Resolution of Sigma-Fields for Multiparticle Finite-State Action Evolutions with Infinite Past

Abstract

For multiparticle finite-state action evolutions, we prove that the observation $\sigma$-field admits a resolution involving a third noise which is generated by a random variable with uniform law. The Rees decomposition from semigroup theory and the theory of infinite convolutions are utilized in our proofs.

Appendices

Appendix A: The Semigroup Consisting of Mappings

Proof of Proposition 1.11

(i) Let us define K by (1.18) and prove that K is a minimal ideal of S. For $f \in K$ and $g,h \in S$, we have $m_S \le \#(gfhV) \le \#(fV) = m_S$, which shows that K is an ideal of S. Let $I \subset K$ be an ideal of S. Let $f \in K$ and $g \in I$. Since $fgf|_{fV}$ is a permutation of fV, there exists an integer $q \ge 1$ such that $(fgf)^q$ is identity on fV, which implies $(fgf)^q f = f$. Hence,

$$\begin{aligned} f = (fgf)^q f = fg(f(fgf)^{q-1}f) \in SIS \subset I , \end{aligned}$$

(6.1)

which shows $I=K$, and thus K is the kernel of S.

(ii) This is obvious.

(iii) Let e be a primitive idempotent of S. Since K is completely simple by Proposition 1.6, we may take $f \in E(K)$. Then, $efe \in SKS \subset K$. Since $efe|_{efeV}$ is a permutation of efeV, there exists an integer $q \ge 1$ such that $(efe)^q$ is identity on efeV, which yields $(efe)^{q+1} = efe$. If we write $g := (efe)^{2q}$, we obtain $eg=ge=g \in E(K)$, which implies $g=e$ by primitivity. Thus, we obtain $e \in E(K)$. The converse is obvious since all idempotents of K are primitive.

(iv) Let $f \in Se = LG$ and take $(x,g) \in L \times G$ such that $f = xg$. Since $g^{-1}f = e$ and $fe = f$, we have [$fv=fw$ $\iff$ $ev=ew$] for all $v,w \in V$, which shows $\pi (f)=\pi (e)$.

Conversely, let $f \in S$ be such that $\pi (f)=\pi (e)$. Then, $\#(fV) = \#(eV) = m_S$, so that $f \in K$. Let $f = xgy$ with $(x,g,y) \in L \times G \times R$. Since $\pi (y) = \pi (gy) = \pi (ef) = \pi (f) = \pi (e)$ and $ye = e$, we obtain $y = e$, so $f \in Se$.

(v) Let $f \in eS = GR$ and take $(g,y) \in G \times R$ such that $f = gy$. Then, $fV = efV \subset eV$. Since $\#(fV) = \#(eV) = m_S$, we have $fV=eV$.

Conversely, let $f \in S$ be such that $fV=eV$. Then, $\#(fV) = \#(eV) = m_S$, so that $f \in K$. Take $(x,g,y) \in L \times G \times R$ such that $f = xgy$. Note that $fe = xgye = xg$ and $x = feg^{-1} = fg^{-1}$. Since $xV = fg^{-1}V \subset fV = eV$, we have $xV = eV$. On the one hand, since e is identity on $xV = eV$, we have $exv = xv$ for $v \in V$. On the other hand, since $ex = e$, we have $exv = ev$ for $v \in V$. We now obtain $x=e$, so $f \in eS$.

(vi) This is immediate from (iv) and (v), since $G = Se \cap eS$. $\square$

Appendix B: Another Example

Let

$$\begin{aligned} V = \!\left\{ \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} : a_1,a_2,a_3 \in \{ -1,1 \} \right\} . \end{aligned}$$

(7.1)

Let $D = \{ (1,0),(-1,0),(0,1),(0,-1) \}$ and set

$$\begin{aligned} S = \!\left\{ \begin{pmatrix} a_{11} &{} a_{12} &{} 0 \\ a_{21} &{} a_{22} &{} 0 \\ 0 &{} 0 &{} b \end{pmatrix} : (a_{11},a_{12}) \in D , \ (a_{21},a_{22}) \in D , \ b \in \{ -1,1 \} \right\} . \end{aligned}$$

(7.2)

Note that S is a finite semigroup with respect to the usual matrix product. In fact,

$$\begin{aligned} \begin{pmatrix} 1&0 \end{pmatrix} \begin{pmatrix} \pm 1 &{} 0 \\ a_{21} &{} a_{22} \end{pmatrix} = \begin{pmatrix} \pm 1&0 \end{pmatrix} , \quad \begin{pmatrix} 1&0 \end{pmatrix} \begin{pmatrix} 0 &{} \pm 1 \\ a_{21} &{} a_{22} \end{pmatrix} = \begin{pmatrix} 0&\pm 1 \end{pmatrix} \end{aligned}$$

