joins two vertices both in either $V_{+}$ or $V_{-}$ and every edge of the
graph labeled $-$ joints a vertex in $V_{+}$ to one in $V_{-}$.

###### Proof of Proposition 12.16.

Suppose $\Sigma_{P}$ contained a component $K$ which is balanced. Then from
the theorem of Harary, there is a partition of vertices of $K$ into $V_{+}(K)$
and $V_{-}(K)$. Let $f=(f_{1},\ldots,f_{n})\in{\mathbb{R}}^{n}$ be defined by

$f_{i}=\begin{cases}1&v_{i}\in V_{+}(K)\\\ -1&v_{i}\in V_{-}(K)\\\
0&v_{i}\notin K\end{cases}$

Recall that an edge of $\Sigma$ labeled $+$ corresponded to $e_{i}-e_{j}$,
meaning $f_{i}=f_{j}=\pm 1$, so $\langle f,e_{i}-e_{j}\rangle=0$. Similarly,
an edge of $\Sigma$ labeled $-$ corresponds to $\pm(e_{i}+e_{j})$, with
$f_{i}=-f_{j}$, so $\langle f,\pm(e_{i}+e_{j})\rangle=0$. Then one has that
$\langle f,\alpha\rangle=\langle-f,\alpha\rangle=0$ for all $\alpha\in P$. In
other words, $K_{P}^{\mathrm{rt}}$ lies in $f^{\perp}$, so cannot be full-
dimensional.

In the other direction, suppose every connected component of $\Sigma_{P}$
contains an unbalanced cycle. To show that $K_{P}^{\mathrm{rt}}$ is full-
dimensional it suffices to show there are $k$ edges corresponding to $k$
linearly independent elements of $P$ in each connected component of
$\Sigma_{P}$ with $k$ vertices. Suppose $K$ is a connected component of
$\Sigma_{P}$ with $k$ vertices. A spanning tree of $K$ has $k-1$ edges
corresponding to $k-1$ linearly independent elements of $P$. Adding any other
edge of $K$ would create a cycle. Imagine that an edge $e$ is added to $K$ to
create a cycle. The only way this edge could fail to correspond to a $k$th
linearly independent element of $P$ is if the edge “completes” both the
unbalanced cycle and a balanced cycle. However, this would mean the first
$k-1$ edges formed a cycle, which is impossible. Thus, $K$ contains edges
corresponding to $k$ linearly independent elements of $P$, so $\Sigma_{P}$ has
$n$ such edges, so $K_{P}^{\mathrm{rt}}$ is full-dimensional. ∎

Note that the $f$ found in the proof of Proposition 12.16 is an ideal such
that $-f$ is also an ideal. In other words, the proof also shows that when
$K_{P}^{\mathrm{rt}}$ is not full-dimensional, $K_{P}^{\mathrm{wt}}$ is not
pointed.

######  Proposition 12.18.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset.
$K_{P}^{\mathrm{wt}}$ is pointed if and only if no connected component of
$\widehat{G}(P)$ is isotropic.

###### Proof.

Suppose $J\subset\widehat{G}(P)$ is a connected component of $\widehat{G}(P)$
which is isotropic. Then $-J$ must also be a connected component of
$\widehat{G}(P)$, which is isotropic and thus an ideal. Consequently,
$K_{P}^{\mathrm{wt}}$ is not pointed.

In the other direction, suppose $J$ and $-J$ are both isotropic ideals in
$\widehat{G}(P)$, i.e. $K_{P}^{\mathrm{wt}}$ is not pointed. Suppose $i\in J$
(so $-i\in-J$) and $i<j$. Then, from the definition of $\widehat{G}(P)$, one
must have $-j<-i$, meaning $-j\in-J$, so $j\in J$. Thus $J$ is an order
filter. An ideal which is also an order filter must be an entire connected
component of the Hasse diagram, meaning $\widehat{G}(P)$ has a connected
component which is isotropic, namely $J$. ∎

######  Proposition 12.19.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset. If
$\widehat{G}(P)$ contains an isotropic connected component, then $\Sigma_{P}$
contains a balanced connected component.

###### Proof.

Suppose $J\subset\widehat{G}(P)$ is an isotropic connected component. Then
$\Sigma_{P}$ has a connected component whose vertices are $|i|$ for $i\in J$,
call it $K$. Then the vertices of $K$ can be partitioned into $V_{+}$ and
$V_{-}$ by the following rule:

$V_{+}=\\{i\colon i\in J\\}\quad\text{and}\quad V_{-}=\\{i\colon-i\in J\\}.$

Checking the definition of $\widehat{G}(P)$, one sees that edges labeled $+$
in $\Sigma_{P}$ correspond to covering relations between two positive or two
negative elements of $\widehat{G}(P)$ and edges labeled $-$ correspond to
covering relations between one positive and one negative element in
$\widehat{G}(P)$. In other words, the $V_{+}$ and $V_{-}$ coming from $J$
satisfy the requirements of Theorem 12.17, so $K$ is a balanced component of
$\Sigma_{P}$. ∎

As an example, consider
$P=\\{+e_{1}+e_{2},+e_{2}-e_{3},-e_{3}-e_{4},+e_{1}-e_{4}\\}$, with
$\Sigma_{P}$ and $\widehat{G}(P)$ are shown in Figure 16. One sees that
$K_{P}^{\mathrm{rt}}$ is not full-dimensional as $\Sigma_{P}$ contains a
balanced cycle. Additionally, one sees that $\widehat{G}(P)$ contains
connected components which are isotropic. The corresponding partition of the
vertices of $\Sigma_{P}$ is $\\{1,4\\}\sqcup\\{2,3\\}$ and one sees that all
elements of $P$ lie in the hyperplane perpendicular to $(1,-1,-1,1)$, meaning
both $(1,-1,-1,1)$ and $(-1,1,1,-1)$ are ideals, so $K_{P}^{\mathrm{wt}}$ is
not pointed.

$1$$2$$3$$4$$-$$+$$-$$+$ (a) $\Sigma_{P}$

$-3$$1$$-2$$4$$2$$-3$$3$$-1$ (b) $\widehat{G}(P)$

Figure 16: $\Sigma_{P}$ and $\widehat{G}(P)$ for
$P=\\{+e_{1}+e_{2},+e_{2}-e_{3},-e_{3}-e_{4},+e_{1}-e_{4}\\}$

######  Proposition 12.20.

Suppose $P\subset\Phi_{B_{n}}$ is a signed poset. The cones
$K_{P}^{\mathrm{rt}}$ and $K_{P}^{\mathrm{wt}}$ are simplicial if and only if
$\Sigma_{P}$ does not contain a balanced cycle or two unbalanced cycles joined
by a path.

###### Proof.

From Theorem 12.15, one has that $K_{P}^{\mathrm{rt}}$ is simplicial if and
only if $\Gamma(\Sigma_{P})$ contains no dependent sets, as dependent sets in
$\Gamma(\Sigma_{P})$ correspond precisely to dependent sets of extreme rays of
$K_{P}^{\mathrm{rt}}$. Therefore, $K_{P}^{\mathrm{rt}}$ is simplicial if and
only if $\Gamma(\Sigma_{P})$ contains no circuits, i.e. no balanced cycles and
no pairs of unbalanced cycles joined by a path. One then has that this
characterizes when $K_{P}^{\mathrm{wt}}$ is simplicial as well, since
$K_{P}^{\mathrm{rt}}$ and $K_{P}^{\mathrm{wt}}$ are either both simplicial or
both not simplicial as they are dual cones. ∎

######  Corollary 12.21.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset. Its root
cone, $K_{P}^{\mathrm{rt}}$, and weight cone $K_{P}^{\mathrm{wt}}$ are
pointed, full-dimensional and simplicial if and only if

  * •

every connected component of $\Sigma_{P}$ contains and unbalanced cycle, and

  * •

$\Sigma_{P}$ contains a balanced cycle nor two unbalanced cycles joined by a
path.

###  13 Two Rational Functions

Recall from the introduction that associated to each signed poset is a pair of
rational functions which are sums over the linear extensions. For
$P\subset\Phi_{B_{n}}$, define

$\displaystyle\Psi_{P}(\mathbf{x})$
$\displaystyle=\sum_{w\in\mathcal{L}(P)}w\left(\dfrac{1}{(x_{1}-x_{2})(x_{2}-x_{3})\cdots(x_{n-1}-x_{n})x_{n}}\right)\quad\text{and}$
$\displaystyle\Phi_{P}(\mathbf{x})$
$\displaystyle=\sum_{w\in\mathcal{L}(P)}w\left(\dfrac{1}{x_{1}(x_{1}+x_{2})\cdots(x_{1}+x_{2}+\cdots+x_{n})}\right),$

and for ${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$, define

$\displaystyle\Psi^{\scriptscriptstyle\vee}_{P^{\scriptscriptstyle\vee}}(\bm{x})$
$\displaystyle=\sum_{w\in\mathcal{L}(P)}w\left(\dfrac{1}{(x_{1}-x_{2})(x_{2}-x_{3})\cdots(x_{n-1}-x_{n})2x_{n}}\right)=\dfrac{1}{2}\Psi_{P}(\bm{x})\quad\text{and}$
(2)
$\displaystyle\Phi^{\scriptscriptstyle\vee}_{P^{\scriptscriptstyle\vee}}(\bm{x})$
$\displaystyle=\sum_{w\in\mathcal{L}(P)}w\left(\dfrac{1}{x_{1}(x_{1}+x_{2})\cdots(x_{1}+\cdots+x_{n-1})(\frac{1}{2}x_{1}+\frac{1}{2}x_{2}+\cdots+\frac{1}{2}x_{n})}\right)$
(3) $\displaystyle=2\Phi_{P}(\bm{x}).$

These functions are the analogues of the $\Psi$ and $\Phi$ considered in [31],
[9] and [21] and discussed in Section 2 above. Note that when
$P\subset\Phi_{B_{n}}$ is a type A signed poset and
$P^{\prime}=P\cap\Phi_{A_{n-1}}$ the corresponding poset, one has
$\Phi_{P}(\bm{x})=\Phi_{P^{\prime}}(\bm{x})$, where $\Phi_{P}$ is calculated
with the type B definition and $\Phi_{P^{\prime}}$ is calculated with the type
A definition, since $P$ and $P^{\prime}$ have the same ideals by Proposition
12.6 and thus the same linear extensions.

As in [9], one can understand these functions as the valuation $s(-;\bm{x})$
from Section 5.1.

######  Theorem 13.1.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset. Then

$\displaystyle\Psi_{P}(\bm{x})$ $\displaystyle=s(K_{P}^{\mathrm{rt}};\bm{x})$
$\displaystyle=\int_{K_{P}^{\mathrm{rt}}}e^{-\langle\bm{x},u\rangle}\,du$
$\displaystyle\Psi^{\scriptscriptstyle\vee}_{{P^{\scriptscriptstyle\vee}}}(\bm{x})$
$\displaystyle=s^{\scriptscriptstyle\vee}(K_{P}^{\mathrm{rt}};\bm{x})$
$\displaystyle=\int_{K_{P}^{\mathrm{rt}}}e^{-\langle\bm{x},u\rangle}\,\dfrac{1}{2}du$
$\displaystyle\Phi_{P}(\bm{x})$ $\displaystyle=s(K_{P}^{\mathrm{wt}};\bm{x})$
$\displaystyle=\int_{K_{P}^{\mathrm{wt}}}e^{-\langle\bm{x},u\rangle}\,du$
$\displaystyle\Phi^{\scriptscriptstyle\vee}_{{P^{\scriptscriptstyle\vee}}}(\bm{x})$
$\displaystyle=s^{\scriptscriptstyle\vee}(K_{P}^{\mathrm{wt}};\bm{x})$
$\displaystyle=\int_{K_{P}^{\mathrm{wt}}}e^{-\langle\bm{x},u\rangle}\,2du$

where $du$ is Lebesgue measure.

The measure used for calculating $\Psi^{\scriptscriptstyle\vee}$ and
$\Phi^{\scriptscriptstyle\vee}$ varies to give the parallelopiped spanned by
the simple roots and fundamental dominant weights, respectively, volume one.

###### Proof.

In all cases, the proof proceeds by induction on the number of pairs
$\\{i,j\\}\subset\pm[n]$ such that $i$ and $j$ are incomparable in
$\widehat{G}(P)$. The proof for $\Psi_{P}$ and $\Phi_{P}$ is given here; the
proofs for $\Psi^{\scriptscriptstyle\vee}_{{P^{\scriptscriptstyle\vee}}}$ and
$\Phi^{\scriptscriptstyle\vee}_{P}$ follow (2) and (3). In the base case,
suppose all pairs $\\{i,j\\}$ are comparable, i.e. $\widehat{G}(P)$ is a
chain. Then $\mathcal{L}(P)$ must consist of a single signed permutation, call
it $w$.

Applying Proposition 5.5 in each case, one has

$s(K_{P}^{\mathrm{rt}};\bm{x})=|w(\alpha_{1})\wedge\cdots\wedge
w(\alpha_{n})|\prod_{i=1}^{n}\langle\bm{x},w(\alpha_{i})\rangle^{-1}\\\
=w\left(\dfrac{1}{(x_{1}-x_{2})\cdots(x_{n-1}-x_{n})x_{n}}\right)=\Psi_{P}(\bm{x})\\\
s(K_{P}^{\mathrm{wt}};\bm{x})=|w(\mu_{1})\wedge\cdots\wedge
w(\mu_{n})|\prod_{i=1}^{n}\langle\bm{x},w(\mu_{i})\rangle^{-1}\\\
=w\left(\dfrac{1}{x_{1}(x_{1}+x_{2})\cdots(x_{1}+\cdots+x_{n})}\right)=\Phi_{P}(\bm{x})$

where $|v_{1}\wedge\cdots\wedge v_{n}|$ is the volume of the parallelopiped
they form.

For the induction step, suppose $\\{i,j\\}\subset\pm[n]$ are incomparable in
$\widehat{G}(P)$. Let

$P_{i<j}=\overline{P\cup\\{\mathrm{sgn}(i)e_{i}-\mathrm{sgn}(j)e_{j}\\}}^{PLC}\quad\text{and}\quad
P_{j<i}=\overline{P\cup\\{\mathrm{sgn}(j)e_{j}-\mathrm{sgn}(i)e_{i}\\}}^{PLC}.$

(If $i=j$, one needs to divide $\mathrm{sgn}(i)e_{i}-\mathrm{sgn}(j)e_{j}$ and
$\mathrm{sgn}(j)e_{j}-\mathrm{sgn}(i)e_{i}$ by $2$ in type B.) The only way
$P_{i<j}$ could fail to be a signed poset is if
$\mathrm{sgn}(j)e_{j}-\mathrm{sgn}(i)e_{i}\in P$, a contradiction since it
would mean $i$ and $j$ were comparable in $\widehat{G}(P)$. A symmetric
argument means $P_{j<i}$ is also a signed poset. Next, observe that, by
construction,

$\mathcal{L}(P)=\mathcal{L}(P_{i<j})\sqcup\mathcal{L}(P_{j<i}),$

meaning

$\displaystyle\Psi_{P}(\bm{x})$
$\displaystyle=\Psi_{P_{i<j}}(\bm{x})+\Psi_{P_{j>i}}(\bm{x})$
$\displaystyle\Phi_{P}(\bm{x})$
$\displaystyle=\Phi_{P_{i<j}}(\bm{x})+\Phi_{P_{j>i}}(\bm{x})$

Recall that $s(-;\bm{x})$ is a valuation, so one wants to write
$K_{P}^{\mathrm{rt}}$ and $K_{P}^{\mathrm{wt}}$ as a sum of
$K^{\mathrm{rt}}_{P_{i<j}}$, $K^{\mathrm{rt}}_{P_{j<i}}$,
$K_{P_{i<j}}^{\mathrm{wt}}$ and $K_{P_{j<i}}^{\mathrm{wt}}$ and apply the
induction assumption to compute $s(-;\bm{x})$ for each of these cones and use
that to compute $s(K_{P}^{\mathrm{rt}};\bm{x})$ and
$s(K_{P}^{\mathrm{wt}};\bm{x})$. To that end, define a set
$P_{i=j}=\overline{P\cup\\{\mathrm{sgn}(i)e_{i}-\mathrm{sgn}(j)e_{j},\mathrm{sgn}(e_{j})e_{j}-\mathrm{sgn}(i)e_{i}\\}}^{PLC}$.
The set $P_{i=j}$ is of course, not a signed poset, but the definitions of
root cone and weight cone still make sense.

Next, observe that

$\displaystyle K_{P}^{\mathrm{rt}}$
$\displaystyle=K^{\mathrm{rt}}_{P_{i<j}}\cap K^{\mathrm{rt}}_{P_{j<i}}$
$\displaystyle K^{\mathrm{rt}}_{P_{i=j}}$
$\displaystyle=K^{\mathrm{rt}}_{P_{i<j}}\cup K^{\mathrm{rt}}_{P_{j<i}}$
$\displaystyle K_{P_{i=j}}^{\mathrm{wt}}$
$\displaystyle=K_{P_{i<j}}^{\mathrm{wt}}\cap K_{P_{j<i}}^{\mathrm{wt}}$
$\displaystyle K_{P}^{\mathrm{wt}}$
$\displaystyle=K_{P_{i<j}}^{\mathrm{wt}}\cup K_{P_{j<i}}^{\mathrm{wt}}.$

One then has by the valuative property that

$\displaystyle s(K^{\mathrm{rt}}_{P_{i=j}};\bm{x})$
$\displaystyle=s(K^{\mathrm{rt}}_{P_{i<j}};\bm{x})+s(K^{\mathrm{rt}}_{P_{j<i}};\bm{x})-s(K_{P}^{\mathrm{rt}};\bm{x})$
$\displaystyle s(K_{P}^{\mathrm{wt}};\bm{x})$
$\displaystyle=s(K_{P_{i<j}}^{\mathrm{wt}};\bm{x})+s(K_{P_{j<i}}^{\mathrm{wt}};\bm{x})-s(K_{P_{i=j}}^{\mathrm{wt}};\bm{x})$

Since $K^{\mathrm{rt}}_{P_{i=j}}$ is not pointed and
$K_{P_{i=j}}^{\mathrm{wt}}$ is not full-dimensional,

$s(K^{\mathrm{rt}}_{P_{i=j}};\bm{x})=s(K_{P_{i=j}}^{\mathrm{wt}};\bm{x})=0.$

Then, applying the induction assumption one has

$\begin{split}s(K_{P}^{\mathrm{rt}};\bm{x})=\Psi_{P_{i<j}}(\bm{x})+\Psi_{P_{j<i}}(\bm{x})=\Psi_{P}(\bm{x})\\\
s(K_{P}^{\mathrm{wt}};\bm{x})=\Phi_{P_{i<j}}(\bm{x})+\Phi_{P_{j<i}}(\bm{x})=\Phi_{P}(\bm{x})\end{split}$

completing the proof. ∎

Together with Proposition 6.14, one now has reason to believe that the cone
and semigroup perspective of [9] and [21] will bear fruit in the signed poset
case.

######  Corollary 13.2.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset. Both
$\Psi_{P}$ and $\Phi_{P}$ (resp.
$\Psi^{\scriptscriptstyle\vee}_{P^{\scriptscriptstyle\vee}}$ and
$\Phi^{\scriptscriptstyle\vee}_{P^{\scriptscriptstyle\vee}}$) vanish when
$\widehat{G}(P)$ has an isotropic component or, equivalently, $\Sigma_{P}$ has
a balanced component.

