groups-rings-fields

[glacial oak]

[rustic crown]

me like the blue color tho

[frigid lark]

Det is evolving?

[rustic crown]

nuu >.<

det has everstone :3

[frigid lark]

but glaceon or vaporeon is blue

[tribal furnace]

me too

[rustic crown]

me no blue anymore

[summer path]

det still has blue dot

[rustic crown]

[solar shore]

could someone assist me in proving this?

[formal ermine]

which direction are you having trouble with

[solar shore]

mm the book's haphazardly stated that it uses induction

i think i can get the direction of where if the product is cyclic, then they're all relatively prime

but im not sure where induction really comes into play here

[tender wharf]

theyre probably inducting on n

[solar shore]

mm i see

i wasn't sure if i was inducting on i and j or just n

ah wait

i get it now

i think'

[formal ermine]

isn't this just like

permutation is a bijection

[sinful mirage]

does anyone know a text, where the relation between Lie algebra cohomology and Lie group cohomology IN LOW dimensions is discussed?
it is important for me in low dimension,because there the corresponding cohomology groups can be easily defined via cocycles and hands-on maps,etc, not ext/tor functors and bar resolutions, which are too abstract for me right now

[south patrol]

Homeomorphism lol

[formal ermine]

yeah but like

Gal(L/K) is a topo group

[south patrol]

Sure

[formal ermine]

so multiplication is continous

permutation is a bijection

[south patrol]

Oh okay I misread lol like

[formal ermine]

and its inverse is also just a permutation

[south patrol]

Yes that is true

[formal ermine]

so its also continuous

[south patrol]

lol

[formal ermine]

just wanting to make sure I understand this notation

when we say R_p or R_l or whatever we mean the p or l adics over that ring?

[agile burrow]

Can you give more context? I haven't heard of p-adics over an arbitrary ring

It could be localization at a prime ideal

[empty rose]

try integrating the channels instead

[lethal dune]

Why do i have such a hideous color

[kind jacinth]

im not sure what i should do in part b. should i understand it as "group action 3Z -> Z with action given by addition thus (n, 3m) = n+3m so then it only has 1 orbit? or is it "group action Z->3Z (which i think makes more sense since 3z is a subgroup of Z?"? but then i dont know how to find the orbits

[south patrol]

It says 3Z acts on Z

But also like the notation "group action 3Z -> Z" is uh smth I would avoid

[lethal dune]

Stay muted

[kind jacinth]

in that case id have (n, 3m) = n + 3m as the action therefore it would only have 1 orbit right? I could say orbit us 0

wait no orbit 0 would give {0, 3, 6, ...} so i think the orbits would be 0, 1, 2?

[south patrol]

What is an orbit?

You seem to have confused the concept w smth else

[chilly ocean]

orbit is like coset

[kind jacinth]

I have an action G x X -> X and my orbits are the elements in X that would give unique sets of G using a group action? so for example 3Z -> Z6 with group action (3n, m) = (3n+m)mod 6. Z6 being {0, 1, 2, 3, 4, 5} orbits 0 and 3 gives same set, orbits 1, 4 give same thing, and orbits 5, 2 give same set

so then i can say the orbits are 0, 1, 5 in the example above

[delicate orchid]

Orbits are sets, you can say those are orbit representatives though

[wooden ember]

why everybody muted

[delicate orchid]

The orbits themselves would be {0, 3}, {1, 4}, {2, 5}

[kind jacinth]

ohhh i see. ok then for the exercise I sent the orbits would be {3k}, {3k-1}, {3k-2} ?

[delicate orchid]

Let me have a look

[formal ermine]

#changelog

[chilly ocean]

The orbits would be 3Z, 3Z+1 and 3Z+2.

[wooden ember]

oh lmao it's the "punishment" ok

[kind jacinth]

Yeah should have done addition there not subtraction. and the cosets are 0+H, 1+H, 2+H?

[delicate orchid]

yeah should have addition not subtraction

No your notation was just wrong

[chilly ocean]

Yes

[delicate orchid]

But those are the cosets

[chilly ocean]

how to do backflip

[delicate orchid]

Spin real fast like

[kind jacinth]

ok so then i have my cosets same as my orbits since H = 3Z. in part (c) they just want me to say my cosets and orbits are equal to n given that H = nZ? and im not sure about (d)

are cosets and orbits always the same? or its just in this case?

