text:
\section{Introduction}
\label{sec-introduction}

   For a matroid $M$ on $[n]$, a \emph{matroid base polytope} $Q(M)$ is
   the polytope in ${\mathbb R}^n$ whose vertices are the incidence vectors of the bases of $M$.
   The polytope $Q(M)$ is a face of a \emph{matroid polytope} first studied by
   Edmonds~\cite{Edmonds}, whose vertices are the incidence vectors of \emph{all}
   independent sets in $M$.
   In this paper, we study flags of faces and the $\mathbf{c}\mathbf{d}$-index
   of matroid base polytopes.

   It is known that a face $\sigma$ of a matroid base polytope is
   the matroid base polytope $Q(M_\sigma)$ for some matroid $M_\sigma$ on $[n]$
   (see \cite{FeichtnerSturmfels} and Section~\ref{sec-matroid-base-polytopes} below).
   We show that $M_\sigma$ can be described using
   equivalence classes of factor-connected flags of subsets of $[n]$.
   As a result, one can describe faces of $Q(M)$ in terms of matroid data:
   \begin{theorem}
   [Theorem~\ref{theorem-posets-for-faces-of-matroid-polytopes}]

      Let $M$ be a connected matroid on a ground set $[n]$.
      For a face $\sigma$ of the matroid base polytope $Q(M)$, one can associate a poset $P_\sigma$
      defined as follows:
      \begin{enumerate}[(i)]
         \item
         the elements of $P_\sigma$ are the connected components of $M_\sigma$, and
         \item
         for distinct connected components $C_1$ and $C_2$ of $M_\sigma$,
         $C_1 < C_2$ if and only if
         $$
         C_2 \subseteq S \subseteq [n] \text{ and }
         \sigma \subseteq H_S \text{ implies } C_1 \subseteq S,
         $$
         where $H_S$ is the hyperplane in ${\mathbb R}^n$ defined by $\sum_{e \in S} x_e = r(S)$.
      \end{enumerate}
   \end{theorem}

   The $\mathbf{c}\mathbf{d}$-index $\Psi(Q)$ of a polytope $Q$, a polynomial in the noncommutative variables
   $\mathbf{c}$ and $\mathbf{d}$, is a very compact encoding of the flag numbers of a polytope $Q$~\cite{BayerKlapper}.
   Ehrenborg and Readdy~\cite{EhrenborgReaddy} express the $\mathbf{c}\mathbf{d}$-indices of a prism, a pyramid, and
   a bipyramid of a polytope $Q$ in terms of $\mathbf{c}\mathbf{d}$-indices of $Q$ and its faces.
   Also, the $\mathbf{c}\mathbf{d}$-index of zonotopes, a special class of polytopes, is well-understood
   \cite{BilleraEhrenborgReaddy97, BilleraEhrenborgReaddy98}.
   Generalizing the formulas of the $\mathbf{c}\mathbf{d}$-indices of a prism and a pyramid of a polytope,
   we show how the $\mathbf{c}\mathbf{d}$-index of a polytope can be expressed when a polytope is split by a hyperplane
   in Section~\ref{sec-cd-index}.
   %

   In Section~\ref{sec-hyperplane-splits},
   we find the conditions when a matroid base polytope is split into two
   matroid base polytopes by a hyperplane:
   \begin{theorem}
   [Theorem~\ref{matroid-base-polytope-split}]
      Let $M$ be a rank $r$ matroid on $[n]$ and $H$ be a hyperplane in ${\mathbb R}^n$
      given by $\sum_{e \in S} x_e = k$.
      Then $H$ decomposes $Q(M)$ into two matroid base polytopes if and only if
      \begin{enumerate}[(i)]
         \item
         $r(S) > k$ and $r(S^c) > r-k$,
         \item
         if $I_1$ and $I_2$ are $k$-element independent subsets of $S$ such that
         $(M/I_1)|_{S^c}$ and $(M/I_2)|_{S^c}$ have rank $r-k$, then
         $(M/I_1)|_{S^c} = (M/I_2)|_{S^c}.$
      \end{enumerate}
   \end{theorem}

   We apply this theorem to the $\mathbf{c}\mathbf{d}$-index of matroid base polytopes for rank $2$ matroids
   in Section~\ref{sec-rank-2-matroids}.

   Throughout this paper, we assume familiarity with the basic concepts of matroid theory.
   For further information, we refer the readers to \cite{Oxley}.

\section{Matroid base polytopes}
\label{sec-matroid-base-polytopes}

   This section contains the description of faces of matroid base polytopes.
   In particular, we associate a poset to each face of a matroid base polytope.

   We start with a precise characterization of matroid base polytopes.
   Let $\mathcal{B}$ be a collection of $r$-element subsets of $[n]$.
   For each subset $B = \{ b_1, \dots, b_r \}$, let
   $$
   e_B = e_{b_1} + \cdots + e_{b_r} \in {\mathbb R}^n ,
   $$
   where $e_i$ is the $i$th standard basis vector of ${\mathbb R}^n$.
   The collection $\mathcal{B}$ is represented by the convex hull of these points
   $$
   Q(\mathcal{B}) = \mathrm{conv} \{ e_B : B \in \mathcal{B} \}.
   $$
   This is a convex polytope of dimension $\le n-1$
   and is a subset of the $(n-1)$-simplex
   $$
   \Delta_n = \{ (x_1, \dots, x_n) \in {\mathbb R}^n : x_1 \ge 0,
         \dots, x_n \ge 0, x_1 + \cdots + x_n = r \}.
   $$
   Gelfand, Goresky, MacPherson, and Serganova~\cite[Thm. 4.1]{GelfandGoreskyMacPhersonSerganova}
   show the following characterization of matroid base polytopes.
   \begin{theorem}
   \label{thm-matroid-polytopes}
      $\mathcal{B}$ is the collection of bases of a matroid if and only if
      every edge of the polytope $Q(\mathcal{B})$ is parallel to
      a difference $e_\alpha - e_\beta$ of two distinct standard basis vectors.
   \end{theorem}

   For a rank $r$ matroid $M$ on a ground set $[n]$
   with a set of bases $\mathcal{B}(M)$,
   the polytope $Q(M) := Q(\mathcal{B}(M))$ is called
   the \emph{matroid base polytope} of $M$.

   By the definition, the vertices of $Q(M)$ represent the bases of $M$.
   For two bases $B$ and $B'$ in $\mathcal{B}(M)$,
   $e_B$ and $e_{B'}$ are connected by an edge if and only if
   $e_B - e_{B'} = e_\alpha - e_\beta$ for some $\alpha, \beta \in [n]$.
   Since the latter condition is equivalent to $B \setminus B' = \{ \alpha \}$
   and $B' \setminus B = \{ \beta \}$, the edges of $Q(M)$ represent
   the basis exchange axiom.
   The basis exchange axiom gives the following equivalence relation
   on the ground set $[n]$ of the matroid $M$:
   $\alpha$ and $\beta$ are \emph{equivalent} if there exist bases $B$ and $B'$ in $\mathcal{B}(M)$
   with $B \setminus B' = \{ \alpha \}$ and $B' \setminus B = \{ \beta \}$.
   The equivalence classes are called the \emph{connected components} of $M$.
   The matroid $M$ is called \emph{connected} if it has only one connected
   component.
   Feichtner and Sturmfels~\cite[Prop. 2.4]{FeichtnerSturmfels} express the dimension of
   the matroid base polytope $Q(M)$ in terms of the number of connected
   components of $M$.

   \begin{proposition}
   \label{dim-of-matroid-polytope}
      Let $M$ be a matroid on $[n]$.
      The dimension of the matroid base polytope $Q(M)$ equals $n - c(M)$, where
      $c(M)$ is the number of connected components of $M$.
   \end{proposition}

   Theorem~\ref{thm-matroid-polytopes} implies that every face of a matroid base polytope
   is also a matroid base polytope.
   For a face $\sigma$ of $Q(M)$, let $M_\sigma$ denote the matroid on $[n]$
   whose matroid base polytope is $\sigma$.
   For $\omega \in {\mathbb R}^n$, let $M_\omega$ denote the matroid whose bases
   $\mathcal{B}(M_\omega)$ is the collection of bases of $M$ having minimum $\omega$-weight.
   Then $Q(M_\omega)$ is the face of $Q(M)$ at which the linear form
   $\sum_{i=1}^n \omega_i x_i$ attains its minimum.
   Let ${\mathcal F}(\omega)$ denote the unique flag of subsets
   $$
   \{ \emptyset =: S_0 \subseteq S_1 \subseteq \cdots \subseteq S_k \subseteq S_{k+1} := [n] \}
   $$
   for which $\omega$ is constant on each set $S_i \setminus S_{i-1}$
   and $\omega|_{S_i \setminus S_{i-1}} < \omega|_{S_{i+1} \setminus S_i}$.
   Ardila and Klivans~\cite{ArdilaKlivans}
   show that $M_\omega$ depends only on ${\mathcal F}(\omega)$, and hence
   one can call it $M_{{\mathcal F}}$.
   They also give the following description of $M_{\mathcal F}$.

   \begin{proposition}
   \emph{\cite[Prop. 2]{ArdilaKlivans}}
   \label{bases-of-M-F}
      Let $M$ be a matroid on $[n]$ and ${\mathcal F}$ be a flag of subsets
      $$
      \{ \emptyset =: S_0 \subseteq S_1 \subseteq \cdots \subseteq S_k \subseteq S_{k+1} := [n] \},
      $$
      then

      $$
      M_{{\mathcal F}} = \bigoplus_{i=1}^{k+1} (M |_{S_i}) / S_{i-1},
      $$
      where $M|_S$ is the restriction of $M$ on $S$ and $M/S$ is the contraction of $M$ on $S$.
   \end{proposition}

   A flag
   ${\mathcal F} = \{ \emptyset =: S_0 \subseteq S_1 \subseteq
   \cdots \subseteq S_k \subseteq S_{k+1} := [n] \}$
   is called \emph{factor-connected} (with respect to $M$) if the matroids
   $(M|_{S_{i}})/ S_{i-1}$ are connected for all $i = 1, \dots, k+1$.
   If ${\mathcal F}$ is factor-connected, then the connected components of $M_{\mathcal F}$ are
   $S_1 \setminus S_0, S_2 \setminus S_1, \dots, S_{k+1} \setminus S_k$.
   Proposition~\ref{dim-of-matroid-polytope} and Proposition~\ref{bases-of-M-F}
   show that the dimension of $Q(M_{{\mathcal F}})$ is $n-k-1$
   when ${\mathcal F}$ is factor-connected.

   Since $Q(M_1 \oplus M_2) = Q(M_1) \times Q(M_2)$, it is enough to
   restrict to attention to connected matroid $M$.
   For a connected matroid $M$ on $[n]$, facets of $Q(M)$ correspond to factor-connected flags of the form
   $\{ \emptyset \subseteq S \subseteq [n] \}$.
   Feichtner and Sturmfels~\cite{FeichtnerSturmfels} show that there are two types of facets of $Q(M)$:
   \begin{enumerate}[(i)]
      \item
      a facet corresponding to a factor-connected flag $\{ \emptyset \subseteq F \subseteq [n] \}$
      for some flat $F$ of $M$ (in this case, the facet is called a \emph{flacet}),
      \item
      a facet corresponding to a factor-connected flag $\{ \emptyset \subseteq S \subseteq [n] \}$
      for an $(n-1)$-subset $S$ of $[n]$.
   \end{enumerate}

   %

   %
   \begin{lemma}
   \label{switching-components}
      Let $M$ be a connected matroid on $[n]$ and
      $$
      {\mathcal F} = \{ \emptyset =: S_0 \subseteq S_1 \subseteq \cdots \subset
      S_k \subseteq S_{k+1} := [n] \}
      $$
      a factor-connected flag with respect to $M$.
      Then the matroid $(M|_{S_{j+1}})/S_{j-1}$ has at most two connected components
      for $1 \le j \le k$.
      \begin{enumerate}[(i)]
         \item
         If it has one connected component, the flag
         $$
         \mathcal{G} = \{ \emptyset =: S_0 \subseteq \cdots \subseteq S_{j-1}
         \subseteq S_{j+1} \subseteq \cdots \subseteq S_{k+1} := [n] \}
         $$
         is factor-connected and $Q(M_{\mathcal{G}})$ covers $Q(M_{{\mathcal F}})$
         in the face lattice of $Q(M)$.
         \item
         If it has two connected components, then they are $S_j \setminus S_{j-1}$ and
         $S_{j+1} \setminus S_j$.
         Moreover, the flag
         $$
         {\mathcal F}' = \{ \emptyset =: S_0 \subseteq \cdots \subseteq S_{j-1}
         \subseteq S'_j \subseteq S_{j+1} \subseteq \cdots \subseteq S_{k+1} := [n] \},
         $$
         where $S'_j = S_{j-1} \cup (S_{j+1} \setminus S_j)$, is factor-connected
         and $Q(M_{{\mathcal F}'}) = Q(M_{{\mathcal F}})$.
         In this case, ${\mathcal F}$ and ${\mathcal F}'$ are said to be \emph{adjacent}.
      \end{enumerate}
   \end{lemma}

   \begin{proof}
      For $j = 1, \dots, k$, we have
      $(M|_{S_{j+1}})/S_j = [(M|_{S_{j+1}})/S_{j-1}]/(S_j\setminus S_{j-1})$
      and $(M|_{S_j})/S_{j-1} = [(M|_{S_{j+1}})/S_{j-1}]|_{S_j}$.
      The first assertion is obtained from \cite[Proposition 4.2.10]{Oxley}, and
      the other assertions follow from \cite[Proposition 4.2.13]{Oxley}
      and Proposition~\ref{bases-of-M-F}.
   \end{proof}

