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 $\operatorname {Arf}(C_{n,1}(K))=0$.

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.

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

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

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

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 $y_j= {1}/{2n(8j +3)}$.

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.

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 $x_j= {1}/{2n(8(N+j)+1)}$.

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.

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.

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 $H_1(E_n(K))$.

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

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

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. There is a standard degree-one map from $E(K) \to E(U)$ that realises the homology equivalence $E(K)\to S^{1}$ and is the identity map on the boundary; see e.g. [Reference Miller and PowellMP19, Construction 7.1] for the details. As submanifolds of $S^{3}$, both $E(K)$ and $E(U)$ have stably trivial tangent bundles and can thus be stably framed. So this degree-one map can be given a choice of normal structure by Remark 4.3. Now extend this degree-one normal map to the Dehn filling.

We construct the homotopy equivalence $\bar {f}$. By construction, $f\colon S^{3}_n(K)\rightarrow S^{3}_n(U)=L(n,1)$ sends a meridian of $K$ to a meridian of the unknot $U$. We also have a handle decomposition of $X_n(K)$ relative to its boundary consisting of a $2$-handle, attached along a meridian of $K$, and a $4$-handle. Define $\bar {f}\colon X_n(K)\to X_n(U)=D_n$ by mapping the $2$-handles and the $4$-handles homeomorphically to each other. To do this, first note that the attaching circle of the 2-handle of $X_n(U)$ is a meridian of $U$, so the attaching circle of the 2-handle of $X_n(K)$ is sent to the attaching circle of the 2-handle of $X_n(U)$. Also the framings agree, so we can extend over the 2-handle. Then note that attaching the 2-handles undoes the Dehn surgeries, converting both $S^{3}_n(K)$ and $S^{3}_n(U)$ to $S^{3}$. The 4-handles are attached to these copies of $S^{3}$. Since every orientation-preserving homeomorphism of $S^{3}$ is isotopic to the identity, we may extend the map over the 4-handles. Observe that $\bar {f}$ is a homotopy equivalence by Whitehead's theorem.

Proof. Fix a cell complex $A\cong S^{1}\times S^{1}$ and extend it to a cell complex $B\cong S^{1}\times D^{2}$. Denote by $Y(K)$ and $Y(U)$ any fixed choice of cell structures for $E(K)$ and $E(U)$ (we will make quite specific choices later). For each of $J=U, K$, let $M(J)$ denote the cell complex obtained as the mapping cylinder of a map $A\to Y(J)$ which is a cellular approximation to the inclusion $S^{1}\times S^{1}\to E(J)$.

Each of these spaces has a $\mathbb {Z}/n$ cover, determined by $\pi _1(L(n,1))\cong \mathbb {Z}/n$ and composing with, where appropriate, the map $f$. Choose a lift of the cell structure on $A$ to the $\mathbb {Z}/n$ cover, and denote this by $\widetilde{A}$. Extend this lift to $B$, $M(J)$ and $M(U)$ and write $\widetilde{B}$, $\widetilde {M(J)}$ and $\widetilde {M(U)}$ for the corresponding covering spaces. Write $\widetilde{f}$ for a lift of $f$ to the $\mathbb {Z}/n$ covers.

In this diagram we use cellular chain complexes and $\mathscr {C}$ denotes taking an algebraic mapping cone. Strictly, we have replaced the maps $\widetilde{f}$ and $\widetilde{f}|_{\widetilde {M(K)}}$ by cellular approximations. This is thus a diagram in the category of finite, finitely generated, based $\mathbb {Z}[\mathbb {Z}/n]$-coefficient chain complexes. All complexes in the lower sequence are acyclic so by the multiplicativity of torsion under such exact sequences, and the fact that the identity map has vanishing torsion, we obtain $\tau (f)=\tau (\widetilde{f}_*)=\tau ((\widetilde{f}|_{\widetilde {M(K)}})_*)\in \operatorname {Wh}(\mathbb {Z}/n)$.

