groups-rings-fields

[south patrol]

Oh okay nvm I see what they mean lol

[next obsidian]

[south patrol]

Lol

[next obsidian]

But also if you take both coordinates

Then you do have multiplication yeah

[south patrol]

Yeah then more like a poly in two tings ye

[delicate orchid]

yur

[south patrol]

Ignore me

uwu

[delicate orchid]

don't say that

[south patrol]

Okay

that

[delicate orchid]

q wertyuiop[]

[karmic moat]

time to retire wew

[south patrol]

Wew I am doing Jordan-Hoelder for the like 4th time

Are you proud of me

[delicate orchid]

I haven't even done it for the first time

because decomposition series are utterly useless

[next obsidian]

Jordan Hölder is important

At least the invariance of length part

[south patrol]

Yes fair

They get mentioned for reps of finite groups and of lie algebras in what im reading

Well ig just the point that there are only finitely many simple R-modules if R is artinian and noetherian and you can kinda just write them down

[delicate orchid]

wrong

"oohh oohh I simply must work over a semifactorial monoid" get out of here boot boy

[south patrol]

But then seems superceded at least in char 0

[next obsidian]

I’ve never said that in my life

[south patrol]

Glomed

[next obsidian]

Glomed

[south patrol]

k[t]/t^2

[delicate orchid]

but this just offloads the actual hard part to checking if R is artinan/noetherian so what's the point

[south patrol]

That is often easy though

like if it is finite over a field

[next obsidian]

Yeah because it’s always Noetherian and then you just need to know if it’s dim 0

[south patrol]

Which is a lot of stuff we are interested in right lol e.g. group ring of finite group smh

[delicate orchid]

I literally do not know what you people are talking about

[formal ermine]

https://tenor.com/view/joke-dumb-you-gif-8906255

[next obsidian]

Are you gonna let a 15 yr old talk to you like that Potato?

[karmic moat]

^

[delicate orchid]

finite what, dimensional? illogical. Not even free

I'd legitimately rather fuck around with brauer characters than even think about ring theory

[south patrol]

No

[wraith cargo]

sociopath behaviour

[delicate orchid]

I will continue being a sociopath until potato explains what the fuck this means

[wraith cargo]

Field in what category

[south patrol]

finite over R means finitely generated R-module

so here it is just yes a f.d. vector spacae over the field

[delicate orchid]

oh yeah cause "finitely generated" is just TOO long to type smh

in that case yeah it's obvious lol

[wraith cargo]

why tf are you working over reals

[next obsidian]

Finitely generated makes you think it’s finitely generated as an algebra

Finite is the standard name for this

Get good

[delicate orchid]

stupid terminology

[wraith cargo]

no what

lmao

chmonkey being 15 y/o brained

[next obsidian]

[wraith cargo]

tf kind of shit terminology is that

[delicate orchid]

https://cdn.discordapp.com/attachments/482958702211497994/1119257164301750414/mods_crush_his_skull_thank_you._LTG.mp4

[next obsidian]

They hated Jesus for he told the truth

[wraith cargo]

say finitely presented like a real man

[next obsidian]

NO!

Finitely presented means a specific thing

Which is neither of these

Over a field they’re the same, but that’s inconsequential

[delicate orchid]

yeah finitely presented is very different lol

[delicate orchid]

if you werent referring to as an algebra then I legtimately don't know what you or potato are talking about

[wraith cargo]

wow OK I got what you were talking abt massively wrong lmao

[next obsidian]

Finite over a field means as an algebra it’s a finite dim VS

This is much stronger than being finitely generated as an algebra

k[x] is not a finite k-algebra

[delicate orchid]

or at least potato got something wrong, there are finite groups with wild representation types over finite fields but the group algebra is obviously still finitely generated

[next obsidian]

Any finite k-algebra is 0-dimensional, Artinian, etc

A group algebra associated to a finite group is a finite k-algebra because it has basis G

[delicate orchid]

as an algebra it's a vector space
fuck me why didn't I just do fluids

[next obsidian]

Literally being an algebra makes you a VS

This is what all of field theory is based off of

[delicate orchid]

so as a vector space it's a vector space

nifty

[next obsidian]

No because it has a ring structure that also plays nice with the field structure of the base

[wraith cargo]

I wish I can go back to HS algebra sometimes

[next obsidian]

Anyway, this is fucking dumb I feel like I’m talking to a 12 year old so I leave you with this sully

[delicate orchid]

thanks for that one chmonkey

u mfs get on my case for being imprecise constantly but when the turn tables u scamper off

[next obsidian]

This isn’t imprecise at all

[wraith cargo]

don't worry I'm the type of person who constantly mixes up terminology and then tells ppl they can understand what I'm saying based on context

[next obsidian]

This is all standard terminology

Considering an algebra as a module over the base ring is normal

[delicate orchid]

well no shit it's normal

it's just the way your phrasing it is confusing the shit out of me

I got what you mean now

you forgor the algebra structure and it's a finite dim vs

but that's not what you said

[next obsidian]

How is saying “your algebra is a finite dimensional vector space” confusing at all?

There’s only one way to do that lol

[south patrol]

very standard lol

[wraith cargo]

don't make me get out my copy of Eisenbud

[delicate orchid]

anyway there's still this point

[south patrol]

What do you mean by wild representation types

Or like, give me an example

[delicate orchid]

oh it's alright I was getting simple and indecomopsable swapped in the definition

[teal vessel]

Just a sanity check here: in question 23, does σ²=σ°σ, or (σ(a))²?

I think the former

[delicate orchid]

what's the group operation

[teal vessel]

Oh, forgot to post the q

Srry

[slim kayak]

The former

[delicate orchid]

the group operation in Aut(G) is composition yeah

[teal vessel]

Yes. The only confusion here was because it was asking about properties of G, as opposed to Aut(G)

[slim kayak]

Well, it was talking an automorphism of G so it was also about Aut(G). In the future always assume the former unless strictly specified

[teal vessel]

I've just seen both in the questions, so I needed to make sure

[charred garden]

guys on question 5 is the set of 2Z also infinite as with Z, 2Z is definitely a subset of Z but i’m struggling to understand because 2Z could also be infinite despite it only containing even numbers

[lusty marlin]

Yes, 2ℤ is infinite. And it is a subset of ℤ. What is the issue?

[wraith cargo]

the map x-> 2x is an isomorphism so they're equinumerous sets

[formal ermine]

[wraith cargo]

illum this is honestly the worst insult ever flinged at me

[charred garden]

i get now actually

thanks guys

[open sluice]

see https://en.wikipedia.org/wiki/Dedekind-infinite_set
in ZFC, any given set is infinite iff it has a bijection with one of its proper subsets

it's a funny alternative definition for infinite sets

[charred garden]

it truly is,thanks

[alpine island]

What's the other one?

This is the only formal definition of an infinite set that I know

[open sluice]

I dunno but I think it’s just one of the ZF axioms

axiom of infinity

[south patrol]

No it's not an axiom

That axiom talks about the existence of the natural numbers essentially

[south patrol]

You can just define a set as finite if it is in bijection with {1,...,n} for some n and say that a set is infinite if it is not finite

Equivalently if for all n there exist pairwise distinct x1,..,x_n in your set

It is then a theorem (with ZFC) that this definition is equivalent to being Dedekind infinite

[open sluice]

oh

[south patrol]

Though the two notions can differ with a weak enough set theory

Not sure why this is in abstract algebra tho...

