Proof. These claims are [Reference Juhász29, Proposition 9.18] (whose proof is attributed to Yi Ni), [Reference Juhász30, Theorem 1.4] and [Reference Juhász30, Theorem 9.7], respectively.

Proof. Since $(M,\gamma )$ is taut and $\dim \mathit {SFH}(M,\gamma )<4$ , [Reference Juhász31, Corollary 2.2] says that $(M,\gamma )$ is horizontally prime (see [Reference Juhász31, Definition 1.7]). If $(M,\gamma )$ is also reduced, meaning that it does not contain an essential product annulus, and if it is not one of the forbidden products, then [Reference Juhász32, Theorem 3] says that

$$\begin{align*}\dim \mathit{SFH}(M,\gamma) \geq \tfrac{1}{2} b_1(\partial M) + 1. \end{align*}$$

By hypothesis, this is not the case, so since $(M,\gamma )$ is not such a product, it is not reduced. (The products were not excluded in the statement of [Reference Juhász32, Theorem 3], but the proof assumes that there are no essential product disks in $(M,\gamma )$ , which by [Reference Juhász32, Lemma 2.13] holds if and only if $(M,\gamma )$ is not one of these products. See [Reference Ghosh and Li19, Remark 5.10].)

Proof. For any balanced sutured manifold $(M,\gamma )$ , a choice of homology orientation for the pair $(M,R_-(\gamma ))$ gives rise to an absolute lift of the relative $\mathbb {Z}/2\mathbb {Z}$ -grading on $\mathit {SFH}(M,\gamma )$ , and therefore to a well-defined Euler characteristic

$$\begin{align*}\chi(\mathit{SFH}(M,\gamma,\mathfrak{s}))\in \mathbb{Z}\end{align*}$$

for each $\mathfrak {s}\in \operatorname {Spin}^c(M,\gamma )$ , as described in [Reference Friedl, Juhász and Rasmussen15]. Fixing an $H_1(M)$ -affine isomorphism

$$\begin{align*}\iota: \operatorname{Spin}^c(M,\gamma)\to H_1(M),\end{align*}$$

these Euler characteristics can be packaged as an element

$$\begin{align*}\tau(M,\gamma) = \sum_{\mathfrak{s}\in\operatorname{Spin}^c(M,\gamma)} \chi(\mathit{SFH}(M,\gamma,\mathfrak{s})) \cdot \iota (\mathfrak{s})\end{align*}$$

of the group ring $\mathbb {Z}[H_1(M)]$ .

Let us write $E_K = S^3 \setminus N(K)$ for convenience. Then

$$\begin{align*}\tau(E_K,\gamma_0) = 0 \end{align*}$$

as shown in [Reference Friedl, Juhász and Rasmussen15, Example 8.1], which means that

$$\begin{align*}\chi(\mathit{SFH}(E_K,\gamma_0, \mathfrak{s})) = 0 \end{align*}$$

for each $\mathfrak {s} \in \operatorname {Spin}^c(E_K,\gamma _0)$ . In particular, $\dim \mathit {SFH}(E_K,\gamma _0,\mathfrak {s})$ is always even.

Since K is nontrivial, its complement $E_K$ is irreducible. Thus, if we let

$$\begin{align*}S = \{ \mathfrak{s} \in \operatorname{Spin}^c(E_K,\gamma_0) \mid \mathit{SFH}(E_K,\gamma_0,\mathfrak{s}) \not\cong 0 \}, \end{align*}$$

then [Reference Friedl, Juhász and Rasmussen15, Theorem 1.4] tells us that for all $\alpha \in H_2(E_K,\partial E_K;\mathbb {R})$ , we have

$$\begin{align*}\max_{\mathfrak{s},\mathfrak{t} \in S} \langle \mathfrak{s}-\mathfrak{t}, \alpha \rangle = x^s(\alpha), \end{align*}$$

where $x^s$ is the sutured Thurston norm on $(E_K,\gamma _0)$ . If $\alpha $ is the class of a Seifert surface for K, with genus $g= g(K) \geq 1$ , then we compute by [Reference Friedl, Juhász and Rasmussen15, Lemma 7.3] that

$$\begin{align*}x^s(\alpha) = x(\alpha) = 2g-1, \end{align*}$$

and since this is nonzero, there must be two different $\operatorname {Spin}^c$ structures $\mathfrak {s}$ on $(E_K,\gamma _0)$ , each pairing differently with $\alpha $ , for which $\mathit {SFH}(E_K,\gamma _0,\mathfrak {s})$ is nonzero. But then $\mathit {SFH}(E_K,\gamma _0)$ has dimension at least two in each of these two $\operatorname {Spin}^c$ structures, so we conclude that

$$\begin{align*}\dim \mathit{SFH}(E_K,\gamma_0) \geq 4, \end{align*}$$

as desired.

