---
abstract: 'In this paper, we collect various structural results to determine when an integral homology $3$–sphere bounds an acyclic smooth $4$–manifold, and when this can be upgraded to a Stein embedding. In a different direction we study whether smooth embedding of connected sums of lens spaces in $\mathbb{C}^2$ can be upgraded to a Stein embedding, and determined that this never happens.'
address:
- |
    Department of Mathematics\
    Georgia Institute of Technology\
    Atlanta\
    Georgia
- |
    Department of Mathematics\
    University of Alabama\
    Tuscaloosa\
    Alabama
author:
- 'John B. Etnyre'
- Bülent Tosun
title: Homology spheres bounding acyclic smooth manifolds and symplectic fillings
---

Introduction
============

The problem of embedding one manifold into another has a long, rich history, and proved to be tremendously important for answering various geometric and topological problems. The starting point is the Whitney Embedding Theorem: every compact $n$–dimensional manifold can be smoothly embedded in $\mathbb {R}^{2n}$. In this paper we will focus on smooth embeddings of 3–manifolds into $\R^4$ and embeddings that bound a convex symplectic domain in $(\R^4, \omega_{std})$. One easily sees that given such an embedding of a (rational) homology sphere, it must bound a (rational) homology ball. Thus much of the paper is focused on constructing or obstructing such homology balls.

Smooth embeddings
-----------------

In this setting, an improvement on the Whitney Embedding Theorem, due to Hirsch [@Hirsch:embedding] (also see Rokhlin [@Rohlin:3manembedding] and Wall [@Wall:embedding]), proves that every $3$–manifold embeds in $\mathbb{R}^5$ smoothly. In the smooth category this is the optimal result that works for all $3$–manifolds; for example, it follows from a work of Rokhlin that the Poincaré homology sphere $P$ cannot be embeded in $\mathbb{R}^4$ smoothly. On the other hand in the topological category one can always find embeddings into $\R^4$ for any integral homology sphere by Freedman’s work [@Freedman:4manifolds]. Combining the works of Rokhlin and Freedman for $P$ yields an important phenomena in $4$–manifold topology: there exists a closed oriented non-smoothable $4$–manifold — the so called $E_8$ manifold. In other words, the question of [*when does a $3$–manifold embeds in $\mathbb{R}^4$ smoothly*]{} is an important question from the point of smooth $4$–manifold topology. This is indeed one of the question in the Kirby’s problem list (Problem $3.20$) [@Kirby:problemlist]. Since the seminal work of Rokhlin in 1952, there has been a great deal of progress towards understanding this question. On the constructive side, Casson-Harrer [@CH], Stern, and Fickle [@Fickle] have found many infinite families of integral homology spheres that embeds in $\mathbb{R}^4$. On the other hand techniques and invariants, mainly springing from Floer and gauge theories, and symplectic geometry [@FS:R; @Manolescu:T; @OS:grading], have been developed to obstruct smooth embeddings of $3$–manifolds into $\mathbb{R}^4$. It is fair to say that despite these advances and lots of work done in the last seven decades, it is still unclear, for example, which Brieskorn homology spheres embed in $\mathbb{R}^4$ smoothly and which do not.

A weaker question is whether an integral homology sphere can arise as the boundary of an acyclic $4$–manifold? Note that a homology sphere that embeds in $\mathbb{R}^4$ necessarily bounds an acyclic manifold, and hence is homology cobordant to the $3$–sphere. Thus a homology cobordism invariant could help to find restrictions, and plenty of such powerful invariants has been developed. For example, for odd $n$, $\Sigma(2,3,6n-1)$ and $\Sigma(2,3,6n+1)$ have non-vanishing Rokhlin invariant. For even $n$, $\Sigma(2,3,6n-1)$ has $R=1$, where $R$ is the invariant of Fintushel and Stern, [@FS:R]. Hence none of these families of homology spheres can arise as the boundary of an acyclic manifold. On the other hand, for $\Sigma(2,3,12k+1)$ all the known homology cobordism invariants vanish. Indeed, it is known that $\Sigma(2,3,13)$ [@AKi] and $\Sigma(2,3,25)$ [@Fickle] bound contractible manifolds of Mazur type. Motivated by the questions and progress mentioned above and view towards their symplectic analogue, we would like to consider some particular constructions of three manifolds bounding acyclic manifolds.