[delicate orchid]

No, they want you to do parts a and b for H = nZ instead of just H = 3Z

[chilly ocean]

Yes for (c) and for (d) they want you to show that if H is a subgroup of (G,+) then the action defined by the + operation holds the properties we saw previously (equality in cosets and orbits)

[delicate orchid]

No. Consider the action of D_8 on the corners of a square. How can D_8 form cosets of the corners of a square when the corners aren’t even a group

In a very specific group action they are the same though

Same with conjugacy classes - they’re the orbits of a group acting on itself by conjugation

[delicate orchid]

I'd like to correct this

if you have a group action GxX -> X and define an equivalence relation on X by x ~ y if and only if x and y are in the same orbit then the classes in X/~ are obviously the orbits

which is sort of analogous to cosets

so they're not completely distinct ideas

[kind jacinth]

i see, thanks. so I think i was able to do (c) but for (d) im not sure how to generalize it.

because what I get in (C) is nZ, nZ+1, nZ+2, ..., nZ+(n-1)

i dont see how it relates to (d) since G can be any group

[delicate orchid]

yes that's right

so for a general group G and subgroup H what do you think the appropriate choice for the group action HxG -> G would be?

have a guess if you don't know

[kind jacinth]

addition should be my group action

[delicate orchid]

yeah it was addition when you were in Z, but what does it mean to do h+g in an arbitrary group?

[kind jacinth]

binary operation based on the groups operation?

[delicate orchid]

yes!

so we just set (h, g) to be hg

[kind jacinth]

but wouldnt this imply that any binary operation would result in cosets = orbits as long as H is a subgroup of G?

[delicate orchid]

no

I could as easily define (h, g) = h^{-1}gh

it's just that specific action

although out of the 4 I'm thinking of rn 2 do just produce the left/right cosets

so if you round up then yes they all do

[kind jacinth]

but I still dont see the relation of this with (c). so in that case C would be g^n instead of gn? because the current thing in C is addition specific

[delicate orchid]

you probably can't see a relation of that to (c) because it's a completely different action that's irrelevant to the question

I only provided it as an example of an action HxG -> G that doesn't give cosets as orbits

[kind jacinth]

so is everything they want us to do is instead of giving the general addition, set (h, g) to hg (which implies all binary operations) and thats it?

[chilly ocean]

suppose we have a galois extension F|K and two intermediate fields L1, L2. Is it possible to express Gal(F | L1L2) in terms of Gal(F | L1) and Gal(F | L2) ?

[south patrol]

Intersection

This is clear from the Galois correspondence or because a map F -> F fixes L1L2 (pointwise) if and only if it fixes L1 and L2

[kind jacinth]

if 4 is fixed then is cardinality of G = cardinality of H?

or they mean element 4 is fixed on H which is why its S3?

[fleet pelican]

Yeah. It is the elements of S_4 which don't move 4 around. I.e. "fix 4"

[kind jacinth]

alright i see. So I know that the number of left cosets is going to be 4 (because 4!/3! = 4) but is there a quick way i can find these cosets?

[fleet pelican]

yes

[kind jacinth]

my subset S3 = {e, (1 2), (1 3), (2, 3), (1 2 3), (1 3 2)} dont wanna do all elements of G not in S3 to find the cosets

[fleet pelican]

can you find three elements not in H that all give different cosets?

think more abstractly, not concretely

remember how we got H

[kind jacinth]

i think (1 4), (2 4), (3 4) would give different things?

[fleet pelican]

That sounds correct. Can you justify that?

[kind jacinth]

im just thinking of the square with 4 vertices and performing the rotations above would give me something else. Not sure if this is the correct intuition.

[fleet pelican]

What is the stabilizer subgroup of each?

[kind jacinth]

performing a rotation 4 times

or 2 reflections

[fleet pelican]

What in S_4 is the stabilizer of each coset of H

[kind jacinth]

the identity

[kind jacinth]

dont think its right though because (3 4) (1 2 3 4) gives (1 2 4) but so does (1 4)(1 2)

[fleet pelican]

how did you get (1234)?

(1234) along with any transposition will generate all of S_4 if I'm not mistaken

[tribal moss]

(1234) and (13) only generate D_8.

[formal ermine]

yeah that only works if your permutation order (or whatever it's called) is prime

a p cycle and a transposition generate S_p

[fleet pelican]

oh true rip. at any rate, he shouldn't get (1234) in a coset of S_3

[tribal moss]

It has to be in some coset.

