Semi-polytope decomposition of a Generalized permutohedron

TL;DR

This paper presents a method to decompose generalized permutohedra into semi-polytopes, providing a clearer understanding of their structure.

Contribution

It introduces an explicit decomposition technique for generalized permutohedra into semi-polytopes, which was not previously detailed.

Findings

Explicit semi-polytope decomposition of generalized permutohedra

Enhanced understanding of generalized permutohedron structure

Potential applications in combinatorics and polyhedral theory

Abstract

In this short note we show explicitly how to decompose a generalized permutohedron into semi-polytopes.

Figures

Equations

$$P_G(y_1,\dots ,y_m):=y_1{\Delta }_{I_1}+\cdots +y_m{\Delta }_{I_m},$$

$$P_G^{-}(y_1,\dots ,y_m):=\{x\in {\mathbb{R}}^n|x+{\Delta }_{[n]}\subseteq P_G\}.$$

$${\Delta }_{I,j}^{\ast }={\Delta }_I\setminus {\Delta }_{I\setminus j}$$

Full text

1 Introduction

Given a polytope, assume we have disjoint open cells whose closures sum up to be the entire polytope. A question of naturally assigning each of the remaining points (possibly in multiple closures) to a cell has appeared in [1] for studying regular matroids and zonotopes and in [2] for studying h-vectors and $Q$-polytopes. In other words, we are trying to determine ownership of lattice points on boundaries of multiple polytopes. In this note, we study a more general case of doing the same for a Generalized permutohedron, a polytope that can be obtained by deforming the usual permutohedron. We will show explicitly how to construct a semi-polytope decomposition of a trimmed generalized permutohedron.

2 Generalized permutohedron $P_{G}$ and its fine mixed subdivision

Let $\Delta_{[n]}=\operatorname{ConvexHull}(e_{1},\ldots,e_{n})$ be the standard coordinate simplex in $\mathbb{R}^{n}$. For a subset $I\subset[n]$, let $\Delta_{I}=\operatorname{ConvexHull}(e_{i}|i\in I)$ denote the face of $\Delta_{[n]}$. Let $G\subseteq K_{m,n}$ be a bipartite graph with no isolated vertices. Label the vertices of $G$ by $1,\ldots,m,\bar{1},\ldots,\bar{n}$ and call $1,\ldots,m$ the left vertices and $\bar{1},\ldots,\bar{n}$ the right vertices. We identify the barred indices with usual non-barred cases when it is clear we are dealing with the right vertices. For example when we write $\Delta_{\{\bar{1},\bar{3}\}}$ we think of it as $\Delta_{\{1,3\}}$. We associate this graph with the collection $\mathcal{I}_{G}$ of subsets $I_{1},\ldots,I_{m}\subseteq[n]$ such that $j\in I_{i}$ if and only if $(i,\bar{j})$ is an edge of $G$. Let us define the polytope $P_{G}(y_{1},\ldots,y_{m})$ as:

[TABLE]

where $y_{i}$ are nonnegative integers. This lies on a hyperplane $\sum_{i\in[n]}x_{i}=\sum_{j\in[m]}y_{j}$. An example of a coordinate simplex $\Delta_{[3]}$, a bipartite graph $G$ and a generalized permutohedron $P_{G}(1,2,3)$ is given in Figure 1.

Definition 2.1** ([3], Definition 14.1).**

Let $d$ be the dimension of the Minkowski sum $P_{1}+\cdots+P_{m}$. A Minkowski cell in this sum is a polytope $B_{1}+\cdots+B_{m}$ of dimension $d$ where $B_{i}$ is the convex hull of some subset of vertices of $P_{i}$. A mixed subdivision of the sum is the decomposition into union of Minkowski cells such that intersection of any two cells is their common face. A mixed subdivision is fine if for all cells $B_{1}+\cdots+B_{m}$, all $B_{i}$ are simplices and $\sum dimB_{i}=d$.

All mixed subdivisions in our note, unless otherwise stated, will be referring to fine mixed subdivisions. We will use the word cell to denote the Minkowski cells. Beware that our cells are all closed polytopes.

