Extension complexities of Cartesian products involving a pyramid

TL;DR

This paper proves that for the Cartesian product of two polytopes, if one is a pyramid, the extension complexity of the product equals the sum of the individual complexities, resolving an open question in polytope theory.

Contribution

It establishes that the extension complexity of the Cartesian product of a pyramid and another polytope is additive, confirming the conjecture in this specific case.

Findings

Extension complexity of the product equals sum when one polytope is a pyramid.

Addresses an open question in polytope extension complexity.

Provides a new understanding of Cartesian products involving pyramids.

Abstract

It is an open question whether the linear extension complexity of the Cartesian product of two polytopes P, Q is the sum of the extension complexities of P and Q. We give an affirmative answer to this question for the case that one of the two polytopes is a pyramid.

Equations

$$\mathrm{xc}(P\times Q)=\mathrm{xc}(P)+\mathrm{xc}(Q)$$

$$P$$

$$Q$$

$$P\times Q$$

$$⟨a_i^Q,y⟩\le b_i^{Q}i=1,\dots ,m_Q\}$$

$$=\mathrm{conv}\{(v_i^P,v_j^Q):i\in [n_P],$$

$$r_{+}(A)\ge r_{+}(S)+r_{+}(T^{\prime })+1$$

$$r_{+}(A)\le r_{+}(S)+r_{+}(T^{\prime }),$$

Full text

1 Introduction

For a non-empty polytope $P$, the linear extension complexity of $P$ is defined as the smallest number of facets of any polytope that can be affinely projected onto $P$, and is denoted by $\operatorname{xc}(P)$. Given any non-empty polytopes $P$ and $Q$, one can easily observe that $\operatorname{xc}(P\times Q)\leq\operatorname{xc}(P)+\operatorname{xc}(Q)$, while it is an open question whether this inequality actually holds as an equality, i.e., whether

[TABLE]

holds in general. This question has been asked at several occasions (see, e.g., [3, Conj. 1] or [5, Prob. 3]) but it seems that the most general case in which is it known that (1) holds is when one of the two polytopes is a simplex. The latter fact has been observed by several authors and can be explicitly found in [3, Cor. 10]. In this note, we prove that (1) holds whenever one of the two polytopes is a pyramid (in Section 2 we recall the definition of a pyramid):

Theorem 1**.**

Let $P,Q$ be non-empty polytopes such that one of the two polytopes is a pyramid. Then we have $\operatorname{xc}(P\times Q)=\operatorname{xc}(P)+\operatorname{xc}(Q)$.

While pyramids are still very special polytopes, with respect to linear extensions they are closely related to their bases, which can be arbitrary polytopes. Indeed, given a pyramid $P$ with base $B$ it is easy to see that $\operatorname{xc}(P)=\operatorname{xc}(B)+1$ holds. Thus, although our proof crucially exploits the structure of Cartesian products involving a pyramid, we hope that our result opens doors for further generalizations.

In the next section, we discuss basic ingredients needed for the proof of Theorem 1 while the proof itself is given in Section 3.

2 Preliminaries

A polytope $P\subseteq\mathbb{R}^{d}$ is called a pyramid with base $B\subseteq\mathbb{R}^{d}$ and apex $v\in\mathbb{R}^{d}$ if $P=\operatorname{conv}(B\cup\{v\})$ and $v$ is not contained in the affine hull of $B$. Note that $v$ is contained in every facet of $P$ except for one which contains all remaining vertices of $P$.

Let $P=\{x\in\mathbb{R}^{d}:\langle a_{i},x\rangle\leq b_{i}\ i=1,\dotsc,m\}=\operatorname{conv}\{v_{1},\dotsc,v_{n}\}$ for some $a_{1},\dotsc,a_{m}\in\mathbb{R}^{d}$, $b_{1},\dotsc,b_{m}\in\mathbb{R}$, and $v_{1},\dotsc,v_{m}\in\mathbb{R}^{d}$, where $\langle\cdot,\cdot\rangle$ denotes the Euclidean scalar product of $\mathbb{R}^{d}$. Then the matrix $S\in\mathbb{R}^{m\times n}_{\geq 0}$ defined via $S_{i,j}:=b_{i}-\langle a_{i},v_{j}\rangle$ is called a slack matrix of $P$. A well-known result of Yannakakis [6] states that the linear extension complexity of $P$ is equal to the nonnegative rank of $S$, which is defined as the smallest number $r_{+}(S)$ such that $S$ can be written as the sum of $r_{+}(S)$ nonnegative rank-one matrices. The nonnegative rank $r_{+}(S)$ of a polytope is indeed well defined despite the fact its definition relies on the slack matrix $S$ which, in turn, is defined by a particular linear description of $P$. This follows from the fact that $r_{+}(S)$ neither depends on the scaling of the constraints used to describe $P$ nor on the potential presence of redundant constraints.

