groups-rings-fields

[tender wharf]

0 polynomial has 0 constant term

[chilly radish]

A is algebraic so Q(A)=Q[A]

You don't need to think about rational functions

Just polynomials

[lone niche]

I know. Look at the proof they wanted to understand, which is why I mentioned rational functions.

[forest canopy]

As an additional challenge (maybe you already know this): Try proving that the kernel of any ring homomorphism is an ideal, it's a very useful result for questions like these. (For your example just regard the homomorphism mapping any polynomial to it's constant part)

well i guess your example is not too difficult to do directly but it's still useful to know lol

[tender wharf]

conversely every ideal is the kernel of some ring homomorphism (there's an obvious choice) so that the study of ideals is really the study of ring homomorphisms

[arctic dune]

not sure if Lie algebras go in here, but I think it fits better than advanced algebra

the second image is the official answer for (ii)

Why does $\rho (h)$ having three distinct eigenvalues necessarily mean that $V\cong V_2$?

[cloud walrus]

thou_art_an_egg

[arctic dune]

I understand that with three distinct eigenvalues, its eigenspace will have dimension 3 and so its eigenvectors span $V$, but I don't see why this necessarily implies $V\cong V_2$.

[cloud walrus]

thou_art_an_egg

[chilly radish]

Oops, my bad

[tidal torrent]

hey random question but

would U(R) necessarily equal R

let me pull up the question rq

specifically question #22 (very bottom)

[lone niche]

Why do you suppose U(R)=R?

[tidal torrent]

if R is a commutative ring then wouldnt that imply that U(R) is just all of R

[lone niche]

I think there might be some confusion about the definition of unit

Do you remember the definition of unit?

Is 0 a unit?

[tidal torrent]

dont think 0 is

[lone niche]

Now consider the integers Z. Z is commutative. Is 2 a unit?

[tidal torrent]

it would be 1 and -1 right

wait actually

[tidal torrent]

would a unit mean an inverse under two sets of binary operations?

i havent covered much ring theory

[lone niche]

For multiplication.

[tidal torrent]

oh ok

[crystal vale]

I want to show that if in the abelian group, a and b has order m and n then there is element of order lcm (m,n).

Now first I showed that if gcd(m,n) = 1 then it is true.

Let a and b have the order m and n respectively such that m and n are relative prime.

Now ab has a finite order because of abelian we can show it.

Now let ab have order t then t divides mn.
and (ab)^t = e implies b^t = a^-t. But | a^-t| divides both m and n, therefore a^-t = e, and b^t =e.

Thus, m and n divides t thus mn divides t.
It shows that t = mn.

Yes maybe it is not so good proof.

Now I want to show it in general case, if a and b has order m and n and their gcd is d.

Then (a^d)(b^d) has order mn/(d^2)

And a^(m/d) has order d

Any idea?

[crystal vale]

To prove that infinite group is cyclic if and only if it is isomorphic to every proper subgroup. Here {e} is not a proper subgroup.

First direction: let G be an infinite cyclic group then G is isomorphic to Z then we can use the structure of subgroups of Z.

Second direction: So G is infinite then if there is g such that < g> = G then we are done.

Let there be g≠e in G such that < g > ≠ G so and since G is isomorphic to its proper subgroup so G is isomorphic to < g >, then by isomorphism G is infinite cyclic group

[crystal vale]

if p is prime then the non-zero elements of Z/pZ form a group of order p-1 under multiplication.

I know it makes a group but how can I show that it makes a group of order p-1?

[topaz solar]

Count them

[crystal vale]

Actually question is if a is relatively prime to p then | < a > | = p-1

[topaz solar]

Yes, the units are a cyclic group

This excludes a = 1 of course

[crystal vale]

But that comes from field theory

[topaz solar]

So?

[crystal vale]

No I don't want to use field theory here

[topaz solar]

But that’s exactly what you’re doing

The units are the things which make a group under multiplication

[crystal vale]

Yes

[tough raven]

That's false; you can take a = 1 for example. If you mean that |<a>| divides p-1, you can use Lagrange's theorem for groups or Fermat's little theorem to conclude that. If you mean that |<a>| = p-1 for some a, well then, you need that result (existence of primitive root modulo p) whether you count it as field theory or number theory or anything else.

[tough raven]

This is kind of cheating, but: for every prime p dividing lcm(m, n), let k_p, l_p be the number of times that p divides m and n (i.e. p^k_p divides m but p^{k_p+1} does not, etc.). Then lcm(m, n) = \prod_p p^{max(k_p, l_p)}. Now, |a^{m/p^k_p}| = p^k_p and |b^{n/p^l_p}| = p^l_p, so we can find a group element x_p of order p^{max(k_p, l_p)}.
You can now apply your proof of the coprime case (inductively) to \prod_p x_p.

[topaz solar]

I’d say calling 1 coprime is kinda

[tough raven]

I'd say you have to do it to be consistent.

The set of common divisors of 1 and p is {-1, 1}. The ideal generated by 1 and p is the principal ideal (1). The set of numbers coprime to p wouldn't be invariant under the "equal modulo p" equivalence relation if 1 wasn't included.

[tough raven]

But in fairness, you can take a = -1 instead (unless p = 2, 3 - incidentally I now realise I was wrong and |a| = p-1 for all a if p = 2).

[topaz solar]

Truly a small number moment

[crystal vale]

Yes it is wrong because group of units of Z/pZ is isomorphic to Z/(p-1)Z

[crystal vale]

So for every prime p we get an element such that p^{ max(k_p, l_p)} now since we had proof for co-prime case so we can do for distinct primes.

Thank you ❤️

If H and K are subgroups of group G, then [H v K: H ]≥[K: H intersection K].

They define H v K as a subgroup generated by H union K.

Now if I define mapping between set of all right coset of H and intersection K in K and set of right cosets of H in H v K.
Let f be the mapping such that ( H and K)k-> (H)k.

It is well defined because if ( H and K )k_1 = (H and K)k_2 implies k_1k_2^-1 in H implies Hk_1 = Hk_2.

Now it is injective, if (H)k_1 = (H)k_2 then k_1k_2^-1 in H implies k_1k_2^-1 in H and K therefore (H and K)k_1 = (H and K)k_2.

Hence because of injective [K: H and K]≤[H v K : H].

Is it correct?

[tough raven]

Yes! In fact, you can exactly characterise the range of the mapping f and prove that [K : H \cap K] = [HK : H], where HK = {hk | h in H, k in K} (this may just be a subset of G, but it's a union of right cosets of H, so "number of right cosets of H in HK" is well-defined and that's what I mean by [HK : H] on the RHS).

[crystal vale]

Yes cardinality of range is [HK : H], thank you ❤️

If I have a finite generator of a group and there is some relation between it, what are the ways to count the element of that group?

[tough raven]

You mean something like taking a group with presentation <g, h, k | g^3 = h^2 = gkghk^{-1}> and computing its order?

I don't think there are any general strategies for doing this. In fact, it's undecidable whether such a group is trivial (i.e. you cannot even in theory write a computer program to take a presentation and even figure out whether the group is trivial, much less its order).

[knotty badger]

Yeah it’s like… there are some special cases that you can do this for, but definitely no general strategy

[topaz solar]

good luck even verifying it's finite

[knotty badger]

Presentations help you define group homs out of a group

But not really into a group

[crystal vale]

Yes

[topaz solar]

I'd say its good for maps into a group comparing generators though

anyhow

[tough raven]

https://en.wikipedia.org/wiki/Group_isomorphism_problem this is mentioned here for example (I thought there was a Wikipedia article just about deciding whether a group is trivial but I guess not 🤷).

[topaz solar]

word problem

[knotty badger]

Wait wdym

[tough raven]

Not quite the same thing. That asks whether two words in the generator are equal. Triviality would require a yes/no answer in finite time to whether all pairs of words are equal.

[crystal vale]

So if I have some relation between generators then I need to do computation

[topaz solar]

something something homomorphisms are exactly the maps preserving, but not necessarily reflecting, atomic formulae

[knotty badger]

Um

What does that mean….?

