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\documentclass[finalversion]{FPSAC2018}
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\theoremstyle{plain}%{theorem}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
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\newcommand{\rrem}[1]{{\bf \color{red} Ron says:} {\it\color{red} #1}}
\newcommand{\drem}[1]{{\bf \color{blue} Danny says:} {\it\color{blue} #1}}
\newcommand{\erem}[1]{{\bf \color{green} Eran says:} {\it\color{green} #1}}



%%%%%%%%%%%%%%%%%%%%%% our macros %%%%%%%%%%%%%%%%%%%%%
\newcommand{\cyc}{\mathrm{Cyc}}
\newcommand{\lk}{\mathrm{lk}}
\newcommand{\anst}{\mathrm{ast}}
\newcommand{\aff}{\mathrm{Aff}}
\newcommand{\conv}{\mathrm{conv}}
\newcommand{\vertices}{\mathrm{Vert}}
\newcommand{\st}{\mathrm{st}}
\newcommand{\stc}{\mathsf{SC}}
\newcommand{\nc}{\mathsf{NC}}
\newcommand{\ncp}{\mathsf{NCP}}
\newcommand{\cC}{\mathcal{C}}
\newcommand{\cL}{\mathcal{L}}
\newcommand{\lex}{\mathrm{Lex}}

\newcommand{\set}[2]{\left\{ #1\,\middle|\, #2\right\}}
\newcommand{\sizeset}[2]{\left|\set{#1}{#2}\right|}
\newcommand{\gc}{g^c}
\newcommand{\hc}{h^c}
\newcommand{\hsc}{h^{sc}}
\newcommand{\gsc}{g^{sc}}
%\newcommand{\dhalf}[1]{\left\lfloor\frac{#1}{2}\right\rfloor}
\newcommand{\dhalf}[1]{\left\lfloor #1/2\right\rfloor}
\newcommand{\duhalf}{\left\lceil\frac{d}{2}\right\rceil}

\newcommand{\R}{\mathbb{R}}

\newcommand{\mw}{{\sf MW}}

\def\mchoose#1#2{\ensuremath{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{#1}{#2}\right)\kern-.3em\right)}}

\newcommand{\vfig}[2]{{\raisebox{.1em}{$#1$}\left/\raisebox{-.1em}{$#2$}\right.}}

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%% define your title in the usual way
\title{On the cone of $f$-vectors of cubical polytopes}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
\author{Ron
M.~Adin\thanks{\href{mailto:radin@math.biu.ac.il}{radin@math.biu.ac.il}}\addressmark{1},
Daniel
Kalmanovich\thanks{\href{mailto:daniel.kalmanovich@gmail.com}{daniel.kalmanovich@gmail.com}}\addressmark{2},
and Eran
Nevo\thanks{\href{mailto:nevo.eran@gmail.com}{nevo.eran@gmail.com}}\addressmark{2}}

%% then use \addressmark to match authors to institutions here
\address{
\addressmark{1}Department of Mathematics, Bar-Ilan University, Ramat-Gan, Israel \\ 
\addressmark{2}Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Jerusalem, Israel}%\\
%\addressmark{3}Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Jerusalem, Israel}

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\received{November 14, 2017}

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\revised{April 1, 2018}

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\abstract{%
	What is the minimal closed cone containing all $f$-vectors of cubical\break $d$-polytopes?
	We construct cubical polytopes showing that this cone, expressed in the cubical $g$-vector coordinates, contains the nonnegative $g$-orthant,
	thus verifying one direction of the Cubical Generalized Lower Bound Conjecture of Babson, Billera and Chan. Our polytopes also show that a natural cubical analogue of the simplicial Generalized Lower Bound Theorem does not hold.
}

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%\resume{\lipsum[2]}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{cubical polytope, cubical $g$-vector}

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\addbibresource{85-Adin-Kalmanovich-Nevo.bib}

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\begin{document}

\maketitle
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\setcounter{footnote}{0}

\section{Introduction}
Understanding the possible face numbers of polytopes, and of subfamilies of interest, is a fundamental question, dealt with since antiquity.
The celebrated $g$-theorem, conjectured by McMullen~\cite{McMu71} and proved by Stanley~\cite{Stan80} (necessity) and by Billera and Lee~\cite{BillL81} (sufficiency), characterizes the $f$-vectors of simplicial polytopes.
Here we consider $f$-vectors of cubical polytopes;
a $d$-polytope $Q$ is \emph{cubical} if all its facets are combinatorially isomorphic to the $(d-1)$-cube.
Adin~\cite{Adin96} proved analogues of the Dehn--Sommerville relations for cubical polytopes,
thus encoding the $f$-vector of $Q$ by its (long) cubical $g$-vector
\[g^c(Q)=\left(g_1^c(Q), g_2^c(Q), \dots, g_{\dhalf{d}}^c(Q)\right)\]
(with the constant $g_0^c(Q)=2^{d-1}$ omitted).
The construction of neighborly cubical\break $d$-polytopes by Joswig and Ziegler~\cite{JoswZ00}, where the number of vertices varies, shows that the linear span of the vectors $g^c(Q)$, over all cubical $d$-polytopes, is the entire vector space $\mathbb{R}^{\lfloor d/2\rfloor}$.
Adin~\cite[Question 2]{Adin96} asked whether $g^c(Q)$ is always in the nonnegative orthant, and Babson, Billera and Chan~\cite[Conjecture 5.2]{BabsBC97} conjectured further that the minimal closed cone $\mathcal{C}_d$ containing all the vectors $g^c(Q)$ corresponding to cubical $d$-polytopes is exactly this nonnegative orthant $\mathcal{A}_d$.

Denote by $e_i$ the $i$-th unit vector in $\mathbb{R}^{\lfloor d/2\rfloor}$. Stacked cubical polytopes show that the ray spanned by $e_1$ is in $\mathcal{C}_d$, and neighborly cubical polytopes show that the ray spanned by $e_{\lfloor d/2\rfloor}$ is in $\mathcal{C}_d$.
Our main result is that all the rays spanned by the vectors $e_i$ are in $\mathcal{C}_d$:

\begin{theorem}\label{thm:cones}
	$\mathcal{A}_d \subseteq \mathcal{C}_d$.
\end{theorem}

The conjecture of Adin and Babson--Billera--Chan is
\begin{conjecture}\label{conj:cones}
	$\mathcal{A}_d = \mathcal{C}_d$.
\end{conjecture}

An analogue of Theorem~\ref{thm:cones} was previously known for the much wider class of PL cubical spheres~\cite[Theorem 5.7]{BabsBC97}.
Also, Conjecture~\ref{conj:cones} holds for $d\leq 5$, by combining the constructions above with Steve Klee's result \cite[Prop.3.7]{Klee11} asserting that $g^c_k(Q)\geq 0$ for any cubical polytope $Q$ of dimension $2k+1$.

Sanyal and Ziegler~\cite{SanyZ10} showed how to construct, from any simplicial $(d-2)$-polytope $P$ on $n-1$ vertices and a total order $v_1<v_2<\ldots <v_{n-1}$ on its vertices, a cubical\break $d$-polytope $Q=Q(P,<)$ on $2^n$ vertices; it is the projection of a deformed $n$-cube in $\mathbb{R}^n$ onto the last $d$ coordinates.
Further, they showed that if $P$ is $k$-neighborly then
the\break $k$-skeleton of $Q$ is isomorphic to the $k$-skeleton of the $n$-cube.
We apply their construction to the case where $P_n$ is the $k$-neighborly $k$-stacked $(d-2)$-polytope on $n-1$ vertices constructed by McMullen and Walkup~\cite{McMuW71}, with $1\leq k\leq \lfloor\frac{d-2}{2}\rfloor$, and with a suitable total order $<$. Analyzing the cubical $g$-vectors of the resulting polytopes $Q(k,d,n)=Q(P_n,<)$, as $n$ tends to infinity, gives~\autoref{thm:cones}. See Theorem~\ref{t:gc_Q} and Corollary~\ref{t:asymptotics} for the exact values and asymptotic behavior of the cubical $g$-vectors.

The generalized lower bound theorem for simplicial polytopes (GLBT), conjectured by McMullen and Walkup~\cite{McMuW71} and proved by Murai and Nevo~\cite{MuraN13}, asserts that for\break $1\le k<\dhalf{d}$, a simplicial $d$-polytope $P$ is $k$-stacked if and only if $g_{k+1}(P)=0$.
The polytopes $Q(k,d,n)$ demonstrate that the natural cubical analogue of the GLBT is false:
\begin{theorem}
	For any $k\ge 1$ and $n\ge d\ge 2k+2$, we have $\gc_{k+2}(Q(k,d,n))=0$, and $Q(k,d,n)$ is not cubical $(k+1)$-stacked.
\end{theorem}

This is an extended abstract. For the complete paper, see~\cite{AKNarxiv}.

