\subsection{Facets}
\label{sec:facet}
%Here we consider the dimension and facets of a directed edge polytope.
%To describe dimensions, we define \cref{condition:fuga}, and show
%\cref{thm:dim}. 
%To characterize facets,
%we introduce \cref{condition:piyo,condition:hogera},
%and show \cref{thm:facet}.
Let $R$ be a lluf subquiver of $Q$ so that both $\DE(R)$
and $\DE(Q)$ are contained in $\RR^{Q_{0}}$.
In order to prove
\cref{thm:facet}, we would like to know when
$\DE(R)$ is a face of $\DE(Q)$ and $\dim (\DE(R))=\dim(\DE(Q))-1$.

We first obtain the following relation between the coconnectivities of
$Q$ and $R$ by the dimension condition.

By  \cref{thm:dim}, we have the following:
\begin{lemma}
 \label{lem:coconnectivity}
 Let $R$ be a lluf subquiver of $Q$. If $\DE(R)$ is a facet of $\DE(Q)$,
 then $c(R)=c(Q)$ or $c(R)=c(Q)-1$.
% Thus $\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}$ or 
% $\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}+1$. 
\end{lemma}




%% \begin{proof}
%%  When $Q$ has a rank function, so does $R$. And we have
%%  \[
%%  c(R)-1 = \dim(\DE(R)) = \dim(\DE(Q))-1 = c(Q)-2
%%  \]
%%  by \cref{thm:dim}, or $c(R)=c(Q)-1$.
%%  If $Q$ does not have a rank function, $\dim(\DE(Q))=c(Q)$, and we have
%%  $c(R)=c(Q)$ or $c(R)=c(Q)-1$, depending on the existence of a rank
%%  function on $R$.
%% \end{proof}

If a facet $\DE(R)$  of $\DE(Q)$
satisfies
%$\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}$,
$c(R)=c(Q)$,
then $Q$ has no rank function but $R$ has a rank function.
In this case, we obtain \cref{lem:facetispiyo},
which implies nessesary condition of \cref{thm:facet}
for the case where $c(R)=c(Q)$:
\begin{lemma}
 \label{lem:facetispiyo}
 Let $R$ be a lluf subquiver of $Q$ with
 $c(R)=c(Q)$. 
% $\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}$. 
 If $\DE(R)$ is a facet of $\DE(Q)$, then
 the subquiver $R$ has a rank function
 $\rho\in \RR^{Q_0}$ such that
 \[
 (\rho(v)-\rho(w)+1)(\rho(v')-\rho(w')+1)> 0.
  \]
 for $(v,w),(v' ,w') \in Q_1\setminus R_1$ 
 and $Q$ does not have a rank function.
\end{lemma}
%% \begin{proof}
%%  Since $c(R)=c(Q)$ and $\DE(R)$ is a facet of $\DE(Q)$,
%%  \[
%%   c(Q) \ge \dim(\DE(Q)) = \dim(\DE(R))+1 \ge c(R)=c(Q),
%%  \]
%%   $Q$ does not have a rank function and
%%   $R$ has a rank function by \cref{thm:dim}. We also have
%%  $\dim(\DE(Q))=c(R)$ and $\dim(\DE(R))=c(R)-1$.
%%   Since $\numof{\pi_{0}(Q)}=\numof{\pi_{0}(R)}$,
%%   $v$ and $w$ are in the same connected component of $R$
%%   for each $(v,w)\in Q_1$.
%%   Hence $V_{R}$ contains $\varepsilon_{(v,w)}$
%%   for all $(v,w)\in Q_1$. In other words, $\DE(Q)$ is a convex polytope
%%  in $V_{R}$.

%%  Let $\rho$ be a rank function of $R$.
%%  Fix $(v_0,w_0)\in R_1$ and consider the hyperplane
%%  $H_{\rho}=\rho^{\perp}+\varepsilon_{(v_{0},w_{0})}$.
%%  Then $H_{\rho}\cap V_{R}$ is an affine space of dimension $c(R)-1$
%%   which contains $\DE(R)$. It means that $H_{\rho}\cap V_{R}$ is a
%%  supporting hyperplane of $\DE(R)$ in $V_{R}$.
%%  Thus we have
%%     \begin{align*}
%%     \forall (v,w)\in Q_1,&\quad
%%     \Braket{\rho, \varepsilon_{(v,w)}-\varepsilon_{(v_0,w_0)}}=\rho(v)-\rho(w)+1\geq 0
%%     \intertext{or}
%%     \forall (v,w)\in Q_1,&\quad
%%     \Braket{\rho, \varepsilon_{(v,w)}-\varepsilon_{(v_0,w_0)}}=\rho(v)-\rho(w)+1\leq 0.
%%   \end{align*}

