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\usepackage{lipsum}

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\title[The Integer Decomposition Property in Smooth Polytopes]{
\texorpdfstring{The Integer Decomposition Property in \\ Smooth Polytopes}{The Integer Decomposition Property in Smooth Polytopes}    
}

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%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
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\author{\texorpdfstring{Juliana Curtis \thanks{\href{mailto:jacca@uw.edu}{jacca@uw.edu}.}}{Juliana Curtis}
%Longer names me was partially supported by Grant 2017.11.14.$\partial$\;supp.}\addressmark{1}
}

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\address{\addressmark{1}Department of Mathematics, University of Washington, Seattle, Washington, USA}

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%% put your English abstract here, or comment this out if you don't have one yet
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\abstract{
    Tadao Oda conjectured that every smooth polytope, in any dimension, has the Integer Decomposition Property. 
    %Beck et al.\ previously showed that smooth polytopes which are both 3-dimensional and centrally symmetric have this property. 
    In this paper, we show this result for subclasses of polytopes: smooth combinatorial cubes of any dimension and 3-dimensional pseudo-prisms. 
}

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%\resume{\lipsum[2]}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{polytope, smooth, IDP}

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\begin{document}

\maketitle
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Introduction}

%Polytopes are higher dimensional analogues of polygons and are objects of study in many disciplines. In this paper, we will examine lattice polytopes, those with vertices in $\Z^d$, which are fundamental to our study of the following characteristics. 

%: a \textit{polytope} is a bounded intersection of finitely many closed halfspaces in $\R^d$. A $d$-dimensional polytope is equipped with \textit{faces} of each dimension less than or equal to $d$, together forming its face poset via inclusion. We refer to the $(d-1)$-faces as \textit{facets}, the $1$-faces as \textit{edges}, and the $0$-faces as \textit{vertices}. A \textit{lattice polytope} is one whose vertices are all lattice points, i.e. points in $\Z^d$. In this paper, all polytopes discussed will be lattice polytopes, unless stated otherwise.

For lattice polytopes $P$ and $Q$, we say that $(P,Q)$ has the \textit{Integer Decomposition Property}, or that it \textit{is} \textit{IDP}, if every lattice point in the Minkowski sum $P + Q = \{p + q : p \in P, q \in Q\}$ can be written as the sum of a lattice point in $P$ and a lattice point in $Q$. For a single polytope $P$, we say that $P$ is \textit{IDP} when $(P, kP)$ is IDP for all positive integers $k$. IDP polytopes are directly related to Ehrhart theory, and are of great interest in commutative algebra and the study of toric varieties, as well as being of use in integer programming.
%Polytopes which are IDP are related to toric varieties and are broadly of interest in toric geometry and Ehrhart theory, as well as being of use in integer programming. 
In general, it is an open question to characterize when a polytope is IDP. 

While being IDP is a global property of a polytope, we are interested in its relationship to the more local notion of smoothness. First, a $d$-dimensional polytope is \textit{simple} if each vertex is contained in exactly $d$ edges (and so also exactly $d$ facets). We define the \textit{primitive edge directions} of a vertex to be the smallest lattice directions along all its adjacent edges, and then say that a $d$-dimensional polytope is \textit{smooth} if it is simple and if the primitive edge directions at each vertex form a basis for the integer lattice $\Z^d$. 

In 1997, Tadao Oda made the following conjecture, documented in \cite{oda2008problems}, which remains unproven.

\begin{conjecture} [Oda's Conjecture]
    All smooth polytopes are IDP.
\end{conjecture}

%despite attracking consid interest, very little progress has been made

This problem is motivated by its relationship to the study of toric varieties, as there is a correspondence between smooth polytopes and ample divisors of smooth toric varieties. Suppose that $X_\Sigma$ is a smooth projective toric variety, and that $\mathcal{L}$ is an ample line bundle on it. Oda's Conjecture is equivalent to the statement that the embedding of $X_\Sigma$ given by $\mathcal{L}$ is projectively normal, or in other words, that the multiplication map
$$
    H^0 ( X_\Sigma, \mathcal{L}) \otimes ... \otimes H^0 ( X_\Sigma, \mathcal{L}) \to 
    H^0 ( X_\Sigma, \mathcal{L}^{\otimes k}),
$$
is surjective (\cite{hibi2008mini}). 
%There is considerable interest in algebraic geometry to prove this conjecture, and the geometric formulation may be more tractable.


The interest in the consequences of smoothness is not limited to the IDP, but includes stronger properties such as the existence of a unimodular covering or triangulation.  Indeed, there is a hierarchy of properties, cataloged in \cite{Haase_2021}, of which the IDP is the weakest. And despite attracting considerable interest, including as the subject of an Oberwolfach mini-workshop in 2007, Oda's conjecture remains open, even in three dimensions. As such, even partial or computational results relating to any such properties are of interest, as in \cite{gubeladze2012convex,firla1999hilbert,Lundman_2012,Bogart_2015,haase2010generating}.

In particular, recent progress was made towards proving the conjecture in \cite{2019BecksmoothCS3D}, where Beck et al.\ showed that 3-dimensional, centrally symmetric, smooth polytopes are IDP. In this paper, we contribute other sufficient conditions for smooth polytopes to be IDP. 

%Let $P$ be a $d$-dimensional polytope. We say that $P$ is a \textit{pseudo-prism} if its face poset is isomorphic to that of a prism, $Q \times [0,1]$ for some $(d-1)$-dimensional polytope $Q$, and whose faces corresponding to $Q \times \{0\}$ and $Q \times \{1\}$ are parallel. 


We define a $d$-dimensional \textit{combinatorial cube} to be a polytope whose face poset is in bijection with the face poset of the unit cube, $[0,1]^d$. Then, we prove the following.

% Repeated Theorem
\newtheorem*{ThmSmoothCubesAreIDP_statedEarly}{\Cref{CorSmoothCubesAreIDP}}
\begin{ThmSmoothCubesAreIDP_statedEarly}
    Smooth combinatorial cubes of any dimension are IDP.
\end{ThmSmoothCubesAreIDP_statedEarly}

In the process of proving this, we also define a \textit{pseudo-prism}, a more general class of polytopes with specific characteristics reminiscent of a prism, and show the following.

% Repeated Theorem
\newtheorem*{Thm3DPseudoPrismsIDP_statedEarly}{\Cref{ThmPseudoprism3DareIDP}}
\begin{Thm3DPseudoPrismsIDP_statedEarly}
    Every smooth, 3-dimensional pseudo-prism is IDP.
\end{Thm3DPseudoPrismsIDP_statedEarly}

% We begin in Section 2 with careful definitions of subspace parallelism, a review of known results of IDP polytopes, and precise notation for combinatorial cubes. Then in Section 3, we prove the intermediate result, \Cref{TheoremMain2ParallelFaces}, that every smooth combinatorial cube has two parallel faces. Finally in Section 4, we build to our main result, \Cref{CorSmoothCubesAreIDP}. 

% To do this, we briefly review known results about IDP polytopes in Section 2, as well as familiarize ourselves with combinatorial cubes. Then in Section 3, we prove the intermediate result that every smooth combinatorial cube has two parallel faces in \Cref{TheoremMain2ParallelFaces}. In Section 4, we build to our main result, \Cref{CorSmoothCubesAreIDP}. 