[crystal vale]

Actually the question hint that group is isomorphic to Q_8 but what if there is no hint then ?

[tough raven]

An explicit group with unsolvable word problem 👀

[tough raven]

Depending on the presentation, you might be able to figure it out. For example, if you can simplify the presentation to <r, s | r^n = s^2 = (sr)^2 = 1>, you know you have a dihedral group. (Or if you get the same with extra relations, you can express it as a known quotient of the dihedral group.) Or, if you can show that the relations imply that all the generators commute, then you have a finitely generated abelian group, and there's a probably a way to explicitly write down what it is in terms of the presentation.
But IG for an arbitrary presentation it might not be possible to do any better than guess (and get stuck if your guess isn't right).

Textbook hints, making exercises not unsolvable since 1842.

[topaz solar]

if you have a map f:A -> B and an atomic formula \phi(\bar x), then A |= \phi(\bar a) => B |= \phi(f(\bar a)) iff f is a homomorphism (by definition)

[knotty badger]

What is an atomic formula?

[topaz solar]

something like R(t) for a relation symbol R, term t

[knotty badger]

I’ve heard something like this in logic, but not really in group theory

Hm…

I still don’t quite understand

[tough raven]

Do you mean that it's good for checking whether a map specified on generators is a group homomorphism?

[topaz solar]

for groups, you literally only have like

a*b = ab

[knotty badger]

Huh…

Oh, wait

I think I get what you mean

So it’s like

[topaz solar]

like a map from generators to generators

[knotty badger]

It preserves equations?

[crystal vale]

Okay thank you ❤️

[knotty badger]

Right, but then you’re doing a map from generators

[topaz solar]

does a bit more, generally, but yeah

[knotty badger]

And presentations help with that

[topaz solar]

I did say this

[knotty badger]

I’m confused

I said originally that presentations help you map out of a group

But not really into a group

And then… you seem to be saying something about how you can map from generators to generators

[knotty badger]

But that’s an example of this, right?

[topaz solar]

yeah, into the group

[knotty badger]

But the presentation of the codomain doesn’t really help you map into it

It’s the presentation of the domain which does this

[topaz solar]

my face when verification

[knotty badger]

The presentation of the codomain defines the group, of course

And so you use it to verify that a map out of the domain is a group hom

But you’d do that regardless of what the codomain was

In all cases, when mapping out of a presented group, you check the relations hold in the codomain

The codomain doesn’t need to be presented for this to be true

What the presentation does is help you map out of the group - you need not define the group hom on the entire group, merely the generators, subject to relations

In particular, I don’t see how the presentation helps you map from a non-presented group to a presented group

[topaz solar]

the point was verifying an arbitrary function is a homomorphism

[knotty badger]

Beyond giving a definition of the codomain

[knotty badger]

Yes, but my point is that you always do this when mapping out of a presented group

It’s not really anything to do with the codomain

You always check that the images of the generators satisfy the relations

[topaz solar]

thats in the codomain

[knotty badger]

Yes, so the presentation of the codomain gives a definition of the codomain

That’s the only sense in which it helps you map into the group

Whereas the presentation of the domain allows you to specify an entire group hom with a lot less information

[topaz solar]

I give up

[knotty badger]

And if you’d defined the codomain a different way, e.g. as a subgroup of GL(V) or something, you’d use that definition

It’s just that… the universal property for presentations is about mapping out, not mapping in

[topaz solar]

oh boy universal properties, here we go

but you also immediately, from the presentation, get subgroups, groups which surject onto it in a sorta lattice above it topped by F(gens.), things like that

[knotty badger]

Which subgroups do you mean?

[topaz solar]

there are more ways to see math than your categorical proselytizing

[knotty badger]

And - which surjections?

I… I think that’s an unfair accusation

I wasn’t proselytising anything

[topaz solar]

for one, subgroups generated by subsets of generators

now we have a lattice of those which decorate P(gen)

[knotty badger]

Yep, but you can do that for any group, right?

You pick some elements, and find the subgroup generated by them

[topaz solar]

indeed, but these are special as we know it covers everything

these generate G after all

[knotty badger]

Wait, covers everything?

I thought you said a subset of generators

[topaz solar]

yeah

[knotty badger]

You mean all the generators?

[topaz solar]

there is a subgroup for each subset of generators

[knotty badger]

Yes

[topaz solar]

i.e. we have a monotone P(gen) -> Subgroups of G

and it covers G

[knotty badger]

I see I see

[topaz solar]

so for any x, theres a subset which has it in it

ok obviously, take the whole thing

[knotty badger]

This is because it includes G itself?

[topaz solar]

well, finite subsets are enough

[knotty badger]

Ah, that is neat!

[topaz solar]

even if its not finitely generated

[knotty badger]

Mhm mhm, i see that

[topaz solar]

and this is immediately obvious, in fact

since x can only be a finite product of generators

[knotty badger]

Though - these aren’t homomorphisms into G, right? Unless you’re talking about the inclusion maps?

[topaz solar]

indeed there are inclusions

now, lets say we pick our presentation

G is <S|R>

[knotty badger]

So it’s like - for any group G, presented or not, you can take subsets of G and produce subgroups that’s been generated by them

But a presentation gives a canonical choice of subsets to try out - subsets of generators

Is that about right?

[topaz solar]

R is the set of relations, and lets deductively close R, so anything you can derive from them is in it

I'm discarding niceness like finiteness etc, but it means everything is there

[knotty badger]

Wait wdym

Are we considering finite subsets, or all subsets?

[topaz solar]

for any subset R' of R, we have a surjection <S|R'> -> <S|R>

[knotty badger]

Ah, yes!

[topaz solar]

F(S) is the empty

[knotty badger]

Right, yeah

[knotty badger]

And - we know this works cause of how <S|R’> is defined

[topaz solar]

this, in fact, essentially describes all surjections onto it generated by the set S

[knotty badger]

Mhm mhm

[topaz solar]

so its every surjection, excluding those from doing S u {x} and adding x = 1

[knotty badger]

Wait, it’s every surjection?

I’m not sure what you mean

[topaz solar]

every surjection generated by S elements

[knotty badger]

Right, I completely agree with that

[topaz solar]

If you want more, add redundant generators

[knotty badger]

Sure - or choose a different group altogether

[topaz solar]

well its every surjection onto our original G

[knotty badger]

Yes

I suppose im saying that - there are other surjections that occur

Onto G

That might involve a group wholly unrelated to S

[topaz solar]

those all occur from expanded presentations, i.e. S u (x_i)_{i<\kappa}

[knotty badger]

Oh, they do?

How do we know that?

[topaz solar]

S is not a set of elements

[knotty badger]

Oh…?

I thought it was a set

[topaz solar]

its a list counting how many generators we have

[knotty badger]

I don’t quite understand…

[topaz solar]

<S|> = F(S)

S, on its own, does not actually say anything

all the structure is in R

[knotty badger]

So S is not a set?

[topaz solar]

S is just a cardinality, and a collection of labels

its a set obviously

[knotty badger]

Ah, so an isomorphism class of sets you mean

[knotty badger]

Right, well

[topaz solar]

but it is not a "set of elements"

[knotty badger]

You did say this

Oh, hmm

[topaz solar]

its more like, its a set of symbols

[knotty badger]

Don’t all sets contain elements…?

[topaz solar]

names for generators

generator 1, generator 2, etc

[knotty badger]

Right - I haven’t really heard this terminology before, so it was confusing

I was under the impression that sets just contain elements

[topaz solar]

So every surjection occurs in this way

[knotty badger]

And nothing more complicated than that

[topaz solar]

because we can just add "unused" symbols

[knotty badger]

So - how do we know this?

[topaz solar]

take a surjection

[knotty badger]

Mhm

[topaz solar]

pick a generating set on the big one

namely, the whole group

[knotty badger]

Uh

Ok hang on

We have a surjection K -> G

We pick a generating set for K?

[topaz solar]

K = <K|R(K)>

[knotty badger]

Sure sure

[topaz solar]

right?

