groups-rings-fields

[rapid cave]

!

S2 is abelian

[fiery dirge]

(1,2) = (2,1)

Yeah

Then the last 2 pairs aren't abelian

[rapid cave]

yes

so you have two pairs to check if they are isomorphic

[fiery dirge]

Okay so possible matches are 1st and 2nd and 3rd and 4th

[rapid cave]

yes

[fiery dirge]

Alr

What next?

[rapid cave]

1 and 2

what more can you say about them?

specifically 2

[fiery dirge]

Hmm

S2 is cyclic

[rapid cave]

yes

[fiery dirge]

But Z/37Z* probably isn't

And same for Z/10Z*

I don't know how to check if they're cyclic to be honest

[rapid cave]

they are both not cyclic

you can:

check the orders of the elements

[fiery dirge]

So for Z18 possible orders are 1, 2, 3, 6, 9, 18, for S2 elements have order 1

Unsure about the Z/nZ*

[rapid cave]

s2 is 1 and 2

[fiery dirge]

Wait but how?

[rapid cave]

s2 has two elements

the identity and (1,2)

(1,2)^2 = e

[fiery dirge]

Ohh yeah 👍

[fiery dirge]

Can you help me with this?

Is it a^n

Where a is the element we are checking

[rapid cave]

yeah

so this requires some checking

to understand what orders may arise in 2, notice 4 divides 36 so we may have an element of order 4

[fiery dirge]

How the hell do I calculate 35^37 or any element ^37 for that matter

[rapid cave]

^36

and you start multiplying

and reducing mod 37 after each multiplication to keep things small

[tribal moss]

A3 is actually isomorphic to C3, which is abelian.

[rapid cave]

oh yeah xD

but D12 isn't

[tribal moss]

Good point.

[rapid cave]

but notice the first group has no element of order 4

do you see why?

[fiery dirge]

Why 36?

[rapid cave]

phi(37)=36

[rapid cave]

.

[fiery dirge]

Why Z18 doesn't have element of order 4?

Because 18 isn't divisible by 4?

[tribal moss]

That's a good reason.

[rapid cave]

yep

[fiery dirge]

But what would happen in their product

All the gaps would fill in

?

[rapid cave]

no

[fiery dirge]

So we would have element of order 4 in the product

Oh

[tribal moss]

In a product group, the order of (a,b) is the least common multiple of the order of a and the order of b.

[rapid cave]

A element (a,b) in the product has order lcm(ord(a),ord(b))

Also how I arrived at 4,
I tried breaking these two into a product of cyclic groups

you can always do it for abelian groups

from this decomposition you can see mismatches like this

[fiery dirge]

Okay, how do we compare them to the 2nd product?

[rapid cave]

the 2nd has an element of order 4 (6, e)

[tribal moss]

Looking for elements of order 4 worked great so far ...

[cloud walrus]

ExpertEsquieESQUIE

[fiery dirge]

Hmm phi(4) = phi(2^2) = 2^2 - 2 = 2.

[rapid cave]

true

[fiery dirge]

But the order of Z2 x Z2 is 4

How can they be isomoprhic then

[rapid cave]

oh I wanted to say Phi_10 is Z2xZ2

mb

[fiery dirge]

Yup that makes sense

[rapid cave]

this is false

Phi_10 is Z4

mb

[fiery dirge]

Okay so my biggest issue with groups in general is that, finding orders of elements and also finding how many elements of each order there is

[rapid cave]

the second thing is annoying

because you need exact order

[fiery dirge]

Oh ☠️

[rapid cave]

returning to the question

so I made a mistake and order 4 doens't help us

but actually Phi_37 \cong C36

[fiery dirge]

Can you please tell me what is your thinking process? What are we trying to do after figuring out abelian/cyclic

[rapid cave]

orders

[fiery dirge]

We can either count all orders of elements or count number of elements of certain order?

[rapid cave]

so I though the first didn't have an element of order 4, but I was wrong

[rapid cave]

both are possible, the first one is signficantly easier

look at 5 in Phi_37

5 * 5 * 5 = 125 = 14 (mod 37)

14 * 14 = 196 = 11 (mod 37)

11 * 11 = 121 = 10 (mod 37)

10 * (11 * 11) * 10 = 110 * 110 = (-1)^2 = 1 (mod 37)

I calculated 5^3, 5^6, 5^12, 5^18, 5^36 mod 37

basically I check the order of 5 in (Z/37Z)* is 36

[rapid cave]

so this is true

@fiery dirge are you with me?

[fiery dirge]

I'm here

[rapid cave]

do you see why Phi_37 is cyclic?

or at least has an element of order 36

[fiery dirge]

Because of this?

The entire group can be generated with a single element

[rapid cave]

yes

[fiery dirge]

But 37 is an odd prime so it does make sense for it to be cyclic

[rapid cave]

yes ofc

but I checked manually xD

you really shouldn't do this

just use theorems

[fiery dirge]

Which ones?

[rapid cave]

what you have learned

in general

avoid manual computation if possible

unless you want to understand things more concretely

but usually its just a pain

[fiery dirge]

I know elementary number theory like euclidean algorithm, cvt, eulers totient function. Then for groups nothing special, subgroups, normal groups, group actions, isomorphisms, first/second theory of isomorphism etc.

I kinda skipped some topics in between but yeah

[rapid cave]

but you will learn things in the future

its fine

[fiery dirge]

In the next course yeah

This one is pretty much over with rings/fields and it's pretty introductory there

[rapid cave]

returning to the question

we have an element of order 36 in the 2nd group

but we don't have one in the first group

so they are not isomorphic

do you agree?

[fiery dirge]

Yeah

[rapid cave]

so lets move to the 3rd and 4th

are they isomorphic?

[tribal moss]

(The relevant facts to know for (Z/37Z)^× being cyclic are (a) 37 is prime, (b) the integers modulo a prime are a field, (c) the multiplicative group of a finite field is always cyclic).

[tribal moss]

Um, I think (1 in Z18, 3 in Phi10) ought to have order 36 too?

[rapid cave]

fuck I hate this

xD

late night math be like

@tribal moss can you take over?

I am tired

[proud vigil]

this is so funny to me

[rapid cave]

I wanna sleep

[tribal moss]

In fact,
$$ \bZ_{18} \times \Phi_{10} = C_{18} \times C_4 = C_9 \times C_2 \times C_4$$
and
$$ \Phi_{37} \times \mathbb{S}2 = C{36} \times C_2 = C_9 \times C_4 \times C_2$$

[fiery dirge]

D12 has elements of order 24, 12, 8, 6, 4, 3, 2, 1.
A3 is 3!/2 = 3
Z/14Z* has order phi(14) = phi(2) phi(7) = 6.
A4 is 4!/2 = 12

[rapid cave]

🙏

[cloud walrus]

Troposphere

[fiery dirge]

No worries I'll go to sleep as well, we can pause here

[rapid cave]

ok

thanks

xD

cya

[tribal moss]

I'm on the way to bed too.
But my conclusion above is that the first two groups are indeed isomorphic.

For the last two groups I think looking for elements of order 4 actually ought to work :-)

[fiery dirge]

Bye

[earnest delta]

Z_(1-i) is not maximal ideal of Z[i]

[tribal moss]

What does Z_(1-i) mean? If it's the principal ideal of Z[i] generated by 1-i, then that definitely is maximal.

[rapid cave]

localization at 1-i
/j

[thorn jay]

ah yes 1-i the famulous element of Z

[rapid cave]

xD

it was written with a subscript. So it took it literally

[thorn jay]

i mean im sure you could make it work lol

[earnest delta]

How did you check it?

It is isomorphic to z_2

They should be maximal

[rapid cave]

yep

[tribal moss]

Intuitively: The ideal <1-i> consists of exactly those a+bi in Z[i] where a+b is even.
If we have a strictly larger ideal, then it must contain something outside <1-i>, say c+di where c+d is odd.
Then (c+1)+di is also in the ideal, because it is in <1-i>, and the difference between those elements is 1, which is again in the ideal. And now the larger ideal is the entire ring.

Also note that <1+i> and <1-i> are the same ideal, since (1-i)·i = 1+i and (1+i)·(-i) = 1-i.

[astral ivy]

Let G be the subgroup of S10 generated by the permutations (4, 9)(1, 3, 5, 7)(2, 8, 10), acting on the set [10] = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 as a permutation group. Determine the orbits and stabilizers of the elements 1, 6, and 10, as well as the fixed set of a = (1, 5)(3, 7)(2, 8, 10).

How do I think about this?

[tribal moss]

This is just a "remember the definitions" exercise. There's no particular cleverness to employ here -- just apply the definitions.

[astral ivy]

So umm stabilizers are only those that fix them in place?

So for example stab 1 = (4,9)(2,8,10)?

I'm not sure about orbit tho, is 8t just (1,3,5,7)?

[tribal moss]

If you need to ask that, go back in your text or course notes to find and reread the definition of "stabilizer" rather than relying on randos on the internet :-)

Also, I'd read the problem statement very closely to make sure whether (4 9)(1 3 5 7)(2 8 10) is meant as a single permutation that generates G, or three different permutations that generate G together. You get slightly different groups out of those two cases.
(I get suspicious because you wrote "the permutations" in plural, but without commas or "and"s between the three cycles, so either one or the other must be a typo...)

Usually one doesn't write commas between the elements of a cycle in cycle notations -- do those commas really appear in your original?