Although not needed for this work, the interested reader may consider the surveys [4, 1] and the book chapter [2, Chap. 4] as excellent sources for background information and recent developments on linear extended formulations.

In our proof, we make use of two simple facts about decompositions into nonnegative rank-one matrices: Let $S=R^{1}+\dotsb+R^{k}$ where $R^{1},\dotsc,R^{k}$ are nonnegative rank-one matrices and suppose that $S_{i,j}=0$ holds. First, since all $R^{\ell}$ are nonnegative, this implies $(R^{\ell})_{i,j}=0$ for all $\ell$. Second, since all $R^{\ell}$ have rank one, for every pair of indices $(i^{\prime},j^{\prime})$ and every $\ell$ we must have $(R^{\ell})_{i^{\prime},j}=0$ or $(R^{\ell})_{i,j^{\prime}}=0$.

Given two polytopes $P,Q$ with

[TABLE]

one immediately obtains

[TABLE]

Thus, if $S\in\mathbb{R}^{m_{P}\times n_{P}}_{\geq 0}$ and $T=[t_{1}\dotsb t_{n_{Q}}]\in\mathbb{R}^{m_{Q}\times n_{Q}}_{\geq 0}$ are slack matrices of $P$ and $Q$, respectively, then the matrix

S$$S$$\dotsb$$S$$t_{1}\dotsb t_{1}$$t_{2}\dotsb t_{2}$$\dotsb$$t_{n_{Q}}\dotsb t_{n_{Q}}$$\in\mathbb{R}^{(m_{P}+m_{Q})\times(n_{P}\cdot n_{Q})}_{\geq 0}

is a slack matrix of $P\times Q$, where $t_{1},\dotsc,t_{n_{Q}}\in\mathbb{R}^{m_{Q}}_{\geq 0}$ denote the columns of $T$. The columns of the above slack matrix correspond, from left to right, to the vertices $(v_{1}^{P},v_{1}^{Q}),(v_{2}^{P},v_{1}^{Q}),\ldots,(v_{n_{p}}^{P},v_{1}^{Q}),(v_{1}^{P},v_{2}^{Q}),(v_{2}^{P},v_{2}^{Q}),\ldots,(v_{n_{p}}^{P},v_{n_{Q}}^{Q})$. Moreover, the first block of rows correspond to the constraints of $P$ and the second block of rows to the constraints of $Q$.

3 Proof of Theorem 1

We may assume that $Q$ is a pyramid. First, note that there exists a slack matrix $S\in\mathbb{R}^{m_{P}\times n_{P}}_{\geq 0}$ of $P$ such that every row contains at least one entry being zero. Indeed, every row containing no entry being zero corresponds to a redundant inequality and hence can be removed from the description of $P$. Second, by assuming that the description of $Q$ does not contain any redundant inequalities, the slack matrix $T\in\mathbb{R}^{m_{Q}\times n_{Q}}$ of $Q$ has the form

T=$$T^{\prime}$$\mathbb{O}$$\mathbb{O}$$1

where $T^{\prime}\in\mathbb{R}^{(m_{Q}-1)\times(n_{Q}-1)}_{\geq 0}$. Thus, the matrix $A\in\mathbb{R}^{(m_{P}+m_{Q})\times(n_{P}\cdot n_{Q})}_{\geq 0}$ defined via

