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We leave it as an easy exercise to verify that this particular graph is rigid.
In particular, it implies that
$|\operatorname{Supp}(u)\cap\operatorname{Supp}(v_{1},\dots,v_{k})|=1$
contradicting the assumption on $u$. This establishes (c) in the final case
and concludes the proof.∎

Finally we are ready to show that the Working Conditions imply rigidity.

######  Theorem 4.13.

If $\Gamma$ is a disjoint union of linear plumbings satisfying the Working
Conditions, then $\Gamma$ is rigid.

###### Proof.

We prove this by induction on the number of vertices in $\Gamma$. For the base
case we assume that $\Gamma$ contains at most two vertices. It follows from
Condition II that $\Gamma$ must be one of the following five graphs in this
case:

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}\pgfsys<EMAIL_ADDRESS>
Each of these is easily checked to be rigid, giving the base case.

Now suppose that $\Gamma$ contains at least three vertices. We will see that
such a $\Gamma$ must contain a chain of twos. If $\Gamma$ has at least three
vertices in one connected component, then Condition III implies that at least
one of these vertices must have adjusted weight zero which must be contained
in a chain of twos. On the other hand if $\Gamma$ contains only connected
components with at most two vertices, then Condition II implies that only one
of these components contains a vertex with $w^{\prime}(v)>1$. Thus all but one
of these components will take the following form

$\leavevmode\hbox to40.12pt{\vbox
to23.49pt{\pgfpicture\makeatletter\hbox{\hskip-22.61975pt\lower-5.55522pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}}{{}}
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
{{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{25.95276pt}{8.1587pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$2$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
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}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$2$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
{{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{25.95276pt}{-2.22221pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\bullet$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
{{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-2.22221pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\bullet$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} {
{}{}{}}{}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{ {}{}{}}{}{ {}{}{}}
{{{{{}}{ {}{}}{}{}{{}{}}}}}{}{{{{{}}{
{}{}}{}{}{{}{}}}}}{{}}{}{}{}{}{}{{{}{}}}{}{{}}{}{}{}{{{}{}}}{}\pgfsys@moveto{34.48576pt}{0.0pt}\pgfsys@lineto{50.87251pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
{{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{42.67914pt}{3.533pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\scriptstyle{}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>
Thus we can assume that $\Gamma$ contains a chain of twos.

We claim further that $\Gamma$ contains a chain of twos $v_{1},\dots,v_{k}$
such that for every other vertex $v$ we have

$|\operatorname{Supp}(v)\cap\operatorname{Supp}(v_{1},\dots,v_{k})|\leq 1.$

Let $u_{1},\dots,u_{\ell}$ be a chain of twos and suppose there is a vertex
$u$ not contained in the chain of twos $u_{1},\dots,u_{\ell}$ with
$|\operatorname{Supp}(u)\cap\operatorname{Supp}(u_{1},\dots,u_{\ell})|\geq 2$.
Note that if no such $u$ exists then we are done. By Lemma 4.12, we see that
$u$ is the unique vertex with $w^{\prime}(u)>1$ and that
$|\operatorname{Supp}(u)\cap\operatorname{Supp}(u_{1},\dots,u_{\ell})|\geq 3$.
Moreover, $\Gamma$ must contain some other chain of twos $v_{1},\dots,v_{k}$.
Now, suppose there is a vertex $v$ not contained in the chain of twos
$v_{1},\dots,v_{k}$ such that
$|\operatorname{Supp}(v)\cap\operatorname{Supp}(v_{1},\dots,v_{k})|\geq 2$.
Again by Lemma 4.12, we have that $v$ is the unique vertex with
$w^{\prime}(v)>1$ and that

$|\operatorname{Supp}(v)\cap\operatorname{Supp}(v_{1},\dots,v_{k})|\geq 3.$

In particular, we conclude that $u=v$. This is not possible, since by
Condition II, we have that $w(u)\leq 5$, and
$|\operatorname{Supp}(u)\cap\operatorname{Supp}(u_{1},\dots,u_{\ell})|\geq 3$
and Lemma 4.7 imply that

$|\operatorname{Supp}(v)\cap\operatorname{Supp}(v_{1},\dots,v_{k})|\leq 2.$

Hence, we conclude that
$|\operatorname{Supp}(v)\cap\operatorname{Supp}(v_{1},\dots,v_{k})|\leq 1$ for
all vertices $v$ not contained in the chain of twos $v_{1},\dots,v_{k}$.

