Lifting properties of maximal lattice-free polyhedra

Abstract

We study the uniqueness of minimal liftings of cut-generating functions obtained from maximal lattice-free polyhedra. We prove a basic invariance property of unique minimal liftings for general maximal lattice-free polyhedra. This generalizes a previous result by Basu et al. (Math Oper Res 37(2):346–355, 2012) for simplicial maximal lattice-free polytopes, thus completely settling this fundamental question about lifting for maximal lattice-free polyhedra. We further give a very general iterative construction to get maximal lattice-free polyhedra with the unique-lifting property in arbitrary dimensions. This single construction not only obtains all previously known polyhedra with the unique-lifting property, but goes further and vastly expands the known list of such polyhedra. Finally, we extend characterizations from Basu et al. (2012) about lifting with respect to maximal lattice-free simplices to more general polytopes. These nontrivial generalizations rely on a number of results from discrete geometry, including the Venkov-Alexandrov-McMullen theorem on translative tilings and characterizations of zonotopes in terms of central symmetry of their faces.

Appendices

Appendix 1: Proofs of Propositions 1.1 and 1.2

Proof of Proposition 1.1

First we show that the infimum in (1.7) is attained.

Consider the case of a bounded $B$. For a sufficiently large $N \in \mathbb {N}$ the Euclidean ball of radius $N$ centered at $o$ contains $B$. It follows that $\phi _{B-f}(r) \ge \frac{1}{N} \Vert r\Vert$ for every $r \in \mathbb {R}^n$, where $\Vert \,\cdot \,\Vert$ is the Euclidean norm. Consequently, $\phi _{B-f}(r+w) \ge \frac{1}{N} \Vert r + w \Vert \ge \frac{1}{N} ( \Vert w\Vert - \Vert r\Vert ) > \phi _{B-f}(r)$ for $r \in \mathbb {R}^n$ and $w \in \mathbb {Z}^n$ whenever $w$ fulfills $\Vert w\Vert > (N \phi _{B-f}(r) +\Vert r\Vert )$. It follows that $\inf _{w \in \mathbb {Z}^n} \phi _{B-f}(r + w)$ is attained for some of finitely many vectors $w \in \mathbb {Z}^n$ satisfying $\Vert w\Vert \le (N \phi _{B-f}(r) + \Vert r\Vert )$.

Let us switch to the case that $B$ is unbounded. It is known that the recession cone of $B$ is a linear space spanned by rational vectors; see see [5, 22] and [2]. Up to appropriate unimodular transformations, we can assume that $B$ has the form $B = B' \times \mathbb {R}^k$, where $k \in \{1,\ldots ,n-1\}$ and $B'$ is a bounded maximal lattice-free set in $\mathbb {R}^{n-k}$. We denote by $\phi _{B'-f}$ the gauge-function of $B'$; it is well known that $\phi _{B-f}((r',r'')) = \phi _{B'-f}(r')$ for all $(r', r'') \in \mathbb {R}^{n-k}\times \mathbb {R}^{k}$. Thus, it suffices to apply the assertion of the bounded case to $B'$ to get the assertion for an unbounded $B$.

It remains to prove the assertion on polynomial-time computability. Assume that $n \in N$ is fixed and that $f$ and $a_i$ ($i \in I$) are rational vectors, whose components are given as the input in standard binary encoding. We have

$$\begin{aligned} \phi _{B-f}^*(r):=&\min _{w \in \mathbb {Z}^n} \phi _{B-f} (r + w) \nonumber \\ =&\min _{w \in \mathbb {Z}^n} \max _{i\in I} a_i\cdot (r + w) \nonumber \\ =&\min \left\{ \rho \ge 0 \,:\, \rho \ge a_i(r+w) \ \forall i \in I, \ w \in \mathbb {Z}^n \right\} . \end{aligned}$$

(6.1)

Expression (6.1) defines a mixed-integer linear program with rational coefficients with $n$ integer variables (the components of $w$) and one real variable (the value $\rho$). Since $n$ is fixed, Lenstra’s algorithm [21] can be used to determine (6.1) in polynomial time.$\square$

Proof of Proposition 1.2

It was established in  [4] that for every $r\in \mathbb {R}^n$ such that $r+ f \in R(f,B)$, $\pi (r) = \phi _{B-f}(r)$ for every minimal lifting $\pi$ of $\phi _{B-f}$. Moreover, it is not difficult to see that every minimal lifting is periodic with respect to $\mathbb {Z}^n$, i.e., $\pi (r) = \pi (r+w)$ for every $r\in \mathbb {R}^n$ and $w \in \mathbb {Z}^n$. If $\phi _{B-f}$ has a unique lifting, then $R(f,B) + \mathbb {Z}^n = \mathbb {R}^n$. Therefore, for any $r$, there exists $w \in \mathbb {Z}^n$ such that $r + w + f \in R(f,B)$ and thus $\pi (r) = \pi (r + w) = \phi _{B-f}(r+w) \ge \phi _{B-f}^*(r)$ for every minimal lifting $\pi$, thus establishing that $\phi _{B-f}^*$ is a minimal lifting.