[tribal moss]

There are other permutations that fix 1, for example (2 8 10) -- which is in G no matter how we interpret the problem, because (2 8 10) = ((4 9)(1 3 5 7)(2 8 10))^4.
And definitely also the identity :-)

However, (4 9)(2 8 10) is only in G if we consider it to be generated by the three permutations (4 9) and (1 3 5 7) and (2 8 10) separately.

So it looks like you also need to go back and revise the definition of "generated by".

[thorn jay]

it gives you interpretations of some object in another

so studying homomorphisms is studying the relations between things

[tribal moss]

It is extremely rare to say "these two things are homomorphic".
What's interesting is the homomorphISMS that connect them, not whether some random homomorphism exists.
For example, given any two groups G and H there's always a homomophism from G to H -- at least the one that sends everything to the identity! -- but it's still interesting to study the homomorphisms themselves.

[thorn jay]

apples 🍎

[tribal moss]

"Homomorphic image of" is stronger than *"homomorphic to" would be.
In this context it means there is a surjective homomorphism from a finitely generated free module onto the module we start with.

[astral ivy]

It's 3 of them yeah

[north stirrup]

Anyone who wanna talk or have discussions on set theory algebra including group theory etc, linear algebra, propositional logic or predicate calculus
i just like to discuss and maybe learn alot by exchanging opinions

[knotty badger]

It might be better if you mention a specific topic and/or question you have

[thorn jay]

this is like asking for discussions about balf of what you learn in first year undergrad lol

homomorphic image of M means that it is the image of a homomorphism from M to some other module

i.e. isomorphic to a quotient of M

[astral ivy]

How do you discuss something already accepted/proven to be true like definitions or axioms?

[kind temple]

i think they just mean talk about those already accepted notions and develop intuition for them, rather than discuss how they should be changed etc.

[kind temple]

yea, it seems like this works best if you start the conversation with something you are confused on or interested in. somebody will likely respond

[astral ivy]

There's no developing intuition, algebra notation is just fucked up 😭🙏

Makes analysis look easy in comparison

[kind temple]

i disagree with this, but yea lol, it certainly takes a different kind of intuition

[thorn jay]

hard disagree lma

[fading acorn]

there's developing intuition on algebra because existence of algebraic geometry /s

[twilit wraith]

If only we taught them permutations for three years

[karmic moat]

this all sounds like teacher and/or student issue not an issue with the field itself

[tardy hedge]

well in intro algebra classes you quickly see concrete examples that we're all familiar with like Z, polynomials, matrices.. isn't that enough motivation ?

[karmic moat]

in my experience it's the opposite lol

with bad teachers, students dont understand the point of derivatives/integrals

my friend is a TA for calculus classes so I sit in the help room with him and i've seen this plenty

[knotty badger]

hm it wasn't really for me

i still suck at algebra

[proud vigil]

I feel like invertible functions are everywhere in math to the point that it makes sense to study them in that context

[tardy hedge]

dude at my school i stg at least 90% of the students in calc class have barely a motivation for any of it

im a TA for the tutorials

engineering calc

[quiet pelican]

The motivation for algebra is that it’s cute and fun /hj

[thorn jay]

SO TRUE

[manic pollen]

getting my cheeks clapped in algebra rn

missed one lecture and apparently they finished all of the isomorphisms theorems

[tardy hedge]

That made me lol

[thorn jay]

which ones do you guys cover?

[tardy hedge]

Yea also is it a group theory class or what

Modules?

[thorn jay]

Im guessing groups

[manic pollen]

apparently there is 4, i dont know because i wasnt there 😭

[manic pollen]

first course in algebra

us uni

so the curriculum is like

groups rings fields

[tardy hedge]

The Classic

The Signature

[thorn jay]

ah yeah

first iso, second iso, third iso and correspondence theorem!

correspondence theorem my beloved

[manic pollen]

hopefully ill love it as much as you do

[tardy hedge]

You won’t

[thorn jay]

probably not no

given that a huge part of the coherent conditions stuff relies on the correspondence theorem and third isomorphism theorem

[manic pollen]

good to know

[toxic sapphire]

I don't think I need help on this problem right now but I think there's a typo and I'm not certain what my professor meant

"Let y != <x>"

at first I thought he meant to assume y is not in the group generated by x, so I did (a) with that assumption. but isn't that contradictory to (b)?

nevermind.

[candid patrol]

It means that y € G \ <x>

Which is not empty because G has order 8

[toxic sapphire]

right yeah I got it now

thanks though

[candid patrol]

np

[earnest delta]

Order 2 elements in S5

Partition will be

2+2+1

2+1+1+1

2+3

What should I do next?

[tribal moss]

"2+3" doesn't correspond to an element of order 2.

[earnest delta]

Why sir?

Ohh LCM would be 6 so?

Thanks

[earnest delta]

I got 25

[tribal moss]

Oh, you were being asked to count them? Yes, that sound right -- 15 of one kind, 10 of the other.

[earnest delta]

Thanks tropi🥰

I was taking 2-3 by mistake

[rapid cave]

No

Yes

But why would you take more unnecessary generators

[slow bison]

real

[thorn jay]

this lemma is basically one of the defining features of free modules

and informally explains why finitely generated modules are so nice in many ways; finitely generated free modules are often nice and those properties carry over

mooodssss

thank you :3

[alpine plank]

Quotients inherit some properties from the original module (ex:being Noetherian, Artinian)

[thorn jay]

when you want to prove something holds for all finitely generated modules, what you then usually do is prove it for free modules and show that, if it holds for a module, it holds for any quotient of it

[thorn jay]

e.g. properties which depend on the lattice structure of the the set of submodules

[alpine plank]

Yeah literally can't think of anything that isn't basically these two

[thorn jay]

generally, equations which hold

[alpine plank]

Oh right, that's very important

[thorn jay]

so for example if M is annihilated by some subset S ⊂ R then any quotient of M is

definitely not anything

f.e. every free module is flat but not every module

or projective

or, very silly, every free module is free but not every module is free :P

[alpine plank]

For a different example, In topology to show something is connected/compact it's enough to show that it is a continuous image of a connected/compact space

[dull ginkgo]

It more matters that it’s a quotient of a free module

Which is quite literally like the “module presentation” of it analogous to a group presentation

Where the kernel represents the presentation

Arguably it’s a case of the former

[limber sequoia]

concretely this means it's the set of all words of the generators together with rules that say when a sentence is equal to e.

[dull ginkgo]

it allows you to describe the structure of the module by how the generators relate just like a group presentation

Whoever decided to use the words group presentation and group representation needs to be retconned to be hit in the head with a dirt transfer shovel

[limber sequoia]

but this is algebra so it's common to say things with homomoprhisms instead of concretely.

[thorn jay]

funnily enough this naming convention isn't even category theory-pilled

it's widely used in universal algebra (and universal algebraists and category theorists had slight beef lol)

[quiet pelican]

A group representation is when you present a group again 🙃

[dull ginkgo]

I usually just say “group action” when a group maps into the automorphism group of something and call it a day

[karmic moat]

Based

[dull ginkgo]

Monoid action when a monoid maps into the Endomorphism monoid

[thorn jay]

lmao

[dull ginkgo]

The frantic edit of that message before someone vaporizes me

[thorn jay]

i had my keyboard ready to shoot

[dull ginkgo]

fear

Isn’t there a theorem that says you can decompose a finite vector space rep into invariant subobjects or something like that

[quiet pelican]

You can decompose it as a sum of reps

[dull ginkgo]

isn’t that just like Jordan normal form

[quiet pelican]

(Assuming characteristic of the underlying field doesn’t divide the group)

[thorn jay]

yes

[dull ginkgo]

can you do that for actions by a finitely generated PID module

[quiet pelican]

the main content is that you can do it simultaneously for all eles of the group

[thorn jay]

if the characteristic of k does not divide |G|, then k[G] is semisimple, so the k-reps can be decomposed as a direct sum of irreducible/simple ones

[dull ginkgo]

oh that is much less complicated than I thought

Monoid rings ftw I guess lol

[thorn jay]

you like thinking in monoids? I like you

me too

[dull ginkgo]

I had an autistic urge to see when taking the monoid ring functor for a monoid preserves Noetherianness

Like for N, N^2, etc by Hilbert basis

I wonder if applying that to F^2[M] actually gives criteria on M

Since the ideal lattice, and thus the principal ideal lattice, which you can pull back to the divisibility relation, would be Noetherian

Enough of that back to mind melting engineering work

[tardy hedge]

what do u do

are u in school or are u working?

[dull ginkgo]

Some continuum mechanics stuff

School lol

[tardy hedge]

undergrad?

[dull ginkgo]

Yeah

[earnest delta]

hints?

[candid patrol]

Do you know the cycle décomposition for a permutation

[harsh gale]

Lagrange theorem?

[earnest delta]

Then all option satisfied

[candid patrol]

It doesn’t help

[thorn jay]

a nunber is the order of some element in Sn iff it is the lcm of a collection of positive numbers that add up to n

[earnest delta]

2+3

[thorn jay]

yes

[languid wagon]

Is there a version of Mackey’s irreducibility criterion that applies to locally compact groups? The versions I seem to find online all pertain to finite groups https://dec41.user.srcf.net/h/II_L/representation_theory/12

For the particular case I’m interested in G isn’t compact, the induced representation is infinite dimensional, and I’m inducing from a character

[limber sequoia]

you should try in the #advanced-algebra channel

[languid wagon]

Okay will do thanks! I’m guessing there isn’t a dedicated channel for representation theory?