Now, take such chain of twos $v_{1},\dots,v_{k}$ and assume that $v_{i}$ is
embedded as $v_{i}=e_{i}-e_{i-1}$ for each $i=1,\dots,k$. We break down the
proof into three cases.

######  Case.

Both $v_{1}$ and $v_{k}$ are leaves.

In this case, $v_{1},\dots,v_{k}$ forms a connected component of $\Gamma$ and
for every other vertex we have
$\operatorname{Supp}(v)\cap\operatorname{Supp}(v_{1},\dots,v_{k})=\varnothing$.
Let $\Gamma^{\prime}$ be the plumbing obtained by deleting this component.
Note that $\Gamma^{\prime}$ also satisfies the Working Conditions. The
embedding of $\Gamma$ restricts to give an embedding of $\Gamma^{\prime}$. By
the inductive hypothesis $\Gamma^{\prime}$ is rigid and hence the induced
embedding is standard. It follows that the original embedding of $\Gamma$ was
standard.

######  Case.

Exactly one of $v_{1}$ and $v_{k}$ is a leaf.

In this case, we can assume that there is a vertex $v_{k+1}$, which is
adjacent to $v_{k}$. This vertex must take the form
$v_{k+1}=-e_{k}+v^{\prime}$ where $v^{\prime}\cdot e_{0}=\dots=v^{\prime}\cdot
e_{k}=0$. For all other vertices we must have
$\operatorname{Supp}(v)\cap\operatorname{Supp}(v_{1},\dots,v_{k})=\varnothing$.
Now, let $\Gamma^{\prime}$ be the plumbing obtained by deleting
$v_{1},\dots,v_{k}$ and decreasing the weight of $v_{k+1}$ by one. Note that
$\Gamma^{\prime}$ also satisfies the Working Conditions. The plumbing
$\Gamma^{\prime}$ inherits an embedding in $\mathbb{Z}^{n}$ obtained by
deleting $v_{1},\dots,v_{k}$ and sending $v_{k+1}$ into $\mathbb{Z}^{n}$ as
$v^{\prime}$. By the inductive hypothesis $\Gamma^{\prime}$ is rigid and hence
the induced embedding is standard. It follows that the original embedding of
$\Gamma$ was standard.

######  Case.

Neither $v_{1}$ nor $v_{k}$ is a leaf.

In this case, there are vertices $v_{0}$ and $v_{k+1}$ adjacent to $v_{1}$ and
$v_{k}$ respectively. These must embed as $v_{0}=e_{0}+v^{\prime}$ and
$v_{k+1}=-e_{k}+v^{\prime\prime}$, where $v^{\prime}$ and $v^{\prime\prime}$
satisfy $v^{\prime}\cdot e_{0}=\dots=v^{\prime}\cdot e_{k}=0$ and
$v^{\prime\prime}\cdot e_{0}=\dots=v^{\prime\prime}\cdot e_{k}=0$. Take
$\Gamma^{\prime}$ to be the plumbing obtained by deleting $v_{1},\dots,v_{k}$
and decreasing the weights on $v_{0}$ and $v_{k+1}$ by one. Note that
$\Gamma^{\prime}$ also satisfies the Working Conditions. This inherits an
embedding in $\mathbb{Z}^{n}$ obtained by deleting $v_{1},\dots,v_{k}$ and
embedding $v_{0}$ and $v_{k+1}$ as $v^{\prime}$ and $v^{\prime\prime}$. By the
inductive hypothesis $\Gamma^{\prime}$ is rigid and hence the induced
embedding is standard. It follows that the original embedding of $\Gamma$ was
standard. This completes the induction step. ∎

This allows us to conclude with the proof of Theorem 2.7, which forms the
final ingredient in the proof of Theorem 1.1.