% To do this, we need the intermediate result that all smooth combinatorial cubes have two parallel faces in \Cref{TheoremMain2ParallelFaces}. So 




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Preliminaries
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Preliminaries}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Subspace parallelism
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Subspace parallelism} 

% Recall that a hyperplane is an affine subspace $H$ in $\R^d$ that can be written as the solution space $H = \{ x : A x = b \}$ for some $r \times d$ matrix $A$ and $r$-dimensional vector $b$, where $r = \codim(H) = \rank(A)$. In this paper, we need only determine whether two hyperplanes of the same dimension  in $\Rd$ are parallel.

% The solution space $\{ x: A x = 0 \}$ is the unique linear subspace that is \textit{parallel} to $H$, which we denote by $\lin(H)$; it is the translation of $H$ that contains the origin. We say that two hyperplanes $H_1$ and $H_2$ (of the same dimension) are \textit{parallel} to each other when $\lin(H_1) = \lin(H_2)$. We note that whether two affine spaces are parallel is preserved under linear transformations.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Properties of IDP Polytopes and Unimodular Transformations
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Properties of smooth and IDP polytopes} 

In this paper, we will explore the structure imposed on polytopes when they are smooth; in particular, we will be concerned with whether two faces of a polytope are parallel. In general, every $k$-face $F$ of a polytope lies in a unique $k$-dimensional affine subspace, which we denote with $\aff(F)$. We use $\lin(F)$ to denote the linear subspace parallel to $\aff(F)$, and then two faces $F$ and $G$ of a polytope are \textit{parallel} when $\lin(F) = \lin(G)$. 

This additional structure that we find in certain classes of smooth polytopes will allow us to consider the integer decomposition property. We will employ the following well-known facts.

\begin{proposition} \label{PropIDPBasics} 
    Basic IDP properties:
    \begin{enumerate} 
        \item [(a)] \label{PropIDPBasics_a} Let $P$ be a $d$-dimensional polytope. Then, $(P, kP)$ is IDP for all integers $k \geq d-1$. 
        \item [(b)] All polygons are IDP. 
    \end{enumerate}
\end{proposition} 

\begin{proof}
    A proof of (a) can be found in \cite[Theorem 2.2.12]{cox2011toric}, and was originally proved in \cite{viet1997normal}. The result of (b) follows from (a), or alternatively is shown in \cite{koelman1993generators}.
\end{proof}

We also have the following useful characterization of being IDP, which we use repeatedly.

\begin{proposition} [IDP Equivalence] \label{PropIDPEquiv}
    Let $P$ and $Q$ be polytopes. For a lattice point $a \in \Zd$, define
    $$
        R_a = P \cap (a + (-Q)).
    $$
    Then, $(P, Q)$ is IDP if and only if for all $a$, $R_a$ contains a lattice point whenever it is non-empty.
\end{proposition}


In this paper, we rely heavily on the use of \textit{unimodular transformations}, which are linear maps $\Rd \to \Rd$ which send the lattice $\Z^d$ to itself. Equivalently, a linear transformation is unimodular if and only if the determinant of the matrix representing it is $\pm 1$.
Importantly, for every lattice basis, there exists a unimodular transformation which sends it to the standard basis, and follows that unimodular transformations preserve the IDP.

%Every polytope $P$ is equipped with a \textit{normal fan}, $N(P)$, and if a polytope $Q$ has the same normal fan as $P$, we say $P$ and $Q$ are \textit{Minkowski equivalent}. 

It will be relevant to consider when two polytopes are \textit{Minkowski equivalent}, and we will use the following characterization of this property.

\begin{proposition} \label{PropMEquivWhenAllFacetsParallel}
    Polytopes $P$ and $Q$ are Minkowski equivalent if and only if there is a bijection between their face posets such that all corresponding facets are parallel to each other.
\end{proposition}

Since linear transformations preserve the parallelism of affine subspaces, it follows that they also preserve Minkowski equivalence. We will particularly apply this fact to unimodular transformations.

% \begin{proposition} \label{PropUniModPreserveMEquiv}
%     Unimodular transformations preserve Minkowski equivalence.
% \end{proposition} 

There are many geometric implications of being Minkowski equivalent; in particular, we have the following.

\begin{lemma} \label{LemmaMEquivDisjointHyperplane}
    Suppose polytopes $P$ and $-Q$ in $\Rd$ are Minkowski equivalent. If $P$ and $Q$ are disjoint, then there is a hyperplane which separates them that is parallel to a facet of $P$.
\end{lemma}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Combinatorial Cubes
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Combinatorial cubes} 

Consider $C$, an arbitrary smooth $d$-dimensional combinatorial cube. Importantly, each face of a combinatorial cube is itself a combinatorial cube of smaller dimension.

In order to discuss the faces of $C$, we will employ the bijection between the face posets of $C$ and the $d$-dimensional unit cube, $[0,1]^d$. We can label the faces of the unit cube as the $F_I^J$, where $I$ and $J$ are every pair of disjoint subsets of $[d]$. Then we define
$$
    F _I ^J := 
        \Bigg \{ 
            (x_1, x_2, ..., x_d) \in [0,1]^d : 
                x_k = 
                    \begin{cases} 
                        0 & \text{ if } k \in I \\ 
                        1 & \text{ if } k \in J
                    \end{cases}
        \Bigg \},
$$
so that $F_I^J$ is a $(d-(|I|+|J|))$-dimensional face of the unit cube. Via the bijection, we reuse this face labeling for the \textit{faces of} $C$, so that the $F_I^J$ uniquely identify every face of $C$. 

%When it is unambiguous, we will omit the empty set as well as any brackets and commas when writing the sets $I$ and $J$, for example, we'll write $F_{\es}^{\{2,3\}} = F^{23}$.



With this notation, we can also easily recover the following relationships between the faces of a cube.
\begin{proposition}\label{prop:oldnotationcontainment}
Basic properties of faces of cubes.
    \begin{enumerate} 
        \item[(a)] The face $F _{I_1} ^{J_1}$ contains the face $F _{I_2} ^{J_2}$ if and only if $I_1 \sbe I_2$ and $J_1 \sbe J_2$.

        \item[(b)] The intersection of two faces $F _{I_1} ^{J_1}$ and $F _{I_2} ^{J_2}$ is the face $F_{I_1 \cup I_2}^{J_1 \cup J_2}$.  
    \end{enumerate}
\end{proposition}

However, when it is appropriate, we prefer a more concise notation for $F_I^J$: we denote elements of $J$ with a bar, rather than using the superscript. Instead of writing $F_{\{y\}}^{\{x, z\}}$, we may write $F_{\bar x y \bar z}$.

