Generalized crested products of Markov chains

Daniele D’Angeli Daniele D’Angeli. Department of Mathematics, Technion–Israel Institute of Technology, Technion City, Haifa 32 000. dangeli@tx.technion.ac.il and Alfredo Donno Alfredo Donno. Dipartimento di Matematica “G. Castelnuovo”, Sapienza Università di Roma, Piazzale A. Moro, 2, 00185 Roma, Italia. donno@mat.uniroma1.it

July 9, 2019

Abstract.

We define a finite Markov chain, called generalized crested product, which naturally appears as a generalization of the first crested product of Markov chains. A complete spectral analysis is developed and the $k$ -step transition probability is given. It is important to remark that this Markov chain describes a more general version of the classical Ehrenfest diffusion model.
As a particular case, one gets a generalization of the classical Insect Markov chain defined on the ultrametric space. Finally, an interpretation in terms of representation group theory is given, by showing the correspondence between the spectral decomposition of the generalized crested product and the Gelfand pairs associated with the generalized wreath product of permutation groups.

Key words and phrases:

Reversible Markov chain, generalized crested product, Insect Markov chain, spectral theory, Gelfand pairs.

Mathematics Subject Classification (2010): 60J10, 05C50, 06A07, 20B25.

1. Introduction

This paper deals with the study of a finite Markov chain, called generalized crested product, which is defined on the product of finite spaces. The generalized crested product is a generalization of the Markov chains introduced and studied in [6] and [7]. More precisely, in [6] the first crested product of Markov chains is defined, inspired by the analogous definition for association schemes [2], as a sort of mixing between the classical crossed and nested products of Markov chains and it contains, as a particular case, the so called Insect Markov chain introduced by Figà-Talamanca in [10] in the context of Gelfand pairs theory. In [7] a Markov chain on some special structures called orthogonal block structures is introduced. If the orthogonal block structure is a poset block structure, then that Markov chain can also be defined starting from a finite poset $(I, \leq)$ and it can be interpreted as a slightly different generalization of the classical Insect Markov chain and of the associated Gelfand pairs theory. We noticed that, for some particular posets, the Markov chain in [7] has the same spectral decomposition as the first crested product of Markov chains despite the corresponding operators do not coincide. So it is natural to ask if it is possible to define a Markov chain containing the first crested product as a particular case, giving rise to an Insect Markov chain for a particular choice of the parameters involved. This is the reason why this paper aims at introducing a new Markov chain which can be seen as a modification of the Markov chain on the orthogonal block structure and the natural generalization of the first crested product of Markov chains. The idea is to take a finite poset $(I, \leq)$ and a family of Markov operators $P_{i}$ defined on finite sets $X_{i}$ indexed by the elements of the poset. Then we consider the sum, over $I$ , of tensor products of Markov chains reflecting, in some sense, the poset hierarchy structure (see Definition 3.2). A necessary and sufficient condition to have reversibility of the Markov chain is proven in Theorem 3.4. In Theorem 3.5, we give a complete spectral analysis of this Markov chain and we show in Proposition 3.8 that it coincides with the first crested Markov chain when the poset $(I, \leq)$ satisfies some particular properties. A formula for the $k$ -step probability is given in Section 3.4. Moreover, we introduce in Section 4 an Insect Markov chain on the product $X = \prod_{i \in I} X_{i}$ , naturally identified with the last level of a graph $T$ which is the generalization of the rooted tree. This Insect Markov chain is obtained from the generalized crested product of Markov chains for a particular choice of the operators $P_{i}$ , i.e. $P_{i} = J_{i}$ , where $J_{i}$ is the uniform operator on the set $X_{i}$ . If the poset $(I, \leq)$ is totally ordered, this Insect Markov chain coincides with the classical Insect Markov chain [10]. In Section 5 we highlight the correspondence with the Gelfand pairs theory (for a general theory and applications see [8]): taking the generalized wreath product of permutation groups [3] associated with $(I, \leq)$ and the stabilizer of an element of $X$ under the action of this group, one gets a Gelfand pair [7], and the decomposition of the action of the group on the space $L (X)$ into irreducible submodules is the same as the spectral decomposition of the Insect Markov chain associated with $(I, \leq)$ . This allows to study many examples of Gelfand pairs only by using basic tools of linear algebra.
It is important to remark that the generalized crested product can be seen as a generalization of a classical diffusion model, the Ehrenfest model, as well as of the $(C, N)$ -Ehrenfest model described in [6] (see [9] and [11] for more examples and details).

2. Preliminaries

We recall in this section some basic facts about finite Markov chains (see, for instance, [4]). Let $X$ be a finite set, with $| X | = m$ . Let $P = (p (x, y))_{x, y \in X}$ be a stochastic matrix, so that

\sum x \in X p (x_{0}, x) = 1,

for every $x_{0} \in X$ . Consider the Markov chain on $X$ with transition matrix $P$ . By abuse of notation, we will denote by $P$ this Markov chain as well as the associated Markov operator.