Suppose $\phi _{B-f}$ does not have a unique minimal lifting. This implies there are at least two distinct minimal liftings and so there must exist a minimal lifting $\pi$ that is different from the lifting $\phi _{B-f}^*$. However, we show below that $\pi \le \phi _{B-f}^*$. Thus, $\phi _{B-f}^*$ is not a minimal lifting.

To show that $\pi \le \phi _{B-f}^*$, consider any $r \in \mathbb {R}^n$. It is well-known that $\pi \le \phi _{B-f}$ because $\pi$ is a minimal lifting. By Theorem 1.1, there exists $w \in \mathbb {Z}^n$ such that $\phi _{B-f}^*(r) = \phi _{B-f}(r + w)$. By the $\mathbb {Z}^n$-periodicity of $\pi$, we have $\pi (r) = \pi (r+w) \le \phi _{B-f}(r+w) = \phi _{B-f}^*(r)$.$\square$

Appendix 2: Proof of Theorem 6.2

Let $P\subseteq \mathbb {R}^n$ be an $n$-dimensional centrally symmetric polytope with centrally symmetric facets. Let $G$ be any $(n-2)$-dimensional face of $P$. The belt corresponding to $G$ is the set of all facets which contain a translate of $G$ or $-G$. Observe that every centrally symmetric polytope $P$ with centrally symmetric facets has belts of even size greater than or equal to 4.

A zonotope is a polytope given by a finite set of vectors $V=\{v_1, \ldots , v_k\} \subseteq \mathbb {R}^n$ in the following way: $Z(V) := \{\lambda _1v_1 + \cdots + \lambda _kv_k: -1 \le \lambda _i \le 1 \quad \forall i = 1, \ldots , k\}.$ We recall that $F(P,u)$ denotes the face of points in $P$ maximizing the linear function $x \mapsto u \cdot x$. The following simple lemma is well-known.

Lemma 6.6

Let $n \in \mathbb {N}$. Let $V$ be a nonempty finite subset of $\mathbb {R}^n$ and let $u \in \mathbb {R}^n$. Then the face $F(Z(V),u)$ of the zonotope $Z(V)$ coincides, up to a translation, with the zonotope $Z\bigl ( \left\{ v \in V \,:\, u \cdot v = 0 \right\} \bigr )$.

Proof

By the Minkowski additivity of the functional $F(\,\cdot \,,u)$, defined by (4.1), we get $F(Z(V),u) = \sum _{v \in V} F([-v,v],u).$ It is straightforward to verify that for every $v \in V$ one has

$$\begin{aligned} F([-v,v],u):= \left\{ \begin{array}{ll} \{-v \} &{} \text {if} \, u \cdot v < 0, \\ {[}-v,v{]} &{} \text {if} \, u \cdot v = 0, \\ \{v\} &{} \text {if} \, u \cdot v > 0. \end{array}\right\} \end{aligned}$$

Putting these observations together, we have the assertion.$\square$

The latter lemma shows that every face of a zonotope is a zonotope (and, thus, centrally symmetric). The following lemma deals with belts of zonotopes. Each belt of the cube $[-1,1]^n$ consists of exactly four facets. The following theorem shows that the latter property essentially characterizes cubes within all zonotopes.

Theorem 6.7

Let $n \in \mathbb {N}$, $n \ge 3$. Let $V$ be a finite set linearly spanning $\mathbb {R}^n$ and such that each belt of the $n$-dimensional zonotope $Z(V)$ consists of exactly four facets. Then $Z(V)$ is the image of the $n$-dimensional cube $[-1,1]^n$ under a invertible linear transformation.

Proof

Choose a basis $b_1,\ldots ,b_n$ of $\mathbb {R}^n$ consisting of vectors in $V$. It suffices to show that every vector of $V$ is parallel to some vector of $\{b_1,\ldots ,b_n\}$. After a change of coordinates in $\mathbb {R}^n$ we can assume that $b_1,\ldots ,b_n$ is the standard basis $e_1,\ldots ,e_n$.