[knotty badger]

Yeah that makes perfect sense

It’s the canonical presentation of K

[topaz solar]

pick any generators of choice for G, call it S

[knotty badger]

Mhm

[topaz solar]

by surjectivity of the map, and choice, there is some element k_s in K for each s in S, and f(k_s) = s

[knotty badger]

Right - that makes sense

You’re taking a preimage of each generator

Which may not be distinct, but that’s ok

Wait, no

[topaz solar]

well

[knotty badger]

They’re definitely distinct

Lol

So - we have an |S|-sized subset of K

[topaz solar]

f(k_s) generates G, so just order them at the front of an order on K

[knotty badger]

Wait, order them?

[topaz solar]

yeah

[knotty badger]

I didn’t realise that was necessary

But sure, ok

[topaz solar]

probably isn't but I'm thinking of it

yeah this is completely unnecessary

I have not slept

[knotty badger]

Right…

We can pick this up later if you want?

[topaz solar]

anyway, f[K] generates G obviously

[knotty badger]

Well, f(K) = G, right?

That’s the surjectivity hypothesis

[topaz solar]

and we can just add redundant symbols for each point in K that isn't a k_s

[knotty badger]

Mhm mhm - but won’t these satisfy their own relations too?

[topaz solar]

sure, we can add those to our presentation

[knotty badger]

I see - I think I was misunderstanding your statement, then

[topaz solar]

if you want all our redundant symbols to have the relation "x=1" instead then like

[knotty badger]

Then your statement makes perfect sense to me

[knotty badger]

But this one does not

If you allow arbitrary relations on redundant symbols, im happy

[topaz solar]

if k_s generates K, we can call it a day

[knotty badger]

Mhm

[topaz solar]

because like, you're already where you don't need any extra relations to make surjection

[knotty badger]

I don’t imagine this will be true in general, though

[topaz solar]

indeed

[knotty badger]

After all, you have a surjection G x H -> G for any group H!

[topaz solar]

indeed

so take some k not generated by these

[topaz solar]

yeah we probably wanna allow "x = y" between em rather than just "x=1" I'm too tired

[knotty badger]

That’s alright

[topaz solar]

or some similar such

[knotty badger]

Mhm mhm

[topaz solar]

point being, the relation gives so and so data about surjections onto it

[knotty badger]

Yes, so I think I can summarise what’s happened as follows:

[topaz solar]

and you get em all among groups of a bounded size, basically

[knotty badger]

If X is an arbitrary group, then there’s a nice description of homomorphisms <S|R> -> X. Namely, functions from S to X which satisfy the relations R uniquely determine such homomorphisms

However, there’s not a nice description X -> <S|R>, beyond the definition of one

It’s still true, though, that expressing a group as a presentation gives a large family of homomorphisms Y -> <S|R> for some groups Y that depend on S and R - subgroups and quotients, as you described.

Is that fair?

[topaz solar]

it gives all the surjections, assuming you know what maps to what generator

[knotty badger]

Mhm - that fits into my last paragraph

Since all surjections are quotients

[topaz solar]

because the relations exactly describe, like the positive atomic diagram

[knotty badger]

Not sure what that is, but that’s ok

[knotty badger]

The middle part is what I meant when I was talking about universal properties

I still think it’s unfair to call that proselytising, but I can also understand you’re tired

I’d kindly request you avoid calling it that in the future - im not trying to dismiss your maths or anything, I don’t tend to ask questions in bad faith

[topaz solar]

this one in particular isn't bad, but you don't always need to present it as if the universal properties are the end-all-be-all way to see things

[knotty badger]

I didn’t!

In fact, I don’t think they are

But I think they’re a useful perspective

In general, the lesson I learned from category theory is not that there’s One True Perspective on things; instead, it’s useful to have multiple perspectives, so long as you have ways to translate between them

This applies to the categorical perspective itself, too - it’s not the One True Perspective on anything

[topaz solar]

its a useful tool

[knotty badger]

Exactly! That’s how I think of it

I do think category theorists in general can get a bit… overzealous

But I don’t try to be like that

[topaz solar]

Seeing yoneda mentioned while trying to solve a linear system seems a bit much

[knotty badger]

Oh? What do you mean?

[topaz solar]

#linear-algebra message this. Its not like it's a long drawn out spiel, but it's ultimately like a

"ok, but why?"

[knotty badger]

Ah, well that is one of my favourite applications of yoneda

So, the reason why is that I find it cool!

But again - I don’t think yoneda is the One True Perspective on row and column operations

I just find it neat that such an abstract lemma has such a concrete corollary

[topaz solar]

It seems less an "application" and more "it just so happens you can make a category where this can be shown as such"

[knotty badger]

I mean, I actually gave a talk about this

[topaz solar]

Theorems aren't really a "why," ever

[knotty badger]

It’s possible to prove this using the idea of yoneda, without actually needing to introduce categories, functors or natural transformations

[topaz solar]

theyre the "what"

[knotty badger]

Hmm, I think id disagree

For me, maths often tells me “why” something is true

Not merely “what” is true

[knotty badger]

Which is what I did, and it was quite well received

[topaz solar]

if its without categories I do wonder just how yoneda it is

[knotty badger]

Very much so!

It’s essentially the proof of yoneda for that category

But fully unfolded so that you don’t need any categorical language

And it’s actually pretty short too

[topaz solar]

Not much you can say about "why" choice is/isn't true, or why Darboux's theorem is true

a proof can show "how"

[knotty badger]

You’ve given me examples of places where maths doesn’t show “why”

This doesn’t mean that maths never shows “why”

[topaz solar]

Seems like quite a complication

[knotty badger]

Would you like to see it?

[topaz solar]

I don't think it ever does

[knotty badger]

Well, I think learning category theory has shown me the “why” on a lot of things

For me, a proof is a reason why something is true, after all

Otherwise, we’d only need lists of already-proven statements

[topaz solar]

Proofs are more like a mechanism, how something occurs, what it does

[tough raven]

It's really just RowOp(A) = RowOp(Id A) = RowOp(Id) A =: EA.

[knotty badger]

Well, that’s not how they feel to me

[knotty badger]

Yes, exactly!

[topaz solar]

Knowing how to make an epoxide doesn't tell you why it's stable

[knotty badger]

That’s the proof I used

[knotty badger]

Again, giving an non-example of “why” doesn’t mean “why” never occurs

[topaz solar]

I'm saying yoneda-fication is a complication of the proof

[tough raven]

Calling it an instance of Yoneda's lemma really just amounts to recognising that this proof and the proof of Yoneda are strikingly similar.

[knotty badger]

But that is the proof with yoneda

Fully unfolded

You show that row operations satisfy RowOp(AM) = RowOp(A) M

And then you use the trick that A = IA

That’s it

[topaz solar]

it's less "proven with yoneda," more "if I draw a category and used yoneda, I get this same proof as if I just did it"

[knotty badger]

This follows from the definition of matrix multiplication

[knotty badger]

I think we’re just arguing over small details at that point

[topaz solar]

its an unnecessary overhead to call it yoneda

[knotty badger]

Well, it’s just a name

[topaz solar]

it complicates matters, obfuscates

[knotty badger]

You could call it “Bob” and it’d amount to the same thing

I don’t think it obfuscates as much as you claim

[tough raven]

Noticing patterns across different existing proofs can be helpful if you can formalise a generalisation which you can then apply in other contexts where it wasn't obvious that the same technique was applicable.

[knotty badger]

And indeed, based on the feedback I received on my talk

It didn’t obfuscate anything!

[knotty badger]

Though the way I did it was actually with column operations, then transposing

[tough raven]

It's not a harmful overhead though.

[topaz solar]

I think its not too far from being one, since it gives the impression of a handful of matrix manipulations being deep

[knotty badger]

Well, what’s deep is subjective

[tough raven]

And it's interesting to some people (myself included), so might as well let them notice it and be excited about it...

[knotty badger]

Indeed, lots of people found it interesting!

[tough raven]

Call it "not deep" if you want, but people should also have a way to say "this can be seen as a special case of Yoneda's lemma" without being interpreted as claiming that that's deep.