######  Corollary 13.3.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset such that
$K_{P}^{\mathrm{rt}}$ and $K_{P}^{\mathrm{wt}}$ are pointed, full-dimensional
and simplicial. Then

$\displaystyle\Psi_{P}(\bm{x})$
$\displaystyle=\dfrac{1}{2}\Psi^{\scriptscriptstyle\vee}_{P^{\scriptscriptstyle\vee}}(\bm{x})\stackrel{{\scriptstyle(1)}}{{=}}|u_{1}\wedge\cdots\wedge
u_{n}|\prod_{i=1}^{n}\langle\bm{x},u_{u}\rangle^{-1}\stackrel{{\scriptstyle(3)}}{{=}}2^{k}\prod_{i=1}^{n}\langle\bm{x},u_{u}\rangle^{-1}\text{
and }$ $\displaystyle\Phi_{P}(\bm{x})$
$\displaystyle=2\Psi^{\scriptscriptstyle\vee}_{P^{\scriptscriptstyle\vee}}(\bm{x})\stackrel{{\scriptstyle(2)}}{{=}}|J_{1}\wedge\cdots\wedge
J_{n}|\prod_{i=1}^{n}\langle\bm{x},J_{i}\rangle^{-1}\stackrel{{\scriptstyle(4)}}{{=}}2^{n-k}\prod_{i=1}^{n}\langle\bm{x},J_{i}\rangle^{-1},$

where $u_{1},\ldots,u_{n}$ are the extreme rays of $K_{P}^{\mathrm{rt}}$,
$J_{1},\ldots,J_{n}$ are the extreme rays of $K_{P}^{\mathrm{wt}}$ and $k$ is
the number of (non-loop) cycles in the Hasse diagram of $P$.

###### Proof.

Equalities ($1$) and ($2$) are consequences of Corollary 12.21. Equality ($3$)
is the result of combining Theorem 13.1 with [64, Lemma 8A.2]. Recalling that
$K_{P}^{\mathrm{rt}}$ and $K_{P}^{\mathrm{wt}}$ are dual cones and the
definition of coroot (see Definitions 7.3 and 7.4), one sees that

$(u_{i})(J_{i})^{\top}=2I_{n},$

since $\langle u_{i},J_{k}\rangle=2\delta_{ik}$ (for some indexing of the
$u_{i}$ and $J_{k}$). Therefore,

$|J_{1}\wedge\cdots\wedge J_{n}|=\dfrac{2^{n}}{|u_{1}\wedge\cdots\wedge
u_{n}|},$

giving equality ($4$). ∎

Corollary 13.3 is an analogue of [9, Proposition 3.2(1)], which shows that for
a poset, the numerator of $\Psi_{P}$ was $1$ for a tree and $0$ for a forest,
covering the two cases where $K_{P}^{\mathrm{rt}}$ is simplicial, full-
dimensional and pointed. However, instead of being an indicator of
connectedness, it counts the number of cycles in the Hasse diagram.

For example, consider the poset $P$ whose Hasse diagram and Fischer poset are
shown in Figure 17.

$1$$2$$3$$+$$-$$-$$+$$+$$+$ (a) Hasse diagram

$1$$3$$-2$$2$$-1$$-3$ (b) $\widehat{G}(P)$

Figure 17: Hasse diagram and $\widehat{G}(P)$ for
$P=\\{e_{1}+e_{2},e_{1}-e_{2},e_{1}+e_{3},e_{1},e_{3})\\}$

The extreme rays of $K_{P}^{\mathrm{rt}}$ are $(1,-1,0),(0,-1,1),(1,0,1)$ and
the extreme rays of $K_{P}^{\mathrm{wt}}$ are $(1,1,1),(1,-1,-1),(-1,-1,1)$.
Applying Proposition 5.5, one has that

$\Psi_{P}(\bm{x})=\left|\det\begin{pmatrix}1&0&1\\\ -1&-1&0\\\
0&1&1\end{pmatrix}\right|\dfrac{1}{(x_{1}-x_{2})(x_{3}-x_{2})(x_{1}+x_{3})}=\dfrac{2}{(x_{1}-x_{2})(x_{3}-x_{2})(x_{1}+x_{3})},$

and

$\Phi_{P}(\bm{x})=\left|\det\begin{pmatrix}1&-1&1\\\ -1&-1&1\\\
-1&1&1\end{pmatrix}\right|\dfrac{1}{(x_{1}+x_{2}+x_{3})(x_{1}-x_{2}-x_{3})(-x_{1}-x_{2}+x_{3})}\\\
=\dfrac{4}{(x_{1}+x_{2}+x_{3})(x_{1}-x_{2}-x_{3})(-x_{1}-x_{2}+x_{3})}.$

On the other hand, the second half of Corollary 13.3 is a generalization of
Theorem 1.2, which computed $\Phi_{P}$ for forests. For example, consider the
forest $F$ in Figure 18(a) and its embedding as a signed poset in Figure
18(b).

$1$$2$$3$$4$ (a) A forest $F$

$1$$2$$4$$3$$+$$-$$-$$+$$+$$+$ (b) $F$ embedded as a signed poset, $F_{B}$

Figure 18: A forest and its embedding as a signed poset

Embedding a poset as a signed poset preserves its ideals and linear extensions
and therefore preserves $\Phi$. In particular, this means that $\det(J_{i})$,
where $J_{i}$ runs over the connected non-extensible ideals (i.e. all
connected ideals in this case) will be $\pm 1$, so

$\Phi_{F}(\bm{x})=\Phi_{F_{B}}(\bm{x})=\left|\det\begin{pmatrix}1&0&0&0\\\
1&1&0&0\\\ 0&0&1&0\\\
0&0&1&1\end{pmatrix}\right|\dfrac{1}{x_{1}x_{3}(x_{1}+x_{2})(x_{3}+x_{4})}=\prod_{j\in
F}\dfrac{1}{\sum_{i<_{F}j}x_{i}}.$

##  Chapter The Root Cone Semigroup

This chapter considers $K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}$, the
semigroup associated to the root cone in types $B$ and $C$. After discussion
of the generators, Section 14 will consider the toric ideal of an oriented
signed graph, but in a more general setting from which one can recover the
signed poset, digraph and poset and graph cases. Section 15 will give a
sufficient condition for the semigroup ring to be a complete intersection,
enabling the computation of $\Psi_{P}(\bm{x})$ and
$\Psi^{\scriptscriptstyle\vee}_{{P^{\scriptscriptstyle\vee}}}(\bm{x})$ via
Proposition 6.14.

Recall that $\widehat{G}(P)$ is equipped with an involution, $\iota$, sending
$i$ to $-i$. This involution will make repeated appearances throughout this
chapter.

######  Proposition 13.1.

The semigroups $K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{B}$ and
$K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{C}$ are generated by the elements of
$P$ and ${P^{\scriptscriptstyle\vee}}$ corresponding to orbits of edges of
$\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$,
respectively, under the involution.

###### Proof.

In both cases, it suffices to show that every element of $P$ and
${P^{\scriptscriptstyle\vee}}$ lies in the semigroup generated by the orbits
of edges of $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. First, suppose $\pm e_{i}\pm
e_{j}\in P$ with $i\neq j$. It is also an element of
${P^{\scriptscriptstyle\vee}}$. Then, by definition, $\pm i<\mp j$ in both
$\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.
Summing the elements corresponding to edges (see Table 1) along a chain from
$i$ to $-j$ gives $\pm e_{i}\pm e_{j}$ in each case. Next, suppose $\pm
e_{i}\in P$. Then $\pm 2e_{i}\in{P^{\scriptscriptstyle\vee}}$. By definition,
$\pm i<0<\mp i$ in $\widehat{G}_{B}(P)$. Summing the elements corresponding to
edges along a chain from $\pm i$ to $0$ gives $\pm e_{i}$. On the other hand,
since $\pm 2e_{i}\in{P^{\scriptscriptstyle\vee}}$, by definition one has $\pm
i<\mp i$ in $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ and adding the
elements corresponding to edges along a chain from $\pm i$ to $\mp i$ gives
$\pm 2e_{i}$. ∎

This generating set is in fact minimal. All edges of $\widehat{G}_{B}(P)$
correspond to edges of the Hasse diagram of $P$, which in turn correspond to
the extreme rays of the root cone. It may be that
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ has “extra” edges relative to
the Hasse diagram, namely $\delta i\to-\epsilon j$ and $\epsilon j\to-\delta
i$, should the Hasse diagram contain loops at $i$ and $j$. However, $\delta
e_{i}+\epsilon e_{j}$ is _not_ in the semigroup generated by $\delta 2e_{i}$
and $\epsilon 2e_{j}$, so the extra edge in the Fischer poset really does
correspond to a semigroup generator. Proposition 13.1 generalizes (via the
embedding from Section 11) the result in type $A$ (Propositions 5.1 and 7.1 of
[9]), where $K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{A}$ was generated by the
roots corresponding to the covering relations of a poset.

###  14 The Toric Ideal of an Oriented Signed Graph

Recall from Section 8 that one can view a signed poset as an oriented signed
graph. This section will consider the generators of the toric ideal associated
to an oriented signed graph and then use that viewpoint to understand the
generators of the toric ideals associated to posets, directed graphs, graphs
and signed posets. In other words, we will describe the toric ideals for _all_
affine semigroup rings in $k[t_{1}^{\pm 1},\ldots,t_{n}^{\pm 1}]$ generated by
monomials of the form $1,t_{i}^{\pm 1},t_{i}^{\pm 2},t_{i}^{\pm 1}t_{j}^{\pm
1}$.

Suppose $\Sigma$ is a signed graph and $\tau$ a bidirection of $\Sigma$. (Here
$\Sigma$ is allowed to have multiple edges as well as self-loops and half
edges.) The vectors $\tau(i,e)e_{i}+\tau(j,e)e_{j}$ as $e=(i,j)$ runs over the
edges of $\Sigma$ (with a half-edge $e=(i,-)$ giving $\tau(i,e)e_{i}$)
generate a semigroup contained in ${\mathbb{Z}}^{|V(\Sigma)|}$.

######  Definition 14.1.

Suppose $\Sigma$ is a signed graph and $\tau$ a bidirection of $\Sigma$.
Define a polynomial ring $S_{\Sigma}=k[U_{e}]$ where $e$ runs over the
(bidirected) edges of $\Sigma$. The _toric ideal_ of $\Sigma$ is the kernel of
the map $\varphi\colon k[U_{e}]\to k[t_{1}^{\pm 1},\ldots,t_{n}^{\pm 1}]$
defined by

$\varphi(U_{e})=\begin{cases}t_{i}^{\tau(i,e)}t_{j}^{\tau(j,e)}&e=(i,j),\quad
i\neq j\\\ t_{i}^{\tau(i,e)}&e=(i,-)\text{ a half edge}\\\
t_{i}^{2\tau(i,e)}&e=(i,i)\text{ a negative loop}\\\ 1&e=(i,i)\text{ a
positive loop}\end{cases}$

The convention will be to put a bar over negative coordinates to save space.

As an example, consider the oriented signed graph in Figure 19.

$1$$2$$3$$+$$-$$+$$-$$+$$-$$+$$-$$-$$-$$+$ Figure 19: An oriented signed graph

In this case,

$S_{\Sigma}=k[U_{1},U_{2},U_{\bar{3}},U_{1\bar{2}},U_{2\bar{3}},U_{1\bar{3}}]$

and $\varphi$ is defined by

$\displaystyle U_{1}$ $\displaystyle\mapsto 1$ $\displaystyle U_{2}$
$\displaystyle\mapsto t_{2}$ $\displaystyle U_{\bar{3}}$ $\displaystyle\mapsto
t_{3}^{-2}$ $\displaystyle U_{1\bar{2}}$ $\displaystyle\mapsto
t_{1}t_{2}^{-1}$ $\displaystyle U_{2\bar{3}}$ $\displaystyle\mapsto
t_{2}t_{3}^{-1}$ $\displaystyle U_{1\bar{3}}$ $\displaystyle\mapsto
t_{1}t_{3}^{-1}$

The toric ideal $I_{\Sigma}$ is then
$(U_{1\bar{2}}U_{2\bar{3}}-U_{1\bar{3}},U_{2}^{2}U_{\bar{3}}-U_{2\bar{3}}^{2},U_{1}-1)$.
(See Figure 20 for an example of the computation with Macaulay2 [30] and 4ti2
[25].)

⬇

i1 : loadPackage "FourTiTwo";

A = transpose matrix {{1,0,0},{0,1,0},{0,0,-1},{1,-1,0},{0,1,-1},{1,0,-1}};

S = QQ[U1,U2,Um3,U1m2,U2m3,U1m3,Degrees=>entries transpose A];

I = toricGroebner(A,S)

3 6

o2 : Matrix ZZ <\--- ZZ

i4 : using temporary file name /tmp/M2-22805-3

\-------------------------------------------------

4ti2 version 1.3.2, Copyright (C) 2006 4ti2 team.

4ti2 comes with ABSOLUTELY NO WARRANTY.

This is free software, and you are welcome

to redistribute it under certain conditions.

For details, see the file COPYING.

\-------------------------------------------------

Using 64 bit integers.

4ti2 Total Time: 0.00 secs.

o4 = ideal (- U1*Um3 + U1m3, - U1 + U2*U1m2, - U2*Um3 + U2m3)

o4 : Ideal of S

Figure 20: Computing a toric ideal with Macaulay2

Understanding the generators of the toric ideal of an oriented signed graphs
requires the notion of the signed covering of an oriented signed graph.

######  Definition 14.2.

Given a signed graph $\Sigma=(\Gamma,\sigma)$ oriented by $\tau$, define a
directed graph $\widetilde{\Sigma}$ whose vertices are
$V(\Sigma)\times\\{\pm\\}$ and whose edges are determined by the edges of
$\Sigma$ according to Table 4. The graph $\widetilde{\Sigma}$ is the _signed
covering_ of $\Sigma$.

edge in $\Sigma$ | edge(s) in $\widetilde{\Sigma}$
---|---
$i$$j$$+$$-$ |  $i\to j$, $-j\to-i$
$i$$j$$+$$+$ |  $i\to-j$, $j\to-i$
$i$$j$$-$$-$ |  $-i\to j$, $-j\to i$
$i$$+$$+$ | $i\to-i$
$i$$-$$-$ | $-i\to i$
$i$$+$$-$ | self-loops at $i$, $-i$
$i$$+$ | half edges $i\to$, $\to-i$
$i$$-$ | half edges $-i\to$, $\to i$
Table 4: Correspondence between edges in $\Sigma$ and $\widetilde{\Sigma}$

When $\widetilde{\Sigma}$ contains half edges, it will be convenient to
imagine an extra vertex $0$ as the other end of the half edges. With this
extra vertex, when $P\subset\Phi_{B_{n}}$ is a signed poset,
$\widehat{G}_{B}(P)$ is the signed cover of a signed graph.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is also the signed cover of a
signed graph, with or without the dummy $0$, as there are no half edges. (See
Definition 8.4 for the definitions of $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.) Like $\widehat{G}(P)$, a
signed covering $\widetilde{\Sigma}$ is equipped with an involutive digraph
anti-automorphism sending $i$ to $-i$ (and fixing $0$). Figure 21 shows the
signed covering of the oriented signed graph in Figure 19.

$1$$2$$3$$-3$$-2$$-1$$0$ Figure 21: Signed covering of the oriented signed
graph in Figure 19

######  Definition 14.3.

Suppose $\Sigma$ is a signed graph and $\widetilde{\Sigma}$ is its signed
covering. Consider a cycle $C$ in $\widetilde{\Sigma}$ and orient it in some
way (i.e. choose a direction in which to traverse $C$). This partitions the
edges of $C$ into $W\sqcup A$, where $W$ is the set of edges such that the
orientation is consistent with the direction of the edge in
$\widetilde{\Sigma}$ and $A$ consists of the edges oriented opposite their
direction in $\widetilde{\Sigma}$. Say $C$ is _fixed orientation-wise_ by the
involution if $\iota(C)=C$ as edges and $W_{C}=W_{\iota(C)}$ and
$A_{C}=A_{\iota(C)}$.

For example, consider the cycle $3\to 1\to 2\to 0\to-2\to-1\to-3\to 3$ in the
graph in Figure 21. Then

$W_{C}=\\{(1,2),(2,0),(0,-2),(-2,-1),(-3,3)\\}\quad\text{and}\quad
A_{C}=\\{(3,1),(-1,-3)\\}.$

While this cycle is fixed _edgewise_ by the involution, $\iota(C)$ is oriented
in the opposite direction: $-3\to-1\to-2\to 0\to 2\to 1\to 3\to-3$. Thus
$W_{\iota(C)}=A_{C}$ and $A_{\iota(C)}=W_{C}$, so $C$ is _not_ fixed
orientation-wise. On the other hand, consider the signed covering in Figure 22
(which is actually $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ for
${P^{\scriptscriptstyle\vee}}=\\{+e_{1}-e_{2},+e_{1}+e_{2},+2e_{1}\\}$). It
has a single cycle $1\to 2\to-1\to-2\to 1$, which _is_ fixed orientation-wise
by the involution, with $W_{C}=W_{\iota(C)}=\\{(1,2),(2,-1)\\}$ and
$A_{C}=A_{\iota(C)}=\\{(1,-2),(-2,-1)\\}$.

$1$$2$$-2$$-1$ Figure 22: Another signed covering

######  Lemma 14.4.

Suppose $\widetilde{\Sigma}$ is the signed covering of an oriented signed
graph $\Sigma$. Suppose $C$ is a cycle in $\widetilde{\Sigma}$ fixed
orientation-wise by the involution. Then if $e\in W_{C}$ is an edge,
$\iota(e)\in A_{C}$.

###### Proof.

Suppose $e\in W_{C}$ and $\iota(e)\in W_{C}$. There are two cases, when $e$
involves $0$ and when it doesn’t. First, suppose $e$ does not have $0$ as an
endpoint. Then orienting $C$, one has

$\cdots\rightarrow i\stackrel{{\scriptstyle
e}}{{\rightarrow}}j\rightarrow\cdots\rightarrow-j\stackrel{{\scriptstyle\iota(e)}}{{\rightarrow}}-i\rightarrow\cdots$

Applying the involution gives

$\cdots\rightarrow-i\rightarrow-j\rightarrow\cdots\rightarrow j\rightarrow
i\rightarrow\cdots$

so $e,\iota(e)\in A_{\iota(C)}$, meaning $C$ was not fixed orientation-wise by
the involution.

On the other hand, suppose $e$ does have $0$ as an endpoint. Without loss of
generality, one may assume one has

$\cdots\rightarrow i\stackrel{{\scriptstyle
e}}{{\rightarrow}}0\stackrel{{\scriptstyle\iota(e)}}{{\rightarrow}}-i\rightarrow\cdots$

Applying the involution gives

$\cdots\rightarrow-i\rightarrow 0\rightarrow i\rightarrow\cdots,$

so $e,\iota(e)\in A_{\iota(C)}$, meaning $C$ was not fixed orientation-wise by
the involution. ∎

If $C$ is a cycle in $\widetilde{\Sigma}$, it gives rise to a relation in the
semigroup:

$\sum_{\alpha\in W(C)}\alpha=\sum_{\alpha\in A(C)}\alpha.$ (4)

Recall that $\varphi$ was the map from from $S_{\Sigma}=k[U_{e}]$ to the
Laurent polynomial ring sending $U_{ij}$ to
$t_{i}^{\tau(i,e)}t_{j}^{\tau(j,e)}$ for the edge $e=(i,j)$ (see Definition
14.1).

######  Theorem 14.5.

Suppose $\Sigma$ is an oriented signed graph. The toric ideal
$I_{\Sigma}=\ker\varphi$ is generated by _cycle binomials_

$U(C)=\prod_{e\in W(C)}U_{e}-\prod_{e\in A(C)}U_{e},$

where $C$ runs over the cycles of $\widetilde{\Sigma}$ not fixed by the
involution.