Proof. Let us check that the hypotheses of Theorem 2.6 are met. First, note that $S^3(F)$ is not one of the excluded products, since $R_+(\gamma )= F\times \{1\}$ is not an annulus or pair of pants. Next, we have that

(3.1) $$ \begin{align}H_2(S^3(F)) \cong \tilde{H}^0(F) \cong 0\end{align} $$

by Alexander duality. We also know that $S^3(F)$ is irreducible (in fact, this sutured manifold is taut, per Remark 2.2) and that

$$\begin{align*}\dim \mathit{SFH}(S^3(F)) = \dim \widehat{\mathit{HFK}}(K,g) = 2 \end{align*}$$

by (2.2). Note that $g\geq 1$ since the unknot is not nearly fibered. Therefore,

$$\begin{align*}\dim \mathit{SFH}(M,\gamma) =2\leq 2g = \tfrac{1}{2}b_1(\partial M),\end{align*}$$

and so Theorem 2.6 provides an essential product annulus $A\subset (M,\gamma )=S^3(F)$ .

Let us suppose for a contradiction that both boundary components

$$\begin{align*}A_\pm\subset R_\pm(\gamma)\end{align*}$$

of A are boundary-parallel in their respective surfaces. We recover the knot complement

$$\begin{align*}E_K = S^3 \setminus N(K)\end{align*}$$

from M by gluing $R_+(\gamma )$ to $R_-(\gamma )$ by some homeomorphism, and we can assume that this gluing map sends $A_+$ to $A_-$ since these curves are boundary-parallel in $R_\pm (\gamma )$ , respectively. Then A becomes a torus $T \subset E_K$ which meets F in a boundary-parallel circle.

We first claim that T is incompressible in $E_K$ . Indeed, its fundamental group is spanned by a longitude $\lambda $ of K and the image c of a curve

$$\begin{align*}\{\mathrm{pt}\} \times [-1,1] \subset S^1 \times [-1,1] \cong A, \end{align*}$$

which is homologically essential in $E_K$ since it is dual to F. If some product $\lambda ^i c^j$ is nullhomotopic in $E_K$ , then its homology class satisfies

$$\begin{align*}0 = [\lambda^ic^j]\cdot F = j,\end{align*}$$

so it is a power $\lambda ^i$ of the longitude of K, but then $i=0$ since K is a nontrivial knot in $S^3$ . Therefore, $\lambda ^i c^j$ is nullhomotopic in T as well.

We next claim that T is not boundary-parallel. Indeed, if it were, then T and $\partial E_K$ would cobound a thickened torus intersecting F in a properly embedded annulus, in which case cutting $E_K$ back open along F would give a thickened annulus in $(M,\gamma )$ which is the trace of an isotopy between A and $\gamma $ that keeps $\partial A$ in $R(\gamma )$ at all times. But A is essential, which by definition implies that no such isotopy exists – a contradiction.

We have shown that under these circumstances, K must be a satellite knot, and the torus T splits its exterior into two pieces: the exterior $E_C$ of the companion C and the exterior $E_P$ of the pattern $P\subset S^1\times D^2$ . But then T splits the Seifert surface F into two pieces as well, one of which is an annulus in $E_P$ cobounded by the image of $A_\pm $ and the boundary $\partial F$ . This annulus gives an isotopy of the pattern P into T, where it is identified with a longitude of C, so P must be a cable pattern with winding number one. But this means that P is isotopic to the core of $S^1\times D^2$ , so T is boundary-parallel, and we have a contradiction. We conclude that $A_\pm $ cannot both be boundary-parallel, as desired.

Proof. Forgetting about the sutures, note that the Seifert surface complement

$$\begin{align*}S^3(F)\cong S^3\setminus \operatorname{int}(F\times[-1,1])\end{align*}$$

can be recovered from $M_\alpha $ by gluing a thickened annulus $N(A')$ along $\gamma _A$ by a map which identifies $\partial A'$ with $s(\gamma _A).$ The Mayer–Vietoris sequence associated to the decomposition

$$\begin{align*}S^3(F) \cong M_\alpha \cup_{\gamma_A} N(A') \end{align*}$$

with coefficients in R (which we momentarily suppress for convenience) reads in part

$$\begin{align*}\underbrace{H_2( \gamma_A)}_{\cong 0} \to H_2(M_\alpha) \oplus \underbrace{H_2(N(A'))}_{\cong 0} \to \underbrace{H_2(S^3(F))}_{\cong 0} \to H_1( \gamma_A) \to H_1(M_\alpha) \oplus H_1(N(A')). \end{align*}$$

We have that

$$\begin{align*}H_2(S^3(F); R) \cong \tilde{H}^0(F;R) \cong 0,\end{align*}$$

by Alexander duality, so the leftmost portion of the sequence tells us that $H_2(M_\alpha ;R) \cong 0$ , proving the first claim.

