2.1 Knot Floer homology and L-space knots

Let $K$ denote a (doubly pointed) knot in a rational homology sphere $Y$ . Let $N(K)$ denote an open tubular neighborhood of $K$ , $X=Y\smallsetminus N(K)$ the knot exterior, and $\unicode[STIX]{x1D707}\subset \unicode[STIX]{x2202}X$ a meridian of $K$ . Let $\operatorname{Spin}^{c}(Y)$ denote the set of $\operatorname{spin}^{c}$ structures on $Y$ and $\operatorname{Spin}^{c}(X,\unicode[STIX]{x2202}X)$ the set of relative $\operatorname{spin}^{c}$ structures on $(X,\unicode[STIX]{x2202}X)$ . They are torsors over the groups $H_{1}(Y;\mathbb{Z})$ and $H_{1}(X;\mathbb{Z})$ , respectively. Let $\operatorname{Spin}^{c}(K)$ denote the set of orbits in $\operatorname{Spin}^{c}(X,\unicode[STIX]{x2202}X)$ under the action by $[\unicode[STIX]{x1D707}]$ . It forms a torsor over $H_{1}(X;\mathbb{Z})/[\unicode[STIX]{x1D707}]\approx H_{1}(Y;\mathbb{Z})$ , and there exists a pair of torsor isomorphisms $i_{v},i_{h}:\operatorname{Spin}^{c}(K)\rightarrow \operatorname{Spin}^{c}(Y)$ .

We work with the hat-version of knot Floer homology with $\mathbb{Z}$ coefficients, graded by $s\in \operatorname{Spin}^{c}(K)$ :

$$\begin{eqnarray}\widehat{HFK}(K)=\bigoplus _{s}\widehat{HFK}(K,s).\end{eqnarray}$$

There exists a further pair of gradings

$$\begin{eqnarray}a,m:\widehat{HFK}(K,s)\rightarrow \mathbb{Z}\end{eqnarray}$$

on each summand, the Alexander and Maslov gradings. An element is homogeneous if it is homogeneous with respect to both gradings. The group $\widehat{HFK}(K)$ comes equipped with differentials $\widetilde{d}_{v}$ , $\widetilde{d}_{h}$ that preserve the $\operatorname{Spin}^{c}(K)$ -grading, respectively lower $m$ and $m-2a$ by one, and respectively raise and lower $a$ . They are invariants of $K$ , and their homology calculates $\widehat{HF}(Y,i_{v}(s))$ and $\widehat{HF}(Y,i_{h}(s))$ , respectively. The manifold $Y$ is an L-space if $\widehat{HF}(Y,t)\approx \mathbb{Z}$ for all $t\in \operatorname{Spin}^{c}(Y)$ .

Definition 2.1. For a knot $K$ in an L-space $Y$ and $s\in \operatorname{Spin}^{c}(K)$ , the group $\widehat{HFK}(K,s)$ is a positive chain if it admits a homogeneous basis $x_{1},\ldots ,x_{2n+1}$ such that, for all $k$ ,

$$\begin{eqnarray}\widetilde{d}_{v}(x_{2k})=\pm x_{2k-1},\quad \widetilde{d}_{h}(x_{2k})=\pm x_{2k+1},\quad \widetilde{d}_{v}(x_{2k-1})=\widetilde{d}_{h}(x_{2k-1})=0.\end{eqnarray}$$

The group $\widehat{HFK}(K)$ consists of positive chains if $\widehat{HFK}(K,s)$ is a positive chain for all $s$ .

For example, a positive chain with $2n+1=7$ generators takes the form shown here:

Each arrow represents a component of the differential and is an isomorphism between the groups it connects. Similarly, a negative chain is the dual complex to a positive chain with respect to the defining basis. Reversing the arrows above gives an example of a negative chain.

Note that the requirement that the positive chain basis elements are homogeneous does not appear in [Reference Rasmussen and RasmussenRR17, Definition 3.1]. For example, we could alter the positive chain basis displayed above to an inhomogeneous one by replacing $x_{2}$ by $x_{2}+x_{1}$ . However, homogeneity is an intended property of a positive chain basis in the literature. Moreover, an inhomogeneous positive chain basis gives rise to a homogeneous one by replacing each basis element by its homogeneous part of highest bigrading. Therefore, our definition is no more restrictive, and its precision is more convenient when we invoke it in the proofs of Lemmas 2.3 and 2.4.

Let $\unicode[STIX]{x1D706}\subset \unicode[STIX]{x2202}X$ denote the rational longitude of $K$ , the unique slope that is rationally null-homologous in $X$ . Note that $\unicode[STIX]{x1D707}\neq \unicode[STIX]{x1D706}$ , since $Y$ is a rational homology sphere. Another slope $\unicode[STIX]{x1D6FC}\subset \unicode[STIX]{x2202}X$ is positive or negative according to the sign of $(\unicode[STIX]{x1D707}\cdot \unicode[STIX]{x1D706})(\unicode[STIX]{x1D706}\cdot \unicode[STIX]{x1D6FC})(\unicode[STIX]{x1D6FC}\cdot \unicode[STIX]{x1D707})$ , for any orientations on these curves. The knot $K$ is a positive or a negative L-space knot if it has an L-space surgery slope of that sign. The following theorem characterizes positive L-space knots in L-spaces in terms of their knot Floer homology. Ozsváth and Szabó originally proved it for the case of knots in $S^{3}$ [Reference Ozsváth and SzabóOS05, Theorem 1.2]. Boileau, Boyer, Cebanu, and Walsh promoted a significant component of their result to knots in rational homology spheres [Reference Boileau, Boyer, Cebanu and WalshBBCW12]. Building on it, Rasmussen and Rasmussen established the definitive form of the result that we record here: see [Reference Rasmussen and RasmussenRR17, Lemmas 3.2, 3.3, and 3.5], including the proofs of these results.

