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Embedding spheres in knot traces

Published online by Cambridge University Press:  20 October 2021

Peter Feller
Department of Mathematics, ETH Zürich, Switzerland
Allison N. Miller
Department of Mathematics, Rice University, Houston, TX, USA
Matthias Nagel
Department of Mathematics, ETH Zürich, Switzerland
Patrick Orson
Department of Mathematics, Boston College, Chestnut Hill, USA
Mark Powell
Department of Mathematical Sciences, Durham University, UK
Arunima Ray
Max-Planck-Institut für Mathematik, Bonn, Germany
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The trace of the $n$-framed surgery on a knot in $S^{3}$ is a 4-manifold homotopy equivalent to the 2-sphere. We characterise when a generator of the second homotopy group of such a manifold can be realised by a locally flat embedded $2$-sphere whose complement has abelian fundamental group. Our characterisation is in terms of classical and computable $3$-dimensional knot invariants. For each $n$, this provides conditions that imply a knot is topologically $n$-shake slice, directly analogous to the result of Freedman and Quinn that a knot with trivial Alexander polynomial is topologically slice.

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This is an Open Access article, distributed under the terms of the Creative Commons Attribution-NonCommercial licence (, which permits noncommercial re-use, distribution, and reproduction in any medium, provided the original article is properly cited. Written permission must be obtained prior to any commercial use. Compositio Mathematica is © Foundation Compositio Mathematica.
© 2021 The Author(s)

1. Introduction

Question Let $M$ be a compact topological $4$-manifold and let $x \in \pi _2(M)$. Can $x$ be represented by a locally flat embedded $2$-sphere?

Versions of this fundamental question have been studied by many authors, such as [Reference Kervaire and MilnorKM61, Reference TristramTri69, Reference RohlinRoh71, Reference Hsiang and SzczarbaHS71]. The seminal work of Freedman and Quinn [Reference FreedmanFre82, Reference Freedman and QuinnFQ90] provided new tools with which to approach this problem. In independent work of Lee and Wilczyński [Reference Lee and WilczyńskiLW90, Theorem 1.1] and Hambleton and Kreck [Reference Hambleton and KreckHK93, Theorem 4.5], the methods of topological surgery theory were applied to provide a complete answer for simply connected, closed 4-manifolds, in the presence of a natural fundamental group restriction. That is, they classified when an element of the second homotopy group of such a 4-manifold can be represented by a locally flat embedded sphere whose complement has abelian fundamental group. Lee and Wilczyński [Reference Lee and WilczyńskiLW97] later generalised their theorem to apply to simply connected, compact $4$-manifolds with homology sphere boundary. In this article, we expand our understanding to another general class of 4-manifolds with boundary.

Our main result is an answer to the sphere embedding question for $4$-manifolds called knot traces, with $x$ a generator of the second homotopy group. Let $\nu K$ be an open tubular neighbourhood of a knot $K$ in $S^{3}$ and let $n$ be an integer. The $n$-framed knot trace $X_n(K)$ is the smooth 4-manifold obtained by attaching a 2-handle $D^{2}\times D^{2}$ to the 4-ball along $\nu K\subset S^{3}$, using framing coefficient $n$ and smoothing corners. The boundary of $X_n(K)$ is the $n$–framed surgery $S_n^{3}(K) := (S^{3} {{\smallsetminus }} \nu K) \cup D^{2} \times S^{1}$, where $\partial D^{2} \times \{1\}$ is attached to the $n$-framed longitude of $K$.

Theorem 1.1 Let $K$ be a knot in $S^{3}$ and let $n$ be an integer. A generator of $\pi _2(X_n(K))$ can be represented by a locally flat embedded $2$-sphere whose complement has abelian fundamental group if and only if:

  1. (i) $H_1(S_n^{3}(K);\mathbb {Z}[\mathbb {Z}/n])=0$; or equivalently for $n \neq 0$, $\prod _{\{\xi \mid \xi ^{n}=1\} }\Delta _K(\xi ) =1$;

  2. (ii) $\operatorname {Arf}(K)=0$; and

  3. (iii) $\sigma _K(\xi ) =0$ for every $\xi \in S^{1}$ such that $\xi ^{n}=1$.

The theorem is stated in terms of some well-known knot invariants that arise from a Seifert matrix $V$ for $K$. The Alexander polynomial $\Delta _K(t) \in \mathbb {Z}[t,t^{-1}]$ can be most quickly defined as $\det (tV-V^{T})$. Then the Arf invariant of $K$ in $\mathbb {Z}/2$ is $0$ if $\Delta _K(-1) \equiv \pm 1 \mod {8}$ and is $1$ if $\Delta _K(-1) \equiv \pm 3 \mod {8}$. Finally the Tristram–Levine signature, for $\xi \in S^{1} \subseteq \mathbb {C}$, is the signature of the Hermitian matrix $(1-\xi )V + (1-\overline {\xi })V^{T}$. These invariants are independent of the choice of Seifert surface and Seifert matrix, and they are straightforward to compute from a knot diagram.

For certain choices of $n$, there are logical dependencies among the conditions (i), (ii), and (iii) above. When $n=0$, condition (i) states that $H_1(S^{3}_0(K); \mathbb {Z}[\mathbb {Z}])=0$, which is equivalent to $\Delta _K(t)=1$, which in turn implies both conditions (ii) and (iii). When $n=\pm 1$, conditions (i) and (iii) are automatically satisfied.

When $n \neq 0$, condition (i) is equivalent to the condition that $\Sigma _{|n|}(K)$, the $n$-fold cyclic branched cover of $S^{3}$ with branching set $K$, is an integral homology sphere. This is due to the classical formula $|H_1(\Sigma _{|n|}(K);\mathbb {Z})| = \prod _{\{\xi \mid \xi ^{n}=1\} }\Delta _K(\xi )$ due to [Reference GoeritzGoe34, Reference FoxFox56]. When $n \neq 0$ is even, condition (i) implies condition (ii), as follows. The expression $\prod _{\{\xi \mid \xi ^{n}=1\} }\Delta _K(\xi )$ equals the resultant $\operatorname {Res}(\Delta _K(t), t^{n}-1) \in \mathbb {Z}$. Whenever $m$ divides $n$, $t^{m}-1$ divides $t^{n}-1$, and so by the characterising properties of resultants we have

\[ \operatorname{Res}(\Delta_K(t), t^{n}-1) = \operatorname{Res}(\Delta_K(t), t^{m}-1) \cdot \operatorname{Res}(\Delta_K(t),(t^{n}-1)/(t^{m}-1)). \]

Since $\operatorname {Arf}(K)=1$ implies that $\operatorname {Res}(\Delta _K(t), t^{2}-1) = \Delta _K(-1) \neq 1$, it follows that $\operatorname {Arf}(K)=1$ further implies that $\operatorname {Res}(\Delta _K(t), t^{n}-1)= \prod _{\{\xi \mid \xi ^{n}=1\} }\Delta _K(\xi )$ is not equal to 1 for $n$ even.

Throughout the paper we will assume knots are oriented in order to make various constructions in the standard way. However, none of the conditions (i), (ii), or (iii) depends on a given orientation for the knot, and so the characterisation provided by Theorem 1.1 is independent of this choice.

The remainder of the introduction proceeds as follows. In § 1.1, we discuss the applications of Theorem 1.1 to the study of whether a knot is shake slice. In § 1.2 we give a quick proof of the main theorem for the case $n=0$, and then some further results that we have obtained when $n=\pm 1$. In § 1.3 we outline the topological surgery theory strategy we use to prove our main result.

1.1 Shake slice knots

The embedding question for a generator of the second homotopy group of a knot trace is of interest via the lens of knot theory.