We now choose convenient cell structures for $E(K)$ and $E(U)$. Let $Y(K)$ and $Y(U)$ be the cell structures associated to Wirtinger presentations of the respective knot groups (see e.g. proof of [Reference TuraevTur01, Theorem 16.5]), where for $K$ we choose an arbitrary such presentation and for $U$ we choose the presentation of $\mathbb {Z}$ with no relations. By [Reference TuraevTur01, Lemma 8.4], the Whitehead torsions $\tau (M(J),Y(J))=1$ for each of $J=K,U$, so we have that $\tau ((\widetilde{f}|_{\widetilde {M(K)}})_*)$ may be computed by a cellular map $F\colon \widetilde {Y(K)}\to \widetilde {Y(U)}$ representing $\widetilde{f}|_{\widetilde {E(K)}}$. This map is given by

where we are writing $\mathbb {Z}[\mathbb {Z}/n]=\mathbb {Z}[t, t^{-1}]/(t^{n}-1)$, $m+1$ is the number of generators in the Wirtinger presentation for $\pi _1(K)$, and $\partial _2(t)$ is determined by the relations in that presentation. Up to basis change, and a simple homotopy equivalence, the algebraic mapping cone $\mathscr {C}(F)$ is

where $\partial '_2(t)$ is the effect of deleting a row from the matrix $\partial _2(t)$, after the basis change. We then compute that $\tau ((\widetilde{f}|_{\widetilde {M(K)}})_*)=\tau (F)=\big [ \partial '_2(t)\big ]\in \operatorname {Wh}(\mathbb {Z}/n)$.

When $G$ is an abelian group, the determinant map $\det \colon K_1(\mathbb {Z}[G])\to (\mathbb {Z}[G])^{\times }$ is a split surjection (see e.g. [Reference MilnorMil66, p. 359]), and for any finite cyclic $G$ the kernel of this map vanishes [Reference OliverOli88, Theorem 5.6]. So the determinant gives an isomorphism $K_1(\mathbb {Z}[\mathbb {Z}/n])\cong (\mathbb {Z}[\mathbb {Z}/n])^{\times }$. The matrix $\partial _2'(t)$ is a presentation matrix for the Alexander module of $K$ when considered over the ring $\mathbb {Z}[t,t^{-1}]$. From this, and using the left regular representation of $\mathbb {Z}[\mathbb {Z}/n]$, we may consider $\partial '_2$ as an $mn\times mn$ matrix over $\mathbb {Z}$, such that the absolute value of the determinant $|{\det (\partial '_2)}|$ is the order of the first homology of the $n$-fold branched cover $H_1(\Sigma _n(K);\mathbb {Z})$. This, in turn, is the order of $H_1(S^{3}_n(K);\mathbb {Z}[\mathbb {Z}/n])$, which we have assumed to be $1$. Thus, as an element of $K_1(\mathbb {Z}[\mathbb {Z}/n])\cong (\mathbb {Z}[\mathbb {Z}/n])^{\times }$, we have that $\tau (F)=\det (\partial '_2(t))=\pm 1$, under this isomorphism. Finally, both $+1$ and $-1$ become the trivial element on passage to the Whitehead group $\operatorname {Wh}(\mathbb {Z}/n)$, so we obtain the desired result.

Proof. Write $f \colon S_n^{3}(K) \to L(n,1)$ for the degree-one normal map obtained in Lemma 4.4. By Remark 4.3, if we can show there is a degree-one map $F\colon W\to L(n,1)\times I$ describing a stably framed cobordism over $L(n,1)$ from $(S_n^{3}(K),f)$ to $(L(n,1),\operatorname {Id})$, we will be done. (Note, we may change the choice of stable framing on $S_n^{3}(K)$ during the course of the proof.)