Fine Minkowski cells can be described by spanning trees of $G$. When we are looking at a fixed generalized permutohedron $P_{G}(y_{1},\ldots,y_{m})$, we will use $\prod_{J}$ to denote $y_{1}\Delta_{J_{1}}+\cdots+y_{m}\Delta_{J_{m}}$ where $J=(J_{1},\ldots,J_{m})$. We say that $J$ is a tree if the associated bipartite graph is a tree.

Lemma 2.2** ([3], Lemma 14.7).**

Each fine mixed cell in a mixed subdivision of $P_{G}(y_{1},\cdots,y_{m})$ has the form $\prod_{T}$ such that $T$ is a spanning tree of $G$.

An example of a fine mixed subdivision of the polytope considered in Figure 1 is given in Figure 2.

We can say a bit more about the lattice points in each $\prod_{T}$:

Proposition 2.3** ([3], Proposition 14.12).**

Any lattice point of a fine Minkowski cell $\prod_{T}$ in $P_{G}(y_{1},\cdots,y_{m})$ is uniquely expressed (within $\prod_{T}$) as $p_{1}+\cdots+p_{m}$ where $p_{i}$ is a lattice point in $y_{i}\Delta_{T_{i}}$.

3 Semi-polytope decomposition

A mixed subdivision of $P_{G}$ divides the polytope into cells. In this section, we show that from a mixed subdivision of $P_{G}$, one can obtain a way to decompose the set of lattice points of $P_{G}^{-}$.

Definition 3.1** ([3], Definition 11.2).**

The trimmed generalized permutohedron $P_{G}^{-}$ is defined as:

[TABLE]

This is a more general class of polytopes than generalized permutohedra $P_{G}(y_{1},\ldots,y_{m})$. With a slight abuse of notation, we will let $I\setminus j$ stand for $I\setminus\{j\}$.

Definition 3.2** ([3], Theorem 11.3).**

The coordinate semi-simplices are defined as

[TABLE]

for $j\in I\subseteq[n]$.

For each cell $\prod_{T}$, we are going to turn it into a semi-polytope of the form $y_{1}\Delta_{J_{1},j_{1}}^{*}+\cdots+y_{m}\Delta_{J_{m},j_{m}}^{*}$. This will involve deciding which cell takes ownership of the lattice points on several cells at the same time.

We denote the point $((m-1)c+\sum_{i}y_{i},-c,\ldots,-c)$ for $c$ sufficiently large as $\infty_{1}$. For a facet of a polytope, we say that it is negative if the defining hyperplane of the facet (inside the space $\sum_{i\in[n]}x_{i}=\sum_{j\in[m]}y_{j}$ which the polytope lies in) separates the point $\infty_{1}$ and the interior of the polytope. Otherwise, we say that it is positive. We will say that a point of a polytope is good if it is not on any of the positive facets of the polytope.

Lemma 3.3**.**

Fix $T$, a spanning tree of $G\subseteq K_{m,n}$. Let $T_{i}$ be the set of neighbors of $i$. For each $i$ such that $\bar{1}\not\in T_{i}$, there exists a unique element $t_{i}$ in $T_{i}$ such that there exists a path to $\bar{1}$ not passing through $i$.

Proof.

There exists such an element since $T$ is a spanning tree of $T$. There cannot be more than one such element since otherwise, we get a cycle in $T$. ∎

In cases where $T_{i}$ does contain $\bar{1}$, we set $t_{i}$ to be $\bar{1}$.

Lemma 3.4**.**

Let $\prod_{T}$ be a fine mixed cell. Removing the positive facets gives us $\sum_{i}\Delta^{*}_{T_{i},t_{i}}$.

To prove this, we first introduce a tool that will be useful for identifying which hyperplanes the facets lie on. Let $\prod_{T}$ be a fine mixed cell so $T$ a spanning tree. For any edge $e$ of $T$ that is not connected to a leaf on the left side, $T\setminus e$ has two components. Let $I_{e}$ denote the set of right vertices of a component that contains $\bar{1}$. Let $c_{e}$ be the sum of $y_{i}$’s for left vertices contained in that component. Notice that $I_{e}$ cannot be $[n]$ since otherwise $e$ would have a leaf as its left endpoint.

