\section{Sketch of proof}
\label{sec:ske}
Here we give a sketch of proofs of \Cref{thm:dim,thm:facet}.
We omit the details.
See \cite{2203.14521} for the details.

\subsection{Dimension}
\label{sec:dim}

%% Here is our main object of study.

%% \begin{definition}
%% Let $Q$ be a quiver.
%% The vector space of maps from $Q_0$ to $\RR$ is denoted by $\RR^{Q_0}$.
%% It is equipped with an inner product $\Braket{\ ,\ }$ defined by
%% \begin{align*}
%%  \Braket{\rho,\delta}=\sum_{v\in Q_0} \rho(v)\delta(v)
%% \end{align*}
%% for $\rho,\delta\in \RR^{Q_0}$.
%% For a subset $V\subset Q_0$,
%% we define an element $\kappa_{V} \in \RR^{Q_0}$ by
%% \begin{align*}
%%  \kappa_{V}(v)
%% =
%% \begin{cases}
%% 1&(v\in V) \\
%% 0&(v\not\in V)
%%  \end{cases}
%% \end{align*}
%% The set
%%  $\Set{\kappa_{\set{v}}|v\in Q_0}$ is
%%  a standard basis for the vector space $\RR^{Q_0}$. 
%% For $(v,w)\in Q_0\times Q_0$, 
%% we define the vector $\varepsilon_{(v,w)}$ by
%% \begin{align*}
%% \varepsilon_{(v,w)}= \kappa_{\set{v}} - \kappa_{\set{w}}.
%% \end{align*}
%% Define a convex polytope in $\RR^{Q_{0}}$ by
%% \begin{align*}
%%  \DE(Q)=\conv\Set{\varepsilon_{(v,w)}|(v,w)\in Q_1}.
%% \end{align*}
%%  This is called 
%%  the \defit{directed edge polytope} of $Q$.
%% % We define $\dim(\DE(Q))$ to be
%% % the dimension of the polytope $\DE(Q)$.
%% \end{definition}

%% \begin{remark}
%%  \label{remark:incidence_matrix}
%%  For a quiver $Q$, define a map
%% \[
%%  \delta: Q_{0}\times Q_{1} \longrightarrow \{-1,0,1\}
%% \]
%%  by 
%%  \[
%%  \delta(v,e) = 
%%  \begin{cases}
%%   1 & (e=(v,w) \text{ for some } w\in Q_{0}) \\
%%   -1 & (e=(w,v) \text{ for some } w\in Q_{0}) \\
%%   %% 1 & (e=(w,v) \text{ for some } w\in Q_{0}) \\
%%   %% -1 & (e=(v,w) \text{ for some } w\in Q_{0}) \\
%%   0 & (\text{ otherwise}).
%%  \end{cases}
%%  \]
%%  The matrix obtained from this map by choosing appropriate total orders
%%  of $Q_{0}$ and $Q_{1}$ is called the \emph{incidence matrix} of
%%  $Q$ and is denoted by $I(Q)$.
%%  Note that the vectors $\varepsilon_{(v,w)}$ are column vectors of
%%  $I(Q)$ and the directed edge polytope $\DE(Q)$ is the convex hull of these
%%  column vectors.
%% \end{remark}

%% It is easy to determine the vertices of $\DE(Q)$. For a convex polytope
%% $P$, let us denote the set of vertices of $P$ by $\ver(P)$.

%% \begin{lemma}
%%  \label{lem:vertices}
%%  For a quiver $Q$, we have
%%  \[
%%  \ver(\DE(Q)) = \Set{\varepsilon_{(v,w)}| (v,w)\in Q_{1}}.
%%  \]
%% \end{lemma}


Here we give a sketch of proofs of \Cref{thm:dim}.
Note that even if $Q$ is connected, a subquiver $R$ representing a face
of $\DE(Q)$ may not be connected.
It turns out that the number of connected components is closely related
to the dimension of $\DE(R)$.  
In fact, an upper bound is given by the
coconnectivity $c(R)$.