Definition 1.2 A knot $K$ is $n$-shake slice if a generator of $\pi _2(X_n(K)) \cong \mathbb {Z}$ can be realised by a locally flat embedded 2-sphere $S$. We say a knot $K$ is $\mathbb {Z}/n$-shake slice if in addition $\pi _1(X_n(K) {{\smallsetminus }}S) \cong \mathbb {Z}/n$.

For every $n \in \mathbb {Z}$, the fundamental group of the complement of an embedded sphere generating $\pi _2(X_n(K))$ abelianises to $\mathbb {Z}/n$, so our condition that $\pi _1(X_n(K)) \cong \mathbb {Z}/n$ is just a more specific way to express the abelian condition on the fundamental group.

Theorem 1.1 can be viewed as a characterisation of when a knot $K$ is $\mathbb {Z}/n$-shake slice. Although the term ‘$n$-shake slice’ was not coined until much later, classical obstructions to being $n$-shake slice were already obtained in the 1960s. In the course of proving Theorem 1.1 we obtained several new proofs of these classical results. Robertello [Reference RobertelloRob65] showed that the Arf invariant obstructs $K$ from being $n$-shake slice, for every $n$. We give a new proof of Robertello's result in Proposition 6.5 for all $n$, and we outline a second new proof in Remark 4.7 for even $n$. Both proofs are different from Robertello's. Saeki [Reference SaekiSae92] has yet another proof in the smooth category that uses the Casson invariant.

Tristram [Reference TristramTri69] showed that the signatures $\sigma _K(\xi _p^{m})$, for $p$ a prime power dividing $n$, obstruct $K$ being $n$-shake slice. We provide a different proof in § 3 that also explains Tristram's theorem in the context of our results. Our proof is similar to that sketched by Saeki in [Reference SaekiSae92, Theorem 3.4].

For $n=0$, it remains unknown in both the smooth and topological categories whether every $0$-shake slice knot is slice. An immediate consequence of our theorem is that a knot is $\mathbb {Z}$-shake slice if and only if it is $\mathbb {Z}$-slice: both correspond to Alexander polynomial one.

Now we describe some further consequences of Theorem 1.1 that are proved as corollaries in § 2. First, we provide new examples of the difference between the smooth and topological categories. Recall that a knot is smoothly $n$-shake slice if a generator of $\pi _2(X_n(K)) \cong \mathbb {Z}$ can be realised by a smoothly embedded 2-sphere $S$.

Corollary 2.3 For every $n>0$ there exist infinitely many knots that are $n$-shake slice but neither smoothly $n$-shake slice nor $($topologically$)$ slice. These knots may be chosen to be distinct in concordance.

We then show that being $n$-shake slice for infinitely many $n \in \mathbb {Z}$ does not imply slice.

Corollary 2.4 There exist infinitely many non-slice knots, each of which is $n$-shake slice for infinitely many $n \in \mathbb {Z}$. Moreover, these knots may be chosen to be distinct in concordance.

A question of Hedden [Max16] asks whether the concordance class of a knot must be determined by the infinite tuple of homology cobordism classes $(S^{3}_{p/q}(K))_{p/q \in \mathbb {Q}}$. We provide evidence for a negative answer to this question as follows, in what we believe is the first example of non-concordant knots with the property that infinitely many of their integer surgeries are homology cobordant.

Corollary 2.5 There exist infinitely many knots $\{K_i\}$, mutually distinct in concordance, and an infinite family of integers $\{n_j\}$ such that $S^{3}_{n_j}(K_i)$ is homology cobordant to $S^{3}_{n_j}(U)$ for all $i, j\in \mathbb {Z}$.

We remark that $0$ is not an element of our family of integers $\{n_j\}$: $S^{3}_0(K)$ is homology cobordant to $S^{3}_0(U)$ if and only if $K$ is slice in an integral homology 4-ball (which, incidentally, implies that $S^{3}_n(K)$ is homology cobordant to $S^{3}_n(U)$ for all $n \in \mathbb {Z}$). It is currently an open question in both categories whether there exists a non-slice knot that is slice in a homology ball.

We then show that, for most $m$ and $n$, the $m$-shake slice and $n$-shake slice conditions are independent.

Corollary 2.8 If $m$ does not divide $n$ then there exist infinitely many knots which are $n$-shake slice but not $m$-shake slice. These knots may be chosen to be distinct in concordance.

The difference between $n$-shake slice and $\mathbb {Z}/n$-shake slice can also be investigated using Theorem 1.1. For composite $n$ the theorem says there are extra signatures away from the prime power divisors of $n$ whose vanishing is necessary for an $n$-shake slice knot to be moreover $\mathbb {Z}/n$-shake slice; cf. [Reference SaekiSae92, Proposition 3.7]. Consider that every slice knot is $n$-shake slice for all $n$, but according to Cha and Livingston [Reference Cha and LivingstonCL04], for any choice of composite $n$ and root of unity $\xi _n$, there exist slice knots $K$ with $\sigma _K(\xi _n)\neq 0$. We will prove that examples of this sort are not peculiar to composite $n$.

Corollary 2.10 For all $n \neq \pm 1$, there exists a slice, and therefore $n$-shake slice, knot that is not $\mathbb {Z}/n$-shake slice.

1.2 The cases $n=0$ and $n=\pm 1$

The cases $n=0$ and $n=\pm 1$ for Theorem 1.1 are special, in that they can be proved relatively quickly by appealing directly to results of Freedman and Quinn. We provide a quick proof for $n=0$ now. Recall that when $n=0$ the three conditions in Theorem 1.1 reduce to $\Delta _K(t)=1$.

Example 1.3 (The case $n=0$)

First assume a generator of $\pi _2(X_0(K))$ is represented by an embedded sphere whose complement has fundamental group $\mathbb {Z}$. Since $n=0$, the normal bundle of this sphere is trivial. Perform surgery on $X_0(K)$ along this 2-sphere to obtain a pair $(V, S_0^{3}(K))$ where $\pi _2(V)=0$, and $\pi _1(V)\cong \mathbb {Z}$, generated by a meridian of $K$. Now attach a 2-handle to a meridian in the boundary. The cocore of the 2-handle is a slice disc $D'$ for $K$ in a homotopy 4-ball $B'$ such that $\pi _1(B' \smallsetminus D') \cong \mathbb {Z}$. Hence $\Delta _K(t)=1$.

Now assume $\Delta _K(t)=1$. By [Reference FreedmanFre84, Theorem 7] and [Reference Freedman and QuinnFQ90, 11.7B] (see also [Reference Garoufalidis and TeichnerGT04, Appendix]), $K$ has a slice disc $D$ in $D^{4}$ with $\pi _1(D^{4} \smallsetminus D) \cong \mathbb {Z}$. Cap this disc off with the core of the 2-handle to obtain the desired sphere in $X_0(K)$. This completes the proof of Theorem 1.1 for $n=0$.

To see a similarly quick proof for $n=\pm 1$, the reader is invited to skip ahead to Example 8.1. In this case, we rely on the result of Freedman that as $S^{3}_{\pm 1}(K)$ is an integer homology sphere, it must bound a contractible 4-manifold. When $n=\pm 1$ we have obtained a wholly different proof of Theorem 1.1 using Seifert surface constructions. These methods, detailed in § 8, lead to the two results described next.

For each $n$, one can measure how far a knot $K$ is from being $n$-shake slice by considering the minimal genus of a locally flat embedded surface generating $H_2(X_n(K))$. This minimum is called the (topological) $n$-shake genus $g_{\operatorname {sh}}^{n}(K)$. For $n=1$ we have a precise understanding of this invariant.