More concretely, it is definitely in the coset (1234) S_3.

You're probably thinking of (1234) not being in any conjugate of S_3.

[formal ermine]

someone should rename this channel to "how do i solve for x"

[tribal moss]

That's #prealg-and-algebra.

[pastel cliff]

"find x"

"it's right there"

[formal ermine]

this one girl in my pe class was wearing a shirt with that

lmao

[tribal moss]

Oh, I see what you meant to say: (34)(1234) is not in the coset (34)S_3, because (1234) is not in S_3.

[coral shale]

hmmm

because left multiplication is injective, say

[tribal moss]

A better intuition would be: All permutations σ in the coset (4 a) S_3 satisfy σ(4)=a, and onversely the coset contains every permutation with that property.

So you can identify which coset a permutation σ is in just by looking at σ(4).

[chilly ocean]

suppose we have a galois extension $F|K$ and two intermediate fields L1, L2. Is it possible to express Gal($F | L_1 \cap L_2$) in terms of Gal($F | L_1$) and Gal($F | L_2$) ?

[cloud walrus]

matcool473

[coral shale]

https://i.imgur.com/Ygun9e8.png

this right

idk myself, might need normality on some of those to make it work

very familiar diagram

[chilly ocean]

yes this is the diagram

we know that all extensions are galois

[coral shale]

Making a guess: I think the latter 2 are just subgroups (maybe normal), and only when the intersection = K do you have a product

[formal ermine]

isn't it like the subgroup generated by the product of gal f/l1 and gal f/l2 or something

yeah pretty sure it is

[chilly ocean]

i think its the smallest group that inclused Gal(F|L1) and Gal(F|L2)

is there a specific name for such conctruction between two groups?

[boreal inlet]

lmao the channel names threw me off so bad

[formal ermine]

the group generated by their product lol

the proof shouldn't be too involved (maybe 1-2 lines)

it's a fun little exercise

[chilly ocean]

ok great

product in the sense of multiplying their elements right

[formal ermine]

yes

[kind jacinth]

woops ur right. ok (1 4) (2 4) (3 4) are cosets and now we would need to find 1 more coset. actually we dont as thats just the identity.

[fleet pelican]

what do you think the stabilizers are

[kind jacinth]

stabilizer so if i do (1 4)*x id need to get (1 4) back. x should be the identity?

[fleet pelican]

use @tribal moss 's comment to help you think about this

[fleet pelican]

incorrect, you do not need (14) back, you just need that (14)x is in (14)H

[coral shale]

dunt ping to say its theirs

u can no ping reply to the actual comment

[kind jacinth]

then isnt the stabilizer for each coset anything in H?

[fleet pelican]

H only stabilizes H

[kind jacinth]

then for the coset (1 4)H i can have (2 3) as a stabilizer? or anything not involving 1 and 4? i am not sure honestly.

actually, wouldn't the identity in S4 work? (1)*(1 4)H would give back (1 4)H

i dont see why this is incorrect

[fleet pelican]

I think you need to review the definition of stabilizer?

[kind jacinth]

stab(H) = {g element of G: g*H = H} right?

[fleet pelican]

what is stab(xH)?

[kind jacinth]

element of G such that g*xH = xH

[fleet pelican]

so what is stabilizer of H?

in our example

and what is stabilizer (14)H, in our example?

remember, stab(H) is a subgroup

as is stab (xH)

also recall, |stab(H)| = |G|/|orbit(H)|

[kind jacinth]

so then the subgroup consists of 6 elements?

but im not sure how to come up with them

[kind jacinth]

stabilizer of H is H?

and of xH it would be xH

[kind jacinth]

if i do (1 2)*(1 2) (being elements of g and H) id get (1 2) and if i do (23)(12) id get (1 3) which is also in H. I think this would work for all cases

[charred crescent]

would anyone mind reading through this and telling me if i went astray?

[fleet pelican]

that is not correct. xH isn't a subgroup.

[celest furnace]

I think well defined is wrong here shook

Cuz it doesn’t depend on H cap K is trivial

Should be that h h’^{-1} is in both H and K so is the identity and similar for the k’s

Ah! It’s because you were trying to show that if the inputs are equal then the outputs are equal, but started with the outputs being equal

[charred crescent]

so i have to show that g = g' => p(g) = p(g') yes?