[topaz solar]

fair

[tough raven]

I like how both of you reacted with BTW

[knotty badger]

Lol I didn’t notice that

But I mean… yeah, what you said is exactly right

[tough raven]

Anyway 🥱 let's have some actual group, ring or field theory in here

[topaz solar]

I reacted with a in the same sense that FLT proves the 4th root of 2 is irrational

[knotty badger]

Wait, you can do that non-circularly?

[tough raven]

Yes, I realise you probably meant to imply very different things by the reaction.

[topaz solar]

(which that one is a bit lame since those simple equations being the way they are is a step in the proof)

[knotty badger]

I do think people tend to overestimate how hard yoneda is

[topaz solar]

It's not FLT because it's hard,it's because its a sledgehammer

[knotty badger]

And I think that’s fun!

[topaz solar]

oh boy wreath products

[delicate orchid]

https://tenor.com/view/monkey-monke-neuron-activation-neuron-activation-gif-20850506

I have been awoken

[topaz solar]

wew I finally used a theorem about wreath products are you proud

(restricted wreath, that is)

[delicate orchid]

which one

restricted wreath products are just wreath products because all sets are finite

[topaz solar]

so based, so forward thinking

[topaz solar]

uh one from a 2023 arxiv, which is based on like amenability

[tough raven]

Products = coproducts
co = 1
coconut = nut

Now buy a pack of nuts, put it in a box and recite this proof to cause an explosion.

[delicate orchid]

thanks for the mystical spellcasting tips and tricks

[topaz solar]

basically one exploiting some DST nonsense to show some representation theoretic MALARKY for wreaths

[delicate orchid]

I know about representation theoretic malarky for wreaths but considering you're actually making a difference between an underlying direct product/direct sum does not fill me with hope

[topaz solar]

lamplighter moment

[delicate orchid]

.... who?

[topaz solar]

uhhh C_2 \wr G shenanigans

anyway I used stupid wreath theorem #7675776576456454 to prove something

:3 I am now a wewist

[delicate orchid]

so... permutation matrices on a F_2-vector space...

[hl2 scientist voice] fascinating....

[topaz solar]

Basically tbh tbh

Anyhow, there’s relevance to the stuff I do because maps which are close to being reps are nice on some groups

[delicate orchid]

what on earth do you mean "close to being reps"

[topaz solar]

But these are INFINITE groups so

[topaz solar]

Stuff like, uh, the specific flavor here is like, making \sup_{x,y} || f(xy)-f(x)f(y) || small for unitary reps

[delicate orchid]

I see

so they're homomorphisms up to floating point precision

[topaz solar]

Yeah basically

[delicate orchid]

interesting, never seen that before but I'm not sure why

cause it's a natural thing to think about

[topaz solar]

There’s variations where you ask that a map on generators is close to satisfying so and so relationd

[coral spindle]

Sounds interesting

[topaz solar]

Which this is not uniform in x, y, and is free group quotient-y

And has wildly different behavior

(Z^2 is uniformly nice, but if we do this generator approach we’re screwed)

Because the goal is “it’s close to being a homomorphism” -> it’s actually close to a representation

and it’s relevant to like, quantum checking stuff since verifying on generators is way easier then over the whole group

[delicate orchid]

see I don't see the difference there

a representation is a homomorphism

[topaz solar]

Oh that’s the same

The difference is like

[topaz solar]

Make this small, then we have a rep g with \sup_x || f(x) - g(x) || small

So it’s close to a rep uniformly, if it’s close to being a rep

[delicate orchid]

I don't follow. What's g

oh

[topaz solar]

some representation

There’s a bunch of analogues, sofic and hyperlinear groups are a similar class of things

But the one I’m working on is the above one

[delicate orchid]

ah now I see the difference

[topaz solar]

(There’s also p-adic versions somehow, with ultra metric trickery doing a lot)

[delicate orchid]

aprori there's nothing stopping you from constructing such an f that's miles away from any actual representation

[topaz solar]

Indeed

And F_2 you can make explicit maps so that you get like

1/n error in an n-dim rep

But it’s 1/10 away from a real rep

And you can do this with some fundamental groups of surfaces

Compact surface groups, that’s what it was

[delicate orchid]

ok that's uhhh what's the word I'm looking for uhhhh HORRIFIC?

[topaz solar]

yeah

Basically, the things where it fails however are: F_2, Tarski monsters, and like

Things which contain F_2 in a measure-y sense of orbits

[delicate orchid]

absolutely zero idea what that means

markov chain ass sentence

[topaz solar]

yeah neither do I but I just accept it as “F_2 is always the problem”

Even if it’s not a subgroup

Anyway, this is obviously a very stinky property

Because it doesn’t tell you anything about the rep

The generator based or approximability ones you have a bit of structure in limits but uh

This uniform one? Nothin

[crystal vale]

Let G be an abelian group of order 2n, where n is odd, show that there is only one element of order 2.

I used the normal subgroup concept, is there another way to prove it?

[delicate orchid]

sylow

[topaz solar]

Well, if we ask that this is small uniformly in all finite dimensions for unitary reps, we get a lot of interesting groups with this property

[delicate orchid]

all sylow subgroups are normal, hence there's only one of each. And the sylow 2-subgroup is size 2, hence there is a unique involution

of course this is a stupid way of proving it and there's probably a far more elementary way but that would require me to think

[topaz solar]

Like, lamplighters or SL(n, A) for n\geq 3, A a localization of a ring of integers in a finite extension of Q

Well, the ones I care about are even nicer, in fact

[knotty badger]

I suppose - if there was more than one element of order 2, then you’d have a subgroup of size 4

And that violates Lagrange’s theorem

[topaz solar]

If we ask uniformity in every dimension, there’s basically only 1 class of known examples: amenable

[delicate orchid]

yeah there we go that's a much easier way of doing it

[knotty badger]

Cauchy’s theorem would guarantee that you have at least 1 element of order 2

[topaz solar]

Which, what I think I’ve just proven, if we have some asymptotic decay estimates, they’re the only ones

[hidden wind]

reminds me i was idly browsing through the norwegian state library archive the other day (as one does), and found some early 20th century article by thoralf skolem (!) on the mathematics of sylow

i should read that

[topaz solar]

Though this depends on correctness of another paper etc

So somehow silly goofy combinatorial stuff = this density stuff of maps that are almost reps

????

I have no idea what’s going on outside that decay estimate though

I barely know what’s going on under it

[crystal vale]

Thank you

[terse crystal]

There is a kind of “product” of symmetric groups, like product of Sa and Sb, being a subgroup of Sab, by dividing ab into a many blocks of length b, Sa permutes those blocks. What is this product called?

[tender wharf]

for more elementariness we can even use cauchy's theorem for finite abelian groups which does not require sylow to prove

[coral spindle]

This sounds like the wreath product to me

[terse crystal]

Wreath product? Thank you I am searching it

[delicate orchid]

wreath product mentioned TWICE in the same day you love to see it

[coral spindle]

How are you gonna survive wew, you might just explode with happiness

[delicate orchid]

It's ok I'll temper my elation by thinking about Brauer nonsense

[coral spindle]

How are you gonna survive wew, you might just cry so much you'll wither into a tiny human raisin

[delicate orchid]

important question

"sub G-set" or "G-subset"

vote now on your phones

[coral spindle]

I love them both, it's like choosing between my children

I think I find that "G-subset" sounds more natural even if I think that "sub G-set" is more pedantically correct

[delicate orchid]

I agree

"subobject in G-set" it is

[terse crystal]

Oh yes it’s wreath product,thanks 😊

[coral spindle]

1000 years jail

[topaz solar]

subtract 1, out of spite and despair

[marsh scaffold]

Hey @topaz solar (sorry for pinging if you think this is irrelevant) but i actually found a close enough reason on how this equivalence relationship came to be , it's i believe how brauer proved frobenius' theorem

After reading that paper it actually makes much more sense

Also like it's been quite a while ......lol...

