groups-rings-fields

[chilly ocean]

All understood thanks a lot!

[sturdy mirage]

yes, i'm not familiar with the powers you are using, r^m1 means r raised to the power of m1?

[tough raven]

I think it means m1 r's

No wait

[sturdy mirage]

the way i'm used to writing cycles in $S_n$ is.. $\sigma = ( r_1 r_2 \cdots r_k)$, and you then get
$\sigma^2 = (r_1 r_3 r_5 \cdots r_k r_2 r_4 \cdots r_{k-1})$ if $k$ is odd,
$\sigma^2 = (r_1 r_3 r_5 \cdots r_{k-1})(r_2 r_4 \cdots r_k)$ if $k$ is even.

[cloud walrus]

reking

[chilly ocean]

oh sorry m1 is the power or r1

[sturdy mirage]

but what is r1? you just wrote r up there

[chilly ocean]

rectified 🙂

should've written that in latex indeed

[chilly ocean]

that sort of makes sense

[sturdy mirage]

so the ri's are elements of a group, and mi's are integers?

[chilly ocean]

but in my case the cycle shape is given by the ri's

So

the ri is an integer denoting a cycle length and mi denote how many times that cycle of length ri occurs

if that makes sense

[sturdy mirage]

maybe it means something different in square brackets, than the normal paranthesis that i'm used to

[chilly ocean]

it means the cycle shape

[tough raven]

So n = r1 m1 + ... + rk mk?

[chilly ocean]

exactly this

[cloud walrus]

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

[nova plank]

Well, if ker(theta) = F[x] then your theta(f(x)) =0 for all f(x)in F[x]. So does theta then satisfy the other conditions?

Well, do you require that theta(1) = 1 as part of your defn of homomorphism?

Then you cannot assume that because theta(f(x)) = 0 is a counterexample.

In fact, when ker(theta) = F[x] then you always have I as a subset of ker(theta)

Since I is a subset of F[x]

Then you cannot have ker(theta) = F[x]

Because ker(theta) = F[x] means theta(f(x)) = 0 for all f(x) in F[x]

But that's unrelated to the ideal

[tough raven]

Basically the condition I subset ker(theta) does nothing to prevent ker(theta) = full ring, and may even make it more likely.

[nova plank]

In fact you can say that if I is not a subset of ker(theta) then ker(theta) ≠ full ring

[molten silo]

How do i right 17 has a product of irreducible elements in Z(root(-11))

[chilly ocean]

write*

[molten silo]

That is

[rustic crown]

what are the elements with norm 17?

[chilly ocean]

a^2 +11b^2 = 17

I guess you have to look at that

[rustic crown]

yep

[hot lake]

but the ring of integers of Q(root(-11)) isn't Z[root(-11)]

[rustic crown]

Z[sqrt(-11)] is still a ring though

[hot lake]

yeah but I'm having a hard time finding an element of norm 17 in it

[rustic crown]

there aren't any

[molten silo]

its not possible is it

[hot lake]

okay so 17 = 17 I guess

[rustic crown]

b^2 can't be 0, can't be 1, >= 4 is too big

[chilly ocean]

exactly

and 17 isnt a perfect square

so no a for which a^2 = 17

Can there be an even permutations in the symmetric group S_n of odd order?

My intuition says no

[rustic crown]

S_n always has even order for n>=2

[hot lake]

the identity is an even permutation and has odd order ?

[rustic crown]

ahh, i read the associativity wrong

[chilly ocean]

True i was wondering if there were any other than the trivial identity

[hot lake]

any 3-cycle

any 5-cycle

[chilly ocean]

so a cycle of length k is even iff k is odd, got it

[hot lake]

well time to pick permutations at random and look at their squares

[chilly ocean]

I can't find a pattern

[rustic crown]

if you think about it, if sigma has odd order then sgn(sigma^odd) = 1 = (sgn(sigma))^odd, so sgn(sigma) = 1.

[chilly ocean]

I noticed that transpositions have length 2 right

so sigma^2 = 1 iff sigma is a transposition

[hot lake]

wrong

[chilly ocean]

[hot lake]

there are permutations of order 2 or 1 that aren't transpositions

[chilly ocean]

idk why i've written order

corrected it

[rustic crown]

this is still false... sigma = id

[chilly ocean]

So sigma is even

[rustic crown]

yep a permutation of odd order is automatically even

[golden pasture]

yup

one fancier proof is disc is 11, and x^2+11 is irreducible in finite field with 17 elements, and you're done

[hot lake]

yeah odd permutations of odd order are impossible but everything else is possible

[chilly ocean]

Got it

[molten silo]

@golden pasture the question says "Show that the element 17 of Z(√−11) is not irreducible by giving a factorisation
into irreducible elements. Justify your work."

so confused

[rustic crown]

17 is irreducible

[molten silo]

The question must be wrong

seems impossible

[rustic crown]

could you confirm if its Z[sqrt(-11)] or ring of integers of Q(sqrt(-11))?

[golden pasture]

yea it isnt possible

it is irred in the ring of integers

[hot lake]

even in the ring of integers it's irreducible

[golden pasture]

hence irred in Z[sqrt(-11)]

[hot lake]

so yeah the question is pretty wrong

[vestal snow]

Can someone explain this to me?

I understand why A = k[x,y]

But after that I'm pretty lost

[hot lake]

what's U

it's saying that Spec k[x,y] has a point corresponding to the ideal (x,y) of k[x,y]

but U doesn't seem to have it so U can't be Spec k[x,y]

[vestal snow]

U is A^2 - {(0,0)}

[vestal snow]

How do we know that if $A \cong k[x,y]$, then the induced isomorphism $Spec(A) \cong Spec(k[x,y])$ sends the ideal (x,y) to (x,y)?

[cloud walrus]

Have a Banana, Bitch

[hot lake]

if U is an affine scheme then it has to be isomorphic to Spec R with R = the ring of global sections of U, and the topology of U has to be the Zariski topology of Spec R, and the rings Gamma(V,Ok²|U) have to be the correspondings localizations of R

[vestal snow]

Okay so let's assume that U is affine

then U = Spec(A) for some ring A

And we work out that A is isomorphic to k[x,y] as a ring

[hot lake]

yeah

yeah

[vestal snow]

Therefore, Spec(A) is isomorphic to Spec(k[x,y]) as a sheaf

[hot lake]

I think your question is why can we assume that this isomorphism can be the identity

but uh

Idk what I would say to that

yeah Spec(A) is isomorphic to Spec k[x,y] as a sheaf

[vestal snow]

consider the image of (x,y) under this isomorphism

[hot lake]

the (x,y) from where

[vestal snow]

Spec k[x,y]

[hot lake]

that's a maximal ideal of k[x,y] so it corresponds to closed point of Spec k[x,y]

[vestal snow]

yes

So it must map to a closed point of Spec(A)

[hot lake]

yeah

[vestal snow]

but why must it map to (x,y)?