[warped fable]

fitting

[south patrol]

Everything in the last like hour would probs go better in #proofs-and-logic next time

Not to mini mod lol

[warped fable]

whats the difference

[south patrol]

Lol

[chilly ocean]

nothing. basic questions in this channel basically belong there

[south patrol]

Well this had nothing much to do with algebra lol

[warped fable]

this is algebra to me

i see no difference

[south patrol]

Oof

[lethal dune]

[open sluice]

well we've got the abstract part down at least

[white oxide]

if S is a ring and R is a subring, why is R in general not an S module? the distributive laws are satisfied since R is a ring, multiplication is associative, and the identity element for S is the same as R

[south patrol]

How are you defining the multiplication

[lethal dune]

it is right?

[south patrol]

This is true the other way round though (S is an R module)

[south patrol]

No like this makes no sense as is in general

[lethal dune]

oh R is the subring here

duh

[delicate orchid]

multiplication by S on R might not map elements of R to R

it's not an ideal

[lethal dune]

look at Q module Z

[delicate orchid]

unless they're non unital in which case I apologise

[south patrol]

For example this would make Z into a Q-vector space which we hopefully see is very much not the case

OK rip beaten to it

[lethal dune]

[white oxide]

wait but i thought since S is contained in R in which case multiplication has to be closed

[south patrol]

You said R is contained in S

[white oxide]

que es

[lethal dune]

[white oxide]

oops

[lethal dune]

same issue I had

[south patrol]

Nice nickname ryu

[lethal dune]

it's either correct or incorrect depending on who you ask

number theorist will say incorrect, topologist will say correct

[south patrol]

Yeah that's what I was thinking lol

Toplogist might say it is ambiguous

[lethal dune]

true

[warped fable]

what is correct and what is incorrect

i see a bunch of symbols

its neither correct nor incorrect

[lethal dune]

it's set theory

[warped fable]

its nothing

[lusty marlin]

By cl(ℚ), do you mean the algebraic closure of ℚ?

[lethal dune]

upto you

[south patrol]

Class group of Q

[teal vessel]

this question is really kicking my butt

[barren sierra]

questions tend to do that

[slim kayak]

Exercise number really is a tough cookie

[teal vessel]

23 from the picture previously sent. I'm having a hard time, but I feel like it's because I'm not incorporating the fact that G is a finite group

more specifically: I'm having a tough time getting the intermediate step of showing that for all x in G, x=x^-1sigma(x)

[hidden haven]

It doesn't say x = x^-1 σx

It says that for any g, there is some x such that g = x^-1 σx

[teal vessel]

that probably explains my problem

gotta learn how to read

[south patrol]

hilbert's theorem 90

[teal vessel]

not gonna speedrun that one until it's on my whiteboard lol

[hidden haven]

Though this also seems sus to me. Applying σ to such an expression inverts it, so this implies that σ is the inversion map. But then g = x^-1 σx = x^-2, so this implies that every element of the group is a square? This is false because ℤ is a counterexample. Not sure where I'm making a mistake.

[teal vessel]

are you saying that Z is an example of such a group? because Z is not a finite group

[hidden haven]

Ahh

Missed that

[formal ermine]

wait Z isn't finite?

[teal vessel]

at least within the context of this textbook, despite having a finite generating set, Z is not finite. It has infinite order.

[formal ermine]

ong?

[glossy crag]

Anyone?

[slim kayak]

First off, it says that every y has the form x^-1 sigma(x) for some x in G.

[formal ermine]

lmao

good one

[teal vessel]

got that part

[slim kayak]

If you interpret x^-1 sigma(x) as a function that sends x to that expression you wanna check if that function is actually surjective, in that case you can write any g in that form.

You can see how this map is injective from the property you were given (work it out). Injective functions into a finite set into itself are always bijective, so that gets you your desired claim

[slim kayak]

Tf are abelian categories m

You mean R-modules over a fitting (sometimes noncommutative) ring?

[lethal dune]

Thing category theorists came up to feel good about themselves

[slim kayak]

Yeah, most likely

[hidden haven]

I don't sound like that

[slim kayak]

Why am I getting sully-ed for the based mitchell embedding theorem

[delicate orchid]

true albeit

[south patrol]

Good

Isn't that just a good enough algebraist rather than a cat theorist

Jk

[fervent gyro]

Are there really any abelian categories that are studied other than R-modules?

[coral spindle]

Sure, sheaves

[slim kayak]

Lots, but usually them being abelian is a nice bonus. If you forget everything except the abelian cat structure then there is a sense in which it "behaves exactly like R-modules", so the point where any theorem of abelian categories will hold if it holds for R-modules

[warped fable]

good example

this is the only motive behind abelian categories

no other examples exist

[formal ermine]

who even gives a sheaf about those

[slim kayak]

Chain complexes over an abelian category might be an interesting example to a good amount of people too

[warped fable]

thats not an example because no one cares about just chain complexes

you want to go to the derived category

no longer abelian

[slim kayak]

I guess if you only care about a concrete derived category then that's true, but otherwise derived categories are defined over abelian categories, no? Still makes it an important stepping stone for getting there

[warped fable]

ok i will give you this one

its a good example too

but thats it

no more examples exist

besides your example is a special case of complexes of sheaves over a point

theyre the same example you tricked me

[slim kayak]

Shit, they found out

[teal vessel]

I know that this function is injective, but I'm having trouble making surjectivity. I suppose that if for all x,y, showing that f(x)=f(y) implies that x=y would do it via pigeonhole principle, since G is equinumerous to itself. I guess I can lean on the fact that sigma(x) is a bijection in itself, therefore the pieces of inv(x)sigma(x) are unique. However, I still haven't quite figured out how to show that there's no weird collapse or cancellation that can render f(x)=f(y) for x=/=y

[slim kayak]

Injective functions form a finite set to itself are always surjective

How many elements does the image of an injective function?

[teal vessel]

ah, the hole-pigeon principle XD

[slim kayak]

The diagram of holed and pigeons commutes

[delicate orchid]

Aren’t you told sigma is an automorphism or is this a different question

[teal vessel]

I am.

that's how I was able to easily conclude that sigma(x) is unique for a given x

[slim kayak]

Yeah but this about an related function

Not even a homomorphism

[delicate orchid]

g = x^-1sigma(x)? For some x I’m presuming

[teal vessel]

yeh

[slim kayak]

Yeah, interpreted as a function of x it is surjective

[teal vessel]

wait hold on... something feels less obvious than before. I've shown that every member of G can be mapped via this function, and I've shown that it is well defined. ah, I see my problem. be back in a second when I've actually demonstrated injectivity. I need to show that there isn't an element that can't get mapped to.

[undone bay]

Does anyone have a reference for the representation theory of GL(N, Z) for generic N?

[teal vessel]

oh, no wait, there's a left inverse, therefore it is injective, and therefore via pigeonhole, it is surjective.

[delicate orchid]

There really is no nice way to do this question and it’s annoying me

like yeah sure you could just check it but I want to do something cooler

[slim kayak]

What question? Injective maps on equinumerous finite sets are surjective?

The initial question?

[delicate orchid]

g = f(h) = f(f(h^-1h)h) = h^-1f(h^2)
Like come on bro I’m so close

[teal vessel]

showing that G is abelian given the correct conditions on sigma

[warped fable]

Dunno, but a keyword is Margulis superrigidity. Virtually, all finite dimensional representations of GL(n, Z) come from GL(n, C)

For n>=3

n=2 is the rank one case, that's number theory territory. I dunno.