Our first result is the following, which follows by adapting a method of Fickle.

\[main1\] Let $K$ be a knot in the boundary of an acyclic, respectively rationally acyclic, $4$–manifold $W$ which has a genus one Seifert surface $F$ with primitive element $[b]\in H_1(F)$ such that the curve $b$ is slice in $W.$ If $b$ has $F$–framing $s$, then the homology sphere obtained by $\frac{1}{(s\pm 1)}$ Dehn surgery on $K$ bounds an acyclic, respectively rationally acyclic, $4$–manifold.

Fickle [@Fickle] proved this theorem under the assumption that $\partial W$ was $S^3$ and $b$ was an unknot, but under these stronger hypothesis he was able to conclude that the homology sphere bounds a contractible manifold.

Fintushel and Stern conjectured, see [@Fickle], the above theorem for $\frac{1}{k(s\pm 1)}$ Dehn surgery on $K$, for any $k\geq 0$. So the above theorem can be seen to verify their conjecture in the $k=1$ case.

As noted by Fickle, if the conjecture of Fintushel and Stern is true then all the $\Sigma(2,3,12k+1)$ will bound acyclic manifolds since they can be realized by $-1/2k$ surgery on the right handed trefoil knot that bounds a Seifert surface containing an unknot for which the surface gives framing $-1$.

Notice that if $b$ is as in the theorem, then the Seifert surface $F$ can be thought of as obtained by taking a disk around a point on $b$, attaching a 1–handle along $b$ (twisting $s$ times) and then attaching another 1–handle $h$ along some other curve. The proof of Theorem \[main1\] will clearly show that $F$ does not have to be embedded, but just ribbon immersed so that cutting $h$ along a co-core to the handle will result in a surface that is “ribbon isotopic" to an annulus. By ribbon isotopic, we mean there is an 1-parameter family of ribbon immersions between the two surfaces, where we also allow a ribbon immersion to have isolated tangencies between the boundary of the surface and an interior point of the surface.

Consider the (zero twisted) $\pm$ Whitehead double $W_\pm(K_p)$ of $K_p$ from Figure \[exampleKp\]. In [@Cha07], Cha showed that $K_p$ is rationally slice. That is $K_p$ bounds a slice disk in some rational homology $B^4$ with boundary $S^3$. (Notice that $K_1$ is the figure eight knot originally shown to be rationally slice by Fintushel and Stern [@FintushelStern84].) Thus Theorem \[main1\] shows that $\pm 1$ surgery on $W_\pm(K_p)$ bounds a rationally acyclic 4–manifolds. This is easy to see as a Seifert surface for $W_\pm(K_p)$ can be made by taking a zero twisting ribbon along $K_p$ and plumbing a $\pm$ Hopf band to it.

[exampleKp]{} (44.5, 51)[$-p$]{} (48,6)[$p$]{}

Moreover, from Fickle’s original version of the theorem, $\pm\frac12$ surgery on $W_\pm(K_p)$ bounds a contractible manifold.

We can generalize this example as follows. Given a knot $K$, we denote by $R_m(K)$ the $m$-twisted ribbon of $K$. That is take an annulus with core $K$ such that its boundary components link $m$ times. Now denote by $P(K_1, K_2, m_1, m_2)$ the plumbing of $R_{m_1}(K_1)$ and $R_{m_2}(K_2)$. If the $K_i$ are rationally slice then $\frac{1}{m_i\pm 1}$ surgery on $P(K_1, K_2, m_1, m_2)$ yields a manifold bounding a rationally acyclic manifolds; moreover, if the $K_i$ are slice in some acyclic manifold, then the result of these surgeries will bound an acyclic manifold.