Proposition 8.7 For every knot $K$ there exists a locally flat embedded torus in $X_1(K)$ that generates $H_2(X_1(K))$ and has simply connected complement. In particular,

\[ g^{1}_{\operatorname{sh}}(K) = \operatorname{Arf}(K) \in \{0,1\}. \]

Now for a slice knot $K$, and for each $n$, a slice disc capped off by the core of the 2-handle in $X_n(K)$ geometrically intersects the cocore once. This suggests a way to measure of how far an $n$-shake slice knot $K$ is from being slice, by taking the minimum over all embedded spheres $S$ generating $\pi _2(X_n(K))$ of the geometric intersection number of $S$ with the cocore of the 2-handle of $X_n(K)$. We call this minimum the $n$-shaking number of $K$.

Since the algebraic intersection of $S$ with the cocore of the 2-handle of $X_n(K)$ is algebraically one, the shaking number measures the difference between geometric and algebraic intersection numbers. For closed surfaces, this difference was investigated in [Reference Morgan and SzabóMS99, Reference StrleStr03, Reference HorvatHor15, Reference SunukjianSun10].

For a knot $K$, ${g^{\operatorname {top}}_4}(K)$ denotes the (topological) 4-genus or slice genus, the minimal genus among compact, oriented, locally flat surfaces in $D^{4}$ with boundary $K$. The (topological) $\mathbb {Z}$-slice genus ${g_4^{\mathbb {Z}}}(K)$ is the minimal genus among such surfaces whose complement has infinite cyclic fundamental group. Computable upper bounds for ${g_4^{\mathbb {Z}}}(K)$ are discussed in [Reference Feller and LewarkFL18], and include $2{g_4^{\mathbb {Z}}}(K) \leq \deg (\Delta _K)$ [Reference FellerFel16]. For $n=1$, as well as proving again that a knot with vanishing $\operatorname {Arf}$ invariant is $1$-shake slice, the Seifert surface method provides explicit upper bounds on the $1$-shaking number.

Proposition 8.8 For a knot $K$ with $\operatorname {Arf}(K)=0$ we have

\[ 2{g^{\operatorname{top}}_4}(K)+1\leq 1\text{-shaking number of K}\leq 2{g_4^{\mathbb{Z}}}(K)+1. \]

In particular, for each integer $k \geq 0$ there exists a $1$-shake slice knot $K_k$ such that the $1$-shaking number of $K_k$ is exactly $2k+1$.

Note that for every $n$, the $n$-shaking number is always odd since the algebraic intersection of the generator of $\pi _2(X_n(K))$ with the cocore of the 2-handle is $1$. Thus this is a complete realisation result for $1$-shaking numbers.

1.3 Proof outline

Having already proved Theorem 1.1 for the case of $n=0$ above, we now restrict to $n\neq 0$. The components of the ‘only if’ direction are proven respectively in Proposition 3.3, Proposition 6.5, and Proposition 3.5. As mentioned previously, with different approaches Proposition 6.5 is due to Robertello [Reference RobertelloRob65] and Proposition 3.5 is due to Tristram [Reference TristramTri69] (see also [Reference SaekiSae92]).

Now we will outline the proof of the ‘if’ direction. Given a group $\pi$ and closed $3$-manifolds $M_1$ and $M_2$, together with homomorphisms $\varphi _i\colon \pi _1(M_i)\to \pi$, we say a cobordism $W$ from $M_1$ to $M_2$ is over $\pi$ if $W$ is equipped with a map $\pi _1(W)\to \pi$ restricting to the given homomorphisms on the boundary.

The key idea of our proof is that for a fixed knot $K$, a generator of $\pi _2(X_n(K))$ can be represented by a locally flat embedded 2-sphere $S$ with $\pi _1(X_n(K) {{\smallsetminus }}S) \cong \mathbb {Z}/n$ if and only if there exists a homology cobordism $V$ from $S^{3}_n(K)$ to the lens space $L(n,1)$ over $\mathbb {Z}/n$, extending standard maps $\pi _1(S^{3}_n(K))\to \mathbb {Z}/n$ and $\pi _1(L(n,1))\to \mathbb {Z}/n$, such that $\pi _1(V) \cong \mathbb {Z}/n$ and $V \cup _{L(n,1)} D_n$ is homeomorphic to $X_n(K)$, where $D_n$ denotes the $D^{2}$-bundle over $S^{2}$ with euler number $n$.

The proof of the ‘if’ direction involves constructing such a cobordism $V$ when the list of invariants in Theorem 1.1 vanish. Here is an outline.

  1. (1) Show there exists a cobordism $W$ between $S^{3}_n(K)$ and the lens space $L(n,1)$ and a map $W \to L(n,1)\times [0,1]$ that restricts to a degree-one normal map (Definition 4.2) $f\colon S^{3}_n(K) \to L(n,1) \times \{0\}$ and the identity map $L(n,1) \to L(n,1) \times \{1\}$ (see § 4). This uses the assumption that $\operatorname {Arf}(K)=0$ when $n$ is even and no assumptions when $n$ is odd.

  2. (2) Use the computation of the simple surgery obstruction groups $L_4^{s}(\mathbb {Z}[\mathbb {Z}/n])$ in terms of multisignatures to show that we can replace $W$ with a homology cobordism $V$ between $S^{3}_n(K)$ and $L(n,1)$ over $\mathbb {Z}/n$, with $V$ homotopy equivalent to $L(n,1) \times I$ (see § 5). This uses the assumptions that $H_1(S_n^{3}(K);\mathbb {Z}[\mathbb {Z}/n])=0$ and $\sigma _K(\xi _n^{m}) =0$ for all $m$.

  3. (3) Let $X:= V \cup _{L(n,1)} D_n$. Note that a generator for $\pi _2(X)$ is represented by an embedded sphere in $D_n$. Use Boyer's classification (Theorem 7.1) to conclude that $X$ is homeomorphic to $X_n(K)$ (see § 6). This uses the assumption that $\operatorname {Arf}(K)=0$ when $n$ is odd and no additional assumptions when $n$ is even. More precisely, according to the classification, $X$ is homeomorphic to $X_n(K)$ automatically when $n$ is even, and if and only if $\operatorname {ks}(X)=\operatorname {ks}(X_n(K))=0$ when $n$ is odd; the latter equality follows since $X_n(K)$ is smooth. We show in Proposition 6.8 that $\operatorname {Arf}(K)=\operatorname {ks}(X)$.

An interesting aspect of the proof (of both the ‘if’ and ‘only if’ directions) of Theorem 1.1 is that the Arf invariant appears in different places for $n$ odd and $n$ even. However, in each case its vanishing is required.

Recent work of Kim and Ruberman [Reference Kim and RubermanKR20] uses similar techniques to prove the existence of topological spines in certain 4-manifolds. Their argument is in some ways structurally quite similar to ours, since both works follow a surgery theoretic strategy. There is no overlap between our results: every knot trace $X_n(K)$ admits a PL-spine consisting of the cone on $K$ union the core of the attached 2-handle, as is crucially used in [Reference Kim and RubermanKR20]. Moreover, there is a key difference: Kim and Ruberman have flexibility in their choice of a second 3-manifold, whereas we have a fixed choice of $S^{3}_n(K)$ and $L(n,1)$.


From § 3 onwards, we assume for convenience that $n>0$. The case of $n=0$ was proved in Example 1.3. When $n<0$, the argument is the same as for $-n$. Throughout, manifolds are compact and oriented, and knots are oriented.

2. Corollaries to Theorem 1.1

Before embarking on the main work of proving Theorem 1.1, we use it to prove several knot theoretic corollaries. For a knot $K$ in $S^{3}$, let $C_{n,1}(K)$ denote the $(n,1)$-cable of $K$.

Corollary 2.1 Let $K$ be any knot and let $n$ be an integer. Suppose that $n$ is even or ${\operatorname {Arf}(K)=0}$. Then $C_{n,1}(K)$ is $\mathbb {Z}/n$-shake slice.

Proof. We use the formulae for the Alexander polynomial and signatures of a satellite knot, due respectively to Seifert [Reference SeifertSei50] and Litherland [Reference LitherlandLit79], to verify the conditions of Theorem 1.1 for $C_{n,1}(K)$.

