# Monodromy factorizations of Seifert fibered spaces

Victoria Quijano

###### Abstract.

We present relations in the mapping class monoid of $S_{0,0}^{n}$ between
products of boundary parallel twists and those involving only non boundary
parallel twists. These are of particular interest because each element gives
an open book decomposition of a contact 3-manifold, and different
factorizations of the same mapping class give potentially distinct symplectic
fillings with diffeomorphic boundaries. We apply these results in order to
compute the Euler characteristics of fillings resulting from different
factorizations, present plumbing graphs for the fillings given by products of
boundary parallel twists, and provide sharp bounds on the Euler
characteristics for fillings arising from such relations.

##  1\. Introduction

In 2010, Wendl proved that, given a planar open book decomposition of a
contact 3-manifold, each minimal symplectic filling is given by a
factorization of its monodromy [Wen10]. This is particularly relevant to a
process called monodromy substitution, in which one filling is replaced by
another in an ambient 4-manifold. Monodromy substitution preserves the
symplectic structure on the ambient 4-manifold, but in order to apply it using
a relation between factorizations, we must be able to identify the filling
corresponding to one of the sides of the relation embedded inside an existing
4-manifold. It is useful to consider relations involving a factorization that
corresponds to a symplectic plumbing, because in order to find a plumbing
embedded in a 4-manifold, we need only to find a collection of symplectic
surfaces that intersect as described by the plumbing graph.

In this article, we consider relations of a specific type in order to identify
alternative symplectic fillings of the boundaries of plumbings with star-
shaped graphs, called Seifert fibered spaces. In particular, Endo and Gurtas
[EG08] show that monodromy substitution using the lantern relation correlates
to a rational blowdown, a process which has been shown to be symplectic by
Symington [Sym98]. Furthermore, when monodromy subsitution is performed using
a plumbing with a star-shaped graph, the process corresponds to star surgery,
a generalization of the rational blowdown that was introduced and shown to be
symplectic by Karakurt and Starkston [KS16]. We present relations that
generalize the lantern relation, which is defined for
$\text{Mod}(S_{0,0}^{4})$, to the mapping class monoids of planar surfaces
with more boundary components. Margalit and McCammond [MM09] describe a well-
defined correlation between convex twists on braids and Dehn twists on
punctured spheres with exactly one boundary component. In order to prove
relations on $\text{Mod}(S_{0,0}^{n})$, it suffices to show that the
corresponding braids, as defined by [MM09] after the interior boundary
components are capped by once-punctured disks, are isotopic and that the
multiplicity of each boundary component is the same under both sides of the
relation (see [PS21]). We use these results to consider possible relations and
discuss the completeness of each list of relations.

In particular, we focus on relations on the mapping class monoid of
$S_{0,0}^{n}$, for $n\geq 5$, of the form $T_{b_{1}}^{a_{1}}\cdots
T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots T_{\alpha_{k}}$ where each
$\alpha_{i}$ is a non boundary parallel curve. We present restrictions on the
exponents $a_{i}$ in order for such a relation to exist. Applying these
results, we discuss the graphs of plumbings whose contact boundaries do not
have alternative minimal symplectic fillings, concluding that a plumbing given
by a star-shaped graph with a central vertex of self-intersection number $-n$
and at most $n-5$ other vertices, all with self-intersection number $-2$ does
not have any alternative minimal symplectic fillings. We also prove a
generalization of the daisy relation introduced in [GM13], describe the
plumbing graphs associated with relations of this form, and compute the Euler
characteristics of the symplectic fillings associated with each side of the
relation in terms of $n$. Additionally, we consider relations of the specified
form for particular small values of $n$. A complete set of relations for the
$n=5$ case is presented using the generalization of the daisy relation. For
$n=6$ and $n=7$, we prove additional relations that are not special cases of
this generalization, and we show that the list of relations presented for
$n=6$ is complete as well. For each of these sets of relations, we present the
associated plumbing graphs and compute the Euler characteristics of the
fillings associated with each side of the relation. We also present sharp
bounds on the Euler characteristics of plumbings associated with relations of
the specified form.

The organization of this article is as follows. We begin with an introduction
to the background material and notation that will be used in Section 2. In
Section 3, we present products of boundary parallel twists that do not have
alternative factorizations and the graphs of plumbings that correspond to
these products. In Section 4, we present the generalization of the daisy
relation and the list of relations for $n=5,6,$ and 7 and discuss the
associated Euler characteristics and plumbing graphs. Finally, in Section 5,
we discuss bounds on the Euler characteristics of plumbings associated with
relations of the specified form and prove the completeness of the list of
relations presented for the $n=5$ and $n=6$ cases.

Acknowledgments. The author would like to thank Laura Starkston for providing
the initial question that led to this project and for her patience, support,
and guidance along the way.

##  2\. Background

Take $S_{g,n}^{b}$ to be the orientable surface of genus $g$ with $n$
punctures and $b$ boundary components. Recall that the mapping class group of
$S_{g,n}^{b}$ contains the isotopy classes of homeomorphisms from that surface
to itself which fix the boundary pointwise and send every puncture in the
domain to a puncture in the image.

In particular, we can use homeomorphisms called Dehn twists to represent
elements of the mapping class group of a surface.

######  Definition 2.1.

Consider the annulus $A=S^{1}\times[0,1]$ and its embedding into the
$(\theta,r)$-plane by the map $(\theta,t)\mapsto(\theta,t+1)$. Induce an
orientation onto $A$ using the standard orientation of the plane. Consider the
map $T:A\to A$ defined by $T(\theta,t)=(\theta+2\pi t,t)$.

Given an oriented surface $S$, and $\alpha$ a simple closed curve in $S$, let
$N$ be a regular neighborhood of $\alpha$. Let $\phi:A\to N$ be an
orientation-preserving homeomorphism of the annulus.

Then the positive Dehn twist about $\alpha$ is the homeomorphism
$T_{\alpha}:S\to S$ defined by

$T_{\alpha}(x)=\begin{cases}\phi\circ T\circ\phi^{-1}(x)&{\text{if }}x\in N\\\
x&{\text{if }}x\in S/N\end{cases}$

A negative Dehn twist about $\alpha$, $T^{-1}_{\alpha}:S\to S$, is defined by
replacing $T$ with $T^{-1}$ in the piecewise function. Note that
$T^{-1}(\theta,t)=(\theta-2\pi t,t)$ and that $T_{\alpha}^{-1}$ is indeed the
inverse of $T_{\alpha}$.

Figure 1. The image on the right shows the image of $\gamma$ after applying
the positive Dehn twist $\psi$ to the annulus.

Dehn twists are well-defined on isotopy classes of curves, so we can use
$T_{\alpha}$ to denote the positive Dehn twist along the isotopy class of
$\alpha$. Furthermore, the mapping class group of any surface is generated by
Dehn twists over finitely many simple closed curves [FM12]. Therefore, we can
write any element of the mapping class group as a product of Dehn twists.

The mapping class monoid of a surface is the monoid generated by all positive
Dehn twists about isotopy classes of essential simple closed curves and
boundary parallel simple closed curves on that surface. The mapping class
monoids of surfaces are of particular interest because, given a surface and a
product of Dehn twists in its mapping class monoid, we can build a
4-dimensional manifold using a Lefschetz fibration.

######  Definition 2.2.

A Lefschetz fibration on a 4-manifold $X$ is a map $\pi:X\to B$, such that $B$
is a compact, connected, oriented surface, $\pi$ has finitely many critical
values $t_{1},\ldots,t_{n}$ in the interior of $B$, and in a neighborhood
around $t_{i}$ we can assign complex coordinates $(z_{1},z_{2})$ such that, in
these coordinates, $\pi(z_{1},z_{2})=z_{1}^{2}+z_{2}^{2}$.

Furthermore, we can consider the fiber, or preimage, of a point $c\in B$. For
all $c\in B$, if $c$ is not a critical point of $\pi$, then the fiber
$\pi^{-1}(c)$ is diffeomorphic to a compact surface. Furthermore, there is a
single such surface for all $c\in B$ that are not critical points, and this is
called the general fiber [Ful03]. A vanishing cycle in a Lefschetz fibration
is the isotopy class of a curve in the general fiber which is collapsed in the
fiber of a critical point.

We will now consider the properties of a Lefschetz fibration.

######  Definition 2.3.

Given a 3-manifold $M$, an open book decomposition of $M$ is a pair $(B,\pi)$
such that $B$ is an oriented link and $\pi:M\backslash B\to S^{1}$ is a
fibration of $M\backslash B$ such that, for all $\theta\in S^{1}$,
$\pi^{-1}(\theta)$ is the interior of a compact surface $S_{\theta}$ in $M$
and $\partial S_{\theta}=B$.

$B$ is called the binding of the open book, and each $S_{\theta}$ is called a
page.

Conversely, we can begin with an open book and use it to build a 3-manifold.

######  Definition 2.4.

An abstract open book is a pair $(\Sigma,\phi)$ such that $\Sigma$ is an
oriented, compact surface with boundary and $\phi:\Sigma\to\Sigma$ is a
diffeomorphism which acts as the identity on a neighborhood of
$\partial\Sigma$. $\phi$ is called the monodromy of the open book.

Two abstract open books $(\Sigma_{1},\phi_{1})$ and $(\Sigma_{2},\phi_{2})$
are considered equivalent if there exists a diffeomorphism
$h:\Sigma_{1}\to\Sigma_{2}$ such that $\phi_{1}\circ h=h\circ\phi_{2}$.

Given a Lefschetz fibration $\pi:X\to B$, we can obtain an open book
decomposition of the boundary of $X$ [OS04]. In this decomposition, the pages
are the fibers of the points $x\in\partial B$. In order to obtain the binding,
we consider the disjoint union of the boundary components of the fibers.
Because each fiber is a surface, its boundary will be diffeomorphic to the
union of $n$ copies of $S^{1}$. The number of boundary components of the
general fiber does not change when one of the vanishing cycles is collapsed,
so all fibers have the same value for $n$. Therefore, taking the union of the
boundaries of each fiber in the Lefschetz fibration will yield
$\sqcup_{n}S^{1}\times B$.

The following result describes how to obtain the monodromy of the open book
decomposition of the boundary of $X$.

######  Theorem 2.5 ([OS04]).

Let $X$ be a 4-manifold, and let $\pi:X\to D^{2}$ be a Lefschetz fibration on
$X$ with $n$ singularities. Then the corresponding abstract open book
$(F,\phi)$ has monodromy $\phi=T_{\gamma_{1}}\cdots T_{\gamma_{n}}$, where $F$
is the general fiber and each $\gamma_{i}$ is a vanishing cycle in $F$.

