# Ω 17

Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.

This is the eighth post in the Cartesian frames sequence. Here, we define new versions of the sum and tensor of Cartesian frames that can delete spurious possible environments from the frame.

## 1. Motivating Examples

The sum  of  and  is supposed to represent an agent that can do anything either  or  can do, while the tensor, , is supposed to represent an agent that can do anything  and  can do working freely together on a team. However, sometimes these operations produce Cartesian frames with more environments than we would expect.

Consider two players, Alice and Bob, in a prisoner's dilemma. We will let  be the space of utilities for Alice. The Cartesian frame for Alice looks like

,

where the top row represents Alice cooperating, and the left column represents Bob cooperating.

Let  represent Alice committed to cooperating, and let  represent Alice committed to defecting. Since the real Alice can either cooperate or defect, one might expect that Alice () would equal the sum () of Alice cooperating with Alice defecting. However,

.

The last two columns are spurious environments that represent Bob copying Alice's move, and Bob doing the opposite of Alice's move. However, since Bob cannot see Alice's move, Bob should not be able to implement these policies: Bob can only choose to cooperate or defect.

Next, consider a unilateralist's curse game where two players each have access to a button that destroys the Earth. If either player pushes the button, the Earth is destroyed. Otherwise, the Earth is not destroyed. , where  represents the world being destroyed and  represents the world not being destroyed.

Here, both players have the Cartesian frame

,

where the first row represents pressing the button, and the first column represents the other player pressing the button.

The two players together can be expressed with the Cartesian frame

,

where the rows in order represent: both players pressing the button; the first player pressing the button; the second player pressing the button; and neither player pressing the button.

One might expect that  would be , but in fact,

.

The second possible environment, in which the earth is just destroyed regardless of what the players do, is spurious.

In both of the above examples, the spurious environments are only spurious because of our interpretation. In the prisoner's dilemma case,  would be correct if Alice and Bob were playing a modified dilemma where Bob can see Alice's choice. In the unilateralist's curse example,  would be correct if there were three people playing the game. The problem is that the  and  operations do not see our interpretation, and so include all possible environments.

## 2. Deleting Spurious Environments

We introduce two new concepts, called sub-sum and sub-tensor, which represents sum and tensor with some (but not too many) spurious environments removed.

Definition: Let , and let . A sub-sum of C and D is a Cartesian frame of the form , where  and  is  restricted to , such that  and , where  is  restricted to  and  is  restricted to . Let  denote the set of all sub-sums of  and .

Definition: Let , and let . A sub-tensor of  and  is a Cartesian frame of the form , where  and  is  restricted to , such that  and , where  and  are given by  and . Let  denote the set of all sub-tensors of  and .

Thus, we define  and  to be sets of Cartesian frames that can be obtained by deleting columns from from  and , respectively, but we have an extra restriction to ensure that we do not delete too many columns.

We will discuss later how to interpret the extra restriction, but first let us go back to our above examples.

If  and , then  has 7 elements:

The 9 Cartesian frames that can be obtained by deleting columns from  that are not in  are:

.

The Cartesian frames above in  are exactly those with all four entries, , and . This is because the extra restriction to be in  is exactly that if you delete the bottom row, you get an object biextensionally equivalent to , and if you delete the top row, you get an object biextensionally equivalent to .

Similarly, from the unilateralist's curse example,

.

It is easy to see that there are no other Cartesian frames in , since there are only four subsets of the two element environment of , and the Cartesian frames corresponding to the other two subsets do not have 1 in their image, so we cannot build anything biextensionally equivalent to  out of them.

Conversely, let  and  both be , and notice that if

,

then  and  are both two-by-two matrices with a single  entry and three  entries, and so must be isomorphic to . Similarly, if

,

then  and  are both two-by-four matrices with a single  entry and seven  entries, and so must by biextensionally equivalent to .

## 3. Properties of Sub-Sums and Sub-Tensors

3.1. Sub-Sums and Sub-Tensors Are Commutative

Claim: For any Cartesian frames  and , there is a bijection between  and  that preserves Cartesian frames up to isomorphism. Similarly, there is a bijection between  and  that preserves Cartesian frames up to isomorphism.

Proof: Trivial.

3.2. Tensors Need Not Be Sub-Tensors

For any Cartesian frames  and  with nonempty environments, we have that . However, sometimes . Indeed, sometimes .

