arithmetic of finite ordered sets: cancellation of exponents, i

Order 16: 313–333, 1999.© 2000Kluwer Academic Publishers. Printed in the Netherlands.

313

Arithmetic of Finite Ordered Sets: Cancellation ofExponents, I

RALPH McKENZIEDepartment of Mathematics, Vanderbilt University, Nashville, TN 37240, U.S.A.

(Received: 23 March 1999; accepted in revised form: 14 March 2000)

Abstract. Garrett Birkhoff conjectured in 1942 that whenA,B,P are finite posets satisfyingAP ∼=BP, thenA ∼= B. We show that this is true in caseP is dismantlable to each of its points, orP isconnected and each ofA andB is dismantlable to each of its covering pairs.

Mathematics Subject Classification (1991):06A07.

Key words: cancellation of exponents, exponentiation, posets.

1. Introduction

This paper concerns a general operation on ordered sets (posets, partially orderedsets). Given two posetsA andB,AB denotes the set of all order preserving (mono-tone) functions fromB into A, construed as a poset in which the order is definedpointwise,f ≤ g iff f (x) ≤ g(x) for all x ∈ B. This operation, called exponenti-ation, was defined and studied by Birkhoff [3, 4]. In [4], Birkhoff conjectured thatwhenA,B, andP are finite, then the equationAP ∼= BP always entailsA ∼= B.This conjecture remains open, although positive results have been obtained underseveral different sets of hypotheses. The papers [1, 2, 6, 7, 9, 10, 13] and [12] arerepresentative of the literature on this and related problems regarding exponentia-tion of posets. The last paper contains a bibliography that was rather complete asof 1982.

In this paper we supply further evidence suggesting that Birkhoff’s conjecturemay be true. We prove thatA ∼= B is forced whenP is directly indecomposableand the isomorphismφ: AP ∼= BP and its inverse have the Point Covering Property(which will be defined in the third section). For example, if either (i)P is dismant-lable down to each of its one-element subsets; or (ii)A is dismantlable down eachof its two-element subsets{u, v} wherev coversu, andB has the same property –thenφ and its inverse will possess the Point Covering Property.

Condition (i) holds ifP is a fence, or more generally, is a tree-order, i.e., if thecovering graph ofP is a tree (connected and without cycles). Also, condition (i)holds if P has a least or largest element. Condition (ii) holds if bothA andB aretree-orders, or if one of them has a least or largest element.

314 RALPH McKENZIE

We prove thatA ∼= B is forced whenP is directly indecomposable and everymaximal element ofA has maximum height inA. All these results are in the finalsection of the paper.

The implicationAP ∼= BP⇒ A ∼= B was proved in Bergman et al. [2] whenP isany linear order andA andB are arbitrary (they may be infinite). The cancellationof P was proved in Jónsson and McKenzie [12] under an assortment of differenthypotheses. WhenA,B,P are finite and connected, their results amount to thevalidity of Birkhoff’s conjecture in the cases where eitherP or A has a least ora largest element. The methods of this paper seem to overlap only a little withthe methods they used, although their results on the Birkhoff conjecture for finiteposets are corollaries of the results in this paper.

Just as Jónsson’s productive method for proving cancellation of exponents evolv-ed from considering the very special caseP = 2, the ideas in this paper werediscovered while considering the special case whereP is the four-element fence.It is an open question whether Birkhoff’s conjecture is valid when the exponent isthe four-element crown.

2. Finite Posets

We now gather a few basic definitions and results about finite posets which we shallneed later. We use boldface capital letters to denote posets throughout the paper,thusP= 〈P,≤〉whereP is the set of elements (or points) ofP. Thedualof a posetP = 〈P,≤〉 is the posetP∂ = 〈P,≥〉. The covering relation inP is denoted by≺.Thusx ≺ y denotes thatx < y and the intervaly/x = {u : x ≤ u ≤ y} is identicalto the set{x, y}. For a subsetX of P we writeX↑ for the set{y : for somex ∈X, x ≤ y}, andX↓ for the set{y : for somex ∈ X, y ≤ x}; and ifX = {x} then wewrite x↑ in place ofX↑, andx↓ in place ofX↓. An elementx ∈ P is irreducibleiff either x↑\{x} has a smallest elementx◦, or x↓\{x} has a largest elementx◦. Inthe first case, we say thatx is meet-irreduciblewith unique upper coverx◦. In thesecond case,x is join-irreduciblewith unique lower coverx◦.

A subposetQ of a posetP is a subsetQ of P together with the restriction ofthe order relation onP to Q. Frequently, we writeQ, rather thanQ, to denotethis subposet. The set of join-irreducible elements inP will be denotedJ (P) andif nonvoid, the corresponding subposet will be denotedJ(P). We remark that ourdefinition ofJ(P) differs from that in Jónsson and McKenzie [12].

Thediagonalin AP is the set of constant functions belonging toAP . Fora ∈ A,the constant function with valuea is denoted〈a〉, and called adiagonal elementofAP. (In this usage, the domain of the function〈a〉 is understood to beP .)

A finite posetP is dismantlable to its subsetX if P can be written asX ∪{y1, . . . , yn} wheren ≥ 0, y1, . . . , yn are distinct and not inX, and each elementyi is irreducible in the subposetX ∪ {yi, . . . , yn}. A finite posetP is dismantlableiff it is dismantlable to some singleton subset.

ARITHMETIC OF FINITE ORDERED SETS 315

A retractionof a posetP onto a subposetQ is a monotone mapπ :P→ Q suchthatπ restricted toQ is the identity map. A monotone mapφ :P → Q, Q ⊆ P,is increasingif for all x ∈ P, φ(x) ≥ x, decreasingif for all x ∈ P , φ(x) ≤ x.Suppose thatb is a joint-irreducible element inP, with subcovera. Then the mapπ onP defined byπ(b) = a, π(x) = x for x 6= b, is a decreasing retraction ofPonto its subposetP \{b}. If, instead,b is a meet-irreducible element, then there isan analogous increasing retraction ofP ontoP \{b}.

Using these ideas, it is easy to show that a finite posetP is dismantlable toX ⊆ P iff there is a chain of subsetsP = P0 ⊇ P1 ⊇ · · · ⊇ Pn = X and mapsπisuch that fori < n, πi is an increasing, or decreasing, retraction ofPi ontoPi+1.From this, it follows, with a little work, that ifP is a finite poset, retractible onto itssubposetQ, then ifP is dismantlable, so isQ; and ifP is dismantlable to{q} ⊆ Q,thenQ is dismantlable to{q}.

3. Local Behaviour ofφ

Throughout this section and the next, we assume thatA,B andP are finite.

DEFINITION 3.1. Forf, g ∈ AP andp ∈ P , we writef <p g to denote thatf < g andf (x) = g(x) for all x ∈ P \{p}. We writef ≺p g to denote thatf <p g andf ≺ g, i.e.,g coversf in AP.

If u ≺ v in A andp ∈ P , we useCp(u, v) to denote the set of all pairs(f, g)of elements ofAP such thatf ≺p g and(f (p), g(p)) = (u, v). In contexts whereseveral exponents are being considered, we may writeCP

p(u, v), and if both baseand exponent need to be identified, we writeCP

A,p(a, b).

LEMMA 3.1. (i) If f ≺ g in AP, there exists a unique pointp ∈ P such thatf ≺p g.

(ii) Suppose thatf < g in AP andf (p) ≤ u ≺ v ≤ g(p). Then there exists(h, k) ∈ Cp(u, v) such thatf ≤ h ≺p k ≤ g.

Proof. Suppose thatf < g in AP, and choose any pointp for which f (p) <g(p). Defineα andβ so thatα(x) = g(x) for x > p andα(x) = f (x) otherwise,β(x) = g(x) for x ≥ p andβ(x) = f (x) otherwise. Thus,α, β ∈ AP and

f ≤ α <p β ≤ g.Moreover, iff (p) ≤ a ≤ g(p), then the functionδa which agrees withf atx /≥p,agrees withg atx > p, and hasδa(p) = a, clearly belongs toAP and to the intervalβ/α.

To prove (i), suppose that in fact,f ≺ g. The displayed formula implies thatf = α, g = β. If f (p) ≤ a ≤ g(p) then δa ∈ g/f = {f, g} implies a ∈{f (p), g(p)}. Hencef (p) ≺ g(p), andf ≺p g.