Since $C_{n,1}(U)=U$ we have $\Delta _{C_{n,1}(K)}(t)= \Delta _K(t^{n})$ and $\sigma _{C_{n,1}(K)}(\omega )= \sigma _K(\omega ^{n})$ for all $\omega \in S^{1}$. Letting $\omega _n$ denote a primitive $n$th root of unity, we therefore have for $1 \leq k \leq n$ that

\[ \sigma_{C_{n,1}(K)}(\omega_n^{k})= \sigma_K(\omega_n^{nk})= \sigma_K(1) =0. \]

We have that

\[ \prod_{k=1}^{n} \Delta_{C_{n,1}(K)}(\omega_n)= \prod_{k=1}^{n} \Delta_{K}(\omega_n^{n}) =1. \]

Levine [Reference LevineLev66, Proposition 3.4] showed that $\operatorname {Arf}(J)=0$ if and only if $\Delta _J(-1)\equiv \pm 1\ {\rm mod}\ 8$, and so since

\[ \Delta_{C_{n,1}(K)}({-}1)= \Delta_{K}(({-}1)^{n})= \begin{cases} \Delta_K({-}1) & n \text{ odd}, \\ 1 & n \text{ even} \end{cases} \]

we obtain as desired that $\operatorname {Arf}(C_{n,1}(K))=0$.

Remark 2.2 Gordon [Reference GordonGor83] observed that for any knot $K$

\[ S^{3}_n(C_{n,1}(K))= L(n,1)\, \#\, S^{3}_{1/n}(K). \]

This gives a slightly more direct argument that $C_{n,1}(K)$ is $\mathbb {Z}/n$-shake slice whenever $n$ is even or $\operatorname {Arf}(K)=0$, as follows. Let $C$ be the contractible 4-manifold with boundary $S^{3}_{1/n}(K)$ guaranteed by [Reference FreedmanFre82, Theorem 1.4$'$], and define $V$ to be the boundary connected sum of $L(n,1) \times I$ and $C$. The manifold $V$ is now a homology cobordism from $S^{3}_n(C_{n,1}(K))$ to $L(n,1)$ that is homotopy equivalent to $L(n,1) \times I$. This allows one to skip the work of §§ 4 and 5 constructing a homology cobordism and go straight to proving that $V \cup D_n$ is homeomorphic to $X_n(C_{n,1}(K))$ using the results we prove in § 6 and Boyer's classification (Theorem 7.1), which when $n$ is odd requires $\operatorname {Arf}(C_{n,1}(K))= \operatorname {Arf}(K)=0$.

We can use Corollary 2.1 to obtain many new examples of $n$-shake slice knots.

Corollary 2.3 For every $n \neq 0$ there exist infinitely many topological concordance classes of knots that are $n$-shake slice but not smoothly $n$-shake slice.

Proof. In both the smooth and topological categories a knot $K$ is $n$-shake slice if and only if $-K$ is $(-n)$-shake slice. So it suffices to show the $n>0$ case as follows.

We can use the description of $S^{3}_n(C_{n,1}(K))$ from Remark 2.2 to show that $C_{n,1}(K)$ is often not smoothly $n$-shake slice. Given an integer homology sphere $Y$, Ozsváth and Szabó associate a so-called $d$-invariant $d(Y) \in \mathbb {Q}$, with the property that $d(Y)=0$ if $Y$ bounds a rational homology ball [Reference Ozsváth and SzabóOS03a]. We show that if $C_{n,1}(K)$ is smoothly $n$-shake slice for some $n>0$, then ${d(S^{3}_{1}(K))=0}$ as follows.

Suppose that $C_{n,1}(K)$ is smoothly $n$-shake slice via a sphere $S$ in $X_n(J)$. The exterior of $S$ can be quickly confirmed to be a smooth homology cobordism between $S^{3}_n(J)$ and $L(n,1)$; see Lemma 3.1. Therefore $S^{3}_n(C_{n,1}(K))= L(n,1)\, \#\, S^{3}_{1/n}(K)$ and $S^{3}_n(U)= L(n,1)$ are homology cobordant via some smooth $W$. By summing $W$ with $-L(n,1) \times I$ along $D^{4} \times I \subset W$, we further obtain that $S^{3}_{1/n}(K)$ is smoothly rationally homology cobordant to $S^{3}$. Therefore, $d(S^{3}_{1/n}(K))= d(S^{3})=0$. Furthermore, since $n >0$ we have $d(S^{3}_{1/n}(K))= d(S^{3}_1(K))$ by [Reference Ni and WuNW15, Proposition 1.6].

Now for each $j \in \mathbb {N}$ let $K_j:= T_{2,8j+1}$. Note that since $K_j$ is alternating and the ordinary signature $\sigma _{K_j}(-1) <0$, [Reference Ozsváth and SzabóOS03b, Corollary 1.5] implies that $d(S^{3}_1(K)) \neq 0$. So $C_{n,1}(K_j)$ is not smoothly $n$-shake slice, despite being $\mathbb {Z}/n$-shake slice by Corollary 2.1. One can use Litherland's satellite formula [Reference LitherlandLit79] to compute that the first jump of the Tristram–Levine signature function of $C_{n,1}(K_j)$ occurs at $e^{2 \pi i \theta _j}$, where $\theta _j= {1}/{2n(8j+1)}$. Therefore the knots $C_{n,1}(K_j)$ are distinct in concordance.

In another direction, we are able to show that there exist non-slice knots which are nevertheless $n$-shake slice for infinitely many $n \in \mathbb {Z}$. It is presently unknown in either category whether $0$-shake slice implies slice, and so the question of whether being $n$-shake slice for all $n \in \mathbb {Z}$ implies slice appears both interesting and difficult.

Corollary 2.4 There exist knots that are $\mathbb {Z}/n$-shake slice for every prime power $n \in \mathbb {Z}$, but are not slice. Moreover, these knots may be chosen to represent infinitely many concordance classes.

Proof. Let $J$ be a knot with Alexander polynomial equal to the $m\textrm {th}$ cyclotomic polynomial, where $m$ is divisible by at least 3 distinct primes. Since $J \#-J$ does not have trivial Alexander polynomial, there exist infinitely many non-concordant knots sharing its Seifert form [Reference KimKim05]. We show that any such knot $K$ is $\mathbb {Z}/n$-shake slice for all prime powers $n$.

By [Reference LivingstonLiv02], we have $|H_1(\Sigma _{|n|}(J))|=1$, so condition (i) of Theorem 1.1 follows immediately:

\[ \prod_{\{\xi \mid \xi^{n}=1\} }\Delta_K(\xi) = |H_1(\Sigma_{|n|}(K))|= |H_1(\Sigma_{|n|}(J))|^{2}=1. \]

For conditions (ii) and (iii), observe that since $K$ shares a Seifert form with the knot $J \#-J$, we have that $\operatorname {Arf}(K)=0$ and for every $\omega \in S^{1}$ we have

\[ \sigma_{\omega}(K)=\sigma_{\omega}(J)+ \sigma_{\omega}({-}J)=0. \]

We remark that it is open whether there is a smoothly non-slice knot that is smoothly $n$-shake slice for infinitely many $n$. The smooth analogue of the next result is also open.

Corollary 2.5 There exist infinitely many knots $\{K_i\}$, mutually distinct in concordance, and an infinite family of integers $\{n_j\}$ such that $S^{3}_{n_j}(K_i)$ is homology cobordant to $S^{3}_{n_j}(U)$ for all $i, j\in \mathbb {N}$.

Proof. This follows immediately from Corollary 2.4 and Lemma 3.1.

We are also able to show that homology cobordism of the $n$-surgeries is often not enough to determine that a knot is $n$-shake slice, as follows.