   Two factor-connected flags ${\mathcal F}$ and ${\mathcal F}'$ of the same length are said to be \emph{equivalent}
   if there is a sequence of factor-connected flags
   $$
   {\mathcal F} = {\mathcal F}_0, {\mathcal F}_1, \dots, {\mathcal F}_r = {\mathcal F}'
   $$
   such that ${\mathcal F}_i$ is adjacent to ${\mathcal F}_{i-1}$ for $i = 1, \dots, r$.
   We write ${\mathcal F} \sim {\mathcal F}'$ when factor-connected flags ${\mathcal F}$ and ${\mathcal F}'$ are equivalent.

   \begin{lemma}
   \label{switch-connected-case}
      Let $M$ be a connected matroid on $[n]$ and $X, Y$ disjoint subsets of $[n]$
      with $X \cup Y = [n]$.
      Then $\mathcal{B}((M|_X) \oplus (M/X))$ and $\mathcal{B}((M|_Y) \oplus (M/Y))$
      are disjoint.
   \end{lemma}

   \begin{proof}
      Since $X \cap Y = \emptyset$ and $X \cup Y = [n]$,
      one has $r(X) + r(Y) \ge r(M)$.
      If $r(X) + r(Y) = r(M)$, then $|B \cap X| = r(X)$ and $|B \cap Y| = r(Y)$
      for all $B \in \mathcal{B}(M)$, and hence $M = M|_X \oplus M|_Y$.
      But then $M$ is not connected.
      Thus $r(X) + r(Y) > r(M)$.
      Therefore there is no base which has $r(X)$ elements in $X$ and $r(Y)$ elements in $Y$,
      i.e., $\mathcal{B}((M|_X) \oplus (M/X))$ and $\mathcal{B}((M|_Y) \oplus (M/Y))$
      are disjoint.

      %

   \end{proof}

   The following proposition shows that the equivalence classes of factor-connected
   flags characterize faces of a matroid base polytope.

   \begin{proposition}
   \label{equivalence-classes-of-fcf}
      Let $M$ be a connected matroid on $[n]$.
      If ${\mathcal F}$ and ${\mathcal F}'$ are two factor-connected flags of subsets of $[n]$ given by
      $$
      \begin{aligned}
         {\mathcal F} =& \{ \emptyset =: S_0 \subseteq S_1 \subseteq \cdots
                  \subseteq S_k \subseteq                  S_{k+1} := [n] \},\\
         {\mathcal F}' =& \{ \emptyset =: T_0 \subseteq T_1 \subseteq \cdots
                  \subseteq T_l \subseteq T_{l+1} := [n] \},
      \end{aligned}
      $$
      then $M_{{\mathcal F}} = M_{{\mathcal F}'}$ if and only if
      ${\mathcal F}$ and ${\mathcal F}'$ are equivalent.
   \end{proposition}

   \begin{proof}
      If ${\mathcal F} \sim {\mathcal F}'$, then $M_{\mathcal F} = M_{{\mathcal F}'}$
      from Lemma~\ref{switching-components}.

      For the other direction, suppose that $M_{{\mathcal F}} = M_{{\mathcal F}'}$.
      Then ${\mathcal F}$ and ${\mathcal F}'$ have the same length since $\dim Q(M_{{\mathcal F}}) = n-k-1$
      and $\dim Q(M_{{\mathcal F}'}) = n-l-1$.
      Without loss of generality, we may assume that $S_1 \ne T_1$ and $S_k \ne T_k$.

      We will use induction on $k$.
      We claim that
      \begin{eqnarray}
      \label{difference-are-the-same}
         T_1 = S_m \setminus S_{m-1} \text{ for some } m > 1.
      \end{eqnarray}
      This claim is going to be used both in the base case and the inductive step.
      Let $m$ be the smallest index such that $T_1 \cap S_m \ne \emptyset$
      and let $\alpha \in T_1 \cap S_m$.
      Suppose that $T_1$ is not contained in $S_m$, i.e.,
      there is an element $\beta \in T_1 \setminus S_m$.
      Since $M|_{T_1}$ is connected, there are
      $B'_1, \widetilde{B}'_1 \in \mathcal{B}(M|_{T_1})$ such that $\alpha \in B'_1$ and
      $\widetilde{B}'_1 = (B'_1 \setminus \{ \alpha \}) \cup \{ \beta \}$.
      Choose $B'_j \in \mathcal{B}((M|_{T_j})/T_{j-1})$ for $j = 2, \dots, k+1$.
      Then
         $B' = B'_1 \cup B'_2 \cup \cdots \cup B'_{k+1}$

         and

         $\widetilde{B}' = \widetilde{B}'_1 \cup B'_2 \cup \cdots \cup B'_{k+1}$
      are bases of $M_{{\mathcal F}'}$.
      But either $B'$ or $\widetilde{B}'$ is not a base of $M_{{\mathcal F}}$ since
      $$
      |\widetilde{B}' \cap (S_m \setminus S_{m-1})| = |B' \cap (S_m \setminus S_{m-1})|-1.
      $$
      This contradicts the assumption $M_{\mathcal F} = M_{{\mathcal F}'}$.
      Therefore $T_1 \subseteq S_m$.

      Now suppose $(S_m \setminus S_{m-1}) \setminus T_1 \ne \emptyset$, i.e.,
      there is $\gamma \in (S_m \setminus S_{m-1}) \setminus T_1$.
      Since $(M|_{S_m})/S_{m-1}$ is connected, there are
      bases $B_m , \widetilde{B}_m$ of $(M|_{S_m})/S_{m-1}$ satisfying
      $\alpha \in B_m$ and $\widetilde{B}_m = (B_m \setminus \{ \alpha \}) \cup \{ \gamma \}$.
      If $B_j \in \mathcal{B}((M|_{S_j})/S_{j-1})$ for all $j \ne m$,
      then
      $$
      \begin{aligned}
         B &= B_1 \cup \cdots \cup B_{k+1}
         \textrm{ and}\\
         \widetilde{B} &= B_1 \cup \cdots \cup B_{m-1} \cup \widetilde{B}_m \cup B_{m+1} \cup \cdots \cup B_{k+1}
      \end{aligned}
      $$
      are contained in $\mathcal{B}(M_{\mathcal F})$.
      But at least one of them is not in $\mathcal{B} (M_{{\mathcal F}'})$ because
      $|\widetilde{B} \cap T_1| = |B \cap T_1|-1$.
      This is impossible because $M_{\mathcal F} = M_{{\mathcal F}'}$.
      Therefore
      $T_1 = S_m \setminus S_{m-1}$.

      \emph{Base case}: $k = 1$.
      Equation~(\ref{difference-are-the-same}) shows that $T_1 = S_2 \setminus S_1$.
      Since $T_1 \cup S_1 = [n]$,
      $M$ is not connected by Lemma~\ref{switch-connected-case}.
      Thus, ${\mathcal F}$ and ${\mathcal F}'$ are equivalent by Lemma~\ref{switching-components}(ii).

      \emph{Inductive step}.
      Now suppose $k > 1$.
      By Equation~(\ref{difference-are-the-same}), $T_1 = S_m \setminus S_{m-1}$ for some $m > 1$.
      Let $\widetilde{{\mathcal F}}$ be a flag
      $$
      \emptyset =: S_0 \subseteq S_1
      \subseteq S_1 \cup T_1 \subseteq S_2 \cup T_1 \subseteq \cdots
      \subseteq \underbrace{S_{m-1} \cup T_1}_{= S_m} \subseteq S_{m+1} \cdots \subseteq S_{k+1} := [n].
      $$

      We claim that $M|_{S_j \cup T_1} = M|_{S_j} \oplus M|_{T_1}$ for $j = 1, \dots, m-1$.
      Let $B_j \in \mathcal{B}(M|_{S_j})$ and $B'_1 \in \mathcal{B}(M|_{T_1})$.
      Since
      $$
      M_{\mathcal F} = \oplus_{i=1}^{k+1} (M|_{S_i})/S_{i-1}
      = \oplus_{i=1}^{k+1} (M|_{T_i})/T_{i-1} = M_{{\mathcal F}'}
      $$
      and $M|_{T_1} = (M|_{S_m})/S_{m-1}$,
      there is a base $B \in \mathcal{B}(M_{\mathcal F})$ such that $B \cap S_j = B_j$
      and $B \cap T_1 = B'_1$.
      Thus $B_j \cup B'_1 \in \mathcal{B}(M|_{S_j \cup T_1})$,
      and hence
      $$
      r(M|_{S_j \cup T_1}) = r(M|_{S_j}) + r(M|_{T_1}).
      $$
      By \cite[Lemma 3]{MartinReiner}, we have
      $M|_{S_j \cup T_1} = M|_{S_j} \oplus M|_{T_1}$.
      Therefore
      $$
      \begin{aligned}
         (M|_{S_1 \cup T_1})/S_1 &= [M|_{T_1} \oplus M|_{S_1}]/S_1\\

                                 &= (M|_{S_m})/S_{m-1}
      \end{aligned}
      $$
      and
      $$
      \begin{aligned}
         (M|_{S_i \cup T_1})/(S_{i-1} \cup T_1)
         &= [M|_{T_1} \oplus M|_{S_i}]/(S_{i-1} \cup T_1))\\
         &= (M|_{S_i})/S_{i-1}
      \end{aligned}
      $$
      for $i = 2, \dots, m-1$.
      Thus $\widetilde{{\mathcal F}}$ is factor-connected and $M_{\widetilde{{\mathcal F}}} = M_{{\mathcal F}}$.
      By the induction assumption, $\widetilde{{\mathcal F}} \sim {\mathcal F}$.
      Similarly, one can show $S_1 = T_l \setminus T_{l-1}$ for some $l > 1$
      and the chain $\widetilde{{\mathcal F}}'$ given by
      $$
      \emptyset =: T_0 \subseteq T_1
      \subseteq S_1 \cup T_1 \subseteq S_1 \cup T_2 \subseteq \cdots
      \subseteq \underbrace {S_1 \cup T_{l-1}}_{= T_l} \subseteq T_{l+1} \cdots \subseteq T_{k+1} := [n]
      $$

      is factor-connected and $M_{\widetilde{{\mathcal F}}'} = M_{{\mathcal F}'}$.
      Thus the induction assumption gives $\widetilde{{\mathcal F}}' \sim {\mathcal F}'$.

      By the induction assumption again, we have $\widetilde{{\mathcal F}} \sim \widetilde{{\mathcal F}}'$
      and hence ${\mathcal F}$ and ${\mathcal F}'$ are equivalent.
   \end{proof}

   If $M$ is a matroid on $[n]$ and $S$ is a subset of $[n]$,
   then the hyperplane $H_S$ defined by $\sum_{e \in S} x_e = r(S)$
   is a supporting hyperplane of $Q(M)$ and $Q(M) \cap H_S$ is
   the matroid base polytope for $(M|_S) \oplus (M/S)$.
   The next lemma tells us when a face of $Q(M)$ is contained in $H_S$.

   \begin{lemma}
   \label{lem-set-in-flag}
      Let $M$ be a matroid on $[n]$ and $S$ be a subset of $[n]$.
      A face $\sigma$ of $Q(M)$ is contained in $H_S$ if and only if
      there is a factor-connected fla
      $$
      {\mathcal F} = \{ \emptyset =: S_0 \subseteq S_1 \subseteq \cdots
              \subseteq S_k \subseteq S_{k+1} := [n] \}
      $$
      containing $S$ and $\sigma = Q(M_{\mathcal F})$.
   \end{lemma}

   \begin{proof}
      If there is a flag ${\mathcal F}$ containing $S$ and $\sigma = Q(M_{\mathcal F})$,
      then every base of $M_{\mathcal F}$ is contained in $\mathcal{B}((M|_S) \oplus (M/S))$
      and hence $\sigma$ is contained in $H_S$.