###### Proof of Theorem 2.7.

By Lemma 3.6, the manifold $-\natural X(p_{i},p_{i}-q_{i})$ is plumbed
according to a dual plumbing $P^{*}$ satisfying the Working Conditions. By
Theorem 4.13, this implies the plumbing $P^{*}$ is rigid and hence the
intersection form of $-\natural X(p_{i},p_{i}-q_{i})$ is also rigid. ∎

##  5\. Lens spaces not bounding small definite manifolds

In this section, we use ideas from the previous sections to prove Theorem 1.4
which asserts the existence of lens spaces that do not admit a “small”
negative-definite filling with either orientation. As in previous sections we
use $\mathbb{Z}^{N}$ to denote the standard positive definite diagonal lattice
$(\mathbb{Z}^{N},\langle 1\rangle^{N})$.

Throughout this section we will use $L_{m,n}$ to denote the lens space
$L(p,q)$, where $p/q$ is given by the continued fraction expansion

$\frac{p}{q}=\left[[9]^{n},[3,2,2,2,2]^{m}\right]^{-}$

and $m$ and $n$ are positive integers. The examples in Theorem 1.4 will be
obtained by choosing suitable examples among the $L_{m,n}$. By (3.2), we see
that $p/(p-q)$ satisfies the continued fraction

(5.1) $\frac{p}{p-q}=\left[2,[2,2,2,2,2,2,3]^{n},[7]^{m-1},6\right]^{-}.$

First we obtain a lower bound on any smooth negative-definite filling of
$-L_{m,n}$.

######  Proposition 5.1.

Let $X$ be a smooth negative-definite filling of $-L_{m,n}$, then
$b_{2}(X)\geq m-n+1$.

###### Proof.

Let $X$ be a smooth negative-definite filling of $-L_{m,n}$. We construct a
smooth closed positive-definite 4-dimensional manifold
$-X\cup_{L(p,q)}-X(p,q)$. By Donaldson’s diagonalization theorem this has
diagonalizable intersection form and so we obtain a map of lattices

$-Q_{X(p,q)}\hookrightarrow\mathbb{Z}^{N},$

where $N=b_{2}(X(p,q))+b_{2}(X)$. The intersection form of $-X(p,q)$ is given
by the tuple:

$\left([9]^{n},[3,2,2,2,2]^{m}\right).$

The sublattice corresponding to the subchain given by $[3,2,2,2,2]^{m}$ is
rigid; one can check that this satisfies the Working Conditions from Section 3
and so is rigid by Theorem 4.13. By the bound in Remark 4.5(a) this implies
that $N\geq 6m+1$ (the graph corresponding to $[3,2,2,2,2]^{m}$ has a sum of
weights $11m$ and $5m-1$ edges). Since $b_{2}(X(p,q))=n+5m$, this gives

$b_{2}(X)=N-b_{2}(X(p,q))\geq m-n+1,$

which is the desired bound on $b_{2}(X)$. ∎

Next we obtain a lower bound on any smooth negative-definite filling of
$L_{m,n}$. The argument is similar although a little more delicate.

######  Proposition 5.2.

Let $X$ be a smooth negative-definite filling of $L_{m,n}$, then $b_{2}(X)\geq
n-1$.

###### Proof.

Let $X$ be a smooth negative-definite filling of $L_{m,n}$. By the fact that
$-L_{m,n}\cong L(p,p-q)$ and (5.1), the manifold $-X(p,p-q)$ is plumbed along
a linear graph with weights given by the tuple

$\left(2,[2,2,2,2,2,2,3]^{n},[7]^{m-1},6\right).$

In particular, we see that $b_{2}(X(p,p-q))=m+7n+1$. We construct a smooth
closed positive-definite 4-dimensional manifold $-X\cup_{L(p,p-q)}-X(p,p-q)$.
By Donaldson’s diagonalization theorem this has diagonalizable intersection
form and so we obtain a map of lattices

