Eight Definitions of Observability

[Scott Garrabrant]

This is the eleventh post in the Cartesian frames sequence. Here, we compare eight equivalent definitions of observables, which emphasize different philosophical interpretations.

Throughout this post, we let $C=(A,E,\cdot )$ be a Cartesian frame over a nonempty set $W$, we let $V=\{S_1,\dots ,S_n\}$ be a finite partition of $W$, and we let $v:W\to V$ send each element of $W$ to its part in $V$.

The condition that $V$ is finite is an important one. Many of the definitions below can be extended to infinite partitions, and the theory of observability for infinite partitions is probably nice, but we are not discussing it here. The condition that $W$ is nonempty is just ruling out some degenerate cases

1. Definition from Subsets

The definitions in this post will talk about when a finite partition $V$ of $W$ is observable in $C$. This will make some of the definitions more elegant, and it is easy to translate back and forth between the new definitions of the observability of a finite partition and the old definitions of the observability of a subset.

Definition: We say $C$’s agent can observe a finite partition $V$ of $W$ if for all parts $S_i\in V$, $S_i\in \text{Obs}\left(C\right)$. We let ${\text{Obs}}^{\prime }\left(C\right)$ denote the set of all finite partitions of $W$ that are observable in $C$.

Claim: For any nonempty strict subset $S\subset W$, $C$’s agent can observe $S$ if and only if $C\text{'s}$agent can observe $\{S,(W\setminus S\left)\right\}$.

Proof: If $C$‘s agent can observe $\{S,(W\setminus S\left)\right\}$, then clearly $C$’s agent can observe $S$. If $C\text{'s}$agent can observe $S$, then since observability is closed under complements, $C$’s agent can observe $W\setminus S$, and so can observe $\{S,(W\setminus S\left)\right\}$. $\square$

1.1. Example

In “Introduction to Cartesian Frames,” we gave the example of an agent that can choose between unconditionally carrying an umbrella, unconditionally carrying no umbrella, carrying an umbrella iff it’s raining, and carrying an umbrella iff it’s sunny:

$C_0=\begin{array}{cc}& \begin{array}{cc}r& s\end{array}\\ \begin{array}{c}u\\ n\\ u\leftrightarrow r\\ u\leftrightarrow s\end{array}& \left(\begin{array}{cc}ur& us\\ nr& ns\\ ur& ns\\ nr& us\end{array}\right)\end{array}$

Here, $\text{Obs}\left(C_0\right)=\left\{\right\{\},\{ur,nr\},\{us,ns\},W\}$, so the partition $V=\{R,S\}$ is observable in $C_0$, where $R=\{ur,nr\}$ and $S=\{us,ns\}$.

As we go through the definitions in this post, we will repeatedly return to $C_0$ and show how to understand $C_0$’s observables in terms of our new definitions.

Before presenting fundamentally new definitions, we will modify our two old definitions to be about finite partitions instead of subsets.

2. Conditional Policies Definition

Definition: We say that $C$’s agent can observe a finite partition $V$ of $W$ if for all functions $f:V\to A$, there exists an element $a_f\in A$ such that for all $e\in E$, $f\left(v\right(a_f\cdot e\left)\right)\cdot e=a_f\cdot e$.

Claim: This definition is equivalent to the definition from subsets.

Proof: We work by induction on the number of parts in $V$. Since $W$ is nonempty, $V$ has at least one part. If $V=\left\{W\right\}$ has one part, we clearly have that $C$‘s agent can observe $V$ under the definition from subsets. For the conditional policies definition, we also have that $C$’s agent can observe $V$, since we can take $a_f=f\left(W\right)$, and thus, for all $e\in E$,

$$\begin{array}{cc}f\left(v\right(a_f\cdot e\left)\right)\cdot e& =f\left(W\right)\cdot e\\ & =a_f\cdot e.\end{array}$$

If $V=\{S_1,\dots ,S_n\}$ has $n$ parts, consider the partition $V^{\prime }=\{S_1\cup S_2,S_3,\dots ,S_n\}$ which unions together the first two parts $S_1$ and $S_2$ of $V$. Let $v^{\prime }:W\to V^{\prime }$ send each element of $W$ to its part in $V^{\prime }$.

First, assume that $C$’s agent can observe $V$ according to the definition from subsets. Then, since observability of subsets is closed under unions, $C$’s agent can also observe $V^{\prime }$ under the definition from subsets, and thus also under the conditional policies definition.

Given a function $f:V\to A$, let $f^{\prime }:V^{\prime }\to A$ be given by $f^{\prime }(S_1\cup S_2)=f\left(S_2\right)$, and $f^{\prime }\left(S_i\right)=f\left(S_i\right)$ on all other inputs. Since $C$’s agent can observe $V^{\prime }$ under the conditional policies definition, we can let $a_{f^{\prime }}$ be such that for all $e\in E$, $f^{\prime }\left(v^{\prime }\right(a_{f^{\prime }}\cdot e\left)\right)\cdot e=a_{f^{\prime }}\cdot e$.

Choose an $a_f\in A$ such that $a_f\in \text{if}(S_1,f(S_1),a_f^{\prime })$, which we can do because $S_1$ is observable in $C$. Observe that for all $e\in E$, we have that if $a_f\cdot e\in S_1$, then

$$\begin{array}{cc}f\left(v\right(a_f\cdot e\left)\right)\cdot e& =f\left(S_1\right)\cdot e\\ & =a_f\cdot e,\end{array}$$

if $a_f\cdot e\in S_2$, we have $a_f\cdot e=a_{f^{\prime }}\cdot e$, and thus

$$\begin{array}{cc}f\left(v\right(a_f\cdot e\left)\right)\cdot e& =f\left(S_2\right)\cdot e\\ & =f^{\prime }(S_1\cup S_2)\cdot e\\ & =f^{\prime }\left(v^{\prime }\right(a_{f^{\prime }}\cdot e\left)\right)\cdot e\\ & =a_{f^{\prime }}\cdot e\\ & =a_f\cdot e,\end{array}$$

and finally if $a_f\cdot e\in S_i$ for some $i\ne 1,2$, we still have have $a_f\cdot e=a_{f^{\prime }}\cdot e$, and thus

$$\begin{array}{cc}f\left(v\right(a_f\cdot e\left)\right)\cdot e& =f\left(S_i\right)\cdot e\\ & =f^{\prime }\left(S_i\right)\cdot e\\ & =f^{\prime }\left(v^{\prime }\right(a_{f^{\prime }}\cdot e\left)\right)\cdot e\\ & =a_{f^{\prime }}\cdot e\\ & =a_f\cdot e.\end{array}$$

Thus, $C$’s agent can observe $V$ according to the conditional policies definition.

Conversely, if $C$‘s agent can observe $V$ according to the conditional policies definition, then to show that $C$’s agent can observe $V$ according to the definition from subsets, it suffices to show that the agent can observe $S_i$ for all $S_i\in V$. Thus, we need to show that for any $a_0,a_1\in A$, there exists an $a_2\in A$ with $a_2\in \text{if}(S_i,a_0,a_1)$.

