Cutting convex polytopes by hyperplanes

Cutting a polytope is a very natural way to produce new classes of interesting polytopes. Moreover, it has been very enlightening to explore which algebraic and combinatorial properties of the orignial polytope are hereditary to its subpolytopes obtained by a cut. In this work, we put our attention to all the seperating hyperplanes for some given polytope (integral and convex) and study the existence and classification of such hyperplanes. We prove the exitence of seperating hyperplanes for the order and chain polytopes for any finite posets that are not a single chain; prove there are no such hyperplanes for any Birkhoff polytopes. Moreover, we give a complete seperating hyperplane classification for the unit cube and its subpolytopes obtained by one cut, together with some partial classification results for order and chain polytopes.


Introduction
Let P ⊂ R n be a convex polytope of dimension d and ∂P its boundary. If H ⊂ R n is a hyperplane, then we write H (+) and H (−) for the closed half-spaces of R d with H (+) ∩H (−) = H. We say that H cuts P if H∩(P \∂P) = ∅ and if each vertex of the convex polytopes P ∩ H (+) and P ∩ H (−) is a vertex of P. When H ∩ (P \ ∂P) = ∅, it follows that H cuts P if and only if, for each edge e = conv({v, v ′ }) of P, where v and v ′ are vertices of P, one has H ∩ e ⊂ {v, v ′ }. Cutting a polytope is a very natural way to produce new classes of interesting polytopes. For example, the hypersimplices are obtained from cutting the unit cube by hyperplanes of the form x 1 + · · · + x n = k, k + 1, for some integer 0 ≤ k < n, which is a class of very interesting and well-studied polytopes (see for example [Stan1], [LamP] and [L]). A similar class of interesting polytopes obtained from cutting permutahedrons and in general any graphical zonotopes are studied in [LP]. In general, it is a very interesting problem to explore which algebraic and combinatorial properties of P are hereditary to P ∩ H (+) and P ∩ H (−) . For example, in [HLZ] the study on separating hyperplanes of the edge polytope P G of a finite connected simple graph G is achieved and it is shown that P G is normal if and only if each of P G ∩ H (+) and P G ∩ H (−) is normal.
In this paper, we look at the problem from another perspective, focusing more on the hyerplane that cuts the polytope. We are interested in the exitence and classification of such hyperplanes. Let us make it more precise what we mean by a "cut".

The unit cube
Let [0, 1] d ⊂ R d be the unit cube with d ≥ 2. In the study of its separating hyperplane H it is assumed that H passes through the origin of R d . First of all, we discuss the question when a hyperplane H of R d passing through the origin H : a 1 x 1 + · · · + a d x d = 0, (1.1) where each a i ∈ Q, is a separating hyperplane of [0, 1] d . where each ε j ∈ {0, 1}. Suppose that there exists p and q with a p > 0 and a q < 0 and that all nonzero coefficients of H have the same absolute value. Then we may assume that H : x 1 + · · · + x s − x s+1 − · · · − x s+t = 0, where s > 0, t > 0 and s + t ≤ d. Moreover, since s > 0 and t > 0, it follows that H ∩ (0, 1) d = ∅. Hence H is a separating hyperplane of [0, 1] d .
("Only if") If every coefficient a i of (1.1) is nonnegative, then H ∩ [0, 1] d consists only of the origin. Hence H cannot be a separating hyperplane of [0, 1] d . Thus there exists p and q with a p > 0 and a q < 0. Now, suppose that there exist i = j with a i = 0, a j = 0 and |a i | = |a j |. Let, say, |a i | < |a j |. Let e is the edge defined by x i = 1 and x k = 0 for all k with k ∈ {i, j}. If a i a j < 0, then 0 < −a i /a j < 1 and v = (v 1 , . . . , v n ) ∈ e with v j = −a i /a j belongs to H. Thus H cannot be a separating hyperplane of [0, 1] d . Hence a i a j > 0. In particular |a p | = |a q |. Let 1 ≤ k ≤ d with a k > 0. Then, since a k a q < 0, it follows that |a k | = |a q |. Similarly if a k < 0, then |a k | = |a p |. Consequently, |a k | = |a p | (= |a p |) for all k with a k = 0, as desired. Now, by virtue of Lemma 1.1, it follows that a separating hyperplane of [0, 1] d passing through the origin is of the form we can work with a separating hyperplanes of [0, 1] d of the form Finally, the equation (1.3) can be rewritten as In [0, 1] d ∩ H (−) , again by replacing x i with 1 − x i for 1 ≤ i ≤ t, it follows that, since 0 < k − ℓ < k, each of the subpolytopes [0, 1] d ∩ H (±) is, up to unimodular equivalence, of the form We now turn to the problem of finding a separating hyperplane of (1.5). We say that a separating hyperplane of (1.5) is a second separating hyperplane of [0, 1] d following (1.4).