Moreover, the map $H_1( \gamma _A;R) \to H_1(N(A');R)$ sends the class $[s(\gamma _A)]$ to

$$\begin{align*}[\partial A']=0\in H_1(N(A');R).\end{align*}$$

Since the rightmost map in the sequence is injective, and

$$\begin{align*}r\cdot[s(\gamma_A)]\neq 0 \in H_1( \gamma_A;R) \textrm{ for all } r\in R\setminus \{0\},\end{align*}$$

it follows that

$$\begin{align*}r\cdot[s(\gamma_A)]\neq 0 \in H_1( M_\alpha;R) \textrm{ for all } r\in R\setminus \{0\}. \end{align*}$$

Note that the meridian $\mu _\alpha $ of $\alpha $ and the sutures $s(\gamma _A)$ cobound the pair of pants $R_+(\gamma _\alpha )\subset \partial M_\alpha .$ It follows that

$$\begin{align*}[\mu_\alpha] = \pm [s(\gamma_{A})]\in H_1(M_\alpha; R),\end{align*}$$

and therefore that

(4.3) $$ \begin{align}r\cdot[\mu_\alpha]\neq 0 \in H_1( M_\alpha;R) \textrm{ for all } r\in R\setminus \{0\}. \end{align} $$

To prove the second claim, note that $M_A$ is recovered from $M_\alpha $ by gluing back the neighborhood $N(\alpha )$ along the annular neighborhood $N(\mu _\alpha )$ of $\mu _\alpha $ ,

(4.4) $$ \begin{align} M_A \cong M_\alpha \cup_{N(\mu_\alpha)} N(\alpha). \end{align} $$

Let us consider the Mayer–Vietoris sequence corresponding to this decomposition. Since

$$\begin{align*}H_2(M_\alpha;R)\cong H_2(N(\alpha);R)\cong H_1(N(\alpha);R)\cong 0,\end{align*}$$

the portion of the sequence beginning at $H_2(M_\alpha ;R) \oplus H_2(N(\alpha );R)$ has the form

$$\begin{align*}0 \to H_2(M_A;R) \to \underbrace{H_1(N(\mu_\alpha);R)}_{\cong R} \to H_1(M_\alpha;R), \end{align*}$$

with $H_1(N(\mu _\alpha );R)$ generated by the class $[\mu _\alpha ]$ . Then it follows from (4.3) that the rightmost map is injective, and we conclude by exactness that $H_2(M_A;R) \cong 0$ , as desired.

Proof. We know that $(M_\alpha ,\gamma _\alpha )$ is a taut balanced sutured manifold, with $H_2(M_\alpha ;\mathbb {Z})=0$ by Lemma 4.2, and its boundary $\partial M_\alpha $ is a connected genus-2 surface. Then

$$\begin{align*}\dim \mathit{SFH}(M_\alpha,\gamma_\alpha) = 2 \end{align*}$$

by (4.2), so Theorem 2.6 provides the desired annulus.

Proof. Suppose that $B_+$ is isotopic in $\partial M_\alpha $ to $\mu _\alpha $ but $B_-$ is not. From the discussion above, $B_-$ must then be isotopic in $\partial M_\alpha $ to a component of $s(\gamma _A)$ . Recall that $M_A$ is obtained from $M_\alpha $ by gluing back a thickened disk (namely, the neighborhood $N(\alpha )$ ) along a neighborhood of the meridian $\mu _\alpha $ , as in (4.4). It follows that under the inclusion

$$\begin{align*}M_\alpha\hookrightarrow M_A,\end{align*}$$

the boundary component $B_+$ of the annulus B gets capped off with a disk D, so that

$$\begin{align*}B\cup D\subset M_A\end{align*}$$

is a disk bounded by the curve $B_-\subset \partial M_A$ . This disk then gives rise under the inclusion

$$\begin{align*}M_A\hookrightarrow M_F\end{align*}$$

to a disk in $M_F$ bounded by the image

$$\begin{align*}B_-\subset T_-\subset \partial M_F.\end{align*}$$

But $B_-$ is isotopic in $T_-$ to the boundary component $A_-$ of the annulus A, which by Lemma 3.4 is homologically essential. The fact that this curve bounds a disk in $M_F$ then contradicts the fact that the torus $T_-$ is incompressible, as shown in Lemma 3.3.

