Face Structure and Facets

In this section, we study the face structure ofO(P,λ). As it turns out, the faces of marked order polyhedra correspond to certain partitions of the underlying posetP. Our goal is to characterize those partitions combinatorially. We associate to each pointxinO(P,λ)a partitionπ_x ofP, that will suffice to describe the minimal face ofO(P,λ)containingx. The partitions that are obtained in this way from points of the polyhedron will then—ordered by refinement—capture the polyhedrons face structure.

Definition 6.2.1. LetQ = O(P,λ) be a marked order polyhedron. To eachx ∈ Q we associate a partitionπ_x ofP induced by the transitive closure of the relation

p ∼_x q if x_p =x_q andp,qare comparable.

6.2. Face Structure and Facets We may think ofπ as being obtained by first partitioningPinto blocks of constant values underxand then splitting those blocks into connected components with respect to the Hasse diagram ofP.

Given any partitionπ ofP, we call a blockB ∈π free ifP^∗∩B =œand denote byπ˜ the set of all free blocks ofπ. Note that anyx ∈ O(P,λ)is constant on the blocks ofπ_x and the values on the non-free blocks ofπ_x are determined byλ.

Letx ∈Q be a point of a polyhedron. We denote the minimal face ofQ containingx byF_x. Hence,F_x is the unique face havingxin its relative interior. Equivalently,F_x is the intersection of all faces ofQ containingx.

Proposition 6.2.2. Let x ∈ Q = O(P,λ) be a point of a marked order polyhedron with associated partitionπ =π_x. We have

F_x = {y ∈Q :yis constant on the blocks of π } and dimF_x = |π˜|.

Proof. Forp <qinP letH_p<q = ∂H_p<q⁺ be the hyperplane defined byx_p =x_q inR^P. The minimal face of a pointx ∈Q is then given by

F_x =Q ∩ ⋂

p<q, xp=xq

H_p<q.

A pointy ∈Q satisfiesy_p =y_q for allp <qwithx_p =x_q if and only ify is constant on the blocks ofπ_x. Thus,F_x is indeed given by ally ∈Q constant on the blocks ofπ_x.

To determine the dimension ofF_x, we consider its affine hull aff(F_x). It is obtained by intersecting the affine hull ofQ with allH_p<q such thatx_p =x_q. The affine hull ofQ itself is the intersection of allH_a fora∈P^∗and allH_p<q such thaty_p =y_q for ally ∈Q. Putting these facts together, we have

aff(F_x)= ⋂

a∈P^∗

H_a ∩ ⋂

p<q, yp=yq∀y∈Q

H_p<q ∩ ⋂

p<q, xp=xq

H_p<q = ⋂

a∈P^∗

H_a ∩ ⋂

p<q, xp=xq

H_p<q.

This is exactly the set of allyconstant on the blocks ofπ_x and satisfyingy_a =λ(a)for alla ∈P^∗. Suchy are uniquely determined by values on the free blocks ofπ_x and thus dim(F_x)= |π˜_x|as desired.

Corollary 6.2.3. If λ is a strict marking onP, the dimension of O(P,λ) is equal to the number of unmarked elements inP.

Proof. Since all coordinates inP^∗are fixed byλ, we always have dimO(P,λ) ≤ P˜

. Ifλis strict, there is a pointx ∈ O(P,λ)such thatx_p <x_q wheneverp <qby Proposition6.1.2.

Hence,π_x is the partition ofP into singletons and dimF_x = |π˜_x| = P˜

. We conclude that F_x = O(P,λ), soxis a relative interior point and the marked order polyhedron has the desired dimension.

Figure 6.2.: The face partitions of the marked order polytope in Example6.1.3.

Corollary 6.2.4. Letx ∈Q =O(P,λ)be a point of a marked order polyhedron. Fory ∈Q we havey ∈F_x if and only if π_x is a refinement of π_y.

