Markov Chains on Graded Posets

We consider two types of discrete-time Markov chains where the state space is a graded poset and the transitions are taken along the covering relations in the poset. The first type of Markov chain goes only in one direction, either up or down in the poset (an up chain or down chain). The second type toggles between two adjacent rank levels (an up-and-down chain). We introduce two compatibility concepts between the up-directed transition probabilities (an up rule) and the down-directed (a down rule), and we relate these to compatibility between up-and-down chains. This framework is used to prove a conjecture about a limit shape for a process on Young’s lattice. Finally, we settle the questions whether the reverse of an up chain is a down chain for some down rule and whether there exists an up or down chain at all if the rank function is not bounded.


Introduction
A Hasse walk is a walk along the covering relations in a poset [6,5].In [2], Eriksson and Sjöstrand discussed how several famous models of stochastic processes can be regarded as random Hasse walks on Young's lattice, either walks that go steadily upwards (e.g.Simon's model of urban growth) or that alternately go up and down (e.g. the Moran model in population genetics).The aim of the present paper is to develop a general framework of such unidirected and alternatingly directed random Hasse walks on graded posets.
Let I ⊆ Z be a (possibly infinite) interval of the integers.An I-graded poset Ω is a countable (or finite) poset together with a surjective map ρ : Ω → I, called the rank function, such that • u < v implies ρ(u) < ρ(v), and where u ⋖ v means that v covers u, that is u < v but there is no w with u < w < v.We can partition Ω = i∈I Ω i into its level sets Ω i = ρ −1 (i).
We will now describe two types of stochastic processes on Ω.

Up chains and down chains
An assignment of a probability T (u → v) to any pair (u, v) ∈ Ω × Ω is an up rule if for any u with non-maximal rank.Analogously, it is a down rule if • T (u → v) > 0 ⇔ u ⋗ v, and v∈Ω T (u → v) = 1 for any u with non-minimal rank.Example 1. Figure 1 shows an example of a [0, 2]-graded poset with two sets of probabilities forming an up rule (left) and a down rule (right).Up rules and down rules define classes of Markov chains on Ω.We will consider such Markov chains (X i ) i∈J on Ω for any (possibly infinite) subinterval J ⊆ I. Definition 1.Let U be an up rule and let D be a down rule on an I-graded poset Ω.A Markov chain (X i ∈ Ω i ) i∈J is a U -chain (with time interval J) if Prob(X i = u and for any non-maximal i ∈ J and any u, v ∈ Ω. Analogously, it is a D-chain if Prob(X i = u and for any non-minimal i ∈ J and any u, v ∈ Ω. A U -or D-chain is maximal if its time interval is I, and it is positive if Prob(X i = u) > 0 for any i ∈ J and any u ∈ Ω i .
In this paper we shall examine when an up rule and a down rule are compatible with each other.We shall distinguish between a weaker and a stronger notion of compatibility.Definition 2. An up rule U and a down rule D on Ω are compatible if there is a maximal U -chain (X i ∈ Ω i ) i∈I and a maximal D-chain (Y i ∈ Ω i ) i∈I such that, for any i ∈ I, the random variables X i and Y i are equally distributed.
U and D are strongly compatible if there is a Markov chain that is both a maximal U -chain and a maximal D-chain.
Example 2. Let U and D be the up and down rules depicted in Figure 2. Clearly, U and D are compatible -just assign the probability 1/2 to each element in the poset.But they are not strongly compatible since the probability of going diagonally is 3/4 for the up rule and 1/4 for the down rule.In order to study probability distributions on the poset Ω we introduce some notation.
Let ℓ 1 (Ω) denote the Banach space of real-valued functions f on Ω such that the norm denote the subspace consisting of functions with support on the level set Ω i and let denote the set of probability distributions on Ω i .
An up or down rule T induces a linear operator (which, by abuse of notation, also is called T ) on ℓ 1 (Ω) defined by Stepwise application of this operator defines sequences of probability distributions with support on one level set at a time, as follows.
Definition 3. Let U be an up rule and let D be a down rule on an I-graded poset Ω.A sequence (π i ∈ M (Ω i )) i∈J for some (possibly infinite) interval J ⊆ I is a U -sequence if U π i = π i+1 for any non-maximal i ∈ J, and it is a D-sequence if Dπ i = π i−1 for any non-minimal i ∈ J.
A U -or D-sequence (π i ) i∈J is positive if π i (u) > 0 for any i ∈ J and any u ∈ Ω i .
Clearly, there is a one-to-one correspondence between U -sequences and U -chains and between D-sequences and D-chains.
Observation 1. U and D are compatible if and only if there exists a Usequence that is also a D-sequence.