      For the converse, suppose $\sigma$ is contained in $H_S$.
      Since $\sigma$ is a face of $Q(M)$, there is a factor-connected flag
      $$
      {\mathcal F}' = \{ \emptyset =: T_0 \subseteq T_1 \subseteq \cdots
               \subseteq T_k \subseteq T_{k+1} := [n] \},
      $$
      such that $\sigma = Q (M_{{\mathcal F}'})$ and $k = n-\dim \sigma -1$.
      We claim that
      $T_i \setminus T_{i-1}$ is contained in either $S$ or $[n]\setminus S$ for $i = 1, \dots, k+1$.
      Suppose there are $\alpha \in (T_i \setminus T_{i-1}) \cap S$ and $\beta \in (T_i \setminus T_{i-1}) \setminus S$.
      Since $(M|_{T_i})/T_{i-1}$ is connected, there exist bases $B_1$ and $B_2$ of $M_{{\mathcal F}'}$
      with $\alpha \in B_1$ and $B_2 = (B_1 \setminus \{ \alpha \}) \cup \{ \beta \}$.
      Then $|B_1 \cap S| > |B_2 \cap S|$ and so $\sigma$ is not contained in $H_S$, which is a contradiction.

      Construct a flag ${\mathcal F}$ as follows:
      $S_j = S_{j-1} \cup (T_l \setminus T_{l-1})$, where $l$ is the smallest index such that
      \begin{enumerate}[(i)]
         \item
         $T_l \setminus T_{l-1} \subseteq S \setminus S_{j-1}$ if $S \setminus S_{j-1} \ne \emptyset$,
         \item
         $T_l \setminus T_{l-1} \subseteq [n]\setminus S_{j-1}$ if $S \setminus S_{j-1} = \emptyset$.
      \end{enumerate}
      By the construction of ${\mathcal F}$, $S$ is contained in ${\mathcal F}$.
      Since $\sigma$ is contained in $H_S$,
      $$
      M_{\mathcal F} = \oplus_{i=1}^{k+1} (M|_{S_i})/S_{i-1} = \oplus_{i=1}^{k+1} (M|_{T_i})/T_{i-1} =  M_{{\mathcal F}'}
      $$
      and hence $\sigma = Q(M_{{\mathcal F}})$.
   \end{proof}

   For a face $\sigma$ of $Q(M)$, let $L_\sigma$ be the poset of all subsets of $[n]$
   which are contained in some factor-connected flag ${\mathcal F}$ with $\sigma = Q(M_{\mathcal F})$
   ordered by inclusion.
   The following lemma shows that $L_\sigma$ is a lattice.

   \begin{lemma}
   \label{lem-closed-under-intersection}
      Let $M$ be a matroid on $[n]$ and $S, T \subseteq [n]$.
      If $Q(M)$ is contained in $H_S$ and $H_T$, then it is also contained in $H_{S \cap T}$.
   \end{lemma}

   \begin{proof}
      Since $Q(M)$ is contained in $H_S$ and $H_T$, we have
      $$
      r(S \cup T) + r(S \cap T) = r(S) + r(T)
      $$
      (see \cite[Lemma 1.3.1]{Oxley}).

      Suppose that $Q(M)$ is not contained in $H_{S \cap T}$, i.e., there is a base $B$
      of $M$ such that $|B \cap (S \cap T)| < r(S \cap T)$.
      Then
      $$
      \begin{aligned}
         |B \cap (S \cup T)| &= |B \cap S| + |B \cap T| - |B \cap (S \cap T)|\\
                             &> r(S) + r(T) - r(S \cap T) = r(S \cup T)
      \end{aligned}
      $$
      which is impossible.
      Therefore $Q(M)$ is contained in $H_{S \cap T}$.
   \end{proof}
   %

   Since $L_\sigma$ is a sublattice of the Boolean lattice $\mathcal{B}_n$,
   it is distributive.
   The fundamental theorem for finite distributive lattices~\cite{StanleyI}
   shows that there is a finite poset $P_\sigma$
   for which $L_\sigma$ is the lattice of order ideals of $P_\sigma$.
   The following theorem describes the poset $P_\sigma$ in terms of matroid data.

   \begin{theorem}
   \label{theorem-posets-for-faces-of-matroid-polytopes}
      Let $M$ be a matroid on $[n]$ and $\sigma$ be a face of $Q(M)$.
      Then $L_\sigma$ is the lattice of order ideals of $P_\sigma$,
      where $P_\sigma$ is the poset defined as follows:
      \begin{enumerate}[(i)]
         \item
         The elements of $P_\sigma$ are the connected components of $M_\sigma$, and
         \item
         for distinct connected components $C_1$ and $C_2$ of $M_\sigma$,
         $C_1 < C_2$ if and only if
         $$
         C_2 \subseteq S \subseteq [n] \text{ and }
         \sigma \subseteq H_S \text{ implies } C_1 \subseteq S.
         $$
      \end{enumerate}
   \end{theorem}

   Note that $P_\sigma$ is a well-defined poset.
   Reflexivity and transitivity are clear.
   Suppose $C_1$ and $C_2$ are distinct connected components of $M_\sigma$ with $C_1 < C_2$.
   Consider a minimal subset $S$ such that $C_2 \subseteq S$ and $\sigma \subseteq H_S$.
   $C_1 < C_2$ implies $C_1$ is contained in $S$. Since $C_2$ is a connected component of $M_\sigma$,
   Lemma~\ref{lem-set-in-flag} implies that there is a flag containing $S$ and $S \setminus C_2$
   such that $Q(M_{\mathcal F}) = \sigma$.
   Since $C_1 \subseteq S \setminus C_2$, $\sigma \subseteq H_{S \setminus C_2}$,
   and $C_2 \nsubseteq S \setminus C_2$, we have $C_2 \nless C_1$.

   \begin{example}
      Let $M_{2,1,1}$ be the rank $2$ matroid on $[4] = \{ 1, 2, 3, 4 \}$
      whose unique non-base is  $12$ and let $\sigma$ be an edge of $Q(M_{2,1,1})$
      connecting $e_{14}$ and $e_{24}$.
      Then the connected components of $M_\sigma$ are $12$, $3$ and $4$.
      Since $\{ 1, 2, 3, 4 \}$ is the only subset $S$ containing $\{ 3 \}$
      such that $\sigma \subseteq H_S$, $12 < 3$ and $4 < 3$ in $P_\sigma$.
      One can see that there are no other relations in $P_\sigma$.
      Figure~\ref{fig-Face-poset} is the proper part of the face poset of $Q(M_{2,1,1})$
      whose faces (shown on the right-hand side of each box) are labeled by corresponding posets
      (shown on the left-hand side of each box) and $P_\sigma$ is shown in the shaded box.
   \end{example}

   \begin{figure}[!t]

      \begin{center}
         \includegraphics[width=0.9\textwidth]{FacePoset_paper}

         \begin{caption}
            {The proper part of the face poset of $Q(M_{2,1,1})$}
            \label{fig-Face-poset}
         \end{caption}
      \end{center}
   \end{figure}

   \begin{proof}[Proof of Theorem~\ref{theorem-posets-for-faces-of-matroid-polytopes}]
      Let $S$ be a set in $L_\sigma$.
      By Lemma~\ref{lem-set-in-flag}, there is a factor-connected flag
      $$
      {\mathcal F} = \{ \emptyset =: S_0 \subseteq S_1 \subseteq \cdots
              \subseteq S_k \subseteq S_{k+1} := [n] \}
      $$
      such that $M_\sigma = M_{\mathcal F}$ and $S_i = S$ for some $i$.
      Then $S = \cup_{j=1}^i (S_j \setminus S_{j-1})$.
      Suppose $S_l \setminus S_{l-1} < S_j \setminus S_{j-1}$ in $P_\sigma$ for some $j \le i$.
      Since $\sigma \subseteq H_S$, $S_l \setminus S_{l-1} \subseteq S$ and
      hence $S$ is an order ideal of $P_\sigma$.

      Conversely, suppose $T$ is an order ideal of $P_\sigma$.
      Let $U$ be the intersection of all subsets $\widetilde{T}$ such that $\sigma \subseteq H_{\widetilde{T}}$
      and $T \subseteq \widetilde{T}$.
      By Lemma~\ref{lem-closed-under-intersection}, $U$ lies in $L_\sigma$.
      Suppose $T \ne U$.
      By Lemma~\ref{lem-set-in-flag}, there is a factor-connected flag
      $$
      {\mathcal F} = \{ \emptyset =: U_0 \subseteq U_1 \subseteq \dots
              \subseteq U_k \subseteq U_{k+1} := [n] \}
      $$
      such that $\sigma = Q(M_{\mathcal F})$ and $U = U_j$ for some $j$.
      Since $T \ne U$ and $T$ is an order ideal,
      we may choose ${\mathcal F}$ so that $U_j \setminus U_{j-1} \nsubseteq T$.
      Then $U_{j-1} \in L_\sigma$ and $T \subseteq U_{j-1}$
      which is a contradiction.
   \end{proof}

   The posets $P_\sigma$ coincide with the posets obtained from preposets corresponding to normal cones
   of the matroid base polytope $Q(M)$ studied by Postnikov, Reiner, and Williams~\cite{PostnikovReinerWilliams}.

   Billera, Jia, and Reiner~\cite{BilleraJiaReiner} define posets
   related to bases of matroids:
   For each $e \in B \in \mathcal{B}(M)$ the \emph{basic bond for $e$ in $B$}
   is the set of $e' \in [n]\setminus B$ for which $(B \setminus \{ e \}) \cup \{ e' \}$ is another
   base of $M$.
   Dually, for each $e' \in [n]\setminus B$ the \emph{basic circuit for $e$ in $B$} is
   the set of $e \in B$ for which $(B \setminus \{ e \}) \cup \{ e' \}$ is another base of $M$.
   Since $e'$ lies in the basic bond for $e$ if and only if
   $e$ lies in the basic circuit of $e'$,
   one can define the poset $P_B$ to be the poset defined on $[n]$
   where $e <_{P_B} e'$ if and only if $e'$ is in the basic bond for $e$.
   The next proposition shows that our poset $P_\sigma$ is the same as
   the poset $P_B$ of Billera, Jia, and Reiner if $\sigma$ is a vertex $e_B$.

   \begin{proposition}
      If $B$ is a base of a matroid $M$ on $[n]$ and $\sigma$ is a vertex $e_B$ of $Q(M)$, then
      $P_\sigma = P_B$.
   \end{proposition}

   \begin{proof}
      Since all connected components of $M_\sigma$ are singletons,
      $P_\sigma$ and $P_B$ have the same elements.
      If $e_1, e_2 \in B$, then they are not comparable in $P_\sigma$ since
      $e_B \subseteq H_{e_1}$ and $e_B \subseteq H_{e_2}$.
      Also, $e_1, e_2 \notin B$ are not comparable in $P_\sigma$
      because $e_B \subseteq H_{B \cup \{ e_1 \}}$ and $e_B \subseteq H_{B \cup \{ e_2 \}}$.
      Now assume $e_1 \in B$ and $e_2 \notin B$.
      We claim that $e_1 <_{P_\sigma} e_2$ if and only if $B' := (B \setminus \{ e_1 \}) \cup \{ e_2 \} \in \mathcal{B}(M)$,
      i.e., $e_1 <_{P_B} e_2$.
      Assume $e_1 <_{P_\sigma} e_2$, i.e., $e_2 \in S$ and $\sigma \subseteq H_S$ implies $e_1 \in S$.
      Suppose $B'$ is not a base of $M$.
      Then $r(B') = r(M) -1$.
      Thus we have $e_2 \in B', \sigma \subseteq H_{B'}$, but $e_1 \notin B'$, which is a contradiction.
      Therefore $B'$ is a base of $M$.
      Conversely, assume $B' \in \mathcal{B}(M)$.
      Let $S$ be a subset of $[n]$ such that $e_2 \in S$ and $\sigma \subseteq H_S$, i.e., $|B \cap S| = r(S)$.
      If $e_1 \notin S$, then $|B' \cap S| = |B \cap S| + 1 = r(S) + 1$, which is impossible.
      Thus $e_1 <_{P_\sigma} e_2$.
   \end{proof}

\section{The $\mathbf{c}\mathbf{d}$-index}
\label{sec-cd-index}

   In this section, we define the $\mathbf{c}\mathbf{d}$-index for Eulerian posets and give
   the relationship among $\mathbf{c}\mathbf{d}$-indices of polytopes when a polytope is split by a hyperplane.

   Let $P$ be a graded poset of rank $n+1$ with the rank function $\rho$.
   For a subset $S$ of $[n]$, define $f_P (S)$ to be the number of chains of $P$
   whose ranks are exactly given by the set $S$.
   The function $f_P : 2^{[n]} \to {\mathbb N}$ is called the \emph{flag f-vector} of $P$.
   The \emph{flag h-vector} of $P$ is defined by the identity
   $$
   h_P (S) = \sum_{T \subseteq S} (-1)^{|S \setminus T|} \cdot f_P(T).
   $$
   Since  this identity is equivalent to the relation
   $$
   f_P (S) = \sum_{T \subseteq S} h_P (T),
   $$
   the flag $f$-vector and the flag $h$-vector contain the same information.