Definition 2.1.

The Markov chain $P$ is reversible if there exists a strict probability measure $π$ on $X$ such that

π (x) p (x, y) = π (y) p (y, x),

for all $x, y \in X$ .

If this is the case, we say that $P$ and $π$ are in detailed balance [1].

Define a scalar product on $L (X) = {f : X ⟶ C}$ as

⟨ f_{1}, f_{2} ⟩_{π} = \sum x \in X f_{1} (x) ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ f_{2} (x) π (x),

for all $f_{1}, f_{2} \in L (X)$ , and the linear operator $P : L (X) ⟶ L (X)$ as

(1)

(P f) (x) = \sum y \in X p (x, y) f (y) .

It is easy to verify that $π$ and $P$ are in detailed balance if and only if $P$ is self-adjoint with respect to the scalar product $⟨ \cdot, \cdot ⟩_{π}$ . Under these hypotheses, it is known that the matrix $P$ can be diagonalized over the reals. Moreover, $1$ is always an eigenvalue of $P$ and for any eigenvalue $λ$ one has $| λ | \leq 1$ .
Let ${λ_{z}}_{z \in X}$ be the eigenvalues of the matrix $P$ , with $λ_{z_{0}} = 1$ . Then there exists an invertible unitary real matrix $U = (u (x, y))_{x, y \in X}$ such that $P U = U Δ$ , where $Δ = (λ_{x} δ_{x} (y))_{x, y \in X}$ is the diagonal matrix whose entries are the eigenvalues of $P$ . This equation gives, for all $x, z \in X$ ,

(2)

\sum y \in X p (x, y) u (y, z) = u (x, z) λ_{z} .

Moreover, we have $U^{T} D U = I$ , where $D = (π (x) δ_{x} (y))_{x, y \in X}$ is the diagonal matrix of coefficients of $π$ . This second equation gives, for all $y, z \in X$ ,

(3)

\sum x \in X u (x, y) u (x, z) π (x) = δ_{y} (z) .

It follows from (2) that each column of $U$ is an eigenvector of $P$ , and from (3) that these columns are orthogonal with respect to the product $⟨ \cdot, \cdot ⟩_{π}$ .

Proposition 2.2.

The $k$ -th step transition probability is given by

(4)

p^{(k)} (x, y) = π (y) \sum z \in X u (x, z) λ_{z}^{k} u (y, z),

for all $x, y \in X$ .

Proof.

The proof is a consequence of (2) and (3). In fact, the matrix $U^{T} D$ is the inverse of $U$ , so that $U U^{T} D = I$ . In formulæ, we have

\sum y \in X u (x, y) u (z, y) = \frac{1}{π (z)} Δ_{z} (x) .

From the equation $P U = U Δ$ we get $P = U Δ U^{T} D$ , which gives

p (x, y) = π (y) \sum z \in X u (x, z) λ_{z} u (y, z) .

Iterating this argument we obtain $P^{k} = U Δ^{k} U^{T} D$ , which is the assertion. ∎

Recall that there exists a correspondence between reversible Markov chains and weighted graphs.

Definition 2.3.

A weight on a graph $G = (X, E)$ is a function $w : X \times X ⟶ [0, + \infty)$ such that

$w (x, y) = w (y, x)$ ;
$w (x, y) > 0$ if and only if $x \sim y$ .

If $G$ is a weighted graph, it is possible to associate with $w$ a stochastic matrix $P = (P (x, y))_{x, y \in X}$ on $X$ by setting

p (x, y) = \frac{w (x, y)}{W (x)},

with $W (x) = \sum_{z \in X} w (x, z)$ . The corresponding Markov chain is called the random walk on $G$ . It is easy to prove that the matrix $P$ is in detailed balance with the distribution $π$ defined, for every $x \in X$ , as

π (x) = \frac{W (x)}{W},

with $W = \sum_{z \in X} W (z)$ . Moreover, $π$ is strictly positive if $X$ does not contain isolated vertices. The inverse construction can be done. So, if we have a transition matrix $P$ on $X$ which is in detailed balance with the probability $π$ , then we can define a weight $w$ as $w (x, y) = π (x) p (x, y)$ . This definition guarantees the symmetry of $w$ and, by setting $E = {{x, y} : w (x, y) > 0}$ , we get a weighted graph.
There are some important relations between the weighted graph associated with a transition matrix $P$ and its spectrum $σ (P)$ . In fact, it is easy to prove that the multiplicity of the eigenvalue 1 of $P$ equals the number of connected components of $G$ . Moreover, the following propositions hold.