%For $\rho, \epsilon \in \RR^{Q_0}$,
%We define $\rho^{\perp}$ and $\rho^{\perp}+\epsilon$ by
%\begin{align*}
%\rho^{\perp}&=\Set{\delta\in\RR^{Q_0}| \Braket{\rho,\delta}=0},\\
%\rho^{\perp}+\epsilon&=\Set{\delta+\epsilon|\delta\in \rho^\perp}.
%\end{align*}
%First we discuss the upper bound of the dimension.
\begin{lemma}
 \label{lemma:ocs:constant}
 Define a vector subspace $V_{Q}$ of $\RR^{Q_{0}}$ by
 \[
  V_{Q} = \bigcap_{R\in\pi_{0}(Q)} \kappa_{R_{0}}^{\perp},
 \]
 where $\kappa_{R_{0}}^{\perp}$ is
 the hyeperplane orthogonal to
 $\kappa_{R_{0}}=\sum_{v\in R_0}\ee_v$ in $\RR^{n}$.
 Then $\DE(Q)\subset V_{Q}$ and we have
 $\dim \DE(Q) \le \numof{Q_{0}}-\numof{\pi_{0}(Q)}=c(Q)$.
\end{lemma}

%% \begin{proof}
%%  Let $R$ be a connected component of $Q$.
%%  We show that $\DE(Q)$ is contained in the hyperplane
%%  $\kappa_{R_{0}}^{\perp}$. 
%%  For each edge $(v,w)$ in $Q$, either $\{v,w\}\subset R_{0}$ or
%%  $\{v,w\}\cap R_{0}=\emptyset$. In the former case,
%%  $\langle \kappa_{R_{0}},\varepsilon_{(v,w)}\rangle=1-1$ and in the
%%  latter case, 
%%  $\langle \kappa_{R_{0}},\varepsilon_{(v,w)}\rangle=0-0$.
%%  Thus $\DE(Q)\subset \kappa_{R_{0}}^{\perp}$.

%%  Since the hyperplanes defined by the vertex sets of connected
%%  components intersect transversally, we have
%%  \[
%%   \dim \DE(Q) \le \dim \left(\bigcap_{R\in\pi_{0}(Q)}
%%  \kappa_{R_{0}}^{\perp}\right) = \dim \RR^{Q_{0}} 
%%  - \numof{\pi_{0}(Q)} = c(Q). 
%%  \]
%% \end{proof}
%\begin{proof}
%  By definition,
%it follows that
%\begin{align*}
%  \Braket{\kappa_{\check Q_0},\varepsilon_{(v,w)}}
%  =
%\begin{cases}
%  1-1=0 & (v,w,\in \check Q_0)\\
%0-0=0 & (v,w,\not\in \check Q_0),
%\end{cases}
%\end{align*}
%which implies the lemma.
%\end{proof}
%We obtain the following 
%as a corollary to \cref{lemma:ocs:constant}
%\begin{cor}
%  \label{lemma:dimen:upperbound:general}
%  For a quiver $Q$, $\dim(\DE(Q))\leq\rank(Q)$.
%\end{cor}

It turns out that $\dim \DE(Q)$ varies
depending on the existence of a rank function, since such a function
defines another hyperplane that contains $\DE(Q)$.

\begin{definition}
  \label{def:rankfunc}
 For a quiver $Q$, a map $\rho\colon Q_0\to \RR$ is called a
 \emph{rank function} of $Q$ if it satisfies $\rho(v)+1=\rho(w)$ for
 each edge $(v,w)\in Q_{1}$.
\end{definition}
\begin{remark}
 \label{prop:fuga}
 The following are equivalent for a quiver $Q$:
 \begin{enumerate}
  \item
       $Q$ has a rank function $\rho$.
  \item
       $Q$ is asymmetric and satisfies
       \begin{align*}
	\numof{\Set{t|(v_t,v_{t+1})\in Q_1}}
	=\numof{\Set{t|(v_{t+1},v_t)\in Q_1}}
       \end{align*}
       for each undirected closed walk
       $(v_0,v_1,\ldots,v_n)$ in $Q$.
\item
     $Q$ is asymmetric and satisfies
     \begin{align*}
      \numof{\Set{t|(v_t,v_{t+1})\in Q_1}}
      =\numof{\Set{t|(v_{t+1},v_t)\in Q_1}}
     \end{align*}
     for each undirected cycle
     $(v_0,v_1,\ldots,v_n)$ in $Q$.
  \item
       $Q$ is the Hasse diagram of a graded poset $(Q_0,\leq)$ with the rank
       function $\rho$.
%       where a poset is said to be graded if there exists the rank function
%  $\rho\colon Q_0\to \ZZ$ such that $\rho(v)+1=\rho(w)$ for $v$ covered by $w$.
 \end{enumerate}
\end{remark}