To prove (ii), suppose thatf (p) ≤ u ≺ v ≤ g(p). It is clear thatf ≤ δu ≺pδv ≤ g. 2

316 RALPH McKENZIE

POINT COVERING PROPERTY. Suppose thatφ :AP ∼= BP. Fora ≺ b in A andp ∈ P , we say thatφ possesses the point covering property at(p, a, b) providedthere existsq ∈ P such that for all(f, g) ∈ Cp(a, b), φ(f ) ≺q φ(g). We saythatφ possesses thepoint covering propertyprovided it possesses the property at(p, a, b) for all a ≺ b in A and for allp ∈ P .

LEMMA 3.2. Suppose thatφ :AP ∼= BP, a ≺ b in A andp ∈ P . If Cp(a, b) is aconnected subposet ofAP×AP, thenφ has the point covering property at(p, a, b).

Proof.Suppose thatCp(a, b) is connected. For each(f, g) ∈ Cp(a, b), we havethat φ(f ) ≺ φ(g) in BP and so there must exist a uniqueq ∈ P , dependingpossibly on(f, g), so thatφ(f ) ≺q φ(g) in BP. We must prove thatq is inde-pendent of(f, g). Obviously, sinceCp(a, b) is connected, it suffices to prove thatwhenever(f, g), (k, `) ∈ Cp(a, b) andf ≤ k, g ≤ ` then there existsq withφ(f ) ≺q φ(g) andφ(k) ≺q φ(`). So assume that(f, g), (k, `) ∈ Cp(a, b), f ≤k, g ≤ ` andφ(f ) ≺q φ(g) andφ(k) ≺q ′ φ(`). We want to show thatq = q ′;i.e., thatφ(k)(q) 6= φ(`)(q). So assume thatφ(k)(q) = φ(l)(q). In this case,φ(k)(q) ≥ φ(g)(q). Also, if x ∈ P \{q}, thenφ(k)(x) ≥ φ(f )(x) = φ(g)(x).Henceφ(k) ≥ φ(g). But k /≥g, sincek(p) = f (p) < g(p). 2DEFINITION 3.2. Givenp ∈ P, a ≤ c ∈ A, we definej (p, a, c) to be thefunction inAP which takes valuec at all pointsx ≥ p and takes valuea at all otherpoints.

LEMMA 3.3. Suppose thatφ :AP ∼= BP and φ(〈u〉) = 〈v〉, so thatφ inducesisomorphismsφu : (u↑)P ∼= (v↑)P andφd : (u↓)P ∼= (v↓)P. The isomorphismsφuandφd and their inverses have the point covering property.

Proof. Suppose, for instance thatp ∈ P anda ≺ b in u↑. Write U for theposetu↑. Now the setCU,p(a, b) has a least element,(j ′, j (p, u, b)), wherej ′agrees withj (p, u, b) except atp wherej ′(p) = a; hence the set is connected.Consequently,φu has the point covering property, by Lemma 3.2. 2LEMMA 3.4. Suppose thatP is dismantlable to{p}. Then for eacha ≺ b in A,the setCp(a, b) is a connected subset ofAP× AP.

Proof.We proceed by induction on|P |. The lemma is trivial if|P | = 1, for thenCPp(a, b) is a singleton. SinceP is dismantlable to{p}, there is a pointq ∈ P \{p}

such thatP′ = P\{q} is dismantlable to{p} and q either is join-irreducible ormeet-irreducible. Assume thatq is join-irreducible with unique subcoverq0. (Theother case is analogous.) Defineπ to be the decreasing retraction ofP onto P′mappingx 7→ x for x 6= q andq 7→ q0. We have a decreasing retractionπ̂ , withπ̂(f ) = f ◦ π , of AP onto its subposet consisting of all functionsf such thatf (q) = f (q0), and this subset is order-isomorphic toAP′.

First assume thatp 6= q0. Thenπ̂ × π̂ restricted toCPp(a, b) is a decreasing

retraction ofCPp(a, b) onto a subset of itself isomorphic toCP′

p (a, b). By the in-


ductive assumption, the latter set is connected, and the connectivity ofCPp(a, b)

follows from this.Next, suppose thatp = q0. In this case,π̂ × π̂ restricted toCP

p(a, b) is amonotone decreasing map ofCP

p(a, b) onto a subposet ofAP × AP that is order-

isomorphic toCP′p (a, b) (and not a subset ofCP

p(a, b)). The connectivity ofCPp(a, b)

follows again, in the same way as in the first case. 2LEMMA 3.5. Suppose thatA is dismantlable to the set{a, b} wherea ≺ b. Thenfor all p ∈ P , the setCP

p(a, b) is a connected subset ofAP× AP.Proof. We proceed by induction on|A|. The base case,|A| = 2, is trivial.

For in this case,Cp(a, b) has a least element(j ′, j (p, a, b)). (See the proof ofLemma 3.3.)

For the inductive step, there is somec ∈ A\{a, b} such thatA ′ = A\{c} isdismantlable to{a, b} andc has a unique subcover or unique cover. It suffices toconsider the first case. (The argument in the second case is dual.) Thus letc0 bethe unique subcover ofc. We have the decreasing retractionσ of A onto A ′ withσ(c) = c0. The functionσ̂ (f ) = σ ◦ f is a decreasing retraction ofAP onto itssubposet(A ′)P. Thenσ̂ × σ̂ projectsCP

A,p(a, b) onto its subsetCPA′,p(a, b). This

works becausec /∈ {a, b}. The proof is concluded just as for the first case in thelast lemma. 2THEOREM 3.1. Suppose thatφ :AP ∼= BP. If P is dismantlable to each of itspoints, or ifA is dismantlable to each two-element set{a, a′} ⊆ A in whicha ≺ a′,thenφ has the point covering property.

Proof. It follows by Lemmas 3.2, 3.4, 3.5. 2Next we introduce the Diagonal Covering Property. Recall that fora ∈ A, we

use〈a〉 to denote the constant function fromP to {a}.DEFINITION 3.3. Suppose thatφ :AP ∼= BP anda ≺ a′ in A andφ(〈a〉) = 〈b〉.We say that the intervala′/a in A is φ-flat if φ(〈a′〉) = 〈b′〉 for someb′ ∈ B withb ≺ b′.

We say thata′/a is φ-twistedprovided the following hold. There arec ∈ A andd ∈ B such thatc ≥ a′ and:

(1) φ(〈c〉) = 〈d〉.(2) The interval subposetsc/a in A andd/b in B are both isomorphic to the poset

2P in such a fashion that the isomorphism whichφ produces between the inter-val subposets〈c〉/〈a〉 in AP and〈d〉/〈b〉 in BP is the one naturally induced bythe automorphism(x, y) 7→ (y, x) of P× P. More precisely, we have isomor-phismsαc : c/a ∼= 2P andβd : d/b ∼= 2P such that for allg = φ(f ) ∈ BP with〈a〉 ≤ f ≤ 〈c〉, and for all(x, y) ∈ P × P ,[

αc(f (x))](y) = [βd(g(y))](x). (1)

318 RALPH McKENZIE

DIAGONAL COVERING PROPERTY. An isomorphismφ :AP ∼= BP will be saidto have thediagonal covering propertyprovided that every covera′/a in A, whereφ(〈a〉) = 〈b〉 is a diagonal element, is eitherφ-flat orφ-twisted.

THEOREM 3.2. If P is connected and directly indecomposable andA,B,P arefinite, andφ :AP ∼= BP, thenφ has the diagonal covering property.

The remainder of this section is devoted to a proof of Theorem 3.2.In the next lemma,J(a↑), wherea is a non-maximal element ofA, denotes

the subposet ofA consisting of those elements which are join-irreducible in thesubposeta↑. Recall the definition ofj (p, x, y) in Definition 3.2.

LEMMA 3.6 (Duffus and Rival [6]).Letf ∈ AP and a be a non-maximal elementof A. Thenf is a join-irreducible element of the subposet〈a〉↑ of AP iff f =j (p, a, c) for somep ∈ P andc ∈ A wherec is join-irreducible in the poseta↑.ThusJ(〈a〉↑) ∼= P∂ × J(a↑), whereP∂ is the dual ofP.