[thorn jay]

no, falls under advanced algebra

but plenty of people who know rep theory here

[prisma ibex]

sort of but you need extra care with what "Ind" means and extra assumptions on W

especially in the noncompact case it's not actually obvious what "irreducible constituent" means for the kinds of induced representations which appear; representations need not decompose into direct sums of irreducible representations and you often run into representations with continuous spectra and need to work with direct integrals rather than direct sums

this paper proves a reasonably general version of Mackey's criterion for locally compact groups if you want to see what sorts of analytic difficulties are involved https://msp.org/pjm/1998/184-1/pjm-v184-n1-p06-p.pdf

[limber sequoia]

The #advanced-algebra is the "real" algebra channel. This here channel was made to separate discussions at the level of first-second encounter with the topic.

[toxic sapphire]

I'm not sure how to approach this problem. should I be looking to construct an explicit even permutation that s_2 is the conjugate of s_1 by?

[rapid cave]

I think you should try to repeat the proof of what elements of Sn are conjugate

Or you can try to say that if they are conjugate in Sn by some g then g is actually in An

Under the conditions in (a)

[toxic sapphire]

hmm actually I think I'll sleep now and attempt this again tomorrow

[languid wagon]

Thanks! That's for systems of imprimitivity though, I don't see any theorems for irreducibility criteria there (a quick ctrl-F only shows two matches for irreducible at the end of the paper). Unless a system of imprimitivity implies irreducibility?

[tribal moss]

You can isolate the detail work in a lemma asking: Given some permutation sigma, when does there exist an odd tau such that tau·sigma·tau^-1 = sigma?
Once you have the answer to that (namely: ||exactly when sigma contains at least one even cycle or two cycles of the same odd length||) getting to the required statements about conjugacy classes can proceed at a higher level of abstraction.

[mint seal]

fun little problem for anyone who enjoys a nice diagram chase. Found in a book of old qual exam problems; this is one of the easier ones

[rocky cloak]

You don't even need to do any diagram chasing. But cute problem nonetheless

[mint seal]

hm, can it be avoided? I chose an a in A_2 and did a bit of chasing. Or maybe there's a more specific technique called chasing that this isnt?

[rocky cloak]

Like you don't have to think about elements at all is what I mean.

||beta1 is epi, so beta2=0, so beta3 is mono, so alpha1 is epi, so alpha2=0, so alpha3 is mono||

Maybe that still qualifies as chasing idk, but you never have to consider elements (or generalized elements, or pullbacks or nothing)

[elfin wraith]

I feel like I’d still count that as chasing, it’s just slightly smarter chasing

[mint seal]

that is pretty much what I did yeah, only I guess you're using some general fact to get alpha1 being epi?

[quiet pelican]

||beta3 is epi (as phi is) and mono (as proven), so is an iso, so we can invert it and alpha1 is composition of two epis (phi and beta3^-1)|| is how I’d argue it

[mint seal]

oh, I like that, very nice

[thorn jay]

kinda fun!

[thorn jay]

||phi is surjective so b3 is surjective. But b1 is surjective so ker b2 = B1, which means that b3 is an isomorphism. Therefore a1 is surjective, so same reasoning a3 is injective.||

[thorn jay]

this

[thorn jay]

yee

[tribal moss]

Hmm, I recall warnings that epi + mono don't necessarily imply iso in general, so this argument must be more specific to groups than it sounds like at first.

[quiet pelican]

Yeah it doesn’t in general, but for groups, mono and epi are equivalent to surjective and injective
(For epi this is fairly non-trivial, but for mono this is easy)

[thorn jay]

I wonder what algebraic theories have epi => surjective

[quiet pelican]

Same proof works for this diagram in an abelian category too

[tribal moss]

Yeah, it just struck me that the problem statement said "injective" and "surjective", and switching to "mono" and "epi" when phrasing the argument could give the impression that the argument suddenly works in far wider generality than it does.

[quiet pelican]

It works in any abelian category so I feel it’s at least somewhat justified

[knotty badger]

Yeah every abelian category is balanced, and this is typically where one talks about exact sequences

[ivory ore]

any two semisimple decompotions are equivalent, which is easy to state as perhaps a corollay of jordan-holder but what do i do for infite cases?

[rocky cloak]

If S is a simple module, then Hom(S, M) is an End(S)-vector space.

So it just reduces to a dimension argument

[ivory ore]

nice

i mean it's division ring right?

[rocky cloak]

End(S) is a division ring yeah

[ivory ore]

schur's lemma comes 2 sections later, prolly will learn about vector spaces over it then

[rocky cloak]

You can avoid linear algebra as well if you like.

Just say M = Sum_i M_i = Sum_i N_i
with M_i and N_i sums of copies of some simple module S_i.

Then the identity on M can be thought of as a map between these sums. Since there are no maps from M_i to N_j for i different from j M_i must be a subset of N_i. Then similarly interchanging M_i and N_i they must be equal.

[ivory ore]

right yeah, the fact that the blocks can be different is not a problem since there are no non-trivial maps between them

[bitter locust]

is there a generalization of the "divisibility <=> roots" from polynomial rings for general rings?

like a polynomial having a factor of (x - 2) means 2 is a root

can this somehow be reintetpreted to hold for other rings?

[knotty badger]

probably using ideals, i imagine?

this is kind of a starting point for algebraic geometry i think

[bitter locust]

yeah that's what im thinking, prime ideals or somethingn

[knotty badger]

mhm

in this case, the map $R \to R / I$ for $I$ an ideal can be viewed as a generalised kind of "evaluation"

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

if you use the ideal generated by $(x - 2)$ in a polynomial ring, you recover the factor/remainder theorem

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[bitter locust]

ah i see now

[knotty badger]

(also fyi i am very new to AG so take anything i say with a grain of salt, and feel free to leave if i start talking nonsense)

[bitter locust]

so for example
f = 28 from Z and an ideal (7)
on one hand 7 divides 28
on the other hand evaluating 28 at the ideal (7) through Z -> Z_7 yields 0

[knotty badger]

yeah

a big part of AG is trying to view elements of a ring as functions on some space

lots of rings, like polynomial rings, arise this way

[bitter locust]

that makes much more sense now than a few years ago

[knotty badger]

a few years ago?

[bitter locust]

i remember watching some alg geo vids and stuff and was so confused when they mentioned using ring elements as functions

iirc i was new to abstract algebra back then

[knotty badger]

i see i see

from what i understand, a lot of classical AG is about studying solutions to polynomial equations

and what's interesting about polynomials (at least to me) is that you can interpret them in any ring, provided you know how to interpret the coefficients

like the equation "x^2 + y^2 = 1" makes sense in any ring, so every ring has its own version of the circle

[mint seal]

indeed, though for alg geo they typically work over algebraically closed fields, so that all polynomial systems have solutions sets

[knotty badger]

yeah that part i still don't really get

but the circle thing is cool to me

and i guess one of the hopes of AG is that solutions of the same equation across different rings "interact" in a nice way

[mint seal]

well, a polynomial like x^2 + 1 has no roots over R

[bitter locust]

oh that you can plug elements of any ring into such polynomial equations?

[knotty badger]

so that knowing solutions in one ring gives you info about solutions in another ring

[knotty badger]

yeah!

[bitter locust]

i see

[knotty badger]

given any ring $R$, you could define $S^1_R = {(a, b) \in R^2 \mid a^2 + b^2 = 1}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

this is what i mean by "every ring has its own version of the circle"

(in fact, since the circle is an algebraic group, meaning the group operation can be expressed with polynomials, S^1_R is always a group)

[bitter locust]

thats cool

[knotty badger]

when R is the real numbers, you can interpret this geometrically as an actual circle

when R is the rationals, this gives you info about pythagorean triples

[bitter locust]

also technically every group would have its own "taxicab circle" since thatd be x + y = 1

[mint seal]

you want 1 to be a multiplicative identity, though

[bitter locust]

oh shouldve written 0

oh

nvm

[knotty badger]

even something like looking at diophantine equations modulo n is an instance of this

you interpret the polynomial equation in the ring Z / n Z

and use information about the solutions in that ring to help you determine solutions in the ring Z

https://en.wikipedia.org/wiki/Faltings's_theorem

this is also a very cool instance of this concept

where solutions over C interact with solutions over Q in a nice way

from what i can tell, the fact that the same polynomial can be interpreted in multiple rings is pretty core to classical AG

[bitter locust]

makes sense, ik you often end up working with a ring R and then also with its polynomial ring

or the underlying ring

[knotty badger]

then ig modern AG is trying to extend this to non-polynomial rings and allowing geometric intuition to happen

and a big part of that seems to be "localization"

[bitter locust]

ah yes that

still dont understand it that well

but ik its a kind of "zooming in" in some cases

[elfin wraith]

Mfw non commutative alg geo

[noble nexus]

localization tends to be a little unintuitive because when you localize the ring in general gets larger

but if you think in terms of ideals the ring gets "smaller" which is whats important

and the intuition from functions

[knotty badger]

i have some intuition for it if you'd like to hear

[bitter locust]

sure thatd be nice

[tardy hedge]

what is this intuition?