Consider the oriented signed graph in Figure 19, whose signed covering is in
Figure 21. Recall from the start of this section that the toric ideal is

$I_{\Sigma}=(U_{1\bar{2}}U_{2\bar{3}}-U_{1\bar{3}},U_{2}^{2}U_{\bar{3}}-U_{2\bar{3}}^{2},U_{1}-1).$

The signed covering $\widetilde{\Sigma}$ has five orbits of cycles not fixed
orientation-wise by the involution. Their cycle binomials are:

$U_{1}-1,\quad U_{1\bar{2}}U_{2\bar{3}}-U_{1\bar{3}},\quad
U_{1\bar{2}}^{2}U_{2}^{2}U_{\bar{3}}-U_{1\bar{3}}^{2},\quad
U_{2}^{2}U_{\bar{3}}-U_{2\bar{3}}^{2},\quad
U_{1\bar{2}}U_{2}^{2}U_{\bar{3}}-U_{2\bar{3}}U_{1\bar{3}}.$

One needs to check that these five binomials generate $I_{\Sigma}$ and not
some larger ideal. One can use Macaulay2 to find the following relations which
show that the five cycle binomials do generate $I_{\Sigma}$ by writing the two
“extra” generators in terms of the other three generators:

$\begin{split}U_{2\bar{3}}(U_{1\bar{2}}U_{2\bar{3}}-U_{1\bar{3}})+U_{1\bar{2}}(U_{\bar{3}}U_{2}^{2}-U_{2\bar{3}}^{2})&=U_{1\bar{2}}U_{2}^{2}U_{\bar{3}}-U_{2\bar{3}}U_{1\bar{3}}\\\
(-U_{1\bar{2}}U_{2\bar{3}}-U_{1\bar{3}})(U_{1\bar{3}}-U_{1\bar{2}}U_{2\bar{3}})+U_{1\bar{2}}^{2}(U_{\bar{3}}U_{2}^{2}-U_{2\bar{3}}^{2})&=U_{1\bar{2}}^{2}U_{\bar{3}}U_{2}^{2}-U_{1\bar{3}}^{2}.\end{split}$

######  Proposition 14.6.

Suppose $C$ is a cycle in $\widetilde{\Sigma}$ fixed orientation-wise by the
involution. Then its cycle binomial is zero.

###### Proof.

_A priori_ , one has that $\iota(W_{C})=A_{\iota(C)}$ and
$\iota(A_{C})=W_{\iota(C)}$. If a cycle $C$ is fixed by the involution, one
has that $W_{C}=W_{\iota(C)}$ by definition, but this means
$W_{\iota(C)}=\iota(A_{\iota(C)})$ so the two sums in (4) are over the same
set, meaning the cycle binomial is $0$. ∎

It will occasionally be convenient to think of _all_ cycle binomials as
generating the toric ideal. This is acceptable since the proposition shows the
extra cycle binomials are $0$.

###### Proof of Theorem 14.5.

Toric ideals are generated by binomials $U^{\alpha}-U^{\beta}$ such that
$\alpha$ and $\beta$ have disjoint support. (See [58, Corollary 4.3 ].) One
must then show that if such a $U^{\alpha}-U^{\beta}$ is in $I_{\Sigma}$, it is
in the ideal generated by the cycle binomials of $\widetilde{\Sigma}$.

Suppose $U^{\alpha}-U^{\beta}$ is in $I_{\Sigma}$ and $U^{\alpha}$ and
$U^{\beta}$ have disjoint support. The binomial corresponds to a relation

$\sum a_{\delta i,\epsilon j}(\delta e_{i}-\epsilon e_{j})=\sum b_{\delta
i\epsilon j}(\delta e_{i}-\epsilon e_{j}),$

where $\delta,\epsilon\in\\{\pm\\}$ and, for ease of notation, one takes
$e_{0}=0$. One builds a digraph on vertex set $\pm[n]$ as follows. For each
term $\delta e_{i}-\epsilon e_{j}$ in the left hand sum, take $a_{\delta
i,\epsilon j}$ copies of the directed edge $\delta i\to\epsilon j$ and
$a_{\delta i,\epsilon j}$ copies of $-\epsilon j\to-\delta i$. For each term
$\delta e_{i}-\epsilon e_{j}$ in the right hand sum, take $b_{\delta
i,\epsilon j}$ copies of $\delta i\leftarrow\epsilon j$ and $b_{\delta
i,\epsilon j}$ copies of $-\epsilon j\leftarrow-\delta i$. In other words, one
gets $a_{\delta i,\epsilon j}$ edges with orientation coinciding with that of
$\widetilde{\Sigma}$ and $b_{\delta i,\epsilon j}$ edges with orientation
opposite that of $\widetilde{\Sigma}$.

One has constructed a digraph where the in-degree equals the out-degree at
each vertex, so one has a collection of cycles, possibly with multiplicity. By
construction, each edge and its image under the involution are oriented the
same way relative to their direction in $\widetilde{\Sigma}$. Consequently, by
Lemma 14.4, none of these cycles can be fixed orientation-wise by the
involution. Furthermore, if $C$ is a cycle fixed _edgewise_ by the involution,
one will have an even number of copies of $C$.

Call these pairs of cycles (possibly with multiplicity) $C_{1},\ldots,C_{k}$.
Induct on $k$. If $k=1$, then $U^{\alpha}-U^{\beta}$ was itself a cycle
binomial. Suppose that binomials with no common factors corresponding to $k-1$
pairs of cycles lie in the ideal generated by the cycle binomials. Then by
induction one has

$U^{\alpha}-U^{\beta}=\prod_{i=1}^{k}U_{W(C_{i})}-\prod_{i=1}^{k}U_{A(C_{i})}\\\
=(U_{W(C_{1})}-U_{A(C_{1})})\prod_{i=2}^{k}U_{W(C_{i})}+U_{A(C_{1})}\left(\prod_{i=2}^{k}U_{W(C_{i})}-\prod_{i=2}^{k}U_{A(C_{k})}\right),$

where $U_{W(C_{i})}=\prod_{e\in W(C_{i})}U_{e}$ and $U_{A(C_{i})}=\prod_{e\in
A(C_{i})}$. Consequently, $U^{\alpha}-U^{\beta}$ lies in the ideal generated
by the cycle binomials. Thus one has that the toric ideal is generated by the
cycle binomials. ∎

####  14.1 The toric ideal of the root cone semigroup

As noted earlier, examining the definitions of $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ (Definition 8.4), one sees
that $\widehat{G}_{B}(P)$ and $\widehat{G}_{C}(P)$ are both the signed
coverings of some oriented signed graph (at least when one anchors the half
edges to a $0$ in the case of $\widehat{G}_{B}(P)$). Recall from Proposition
13.1 that the semigroups $K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{B}$ and
$K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{C}$ are generated by the edges of
these oriented signed graphs. Denote the two semigroup rings
$k[K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{B}]$ and
$k[K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{C}]$ by $R_{P}^{\mathrm{rt}}$ and
$R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}$, respectively. In this case,
Theorem 14.5 gives the following.

######  Corollary 14.7.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset. Let
$S^{\mathrm{rt}}_{P}=k[U_{\delta i,\epsilon j}]$ and
$S^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}=k[U_{\delta i,\epsilon j}]$,
where the pairs $(\delta i,\epsilon j)$ run over orbits under the involution
of edges $\delta i\to\epsilon j$ in $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, respectively. Then the toric
ideals $I^{\mathrm{rt}}_{P}$ and
$I^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}$ are generated by the cycle
binomials corresponding to cycles in $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, respectively, not fixed
orientation-wise by the involution and

$R_{P}^{\mathrm{rt}}=S^{\mathrm{rt}}_{P}/I^{\mathrm{rt}}_{P}\quad\text{and}\quad
R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}=S^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}/I^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}.$

$1$$4$$0$$-4$$-1$$2$$-3$$3$$-2$ (a) $\widehat{G}_{B}(P)$

$1$$4$$-4$$-1$$2$$-3$$3$$-2$ (b)
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$

Figure 23: $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ for
$P=\\{+e_{1}-e_{2},+e_{1}+e_{3},+e_{1}-e_{3},+e_{1}+e_{2},+e_{1}-e_{4},+e_{1}+e_{4},+e_{1},+e_{4}\\}$

Consider
$P=\\{+e_{1}-e_{2},+e_{1}+e_{3},+e_{1}-e_{3},+e_{1}+e_{2},+e_{1}-e_{4},+e_{1}+e_{4},+e_{1},+e_{4}\\}$.
Figure 23 shows $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. Then

$S^{\mathrm{rt}}_{P}=k[U_{12},U_{23},U_{1\bar{3}},U_{14},U_{4\bar{4}}]\quad\text{and}\quad
S^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}=k[U_{12},U_{23},U_{1\bar{3}},U_{14},U_{40}].$

Both $\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$
have a cycle $1\to 2\to 3\to-1\to-2\to-3\to 1$ fixed orientation-wise by the
involution, so its cycle binomial is $0$. Figure 24 shows the cycles
constructed in the proof of Theorem 14.5.

$1$$4$$0$$-4$$-1$$2$$-3$$3$$-2$ (a) $\widehat{G}_{B}(P)$

$1$$4$$-4$$-1$$2$$-3$$3$$-2$ (b)
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$

Figure 24: The directed (multi-)graph from the proof of Theorem 14.5

In both $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, there is a single pair
$(C,\iota(C))$, so the toric ideals are principal. One then has

$I^{\mathrm{rt}}_{P}=(U_{12}U_{23}U_{1\bar{3}}-U_{14}^{2}U_{40}^{2})\quad\text{and}\quad
I^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}=(U_{12}U_{23}U_{1\bar{3}}-U_{14}^{2}U_{4}).$

####  14.2 Toric ideals of posets and directed graphs

Another application of Theorem 14.5 is to posets and directed graphs. The
semigroup associated to a directed graph is generated by $e_{i}-e_{j}$ for
each edge $i\to j$. When the directed graph is the Hasse diagram of a poset,
this is the root cone semigroup in type A of [9]. Using a similar argument to
the proof of Theorem 14.5 (but simpler due to having the Hasse diagram rather
than the signed covering), Boussicault, Féray, Lascoux and Reiner showed in
[9, Proposition 8.1] that, given a poset, its toric ideal is generated by
binomials corresponding to cycles in the Hasse diagram. Their argument easily
generalizes to all directed graphs. Earlier, Gitler, Reyes and Villarreal
obtained the same result for directed graphs [27, Proposition 4.3] using a
result of Sturmfels [57, §5].

To see how Theorem 14.5 works for directed graphs, first observe that in a
signed graph $\Sigma$ with an orientation coming from a directed graph $G$,
all edges are signed $+$, meaning all cycles are balanced. Furthermore, the
signed covering of $\Sigma$ is $G\sqcup\iota(G)$ and a cycle in $G$
corresponds to a pair of cycles $(C,\iota(C))$ in $G\sqcup\iota(G)$, and no
cycle in the signed covering is fixed orientation-wise by the involution.
Consequently, the generating set from Theorem 14.5 is the same as the
generating set from [9] and [27].

As an example, consider the digraph (and poset, depending on perspective)
shown in Figure 25(a). The associated oriented signed graph is shown in Figure
25(b) and its signed covering in Figure 25(c).

|
---|---

$1$$2$$3$$4$$5$$6$ (a) digraph/poset

$2$$4$$6$$1$$3$$5$$+$$-$$+$$-$$+$$-$$+$$-$$+$$-$$+$$-$$+$$-$ (b) associated
oriented signed graph

$1$$2$$3$$4$$5$$6$$-2$$-1$$-3$$-4$$-5$$-6$ (c) signed covering

Figure 25: A digraph, its associated oriented signed graph and signed covering

Then
$S_{G}=k[U_{1\bar{2}},U_{1\bar{3}},U_{2\bar{4}},U_{3\bar{4}},U_{3\bar{5}},U_{4\bar{6}},U_{5\bar{6}}]$
and, according to Theorem 14.5, the toric ideal is

$I_{G}=(U_{1\bar{2}}U_{2\bar{4}}-U_{1\bar{3}}U_{3\bar{4}},U_{3\bar{4}}U_{4\bar{6}}-U_{3\bar{5}}U_{5\bar{6}},U_{1\bar{2}}U_{2\bar{4}}U_{4\bar{6}}-U_{1\bar{3}}U_{3\bar{5}}U_{5\bar{6}}).$

####  14.3 Toric ideals of graphs

Suppose $G$ is a (unsigned) graph with no loops or multiple edges. Associated
to $G$ is the semigroup generated by the vectors $e_{i}+e_{j}$ where $(i,j)$
runs over all edges of $G$. This semigroup then gives rise to a toric ideal
$I_{G}$. Of course, this is the same semigroup as one obtains from an oriented
signed graph $\Sigma$ where $G$ is $\Sigma$’s underlying graph, all edges are
signed $-$ and $\Sigma$ is oriented with all incidences $+$. Consequently,
Theorem 14.5 gives a set of generators for $I_{G}$. This generating set lies
between two other known generating sets.

######  Definition 14.8.

A sequence $v_{0},\ldots,v_{n}$ of (not necessarily distinct) vertices in a
graph $G$ is a _closed walk_ if $v_{0}=v_{n}$ and there is an edge between
$v_{i}$ and $v_{i+1}$ for $i=0,\ldots,n-1$. It is an _even closed walk_ if $n$
is even. Let $S_{G}=k[U_{e}]$ where $e$ runs over the edges of $G$. Associated
to an even closed walk $w=\\{v_{0},\ldots,v_{2k}\\}$ is a binomial

$T_{w}=U_{v_{0}v_{1}}U_{v_{2}v_{3}}\cdots
U_{v_{2k-2}v_{2k-1}}-U_{v_{1}v_{2}}U_{v_{3}v_{4}}\cdots U_{v_{2k-1}v_{2k}}$

######  Theorem 14.9 (Villarreal, [61, Proposition 3.1]).

Suppose $G$ is a graph. Then its toric ideal has this description:

$I_{G}=(T_{w}\colon w\text{ an even closed walk in }G).$

This generating set and the generating set from Theorem 14.5 do not
necessarily coincide. Consider the graph in Figure 26. According to Theorem
14.9, the toric ideal is

$(U_{12}U_{13}U_{45}-U_{23}U_{14}U_{15},U_{18}U_{67}-U_{16}U_{78},U_{12}U_{13}U_{45}U_{18}U_{67}-U_{23}U_{14}U_{15}U_{78}U_{16}).$

On the other hand, according to Theorem 14.5, it is

$(U_{12}U_{13}U_{45}-U_{23}U_{14}U_{15},U_{18}U_{67}-U_{16}U_{78}).$

One has that

$U_{12}U_{13}U_{45}(U_{67}U_{18}-U_{16}U_{78})-U_{16}U_{78}(U_{23}U_{14}U_{15}-U_{12}U_{13}U_{45})\\\
=U_{12}U_{13}U_{45}U_{18}U_{67}-U_{23}U_{14}U_{15}U_{78}U_{16},$

so the two ideals really are equal.

$1$$6$$5$$4$$3$$2$$8$$7$ (a) A graph $G$

$-2$$2$$-3$$3$$-4$$4$$-5$$5$$-6$$6$$-7$$7$$-8$$8$$1$$-1$ (b) The signed
covering of $G$

Figure 26: A graph and its signed covering

Since the signed covering is bipartite, any cycle will correspond to an even
closed walk in $G$. Therefore, the generating set of Theorem 14.5 is contained
in the generating set of Theorem 14.9.

######  Definition 14.10.

If $I\subset k[x_{1},\ldots,x_{k}]$ is a toric ideal, a _primitive binomial_
is a binomial $\bm{x}^{u^{+}}-\bm{x}^{u^{-}}\in I$ such that there does not
exist an $\bm{x}^{v^{+}}-\bm{x}^{v^{-}}\in I$ with $u\neq v$ and
$\bm{x}^{v^{+}}|\bm{x}^{u^{+}}$ and $\bm{x}^{v^{-}}|\bm{x}^{u^{-}}$. The set
of primitive binomials is called the _Graver basis_ and generates the toric
ideal.

See [58] for more on the Graver basis. Ohsugi and Hibi described the Graver
basis of the toric ideal of a graph in [46].

######  Theorem 14.11 (Ohsugi and Hibi, [46, Lemma 3.2]).

Let $G$ be a finite connected graph. If $f\in I_{G}$ is primitive, it is
$f=T_{\Gamma}$ where $\Gamma$ is one of the following even closed walks:

  1. (i)

$\Gamma$ is an even cycle

  2. (ii)

$\Gamma$ is a pair of odd cycles having exactly one common vertex

  3. (iii)

$\Gamma=(C_{1},\Gamma_{1},C_{2},\Gamma_{2})$ where $C_{1}$ and $C_{2}$ are odd
cycles having no common vertex and where $\Gamma_{1}$ and $\Gamma_{2}$ are
walks of $G$, both of which connect a vertex $v_{1}$ in $C_{1}$ to a vertex
$v_{2}$ of $C_{2}$.

A moment’s thought will show that an even closed walk $\Gamma$ corresponding
to a primitive binomial corresponds to a cycle not fixed orientation-wise by
the involution. However, the converse is not true. Consider the graph in
Figure 27. The cycle $1\to-2\to 3\to-1\to 2\to-4\to 5\to-6\to 4\to-3\to 1$ in
the signed covering is not fixed orientation-wise by the involution, but does
not project to an even closed walk of any of the forms in Theorem 14.11. The
binomial corresponding to this cycle is
$U_{12}U_{13}U_{24}U_{56}U_{34}-U_{23}U_{12}U_{45}U_{46}U_{13}$, which is
clearly not primitive as the two terms are not relatively prime. Cancelling
$U_{12}U_{13}$ from both terms gives the primitive binomial
$U_{24}U_{56}U_{34}-U_{23}U_{45}U_{46}$, corresponding to the even closed walk
$2\to 4\to 6\to 5\to 4\to 3\to 2$.

$1$$2$$3$$4$$6$$5$ (a) $G$

$1$$-2$$3$$-1$$2$$-3$$4$$-5$$6$$-4$$5$$-6$ (b) Signed covering of $G$

Figure 27: A graph for which the generating sets from Theorem 14.5 and Theorem
14.11 differ

The following proposition summarizes the results of this section, linking the
generating sets of Theorems 14.5, 14.9, and 14.11.

######  Proposition 14.12.

Suppose $G$ is a finite connected graph. There are then three generating sets,
$S_{1}\subset S_{2}\subset S_{3}$ for the toric ideal $I_{G}$:

  * •

$S_{1}=\\{T_{w}\colon w\text{ an even closed walk in the forms of
Theorem~{}\ref{thm:ohsugihibi}}\\}$

  * •

$S_{2}=\\{U(C)\colon C\text{ a cycle in the signed covering}\\}$

  * •

$S_{3}=\\{T_{w}\colon w\text{ an even closed walk}\\}$.

###  15 Some Complete Intersection Rings $R_{P}^{\mathrm{rt}}$

Now that one understands the toric ideal, one can turn one’s attention to the
question of computing $\Psi_{P}$ and $\Psi_{P}^{\scriptscriptstyle\vee}$ via
complete intersection presentations. This section gives factorizations of

$\displaystyle\Psi_{P}(\bm{x})$
$\displaystyle=\sum_{w\in\mathcal{L}(P)}w\left(\dfrac{1}{(x_{1}-x_{2})(x_{2}-x_{3})\cdots(x_{n-1}-x_{n})x_{n}}\right)\quad\text{and}\quad$
$\displaystyle\Psi^{\scriptscriptstyle\vee}_{{P^{\scriptscriptstyle\vee}}}(\bm{x})$
$\displaystyle=\sum_{w\in\mathcal{L}(P)}w\left(\dfrac{1}{(x_{1}-x_{2})(x_{2}-x_{3})\cdots(x_{n-1}-x_{n})2x_{n}}\right)$

for a certain class of signed posets using Proposition 6.14.