%%  It remains to show that these values are nonzero if 
%%  $(v,w)\in Q_{1}\setminus R_{1}$.
%%   If $\rho(v)-\rho(w)=1$,
%%   then $H_{\rho}\cap V_{R}$ contains $\varepsilon_{(v,w)}$.
%%   Since $\DE(R)$ is a facet  of $\DE(Q)$,
%%   it follows that
%%   $\DE(R)$ contains $\varepsilon_{(v,w)}$,
%%   which implies $(v,w)\in R_1$.
%% \end{proof}



If a facet $\DE(R)$  of $\DE(Q)$
%satisfies $\numof{\pi_{0}(R)}<\numof{\pi_{0}(Q)}$,
satisfies $c(R)>c(Q)$,
then both $Q$ and $R$ have a rank function
or both has no rank function.
In this case, we have the following:
\begin{lemma}
  \label{lem:facetisfull}
  Let $R$ be a subquiver of $Q$
  with
  $c(R)=c(Q)-1$.
%$\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}+1$.
  If $\DE(R)$ is a facet of $\DE(Q)$,
%  then each connected component of $R$
%  is a full subquiver of $Q$.
  then $R$ is a component-wise full subquiver of $Q$.
\end{lemma}
%% \begin{proof}
%%  Let $n=\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}+1$ and
%%  $R^{(1)},\ldots, R^{(n)}$ be the complete list of
%%  connected components of $R$ so that
%%  \[
%%   V_{R} = \bigcap_{k=1}^{n} \kappa_{R^{(k)}}^{\perp}.
%%  \]
%%  Let $(v,w)\in Q_1$ satisfy $v,w\in R^{(k)}_0$ for some $k$.
%%  It suffices to show that $\varepsilon_{(v,w)}\in \DE(R)$, since it is
%%  equivalent to $(v,w)\in R_{1}$ by \cref{lem:vertices}.

%%  By \cref{lemma:ocs:constant}, $V_{R}$ is a $c(R)$-dimensional affine
%%  space, which is a hyperplane in $V_{Q}$ by the assumption
%%  $c(R)=c(Q)-1$.  
%%  We have $\varepsilon_{(v,w)} \in V_{R}$, since
%%  \begin{align*}
%%   \kappa_{R^{(i)}_{0}}(\varepsilon_{(v,w)}) & =
%%   \begin{cases}
%%    1-1 & (i=k) \\
%%    0-0 & (i\neq k)
%%   \end{cases} \\
%%   & = 0
%%  \end{align*}
%%  for any $i=1,\ldots,n$.

%% % For all $v,w\in R^{(k)}_0$,
%% %  \begin{align*}
%% %    \braket{\kappa_{R^{(i)}_0},\varepsilon_{(v,w)}}=0.
%% %  \end{align*}
%% %  Let $A$ be the intersection of 
%% %  $\kappa_{R^{(k)}_0}^\perp$ for all $k$.
%% %  Then $A$ is an affine space of dimension $c(Q)$,
%% %  and $A$ contains $\varepsilon_{(v,w)}$.
  
%%   If $R$ does not have a rank function,
%%   then $\DE(R)$ is a $c(R)$-dimensional polytope
%%   contained in $V_{R}$. In other words, $V_{R}$ is a supporting
%%  hyperplane of $\DE(R)$ in $V_{Q}$, which implies that
%%   $\varepsilon_{(v,w)}\in \DE(R)$. 
  
%%  Suppose that $R$ has a rank function.
%%  $\DE(R)$ is of dimension $c(R)-1$ by \cref{thm:dim}.
%%  Since $\DE(R)$ is a facet of $\DE(Q)$,
%%  \[
%%   \dim(\DE(Q)) = \dim(\DE(R))+1 = c(R) = c(Q)-1.
%%  \]
%%  By \cref{thm:dim}, $Q$ also has a rank function, which is denoted by  
%%  $\rho$.
%%  It defines a
%%  hyperplane 
%%  \[
%%   H_{\rho} = \rho^{\perp} + \varepsilon_{(v_{0},w_{0})}
%%  \]
%%  in $\RR^{Q_{0}}$ for some $(v_{0},w_{0})\in Q_{1}$.
%%  We may choose $(v_{0},w_{0})\in R_{1}$.
%%  By \cref{lemma:ocs:rank}, $\DE(R)$ is contained in $H_{\rho}$, hence in
%%  $H_{\rho}\cap V_{R}$.
 