Proof. By Proposition6.2.2,y ∈F_x if and only ifyis constant on the blocks ofπ_x. Lety be constant on the blocks ofπ_x. Any blockBofπ_x is connected with respect to the Hasse diagram ofP andy takes constant values onB, henceBis contained in a block ofπ_y by construction andπ_x is a refinement ofπ_y. Now lety ∈Qwithπ_x being a refinement of π_y. We conclude thaty is constant on the blocks ofπ_x, since it is constant on the blocks ofπ_yandπ_x is a refinement ofπ_y.

Corollary 6.2.5. Given any two pointsx,y ∈ O(P,λ), we haveF_y ⊆ F_x if and only if π_x is a refinement of π_y. In particular F_y =F_x if and only if π_y =π_x.

Hence, the partition of O(P,λ)into relative interiors of its faces is the same as the partition given byx ∼ y ifπ_x = π_y and we can associate to each non-empty face F a partitionπ_F withπ_F =π_x for anyxin the relative interior ofF. We call a partitionπ ofP aface partitionof(P,λ)ifπ =π_F for some non-empty face of O(P,λ). We arrive at the following description of face lattices of marked order polyhedra.

Corollary 6.2.6. LetQ = O(P,λ)be a marked order polyhedron. The poset F (Q) \ {œ}

of non-empty faces of Q is isomorphic to the induced subposet of the partition lattice onP given by all face partitions of (P,λ).

For the marked order polytope from Example6.1.3, we illustrated the face partitions in Figure6.2. The free blocks are highlighted in blue round shapes, non-free blocks in red angular shapes. We see that the dimensions of the faces are given by the numbers of free blocks in the associated face partitions and that face inclusions correspond to refinements of partitions.

In order to characterize the face partitions of a marked poset(P,λ)combinatorially, we introduce some properties of partitions ofP.

6.2. Face Structure and Facets Definition 6.2.7. Let(P,λ)be a marked poset. A partitionπ ofP isconnected if the blocks ofπ are connected as induced subposets ofP. It isP-compatible, if the relation≤ defined onπ as the transitive closure of

B ≤C if p ≤qfor somep ∈B,q ∈C

is anti-symmetric. In this case≤ is a partial order onπ. AP-compatible partitionπ is called(P,λ)-compatible, if whenevera ∈B∩P^∗ andb ∈C ∩P^∗for some blocksB ≤C, we haveλ(a) ≤ λ(b).

Remark 6.2.8. Whenever a partitionπ of a posetP isP-compatible, it is alsoconvex. That is, fora <b < c withaandc in the same blockB ∈π, we also haveb ∈B, since otherwise the blocks containingaandbwould contradict the relation on the blocks being anti-symmetric. This implies that the blocks in a connected,P-compatible partition are not just connected as induced subposets ofP but even connected as induced subgraphs of the Hasse diagram ofP.

Proposition 6.2.9. Let(P,λ)be a marked poset. A(P,λ)-compatible partitionπof Pgives rise to a marked poset (P/π,λ/π) whereP/π is the poset of blocks inπ,(P/π)^∗ = π \π˜ andλ/π: (P/π)^∗ →Ris defined by(λ/π)(B)=λ(a)for anya ∈B∩P^∗. Furthermore, the quotient mapP →P/π defines a map(P,λ) → (P/π,λ/π)of marked posets.

Proof. Sinceπ isP-compatible, the blocks ofπ form a posetP/π as in Definition6.2.7.

Since π is (P,λ)-compatible, we have λ(a) = λ(b) whenevera,b ∈ B ∩P^∗ for some non-free blockB ∈π. Hence, the mapλ/π is well-defined. It is order-preserving by the definition of(P,λ)-compatibility. Furthermore, we have a commutative diagram

P P^∗ R

P/π (P/π)^∗ R.

λ

λ/π

Thus, we have a quotient map(P,λ) → (P/π,λ/π).