Indeed, if we let $f:V\to A$ send $S_i$ to $a_0$, and send all other inputs to $a_1$, then we can take an $a_f$ such that for all $e\in E$, $f\left(v\right(a_f\cdot e\left)\right)\cdot e=a_f\cdot e$. But then, if $a_f\cdot e\in S_i$, then

$$\begin{array}{cc}{a}_f\cdot e& =f\left(v\right(a_f\cdot e\left)\right)\cdot e\\ & =f\left(S_1\right)\cdot e\\ & =a_0\cdot e,\end{array}$$

and otherwise,

$$\begin{array}{cc}{a}_f\cdot e& =f\left(v\right(a_f\cdot e\left)\right)\cdot e\\ & =a_1\cdot e.\end{array}$$

Thus, $C$’s agent can observe $V$ according to the definition from subsets. $\square$

2.1. Example

Let $C_0=(A,E,\cdot )$ be defined as in the §1.1 example, with $R=\{ur,nr\}$, $S=\{us,ns\}$, and $V=\{R,S\}$.

$A=\{u,n,u\leftrightarrow r,u\leftrightarrow s\}$ is a four-element set, and $V=\{R,S\}$ is a two-element set, so there are sixteen functions $f:V\to A$. For each function, there is a possible agent $a_f\in A$ that satisfies $f\left(v\right(a_f\cdot e\left)\right)\cdot e=a_f\cdot e$ for all $e\in E$. We can illustrate the sixteen functions and the corresponding $a_f\in A$ in a sixteen-row table:

$f\left(R\right)$	$f\left(S\right)$	$a_f$
$u$	$u$	$u$
$u$	$n$	$u\leftrightarrow r$
$u$	$u\leftrightarrow r$	$u\leftrightarrow r$
$u$	$u\leftrightarrow s$	$u$
$n$	$u$	$u\leftrightarrow s$
$n$	$n$	$n$
$n$	$u\leftrightarrow r$	$n$
$n$	$u\leftrightarrow s$	$u\leftrightarrow s$
$u\leftrightarrow r$	$u$	$u$
$u\leftrightarrow r$	$n$	$u\leftrightarrow r$
$u\leftrightarrow r$	$u\leftrightarrow r$	$u\leftrightarrow r$
$u\leftrightarrow r$	$u\leftrightarrow s$	$u$
$u\leftrightarrow s$	$u$	$u\leftrightarrow s$
$u\leftrightarrow s$	$n$	$n$
$u\leftrightarrow s$	$u\leftrightarrow r$	$n$
$u\leftrightarrow s$	$u\leftrightarrow s$	$u\leftrightarrow s$

Since there is an $a_f\in A$ for each function, $C_0$’s agent can observe $V$ according to the conditional policies definition.

3. Additive Definitions

Next, we give an additive definition of observables. This is a version of our categorical definition of observables from “Controllables and Observables, Revisited,” modified to be about finite partitions.

Definition: We say $C$’s agent can observe a finite partition $V=\{S_1,\dots ,S_n\}$ of $W$ if there exist $C_1,\cdots C_n$, Cartesian frames over $W$, with $C_i◃{\perp }_{S_i}$ such that $C\simeq C_1\&\dots \&C_n$.

This can also be strengthened to a constructive version of the additive definition, which we will call the assuming definition.

Definition: We say $C$’s agent can observe a finite partition $V=\{S_1,\dots ,S_n\}$ of $W$ if $C\simeq {\text{Assume}}_{S_1}\left(C\right)\&\dots \&{\text{Assume}}_{S_n}\left(C\right)$.

Claim: These definitions are equivalent to each other and the definitions above.

Proof: We assume that $n\ge 2$, and that $A$ is nonempty. The case where $n=1$ and the case where $A=\left\{\right\}$ are trivial.

If $C$’s agent can observe $V$ according to the assuming definition of observables, then it can also clearly observe $V$ according to the additive definition, since ${\text{Assume}}_{S_1}\left(C\right)◃{\perp }_{S_1}$.

Next, assume that $C$‘s agent can observe $V$ according to the additive definition. We will show that $C$’s agent can observe $S_1$. Consider the pair of Cartesian frames $C_1$ and $C_2\&\dots \&C_n$. Observe that $C_1◃{\perp }_{S_1}$ and that $C_2\&\dots \&C_n◃{\perp }_{W\setminus S_1}$, and that $C\simeq C_1\&(C_2\&\dots \&C_n)$. Thus, $S_1$ is observable in $C$. Symmetrically, $S_i$ is observable in $C$ for all $i=1,\dots n$, and thus $V$ is observable in $C$ according to the definition from subsets.

Finally, assume that $C$’s agent can observe $V$ according to the conditional policies definition (and also the definition from subsets). We will show that $C\simeq C_1\&\dots \&C_n$, where $C_i={\text{Assume}}_{S_i}\left(C\right)$.

We have $C_1\&\dots \&C_n=(A^n,E_1\bigsqcup \cdots \bigsqcup E_n,\star )$, where $C_i=(A,E_i,{\cdot }_i)$, and $\star$ is given by $(a_1,\dots ,a_n)\star e=a_i\cdot e$, where $e\in E_i$.

First observe that for every $e\in E$, there is a unique $i\in \{1,\dots ,n\}$ such that $e\in E_i$. This is because there exists an $a_0\in A$, and from the definition from subsets, $C$’s agent can observe each $S_i$, and so given an $e\in E$, if $a_0\cdot e\in S_i$, it must be the case that for all $a\in A,$ $a\cdot e\in S_i$. Thus, we have that that $E=E_1\bigsqcup \cdots \bigsqcup E_n$.

We construct $(g_0,h_0):(A^n,E,\star )\to C$ and $(g_1,h_1):C\to (A^n,E,\star )$ which compose to something homotopic to the identity in each order. Let $g_1:A\to A^n$ be the diagonal, given by $g_1\left(a\right)=(a,\dots ,a)$. Let $h_0$ and $h_1$ be the identity on $E$. Let $g_0$ be given by $g_0(a_1,\dots ,a_n)=a_f$, where $f:V\to A$ is given by $f\left(S_i\right)=a_i$, and $a_f$ satisfies $f\left(v\right(a_f\cdot e\left)\right)\cdot e=a_f\cdot e$ for all $e\in E$, which is possible by the conditional policies definition.