% \begin{figure}
%   \centering
%   \includegraphics[width=0.5\textwidth]{plot}
%   \caption{A plot of a function.}
%   \label{fig:plot}
% \end{figure}

\begin{figure}[H]
    \centering
    \begin{subfigure}[t]{0.46\textwidth}
        \resizebox{\linewidth}{!}{
            \includegraphics{tikz_diagrams/cubeview2.pdf}
        }
    \end{subfigure}
    \hfill
    \begin{subfigure}[t]{0.46\textwidth}
        \resizebox{\linewidth}{!}{
            \includegraphics{tikz_diagrams/cubeview1.pdf}
        }
    \end{subfigure}
    \caption{The 3-dimensional unit cube, viewed from the `inside' and `outside.'}
    \label{fig:standard3dcube}
\end{figure}

% \begin{figure}[H]
%     \centering
%     \begin{subfigure}[t]{0.45\textwidth}
%         \resizebox{\linewidth}{!}{
%             \subimport{./tikz_diagrams/}{cubeview2.tex}
%         }
%     \end{subfigure}
%     \hfill
%     \begin{subfigure}[t]{0.45\textwidth}
%         \resizebox{\linewidth}{!}{
%             \subimport{./tikz_diagrams/}{cubeview1.tex}
%         }
%     \end{subfigure}
%     \caption{The 3-dimensional unit cube, viewed from the `inside' and `outside.'}
%     \label{fig:standard3dcube}
% \end{figure}


For example, as in \Cref{fig:standard3dcube}, when $C$ is 3-dimensional the six (2-dimensional) facets of $C$ are $F_1$, $F_{\bar{1}}$, $F_2$, $F_{\bar{2}}$, $F_3$, $F_{\bar3}$. As in \Cref{prop:oldnotationcontainment}, $F_1$ contains the 1-dimensional faces $F_{12}, F_{13}, F_{1 \bar 2}, F_{1 \bar3}$, and the intersection of the 2-dimensional faces $F_1$ and  $F_{\bar 2}$ is the 1-dimensional face $F_{1 \bar2}$.

% and the twelve 1-dimensional faces are $F_{12}$, $F_{1 \bar2}$, $F_{\bar1 2}$, $F_{\bar1 \bar2}$, $F_{13}$, $F_{1 \bar3}$, $F_{\bar1 3}$, $F_{\bar1 \bar3}$, $F_{23}$, $F_{2 \bar3}$, $F_{\bar2 3}$, $F_{\bar2 \bar3}$

In cubes of all dimensions there is a notion of opposite facets, and our notation indicates this: We say that two facets $F_x$ and $F_{\bar x}$ of $C$ are \textit{opposite} each other. Since faces of combinatorial cubes are themselves cubes, we have that $F_{x y}$ and $F_{x \bar y}$ are \textit{opposite} each other within $F_x$. In a full-dimensional cube, two facets can be parallel only if they are opposite.

As $C$ is smooth, the primitive edge directions at each vertex of $C$ span the integer lattice, and there always exists a unimodular transformation which sends these directions to the standard basis vectors. Thus, since unimodular transformations and translations preserve the IDP and subspace parallelism, we may assume that one corner of $C$ lies at the origin and has primitive edge directions along the coordinate axes, $e_1, e_2, ..., e_d$. We note, however, that not all of the faces of $C$ will inherit this placement.

%In general, combinatorial cubes in this paper will be assumed to have this property, though we note that not all of the faces of $C$ will inherit it. 

%In particular, we call a facet $F_I^J$ of $C$ \textit{primary} if $J = \es$, that is, when it lies entirely within a coordinate (linear) subspace, and so contains the origin. We note for future use that a primary face $F_I$ lies in $\spann( \{e_k : k \in [d], k \notin I \})$.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Section: Parallelism in combinatorial cubes
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Parallelism in combinatorial cubes}



% \subsection{Properties of parallel subspaces} 

% We begin by distinguishing a few basic results of purely linear algebra which will be useful.

% \begin{lemma} \label{LemmaH1H2Parallel}
%     Suppose that $H_1$ and $H_2$ are $k$-dimensional affine subspaces in $\Rd$ and that each contains two $(k-1)$-dimensional, non-parallel, affine subspaces: $F_1, G_1 \sbe H_1$ and $F_2, G_2 \sbe H_2$. Suppose also that that $F_1$ is parallel to $F_2$ and $G_1$ is parallel to $G_2$, as pictured in \Cref{fig:lemma3_1}. Then, $H_1$ is parallel to $H_2$.
% \end{lemma}

% \begin{figure}[H]
%     \centering
%         \resizebox{0.6\linewidth}{!}{
%             \subimport{./tikz_diagrams/}{lemma3_1.tex}
%         }
%     \caption{In $\R^3$, $H_1$ and $H_2$ are planes while $F_1, F_2, G_1$, and $G_2$ are lines.}
%     \label{fig:lemma3_1}
% \end{figure}

% \begin{proof}
%     Since $F_1$ is not parallel to $G_1$, it must be that $\lin(F_1) \neq \lin(G_1)$, and so for dimensional reasons, 
%     $$
%         \spann(\lin(F_1), \lin(G_1)) = \lin(H_1).
%     $$
%     Similarly, 
%     $$
%         \spann(\lin(F_2), \lin(G_2)) = \lin(H_2).
%     $$
%     By hypothesis, $\lin(F_1) = \lin(F_2)$ and $\lin(G_1) = \lin(G_2)$, so it follows that
%     $$
%         \spann(\lin(F_1), \lin(G_1)) = \spann(\lin(F_2), \lin(G_2)).
%     $$ 
%     Thus, $\lin(H_1) = \lin(H_2)$, so $H_1$ and $H_2$ are parallel.
% \end{proof}

% % ``Degenerate'' case
% \begin{corollary} \label{CorDegenerateCase}
%      Let $C$ be a d-dimensional smooth combinatorial cube for $d \geq 3$, and let $x,y,z \in [d]$ be distinct. Suppose that the pair of faces $(F_{x z},F_{x \bar z})$ are parallel and the pair of faces $(F_{y z},F_{y \bar z})$ are parallel. Then, $F_z$ is parallel to $F_{\bar z}$.
% \end{corollary} 
% \begin{proof}
%     As $F_{x z}, F_{y z} \sbe F_z$ and $F_{x \bar z}, F_{y, \bar z} \sbe F_{\bar z}$, by a direct application of \Cref{LemmaH1H2Parallel}, $F_z$ is parallel to $F_{\bar z}$.
% \end{proof}

% % \begin{lemma} \label{LemmaIntersectionParallel}
% %     Let $F$ and $G$ be two non-parallel, $(d-1)$-dimensional affine hyperplanes in $\Rd$, with $d \geq 3$. Suppose that they contain $(d-2)$-dimensional affine subspaces $F'$ and $G'$, respectively, which are parallel to each other. Then, the intersection of $F$ and $G$ is parallel to $F'$ (and $G'$) also.
% % \end{lemma}

% % \begin{proof}
% %     Let $H = F \cap G$, nonempty since $F$ and $G$ must intersect. That $F'$ and $G'$ are parallel means that $\lin(F') = \lin(G')$, so let $L$ be this $(d-2)$-dimensional linear subspace. Then, to show that $H$ is parallel to $F'$ (and $G'$), we must show that $\lin(H) = L$.

% %     First, since $H = F \cap G$, we have that $\lin(H) = \lin(F \cap G) = \lin(F) \cap \lin(G)$.

% %     Then, since $F' \sbe F$ and $G' \sbe G$, it follows that $L = \lin(F') \sbe \lin(F)$ and $L = \lin(G') \sbe \lin(G)$. Thus, $L \sbe \lin(F) \cap \lin(G) = \lin(H)$.