Recall Theorem 1.1, where Greene showed that, for a strongly planar poset,

$\Psi_{P}(\bm{x})=\dfrac{\prod_{\rho}(x_{\min(\rho)}-x_{\max(\rho)})}{\prod_{i\lessdot
j}(x_{i}-x_{j})},$

where $\rho$ runs over all the regions enclosed by the Hasse diagram of $P$
and $i\lessdot j$ over the covering relations of the poset. [8] showed that a
similar factorization occurs for posets that are “gluings of diamonds along
chains” and, in particular, for strongly planar posets, recovering Greene’s
result. In [9], Boussicault, Féray, Lascoux and Reiner gave an algebraic
explanation for the disconnecting chains of [8] by showing certain posets
(including strongly planar posets) are complete intersections by constructing
a regular sequence using the operation of opening/closing a notch.

The situation for signed posets is quite similar. We will construct a regular
sequence by opening/closing signed notches, culminating in Theorem 15.12.

######  Theorem 15.12.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset such that
$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$)
has an embedding in ${\mathbb{R}}^{2}$ which is strongly planar. Then

  * •

$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$)
has an embedding that is both centrally symmetric and strongly planar,

  * •

$R_{P}^{\mathrm{rt}}$ (resp. $R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}$)
is a complete intersection, with $I_{P}$ (resp.
$I_{{P^{\scriptscriptstyle\vee}}}$) generated by the cycle binomials of the
cycles in $\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}(P)$) defining the faces
of the graph, and

  * •

one has

$\displaystyle\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})$
$\displaystyle=\dfrac{\prod_{\rho}(1-\bm{x}^{\rho})}{\prod_{e}(1-x_{a}^{\delta}x_{b}^{-\epsilon})}\quad\text{and}$
$\displaystyle\mathrm{Hilb}(R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}},\bm{x})$
$\displaystyle=\dfrac{\prod_{\rho}(1-\bm{x}^{\rho})}{\prod_{e}(1-x_{a}^{\delta}x_{b}^{-\epsilon})},$

where $\rho$ runs over all regions enclosed by $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}(P)$) not fixed by the involution, $e$ runs over the orbits of
edges $\delta a\to\epsilon b$ ($a<b$) in $\widehat{G}_{C}(P)$, $x_{0}$ is
taken to be $1$ and

$\bm{x}^{\rho}=x_{\min(\rho)}^{\mathrm{sgn}(\min(\rho))}x_{\max(\rho)}^{-\mathrm{sgn}(\max(\rho))}.$

The next several sections will build up to the proof of Theorem 15.12 in
Section 15.3, where what it means for a signed poset to be strongly planar
will be defined.

  * •

Section 15.1 will reduce to the case of signed posets consisting of a single
biconnected component (see Definition 15.1 and Proposition 15.2) and whose
root cones are full-dimensional (see Propositions 15.3 and 15.4).

  * •

Section 15.2 defines the notion of a signed notch (see Definition 15.5),
explains what it means to open/close a signed notch and what impact that has
on the root cone semigroup ring (see Proposition 15.9).

  * •

Section 15.3 defines strongly planar posets (see Definition 15.10), proves a
number of propositions and lemmas regarding strongly planar posets and then
proves Theorem 15.12.

  * •

After the proof, Section 15.4 will explain how Theorem 15.12 can be used to
compute $\Psi_{P}$ and
$\Psi^{\scriptscriptstyle\vee}_{P^{\scriptscriptstyle\vee}}$.

####  15.1 A few reductions

The first reduction will be to the case of signed posets consisting of a
single biconnected component.

######  Definition 15.1.

Let $P\subset\Phi_{B_{n}}$ and
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$ be signed posets. Let
$A\subset P$ (resp. ${P^{\scriptscriptstyle\vee}}$) be the elements
corresponding to the edges of $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$). Say two elements of $A$ are
_cycle equivalent_ if there is a cycle in $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}(P)$) not fixed orientation-wise by the involution passing
through an edge corresponding to each element. Taking the transitive closure
gives an equivalence relation. Combine equivalence classes lying in the same
orbit of the involution. This partitions the edges of $\widehat{G}_{B}(P)$
(rep. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) into the _biconnected
components_ of $P$ (resp. ${P^{\scriptscriptstyle\vee}}$). Each biconnected
component corresponds to a signed poset, which will also be called the
biconnected components of $P$ (resp. ${P^{\scriptscriptstyle\vee}}$).

Consider
$P=\\{+e_{1}-e_{2},+e_{1}+e_{2},+e_{1}-e_{3},+e_{1}+e_{3},+e_{1}-e_{4},+e_{1}-e_{5},+e_{1}-e_{6},+e_{2}-e_{3},+e_{2}-e_{4},+e_{2}-e_{5},+e_{2}-e_{6},+e_{4}-e_{6},+e_{5}-e_{6},+e_{1}\\}$.
Figure 28 shows $\widehat{G}(P)$ with the edges of one biconnected component
solid and the edges of the other dashed.

$1$$2$$3$$-3$$-2$$-1$$5$$4$$6$$-5$$-4$$-6$$0$ (a) $\widehat{G}_{B}(P)$

$1$$2$$3$$-3$$-2$$-1$$5$$4$$6$$-5$$-4$$-6$ (b)
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$

Figure 28: $\widehat{G}(P)$ for
$P=\\{+e_{1}-e_{2},+e_{1}+e_{2},+e_{1}-e_{3},+e_{1}+e_{3},+e_{1}-e_{4},+e_{1}-e_{5},+e_{1}-e_{6},+e_{2}-e_{3},+e_{2}-e_{4},+e_{2}-e_{5},+e_{2}-e_{6},+e_{4}-e_{6},+e_{5}-e_{6},+e_{1}\\}$
with biconnected components indicated by dashed and solid lines.

######  Proposition 15.2.

Let $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) be a signed poset and
$P_{1},\ldots,P_{k}$ (resp.
$P^{\scriptscriptstyle\vee}_{1},\ldots,P^{\scriptscriptstyle\vee}_{k}$) its
biconnected components. Then

$R_{P}^{\mathrm{rt}}\cong R_{P_{1}}^{\mathrm{rt}}\otimes_{k}\cdots\otimes
R_{P_{k}}^{\mathrm{rt}}\quad\text{and}\quad
R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}\cong
R_{P^{\scriptscriptstyle\vee}_{1}}^{\mathrm{rt}}\otimes_{k}\cdots\otimes_{k}R_{P^{\scriptscriptstyle\vee}_{k}}^{\mathrm{rt}}$

and, as a consequence,

$\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})=\prod_{i=1}^{k}\mathrm{Hilb}(R_{P_{i}}^{\mathrm{rt}},\bm{x})\quad\text{and}\quad\mathrm{Hilb}(R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}},\bm{x})=\prod_{i=1}^{k}\mathrm{Hilb}(R_{P^{\scriptscriptstyle\vee}_{i}}^{\mathrm{rt}},\bm{x}).$

###### Proof.

Recall that $R_{P}^{\mathrm{rt}}=S_{P}/I_{P}$ and
$R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}=S_{{P^{\scriptscriptstyle\vee}}}/I_{{P^{\scriptscriptstyle\vee}}}$,
where $I_{P}$ and $I_{{P^{\scriptscriptstyle\vee}}}$ are generated,
respectively, by the cycle binomials of cycles of $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ not fixed by the involution.
Every edge in $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ lies in a unique biconnected
component and is thus associated to a unique $P_{i}$ (resp.
$P^{\scriptscriptstyle\vee}_{i}$). Consequently,
$S_{P}\cong\bigotimes_{\ell=1}^{k}S_{P_{\ell}}$ and
$S_{P^{\scriptscriptstyle\vee}}\cong\bigotimes_{\ell=1}^{k}S_{P^{\scriptscriptstyle\vee}_{\ell}}$.
Furthermore, each cycle not fixed by the involution lies wholly in a single
biconnected component. Consequently, $I_{P}=\bigoplus_{\ell=1}^{k}I_{P_{k}}$
and
$I_{{P^{\scriptscriptstyle\vee}}}=\bigoplus_{\ell=1}^{k}I_{P^{\scriptscriptstyle\vee}_{k}}$.
Then

$R_{P}^{\mathrm{rt}}\cong
S_{P}/I_{P}=S/\oplus_{\ell=1}^{k}I_{P_{k}}\cong\bigotimes_{\ell=1}^{k}S_{P_{\ell}}/I_{P_{\ell}}\cong
R_{P_{1}}^{\mathrm{rt}}\otimes_{k}\cdots\otimes_{k}R_{P_{k}}^{\mathrm{rt}}$

and

$R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}\cong
S_{P^{\scriptscriptstyle\vee}}/I_{{P^{\scriptscriptstyle\vee}}}=S/\oplus_{\ell=1}^{k}I_{P^{\scriptscriptstyle\vee}_{k}}\cong\bigotimes_{\ell=1}^{k}S_{P^{\scriptscriptstyle\vee}_{\ell}}/I_{P^{\scriptscriptstyle\vee}_{\ell}}\cong
R_{P^{\scriptscriptstyle\vee}_{1}}^{\mathrm{rt}}\otimes_{k}\cdots\otimes_{k}R_{P^{\scriptscriptstyle\vee}_{k}}^{\mathrm{rt}},$

as claimed. ∎

For the remainder of this chapter, it will be assumed that all signed posets
consist of a single biconnected component. In particular, one may assume that
$\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ have
no vertices of degree one.

Furthermore, signed posets for which $K_{P}^{\mathrm{rt}}$ is not full-
dimensional need not be considered here, as they reduce to work of [9], as is
now explained.

Suppose $P$ is a signed poset such that $K_{P}^{\mathrm{rt}}$ is not full-
dimensional. There are two possibilities:

  * •

$P$ can be relabeled so that $P\subset\Phi_{B_{n-1}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n-1}})$, i.e. one should think of
$K_{P}^{\mathrm{rt}}\subset{\mathbb{R}}^{n-1}$ rather than
$K_{P}^{\mathrm{rt}}\subset{\mathbb{R}}^{n}$.

  * •

The signed graph $\Sigma_{P}$ underlying the Hasse diagram of $P$ is balanced.
(See Definition 8.3 for the definition of the Hasse diagram of a signed poset
and Definition 12.12 for the notion of a balanced signed graph.)

While the first case is straightforward, the second requires a little care.

######  Proposition 15.3.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset such that
$\Sigma_{P}$ is balanced. Then $P$ (resp. ${P^{\scriptscriptstyle\vee}}$) is
isomorphic to a poset $P^{\prime}\subset\Phi_{A_{n-1}}\cap\Phi_{B_{n}}$ (resp.
$P^{\prime}\subset\Phi_{A_{n-1}}\cap\Phi_{C_{n}}$).

###### Proof.

One may assume $\widehat{G}(P)$ consists of a single biconnected component.
Consequently, $\Sigma_{P}$ must be connected. Since $\Sigma_{P}$ is balanced,
the vertices can be partitioned into sets $V^{+}$ and $V^{-}$ such that edges
labeled $+$ join a vertex in $V^{+}$ to one in $V^{-}$ and edges labeled $-$
join two vertices in either $V^{+}$ or $V^{-}$. (This is Theorem 12.17.) Note
that since $\Sigma_{P}$ is balanced, it cannot contain any self-loops.
Therefore, there is no $\pm i<\mp i$ in $\widehat{G}(P)$. (In particular,
should one be considering $\widehat{G}_{B}(P)$, $0$ must lie in its own
connected component.) Then $\widehat{G}(P)$ consists of two isotropic
components (plus $0$ in $\widehat{G}_{B}(P)$). One can then read off a signed
permutation sending $P$ to some $P^{\prime}\subset\Phi_{A_{n-1}}$ (it is the
permutation flipping signs so that all vertices in each isotropic component
have the same sign), as desired. ∎

Next, one checks that, given an $\Phi_{A_{n-1}}$-poset, its root cone
semigroup does not change when one switches among the $\Phi_{A_{n-1}}$,
$\Phi_{B_{n}}$ and $\Phi_{C_{n}}$ root lattices.

######  Proposition 15.4.

Suppose $P\subset\Phi_{A_{n-1}}$ is a poset. Then the semigroups
$K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{A}$, $K_{P}^{\mathrm{rt}}\cap
L^{\mathrm{rt}}_{B}$ and $K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{C}$ are
equal.

###### Proof.

First, begin by observing that $K_{P}^{\mathrm{rt}}\cap
L^{\mathrm{rt}}_{A}\subset K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{B}$ and
$K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{C}$. Suppose $\alpha\in
K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{B}$. Then $\alpha$ can be written as
a positive integer linear combination of elements of $P$ corresponding to the
covering relations in $\widehat{G}_{B}(P)$. However, these elements correspond
precisely to the covering relations in the Hasse diagram of $P$, which are a
minimal generating set for $K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{A}$ (see
[9, Proposition 7.1]). Therefore, $\alpha\in K_{P}^{\mathrm{rt}}\cap
L^{\mathrm{rt}}_{A}$, so $K_{P}^{\mathrm{rt}}\cap
L^{\mathrm{rt}}_{A}=K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{B}$. The same
argument shows $K_{P}^{\mathrm{rt}}\cap
L^{\mathrm{rt}}_{A}=K_{P}^{\mathrm{rt}}\cap L^{\mathrm{rt}}_{C}$. ∎

As a consequence of the last two propositions, if $P$ is a signed poset and
$K_{P}^{\mathrm{rt}}$ is not full-dimensional, its root cone semigroup ring
can be understood using the results of [9]. If $\widehat{G}(P)$ is
disconnected, it either has multiple biconnected components or consists of a
single biconnected component made up of two isotropic connected components,
meaning $K_{P}^{\mathrm{rt}}$ is not full-dimensional. Consequently, for the
remainder of the chapter it will be assumed that $\widehat{G}(P)$ is connected
and consists of a single biconnected component.

####  15.2 Signed Notches

In [9], a _$\vee$ -shaped notch_ (resp. a _$\wedge$ -shaped notch_) in a poset
was defined to be a triple of elements $(a,b,c)$ such that $a\lessdot
b,a\lessdot c$ (resp. $a\gtrdot b,a\gtrdot c$) and $b$ and $c$ are in
different connected components of $P\smallsetminus P_{\leq a}$ (resp.
$P\smallsetminus P_{\geq a}$). One defines a signed notch as a pair of notches
in this sense.

######  Definition 15.5.

Given a signed poset $P$, a _signed notch_ in $\widehat{G}(P)$ is a pair
$(a,b,c)$ and $(-a,-b,-c)$ such that both $(a,b,c)$ and $(-a,-b,-c)$ are
notches in $\widehat{G}(P)$ and, if in $\widehat{G}_{B}(P)$, neither $b$ nor
$c$ is $0$.

Consider again the signed poset $P$ from Figure 28. Together $(2,3,5)$ and
$(-2,-3,-5)$ form a signed notch.

If $(a,b,c)$ is a notch in $\widehat{G}(P)$, it follows from the properties of
the involution that $(-a,-b,-c)$ is also a notch in the type $A$ sense, but
facing the other way. Consequently, one may assume that $(a,b,c)$ is a
$\vee$-shaped notch and $(-a,-b,-c)$ is a $\wedge$-shaped notch. One may also
always assume, without loss of generality, that
$\mathrm{sgn}(b)=\mathrm{sgn}(c)$, by relabeling the elements of
$\widehat{G}(P)$ as necessary.

One next wants to show that closing a signed notch in either
$\widehat{G}_{B}(P)$ or $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ gives
the $\widehat{G}_{B}$ or $\widehat{G}_{C}$ associated to some other signed
poset. If $(a,b,c)$ and $(-a,-b,-c)$ form a signed notch in $\widehat{G}(P)$,
the poset $\widehat{G}(P)/\\{b\equiv c,-b\equiv-c\\}$ is the poset obtained by
_closing the notch_. It will often be useful to think of this operation as
closing the type $A$ notch $(a,b,c)$ and then the notch $(-a,-b,-c)$ (or vice
versa). To legitimize this point of view, one needs to show the following:

  * •

$(-a,-b,-c)$ remains a notch in $\widehat{G}(P)/\\{b\equiv c\\}$ (Proposition
15.6)

  * •

There is a signed poset $P^{\prime}$ such that
$\widehat{G}(P^{\prime})=\widehat{G}(P)/\\{b\equiv c,-b\equiv-c\\}$. Such a
$P^{\prime}$ will be said to have been obtained from $P$ by _closing the
signed notch_. (Propositions 15.7 and 15.8)

The result of closing the signed notch formed by $(2,3,5)$ and $(-2,-3,-5)$ in
$P$ shown in Figure 29.

$1$$2$$3\equiv 5$$-3\equiv-5$$-2$$-1$$4$$6$$-4$$-6$$0$ (a)
$\widehat{G}_{B}(P)/\\{3\equiv 5,-3\equiv-5\\}$

$1$$2$$3\equiv 5$$-3\equiv-5$$-2$$-1$$4$$6$$-4$$-6$ (b)
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})/\\{3\equiv 5,-3\equiv-5\\}$

Figure 29: $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ from Figure 28 after closing
the signed notch $(2,3,5)$ and $(-2,-3,-5)$.

######  Proposition 15.6.

If $(a,b,c)$ and $(-a,-b,-c)$ form a signed notch in $\widehat{G}(P)$,
$(-a,-b,-c)$ remains a notch in $\widehat{G}(P)/\\{b\equiv c\\}$.

###### Proof.

One knows from [9, Definition 8.5] that $\widehat{G}(P)/\\{b\equiv c\\}$ is a
poset and thus so is $(\widehat{G}(P)/\\{b\equiv c\\})/\\{-b\equiv-c\\}$, as
long as $(-a,-b,-c)$ remains a notch in $\widehat{G}(P)/\\{b\equiv c\\}$. The
only way in which $(-a,-b,-c)$ can fail to be a notch in
$\widehat{G}(P)/\\{b\equiv c\\}$ is if $-b$ and $-c$ lie in the same connected
component of $(\widehat{G}(P)/\\{b\equiv
c\\})\smallsetminus(\widehat{G}(P)/\\{b\equiv c\\})_{\geq-a}$. In that case,
there is a path from $-b$ to $-c$ avoiding $(\widehat{G}(P)/\\{b\equiv
c\\})_{\geq-a}$. However, since $(-a,-b,-c)$ was a notch in $\widehat{G}(P)$,
this path cannot lift to a path in $\widehat{G}(P)$, so it must pass through
the vertex $\\{b\equiv c\\}$. Therefore at least one of $b$ and $c$ lies in
$\widehat{G}(P)_{\geq-a}$. Without loss of generality, assume
$b\in\widehat{G}(P)_{\geq-a}$. But then $b\equiv c\geq-a$ in
$\widehat{G}(P)/\\{b\equiv c\\}$, a contradiction. Thus, one must have that
$(-a,-b,-c)$ is a notch in $\widehat{G}(P)/\\{b\equiv c\\}$.

∎

With Proposition 15.6 in hand, it is easier to address $\widehat{G}_{B}(P)$
and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ separately, as the next
two propositions.

######  Proposition 15.7.

If $(a,b,c)$ and $(-a,-b,-c)$ form a signed notch in $\widehat{G}_{B}(P)$,
there is a signed poset $P^{\prime}\subset\Phi_{B_{n}}$ such that
$\widehat{G}_{B}(P^{\prime})=\widehat{G}_{B}(P)/\\{b\equiv c,-b\equiv-c\\}$.