Proposition 2.4.

Let $G = (X, E, w)$ be a finite connected weighted graph and denote $P$ the corresponding transition matrix. Then the following are equivalent:

$G$ is bipartite;
the spectrum $σ (P)$ is symmetric;
$- 1 \in σ (P)$ .

Definition 2.5.

Let $P$ be a stochastic matrix. $P$ is ergodic if there exists $n_{0} \in N$ such that

p^{(n_{0})} (x, y) > 0, for all x, y \in X .

Proposition 2.6.

Let $G = (X, E)$ be a finite graph. Then the following conditions are equivalent:

$G$ is connected and not bipartite;
for every weight function on $G$ , the associated transition matrix $P$ is ergodic.

So we can conclude that a reversible transition matrix $P$ is ergodic if and only if the eigenvalue 1 has multiplicity one and $- 1$ is not an eigenvalue. Note that the condition that 1 is an eigenvalue of $P$ of multiplicity one is equivalent to require that the probability $P$ is irreducible, according with the following definition.

Definition 2.7.

A stochastic matrix $P$ on a set $X$ is irreducible if, for every $x_{1}, x_{2} \in X$ , there exists $n = n (x_{1}, x_{2})$ such that $p^{(n)} (x_{1}, x_{2}) > 0$ .

3. Generalized crested product

3.1. Definition

Let $(I, \leq)$ be a finite poset, with $I = {1, \dots, n}$ . The following definitions are given in [3].

Definition 3.1.

A subset $J \subseteq I$ is said

ancestral if, whenever $i > j$ and $j \in J$ , then $i \in J$ ;
hereditary if, whenever $i < j$ and $j \in J$ , then $i \in J$ ;
a chain if, whenever $i, j \in J$ , then either $i \leq j$ or $j \leq i$ ;
an antichain if, whenever $i, j \in J$ and $i \neq j$ , then neither $i \leq j$ nor $j \leq i$ .

Given an element $i \in I$ , we set $A (i) = {j \in I : j > i}$ to be the ancestral set of $i$ and $A [i] = A (i) ⊔ {i}$ . Analogously we set $H (i) = {j \in I : j < i}$ to be the hereditary set of $i$ and $H [i] = H (i) ⊔ {i}$ . For a subset $J \subseteq I$ we put $A (J) = ⋃_{j \in J} A (j)$ , $A [J] = ⋃_{j \in J} A [j]$ , $H (J) = ⋃_{j \in J} H (j)$ and $H [J] = ⋃_{j \in J} H [j]$ . Moreover, we denote by $S$ the set of the antichains of $(I, \leq)$ and we set $¯ ¯¯ ¯ S = {j \in I : A (j) = \emptyset}$ . It is clear that $¯ ¯¯ ¯ S$ and the empty set $\emptyset$ belong to $S$ . Note that $A (\emptyset) = H (\emptyset) = A [\emptyset] = H [\emptyset] = \emptyset$ .

For each $i \in I$ , let $X_{i}$ be a finite set, with $| X_{i} | = m_{i}$ , so that we can identify $X_{i}$ with the set ${0, 1, \dots, m_{i} - 1}$ . Moreover, let $P_{i}$ be an irreducible Markov chain on $X_{i}$ and let $p_{i}$ be the corresponding transition probability. We also denote by $P_{i}$ the associated Markov operator $P_{i} : L (X_{i}) ⟶ L (X_{i})$ defined as in (1). Let $I_{i}$ be the identity matrix of size $m_{i}$ and set:

J_{i} = \frac{1}{m_{i}} ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 1 & 1 & \dots & 1 1 & ⋱ & ⋮ ⋮ & ⋱ & ⋮ 1 & \dots & \dots & 1 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ .

We also denote by $I_{i}$ and $J_{i}$ the associated Markov operator on $L (X_{i})$ , that we call the identity and the uniform operator, respectively. We are going to define a Markov chain on the set $X = X_{1} \times \dots \times X_{n}$ .

Definition 3.2.

Let $(I, \leq)$ be a finite poset and let ${p_{i}^{0}}_{i \in I}$ be a probability distribution on $I$ , i.e. $p_{i}^{0} > 0$ for every $i \in I$ and $\sum_{i = 1}^{n} p_{i}^{0} = 1$ . The generalized crested product of the Markov chains $P_{i}$ defined by $(I, \leq)$ and ${p_{i}^{0}}_{i \in I}$ is the Markov chain on $X$ whose associated Markov operator is

(5)

P = \sum i \in I p_{i}^{0} ⎛ ⎝ P_{i} \otimes ⎛ ⎝ ⨂ j \in H (i) U_{j} ⎞ ⎠ \otimes ⎛ ⎝ ⨂ j \in I ∖ H [i] I_{j} ⎞ ⎠ ⎞ ⎠ .