Let $Q$ have a rank function $\rho$.
Since $Q_0=[n]$,
 we regard a rank function as a vector in $\RR^n$.
The affine hyperplane 
$\rho^{\perp} + \varepsilon_{(v,w)} = \Set{\delta+\varepsilon_{(v,w)}|\delta\in \rho^\perp}$ 
is independent of choice of an egde $(v,w) \in Q_1$.
We define $ H_{\rho}$ to be $\rho^{\perp} + \varepsilon_{(v,w)}$
for some edge $(v,w) \in Q_1$.
The hyperplane  $H_{\rho}$ contains  $\DE(Q)$, and
is transversal to the hyperplanes defined by
connected components of $Q$.
We have the following upper bound of $\dim\DE(Q)$. 
%% \begin{lemma}
%%  \label{lemma:ocs:rank}
%%  When $Q$ has a rank function $\rho$, choose an edge $(v,w)\in Q_{1}$
%%  and define a hyperplane $H_{\rho}$ in
%%  $\RR^{Q_{0}}$ by
%%  \[
%%   H_{\rho} = \rho^{\perp} + \varepsilon_{(v,w)} =
%%  \Set{\delta+\varepsilon_{(v,w)}|\delta\in \rho^\perp}.
%%  \]
%%  Then this is independent of the choice of an edge $(v,w)$ and contains
%%  $\DE(Q)$. 
%% \end{lemma}

%% \begin{proof}
%%  For edges $(v,w)$ and $(v',w')$ in $Q$, we have
%%  \[
%%  \langle \rho, \varepsilon_{(v,w)}\rangle = \rho(v)-\rho(w)  = -1 =
%%  \rho(v')-\rho(w') = 
%%  \langle \rho, \varepsilon_{(v',w')}\rangle,
%%  \]
%%  which implies that
%%  $\varepsilon_{(v,w)}-\varepsilon_{(v',w')}\in\rho^{\perp}$ and  
%%  $\rho^{\perp}+\varepsilon_{(v,w)}=\rho^{\perp}+\varepsilon_{(v',w')}$.
%%  It also implies that all vertices of $\DE(Q)$ are contained in
%%  $H_{\rho}$ and thus $\DE(Q)\subset H_{\rho}$.
%% \end{proof}

%% The hyperplane $H_{\rho}$ is transversal to the hyperplanes defined by
%% connected components of $Q$. And we have the following upper bound of
%% $\dim\DE(Q)$. 

\begin{lemma}
 \label{cor:dim:upperbound}
 If $Q$ has a rank function, then $\dim (\DE(Q))\le c(Q)-1$.
\end{lemma}

%% The choice of the term ``rank function'' is justified by the following
%% fact. 

%By definition,
%if $Q$ satisfies \Cref{condition:fuga} with $\rho$,
%then each subquiver of $Q$ also satisfies \Cref{condition:fuga}. 
%It is easy to show the following:
%% \begin{proposition}
%%  \label{prop:fuga}
%%  The following are equivalent for a quiver $Q$:
%%  \begin{enumerate}
%%   \item
%%        $Q$ has a rank function $\rho$.
%%   \item
%%        $Q$ is asymmetric and satisfies
%%        \begin{align*}
%% 	\numof{\Set{t|(v_t,v_{t+1})\in Q_1}}
%% 	=\numof{\Set{t|(v_{t+1},v_t)\in Q_1}}
%%        \end{align*}
%%        for each undirected closed walk
%%        $(v_0,v_1,\ldots,v_n)$ in $Q$.
%% \item
%%      $Q$ is asymmetric and satisfies
%%      \begin{align*}
%%       \numof{\Set{t|(v_t,v_{t+1})\in Q_1}}
%%       =\numof{\Set{t|(v_{t+1},v_t)\in Q_1}}
%%      \end{align*}
%%      for each undirected cycle
%%      $(v_0,v_1,\ldots,v_n)$ in $Q$.
%%   \item
%%        $Q$ is the Hasse diagram of a graded poset $(Q_0,\leq)$ with rank
%%        function $\rho$.
%% %       where a poset is said to be graded if there exists the rank function
%% %  $\rho\colon Q_0\to \ZZ$ such that $\rho(v)+1=\rho(w)$ for $v$ covered by $w$.
%%  \end{enumerate}
%% \end{proposition}