[hidden wind]

vibing with Burnside

[coral spindle]

What a lovely diagram

[knotty badger]

Oh this is a lovely way to think about normality, I think

[hidden wind]

[delicate orchid]

God damn this is swag

cheeky hyperbolic group

[hidden wind]

compare my viz of klein’s icosahedral invariant with fig. 13

mwahahaha

(each of the regions in burnside’s diagram are conformally equivalent with a complex half-plane)

[rain grove]

How to prove this?

Is this ok:

<x> ,<x'> subgroups of Z_n of order k. From the theorem before subgroup of Z_n of order k is exactly when k|n. So for both x and x' I can write: kx = n and kx'=n (in Z) and in Z it follows that x = x'. So the subgroups are generated with the same element, so they are the same.

[crystal vale]

Why do you write kx= n and kx' = n ?

[rain grove]

Because I get the result x' = x and im done

[tardy hedge]

Oh HELLLLL nahh!!!

[rain grove]

Wait this is for sure wrong... cause in Z_6, <2> and <4> are both order 3 but 3x2 = 6 and 3x4!=6

ahh idk

[delicate orchid]

they're both 0 mod 6 idk what ur smoking

what are we doing?

[rain grove]

That in Z_n there is only one group of order k

[delicate orchid]

yeah there's one for each divisor

[rain grove]

I already proved this: k|n <=> there is a subgroup in Z_n of order k

I need to prove that its the only one

[delicate orchid]

right so you're stuggling with uniqness

[rain grove]

yes

[delicate orchid]

<x>, <x'> both subgroups of order k in Z/nZ. See if you can show that x = mx' for some m coprime to n

alternatively show that they're both equal to <n/k>

[rain grove]

I did this:
Because Z_n is cyclic I can write x = n*1 for some n. let m, m' smallest positive integers that x = m and x' = m'. then I can do m = qm' + r... x = qm' + r = qx' + r where 0<=r<m'. Because x element of <x> so is qx' + r ...

and I wanted to conclude that qx' element of <x> and r element of <x> but because smallest integer to be in <x> is m and r is smaller than m it must be 0, so x = qx'

but r is smaller than m' and not m

And also I don't think from qx' + r in <x> follows qx' and r in <x> so im wrong there too probably

[delicate orchid]

any subgroup of Z/nZ is of the form <k> for some k that divides n - this can be seen as if we chose <m> , then there are integers a, b such that gcd(m, n) = am+nb, so <m> = <gcd(n, m)>

I tried to understand your argument for a solid 10 minutes and I still don't get it

why are you writing x as n*1?

[rain grove]

What

[rain grove]

I mean its wrong don't worry about it

[rain grove]

How do you conclude ... so <m> = <gcd(n,m)>

[delicate orchid]

reduce what I said mod n and you get that gcd(m, n) = am, so gcd(m, n) is in <m>

if it didn't generate all of <m>, that would imply that there was an integer smaller than gcd(m, n) that divides both m and n

[rain grove]

Im sorry but can you show me the proof where we have two subgroups and we shows they are the same

[delicate orchid]

I just did

I'm kinda done thinking about this ngl

[rain grove]

ahh

Ok

Im trying to do this for 3 days

and can't

[delicate orchid]

Idve Googled it after an hour ngl

[rain grove]

lol

[delicate orchid]

Better things to spend your finite life doing

[rain grove]

you think I didn't

[delicate orchid]

You’re expecting me to believe there aren’t any clear proofs of this on the internet

[rain grove]

yes

[delicate orchid]

https://math.stackexchange.com/questions/2930733/finding-all-the-subgroups-of-a-cyclic-group this answer is fantastic

Even better than my hack fraud proof using bezout’s

There’s a 9 page document entirely about cyclic groups https://kconrad.math.uconn.edu/blurbs/grouptheory/cyclicgp.pdf see theorem 3.6

[rain grove]

Okk nice 3.6 saved me from || kissing myself||

[delicate orchid]

same proof as me

maybe it wasn't so fraudulent after all

[rain grove]

Ok is there an easier version for Z_n tho?

[spice whale]

mwah

[delicate orchid]

....

all cyclic groups are Z_n (or Z (which is just Z_0))

that's all of them

[rain grove]

OOO omg

[south patrol]

Z_n is only topologically cyclic

[delicate orchid]

Someone ban this guy

[coral spindle]

And why would you call a prime n???!?!?!?!

[south patrol]

Sorry for using normal notation

[delicate orchid]

Does he know? Oh no no no no

[delicate orchid]

Do u know how much will power it takes me to not write C_n

[south patrol]

Lmao

C_n to me is just a group theorist thing

[coral spindle]

shut it nerd

I can construct Z_n I don't CARE

[south patrol]

Construct it then

[coral spindle]

No

I imagine it's the usual construction with Z/n ← Z/n^2 ← Z/n^3 ...

But for any number n

just a guess

[south patrol]

Yeah

I mean these are completions of Z with respect to the ideals (n) innit

[coral spindle]

Ye

[delicate orchid]

https://tenor.com/view/yeah-dats-what-i-woulda-gif-18253305946678612829

[heady acorn]

suppose m is a proper nonmaximal ideal of a ring A. is it always true that there exists a in A\m, such that m is properly contained in m+(a) and m+(a) is properly contained in A?

(a) is the principal ideal generated by a

[south patrol]

Yes

Indeed since m is non-maximal, there's an ideal n with proper inclusions m < n < A

Now any a in n will work!

[heady acorn]

n\m ig? and thanks! i get it now

[south patrol]

yes sorry lol

[topaz solar]

Horrible

[topaz solar]

@delicate orchid no I’m just an idiot, I blame \wr for my agony

[delicate orchid]

how has blud done this

and don't give me the details using a shit ton of words I have to google just. Da overview

[topaz solar]

For A \wr G, I can move A around by multiplication by A pretty freely. I made a mistake and moved an A by the G action in the middle of it, which is not “safe”

[delicate orchid]

I see

now u... got me thinking

nah nvm what I was considering is a triviality

[topaz solar]

What is it

Because I very clearly am not the \wr expert

[delicate orchid]

S a p-group, fusion system given by a wreath product on S

completely unrelated

[topaz solar]

And, to clarify, it’s the restricted wreath

[delicate orchid]

it's essential rank 0 and thus INCREDIBLY boring

[topaz solar]

Because it definitely doesn’t work unrestricted

[topaz solar]

Average thing I think about

[delicate orchid]

like I completely solved the representation theory of them back in 2022 levels of boring

[topaz solar]

Based

[delicate orchid]

then found out two mexican guys were also doing the same thing

[topaz solar]

But are they as based? Didn’t think so bub

[delicate orchid]

feel that? that's true

[tardy hedge]

Lmao why is this so funny

[crystal vale]

1. N is normal in G
2. | G: N| is finite
3. H< G
4. | H | finite
5. | G: N| and | H | are relatively prime numbers.

Now I want to prove H < N.

Take | H | = n and | G : N | = m, since N is normal so the Quotient group by N makes sense.

Let h in H then h^n = e and (h+N)^m = e + N.

Now use m and n are relative prime so we get h+N = N implies h in N.

Thus H < N
Is it correct?

[rotund aurora]

can you classify the 1 dimensional Lie groups (over R or C)?

and should they be locally isomorphic? (for example, R and R/Z are locally isomorphic)

[rotund aurora]

The Lie algebra in all those cases should be R, or C if the Lie groups are over C. Right? So they should be locally isomorphic?

[junior badge]

I'm given a Noetherian module M. I want to show R/ann M is Noetherian.

Suppose for a contradiction that I_0 ⊂ I_1 ⊂ ... is an infinite chain in R/ann M, all containments proper. This gives a corresponding chain in R, I_0 + ann M ⊂ I_1 + ann M ⊂ ..., all containments proper. This allows us to produce a chain of submodules of M:

(I_0 + ann M)M ⊂ (I_1 + ann M)M ⊂ ... or I_0M ⊂ I_1M ⊂ ...

However, I dunno if the containments are proper here. I've tried proving it, using the generators for M and some version of Nakayama. But yeah, kinda stuck here. Am I on the right track?