Consider the closed 3-manifold

\[ N_K:= E(K) \cup_{S^{1} \times S^{1}} S^{1} \times S^{1} \times I \cup_{S^{1} \times S^{1}} -E(U), \]

glueing so that the result is homeomorphic to the 0-framed surgery $S^{3}_0(K)$, but we decompose it in this way for later. Choose a framing of the tangent bundle of $N_K$ that is a product framing on $S^{1} \times S^{1} \times I$. This determines an element of $[N_K,g]\in \Omega _3^{\operatorname {fr}}(S^{1} \times D^{2}\times I)$, where the map $g\colon N_K\to S^{1} \times D^{2} \times I = E(U) \times I$ has image

\[ \partial (S^{1} \times D^{2} \times I) = S^{1} \times D^{2} \cup S^{1} \times S^{1} \times I \cup{-} S^{1} \times D^{2}, \]

and is obtained by glueing together the standard degree-one map $E(K) \to E(U) = S^{1} \times D^{2}$, the identity on $S^{1} \times S^{1} \times I$, and the canonical identification $E(U) \to S^{1} \times D^{2}$.

We have $\Omega _3^{\operatorname {fr}}(S^{1} \times D^{2}\times I) \cong \Omega ^{\operatorname {fr}}_3(S^{1})$. The map from $S^{1}$ to a point induces a split short exact sequence $0\to \widetilde{\Omega }_3^{\operatorname {fr}}(S^{1})\to \Omega ^{\operatorname {fr}}_3(S^{1})\to \Omega ^{\operatorname {fr}}_3\to 0$. Thus $\Omega ^{\operatorname {fr}}_3(S^{1})\cong \Omega ^{\operatorname {fr}}_3\oplus \Omega ^{\operatorname {fr}}_2$, where we have used the isomorphism $\widetilde{\Omega }^{\operatorname {fr}}_3(S^{1})\cong \widetilde{\Omega }^{\operatorname {fr}}_2(S^{0})\cong \Omega ^{\operatorname {fr}}_2$, coming from the fact that $\Omega _*^{\operatorname {fr}}(-)$ is a generalised homology theory and thus satisfies the suspension axiom. Under this isomorphism, the projection of $[N_K,g]$ to $\Omega _3^{\operatorname {fr}}\cong \mathbb {Z}/24$ is given by the framed bordism class of $N_K$. Since any element in $\Omega _3^{\operatorname {fr}}\cong \mathbb {Z}/24$ can be realised by a suitable framing on $S^{3}$, the framing of $N_K$ can be modified in a small neighbourhood until the element in $\Omega _3^{\operatorname {fr}}\cong \mathbb {Z}/24$ vanishes; see e.g. [Reference Cha and PowellCP14, p. 13] for details.

The map from $\Omega _2^{\operatorname {fr}}\cong \widetilde{\Omega }_2^{\operatorname {fr}}(S^{0})\cong \widetilde{\Omega }_3^{\operatorname {fr}}(S^{1})\to \Omega _3(S^{1})$ is given by applying the reduced suspension, sending the class of $f\colon F^{2}\to \operatorname {pt}$ to $f\times \operatorname {Id}\colon F\times S^{1}\to S^{1}$ in $\Omega _3(S^{1})$, and framing $F\times S^{1}$ via the product framing with the trivial framing on $S^{1}$. Under the projection $\Omega _3(S^{1})\to \Omega _2^{\operatorname {fr}}$, the class $[N_K,g]$ is thus mapped to a framed surface $F$ given by taking the transverse preimage of a regular point in $S^{1}$ under $g$. Recall that $g$ was obtained using the standard degree-one normal maps $E(K)\to E(U)$ and $E(U)\to E(U)$. Under these, generic transverse point preimages are Seifert surfaces, $F_K$ and $F_U$, respectively for the knots $K$ and $U$. Thus the projection to $\Omega _2^{\operatorname {fr}}$ is the Arf invariant of the framed surface $F=F_K \cup (\{\operatorname {pt}\} \times S^{1} )\cup - F_U$. Since we assumed that $\operatorname {Arf}(K)=0$ this component already vanishes in $\Omega ^{\operatorname {fr}}_2$. So overall we obtain $[N_K,g]=0\in \Omega _3^{\operatorname {fr}}(S^{1} \times D^{2}\times I)$

Write $W'$ for a framed null bordism of $(N_K,g)$ over $S^{1} \times D^{2}\times I$. Now attach $D^{2}\times S^{1} \times I$ to $S^{1} \times S^{1} \times I$ such that $D^{2}\times S^{1} \times \{t\}$ attaches to $S^{1} \times S^{1} \times \{t\}$ with the $n$-framing, and similarly attach the $n$-framed $D^{2}\times S^{1} \times I$ to the codomain $S^{1}\times D^{2} \times I$. This yields a map $F\colon W\to L(n,1)\times I$ that describes a framed cobordism from $(S_n^{3}(K),f)$ to $(L(n,1),\operatorname {Id})$, over $L(n,1)$. To see that the cobordism is framed, note that we can glue the two framings together along the product framing on both pieces $W'$ and $D^{2} \times S^{1} \times I$.