Remark 3.3.

The generalized crested product can be seen as a generalization of the classical diffusion Ehrenfest model. This classical model consists of two urns numbered 0, 1 and $n$ balls numbered $1, \dots, n$ . A configuration is given by a placement of the balls into the urns. Note that there is no ordering inside the urns. At each step, a ball is randomly chosen (with probability $1 / n$ ) and it is moved to the other urn. In [6] we generalized it to the $(C, N)$ -Ehrenfest model. Now put $| X_{i} | = m$ , for each $i = 1, \dots, n$ : then we have the following interpretation of the generalized crested product. Suppose that we have $n$ balls numbered by $1, \dots, n$ and $m$ urns. Let $(I, \leq)$ be a finite poset with $n$ elements. At each step, we choose a ball $i$ according with a probability distribution $p_{i}^{0}$ : then we move it to another urn following a transition probability $P_{i}$ and all the other balls numbered by indices $j$ such that $j \leq i$ in the poset $(I, \leq)$ are moved uniformly to a new urn. The balls corresponding to all the other indices are not moved.

From now on, we suppose that each $P_{i}$ is in detailed balance with the probability measure $σ_{i}$ .

Theorem 3.4.

The generalized crested product is reversible if and only if $P_{k}$ is symmetric for every $k \in I ∖ ¯ ¯¯ ¯ S$ , i.e. $p_{k} (x_{k}, y_{k}) = p_{k} (y_{k}, x_{k})$ , for all $x_{k}, y_{k} \in X_{k}$ . If this is the case, $P$ is in detailed balance with the strict probability measure $π$ on $X$ given by

π (x_{1}, \dots, x_{n}) = \frac{\prod_{i \in ¯ ¯ ¯ S} σ_{i} (x_{i})}{\prod_{i \in I ∖ ¯ ¯ ¯ S} m_{i}} .

Proof.

We start by proving that the condition $σ_{k} = \frac{1}{m_{k}}$ , for each $k \in I ∖ ¯ ¯¯ ¯ S$ , is sufficient. Consider two elements $x = (x_{1}, \dots, x_{n})$ and $y = (y_{1}, \dots, y_{n})$ in $X$ . First, suppose that there exists $j \in ¯ ¯¯ ¯ S$ such that $x_{j} \neq y_{j}$ and $x_{i} = y_{i}$ for all $i \in ¯ ¯¯ ¯ S ∖ {j}$ . In this case, we have

$π (x) P (x, y)$	$=$	$\frac{\prod_{i \in ¯ ¯ ¯ S} σ_{i} (x_{i})}{\prod_{i \in I ∖ ¯ ¯ ¯ S} m_{i}} \cdot p_{j}^{0} \cdot \frac{p_{j} (x_{j}, y_{j})}{\prod_{i \in H (j)} m_{i}} \cdot \prod i \in I ∖ H [j] δ_{i} (x_{i}, y_{i})$
	$=$	$σ_{j} (x_{j}) p_{j} (x_{j}, y_{j}) \cdot p_{j}^{0} \cdot \frac{\prod_{i \in ¯ ¯ ¯ S ∖ {j}} σ_{i} (x_{i})}{\prod_{i \in I ∖ ¯ ¯ ¯ S} m_{i}} \cdot \frac{\prod_{i \in I ∖ H [j]} δ_{i} (x_{i}, y_{i})}{\prod_{i \in H (j)} m_{i}}$
	$=$	$σ_{j} (y_{j}) p_{j} (y_{j}, x_{j}) \cdot p_{j}^{0} \cdot \frac{\prod_{i \in ¯ ¯ ¯ S ∖ {j}} σ_{i} (y_{i})}{\prod_{i \in I ∖ ¯ ¯ ¯ S} m_{i}} \cdot \frac{\prod_{i \in I ∖ H [j]} δ_{i} (y_{i}, x_{i})}{\prod_{i \in H (j)} m_{i}}$
	$=$	$\frac{\prod_{i \in ¯ ¯ ¯ S} σ_{i} (y_{i})}{\prod_{i \in I ∖ ¯ ¯ ¯ S} m_{i}} \cdot p_{j}^{0} \cdot \frac{p_{j} (y_{j}, x_{j})}{\prod_{i \in H (j)} m_{i}} \cdot \prod i \in I ∖ H [j] δ_{i} (y_{i}, x_{i})$
	$=$	$π (y) P (y, x) .$

If we suppose that there exist $j_{1}, j_{2} \in ¯ ¯¯ ¯ S$ such that $x_{j_{h}} \neq y_{j_{h}}$ , for $h = 1, 2$ , then $P (x, y) = P (y, x) = 0$ and there is nothing to prove.