Furthermore, if two abstract open books are equivalent, then their
corresponding 3-manifolds are diffeomorphic. In particular, given a surface
$F$ with two different factorizations of an element of its mapping class
monoid, $\tau$ and $\tau^{\prime}$, the open books $(F,\tau)$ and
$(F,\tau^{\prime})$ are equivalent. Let $X$ and $X^{\prime}$ be the fillings
of $(F,\tau)$ and $(F,\tau^{\prime})$, respectively. Because the open books
are equivalent, $\partial X$ and $\partial X^{\prime}$ are diffeomorphic.
Since $X$ and $X^{\prime}$ are not necessarily diffeomorphic, this gives us
two potentially distinct 4-manifolds with diffeomorphic boundaries. In fact,
we will see that the fillings associated with the relations presented in later
sections are distinct by computing their Euler characteristics.

A Lefschetz fibration to the disk is called _allowable_ if all of its
vanishing cycles are essential curves. In this article, all Lefschetz
fibrations we consider will be allowable, so we will typically omit this
adjective. A correspondence between allowable Lefschetz fibrations and a
particularly nice kind of symplectic fillings called Stein fillings was
established in [LP01] (see also an alternative proof in [AO01]). A
correspondence between contact structures and open book decompositions up to
stabilizations was established by [TW75, Gir02]. Given an allowable Lefschetz
fibration, the compatible symplectic filling structure induces a contact
manifold on the boundary, and under the correspondences, this contact manifold
is compatible with the open book induced on the boundary of the Lefschetz
fibration.

We will be restricting our consideration to the case where the general fiber
of a Lefschetz fibration has genus 0. In part, this is because it is a simpler
starting point in studying mapping class monoids, and it is also partly
motivated by the stronger implications for symplectic fillings. We say that an
open book decomposition is _planar_ if the pages are genus zero. In this case,
all (minimal) symplectic fillings, not only Stein fillings, have corresponding
Lefschetz fibrations. Moreover, every (minimal) symplectic filling has a
Lefschetz fibration such that the open book decomposition induced on its
boundary is the fixed planar open book that you started with, as shown in the
following theorem of Wendl.

######  Theorem 2.6 (Wendl [Wen10]).

If $(Y,\xi)$ is a contact $3$-manifold supported by a planar open book
$\mathcal{OB}$, then, up to symplectic deformation, every minimal symplectic
filling $(W,\omega)$ of $(Y,\xi)$ is supported by a Lefschetz fibration which
induces $\mathcal{OB}$ on its boundary.

As a consequence, if we have a contact manifold supported by a planar open
book, then every symplectic filling corresponds to a factorization of the
monodromy of that particular planar open book.

######  Definition 2.7.

Let $F$ be a closed surface and let $\pi:X\to F$ be a $D^{2}$-bundle over $F$.
Given two disjoint disks in $F$, $D_{1}$ and $D_{2}$, we can plumb $X$ at
$D_{1}$ and $D_{2}$ by identifying $D_{1}\times D^{2}$ with $D_{2}\times
D^{2}$ using a map that interchanges the factors but preserves product
structures.

A plumbing is a manifold obtained by finitely many applications of this
procedure.

Alternatively, we can think of a plumbing as a 4-dimensional thickening of
some number of surfaces that are connected by some number of intersections,
where any two surfaces must intersect each other at a single point
transversally (see [GS99]). In the case that all surfaces involved are spheres
or disks, we can build a plumbing graph in the following way:

Represent each sphere with a vertex. For any pair of spheres that intersect
each other at a point, draw an edge between their corresponding vertices.
Finally, label the vertex corresponding to each sphere with the number $-k$,
where $k$ is the number of intersections this sphere has with other surfaces.
This is called the self-intersection number of the vertex. By convention, we
will consider vertices with no label to have a self-intersection number of
$-2$.

Furthermore, a Lefschetz fibration with vanishing cycles that are pairwise
disjoint builds a plumbing. The collection of surfaces corresponding to this
plumbing can be obtained by contracting all vanishing cycles to a point
[GM13].

Figure 2.

######  Example 2.8.

Consider the Lefschetz fibration given by taking the surface and the product
of twists
$T_{b_{1}}^{2}T_{b_{2}}T_{b_{3}}T_{b_{4}}T_{b_{5}}$ on the surface
$S_{0,0}^{5}$. The figure on the left of Figure 2 shows the vanishing cycles
as colored curves. Collapsing each curve to a point results in the figure in
the middle, which completely describes the plumbing that is built by this
Lefschetz fibration. On the right, we present the associated plumbing graph.

##  3\. Monodromies with a unique positive factorization

We will now present monodromy factorizations on surfaces of the form
$S_{0,0}^{n}$ that have no alternative factorizations in the mapping class
monoid of the surface. We will focus on mapping class elements of the form
$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}$, using the notation
described below.

Choose an embedding of $S_{0,0}^{n}$ into $\mathbb{R}^{2}$ such that the
boundary components are arranged convexly and label its boundary components as
in Figure 3.

Figure 3.

Let $T_{b_{i}}$ denote the positive Dehn twist over a simple closed curve that
is parallel to the boundary component $b_{i}$. Additionally, let
$T_{b_{i_{1}},\ldots,b_{i_{m}}}$ denote the positive Dehn twist over the
convex simple closed curve containing the boundary components
$b_{i_{1}},\ldots,b_{i_{m}}$ when the interior boundary components are
arranged convexly. Note that this is well-defined because there is one such
curve up to isotopy. When discussing products of twists, we will use function
notation; that is, given a product of twists $T_{\alpha_{1}}\cdots
T_{\alpha_{m}}$, $T_{\alpha_{m}}$ is applied first, and this pattern continues
in descending order of the index on the $\alpha_{i}$’s.

Furthermore, the following lemma will allow us to use braids to compare
products of Dehn twists.

######  Lemma 3.1 (Margalit and McCammond [MM09]).

Let $D_{A}$ be a convexly punctured disk. In the mapping class group of
$D_{A}$, every convex swing over a set of punctures $B$ is equivalent to the
Dehn twist over the convex curve whose interior contains exactly the punctures
in $B$. Additionally, every Dehn twist over a convex curve $b$ is equivalent
to the convex swing over the points $B$ contained in the interior of $b$.

Using this lemma, we can use braids to represent products of Dehn twists on
$S_{0,0}^{n}$. In particular, we can push them into a semicircle with $b_{1}$
at the top and $b_{n-1}$ at the bottom, then cap each interior boundary
component with a once-punctured disk. This will result in a convexly punctured
disk. Then we can use a braid to represent the movement of the boundary
components through time as each swing is performed.

Additionally, the multiplicity of a boundary component $b_{i}$ under a product
of Dehn twists $T_{\alpha_{1}}\cdots T_{\alpha_{k}}$ is the number of curves
$\alpha_{j}$ that contain $b_{i}$ in their interiors.

In particular, we will use braids and multiplicity to compare products of Dehn
twists using the following lemma.

######  Lemma 3.2.

Two products of Dehn twists over convex curves in $S_{0,0}^{n}$ are equivalent
if and only if their associated braids are isotopic and each interior boundary
components have the same multiplicity under each product of Dehn twists.

###### Proof.

If we have two products of Dehn twists on the sphere with boundary that are
equivalent, then capping each of the interior boundary components with a
punctured disk and performing each product of twists will yield the same
braid. Furthermore, the first half of the proof of Lemma 2.3 in [PS21] shows
that multiplicity is well-defined under any element of the mapping class
group, so the multiplicities of each boundary component under each product
must be equal as well.

The proof of Lemma 2.3 in [PS21] also shows that, if two products of Dehn
twists have isotopic braids and each boundary component has the same
multiplicity, then the two products of Dehn twists are equivalent in the
mapping class group. This completes the argument. ∎

We will use this result to present new relations and disprove the existence of
certain types of relations. In order to prove our first result about relations
that do not exist, we will use the following lemma.

######  Lemma 3.3.

Given a product of Dehn twists over a collection of curves
$\\{\gamma_{1},\ldots,\gamma_{k}\\}$ on the surface $S_{0,0}^{n}$, the linking
number between any two strands $b_{x},b_{y}$ in the corresponding braid is
equal to $-m$, where $m$ is the number of curves $\gamma_{i}$ that contain
both $b_{x}$ and $b_{y}$ in their interior.

###### Proof.

Let $\gamma$ be a simple closed curve that contains at least two boundary
components. We claim that the braid corresponding to $T_{\gamma}$ will result
in a linking number of $-1$ for each pair $b_{i},b_{j}$ that are both
contained in the interior of $\gamma$ and a linking number of $0$ for any
other pair of boundary components.

We will first consider the case where $\gamma$ is convex.

Let $\\{b_{a_{i}},\ldots,b_{a_{k}}\\}$ be the set of such boundary components.
If we cap all other boundary components with a disk, the braid corresponding
to $T_{\gamma}$ will be as shown in Figure 5. From the first section of this
braid, we see that $b_{a_{k}}$ has two crossings of the value $-1$ with all
other strands, and it does not cross any other strands later in the braid.
Thus, the linking number between $b_{a_{k}}$ and any other interior boundary
component in $B$ is $\frac{-1-1}{2}=-1$.

Figure 4.

Figure 5.

Then $b_{a_{k-1}}$ crosses similarly over all $b_{a_{i}}$ such that $i<k-1$,
giving a linking number of $-1$ between $b_{a_{k-1}}$ and any other interior
boundary component above it. Furthermore, we already know that $b_{a_{k}}$ has
a linking number of $-1$ with $b_{a_{k-1}}$, so $b_{a_{k-1}}$ has a linking
number of $-1$ with every other element of $\gamma$. Continuing inductively,
we see that, for any choice of $i$ and $j$, $b_{a_{i}}$ and $b_{a_{j}}$ will
have a linking number of $-1$.

Now we will consider the linking number between a boundary component in
$\gamma$ and another outside of $\gamma$.

Let $b_{j}$ be the boundary component not in $\gamma$, and let $b_{a_{i}}$ be
given. Then there are three possible positions for $b_{j}$, as shown in Figure
5: above all $b_{a_{i}}$, between the $b_{a_{i}}$, and below all $b_{a_{i}}$.
If $b_{j}$ is above or below all boundary components in $b_{j}$, we see that
it has no crossings and therefore has a linking number of 0 with $b_{a_{i}}$.

In the case where $a_{1}<j<a_{k}$, $b_{j}$ may be above or below $b_{a_{i}}$.
In the case where it is below, we see that it has no crossings with
$b_{a_{i}}$ and therefore has a linking number of 0 with $b_{a_{i}}$. In the
case where it is above $b_{a_{i}}$ but below $b_{a_{1}}$, we see that it has
two crossings with $b_{a_{i}}$: one with a value of $-1$ and the other with a
value of $1$. Therefore, the linking number will be $\frac{1-1}{2}=0$.