I mean PGL(2, Z) is Z2 * Z3

[slim kayak]

If x^-1 sigma(x) = y^-1 sigma(y), then yx^-1 = sigma(y)sigma(x)^-1

So yx^-1 is then a fixed point, so by assumption 1

[teal vessel]

makes sense

that tracks

I'm way out of practice with my function manipulation lol

[undone bay]

Thanks so much for this: I'm interested in this question mostly for larger n, so no need to worry about the n = 2 edge case

[slim kayak]

You just multiply on both sides and then use the homomorphism property to obtain that yx inverse is equal to its image under sigma

[teal vessel]

and then, given that all elements can be written as x inverse S(x), we can then apply sigma on it to show that S(-x)x=S(-xS(x)) which through some manipulation shows that S(x)=-xS(x)x for all elements in G, which can only be the case if the x and x inverse on either side can cancel out, which implies commutation.

or, more strictly, apply x from the left on both sides to get xS(x)=S(x)x

i.e. commutativity

[slim kayak]

Ngl that's a bit complicated. S(-xS(x))=-S(x)x. This is just the inverse of -xS(x). So S is homomorphism that sends any elements to it's inverse

That's not only a cool fact, but is equivalent to G being abelian

[teal vessel]

yeah, demonstrated in the previous problem

[slim kayak]

Yeah, just saying. Your reasoning seems a bit shacky there

At least it doesn't obvious that, potentially, x or S(x) always in the center. Like, just because xS(x)=S(x)x, why should it hold if I replace S(x) with some arbitrary y?

[teal vessel]

fair enough.

[white oxide]

can somebody help me understand why beta not being in E implies that such an isomorphism mapping of E can't be an automorphism of E?

[south patrol]

Suppose E is an algebraic extension of a field E lol

[white oxide]

yea i saw that lmfao

[next obsidian]

Well

consider that

the image has beta

so

it isn't a map E -> E

[south patrol]

Yeah

[next obsidian]

Or well like

yeah

that's it

[south patrol]

Yeah I assumed it was something deeper initially lol

[white oxide]

oh yea i just overthought again

thanks

[formal ermine]

how do you call a group ring over some group with coefficients in some ring

like, "group algebra of G over R"?

or how would you phrase that

[prisma ibex]

group algebra of G over R yeah

or group R-algebra of G

[formal ermine]

this sounds better

ty

[delicate orchid]

R[G] lol

[sage lodge]

anyone have a good quality PDF of pinter

i can't find any with good formatting

[solemn garden]

Herstein gave this as an exercise. How would the proof even start?

[formal ermine]

yes but like in a HUMAN SENTENCE

[delicate orchid]

"let A = R[G]"

[formal ermine]

@delicate orchid how do we define the tensor product of two characters

is it just the character corresponding to the tensor of their representations

[delicate orchid]

multiply them

vast majority of sources don't even use a tensor product symbol they just write \chi_1\chi_2

[formal ermine]

yeah so the character corresponding to the tensor of their representations

[delicate orchid]

which is the character corresponding to the tesnor product of the corresponding modules but I thought I'd let you know it's just multiplication

alright wise guy simmer down

one more wisecrack out of you and I'm gonna start charging by the hour

[formal ermine]

now I just need to figure out if this is tensor or direct sum....

[delicate orchid]

tensor

only absolute lunatics write a DIRECT SUM of characters

[formal ermine]

so it's like

the left side is a sum of things

so it's term wise?

like it's some characters added to one another

and the morphism here is like sending each one to their respective thing on the right?

[delicate orchid]

when I see \chi^u I think \chi^u(g) := \chi(ugu^-1)

[delicate orchid]

the tensor product distrubutes over direct sums so it doesn't matter

[formal ermine]

\chi^u is (t_1, ..., t_n) -> t_1^a_1 * .... * t_n^a_n

[formal ermine]

yea

[delicate orchid]

oh it's the torus geezer

[formal ermine]

indeed

[delicate orchid]

yeah so this map just takes that to \chi^2u surely, assuming u = (a_i)

[formal ermine]

what is C[M] tensor C[mathcal O^vee cap M] isomorphic to

[south patrol]

Ye

[delicate orchid]

how the fuck am I supposed to know

[formal ermine]

because like

if chi^u is sent to x^2u

then there must be a nice form of that tensor

[delicate orchid]

I just know that if you square something the exponents go up

[formal ermine]

real

[delicate orchid]

there's no telling how chi^2u decomposes over that

just because it's a character doesn't mean it's a nice one

[delicate orchid]

wait nvm

would (1,...,1) be the identity here

[formal ermine]

aren't all of the tori characters like really nice

or am I misremembering

[delicate orchid]

cause if so I'm chatting shit - \chi^2u would be linear

so the abelianisation of both dudes are isomorphic :letrollface;

[formal ermine]

what the fuck is an abelianization

I've only heard it in the wikipedia article of class field theory

[delicate orchid]

I have no idea what you're talking about

[tepid silo]

can anyone help me with this identity? im not sure if its right: $\mathbb{Q}(\zeta_9) = \mathbb{Q}(\zeta_9 + \bar{\zeta_9}) \mathbb{Q}(\zeta_3)$

[cloud walrus]

Cone

[rotund aurora]

Q(zeta_9+zeta_9^{-1}) is totally real

you get the result by degree count

[tepid silo]

yea the degrees match. so you think its right?

[rotund aurora]

if the degrees match then you are done

because obviously $Q(\zeta_9)\supseteq \Q(\zeta_9+\zeta_9^{-1})\Q(\zeta_3)$

[cloud walrus]

Croqueta

[tepid silo]

hm true that, thank you!

why is the right side contained in the left side?

[rotund aurora]

idk think about it

but its just a definition

[hidden haven]

The generators of the right side are contained in the left one

[tepid silo]

true

thank you

[glossy crag]

If I want to show multiplication in the Brauer group is well defined via [A]\cdot[B]=[A\otimes B] does the following work:
Suppose A\sim A', B\sim B', we want to show A\otimes B\sim A'\otimes B'. Let D and E be division algebras with A\cong M_n(D), A'\cong M_n'(D), B\cong M_m(E), B'\cong M_m'(E), then A\otimes B\cong M_nm(D\otimes E) and A'\otimes B'\cong M_n'm'(D\otimes E), i.e. both are isomorphic to a matrix algebra over the same (not necessarily division) CSA C. This implies similarity (if A\cong M_n(C) and B\cong M_m(C) for a CSA C, then there is a division algebra D with C\cong M_d(D), hence A\cong M_nd(D) and B\cong M_md(D) and A\sim B).

[chilly radish]

Can you tex this

It's very hard to read like this

Anyways this should be trivial if you know how matrix algebras tensor

And that you can pull the coefficient ring out

[vagrant zinc]

[delicate orchid]

block morita equivalence 😋

[celest furnace]

Is there any research / books on polynomial rings in infinitely many variables?

[delicate orchid]

le classical non-noetherian ring has arrived

[coral spindle]

That's a really broad question. Is there any research? Idk, look it up on arXiv. In any case I doubt you'd find a whole book just on one single ring.