For example, if  and  both have nonempty image, but there are no morphisms from  to , then  has no environments, and it is easy to see that  must be empty.

3.3. Sub-Sums and Sub-Tensors Are Superagents

Claim: For any Cartesian frames  and , and for any , we have  and . Similarly, for any , we have  and .

Proof: Let  and . First, we show  using the currying definition of subagent. Observe . Consider the Cartesian frame  over , where  is given by . Observe that the definition of sub-sum says that , so . Therefore, , and by commutativity, we also have  .

Similarly, we show  using the currying definition of subagent. Observe that . Consider the Cartesian frame  over , where  is given by . Observe that the definition of sub-tensor says that . Therefore, , and by commutativity, we also have

Observe that in the above proof, the Cartesian frame over  we constructed to show that sub-sums are superagents had a singleton environment, and the Cartesian frame over  we constructed to show that sub-tensors are superagents had a surjective evaluation function. The relevance of this observation will become clear later.

3.4. Biextensional Equivalence

As we do with most of our definitions, we will now show that sub-sums and sub-tensors are well-defined up to biextensional equivalence.

Claim: Given two Cartesian frames over  and , and given any , we have that for all  and , there exists a , with .

Proof: Let , let , and let  be an element of , so , and  if , and  if .

The fact that  tells us that for , where  with  given by  and .

For , let  satisfy . Thus, there exist morphisms , and , such that , and  are all homotopic to the identity.

We define a function  by . Then, we define  by , and let , where  if , and  if . We need to show , and that .

To show that , we will construct a pair of morphisms  and  that compose to something homotopic to the identity in both orders. We define  by  if , and  if . We similarly define  by  if , and  if . We define  by , which is clearly a function into , by the definition of . Further,  is surjective, and thus has a right inverse. We choose  to be any right inverse to , so  for all .

To see that  is a morphism, observe that for , and , if  then

To see that  is a morphism, consider an arbitrary  and , and let . Then, if , we have

To see that  is homotopic to the identity on , observe that for all  and , we have that if ,

Similarly, to see that  is homotopic to the identity on , observe that for all  and , we have that if ,

Thus, .

To see , we need to show that , where  with  given by  and . It suffices to show that , since .

For , we construct morphisms , and . We define , and .

To see that  is a morphism, observe that for all  and , we have

and to see that  is a morphism, observe that for all , and , we have

To see  is homotopic to the identity on , observe that for all  and , we have

and similarly, to show that  is homotopic to the identity on , observe that for all  and , we have

Thus, we have that , completing the proof.

We have a similar result for sub-tensors, whose proof directly mirrors the proof for sub-sums:

Claim: Given two Cartesian frames over  and , and any , we have that for all  and , there exists a , with .

Proof: Let , let , and let  be an element of , so , and

The fact that  tells us that for , where  with  given by

For , let  satisfy . Thus, there exist morphisms , and , such that , and  are all homotopic to the identity.

We define a function  by . This function is well-defined, since  and .

Then, we define  by , and let , where

We need to show that , and that .

To show that , we will construct a pair of morphisms  and  that compose to something homotopic to the identity in both orders. We define  by , and we similarly define  by . We define  by , which is clearly a function into , by the definition of . Further,  is surjective, and thus has a right inverse. We choose  to be any right inverse to , so  for all .

To see  is a morphism, observe that for , and , we have

To see that  is a morphism, consider an arbitrary  and , and let . Then, we have:

To see that  is homotopic to the identity on , observe that for all  and , we have:

Similarly, to see that  is homotopic to the identity on , observe that for all  and , we have:

Thus, .

To see , we need to show that , where  with  given by

It suffices to show that , since .

For , we construct morphisms , and . We define  and . We define  by , and similarly define  by .

To see that  is a morphism, observe that for all  and , we have:

To see that , and  are morphisms is similar.

To see  is homotopic to the identity on , observe that for all  and , we have

and seeing that , and  are homotopic to the identity is similar.

Thus, we have that , completing the proof.

In our next post, we will use sub-sum and sub-tensor to define additive subagents, which are like agents that have committed to restrict their class of options; and multiplicative subagents, which are like agents that are contained inside other agents. We will also introduce the concept of sub-environments.

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