Furthermore, any two boundary components not in $\gamma$ will have a linking
number of $0$ with each other because their strands will stay in place.

Thus, under $T_{\gamma}$, any two boundary components in the interior of
$\gamma$ will have a linking number of $-1$ and any other pair of boundary
components will have a linking number of 0.

We will now show that this also holds in the case that $\gamma$ is not convex.
Let $\gamma$ be a nonconvex simple closed curve that contains at least two
boundary components, labeled $b_{a_{1}},\ldots b_{a_{k}}$. We claim that there
exists a homeomorphism that takes $\gamma$ to a convex curve $\gamma^{\prime}$
which contains exactly the same boundary components. First, cut along $\gamma$
to obtain two planar subsurfaces. Then take a homeomorphism to reshape the
subsurface containing $b_{a_{1}},\ldots b_{a_{k}}$ so that it is convex, while
returning each boundary component in its place. We can similarly take a
homeomorphism on the other subsurface so that the image of $\gamma$ has the
same shape as the outer boundary of the first subsurface and returning each of
the original boundary components to its place.

Then glue the subsurfaces back together on the images of $\gamma$ under each
homeomorphism. We claim that this operation is a homeomorphism on the whole
surface. It is a homeomorphism on each of its subsurfaces, so we need only to
ensure that the process of gluing them together is continuous. If it is not
continuous along the image of $\gamma$, we can apply an isotopy to a
neighborhood of the image of $\gamma$ on one of the subsurfaces to obtain a
new function that is continuous. Thus, we have a homeomorphism of the surface
that takes a non convex curve $\gamma$ containing $\\{b_{a_{1}},\ldots
b_{a_{k}}\\}$ to a convex curve $\gamma^{\prime}$ containing
$\\{b_{a_{1}},\ldots b_{a_{k}}\\}$ and fixes the boundary pointwise.

Therefore, there also exists a homeomorphism $\phi$ that takes
$\gamma^{\prime}$ to $\gamma$. Furthermore, we can choose $\phi$ such that it
fixes the boundary pointwise. Additionally, because $\phi$ is a homeomorphism,
$T_{\gamma}=T_{\phi(\gamma^{\prime})}=\phi T_{\gamma^{\prime}}\phi^{-1}$ by
Fact 3.7 in [FM12]. Because $\phi$ fixes the boundary pointwise, $\phi$ and
$\phi^{-1}$ each belong to mapping classes in $\text{Mod}(S_{0,0}^{n})$. We
will now consider the braid corresponding to $T_{\gamma}=\phi
T_{\gamma^{\prime}}\phi^{-1}$. Because $\phi$ belongs to a mapping class, it
is isotopic to a product of Dehn twists over curves on $S_{0,0}^{n}$, call it
$\tau$. Additionally, $\phi^{-1}$ is isotopic to $\tau^{-1}$. Note that each
Dehn twist $T_{\alpha}$ appears in $\tau$ exactly as many times as
$T_{\alpha}^{-1}$ appears in $\tau^{-1}$.

Now we will consider two arbitrary boundary components $b_{i}$ and $b_{j}$. By
the convex case proven above, we know that the linking number for $b_{i}$ and
$b_{j}$ from $\tau$ is equal to $-m^{\prime}$, where $m^{\prime}$ is the
number of curves involved in twists in $\tau$ that contain both $b_{i}$ and
$b_{j}$. Additionally, because negative Dehn twists correspond to the inverse
of a swing on the boundary components in its interior, the linking number
between $b_{i}$ and $b_{j}$ under $\tau^{-1}$ is $m^{\prime}$. Therefore, if
$b_{i}$ and $b_{j}$ are both contained in $\gamma^{\prime}$, then their
linking number under the twist over $\gamma$ is
$(-m^{\prime})+(-1)+m^{\prime}=-1$. Otherwise, it is
$(-m^{\prime})+0+m^{\prime}=0$. Therefore, for a non convex curve $\gamma$,
under $T_{\gamma}$, any pair of boundary components both contained in $\gamma$
will have a linking number of $-1$, while all other pairs will have a linking
number of $0$. ∎

We are now equipped to present the following result.

######  Theorem 3.4.

Let $n\geq 5$, and consider the mapping class monoid of $S^{n}_{0,0}$. There
does not exist a relation between

$T_{b_{1}}\cdots T_{b_{n}}$

and any product of twists over non-boundary parallel curves.

###### Proof.

We proceed by contradiction. Assume that there is a product of Dehn twists
over non boundary parallel curves that is equivalent to $T_{b_{1}}\cdots
T_{b_{n}}$. Then there exists a set of non boundary parallel simple closed
curves $B=\\{\alpha_{1},\ldots,\alpha_{k}\\}$ such that $T_{b_{1}}\cdots
T_{b_{n}}=T_{\alpha_{1}}\cdots T_{\alpha_{k}}$

We know that the braid corresponding to $T_{b_{1}}T_{b_{2}}\cdots
T_{b_{n-1}}T_{b_{n}}$ is equivalent to the braid corresponding to $T_{b_{n}}$.
Note that this braid will be as shown in Figure 6.

Figure 6.

Additionally, the linking number between any two interior boundary components
is $-1$. Since the linking number is a braid invariant, this means that each
pair of interior boundary components must have a linking number of $-1$ under
the product of twists over curves in $B$. Therefore, by Lemma 3.2, for any two
interior boundary components $b_{i}$ and $b_{j}$, there must be exactly one
curve $\alpha_{k}\in B$ such that both $b_{i}$ and $b_{j}$ are contained in
its interior.

Furthermore, we see that, under $T_{b_{1}}T_{b_{2}}\cdots
T_{b_{n-1}}T_{b_{n}}$, each boundary component has a multiplicity of 2.
Therefore, for every interior boundary component $b_{i}$, there must exist
exactly two curves $\alpha_{j},\alpha_{k}\in B$ such that $b_{i}$ is contained
in their interiors.

We claim that, for $n\geq 5$, it is not possible to satisfy both conditions.

We will model the situation using a graph. Build a graph $G$ such that each
vertex $v_{i}$ represents the interior boundary component $b_{i}$ in
$S_{0,0}^{n}$.

We will add edges, and colors of edges, to this graph in the following way:
There exists an edge between vertices $v_{i}$ and $v_{j}$ if and only if
$b_{i}$ and $b_{j}$ have a linking number of $-1$. All edges resulting from a
particular curve $\alpha_{i}$ will be the same color, with a distinct color
for each $\alpha_{i}$.

Note that the subgraph of $G$ containing all edges of a particular color, and
their adjacent vertices, must be a complete subgraph. This is because, for any
two boundary components $b_{i},b_{j}$ contained in a curve $\alpha_{k}$, the
edge between them will be the color corresponding to $\alpha_{k}$. There is an
edge between them because $\alpha_{k}$ is the only curve containing them both,
so they have linking number $-1$ resulting from the twist over $\alpha_{k}$.
Additionally, we note that the number of colors of edges adjacent to a vertex
is equal to the multiplicity of the corresponding boundary component.

Then, in order to have a relation with $T_{b_{1}}T_{b_{2}}\cdots
T_{b_{n-1}}T_{b_{n}}$, we must have a complete graph. Equivalently, each
interior boundary component must have a linking number of $-1$ with every
other interior boundary component. Additionally, every vertex must have
adjacent edges of exactly two distinct colors (i.e. all interior boundary
components have a multiplicity of 2). We will show this is not possible for
$n\geq 5$.

Let $n\geq 5$ and consider the complete graph with $n-1$ vertices, $K_{n-1}$.
Let $v_{1}\in K_{n-1}$ be given. We know that $v_{1}$ must have adjacent edges
of exactly two distinct colors, call them blue and green. Then $v_{1}$ is part
of two complete subgraphs, one with blue edges and one with green edges. Since
$n-1\geq 4$, at least one of the subgraphs must contain two vertices other
than $v_{1}$. Without loss of generality, assume it is the blue subgraph. Let
$v_{2},v_{3}$ be distinct vertices in the blue subgraph and let $v_{4}$ be a
vertex in the green subgraph such that none are equal to $v_{1}$, as
represented in Figure 9.

Because the graph is complete, we know that there must be an edge between
$v_{4}$ and $v_{2}$, and it must be a third color, call it red. This is shown
in Figure 9.

There must also be an edge between $v_{4}$ and $v_{3}$, and it cannot be blue
or green because the two vertices are not both in either subgraph.
Furthermore, it cannot be red because the edge between $v_{2}$ and $v_{3}$ is
blue, so the red subgraph would not be complete. Thus, it must be a fourth
color, call it black. The result of this is shown in Figure 9.

Now we see that $v_{4}$ is adjacent to edges of three distinct colors, which
contradicts the original conditions. Therefore, such a graph is not possible.
∎

Figure 7.

Figure 8.

Figure 9.

Translating the implications of Theorem 3.4 for symplectic fillings, we obtain
as a corollary, the following result, which was originally proven by McDuff
[McD90, Theorem].

######  Corollary 3.5.

For $n\geq 5$, the contact manifold $(Y,\xi)=(L(n,1),\xi_{std})$ has a unique
minimal symplectic filling up to symplectic deformation, given by the plumbing
whose graph is a single vertex decorated by self-intersection number $-n$.
($L(n,1)$ denotes the lens space, and $\xi_{std}$ is the standard contact
structure obtained as a quotient under the $\mathbb{Z}_{n}$ action from the
standard contact structure on $S^{3}$.)

###### Proof.

The open book decomposition with page $\Sigma^{n}_{0,0}$ and monodromy
$T_{b_{1}}\cdots T_{b_{n}}$ supports the contact manifold $(L(p,1),\xi_{std})$
by [PVHM10]. By Wendl’s Theorem 2.6, any minimal symplectic filling of
$(L(p,1),\xi_{std})$ is supported by a Lefschetz fibration which corresponds
to a factorization of $T_{b_{1}}\cdots T_{b_{n}}$ into positive Dehn twists.
By Theorem 3.4, there are no other such positive factorizations, so the only
minimal symplectic filling is the one corresponding to the Lefschetz fibration
with vanishing cycles $b_{1},\dots,b_{n}$. The construction of Gay and Mark
[GM13] shows that the symplectic filling with this Lefschetz fibration is the
plumbing corresponding to the single vertex with self-intersection number
$-n$, which is a single disk bundle over a $2$-sphere with self-intersection
number $-n$. ∎

Next we give some generalizations of this result. As we will see in the next
section, there does exist a relation of the form $T_{b_{1}}\cdots
T_{b_{i}}^{n-3}\cdots T_{b_{n}}=T_{\alpha_{1}}\cdots T_{\alpha_{k}}$ in the
mapping class monoid of $S_{0,0}^{n}$. However, here we prove that if the
exponent $n-3$ is replaced by any smaller number, there is no relation of that
form.