Suppose now that $x_{i} = y_{i}$ for every $i \in ¯ ¯¯ ¯ S$ and there is $j \in I ∖ ¯ ¯¯ ¯ S$ such that $x_{j} \neq y_{j}$ . We have

$π (x) P (x, y)$	$=$	$\frac{\prod_{i \in ¯ ¯ ¯ S} σ_{i} (x_{i})}{\prod_{i \in I ∖ ¯ ¯ ¯ S} m_{i}} \cdot ⎛ ⎝ \sum i \in A (j) p_{i}^{0} \cdot \frac{p_{i} (x_{i}, y_{i}) \prod_{h \in I ∖ H [i]} δ_{h} (x_{h}, y_{h})}{\prod_{h \in H (i)} m_{h}}$
	$+$	$p_{j}^{0} \cdot \frac{p_{j} (x_{j}, y_{j}) \prod_{h \in I ∖ H [j]} δ_{h} (x_{h}, y_{h})}{\prod_{h \in H (j)} m_{h}})$
	$=$	$\frac{\prod_{i \in ¯ ¯ ¯ S} σ_{i} (y_{i})}{\prod_{i \in I ∖ ¯ ¯ ¯ S} m_{i}} \cdot ⎛ ⎝ \sum i \in A (j) p_{i}^{0} \cdot \frac{p_{i} (y_{i}, x_{i}) \prod_{h \in I ∖ H [i]} δ_{h} (y_{h}, x_{h})}{\prod_{h \in H (i)} m_{h}}$
	$+$	$p_{j}^{0} \cdot \frac{p_{j} (y_{j}, x_{j}) \prod_{h \in I ∖ H [j]} δ_{h} (y_{h}, x_{h})}{\prod_{h \in H (j)} m_{h}})$
	$=$	$π (y) P (y, x) .$

On the other hand, we show that the condition $σ_{k} = \frac{1}{m_{k}}$ , for each $k \in I ∖ ¯ ¯¯ ¯ S$ , is necessary. Suppose that the equality $π (x) P (x, y) = π (y) P (y, x)$ holds for all $x, y \in X$ . Let $i \in ¯ ¯¯ ¯ S$ : by irreducibility, we can choose $x, y \in X$ such that $x_{i} \neq y_{i}$ and $p_{i} (x_{i}, y_{i}) \neq 0$ . Let $x_{j} = y_{j}$ for every $j \in ¯ ¯¯ ¯ S ∖ {i}$ . We have

π (x) P (x, y) = π (y) P (y, x) ⟺ π (x) p_{i} (x_{i}, y_{i}) = π (y) p_{i} (y_{i}, x_{i}) .

This gives

(6)

\frac{π (x)}{π (y)} = \frac{p_{i} (y_{i}, x_{i})}{p_{i} (x_{i}, y_{i})} = \frac{σ_{i} (x_{i})}{σ_{i} (y_{i})} .

Let $¯ ¯ ¯ x \in X$ such that ${¯ ¯ ¯ x}_{j} = y_{j}$ for each $j \in H (i)$ and ${¯ ¯ ¯ x}_{j} = x_{j}$ for each $j \in I ∖ H (i)$ . Proceeding as above we get

(7)

\frac{π (¯ ¯ ¯ x)}{π (y)} = \frac{p_{i} (y_{i}, x_{i})}{p_{i} (x_{i}, y_{i})} = \frac{σ_{i} (x_{i})}{σ_{i} (y_{i})},

so that (6) and (7) imply $π (x) = π (¯ ¯ ¯ x)$ , i.e. $π$ does not depend on the coordinates corresponding to indices in $H (i)$ . Let $j \in H (i)$ and let $x^{'} \in X$ such that $x_{h}^{'} \neq x_{h}$ for each $h \in H [j]$ and $x_{k}^{'} = x_{k}$ for each $k \in I ∖ H [j]$ . The condition $π (x) P (x, x^{'}) = π (x^{'}) P (x^{'}, x)$ reduces to

(8)

P (x, x^{'}) = P (x^{'}, x),

since $x$ and $x^{'}$ differ only for indices in $H (i)$ and so $π (x) = π (x^{'})$ . Observe that $x_{ℓ} = x_{ℓ}^{'}$ for each $ℓ \in I ∖ H [j]$ and so the summands corresponding to these indices are equal in both sides of (8). Moreover, for each $k \in H (j)$ , one has $j \in I ∖ H [k]$ and so the summands corresponding to these indices are 0 in both sides of (8), since $x_{j}^{'} \neq x_{j}$ . Hence, (8) reduces to $p_{j} (x_{j}, x_{j}^{'}) = p_{j} (x_{j}^{'}, x_{j})$ , what implies $σ_{j} (x_{j}) = σ_{j} (x_{j}^{'})$ and so the hypothesis of irreducibility guarantees that $σ_{j}$ is uniform on $X_{j}$ . This completes the proof. ∎