(7.3)

for $(a_{21},a_{22}) \in D$, etc. We regard an element of S as a map of V into itself with respect to the usual matrix product. Set

$$\begin{aligned} s_0 = \begin{pmatrix} 1 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} -1 \end{pmatrix} , \quad s_1 = \begin{pmatrix} -1 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} -1 \end{pmatrix} , \quad s_2 = \begin{pmatrix} 0 &{} 1 &{} 0 \\ 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 \end{pmatrix} , \quad \end{aligned}$$

(7.4)

and

$$\begin{aligned} g = \begin{pmatrix} 0 &{} 1 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} -1 \end{pmatrix} , \end{aligned}$$

(7.5)

so that $s_0,s_1,s_2,g \in S$. Let $\mu = (\delta _{s_0} + \delta _{s_1} + \delta _{s_2} + \delta _{g})/4$ be the uniform law on $\mathcal {S}(\mu ) = \{ s_0,s_1,s_2,g \}$. It is easy to see that $\mathcal {S}(\mu )$ generates S, i.e., $S = \bigcup _{n=1}^{\infty } \{ s_0,s_1,s_2,g \}^n$. Let us apply Propositions 1.8, 1.11 and 1.12.

If we write

$$\begin{aligned} A = \!\left\{ \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \in V : a_1=a_2 \right\} , \quad B = \!\left\{ \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \in V : a_1=-a_2 \right\} , \end{aligned}$$

(7.6)

then

$$\begin{aligned} {\left\{ \begin{array}{ll} s_0A=A \\ s_0B=B \end{array}\right. } \quad {\left\{ \begin{array}{ll} s_1A=B \\ s_1B=A \end{array}\right. } \quad {\left\{ \begin{array}{ll} s_2A=A \\ s_2B=B \end{array}\right. } \end{aligned}$$

(7.7)

and $gV=gA=gB=A$, and hence we see that $m_\mu = m_S = \#(A) = \#(B) = 4$.

Set

$$\begin{aligned} S_+ = \!\left\{ \begin{pmatrix} a_{11} &{} a_{12} &{} 0 \\ a_{21} &{} a_{22} &{} 0 \\ 0 &{} 0 &{} b \end{pmatrix} \in S : b=1 \right\} , \quad S_- = \!\left\{ \begin{pmatrix} a_{11} &{} a_{12} &{} 0 \\ a_{21} &{} a_{22} &{} 0 \\ 0 &{} 0 &{} b \end{pmatrix} \in S : b=-1 \right\} . \end{aligned}$$

(7.8)

Since $\mathcal {S}(\mu ) \subset S_-$, we have $\mathcal {S}(\mu ^2) = \mathcal {S}(\mu ) \mathcal {S}(\mu ) \subset S_- S_- = S_+$. Since $\mathcal {S}(\mu ^2)$ generates $S_+$ and contains the identity map, we see that the left or right random walk on $S_+$ whose steps have law $\mu ^2$ is aperiodic, whereas the random walk on S whose steps have law $\mu$ is not. Hence, we obtain $p=2$, and consequently the sequence $\{ \mu ^{2n} \}_{n=1}^{\infty }$ converges to $\eta$.

Let $K = \mathcal {S}(\nu )$ and $K_+ = \mathcal {S}(\eta )$ denote the kernels of S and $S_+$, respectively. Then,

$$\begin{aligned} K =&\{ s \in S : \#(sV) = 4 \} = \!\left\{ \begin{pmatrix} a_{11} &{} 0 &{} 0 \\ a_{21} &{} 0 &{} 0 \\ 0 &{} 0 &{} b \end{pmatrix}, \begin{pmatrix} 0 &{} a_{11} &{} 0 \\ 0 &{} a_{21} &{} 0 \\ 0 &{} 0 &{} b \end{pmatrix} \in S \right\} , \end{aligned}$$

(7.9)

$$\begin{aligned} K_+ =&\{ s \in S_+ : \#(sV) = 4 \} = \!\left\{ \begin{pmatrix} a_{11} &{} 0 &{} 0 \\ a_{21} &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix}, \begin{pmatrix} 0 &{} a_{11} &{} 0 \\ 0 &{} a_{21} &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix} \in S \right\} . \end{aligned}$$

(7.10)

Set

$$\begin{aligned} e := g^2 = \begin{pmatrix} 0 &{} 1 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix} \in S_+ , \end{aligned}$$

(7.11)

which is an idempotent of $S_+$. Since $\#(eV) = 4$, we have $e \in K_+$. Let us determine the Rees decompositions $K=LGR$ and $K_+=LHR$ at $e \in E(K_+)$. Set

$$\begin{aligned} f = \begin{pmatrix} 0 &{} -1 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix} , \quad k = \begin{pmatrix} 1 &{} 0 &{} 0 \\ 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix} . \end{aligned}$$

(7.12)

Then, we have

$$\begin{aligned} L =&\{ s \in E(K_+) : \pi (s) = \pi (e) \} = \{ e,f \} , \end{aligned}$$