Similarly, $K$ is a negative L-space knot if and only if $\widehat{HFK}(K)$ consists of negative chains. The reason amounts to the behavior of Dehn surgery and knot Floer homology under mirroring.

2.2 Calculating the invariants from a Heegaard diagram

The invariants can be calculated from any doubly pointed Heegaard diagram $D=(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},z,w)$ of $K$ after making some additional analytic choices. Here, as usual, $\unicode[STIX]{x1D6F4}$ denotes a closed, oriented surface of some genus $g$ ; $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ are $g$ -tuples of homologically linearly independent, disjoint, simple closed curves in $\unicode[STIX]{x1D6F4}$ ; and the two basepoints $w$ and $z$ lie in the complement of the $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ curves on $\unicode[STIX]{x1D6F4}$ . The curve collections induce tori $\mathbb{T}_{\unicode[STIX]{x1D6FC}},\mathbb{T}_{\unicode[STIX]{x1D6FD}}$ in the $g$ -fold symmetric product $\text{Sym}^{g}(\unicode[STIX]{x1D6F4})$ . The underlying group of the Floer chain complex $\widehat{CFK}(D)$ is freely generated by $\mathfrak{T}(D)=\mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}}$ . The elements of $\mathfrak{T}(D)$ fall into equivalence classes in one-to-one correspondence with $\operatorname{Spin}^{c}(K)$ : two elements $x,y$ lie in the same equivalence class if and only if the set $\unicode[STIX]{x1D70B}_{2}(x,y)$ of homotopy classes of Whitney disks from $x$ to $y$ is non-empty. Write $\mathfrak{T}(D,s)$ for the equivalence class corresponding to $s\in \operatorname{Spin}^{c}(K)$ and $\widehat{CFK}(D,s)$ for the subgroup of $\widehat{CFK}(D)$ generated by the elements in $\mathfrak{T}(D,s)$ . Each Whitney disk $\unicode[STIX]{x1D719}\in \unicode[STIX]{x1D70B}_{2}(x,y)$ has a pair of multiplicities $n_{z}(\unicode[STIX]{x1D719})$ , $n_{w}(\unicode[STIX]{x1D719})$ and a Maslov index $\unicode[STIX]{x1D707}(\unicode[STIX]{x1D719})$ . The Alexander and Maslov gradings on $\widehat{CFK}(D,s)$ are characterized up to an overall shift by the relations

$$\begin{eqnarray}a(x)-a(y)=n_{z}(\unicode[STIX]{x1D719})-n_{w}(\unicode[STIX]{x1D719}),\quad m(x)-m(y)=\unicode[STIX]{x1D707}(\unicode[STIX]{x1D719})-2n_{w}(\unicode[STIX]{x1D719})\end{eqnarray}$$

for all $x,y\in \mathfrak{T}(D,s)$ and $\unicode[STIX]{x1D719}\in \unicode[STIX]{x1D70B}_{2}(x,y)$ . There exist endomorphisms $d_{0},d_{v},d_{h}$ of $\widehat{CFK}(D,s)$ defined on generators $x\in \mathfrak{T}(D,s)$ by the same general prescription:

$$\begin{eqnarray}d(x)=\mathop{\sum }_{y,\unicode[STIX]{x1D719}}\#\widehat{{\mathcal{M}}}(\unicode[STIX]{x1D719})\cdot y.\end{eqnarray}$$

Here $y$ ranges over $\mathfrak{T}(D,s)$ ; $\unicode[STIX]{x1D719}$ ranges over the elements of $\unicode[STIX]{x1D70B}_{2}(x,y)$ with Maslov index $\unicode[STIX]{x1D707}(\unicode[STIX]{x1D719})=1$ and a constraint on the multiplicities $n_{w}(\unicode[STIX]{x1D719})$ , $n_{z}(\unicode[STIX]{x1D719})$ ; and $\#\widehat{{\mathcal{M}}}(\unicode[STIX]{x1D719})$ is the count of pseudo-holomorphic representatives of $\unicode[STIX]{x1D719}$ . The specific constraints on the multiplicities are $n_{w}(\unicode[STIX]{x1D719})=n_{z}(\unicode[STIX]{x1D719})=0$ for $d=d_{0}$ ; $n_{w}(\unicode[STIX]{x1D719})=0,n_{z}(\unicode[STIX]{x1D719})>0$ for $d=d_{v}$ ; and $n_{w}(\unicode[STIX]{x1D719})>0,n_{z}(\unicode[STIX]{x1D719})=0$ for $d=d_{h}$ . The maps $d_{0},d_{0}+d_{v},d_{0}+d_{h}$ are all differentials. The differential $d_{0}+d_{v}$ lowers $m$ by one and is filtered with respect to $a$ . The differential $d_{0}+d_{h}$ lowers $m-2a$ by one and is filtered with respect to $-a$ . The differential $d_{0}$ is the $a$ -filtration-preserving component of each. The groups $\widehat{HFK}(K,s)$ , $\widehat{HF}(Y,i_{v}(s))$ , and $\widehat{HF}(Y,i_{h}(s))$ are the homology groups of $\widehat{CFK}(D,s)$ with respect to $d_{0}$ , $d_{0}+d_{v}$ , and $d_{0}+d_{h}$ , respectively, and the Alexander and Maslov gradings on $\widehat{CFK}(D,s)$ descend to the respective gradings on $\widehat{HFK}(K,s)$ . The maps $d_{0}+d_{v}$ and $d_{0}+d_{h}$ induce the differentials $\widetilde{d}_{v}$ and $\widetilde{d}_{h}$ on $\widehat{HFK}(K,s)$ , respectively.

The Alexander and Maslov gradings enable us to constrain the existence of pseudo-holomorphic disks for L-space knots.

