This is the seventh post in the Cartesian frames sequence.

Here, we introduce three new binary operations on Cartesian frames, and discuss their properties.

1. Tensor

Our first multiplicative operation is the tensor product, $\otimes$.

One way we can visualize our additive operations from before, $\oplus$ and $\&$, is to imagine two robots (say, a mining robot $\text{Agent}\left(C\right)$ and a drilling robot $\text{Agent}\left(D\right)$) that have an override mode allowing an AI supervisor to take over that robot's decisions.

• $C\oplus D$ represents the supervisor deciding which robot to take control of, then selecting that robot's action. (The other robot continues to run autonomously.)
• $C\&D$ represents something in the supervisor's environment (e.g., its human operator) deciding which robot the supervisor will take control of. Then the supervisor selects that robot's action (while the other robot runs autonomously).

$C\otimes D$ represents an AI supervisor that controls both robots simultaneously. This lets $\text{Agent}(C\otimes D)$ direct $\text{Agent}\left(C\right)$ and $\text{Agent}\left(D\right)$ to work together as a team.

Definition: Let $C=(A,E,\cdot )$ and $D=(B,F,\star )$ be Cartesian frames over $W$. The tensor product of $C$ and $D$, written $C\otimes D$, is given by $C\otimes D=(A\times B,\text{hom}(C,D^{\ast }),\diamond )$, where $\text{hom}(C,D^{\ast })$ is the set of morphisms $(g,h):C\to D^{\ast }$ (i.e., the set of all pairs $(g:A\to F,h:B\to E)$ such that $b\star g\left(a\right)=a\cdot h\left(b\right)$ for all $a\in A$, $b\in B$), and $\diamond$ is given by $(a,b)\diamond (g,h)=b\star g\left(a\right)=a\cdot h\left(b\right)$.

Let us meditate for a moment on why this definition represents two agents working together on a team. The following will be very informal.

Let Alice be an agent with Cartesian frame $C=(A,E,\cdot )$, and let Bob be an agent with Cartesian frame $D=(B,F,\star )$. The team consisting of Alice and Bob should have agent $A\times B$, since the team's choices consist of deciding what Alice does and also deciding what Bob does.

The environment is a bit more complicated. Starting from Alice, to construct the team, we want to internalize Bob's choices: instead of just being choices in $A$'s environment, Bob's choices will now be additional options for the team $A\times B$.

To do this, we want to first see Bob as being embedded in Alice's environment. This embedding is given by a function $h:B\to E$, which extends each $b\in B$ to a full environment $e\in E$. We will view Alice's possible environments as being constructed by combining a choice by Bob (that is, a $b\in B$) with a function from Bob's choices to possible environments ($h:B\to E$). Then, we will move the $B$ part across the Cartesian boundary into the agent.

Now, the agent looks like $A\times B$, while the environment looks like $B\to E$. However, we must have been able to do this starting from Bob as well, so a possible environment can also be viewed as function $g:A\to F$.

Since we should get the same world regardless of whether we think of the team as starting with Alice or with Bob, these functions $g$ and $h$ should agree with each other. This looks a bit like currying. The environment for an Alice-Bob team should be able to take in a Bob to create an environment for Alice, and it should also be able to take in an Alice to create an environment for Bob.

1.1. Example

We will illustrate this new operation using a simple formal example.

Jack, Kate, and Luke are simultaneously casting votes on whether to have a party. Each agent can vote for or against the party. The possible worlds are encoded as strings listing which people vote for the party, $W=\{\epsilon ,\text{J},\text{K},\text{L},\text{JK},\text{JL},\text{KL},\text{JKL}\}$. Jack's perspective is given by the frame

$C_J=\left(\begin{array}{cccc}\text{J}& \text{JK}& \text{JL}& \text{JKL}\\ \epsilon & \text{K}& \text{L}& \text{KL}\end{array}\right)$,

Kate's perspective is given by the frame

$C_K=\left(\begin{array}{cccc}\text{K}& \text{JK}& \text{KL}& \text{JKL}\\ \epsilon & \text{J}& \text{L}& \text{JL}\end{array}\right)$,

and Luke's perspective is given by the frame

$C_L=\left(\begin{array}{cccc}\text{L}& \text{JL}& \text{KL}& \text{JKL}\\ \epsilon & \text{J}& \text{K}& \text{JK}\end{array}\right)$.