(7.13)

$$\begin{aligned} R =&\{ s \in E(K_+) : sV = eV \} = \{ e,k \} , \end{aligned}$$

(7.14)

$$\begin{aligned} H =&\{ s \in K_+ : \pi (s) = \pi (e) , \ sV=eV \} = \{ e,-g \} , \end{aligned}$$

(7.15)

$$\begin{aligned} G =&\{ s \in K : \pi (s) = \pi (e) , \ sV=eV \} = \{ e,-e,g,-g \} . \end{aligned}$$

(7.16)

We now see that we may choose $\gamma = g$, so that $C = \{ e,g \}$ and $G = CH$. We have the following multiplication tables (the table of ab for a and b):

(7.17)

Note that $-f = fg(-g)$ and $-k = g(-g)k$.

Let us compute $\eta ^L$. Let $\eta ^L = \alpha \delta _e + \beta \delta _f$ for some $\alpha ,\beta >0$ with $\alpha + \beta = 1$. Note that

$$\begin{aligned} \mu * \eta ^L * \omega _H =&\mu * \eta ^L * \omega _H * \eta ^R * \omega _H = \mu * \eta * \omega _H \end{aligned}$$

(7.18)

$$\begin{aligned} =&\eta ^L * \delta _g * \omega _H * \eta ^R * \omega _H = \eta ^L * \delta _g * \omega _H . \end{aligned}$$

(7.19)

On the one hand, we have

$$\begin{aligned} \eta ^L * \delta _g * \omega _H = \!\left( \alpha \delta _e + \beta \delta _f \right) * \delta _g * \omega _H . \end{aligned}$$

(7.20)

On the other hand, by (7.17), we have

$$\begin{aligned} \mu * \eta ^L&= \frac{1}{4} \!\left( \delta _{s_0} + \delta _{s_1} + \delta _{s_2} + \delta _{g} \right) * \!\left( \alpha \delta _e + \beta \delta _f \right) \nonumber \\&= \frac{3\alpha + 2\beta }{4} \delta _g + \frac{\alpha + \beta }{4} \delta _{fg} + \frac{\beta }{4} \delta _{(-f)} . \end{aligned}$$

(7.21)

Since $-f = fg(-g)$ and $-g \in H$, we have $\delta _{-f} * \omega _H = \delta _{fg} * \omega _H$, so

$$\begin{aligned} \mu * \eta * \omega _H = \mu * \eta ^L * \omega _H = \!\left( \frac{3\alpha + 2\beta }{4} \delta _e + \frac{\alpha + 2\beta }{4} \delta _f \right) * \delta _g * \omega _H . \end{aligned}$$

(7.22)

Hence, we obtain $\alpha = 2/3$ and $\beta = 1/3$, so that

$$\begin{aligned} \eta ^L = \frac{2}{3} \delta _e + \frac{1}{3} \delta _f . \end{aligned}$$

(7.23)

By a similar argument, using the identities $-k = g(-g)k$, $\omega _H * \delta _g = \delta _g * \omega _H$ and

$$\begin{aligned} \omega _H * \eta ^R * \mu = \delta _g * \omega _H * \eta ^R , \end{aligned}$$

(7.24)

we obtain

$$\begin{aligned} \eta ^R = \frac{2}{3} \delta _e + \frac{1}{3} \delta _k . \end{aligned}$$

(7.25)

Note that $eV = \{ v_1,v_2,v_3,v_4 \}$ and $fV = \{ fv_1,fv_2,fv_3,fv_4 \}$, where

$$\begin{aligned} v_1 = \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} , \quad v_2 = \begin{pmatrix} -1\\ -1\\ 1 \end{pmatrix} , \quad v_3 = \begin{pmatrix} 1\\ 1\\ -1 \end{pmatrix} , \quad v_4 = \begin{pmatrix} -1\\ -1\\ -1 \end{pmatrix} . \end{aligned}$$

(7.26)

By (3.1) and (3.2), we see that

$$\begin{aligned} W_\mu = \{ (x^1,x^2,x^3,x^4) : \text {a permutation of } (v_1,v_2,v_3,v_4) \text { or } (fv_1,fv_2,fv_3,fv_4) \} \end{aligned}$$

(7.27)

and

$$\begin{aligned} eW_\mu = \{ (x^1,x^2,x^3,x^4) : \text {a permutation of } (v_1,v_2,v_3,v_4) \} . \end{aligned}$$

(7.28)

We have the following multiplication table (the table of sv for s and v):

(7.29)

From this table, we see that we may take a set W as

$$\begin{aligned} W = \{ (x^1,x^2,x^3,x^4) : x^4=v_4 \text { and } (x^1,x^2,x^3) \text {is a permutation of } (v_1,v_2,v_3) \} , \end{aligned}$$

(7.30)

which is a minimal subset of $W_\mu$ such that $eW_\mu = GW$.