A characterization of quasipositive two-bridge knots

Suppose that $p/q=[a_{1},a_{2},\ldots,a_{k}]^{-}$, where $a_{i}\geq 2$ is even for all $1\leq i\leq k$. Note that $L(p,q)$ can be obtained by surgery on a chain of unknots with framings $-a_{1},-a_{2},\ldots,-a_{k}$, respectively. Since $a_{i}\geq 2$ is even for all $1\leq i\leq k$, we can Legendrian realize each unknot in the chain, with respect to the standard contact structure, so that the rotation number of each Legendrian unknot is zero. It follows by [14, Proposition 2.3] that the first Chern class of the Stein fillable (and hence tight) contact structure on $L(p,q)$ obtained by Legendrian surgery on the resulting Legendrian link is zero.

To prove the only if direction, suppose that $p/q=[a_{1},a_{2},\ldots,a_{k}]^{-}$, where $a_{i}\geq 2$ for all $1\leq i\leq k$ and $a_{j}$ is odd for some $1\leq j\leq k$. Let $\xi$ be the Stein fillable contact structure on $L(p,q)$ obtained by Legendrian surgery along an arbitrary Legendrian realization $\mathcal{L}$ of the chain of unknots with smooth framings $-a_{1},-a_{2},\ldots,-a_{k}$, respectively. Let $\overline{\mathcal{L}}$ be the Legendrian link obtained from $\mathcal{L}$ by taking the mirror image of each component and let $\overline{\xi}$ be the Stein fillable contact structure on $L(p,q)$ obtained by Legendrian surgery along $\overline{\mathcal{L}}$. It follows that rotation number of each component of $\overline{\mathcal{L}}$ is the negative of the rotation number of the corresponding component of $\mathcal{L}$ and hence $\overline{\xi}$ is obtained from $\xi$ by reversing the orientation of the contact planes. By Honda’s classification [16, Theorem 2.1] of tight contact structures on $L(p,q)$, the contact structures $\xi$ and $\overline{\xi}$ are not isotopic because of the assumption that $a_{j}$ is odd for some $1\leq j\leq k$. Note that non-isotopic tight contact structures on $L(p,q)$ are non-homotopic [16, Proposition 4.24]. This implies that $c_{1}(\xi)\neq 0$ since according to Gompf [14, Corollary 4.10], an oriented plane field in any closed oriented $3$-manifold is homotopic to itself with reversed orientation if and only if it has trivial first Chern class. ∎

The double cover of $S^{3}$ branched along $K(p,q)$ is $L(p,q)$. We can make $K(p,q)$ transverse to the standard contact structure $\xi_{st}$ by isotopy. Let $(L(p,q),\xi)$ be the contact double cover of $(S^{3},\xi_{st})$ branched along the transverse link $K(p,q)$. Suppose that $K(p,q)$ is quasipositive. By the work of Plamenevskaya [26, Proposition 1.4 and Lemma 5.1], we conclude that $\xi$ is Stein fillable and moreover $c_{1}(\xi)=0$. It follows by Proposition 1.6 that the negative continued fraction of $p/q$ must be even. ∎

A regular continued fraction of $p/q$ is defined by

$$\frac{p}{q}=[c_{1},c_{2},\ldots,c_{2m+1}]=c_{1}+\cfrac{1}{c_{2}+\cfrac{1}{\ddots+\cfrac{1}{c_{2m+1}}}}\;,\qquad c_{i}>0\text{ for all $1\leq i\leq 2m+1$}.$$

Note that there is always a regular continued fraction of $p/q$ of odd length, as above. To see this, suppose that $p/q$ has a regular continued fraction of even length, i.e. $\frac{p}{q}=[c_{1},c_{2},\ldots,c_{2m}]$ with $c_{i}>0$. If $c_{2m}=1$, then

$$\frac{p}{q}=[c_{1},c_{2},\ldots,c_{2m}]=[c_{1},c_{2},\ldots,c_{2m-2},1+c_{2m-1}]$$

and otherwise,

$$\frac{p}{q}=[c_{1},c_{2},\ldots,c_{2m}]=[c_{1},c_{2},\ldots,c_{2m-1},-1+c_{2m},1].$$

Using an odd length regular continued fraction $\frac{p}{q}=[c_{1},c_{2},\ldots,c_{2m+1}]$, we define the two bridge link $K(p,q)$ as depicted in Figure 1, where the integer inside each box denotes the signed number of half twists to be inserted.

