10.2.2: Simple example---Finite causal operator is represented by matrix

Define the deterministic causal map $\phi_{1,2} : \Omega_1 \to \Omega_2 $ such that

Then, by (10.11), we get the deterministic dual causal operator ${\Phi}^*_{1,2}:{\cal M}_{}(\Omega_1) \to {\cal M}_{}(\Omega_2) $ such that

where $\delta_{(\cdot)}$ is the point measure. Also, the deterministic causal operator $\Phi_{1,2}:L^\infty(\Omega_2) \to L^\infty (\Omega_1) $ is defined by

Example 10.8 [Dual causal operator, causal operator] Recall Remark 2.13, that is, if $\Omega$ $(=\{1,2,...,n \})$ is finite set ( with the discrete metric $d_D$ and the counting measure $\nu$,), we can consider that

For example, put $\Omega_1=\{ \omega_1^1, \omega_1^2, \omega_1^3 \}$ and $\Omega_2=\{ \omega_2^1 , \omega_2^2\}$. And define $\rho_1 (\in {\cal M}_{+1}(\Omega_1))$ such that

Then, the dual causal operator ${\Phi}^*_{1,2}: {\cal M}_{+1}(\Omega_1) \to {\cal M}_{+1}(\Omega_2)$ is represented by

and, consider the identification:${\cal M}(\Omega_1) {\; \approx \;} {\mathbb C}^3$, ${\cal M}(\Omega_2) {\; \approx \;} {\mathbb C}^2$, That is,

Then, putting

write, by matrix representation, as follows.

Next, from this dual causal operator ${\Phi}^*_{1,2}: {\cal M}_{}(\Omega_1) \to {\cal M}_{}(\Omega_2)$, we will construct a causal operator $\Phi_{1,2}: C_0(\Omega_2) \to C_0(\Omega_1)$. Consider the identification:${C_0}(\Omega_1) {\; \approx \;} {\mathbb C}^3$, ${C_0}(\Omega_2) {\; \approx \;} {\mathbb C}^2$, that is,

Let $f_2 \in C_0(\Omega_2)$, $f_1 = \Phi_{1,2} f_2$. Then, we see

Therefore, the relation between the dual causal operator ${\Phi}^*_{1,2}$ and causal operator $\Phi_{1,2}$ is represented as the the transposed matrix.

Example 10.9[Deterministic dual causal operator, deterministic causal map, deterministic causal operator] Consider the case that dual causal operator ${\Phi}^*_{1,2}: {\cal M}(\Omega_1) ({\approx} {\mathbb C}^3) \to {\cal M}(\Omega_2) ({\approx} {\mathbb C}^2) $ has the matrix representation such that

In this case, it is the deterministic dual {{causal operator}}. This deterministic causal operator $\Phi_{1,2}: C_0(\Omega_2) \to C_0(\Omega_1)$ is represented by

with the deterministic causal map $\phi_{1,2}: \Omega_1 \to \Omega_2$ such that

Let $(T,\le)$ be a finite tree (in Chapter 14, we discuss the infinite case), i.e., a tree like semi-ordered finite set such that "$t_1 \le t_3$ and $t_2 \le t_3$" implies "$t_1 \le t_2$ or $t_2 \le t_1$"

Assume that there exists an element $t_0 \in T$, called the root of $T$, such that $t_0 \le t$ ($\forall t \in T$) holds.

Put $T^2_\le = \{ (t_1,t_2) \in T^2{}: t_1 \le t_2 \}$. An element $t_0 \in T$ is called a root if $t_0 \le t$ ($\forall t \in T$) holds. Since we usually consider the subtree $T_{t_0}$ $(\subseteq T)$ with the root $t_0$, we assume that the tree has a root. In this chapter, assume, for simplicity, that $T$ is finite (though it is sometimes infinite in applications).

For simplicity, assume that $T$ is finite, or a finite subtree of a whole tree. Let $T$ $(= \{ 0, 1, ..., N \})$ be a tree with the root $0$. Define the parent map $\pi{}: T \setminus \{ 0 \} \to T$ such that $\pi (t) = \max \{ s \in T{}: s < t \}$. It is clear that the tree $(T\equiv \{ 0,1,..., N\} , \le)$ can be identified with the pair $(T\equiv \{ 0,1,..., N\} , \pi: T \setminus \{ 0 \} \to T)$. Also, note that, for any $t \in T \setminus \{ 0 \}$, there uniquely exists a natural number $h(t)$ $($called the height of $t$ $)$ such that $\pi^{h(t)} (t) = 0 $. Here, $\pi^2 (t) = \pi (\pi (t))$, $\pi^3 (t) = \pi (\pi^2 (t))$, etc. Also, put $\{ 0,1,..., N \}^2_{{}_{\le}} $ $=$ $\{ (m,n) \; | \; 0 \le m \le n \le N \} $. In Fig.10.2, see the root $t_0$, the parent map: $\pi(t_3)=\pi(t_4)=t_2$, $\pi(t_2)=\pi(t_5)=t_1$, $\pi(t_1)=\pi(t_6)=\pi(t_7)=t_0$

Definition 10.10 [Sequential causal operator; Heisenberg picture of causality] The family $\{ \Phi_{t_1,t_2}{}: $ $ \overline{\mathcal A}_{t_2} \to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $\Big($ or, $\{$ $\overline{\mathcal A}_{t_2} \overset{\Phi_{t_1,t_2}}\to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $\Big)$ is called a sequential causal operator, if it satisfies that

(i): [pre-dual sequential causal operator : Schrödinger picture of causality ] The sequence $\{ ({\Phi}_{t_1,t_2})_*{}: $ $ (\overline{\mathcal A}_{t_1})_* \to (\overline{\mathcal A}_{t_1})_* \}_{(t_1,t_2) \in T^2_{\leqq}}$ is called a pre-dual sequential causal operator of $\{ {\Phi_{t_1,t_2}}{}: $ ${ \overline{\mathcal A}_{t_2}} \to {\overline{\mathcal A}{t_1}} \}_{(t_1,t_2) \in T^2_{\leqq}}$

(ii): [Dual sequential causal operator : Schrödinger picture of causality ] A sequence $\{ {\Phi}^*_{t_1,t_2}{}: $ $ {\mathcal A}^\ast_{t_1} \to {\mathcal A}^\ast_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ is called a dual sequential causal operator of $\{ \Phi_{t_1,t_2}{}: $ ${ \overline{\mathcal A}_{t_2}} \to {\overline{\mathcal A}_{t_1}} \}_{(t_1,t_2) \in T^2_{\leqq}}$.