Swapping the roles of $B_+$ and $B_-$ leads to the same contradiction, so let us now assume that the curves $B_\pm $ are both are isotopic in $\partial M_\alpha $ to $\mu _\alpha $ . In this case, reversing the decompositions $S^3(K) \stackrel {F}{\leadsto } S^3(F) \stackrel {A}{\leadsto } (M_\alpha ,\gamma _\alpha )$ , we can glue $B_+$ to $B_-$ to turn the annulus B into a closed, embedded surface $\Sigma _B$ in $S^3(K)$ that meets F transversely in a single boundary-parallel curve. Then $\Sigma _B$ must be a torus, since if it were a Klein bottle, it could not embed in $S^3(K) \subset S^3$ ; as a torus in $S^3$ , it must bound a solid torus $V_B$ on one side or the other.

If $V_B \subset S^3$ were contained in the knot complement $S^3(K)$ , then

$$\begin{align*}V_B \cap F \subset V_B \end{align*}$$

would be a properly embedded, punctured torus (consisting of F minus a collar neighborhood of its boundary) in the solid torus $V_B$ ; but then it must compress inside $V_B$ and hence in $S^3(K)$ , contradicting the incompressibility of F. Thus, $V_B$ must not lie entirely in $S^3(K)$ , and this means that it must contain $\partial \big (S^3(K)\big ) = \partial N(K)$ as well as the knot K. We now argue exactly as in the proof of Lemma 3.1: the torus $\partial V_B = \Sigma _B$ must be incompressible in $S^3(K)$ , realizing K as a satellite knot, but then the annulus $F \cap V_B$ provides an isotopy from K to its companion knot, so the satellite pattern must have been trivial. This means that $\Sigma _B = \partial V_B$ is boundary-parallel in $S^3(K)$ . Decomposing again along F and then A, we conclude that our original annulus B must have been parallel to an annular neighborhood of $\mu _\alpha $ in $\partial M_\alpha $ . But this contradicts the claim from Lemma 4.3 that B is essential, so we are done.

Proof. Let us orient B as well as its boundary curves $B_+$ and $B_-$ so that

$$\begin{align*}\partial B= B_+ \sqcup -B_-.\end{align*}$$

Recall from Lemma 4.2 that $H_2(M_A) = 0$ . Therefore, the long exact sequence of the pair $(M_A,\partial M_A)$ reads in part

$$\begin{align*}0 \to H_2(M_A,\partial M_A) \xrightarrow{\partial_*} H_1(\partial M_A) \to H_1(M_A). \end{align*}$$

If B is nonseparating in $M_A$ , then it is nonzero in $H_2(M_A,\partial M_A)$ . It then follows from the exact sequence above that the class $[\partial B]$ is nonzero in $H_1(\partial M_A)$ , and hence that

(4.5) $$ \begin{align}[B_+] \neq [B_-]\in H_1(\partial M_A).\end{align} $$

Let us suppose for a contradiction that this is the case.

As discussed before Lemma 4.4, $B_+$ and $B_-$ are each isotopic in $\partial M_\alpha $ either to components of the sutures $s(\gamma _A)$ or to a meridian of the arc $\alpha $ , as unoriented curves. We ruled out the latter possibility in Lemma 4.4. Therefore, when viewed as curves in $\partial M_A$ , $B_\pm $ are each isotopic to components of $s(\gamma _A)$ (and are thus core circles of $R_\pm (\gamma _A)$ ). In particular, $B_+$ and $B_-$ are isotopic to one another as unoriented curves in $\partial M_A$ . Given (4.5), it must therefore be the case that $B_+$ and $B_-$ are parallel, oppositely oriented curves in $\partial M_A$ .

Forgetting their orientation, these curves cobound an annulus in $\partial M_A$ , whose union with B is then a Klein bottle

$$\begin{align*}\Sigma \subset M_A.\end{align*}$$

Since $M_A$ is orientable, the Klein bottle $\Sigma $ must be one-sided and in particular nonseparating. This implies that the mod- $2$ intersection pairing

$$\begin{align*}H_1(M_A,\partial M_A;\mathbb{Z}/2\mathbb{Z}) \times H_2(M_A;\mathbb{Z}/2\mathbb{Z}) \to \mathbb{Z}/2\mathbb{Z} \end{align*}$$

is nonzero. But this contradicts the fact that $H_2(M_A;\mathbb {Z}/2\mathbb {Z})=0$ , by Lemma 4.2. Therefore, $[B_+] = [B_-]$ , and then B has the desired properties.