   For $S \subseteq [n]$, define the \emph{noncommutative} $\mathbf{a}\mathbf{b}$-monomial $u_S = u_1 \dots u_n$,
   where
   $$
   u_i = \left\{
   \begin{array}{ll}
      \mathbf{a} & \text{if } i \notin S, \\
      \mathbf{b} & \text{if } i \in S.
   \end{array}
   \right.
   $$
   The $\mathbf{a}\mathbf{b}$-\emph{index} of the poset $P$  is defined to be the sum
   $$
   \Psi (P) = \sum_{S \subseteq [n]} h_P (S) \cdot u_S.
   $$

   An alternative way of defining the $\mathbf{a}\mathbf{b}$-index is as follows.
   For a chain
   $$
   c := \{ \hat{0} < x_1 < \cdots < x_k < \hat{1} \},
   $$
   we give a \emph{weight} $w_P (c) = w(c) = z_1 \cdots z_n$, where
   $$
   z_i = \left\{
   \begin{array}{ll}
      \mathbf{b} & \text{if } i \in \{ \rho(x_1), \dots, \rho(x_k) \},\\
      \mathbf{a} - \mathbf{b} & \text{otherwise.}
   \end{array}
   \right.
   $$
   Define the $\mathbf{a}\mathbf{b}$-\emph{index} of the poset $P$ to be the sum
   $$
   \Psi (P) = \sum_c w(c),
   $$
   where the sum is over all chains $c = \{ \hat{0} < x_1 < \cdots < x_k < \hat{1} \}$
   in $P$.
   Recall that a poset $P$ is \emph{Eulerian} if its M\"{o}bius function $\mu$
   is given by $\mu(x, y) = (-1)^{\rho(y) - \rho(x)}$
   (see \cite{StanleyI} for more details).
   One important class of Eulerian posets is face lattices of convex polytopes
   (see \cite{Lindstrom,Rota}).
   Fine conjectured and Bayer and Klapper~\cite{BayerKlapper} proved that
   the $\mathbf{a}\mathbf{b}$-index of an Eulerian poset $P$ can be written uniquely
   as a polynomial of $\mathbf{c} = \mathbf{a} + \mathbf{b}$ and $\mathbf{d} = \mathbf{a}\mathbf{b} + \mathbf{b}\mathbf{a}$.
   When the $\mathbf{a}\mathbf{b}$-index can be written as a polynomial in $\mathbf{c}$ and $\mathbf{d}$, we call
   $\Psi (P)$ the $\mathbf{c}\mathbf{d}$-\emph{index} of $P$.
   We will use the notation $\Psi(Q)$ for the $\mathbf{c}\mathbf{d}$-index of the face poset of
   a convex polytope $Q$.

   Let $v$ be a vertex of a polytope $Q$ and let $l(x) = c$ be a supporting hyperplane
   of $Q$ defining $v$.
   The \emph{vertex figure} $Q/v$ of $v$ is defined by
   $$
   Q/v = Q \cap \{ l(x) = c+ \delta \}
   $$
   where $\delta$ is an arbitrary small positive number.
   The polytope $Q/v$ depends on the choice of $l, c$, and $\delta$, but
   it is well-known that its combinatorial type is independent of these variables.
   For a face $\sigma$ of $Q$, the \emph{face figure} $Q/\sigma$ of $\sigma$ is defined by
   $$
   Q/\sigma = ( \dots (( Q / \sigma_0 ) / \sigma_1 ) \dots ) / \sigma_k ,
   $$
   where $\sigma_0 \subseteq \sigma_1 \subseteq \dots \subseteq \sigma_k = \sigma$
   is a maximal chain with $\dim \sigma_i = i$.
   For faces $\sigma$ and $\tau$ of $Q$ with $\sigma \subseteq \tau$,
   the face lattice of the face figure $\tau / \sigma$ is the interval $[\sigma, \tau]$.

   Ehrenborg and Readdy~\cite[Prop. 4.2]{EhrenborgReaddy} give formulas for the $\mathbf{c}\mathbf{d}$-index
   of a pyramid, a prism and a bipyramid of a polytope.

   \begin{proposition}
   \label{prop-pyramid-and-prism}
      Let $Q$ be a polytope. Then
      $$
      \begin{aligned}
         \Psi (\mathrm{Pyr} (Q))   &= \frac 1 2 \left[ \Psi(Q) \cdot \mathbf{c} + \mathbf{c} \cdot \Psi (Q)
                              + \sum_{\sigma}\Psi (\sigma) \cdot \mathbf{d} \cdot \Psi (Q/\sigma)
                              \right],\\
         \Psi (\mathrm{Prism} (Q)) &= \Psi (Q) \cdot \mathbf{c} +
                              \sum_{\sigma} \Psi (\sigma) \cdot \mathbf{d} \cdot \Psi (Q/\sigma),\\
         \Psi (\mathrm{Bipyr} (Q)) &= \mathbf{c} \cdot \Psi (Q) +
                              \sum_{\sigma} \Psi (\sigma) \cdot \mathbf{d} \cdot \Psi (Q/\sigma),
      \end{aligned}
      $$
      where the sums are over all proper faces $\sigma$ of $Q$.
   \end{proposition}

   Note that the $\mathbf{c}\mathbf{d}$-index of $\mathrm{Bipyr} (Q)$ is obtained from the $\mathbf{c}\mathbf{d}$-index of $\mathrm{Prism} (Q)$
   because $\mathrm{Bipyr} (Q)$ is the dual of the prism of the dual of $Q$
   and the $\mathbf{c}\mathbf{d}$-index of the dual polytope is obtained by writing every $\mathbf{a}\mathbf{b}$-monomial in reverse order
   (see~\cite{EhrenborgReaddy} for details).

   Let $Q$ be a polytope in ${\mathbb R}^n$.
   Let $H$ be a hyperplane in ${\mathbb R}^n$ defined by $l(x) = c$ and
   $H^+$ (resp. $H^-$) be the closed halfspace $l(x) \ge c$ (resp. $l(x) \le c$).

   For simplicity, let $Q^+ := Q \cap H^+$, $Q^- = Q \cap H^-$, and $\widehat{Q} := Q \cap H$.
   Also we use the notations $\sigma^+ := \sigma \cap H^+$, $\sigma^- := \sigma \cap H^-$, and
   $\hat{\sigma} := \sigma \cap H$ for a face $\sigma$ of $Q$.
   The following theorem provides the relationship among $\mathbf{c}\mathbf{d}$-indices of polytopes
   $Q$, $Q^+$, $Q^-$ and faces of $\widehat{Q}$ when a polytope $Q$ is split by a hyperplane.

   \begin{theorem}
   \label{cd-index-hyperplane-split-case}
      Let $Q$ be a polytope in ${\mathbb R}^n$ and $H$ be a hyperplane in ${\mathbb R}^n$ intersecting
      the interior of $Q$.
      Then the following identity holds:
      $$
      \Psi (Q) = \Psi (Q^+) + \Psi (Q^-) - \Psi (\widehat{Q}) \cdot \mathbf{c} -
      \sum_\sigma \Psi (\hat{\sigma}) \cdot \mathbf{d} \cdot \Psi ( \widehat{Q} / \hat{\sigma} ),
      $$
      where the sum is over all proper faces $\sigma$ of $Q$ intersecting both
      open halfspaces $H^+ \setminus H$ and $H^- \setminus H$ nontrivially.
   \end{theorem}

   \begin{proof}
      Let
      $$
      c = \{ \emptyset = \sigma_0 < \sigma_1 < \cdots < \sigma_k < \sigma_{k+1} = Q \}
      $$
      be a chain of faces of $Q$.
      If there is a face $\sigma_i$ which meets $H^+ \setminus H$
      (resp. $H^- \setminus H$) but does not meet $H^- \setminus H$
      (resp. $H^+ \setminus H$),
      then
      $$
      c^+ = \{ \emptyset = \sigma_0^+ < \sigma_1^+ < \cdots < \sigma_k^+ < \sigma_{k+1}^+ = Q^+ \}
      $$
      $$
      (\text{resp. }c^- = \{ \emptyset = \sigma_0^- < \sigma_1^- < \cdots < \sigma_k^- < \sigma_{k+1}^- = Q^- \})
      $$
      is a corresponding chain in the face poset of $Q^+$ (resp. $Q^-$) and
      $w(c) = w(c^+)$ (resp. $w(c) = w(c^-)$).

      Now suppose that either $\sigma_i$ is contained in $H$ or intersects both
      $H^+ \setminus H$ and $H^- \setminus H$ for all $i = 0, \dots , k+1$.
      Let $s$ be the smallest index such that $\sigma_i \nsubseteq H$ and
      $\hat{c}$ be the chain in the face poset of $\widehat{Q}$ defined by
      $$
      \hat{c} := c \cap H = \{ \emptyset = \sigma_0 < \hat{\sigma}_1 < \cdots
      < \hat{\sigma}_{k+1} = \widehat{Q} \}.
      $$

      \begin{enumerate}
         \item
         Consider a chain $c^+$ in the face poset of $Q^+$ such that
         $c^+ \cap H$ and $\hat{c}$ are the same.
         Then $c^+$ is one of the following chains for some $j$ with $s \le j \le k+1$:
         $$
         \begin{aligned}
            c^+_1 &= \{ \emptyset < \hat{\sigma}_1 < \cdots < \hat{\sigma}_{j-1}
                     < \sigma_j^+ < \cdots < \sigma_k^+ < \sigma_{k+1}^+ = Q^+ \},\\
            c^+_2 &= \{ \emptyset < \hat{\sigma}_1 < \cdots < \hat{\sigma}_{j-1}
                     < \hat{\sigma}_j < \sigma_j^+ < \cdots < \sigma_k^+ < \sigma_{k+1}^+ = Q^+ \}.
         \end{aligned}
         $$
         Let $\sigma := \sigma_j$.
         Then $\sigma$ meets both open halfspaces $H^+ \setminus H$ and $H^- \setminus H$ nontrivially.
         Let
         $\hat{c}_1 := \{ \emptyset < \hat{\sigma}_1 < \cdots < \hat{\sigma}_{j-1} < \hat{\sigma} \}$ and
         $\hat{c}_2 := \{ \hat{\sigma} < \hat{\sigma}_{j+1} < \cdots < \hat{\sigma}_k < \widehat{Q} \}$.
         There are two cases:
         \begin{enumerate}
            \item
            The first case is when $s \le j \le k$.
            Then the sum of the weights of the chains $c^+_1$ and $c^+_2$ is given by
            $$
            w(c^+_1) + w(c^+_2) = w_{[\emptyset, \hat{\sigma} ]}(\hat{c}_1)\cdot \mathbf{a} \cdot \mathbf{b} \cdot
            w_{[\hat{\sigma}, \widehat{Q}]}(\hat{c}_2).
            $$
            Note that this case does not occur when $s = k+1$.
            \item
            The second case is when $j = k+1$.
            Then the sum of the weights of the chains is
            $$
            w(c^+_1) + w(c^+_2) = w(\hat{c}) \cdot \mathbf{a}.
            $$
         \end{enumerate}
         \item
         Consider a chain $c^-$ in the face poset of $Q^-$ such that
         $c^- \cap H$ and $\hat{c}$ are the same.
         Then $c^-$ is one of the following chains for some $j$ with $s - 1 \le j \le k+1$:
         $$
         \begin{aligned}
            c^-_1 &= \{ \emptyset < \hat{\sigma}_1 < \cdots < \hat{\sigma}_j
                     < \sigma_{j+1}^- < \cdots < \sigma_k^- < \sigma_{k+1}^- = Q^- \},\\
            c^-_2 &= \{ \emptyset < \hat{\sigma}_1 < \cdots < \hat{\sigma}_j
                     < \sigma_j^- < \sigma_{j+1}^- < \cdots < \sigma_k^- < \sigma_{k+1}^- = Q^- \}.
         \end{aligned}
         $$
         Let $\sigma := \sigma_j$ and define chains $\hat{c}_1$ and $\hat{c}_2$ as in Case (1).
         If $j \ge s$, then $\sigma$ meets both $H^+ \setminus H$ and $H^- \setminus H$.
         There are three cases:
         \begin{enumerate}
            \item
            The first case is when $s \le j \le k$.
            Then the sum of the weights of the chains $c^-_1$ and $c^-_2$ is given by
            $$
            w(c^-_1) + w(c^-_2) = w_{[\emptyset, \hat{\sigma} ]}(\hat{c}_1)\cdot \mathbf{b} \cdot \mathbf{a} \cdot
            w_{[\hat{\sigma}, \widehat{Q}]}(\hat{c}_2).
            $$
            Note that this case does not occur when $s = k+1$.
            \item
            The second case is when $j = k+1$.
            Then the only possible chain is $c^-_2$ and its weight is
            $$
            w(c^-_2) = w(\hat{c}) \cdot \mathbf{b}.
            $$
            \item
            The third case is when $j = s - 1$.
            Since $\sigma_{s-1} \subseteq H$, we have $\hat{\sigma}_{s-1} = \hat{\sigma}^-_{s-1}$.
            Therefore the only possible chain is $c^-_1$ and its weight is
            $$
            w(c^-_1) = w_{[\emptyset, \hat{\sigma} ]}(\hat{c}_1) \cdot \mathbf{b} \cdot (\mathbf{a} - \mathbf{b}) \cdot
            w_{[\hat{\sigma}, \widehat{Q}]}(\hat{c}_2) = w(c).
            $$
         \end{enumerate}
      \end{enumerate}
      Now summing over all such chains in the face posets of $Q^+$ and $Q^-$, we obtain
      $$
      \sum_{c^+} w(c^+) + \sum_{c^-} w(c^-) =
         \sum_{\hat{c}_1, \hat{c}_2} w_{[\emptyset, \hat{\sigma} ]}(\hat{c}_1)\cdot \mathbf{d} \cdot
         w_{[\hat{\sigma}, \widehat{Q}]}(\hat{c}_2)
         + w(\hat{c}) \cdot \mathbf{c} + w(c),
      $$
      where the first (resp. second) sum is over all chains $c^+$ (resp. $c^-$) in the face poset of
      $Q^+$ (resp. $Q^-$) such that $c^+ \cap H = \hat{c}$ (resp. $c^- \cap H = \hat{c}$),
      and the third sum is over all pairs of chains $\hat{c}_1, \hat{c}_2$ obtained above
      such that $\hat{c}_2$ is nontrivial.
      Finally, summing over all chains in the face poset of $Q$,
      we finish the proof.
   \end{proof}
   %