[**Symplectic embeddings.**]{} Another way to build examples of integral homology spheres that bound contractible manifolds is via the following construction. Let $K$ be a slice knot in the boundary of a contractible manifold $W$ (e.g. $W=B^4$), then $\frac{1}{m}$ Dehn surgery along $K$ bounds a contractible manifold. This is easily seen by removing a neighborhood of the slice disk from W (yielding a manifold with boundary 0 surgery on $K$) and attaching a 2–handle to a meridian of $K$ with framing $-m$. With this construction one can find examples of three manifolds modeled on not just Seifert geometry, for example $\Sigma(2,3,13)$ is the result of $1$ surgery on Stevedore’s knot $6_1$ but also hyperbolic geometry, for example the boundary of the Mazur cork is the result of $1$ surgery on the pretzel knot $P(-3,3,-3)$, which is also known as $\overline {9}_{46}$. See Figure \[fig:smooth\].

![On the left is the 3-manifold $Y_{m,n}$ described as a smooth $\frac{1}{m}$ surgery on the slice knot $P(3,-3,-n)$ for $n\geq 3$. On the right is the contractible Mazur-type manifold $W_{m,n}$ with $\partial W_{m,n}\cong Y_{m,n}$. Note the $m=1,~n=3$ case yields the original Mazur manifolds (with reversed orientation).[]{data-label="fig:smooth"}](RegSlicetoContractible.pdf){width="16cm"}

We ask the question of when $\frac{1}{m}$ surgery on a slice knot produces a [*Stein contractible*]{} manifold. Here there is an interesting asymmetry not seen in the smooth case.

\[regslice\] Let $L$ be a Legendrian knot in $(S^3, \xi_{std})$ that bounds a regular Lagrangian disc in $(B^4, w_{std})$. Contact $(1+\frac{1}{m})$ surgery on $L$ (so this is smooth $\frac{1}{m}$ surgery) is the boundary of a contractible Stein manifolds if and only if $m>0$.

This result points out an interesting angle on a relevant question in low dimensional contact and symplectic geometry: which compact contractible 4-manifolds admit a Stein structure? In [@MT:pseudoconvex] the second author and Mark found the first example of a contractible manifold without Stein structures with either orientation. This manifold is a Mazur-type manifold with boundary the Brieskorn homology sphere $\Sigma(2,3,13)$. A recent conjecture of Gompf remarkably predicts that Brieskorn homology sphere $\Sigma(p,q,r)$ can never bound acyclic Stein manifolds. It is an easy observation that $\Sigma(2,3,13)$ is the result of smooth $1$ surgery along the stevedore’s knot $6_1$. The knot $6_1$ is not Lagrangian slice, and indeed if Gompf conjecture is true, then by Theorem \[regslice\] $\Sigma(2,3,13)$ can never be obtained as a smooth $\frac{1}{n}$ surgery on a Lagrangian slice knot for any natural number $n$. Motivated by this example, Theorem \[regslice\], and Gompf’s conjecture we make the following weaker conjecture.

No non-trivial Brieskorn homology sphere $\Sigma(p,q,r)$ can be obtained as smooth $\frac{1}{n}$ surgery on a regular Lagrangian slice knot.

On the other hand as in Figure  \[fig:smooth\] we list a family of slice knots, that are regular Lagrangian slice because they bound decomposable Lagrangian discs and by [@CET] decomposable Lagrangian cobordisms/fillings are regular. We explicitly draw the contractible Stein manifolds these surgeries bound in Figure \[fig:stein\].