[knotty badger]

so the way i think of localization geometrically is as an identification of functions

if you zoom into some local region, then two functions might appear to behave the same, even if they don't globally

[noble nexus]

consider the ring of functions on the real number line (under pointwise multiplication). Then define the ring of "germs of functions" at 0, which functions defined on an open neighborhood of 0 but we say two functions are equivalent of they agree on some smaller open set

[knotty badger]

you're identifying things that have the same local behaviour, essentially

[noble nexus]

you can check that this ring of germs is indeed a ring

importantly, notice that the functions with f(0)=0 form an ideal

and if f is not in this ideal (a function does not vanish at 0) then 1/f must be defined on a neighborhood of 0

by continuity

thus f is invertible in this ring

so the ring of germs is a local ring with a single maximal ideal of functions vanishing at that point

in algebraic geometry its the same idea

prime ideals <-> points
elements of R <-> functions on the prime ideals
elements of a prime ideal p <-> functions vanishing at p

[rapid cave]

This is actually very nice

[noble nexus]

if you localize at a prime p, you are inverting all elements not in p, which is basically inverting every function that doesn't vanish at p

then in the local ring, the prime p is maximal and everything not in p is a unit

[tardy hedge]

Thanks

[bitter locust]

both of those intuitions are quite nice tbh

[noble nexus]

this association is easier to understand if you've been through varieties or if you've seen gelfand duality in functional analysis

[elfin wraith]

Im struggling to parse this, what is the ring of germs defined to be?

[noble nexus]

the elements are pairs (f,U) where U is an open set containing 0 and f is a function from U to (say) the reals

[elfin wraith]

I do know the intuition you’re building, I remeber my UG advisor talking about this at some point but I’m not following the specifics

[noble nexus]

but we put an equivalence relations where (f,U) ~ (g,V) if there is some open subset W contained in the intersection of U and V such that f|_W = g|_W

its a fairly natural object to think about when talking about differentability for example, since the derivative can be defined on an arbitrarily small open set and if two functions restrict to be the same on a small open set they have the same derivative at that point

[elfin wraith]

Ah I see that makes sense

I’ve always wondered what geometers were talking about with germs

[noble nexus]

evidently you use germs to define the tangent space of manifolds, and modeled on that you can actually define the tangent space of objects in algebraic geometry as well

[tardy hedge]

thats cool cuz im taking diff geo this term

[noble nexus]

most differential geometry sources won't actually use germs since they'll kinda cheat

basically to define the tangent space at a point of a manifold, you want a vector space of "possible directions" at that point

and the way you do that is by looking at the vector space of "directional derivative operators"

if you denote by $C_p^\infty$ the ring of smooth germs at a point $p$ (whatever smooth means, you'll learn it)

[cloud walrus]

Blake

[noble nexus]

then a directional derivative operator is a linear map $d:C_p^\infty\to\mathbb{R}$ such $d(fg)=d(f)g(p)+f(p)d(g)$

[cloud walrus]

Blake

[elfin wraith]

Oh yeah my alggeo course did actually do that, just implicitly I suppose

[noble nexus]

so for example in R^n you have the partial derivative operators $\partial_i|_p$ defined by $(\partial_i|_p)f := (\partial_i f)(p)$

[cloud walrus]

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

[noble nexus]

the way that most differential geometry courses cheat is that they ignore germs and instead define a derivative on the space $C^\infty(M)$ of smooth functions on the entire manifold

[cloud walrus]

Blake

[elfin wraith]

My alggeo course tried to minimise the amount of topology in it to a ridiculous level that just made everything harder and worse for some reason I still don’t understand

[noble nexus]

b/c on a manifold any germ can be extended to the entire manifold if you muck around with topology a bit

in some way

[quiet pelican]

They’re trying to infect the rest of maths

[elfin wraith]

Those filthy agriculturalists

It is the area of topology I have a bit of a blind spot for though, like I should really learn about sheaves at some point, I had a hard time with Wibel because I didn’t know about them

[knotty badger]

sheaves are a cool idea imo

[elfin wraith]

I very barely missed out on a sheaf theory course at my UG which kinda sucked, that would’ve been a very interesting and useful class

[thorn jay]

theyre starting to feel like the "true" geometric objects to me

lol

[knotty badger]

in what way

[elfin wraith]

From a very good topologist who doesn’t believe in exams no less (my dream course)

[thorn jay]

well sheaves are in essence "thickenings" of a topological space

[knotty badger]

[thorn jay]

which makes it "more geometric", like idk

[knotty badger]

i don't understand what this even means

[noble nexus]

you add geometry through a sheaf of functions p much

[thorn jay]

they add more information

yeah

[noble nexus]

since for example real manifolds, complex manifolds, complex analytic varieties, algebraic varieties etc are all just topological spaces with certain sheaves of functions

[knotty badger]

uh

hm

[noble nexus]

and u can do other random things like look at piecewise linear manifolds

[knotty badger]

@.@

[thorn jay]

the prime motivation for this viewpoint are of course schemes and manifolds for me

[noble nexus]

things didn't really click to me till I saw both the alg geo variety and manifold perspective on sheaves

[manic pollen]

anyone have suggestions for books to read that follows closely to my lecturer's progression of topics. (US Univ, first course in algebra)

all the books ive looked into looks very different, and im looking for a book because he posts lecture notes so late...

[thorn jay]

things didnt click for me with schemes until i went through the motions of seeing how nice regular functions on affine varieties were

[knotty badger]

could you explain a little more?

[thorn jay]

well the Zariski topology does not give much rigidity to the points in the spectrum, i.e. doesnt carry much geometric information, because there are so little closed sets in contrast to, say the topology on the real line

this means for example that theres only a contravariant functor to Top, rather than a duality with a subcategory of it

[elfin wraith]

I’m on a train with shit service so I can’t open these, but going from set theory to category theory in 4 lectures seems quick

If I had to guess I’d say look at Aluffis chapter 0, thats a book which does category theory from the start so it’s possibly similar

[thorn jay]

if you add the structure sheaves onto the spectra, though, that for example carries enough additional information (which is also geometric in some sense, as they are to be interpreted as continuous functions with some local "niceness" property, e.g. looking locally like a quotient of polynomials in the case of classical AG) to make a duality appear

[rocky cloak]

The section seems more about homomorphisms and kernels and so on, as opposed to defining what a category is.

[knotty badger]

If I’m understanding you correctly, are you saying something like “geometry is when spaces with local structure/data”?

[thorn jay]

ges

yes

[knotty badger]

Interesting

[thorn jay]

some data or information that links points together in any way

[knotty badger]

And that is what a sheaf is meant to capture..

[thorn jay]

exactly :3 that is what i am realizing

[knotty badger]

Ok I understand this

And that even makes sense for manifolds

If I interpret local structure as a local diffeo to R^n

[thorn jay]

yis, that is what the sheaf definition of a manifold captures

[elfin wraith]

Ah fair, that seems much more standard

[knotty badger]

Which isn’t exactly sheafy but still geometric

[knotty badger]

The sheaf def is more about the algebra of smooth functions right

[thorn jay]

yeah, but it carries the same idea

the precise definition is Man1fold's bio lmao

[knotty badger]

Mhm

To be clear I still prefer the standard def

But I have a bit more appreciation for the algebra def

[manic pollen]

yea, one other thing is that he focuses a lot on using a word problem to represent a group, which i couldnt find in some of these books ive looked into

[thorn jay]

that makes sense, stuff like local coordinates seem easier to intuit

[noble nexus]

the sheaf definition of a manifold is more helpful for understanding sheaves than it is for understanding manifolds

[thorn jay]

hahaha

[elfin wraith]

I’m not sure what you mean by that, do you just mean that you often use the generators and relations way to define a group? That is pretty standard

I’m almost home I’ll check your notes when I’m back and see if I can be of more help lol

[manic pollen]

yea, but for some reason i just couldnt find it in the books prescribed in the syllabus (Artin and hungerford)

i might just be looking for it in the wrong place

[manic pollen]

thanks

[elfin wraith]

Ok I’ve had a look over it and my possibly unhelpful advice is to just not worry about it too much. You should definitely try to understand the section on the word problem and stuff because it’s not super important

I think most books when they introduce group presentations will present the same material, but those lecture notes are the most in depth and explicit I’ve ever seen, in practice people are usually a lot more fast and loose with the technicalities of formal symbols and words etc

[white oxide]

Silly question but here is minimal taken to be the minimal defined in a set theoretic context

Aka if p_j is minimal there is no p_i with p_i \subset p_j

[tardy hedge]

yea

[manic pollen]

mmm, ic. ive more or less understood word problems(i think at least). yea this prof makes damn good lecture notes

the problem is that he upload them quite slow

so we are currently missing 3 lecture notes

and since i missed a lecture a while ago

i can no longer understand what the hell they are talkijg about anymore

im just worried of falling too far behind

[rocky cloak]

Unless there also is a pretty good description of what's going on in class I'm not sure having a book will help catch up necessarily.

You can try to ask the lecturer where in the book the relevant material is or you can try to see if you have some dilligent colleagues who will share their notes from the relevant classes.