###### Proof.

From Proposition 15.6, one knows that $\widehat{G}_{B}(P)/\\{b\equiv c\\}$ has
a notch $(-a,-b,-c)$. Closing the second notch $(-a,-b,-c)$ one obtains

$\widehat{G}_{B}(P^{\prime})=(\widehat{G}_{B}(P)/\\{b\equiv
c\\})/\\{-b\equiv-c\\}\\\ =(\widehat{G}_{B}(P)/\\{-b\equiv-c\\})/\\{b\equiv
c\\}=\widehat{G}_{B}(P)/\\{b\equiv c,-b\equiv-c\\},$

as the resulting poset is independent of the order in which the notches are
closed. Since $\mathrm{sgn}(b)=\mathrm{sgn}(c)$, this poset has an involution
$\pm i\mapsto\mp i$ such that $i\to j$ is sent to $-j\to-i$, since
$\widehat{G}_{B}(P)$ had an involution with this property.

To show that $\widehat{G}_{B}(P^{\prime})=\widehat{G}_{B}(P)/\\{b\equiv
c,-b\equiv-c\\}$ is really associated to a signed poset $P^{\prime}$, one must
show that $\widehat{G}_{B}(P^{\prime})$ has the property that if $i<-i$, then
$i<0<-i$. Suppose $i<-i$ in $\widehat{G}_{B}(P^{\prime})$. It suffices to
consider only the case where $i\not<-i$ in $\widehat{G}_{B}(P)$. Then, without
loss of generality, one may assume that $b<-i$ and $i<c$. First, consider the
case where $a\neq 0$. One then has the situation depicted in Figure 30, where
the solid lines are edges in the Hasse diagram of $\widehat{G}_{B}(P)$ and the
dashed lines are chains.

$i$$c$$a$$b$$-b$$-a$$-c$$-i$ Figure 30: After closing two notches in the proof
of Proposition 15.7

This gives a path from $b$ to $c$, via $-i,-c,-a,-b$ and $i$. Since $(a,b,c)$
is a notch, this path must intersect $\widehat{G}_{B}(P)_{\leq a}$ at some
$d$. There are a few cases.

  1. (a)

Suppose $-c\leq d\leq-i$. Then $-c<c$ in $\widehat{G}_{B}(P)$. Therefore,
$-c<0<c$. Then in $\widehat{G}_{B}(P^{\prime})$, one has $b\equiv c<-i$ and
$i<-b\equiv-c$, so since $-c<0<c$ one has $i<-b\equiv-c<0<b\equiv c<-i$, as
desired.

  2. (b)

Suppose $i\leq d\leq-b$. Then in $\widehat{G}_{B}(P)$ one has $i\leq d<a<-i$,
so $i<0<-i$ and this relation is preserved in $\widehat{G}_{B}(P^{\prime})$.

  3. (c)

Suppose $i\leq d<c$. Then one has $i\leq d<c<-i$ in $\widehat{G}_{B}(P)$, so
$i<0<-i$ and this relation is preserved in $\widehat{G}_{B}(P^{\prime})$.

  4. (d)

Suppose $-a=d<a$. Then $i<-i$ in $\widehat{G}_{B}(P)$, so $i<0<-i$ and this
relation is preserved in $\widehat{G}_{B}(P^{\prime})$.

In the other case, suppose $a=0$. Then $-b<0<b$. Since $-i>b$, $i<-b$, so
$i<-i$ in $\widehat{G}_{B}(P)$, so $i<0<-i$ and this relation is preserved in
$\widehat{G}_{B}(P^{\prime})$. ∎

######  Proposition 15.8.

If $(a,b,c)$ and $(-a,-b,-c)$ form a signed notch in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, there is a signed poset
$P^{\prime{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$ such that
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})=\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})/\\{b\equiv
c,-b\equiv-c\\}$.

###### Proof.

From Proposition 15.6, one knows that
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})/\\{b\equiv c\\}$ has a notch
$(-a,-b,-c)$. Closing the second notch $(-a,-b,-c)$ one obtains

$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})=(\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})/\\{b\equiv
c\\})/\\{-b\equiv-c\\}\\\
=(\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})/\\{-b\equiv-c\\})/\\{b\equiv
c\\}=\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})/\\{b\equiv
c,-b\equiv-c\\},$

as the resulting poset is independent of the order in which the notches are
closed. Since $\mathrm{sgn}(b)=\mathrm{sgn}(c)$, this poset has an involution
$\pm i\mapsto\mp i$ such that $i\to j$ is sent to $-j\to-i$, since
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ had an involution with this
property.

It remains to show that $\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$
has the property that if $i<-i$ and $j<-j$ (possibly after relabelling) that
$i<-j$ and $j<-i$. Closing a notch introduces relations and does not remove
any relations. There are then two cases:

  1. 1.

Suppose $i<-i$ in $\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$, but
not in $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ and $j<-j$ in both
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ and
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$. Then, without loss of
generality, suppose $i<b$ and $c<-i$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. Then $-b<-i$ and $i<-c$. Then
one has the situation in $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$
depicted in Figure 31, where the dashed edges are chains and the solid edges
are edges in the Hasse diagram of
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.

$i$$-c$$-b$$a$$b$$c$$-a$$-i$ Figure 31:
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ in case 1 of the proof of
Proposition 15.8

Since $(a,b,c)$ is a notch in $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$,
one must have that the path from $b$ to $c$ via $i$ must pass through
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})_{\leq a}$ at some $d$. There
are a few cases

    1. (1.a)

Suppose $d$ lies in the chain from $b$ to $i$ or the chain from $i$ to $-c$.
Then $i<a$, so $i<-i$ in $\widehat{G}(P)$, a contradiction.

    2. (1.b)

Suppose $d<a$ and $-b\leq d<-i$. Then, $-b<d<a<b$. Consequently, if $j<-j$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ and $b\neq j$, one has that
$-b<-j$ and $j<b$ in both $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ and
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$. Then, in
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$, one has $j<b\equiv
c<-i$ and $i<-b\equiv-c<-j$, as required.

    3. (1.c)

Suppose $d=-a$ and $-a<a$. Then in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ one has that $i<-i$, a
contradiction.

    4. (1.d)

Suppose $d=-c$ and $-c<a$. Then
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is as in Figure 32. Then
$i<-i$ in $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, a contradiction.

$-c$$i$$a$$-a$$c$$-i$$b$$-b$ Figure 32:
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ in case 1(d) in the proof of
Proposition 15.8

  2. 2.

Suppose $i<-i$ and $j<-j$ in
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$, but neither relation
exists in $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. There are again two
cases (up to relabeling the vertices).

    1. (2.a)

Suppose $i<b$, $c<-i$ and $j<c$, $-j>b$. Then in
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$, one has $i<b\equiv
c<-j$ and $j<b\equiv c<-i$, as required.

    2. (2.b)

Suppose $i,j<b$ and $c<-i,-j$. Without loss of generality, suppose $i<j$.
Consequently $-j<-i$. Then in
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$, one has $i<j<-j<-i$,
giving the required relations.

Thus, this new poset is $\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$
for some signed poset $P^{\prime{\scriptscriptstyle\vee}}$. ∎

Now that one understands what it means to close a notch in a signed poset, the
natural question is what effect this maneuver has on the semigroup ring.

######  Proposition 15.9.

Let $P\subset\Phi_{B_{n}}$ be a signed poset and $(a,b,c)$ and $(-a,-b,-c)$
form a signed notch in $\widehat{G}_{B}(P)$. Let $P^{\prime}$ be the signed
poset obtained from $P$ by closing this notch. Then

$R_{P^{\prime}}^{\mathrm{rt}}\cong\begin{cases}R_{P}^{\mathrm{rt}}/(t_{a}^{\delta}t_{b}^{\epsilon}-t_{a}^{\delta}t_{c}^{\epsilon})&\text{if
}a\neq 0\\\ R_{P}^{\mathrm{rt}}/(t_{b}^{\epsilon}-t_{c}^{\epsilon})&\text{if
}a=0\\\ \end{cases}$

where $\delta=\mathrm{sgn}(a)$ and $\epsilon=-\mathrm{sgn}(b)$.

Let ${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$ be a signed poset and
$(a,b,c)$ and $(-a,-b,-c)$ form a signed notch in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. Let
$P^{\prime{\scriptscriptstyle\vee}}$ be the signed poset obtained from
${P^{\scriptscriptstyle\vee}}$ by closing this notch. Then

$R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}\cong
R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}/(t_{a}^{\delta}t_{b}^{\epsilon}-t_{a}^{\delta}t_{c}^{\epsilon}),$

where $\delta=\mathrm{sgn}(a)$ and $\epsilon=-\mathrm{sgn}(b)$.

To see how this works, consider once again
$P=\\{+e_{1}-e_{2},+e_{1}+e_{2},+e_{1}-e_{3},+e_{1}+e_{3},+e_{1}-e_{4},+e_{1}-e_{5},+e_{1}-e_{6},+e_{2}-e_{3},+e_{2}-e_{4},+e_{2}-e_{5},+e_{2}-e_{6},+e_{4}-e_{6},+e_{5}-e_{6},+e_{1}\\}$
shown in Figure 28 (page 28). Closing the notch $(2,3,5)$ and $(-2,-3,-5)$
gives the poset shown in Figure 29 (page 29),
$Q=\\{+e_{1}-e_{2},+e_{1}+e_{2},+e_{1}+e_{3},+e_{1}-e_{3},+e_{1}-e_{4},+e_{1}-e_{6},+e_{2}-e_{3},+e_{2}-e_{4},+e_{3}-e_{6},+e_{4}-e_{6}\\}$
(with $3\equiv 5$ and $3\equiv-5$ being renamed $3$ and $-3$, respectively).
In type $B$, one has
$S^{\mathrm{rt}}_{P}=k[U_{12},U_{23},U_{\bar{1}3},U_{24},U_{25},U_{46},U_{56},U_{10}]$.
Define a map $\psi\colon S^{\mathrm{rt}}_{P}\to R_{Q}^{\mathrm{rt}}$ by

$\begin{array}[]{llll}U_{12}\mapsto t_{1}t_{2}^{-1}&U_{23}\mapsto
t_{2}t_{3}^{-1}&U_{\bar{1}3}\mapsto t_{1}t_{3}&U_{24}\mapsto
t_{2}t_{4}^{-1}\\\ U_{25}\mapsto t_{2}t_{3}^{-1}&U_{46}\mapsto
t_{4}t_{6}^{-1}&U_{56}\mapsto t_{3}t_{6}^{-1}&U_{10}\mapsto t_{1}\end{array}$

Then

$\displaystyle R_{Q}^{\mathrm{rt}}$ $\displaystyle\cong
S^{\mathrm{rt}}_{P}/\ker\psi$ $\displaystyle\cong
S^{\mathrm{rt}}_{P}/(U_{24}U_{46}-U_{25}U_{56},U_{12}U_{\bar{1}3}U_{25}-U_{10}^{2},U_{23}-U_{25})$
$\displaystyle=S^{\mathrm{rt}}_{P}/(I^{\mathrm{rt}}_{P}+(U_{23}-U_{25}))$
$\displaystyle\cong(S^{\mathrm{rt}}_{P}/I^{\mathrm{rt}}_{P})/(\overline{U}_{23}-\overline{U}_{25})$
$\displaystyle\cong R_{P}^{\mathrm{rt}}/(t_{2}t_{3}^{-1}-t_{2}t_{5}^{-1})$

###### Proof of Proposition 15.9.

Recall the definitions of $S^{\mathrm{rt}}_{P}$,
$S^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}$ and of the toric ideals
$I^{\mathrm{rt}}_{P}$ and $I^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}$
from Section 14.1. Let $P^{\prime}$ and $P^{\prime{\scriptscriptstyle\vee}}$
be the signed posets obtained by closing notches in $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, respectively. One defines
maps from $S^{\mathrm{rt}}_{P}$ and
$S^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}$ to the semigroup rings
$R_{P^{\prime}}^{\mathrm{rt}}=k[K_{P^{\prime}}^{\mathrm{rt}}\cap
L^{\mathrm{rt}}_{B}]$ and
$R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}=k[K_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}\cap
L^{\mathrm{rt}}_{C}]$.

Define a map $\psi\colon S^{\mathrm{rt}}_{P}\to R_{P^{\prime}}^{\mathrm{rt}}$
by

$U_{\delta i,\epsilon
j}\mapsto\begin{cases}t_{i}^{\delta}t_{j}^{-\epsilon}&i,j\neq c,0\\\
t_{i}^{\delta}t_{b}^{-\epsilon}&j=c,i\neq 0\\\
t_{b}^{\delta}t_{j}^{-\epsilon}&i=c\\\ t_{j}^{-\epsilon}&i=0,j\neq c\\\
t_{b}^{-\epsilon}&i=0,j=c\end{cases}$

and define a map $\psi^{\scriptscriptstyle\vee}\colon
S^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}\to
R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}$ by

$U_{\delta i,\epsilon
j}\mapsto\begin{cases}t_{i}^{\delta}t_{j}^{-\epsilon}&i,j\neq c\\\
t_{b}^{\delta}t_{j}^{-\epsilon}&i=c\\\
t_{i}^{\delta}t_{b}^{-\epsilon}&j=c\end{cases}$

One shows that $\ker\psi=I_{P}+(U_{ab}-U_{ac})$ and
$\ker\psi^{\scriptscriptstyle\vee}=I_{{P^{\scriptscriptstyle\vee}}}+(U_{ab}-U_{ac})$.
The argument is the same for $P$ and ${P^{\scriptscriptstyle\vee}}$. To
simplify notation, in the remainder of the proof, write $J$ for the kernel,
$I$ for the toric ideal, let the variables of the polynomial ring $S$ be
$U_{ij}$ and use $\widehat{G}(P)$ to denote $\widehat{G}_{B}(P)$ or
$\widehat{G}_{C}(P)$, as applicable. Thus, one needs to show that
$J=I+(U_{ab}-U_{ac})$.

Recall that $I$ is generated by the cycle binomials of cycles in
$\widehat{G}(P)$ not fixed by the involution. Let $U(C)$ be such a cycle
binomial. The definition of $\varphi$ (resp. $\psi$) ensures that $U(C)\in J$,
so $I\subset J$. Both $U_{ab}$ and $U_{ac}$ have the same image in
$R_{P^{\prime}}^{\mathrm{rt}}$, so $(U_{ab}-U_{ac})\subset J$. Thus
$I+(U_{ab}-U_{ac})\subset J$.

Let $P^{+}$ be the directed graph that has the same vertices and edges as
$\widehat{G}(P^{\prime})$, but with the edges $-b\equiv-c\to-a$ and $a\to
b\equiv c$ doubled. The argument in the proof of Theorem 14.5 shows that $J$
is generated by the cycle binomials corresponding to cycles of $P^{+}$ not
fixed by the involution. (See Figure 33 for $P^{+}$ for the
$\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ from
Figure 29.)

$1$$2$$3\equiv 5$$-3\equiv-5$$-2$$-1$$4$$6$$-4$$-6$$0$ (a) $P^{+}_{C}$

$1$$2$$3\equiv 5$$-3\equiv-5$$-2$$-1$$4$$6$$-4$$-6$ (b) $P^{+}_{C}$

Figure 33: $P^{+}_{B}$ and $P^{+}_{C}$ for $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ from Figure 29

One then needs to show that for every cycle of $P^{+}$ not fixed by the
involution, its circuit binomial lies in $I+(U_{ab}-U_{ac})$. There are a
number of cases.

  1. 1.

Suppose $C$ is a cycle in $P^{+}$ not fixed by the involution passing through
neither $b\equiv c$ nor $-b\equiv-c$. Then $C$ lifts to a cycle in
$\widehat{G}(P)$, meaning $U(C)\in I_{P}$.

  2. 2.

Suppose $C$ is a cycle in $P^{+}$ not fixed by the involution, but passing
through at least one of $b\equiv c$ and $-b\equiv-c$. One can partition the
edges incident to $b\equiv c$ and $-b\equiv c$ into $E_{b}\sqcup E_{c}$
according to whether they lift to edges incident to either $b$ or $-b$ or to
either $c$ or $-c$ in $\widehat{G}(P)$.

    1. (2.a)

Suppose the edges of $C$ incident to $b\equiv c$ or $-b\equiv-c$ all lie in
one of $E_{b}$ and $E_{c}$. Then $C$ lifts to a cycle in $\widehat{G}(P)$, so
$U(C)\in I$.

    2. (2.b)

Suppose $C$ contains an edge from both $E_{b}$ and $E_{c}$. Assume $C$ has an
edge from each of $E_{b}$ and $E_{c}$ incident to $b\equiv c$. The other case
is symmetric.

Since $(a,b,c)$ was a notch in $\widehat{G}(P)$, one has that $b$ and $c$ lie
in different connected components of
$\widehat{G}(P)\smallsetminus\widehat{G}(P)_{\leq a}$, so $C$ must pass
through at least one vertex $d\leq a$. Let $\pi_{da}$ be a saturated chain
between $d$ and $a$. Let $C_{b}$ be the cycle in $P^{+}$ that follows $C$ from
$b\equiv c$ to d, then $\pi_{da}$ from $d$ to $a$ and finishes along the edge
in $E_{b}$ between $a$ and $b\equiv c$. Let $C_{c}$ be the cycle in $P^{+}$
that follows $C$ from $b\equiv c$ to $d$, then $\pi_{da}$ from $d$ to $a$ and
finishes along the edge in $E_{c}$ from $a$ to $b\equiv c$.

      1. (2.b.i)

Suppose $C_{b}$ and $C_{c}$ both lift to cycles in $\widehat{G}(P)$ and, with
$C_{b}$ and $C_{c}$ are oriented so they traverse $\pi_{da}$ in opposite
directions. One has

$U(C)=U(C_{b})\left(\prod_{e\in
W(\pi_{dc})}U_{e}\right)+U(C_{c})\left(\prod_{e\in A(\pi_{bd})}U_{e}\right)\\\
+(U_{ab}-U_{ac})\left(\prod_{e\in W(\pi_{dc})}U_{e}\right)\left(\prod_{e\in
A(\pi_{bd})}U_{e}\right)\left(\prod_{e\in W(\pi_{da})}U_{e}\right),$

where $\pi_{dc}$ is the portion of $C_{c}$ between $d$ and $c$ and $\pi_{bd}$
is the portion of $C_{b}$ between $d$ and $b$. Note that it is possible one of
$C_{b}$ and $C_{c}$ is fixed by the involution, in which case its cycle
binomial is zero since $W(C)=A(C)$. Thus, $U(C)\in I+(U_{ab}-U_{ac})$, as
desired.

      2. (2.b.ii)

Now suppose at least one of $C_{b}$ and $C_{c}$ does not lift to a cycle in
$\widehat{G}(P)$. One can use the previous argument to show both $U(C_{b})$
and $U(C_{c})$ lie in $I_{P}+(U_{ab}-U_{ac})$. Without loss of generality,
suppose $C_{b}$ does not lift to a cycle in $\widehat{G}(P)$. (If neither
lifts, one repeats the following argument with each cycle.) Since $C_{b}$ does
not lift to a cycle in $\widehat{G}(P)$, it must pass through $-b\equiv-c$,
along one edge in $E_{b}$ and one in $E_{c}$. Since $(-a,-b,-c)$ was a notch
in $\widehat{G}(P)$, there is $d^{\prime}\geq-a\in P^{+}$ which $C_{b}$ passes
through. Then choose a saturated chain between $d^{\prime}$ and $-a$, call it
$\pi_{da}^{\prime}$, with the proviso that if $\pi_{da}^{\prime}$ includes $a$
and $b\equiv c$, it includes the edge in $C$. One can then construct $C_{bb}$
and $C_{bc}$ as in the previous case and the same argument shows that
$U(C_{b})\in I_{P}+(U_{ab}-U_{ac})$. (However, if $\pi_{da}^{\prime}$
coincides with $C$, it will be the case that one of $C_{bb}$ and $C_{cc}$ is
not genuinely a cycle, rather one will have a chain from (say) $b$ to
$d^{\prime}$ and then return to $b$ down the same chain, in which case
$U(C_{bb})=0$.)

Thus, $J=I+(U_{ab}-U_{ac})$.