%%  Since $\rho$ is a rank function on $Q$, we have
%%  $\varepsilon_{(v,w)}\in H_{\rho}$ for $(v,w)\in Q_{1}$.
%%  If $v,w\in R_{1}^{(k)}$ for some $k$, we also have
%%  $\varepsilon_{(v,w)}\in V_{R}$, and hence
%%  $\varepsilon_{(v,w)} \in H_{\rho}\cap V_{R}$.
%%  Note that $H_{\rho}$ and $V_{R}$ intersect transversally, and we have 
%%  \[
%%   \dim (H_{\rho}\cap V_{R}) = \dim(V_{R})-1 = c(R)-1=\dim(\DE(R)).
%%  \]
%%  In other words, $H_{\rho}\cap V_{R}$ is the affine hull of $\DE(R)$ in
%%  $H_{\rho}$ and it should be the supporting hyperplane of $\DE(R)$ in
%%  $H_{\rho}$, since $\DE(R)$ is a facet of $\DE(Q)$.
%%  It implies that $\varepsilon_{(v,w)}\in \DE(R)$.
%% \end{proof}
If $Q/R$ has a directed cycle,
then we have a pair $(R^{1},R^{2})$ of connented components of $R$
with an edge of $Q$ connecting $R^1$ to $R^2$
and  an edge of $Q$ connecting $R^2$ to $R^1$.
The hyperplanes corrsponding to those edges
separates the connected components  $R^{1},R^{2}$.
Hence we have the following:
\begin{lemma}
\label{lem:facetisacyclic}
  Let $R$ be a subquiver of $Q$
  with
  $c(R)=c(Q)-1$.
%  $\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}+1$.
  If $\DE(R)$ is a facet of $\DE(Q)$,
  then $Q/R$ is acyclic.
\end{lemma}
\Cref{lem:facetisacyclic,lem:facetisfull} imply
nessesary condition of \cref{thm:facet}
for the case where $c(R)+1=c(Q)$
%% \begin{proof}
%% % By assumption, $\numof{\pi_{0}(R)}=\numof{\pi_{0}(Q)}+1$.
%%  Denote
%%  \begin{align*}
%%   \pi_{0}(Q) & =\{Q^{(1)},\ldots, Q^{(n-1)}\} \\
%%   \pi_{0}(R) & =\{R^{(1)},\ldots, R^{(n)}\}.
%%  \end{align*}
%%  Without loss of generality, we may assume that $R^{(i)}\subset Q^{(i)}$
%%  for $i=1,\ldots,n-2$ and $R^{(n-1)}\cup R^{(n)}\subset Q^{(n-1)}$.
%%  Let us denote $R'=R^{(n-1)}\cup R^{(n)}$ and $Q'=Q^{(n-1)}$.
%%  We should have
%%  $R^{(i)}=Q^{(i)}$ for $i=1,\ldots,n-2$ and $\DE(R')$
%%  is a facet of $\DE(Q')$, since $\dim(\DE(R))=\dim(\DE(Q))-1$.
%%  It implies that $Q/R$ is a union of $(n-2)$
%%  quivers consisting of a single vertex and
%%  $Q'/R'$.

%%  If $Q/R$ were to have a directed cycle, it should be contained in
%%  $Q'/R'$. By \cref{lem:facetisfull},
%%  both $R^{(n-1)}$ and $R^{(n)}$ are full subquivers of $Q'$.
%%  Since $R$ is lluf, such a directed cycle contains edges
%%  $(v,w),(v',w')\in Q'_{1}$ with $v,w'\in R^{(n-1)}_{0}$ and
%%  $w,v'\in R^{(n)}_{0}$.
%%  Then we have
%%   \begin{align}
%%    \Braket{\kappa_{R^{(n-1)}_0},\varepsilon_{(v,w)}} &=1 \label{align:1} \\
%%     \Braket{\kappa_{R^{(n-1)}_0},\varepsilon_{(v',w')}}
%%    &=-1. \label{align:-1} 
%%   \end{align}
%% % and the hyperplane $\kappa_{R^{(n-1)}_{0}}^{\perp}$ is not a supporting
%% % hyperplane of $\DE(R')$. 

