Additive and Multiplicative Subagents

This is the ninth post in the Cartesian frames sequence. Here, we refine our notion of subagent into additive and multiplicative subagents. As usual, we will give many equivalent definitions.

The additive subagent relation can be thought of as representing the relationship between an agent that has made a commitment, and the same agent before making that commitment. The multiplicative subagent relation can be thought of as representing the relationship between a football player and a football team.

Another way to think about the distinction is that additive subagents have fewer options, while multiplicative subagents have less refined options.

We will introduce these concepts with a definition using sub-sums and sub-tensors.

1. Definitions of Additive and Multiplicative Subagent

1.1. Sub-Sum and Sub-Tensor Definitions

Definition: is an additive subagent of , written , if there exists a and a with .

Definition: is a multiplicative subagent of , written , if there exists a and with .

These definitions are nice because they motivate the names “additive” and “multiplicative.” Another benefit of these definitions is that they draw attention to the Cartesian frames given by C′. This feature is emphasized more in the below (clearly equivalent) definition.

1.2. Brother and Sister Definitions

Definition: is called a brother to in if for some . Similarly, is called a sister to in if for some .

E.g., one “sister” of a football player will be the entire rest of the football team. One “brother” of a person that precommitted to carry an umbrella will be the counterfactual version of themselves that instead precommitted to not carry an umbrella.

This allows us to trivially restate the above definitions as:

Definition: We say if has a brother in and if has a sister in .

Claim: This definition is equivalent to the ones above.

Proof: Trivial.

1.3. Committing and Externalizing Definitions

Next, we will give the committing definition of additive subagent and an externalizing definition of multiplicative subagent. These definitions are often the easiest to work with directly in examples.

We call the following definition the “committing” definition because we are viewing as the result of making a commitment (up to biextensional equivalence).

Definition: Given Cartesian frames and over , we say if there exist three sets , , and , with , and a function such that and , where and are given by and .

Claim: This definition is equivalent to the sub-sum and brother definitions of .

Proof: First, assume that has a brother in . Let , and let . Let be brother to in . Let be such that and . Then, if we let , let , let , and let , we get , where , and by the definition of sub-sum, , where .

Conversely, let , , and be arbitrary sets with , and let . Let , and let , where and . We want to show that has a brother in . It suffices to show that has a brother in , since sub-sum is well-defined up to biextensional equivalence. Indeed, we will show that is brother to in , where is given by .

Observe that , where is given by

if , and is given by

otherwise. Consider the diagonal subset given by . Observe that the map is a bijection from to . Observe that if we restrict to , we get given by . Thus with the isomorphism coming from the identity on , and the bijection between and .

If we further restrict to or , we get and respectively, given by and . Thus and , with the isomorphisms coming from the identities on and , and the bijection between and .

Thus , and , so is brother to in , so has a brother in .

Next, we have the externalizing definition of multiplicative subagent. Here, we are viewing as the result of sending some of its decisions into the environment (up to biextensional equivalence).

Definition: Given Cartesian frames and over , we say if there exist three sets , , and , and a function such that and , where and are given by and .

Claim: This definition is equivalent to the sub-tensor and sister definitions of .

Proof: First, assume that has a sister in . Let , and let . Let be sister to in . Let be such that and . Then, if we let , let , let , and let

we get , where , and by the definition of sub-tensor, , where .

Conversely, let , , and be arbitrary sets, and let . Let , and let , where . We will assume for now that at least one of and is nonempty, as the case where both are empty is degenerate.

We want to show that has a sister in . It suffices to show that has a sister in , since sub-tensor is well-defined up to biextensional equivalence. Indeed, we will show that is sister to in , where is given by .

Observe that , where is given by

For every , there is a morphism , where is given by , and is given by . This is clearly a morphism. Consider the subset given by . Observe that the map is a bijection from to . (We need that at least one of and is nonempty here for injectivity.)

If we restrict to , we get given by . Thus, , with the isomorphism coming from the identity on , and the bijection between and .

To show that , we need to show that and , where and are given by

Indeed, and , so and , with the isomorphisms coming from the identities on and , and the bijection between and .

Thus , and , so is sister to in , so has a sister in .

Finally, in the case where and are both empty, , and either or , depending on whether is empty. It is easy to verify that , since , taking the two subsets of the singleton environment in yields and as candidate sub-tensors, and both are valid sub-tensors, since either way, the conditions reduce to .

Next, we have some definitions that more directly relate to our original definitions of subagent.

1.4. Currying Definitions

Definition: We say if there exists a Cartesian frame over with , such that .

Claim: This definition is equivalent to all of the above definitions of .

Proof: We show equivalence to the committing definition.

First, assume that there exist three sets , , and , with , and a function such that and , where and are given by and .

Let , and let and compose to something homotopic to the identity in both orders.

We define , a Cartesian frame over , by , where is given Observe that , where is given by

To show that , we construct morphisms and that compose to something homotopic to the identity in both orders. Let and both be the identity on . Let be given by , and let be given by .

We know is a morphism, since for all and , we have

We also have that is a morphism, since for all and , we have

Observe that and clearly compose to something homotopic to the identity in both orders, since and are the identity on .

Thus, , and .

Conversely, assume , with . We define and . We define by , where .

Let be given by . Since , we have . Thus, . Unpacking the definition of , we get , where is given by , which is isomorphic to , where is given by . Thus and , as in the committing definition.

Definition: We say if there exists a Cartesian frame over with , such that .

Claim: This definition is equivalent to all of the above definitions of .