Up-and-down processes
Next we turn to alternatingly directed Markov chains on Ω.
Given an up rule U and a down rule D, we define a UD-chain as a Markov chain with state space Ω and with transitions induced by U and D alternately.Definition 4. Let U be an up rule and let D be a down rule on an I-graded poset Ω.A UD-chain is a Markov chain (X (0) , X (1) , . . . ) on Ω such that Prob(X (t) = u and Figure 3 depicts the level sets Ω 1 = {u 1 , u 2 } and Ω 2 = {v 1 , v 2 , v 3 } of the poset Ω in Figure 1 with the same up rule U and down rule D. The up rule U and the down rule D define a UD -chain with state space Ω 1 ∪ Ω 2 .

Results
Our first result relates the UD -process to the compatibility of U -and Dchains.We state this as two theorems, one about strong compatibility and one about compatibility.These theorems follow almost immediately from the definitions, but nevertheless they can be powerful tools.In Section 2 we use Theorem 2 to prove a conjecture of Eriksson and Sjöstrand [2] about the limit shape of a Markov process on Young's lattice.

Theorem 1.
Let Ω be an I-graded poset with an up rule U and a down rule D, and let (X i ∈ Ω i ) i∈I be a Markov chain.The following are equivalent.(a) (X i ) i∈I is both a U -chain and a D-chain (and hence U and D are strongly compatible).
(b) For any adjacent levels i, i + 1 ∈ I it holds that X i , X i+1 , X i , X i+1 , . . . is a UD-chain.
Proof.This follows directly from the definitions.
In other words, Theorem 1 says that if and only if random variables can be defined on each level set such that they correspond both to a Markov chain following the up rule and a Markov chain following the down rule, they also correspond to all Markov chains following the up and down rule alternatingly.
Theorem 2. Let Ω be an I-graded poset with an up rule U and a down rule D, and let (π i ∈ M (Ω i )) i∈I be a sequence of probability distributions.The following are equivalent.(b) For any adjacent levels i, i + 1 ∈ I it holds that (π i + π i+1 )/2 is a stationary distribution of the UD-process.
Proof.Note that (π i +π i+1 )/2 is a stationary distribution of the UD -process if and only if U π i = π i+1 and Dπ i+1 = π i .Now the theorem follows directly from the definitions.
In words, Theorem 2 says that if and only if a given sequence of probability distributions is both a U -sequence and a D-sequence, the average (in ℓ 1 (Ω)) of two adjacent such distributions is a stationary distribution of the UD -process.
The second part of this paper will answer the following questions that arise naturally from the notion of compatibility.Q1.Given an up rule U and a U -chain (X i ), is there a down-rule D such that (X i ) is a D-chain?
Q2.Given an up or down rule T , is there a maximal T -chain?
The answer to Q1 is that every positive U -chain is a D-chain for some D. This is Theorem 3 in Section 3 and we give a constructive proof using Bayesian updating of probability distributions between adjacent levels.
The answer to Q2 is more complex and is presented as three theorems in Section 4. If all level sets Ω i are finite, it turns out that there always exists a T -chain, but not necesserily a positive one.The proofs rely on topological arguments and require the axiom of choice.