Proof. It is clear that ifc is join-irreducible ina↑ with unique subcoverc◦ in a↑thenj (p, a, c) has in〈a〉↑ the unique subcover which is the functionf such thatf (p) = c◦, andf agrees withj (p, a, c) at all pointsx 6= p. Thus,j (p, a, c) isjoin-irreducible in〈a〉↑. Conversely, letf be join-irreducible in〈a〉↑ with uniquesubcoverf ′ in that poset. Thenf ′ ≺p f at somep ∈ P . Definec = f (p). Now〈a〉 ≤ j (p, a, c) ≤ f and j (p, a, c) /≤ f ′, hencef = j (p, a, c). It is easy tosee thatc must be join-irreducible ina↑ with unique subcoverf ′(p). 2

Proof of Theorem 3.2.Assume thatφ(〈a〉) = 〈b〉 anda ≺ a′. Then we haveφ(J (〈a〉↑)) = J (〈b〉↑) and this is an isomorphism of posets. By Lemma 3.6,there is an isomorphismφ◦ :P∂ × J(a↑) ∼= P∂ × J(b↑) such thatφ(j (p, a, u)) =j (q, b, v) iff φ◦(p, u) = (q, v) whenp ∈ P, u ∈ J (a↑). Let J ′ be the connectedcomponent ofJ(a↑) containing the elementa′. Thenφ◦(P × J ′) = P ×K ′ whereK ′ is a connected component ofJ(b↑). We haveφ◦: P∂×J′ ∼= P∂×K ′. Since theseposets are connected andP is directly indecomposable, by the refinement theoremof Hashimoto [11], there are only two possibilities.

Case (i): There are isomorphismsσ :P ∼= P and τ : J′ ∼= K ′ such thatφ(j (p, a, u)) = j (σ (p), b, τ(u)) for p ∈ P, u ∈ J ′. In this case,φ(j (p, a, a′)) =j (σ (p), b, b′) for all p, whereb′ = τ(a′). Then

〈a′〉 =∨p∈P

j (p, a, a′), 〈b′〉 =∨q∈P

j (q, b, b′)

and it is clear thatφ(〈a′〉) = 〈b′〉. These joins in the posets〈a〉↑ and 〈b〉↑ areeasily verified, considering the definition of the functionsj (p, x, y). Incidentally,it follows that b ≺ b′, since for example,〈a〉 ≺ j (p, a, a′) if p is chosen to bemaximal inP.


Case (ii): There exists a posetE and there exist isomorphismsα: J′ ∼= P∂ × Eandβ :K ′ ∼= P∂ × E, such thatφ(j (p, a, u)) = j (q, b, v) holds precisely whenfor somey ∈ E,α(u) = (q, y) andβ(v) = (p, y).

Now α(a′) = (m, e) wherem is maximal inP ande is minimal inE, sincea′is minimal in J (a↑). Writing vp for β−1(p, e), we have thatφ(j (p, a, a′)) =j (m, b, vp) for p ∈ P . Since 〈a′〉 = ∨

p j (p, a, a′) in 〈a〉↑, it is clear that

φ(〈a′〉) = j (m, b, d) for somed, where in factd =∨p vp in B.Sincem is maximal inP, it immediately follows that

d/b ∼= j (m, b, d)/〈b〉φ−1

∼= 〈a′〉/〈a〉 = (a′/a)P ∼= 2P.

The composition of these maps is an isomorphismβd : d/b ∼= 2P. Forb ≤ y ≤ d,andp ∈ P , we have that[

βd(y)](p) = 1 ↔ [

φ−1(j (m, b, y))](p) = a′

↔ φ−1(j (m, b, y)) ≥ j (p, a, a′)↔ j (m, b, y) ≥ j (m, b, vp)↔ y ≥ vp.

Also, write b′ = β−1(m, e) whereα(a′) = (m, e); and for anyp ∈ P , putup = α−1(p, e). Then we have thatφ(j (m, a, up)) = j (p, b, b′). Note thatum =a′ and thusφ(j (m, a, a′)) = j (m, b, b′), showing thatb ≺ b′. Since〈b′〉 is thejoin in 〈b〉↑ of the j (p, b, b′), then clearly we haveφ(j (m, a, c)) = 〈b′〉, wherec =∨p up (and this join exists inA).

Now we have

c/a ∼= j (m, a, c)/〈a〉φ∼= 〈b′〉/〈b〉 = (b′/b)P ∼= 2P.

This yields an isomorphismαc : c/a ∼= 2P so that fora ≤ x ≤ c andp ∈ P ,[αc(x)](p) = 1↔ x ≥ up.

Recall that foru ∈ J ′, v ∈ K ′, andp, q ∈ P , φ(j (p, a, u)) = j (q, b, v)

is equivalent to: for somey ∈ E,α(u) = (q, y) andβ(v) = (p, y). Consider-ing our definitions ofuq and vp, this means thatφ(j (p, a, uq)) = j (q, b, vp).Furthermore, sincec =∨q uq andd =∨p vp, it follows that

〈c〉 =∨p,q

j (p, a, uq) and 〈d〉 =∨p,q

j (q, b, vp),

these joins being taken in〈a〉↑, respectively〈b〉↑. So we have thatφ(〈c〉) = 〈d〉.Also, a′ = um ≤ c andb′ = vm ≤ d.

To complete the proof thata′/a is φ-twisted in case (ii), letg = φ(f ) where〈a〉 ≤ f ≤ 〈c〉 and〈b〉 ≤ g ≤ 〈d〉. Letp, q ∈ P . We have to show that[

αc(f (q))](p) = [βd(g(p))](q).

320 RALPH McKENZIE

We have already noted that fora ≤ x ≤ c, [αc(x)](p) = 1 iff x ≥ up; conse-quently,[

αc(f (q))](p) = 1⇔ f (q) ≥ up ⇔ f ≥ j (q, a, up)⇔ g ≥ j (p, b, vq);

and of course, the analogous computation shows that[βd(g(p))

](q) = 1⇔ g ≥ j (p, b, vq). 2

4. Details of Flat and Twisted Covers

In this section, we assume thatA,B,P are finite andφ :AP ∼= BP. We establishsome elementary facts which will be used in the next section.

LEMMA 4.1 (The2P lemma).

(i) Suppose thata ≺ a′ in A. Then the interval〈a′〉/〈a〉 in AP is equal to(a′/a)P,wherea′/a is the interval subposet ofA, and is isomorphic to2P. The mapκ : 〈a′〉/〈a〉 → 2P given by

[κ(g)](x) ={

1, if g(x) = a′,0, otherwise

is an isomorphism. Furthermore, any maximal chainf0 ≺ f1 ≺ · · · ≺ fn in〈a′〉/〈a〉 can be realized by an enumerationp1, . . . , pn of P , where|P | = n,and, for eachi, {p1, . . . , pi} is a filter inP , by defining

fi(x) ={a′, if x = pk, k ≤ i,a, otherwise.

(Every maximal chain has lengthn = |P |.)(ii) Suppose thatα < α′ in AP (or in BP), T ⊆ P , andα(r) ≺ α′(r), for all

r ∈ T , whileα(r) = α′(r) for all r ∈ P \T . Any maximal chainf0 ≺ f1 ≺· · · ≺ fm in α′/α can be realized by an enumerationp1, . . . , pm of T , where|T | = m, by defining

fi(x) ={α′(x), if x = pk, k ≤ i,α(x), otherwise.

(iii) If 〈a〉 < α′ in AP (or in BP) and a ≺ α′(r), for all r ∈ P , and if P isconnected, thenα′ = 〈a′〉, for somea′ � a.

(iv) Let c/a be a cover inA with φ(〈c〉) = 〈d〉 and φ(〈a〉) = 〈b〉. Thend/bis a cover inB and there is an automorphismτ [d/b] = τ of P defined byφ(j (r, a, c)) = j (τ(r), b, d), for r ∈ P . Furthermore, forf ∈ 〈c〉/〈a〉 andfor r ∈ R, f (r) = c is equivalent to[φ(f )](τ(r)) = d.