######  Theorem 3.6.

Let $n\geq 5$, and consider the mapping class monoid of $\Sigma^{n}_{0,0}$.
Then for any $0<k<n-3$, there does not exist a relation between

$T_{b_{1}}\cdots T_{b_{i}}^{k}\cdots T_{b_{n}}$

and any product of twists over non-boundary parallel curves.

###### Proof.

Assume for the sake of contradiction that there does exist such a relation.
Then we know there exist non boundary parallel curves $\gamma_{i}$ such that

$T_{b_{1}}\cdots T_{b_{i}}^{k}\cdots T_{b_{n}}=T_{\gamma_{1}}\cdots
T_{\gamma_{l}}$

As in the proof of Theorem 3.2, we will represent the product of twists on the
right hand side using a graph $G$. Let there be one vertex in $G$ for each
interior boundary component, and let there be an edge between any two vertices
if they are contained within one of the curves $\gamma_{i}$. Designate a
distinct color for each $\gamma_{i}$, and color the edges of the graph so that
the colors correspond to the curve they represent.

In particular, this means that there are $n-1$ vertices in $G$. We know that
$n\geq k+4$, so $n-1\geq k+3$.

Thus, there are at least $k+3$ vertices in $G$.

Additionally, because there must be exactly one curve containing any pair of
boundary components in the collection $\gamma_{1},\cdots,\gamma_{l}$, we know
that $G$ is a complete graph.

Now we consider $b_{i}$. Because $G$ is complete and has at least $k+3$
vertices, $b_{i}$ is incident to at least $k+2$ edges.

Additionally, we note that the multiplicity of $b_{i}$ under the product of
twists on the left hand side is $k+1$. In particular, this means that the set
of edges incident to $b_{i}$ must contain edges of exactly $k+1$ distinct
colors. Thus, by the pigeonhole principle, there exist at least two edges
adjacent to $b_{i}$ that are of the same color, call it blue.

This means that there is a complete subgraph with blue edges and at least
three vertices. Call the other vertices $b_{x}$ and $b_{y}$.

Let $b_{j}$ be a vertex not in this subgraph. We know such a vertex exists
because $b_{i}$ is incident to at least one edge that is a color other than
blue.

Because $G$ is complete, there must be edges between $b_{j}$ and each vertex
of the blue subgraph, and these edges cannot be blue. Because a subgraph of
any color must be complete, and these edges are not blue, each one must be of
a distinct color from the others.

Thus, $b_{j}$ has a multiplicity of at least 3, which produces a
contradiction. ∎

Combining Theorem 3.6 with Wendl’s Theorem 2.6, we immediately obtain the
following corollary.

######  Corollary 3.7.

Let $(Y,\xi)$ be the contact boundary of the symplectic plumbing corresponding
to the graph shown below, where $k<n-3$.

Then this symplectic plumbing is the unique minimal symplectic filling of
$(Y,\xi)$ up to symplectic deformation.

We will now present a further generalization of this result. We proved that
products of boundary-parallel Dehn twists on a surface with a particular
collection of exponents on $T_{b_{i}}$ would not yield a relation with twists
over non boundary parallel curves. We will now prove that, given a product of
boundary parallel Dehn twists, if the sum of the exponents is less than or
equal to the sum of the particular collection of exponents presented in
Theorem 3.7, then there is also no relation with a product of non boundary
parallel twists.

######  Theorem 3.8.

Let $n\geq 5$ and consider the mapping class monoid of $S_{0,0}^{n}$. Given a
product of boundary parallel twists of the form

$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}$

where each $a_{i}$ is a positive integer, if $1+\sum_{i=1}^{n-1}a_{i}\leq
2n-5$, then there is no relation between the given product and any product of
twists over non boundary parallel curves.

###### Proof.

We will again model the situation using a graph, as described in the proof of
Theorem 3.4. We claim that, for the complete graph with $n-1$ vertices
$b_{i}$, there is no way to partition the edges by color into strict complete
subgraphs such that $1+\sum a_{i}\leq 2n-5$, where $a_{i}$ is the number of
distinct colors represented by the edges incident to the vertex $b_{i}$ minus
1.

We proceed by induction on $n$, with the base case $n=5$. In this case, we
will consider all possible ways to partition the edges of the complete graph
with 4 vertices, $K_{4}$ into complete subgraphs.

The largest possible strict complete subgraph has three vertices. In the case
that three of the edges are part of a complete subgraph with three vertices,
up to relabeling and using arbitrary colors, we have the graph shown in Figure
10.

Then $1+\sum a_{i}=1+(2-1)+(3-1)+(2-1)+(2-1)+(2-1)=1+1+2+1+1+1=6>2(5)-5$

The only other possible case is to color each edge differently. Then each
$a_{i}$ would be $4-1=3$, so $1+\sum a_{i}=1+4(3)=13>2(5)-5$.

Thus, the claim holds for $n=5$.

We now proceed with the inductive step. Assume that, for some $k=n$ such that
$k\geq 5$, the hypothesis holds. In particular, for $n=k$, there does not
exist a partition of the edges of $K_{n-1}$ into complete subgraphs such that
$1+\sum a_{i}\leq 2n-5$, with $a_{i}$ as defined above. We will show that this
is true for $n=k+1$.

Assume for the sake of contradiction that there does exist such a partition by
color on $K_{k}$ into complete strict subgraphs such that $1+\sum a_{i}\leq
2(k+1)-5=2k-3$.

We claim that, given such a partition on $K_{k}$, there must exist at least
one vertex $b_{j}$ such that the edges incident to $b_{j}$ represent at least
three distinct colors.

In the case that all subgraphs have exactly two vertices, all edges incident
to any particular vertex must be of distinct colors. Since the graph is
complete, all vertices are incident to exactly $k-1$ vertices. Therefore, the
edges incident to any $b_{i}$ represent $k-1$ vertices. Since $k\geq 5$, this
means that all vertices are incident to edges of at least four distinct
colors.

In the case that there is at least one subgraph with three vertices,
$v_{1},v_{2},$ and $v_{3}$, we consider a vertex outside of that subgraph,
$v_{4}$. Because $v_{4}$ is not part of the subgraph, the edge between $v_{1}$
and $v_{4}$ cannot be the same color as the subgraph containing $v_{1},v_{2},$
and $v_{3}$. Similarly, the edge between $v_{2}$ and $v_{4}$ and that between
$v_{3}$ and $v_{4}$ must each have their own distinct color, so $v_{4}$ has is
incident to edges that represent at least three distinct colors.

Thus, in our partition of $K_{k}$, we can choose a vertex $b_{j}$ that is
incident to edges of at least three distinct colors. In this case, $a_{j}\geq
3-1=2$.

Consider the graph resulting from removing $b_{j}$ and the edges incident to
it. Because removing a vertex from a complete graph $K_{m}$ gives the complete
graph $K_{m-1}$, we know that the resulting graph, and the subgraphs in the
partition, are complete.

Then we have a complete graph with $k-1$ vertices
$b_{1}^{\prime},\ldots,b_{k-1}^{\prime}$.

In the case that the associated coloring of the edges of this new graph are
all one color, we know that all other subgraphs of $K_{k}$ must be edges from
$b_{j}$ to vertices in this subgraph with $k-1$ vertices. This means that
$a_{j}=k-2$, and $a_{i}=1$ for all $i\neq j$ so $1+\sum
a_{i}=1+k-2+(k-1)(1)=2k-2>2k-3$. Thus, this is not possible. Therefore, the
associated partition of the resulting graph will still contain strict complete
subgraphs.

Additionally, $a_{j}$ is not part of the sum $1+\sum a_{i}^{\prime}$, and,
given a vertex $b_{i}$ and its associated vertex $b_{m}^{\prime}$ in
$K_{k-1}$, $a_{i}\geq a_{m}^{\prime}$.

Thus, $1+\sum a_{i}^{\prime}\leq(1+\sum a_{i})-a_{j}\leq 2k-3-2=2k-5$

This contradicts the inductive hypothesis, so the hypothesis must also be true
for $n=k+1$.

Therefore, for the complete graph with $n-1$ vertices $b_{i}$, there is no way
to partition the edges by color into strict complete subgraphs such that
$1+\sum a_{i}\leq 2n-5$.

This implies that, given $a_{i}$ such that $1+\sum a_{i}\leq 2n-5$ and any
combination of non boundary parallel curves $\gamma_{i}$ such that
$T_{\gamma_{1},\cdots,\gamma_{m}}$ corresponds to a braid with a pairwise
linking number of $-1$ for all interior boundary components, it is not
possible for the multiplicity of all interior boundary components to be equal
under both $T_{\gamma_{1}}\cdots T_{\gamma_{m}}$ and $T_{b_{1}}^{a_{1}}\cdots
T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}$.

Therefore, given $a_{i}$ satisfying $1+\sum a_{i}\leq 2n-5$, there is no
relation between

$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}$

and any product of twists over non boundary parallel curves. ∎

We now consider the plumbing graph associated with a product of boundary
parallel twists as described in Theorem 3.8. By the construction of plumbing
graphs, the graph for any such factorization is star shaped and the central
vertex has self-intersection number $-n$. Each factor $T_{b_{i}}^{a_{i}}$, for
$i\leq n-1$, corresponds to a linear subgraph with $a_{i}-1$ vertices, each
with self intersection number $-2$, where exactly one of these vertices is
adjacent to the central vertex. Additionally, all such subgraphs are disjoint.
This means that there are $\sum_{i=1}^{n-1}(a_{i}-1)$ vertices with self-
intersection number $-2$. Since $1+\sum a_{i}\leq 2n-5$, we can conclude that
$\sum_{i=1}^{n-1}(a_{i}-1)\leq n-5$, so there must be at most $n-5$ vertices
other than the central vertex. , note that, for $n\geq 5$, every star shaped
graph with a central vertex of self-intersection number $-n$ and at most $n-5$
other vertices, each with self intersection number $-2$, describes a plumbing
given by a monodromy factorization of the form described in Theorem 3.8.
Combining this with Wendl’s Theorem 2.6, we obtain the following corollary.

######  Corollary 3.9.

Let $(Y,\xi)$ be the contact boundary of the symplectic plumbing corresponding
to a star-shaped graph with at most $n-4$ vertices, where the central vertex
has self-intersection number $-n$ and all others have self-intersection number
$-2$. Then this symplectic plumbing is the unique minimal symplectic filling
of $(Y,\xi)$ up to symplectic deformation.