Corollary 2.6 For each odd $n \in \mathbb {N}$, there exists $K$ such that $S^{3}_n(K)$ and $S^{3}_n(U)$ are topologically homology cobordant but $K$ is not $n$-shake slice. In fact, for each odd $n$ there exist knots representing infinitely many concordance classes that satisfy this.

Proof. By the proof of Theorem 1.1, if $n$ is odd and $K$ is a knot with $\prod _{\{\xi \mid \xi ^{n}=1\} }\Delta _K(\xi ) =1$ and $\sigma _K(\xi )=0$ for all $n$th roots of unity $\xi$, then $S^{3}_n(K)$ and $S^{3}_n(U)= L(n,1)$ are homology cobordant. As discussed in the proof of Corollary 2.3, for every knot $J$ the knot $K= C_{n,1}(J)$ satisfies these conditions. However, if $\operatorname {Arf}(J) \neq 0$, then since $\operatorname {Arf}(K)= \operatorname {Arf}(J)$ we obtain that $C_{n,1}(J)$ is not $n$-shake slice (or even, for the cognoscenti, $n$-shake concordant to the unknot). Therefore the set $\{C_{n,1}(T_{2, 8j+3}\}_{j\geq 1}$ is an infinite collection of such knots, all distinguished in concordance by the first jump of their Tristram–Levine signature functions, which occur at $e^{2 \pi i y_j}$, where $y_j= {1}/{2n(8j +3)}$.

We remark that in the examples of Corollary 2.6, $S^{3}_n(K)$ and $S^{3}_n(U)$ are homology cobordant not just with integer coefficients, but also with $\mathbb {Z}[\mathbb {Z}/n]$-coefficients.

We also compare the difference between $m$-shake slice and $n$-shake slice for $m\neq n$.

Corollary 2.7 If $m \mid n$ and $K$ is $\mathbb {Z}/n$-shake slice, then $K$ is $\mathbb {Z}/m$-shake slice.

Proof. First, note that if $n=0$ and $K$ is $\mathbb {Z}/n$-shake slice then $\Delta _K(t)=1$, and so the conditions (i), (ii), and (iii) of Theorem 1.1 are satisfied for all $m \in \mathbb {Z}$.

So assume $n \neq 0$. Since $\prod _{\{\xi \mid \xi ^{m}=1\} }\Delta _K(\xi )$ divides $\prod _{\{\xi \mid \xi ^{n}=1\} }\Delta _K(\xi )$ and both are integers, if the criterion (i) holds for $n$, then it also holds for $m$. The signature and Arf invariant conditions are immediate.

Corollary 2.8 If $m$ does not divide $n$ then there exist infinitely many knots that are $n$-shake slice but not $m$-shake slice. These knots may be chosen to be distinct in concordance.

Proof. Let $q$ be a prime power which divides $m$ but not $n$. Let $K$ be any knot with $\operatorname {Arf}(K)=0$ and $\sigma _{K}(e^{2 \pi ik/q}) \neq 0$ for all $k=1, \dots , q-1$. Such knots are easy to find, for example by taking $K=T_{2, 8N+1}$ for sufficiently large $N$. Now let $J=C_{n,1}(K)$, and note that since $\operatorname {Arf}(K)=0$ Corollary 2.1 tells us that $J$ is $n$-shake slice. However, $\sigma _{C_{n,1}(K)}(e^{2 \pi i/q})= \sigma _{K}(e^{2 \pi in/q}) \neq 0$, since $q$ does not divide $n$. So $J$ is not $m$-shake slice.

We can obtain $\{J_j\}_{j \in \mathbb {N}}$ representing infinitely many concordance classes of $n$-shake but not $m$-shake slice knots by letting $J_j=C_{n,1}(T_{2,8(N+j)+1})$ for sufficiently large $N$. As in the proof of Corollary 2.3, these knots are distinguished in concordance by the first jump of the Tristram–Levine signature, which occurs for $J_j$ at $e^{2 \pi i x_j}$, where $x_j= {1}/{2n(8(N+j)+1)}$.

Remark 2.9 For many pairs $(n,m)$ such that $m$ does not divide $n$, one can also find examples of $n$-shake slice knots which are not $m$-shake slice by considering certain linear combinations of $(2,2k+1)$ torus knots. For example, one can verify that $K_n:=6T_{2,4n+1} \# - 4T_{2, 6n+1}$ satisfies the conditions of Theorem 1.1 for $n$, and hence is $\mathbb {Z}/n$-shake slice. Additionally, computation of Tristram–Levine signatures shows that if $m$ does not divide $2n$, then $K_n$ is not $m$-shake slice.

Finally, we show in almost all cases that $\mathbb {Z}/n$-shake slice is a strictly stronger condition than $n$-shake slice.

Corollary 2.10 A knot is $(\pm 1)$-shake slice if and only if it is $\mathbb {Z}/1$-shake slice. For all other $n$ there exists a slice, and therefore $n$-shake slice, knot that is not $\mathbb {Z}/n$-shake slice.

Proof. The ‘if’ direction of the first sentence is obvious. For the other direction, observe that an embedded sphere $S$ representing a generator of $\pi _2(X_{\pm 1}(K))$ has a normal bundle $\nu S$ with euler number $\pm 1$ (see Lemma 6.1). Since $\nu S \to S$ is a homotopy equivalence, $\nu S$ is simply connected. Since the euler number is $\pm 1$, $\partial (D\nu S) \cong S^{3}$, where $D\nu S$ denotes the disc bundle of the normal bundle. We know that $X_{\pm 1}(K)$ is simply connected. Apply the Seifert–Van Kampen theorem to the decomposition

\[ X_{{\pm} 1}(K) = (X_{{\pm} 1}(K) {{\smallsetminus}}D\nu S) \cup_{S^{3} \times (1,\infty)} \nu S \]

to deduce that $\pi _1(X_{\pm 1}(K) {{\smallsetminus }}D\nu S) \cong \pi _1(X_{\pm 1}(K) {{\smallsetminus }}S) \cong \{1\} \cong \mathbb {Z}/1$. This proves the reverse direction.

For the second sentence, the knot $K=4_1 \# 4_1$ is slice and hence $n$-shake slice for all $n$. But $K$ has $|H_1(\Sigma _n(K))| \neq 1$ for all $n>1$, and hence is not $\mathbb {Z}/n$-shake slice for $|n|>1$ by condition (iii) of Theorem 1.1. Recall that when $n=0$, the conditions of Theorem 1.1 reduce to triviality of the Alexander polynomial. But since $\Delta _K(t)=(t^{2}-3t+1)^{2}$, this knot $K$ is not $\mathbb {Z}/0$-shake slice either.

When $n=\pm 1$, the conditions (i) and (iii) of Theorem 1.1 are automatically satisfied, so Corollary 2.10 immediately gives the following.

Corollary 2.11 A knot $K$ is $(\pm 1)$-shake slice if and only if it is $\mathbb {Z}/ (\pm 1)$-shake slice if and only if $\operatorname {Arf}(K)=0$.

If a knot is $n$-shake slice, then it has vanishing Arf invariant [Reference RobertelloRob65], so Corollary 2.11 immediately gives the following.

Corollary 2.12 If a knot $K$ is $n$-shake slice for some integer $n$, then it is $(\pm 1)$-shake slice.

Changing the orientation on an $n$-trace $X_n(K)$ results in the trace $X_{-n}(-K)$, and so $K$ is $n$-shake slice if and only if $-K$ is $(-n)$-shake slice. Surprisingly, the conditions in Theorem 1.1 show that an even stronger symmetry holds, as in the following corollary.

Corollary 2.13 A knot $K$ is $\mathbb {Z}/n$-shake slice if and only if it is $\mathbb {Z}/(-n)$-shake slice.