Proof: We show equivalence to the externalizing definition.

First, assume there exist three sets , , and , and a function such that and , where and are given by .

Let , and let and compose to something homotopic to the identity in both orders.

We define , and we define , a Cartesian frame over , by , where is given by if and , and otherwise. Clearly, , since for any , if we let , we have .

Observe that for all , , and , if and , then

and on the other hand, if or , we also have .

Thus, we have that , where is given by

To show that , we construct morphisms and that compose to something homotopic to the identity in both orders. Let and both be the identity on . Let be given by , and let be given by .

We know is a morphism, since for all and ,

We also have that is a morphism, since for all and , we have

Observe that and clearly compose to something homotopic to the identity in both orders, since and are the identity on .

Thus, , where .

Conversely, assume , with . Let , let , and let . Let be given by , where and .

Thus , where is given by . All that remains to show is that , where . Let .

We construct morphisms and that compose to something homotopic to the identity in both orders. Let and be the identity on . Let be given by . Since is surjective, it has a right inverse. Let be any choice of right inverse of , so for all .

We know is a morphism, since for all and ,

To see that is a morphism, given and , let , and observe

and clearly compose to something homotopic to the identity in both orders, since and are the identity on . Thus , completing the proof.

Consider two Cartesian frames and , and let be a frame whose possible agents are and whose possible worlds are . When is a subagent of , (up to biextensional equivalence) there exists a function from , paired with , to .

Just as we did in “Subagents of Cartesian Frames” §1.2 (Currying Definition), we can think of this function as a (possibly) nondeterministic function from to , where represents the nondeterminism. In the case of additive subagents, is a singleton, meaning that the function from to is actually deterministic. In the case of multiplicative subagents, the (possibly) nondeterministic function is surjective.

Recall that in “Sub-Sums and Sub-Tensors” §3.3 (Sub-Sums and Sub-Tensors Are Superagents), we constructed a frame with a singleton environment to prove that sub-sums are superagents, and we constructed a frame with a surjective evaluation function to prove that sub-tensors are superagents. The currying definitions of and show why this is the case.

1.5. Categorical Definitions

We also have definitions based on the categorical definition of subagent. The categorical definition of additive subagent is almost just swapping the quantifiers from our original categorical definition of subagent. However, we will also have to weaken the definition slightly in order to only require the morphisms to be homotopic.

Definition: We say if there exists a single morphism such that for every morphism there exists a morphism such that is homotopic to .

Claim: This definition is equivalent to all the above definitions of .

Proof: We show equivalence to the committing definition.

First, let and be Cartesian frames over , and let be such that for all , there exists a such that is homotopic to . Let .

Let , let , and let . Let be given by . We already have , and our goal is to show that where is given by .

We construct and that compose to something homotopic to the identity in both orders.

We define by . is surjective, and so has a right inverse. We let be any right inverse to , so for all . We let be given by .

Defining will be a bit more complicated. Given an , let be the morphism from to , given by and . Let be such that is homotopic to . We define by .

We trivially have that is a morphism, since for all and ,

To see that is a morphism, consider and , and define and as above. Then,

We trivially have that is homotopic to the identity, since is the identity on . To see that is homotopic to the identity on , observe that for all and , defining and as above,

Thus , and according to the committing definition.

Conversely, let , , and be arbitrary sets with , let , and let and , where and are given by and .

Let and compose to something homotopic to the identity in both orders, and let and compose to something homotopic to the identity in both orders. Let be given by is the embedding of in and is the identity on . is clearly a morphism.

We let .

Given a , our goal is to construct a such that is homotopic to .

Let , let , and let . Let be given by . Let be given by . This is clearly a morphism, since for all and ,

To see that is homotopic to , we just need to check that is a morphism. Or, equivalently, that , since is the identity, and .

Indeed, for all and ,

Thus is homotopic to , completing the proof.

Definition: We say if for every morphism , there exist morphisms and such that , and for every morphism , there exist morphisms and such that .

Before showing that this definition is equivalent to all of the above definitions, we will give one final definition of multiplicative subagent.

1.6. Sub-Environment Definition

First, we define the concept of a sub-environment, which is dual to the concept of a sub-agent.

Definition: We say is a sub-environment of , written , if .

We can similarly define additive and multiplicative sub-environments.

Definition: We say is an additive sub-environment of , written , if . We say is an multiplicative sub-environment of , written , if .

This definition of a multiplicative sub-environment is redundant, because the set of frames with multiplicative sub-agents is exactly the set of frames with multiplicative sub-environments, as shown below:

Claim: if and only if .

Proof: We prove this using the externalizing definition of .

If , then for some , , , and , we have and , where and are given by and .

Observe that and , where and are given by and . Taking , , , and , this is exactly the externalizing definition of , so .

Conversely, if , then , so .

We now give the sub-environment definition of multiplicative subagent:

Definition: We say if and . Equivalently, we say if and .

Claim: This definition is equivalent to the categorical definition of .

Proof: The condition that for every morphism , there exist morphisms and such that , is exactly the categorical definition of .

The condition that for every morphism , there exist morphisms and such that , is equivalent to saying that for every morphism , there exist morphisms and such that . This is the categorical definition of .

Claim: The categorical and sub-environment definitions of are equivalent to the other four definitions of multiplicative subagent above: sub-tensor, sister, externalizing, and currying.

Proof: We show equivalence between the externalizing and sub-environment definitions. First, assume that and are Cartesian frames over with and .

We define , , and . We define