbb {Z}[t^{\pm 1}])$ is the homomorphism induced by the inclusion, ${\textrm {PD}}^{-1} \colon H_2(X,\partial X;\mathbb {Z}[t^{\pm 1}]) \to H^2(X;\mathbb {Z}[t^{\pm 1}])$ is the inverse of the Poincaré duality isomorphism, and $\textrm {ev} \colon H^2(X;\mathbb {Z}[t^{\pm 1}]) \to \overline {{\textrm {Hom}}(H_2(X;\mathbb {Z}[t^{\pm 1}]),\mathbb {Z}[t^{\pm 1}])}$ is the evaluation homomorphism. The equivariant intersection form is sesquilinear and Hermitian.

We will occasionally also consider the relative pairing

\begin{align*} \lambda _X^\partial \colon H_2(X;\mathbb {Z}[t^{\pm 1}]) \times H_2(X,\partial X;\mathbb {Z}[t^{\pm 1}]) &\to \mathbb {Z}[t^{\pm 1}] \\ (x,y) &\mapsto \textrm {ev}({\textrm {PD}}^{-1}(y))(x). \end{align*}

Finally, we record the relation between the equivariant intersection form and the Blanchfield form. This is a well-known statement (see for example [Reference Conway and PowellCP23, Remark 2.4 and Proposition 3.5] for a reference involving conventions that match ours) and so we omit the proof.

Proposition 2.3. Let $W$ be a $4$ -manifold with $\pi _1(W) \cong \mathbb {Z}$ , connected boundary, $\pi _1(\partial W ) \to \pi _1(W)$ surjective, and $H_1(\partial W^\infty )$ a $\mathbb {Z}[t^{\pm 1}]$ -torsion module. Any matrix $A$ representing $\lambda _W$ pre-sents the Blanchfield form ${\textrm {Bl}}_{\partial W}$ , meaning that ${\textrm {Bl}}_{\partial W}$ is isometric to the linking form

\begin{align*} \mathbb {Z}[t^{\pm 1}]^n/A^T\mathbb {Z}[t^{\pm 1}]^n \times \mathbb {Z}[t^{\pm 1}]^n/A^T\mathbb {Z}[t^{\pm 1}]^n &\to \mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}] \\ ([x],[y]) &\mapsto -x^TA^{-1}\overline {y}, \end{align*}

where $n$ denotes the size of $A$ and the inverse is taken over $\mathbb {Q}(t).$

It is worth noting that the sign in this proposition depends on various conventions that might differ depending on the article. For example, no such sign appears in [Reference Borodzik and FriedlBF15, § 1.3 and Theorem 2.6], whereas [Reference Cha, Orr and PowellCOP20, Lemma 10.2] contains the sign but has all instances of $A$ replaced by $A^T$ .

3. Existence of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$

The goal of this section is to prove Theorem1.1 from the introduction which lists criteria for a knot $K$ that are equivalent to it being $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ .

Let $N$ be a simply-connected $4$ -manifold with boundary $S^3$ and let $D \subset N$ be a $\mathbb {Z}$ -disk with boundary a knot $K \subset S^3$ . We use $N_D:=N \setminus \nu (D)$ to denote the exterior of $D$ and recall that $\partial N_D=M_K$ is the result of $0$ -framed surgery along $K$ . The next lemma lists some algebro-topological properties of $\mathbb {Z}$ -disk exteriors.

Lemma 3.1. Let $N$ be a simply-connected $4$ -manifold with boundary $S^3$ . Given a $\mathbb {Z}$ -disk $D \subset N$ with boundary $K$ , the following assertions hold:

  1. (1) the $\mathbb {Z}[t^{\pm 1}]$ -module $H_2(N_D;\mathbb {Z}[t^{\pm 1}])$ is free of rank $b_2(N)$ ;

  2. (2) if $A(t)$ is a matrix representing $\lambda _{N_D}$ , then $A(1)$ represents the intersection form of $N$ ; and

  3. (3) if $A(t)$ is a matrix representing $\lambda _{N_D}$ , then the transpose $A^T(t)$ presents the Alexander module $H_1(M_K;\mathbb {Z}[t^{\pm 1}])$ .

Proof. The first assertion is proved in [Reference Conway and PowellCP23, Claim 4, p. 43] for $N=D^4$ but the proof generalizes to other simply-connected $4$ -manifolds with boundary $S^3$ . The second assertion follows from [Reference Conway and PowellCP23, Lemma 5.10]. The third assertion is proved in [Reference Conway and PowellCP23, Lemma 3.2 (4)].

Next, we restrict to $\mathbb {Z}$ -disks in $N=(\mathbb {C}P^2)^\circ$ with boundary a knot $K$ and recall that $\Delta _K$ denotes the unique symmetric representative of the Alexander polynomial of a knot $K$ that evaluates to $-1$ at $t=1$ . The next lemma shows that only one Hermitian form $(H,\lambda )$ over $\mathbb {Z}[t^{\pm 1}]$ can arise as the equivariant intersection form of such a $\mathbb {Z}$ -disk exterior.

Proposition 3.2. If $D \subset (\mathbb {C}P^2)^\circ$ is a $\mathbb {Z}$ -disk with $\partial D=K$ , then the equivariant intersection form $\lambda _{N_D}$ is represented by the size one matrix $(-\Delta _K)$ .

Proof. The first item of Lemma 3.1 implies that $H_2(N_D;\mathbb {Z}[t^{\pm 1}]) \cong \mathbb {Z}[t^{\pm 1}]$ and the third item of Lemma 3.1 implies that $\lambda _{N_D}$ is represented by a size one Hermitian matrix whose determinant is equal (up to multiplication by units) to the Alexander polynomial of $K$ . Since $\lambda _{N_D}$ is Hermitian, such a matrix must therefore be of the form $(f(t))$ where $f(t)$ is a symmetric representative of this Alexander polynomial. There are two such representatives, namely $\pm \Delta _K$ . Since the second item of Lemma 3.1 ensures that $A(1)$ is a matrix for the intersection form of $(\mathbb {C}P^2)^\circ$ , i.e. $A(1)=(1)$ , we deduce that $A(t)=(-\Delta _K)$ . This concludes the proof of the proposition.

We can now prove Theorem1.1 from the introduction. We emphasize again that this result can readily be deduced from Fogel’s work [Reference FogelFog94] and that the equivalence $(2) \Leftrightarrow (4)$ is in fact [Reference FogelFog94, Theorem]. Our reason for nevertheless outlining the proof of $(4) \Rightarrow (1)$ is to build intuition for the related, but more detailed, construction of § 4.

Theorem 1.1. Given a knot $K$ , the following assertions are equivalent:

  1. (1) $K$ is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ ;

  2. (2) the Blanchfield form ${\textrm {Bl}}_K$ is presented by $-\Delta _K(t)$ ;

  3. (3) $K$ can be converted into an Alexander polynomial one knot by switching a single positive crossing to a negative crossing; and

  4. (4) $K$ can be converted into an Alexander polynomial one knot by a single positive generalized crossing change.

Proof. The implication $(1) \Rightarrow (2)$ follows from the combination of Propositions 2.3 and 3.2. The implication $(2) \Rightarrow (3)$ is due to Fogel [Reference FogelFog94, Theorem] (see also [Reference Borodzik and FriedlBF14, Theorem 5.1]), and $(3) \Rightarrow \,(4)$ is immediate. The fact that $(4) \Rightarrow (1)$ is well known and will be discussed in detail in the next section, but we outline the proof briefly. Assume that $K$ can be transformed into an Alexander polynomial one knot $J$ by a generalized crossing change. The generalized crossing change is performed by surgering $(S^3,K)$ along a $(-1)$ -framed unknotted curve $\gamma \subset S^3$ that is disjoint from $K$ and nullhomologous in $E_K$ . It follows that $K \subset S^3$ and $J \subset S^3$ are concordant in $W:=(S^3 \times I)\# {\mathbb {C}} P^2$ via a concordance $C$ . Since $J$ has Alexander polynomial one, the work of Freedman ensures that $J$ bounds a $\mathbb {Z}$ -disk $\Delta \subset D^4$ [Reference FreedmanFre82]. Delaying a thorough discussion of orientations to § 4, the required disk $D \subset (\mathbb {C}P^2)^\circ$ is obtained by using $\Delta$ to cap off the concordance:

\begin{equation*}((\mathbb {C}P^2)^\circ ,D) := (-((S^3 \times I)\# {\mathbb {C}} P^2),C) \cup _{(S^3,J)} (D^4,\Delta ).\end{equation*}

It remains to show that the complement of $D$ , denoted $N_D$ , has fundamental group $\mathbb {Z}$ . By construction of the disk $D$ , we know that $(\mathbb {C}P^2)^\circ _D=-W_C \cup _{M_J} N_\Delta$ where $W_C$ (respectively, $N_\Delta$ ) denotes exterior of $C \subset W$ (respectively, of $\Delta \subset D^4$ ). The claim now follows from a van Kampen argument using the fact that $\gamma$ is nullhomologous in $E_K \subset M_K$ and that $\Delta$ is a $\mathbb {Z}$ -disk.

4. Crossing change $\mathbb {Z}$ -disks

The proof of Theorem 1.1 shows that if a single generalized positive crossing change turns $K$ into a knot that bounds a $\mathbb {Z}$ -disk in $D^4$ , then $K$ bounds a $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ . We consider this construction in more detail (§ 4.1), and study the exterior of such disks (§ 4.2) and their further properties (§ 4.3).

4.1 The definition of generalized crossing change $\mathbb {Z}$ -disks

We describe in more detail how a generalized positive crossing change leads to a $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ , provided the knot resulting from the crossing change has Alexander polynomial one. Before doing so, we introduce some notation that will be used throughout this section.

Notation 4.1. The following notation is introduced to ensure that we are comparing disks in a fixed copy of $(\mathbb {C}P^2)^\circ$ . Equivalences and isotopies of disks are henceforth understood to take place in this fixed model:

  1. a copy of $D^4$ with boundary $S^3$ ;

  2. a $(-1)$ -framed oriented unknot $c \subset S^3$ ;

  3. an orientation-reversing homeomorphism $h \colon S_{-1}^3(c) \xrightarrow {\cong } S^3$ ;

  4. a copy of $(\mathbb {C}P^2)^\circ$ which we think of as

    \begin{equation*}(\mathbb {C}P^2)^\circ =( -((S^3 \times [0,1]) \cup _{c \times \lbrace 1 \rbrace } h^2)) \cup _h D^4;\end{equation*}
    here $h^2$ denotes a $2$ -handle whose core we denote by $\textrm {core}(h^2)$ and whose cocore we denote by $\textrm {cocore}(h^2).$ We think of $\mu _c:=\partial \textrm {cocore}(h^2)$ both as a curve in $S^3 \times \lbrace 1 \rbrace$ and in $S^3_{-1}(c)$ ; and
  5. an oriented knot $K \subset S^3$ ;

We reverse the orientation of $(S^3 \times [0,1]) \cup h^2$ so that $S^3 \times \lbrace 0 \rbrace \subset -((S^3 \times [0,1]) \cup h^2)$ has the same orientation as the usual one of $S^3$ , thought of as the boundary of $D^4$ . Recall that we orient the boundary of a manifold according to the ‘outwards normal vector first’ rule so that

\begin{equation*}\partial (S^3 \times [0,1]) =-S^3\times \{ 0\} \sqcup S^3 \times \{ 1 \}.\end{equation*}

For this reason, we frequently think of $K$ as a subset of $S^3\times \{ 0\} \subset -((S^3 \times [0,1]) \cup h^2).$

Remark 4.2. Given a simple closed curve $\eta \subset S^3$ and an ambient isotopy $(f_t \colon S^3 \to S^3)_{t \in [0,1]}$ with $f_0={\textrm {id}}_{S^3}$ , we think of the trace $C_{\eta ,f}$ of $(f_t|_{\eta })_{t \in [0,1]}$ as a subset of $-(S^3 \times [0,1])$ with boundary

\begin{equation*}\partial C_{\eta ,f} =\eta \sqcup -f(\eta ) \subset S^3\times \{ 0\} \sqcup - S^3 \times \{ 1 \}.\end{equation*}

Notation 4.3. We also fix the following notation, which we will use frequently when working with $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ that arise from a single positive generalized crossing change:

  1. an oriented curve $\gamma \subset E_K$ that is unknotted in $S^3$ with $\ell k(\gamma ,K)=0$ such that a generalized positive crossing change along $\gamma$ results in an Alexander polynomial one knot;

  2. an ambient isotopy $(f_t \colon S^3 \to S^3)_{t \in [0,1]}$ with $f_0={\textrm {id}}_{S^3}$ and $f_1(\gamma )=c$ as oriented curves; write

    • $\circ$ $C_{\gamma ,f} \subset -((S^3 \times [0,1]) \cup _{c \times \lbrace 1 \rbrace } h^2)$ for the trace of the isotopy $(f_t|_{\gamma })_{t \in [0,1]}$ ,

    • $\circ$ $C_{K,f} \subset -((S^3 \times [0,1]) \cup _{c \times \lbrace 1 \rbrace } h^2)$ for the trace of the isotopy $(f_t|_K)_{t \in [0,1]}$ ,

    • $\circ$ $K' \subset S^3_{-1}(c)$ for the knot $f_1(K) \subset S^3$ viewed as a subset of $S^3_{-1}(c)$ , and

    • $\circ$ $J:=h(-K') \subset S^3$ for the image of $K' \subset S^3_{-1}(c)$ under the orientation-reversing homeomorphism $h \colon S^3_{-1}(c) \xrightarrow {\cong } S^3$ ;

  3. a $\mathbb {Z}$ -disk $\Delta \subset D^4$ with boundary the knot $J \subset S^3$ .

Note that, in the topological category, a knot bounds a $\mathbb {Z}$ -disk in $D^4$ if and only if it has Alexander polynomial one [Reference FreedmanFre82]; this disk is unique up to topological isotopy rel. boundary [Reference Conway and PowellCP21]. Consequently, in the topological category, there is no ambiguity in choosing a disk $\Delta$ with boundary $J$ . On the other hand, the existence of a smoothly embedded disk $\Delta$ with boundary $J$ is not a given and, if such a disk exists, it may not be unique [Reference AkbulutAkb91, Reference HaydenHay20].

Definition 4.4. Given a curve $\gamma \subset E_K$ , an isotopy $(f_t \colon S^3 \to S^3)_{t \in [0,1]}$ , and a $\mathbb {Z}$ -disk $\Delta \subset D^4$ as in Notation 4.3, the generalized crossing change $\mathbb {Z}$ -disk $D_{\gamma }$ is

\begin{equation*} ((\mathbb {C}P^2)^\circ ,D_{\gamma })=( - ((S^3 \times [0,1]) \cup _{c \times \lbrace 1 \rbrace } h^2),C_{K,f} ) \cup _h (D^4,\Delta ).\end{equation*}

The following proposition shows that, up to isotopy rel. boundary, the disk $D_\gamma$ only depends on the knot $K$ and on the surgery curve $\gamma$ .

Proposition 4.5. Let $K$ be a knot and let $\gamma \subset E_K$ be a surgery curve as in Notation 4.3 . Up to isotopy rel. boundary, a generalized crossing change $\mathbb {Z}$ -disk $D_\gamma$ with boundary $K$ depends neither on the choice of the ambient isotopy $(f_t)_{t \in [0,1]}$ nor on the $\mathbb {Z}$ -disk $\Delta$ .

Proof. Fix two ambient isotopies $(f_t)_{t \in [0,1]},(g_t)_{t \in [0,1]}$ of $S^3$ with $f_0={\textrm {id}}_{S^3}=g_0$ and $f_1,g_1$ taking $\gamma$ to $c$ . During this proof, we write $D_{\gamma ,f}$ and $D_{\gamma ,g}$ for the the resulting generalized crossing change $\mathbb {Z}$ -disks. Although not included in the notation, we note that $D_{\gamma ,f}$ and $D_{\gamma ,g}$ a priori also depend on the disks used to cap off the traces $C_{K,f}$ and $C_{K,g}$ . In a nutshell, the idea of the proof is to use the Smale conjecture [Reference HatcherHat83] to show that the isotopies $(f_t)_{t \in [0,1]}$ and $(g_t)_{t \in [0,1]}$ are isotopic and to use this isotopy of isotopies, together with the fact that $\mathbb {Z}$ -discs in $D^4$ are unique up to isotopy rel. boundary [Reference Conway and PowellCP23], to prove that $D_{\gamma ,f}$ and $D_{\gamma ,g}$ are isotopic. We now expand on this outline.

Our goal is to define a $1$ one-parameter family of rel. boundary homeomorphisms $\Phi _s \colon (\mathbb {C}P^2)^\circ \to (\mathbb {C}P^2)^\circ$ with $\Phi _0={\textrm {id}}_{(\mathbb {C}P^2)^\circ }$ and $\Phi _1$ taking one disk to another.

Claim 1. If $(f_t)$ and $(g_t)$ are isotopic through a two-parameter family $(h_{t,s} \colon S^3\times \{ t \}\times \{ s \}\to S^3)_{t,s \in I}$ with

\begin{equation*}h_{t,0}=f_t, \quad h_{t,1}=g_t, \quad h_{0,s}={\textrm {id}}_{S^3}, \quad h_{1,s}(\gamma )=c,\end{equation*}

then $D_{\gamma ,f}$ and $D_{\gamma ,g}$ are isotopic rel. boundary in $(\mathbb {C}P^2)^\circ$ .

In what follows we will sometimes think of $\mathbf {f}:=(f_t)$ and $\mathbf {g}:=(g_t)$ as paths in the space of self-homeomorphisms ${\textrm {Homeo}}(S^3)$ . The assumptions of the claim are then that there is a homotopy $\mathbf {h} \colon I \times I \to {\textrm {Homeo}}(S^3)$ with $\mathbf {h}_0=\mathbf {f}, \mathbf {h}_1=\mathbf {g}$ as well as $\mathbf {h}_s(0)={\textrm {id}}_{S^3}$ and $\mathbf {h}_s(1)(\gamma )=c$ for every $s \in I$ .

Proof of Claim 1 . We have written $(\mathbb {C}P^2)^\circ$ as the union of the three pieces $S^3 \times I$ , $h^2$ , and $D^4$ :

\begin{equation*} (\mathbb {C}P^2)^\circ = ( -((S^3 \times I) \cup _{c \times \lbrace 1 \rbrace } h^2)) \cup _{h} D^4. \end{equation*}

We construct a one-parameter family of self-homeomorphisms

\begin{equation*} \Phi _s \colon (\mathbb {C}P^2)^\circ \rightarrow (\mathbb {C}P^2)^\circ ,\end{equation*}

with the following properties:

  1. (1) $\Phi _0$ is the identity;

  2. (2) $\Phi _s$ is the identity on $\partial (\mathbb {C}P^2)^\circ$ for all $s$ ;

  3. (3) $\Phi _s$ preserves each of the pieces $S^3 \times I$ , $h^2$ , and $D^4$ setwise for all $s$ ; and

  4. (4) $\Phi _1(C_{K,f}) = C_{K,g}$ .

This suffices to prove the claim. Indeed, $(\Phi _s)_{s \in [0,1]}$ tautologically constitutes an ambient isotopy of $(\mathbb {C}P^2)^\circ$ sending the disk $D_{f, \gamma }$ to the disk $\Phi _1(D_{f, \gamma })$ . Due to the last two conditions above, $\Phi _1(D_{\gamma ,f})$ and $D_{\gamma ,g}$ coincide in $S^3 \times I$ . Hence $\Phi _1(D_{\gamma ,f})$ and $D_{\gamma ,g}$ differ only by a choice of capping $\mathbb {Z}$ -disk in $D^4$ ; any two such $\mathbb {Z}$ -disks are topologically isotopic rel. boundary [Reference Conway and PowellCP21].

We build $\Phi _s$ by first defining a one-parameter family of self-homeomorphisms of $S^3 \times I$ satisfying the desired conditions. We then extend this family to a one-parameter family of self-homeomorphisms of $(S^3 \times I) \cup _{c \times \lbrace 1 \rbrace } h^2$ , and then successively to a one-parameter family of self-homeomorphisms of all of $(\mathbb {C}P^2)^\circ$ . The construction of $\Phi _s$ on $S^3 \times I$ is immediate from the hypotheses of the claim by defining

\begin{equation*} (\Phi _s)|_{S^3 \times I} = \mathbf {h}_s \circ \mathbf {h}_0^{-1} = \mathbf {h}_s \circ \mathbf {f}^{-1}. \end{equation*}

Here, we briefly abuse notation by considering $\mathbf {f}$ as a map from $S^3 \times I$ to itself sending $(x, t)$ to $(f_t(x), t)$ , and likewise for $\mathbf {h}_s$ . This is clearly consistent with the desired conditions.