Proposition 6.2.10. Every face partitionπ_F of (P,λ)is(P,λ)-compatible, connected and the induced marking on(P/π_F,λ/π_F)is strict.

Proof. LetF be a non-empty face ofO(P,λ). It is obvious thatπ_F is connected by con-struction, since it is given by the transitive closure of a relation that only relates pairs of comparable elements. To verify thatπ_F isP-compatible, we need to check that the induced relation ≤ on the blocks of π_F is anti-symmetric. Assume we have blocks B,C ∈π_F such thatB ≤C andC ≤ B. SinceB ≤C, there is a finite sequence of blocks

B =X1,X2,. . .,X_k,X_k+1 =C such that fori = 1,. . .,kthere are somep_i ∈X_i,q_i ∈X_i+1

withp_i ≤ q_i. Take anyx in the relative interior of F, thenx_pi ≤ x_qi fori = 1,. . .,k and sincex is constant on the blocks ofπ_F, we havex_qi = x_pi+1

fori = 1,. . .,k−1. To

summarize, we have

x_p1 ≤ x_q1 =x_p2 ≤ x_q2 = · · · ≤ · · ·=x_pk ≤ x_qk. (6.1)

Hence, the constant valuex takes onB is less than or equal to the constant valuextakes onC. Since we also haveC ≤ B, we conclude thatx takes equal values on the blocks BandC. From (6.1) we conclude thatx takes equal values on all blocksX_i. From the definition ofπ_x =π_F it follows that the blocksX_iare in fact all equal, in particularB =C and the relation is anti-symmetric.

To see thatπ_F is(P,λ)-compatible, letB,C ∈π be non-free blocks withB ≤C. By the same argument as above, we know that anyx ∈F has constant value onBless than or equal to the constant value onC, soλ(a) ≤ λ(b)for markeda∈B,b ∈C. Ifλ(a)=λ(b) we haveB =C, by the same argument as above, so the induced marking is strict.

Given any partitionπ ofP, we can define a polyhedronF_π contained inO(P,λ)by F_π = {y ∈Q :y is constant on the blocks ofπ}.

Ifπ =π_F is a face partition of(P,λ), we haveF_π =F by Proposition6.2.2. However,F_π is not a face for all partitionsπ ofP.

As long as π is (P,λ)-compatible, we can show that the polyhedron F_π is affinely isomorphic to the marked order polyhedron O(P/π,λ/π). The isomorphism will be induced by the quotient mapP → P/π. Our first step is to verify that this induced map is indeed an injection.

Lemma 6.2.11. Let f : (P,λ) → (P^′,λ^′)be a map of marked posets. If f is surjective, the induced map f^∗: O(P^′,λ^′) → O(P,λ)is injective.

Proof. Letx,y ∈ O(P^′,λ^′)such that f^∗(x) = f^∗(y). Given anyp ∈P^′we need to show x_p =y_p. Since f is surjective,p = f(q)for someq ∈P and thus

x_p =x_f(q) = f^∗(x)_q = f^∗(y)_q =y_f(q) =y_p.

Proposition 6.2.12. Let(P,λ)be a marked poset andπ a(P,λ)-compatible partition. The quotient mapq: (P,λ) → (P/π,λ/π)induces an injection

q^∗: O(P/π,λ/π)↪−→ O(P,λ) with imageq^∗(O(P/π,λ/π))=F_π.

Proof. By Lemma6.2.11we know thatq^∗is an injection. Hence, we only need to verify thatF_π is the image ofq^∗. The image is contained inF_π, since wheneverpandp^′are in the same blockB ∈π, we have

q^∗(x)_p =x_q(p) =x_B =x_q(p^′) =q^∗(x)_p^′.

Hence, allq^∗(x)are constant on the blocks ofπ. Conversely, given any pointy ∈ O(P,λ) constant on the blocks ofπ, we obtain a well defined mapx: P/π → Rsending each block to the constant valuey_p for allpin the block. This map is a pointx ∈ O(P/π,λ/π) mapped toybyq^∗.