To finish the proof, one sees that

$\displaystyle R_{P^{\prime}}^{\mathrm{rt}}$ $\displaystyle\cong
S_{P}/\ker\psi$ $\displaystyle\cong S_{P}/(I_{P}+(U_{ab}-U_{ac}))$
$\displaystyle\cong(S_{P}/I_{P})/(\bar{U}_{ab}-\bar{U}_{ac})$
$\displaystyle\cong
R_{P}^{\mathrm{rt}}/(t_{a}^{\mathrm{sgn}(a)}t_{b}^{-\mathrm{sgn}(b)}-t_{a}^{\mathrm{sgn}(a)}t_{c}^{-\mathrm{sgn}(c)}),$

where $t_{a}=1$ if $a=0$, and

$\displaystyle R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}$
$\displaystyle\cong
S_{{P^{\scriptscriptstyle\vee}}}/\ker\psi^{\scriptscriptstyle\vee}$
$\displaystyle\cong
S_{{P^{\scriptscriptstyle\vee}}}/(I_{{P^{\scriptscriptstyle\vee}}}+(U_{ab}-U_{ac}))$
$\displaystyle\cong(S_{{P^{\scriptscriptstyle\vee}}}/I_{{P^{\scriptscriptstyle\vee}}})/(\bar{U}_{ab}-\bar{U}_{ac})$
$\displaystyle\cong
R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}/(t_{a}^{\mathrm{sgn}(a)}t_{b}^{-\mathrm{sgn}(b)}-t_{a}^{\mathrm{sgn}(a)}t_{c}^{-\mathrm{sgn}(c)}).$

∎

####  15.3 Strongly Planar Signed Posets

Having looked at the notion of opening notches and its impact on
$R_{P}^{\mathrm{rt}}$, one is ready to show that a certain class of signed
posets have $R_{P}^{\mathrm{rt}}$ a complete intersection, implying that the
numerator of $\Psi_{P}$ factors.

######  Definition 15.10.

A poset is said to be _strongly planar_ if, after the addition of a maximal
element $\hat{1}$ and a minimal element $\hat{0}$, there is an embedding of
its Hasse diagram in ${\mathbb{R}}^{2}$ that is planar and has the property
that if $a<_{P}b$, the $y$-coordinate of $a$ is smaller than that of $b$. A
signed poset $P\subset\Phi_{B_{n}}$ (resp. $P\subset\Phi_{C_{n}}$) will be
said to be _strongly planar_ if $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) is strongly planar.

As an example, consider $P=\\{+e_{1}+e_{2},+e_{1},+e_{2}\\}$. Figure 34 shows
$\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. One
sees that $\widehat{G}_{B}(P)$ is strongly planar, but
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is not.

$0$$-1$$-2$$1$$2$ (a) $\widehat{G}_{B}(P)$

$-1$$-2$$1$$2$ (b) $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$

Figure 34: $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ for
$P=\\{+e_{1}+e_{2},+e_{1},+e_{2}\\}$

######  Definition 15.11.

An embedding of $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) in ${\mathbb{R}}^{n}$ is said
to be _centrally symmetric_ if it is fixed by the map $(x,y)\mapsto(-x,-y)$.

The main result of this section will be the following.

######  Theorem 15.12.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset such that
$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$)
has an embedding in ${\mathbb{R}}^{2}$ which is strongly planar. Then

  * •

$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$)
has an embedding that is both centrally symmetric and strongly planar,

  * •

$R_{P}^{\mathrm{rt}}$ (resp. $R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}$)
is a complete intersection, with $I_{P}$ (resp.
$I_{{P^{\scriptscriptstyle\vee}}}$) generated by the cycle binomials of the
cycles in $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) defining the faces of the
graph, and

  * •

one has

$\displaystyle\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})$
$\displaystyle=\dfrac{\prod_{\rho}(1-\bm{x}^{\rho})}{\prod_{e}(1-x_{a}^{\delta}x_{b}^{-\epsilon})}\quad\text{and}$
$\displaystyle\mathrm{Hilb}(R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}},\bm{x})$
$\displaystyle=\dfrac{\prod_{\rho}(1-\bm{x}^{\rho})}{\prod_{e}(1-x_{a}^{\delta}x_{b}^{-\epsilon})},$

where $\rho$ runs over all regions enclosed by $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}(P)$) not fixed by the involution, $e$ runs over the orbits of
edges $\delta a\to\epsilon b$ ($a<b$) in $\widehat{G}_{C}(P)$, $x_{0}$ is
taken to be $1$ and

$\bm{x}^{\rho}=x_{\min(\rho)}^{\mathrm{sgn}(\min(\rho))}x_{\max(\rho)}^{-\mathrm{sgn}(\max(\rho))}.$

$0$$-2$$-1$$-3$$-4$$1$$4$$3$$2$$\rho$ (a) $\widehat{G}_{B}(P)$

$-2$$-1$$-3$$-4$$1$$4$$3$$2$$\rho$ (b)
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$

Figure 35: A signed poset with $\widehat{G}_{B}(P)$ and
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ both strongly planar

The proof of Theorem 15.12 is quite involved and requires a number of
propositions. First, the idea of the proof is illustrated using the posets in
Figure 35. The proof is by induction on the number of orbits of regions
enclosed by $\widehat{G}(P)$ under the involution. Proposition 15.14 will show
that one can find at least one “rightmost” region in $\widehat{G}(P)$, such as
the region marked by $\rho$ in Figure 35. Lemma 15.17 gives that the region
$\rho$ is not fixed by the involution.

$0$$-2$$-1$$-5$$-3$$-4$$1$$5$$4$$3$$2$ (a) $\widehat{G}_{B}(P^{\prime})$

$-2$$-1$$-5$$-3$$-4$$1$$5$$4$$3$$2$ (b)
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}}^{\prime})$

Figure 36: Having opened one notch to obtain $P^{\prime}$

Lemma 15.19 guarantees that $\widehat{G}(P)$ can be obtained by closing a
notch along the left border of $\rho$. Figure 36 shows the
$\widehat{G}(P^{\prime})$ obtained by opening a notch along $(-2,-1)$ and
$(1,2)$ in $\widehat{G}(P)$. Proposition 15.9 gives that

$R_{P}^{\mathrm{rt}}=R_{P^{\prime}}^{\mathrm{rt}}/(x_{1}x_{2}^{-1}-x_{5}x_{2}^{-1})\quad\text{and}\quad
R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}=R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}/(x_{1}x_{2}^{-1}-x_{5}x_{2}^{-1}).$

Typically, one thinks of $R_{P^{\prime}}^{\mathrm{rt}}$ as graded by
${\mathbb{Z}}^{5}$ and $R_{P}^{\mathrm{rt}}$ as graded by ${\mathbb{Z}}^{4}$.
One can alter the grading of $R_{P^{\prime}}^{\mathrm{rt}}$ so that $\deg
x_{1}=\deg x_{5}=(1,0,0,0)$, so that $R_{P^{\prime}}^{\mathrm{rt}}$ is also
graded by ${\mathbb{Z}}^{4}$ and $R_{P}^{\mathrm{rt}}$ is a quotient of
$R_{P^{\prime}}^{\mathrm{rt}}$ by a homogeneous ideal. Keeping the altered
grading in mind, one then has from Proposition 6.12 that

$\begin{split}\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})&=(1-x_{1}x_{2}^{-1})\mathrm{Hilb}(R_{P^{\prime}}^{\mathrm{rt}},\bm{x})|_{x_{5}=x_{1}}\quad\text{and}\quad\\\
\mathrm{Hilb}(R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}},\bm{x})&=(1-x_{1}x_{2}^{-1})\mathrm{Hilb}(R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}},\bm{x})|_{x_{5}=x_{1}}\end{split}$
(5)

$\widehat{G}_{B}(P^{\prime})$ and
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$ each have two
biconnected components, associated to signed posets $P_{1}$ and $P_{2}$, shown
in Figure 37.

|
---|---
|

$0$$-2$$-5$$-3$$-4$$2$$3$$4$$5$ (a) $\widehat{G}_{B}(P_{1})$

$-2$$-5$$-3$$-4$$2$$3$$4$$5$ (b)
$\widehat{G}_{C}(P^{\scriptscriptstyle\vee}_{1})$

$0$$2$$-2$$1$$-1$ (c) $\widehat{G}_{B}(P_{2})$

$2$$-2$$1$$-1$ (d) $\widehat{G}_{C}(P^{\scriptscriptstyle\vee}_{2})$

Figure 37: $P^{\prime}$ broken into biconnected components

Both $P_{1}$ and $P^{\scriptscriptstyle\vee}_{1}$ have a single biconnected
component and principal toric ideals generated by
$U_{\bar{3}4}U_{\bar{2}3}-U_{4\bar{5}}U_{\bar{2}5}$. Then one has

$\begin{split}\mathrm{Hilb}(R_{P_{1}}^{\mathrm{rt}},\bm{x})&=\dfrac{(1-x_{4}x_{2}^{-1})}{(1-x_{4}x_{3}^{-1})(1-x_{3}x_{2}^{-1})(1-x_{4}x_{5}^{-1})(1-x_{5}x_{2}^{-1})}\quad\text{and}\\\
\mathrm{Hilb}(R_{P^{\scriptscriptstyle\vee}_{1}}^{\mathrm{rt}},\bm{x})&=\dfrac{(1-x_{4}x_{2}^{-1})}{(1-x_{4}x_{3}^{-1})(1-x_{3}x_{2}^{-1})(1-x_{4}x_{5}^{-1})(1-x_{5}x_{2}^{-1})}.\end{split}$
(6)

Looking at Figures 37(c) and 37(d), one sees that an additional notch can be
opened in $\widehat{G}_{B}(P_{2})$, but not in
$\widehat{G}_{C}(P_{2}^{\scriptscriptstyle\vee})$. Opening the notch in
$\widehat{G}_{B}(P_{2})$ results in a signed poset $P_{3}$ shown in Figure 38.

$2$$0$$-2$$1$$3$$-3$$-1$ Figure 38: $\widehat{G}_{B}(P_{3})$

Both $\widehat{G}_{B}(P_{3})$ and
$\widehat{G}_{C}(P_{2}^{\scriptscriptstyle\vee})$ each have only one
biconnected component and a single cycle. This gives

$\displaystyle\mathrm{Hilb}(R_{P_{2}}^{\mathrm{rt}},\bm{x})$
$\displaystyle=(1-x_{1})\mathrm{Hilb}(R_{P_{3}}^{\mathrm{rt}},\bm{x})|_{x_{3}=x_{1}}$
$\displaystyle=\left.\dfrac{(1-x_{1})(1-x_{1}x_{3})}{(1-x_{1}x_{2}^{-1})(1-x_{2}x_{3})(1-x_{3})(1-x_{1})}\right|_{x_{3}=x_{1}}$
(7)
$\displaystyle=\dfrac{1-x_{1}^{2}}{(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2})(1-x_{1})}$

and

$\mathrm{Hilb}(R_{P_{2}^{\scriptscriptstyle\vee}}^{\mathrm{rt}},\bm{x})=\dfrac{(1-x_{1}^{2})}{(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2})}$
(8)

Since $P_{1}$ and $P_{2}$ are the biconnected components of $P^{\prime}$, from
Proposition 15.2, one has that
$R_{P^{\prime}}^{\mathrm{rt}}=R_{P_{1}}^{\mathrm{rt}}\otimes
R_{P_{2}}^{\mathrm{rt}}$ and
$R_{{P^{\scriptscriptstyle\vee}}^{\prime}}^{\mathrm{rt}}=R_{P_{1}^{\scriptscriptstyle\vee}}^{\mathrm{rt}}\otimes
R_{P_{2}^{\scriptscriptstyle\vee}}^{\mathrm{rt}}$, meaning

$\mathrm{Hilb}(R_{P^{\prime}}^{\mathrm{rt}},\bm{x})=\mathrm{Hilb}(R_{P_{1}}^{\mathrm{rt}},\bm{x})\mathrm{Hilb}(R_{P_{2}}^{\mathrm{rt}},\bm{x})\quad\text{and}\\\
\mathrm{Hilb}(R_{{P^{\scriptscriptstyle\vee}}^{\prime}}^{\mathrm{rt}},\bm{x})=\mathrm{Hilb}(R_{P^{\scriptscriptstyle\vee}_{1}}^{\mathrm{rt}},\bm{x})\mathrm{Hilb}(R_{P_{2}^{\scriptscriptstyle\vee}}^{\mathrm{rt}},\bm{x}).$

Combining (5), (6), (7) and (8) gives

$\begin{split}\mathrm{Hilb}&(R_{P}^{\mathrm{rt}},\bm{x})=(1-x_{1}x_{2}^{-1})\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})|_{x_{5}=x_{1}}\\\
=&(1-x_{1}x_{2}^{-1})\mathrm{Hilb}(R_{P_{1}}^{\mathrm{rt}},\bm{x})\mathrm{Hilb}(R_{P_{2}}^{\mathrm{rt}},\bm{x})|_{x_{5}=x_{1}}\\\
=&\left.\dfrac{(1-x_{1}x_{2}^{-1})(1-x_{4}x_{2}^{-1})(1-x_{1}^{2})}{(1-x_{4}x_{3}^{-1})(1-x_{3}x_{2}^{-1})(1-x_{4}x_{5}^{-1})(1-x_{5}x_{2}^{-1})(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2})(1-x_{1})}\right|_{x_{5}=x_{1}}\\\
=&\dfrac{(1-x_{4}x_{2}^{-1})(1-x_{1}^{2})}{(1-x_{4}x_{3}^{-1})(1-x_{3}x_{2}^{-1})(1-x_{4}x_{1}^{-1})(1-x_{1}x_{2}^{-1})(1-x_{2}x_{4})(1-x_{1}x_{2})(1-x_{1})}.\end{split}$

and

$\begin{split}\mathrm{Hilb}&(R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}},\bm{x})=(1-x_{1}x_{2}^{-1})\mathrm{Hilb}(R_{{P^{\scriptscriptstyle\vee}}^{\prime}}^{\mathrm{rt}},\bm{x})|_{x_{5}=x_{1}}\\\
=&(1-x_{1}x_{2}^{-1})\mathrm{Hilb}(R_{P^{\scriptscriptstyle\vee}_{1}}^{\mathrm{rt}},\bm{x})\mathrm{Hilb}(R_{P_{2}^{\scriptscriptstyle\vee}}^{\mathrm{rt}},\bm{x})|_{x_{5}=x_{1}}\\\
=&\left.\dfrac{(1-x_{1}x_{2}^{-1})(1-x_{4}x_{2}^{-1})(1-x_{1}^{2})}{(1-x_{4}x_{3}^{-1})(1-x_{3}x_{2}^{-1})(1-x_{4}x_{5}^{-1})(1-x_{5}x_{2}^{-1})(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2})}\right|_{x_{5}=x_{1}}\\\
=&\dfrac{(1-x_{1}x_{2}^{-1})(1-x_{4}x_{2}^{-1})(1-x_{1}^{2})}{(1-x_{4}x_{3}^{-1})(1-x_{3}x_{2}^{-1})(1-x_{4}x_{1}^{-1})(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2})}\\\
=&\dfrac{(1-x_{4}x_{2}^{-1})(1-x_{1}^{2})}{(1-x_{4}x_{3}^{-1})(1-x_{3}x_{2}^{-1})(1-x_{4}x_{1}^{-1})(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2}^{-1})(1-x_{1}x_{2})}.\end{split}$

The edges which correspond to the notches that are opened form a set of
_disconnecting chains_ splitting $\widehat{G}(P)$ into biconnected components.

######  Definition 15.13.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset. A
_disconnecting chain_ of $\widehat{G}(P)$ is a chain $c_{1}\lessdot
c_{2}\lessdot\dots\lessdot c_{k}$ such that removing
$c_{1}\lessdot\cdots\lessdot c_{k}$ and $-c_{k}\lessdot\cdots\lessdot-c_{1}$
breaks $\widehat{G}(P)$ into three connected components.

The first step in the proof is to locate a region to work with.

######  Proposition 15.14.

Suppose $Q$ is a strongly planar poset whose Hasse diagram is connected. Then
$Q$ has a region $\rho$ such that any vertex other than the maximum and
minimum element of $\rho$ in the right border of $\rho$ is in the border of no
other region. Such a region is called a _rightmost region_.

By considering a strongly planar poset, one has the luxury of assuming a given
strongly planar embedding, allowing sensible notions of “left” and “right”.

######  Definition 15.15.

Suppose $Q$ is a strongly planar poset. The _rightmost cover_ of $x\in Q$ is
the $y$ that covers $x$ such that if one traversed a small circle around $x$
counterclockwise (starting from the bottom, say) one passes the edge leading
to $y$ last. The _rightmost lower cover_ is the $z\lessdot x$ such that the
edge leading to $z$ is encountered first when traveling clockwise from the top
of the circle.

###### Proof of Proposition 15.14.

Without loss of generality, one may assume that every vertex in $Q$ has degree
at least two, since one makes this assumption of $\widehat{G}(P)$. Since $Q$
is strongly planar, the poset $\widehat{Q}$ obtained by adding a $\hat{0}$ and
$\hat{1}$ to $Q$ is also planar. Construct a saturated chain $\hat{0}\lessdot
c_{1}\lessdot\cdots\lessdot c_{k}\lessdot\hat{1}$ such that $c_{i+1}$ is the
rightmost cover of $c_{i}$ for all $i$ and $c_{1}$ is the rightmost cover of
$\hat{0}$.

Let $a=c_{i}$, where $i$ is maximal such that $c_{i}$ is covered by at least
two elements. (Since $c_{1}$ is a minimal element of $Q$, and $Q$ has been
assumed to have no vertices of degree 1, it has at least two covers, so such
an $a$ exists.) Construct a chain $a\lessdot b_{1}\lessdot\cdots\lessdot
b_{\ell}\lessdot\hat{1}$ where $b_{1}$ is the cover of $a$ that is rightmost
but one and $b_{i+1}$ is the rightmost cover of $b_{i}$.

Claim:  $c_{k}$ is the rightmost lower cover of $\hat{1}$ in $\widehat{Q}$.

Suppose not and $d\lessdot\hat{1}$ is to the right of $c_{k}$. Then $d$ must
lie above some minimal element of $Q$, say $f$. Since $Q$ is strongly planar,
a maximal chain from $f$ to $d$ must intersect $c_{1}\lessdot\cdots\lessdot
c_{k}$. Let $j$ be maximal such that $c_{j}$ is in this chain from $f$ to $d$.
Since $d\neq c_{k}$, one must have that $j\neq k$. Since $d$ is to the right
of $c_{k}$, the chain must continue along the rightmost cover of $c_{j}$.
However, this is $c_{j+1}$, contradicting the maximality of $j$.

Claim:  $b_{\ell}=c_{k}$.

Suppose not. Then $c_{k}$ has degree $\geq 2$ and it was maximal in $Q$, so it
must cover some $d\neq c_{k-1}$. Then $Q$ has a minimal element $a^{\prime}$
such that $a^{\prime}<d$. As $Q$ is strongly planar, a saturated chain from
$a^{\prime}$ to $d$ must include either some $b_{i}$ or $c_{j}$, $j<k$. If the
chain includes $b_{i}$, then there must be some $m$ where $b_{m+1}$ is not the
rightmost cover of $b_{m}$, a contradiction. If $c_{j}$ is in the chain, but
no $b_{i}$, then, since $d\neq c_{k}$, there must be an $i$ such that
$c_{n}>a$ and $c_{n}$ has more than two covers, contradicting that $a=c_{i}$
was maximal with this property.

Let $\rho$ be the region enclosed by the $b_{i}$ and $c_{j}$. Note that
$b_{\ell}=c_{k}$ need not be the maximal element of $\rho$. However, by
construction, $a$ will be the minimal element of $\rho$.