[hot lake]

and there should be a closed subset of Spec A called V(x) and one called V(y) such that V(x) intersected with V(y) is that point

no I think you have to pick x and y in the ring of sections of U first

then map them into k[x,y]

[vestal snow]

I don't get it

[hot lake]

U has a point whose closure is basically that point + {(x,0) for nonzero x}

so it should correspond to some point in Spec k[x,y]

[vestal snow]

We should probably make this stuff more precise

[hot lake]

U = the set of prime ideals of k[x,y] with the prime ideal (x,y) removed, right ?

[vestal snow]

Yes

[hot lake]

so the ideal (y) is a point of U

so if U is an affine scheme Spec A

there should be a point of Spec A corresponding to (y)

[vestal snow]

yes

[hot lake]

I think our whole problem reduces that we magically know that A has to be isomorphic to k[x,y]

but we don't know why we can assume that this isomorphism can be literally the identity

[vestal snow]

Yeah

That's exactly it

[hot lake]

and there is nothing stopping us from mistakenly picking the isomorphism that swaps x with y

for example

and I'm not sure that would "break" anything

... it's so hard to articulate stuff

so uh

I don't think I have a quick answer right now sorry

[vestal snow]

It's okay

Thanks for the help

[hot lake]

I think it comes down to if you have an affine scheme Spec R there is a unique isomorphism from Spec R to the ring of global sections that makes it so the functions vanish on the right points

[chilly ocean]

We need (f(x) + g(x))^2 = f(x)^2 +g(x)^2

[delicate bloom]

expand the left side and simplify maybe

[next obsidian]

It has to do with the nature of the isomorphism of Spec A to Spec Gamma(O_Spec A(Spec A))

Edit: I thought harder and it’s not very clear to me either lol

I think

[chilly ocean]

Tried that

doesn't give something nice

[delicate bloom]

don't you get 2f(x)g(x)=0

[chilly ocean]

Right I did that by multiplying the polynomials out with their coefficients

[chilly ocean]

but yeah

[delicate bloom]

this must hold for all f(x), g(x) right?

[chilly ocean]

Correct

Therefore

[delicate bloom]

might as well pick something simple like f(x)=g(x)=1

[chilly ocean]

Right so we shall look at specific functions

So for p = 2 that holds

[delicate bloom]

yeah

our choice works for every Z_p so we're done

[chilly ocean]

Right so its a ring homomorphism for every prime?

[delicate bloom]

lol no

f(x)=g(x)=1 was chosen carefully, although simple to work in every Z_p

and that reduced the criteria to needing 2=0

this is only true in Z_2 though

[chilly ocean]

Ok right, the choice of the functions was true for every Z_p. Understood

[delicate bloom]

cool

[chilly ocean]

Thanks

[delicate bloom]

yw

[golden pasture]

The degree of 2f(x)g(x) must equal deg(2)+deg(f(x))+deg(g(x))

but then deg(0)=-infty

soooo

[delicate bloom]

so g(x)=0 or f(x)=0 😌

[golden pasture]

i was so tripped up initially cuz i was like

b...but frobenius dont lift!

damnit people using Z_p for F_p

[delicate bloom]

lol I almost asked but then I figured I would just assume

question would be silly for Z_p haha

[golden pasture]

F_p[[t]] > Z_p

but also Z_p is perfect

[delicate bloom]

I like carrying when I add

[golden pasture]

🤔

hard choice

mero cohomology simp

[delicate bloom]

lol idk about that

[golden pasture]

(https://ncatlab.org/nlab/show/carrying)

[delicate bloom]

well I know what you're referring to

I just know 0 cohomology haha

[golden pasture]

there is your motivation to learn :D

[delicate bloom]

I just downloaded this paper the other day

ngroupoid said it was a decent intro to cohomology haha

time to find out how memed on I am

[golden pasture]

[prisma ibex]

wait which one lmfao

[delicate bloom]

https://www.jstor.org/stable/3072368

[prisma ibex]

oh this thing lmfao

yea it's an okay intro lol

obviously not a place to learn it properly

[delicate bloom]

haha ofc

[delicate orchid]

I'm reading something on the spectrum of maximal ideals in algebraic geometry and they've used this map and I cannot for the life of me see how it's well defined

cause surely if you take an maximal ideal in R[X, Y] like (X-1, Y-2) this would map to (1, 2) but the same ideal can be easily written as (Y-2, X-1) because the ring is commutative and it would map to (2, 1), so we have two ideals that are equal but with differing images

ping me if you've got any idea about this cause I've been stuck on this for a while now

[delicate bloom]

the labels aren't commutative, you can't swap out X,Y for Y,X

[delicate orchid]

oh that's so obvious once you've said it

[delicate bloom]

haha cool

[delicate orchid]

thanks lmao

[delicate bloom]

I don't even know what M-spec(R) is

[delicate orchid]

spectrum of maximal ideals

using notation from my uni so it might not be universal

[delicate bloom]

idk what that is, like you make a ring out of all the maximal ideals I guess?

[delicate orchid]

it's just the set of all maximal ideals of R

[delicate bloom]

oh ok

[molten silo]

Find two non-trivial subrings of Z_22 anyone got any ideas

[delicate orchid]

wait is that Z/22Z or 22Z

[molten silo]

second one

[delicate orchid]

multiples of 22 work then

wait 22Z isn't a ring it's an ideal

[molten silo]

?

[delicate orchid]

1 isn't in 22Z so it can't be a ring

but if you want subideals 44Z and 88Z work

[molten silo]

it can in some definitions

[delicate orchid]

oh sure if you're working with non-unital rings then go ahead

[delicate orchid]

oh apparently that is Z/22Z

[molten silo]

i see that now

[delicate orchid]

never seen that notation before so I googled it

uhh in that case any divisor "n" of 22 will form a subring Z/nZ or Z_n in your notation

[molten silo]

ahh

ok

[molten silo]

ψ : Z_5 → Z_10 , a → ψ(a) = 2a is this a ring homomorphism

same notation as above

[woven delta]

Yes

[molten silo]

how?

[woven delta]

Well I guess you need to map 1 to 1

So no

Sorry I read that initially as group

[sturdy mirage]

is ψ(1+1) equal to ψ(1) + ψ(1)?

well yea

and then you have to check that for all numbers not just for 1

and for multiplication too

[molten silo]

i did that it didnt work

false for 2*3

[sturdy mirage]

for example ψ(1*1)

or that

if it doesnt work , then its not a homomorphism

[molten silo]

Im trying to create a homomorphism between these sets

[scarlet estuary]

both under +?