3.2. Spectral analysis

The next step is to study the spectral decomposition of the operator $P$ . Suppose that $L (X_{i}) = ⨁_{j_{i} = 0}^{r_{i}} V_{j_{i}}^{i}$ is the decomposition of $L (X_{i})$ into eigenspaces of $P_{i}$ and that $λ_{j_{i}}$ is the eigenvalue corresponding to $V_{j_{i}}^{i}$ . The eigenspace $V_{0}^{i}$ is the space of the constant functions over $X_{i}$ : under our hypothesis of irreducibility, we have $dim (V_{0}^{i}) = 1$ .
For every antichain $S = {i_{1}, \dots, i_{k}} \in S$ , define

J_{S} := {j - = (j_{i_{1}}, \dots, j_{i_{k}}) : j_{i_{h}} \in {1, \dots, r_{i_{h}}}}

and, for $S \in S$ and $j - \in J_{S}$ , we put

V_{S, j -} := V_{j_{i_{1}}}^{i_{1}} \otimes \dots \otimes V_{j_{i_{k}}}^{i_{k}} .

Moreover, we set

(9)

W_{S, j -} := V_{S, j -} \otimes ⎛ ⎝ ⨂ i \in A (S) L (X_{i}) ⎞ ⎠ \otimes ⎛ ⎝ ⨂ i \in I ∖ A [S] V_{0}^{i} ⎞ ⎠

and

(10)

λ_{S, j -} = k \sum h = 1 p_{i_{h}}^{0} λ_{j_{i_{h}}} + \sum i \in I ∖ A [S] p_{i}^{0} .

Theorem 3.5.

Let $(I, \leq)$ be a finite poset and let $P$ be the generalized crested product defined in (5). The decomposition of $L (X)$ into eigenspaces for $P$ is

L (X) = ⨁ S \in S ⎛ ⎝ ⨁ j - \in J_{S} W_{S, j -} ⎞ ⎠ .

Moreover, the eigenvalue associated with $W_{S, j -}$ is $λ_{S, j -}$ .

Proof.

Let $φ$ be a function in $W_{S, j -}$ . We can represent $φ$ as the tensor product $φ_{1} \otimes \dots \otimes φ_{n}$ , with $φ_{i} \in V_{j_{i}}^{i}$ if $i \in S$ , $φ_{i} \in L (X_{i})$ if $i \in A (S)$ , and $φ_{i} \in V_{0}^{i}$ if $i \in I ∖ A [S]$ . We have to show that $(P φ) (x) = λ_{S, j -} φ (x)$ for every $x = (x_{1}, \dots, x_{n}) \in X$ . One has:

	$(P φ) (x)$	$=$	$\sum (y_{1}, \dots, y_{n}) \in X \sum i \in I p_{i}^{0} ⎛ ⎝ p_{i} (x_{i}, y_{i}) φ_{i} (y_{i}) \frac{\prod_{j \in I ∖ H [i]} δ_{j} (x_{j}, y_{j})}{\prod_{j \in H (i)} m_{j}} \prod j \neq i φ_{j} (y_{j}) ⎞ ⎠$
		$=$	$\sum i \in I p_{i}^{0} \sum y_{j} : j \in H [i] \frac{p_{i} (x_{i}, y_{i}) φ_{i} (y_{i})}{\prod_{j \in H (i)} m_{j}} \prod j \in I ∖ H [i] φ_{j} (x_{j}) \prod j \in H (i) φ_{j} (y_{j}) .$

Observe that, if $i \in S$ , then $H (i) \subseteq I ∖ A [S]$ and so $φ_{j}$ is constant for every $j \in H (i)$ . Suppose $i = i_{h}$ , for some $h \in {1, \dots, k}$ . Hence, the term of $P$ corresponding to $i_{h}$ is

p_{i_{h}}^{0} \prod j \in H (i_{h}) m_{j} \cdot \sum y_{i_{h}} \frac{p_{i_{h}} (x_{i_{h}}, y_{i_{h}}) φ_{i_{h}} (y_{i_{h}})}{\prod_{j \in H (i_{h})} m_{j}} \prod j \neq i_{h} φ_{j} (x_{j}) = (p_{i_{h}}^{0} λ_{j_{i_{h}}}) φ (x) .