[vague granite]

Suppose you have a field F. Let F↪️F[u] be a ring extension such that u is invertible. Is F[u] a field then?

[tough raven]

No: consider F[X, X^{-1}] inside F(X).

Also F[X]/(p), provided p is reducible but has invertible constant term.

[tough raven]

On the other hand, if F[u] is an integral domain which is finite-dimensional over F, it's a field.

[vague granite]

That’s what I was thinking since F[X] localised at {1,X,X^2,…} has more than one element in its spectrum

I’m pretty sure it’s just the points which don’t have a root at 0

[hazy lion]

if F is a field then yes, Spec F[X]_X is identifiable with the monic polynomials indivisible by X

[rain grove]

Why is this true?

just that part, why is the order m/gcd(k,m)

[tough raven]

You need to show that for any n, (g^k)^n = 1 iff m/gcd(k, m) divides n.

[rain grove]

and why from this it follows that order of <g^k> is m/(k,m)?

[tough raven]

Well, assuming you have shown that, the smallest positive integer n such that (g^k)^n = 1 is the smallest positive integer divisible by m/gcd(k, m), and I think you can guess what the latter is.

[rain grove]

Ok im sorry I don't get it...
Why you said any n before and now its smallest positive n

And i don't guess what the latter is 😩

[tough raven]

OK, my bad.

What's your definition of order of an element of a group?

[rain grove]

smallest n >= 0 that g^n = 1 for some g in G. Then n is order of element g

[tough raven]

Right.

(Should be smallest n > 0 BTW.)

[rain grove]

Ohh yeah it is n >0 mb

[tough raven]

So you can further prove that if n is the order of g, then for any integer m, g^m = 1 iff n divides m.

[rain grove]

Yeah I proved this

[tough raven]

So using this, in your problem, what is the condition on n for (g^k)^n to be 1?

[rain grove]

smallest positive integer?

[tough raven]

No, for any integer n.

(g^k)^n = 1 iff ________

[rain grove]

Well If i look atthat statement: if n is the order of g, then for any integer m, g^m = 1 iff n divides m.

If x is order of g^k then for any n, (g^k)^n <=> x | n

but idk what x is

[tough raven]

Try to use the order of g (which is called m in your problem).

[knotty badger]

Ooh huh, I suppose one could call that a universal property…

[rain grove]

Don't i just get: if m is order of g then for any n, g^n <=> m | n

[coral spindle]

g^n <=> m | n
On the left of the equivalence you have an element of the group, and on the right you have a proposition. This is meaningless.

You mean g^n = 1 <=> m | n

[rain grove]

yeah thats what I mean

I forgot to write that mb

[knotty badger]

Ah, so you want the order of g^k given the order of g, hmm

[coral spindle]

Pseudo, don't spoil this question.

[knotty badger]

Ok! I’ll think about it on my own though

I wasn’t planning on spoiling it, dw

Ok cool, you can indeed solve this using yoneda if you want

That’s neat!

[rain grove]

m is order of g and also g^k.

So looking at this then: ...then for any integer x, (g^k)^x = 1 <=> m divides x

So I need to find a integer x that divides m? Can't that be any number that divides m, why is gcd(k,m) special?

[tough raven]

Yes, and (g^k)^n = 1 <=> ???

[rain grove]

(g^k)^n = 1 <=> m | nk ?

I don't understand

[knotty badger]

Can I make a suggestion?

[rain grove]

PLEASE

[knotty badger]

So, given $g \in G$, $\text{ord}(g)$ is defined to be the smallest integer $k$ such that $g^k = 1$, right?

I.e. it is $\min { k > 0 \text{ such that } g^k = 1}$

[cloud walrus]

Pseudonium

[rain grove]

Yes

[knotty badger]

We also have an additional property - that $\text{ord}(g) | n \iff g^n = e$

[cloud walrus]

Pseudonium

[rain grove]

Yes

[knotty badger]

So far, you’ve been trying to attack $\text{ord}(g^k)$ with this definition, as far as I can tell.

My suggestion would be to instead start from the additional property

[cloud walrus]

Pseudonium

[knotty badger]

Is it clear what I mean by that?

[rain grove]

Maybe i got some ideas

wait

[knotty badger]

What I mean is - start from $\text{ord}(g^k) | n$, and see what you can deduce

[cloud walrus]

Pseudonium

[rain grove]

ok i give up... so from ord(g^k) | n, I can deduce that g^n = 1

nothing else

[knotty badger]

Actually, there’s more! Try applying the additional property of order

I don’t think g^n = 1 is correct

[rain grove]

order of (g^k): m is smallest positive integer that (g^k)^m = 1.
and ord(g^k) | n => (g^k)^n = 1

I don't see how to conenct them together

[knotty badger]

Yes, so actually we have $\text{ord}(g^k) | n \iff (g^k)^n = 1$, right?

[cloud walrus]

Pseudonium

[rain grove]

Yes

[knotty badger]

And then - what can you deduce further?

[rain grove]

Idk im trying to use the proprety of the order but idk how

[knotty badger]

So - we can deduce $g^{kn} = 1$, right

[cloud walrus]

Pseudonium

[rain grove]

yeah

[knotty badger]

Then - use the property of $\text{ord}(g)$

[cloud walrus]

Pseudonium

[rain grove]

how?

[knotty badger]

Well, what is the property of $\text{ord}(g)$?

[cloud walrus]

Pseudonium

[rain grove]

its the smallest positive integer that $g^{ord(g)} = 1$

[cloud walrus]

OHHELLNAH

[knotty badger]

I mean the other one

[knotty badger]

This

Do you see how we can apply that?

[rain grove]

ord(g) | kn

[knotty badger]

Indeed! So, just to recap

We’ve done $\text{ord}(g^k) | n \iff (g^k)^n = 1 \iff g^{kn} = 1 \iff \text{ord}(g) | kn$

[cloud walrus]

Pseudonium

[rain grove]

nice

[knotty badger]

I think this could help you prove the statement you want

Note that we’ve actually reduced this to a problem about integers

We have that $a | n \iff b | kn$

[cloud walrus]

Pseudonium

[knotty badger]

And you want to show $a = \frac{b}{\text{gcd}(k, b)}$

[cloud walrus]

Pseudonium

[rain grove]

What is a and b?

So a and b can eb any unrelated numbers?

[knotty badger]

Here, $a = \text{ord}(g^k)$

[cloud walrus]

Pseudonium

[knotty badger]

And $b = \text{ord}(g)$

[cloud walrus]

Pseudonium

[rain grove]

But if i choose a then b is already chosen no? Because both depend on g

so they are not any a and b?

[knotty badger]

Well, what I mean is that

It is true that, for all naturals a and b, if you have a | n iff b | kn, then a = b/gcd(k, b)

At this point, we have reduced it to a number theory problem, not a group theory one

[knotty badger]

So you are correct that, once you know ord(g), then you know ord(g^k)

But what I mean is - you can actually forget about the group theory at this point, and focus on the number theory

Of course, if you’re not confident with the number theory, then this strategy may not be a good one for you to try

[rain grove]

I get the transition but to me It feels like there could be problems because in "group theory" our "a" and "b" were kinda related and then in number theory they are any numbers a and b

[knotty badger]

Mhm - though even in the number theory, you can prove that a and b are actually related

[hazy lion]

if b=ord(g) then you can just look at the cyclic group Z/bZ and model it canonically on the elements {0, 1, ..., b-1}

[knotty badger]

You need not use a and b if that’s confusing, you can stick with ord(g^k) and ord(g)

What we have is that $\text{ord}(g^k) | n \iff \text{ord}(g) | kn$

[cloud walrus]

Pseudonium

[rain grove]

Yeah ill try proving it

[knotty badger]

Ping me if you get stuck

I suppose what’s neat about the universal property of $\text{ord}(g)$ is that

The definition gives you $g^n = e \implies \text{ord}(g) \leq n$

But the universal property strengthens this to $g^n = e \implies \text{ord}(g) | n$, and moreover you can replace $\implies$ with an $\iff$

[cloud walrus]

Pseudonium

[knotty badger]

Very much like what happens for the universal property of gcd

[crystal vale]

Where may I learn Introductory category?