To see that $(g_1,h_1)$ is a morphism, observe that for all $a\in A$ and $e\in E$,

$$\begin{array}{cc}{g}_1\left(a\right)\star e& =(a,\dots ,a)\star e\\ & =a\cdot e\\ & =a\cdot h_1\left(e\right).\end{array}$$

To see that $(g_0,h_0)$ is a morphism, observe that for all $(a_1,\dots ,a_n)\in A$, and $e\in E$, if we let $f:V\to A$ be given by $f\left(S_i\right)=a_i$, we have

$$\begin{array}{cc}{g}_0(a_1,\dots ,a_n)\cdot e& =f\left(v\right(g_0(a_1,\dots ,a_n)\cdot e\left)\right)\cdot e\\ & =f\left(S_i\right)\cdot e\\ & =a_i\cdot e\\ & =(a_1,\dots ,a_n)\star e\\ & =(a_1,\dots ,a_n)\star h_0\left(e\right),\end{array}$$

where $i$ is such that $e\in E_i$. The fact that $(g_0,h_0)$ and $(g_1,h_1)$ compose to something homotopic to the identity in both orders follows from the fact that $h_0\circ h_1$ and $h_1\circ h_0$ are the identity on $E$. Thus, $C\simeq {\text{Assume}}_{S_1}\left(C\right)\&\dots \&{\text{Assume}}_{S_n}\left(C\right)$, and so $V$ is observable in $C$ according to the assuming definition. $\square$

3.1. Example

Let $C_0$ be defined as in the previous examples, with $R=\{ur,nr\}$ and $S=\{us,ns\}$. By the assuming definition, there exist two frames

$C_1={\text{Assume}}_R\left(C_0\right)=\begin{array}{cc}& \begin{array}{c}r\end{array}\\ \begin{array}{c}u\\ n\end{array}& \left(\begin{array}{c}ur\\ nr\end{array}\right)\end{array}$

and

$C_2={\text{Assume}}_S\left(C_0\right)=\begin{array}{cc}& \begin{array}{c}s\end{array}\\ \begin{array}{c}u\\ n\end{array}& \left(\begin{array}{c}us\\ ns\end{array}\right)\end{array}$

such that $C_0\simeq C_1\&C_2$.

This example both illustrates the idea behind the additive definitions, and shows the construction used in the assuming definition. This is also the same example we provided to illustrate products of Cartesian frames in “Additive Operations on Cartesian Frames.”

Another way of thinking about the additive definition of observables: Recall “Committing, Assuming, Externalizing, and Internalizing” §3.2 (Committing and Assuming Can Be Defined Using Lollipop and Tensor), where we saw that ${\text{Assume}}_S\left(C\right)\cong 1_S\otimes C$. This means that (up to isomorphism) we can restate $C_0\simeq C_1\&C_2$ as $C_0\simeq (1_R\otimes C_0)\&(1_S\otimes C_0)$, i.e.,

$\begin{array}{c}\left(\begin{array}{cc}ur& us\\ nr& ns\\ ur& ns\\ nr& us\end{array}\right)\end{array}\simeq \begin{array}{cc}& \left(\begin{array}{cc}ur& nr\end{array}\right)\end{array}\otimes \begin{array}{c}\left(\begin{array}{cc}ur& us\\ nr& ns\\ ur& ns\\ nr& us\end{array}\right)\end{array}\&\begin{array}{cc}& \left(\begin{array}{cc}us& ns\end{array}\right)\end{array}\otimes \begin{array}{c}\left(\begin{array}{cc}ur& us\\ nr& ns\\ ur& ns\\ nr& us\end{array}\right)\end{array}$.

This (equivalent) framing makes it easier to keep track of what “assuming” is doing categorically, so that we can see what interfaces between frames we are relying on when we say that something is “observable” using an additive definition.

4. Multiplicative Definitions

Our multiplicative definitions will depend on a notion of agents being powerless outside of a subset.

4.1. Powerless Outside of a Subset

Definition: Given a subset $S$ of $W$, we say that $C$’s agent is powerless outside $S$ if for all $e\in E$, and all $a_0,a_1\in A$, if $a_0\cdot e\notin S$, then $a_0\cdot e=a_1\cdot e$.

To say that $C$‘s agent is powerless outside $S$ is to say that the if the world is at all dependent on $C$’s agent, then the world must be in $S$.

Here are some lemmas about being powerless outside of a subset, which we will use later.

Lemma: If $C$‘s agent is powerless outside $S$ and $T\supseteq S$, then $C$’s agent is powerless outside $T$.

Proof: Trivial. $\square$

Lemma: If $C$ and $D$‘s agents are both powerless outside $S$, then $C\otimes D$’s agent is powerless outside $S$.

Proof: Let $D=(B,F,\star )$, and let $C\otimes D=(A\times B,\text{hom}(C,D^{\ast }),\diamond )$. Consider some $(a_0,b_0),(a_1,b_1)\in A\times B$ and $(g,h)\in \text{hom}(C,D^{\ast })$. We will use the fact that if $a_0\cdot h\left(b_0\right)\notin S$ then $a_0\cdot h\left(b_0\right)=a_1\cdot h\left(b_0\right)$, and the fact that if $b_0\star g\left(a_1\right)\notin S$ then $b_0\star g\left(a_1\right)=b_1\star g\left(a_1\right)$. Observe that if $(a_0,b_0)\diamond (g,h)\notin S$, then

$$\begin{array}{cc}(a_0,b_0)\diamond (g,h)& =a_0\cdot h\left(b_0\right)\\ & =a_1\cdot h\left(b_0\right)\\ & =b_0\star g\left(a_1\right)\\ & =b_1\star g\left(a_1\right)\\ & =(a_1,b_1)\star (g,h).\end{array}$$

$\square$

Now, we are ready for our first truly new definition of the observability of a finite partition.

4.2. Multiplicative Definitions of Observables

Definition: We say that $C$‘s agent can observe a finite partition $V=\{S_1,\dots ,S_n\}$ of $W$ if $C\simeq C_1\otimes \cdots \otimes C_n$, where each $C_i$’s agent is powerless outside $S_i$.

Again, we also have a constructive version of this definition:

Definition: We say that $C$’s agent can observe a finite partition $V=\{S_1,\dots ,S_n\}$ of $W$ if $C\simeq C_1\otimes \cdots \otimes C_n$, where $C_i={\text{Assume}}_{S_i}\left(C\right)\&1_{T_i}$, where $T_i=\left(W\setminus S_i\right)\cap \text{Image}\left(C\right)$.

Claim: These definitions are equivalent to each other and equivalent to the definitions above.

Proof: First, observe that if $C$’s agent can observe $V$ according to the constructive version of the multiplicative definition, it can also observe $V$ according to the nonconstructive version of the multiplicative definition, since the agent of ${\text{Assume}}_{S_i}\left(C\right)\&1_{T_i}$ is clearly powerless outside $S_i$.

Next, we show that if $C$‘s agent can observe $V$ according to the nonconstructive multiplicative definition, it can also observe $V$ according to the definition from subsets. Let $C\simeq D=C_1\otimes \cdots \otimes C_n$, where each $C_i$‘s agent is powerless outside $S_i$. It suffices to show that $D$‘s agent can observe $V$, since the definition from subsets is equivalent to the additive definition, and thus closed under biextensional equivalence. Thus, it suffices to show that $D$‘s agent can observe $S_i$ for all $i=1,\dots ,n$. We will show that $D$’s agent can observe $S_1$, and the rest will follows by symmetry.