On the other hand, if $i \in I ∖ A [S]$ , then $S \subseteq I ∖ H [i]$ and so the identity operator $I_{j}$ , for $j \in S$ , acts on the space orthogonal to $V_{0}^{j}$ and $P_{i}$ acts on $V_{0}^{i}$ . Note that $H (i) \subseteq I ∖ A [S]$ , so that the term corresponding to the index $i$ is

			$p_{i}^{0} \prod j \in H (i) m_{j} \cdot \sum y_{i} \frac{p_{i} (x_{i}, y_{i}) φ_{i} (y_{i})}{\prod_{j \in H (i)} m_{j}} \prod j \neq i φ_{j} (x_{j})$
		$=$	$p_{i}^{0} \prod j \in H (i) m_{j} \cdot \frac{φ_{i} (y_{i})}{\prod_{j \in H (i)} m_{j}} \prod j \neq i φ_{j} (x_{j}) = p_{i}^{0} φ (x) .$

Finally, if $i \in A (S)$ , then there exists $k \in S$ such that $k \in H (i)$ . In particular, $φ_{k}$ is orthogonal to $V_{0}^{k}$ , i.e. $\sum_{y_{k} \in X_{k}} φ_{k} (y_{k}) = 0$ and so the term corresponding to the index $i$ is

			$p_{i}^{0} \sum y_{j} : j \in H [i] \frac{p_{i} (x_{i}, y_{i}) φ_{i} (y_{i})}{\prod_{j \in H (i)} m_{j}} \prod j \in I ∖ H [i] φ_{j} (x_{j}) \prod j \in H (i) φ_{j} (y_{j})$
		$=$	$p_{i}^{0} ⎛ ⎜ ⎝ \sum y_{j} : k \neq j \in H [i] \frac{p_{i} (x_{i}, y_{i}) φ_{i} (y_{i})}{\prod_{k \neq j \in H (i)} m_{j}} \prod j \in I ∖ H [i] φ_{j} (x_{j}) \prod k \neq j \in H (i) φ_{j} (y_{j}) ⎞ ⎟ ⎠ \cdot \frac{1}{m_{k}} \sum y_{k} \in X_{k} φ_{k} (y_{k})$
		$=$	$0.$

Hence

(P φ) (x) = ⎛ ⎝ k \sum h = 1 p_{i_{h}}^{0} λ_{j_{i_{h}}} + \sum i \in I ∖ A [S] p_{i}^{0} ⎞ ⎠ φ (x) .

and the claim is proven. ∎

Corollary 3.6.

If $P_{i}$ is ergodic for each $i \in I$ , then $P$ is ergodic.

Proof.

The expression of the eigenvalues of $P$ given in (10) ensures that the eigenvalue 1 is obtained with multiplicity one and the eigenvalue $- 1$ can never be obtained. ∎

We are able now to provide the matrices $U, D$ and $Δ$ associated with $P$ . For every $i$ , let $U_{i}$ , $D_{i}$ and $Δ_{i}$ be the matrices of eigenvectors, of the coefficients of $σ_{i}$ and of eigenvalues for the probability $P_{i}$ , respectively. Recall the identification of $X_{i}$ with the set ${0, 1, \dots, m_{i} - 1}$ .

Proposition 3.7.

The matrices $U, D$ and $Δ$ have the following form:

$U = \sum_{S \in S} (⨂_{i \in S} (U_{i} - A_{i})) \otimes (⨂_{i \in I ∖ A [S]} A_{i}) \otimes (⨂_{i \in A (S)} I_{i}^{σ_{i} - n o r m})$ , where

$I_{i}^{σ_{i} - n o r m} = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} \frac{1}{\sqrt{σ_{i} (0)}} \frac{1}{\sqrt{σ_{i} (1)}} ⋱ \frac{1}{\sqrt{σ_{i} (m_{i} - 1)}} \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ .$

By $A_{i}$ we denote the matrix of size $m_{i}$ whose entries on the first column are all 1 and the remaining ones are 0.
$D = ⨂_{i \in I} D_{i}$ .
$Δ = \sum_{i \in I} p_{i}^{0} Δ_{i} \otimes (⨂_{j \in I ∖ H [i]} I_{j}) \otimes (⨂_{j \in H (i)} J_{j}^{d i a g})$ , where $J_{j}^{d i a g}$ is the diagonal matrix $⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 10 ⋱ 0 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$ of size $m_{j}$ .

Proof.

Let us start by proving the statement for the matrix $U$ . By construction, each column of $U$ is an eigenvector of $P$ . Let us show that the rank of $U$ is maximal. Fix $S \in S$ . Then the matrix

(11)