$-Q_{X(p,q)}\hookrightarrow\mathbb{Z}^{N},$

where $N=b_{2}(X(p,p-q))+b_{2}(X)$. We now separate the vertices in the
plumbing graph of $-X(p,p-q)$ into two groups. We will use
$v_{1},\dots,v_{7n}$ to denote the first $7n$ vertices, that is the vertices
corresponding to the tuple:

$\left(2,[2,2,2,2,2,2,3]^{n-1},2,2,2,2,2,2\right)$

and $u_{0},\dots,u_{m}$ to denote the remaining $m+1$ vertices, that is the
vertices corresponding to the tuple:

$\left(3,[7]^{m-1},6\right).$

Now, by Theorem 4.13 the plumbing corresponding to the vertices
$v_{1},\dots,v_{7n}$ is rigid and thus up to automorphisms of $\mathbb{Z}^{N}$
admits a unique embedding in a diagonal lattice. Moreover, using the formula
derived in Remark 4.5(a) this unique embedding uses $8n$ distinct orthogonal
unit vectors. Thus we can assume that the first $7n$ vertices are embedded in
a standard way $v_{1},\dots v_{7n}\subseteq\langle e_{1},\dots,e_{8n}\rangle$.
However, for every vector $v$ in the orthogonal complement

$\langle v_{1},\dots v_{7n}\rangle^{\bot}\subseteq\langle
e_{1},\dots,e_{8n}\rangle$

must have length satisfying $\lVert v\rVert^{2}\geq 8$.444One can see that the
orthogonal complement of $v_{1},\dots v_{7n}$ standardly embedded in $\langle
e_{1},\dots,e_{8n}\rangle$ is isomorphic to the linear lattice corresponding
to the linear graph with vertices  $([9]^{n-1},8).$ By equation (3.1) in the
proof of Lemma 3.4, every non-trivial element of this linear lattice satisfies
$\lVert v\rVert^{2}\geq 8$.. Since vertices $u_{1},\dots,u_{m}$ satisfy
$\lVert u_{i}\rVert^{2}\leq 7$ and are orthogonal to $v_{1},\dots v_{7n}$,
they cannot pair nontrivially with any of $e_{1},\dots,e_{8n}$. This implies
that the embeddings of $u_{1},\dots,u_{m}$ must use at least $m$ additional
unit vectors, and we have that $N\geq 8n+m$. Thus,

$b_{2}(X)=N-b_{2}(X(p,p-q))\geq 8n+m-(m+7n+1)=n-1,$

which is the required bound. ∎

With these bounds in hand we exhibit the examples for Theorem 1.4.

###### Proof of Theorem 1.4.

We take $L_{k}$ to be the lens space $L_{m,n}$ with $m=2k$ and $n=k+1$. By
Proposition 5.1, we see that a negative-definite filling $X$ of $-L_{k}$
satisfies $b_{2}(X)\geq m-n+1=k$ and, by Proposition 5.2, a negative-definite
filling $X$ of $L_{k}$ satisfies $b_{2}(X)\geq n-1=k$. ∎

##  6\. Embedding lens spaces in 4-manifolds

We now turn our attention to embedding lens spaces in 4-manifolds.

######  Proposition 6.1.

Let $L_{k}$ be a lens space as given in Theorem 1.4 and suppose that $L_{k}$
embeds in a smooth 4-manifold $X$ with $|\sigma(X)|\geq b_{2}(X)-2$, then
$k\leq b_{2}(X)$.