Assume to the contrary, that there exists a vector $a = (\alpha _1,\alpha _2,\ldots ,\alpha _n) \in V$ which is not parallel to any vector of the basis $e_1,\ldots ,e_n$. Thus, at least two of its components $\alpha _1,\ldots ,\alpha _n$ are nonzero. Without loss of generality let $\alpha _1 \ne 0$ and $\alpha _2 \ne 0$. Let $W:= V \cap (\{0\}^2 \times \mathbb {R}^{n-2})$. We have $e_3,\ldots ,e_n \in W$ and $e_1,e_2, a \in V {\setminus } W$. Choose a nonzero vector $u' = \mathbb {R}^2 \times \{0\}^{n-2}$ such that $u'$ is not orthogonal to any vector from $V {\setminus } W$ (e.g., one can choose $u'= (1,\varepsilon ,0,\ldots ,0)$, where $\varepsilon >0$ is small). By Lemma 6.6, the face $G:=F(Z(V),u')$ is a translation of $Z(W)$. By the choice of $W$, the zonotope $Z(W)$ is $(n-2)$-dimensional. We analyze the belt of $Z(V)$ determined by the $(n-2)$-dimensional face $G$.

We shall construct a number of facets $F(Z(V),u)$ with $u \in \mathbb {R}^2 \times \{0\}^{n-2}$ belonging to the belt generated by $G$. In view of Lemma 6.6, for $u=e_1$ the face $F( Z(V), u)$ contains a translation of $Z( \{e_2\} \cup W)$. Similarly, for $u= e_2$ the face $F(Z(V),u)$ contains a translation of $Z(\{e_1\} \cup W)$. For a nonzero vector $u \in \mathbb {R}^2 \times \{0\}^{n-2}$ orthogonal to $a$ (say, for $u=(-\alpha _2,\alpha _1,0,\ldots ,0)$) the face $F(Z(V),u)$ contains a translation of $Z(\{a\} \cup W)$. Since the zonotopes $Z(\{e_1\} \cup W)$, $Z(\{e_2\} \cup W)$ and $Z(\{a\} \cup W)$ are $(n-1)$-dimensional, we see that for all three choices of $u$ above, the face $F(Z(V),u)$ is actually a facet. The latter shows that the six distinct facets $F(Z(V),u)$ with $u \in \{\pm e_1,\pm e_2, \pm (-\alpha _2,\alpha _1,0,\ldots ,0) \}$ belong to the belt generated by $G$. The latter is a contradiction to the assumptions on $Z(V)$.$\square$

Theorem 6.8

(McMullen  [25]) Let $n \in \mathbb {N}$, $n \ge 3$, and let $S\subseteq \mathbb {R}^n$ be an $n$-dimensional spindle with centrally symmetric facets. Then $S$ is the image of the $n$-dimensional hypercube under an invertible affine transformation.

Proof

Since all facets of $S$ are centrally symmetric, by the Alexandrov-Shephard theorem (see [24] for a short proof), the polytope $S$ itself is also centrally symmetric. Without loss of generality, we assume that $S$ is symmetric in the origin. Let $a$ and $-a$ be the apexes of the spindle $S$.

We first show that every belt of $S$ is of length 4. Let $G$ be an arbitrary $(n-2)$-dimensional face of $S$ and consider the belt of $S$ associated with $G$. Since $S$ is centrally symmetric, each belt is even length, i.e., of length $2 k$ where $k \ge 2$. There are $k$ facets $F_1, \ldots , F_{k }$ involved in this belt that contain $a$; the remaining $k$ facets contain $-a$. We project $S$ onto the two-dimensional space perpendicular to $G$ to get a polygon $P$. The facets $F_1, \ldots , F_{k}$ are all projected onto $k$ distinct edges of the polygon $P$. Moreover, observe the projection of $a$ is contained in all these edges. Since $P$ is two-dimensional, intersection of more than three edges of $P$ is empty. Hence $k \le 2$ and since we also have $k \ge 2$, we get $k=2$.

