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ASC Report No. 39/2012

Congruences of Convex Algebras

A. Sokolova, H. Woracek

Institute for Analysis and Scientific Computing

Vienna University of Technology TU Wien

www.asc.tuwien.ac.at ISBN 978-3-902627-05-6

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c Alle Rechte vorbehalten. Nachdruck nur mit Genehmigung des Autors.

ASCTU WIEN

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Congruences of Convex Algebras

Ana Sokolova1 , and Harald Woracek2

Abstract

We provide a full description of congruence relations of convex, positiveconvex, and totally convex algebras. As a consequence of this result weobtain that finitely generated convex (positive convex, totally convex)algebras are finitely presentable. Convex algebras, in particular positiveconvex algebras, are important in the area of probabilistic systems. Theyare the Eilenberg-Moore algebras of the subdistribution monad.

AMS MSC 2010: Primary 08A30, 52A20, 18C20. Secondary 08A62, 18C05

Keywords: convex algebra, congruence relation, finitely presentable

1 Introduction

In this paper we present a study of the equational classes CA, PCA, and TCA,of convex, positive convex, and totally convex algebras. We describe all con-gruence relations of such algebras. Knowing the congruences, we obtain thatfinitely generated convex (positive convex, totally convex) algebras are finitelypresentable.

A convex algebra is an algebra (with nonempty carrier set) with an infinite

set of operations of arbitrary positive arities providing convex combinations ofthe arguments, which satisfy two axioms (axiom schemes): (1) the projectionaxiom stating that a convex combination with a single coefficient equal to 1equals the identity map, and (2) the baricenter axiom stating that a convexcombination of convex combinations equals the convex combination with suit-ably multiplied and summed coefficients. Positive convex and totally convexalgebras are defined in a similar way from larger convex structures (sub-convexcombinations for positive convex algebras and linear combinations with coeffi-cients whose absolute values are sub-convex for totally convex algebras). Theprecise definitions and details follow in Section 3.

Examples of convex algebras are provided by convex subsets of a vector spaceover the scalar field R (a subset of a vector space is convex, if it contains witheach two points the whole line segment connecting them): IfK is such, then K isa convex algebra with the operations inherited from the vector space. However,these examples do not exhaust the class CA; the major obstacle being possiblefailure of cancellation laws in general convex algebras. Examples of positiveconvex algebras are provided by convex subsets of a vector space over R whichcontain the zero vector. Examples of totally convex algebras are provided byconvex subsets of a vector space R which are symmetric around the zero vector.

Convex algebras appear in a categorical context. To explain this, e.g. for thePCA-situation, consider the category Vec+1 whose objects are regularly ordered

1University of Salzburg, email: anas@cs.uni-salzburg.atResearch supported by Austrian Science Fund (FWF) grant V00125.2Vienna University of Technology, email: harald.woracek@tuwien.ac.at

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and topological questions, see, e.g., [PR85, BK93, Ke98, Pu01a, Pu01b, Pu03]

and the references therein. A far reaching generalization, namely the conceptof convexity theories, has been developed in a series of papers involving severalauthors which started with [Ro94], and went on (at least) till [Ro01].

Previous work which is the closest to our approach is [PR90, Ke99, Ke00],where congruence relations in (infinitary or p-) totally convex modules (alge-bras) are studied. Some parts of our results read similarly and several geometricideas employed there can also be used in the present setting. In order to preventconfusion concerning terminology, let us note explicitly that in the literaturealgebras in CA (PCA or TCA) are also called finitely (positively/totally) con-vex modules. The term finitely thereby refers to the fact that they carryonly finitary operations. However, in the present paper we stick to the purelyalgebraic setting and do not touch upon the possibility of allowing infinitaryoperations. Hence, we omit the prefix finitely from the notation. Moreover,

we also choose the term algebra over module since it has been used in recentwork regarding positive convex algebras [Do06, Do08, SS11].

The structure of the paper is as follows. After the introduction, we recallsome notions and facts from convex geometry in Section 2. In Section 3, wepresent the equational classes of convex, positive convex and totally convex al-gebras. After collecting some basic facts, we investigate the relationship betweenCA, PCA, and TCA. Interestingly, it turns out that CA and PCA are in essencejust the same, whereas TCA carries a significantly stronger structure. Section 4is the core of the paper. There we formulate and prove Theorems 4.3 and 4.4that describe the congruence relations on a polytope K in euclidean space. Itturns out that a congruence on K is fully determined by two ingredients: (1)a family of linear subspaces, describing the congruence classes in the interior

of K and in the interior of each of its lower dimensional facets; (2) a graph,describing how the interiors of K and each of its facets are related to the lowerdimensional facets forming the respective boundary. Finally, in Section 5, wegive the already mentioned application, and show that in each of the algebraiccategories corresponding to CA, PCA, and TCA, the notions finitely generatedand finitely presentable coincide.

Our basic reference concerning terminology and results of universal algebra isthe (old but still excellent) book [Gra68], or the more recent treatment [BS00].For the (few) notions from category theory which are used in this paper, werefer the reader to [La98]. Our standard reference concerning convex geometryis [Gru03].

2 Preliminaries from convex geometryBasic universal algebra and euclidean topology notions and results will be re-called when they are needed. In this section we explicitly recall some definitionsand results from convex geometry. We give even proofs of simple properties, toset the mood for what follows in the paper.

We start with the definition of a convex set.

Definition 2.1. A subset C of a vector space V over the field R is convexif for all x, y C and any scalar [0, 1] it holds that x + (1 )y C.Geometrically, this means that C contains, together with each two points, thewhole line segment connecting them.

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Convex sets are mapped by linear functions to convex sets. The following

simple property shows that convexity is the same as being closed under arbitraryconvex linear combinations.

Lemma 2.2. LetC be a subset of a vector space V overR. Then C is convexif and only if

ni=1

pixi C

for all n N+, all xi C, and all pi [0, 1] with i = 1, . . . , n such thatni=1pi = 1.

Proof. It is clear that being closed under convex linear combinations impliesconvexity. The other direction is proved by induction. Let C be a convex subsetand let x

i C, p

i [0, 1], for i = 1, . . . , n be given such that ni=1pi = 1. Ifn = 1 or pi = 1 for some i, the statement is trivial. If n = 2, the statement

holds by convexity. Assume that n > 2, p1 = 1, and the statement holds forn 1. Then

ni=1

pixi = p1x1 + (1 p1) ni=2

pi1 p1

xi

andn

i=2pi

1p1= 1. Hence, from the inductive hypothesis

ni=2

pi1p1

xi C

and by convexity alson

i=1pixi C.

The following subsets, associated with a finite subset Y ofV, play an importantrole:

span Y :=yY yy | y R,dir Y :=

yY

yy | y R,yY

y = 0

,

affY :=

yY

yy | y R,yY

y = 1

,

co Y :=

yY

yy | y [0, 1],yY

y = 1

,

co Y :=

yY

yy | y (0, 1],yY

y = 1

.