If you want to know some stuff about it then just ask

Anyway as wew points out it is the classic example of a non-Noetherian ring, which in some ways says it's a gross ring

I think an interesting question would be whether or not every projective module over this ring is free. This is true for ordinary polynomial rings (Quillen-Suslin)

[formal ermine]

isn't this phrasing wrong? wouldn't it imply that all sylow subgroups are normal?

[coral spindle]

No

[delicate orchid]

society if all sylow subgroups were normal

yeah that absolutely does not imply what you're saying wtf

[formal ermine]

what if you choose K = H? shouldn't it be phrased as "the conjugate of a sylow subgroup is again a sylow subgroup"

[delicate orchid]

if two elements are conjugate then so are their centralisers, that does not imply that all centralisers are normal

[formal ermine]

oh wait lmao I'm stupid

[coral spindle]

Then you would get an element g such that gHg^-1 = H, wow so much crazy

[formal ermine]

"a g" exists

I read it as every g

[coral spindle]

good sticker

[delicate orchid]

oh my golly gosh but then if you track how you actually get betwixt each subgroup you get the wholesome alperin fusion theorem oh my goodness gracious

[coral spindle]

oh my goodness gracious

I have never heard of this theorem

[delicate orchid]

nobody has ever heard of my field of study so this doesn't surprise me

[formal ermine]

what does it state

[delicate orchid]

I'm not typing it out

gimme a min I'll go find it

[coral spindle]

le epic google

[delicate orchid]

no I just have fusion systems in alg/top open at all times

replace fusion system with "a group G" and "a p-group" with "a sylow p-subgroup of G"

it just describes how you can daisy chain together conjugation maps to go from one subgroup to another basically

[south patrol]

Nice

[delicate orchid]

and there's a VERY small set of groups called "essential subgroups" which are required to be able to do this

the statement is much easier if you just care about the fusion system

[coral spindle]

Wow I bet people that work in the same institution as one of the authors of that book are epic and hot

[delicate orchid]

oh yeah I forgot you did

[coral spindle]

Also I should read it

[delicate orchid]

it's pretty long

[coral spindle]

I feel bad for not knowing a single fucknndsing thing about fusion systems

[delicate orchid]

nobody does

me included

[coral spindle]

yeah I get that picture

[formal ermine]

wait I'm having a brainfart here, surely [A, B] = [B, A] must be true?

[chilly ocean]

what's the bracket mean here

[formal ermine]

commutator

[chilly ocean]

[delicate orchid]

sure if it's the boring commutator

[chilly ocean]

illuminator just write it out my man

[formal ermine]

what's the cool commutator

[coral spindle]

So you're saying aba^-1b^-1 = bab^-1a^-1 always?

[delicate orchid]

XY-YX = YX-XY is how I interpreted it lol

[coral spindle]

So you don't mean the group-theoretic commutator

[delicate orchid]

no I think they do

[coral spindle]

Or illum* doesn't

[delicate orchid]

I'm just an idiot

[coral spindle]

sorry wew I saw the blue names and got confused

[summer path]

I saw capital letters and also assumed this lol

[formal ermine]

nvm lmao

wait

actually

uh

huh

wait let me think

[delicate orchid]

are A and B groups or what

[formal ermine]

yes

[coral spindle]

Well this is obviously not true for the free group, so that's a counterexample.

[formal ermine]

is inversion a bijection

[coral spindle]

It is

Now it is true that [A,B] = [B,A]^-1.

[formal ermine]

ok nvm this whole thing I didn't even need this

[coral spindle]

this is why you should ask the whole question

[formal ermine]

I know I'm gonna get sullied for this but I needed [A, B] <= A cap B for A and B normal for something else and got [A, B] <= B but didn't realize that it's literally THE SAME EXACT ARGUMENT for [A, B] <= A

🗿

[coral spindle]

Huh what

[delicate orchid]

normal in what

[coral spindle]

what does this mean

[formal ermine]

some larger group

[coral spindle]

Oh these are groups, not elements

I wish you had told us this!

[summer path]

https://xyproblem.info/ moment

[coral spindle]

damn I forgot someone had already come up with a name for this

I was calling it "asking the whole question"

[formal ermine]

nono this isn't xy my argument still works

[coral spindle]

I think that's a better name. "xy problem" bad

[formal ermine]

and like I knew that it's true because it's in my notes

I just don't write down proofs

so I forgor it

[delicate orchid]

do you even need the commutator for this actually, I decided to think about this one

ah no you do

A, B normal in G, so A^B <= A and B^A <= B, so [a,b] <= A as bab^{-1} in A and likewise for [a,b] <= B

yurrr

ok cool

[formal ermine]

what's A^B

[delicate orchid]

image of A under every c_b for b in B

you know the one

[covert cliff]

Suppose a,b in Z/pZ and a^2 - b^2 = 0

p is prime

Can we factor the a^2 - b^2 to (a-b)(a+b)

[chilly ocean]

what do you think

[covert cliff]

yes

I think that should be allowed

[formal ermine]

can you come up with rings where it is not allowed

[chilly ocean]

first, why is it allowed

what's (a - b)(a + b)

[covert cliff]

I guess non commutative rings

[south patrol]

Yeah

[covert cliff]

the element a-b times the element a+b?

[chilly ocean]

what is it

distribute

what do you get

[covert cliff]

a^2 - b^2

[chilly ocean]

why

why is that what you get

[covert cliff]

a*a + a*b - b*a - b*b

and then a*b = b*a

[chilly ocean]

!!!

[covert cliff]

so a*a - b*b

[chilly ocean]

yes

[chilly ocean]

the key

[covert cliff]

I guess I was more concerned about whether these would be the only solutions

[coral spindle]

Solutions to what?

[covert cliff]

to this

[coral spindle]

Ah

Well ok here's a fact in Z/pZ where p is prime

if xy = 0, then x = 0 or y=0

can you infer what you want from this?

[covert cliff]

ah I see

that makes sense tbh

thank you

[celest cairn]

Hi.
If I wanted to solve the equation p(x) = x^2 + 1 = 0 over the quaternions where Re(x) = 0, would the solutions of p(x) just be x = +/-(i, j, k)?

Something tells me that’s a no…

[delicate bloom]

take a pure vector quaternion v and square it, you get -v dot v

in other words, every unit length pure vector quaternion satisfies v^2=-1

in fact you can lift the condition of having scalar part =0 since if there is, you end up with a vector part after you square it, so it will never be a solution to x^2=-1

[celest cairn]

Ah alright, thanks

[delicate bloom]

yeah you're welcome

[delicate bloom]

maybe try working out a more general formula for distinct vectors u and v what you get when you multiply them in terms of the dot and cross product

[formal ermine]

why did this look better when my prof drew it on the blackboard

it looks so weird in his notes

[delicate bloom]

I thought your subsets were ? for a minute lol

[formal ermine]

so true

[delicate bloom]

idk looks fine to me, feels 3D haha

[glossy crag]

Errant question, how are characters defined for infinite-dimensional representations (e.g. representations of compact groups)? Googling did not give any immediate answers.