Since Luke's environment can be thought of as the team consisting of Jack and Kate, one might expect that $C_J\otimes C_K\cong C_L^{\ast }$. Indeed, we will show this is the case.

Let $C_J=(A,E,\cdot )$, and let $C_K=(B,F,\star )$. We label the elements of $A$, $E$, $B$, and $F$ as follows:

$C_J=\begin{array}{cc}& \begin{array}{cccc}{e}_{\epsilon }& e_K& e_L& e_{KL}\end{array}\\ \begin{array}{c}{a}_J\\ a_{\epsilon }\end{array}& \left(\begin{array}{cccc}\text{J}& \text{JK}& \text{JL}& \text{JKL}\\ \epsilon & \text{K}& \text{L}& \text{KL}\end{array}\right)\end{array}$, and $C_K=\begin{array}{cc}& \begin{array}{cccc}{f}_{\epsilon }& f_J& f_L& f_{JL}\end{array}\\ \begin{array}{c}{b}_K\\ b_{\epsilon }\end{array}& \left(\begin{array}{cccc}\text{K}& \text{JK}& \text{KL}& \text{JKL}\\ \epsilon & \text{J}& \text{L}& \text{JL}\end{array}\right)\end{array}$.

We will first enumerate all of the morphisms from $C_J$ to $C_K^{\ast }$. A morphism $(g,h):C_J\to C_K^{\ast }$ consists of a function $g:A\to F$ and a function $h:B\to E$. There are 16 functions from $A$ to $F$ and 16 functions from $B$ to $E$, but most of the 256 pairs do not form morphisms.

Let us break the possibilities into cases based on $g\left(a_J\right)$. Observe that $b_K\star g\left(a_J\right)=a_J\cdot h\left(b_K\right)$: the possible worlds where (from Kate's perspective) Kate votes for the party and Jake-interfacing-with-Kate's-perspective votes for the party too, are the same as the possible worlds where (from Jake's perspective) Jake votes for the party and Kate-interfacing-with-Jake's-perspective does too. These possible worlds must have a $\text{J}$ in them, so $g\left(a_J\right)$ must be either $f_J$ or $f_{JL}$.

If $g\left(a_J\right)=f_J$, then

so $h\left(b_K\right)=e_K$. Similarly,

so $h\left(b_{\epsilon }\right)=e_{\epsilon }$, and

so $g\left(a_{\epsilon }\right)=f_{\epsilon }$.

Similarly, if $g\left(a_J\right)=f_{JL}$, then $h\left(b_K\right)=e_{KL}$, $h\left(b_{\epsilon }\right)=e_L$, and $g\left(a_{\epsilon }\right)=f_L$.

Thus, there are only two candidate morphisms:

• The first, which we will call ${\varphi }_{\epsilon }=(g_{\epsilon },h_{\epsilon })$, is given by $g_{\epsilon }\left(a_{\epsilon }\right)=f_{\epsilon }$, $g_{\epsilon }\left(a_J\right)=f_J$, $h_{\epsilon }\left(b_{\epsilon }\right)=e_{\epsilon }$, and $h_{\epsilon }\left(b_K\right)=e_K$.
• The second, ${\varphi }_L=(g_L,h_L)$, is given by $g_L\left(a_{\epsilon }\right)=f_L$, $g_L\left(a_J\right)=f_{JL}$, $h_L\left(b_{\epsilon }\right)=e_L$, and $h_L\left(b_K\right)=e_{KL}$.

It is easy to see that these are both indeed morphisms, by checking the definition of morphism on all four pairs in $A\times B$.