Here of course $\mathbb {Z}/n \cong \mathbb {Z}/(-n)$, but the 4-manifolds $X_n(K)$ and $X_{-n}(K)$ are generally different. In particular, we know of no reason to believe that a knot is $n$-shake slice if and only if it is $(-n)$-shake slice.

For any $n$, it remains unknown in both the smooth and topological category whether $n$-shake slice knots have $n$-shake slice connected sum. As the invariants involved in conditions (i), (ii), and (iii) of Theorem 1.1 are all additive under connected sum we have the following immediate corollary.

Corollary 2.14 If $K$ and $J$ are $\mathbb {Z}/n$-shake slice, then so is $K \#J$.

3. Obstructions to $\mathbb {Z}/n$-shake sliceness

In this section we prove that the conditions listed in Theorem 1.1 on the knot signatures and the Alexander polynomial are indeed necessary conditions for a knot to be $\mathbb {Z}/n$-shake slice. Some of these results have been shown by Tristram [Reference TristramTri69] and Saeki [Reference SaekiSae92]; we include our own proofs for completeness, for the convenience of the reader, and to introduce the identification of certain Atiyah–Singer/Casson–Gordon signatures with Tristram–Levine knot signatures that will be needed later on. As stated in the conventions section, we henceforth assume that $n>0$.

Lemma 3.1 If a knot $K$ is $n$-shake slice via an embedded sphere $S$ in $X_n(K)$ then $W:= X_n(K) \smallsetminus \nu (S)$ is a homology cobordism from $S^{3}_n(K)$ to $L(n,1)$. If $K$ is further $\mathbb {Z}/n$-shake slice via $S$, then $\pi _1(W) \cong \mathbb {Z}/n$.

Proof. The first statement follows from computation using the Mayer–Vietoris sequence for $X_n(K)= W \cup \nu (S)$. The second statement is immediate from the definition.

Remark 3.2 In this section, we only use the hypothesis that the manifolds $S^{3}_n(K)$ and $L(n,1)$ are homology cobordant via a homology cobordism $W$ with $\pi _1(W) \cong \mathbb {Z}/n$. This hypothesis is sufficient to establish conditions (i) and (iii) of Theorem 1.1 However, one cannot make this weaker assumption, at least when $n$ is odd, if we wish to conclude that $\operatorname {Arf}(K)=0$. In particular, for every knot $K$ we know that $S^{3}_1(K)$ is homology cobordant to $L(1,1)=S^{3}$ by a simply connected cobordism [Reference FreedmanFre82, Theorem 1.4$'$] (see also [Reference Freedman and QuinnFQ90, 9.3C]), but of course some knots have Arf invariant 1. On the other hand, the existence of a smooth homology cobordism would be enough to imply that $\operatorname {Arf}(K)=0$ [Reference SaekiSae92].

Proposition 3.3 Let $K$ be a knot such that $S^{3}_n(K)$ and $L(n,1)$ are homology cobordant via a homology cobordism $W$ with $\pi _1(W) \cong \mathbb {Z}/n$. Then $H_1(S^{3}_n(K); \mathbb {Z}[\mathbb {Z}/n])=0= H_1(\Sigma _n(K);\mathbb {Z})$.

Proof. We will show that the $n$-sheeted cyclic cover $Y_n:=(S^{3}_n(K))_n$ of $S^{3}_n(K)$ has trivial integral homology. Since $H_1(Y_n;\mathbb {Z})=0$ implies that $H_1(S^{3}_n(K); \mathbb {Z}[\mathbb {Z}/n])=0$, this will imply the first part of our desired result. By the universal coefficient theorem, it suffices to show that $H_1(Y_n ;\mathbb {F}_p)=0$ for all primes $p$.

Let $p$ be a prime. For each $k \in \mathbb {N}$ and space $X$, let $b_k^{p}(X)$ denote the dimension of $H_k(X; \mathbb {F}_p)$ as a $\mathbb {F}_p$-vector space. Let $\widetilde {W}$ be the $\mathbb {Z}/n$-cover of $W$, and observe that since $\widetilde {W}$ is simply connected we have $H_1(\widetilde {W}; \mathbb {F}_p)=0$. It follows that

\[ 0=n \cdot \chi(W)= \chi(\widetilde {W})= 1+ b_2^{p}(\widetilde {W})- b_3^{p}(\widetilde {W}). \]

Here the first equality holds since $W$ has the same Euler characteristic as a closed 3-manifold. By considering the long exact sequence of the pair $(W, \partial W)$, we obtain that

\[ H^{3}(\widetilde {W}; \mathbb{F}_p) \cong H_1(\widetilde {W}, \partial \widetilde {W}; \mathbb{F}_p) \cong \mathbb{F}_p, \]

and hence that $b_2^{p}(\widetilde {W})=0$. It then follows from the same long exact sequence that $b_1^{p}(\partial \widetilde {W})= b_1^{p}( Y_n \sqcup S^{3})=0$, and so we have established the first claim.

Let $E(K)$ denote the exterior of the knot $K$ in $S^{3}$ and $\mu _K$ and $\lambda _K$ denote the meridian and longitude respectively. By definition the manifold $S^{3}_n(K)= E(K) \cup (S^{1} \times D^{2})$, where $\{\operatorname {pt}\} \times \partial D^{2}$ is identified with $n \mu _K + \lambda _K$ and $S^{1} \times \{\operatorname {pt}\}$ with $\lambda _K$. Therefore, we have that

\[ Y_n= ( E(K) \cup (S^{1} \times D^{2}))_n= E_n(K) \cup (S^{1} \times D^{2}), \]

where $\{\operatorname {pt}\} \times \partial D^{2}$ is identified with $\widetilde {\mu _K^{n}}+ \widetilde{\lambda }_K$ and $S^{1} \times \{\operatorname {pt}\}$ with $\widetilde{\lambda }_K$, where $\widetilde{\cdot }$ denotes a lift to the $n$-fold cover $E_n(K)\to E(K)$.

We also have that

\[ \Sigma_n(K)= E_n(K) \cup (S^{1} \times D^{2}) \]

where $\{\operatorname {pt}\} \times \partial D^{2}$ is identified with $\widetilde {\mu _K^{n}}$ and $S^{1} \times \{\operatorname {pt}\}$ with $\widetilde{\lambda }_K$. Since $\widetilde{\lambda }_K$ is null-homologous in $E_n(K)$, we see that $H_1(\Sigma _n(K))$ and $H_1(Y_n)$ are isomorphic quotients of $H_1(E_n(K))$.

We extract the next statement from the proof of Proposition 3.3 for later use.

Corollary 3.4 Let $K$ be a knot such that $S^{3}_n(K)$ and $L(n,1)$ are homology cobordant via a homology cobordism $W$ with $\pi _1(W) \cong \mathbb {Z}/n$ and let $\widetilde {W}$ denote the $\mathbb {Z}/n$-cover of $W$. Then $H_2(\widetilde {W};\mathbb {F}_p)=0$ for every prime $p$.

We would now like to prove the following. Note that the first statement implies that if a knot $K$ is $n$-shake slice then $\sigma _{\omega }(K)=0$ for every $q\textrm {th}$ root of unity $\omega$, where $q$ is a prime power dividing $n$. This was originally proved using different methods by Tristram [Reference TristramTri69].

Proposition 3.5 [Reference TristramTri69]

Let $K$ be a knot such that $S^{3}_n(K)$ and $L(n,1)$ are homology cobordant via a homology cobordism $W$. Then $\sigma _{\omega }(K)=0$ for every $q{th}$ root of unity $\omega$, where $q$ is a prime power dividing $n$. If $\pi _1(W) \cong \mathbb {Z}/n$, then $\sigma _{\omega }(K)=0$ for every $n{th}$ root of unity $\omega$.