%% \begin{proof}
%%  Suppose $Q$ has a rank function $\rho$. Then $Q$ cannot have a pair
%%  $(v,w)$ of vertices with $v\to w$ and $v\leftarrow w$ and hence is
%%  asymmetric.  
%%  Let $(v_{0},\ldots,v_{n})$ be an undirected closed walk or cycle in $Q$.
%%  Then
%%  \[
%%   0 = \rho(v_{0})-\rho(v_{n}) = \sum_{t\in\Set{t|(v_{t},v_{t+1})\in
%%  Q_{1}}} 1 + \sum_{t\in\Set{t|(v_{t+1},v_{t})\in Q_{1}}}(-1)
%%  \]
%%  and we have
%%  \[
%%  \numof{\Set{t|(v_{t},v_{t+1})\in Q_{1}}} = \numof{\Set{t|(v_{t+1},v_{t})\in
%%  Q_{1}}}. 
%%  \]

%%  Conversely, suppose that the second condition is
%%  satisfied. Choose a vertex $v_{0}$. For a vertex $w\in Q_{0}$, choose
%%  an undirected walk $(v_{0},\ldots,v_{n}=w)$ from $v_{0}$ to $w$ and
%%  define
%%  \[
%%  \rho(w) = \numof{\Set{t|(v_{t},v_{t+1})\in Q_{1}}} -
%%  \numof{\Set{t|(v_{t+1},v_{t})\in Q_{1}}}. 
%%  \]
%%  The second condition guarantees that this is independent of the choice
%%  of a walk.

%%  Since any undirected closed walk can be decomposed into undirected
%%  cycles, the second and the third conditions are equivalent. 
%%  Finally the first and the fourth conditions are equivalent by definition.
%% \end{proof}

In order to obtain lower bounds of $\dim\DE(Q)$,
we consider the case of a quiver whose underlying graph is acyclic.
 Then the dimension of the vector space spanned by
 $\set{\varepsilon_{(v,w)}|(v,w)\in Q_{1}}$ is 
 given by $c(Q)$.
 Thus we obtain
 \[
 \dim \DE(Q) = \dim
 \aff\ideal{\varepsilon_{(v,w)}}{(v,w)\in Q_{1}} = c(Q) -1,  
 \]
 where $\aff$ denotes the affine hull.
Hence we have the following.
 %\end{lemma}
%
%We define $\corank(Q)$ to be the number of weakly connected components of $Q$.
%We also define  $\rank(Q)$ to be $\rank(Q)=\numof{Q_0}-\corank(Q)$.
%
%\subsection{Main Theorems}
%First we discuss the dimension of a directed edge polytope.
%For a quiver $Q$,
%we define
%\cref{condition:fuga}
%as follows:
%\begin{condition}
%\label{condition:fuga}
%There exists $\rho\in\RR^{Q_0}$
%such that
%\begin{align*}
%  (v,w)\in Q_1 \implies \rho(v)+1=\rho(w).
%\end{align*}
%\end{condition}
%
%Next we discuss the lower bound of the dimension.
%\begin{lemma}
%\label{lem:dim:forest}
%If the underlying graph of an asymmetric quiver $Q$ is acyclic,
%then the dimension of the vector (resp.\ affine) space
%spanned by $\Set{\varepsilon_{(v,w)}|(v,w)\in Q_1}$ is $\rank(Q)$
%(resp.\ $\rank(Q)-1$).
%\end{lemma}
%
%% \begin{lemma}
%%  \label{lem:dim:forest}
%%  Let $F$ be a quiver whose underlying graph is acyclic.
%%  Then the dimension of the vector space spanned by
%%  $\set{\varepsilon_{(v,w)}|(v,w)\in F_{1}}$ is 
%%  given by $c(F)$.
%%  Thus we obtain
%%  \[
%%  \dim \DE(F) = \dim
%%  \aff\ideal{\varepsilon_{(v,w)}}{(v,w)\in F_{1}} = c(F) -1,  
%%  \]
%%  where $\aff$ denotes the affine hull.
%% \end{lemma}

%% \begin{proof}
%%  Recall that $I(F)$ is the incidence matrix of $F$. Then the dimension
%%  of the vector space spanned by 
%%  $\Set{\varepsilon_{(v,w)}|(v,w)\in F_{1}}$ is $\rank I(F)$.
%%  By the additivity of the rank of the incidence matrix with respect to
%%  disjoint unions, it suffices to prove that
%%  $\rank I(F)=\numof{F_{0}}-1$, when $F$ is connected.