We finally also note that the map $F\colon W\to L(n,1)$ is degree one. This can be computed by considering the naturality of the long exact sequences in homology of the pairs $(W,\partial W)$ and $(L(n,1) \times I,L(n,1) \times \{0,1\})$, and the fact that both components of $\partial F$ are already known to be degree one.

Proof. Write $f \colon S_n^{3}(K) \to L(n,1)$ for the degree-one normal map obtained in Lemma 4.4. Again, by Remark 4.3, the objective is to show there is a degree-one map $F\colon W\to L(n,1)\times I$ describing a stably framed cobordism over $L(n,1)$ from $(S_n^{3}(K),f)$ to $(L(n,1),\operatorname {Id})$.

We show that there is a choice of stable framing for $S_n^{3}(K)$ such that $[S_n^{3}(K),f] - [L(n,1),\operatorname {Id}] =0 \in \Omega ^{\operatorname {fr}}_3(L(n,1))$.

Consider the Atiyah–Hirzebruch spectral sequence for $\Omega _*^{\operatorname {fr}}(-)$, with $E^{2}$ page $E^{2}_{p,q}= H_p(L(n,1);\Omega ^{\operatorname {fr}}_q)$ and converging to $\Omega _{p+q}^{\operatorname {fr}}(L(n,1))$. By inspection of the differentials, this sequence collapses already at the $E^{2}$ page, so the groups $H_p(L(n,1);\Omega ^{\operatorname {fr}}_q)$ form iterated graded quotients for a filtration of $\Omega ^{\operatorname {fr}}_3(L(n,1))$. We note that by the Pontryagin–Thom theorem, we have $\Omega _q^{\operatorname {fr}}= \mathbb {Z},\mathbb {Z}/2,\mathbb {Z}/2,\mathbb {Z}/24$ when $q=0,1,2,3$ respectively.

We analyse the groups on the $E^{2}$ page in turn by considering the CW decomposition of $L(n,1)$ with a single $p$-cell $e^{p}$ for each of $p=0,1,2,3$ and recalling that the $E^{1}$ page is given by

\[ E^{1}_{p,q}=C^{\text{cell}}_p(L(n,1);\mathbb{Z})\otimes\Omega^{\operatorname{fr}}_{q}\cong \widetilde{\Omega}^{\operatorname{fr}}_{q}(S^{0}) \cong \widetilde{\Omega}_{p+q}^{\operatorname{fr}}(e^{p}/\partial e^{p})\cong \Omega_{p+q}^{\operatorname{fr}}(X^{(p)},X^{(p-1)}) \]

where $X^{(p)}$ denotes the $p$-skeleton of $L(n,1)$. The groups on the $E^{2}$ page can be considered as a sequence of obstructions to finding a framed null-bordism of $S^{3}_n(K)\sqcup -L(n,1)$ over $L(n,1)$ so we analyse these obstructions in turn.

At $q=0$, a representative class for $[S_n^{3}(K),f] - [L(n,1),\operatorname {Id}]\in \Omega _3^{\operatorname {fr}}(L(n,1))$ is given by a cycle in $E^{1}_{3,0}=\Omega ^{\operatorname {fr}}_3(X^{(3)},X^{(2)})$. The resulting class in $E^{2}$ vanishes if we can do surgery on the disjoint union $S_n^{3}(K)\sqcup - L(n,1)$ in a way that respects the degree-one normal maps to $L(n,1)$ and such that the effect of surgery maps to the $2$-skeleton of $L(n,1)$. In other words we seek to compatibly stably frame the connected sum $S_n^{3}(K)\#(- L(n,1))$. The obstruction to doing this is the difference of $f_*[S_n^{3}(K)]-\operatorname {Id}_*[L(n,1)]\in H_3(L(n,1);\Omega _0^{\operatorname {fr}})$, which vanishes because the maps $f$ and $\operatorname {Id}$ are degree one.