[fleet pelican]

yes

[celest furnace]

Also I think your injective proof uses circular reasoning too

Phi(g) = e implies g = e seems sketch

[charred crescent]

well it's using the fact that ker(phi) is trivial implies injective

at least that was my aim

[fleet pelican]

you have to use it explicitly.

[celest furnace]

I think it’s cleaner to just do it directly

[charred crescent]

what do you mean?

so let g,g' in G s.t phi(g) = phi(g') and show that this implies g = g'

[celest furnace]

Yes

[charred crescent]

okay, thank you for your help y'all

[celest furnace]

Actually your proof is right but needs one more step

You should decompose g = hk (which exists clearly) then you get

phi(g) = (h,k) = (e,e)

Then you can use that (a,b) = (c,d) iff a = c and b = d

Technically you did this but I think it would help to state it explicitly

[charred crescent]

this is in reference to the injective part?

[celest furnace]

Yes

[fleet pelican]

but you first need well-defined

[celest furnace]

Homomorphism looks great

[south patrol]

lol i think it's easier to just map H x K -> G

well much easier

[celest furnace]

Potato how are you talking while muted

Honorable shenanigans

[charred crescent]

okay so im still a little unsure as to what i still need to show regarding injectivity

[celest furnace]

Injective is phi(a) = phi(b) implies a = b

Equivalently for groups that given a in ker(phi) a = e

You need the implication going in the right way

[charred crescent]

so i should not use the phi(g) = e argument?

[celest furnace]

You can

But it’s easier to make mistakes

[charred crescent]

okay fair

as for the well-defined part:
let g,g' in G such that g = g'.
then g = hk and g'=h'k' for some h,h' in H, k,k' in K
so g = g' => hk = h'k' => (h,k) = (h',k')
thus phi(g) = phi(g'), hence phi is well defined

is this sufficient?

[sonic coral]

noooo my abelian grapes

[celest furnace]

No because in the second implication you are assuming it’s well defined

To get that implication you need to show that h=h’ and k=k’

Pretty much if you ever haven’t used a given something is probably wrong ( in this case, you didn’t use that H cap K is trivial)

[charred crescent]

okay so i must somehow use H cap K is trivial to show h = h' and k = k'

let me try that

[celest furnace]

Yes

[charred crescent]

ah okay how about this:
let g,g' in G such that g = g'
then g = hk and g' = h'k' for some h,h' in H, k,k' in K
then hk = h'k' => k(k')^{-1} = h^{-1}h'
because H cap K is trivial, k(k')^{-1} = h^{-1}h' => h^{-1}h' = e and k(k')^{-1} = e
thus h = h' and k = k', so (h,k) = (h',k'), thus phi(g) = phi(g')

[celest furnace]

yes

good start

[charred crescent]

i hit enter prematurely lol

[celest furnace]

yep that's what i had in mind

[charred crescent]

okay cool

thank you for that, it makes much more sense now

[kind jacinth]

Thats true. the stabilizer of xH should still be H?

it needs to be a subgroup of 6 elements. I can not think of somthing else that could work

[south patrol]

lol

[pastel cliff]

muted potat

[kind jacinth]

nvm the inverse can work. so (1 4)H(4 1)

[sinful mirage]

II am trying to prove this. Can someone help me see why the second square commutes?

this is my trial. I proved that first square commutes

[fleet pelican]

I believe there is a theorem which states that the stabilizers in an orbit are all cojugate

[prisma wren]

how does y = ax²+bx+c work?

[cloud walrus]

memeke

[formal ermine]

you mod k[x, y] by (x^3, y^2)

[sinful mirage]

G_3

[lethal dune]

ϕ_3 is an isomorphism

[sinful mirage]

but i don't see how this helps

[fleet pelican]

I don't know what you still need to prove?

[sinful mirage]

yes,but how does this help?

I need to prove that the square commutes

this means $\phi_3 \circ f_2= \pi \circ \phi_2$

[fleet pelican]

that projection->iso is same as iso-> projection?

[lethal dune]

check that $G_2\rightarrow[f_3] G_3 \rightarrow{\phi_3} G_2/f_1(G_1)$

[cloud walrus]

ProphetX

[lethal dune]

bruh

[pastel cliff]

khaki ryu

[sinful mirage]

maybe I am confusing what these maps mean,idk

they don't seem to be equal

[lethal dune]

anyway you can show ϕ3 ∘f3= π by definition

[sinful mirage]

what si f_3?