In [34], Tanaka uses the notation $\mathcal{C}(c_{1},c_{2},\ldots,c_{2m+1})$, with each $c_{i}>0$, to describe a two-bridge knot. By comparing [34, Figure 6], for example, with our Figure 1, we see that our definition of $K(p,q)$ is the mirror image of the one described by Tanaka, but agrees with the one described by Lisca [18].

According to [34, Proposition 3.1], the two-bridge knot $\mathcal{C}(c_{1},c_{2},\ldots,c_{2m+1})$, which is the mirror image of our $K(p,q)$ defined above, is strongly quasipositive as long as $c_{i}$ is even for each even index $2\leq i\leq 2m$. To express this condition in terms of negative continued fractions, we just describe how to convert a given regular continued fraction of any $p/q$ to the negative continued fraction of $p/(p-q)$ and vice versa in Lemma 2.1. This finishes the proof of Theorem 1.1, since by Lemma 2.1, we can easily deduce that the negative continued fraction of $p/(p-q)$ is even if and only if $p/q$ has a regular continued fraction of odd length where each even indexed coefficient is even. ∎

The negative continued fraction of $p/q$ can be obtained from a given regular continued fraction of $p/q$ by a straightforward induction argument, whereas the negative continued fraction of $p/(p-q)$ is obtained from that of $p/q$ by the Riemenschneider’s point diagram method [27]. ∎

Rudolph [28] showed that quasipositive links arise as the transverse intersection of $S^{3}\subset\mathbb{C}^{2}$, with a complex curve. Therefore, the quasipositivity assumption implies that $K$ bounds a complex curve in $B^{4}$. Since $K$ is assumed to be smoothly slice as well, there is a smooth disk in $B^{4}$ with boundary $K$. But by the ”local Thom conjecture” [17, Corollary 1.3] , the complex curve minimizes genus, so the slice disk can be assumed to be complex. Hence, the analytic double cover of $B^{4}$ equipped with its standard complex structure, branched along this complex slice disk, is a rational homology ball Stein filling of $(Y,\xi)$. It is well-known that the double cover is a rational homology ball and it is in fact Stein by [19, Theorem 3].∎

Lisca [18] showed that $L(p,q)$ bounds a rational homology ball if and only if $p/q$ belongs to a certain subset $\mathcal{R}$ of the set of positive rational numbers. Lisca also showed that, for odd $p$, any two-bridge knot $K(p,q)$ is smoothly slice if and only if $p/q\in\mathcal{R}$. By definition, the set $\mathcal{R}$ contains a subset, denoted by $\mathcal{O}$ here, which consists of $p/q>0$ such that $p=m^{2}$ (for odd $m$), and $q=mh-1$ where $0<h<m$ and $(m,h)=1$.

Suppose that $p/q\in\mathcal{R}\setminus\mathcal{O}$ and $K(p,q)$ is a smoothly slice, quasipositive two-bridge knot. We can isotope $K(p,q)$ to be transverse to the standard contact structure $\xi_{st}$ in $S^{3}$. Let $(L(p,q),\xi)$ be the double cover of $(S^{3},\xi_{st})$ branched along the transverse knot $K$. According to Proposition 1.7, $(L(p,q),\xi)$ admits a rational homology ball Stein filling, which in turn, implies that $\xi$ is isomorphic to the canonical contact structure $\xi_{can}$, because no virtually overtwisted lens space has a rational homology ball symplectic filling by the work of Golla and Starkston [13]. (See also [10, Lemma 1.5], [11, Proposition 11]). This gives a contradiction to the fact that $(L(p,q),\xi_{can})$ admits a rational homology ball symplectic filling if and only if $p/q\in\mathcal{O}$, as shown by Lisca [18, Corollary 1.2(c)].

Now suppose that $p/q\in\mathcal{O}$, i.e. $p=m^{2}$, for odd $m>1$ and $q=mh-1$ where $0<h<m$ and $(m,h)=1$. If $h$ is even, then $q=mh-1$ is odd and $K(p,q)$ is non-quasipositive by Corollary 1.4. On the other hand, if $h$ is odd, then $q=mh-1$ is even but $q^{\prime}=m(m-h)-1$ is odd and $qq^{\prime}\equiv 1(\mbox{mod}\;m^{2}).$ Since $K(p,q)$ is isotopic to $K(p,q^{\prime})$ [5, Chapter 12], we conclude that $K(p,q^{\prime})$ and hence $K(p,q)$ is non-quasipositive, again by Corollary 1.2. ∎