%%  By the acyclicity assumption, the underlying graph of $F$ is a tree. 
%%  Let $n=\numof{F_{0}}$. We may choose an ordering
%%  $F_{0}=\{v_{1},\ldots,v_{n}\}$ in such a way that the underlying graph
%%  of the full subquiver $F^{(i)}$ with vertices $\{v_{i},\ldots,v_{n}\}$ is 
%%  a tree for each $i=1,\ldots,n-1$.
%%  In other words, $v_{i}$ is connected to a vertex in
%%  $\{v_{i+1},\ldots,v_{n}\}$ by a unique edge for each
%%  $i=1,\ldots,n-1$. 
%%  With this ordering, $I(F^{(i)})$ is of the form
%%  \[
%%   I(F^{(i)}) = 
%%  \begin{pmatrix}
%%   \pm 1 & 0 & \cdots & 0 \\
%%   0 & \\
%%   \vdots & \\
%%   0 & \\
%%   \mp 1 & & I(F^{(i+1)})& \\
%%  0 & \\
%%   \vdots & \\
%%   0 & 
%%  \end{pmatrix}
%%  \]
%%  and we obtain $\rank I(F)=n-1=\numof{F_{0}}-1$ by induction.

%%  Since the origin is not contained in the affine hull, we have
%%  \begin{align*}
%%  \dim\conv\ideal{\varepsilon_{(v,w)}}{(v,w)\in F_{1}} & = \dim
%%  \aff\ideal{\varepsilon_{(v,w)}}{(v,w)\in F_{1}} \\
%%   & = c(F)-1.  
%%  \end{align*}
%% \end{proof}
%\begin{proof}
%  For each weakly connected component $\check Q$ of $Q$,
%  we have the following:
%  Let $\numof{\check Q_0}=n$.
%  Since the underlying graph of $\check Q$ is a tree,
%  we can assume that $Q_0=\Set{v_1,\ldots,v_n}$
%  satisfies the following:
%  the underlying graph of the full subquiver of $\check Q$ with respect to 
%  $\Set{v_i,\ldots,v_n}$
%  is a tree for each $i=1,\ldots,n-1$.
%  For each $i=1,\ldots,n-1$,
%  there exists a unique edge $a_i$ connecting $v_i$ 
%  and a vertex in $\Set{v_{i+1},\ldots,v_n}$.
%  It is easy to show that
%  the rank of the incidence matrix of $\check Q$
%  with respect to these orderings
%  is full.
%  Hence the rank of the incidence matrix of $\check Q$
%  is $\rank(\check Q)$.
%
%  Since the rank of the incidence matrix of $\check Q$ is $\rank(\check Q)$
%  for each weakly connected component,
%  it follows that
%  the rank of the incidence matrix of $Q$ is $\rank(Q)$,
%  which implies the lemma.
%\end{proof}

%Since a quiver  $Q$
%has a spanning polyforest in $Q$,
%that is a weakly asymmetric subquiver whose underlying graph is acyclic,
%we obtain the following:
\begin{lemma}
\label{lem:dim:lowerbound}
 For any quiver $Q$, we have $\dim(\DE(Q))\geq c(Q)-1$.
\end{lemma}

%% \begin{proof}
%%   Choose a spanning polyforest $F$ in $Q$.
%%  Then we have
%%  \[
%%  \dim (\DE(Q)) \ge \dim (\DE(F)) = c(F)-1 \ge c(Q)-1,
%%  \]
%%  since $R_{0}=Q_{0}$.
%% %  If follows from \cref{lem:dim:forest}
%% %  that $\dim(\DE(F))\geq\rank(Q)-1$.
%% %  Since $\DE(F)\subset \DE(Q)$,
%% %  $\dim(\DE(Q))\geq\rank(Q)-1$.
%% \end{proof}

%Hence, e.g., an asymmetric quiver without undirected cycles
%has a rank function.