We now extend $\Phi _s$ over the $2$ -handle attachment. Note $\Phi _s$ maps $c \times \{1\}$ to itself for all $s$ ; without loss of generality, we may assume $\Phi _s$ preserves a tubular neighborhood $\overline {\nu }(c)$ of $c \times \{1\}$ in $S^3 \times \{1\}$ . Denote the induced one-parameter family of self-diffeomorphisms of $\overline {\nu }(c)$ by $\phi _s \colon \overline {\nu }(c) \rightarrow \overline {\nu }(c)$ . Note that $\phi _0={\textrm {id}}_{\overline {\nu }(c)}$ because $\Phi _0={\textrm {id}}_{S^3}.$ For convenience of notation, we insert a collar $\overline {\nu }(c) \times I$ between $S^3 \times \{1\}$ and the $2$ -handle $h^2$ . We thus consider the $2$ -handle attachment as being $(S^3 \times I) \cup (\overline {\nu }(c) \times I) \cup h^2$ , where $\overline {\nu }(c) \subseteq S^3 \times \{1\}$ is glued to $\overline {\nu }(c) \times \{0\}$ via the identity and $h^2$ is attached along $\overline {\nu }(c) \times \{1\}$ . Extend $\Phi _s$ over $\overline {\nu }(c) \times I$ by defining

\begin{equation*} (\Phi _s)|_{\overline {\nu }(c) \times I} (x, t) = (\phi _{(1-t)s}(x), t). \end{equation*}

Note that $\Phi _s$ is the identity on $\overline {\nu }(c) \times \{1\}$ for all $s$ . Hence we may identically extend

\begin{equation*} (\Phi _s)|_{h^2} = {\textrm {id}}. \end{equation*}

This produces the desired extension of $\Phi _s$ over the $2$ -handle attachment.

Finally, we observe that $\Phi _s$ may be extended over the capping $4$ -ball $D^4$ . Indeed, our definition of $\Phi _s$ so far gives a one-parameter family $\psi _s$ of self-diffeomorphisms of $S^3 = \partial D^4$ . This extends to a one-parameter family of self-homeomorphisms of $D^4$ via the usual radial extension/Alexander trick:

\begin{equation*} (\Phi _s)|_{D^4}(x, r) = (\psi _s(x), r), \end{equation*}

where $x \in S^3$ and $r$ is the radial coordinate of $D^4$ . This completes the proof of the claim.

We now prove that the assumption of Claim1 always holds.

Claim 2. There is a homotopy $\mathbf {h} \colon I \times I \to {\text {Homeo}}(S^3)$ with $\mathbf {h}_0=\mathbf {f}, \mathbf {h}_1=\mathbf {g}$ as well as $\mathbf {h}_s(0)={\textrm {id}}_{S^3}$ and $\mathbf {h}_s(1)(\gamma )=c$ for every $s \in I$ .

Note that $h$ is a free homotopy, not a homotopy rel. endpoints.

Proof of Claim 2. We start with an observation whose proof is left to the reader. Fix a path of homeomorphisms $\boldsymbol {\eta } \colon I \to {\textrm {Homeo}}(S^3)$ with $\eta _0={\textrm {id}}_{S^3}$ .

If $\mathbf {r} \colon I \to {\textrm {Homeo}}(S^3)$ is a path of homeomorphisms with $r_0={\textrm {id}}_{S^3}$ and $r_t(c)=c$ for every $t$ , then the path $\boldsymbol {\eta }$ is freely homotopic to the path $\boldsymbol {\eta * r}$ given by

\begin{equation*} (\eta * r)_t:= \begin{cases} \eta _{2t}, &t \in [0,\tfrac {1}{2}], \\ r_{2t-1} \circ \eta _1, & t \in [\tfrac {1}{2},1], \end{cases} \end{equation*}

through a homotopy $\boldsymbol {\varphi } \colon I \times I \to {\textrm {Homeo}}(S^3)$ with $\boldsymbol {\varphi }_0=\boldsymbol {\eta },\boldsymbol {\varphi }_1=\boldsymbol {r * \eta }$ as well as $\varphi _s(0)=\eta _0={\textrm {id}}_{S^3}$ and $\varphi _s(1)(c)=r_s(c)=c$ for every $s \in I$ .

We now begin the proof of the claim proper and assert that, without loss of generality, we can assume that $f_1$ and $g_1$ agree pointwise on a tubular neigbhorhood $\overline {\nu }(\gamma )$ of $\gamma$ . Let $\mathbf {r} \colon I \to {\textrm {Homeo}}(S^3)$ be a path of homeomorphisms with $r_0={\textrm {id}}_{S^3}$ and $r_1 \circ f_1|_{\overline {\nu }(\gamma )} =g_1|_{\overline {\nu }(\gamma )}$ . The observation implies that $\mathbf {f}$ is homotopic to $\boldsymbol {f * r}$ through a homotopy $\boldsymbol {\varphi }$ with $\varphi _0(s)={\textrm {id}}_{S^3}$ and $\varphi _1(s)(c)=c$ for every $s \in I$ . Since $(f*r)_1|_{\overline {\nu }(\gamma )}=(r_1 \circ f_1)|_{\overline {\nu }(\gamma )}=g_1|_{\overline {\nu }(\gamma )}$ , this concludes the proof of the assertion.

Next we assert that, without loss of generality, we can assume that $f_1$ and $g_1$ agree on the union of $\overline {\nu }(\gamma )$ with a closed tubular neigbhorhood of a disk bounded by $\gamma$ . Let $D \subset S^3$ be a smoothly embedded disk bounded by $\gamma$ intersecting $\overline {\nu }(\gamma )$ only in a collar of $\partial D$ . Initially $f_1(D)$ and $g_1(D)$ are disks in $S^3$ with boundary $c$ that may be distinct. An application of Schoenflies’s theorem shows that these disks are ambiently isotopic rel. boundary via an ambient isotopy $(r_t \colon S^3 \to S^3)_{t \in [0,1]}$ with $r_0={\textrm {id}}_{S^3}$ and $r_1 \circ f_1|_D=g_1|_D$ . This can be extended so that $r_1 \circ f_1|_{\overline {\nu }(D)}=g_1|_{\overline {\nu }(D)}$ . The assertion again follows from the observation at the beginning of the proof.

We now assert that, without loss of generality, we can assume that $f_1$ and $g_1$ agree pointwise on the whole of $S^3$ . Presently, the homeomorphisms $f_1$ and $g_1$ agree pointwise on $\overline {\nu }(D)$ but might not agree on the closed ball $B=S^3 \setminus f_1(\nu (D))$ . They do, however, agree on $\partial B$ , so Alexander’s trick ensures the existence of a family $(\widetilde {r}_t \colon B \to B)_{t \in [0,1]}$ of rel. boundary homeomorphisms with $\widetilde {r}_0={\textrm {id}}_B$ and $\widetilde {r}_1 \circ f_1|_B=g_1|_B.$ Since $\widetilde {r}_t|_{\partial B}={\textrm {id}}_{\partial B}$ for every $t \in [0,1]$ , we can extend $\widetilde {r}_t$ by the identity on $S^3 \setminus \textrm {Int}(B)$ resulting in a path $\mathbf {r} \colon I \to {\textrm {Homeo}}(S^3)$ with $r_0={\textrm {id}}_{S^3},r_1 \circ f_1=g_1$ and $r_1(s)(c)=c$ for every $s \in I$ . The assertion again follows from the observation.

We now prove Claim2 for paths of homeomorphisms $\mathbf {f}=(f_t)$ and $\mathbf {g}=(g_t)$ with $f_1=g_1$ . Consider the loop in $[\boldsymbol {f \cdot \overline {g}}] \in \pi _1({\textrm {Homeo}}(S^3),{\textrm {id}}_{S^3})$ :

\begin{equation*} (f \cdot \overline {g})_t =\begin{cases} f_{2t},& t \in [0,\tfrac {1}{2}], \\ g_{2t-1},& t \in [\tfrac {1}{2},1]. \end{cases} \end{equation*}

If $[\boldsymbol {f \cdot \overline {g}}]=1$ , then $f,g$ are homotopic rel. $f_0=g_0={\textrm {id}}_{S^3}$ and $f_1=g_1$ through a homotopy $\mathbf {h}$ . This homotopy satisfies the conditions of the claim because $h_s(0)=f_0={\textrm {id}}_{S^3}$ and $h_s(1)(\gamma )=f_1(\gamma )=c.$

Assume now that $[\boldsymbol {f \cdot \overline {g}}] \neq 1 \in \pi _1({\textrm {Homeo}}(S^3),{\textrm {id}}_{S^3})$ . The Smale conjecture [Reference HatcherHat83] gives the isomorphism $\pi _1({\textrm {Homeo}}(S^3),{\textrm {id}}) \cong \pi _1(O(4)) \cong \mathbb {Z}_2$ , where the generator is represented by the loop of homeomorphisms $\mathbf {r}=(r_t)_{t \in [0,1]}$ with $r_t$ the rotation of angle $2\pi t$ about an axis fixing $c$ setwise.

Since $r_0={\textrm {id}}_{S^3}$ and $r_t(c)=c$ for every $t$ , we can use the observation at the beginning of the proof of this claim to reduce ourselves to proving that $[\boldsymbol {(f * r) \cdot \overline {g}}] = 1\in \pi _1({\textrm {Homeo}}(S^3),{\textrm {id}}_{S^3})$ : thanks to the previous paragraph, replacing $\mathbf {f}$ by $\mathbf {f *r}$ would then give the desired outcome.

We now prove that $[\boldsymbol {(f * r) \cdot \overline {g}}] = 1\in \pi _1({\textrm {Homeo}}(S^3),{\textrm {id}}_{S^3})$ . This is equivalent to proving that $[\boldsymbol {\overline {g} \cdot (f*r)}] = 1 \in \pi _1({\textrm {Homeo}}(S^3),f_1)$ . Consider the nontrivial loop $\boldsymbol {r \circ f_1}$ of homeomorphisms based at $f_1$ and defined by $(r \circ f_1)_t:=r_t \circ f_1$ . Note that $\boldsymbol {\overline {g} \cdot (f*r)} \simeq \boldsymbol {(\overline {g} \cdot f) \cdot (r \circ f_1)}.$ Since neither loop is nullhomotopic and $\pi _1({\textrm {Homeo}}(S^3),f_1)\cong \mathbb {Z}_2$ , the conclusion follows.

Claim2 implies that the paths of homeomorphisms $\mathbf {f}$ and $\mathbf {g}$ are homotopic via a homotopy satisfying the hypotheses of Claim1 and thus that $D_{\gamma ,f},D_{\gamma ,g}$ are isotopic rel. boundary.

The orientations of $c$ and $\gamma$ are essential in the definition of $D_{\gamma }$ . The two disks $D_{\gamma }$ and $D_{-\gamma }$ are related by a homeomorphism of $(\mathbb {C}P^2)^\circ$ inducing the map $-1$ on $H_2((\mathbb {C}P^2)^\circ )=\mathbb {Z}$ , but in general need not be isotopic rel. boundary.

4.2 The exterior of a generalized crossing change $\mathbb {Z}$ -disk

The goal of this section is to describe a decomposition of the exterior of a generalized crossing change $\mathbb {Z}$ -disk and its infinite cyclic cover.

Notation 4.6. Continuing with Notation 4.3, we write $M_{K'}$ for the $0$ -surgery on $K' \subset S^3_{-1}(c)$ and $E_{K'} \subset S^3_{-1}(c)$ for its exterior. Since $J:=h(K')$ , observe that our fixed homeomorphism $h \colon S^3_{-1}(c) \xrightarrow {\cong } S^3$ induces homeomorphisms $E_{K'} \xrightarrow {\cong } E_J$ and $M_{K'} \xrightarrow {\cong } M_J$ that we also denote by $h$ .

For convenience, we write the exterior of the trace $C_{K,f}$ of the isotopy $(f_t|_K)_{t \in [0,1]}$ as

\begin{equation*} W := (S^3 \times I) \setminus \nu (C_{K, f}). \end{equation*}

This way, the exterior $N_{D_\gamma }$ of a generalized crossing change $\mathbb {Z}$ -disk decomposes into three pieces, where (suppressing orientations for ease of notation)

\begin{equation*} N_{D_\gamma } =W \cup h^2 \cup N_\Delta . \end{equation*}

Here, we initially view the union as being formed from left to right, in which case the handle $h^2$ is attached to $W$ along the curve $c \subset S^3 \times \{1\}$ and $W \cup h^2$ is attached to $N_\Delta$ using the homeomorphism $h \colon \partial (W \cup h^2) \supset M_{K'} \xrightarrow {\cong } M_J =\partial N_\Delta$ . We may also consider the union as being formed from right to left, in which case $h^2$ is attached to $N_\Delta$ via the cocore curve $\mu _c$ . Note, in the latter situation, we view $W$ as then being attached along the subset of its boundary given by $E_{f_1(K)} \subset S^3 \times \{1\}$ .

Now fix a basepoint $z \in E_K$ and consider the infinite cyclic cover $p \colon N_{D_\gamma }^\infty \to N_{D_\gamma }$ corresponding to the isomorphism $\pi _1(N_{D_\gamma }, z) \cong \mathbb {Z}$ . Denote the restricted covers by

\begin{equation*} W^\infty :=p^{-1}(W) \quad \text {and} \quad N_\Delta ^\infty :=p^{-1}(N_\Delta ), \end{equation*}

so that the infinite cyclic cover of $N_{D_\gamma }$ decomposes as

\begin{equation*} N_{D_\gamma }^\infty = W^\infty \cup \bigcup _{j \in \mathbb {Z}} t^j \tilde {h}^2 \cup N_\Delta ^\infty . \end{equation*}

The inclusion maps $W \hookrightarrow N_{D_\gamma }$ and $N_\Delta \hookrightarrow N_{D_\gamma }$ induce isomorphisms on first homology, so the restricted covers $W^\infty$ and $N_\Delta ^\infty$ are homeomorphic (as manifolds) to the infinite cyclic covers of $W$ and $N_{\Delta }$ defined using some (potentially different) choice of basepoint. As before, we may view the above union as being formed from left to right or right to left, with the handles $t^j \tilde {h}^2$ being attached along $t^j \tilde {c}$ or $t^j \tilde {\mu }_c$ , respectively. Note that, in the right-to-left case, the final piece $W^\infty$ is then attached along the subset

\begin{equation*} E_{f_1(K)}^\infty := p^{-1}(E_{f_1(K)}) \end{equation*}

of its boundary. We argue that $N_{D_\gamma }^\infty$ deformation retracts onto the subset of $\bigcup _{j \in \mathbb {Z}} t^j \tilde {h}^2 \cup N_\Delta ^\infty$ as this will be helpful for later homological calculations.

Remark 4.7. We have a retract from $W$ onto $E_{f_1(K)}$ given by the map

\begin{equation*} r(x, t) = (f_1 \circ f_t^{-1}(x), 1). \end{equation*}

This is readily seen to be a deformation retract via the homotopy

\begin{equation*} H_s(x, t) = (f_{t + s(1-t)} \circ f_t^{-1}(x), t + s(1-t)). \end{equation*}

It follows that $W^\infty$ deformation retracts onto the subset $E^\infty _{f_1(K)}$ , and thus that $N_{D_\gamma }^\infty$ deformation retracts onto the subset

\begin{equation*} \bigcup _{j \in \mathbb {Z}} t^j \tilde {h}^2 \cup N_\Delta ^\infty . \end{equation*}

4.3 A preferred basis for $H_2(N_{D_\gamma }^\infty )$

We saw in Lemma 3.1 that, for any $\mathbb {Z}$ -disk $D \subset (\mathbb {C}P^2)^\circ$ with boundary $K$ , the $\mathbb {Z}[t^{\pm 1}]$ -module $H_2(N_D^\infty )$ is free of rank one. In this section, when $D$ is a generalized crossing change $\mathbb {Z}$ -disk, we describe a surface in the infinite cyclic cover of this disk exterior that freely generates this module.

Construction 4.8 (Bases for $H_2(N_{D_\gamma }^\infty )$ and $H_2(N_{D_\gamma }^\infty ,M_K^\infty )$ ). Given a curve $\gamma \subset E_K$ , an ambient isotopy $(f_t \colon S^3 \to S^3)_{t \in [0,1]}$ , and a $\mathbb {Z}$ -disk $\Delta \subset D^4$ as in Notation 4.3, we construct a surface $\Sigma \subset N_{D_\gamma }^\infty$ that freely generates $H_2(N_{D_\gamma }^\infty )\cong \mathbb {Z}[t^{\pm 1}]$ . As in § 4.2, we set

\begin{equation*}W:=(S^3 \times I) \setminus \nu (C_{K, f})\end{equation*}

and write $p \colon N_{D_\gamma }^\infty \to N_{D_\gamma }$ for the infinite cyclic cover so that $N_{D_\gamma }^\infty$ decomposes as

\begin{equation*}N_{D_\gamma }^\infty =\biggl ( {-}\biggl ( W^\infty \cup \bigcup _{j \in \mathbb {Z}} t^j \widetilde {h}^2\biggr )\biggr ) \cup N_\Delta ^\infty .\end{equation*}

Recall from Notation 4.3 that $\mu _c$ denotes $\partial \textrm {cocore}(h^2)$ . Thus the loop $\widetilde {\mu }_c \subset M_{J}^\infty =\partial N_\Delta ^\infty$ bounds $\textrm {cocore}(\widetilde {h}^2)$ . On the other hand, since $H_1(M_J^\infty )=0$ , this loop bounds a surface $S \subset M_J^\infty$ . One can assume that $S \cap \textrm {core}(h^2)=\emptyset$ . The union of these two surfaces gives rise to the closed surface

\begin{equation*}\Sigma :=S \cup _{\widetilde {\mu }_c} \textrm {cocore}(\widetilde {h}^2) \subset N_{D_\gamma }^\infty .\end{equation*}

Generically, we can assume that the surface $S\subset M_J^\infty$ projects down to an immersed surface in $M_J$ . As we mentioned in Remark 4.7, $W^\infty$ deformation retracts onto $E_{f_1(K)}^\infty$ , and since $H_2(N_\Delta ^\infty )=0$ (recall Lemma 3.1) a Mayer–Vietoris argument therefore shows that

\begin{equation*}H_2(N_{D_\gamma }^\infty ) \cong H_2\biggl (\bigcup _{j \in \mathbb {Z}} t^j \widetilde {h}^2 \cup N_\Delta ^\infty \biggr ) \cong \mathbb {Z}[t^{\pm 1}],\end{equation*}

is freely generated by $[\Sigma ]$ . As explained for example in [Reference Conway and PowellCP23] proof of Lemma 3.2], the universal coefficient spectral sequence implies that the following composition is an isomorphism:

\begin{equation*}\textrm {ev} \circ {\textrm {PD}}^{-1} \colon H_2(N_{D_\gamma },M_K;\mathbb {Z}[t^{\pm 1}]) \xrightarrow {\cong } \overline {{\textrm {Hom}}_{\mathbb {Z}[t^{\pm 1}]}(H_2(N_{D_\gamma };\mathbb {Z}[t^{\pm 1}]),\mathbb {Z}[t^{\pm 1}])}=:H_2(N_{D_\gamma };\mathbb {Z}[t^{\pm 1}])^*. \end{equation*}

It follows that $H_2(N_{D_\gamma }^\infty ,M_K^\infty )\cong \mathbb {Z}[t^{\pm 1}]$ is generated by the homology class of a surface Poincaré dual to $\Sigma$ , for example the class of

\begin{equation*}\widehat {\textrm {core}}(\widetilde {h}^2):=\textrm {core}(\widetilde {h}^2) \cup _{\widetilde {c} \times \lbrace 1 \rbrace } \widetilde {C}_{\gamma ,f}.\end{equation*}

Here recall from Notation 4.3 that $\widetilde {C}_{\gamma ,f}$ denotes the lift of the trace of the isotopy from $\gamma$ to $c$ . Details on why $[\Sigma ]$ and $\widehat {\textrm {core}}(\widetilde {h}^2)$ are Poincaré dual will be given during the proof of Proposition 4.10.

Summarizing the outcome of this construction, we have

\begin{align*} H_2(N_{D_\gamma }^\infty )&\cong \mathbb {Z}[t^{\pm 1}] \langle [\Sigma ] \rangle , \\ H_2(N_{D_\gamma }^\infty ,M_K^\infty )&\cong \mathbb {Z}[t^{\pm 1}] \langle [\widehat {\textrm {core}}(\widetilde {h}^2)]\rangle . \end{align*}

Note for later use that $\partial (\widehat {\textrm {core}}(\widetilde {h}^2))=\widetilde {\gamma }$ .

Construction 4.9 (A basis for $H_2((\mathbb {C}P^2)^\circ )$ ). We define a generator

\begin{equation*}x_{{\mathbb {C}} P^2} \in H_2((\mathbb {C}P^2)^\circ ),\end{equation*}

by capping off $\textrm {cocore}(h^2)$ with any immersed surface $F \looparrowright E_J \subset S^3_{-1}(c)$ with boundary $\mu _c$ . Here, as previously, we are using the decomposition $(\mathbb {C}P^2)^\circ = (-( (S^3 \times [0,1]) \cup _{c \times \lbrace 1 \rbrace } h^2) ) \cup D^4$ .

Given a generalized crossing change $\mathbb {Z}$ -disk $D_\gamma$ with boundary $K$ , the next proposition describes some further properties of the homology classes $[\widehat {\textrm {core}}(\widetilde {h}^2)] \in H_2(N_{D_\gamma }^\infty ,M_K^\infty )$ and $[\widetilde {\gamma }] \in H_1(M_K^\infty )$ .