![Stein contractible manifold with $\partial X_{m,n}\cong Y_{m,n}$.[]{data-label="fig:stein"}](RegSlicetoContractibleStein.pdf){width="12cm"}

A related embedding question is the following: when does a lens space L(p,q) embeds in $\mathbb{R}^4$ or $S^4$? Two trivial lens spaces, $S^3$ and $S^1\times S^2$ obviously have such embeddings. On the other hand, Hantzsche in $1938$ [@Hantzsche] proved, by using some elementary algebraic topology that if a $3$–manifold $Y$ embeds in $S^4$, then the torsion part of $H_1(Y)$ must be of the form $G \oplus G$ for some finite abelian group $G$. Therefore a lens space $L(p,q)$ for $|p|>1$ never embeds in $S^4$ or $\mathbb{R}^4$. For punctured lens spaces, however the situation is different. By combining the works of Epstein [@Epstein] and Zeeman [@Zeeman], we know that, a punctured lens space $L(p,q)\setminus B^3$ embeds in $\mathbb{R}^4$ if and only if $p>1$ is odd. Note that given such an embedding a neighborhood of $L(p,q)\setminus B^3$ in $\mathbb{R}^4$ is simply $(L(p,q)\setminus B^3)\times [-1,1]$ a rational homology ball with boundary $L(p,q)\# L(p,p-q)$ (recall $-L(p,q)$ is the same manifold as $L(p,p-q)$).

One way to see an embedding of $L(p,q)\# L(p,p-q)$ into $S^4$ is as follows: First, it is an easy observation that if $K$ is a doubly slice knot (that is there exists a smooth unknotted sphere $S\subset S^4$ such that $S\cap S^3=K$), then its double branched cover $\Sigma_2(K)$ embeds in $S^4$ smoothly. Moreover by a known result of Zeeman $K \# m(K)$ is a doubly slice knot for any knot $K$ (here $m(K)$ is the mirror of $K$). It is a classic fact that $L(p,q)$ is a double branched cover over the the 2-bridge knot $K(p,q)$ (this is exactly where we need $p$ to be odd, as otherwise $K(p,q)$ is a link). In particular $L(p,q)\#L(p,p-q)$, being double branched cover of doubly slice knot $K(p,q)\#m(K(p,q))$, embeds in $S^4$ smoothly. On the other hand, Fintushel-Stern [@FS:lensspace] and independently Gilmer-Livingston [@GilmerLivingston] showed this is all that could happen. That is they proved that $L(p,q)\#L(p,q')$ embeds in $S^4$ if and only if $L(p,q')=L(p,p-q)$ and $p$ is odd. In particular for $p$ odd, $L(p,q)\#L(p,p-q)$ bounds a rational homology ball in $\mathbb{R}^4$. A natural question in this case is to ask whether any of this smooth rational homology balls can be upgraded to be Symplectic or Stein submanifold of $\mathbb{C}^2$. We prove that this is impossible.

\[main2\] No contact structure on $L(p,q)\#L(p,p-q)$ has a symplectic filling by a rational homology ball. In particular, $L(p,q)\#L(p,p-q)$ cannot embed in $\mathbb{C}^2$ as the boundary of exact symplectic submanifold in $\mathbb{C}^2$.

Donald [@Donald] generalized Fintushel-Stern and Gilmer-Livingston’s construction further to show that for $L=\#_{i=1}^h L(p_i,q_i)$, the manifold $L$ embeds smoothly in $\mathbb{R}^4$ if and only if there exists $Y$ such that $L\cong Y\#-Y$. Our proof of Theorem \[main2\] applies to this generalization to prove none of the sums of lens spaces which embed in $\mathbb{R}^4$ smoothly can bound an exact symplectic manifold in $\mathbb{C}^2.$

To prove this theorem we need a preliminary result of independent interest.

\[prop\] If a symplectic filling $X$ of a lens space $L(p,q)$ is a rational homology ball, then the induce contact structure on $L(p,q)$ is a universally tight contact structure $\xi_{std}$.