Next, for $q=1$ and $q=2$ homology computations using that $n$ is odd show that both $H_2(L(n,1);\Omega ^{\operatorname {fr}}_1) = 0$ and $H_1(L(n,1);\Omega ^{\operatorname {fr}}_2)=0$, so there is no obstruction here. Thus we may assume we have done surgery on $S_n^{3}(K)\sqcup (-L(n,1))$ over $L(n,1)$ to obtain some closed, connected $Y$ together with a degree-one normal map $g\colon Y\to L(n,1)$, where the map $g\colon Y\to L(n,1)$ has codomain the $0$-skeleton, and $[Y,g]=[S_3(K),f)]-[L(n,1),\operatorname {Id}]\in \Omega _3^{\operatorname {fr}}(L(n,1))$.

Finally, for $q=3$, the last $E^{2}$ page obstruction is given by a class in $E^{1}_{0,3}=\Omega _3^{\operatorname {fr}}(X^{(0)},\emptyset )$ which is equal to $[Y]\in \Omega _3^{\operatorname {fr}}\cong \mathbb {Z}/24$. By framed bordism invariance, this is equal to $[S_n^{3}(K)]-[L(n,1)]\in H_0(L(n,1);\Omega ^{\operatorname {fr}}_3)\cong \Omega _3^{\operatorname {fr}}\cong \mathbb {Z}/24$. Similarly to the proof of Lemma 4.6, the framing near a point in $S_n^{3}(K)$ may be modified in a small neighbourhood to force vanishing of this obstruction.

As the various representative elements of $[S_n^{3}(K),f] - [L(n,1),\operatorname {Id}]\in \Omega _3^{\operatorname {fr}}(L(n,1))$ can be made to vanish on the $E_2$ page, this class vanishes in $\Omega ^{\operatorname {fr}}_3(L(n,1))$, and we obtain the required bordism. Similarly to the proof of Lemma 4.6, this bordism over $L(n,1)$ is seen to be a degree one normal bordism.

Proof. When $n$ is odd, we use Lemma 4.8, and when $n$ is even we use Lemma 4.6, to obtain a (smooth) degree-one normal bordism $W$. Now perform (smooth) 1-surgeries on the interior of $W$ to modify the fundamental group to $\mathbb {Z}/n$ while still retaining a degree-one normal map. We abuse notation and continue to write $F\colon W\to L(n,1)\times I$ for this degree-one map, so that now $\pi _1(W) \cong \mathbb {Z}/n$. Next take connected sums with the degree-one normal map $E_8\to S^{4}$ (or the oppositely oriented version $-E_8\to S^{4}$), in order to kill the signature of $W$. Here $E_8$ denotes the $E_8$ manifold, that is, a closed, simply connected topological $4$-manifold with intersection form given by the $E_8$ matrix. The result may no longer be smooth, but now has all the desired properties.

Proof. The quadratic form is non-singular because $H_1(S_n^{3}(K);\mathbb {Z}[\mathbb {Z}/n]) = 0$, by assumption. We claim this quadratic form is moreover simple (with respect to some stable $\mathbb {Z}[\mathbb {Z}/n]$ basis). Indeed, the manifold with boundary $(W,S_n^{3}(K)\sqcup -L(n,1))$ is a simple Poincaré pair [Reference Kirby and SiebenmannKS77, Essay III, Theorem 5.13], so it remains to check that the $\mathbb {Z}[\mathbb {Z}/n]$-homology equivalence on the boundary

\[ f\sqcup \operatorname{Id}\colon S^{3}_n(K)\sqcup{-}L(n,1)\to L(n,1)\sqcup{-}L(n,1) \]

has vanishing Whitehead torsion. But the identity map has vanishing Whitehead torsion, and so does $f$ by Lemma 4.5.