[tribal moss]

One way to think of it is: We have concluded that the coset (14)S3 consists of exactly those permutations that send 4 to 1. What are the elements g such that if h(4)=1 then gh(4) is also 1?

[fleet pelican]

what kind of map is f_2 (remember the sequence is exact)

[sinful mirage]

f_2 is surjective.

[sinful mirage]

this is the unviersal property of the quotients

yeah makes sense

i see now

thanks

[chilly ocean]

3x^2+6x-9
how do I factor this quickly?

[queen quarry]

Divide by 3
x²+2x-3 = (x+p)(x+q) = x² + (p+q)c + pq
=> pq = -3, p+q = 2
Therefore p=-1, q=3 is a solution so
3x²+6x-9 = 3(x-1)(x+3)

I have trouble understanding this proof. Q^# denotes the set of all nonidentity elements of Q.

Since we choose Q as the smallest counterexample, then for all 1=/=g in Q we have g is not in <A, A^g>.

What I don't understand is why does that mean that [[g,A],A] = 1

Can someone give me an idea? Since g is not in <A, A^g> then g is not contained in [g,A] for all g=/=1 but I don't see how to go on from here

[formal ermine]

#prealg-and-algebra

[fleet pelican]

not entirely sure how the notation here is being used

A^g, <A,A^g>, [g, A]

[warped fulcrum]

I am going through Aluffi and there's a problem I'd skipped which I'm trying to have a 2nd go at rn:

Proposition:

There exists a surjective homomorphism $\sigma: \bZ * \bZ \to C_2 * C_3$ where coproducts are considered in Grp.

I avoid denoting many (intermediate) hmorphisms by symbols because otherwise I'd quickly run out of symbols to use.
It is fairly trivial to construct a homomorphism, but I'm having some trouble trying to show that it's surjective. My approach has been to consider homomorphisms $\varphi: \bZ \to C_2 * C_3$ given in the obvious way by composing a (surjective) homomorphism $\bZ \to C_2$ with the (injective) inclusion map $i: C_2 \to C_2 * C_3$, and similarly for $C_3$. Then by applying the universal property on $\bZ * \bZ$, the existence of a homomorphism $\sigma: \bZ * \bZ \to C_2 * C_3$ (unique given the chosen maps) is immediate. I suspect this is the correct homomorphism, but I'm struggling to show that it is surjective. It doesn't seem like I can actually conclude anything about the (surjectivity or injectivity) of the maps $\varphi: \bZ \to C_2 * C_3$, even though they seem to be the most natural choice. Am I on the right track? Any hints?

[cloud walrus]

humefanboy

[lethal dune]

note that Z →C2 and Z → C3 are surjective, can you use that somehow

[warped fulcrum]

yeah but I'm composing them with injective inclusion maps

I'm not immediately using Z to C2 or Z to C3, I'm using Z to C2 * C3, it doesn't seem like I can say much about that

[lethal dune]

you can try to use explicit definition of G*H

[warped fulcrum]

I have not covered that yet

this is supposed to be before the explicit definition which I believe is covered in chapter 8

I'm still at chapter 3

[lethal dune]

I see

one more thing you can try showing is it's epi (in categorical sense)

Say you have f,g: C2*C3 → X then this will give you 2 maps from Z to X through C2 and C3 and from which you can claim f=g showing epi

in Grp epi=> surjective

try to see if you can fill the details

@warped fulcrum

[warped fulcrum]

epi is equivalent to surj hmorphism in Grp right?

[lethal dune]

well that I'm not too sure but they are in Ab

let me think

[formal ermine]

yes

[lethal dune]

the proof I was thinking doesn't work

[formal ermine]

I remember seeing the nlab article for this lol

[elder wave]

it's not trivial

[south patrol]

Yeah it is lol skill issue

jk

Uhh

[elder wave]

well ok the direction ryu assumed is the harder one

[warped fulcrum]

lol

[empty rose]

...$*$ is coproduct right? i feel like i remember that but idk if i'm just wrong

[cloud walrus]

bee

[empty rose]

(as in, in Grp)

[elder wave]

important tex

[lethal dune]

yeah didn't realise it stops at epi step

[warped fulcrum]

yes

[coral spindle]

Yeah, but they call it a "free product" usually

[elder wave]

🦅

[empty rose]

(yeah i knew that)

[coral spindle]

OK

[lethal dune]

amalgamated product by {e}

so what should be a categorical way to solve it

[south patrol]

Isn't this true whenever the forgetful functor has a left adjoint

Or am I smoking

[lethal dune]

use explicit construction and call it done

[south patrol]

Idk I did smth like this a while back lol

[lethal dune]

yes the Free functor

[queen quarry]

A^g := {g^-1 a g | a in A}
<A, A^g> is the group generated by A and A^g
[g,A] is the commutator, i.e.
[g,A] := {g^-1 a^-1 g a | a in A}

[south patrol]

wait so which direction didn you want to prove

which is the hard one

oh epi => surjective ?