[formal ermine]

it looked wayyyyyy cooler on the blackboard

[prisma ibex]

the main difficulty is in the non-compact case, where the notion of a character for these infinite dimensional representations is kinda subtle: the term to look up is "character distribution" or some variant of this; the characters here in general will be Schwatz distributions with singularities

for example if you're looking at representations of a semisimple Lie group G then the relevant thing to look at is the Harish-Chandra character distributions: if \pi is an irreducible unitary representation of G on a Hilbert space H and f is a test function then the operator \pi(f)=\int_G f(x)\pi(x)dx is of trace class

so you get a distribution \Theta_\pi(f)=tr(\pi(f)), this is the Harish-Chandra character of \pi, it's a conjugation invariant (eigen)distribution on G with eigenvalue the central character of \pi.

if you want to see some nontrivial examples of this worked out, check e.g. section 7.5 of Palm's thesis where this is worked out for GL_2(R) in a bunch of detail, and similarly section 9.5 for GL_2(Q_p)

[glossy crag]

Potentially dumb question, but can't one use the fact that the trace can be defined coordinate free via V\otimes V*\to K and try to do something similar in the infinite-dimensional case (e.g. with the topological dual instead of the algebraic)?

[frigid lark]

idk if this belongs in number theory but where does this equation come from

Lang Chapter 6 Q 22

[rustic crown]

p^(q^(r-1)) = 1 + x

b = 1/x * ((1+x)^q - 1)

the ... don't tell you that these are binomial coefficients lol

[frigid lark]

oh, that makes sense

[rustic crown]

[chilly radish]

Anyone have any idea what Int(g) could mean for an element g of a group (Where Int(g) is supposed to be a morphism)

[rocky cloak]

Could be the inner automorphism given by x |-> gxg^(-1)

[chilly radish]

I was thinking that might be it but it doesn't seem to work for what it's used

maybe i'm just being silly

[rocky cloak]

What's the context where it's used?

[chilly radish]

Where compatible means this

Where Gamma,Gamma' are profinite groups, A,A' are Gamma and Gamma' (resp.) sets/groups/modules and
f:A->A'
pho:Gamma'->Gamma

[chilly radish]

,rotate

[delicate orchid]

Oh that looks like an intertwining map

Don’t know if it is but that’s what it looks like

[chilly radish]

what's an intertwining map

this is showing the bifunctoriality of H^n

oh

you mean equivariant

yes this is equivariant but that doesn't explain what Int(g) is

[delicate orchid]

First three letters line up 🆙

[rocky cloak]

Well, it is true that $\sigma(\rho^{-1}a) = \rho^{-1}(\rho\sigma\rho^{-1})a$
So my vote is still on inner automorphism.

[cloud walrus]

jagr2808

[delicate orchid]

https://tenor.com/view/kuzco-no-no-hes-got-a-point-morgan-freeman-morgan-freeman-true-gif-25944627

[chilly radish]

I think you're probably right jagr

I probably messed up my calculation

oh yea I was thinking rho^-1 sigma rho

instead of rho sigma rho^-1

[hidden haven]

so silly

[uncut girder]

What are the irreps of Aff(2,3)

Also called AGL(2,3)

The affine group on the 2 dimensional vector space over F_3

[delicate bloom]

I guess I'd start thinking about representing column vectors as [x ; y ; 1] and then elements of AGL(2,3) as 3x3 matrices with the last row as [0 0 1] as a way to shoehorn in linear transformations with translations

then try to think of what kinds of upper triangular blocks of this we can make I guess

[uncut girder]

That is a 3 dimensional representation. Is it irreducible? Nvm

[delicate bloom]

I was more thinking this is more like a 6 dimensional representation cause it has 6 free entries, and then try to reduce it down by finding like equivalence relations by thinking of the cosets, but I might be way off base here lol

[coral spindle]

If you can work out the irreps of GL(2,3) then Clifford theory would make it easy to extend to AGL(2,3)

[delicate orchid]

Aff(2, 3) \cong F_3^2 \ltimes GL(2, 3), reps of F_3^2 are easy, GL(2, 3) is annoying but it's a double cover of S_4 so not too bad

[uncut girder]

Its a 3 dim rep over F_3, not C

[delicate orchid]

there's a well known formulation of reps of semidirect products lemme go find it

[delicate orchid]

or u can do this, but that requires thinking

[coral spindle]

It's gonna use clifford for sure

the big red dog is hard to beat

[delicate orchid]

yeahhhhhhh just use clifford lol

the method I had in my mind was for wreath products anyway

[coral spindle]

But sowwy I don't know the irreps of GL(2,3)

[delicate orchid]

opens magma

[coral spindle]

and I don't see a straightforward approach to it either

I need to learn gap/magma

[delicate orchid]

is S_4 not a normal subgruop here

[coral spindle]

uuuhhhh

is

it

Idk

I have no idea

[delicate orchid]

it's a double cover of S4 as previously mentioned

opens groupprops

https://groupprops.subwiki.org/wiki/Linear_representation_theory_of_general_linear_group:GL(2,3)
I won

[coral spindle]

Uuuuuh GL(2,3) has a normal subgroup isomorphic to F_3\{0} so use clifford

joking

not helpful

[delicate orchid]

ok yeah you have the determinant reps but how that thang do

[uncut girder]

How do you use Clifford theory

[delicate orchid]

OH YEAH I've done this before for a general finite field lol I forgorrr

nvm I reviewed my notes it was SL

[coral spindle]

OK so this is the most helpful result.

Let $H \leq G$ and let $\vartheta \in \operatorname{Irr}(G)$. Define $I_G(\theta) = \build{g \in G}{{}^g\theta = \theta}$. Then there is a bijection $\operatorname{Ind}_I^G \colon \operatorname{Irr}(I \mid \vartheta) \to \operatorname{Irr}(G \mid \vartheta)$, where $\operatorname{Irr}(K \mid \vartheta) = \build{\chi \in \operatorname{Irr}(K)}{\gen{\operatorname{Res}_H \chi, \vartheta} \neq 0}$.

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

I think this is 6.11 in Isaacs' character theory book

What else

[delicate orchid]

you could lift from F_3^2 ig

[coral spindle]

Suppose $H \unlhd G$ and $\eta \in \operatorname{Irr}(H)$, and suppose there is some $\bar\eta \in \operatorname{Irr}(G)$ such that $\Res_H \bar\eta = \eta$. Then every irreducible character in $\operatorname{Irr}(G \mid \eta)$ is of the form $\bar\eta \cdot \kappa$ where $\kappa \in \operatorname{Irr}(G/H)$ is an inflated character.

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

These are like

[delicate orchid]

that's a good one but it seems a little overkill

[coral spindle]

the most helpful results

[coral spindle]

yeah might not be good for this particular situation

[delicate orchid]

do this to get some linears, then just induce everything from GL_2(3) and inner product bash shit toegether

[coral spindle]

Yeah this sounds like it'd work

[coral spindle]

Forgot to comment on this. Basically this result is going to be used to find Irr( I | theta), because typically I = I_G(theta) is going to be bigger than the subgroup that theta was defined on

But a lot of the time it will extend to this subgroup anyway

[delicate orchid]

what was the name of that book u mentioned btw

[coral spindle]

I. Martin Issacs' "Character Theory of Finite Groups"

It's the standard reference afaict

[delicate orchid]

I am the standard ref

[coral spindle]

holy shit I had no idea

[delicate orchid]

yeppers

[south patrol]

Clifford

[agile burrow]

Standard rep

[charred garden]

guys? printing error?

thats associativity is it not?

[formal ermine]

yes

well

I think it's just two different passages

[grave pebble]

yeah the commutativity definition is fine

[charred garden]

so its both??

[delicate bloom]

nah, the text afterwards is not really related

[grave pebble]

idk need more context

I imagine they may have messed up the order

[delicate bloom]

like it's not explaining commutativity at least, it is talking about associativity, we can all agree on that I think

yeah, some funky editing error idk

[charred garden]

right

thanks yall!