[molten silo]

doesnt work for multiplication

[scarlet estuary]

oh theyre fields?

wait im confused

what structures do they have

[delicate orchid]

Those two sets aren’t homomorphic, the right contains a multiplicative identity the left does not - their structure is different as fields/rings

[sturdy mirage]

i dont think a homomorphism is defined between sets

[molten silo]

z_7 uses modulo 7

[sturdy mirage]

they need to be rings, groups, or something

[delicate orchid]

As additive groups you can get away with it

[scarlet estuary]

my question is

are they groups

rings

fields

???

and if so, under what operations?

[molten silo]

must be rings

[delicate orchid]

I’m pretty sure Z_7 is F7

So fields

[scarlet estuary]

but {0, 2, 4, ...} is not a field

[delicate orchid]

In which case what I said here is the result

[scarlet estuary]

at least under standard *

theres no identity

unless its like

{0, 2, 4, ...} under multiplication mod 7

in which case theyre already the same structure lmao

the homomorphism is the identity

[delicate orchid]

a*b = 1/4(ab) could work maybe?

[scarlet estuary]

can you give the full question/context? @molten silo

since clearly somethings missing

as stated, assuming these are meant to be rings under standard + and *, they are not homomorphic

[molten silo]

The second set is modulo 14

[scarlet estuary]

then it has no multiplicative identity.

whereas the first set does

so they cant be homomorphic.

[delicate orchid]

But can we make it have one

[scarlet estuary]

(since homomorphisms map identities to identities)

[molten silo]

i think it does

[scarlet estuary]

?

what do you think its identity is

[molten silo]

8

[delicate orchid]

You’d need a unit in Z_14 to have an identity assuming closure

8*4 = 2

[scarlet estuary]

8 * 4 = 24

[delicate orchid]

Close enough

[scarlet estuary]

the sheer genius of #groups-rings-fields

[delicate orchid]

It’s 1am bro

[magic owl]

maybe leave this to nami

[delicate orchid]

WOW

Haha I’m going to edit it and ruin ur pin

[sturdy mirage]

BAN HIM

[scarlet estuary]

your contribution to mathematics has been eternalized

fine

[sturdy mirage]

no edits

[scarlet estuary]

edit that

[delicate orchid]

Ok I’ll give it a go

Multiplication is a spook bro

[scarlet estuary]

er wait though

did i just brain fart

[hidden haven]

They're just additive groups? Then Z/7Z ≅ 2Z/14Z

[scarlet estuary]

no wait

[delicate orchid]

They’re rings

[scarlet estuary]

@molten silo sorry im being a dumbass

youre right

[hidden haven]

Oh ok

[scarlet estuary]

8 is a * identity

here

[sturdy mirage]

8*3 mod 14 = 10 ?

[molten silo]

yes

[scarlet estuary]

but they still dont have the same structure

[sturdy mirage]

oh

3 isnt in it

[hidden haven]

ohh

[sturdy mirage]

my god

[scarlet estuary]

lmao

i wasnt thinking

sorry

[molten silo]

so do you know the homomorphism?

[scarlet estuary]

hm

[molten silo]

does the idenity map to the idenityt

[delicate orchid]

I’m having an existential crisis rn

[scarlet estuary]

well 1 -> 8 should surely be enough

to derive the rest

[hidden haven]

1 has to keep to 8

[molten silo]

i see

danker

[delicate orchid]

How tf is 8 the identity is this some weird quadratic residue number theory thing bro I can’t cope with this

[sturdy mirage]

and 0 has to map to 0

i think?

[hidden haven]

Ye

[molten silo]

times by 8

[delicate orchid]

I mean I believe it’s the identity because it is but how

[scarlet estuary]

0 |-> 0

1 |-> 8

2 = 1 + 1 |-> 8 + 8

and so on

[sturdy mirage]

so the homomorphism is just T(x) = 8*x (mod 14) ?maybe

[molten silo]

yes

[scarlet estuary]

since 1 generates all of Z_7

[molten silo]

this makes sense

[scarlet estuary]

so knowing where 1 goes is sufficient

[molten silo]

i never new that

[hidden haven]

8*2 = 2 so 8*2n = 2n

[sturdy mirage]

isnt T(x) = 0 also a homomorphism technically

[scarlet estuary]

anyway sorry for that brain fart earlir

that was

monumentally stupid

[delicate orchid]

Wait is it because 8/2 and 14/2 are Coprime and we’re working in a quotient ring of 2Z?

Is that it???

[molten silo]

make the mutiplication table of 0, 2, 4, 6, 8, 10, 12 modulo 14

[hidden haven]

I don't think so

[delicate orchid]

I’m trying to generalise it to arbitrary rings in my head

[hidden haven]

You just check that 8 acts as identity on the generator

[delicate orchid]

I mean that works yeah but like

Ok so if you have 3Z/12Z = {0, 3, 6, 9} would that make 9 the identity cause 9/3 and 12/3 are coprime and 9*3 = 27 = 3

Ah yes that’s it

Delicious

That only works for rings nZ/mZ where n divides m hmmmm

I suppose if n and m are coprime you just get a rearrangement of Z/mZ

Ok I’m satisfied with that rant I get it now

[hidden haven]

14/2 and 10/2 are also coprime

[delicate orchid]

But 10/2 isn’t in 2Z

[hidden haven]

Wait so what is your condition

[sturdy mirage]

14/2 and 12/2 then?

[delicate orchid]

Uhhh lemme actually formulate this

Nvm it doesn’t work

If you take an element of x of nZ/mZ where n divides m such that x/n and m/n are coprime then x is a unit

not the identity

[hidden haven]

Ye for being the identity you just get an equation

[delicate orchid]

well at least that was fun to think about

[hidden haven]

[sturdy mirage]

wait does it also work with the homomorphism T(x) = 12x?

with 12 as the identity

[delicate orchid]

anyway for my next trick I'm going to define the wew ring where 4*8 = 24 is it's defining property

[delicate orchid]

it'll work with any unit I think?

[hidden haven]

12 isn't the identity tho

[delicate orchid]

it can be whatever it wants...