%We have the following equation for dimension of a direct edge polytope,
%which is proved in \cref{sec:proof}.
%\begin{theorem}
%  For a quiver $Q$,
%\begin{align*}
%\dim(\DE(Q))=
%  \begin{cases}
%\rank(Q)-1 &(\text{$Q$ satisfies \cref{condition:fuga}})\\
%\rank(Q)  &(\text{otherwise}).
%\end{cases}
%\end{align*}
%\end{theorem}
%

%
%\begin{lemma}
%\label{lemma:ocs:rank}
%Let $Q$ be a quiver.
%If $\rho \in \RR^{Q_0}$ satisfies
%$\rho(v)+1=\rho(w)$ for all $(v,w)\in Q_1$,
% then
%$\DE(Q)\subset \rho^\perp + \varepsilon_{(v,w)}$
%for $(v,w) \in \RR^{Q_0}$.
%\end{lemma}
%\begin{proof}
%  For each $(v,w)\in Q_1$,
%  it follows from direct calculation that
%  $\Braket{\rho,\varepsilon_{(v,w)}}=\rho(v)-\rho(w)=-1$.
%  Hence, for  $(v,w), (v',w')\in Q_1$,
%  we have $\Braket{\rho,\varepsilon_{(v',w')}-\varepsilon_{(v,w)}}=0$,
%  which implies
%  $\varepsilon_{(v',w')} \in \rho^\perp + \varepsilon_{(v,w)}$.
%  Since $\rho^\perp + \varepsilon_{(v,w)}$ contains
%  all vertices of $\DE(Q)$ 
%  the space also contains the convex hull $\DE(Q)$.
%\end{proof}

%We have the following as a corollary to 
%\cref{lemma:ocs:constant,lemma:ocs:rank}
%\begin{cor}
%  \label{cor:dim:upperbound}
%  If $Q$ satisfies \cref{condition:fuga},
%  then $\dim(\DE(Q))\leq\rank(Q)-1$.
%  If $Q$ does not satisfy \cref{condition:fuga},
%  then $\dim(\DE(Q))\leq\rank(Q)$.  
%\end{cor}

For those quivers that do not have rank functions, we have the following
lower bound.

\begin{lemma}
  \label{lem:dim:lowerbound1}
  If $Q$ does not have a rank function,
  then
  $\dim(\DE(Q))\geq c(Q)$.
\end{lemma}

%% \begin{proof}
%%  Let $F$ be spanning polyforest of $Q$. Then $F_{0}=Q_{0}$ and
%%  $\numof{\pi_{0}(F)} = \numof{\pi_{0}(Q)}$.
%%  By \cref{lem:dim:forest}, we have $\dim(\DE(F)) = c(F)-1=c(Q)-1$.