M^{S} := (⨂ i \in S (U_{i} - A_{i})) \otimes ⎛ ⎝ ⨂ i \in I ∖ A [S] A_{i} ⎞ ⎠ \otimes ⎛ ⎝ ⨂ i \in A (S) I_{i}^{σ_{i} - n o r m} ⎞ ⎠

has rank $\prod_{j \in S} (m_{j} - 1) \prod_{j \in A (S)} m_{j}$ if $S \neq \emptyset$ . If $S = \emptyset$ , then $M^{S} = \otimes_{i \in I} A_{i}$ has rank 1. Moreover, eigenvectors arising from different $S$ are independent because they belong to subspaces of $L (X)$ which are orthogonal with respect to the scalar product $⟨ \cdot, \cdot ⟩_{π}$ .
First, let us show that, if $S \neq S^{'}$ , then the sets of indices corresponding to the non zero columns of $M^{S}$ and $M^{S^{'}}$ are disjoint. Note that $S \neq S^{'}$ implies $I ∖ A [S] \neq I ∖ A [S^{'}]$ . Hence, we can assume without loss of generality that there exists $h \in I ∖ A [S]$ such that either $h \in S^{'}$ or $h \in A (S^{'})$ . Suppose $h \in S^{'}$ , and put $M_{k}^{S} := ⨂_{j \geq k} M_{j}^{S}$ , where $M_{j}^{S} = U_{j} - A_{j}, A_{j}$ or $I_{j}^{σ_{j} - n o r m}$ according with (11). Under our assumption $M_{h}^{S} = A_{h}$ and $M_{h}^{S^{'}} = U_{h} - A_{h}$ , so that our claim is true for $M_{h}^{S}$ and $M_{h}^{S^{'}}$ . Then the same property can be deduced for $M^{S}$ and $M^{S^{'}}$ . Suppose now that $h \in A (S^{'})$ . Then there is $h^{'} \in S^{'}$ such that $h \in A (h^{'})$ . We claim that $h^{'} \in I ∖ A [S]$ . In fact if $h^{'} \in S$ then $h \in A (S)$ , which is absurd. If $h^{'} \in A (S)$ then $h \in A (S)$ , a contradiction again. Hence, there exists an index $h^{'} \in S^{'}$ such that $h^{'} \in I ∖ A [S]$ and from (11) we deduce that the claim is true for $M^{S}$ and $M^{S^{'}}$ .
Hence, we deduce from Theorem 3.5 that the rank of $U$ is $1 + \sum_{S \neq \emptyset} \prod_{j \in S} (m_{j} - 1) \prod_{j \in A (S)} m_{j} = \prod_{j \in I} m_{j}$ and so it is maximal.

In order to get the diagonal matrix $D$ , whose entries are the coefficients of $π$ , it suffices to consider the tensor product of the corresponding matrices associated with the probability $P_{i}$ , for every $i = 1, \dots, n$ .
Finally, to get the matrix $Δ$ of eigenvalues of $P$ it suffices to replace, in the expression of the matrix $P$ , the matrix $P_{i}$ by $Δ_{i}$ and the matrix $J_{i}$ by the corresponding diagonal matrix $J_{i}^{d i a g}$ . ∎

3.3. The case of the first crested product

In [6] the definition of the first crested product of Markov chains is given. More precisely, considering the product $X_{1} \times \dots \times X_{n}$ and a partition

(12)

{1, \dots, n} = C ∐ N

of the set ${1, \dots, n}$ , given a probability distribution ${p_{i}^{0}}_{i = 1}^{n}$ on ${1, \dots, n}$ , the first crested product of the Markov chains $P_{i}$ with respect to the partition (12) is defined as the Markov chain on $X_{1} \times \dots \times X_{n}$ whose transition matrix is

	$P$	$=$	$\sum i \in C p_{i}^{0} (I_{1} \otimes \dots \otimes I_{i - 1} \otimes P_{i} \otimes I_{i + 1} \otimes \dots \otimes I_{n})$
		$+$	$\sum i \in N p_{i}^{0} (I_{1} \otimes \dots \otimes I_{i - 1} \otimes P_{i} \otimes J_{i + 1} \otimes \dots \otimes J_{n}) .$

We want to show in this section that, if the poset $(I, \leq)$ satisfies some special conditions, then the generalized crested product defined in (5) reduces to the first crested product. We denote by $⪯$ the usual ordering of natural numbers.

Proposition 3.8.

Suppose that $(I, \leq)$ satisfies the following property: given $i$ such that $H (i) \neq \emptyset$ , then $j \in H (i)$ if and only if $i ≺ j$ . Then the first crested product of Markov chains is obtained by the operator defined in (5) by putting:

N = {i : H (i) \neq \emptyset} and C = {1, \dots, n} ∖ N .

Proof.

The partition ${1, \dots, n} = C ⊔ N$ , with

N = {i : H (i) \neq \emptyset} and C = {1, \dots, n} ∖ N,

gives:

	$P$	$=$	$\sum I \in C p_{i}^{0} I_{1} \otimes \dots \otimes I_{i - 1} \otimes P_{i} \otimes I_{i + 1} \dots \otimes I_{n}$
		$+$	$\sum I \in N p_{i}^{0} I_{1} \otimes \dots \otimes I_{i - 1}$