% %     Since $\dim(H) = \dim(\lin(H)) \leq d-2$, it must be that $L = \lin(H)$. 
% %     %\red{(I think that making F and G be non-parallel, $\dim H$ must be exactly $d-2$?) but $\leq$ is fine here)}
% % \end{proof}

% \begin{lemma} \label{LemmaIntersectionParallel}
%     Let $H_1$ and $H_2$ be two non-parallel, $(d-1)$-dimensional affine hyperplanes in $\Rd$, with $d \geq 3$. Suppose that they contain $(d-2)$-dimensional affine subspaces $F_1$ and $F_2$, respectively, which are parallel to each other. Then, the intersection of $H_1$ and $H_2$ is parallel to $F_1$ (and also $F_2$).
% \end{lemma}

% \begin{figure}[H]
%     \centering
%         \resizebox{0.5\linewidth}{!}{
%             \subimport{./tikz_diagrams/}{lemma3_3.tex}
%         }
%     \caption{In $\R^3$, $H_1$ and $H_2$ are planes while $F_1$, $F_2$, and $G$ are lines. }
%     \label{fig:lemma3_3}
% \end{figure}

% \begin{proof}
%     Let $G = H_1 \cap H_2$, nonempty since $H_1$ and $H_2$ must intersect, so we have that $F_1, G \sbe H_1$ and $F_2, G \sbe H_2$. Then by the contrapositive of \Cref{LemmaH1H2Parallel}, because $H_1$ is not parallel to $H_2$, it must be that $G$ is parallel to $F_1$ (and so also $F_2$). 
% \end{proof}

% \begin{corollary} \label{CorThreeParallelImpliesFourth}
%     Let $C$ be a d-dimensional smooth combinatorial cube with $d \geq 3$. Fix $x,y \in [d]$ with $x \neq y $, and consider the four $(d-2)$-dimensional faces $F_{xy}$, $F_{x \bar y}$, $F_{\bar x y}$, $F_{\bar x \bar y}$. If three of them are parallel to each other, then so is the fourth. 
%     % obfuscating way to write it:$F_{I_i}^{J_i}$, $i = 1,...,4$, where each $I_i \cup J_i = \{x, y\}$.
% \end{corollary}

% \begin{proof}
%     Suppose that of the four faces $F_{x y}$, $F_{x \bar{y}}$, $F_{\bar{x} y}$, $F_{\bar{x} \bar{y}}$, the first three are parallel. 

%     We see that the fourth face, $F_{\bar x \bar y}$, is precisely the intersection of the facets $F_{\bar x}$ and $F_{\bar y}$. But $F_{\bar x}$ contains $F_{\bar x y}$, and $F_{\bar y}$ contains $F_{x \bar y}$, which are parallel to each other. So, by \Cref{LemmaIntersectionParallel}, $F_{\bar x \bar y}$ is also parallel to $F_{x \bar y}$ and $F_{\bar x y}$ (and so also $F_{x y}$).
% \end{proof}

% %\begin{lemma}
% %    Let $C$ be a d-dimensional smooth combinatorial cube. Fix $x,y \in [d]$ with $x \neq y $, and consider the four $(d-2)$-dimensional faces $F_{xy}, F_x^y, F_y^x, F^{xy}$. If three of them are parallel to each other, then $C$ has two facets which are parallel.
% %\end{lemma}

% %\begin{proof}
% %    Suppose without loss of generality that $x=1$ and $y=2$ and that the three faces $F_12, F^1_2, F^{1 2}$ are parallel. 
% %\end{proof}

In general, a combinatorial cube may be irregular in shape, as in \Cref{fig:wonkycube}. 

\begin{figure}[ht]
    \centering
        \resizebox{0.46\linewidth}{!}{
            \includegraphics{tikz_diagrams/wonkycube.pdf}
        }
    \caption{A 3-dimensional combinatorial cube with no parallel faces.}
    \label{fig:wonkycube}
\end{figure}

However, we will show that smoothness imposes restrictions on the relative positions of the facets of a cube, in the following theorem.

\begin{theorem} \label{TheoremMain2ParallelFaces}
    Let $C$ be a $d$-dimensional smooth combinatorial cube in $\Rd$ with $d \geq 2$. Then, $C$ has two parallel facets. 
\end{theorem}


%Now, we are ready to examine the parallelism of faces in combinatorial cubes. 

To build intuition, let us observe the consequences of assuming smoothness of a cube in two dimensions: Suppose that $C$ is a smooth quadrilateral. 

As above, we can take one corner of $C$ to be at the origin, with primitive edge directions $\{ (1, 0), (0, 1) \}$. Then, since $C$ is smooth, the primitive edge directions at each of the four vertices must span the integer lattice $\Z^2$, so they must be of the form given in \Cref{fig:parallelsides2d}, for some integers $m$ and $n$.

\begin{figure}[ht]
    \centering
        \resizebox{0.5\linewidth}{!}{
            \includegraphics{tikz_diagrams/parallelsides2d.pdf}
        }
    \caption{2-dimensional cube, with primitive edge directions at each vertex.}
    \label{fig:parallelsides2d}
\end{figure}

Then, since $\{ (-1, -n), (-m, -1) \}$ must span $\Z^2$, it must be that 
$$
    \det \begin{pmatrix}
            -1 & -m\\
            -n & -1
         \end{pmatrix}
    = \pm 1.
$$

It follows easily that one of $n, m$ must be 0, resulting in either vertical or horizontal edges which are parallel.

The higher dimensional result follows inductively, with an examination by cases shown in three dimensions in \Cref{fig:parallelproof}.



% Next, we prove the following lemma, which is a technical tool used in the main theorem.


% \begin{lemma} \label{LemmaCaseA}
%     Let $C$ be a smooth combinatorial cube of dimension $d \geq 3$ and suppose that all smooth combinatorial cubes of dimension smaller than $d$ have two parallel facets. Fix $x,y \in [d]$ with $x \neq y $, and suppose that the four  $(d-2)$-dimensional faces $F_{xy}, F_{x \bar y}, F_{\bar x y}, F_{\bar x \bar y}$ are all parallel to one another. Then, either $F_x$ and $F_{\bar x}$ are parallel or $F_y$ and $F_{\bar y}$ are parallel.
% \end{lemma}

% \begin{proof}
%     Suppose without loss of generality that $x = 1$ and $y = 2$.
    
%     We proceed by induction on $d$. When $d=3$, consider the facet of the cube $F_3$. It has a pair of parallel faces by hypothesis, either $(F_{1 3}, F_{\bar1 3})$ or $(F_{2 3}, F_{\bar2 3})$. By \Cref{CorDegenerateCase}, either $F_1$ and $F_{\bar 1}$ are parallel or $F_2$ and $F_{\bar 2}$ are parallel, respectively.

%     Suppose $d > 3$, and that the result holds for all smaller dimensional cubes. As in the base case, the facet of the cube $F_3$ has a pair of parallel faces $(F_{3 x}, F_{3 \bar x})$ for some $x \neq 3$. If $x = 1$ or $x = 2$, then by \Cref{CorDegenerateCase}, either $F_1$ and $F_{\bar 1}$ are parallel or $F_2$ and $F_{\bar 2}$ are parallel, respectively. So, suppose that $x \neq 1,2,3$; without loss of generality, let $x = 4$.