6.2. Face Structure and Facets The previous proposition tells us, that whenever we have a(P,λ)-compatible partition π, the marked order polyhedronO(P/π,λ/π)is affinely isomorphic to the polyhedron F_π ⊆ O(P,λ)via the embeddingq^∗induced by the quotient map. From now on, we refer to affine isomorphisms arising this way as thecanonical affine isomorphismO(P/π,λ/π) F_π.

Corollary 6.2.13. For every non-empty face F of a marked order polyhedronO(P,λ)we have a canonical affine isomorphismO(P/π_F,λ/π_F) F.

We are now ready to state and prove the characterization of face partitions of marked posets.

Theorem 6.2.14. A partitionπ of a marked poset(P,λ)is a face partition if and only if it is (P,λ)-compatible, connected and the induced marking on(P/π,λ/π)is strict.

Proof. The fact that face partitions satisfy the above properties is the statement of Propo-sition6.2.10. Now letπ be a partition ofPthat is(P,λ)-compatible, connected and induces a strict markingλ/π. By Proposition6.1.2, there is a pointz ∈ O(P/π,λ/π)such that z_B < z_C wheneverB < C. Letx ∈ R^P be the point in the polyhedron F_π ⊆ O(P,λ) obtained as the image ofzunder the canonical affine isomorphismO(P/π,λ/π) ∼−→F_π. We claim thatπ =π_x, soπ is a face partition. Sincexis constant on the blocks ofπ and π is connected, we know thatπ is a refinement ofπ_x. Now assume that the equivalence relation∼_x definingπ_x relates elements in different blocks ofπ. In this case, there are blocksB ,C ofπ with elementsp ∈B,q ∈C such thatx_p =x_q andp <q. This implies thatz_B =z_C andB <C, a contradiction to the choice ofz. Hence,π =π_x andπ is a face partition of(P,λ).

Remark 6.2.15. To decide whether a given partitionπ of a marked poset(P,λ)satisfies the conditions in Theorem6.2.14, it is enough to know the linear order onλ(P^∗). The exact values of the marking are irrelevant. Hence, the face lattice ofO(P,λ)is determined solely by discrete, combinatorial data. In fact, since the directions of facet normals do not depend on the values ofλ, we can conclude that the normal fanN (O(P,λ))is determined by this combinatorial data. However, the affine isomorphism type ofO(P,λ)does depend on the exact values ofλ.

Example 6.2.16. We construct a continuous family(Q_t)_t∈[0,1]of marked order polytopes, whose underlying marked posets all yield the same combinatorial data in the sense of Remark6.2.15, butQ_s andQ_t are affinely isomorphic if and only ifs = t. Let(P,λ_t)be the marked poset shown in Figure 6.3a. Lettingt vary in[0, 1], we obtain for eacht a different affine isomorphism type, since two of the vertices ofQ_t will move, while the other three stay fixed and are affinely independent as can be seen in Figure6.3b. However, allQ_t share the same normal fan and are in particular combinatorially equivalent. ♢ We continue our study of the face structure of marked order polyhedra by having a closer look at facets. Since inequalities in the description of marked order polyhedra come from covering relations in the underlying poset, we expect a correspondence of

0 p

q 4

1+t

3

(a) the marked poset(P,λ_t)

x_p x_q

Q_t

3 4

1+t

(b) the polytopeQt =O(P˜ ,λt)

Figure 6.3.: The marked poset (P,λ_t)from Example 6.2.16 and the associated marked order polytopeQ_t =O(˜P,λ_t).

facets to certain covering relations. If the marked poset satisfies a certain regularity condition, the facets are indeed in bijection with the covering relations. Hence, if we can change the underlying poset of a marked order polyhedron to a regular one, without changing the associated polyhedron, we obtain an enumeration of facets. We start by modifying an arbitrary marked poset to a strict one by contracting constant intervals.