Let $C_1=(A_1,E_1,{\cdot }_1)$, and let $D_1=(B_1,F_1,{\star }_1)=C_2\otimes \cdots \otimes C_n$. We start by showing that $D_1$‘s agent is powerless outside $W\setminus S_1$. We have that the agents of $C_2,\dots ,C_n$ are all powerless outside $W\setminus S_1$, since being powerless outside something is closed under supersets. Thus we have that $D_1$’s agent is powerless outside $W\setminus S_1$, since being powerless outside $W\setminus S_1$ is closed under tensor.

Thus, we have $D=(A_1\times B_1,\text{hom}(C,D^{\ast }),\diamond )=C_1\otimes D_1$, with $C_1$‘s agent powerless outside $S_1$ and $D_1$‘s agent powerless outside $W\setminus S_1$. Given an arbitrary $(a_1,b_1),(a_2,b_2)\in A_1\times B_1$ , we will show that $(a_1,b_2)\in \text{if}(S_1,(a_1,b_1),(a_2,b_2\left)\right)$, and thus show that $D$’s agent can observe $S_1$.

It suffices to show that for all $(g,h):C\to D^{\ast }$, if $(a_1,b_2)\diamond (g,h)\in S_1$, then $(a_1,b_2)\diamond (g,h)=(a_1,b_1)\diamond (g,h)$, and if $(a_1,b_2)\diamond (g,h)\notin S_1$, then $(a_1,b_2)\diamond (g,h)=(a_2,b_2)\diamond (g,h)$. Indeed, if $(a_1,b_2)\diamond (g,h)\in S_1$, then, since $D_1$’s agent is powerless outside $W\setminus S_1$, we have

$$\begin{array}{cc}(a_1,b_2)\diamond (g,h)& =b_2{\star }_{1}g\left(a_1\right)\\ & =b_1{\star }_{1}g\left(a_1\right)\\ & =(a_1,b_1)\diamond (g,h).\end{array}$$

Similarly, if $(a_1,b_2)\diamond (g,h)\notin S_1$, then, since $C_1$’s agent is powerless outside $S$, we have

$$\begin{array}{cc}(a_1,b_2)\diamond (g,h)& =a_1{\cdot }_{1}h\left(b_2\right)\\ & =a_2{\cdot }_{1}h\left(b_2\right)\\ & =(a_2,b_2)\diamond (g,h).\end{array}$$

Thus, $D$‘s agent can observe $S_1$, so $C$’s agent can observe $V$ according to the definition from subsets.

Finally, we assume that $C$‘s agent can observe $V$ according to the assuming definition, and show that $C$’s agent can observe $V$ according to the constructive version of the multiplicative definition.

We work by induction on $n$, the number of parts. The case where $n=1$ is trivial. Let $C\simeq {\text{Assume}}_{S_1}\left(C\right)\&\dots \&{\text{Assume}}_{S_n}\left(C\right)$. Thus, we also have that $C\simeq {\text{Assume}}_{S_1\cup S_2}\left(C\right)\&{\text{Assume}}_{S_3}\left(C\right)\&\dots \&{\text{Assume}}_{S_n}\left(C\right)$, and so by induction, we have that $C\simeq \left({\text{Assume}}_{S_1\cup S_2}\right(C\left)\&1_{T_1\cap T_2}\right)\otimes C_3\otimes \cdots \otimes C_n$, where $C_i$ and $T_i$ are as in the constructive multiplicative definition. Thus, it suffices to show that

$$\begin{array}{cc}{\text{Assume}}_{S_1\cup S_2}\left(C\right)\&1_{T_1\cap T_2}& \simeq C_1\otimes C_2\\ & =\left({\text{Assume}}_{S_1}\right(C\left)\&1_{T_1}\right)\otimes \left({\text{Assume}}_{S_2}\right(C\left)\&1_{T_2}\right).\end{array}$$

First, observe that we have $C\simeq D_1\&D_2\&D_3$, where $D_1={\text{Assume}}_{S_1}$(C), $D_2={\text{Assume}}_{S_2}\left(C\right)$, and $D_3={\text{Assume}}_{S_3}\left(C\right)\&\dots \&{\text{Assume}}_{S_n}\left(C\right)$. Let $D_i=(B_i,F_i,{\star }_i)$. Let $R_i=\text{Image}\left(D_i\right)$.

Observe that $T_1=R_2\cup R_3$, $T_2=R_1\cup R_3$, and $T_1\cup T_2=R_3$, and observe that ${\text{Assume}}_{S_1\cup S_2}\left(C\right)\simeq D_1\&D_2$. Thus it suffices to show that $\left(D_1\&1_{R_2\cup R_3}\right)\otimes \left(D_2\&1_{R_1\cup R_3}\right)\simeq D_1\&D_2\&1_{R_3}$.

Let $D_1\&1_{R_2\cup R_3}=(B_1,F_1\bigsqcup R_2\bigsqcup R_3,{\bullet }_1)$, let $D_2\&1_{R_1\cup R_3}=(B_2,F_2\bigsqcup R_1\bigsqcup R_3,{\bullet }_2)$, and let $D_1\&D_2\&1_{R_3}=(B_1\times B_2,F_1\bigsqcup F_2\bigsqcup R_3,{\bullet }_3)$ where ${\bullet }_1$, ${\bullet }_2$, and ${\bullet }_3$ are all given by $b{\bullet }_{i}f=b{\star }_{1}f$ if $f\in F_1$, $b{\bullet }_{i}f=b{\star }_{2}f$ if $f\in F_2$, and $b{\bullet }_{i}f=f$ otherwise.

Let $H=\text{hom}(B_1\&1_{R_2\cup R_3},(D_2\&1_{R_1\cup R_3}{)}^{\ast })$. Let $\left(D_1\&1_{R_2\cup R_3}\right)\otimes \left(D_2\&1_{R_1\cup R_3}\right)=(B_1\times B_2,H,{\bullet }_4)$), where

$$\begin{array}{cc}(b_1,b_2){\bullet }_4(g,h)& =b_1{\bullet }_{1}h\left(b_2\right)\\ & =b_2{\bullet }_{2}g\left(b_1\right).\end{array}$$

Observe that for any $f_1\in F_1$, there is a $(g_{f_1},h_{f_1})\in H$, given by $g_{f_1}\left(b_1\right)=b_1{\cdot }_1{f}_1$ and $h_{f_1}\left(b_2\right)=f_1$. This is clearly a morphism, since

$$\begin{array}{cc}{b}_1{\bullet }_1{h}_{f_1}\left(b_2\right)& =b_1{\bullet }_1{f}_1\\ & =b_1{\cdot }_1{f}_1\\ & =g_{f_1}\left(b_1\right)\\ & =b_2{\bullet }_2{g}_{f_1}\left(b_1\right).\end{array}$$

Similarly, for any $f_2\in F_2$, there is a morphism $(g_{f_2},h_{f_2})\in H$ given by $g_{f_2}\left(b_1\right)=f_2$ and $h_{f_2}\left(b_2\right)=b_2{\cdot }_2{f}_2$. Finally, for any $r\in R_3$, there is a morphism $(g_r,h_r)\in H$, given by $g_r\left(b_1\right)=h_r\left(b_2\right)=r$, which is also clearly a morphism.