Proposition 4.10. Let $K$ be a knot and let $D_\gamma \subset (\mathbb {C}P^2)^\circ$ be a generalized crossing change $\mathbb {Z}$ -disk with $\partial D_\gamma =K$ . The basis elements $[\Sigma ] \in H_2(N_{D_\gamma }^\infty )$ and $[\widehat {\textrm {core}}(\widetilde {h}^2)] \in H_2(N_{D_\gamma }^\infty ,M_K^\infty )$ from Construction 4.8 satisfy the following properties.

  1. (1) The connecting homomorphism

    \begin{equation*}\partial _{D_\gamma } \colon H_2(N_{D_\gamma }^\infty ,M_K^\infty ) \to H_1(M_K^\infty ),\end{equation*}
    is entirely determined by the equality $\partial _{D_\gamma }([\widehat {\textrm {core}}(\widetilde {h}^2)])=[\widetilde {\gamma }]$ .
  2. (2) The classes $[\widehat {\textrm {core}}(\widetilde {h}^2)]$ and $[\Sigma ]$ are Poincaré dual,

    \begin{equation*}\textrm {ev} \circ \textrm {PD}^{-1} ([\widehat {\textrm {core}}(\widetilde {h}^2)])=[\Sigma ]^*.\end{equation*}
  3. (3) The homology class $[\widetilde {\gamma }] \in H_1(M_K^\infty )$ belongs to

    (4.1) \begin{equation} \mathcal {G}_K:= \left \lbrace x \in H_1(M_K^\infty ) \ \Big | \ x \text { is a generator and } {\textrm {Bl}}_K(x,x) =\frac {1}{\Delta _K} \right \rbrace . \end{equation}
  4. (4) Under the composition

    \begin{equation*} H_2(N_{D_\gamma }^\infty ) \xrightarrow {p_*} H_2(N_{D_\gamma }) \xrightarrow {\phi _{D_\gamma },\cong } H_2((\mathbb {C}P^2)^\circ ),\end{equation*}
    of the projection and inclusion induced maps, the class $[\Sigma ]$ is mapped to the preferred generator $x_{{\mathbb {C}} P^2}$ from Construction 4.9 .

Proof. We first show that $[\Sigma ]$ and $[\widehat {\textrm {core}}(\widetilde {h}^2)]$ are Poincaré dual as was already alluded to in Construction4.8. Since $H_2(N_{D_\gamma }^\infty )$ is freely generated by $[\Sigma ]$ , it suffices to show that evaluating $\textrm {ev} \circ {\textrm {PD}}^{-1}([\widehat {\textrm {core}}(\widetilde {h}^2)])$ and $[\Sigma ]^*$ on $[\Sigma ]$ leads to the same outcome. This follows because $\Sigma =S \cup _{\widetilde {\mu }_c} \textrm {cocore}(\widetilde {h}^2)$ and $\widehat {\textrm {core}}(\widetilde {h}^2)=\textrm {core}(\widetilde {h}^2) \cup _{\widetilde {c}} \widetilde {C}_{\gamma ,f}$ are geometrically dual (the intersection point occurs at the intersection of $\textrm {cocore}(\widetilde {h}^2)$ and $\textrm {core}(\widetilde {h}^2)$ ), and thanks to the definition of the equivariant intersection form

\begin{equation*} \langle [\Sigma ]^*,[\Sigma ] \rangle =1 =\lambda _{N_{D_\gamma }}^\partial ([\Sigma ],[\widehat {\textrm {core}}(\widetilde {h}^2)]) = \langle \textrm {ev} \circ {\textrm {PD}}^{-1}([\widehat {\textrm {core}}(\widetilde {h}^2)]),[\Sigma ]\rangle .\end{equation*}

This proves the second assertion.

The first assertion follows readily: since $\widehat {\textrm {core}}(\widetilde {h}^2)$ freely generates $H_2(N_{D_\gamma }^\infty ,M_K^\infty ) \cong \mathbb {Z}[t^{\pm 1}]$ and $\partial (\widehat {\textrm {core}}(\widetilde {h}^2))=\widetilde {\gamma }$ , we deduce that $\partial _{D_\gamma }([\widehat {\textrm {core}}(\widetilde {h}^2)])=[\widetilde {\gamma }]$ ; this equality determines $\partial _{D_\gamma }$ because its domain is freely generated by $[\widehat {\textrm {core}}(\widetilde {h}^2)]$ .

We move on to the third assertion. First, the class $[\widetilde {\gamma }]$ generates the Alexander module because $\partial _{D_\gamma }$ is surjective and takes the generator of $H_2(N_{D_\gamma }^\infty ,M_K^\infty )$ to $[\widetilde {\gamma }]$ . In order to show that ${\textrm {Bl}}_K([\widetilde {\gamma }],[\widetilde {\gamma }])= {1}/{\Delta _K}$ , we first record a more detailed formulation of Proposition 2.3.

Claim 3. Let $W$ be a $4$ -manifold with $\pi _1(W) \cong \mathbb {Z}$ , connected boundary, and $\pi _1(\partial W ) \to \pi _1(W)$ surjective. If $H_1(\partial W^\infty )$ is $\mathbb {Z}[t^{\pm 1}]$ -torsion, then, for any $\mathbb {Z}[t^{\pm 1}]$ -basis $e_1,\ldots ,e_n$ of $H_2(W;\mathbb {Z}[t^{\pm 1}])$ with dual basis $e_1^*,\ldots ,e_n^*$ of $ H_2(W;\mathbb {Z}[t^{\pm 1}])^*$ , we have

\begin{equation*}{\textrm {Bl}}_{\partial W} (\partial ({\textrm {PD}} \circ \textrm {ev}^{-1} (e_i^*)),\partial ({\textrm {PD}} \circ \textrm {ev}^{-1} (e_j^*)) )\equiv -\lambda _W(e_i,e_j)^{-1} \in \mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}],\end{equation*}

where $\partial \colon H_2(W,\partial W;\mathbb {Z}[t^{\pm 1}]) \to H_1(\partial W;\mathbb {Z}[t^{\pm 1}])$ denotes the connecting homomorphism in the long exact sequence of the pair $(W,\partial W)$ , and the inverse is taken over $\mathbb {Q}(t)$ .

We saw in the second assertion that the isomorphism ${\textrm {PD}} \circ \textrm {ev}^{-1}$ takes the dual class $[\Sigma ]^*$ to $[\widehat {\textrm {core}}(\widetilde {h}^2)] \in H_2(N_D,M_K;\mathbb {Z}[t^{\pm 1}])$ . Since $\partial _{D_\gamma }([\widehat {\textrm {core}}(\widetilde {h}^2)])=[\widetilde {\gamma }]$ , the claim now implies that

\begin{equation*}{\textrm {Bl}}_K([\widetilde {\gamma }],[\widetilde {\gamma }]) =-\lambda _{N_{D_\gamma }}([\Sigma ],[\Sigma ])^{-1} =-\left (\frac {1}{-\Delta _K}\right ) =\frac {1}{\Delta _K} \in \mathbb {Q}(t)/\mathbb {Z}[t^{\pm 1}]. \end{equation*}

For the penultimate equality, since $H_2(N_{D_\gamma }^\infty ) \cong \mathbb {Z}[t^{\pm 1}]$ , Proposition 3.2 implies that $\lambda _{N_{D_\gamma }}(x,x)=-\Delta _K$ for any generator $x$ of $H_2(N_{D_\gamma }^\infty )$ , for example $x=[\Sigma ]$ . This concludes the proof of the third assertion.

The fourth assertion essentially follows from the definitions of $[\Sigma ]$ and $x_{{\mathbb {C}} P^2}$ : the composition

\begin{equation*}H_2(N_{D_\gamma }^\infty ) \xrightarrow {p_*} H_2(N_{D_\gamma }) \xrightarrow {\phi _{D_\gamma },\cong } H_2((\mathbb {C}P^2)^\circ),\end{equation*}

maps the surface $\Sigma =S \cup _{\widetilde {\mu }_c} \textrm {cocore}(\widetilde {h}^2)$ to a surface in $(\mathbb {C}P^2)^\circ$ of the form $\textrm {cocore}(h^2) \cup _{\mu _c} F$ , where $F \looparrowright E_J \subset S^3$ is an immersed surface with boundary $\mu _c$ ; as mentioned in Construction4.9, such a closed immersed surface represents the generator $x_{{\mathbb {C}} P^2} \in H_2((\mathbb {C}P^2)^\circ )$ .

The next proposition uses the work of Fogel [Reference FogelFog94] and Borodzik and Friedl [Reference Borodzik and FriedlBF14] to show that elements of $\mathcal {G}_K$ can be realized by crossing change $\mathbb {Z}$ -disks. While several of the ideas go back to Fogel, we reference the work of Borodzik and Friedl as our set-up is closer to theirs.

Proposition 4.11. For every knot $K$ that is $\mathbb {Z}$ -slice in $(\mathbb {C}P^2)^\circ$ and every $x \in \mathcal {G}_K$ , there exists a crossing change $\mathbb {Z}$ -disk $D_\gamma$ with boundary $K$ such that $[\widetilde {\gamma }]=x$ .

Proof. The work of Borodzik and Friedl implies that, for every $x \in \mathcal {G}_K$ , there exists an embedded disk $\mathcal {D} \subset S^3$ that intersects $K$ in two transverse intersections of opposite signs, such that the lift of $\gamma :=\partial \mathcal {D}$ to the infinite cyclic cover satisfies $[\widetilde {\gamma }]=x$ and such that a positive crossing change using $\gamma$ leads to an Alexander polynomial one knot [Reference Borodzik and FriedlBF14, Lemma 5.5 and Proof of Theorem 5.1]. In particular, after choosing an ambient isotopy $(f_t)_{t \in [0,1]}$ of $S^3$ with $f_1(\gamma )=c$ , we see that every $x \in \mathcal {G}_K$ leads to a crossing change $\mathbb {Z}$ -disk with $[\widetilde {\gamma }]=x$ .

We include some more details on how to apply [Reference Borodzik and FriedlBF14]. First, note that Borodzik and Friedl work with a representative of the Alexander polynomial that evaluates to $1$ at $t=1$ [Reference Borodzik and FriedlBF14, Example 2.3] whereas for us, $\Delta _K(1)=-1$ . Next, we note that in [Reference Borodzik and FriedlBF14, § 1.3] the authors use the opposite sign convention than ours in their definition of a matrix presenting a linking form: contrarily to (1.1), no minus sign appears.

We now give the argument proper. Proposition 4.10 ensures that ${\textrm {Bl}}_K(x,x)=\mbox {${1}/{\Delta _K}$}=\mbox {${-1}/({-\Delta _K})$}$ . We apply [Reference Borodzik and FriedlBF14, Lemma 5.5] (in their notation, we take $n=1$ and $p_{11}(t)=-1$ ): the outcome is a based embedded disk $\mathcal {D} \subset S^3$ that intersects $K$ in two transverse intersections of opposite signs and whose boundary $\gamma =\partial \mathcal {D}$ lifts to a loop that represents $x$ and has equivariant self-linking $ {-1}/({-\Delta _K})=\mbox {${1}/{\Delta _K}$} \in \mathbb {Q}(t)$ . The argument in [Reference Borodzik and FriedlBF14, Proof of Theorem 5.1] shows that the knot obtained by the resulting positive crossing change has Alexander polynomial one. In their notation we take $A(t)$ to be the size one matrix $(\Delta _K)$ (as this presents the Blanchfield pairing with their sign conventions) and $\varepsilon _1=A_{11}(1)=\Delta _K(1)=-1$ .

The next remark describes the relation between the homology of $N_D^\infty$ and the homology of $N_D$ . For this, we write $\mathbb {Z}_{\textrm {aug}}$ for the $\mathbb {Z}[t^{\pm 1}]$ -module whose underlying abelian group is $\mathbb {Z}$ and where the action is by $p(t) \cdot n:=p(1)n$ for $p(t) \in \mathbb {Z}[t^{\pm 1}]$ and $n \in \mathbb {Z}$ . In other words, $\mathbb {Z}_{\textrm {aug}}$ has the $\mathbb {Z}[t^{\pm 1}]$ -module induced by the augmentation map $\textrm {aug} \colon \mathbb {Z}[t^{\pm 1}] \to \mathbb {Z},p(t) \mapsto p(1).$

Remark 4.12. Given an infinite cover $p \colon X^\infty \to X$ associated to an epimorphism $\varphi \colon \pi _1(X) \twoheadrightarrow \mathbb {Z}$ , it is customary to identify $H_*(X;\mathbb {Z}_{\textrm {aug}}):=H_*(X;\mathbb {Z}[t^{\pm 1}] \otimes _{\mathbb {Z}[t^{\pm 1}]} \mathbb {Z}_{\textrm {aug}})$ with $H_*(X)$ . This isomorphism is induced by the projection $p$ and, in particular, the following diagram commutes.

The map labeled $\psi$ maps $[\sigma ] \otimes n$ to $[\sigma \otimes n]$ and its kernel and cokernel can be analyzed using the universal coefficient spectral sequence. When $X=N_D$ is a $\mathbb {Z}$ -disk exterior and $k=2$ , this map is an isomorphism [Reference Conway and PowellCP23, proof of Lemma 5.10].

The upshot of this discussion is that will frequently use the identification

\begin{equation*}p_* \circ \psi \colon H_2(N_D;\mathbb {Z}[t^{\pm 1}]) \otimes _{\mathbb {Z}[t^{\pm 1}]} \mathbb {Z}_{\textrm {aug}} \xrightarrow {\cong } H_2(N_D)\end{equation*}

and, given $x \in H_*(N_D^\infty )=H_k(N_D;\mathbb {Z}[t^{\pm 1}])$ , we will identify $x \otimes 1 \in H_2(N_D;\mathbb {Z}[t^{\pm 1}]) \otimes _{\mathbb {Z}[t^{\pm 1}]} \mathbb {Z}_{\textrm {aug}}$ with $p_*(x) \in H_2(N_D)$ .

For example, using these identifications, the fourth item of Proposition 4.10 states that

\begin{equation*}\phi _{D_\gamma }([\Sigma ] \otimes 1)=x_{{\mathbb {C}} P^2}.\end{equation*}

5. The classification of $\mathbb {Z}$ -disks

The goal of this section is to prove Theorem1.2 from the introduction. After introducing some notation, we extract some results from [Reference Conway and PowellCP23] that provide the main ingredients for the proof.

Notation 5.1. In what follows, given a $\mathbb {Z}$ -disk $D \subset (\mathbb {C}P^2)^\circ$ with boundary a knot $K$ , we use the composition $\textrm {ev} \circ {\textrm {PD}}^{-1}$ of the Poincaré duality isomorphism with evaluation to identify $H_2(N_D,M_K;\mathbb {Z}[t^{\pm 1}])$ with

\begin{equation*}H_2(N_D;\mathbb {Z}[t^{\pm 1}])^*:= \overline {{\textrm {Hom}}_{\mathbb {Z}[t^{\pm 1}]}(H_2(N_D;\mathbb {Z}[t^{\pm 1}]),\mathbb {Z}[t^{\pm 1}])}.\end{equation*}

The fact that $\textrm {ev}$ is an isomorphism can be seen using the universal coefficient spectral sequence; we refer to [Reference Conway and PowellCP23, proof of Lemma 3.2] for the details.

Given a basis $e$ of $H_2(N_D;\mathbb {Z}[t^{\pm 1}]) \cong \mathbb {Z}[t^{\pm 1}]$ , we write $e^* \in H_2(N_D;\mathbb {Z}[t^{\pm 1}])^*$ for the dual basis. From now on, we use $\textrm {ev} \circ {\textrm {PD}}^{-1}$ to view $e^*$ as an element of $H_2(N_D,M_K;\mathbb {Z}[t^{\pm 1}])$ . We also write

\begin{equation*}\phi _D \colon H_2(N_D) \to H_2\big(\big(\mathbb {C}P^2\big)^\circ \big)\end{equation*}

for the inclusion induced map. Continuing with the notation from Remark 4.12, we note that any $\sigma \in H_2(N_D;\mathbb {Z}[t^{\pm 1}])$ determines an element $\phi _D(\sigma \otimes 1) \in H_2(N_D).$

The next proposition uses results from [Reference Conway and PowellCP23] and [Reference Orson and PowellOP22] to formulate a criterion ensuring that two $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ are isotopic rel. boundary.

Proposition 5.2. Assume that $D_0,D_1 \subset (\mathbb {C}P^2)^\circ$ are $\mathbb {Z}$ -disks with boundary a knot $K$ . The following assertions are equivalent:

  1. (i) the disks $D_0$ and $D_1$ are isotopic rel. boundary;

  2. (ii) there are generators $e_0 \in H_2(N_{D_0}^\infty )$ and $e_1 \in H_2(N_{D_1}^\infty )$ such that, for some $k \in \mathbb {Z}$ ,

    \begin{align*} \phi _{D_0}(e_0 \otimes 1) & =\phi _{D_1}(e_1 \otimes 1) \in H_2({\mathbb {C}} P^2), \\ t^k\partial _{D_0}(e_0^*)&=\partial _{D_1}(e_1^*) \in H_1(E_K^\infty ). \end{align*}

Proof. We start with a claim that we will be using again later on.

Claim 4. Assume that $\Phi \colon ((\mathbb {C}P^2)^\circ ,D_0) \to ((\mathbb {C}P^2)^\circ ,D_1)$ is a homeomorphism rel. boundary. For any generators $x_0 \in H_2(N_{D_0}^\infty )$ and $x_1 \in H_2(N_{D_1}^\infty )$ , there is a sign $\varepsilon =\pm 1$ so that

\begin{align*} \phi _{D_0}(x_0 \otimes 1) & =\textrm {sign}(\Phi ) \varepsilon \phi _{D_1}(x_1 \otimes 1), \\ \partial _{D_0}(x_0^*) & =\varepsilon t^k \partial _{D_1}(x_1^*), \end{align*}

where $\textrm {sign}(\Phi )=1$ if $\Phi \colon {\mathbb {C}} P^2 \to {\mathbb {C}} P^2$ induces the identity on $H_2$ and $\textrm {sign}(\Phi )=-1$ otherwise.

Proof of Claim 4. Restricting $\Phi$ to the disk exteriors, lifting the outcome to the covers, and taking the induced map $\widetilde {\Phi }_*$ on second homology, we obtain a $\mathbb {Z}[t^{\pm 1}]$ -isomorphism. As these modules are free of rank one, for any generators $x_0 \in H_2(N_{D_0}^\infty )$ and $x_1 \in H_2(N_{D_1}^\infty )$ there is a $\varepsilon =\pm 1$ and a $k \in \mathbb {Z}$ so that

\begin{equation*} \widetilde {\Phi }_*(x_0)=\varepsilon t^kx_1.\end{equation*}

Apply the projection induced map $(p_1)_* \colon H_2(N_{D_1}^\infty ) \to H_2(N_{D_1})$ to obtain $\Phi \circ (p_0)_*(x_0)=p_1(\varepsilon t^k x_1).$ Recalling the notation from Remark 4.12 we rewrite this as $\Phi (x_0 \otimes 1)=\varepsilon t^k x_1 \otimes 1.$ By definition of the augmentation map, this can be rewritten as $\Phi (x_0 \otimes 1)=\varepsilon x_1 \otimes 1.$ Apply $\phi _{D_1}$ to obtain the first claimed equality:

\begin{equation*} \varepsilon \phi _{D_1}(x_1 \otimes 1) =\phi _{D_1} \circ \Phi (x_0 \otimes 1)= \textrm {sign}(\Phi ) \phi _{D_0}(x_0 \otimes 1). \end{equation*}

For the second equality, since $\Phi$ restricts to the identity on the boundary, we have

\begin{equation*} \partial _{D_0}(x_0^*) =\partial _{D_1} \circ \widetilde {\Phi }_*(x_0^*) =\varepsilon t^k \partial _{D_1}(x_1^*). \end{equation*}

This concludes the proof of Claim4.

We now prove the (i) $\Rightarrow$ (ii) direction of the proposition. Assume that $\Phi \colon ((\mathbb {C}P^2)^\circ ,D_0) \to ((\mathbb {C}P^2)^\circ ,D_1)$ is an ambient isotopy rel. boundary and pick generators $x_0 \in H_2(N_{D_0}^\infty )$ and $x_1 \in H_2(N_{D_1}^\infty )$ . Since $\Phi \colon (\mathbb {C}P^2)^\circ \to (\mathbb {C}P^2)^\circ$ induces the identity on $H_2$ , Claim4 states that there is a sign $\varepsilon =\pm 1$ so that

\begin{align*} \phi _{D_0}(x_0 \otimes 1) & =\varepsilon \phi _{D_1}(x_1 \otimes 1), \\ \partial _{D_0}(x_0^*) & =\varepsilon t^k \partial _{D_1}(x_1^*). \end{align*}

It follows that the generators $e_0:=x_0$ and $e_1:=\varepsilon x_1$ satisfy the relations stated in (ii).

Next we prove the converse, namely we assume that there are $\mathbb {Z}[t^{\pm 1}]$ -generators $e_0 \in H_2(N_{D_0}^\infty )$ and $e_1 \in H_2(N_{D_1}^\infty )$ as in (ii) and prove that the disks are isotopic rel. boundary. The idea is as follows: the first equation in (ii) is used to show that the disks are equivalent rel. boundary; the second condition is used to upgrade this equivalence to an isotopy.