%     Since $F_{12}, F_{1 \bar 2}, F_{\bar 1 2}, F_{\bar 1 \bar 2}$ are all parallel, their intersections with $F_3$ are also parallel, i.e. $F_{123},$ $F_{1 \bar 2 3},$ $F_{\bar 1 2 3},$ $F_{\bar 1 \bar 2 3}$ are all parallel to each other. Then by the inductive hypothesis, $F_3$ has a pair of parallel faces, which is either $(F_{13}, F_{\bar1 3})$ or $(F_{2 3}, F_{\bar 2 3})$. So suppose that the former are parallel.

%     Then, in $F_3$, there are pairs of parallel faces $(F_{13}, F_{\bar1 3})$ and $(F_{3 4}, F_{3 \bar4})$. It follows that $F_{1 3 4}$ and $F_{\bar1 3 \bar4}$ are parallel. Since $F_1$ contains both $F_{1 3 4}$ and $F_{1 2}$, and $F_{\bar1}$ contains both $F_{\bar1 3 \bar4}$ and $F_{\bar1 2}$, we get that 
%     $$
%         \lin(F_1) = \spann \big( \lin(F_{1 3 4}), \lin(F_{1 2}) \big) 
%                   = \spann \big( \lin(F_{\bar1 3 \bar4}), \lin(F_{\bar1 2}) \big) = \lin(F_{\bar 1}).
%     $$
%     Thus, $F_1$ and $F_{\bar1}$ are parallel. 
    
%     If instead we suppose that $(F_{2 3}$ and $F_{\bar 2 3})$ are parallel, we similarly get that $F_2$ and $F_{\bar2}$ are parallel. Therefore, we conclude that either $F_1$ and $F_{\bar1}$ are parallel or $F_2$ and $F_{\bar2}$ are parallel.
% \end{proof}








% \begin{proof}

%     We proceed by induction on $d$. \Cref{TheoremBaseCase2ParallelFaces} gives us our base case when $d=2$, so let $d \geq 3$ and inductively assume that each smaller dimensional such cube has two parallel facets. As usual, we may assume that one corner of $C$ lies at the origin and has primitive edge directions $e_1, e_2, ..., e_d$. A priori, a smooth 3-dimensional combinatorial cube might look as in \Cref{fig:wonkycube}, before any parallelism properties of the inductive hypothesis have been applied.

%     Consider the $d$ primary facets of the cube, $F_1, F_2, ..., F_d$, each of which is a $(d-1)$-dimensional cube itself. By our inductive hypothesis, each of them contains a pair of parallel $(d-2)$-faces. So, let us say that for each $i \in [d]$, $F_i$ contains a pair of parallel faces $(F_{i k_i}, F_{i \bar{k_i}})$ for some $k_i \in [d]$. Notice that one face of each pair, $F_{i k_i}$, is primary, and the other, $F_{i \bar{k_i}}$, is non-primary.

%     We first consider a straightforward case: Suppose that there are distinct indices $i,j \in [d]$ such that $k_i = k_j =: k$, as in \Cref{fig:parallelproof}\subref{subfig:straightforward}. (Note that $k \neq i,j$.) Then, by \Cref{CorDegenerateCase}, it must be that $F_k$ and $F_{\bar k}$ are parallel.

%     Now, assume the opposite, i.e. that all of the $k_i$'s are distinct, as in \Cref{fig:parallelproof}\subref{subfig:pinwheel}. It follows that for each $k \in [d]$ that there is a unique $i$ such that $k_i = k$.

%     Consider the non-primary facets of the cube, $F_{\bar1}, F_{\bar2}, ..., F_{\bar d}$. By the inductive hypothesis, they also contain pairs of parallel $(d-2)$-faces. So, let us say that for each $i \in [d]$, $F_{\bar i}$ contains the pair of parallel faces $(F_{\bar i {m_i}}, F_{\bar i \bar{m_i}})$ for some $m_i \in [d]$. 

%     We must now be careful to consider different cases. 

%     \vspace{2mm} \noindent \textbf{Case (a)}: There is a non-primary facet $F_{\bar j}$ such that in the primary facet $F_{m_j}$, the pair $(F_{j m_j}, F_{\bar j {m_j}})$ are parallel. 

%     Without loss of generality, let us take $j = 1$ and $m_j = 2$, so that $F_{\bar1}$ has parallel pair $(F_{\bar1 2}, F_{\bar1 \bar2})$ and $F_2$ has parallel pair $(F_{1 2}, F_{\bar1 2})$. Thus, the three faces $F_{1 2}$, $ F_{\bar1 2}$, and $F_{\bar1 \bar2}$ are all parallel. By \Cref{CorThreeParallelImpliesFourth}, $F_{1 \bar2}$ is parallel to them as well. 
    
%     We see that this case is actually impossible if $d = 3$: Reconsidering the primary facets, the above tells us that $F_1$ has parallel pair $(F_{1 2}, F_{1 \bar2})$ and $F_2$ has parallel pair $(F_{1 2}, F_{\bar 1 2})$. So it is impossible for $k_1, k_2, k_3$ to be distinct, as the parallel faces in $F_3$ must either be $(F_{1 3}, F_{\bar1 3})$ or $(F_{2 3}, F_{\bar2 3})$. So, suppose that $d \geq 4$. Then, by \Cref{LemmaCaseA}, either $F_1$ and $F_{\bar1}$ are parallel or $F_2$ and $F_{\bar2}$ are parallel.






%     \vspace{2mm} \noindent \textbf{Case (b)}: For every $i \in [d]$,  $F_{\bar i {m_i}}$ is not one of the parallel faces in $F_{m_i}$.  
    
%     Without loss of generality, let us again take $F_{\bar1}$ to have parallel pair $(F_{\bar1 2}, F_{\bar1 \bar2})$. We will further subdivide into more cases, and this division has a geometric interpretation shown in \Cref{fig:parallelproof}\subref{subfig:casebi} and \Cref{fig:parallelproof}\subref{subfig:casebii}, respectively. 

%     \vspace{2mm} \noindent \textbf{Case (b)(i)}: $\lin(F_{\bar1 2}) = \spann(e_3, e_4, ..., e_d)$. 
    
%     By assumption, there is an $x \neq 2$ such that in $F_x$, the faces $F_{1x}$ and $F_{\bar1 x}$ are parallel. Without loss of generality, we can take $x = 3$. Then, since $F_{13}$ and $F_{\bar1 3}$ are parallel and $F_{13}$ is primary, 
%     $$
%         \lin(F_{\bar1 3}) = \lin(F_{13}) = \spann(e_2, e_4, e_5 ..., e_d).
%     $$    
%     But since $\lin(F_{\bar1 2})$ and $\lin(F_{\bar1 3})$ are both in $\lin(F_{\bar1})$ and not equal to each other, for dimensional reasons 
%     $$
%         \spann( \lin(F_{\bar1 2}), \lin(F_{\bar1 3}) ) = \spann{(e_2, e_3, ..., e_d)} = \lin(F_{\bar1}),
%     $$
%     and since the middle expression is precisely $\lin(F_1)$, we have that $F_1$ and $F_{\bar1}$ are parallel.