Recall that every lens space admits a unique contact structure $\xi_{std}$ that is tight when pulled back the covering space $S^3$. Here we are not considering an orientation on $\xi_{std}$ when we say it is unique. On some lens spaces the two orientations on $\xi_{std}$ give the same oriented contact structure and on some they are different.

After completing a draft of this paper, the authors discovered that this result was previously proven by Fossati [@Fossati19pre] and Golla and Starkston [@GollaStarkston19pre]. As the proof we had is considerably different we decided to present it here.

[**Acknowledgements:**]{} We are grateful to Agniva Roy for pointing out the work of Fossati and of Golla and Starkston. The first author was partially supported by NSF grant DMS-1906414. Part of the article was written during the second author’s research stay in Montreal in Fall 2019. This research visit was supported in part by funding from the Simons Foundation and the Centre de Recherches Mathmatiques, through the Simons-CRM scholar-in-residence program. The second author is grateful to CRM and CIRGET, and in particular to Steve Boyer for their wonderful hospitality. The second author was also supported in part by a grant from the Simons Foundation (636841, BT)

Bounding acyclic manifolds
==========================

We now prove Theorem \[main1\]. The proof largely follows Fickle argument from [@Fickle], but we repeat it here for the readers convince (and to popularize Fickle’s beautiful argument) and to note where changes can be made to prove our theorem.

Suppose the manifold $\partial W$ is given by a surgery diagram $D$. Then the knot $K$ can be represented as in Figure \[TheKnot\]. There we see in grey the ribbon surface $F$ with boundary $K$ and the curve $b$ on the surface.

[TheKnot]{} (48, 68)[$D$]{} (103, 15)[$K$]{} (62, 15)[$b$]{} (11, 10)[$-s+1$]{} (90, 33)[$0$]{}

The result of $\frac{1}{s-1}$ surgery on $K$ is obtained by doing $0$ surgery on $K$ and $(-s+1)$ surgery on a meridian as shown in Figure \[TheKnot\]. (The argument for $\frac{1}{s+1}$ surgery is analogous and left to the reader.) Now part of $b$ is the core of one of the 1–handles making up $F$. So we can handle slide $b$ and the associated 1–handle over the $(-s+1)$ framed unknot to arrive at the left hand picture in Figure \[Step1\]. Then one may isotope the resulting diagram to get to the right hand side of Figure \[Step1\].

[Step1]{} (30, 26)[$D$]{} (84, 26)[$D$]{} (48, 11)[$0$]{} (100, 11)[$0$]{} (11, 6)[$-s+1$]{} (63, 6)[$-s+1$]{} (14.5, 17.5)[$1$]{}

We now claim the left hand picture in Figure \[Step2\] is the same manifold as the right hand side of Figure \[Step1\]. To see this notice that the green part of the left hand side of Figure \[Step2\] consists of two 0-framed knots. Sliding one over the other and using the new 0-framed unknot to cancel the non-slid component results in the right hand side of Figure \[Step1\].

Before moving forward we discuss the strategy of the remainder of the proof. The left hand side of Figure \[Step2\] represents the 3-manifold $M$ obtained from $\partial W$ by doing $\frac{1}{s-1}$ surgery on $K$. We will take $[0,1]\times M$ and attach a 2–handle to $\{1\}\times M$ to get a 4-manifold $X$ with upper boundary $M'$ so that $M'$ is obtained from $W$ by removing a slice disk $D$ for $b$. Since $W$ is acyclic, the complement of $D$ will be a homology $S^1\times D^3$. Let $W'$ denote this manifold. Attaching $X$ upside down to $W'$ (that is attaching a 2–handle to $W'$) to get a 4–manifold $W''$ with boundary $-M$. Since $-M$ is a homology sphere, we can easily see that $W''$ is acyclic. Thus $-W''$ is an acyclic filling of $M$.

Now to see we can attach the 2–handle to $[0,1]\times M$ as described above, we just add a 0-framed meridian to the new knot unknot on the left hand side of Figure \[Step2\]. This will result in the diagram on the right hand side of Figure \[Step2\].