Define an isomorphism $F \colon H_2(N_{D_0};\mathbb {Z}[t^{\pm 1}]) \to H_2(N_{D_1};\mathbb {Z}[t^{\pm 1}])$ by setting $F(e_0):=t^{-k}e_1$ and extending $\mathbb {Z}[t^{\pm 1}]$ -linearly. Note that $F$ is automatically an isometry because

\begin{align*} \lambda _{N_{D_0}}(e_0,e_0) & =-\Delta _K,\\ \lambda _{N_{D_1}}(F(e_0),F(e_0)) & = \lambda _{N_{D_1}}(t^{-k}e_1,t^{-k}e_1) = \lambda _{N_{D_1}}(e_1,e_1) =-\Delta _K. \end{align*}

Here we used the fact that $\lambda _{N_{D_i}}(x,x)=-\Delta _K$ for any generator $x \in H_2(N_{D_i};\mathbb {Z}[t^{\pm 1}]) \cong \mathbb {Z}[t^{\pm 1}]$ ; this is a consequence of Lemma 3.1.

Using the identification $H_2(N_{D_i};\mathbb {Z}[t^{\pm 1}])^* \cong H_2(N_{D_i},M_K;\mathbb {Z}[t^{\pm 1}])$ for $i=0,1$ , we think of the dual isomorphism as a map $F^* \colon H_2(N_{D_1},M_K;\mathbb {Z}[t^{\pm 1}]) \to H_2(N_{D_0},M_K;\mathbb {Z}[t^{\pm 1}])$ . Applying the connecting homomorphism $\partial _{D_1}$ to the equality $(F^*)^{-1}(e_0^*)=t^{-k}e_1^*$ , we obtain

\begin{equation*}\partial _{D_1} \circ (F^*)^{-1}(e_0^*)=t^{-k}\partial _{D_1}(e_1^*) =\partial _{D_0}(e_0^*).\end{equation*}

In the second equality, we used our assumption that $t^k\partial _{D_0}(e_0^*)=\partial _{D_1}(e_1^*)$ .

Since $F$ is an isometry that satisfies $\partial _{D_1} \circ (F^*)^{-1}=\partial _{D_0}$ , we can apply [Reference Conway and PowellCP23, Theorem1.3] to deduce that there is an equivalence rel. boundary $\Phi \colon ((\mathbb {C}P^2)^\circ ,D_0) \cong ((\mathbb {C}P^2)^\circ ,D_1)$ realizing  $F$ .

It remains to verify that the homeomorphism $\Phi$ is isotopic to the identity. As shown by [Reference Conway and PowellCP23, Lemma 5.10], any isometry $G \colon \lambda _{N_{D_0}} \cong \lambda _{N_{D_1}}$ induces a self-isomorphism $G_{\mathbb {Z}}$ on $H_2((\mathbb {C}P^2)^\circ )$ that satisfies $G_{\mathbb {Z}} \circ \phi _{D_0}=\phi _{D_1}\circ (G \otimes _{\mathbb {Z}[t^{\pm 1}]} {\textrm {id}}_{\mathbb {Z}_{\textrm {aug}}})$ . The same reference states that if $G$ is induced by a homeomorphism $\Psi \colon ((\mathbb {C}P^2)^\circ ,D_0) \xrightarrow {\cong } ((\mathbb {C}P^2)^\circ ,D_1)$ , then $\Psi _*=G_{\mathbb {Z}}$ .

Returning to our setting, we endow $H_2((\mathbb {C}P^2)^\circ )$ with the basis $\phi _{D_0}(e_0 \otimes 1)$ and note that

\begin{equation*}F_{\mathbb {Z}} \circ \phi _{D_0}(e_0 \otimes 1) =\phi _{D_1} \circ (F \otimes _{\mathbb {Z}[t^{\pm 1}]} {\textrm {id}}_{\mathbb {Z}_{\textrm {aug}}})(e_0 \otimes 1) =\phi _{D_1}(t^{-k}e_1 \otimes 1) =\phi _{D_1}(e_1 \otimes 1) =\phi _{D_0}(e_0 \otimes 1).\end{equation*}

In the last equality we used the assumption from (ii). This calculation implies that the rel. boundary homeomorphism $\Phi \colon (\mathbb {C}P^2)^\circ \to (\mathbb {C}P^2)^\circ$ induces the identity on $H_2$ . Work of Orson and Powell [Reference Orson and PowellOP22, Corollary C] now implies that $\Phi$ is isotopic rel. boundary to the identity. This concludes the proof that $D_0$ and $D_1$ are isotopic rel. boundary.

The next result applies the criterion from Proposition 5.2 to generalized crossing change $\mathbb {Z}$ -disks.

Proposition 5.3. Assume that $D_0,D_1 \subset (\mathbb {C}P^2)^\circ$ are generalized crossing change $\mathbb {Z}$ -disks for a knot $K$ , with respective surgery curves $\gamma _0,\gamma _1 \subset E_K$ . Let $\widetilde {\gamma }_0,\widetilde {\gamma }_1$ be the lifts of $\gamma _0,\gamma _1$ to the infinite cyclic cover. The following assertions are equivalent:

  1. (i) the disks $D_0$ and $D_1$ are isotopic rel. boundary;

  2. (ii) the homology classes $[\widetilde {\gamma }_0],[\widetilde {\gamma }_1] \in \mathcal {G}_K \subset H_1(E_K^\infty )$ satisfy $[\widetilde {\gamma }_0]=t^k [\widetilde {\gamma }_1]$ for some $k \in \mathbb {Z}$ .

Proof. For $i=0,1,$ consider the generator $e_i:=[\Sigma _i]$ for $H_2(N_{D_i}^\infty )$ from Construction4.8 and recall from Proposition 4.10 that this generator satisfies $\partial _{D_i}(e_i^*)=[\widetilde {\gamma }_i]$ . We also noted in Remark 4.12 that $\phi _{D_i}(e_i \otimes 1)=x_{{\mathbb {C}} P^2}$ , the preferred homology class $x_{{\mathbb {C}} P^2} \in H_2((\mathbb {C}P^2)^\circ )$ from Construction4.9.

We can now prove the (ii) $\Rightarrow$ (i) direction: we have $\phi _{D_0}(e_0 \otimes 1)=x_{{\mathbb {C}} P^2}=\phi _{D_1}(e_1 \otimes 1)$ and $\partial _{D_0}(e_0^*)=[\widetilde {\gamma }_0]=t^k [\widetilde {\gamma }_1]=t^k \partial _{D_1}(e_1^*)$ and therefore Proposition 5.2 implies that $D_0$ and $D_1$ are isotopic rel. boundary.

Next we consider the (i) $\Rightarrow$ (ii) direction. Assume the disks are isotopic rel. boundary via a homeomorphism $\Phi \colon ((\mathbb {C}P^2)^\circ ,D_0) \xrightarrow {\cong } ((\mathbb {C}P^2)^\circ ,D_1)$ . Claim4 shows that, for any generators $x_0 \in H_2(N_{D_0}^\infty )$ and $x_1 \in H_2(N_{D_1}^\infty )$ , there is a sign $\varepsilon =\pm 1$ so that

\begin{align*} \varepsilon \phi _{D_1}(x_1 \otimes 1) & = \phi _{D_0}(x_0 \otimes 1), \\ \partial _{D_0}(x_0^*) & =\varepsilon t^k \partial _{D_1}(x_1^*). \end{align*}

Since $\phi _{D_0}(e_0 \otimes 1)=x_{{\mathbb {C}} P^2}=\phi _{D_1}(e_1 \otimes 1)$ , if we apply these equalities to $x_0=e_0$ and $x_1=e_1$ , we deduce that $\varepsilon =1$ . It follows that $ [\widetilde {\gamma }_0] =\partial _{D_0}(e_0^*) =t^k \partial _{D_1}(e_1^*) = t^k[\widetilde {\gamma }_1],$ as required. This concludes the proof of the proposition.

We now prove the first main result of this section.

Theorem 5.4. Let $K$ be a knot. Every $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ with boundary $K$ is isotopic rel. boundary to a crossing change $\mathbb {Z}$ -disk $D_\gamma$ .

Proof. Let $D \subset (\mathbb {C}P^2)^\circ$ be a $\mathbb {Z}$ -disk with boundary $K$ . We are going to prove that $D$ is isotopic rel. boundary to a crossing change $\mathbb {Z}$ -disk. Fix a generator $y \in H_2(N_D^\infty )$ . The idea of the proof is that the work of Fogel [Reference FogelFog94] and Borodzik and Friedl [Reference Borodzik and FriedlBF14] (as recast in Proposition 4.11) implies that the projection of a loop representing $\partial _D([y^*]) \in H_1(M_K^\infty ) \cong H_1(E_K^\infty )$ will give the required surgery curve.

Use $p \colon N_{D^\infty } \to N_D$ to denote the covering map and consider the portion

(5.1) \begin{equation} \cdots \to H_2(N_D;\mathbb {Z}[t^{\pm 1}]) \xrightarrow {\cdot (t-1)} H_2(N_D;\mathbb {Z}[t^{\pm 1}]) \xrightarrow {p_*} H_2(N_D) \to H_1(N_D;\mathbb {Z}[t^{\pm 1}]) \to \cdots , \end{equation}

of the exact sequence induced by the short exact sequence of coefficients

\begin{equation*}0 \to \mathbb {Z}[t^{\pm 1}] \xrightarrow {\cdot (t-1)} \mathbb {Z}[t^{\pm 1 }] \to \mathbb {Z}_{\textrm {aug}} \to 0.\end{equation*}

In the third term of (5.1), we identified $H_2(N_D)$ with $H_2(N_D;\mathbb {Z}_{\textrm {aug}})$ as explained in Remark 4.12. This sequence is sometimes called the Milnor exact sequence because of its use in [Reference MilnorMil68].

Since $H_1(N_D;\mathbb {Z}[t^{\pm 1}]) =H_1(N_D^\infty )=0$ , the projection induced map $p_* \colon H_2(N_D^\infty ) \to H_2(N_D)$ is surjective and therefore $p_*(y) \in H_2(N_D)\cong \mathbb {Z}$ is a generator. Since $\phi _D \colon H_2(N_D) \to H_2((\mathbb {C}P^2)^\circ )$ is an isomorphism, it follows that $\phi _D(p_*(y))=\pm x_{{\mathbb {C}} P^2} \in H_2((\mathbb {C}P^2)^\circ ) \cong \mathbb {Z}$ . Picking the negative of $y$ if necessary, we can therefore assume that $\phi _D(p_*(y))= x_{{\mathbb {C}} P^2}$ .

Using the connecting homomorphism $\partial _D$ in the long exact sequence of the pair $(N_D^\infty ,M_K^\infty )$ , we see that the generator $y^* \in H_2(N_D^\infty ,M_K^\infty )$ determines a generator $x:=\partial _D([y^*]) \in H_1(M_K^\infty )$ ; this uses the fact that $\partial _D$ is surjective. Since $y$ generates $H_2(N_D^\infty ) \cong \mathbb {Z}[t^{\pm 1}]$ , Proposition 3.2 shows that $\lambda _{N_D}(y,y)=-\Delta _K$ . Claim3 implies that

\begin{equation*}{\textrm {Bl}}_K(x,x) \equiv -\lambda _{N_D}(y,y)^{-1}=\frac {1}{\Delta _K},\end{equation*}

and therefore $x \in \mathcal {G}_K$ . As mentioned in Proposition 4.11, the work of Borodzik and Friedl implies that there is a crossing change $\mathbb {Z}$ -disk $D_\gamma$ with boundary $K$ and $[\widetilde {\gamma }]=x \in \mathcal {G}_K$ .

We show that $D_\gamma$ is isotopic rel. boundary to the $\mathbb {Z}$ -disk $D$ that we started with. Consider the generator $[\Sigma ] \in H_2(N_{D_\gamma };\mathbb {Z}[t^{\pm 1}])$ from Construction4.8. Recall from Proposition 4.10 that $\partial _{D_\gamma }([\Sigma ]^*)=[\widetilde {\gamma }]$ and $\phi _{D_\gamma }([\Sigma ])=x_{{\mathbb {C}} P^2}$ . By construction of $y$ , we also have $\partial _D(y^*)=[\widetilde {\gamma }]$ and $\phi _D([y])=x_{{\mathbb {C}} P^2}$ . Proposition 5.2 now implies that $D$ and $D_\gamma$ are isotopic rel. boundary.

Combining these results, we obtain Theorem1.2 from the introduction.

Theorem 1.2. Let $K$ be a knot.

  1. (1) Every $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ with boundary $K$ is isotopic rel. boundary to a crossing change $\mathbb {Z}$ -disk.

  2. (2) Two crossing change $\mathbb {Z}$ -disks with boundary $K$ are isotopic rel. boundary if and only if the homology classes of the lifts of their surgery curves agree in the Alexander module $H_1(E_K^\infty )$ up to multiplication by $t^k$ for some $k \in \mathbb {Z}$ .

  3. (3) Mapping a crossing change $\mathbb {Z}$ -disk with boundary $K$ to the homology class of the lift of its surgery curve defines a bijection

    \begin{equation*}\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ ) \xrightarrow {\approx } \mathcal {G}_K/\lbrace t^k \rbrace _{k \in \mathbb {Z}}.\end{equation*}

Proof. The first item is proved in Theorem5.4 and the second in Proposition 5.3. The third item is also a consequence of these results as we now explain. The combination of these results shows that the map $\mathcal {D}_{\mathbb {Z}}(K,(\mathbb {C}P^2)^\circ ) \to \mathcal {G}_K/\lbrace t^k \rbrace _{k \in \mathbb {Z}}$ is well defined and injective: Theorem5.4 shows that every $\mathbb {Z}$ -disk is isotopic rel. boundary to a crossing change $\mathbb {Z}$ -disk and Proposition 5.3 states that this disk is uniquely determined by the homology class of the lift of the surgery curve. Surjectivity is a consequence of Proposition 4.11.

Remark 5.5. All the results in this section can be modified to involve equivalence rel. boundary instead of isotopy rel. boundary: in Propositions 5.2 and 5.3 and Theorem 1.2, one substitutes multiplication (and modding out) by $t^k$ with multiplication (and modding out) by $\pm t^k$ . We leave the details to the reader, but note the key point: an equivalence between $\mathbb {Z}$ -surfaces in $(\mathbb {C}P^2)^\circ$ need not induce the identity on $H_2((\mathbb {C}P^2)^\circ )$ ; indeed it can also induce multiplication by $-1$ .

6. Units in rings of quadratic integers

We have reduced the enumeration of $\mathbb {Z}$ -disks in $({\mathbb {C}} P^2)^\circ$ with boundary a knot $K$ to understanding (a quotient of) the group $U(\mathbb {Z}[t^{\pm 1}]/(\Delta _K))$ . Here, recall that

\begin{equation*}U(\mathbb {Z}[t^{\pm 1}]/(\Delta _K))=\lbrace x \in \mathbb {Z}[t^{\pm 1}]/(\Delta _K) \mid x\overline {x}=1 \rbrace .\end{equation*}

Studying this group involves some number theory. In § 7, we re-interpret $U(\mathbb {Z}[t^{\pm 1}]/(\Delta _K))$ in terms of the localization of a ring of quadratic integers $\mathbb {Z}[\omega ]$ . This latter object will thus be the focus of the present section. Section 6.1 collects some definitions and facts concerning $\mathbb {Z}[\omega ]$ , whereas § 6.2 describes the group of units of $\mathbb {Z}[1/n, \omega ]$ for any nonzero $n \in \mathbb {Z}$ .

6.1 Rings of quadratic integers

We recall some facts about rings of quadratic integers. Further discussion of these can be found in many texts on algebraic number theory such as [Reference NeukirchNeu99]. We note that the following results admit generalizations to rings of (not necessarily quadratic) integers, but we will not use these here.

Definition 6.1. Given a square free integer $d$ , let

(6.1) \begin{equation} \omega = \begin{cases} \sqrt {d} & \text {if } d \equiv 2, 3 \bmod 4, \\ \dfrac {1 + \sqrt {d}}{2} & \text {if } d \equiv 1 \bmod 4, \end{cases} \end{equation}

We refer to $\mathbb {Z}[\omega ]$ as a ring of quadratic integers.

The reason for this terminology is as follows: the field $\mathbb {Q}(\sqrt {d})$ is typically called a quadratic number field, and $\mathbb {Z}[\omega ]$ is the integral closure of $\mathbb {Z}$ in $\mathbb {Q}(\sqrt {d})$ ; see for example [Reference Stewart and TallST16, § 3.1]. The next result collects some properties of this ring and its ideals. For a proof of the first assertion we refer to [Reference Stewart and TallST16, § 5.2]; for the second we refer to [Reference Stewart and TallST16, § 9.3].

Theorem 6.2 (Finiteness of the class group). Given a square free integer $d$ and $\omega$ as in (6.1), we have the following.

  1. (1) The ring of integers $\mathbb {Z}[\omega ]$ is a Dedekind domain: every ideal in $\mathbb {Z}[\omega ]$ admits a unique prime factorization.

  2. (2) The ring of integers $\mathbb {Z}[\omega ]$ has finite class number: there exists some $h \in \mathbb {N}$ such that $\mathfrak {p}^h$ is principal for any ideal $\mathfrak {p}$ in $\mathbb {Z}[\omega ]$ .

The following result describes the units of $\mathbb {Z}[\omega ]$ and is known as Dirichlet’s Unit Theorem; see for example [Reference Stewart and TallST16, Appendix B] for a proof.

Theorem 6.3 (Dirichlet’s Unit Theorem). Given a square free integer $d$ and $\omega$ as in (6.1), we have the following.

  1. (1) If $d \gt 0$ , then $\mathbb {Z}[\omega ]^\times \cong \{\pm 1\} \times \mathbb {Z}$ . In particular, there exists a unit $u \in \mathbb {Z}[\omega ]^\times$ , called a fundamental unit, such that every unit in $\mathbb {Z}[\omega ]$ is of the form $\pm u^i$ for some $i \in \mathbb {Z}$ .

  2. (2) If $d \lt 0$ , then $\mathbb {Z}[\omega ]^\times$ is finite. More precisely:

    1. (a) if $d = -1$ , then there are four units, given by $\pm 1$ and $\pm i$ ;

    2. (b) if $d = -3$ , then there are six units, given by $\pm 1$ and $(\pm 1 \pm \sqrt {-3})/2$ ; and

    3. (c) if $d \lt -3$ , then the only units are $\pm 1$ .

In this paper, we will be most concerned with the case where $d \equiv 1$ mod $4$ . In this setting, the next result describes the factorization of ideals $(p)$ in $\mathbb {Z}[\omega ]$ for $p$ a prime in $\mathbb {Z}$ . See for example the discussion surrounding [Reference NeukirchNeu99, Proposition 8.5] for a proof.

Theorem 6.4 (Splitting criterion). Assume $d \equiv 1 \bmod 4$ is square free, and let $\omega$ be as in (6.1). Let $p$ be an odd prime in $\mathbb {Z}$ . Then $(p)$ factors into prime ideals in $\mathbb {Z}[\omega ]$ as follows:

  1. (1) $(p)$ prime factors into the form $\mathfrak {p}^2$ if and only if $p$ divides $d$ ;

  2. (2) $(p)$ prime factors into the form $\mathfrak {p}_1 \mathfrak {p_2}$ if and only if $p$ does not divide $d$ and $d$ is a quadratic residue mod $p$ ; and

  3. (3) $(p)$ remains a prime ideal in $\mathbb {Z}[\omega ]$ if and only if $p$ does not divide $d$ and $d$ is not a quadratic residue mod $p$ .

In the special case $p = 2$ , the quadratic residue conditions for possibilities $(2)$ and $(3)$ are replaced by the conditions $d \equiv 1 \bmod 8$ and $d \equiv 5 \bmod 8$ , respectively.

The above possibilities are referred to as $p$ ramifying, splitting, and remaining inert.

6.2 The units of $\mathbb {Z}[1/n, \omega]$

While Dirichlet’s Unit Theorem describes the group of units of $\mathbb {Z}[\omega ]$ , in § 7 we will be interested in the localization of $\mathbb {Z}[\omega ]$ at a particular integer $n$ . This section is thus devoted to describing the units of $\mathbb {Z}[1/n, \omega ]$ for any nonzero $n \in \mathbb {Z}$ . The answer will follow promptly from Dirichlet’s Unit Theorem once we recall the notion of the saturation of a multiplicatively closed subset.

Definition 6.5. The saturation of a multiplicatively closed subset $S \subset R$ of an integral domain is defined by

\begin{equation*} \widetilde {S} = \{ x \in R \text { such that } x \text { divides } s \text { for some } s \in S \}. \end{equation*}

The next lemma is essentially the observation that if we invert $S$ , then every element of $S$ gives us a new unit in $S^{-1}R$ and, more generally, every divisor of every element of $S$ becomes a unit in $S^{-1}R$ .

Lemma 6.6. Given a multiplicatively closed subset $S \subset R$ of an integral domain, the group of units of $S^{-1}R$ is multiplicatively generated by the saturation of $S$ .