[hidden haven]

[molten silo]

Anyone recommend any books for abstract algebra

[delicate orchid]

there'll probably be some suggestions already in #book-recommendations

[molten silo]

k

[maiden ocean]

Context: G is the Galois group of an infinite Galois extension K of k

I'm not sure I understand what's going on with the fundamental open neighborhood, or how H hitting the coset sigma U_M for all M we have sigma in the closure of H

[oblique river]

The U_M form a basis of open neighborhoods of the identity

and so the cosets sigma U_M form a basis of open neighborhoods of sigma

so if you think of this topologically -- H has nontrivial intersection with every open neighborhood of sigma is exactly what it means for sigma to be in the closure of H

maybe that doesn't entirely answer the first part of your question

[maiden ocean]

I think I see

the U_M are the kernels of the projections

right

[oblique river]

yeah

sorry if that wasn't super clear though i can try and elucidate some part of it

[maiden ocean]

I think I get it

I just wasnt realizing that the U_M being a basis for 1 implies sigma U_M are a basis for sigma

but that makes sense I think

[oblique river]

ah yeah, that's sort of a topological group thing

part of the definition of a topological group is that (left) multiplication by some fixed element is a homeomorphism of the group

so x --> sigma*x

and the image of a neighborhood basis will be a neighborhood basis cuz homeos preserve that kind of thing

[maiden ocean]

Right

Ok that makes sense

[oblique river]

awesome :)

[maiden ocean]

thank you

this is pretty cool

[oblique river]

no problem!

I think so, too

in some sense, algebraic number theory (and the langlands program) is all about just trying to understand the structure of the absolute galois group of Q

[maiden ocean]

I am definitely enjoying this more than I enjoyed finite galois theory lol

[oblique river]

nice! what reference are you using?

[maiden ocean]

szamuely

galois groups and fundamental groups

I am in a reading group its very neat

[oblique river]

oh cool! that's a great book

well, i've only read part of it

but i liked it and the content is good and ive heard good things about it

reading groups are great

[golden pasture]

:o nice

[bleak abyss]

Esp abelian ones

[stone fulcrum]

If your reading group commutes, you know you have a good one

[peak root]

ooh that's so nice, I kinda wish I was in one

might be doing a reading course through uni next year though

[chilly ocean]

whats a reading group?

[paper flint]

A group of people who informally select some book/literature to work through together, and occasionally gather in some way to discuss/clarify.

[tough raven]

Oh

[tough raven]

I legit thought this was meaningful

🤡

[vestal snow]

Can someone help me out with this problem?

[cloud walrus]

bim

[tulip crane]

hi

how can i prove this?

by the definition of exact if i supose 1) then exists a function $\psi: M \rightarrow M'$ such tha $\psi \circ f=1 $,because $f$ is inyective, but how can i know thar $\psi$ is homomorfism?

[cloud walrus]

r2h

[carmine fossil]

I think you can show that by constructing phi

phi(a)=f^-1(a) if a in range
and phi(a)=e if a is not in range

[tulip crane]

thanks

[sturdy mirage]

Is this correct? ie. the degree of the extension field over Q is 8?
$$\left[\mathbb{Q}(i, \sqrt[4]{3}) : \mathbb{Q}\right] = 8$$

[cloud walrus]

reking

[hidden haven]

yes

[sturdy mirage]

thanks. just seeing if i understodo this dumb example i made up myself

[molten silo]

Can anyone explain why Z[√−3] is not a euclidean domain

[hot lake]

it's not a euclidean domain because it's not a unique factorisation domain

[south storm]

Hey I’m reading about groups of orders p^n, for odd primes p and I keep seeing the trend that for sufficiently large p, the amount of groups of orders p^n is a polynomial of p,gcd(p-1,q_1),gcd(p-1,q_2)... for some natural numbers q_i. Is this a theorem? If yes how is it proved? If someone would have some papers they could link that would be helpful.

[vestal snow]

Can someone explain how 4.4.6 can be viewed as described in 4.4.7?

The way I'm understanding this is that P^1 is {(x)} U {(t)} U V where V is the set of equivalance classes of prime ideals under the equivalence relation (x-a) ~ (t-1/a)

[hot lake]

if you have been introduced to the projective line before, they usually do it with projective coordinates [x,y]

your point (x) would correspond to [0,1]

and (x-a) to [a,1]

which is equivalent to [1,1/a]

meanwhile, (t) corresponds to [1,0]

and (t-b) to [1,b]

and so (x-a) and (t-1/a) correspond to the same projective coordinates, as is required

and any point [a,b] with x and y not both zero, comes from (x-a/b) or (y-b/a), possibly both

[vestal snow]

That's what I figured. Thanks

[maiden ocean]

etale time

[tulip crane]

what is an example of exact secuence that doesnt split?

[maiden ocean]

0 -> Z -> Z -> Z/2 -> 0 where the first map sends x to 2x and the second is the quotient

by the splitting lemma this is equivalent to the first map not having an inverse on the entire codomain

[chilly ocean]

[gritty sparrow]

I think both are even permutations, does that give you a hint?

[gritty sparrow]

See, here is the thing if K contains Fp^2, then it already is the splitting field, but it may just so happen that Fp is a splitting field itself. Eg:p=17. If p=1 mod 8, then( Fp)* has a cyclic subgrp of order 8, this subgroup consists of exactly the roots of the equation t^8-1, hence t^4+1 will split in Fp

[uncut girder]

MOTH

[maiden ocean]

Hi

[uncut girder]

Are you reading Szamuely's book?

[maiden ocean]

ya

[uncut girder]

Yayayayyayayayyay

[maiden ocean]

its cool

I am doing the exercises for ch 1 today

[uncut girder]

I've been reading it for a while

I just gave lectures on the end of chapter 2

[maiden ocean]

epic

[uncut girder]

https://youtu.be/usnQ2BuBM6o

[maiden ocean]

the finite etale stuff is neat because ive seen a similar formulation of normal covering theory so its helping me understand that as well

BASED

I will watch this once I get to that part of the book

[uncut girder]

Every section of Szamuely's book be like: this is another equivalence of categories

[maiden ocean]

lul

dieck puts it to shame though

in terms of density of categorical memes

[steady axle]

When we say that a general nth degree polynomial for n>4 is not solvable by radicals do we assume that the polynomial's splitting field has galois group S_n?

[prisma ibex]

not necessarily

any non-solvable Galois group will suffice

[steady axle]

yea what I meant to ask was how do we say this:

[hidden haven]

is "general" nth degree polynomial defined in the book?