We refer to co Y as the closed convex hull ofY and co Y as the open convex hullor the interior of co Y. We will see later that this choice of terminology is indeedjustified, cf. Lemma 2.5. The linear span span Y is the smallest vector subspacethat contains Y. Moreover, we refer to affY as the affine space generated byY and dir Y as the directions of affY. Note that dir Y is a vector subspace.Clearly, for each finite set Y,

co Y co Y affY span Y and dir Y span Y.

If Y contains only one element, then co Y = co Y = affY = Y and dir Y = {0}.If |Y| 2, then co Y co Y affY and dir Y = {0}.

First, some simple geometric properties of these sets.

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Lemma 2.3. Let Y be a finite subset of a vector space V overR. Then the

following hold:(i) For each z affY we have

affY = z + dir Y .

(ii) For each z affY, we have

dir Y =

w z | w affY

.

(iii) We havedir Y =

w z | z, w affY

.

(iv) And, for every y0 Y,

dir Y = span{y y0 | y Y .

Proof. Let z affY be given, and write z =

yY yy with

yY y = 1.Given x affY, write x =

yY yy with

yY y = 1. Then

x z =yY

(y y)y ,

and

yY(y y) = 0. This means that x z dir Y, and we see that theinclusion in item (i) holds. All other inclusions in item (i) - item (iii) followin the same way. Item (iv) is straightforward by the definition of dir Y.

In the situation that V = Rn some important topological properties hold. These

are expressed in the following two lemmas.

Lemma 2.4. LetY be a finite subset ofRn. Thenco Y is compact and convex.Moreover, co Y is the closure of co Y.

Proof. It is easy to check by the definitions that co Y is convex. Recall that aset is compact if it is closed and bounded. Moreover, continuous functions mapa compact set to a compact set. The set

= {(y)yY R|Y| | y [0, 1],

yY

y = 1}

is compact, the function (y)yY

yY yy is continuous, and it maps to

co Y. Hence, co Y is compact.

Since co Y co Y and co Y is closed, we get Clos( co Y) co Y. For theopposite inclusion, take an element y co Y. Write Y = {y1, . . . , yn}, andy =

ni=1 iyi with i [0, 1] and

ni=1 i = 1. We will construct a sequence

of elements in co Y whose limit is y, showing that y Clos( co Y). Without lossof generality, we may assume that n > 0 and 1, . . . , k are all coefficients thatare equal to 0. Then for y =

ni=1 iyi where (0,

nk

) and

i :=

i = 1, . . . , ki i = k + 1, . . . , n 1n k i = n

we have y co Y. The elements y tend to y when tends to 0.

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Lemma 2.5. Let Y be a finite subset ofRn with |Y| 2. Then co Y is open

considered as a subset of affY.Proof. Let Y = {y0, . . . , ym} Rn with m 1. From Lemma 2.3, for eachz affY, affY = z + span{yk y0 | k = 1, . . . , m}. Let {b1, . . . , bl} bea maximal linearly independent subset of {yk y0 | k = 1, . . . , m}. ThenaffY = z + span{b1, . . . , bl}.

Choose (if necessary) bl+1, . . . , bn such that {b1, . . . , bn} is a basis ofRn andconsider the norm

||x|| := maxi=1,...,n

|i|, x Rn ,

where i are the unique coefficients in x =n

i=1 ibi.For each > 0, the set U = {x Rn | ||x|| < } is open with respect to the

euclidean topology in Rn (by equivalence of norms2). Thus, for each z affY,

the set (z + U) affY is an open subset of affY. We have

(z + U) affY = z + {l

i=1

ibi | |i| < } ,

where is trivial and follows since {b1, . . . , bn} is linearly independent.Assume now that z co Y, and write z =

mi=0 iyi with i (0, 1). This

is always possible since |Y| 2 and hence no i equals 1. Set

:=1

mmin({i | i = 0, . . . , m} {1 i | i = 0, . . . , m}) .

Then > 0 and it is not difficult to check that the choice of guarantees that

(z + U) affY co Y.

Hence we have found an open set in aff Y containing z and being contained inco Y. Since z co Y was arbitrary, this shows that co Y is open as a subset ofaffY.

Lemma 2.5 implies an alternative characterization of dir Y.

Lemma 2.6. Let Y be a finite subset ofRn. Then we have

dir Y = span{y z | y co Y}, z affY ,

and

dir Y = span{y2 y1 | y1, y2 co Y} .Proof. If |Y| = 1, we have dir Y = {0} and co Y = affY = Y. Hence, the statedequalities hold in this case. Moreover, the second asserted equality will followimmediately once the first is shown, and the inclusion in the first one istrivial.

Assume that |Y| 2, and let z affY be given. Since span{y z | y co Y} dir Y and both are vector subspaces, we get that span{y z | y co Y}is a subspace of dir Y. Note that no proper subspace of dir Y contains a non-empty open subset of dir Y (think as an illustration of a line in a plane and

2 maxi=1,...,n

|i| n

i=12i

12 n max

i=1,...,n|i|.

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a full open disc in this plane). In order to show the needed equality, we will

show that span{y z | y co Y} does contain a non-empty open set in dir Y.Consider the translation map Tz : x x z. It is a homeomorphism, i.e.,a continuous bijective map whose inverse is also continuous. Homeomorphismsmap open subsets onto open subsets. Note that Tz maps affY onto dir Y. Asa consequence, also by Lemma 2.5, Tz( co Y) is a non-empty open subset ofdir Y. Clearly, Tz( co Y) = {y z | y co Y} span{y z | y co Y}.

Let C Rn be convex. A point e C is called an extremal point if

e = tx + (1 t)y with x, y C, t (0, 1) x = y = e .

Geometrically, this means that e does not lie in the interior of any line segmentwith endpoints in C. We denote the set of all extremal points of C by ext C.

It is an important fact that compact convex sets can be recovered from theirextremal points. The Kren-Milman theorem states in a very general contextthat each compact convex set is the closed convex hull of its extremal points,see, e.g., [Ru91, 3.23]. The version of this theorem for subsets C ofRn, and thisis what we use here, can be found in [Gru03, 2.4.5].

We mainly deal with a certain kind of geometric objects called polytopes.

Definition 2.7. Let K be a subset of the euclidean space Rn. The set K is apolytope if it is of the form K = co Y for some finite set Y Rn.

A fundamental example of a polytope is a simplex.

Example 2.8 (A d-dimensional simplex). Let a Rn, and let {u1, . . . , ud} bea linearly independent subset ofRn. Then the polytope

K := co

{a} {a + ui | i = 1, . . . , d}

is called a d-dimensional simplex.For instance, for d = n = 3 and

a :=

00

0

, u1 :=

10

0

, u2 :=

01

0

, u3 :=

00

1

, (2.1)

we obtain the pyramid having the triangle with corner points (0, 0, 0), (1, 0, 0),(0, 1, 0) as its base and the point (0, 0, 1) as its apex.

Another concrete example of a polytope is an octahedron.Example 2.9 (A d-dimensional octahedron). Let a Rn, and let {u1, . . . , ud}be a linearly independent subset ofRn. Then the polytope

K := co

{a + ui | i = 1, . . . , d} {a ui | i = 1, . . . , d}

is called a d-dimensional octahedron.