Proof.To prove (i), note that the first two statements are obvious. Now letf0 ≺f1 ≺ · · · ≺ fn be a maximal chain in〈a′〉/〈a〉. There are elementsp1, . . . , pnin P so thatfi ≺pi+1 fi+1, for i < n. Sincef0 = 〈a〉; this means that for 1≤j ≤ k < i ≤ n, fk(pj ) ≥ fj(pj ) = a′ andfk(pi) ≤ fi−1(pi) = a, and forx ∈ P \{p1, . . . , pk},

a = f0(x) = f1(x) = · · · = fk(x).Sincefn = 〈a′〉, it is obvious thatn = |P | andp1, . . . , pn is an enumeration ofP .It is also clear thatfi(pj ) = a′ iff j ≤ i. If 1 ≤ i < j ≤ n thenfi(pi) = a′, whilefi(pj ) = a – consequently,pj /≥pi . Thus{p1, . . . , pi} is a filter inP.

To prove (ii), letf0, . . . , fn be any maximal chain inα′/α. The argument for (i)produces the desired conclusion here, although we cannot claim that{p1, . . . , pi} isa filter in T. Note that it does not necessarily follow in (ii) thatα′/α is isomorphicto 2T (whereT = 〈T,≤〉). For example, takeP = 〈{0,1, . . . , n − 1}, ≤〉, A =〈{0,1, . . . , n},≤〉, T = P, α(x) = x, α′(x) = x + 1.

For (iii), sinceP is connected, it suffices to prove thatα′(r) = α′(s), if r < s.But a ≺ α′(r) ≤ α′(s) (α′ is monotone), anda ≺ α′(s); and the desired conclusionis obvious.

To prove (iv), note that the poset〈c〉/〈a〉 is isomorphic to both〈d〉/〈b〉 and tothe poset of monotone functions fromP into {b, d} (ordered byb < d). All the setsare finite, and thus all members of〈d〉/〈b〉 belong to the subset of functions withrange included in{b, d}. This clearly implies thatb ≺ d.

The mapφ restricted to join-irreducible elements of〈c〉/〈a〉 is an isomorphismof J(〈c〉/〈a〉) with J(〈d〉/〈b〉). By Lemma 3.6, we have a permutationτ of P withφ(j (r, a, c)) = j (τ(r), b, d), for r ∈ P . Thenτ is monotone becauser < r ′ iffj (r ′, a, c) < j (r, a, c) iff φ(j (r ′, a, c)) < φ(j (r, a, c)) iff τ(r) < τ(r ′).

For 〈a〉 ≤ f ≤ 〈c〉 and forr ∈ R, f (r) = c iff f ≥ j (r, a, c) and[φ(f )](τ(r))= d iff φ(f ) ≥ j (τ(r), b, d). This gives the final assertion in (iv). 2

Now we turn to consideration of twisted covers.

LEMMA 4.2 (The2P×P lemma).Suppose thata′/a is aφ-twisted cover inA. Thuswe haveφ(〈a〉) = 〈b〉 and φ(〈c〉) = 〈d〉, wherea′ ≤ c, and we have isomor-phismsαc : c/a ∼= 2P and βd : d/b ∼= 2P, and formula(1) holds. Letf, f ′, fi ∈〈c〉/〈a〉, φ(f ) = g, φ(f ′) = g′, φ(fi) = gi.

(i) If for somep, f0 ≺p f1 ≤ f2 ≺p f3, theng0 ≺q g1 ≤ g2 ≺q ′ g3 for someq 6= q ′. Similarly, if g0 ≺q g1 ≤ g2 ≺q g3 thenf0 ≺p f1 ≤ f2 ≺p′ f3 forsomep 6= p′. If f0 ≺p f1, f0 ≺p f2, g0 ≺q q1, g0 ≺q g2 for somep, q ∈ P ,thenf1 = f2.

(ii) Suppose thatp ∈ P and f <p f′. Then there is a setT ⊆ P such that

g(t) ≺ g′(t) for all t ∈ T , while g(r) = g′(r) for all r ∈ P \T . Further,T = I ∩ F for some filterF and idealI in P.

322 RALPH McKENZIE

(iii) Suppose thatp ∈ P andf <p f ′, f (p) = a, f ′(p) = c. Theng(r) ≺ g′(r)for all r ∈ P . Furthermore, ifg = g0 ≺ g1 ≺ · · · ≺ gn = g′ is any maximalchain ing′/g, then there is an enumerationp1, . . . , pn ofP , |P | = n, so thatfor all i < n, gi ≺pi+1 gi+1, and{p1, . . . , pi} is a filter in P.

(iv) Suppose thata′′ ∈ A anda ≺ a′′ ≤ c. Thenφ(〈a′′〉) = j (q, b, d) for somemaximal elementq ∈ P .

Proof.For (i), let us assume thatf0 ≺p f1 ≤ f2 ≺p f3. Thus

αc(f0(p)) ≺ αc(f1(p)) ≤ αc(f2(p)) ≺ αc(f3(p)),

and there must existq, q ′ ∈ P such that

αc(f0(p)) ≺q αc(f1(p)) ≤ αc(f2(p)) ≺q ′ αc(f3(p)).

Since these four functions belong to2P, thenq 6= q ′; further, [αc(f0(p))](q) 6=[αc(f1(p))](q), and [αc(f2(p))](q ′) 6= [αc(f3(p))](q ′). By Definition 3.3, for-mula (1), we then have[βd(g0(q))](p) 6= [βd(g1(q))](p) and[βd(g2(q

′))](p) 6=[βd(g3(q

′))](p). These formulas display thatg0(q) 6= g1(q), g2(q′) 6= g3(q

′),implying thatg0 ≺q g1, g2 ≺q ′ g3.

The proof that iff0 ≺p f1 andf0 ≺p f2 andf1 6= f2 theng0 ≺q g1 andg0 ≺q ′ g3 whereq 6= q ′, is similar.

To prove (ii), letf = f0 ≺ f1 ≺ · · · ≺ fm = f ′ be a maximal chain inf ′/f .Thusfi ≺p fi+1, for all i < m. Hence, wheregi = φ(fi), we have by statement(i), that there are distinct elementsq1, . . . , qm in P such thatgi ≺qi+1 gi+1, for alli < m. It follows thatg′(x) � g(x) for x ∈ T = {q1, . . . , qm}, andg′(x) = g(x)for x ∈ P \T . By formula (1),T is identical with the set ofx ∈ P for which[αc(f (p))](x) = 0 and[αc(f ′(p))](x) = 1.

To prove (iii), note that we have the assumptions of (ii) and moreover,f (p) = aandf ′(p) = c, so thatαc(f (p)) = 〈0〉 andαc(f ′(p)) = 〈1〉. In this case,T = P .Now let g = g0 ≺ g1 ≺ · · · ≺ gn = g′ be any maximal chain, saygi ≺pi+1 gi+1

for i < n. It is clear thatp1, . . . , pn is a one-to-one enumeration ofP . Putfi =φ−1(gi). Sincef = f0 ≤p fi, andp1, . . . , pi are precisely the points at whichgandgi differ, then it is clear from Equation (1) that{p1, . . . , pi} is exactly the setof pointsx for which [αc(fi(p))](x) = 1; thus it is a filter.

To prove (iv), choose any maximalp ∈ P . Then 〈a〉 ≺ j (p, a, a′′) ≤ 〈c〉,and φ(j (p, a, a′′)) = j (q, b, b′′) for some maximalq ∈ P and whereb ≺b′′ ≤ d. Now Equation (1) implies thatαc(a′′) = j (q,0,1), and thatφ(〈a′′〉) =j (q, b, d). 2


5. Cancellation of Exponents

This section is devoted to a proof of the next theorem.

THEOREM 5.1. Suppose thatA,B,P are finite connected ordered sets andφ :AP ∼= BP, whereφ andφ−1 have the point covering property and the diagonalcovering property, both considered as isomorphisms betweenAP and BP, and asisomorphisms between(A∂)P

∂

and(B∂)P∂

. ThenA ∼= B.

To prove the theorem, we follow a method invented by B. Jónsson about 1977in order to prove that the two-element chain cancels as exponent. The hypothesesof Theorem 5.1 are assumed throughout the section.

DEFINITION 5.1. We defineR(φ) to be the set of alla ∈ A such thatφ(〈a〉) =〈b〉 for someb ∈ B, and we defineR(φ−1) ⊆ B in a similar manner.

We put0(φ) equal to the set of alla ∈ A such thatφ(〈a〉) = g ∈ R(φ−1)P and,wheref is the function defined byφ(〈f (p)〉) = 〈g(p)〉 for all p ∈ P , it is the casethatφ(f ) = 〈b〉 for an elementb ∈ B, which will be written asb = γ (a).