%%  Since the underlying graph of
%%  $F$ is acyclic, $F$ has a rank function $\rho$.
%%  Let $H_{\rho}$ be the hyperplane in \cref{lemma:ocs:rank}. It is
%%  given by
%%  \[
%%   H_{\rho} = \rho^{\perp} + \varepsilon_{(v_{0},w_{0})}
%%  \]
%%  for an edge $(v_{0},w_{0})$ in $F$.
%%  Since $Q$ does not have a rank function, there exists an edge $(v,w)$
%%  in $Q$ such that $\rho(v)+1\neq \rho(w)$. Then we have
%%  \begin{align*}
%%   \langle \rho, \varepsilon_{(v,w)}-\varepsilon_{(v_{0},w_{0})}\rangle & =
%%   \rho(v)-\rho(w) - (\rho(v_{0})-\rho(w_{0})) \\
%%   & = \rho(v)-\rho(w)+1 \neq 0,
%%  \end{align*}
%%  which implies that the vertex
%%  $\varepsilon_{(v,w)}$ is not contained in the hyperplane $H_{\rho}$.
%%  It is not contained in any one of hyperplanes of the form
%%  $\kappa_{G_{0}}$ for a connected component $G$
%%  of $F$, either. In other words,
%%  $\varepsilon_{(v,w)}\not\in H_{\rho}\cap V_{F}$.
%%  Since $ \DE(F) \subset H_{\rho} \cap V_{F}$, 
%%  we have $\dim(\DE(Q)) \ge \dim(\DE(F))+1 = c(Q)$.
%% \end{proof}
%\begin{proof}
%  Let $F$ be a spanning polyforest in $Q$.
%  Since the underlying graph of $F$ is acyclic,
%  we have $\rho\in \RR^{Q_0}$ satisfying
%  \begin{align*}
%    (v,w)\in F_1\implies \rho(v)+1=\rho(w).
%  \end{align*}
%  Let $(v_0,w_0)\in F_1$.
%  By \cref{lemma:ocs:constant,lemma:ocs:rank},
%  $\DE(F)$ is a $(\rank(Q)-1)$-dimensional polytope contained by
%  $\rho^\perp + \varepsilon_{(v_0,w_0)}$
%  and $\kappa_{\check Q_0}^\perp$ for all weakly connected component of $Q$.
%  Since $Q$ does not satisfy \cref{condition:fuga},
%  there exists $(v,w)\in Q_1$ such that
%  $\rho(v)+1\neq\rho(w)$.
%  Hence 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 \neq 0,
%  \end{align*}
%  which implies
%  $\rho^\perp + \varepsilon_{(v_0,w_0)}$
%  does not contain $\varepsilon_{(v,w)}$.
%  Let $S$ be 
%  the intersection of 
%  $\rho^\perp + \varepsilon_{(v_0,w_0)}$
%  and $\kappa_{\check Q_0}^\perp$ for all weakly connected component of $Q$.
%  Since $S$ 
%  is an affine space of dimension $\rank(Q)-1$ such that
%  $S$ contains $\DE(F)$ and $S$ does not contain $\DE(Q)$,
%  we have $\dim(\DE(Q))\geq\rank(Q)$.
%\end{proof}

Now \cref{thm:dim} follows from 
\cref{cor:dim:upperbound,lem:dim:lowerbound,lemma:ocs:constant,lem:dim:lowerbound1}. 


%% \begin{lemma}
%% \label{lem:dim:tree}
%% Let $Q$ be asymmetric and weakly connected. 
%% If the underlying graph of $Q$ is acyclic,
%% then the dimension of the vector (resp.\ affine) space
%% spanned by $\Set{\varepsilon_{(v,w)}|(v,w)\in Q_1}$ is $\numof{Q_0}-1$
%% (resp.\ $\numof{Q_0}-2$).
%% \end{lemma}
%% \begin{proof}
%% Let $\numof{Q_0}=n$.
%% Since the underlying graph of $Q$ is a tree,
%% we can assume that $Q_0=\Set{v_1,\ldots,v_n}$ satisfies the following:
%% the underlying graph of the full subquiver of $Q$ with respect to 
%% $\Set{v_i,\ldots,v_n}$
%% is a tree for each $i=1,\ldots,n-1$.
%% For each $i=1,\ldots,n-1$,
%% there exists a unique edge $a_i$ connecting $v_i$ 
%% and a vertex in $\Set{v_{i+1},\ldots,v_n}$.
%% It is easy to show that
%% the rank of the incidence matrix of $Q$ with respect to these orderings
%% is full.
%% Hence we have the lemma.
%% \end{proof}
%% Since a weakly connected quiver $Q=(Q_0,Q_1)$
%% has
%% a spanning polytree in $Q$,
%% that is a weakly connected asymmetric subquiver whose
%% underlying graph is acyclic,
%% we obtain the following:
%% \begin{lemma}
%% \label{lem:dim:lowerbound}
%% If $Q=(Q_0,Q_1)$ is weakly connected,
%% then the dimension of $P(Q)$ is greater than or equal to $\numof{Q_0}-2$.
%% \end{lemma}
%% \begin{proof}
%%   Let $T=(T_0,T_1)$ be a spanning polytree in $Q=(Q_0,Q_1)$.
%%   By \cref{lem:dim:tree},
%%   we have the dimension of $P(T)$ is greater than or equal to $\numof{Q_0}-2$.
%%   Since $P(T)\subset P(Q)$,
%%   the dimension of $P(Q)$ is greater than or equal to $\numof{Q_0}-2$.
%% \end{proof}


%\Cref{thm:dim} follows from
%\cref{cor:dim:upperbound,lem:dim:lowerbound,lem:dim:lowerbound1}.