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For instance, if d = n = 3 and a, u1, u2, u3 again as in (2.1), we obtain a

regular octahedron with center at the origin.3

(2.2)

Polytopes can be defined in several equivalent ways. The definition used in[Gru03] is presented in the next lemma. The fact that this definition is equivalentto the one above, i.e., the proof of the lemma is, in essence, a consequence ofthe Kren-Milman theorem; we skip the details.

Lemma 2.10. A subset K Rn is a polytope if and only if K is compact,convex, and the set ext K of its extremal points is finite.

Note that if K is a polytope and K = co Y for some finite set Y, thenext K Y and K = co(ext K).

Remark 2.11. The mentioned concrete examples, the simplex (2.1) and theoctahedron (2.2), are of particular interest in the present context. They arethe free algebras with 3 generators in the equational classes PCA and TCA,respectively (similarly as the standard simplex (1.1) is in CA). This fact (ofcourse for dimension n instead of 3), together with the results of Section 3 below,shows that understanding polytopes as convex algebras suffices to understandall finitely generated algebras in CA, PCA, and TCA.

3 The equational classes CA, PCA, and TCA

In this section we investigate the three convexity theories of convex, positiveconvex, and totally convex algebras and their induced equational classes. Tostart with, let us recall the definitions.

Definition 3.1. A convex algebra is an algebra (with nonempty carrier set) oftype

Tca :=

(pi)ni=1 R

n | n N+, p1, . . . , pn 0,ni=1

pi = 1

,

which satisfies (we denote by f(pi)

n

i=1

, (pi)ni=1

Tca

, the operations of the alge-bra)

(1) The projection axiom:

f(ij)ni=1(x1, . . . , xn) = xj , n N+, j = 1, . . . , n ,

where ij denotes the Kronecker-delta

ij :=

1 , i = j

0 , i = j

3Picture source: http://en.wikipedia.org/wiki/Octahedron

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(2) The baricenter axiom:

f(pi)ni=1

f(p1j)mj=1(x1, . . . , xm), . . . , f (pnj)mj=1(x1, . . . , xm)

=

= f(

ni=1 pipij)

mj=1

(x1, . . . , xm),

whenever n, m N+, (pi)ni=1 Tca, and (pij)mj=1 Tca, i = 1, . . . , n.

The operation appearing on the right hand side of the baricenter axiom is well-defined since

mj=1

ni=1

pipij

=

ni=1

pi

mj=1

pij

, (3.1)

and hence (n

i=1pipij)mj=1 Tca.

By CA we denote the equational class of convex algebras.

Definition 3.2. A positive convex algebra is an algebra (with nonempty carrierset) of type

Tpca :=

(pi)ni=1 R

n | n N+, p1, . . . , pn 0,ni=1

pi 1

,

which satisfies the projection axiom and the baricenter axiom, where in thelatter (pi)

ni=1 and (pij)

mj=1 vary through Tpca.

We denote the equational class of all positive convex algebras as PCA.

Definition 3.3. A totally convex algebra is an algebra (with nonempty carrierset) of type

Ttca

:= (pi)ni=1 Rn | n N+,n

i=1

|pi| 1,

which satisfies the projection axiom and the baricenter axiom, where in thelatter (pi)

ni=1 and (pij)

mj=1 vary through Ttca.

We denote the equational class of all positive convex algebras as TCA.

Note that, again because of (3.1), the operation appearing on the right handside of the baricenter axiom is always well-defined, i.e., is of type Tpca or Ttca,respectively (to see this for Ttca, use the triangle inequality).

It is obvious that each positive convex algebra can be considered as a convexalgebra. To be precise, if P, (f)Tpca is a positive convex algebra, thenP, (f)Tca is a convex algebra. Similarly: If P, (f)Ttca is a totallyconvex algebra, then P, (f)Tpca is a positive convex algebra, and in turn

P, (f)Tca is a convex algebra. Due to this fact, many results can immedi-ately be transferred from CA to PCA and TCA.Another interesting fact, which we shall explain in the sequel, is that CA and

PCA are essentially the same from an algebraic viewpoint, whereas TCA carriesmore structure than CA (compare Proposition 3.6 with Proposition 3.8 below).

When working with algebras of one of the types Tca, Tpca, or Ttca, it ispractical (and customary) to write operations as formal sums and/or to usevector notation:

f(pi)ni=1(x1, . . . , xn) =ni=1

pixi = (p1, . . . , pn)

x1...

xn

.

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With this notation the projection axiom writes as

(0, . . . , 1

j-th place

, . . . , 0)

x1...xn

= xj ,and the baricenter axiom as

(p1, . . . , pn)

p11 . . . p1m... ...

pn1 . . . pnm

x1...

xm

=

= (p1, . . . , pn)p11 . . . p1m

.

..

.

..pn1 . . . pnm

x1.

..xm .

In the next lemma we provide some simple but useful identities which followfrom the projection and baricenter axioms. For the convenience of the reader,we provide an explicit proof. In the setting ofTCA (with infinitary operations)these identities were shown in [PR84, Theorem 2.4] using a different proof.

Lemma 3.4. For items (i)(iii), let P be an algebra in any of the classesCA,PCA, orTCA. For items (iv) and (v), assume that P belongs to PCA orTCA.Let c stand for ca, pca, tca, if P is inCA, PCA, TCA, respectively.

(i) The above introduced sum notation already suggests that operations are in

a sense commutative. They indeed are, we have

f(pi)ni=1(x1, . . . , xn) = f(p(i))ni=1(x(1), . . . , x(n)),

whenever n N+, is a permutation of {1, . . . , n}, and (pi)iI Tc.

(ii) The extended projection law:

f(pi)ni=1(x1, . . . , xn) = f(pik )mk=1

(xi1 , . . . , xim),

holds whenever (pi)ni=1 Tc and i1, . . . , im satisfy

i1 < < im, {i1, . . . , im} {i {1, . . . , n} | pi = 0}.

(iii) Whenever (pi)ni=1 Tc and x P, we have

f(pi)ni=1(x , . . . , x) = f(

ni=1 pi)

(x).

(iv) The elements

f(0)ni=1(x1, . . . , xn), n N+, x1, . . . , xn P,

all coincide. We denote this element as 0P.

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(v) The element 0P plays the role of a zero element: Let (pi)ni=1 Tc and let

i1, . . . , im satisfyi1 < < im, {i1, . . . , im} {i {1, . . . , n} | xi = 0P}.

Then

f(pi)ni=1(x1, . . . , xn) =

f(pik )

mk=1

(xi1 , . . . , xim) , m 1

0P , m = 0.

Proof. We denote, here and in the sequel, by In the n n identity matrix andby 0n,m the n m zero matrix.

(i) Set pij := i,1(j) and compute

f(p(i))

n

i=1x(1), . . . , x(n) ==f(p(i))ni=1

f(p1j)nj=1(x1, . . . , xn), . . . , f (pnj)nj=1(x1, . . . , xn)

=

=f(

ni=1 p(i)pij)

nj=1

(x1, . . . , xn) = f(pj)nj=1(x1, . . . , xn).