      %

   \begin{remark}
      The formula for the prism of a polytope in
      Proposition~\ref{prop-pyramid-and-prism} is a special case of
      Theorem~\ref{cd-index-hyperplane-split-case},
      since in this case
      $$
      \begin{aligned}
         Q = \mathrm{Prism}(Q') &\simeq Q' \times \left[0, 0.5 \right] = Q^- \\
                        &\simeq Q' \times \left[0.5, 1 \right] = Q^+
      \end{aligned}
      $$
      and $Q' \simeq Q' \times \{ 0.5 \} = \widehat{Q}$.
      Also, the formula for the pyramid of a polytope is obtained from
      Theorem~\ref{cd-index-hyperplane-split-case}
      by considering $Q = \mathrm{Bipyr} (Q')$ split by the hyperplane containing $Q'$:
      in this case, $Q^+ = Q^- = \mathrm{Pyr} (Q')$ and there are no proper faces of $\mathrm{Bipyr} (Q')$
      intersecting both open halfspaces nontrivially.
   \end{remark}

\section{Hyperplane splits of a matroid base polytope}
\label{sec-hyperplane-splits}

   In this section, we define hyperplane splits of a matroid base polytope
   and give conditions when they occur.

   For a matroid $M$ on $[n]$, a \emph{hyperplane split} of the matroid base polytope $Q(M)$ is a decomposition
   $Q(M) = Q(M_1) \cup Q(M_2)$ where
   \begin{enumerate}[(i)]
      \item
      $M_1$ and $M_2$ are matroids on $[n]$, and
      \item
      the intersection $Q(M_1) \cap Q(M_2)$ is a proper face of
      both $Q(M_1)$ and $Q(M_2)$.
   \end{enumerate}

   Let $\sum_{i=1}^n a_i x_i = b$ be an equation defining corresponding hyperplane $H$.
   Since the intersection $Q(M_1) \cap Q(M_2)$ is a matroid base polytope contained in $H$ and
   its edges are parallel to $e_i - e_j$ for some $i \ne j$,
   the only constraints on the normal vector $(a_1, a_2, \dots, a_n)$ of $H$
   are of the form $a_i = a_j$.
   Using the fact that $Q(M)$ is a subset of a simplex defined by $\sum_{i=1}^n x_i = r(M)$
   and scaling the right hand side $b$, one can assume that $H$ is defined by
   $\sum_{e \in S} x_e = k$ for some subset $S$ of $[n]$ and some positive integer $k < r(M)$.

   The following result characterizes when hyperplane splits occur.

   \begin{theorem}
   \label{matroid-base-polytope-split}
      Let $M$ be a rank $r$ matroid on $[n]$ and $H$ a hyperplane defined by
      $\sum_{e \in S} x_e = k$.
      Then $H$ gives a hyperplane split of $Q(M)$
      if and only if the following are satisfied:
      \begin{enumerate}[(i)]
         \item
         $r(S) > k$ and $r(S^c) > r-k$,
         \item
         if $I_1$ and $I_2$ are $k$-element independent subsets of $S$ such that
         $(M/I_1)|_{S^c}$ and $(M/I_2)|_{S^c}$ have rank $r-k$, then
         $(M/I_1)|_{S^c} = (M/I_2)|_{S^c}.$

      \end{enumerate}
   \end{theorem}

   \begin{remark}
      Note that if $I$ is a $k$-element independent subset of $S$ and
      $J$ is an $(r-k)$-element independent subset of $S^c$, then
      $I$ is a base for $(M/J)|_S$ if and only if $J$ is a base for $(M/I)|_{S^c}$.
      Therefore the condition (ii) can be replaced with the following condition for $S^c$:
      \begin{enumerate} \it
         \item[(ii$\,'$)]
         if $J_1$ and $J_2$ are $(r-k)$-element independent subsets of $S^c$ such that
         $(M/J_1)|_S$ and $(M/J_2)|_S$ have rank $k$, then
         $(M/J_1)|_S = (M/J_2)|_S$.
      \end{enumerate}
   \end{remark}

   \begin{remark}
      Herrmann and Joswig \cite{HerrmannJoswig} study the splits of general polytopes.
      For the entire hypersimplex, Theorem~\ref{matroid-base-polytope-split} reduces to \cite[Lemma 38]{HerrmannJoswig}.
   \end{remark}

   \begin{proof}[Proof of Theorem~\ref{matroid-base-polytope-split}]
      The condition (i) is equivalent to the condition that $H$ intersects the interior of $Q(M)$ nontrivially.

      Define $\mathcal{B}_k = \{ B \in \mathcal{B}(M) : | B \cap S | = k \}$.
      We will show that the condition (ii) holds if and only if
      $\mathcal{B}_k$ is a collection of bases of some matroid.
      Then the assertion follows from Theorem~\ref{thm-matroid-polytopes}.

      Suppose that the condition (ii) is true.
      Choose any bases $B_1$ and $B_2$ in $\mathcal{B}_k$ and $x \in B_1 \setminus B_2$
      (without loss of generality, we may assume $x \in B_1 \cap S$).
      Let $I_i = B_i \cap S$ and $J_i = B_i \setminus S$ for $i = 1,2$.
      Then the condition (ii) implies that there is a base $B = I_2 \cup J_1$ in $\mathcal{B}_k$.
      Since $B_1, B \in \mathcal{B}$, there is $y \in B \setminus B_1 \subseteq I_2 \subseteq B_2$
      such that $B_3 = (B \setminus \{ x \}) \cup \{ y \} \in \mathcal{B}$.
      Since $y \in I_2 \subseteq S$, $B_3 \in \mathcal{B}_k$.
      Thus $\mathcal{B}_k$ forms a collection of bases of a matroid.

      Conversely suppose that $\mathcal{B}_k$ is a collection of bases of some matroid.
      Let $I_1$ and $I_2$ be $k$-element independent subsets of $S$ such that
      $(M/I_1)|_{S^c}$ and $(M/I_2)|_{S^c}$ have rank $r-k$.
      Choose $J_1 \in \mathcal{B}((M/I_1)|_{S^c})$ and $J_2 \in \mathcal{B}((M/I_2)|_{S^c})$.
      Then $B_1 = I_1 \cup J_1$ and $B_2 = I_2 \cup J_2$ are bases for $B$.
      We claim that $I_2 \cup J_1$ is also a base of $M$: this implies
      $\mathcal{B}((M/I_1)|_{S^c}) \subseteq \mathcal{B}((M/I_2)|_{S^c})$ and
      (ii) follows by symmetry.
      We use induction on the size of $I_1 - I_2$.

      \emph{Base Case}:
         If $|I_1 \setminus I_2| = 0$, we have $I_2 \cup J_1 = B_1 \in \mathcal{B}$.

      \emph{Inductive Step}:
         Suppose $|I_1 \setminus I_2| = l$ for some $l \le k$.
         Choose an element $x \in I_1 \setminus I_2 \subseteq B_1 \setminus B_2$.
         Since $\mathcal{B}_k$ forms a matroid, there exist $y \in I_2 \setminus I_1$ such that
         $B_3 = (B_1 \setminus \{ x \}) \cup \{ y \} \in \mathcal{B}_k \subseteq \mathcal{B}$.
         Since $B_3 = [(I_1 \setminus \{ x \}) \cup \{ y \}] \cup J_1$, we have $|(B_3 \cap S) \setminus I_2| = l-1$
         and the induction hypothesis implies $I_2 \cup J_1 \in \mathcal{B}$.
   \end{proof}

\section{Rank $2$ matroids}
\label{sec-rank-2-matroids}

   In this section we study the $\mathbf{c}\mathbf{d}$-index of a matroid base polytope
   when a matroid has rank $2$.

   A (loopless) rank $2$ matroid $M$ on $[n]$ is determined up to isomorphism by the composition
   $\alpha(M)$ of $n$ that gives the sizes $\alpha_i$ of its parallelism classes.

   %

   %

   %

   %
   %

   %

   %
   Let $\alpha := (\alpha_1, \alpha_2, \dots ,\alpha_k)$ be a composition of $n$ with the length $\ell(\alpha) = k$
   and let $M_\alpha$ be the corresponding rank $2$ matroid on $[n]$.
   When $k = 2$, we have
   $$
   Q(M_\alpha) \cong Q(U_{1,\alpha_1}) \times Q(U_{1, \alpha_2})
   \cong \Delta_{\alpha_1} \times \Delta_{\alpha_2},
   $$
   where $U_{r,n}$ is the uniform matroid of rank $r$ on $n$ elements
   and $\Delta_n$ is an $(n-1)$-dimensional simplex.

   For two weak compositions (i.e., compositions allowing $0$ as parts)
   $\alpha$ and $\beta$ of the same length, we define
   $\beta \le \alpha$ if $\beta_i \le \alpha_i$ for all $i = 1, 2, \dots, \ell(\alpha)$.
   Let $\bar{\beta}$ be the composition obtained from $\beta$ by deleting $0$ parts.
   If $\alpha = (2,4,0,6,7)$ and $\beta = (1,3,0,6,3)$, then $\beta < \alpha$ and
   $\bar{\beta} = (1,3,6,3)$.

   When $M$ has rank $2$, Theorem~\ref{matroid-base-polytope-split}
   can be rephrased in the following way.

   \begin{corollary}
   \label{matroid-base-polytope-split-rank-two}
      Let $M$ be a rank $2$ matroid on $[n]$ and $H$ be a hyperplane defined by
      $\sum_{e \in S} x_e = 1$.
      Then $H$ gives a hyperplane split of $Q(M)$ if and only if
      $S$ and $S^c$ are both unions of at least two parallelism classes of $M$.
   \end{corollary}
   %

   %

   %

   %

   \begin{question}
   \label{matroidal-version-of-cd-index-hyperplane-split-case}
      When $Q(M_1) \cup Q(M_2)$ is a hyperplane split of $Q(M)$ with a corresponding
      hyperplane $H$, can Theorem~\ref{cd-index-hyperplane-split-case} be restated
      in terms of matroids?
   \end{question}

   %

   %

   %

   %

   If $M$ has rank $2$, then one can rephrase Theorem~\ref{cd-index-hyperplane-split-case}
   in terms of matroids (see Proposition~\ref{cd-index-hyperplane-split-case-rank-two} below), but
   Question~\ref{matroidal-version-of-cd-index-hyperplane-split-case} is open for higher ranks.