Applications
We will apply the results to processes on two particular posets: (i) Young's lattice; and (ii) the d-dimensional nonnegative integer lattice.
In [2], Eriksson and Sjöstrand study stochastic processes on Young diagrams of integer partitions, in particular their limit shapes.In the framework of this paper, these processes are up processes and up-and-down processes, respectively, where the underlying poset is Young's lattice.A motivation for this work is a limit shape conjecture in [2] for a certain up-and-down process (described in Section 1.5 below).Using the results in the current paper, in particular Theorem 2, we will prove this conjecture.This is done in Corollary 2 in Section 2.
Applications of our theory to the d-dimensional nonnegative integer lattice are presented in Section 3.1.

An application of Theorem 2 to a process on Young's lattice
One of the processes studied in [2], called derow-row(µ), is a UD -chain on Young's lattice.We will first introduce some notation for integer partitions and Young diagrams and then describe the up rule row(µ) and the down rule derow used in this process.

Notation
An introduction to the theory of integer partitions can be found in [1].
With P n we mean the set of all partitions of the positive integer n.For λ ∈ P n , we write |λ| = n.Denote the parts of the partition by λ An integer partition λ can be represented by a Young diagram drawn as left-aligned rows of squares in the first quadrant such that the ith row from the bottom has length λ i .The diagrams in the first three levels of Young's lattice can be seen in Figure 4.With λ ∈ P n we mean either the partition or its corresponding Young diagram.

Description of the process
To describe the UD -chain derow-row(µ), we need a tool introduced in [2], used to associate squares with row-lengths and row-lengths with corners in a Young diagram.Definition 5. Consider some given Young diagram λ.For any of its squares s let κ(s) denote the length of the row to which s belongs.If κ is a row length, let ω(κ) and ι(κ) denote the unique outer corner and inner corner, respectively, for which the row coordinate is κ: Consider a current Young diagram λ.The action of the down rule derow is defined by choosing a non-empty row i uniformly at random and removing the corresponding outer corner ω(λ i ).
The action of the up rule row(µ) is defined as follows: With probability µ create a new row of length 1. Otherwise make a uniformly random choice of a row i among the N (λ) non-empty rows and insert a new square at the corresponding inner corner ι(λ i ).
The process derow-row(µ) is the up-and-down process on P n (for some n ≥ 2) using these up and down rules.

The stationary distributions
The following result is a refinement of equations ( 12) and (13) in [2] for the stationary distributions of the process derow-row(µ).Lemma 1.For any 0 < µ < 1, the stationary distribution over the partitions in P i ∪ P i+1 in the process row-derow(µ) is Proof.Up to a normalization constant, π UD i (λ) is given in ( 12) and ( 13) in [2], namely, for some constant c i (independent of λ).It follows that c i+1 /(1 − µ) = c i for each i, and since Combining the lemma with Theorem 2 yields a formula for the distributions of the U -chain induced by the up rule row(µ).
Corollary 1.The up rule row(µ) and the down rule derow are compatible and the distribution of the row(µ) process on P i is given by the probability function In fact the derivation of equations ( 12) and ( 13) in [2] reveals that the stationary derow-row(µ) process is reversible, and hence, by Theorem 1, row(µ) and derow are strongly compatible.