Proof. We first claim that the elements of $\widetilde {S}$ are units in $S^{-1}R$ . Indeed, let $x_1 \in \widetilde {S}$ , so that $x_1 x_2 = s$ for some $s \in S$ . Then $x_1$ and $x_2$ are units in $S^{-1} R$ , with inverses $x_2/s$ and $x_1/s$ , respectively. Conversely, suppose that $r_1/s_1$ is a unit in $S^{-1}R$ . Then we have that $r_1/s_1 \cdot r_2/s_2 = 1$ for some $r_2/s_2 \in S^{-1}R$ . But then $r_1 r_2 = s_1 s_2$ , whereupon $r_1$ and $r_2$ are elements of $\widetilde {S}$ .

The ring $\mathbb {Z}[1/n, \omega ]$ corresponds to $S^{-1}R$ in the case $R=\mathbb {Z}[\omega ]$ and $S=\{n^k\}$ . Lemma 6.6 thus gives a set of generators for $\mathbb {Z}[1/n, \omega ]^\times$ , consisting of the set of units of $\mathbb {Z}[\omega ]$ together with the saturation (in $\mathbb {Z}[\omega ]$ ) of the set $\{n^k\}$ . Since the former is given by Dirichlet’s Unit Theorem, the principal difficulty is to understand the latter. If $\mathbb {Z}[\omega ]$ is a unique factorization domain (UFD), then the saturation of $\{n^k\}$ may be computed by factoring $n$ in $\mathbb {Z}[\omega ]$ and taking all possible products of these prime factors. In general, however, $\mathbb {Z}[\omega ]$ is not a UFD and so the saturation of $\{n^k\}$ cannot be computed purely from the factorization of $n$ .

Instead, we have the following criterion based on the factorization of the ideal $(n)$ .

Lemma 6.7. Let $R$ be a Dedekind domain and fix $n \in R$ . Then $x \in R$ is in the saturation of $\{n^k\}$ if and only if $(x)$ is a product of prime ideals occurring in the factorization of $(n)$ .

Proof. Suppose that $x$ is in the saturation of $\{n^k\}$ . Then $x$ divides some element $n^k$ ; this implies that $(x)$ divides $(n^k) = (n)^k$ as an ideal in $R$ . By unique prime factorization of ideals, this shows that $(x)$ must be a product of prime ideals occurring in the factorization of $(n)$ . Conversely, suppose that $(x)$ is a product of prime ideals occurring in the factorization of $(n)$ . Clearly, for $k$ sufficiently large, the factorization of $(n)^k$ then contains the factorization of $(x)$ . Hence $(x)$ divides $(n^k)$ as an ideal, which implies that $x$ divides $n^k$ as an element of $R$ .

Remark 6.8. Lemma 6.7 gives a recipe for understanding the saturation of $\{n^k\}$ in $\mathbb {Z}[\omega ]$ as follows.

  1. (1) Factor $(n)$ into prime ideals in $\mathbb {Z}[\omega ]$ .

  2. (2) Consider all possible products of prime ideals occurring in this factorization and determine which of these products are principal.

  3. (3) For each such principal ideal, choose a generating element. Up to multiplication by units in $\mathbb {Z}[\omega ]$ , this procedure gives all elements in the saturation of $\{n^k\}$ .

In general, the algorithm of Remark 6.8 is difficult to carry out explicitly, as the second step involves computing products of prime ideals and deciding which are principal. Note, however, that due to Theorem6.2 we have a particular set of such products which are always principal. Let

\begin{equation*}(n) = \mathfrak {p}_1^{\alpha _1} \mathfrak {p}_2^{\alpha _2} \cdots \mathfrak {p}_r^{\alpha _r},\end{equation*}

be the factorization of $(n)$ into prime ideals in $\mathbb {Z}[\omega ]$ . If the class number of $\mathbb {Z}[\omega ]$ is $h$ , then by Theorem6.2 each $\mathfrak {p}_i^h$ is principal. Write

\begin{equation*} \mathfrak {p}_i^h = (x_i). \end{equation*}

By Lemma 6.7, $x_i$ is in the saturation of $\{n^k\}$ and is hence a unit in $\mathbb {Z}[1/n, \omega ]$ . We thus obtain a subgroup of $\mathbb {Z}[1/n, \omega ]^\times$ generated by the set of $x_i$ together with $\mathbb {Z}[\omega ]^\times$ . The next proposition computes the rank of this subgroup and uses it to calculate the rank of $\mathbb {Z}[1/n, \omega ]^\times$ .

Proposition 6.9. Let $d$ be square free and $\omega$ as in (6.1). Let $n$ be any nonzero integer and

\begin{equation*}(n) = \mathfrak {p}_1^{\alpha _1} \mathfrak {p}_2^{\alpha _2} \cdots \mathfrak {p}_r^{\alpha _r},\end{equation*}

be the factorization of $(n)$ into prime ideals in $\mathbb {Z}[\omega ]$ . Denote the class number of $\mathbb {Z}[\omega ]$ by $h$ , so that $\mathfrak {p}_i^h = (x_i)$ is principal for each $i$ . Then the following hold.

  1. (1) The subgroup of $\mathbb {Z}[1/n, \omega ]^\times$ generated by the $x_i$ and $\mathbb {Z}[\omega ]^\times$ has the presentation

    (6.2) \begin{equation} \mathbb {Z}[\omega ]^\times \times \langle x_1, x_2, \ldots , x_r \rangle . \end{equation}
  2. (2) We have ${\textrm {rk}}(\mathbb {Z}[1/n, \omega ]^\times ) = \textrm {rk}(\mathbb {Z}[\omega ]^\times \times \langle x_1, x_2, \ldots , x_r \rangle )$ .

In particular,

\begin{equation*} \textrm {rk}(\mathbb {Z}[1/n, \omega ]^\times ) = \textrm {rk}(\mathbb {Z}[\omega ]^\times ) + r = \begin{cases} 1 + r & \text {if } d \gt 0, \\ r & \text {if } d \lt 0. \end{cases} \end{equation*}

Proof. To establish the first claim, we check that the $x_i$ are multiplicatively linearly independent and that there are no nontrivial relations between the $x_i$ and $\mathbb {Z}[\omega ]^\times$ . Suppose we had a linear relation amongst the $x_i$ and $u \in \mathbb {Z}[\omega ]^\times$ , say

\begin{equation*} u x_1^{a_1} \cdots x_q^{a_q} = x_{q+1}^{a_{q+1}} \cdots x_r^{a_r}, \end{equation*}

with all $a_i \geq 0$ . Then

\begin{equation*} \mathfrak {p}_1^{h a_1} \cdots \mathfrak {p}_q^{h a_q} = \mathfrak {p}_{q+1}^{h a_{q+1}} \cdots \mathfrak {p}_r^{h a_r}, \end{equation*}

contradicting unique prime factorization of ideals. This shows that the $x_i$ are linearly independent, even modulo multiplication by any unit in $\mathbb {Z}[\omega ]$ .

We now prove the second claim by relying on the following fact: if $H \leq G$ is a subgroup of an abelian group $G$ and there is an integer $n$ such that $x^n \in H$ for all $x \in G$ , then $H$ and $G$ have the same rank. We will therefore show that that if $x \in \mathbb {Z}[1/n, \omega ]^\times$ , then $x^h$ lies in (6.2), where $h$ is the class number of $\mathbb {Z}[\omega ]$ . Without loss of generality, we may assume that $x$ is an element of the saturation of $\{n^k\}$ in $\mathbb {Z}[\omega ]$ , as these generate $\mathbb {Z}[1/n, \omega ]^\times$ . Then $x$ divides some $n^k$ , which implies that $(x)$ divides $(n^k)$ as an ideal in $\mathbb {Z}[\omega ]$ . But by prime factorization of ideals, this means that $(x)$ is a product of powers of $\mathfrak {p}_1, \ldots , \mathfrak {p}_r$ . Raising both sides to the $h$ th power, we see that $(x)^h=(x^h)$ is a product of powers of the $(x_i)$ . Hence (up to multiplication by a unit in $\mathbb {Z}[\omega ]$ ) we have that $x^h$ is a product of powers of the $x_i$ , as desired.

The final claim of the proposition follows from the observation that

\begin{equation*} \textrm {rk}(\mathbb {Z}[\omega ]^\times \times \langle x_1, x_2, \ldots , x_r \rangle ) = \textrm {rk}(\mathbb {Z}[\omega ]^\times ) + r = \begin{cases} 1 + r & \text {if } d \gt 0, \\ r & \text {if } d \lt 0. \end{cases} \end{equation*}

Here, we have used the fact that the rank of $\mathbb {Z}[\omega ]^\times$ is zero or one, depending on whether $d \lt 0$ or $d \gt 0$ ; recall Theorem6.3.

7. Enumerating unitary units

We are now ready to study the group $U(\mathbb {Z}[t^{\pm 1}]/(\Delta _K))$ when the Alexander polynomial $\Delta _K$ is quadratic. We structure our casework according to the possible Alexander polynomials which arise. In more detail, we set

\begin{equation*} \Delta _n = nt^2 - (2n + 1) t + n,\end{equation*}

since, as $n$ varies over $\mathbb {Z}$ , this encompasses (up to multiplication by $\pm t^k, k \in \mathbb {Z}$ ) all possible quadratic Alexander polynomials. For each of the above Alexander polynomials, the number of $\mathbb {Z}$ -disks for any corresponding knot is deduced by considering the ring

\begin{equation*} \Lambda _n = \mathbb {Z}[t^{\pm 1}]/(\Delta _n). \end{equation*}

More precisely, we saw in Theorem1.2 that the number of $\mathbb {Z}$ -disks in $({\mathbb {C}} P^2)^\circ$ for a knot with polynomial $\Delta _n$ is in bijection with $U_n/\{t^k\}_{k \in \mathbb {Z}}$ , where $U_n$ is the group of unitary units of $\Lambda _n$ :

\begin{equation*} U(\Lambda _n)=\lbrace p \in \Lambda _n \mid p\overline {p}=1 \rbrace .\end{equation*}

Here, recall that $\overline {p}$ is the conjugate of $p$ formed by precomposing with the involution $t \mapsto t^{-1}$ .

We will primarily be interested in determining whether $U_n/\{t^k\}_{k \in \mathbb {Z}}$ has zero or positive rank. In the case that the group is finite, we will make an explicit count of the number of unitary units; in the case that the group is infinite, we compute the rank. The answer will involve

\begin{equation*} \Omega (n):=\# \lbrace \text {positive primes dividing } n \rbrace .\,\end{equation*}

The next theorem is our main number-theoretic result and is Theorem1.7 from the introduction.

Theorem 1.7. For every $n \in \mathbb {Z}$ , the group $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ can be described as follows.

  1. (1) The group $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ is finite precisely when $n = 2, 1, 0, -1$ , or $-p^k$ for a prime $p$ and

    \begin{equation*} U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}} \cong \begin{cases} \lbrace 1 \rbrace & \text {if $n=-1, 0$}, \\ \mathbb {Z}/2\mathbb {Z} & \text {if $n=1,2$ or $n=-p^k$ with $k$ odd}, \\ \mathbb {Z}/4\mathbb {Z} & \text {if $n=-p^k$ with $k$ even}. \\ \end{cases} \end{equation*}
  2. (2) In all other cases, $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ is infinite and

    \begin{equation*} {\textrm {rk}} ( U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}) =\begin{cases} \Omega (n) & \text {if $\Delta _n$ is irreducible and $n\gt 0$}, \\ \Omega (n)-1 & \text {if $\Delta _n$ is irreducible and $n\lt 0$ or if $\Delta _n$ is reducible.} \end{cases} \end{equation*}

Proof. Proposition7.12 is concerned with the cases where $\Delta _n$ is irreducible and the count is infinite. Propositions 7.14 and 7.19 carry out the count when $\Delta _n$ is irreducible and the group is finite $(n=-1,0,1,-p^k$ with $p$ prime). Proposition 7.25 considers the case where $\Delta _n$ is reducible leaving only the case where $\Delta _n$ is reducible and the group is finite ( $n=2$ ) which is studied in Proposition 7.26.

Remark 7.1. We collect some remarks on this theorem.

  1. The polynomial $\Delta _n$ is reducible if and only if the discriminant $4n + 1$ of $\Delta _n$ is a perfect square. We also note that all $\Delta _n$ with $n \lt 0$ are irreducible and that, for $n \gt 0$ , the polynomial $\Delta _n$ is reducible precisely when $n = m(m+1)$ for some $m$ .

  2. The cases in which $(2)$ give a rank less than or equal to zero correspond precisely to the cases in $(1)$ . We use the convention that strictly negative rank should be interpreted as zero rank; we write this for simplicity of the theorem statement.

Before embarking on the proof of Theorem1.7, we give some basic examples of distinct elements in $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ for various values of $n$ , both for concreteness and because of their relevance to topological applications. We first show that $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ is at least of cardinality two except in the cases $n = 0$ and $n = -1$ .

Proposition 7.2. The classes of $+1$ and $-1$ in $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ are distinct except in the cases $n = 0$ and $n = -1$ .

Proof. Suppose that the classes of $+1$ and $-1$ in $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ coincide. Multiplying through by powers of $t$ as necessary, we then have the congruence of polynomials

\begin{equation*} t^a \cdot (+1) \equiv t^b \cdot (- 1) \bmod \Delta _n. \end{equation*}

Clearly, this forces the leading coefficient of $\Delta _n$ to be $\pm 1$ . Moreover, the roots of $\Delta _n$ must lie on the unit circle since $\Delta _n$ divides $t^a + t^b$ ; this rules out the case $n = 1$ . On the other hand, when $n = -1$ it is indeed the case that

\begin{equation*} t^3 \equiv -1 \bmod \Delta _{-1}, \end{equation*}

since $\Delta _{-1} = - t^2 + t - 1$ divides $t^3 + 1$ . In the case $n = 0$ , the ring $\Lambda _n = \mathbb {Z}[t^{\pm 1}]/(1)$ is trivial and the claim is evident.

Remark 7.3. Using Proposition 7.2 and Theorem 1.7, it is straightforward to determine the group

\begin{equation*} U(\Lambda _n)/\{\pm t^k\}_{k \in \mathbb {Z}}, \end{equation*}

discussed in Remark 1.8. Explicitly, consider the subgroup $\{\pm 1\}$ of $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ . If $n = -1$ , then $+1$ and $-1$ actually constitute the same class and hence this subgroup is trivial; in this case $U(\Lambda _n)/\{\pm t^k\}_{k \in \mathbb {Z}} = U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ . Otherwise, we have that $U(\Lambda _n)/\{\pm t^k\}_{k \in \mathbb {Z}}$ is a quotient of $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ by a subgroup of order two. The groups enumerated in Theorem 1.7 then necessarily give those listed in Remark 1.8. In the case of positive rank, it is clear that the rank remains unchanged by this further quotient.

Remark 7.4. It is not difficult to generalize Proposition 7.2 to other Alexander polynomials. Indeed, let $\Delta$ be any (not necessarily quadratic) Alexander polynomial and $\Lambda := \mathbb {Z}[t^{\pm 1}]/(\Delta )$ . Then the classes of $+1$ and $-1$ in $U(\Lambda )/\{t^k\}_{k \in \mathbb {Z}}$ coincide if and only if $\Delta$ (up to multiplication by $t^k$ ) divides $t^\ell + 1$ for some $\ell$ .

The following presents a slightly less trivial set of distinct elements in the case that $n = -k^2$ for some $k \gt 1$ .

Proposition 7.5. Let $n = - k^2$ for $k \gt 1$ . Then the classes of

\begin{equation*} \{\pm 1, \pm k(t - 1)\}, \end{equation*}

are distinct elements of $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ .

Proof. We have already seen in Proposition 7.2 that (the classes of) $\pm 1$ represent distinct unitary units in $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ . It is clear that (the class of) $k(t - 1)$ is a unitary unit, since

\begin{equation*} k(t-1) \cdot k(t^{-1} - 1) = k^2(-t +2 - t^{-1}) = nt - 2n + nt^{-1}. \end{equation*}

Up to multiplication by powers of $t$ , this is congruent to $1$ modulo $\Delta _n = nt^2 - (2n + 1)t + n$ . Furthermore, we claim that $k(t - 1)$ is not in the same class as $\pm 1$ . Indeed, suppose this were the case. Multiplying through by powers of $t$ as necessary, we then have the congruence of polynomials

\begin{equation*} t^a \cdot k(t-1) \equiv t^b \cdot (\pm 1) \bmod \Delta _n. \end{equation*}

Since the leading coefficient of $\Delta _n$ is greater in absolute value than the leading coefficient of $t^a \cdot k(t - 1) - t^b \cdot (\pm 1)$ , it is clear that this is impossible. The case of $- k(t - 1)$ is analogous. Note that $k(t - 1)$ and $- k(t-1)$ are not in the same class, since this would imply $-1$ lies in the trivial class.

Remark 7.6. According to Theorem 1.7, the four elements presented in Proposition 7.5 constitute the entirety of $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ if $n$ is $- p^k$ for $k$ even. In the general case of Proposition 7.5, these four elements are still distinct, but do not necessarily constitute all of $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ .

7.1 Strategy of the proof in the irreducible case

The proof when $\Delta _n$ irreducible is significantly longer than in the reducible case. We therefore begin with a brief outline of this former case. Fix $n \in \mathbb {Z}$ . Throughout this section, let

\begin{equation*} a = \dfrac {(2n+1) + \sqrt {4n+1}}{2} \quad \text {and} \quad \xi = \dfrac {1 + \sqrt {4n+1}}{2}. \end{equation*}

We assume throughout the section that $\Delta _n$ is irreducible. The first step of the proof will be to construct a ring isomorphism

\begin{equation*} f \colon \Lambda _n = \mathbb {Z}[t^{\pm 1}]/(\Delta _n) \rightarrow \mathbb {Z}[1/n, \xi ]. \end{equation*}

Both the domain and the codomain of $f$ are equipped with conjugation involutions giving rise to norm maps.

Definition 7.7. The conjugation map $\overline {\cdot }$ on $\mathbb {Z}[t^{\pm 1}]$ is defined by sending $p(t)$ to $p(t^{-1})$ ; this descends to a conjugation map on $\mathbb {Z}[t^{\pm 1}]/(\Delta _n)$ . We thus obtain a group homomorphism

\begin{align*} N_{\Lambda } \colon \Lambda _n^\times &\to \Lambda _n^\times \\ p &\mapsto p\overline {p}. \end{align*}

The group of unitary units in $\Lambda _n$ is defined to be $U(\Lambda _n)=\ker (N_{\Lambda })$ . We likewise have a conjugation map $\overline {\cdot }$ on $\mathbb {Z}[1/n, \xi ]$ which is trivial on the subring $\mathbb {Z}[1/n]$ and satisfies

\begin{equation*}\overline {\xi }=\frac {1-\sqrt {4n+1}}{2}.\end{equation*}

This defines a group homomorphism:

\begin{align*} N \colon \mathbb {Z}[1/n, \xi ]^\times &\to \mathbb {Z}[1/n]^\times \\ x &\mapsto x\overline {x}. \end{align*}

The group of unitary units in $\mathbb {Z}[1/n, \xi ]$ is defined to be $U(\mathbb {Z}[1/n, \xi ]) = \ker (N)$ .

The following proposition constructs $f$ and shows that it intertwines the conjugation involution on $\Lambda _n$ with the conjugation involution on $\mathbb {Z}[1/n, \xi ]$ . This will show that $f$ gives an isomorphism between $U(\Lambda _n)$ and $U(\mathbb {Z}[1/n, \xi ])$ .

Proposition 7.8. Assume $\Delta _n$ is irreducible. Extending the assignment $t \mapsto a/n$ to a ring homomorphism defines a surjection

\begin{equation*}f \colon \mathbb {Z}[t^{\pm 1}] \twoheadrightarrow \mathbb {Z}[1/n, \xi ],\end{equation*}

with kernel $(\Delta _n)$ . In particular, this assignment defines an isomorphism

\begin{equation*}f \colon \Lambda _n \xrightarrow {\cong } \mathbb {Z}[1/n, \xi ].\end{equation*}

Moreover, $f$ intertwines conjugation on $\Lambda _n$ with conjugation on $\mathbb {Z}[1/n, \xi ]$ .

Proof. First note that since $a$ and $\xi$ differ by an integer, it is clear that $\mathbb {Z}[1/n, a] = \mathbb {Z}[1/n, \xi ]$ and therefore $f$ does indeed take values in $\mathbb {Z}[1/n, \xi ]$ . Next, observe that since $a/n \cdot \bar {a}/n = 1$ , extending $f$ as a ring homomorphism implies that $f(t^{-1})=\overline {a}/n$ ; here note that $\bar {a}/n = 2 + 1/n - a/n$ is indeed an element of $\mathbb {Z}[1/n, a]$ . This also shows that $f$ intertwines conjugation on $\mathbb {Z}[t^{\pm 1}]$ with conjugation on $\mathbb {Z}[1/n, \xi ]$ . The fact that $\ker (f)=(\Delta _n)$ follows from the claim that $\Delta _n$ is the minimal integral polynomial relation satisfied by $a/n$ ; this holds since $\Delta _n$ is irreducible. To see that $f$ is surjective, note that $f(nt) = a$ and $f(t + t^{-1} - 2) = 1/n$ , so $a$ and $1/n$ are both in ${\textrm {im}}(f)$ .