%The paper is organized as follows. Basic definitions and notation are are given in~\autoref{sec:preliminaries}. In~\autoref{sec:MW} we define our variant of the McMullen--Walkup polytopes. A sketch of the Sanyal--Ziegler construction is given in~\autoref{sec:SZ}. In~\autoref{sec:Q} we construct the polytopes $Q(k,d,n)=Q_n$ mentioned above and analyze their cubical $g$-vector. In~\autoref{sec:stack} we show that a cubical GLBT fails, by showing that $Q_n$ is not $(k+1)$-stacked. The final~\autoref{sec:concluding_remarks} is devoted to concluding remarks.


\acknowledgments{
The authors thank Michael Joswig and Isabella Novik for helpful comments.
The research of Adin was supported by an MIT-Israel MISTI grant. He also thanks the Israel Institute for Advanced Studies, Jerusalem, for its hospitality during part of the work on this paper.
The research of Kalmanovich and Nevo was partially supported by Israel Science Foundation grant ISF-1695/15 and by grant 2528/16 of the ISF-NRF Singapore joint research program.
This work was also partially supported by the National Science Foundation
under Grant No.\ DMS-1440140, while Nevo was in residence at the
Mathematical Sciences Research Institute in Berkeley, California, during
the Fall 2017 semester.}











\section{Preliminaries}\label{sec:preliminaries}

The purpose of this section is mainly to set the notation that we will use throughout the paper. For undefined terminology we refer the reader to~\cite{Zieg95}. A {\bf $d$-dimensional polytope} $P$ is the convex hull of a finite set of points in $\R^d$ which affinely span $\R^d$. A (proper) {\bf face} $\sigma$ of $P$ is the intersection of $P$ with one of its supporting hyperplanes, the {\bf dimension} $\dim\sigma$ of $\sigma$ is then the dimension of the affine span of that intersection. The faces of dimensions $0$, $1$, and $d-1$ are called {\bf vertices}, {\bf edges}, and {\bf facets}, respectively. The empty set $\emptyset$ and the polytope $P$ itself are called {\bf trivial faces} and have dimensions $-1$ and $d$, respectively. We will abbreviate and write $d$-polytope and $i$-face to denote dimension.
%The number of $k$-faces of $P$ is denoted by $f_k(P)$, and the {\bf $f$-vector} of $P$ is $f(P)=\left(f_0(P),f_1(P),\dots ,f_{d-1}(P)\right)$. The {\bf $f$-polynomial} of $P$ is defined by
%\[
%f(P,t)=\sum_{i=0}^d f_{i-1}(P)t^i,
%\]
%where $f_{-1}(P) = 1$.
We denote by $\vertices(P)$ the set of vertices of $P$, and for a vertex
$v\in\vertices(P)$, we denote by $\vfig{P}{v}$ the {\bf vertex figure} of
$P$ at $v$, that is, $\vfig{P}{v}$ is a $(d-1)$-polytope obtained as the
intersection of $P$ with a hyperplane which strictly separates $v$ from
$\vertices(P)\setminus\{v\}$; the face lattice of $\vfig{P}{v}$ does not
depend on the separating hyperplane chosen.

A {\bf polytopal complex} $K$ is a finite collection of polytopes in $\R^d$ such that
\begin{enumerate}[(i)]
	\item the empty polytope is in $K$,
	\item if $P\in K$ then all faces of $P$ are also in $K$,
	\item if $P,Q\in K$ then $P\cap Q$ is a face of both $P$ and $Q$.
\end{enumerate}
The {\bf dimension} $\dim K$ of $K$ is the maximum of $\dim P$ over all $P\in K$; we say that $K$ is a $\dim K$-complex. %to denote its dimension,
The elements in $K$ are called {\bf faces}; the faces of dimension $\dim K$ are called {\bf facets}.
For $F\in K$ we define the (open) {\bf star} of $F$ and the {\bf antistar} of $F$, respectively, by
\begin{align*}
\st_F(K)&=\set{G\in K}{F\subseteq G},\\
\anst_F(K)&=\set{G\in K}{F\nsubseteq G}.
\end{align*}
The number of $i$-faces in $K$ is denoted by $f_i(K)$, and the {\bf $f$-vector} of $K$ is $f(K)=\left(f_0(K),f_1(K),\dots ,f_{\dim K}(K)\right)$. The {\bf $f$-polynomial} of $K$ is defined by
\[
f(K,t)=\sum_{i=0}^{\dim K+1} f_{i-1}(K)t^i,
\]
where $f_{-1}(K) = 1$.

For a polytope $P$ we denote by $\langle P\rangle$ the complex of all faces of $P$. The {\bf boundary complex} $\partial P$ is the complex formed by all the proper faces of $P$, that is $\partial P=\langle P\rangle\setminus \{P\}$. We also define the $f$-vector and $f$-polynomial of $P$ by $f(P) = f(\partial P)$ and $f(P,t) = f(\partial P,t)$.
We use $\lk_v(P)$ to denote the boundary complex $\partial \left(\vfig{P}{v}\right)$ of the vertex figure of $P$ at $v$.


\subsection{Simplicial complexes and polytopes}

A {\bf simplicial complex} is a polytopal complex in which all polytopes are simplices. Let $K$ be a simplicial $(D-1)$-complex; the {\bf $h$-polynomial} of $K$ is defined by
\begin{equation*}
\begin{aligned}
h(K,t)
&=h_0(K)+h_1(K)t+\dots +h_D(K)t^D \\
&:=(1-t)^D\cdot f\left(K,\frac{t}{1-t}\right),
\end{aligned}
\end{equation*}
and the {\bf $h$-vector} of $K$ is $(h_0(K),\dots ,h_D(K))$. If
$K=\partial P$ for a simplicial $D$-polytope $P$ then the {\bf Dehn--Sommerville relations} assert that $h_i(K)=h_{D-i}(K)$ for any $0\le i\le D$.
The corresponding {\bf $g$-vector} $\left(g_0(K),\dots ,g_{\dhalf{D}}(K)\right)$ of $K$ is then defined by
\begin{equation*}
\begin{aligned}
g_0(K)&= h_0(K) = 1,\\
g_i(K)&=h_i(K)-h_{i-1}(K),\quad\text{for all}\quad 1\leq i\leq\dhalf{D}.
\end{aligned}
\end{equation*}

For two simplicial complexes $K$ and $L$ we define the {\bf join} $K\ast L$ to be the simplicial complex whose simplices are the disjoint unions of simplices of $K$ and simplices of $L$.

A polytope is {\bf simplicial} if each of its proper faces is a simplex.
For a simplicial polytope $P$ we write $h(P,t)$ to mean $h(\partial P,t)$, and similarly $h_i(P):=h_i(\partial P)$ and $g_i(P):=g_i(\partial P)$.

A simplicial $d$-polytope $P$ is called {\bf $k$-stacked} if $P$ has a triangulation in which there are no interior faces of dimension less than $d-k$. A simplicial polytope $P$ is called\break {\bf $k$-neighborly} if each subset of at most $k$ vertices forms the vertex set of a face of $P$. We denote by $C(d,n)$ the {\bf cyclic $d$-polytope with $n$ vertices}:
$$C(d,n) := \conv\left\{x(t_1), x(t_2)\ldots, x(t_n)\right\},$$
where $t_1 < t_2 <\dots < t_n$ and $x(t) := \left(t,t^2,\dots ,t^d\right)$ is the moment curve in $\R^d$.