We show that these are in fact all of the morphisms in $H$. Indeed, let $(g,h)$ be a morphism in $H$, let $b_1$ be an element of $B_1$, and let $b_2$ be an element of $b_2$. Let

$$\begin{array}{cc}r& =b_2{\bullet }_{2}g\left(b_1\right)\\ & =b_1{\bullet }_{1}h\left(b_2\right).\end{array}$$

If $r\in R_3$, then $g\left(b_1\right)=h\left(b_2\right)=r$, so given any $b_1^{\prime }\in B_1$,

$$\begin{array}{cc}{b}_2{\bullet }_{2}g\left(b_1^{\prime }\right)& =b_1^{\prime }{\bullet }_{1}h\left(b_2\right)\\ & =r\end{array}$$

$\in R_3$, and so

$$\begin{array}{cc}g\left(b_1^{\prime }\right)& =b_2{\bullet }_{2}g\left(b_1^{\prime }\right)\\ & =r.\end{array}$$

Similarly, for any $b_2^{\prime }\in B_2$, $h\left(b_2^{\prime }\right)=r$ and so $(g,h)=(g_r,h_r)$.

If $r\in R_1$, then $g\left(b_1\right)=r$, and $h\left(b_2\right)\in F_1$. Let $f_1=h\left(b_2\right)$. Given any $b_1^{\prime }\in B_1$,

$$\begin{array}{cc}{b}_2{\bullet }_{2}g\left(b_1^{\prime }\right)& =b_1^{\prime }{\bullet }_{1}h\left(b_2\right)\\ & =b_1^{\prime }{\bullet }_1{f}_1\end{array}$$

$\in R_1$, so

$$\begin{array}{cc}g\left(b_1^{\prime }\right)& =b_2{\bullet }_{2}g\left(b_1^{\prime }\right)\\ & =b_1^{\prime }{\bullet }_1{f}_1\\ & =b_1^{\prime }{\cdot }_1{f}_1.\end{array}$$

Given any $b_2^{\prime }\in B_2$,

$$\begin{array}{cc}{b}_1{\bullet }_{1}h\left(b_2^{\prime }\right)& =b_2^{\prime }{\bullet }_{2}g\left(b_1\right)\\ & =b_2^{\prime }{\bullet }_{2}r\\ & =r\end{array}$$

$\in R_1$, and so

$$\begin{array}{cc}h\left(b_2^{\prime }\right)& =b_1{\bullet }_{1}h\left(b_2^{\prime }\right)\\ & =r.\end{array}$$

Thus, $(g,h)=(g_{f_1},h_{f_1})$.

Finally, if $r\in R_2$, we similarly have $(g,h)=(g_{f_2},h_{f_2})$, where $f_2=g\left(b_1\right)\in F_2$.

We construct a pair of morphisms

$$\begin{array}{c}(g_0,h_0):(B_1\times B_2,H,{\bullet }_4)\to (B_1\times B_2,F_1\bigsqcup F_2\bigsqcup R_3,{\bullet }_3)\end{array}$$

and

$$\begin{array}{c}(g_1,h_1):(B_1\times B_2,F_1\bigsqcup F_2\bigsqcup R_3,{\bullet }_3)\to (B_1\times B_2,H,{\bullet }_4),\end{array}$$

by letting $g_0$ and $g_1$ be the identity on $B_1\times B_2$, letting $h_0:F_1\bigsqcup F_2\bigsqcup R_3\to H$ be given by $h_0\left(f\right)=(g_f,h_f)$ as above. Since we have shown that $h_0$ is surjective, we let $h_1$ be any right inverse to $h_0$. It is easy to show that both of these are morphisms by the construction of $(g_f,h_f)$, and they compose to something homotopic to the identity in both orders since $g_0\circ g_1$ and $g_1\circ g_0$ are the identity of $B_1\times B_2$.

Thus $\left(D_1\&1_{R_2\cup R_3}\right)\otimes \left(D_2\&1_{R_1\cup R_3}\right)\simeq D_1\&D_2\&1_{R_3}$, so $C$’s agent can observe $V$ according to the constructive multiplicative definition, completing the proof. $\square$

You may have noticed that the last part of the proof would have been much simpler if $\otimes$ distributed over $\&$, but $\otimes$ does not in general distribute over $\&$. ($\otimes$ distributes over $\oplus$ and $⅋$ distributes over $\&$.)

In this case, however, $\otimes$ does distribute over $\&$. I do not plan on going over it now, but there is actually an interesting relationship between observables and cases where $\otimes$ distributes over $\&$.

4.3. Example

Let $C_0$ be defined as in the previous examples, with $R=\{ur,nr\}$ and $S=\{us,ns\}$. Let $T_X=\left(W\setminus X\right)\cap \text{Image}\left(C_0\right)$, so that $1_{T_R}=1_S$ and $1_{T_S}=1_R$. By the multiplicative definitions of observables, there then exist two frames

$C_1={\text{Assume}}_R\left(C\right)\&1_S=\begin{array}{cc}& \begin{array}{ccc}r& us& ns\end{array}\\ \begin{array}{c}r\to u\\ r\to n\end{array}& \left(\begin{array}{ccc}ur& us& ns\\ nr& us& ns\end{array}\right)\end{array}$

and

$C_2={\text{Assume}}_S\left(C\right)\&1_R=\begin{array}{cc}& \begin{array}{ccc}s& ur& nr\end{array}\\ \begin{array}{c}s\to u\\ s\to n\end{array}& \left(\begin{array}{ccc}us& ur& nr\\ ns& ur& nr\end{array}\right)\end{array}$

such that $C_0\simeq C_1\otimes C_2$.

Here, $C_1$ is an agent that treats the “makes decisions when it’s sunny” part of itself as though it were an external process. Similarly, $C_2$ externalizes its ability to make decisions when it’s rainy.

This example illustrates both multiplicative definitions, and also shows the construction used in the constructive multiplicative definition.

Appealing again to the fact that ${\text{Assume}}_S\left(C\right)\cong 1_S\otimes C$, we also have the option of restating $C_0\simeq C_1\otimes C_2$ here as $C_0\simeq \left(\right(1_R\otimes C_0\left)\&1_S\right)\otimes \left(\right(1_S\otimes C_0\left)\&1_R\right)$. In words, this says that $\text{Agent}\left(C_0\right)$ is (biextensionally equivalent to) a team consisting of:

1. that very agent, picking an action after the environment either (a) gives it a promise it will rain or (b) makes it powerless and doesn’t rain; and

2. that very agent, picking an action after the environment either (a) gives it a promise it won’t rain or (b) makes it powerless and rains.

4.4. Updatelessness

The relationship between observables’ additive and multiplicative definitions is interesting. You can think of the additive definition as updateful, while the multiplicative definition is updateless.

The $C_i$ in the additive definition are basically given a promise that the world will end up in $S_i$. The $C_i$ in the multiplicative definition, however, are instead given a promise that their choices have no effect on worlds outside of $S_i$.