   After the relabeling, one may assume that a rank $2$ matroid $M$ has parallelism classes $P_1, P_2, \dots, P_k$
   and $S = \cup_{i = 1}^m P_i$ for some $m$.
   In this case, one can restate Theorem~\ref{cd-index-hyperplane-split-case} in terms of matroids as follows.

   \begin{proposition}
   \label{cd-index-hyperplane-split-case-rank-two}
      Let $M$ be a rank $2$ matroid on $[n]$ with at least four parallelism classes $P_1, P_2, \dots, P_k$
      and $S = \cup_{i = 1}^m P_i$ for some $m$ satisfying $2 \le m \le k-2$.
      Then the hyperplane $H$ defined by $\sum_{e \in S} x_e = 1$ gives a hyperplane split
      $Q(M_1) \cup Q(M_2)$ of $Q(M)$ where $M_1$ is a matroid with parallelism classes $S, P_{m+1}, \dots, P_k$
      and $M_2$ is a matroid whose parallelism classes are $P_1, \dots, P_m, S^c$.
      Moreover,
      $$
      \begin{aligned}
         \Psi (Q(M)) =& \Psi (Q(M_1)) + \Psi (Q(M_2)) - \Psi (\Delta_{|S|} \times \Delta_{n-|S|}) \cdot \mathbf{c} \\
                      &- \sum_T \Psi (\Delta_{|T \cap S|} \times \Delta_{|T \setminus S|}) \cdot \mathbf{d} \cdot \Psi (\Delta_{n-|T|}),
      \end{aligned}
      $$
      where the sum is over all proper subsets $T$ of $[n]$ such that
      $M|_T$ has at least four parallelism classes and
      $S$ and $S^c$ both contain at least two parallelism classes of $M|_T$.
   \end{proposition}

   \begin{proof}
      Note that there is no flacet (i.e., facet corresponding to a flat of $M$)
      of $Q(M)$ which intersects both open halfspace given by $H$ nontrivially
      since every flacet of $Q(M)$ corresponds to a base set of the form
      $$
      \{ B \in \mathcal{B}(M) : |B \cap P_i| = 1  \}
      $$
      for some $i$.
      (The corresponding flacet has empty intersection with the open halfspace $H^- \setminus H$ if $i \le m$
      while it has empty intersection with $H^+ \setminus H$ otherwise.)
      If $\sigma$ is a face of $Q(M)$ which has nonempty intersection with both open halfspaces given by $H$,
      then $\sigma$ is the intersection of some facets of $Q(M)$ which are not flacets.
      Since each facet of $Q(M)$ which is not a flacet corresponds to the deletion of an element of $[n]$,
      $\sigma$ corresponds to a matroid $M|_T$ for some subset $T$ of $[n]$.
      Also, the upper interval $[\sigma \cap H, Q(M) \cap H]$ $(\cong [\sigma, Q(M)])$ is isomorphic to
      the Boolean algebra of order $n - |T|$,
      and hence $\Psi[(Q(M) \cap H) / (\sigma \cap H)] = \Psi (\Delta_{n-|T|})$.
      Finally, $\sigma$ has nonempty intersection with both open halfspaces given by $H$ if and only if
      $M|_T$ has at least four parallelism classes and
      $S$ and $S^c$ both contain at least two parallelism classes of $M|_T$.
      Now, the result follows from Theorem~\ref{cd-index-hyperplane-split-case}.
   \end{proof}

   The following corollary, which is obtained from Corollary~\ref{matroid-base-polytope-split-rank-two}
   and Proposition~\ref{cd-index-hyperplane-split-case-rank-two},
   expresses the $\mathbf{c}\mathbf{d}$-index of a matroid base polytope
   of a rank $2$ matroid $M$
   in terms of $\mathbf{c}\mathbf{d}$-indices of matroid base polytopes of matroids corresponding to compositions of length $\le 3$.
   For simplicity, we use the following notations:
   $$
   \begin{aligned}
      \lambda(\alpha, i) &= \left( \sum_{j=1}^{i-1} \alpha_j, \alpha_i, \sum_{j=i+1}^{\ell(\alpha)} \alpha_j \right)
                         &\text{ for } 2 \le i \le \ell(\alpha) - 1, \\
      \mu(\alpha, i)     &= \left(\sum_{j=1}^{i} \alpha_j, \sum_{j=i+1}^{\ell(\alpha)} \alpha_j \right)
                         &\text{ for } 1 \le i \le \ell(\alpha) - 1.
   \end{aligned}
   $$
   If $\alpha = (2,4,0,6,7)$, then
   $\lambda(\alpha, 4) = (6,6,7)$ and $\mu(\alpha, 4) = (12,7)$.
   For a rank $2$ matroid $M$ with parallelism classes $P_1, \dots, P_k$,
   define $M_i$ be the rank $2$ matroid with two parallelism classes $P_1 \cup \dots \cup P_i$ and $P_{i+1} \cup \dots \cup P_k$.
   Note that $(M_\alpha)_i$ is isomorphic to $M_{\mu(\alpha, i)}$.

   \begin{corollary}
   \label{cd-index-of-rank-2-matroids}
      Let $\alpha$ be a composition of $n$ with at least three parts and
      $M_\alpha$ be the corresponding rank $2$ matroid on $[n]$.
      Then the $\mathbf{c}\mathbf{d}$-index of $Q(M_\alpha)$ can be expressed as follows:
      $$
      \begin{aligned}
         \Psi(Q(M_\alpha)) =& \sum_{i = 2}^{\ell(\alpha)-1} \Psi(Q(M_{\lambda(\alpha, i)}))
                              - \left( \sum_{i=2}^{\ell(\alpha)-2}
                              \Psi(\Delta_{\mu(\alpha,i)_1} \times \Delta_{\mu(\alpha,i)_2}) \right) \cdot \mathbf{c} \\
                            &- \sum_{\substack{\beta < \alpha\\ \ell(\bar{\beta}) \ge 4}} \prod_{j=1}^{\ell(\alpha)} {\binom{\alpha_j}{\beta_j}}
                             \left( \sum_{i=2}^{\ell(\bar{\beta})-2}
                             \Psi(\Delta_{\mu(\bar{\beta}, i)_1} \times \Delta_{\mu(\bar{\beta},i)_2}) \right)
                             \cdot \mathbf{d} \cdot \Psi(\Delta_{n-|\bar{\beta}|}).
      \end{aligned}
      $$

   \end{corollary}

   %
   \begin{proof}

      For simplicity, let $M := M_\alpha$ and $k := \ell (\alpha)$.
      After the relabeling, one may assume that the matroid $M$ has parallelism classes $P_1, \dots, P_k$
      with $|P_i| = \alpha_i$ for all $i = 1, \dots, k$.
      We first claim that
      $$
      \begin{aligned}
      \Psi(Q(M)) = & \sum_{i = 2}^{k-1} \Psi(Q(M_{\lambda(\alpha, i)}))
                     - \left( \sum_{i=2}^{k-2} \Psi(\Delta_{\mu(\alpha,i)_1} \times \Delta_{\mu(\alpha,i)_2}) \right) \cdot \mathbf{c} \\
                   &- \sum_S \left( \sum_{i=2}^{p(S)-2}
                             \Psi(Q((M|_S)_i) \right) \cdot \mathbf{d} \cdot \Psi(\Delta_{n-|S|}).
      \end{aligned}
      $$
      where the sum in the second line runs over all proper subsets $S$ of $[n]$
      such that $M|_S$ has at least four parallelism classes,
      and $p(S)$ is the number of parallelism classes of $M|_S$.
      We will use induction on $k$.

      \emph{Base Case}:
         If $k=3$, then the result is clear.

      \emph{Inductive Step}:
         By Corollary~\ref{matroid-base-polytope-split-rank-two},
         the hyperplane $H$ defined by
         $$
         \sum_{e \in P_1 \cup \dots \cup P_{k-2}} x_e = 1
         $$
         gives a hyperplane split $Q(M^+) \cup Q(M^-)$ of $Q(M)$ where
         $M^+$ is the matroid whose parallelism classes are $P_1, \dots, P_{k-2}, P_{k-1} \cup P_k$
         and the matroid $M^-$ has parallelism classes $P_1 \cup \dots \cup P_{k-2}, P_{k-1}, P_k$.

         Since $M^-$ is isomorphic to $M_{\lambda(\alpha, k-1)}$,
         Proposition~\ref{cd-index-hyperplane-split-case-rank-two} implies
         $$
         \begin{aligned}
            \Psi (Q(M)) =& \Psi (Q(M^+)) + \Psi (Q(M_{\lambda(\alpha, k-1)}))
                           - \Psi (\Delta_{\mu(\alpha, k-2)_1} \times \Delta_{\mu(\alpha, k-2)_2}) \cdot \mathbf{c} \\
                         &- \sum_T \Psi (Q((M|_T)_{p(T)-2})) \cdot \mathbf{d} \cdot \Psi (\Delta_{n-|T|}),
         \end{aligned}
         $$
         where the sum is over all proper subsets $T$ of $[n]$ such that
         $M|_T$ has at least four parallelism classes, two of which are subsets of
         $P_{k-1}$ and $P_k$ respectively.
         Since
         $M^+$ has $k-1$ parallelism classes, the induction hypothesis implies the claim.

      Since $M|_S$ corresponds to the composition
      $$\beta_T := (|P_1 \cap S|, |P_2 \cap S|, \dots, |P_k \cap S|) < \alpha$$
      and there are
      $$\prod_{j=1}^k \binom{\alpha_j}{\beta_j}$$
      subsets corresponding to $\beta < \alpha$,
      we finish the proof.
   \end{proof}

   Purtill~\cite{Purtill} shows that the $\mathbf{c}\mathbf{d}$-index of the simplex $\Delta^n$ is
   the $(n+1)$-st Andr\'{e} polynomial.
   Using the formula for the $\mathbf{c}\mathbf{d}$-index of a product of two polytopes given
   by Ehrenborg and Readdy~\cite{EhrenborgReaddy}, one can calculate
   terms in the second and the third sums in Corollary~\ref{cd-index-of-rank-2-matroids}.
   We still don't have a simple interpretation for the $\mathbf{c}\mathbf{d}$-index for $Q(M_\alpha)$
   when $\alpha$ has three parts.
   The first few values of the $\mathbf{c}\mathbf{d}$-index for $Q(M_\alpha)$, where $\alpha$ has three parts,
   are displayed in Table~\ref{table-cd-index-for-indecomposable-matroids}.

   \begin{table}[!ht]
   \begin{center}
      \begin{tabular}{|c|l|}
         \hline
         $\alpha$    & $\Psi (Q(M_\alpha))$ \\
         \hline \hline
         $(1,1,1)$ & $\mathbf{c}^2 + \mathbf{d}$ \\
         \hline
         $(2,1,1)$ & $\mathbf{c}^3 + 3 \mathbf{c} \mathbf{d} + 3 \mathbf{d} \mathbf{c}$ \\
         \hline
         $(3,1,1)$ & $\mathbf{c}^4 + 4 \mathbf{c}^2 \mathbf{d} + 8 \mathbf{c}\mathbf{d}\mathbf{c} + 5 \mathbf{d} \mathbf{c}^2 + 7 \mathbf{d}^2$ \\
         $(2,2,1)$ & $\mathbf{c}^4 + 5 \mathbf{c}^2 \mathbf{d} + 10 \mathbf{c}\mathbf{d}\mathbf{c} +  6 \mathbf{d} \mathbf{c}^2 + 10 \mathbf{d}^2$ \\
         \hline
         $(4,1,1)$ & $\mathbf{c}^5 + 5 \mathbf{c}^3 \mathbf{d} + 13 \mathbf{c}^2 \mathbf{d} \mathbf{c} + 15 \mathbf{c} \mathbf{d} \mathbf{c}^2 + 20 \mathbf{c} \mathbf{d}^2 + 7 \mathbf{d} \mathbf{c}^3 + 18 \mathbf{d} \mathbf{c} \mathbf{d} + 22 \mathbf{d}^2 \mathbf{c}$ \\
         $(3,2,1)$ & $\mathbf{c}^5 + 6 \mathbf{c}^3 \mathbf{d} + 17 \mathbf{c}^2 \mathbf{d} \mathbf{c} + 20 \mathbf{c} \mathbf{d} \mathbf{c}^2 + 28 \mathbf{c} \mathbf{d}^2 + 9 \mathbf{d} \mathbf{c}^3 + 26 \mathbf{d} \mathbf{c} \mathbf{d} + 33 \mathbf{d}^2 \mathbf{c}$ \\
         $(2,2,2)$ & $\mathbf{c}^5 + 7 \mathbf{c}^3 \mathbf{d} + 21 \mathbf{c}^2 \mathbf{d} \mathbf{c} + 24 \mathbf{c} \mathbf{d} \mathbf{c}^2 + 36 \mathbf{c} \mathbf{d}^2 + 10 \mathbf{d} \mathbf{c}^3 + 34 \mathbf{d} \mathbf{c} \mathbf{d} + 42 \mathbf{d}^2 \mathbf{c}$ \\
         \hline
         $(5,1,1)$ & $\mathbf{c}^6 + 6 \mathbf{c}^4 \mathbf{d} + 19 \mathbf{c}^3 \mathbf{d} \mathbf{c} + 29 \mathbf{c}^2 \mathbf{d} \mathbf{c}^2 + 38 \mathbf{c}^2 \mathbf{d}^2 + 24 \mathbf{c} \mathbf{d} \mathbf{c}^3 + 60 \mathbf{c} \mathbf{d} \mathbf{c} \mathbf{d}$ \\
                   & \qquad $+ 72 \mathbf{c} \mathbf{d}^2 \mathbf{c} + 9 \mathbf{d} \mathbf{c}^4 + 33 \mathbf{d} \mathbf{c}^2 \mathbf{d} + 65 \mathbf{d} \mathbf{c} \mathbf{d} \mathbf{c} + 47 \mathbf{d}^2 \mathbf{c}^2 + 64 \mathbf{d}^3$ \\