Either way asking the lecturer what the best way for you to catch up is a good idea

[manic pollen]

the past few lectures have been about the isomorphism theorems, orbit stabilizer theorem

and the rest i did not catch

[manic pollen]

yea that sounds smart

[rocky cloak]

These sound like subjects you should be able to find in any algebra book

[manic pollen]

im worried about the order that its presented in, cuz the lecture order and the order that it is introduced in the books are different

is it normal to skip chapters while reading?

for example in one book groups is one of the last topics presented

while it is first topic in the course

[karmic moat]

I don't think it's a cause of worry for a prof to skip around a textbook

Most textbooks aren't meant to be read as novels anyway (as in, cover to cover in order)

[rocky cloak]

Yeah totally normal.

People learn / lecturer teach in whatever order they prefer and might not use everything a book has to offer

[manic pollen]

ah

i just always feel a tinge of unease whenever i skip a chapter

[manic pollen]

ic

im just pretty used to going in order of the table of contents

[rocky cloak]

Some books have these fun diagrams of which chapters use theory from which other chapters to tell what not to skip

[thorn jay]

those are so good

they look pretty too

[karmic moat]

Vakil's diagrams are always funny because he handwrites them

So they look like xkcd

[thorn jay]

🔥 🔥

[manic pollen]

ic ill try to find these diagrams

thanks yall

[rocky cloak]

Not every book has fancy diagram, but they should hopefully have some text about the intended reader / how to follow the book at the start

[delicate orchid]

these chapter names...... they really shoulda gone into comedy

[karmic moat]

how do you even read that one

each time i remove a block i think the whole thing falls over

[delicate orchid]

prerequisites support the chapters above

[karmic moat]

maybe i suck at jenga

[quiet pelican]

What if your dependency graph isn’t planar

[karmic moat]

write two volumes

[delicate orchid]

good thing this one is

look at em go!

[karmic moat]

ohhh i see it now

alternatively you dont even have to add a dependency graph

and just leave it up to the lecturer/learner to struggle to find it themselves

[quiet pelican]

Me making a dependency graph that needs 6 volumes for 30 chapters

[karmic moat]

did classification of finite simple groups have any dependencies within it

[delicate orchid]

I don't think anyone has ever actually seen the classification

it is lost in the tombs of the vatican 💔

[karmic moat]

ah yeah just the usual coffee table book

[quiet pelican]

I wouldn’t be surprised either way tbh
Like I could see someone being crazy enough to print out the tens of thousands of pages of papers
But equally I could see no one being that crazy about CFSG

[karmic moat]

I don't even recognize most of these groups

[quiet pelican]

It isn’t unreasonable to spend a couple hundred pages defining them all if you’re aiming at a strong undergrad, and you’re not just beelining from one to the next

[rocky cloak]

I recognize cyclic, alienating, Tits, Mathieu, monster and baby monster.

Would have to look up which of lie type is which and I don't know the other sporadics

[delicate orchid]

I recognise all of the sporadics and most of the lie types

[elfin wraith]

Get a load of this guy

[delicate orchid]

that's not a good thing

[elfin wraith]

I’m aware, that much group theory is bad for the soul

[karmic moat]

how often do you use the sporadics?

[delicate orchid]

they appear quite often in my area

[karmic moat]

oh damn

[delicate orchid]

like, Aut(M22) = M22 : 2 was used in a counter example to a conjecture recently

M22 or M24 I forget which

it was M22 I'm not a fraud

[karmic moat]

what was the conjecture?

[delicate orchid]

now if I tell you this I will dox myself but I'll do it anyway

[karmic moat]

oh you don't have to lol

i won't understand it anyway

just curious

[delicate orchid]

it was about certain representations being subreps of the regular representation

[karmic moat]

oh okay i did understand that

[delicate orchid]

it was quite silly in the end - it was a disproof by the pidgeonhole principle

there were simply MORE of the certain reps than there were subreps of the regular

which is absolutely digusting and it's not a surprise you need something as disgusting as Aut(M22) to do it

[karmic moat]

does M22 show up naturally anywhere

[delicate orchid]

M23 and M24 appear here https://en.wikipedia.org/wiki/Binary_Golay_code

which was used as part of the data compression for the transmission protocol in the Voyager space craft

[karmic moat]

ohh interesting

that's cool actually

[manic pollen]

will note, thanks 👍

[delicate orchid]

not data compression sorry, the error checking

[karmic moat]

oh well see now it's really lame

jk that's still pretty cool

[delicate orchid]

https://tenor.com/view/rizzler-costco-rizz-rizz-face-the-rizzler-gif-12462530958478786405

fuck it says that fun fact on the wikipedia page now I look like a HACK

[elfin wraith]

How do you even find that? GAP go burr?

[delicate orchid]

yeah

although the sporadics are usually a good place to check for "weird shit"

so probably not too much burr

[elfin wraith]

Yeah same thing as the unit conjecture then I guess

[delicate orchid]

is that mf still open

[elfin wraith]

Narrow it down to one of two families of groups and just abuse a computer

[elfin wraith]

Nope, counter example in 21

[karmic moat]

also pigeonhole counterexample?

[delicate orchid]

https://tenor.com/view/mini-oso-luna-tabby-lunaistabby-gif-220644464987255023

[elfin wraith]

The zero divisor and idempotent conjectures are still open though, the zero divisor conjecture in particular seems quite interesting

[elfin wraith]

No just an explicit counter example found by GAP

[delicate orchid]

nah the Kaplansky conjectures are actually like, complicated

[karmic moat]

is GAP like macaulay2 for group theory

[delicate orchid]

since you know about the regular rep you'll understand the statement of them https://en.wikipedia.org/wiki/Kaplansky's_conjectures

[karmic moat]

oh interesting

[elfin wraith]

Why do you need to know about representations to understand them? Just because of group rings?

[delicate orchid]

yeah

[elfin wraith]

Shameless plug if you’re interested in them

I believe my name is on that just be nice and don’t dox me

[delicate orchid]

although this is still super surprising to me personally. Like taking group rings is an adjoint to the forgetful functor and taking groups of units is the other adjoint. Can we just fuckin get a grip it can't be that hard.

[elfin wraith]

I only ask because I know piss all representation theory lol

[karmic moat]

these statements look really innocent in nature lol

[elfin wraith]

The zero divisor conjecture is particularly interesting because it’s like wrapped up in a web of other conjectures lol

[delicate orchid]

representation theory will tell you about the indecomposable and simple ideals but nothing about the elements directly like the conjectures ask

[elfin wraith]

Like it’s very strongly related to the Atiyah conjecture

[quiet pelican]

Apparently I’ve started recognising names at least in my little corner of maths
Save me

[delicate orchid]

you are beyond salvation

[elfin wraith]

Does Gardham work on GGT stuff?

[quiet pelican]

He’s someone that was recommended to me to look at for PhDs yeah
But he’s not taking any this year 😢

[elfin wraith]

Sad times, is he still at Bonn?

[quiet pelican]

Yes

[elfin wraith]

You don’t want to go there anyway (they rejected me)

[thorn jay]

i wont trust me 🙏 🙏

[karmic moat]

the only group ring i care about is the hopf algebra of algebraic groups/group schemes amen 🙏

anyway still plenty of rep theory in them i mean the comorphisms come from the regular representations

[elfin wraith]

For context that was a 5 page report I had to write for my topics in ring and rep course on non com rings, so it could be more detailed but I was working to some strict length limits and under the assumption that everyone in the class had to understand it

[karmic moat]

last thursday or something i spent a long long time computing stuff about SL(3, C) lol

just to see where it all came from

[delicate orchid]

the only group I'm scheming with is my lawyer adviced me I do not finish this joke

[karmic moat]

computing all the root systems and weight spaces and nonsense

[delicate orchid]

SL(3, C) is the degree uhhhhhh

3? symmetric polynomials?

how does ts mf work again

[karmic moat]

i thought SL(3, C) was just the special linear group of GL(3, C)

[delicate orchid]

I mean the associated lie algebra

[karmic moat]

3x3 traceless matrices

[delicate orchid]

yeah

ok what do I actually mean

there's some way you can get the reps of sl(n, C) out of looking at the action of the symmetric polynomials of degree something in some number of variables

[karmic moat]

oh interesting

what about other fields?

[delicate orchid]

no clue

[karmic moat]

like if we loosen C to alg closed fields, or char 0, or perfect

darn

[delicate orchid]

I don't care for anything with infinite cardinality

Z is an exception

[karmic moat]

i don't care for char p >0

i'm happy in my little world of C

[thorn jay]

you should start doing tame congruence theory

[thorn jay]

do you have these ready or do you make them on the fly

[karmic moat]

AAAAAH AAAAAAH

[delicate orchid]

make them on the fly

[karmic moat]

get that UGLY thing from the bottom away from me

[twilit wraith]

My favorite field is the one with cardinality Beth omega 1

[thorn jay]

amazing

[delicate orchid]

who tf is beth

[twilit wraith]

Who knows bruh

[karmic moat]

comes from hebrew apparently

[rocky cloak]

When Latin fails you reach for Greek, when Greek fails Japanese, Cyrillic and Hebrew start fighting it out.