Our strategy in proving Proposition 3.5 will be to relate the Tristram–Levine signature of $K$ at the $n\textrm {th}$ roots of unity to the Atiyah–Singer/Casson–Gordon signatures [Reference Atiyah and SingerAS68, Reference Casson and GordonCG78] of the 3-manifold $S^{3}_n(K)$. We will then use the $n$-fold cyclic cover of the hypothesised homology cobordism $W$ between $S^{3}_n(K)$ and $L(n,1)$, capped off in a certain nice way, to compute these Casson–Gordon signatures. We therefore recall the definition of the Casson–Gordon signatures of a 3-manifold, as given in [Reference Casson and GordonCG78]. For every $n \in \mathbb {N}$ we think of the cyclic group $\mathbb {Z}/n$ as coming with a canonical multiplicative generator $t$.

To a closed oriented 3-manifold $Y$ and a map $\phi \colon H_1(Y) \to \mathbb {Z}/n$, we wish to associate $\sigma _k(Y, \phi ) \in \mathbb {Q}$ for $k=1, \dots , n-1$. Let $\widetilde{Y} \to Y$ be the covering induced by $\phi$, and note that there is a canonical covering transformation $\tau$ of $\widetilde{Y}$ corresponding to $t \in \mathbb {Z}/n$.

Now suppose there exists a $\mathbb {Z}/n$ branched covering of 4-manifolds $\widetilde{Z} \to Z$, branched over a surface $F$ contained in the interior of $Z$, and such that $\partial (\widetilde{Z} \to Z)=(\widetilde{Y} \to Y)$. Suppose in addition that the covering transformation $\widetilde{\tau }\colon \widetilde{Z}\to \widetilde{Z}$ that induces rotation through $2 \pi /n$ on the fibres of the normal bundle of $\widetilde{F}$ is such that it restricts on $\widetilde{Y}$ to $\tau$. An explicit construction as given, for example, in the proof of [Reference Casson and GordonCG78, Lemma 3.1], shows that such a branched cover does always exist. Then $H_2(\widetilde{Z};\mathbb {C})$ decomposes as $\bigoplus _{k=0}^{n-1} V_k$, where $V_k$ is the $\xi _n^{k}$-eigenspace of the $\widetilde{\tau }$-induced action on second homology, and as before $\xi _n := e^{2 \pi i/n}$. Let $\sigma _k(\widetilde{Z})$ denote the signature of the intersection form of $\widetilde{Z}$ restricted to $V_k$. Define, for $k=1, \dots , n-1$, the signature defect

(3.6)\begin{equation} \sigma_k(Y, \phi) := \sigma(Z)- \sigma_k(\widetilde{Z}) - \frac{2 ([F]\cdot [F]) k (n-k)}{n^{2}}. \end{equation}

Casson and Gordon [Reference Casson and GordonCG78] used the Atiyah–Singer $G$-signature theorem [Reference Atiyah and SingerAS68] to show that $\sigma _k(Y, \phi )$ is an invariant of the pair $(Y,\phi )$ for each $0< k< n$.

Proposition 3.7 Let $K$ be a knot and $W$ be a cobordism between $S^{3}_n(K)$ and $L(n,1)$ over $\mathbb {Z}/n$. Let $\xi _n=e^{2 \pi i/n}$ and $1 \leq k \leq n-1$. Then

\[ \sigma_{\xi_n^{k}}(K)= \sigma_k(\widetilde {W})- \sigma(W), \]

where $\sigma _k(\widetilde {W})$ is the signature of the intersection form of the $\mathbb {Z}/n$-cover of $W$ induced by the map $H_1(W) \to \mathbb {Z}/n$ when restricted to the $\xi _n^{k}$-eigenspace of the action of the generator of the group of deck transformations on $H_2(\widetilde {W}, \mathbb {C})$.

Proof. Our proof follows from computing $\sigma _k(S^{3}_n(K), \phi )$, where $\phi \colon H_1(S^{3}_n(K)) \to \mathbb {Z}/n$ is the canonical map sending the meridian $\mu _K$ to $1 \in \mathbb {Z}/n$, in two different ways.

First, observe that in this setting the surgery formula of [Reference Casson and GordonCG78, Lemma 3.1] is particularly simple and reduces to

(3.8)\begin{equation} \sigma_k(S^{3}_n(K), \phi)= 1- \sigma_{\xi_n^{k}}(K)-\frac{2k(n-k)}{n}. \end{equation}

Second, let $\widetilde {W} \to W$ be the $n$-fold cyclic cover. We have that

\[ \partial(\widetilde {W} \to W)= (\widetilde{S^{3}_n(K)} \to S^{3}_n(K)) \sqcup (S^{3} \to L(n,1)). \]

Now let $X_n(U)$ denote the $n$-trace of the unknot, i.e. the disc bundle $D_n$ over $S^{2}$ with euler number $n$. Let $S$ be the $n$-framed embedded 2-sphere in $X_n(U)$. There is an $n$-fold cyclic branched cover $\widetilde{X}_S$ of $X_n(U)$ along $S$, which restricts on the boundary to the same (unbranched) cover $S^{3} \to L(n,1)$ we saw above. Note that $\widetilde{X}_S$ is a punctured $\mathbb {CP}^{2}$.

We can therefore use $Z:= W \cup _{L(n,1)} X_n(U)$ and $\widetilde{Z}= \widetilde {W} \cup _{S^{3}} \widetilde{X}_S$ to compute $\sigma _k(S^{3}_n(K), \phi )$ using (3.6). Note that $H_2(\widetilde{Z}) \cong H_2(\widetilde {W}) \oplus \mathbb {Z}$, where the generator of the $\mathbb {Z}$ summand is represented by the lift of $S$ and hence has self-intersection $+1$, intersects trivially with all elements of $H_2(\widetilde {W})$, and is preserved under the action of the covering transformation so lies in the $1$-eigenspace. Consequently, for $0< k< n$ we have $\sigma _k(\widetilde{Z})=\sigma _k(\widetilde {W})$ and hence

(3.9)\begin{align} \sigma_k(S^{3}_n(K), \phi)&= \sigma(Z)- \sigma_k(\widetilde{Z}) - \frac{2 ([S] \cdot [S]) k(n-k)}{n^{2}} \nonumber\\ &= (\sigma(W)+1)- \sigma_k(\widetilde {W}) - \frac{2nk(n-k)}{n^{2}}. \end{align}

Since $\sigma _k(S^{3}_n(K), \phi )$ is well defined, by comparing the formulae of (3.8) and (3.9) we obtain as desired that

\[ \sigma_{\xi_n^{k}}(K)= \sigma_k(\widetilde {W})- \sigma(W). \]

Versions of the next lemma have appeared in many places, the earliest of which we know of is [Reference LevineLev94, Lemma I.4.3 and II.3.2]. See also [Reference Cochran, Orr and TeichnerCOT03, Proposition 2.10].

Lemma 3.10 Let $i \colon X \to Y$ be a map of $($spaces homotopy equivalent to$)$ finite CW complexes that induces isomorphisms $i_* \colon H_k(X; \mathbb {Z}) \to H_k(Y;\mathbb {Z})$ for all $k$. Suppose $\varepsilon \colon H_1(Y;\mathbb {Z}) \to \mathbb {Z}/{q}$ is a surjective map inducing $\mathbb {Z}/q$-covers $\widetilde{Y} \to Y$ and $\widetilde{X} \to X$.

If $q$ is a prime power, then the induced map

\[ \widetilde{i}_* \colon H_k(\widetilde{X}; \mathbb{Q}) \to H_k(\widetilde{Y};\mathbb{Q}) \]

is an isomorphism for all $k$.