Proof. Definition 2.1 and the paragraph preceding it show that the positive chain basis satisfies

$$\begin{eqnarray}a(x_{2k})-a(x_{2k-1})=b_{k},\quad m(x_{2k})-m(x_{2k-1})=1\end{eqnarray}$$

and

$$\begin{eqnarray}a(x_{2k})-a(x_{2k+1})=-c_{k},\quad m(x_{2k})-m(x_{2k+1})=1-2c_{k},\end{eqnarray}$$

for some positive integers $b_{k},c_{k}$ , $k=1,\ldots ,n$ . In particular, the Maslov and Alexander gradings of the $x_{l}$ both decrease with the index $l$ . The generators $x,y\in \mathfrak{T}(D,s)$ are homogeneous with respect to both gradings, as are the positive chain basis elements, so it follows that $x=\pm x_{i}$ and $y=\pm x_{j}$ for some $i$ , $j$ , and choices of sign.

The assumption that $\#{\mathcal{M}}(\unicode[STIX]{x1D719})\neq 0$ implies that $n_{w}(\unicode[STIX]{x1D719}),n_{z}(\unicode[STIX]{x1D719})\geqslant 0$ , and at least one inequality is strict, since $d_{0}=0$ . Thus,

(1) $$\begin{eqnarray}a(x_{i})-a(x_{j})=n_{z}(\unicode[STIX]{x1D719})-n_{w}(\unicode[STIX]{x1D719})\quad \text{and}\quad m(x_{i})-m(x_{j})=1-2n_{w}(\unicode[STIX]{x1D719})\leqslant 1.\end{eqnarray}$$

In the case of equality $m(x_{i})-m(x_{j})=1$ , then $j=i-1$ , and either $i=2k$ , which gives the desired conclusion, or else $j=2k$ , which we must rule out. In this case, $n_{w}(\unicode[STIX]{x1D719})=0$ and $n_{z}(\unicode[STIX]{x1D719})>0$ . Since $Y$ is a rational homology sphere, it follows that $\unicode[STIX]{x1D719}\in \unicode[STIX]{x1D70B}_{2}(x,y)$ is the unique disk with $\unicode[STIX]{x1D707}(\unicode[STIX]{x1D719})=1$ , so the coefficient on $y$ in $\widetilde{d}_{v}(x)$ is $\#{\mathcal{M}}(\unicode[STIX]{x1D719})\neq 0$ . However, this implies that $\widetilde{d}_{v}(x_{2k+1})\neq 0$ , which violates the form of the positive chain.

Otherwise, $m(x_{i})-m(x_{j})\leqslant -1$ , so $i<j$ . We have

(2) $$\begin{eqnarray}a(x_{i})-a(x_{j})=\mathop{\sum }_{l=i}^{j-1}a(x_{l})-a(x_{l+1}),\quad m(x_{i})-m(x_{j})=\mathop{\sum }_{l=i}^{j-1}m(x_{l})-m(x_{l+1}).\end{eqnarray}$$

Each term in the second sum is odd, and their total sum is odd, so it contains an odd number of terms. The terms of the first sum alternately take the form $-b_{k}$ and $-c_{k}$ , while the terms of the second sum alternately take the form $-1$ and $1-2c_{k}$ . Let $b$ denote the sum of the values $b_{k}$ that appear and $c$ the sum of the $c_{k}$ that appear. All values $b_{k}$ are positive, so $b\geqslant 0$ , and $b=0$ only if $i$ is even and $j=i+1$ , and $b=1$ only if $i$ is odd and $j=i+1$ . The first sum in (2) equals $-b-c$ , while the second sum equals $1-2c$ or $-1-2c$ , according to whether $i$ is even or odd. Comparing with the second sum in (1) gives $n_{w}(\unicode[STIX]{x1D719})=c$ or $c+1$ , according to whether $i$ is even or odd. Comparing with the first sum in (1) then gives $n_{z}(\unicode[STIX]{x1D719})=-b$ or $-b+1$ , according to whether $i$ is even or odd. If $i$ is even, then $0\leqslant n_{z}(\unicode[STIX]{x1D719})=-b$ gives $n_{z}(\unicode[STIX]{x1D719})=0$ and $j=i+1$ , which gives the desired conclusion. If $i$ is odd, then $0\leqslant n_{z}(\unicode[STIX]{x1D719})=-b+1$ again gives $n_{z}(\unicode[STIX]{x1D719})=0$ and $j=i+1$ . However, in this case it follows as before that $y$ appears with coefficient $\#{\mathcal{M}}(\unicode[STIX]{x1D719})\neq 0$ in $\widetilde{d}_{h}(x)$ , which once again violates the form of the positive chain.◻

2.3 $(1,1)$ knots

Now assume that $D$ is a $(1,1)$ diagram, so $\unicode[STIX]{x1D6F4}$ has genus one. In this case, the differentials on $\widehat{CFK}(D)$ admit an explicit description that requires no analytic input, owing to the Riemann mapping theorem and the correspondence between holomorphic disks in the Riemann surface $\unicode[STIX]{x1D6F4}=\text{Sym}^{1}(\unicode[STIX]{x1D6F4})$ and its universal cover $\mathbb{C}$ . Specifically, the conditions that $\unicode[STIX]{x1D719}\in \unicode[STIX]{x1D70B}_{2}(x,y)$ and $\unicode[STIX]{x1D707}(\unicode[STIX]{x1D719})=1$ imply that $\#\widehat{{\mathcal{M}}}(\unicode[STIX]{x1D719})=\pm 1$ for any choice of analytic data. Furthermore, these conditions are met if and only if the image of $\unicode[STIX]{x1D719}$ lifts under the universal covering map $\unicode[STIX]{x1D70B}:\mathbb{R}^{2}\rightarrow \unicode[STIX]{x1D6F4}$ to a bigon cobounded by lifts of $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ . See [Reference Goda, Matsuda and MorifujiGMM05] and [Reference Ozsváth and SzabóOS04, pp. 89–96] for more details.