##  4\. Relations in the planar monoid

In this section we consider situations in which there exist nontrivial
relations between different factorizations of positive Dehn twists. In
particular, in order to admit a plumbing, a product of Dehn twists must
involve only disjoint curves, and we are interested in relations between
factorizations that admit plumbings and those that do not. For the scope of
this project, we will focus on relations between products of the form
$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}$, where each $a_{i}$
is a positive integer, and products of twists over convex, non boundary
parallel curves.

Figure 10.

###  4.1. Relations for general $S^{n}_{0,0}$

We start by giving relations that hold in the planar mapping class group for
an arbitrary number of boundary components.

######  Theorem 4.1.

Consider the mapping class monoid of $S_{0,0}^{n}$. For all $2\leq i<n-1$ the
following relation holds:

$[T_{b_{1}}^{n-i-1}T_{b_{2}}^{n-i-1}\cdots
T_{b_{i}}^{n-i-1}][T_{b_{i+1}}^{n-3}\cdots
T_{b_{n-1}}^{n-3}]T_{b_{n}}=T_{b_{1},b_{2},\ldots,b_{i}}[T_{b_{i+1},b_{i}}\cdots
T_{b_{i+1},b_{1}}]\cdots[T_{b_{n-1},b_{n-2}}\cdots T_{b_{n-1},b_{1}}]$

Note that this relation generalizes the _daisy relation_ defined in [EMVHM11].

###### Proof.

We will first show that the multiplicities are equal. Let $b_{j}$ be an
interior boundary component.

If $j\leq i$, then on the left hand side, the twists in the product involving
$b_{j}$ are $T_{b_{j}}^{n-i-1}$ and $T_{b_{n}}$. All other components of the
product involve twists over curves that do not contain $b_{j}$. Thus, the
multiplicity of $b_{j}$ is $n-i-1+1=n-i$ on the left hand side. On the right
hand side, the twists involving $b_{j}$ are $T_{b_{1},b_{2},\ldots,b_{i}}$ and
the $T_{b_{j},b_{k}}$ contained in $[T_{b_{k-1},b_{k}}\cdots T_{b_{1},b_{k}}]$
for each $k>i$. We know that there are $n-i-1$ values of $k$ that are greater
than $i$, so the multiplicity of $b_{j}$ is $n-i-1+1=n-i$.

If $j>i$, then on the left hand side, then the twists in the product involving
$b_{j}$ are $T_{b_{j}}^{n-3}$ and $T_{b_{n}}$. Therefore, the multiplicity of
$b_{j}$ on this side is $n-3+1=n-2$. On the right hand side, twists over all
curves containing $b_{j}$ and exactly one other interior boundary component
are included. Since there are exactly $n-2$ such curves, the multiplicity of
$b_{j}$ is $n-2$.

Thus, the multiplicities are equal.

Now we will show that the corresponding braids are isotopic.

We will begin by finding the braid corresponding to the left hand side. We
note that all twists except for $T_{b_{n}}$ are boundary parallel. Because a
twist along a boundary parallel curve corresponds to a braid that demonstrates
a swing on the single puncture contained within that curve, it will not add
any crossings to the braid. Therefore, the braid corresponding to the left
hand side is exactly the braid corresponding to just $T_{b_{n}}$.

$T_{b_{n}}$ is represented by a swing on all interior boundary components, as
shown in Figure 12.

We will now find the braid corresponding to the right hand side.

We note that each portion in brackets is of the form $[T_{b_{j},b_{j-1}}\cdots
T_{b_{j},b_{1}}]$ and goes from $j=n-1$ to $j=i+1$.

We will consider the braid corresponding to $T_{b_{j},b_{k+1}}T_{b_{j},b_{k}}$
for any $k<j$. Performing the required swings gives us the braid in Figure 12.

Cancelling out crossings that are inverses of each other gives an isotopy that
leads to the braid in Figure 14. This is isotopic to the braid in Figure 14.

We will apply the same isotopy inductively to the braid corresponding to
$[T_{b_{j},b_{j-1}}\cdots T_{b_{j},b_{1}}]$ for arbitrary $j$. We know that
$[T_{b_{j},b_{j-1}}\cdots T_{b_{j},b_{1}}]$ is represented by the braid in
Figure 16, and applying the isotopy yields the braid in Figure 16.

Thus, $[T_{b_{i+1},b_{i}}\cdots
T_{b_{i+1},b_{1}}]\cdots[T_{b_{n-1},b_{n-2}}\cdots T_{b_{n-1},b_{1}}]$ is
represented by the braid shown in Figure 18.

Furthermore, $T_{b_{1},\cdots,b_{i}}$ corresponds to the braid in Figure 18.

Putting these together in the order they are shown on the right hand side
yields the braid in Figure 19, which is equivalent to the braid corresponding
to the left hand side.

Thus, the relation holds. ∎

Figure 11.

Figure 12.

Figure 13.

Figure 14.

Figure 15.

Figure 16.

Figure 17.

Figure 18.

Figure 19.

######  Corollary 4.2.

A symplectic plumbing with the following graph

has Euler characteristic

$n^{2}-5n+6+2i-i^{2}$

and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is

$3-n+(n-i-1)(i-1)+\frac{(n-i-1)(n-i)}{2}$

###### Proof.

Consider the product of twists on the left hand side of the relation given in
Theorem 4.1:

$[T_{b_{1}}^{n-i-1}T_{b_{2}}^{n-i-1}\cdots
T_{b_{i}}^{n-i-1}][T_{b_{i+1}}^{n-3}\cdots T_{b_{n-1}}^{n-3}]T_{b_{n}}$

We see that, when considering the curves over which we are twisting and the
corresponding fiber after all vanishing cycles have been compressed to a
point, there is a sphere corresponding to the interior of $S_{0,0}^{n}$. It is
connected to $n$ spheres, each corresponding to a curve parallel to a boundary
component. Thus, in the plumbing graph there is a vertex with a value of $-n$.

We note that, for a boundary component $b_{j}$ for which the product of twists
involved $T_{b_{j}}^{m}$, the fiber with all vanishing cycles collapsed will
contain $m-1$ spheres and one disk corresponding to that particular boundary
component. Thus, in the plumbing graph, $T_{b_{j}}^{m}$ will result in a
branch connected to the first vertex with $m-1$ vertices, each with a value of
$-2$.

Therefore, for each boundary component $b_{j}$ such that $T_{b_{j}}^{n-i-1}$
is included, we will have a branch from the first vertex with $n-i-2$
vertices. Note that there are exactly $i$ of these boundary components, so we
will have $i$ such branches. Similarly, we will have $n-i-1$ branches with
$n-4$ vertices each.

Therefore, we see that the left hand side of the relation corresponds to the
plumbing graph. We will now compute the Euler characteristic of the 4-manifold
built by a Lefschetz fibration using this product of Dehn twists.

The Euler characteristic of a 4-manifold $X$ given by the Lefschetz fibration
$\pi:X\to D^{2}$ with general fiber $F$ is equal to $\chi(D^{2})\chi(F)+k$,
where $k$ is the number of vanishing cycles in the fiber [OS04]. Recall that,
when defining a Lefschetz fibration using a product of positive Dehn twists,
each Dehn twist gives a distinct vanishing cycle in the fiber.

Furthermore, the Euler characteristic of the surface $S_{g,b}^{n}$ is
$2-2g-(b+n)$ [FM12]. Thus, $\chi(S_{0,0}^{n})=2-2(0)-(0+n)=2-n$ and
$\chi(D^{2})=2-2(0)-(1+0)=2-1=1$, so $\chi(D^{2})\chi(S_{0,0}^{n})=2-n$

Therefore, the Euler characteristic of the symplectic filling corresponding to
a product of Dehn twists on $S_{0,0}^{n}$ is equal to
$\chi(D^{2})\chi(S_{0,0}^{n})+k=2-n+k$, where $k$ is the number of Dehn
twists.

Furthermore, we note that, for $1\leq j\leq i$, $T_{b_{j}}$ has an exponent of
$n-i-1$. Since there are $i$ total boundary components in this collection, we
see that there are $i(n-i-1)$ Dehn twists in the first set of brackets.
Similarly, there are $n-i-1$ boundary components $b_{j}$ for which $T_{b_{j}}$
has an exponent of $n-3$ and finally, $T_{b_{n}}$ has an exponent of 1.
Therefore, in total there are $i(n-i-1)+(n-i-1)(n-3)+1$ Dehn twists on the
left hand side.

Thus, the Euler characteristic of the resulting filling is

$2-n+[i(n-i-1)+(n-i-1)(n-3)+1]=n^{2}-5n+6+2i-i^{2}$

For the right hand side of the relation, we see that there is one Dehn twist
outside of the brackets. For each set of twists inside brackets
$[T_{(b_{j},b_{j-1})}\cdots T_{(b_{j},b_{1})}]$, we see that there are $j-1$
twists. Thus, the total number of twists is

$1+\sum_{j=i+1}^{n-1}(j-1)$

Adjusting the index on this sum and evaluating the sum gives the result

$1+\sum_{j=1}^{n-i-1}(j+i-1)=1+\sum_{j=1}^{n-i-1}(i-1)+\sum_{j=1}^{n-i-1}j=1+(n-i-1)(i-1)+\frac{(n-i-1)(n-i)}{2}$

Therefore, the Euler characteristic of the associated filling is

$2-n+\left[1+(n-i-1)(i-1)+\frac{(n-i-1)(n-i)}{2}\right]=3-n+(n-i-1)(i-1)+\frac{(n-i-1)(n-i)}{2}$

as desired. ∎

###  4.2. Relations for $S^{n}_{0,0}$ for small values of $n$

We skip the cases where $n<5$ because a full set of relations has been defined
for $S_{0,0}^{n}$ for each $n<5$. In particular, for $n=2$ and $n=3$,
$S_{0,0}^{n}$ contains only boundary parallel curves, and the Lantern
Relation, which is defined for $S_{0,0}^{4}$, completely describes the
relations between elements of its mapping class monoid of the form we are
considering. For the values $n=5,6,7$, we will present relations on the
mapping class monoid of $S_{0,0}^{n}$ and discuss the implications of these
relations on associated plumbing graphs and the Euler characteristics of the
resulting fillings.