Limiting objects
The original motivation for the current work has been the study of limiting objects, in particular limit shapes of Young diagrams under stochastic processes as initiated in [2].As a corollary to Theorem 2, we can now prove a limit shape conjecture in [2] (Conjecture 2).
First, let us remind ourselves what is meant by a scaling of a Young diagram.As the number of squares n grows we need to rescale the diagram to achieve any limiting behaviour.Following [2] and [7], a diagram is rescaled using a scaling factor a n > 0 such that all row lengths are multiplied by 1/a n and all column heights are multiplied by a n /n, yielding a constant diagram area of 1.
Corollary 2. Let µ n log(µ n n) → 0 and µ n n → ∞ as n → ∞ and choose the scaling a n = 1/µ n .Then the stationary distribution of the derow-row(µ n ) process has the limit shape y(x) = e −x .
Proof.By Corollary 1, the stationary distribution for derow-row(µ) equals the distribution for row(µ).By Theorem 4 in [2], the limit shape for this distribution with µ = µ n is y = e −x as long as µ n log(µ n n) → 0 and µ n n → ∞ as n → ∞.Therefore, under these conditions the limit shape for derow-row(µ n ) must also be y = e −x .
2.1 The asymptotics of µ n in derow-row(µ n ) Theorem 4 in [2] has the conditions µ n n → ∞ and µ n log(µ n n) → 0 as n → ∞.After scaling a Young diagram with scaling factor a n = 1/µ n , each square will have width 1/a n = µ n and height a n /n = 1 µnn .In order for the boundary of the Young diagram (which is the object whose limiting behaviour one considers) to resolve properly as n → ∞, we must at least have However, Theorem 4 in [2] uses the stronger assumption µ n log(µ n n) → 0 as n → ∞.It is still to be investigated whether this can be relaxed in Corollary 2.

A generalization of limiting objects
One may also generalize the concepts of limiting object and limit distribution.In order to be able to talk about a generic process having a limiting object and a limit distribution, let S be a separable metric space, called a limit space, with the metric d S .
For n = 0, 1, . . .let R n be the countable set of states reachable after n steps, and let f n : R n → S be a "scaling" function.A stochastic process (X n ∈ R n ) ∞ n=0 having a limit distribution (with respect to the metric d S ) corresponds to the convergence of the random variables {f n (X n )} in distribution to a random variable X ∈ S; the distribution of X is the limit distribution.
Further, if the limit distribution has all probability mass concentrated in a single point in S, in other words, if X is constant, then this point is the limiting object of the process (in which case we may equivalently talk about convergence in probability of {f n (X n )} to the limiting object).
In this paper, we have made this concept tangible by studying processes on Young's lattice, where the limit space is the set of decreasing functions in the first quadrant with integral 1.In the next section, we will study processes on the d-dimensional nonnegative integer lattice N d where the limit space is the set of nonnegative real d-tuples adding to 1.We see a further possible application to this generalization in the study of limits of permutation sequences (see for example [3]).In this case S is the set of Lebesgue measurable functions [0, 1] 2 → [0, 1] with certain properties.
3 Given an up rule, is there a compatible down rule?
In this section we will prove the existence of a down rule compatible with a given up rule.The proof is constructive and we will use the construction in some examples.
Theorem 3. Let U be an up rule on an I-graded poset Ω, and let (X i ) i∈I be a U -chain.Then (X i ) is a D-chain for some down rule D on Ω if and only if Prob(X i = u) = 0 implies Prob(X i+1 = v) = 0 for any i, i + 1 ∈ I and any In particular this holds when (X i ) is positive.
Proof.By Theorem 1, (X i ) is a D-chain if and only if X i , X i+1 , X i , X i+1 , . . . is a UD -chain for any i ∈ I. Since X i , X i+1 , X i , X i+1 , . . . is a reversible Markov chain (see any text book on Markov chains, for instance [4]), it is a UD -chain if and only if Prob( for any u, v ∈ Ω.A D satisfying that equation can be chosen by letting but this is possible only unless Prob(X i = u) = 0 and Prob(X i+1 = v) > 0 for some u ⋖ v or Prob(X i = u) > 0 and Prob(X i+1 = v) = 0 for some u ⋖ v.The latter is impossible since Prob(X i+1 = v) ≥ Prob(X i = u)U (u → v).

Processes on the d-dimensional nonnegative integer lattice
In this section we will demonstrate two applications of Theorem 3 on the d-dimensional nonnegative integer lattice N d .We will construct down rules compatible with given up rules.We shall also see examples of processes both with and without a limiting object.