Notice that whena ∈ 0(φ) then γ (a) ∈ 0(φ−1). Our goal is to prove that0(φ) = A and0(φ−1) = B. It will then be obvious thatγ is an isomorphismbetweenA andB. The first (and easiest) step is to show that0(φ) 6= ∅. The second(and final) step is to prove that whenevera ≺ a′ thena ∈ 0(φ) iff a′ ∈ 0(φ).SinceP is connected, it will then follow thatA = 0(φ).

We need the concept of height. Theheightof a posetP, denoted ht(P), is thegreatest cardinality of a chain inP, minus one. We have assumed thatA,B andPare finite and connected, but the validity of the next lemma requires only thatP beconnected andAP be finite.

LEMMA 5.1 (Birkhoff [5]). ∅ 6= R(φ) ⊆ 0(φ).Proof.Since every covering inAP is of the formf ≺p g with f (p) ≺ g(p) (for

somep ∈ P ), it follows that forf < g in AP,

ht(g/f ) ≤∑p∈P

ht(g(p)/f (p)) ≤ |P | · ht(A).

Since ht(g(p)/f (p)) ≤ ht(A), it is clear that the equality ht(g/f ) = |P |·ht(A) canhold only if ht(g(p)/f (p)) = ht(A) for all p ∈ P . This latter condition impliesthatg(p) is maximal inA andf (p) is minimal inA for all p. This in turn impliesthatf andg are constant, sinceP is connected. Thus ht(g/f ) = |P | ·ht(A) impliesthatg = 〈c〉, f = 〈a〉 where ht(c/a) = ht(A).

On the other hand, there do exista, c ∈ A such that ht(c/a) = ht(A). Leta = a0 < a1 < · · · < an = c be a chain wheren = ht(A). Let C be thesubposet ofA formed on the set{a0, a1, . . . , an}. Let f0, . . . , fm be any maximalchain in the posetCP. Clearly,f0 = 〈a〉 andfm = 〈c〉. For eachp ∈ P , there

324 RALPH McKENZIE

must ben distinct indicesi0 < i1 < · · · < in−i < m such thatfij ≺p fij+1 andfij (p) = aj , fij+1(p) = aj+1. Clearly, we have here that ht(〈c〉/〈a〉) = |P | ·ht(A).

We have now established that

|P | · ht(A) = ht(AP) = ht(BP) = |P | · ht(B);from which it follows that ht(A) = ht(B). Also, we have proved that forf < g

in AP (or in BP), ht(g/f ) = ht(AP) holds if and only iff = 〈x〉, g = 〈y〉 whereht(y/x) = ht(A).

So now, letc/a be any interval inA of height equal to ht(A). Then

ht(φ〈c〉)/φ(〈a〉)) = ht(〈c〉/〈a〉) = ht(BP),

implying (as we saw above) that the elementsφ(〈c〉) and φ(〈a〉) are diagonalelements ofBP – i.e.,{a, c} ⊆ R(φ). 2

In view of the above lemma, in order to prove that0(φ) = A (which willconclude the proof of Theorem 5.1), it remains to prove that whena ≺ a′ in Athena ∈ 0(φ) iff a′ ∈ 0(φ). We shall explicitly prove thata ∈ 0(φ) impliesa′ ∈ 0(φ). We will not prove thata′ ∈ 0(φ) impliesa ∈ 0(φ), as it follows by adual argument. (Our assumptions are self-dual.) Throughout, we shall refer to thetwo covering properties as the PCP and the DCP.

ASSUMPTIONS. Thus, for the remainder of this section, we assume thata, a′ ∈ Aare fixed,a ≺ a′ anda ∈ 0(φ). We putβ = φ(〈a〉). Then we takeα ∈ AP to bethe function such thatφ(〈α(p)〉) = 〈β(p)〉, and we haveφ(α) = 〈b〉 = 〈γ (a)〉.Also, we defineλ = φ(〈a′〉).

Our goal is to prove thata′ ∈ 0(φ). We first show that this conclusion followsunless it is the case that for allp ∈ P, λ(p)/β(p) ⊆ R(φ−1).

LEMMA 5.2. (i) The intervalλ/β in BP is a distributive lattice isomorphic to2P,and for eachp ∈ P , the intervalλ(p)/β(p) in B is a distributive lattice.

(ii) Suppose thatδ ∈ λ/β. Thenδ is a join-irreducible element of the intervalsubposetλ/β of BP iff there isp ∈ P such that:(a) δ(p) is join irreducible inλ(p)/β(p); (b) δ(q) = β(q) for all q /≥p; and (c) for all y > p, δ(y) is a minimalcommon upper bound ofβ(y) and all elementsδ(x) wherex < y.

CAUTION. We cannot conclude that as lattices,λ/β is embedded as a sublatticein∏p λ(p)/β(p); i.e., that joins and meets are taken coordinatewise.

Proof. To prove (i), note first thatλ/β ∼= 〈a′〉/〈a〉. By Lemma 4.1, this isisomorphic with2P. Now letp ∈ P . Definef, g ∈ BP to be the functions suchthat,f (q) = β(q) for all q />p andf (q) = λ(q) for all q > p; g >p f andg(p) = λ(p). The intervalg/f is isomorphic to the intervalλ(p)/β(p); but it isalso isomorphic byφ−1 to a sub-interval of the interval〈a′〉/〈a〉 in AP. Now (i)follows from the fact that every interval in2P is a distributive lattice.


For (ii), let us first assume thatδ satisfies the conditions (a), (b), (c). Letµ ≺p δbe the function such thatµ(p) is the subcover ofδ(p) aboveβ(p). Clearly,µ ∈ BP

(by (a) and (b)), and it is easy to see thatµ is the unique subcover ofδ aboveβ;i.e.,δ is join-irreducible inβ↑. Indeed, suppose thatβ ≤ f < δ. Let q be minimalsuch thatf (q) < δ(q). We have thatf (q) ≥ β(q) and for allx < q, f (q) ≥f (x) = δ(x), and alsof (q) ≤ δ(q). Conditions (b) and (c) yield thatq = p. Thusβ(p) ≤ f (p) < δ(p), and by (a), this means thatf ≤ µ.

To prove the converse, let us now assume thatδ is join-irreducible inλ/β. Letβ ≤ µ ≺ δ, and say,µ ≺p δ. The function〈β(x) : x /≥p〉 ∪ 〈δ(x) : x ≥ p〉 is inthe intervalδ/β and not≤ µ, hence it isδ, implying thatδ(x) = β(x) = µ(x) forx /≥p. If β(p) ≤ d < δ(p), then the function which equalsd at p and equalsδelsewhere is monotone, and must be≤ µ. Hence,µ(p) is the unique subcover ofδ(p) aboveβ(p). We have now established (a) and (b) forδ. For (c), suppose thatwe did haveq > p and some elements ∈ B with δ(q) > s ≥ β(q) ands ≥ δ(x)for all x < q. Then the functionν obtained by changing the value ofδ at q to sis monotone and satisfiesβ ≤ ν <q δ andν /≤µ. But this is a contradiction. Thussuchq ands do not exist. 2

Recall thatj (p, a, a′) is the function taking the valuea′ at all x ≥ p and thevaluea at all x /≥p. If p is a maximal element ofP, then〈a〉 ≺ j (p, a, a′). In thiscase, for someq ∈ P, β ≺q φ(j (p, a, a′)) ≤ λ. Note thatβ(q) ∈ R(φ−1), sincea ∈ 0(φ). In fact,φ−1(〈β(q)〉) = 〈α(q)〉.LEMMA 5.3. Letp be maximal inP andβ ≺q δ = φ(j (p, a, a′)). If δ(q)/β(q)is not aφ−1-flat cover, thena′ ∈ 0(φ).

Proof. We begin with an application of the fact thatφ−1 has the PCP. Wehave that(β, δ) ∈ Cq(β(q), δ(q)). Also, j (q, β(q), δ(q)) is a join-irreducibleelement of〈β(q)〉↑ with unique subcover, sayg, above〈β(q)〉; and we have that(g, j (q, β(q), δ(q))) ∈ Cq(β(q), δ(q)). Sinceφ−1(β) ≺p φ−1(δ), it follows that

〈α(q)〉 ≤ φ−1(g) ≺p φ−1(j (q, β(q), δ(q))).

