quotient agents i

okay fine i'll divide posts into parts but they're still going to be 2000+ words each

Cruelly, Tetraspace is forcing me to figure out what quotient agents are. So we’re going to work through that, together. First of all: why did I even find the idea of promising a quotient agent post funny? Today we will address that question, and tomorrow we will try to actually figure out what a quotient agent is.

Let’s get this idea down. We will begin with subobjects. What’s your favorite set? Mine is X = {a, b, A, B}. That’s a good set. An example of a subset of X is {a, B}. The set X has sixteen subsets, from the empty set, to the entirety of X, to the four singletons, six unordered pairs, and four unordered triples. In general, this number is given by

because to determine a subset you get to pick, for each element of X, whether it is included in the subset.

X was a boring favorite set. My new favorite set is Z. Some examples of subsets of Z are the odd integers, the even integers, and the threeven integers.

You understand subsets, those are easy.

A (binary) relation on X is a function

where if R(a, b) = 1, we say a ~ b (pronounced “a is related to b (under the relation R)”), and if R(a, b) = 0, we say a ≁ b (pronounced “a is not related to b (under the relation R)”). An equivalence relation is furthermore reflexive, symmetric, and transitive. An example of an equivalence relation on X is the relation where a ~ A and b ~ B, but a ≁ b, a ≁ B, A ≁ b, and A ≁ B—they’re related if they’re the same letter, but not if they’re different letters.

A partition of X is a collection of nonempty subsets of X such that each element of X appears in exactly one of the subsets. An example of a partition of X is { {a, A}, {b, B} }. If you look closely, you might notice a certain sort of similarity in character between this example of a partition and our earlier example of an equivalence relation. This is because, in this example, we partitioned X by splitting it into the equivalence classes of X under that relation—or, equivalently, because we said that two elements of X are related if and only if they are elements of the same subset in our partition. In general, every equivalence relation has an associated partition and every partition has an associated equivalence relation—they are, essentially, the same concept seen from different angles.

The set X has fifteen partitions and fifteen equivalence relations, which can (as just described) be placed into a natural one-to-one correspondence. According to Wolfram Mathworld, this number is in general given by

where 𝕴 is the imaginary part function, which is given by 𝕴(a+bi) = b¹¹ Or at least, it’s given by that if a and b are real. If they’re complex, that should be expanded to 𝕴(a+bi) = 𝕴(a) + 𝕽(b), where 𝕽 is the real part function, which is given by 𝕽(a+bi) = a when a and b are real or 𝕽(a+bi) = 𝕽(a) - 𝕴(b) when they’re complex..

The astute among you may, upon inspecting this formula, notice that

This is because the formula is a product where one of the factors is the imaginary part of a real number. As someone who took a Complex Analysis once, I suspect that there’s probably meant to be an i in one or more of the exponents, but that is the only answer I have for you.

Examples of partitions of Z include {evens, odds} and {threevens, unthreevens}.

Okay, let’s bring back our equivalence relation ~²² Earlier I called the relation R and had ~ be a separate symbol but I don’t care about that anymore you can tell what I mean probably. on X whose equivalence classes are given by the partition P = { {a, A}, {b, B} }. We are going to introduce a piece of corresponding notation and terminology that, it seems to me, might be a little odd initially. We write

and we say “the quotient set of X (under ~) is P”, or sometimes “X modulo ~ is P”.

I think there would be no reason to use this terminology if we were just concerned with sets—it’s just that understanding what’s going on with the sets makes other cases easier. Let’s go back to Z.