A limiting object on N d
As an analogy to the concepts of limit shapes for birth-and birth-anddeath processes on Young diagrams we will here define a limiting object for processes on N d .
If we divide the coordinates of a lattice point reached in an up process on N d by the number of steps taken in the process, the result is a point in the Under this scaling we define a limiting object as follows.Definition 6.For an up process on N d , let X n = (X 1 n , . . ., X d n ) be the lattice point after n steps.A point x ∈ ∆ d−1 is a limit point of the up process if for any ε > 0 we have lim n→∞ Prob 1 n X n − x < ε = 1.
For n ∈ N, let p n : N d n → [0, 1] be the probability function on N d n induced by the up rule U .

An up process on N d with a limiting object
Consider the up rule U on N d governed by the constant transition probabilities U (x → x (i) ) = ν i .We may use Theorem 3 to conclude that there is a down rule D compatible with U .Let us construct it!First of all, the probability to reach x = (x 1 , . . ., x d ) (after n = |x| steps in the up process induced by U ) is clearly As a consequence, We use (4) to compute the transition probability Here we see that an up rule employing the degree of freedom parameters ν 1 , . . ., ν d has a compatible down rule with no such degree of freedom present.As we saw in Corollary 1, this is also the case with the up rule in the process row(µ) having a compatible down rule derow, void of the degree of freedom parameter µ.
Proposition 1.The up process on N d using the up rule U (x → x (i) ) = ν i has the limit point (ν 1 , . . ., ν d ).