I think the updateless factorization is better, and thus prefer the multiplicative definition in spite of the fact that it is more complicated.

When an updateless agent observes something, it becomes the version of itself that only affects the worlds in which it makes that observation. When an updateful agent observes something, we assume that all the worlds in which it does not make that observation do not exist. The fact that the additive and multiplicative definitions above are equivalent illustrates the equivalence of the updateful and updateless views in the simple cases where there is true observation. However, they diverge as soon as you want to try to approximate observation. The updateless view approximates better, as it makes sense to think of a subagent that has only a very small effect on worlds in which it does not make the observation that it makes.

Also, note that the $C_i$ in the additive definition are not subagents of $C$, but they are additive sub-environments. The $C_i$ in the multiplicative definition are multiplicative subagents of $C$.

5. Internalizing-Externalizing Definitions

Next, we have the nonconstructive internalizing-externalizing definition of observables.

Definition: We say that $C$’s agent can observe a finite partition $V$ of $W$ if either $A=\left\{\right\}$ or $C$ is biextensionally equivalent to something in the image of ${\text{External}}_V\circ {\text{Internal}}_V$.

Again, we have a constructive version of this definition.

Definition: We say that $C$’s agent can observe a finite partition $V$ of $W$ if either $A=\left\{\right\}$ or $C\simeq {\text{External}}_V\left({\text{Internal}}_V\right(C\left)\right)$.

Claim: These definitions are equivalent to each other and to the definitions above.

Proof: The case where $A=\left\{\right\}$ is trivial, so we assume that $A$ is nonempty. Clearly if $C$’s agent can observe $V$ under the constructive internalizing-externalizing definition, then $V$ is also observable in $C$ under the non-constructive version.

Next, assume that $C$ is in the image of ${\text{External}}_V\circ {\text{Internal}}_V$ (up to biextensional equivalence). Recall that the image of ${\text{Internal}}_V$ up to biextensional equivalence is exactly those Cartesian frames $(B,F,\star )$ such then $F$ is nonempty and for all $f_0,f_1\in F$ and $b\in B,$ we have $v(b\star f_0)=v(b\star f_1)$. Thus, $C\simeq {\text{External}}_V(B,F,\star )$, where $(B,F,\star )$ is of this form. Let $v_B:B\to V$ send each element $b\in B$ to the unique $v_b\in V$ such that $v(b\star f)=v_b$ for all $f\in F$, and let $V_B$ be the image of $v_B$. Then, ${\text{External}}_V(B,F,\star )=(B/X,X\times F,\diamond )$, where $X=\left\{\right\{b\in B|v_B\left(b\right)=v^{\prime }\}|v^{\prime }\in V_B\}$, and $q\diamond (x,f)=q\left(x\right)\star f$.

Let $V_B=\{v_1,\dots ,v_m\}$, and let $B_i=\{b\in B|v_B(b)=v_i\}$. Then, we clearly have that ${\text{External}}_V(B,F,\star )\cong (B_1\times \cdots \times B_m,V_B\times F,\bullet )$, where $(b_1,\dots ,b_m)\bullet (v_i,f)=b_i\star f$. But this is clearly isomorphic to $D_1\&\dots \&D_m$, where $D_i=(B_i,F,{\star }_i)$, where $b{\star }_{i}f=b\star f$. Thus, $C$’s agent can observe $V$ according to the nonconstructive additive definition of observables.

Finally, we assume that $C$‘s agent can observe $V$ according to the nonconstructive additive definition of observables, and we show that $C$’s agent can observe $V$ according to the constructive internalizing-externalizing definition. Let $C\simeq C_1\&\dots \&C_n$, where $C_i◃{\perp }_{S_i}$. Let $C_i=(A_i,E_i,{\cdot }_i)$, and without loss of generality, let $C=C_1\&\dots \&C_n=(A,E,\cdot )$, where $A=A_1\times \cdots \times A_n$ and $E=E_1\bigsqcup \cdots \bigsqcup E_n$.

First, we show that ${\text{Internal}}_V\left(C\right)\simeq C_1\oplus \cdots \oplus C_n$. Let $C_1\oplus \cdots \oplus C_n=(A_1\bigsqcup \cdots \bigsqcup A_n,E_1\times \cdots \times E_n,\star )$. Observe that (since $A$ is nonempty), ${\text{Internal}}_V\left(C\right)\cong (A\times F,B/F,{\star }^{\prime })$, where $F=\{E_1,\dots ,E_n\}$, where $(a,f)\star q=a\cdot q\left(f\right)$.

We construct

$$\begin{array}{c}(g_0,h_0):(A_1\bigsqcup \cdots \bigsqcup A_n,E_1\times \cdots \times E_n,\star )\to (A\times F,B/F,{\star }^{\prime })\end{array}$$

and

$$\begin{array}{c}(g_1,h_1):(A\times F,B/F,{\star }^{\prime })\to (A_1\bigsqcup \cdots \bigsqcup A_n,E_1\times \cdots \times E_n,\star )\end{array}$$

as follows. Let $g_1\left(\right(a_1,\dots ,a_n),E_i)=a_i$. Let $g_0\left(a_i\right)=\left(\right(a_1,\dots ,a_i,\dots ,a_n),E_i)$, where $a_i\in A_i$, and $a_j\in A_j$ is chosen arbitrarily for $j\ne i$. Let $h_0\left(q\right)=\left(q\right(E_1),\dots ,q(E_n\left)\right)$, and $h_1(e_1,\dots ,e_n)=q$, where $q\left(E_i\right)=e_i$. Clearly, $h_0$ and $h_1$ are inverses.

To see that $(g_0,h_0)$ is a morphism, observe that for all $a_i\in A_1\bigsqcup \cdots \bigsqcup A_n$ and $q\in B/F$, we have

$$\begin{array}{cc}{g}_0\left(a_i\right){\star }^{\prime }q& =\left(\right(a_1,\dots ,a_i,\dots ,a_n),E_i){\star }^{\prime }q\\ & =(a_1,\dots ,a_i,\dots ,a_n)\cdot q\left(E_i\right)\\ & =a_i{\cdot }_{i}q\left(E_i\right)\\ & =a_i\star \left(q\right(E_1),\dots ,q(E_n\left)\right)\\ & =a_i\star h_0\left(q\right),\end{array}$$

where $a_i\in A_i$.

To see that $(g_1,h_1)$ is a morphism, observe that for all $\left(\right(a_1,\dots ,a_n),E_i)\in A\times F$, and for all $(e_1,\dots e_n)\in E_1\times \cdots \times E_n$, we have

$$\begin{array}{cc}{g}_1\left(\right(a_1,\dots ,a_n),E_i)\star (e_1,\dots ,e_n)& =a_i\star (e_1\dots ,e_n)\\ & =a_i{\cdot }_i{e}_i\\ & =(a_1,\dots ,a_n)\cdot e_i\\ & =(a_1,\dots ,a_n)\cdot h_1(e_1,\dots ,e_n)\left(E_i\right)\\ & =\left(\right(a_1,\dots ,a_n),E_i)\star h_1(e_1,\dots ,e_n).\end{array}$$

It is clear that $(g_0,h_0)\circ (g_1,h_1)$ and $(g_1,h_1)\circ (g_0,h_0)$ are both homotopic to the identity, since $h_0\circ h_1$ and $h_1\circ h_0$ are both the identity.