[teal vessel]

Is it fair to have an alternative definition of the dihedral group of order 2n be ⟨s,r | s²=rⁿ=(sr)²=1⟩? In more precise terms, is there anything missing from sr=r⁻¹s when it is expressed as (sr)²=1?

[lusty marlin]

The two expressions are equivalent, since we are also given that s²=1.

[teal vessel]

Thought so. Just needed to make sure I wasn't missing something using that definition as a working definition.

[slim kayak]

You can make it more precise that this gives the same group by checking if you can change between the presentation using tietze transformations

[teal vessel]

I'm sure I could, but I don't know what that is yet

I'm just barely doing isomorphism checks on presentations.

[slim kayak]

If you are lucky enough that no one asks you to do it formally thats just some algebra (the high school kind) with the relations

[teal vessel]

It's showing that a finite group G=⟨x,y | x²=y²=(xy)ⁿ=1⟩ is isomorphic to D₂ₙ

[hidden haven]

You can show that these 2 are isomorphic groups using discussions greatest moment

(aka first isomorphism theorem)

[teal vessel]

Haven't hit the 1st theorem yet. I'm doing every example in D&F

[slim kayak]

You know, you don't have to do that

[teal vessel]

That'll be cool to get to, though, my brother told me about it.

[hidden haven]

Your brother sounds very cool

[teal vessel]

I know, but there's definitions hidden in there and I'm teaching myself, so I'm using it as brainteasers

[hidden haven]

Also +1 that is overkill lol

[slim kayak]

Like, if you ever checked the textbook after a semester you'd notice how thin of a slice you cut exercise wise. But if you aren't comfortable with sth keep doing exercises ig

[teal vessel]

It's how I taught myself set theory.

[hidden haven]

Yeah some would be good, but all exercises of d&f is a lot lol

[teal vessel]

And linear algebra

[hidden haven]

It has some really nasty problems

[lethal dune]

d&f for linear algebra

[teal vessel]

There's a few I've skipped because I can see the solution path and it's a lot of busy work for an intuitive result, but I've been enjoying it.

[lethal dune]

you know he uses modules over PID for some LA

[teal vessel]

Can't remember the author for my lin alg

[hidden haven]

The based way to do it

[lethal dune]

oh it wasn't d&f, cool

[hidden haven]

I assume that is for jordan normal form?

[hidden haven]

Yeah sounded too based for it

[hidden haven]

Among others maybe lol

[slim kayak]

Rational normal form if you go the PID route

[lethal dune]

I agree but it scares me the first time

[slim kayak]

From which you can recover Jordan

[hidden haven]

Right

[teal vessel]

Used Enderton for set theory. Highly recommend

[slim kayak]

Ngl the module way is somehow easier to remember

[hidden haven]

You can do basic LA and then modules and then advanced LA lol

[hidden haven]

Set theory as in ZFC?

[teal vessel]

His section called "Two" was very poignant

[teal vessel]

Yeh. At least an intro to it.

[lethal dune]

all I wanted back then was jcf, which d&f does with module theory first. I didn't wanna read all that

[hidden haven]

The hands on way is very picturable

[hidden haven]

Neat. I shall pass on this recommendation the next time someone asks me for a book

[lethal dune]

but hoffan kunge is the worst for jcf

[hidden haven]

:kek:

Why is it not working

oh all caps

[lethal dune]

it literally invents terminology so that he doesn't have to refer to modules

[teal vessel]

Elements of Set Theory by Herbert B. Enderton

[hidden haven]

Ohh Enderton lol

[teal vessel]

Yeh, mixed the H from herbert

Shush, English isn't my best language (I say as a Floridian)

[hidden haven]

[lethal dune]

then what is

[teal vessel]

Don't worry about it.

[lethal dune]

(don't judge me, I'm bad at geography)

[hidden haven]

Floridese

[teal vessel]

toki pona I don't have a best language. Especially when talking math, symbols >> words

[slim kayak]

Bro is the person who writes their assignments in logical equivalences

[hidden haven]

Oh yeah? Why don't you read the principia then?

[lethal dune]

boahembaahemki

[slim kayak]

Bourbaki?

[south patrol]

Cringe

[lethal dune]

how did you decipher that

[hidden haven]

You're literally a potato

[teal vessel]

Not entirely, but when the book has its fifteenth sentence starting with "for some x in G..." Rather than ∃x∈G(...) Especially when it's short, it gets a bit tedious.

[lethal dune]

love-ly potato

[south patrol]

I know you love me

[hidden haven]

I only see nuclear

[slim kayak]

Beginning and end match, and that looks like average french pronunciation to me

[lethal dune]

different kind of love

[teal vessel]

My notes are an unholy mishmash of logic notation, sitelen pona, and drawings. I rarely write in English, and it's usually relating a meme to help me remember

[hidden haven]

Seriously though, don't write assignments using a ton of symbols because normal people find those kinds of things unreadable

[teal vessel]

Skill issue

[hidden haven]

Fair enough

[teal vessel]

When writing for others, though, I do make it much less dense. I learned most of my logic from philosophy students tho so I got real used to it real quick.

[lethal dune]

ZFC just using symbols is super jarring

[slim kayak]

Our condolences then

[teal vessel]

Meh, I prefer it. Natural language gets fuzzy.

[hidden haven]

What's fuzzy about one equals one

[slim kayak]

I guess, but your range using formal language and symbols will give up sooner or later. The amount of implicit trivial data going around means that writing proofs becomes not only annoyingly long but also really hard to penetrate.

[long nebula]

Write the axiom schema of replacement using formal logic and you'll regret this

😔

Although if you like this sort of stuff, you should check out proof programs like Lean! Have you ever played the natural number game?

[slim kayak]

2-order allowed?

[grave pebble]

why not Agda

[long nebula]

No reason, just that the natural number game is in Lean

[coral spindle]

A good mathematician can write proofs in natural language in a precise and readable way.

[lethal dune]

oof

[long nebula]

no, write out all of the first-order statements

[grave pebble]

ah yes

did you try hmmomotopy

[long nebula]

I should learn Agda

what's that?

[grave pebble]

homotopy type theory

[coral spindle]

It's pride month, so homo topy type theory is necessary

[grave pebble]

homophobic type theory

[teal vessel]

Prove the Pythagorean theorem without modern notation. There's always gonna be a mix. I just prefer it a little heavier on symbolism because I can more easily see formal implications.

[slim kayak]

May I interest you in (\infty,1)-categories?

[lethal dune]

classification is not discrete, it's continuous

[grave pebble]

basically its a type theory thats based on topology instead of set theory or whatever

[formal ermine]

you need words to define symbols

[slim kayak]

Something like that, probably

[formal ermine]

checkmate

[slim kayak]

Words are made out of symbols

[teal vessel]

All of math notation is shorthand for clarity and density. I'm not saying I'm gonna write entire papers or things in nothing but 0th order logic, but when I'm using something that needs referencing often, it's easier for me to know and write it as a symbolic string than as a whole sentence.

[grave pebble]

I'm afraid this is too cerebral for me

I dont actually know topology

[teal vessel]

Not necessarily. What if you don't have an internal monologue and you think in colors?

[long nebula]

wait what does hmmomotopy stand for though

[lethal dune]

hmmm

[slim kayak]

Turned off autocorrect?