An up process on N d without a limiting object
We will now consider an up process on N d starting at (0, . . ., 0) induced by an up rule that depends on the current level.
Proposition 2. The process on N d induced by the up rule U governed by the transition probabilities has a uniform distribution on N d n , for all n ≥ 0.
Proof.We prove this by induction over the number of steps n in the process.
First of all we observe that |N d n | = d+n−1 d−1 .For n = 0, we have a trivial uniform distribution.For n ≥ 0, assume that the distribution on N d n under this process is uniform, i.e., p n We want to prove that the distribution over x can be reached from any of the elements x (1) , . . ., x (d) ∈ N d n .By (5), and by the induction hypothesis, p n (x (j) ) = d+n−1 d−1 −1 , so the probability , so the distribution is uniform also on N d n+1 , and the result follows by induction.
Since the distribution is uniform, this process cannot have a limiting object.
As in Section 3.1.4,let us use (4) to construct the down rule compatible with the up rule in Theorem 2. We get the probability 4 Given an up or down rule T , is there a maximal T -chain?
Let T be an up or down rule on an I-graded poset Ω. Recall that there is a one-to-one correspondence between T -chains and T -sequences, so questions about the existence of maximal T -chains are equivalent to questions about the existence of maximal T -sequences.
If I has a minimal element m there obviously exists a maximal Usequence for a given up rule U : Just choose any probability distribution π m ∈ M (Ω m ) on level m, and the up rule U will induce a U -sequence (U i−m π m ) i∈I .However, if I has no lower bound it is not obvious whether there exists a maximal U -sequence, and, by symmetry, if I has no upper bound it is not obvious whether there exists a maximal D-sequence for a given down rule D.Here is an example where no such D-sequence exists.
Example 3. Let Ω be the two-dimensional integer lattice Z 2 with the partial order (x, y) ≤ (x ′ , y ′ ) if x ≤ x ′ and y ≤ y ′ .It is Z-graded by ρ(x, y) = x + y.Let D be the down rule with probability 1/2 for each edge in the Hasse diagram.If we start at any element of high rank n and follow down edges randomly according to the down rule, the chance of hitting any particular element of rank zero is very small; it tends to zero as n grows.Thus, in a maximal D-sequence any rank-zero element must be given the probability zero, which is impossible.
The phenomenon in the above example cannot happen if the level sets Ω i are finite, and, as we will see in Theorem 6, in this case there is always a maximal T -sequence.As the following example reveals, however, the existence of a positive maximal T -sequence is not guaranteed.
Example 4. Look at the two-dimensional nonnegative integer lattice N 2 with the down rule given by the probability 1/2 at every edge in the Hasse diagram, except for the leftmost and rightmost edges which must have probability one; see Figure 5.If we start at any element of high rank n and follow down edges randomly according to the down rule, with very high probability we will walk into the left or right border before we reach level 2. Thus, the probability of reaching the middle element (1, 1) at level 2 tends to zero as n grows.This means that every D-chain (X i ) must have Prob(X 2 = (1, 1)) = 0.
We will present three theorems about the existence of T -sequences and positive T -sequences.Their proofs all depend on the following lemma.Proof.Without loss of generality, we may assume that T = U is an up rule.
Let Π be the product space Π = i∈I C i .For any non-maximal Let us first show that the set Π k is closed.Clearly, Π k is homeomorphic to the product of i∈I\{k,k+1} C i and the graph G ⊆ C k × C k+1 of the restriction of U to C k ∩ U −1 C k+1 .The map U has operator norm 1, so it is continuous.Hence, the preimage U −1 C k+1 is closed and so is C k ∩ U −1 C k+1 .It follows that the graph G is closed and hence Π k is closed, being homeomorphic to a product of closed sets.By Tychonoff's theorem Π is compact, and therefore the closed subsets Π k are compact, and so is the intersection Λ of all Π k .Clearly, Λ is precisely the set of maximal U -sequences, so our task is to show that Λ is nonempty.To that end, suppose it is empty and define , let π ′ i be an arbitrary element in C i .The sequence (π ′ i ) i∈I so obtained is a point outside the union k∈F U k , which contradicts the fact that {U k } k∈F covers Π.Hence, our supposition that Λ is empty is false.
Our first existence theorem states that if there exists a T -sequence for any finite subinterval of I, and these sequences are uniformly bounded in a certain sense, then there exists a maximal T -sequence.
Theorem 4. Let T be an up or down rule on an I-graded poset Ω. Suppose there is a sequence ( bi ∈ ℓ 1 (Ω i )) i∈I such that for any m ≤ n there exists a T -sequence (π i ) i∈[m,n] with π i ≤ bi for any i ∈ [m, n].Then there exists a maximal T -sequence.
Proof.For any i ∈ I, let C i = {π ∈ M (Ω i ) : π ≤ bi }.If we can show that C i is compact, the theorem will follow from Lemma 2. We can write Since M (Ω i ) is closed it suffices to show that L i is compact.By the dominated convergence theorem L i is homeomorphic to the product space u∈Ω i [0, bi (u)] which is compact by Tychonoff's theorem.
For the existence of a positive maximal T -sequence, it is not enough to assume the existence of finite T -sequences that are uniformly bounded from above; they must be uniformly bounded from above and from below simultaneously!Let ℓ + 1 (Ω i ) = {f ∈ ℓ 1 (Ω i ) : f (u) > 0 for any u ∈ Ω i } denote the set of strictly positive ℓ 1 -functions on Ω i .Theorem 5. Let T be an up or down rule on an I-graded poset Ω.
Suppose there are sequences ( bi ∈ ℓ + 1 (Ω i )) i∈I and ( bi ∈ ℓ + 1 (Ω i )) i∈I such that for any m ≤ n in I there exists a T -sequence (π i ) i∈[m,n] with bi ≤ π i ≤ bi for any i ∈ [m, n].Then there exists a positive maximal T -sequence.
Proof.The proof is completely analogous to that of Theorem 4, but with In most combinatorial applications, the level sets Ω i are finite.In that case, the uniform upper bound ( bi ) in the assumption in Theorem 4 automatically exists.Furthermore, the requirement of uniformicity of the lower bound ( bi ) in Theorem 5 can be relaxed.
The following theorem is stated for an up rule, but the dual statement for a down rule is of course equivalent.Theorem 6.Let U be an up rule on an I-graded poset Ω with finite level sets Ω i .Then there exists a maximal U -sequence, and there exists a positive maximal U -sequence if and only if there is a sequence (b i : Ω i → (0, ∞)) i∈I with the property that for any m ≤ n in I there is a distribution π m,n ∈ M (Ω m ) such that U n−m π m,n ≥ b n .
Proof.Define a sequence ( bi ∈ ℓ 1 (Ω i )) i∈I by letting bi (u) = 1 for any u ∈ Ω i .Now Theorem 4 guarantees the existence of a maximal U -sequence.
Next, suppose there is a sequence (b i : Ω i → (0, ∞)) i∈I and distributions (π m,n ∈ M (Ω m )) m≤n with the property given in the theorem.Let (γ i ) i∈I be positive numbers adding to one, and for each i ∈ I put bi = γ i b i .Now, fix m ≤ n in I. Let Γ = i∈[m,n] γ i and define Since U is a linear operator, for any j ∈ [m, n] we have Theorem 6 guarantees the existence of a maximal T -sequence if all level sets are finite, but in general it is not possible to extend a T -sequence for a subinterval J ⊂ I to a maximal T -sequence.For example, consider the poset in Figure 6 with a defined down rule D. The distributions π 0 ∈ M (Ω 0 ) and π 1 ∈ M (Ω 1 ) given by π 0 ( 0) = 1, π 1 (s 1 ) = 1/4 and π 2 (s 2 ) = 3/4 constitute a D-chain, but this D-chain can obviously not be extended to the top level.