[rapid cave]

I have never seen cyrillic and japanese in math

[elfin wraith]

You need to start getting more esoteric in your interests

[karmic moat]

i recall seeing the hiragana "na" somewhere

[rocky cloak]

yo for yoneda embedding I've seen

[karmic moat]

oh yeah yo sorry not na

[rocky cloak]

よ

[karmic moat]

chinese is such an untapped resource... thousands of characters... no more overlaps in notation

[elfin wraith]

Learning pinyin to maximise the horrific state of your notation

[karmic moat]

pinyin aint that bad

at least not for mandarin idk about the other dialects

[rapid cave]

I cannot wait to show there is an isomorphism between 数 and 学

[elfin wraith]

How long until we get Aramaic in maths

[karmic moat]

gonna write my complaint to ea-nasir and put it on arxiv

[rocky cloak]

Then there's this
https://en.m.wikipedia.org/wiki/Dirac_comb

[rapid cave]

Cursed

[karmic moat]

yeah whatever

[charred iris]

Well cyrillic you wouldn't see much since a lot of its letters look like latin or greek ones already

[elfin wraith]

Flashbacks to doing a physics degree, make it stop

[south patrol]

good

[limber sequoia]

א ב ג ד ...

I like using ж sometimes when I feel bored

[tough raven]

I believe that's not a group ring except for diagonalisable algebraic groups...

[tough raven]

Do you mean Schur--Weyl duality?

[karmic moat]

i likely misunderstood then definition then

[soft tiger]

How would you prove that the first bullet point implies the second

[solemn garden]

you could prove that the first one is equivalent to normality of N, and then do the same thing for the second

[tough raven]

Hint: ||Note that (gh)N = g(hN). What do you know about hN?||

[lucid shadow]

repost here

[agile ingot]

Hello, this might be a elementary question but I encountered while studying cosets. So, what does it mean for a operation to be well-defined? I know it means for different representative of the inputs it must have the same output but
Like, take (uK)(vK)=(uv)K

if i take u_1 K = uk and v_1 K = vK
isnt (uv)K=(uK)(vK)=(u_1K)(v_1 K) = (u_1 v_1)K

this seems perfectly valid as well? am i missing somthing?

[velvet hull]

a map is said to be well-defined if a = b implies that f(a) = f(b)

it is a very subtle mistake that can be made while constructing homomorphisms

for instance I could erroniously claim that Z/2Z surjects onto Z/3Z, by sending 1 to 1.
Indeed if you checked for yourself that this function does satisfy the group homomorphism property; but the mistake was made in assuming that this was a function to begin with

[agile ingot]

so if my operation is a function then what i just wrote is fine yeah?

[velvet hull]

no, well-definedness is exactly a problem that arises with functions

you have to make sure what you have is a function to begin with

and that's what well-definedness is concerned with

[agile ingot]

but to verify its a function you have to verify its well defined, so it seems circular

[velvet hull]

no, well-definedness is how you verify it is a function, there is no circularity here

[twilit wraith]

well what you initally define is a relation

usually you can just handwave the fact that it is indeed a function but with cosets it requires that you verify

[velvet hull]

we don't "care" about well-definedness

we just care that it's a function

[agile ingot]

i see now

ok

got it

[ivory ore]

is the hypothesis of locality a direct consequence of semisimplicity(the maps being either unit or nilpotent(which forms the unique max ideal i suppose))

does the locality imply semisimplicity at least for finite length

[delicate orchid]

If so that’s a connection I was unaware of. I’ll go and find what I’m thinking of at some point today

[rocky cloak]

For finite length modules, having local endomorphism ring is equivalent to being indecomposable.

You seem to maybe be mixing something in your first sentence, or maybe I just dont understand what youre saying. But if M is semisimple, then each simple summand will have local endomorphism ring, since division rings are local.

Endomorphisms of semisimple modules arent either nilpotent or unit. Maybe youre thinking of Fittings lemma for indecomposable finite length modules?

[tribal moss]

Unless you know that more about K (namely that it is normal), it is not automatically true that (uv)K and (u1v1)K are the same coset.
For example, working in S3 -- I hope you know permutation notation! -- we could take K={e,(2 3)}. Then (1 2)K = (1 2 3)K = {(1 2), (1 2 3)}.
But if we take u=v=(1 2) and u1=v1=(1 2 3), we get uv = e but u1v1 = (1 3 2), and those do not have the same coset.

isnt (uv)K=(uK)(vK)=(u_1K)(v_1 K) = (u_1 v_1)K
The middle equals sign assumes that you already know that (uK)(vK) is an operation on cosets whose results only depends on what they are as sets. And that's what you're trying to argue for in the first place.
This is potentially confusing, because there is actually a multiplication operation on subsets of the group that only depends on the sets, namely AB = { ab | a in A, b in B}. However, if that is the operation you're thinking of, then
(uK)(vK)=(uv)K
stops being a definition of anything, but is just a claim about already-defined concepts which might turn out to be false. It happens to be true if K is a normal subgroup, but that is not obvious and needs an actual proof.

[tribal moss]

"verify it is well-defined" is a conventional but somewhat misleading phrasing. There's no such thing as "a function that is not well-defined". What you should actually think of is "verify that such-and-such proposed definition manages to define a function at all". It's a property of the definition rather than a property of the function (since there is no function until the definition works).

[dim wagon]

i copied this as part of my notes in a lecture, can i check what it means to embed something into an S-module?

I thought maybe it just meant that something can be made into an S-module but that wouldnt make sense since i can just make S-action trivially so what is it trying to tell me here

[tribal moss]

What does N mean here? Some arbitrary R-module?

[dim wagon]

${}_RN$

[tribal moss]

Vibe word, technical meaning very context-dependent.

[cloud walrus]

somethingwrong

[dim wagon]

a left R-module

[agile ingot]

I appreciate your explanation.

[dim wagon]

"We can interpret the above theorem as ker j being the largest obstruction such that N can be
embedded into an S-module." I also copied this down but i dont know if it helps

[tribal moss]

Huh, I would have read the subscript R as attaching to the tensor-product symbol rather than to N. I may be out of my depth here, sorry.

[dim wagon]

oh no dont worry about it, but for my course, we would usually write something like $M_R$ and ${}_RN$ to mean that $M$ is a right $R$-module and $N$ is a left $R$-module, then write $M\otimes_R N$ to mean the tensor product of $M$ and $N$ over $R$ so the $R$ is indeed attached to the tensor product

[cloud walrus]

somethingwrong

[tribal moss]

"A embeds into B" means "A is isomorphic to a subthing of B", or equivalently, "there is an injective homomorphism from A to B".
This ought to work the same for groups, rings, S-modules, etc.

[rocky cloak]

Any S-module is also an R-module through phi.

So embedding into an S-module means that it's (isomorphic to) an R-submodule of an S-module.

Really a stronger statement is true, that if M is any S-module, then for any R-linear map
N -> M
there is a unique S-linear map S(x)N -> M that it factors through (using j)

[elfin wraith]

Yeah I’m pretty sure I made a mistake in my application because my rejection was worded very strangely and my friend who did practically the same as me in the degree is now studying there lol

Oopsies

I’m not too fussed but I do mourn all of the kölsch I could’ve been drinking

Nah just Bonn outside of the UK, I looked at Essen but it was in German and my German is not that great lol

No I’m aware lol, it cost me quite a lot of money fucking up the application, Bonn would be free and it’s a pretty cheap place to live

But so it goes, I’m at a really good uni in the UK now anyway, only behind oxbridge for maths and pretty much on par with imperial (but without London living costs) so it’s all good

I was going to apply to EPFL but I just couldn’t justify the cost or risk speaking French

Oh shit yeah you’re Swiss aren’t you? A lot of German unis will charge you too right

Yeah TUM is a no go if you have to pay tuition lol, on top of the cost of Munich

Is it all UEZ countries that get it free not just EU?

Brexit continues to ruin my life

EPFL would be great, amazing uni and probably the most beautiful campus in the world

No but just from all the pictures and stuff I saw when I was thinking about applying

Brother it’s on lake Geneva

Though I did go on a date with a girl from Lausanne and the only positive thing she had to say about the place was there’s good illegal raves lol

A few rows of buildings is whatever, I go to uni in the West Midlands 🥀

[azure hull]

yea looks good same arguments work for descending chains any descending chain in a submodule K is a descending chain in M and any descending chain in M/K corresponds (via the correspondence theorem) to a descending chain of submodules of M containing K If M is Artinian those chains stabilize, so K and M/K are Artinian

yea vector space over F is precisely an F module where F is a field more generally an R module is like a vector space but over a ring R instead of a field When R = F is a field every nonzero scalar is invertible which gives lot more structure
this invertibility has powerful structure every vector space has a basis,any linearly independent set can be extended to a basis,dimension is well defined (all bases have the same cardinality), and every subspace has a complement

[thorn jay]

genuinely what do you think 😭

yes, we call it that because conventially we call modules over a field vector spaces

so its just to accentuate the fact that F is a field and we have a vectorspace

type shiii

nice

[thorn jay]

yes

[quiet pelican]

Pedantic note: I wouldn’t use the notation K/N unless K contains N

[thorn jay]

though it does make sense as the image of K under the natural projection

[quiet pelican]

(K + N)/N

[thorn jay]

or π(K) where π: M → M/N

[quiet pelican]

K + N is the minimal module containing K and N

It’s also the elementwise sum of K and N

Yeah

Correct

[thorn jay]

this one right

[balmy wraith]

off topic, but i love your pfp

[delicate orchid]

thanks!

[tardy hedge]

you're welcome

[thorn jay]

this is the perfect joke

[tardy hedge]

dude ikr

[tough raven]

It would be if it consisted of 6 messages.