######  Proposition 4.3.

The following are relations in the mapping class monoid of $S^{5}_{0,0}$:

  1. (1)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}^{2}T_{b_{5}}=[T_{b_{1},b_{2}}][T_{b_{2},b_{3}}T_{b_{1},b_{3}}][T_{b_{3},b_{4}}T_{b_{2},b_{4}}T_{b_{1},b_{4}}]$

  2. (2)

$T_{b_{1}}T_{b_{2}}T_{b_{3}}T_{b_{4}}^{2}T_{b_{5}}=[T_{b_{3},b_{4}}T_{b_{2},b_{4}}T_{b_{1},b_{4}}]T_{b_{1},b_{2},b_{3}}$

###### Proof.

Note that relation 1 follows directly from Theorem 4.1 by taking $n=5$ and
$i=2$. Similarly, 2 follows from Theorem 4.1 by taking $n=5$ and $i=3$. ∎

######  Corollary 4.4.

  * i.

A symplectic plumbing with the graph in Figure 21 has Euler characteristic 3
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 3.

  * ii.

A symplectic plumbing with the graph in Figure 21 has Euler characteristic 6
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 1.

Figure 20.

Figure 21.

Note that both of these statements follow directly from Corollary 4.2.

######  Proposition 4.5.

The following are relations in the mapping class monoid of $S^{6}_{0,0}$:

  1. (1)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{3}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}=T_{b_{2},b_{1}}[T_{b_{3},b_{2}}T_{b_{3},b_{1}}][T_{b_{4},b_{3}}T_{b_{4},b_{2}}T_{b_{4},b_{1}}][T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{5},b_{2}}T_{b_{5},b_{1}}]$

  2. (2)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}=T_{b_{1},b_{2},b_{3}}[T_{b_{4},b_{3}}T_{b_{4},b_{2}}T_{b_{4},b_{1}}][T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{5},b_{2}}T_{b_{5},b_{1}}]$

  3. (3)

$T_{b_{1}}T_{b_{2}}T_{b_{3}}T_{b_{4}}T_{b_{5}}^{3}T_{b_{6}}=T_{b_{1},b_{2},b_{3},b_{4}}[T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{5},b_{2}}T_{b_{5},b_{1}}]$

  4. (4)

$T_{b_{1}}^{2}T_{b_{2}}T_{b_{3}}^{2}T_{b_{4}}^{2}T_{b_{5}}^{2}T_{b_{6}}=T_{b_{3},b_{4}}T_{b_{1},b_{2},b_{4}}T_{b_{4},b_{5}}T_{b_{2},b_{3},b_{5}}T_{b_{1},b_{5}}T_{b_{1},b_{3}}$

  5. (5)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}T_{b_{5}}^{2}T_{b_{6}}=T_{b_{2},b_{3}}T_{b_{1},b_{3}}T_{b_{3},b_{4},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}T_{b_{1},b_{2},b_{4}}$

  6. (6)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}T_{b_{4}}^{2}T_{b_{5}}^{2}T_{b_{6}}=T_{b_{1},b_{2},b_{3}}T_{b_{3},b_{4},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}T_{b_{2},b_{4}}T_{b_{1},b_{4}}$

  7. (7)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{3}T_{b_{4}}^{2}T_{b_{5}}^{3}T_{b_{6}}=T_{b_{2},b_{3}}T_{b_{1},b_{3}}T_{b_{3},b_{4}}T_{b_{4},b_{5}}T_{b_{3},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}T_{b_{1},b_{2},b_{4}}$

###### Proof.

Note that relations 1, 2, and 3 follow directly from Theorem 4.1.

Now we prove the remaining relations. The braid corresponding to the left hand
side of each relation is shown in Figure 22.

Figure 22.

Figure 23 shows the braid corresponding to the right hand side of relation 4
and two steps in the isotopy between that braid and the braid corresponding to
the left hand side of the relation. Since the two braids are isotopic and each
boundary component has the same multiplicity under the products of twists on
both sides, we conclude that 4 holds.

Similarly, note that the multiplicity of each boundary component is the same
under both sides of relations 5, 6, and 7. Furthermore, Figures 24, 25, and 26
show the braid corresponding to the right hand side of relations 5, 6, and 7,
respectively, and demonstrate some of the steps in each isotopy between the
braids corresponding to the right and left sides of each relation.

Thus, for relations 5, 6, and 7, the multiplicity of each boundary component
is the same on both sides and the corresponding braids for each side are
isotopic, so these relations hold. ∎

Figure 23.

Figure 24.

Figure 25.

Figure 26.

Determining the plumbing and Euler characteristics of the Lefschetz fibrations
corresponding to these relations as in the proof of Corollary 4.2, we obtain
the following corollary.

######  Corollary 4.6.

  * i.

A symplectic plumbing with the graph in Figure 30 has Euler characteristic 12
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 6.

  * ii.

A symplectic plumbing with the graph in Figure 30 has Euler characteristic 9
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 4.

  * iii.

A symplectic plumbing with the graph in Figure 30 has Euler characteristic 4
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 1.

  * iv.

A symplectic plumbing with the graph in Figure 30 has Euler characteristic 6
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 2.

Figure 27.

Figure 28.

Figure 29.

Figure 30.

######  Proposition 4.7.

The following are relations in the mapping class monoid of $S^{7}_{0,0}$:

  1. (1)

$T_{b_{1}}^{4}T_{b_{2}}^{4}T_{b_{3}}^{4}T_{b_{4}}^{4}T_{b_{5}}^{4}T_{b_{5}}^{4}T_{b_{7}}=T_{b_{1},b_{2}}[T_{b_{3},b_{2}}T_{b_{3},b_{1}}][T_{b_{4},b_{3}}T_{b_{4},b_{2}}T_{b_{4},b_{1}}][T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{5},b_{2}}T_{b_{5},b_{1}}][T_{b_{6},b_{5}}T_{b_{6},b_{4}}T_{b_{6},b_{3}}T_{b_{6},b_{2}}T_{b_{6},b_{1}}]$

  2. (2)

$T_{b_{1}}T_{b_{2}}T_{b_{3}}T_{b_{4}}T_{b_{5}}T_{b_{6}}^{4}T_{b_{7}}=T_{b_{1},b_{2},b_{3},b_{4},b_{5}}[T_{b_{6},b_{5}}T_{b_{6},b_{4}}T_{b_{6},b_{3}}T_{b_{6},b_{2}}T_{b_{6},b_{1}}]$

  3. (3)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{3}T_{b_{4}}^{4}T_{b_{5}}^{4}T_{b_{6}}^{4}T_{b_{7}}=T_{b_{1},b_{2},b_{3}}[T_{b_{4},b_{3}}T_{b_{4},b_{2}}T_{b_{4},b_{1}}][T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{5},b_{2}}T_{b_{5},b_{1}}][T_{b_{6},b_{5}}T_{b_{6},b_{4}}T_{b_{6},b_{3}}T_{b_{6},b_{2}}T_{b_{6},b_{1}}]$

  4. (4)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}^{2}T_{b_{5}}^{4}T_{b_{6}}^{4}T_{b_{7}}=T_{b_{1},b_{2},b_{3},b_{4}}[T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{5},b_{2}}T_{b_{5},b_{1}}][T_{b_{6},b_{5}}T_{b_{6},b_{4}}T_{b_{6},b_{3}}T_{b_{6},b_{2}}T_{b_{6},b_{1}}]$

  5. (5)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}T_{b_{5}}^{3}T_{b_{6}}^{3}T_{b_{7}}=T_{b_{1},b_{2},b_{3},b_{4}}T_{b_{4},b_{5},b_{6}}T_{b_{3},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{3},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}$

  6. (6)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{2}T_{b_{4}}T_{b_{5}}^{2}T_{b_{6}}^{2}T_{b_{7}}=T_{b_{2},b_{3}}T_{b_{1},b_{3}}T_{b_{3},b_{4},b_{5},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}T_{b_{1},b_{2},b_{4}}$

  7. (7)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}T_{b_{4}}^{3}T_{b_{5}}^{2}T_{b_{6}}^{3}T_{b_{7}}=T_{b_{4},b_{5}}T_{b_{1},b_{2},b_{3},b_{5}}T_{b_{5},b_{6}}T_{b_{3},b_{4},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{2},b_{4}}T_{b_{1},b_{4}}$

  8. (8)

$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}^{3}T_{b_{5}}T_{b_{6}}^{3}T_{b_{7}}=T_{b_{4},b_{3}}T_{b_{4},b_{2}}T_{b_{4},b_{1}}T_{b_{4},b_{5},b_{6}}T_{b_{6},b_{3}}T_{b_{6},b_{2}}T_{b_{6},b_{1}}T_{b_{1},b_{2},b_{3},b_{5}}$

  9. (9)

$T_{b_{1}}^{3}T_{b_{2}}^{2}T_{b_{3}}^{3}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}^{4}T_{b_{7}}=T_{b_{3},b_{4}}T_{b_{1},b_{2},b_{4}}T_{b_{4},b_{5}}T_{b_{2},b_{3},b_{5}}T_{b_{1},b_{5}}T_{b_{5},b_{6}}T_{b_{4},b_{6}}T_{b_{3},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{1},b_{3}}$

  10. (10)

$T_{b_{1}}^{3}T_{b_{2}}^{2}T_{b_{3}}^{3}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}^{4}T_{b_{7}}=T_{b_{4},b_{3}}T_{b_{1},b_{2},b_{4}}T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{5},b_{2}}T_{b_{5},b_{1}}T_{b_{5},b_{6}}T_{b_{4},b_{6}}T_{b_{2},b_{3},b_{6}}T_{b_{1},b_{6}}T_{b_{1},b_{3}}$

  11. (11)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{3}T_{b_{4}}^{2}T_{b_{5}}^{3}T_{b_{6}}^{4}T_{b_{7}}=T_{b_{2},b_{3}}T_{b_{1},b_{3}}T_{b_{3},b_{4},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}T_{b_{5},b_{6}}T_{b_{1},b_{2},b_{4}}T_{b_{4},b_{6}}T_{b_{3},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}$

  12. (12)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{4}T_{b_{4}}^{2}T_{b_{5}}^{3}T_{b_{6}}^{3}T_{b_{7}}=T_{b_{2},b_{3}}T_{b_{1},b_{3}}T_{b_{3},b_{4}}T_{b_{1},b_{2},b_{4}}T_{b_{4},b_{5},b_{6}}T_{b_{3},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{3},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}$

  13. (13)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{2}T_{b_{4}}^{3}T_{b_{5}}^{4}T_{b_{6}}^{3}T_{b_{7}}=T_{b_{4},b_{5}}T_{b_{3},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}T_{b_{5},b_{6}}T_{b_{3},b_{4},b_{5}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{2},b_{4}}T_{b_{1},b_{4}}T_{b_{1}b_{2}b_{3}}$

  14. (14)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{3}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}^{3}T_{b_{7}}=T_{b_{3},b_{2}}T_{b_{3},b_{1}}T_{b_{5},b_{4}}T_{b_{5},b_{3}}T_{b_{1},b_{2},b_{5}}T_{b_{5},b_{6}}T_{b_{3},b_{4},b_{6}}T_{b_{6},b_{2}}T_{b_{6},b_{1}}T_{b_{4},b_{2}}T_{b_{4},b_{1}}$

  15. (15)

$T_{b_{1}}^{4}T_{b_{2}}^{3}T_{b_{3}}^{2}T_{b_{4}}^{3}T_{b_{5}}^{2}T_{b_{6}}^{4}T_{b_{7}}=T_{b_{2},b_{3},b_{4}}T_{b_{1},b_{4}}T_{b_{1},b_{3}}T_{b_{1},b_{2}}T_{b_{4},b_{5},b_{6}}T_{b_{3},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{3},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}$

  16. (16)

$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{3}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}^{3}T_{b_{7}}=T_{b_{1},b_{2},b_{3}}T_{b_{3},b_{4}}T_{b_{2},b_{4}}T_{b_{1},b_{4}}T_{b_{4},b_{5},b_{6}}T_{b_{3},b_{6}}T_{b_{2},b_{6}}T_{b_{1},b_{6}}T_{b_{3},b_{5}}T_{b_{2},b_{5}}T_{b_{1},b_{5}}$

###### Proof.

Note that relations 1, 2, 3, and 4 follow directly from Theorem 4.1.