If not then this is just way of saying that the claim "all degree n polynomials are solvable" is false whenever n>=5

So you can pick the counterexample

[steady axle]

perhaps it is assumed to be irreducible but still that does not guarantee galois group S_n

[hidden haven]

hmm weird, then the statement seems to say that any polynomial of that form is not solvable

[steady axle]

what I don't understand is how this is true

[hidden haven]

but x^n-2 for example is always solvable

pretty sure its just badly phrased and wants to say that in general, polynomials of degree >4 wont be solvable, but stated that no polynomial of degree >4 will be solvable since all polynomials have that form

[steady axle]

perhaps by 'general' it is meant that galois group of splitting field S_5

[hidden haven]

yeah possible

[hot lake]

the way they define things it's alright

and the general nth degree polynomial has galois group Sn

[hidden haven]

wait arent all degree n polynomials general by that definition

[hot lake]

if they take the coefficient field to be k(s1...sn) instead of k(t1...tn)

[steady axle]

what are the t_i s

[hot lake]

if they take the coefficient field to be k(t1...tn) then it's a very moment

I think they are supposed to be indeterminates

[hidden haven]

ohhh

I was thinking t_i s are elements of some algebraic extension

[steady axle]

yeah me too

[hidden haven]

wait

[steady axle]

I think I get it now

[hidden haven]

book says general nth degree polynomial over k

wot

[steady axle]

its the field of rational functions

[hidden haven]

[steady axle]

that makes sense now

[hidden haven]

so polynomial is not actually over k then

[hot lake]

no it's not actually over k

[steady axle]

its over k (symmetric polys)

[hidden haven]

oh multivariable over k

and when you take galois group you only treat x as the variable

seems like a very weird definition

[tough raven]

Does the quotient Z^2 of F_2 split?

Here F_2 = free group with two generators
Z^2 = well, you know, abelian version of F_2

[woven delta]

@tough raven https://en.wikipedia.org/wiki/Nielsen–Schreier_theorem

if it were to split then there would be a rank 2 finitely generated abelian subgroup of F_2

[tough raven]

Nice

Thanks!

[neat valley]

I'm trying to calculate the quotient group $H/K$ where $H$ and $G$ are (free abelian) subgroups of the free abelian group generated by $a_1,a_2,a_3,a_4,a_5,a_6$ given like so:
\[H=\langle a_1-a_2+a_4,\,a_1-a_3+a_5,\,a_2-a_3+a_6\rangle\]
and
\[K=\langle a_4-a_5+a_6\rangle.\]
For context, I'm trying to learn simplicial homology and my algebra knowledge, specifically regarding taking subgroups, is pretty shaky.

[cloud walrus]

Isaiah

[neat valley]

I understand $H/K$ is isomorphic to $\bZ^2$, but I want a better argument than ``the top has three generators and the bottom has one generator'', since obviously that doesn't always hold. Can someone give me a general outline of how I would go about calculating this? Thanks!

[cloud walrus]

Isaiah

[hidden haven]

H is isomorphic to Z³ by sending the 3 generators to (1,0,0), (0,1,0), (0,0,1) respectively, and the generator of K maps to (1,-1,1) under this isomorphism. Try using your visual intuition about Z³ to guide yourself from here (alternatively ||use the first isomorphism theorem directly. This step can also become simpler if you take the isomorphism to Z³.||)

||for the first isomorphism theorem you can take the map from Z³ to Z² which removes the last coordinate the first 2 generators and maps (0,0,1) to (-1,1). The kernel of this is exactly the image of K under the isomorphism to Z³||

[neat valley]

Ahhh, I see, thank you so much!

[hidden haven]

[cloud walrus]

squirtlespoof

[hidden haven]

Try to figure out the order of the frobenius automorphism

Show that it is ≥ the order of the group

You apply that automorphism to K

[gritty sparrow]

The frobenius map here is not a->a^p

[hidden haven]

What you showed is that the given automorphism fixes F

But you need its order as it acts on K

That's not the order

[gritty sparrow]

The map is a->a^(p^n), look carefully to see at what point you get back to a guaranteed

[hidden haven]

Yes

So you get x^(p^n)^(m/n) = x^p^m = x for all x

But from here, can you justify that order can't be smaller?

Sorry yes

Yes we get x^p^k for some k < m

But then what's the problem with that?

It is a power of an automorphism

[gritty sparrow]

What isn’t?

[hidden haven]

So will be an automorphism

You have to show that x ↦ x^p^k can't be the identity automorphism if k<m

[gritty sparrow]

Imagine x^p^k=x for all x where k<m. I claim this is impossible, why do you think so?

So?

[hidden haven]

So?

[gritty sparrow]

🍯

[hidden haven]

Saketh fuck off this is my help

[gritty sparrow]

🍯 🦞

[hidden haven]

(F)^x?

What does that mean

oh

Yeah so you get that K has more elements than p^k

what's the contradiction?

hint ||how many roots can a polynomial have?||

Yes, because they'd all be roots of x^p^k - x

So you get the order of the automorphism is exactly m/n, and you know the order of G(K/F)?

Yep

GMT + 5:30

What about you?

Oh west coast?

oof

[gritty sparrow]

Lmao you have gmt zones memorised?

[hidden haven]

No I know you have GMT-7 too

[gritty sparrow]

You memorized it just for me? I’m so touched

[hidden haven]

But it sounds cooler to ask west coast than to say "oh same as saketh then"

[gritty sparrow]

Same as saketh is a maxima for coolness

[hidden haven]

🤡

a maxima

[gritty sparrow]

Sorry, maximum

[hidden haven]

You can use the first part

Every theta generates an intermediate field

Try to prove that infinitely many distinct ones exist

alternatively you can use the theorem that any field extension is primitive iff it has finitely many intermediate extensions, if you've seen this

L is infinite

F is infinite

And L is an extension

[gritty sparrow]

Also you should show [L:F]>p if you haven’t

Ye, i mean that it was smth you needed to show for the proof to work properly, it isn’t too hard

[hidden haven]

I see

[gritty sparrow]

Well just try by the first method then

[hidden haven]

Hint for choosing the infinite family of thetas so that each gives a unique intermediate extension is ||choose them such that if any 2 distinct thetas are there in a single extension, so are t_1 and t_2, which would mean the extension is all of L, which can't be generated by any one theta||

F_0 may be finite

Yeah you are adjoining t_1^p and t_2^p

Infinite fields can also have char ≠ 0

eg algebraic closure of F_p

Yep

Hold on

Not L\F

F

Well because F is infinite

Like t_1^(ap) is an element for each integer a

Or a more useful fact

F(x^p, y^p) is isomorphic to F(x, y) if x and y are transcendental

Via the map x ↦x^p and y ↦ y^p

theta is in L\F

Oh shit we were using theta for different things

I was using theta for t1+ct2 where c is anything in F

[hidden haven]

Because you want this property

Why are you doing that?