%     \vspace{2mm} \noindent \textbf{Case (b)(ii)}: $\lin(F_{\bar1 2}) \neq \spann(e_3, e_4, ..., e_d)$. 
    
%     It still holds that 
%     $$
%         \lin(F_{\bar1 \bar2}) = \lin(F_{\bar1 2}) \sbe \lin(F_2) = \spann(e_1, e_3, e_4, ..., e_d).
%     $$
%     By assumption, there is a $y \neq 1$ such that the facet $F_y$ has the pair of parallel faces $(F_{2 y}, F_{\bar2 y})$. Without loss of generality, we can take $y = 3$, and then $\lin(F_{\bar2 3}) = \lin(F_{2 3}) = \spann(e_1, e_4, e_5, ..., e_d)$.

%     Since $\dim(F_{\bar1 \bar2}) = d-2$, we have that 
%     $$
%         \lin(F_{\bar1 \bar2}) = \spann(v_1, v_2, ..., v_{d-2})
%     $$ 
%     where each $v_k \in \spann(e_1, e_3, e_4, ..., e_d)$. However, we observe two things: First, that these vectors can't be such that $\lin(F_{\bar1 \bar2}) = \spann(e_3, e_4, ..., e_d)$ by assumption. Second, we can't have $\lin(F_{\bar1 \bar2}) = \spann( \{e_1, e_3, e_4, ..., e_d\} \sm \{e_p\})$ for any $p \in \{3, 4, ..., d\}$, otherwise $\lin(F_{1 \bar2})$ would equal $\lin(F_{p \bar2})$, collapsing the face $F_{\bar2}$. So in particular, we know that there is at least one $v_k$ which can be written as a linear combination of $e_1, e_3, e_4, ..., e_d$ with a nonzero coefficient for $e_3$. Then,
%     \begin{align*}
%         \spann(\lin(F_{\bar1 \bar2}), \lin(F_{\bar2 3})) 
%             &= \spann(e_1, e_4, e_5, ..., e_d, v_1, v_2, ..., v_{d-2}) \\
%             &= \spann(e_1, e_3, e_4, ..., e_d) \\
%             &= \lin(F_2).
%     \end{align*}
%     Lastly, since $F_{\bar1 \bar2}, F_{\bar2 3} \sbe F_{\bar2}$, we have that $\spann(\lin(F_{\bar1 \bar2}), \lin(F_{\bar2 3}))= \lin(F_{\bar2})$. So putting these together, we get that $\lin(F_2) = \lin(F_{\bar2})$, and thus that $F_2$ and $F_{\bar2}$ are parallel.

%     Therefore, in all cases, there are two facets of $C$ which are parallel.
% \end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Section: IDP stuff
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{IDP in prisms, pseudo-prisms, and cubes}

In the process of showing that combinatorial cubes of every dimension are IDP, we produce several intermediate results about a more general class of polytopes.

A $d$-dimensional \textit{prism} is a polytope which is affinely equivalent to a polytope $Q \times [0,1]$ for some $(d-1)$-dimensional polytope $Q$. This yields \textit{top} and \textit{bottom} facets, $Q \times \{1\}$ and $Q \times \{0\}$, respectively, which are parallel to each other. Thus, there are edges between the corresponding vertices of the top and bottom, and so all other facets of a prism, the \textit{side} facets, are themselves $(d-1)$-dimensional prisms, whose tops and bottoms are $(d-2)$-faces of the top and bottom of the original prism. 

We define a \textit{pseudo-prism} to be a polytope whose face lattice is isomorphic to that of a prism and whose corresponding top and bottom facets are parallel. Observe that this means that the top and bottom faces of a pseudo-prism have the same face poset, so in particular have the same number of vertices. Further note that by \Cref{TheoremMain2ParallelFaces}, every $d$-dimensional smooth combinatorial cube is a pseudo-prism, where $Q$ is a $(d-1)$-dimensional smooth combinatorial cube.

It is convenient to visualize pseudo-prisms whose top and bottom facets are parallel to the coordinate plane $\{(x_1, ..., x_d) : x_d = 0\}$. Again, since unimodular transformations and translations preserve the IDP, Minkowski equivalence, and subspace parallelism, we may assume that pseudo-prisms have tops and bottoms parallel to this coordinate plane, and further that smooth pseudo-prisms have one vertex at the origin, with primitive edge directions $e_1, e_2, ..., e_d$.


\begin{figure}[H]
    \centering
    \begin{subfigure}[t]{0.44\textwidth}
        \resizebox{\linewidth}{!}{
            \includegraphics{tikz_diagrams/straightforwardcase.pdf}
        }
        \caption{The faces $F_{1 \bar3}$, $F_{13}$, and $F_{\bar1 3}$ are all parallel to each other.}
        \label{subfig:straightforward}
    \end{subfigure}
    \hfill
    \begin{subfigure}[t]{0.44\textwidth}
        \resizebox{\linewidth}{!}{
            \includegraphics{tikz_diagrams/pinwheel.pdf}
        }
        \caption{The pairs of faces $(F_{12}, F_{1 \bar2})$, $(F_{13}, F_{\bar1 3})$, and $(F_{23}, F_{2 \bar3})$ are parallel.}
        \label{subfig:pinwheel}
    \end{subfigure}

% \end{figure}
% \begin{figure}[ht]\ContinuedFloat
%     \centering
    \begin{subfigure}[t]{0.44\textwidth}
        \resizebox{\linewidth}{!}{
            \includegraphics{tikz_diagrams/case2bi.pdf}
        }
        \caption{The pairs of faces $(F_{\bar1 2}, F_{\bar1 \bar2})$, $(F_{1 \bar3}, F_{\bar1 \bar3})$, and $(F_{\bar2 3}, F_{\bar2 \bar3})$ are also parallel, and \\ \centering{$\lin(F_{\bar1 2}) = \spann(e_3)$.}}
        \label{subfig:casebi}
    \end{subfigure}
    \hfill
    \begin{subfigure}[t]{0.44\textwidth}
        \resizebox{\linewidth}{!}{
            \includegraphics{tikz_diagrams/case2bii.pdf}
        }
        \caption{The pairs of faces $(F_{\bar1 2}, F_{\bar1 \bar2})$, $(F_{1 \bar3}, F_{\bar1 \bar3})$, and $(F_{\bar2 3}, F_{\bar2 \bar3})$ are also parallel, and  \\ \centering{$\lin(F_{\bar1 2}) \neq \spann(e_3)$.}}
        \label{subfig:casebii}
    \end{subfigure}
    
    \caption{A 3-dimensional combinatorial cube, with various parallel faces.}
    \label{fig:parallelproof}
\end{figure}



It is clear that the top and bottom facets of $d$-dimensional prisms are translations of the same $(d-1)$-dimensional polytope. While this is not true in general for pseudo-prisms, there is a strong relationship between their top and bottom facets, which we give in the following lemma. 

\begin{lemma} \label{LemmaTopBottomComboPrismsMinkowskiEquiv}
    The top and bottom of a pseudo-prism are Minkowski equivalent. 
\end{lemma}

% \begin{proof}
%     Let $P$ be a pseudo-prism of dimension $d$ with bottom facet $B$ and top facet $T$, which lie in the planes $\{(x_1, x_2, ..., x_d) : x_d = 0\}$ and $\{(x_1, x_2, ..., x_d) : x_d = h\}$ for some integer $h$, respectively. 