General homogeneous Markov chains
Finally, we shall use Theorem 4 to prove a more general result about the existence of Markov chains with time interval Z.As far as we know, this has not been treated before.
Let S be a countable (or finite) state space and let P be a transition function on S, that is, an assignment of a probability P (s → s ′ ) to each pair (s, s ′ ) ∈ S × S such that s ′ ∈S P (s → s ′ ) = 1 for any s ∈ S. For any initial (random) state X 0 , there is a unique Markov chain X 0 , X 1 , . . .with transition function P , but is there a Markov chain . . ., X −1 , X 0 , X 1 , . . .with time interval Z? The following "homogeneous" version of Theorem 4 answers that question.

1 Figure 1 :
Figure 1: An up rule (left) and a down rule (right) on some Ω.

Figure 2 :
Figure 2: The up rule (bent arrows) and the down rule (straight arrows) are compatible but not strongly compatible.
(a) (π i ) i∈I is both a U -sequence and a D-sequence (and hence U and D are compatible).

3. 1 . 1
The lattice N d For a positive integer d, let (N d , ≤) be the poset of nonnegative integer dtuples (x 1 , . . ., x d ) ordered component-wise, i.e. for x = (x 1 , . . ., x d ), y = (y 1 . . ., y d ) ∈ N d , we have x ≤ y if x i ≤ y i for all i = 1, . . ., d.For x, y ∈ N d , we have x ∨ y = (max(x 1 , y 1 ), . . ., max(x d , y d )) and x ∧ y = (min(x 1 , y 1 ), . . ., min(x d , y d )), so N d is a lattice, which we will refer to as the d-dimensional nonnegative integer lattice.Obviously, N d is graded with the rank function |x| = d j=1 x j and for n ≥ 0, the level set N d n = {x ∈ N d : |x| = n} is the set of weak compositions of n.

1 Figure 5 :
Figure 5: The down rule on N 2 in example 4.
so the family of all sets U k is an open cover of Π.Since Π is compact there is an open subcover {U k } k∈F , where F is a finite set of non-maximal elements in I. Choose m and n in I so that m ≤ k < n for any k ∈ F .Now, by the assumption in the theorem there is a U j−m π) j∈[m,n] is a T -sequence such that U j−m π ≥ bj for any j ∈ [m, n].Theorem 5 now yields the existence of a positive maximal Usequence.