We next show that all faces of $S$ are centrally symmetric. To do this, we first show that every $n-2$-dimensional face $G$ is centrally symmetric (for $n=3$ this is clear). For $i \in \{1,2\}$, by $c_i$ we denote the center of symmetry of $F_i$. Then $G$ has the form $F_i \cap F_j$ or $(-F_i) \cap (-F_j)$ or $F_i \cap (-F_j)$ with appropriate $i,j$ satisfying $\{i,j\} = \{1,2\}$. Consider the case $G = F_i \cap F_j = F_1 \cap F_2$. The symmetry of $F_1$ implies that $2 c_1 - G$ (the reflection of $G$ with respect to $c_1$) is a face of $F_1$. Since $a \in G$, the face $2 c_1 - G$ does not contain $a$. The face $G$ is contained in exactly two facets of $S$, both belonging to the belt $\{F_1,F_2,-F_1,-F_2\}$ generated by $G$. The facet $-F_1$ cannot contain $2 c_1 - G$, because $-F_1$ is opposite to $F_1$ and thus does not share any nonempty face with $F_1$. The facet $F_2$ of $S$ cannot contain $2 c_1 - G$, because $F_2$ contains $a$, while $2 c_1 - G$ does not contain $a$. It follows that the facet $-F_2$ contains $2 c_1 - G$. Then the reflection $-2c_2 - (2 c_1 - G)$ of $2c_1 -G$ with respect to the center $-c_2$ of $-F_2$ is a facet of $-F_2$. On the other hand, the reflection $-G$ of $G$ with respect to the center $o$ of $S$ is a face of $S$ which does not contain $a$. Hence $-G$ is a facet of $-F_2$. We have shown that $2 c_1 -G$, $-2 c_2 - 2 c_1 + G$ and $-G$ are facets of $-F_2$. Since all these facets of $F_2$ are parallel, two of them must coincide. We cannot have $2 c_1 - G = -G$, since this would imply $c_1 = o$ and, by this, $\hbox {relint}(F_1) \cap \hbox {int}(S) \ne \emptyset$, which is a contradiction. Consequently, $2 c_1 - G = -2 c_2 - 2 c_1 + G$ or $-2 c_2 - 2 c_1 + G = -G$, where each of the two equalities implies that $G$ is centrally symmetric. The case $G = (-F_i) \cap (-F_j)$ is completely analogous to the case $G = F_i \cap F_j$.

Let us switch to the case $G = F_i \cap (-F_j)$ with $\{i,j\} = \{1,2\}$. Without loss of generality, let $G = F_1 \cap (-F_2)$. The face $G$ of $S$ contains neither $a$ nor $-a$. The same also holds for the face $-G$ of $S$. The reflection $2 c_1 - G$ of $G$ with respect to the center $c_1$ of $F_1$ is a facet of $F_1$. Then $2 c_1 - G$ is not a facet of $-F_1$, because $2 c_1 - G$ is a facet of $F_1$, while $F_1$ and $-F_1$ are opposite facets of $S$. Thus, $2 c_1 - G$ is a facet of $F_2$ or $-F_2$. If $2c_1 - G$ is a facet of $F_2$, then also $2 c_2 - (2 c_1 - G)$ is a facet of $F_2$. It follows that $2 c_1 - G, 2 c_2 - 2c_1 + G$ and $-G$ are facets of $F_2$. Again, since they are all parallel, two of them must coincide. Coincidence of $2 c_1 - G$ and $-G$ implies $c_1 =o$ and yields a contradiction. Coincidence of any two other of these three facets of $F_2$ implies that $G$ is centrally symmetric. In the case that $2 c_1 - G$ is a facet of $-F_2$, we get that $- 2 c_2 - (2 c_1 - G)$ is a facet of $-F_2$. Thus, $G$, $-2 c_2- G$ and $- 2 c_2 - 2 c_1 + G$ are facets of $-F_2$. Coincidence of $G$ and $- 2 c_2 - 2 c_1 + G$ implies $c_1 = c_2$, yielding $\hbox {relint}(F_1) \cap \hbox {relint}(F_2) \ne \emptyset$, which is a contradiction. Coincidence of any other of these three facets of $S$ implies that $G$ is centrally symmetric.

It follows that every $(n-2)$-dimensional face of $S$ is centrally symmetric. Therefore, by a theorem of McMullen  [23], in the case $n \ge 4$, every face of $S$ is centrally symmetric (in the case $n=3$ this is clear from the assumptions). Consequently, all 2-dimensional faces of $S$ are centrally symmetric and, by this, $S$ is a zonotope; see, for example, [28, Theorem 3.5.1]. Since $S$ is a zonotope whose belts are length 4, by Lemma 6.6, $S$ is the image of the $n$-dimensional hypercube under an invertible affine transformation.$\square$

Theorem 6.8 was communicated to us by Peter McMullen via personal email. We include a complete proof here as the result does not appear explicitly in the literature. The above proof is based on a proof sketch by Prof. McMullen.

We now state the celebrated Venkov-Alexandrov-McMullen theorem on translative tilings.

Theorem 6.9

(Venkov-Alexandrov-McMullen; see [17, Theorem 32.2]) Let $P$ be a compact convex set with nonempty interior that translatively tiles $\mathbb {R}^n$. Then the following assertions hold:

• (a) $P$ is a centrally symmetric polytope.

• (b) All facets of $P$ are centrally symmetric.

• (c) Every belt of $P$ is either length 4 or 6.

Proof of Theorem 6.2

We only need to consider the case $n \ge 3$. The assertion follows directly from Theorem 6.9 (assertions (a) and (b)) and Theorem 6.8.$\square$