Proof. Let $\alpha \colon H_1(Y) \to \operatorname {GL}_q(\mathbb {Q})$ be the map obtained by composing $\varepsilon$ with the regular representation

\begin{align*} \mathbb{Z}/q &\to \operatorname{GL}_q(\mathbb{Q}) \\ k &\mapsto \left[ \begin{array}{ccccc} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ 1 & 0 & 0 & \dots & 0 \end{array} \right]^{k}. \end{align*}

As usual, this regular representation endows $\mathbb {Q}^{q}$ with the structure of a free $\mathbb {Q}[\mathbb {Z}/q]$-module of rank one.

From the work of Friedl and Powell [Reference Friedl and PowellFP12, Proposition 4.1] (applied with, in their notation, $H=\{1\}$) we have that

\[ i_* \colon H_*(X; \mathbb{Q}^{q}) \to H_*(Y; \mathbb{Q}^{q}) \]

is an isomorphism. Then for $Z\in \{X,Y\}$, we have natural identifications

\[ H_*(Z; \mathbb{Q}^{q})= H_*(C_*(\widetilde{Z}, \mathbb{Q}) \otimes_{\mathbb{Q}[\mathbb{Z}/q]} \mathbb{Q}^{q})=H_*(C_*(\widetilde{Z}, \mathbb{Q}) \otimes_{\mathbb{Q}[\mathbb{Z}/q]} \mathbb{Q}[\mathbb{Z}/q])=H_*(\widetilde{Z}, \mathbb{Q}), \]

and the desired result follows.

Proof of Proposition 3.5. Let $\widetilde {W}$ denote the $\mathbb {Z}/n$ cover of $W$. Note that since $W$ is a homology cobordism between $S^{3}_n(K)$ and $L(n,1)$ we have that $H_2(W)=0$ and so certainly $\sigma (W)=0$.

The case when $\pi _1(W)\cong \mathbb {Z}/n$ follows quickly: Corollary 3.4 tells us that $H_2(\widetilde {W}; \mathbb {F}_p)=0$ for all primes $p$, and hence that $H_2(\widetilde {W}; \mathbb {Z})=0$ and so $H_2(\widetilde {W}; \mathbb {C})=0$. Therefore, by Proposition 3.7 we have for $k=1, \ldots , n-1$ that

\[ \sigma_{\xi_n^{k}}(K)= \sigma_k(\widetilde {W})- \sigma(W)=0-0=0. \]

So we now assume only that $W$ is a homology cobordism, with no condition on the fundamental group. Let $q$ be a prime power dividing $n$, and let $1 \leq k \leq q-1$ be relatively prime to $q$. Let $\phi \colon H_1(S^{3}_n(K)) \to \mathbb {Z}/q$ be the map sending the class $[\mu _K]$ of the meridian of $K$ to $+1 \in \mathbb {Z}/q$. We now argue exactly as in the proof of Proposition 3.7 to show that

\[ (\sigma(W)+1)- \sigma_k(\widetilde {W}) - \frac{2k(q-k)}{q}= \sigma_k(S^{3}_n(K), \phi)= 1- \sigma_{\xi_q^{k}}(K)-\frac{2k(q-k)}{q} \]

and hence, since $H_2(W; \mathbb {Q})=0$, that

\[ \sigma_{\xi_q^{k}}(K)= \sigma_k(\widetilde {W})- \sigma(W)= \sigma_k(\widetilde {W}). \]

But by Lemma 3.10, since the inclusion induced map

\[ i_* \colon H_*(L(n,1); \mathbb{Z}) \to H_*(W; \mathbb{Z}) \]

is an isomorphism and $q$ is a prime power, we have that

\[ \widetilde{i}_* \colon H_*(\widetilde{L(n,1)}; \mathbb{Q}) \to H_*(\widetilde {W}; \mathbb{Q}) \]

is also an isomorphism. But since $\widetilde {L(n,1)}$ is itself a lens space (or $S^{3}$ if $n=q$), we have that $H_2(\widetilde {L(n,1)}; \mathbb {Q})=0$ and so $H_2(\widetilde {W}; \mathbb {Q})=0$ as well. Thus the $\xi _n^{k}$-eigenspace $V_k=0$, and so as desired

\[ \sigma_{\xi_q^{k}}(K)= \sigma_k(\widetilde {W})=0. \]

4. Setting up the surgery problem

We will use surgery theory to construct the exterior of the desired embedded sphere in an $n$-trace. We will eventually apply surgery in the topological category, but our initial input manifolds will be smooth. We thus now recall the input to a surgery problem in the smooth category. There is an analogous theory in the topological category, and we will discuss this below at the point when it becomes necessary.

Definition 4.1 Given a smooth $m$-manifold $X$, the tangent bundle is classified up to isomorphism by a homotopy class of maps $\tau _X\colon X\to \operatorname {BO}(m)\subset \operatorname {BO}$. The unique stable bundle $\nu _X\colon X\to \operatorname {BO}$ such that $\tau _X\oplus \nu _X\colon X\to \operatorname {BO}$ is null-homotopic is called the stable normal bundle of $X$. The manifold $X$ can be stably framed if $\nu _X$ is null-homotopic, and a choice of null homotopy is called a stable framing. A choice of stable framing for $X$ is determined by a choice of stable trivialisation of the tangent bundle,that is, a choice of $k$ and vector bundle isomorphism $TX\oplus \underline {\mathbb {R}}^{k}\cong \underline {\mathbb {R}}^{m+k}$.

Recall that an oriented $m$-manifold $X$ with (possibly empty) boundary has a fundamental class, denoted $[X,\partial X]\in H_m(X,\partial X;\mathbb {Z})$, capping with which induces (twisted) Poincaré–Lefschetz duality isomorphisms $H^{m-k}(X,\partial X;\mathbb {Z}[\pi _1(X)]) \xrightarrow {\cong } H_{k}(X;\mathbb {Z}[\pi _1(X)])$ and $H^{m-k}(X;\mathbb {Z}[\pi _1(X)]) \xrightarrow {\cong } H_{k}(X,\partial X;\mathbb {Z}[\pi _1(X)])$ for every $k$.

Definition 4.2 A map $(f,\partial f)\colon (X,\partial X)\to (Y,\partial Y)$ of smooth oriented $m$-manifolds with (possibly empty) boundary is called degree one if $f_*([X, \partial X])=[Y,\partial Y]$. Given a degree-one map $f$, a normal structure is an isomorphism of stable bundles $\nu _X\simeq \nu _Y\circ f$. A degree-one map with choice of normal structure is called a degree-one normal map. We will often write $(X,f)$ for the data of a degree-one normal map (suppressing the choice of stable bundle isomorphism).

For a topological space $Y$, a bordism $Z$ between closed $m$-manifolds $X$ and $X'$ is over Y if there is a proper map $F\colon Z\to Y\times I$ such that $F(X)\subset Y\times \{0\}$ and $F(X')\subset Y\times \{1\}$. If $Y$ and $Z$ are smooth oriented $m$-manifolds and the map $F$ is a degree-one normal map, then we call $(Z,F)$ a degree-one normal bordism from $(X,F|_X)$ to $(X',F|_{X'})$.

Remark 4.3 Given a degree-one map $(f,\partial f)\colon (X,\partial X)\to (Y,\partial Y)$, if $\nu _Y$ is null-homotopic, then so is $\nu _Y\circ f$. So $f$ admits a normal structure if and only if $X$ can be stably framed.

Furthermore, we will sometimes be interested in picking a normal structure on $f$ that is compatible with a given one on $\partial f$. To understand this, suppose we are given a choice of stable framing on $Y$. This induces a choice of stable framing on $\partial Y$. Suppose we have a degree-one map $\partial f\colon \partial X\to \partial Y$. A choice of normal structure on $\partial f$ is equivalent to a choice of stable framing on $\partial X$. Suppose such a choice has been made. Then a degree-one map $(f,\partial f)\colon (X,\partial X)\to (Y,\partial Y)$ admits a normal structure inducing the given one on the boundary if and only if $X$ can be stably framed compatibly with $\partial X$.

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