Note that, for each of the remaining relations, the multiplicity of each
boundary component is the same on both sides. Therefore, in order to prove
each relation, it suffices to show that the corresponding braids are isotopic.

Figure 31 shows the braid corresponding to any product of boundary twists of
the form $T_{b_{1}}^{a_{1}}\cdots T_{b_{6}}^{a_{6}}T_{b_{7}}$. Note that the
left hand side of each relation is of this form.

Figures 32 through 43 show the braids corresponding to the right hand side of
each relation in ascending numerical order. Each is isotopic to the braid in
Figure 31. Verification of the existence each isotopy is left to the reader.

From this, we conclude that each relation holds. ∎

Figure 31.  Figure 32.  Figure 33.  Figure 34.  Figure 35.  Figure 36.  Figure
37.  Figure 38.  Figure 39.  Figure 40.  Figure 41.  Figure 42.  Figure 43.

######  Corollary 4.8.

  * i.

A symplectic plumbing with the graph in Figure 47 has Euler characteristic 16
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 10.

  * ii.

A symplectic plumbing with the graph in Figure 47 has Euler characteristic 5
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 1.

  * iii.

A symplectic plumbing with the graph in Figure 47 has Euler characteristic 17
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 8.

  * iv.

A symplectic plumbing with the graph in Figure 47 has Euler characteristic 12
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 5.

  * v.

A symplectic plumbing with the graph in Figure 50 has Euler characteristic 9
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 3.

  * vi.

A symplectic plumbing with the graph in Figure 50 has Euler characteristic 14
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 6.

  * vii.

A symplectic plumbing with the graph in Figure 50 has Euler characteristic 14
and can be replaced by a different symplectic filling with the same contact
boundary whose Euler characteristic is 6.

Figure 44.

Figure 45.

Figure 46.

Figure 47.

Figure 48.

Figure 49.

Figure 50.

##  5\. On the completeness of our lists of relations

We will now discuss the minimal and maximal Euler characteristic for the
filling of a plumbing corresponding to a product of twists of the form
$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}$ which is equivalent
to a product of non-boundary parallel twists.

Because the Euler characteristic increases as the number of Dehn twists
increases, we can consider the relations that involve the maximum and minimum
number of boundary-parallel twists.

In particular, by Theorem 3.8, for a product of twists
$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}$, if
$1+\sum_{i=1}^{n-1}a_{i}\leq 2n-5$, then there is no relation involving this
product and some product of non boundary parallel twists. Equivalently, if
there are $2n-5$ or fewer twists, no such relation exists. Thus, the minimum
number of Dehn twists must be $2n-4$ or greater.

Additionally, taking $i=n-2$ in Theorem 4.1 gives us a relation between

$[T_{b_{1}}\cdots T_{b_{n-2}}][T_{b_{n-1}}^{n-3}]T_{b_{n}}$

and a product of twists over non boundary parallel curves.

We note that the number of Dehn twists in this product is
$n-2+n-3+1=2n-5+1=2n-4$. Since a relation involving this number of Dehn twists
exists, we can say that this is the minimum number of boundary-parallel Dehn
twists needed for a nontrivial relation of the specified form.

The Euler characteristic of the symplectic plumbing given by this product of
twists is $2-n+2n-4=n-2$ and the plumbing graph is as shown in Figure 47. This
is the minimal possible Euler characteristic of a plumbing for which there
exists a different symplectic filling given by a relation of the form we are
considering.

Furthermore, the maximal Euler characteristic for such a plumbing can be found
by taking the highest number of boundary-parallel Dehn twists represented by
any relation of the specified form.

Consider a relation

$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots
T_{\alpha_{m}}$

where each $\alpha_{i}$ is some non boundary parallel curve. Since the left
hand side includes exactly one twist over the outer boundary, the braid
corresponding to both sides of the relation must have a linking number of $-1$
for all pairs of boundary components. By Lemma 3.2, no two boundary components
can both be contained in multiple $\alpha_{i}$’s. Therefore, the maximum
multiplicity of any boundary component $b_{i}$ is $n-2$. This is because we
can twist over each curve containing exactly $b_{i}$ and one other interior
boundary component. This results in a collection of $\binom{n-1}{2}$ non
boundary parallel curves on the right hand side.

In fact, using Theorem 4.1 and choosing $i=2$, we have

$[T_{b_{1}}^{n-3}T_{b_{2}}^{n-3}][T_{b_{3}}^{n-3}\cdots
T_{b_{n-1}}^{n-3}]T_{b_{n}}=T_{b_{1},b_{2}}[T_{b_{3},b_{2}}T_{b_{3},b_{1}}]\cdots[T_{b_{n-1},b_{n-2}}\cdots
T_{b_{n-1},b_{1}}]$

Note that the choice of curves on the right hand side is exactly the
collection of $\binom{n-1}{2}$ curves described above. Thus, there exists such
a relation that gives the maximum possible multiplicity to each interior
boundary component.

On the left hand side, the total number of Dehn twists is $(n-3)(n-1)+1$, so
the maximum Euler characteristic of a plumbing given by a relation of the
specified form is $2-n+(n-3)(n-1)+1=n^{2}-5n+6$.

######  Proposition 5.1.

Up to isometry, the relations described in Proposition 4.3 are the only
relations on the mapping class monoid of $S_{0,0}^{5}$ which are of the form

$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots
T_{\alpha_{m}}$

where each $a_{i}$ is a positive integer and each $\alpha_{i}$ is a convex,
non boundary parallel simple closed curve.

###### Proof.

Assume that $T_{b_{1}}^{a_{1}}\cdots
T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots T_{\alpha_{m}}$ is a
relation on the mapping class monoid of $S_{0,0}^{5}$. We know that the braid
corresponding to the left hand side must be as in Figure 51.

Figure 51.

In order for the relation to hold, we know this braid must also correspond to
the right hand side. Thus, we see that the pairwise linking numbers must all
be equal to $-1$. By Lemma 3.2, this means that, for any two boundary
components there must be exactly one curve $\alpha_{i}$ that contains both
boundary components in its interior. Furthermore, because the $\alpha_{i}$’s
are not boundary parallel, each must contain at least 2 interior boundary
components and no more than 3 interior boundary components.

In the case where there is a convex curve $\alpha_{i}$ that contains exactly
three interior boundary components, up to isometry, there is only one such
arrangement for a curve. Such a curve must contain three adjacent boundary
components. Without loss of generality, assume they are $b_{1},b_{2}$, and
$b_{3}$. Any other curves included in the collection of $\alpha_{i}$’s may not
contain more than one of these boundary components. Furthermore, we still need
a linking number of $-1$ between $b_{4}$ and each of $b_{1},b_{2}$, and
$b_{3}$. Therefore, we must also twist over each curve that contains $b_{4}$
and exactly one of the other interior boundary components. We may not include
any additional curves because we now have a linking number of $-1$ for all
pairs of interior boundary components. We note that Relation 1 in Proposition
4.3 describes this case.

In the case where none of the $\alpha_{i}$’s contain more than two boundary
components, we must include the $\binom{4}{2}$ curves that each contain
exactly two interior boundary components in the twists on the right hand side.
This describes the case covered by Relation 2 in Proposition 4.3.

Since these are the only two cases, we conclude that Proposition 4.3 presents
all relations of the specified form, up to isometry. ∎

We present a similar proposition for $S_{0,0}^{6}$:

######  Proposition 5.2.

Up to isometry, Proposition 4.5 presents all relations on the mapping class
monoid of $S_{0,0}^{6}$ of the form

$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots
T_{\alpha_{m}},$

where each $a_{i}$ is a positive integer and each $\alpha_{i}$ is a convex,
non boundary parallel curve on $S_{0,0}^{6}$.

Figure 52.

Figure 53.

Figure 54.

Figure 55.

###### Proof.

Using the same reasoning as above, we will consider all possible collections
of non boundary parallel curves $\alpha_{i}$ on $S_{0,0}^{6}$ such that for
any two boundary components, there is exactly one curve that contains both in
its interior.

Since a curve that contains all 5 interior boundary components is parallel to
the outer boundary, any $\alpha_{i}$ must contain at most 4 interior boundary
components.

In the case where there is one curve that contains four boundary components,
we can assume without loss of generality that $b_{5}$ is not contained in this
curve. Then we see that we must additionally include all $\binom{4}{2}$ curves
that contain $b_{5}$ and exactly one other interior boundary component. Thus,
up to isometry, Relation 3 in Proposition 4.5 describes this case.

Now we will consider the case where at least one curve $\alpha_{i}$ contains
exactly three interior boundary components. Up to isometry, there are two
possibilities: the boundary components are all adjacent or they are not.

In the case that they are not adjacent, we have the curve shown in Figure 54.

There are three possible cases for the remaining curves: there is another
curve with three non adjacent boundary components, there is a curve with three
adjacent boundary components, and there are no other curves with three
boundary components.

In the case that all other curves $\alpha_{i}$ contain two boundary
components, we get a collection of curves as represented by Relation 7.

In the case that there is another curve with three non adjacent boundary
components, up to isometry, we will have the curves shown in Figure 54. Then
all other curves $\alpha_{i}$ must contain exactly two boundary components
each. This case is represented by Relation 4.