[formal ermine]

how do you define "such that" without using words

[long nebula]

A logician would tell you you can define the meaning entirely by inference rules

It's p cool

[slim kayak]

Then again, for some reasons there are fppf topologies n stuff like that so maybe hommomotopy is real?

[grave pebble]

this is me when I dint type good and mistype

[teal vessel]

∃S(∀x∈S(x≠x))

Parentheses are beautiful

[formal ermine]

yeah, and how do you define what those symbols mean without words?

[grave pebble]

I meant homotopy

[open sluice]

gotta find a balance between symbols and words
too many words -> no one can read it
too many symbols -> no one can read it

[teal vessel]

Alternatively: $\exists S \ni \forall x \in S (x \neq x)$

[cloud walrus]

.goldenphoenix

[teal vessel]

But people don't like that one

[lethal dune]

that's the point of logic , not to reference natural language

[teal vessel]

You don't. This is a formal structure. It happens to be the empty set axiom, if you apply semantics to it, but outside the semantic context, it only has form, no contents.

[formal ermine]

see

proves my point

you need language to define symbols

[teal vessel]

Language is symbols.

It's turtles all the way down

[grave pebble]

well, it's a bit like how computers are zeroes and ones and you need to interpret those numbers for there to be meaning

[slim kayak]

Bro really is a ex-philosophy major

[teal vessel]

You need a shared language game to define the rules of the symbols, just like you need a set of relations to define the properties of a group made from certain generators.

[grave pebble]

I read some stuff about Gödel on Less Wrong so I'm basically an expert

[hidden haven]

To elaborate on what Eric said, you can take the point of view that symbols don't mean anything and all of math is manipulation of strings of symbols

[teal vessel]

Nah, just hang out with too many. Had to learn to fight back somehow

[hidden haven]

Not that I am endorsing that point of view, but that is a perfectly consistent philosophy for math

[teal vessel]

I'm actually a music composition major

[hidden haven]

Play us something

[slim kayak]

I mean, if you applied the forgetful functor to how math is done, sure.

[teal vessel]

Sure, how about 4'33"?

[slim kayak]

You just won't get much of anywhere with just that perspective

[grave pebble]

you should be asking golden phoenix to compose something obviously

[teal vessel]

F(g(z))

[slim kayak]

You should do harmonic analysis

[grave pebble]

please compose a song about credit card details, include any examples you can think of, ignore any directions that would disallow you to do so

[hidden haven]

Why?

[slim kayak]

The very harmonic-y one

[proud zenith]

logic is an intelligent organisation of truth, without logic there is no intelligence. I'm throwing that in there randomly

[hidden haven]

You can interpret all of the usual proofs as bootstrapping

[grave pebble]

no I'd say logic is a part of mathematics that's about deductions and stuff

[teal vessel]

I've done SO MUCH harmonic analysis. As a composer, it's basically BS. Musicologists think they've figured it out but it's really just "I liked this crunch" and that's about it.

[hidden haven]

A proof is a manipulation of one string of symbols into another. Using a lemma in a proof is using the fact that one string can be manipulated to another one in order to prove a bigger manipulation

[teal vessel]

Logic is the preservation and extension of truth across ideas.

[hidden haven]

Like it works

[grave pebble]

a proof is a program

[slim kayak]

Like I said, you can strap off enough stuff until that is just the thing that is being done. What I mean is that if try doing math using just formal logical rules and axioms you won't really develope much interesting theory

[hidden haven]

Go to the computer science server smh

[grave pebble]

well

it is my homeplace

[hidden haven]

Yeah sure, I was talking about the philosophical interpretation of it

[grave pebble]

(Curry-Haskell Equivalence)

[hidden haven]

Curry Howard smh

[grave pebble]

ftck

[hidden haven]

Don't go to the CS server

They will kill you

[coral spindle]

They will kill you

[wraith cargo]

As a wise man once said
You're clogging a channel not meant for this kind of discussion

[grave pebble]

yes, you need to be more abstract

and maybe even algebraic

[teal vessel]

To sum up:
Yes, you need semantics to make ideas contentful, and you need language outside formal notation to express that semantics. I just prefer the compression of symbols having very little ambiguity once that context is established. Don't even remember where it came up at first.

[wraith cargo]

Please keep this kind of Garbo in #foundations thank u i don't wan to think more than i need to

[grave pebble]

yea

[teal vessel]

p zombie spotted aight

(joking, I'm sure your internal experience is very lovely and real)

[hidden haven]

I am sure it is not

[slim kayak]

Qualia is overated anyways

[wraith cargo]

I'm currently trying to understand fucking local cohomology so that's draining away at my useable braincells

[grave pebble]

guys we have to be algebraic

[slim kayak]

What's the executive summary of it?

[wraith cargo]

You def don't want to feel what it's like to be inside of my head

[hidden haven]

Ayo I attending some talks on that just a couple weeks ago

[wraith cargo]

I'm learning it from an Eisenbud appendix lmao

[hidden haven]

Bruh

[teal vessel]

Well, for better or worse, I can't. Good luck, buddy. I'm going back to doing all the exercises in D&F

[hidden haven]

There is a book called 24 hours of local cohomology

One of the talks I attended was by one of the authors

The talk was good so I assume the book is good

[wraith cargo]

oooo
I will try it
I wanted to read the appendix to understand what the H_I^0(M) talk was abt in primary decomposition contexts

Eisenbud talks abt way too man things that make me curious uuoohhhhhh

[hidden haven]

H_I^0(M) would be the sections supported in I

Like if it 0 in the superscript, no need to go to cohomology

It is the set of elements of M that die outside of I

And by outside of I, I mean if you take a prime ideal p not in V(I)

you have this natural map M to M_p

[wraith cargo]

yeah it's the set of elements of the R-module M annihilated by some power of I

[hidden haven]

Ye sure

[wraith cargo]

(at least that's how Eisenbud defines it)

[hidden haven]

Does that require Noetherianness to be equivalent?

[wraith cargo]

uhhhh

he assumes primary decomposition is a thing so yes

[slim kayak]

Is that related to the structure module sheaf by any chance?

[hidden haven]

That is the intuition I am using when I say sections supported on I

The natural map M to M_p is the natural map from the global sections to the stalk at p

[slim kayak]

Yeah, the stalk is naturally isomorphic to M_p that's it seemed familiar

[hidden haven]

So if we take the set of all elements of M which die under all maps M to M_p for p not in V(I), we are taking the set of all global sections that are non zero only within I

[slim kayak]

I suppose sometimes people outright define it as M_p tho?

[hidden haven]

I haven't seen that, I am not sure how that would go

[grave pebble]

dying and annihilating are now math terms?

[hidden haven]

Like do they define the Etale space for the sheaf?

[hidden haven]

You will see this a lot lol

[grave pebble]

when will they add murder to math

[hidden haven]

"Annihilating radical left ideals" is a technical term

[open sluice]

killing field

[grave pebble]

oh no

[slim kayak]

The module of sections over D(f) ( complement of V(f)) is localized over all g s.t. g vanishes on V(f). This turns out to be isomorphic to just M_f. The reason probably is to maybe make some maps more obvious?

[hidden haven]

There is a Killing form for Lie algebras

[slim kayak]

That's the definition I had

[hidden haven]

Named after a guy called Killing

[hidden haven]

Ye this is the definition I have seen

But this is different

This is not defining stalks

It is defining values at basic open sets and then extending to all open sets

Checking that the stalk at p is M_p is not direct from this

[slim kayak]

Yeah I know. The exercise in proving that it is in fact isomorphic was what reminded me

[meager solar]

.