Now, we have that ${\text{Internal}}_V(C_1\&\dots \&C_n)\simeq C_1\oplus \cdots \oplus C_n$, and so we also have dually that ${\text{External}}_V(C_1\oplus \cdots \oplus C_n)\simeq C_1\&\dots \&C_n$. Thus, $C\simeq {\text{External}}_V\left({\text{Internal}}_V\right(C\left)\right)$. $\square$

The thing that is going on here is that when $C$ internalizes $V$, the agent of $C$ then has the full ability to choose how $V$ goes (among ways of $V$ going that were possible in $C$). ${\text{Internal}}_V\left(C\right)$ might have other choices than just choosing how $V$ goes. If it does, then it can freely entangle those other choices with the choice of $V$ however it wants.

When $C$ then externalizes $V$, it loses all control over $V$. However, it preserves the ability to entangle all of its other choices with the way that $V$ goes. This ability for the agent to entangle its choices with $V$ is exactly what it means to say ”$V$ is observable.”

5.1. Example

Let $C_0$ be defined as in the previous examples, with $V=\left\{\right\{ur,nr\},\{us,ns\left\}\right\}$. By the internalizing-externalizing definitions, there exists a frame

${\text{Internal}}_V\left(C_0\right)\cong \begin{array}{cc}& \begin{array}{}\end{array}\\ \begin{array}{c}(u,r)\\ (u,s)\\ (n,r)\\ (n,s)\\ (u\leftrightarrow r,r)\\ (u\leftrightarrow r,s)\\ (u\leftrightarrow s,r)\\ (u\leftrightarrow s,s)\end{array}& \left(\begin{array}{c}ur\\ us\\ nr\\ ns\\ ur\\ ns\\ nr\\ us\end{array}\right)\end{array}$,

which is biextensionally equivalent to

$C_1=\begin{array}{cc}& \begin{array}{}\end{array}\\ \begin{array}{c}ur\\ nr\\ us\\ ns\end{array}& \left(\begin{array}{c}ur\\ nr\\ us\\ ns\end{array}\right)\end{array}$.

We then have that

${\text{External}}_V\left(C_1\right)\cong \begin{array}{cc}& \begin{array}{cc}r& s\end{array}\\ \begin{array}{c}(r\to ur,s\to us)\\ (r\to nr,s\to ns)\\ (r\to ur,s\to ns)\\ (r\to nr,s\to ur)\end{array}& \left(\begin{array}{cc}ur& us\\ nr& ns\\ ur& ns\\ nr& us\end{array}\right)\end{array}$,

which is isomorphic to $C_0$.

This example illustrates both internalizing-externalizing definitions, and also shows the construction used in the constructive definition.

In our next post, we’ll conclude the sequence by showing how to formalize agents that learn and act over time using Cartesian frames.

[Ramana Kumar]

How is this supposed to work (focusing on the $h=h_{f_1}$ claim specifically)?

and so

$$\begin{array}{cc}h\left(b_2^{\prime }\right)& =b_1{\bullet }_{1}h\left(b_2^{\prime }\right)\\ & =r.\end{array}$$

Thus, $(g,h)=(g_{f_1},h_{f_1})$.

Earlier, $h_{f_1}$ was defined as follows:

given by $g_{f_1}\left(b_1\right)=b_1{\cdot }_1{f}_1$ and $h_{f_1}\left(b_2\right)=f_1$

but there is no reason to suppose $f_1=r$ above.

[Rohin Shah]

The problem is a bit earlier actually:

$h\left(b_2^{\prime }\right)=b_1{\bullet }_{1}h\left(b_2^{\prime }\right)$

This isn’t true, because ${\bullet }_1$ doesn’t just ignore $b_1$ here (since $r\in R_1$). I think the route is to say “Let $h\left(b_2^{\prime }\right)=f_2$ . Then ${\bullet }_1$ must treat $f_1$ and $f_2$ identically, meaning that either they are equal, or the frame is biextensionally equivalent to one where they are equal.”

[Ramana Kumar]

Using the idea we talked about offline, I was able to fix the proof—thanks Rohin!
Summary of the fix:
When $D_1$ and $D_2$ are defined, additionally assume they are biextensional (take their biextensional collapse), which is fine since we are trying to prove a biextensional equivalence. (By the way this is why we can’t take $b_1=b_2$, since we might have $A\supseteq B_1\ne B_2\subseteq A$ after biextensional collapse.) Then to prove $h=h_{f_1}$, observe that for all $b\in B_1$, $b{\bullet }_{1}h\left(b_2^{\prime }\right)=b{\bullet }_{1}h\left(b_2\right)$ which means $b{\star }_{1}h\left(b_2^{\prime }\right)=b{\star }_1{f}_1$, hence $h\left(b_2^{\prime }\right)=f_1$ since a biextensional frame has no duplicate columns.

[Ramana Kumar]

this is clearly isomorphic to $D_1\&\dots \&D_m$, where $D_i=(B_i,F,{\star }_i)$, where $b{\star }_{i}f=b\star f$. Thus, $C$’s agent can observe $V$ according to the nonconstructive additive definition of observables.

I think this is only true if $V_B$ partitions $W$, or, equivalently, if $v_B$ is surjective. This isn’t shown in the proof. Is it supposed to be obvious?

EDIT: may be able to fix this by assigning any $s\in V$ that is not in $V_B$ to the frame $\top$ so it is harmless in the product of $D_i$s—I will try this.

[Ramana Kumar]

and observe that ${\text{Assume}}_{S_1\cup S_2}\left(C\right)\cong D_1\&D_2$

This cannot be true. I can prove in general ${Assume}_{S_1\cup S_2}\left(C\right)≆{Assume}_{S_1}\left(C\right)\&{Assume}_{S_2}\left(C\right)$ whenever $|Agent\left(C\right)|>1$ by observing that the agent cardinalities on each side differ.

[Scott Garrabrant]

Yep, changed it to $\simeq$.

[Ramana Kumar]

I haven’t yet figured out why it’s true under $\simeq$ - I’ll keep trying, but let me know if there’s a quick argument for why this holds. (Default next step for me would be to see if I can restrict attention to the world $S_1\cup S_2$ then do something similar to my other comment.)

[Scott Garrabrant]

I am confused, why is it not identical to your other comment?

[Ramana Kumar]

Because $S_1$ and $S_2$ are not a partition of the world here.

EDIT: but what we actually need in the proof is ${Assume}_{S_1}\left(C\right)\&{Assume}_{S_2}\left(C\right)\&\cdots \simeq {Assume}_{S_1\cup S_2}\left(C\right)\&\dots$ where the $\dots$ do result in a partition, so I think this will work out the same as the other comment. I’m still not convinced about biextensional equivalence between the frames without the rest of the product.