%%  If $R'$ does not have a rank function, this contracts to the fact that    
%%  $V_{R'}=\kappa_{R^{(n-1)}_{0}}^{\perp}\cap\kappa_{R^{(n)}_{0}}^{\perp}$
%%  is a supporting hyperplane of $\DE(R')$ in $V_{Q'}$,
%%  as we have seen in the proof of
%%  \cref{lem:facetisfull}.

%%  If $R'$ has a rank function, so does $Q'$ as is shown in the proof of
%%  \cref{lem:facetisfull}. Let $\rho$ be a rank 
%%  function of $Q'$. Then, again by the proof of \cref{lem:facetisfull},
%%  $H_{\rho}\cap V_{R'}$ is a supporting hyperplane of $\DE(R')$, which
%%  contradicts to (\ref{align:1}) and (\ref{align:-1}). 

%%  Hence $Q/R$ is acyclic.
%% \end{proof}  



%% One of sufficient conditions for $\DE(R)$ being a facet is the
%% acyclicity of the quiver $Q/R$ obtained from $Q$ by ``contracting''
%% $R$. 

%% \begin{definition}
%%  \label{def:contraction}
%%  Let $R$ be a subquiver of a quiver $Q$ such that each
%%  connected component of $R$ is a full subquiver of
%%  $Q$. Define an equivalence relation $\sim$ on
%%  $Q_{0}$ by 
%%  \[
%%   v \sim w \Longleftrightarrow \text{$v$ and $w$
%%  are connected by an undirected walk in $R$}.
%%  \]
%%  The equivalence class of $v\in Q_{0}$ is denoted by $[v]$.
%%  Define a quiver $Q/R$ by 
%%  \begin{align*}
%%   (Q/R)_{0} & = Q_{0}/_{\sim} \\
%%   (Q/R)_{1} & = \Set{([v],[w]) | (v,w)\in Q_{1}\setminus R_{1}}. 
%%  \end{align*}
%% \end{definition}
%% %\begin{definition}
%% % \label{def:contraction}
%% %For a subquiver $\check Q$ of $Q$
%% %such that
%% %each connected component of $\check Q$ is a full subquiver of $Q$,
%% %we define a quiver $Q/\check Q$ to be the quiver $(Q_0/\check Q_0,Q_1/\check Q_1)$
%% %obtained by
%% %\begin{align*}
%% %  Q_0/\check Q_0 &= \Set{[v]|v\in Q_0},\\
%% %  Q_1/\check Q_1 &= \Set{([v],[w])|(v,w)\in Q_1\setminus \check Q_1}, 
%% %\end{align*}
%% %where
%% %$[v]=\Set{w|\text{$v$ and $w$ is weakly connected in $(Q_0,\check Q_1)$}}$.
%% %\end{definition}

%% Roughly speaking, $Q/R$ is the quiver obtained from $Q$ by
%% collapsing each connected component of $R$ to a point.

By the calculation of the supporting hyperplanes,
we have \cref{lem:hogeraisface,lem:piyoisface}.
\begin{lemma}
 \label{lem:hogeraisface}
 Let $R$ be
 a component-wise full subquiver of $Q$.
% a lluf proper subquiver of a quiver $Q$ such that
% each connected component of $R$ is a full subquiver of $Q$.
 If $Q/R$ is acyclic, then $\DE(R)$ is a face of
 $\DE(Q)$. 
\end{lemma}

%% \begin{proof}
%%  Let us denote the connected components of $R$ by
%%  \[
%%   \pi_{0}(R) = \{R^{(1)},\ldots, R^{(n)}\}.
%%  \]
%%  Since $Q/R$ is acyclic, we may assume that, if
%%  $v\in R^{(i)}_{0}$ and $w\in R^{(j)}_{0}$ are connected
%%  by an edge in $Q$, then $i< j$.

%%  Denote $C_{k}= \bigcup_{i=1}^{k} R^{(i)}_{0}$.
%%  Then, for $v,w\in Q_{0}$,
%%  \begin{align*}
%%   \langle \kappa_{C_{k}},\varepsilon_{(v,w)}\rangle & =  
%%   \kappa_{C_{k}}(v) - \kappa_{C_{k}}(w) \\
%%   & = 
%%   \begin{cases}
%%    1 & (v\in C_{k} \text{ and } w\not\in C_{k}) \\
%%    -1 & (v\not\in C_{k} \text{ and } w\in C_{k}) \\
%%    0 & (\text{otherwise}).
%%   \end{cases}
%%  \end{align*}
%%  By our choice, the second case does not occur and we have
%%  $\langle\kappa_{C_{k}},\varepsilon_{(v,w)}\rangle\ge 0$.
%%  In other words, the orthogonal complement $\kappa_{C_{k}}^{\perp}$ is a
%%  supporting hyperplane of 
%%  $\DE(Q)$ and thus $\DE(Q)\cap\kappa_{C_{k}}^{\perp}$ is a face of
%%  $\DE(Q)$ for each $k$.

