Additive and Multiplicative Subagents

by Scott Garrabrant12 min read6th Nov 20207 comments

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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