Assuming now that $D$ is reduced, we have $d_{0}=0$ and $\widehat{HFK}(K,s)\approx \widehat{CFK}(D,s)$ . We proceed to analyze the differentials $\widetilde{d}_{h}$ and $\widetilde{d}_{v}$ . Without loss of generality, we henceforth fix $\unicode[STIX]{x1D6F4}=\mathbb{R}^{2}/\mathbb{Z}^{2}$ , $z=\unicode[STIX]{x1D70B}(\mathbb{Z}^{2})$ , a horizontal line $\widetilde{\unicode[STIX]{x1D6FC}}\subset \mathbb{R}^{2}\smallsetminus \mathbb{Z}^{2}$ , and $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D70B}(\widetilde{\unicode[STIX]{x1D6FC}})$ . After a further homeomorphism, we may assume that $D$ takes the form shown in Figure 2. Observe that the embedded bigons in $D$ are the obvious ones cobounded by the $\unicode[STIX]{x1D6FC}$ with the ‘rainbow’ arcs of $\unicode[STIX]{x1D6FD}$ above $w$ and below $z$ . Coherence is the condition that when the $\unicode[STIX]{x1D6FD}$ curve is oriented, all of the rainbow arcs around a fixed basepoint orient the same way. This makes it easy to check coherence from a diagram in standard form.

Let $H^{\pm }$ denote the upper and lower half-planes bounded by $\widetilde{\unicode[STIX]{x1D6FC}}$ . Choose a lift to $\widetilde{\unicode[STIX]{x1D6FC}}$ of any point in $\mathfrak{T}(D,s)$ . There exists a unique lift $\widetilde{\unicode[STIX]{x1D6FD}}_{s}\subset \mathbb{R}^{2}$ that passes through it. The points of $\widetilde{\unicode[STIX]{x1D6FC}}\cap \widetilde{\unicode[STIX]{x1D6FD}}_{s}$ are in one-to-one correspondence with $\mathfrak{T}(D,s)$ under $\unicode[STIX]{x1D70B}$ , and there are an odd number $2n+1$ of them, since $\widetilde{\unicode[STIX]{x1D6FC}}\cdot \widetilde{\unicode[STIX]{x1D6FD}}_{s}=\unicode[STIX]{x1D712}(\widehat{CFK}(D,s))=1$ with appropriate orientations on the curves. The curve $\widetilde{\unicode[STIX]{x1D6FD}}_{s}$ meets each half-plane in one closed ray and $n$ compact arcs. These compact arcs cobound positive bigons $D_{k}^{+}\subset H^{+}$ and negative bigons $D_{k}^{-}\subset H^{-}$ with $\widetilde{\unicode[STIX]{x1D6FC}}$ for $k=1,\ldots ,n$ . A positive bigon attains a local maximum, and its image under $\unicode[STIX]{x1D70B}$ is the top of one of the rainbow arcs above $w$ . Consequently, it contains a lift of the bigon in $D$ cobounded by that rainbow arc with $\unicode[STIX]{x1D6FC}$ . Thus, $n_{w}(D_{k}^{+})>0$ , and similarly $n_{z}(D_{k}^{-})>0$ , for all $k$ . Lastly, the positive bigons go from their leftmost corner to their rightmost corner, and vice versa for the negative bigons.

Drawing on the proof of Lemma 2.4, we make a definition and prove a converse.

The bottom right picture in Figure 3 displays a positive graphic curve. The reason for the terminology is that a curve $\widetilde{\unicode[STIX]{x1D6FD}}\subset \mathbb{R}^{2}$ is graphic if and only if there exists a homeomorphism of $\mathbb{R}^{2}$ taking $(\widetilde{\unicode[STIX]{x1D6FC}},\widetilde{\unicode[STIX]{x1D6FD}})$ to the pair consisting of the $x$ -axis and the graph of an odd-degree polynomial with all roots real and distinct. Its sign, positive or negative, is minus the sign of the leading coefficient of the polynomial.

Figure 3. Constructing a positive coherent diagram of $P(-2,3,7)$ , the inclusion of $K(7,4,2)$ into $S^{3}$ . The $\unicode[STIX]{x1D6FC}$ curve is represented by a horizontal arc and the $\unicode[STIX]{x1D6FD}$ curve by a vertical arc in the top left. Moving from left to right illustrates the procedure described in the proof of Theorem 1.2. At the top are diagrams on $\unicode[STIX]{x1D6F4}$ and at the bottom their lifts to $\mathbb{R}^{2}$ . The bigons at the bottom right are shaded for emphasis.

Proof. If $\widehat{HFK}(K,s)$ is a positive chain, then Lemma 2.4 parts (2) and (3) show that $\widetilde{\unicode[STIX]{x1D6FD}}_{s}$ is positive graphic. Conversely, suppose that $\widetilde{\unicode[STIX]{x1D6FD}}_{s}$ is positive graphic. Label the points of intersection in $\widetilde{\unicode[STIX]{x1D6FC}}\cap \widetilde{\unicode[STIX]{x1D6FD}}_{s}$ by $x_{1},\ldots ,x_{2n+1}$ in the order they occur along $\widetilde{\unicode[STIX]{x1D6FC}}$ . The only bigons cobounded by $\widetilde{\unicode[STIX]{x1D6FC}}$ and $\widetilde{\unicode[STIX]{x1D6FD}}_{s}$ are the positive and negative bigons. This latter fact can be verified using the equivalent definition of a graphic curve mentioned after Definition 2.5. Therefore,

$$\begin{eqnarray}\widetilde{d}_{v}(x_{2k})=\unicode[STIX]{x1D6FF}_{k}\cdot x_{2k-1},\quad \widetilde{d}_{h}(x_{2k})=\unicode[STIX]{x1D716}_{k}\cdot x_{2k+1},\quad \widetilde{d}_{v}(x_{2k-1})=\widetilde{d}_{h}(x_{2k-1})=0,\end{eqnarray}$$