%%  We claim that
%%  \[
%%   \DE(Q)\cap \bigcap_{i=1}^{n} \kappa_{C_{k}}^{\perp} = \DE(R)
%%  \]
%%  or
%%  \[
%%  \Set{\varepsilon_{(v,w)}|(v,w)\in Q_{1}}
%%  \cap \bigcap_{i=1}^{n} \kappa_{C_{k}}^{\perp} =
%%  \Set{\varepsilon_{(v,w)}|(v,w)\in R}. 
%%  \]
%%  If $(v,w)\in R_{1}$, 
%%  $\langle \kappa_{C_{k}},\varepsilon_{(v,w)}\rangle =0$ for all
%%  $k$ by the previous calculation. Conversely, suppose that
%%  $(v,w)\in Q_{1}\setminus R_{1}$ with $v\in R^{(i)}_{0}$
%%  and $w\in R^{(j)}_{0}$. By assumption, $i<j$, which implies that
%%  \[
%%   \langle \kappa_{C_{i}},\varepsilon_{(v,w)}\rangle = 1-0=1\neq 0
%%  \]
%%  and we have $\varepsilon_{(v,w)}\not\in \kappa_{C_{i}}^{\perp}$.
%% \end{proof}

%Another sufficient condition for being a face is the following.

\begin{lemma}
  \label{lem:piyoisface}
  Let $R$ be a lluf proper subquiver of a quiver $Q$.
  If $R$ has a rank function $\rho\in \RR^{Q_0}$ such that 
  \begin{align*}
    (\rho(v)-\rho(w)+1)(\rho(v')-\rho(w')+1)> 0
  \end{align*}
 for any $(v,w),(v',w')\in Q_{1}\setminus R_{1}$,
 then $\DE(R)$ is a face of $\DE(Q)$.
\end{lemma}
%% \begin{proof}
%%  Suppose a lluf subquiver $R$ of $Q$
%%  has such a rank function $\rho$.
%%  Fix $(v_0,w_0)\in R_1$ and consider the hyperplane
%%  \[
%%   H_{\rho} = \rho^{\perp}+\varepsilon_{(v_{0},w_{0})}.
%%  \]
%%  We claim that $H_{\rho}$ is a supporting hyperplane of $\DE(Q)$.
 
%%  For $(v,w)\in Q_1$, we have
%%  \begin{align*}
%%   \Braket{\rho, \varepsilon_{(v,w)}-\varepsilon_{(v_0,w_0)}} = 
%%   \rho(v)-\rho(w) - \rho(v_{0})+\rho(w_{0}) = \rho(v)-\rho(w)+1.
%%  \end{align*}
%%  Note that $\Braket{\rho, \varepsilon_{(v,w)}-\varepsilon_{(v_0,w_0)}}=0$
%%  for $(v,w)\in R_{1}$.
%%  By our assumption on $\rho$ we have
%%  $\Braket{\rho, \varepsilon_{(v,w)}-\varepsilon_{(v_0,w_0)}}\ge 0$
%%  for any $(v,w)\in Q_{1}$ or
%%  $\Braket{\rho, \varepsilon_{(v,w)}-\varepsilon_{(v_0,w_0)}}\le 0$
%%  for any $(v,w)\in Q_{1}$
%%  and $H_{\rho}$ is a supporting hyperplane of $\DE(Q)$.

%%  It remains to show that $\DE(Q)\cap H_{\rho}=\DE(R)$.
%%  Again by the assumption on $\rho$,
%%   \begin{align*}
%%     \Braket{\rho, \varepsilon_{(v,w)}-\varepsilon_{(v_0,w_0)}}=
%%     \rho(v)-\rho(w)+1=0 \iff  (v,w)\in R_{1}
%%   \end{align*}
%%   for $(v,w)\in Q_1$.
%%   Hence the hyperplane
%%   $H_{\rho}$ contains $\DE(R)$,
%%   but the hyperplane does not contain $ \varepsilon_{(v,w)}$
%%   for any $(v,w)\in Q_1\setminus R_1$.
%%   Hence the face $\DE(Q)\cap H_{\rho}$ coincides with $\DE(R)$.
%% \end{proof}


%% Now we are ready to prove \cref{thm:facet,thm:face}.