So far we have more or less just been thinking of Z as a set. I am now going to ask you to think of it as a (commutative) rng. This means it has two binary operations, + and •, where (Z, +) is an abelian group, (Z, •) is a (commutative) semigroup (that is, an associative magma), and • distributes over +. A rng is exactly like a ring³³ Examples of rings include Z, Q, R, H (but not O, which is merely a not-necessarily-associative ring (usually “nonassociative ring” for short but that’s technically incorrect unless you actually add an axiom necessitating nonassociativity instead of just dropping the associativity axiom)), every field (as a field is just a commutative division ring), matrices, Clifford algebras, and the singleton with two binary operations. The singleton with two binary operations is maybe more commonly known as the zero ring 0 = {0}. except that a ring must have a multiplicative identity (usually denoted “1”), I am calling it a rng entirely so that I can get away with calling things subrngs even if they don’t include one. It doesn’t matter much if you don’t have the idea of a rng down perfectly, the essential point here is mostly that Z has some additional structure beyond being just a set.

Okay. Earlier we talked about various subsets of the integers: the even numbers, the odd numbers, the threeven numbers, and the unthreeven numbers. Some of these, but not others, are furthermore subrngs of the integers. Let’s explain with the example of the evens.

First, we are going to start writing the evens as 2Z, because

I think this notation basically makes sense. Now, there are three crucial things you need to check: first, you need to check that adding or multiplying two even numbers still gives you an even number. Second, you need to check that we didn’t, for example, include 2 but forget to include -2⁴⁴ The positive evens are merely a rg. Or maybe a rg still has to have zero, so the nonnegative evens are a rg, but the positive rings are actually only a semiring.. Third, at least one even number exists⁵⁵ There’s an argument that it would be more natural to phrase this as “the evens contain the additive identity”, because in a sense the point of this condition is that the empty set fails to be a group under addition, and the reason it fails to be a group is the lack of identity.

If we merely asked for addition to be a commutative associative quasigroup, then the empty rng would be permissible. But nothing else would change, because nonempty associative quasigroups are necessarily groups.. Addition isn’t going to suddenly stop being commutative or something, so as long as we check those two things, we know 2Z is a subrng.

On the other hand, the odd integers 2Z+1 are not a subrng of the integers, because 1+1 = 2 and therefore + isn’t a binary operation on the odd integers. You can’t look at the odd integers all on their own, and forget the even integers, and start adding them together and multiplying. But if you’re looking at the even integers, you can get away with forgetting the odds—because 2Z is a subrng of Z, and 2Z+1 isn’t. Similarly, the threevens 3Z are a subrng, but the unthreevens 3Z±1 aren’t.

Much like quotient sets, there is also an idea of a quotient rng. And like how 2Z is a subrng but 2Z+1 isn't (even though they’re both subsets of Z), most partitions of Z aren’t quotient rngs.

Let’s start with {2Z, 2Z+1}, which is the partition corresponding to the parity relation ≡₂, where a≡₂b iff a and b have the same parity⁶⁶ This is equivalent to saying that a≡₂b iff 2 is a factor of (a-b), which might be helpful if you’re constructing math from the ground up and don’t have a definition of “parity” on hand. (either both are even or both are odd). So, when we’re looking at a prospective subrng, we have a collection of elements of the ring, and it’s obvious how to add them together—we just might get a sum that isn’t an element of the prospective subrng.

There are a few ideas that all seem to me like they produce the operations which I already know are the ones we want. We could define set addition with something like

and get the right addition. But that approach doesn’t quite work for multiplication, since it would give the result that 2Z•2Z = 4Z. But we can notice that 4Z is a subset of 2Z (and therefore is disjoint with 2Z+1), and then say that, for the purposes of putting our operations on our partition, that’s good enough.

The more normal way to define this is to define the notation [a]₂ := {n∈Z : n≡₂a}, where [a]₂ is read as “the equivalence class of a (under ≡₂)”. And then we can just say [a]₂+[b]₂=[a+b]₂ and [a]₂•[b]₂=[a•b]₂. To make sure this definition works for a given equivalence relation, you need to check that [a+b]₂ and [a•b]₂ are always the same set, regardless of which a and b we choose from a given equivalence class—which is an equivalent condition to the subset thing we noticed in the previous paragraph.

Now, maybe this part will help explain the “quotient” term: we don’t, generally, write this quotient rng as Z/≡₂. We write it as

which is usually read “Z mod 2Z” or “the integers modulo 2”. This is sort of like division in some ways, which is where the notation and the term “quotient” comes from. It might help a little if I say the term “remainder”.