The next result is a direct consequence of Proposition 7.8 and reduces the calculation of the rank ${\textrm {rk}}(U(\Lambda _n))$ to the calculation of $\textrm {rk}(\mathbb {Z}[1/n, \xi ]^\times )$ and ${\textrm {rk}}({\textrm {im}}(N))$ .

Proposition 7.9. The isomorphism $f$ intertwines the norms on $\Lambda _n$ and $\mathbb {Z}[1/n, \xi ]$ and therefore fits into the following commutative diagram of groups.

In particular, the following equality holds:

\begin{equation*} {\textrm {rk}} (U(\Lambda _n))={\textrm {rk}}(\mathbb {Z}[1/n,\xi ]^\times )-{\textrm {rk}}({\textrm {im}}(N)).\end{equation*}

Proof. The fact that $f$ intertwines the norms follows because $f$ is a homomorphism of rings with involutions:

\begin{equation*}N(f(p))=f(p)\overline {f(p)}=f(p\overline {p})=f(N_{\Lambda }(p)).\end{equation*}

The remaining assertions follow from Proposition 7.8.

In the following sections, we calculate the rank of the groups involved in Proposition 7.9. Once these ranks have been calculated, leading to a calculation of ${\textrm {rk}}(U(\Lambda _n))$ , it is then not difficult to deduce the rank of $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ by studying the image of $\{ t^k \}_{k \in \mathbb {Z}}$ under $f$ .

7.2 Rank calculations when $\Delta _n$ is irreducible

Following the discussion of the previous subsection, we now calculate $\textrm {rk}(\mathbb {Z}[1/n, \xi ]^\times )$ and ${\textrm {rk}}({\textrm {im}}(N))$ . We begin with the former. Proposition 6.9 shows how to calculate $\textrm {rk}(\mathbb {Z}[1/n, \omega ]^\times )$ when $\mathbb {Z}[\omega ]$ is a ring of quadratic integers. However, Proposition 6.9 is not directly applicable to our situation: since $4n + 1$ is not necessarily square free, $\xi$ may not be of the form described in (6.1). To rectify this, let

\begin{equation*} 4n + 1 = c^2 d, \end{equation*}

for $c \gt 0$ and $d$ a square free integer, and define

\begin{equation*} \omega = \dfrac {1 + \sqrt {d}}{2}. \end{equation*}

Note that $\omega$ is of the form described in (6.1), whereas

\begin{equation*} \xi = \dfrac {1 + \sqrt {4n + 1}}{2} = \dfrac {1 + c\sqrt {d}}{2}. \end{equation*}

It is readily checked that $\mathbb {Z}[\xi ]$ is a subring of $\mathbb {Z}[\omega ]$ , and hence that $\mathbb {Z}[1/n, \xi ]^\times$ is a subgroup of $\mathbb {Z}[1/n, \omega ]^\times$ . Our strategy will thus be to prove that the ranks of $\mathbb {Z}[1/n, \xi ]^\times$ and $\mathbb {Z}[1/n, \omega ]^\times$ coincide, and then compute the latter using Proposition 6.9.

Recall that $\Omega (n)$ denotes the number of (positive) distinct primes dividing $n$ in $\mathbb {Z}$ .

Proposition 7.10. If $\Delta _n$ is irreducible, then

\begin{equation*} \textrm {rk}(\mathbb {Z}[1/n, \xi ]^\times ) = \textrm {rk}(\mathbb {Z}[1/n, \omega ]^\times ) = \begin{cases} 1 + 2\Omega (n)& \text {if } n \gt 0, \\ 2\Omega (n) & \text {if } n \lt 0. \end{cases} \end{equation*}

Proof. We establish the claimed equalities in succession. For the first equality, it suffices to prove there exists a natural number $g$ such that if $x_1 \in \mathbb {Z}[1/n, \omega ]^\times$ , then $x_1^g \in \mathbb {Z}[1/n, \xi ]^\times$ . Without loss of generality, we may assume that $x_1$ is an element of the saturation of $\{n^k\}$ in $\mathbb {Z}[\omega ]$ , since these generate $\mathbb {Z}[1/n, \omega ]^\times$ ; recall Lemma 6.6. Then $x_1 x_2 = n^k$ for some $x_2 \in \mathbb {Z}[\omega ]$ and $k \in \mathbb {N}$ .

Consider the quotient ring $\mathbb {Z}[\omega ]/(c)$ . Since $\mathbb {Z}[\omega ]$ is a free $\mathbb {Z}$ -module of rank two (generated by $1$ and $\omega$ ), it is clear that $\mathbb {Z}[\omega ]/(c) \cong (\mathbb {Z}/c\mathbb {Z}) \oplus (\mathbb {Z}/c\mathbb {Z})$ . In particular, the group of units in $\mathbb {Z}[\omega ]/(c)$ is finite, being a subset of a finite ring. We claim that $g$ may be taken to be the order of this group. Rearranging the equation $4n + 1 = c^2d$ and exponentiating gives $(4n)^k = (c^2d - 1)^k$ . Expanding this latter equality and using the fact that $x_1 x_2 = n^k$ gives a linear combination of $x_1$ and $c$ which is equal to $1$ . This means that the class of $x_1$ is a unit in $\mathbb {Z}[\omega ]/(c)$ . Thus $x_1^g = 1$ in $\mathbb {Z}[\omega ]/(c)$ ; that is, $x_1^g \equiv 1 \bmod c \mathbb {Z}[\omega ]$ . A similar statement holds for $x_2^g$ .

We have thus shown that $x_1^g \in 1 + c \mathbb {Z}[\omega ]$ . However, by direct inspection, we clearly have $1 + c\mathbb {Z}[\omega ] \subseteq \mathbb {Z} + c\mathbb {Z}[\omega ] \subseteq \mathbb {Z}[\xi ]$ , so $x_1^g \in \mathbb {Z}[\xi ]$ . Since the same holds for $x_2^g$ , we conclude that $x_1^g$ is an element of the saturation of $\{n^k\}$ in $\mathbb {Z}[\xi ]$ , and hence a unit in $\mathbb {Z}[1/n, \xi ]$ by Lemma 6.6. This completes the proof of the first equality.

We now apply Proposition 6.9 to establish the second equality. Let $n=p_1^{\alpha _1} \cdots p_{\Omega (n)}^{\alpha _{\Omega (n)}}$ be the prime factorization of $n$ in $\mathbb {Z}$ . Note that since $4n + 1 = c^2 d$ , we have that $c$ is odd and hence $c^2 \equiv 1 \bmod 4$ . It follows that $d \equiv 1 \bmod 4$ and we may apply the splitting criterion (Theorem6.4) to determine the factorization of each $(p_i) \subset \mathbb {Z}[\omega ]$ . Clearly, none of the $p_i$ divide $4n+1$ . The equality $4n + 1 = c^2 d$ shows that $d \equiv 1/c^2 \bmod p_i$ , so $d$ is a quadratic residue mod $p_i$ . Hence, by Theorem6.4, each $(p_i) \subset \mathbb {Z}[\omega ]$ decomposes as a product of two distinct prime ideals in $\mathbb {Z}[\omega ]$ . (In the special case $p_i = 2$ , we have $d \equiv 1 \bmod 8$ since necessarily $c^2 \equiv 1 \bmod 8$ .)

Proposition 6.9 applied with $r=2 \Omega (n)$ then shows that

\begin{equation*} \textrm {rk}(\mathbb {Z}[1/n, \omega ]^\times ) = \begin{cases} 1 + 2\Omega (n)& \text {if } n \gt 0, \\ 2\Omega (n) & \text {if } n \lt 0, \end{cases} \end{equation*}

as desired.

We now carry out the last step in our calculation of ${\textrm {rk}}(U(\Lambda _n))$ by determining ${\textrm {rk}}({\textrm {im}}(N))$ .

Proposition 7.11. If $\Delta _n$ is irreducible, then

\begin{equation*}{\textrm {rk}}({\textrm {im}}(N)) = {\textrm {rk}}(\mathbb {Z}[1/n]^\times )=\Omega (n).\end{equation*}

Proof. We establish the claimed equalities in succession. Note that, by Lemma 6.6, $\mathbb {Z}[1/n]^\times$ is generated by the primes in $\mathbb {Z}$ dividing $n$ . (Since $\mathbb {Z}$ is a UFD, it is clear that the saturation of $\{n^k\}$ in $\mathbb {Z}$ is generated by the prime factors of $n$ .) If $p$ is such a prime, then $p^2$ lies in the image of $N$ , since by definition

\begin{equation*} N(p) = p^2. \end{equation*}

It follows that ${\textrm {rk}}({\textrm {im}}(N))={\textrm {rk}}(\mathbb {Z}[1/n]^\times )$ .

To prove the second equality, observe that

\begin{equation*} \mathbb {Z}[1/n]^\times \cong \lbrace \pm 1 \rbrace \times \langle p_1,\ldots ,p_{\Omega (n)}\rangle, \end{equation*}

where $n = p_1^{\alpha _1} \cdots p_{\Omega (n)}^{\alpha _{\Omega (n)}}$ is the prime factorization of $n$ in $\mathbb {Z}$ . Indeed, the linear independence of $p_1, \ldots , p_{\Omega (n)}$ follows immediately from uniqueness of prime factorization. Hence ${\textrm {rk}}(\mathbb {Z}[1/n]^\times )=\Omega (n)$ , as desired.

We now prove the main result of this section.

Theorem 7.12. If $\Delta _n$ is irreducible, then

\begin{equation*} \textrm {rk}(U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}) = \begin{cases} \Omega (n) & \text {if } n \gt 0, \\ \Omega (n) - 1& \text {if } n \lt 0. \end{cases} \end{equation*}

In particular, $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ is finite if and only if either $n=-1,0,1$ or $-p^k$ with $p$ prime.

Proof. Proposition 7.9 shows that

\begin{equation*} {\textrm {rk}} (U(\Lambda _n))={\textrm {rk}}(\mathbb {Z}[1/n,\xi ]^\times )-{\textrm {rk}}({\textrm {im}}(N)).\end{equation*}

Combining Proposition 7.10 with Proposition 7.11 we deduce that

\begin{equation*} \textrm {rk}(U(\Lambda _n))= \begin{cases} \Omega (n)+1 & \text {if } n \gt 0, \\ \Omega (n)& \text {if } n \lt 0. \end{cases} \end{equation*}

It remains to mod out by $\{t^k\rbrace$ . We claim that if $n\neq -1$ , then $\{t^k\}_{k \in \mathbb {Z}}$ is a rank-one subgroup of $U(\Lambda _n)$ . Suppose not. Then there is some $k$ such that $t^k \equiv 1 \bmod \Delta _n$ in $\mathbb {Z}[t^{\pm 1}]$ . This means that $t^k - 1 = \Delta _n \cdot p(t)$ for some Laurent polynomial $p(t)$ . This is only possible if $\Delta _n$ is monic, which shows $n = \pm 1$ . In the case that $n = 1$ , an explicit calculation of $U(\Lambda _n)$ shows that $t$ is of infinite order; see Proposition 7.14 below. This concludes the proof of the claim.

The claim shows that, as long as $n \neq -1$ , quotienting out $U(\Lambda _n)$ by $\{t^k\}_{k \in \mathbb {Z}}$ decreases the rank by one, which immediately gives the theorem. On the other hand, if $n = -1$ , then $\Omega (n)=0$ , so $U(\Lambda _n)$ is already finite and the desired equality is trivial.

7.3 The cases where $\Delta _n$ is irreducible and the count is finite

Throughout this section, we continue assuming that $\Delta _n$ is irreducible. In Theorem7.12 we determined the cases in which the number of disks is infinite; we now focus on the cases in which the count is finite. We first treat the cases $n=-1,0,1$ where the computation of $U(\Lambda _n)$ is immediate, before moving on to the case $n=-p^k$ which requires more involved arguments. It will be helpful to keep in mind the norm map on the set of ideals of $\mathbb {Z}[\omega ]$ .

Definition 7.13. If $\mathfrak {p}$ is an ideal in $\mathbb {Z}[\omega ]$ , we define the ideal norm of $\mathfrak {p}$ by

\begin{equation*}N(\mathfrak {p}):= |\mathbb {Z}[\omega ]/\mathfrak {p}|.\end{equation*}

We refer to [Reference Stewart and TallST16, § 5.3] for further details. Note that this is a multiplicative map from the set of ideals in $\mathbb {Z}[\omega ]$ to $\mathbb {N}$ and that if $\mathfrak {p} = (x)$ is principal, then $N(\mathfrak {p}) = |N(x)|$ . (Here, by $N(x)$ we mean the norm of $x$ constructed as in Definition 7.7.)

We now begin our analysis of $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ in the finite case.

Proposition 7.14. For $n=-1,0,1$ , the group $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ can be described as follows:

\begin{equation*} U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}} \cong \begin{cases} 1 & \text {if } n=-1,0, \\ \mathbb {Z}/2\mathbb {Z}& \text {if } n=1. \end{cases} \end{equation*}

Proof. First let $n = -1$ , in which case $\xi = (1 + \sqrt {-3})/2$ . The ring $\mathbb {Z}[\xi ]$ in this setting is known as the ring of Eisenstein integers; we have

\begin{equation*} \mathbb {Z}[\xi ]^\times = \{\pm 1\} \times \{1, \xi , \xi ^2\}, \end{equation*}

as discussed in the case $d = -3$ of Theorem6.3. Since $\xi \overline {\xi } = 1$ , each of these units is unitary. Using the homomorphism $f$ from Proposition 7.8, we compute that

\begin{equation*}f(t)=\dfrac {a}{n} = \dfrac {1 - \sqrt {-3}}{2} = - \xi ^2.\end{equation*}

It readily follows that $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ is trivial. Next we assume that $n = 0$ . In this case, the ring $\Lambda _n =\mathbb {Z}[t^{\pm 1}]/(1)$ is trivial and therefore so is the group of units.

We conclude with the case $n=1$ , in which case $\xi = (1 + \sqrt {5})/2$ . Then $\xi$ itself is a fundamental unit for $\mathbb {Z}[\xi ]$ and $\mathbb {Z}[\xi ]^\times = \{ \pm \xi ^k \}$ ; see [Reference NeukirchNeu99, § 7 (Exercise 2)]. Since $\xi \overline {\xi } = -1$ , the unitary units in $\mathbb {Z}[\xi ]^\times$ are given by $\{\pm \xi ^{2k}\}$ . Recalling from Proposition 7.8 that $f(t)=a/n$ , the fact that $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}} \cong \mathbb {Z}/2\mathbb {Z}$ now follows from the following calculation:

\begin{equation*}f(t)=\dfrac {a}{n} = \dfrac {3 + \sqrt {5}}{2} = \xi ^2. \end{equation*}

This concludes the proof of the proposition.

Now let $n=-p^k$ . Our main goal will be to explicitly determine $\mathbb {Z}[1/n, \xi ]^\times$ . However, it will be easier to first analyze the supergroup $\mathbb {Z}[1/n, \omega ]^\times$ , since Remark 6.8 can be utilized in this setting. In Lemmas 7.15 and 7.16, we carry out the procedure of Remark 6.8 to determine the structure of $\mathbb {Z}[1/n, \omega ]^\times$ . In Lemmas 7.17 and 7.18, we use our computation of $\mathbb {Z}[1/n, \omega ]^\times$ to subsequently calculate $\mathbb {Z}[1/n, \xi ]^\times$ . It is then straightforward to determine the unitary units and quotient out by $\{t^k\}_{k \in \mathbb {Z}}$ ; this is done in Proposition 7.19.

The first step of Remark 6.8 is to factor $(n)$ into prime ideals in $\mathbb {Z}[\omega ]$ . Lemma 7.15 carries this out and records some identities involving this factorization which will be useful later.

Lemma 7.15. The ideals

\begin{equation*} \mathfrak {p}_1 = (p, \xi ) \quad \text {and} \quad \mathfrak {p}_2 = (p, \overline {\xi }), \end{equation*}

of $\mathbb {Z}[\omega ]$ are prime and satisfy the relations

(7.1) \begin{equation} \mathfrak {p}_1^k = (\xi ), \quad \mathfrak {p}_2^k = (\overline {\xi }), \quad \text {and} \quad \mathfrak {p}_1 \mathfrak {p}_2 = (p). \end{equation}

In particular, $(n) = (-p^k) = \mathfrak {p}_1^k \mathfrak {p}_2^k$ .

Proof. It will be helpful to keep in mind that $\xi = (1 + \sqrt {4n + 1})/2$ satisfies

\begin{equation*} \xi ^2 = n + \xi = -p^k + \xi , \quad \xi + \overline {\xi } = 1, \quad \text {and} \quad \xi \overline {\xi } = -n = p^k. \end{equation*}

We first show $\mathfrak {p}_1 \mathfrak {p}_2 = (p)$ . We compute

(7.2) \begin{equation} \mathfrak {p}_1 \mathfrak {p}_2 = (p^2, p \overline {\xi }, p \xi , \xi \cdot \overline {\xi }) = (p^2, p \overline {\xi }, p \xi , p^k). \end{equation}

This is equal to the ideal $(p)$ . Indeed, all four generators of the right-hand side of (7.2) are multiples of $p$ ; conversely, $p$ is the sum of the second and third generators of (7.2). It additionally follows that $N(\mathfrak {p}_1) N(\mathfrak {p}_2) = N(p) = p^2$ ; hence $N(\mathfrak {p}_1) = N(\mathfrak {p}_2) = p$ and so $\mathfrak {p}_1$ and $\mathfrak {p}_2$ are prime. Moreover, $\mathfrak {p}_1$ and $\mathfrak {p}_2$ are distinct since (for example) the fact that $\xi + \overline {\xi } = 1$ shows $\mathfrak {p}_1 + \mathfrak {p}_2 = \mathbb {Z}[\omega ]$ . Note that the factorization $(p) = \mathfrak {p}_1 \mathfrak {p}_2$ is precisely the outcome of the splitting criterion applied to $(p)$ ; recall Theorem6.4.

We now prove $\mathfrak {p}_1^k = (\xi )$ . We begin by inductively showing that $\mathfrak {p}_1^i = (p^i, \xi )$ for all $i \leq k$ . Indeed, assuming the inductive hypothesis, observe

(7.3) \begin{equation} \mathfrak {p}_1^i = \mathfrak {p}_1^{i - 1} \mathfrak {p}_1= (p^{i-1}, \xi )(p, \xi ) = (p^i, p\xi , p^{i-1}\xi , \xi ^2) = (p^i, p\xi , p^{i-1}\xi , - p^k + \xi ). \end{equation}

We claim that the right-hand side of (7.3) is precisely $(p^i, \xi )$ . Indeed, to see (7.3) is contained in $(p^i, \xi )$ , note that the first generator of (7.3) is $p^i$ itself, the second and third are multiples of $\xi$ , and the last is a linear combination of $p^i$ and $\xi$ . Conversely, to see $(p^i, \xi )$ is contained in (7.3), note that $p^i$ is the first generator of (7.3), while $\xi$ is a linear combination of the first and fourth generators. (In both directions, we use the fact that $i \leq k$ .) In the special case $i = k$ , we moreover have

\begin{equation*} \mathfrak {p}_1^k = (p^k, \xi) = (\xi \overline {\xi }, \xi ) = (\xi ), \end{equation*}

as desired. The claim $\mathfrak {p}_2^k = (\overline {\xi })$ is similar.

According to the algorithm of Remark 6.8, we now need to determine which ideals of the form $\mathfrak {p}_1^{a_1} \mathfrak {p}_2^{a_2}$ are principal. The result is recorded in Lemma 7.16 and establishes the structure of $\mathbb {Z}[1/n, \omega ]^\times$ .

Lemma 7.16. Let $s$ be the least natural number such that $\mathfrak {p}_1^s$ and $\mathfrak {p}_2^s$ are principal Footnote 1 and write

\begin{equation*} \mathfrak {p}_1^s = (y) \quad \text {and} \quad \mathfrak {p}_2^s = (\overline {y}). \end{equation*}

Then the group $\mathbb {Z}[1/n, \omega ]^\times$ is isomorphic to

(7.4) \begin{equation} \mathbb {Z}[\omega ]^\times \times \langle p, y, \overline {y} \ | \ p^s = y \overline {y} \rangle . \end{equation}

Proof. We begin by determining which ideals of the form $\mathfrak {p}_1^{a_1} \mathfrak {p}_2^{a_2}$ are principal. Suppose $(x) = \mathfrak {p}_1^{a_1} \mathfrak {p}_2^{a_2}$ for $a_1, a_2 \geq 0$ and $x \in \mathbb {Z}[\omega ]$ . We assert that, up to multiplication by a unit in $\mathbb {Z}[\omega ]$ , $x$ is a product of $p$ , $y$ , and $\overline {y}$ . Indeed, assume $a_1 \geq a_2$ and write $a_1 - a_2 = gs + h$ for $0 \leq h \lt s$ . Then

\begin{equation*} (x) = \mathfrak {p}_1^{a_1 - a_2} (\mathfrak {p}_1 \mathfrak {p}_2)^{a_2} = \mathfrak {p}_1^{gs + h} (\mathfrak {p}_1 \mathfrak {p}_2)^{a_2} = (y)^g \mathfrak {p}_1^h (p)^{a_2}, \end{equation*}

where we have used $\mathfrak {p}_1^s = (y)$ and $\mathfrak {p}_1 \mathfrak {p}_2 = (p)$ . It is straightforward to check that the product of a principal ideal with a non-principal ideal is non-principal.Footnote 2 Hence $\mathfrak {p}_1^h$ is principal. But then the minimality of $s$ implies $h = 0$ , which shows that $x = uy^gp^{a_2}$ for some unit $u \in \mathbb {Z}[\omega ]^\times$ , as desired. The case $a_2 \geq a_1$ is similar.