for all $k$ and some $\unicode[STIX]{x1D6FF}_{k},\unicode[STIX]{x1D716}_{k}\in \{-1,0,+1\}$ . A value $\unicode[STIX]{x1D6FF}_{k}$ or $\unicode[STIX]{x1D716}_{k}$ is 0 if and only if the corresponding bigon contains both $z$ and $w$ basepoints. Since

$$\begin{eqnarray}H_{\ast }(\widehat{HFK}(K,s),\widetilde{d}_{v})\approx \widehat{HF}(Y,i_{v}(s))\approx \mathbb{Z}\quad \text{and}\quad H_{\ast }(\widehat{HFK}(K,s),\widetilde{d}_{h})\approx \widehat{HF}(Y,i_{h}(s))\approx \mathbb{Z},\end{eqnarray}$$

it follows that $\unicode[STIX]{x1D6FF}_{k},\unicode[STIX]{x1D716}_{k}\neq 0$ for all $k$ , and $\widehat{HFK}(K,s)$ is a positive chain. The corresponding statements for a negative chain and negative coherent diagram follow as well.◻

3.1 1-bridge braids are L-space knots

We prepare by describing a reduced $(1,1)$ -diagram of a 1-bridge braid in a rational homology sphere.

Let $\unicode[STIX]{x1D6FE}\cup \unicode[STIX]{x1D6FF}\subset S^{1}\times D^{2}$ denote a 1-bridge braid. Identify $\unicode[STIX]{x2202}(S^{1}\times D^{2})$ with the flat torus $\unicode[STIX]{x1D6F4}$ in such a way that meridians $\unicode[STIX]{x1D703}\times \unicode[STIX]{x2202}D^{2}$ lift to horizontal lines and longitudes $S^{1}\times x$ lift to vertical lines under $\unicode[STIX]{x1D70B}:\mathbb{R}^{2}\rightarrow \unicode[STIX]{x1D6F4}$ . Isotope $\unicode[STIX]{x1D6FE}$ rel endpoints into a geodesic on $\unicode[STIX]{x1D6F4}$ , so that $\unicode[STIX]{x1D70B}^{-1}(\unicode[STIX]{x1D6FE})$ consists of parallel line segments in $\mathbb{R}^{2}$ of some common slope $s(\unicode[STIX]{x1D6FE})\in P^{1}(\mathbb{R})$ . Orient $\unicode[STIX]{x1D6FE}$ and write $\unicode[STIX]{x2202}\unicode[STIX]{x1D6FE}=z-w$ so that in each lift of $\unicode[STIX]{x1D6FE}$ to $\mathbb{R}^{2}$ , the lift of $z$ lies below that of $w$ .

Form a genus-one rational homology sphere $Y$ by gluing a solid torus $V$ to $S^{1}\times D^{2}$ along their boundaries, and let $K$ denote the image of $\unicode[STIX]{x1D6FE}\cup \unicode[STIX]{x1D6FF}$ in $Y$ . Let $\unicode[STIX]{x1D6FC}$ denote a meridian of $S^{1}\times D^{2}$ and let $\unicode[STIX]{x1D6FD}_{0}$ denote the curve to which the meridian of $V$ attaches, which we take to be a geodesic. In meridian-longitude coordinates, $\unicode[STIX]{x1D6FD}_{0}$ has some slope $p/q\neq 0/1$ , and $Y$ is homeomorphic to $L(p,q)$ if $p\neq 1$ and $S^{3}$ otherwise. Assume that $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}_{0}$ are disjoint from $\unicode[STIX]{x2202}\unicode[STIX]{x1D6FE}$ . Isotope $\unicode[STIX]{x1D6FD}_{0}$ by a finger-move along $\unicode[STIX]{x1D6FE}$ into a curve $\unicode[STIX]{x1D6FD}_{1}$ disjoint from $\unicode[STIX]{x1D6FE}$ . Observe that $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}_{1},z,w)$ is a doubly pointed Heegaard diagram for $K\subset Y$ . Put it into reduced form by further isotoping $\unicode[STIX]{x1D6FD}_{1}$ so as to eliminate all bigons that do not contain basepoints, one by one, resulting in a curve $\unicode[STIX]{x1D6FD}_{2}$ . Let $D=(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}_{2},z,w)$ denote the resulting reduced diagram of $K\subset Y$ . Figure 3 exhibits this procedure for the inclusion of the 1-bridge braid $K(7,4,2)$ into $S^{3}$ , the pretzel knot $P(-2,3,7)$ . The notation derives from the braid presentation of 1-bridge braids; see [Reference BergeBer91, Reference GabaiGab90, Reference WuWu04]. The $1$ -bridge braid $K(\unicode[STIX]{x1D714},b,m)$ denotes the closure of the braid word $(\unicode[STIX]{x1D70E}_{b}\unicode[STIX]{x1D70E}_{b-1}\cdots \unicode[STIX]{x1D70E}_{1})(\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D714}-1}\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D714}-2}\cdots \unicode[STIX]{x1D70E}_{1})^{m}$ in $S^{1}\times D^{2}$ , where the $\unicode[STIX]{x1D70E}_{i}$ are the generators of the braid group  $B_{w}$ . Note that $K(\unicode[STIX]{x1D714},b,m)$ and $K(\unicode[STIX]{x1D714},b,m+\unicode[STIX]{x1D714})$ differ by a full twist, so there is a diffeomorphism of the solid torus that carries one to the other. Therefore, we may assume that $0\leqslant m<w$ .

A pair of distinct slopes $s_{1},s_{2}\in P^{1}(\mathbb{Q})$ determines an interval $[s_{1},s_{2}]\subset P^{1}(\mathbb{Q})$ oriented from $s_{1}$ to $s_{2}$ in the counterclockwise sense; thus, $[s_{1},s_{2}]\cup [s_{2},s_{1}]=P^{1}(\mathbb{Q})$ and $[s_{1},s_{2}]\cap [s_{2},s_{1}]=\{s_{1},s_{2}\}$ .