Proof. Lemma 4.5 implies that decomposing $(M_\alpha ,\gamma _\alpha )$ along the product annulus B produces a disconnected balanced sutured manifold

$$\begin{align*}(M_\alpha,\gamma_\alpha) \stackrel{B}{\leadsto} (M_2,\gamma_2) \sqcup (M_3,\gamma_3), \end{align*}$$

where we have labeled the components so that $(M_2,\gamma _2)$ has two sutures and $(M_3,\gamma _3)$ has three, as depicted in Figure 3.

Figure 3 We decompose $(M_A,\gamma _A)$ along B to obtain $(M_2,\gamma _2)\sqcup (M',\gamma ')$ . Removing $\alpha $ and adding a meridional suture produces $(M_2,\gamma _2) \sqcup (M_3,\gamma _3)$ , which is also the result of decomposing $(M_\alpha ,\gamma _\alpha )$ along B.

Indeed, in $M_A$ , the components of $\partial B$ are core circles of the annuli $R_+(\gamma _A)$ and $R_-(\gamma _A)$ , so decomposing $(M_A,\gamma _A)$ along the separating B produces a disjoint union of two sutured manifolds, with two sutures each,

$$\begin{align*}(M_A,\gamma_A) \stackrel{B}{\leadsto} (M_2,\gamma_2) \sqcup (M',\gamma'). \end{align*}$$

One of these components is disjoint from the arc $\alpha $ , so we label it $(M_2,\gamma _2)$ . We then remove a tubular neighborhood of $\alpha $ from the other component $(M',\gamma ')$ and add a meridional suture $\mu _\alpha $ to get $(M_3,\gamma _3)$ .

Since the components of $\partial B$ are homologically essential in $R(\gamma _A)$ , we have that

$$ \begin{align*} \mathit{SFH}(M_\alpha,\gamma_\alpha) &\cong \mathit{SFH}((M_2,\gamma_2) \sqcup (M_3, \gamma_3)) \\ &\cong \mathit{SFH}(M_2,\gamma_2) \otimes \mathit{SFH}(M_3,\gamma_3), \end{align*} $$

by Theorem 2.4. Since the left side is 2-dimensional, per (4.2), it follows that

$$\begin{align*}\dim\mathit{SFH}(M_i,\gamma_i) =1 \end{align*}$$

for some $i\in \{2,3\}$ . Then Theorem 2.1 tells us that the corresponding $(M_i,\gamma _i)$ is a product sutured manifold (note that $(M_i,\gamma _i)$ is taut since $(M_\alpha ,\gamma _\alpha )$ is taut, per Remark 2.5).

Suppose first that $(M_2,\gamma _2)$ is a product sutured manifold. Since $\partial M_2$ is a torus and the sutures $s(\gamma _2)$ consist of two parallel essential curves on this torus, $R_+(\gamma _2)$ is an annulus, and so there is a homeomorphism

$$\begin{align*}(M_2,\gamma_2) \cong \big((S^1\times I) \times [-1,1], (S^1\times \partial I) \times [-1,1]\big). \end{align*}$$

But if this is the case, then B could have been isotoped onto the component of $\gamma _\alpha $ which became a component of $\gamma _2$ , by an isotopy keeping $\partial B$ in $R(\gamma _\alpha )$ at all times. This contradicts the fact that B is essential.

It follows that $(M_3,\gamma _3)$ is a product sutured manifold. Since $R_+(\gamma _3)$ is a pair of pants P, we have that

$$\begin{align*}(M_3,\gamma_3) \cong (P\times[-1,1], \partial P \times [-1,1]). \end{align*}$$

One component of the sutures $s(\gamma _3)$ is a meridian $\mu _\alpha $ of $\alpha $ , and $(M',\gamma ')$ is recovered by gluing back a thickened disk $D^2\times I$ along an annular neighborhood of this meridian. The meridian $\mu _\alpha $ corresponds to a certain boundary component of P. Letting

$$\begin{align*}S^1\times [0,1] = P\cup D^2\end{align*}$$

be the annulus formed by capping off this boundary component with a disk, we then have the identification

$$\begin{align*}(M',\gamma') \cong \big((S^1\times [0,1]) \times [-1,1], (S^1\times \{0,1\}) \times [-1,1]\big), \end{align*}$$

where the arc $\alpha \subset M'$ is given by

$$\begin{align*}\alpha = \{\mathrm{pt}\} \times [-1,1] \subset M' \end{align*}$$

for some point

$$\begin{align*}\mathrm{pt} \in D^2\subset (S^1\times [0,1]),\end{align*}$$

as depicted in Figure 4.