[Ramana Kumar]

And it seems we do actually need ${Assume}_{S_1}\left(C\right)\&{Assume}_{S_2}\left(C\right)\simeq {Assume}_{S_1\cup S_2}\left(C\right)$ in the proof to justify:

Thus it suffices to show that $\left(D_1\&1_{R_2\cup R_3}\right)\otimes \left(D_2\&1_{R_1\cup R_3}\right)\simeq D_1\&D_2\&1_{R_3}$.

Without it, we have to show $\left(D_1\&1_{R_2\cup R_3}\right)\otimes \left(D_2\&1_{R_1\cup R_3}\right)\simeq {Assume}_{S_1\cup S_2}\left(C\right)\&1_{R_3}$ instead.

[Ramana Kumar]

UPDATE: I was able to prove ${Assume}_{S_1}\left(C\right)\&{Assume}_{S_2}\left(C\right)\simeq {Assume}_{S_1\cup S_2}\left(C\right)$ in general whenever $S_1$ and $S_2$ are disjoint and both in $Obs\left(C\right)$, with help from Rohin Shah, following the “restrict attention to world $S_1\cup S_2$” approach I hinted at earlier.

[Ramana Kumar]

where $D_1={\text{Assume}}_{S_1}$(C), $D_1={\text{Assume}}_{S_1}\left(C\right)$

Presumably two of those indices should be $2$

[Scott Garrabrant]

Fixed, thanks.

[Ramana Kumar]

Let $C\simeq {\text{Assume}}_{S_1}\left(C\right)\&\dots \&{\text{Assume}}_{S_n}\left(C\right)$. Thus, we also have that $C\simeq {\text{Assume}}_{S_1\cup S_2}\left(C\right)\&{\text{Assume}}_{S_3}\left(C\right)\&\dots \&{\text{Assume}}_{S_n}\left(C\right)$

I’m not seeing why this follows. I’ll look for a counterexample, but in the meantime maybe there’s a simple explanation for why we can combine the product of two assumes as an assume of the union? (I think the only relevant assumption in this context is that the $S_i$s partition the world; but I might be missing some other important assumption.)

EDIT: I can see how maybe this will follow from the definition of observability of a partition from subsets (which we are also assuming) and the fact that $Obs$ is closed under union… will try to figure that out. -- Yep I think this works out. Sorry for the confusion.

[Ramana Kumar]

Let $C_1=(A_1,E_1,{\cdot }_1)$, and let $D_1=(B_1,F_1,{\star }_1)=C_1\otimes \cdots \otimes C_n$

Is $D_1$ supposed to be $C_2\otimes \dots$ here, rather than including $C_1$?

[Scott Garrabrant]

Fixed, thanks.

[Ramana Kumar]

Next, assume that $C$‘s agent can observe $V$ according to the additive definition. We will show that $C$’s agent can observe $S_1$.

I might be misunderstanding this, but the proof suggests you’re actually assuming the assuming definition here, not the additive definition. In which case we may be missing the proof of implication of any of the other definitions from the additive definition.

[Scott Garrabrant]

I think I fixed it. Thanks.

[Ramana Kumar]

Definition: We say that $C$’s agent can observe a finite partition $V$ of $W$ if for all functions $f:V\to A$, there exists an element $a_f\in A$ such that for all $e\in E$, $f\left(v\right(a_f\cdot e\left)\right)\cdot e=a_f\cdot e$.

Claim: This definition is equivalent to the definition from subsets.

This doesn’t hold in the degenerate case $W=\varnothing$, since then we have an empty function $f$ but no elements of $A$. (But the definition from subsets holds trivially.)

[Scott Garrabrant]

This was annoying to fix, so I just made $W$ nonempty in the intro to the post.

[adamShimi]

This is the first post in the sequence that I fully read since the Introduction. So I’m not going to be able to say anything really useful about the proofs. Still, I was curious about the philosophical aspects of these definitions, so I read this post anyway.

That being said, I still think that I understood some part of the definitions, after checking terms from previous posts. My handwavy understanding of your definitions is

• The definitions about subset and conditional policy just rephrase that an observable is something on which the agent can condition it’s own policy. So the agent can observe the partition if it can condition policy on the set of the partition in which it finds itself.

• The additive definitions say that that the agent can observe the partition if the cartesian frame can be decomposed into mutually exclusive cartesian frames, one for each set of the partition, in which the agent acts as if it is in a world of this set.

• The multiplicative definitions say that the agent can observe the partition if the cartesian frame can be decomposed into a product of cartesian frames, one for each subset of the partition, such that the agent is unable to impact the world if it is outside its subset of the partition. The interpretation of the product is that there’s a supervisor agent that control all agents at the same time, and so here, it controls all of them until the observation, after which he morally only controls the one in the observed set (because the other are powerless).

• The internalizing-externalizing definitions say the agent can observe the partition if the cartesian frame can be decomposed into the composition of making the agent able to choose in which set of V it is, but then removing this choice from it, which amount to letting it condition on V, without actually giving it the power to do so.

Is there something really wrong here?

Also, I’m curious if you have an interpretation of the differences between internalizing-externalizing definitions and the others, just like your section on updatelessness compared additive and multiplicative definitions. (Really cool section philosophically, by the way!)

[Scott Garrabrant]

Seems right, except I don’t use the word “product” for the multiplicative definition.

I don’t have much to say about the internalizing-externalizing definition philosophically. One thing to say is that I think the condition that ${\text{External}}_S\left(C\right)$ observes $S$ is a weaker notion of observability, that might actually agree with philosophical intuition more, and the internalizing-externalizing definition might be easier to interpret if you are thinking in terms of this condition.

[Ramana Kumar]

Let $H=\text{hom}(B_1\&1_{R_2\cup R_3},(D_2\&1_{R_1\cup R_3}{)}^{\ast })$

nit: $B_1$ should be $D_1$ here

and let $b_2$ be an element of $b_2$.

and the second $b_2$ should be $B_2$. I think for these $b_1$ and $b_2$ to exist you might need to deal with the $A=\varnothing$ case separately (as in Section 5). (Also couldn’t you just use the same $b$ twice?)

[Ramana Kumar]

Indeed I think the $A=\varnothing$ case may be the basis of a counterexample to the claim in 4.2. I can prove for any (finite) $W$ with $|W|>1$ that there is a finite partition $V$ of $W$ such that $C$’s agent observes $V$ according to the assuming definition but does not observe $V$ according to the constructive multiplicative definition, if I take $C=null.$

[Ramana Kumar]

I presume the fix here will be to add an explicit $A=\varnothing$ escape clause to the multiplicative definitions. I haven’t been able to confirm this works out yet (trying to work around this), but it at least removes the $null$ counterexample.

[Ramana Kumar]

With the other problem resolved, I can confirm that adding an $A=\varnothing$ escape clause to the multiplicative definitions works out.