\subsection{Cubical complexes and polytopes}

A {\bf cubical complex} is a polytopal complex in which all polytopes are combinatorially isomorphic to cubes. Let $Q$ be a cubical $(d-1)$-complex, the {\bf short cubical $h$-polynomial} is defined by
\[\hsc(Q,t)=
\sum_{i=0}^{d-1} \hsc_i(Q) t^i
= \sum_{j=0}^{d-1} f_j(Q) (2t)^j (1-t)^{d-1-j}.
\]
When $Q$ is clear from the context, we may sometimes drop it from the notation, as we do in the following few definitions.
%The {\bf short cubical $h$-vector} $(\hsc_0,\hsc_1,\dots ,\hsc_{d-1})$ satisfies
%\begin{equation*}
%\hsc_i=\sum_{j=0}^i\binom{d-1-j}{d-1-i}(-1)^{i-j}2^jf_j,\quad\text{for}\quad 0\leq i\leq d-1.
%\end{equation*}
The {\bf (long) cubical $h$-vector} $(\hc_0,\hc_1,\dots ,\hc_d)$ is defined by
\begin{align*}
\hc_0&=2^{d-1},\\
\hsc_i&=\hc_i+\hc_{i+1},\quad\text{for}\quad 0\leq i\leq d-1,
\end{align*}
and the corresponding ({\bf short} and {\bf long}) {\bf cubical $g$-vector}s are defined, as in the simplicial case, by
\begin{align*}
\gsc_0=\hsc_0=f_0,& &  \gsc_i&=\hsc_i -\hsc_{i-1}\quad\text{for } 1\leq i\leq\dhalf{(d-1)}; \\
\gc_0=\hc_0=2^{d-1},& &  \gc_i&=\hc_i -\hc_{i-1}\quad\text{for } 1\leq i\leq\dhalf{d}.
\end{align*}

A polytope is {\bf cubical} if each of its proper faces is combinatorially a cube.
Adin~\cite{Adin96} showed that any cubical $d$-polytope $Q$ satisfies an
analogue of the Dehn--Sommerville relations: $\hc_i(Q)=\hc_{d-i}(Q)$ for all $0\le i\le d$.

In analogy with the simplicial case, \cite{BabsBC97} defined cubical neighborliness and cubical stackedness: a cubical $d$-polytope is {\bf $k$-neighborly} if it has the $k$-skeleton of a cube (of some dimension);
it is {\bf $k$-stacked} if it has a cubical subdivision with no interior faces of dimension less than $d-k$.

Each vertex figure in a cubical $d$-polytope $P$ is a simplicial $(d-1)$-polytope;  we finish this section with the relation known as {\bf Hetyei's observation}:
\begin{equation}\label{eq:Hetyei}
\hsc(P,t):=\hsc(\partial P,t)=\sum_{v\in \vertices(P)}h(\lk_v(P),t).
\end{equation}
%Note that Adin's linear relations on $\hc(P)$ easily follow from the
It shows that the cubical Dehn--Sommerville relations follow from the simplicial ones.


\section{The McMullen--Walkup polytopes}\label{sec:MW}

In section 3 of~\cite{McMuW71}, McMullen and Walkup describe the construction of $k$-neighborly $k$-stacked simplicial $D$-polytopes with $N$ vertices, for any set of parameters satisfying $2\leq 2k\leq D < N$. Their construction takes a $k$-neighborly $2k$-polytope $C$ with $N-D+2k$ vertices (e.g. the cyclic $2k$-polytope with $N-D+2k$ vertices), and a $(D-2k)$-simplex $T$, both lying in $\R^D$ in such a way that the relative interior of $T$ intersects the affine hull $\aff (C)$ in a vertex $x$ of $C$. Then the convex hull $\conv (C\cup T)$ is the desired polytope. We define a slightly more general construction.
\begin{definition}\label{def:MW}
	Let $2 \leq K \leq D < N$. Let $C=C(K,N-D+K)$ be the cyclic\break $K$-polytope with $N-D+K$ vertices, and let $T$ be a $(D-K)$-simplex, both lying in $\R^D$ in such a way that the relative interior of $T$ intersects $\aff (C)$ in a vertex $x$ of $C$. The polytope $\conv({C\cup T})$ is a $D$-dimensional simplicial polytope with $N$ vertices, denoted $\mw(K,D,N;x)$.
\end{definition}

%The faces of $\mw(K,D,N;x)$ are of two types:
%\begin{enumerate}[(a)]
%	\item the convex hull of $T$ and a face of $C$ that contains $x$,
%	\item the convex hull of a proper face of $T$ and a face of $C$ that does not contain $x$.
%\end{enumerate}
The boundary complex of $\mw(K,D,N;x)$ is thus described by
\begin{equation}\label{eq:P}
\partial \mw(K,D,N;x)=\langle T\rangle \ast \lk_x(C)\bigcup_{\partial T\ast \lk_x(C)}\partial T\ast \anst_x(C).
\end{equation}

McMullen and Walkup have shown that $\mw(2k,D,N;x)$ is $k$-neighborly as well as $k$-stacked, thus satisfying
\begin{equation}\label{eq:g_of_P}
g_i(\mw(2k,D,N;x))=\begin{cases}
0 & i> k,\\
\mchoose{N-D-1}{i}=\binom{N-D+i-2}{i} & i\leq k.
\end{cases}
\end{equation}

In fact, the proof that $\mw(2k,D,N;x)$ is $k$-neighborly and $k$-stacked given in~\cite[p.~269]{McMuW71} shows:

\begin{lemma}\label{lem:gMW}
	The polytope $\mw(K,D,N;x)$ is $\dhalf{K}$-neighborly and $\dhalf{K}$-stacked. In particular,
	\begin{equation}
	g_i(\mw(2k-1,D,N;x))=\begin{cases}
	0 & i> k-1,\\
	\mchoose{N-D-1}{i}=\binom{N-D+i-2}{i} & i\leq k-1.
	\end{cases}
	\end{equation}
\end{lemma}

The vertices of $C$ come with a natural order $v_1<v_2<\dots v_{N-D+K}$ (inherited from the order of the parameters $t_1 < t_2 <\dots < t_{N-D+K}$ in the definition of $C$). We will take $x$ to be the last vertex in that ordering,
%which determines the combinatorial type of
denoting the resulting polytope simply by $\mw(K,D,N)$.
Removing $x=v_{N-D+K}$, we extend the order $v_1<\dots<v_{N-D+K-1}$ of the remaining vertices of $C$  to an order $v_1<\dots<v_{N-D+K-1}<v_{N-D+K}<\dots<v_N$ of the vertices of $\mw(K,D,N)$, where $v_{N-D+K},\dots,v_N$ are the vertices of the $(D-K)$-simplex $T$.
We will use the following result:

\begin{lemma}\label{lem:vlink_of_MW}
	$\vfig{\mw(2k,D,N)}{v_1}$ is combinatorially isomorphic to $\mw(2k-1,D-1,N-1)$.
\end{lemma}
\begin{proof}[Proof sketch]
	For $C=C(2k,N-D+2k)$ denote $C'=\vfig{C}{v_1}$, and note that\break $C'\cong C(2k-1,N-D+2k-1)$. Applying the construction in Definition~\ref{def:MW} with $C'$ produces an $\mw(2k-1,D-1,N-1)$ with boundary complex
	\begin{equation}\label{eq:boundery_complex_C'}
	\langle T\rangle \ast \lk_x(C')\bigcup_{\partial T\ast \lk_x(C')}\partial T\ast \anst_x(C').
	\end{equation}
Now one shows that the complex above is equal to $\lk_{v_1}(\mw(2k,D,N))$.
%	Let $F$ be a face of $\mw(2k,D,N)$ containing $v_1$.
%	If $F$ contains $T$
%	then $F\setminus\{v_1\}$ is in $\langle T\rangle \ast \lk_x(C')$,
%	and if $F$ does not contain $T$
%	then $F\setminus\{v_1\}$ is in $\partial T\ast \anst_x(C')$.
%	Thus $\lk_{v_1}(\mw(2k,D,N))$ is contained in the complex~\eqref{eq:boundery_complex_C'}.
%	Similarly, for the other direction, take a face $F'$ in the complex~\eqref{eq:boundery_complex_C'} and observe that $F'\cup\{v_1\}$ is in the boundary complex of $\mw(2k,D,N)$, as described in~\eqref{eq:P}.
\end{proof}


\section{The Sanyal--Ziegler construction}\label{sec:SZ}

We give a very brief sketch of the construction, focusing on the combinatorial description of links of vertices. The reader is prompted to confer with the paper~\cite{SanyZ10}, or with Sanyal's diploma thesis~\cite{Sany05}.

Let $(P,<)$ be a simplicial $(d-2)$-polytope with $n-1$ vertices. Label the vertices $v_1,\dots,v_{n-1}\in\R^{d-2}$ according to the given order $v_1<v_2<\dots <v_{n-1}$, and assume that the vertices are in general position, i.e., no $d-1$ vertices of $P$ lie on a hyperplane. We start by defining the lexicographic diamonds of $P$.