[twilit wraith]

Am I right that if G can be given as a left conjugation of K then K can be given as a right conjugation of G

It feels obvious but im worried

[kind temple]

if G and K are subgroups of some larger group and G = x K x^-1, then K = x^-1 G x

is this what you mean?

[twilit wraith]

Yes

Im like 99% sure it is but maybe im not thinking about it hard enough

[ivory ore]

right, sorry i brought up semisimplicity for no reason at all.

[twilit wraith]

What the hell is a left regular representation

They use the word afford but idk what it means

[prisma ibex]

this is such a silly detail to get stuck on

[twilit wraith]

Am I just supposed to know whether or not the word has a formal definition in the context of group theory before knowing the definition

[blissful lantern]

bro, this field doesnt make any sense to me

im getting fried in my first algebra course

But im lowk addicted to this feeling of being stupid, then figuring it out

[prisma ibex]

[twilit wraith]

one more sully and i might just get it

[prisma ibex]

anyways what are you still confused about

[twilit wraith]

nothing i just read a little more

[prisma ibex]

excellent

[twilit wraith]

apparently affords just associates a homomorphism into the relevant symmetric group with the considered action

but it was defined in a previous chapter i skimmed over hence my confusion

[prisma ibex]

affords is not a formal definition lol

[sly crescent]

Are the elements of a tensor product of matrix groups Kronecker products of the elements of the multiplicands?

[prisma ibex]

but yes left regular/right regular representations are defined in various contexts by the left/right multiplcation action of a group on itself

[twilit wraith]

i see

so the left regular representation is a map from G into its symmetric group that maps g to its left multiplication action

at least thats what im understanding

[tribal moss]

Just a (somewhat archaic) way to state that such-and-such representation exists.

[mighty spade]

note sense 2

[karmic moat]

Affording representations? In this economy?

[tardy hedge]

it be like that

[twilit wraith]

Gotcha

Seems like induces is an alternate and in my opinion more understandable word to use for that

So ill just say that I guess

[dim wagon]

this is a portion of the proof i copied to show that 1 implies 2

Let $Y' \subseteq Y$ with $Y = Y' \oplus \alpha(X)$. For $z \in Z$, there exist $y \in Y$ such that $\beta(y)=z$ since $\beta$ is surjective. Write $y=a+b$ with $a \in Y'$, $b \in \alpha(X)$, and define $\gamma:Z \to Y$ by $\gamma(z)=a$. This is well-defined since if $\beta(y)=z=\beta(y')$ with $y'=a'+b'$, then $0=\beta(y-y')$ so $y-y' \in \alpha(X)$. Since $y-y'=(a-a')+(b-b')$ and the decomposition is unique, we get $a=a'$

[cloud walrus]

somethingwrong

[dim wagon]

actually, we don't really need to show well-defineness right? we can just make a choice of any y since we are just trying to define the function gamma

[naive lance]

whats a "positive Energy representation of the Lorentzgroup" i know what a grouprep is its just about the energy part

[lime badge]

I have a very silly question

Let's say I have a ring R and I consider the opposite ring R^op. I'll denote the multiplication in the opposite ring by •

Is the right way to define a ring homomorphism φ: R^op --> S as φ(r • s) = φ(s)φ(r) or φ(r • s) = φ(r)φ(s)?

[vapid vale]

is there a good, elementary, non-casework based way to communicate (convicingly, but perhaps imprecisely if needed) that the finite subgroups of SO(3) must be the platonic solids? i mean maybe you can start by considering the orbit of some point on the sphere and take its convex hull and maybe conclude regularity of that polyhedron (or polygon)

but that seems kind of frustrating to communicate that sort of object, and even then its not very easy to convince someone that this group action must preserve the right properties

[rocky cloak]

A ring homomorphism should preserve the order of multiplication, so you'd want
phi(r * s) = phi(r) phi(s).

Otherwise what you would have would be a homomorphism R -> S

[vapid vale]

i guess with that explanation i either just handwave at this convex hull and try to let the person believe its true, or go a little longer than i'd like if i am to be rigorous. so i'm trying to find a middle ground

and i just dont know the easy way to see it

[lime badge]

Ok perfect, thanks!

[tribal moss]

Well, it's not quite true, because there are also finite subgroups isomorphic to the dihedral groups.
(Or alternatively, you need to work with an extended sense of "Platronic solid").

[vapid vale]

yes i do mean the extended sense

or not "the". an extended sense which includes those

[rocky cloak]

I mean, I think this falls under case-work, but is somewhat similar to what you're describing.

https://staff.fnwi.uva.nl/r.r.j.bocklandt/notes/kleinian.pdf

The key idea is that each rotation is about some axis, and you can think about the orbits of axes by the action of G.

I think the intersection elementary and non-casework would be impossible. But I'm quite interested if there is a more abstract solution without (or at least hiding) the case work.

[vapid vale]

this is sort of the argument i was trying to avoid 😭

[rocky cloak]

Like if one could prove the McKay correspondence without classifying the subgroups first, then one could defer to the classification of semisimple lie algebras (or root systems or rep finite quivers or what have you)

[vapid vale]

i feel like there is some way in which you take the convex hull of the orbit of an element, and pretty easily convince (sweeping details under the rug) someone the required facts about the meeting of edges – you wouldn't like, prove the nature of the platonic solids, but if you already knew the precise solids which met the criteria, you'd be done

i think that sort of exposition still requires some weird argumentation though. like its not very obvious how the group acts on an edge

[tribal moss]

The trouble is if you just take a random vector, then the convex hull of its orbit won't be a Platonic solid.

That requires you to have randomly chosen either vertex or a face midpoint (in which case you get the dual polyhedron instead, of course).

[rocky cloak]

You probably want to start with a vector that is the axis of one of the rotations

Some care must be taken for the dihedral and cyclic group, but otherwise I think that should give the right figure

[tribal moss]

Yeah, and in order to avoid picking an edge midpoint, choose a vector that's fixed by a group element of maximal order.

[vapid vale]

righ

t

[rocky cloak]

But how to prove that this works without first establishing what all the groups are though...

[vapid vale]

i suppose those can just be argued separately

[vapid vale]

yeah idk

my hope was that there'd be a slick way to do it

[tribal moss]

Hmm, if you do pick the axis of an element of maximal order and are okay with treating dihedral and cyclic groups specially, the polyhedron you get is a tetrahedron, octahedron, or icosahedron -- in each case with faces that are equilateral triangles distinguished only by how many meet at each corner. There seems to be some hope that you can treat those three cases in parallel much of the way.

[vapid vale]

hm yeah

lol this may not be the best

[rocky cloak]

So take an element of maximal order and pick the unit vector for its axis of rotation (with some orientation).

If all rotations fix this axis you're dohedral or cyclic. Otherwise you will get a meaningful convex hull.

It will obviously have vertex transitivity.

Then something something, the edges connected to the original vertex are cycle around by the original permutation you started with. (Maybe something about the closest vertecies giving edges and something using maximal order).

This should then give both edge transitivity and face transitivity.

[elfin wraith]

It’s a combinatorics problem, there’s 100 cases and much suffering guaranteed

[rocky cloak]

So let the original point be v and let w be the closets point in the orbit. Let g be the element of maximal order, then the claim is that the edges incident to v are exactly
(v, g^i w).

Say there is a u with (v, u) an edge. Since the distance between w and u must be at least the distance between v and w I feel like one should be able to argue this is impossible, using convexity of the sphere or whatever.

[south patrol]

Just give it to an undergrad to boost their CV

[rocky cloak]

So then G would need to be a subgroup of the symmetry group of a platonic solid, and then I guess you have to check the subgroups manually

[tribal moss]

It's not that bad. We know the correct conclusion is that there are at two infinite families and three sporadic cases (and at best the three sporadic cases can be considered a nice finite family). All we need is to find an argument that reflects that structure.

[quiet pelican]

Iirc there’s a nice way to do this with ADE diagrams that one of the lectures at my summer research briefly talked about

[rocky cloak]

There's a nice bijection with ADE diagrams given by the McKay correspondence, but I'm not sure you can use this to classify the subgroups

If you can I'm very interested, but all the proofs I've seen are essentially: take the finite subgroups, compute their (reduced) McKay graphs, OMG it's the ADE diagrams!!!

[tardy hedge]

lol

why is jagr still not jagr

[rocky cloak]

Illusion!

[tardy hedge]

wow welcome back jagr

long time

[rocky cloak]

Yeah, took a break for a while

[vapid vale]

this is meant to be extremely elementary lol

[rocky cloak]

Get down my screentime you know

[vapid vale]

so ADE diagrams would unfortunately not cut it (and in particular i think using McKay is like invoking magic)

[rocky cloak]

Yeah, I'm not sure what would be a good connection there anyway.

Like if you let G act on the polynomial ring S = C[x, y], then a connection is that the skew group algebra S#G is equal to the preprojective algebra of the corresponding diagram.

[vapid vale]

think i'd probably just forgo this whole thing lol. it was intended to be the last section of an expository thing on group actions and i was hoping for a fun little punchline to bring in the platonic solids. so i was hoping for some argument where i didn't need to be that careful

[rocky cloak]

Are you unhappy with an argument like this though?

[vapid vale]

explaining group actions throgh noncommutative crepant resolutions

[south patrol]

Matrix factorisations.

[rocky cloak]

I think it works, like u would need to ly between w and gw say, and then if a triangle has two long legs it's height will be more than if it had shorter legs...