with each a i ∈ Q, be a separating hyperplane of [0, 1] d passing through v. In (1.6) replace x j with 1 − x j for j ∈ J, and the hyperplane is a separating hyperplane of [0, 1] d passing through the origin. It then follows from Lemma 1.1 that all nonzero coefficients of (1.7) have the same absolute value. Thus each of a i 's and a j ' belongs to {0, ±1}. It turns out that the equation (1.6) is H : Theorem 1.4. A hyperplane H ′ of (1.8) is a second separating hyperplane of [0, 1] d following (1.4) if and only if one of the following conditions is satisfied: Proof. Let P ⊂ R d denote the subpolytope (1.5) of [0, 1] d . Then a hyperplane H ′ of (1.8) is a second separating hyperplane of [0, 1] d following (1.4) if and only if one has H ′ ∩ [0, 1] d ⊂ P.
Hence v ∈ P.
Corollary 1.5. Let be a separating hyperplane of [0, 1] d . Then a hyperplane is a second separating hyperplane of [0, 1] d following (1.9) if and only if one of the following conditions is satisfied:

Order and chain polytopes
A maximal chain is a saturated chain such that x j 1 is a minimal element and x j ℓ is a maximal element of the poset.
The order polytope of P is the convex polytope O(P ) ⊂ R d which consists of those (a 1 , . . . , a d ) ∈ R d such that 0 ≤ a i ≤ 1 for every 1 ≤ i ≤ d together with The chain polytope of P is the convex polytope C(P ) ⊂ R d which consists of those (a 1 , . . . , a d ) ∈ R d such that a i ≥ 0 for every 1 ≤ i ≤ d together with The number of vertices of O(P ) is equal to that of C(P ). Moreover, the volume of O(P ) and that of C(P ) are equal to e(P )/d!, where e(P ) is the number of linear extensions of P ([Stan2, Corollary 4.2]). It also follows from [Stan2] that the facets of O(P ) are the following: , where x j covers x i , and that the facets of C(P ) are the following: • x i = 0 for all x i ∈ P ; • x i 1 + · · · + x i k = 1, where x i 1 < · · · < x i k is a maximal chain of P . Moreover, we have the following descriptions for vertices, which will be used frequently in this section.

Lemma 2.1 ([Stan2]).
(1) Each vertex of O(P ) is ρ(I) such that I is a poset ideal of P ; (2) each vertex of C(P ) is ρ(A) such that A is an antichain of P .
2.1. Existence of seperating hyerplanes for order and chain polytopes. In this subsection, we study the existence of separating hyperplanes of order polytopes and chain polytopes (see Theorem 2.6). First we need an explicit description of edges in terms of vertices.
Conversely, suppose that I = ∅ and J = P and that P is connected. Let x i 1 , . . . , x iq be the maximal elements of P and A i j the set of those elements y ∈ P with y < Clearly h(ρ(P )) = h(ρ(∅)) = 0. Let I be a poset ideal of P with I = ∅ and I = P . What we must prove is h(ρ(I)) > 0. To simplify the notation, suppose that . Then J is a poset ideal of P and h(ρ(J)) ≤ h(ρ(I)). We claim h(ρ(J)) > 0. One has h(ρ(J)) ≥ 0. Moreover, h(ρ(J)) = 0 if and only if no z ∈ J belongs to A i r+1 ∪ · · · ∪ A iq . Now, since P is connected, if follows that there exists z ∈ J with z ∈ A i r+1 ∪ · · · ∪ A iq . Hence h(ρ(J)) > 0. Thus h(ρ(I)) > 0, as desired. Now, suppose that P is connected. If there exist x, x ′ ∈ A and y, y ′ ∈ B with x < y and y ′ < x ′ , then P cannot be connected. We assume y < x if x ∈ A and y ∈ B are comparable. For each x i ∈ A we write a i for the number of elements y ∈ B with y < x i . For each x j ∈ B we write b j for the number of elements z ∈ A with x j < z.