Figure 4 Left, the product sutured manifold $(M',\gamma ')$ , together with the arc $\alpha $ . Right, the same manifold with $\alpha $ isotoped into $\partial M'$ .

The portion of $\partial M'$ which came from the annulus B (i.e., which was in the interior of $M_A$ ) is contained in a tubular neighborhood $N \subset \partial M'$ of one of the two components of $\gamma '$ – let us say the component

$$\begin{align*}(S^1\times\{0\})\times[-1,1].\end{align*}$$

Since $\alpha $ is disjoint from B, we can isotope this arc into $\partial M' \setminus N$ while keeping its endpoints fixed, so that it meets the other component

$$\begin{align*}(S^1\times \{1\})\times[-1,1]\end{align*}$$

of $\gamma '$ in an arc $\{\mathrm {pt}\} \times [-1,1]$ , as indicated in Figure 4. Gluing $(M',\gamma ')$ and $(M_2,\gamma _2)$ back together to form $(M_A,\gamma _A)$ , this gives an isotopy in $M_A$ which fixes the endpoints of $\alpha $ while carrying $\alpha $ to an arc in $\partial M_A$ which meets the sutures $s(\gamma _A)$ in one point.

Proof of Proposition 4.1.

Viewing $\alpha $ as an arc in $M_A$ , Lemma 4.6 says that we can isotope it rel its endpoints to lie in $\partial M_A$ , so that it meets the sutures $s(\gamma _A)$ transversely in a single point, as shown on the left side of Figure 5.

Figure 5 Viewing $M_A$ as a submanifold of $M_F$ , the arc $\alpha \subset \partial M_A$ lies in a pushoff of the annulus A. On the right, we see the region swept out by the isotopy of $\alpha $ into A.

Recall that $M_A$ was formed from $M_F$ by removing the interior of a tubular neighborhood $A \times [-1,1]$ , where the original cabling annulus A is identified as $A\times \{0\}$ . We can arrange the interval coordinate so that $\alpha \subset A\times \{1\}$ , and then the desired isotopy is simply $\phi _t(x) = (x,1-t)$ for $x\in \alpha $ .

Proof. As defined in §4, the manifold $(M_A,\gamma _A)$ is obtained from $(M_F,\gamma _F)$ by decomposing along the cabling annulus A provided in Lemma 3.4,

$$\begin{align*}(M_F,\gamma_F) \stackrel{A}{\leadsto} (M_A,\gamma_A). \end{align*}$$

Recall from Definition 3.2 that

$$\begin{align*}(M,\gamma) = S^3(F)\end{align*}$$

can be recovered from $M_F$ by removing a neighborhood $N(\alpha )$ of the arc $\alpha $ , where the suture $s(\gamma )$ is identified with a meridian $\mu _\alpha $ of $\alpha $ . By Proposition 4.1, we can assume that $\alpha \subset A$ . Therefore, when we remove a neighborhood of $\alpha $ from $M_F$ to form $S^3(F)$ , what remains of the cabling annulus A is a product disk $D\subset S^3(F)$ . Thus, $(M_A,\gamma _A)$ can alternatively be obtained via the product disk decomposition

$$\begin{align*}S^3(F) \stackrel{D}{\leadsto} (M_A,\gamma_A), \end{align*}$$

as indicated in Figure 6.

Figure 6 A schematic picture which shows that decomposing $(M_F,\gamma _F)$ along the cabling annulus A is the same as first removing a neighborhood of $\alpha \subset A$ and then decomposing along the product disk D.

This shows in particular that $M_A$ is a subset of $S^3$ , as

$$\begin{align*}M_A\subset S^3(F)\subset S^3.\end{align*}$$

Since $M_A$ has torus boundary, it follows that $M_A$ can be identified with the complement of a knot $C\subset S^3$ and moreover that we have an identification of sutured manifolds,

$$\begin{align*}(M_A,\gamma_A)\cong (S^3\setminus N(C),\gamma_r),\end{align*}$$

where the sutures $s(\gamma _r)$ are a union of two parallel oppositely oriented curves of slope r, with respect to the Seifert framing of C. Furthermore, we have

(5.1) $$ \begin{align} \mathit{SFH}(S^3\setminus N(C),\gamma_r) \cong \mathit{SFH}(S^3(F))\cong\mathbb{Q}^2, \end{align} $$

by Theorem 2.3. It remains to determine the slope r and the knot C.