[hidden haven]

Well if you naively use this reasoning you gotta justify why all those elements are distinct

So it's better to use the fact that F(x,y) ≅ F(x^p, y^p)

And cite the latter's infinitude

Or idk actually that works too

Nope

In part 1 you already showed that every element has degree p over F

But now you have to show that all t1 + ct2 generate distinct extensions

And there are infinitely such elements so you'll get an infinite family of extensions

y is in it

so cy is

so x+cy - cy is

[tough raven]

Why are you not sleeping

[hidden haven]

All nighter for gal thy assignment

[tough raven]

Oh assignment

Then do what you have to do

Get sleep before test
Or risk falling asleep 20 minutes before

[hidden haven]

They make you redo problems?

I see

[chilly ocean]

which part are you struggling with

Okay you should be careful here

An isomorphism

Is a ring homomorphism that has an inverse, such that the inverse is also a ring homomorphism

Don’t get tricked into thinking that because I can make an inverse “set theoretic” function

That I will have a ring isomorphism

Wait what conditions do you have on a b c d?

And what rings are we going to and from?

What field is it

Okay I think you should go figure out what t rings are

Oops

The rings*

Well I think you are missing some stuff

Because if I take Q

[rustic crown]

okay what does this map do to the elements in F?

you only told me were t goes.

[chilly ocean]

I can pick abcd that this seems to not be a homomorphism

[glossy wing]

how do I find the order of this matrix?

[rustic crown]

yea but what is the map?

you're not sending an element x in F(t) to ax+b/cx+d

only t goes to at+b/ct+d

wait i think this problem has been done before in this channel

found it!

[chilly ocean]

can someone help me see why an inclusion of dynkin diagrams induces an inculsion on root systems

The nodes should be the simple roots?

[rustic crown]

so how do we get that map?

you start with the inclusion F --> F(t)
now using universal property of polynomial rings, extend this to
F[t] --> F(t) sending t to some u.
but since all non-zero things go to invertible elements, by property of fraction field, you get the map F(t) --> F(t) sending t to some u.

this map only sends t to that...
the map phi(x) = ax+b/cx +d is completely different

i'm just saying at+b/ct+d is just any other invertible element in F(t)... you'll get a ring homomorphism even if you send t to say t^2

and that's not the reason why ab = bc doesn't work.

if you send t to 1, then the you get a map F[t] --> F(t) which has t-1 in the kernel!!! so its a non-zero element but it doesn't get sent to an invertible element, and so i can't extend this to a map of F(t) --> F(t)

[rustic crown]

okay wait maybe i'm wrong... but why t |--> 1 cannot be true?

okay so first do you get what exactly this map is?

F(t) --> F(t) which sends elements of F to itself and send t to at+B/ct+d and everything else is determined by the fact it should be a ring hom.

ah sorry for that mistake

if you're used to using universal properties, then there is no work to do... if not... then you need to verify everything by hand

also i think that should evaluate to a/c and not 1

in the second line of display math it should be (act + ad)/(c^2t + bc^2/a) if you mutiplied both numerator and denominator by c

because any map between two fields is automatically injective

you just need to see that if ad = bc then you're mapping t to some element e in the field

but then both t and e would be mapped to e

because the map is an F-hom

yea also need to see if c is 0 and all that.

nope

elements of a field either 0 or units

so any to non-zero elements are automatically coprime

yea?

okie so what is the leading coefficient of phi_n and phi'_n?

what about f(t) and F(t)?

😄

[glossy wing]

how can I show that for each positive integer $n$ we have $n=\sum_{d\mid n}\phi(d)$?

[cloud walrus]

notsushY

[rustic crown]

how many elements are there in G?

[glossy wing]

it doesn't say

[rustic crown]

i mean say G is a cyclic group of order n... then it has n elements

can you try to count this in another way?

[carmine fossil]

Basically you divide the set {1,2...n} into collections in such a way that gcd(k,n) is constant across a collection as k varies over the elements

That is one approach

For example if n=6
You have your collections as
{1,5},{2,4},{3} and {6}

No of elements in these collections are phi(6),phi(6/2),phi(6/3) and phi(6/6)

[rustic crown]

if you wanna make this pleasing to the eye, write out the fraction 1/n, 2/n,..., n/n and reduce them until the numerator and denominator have no common factors. now there are phi(d) fractions with denominator d namely c/d where c and d are coprime!

[glossy wing]

hmm

[rustic crown]

but since you posted that other result, this is probably what they want you to go for?

[sturdy mirage]

The number of elements in G is equal to (the number of elements of order 1) + (the number of elements of order 2) + .... + (the number of elements of order n)

[hidden haven]

Not necessarily, x⁴-2 over Q has splitting field which isn't generated by a single root

Let L be the splitting field of f, then you have L/K(u)/K

Not always

The second one is always 4

Because u is the root of an irreducible quartic over K

So if L/K is the splitting field of f

Then what can you say about G(L/K)?

Yeah but do you know anything about the group itself (not the size), based on the fact that f is quartic?

Yes, it's isomorphic to a subgroup of S4, based on its action on the roots of f

Now use fundamental theorem of gal thy

You have L/K(u)/K

K(u) corresponds to a subgroup of G(L/K), specifically to G(L/K(u))

When does K(u)/K have intermediate extensions?

Ok before that try to figure out what G(L/K(u)) could be

No, an automorphism acts on the field

But this action restricts to the set of roots of f

Because roots of f map to roots of f

And this action on the roots of f is faithful

So on those 4 roots, you can see how an automorphism sigma acts, and identify it with the corresponding element of S_4

Not necessarily

Faithful doesn't imply transitive

It just means that no 2 automorphisms act the same way on the roots of f

ie action on the roots completely determines the automorphism

Because the roots generate the field extension

In general this is an important fact to keep in mind, that the Galois group of a polynomial of degree n is always a subgroup of S_n

So now, you have to use the fundamental theorem

oh lol

You want to look at intermediate fields, so it's gonna be useful to have, instead of a list of subgroups of G(L/K), the exact poset representation of it

(but remove the duplicates, just write 1 Z/2Z instead of listing all because that's too much)

So among the 5 subgroups you listed try to order them by containment (I'm trusting you on there being 5 idk)

So D_4 contains Z/4Z

S4 contains A4 and D4 independently since their orders don't divide each other

Yeah S4 contains all

And I guess D4 contains Z/2Z x Z/2Z too

It should

A4 contains Z/2Z x Z/2Z

I think you might have multiple copies of that, some in A4 and some in D4

Just check this

You're missing subgroups I think

In particular S3

Yeah that too

C3 as well

A3 is just C3

Best way is too look up S4 subgroup lattice on google

Which one are you looking at?

Yeah it's mostly because duplicates

Ok so you know that order of G(L/K) is a multiple of 4

That means it can be one of the 6 things in the top left section

In the one I've sent

Do you see that?