Note that Proposition 3.1 does not characterize the sign of the coherence in terms of $s(\unicode[STIX]{x1D6FE})$ . Theorem 3.2 does so in terms of the slope interval, and its proof relies on Proposition 3.1.

We proceed to sharpen the statement of Theorem 1.4. To do so, we introduce the notion of a strict 1-bridge braid and its basic invariants: the winding number and slope interval. Choose the lift of $\unicode[STIX]{x1D6FE}$ with one endpoint at $(0,0)$ . Its other endpoint takes the form $(t,\unicode[STIX]{x1D714})\in \mathbb{R}\times \mathbb{Z}$ . We have $\unicode[STIX]{x1D714}>0$ by our convention on the basepoints, and $t$ is not a proper divisor of $\unicode[STIX]{x1D714}$ . Moreover, $\unicode[STIX]{x1D714}$ is an invariant of the isotopy type of $K$ , since $[K]=\unicode[STIX]{x1D714}\cdot [S^{1}\times x]\in H_{1}(S^{1}\times D^{2};\mathbb{Z})$ . This is the winding number $\unicode[STIX]{x1D714}(K)$ .

Consider the sweep-out of line segments with one endpoint at $(0,0)$ and the other endpoint varying along line $y=\unicode[STIX]{x1D714}$ . There exists a maximal slope interval $I(\unicode[STIX]{x1D6FE})\subset P^{1}(\mathbb{Q})$ containing $s(\unicode[STIX]{x1D6FE})$ and so that the sweep-out of line segments through slopes in $I(\unicode[STIX]{x1D6FE})$ contains no lattice point in its interior. If $\unicode[STIX]{x1D714}=1$ , then $I(\unicode[STIX]{x1D6FE})=P^{1}(\mathbb{Q})\smallsetminus \{0\}$ , and otherwise $I(\unicode[STIX]{x1D6FE})=[s_{-}(\unicode[STIX]{x1D6FE}),s_{+}(\unicode[STIX]{x1D6FE})]$ with $s_{-}(\unicode[STIX]{x1D6FE})\leqslant s_{+}(\unicode[STIX]{x1D6FE})$ . The sweep-out descends to an isotopy through 1-bridge braids with slopes in the interior of $I(\unicode[STIX]{x1D6FE})$ . If a line segment of slope $s_{\pm }(\unicode[STIX]{x1D6FE})$ has endpoint $(q,\unicode[STIX]{x1D714})\in \mathbb{Z}^{2}$ , then write $(q,\unicode[STIX]{x1D714})=(dq^{\prime },d\unicode[STIX]{x1D714}^{\prime })$ , where $d=\gcd (q,\unicode[STIX]{x1D714})$ . Let $\unicode[STIX]{x1D707}$ denote the meridian of the torus knot $T(q^{\prime },\unicode[STIX]{x1D714}^{\prime })$ and $\unicode[STIX]{x1D706}$ the surface framing, oriented so that $\unicode[STIX]{x1D707}\cdot \unicode[STIX]{x1D706}=+1$ . If $d=1$ , then $K$ is isotopic to the torus knot $T(q,\unicode[STIX]{x1D714})$ , and if $d>1$ , then $K$ is isotopic to the $(d\unicode[STIX]{x1D706}\pm \unicode[STIX]{x1D707})$ cable of $T(q^{\prime },\unicode[STIX]{x1D714}^{\prime })$ , where the sign is positive or negative if the slope of the line segment is $s_{+}(\unicode[STIX]{x1D6FE})$ or $s_{-}(\unicode[STIX]{x1D6FE})$ , respectively. We call such a cable an exceptional cable of a torus knot. Otherwise, if no line segment has endpoint $(q,\unicode[STIX]{x1D714})\in \mathbb{Z}^{2}$ , then we call $K$ a strict 1-bridge braid (and justify the terminology in Corollary 3.3). In this case, there exists a unique $m\in \mathbb{Z}$ so that $m<t<m+1$ , and $s_{-}(\unicode[STIX]{x1D6FE})$ and $s_{+}(\unicode[STIX]{x1D6FE})$ are consecutive terms in the Farey sequence of fractions from $\unicode[STIX]{x1D714}/(m+1)$ to $\unicode[STIX]{x1D714}/m$ whose denominators are bounded by $|m|$ when expressed in lowest terms.

For the following result, assume $p>q\geqslant 0$ are coprime integers, and let $L(1,0)$ denote $S^{3}$ . A simple knot in $L(p,q)$ is a $(1,1)$ knot whose defining arcs are contained in meridian disks in the respective Heegaard solid tori, where the disks’ boundaries meet in $p$ transverse points of intersection. We permit the degenerate case that the the cable arc $\unicode[STIX]{x1D6FE}$ is a simple closed curve and the bridge $\unicode[STIX]{x1D6FF}$ is a point, in which case the simple knot is an unknot. Equivalently, a simple knot is a knot admitting a $(1,1)$ diagram in which all points of intersection between $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ have the same sign. Note that every simple knot is a 1-bridge braid: to see this, push one of the defining arcs onto the Heegaard torus.

For example, the inclusion of $K(7,4,2)$ into $L(p,q)$ is a simple knot if and only if $p/q\in [5/2,3]$ .

Proof of Theorem 3.2.

The reverse directions of (1) and (2) follow from Proposition 3.1 and Theorem 1.2.