[Step2]{} (32, 29)[$D$]{} (87, 29)[$D$]{} (50, 11)[$0$]{} (97, 11)[$0$]{} (45.5, 19)[$0$]{} (26.5, 17.5)[$0$]{} (11, 6.5)[$-s+1$]{} (70, 6)[$-s+1$]{}

We are left to see that the right hand side of Figure \[Step2\] is the boundary of $W$ with the slice disk for $b$ removed. To see this notice that the two green curves in Figure \[Step2\] co-bound an embedded annulus with zero twisting (the grey in the figure) and one boundary component links the $(-s+1)$ framed unknot and the other does not. Sliding the former over the latter results in the left hand diagram in Figure \[Step3\]. Cancelling the two unknots from the diagram results in the right hand side of Figure \[Step3\] which is clearly equivalent to removing the slice disk $D$ for $b$ from $W$.

[Step3]{} (34, 31)[$D$]{} (78, 31)[$D$]{} (26, 3.5)[$0$]{} (26.5, 17.5)[$0$]{} (11, 3)[$-s+1$]{} (92, 6)[$0$]{}

Stein fillings
==============

We begin this section by proving Theorem \[regslice\] concerning smooth $\frac 1 m$ surgery on a Lagrangian slice knot.

We begin by recalling a result from [@CET] that says contact $(r)$ surgery on a Legendrian knot $L$ for $r\in(0,1]$ is strongly symplectically fillable if and only if $L$ is Lagrangian slice and $r=1$. Thus $(1+1/m)$ contact surgery for $m<0$ will never be fillable, much less fillable by a contractible Stein manifold.

We now turn to the $m>0$ case and start by a particularly helpful visualization of the knot $L$ (here and forward $L$ stands both for the knot type and Legendrain knot that realizing the knot type that bounds the regular Lagrangian disk). By [@CET Theorem $1.9$, Theorem $1.10$], we can find a handle presentation of the 4-ball $B^4$ made of one 0–handle, and $n$ cancelling Weinstein $1$– and $2$–handle pairs, and a maximum Thurston-Bennequin unknot in the boundary of the 0–handle that is disjoint from $1$– and $2$–handles such that when the $1$– and $2$–handle cancellations are done the unknot becomes $L$. See Figure \[Handle\].

[Handle2]{} (56, 2)[$L$]{}

Now smooth $1/m$ surgery on $L$ can also be achieved by smooth $0$ surgery (which corresponds to taking the complement of the slice disk) on $L$ followed by smooth $-m$ surgery on its meridian.

As the proof of Theorem 1.1 in [@CET] shows, removing a neighborhood of the Lagrangian disk $L$ bounds from $B^4$ gives a Stein manifold with boundary $(+1)$ contact surgery on $L$ (that is smooth $0$ surgery on $L$). Now since the meridian to $L$ can clearly be realized by an unknot with Thurston-Bennequin invariant $-1$, we can stabilize it as necessary and attach a Stein $2$–handle to it to get a contractible Stein manifold bounding $(1+1/m)$ contact surgery on $L$ for any $m>1$.

For the $m=1$ case we must argue differently. One may use Legendrian Reidemeister moves to show that in any diagram for $L$ as described above the $2$–handles pass through $L$ as shown on the left hand side of Figure \[normalform\].

[NormalForm]{} (74, 7.75)[$-1$]{}

Smoothly doing contact $(1+1/1)$–surgery on $L$ (that is smooth $1$ surgery) is smoothly equivalent to replacing the left hand side of Figure \[normalform\] with the right hand side and changing the framings on the strands by subtracting their linking squared with $L$.

Now notice that if we realize the right hand side of Figure \[normalform\] by concatenating $n$ copies of either diagram in Figures \[sandz\] (where $n$ is the number of red strands in Figure \[normalform\]) then the Thurston-Bennequin invariant of each knot involved in Figure \[normalform\] is reduced by the linking squared with $L$. Thus we obtain a Stein diagram for the result of $(2)$ contact surgery on $L$.