(ii) By commutativity it is enough to consider the case that ik = k, k =1, . . . , m. Then we compute, using the projection and baricenter axioms,

(p1, . . . , pm)

x1...

xm

=(p1, . . . , pm)

Im 0m,nm

x1...

xmxm+1

.

.

.xn

=

=

(p1, . . . , pm)

Im 0m,nm x1...

xn

=

= (p1, . . . , pm, 0, . . . , 0) (p1,...,pn)

x1...

xn

.

(iii) f(pi)ni=1(x , . . . , x) = f(pi)ni=1(f(1)(x), . . . , f (1)(x)) = f(

ni=1 pi)

(x).

(iv) Apply the extended projection law twice to obtain

f(0)ni=1(x1, . . . , xn) = f(0)n+n

i=1 (x1, . . . , xn, x

1, . . . , x

n

) = f(0)n

i=1(x

1, . . . , x

n

).

(v) Consider first the case that m = 0. We have, using (iv),

(p1, . . . , pn)

0P...

0P

= (p1, . . . , pn)

0n,n

0P...

0P

=

=

(p1, . . . , pn)0n,n0P...

0P

= (0, . . . , 0)

0P...

0P

= 0P.

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This remark also clarifies the note made in [Pu03, p.110] immediately after the

definition of positively convex algebra (in which the above stronger form of thebaricenter axiom is required).

3.1 Positive convex algebras vs. convex algebras

In this subsection we make precise what we meant above by claiming that CAand PCA are just the same.

Let P, (f)Tca CA. We call a collection {f| Tpca} of operations onP an operational extension of P, (f)Tca to PCA, if

f = f, Tca, and P, (f)Tpca PCA . (3.4)

Proposition 3.6. Let P, (f)Tca CA. Then the set of all operational

extensions of P, (f)Tca to PCA corresponds bijectively to P. In particular,since P = , P can be operationally extended to a positive convex algebra.

Proof. In the first step we show existence of operational extensions. Choose anelement z P, and define for (pi)

ni=1 Tpca

f(pi)ni=1(x1, . . . , xn) := f(p1,...,pn,p)(x1, . . . , xn, z),

where

p := 1 ni=1

pi.

The extended projection law in P, (f)Tca gives f = f, Tca.Knowing that f

= f

, T

ca, it is clear that the projection axiom holds

in P, (f)Tpca; the operations involved all belong to Tca. To show the PCA-baricenter axiom, let (pi)

ni=1 Tpca and (pij)

mi=1 Tpca, i = 1, . . . , n, be given.

Denote

p := 1 ni=1

pi, pi := 1 mj=1

pij, i = 1, . . . , n .

Then (ij again denotes the Kronecker-delta)

(p1, . . . , pn, p), (pi1, . . . , pim, pi), i = 1, . . . , n , (i,m+1)m+1i=1

all belong to Tca, and hence we may apply the CA-baricenter axiom. This gives

f(pi)

ni=1 f(p1j)mj=1(x1, . . . , xm), . . . , f(pnj)mj=1(x1, . . . , xm) =

= (p1, . . . , pn, p)

f(p1j)mj=1(x1, . . . , xm)...

f(pnj)mj=1(x1, . . . , xm)

z

=

= (p1, . . . , pn, p)

p11 . . . p1m p1...

......

pn1 . . . pnm pn0 . . . 0 1

x1...

xmz

=

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= (p1, . . . , pn, p)p11 . . . p1m p1

.

.....

.

..pn1 . . . pnm pn

0 . . . 0 1

x1...

xmz

=

= n

i=1

pipi1, . . . ,ni=1

pipim,ni=1

pipi + p

x1...

xmz

=

= f(

ni=1 pipij)

mj=1

(x1, . . . , xm).

The last equality holds since, due to (3.1),

ni=1

pipi + p = 1 ni=1

pipi1 + +

ni=1

pipim.Moreover, we have for arbitrary x P

0P = f(0)(x) = f(0,1)(x, z) = z .

Second, assume conversely that operations f, Tpca, with (3.4) are given.Then, by Lemma 3.4(v),

f(pi)ni=1(x1, . . . , xn) = f(p1,...,pn,p)(x1, . . . , xn, 0P) =

= f(p1,...,pn,p)(x1, . . . , xn, 0P), (pi)ni=1 Tpca ,

where p := 1

ni=1pi. Hence, the operations f coincide with the operaions

constructed in the first step with the choice z = 0P.

Finally, the above computation also shows that an operational extensionaccording to (3.4) is uniquely determined by the corresponding element 0P,showing that P indeed corresponds bijectively to the set of all operational ex-tensions.

Next, we show that congruence relations within CA or PCA are just the same.Let P be an algebra in CA or PCA and an equivalence relation on P. Recallthat is a congruence if whenever (xi, x

i) , for i {1, . . . , n} then also

(n

i=1pixi,n

i=1pixi) for (pi)

ni=1 Tca or (pi)

ni=1 Tpca, respectively. By

ConCA P or ConPCA P we denote the sets of all CA- or PCA-congruence relationson P, respectively.

Lemma 3.7. Let P, (f)Tca CA and let {f| Tpca} be an operationalextension of P, (f)Tca to PCA. Moreover, let be an equivalence relationon P. Then ConCAP, (f)Tca if and only if ConPCAP, (f)Tpca.

Proof. Clearly, each PCA-congruence on P is also a CA-congruence. Conversely,assume that ConCA P. Let (pi)

ni=1 Tpca and (xi, x

i) , i = 1, . . . , n,

be given. Again set p := 1 n

i=1pi, so that (p1, . . . , pn, p) Tca. Since ConCA P, it follows that n

i=1

pixi,ni=1

pixi

=

=p1x1 + +pnxn + p 0P , p1x

1 + +pnx

n + p 0P

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3.2 Totally convex algebras vs. convex algebras

In this subsection we characterize those convex algebras which can be opera-tionally extended to totally convex algebras. In view of Proposition 3.6 thisproblem is equivalent to characterizing those positive convex algebras that canbe operationally extended4 to totally convex algebras. It turns out that not ev-ery convex algebra has this property, contrasting the situation for passing fromCA to PCA.

The decisive additional property which allows to operationally extend toTCA is existence of an involutory endomorphism satisfying a particular compu-tation rule. Recall that an endomorphism of an algebra is called involutoryif it is inverse to itself, i.e., if 2 = id.

Proposition 3.8. Let P, (f)Tpca PCA. Then there exists an operationalextension {f| Ttca} of P to TCA if and only if there exists an involutoryPCA-endomorphism of P which has the property that

f(p1,...,pn,q1,...,qn)(x1, . . . , xn, x1, . . . , xn) =

=f(p1,...,pn,q1,...,qn)(x1, . . . , xn, x1, . . . , xn),(3.5)

whenever n N+,

(p1, . . . , pn, q1, . . . , q n), (p1, . . . , p

n, q

1, . . . , q

n) Tpca,

pk qk = pk q

k, k = 1, . . . , n .