\subsection{Lexicographic diamonds}

Let $w_1,\dots ,w_{n-1}\in\R$ be a set of {\bf heights}, and denote by
$V^w=\set{(w_i,v_i)}{1\leq i\leq n-1}\subset\R^{d-1}$ the set of {\bf
lifted vertices}. Let $p=(w_0,v_0)\in\R^{d-1}$ be an arbitrary point with
$w_0\gg |w_i|$ for every $1\leq i\leq n-1$, and consider the
$(d-1)$-polytope $D(P,w)=\conv(V^w,p)$. We call $D(P,w)$ the {\bf diamond} over $P$ with subdivision $w$, noting that, for $w_0$ big enough, the combinatorial type of $D(P,w)$ is independent of the choice of point $p$.

Of special interest are the subdivisions of $P$ induced by the heights $$(w_1,w_2,\dots,w_{n-1})=(\pm h,0,\dots,0)\quad\text{with }h>0.$$
The subdivision induced by $(-h,0,\dots,0)$ is obtained by {\bf pulling} $v_1$, it is a triangulation of $P$, and its cells are the pyramids with apex $v_1$ over facets in $P\cap P_1$ with $P_1=\conv(v_2,\dots,v_{n-1})$.
The subdivision induced by $(h,0,\dots,0)$ is obtained by {\bf pushing} $v_1$, and it consists of the pyramids with apex $v_1$ over facets in $P_1\setminus P$, and one (possibly non-simplex) cell $P_1$.
The {\bf $a$-th lexicographic subdivision} $\lex_a(P)$ of $P$ is obtained by successively pushing the vertices $v_1,\dots,v_{a-1}$, and then pulling $v_a$. That is, pushing $v_1$ creates a subdivision with one non-simplex cell $P_1$, which we replace by its subdivision obtained by pushing $v_2$, which, in turn, has only one non-simplex cell $P_2=\conv\{v_3,\dots ,v_{n-1} \}$, and so on, until we finally replace $P_{a-1}=\conv\{v_a,\dots, v_{n-1} \}$ by its triangulation obtained by pulling $v_a$.

The above iterative procedure amounts to choosing a set of heights $w_1,\dots w_{n-1}$ with
\[
w_1>\dots>w_{a-1}>0>w_a,\quad\text{and}\quad w_{a+1}=\dots =w_{n-1}=0.
\]
The resulting diamond, denoted $D_a$, is called the {\bf $a$-th lexicographic diamond}.
Its vertices are labeled $v_0, v_1, \dots, v_{n-1}$,
with $v_0$ corresponding to the apex $p$.

%Denote by $Q=Q_0\supset Q_1\supset\dots\supset Q_{n-d}$ the chain of simplicial $(d-2)$-polytopes defined by $Q_i=\conv\{v_{i+1},\dots v_{n-1}\}$, and note that $Q_{n-d}$ is the $(d-2)$-simplex.

\begin{remark}
	Note that pushing or pulling a vertex in a simplex has no effect, thus the (possibly) different diamonds
	are $D_a$ with $1\leq a\leq n-d+1$.
\end{remark}


\subsection{The vertex figures of $Q$}

Take a Gale transform $G\in \R^{(n-1)\times(n-d)}$ of $P$ that has the form $G=\begin{bmatrix}
I_{n-d}\\
\overline{G}
\end{bmatrix}$, where $\overline{G}\in\R^{(d-1)\times (n-d)}$. Plugging $\overline{G}$ into the {\bf deformed cube template} (see~\cite[Definition 3.1]{SanyZ10}) produces a combinatorial $n$-cube $C=C_n(\overline{G})$. The projection of $C$ onto the last $d$ coordinates $\pi_d(C)$ is the cubical polytope $Q=Q(P,<)$ mentioned in the introduction.

The following key result from~\cite{SanyZ10}\footnote{Theorem 3.7 in~\cite{SanyZ10} actually contains a typo, having $n-d-1$ instead of the correct value $n-d+1$. Their proof, however, does give the correct value.}
states that each vertex figure of $Q$ is combinatorially equivalent to some diamond $D_a$, and moreover, it tells us which diamond corresponds to a given vertex $v$ of $Q$.
\begin{lemma}[{\cite[Theorem 3.7]{SanyZ10}}]\label{lem:SZ}
	Let $v$ be a vertex of $C$
	%with $\mathsf{eq} v$ given
	labeled
	by $\sigma\in\{+,-\}^n$. Then the vertex figure of $\pi_d(v)$ in $Q$ is isomorphic to $D_a$ with
	$$a=\min\left(\set{i\in [n]}{\sigma_i=+}\bigcup \{n-d+1\}\right).$$
\end{lemma}
%Here, the vertices of $D_a$ are $v_0,v_1,\dots ,v_{n-1}$, where $v_0$ denotes the apex $p$ and $v_1,\dots ,v_{n-1}$ are the vertices of $(P,<)$.
The isomorphism $D_a\cong \vfig{Q}{v}$ is given by:
$v_i$ (in $D_a$) corresponds to the neighbor of $v$ obtained by flipping the $(i+1)$-st coordinate of $v$ $(0 \leq i \leq n-1)$.


\section{The cubical polytopes $Q(k,d,n)$}\label{sec:Q}

Fix positive integers $k \ge 1$ and $n \ge d \ge 2k+2$.
We apply the Sanyal--Ziegler construction to the McMullen--Walkup polytope $P=\mw(2k,d-2,n-1)$, with a total order $<$ on its vertices as described after Lemma~\ref{lem:gMW} above.
The result is a $d$-dimensional cubical polytope $Q=Q(k,d,n)=Q(P,<)$ with $2^n$ vertices.
We now compute its cubical $g$-vector $\gc(Q)$, in stages.


\subsection{Computing $\gsc(Q(k,d,n))$}
%\subsection{Computing $\gc(Q(k,d,n))$}

By Hetyei's observation~\eqref{eq:Hetyei} we have
\begin{equation*}
\hsc_i(Q)=\sum_{v\in \vertices(Q)}h_i(\lk_v(Q))
\qquad (0 \leq i \leq d-1).
\end{equation*}
%and since
%\begin{equation*}
%\hc_i(Q)=(-1)^i2^{d-1}+\sum_{j=0}^{i-1}(-1)^{i-1-j}\hsc_j(Q)
%\qquad (0 \leq i \leq d)
%\end{equation*}
%we have for $1\leq i\leq \lfloor d/2\rfloor$
%\begin{equation}\label{eq:Q_g-vector}
%\begin{aligned}
%\gc_i(Q)
%&=(-1)^i2^d+\sum_{j=0}^{i-1}(-1)^{i-1-j}\gsc_j(Q)\\
%&=(-1)^i2^d+\sum_{v\in \vertices(Q)}\sum_{j=0}^{i-1}(-1)^{i-1-j}g_j(\lk_v(Q))\\
%%FALSE:&=(-1)^i2^d+\sum_{a=1}^{n-d-1}2^{n-a}\left(\sum_{j=0}^{i-1}(-1)^{i-1-j}g_j(D_a)\right)\\
%&=(-1)^i2^d\\
%&\quad+2^{n-1}\left(g_{i-1}(D_1)-g_{i-2}(D_1)+\dots +(-1)^{i-1}g_0(D_1)\right)\\
%&\quad+2^{n-2}\left(g_{i-1}(D_2)-g_{i-2}(D_2)+\dots +(-1)^{i-1}g_0(D_2)\right)\\
%&\vdots\\
%&\quad+2^{d}\left(g_{i-1}(D_{n-d})-g_{i-2}(D_{n-d})+\dots +(-1)^{i-1}g_0(D_{n-d})\right)\\
%&\quad+2^{d}\left(g_{i-1}(D_{n-d+1})-g_{i-2}(D_{n-d+1})+\dots +(-1)^{i-1}g_0(D_{n-d+1})\right).
%\end{aligned}
%\end{equation}
%%\erem{I changed the summation above: two more diamonds types.}
%%The above computation applies generally, next we
Therefore, for $1\leq i\leq \lfloor\frac{d-1}{2}\rfloor$:
\begin{equation}\label{eq:Q_g-vector}
%\begin{aligned}
\gsc_i(Q)
= \sum_{v\in \vertices(Q)} g_i(\lk_v(Q))
= \sum_{a=1}^{n-d} 2^{n-a} g_i(D_a) + 2^{d} g_i(D_{n-d+1}).
%\end{aligned}
\end{equation}
We compute the $g$-vectors of the diamonds $D_a$ at hand, i.e. for our choice of $(P,<)$.