###### Proof.

By hypothesis, any smooth negative-definite filling of $L_{k}$ or $-L_{k}$ has
$b_{2}\geq k$. Suppose that $L_{k}$ smoothly embeds in a $4$-manifold $X$ as a
separating submanifold and that $|\sigma(X)|\geq b_{2}(X)-2$. By changing the
orientation of $X$ if necessary we may assume that $\sigma(X)\leq 2-b_{2}(X)$
and hence that $b_{2}^{+}(X)\leq 1$. The embedding of $L_{k}$ in $X$ gives a
decomposition of $X$ into two pieces $X_{1}$ and $X_{2}$ with $\partial
X_{1}\cong-\partial X_{2}\cong L_{k}$. Moreover, by Novikov additivity and the
fact that a lens space is a rational homology sphere, we have that
$b_{2}^{+}(X)=b_{2}^{+}(X_{1})+b_{2}^{+}(X_{2})$ and
$b_{2}^{-}(X)=b_{2}^{-}(X_{1})+b_{2}^{-}(X_{2})$. Thus, we have that at least
one of $b_{2}^{+}(X_{1})$ or $b_{2}^{+}(X_{2})$ is zero. That is, at least one
of $X_{1}$ or $X_{2}$ is negative-definite. This implies that
$b_{2}(X_{1})\geq k$ or $b_{2}(X_{2})\geq k$ and hence that $b_{2}(X)\geq k$.
∎

Mimicking arguments that can be found in [AGL17], we note that there is no
spin 4-manifold that contains every lens space as a separating submanifold.

######  Proposition 6.2.

Suppose that $L(n,n-1)$ with $n\geq 1$ odd smoothly embeds in a closed spin
4-manifold $X$ as a separating submanifold. Then

$n\leq 9b_{2}(X)+1.$

###### Proof.

Note that since $n$ is odd, the lens space $L(n,n-1)$ admits a unique spin
structure. Cutting $X$ along the embedded $L(n,n-1)$ yields a spin 4-manifold
$Z$ bounding $-L(n,n-1)$. On the other hand, the manifold $X(n,n-1)$ is
negative-definite and spin with $b_{2}(X(n,n-1))=n-1$. Since the spin
structure on $L(n,n-1)$ is unique, the closed 4-manifold
$W=X(n,n-1)\cup_{L(n,n-1)}Z$ is spin. We have

$b_{2}(W)=b_{2}(Z)+n-1\leq
b_{2}(X)+n-1\quad\text{and}\quad|\sigma(W)|=|\sigma(Z)-n+1|\geq n-1-b_{2}(X).$

Thus by Furuta’s theorem [Fur01], we have

$b_{2}(X)+n-1\geq
b_{2}(W)\geq\frac{5}{4}|\sigma(W)|\geq\frac{5}{4}(n-1-b_{2}(X)).$

Rearranging implies that $n\leq 9b_{2}(X)+1$. ∎

###### Proof of Theorem 1.3.

Let $X$ be a smooth compact 4-manifold that contains every lens space as a
smoothly embedded separating submanifold. Proposition 6.2 implies that $X$
cannot be spin. Proposition 6.1 implies that $|\sigma(X)|<b_{2}(X)-2$. However
$b_{2}(X)\equiv\sigma(X)\bmod 2$, so this implies that $|\sigma(X)|\leq
b_{2}(X)-4$.∎

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##  Appendix A Appendix: Computer data

This is the data necessary to justify Lemma 4.8. We consider each of the five
graphs enumerated in the proof and verify that all embeddings restrict to a
standard embedding four the vertices $v_{1}$, $v_{2}$, $v_{3}$ and $v_{4}$.
This shows that the subgraph given by $v_{1}$, $v_{2}$, $v_{3}$ and $v_{4}$ is
rigid. We did this by using the OrthogonalEmbeddings command GAP to find all
possible embeddings up to automorphisms of $\mathbb{Z}^{n}$ [GAP21]. The use
of a computer here is merely for convenience. The calculation is certainly
within the realm of hand calculation, however the details would be
sufficiently onerous and unenlightening that a computer calculation is
preferable.