Proof. Consider first the case that |P| = 1. Then P = {0P} and trivially Pcan be operationally extended to a totally convex algebra. Equally trivially the

identity map is an involutory endomorphism and satisfies (3.5). Hence, inthis case, the asserted equivalence holds. For the rest of the proof we may thusassume that P contains more than one element.

Necessity. Assume that an operational extension {f| Ttca} is given, andconsider the map : P P defined as

(x) = f(1)(x).

Let (pi)ni=1 Ttca, then

f(1)

f(pi)ni=1(x1, . . . , xn)

= f(pi)ni=1(x1, . . . , xn)

= f(pi)ni=1f(1)(x1), . . . , f(1)(xn).This shows that is an endomorphism (even a TCA-endomorphism). Moreover,we have

f(1)(f(1)(x)) = f(1)(x) = x,

i.e. is involutory.In order to show (3.5), it is certainly sufficient to show that

f(p1,...,pn,q1,...,qn)(x1, . . . , xn,x1, . . . , xn) =

=f(p1q1,...,pnqn)(x1, . . . , xn),(3.6)

4Operational extensions of an algebra in CA or PCA to an algebra in TCA are definedanalogously as operational extensions of an algebra in CA to one in PCA.

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whenever (p1, . . . , pn, q1, . . . , q n) Ttca. Note here that

ni=1

|pi qi| ni=1

(|pi| + |qi|) =ni=1

|pi| +ni=1

|qi| 1,

and hence the operation written on the right side is legitimate. To see (3.6), wecompute

(p1, . . . , pn, q1, . . . , q n)

x1...

xnx1

...xn

= (p1, . . . , pn, q1, . . . , q n)

In

In

x1...

xn

=

=

(p1, . . . , pn, q1, . . . , q n)

In

In

x1...xn

= (p1 q1, . . . , pn qn)

x1...

xn

.

Sufficiency. Assume that an involutory PCA-endomorphism on P with (3.5)is given. For each (pi)

ni=1 Ttca we define an operation f(pi)ni=1 as

f(pi)ni=1(x1, . . . , xn) := f(p+1 ,...,p+n ,p

1 ,...,p

n )(x1, . . . , xn, x1, . . . , xn), (3.7)

wherep+ := max{p, 0}, p := min{p, 0}, p R. (3.8)

Note that p = p+ p. Moreover, we have

ni=1

p+i +ni=1

pi =ni=1

|pi|,

and hence the operation in Tpca on the right side of (3.7) is well-defined. If(pi)

ni=1 Tpca, then p

+i = pi and p

i = 0 for all i {1, . . . , n}, and the extended

projection law for P, (f)Tpca gives

f(pi)ni=1(x1, . . . , xn) = f(p1,...,pn,0,...,0)(x1, . . . , xn, x1, . . . , xn)

= f(pi)ni=1(x1, . . . , xn).

It is not difficult to see that the projection axiom holds for P, (f)Ttca; weskip the details. To check the TCA-baricenter axiom, let (pi)

ni=1 Ttca and

(pij)mj=1 Ttca, i = 1, . . . , n, be given. First, compute

f(pij)mj=1(x1, . . . , xm) = f(p+i1,...,p+im,p

i1,...,p

im)(x1, . . . , xm, x1, . . . , xm)

= f(p+i1,...,p+im

,p

i1,...,p

im)(x1, . . . , xm, x1, . . . , xm)

= f(pi1,...,p

im,p+i1,...,p

+im)

(x1, . . . , xm, x1, . . . , xm).

Next, setP+ :=

p+ij

i=1,...,nj=1,...,m

, P :=

piji=1,...,nj=1,...,m

,

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and use the baricenter axiom for P, (f)Tpca to compute

f(pi)ni=1f(p1j)mj=1(x1, . . . , xm), . . . , f(pnj)nj=1(x1, . . . , xm) =

=(p+1 , . . . , p+n , p

1 , . . . , p

n )

P+ P

P P+

x1

.

.

.xnx1

.

.

.xn

=

=

(p+1 , . . . , p

+n , p

1 , . . . , p

n )

P+ P

P P+

x1

.

.

.xnx1

.

.

.xn

=

= n

i=1

p+i p+i1 +

ni=1

pi pi1, . . . ,

ni=1

p+i pi1 +

ni=1

pi p+i1, . . .

x1

..

.xnx1

.

.

.xn

=

= n

i=1

pipi1

+, . . . ,

ni=1

pipi1

, . . .

x1...xnx1

.

.

.xn

where the last equality follows from (3.5) since

ni=1

p+i p+ij +

ni=1

pi pij ni=1

p+i pij +

ni=1

pi p+ij ==

ni=1

pi0,pij0

pipij +ni=1

pi

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(xi, xi) , i = 1, . . . , n, be given. Using the notation (3.8), we obtain from

(3.6) that

f(pi)ni=1(x1, . . . , xn) = f(p+1 ,...,p+n ,p

1 ,...,p

n )(x1, . . . , xn, x1, . . . , xn).

The same equation holds for xi instead of xi. Since is a PCA-congruence,xix

i, and xix

i by the assumptions, we see that indeed

f(pi)ni=1(x1, . . . , xn) f(pi)ni=1(x1, . . . , x

n) .

4 Convex equivalences on polytopes

Let K Rn be a polytope. We consider Kas a convex algebra endowed with theoperations inherited from Rn (as in Proposition 3.10). The following propertyis a direct consequence of the definitions and Lemma 2.2 but it is an importantobservation for what follows.

Lemma 4.1. An equivalence relation on a polytope K is in ConCA K if andonly if it is convex as a subset of K K Rn Rn, with operations definedcomponent-wise.

Throughout the paper, we denote

VK := P(ext K) \ {},

where P(M) denotes the power set of a set M, and consider VK as a join-subsemilattice of the lattice P(ext K). Moreover, we denote by SubRn the setof all linear subspaces ofRn, and consider SubRn as being ordered by inclusion.

In the following definition we introduce the crucial concepts for describingConCA K.

Definition 4.2. Let K Rn be a polytope and let ConCA K. We define amap : VK SubRn by

(Y) = span

x2 x1 | x1, x2 co Y, x1x2

, Y VK.

We define a graph G with vertices VK and (undirected) edges E given by

{Y1, Y2} E ( co Y1 co Y2) = , Y1, Y2 VK.

We denote by the equivalence relation on VK defined as

Y1 Y2 Y1 and Y2 are connected by a path in G.

The equivalence classes of are the connected components of G, see, e.g.,[Di00]. Note that there is always an edge in G connecting a vertex Y withitself, even if |Y| = 1, due to the definition of co Y.

In the following three statements we give a complete description of the con-gruence lattice ConCA K.

Theorem 4.3. LetK be a polytope inRn and let ConCA K. Then

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(i) The map is monotone.

(ii) LetC be a component of the graph G. Then C contains a largest elementwith respect to inclusion. Denoting this largest element by Y(C), we have{Y, Y(C)} E, Y C.

(iii) The relation is a congruence of the join-semilattice VK.