Now, {3Z, 3Z±1}, while a perfectly valid partition, does not form a quotient rng. Because 1 and -1 are equal to 3•0+1 and 3•0-1 respectively, so we have 1-1=0=3•0, which would mean we have to have (3Z±1) + (3Z±1) = 3Z. But we also have 1+1=2=3•1-1, so we would have to have (3Z±1) + (3Z±1) = 3Z±1, and we can’t have both. If you want a quotient rng, you need to use the partition {3Z-1, 3Z, 3Z+1}.

It turns out that there is a one-to-one equivalence between subrngs⁷⁷ More specifically, these subrngs are ideals, which satisfy the slightly stronger condition of being closed under multiplication by any element of the rng, not just the elements in the subset as a subrng requires. This happens to not make any difference if the rng is Z, but you’ll run into issues if you (for example) try to put a multiplication operation on Q/Z, even though Z is a subrng of Q. (Q/Z is a perfectly fine quotient group under addition though).

In group theory, the analogous notion to an ideal is a normal subgroup, which is a little trickier to define—except if you only care about the abelian case, as all subgroups of abelian groups are normal. I’m not sure why ideals aren’t just called normal subrngs—or actually, I do know the etymology, I just think “normal subrng” is more fitting.

The normal subobjects are generally supposed to be the ones you can quotient by. nZ⁸⁸ You can assume n is a natural without losing much, because nZ = (-n)Z. and quotient rngs Z/nZ, and that Z/nZ is the only quotient ring which has nZ as an element, which is part of why the notation Z/nZ works. These are also the only⁹⁹ Don’t forget Z/0Z = Z/0 = Z and Z/1Z = Z/Z = 0, my beloveds.

…Something about “Z/0 = Z” and “Z/Z = 0” might look very wrong to you, but just firmly tell yourself that 0=1 in 0 and everything will make sense.

Do you want something to be angry about? {0} is a subset of Z that is a (unital) ring, but it’s not a subring of Z, because—what’s the funniest way to describe this as working which isn't obviously wrong—{0} fails to be closed under Z’s multiplication operation, because the empty product in Z is 1, whereas in 0, the empty product is 0. subrngs and quotient rngs¹⁰¹⁰ Unless you want to do stuff like bespokely considering Q a quotient ring of Z on the basis that the inclusion of Z into Q is an epimorphism even though it’s not a surjection.

Or on the dumber basis that the elements of Q are quotients of elements of Z. Like, that is why we call it Q… of Z.

I hope that this point, where there is a clear correspondence between subrngs and quotient rngs of Z, helps a least a little to convey the intuition that subobjects and quotient objects are somehow dual notions to each other. They’re basically opposites.

Like, you can see A being a subobject of B as basically meaning that there’s a structure-preserving injection ι from A to B, if you’re willing to identify a with ι(a)—which you should be, since it’s injective, which means A and ι(A) are isomorphic. And you can see A being a quotient object of B as basically meaning that there’s a structure-preserving surjection q from B to A, if you’re willing to identify [b]_~ with q(b), where ~ is the equivalence relation where b₁ ~ b₂ iff q(b₁) = q(b₂), which you should be, since q is surjective and therefore (when reinterpreted in the natural manner¹¹¹¹ Where for a subset X of B, we define q(X) := {q(x) : x∈X}, and we say q(X) is the image of X under q. as a function q: 𝓟(B) → 𝓟(A)) is an isomorphism between the equivalence classes of ~ and the one-element subsets of A. And the one-element subsets of A is basically just A, come on, just use the isomorphism a ↦ {a}.

What I’m saying is that they’re opposites. Which means it’s really funny to suggest that there should be some notion of a quotient agent dual to the common folk notion of a subagent, even though subagents aren’t¹²¹² Usually, cf the subagents post a precise mathematical concept.

Actually figuring out what that notion is, I am leaving for tomorrow. Hopefully Celene or whoever at least sort of gets why that’s supposed to be funny now.