(btw if your book has a section on computing gal groups of quartics there is likely a section on subgroups of S_4 with order a multiple of 4)

Yeah

Oh nice that's the list you sent before

Lol

Im guessing it's somewhere in your book?

So what can you say about G(K(u)/K)

Sorry

G(L/K(u))

[hidden haven]

This one need not be galois

[hidden haven]

Use the fundamental theorem of galois theory

hmm not necessarily

It corresponds to a subgroup of G(L/K) right?

And you know the degree of K(u)/K

So can you say anything about the order of G(L/K(u))?

Yes exactly

Now what can you say about the intermediate fields of K(u)/K

Lets say M

Yeah if it exists

We want to show that it doesn't

[L:M]

Wait sorry ignore that

Yes, now what do those degrees say about what subgroups of G(L/K)

Can you make that more precise?

Yes but I'm not sure what you meant by this statement

Like can you say exactly what the chain is, and what the orders are

K(u) is a field

Or are you writing the corresponding things?

right, so always, L corresponds to e

The correspondence is inclusion reversing

And the correspondence is N ↦ G(L/N) ie the top extension

So your chain of fields
L
K(u)
M
K

The last one is wrong

K(u)/K is 4

Nope still wrong

Look at the correspondence again

Lol

[K(u):K] = index of G(L/K(u))

In G(L/K)

So you don't actually get the order directly

You get the index

So the value there will be |G(L/K)|/4

Similarly G(L/M) will be an intermediate subgroup

Between G(L/K) and G(L/K(u))

Where each thing is index 2 in the previous one

So far we haven't really used the poset thing lol, the only thing we've done is say that you have L/K(u)/K, and if there's an M so that L/K(u)/M/K, then the corresponding groups have order 1\x\2x\4x (using \ to emphasize order reversing nature)

Where x is an unknown

Is this ok?

Yes

And also it's contains G(L/K(u)) (so it's not just index 2, it is index 2 + contains that subgroup)

Like we started with a chain of extensions and we get a chain of subgroups, just with reversed order

Yes

So just see what choices of 4x force you to have that kind of a subgroup chain

What you're trying to show is that if G(L/K) is anything other than 12 or 24, then you are forced to have an intermediate group between G(L/K) and G(L/K(u))

M is not something that exists yet, we figured out that its existence is equivalent to the existence of an intermediate group between those 2

I need to sleep so I'll just type this.
||You have order of the whole gal group = 4x. The possible values are 4, 8, 12, 24 (must divide 24). You need to eliminate 4 and 8.
If 4x were 4, then G(L/K(u)) = e. And any group of order 4 has an order 2 subgroup (which will definitely contain e), so the existence of an intermediate group is guaranteed in this case.
If 4x = 8, then G(L/K) = D4. Then G(L/K(u) which has index 4 must be C2. However you include C2 in D4, you will get an intermediate index 4 group.||

[hidden haven]

The chart wasn't really needed. The whole solution is the above message and this one

[limpid edge]

😩 I still don't understand quotient groups

it's been 2 weeks

[hidden haven]

Np

[limpid edge]

let's take Z/nZ as an example

or actually Z/3Z for simplification

the three cosets for the subgroup 3Z

are

[chilly ocean]

cosets are the numbers that are the same right

mod 3

[limpid edge]

1+3Z

2+3Z

3Z

yeah

[chilly ocean]

I think its good to think about those as classes of abstraction

[limpid edge]

that's why people just call them 0, 1, and 2?

I thought you had to be more formal in mathematics

[chilly ocean]

you have to be formal and you also have to be efficient with notation such that its also not confusing

[limpid edge]

no because I've seen people say

Z/nZ = {0, 1, 2, \dots, n - 1}

[chilly ocean]

yes

[limpid edge]

when in fact it's supposed to be a group of n cosets

[chilly ocean]

correct, but Zn is a special case

where it is clear what the cosets are and there is no confusion

[limpid edge]

oh here
"A quotient group or factor group is a mathematical group obtained by aggregating similar elements of a larger group using an equivalence relation that preserves some of the group structure (the rest of the structure is "factored" out)."

where is the equivalence relation specified

[chilly ocean]

a +nZ = b+nZ iff a-b in nZ

[limpid edge]

yes, I agree that's the equiv. relation for Z/nZ = {0, 1, 2, \dots, n - 1}

but where was it specified in the notation

it wasn't?

[chilly ocean]

As I said, in the case of Zn its clear from the context

[limpid edge]

ah I see

[chilly ocean]

the a+ I = b+I iff a-b in I is the equivalence relation for all quotient groups

for I normal subgroup

[limpid edge]

oh that's more interesting

[chilly ocean]

from my experience somehow the quotient structure has been more clear after I learned rings for some reason

[limpid edge]

noted

can I do this question now

lol

[chilly ocean]

whats G? Galois group?

[limpid edge]

Idk I just copied it from above

[hot lake]

it looks like a Galois group

[limpid edge]

yea there's more context to the question oops

so I probably can not do it

[chilly ocean]

Yeah, but kinda confused you are asking about it lol

No, you can't

because it is not a quotient

[limpid edge]

what does L/K mean here?

[chilly ocean]

L/K is what we call field extension

[limpid edge]

ohh

[chilly ocean]

Where L and K are fields

[limpid edge]

like Q[sqrt2]

[chilly ocean]

sure

Qsqrt2 is an extension of Q

does anyone know why this is the case?

this is page 331 from fulton and harris' representation theory

[glossy wing]

how can i show that $\mathbb{F}_2[x]/\left<x^4+x^3+1\right>\cong \mathbb{F}_2[x]/\left<x^4+x+1\right>$

[cloud walrus]

notsushY

[glossy wing]

with both of them being irredicble in F2

[lavish gale]

i dont understand the last statement? what is it asking me to prove?

'half of the members are even...' is it asking me to show half of the elements of S_n are even?

[chilly ocean]

no, H

[lavish gale]

oo so i hv to show either every member is even or half of its elements are even right?

in H

[chilly ocean]

yes

[lavish gale]

ty

[gusty halo]

thats the definition of a basis for the root system

each root can be expressed as a linear combination of simple roots that either uses only nonnegative or only nonpositive coefficients

(that such a basis actually exists has to be shown though)

ah and maybe you can add that since the simple roots are a basis of the vector space, the coefficients are unique

[chilly ocean]

okay I am okay with this

but this seems to be saying that beta-p alpha is not a root

maybe is it that it means to say its not a root in our basis?