  1. (1)

The plumbing $2$3223$2$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$
admits only one embedding:

$\begin{pmatrix}0&0&0&-1&1&0&0&0&0\\\ \hline\cr-1&0&0&1&0&0&0&1&0\\\
1&-1&0&0&0&0&0&0&0\\\ 0&1&-1&0&0&0&0&0&0\\\ 0&0&1&0&0&-1&0&0&1\\\ \hline\cr
0&0&0&0&0&1&1&0&0\end{pmatrix}$

  2. (2)

The plumbing
$2$3223$3$$2$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$
admits two embeddings:

$\begin{pmatrix}0&0&0&0&0&-1&1&0&0&0\\\ \hline\cr 0&0&-1&0&0&1&0&1&0&0\\\
0&0&1&-1&0&0&0&0&0&0\\\ 0&0&0&1&-1&0&0&0&0&0\\\ 1&-1&0&0&1&0&0&0&0&0\\\
\hline\cr-1&0&0&0&0&0&0&0&1&1\\\ 1&1&0&0&0&0&0&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{11}\begin{pmatrix}0&0&0&-1&1&0&0&0&0&0&0\\\
\hline\cr-1&0&0&1&0&0&0&0&1&0&0\\\ 1&-1&0&0&0&0&0&0&0&0&0\\\
0&1&-1&0&0&0&0&0&0&0&0\\\ 0&0&1&0&0&0&0&-1&0&1&0\\\ \hline\cr
0&0&0&0&0&-1&0&1&0&0&1\\\ 0&0&0&0&0&1&1&0&0&0&0\end{pmatrix}$

  3. (3)

The plumbing
$2$$2$3223$3$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$
admits two embeddings:

$\begin{pmatrix}1&-1&0&0&0&0&0&0&0&0\\\ -1&0&1&0&0&0&0&0&0&0\\\ \hline\cr
1&1&0&-1&0&0&0&0&0&0\\\ 0&0&0&1&-1&0&0&0&0&0\\\ 0&0&0&0&1&-1&0&0&0&0\\\
0&0&0&0&0&1&-1&1&0&0\\\ \hline\cr 0&0&0&0&0&0&1&0&1&1\end{pmatrix}$

$\setcounter{MaxMatrixCols}{11}\begin{pmatrix}0&0&0&-1&1&0&0&0&0&0&0\\\
0&0&0&1&0&-1&0&0&0&0&0\\\ \hline\cr-1&0&0&0&0&1&1&0&0&0&0\\\
1&-1&0&0&0&0&0&0&0&0&0\\\ 0&1&-1&0&0&0&0&0&0&0&0\\\ 0&0&1&0&0&0&0&-1&1&0&0\\\
\hline\cr 0&0&0&0&0&0&0&1&0&1&1\end{pmatrix}$

  4. (4)

The plumbing
$2$$2$3223$4$$2$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$
admits 4 embeddings:

$\setcounter{MaxMatrixCols}{11}\begin{pmatrix}1&-1&0&0&0&0&0&0&0&0&0\\\
-1&0&1&0&0&0&0&0&0&0&0\\\ \hline\cr 1&1&0&0&0&-1&0&0&0&0&0\\\
0&0&0&0&0&1&-1&0&0&0&0\\\ 0&0&0&0&0&0&1&1&0&0&0\\\ 0&0&0&1&-1&0&0&-1&0&0&0\\\
\hline\cr 0&0&0&-1&0&0&0&0&1&1&1\\\ 0&0&0&1&1&0&0&0&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{12}\begin{pmatrix}1&-1&0&0&0&0&0&0&0&0&0&0\\\
-1&0&1&0&0&0&0&0&0&0&0&0\\\ \hline\cr 1&1&0&-1&0&0&0&0&0&0&0&0\\\
0&0&0&1&-1&0&0&0&0&0&0&0\\\ 0&0&0&0&1&1&0&0&0&0&0&0\\\
0&0&0&0&0&-1&0&0&1&1&0&0\\\ \hline\cr 0&0&0&0&0&0&-1&0&-1&0&1&1\\\
0&0&0&0&0&0&1&1&0&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{12}\begin{pmatrix}0&0&0&0&0&-1&1&0&0&0&0&0\\\
0&0&0&0&0&1&0&-1&0&0&0&0\\\ \hline\cr 0&0&-1&0&0&0&0&1&1&0&0&0\\\
0&0&1&-1&0&0&0&0&0&0&0&0\\\ 0&0&0&1&1&0&0&0&0&0&0&0\\\
1&-1&0&0&-1&0&0&0&0&0&0&0\\\ \hline\cr-1&0&0&0&0&0&0&0&0&1&1&1\\\
1&1&0&0&0&0&0&0&0&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{13}\begin{pmatrix}0&0&0&-1&1&0&0&0&0&0&0&0&0\\\
0&0&0&1&0&-1&0&0&0&0&0&0&0\\\ \hline\cr-1&0&0&0&0&1&1&0&0&0&0&0&0\\\
1&-1&0&0&0&0&0&0&0&0&0&0&0\\\ 0&1&1&0&0&0&0&0&0&0&0&0&0\\\
0&0&-1&0&0&0&0&0&0&1&1&0&0\\\ \hline\cr 0&0&0&0&0&0&0&-1&0&-1&0&1&1\\\
0&0&0&0&0&0&0&1&1&0&0&0&0\end{pmatrix}$