Proposition 6.2.17. Given any marked poset (P,λ), the partitionπ induced by the rela-tionsa ∼ pandp ∼b whenever [a,b]is a constant interval containingpyields a strictly marked poset (P/π,λ/π)such that O(P/π,λ/π) F_π = O(P,λ)via the canonical affine isomorphism.

Proof. Letx ∈ O(P,λ)be a point constructed as in the proof of Proposition6.1.2. By construction we havex_p = x_q forp < q if and only if there area,b ∈ P^∗ witha ≤ p <

q ≤ b withλ(a) = λ(b). Thus, we conclude that π_x = π andπ is a face partition of O(P,λ). Since every point of O(P,λ)satisfiesx_a =x_p =x_b whenever[a,b]is a constant interval containingp, we conclude that F_π is indeed the whole polyhedron. Hence, O(P/π,λ/π) F_π = O(P,λ), whereλ/π is a strict marking by Proposition6.2.10.

Definition 6.2.18. Let(P,λ) be a marked poset. A covering relationp ≺ q is called non-redundantif for all marked elementsa,bsatisfyinga≤ qandp ≤b, we havea=b orλ(a) < λ(b). Otherwise the covering relation is calledredundant. The marked poset (P,λ)is calledregular, if all its covering relations are non-redundant.

Apart from the desired correspondence of covering relations and facets, regularity of marked posets implies some useful properties of the marked poset itself.

Proposition 6.2.19. Let (P,λ) be a regular marked poset. The following conditions are satisfied:

i) the markingλis strict,

ii) there are no covering relations between marked elements,

iii) every element inP covers and is covered by at most one marked element.

6.2. Face Structure and Facets

0 p

q 3

2

1

(a) a marked poset(P,λ)

x_p x_q

x_p =x_q

1 2 3

(b) the polytopeO(P˜ ,λ)

Figure 6.4.: The marked poset(P,λ)from Remark6.2.20and the associated marked order polytopeO(˜P,λ). The covering relationp ≺qis redundant.

Proof. i) Whena <bare marked elements ofP, there is some covering relationp ≺q such thata ≤p ≺q ≤b. Sincea ≤qandp ≤b, we haveλ(a)< λ(b)by regularity.

ii) Whenb ≺ais a covering relation between marked elements, we haveλ(a) <λ(b) by choosingp =b,q =ain the regularity condition. This is a contradiction toλ being order-preserving.

iii) Whena,b ≺qfor markeda,b, the regularity condition fora ≤qandb ≤bimplies a = b orλ(a) < λ(b). By the same argument we geta = b orλ(b) < λ(a). We conclude thata =b.

Remark 6.2.20. The conditions in Proposition6.2.19are necessary, but not sufficient for(P,λ)to be regular. The marked poset in Figure6.4asatisfies all three conditions, but the covering relationp ≺qis redundant.

In fact, this example shows that the process described by Fourier in [Fou16, Sec. 3] does not remove all redundant covering relations and hence leads to a notion of regularity that is not sufficient to have facets in correspondence with covering relations.

The same marked poset also serves as a counterexample to the characterization of face partitions in [JS14, Prop. 2.3]. Instead of partitions of P in terms of blocks, they use subposets ofP that have all the elements ofP but only some of the relations. The connected components of the Hasse diagram of such a subposetG give a connected partitionπ_GofP and conversely every connected partitionπ defines a subposetG_π on the elements ofP by havingp ≤ qinG_π if and only ifp ≤qinP andpandqare in the same block ofπ. When(P,λ)is the marked poset in Figure6.4andGis the subposet with p ≤ q as the only non-reflexive relation—i.e.,{p,q} is the only non-singleton block in π_G—the conditions in Proposition 2.3 of [JS14] are satisfied butG does not yield a face of O(P,λ)as can be seen in Figure6.4b.