[sandz]{}

Notice that the diagram clearly describes an acyclic 4–manifolds and moreover the presentation for its fundamental group is the same as for the presentation for the fundamental group of $B^4$ given by the original diagram. Thus the 4–manifolds is contractible.

We now turn to the proof that connected sums of lens spaces can never have acyclic symplectic fillings, but first prove Proposition \[prop\] that says any contact structure on a lens space that is symplectically filled by a rational homology ball must be universally tight.

Let $X$ be a rational homology ball symplectic filling of $L(p,q)$. We show the induces contact structure must be the universally tight contact structure $\xi_{std}$. This will follow from unpacking recent work of Menke [@menke2018jsjtype] where he studies exact symplectic fillings of a contact $3$–manifold that contains a [*mixed torus*]{}.

We start with the set-up. Honda [@Honda:classification1] and Giroux [@Giroux:classification] have classified tight contact structures on lens spaces. We review the statement of Honda in terms of the Farey tessellation. We use notation and terminology that is now standard, but see see [@Honda:classification1] for details. Consider a minimal path in the Farey graph that starts at $-p/q$ and moves counterclockwise to $0$. To each edge in this path, except for the first and last edge, assign a sign. Each such assignment gives a tight contact structure on $L(p,q)$ and each tight contact structures comes from such an assignment. If one assigns only $+$’s or only $-$’s to the edges then the contact structure is universally tight, and these two contact structures have the same underlying plane field, but with opposite orientations. We call this plane field (with either orientation) the the universally tight structure $\xi_{std}$ on $L(p,q)$. All the other contact structures are virtually overtwisted, that is they are tight structures on $L(p,q)$ but become overtwisted when pulled to some finite cover. The fact that at some point in the path describing a virtually overtwisted contact structure the sign must change is exactly the same as saying a Heegaard torus for $L(p,q)$ satisfies Menke’s mixed torus condition.

\[mixed\] Let $(Y, \xi)$ denote closed, co-oriented contact $3$–manifold and let $(W, \omega)$ be its strong (resp. exact) symplectic filling. If $(Y,\xi)$ contains a mixed torus $T$, then there exists a (possibly diconnected) symplectic manifold $(W', \omega')$ such that:

-   $(W', \omega')$ is a strong (rep. exact) symplectic filling of its boundary $(Y',\xi')$.

-   $\partial W'$ is obtained from $\partial W$ by cutting along $T$ and gluing in two solid tori.

-   $W$ can be recovered from $W'$ by symplectic round $1$–handle attachment.

In our case we have $X$ filling $L(p,q)$. Suppose the contact structure on $L(p,q)$ is virtually overtwisted. The theorem above now gives a symplectic manifold $X'$ two which a round 1–handle can be attached to recover $X$; moreover, $\partial X'$ is a union of two lens spaces or $S^1\times S^2$. However, Menke’s more detailed description of $\partial X'$ shows that $S^1\times S^2$ is not possible. We digress for a moment to see why this last statement is true. When one attaches a round 1–handle, on the level of the boundary, one cuts along the torus $T$ and then glues in two solid tori. Menke gives the following algorithm to determine the meridional slope for these tori. That $T$ is a mixed torus means there is a path in the Farey graph with three vertices having slope $r_1, r_2,$ and $r_3$, each is counterclockwise of the pervious one and there is an edge from $r_i$ to $r_{i+1}$ for $i=1,2$. The torus $T$ has slope $r_2$ and the signs on the edges are opposite. Now let $(r_3,r_1)$ denote slopes on the Farey graph that are (strictly) counterclockwise of $r_3$ and (strictly) clockwise of $r_1$. Any slope in $(r_3,r_1)$ with an edge to $r_2$ is a possible meridional slope for the glued in tori, and these are the only possible slopes. Now since our $r_i$ are between $-p/q$ and $0$ we note that if there was an edge from $r_2$ to $-p/q$ or $0$ then $r_2$ could not be part of a minimal path form $-p/q$ to $0$ that changed sign at $r_2$. Thus when we glue in the solid tori corresponding to the round 1–handle attachment, they will not have meridional slope $0$ or $-p/q$ and thus we cannot get $S^1\times S^2$ factors.