(iv) Let C be a component of G and Y(C) its largest element. Then

(Y) = (Y(C)) dir Y for Y C, (4.1)co Y + (Y(C))

co Y(C) = for Y C. (4.2)

SetZ(C) :=

YC

co Y, (4.3)

then the congruence can be recovered from and G as

=

C componentof G

(x1, x2) Z(C) Z(C) : x2 x1 (Y(C))

. (4.4)

Let us point out that reconstructing by means of the formula (4.4) onlyrequires knowledge of the classes of and the values of on the respectivelargest elements of these classes.

Theorem 4.4. Let K be a polytope in Rn. Let be a congruence relationof the join-semilattice VK with the property that each congruence class C of

contains a largest element, say Y(C). Moreover, let

: {Y(C) | C class of } SubRn

be a monotone map such that, for each class C of ,

(Y(C)) dir Y(C),

co Y + (Y(C))

co Y(C) = for Y C.

Then there exists a unique congruence ConCA K such that

= , (Y(C)) = (Y(C)) for C a class of .

This congruence can be computed from and by means of the formula

= C classof

(x1, x2) Z(C) Z(C) : x2 x1 (Y(C)), (4.5)where again Z(C) :=

YC co Y. Its associated function is given as

(Y) =

Y(C)

dir Y for Y C, (4.6)

and the set of edges E of its associated graph G is given as

{Y1, Y2} E

Y1 Y2

co Y1 +

Y([Y1])

co Y2 =

(4.7)

where [Y1] denotes the equivalence class of Y1.

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4.1 Geometric consequences of convexity

We start with a definition of a useful concept of perspective.

Definition 4.8. For each z K we denote by z : [0, 1] Rn Rn the mapdefined as

z(s, x) := sz + (1 s)x, s [0, 1], x Rn.

Based on geometric intuition, we refer to z as the perspective with center z.

Since K is convex and z K, we have z(s, K) K for any s [0, 1].Moreover, z(0, .) is the identity map, and z(1, .) is the constant map withvalue z.

The following observation is simple but important, and we further refer toit as perspective invariance. Intuitively speaking, perspective invariance means

that a congruence class cannot split and distribute over several different classeswhen moved with a perspective.

Lemma 4.9. Let A K be an equivalence class of . Then, for each z Kand s [0, 1], there exists an equivalence class Az,s K of with

z(s, A) Az,s.

K

z

A

z (s,A)

Moreover, perspective invariance implies that each equivalence class of is con-vex.

Proof. Let x1, x2 K with x1x2, z K, and s [0, 1] be given. Since zzand (x, y) sx + (1 s)y is an operation of K CA, we have

z(s, x1), z(s, x2)

= (sz + (1 s)x1, sz + (1 s)x2) .

To deduce that equivalence classes are convex, let x1, x2 K with x1x2, ands [0, 1] be given. Perspective invariance with z := x1 gives

(x1, sx1 + (1 s)x2) =

x1(s, x1), x1(s, x2)

.

Also the next fact is simple but useful.

Lemma 4.10.

(i) Let Y1, Y2 VK be given. Then

z co Y1, x co Y2 : z(s, x) co(Y1 Y2), s (0, 1).

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(ii) Let Y VK be given. Then

z co Y, x co Y : z(s, x) co Y, s [0, 1).

z co Y, x co Y : z(s, x) co Y, s (0, 1].

Proof. For the proof of (i) write

z =yY1

1yy, 1y (0, 1],

yY1

1y = 1,

x =yY2

2yy, 2y (0, 1],

yY2

2y = 1.

Then

z(s, x) = s yY1

1yy

+ (1 s) yY2

2yy

=

=

yY1\Y2

s1yy +

yY1Y2

s1y + (1 s)

2y

y +

yY2\Y1

(1 s)2yy.

All coefficients in these sums are positive and they sum up to 1. Hence,z(s, x) co(Y1 Y2). Item (ii) follows in the same manner.

Let us immediately exploit perspective invariance to show that is monotone.

Proof (of Theorem 4.3, (i)). Let Y1, Y2 VK with Y1 Y2 be given. IfY1 = Y2,there is nothing to prove. Hence, assume that Y1 Y2.

Let w (Y1), and write according to the definition of

w =mk=1

k(xk2 x

k1),

with somek R, x

k1 , x

k2 co Y1, x

k1x

k2 , k = 1, . . . , m .

Set z := 1|Y2\Y1|

yY2\Y1y, and fix s (0, 1). Note that the assumption Y1 Y2

ensures that z is well defined. Perspective invariance gives

z(s, xk1 ) z(s, x

k2), k = 1, . . . , m .

By Lemma 4.10(i), since z co(Y2\Y1) and Y2 = (Y2\Y1)Y1 by the assumption,we have z(s, xk1 ), z(s, xk2 ) co Y2, and hence

w :=mk=1

k

z(s, xk2) z(s, x

k1)

(Y2).

However,z(s, x

k2) z(s, x

k1) = (1 s)(x

k2 x

k1),

and hence w = (1 s)w. Since (Y2) is a vector subspace, it follows that alsow (Y2).

The property shown in the next lemma will be used in several instances.

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Lemma 4.11. Let Y VK, |Y| 2, and let x1, x2 co Y, x1 = x2. Consider

the map : R affY given by(t) := tx2 + (1 t)x1, t R.

Clearly, (R) = aff{x1, x2} is the line containing x1 and x2.5 Then

(r)(s, (t)) =

sr + (1 s)t

, r, t R, s [0, 1]. (4.8)

There exist numbers t < 0 and t+ > 1, such that

1( co Y) = (t, t+), 1(co Y) = [t, t+]

t

0

1

t+

R

affYcoY

x1

x2 (t+)

(t)

Proof. To show (4.8), we compute

(r)(s, (t)) = s(r) + (1 s)(t) =

=s

rx2 + (1 r)x1

+ (1 s)

tx2 + (1 t)x1

=

=

sr + (1 s)t

x2 +

s(1 r) + (1 s)(1 t)

x1 =

=

sr + (1 s)t

x2 +

1 (sr + (1 s)t)

x1 = (sr + (1 s)t).

Note that R and affY inherit the euclidean topology from Rn. Moreover theyalso inherit the euclidean metric from Rn. Also, a line is continuous, i.e., isa continuous function. Consider the set 1( co Y). Since is continuous andco Y is an open subset of affY by Lemma 2.5, this set is an open subset ofR.Since is linear and co Y is convex, cf. Section 2, it is convex. We will nowshow that 1( co Y) is bounded. Note that (t) = x1 + t(x2 x1). Therefore,for denoting the euclidean norm, using the downward triangle inequality,we get

(t) |t(x2 x1) x1| |t| x2 x1 x1.

Hence, for each positive real number R, if |t| > R+x1x2x1 , then (t) > R. This

shows that the inverse image by of a bounded set in aff Y is a bounded setin R. Now, since co Y co Y and the latter is bounded by Lemma 2.4, we get

5Note that actually depends on the points x1 and x2 but we prefer a light, overloadednotation.

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that 1( co Y) is bounded. Note that in R a set is open, convex, and bounded

if and only if it is a finite open interval. Hence1( co Y) = (t, t+)

with some numbers t, t+ R. Since (0) = x1 co Y and (1) = x2 co Y,we must have t < 0 and t+ > 1.