For rational links we follow the orientation convention used in Murasugi’s book [22]. Let $L$ be a rational oriented link of type $(p,q)$. For the mirror image $D^{*}$ of $D$, we have $d^{\pm}(D^{*})=d^{\mp}(D)$ and $\mathbb{r}^{\pm}(D^{*})=\mathbb{r}^{\mp}(D)$. Therefore, without loss of generality we may assume that $q$ is odd and $0<q<p$. Using the notation introduced in [21, Section 3], let

$$\frac{p}{p-q}=[2n_{1,1},\ldots,2n_{1,k_{1}},\;-2n_{2,1},\ldots,-2n_{2,n_{2}},\;\ldots\;,(-1)^{t-1}2n_{t,1},\ldots,(-1)^{t-1}2n_{t,k_{t}}]^{-},$$

where we assume $n_{i,j}>0$.

Let $b$ be the braid index of $L$ and $e=w(\beta)$ be the exponent sum of a $b$-braid $\beta$ representing $L$. By [21, Prop. 4.2 and Thm. 4.3] we have

$$b=t+1+\sum_{i=1}^{t}\sum_{j=1}^{k_{i}}(n_{i,j}-1),\qquad e=\frac{1-(-1)^{t}}{2}+\sum_{i=1}^{t}(-1)^{i-1}\sum_{j=1}^{k_{i}}n_{i,j}.$$

(4.2)

Using the standard properties of rational links and continued fractions, one easily checks that $L$ admits an alternating diagram $D$ shown in Figure 2 where $T_{i}$ are the tangles defined by the braids

$$T_{i}=\begin{cases}\sigma_{1}^{1-2n_{i,1}}\Big{(}\prod_{j=2}^{k_{i}}\sigma_{2}\sigma_{1}^{2-2n_{i,j}}\Big{)}\sigma_{1}^{-1},&\text{$i$ is odd,}\\ \sigma_{2}^{2n_{i,1}-1}\Big{(}\prod_{j=2}^{k_{i}}\sigma_{1}^{-1}\sigma_{2}^{2n_{i,j}-2}\Big{)}\sigma_{2},&\text{$i$ is even,}\end{cases}$$

as depicted in Figure 3 for odd $i$. Note that, for odd (resp. even) $i$, all crossings of $T_{i}$ are positive (resp. negative).

Let $w=w(D)$ and $s=s(D)$ be the writhe and the number of Seifert circles of $D$. Each tangle $T_{i}$ contributes $(-1)^{i-1}w_{i}$ to $w$, where $w_{i}=1+\sum_{j}(2n_{i,j}-1)$. If $i$ is odd (resp. even), then $T_{i}$ contributes $w_{i}-k_{i}$ to $d^{+}$ (resp. to $d^{-}$). We have $s-(d^{+}+d^{-})=1$ (the Euler characteristic of a spanning tree).

Hence

$$s=t+1+\sum_{i=1}^{t}\sum_{j=1}^{k_{i}}2(n_{i,j}-1),\quad w=\frac{1-(-1)^{t}}{2}+\sum_{i=1}^{t}(-1)^{i-1}\sum_{j=1}^{k_{i}}(2n_{i,j}-1),$$

(4.3)

$$d^{+}=\lceil t/2\rceil+\sum_{\text{$i$ odd}}\;\sum_{j=1}^{k_{i}}2(n_{i,j}-1),\qquad d^{-}=\lfloor t/2\rfloor+\sum_{\text{$i$ even}}\;\sum_{j=1}^{k_{i}}2(n_{i,j}-1).$$

(4.4)

By combining (4.2) and (4.3) we obtain

$$s-b=\sum_{i=1}^{t}\sum_{j=1}^{k_{i}}(n_{i,j}-1),\qquad w-e=\sum_{i=1}^{t}(-1)^{i-1}\sum_{j=1}^{k_{i}}(n_{i,j}-1).$$

(4.5)

Recall that $\mathbb{r}^{\pm}=\mathbb{r}^{\pm}(D)$ are defined by

$$\mathbb{r}^{+}+\mathbb{r}^{-}=s-b,\qquad\mathbb{r}^{+}-\mathbb{r}^{-}=w-e.$$

(4.6)

By combining (4.4), (4.5), and (4.6) we obtain

$$2\mathbb{r}^{+}=d^{+}-\lceil t/2\rceil\leq d^{+},\qquad 2\mathbb{r}^{-}=d^{-}-\lfloor t/2\rfloor\leq d^{-}.$$

∎