Thus, $\text{Env}(C_J\otimes C_K)=\{{\varphi }_{\epsilon },{\varphi }_L\}$, and $\text{Agent}(C_J\otimes C_K)=A\times B$, and we can compute $\text{Eval}(C_J\otimes C_K)$ from the definitions of the morphisms. The result is as follows:

$C_J\otimes C_K=\begin{array}{cc}& \begin{array}{cc}{\varphi }_{\epsilon }& {\varphi }_L\end{array}\\ \begin{array}{c}(a_J,b_K)\\ (a_J,b_{\epsilon })\\ (a_{\epsilon },b_K)\\ (a_{\epsilon },b_{\epsilon })\end{array}& \left(\begin{array}{cc}\text{JK}& \text{JKL}\\ \text{J}& \text{JL}\\ \text{K}& \text{KL}\\ \epsilon & \text{L}\end{array}\right)\end{array}$.

This is clearly $C_L^{\ast }$, up to reordering and relabeling rows and columns.

2. Properties of Tensor

Tensor introduces a lot of categorical structure to Chu spaces, in fact giving us a star-autonomous category. This post and the ones to come will be ignoring connections to larger topics in category theory, but only because my time and my familiarity with category theory are limited, not because these connections are unimportant.

I encourage the interested reader to learn more about the structure of Chu spaces on the excellent category theory wiki nLab, beginning with their article on the Chu construction.

2.1. Commutativity, Associativity, and Identity

Claim: $\otimes$ is commutative and associative, and $1$ is the identity of $\otimes$ (up to isomorphism).

Proof: Commutativity is clear from the definition of $\otimes$, once one unpacks the definition of $\text{hom}(C,D^{\ast })$.

To see that 1 is the identity of $\otimes$, let $C=(A,E,\cdot )$, let $1=\left(\right\{b\},W,\star )$, and let $C\otimes 1=(A\times \{b\},\text{hom}(C,1^{\ast }),\diamond )$.

Consider the isomorphism $({\iota }_0,{\iota }_1):C\to C\otimes 1$ given by ${\iota }_0\left(a\right)=(a,b)$ and ${\iota }_1(g,h)=h\left(b\right).$ We need to show that $({\iota }_0,{\iota }_1)$ is a morphism, and that both ${\iota }_0$ and ${\iota }_1$ are bijective. To see that $({\iota }_0,{\iota }_1)$ is a morphism, observe that for all $a\in A$ and $(g,h):C\to 1^{\ast }$,

 Clearly, ${\iota }_0$ is a bijection, so all that remains to show is that ${\iota }_1$ is bijective.

To see that ${\iota }_1$ is injective, observe that if ${\iota }_1(g_0,h_0)={\iota }_1(g_1,h_1)$, then $h_0\left(b\right)=h_1\left(b\right)$, so $h_0=h_1$, and

for all $a\in A$, so $g_0=g_1$.

To see that ${\iota }_1$ is surjective, observe that for every $e\in E$, there exists a morphism $(g_e,h_e):C\to 1^{\ast }$, given by $h_e\left(b\right)=e$ and $g_e\left(a\right)=a\cdot e$. This is clearly a morphism, since 

and ${\iota }_1(g_e,h_e)=e$.

Next, we need to show that $\otimes$ is associative, which will be much more tedious. Let $C_i=(A_i,E_i,\cdot )$. Since we have already established commutativity, it suffices to show that $(C_0\otimes C_1)\otimes C_2\cong (C_0\otimes C_2)\otimes C_1$.

Let $D=(A_0\times A_1\times A_2,F,\star )$, where $F$ is the set of all triples of functions $(g_0:A_1\times A_2\to E_0,g_1:A_0\times A_2\to E_1,g_2:A_0\times A_1\to E_2)$, such that for all $a_i\in A_i$, we have

 and $\star$ is given by

We will show that $(C_0\otimes C_1)\otimes C_2\cong D$, and since the definition of $D$ is symmetric in swapping $C_1$ and $C_2$, it will follow that $(C_0\otimes C_2)\otimes C_1\cong D$, so $(C_0\otimes C_1)\otimes C_2\cong (C_0\otimes C_2)\otimes C_1$.