The assertion, together with Lemmas 6.6 and 6.7, shows that $\mathbb {Z}[1/n, \omega ]^\times$ is generated by $p$ , $y$ , and $\overline {y}$ , together with the units of $\mathbb {Z}[\omega ]$ . The relation $p^s = y \overline {y}$ holds, since

\begin{equation*} y \overline {y} = N(y) = N(\mathfrak {p}_1)^s = p^s. \end{equation*}

It remains to show that there are no other relations. This follows from the same linear independence argument as in Proposition 6.9. Indeed, suppose that we had a positive-exponent linear relation between $p$ , $y$ , and $\overline {y}$ . Passing to ideals, we obtain a positive-exponent linear relation between $(p) = \mathfrak {p}_1 \mathfrak {p}_2$ , $(y) = \mathfrak {p}_1^s$ , and $(\overline {y}) = \mathfrak {p}_2^s$ . By unique factorization of ideals in $\mathbb {Z}[\omega ]$ , it is clear that every linear relation between $(p)$ , $(y)$ , and $(\overline {y})$ is a multiple of the fundamental relation $(p)^s = (y)(\overline {y})$ . This completes the proof.

We have now computed $\mathbb {Z}[1/n, \omega ]^\times$ . The next step is to determine the subgroup $\mathbb {Z}[1/n, \xi ]^\times$ . Clearly, $\mathbb {Z}[1/n, \xi ]^\times$ contains the elements $\pm 1$ , $p$ , $\xi$ , and $\overline {\xi }$ , since these are all in the saturation of $\{n^k\}$ in $\mathbb {Z}[\xi ]$ . (Note that $\xi \overline {\xi } = -n = p^k$ .) Hence $\mathbb {Z}[1/n, \xi ]^\times$ certainly contains the subgroup of $\mathbb {Z}[1/n, \omega ]^\times$ generated by $\pm 1$ , $p$ , $\xi$ , and $\overline {\xi }$ . The structure of this subgroup is given by the following.

Lemma 7.17. The subgroup of $\mathbb {Z}[1/n, \omega ]^\times$ generated by $\pm 1$ , $p$ , $\xi$ , and $\overline {\xi }$ has presentation

(7.5) \begin{equation} \{\pm 1\} \times \langle p, \xi , \overline {\xi } \ | \ \xi \overline {\xi } = p^k\rangle . \end{equation}

Proof. The claim is almost a tautology, but crucially we must verify that $\xi \overline {\xi } = p^k$ is the only relation. To see this, we appeal to the structure of the ambient group $\mathbb {Z}[1/n, \omega ]^\times$ established in Lemma 7.16. Observe that the natural number $s$ from Lemma 7.16 must divide $k$ . Indeed, write $k = gs + h$ with $0 \leq h \lt s$ ; then $(\xi ) = \mathfrak {p}_1^k = \mathfrak {p}_1^{gs}\mathfrak {p}_1^h = (y)^g \mathfrak {p}_1^h$ . Since the product of a principal ideal with a non-principal ideal is non-principal, we have that $\mathfrak {p}_1^h$ is principal. The minimality of $s$ then forces $h = 0$ . We thus see that

\begin{equation*} \xi = u y^{k/s} \end{equation*}

for some unit $u \in \mathbb {Z}[\omega ]^\times$ . The fact that $\xi \overline {\xi } = p^k$ is the only relation then follows from the fact that all relations between $p$ , $y$ , $\overline {y}$ , and $\mathbb {Z}[\omega ]^\times$ are given in (7.4).

We now come to the computation of $\mathbb {Z}[1/n, \xi ]^\times$ .

Lemma 7.18. The unit group $\mathbb {Z}[1/n, \xi ]^\times$ is the subgroup of $\mathbb {Z}[1/n, \omega ]^\times$ given by

\begin{equation*}\mathbb {Z}[1/n, \xi ]^\times = \{\pm 1\} \times \langle p, \xi , \overline {\xi } \ | \ \xi \overline {\xi } = p^k\rangle .\end{equation*}

Proof. It suffices to prove that $\mathbb {Z}[1/n, \xi ]^\times$ is generated by $\pm 1$ , $p$ , $\xi$ , and $\overline {\xi }$ as the conclusion will then follow from Lemma 7.17. As noted previously, it is clear that $\mathbb {Z}[1/n, \xi ]^\times$ contains (7.5), so it suffices to prove the reverse containment. Let $x \in \mathbb {Z}[1/n, \xi ]^\times$ be arbitrary. Since $\mathbb {Z}[1/n, \xi ]^\times \subset \mathbb {Z}[1/n, \omega ]^\times$ , Lemma 7.16 shows that we may write

\begin{equation*} x = u p^g y^h, \end{equation*}

for some $u \in \mathbb {Z}[\omega ]^\times$ and $g, h \in \mathbb {Z}$ . Recall from the proof of Lemma 7.17 that the exponent $s$ of Lemma 7.16 divides $k$ and that

\begin{equation*} u' \xi = y^{k/s}, \end{equation*}

for some $u' \in \mathbb {Z}[\omega ]^\times$ . Write $h = (k/s) t + r$ for $0 \leq r \lt k/s$ . Then

\begin{equation*} x = u p^g y^{(k/s)t + r} = u'' p^g \xi ^t y^r = (u'' y^r) (p^g \xi ^t), \end{equation*}

for $u'' = u (u')^t \in \mathbb {Z}[\omega ]^\times$ . Note that $p^g \xi ^t$ already lies in (7.5). It thus suffices to prove that $u'' y^r$ must be $\pm 1$ . For this, we will utilize the following technical claim.

Claim 5. Let $z$ be an element of $\mathbb {Z}[\omega ]$ with

\begin{equation*} N(z) = p^i, \end{equation*}

for $0 \leq i \lt k$ . Suppose moreover that $p^\ell z \in \mathbb {Z}[\xi ]$ for some $\ell \geq 0$ . Then $i$ is even and $z = \pm p^{i/2}$ .

Proof. Write $z = a + b \omega$ for $a, b \in \mathbb {Z}$ . First, we analyze the consequences of the condition $p^\ell z \in \mathbb {Z}[\xi ]$ , before moving on to the condition $N(z) = p^i$ .

The requirement that $p^\ell z \in \mathbb {Z}[\xi ]$ gives

\begin{equation*} p^\ell a + p^\ell b \omega = e + f \xi , \end{equation*}

for $e, f \in \mathbb {Z}$ . Using the definitions of $\omega$ and $\xi$ , we compute

\begin{equation*} \xi = \dfrac {1 + c\sqrt {d}}{2} = c \left (\dfrac {1 + \sqrt {d}}{2} \right ) - \dfrac {c-1}{2} = c \omega - \dfrac {c-1}{2}. \end{equation*}

Substituting this expression into the right-hand side above and matching coefficients of $\omega$ shows that $p^\ell b = cf$ . Hence $c$ divides $p^\ell b$ . Since $-4 p^k + 1 = c^2 d$ , we have that $c$ and $p$ are coprime; thus $c$ divides $b$ . We deduce that, for some $g \in \mathbb {Z}$ , we have

\begin{equation*} b = gc. \end{equation*}

We now consider that, for $z=a+b\omega$ , our assumption that $N(z)=p^i$ is equivalent to

(7.6) \begin{equation} a^2 + ab + b^2 \left (\dfrac {1-d}{4}\right ) = p^i ,\end{equation}

over $\mathbb {Z}$ . The discriminant of (7.6) (viewed as a quadratic equation in $a$ ) is given by

\begin{equation*} b^2 - 4\left (b^2 \left (\dfrac {1-d}{4}\right ) - p^i\right ) = b^2 d + 4p^i = g^2(-4p^k + 1) + 4p^i, \end{equation*}

where in the last equality we have substituted $b = gc$ and used the fact that $-4 p^k + 1 = c^2 d$ . In order for $N(a + b \omega ) = p^i$ , the discriminant of (7.6) must certainly be non-negative. However, if $0 \leq i \lt k$ , then this is readily seen to be impossible unless $g = 0$ . It follows that $(a, b) = (\pm p^{i/2}, 0)$ , and so $z=a+b\omega =\pm p^{i/2}$ , concluding the proof of the claim.

We now verify that $z = u'' y^r$ satisfies the hypotheses of the claim. Indeed,

\begin{equation*} N(z) = N(y)^r = p^{rs}. \end{equation*}

Since $0 \leq r \lt k/s$ , we have $0 \leq rs \lt k$ . Moreover, recall that $x = z(p^g \xi ^t)$ . Since $x$ and $p^g \xi ^t$ both lie in $\mathbb {Z}[1/n, \xi ]^\times$ , it follows that $z$ does as well. Hence $z$ is equal to an element of $\mathbb {Z}[\xi ]$ divided by some power of $n = -p^k$ . The claim then shows that $z = \pm p^{(rs)/2}$ , which implies

\begin{equation*} u'' y^r = \pm p^{(rs)/2}. \end{equation*}

However, the fact that there are no nontrivial relations between $y$ , $p$ , and $\mathbb {Z}[\omega ]^\times$ in $\mathbb {Z}[1/n, \omega ]^\times$ then forces $z = \pm 1$ . Thus $x=(u'' y^r) (p^g \xi ^t)=\pm p^g \xi ^t$ and we conclude that $x$ lies in the subgroup (7.5), completing the proof.

Putting everything together gives the desired count of units.

Proposition 7.19. For $n=-p^k$ , the group $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ can be described as follows:

\begin{equation*} U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}} \cong \begin{cases} \mathbb {Z}/2\mathbb {Z} & \text {if $k$ is odd, } \\ \mathbb {Z}/4\mathbb {Z} & \text {if $k$ is even.} \end{cases} \end{equation*}

Proof. Lemma 7.18 states that

\begin{equation*} \mathbb {Z}[1/n, \xi ]^\times = \{\pm 1\} \times \langle p, \xi , \overline {\xi } \ | \ \xi \overline {\xi } = p^k\rangle . \end{equation*}

We now determine $U(\Lambda _n)$ by understanding the kernel of $N$ : indeed, recall from Propositions 7.8 and 7.9 that the isomorphism $f$ induces a group isomorphism

\begin{equation*}U(\Lambda _n) \cong U(\mathbb {Z}[1/n, \xi ]) = \ker (N).\end{equation*}

Note that $N(p) = p^2$ , while $N(\xi ) = N(\overline {\xi }) = p^k$ . It follows that there are two possible cases. If $k$ is odd, then $U(\Lambda _n)$ is generated (up to multiplication by $\pm 1$ ) by $\xi /\overline {\xi }$ . If $k$ is even, then $U(\Lambda _n)$ has a pair of natural potential generators, given by $\xi /\overline {\xi }$ and $\xi /p^{k/2}$ . However, the square of the second generator is given by $\xi ^2/p^k = \xi ^2/(\xi \overline {\xi })$ , and is thus equal to the first. Hence in both cases, we have $U(\Lambda _n) = \{\pm 1\} \times \mathbb {Z}$ , but there is a slight difference in the generating element: when $k$ is odd $\xi /\overline {\xi }$ generates, whereas when $k$ is even $\xi /p^{k/2}$ generates. An explicit calculation shows that the isomorphism $f$ satisfies

\begin{equation*} f(t)= \dfrac {2n +1 + \sqrt {4n +1}}{2n} = \dfrac {\xi ^2}{n} = - \dfrac {\xi }{\overline {\xi }}. \end{equation*}

Hence in the odd case $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ is readily seen to be isomorphic to $\mathbb {Z}/2\mathbb {Z}$ , while in the even case it is isomorphic to $\mathbb {Z}/4\mathbb {Z}$ .

7.4 The case where $\Delta _n$ is reducible

We now turn to the case when $\Delta _n$ is reducible. It is straightforward to check that, in this setting,

\begin{equation*} \Delta _n = ((m+1)t - m)(mt - (m+1)), \end{equation*}

where $n = m(m+1)$ and $m \in \mathbb {N}$ ; recall Remark 7.1.

Our overall strategy will be similar to the irreducible case, except that a different ring will take the place of $\mathbb {Z}[1/n, \xi ]$ . The first goal will be to describe the ring $\Lambda _n$ when $\Delta _n$ is reducible. The outcome, which is stated in Proposition 7.23, is that $\Lambda _n$ can be identified with a certain subring of $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ . Thanks to this identification we are able to identify $U(\Lambda _n)$ with a concrete subset of $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ (Proposition 7.24) and calculate its rank (Proposition 7.25). The outcome is that $U(\Lambda _n)/\{ t^k \}$ is finite if and only if $n=2$ , in which case $U(\Lambda _n)/\{ t^k \} \cong \mathbb {Z}/2\mathbb {Z}$ (Proposition 7.26).

Construction 7.20. We construct a ring homomorphism

\begin{equation*}f=f_1 \oplus f_2\colon \mathbb {Z}[t^{\pm 1}] \rightarrow \mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n],\end{equation*}

by extending the following assignments:

\begin{equation*} f_1(t) = \dfrac {m}{m+1} \quad \text {and} \quad f_2(t) = \dfrac {m+1}{m}. \end{equation*}

Note that both $f_1$ and $f_2$ are individually surjective since $f_i(t + t^{-1} - 2)=1/n$ for $i=1,2$ .

Definition 7.21. Endow the ring $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ with the conjugation involution $\overline {(a,b)} = (b, a)$ . This defines a norm map

\begin{align*} N \colon (\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n])^\times &\to (\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n])^\times \\ (a,b) &\mapsto (a,b)\overline {(a,b)} = (ab, ab), \end{align*}

which is readily seen to be a group homomorphism.

The next lemma uses the homomorphism $f$ to describe $\Lambda _n$ as a subring of $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ .

Lemma 7.22. The homomorphism $f$ from Construction 7.20 has kernel $(\Delta _n)$ and thus induces an injective homomorphism

\begin{equation*} f\colon \Lambda _n \hookrightarrow \mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n].\end{equation*}

Moreover, $f$ intertwines the norms on $\Lambda _n$ and $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ and therefore fits into the following commutative diagram of groups.

Proof. A short verification shows that

\begin{equation*} \ker (f_1) = ((m+1)t - m) \quad \text {and} \quad \ker (f_2) = (mt - (m+1)). \end{equation*}

Since $\mathbb {Z}[t^{\pm 1}]$ is a UFD and the polynomials $(m+1)t - m$ and $mt - (m+1)$ have no common factors, we deduce that

\begin{align*} \ker (f) &= \ker (f_1) \cap \ker (f_2) = ( ((m+1)t - m))(mt - (m+1))) = (\Delta _n). \end{align*}

It is straightforward to show that $f$ intertwines conjugation on $\mathbb {Z}[t, t^{-1}]$ with conjugation on $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ . The claim then follows as in the proof of Proposition 7.9.

The following proposition determines the image of $f$ , thus concluding our description of $\Lambda _n$ as a subring of $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ .

Proposition 7.23. The image of $f$ agrees with the following subring of $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ :

\begin{equation*} S = \{(a, b) \in \mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n] \ | \ b - a \in (2m+1) \cdot \mathbb {Z}[1/n] \text { for some } m \in \mathbb {Z}\}. \end{equation*}

Proof. Let $D$ be the diagonal $\{(a,a)\}$ of $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ . Note that $D \subset {\textrm {im}}(f)$ : this follows by observing that $(1/n, 1/n)=f(t + t^{-1} - 2)$ and taking the image of linear combinations of powers of $t + t^{-1} - 2$ .

We claim that, for every $b \in \mathbb {Z}[1/n]$ , we have $(0,b) \in {\textrm {im}}(f)$ if and only if $(0,b) \in S$ . Suppose $(0, b) = f(p)$ for some $p \in \Lambda$ . Then $p \in \ker (f_1)$ , so $p$ is a multiple of $(m+1)t - m$ . Now,

\begin{equation*} f((m+1)t - m) = \left (0, (m+1)\dfrac {m+1}{m} - m \right ) = \left (0, \dfrac {2m+1}{m}\right ). \end{equation*}

Thus, $b$ is $(2m+1)/m$ times an element of $\mathbb {Z}[1/n]$ . As $1/m = (m+1)/n$ , we see that $b$ is $2m+1$ times an element of $\mathbb {Z}[1/n]$ , as desired. Conversely, suppose that $b = (2m+1) \cdot c$ for some $c \in \mathbb {Z}[1/n]$ . Let $p$ be the linear combination of powers of $t + t^{-1} - 2$ mapping to $(c, c) \in D$ . Then $f$ maps $((m+1)t - m) \cdot mp$ to $(0, b)$ , as desired. This concludes the proof of the claim.

We now conclude that ${\textrm {im}}(f)=S$ . Since $D \subset {\textrm {im}}(f)$ , subtracting off $(a, a) \in D$ shows that $(a, b) \in {\textrm {im}}(f)$ if and only if $(0, b - a) \in {\textrm {im}}(f)$ . Thanks to the claim, this is equivalent to $(0,b-a) \in S$ . Since the condition of being in $S$ is clearly invariant under adding elements of $D$ , this is in turn equivalent to $(a,b) \in S$ .

Now that we have described $\Lambda _n$ as a subring of $\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n]$ , we are able to understand its group of unitary units using the injection $f \colon \Lambda \hookrightarrow \mathbb {Z}[1/n]\oplus \mathbb {Z}[1/n]$ and the projection $\textrm {pr}_1 \colon \mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n] \to \mathbb {Z}[1/n]$ onto the first component.

Proposition 7.24. If $\Delta _n$ is reducible, then the composition $\textrm {pr}_1 \circ f$ induces an isomorphism

\begin{equation*} U(\Lambda _n) \cong T := \{x \in \mathbb {Z}[1/n] \ | \ x \text { is a unit in } \mathbb {Z}[1/n] \text { and } x - x^{-1} \in (2m+1) \cdot \mathbb {Z}[1/n]\}. \end{equation*}

Proof. Since the norm on $(\mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n])^\times$ is given by $N(a,b)=(ab,ab)$ , Proposition 7.23 implies that $f$ induces a group isomorphism

\begin{align*} U(\Lambda ) &\cong \{(a,b) \in S^\times \ | \ N(a, b) = (ab, ab) = (1, 1)\} \\ &= \{(x,x^{-1}) \in \mathbb {Z}[1/n] \oplus \mathbb {Z}[1/n] \ | \ x \text { is a unit in } \mathbb {Z}[1/n] \text { and } x - x^{-1} \in (2m+1) \cdot \mathbb {Z}[1/n]\}. \end{align*}

The proposition now follows because $\textrm {pr}_1$ maps this group isomorphically onto $T$ .

We now complete the proof of Theorem1.7 for reducible $\Delta _n$ and $n \neq 2.$

Proposition 7.25. If $\Delta _n$ is reducible, then $\textrm {rk}(U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}) = \Omega (n) - 1$ . In particular, $\textrm {rk}(U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}})$ is finite if and only if $n=2$ .

Proof. We first show that

\begin{equation*} \textrm {rk}(T) = \textrm {rk}(\mathbb {Z}[1/n]^\times ) = \Omega (n). \end{equation*}

The second equality was established in Proposition 7.11. Since we proved in Proposition 7.11 that

\begin{equation*} \mathbb {Z}[1/n]^\times \cong \lbrace \pm 1 \rbrace \times \langle p_1,\ldots ,p_{\Omega (n)} \rangle , \end{equation*}

where $n = p_1^{\alpha _1},\ldots ,p_{\Omega (n)}^{\alpha _{\Omega (n)}}$ , the first equality immediately follows from the claim below.

Claim 6. There exists a $g \in \mathbb {N}$ such that, for each prime divisor $p_i$ of $n$ , we have $p_i^{g} \in T$ .

Proof. We need to show that there exists a $g \in \mathbb {N}$ such that $p_i^g \in T$ for every $i=1,\ldots, r$ . By definition of $T$ , this is equivalent to proving the existence of a $g \in \mathbb {N}$ such that $p_i^g-(1/p_i^g) \in (2m+1)\mathbb {Z}[1/n]$ for each $i$ . Observe that

\begin{equation*} p_i^{g} - \frac {1}{p_i^{g}} = \frac {p_i^{2g} - 1}{p_i^{g}}. \end{equation*}

Note that since $p_i^g$ is always in the saturation of $\{n^k\}$ , we have $1/p_i^g \in \mathbb {Z}[1/n]$ . It thus suffices to show that there exists a $g \in \mathbb {N}$ such that $p_i^{2g} - 1 \in (2m+1)\mathbb {Z}[1/n]$ for each $i$ , i.e. $p_i^{2g} \equiv 1$ mod $2m+1$ . Now, $p_i$ is a divisor of either $m$ or $m+1$ , both of which are coprime to $2m + 1$ . Hence $p_i$ is coprime to $2m+1$ , and we can take $g$ to be the cardinality of $(\mathbb {Z}/(2m+1)\mathbb {Z})^\times$ . This concludes the proof of the claim.