[vapid vale]

oh, didnt see that – thats a nice argument actually. my issue is that the audience is one ostensibly new to group theory (although bright students). so im literally defining and talking about groups through toy examples, and every additional thing i need to define about groups is taking away from the point of the lecture which should be to motivate them

i will have to think about it. i dont want to bloat the talk with definitions

[rocky cloak]

Well, if your giving a talk I think it's fine to not present a very rigorous proof.

Just the fact that these are the finite subgroups of SO(3) is already interesting in it's own right

Not sure you need to go into a detailed proof of it

[vapid vale]

what is the connection? i know you get some categorical equivalences bewteen MCM modules and matrix factorizations, and these kleinian singularities should have finite CM type

[south patrol]

No I mostly just said that as a joke because of being in the same rough domain ig

But I unfortunately do not actually know anything about matrix factorisations lol just it is kind of adjacent to stuff I like

[vapid vale]

yeah fair. i think i can probably find a middle group on what to handwave

thank you for the help :)

[south patrol]

You know more than me lol

[vapid vale]

theyre kewl

[south patrol]

Says u

[prisma ibex]

McKay is a very neat collection of results

[south patrol]

McKay when Mackey shows up

[prisma ibex]

the relation with minimal model conformal field theories is especially fun

[vapid vale]

mukai + mckay :3

[south patrol]

Now is there a positive characteristic variant

[vapid vale]

i think so? probably can’t have characteristic 2,3,5 but there’s some statement

[rocky cloak]

As long as the characteristic is relatively prime to the order of the group I think most things go through

[vapid vale]

oh wait i forgot about cyclic and dihedral groups lol

the lame ones

[rocky cloak]

I was thinking more the McKay graph equals the Gabrial quiver of skew group equals Ausland--Reiten quiver of MCM modules part, not so much whatever specific classification you get

[vapid vale]

i don’t think the MCM AR quiver is the same?

like An should depend on even or odd

[rocky cloak]

What do you mean?

Like in positive characteristic you mean?

[vapid vale]

in general

unless im confused

but i think the AR quiver of the MCM category is just a completley different thing

[rocky cloak]

Like if G is a finite subgroups of SL(2, C) and R = C[[x, y]]^G, then the AR-quiver of MCM R-modules is equal to the McKay quiver of G

This is part of the McKay correspondence

And I think more broadly if G is a finite subgroup of GL(n, C) (without pseudoreflections), then the same should be true if you restrict to MCM modules that appear as summands of the power series ring (viewed as an R-module)

[vapid vale]

hmm

i may be getting confused then

this was in yoshino's book

[rocky cloak]

So looking in Yoshino this R is equal to k[x, y]/(y^2 + x^n) which is not the Kleinian singularity
k[x, y, z]/(x^2 + y^n+1 + z^2)

I guess there is some 1D version that is also indexed by Dynkin diagrams...

[vapid vale]

yeah

oh ok i haven't really read this text

[rocky cloak]

But chapter 10 is what I'm talking about

[vapid vale]

oh i see

[rocky cloak]

These bad boys right here

[vapid vale]

:D

i had only seen the 1d MCM AR-quivers before

there was a paper by iyama and some others which had them

https://arxiv.org/abs/0704.1249

[rocky cloak]

Hmm, so then I'm wondering

How do these guys get their associated Dynkin diagrams

If it's not from the AR quiver...

Seems like it might be alggeo magic...

[vapid vale]

i forgot how this happens lol

i think it is just the same thing, k[[x,y]]/(f) given by finite subgroup of SL(2,k) action

[untold hearth]

Does anyone have an intuitive way to understand the tensor product of two group representations?

[delicate orchid]

rho_{U (x) V}(g) is the kronecker product of rho_U(g) and rho_V(g)

[elfin wraith]

I’m guessing they mean more heuristically than just, tensor the matrices

[delicate orchid]

or g(u (x) v) = gu (x) gv

or just multiply the characters

[delicate orchid]

I don't really know what this means

[elfin wraith]

As in, how to interpret it

[rocky cloak]

I mean, if you want a representation structure on the tensor product then I think
g( u(x)v ) = gu (x) gv
should feel pretty natural.

It's what you get from just multiplying characters.

It's adjoint to Hom_k(V, -), whose action should make sense because it has G-linear maps as invariants.

[untold hearth]

Could you specify this adjunction more explicitly please?

[rocky cloak]

The action of g on f in Hom_k(V, W)
is the map taking v to
g ( f ( g^-1 v )))

And then it's the usual adjunction
Hom_G(U, Hom_k(V, W)) = Hom_G(U (x) V, W)
taking f to
u(x)v |-> f(u)(v)

[azure hull]

fundamental definition of a function f: A \to B requires that for every element a \in A there corresponds a \textbf{unique} element f(a) \in B
proposed construction of the function \gamma: Z \to Y violates this principle until well-definedness is proven right ? procedure is as follows:
For a given z \in Z, select an element y from the preimage set \beta^{-1}({z})
Define \gamma(z) based on this choice of y
source of the problem is that the homomorphism \beta is surjective but not necessarily injectiv this means the preimage set \beta^{-1}({z}) is non-empty but is not guaranteed to be a singleton If there exist distinct elements y_1, y_2 \in \beta^{-1}({z}), the definition is ambiguous ig If the procedure yields a different result for y_1 than for y_2 then the input z would map to multiple outputs and \gamma would fail to be a function.

[kind temple]

consider the collection of paths I -> X under path-homotopy equivalence (so two paths are path-homotopic if there is a homotopy between them such that the endpoints are constant).
this consists of the following data:

• an associative partial binary operation X^I x X^I -> X^I, where two paths f,g are composable if f(1) = g(0),
• each path f has a left and right identity (a kind of relative identity?),
• each path f has an inverse with respect to f's relative identities

are there any structures like this?

[thorn jay]

fundamental groupoid

https://en.m.wikipedia.org/wiki/Fundamental_groupoid

[kind temple]

ah dang. is my pattern recognition still that weak lmao

[thorn jay]

:P

yeah what you described is a groupoid lol

[kind temple]

thanks

[warm needle]

Damn bruh never thought knowing what I need to do to take my socks & shoes off would come in so clutch

[tardy hedge]

Lol

Inverse

[bleak sandal]

Anyone have any hints?

[vapid vale]

how do you imagine an arbitrary element of SO_2(R)?

[tribal moss]

☝️ Officially SO2(R) consists of certain 2×2 matrices, but there are several simpler ways to think of it (that is, groups with a more straightforward definitions that it happens to be isomorphic to). Do you know some of them?

[white oxide]

Could I have a hint on this problem? I'm a little stuck on where to start lol. If $\mathfrak{a} = \bigcap_{i = 1}^n \mathfrak{q}_i$ where $r(\mathfrak{q}_i) = \mathfrak{p}i$ for all $i$, then we have $\mathfrak{a} = \bigcap{i = 1}^n \mathfrak{q}i = \bigcap{i = 1}^n \mathfrak{p}_i = r(\mathfrak{a})$. I am assuming for contradiction that there is an embedded ideal, that is, there is a non minimal element of ${\mathfrak{p}_1, \dots, \mathfrak{p}_n}$, so that $\mathfrak{p}_k \subseteq \mathfrak{p}_j$ for some $k \neq j$.

[cloud walrus]

okeyokay

[white oxide]

I guess one thing I noticed is that this implies that $\mathfrak{q}_k \subseteq \mathfrak{p}_j$, which is not usually the case?

[cloud walrus]

okeyokay

[rapid cave]

what is an embedded ideal?

[white oxide]

Hol up I'll post the book definition in a sec

[karmic moat]

Then in your intersection you can throw out p_j, contradicting uniqueness (theorem 4.5)

[rocky cloak]

If a is radical then it is the intersection of prime ideals, so you can assume the primary ideals in the decomposition are prime

[white oxide]

Wait so we can say that a = \bigcap_{i} p_i is a primary decomposition for a?

I guess this makes sense... the only conditions are that every ideal in the intersection is primary, and prime ideals are primary

and moreover there are some remarks on the page before that saying we can reduce any primary decomposition to a minimal primary decomposition(?), so theorem 4.5 applies

[karmic moat]

Yeah

[karmic moat]

You can just assume it’s a minimal decomposition from the get-go

[white oxide]

Wait I'm a little bit confused about how this contradicts Theorem 4.5 - doesn't it contradict the definition of a minimal primary decomposition? For if $\mathfrak{p}_j \subseteq \mathfrak{p}k$, then $\bigcap{i \neq k} \mathfrak{p}_i \subseteq \mathfrak{p}_j \subseteq \mathfrak{p}_k$, which contradicts the definition of a minimal primary decomposition

[cloud walrus]

okeyokay

[white oxide]

Also if Z[t]/(2, t) isomorphic to Z/2Z via evaluating at zero then reducing mod 2

[karmic moat]

Yeah sorry contradicting minimality

[chilly ocean]

I know you were already informed this is a groupoid, so I just want to add that groupoids can be useful for, say, group theory, not just category theory or alg top. In fact, you can prove that a subgroup of a free group is itself free(whereas the “typical” proof is topological). I don’t have the exact reference for this but it’s somewhere in Ronald Brown’s Topology and Groupoids

[warped wedge]

can someone help with my group theory homework im so tired and so lost idek if what I attempted is right

its super overdue too and idk if im proving things in the right oder