Suppose first that $r=0$ . Then C is the unknot because otherwise, we would have

$$\begin{align*}\dim \mathit{SFH}(S^3\setminus N(C),\gamma_0)\geq 4,\end{align*}$$

by Proposition 2.7, contradicting (5.1). But then $M_F$ is the complement of the $(2,0)$ -cable of the unknot in $S^3$ , which contradicts the fact in Lemma 3.3 that $M_F$ is irreducible.

The above argument shows that $r\neq 0$ . Note that we can identify $(S^3\setminus N(C),\gamma _r)$ as the sutured complement of the core $C' \subset S^3_r(C)$ of r-surgery on C, whose sutures are a union of two oppositely oriented meridians of $C'$ . With this in mind, equation (2.1) becomes

$$ \begin{align*} \widehat{\mathit{HFK}}(S^3_r(C), C') &\cong \mathit{SFH}(S^3_r(C)(C')) \\ &\cong \mathit{SFH}(S^3 \setminus N(C), \gamma_r) \cong \mathbb{Q}^2. \end{align*} $$

Since $r\neq 0$ , the core $C'$ is rationally nullhomologous in $S^3_r(C)$ . It follows that there is a spectral sequence

$$\begin{align*}\mathbb{Q}^2 \cong \widehat{\mathit{HFK}}(S^3_r(C),C') \ \Longrightarrow\ \widehat{\mathit{HF}}(S^3_r(C)) \end{align*}$$

leading to the chain of inequalities

$$ \begin{align*} 1 \leq |H_1(S^3_r(C);\mathbb{Z})| &\leq \dim \widehat{\mathit{HF}}(S^3_r(C)) \\ &\leq \dim \widehat{\mathit{HFK}}(S^3_r(C),C') = 2. \end{align*} $$

We conclude that

(5.2) $$ \begin{align}\dim\widehat{\mathit{HF}}(S^3_r(C))=2,\end{align} $$

as this dimension has the same parity as

$$\begin{align*}\dim\widehat{\mathit{HFK}}(S^3_r(C), C')=2.\end{align*}$$

It also has the same parity as

$$\begin{align*}\chi\big( \widehat{\mathit{HF}}(S^3_r(C)) \big) = |H_1(S^3_r(C);\mathbb{Z})|, \end{align*}$$

which then implies that

(5.3) $$ \begin{align}|H_1(S^3_r(C);\mathbb{Z})| = 2. \end{align} $$

Combining (5.2) and (5.3), we have shown that $S^3_r(C)$ is an L-space. Moreover, if $r=p/q$ with $q\geq 0$ and $\gcd (p,q)=1$ , then $|p|=2$ .

We now recall from [Reference Ozsváth and Szabó54, Proposition 9.6] (see [Reference Hom27, §2] for details) that if $C \subset S^3$ is a nontrivial knot, then r-surgery on C can only be an L-space if

$$\begin{align*}|r| \geq 2g(C)-1.\end{align*}$$

Moreover, if we also have that $r>0$ , then C must additionally be fibered [Reference Ghiggini18, Reference Ni44] and strongly quasipositive [Reference Hedden24]. Note that when C is knotted, we have that

$$\begin{align*}0 < |r| < 1 \leq 2g(C)-1\end{align*}$$

for slopes $r=\pm 2/q$ unless $q=1$ , so there are three cases to consider:

1. 1. C is an unknot and $r=2/q$ for some odd $q\in \mathbb {Z}$ .

2. 2. C is knotted and $r=2$ . Then $S^3_2(C)$ is an L-space, so $g(C) = 1.$ Then C must be the right-handed trefoil since this is the only genus-1, fibered, strongly quasipositive knot in the 3-sphere.

3. 3. C is knotted and $r=-2$ . Then $S^3_{-2}(C)$ is an L-space, so again, $g(C) \leq 1$ . But now C must be the left-handed trefoil since its mirror $\overline {C}$ admits a positive L-space surgery and is therefore the right-handed trefoil, as discussed above.