%% \begin{proof}[Proof of \cref{thm:facet}]
%%  Let $R$ be a lluf subquiver of $Q$ with $\dim(\DE(R))=\dim(\DE(Q))-1$.
 
%%  Suppose that $\DE(R)$ is a facet of $\DE(Q)$. 
%%  By \cref{lem:coconnectivity}, $c(R)=c(Q)$  or $c(R)=c(Q)-1$.
%%  When $c(R)=c(Q)$, \cref{lem:facetispiyo} implies that the
%%  condition (\ref{condition:piyo}) in \cref{thm:facet} holds.
%%  If $c(R)=c(Q)-1$, the condition (\ref{condition:hogera}) in
%%  \cref{thm:facet} follows from  
%%  \cref{lem:facetisfull,lem:facetisacyclic}.  

%%  Conversely, if a lluf subquiver $R$ satisfies 
%%  (\ref{condition:hogera}) in \cref{thm:facet}, then $\DE(R)$ is a face
%%  of $\DE(Q)$ by \cref{lem:hogeraisface}. 
%%  If $R$ satisfies (\ref{condition:piyo}) in \cref{thm:facet}, $\DE(R)$
%%  is a face of $\DE(Q)$ by \cref{lem:piyoisface}. Hence, in both cases,
%%  $\DE(R)$ is a facet.  
%% \end{proof}

%% \begin{proof}[Proof of \cref{thm:face}]
%%  Suppose that $\DE(R)$ is a face of $\DE(Q)$. Then there exists a
%%  descending sequence of subquivers
%% %% \[
%% %%  R=Q^{(n)} \subset Q^{(n-1)} \subset \cdots \subset Q^{(1)}\subset
%% %%  Q^{(0)} =Q 
%% %% \]
%%  \[
%%  Q = Q^{(0)}
%%  \supset Q^{(1)}
%%  \supset \cdots
%%  \supset Q^{(n-1)}
%%  \supset  Q^{(n)}=R
%% \]
%%  such that $\DE(Q^{(i+1)})$ is a facet of $\DE(Q^{(i)})$ for each
%%  $i=0,\ldots,n-1$. 
%%  The rank function of $Q$ serves as a rank function of $Q^{(i)}$ and we have
%%  \[
%%   c(Q^{(i+1)})=\dim(\DE(Q^{(i+1)}))+1 = \dim(\DE(Q^{(i)}))-1+1=c(Q^{(i)})-1.
%%  \]
%%  by \cref{thm:dim}.
%%  It implies by \cref{thm:facet} that each connected component of
%%  $Q^{(i+1)}$ is a full subquiver of $Q^{(i)}$ and that
%%  $Q^{(i)}/Q^{(i+1)}$ is an acyclic quiver for all $i$.
%%  Hence
%%  each connected component of $Q^{(n)}=R$ is a full subquiver of $Q$
%%  and $Q^{(0)}/Q^{(n)}=Q/R$ is acyclic.

%%  The converse follows from \cref{lem:hogeraisface} .
%% \end{proof}



%% %% Assume that $\dim(\check P)+1$
%% %% Let $P(R)$ be a facet of $P(Q)$.
%% %% We have $\dim(P(R))=\dim(P(Q))-1$.
%% %% Since $\dim(P(R))\leq \rank(R)$ and $\rank(Q)-1\leq \dim(P(Q))$,
%% %% $\rank(R)\geq \rank(Q)-2$.
%% %% In the case where $\rank(R)= \rank(Q)-2$,
%% %% $\dim(P(R))=\dim(P(Q))-1$.
%% %% only if $\dim(P(R))=\rank(R)$ and $\dim(P(Q))=\rank(Q)-1$.
%% %% If $\dim(P(Q))=\rank(Q)-1$,
%% %% then $Q$ satisfies \cref{condition:fuga}.
%% %% Since $Q$ satisfies \cref{condition:fuga},
%% %% the subquiver $R$ of $Q$ also satisfies \cref{condition:fuga}.
%% %% Hence we have $\dim(P(R))=\rank(R)-1$.
%% %% Thus,
%% %% if  $P(R)$ is a facet of $P(Q)$,
%% %% then $\rank(R)$ is $\rank(Q)-1$ or $\rank(Q)$.
%% %% Therefore
%% %% \cref{thm:facet}
%% %% follows from
%% %% \cref{lem:piyoisface,lem:hogeraisface,lem:facetisacyclic,lem:facetisfull,lem:facetispiyo}.