Again by Lemma 2.4, co Y is the closure of co Y and hence continuity of implies that 1(co Y) is closed. Since 1( co Y) 1(co Y) and [t, t+] isthe closure of 1( co Y) = (t, t+), we also have [t, t+]

1(co Y).To show the opposite inclusion, let t R with (t) co Y be given. If

t [0, 1], we have (t) co{x1, x2} co Y, and hence t (t, t+). Next,

consider the case that t > 1. Choose t (1, t) and set s := t

t. Then s (0, 1)

and

(t)(s, x1) = s(t) + (1 s)x1 = s(tx2 + (1 t)x1) + (1 s)x1 =

= (st)x2 + (1 st)x1 = tx2 + (1 t

)x1 = (t).

By Lemma 4.10, we have (t) co Y. Thus t (t, t+). Since t was arbitrary

in (1, t), we have (1, t) (t, t+) and hence [1, t] [t, t+] showing t [t, t+].The case that t < 0 is analogous.

4.2 The structure of G

In order to understand the structure of G, we need one preparatory result.

Lemma 4.12. Let Y1, Y2 VK, and assume that Y1 Y2. If {Y1, Y2} E,i.e.,

x1 co Y1, x2 co Y2 : x1x2,

then actuallyx1 co Y1, x

2 co Y2 : x

1x

2.

Proof. If Y1 = Y2, we can always choose x2 := x

1. If |Y1| = 1, also | co Y1| = 1,

and hence the assertion is trivial. Note that the assertion is not trivial in generalif Y1 Y2 as it often happens that then co Y1 co Y2 = , think for example of|Y1| = 2 and |Y2| = 3 when co Y2 is the interior of a triangle and co Y1 is theinterior of one of its sides.

Assume that Y1 Y2, |Y1| > 1, and choose x1 co Y1 and x2 co Y2 withx1x2.

Let x1 co Y1 be given. If x1 = x1, we can take x

2 := x2. Hence, assume

that x

1 = x1. Consider the line containing the points x

1, x1, i.e.,(t) := tx1 + (1 t)x

1, t R.

By Lemma 4.11, we have 1( co Y1) = (t, t+) and 1(co Y1) = [t, t+] with

some t < 0 and t+ > 1. Set s :=1

1t, then s (0, 1) and st + (1 s) 1 = 0.

By (4.8) since (1) = x1 we have

(t)(s, x1) = (st + (1 s) 1) = (0) = x1.

Take x2 := (t)(s, x2). By perspective invariance and the above equality, weget x1x

2. Since (t) co Y1 co Y2 and x2 co Y2, Lemma 4.10(ii) implies

x2 = (t)(s, x2) co Y2, which completes the proof.

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x1

x2

x1

(t)(s, x1)

(t)(s, x2)

x2

(t)

(R)

co Y2

Corollary 4.13. LetY1, Y2, Y3 VK, and assume that {Y1, Y2}, {Y2, Y3} E.If Y2 Y3, then also {Y1, Y3} E.

Proof. Choose x1 co Y1 and x2 co Y2 with x1x2. According to Lemma4.12 there exists x3 co Y3 with x2x3. Transitivity gives x1x3, and we seethat {Y1, Y3} E.

We can now establish the result on the structure of components of G.

Proof (of Theorem 4.3, (ii)). We proceed in three steps.

The first step is to show that

{Y1, Y2} E {Y1, Y1 Y2}, {Y2, Y1 Y2} E

To this end, choose x1 co Y1 and x2 co Y2 with x1x2. By convexity ofequivalence classes, we have

1

2(x1 + x2) xj, j = 1, 2,

and since x1

12 , x2

= 12 (x1 + x2), Lemma 4.10 gives

12 (x1 + x2) co(Y1 Y2)

which shows the above property.

Let C be a component of G. The second step is to show that

Y1, Y2 C Y3 C : Y1 Y2 Y3, {Y1, Y3}, {Y2, Y3} E

To established existence ofY3 with the required properties, we use induction onthe length of a path in G connecting Y1 and Y2.

If Y1 and Y2 can be connected by a path of length 1, i.e., if {Y1, Y2} E,then set Y3 := Y1 Y2. By the property shown in the first step, we have{Y1, Y3}, {Y2, Y3} E. This also implies that Y3 C.

For the inductive step, let m N+ be given, and assume that Y1 and Y2 canbe connected by a path of length m + 1. This means that there exist

Y0 , . . . , Y m+1 VK, Y

0 = Y1, Y

m+1 = Y2 with {Y

k, Y

k+1} E

for k {0, . . . , m}. Clearly, all Yk belong to C. The inductive hypothesisprovides a vertex Y C with

Y1 Ym Y

, {Y1, Y}, {Ym, Y

} E.

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Now, since Ym Y and {Y2, Y

m}, {Y

m, Y

} E, from Corollary 4.13, we get

{Y2, Y

} E. From the property shown in the first step, we get {Y2, Y2 Y}, {Y, Y2 Y} E. Another application of Corollary 4.13, this time since

{Y1, Y}, {Y, Y2 Y

} E and obviously Y Y2 Y

, gives that also {Y1, Y2 Y} E. Thus the element Y3 := Y2 Y

has all required properties. Thisfinishes the proof of the second step.

Let Y C. If there exists Y C with Y Y, the property shown in thesecond step gives an element Y C with Y Y Y and Y Y (sinceY Y Y). Hence, Y is not maximal in C. We conclude that if an elementis maximal in C, then it must also be the largest in C. Since C is finite (thewhole graph is finite), there certainly exists a maximal element. Let Y0 be such.Then, since it is the largest, we have

Y0 YC

Y,

and hence Y0 =

YC Y.Let Y C be given. By the property shown in the second step, applied to

Y and Y0, we obtain Y3 C with

Y Y0 Y3, {Y, Y3}, {Y0, Y3} E.

However, since Y0 is the largest element of C, it follows that Y3 = Y0. Hence,we have shown that {Y, Y0} E.

4.3 Behaviour of inside of co Y

Let Y VK. Our aim is to describe ( co Y co Y). If Y contains only oneelement, then co Y = Y and clearly ( co Y co Y) = Y Y. Hence, assumethat |Y| 2.

The main observation is that an equivalence class of which contains twodifferent points of co Y must already stretch all the way to the boundary ofco Y. This is the analogue of [PR90, Theorem 1.3], where a similar result wasestablished for congruences of totally convex algebras. However, the proof giventhere does not immediately carry over to the presently considered situation,since we are bound to operations in Tca, i.e., true convex combinations.

Lemma 4.14. Letx1, x2 co Y, x1 = x2, with x1x2. Then

aff{x1, x2} co Y [x1],where [x1] denotes the -equivalence class of x1.