%\begin{proposition}\label{prop:D_1_g-vector}
%	For $a=1$ and $0 \leq i\leq \lfloor\frac{d-1}{2}\rfloor$:
%	$$g_i(D_1)=\begin{cases}
%	0,& \text{if } i> k+1;\\
%	\mchoose{n-d}{k},& \text{if } i=k+1;\\
%	\mchoose{n-d}{i},& \text{if } i\leq k.
%	\end{cases}$$
%\end{proposition}
%\begin{proof}
%	Let $B_1=\lex_1(P)$ be the simplicial $(d-2)$-ball triangulating $P$, obtained by pulling $v_1$. Then
%	$$f(D_1,t)=f(B_1,t)+t\cdot f(P,t).$$
%	Note that, strictly speaking, the simplicial complexes actually considered here are $\partial D_1$, $B_1$ and $\partial P$, of dimensions $d-2$, $d-2$ and $d-3$, respectively.
%	The corresponding $h$-polynomial is
%	\[
%	\begin{aligned}
%	h(D_1,t)
%	&= (1-t)^{d-1} \cdot f\left(D_1,\frac{t}{1-t}\right) \\
%	&= (1-t)^{d-1} \cdot \left[ f\left(B_1,\frac{t}{1-t}\right) + \frac{t}{1-t} \cdot f\left(P,\frac{t}{1-t}\right)\right] \\
%	&= h(B_1,t)+t\cdot h(P,t).
%	\end{aligned}
%	\]
%	To compute $h(B_1,t)$, observe that
%	\[
%	f(B_1,t)=(1+t)\cdot f(\anst_{v_1}(P))=(1+t)\cdot \Big( f(P,t)-t\cdot f(\lk_{v_1}(P),t)\Big).
%	\]
%	The $h$-polynomial is therefore
%	\[
%	\begin{aligned}
%	h(B_1,t)
%	&= (1-t)^{d-1} \cdot f\left(B_1,\frac{t}{1-t}\right) \\
%	&= (1-t)^{d-1} \cdot \left(1+\frac{t}{1-t}\right)\cdot \left[ f\left(P,\frac{t}{1-t}\right)-\frac{t}{1-t}\cdot f\left(\lk_{v_1}(P),\frac{t}{1-t}\right)\right]\\
%	&= (1-t)^{d-2}\cdot f\left(P,\frac{t}{1-t}\right)-t\cdot (1-t)^{d-3}\cdot f\left(\lk_{v_1}(P),\frac{t}{1-t}\right)\\
%	&= h(P,t)-t\cdot h(\lk_{v_1}(P),t).
%	\end{aligned}
%	\]
%	Summing up, we have
%	\begin{equation*}
%	h(D_1,t)=(1+t)\cdot h(P,t)-t\cdot h(\lk_{v_1}(P),t)
%	\end{equation*}
%	which yields, for $1\leq i\leq \lfloor\frac{d-1}{2}\rfloor$,
%	\begin{equation}\label{eq:g_of_D_1}
%	g_i(D_1)=g_i(P)+g_{i-1}(P)-g_{i-1}(\lk_{v_1}(P)).
%	\end{equation}
%	Using~\eqref{eq:g_of_P} and Lemmas~\ref{lem:vlink_of_MW} and \ref{lem:gMW} we compute the right-hand side of~\eqref{eq:g_of_D_1} to obtain the claimed result.
%\end{proof}

%\erem{Now the Prop. below DOES give the correct g-entries implied by k-neighborliness: recheck the combinatorial identity below, and correct indexing in the proof.}
\begin{proposition}\label{prop:D_a_g-vector}
	For each $1\leq a\leq n-d+1$ and $0 \leq i\leq \lfloor\frac{d-1}{2}\rfloor$:
	\[
	g_i(D_a)=\begin{cases}
	0,& \text{if } i>k+1;\\
	\mchoose{n-d-a+1}{k},& \text{if } i=k+1;\\
	\mchoose{n-d}{i},& \text{if } i\leq k.
	\end{cases}
	\]
\end{proposition}
\begin{proof}[Proof sketch]
For $a=1$ this follows from the $f$-polynomials identity
\[f(D_1,t)=f(\lex_1 (P),t)+t\cdot f(P,t)=
(1+t)(f(P,t)-t\cdot f(\lk_{v_1}(P),t))+t\cdot f(P,t)
\]
combined with Lemma~\ref{lem:gMW}, Lemma~\ref{lem:vlink_of_MW} and the transformation to $h$-polynomials.

For $a>1$, contract the edge $v_0v_1$ in $P$ to obtain the $(a-1)$-lexicographic diamond $D_{a-1}(P_1)$ over $P_1=\mw(2k,d-2,n-2)$. Use the relation
\[
h(D_a(P),t)=h(D_{a-1}(P_1),t)+t\cdot h(\lk_{v_0v_1}(D_a(P),t))
\]
and iterate by edge contraction in $P_1$ etc. 
\end{proof}


%\begin{proof}
%	Let $v_0$ be the apex of the diamond $D_a$, and let $B_a=\lex_a(P)$.
%	The link condition $\lk_{v_1v_0}(D_a)=\lk_{v_1}(D_a)\cap\lk_{v_0}(D_a)$ holds;
%	denote by $Y_{a-1}(n-1)$ the $(d-2)$-complex obtained from $\partial D_a$ by contracting the edge $v_1v_0$.
%	Now, since
%	\[
%	\lk_{v_1v_0}(D_a)
%	=\lk_{v_1}(D_a)\cap\lk_{v_0}(D_a)
%	=\lk_{v_1}(B_a)\cap\partial P
%	=\lk_{v_1}(\partial P),
%	\]
%	we have by Lemma~\ref{lem:vlink_of_MW}
%	\begin{equation*}
%	h(D_a,t)=h(Y_{a-1}(n-1),t)+t\cdot h(\mw(2k-1,d-3,n-2),t).
%	\end{equation*}
%	Since $Y_{a-1}(n-1)$ is the $(a-1)$-lexicographic diamond over $P_1=\mw(2k,d-2,n-2)$, we may again contract the edge between the first vertex of $P_1$ and the apex $v_0$ to obtain $Y_{a-2}(n-2)$. We do these edge contractions $a-1$ times, and get
%	\begin{equation*}
%	h(D_a,t)=h(Y_1(n-a+1),t)+t\cdot \sum_{j=1}^{a-1}h(\mw(2k-1,d-3,n-1-j),t),
%	\end{equation*}
%	and therefore
%	\begin{equation}
%	g_i(D_a)=g_i(Y_1(n-a+1))+\sum_{j=1}^{a-1}g_{i-1}(\mw(2k-1,d-3,n-1-j)).
%	\end{equation}
%	As $Y_1(n-a+1)$ is the $1$-lexicographic diamond over $\mw(2k,d-2,n-a)$,
%	the result follows from Proposition~\ref{prop:D_1_g-vector}, Lemma~\ref{lem:gMW} and the identity
%	\[
%	\begin{aligned}
%	&\quad \mchoose{n-d-a+1}{i}+\sum\limits_{j=1}^{a-1}\mchoose{n-d-j+1}{i-1} \\
%	&= \mchoose{n-d-a+1}{i}+\sum\limits_{j=1}^{a-1}\left[\mchoose{n-d-j+1}{i}-\mchoose{n-d-j}{i}\right] \\
%	&= \mchoose{n-d}{i}.\qedhere
%	\end{aligned}
%	\]
%\end{proof}

Combining~\eqref{eq:Q_g-vector} with Proposition~\ref{prop:D_a_g-vector},
and noting that $\mchoose{0}{k} = 0$ for $k \ge 1$,
we conclude
\begin{corollary}\label{t:gsc_Q}
	%\begin{equation}
	For each $0 \leq i \leq \lfloor \frac{d-1}{2} \rfloor$,
	%\[
	%\gc_{i}(Q) + \gc_{i+1}(Q) = 0.
	%\]
	\[
	\gsc_i(Q)=\begin{cases}
	0,& \text{if } i>k+1;\\
	{\displaystyle\sum_{a=1}^{n-d}} 2^{n-a} \mchoose{n-d-a+1}{k}% + 2^d \mchoose{0}{k}
	,& \text{if } i=k+1;\\
	2^n \mchoose{n-d}{i},& \text{if } i\leq k.
	\end{cases}
	\]
	%\end{equation}
\end{corollary}