Now that we have calculated ${\textrm {rk}}(U(\Lambda _n))$ , we turn to $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ . Tracing through the definition of $f$ and the isomorphism of Proposition 7.24, we see that $\{t^k\}_{k \in \mathbb {Z}}$ corresponds to the subgroup

\begin{equation*} \left \langle \dfrac {m}{m+1}\right \rangle \leq T. \end{equation*}

Note that since $T \subset \mathbb {Z}[1/n]$ , the image of $t$ has infinite order and so ${\textrm {rk}} (\{t^k\}_{k \in \mathbb {Z}})=1$ . It follows that $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ has rank $\Omega (n) - 1$ , as desired. The last sentence of the proposition follows because if $m \gt 1$ , then $n = m(m+1)$ has at least two distinct prime factors.

Thanks to Proposition 7.25, it only remains to determine the count when $n=2$ .

Proposition 7.26. For $n=2$ , the group $U(\Lambda _n)/\{t^k\}_{k \in \mathbb {Z}}$ is isomorphic to $\mathbb {Z}/2\mathbb {Z}$ .

Proof. We calculate this quotient explicitly under the isomorphism of Proposition 7.25, starting with $T$ and then quotienting out by the image of $\{ t^k \}$ . For $n=2$ , we have that $\mathbb {Z}[1/n]^\times$ is generated (up to multiplication by $\pm 1$ ) by $2$ . Since $2 - 1/2 = 3/2 \in 3 \cdot \mathbb {Z}[1/n]$ , we deduce that $U \cong T = \mathbb {Z}[1/n]^\times$ . Since the isomorphism $\textrm {pr}_1 \circ f$ from Proposition 7.25 satisfies $\textrm {pr}_1 \circ f(t)=1/2$ , the quotient $U/\{t^k\}_{k \in \mathbb {Z}}$ is isomorphic to $\mathbb {Z}/2\mathbb {Z}$ .

8. Examples of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$

The goal of this section is to describe examples of smoothly embedded $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ with a focus on attempting to find smooth representatives for the enumeration from Theorem1.3. Smooth representatives will be constructed using the following observation.

Observation 8.1. Let $\gamma$ be a generalized unknotting curve for $K$ such that applying a positive generalized crossing change to $K$ along $\gamma$ yields a smoothly $\mathbb {Z}$ -slice knot. Then the disk $D_\gamma$ is smoothable.

From now on, as in the introduction, we use $K_n$ to denote the twist knot depicted in Figure 2. We recall that the Alexander polynomial of $K_n$ is

\begin{equation*}\Delta _n:=nt-(2n+1)+nt^{-1}.\end{equation*}

We start with the trefoil $K_{-1}$ , which bounds a unique $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ . This disk can be realized smoothly.

Example 8.2. Theorem 1.3 predicts that the trefoil knot $K_{-1}$ bounds a unique $\mathbb {Z}$ -disk in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary. The fact that the disk can be realized smoothly is well known; we illustrate a crossing change curve $\gamma$ for $K_{-1}$ with $D_\gamma$ smoothly embedded (according to Observation 8.1) in Figure 3.

As discussed in § 4, in order to define $D_\gamma$ the unknotting curve $\gamma$ must be oriented. Reversing orientation on $\gamma$ gives a disk $D_{-\gamma }$ which need not be isotopic rel. boundary to $D_\gamma$ . From this perspective, it may seem counterintuitive that $K_{-1}$ bounds a unique $\mathbb {Z}$ -disk. However, in the case of the trefoil, the curves $\gamma$ and $-\gamma$ are in fact isotopic in the complement of $K_{-1}$ , as illustrated in Figure 4. Thus, the disks $D_{\gamma }$ and $D_{-\gamma }$ are even smoothly isotopic rel. boundary.

Figure 3 A positive crossing change curve (bold) for $K_{-1}$ . By Observation 8.1, the resulting disk in $(\mathbb {C}P^2)^\circ$ can be taken to be smoothly embedded.

Figure 4 From left to right, top to bottom, we illustrate an isotopy from $\gamma$ to $-\gamma$ in the complement of $K_{-1}$ .

Remark 8.3. In Proposition 7.2 we show that if $\Delta _K = \Delta _n$ , then the two disks $D_{\pm \gamma }$ are always distinct except in the cases $n = 0$ and $n = -1$ . In these latter cases, $D_{\pm \gamma }$ are always topologically isotopic rel. boundary (for any $\gamma$ ).

Figure 5 An unoriented unknotting curve (bold) for $K_1$ . The two possible choices of orientation yield the two isotopy rel. boundary classes of $\mathbb {Z}$ -disk for $K_1$ in $(\mathbb {C}P^2)^\circ$ .

Figure 6 Left: two surgery curves (bold) $\gamma _1,\gamma _2$ for $K_{n}$ ( $n=-k^2, k\gt 1$ ). Right: adding a positive twist about $\gamma _2$ transforms $K_n$ into the untwisted negative Whitehead double of $T_{k,k-1}$ .

Figure 7 Leftmost: two surgery curves (bold) $\gamma _1,\gamma _2$ for $K_{-4}$ . From left to right: finding lifts of $\gamma _1, \gamma _2$ in $E_{K_{-4}}^\infty$ . Rightmost: a surgery presentation of $E_{K_{-4}}^\infty$ ; a meridian of a surgery curve (dashed) represents the homology class $\pm t^k$ for some $k$ and choosing which meridian represents $1$ determines a basis for $H_1(E_{K_{-4}}^\infty )$ . For one choice of basis, the pictured lift of $\gamma _1$ represents $\pm (4t-8+4t^{-1})\equiv \mp 1$ while the pictured lift of $\gamma _2$ represents $\pm (2-2t)$ .

Figure 8 Top left: two surgery curves (bold) $\gamma _1,\gamma _2$ for $J_{-k^2}$ , $k\gt 1$ , pictured for $k=\,4$ . The square twist box contains $(k-1)^2-k^2=-2k+1$ full twists, so that the pictured knot is the $-k^2$ -twisted Whitehead double $J_{-k^2}$ . Top row, from left to right: we manipulate the original figure until we obtain the bottom left configuration. The rectangular box indicates the double of a $-(k-1)/k$ twist (compare the third and fourth figures in the top row). Bottom right: we find lifts of $\gamma _1, \gamma _2$ (for some choice of basepoint) to $E_{J_{-k^2}}^\infty$ , suppressing surgery framings. In the surgery diagram of $E_{J_{-k^2}}^\infty$ , the Dehn surgery curves are dashed. For one choice of basis of $H_1(E_{J_{-k^2}}^\infty )$ (taking the meridian of one dashed surgery curve to represent $1$ ), the lift of $\gamma _1$ represents $\pm (k^2 t-2k^2+k^2 t^{-1})\equiv \mp 1$ while the lift of $\gamma _2$ represents $\pm (kt-k)$ .

Example 8.4. Theorem 1.3 predicts that the figure eight knot $K_1$ bounds two $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ up to isotopy rel. boundary. Figure 5 shows an unoriented unknotting curve $\gamma$ for $K_1$ . The two possible orientations of $\gamma$ yield $\mathbb {Z}$ -disks for $K_1$ in $(\mathbb {C}P^2)^\circ$ that (by Proposition 7.2) are not isotopic rel. boundary and hence represent the two possible classes of such disks. Since performing a positive crossing change along $\gamma$ yields the unknot, these two disks are smoothly embedded in $(\mathbb {C}P^2)^\circ$ .

Example 8.5. We now consider the twist knots $K_n$ with $n=-k^2$ and $k\gt 1$ ; the case $k = 1$ has been treated in Example 8.2. Theorem 1.3 predicts that, up to isotopy rel. boundary, $K_n$ bounds four disks when $k$ is a prime power, and infinitely many $\mathbb {Z}$ -disks otherwise. We realize two of these disks smoothly and, for $k=2$ , we realize all four of the disks smoothly.

The left-hand side of Figure 7 shows two generalized unknotting curves $\gamma _1$ and $\gamma _2$ for the knot $K_{-k^2}$ . The fact that $\gamma _1$ is an unknotting curve is evident, but we show explicitly in Figure 7 that a positive generalized crossing change along $\gamma _2$ results in an Alexander polynomial one knot. Specifically, performing a generalized crossing change along $\gamma _2$ yields the negative untwisted Whitehead double of the torus knot $T(k,k-1)$ . The disks $D_{\pm \gamma _1}$ are smoothly embedded by Observation 8.1. However, we can only ascertain that $D_{\pm \gamma _2}$ are smoothly embedded when $k=2$ , since in this case $T(k,k-1)$ is the unknot.

We claim that $D_{\pm \gamma _1}$ and $D_{\pm \gamma _2}$ represent four distinct topological isotopy classes. For this, after fixing a basepoint, we determine lifts of $\pm \gamma _1$ and $\pm \gamma _2$ to $E_{K_{-k^2}}^\infty$ . Figure 7 illustrates this for $k=2$ ; the general case is similar. For an appropriate choice of the generator of the Alexander module $H_1(E_{K_{-k^2}}^\infty )$ , these lifts are given by

\begin{align*} [\pm \widetilde {\gamma }_1]&=\pm (nt-2n+nt^{-1}) \equiv \mp 1, \\ [\pm \widetilde {\gamma }_2]&=\pm (kt-k). \end{align*}

In Proposition 7.5, we show that these are distinct elements of $\mathcal {G}_{K_n}/\{ t^\ell \}_{\ell \in \mathbb {Z}}$ . If $k$ is a prime power, then by Theorem1.3, this realizes all of the $\mathbb {Z}$ -disks for $K_n$ up to isotopy rel. boundary.

We now turn to a slightly different example which also has Alexander polynomial $\Delta _{-k^2}$ .

Example 8.6. Let $n=-k^2$ and $k\gt 1$ . In Example 8.5, we could only ascertain that (in the general case) two of the four $\mathbb {Z}$ -disks bounded by $K_n$ were smoothly embedded. Let $J_n$ instead denote the $n$ -twisted negative Whitehead double of the torus knot $T(-k,k-1)$ . The knots $J_n$ and $K_n$ have the same Alexander polynomial because they are $n$ -twisted Whitehead doubles. Since $J_n$ is a genus one knot with $\Delta _{J_n} = \Delta _{K_n}$ , Theorem 1.3 implies that $J_n$ bounds four isotopy rel. boundary classes of $\mathbb {Z}$ -disks in $(\mathbb {C}P^2)^\circ$ if $k$ is a prime power and infinitely many classes otherwise. We argue that, this time, we can realize the four classes smoothly.

Applying a positive twist to either of the curves $\gamma _1$ or $\gamma _2$ in Figure 8 (top left) transforms $J_n$ into the unknot. As illustrated in Figure 8, with an appropriate choice of the generator of the Alexander module $H_1(E_{K_{-k^2}}^\infty )$ , the lifts of $\pm \gamma _1$ and $\pm \gamma _2$ are again given by

\begin{align*} [\pm \widetilde {\gamma }_1]&=\pm (nt-2n+nt^{-1}) \mp \pm 1, \\ [\pm \widetilde {\gamma }_2]&=\pm (kt-k). \end{align*}

Since $\Delta _{J_n} = \Delta _{K_n}$ , an argument identical to that of Example 8.5 shows that these disks are distinct. Once again, if $k$ is a prime power, then by Theorem 1.3, this realizes all of the $\mathbb {Z}$ -disks for $J_n$ up to isotopy rel. boundary.

Acknowledgment

The authors would like to thank Levent Alpöge for his significant help regarding the number-theoretic aspects of this paper.

Conflicts of interest

None.

Financial support

AC was partially supported by the NSF grant no. DMS–2303674. ID was partially supported by the NSF grant no. DMS–2303823. MM was partially supported by a Stanford Science Fellowship and a Clay Research Fellowship.

Journal information

Compositio Mathematica is owned by the Foundation Compositio Mathematica and published by the London Mathematical Society in partnership with Cambridge University Press. All surplus income from the publication of Compositio Mathematica is returned to mathematics and higher education through the charitable activities of the Foundation, the London Mathematical Society and Cambridge University Press.

Footnotes

1 Note that since $\mathfrak {p}_1$ and $\mathfrak {p}_2$ are conjugate, this value of $s$ is the same for both.

2 In general, let $\mathfrak {a}$ , $\mathfrak {b}$ , and $\mathfrak {c}$ be ideals in an integral domain $R$ . Assume $\mathfrak {a} = (a)$ and $\mathfrak {b} = (b)$ are principal and suppose that $\mathfrak {a} = \mathfrak {b} \mathfrak {c}$ . Then $a \in \mathfrak {b}\mathfrak {c}$ , so $a = bc$ for some $c \in \mathfrak {c}$ . We claim that $\mathfrak {c} = (c)$ . To see this, let $c' \in \mathfrak {c}$ be arbitrary. Then $bc' \in \mathfrak {a}$ , so $bc' = ra$ for some $r \in R$ . But then $bc' = r(bc)$ , which implies $c' = rc$ since $R$ is an integral domain.

References

Akbulut, S., A solution to a conjecture of Zeeman, Topology 30 (1991), 513515.Google Scholar
Borodzik, M. and Friedl, S., On the algebraic unknotting number, Trans. London Math. Soc. 1 (2014), 5784.Google Scholar
Borodzik, M. and Friedl, S., The unknotting number and classical invariants, I, Algebr. Geom. Topol. 15 (2015), 85135.Google Scholar
Boyer, S.. Simply-connected 4-manifolds with a given boundary, Trans. Amer. Math. Soc. 298 (1986), 331357 Google Scholar
Boyer, S., Realization of simply-connected, 4-manifolds with a given boundary, Comment. Math. Helv. 68 (1993), 2047.Google Scholar
Cha, J. C., Orr, K. E. and Powell, M., Whitney towers and abelian invariants of knots, Math. Z. 294 (2020), 519553.Google Scholar
Conway, A., Crowley, D. and Powell, M., Infinite homotopy stable class for 4-manifolds with boundary, Pacific J. Math. 325 (2023), 209237.Google Scholar
Conway, A., Crowley, D., Powell, M. and Sixt, J., Stably diffeomorphic manifolds and modified surgery obstructions, Preprint (2021), arXiv:2109.05632.Google Scholar
Conway, A. and Orson, P., Locally flat simple spheres in, $\Bbb {C}P^2$ , bull, Lond. Math. Soc. 57 (2025), 150163.Google Scholar
Conway, A., Piccirillo, L. and Powell, M., $4$ -manifolds with boundary and fundamental group $\mathbb {Z}$ , Comment. Math. Helv. 100 (2023), 323420.CrossRefGoogle Scholar
Conway, A. and Powell, M., Characterisation of homotopy ribbon discs, Adv. Math. 391 (2021), 107960.CrossRefGoogle Scholar
Conway, A. and Powell, M., Embedded surfaces with infinite cyclic knot group, Geom. Topol. 27 (2023), 739821.Google Scholar
Crowley, D. and Sixt, J., Stably diffeomorphic manifolds and $l_{2q+1}(\Bbb Z[\pi ])$ , Forum Math. 23 (2011), 483538.CrossRefGoogle Scholar
Dai, I., Mallick, A. and Stoffregen, M., Equivariant knots and knot floer homology, J. Topol. 16 (2023), 11671236.Google Scholar
Feller, P. and Lewark, L., On classical upper bounds for slice genera, Selecta Math. (N.S.). 24 (2018), 48854916.Google Scholar
Feller, P. and Lewark, L., Balanced algebraic unknotting, linking forms, and surfaces in three- and four-space, J. Differential Geom. 127 (2024), 213275.Google Scholar
Fogel, M. E., Knots with algebraic unknotting number one, Pacific J. Math. 163 (1994), 277295.Google Scholar
Freedman, M. H., The topology of four-dimensional manifolds, J. Differential Geometry 17 (1982), 357453.Google Scholar
Friedl, S., Nagel, M., Orson, P. and Powell, M., A survey of the foundations of four-manifold theory in the topological category, New York J. Math., to appear. Preprint (2019), arXiv:1910.07372.Google Scholar
Friedl, S. and Powell, M., A calculation of Blanchfield pairings of 3-manifolds and knots, Mosc. Math. J. 17 (2017), 5977.Google Scholar
Hatcher, A. E., A proof of the Smale conjecture, $\textrm{Diff}(S^3) \simeq O(4)$ , Ann. of Math. (2) 117 (1983), 553607.Google Scholar
Hayden, K., Exotic ribbon disks and symplectic surfaces, Preprint (2020), arXiv:2003.13681.Google Scholar
Hayden, K. and Sundberg, I., Khovanov homology and exotic surfaces in the 4-ball, J. Reine Angew. Math. 809 (2024), 217246.Google Scholar
Kjuchukova, A., Miller, A. N., Ray, A. and Sakalli, S., Slicing knots in definite 4-manifolds, Trans. Amer. Math. Soc. 377 (2024), 59055946.Google Scholar
Livingston, C. and Moore, A. H.. Knotinfo: Table of Knot Invariants (2024), https://knotinfo.math.indiana.edu.Google Scholar
Milnor, J. W., Infinite cyclic coverings, in Proc. conf. on the topology of manifolds (Michigan State University, East Lansing, MI, 1967) (Prindle, Weber & Schmidt, Boston, MA, 1968), 115133.Google Scholar
Neukirch, J., Algebraic number theory, Grundlehren der mathematischen Wissenschaften [Fundamental principles of mathematical sciences], vol. 322 (Springer-Verlag, Berlin, 1999), translated from the 1992 German original with a note by Norbert Schappacher and a foreword by G. Harder.CrossRefGoogle Scholar
Orson, P. and Powell, M., Mapping class groups of simply connected 4-manifolds with boundary, Preprint (2022), arXiv:2207.05986.Google Scholar
Powell, M., Twisted Blanchfield pairings and symmetric chain complexes, Q. J. Math. 67 (2016), 715742.Google Scholar
Stewart, I. and Tall, D., Algebraic number theory and Fermat’s last theorem, fourth edition (CRC Press, Boca Raton, FL, 2016).Google Scholar
Stong, R., Simply-connected 4-manifolds with a given boundary, Topology Appl. 52 (1993), 161167.Google Scholar
Figure 0

Figure 1 Top: a positive crossing change from $K$ to $K'$ realized as a (generalized) crossing change about a curve $\gamma$ (bold). Bottom: a generalized positive crossing change about $\gamma$ transforms $K$ into $K'$. In this example, $\gamma$ links $K$ geometrically more than two times (although as always links $K$ zero times algebraically).

Figure 1

Figure 2. Left: the twist knot $K_n$ where the box denotes $n$ full twists. Middle: the figure eight knot $K_1$. Right: the right-handed trefoil $K_{-1}$.

Figure 2

Figure 3 A positive crossing change curve (bold) for $K_{-1}$. By Observation 8.1, the resulting disk in $(\mathbb {C}P^2)^\circ$ can be taken to be smoothly embedded.

Figure 3

Figure 4 From left to right, top to bottom, we illustrate an isotopy from $\gamma$ to $-\gamma$ in the complement of $K_{-1}$.

Figure 4

Figure 5 An unoriented unknotting curve (bold) for $K_1$. The two possible choices of orientation yield the two isotopy rel. boundary classes of $\mathbb {Z}$-disk for $K_1$ in $(\mathbb {C}P^2)^\circ$.

Figure 5

Figure 6 Left: two surgery curves (bold) $\gamma _1,\gamma _2$ for $K_{n}$ ($n=-k^2, k\gt 1$). Right: adding a positive twist about $\gamma _2$ transforms $K_n$ into the untwisted negative Whitehead double of $T_{k,k-1}$.

Figure 6

Figure 7 Leftmost: two surgery curves (bold) $\gamma _1,\gamma _2$ for $K_{-4}$. From left to right: finding lifts of $\gamma _1, \gamma _2$ in $E_{K_{-4}}^\infty$. Rightmost: a surgery presentation of $E_{K_{-4}}^\infty$; a meridian of a surgery curve (dashed) represents the homology class $\pm t^k$ for some $k$ and choosing which meridian represents $1$ determines a basis for $H_1(E_{K_{-4}}^\infty )$. For one choice of basis, the pictured lift of $\gamma _1$ represents $\pm (4t-8+4t^{-1})\equiv \mp 1$ while the pictured lift of $\gamma _2$ represents $\pm (2-2t)$.

Figure 7

Figure 8 Top left: two surgery curves (bold) $\gamma _1,\gamma _2$ for $J_{-k^2}$, $k\gt 1$, pictured for $k=\,4$. The square twist box contains $(k-1)^2-k^2=-2k+1$ full twists, so that the pictured knot is the $-k^2$-twisted Whitehead double $J_{-k^2}$. Top row, from left to right: we manipulate the original figure until we obtain the bottom left configuration. The rectangular box indicates the double of a $-(k-1)/k$ twist (compare the third and fourth figures in the top row). Bottom right: we find lifts of $\gamma _1, \gamma _2$ (for some choice of basepoint) to $E_{J_{-k^2}}^\infty$, suppressing surgery framings. In the surgery diagram of $E_{J_{-k^2}}^\infty$, the Dehn surgery curves are dashed. For one choice of basis of $H_1(E_{J_{-k^2}}^\infty )$ (taking the meridian of one dashed surgery curve to represent $1$), the lift of $\gamma _1$ represents $\pm (k^2 t-2k^2+k^2 t^{-1})\equiv \mp 1$ while the lift of $\gamma _2$ represents $\pm (kt-k)$.