[knotty badger]

Category theory, you mean?

Or something else?

[crystal vale]

Yes

[knotty badger]

Hm…

For a book, I think Leinster’s basic category theory is good

Awodey’s book is also good

And if you know a bit more maths, Riehl’s category theory in context is very good

[crystal vale]

And prerequisite?

[knotty badger]

It’s honestly a little hard to say

How much maths do you know?

Like, a lot of the examples come from undergrad

[crystal vale]

Undergraduate, group theory, ring Theory, topology, analysis

[knotty badger]

But it’s not strictly necessary or anything

[knotty badger]

Ah, this will be good then

Do you know any universal properties?

[crystal vale]

Not exactly, but I heard it a lot

[knotty badger]

I can give an example of one, if you’d like?

[crystal vale]

The property which ones invariant under a different structure?

[knotty badger]

That’s an abstract description of them, but there are plenty of concrete examples

Have you heard about the universal property of product?

[crystal vale]

No

[knotty badger]

Ok so - given sets A and B, we can form the Cartesian product A x B

What it “is” is the collection of ordered pairs (a, b), with a in A, b in B

[crystal vale]

Okay

[knotty badger]

But we can also ask what it “does”, how it relates to other sets, and how to use it

[crystal vale]

Yes

Canonical mapping?

[knotty badger]

In particular, notice the following - if you have a function f : Z -> A x B, then it has to take the form f(z) = (g(z), h(z))

And so you can “unpackage” it to two functions g : Z -> A, and h : Z -> B

[crystal vale]

Yes

[knotty badger]

Conversely, if you have two functions g : Z -> A, and h : Z -> B, you can “package” them into a single function f : Z -> A x B

Defined by f(z) = (g(z), h(z))

[crystal vale]

Yes

[knotty badger]

https://media.discordapp.net/attachments/834837272707727372/1263928509978640454/image.png?ex=66af2267&is=66add0e7&hm=c7a68711feb0ee9dc3dae6e23d2112cd8481911ef01bbe8adc73316803ddac71&

https://media.discordapp.net/attachments/834837272707727372/1263928868088315945/image.png?ex=66af22bc&is=66add13c&hm=0f07e48b99a81b1d5d10e22cee528c1815c4ae02a96da81e9906e47038e66866&

This is what the Cartesian product “does” - it lets you package and unpackage functions

Does that make sense?

[crystal vale]

Yes so what do we get here?

[knotty badger]

Well, that’s the universal property!

It’s an alternative description of maps into the Cartesian product

As an exercise - what does the identity map A x B -> A x B unpackage to?

[crystal vale]

I don't understand which property called universal property

[crystal vale]

Identity map I_A and identity map I_B?

[knotty badger]

I’ve described how A x B relates to all other sets in the “universe”

Namely, that maps Z -> A x B naturally correspond to a pair of maps Z -> A, Z -> B

[knotty badger]

Hmm, not quite - careful about the domain and codomain of the unpackaged maps

[crystal vale]

Other set means?

[knotty badger]

So, Z is any other set

It could be the naturals, it could be the real numbers, it could be the rationals

[crystal vale]

A×B -> A and A×B-> B such that (a,b)->a and (a,b)->b respectively

[knotty badger]

Yes! These are the projections

And you can relate the packaged/unpackaged maps using these projections, in a commutative diagram

[crystal vale]

But what do you mean by unpackage?

[knotty badger]

I mean that - if you have a map Z -> A x B, you can convert it into two maps Z -> A, Z -> B

[crystal vale]

Yes I saw some Commutative diagrams but in general I don't understand the use of that diagram

[knotty badger]

This operation I am calling “unpackaging”

[crystal vale]

Ok got it

[knotty badger]

Well, do you understand the packaging/unpackaging?

[crystal vale]

Yes

[knotty badger]

You can focus on that part!

So, as another example

Given two numbers a and b

[crystal vale]

Yes

[knotty badger]

You can take their minimum, min(a, b)

[crystal vale]

Yes

[knotty badger]

What it “is” is the smaller of the two

We can also ask what it “does”, how it relates to other numbers, what you can use it for

And it’s this - we have that x <= min(a, b) if and only if x <= a, and x <= b

It “packages” together two <= conditions into one

[crystal vale]

Yes

[knotty badger]

And also lets you unpackage

Now, what happens if you set x = min(a, b)? What does this unpackage to?

[crystal vale]

Can you ask me in briefly?

[knotty badger]

So, this is true for all numbers x

In particular, it is true for x = min(a, b)

And - when you do that substitution, what happens?

It’s quite similar to when you “unpackaged” the identity function A x B -> A x B

[crystal vale]

Means change a and b order?

[knotty badger]

No, I think - let me go through it

When we substitute x = min(a, b)

We get that min(a, b) <= min(a, b) if and only if min(a, b) <= a, and min(a, b) <= b

But we know that min(a, b) <= min(a, b) - this is called “reflexivity”

[delicate orchid]

left Ore localisation got me like

[knotty badger]

So, we can deduce that min(a, b) <= a, and min(a, b) <= b

These are quite analogous to the projections A x B -> A, A x B -> B you found earlier

Does that make sense now?

[tough raven]

Another example you may have seen are free groups (or more generally, groups with presentations).
If G is the free group on a set S, then for any group H, any function from S to H extends uniquely to a group homomorphismm from G to H. This is also a universal property.

[crystal vale]

Sorry but what do you mean by substitute x=min(a,b), where do you substitute?

[knotty badger]

So, the expression holds for all numbers x

E.g. it holds for x = 1

So I can replace everywhere I see an “x” with a “1”

To get the statement that 1 <= min(a, b) if and only if 1 <= a, and 1 <= b

[crystal vale]

Yes I read this result

[knotty badger]

We also have 24 <= min(a, b) if and only if 24 <= a, and 24 <= b

[crystal vale]

Yes

[knotty badger]

Or -27283746 <= min(a, b) if and only if -27283746 <= a, and -27283746 <= b

[crystal vale]

Now maybe I am getting

[knotty badger]

We can substitute any number into this

[crystal vale]

Yes

[knotty badger]

But - min(a, b) is itself a number!

So we can substitute x = min(a, b)

[crystal vale]

Yes

Yes

[knotty badger]

To get min(a, b) <= min(a, b) if and only if min(a, b) <= a, and min(a, b) <= b

[crystal vale]

Yes

[knotty badger]

But we know the left-hand side is true, because every number is equal to itself, so in particular is less than or equal to itself

So, we can deduce the right-hand side is true

That is, we can deduce min(a, b) <= a, and min(a, b) <= b

[crystal vale]

Yes

[knotty badger]

I.e. min(a, b) is indeed a lower bound of {a, b}

[crystal vale]

Yes

[knotty badger]

It turns out this idea of “packaging and unpackaging” is common to many notions of product

Products of groups let you package and unpackage group homomorphisms

Products of rings let you package and unpackage ring homomorphisms

Products of metric spaces, and of topological spaces, let you package and unpackage continuous functions

Products of manifolds let you package and unpackage smooth functions

What the product “is” changes from place to place, but what it “does” is always the same - it lets you package and unpackage

Does that make sense?

[crystal vale]

Yes

Got it

[knotty badger]

There are even examples like min(a, b) which don’t look at all like products!

But in general, category theory focuses on what something “does” rather than what something “is”

[crystal vale]

So Cartesian products are universal property

[knotty badger]

How it relates to other things, what you can use it for

[knotty badger]

These are usually called categorical products

[crystal vale]

Okay

[knotty badger]

But yeah - it’s a universal property

This is the kind of thing that category theory studies

If it seems interesting, I’d recommend reading about it

[rain grove]

I think I did it

[knotty badger]

Oh, well done!

[rain grove]

from here

[knotty badger]

There’s a way to simplify your proof, if you’d like to hear?

[rain grove]

Yes

[knotty badger]

So - you successfully show that $a | n \iff \frac{b}{\text{gcd}(b, k)} } | n$, right?