  5. (5)

The plumbing
$2$$3$3223$3$$2$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$
admits 5 embeddings:

$\begin{pmatrix}0&1&-1&0&0&0&0&0&0\\\ 1&-1&0&-1&0&0&0&0&0\\\ \hline\cr
0&1&1&0&0&0&-1&0&0\\\ 0&0&0&0&0&0&1&-1&0\\\ 0&0&0&0&0&0&0&1&-1\\\
0&0&0&0&1&-1&0&0&1\\\ \hline\cr 1&0&0&1&-1&0&0&0&0\\\
0&0&0&0&1&1&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{11}\begin{pmatrix}1&-1&0&0&0&0&0&0&0&0&0\\\
-1&0&0&0&0&0&0&1&1&0&0\\\ \hline\cr 1&1&0&0&-1&0&0&0&0&0&0\\\
0&0&0&0&1&-1&0&0&0&0&0\\\ 0&0&0&0&0&1&-1&0&0&0&0\\\ 0&0&1&-1&0&0&1&0&0&0&0\\\
\hline\cr 0&0&-1&0&0&0&0&0&0&1&1\\\ 0&0&1&1&0&0&0&0&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{12}\begin{pmatrix}1&-1&0&0&0&0&0&0&0&0&0&0\\\
-1&0&0&0&0&0&0&0&0&1&1&0\\\ \hline\cr 1&1&-1&0&0&0&0&0&0&0&0&0\\\
0&0&1&-1&0&0&0&0&0&0&0&0\\\ 0&0&0&1&-1&0&0&0&0&0&0&0\\\
0&0&0&0&1&0&0&-1&1&0&0&0\\\ \hline\cr 0&0&0&0&0&-1&0&1&0&0&0&1\\\
0&0&0&0&0&1&1&0&0&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{12}\begin{pmatrix}0&0&0&0&0&-1&1&0&0&0&0&0\\\
0&0&0&0&0&1&0&-1&0&1&0&0\\\ \hline\cr 0&0&-1&0&0&0&0&1&1&0&0&0\\\
0&0&1&-1&0&0&0&0&0&0&0&0\\\ 0&0&0&1&-1&0&0&0&0&0&0&0\\\
1&-1&0&0&1&0&0&0&0&0&0&0\\\ \hline\cr-1&0&0&0&0&0&0&0&0&0&1&1\\\
1&1&0&0&0&0&0&0&0&0&0&0\end{pmatrix}$

$\setcounter{MaxMatrixCols}{13}\begin{pmatrix}0&0&0&-1&1&0&0&0&0&0&0&0&0\\\
0&0&0&1&0&0&0&-1&0&0&0&1&0\\\ \hline\cr-1&0&0&0&0&0&0&1&0&1&0&0&0\\\
1&-1&0&0&0&0&0&0&0&0&0&0&0\\\ 0&1&-1&0&0&0&0&0&0&0&0&0&0\\\
0&0&1&0&0&0&0&0&-1&0&1&0&0\\\ \hline\cr 0&0&0&0&0&-1&0&0&1&0&0&0&1\\\
0&0&0&0&0&1&1&0&0&0&0&0&0\end{pmatrix}$