In the case that there is a curve containing three adjacent boundary
components, up to isotopy, we have the curves shown in Figure 54. All
remaining curves must contain exactly two boundary components, describing the
collection of curves used in Relation 5.

Now we can consider the case where there is a curve containing exactly three
adjacent boundary components and no curves containing three non adjacent
boundary components.

If this is the only curve with more than two boundary components, all other
curves must contain two boundary components each. Up to isotopy, this is
described by the right-hand side of Relation 2.

If there is another curve with three adjacent boundary components, then up to
isometry we have an arrangement as shown in Figure 55. All remaining curves
must contain exactly two interior boundary components each. Up to isometry,
this collection of curves is described by Relation 6.

Finally, if all curves $\alpha_{i}$ have no more than two boundary components,
we must have all $\binom{6}{2}$ curves that contain exactly two boundary
components each. This case is described by Relation 1.

Since this covers all possible cases, we see that, up to isometry, Proposition
4.5 describes all relations of the specified form. ∎

We suspect that Proposition 4.7 describes all possible relations of the
specified form on the mapping class monoid of $S_{0,0}^{7}$. However, since
they do not include all possible collections of curves $\alpha_{i}$ which
would give the correct pairwise linking numbers, in order to prove this claim,
it must be shown that some of the possible collections would not give a
relation with any product of boundary parallel twists.

We have a weaker claim that we can prove for $S_{0,0}^{7}$:

######  Proposition 5.3.

The plumbing graphs in Corollary 4.8 describe all possible plumbings that
correspond to a relation on the mapping class monoid of $S_{0,0}^{7}$ of the
form

$T_{b_{1}}^{a_{1}}\cdots T_{b_{6}}^{a_{6}}T_{b_{7}}=T_{\alpha_{1}}\cdots
T_{\alpha_{m}},$

where each $a_{i}$ is a positive integer and each $\alpha_{i}$ is a non
boundary parallel curve.

###### Proof.

We will show this by considering a relation $T_{b_{1}}^{a_{1}}\cdots
T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots T_{\alpha_{m}}$ on the
mapping class monoid of $S_{0,0}^{7}$. In order for this relation to hold, we
know that, for any pair of interior boundary components, there must be exactly
one curve $\alpha_{i}$ that contains both in its interior. Furthermore, since
each $\alpha_{i}$ is not boundary parallel, we know that each one must contain
at least 2 and at most 5 boundary components in its interior.

Consider the case where there is an $\alpha_{i}$ that contains exactly 5
boundary components in its interior. Let $b_{j}$ denote the interior boundary
component not contained in $\alpha_{i}$. This means that we must include all
curves that contain $b_{j}$ and exactly one other boundary component and no
additional curves. Therefore, under the right hand side, the multiplicity of
$b_{j}$ is 5 and the multiplicity of all other interior boundary components is
2. Because the multiplicity of each boundary component must be equal under
both sides of the relation, the left hand side must be $T_{b_{1}}\cdots
T_{b_{i}}^{4}\cdots T_{b_{6}}T_{b_{7}}$. The graph for the corresponding
plumbing shown in Figure 47.

Note that the plumbing graph is equivalent under any homeomorphism that
rearranges the boundary components, so this hold regardless of the relative
position of the interior boundary components. This is true for all plumbing
graphs, so we can consider all subsequent cases up to relabeling by any
permutation.

In the case that there is one curve $\alpha_{i}$ that contains exactly 4
boundary components, we can either have the case where all other
$\alpha_{i}$’s contain exactly two boundary components or the cause where
there is a curve $\alpha_{j}$ which contains exactly three boundary
components. In the former case, we see that the boundary components inside of
$\alpha_{j}$ will each have a multiplicity of 3 and those outside will have a
multiplicity of 5. Up to relabeling, this means that the left hand side must
be
$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}^{2}T_{b_{5}}^{4}T_{b_{6}}^{4}T_{b_{7}}$.
Therefore, any relation of this form will correspond to the plumbing graph
shown in Figure 47. In the latter case, we know that $\alpha_{i}$ and
$\alpha_{j}$ must contain exactly one boundary component in common and all
pairs of boundary components not both in $\alpha_{i}$ or $\alpha_{j}$ must be
contained in a curve with exactly those two boundary components. Up to
relabelling, a product of twists over such a collection of curves would be
equivalent to
$T_{b_{1}}^{2}T_{b_{2}}^{2}T_{b_{3}}^{2}T_{b_{4}}T_{b_{5}}^{3}T_{b_{6}}^{3}T_{b_{7}}$,
which corresponds to the plumbing graph in Figure 50.

In the case where all curves contain at most three boundary components in
their interior, we have three cases: there is exactly one such curve, there
are two such curves that have one boundary component in common, and there are
two such curves that do not have any boundary components in common. In the
first case, all other curves must contain exactly two boundary components, and
there must be one that contains any pair of boundary components that are not
both in the first curve. Up to relabelling, the product of twists over these
curves will be equivalent to
$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{3}T_{b_{4}}^{4}T_{b_{5}}^{4}T_{b_{6}}^{4}T_{b_{7}}$,
which corresponds to the plumbing graph in Figure 47. In the case where there
are two curves with three boundary components that have one boundary component
in common, all remaining curves must contain exactly two boundary components,
and this will be equivalent to
$T_{b_{1}}^{3}T_{b_{2}}^{2}T_{b_{3}}^{3}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}^{4}T_{b_{7}}$
up to relabelling. This corresponds to the graph in Figure 50. Finally, if the
two curves do not have any boundary components in common, then each additional
curve must contain exactly one boundary component from each of these curves.
Up to relabelling, this is equivalent to
$T_{b_{1}}^{3}T_{b_{2}}^{3}T_{b_{3}}^{3}T_{b_{4}}^{3}T_{b_{5}}^{3}T_{b_{6}}^{3}T_{b_{7}}$,
which corresponds to the plumbing graph given in Figure 50.

Finally, in the case where all curves contain at most two boundary components,
we must include twists over each of the $\binom{6}{2}$ curves that contain
exactly two interior boundary components. Any product of twists over this
collection of curves that corresponds to the specified braid is equivalent to
$T_{b_{1}}^{4}T_{b_{2}}^{4}T_{b_{3}}^{4}T_{b_{4}}^{4}T_{b_{5}}^{4}T_{b_{6}}^{4}T_{b_{7}}$
and corresponds to the graph in Figure 47.

Since we have considered all possible collections of curves $\alpha_{i}$, we
see that the graphs in Corollary 4.8 represent all possible plumbings that are
built from relations of the specified form on the mapping class monoid of
$S_{0,0}^{7}$. ∎

We can also consider the possible relations of the form
$T_{b_{1}}^{a_{1}}\cdots T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots
T_{\alpha_{m}}$, this time allowing the $\alpha_{i}$’s to be non convex.
Because Lemma 3.2 applies for convex and non convex curves, using the same
combinatorial argument as in the proof of Proposition 4.7, we can conclude
that the plumbing graph and associated Euler characteristics for any relation
of this form must be the same as those associated with a relation involving
only convex curves. Therefore, we must find another way to determine whether
the resulting fillings are distinct from those given in the existing
relations.

######  Question 5.4.

Do there exist relations of the form $T_{b_{1}}^{a_{1}}\cdots
T_{b_{n-1}}^{a_{n-1}}T_{b_{n}}=T_{\alpha_{1}}\cdots T_{\alpha_{m}}$, where at
least one $\alpha_{i}$ is non convex, that cannot be related by a global
diffeomorphism to a relation involving only convex curves?

## References

  * [AO01] Selman Akbulut and Burak Ozbagci.  Lefschetz fibrations on compact Stein surfaces.  Geom. Topol., 5:319–334, 2001.
  * [EG08] Hisaaki Endo and Yusuf Z. Gurtas. Lantern relations and rational blowdowns. Proc. Amer. Math. Soc. 138:1131-1142, 2010.
  * [EMVHM11] Hisaaki Endo, Thomas E. Mark, and Jeremy Van Horn-Morris.  Monodromy substitutions and rational blowdowns.  J. Topol., 4(1):227–253, 2011.
  * [FM12] Benson Farb and Dan Margalit.  A Primer on Mapping Class Groups.  Princeton University Press, 2011.
  * [Ful03] Terry Fuller.  Lefschetz fibrations of 4-dimensional manifolds.  Cubo Mat. Educ., 5(3): 275–294, 2003.
  * [GM13] David Gay and Thomas E. Mark.  Convex plumbings and Lefschetz fibrations.  J. Symplectic Geom., 11(3):363–375, 2013.
  * [Gir02] Emmanuel Giroux.  Géométrie de contact: de la dimension trois vers les dimensions supérieures.  In Proceedings of the International Congress of Mathematicians, Vol. II (Beijing, 2002), pages 405–414. Higher Ed. Press, Beijing, 2002.
  * [GS99] Robert E. Gompf and András I. Stipsicz.  $4$-manifolds and Kirby calculus, volume 20 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 1999.
  * [KS16] Çağri Karakurt and Laura Starkston.  Surgery along star-shaped plumbings and exotic smooth structures on 4-manifolds.  Algebr. Geom. Topol. 16(3):1585-1635, 2016
  * [LP01] Andrea Loi and Riccardo Piergallini.  Compact Stein surfaces with boundary as branched covers of $B^{4}$.  Invent. Math., 143(2):325–348, 2001.
  * [MM09] Dan Margalit and Jon McCammond.  Geometric presentations for the pure braid group.  J. Knot Theory Ramifications, 18(1): 1–20, 2009.
  * [McD90] Dusa McDuff.  The structure of rational and ruled symplectic $4$-manifolds.  J. Amer. Math. Soc., 3(3):679–712, 1990.
  * [OS04] Burak Ozbagci and András I. Stipsicz. Surgery on Contact 3-Manifolds and Stein Surfaces. _Springer-Verlag Berlin Heidelberg_ , 2004.
  * [PS21] Olga Plamenevskaya and Laura Starkston.  Unexpected Stein fillings, rational surface singularities, and plane curve arrangements. preprint, arXiv:2006.06631 [math.GT].
  * [PVHM10] Olga Plamenevskaya and Jeremy Van Horn-Morris.  Planar open books, monodromy factorizations and symplectic fillings.  Geom. Topol., 14(4):2077–2101, 2010.
  * [Sym98] Margaret Symington. Symplectic rational blowdowns. J. Differential Geom., 50:505-518, 1998
  * [TW75] W. P. Thurston and H. E. Winkelnkemper.  On the existence of contact forms.  Proc. Amer. Math. Soc., 52:345–347, 1975.
  * [Wen10] Chris Wendl.  Strongly fillable contact manifolds and $J$-holomorphic foliations.  Duke Math. J., 151(3):337–384, 2010.