In case (1), it follows that

$$\begin{align*}(M_A,\gamma_A) \cong (S^3 \setminus N(U), \gamma_2), \end{align*}$$

since any two choices of $\gamma _{2/q}$ are related by a homeomorphism of the solid torus $S^3 \setminus N(U)$ . We conclude that

$$\begin{align*}M_F \cong S^3 \setminus N(C_{2,4}(U)) \cong S^3 \setminus N(T_{2,4}). \end{align*}$$

Similarly, in case (2), we have that

$$\begin{align*}(M_A,\gamma_A) \cong (S^3 \setminus N(T_{2,3}), \gamma_2) \end{align*}$$

and therefore conclude that

$$\begin{align*}M_F \cong S^3 \setminus N(C_{2,4}(T_{2,3})). \end{align*}$$

This leaves only case (3), in which

$$\begin{align*}(M_A,\gamma_A) \cong (S^3 \setminus N(T_{-2,3}), \gamma_{-2}). \end{align*}$$

Then we have

$$ \begin{align*} M_F &\cong S^3 \setminus N(C_{2,-4}(T_{-2,3})) \\ &\cong -\left( S^3 \setminus N(C_{2,4}(T_{2,3})) \right). \end{align*} $$

But in this case, we can replace K with its mirror $\overline {K}$ , and doing so replaces $M_F$ with $-M_F$ . So again, case (2) applies here, and we are done.

Proof. We can rotate the diagram of the unknot $U=\tau \cup \beta $ on the right side of Figure 8 about a vertical axis, and this preserves the tangle $\tau $ and the linked curve $\kappa $ while replacing the braid $\beta $ with its reverse $r(\beta )$ . It follows that $\tau \cup r(\beta )$ is also unknotted, since it is isotopic to the unknot U. This isotopy also carries $\kappa $ to itself, so up to isotopy, $\kappa $ must lift to both $K_\beta $ and $K_{r(\beta )}$ in the branched double cover of U; hence, $K_\beta \cong K_{r(\beta )}$ .

Proof. In Figure 9, we perform an isotopy of $U = \tau \cup \big (m(\beta )y\big )$ in the complement of $\kappa $ , and we quickly find ourselves with a mirror image (reflecting across the plane of the page) of the diagram used to recover $K_\beta $ . Thus, if $\tau \cup \beta $ is unknotted, then so is $\tau \cup \big (m(\beta )y\big )$ , and the unknot $\kappa $ for $\tau \cup \big (m(\beta )y\big )$ lifts to the mirror of the lift $K_\beta $ of the corresponding knot in the $\tau \cup \beta $ diagram.

Proof. It is straightforward to see that $\tau \cup (y\beta y^{-1})$ is isotopic to $\tau \cup \beta $ in the complement of $\kappa $ , so the lemma follows by induction on a.

Proof of Theorem 6.1.

By Lemma 6.2, it suffices to classify the braids $\beta \in B_3$ such that $U = \tau \cup \beta $ is unknotted and to determine $K = K_\beta $ for each of them.

Supposing that U is an unknot, in Subsection 6.1, we will identify an arc (see Figure 10) that lifts to a knot $\gamma $ in the branched double cover $\Sigma _2(U) \cong S^3$ . We will argue via the cyclic surgery theorem [Reference Culler, McA, Gordon and Shalen12] that $\gamma $ must be an unknot or a torus knot, and we will study various surgeries on $\gamma $ which must be lens spaces or connected sums of lens spaces. In Subsections 6.2 and 6.3, we will study the cases $\gamma \cong U$ and $\gamma \cong T_{p,q}$ separately, proving in Propositions 6.12 and 6.13 that $\beta $ must be one of

$$\begin{align*}x^{-1},\ xy, \text{ or } x^ny^{-1}xy \ (n\in\mathbb{Z}) \end{align*}$$

or

$$\begin{align*}x^3y^{-1}x^2y \text{ or } x^{-3}yx^{-2}, \end{align*}$$

respectively, up to reversal and conjugation by powers of y. Lemmas 6.3 and 6.5 tell us that it is enough to consider these particular braids.

Figure 10 Resolving the topmost crossing in the clasp of $U = \tau \cup \beta $ in several different ways.

After classifying these braids, we devote Subsection 6.4 to determining the knot $K_\beta $ for each of

$$\begin{align*}\beta = x^{-1},\ x^ny^{-1}xy, \text{ or } x^3y^{-1}x^2y. \end{align*}$$

These cases occupy Propositions 6.21, 6.22 and 6.23, respectively, and they recover the knots $5_2$ , $P(-3,3,2n+1)$ and $15n_{43522}$ . The only remaining braids are

$$\begin{align*}\beta = xy = m(x^{-1}) y \end{align*}$$

and

$$\begin{align*}\beta = x^{-3}yx^{-2} = m(x^3y^{-1}x^2y) y, \end{align*}$$

but then Lemma 6.4 says that the corresponding $K_\beta $ are the mirrors of knots which we already found, so the proof is complete.