Lemma 3.5**.**

Let $\prod_{T}$ be a fine mixed cell. For any edge $e$ of $T$ that is not connected to a leaf on the left side, $\prod_{T\setminus e}$ is a facet of $\prod_{T}$ that lies on $\sum_{j\in I_{e}}x_{j}=c_{e}$. If the right endpoint of $e$ is in $I_{e}$, then $\prod_{T}$ lies in half-space $\sum_{j\in I_{e}}x_{j}\geq c_{e}$. Otherwise it lies in $\sum_{j\in I_{e}}x_{j}\leq c_{e}$

Proof.

The dimension difference between $\prod_{T}$ and $\prod_{T\setminus e}$ is at most one, and all endpoints of $\prod_{T\setminus e}$ lie on $\sum_{j\in I_{e}}x_{j}=c_{e}$. If the right endpoint of $e$ is in $I_{e}$, that means we can find a point $x$ using $e$ so that $\sum_{j\in I_{e}}x_{j}>c_{e}$. If not, that means we can find a point $x$ using $e$ so that $\sum_{j\in I_{e}}x_{j}<c_{e}$. ∎

Proof of Lemma 3.4.

If $\prod_{T\setminus e}$ is a positive facet of $\prod_{T}$, then Lemma 3.5 tells us that the right endpoint of $e=(i,\bar{j})$ is in $I_{e}$. From definition of $t_{i}$, we have $\bar{j}=t_{i}$. In other words we are removing sets of form $\Delta_{T_{1}}+\cdots+\Delta_{T_{i}\setminus t_{i}}+\cdots+\Delta_{T_{m}}$. At the end we end up with $\sum_{i}(\Delta_{T_{i}}\setminus\Delta_{T_{i}\setminus t_{i}})$.

∎

Let $\prod_{T}$ and $\prod_{T}^{\prime}$ be two cells inside a mixed subdivision of $P_{G}$ that share a facet $F$. The sign of $F$ in $\prod_{T}$ and the sign of $F$ in $\prod_{T}^{\prime}$ has to be different, since $\infty_{1}$ can be on exactly one side of the defining hyperplane of $F$. This implies that all lattice points of $P_{G}$ are good in at most one cell of $P_{G}$.

Lemma 3.6**.**

$p\in P_{G}^{-}\cap\mathbb{Z}^{n}$* if and only if $p+e_{1}$ is a good point of $P_{G}$.*

Proof.

Having $p\in P_{G}^{-}\cap\mathbb{Z}^{n}$ implies from definition that $p+e_{i}\in P_{G}$ for each $i\in[n]$. Assume for sake of contradiction $p+e_{1}$ is on some positive facet $x_{I}=c_{I}$ with $1\in I$. Any $p+e_{i}$ is either on that facet or is on the same side as $\infty_{1}$. Hence $I=[n]$ and we get a contradiction.

Now look at the case when $p+e_{1}$ is a good point of $P_{G}$. Assume for the sake of contradiction that $p+e_{j}$ is not in $P_{G}$ for some $j\in[n]$. Then $p+e_{1}$ is on a facet of $P_{G}$, whose corresponding hyperplane is given by $x_{I}=c_{I}$ where $1\in I$ and $j\not\in I$. This hyperplane separates the interior of $P_{G}$ with $p+e_{j}$. Since $j\not\in I$, the point $\infty_{1}$ has to be on opposite side of $p+e_{j}$. This is a positive facet and we get a contradiction.

∎

Combining what we have so far, we state the main result of our note on how to do the semi-polytope decomposition with an explicit way to obtain each of the semi-polytopes:

Theorem 3.7**.**

Identify the lattice points of $P_{G}^{-}$ with points not lying on any of the positive facets of $P_{G}$ via the map $p\rightarrow p+e_{1}$ (as in Lemma 3.6). Pick any full mixed subdivision of $P_{G}$. For each cell $\prod_{T}$, construct a semi-polytope by $\sum_{i}\Delta^{*}_{T_{i},t_{i}}$ (where $t_{i}$ is chosen as in Lemma 3.3). Then the (disjoint) union of the semi-polytopes is exactly $P_{G}$ minus the positive facets. Each lattice point of $P_{G}^{-}$ with the above identification is contained in exactly one semi-polytope.