%     The the facets of the top and bottom of $P$ (which are $(d-2)$-faces of $P$), are precisely the intersections of $T$ and $B$ respectively with the side facets of $P$: every facet of $T$ corresponds to the facet of $B$ which is the intersection with the same side facet. Thus, there is a bijection between the facets of $T$ and $B$.
    
%     Letting corresponding facets of the top and bottom be $F_T$ and $F_B$, there is some side facet $F$ of $P$ such that $F_T = T \cap F$, and $F_B = B \cap F$. Then, since $T$ and $B$ are parallel,
%     $$
%         \lin(F_T) = \lin(F) \cap \lin(T) = \lin(F) \cap \lin(B) = \lin(F_B),
%     $$
%     so $F_T$ and $F_B$ are parallel. Thus, since this holds for every pair of corresponding facets, it follows from \Cref{PropMEquivWhenAllFacetsParallel} that $T$ and $B$ are Minkowski equivalent.
% \end{proof}

Supposing that a pseudo-prism $P$ has bottom which lies in $\{(x_1, ..., x_d) : x_d = b\}$ and top which lies in $\{(x_1, ..., x_d) : x_d = b + h\}$ for integers $b$ and $h$, we define \textit{slices} $S_l$ of $P$ as the nonempty intersections
$$
    S_l = P \cap \{(x_1, ..., x_d) : x_d = b + l\}
$$
for heights $l \in [h]$. It is not hard to show that each slice of a smooth pseudo-prism has only integer vertices, and the following lemma is a natural consequence.

\begin{lemma} \label{LemmaSlicesAreMEquiv}
    If $P$ is a smooth pseudo-prism of dimension $d$, then every slice of $P$ is an integer polytope of dimension $d-1$ and is Minkowski equivalent to its bottom (and so also its top). 
\end{lemma}

% \begin{proof}
%     Let $P$ be a smooth pseudo-prism of dimension $d$ with top $T$ and bottom $B$ and let $n$ be the number of vertices each. We will call the edges which connect corresponding vertices in $B$ and $T$ semi-vertical edges: For $i \in [n]$, let $b_i$ be a vertex in $B$, $t_i$ be a vertex in $T$, and $e_i$ the semi-vertical edge connecting $b_i$ and $t_i$. (Since $P$ is a pseudo-prism, the $e_i$ do not intersect each other.)

%     The semi-vertical edges are the only edges at each vertex which have a primitive edge direction with a non-zero $d^{\text{th}}$ coordinate. So consider one such edge, $e_i$: since $P$ is smooth, the primitive edge direction out of $e_i$ must be $(u_1, u_2, ..., u_{d-1}, 1)$ for some integers $u_j$. Then, for each integer height $h' \in [h]$, $e_i$ intersects the hyperplane $\{(x_1, ..., x_d) : x_d = h'\}$ at the integer point $b_i + h'(u_1, u_2, ..., u_{d-1}, 1)$. 
    
%     Since this holds for all semi-vertical edges, we see that a slice of $P$,  $T_{h'} = P \cap \{(x_1, ..., x_d) : x_d = h'\}$ for some integer $h'$, has only integer vertices, and so is $(d-1)$-dimensional polytope which lies in the hyperplane $\{(x_1, ..., x_d) : x_d = h'\}$. 

%     Then, for each $h' \in [h]$, the $d$-dimensional polytope $P \cap \{(x_1, ..., x_d) : b \leq x_d \leq b + h'\}$ is a pseudo-prism with bottom $B$ and top $T_{h'}$. Further, it is smooth, as at every vertex in $T_{h'}$, the primitive edge directions are the same, up to sign, as those of $T$. So by \Cref{LemmaTopBottomComboPrismsMinkowskiEquiv}, $T_{h'}$ is Minkowski equivalent to $B$ (and $T$).

%     Thus, every slice of $P$ is an integer polytope of dimension $d-1$ and is Minkowski equivalent to $B$ (and $T$).
% \end{proof}

%As we saw in \Cref{PropPkPisIDP}, 
Oda's Conjecture has been proved in only two dimensions. However, another two dimensional result that doesn't require smoothness was proved by Hasse et al.\ in 2007.

\begin{theorem}[\cite{haase2007lattice}] \label{TheoremMEquivPolygonsAreIDP}
     Let $P$ and $P'$ be Minkowski equivalent lattice polygons. Then, $(P, P')$ is IDP.
\end{theorem}

An analogous result for dimensions higher than two is known to be untrue in many cases, for instance, any pair $(P, kP)$ where $P$ is not IDP, such as the Reeve tetrahedron. However, the structure of a pseudo-prism lends itself to the following theorem.
%is somewhat conducive to induction

\begin{theorem} \label{TheoremRecursiveIDP}
    Let $P$ and $P'$ be Minkowski equivalent, smooth pseudo-prisms which are $d$-dimensional. Let $\{S_l\}$ and $\{S_m'\}$ be the slices of $P$ and $P'$ respectively, and suppose that for every $l$ and $m$, the pair $(S_l, S_m')$ is IDP. Then, $(P, P')$ is IDP.
\end{theorem}

\begin{proof}
    Assume without loss of generality that the bottom $B$ and top $T$ of $P$ lie in the respective hyperplanes $\{(x_1, ..., x_d) : x_d = 0\}$ and $\{(x_1, ..., x_d) : x_d = t\}$ for some integer $t$, while the bottom $B'$ and top $T'$ of $P'$ lie in the hyperplanes $\{(x_1, ..., x_d) : x_d = b'\}$ and $\{(x_1, ..., x_d) : x_d = t'\}$ for integers $b' > t'$, respectively.
    
    To show that $(P, P')$ is IDP, we will use \Cref{PropIDPEquiv}. Suppose that $R_a = P \cap (a + (-P'))$ is non-empty for some point $a = (a_1, ..., a_d) \in \Z^d$.

    We see that $a + (-P')$ has top $a + (-B)$ which is contained in $\{(x_1, ..., x_d): x_d = a_d - b' \}$ and bottom $a + (-T)$ which is contained in $\{(x_1, ..., x_d): x_d = a_d -t' \}$. All of $a + (-P')$ lies between these two hyperplanes, and all of $P$ lies between $\{(x_1, ..., x_d): x_d = 0 \}$ and $\{(x_1, ..., x_d) : x_d = t\}$. 
    
    In order for $R_a$ to be non-empty, the top or bottom of at least one of $P$ or $a+(-P')$ lies in a hyperplane which is vertically between the top and bottom hyperplanes of the other. So, suppose without loss of generality that $B$ lies between the top and bottom hyperplanes of $a + (-P')$. This means that there is a slice $S_m'$ of $P'$ such that the slice $G := a +(-S_m')$ of $a + (-P')$ lies in the hyperplane $\{(x_1, ..., x_d): x_d = 0\}$. 
    