%\section{A new cubical $g$-vector}
\subsection{Computing $\gc(Q(k,d,n))$}

In order to compute the cubical $g$-vector of $Q$, and in particular $\gc_{k+2}(Q)$, we need the following binomial identity; its proof is omitted here.
\begin{lemma}\label{t:binomial}
	For any integers $k \ge 1$ and $m \ge 0$,
	\[
	\sum_{a=1}^{m} 2^{m-a} \mchoose{m-a+1}{k} %+ \mchoose{0}{k}
	= (-1)^{k+1} + 2^m \sum_{j=0}^{k} (-1)^j \mchoose{m}{k-j}.
	\]
\end{lemma}
%\begin{proof}
%	Fix $k \ge 1$, and denote
%	\[
%	\begin{aligned}
%	L_m &:= \sum_{a=1}^{m} 2^{m-a} \mchoose{m-a+1}{k}
%	= \sum_{b=0}^{m-1} 2^{b} \mchoose{b+1}{k}, \\
%	R_m &:= (-1)^{k+1} + 2^m \sum_{j=0}^{k} (-1)^j \mchoose{m}{k-j}.
%	\end{aligned}
%	\]
%	We will prove that $L_m = R_m$ for all $m \ge 0$, by induction on $m$.
%	
%	For $m=0$, indeed $L_0 = R_0 = 0$.
%	For the induction step, note that
%	$L_{m+1} = L_m + 2^m \mchoose{m+1}{k}$.
%	Thus it suffices to prove that
%	$R_{m+1} = R_m + 2^m \mchoose{m+1}{k}$ as well. Indeed,
%	\begin{equation*}
%	\begin{aligned}
%	R_{m+1} - R_m
%	&= 2^m \sum_{j=0}^{k} (-1)^j \left[ 2\mchoose{m+1}{k-j} - \mchoose{m}{k-j} \right] \\
%	&= 2^m \sum_{j=0}^{k} (-1)^j \left[ \mchoose{m+1}{k-j} + \mchoose{m+1}{k-j-1} \right] \\
%	&= 2^m \left[ \mchoose{m+1}{k} + (-1)^{k-1} \mchoose{m+1}{-1} \right] = 2^m \mchoose{m+1}{k},
%	\end{aligned}
%	\end{equation*}
%	as claimed.
%\end{proof}

\begin{theorem}\label{t:gc_Q}
	%\begin{equation}
	For each $1 \leq i \leq \dhalf{d}$,
	%\[
	%\gc_{i}(Q) + \gc_{i+1}(Q) = 0.
	%\]
	\[
	\gc_i(Q)=\begin{cases}
	0,& \text{if } i>k+1;\\
	2^n {\displaystyle\sum_{j=1}^{i}} (-1)^{j-1} \mchoose{n-d}{i-j} + (-1)^i 2^d,& \text{if } i\leq k+1.
	\end{cases}
	\]
	%\end{equation}
\end{theorem}
\begin{proof}[Proof sketch]
	From the definitions of $\gc$ and $\gsc$, we have
%	\[
%	\begin{aligned}
%	\gc_0(Q) &= 2^{d-1}, \qquad
%	\gsc_0(Q) = 2\gc_0(Q) + \gc_1(Q), \\
%	\gsc_i(Q) &= \gc_i(Q) + \gc_{i+1}(Q)
%	\qquad (1 \leq i \leq \lfloor (d-1)/2 \rfloor)
%	\end{aligned}
%	\]
%	it follows that
	\begin{equation}\label{eq:gc_Q}
	\gc_i(Q) = \sum_{j=1}^{i} (-1)^{j-1} \gsc_{i-j}(Q) + (-1)^i 2^d
	\qquad (1 \leq i \leq \lfloor d/2 \rfloor).
	\end{equation}
	The values of $\gc_i(Q)$ for $i \le k+1$ now follow easily from Corollary~\ref{t:gsc_Q}.
	It also follows that
	\[
	\gc_i(Q) + \gc_{i+1}(Q) = 0 \qquad (i \ge k+2),
	\]
	and all that remains is to show that $\gc_{k+2}(Q) = 0$.
	Indeed, by \eqref{eq:gc_Q} and Corollary~\ref{t:gsc_Q},
	\[
	\begin{aligned}
	\gc_{k+2}(Q)
	%&= \sum_{j=1}^{k+2} (-1)^{j-1} \gsc_{k+2-j}(Q) + (-1)^{k+2} 2^d \\
	%&= \sum_{a=1}^{n-d} 2^{n-a} \mchoose{n-d-a+1}{k}
	%+ 2^n \sum_{j=2}^{k+2} (-1)^{j-1} \mchoose{n-d}{k+2-j} + (-1)^{k} 2^d \\
	&= \sum_{a=1}^{n-d} 2^{n-a} \mchoose{n-d-a+1}{k}
	- \left[ 2^n \sum_{j=0}^{k} (-1)^{j} \mchoose{n-d}{k-j} + (-1)^{k+1} 2^d \right].
	\end{aligned}
	\]
	Using Lemma~\ref{t:binomial} with $m = n-d$ gives, indeed,
	$\gc_{k+2}(Q) = 0$ as claimed.
\end{proof}

\begin{corollary}\label{t:asymptotics}
	Fix $k \ge 1$ and $d \ge 2k+2$, and let $\{Q_n\}_{n=d}^{\infty}=\{Q(k,d,n)\}_{n=d}^{\infty}$. Then
	\begin{equation*}
	\lim_{n \to \infty} \frac{\gc_{k+1}(Q_n)}{2^{n} \mchoose{n-d}{k}} = 1,
	\qquad\text{and}\qquad
	\lim_{n \to \infty} \frac{\gc_i(Q_n)}{2^{n} \mchoose{n-d}{k}} = 0
	\quad (\forall i \ne k).
	\end{equation*}
\end{corollary}

Corollary~\ref{t:asymptotics} shows that the ray spanned by $e_k$
$(2 \leq k \leq \lfloor d/2 \rfloor)$ in $\R^{\lfloor d/2 \rfloor}$ belongs to the closed cone $\mathcal{C}_d$.
Note that this was already known for the ray spanned by $e_{\lfloor d/2 \rfloor}$ because of the existence of neighborly cubical $d$-polytopes, such as $Q(k,2k+2,n)$).
The ray spanned by $e_1$ belongs to this cone because of the existence of stacked cubical $d$-polytopes.
Thus $\mathcal{C}_d$ contains $\mathcal{A}_d$, and \autoref{thm:cones} is proved.


\section{No obvious cubical GLBT}\label{sec:stack}

In~\cite{BabsBC97}, after introducing the definitions of cubical stackedness and cubical neighborliness, the authors show that cubical $1$-stacked $d$-polytopes with at least $n$ vertices exist, for any $n\geq 2^d$ (see~\cite[Corollary 5.6]{BabsBC97}). It is also shown (see~\cite[proof of Proposition 5.5]{BabsBC97}) that if $Q$ is a cubical $k$-stacked $d$-polytope, then $\gc_i(Q)=0$ for $k<i\leq\dhalf{d}$. The converse claim, namely, that $\gc_{k+1}(Q)=0$ implies that $Q$ is cubical $k$-stacked, is false, as our analysis of $Q(k,d,n)$ below shows. This is in apparent contrast with the simplicial GLBT.


\begin{theorem}\label{prop:notStacked}
	The polytope $Q=Q(k,d,n)$ is not cubical $(k+1)$-stacked.
\end{theorem}
\begin{proof}[Proof sketch]
	Assume by contradiction that $Q$ is cubical $(k+1)$-stacked, so $Q$ has some cubulation $Q'$, namely a subdivision into (combinatorial) cubes, without interior\break $(d-k-2)$-faces.
	Let $C_n$ be the deformed $n$-cube that $Q$ is a projection of.
	\begin{lemma}\label{lem:cubulationBySubcubes}
		All faces of $Q'$ must be faces of $C_n$.
	\end{lemma}
	To see this, note that any $1$-dimensional subcomplex of $C_n$
	which is isomorphic to the graph of an $m$-cube, is the
	$1$-skeleton of an $m$-face of $C_n$.
	