For the forward directions, suppose first that $s_{+}(\unicode[STIX]{x1D6FE})\in (0,p/q)$ . Write $s_{+}(\unicode[STIX]{x1D6FE})=r/s$ in lowest terms, $s>0$ . As before, let $\widetilde{\unicode[STIX]{x1D6FE}}$ denote the lift of $\unicode[STIX]{x1D6FE}$ with one endpoint at $(0,0)$ and the other at $(t,\unicode[STIX]{x1D714})\in \mathbb{R}\times \mathbb{Z}$ . Note that $\unicode[STIX]{x1D714}>|r|$ . Let $\widetilde{\unicode[STIX]{x1D6FC}}\subset \mathbb{R}^{2}\smallsetminus \mathbb{Z}^{2}$ denote a horizontal line that separates $(s,r)$ from $(t,\unicode[STIX]{x1D714})$ , and let $\widetilde{\unicode[STIX]{x1D6FD}}_{0}\subset \mathbb{R}^{2}\smallsetminus \mathbb{Z}^{2}$ denote a line of slope $p/q$ that separates $(s,r)$ from $(0,0)$ . Then $\widetilde{\unicode[STIX]{x1D6FC}}$ , $\widetilde{\unicode[STIX]{x1D6FD}}_{0}$ , and $\widetilde{\unicode[STIX]{x1D6FE}}$ intersect in pairs, and since $r/s\in (0,p/q)$ , they cobound a triangle containing $(s,r)$ in its interior. See Figure 5(a).

Figure 4. On the left, a braid with closure the 1-bridge braid $K(7,4,2)\subset S^{1}\times D^{2}$ . The bold points suggest its decomposition into the form $\unicode[STIX]{x1D6FE}\cup \unicode[STIX]{x1D6FF}$ . On the right, a sweep-out of arcs in $\mathbb{R}^{2}$ indicating that $I(\unicode[STIX]{x1D6FE})=[5/2,3]$ . The bold points arise in the proof of Proposition 3.4.

Figure 5. Capturing a lattice point.

Follow the procedure for producing a coherent diagram $D$ for $K\subset Y$ using the curves $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D70B}(\widetilde{\unicode[STIX]{x1D6FC}})$ , $\unicode[STIX]{x1D6FD}_{0}=\unicode[STIX]{x1D70B}(\widetilde{\unicode[STIX]{x1D6FD}}_{0})\subset \unicode[STIX]{x1D6F4}$ . The isotopy from $\widetilde{\unicode[STIX]{x1D6FD}}_{0}$ to $\widetilde{\unicode[STIX]{x1D6FD}}_{1}$ captures $(s,r)$ in a negative bigon and $(t,\unicode[STIX]{x1D714})$ in a positive bigon in $(\mathbb{R}^{2},\widetilde{\unicode[STIX]{x1D6FC}}\cup \widetilde{\unicode[STIX]{x1D6FD}}_{1})$ . See Figure 5(b). These points remain in bigons of the respective types following the isotopy from $\widetilde{\unicode[STIX]{x1D6FD}}_{1}$ to $\widetilde{\unicode[STIX]{x1D6FD}}_{2}$ . It follows that $\widehat{HFK}(K,s)$ is a positive chain of rank greater than one for the $\operatorname{spin}^{c}$ structure $s$ corresponding to the pair of lifts $\widetilde{\unicode[STIX]{x1D6FC}},\widetilde{\unicode[STIX]{x1D6FD}}_{2}$ . Thus, $K$ is not a negative L-space knot, which gives the forward direction of (2). The forward direction of (1) follows the same line of reasoning.

Since a simple knot is both a positive and a negative L-space knot, the forward direction of (3) follows. Finally, taking a 1-bridge presentation of $K$ in which $\unicode[STIX]{x1D6FE}$ has slope $p/q$ results in a diagram $D$ of a simple knot, and the reverse direction of (3) follows.◻

The following result recovers the isotopy classification of 1-bridge braids from Theorem 3.2. Compare [Reference GabaiGab90, Proposition 2.3], which does so in terms of braid parameters.

3.2 Solid torus fillings

Gabai proved in [Reference GabaiGab90] that a knot in the solid torus with a non-trivial solid torus surgery is a 1-bridge braid, and Berge classified them in [Reference BergeBer91]. Berge found that, up to mirroring, $K(7,4,2)$ is the unique strict 1-bridge braid that admits more than one such surgery. Its exterior is known as the Berge manifold. Menasco and Zhang studied 1-bridge braids whose exteriors admit a solid torus filling on the outer torus in [Reference Menasco and ZhangMZ01], and Wu classified them in [Reference WuWu04].

We indicate a line of approach towards Wu’s result using Theorem 3.2. Suppose that $K$ is a 1-bridge braid in $S^{1}\times D^{2}$ and $(p/q)$ -filling on its outer torus is a solid torus; equivalently, the inclusion of $K$ into $L(p,q)$ is a knot with solid torus exterior. A knot in $L(p,q)$ has a solid torus exterior if and only if it is simple and homologous to the oriented core of a Heegaard solid torus. Since $[K]$ equals $\unicode[STIX]{x1D714}$ times a generator of $H_{1}(S^{1}\times D^{2};\mathbb{Z})$ , the uniqueness of genus-one Heegaard splittings of $L(p,q)$ implies that either

(3) $$\begin{eqnarray}\text{(a)}\quad \unicode[STIX]{x1D714}\equiv \pm 1\hspace{0.6em}({\rm mod}\hspace{0.2em}p)\quad \text{or}\quad \text{(b)}\quad \unicode[STIX]{x1D714}\cdot q\equiv \pm 1\hspace{0.6em}({\rm mod}\hspace{0.2em}p);\end{eqnarray}$$

the $2\times 2$ possibilities correspond to the two Heegaard solid tori and the two orientations. Therefore, Theorem 3.2 reduces the problem Wu solved to one about lattice points. However, it appears to require considerable effort to extract Wu’s result from it. Nevertheless, we can quickly derive the following characterization of the Berge manifold.

Wu points out that this result follows from the work of Berge and Gabai, which is an amusing exercise [Reference WuWu04, §4(1)]. By contrast, we deduce Proposition 3.4 from Theorem 3.2 and the uniqueness of genus-one Heegaard splittings of lens spaces.