Proof. Consider the line containing the points x1 and x2, i.e., the map

(t) := tx2 + (1 t)x1, t R,

and let t, t+ be as given by Lemma 4.11. As a first step we show that, for eacht (t, t+), the set

1([(t)] co Y) contains an open neighbourhood of t.Consider first the case that t 12 . Choose s [0, 1) such that st++(1s)

12

=t. This is doable since under the assumption for t we have t+ >

12

. Then by

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where the last equality follows by Lemma 2.6. To show the reverse inclusion,

let u (Y(C)) dir Y be given. Choose x1 co Y, let > 0, and set x2 :=x1 + u. Since u dir Y and x1 affY, we have x1 + span{u} affY. Sinceco Y is an open subset of affY, we can choose > 0 so small that x2 co Y.We have x2 x1 = u (Y(C)), and hence (4.4) implies that x1x2. Thisis enough to conclude that u = 1

u = 1

(x2 x1) (Y).

It remains to prove (4.2). To this end, remember that {Y, Y(C)} E,i.e., there exist x1 co Y, x2 co Y(C), with x1x2. By (4.4), we havex2 x1 (Y(C)), and hence

x2

co Y + (Y(C))

co Y(C).

By now all assertions of Theorem 4.3 are proved.

4.6 The converse construction

In this subsection let a relation and a map as in the statement of Theorem4.4 be given. For each equivalence class C of , we denote its largest elementby Y(C) and set

Z(C) :=YC

co Y.

Moreover, we define a relation on K as

=

C classof (x1, x2) Z(C) Z(C) | x2 x1 (Y(C))

.

The main step is to show:

Lemma 4.17. The relation is a convex congruence onK, i.e., ConCA K,and

= , (Y(C)) = (Y(C)) for C class of .

Proof. First of all, is an equivalence relation as a direct consequence of(Y(C)) being a linear subspace.

Next, notice the following: If x co Y, x co Y, and xx, then x and x

must both belong to the same of the sets Z(C), and hence Y Y.To show that is a convex congruence, let (x1, x

1), (x2, x

2) and s

(0, 1) be given. Choose Yj , Yj , j = 1, 2, such that xj co Yj, x

j co Y

j . Then

Yj Yj , and hence alsoY1 Y2 Y

1 Y

2 .

Let C be the class of which contains Y1 Y2. Lemma 4.10 gives

x := sx1 + (1 s)x2 co(Y1 Y2), x := sx1 + (1 s)x

2 co(Y

1 Y

2 ),

and hence x, x Z(C).Let Cj be the class which contains Yj. Since Yj Y(Cj), it follows that

Y1 Y2 Y(C1)Y(C2), and hence Y(C1)Y(C2) Y(C). Since is monotone,we conclude that

(Y(Cj)) (Y(C)), j = 1, 2.

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Now using (4.4), since ext K is a vertex of G and hence in some connected

component C, and is monotone, we getz(s, x1), z(s, x2)

, s (0, 1).

Letting s tend to 0, z(s, x1) tends to x1 and z(s, x2) to x2 and it follows that(x1, x2) Clos . Hence

0 Clos

and therefore 0 = Clos .

This lemma tells us, in particular, that is closed if and only if = 0.

Proof (of Proposition 5.6).Necessity. Assume that R is a finite subset of K K, such that is generated

by R as a subalgebra of K K CA (PCA or TCA, respectively). Then wehave

=

co R (CA situation )

co

{0K} R

(PCA situation )

co

R (R)

(TCA situation )

where is as in Proposition 3.8. For the PCA or TCA situation, note here thateach PCA or TCA linear combination can be written as a convex combinationby making use of the zero element 0K and the -images of the involved elements,and that for convex combinations the operations coincide with the usual vectoroperations.

In any case, is a polytope in R2n and thus in particular closed.

Sufficiency. Assume that ConCA K, and that is closed as a subset ofRn Rn. We will now show that is finitely generated as a CA-subalgebra ofK K. Set d := n dim (ext K), and choose a linear and surjective map : Rn Rd whose kernel equals (ext K). Note that this means

(x1, x2) ker (x1) = (x2) x2 x1 {x | (x) = 0}

x2 x1 Ker x2 x1 (ext K).

We denote by the diagonal in Rn Rn. Since is closed, by Lemma 5.7 andthe above, we have

= Clos = (K K) ker = (K K) ( )1().

The set K K is a compact (since K is compact) and convex subset ofR2n,and one can easily show unfolding the definition of an extremal point that

ext(K K) = ext K ext K,

so ext(K K) is a finite set. Hence, K K is a polytope in R2n.Now we will employ some non-trivial geometric arguments from [Gru03]. The

diagonal is a linear subspace, and hence can be written as a finite intersectionof halfspaces, i.e., it is a polyhedral set in the sense of [Gru03, 2.6]. Since ()is linear and surjective, the inverse image of a halfspace is again a halfspace.Hence, ( )1() is again a polyhedral set.

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As an intersection of a polytope with a polyhedral set, is a polytope,

cf. [Gru03,

3.1]. Hence, has only finitely many extremal points and =co(ext). This means that the finite set ext generates as a CA-subalgebraof K K.

It remains to consider PCA and TCA congruences. Assume that ConPCAK (or ConTCA K, respectively) and is closed. Then also ConCA K,and hence by what we have proven so far is finitely generated as a CA-subalgebra of K K. In particular, it is therefore also finitely generated as aPCA- (TCA-, respectively) subalgebra of K K.

Example 5.8. Let K Rn be a polytope with | ext K| 2, and let Y0 be anonempty and proper subset of ext K. Consider the join-semilattice congruence on VK defined by specifying its equivalence classes to be

Cy := {y}, y Y0, C0 := VK \ yY0

Cy ,

and the map : {Y(C)|C class of } SubRn defined as

Y(Cy)

:= {0}, y Y0,

Y(C0)

:= dir(ext K) ,

cf. Remark 4.5. Applying Theorem 4.4, we obtain that the relation

:=

(x, x) | x Y0

(x1, x2) Z(C0) Z(C0) | x2 x1 dir(ext K)

is a congruence of K when K is considered as a convex algebra with the usualvector space operations ofRn.

Let y0 Y0. Since y0 is an extremal point, it cannot be an element ofZ(C0). However, it can be approximated by elements from Z(C0): Choose y (ext K) \ Y0, and set x := y + (1 )y0, (0, 1]. Then one can check from

the definitions that (x, y) and lim0(x, y) = (y0, y) , and this showsthat is not closed.

This example applies immediately to the free algebras Fn(CA), n 2, andFn(PCA), n N+, and shows that they contain congruences which are notfinitely generated as subalgebras. For Fn(TCA), choose Y0 symmetric aroundthe zero vector.

6 Conclusion

We fully describe the congruence lattice of a polytope in finite-dimensional eu-clidean space when considered as a convex algebra. This description appliesin particular to the free algebras with a finite number of generators in thisequational class, and hence clarifies the structure of finitely generated alge-bras. In particular, we see that each finitely generated convex algebra is finitelypresented. The proofs are algebraic in their nature and use the geometry ofeuclidean space.

We show that the equational classes of positive- or totally convex algebras(and their respective congruence lattices) are closely related with convex alge-bras. Using this relation, similar structure results for these equational classesfollow.

The presented results have an interpretation in a categorical context, sincethe considered equational classes are the categories of Eilenberg-Moore algebrasassociated with certain monads.

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