         $(4,2,1)$ & $\mathbf{c}^6 + 7 \mathbf{c}^4 \mathbf{d} + 24 \mathbf{c}^3 \mathbf{d} \mathbf{c} + 39 \mathbf{c}^2 \mathbf{d} \mathbf{c}^2 + 52 \mathbf{c}^2 \mathbf{d}^2 + 33 \mathbf{c} \mathbf{d} \mathbf{c}^3 + 86 \mathbf{c} \mathbf{d} \mathbf{c} \mathbf{d}$ \\
                   & \qquad $+ 104 \mathbf{c} \mathbf{d}^2 \mathbf{c} + 12 \mathbf{d} \mathbf{c}^4 + 48 \mathbf{d} \mathbf{c}^2 \mathbf{d} + 98 \mathbf{d} \mathbf{c} \mathbf{d} \mathbf{c} + 72 \mathbf{d}^2 \mathbf{c}^2 + 100 \mathbf{d}^3$ \\
         $(3,3,1)$ & $\mathbf{c}^6 + 7 \mathbf{c}^4 \mathbf{d} + 25 \mathbf{c}^3 \mathbf{d} \mathbf{c} + 42 \mathbf{c}^2 \mathbf{d} \mathbf{c}^2 + 55 \mathbf{c}^2 \mathbf{d}^2 + 36 \mathbf{c} \mathbf{d} \mathbf{c}^3 + 93 \mathbf{c} \mathbf{d} \mathbf{c} \mathbf{d}$ \\
                   & \qquad $+ 114 \mathbf{c} \mathbf{d}^2 \mathbf{c} + 13 \mathbf{d} \mathbf{c}^4 + 52 \mathbf{d} \mathbf{c}^2 \mathbf{d} + 109 \mathbf{d} \mathbf{c} \mathbf{d} \mathbf{c} + 81 \mathbf{d}^2 \mathbf{c}^2 + 112 \mathbf{d}^3$ \\
         $(3,2,2)$ & $\mathbf{c}^6 + 8 \mathbf{c}^4 \mathbf{d} + 30 \mathbf{c}^3 \mathbf{d} \mathbf{c} + 51 \mathbf{c}^2 \mathbf{d} \mathbf{c}^2 + 69 \mathbf{c}^2 \mathbf{d}^2 + 42 \mathbf{c} \mathbf{d} \mathbf{c}^3 + 116 \mathbf{c} \mathbf{d} \mathbf{c} \mathbf{d}$ \\
                   & \qquad $+ 142 \mathbf{c} \mathbf{d}^2 \mathbf{c} + 14 \mathbf{d} \mathbf{c}^4 + 64 \mathbf{d} \mathbf{c}^2 \mathbf{d} + 136 \mathbf{d} \mathbf{c} \mathbf{d} \mathbf{c} + 98 \mathbf{d}^2 \mathbf{c}^2 + 142 \mathbf{d}^3$ \\

         \hline
      \end{tabular}
      \begin{caption}
         {$\mathbf{c}\mathbf{d}$-index for $Q(M_\alpha)$ for composition $\alpha$ with three parts}
         \label{table-cd-index-for-indecomposable-matroids}
      \end{caption}
   \end{center}
   \end{table}

\def$'${$'$}

redpajama_set_name: RedPajamaArXiv


text:
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For Asset Managers, the Right CRM Should Be More than a Customer Database: Does Your CRM Have What It Takes?
June 5, 2018 · AKA Enterprise Solutions · Comments Off on For Asset Managers, the Right CRM Should Be More than a Customer Database: Does Your CRM Have What It Takes?
SIX REASONS YOU NEED A CRM AND WHY MICROSOFT DYNAMICS IS THE RIGHT CHOICE FOR YOU!
November 29, 2016 · Brian Begley, enCloud9 Social/CRM · Comments Off on SIX REASONS YOU NEED A CRM AND WHY MICROSOFT DYNAMICS IS THE RIGHT CHOICE FOR YOU!


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WPS Office
WPS Wins 4-Star Reviews from LAPTOP Magazine and TechWorld
by Joel (KINGSOFT Office)
on Aug 30, 2016 at 16:25 UTC
Brand Representative for WPS Office
Next: Better Value, No Compromise
WPS Office Software, a leading office productivity suite for PC and mobile devices, has earned two four-star editorial reviews from the publications Laptop Magazine and Techworld.
In the WPS Office 2016 Review in Laptop Magazine, the editor praised the office productivity suite for its compatibility with MS Office, its customizable interface, and other unique features. In the review, Laptop noted that "WPS Office does an excellent job of fulfilling its promise to be as much like Microsoft Office as possible while costing less and providing a few more neat features," adding that "WPS Office has a narrow niche and fills it well." Among the "Pros" of the software, the review lists features such as tabbed documents, eye-protection mode, and a user interface that can be either static menus or modern Ribbon, concluding that "WPS Office is the most attractive Office 365 alternative, thanks to its customizable interface and other unique features." The full review is available at: http://www.laptopmag.com/reviews/software/wps-office.
Techworld's review highlights many advantages to using WPS Office's "light yet powerful office suite," including a "speedy installation process"; "real depth" with support for tables, charts, shapes, images, equations, and more; compatibility with SmartArt and WordArt; and an internal reader in the word processor application for opening PDFs. "A seriously impressive and attractive suite of tools," Techworld concludes in awarding its four-star review. The full review is available at: http://www.techworld.com/download/office-business/wps-office-2016-free-v10105652-3330664/.
Learn more at:
http://www.wps.com/press/four-star-editorial-reviews


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NEW EQUIPMENT: Lisa Larkin tries out some of the machines. Photo supplied. Fitness taken to the public PEOPLE WILL not have to join a gym to put on muscle in Lincoln -- the fitness machines have been brought outside to them. A row of machines installed near the Lincoln Events Centre rely on using body weight to provide resis- tance and are available for the public to use. It is the culmination of an idea that came up in 2009 from the Rotary Club of Lincoln. The club finally managed to get funding from the Southern Trust and Selwyn District Council, which enabled the purchase of the equip- ment. With the help of the Concrete Protection & Repair Company and their own manpower, they were put in in January and officially opened last week. Selwyn Mayor Kelvin Coe said at the opening that many people in the community wanted to be able to keep fit, but it was not always easy for them to get to a gym, so it was great to have the equipment in place for people to be able to access locally and for free . Lincoln Rotary president, Farhad Dastgheib, encouraged people to haveagoonit. Enjoy it and encourage your friends and neighbours to come along too -- it is more fun that way. Personal trainer Jacqui Restieaux will host free one-hour sessions, showing people how they can use the equipment to achieve their training objectives, tomorrow at 10.30am and 5.15pm, no booking needed. Primary school rolls ahead The school roll has already had a fillip since the decision, and could have new facilities with the repair bill from earthquake damage higher than replacement costs. THE REPRIEVE of Burnham School from closure will not affect plans for a new primary school in Rolleston, a Ministry of Education official told a Selwyn District Council meeting last week. The Canterbury education renewal programme director for the ministry, Coralanne Child, told the council the ministry was actively pursuing sites for a new primary and high school in Rolleston. Burnham School was propo- sed for closure, but an interim decision last month reversed that. The school roll has already had a fillip since the decision and could have new facilities, with the repair bill from earth- quake damage higher than replacement costs. Though the school and its roll capacity was retained, it would not change the timetable for a new school, Ms Child said. Burnham School would really serve the needs of the military camp, though there was still the choice for parents from Rolles- ton to send their children there. It would not be used to cater for roll growth in Rolleston, though primary schools there were under pressure. A third primary school is not planned until 2016 in Rolleston and a high school is not planned until 2017. There was still public consul- tation to be undertaken, particu- larly for the new high school, she said. People in the community had already expressed a desire to have a stand-alone high school in Rolleston, but the proposition of a Lincoln High School hub in the town was still an option. She said a new primary school for Lincoln was only medium priority, with Rolleston having the more pressing need. Councillor Malcolm Lyall raised concerns about the capacity of Prebbleton School to cater for sustained roll growth. He said that -- with more than 400 pupils -- it was now a city size school in a rural location. Deputy mayor Sarah Walters also had issues with new schools in Rolleston being too big, which could put parents off. Fire-hit to decide BROADFIELD RESIDENTS who were affected by a fire in the area in January will be able to decide how to allocate $10,000 from the Selwyn Road Fire Relief Fund, the Selwyn District Council decided at a meeting last week. Thirty-eight properties suffered damage as a result of the fire. The council decided Broadfield residents who had been affected by the blaze would be able to decide collectively how they wished to use the money. Nearly $20,000 was donated to the fund, established to provide relief for families soon after the fire -- with $5420 set aside for one family who lost their home in the blaze. Of the remaining money, $4101 will remain in the fund and be avail- able for individual applications from residents affected by the fire. The Broadfield community will be invited to attend a meeting in March and be asked to put forward ideas about how the other $10,000 could be used, Mayor Kelvin Coe said. If the $4101 available for indivi- dual applications from residents is not fully allocated, any leftover fund- ing will also go to affected residents to decide how they want to use it.


The hybrid Application HtmlHelper extension is a server-side wrapper for the hybrid Kendo UI Application widget.
It allows you to configure the hybrid Kendo UI Application from server-side code.
The Hybrid UI Application for ASP.NET MVC provides the necessary tools for building native-looking web-based mobile applications. When initialized, the mobile Application modifies the behavior of the hybrid Kendo UI widgets, so that they navigate between the mobile views when the user taps them.
Add an action method for the detail view.
Add the default Hybrid UI View for ASP.NET MVC. The mobile application expects that the immediate child of the application element is a MobileView.
Create a new Details ASP.NET MVC View file under the /Views/Home/ folder.
Initialize the Application inside the Master/Layout page and enable the server navigation.
Notice that the view is returned as partial.
Initialize the mobile application inside the Master/Layout page.
You can reference a hybrid Application instance by using the code from the example below. Once a reference is established, use the hybrid Application API to control its behavior.
//Notice that the casing is important.


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Nielsen: Symphony No. 4
Carl Nielsen's Symphony No. 4, also known as The Inextinguishable, is among the most dramatic symphonies Nielsen composed. Composed in 1916, the symphony was composed against the backdrop of the First World War. It features a battle between two sets of timpani. In this symphony Nielsen set out to capture in music the idea of an inextinguishable life force that runs through all creation.
Half a bar of Carl Nielsen and you know it's him; a characteristic of great composers down the ages, to be sure, but one peculiarly concentrated around nationalist composers of the late 19th and early 20th centuries. Nationalist, not necessarily through any political affiliation but because they sought to renew both the standard musical forms and genres, and their own country's culture, with bold and fresh attempts to marry the two. So successful was Nielsen at this (like Sibelius in Finland), indeed, that for decades afterwards he seemed to cast a shadow over those who came after.
Composer: Carl Nielsen
Artists: Janáček Philharmonic Orchestra, Theodore Kuchar (conductor)
Stream this album: https://brilliant-classics.lnk.to/NielsenSymphonyNo4ID
Symphony No. 4, Op. 29 "The Inextinguishable":
00:00 I. Allegro
11:08 II. Poco allegretto
15:40 III. Poco adagio quasi andante
25:14 IV. Allegro
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Beautiful Classical Music: https://brilliant-classics.lnk.to/BeautifulClassicalMelodiesID
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Thank you for watching this video by Brilliant Classics, we hope you enjoyed it! Don't forget to share it and subscribe to our YouTube channel: https://brilliant-classics.lnk.to/YouTubeID
And visit our channel for the best classical music from the greatest composers like: Bach, Satie, Mozart, Tchaikovsky, Beethoven, Chopin, Haydn, Ravel, Debussy, Verdi, Vivaldi, Handel, Brahms, Liszt, Schubert, Mendelssohn, Rachmaninoff, Wagner, Strauss, Handel, Dvorak and many more! We upload complete albums, music for relaxing, working, studying, meditating, concentrating, instrumental music, opera, violin, classical piano music, sonatas and more!
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My sister's keeper is an automatic monthly donation that is made to help young women in need. Your funds will be used to help women receive job-skills training, business and money management basics and self assertiveness skills education. As well as support women survivors of war and conflicts as they try to progress through life and try to rebuild.