We construct a morphism $({\iota }_0,{\iota }_1)$ from $(C_0\otimes C_1)\otimes C_2$ to $D$ as follows. ${\iota }_0$ is just the identity on $A_0\times A_1\times A_2$. We will let ${\iota }_1(g_0,g_1,g_2)$ be the morphism $(g_2,h):C_0\otimes C_1\to C_2^{\ast }$, where $h:A_2\to \text{hom}(C_0,C_1^{\ast })$ is given by $h\left(a_2\right)=(h_0^{a_2},h_1^{a_2}):C_0\to C_1^{\ast }$, where $h_0^{a_2}\left(a_0\right)=g_1(a_0,a_2)$, and $h_1^{a_2}\left(a_1\right)=g_0(a_1,a_2)$.

First, we need to show that ${\iota }_1$ is well-defined, by showing that $h\left(a_2\right)$ is a morphism from $C_0$ to $C_1^{\ast }$, and that $(g_2,h)$ is a morphism from $C_0\otimes C_1\to C_2^{\ast }$. To see that $h\left(a_2\right)=(h_0^{a_2},h_1^{a_2})$ is a morphism, observe that for $a_0\in A_0$ and $a_1\in A_1$,

To see that $(g_2,h)$ is a morphism, observe for all $(a_0,a_1)\in A_0\times A_1$ and all $a_2\in A_2$,

where $\diamond =\text{Eval}(C_0\otimes C_1)$.

Now that we know ${\iota }_1$ is well-defined, we need to show that $({\iota }_0,{\iota }_1)$ is a morphism. Indeed, for all $(a_0,a_1,a_2)\in A_0,A_1,A_2$, and for all $(g_0,g_1,g_2)\in F$, we have

where $\bullet =\text{Eval}\left(\right(C_0\otimes C_1)\otimes C_2)$.

Finally, to show that $({\iota }_0,{\iota }_1)$ is an isomorphism, we need to show that ${\iota }_0$ and ${\iota }_1$ are bijective. ${\iota }_0$ is trivial, since it is the identity, so it suffices to show that ${\iota }_1$ is bijective.

To see that ${\iota }_1$ is surjective, let $(g,h)$ be a morphism from $C_0\otimes C_1$ to $C_2^{\ast }$, so $g:A_0\times A_1\to E_2$, and $h:A_2\to \text{hom}(C_0,C_1^{\ast })$. Again, let $h\left(a_2\right)=(h_0^{a_2},h_1^{a_2})$. We will define $(g_0,g_1,g_2)$ by $g_2=g$, $g_1(a_0,a_2)=h_0^{a_2}\left(a_0\right)$, and $g_0(a_1,a_2)=h_1^{a_2}\left(a_1\right)$.

We need to show that $(g_0,g_1,g_2)\in F$, by showing that for all $(a_0,a_1,a_2)\in A_0\times A_1\times A_2$, we have

Observe that since $(g,h)$ is a morphism,

where $\star =\text{Eval}(C_0\otimes C_1)$. Also, by the definition of $C_0\otimes C_1$, we have that

and similarly

Thus,

so $(g_0,g_1,g_2)\in F$. Finally, observe that ${\iota }_1(g_0,g_1,g_2)$ is in fact $(g,h)$.

To show that ${\iota }_1$ is injective, assume ${\iota }_1(g_0,g_1,g_2)={\iota }_1(g_0^{\prime },g_1^{\prime },g_2^{\prime })=(g,h)$, and given an $a_2\in A_2$, let $h\left(a_2\right)=(h_0^{a_2},h_1^{a_2})$. Clearly, this means $g_2=g=g_2^{\prime }$. Further, for all $a_0\in A_0$, $a_1\in A_1$, and $a_2\in A_2$,