Theorem 6.2.21. Let (P,λ)be a regular marked poset. The facets of O(P,λ)correspond to the covering relations in(P,λ).

Proof. Since(P,λ)is strictly marked, the dimension ofO(P,λ)is equal to the number of unmarked elements inP. Hence, a facetF corresponds to a(P,λ)-compatible, connected partitionπ ofP such thatλ/π is strict andπ has exactly |P˜| −1 free blocks. We claim

that the number of non-free blocks ofπ is|P^∗|. Assume there are marked elementsa,b in a common blockBofπ. Sinceπ has

P˜

−1 free blocks, at most one unmarked element can be in a non-free block. Since(P,λ)is regular, there are no covering relations between marked elements. Hence, since B is connected as an induced subgraph of the Hasse diagram ofP and contains botha andb, it also contains the only unmarked element p in a non-free block, and we have one of the following four situations: a ≺ p ≺ b, a ≻ p ≻ b, a ≺ p ≻ b ora ≻ p ≺ b. Sincea andb are in the same block, they are identically marked and the first two possibilities contradictλ being strict. The other two possibilities contradict regularity, sincepcovers—or is covered by—more than one marked element. Hence,π has exactly|P^∗|non-free blocks and we conclude thatπ has

|P| −1 blocks overall. Therefore,π consists of|P| −2 singletons and a single connected 2-element block corresponding to a covering relation ofP.

Conversely, letp ≺qbe a covering relation ofP. We claim that the partitionπ with the only non-singleton block{p,q}is a face partition with|P˜| −1 free blocks. Since(P,λ) is regular, it contains no covering relation between marked elements andπ has exactly

|P˜| −1 free blocks. Since{p,q}is the only non-singleton block andp ≺q, the partitionπ is connected andP-compatible. To verify thatπ is(P,λ)-compatible andλ/π is strict, let B,Cbe non-free blocks ofπ witha ∈B∩P^∗andb ∈C∩P^∗such thatB ≤C. WhenB =C, we havea =b andλ(a)= λ(b). WhenB <C, we concludea < bora ≤ q,p ≤b, since {p,q}is the only non-trivial block. In both cases, regularity impliesλ(a)< λ(b).

Now that we established a regularity condition on marked posets that guarantees a bijection between covering relations inP and facets of the marked order polyhedron, we explain how to transform any given marked poset to a regular one.

Proposition 6.2.22. Let (P,λ)be a strictly marked poset. Redundant covering relations inP can be removed successively to obtain a regular marked poset (P^′,λ) with the same associated marked order polyhedronO(P^′,λ)= O(P,λ).

Proof. Letp ≺qbe a redundant covering relation inP. That is, there are marked elements a,bsatisfyinga ≤q,p ≤b andλ(a) ≥λ(b). LetP^′be obtained fromP by removing the covering relationp ≺qfromP. ObviouslyO(P,λ)is contained in O(P^′,λ).

Now letx ∈ O(P^′,λ). To verify thatx is a point ofO(P,λ), we have to showx_p ≤ x_q. Sinceλis a strict marking onP, we can not havea ≤ p. Otherwisea ≤ p ≤ b implies a <b, in contradiction toλ(a) ≥λ(b). Hence, removing the covering relationp ≺qwe still havea ≤^′qinP^′. By the same argumentp ≤^′b. Thus, by the defining conditions of O(P^′,λ), we have

x_p ≤x_b =λ(b) ≤λ(a)=x_a ≤x_q.

Therefore,x ∈ O(P,λ)and we concludeO(P^′,λ)= O(P,λ). This process can be repeated until all redundant covering relations have been removed, resulting in a regular marked poset defining the same marked order polyhedron.

Remark 6.2.23. Note that Proposition6.2.22does not imply, that all covering relations that are redundant in(P,λ)can be removed simultaneously. Removing a single redundant

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