Let us now assume thatδ(q)/β(q) is not φ−1-flat, so that it is aφ−1-twistedcover. (We have just applied our assumption that the DCP holds forφ−1.) Then byLemma 4.2(iv) it follows thatφ−1(〈δ(q)〉) = j (p′, α(q), c) for some maximalp′,whereφ−1(〈d〉) = 〈c〉 for a certaind ≥ δ(q). Thus, the displayed formula aboveexpands to

〈α(q)〉 ≤ φ−1(g) ≺p φ−1(j (q, β(q), δ(q))) ≤ j (p′, α(q), c).This clearly impliesp′ = p and〈α(q)〉 <p φ−1(j (q, β(q), δ(q))).

Recall thatβ ≺q δ. Thus for eachq ′ > q, we haveβ(q ′) = δ(q ′) ≥ δ(q) andso

〈α(q ′)〉 = φ−1(〈β(q ′)〉) ≥ j (p, α(q), c),

326 RALPH McKENZIE

i.e.,α(q ′) ≥ c. Sinceφ(〈α(q ′)〉) = 〈β(q ′)〉 andφ(〈c〉) = 〈d〉, thenβ(q ′) ≥ d forq ′ > q. Let β ′ be the function which agrees withβ except atq, with β ′(q) = d.We have just shown thatβ ′ ∈ BP andβ ′ > β.

Let j◦(q, β(q), d) denote the function equal tod at all q ′ > q and equal toβ(q) elsewhere. From Lemma 4.2(iii) applied to theφ−1-twisted coverδ(q)/β(q),it follows thatφ−1(j (q, β(q), d)) coversφ−1(j◦(q, β(q), d)) at every coordinate.Sinced/β(q) ∼= 2P, there is a maximal chain ind/β(q),

β(q) = uo ≺ u1 ≺ · · · ≺ un = d (n = |P |),with u1 = δ(q). Definef0 = j◦(q, β(q), d) and forn ≥ i > 0, letfi >q f0 withfi(q) = ui. Also, defineg0 = β and, forn ≥ i > 0, letgi >q g0, gi(q) = ui . Thenwe have maximal chains

j◦(q, β(q), d) = f0 ≺q f1 ≺q · · · ≺q fn = j (q, β(q), d), and

β = g0 ≺q g1 ≺q · · · ≺q gn = β ′.By Lemma 4.2(iii), there is an enumerationp1, . . . , pn of P, |P | = n, such thatφ−1(fi) ≺pi+1 φ

−1(fi+1) and{p1, . . . , pi} is a filter inP, for all i < n. By the PCPfor φ−1, φ−1(gi) ≺pi+1 φ

−1(gi+1), for i < n. We conclude thatφ−1(β ′) coversφ−1(β) = 〈a〉 at every coordinate.

Then by Lemma 4.1(iii),φ−1(β ′) = 〈a′′〉wherea′′ � a. Butg1 = δ ≤ β ′, hencej (p, a, a′) ≤ 〈a′′〉, implying thata′′ = a′. Thus we have thatλ = φ(〈a′〉) = β ′.

Now let α′ be the function which agrees withα except atq, with α′(q) = c.Thusφ(〈α′(y)〉) = 〈β ′(y)〉 for all y ∈ P , andα′ ∈ AP. All that remains to finishthe proof of this lemma, is to show thatφ(α′) = 〈b′〉 for someb′ � b. This is donein precisely the same way that we showedφ−1(β ′) covers〈a〉 at every coordinate,by using Lemma 4.2, combined with Lemma 4.1(iii) and the PCP. That argumentis left to the reader. 2LEMMA 5.4. Suppose that for someq ∈ P , the intervalλ(q)/β(q) contains anelement outsideR(φ−1). Then there is a maximal elementp such thatφ(j (p, a, a′))= δ �q β andδ(q)/β(q) is aφ−1-twisted cover. Under this assumption,a′ ∈ 0(φ)by the preceding lemma.

Proof.Let y′ be a minimal element inλ(q)/β(q) outside ofR(φ−1) and choosey with β(q) ≤ y ≺ y′. Thus y ∈ R(φ−1) and y′ /∈ R(φ−1), hencey′/y isφ−1-twisted, by the DCP. Let(g, g′) ∈ Cq(y, y′) ∩ λ/β. (Such a pair exists byLemma 3.1.) Then we haveφ−1(g) ≺p φ−1(g′), say. It follows that(φ−1(g), φ−1(g′)) ∈ Cp(a, a′). Letj◦(p, a, a′) be the unique subcover ofj (p, a, a′)above 〈a〉. Thus j◦(p, a, a′) ≺p j (p, a, a′) ≤ φ−1(g′), j (p, a, a′) /≤φ−1(g), j◦(p, a, a′) ≤ φ−1(g). Definingδ = φ(j (p, a, a′)), δ◦ = φ(j◦(p, a, a′)),the PCP implies thatβ ≤ δ◦ ≺q δ ≤ g′.

Also, it follows by the PCP thatφ−1(j◦(q, y, y′)) ≺p φ−1(j (q, y, y′)) wherej◦(q, y, y′) is the subcover ofj (q, y, y′) above〈y〉. Sincey′/y is φ−1-twisted, this


implies by Lemma 4.2(iv), thatp is maximal. So we havej◦(p, a, a′) = 〈a〉, andβ = δ◦ ≺q δ ≤ g′, δ /≤g – in particular,β(q) ≺ δ(q) ≤ y′, δ(q) /≤ y.

In this situation, withβ(q), y ∈ R(φ−1), if δ(q)/β(q) were φ−1-flat, y′/ywould be forced also to beφ−1-flat, which it’s not. Indeed, suppose that we didhaveφ(〈u〉) = 〈δ(q)〉, α(q) ≺ u. We also haveφ(〈v〉) = 〈y〉, α(q) ≤ v, u /≤ v(sinceδ(q) /≤ y). By Lemma 4.2(iv), applied to the twisted covery′/y, we haveφ(j (x, v, e)) = 〈y′〉 for some maximalx and somee ∈ A. Then〈u〉 ≤ j (x, v, e),sinceδ(q) ≤ y′. Sinceu /≤ v, this impliesP = {x}. But in this case, obviously,R(φ) = A, a contradiction. Thusδ(q)/β(q) is φ−1-twisted, as we hoped. 2ASSUMPTIONS. For the remainder of the argument, we assume that every intervalλ(q)/β(q) in B is contained inR(φ−1), and hence all members ofλ/β in BP belongtoR(φ−1)P .

We writeλ′p = φ(j (p, a, a′)). Thus theλ′p are precisely the join irreducible ele-ments of the distributive latticeλ/β, described in Lemma 5.2. Letλp =φ(j◦(p, a, a′)) be the unique subcover ofλ′p aboveβ. We writeπ(p) for the uniqueq ∈ P such thatλ′p �q λp.

We require five lemmas to prove thatπ is an automorphism ofP.

LEMMA 5.5. Let p ∈ P . If 〈a〉 ≤ f ≺p f ′ ≤ 〈a′〉 thenφ(f ) ≺π(p) φ(f ′).Moreover, there is an automorphismρ of P with ρ(p) = π(p). Thus, in particular,p is maximal iffπ(p) is maximal.

Proof. If f ≺p f ′ in 〈a′〉/〈a〉, then it follows immediately from the PCPthat φ(f ) ≺π(p) φ(f ′), since(f, f ′) and (j◦(p, a, a′), j (p, a, a′)) both belongtoCp(a, a′). Now choose anyf, f ′ ∈ 〈a′〉/〈a〉 with f ≺p f ′ and writeq = π(p),g = φ(f ), g′ = φ(f ′), y = g(q), y′ = g′(q).

Theny′ � y and bothy, y′ ∈ R(φ−1), so this is a flat cover. There existx, x′ ∈A such thatφ(〈x〉) = 〈y〉, φ(〈x′ 〉) = 〈y′〉, and by Lemma 4.1(iv), there is anautomorphismρ of P such that for allχ ∈ 〈x′〉/〈x〉 andp′ ∈ P , φ(χ)(ρ(p′)) =y′ iff χ(p′) = x′. Then choosing any pairh ≺q h′ in the interval〈y′〉/〈y〉, itfollows thatφ−1(h) ≺ρ−1(q) φ

−1(h′). But it also follows by the PCP thatφ−1(h) ≺pφ−1(h′), sinceφ−1(g) ≺p φ−1(g′). Thusp = ρ−1(q), or stated otherwise,ρ(p) =π(p). 2LEMMA 5.6. For eachp ∈ P ,π maps the posetp↑ isomorphically onto the posetπ(p)↑. Thusλ′p(q) � β(q) for all q ≥ π(p) andλ′p(q) = β(q) for q /≥π(p).