A:=$$S$$S$$\dotsb$$S$$S$$t^{\prime}_{1}\dotsb t^{\prime}_{1}$$t^{\prime}_{2}\dotsb t^{\prime}_{2}$$\dotsb$$t^{\prime}_{k}\dotsb t^{\prime}_{k}$$\mathbb{O}$$\mathbb{O}$$\mathbb{O}$$\dotsb$$\mathbb{O}$$1\dotsb 1

is a slack matrix of $P\times Q$, where $t^{\prime}_{1},\dotsc,t^{\prime}_{k}\in\mathbb{R}^{m_{Q}-1}_{\geq 0}$ are the columns of $T^{\prime}$ (here $k=n_{Q}-1$). Recall that we have $\operatorname{xc}(P\times Q)=r_{+}(A)$, $\operatorname{xc}(P)=r_{+}(S)$, and $\operatorname{xc}(Q)=r_{+}(T)$. Furthermore, it is straightforward to check that $r_{+}(T)=r_{+}(T^{\prime})+1$ holds. Thus, it remains to show that

[TABLE]

holds. For the sake of contradiction, let us assume that we have

[TABLE]

i.e., there exists a set $\mathcal{R}$ of nonnegative rank-one matrices in $\mathbb{R}^{(m_{P}+m_{Q})\times(n_{P}\cdot n_{Q})}_{\geq 0}$ with $|\mathcal{R}|\leq r_{+}(S)+r_{+}(T^{\prime})$ whose sum is equal to $A$. Let {\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}} and {\color[rgb]{0,0,0.8}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0.8}\mathcal{R}^{\prime\prime}} denote the set of matrices in $\mathcal{R}$ that have support in the red and blue parts of $A$, respectively.

Claim 1: The sets {\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}} and {\color[rgb]{0,0,0.8}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0.8}\mathcal{R}^{\prime\prime}} form a partition of $\mathcal{R}$ satisfying |{\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}}|=r_{+}(T^{\prime}) and |{\color[rgb]{0,0,0.8}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0.8}\mathcal{R}^{\prime\prime}}|=r_{+}(S).

First, observe that {\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}} and {\color[rgb]{0,0,0.8}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0.8}\mathcal{R}^{\prime\prime}} are disjoint due to the $\mathbb{O}$-block within $A$ that is below the blue $S$-block. Since the red part of $A$ contains $T^{\prime}$ as a submatrix, we must have |{\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}}|\geq r_{+}(T^{\prime}), and since the blue part contains $S$ as a submatrix, we must have |{\color[rgb]{0,0,0.8}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0.8}\mathcal{R}^{\prime\prime}}|\geq r_{+}(S), which yields the claim.

Claim 2: There exists at least one matrix in {\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}} that has support in the green part of $A$.

Since the nonnegative rank of the green submatrix of $A$ is equal to the nonnegative rank of $S$, at least $r_{+}(S)$ matrices in $\mathcal{R}$ must have support in this part. Note that at least one matrix in {\color[rgb]{0,0,0.8}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0.8}\mathcal{R}^{\prime\prime}} has support in the last row of the blue part of $A$ and hence it cannot have support in the green part of $A$. The claim follows since |{\color[rgb]{0,0,0.8}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0.8}\mathcal{R}^{\prime\prime}}|=r_{+}(S).

Claim 3: Let R\in{\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}} and pick exactly one column of each of the $k$ red submatrices of $A$. Then $R$ has support in at least one of these columns.

Suppose the contrary. Then we can pick exactly one column of each of the $k$ red submatrices of $A$ such that $R$ has no support on any of these columns. Restricting to the submatrix formed by these columns, observe that this submatrix is identical to $T^{\prime}$ but can be written as the sum of all matrices in {\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}}\setminus\{R\} and hence r_{+}(T^{\prime})\leq|{\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}}|-1=r_{+}(T^{\prime})-1, a contradiction.

Claim 4: No matrix in {\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}} can have support in the green part of $A$ (a contradiction to Claim 2).

Assume that there is some R\in{\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\mathcal{R}^{\prime}} that has a positive entry $e_{1}$ in the green part of $A$. By our choice of $S$, every of the first $k$ blocks of $A$ contains a column of $A$ in which this row has a zero entry. By the previous claim, $R$ has a positive entry $e_{2}$ in the red part of one of these columns. Restricting $R$ to the two-by-two submatrix containing the entries $e_{1},e_{2}$, it looks as follows (up to swapping its columns):

$e_{1}>0$[math]e_{2}>0$$*

However, there is no rank-one matrix with such a sign pattern. ∎