We claim that, for any antichain C of P with C = A and C = B, one has h(ρ(C)) < q. Let C = A ′ ∪ B ′ with A ′ ⊂ A and B ′ ⊂ B. Since P = A ∪ B is connected and since C is an antichain of P , it follows that x i ∈A ′ a i + x j ∈B ′ b j < q. Thus h(ρ(C)) < q, as desired. Now we ask the question whether there exists a separating hyperplane of an order polytope as well as that of a chain polytope.
Lemma 2.4. Let x i , x j ∈ P with x i = x j and H i,j the hyperplane of R d defined by the equation x i = x j . Then the following conditions are equivalent: i,j \ H i,j . In other words, there exist poset ideals I and J of P with x i ∈ I \ J and x j ∈ J \ I. Thus in particular x i and x j are incomparable in P . Hence (ii) ⇒ (iii) follows.
Suppose (iii). Let I be the poset ideal of P consisting of those y ∈ P with y ≤ x i and J the poset ideal of P consisting of those y ∈ P with y ≤ x j . Since x i and x j are incomparable in P , it follows that x i ∈ J and x j ∈ I. Thus ρ(I) ∈ H By virtue of Lemmata 2.4 and 2.5, it follows immediately that Theorem 2.6. Let P be a finite poset, but not a chain. Then each of the order polytope O(P ) and the chain polytope C(P ) possesses a separating hyperplane.

Description of seperating hyperplanes for order and chain polytopes.
In this subsection, we study the necessary and sufficient conditions such that the following hyperplane H : h(x) = c 1 x 1 + c 2 x 2 + · · · + c d x d = 0 becomes a seperating hyperplane for a centain d-element poset P. This study can be very difficult for general posets. Therefore, we focus on the following three basic posets: disjoint chains; binary trees (assume connected); and zigzag posets (assume connected). Notice that there are no "X" shape in all of the three classes of posets, therefore their chain polytopes and order polytopes are unimodular equivalent ( [HL]). In this subsection, we will focus on order polytopes, and all results are also true for chain polytopes.
First, by the definitions of seperating hyperplanes, together with Lemma 2.1 and Lemma 2.2 about the descriptions of the vertices and edges for order polytopes, we have the following description. (1) there exist two poset ideals I and J such that h(ρ(I)) > 0 and h(ρ(J)) < 0 (getting two nontrivial subpolytopes); (2) h(ρ(I))h(ρ(J)) ≥ 0, for each pair of poset ideals I and J such that (I\J) ∪ (J\I) is connected in P .
We call a pair of poset ideals I and J that does not satisfy the second property in Lemma 2.7 a bad pair for h, i.e., h(ρ(I))h(ρ(J)) < 0 and (I\J) ∪ (J\I) is connected in P . In the rest of this subsection, we will prove most necessary conditions for being a seperating hyperplane by constructing bad pairs. We are looking for posets which have the following property.
Consider the following three properties of the hyperplane H.
Property 2.8. Given a poset P , the following form the necessary and sufficient conditions for H to be a seperating hyperplane for O(P ).
Here we always try to aviod having zero coefficients.
Notice that once Property 2.8 is true for some poset P , we can easily check whether a hyperplane is a seperating hyperplane for O(P ). Moreover, the total number of seperating hyperplane will be 2 #{of min elements in P } . Among the three classes of posets we mentioned: disjoint chains, connected binary trees and connected zigzag posets, only disjoint chains satisfy Property 2.8. We will provide counter examples for the other two posets and give the best possible results under certain conditions. Proposition 2.9. Property 2.8 is true for disjoint chains.