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Extracellular nucleases, which occur in most body fluids of vertebrates, predominantly belong to a family of Ca2+ and Mg2+ (Mn2+)-dependent endonucleases, which was named by its first identified member, Deoxyribonuclease I (DNase I / DNASE1). So far three further members beside DNASE1 were found: DNASE1-like 1-3 (DNASE1L1-3).
Our group characterizes the physiological functions of these nucleases by employing gene-knockout mouse models. Especially the functions of DNASE1, which is highly expressed by glands lining the gastro- and urogenital tract and DNASE1L3 (DNase γ), which is expressed by macrophages and activated B-cells, are of our major interest. Both nucleases occur in serum of mammals and our experiments hint to their contribution in the suppression of anti-nuclear autoimmunity, which is characteristic for a human disease called Systemic Lupus Erythematosus (SLE). Both enzymes display a complementary substrate specificity – DNASE1 efficiently degrades protein-free DNA whereas DNASE1L3 preferentially hydrolyzes protein-complexed DNA (chromatin) – and appear to cooperate in the destruction of exogenous (microorganisms) and endogenous (primary or secondary necrotic cells) DNA. Impairment of these processes might lead to a prolonged and chronic exposure of the immune system to DNA-antigens leading to an increased probability for the breakdown of self-tolerance towards nuclear antigens. Indeed, SLE is regarded as a multigenetic disease caused by defects which impair the removal of dying cells and cell debris in the context of defects leading to a deregulated and hypersensitive immune system.
In summary, we intend to elucidate the role of extracellular nucleases in the removal of nuclear cell debris released under physiological and pathological conditions (necrosis) and the consequences resulting from an impairment of these processes.
Dr. Markus Napirei
Abteilung für Anatomie und Molekulare Embryologie
Detection of anti-nuclear (chromatin) autoantibodies (ANA) in the serum of an ANA-positive mouse by indirect immunofluorescence using NIH3T3 fibroblasts.


We can tell stories. We can facilitate learning. But how do we use stories to facilitate learning? This session is designed to fill that void. We'll start from the beginning – mining your own life and experiences for relevant stories to tell, regardless of the topic you're teaching. Then we'll discuss how to pull from other disciplines, like literature, film, and writing, to help you select, structure and shape stories that can facilitate connections between you, your learners, and the material. You'll also learn about the characteristics of stories that are most useful for facilitating learning. Finally, we'll address how you can help learners find and shape their own stories. You'll leave this session with templates and strategies that will help you apply what you've learned about finding and shaping back on the job.
List strategies for mining your own life and experiences for relevant stories.
Explain how to structure stories so that the intent is clear, the listeners are captivated, and the connection to the content is evident.
Hadiya Nuriddin is the owner of Focus Learning Solutions.
She has nearly 20 years' experience in all levels of learning and development.She specializes in instructional design and development for both corporate and academic environments. She has extensive experience in designing, developing, and delivering both technical and professional development courses. She also designs and develops eLearning courses using Articulate Studio, Articulate Storyline, and Captivate. Hadiya has an MEd in Curriculum Design, an MA in Writing, and a BA in English. She holds the Certified Professional in Learning and Performance (CPLP), the Professional in Human Resources (PHR), and is MBTI certified. She's also earned several technical certifications throughout her career, including Microsoft Certified Professional and Microsoft Office Master Instructor.


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Haunted Mansion Holiday Overview | Disneyland Attractions
Posted on December 6, 2022 January 24, 2023 by Bobby Asen | Leave A Comment
Attraction OverviewDisney AttractionsDisney Holidays
The classic Haunted Mansion is in brighter spirits with The Haunted Mansion Holiday overlay at Disneyland.
Haunted Mansion Details
Disneyland Resort Park
Location in Park
Any Height
Age Interests
Kids, Teens, Adults
Thrill Type
Slow ride, dark, and scary
Open to Park Close
Must Transfer from Wheelchair/ECV
Attraction Overview
Jack Skellington has taken over decorating The Haunted Mansion for the Christmas season as he prepares to take over for Santa Claus. The ride's entry has Jack Skellington sitting outside and a giant countdown sign on the house, letting you know it's officially the holiday season. All the Haunted Mansion attraction locations are changed with overlays that you will only be able to see the first week of September to January 8th each year.
Even the queue line is decorated for Christmas! Photo by Bobby Asen
Once you enter The Haunted Mansion Holiday, you will see that the stretching portrait room has different artwork from the beloved film The Nightmare Before Christmas. The ceiling will burst with an appearance from Jack Skellington and his ghost dog Zero. From there, you will walk into the hallway of artwork featuring paintings of Sally, Jack Skellington, Santa Claus, and Zero that come to life, and you'll find your doom buggy waiting for you to be loaded on the slow-moving attraction.
Haunted Mansion Holiday Entrance. Photo by Bobby Asen
When you get to your doom buggy, you'll notice the citizens of Halloween Town have begun singing their Christmas carols, and you are off to start your journey. You'll see that the regular Haunted Mansion animatronics have been replaced with ones of characters from The Nightmare Before Christmas. One of the significant changes is the famous ballroom scene where instead of just waltzing ghosts, you see that they are celebrating the holidays with a ginormous scented gingerbread house. Each year the design of the gingerbread house changes.
Jack and Sally. Photos by Bobby Asen
Finally you and your doom buggy make it to the graveyard you'll be greeted by Jack Skellington, Sally, Zero, and even Oogie Boogie. Before exiting, Oogie Boogie gives you a ghost present you can only see in the Haunted Mansion mirror from his wheel of gifts. Keep your eyes out for Lock, Shock, and Barrel before you wave goodbye to Halloweentown.
Photo by Bobby Asen
Location & Wait Times
You can find the Haunted Mansion between Pirates of The Caribbean and the beginning of Critter Country. The Haunted Mansion is right in the heart of New Orleans Square, located by The Disneyland Railroad. With this attraction having Doom Buggies constantly running, the wait time is typically moderate.
You may spot Jack and Sally in the are for meet and greets! Photo by Bobby Asen
Queue Experience
The queue for The Haunted Mansion is through the grounds of the home itself outside. You'll find yourself getting a good laugh from many funny engraved tombstones, which give you a preview of what's to come inside. The second half of your wait for this attraction is indoors, including the stretching chamber in the hallway of moving portraits.
Bobby Asen
Bobby has a strong passion for Walt Disney's stories and the Disney Parks. From taking vacations to the parks, to being an Annual Passholder, to working as a cast member for seven years at Magic Kingdom, his passion for Disney continues to grow. After visiting every Disney Park in the world, Bobby knew he wanted to share his experiences and his love for Disney in his writing.
Win Tickets to Night Of A Million Lights Hosted By Give Kids The World Village
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A bison near a warm stream from "Silence and Solitude: Yellowstone's Winter Wilderness" by Tom Murphy.
Written across fresh snow, a line of mouse track collided with an imprint of hawk feather to record a moment when winter stripped nature and story to their basic elements, survival and death. Yet even in this minimalist environment is beautiful abundance that is illustrated in "Silence and Solitude: Yellowstone's Winter Wilderness" by Tom Murphy.
The exquisite excess in the act of survival captivates the imagination of Murphy, a naturalist, photographer and author who has spent more than four decades recording the harsh beauty and tenacious spirit of Yellowstone National Park, one of the few remaining wild places in the United States.
Murphy's curiosity and gratitude for nature are exposed and expressed in his photographs and writing, which has appeared in National Geographic, Newsweek, Audubon and the Corporation for Public Broadcasting. His dedication is evident in the 80 to 100 days he spends within the national park's borders each year, and the five photography books published.
Art and science both originate from a need to understand the world we live in, said Murphy, who draws on both when he photographs Yellowstone in winter, a place of rare beauty and near impossible discoveries now imperiled by climate change.
Anticipated climate change to the Yellowstone ecosystem. The NASA Earth Observatory map by Joshua Stevens.
Reorienting ourselves in place and import within the larger needs of the world was an underlying argument of Thomas Berry (1914 – 2009), a pioneer in the field of religion and ecology who argued that we could not rescue ourselves from ecological crisis without tending to the Earth.
Yellowstone, our oldest national park, has long served to reconnect humans with nature. In 1883, The New York Times described this land of fire and ice as an "almost mystical wonderland." More than a century later, that reverence appears in Murphy's photographs.
Murphy developed an intimacy with wide-open places during his childhood on a cattle ranch in western South Dakota. There he developed a deep love for spacious land, as well as a determination to not spend his lifetime chasing cows, according to his website autobiography. After earning a degree in anthropology, he spent many years combining photography with anthropological fieldwork.
Cahill might think his friend foolhardy to brave the elements in 50-cent socks, but he expressed only admiration for Murphy's ability to tell a story with an image.
In a place where temperatures can dip 50-degrees below freezing, Murphy wrote that he continues to use tripods 99 percent of the time, although they are cold, pinch his fingers, and are heavy and awkward on his skinny shoulders. When traveling through steam in thermal areas, he learned to keep his equipment in plastic bags to reduce damage from the condensation that occurred when going from cold to warm areas. Despite the extreme contrasts between bright snow and dark objects, he said he avoids the additional weight of flash, filter and reflector (which he admits are excellent tools), and chooses to bracket exposures, taking a series of the same image at different camera settings.
Like the land and the animals he photographs, Murphy endures long periods of exertion followed by hours of still observation. His dedication rewards us with a rare and insightful perspective on the tenacious, yet tenuous nature of life.
Thomas Berry on our shared future.
"Silence and Solitude: Yellowstone's Winter Wilderness" (public library) by Murphy, forward by Tim Cahill.
"Your children's Yellowstone will be radically different," The New York Times.
Yellowstone Science: special report on the ecological implications of climate change on the park's ecosystem, published by the National Forest Service, 2015.
Page 28: Scientist Michael Tercek and his daughter.
"Natural beauty at risk: Preparing for change in western parks," by NASA Earth Observatory blog.


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Master of Financial Engineering Program Makes Top 10 List
May 21, 2012 | by Haas News
The Berkeley Master of Financial Engineering (MFE) Program was named among the top 10 quantitative finance programs in the nation for a second time by Advanced Trading magazine.
Berkeley MFE Program also made the top 10 list in 2008, the first year Advanced Trading published its ranking. Advanced Trading is a magazine and Web site that covers stock trading trends and technology strategies. Its list does not rank the elite programs against each other.
A profile of the Berkeley MFE Program notes it is among the few financial engineering programs offered by a business school, teaching students about computational finance within the context of business and economics.
The schools were selected by five judges: Scott R. Burrill, a partner and managing director at Rosenblatt Securities; David Leinweber, founder of two tech firms and a fellow in finance at Haas, who was not allowed to vote for Berkeley-Haas; Michael Levas, founder, senior managing principal, and director of trading for Olympian Capital Management; Unson Allen, VP and global head of recruiting for Knight Capital Group; and Lou Ricci, a headhunter for the Hagan-Ricci Group.
Updated 6.18.2012: The MFE program was also featured in the Financial Times Financial Training Special Report on June 18, 2012.
Read the Advanced Trading Ranking
Posted in Rankings


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Springfield principal prepared to lead
Brittney L. Jackson
Springfield Middle School Principal Grant Bell is looking to the future.
The veteran educator was one of 14 people to recently complete the Superintendents Academy hosted by Tennessee Schools Boards Association (TSBA) and Tennessee Organization of School Superintendents (TOSS), according to a news release issued by TSBA.
"What I really loved about the program is that it was run by superintendents and retired superintendents," Bell said in an interview with The Times on Wednesday.
The 16-day program is not required to be a school superintendent, but it offers training in school law, finance and budget, facilities, school board relations, working with the media and personnel decisions, he said.
Bell decided to take the course to learn about being a superintendent in case the future leads him in that direction, he said.
"I absolutely love leading Springfield Middle," Bell said. "There will be a time in the future where I would want to lead Robertson County Schools."
He added that he loves Robertson County and does not plan to leave the area.
TSBA does a number of superintendent searches in Tennessee and the organization has faced a shortage of qualified candidates in recent years, the release said. The academy was created in 2007 as a solution to the problem, the release said, noting that it provides educators with the training they need to manage a school system.
Bell was the only educator from Robertson County to complete the most recent session. Each of the 14 candidates received certification from the Academy's independent review board made up of educators and business leaders, the release said. They were then evaluated on their work product, class participation and a final oral presentation and interview, it added.
"We are very pleased with the quality of the candidates who completed the course this year," TSBA Executive Director Tammy Grissom said in the release. "A dynamic, innovative and highly-skilled superintendent is crucial to the success of any school system, and based on what we have seen over the past year, we believe these 14 individuals have the tools to meet that challenge."
Reach Brittney Jackson at 615-384-3567 or on Twitter @britt_lacole.


Whether you are looking for enhanced well-being or are weighed down by troubles and concerns and want real solutions for them, we provide a wide range of services specifically tailored to suit your needs to do just that.
This Personality Test is your first step. Rating you against ten different personality traits, it will accurately show you both your strong points - so you can take advantage of them - and your problem areas - those things that are blocking your true potentials and happiness in life.
Fill it out here or contact us to send it to you. We will follow this up with your free Initial Consultation.
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