	
	Each $\lk_v(Q')$ (the simplicial complex whose face lattice is the ideal above the vertex $v$ in the face lattice of $Q'$) is a triangulation of $\vfig{Q}{v}$ with no interior $(d-k-3)$-faces.
	Thus the vertex figure of $v$ in $Q$ --- isomorphic to some diamond $D_a$ --- is $(k+1)$-stacked, and by the GLBT, %(an easy part of it, see \cite[Thm.2.3(ii)]{MuraN13} or \cite[Thm.2.12]{Bagchi-Datta}),
	%from the Murai--Nevo part of the GLBT~\cite{MuraN13},
	% that it has a unique triangulation without interior $(d-k-3)$-faces,
	$\lk_v(Q')$ is the triangulation
	obtained from $D_a$  by inserting all
	$(d-1)$-simplices whose $(d-k-3)$-skeleton is contained in the boundary of the diamond $D_a$.
	(We abuse notation and identify $\partial D_a$ with $\lk_v(Q)$ by the isomorphism given in Lemma~\ref{lem:SZ}.)
	This description allows us to determine, for each vertex $v$ of $Q$, the set of $d$-cubes in $Q'$ that contain $v$. The compatibility condition mentioned above is the requirement that
	%these lists of $d$-cubes are compatible, i.e.,
	if $u$ is a vertex in a $d$-cube from the list of $v$, then the list of $u$ must contain this $d$-cube too. 
%The following series of lemmas shows that the $(k+1)$-stacked triangulations of the diamonds $D_a$ are incompatible, thus such a cubulation $Q'$ cannot exist.
	
%	\begin{lemma}[The operations $\mw$ and $\lex$ commute]\label{lem:MWcommuteLEX}
%		\begin{equation}\label{eq:MWcommuteLEX}
%		\lex_a(P)=\langle T\rangle \ast \lk_x(\lex_a(C))\bigcup_{\partial T\ast \lk_x(\lex_a(C))}\partial T\ast \anst_x(\lex_a(C)).
%		\end{equation}
%	\end{lemma}
%	\begin{proof}
%		Let $P=\mw(2k,d-2,m)$. By the construction of $P$ we have:
%		\begin{itemize}
%			\item $F$ is a face of $P$, with $V(T) \subset V(F)$, iff $(V(F) \setminus V(T) ) \cup {x}$ is the set of vertices of a face $\bar{F}$ of $C$, and
%			\item $F$ is a face of $P$, with $V(T) \not\subset V(F)$, iff $V(F) \setminus V(T)$  is the set of vertices of a face $\bar{F}$ of $C$.
%		\end{itemize}
%		
%		Now, let $F$ be a face added when pulling $v_1$ in $P$. Then $F=v_1\ast F'$, with $F'$ a face of $P$, and $F'$ a face of $P_1=\conv(v_2,...,v_m)$. By the argument above, $\bar{F'}$ is a face of both $C$ and $C_1$, and thus $\bar{F}$ is added when pulling $v_1$ in $C$.
%		Similarly, when pushing $v_1$ in $P$, we add faces $F$ (and $v_1*F$) of $P_1$ which are non-faces of $P$, and so $\bar{F}$ is a face of $C_1$, and a non-face of $C$, and thus is added when pushing $v_1$ in $C$.
%		Reversing the argument shows that faces added when pushing/pulling $v_1$ in $C$ give corresponding added faces in pushing/pulling in $P$.
%		When constructing $\lex_a(P)$ and $\lex_a(C)$ sequentially, we always push/pull $v_1$ in a suitable $\mw$-polytope and the corresponding suitable cyclic polytope $C$, hence \eqref{eq:MWcommuteLEX} follows.
%	\end{proof}
	
Omitting many details, here is a description of 
the facets in the $(k+1)$-stacked triangulation of $D_a$, after introducing some terminology: 
	let $G$ be a subset of vertices of the cyclic polytope $C$. A {\em block} $B\subset G$ is a maximal subset of $G$ (w.r.t inclusion) of consecutive vertices, 
%and $B$ is {\em inner} if $\min B>v_1$ and $\max B<v_{n-d+2k+1}=x$, 
and $B$ is even / odd if its size is. Recall that $p$ denotes the apex of $D_a$.	
	\begin{proposition}
		The sets of $d$ vertices that form the $(d-1)$-simplices of the $(k+1)$-stacked triangulation of $D_a$ are exactly the ones of one of the following two types:
		\begin{enumerate}[{\rm (i)}]
			\item $\{p\}\cup \left\{2k \text{-set in } C\setminus \{x\}\text{ consisting of even blocks}\right\}\cup T$,
			\item $\{a\}\cup \left\{2k \text{-set } F \text{ in } C\setminus \{x\}\text{ consisting of even blocks}, \min F>a\right\}\cup T$.
		\end{enumerate}
	\end{proposition}
	
%	\begin{proof}
%		First, we determine the missing faces of $D_a$.
%		
%		By the Gale evenness criterion, a subset $F$ of $i$ vertices of $C$ forms a face iff it contains at most $2k-i$ inner odd blocks. It follows that a missing face of $C$ is a set of $k+1$ isolated vertices, containing at most one of $\{v_1,x\}$.
%		
%		The polytope $P$ has two types of missing faces: ``small", of size $k+1$, arising from a missing face of $C$ which does not contain $x$; ``large", of size $d-k-1$, equal to $(MF \setminus x, T)$ where $MF$ is a missing face of $C$ which contains $x$.
%		
%		Kalai~\cite{Kalai:missing} and Nagel~\cite{Nagel:missing} showed that if $g_i(P)=0$ for a simplicial $j$-polytope, then the polytope has no missing $l$-faces for $i\le l\le j-i$. Combined with the GLBT, a subset $A$ of $d$ vertices of $D_a$ is a facet in the unique
%		$(k+1)$-stacked triangulation of $D_a$ iff all its subsets $B$ of size $\leq k+2$ are faces of $D_a$. Since $D_a$ is $k$-neighborly, it is enough to determine the missing faces of $D_a$ of sizes $k+1$ and $k+2$.
%		
%		It follows that we are only \erem{Why only them??} interested in missing faces of $\lex_a(P)$ of size $\leq k+2$, and using Lemma~\ref{lem:MWcommuteLEX} we can now determine the missing faces $F$ of $D_a$ of size $\leq k+2$:
%		\begin{itemize}
%			\item $p\in F$, $F\setminus\{p\}$  has size $k+1$, $F\setminus \{p\}$ missing in $P$, $F\setminus \{p\}$ is a face of $\lex_a(P)$.
%			Thus $F\setminus \{p\}$ is missing in $C$, without $x$, and $F\setminus \{p\}$ a face of $\lex_a(C)$.
%			\item $p\notin F$, $F$ has size $k+1$, $F$ missing in $P$ and in $\lex_a(P)$.
%			Thus $F$ is missing face in $C$ and in $\lex_a(C)$, without $x$.
%		\end{itemize}
%		So we have that $F$ is either of type \erem{pf of why (1,2) is the right description is missing. Pullinh/pushing comes into the argument.}
%		\erem{also: add pf that there are no missing faces of size k+2 and w/o p.}
%		\begin{enumerate}[(1)]
%			\item $F=\{p\} \cup F'$, with $F'$ of size $k+1$, consisting of isolated vertices of $C$, $x\notin F'$, and $\min F'=v_a$, or
%			\item $F$ of size $k+1$, $x\notin F$, $F$ consisting of isolated vertices of $C$, $\min F\neq v_a$.
%		\end{enumerate}
%		
%		Let $Y$ be a $d$-set that forms a facet in the $(k+1)$-stacked triangulation of $D_a$. Then $Y$ cannot contain more than $2k+1$ vertices of $C\setminus \{x\}$, because then it would contain two distinct missing faces, one of which must be of type (2). Clearly, $Y$ has to contain at least $2k$ vertices of $C\setminus \{x\}$. So we have that $Y$ is either of the form
%		\begin{itemize}
%			\item $2k$ vertices of $C\setminus \{x\}$, the apex $p$, and all of the vertices of $T$, or
%			\item $2k+1$ vertices of $C\setminus \{x\}$, and all of the vertices of $T$.
%		\end{itemize}
%		
%		If $Y$ has $2k+1$ vertices of $C\setminus \{x\}$, then it contains $k+1$ isolated vertices of $C\setminus \{x\}$, so $\min Y=a$, since $Y$ contains no missing faces of type (2), and $p\notin Y$ since $Y$ contains no missing faces of type (1). These are the $d$-sets of type (ii) in the statement. Otherwise, $Y$ has $2k$ vertices of $C\setminus \{x\}$, and hence must contain $p$ and all vertices of $T$, and since $Y$ does not contain missing faces of type (1) or (2), it must be of type (i).
%	\end{proof}
%	
	