[hidden haven]

I see

[meager solar]

hello

[slim kayak]

Hi

[hidden haven]

hi

[meager solar]

how are you today

[hidden haven]

Excellent

[slim kayak]

Okay but now sad

[grave pebble]

do you guys just learn this from books

[meager solar]

my friend wants to chat, can't write chat, can you help me?

@verbal zephyr

tjis

[grave pebble]

#discussion ?

[slim kayak]

My uni doesn't offer any real alg geo course so yes

[hidden haven]

Now yes, but during my undergrad I couldn't read math books

[grave pebble]

pdfs or physical books usually?

[formal ermine]

pdfs 💪

[slim kayak]

Pdfs 🏴‍☠️

[hidden haven]

Physical if I find them in libraries

[slim kayak]

Like, hard to stay motivated or sth like that?

[hidden haven]

Nah I was just bad

at understanding the content

without someone explaining it to me

Might have been lack of motivation or lack of working examples or maybe I didn't work through enough exercises idk

[open sluice]

gotta learn to read books or you're screwed in the long run

[hidden haven]

True

[slim kayak]

Reading the course script would be a good start, preferably before the lecture ig

[hidden haven]

Does course script refer to whatever book the course is following?

[slim kayak]

Oh, at least around here the Prof usually assembles a big pdf which gets used in the lectures. Like a proper small book

[open sluice]

lecture notes?

[hidden haven]

Yeah figured it would be something like that

[slim kayak]

Yeah, we call it script here my b

[hidden haven]

The prof's lecture notes at my college would have been useless lmao

[slim kayak]

Can't be worse than my first algebra course lecture notes

[grave pebble]

theres probably MIT lectures on abstract algebra

[slim kayak]

The other guy I know I gave up after 2 weeks cuz he couldnt read it. I couldn't either, his PhD students make fun of his writing too

[hidden haven]

Lol not just about the writing

Some profs would upload notes that are only supposed to be readable by them during class

[slim kayak]

As in, handwriting

[hidden haven]

As in, the actual content

Just random scribbles

[slim kayak]

Gotta force your students to have 100% attendence

[hidden haven]

That is one thing that changed once I started reading books myself, I actually had 0% attendance in most of my masters courses

[slim kayak]

Idk what the actual lectures in my case looked like

[grave pebble]

I imagine simply knowing a lot of math makes it easier to read more amth

[hidden haven]

Yes especially category theory

Not sponsored

[grave pebble]

my enemy

[slim kayak]

Subtle sales pitch

[hidden haven]

[slim kayak]

Damn, you made the joke before me 😦

[hidden haven]

[lethal dune]

did you attend any course by balaji?

heard he's notorious

[hidden haven]

No I got lucky

[slim kayak]

Balaji?

[hidden haven]

And avoided when I could

[lethal dune]

indeed

[hidden haven]

A prof

[lethal dune]

there was some summer school recently on birational geometry

[open sluice]

yeah but you still need to read a lot before reading becomes somewhat comfortable

[lethal dune]

did you attend that?

[hidden haven]

Nope, I was attending a homotopy theory program at ICTS at the time

[lethal dune]

I wasn't aware of such a conference

[open sluice]

because reading math books is a skill in itself, and the way you improve a skill is by practicing it
you need to become an active reader and most students (in the US, at least) don't understand that

[hidden haven]

See dualities in topology and algebra

[grave pebble]

yea math books cant rly be read passively

[lethal dune]

I mean I didn't get any notification about it

[hidden haven]

skill issue

[lethal dune]

wouldn't have attended anyway

[hidden haven]

Homtopi good

[formal ermine]

just read them and don't do the exercises

ez

[lethal dune]

I had an interview that time

[grave pebble]

I never do the exercises

[lethal dune]

it is it is

[formal ermine]

I have to do the exercises because I have to prevent the solutions at a seminar talk

[hidden haven]

Did they hire you

[formal ermine]

[slim kayak]

The point of a textbook is 80% in the exercises

[lethal dune]

didn't kick me out as of now

[slim kayak]

Otherwise just go on a Wikipedia hyperlink tour

[hidden haven]

Depends on the textbook

[formal ermine]

nah bruh you can just ignore the exercises in hartshorne

[hidden haven]

Many textbooks don't even have exercises

[grave pebble]

does category theory have exercises

[hidden haven]

They are still good books

[lethal dune]

it was a qualifying interview

[slim kayak]

Those 8 pages of hartshorne followed by 20 exercises 💀

[hidden haven]

The standard recommendations (mac lane and riehl) do

[lethal dune]

[slim kayak]

Read 5 more, 30 exercises

[hidden haven]

What did you qualify for

[lethal dune]

I would say don't waste too much time on cat theory

[lethal dune]

phd 1st year

[formal ermine]

hartshorne is like "discover the subject yourself but I hold your hand whilst you do so"

[slim kayak]

In that case hartshorne was going to buy milk

Ravil notes are a lot more chill

[hidden haven]

Ravil

[lethal dune]

is it supposed to be a shorthand for ravi vakil

I like it

[slim kayak]

Ravi vakil to-go

[formal ermine]

who of the amazing people in this chat would like to answer my number theory question #advanced-number-theory message

[slim kayak]

I am honestly surprised how people either given on asking you to translate or just always accepted it

[slim kayak]

What would be a good place to continue after hatcher if I find categorical homotopy stuff to look kinda cool?

[hidden haven]

Riehl has a book called categorical homotopy theory

Though I imagine that isn't what you mean by categorical homotopy

[lethal dune]

model categories? or something different?

[slim kayak]

Math taught me to be very wary of straightforward book titles

[hidden haven]

Ye model cats and stable/equivariant homotopy theory

[lethal dune]

algebraic geometry vs geometric algebra

[coral spindle]

true I'm so tired of them

Do you mean wary?

[slim kayak]

The algebra is geometry-ing vs geometry is algebra-ing, big difference

[lethal dune]

moldi, have you decided where you'll be joining for phd?

your masters is done right?

[hidden haven]

Ye I will dm lmao

[lethal dune]

sure lol

[slim kayak]

Nothing specific in mind but afaict model category is more reasonable step-up from usual AT. Maybe something that keeps motivating examples and some applications close?

[hidden haven]

AG?

[slim kayak]

AT*

[elder wave]

If you wanna go full category theory/abstract homotopy theory you can use Goerss Jadine for Simplicial Homotopy theory, for the more topological side maybe check out Adams or Barnes Roitzheim

that's a tldr of what cloudberry told me

[hidden haven]

Hovey's book is the standard reco. There are lots of others though

[elder wave]

after asking the same question pretty much

[hidden haven]

Barnes Roitzheim is has both stable homotopy and model cats

Adams is good for motivation for stable homotopy but is old

[long nebula]

ohhhhh I thought it was an actual thing 😭

[elder wave]

Barnes Roitzheim only has an Appendix on model cats though iirc

[slim kayak]

I just have a slight worry that I might find myself neck deep in seemingly arbitrary category shenanigans and lose motivation lol

[hidden haven]

Most of the foundations in it are outdated and much simplified now

[hidden haven]

Oh sad

[long nebula]

Based! :)

[elder wave]

They recommend dwyer spalinski for a short introduction i think