Given linearly independent $v,w\in \mathbb{Z}^{2}$ , let $\unicode[STIX]{x1D6E5}(v,w)$ denote the triangle with vertices $0$ , $v$ , $w$ . It is empty if it contains no lattice points besides its vertices, that is, $\{v,w\}$ is a basis of $\mathbb{Z}^{2}$ .

Proof. Suppose that $K\subset S^{1}\times D^{2}$ is a strict 1-bridge braid with winding number $\unicode[STIX]{x1D714}$ and slope interval $[b/d,a/c]\subset (\unicode[STIX]{x1D714}/(m+1),\unicode[STIX]{x1D714}/m)$ , $m\in \mathbb{Z}$ . Let $X$ denote the exterior of $K$ . Suppose that $(p/q)$ -Dehn filling on the outer torus of $X$ is a solid torus. Thus, $p/q\in [b/d,a/c]$ , and one of the congruences in (3) holds.

If (3)(b) holds, then there must exist $s\in \mathbb{Z}$ so that $\unicode[STIX]{x1D6E5}((s,\unicode[STIX]{x1D714}),(p,q))$ is empty. If $p/q\in (b/d,a/c)$ , then this triangle contains $(c,a)$ if $s\leqslant m$ and $(d,b)$ if $s\geqslant m+1$ . Therefore, we must have $p/q\in \{a/c,b/d\}$ . Furthermore, $s=m$ if $p/q=a/c$ and $s=m+1$ if $p/q=b/d$ .

If instead $p>\unicode[STIX]{x1D714}$ , then it follows that (3)(a) holds, and we obtain $\unicode[STIX]{x1D714}=p-1$ . Furthermore, $(q,\unicode[STIX]{x1D714}+1)=(q,p)$ is in the cone bounded by the rays from $(0,0)$ through $(d,b)$ and $(c,a)$ . Since $m+1\leqslant \unicode[STIX]{x1D714}$ , it follows that $(q,\unicode[STIX]{x1D714}+1)=(m+1,\unicode[STIX]{x1D714}+1)$ . Moreover, $(m+1)/(\unicode[STIX]{x1D714}+1)$ is the mediant of $d/b$ and $c/a$ , meaning that $a+b=\unicode[STIX]{x1D714}+1$ and $c+d=m+1$ .

Hence, if $X$ has three distinct solid torus fillings, then they have slopes $a/c$ , $b/d$ , and $(a+b)/(c+d)$ (in conformity with the cyclic surgery theorem [Reference Culler, Gordon, Luecke and ShalenCGLS87]). We assume this going forward.

Suppose that (3)(b) holds for both $b/d$ and $a/c$ . It follows that all lattice points interior to $\unicode[STIX]{x1D6E5}((m,\unicode[STIX]{x1D714}),(m+1,\unicode[STIX]{x1D714}))$ fall on the rays generated by $(d,b)$ and $(c,a)$ . In particular, $(m,\unicode[STIX]{x1D714}-1)$ is a multiple of exactly one of these two points, meaning that $\unicode[STIX]{x1D714}\equiv 1\hspace{0.6em}({\rm mod}\hspace{0.2em}e)$ for a unique value $e\in \{a,b\}$ . Thus, (3)(a) holds for whichever of $b/d$ and $a/c$ has numerator different from $e$ .

On the other hand, if (3)(a) holds for some $p/q\in \{a/c,b/d\}$ , then it must hold with the sign $+1$ , since otherwise one of $a$ or $b$ divides the other, a contradiction. Since $(m,\unicode[STIX]{x1D714}-1)$ is the unique lattice point in $\unicode[STIX]{x1D6E5}((m,\unicode[STIX]{x1D714}),(m+1,\unicode[STIX]{x1D714}))$ with its $y$ -coordinate, it follows that it is a multiple of $(q,p)$ . Thus, (3)(a) holds for at most one of $a/c,b/d$ .

In total, (3)(a) holds for one of $a/c$ , $b/d$ and (3)(b) holds for the other. Applying the linear map $(x,y)\mapsto (y-x,y)$ exchanges $K$ with its mirror (see [Reference WuWu04, §4]). Thus, we may assume that (3)(a) holds for $a/c$ and (3)(b) for $b/d$ . We have $k\cdot (c,a)=(m,\unicode[STIX]{x1D714}-1)$ for some $k\in \mathbb{Z}$ , $k>0$ . Since $(m+1,\unicode[STIX]{x1D714}+1)=(c,a)+(d,b)$ , we obtain $(m+1,\unicode[STIX]{x1D714}+1)=k\cdot (c,a)+(1,2)$ , and so $(k-1)\cdot (c,a)+(1,2)=(d,b)$ . Hence

$$\begin{eqnarray}1=\left|\begin{array}{@{}cc@{}}d & b\\ c & a\end{array}\right|=\left|\begin{array}{@{}cc@{}}1 & 2\\ c & a\end{array}\right|=a-2c\quad \text{and}\quad 1=\left|\begin{array}{@{}cc@{}}m+1 & \unicode[STIX]{x1D714}\\ d & b\end{array}\right|=\left|\begin{array}{@{}cc@{}}m+1 & \unicode[STIX]{x1D714}+1\\ d & b\end{array}\right|+d=-1+d.\end{eqnarray}$$

We deduce in turn that $d=2$ , $k=2$ , $c=1$ , $a=3$ , $b=5$ , $\unicode[STIX]{x1D714}=7$ , and $m=2$ . Since $(7/3,5/2,3/1,7/2)$ is the Farey sequence of fractions between $\unicode[STIX]{x1D714}/(m+1)$ and $\unicode[STIX]{x1D714}/m$ , the slope interval of $K$ is $[5/2,3/1]$ , and it has winding number $7$ . See Figure 4. These invariants specify $K\simeq K(7,4,2)$ .◻