# Sub-Sums and Sub-Tensors

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*.

- Additive and Multiplicative Subagents by 6 Nov 2020 14:26 UTC; 20 points) (
- 5 Nov 2020 18:41 UTC; 6 points) 's comment on Multiplicative Operations on Cartesian Frames by (

Can subsums and subtensors be defined as diagrams? Tensors not needing to be subtensors sounds to me like subsums/subtensors should be fronter-row citizens than sums/tensors.

In the first claim of 3.4, the last bit of the claim is D′∈C