Proof. We first prove that all three conditions listed in Property 2.8 are necessary for H to be a seperating hyperplane.
(1) By Lemma 2.7 (1), there exists one order ideal I of P, such that h(ρ(I)) > 0. We assume I is connected, otherwise we look at the chain decomposition of P = C 1 ∪ · · · ∪ C r and consider I ∩ C i , for i = 1, . . . , r. At least one of the intersections is nonempty and satisfies h(ρ(I ∩ C i )) > 0. Now back to the case when I is connected. Since I is a chain, there exists a unique minimal element i in I. We claim that c i ≥ 0, where c i is the coefficient of In fact, if c i < 0, I and J = {i} is a bad pair. Actually, here we can assume c i > 0, since in the case c i = 0, we can simplely throw this element away from the poset and look at the new minimal element in the subposet P\{i}. Since the whole I can not have all zero coefficients, we will just assume c i = 0. Similarly, we also have another minimal element j with c j < 0. (2) We first prove that nonzero coefficients of the minimal elements need to have the same absolute value. For example, consider the following poset. Proposition 2.10. For the binary trees, the following are true: (1) Property 2.8 (1) is necessary.
However, all three conditions in Property 2.8 together are not sufficient for a hyperplane to be a seperating hyperplane. Proof.
(1) We want to show that, there exist two minimal elements i and j such that c i > 0 and c j < 0. The argument in the proof for the disjoint union of chains also works here. The key point is that for any connected poset ideal I in the binary tree and one of its minimal element i, I\{i} is still connected in P.
(2) The argument that all the minimal elements have the same absolute value still holds as in the disjoint union of chains. But it is possible that not all elements have the same absolute value. For example. consider the typerplane as the following labelled represented poset, where the label for an element i in P is the coefficient c i in H. We can check that there are no bad pairs for H, thus H is a seperating hyperplane. But not all coefficients in H have the same absolute value.
(3) Now assume all coefficients have the same absolute value, and thus can only take value from {−1, 0, 1} after rescaling. So here we only need to talk about the sign for an element i in P (+ refers to c i = 1 and − refers to c i = −1). Now we want to show that the sign of an element is determined by the sign of its two children. Here "the sign of the child" refers to the sign of the poset ideal generated by that child. In particular, there are exactly six local sign patterns: Notice that 0 appears if and only if its children have a + and a −. For two elements a, b with a common parent d, we have c d = 0, which corresponds to the third tree above. (c) suppose c b > 0 and c a = 0. This indicates that a is larger than some minimal element e with c e < 0. Then by the pair ({e}, < d >) and ({b}, < d >), we have h(ρ(< d >)) = 0, thus c d = −c b , which corresponds to the forth tree above. The fifth and the sixth tree can be obtained in a similar way. (4) Following the above rule will not always result in a separating hyperplane.
For example, consider the hyperplane represented by the following labelled poset.
One can easily check that the above hyperplane follows the six local rules listed above as well the other two conditions in Property 2.8. However, for example, I =< g, e, f > and J =< a, b, c, d, e, f > is a bad pair.
Proposition 2.11. For the zigzag posets, Property 2.8 (1) is not necessary for H to be a seperating hyperplane. However, for any hyperplane H with Property 2.8 (1), the rest two conditions listed in Property 2.8 are necessary and sufficient conditions for H to be a seperating hyperplane.
Proof. The following example is a seperating hyperplane but does not satisfy Property 2.8 (1).
Now assume H is a hyperplane satisfying Property 2.8 (1). We first prove that if H is a seperating hyperplane, then both Property 2.8 (2) and (3) are true.