Claim:  The $c_{j}$ with $c_{j}>a$ are rightmost in $Q$, i.e. they are not in
the left border of any region.

Suppose $c_{j}$ is on the right border of $\rho$ (and is not $\max(\rho)$ or
$\min(\rho)$), that $c_{j}$ is on the left border of some region $\sigma$ and
that $j$ is minimal such that this is the case. Then, since $c_{j+1}$ is the
rightmost cover of $c_{j}$, the edge $(c_{j},c_{j+1})$ must be in the left
border of $\sigma$. Consequently, $c_{j}$ cannot be the minimal element of
$\sigma$. Then, by the minimality of $j$, there is some $d\lessdot c_{j}$ such
that $d$ is to the right of $c_{j-1}$, as depicted in Figure 39

$c_{j}$$c_{j-1}$$d$$c_{j+1}$$\sigma$$\rho$ Figure 39: The situation in the
last claim of the proof of Proposition 15.14

The region $\sigma$ must have a minimum element, call it $f$. Then $f\geq g$
for some $g$ that is minimal in $Q$. Then, since $Q$ is strongly planar and
$c_{1}$ is the rightmost minimal element of $Q$, a saturated chain from $g$ to
$f$ must pass through some $c_{i}$. But then, since $c_{i+1}$ is the rightmost
cover of $c_{i}$, one must have that $f$ is one of the $c$’s, a contradiction.
∎

One can now use the previous result to prove a few lemmas specific to
$\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.

######  Lemma 15.16.

Suppose ${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$ is a
$\Phi_{C_{n}}$-signed poset such that
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is strongly planar. Then there
is at most one $i$ such that $\pm 2e_{i}$ corresponds to an edge in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.

###### Proof.

Suppose ${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$ is a signed poset
such that $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is strongly planar
and $i\lessdot-i$ and $j\lessdot-j$. Then one must have $i<-j$ and $j<-i$.
Then if $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is embedded in the
plane (so that if $a<b$, then $y_{a}<y_{b}$), there must be a path from $i$ to
$-j$ and $j$ to $-i$, but these paths must intersect. For
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ to be planar, they must
intersect at a vertex, say $k$. But then
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ would cease to be the Hasse
diagram of a poset, as the relation $i<-i$ would follow by transitivity from
$i<k$ and $k<-i$. Thus, such a $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$
cannot be strongly planar. ∎

######  Lemma 15.17.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset such that
$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$)
is strongly planar. Then, if $\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}(P)$)
encloses more than one region, a cycle defining a rightmost region of
$\widehat{G}(P)$ is not fixed by the involution.

###### Proof.

Suppose $C$ is a cycle in $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) that encloses a rightmost
region, $\rho$. Suppose $C$ is fixed by the involution. Let $\sigma$ be a
region to the immediate left of $\rho$. The type $B$ and type $C$ cases are
slightly different.

First, suppose $P\subset\Phi_{B_{n}}$. There are two cases.

  1. 1.

Suppose $0$ is not in the left border of $\rho$. Let $e$ be an edge that lies
in the left border of $\rho$ and the right border of $\sigma$. Since $0$ is
not in the left border of $\rho$, an edge and its image under the involution
cannot both be in the left border. Therefore, since $\rho$ is fixed by the
involution, $\iota(e)$ must lie in the right border of $\rho$. But $\iota(e)$
is also in the border of $\iota(\sigma)$, contradicting that $\rho$ is a
rightmost region.

  2. 2.

Suppose $0$ is in the left border of $\rho$. Then if $j$ is in the left
border, $-j$ must also be in the left border, since $\rho$ is fixed by the
involution and $\pm j<0<\mp j$. Then the left border is fixed by the
involution. Since $\rho$ itself is fixed by the involution, this means that
the right border must also be fixed and thus symmetric, a contradiction, since
$0$ is in the left border, not the right.

Next, consider ${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$. There are
two cases.

  1. 1.

Suppose there is no edge of the form $(i,-i)$ in the left border of $\rho$.
Since $\sigma$ is to the left of $\rho$, there is some edge $e$ that lies in
the left border of $\rho$ and the right border of $\sigma$. Since $\rho$ is
fixed by the involution, $\iota(e)$ lies in the border of $\rho$. Since there
is no edge of the form $(i,-i)$ in the left border, it is not symmetric and
$\iota(e)$ cannot be in the left border, so it is in the right border. But,
$\iota(e)$ is in the border of $\iota(\sigma)$, meaning $\rho$ cannot be a
rightmost region, a contradiction.

  2. 2.

Suppose there is an edge $(i,-i)$ in the left border of $\rho$. Then, by the
definition of $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, if $j$ is in
the left border of $\rho$, then $-j$ is also in the left border of $\rho$.
From Lemma 15.16, one knows that the right border of $\rho$ does not include
an edge $(k,-k)$, so there must be some $k$ in the right border such that $-k$
is not in the right border. (In fact, the absence of such an edge means only
one of $k$ and $-k$ is in the right border for any $k$ that is not maximal or
minimal in $C$.) Then $\iota(k)=-k$ is a vertex in $\iota(C)$, which cannot be
$C$, so $C$ is not fixed by the involution.

∎

######  Proposition 15.18.

Suppose $P\subset\Phi_{B_{n}}$ (resp.
${P^{\scriptscriptstyle\vee}}\subset\Phi_{C_{n}}$) is a signed poset. Then
$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}}$)
has a strongly planar embedding if and only if it has a centrally symmetric
strongly planar embedding.

###### Proof.

One needs to show only that if $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) is strongly planar, it has a
centrally symmetric strongly planar embedding. Instead, prove a slightly
stronger statement, namely that if $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) has a strongly planar
embedding, it also has a centrally symmetric strongly planar embedding such
that the outer border of $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) is the same in both the first
strongly planar embedding and the centrally symmetric strongly planar
embedding.

Induct on the number of regions in $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$). There are two base cases.

First, suppose $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) consists of a single region,
bounded by a single cycle. One needs to address type $B$ and type $C$
separately.

  * •

In type $C$, consider $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. Since
it is strongly planar, it has a single maximum and a single minimum, which
must be $i$ and $-i$.

Claim:  There is no $j\neq i$ such that $j<-j$ or $-j<j$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.

Suppose not. Since $\widehat{G}_{C}(P)$ consists of a single cycle, one has
that either $i<j<-j<-i$, in which case one border of
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is fixed by the involution, so
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ must have more than one
region, a contradiction, or $j$ lies in the left border of
$\widehat{G}_{C}(P)$ and $-j$ lies in the right border of
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. In the latter case, one must
have that the left border of $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$
is sent to the right border under the involution, i.e. if $i\lessdot
c_{1}\lessdot\cdots\lessdot c_{k}\lessdot-i$ is the left border, then
$i\lessdot-c_{k}\lessdot\cdots\lessdot-c_{1}\lessdot-i$ is the right border.
However, it is then impossible that $j<-j$ since
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ encloses a single region.

Thus, when $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is strongly planar
and encloses a single region, $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$
can be embedded centrally symmetrically by evenly spacing the vertices around
a circle centered at the origin with $-i$ at the top and $-i$ at the bottom.

  * •

In type $B$, consider $\widehat{G}_{B}(P)$. Since $i<-i$, one must have
$i<0<-i$, meaning either the left or right border of $\widehat{G}_{B}(P)$ is
symmetric about $0$, so it is fixed by the involution. But then
$\widehat{G}_{B}(P)$ must have more than one region, contradicting the
assumption.

Next, consider the second base case where $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) consists of two regions. From
Lemma 15.17, one knows that the rightmost region $\rho$ is not fixed by the
involution. Therefore, the border between the two regions consists of a chain
fixed by the involution and the two regions are exchanged by the involution.
One then has the scenario depicted in Figure 40.

$\rho$$\iota(\rho)$$c$$d$$e$$\iota(e)$$\iota(d)$ Figure 40: Two regions
exchanged by the involution separated by a chain, $c$, fixed by the involution

One sees that $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) (when translated
appropriately) is centrally symmetric and strongly planar.

Now suppose $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) encloses more than two
regions. Then, by Proposition 15.14, it has a rightmost region, call it
$\rho$. The right border of $\rho$ is defined by some chain
$\min(\rho)\lessdot c_{1}\lessdot\cdots\lessdot c_{k}\lessdot\max(\rho)$.
Since $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) is a poset and $\rho$ is
enclosed by a cycle, $k\geq 1$. Since $\rho$ is rightmost, by definition each
edge in its right border lies in no cycle defining a region other than $\rho$.
The same then must be true of the image of the right border of $\rho$ in
$\iota(\rho)$. Deleting $c_{1},\ldots,c_{k},-c_{1},\ldots,-c_{k}$ and the
incident edges gives $\widehat{G}_{B}(P^{\prime})$ (resp.
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$) for a signed poset
$P^{\prime}$

Since $\widehat{G}(P)$ was strongly planar, $\widehat{G}(P^{\prime})$ is
strongly planar. By induction, $\widehat{G}(P^{\prime})$ has a centrally
symmetric strongly planar embedding. The left border of $\rho$ is now part of
the right border of $\widehat{G}(P^{\prime})$, meaning the chain
$c_{1}\lessdot\cdots\lessdot c_{k}$ and be attached to form a new region on
the right of $\widehat{G}(P^{\prime})$ and similarly for
$-c_{k}\lessdot\cdots\lessdot-c_{1}$ on the left. Since $\pm\min(\rho)$ and
$\pm\max(\rho)$ are positioned so as to be centrally symmetric, it is possible
to place the chains in such a way as to preserve central symmetry. Replacing
the chains in $\widehat{G}(P^{\prime})$ gives a centrally symmetric strongly
planar embedding of $\widehat{G}(P)$. ∎

######  Lemma 15.19.

Suppose $P\subset\Phi_{B_{n}}$ (resp. $P\subset\Phi_{C_{n}}$) is a signed
poset such that $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) is centrally symmetric and
strongly planar. Suppose $\rho$ is a rightmost region of $\widehat{G}(P)$ and
$c_{1}\lessdot\cdots\lessdot c_{k}$ is the portion of the left border of
$\rho$ which is part of the right border of some other region(s), with $k\geq
2$ and $c_{k}\neq-c_{k-1}$ (or $c_{1}\neq-c_{2}$). Then $\widehat{G}_{B}(P)$
(resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) is obtained from
$\widehat{G}_{B}(P^{\prime})$ (resp. $\widehat{G}_{C}(P^{\prime})$) for some
signed poset $P^{\prime}$ by closing a signed notch
$(c_{k-1},c_{k},c_{k}^{\prime})$ and $(-c_{k-1},-c_{k},-c_{k}^{\prime})$ or a
signed notch $(c_{2},c_{1},c_{1}^{\prime})$ and
$(-c_{2},-c_{1},-c_{1}^{\prime})$.

###### Proof.

Begin by observing that in the type $B$ case, one of $c_{1}$ and $c_{k}$ is
nonzero, and in the type $C$ case, at least one of $c_{k}\neq-c_{k-1}$ and
$c_{1}\neq-c_{2}$ holds. Without loss of generality, assume $c_{k}\neq 0$ in
$\widehat{G}_{B}(P)$ and $c_{k}\neq-c_{k-1}$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. (The argument in the other
cases is symmetric.)

Construct $\widehat{G}_{B}(P^{\prime})$ (resp. $\widehat{G}_{C}(P^{\prime})$)
by replacing $c_{k}$ by $c_{k}$ and $c_{k}^{\prime}$, with an edge between
$c_{k-1}$ and each of $c_{k}$ and $c_{k}^{\prime}$, with the edges incident to
$c_{k}$ in $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) that are not in the border of
$\rho$ moved to be incident to $c_{k}^{\prime}$. Replace $-c_{k}$ by $-c_{k}$
and $-c_{k}^{\prime}$ and partition the edges incident to $-c_{k}$ in
$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$)
in the same way. By construction, $\widehat{G}_{B}(P^{\prime})$ (resp.
$\widehat{G}_{C}(P^{\prime}c)$) will be centrally symmetric and strongly
planar and have an involution sending $i<j$ to $-j<-i$ (and fixing $0$ in the
type $B$ case). Then to show $P^{\prime}$ is really a signed poset, one needs
to check the conditions in Proposition 8.5 and Proposition 8.6.

First, consider the type $B$ case. Suppose $i<-i$ in
$\widehat{G}_{B}(P^{\prime})$. Since opening a notch does not introduce
relations, one must have that $i<-i$ in $\widehat{G}_{B}(P)$, so $i<0<-i$ in
$\widehat{G}_{B}(P)$. Suppose one does not have $-i>0$ in
$\widehat{G}_{B}(P^{\prime})$. There are two cases where $0<-i$ in
$\widehat{G}_{B}(P)$ could fail to lift to $\widehat{G}_{B}(P^{\prime})$.

  * •

Suppose $-i>c_{k}$ and $0<c_{k}^{\prime}$, but $0\not<-i$.

By construction (keeping in mind that $\widehat{G}_{B}(P)$ is centrally
symmetric and strongly planar), $0$ must be in the right border of $\rho$ and
$c_{k}^{\prime}=\max(\rho)$. However $0$ being in the right border contradicts
central symmetry.

  * •

Suppose $-i>-c_{k}$ and $0<-c_{k}^{\prime}$, but $0\not<-i$.

By construction, one has that $0$ is in the part of the border of
$\iota(\rho)$ in both $\widehat{G}_{B}(P)$ and in
$\widehat{G}_{B}(P^{\prime})$ not shared by any other region. Thus, since
$i<0$ in $\widehat{G}_{B}(P)$, one must have $i<0$ in
$\widehat{G}_{B}(P^{\prime})$, contradicting that $i<-i$ in
$\widehat{G}_{B}(P^{\prime})$.

In the type $C$ case, one must check that if $i<-i$ and $j<-j$ in
$\widehat{G}_{C}(P^{\prime})$, then $i<-j$ and $j<-i$.

First, observe that if $i<-i$ in $\widehat{G}_{C}(P^{\prime})$, then $i<-i$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$. (If $i=c_{k}^{\prime}$ in
$\widehat{G}_{C}(P^{\prime})$, then $c_{k}<-c_{k}$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.) Now suppose that $i<-i$ and
$j<-j$ in $\widehat{G}_{C}(P^{\prime})$. Then $i<-j$ and $j<-i$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ (possibly vacuously if $i=j$
in $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$). Should these relations
fail to lift to $\widehat{G}_{C}(P^{\prime})$, one is in one of the following
two cases.

  1. 1.

Suppose $i\leq c_{k}$ and $-j\geq c_{k}^{\prime}$.

There are several subcases.

    1. (1.a)

Suppose $i<c_{k-1}$. Then $i<-j$ and one is done.

    2. (1.b)

Suppose $i<c_{k}$, but $i\not<c_{k-1}$. In other words, $i$ is on the right
border of $\rho$ and $c_{k}=\max(\rho)$. Since $\rho$ is not fixed by the
involution and is a right region, $-i$ is not on the right border of $\rho$.
(If it were, either $\rho$ would contain an edge $(k,-k)$ and thus not be a
right region, or $i=\min(\rho)$, $-i=\max(\rho)$ and $\rho$ would be cut out
by a cycle fixed by the involution, since $\max(\rho)$ has degree 2, by
construction.) Therefore, one cannot have that $i<-i$ in
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, a contradiction.

    3. (1.c)

Suppose $i=c_{k}$. Then $c_{k}<-c_{k}$, but, by construction of
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$, elements in the border
of $\rho$ other than $c_{1}\lessdot\cdots\lessdot c_{k-1}$ are not in the
border of any other region in $\widehat{G}_{C}(P^{\prime})$ and $-c_{k}$ is in
the border of $\iota(\rho)$, a contradiction, a contradiction.

  2. 2.

Suppose $i\leq c_{k}^{\prime}$ and $-j\geq c_{k}$. In this case, $-j$ must be
on the outer border of $\rho$, as it is not one of
$c_{1}\lessdot\cdots\lessdot c_{k-1}$. Then, if $j<-j$, one has that either
$-c_{k}<c_{k}$ or $-c_{1}<c_{1}$. In the latter case, it follows that
$-c_{k}<c_{k}$. Therefore, without loss of generality, suppose $-j=c_{k}$.
Consider two cases.

    1. (2.a)

There is a chain from $-c_{k}$ to $c_{k}$ which is not fixed by the
involution. This chain and its image form a cycle fixed by the involution,
having $c_{k}$ as its maximum and $-c_{k}$ as its minimum. But then $c_{k}$,
which has degree two, must cover two elements in this cycle, meaning
$c_{k}=\max(\rho)$ and this cycle is the one that cuts out $\rho$. This is a
contradiction, since $\rho$ is not fixed by the involution.

    2. (2.b)

There is a unique chain from $-c_{k}$ to $c_{k}$. By construction of
$\widehat{G}_{C}(P^{\prime})$, this chain passes through $c_{k-1}$ and
$-c_{k-1}$ and extends uniquely to a maximal chain from $\min(\iota(\rho))$ to
$\max(\rho)$. On the other hand, since $i<-i$, there is a chain from $-i$ to
$i$. Since $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is centrally
symmetric and strongly planar, this chain must pass through the chain from
$\min(i\rho))$ to $\max(\rho)$, meaning $-c_{k-1}<-i$ and $i<c_{k-1}$, so
$-c_{k}<-i$ and $i<c_{k}$, as desired.

Lastly, one must show that $(c_{k-1},c_{k},c_{k}^{\prime})$ and
$(-c_{k-1},-c_{k},-c_{k}^{\prime})$ forms a signed notch in
$\widehat{G}(P^{\prime})$. However, this follows from the fact
$c_{1}\lessdot\cdots\lessdot c_{k-1}$ and $-c_{k-1}\lessdot\cdots\lessdot-
c_{1}$ are disconnecting chains in $\widehat{G}_{B}(P^{\prime})$ (resp.
$\widehat{G}_{C}(P^{\prime}))$.

∎

One is finally ready to prove Theorem 15.12.

###### Proof of Theorem 15.12.

Begin by realizing one may assume courtesy of Proposition 15.18 that
$\widehat{G}(P)$ is embedded in ${\mathbb{R}}^{n}$ so as to be centrally
symmetric and strongly planar. Next, recall that a centrally symmetric
strongly planar $\widehat{G}_{B}(P)$ contains at least two regions. If a
centrally symmetric strongly planar
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ has no region not fixed by the
involution, central symmetry guarantees it consists of one region, cut out by
a cycle fixed by the involution. This fixed cycle is the only cycle of
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$, so the toric ideal
$I^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}$ is the zero ideal and
$R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}$ is a complete intersection.

Thus, one may assume $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) has at least one region not
fixed by the involution. Induct on the number of two-element orbits of regions
under the involution of $\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}(P)$).
Since $\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}(P)$) is centrally symmetric
and strongly planar, it has a rightmost region $\rho$ and $\rho$ is not fixed
by the involution.

As a base case, consider the case where $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) has only one orbit. The case
where $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ consists of a single
orbit of one region has already been addressed. Thus, one may assume in both
the $\widehat{G}_{B}(P)$ and $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$
cases that there are precisely two regions, which are exchanged by the
involution. Call them $\rho$ and $\iota(\rho)$, with $\rho$ being the right
region. There are two cases.

  1. 1.

The regions $\rho$ and $\iota(\rho)$ share an edge and so intersect along a
chain. (See Figure 41(a).)

$\rho$$\iota(\rho)$ (a) Case 1

$0$$\rho$$\iota(\rho)$ (b) Case 2

Figure 41: The first two cases of the proof of Theorem 15.12

Then $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$) has three cycles: the cycle
defining $\rho$, its image under the involution, which defines $\iota(\rho)$,
and the cycle consisting of the “outer border” of $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$), i.e. those edges surrounding
only one of $\rho$ and $\iota(\rho)$. This last cycle must be fixed by the
involution, as the involution exchanges the cycles defining $\rho$ and
$\iota(\rho)$ while fixing the border between them. Then $I^{\mathrm{rt}}_{P}$
(resp. $I^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}$) is generated by the
cycle defining $\rho$ and, as the toric ideal is then principal,
$R_{P}^{\mathrm{rt}}$ (resp. $R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}$)
is a complete intersection.