An example of a semi-polytope decomposition of the generalized permutohedron considered in Figure 1 and in Figure 2 is given in Figure 3.

4 Application to Erhart theory

In this section we show how the semi-polytope decomposition can be used in Erhart theory as guided in [3]. Given any subgraph $T$ in $G$, define the left degree vector $ld(T)=(d_{1}-1,\cdots,d_{n}-1)$ and the right degree vector $rd(T)=(d_{1}^{\prime}-1,\cdots,d_{m}^{\prime}-1)$ where $d_{i}$ and $d_{j}^{\prime}$ are the degree of the vertex $i$ and $\bar{j}$ respectively. The raising powers are defined as $(y)_{a}:=y(y+1)\cdots(y+a-1)$ for $a\geq 1$ and $(y)_{0}:=1$.

Corollary 4.1**.**

Fix a fine mixed subdivision of $P_{G}(y_{1},\ldots,y_{m})$ where $y_{i}$’s are nonnegative integers. The number of lattice points in the trimmed generalized permutohedron $P_{G}^{-}(y_{1},\ldots,y_{m})$ equals $\sum_{(a_{1},\ldots,a_{m})}\prod_{i}\frac{(y_{i})_{a_{i}}}{a_{i}!}$ where the sum is over all left-degree vectors of fine mixed cells inside the subdivision.

Proof.

Obtain a semi-polytope decomposition as in Theorem 3.7. Then each lattice point of $P_{G}^{-}(y_{1},\ldots,y_{m})$ is in exactly one semi-polytope. The claim follows since the number of lattice points of a semi-polytope $y_{1}\Delta_{T_{1},t_{1}}^{*}+\cdots+y_{m}\Delta_{T_{m},t_{m}}^{*}$ (thanks to Proposition 2.3, different sum gives a different point) is given by $\prod_{i}\frac{(y_{i})_{|T_{i}|-1}}{(|T_{i}|-1)!}$.

∎

The expression in Corollary 4.1 is called the Generalized Erhart polynomial of $P_{G}^{-}(y_{1},\ldots,y_{m})$ by [3]. As foretold in [3], Theorem 3.7 gives us a pure counting proof of Theorem 11.3 of [3].

Definition 4.2** ([3], Definition 9.2).**

Let us say that a sequence of nonnegative integers $(a_{1},\cdots,a_{m})$ is a $G$-draconian sequence if $\sum a_{i}=n-1$ and, for any subset $\{i_{1}<\cdots<i_{k}\}\subseteq[m]$, we have $|I_{i_{1}}\cup\cdots\cup I_{i_{k}}|\geq a_{i_{1}}+\cdots+a_{i_{k}}+1$.

Theorem 4.3** ([3], Theorem 11.3).**

For nonnegative integers $y_{1},\ldots,y_{m}$, the number of lattice points in the trimmed generalized permutohedron $P_{G}^{-}(y_{1},\ldots,y_{m})$ equals $\sum_{(a_{1},\ldots,a_{m})}\prod_{i}\frac{(y_{i})_{a_{i}}}{a_{i}!}$, where the sum is over all $G$-draconian sequences $(a_{1},\ldots,a_{m})$.

Proof.

Thanks to Corollary 4.1, all we need to do is show that the set of $G$-draconian sequences is exactly the set of left-degree vectors of the cells inside a fine mixed subdivision of $P_{G}$. Lemma 14.9 of [3] tells us that the right degree vectors of the fine cells is exactly the set of lattice points of $P_{G^{*}}^{-}(1,\ldots,1)$ where $G^{*}$ is obtained from $G$ by switching left and right vertices. Then Lemma 11.7 of [3] tells us that the set of $G$-draconian sequences is exactly the set of lattice points of $P_{G^{*}}^{-}(1,\ldots,1)$. ∎

This approach has an advantage that the generalized Erhart polynomial is obtained from a direct counting method, without using any comparison of formulas.