(1) We want to prove that all the nonzero coefficients in any separating hyperplane for a zigzag poset have the same absolute value. First notice that, all the minimal elements have the same absolute value, as proved in Proposition 2.9. Following the same proposition, all the non maximal elements (if nonzero) have the same absolute value. As for the maximal elements, let us has a closer look at the zigzag poset. One maximal element m covers at most two minimal elements p, q. For the case m only covers one minimal element, we have the coefficient c m need to have the same absolute value for the same reason as disjoint chains proved in Proposition 2.9. Now there are two cases when m covers two minimal elements p, q: (a) c p · c q < 0. Let I =< m > be the poset ideal generated by m. Consider the pair I and J, where J = {p} or {q}. We have h(ρ(I)) = 0, which implies |c m | ≤ 1. (b) c p · c q > 0. Say c p = c q = 1. Let n be a maximal element adjacent to m that covers two minimal elements with different signs. For example, Consider the poset ideal I =< m, n >. Similar as the previous case, we have h(ρ(I)) = 0, which still implies |c m | ≤ 1.
(2) Since H satisfies conditions (1) and (2) in Property 2.8, once we fix the signs of all the minimal elements, all elements except those maximal are uniquely determined the same way as the disjoint chains (Proposition 2.9). As for the the maximal elements, they are uniquely determined by the signs of their two children the same as the binary trees (Proposition 2.10). Now we want to prove that any hyperplane h(x) = 0 satisfying the three conditions listed in Property 2.8 is a seperating hyperplane. The condition (1) in Property 2.8 implies condition (1) in Lemma 2.7. Now we want to show that there are no bad pairs. Notice that by the rules descripted above, any connected component has value sum to {1, 0, −1}. In the case I ⊂ J, if h(ρ(I)) < 0, then h(ρ(J)) = h(ρ(I)) + h(ρJ\I) < 1, since h(ρ(J\I)) < 1. Now we claim that for the zigzag poset, the condition that (I\J) ∪ (J\I) is connected, implies that I\J or J\I is empty. Consider a generic connected subposet S = (I\J) ∪ (J\I). We want to show that S ⊂ I or S ⊂ J. If S only has one maximal element, then it is clear that all the elements belong to the same order ideal as the maximal element (either I or J). If there are more than one maximal element, see the following example.
Consider two adjacent maximal elements (here they are a and b in the example). These two maximal elements cover a common minimal element d, because this subposet is connected. Then d belongs to the same poset ideal as both a and b. Therefore, both a and b belong to the same poset ideal. This shows that S belongs to either I or J.

Birkhoff polytopes
Birkhoff polytopes B n are defined to be the convex hull of all n × n nonnegative matrices with row sum and column sum equal to one. These matrices are known as the doubly stochastic matrices. Here we consider an n × n matrix as a n 2vector. Birkhoff polytopes are well-studied polytopes and have many applications, in combinatorial optimization and Bayesian statistics, for example. In this section, we look for seperating hyperplanes for B n (Theorem 3.3).
In the rest of the section, we assume the hyperplanes have the form h(x) = 0, but actually all the results holds for general hyperplanes h(x) = r for any constant r. We start with the following known properties of Birkhoff polytopes B n . Here we use both the one line notation and the cycle notation for a permutation. For example, w = 34256187 is the one line notation for the permutation sending 1 → 3, 2 → 4, 3 → 2, 4 → 5, 5 → 6, 6 → 1, 7 → 8 and 8 → 7. The cycle notation for w is (132456)(78), thus w has two cycles.
(1) dim B n = (n − 1) 2 ; (2) B n has n! vertices, which are all the matrices corresponding to permutations S n ; (3) permutations w and u form an edge in B n if and only if w −1 u has one cycle (excluding the fixed points). [reference?] In particularly, for n = 3, w −1 u has one cycle for any w, u ∈ S 3 . In other words, the skeleton graph for B 3 is the complete graph K 6 . Therefore, there are no seperating hyperplanes for B 3 . Moreover, we have Lemma 3.1. B 4 has no seperating hyperplanes. Now consider the permutation σ 1 = (C 3 ) · · · (C k ). Since it has fewer than k cycles, we have h(x σ 1 ) ≥ 0. Notice that where σ 2 = (325A)C 3 · · · C k and σ 3 = (146B)C 3 · · · C k . One can check that σ 2 and σ 3 are both connected with v. Since σ 2 and σ 3 both have fewer cycles than v, we have h(x (σ 2 ) ) = 0 and h(x σ 3 ) = 0. This is a contradiction, since h(x τ 1 ) > 0 and h(x σ 1 ) ≥ 0.