%We are now ready to complete the proof of the theorem.
	Consider a vertex $v_{\sigma}$ of $Q$ with $\sigma\in\{+,-\}^n$, and with $a=\min\{i\mid \sigma_i=+\} <n-d+1$; then $\vfig{Q}{v_{\sigma}}\cong D_a$. Now $v_{\sigma'}$, where $\sigma'$ is obtained by flipping the $(a+1)$-th coordinate of $\sigma$, is a neighbor of $v_{\sigma}$ with $\vfig{Q}{v_{\sigma'}}\cong D_b$, for some $b>a$. Since $\left\{v_{\sigma},v_{\sigma'}\right\}$ forms an edge of $Q$, it must be contained in some $d$-cube of the cubulation of $Q$, and, by the isomorphism $\vfig{Q}{v_{\sigma}}\cong D_a$ given after Lemma~\ref{lem:SZ}, the $d$-set $F$ of $D_a$ corresponding to this $d$-cube must be of type (ii). But $F$ is not a $d$-set of either type (i) or (ii) in $D_b$, and so the triangulations of $\vfig{Q}{v_{\sigma}}$ and $\vfig{Q}{v_{\sigma'}}$ are incompatible.
%From the uniqueness of the $(k+1)$-stacked triangulations of the diamonds, it follows that there cannot exist another set of (potentially compatible) triangulations of the links of vertices in $Q$, and so $Q$ is not cubical $(k+1)$-stacked.
	%\end{proof}

This contradicts the assumption that $Q'$ exists.
\end{proof}



\section{Concluding remarks}\label{sec:concluding_remarks}

%\subsection{Cubical $k$-stacked $d$-polytopes}
%As mentioned in the proof of Theorem~\ref{prop:notStacked}, since $g_{k+2}(D_a)=0$, the vertex figure $D_a$ has no missing $i$-faces for $\dhalf{d}\leq i\leq d-k-3$, and so $Q(k,d,n)$ is not cubical $j$-stacked for any $k+1\leq j\leq \dhalf{d}$. It is then natural to ask:
%\begin{question}\label{Q:exist?}
%	Let $2\leq k\leq\dhalf{d}-1$.
%	Do cubical $k$-stacked $d$-polytopes with $g^c_k\neq 0$ (and are thus, in particular, not $(k-1)$-stacked) actually exist? %Are there infinitely many combinatorial types of such polytopes?
%\end{question}
%The {\em $k$-elementary} cubical $d$-polytopes $C^d_k$ constructed (for any $0\leq k\leq d$) by Blind and Blind in~\cite{BlinB95} provide a positive answer to Question~\ref{Q:exist?}.
%For $1 \le k \le \lfloor d/2 \rfloor$, $C^d_k$ is $k$-stacked by construction and its $\gc$-vector is
%\[
%g^c_i(C^d_k)
%%= \sum_{j=1}^{k} 2^{d-j} \left[ \binom{j-1}{i-1} - \binom{j-1}{d-i} \right]
%= \sum_{j=1}^{k} 2^{d-j} \binom{j-1}{i-1}
%\qquad (1 \le i \le \lfloor d/2 \rfloor)
%\]
%so that, in particular, $g^c_k(C^d_k) = 2^{d-k} > 0$.

The following question, implicit in~\cite{BabsBC97}, asks for a sequence of cubical $k$-stacked\break $d$-polytopes with $\gc$-vector approaching the ray spanned by $e_k$. It is still unanswered.
\begin{question}\label{Q:existBBC?}
	Let $2\leq k\leq\dhalf{d}-1$.
	Does there exist a sequence of cubical $k$-stacked\break $d$-polytopes such that the $k$-th coordinate of $\gc$ dominates the other coordinates? 
	%with dominant $g^c_k$?
\end{question}



%\subsection{The Cubical Lower Bound Conjecture}
Jockusch studied the lower and upper bound problems for cubical polytopes in~\cite{Jock93}, where he stated a {\em Cubical Lower Bound Conjecture}:
\begin{conjecture}[CLBC, \cite{Jock93}]\label{conj:JCLBC}
	Let $Q$ be a cubical $d$-polytope with $n$ vertices. Then
	$$f_k(Q)\geq\left(2^{d-k}\binom{d}{k}-2^{d-k-1}\binom{d-1}{k}\right)
	\left(\frac{n}{2^{d-1}}-2\right)+2^{d-k}\binom{d}{k}\qquad (1\leq k\leq d-1).$$
%	for all $0\leq k\leq d-1$.
\end{conjecture}

We prove a cubical version of the {\em MPW-reduction} stating that the case $k=1$ of Conjecture~\ref{conj:JCLBC} implies that it holds for all $k$. The case $k=1$ is equivalent to $\gc_2\geq 0$.
%
%\begin{lemma}[Cubical MPW-reduction]\label{lem:cMPW}
%	If the CLBC is true for edges of cubical polytopes then it is true for faces of all dimensions.
%\end{lemma}
%%\begin{proof}
%%	Let $Q$ be cubical $d$-polytope witn $n$ vertices $v_1,\dots ,v_n$, and let $d_i=\deg(v_i)$ be the valency of $v_i$. Due to the distinction between the formulas for $2\leq k\leq d-2$ and $k=d-1$ in the LBT, we write out the details for dimensions $2\leq k\leq d-2$, and remark that the same argument holds in the case $k=d-1$.
%%	
%%	Since the vertex figure $\vfig{Q}{v_i}$ is a simplicial $(d-1)$-polytope with $d_i$ vertices, we have, by the LBT:
%%	\begin{equation}\label{eq:lklbt}
%%	f_{k-1}(\lk_{v_i}(Q))\geq \binom{d-1}{k-1}d_i-\binom{d}{k}(k-1).
%%	\end{equation}
%%	Count in two ways the pairs vertex-in-$k$-face:
%%	$$2^kf_k(Q)=\sum_{i=1}^n f_{k-1}(\lk_{v_i}(Q)),$$
%%	and then plug in~\eqref{eq:lklbt} to get
%%	\begin{align*}
%%	2^kf_k(Q)&\geq \sum_{i=1}^n \binom{d-1}{k-1}d_i-\binom{d}{k}(k-1)\\
%%	&\geq -n\binom{d}{k}(k-1)+\binom{d-1}{k-1}\sum_{i=1}^nd_i.
%%	\end{align*}
%%	Now, use the CLBC for edges:
%%	$$\sum_{i=1}^nd_i\geq (d+1)n-2^d$$
%%	to obtain
%%	\begin{align*}
%%	2^kf_k(Q)&\geq -n\binom{d}{k}(k-1)+\binom{d-1}{k-1}\left((d+1)n-2^d\right)\\
%%	&\geq \left((d+1)\binom{d-1}{k-1}-\binom{d}{k}(k-1)\right)n-2^d\binom{d-1}{k-1}.
%%	\end{align*}
%%	So we have
%%	\begin{align*}
%%	f_k(Q)&\geq \left((d+1)\binom{d-1}{k-1}-\binom{d}{k}(k-1)\right)\frac{n}{2^k}-2^{d-k}\binom{d-1}{k-1}%\\
%%	%&=\left(\binom{d}{k}+\binom{d-1}{k-1}\right)
%%	%\frac{n}{2^{k}}-2^{d-k}\binom{d-1}{k-1}\\
%%	%&=\left(2\binom{d}{k}-\binom{d-1}{k}\right)
%%	%\frac{n}{2^{k}}+\left(2^{d-k}\binom{d-1}{k}-2^{d-k}\binom{d}{k}\right)\\
%%	%&=\left(2^{d-k}\binom{d}{k}-2^{d-k-1}\binom{d-1}{k}\right)
%%	%\frac{n}{2^{d-1}}+\left(2^{d-k}\binom{d}{k}-2\cdot 2^{d-k}\binom{d}{k}+2^{d-k}\binom{d-1}{k}\right)\\
%%	%&=\left(2^{d-k}\binom{d}{k}-2^{d-k-1}\binom{d-1}{k}\right)
%%	%\left(\frac{n}{2^{d-1}}-2\right)+2^{d-k}\binom{d}{k}\\
%%	%&\hfill\qedhere
%%	\end{align*}
%%	It is straightforward to verify that the above right-hand side is equal to the expression in Conjecture~\ref{conj:JCLBC}.
%%\end{proof}
%
%It follows that Conjecture~\ref{conj:JCLBC} is equivalent to:
%\begin{conjecture}[CLBC]\label{conj:CLBC}
%	If $Q$ is a cubical $d$-polytope with $n$ vertices then
%	\begin{equation}\label{eq:CLBC}f_1(Q)\geq\frac{d+1}{2}n-2^{d-1}.\end{equation}
%\end{conjecture}
%Clearly, \eqref{eq:CLBC} is equivalent to $\gc_2(Q)\geq 0$.





%\acknowledgements{\lipsum[20]}

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