    Suppose towards contradiction that $B$ itself does not intersect $G$. 
    As $B$ and $G$ are both $(d-1)$-dimensional polytopes, consider them only in the subspace $\{(x_1, ..., x_d): x_d = 0\}$. Since they are Minkowski equivalent, \Cref{LemmaMEquivDisjointHyperplane} gives that there is a $(d-2)$-dimensional hyperplane $\Tilde{H} \sbe \{(x_1, ..., x_d): x_d = 0 \}$ which separates $B$ and $G$. 
    Further, this Proposition gives us that $\Tilde{H}$ is parallel to a $(d-2)$-face $\Tilde{B}$ of $B$ and so is also parallel to the corresponding face $\Tilde{G}$ of $G$. 

    Since $P$ is a pseudo-prism and $\Tilde{B}$ is a face of $B$, we know that $P$ has a facet $F$ such that $\Tilde{B} = B \cap F$, a $(d-2)$-face of $P$. As $P'$ is Minkowski equivalent to $P$, it has corresponding facet $F'$ which is parallel to $F$, and $\Tilde{G} = G \cap (a + (-F'))$.
    
    In particular, we have that $F$ and $a + (-F')$ are parallel and non-equal since $\Tilde{B}$ and $\Tilde{G}$ are disjoint. Thus, there is a hyperplane $H$ which is parallel to both $F$ and $a + (-F')$ which contains $\Tilde{H}$. However, this $H$ must separate $C$ and $a + (-P')$, contradicting our supposition that $R_a$ is non-empty. Therefore, it must be that $B$ intersects $G$, as in \Cref{fig:slices_overlap}.

    \begin{figure}[H]
        \centering
            \resizebox{0.6\linewidth}{!}{
                \includegraphics{tikz_diagrams/prisms_slice_overlap.pdf}
            }
        \caption{Slices $B$ of $P$ and $G$ of $a + (-P')$.}
        \label{fig:slices_overlap}
    \end{figure}

    By assumption, $(B, S_m')$ is IDP. By \Cref{PropIDPEquiv}, since $B \cap G = B \cap (a +(-S_m'))$ is nonempty, it must contain a lattice point. Thus, $R_a$ contains a lattice point, so we conclude that $(P, P')$ is IDP.
\end{proof}

By employing \Cref{TheoremMEquivPolygonsAreIDP}, we satisfy the required hypotheses of \Cref{TheoremRecursiveIDP} when we restrict to 3-dimensional smooth pseudo-prisms, and we can conclude the following.

\begin{theorem} \label{Theorem3DMEquivPseudoPrismsAreIDP}
    Let $P$ and $P'$ be Minkowski equivalent, smooth, 3-dimensional pseudo-prisms. Then, $(P, P')$ is IDP.
\end{theorem}

% \begin{proof}
%     Suppose that $P$ has bottom $B$ and the top $T$, and $P'$ has bottom $B'$ and top $T'$. That $P$ and $P'$ are Minkowski equivalent means that $B$ and $B'$ are also.

%     As in \Cref{LemmaSlicesAreMEquiv}, let $\{S_l\}$ be the slices of $P$, and $\{S_m'\}$ the slices of $P'$. Since $P$ and $P'$ are Minkowski equivalent, by \Cref{LemmaSlicesAreMEquiv} we have that all slices of each, including their tops and bottoms, are Minkowski equivalent to both $B$ and $B'$, and therefore to each other.

%     Then, by \Cref{TheoremMEquivPolygonsAreIDP}, it must be that for every $l$ and $m$, the pair $(S_l, S_m')$ is IDP. Therefore, \Cref{TheoremRecursiveIDP} gives that $(P, P')$ is IDP.
% \end{proof}

\begin{theorem} \label{ThmPseudoprism3DareIDP}
    Every smooth, 3-dimensional pseudo-prism is IDP.
\end{theorem}

% \begin{proof}
%     Let $P$ be a smooth, 3-dimensional pseudo-prism. By \Cref{PropIDPBasics}(a) it suffices to show that $(P, P)$ is IDP, and clearly $P$ is Minkowski equivalent to itself. Thus by \Cref{Theorem3DMEquivPseudoPrismsAreIDP}, $P$ is IDP. 
% \end{proof}

Combinatorial cubes are highly structured polytopes, and as such are conducive to induction. This allows us to satisfy the hypotheses of \Cref{TheoremRecursiveIDP} in any dimension, yielding the following results.

\begin{theorem} \label{TheoremTwoCubesMEquivIDP}
    Suppose that $C$ and $C'$ are Minkowski equivalent smooth cubes of dimension $d \geq 2$. Then, $(C, C')$ is IDP.
\end{theorem}

% \begin{proof}
%     We proceed by induction on $d$. When $d = 2$, \Cref{TheoremMEquivPolygonsAreIDP} provides the result, so suppose that $d > 2$ and that for pairs $(P, Q)$ of Minkowski equivalent smooth $(d-1)$-cubes that $(P, Q)$ is IDP.

%     Since $C$ is smooth, by \Cref{TheoremMain2ParallelFaces} it has two parallel facets, and so is a pseudo-prism, with top $T$ and bottom $B$. Similarly, $C'$ has top $T'$ and bottom $B'$.
    
%     Let $\{S_l\}$ be the slices of $C$ and $\{S_m'\}$ the slices of $C'$. By \Cref{LemmaSlicesAreMEquiv}, as $B$ is a $(d-1)$-dimensional smooth cube, all slices of $C$ are also $(d-1)$-dimensional smooth cubes which are Minkowski equivalent to it. Similarly, all slices of $C'$ are $(d-1)$-dimensional smooth cubes which are Minkowski equivalent to $B'$. As $C$ and $C'$ are Minkowski equivalent to each other, so are $B$ and $B'$, and so it follows that every slice $S_l$ of $C$ is Minkowski equivalent to every slice $S_m'$ of $C'$.
    
%     Then by our inductive assumption, it must be that for every $l$ and $m$, the pair $(S_l, S_m')$ is IDP. Therefore, by \Cref{TheoremRecursiveIDP}, $(C, C')$ is IDP.
% \end{proof}



\begin{theorem} \label{CorSmoothCubesAreIDP}
    Smooth combinatorial cubes of any dimension are IDP. 
\end{theorem}

% \begin{proof}
%     Suppose $C$ is a smooth combinatorial cube of dimension $d$. For any integer $k$, the dilation $kC$ is Minkowski equivalent to $C$. Thus, by \Cref{TheoremTwoCubesMEquivIDP}, $(C, kC)$ is IDP, so we conclude that $C$ is IDP.
% \end{proof}

\section{Future work}

Smooth polytopes have only been shown to be IDP under strong additional assumptions; in its full generality, Oda’s Conjecture remains open. Following these results, it is natural to try to relax the structure of pseudo-prisms and examine polytopes with some weaker characteristic involving parallelism, or to more deeply explore the geometric consequences of smoothness. Alternatively, it is interesting to explore which assumption in \cite{2019BecksmoothCS3D}, 3-dimensional or centrally symmetric, could be shown to be superfluous. Others have classified all smooth, 3-dimensional polytopes which have small numbers of internal lattice points: In \cite{Lundman_2012}, Lundman expanded on the work in \cite{Bogart_2015}, leaving open ample opportunity for exploration. There are stronger properties than the IDP, e.g. the existence of a unimodular triangulation, which might be associated to these or other classes of polytopes.


\acknowledgements{The author is very grateful to Gaku Liu for his guidance and support in all aspects of this research. }


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