Proof. Claim 1:If x < y, thenπ(x) < π(y) andλ′x(π(y)) > β(π(y)).Indeed, assume thatx < y. Thenλ′x > λ′y and thusλ′x(π(y)) > β(π(y)), since

λ′x(π(y)) ≥ λ′y(π(y)) � λy(π(y)) ≥ β(π(y)).Then with Lemma 5.2, we can conclude thatπ(x) ≤ π(y), since otherwiseλ′x(π(y)) = β(π(y)), as the unique subcover ofλ′x differs from it only atπ(x).

328 RALPH McKENZIE

Sincex < y entails ht(y↑) < ht(x↑), andx is automorphic toπ(x) andy is auto-morphic toπ(y) (by the previous lemma), then we have ht(π(y)↑) < ht(π(x)↑);consequently we have not justπ(x) ≤ π(y), butπ(x) < π(y).

Claim 2: If x ∈ P , λ′x(π(x)) � β(π(x)), i.e.,λx(π(x)) = β(x).If this fails, there is some coverλx �π(x) g ≥ β (sinceλx(z) = β(z) for all

z < π(x) by Lemma 5.2). We haveφ(f ) = g where〈a〉 ≤ f ≺ j◦(x, a, a′). Heref ≺y j◦(x, a, a′) for somey > x, implying thatg ≺π(y) λx (by Lemma 5.5). Butg ≺π(x) λx also, yieldingπ(x) = π(y), and contradicting what we proved in thelast paragraph.

Now we consider a fixedp = pm and putqm = π(pm). For x ∈ P , we shallwrite dx = λ′pm(x), alsobx = β(x), and we writeax andcx for the elements ofA such thatφ(〈ax〉) = 〈bx〉 andφ(〈cx〉) = 〈dx〉 (recalling that all these elementshave been assumed to lie inR(φ−1)). By Lemma 5.2 and the observations above,we have thatdx = bx if x /≥ qm, dqm � bqm , anddπ(x) > bπ(x) if x > pm. Finally, weputτ = τ [dqm/bqm] (see Lemma 4.1(iv)), so that for allx ∈ P ,φ(j (x, aqm, cqm)) =j (τ(x), bqm, dqm).

Claim 3:For allq ≥ qm and for allx ∈ P , φ(j (x, aq, cq)) = j (τ(x), bq , dq).We prove this by induction on ht(q/qm). For q = qm, it is the definition of

τ [dqm/bqm]. Now suppose that we haveqm ≤ q ≺ q ′ and it is true forq. Recallthat dq ′ is a minimal element of the set of all common upper bounds ofbq ′ and{dy : y < q ′} (by Lemma 5.2). Thuscq ′ is a minimal upper bound to the set{aq ′ } ∪ {cy : y < q ′}. It follows that j (x, aq ′ , cq ′) is a minimal common upperbound of〈aq ′ 〉 and{j (x, ay, cy) : y < q ′}. Notice that wheny < q ′ andy /≥ qmthen dy = by and ay = cy and it is automatically true thatφ(j (x, ay, cy)) =j (τ(x), by, dy). Thus we can applyφ and use the induction assumption, to get thatwhereφ(j (x, aq ′ , cq ′)) = g, we have thatg ≤ 〈dq ′ 〉 andg is a minimal commonupper bound of〈bq ′ 〉 and{j (τ(x), by, dy) : y < q ′}. It follows thatg(τ(x)) = dq ′implying, by minimality ofg, thatg = j (τ(x), bq ′ , dq ′), which is the desired result.

Claim 4:π andτ are equal on the setpm↑.To prove this, note first that(λpm, λ

′pm) and(j◦(qm, bqm, dqm), j (qm, bqm, dqm))

both belong toCqm(bqm, dqm) and so, by PCP,τ(pm) = qm = π(pm). Now letx >pm, with π(x) = z > qm. Putb′z = λ′x(z) andφ(〈a′z〉) = 〈b′z〉. By Claim 2,b′z � bz.Defineσ = τ [b′z/bz] (from Lemma 4.1(iv)). We know thatσ(x) = z (replacingpm, τ by x, σ in our argument thatτ(pm) = π(pm)). Whenceφ(j (x, az, a′z)) =j (z, bz, b

′z). Also, λ′x < λ′pm and henceb′z ≤ dz anda′z ≤ cz. From Claim 3, we

have thatφ(j (x, az, cz)) = j (τ(x), bz, dz). Sincej (x, az, a′z) ≤ j (x, az, cz), weobtain thatj (z, bz, b′z) ≤ j (τ(x), bz, dz). Sinceb′z > bz, this implies thatτ(x) ≤z = π(x) = σ(x). Henceσ−1τ(x) ≤ x. This forcesσ−1τ(x) = x, sinceσ−1τ isan automorphism of the finite posetP. Henceτ(x) = σ(x) = π(x) as claimed.

To conclude this proof, let

〈a〉 = f0 ≺p1 f1 ≺p2 · · · ≺pm fm = j (pm, a, a′)


be a maximal chain, where{p1, . . . , pm} = pm↑. Thus withgi = φ(fi), andqi = π(pi), we have the maximal chain

β = g0 ≺q1 g1 ≺q2 · · · ≺qm gm = λ′pm.Since{q1, . . . , qm} = τ(pm ↑) = qm↑, it follows immediately thatλ′pm(x) � β(x)for all x ≥ qm andλ′pm(x) = β(x) for all x /≥ qm. 2DEFINITION 5.2. We now modify and extend the notation employed in the lastproof. We putax = α(x), bx = β(x), as before. For allx ∈ P andy ≥ π(x), weput dxy = λ′x(y). By Lemma 5.6,λ(y) ≥ dxy � by . We definecxy by the formulaφ(〈cxy〉) = 〈dxy〉. We defineτxy = τ [dxy/by].LEMMA 5.7. For p ∈ P andy, y′ ≥ π(p), we haveτpy = τpy ′, andπ = τpy onp↑.

Proof. Let π(p) = q and τpq = τ . In Claim 3 of the proof of the previouslemma, we showed that for ally ≥ q and for allx,

φ(j (x, ay, cpy)) = j (τ(x), by, dpy),and in Claim 4 thatπ = τ onp↑. The displayed equations show thatτ = τpy. 2LEMMA 5.8. For all p,p′ ∈ P andy ≥ π(p), y′ ≥ π(p′) we haveτpy = τp′y ′.

Proof. It suffices, by the connectedness ofP, to prove this in case there is anelementu which is an upper bound of bothp andp′. Assume thatp,p′, y, y′, uare given withp ≤ u, p′ ≤ u, π(p) ≤ y, π(p′) ≤ y′. Now π(u) = v is anelement bounding above bothπ(p) andπ(p′) (by Lemma 5.6). Sinceλ′u ≤ λ′pandλ′u ≤ λ′p′, thenbv ≺ duv ≤ dpv andduv ≤ dp′v, implying dpv = duv = dp′v,since these elements all coverbv. Thus,τpv = τp′v follows from the definitionof these maps. Then ify ≥ π(p) andy′ ≥ π(p′), we have by Lemma 5.7 thatτpy = τpv = τp′v = τp′y ′. 2LEMMA 5.9. Letp, q ∈ P . Thenβ(q) ≺ λ(q), and ifq ≥ π(p), thenλ′p(q) =λ(q) andπ = τpq .