The manifold $X'$ is either connected or disconnected. We notice that it cannot be connected because it is know that any contact structure on a lens space is planar [@Schoenenberger05], and Theorem 1.2 from [@Etnyre:planar] says any filling of a contact structure supported by a planar open book must have connected boundary. Thus we know that $X'$ is, in fact, disconnected. So $X'=X'_1\cup X'_2$ with $\partial X'_i$ a lens space. The Mayer–Vietoris sequence for the the decomposition of $X'$ into $X'_1\cup X'_2$ (glued along an $S^1\times D^2$ in their boundaries) shows that $H_1$ of $X_1'$ or $X_2'$ has rank 1, while both of their higher Betti numbers are 0. But now the long exact sequence for the pair $(X'_i,\partial X'_i)$ implies that $b_1$ must be 0 for both the $X_i'$. This contradiction shows that a symplectic manifold which is rational homology ball and with boundary $L(p,q)$ must necessarily induce the universally tight contact structure on the boundary.

The statement about embeddings follows directly from the statement about symplectic fillings. To prove that result let $X$ be an exact symplectic filling of $L(p,q)\# L(p,p-q)$ that is also a rational homology ball. Observe that there is an embedded sphere in $\partial X$ as it is reducible. Eliashberg’s result in [@CE Theorem $16.7$] says that $X$ is obtained from another symplectic manifold with convex boundary by attaching a 1–handle. Thus $X\cong X_1\natural X_2$ where $X_1$ and $X_2$ are exact symplectic manifolds with $\partial X_1=L(p,q)$ and $\partial X_2=L(p,p-q)$ or $X\cong X'\cup (\text{1--handle})$ where $X'$ is symplectic 4-manifold with the disconnected boundary $\partial X'\cong L(p,q)\sqcup L(p,p-q)$.

As argued above in the proof of Proposition \[prop\] it is not possible to have $X'$ with disconnected boundary being lens spaces and we must be in the case $X\cong X_1\natural X_2$; moreover, since $X$ is a rational homology balls, so are the $X_i$. Moreover, since $X_1$ and $X_2$ are symplectic filling of their boundaries, they induce tight contact structures on $L(p,q)$ and $L(p,p-q)$), respectively.

Proposition \[prop\] says that these tight contact structures must be, the unique up to changing orientation, universally tight contact structures $\xi_{std}$ on $L(p,q)$ and $\xi'_{std}$ on $L(p,p-q)$. Thus we have that $X_1$ and $X_2$ are rational homology balls, and are exact symplectic fillings of $(L(p,q), \xi_{std})$, and $(L(p,p-q), \xi'_{std})$, respectively. In [@Lisca08 Corollary 1.2(d)] Lisca classified all such fillings. According to Lisca’s classification, symplectic rational homology ball fillings of $(L(p,q), \xi_{std})$ are possible exactly when $(p,q)=(m^2, mh-1)$ for some $m$ and $h$ co-prime natural numbers, and similarly for $(L(p,p-q), \xi'_{std})$ exactly when $(p,p-q)=(m^2, mk-1)$ for $m$ and $k$ co-prime natural numbers. Now simple calculation shows that, the only possible value for $m$ satisfying these equations is $m=2$. In particular, we get that $p=4$, but then we must have $\{q, p-q\}=\{1,3\}$, and 3 cannot be written as $2k-1$, for $k$ co-prime to $2$. Thus there is no such $X$.

[10]{}

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