  2. 2.

The regions $\rho$ and $\iota(\rho)$ do not share an edge and instead share a
single vertex. (See Figure 41(b). Note that this case only arises in
$\widehat{G}_{B}(P)$ (where the shared vertex is $0$).

Then the cycle defining $\rho$ and its imagine under the involution, which
defines $\iota(\rho)$, are the only cycles in $\widehat{G}_{B}(P)$, so
$I^{\mathrm{rt}}_{P}$ is a principal ideal and $R_{P}^{\mathrm{rt}}$ is a
complete intersection.

Suppose $\widehat{G}(P)$ has $n>1$ orbits of regions and the result holds for
posets $P^{\prime}$ such that $\widehat{G}(P^{\prime})$ has $k$ orbits of
regions not fixed by the involution. Let $\rho$ be a rightmost region of
$\widehat{G}(P)$. Let $c_{1}\lessdot\cdots\lessdot c_{k}$ be the chain
defining the left border of $\rho$. Note that the removal of
$c_{1}\lessdot\cdots\lessdot c_{k}$ would disconnect $\widehat{G}(P)$.

Let $c_{j_{1}}\lessdot\cdots\lessdot c_{j_{\ell}}$ be the portion of the
border of $\rho$ such that $c_{j_{1}}$ is the minimal vertex in the border
that lies in the border of more than one region and $c_{j_{\ell}}$ is the
maximal vertex in the border that lies in the border of more than one region.
(Equivalently, $c_{j_{1}}$ and $c_{j_{\ell}}$ are minimal and maximal,
respectively, elements in the chain having degree $\geq 3$. See Figure 42.)
Induct on $\ell$.

$c_{j_{1}}$$c_{j_{\ell}}$$c_{1}$$c_{k}$$\rho$ Figure 42: The chain
$c_{j_{1}}\lessdot\cdots\lessdot c_{j_{\ell}}$ in the proof of Theorem 15.12.

Postpone discussing the base case and suppose $\ell\geq 3$.
Claim: $P$ is obtained by closing a notch in some signed poset $P^{\prime}$
such that $\widehat{G}_{B}(P^{\prime})$ (resp.
$\widehat{G}_{C}(P^{\prime{\scriptscriptstyle\vee}})$) has a right region
whose left border is $c_{1}\lessdot\cdots\lessdot c_{k}$ such that
$c_{j_{\ell-1}}$ is the maximal element along the left border that is in the
border of more than one region.

There are three cases.

  1. 1.

Suppose $c_{j_{\ell}-1}\neq-c_{j_{\ell}}$ and $c_{j_{\ell}-1}\neq 0$.

By Lemma 15.19, there is a signed poset $P^{\prime}$ such that $P$ is obtained
from $P^{\prime}$ by closing a signed notch
$(c_{j_{\ell}-1},c_{j_{\ell}}^{\prime},c_{j_{\ell}})$ and
$(-c_{j_{\ell}-1},-c_{j_{\ell}}^{\prime},-c_{j_{\ell}})$.

By construction, $\widehat{G}_{B}(P^{\prime})$ (resp.
$\widehat{G}_{C}(P^{\prime})$) has a right region whose left border is
$c_{1}\lessdot\cdots\lessdot c_{k}$, such that $c_{j_{\ell}-1}$ is the maximal
element along the left border that is in the border of more than one region.
Then $\widehat{G}_{B}(P^{\prime})$ (resp. $\widehat{G}_{C}(P^{\prime})$) also
has a right region whose right border is $-c_{k}\lessdot\cdots\lessdot-c_{1}$,
such that $-c_{j_{\ell}-1}$ is the minimal element along the right border that
is in the border of more than one region.

The result then follows by induction.

  2. 2.

Suppose $c_{\ell-1}=-c_{\ell}$. This case can only arise in type $C$. Then
since $\ell\geq 3$, from Lemma 15.16, one knows that $c_{1}\neq-c_{2}$.

Once again, from Lemma 15.19, there is a signed poset $P^{\prime}$ such that
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is obtained from
$\widehat{G}_{C}(P^{\prime})$ by closing the signed notch comprised of
$(-c_{j_{2}},-c_{j_{1}},-c_{j_{1}}^{\prime})$ and
$(c_{j_{2}},c_{j_{1}},c_{j_{1}}^{\prime})$. By construction,
$\widehat{G}_{C}(P^{\prime})$ has a right region whose left border is
$c_{1}\lessdot\cdots\lessdot\cdot c_{k}$, such that
$c_{j_{2}}\lessdot\cdots\lessdot c_{j_{\ell}}$ is the shortest saturated chain
including all vertices in the left border of this region that lie in the right
border of another region.

The result then follows by induction.

  3. 3.

Suppose $c_{j_{\ell}}=0$. This case only arises in type $B$. Since $\ell\geq
3$, one knows that $c_{j_{1}}\neq 0$.

Then, from Lemma 15.19, one has a signed poset $P^{\prime}\subset\Phi_{B_{n}}$
such that $\widehat{G}_{B}(P)$ is obtained from $\widehat{G}_{B}(P^{\prime})$
by closing a signed notch $(-c_{j_{2}},-c_{j_{1}},-c_{j_{1}}^{\prime})$ and
$(c_{j_{2}},c_{j_{1}},c_{j_{1}}^{\prime})$ By construction,
$\widehat{G}_{B}(P^{\prime})$ has a right region whose right left border is
$c_{1}\lessdot\cdots\lessdot\cdot c_{k}$, such that
$c_{j_{2}}\lessdot\cdots\lessdot c_{j_{\ell}}$ is the shortest saturated chain
including all the vertices in the left border of this region that lie in the
right border of another region.

The result then follows by induction.

For the base case $\ell=2$, address types $B$ and $C$ separately. In type B,
there are two cases.

  1. 1.

Neither $c_{1}$ nor $c_{2}$ is $0$.

From Lemma 15.19, one has a signed poset $P^{\prime}$ such that
$\widehat{G}_{B}(P)$ is obtained from $\widehat{G}_{B}(P^{\prime})$ by closing
a signed notch $(c_{1},c_{2},c_{2}^{\prime})$ and
$(-c_{1},-c_{2},-c_{2}^{\prime})$. Then $\widehat{G}_{B}(P^{\prime})$ has a
rightmost region $\rho$ such that the left border of $\rho$ has precisely one
vertex in the border of another region and similarly for the right border of
$\iota(\rho)$. Then $\rho$ and $\iota(\rho)$ form a biconnected component in
$\widehat{G}(P^{\prime})$.

  2. 2.

One of $c_{1}$ and $c_{2}$ is $0$. Without loss of generality, assume
$c_{1}=0$.

One applies Lemma 15.19 once again, and one has a signed poset $P^{\prime}$
such that $\widehat{G}_{B}(P)$ is obtained from $\widehat{G}_{B}(P^{\prime})$
by closing a signed notch $(c_{1},c_{2},c_{2}^{\prime})$ and
$(-c_{1},-c_{2},-c_{2}^{\prime})$. Then $\widehat{G}_{B}(P^{\prime})$ has a
rightmost region $\rho$ such that the left border of $\rho$ has precisely one
vertex in the border of another region and similarly for the right border of
$\iota(\rho)$. Then $\rho$ and $\iota(\rho)$ form a biconnected component in
$\widehat{G}(P^{\prime})$.

In type C, there are two cases.

  1. 1.

Suppose $c_{1}\neq-c_{2}$.

In this case, Lemma 15.19 applies and there is a signed poset $P^{\prime}$
such that $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ is obtained from
$\widehat{G}_{C}(P^{\prime})$ by closing a signed notch
$(c_{1},c_{2},c_{2}^{\prime})$ and $(-c_{1},-c_{2},-c_{2}^{\prime})$. Then
$\widehat{G}_{C}(P^{\prime})$ has a rightmost region $\rho$ such that the left
border of $\rho$ has precisely one vertex in the border of another region and
similarly for the right border of $\iota(\rho)$. Then $\rho$ and $\iota(\rho)$
form a biconnected component in $\widehat{G}_{C}(P^{\prime})$.

  2. 2.

Suppose $c_{1}=-c_{2}$.

In this case, one has that $(c_{1},c_{2})$ is the only edge of either $\rho$
or $\iota(\rho)$ shared by another region. Since $(c_{1},c_{2})$ is of the
form $(i,-i)$, it must be shared by $\rho$ and $\iota(\rho)$ (and, since
$\widehat{G}(P)$ is strongly planar, by no other region). Consequently, $\rho$
and $\iota(\rho)$ must form an entire biconnected component of
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$.

In each of these cases, the rightmost region $\rho$ and its image under the
involution form a biconnected component of $\widehat{G}_{B}(P)$ (resp.
$\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$). Let $P_{1}$ be this
biconnected component and $P_{2}$ correspond to the rest of
$\widehat{G}_{B}(P)$ (resp. $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$).
Then, by the same argument as in the proof of Proposition 15.2, one has

$R_{P}^{\mathrm{rt}}=R_{P_{1}}^{\mathrm{rt}}\otimes_{k}R_{P_{2}}^{\mathrm{rt}}\quad\text{and}\quad
R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}}=R_{P_{1}}^{\mathrm{rt}}\otimes_{k}R_{P_{2}}^{\mathrm{rt}}.$

One has that $\widehat{G}(P_{2})$ is centrally symmetric and strongly planar
since $\widehat{G}(P)$ was and $\widehat{G}(P_{2})$ has one fewer region not
fixed by the involution than $\widehat{G}(P)$, so the result follows by
induction for $R_{P_{2}}^{\mathrm{rt}}$. One then has

$I^{\mathrm{rt}}_{P}=(U(C_{\rho}))\oplus
I^{\mathrm{rt}}_{P_{2}}=(U(C_{\rho}),U(C_{\sigma_{1}}),\ldots,U(C_{\sigma_{n-1}}))$

and

$I^{\mathrm{rt}}_{{P^{\scriptscriptstyle\vee}}}=(U(C_{\rho}))\oplus
I^{\mathrm{rt}}_{P_{2}}=(U(C_{\rho}),U(C_{\sigma_{1}}),\ldots,U(C_{\sigma_{n-1}})),$

where $\sigma_{1},\ldots,\sigma_{n-1}$ are the orbits of regions of
$\widehat{G}(P_{2})$ not fixed by the involution. Then

$\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})=\mathrm{Hilb}(R_{P_{1}}^{\mathrm{rt}},\bm{x})\mathrm{Hilb}(R_{P_{2}}^{\mathrm{rt}},\bm{x})\quad\text{and}\quad\mathrm{Hilb}(R_{{P^{\scriptscriptstyle\vee}}}^{\mathrm{rt}},\bm{x})=\mathrm{Hilb}(R_{P_{1}}^{\mathrm{rt}},\bm{x})\mathrm{Hilb}(R_{P_{2}}^{\mathrm{rt}},\bm{x}).$

Since, in each case, $I^{\mathrm{rt}}_{P_{1}}$ is a principal ideal and $\rho$
has a single maximum and a single minimum, one knows that

$\mathrm{Hilb}(R_{P_{1}}^{\mathrm{rt}},\bm{x})=\dfrac{(1-\bm{x}^{\rho})}{\prod_{e\in
C_{\rho}}(1-x_{a}^{\delta}x_{b}^{-\epsilon})},$

where the product in the denominator runs over edges $e=\delta a\to\epsilon b$
in the cycle defining $\rho$, and taking $x_{0}=1$ in type $B$. By induction,

$\mathrm{Hilb}(R_{P_{2}}^{\mathrm{rt}},\bm{x})=\dfrac{\prod_{\sigma}(1-\bm{x}^{\rho})}{\prod_{e}(1-x_{a}^{\delta}x_{b}^{-\epsilon})},$

where $\sigma$ runs over orbits of regions enclosed by $\widehat{G}(P_{2})$
not fixed by the involution and $e$ runs over the orbits of edges $\delta
a\to\epsilon b$ $(a<b)$ in $\widehat{G}(P_{2})$.

Then

$\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})=\dfrac{\prod_{\rho}(1-\bm{x}^{\rho})}{\prod_{e}(1-x_{a}^{\delta}x_{b}^{-\epsilon}),}$

where $\rho$ runs over all orbits of regions enclosed by $\widehat{G}(P)$, as
desired. ∎

####  15.4 Computing $\Psi_{P}$

While the proof of Theorem 15.12 proceeds as above, using Proposition 6.14 to
compute $\Psi_{P}$ requires having a regular sequence generating the toric
ideal. This regular sequence is implicit in Theorem 15.12—the cycle binomials
of the cycles cutting out the regions of $\widehat{G}(P)$ form a regular
sequence—though one needs to work in the opposite order to the proof to see
the regular sequence, building the poset up from cycles rather than breaking
it apart into cycles.

Recall the posets $P_{1}$ (Figure 37(a) and 37(b)), $P_{2}$ (Figures 37(c) and
37(d)) and $P_{3}$ (Figure 38) from the previous section. To find a regular
sequence generating $I^{\mathrm{rt}}_{P}$, one starts by noting that

$R_{P_{3}}^{\mathrm{rt}}\cong
k[U_{1},U_{1\bar{2}},U_{23},U_{3}]/(U_{1\bar{2}}U_{23}-U_{1}U_{3}).$

One then closes the notch to obtain $\widehat{G}_{B}(P_{2})$, which has the
following impact on the semigroup ring courtesy of Proposition 15.9:

$\displaystyle R_{P_{2}}^{\mathrm{rt}}$ $\displaystyle\cong
k[U_{1},U_{1\bar{2}}U_{23},U_{3}]/(U_{1\bar{2}}U_{23}-U_{1}U_{3},U_{1}-U_{3})$
$\displaystyle\cong
k[U_{1},U_{1\bar{2}},U_{12}]/(U_{1\bar{2}}U_{12}-U_{1}^{2})$

One sees that $\widehat{G}_{B}(P^{\prime})$ has two biconnected components,
one coming from $P_{1}$ and the other from $P_{2}$.

$R_{P_{1}}^{\mathrm{rt}}\cong
k[U_{\bar{3}4},U_{4\bar{5}},U_{\bar{2}3},U_{\bar{2}5}]/(U_{\bar{3}4}U_{\bar{2}3}-U_{4\bar{5}}U_{\bar{2}5})$

Then

$\displaystyle R_{P^{\prime}}^{\mathrm{rt}}$ $\displaystyle\cong
R_{P_{1}}^{\mathrm{rt}}\otimes R_{P_{2}}^{\mathrm{rt}}$ $\displaystyle\cong
k[U_{1},U_{1\bar{2}},U_{12},U_{\bar{2}3},U_{\bar{3}4},U_{4\bar{5}},U_{\bar{2}5}]/(U_{1\bar{2}}U_{12}-U_{1}^{2},U_{\bar{3}4}U_{\bar{2}3}-U_{4\bar{5}}U_{\bar{2}5}).$

Closing the last notch to obtain $\widehat{G}_{B}(P)$ means

$\displaystyle R_{P}^{\mathrm{rt}}$ $\displaystyle\cong
k[U_{1},U_{1\bar{2}},U_{12},U_{\bar{2}3},U_{\bar{3}4},U_{4\bar{5}},U_{\bar{2}5}]/(U_{1\bar{2}}U_{12}-U_{1}^{2},U_{\bar{3}4}U_{\bar{2}3}-U_{4\bar{5}}U_{\bar{2}5},U_{\bar{2}5}-U_{1\bar{2}})$
$\displaystyle\cong
k[U_{1},U_{1\bar{2}},U_{12},U_{\bar{2}3},U_{\bar{3}4},U_{\bar{1}4}]/(U_{12}U_{1\bar{2}}-U_{1}^{2},U_{1\bar{2}}U_{\bar{1}4}-U_{\bar{3}4}U_{\bar{2}3}).$

Each step in this process results in a regular sequence, so one finishes with
a regular sequence generating $I^{\mathrm{rt}}_{P}$. One has

$\mathrm{Hilb}(R_{P}^{\mathrm{rt}},\bm{x})=\dfrac{(1-x_{1}^{2})(1-x_{2}^{-1}x_{4})}{(1-x_{1})(1-x_{1}x_{2})(1-x_{1}x_{2}^{-1})(1-x_{1}^{-1}x_{4})(1-x_{3}^{-1}x_{4})(1-x_{2}^{-1}x_{3})}.$

Applying Proposition 6.14 gives

$\Psi_{P}(\bm{x})=\dfrac{2x_{1}(x_{4}-x_{2})}{x_{1}(x_{1}+x_{2})(x_{1}-x_{2})(x_{4}-x_{1})(x_{4}-x_{3})(x_{3}-x_{2})}.$
(9)

On the other hand, Figure 43 shows the poset of order ideals of $P$, from
which the linear extensions can be read off.

$(0,0,0,0)$$(0,0,0,1)$$(0,0,1,1)$$(1,0,0,1)$$(1,0,1,1)$$(1,-1,0,1)$$(1,1,1,1)$$(1,-1,1,1)$$(1,-1,-1,1)$
Figure 43: $J(P)$ for $P$ in Figure 35

The linear extensions are:

$\begin{split}\begin{pmatrix}1&2&3&4\\\
4&3&1&2\end{pmatrix},\begin{pmatrix}1&2&3&4\\\
4&3&1&-2\end{pmatrix},\begin{pmatrix}1&2&3&4\\\ 4&1&3&2\end{pmatrix},\\\
\begin{pmatrix}1&2&3&4\\\ 4&1&3&-2\end{pmatrix},\begin{pmatrix}1&2&3&4\\\
4&1&-2&3\end{pmatrix},\begin{pmatrix}1&2&3&4\\\
4&1&-2&-3\end{pmatrix}.\end{split}$

Then

$\Psi_{P}(\bm{x})=\sum_{w\in\mathcal{L}(P)}w\left(\dfrac{1}{(x_{1}-x_{2})(x_{2}-x_{3})(x_{3}-x_{4})x_{4}}\right)=\\\
\dfrac{1}{(x_{4}-x_{3})(x_{3}-x_{1})(x_{1}-x_{2})x_{2}}-\dfrac{1}{(x_{4}-x_{3})(x_{3}-x_{1})(x_{1}+x_{2})x_{2}}\\\
+\dfrac{1}{(x_{4}-x_{1})(x_{1}-x_{3})(x_{3}-x_{2})x_{2})}-\dfrac{1}{(x_{4}-x_{1})(x_{1}-x_{3})(x_{2}+x_{3})x_{2}}\\\
-\dfrac{1}{(x_{4}-x_{1})(x_{1}+x_{2})(x_{2}+x_{3})x_{3}}-\dfrac{1}{(x_{4}-x_{1})(x_{1}+x_{2})(x_{3}-x_{2})x_{3}}\\\
=\dfrac{2(x_{4}-x_{2})}{(x_{4}-x_{3})(x_{3}-x_{2})(x_{4}-x_{1})(x_{1}-x_{2})(x_{1}+x_{2})}\cdot\dfrac{x_{1}}{x_{1}},$

agreeing with (9).

One builds $\widehat{G}_{C}({P^{\scriptscriptstyle\vee}})$ (see Figure 35)
similarly, except one actually arrives at a principal ideal.

$R_{P^{\prime{\scriptscriptstyle\vee}}}^{\mathrm{rt}}\cong
R_{P_{1}^{\scriptscriptstyle\vee}}^{\mathrm{rt}}\otimes
R_{P_{2}^{\scriptscriptstyle\vee}}^{\mathrm{rt}}.$