[cloud walrus]

Pseudonium
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rain grove]

$b | kn \iff \frac{b}{\text{gcd}(b, k)} } | n$

[knotty badger]

You mean $b | kn$, surely

[cloud walrus]

Pseudonium

[rain grove]

Ohh yeah

[cloud walrus]

OHHELLNAH
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[knotty badger]

We get this by chaining the result $a | n \iff b | kn$

[cloud walrus]

Pseudonium

[rain grove]

chaining?

[knotty badger]

As in - $a | n \iff b | kn \iff \frac{b}{\text{gcd}(b, k) } } | n$

[cloud walrus]

Pseudonium
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[knotty badger]

And you can deduce this

[rain grove]

Ok yeah

[knotty badger]

Now, what happens if you substitute $n = a$ here?

[cloud walrus]

Pseudonium

[rain grove]

truth $\iff \frac{b}{\text{gcd}(b, k) } } | a$

[cloud walrus]

OHHELLNAH
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[knotty badger]

Yep! So - we can deduce that $\frac{b}{\text{gcd}(b, k) } } | a$

[cloud walrus]

Pseudonium
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crystal vale]

Okay thank you for all the information ❤️

[knotty badger]

And then - what happens if you substitute $n = \frac{b}{\text{gcd}(b, k) } }$?

[cloud walrus]

Pseudonium
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rain grove]

where?

[knotty badger]

Back in here!

[rain grove]

oh we also get that a divides b/gcd(b,k)

[knotty badger]

Yep! And so they have to be equal

Right?

[rain grove]

omg

nice

ty

[knotty badger]

It’s generally true that if c | n iff d | n, then c = d

This is an instance of a very general result called the “yoneda lemma”, also from category theory

No need to look it up, just a fun tidbit

Note that we never actually had to use the definition of ord(g) or ord(g^k), as the minimal integer where g^ord(g) = e

We merely focused on their “universal property”

How they relate to all other naturals, through divisibility

[rain grove]

hm cool

[knotty badger]

I’m glad you find it cool!

[sonic coral]

challenge: go one day without mentioning category theory

level: impossible

[knotty badger]

Why do you propose this challenge?

[sonic coral]

i just associate you with category theory as much as you mention it to students who are not learning it

it’s not a bad thing or an attack on you just an observation

[knotty badger]

Right right, I think that’s completely fair

But I mean, I mention it cause it’s useful

[dull ginkgo]

WHAT

[knotty badger]

Yep! It totally is

[delicate orchid]

I hate to admit it but it kind of is

[dull ginkgo]

Oh okay I sorta see it

[delicate orchid]

Poset category of Z ordered by divisibility

[dull ginkgo]

Yeah

I wonder if you can go about doing a lot of Galois theory due to independence of characters

I also wonder if a lot of Galois theory stems from group algebras over the automorphism-stabilizer groups

Let F be a field, G be a finite automorphism subgroup and F^G be the fixed point subfield, I wonder if you can prove shit off of F^G[G]

[tardy hedge]

Damn ur being like a mathematician rn^

Dats Cray

[ashen heron]

[tardy hedge]

What

As if youre so confused by what i said

[dull ginkgo]

Let F be a field, then F* (the set of nonzero elements) can be regarded as an abelian group.
The F-vector space V = F^F (set of maps from F to itself) can be equipped as an associative, unital, commutative F-algebra (which I will call a commutative F-ring) by multiplication.

Every nontrivial ring-morphism from a field (simple ring) is injective, so every nontrivial endomorphism of F is injective, and can be forgetful’d to a monoid endomorphism of F*. By independence of characters, End(F)* (excluding the 0 morphism) is a linearly independent set in V.

[mystic ether]

Pseudonym boutta start explaining how sup inf <= inf sup is actually a special instance of limits and colimits commuting or some shit

[knotty badger]

I mean, it is - though it’s them not commuting

[mystic ether]

Yeah ik it’s why I made the joke

But it’s a very obnoxious way to learn that fact lol

[knotty badger]

Well it’s like

You don’t really need limits and colimits for this

Just the universal properties of sup and inf

Riehl has a proof of this fact near the start of CTiC if I recall correctly

[dull ginkgo]

I am really pondering some algebraic crap

[mystic ether]

Yes I’ve read this

[knotty badger]

In general I’ve found that like

It’s harder to recognise something is an example of a general categorical thing, than it is to just… quickly reprove it yourself

I don’t usually remember things like “limits commute” or “right adjoints preserve limits” or anything like that

I’m just familiar enough with cat theory that I can very quickly reprove these whenever I need to, in a specific example, and then later realise “oh I just redid that proof”

Sometimes I even forget the statement of the yoneda lemma, but I can quickly reprove it whenever I need to

[dull ginkgo]

Miz pondering about field stuff again

[dull ginkgo]

I am going to ask about the following:

[tardy hedge]

Yes Sir

[dull ginkgo]

Let $R$ be an infinite integral domain, then is the set of monoid morphisms (characters) from monoid $M$ to $R$ \textbf{algebraically} independent as a subset of $F^M$

[cloud walrus]

The Library of Babble

[dull ginkgo]

actually

[rotund aurora]

you could let R=C, M=Z/2Z. There are two characters, one that sends 1 to -1 and the identity

squaring the first gives you the second

[dull ginkgo]

Ah, thank you

[rotund aurora]

in fact

if you take any finite group, Hom(G, C^x) is isomorphic to G (noncanonically)

[dull ginkgo]

pontryagin bs

jk

anyway

[rotund aurora]

wait you needed G abelian I guess

[coral spindle]

ahdshajf abelian

[dull ginkgo]

Algebraic independence only exists over infinite fields unless you like, say modulo (X^|F| - X) for each of the indeterminates

[coral spindle]

Lol

[dull ginkgo]

afaik

[rotund aurora]

and no, for finite G and abelian it's somewhat easy. Pontryagin duality is much more difficult

[coral spindle]

But Hom(G, C^x) is noncanonically isomorphic to Ab(G)

Which is still cool

[dull ginkgo]

That makes more sense

[rotund aurora]

and also pontryagin duality only gives an isomorphism with the double dual

[dull ginkgo]

I am trying to think of a different way to prove Artin's Lemma

and it turns out there's a lot of structure I didn't realize was there in generality

[south patrol]

Kinda sad it isn't canonical lol

C'est la vie.

[coral spindle]

WEll I mean

[south patrol]

I mean they go in opposite directions

Lol

[coral spindle]

It gives us a beautiful world of Galois theory of characters <3

[south patrol]

Yes

[rotund aurora]

What is that

[coral spindle]

Galois groups act on characters

This tells u things about the reps

over smaller fields

[topaz solar]

Galois stuff is one of like, the 3 flavors of algebra it feels like

[coral spindle]

I was thinking the other day that it really is a fantastic motivation for group theory

You can really see why people got excited about these things

[south patrol]

Interesting take

Hm

To me it feels like more of a smaller area but very intertwined w arithmetic considerations

[topaz solar]

The motivation :3

[coral spindle]

yeah exactly lol

It almost feels like pedagogically, we should start with Galois theory. But of course this would be asking a lot of learners.

[topaz solar]

Well, Galois-y stuff about automorphisms fixing XYZ subsets are certainly a point in model stuff

[south patrol]

Hm is there enough motivation for fields at that point though

I guess you could do it via number theory lol

[rotund aurora]

Isnt it half of Langlands

[south patrol]

Sure but that should fit into like arithmetic stuff and representation theory

rather than like galois theory being the area in itself

[topaz solar]

I said flavor not field :3

[south patrol]

Not much point arguing tho lol

Oh sure lol

Fair

[topaz solar]

Like the methodology, focus, etc

[topaz solar]

Start with polynomials :3

[rotund aurora]

Galois theory can also motivate algebraic geometry, btw

[topaz solar]

it’s almost like orbits = types

[south patrol]

Why care about polynomials

[tough raven]

Different from?

[dull ginkgo]

The standard way using descent\

[tough raven]

Which way is that?