Proof.Let τ be the functionτpq which, as we have seen, is independent ofp, q.From the previous two lemmas, we have thatπ = τ , and soπ is an automorphismof P. Consider any maximal chain

〈a〉 = f0 ≺p1 f1 ≺p2 · · · ≺pn fn = 〈a′〉,where{p1, . . . , pn} = P . Then withgi = φ(fi), qi = π(pi), we have a maximalchain

β = g0 ≺q1 g1 ≺q2 · · · ≺qn gn = λ.Since{q1, . . . , qn} = P and these elements are all distinct, we can conclude thatλ(q) � β(q) for all q. Since forq ≥ π(p), λ′p(q) > β(q), and sinceβ ≤ λ′p ≤ λ,andβ(q) ≺ λ(q), we obtain thatλ′p(q) = λ(q). 2

330 RALPH McKENZIE

LEMMA 5.10. a′ ∈ 0(φ).Proof. Write b′p = λ(p), φ(〈a′p〉) = 〈b′p〉, and letα′ be defined byα′(p) = a′p

for all p. We must show thatφ(α′) = 〈b′〉 for someb′ � b.We claim that wheneverα ≤ f ≺p f ′ ≤ α′ thenφ(f ) ≺π(p) φ(f ′). To see it,

note thatbp ≺ b′p andb′p = dxp whereπ(x) = p. Thusτ [b′p/bp] = π . This impliesthatφ(j◦(p, ap, a′p)) ≺π(p) φ(j (p, ap, a′p)). Sincef (p) = ap andf ′(p) = a′p,then the PCP implies thatφ(f ) ≺π(p) φ(f ′).

Now by Lemma 4.1(ii), any maximal chain inα′/α can be realized as

α = f0 ≺p1 f1 ≺p2 · · · ≺pn fn = α′, |P | = n.This yields a maximal chain

〈b〉 = g0 ≺q1 g1 ≺q2 · · · ≺qn gn = φ(α′)in φ(α′)/〈b〉 with gi = φ(fi) and qi = π(pi). Clearly, this means thatb ≺[φ(α′)](p) for all p ∈ P . Then by Lemma 4.1(iii),φ(α′) = 〈b′〉 whereb ≺ b′. 2

This concludes our proof of Theorem 5.1. We note one useful corollary of the ar-gument, which is also an immediate consequence of Theorem 3.2 and Corollary 7.8in Jónsson and McKenzie [12].

THEOREM 5.2. Suppose thatA,B,P are finite,P is connected and directly inde-composable, andφ : AP ∼= BP. ThenR(φ)↑ ∪ R(φ)↓ ⊆ 0(φ).

Proof. Let φ(〈a〉) = 〈b〉, and writeU,V for the subposetsa↑, b↑ of A, re-spectivelyB. By Lemmas 3.3 and 3.2, the induced isomorphismsφu : UP ∼= VP

andφu : (U∂)P∂ ∼= (V∂)P

∂

, and their inverses have the PCP and the DCP. Henceour proof of Theorem 5.1, applied in this case, yields thatU = 0(φu), which isequivalent toa↑ ⊆ 0(φ).

The inclusiona↓ ⊆ 0(φ) is obtained the same way. 2

6. Concluding Results, Perspective

THEOREM 6.1. Suppose thatφ : AP ∼= BP whereA,B,P are finite and con-nected and eitherP is dismantlable down to each of its points orA and B aredismantlable down to each of their covering pairs. ThenA ∼= B.

COROLLARY 6.1 (Jónsson and McKenzie [12]).Suppose thatφ : AP ∼= BP whereA,B,P are finite and connected and one ofA or P has a least or largest element.ThenA ∼= B.

THEOREM 6.2. Let φ : AP ∼= BP where P is finite, connected and directlyindecomposable, and for every minimal elementa in A, ht(a↑) = ht(A). ThenA ∼= B. In fact,0(φ) = A and0(φ−1) = B.


Proof of Theorem 6.1 and Corollary 6.1.First, expressP as isomorphic toP1×· · · × Pn, a product of directly indecomposable connected posets.

Now assume thatP is dismantlable to each of its points. Each direct factor ofPis a retract ofP and therefore also has this property. ThusP′ = P1×· · ·×Pn−1,Pnare dismantlable to each point, which, in particular, implies thatAP′ is connected.By Theorems 3.1 and 3.2, the isomorphism(AP′)Pn ∼= (BP′)Pn , and its inverse,and their duals (replacingAP′,BP′,Pn by their duals), have the PCP and DCP.Therefore Theorem 5.1 yields thatAP′ ∼= BP′. Continuing in this fashion, we cancelsuccessivelyPn,Pn−1, . . . ,P1 and find thatA ∼= B.

Next, assume thatA andB are dismantlable down to their covering pairs. Thenevery isomorphismψ : AQ ∼= BQ with Q connected has the Point Covering Prop-erty, by Theorem 3.1, as doesψ−1 (and of course, their duals). Further, it is easy tosee that ifψ : AQR ∼= BQR has the PCP, then the naturally correlated isomorphism(AQ)R ∼= (BQ)R has the PCP. Moreover,AQ andBQ are connected becauseA andB are dismantlable. Thus we are able to conclude from Theorems 3.2 and 5.1 thatAP′ ∼= BP′. Then it follows by induction onn thatA ∼= B.

Corollary 6.1 follows from Theorem 6.1, because if a finite posetQ is lower orupper bounded then it dismantles to each of its covering pairs and hence to eachof its points. Also,A has a lower or upper bound, if and only ifAP has a lower,respectively upper bound. To verify the first assertion, assume thatA is finite withleast element 0, and say, 0< a ≺ b in A. To dismantleA to {a, b}, successivelyremove elements ofA\(a↑ ∪ {0}) that have become minimal in this set, until thisentire set is removed. Then remove 0 (which has become meet-irreducible). Nextremove elements ofa↑\(b↑ ∪ {a}) as they become minimal in this set, until thisset has been entirely removed. Finally, remove elements ofb↑ \{b} as they becomeminimal in this set. What is left at the end is{a, b}. If b � 0, the process ofdismantlingA to {0, b} is the same, omitting the first two steps. 2

Proof of Theorem 6.2.Let m be any minimal element ofA. Since ht(m↑) =ht(A), the proof of Lemma 5.1 shows thatm ∈ R(φ). By Theorem 5.2,m↑ ⊆0(φ). Sincem is an arbitrary minimal element, it follows that0(φ) = A. Thusthe mapγ : 0(φ) ∼= 0(φ−1) is an isomorphism ofA with the subposet0(φ−1) ofB. If B 6= 0(φ−1), then clearly,|BP| > |0(φ−1)P| = |AP|, a contradiction. Hence0(φ−1) = B. 2

The results in this paper include everything this author knows about the Birkhoffcancellation problem. Although we have somewhat extended the results on theBirkhoff problem proved in Jónsson and McKenzie [12], which are summarizedin Corollary 6.1 above, our knowledge still seems very limited. In particular, weknow of no isomorphismφ : AP ∼= BP,A,B,P finite and connected,P directly in-decomposable, such that0(φ) 6= A. We remark that, in this situation, the connectedcomponent ofAP containing the diagonal is mapped byφ onto the connectedcomponent ofBP containing the diagonal. This follows from the fact thatR(φ) is

332 RALPH McKENZIE

Figure 1.

non-empty (Lemma 5.1). Our argument to prove Theorem 5.1 at no point involvesthe consideration of elements outside of this connected component. Hence, in orderto conclude that0(φ) = A, it would suffice for the point covering property to bevalid for the elements of this connected component.

The full point covering property, as we have defined it, can indeed fail. To seethis, letA be the poset pictured in Figure 1 and letP be the subposet formed onA\{b}. (P is a four-element crown.) Takef to be the identity function onP andf1

to be the automorphism ofP which switches the two minimal elements and leavesthe maximal elements unchanged. Leth be an automorphism ofP which switchesthe two maximal elements. Note that the connected component ofAP containingf is a two-element set{f, g} where(f, g) ∈ Ca(a, b). Likewise, the componentcontainingf1 is {f1, g1} where(f1, g1) ∈ Ca(a, b), and the component containingf1h is {f1h, g1h} where(f1h, g1h) ∈ Cc(a, b). Let φ be the automorphism ofAP

which switchesf1 andf1h and switchesg1 andg1h and leaves all other elementsfixed. Clearly,φ does not have the point covering property at(a, a, b).

There appear to be two obvious next steps to take in the investigation of Birk-hoff’s conjecture: letPbe the four-element crown, or assume thatP is dismantlable.

Incidentally, the posetB in Figure 1 is dismantlable to each point, but fails todismantle down to the coveringv/u. In fact, B dismantles down to each of itscoverings exceptv/u. It is the smallest example of this type that we know of.

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