[gusty halo]

the root system stays the same independent of which basis of it you choose, kind of like a usual vectorspace

[chilly ocean]

okay maybe this will help explain my confusion for A2 with alpha and beta, beta-alpha is a root

[gusty halo]

I think the sentence in the image you posted is worded a bit confusingly, basically it just says that we know p and q by induction

and the stuff in the parentheses explains why we know p by induction

[chilly ocean]

and the induction is

we are claiming that we know all roots of height m

or level to use this terminology

[hot lake]

p is the largest integer <= mj such that beta - p alpha is a root

and you know whether beta - p alpha is or isn't a root because its level is m-p, which is <= m

[chilly ocean]

just to double check

if beta=sum m_i alpha_i

then the level of beta is sum m_i

[hot lake]

yes

[chilly ocean]

okay so i know then that beta + alpha_j where alpha_j is a simple root is a root of level m+1

[chilly ocean]

i'm still not sure about this

ooooooh

[hot lake]

it has level m+1 and we don't know yet if it's a root or not

[chilly ocean]

are you saying

that we ignore the beta - k alpha_j with k>m_j

and start with beta - p alpha_j

because we already know those roots

as in they are inour list

[hot lake]

the beta - k alpha_j with k > m_j would have a negative coefficient

so either it would be -alphaj, -2alphaj etc ; or it's not a root

and if we are in the first case

[chilly ocean]

okay this confuses me

[hot lake]

then beta would be alphaj

[chilly ocean]

what is wrong with negative coefficent in A2 alpha-beta is a root?

[hot lake]

and so one of the axioms says that beta + alphaj = 2alphaj is not a root

[gusty halo]

this cant be because it has negative and positive coefficients

and alpha and beta are simple

[hot lake]

if beta is a root that is not alphaj then it is a linear combination of SEVERAL positive simple roots

then removing too many alphaj would give you a linear combination with coefficients of mixed signs

and that's not a root

[chilly ocean]

oh my bad

i see what i had wrong now

i thought this top right root was alpha- beta

so we literally just cannot have a root

that is a mixed combination of simple roots

never

[hot lake]

yeah

[gusty halo]

yes by definition

[chilly ocean]

mucho thank you

[chilly ocean]

did you ever figure this out @glossy wing

[sturdy mirage]

each of those fields will have a generator that generates all the units. a homomorphism that maps 0 to 0, and a generator in one field to a generator of the other field will be an isomorphism. does that sound right?

[chilly ocean]

beta+alpha will be a root when q is greater than or equal to 1

so this gives us p greater than the cartan integer for beta alpha

its still not clear to me how we are reconstructing the root system

This is the situation as I understand it

We can read off directly the simple roots and the roots of level 1

Next we will use induction

We assume that we have all the roots of level m

we want to then find all the roots of level m+1

(there can never be a gap, as in we can't have roots of level 2 and roots of level 4 without a root of level 4 -- so this induction would give us all the roots)

So then what we are doing is, for every root beta of level m, we check all the simple roots alpha_j to see if we get a beta+alpha_j as a root

fix alpha=alpha_j

we know that there are not any gaps in an alpha-string through a root, so we consider the alpha string through beta and we want to check if beta + alpha is in this string

that means there must be some string beta -p alpha,.., beta +q alpha

and we want q greater than equal to 1 ( and p must be positive?)

be definition of a root system and fleshed out above we cannot have beta -p alpha being a root with p > m_j

~~ this is where i get confused

we can read the cartan integers off the dynkin diagram

what we are trying to do is show that given just the information in the dynkin diagram, we can find all the roots?

so we need to just check the dynkin diagram and see that the cartan integer for beta-alpha is less than p?

but i am confused on what p is, we know that it must be less than m_j but do we have other conditions on it?

fuck

me

thank you very much @gusty halo

[hot lake]

p is the largest integer such that beta - p alpha is a root, and when they say that "we know p by induction" they mean that the information of who is a root for levels <= m is enough to determine it completely : when beta is not alpha, p is the largest integer <= mj such that beta - p alpha is one of the roots you have found so far, and so is completely determined by the data that you have so far in the procedure

the part I find possibly unclear is explaining why if beta is a positive root then there is a path from 0 to it that can be made entirely through positive roots by adding simple positive roots one at a time

say you have all the lists of positive roots of level up to 4, you have beta = a1 + 2a2 + a3 and you want to know if beta + a2 is a root. To get p, you look at a1+a2+a3. Is it in the list of roots of level 3 ? if no then p=0, if yes then you continue and look at a1+a3. Is it in the list of roots of level 2 ? if no then p=1, if yes then p=2 because it is impossible that a1-a2+a3 is a root.

so p is completely determined by the lists you have made so far

[chilly ocean]

thanks to both of you

[chilly ocean]

Sup. What’s a Lie Group and which are the principal lie groups?

a lie group is a group with a smooth manifold structure making the group operation smooth

for example, the group of all invertible real matrices of your favorite dimension

or the circle

ive never heard of a principal lie group, and google only returns things on principal bundles. is that what you had in mind?

[delicate orchid]

could be lie groups that are principle ideal domains

[magic owl]

maybe they mean it colloquially like what are the most importsnt examples

[chilly ocean]

hmm

i would say the matrix groups, then

probably the easiest to get into

tori as well

[magic owl]

matrix groups and tori for sure

tbh i dont even know any others off the top of my head

[chilly ocean]

the isometry group of any riemannian manifold

[viscid pewter]

1

[chilly ocean]

@chilly ocean Thanks

[chilly ocean]

I would like to know about the groups where we use more, like the Special Unitary group for NxN matrices

Through some searches by me I found some groups where we use a lot in Quantum Field theory, where is my focus

A Lie group is a group object in the category of smooth manifolds

We’ve the SL(n,R), GL(n,R) and the SU(n), but I would like to know if we need to know more

[chilly ocean]

I see

There is a nice theorem that says something that any commutative Lie group is just R^n x T^m

Hmmmmm

But as was said above just look up matrix groups

T^m either is a tensor or something like this?

[magic owl]

there are a lot of good references

for the matrix groups

like a lot

[chilly ocean]

No no SxSx...xS

[magic owl]

S^1*

[chilly ocean]

A product of circles

I see

[magic owl]

virgin torus vs chad power of the sphere spectrum

(jk the latter thing is just the sphere spectrum again)

[chilly ocean]

What are you actually asking btw?

And why are you asking it?

Do you want to just have a list of examples of Lie groups?

If you want to learn about Lie groups the best thing to do is just pick up a book on the subject

[chilly ocean]

Unless you already know about Lie groups this definition is pretty crap

[magic owl]

(to be fair there is a lot to be said about these specific groups and one can basically make a career out of studying them)

[chilly ocean]

Yes

I’d just hate to see

Some undergrad who hasn’t done topology