[warped fulcrum]

No

I want to prove that $\sigma$ is surjective

[cloud walrus]

humefanboy

[warped fulcrum]

the map I constructed above

I don't know how to show it's epi either

I do know the proof epi <=> Surj in Set, I suspect epi => Surj in Grp is going to trivially follow from that

[lethal dune]

yeah the latter statement is not actually that easy to prove lol

[warped fulcrum]

lol

it seems that the phi's I constructed ought to be surjective because if they were the proof works out perfectly

does that pass for a rigorous proof

[empty rose]

hm

ok wait

the phis you constructed aren't surjective, but i think the overall sigma is

it's not actually that hard to just take any element of C2 * C3 and produce an element of Z * Z that gets mapped onto it

[warped fulcrum]

tbh I don't even have the slightest clue what elements of C2 * C3 are supposed to look like, the explicit construction is not given until like chapter 8 of Aluffi I believe

[empty rose]

...hm

ok well it's still definitely surjective, but idk how to prove it purely from the universal property of C2 * C3

[warped fulcrum]

ok I will trust you and wait until I learn the explicit definition lol

[empty rose]

i can explain the basic idea even without the details of the explicit definition

everything in C2 * C3 is some combination of things from C2, and things from C3

things from C2 are produced by the left Z in the Z * Z, things from C3 are produced by the right Z

so overall every combination is produced by Z * Z

but if you try to view this categorically, you don't have like, a homomorphism from C2 to Z
i'm not sure if there's a reasonable way to phrase "surjective" group homomorphisms entirely in terms of arrows to get away from talking about elements of C2 * C3

[warped fulcrum]

oh I see, pretty weird that Aluffi would give it as a problem this early in the book then

[empty rose]

i might just be missing something

[lethal dune]

Aluffi is wild unless there's some easy solution I'm not seeing

[queen quarry]

Does anybody have hints for this?

[queen quarry]

Also [[g,A],A] should be a subgroup of <A, A^g> so then for all nonidentity g: g is not in [[g,A],A] but that's not enough to prove that [[g,A],A] = 1

[feral agate]

I don't really understand why there is an inequality at the arrow instead of equality, isn't the orbit-stabilizer theorem applied here?

because we know that the stabilizer is a subgroup

[fleet pelican]

what is the meanings of the symbols you're using?

[queen quarry]

[x,y] := x^-1 y^-1 x y is the commutator of x and y

A^g := g^-1 A g is the conjugate

[fleet pelican]

i see

so to answer your original question, by your definitions [[g,A],A] is a subset of <A, A^g>, isn't it?

[queen quarry]

That's correct

[fleet pelican]

so ur good then?

[queen quarry]

Not quite, i need to show that [[g,A],A] = 1 for all g in Q but so far I only know that g is not in [[g,A],A] for all g=/=1

I don't see how it's useful that A acts non-trivially on Q

And also I don't understand where the minimality of Q is being used

[charred crescent]

is dummit and foote a good place to start for representation theory?

[fleet pelican]

artin has a great treatment of rep theory

[charred crescent]

okay thanks

[fleet pelican]

so, if something in [[g,A],A] were nonzero

it would be in <A^g,A>

so either you can find a bigger Q, by construction, or you're still in Q.

[chilly ocean]

Hi

How can i start in abstract-algebra

[formal ermine]

I recommend artin

[chilly ocean]

Books?

[formal ermine]

yes, artin's algebra book

[chilly ocean]

Just one book?

[formal ermine]

it's a good starting point

[chilly ocean]

Hmm

And after

?

[formal ermine]

read it first, then ask "and after"

[chilly ocean]

sure my man

thanks

i will search for an PT-BR version

[sonic coral]

thoughts on gallian?

[formal ermine]

no idea

[queen quarry]

What do you mean by that?

[feral agate]

Any ideas on this?

[empty rose]

well we're saying that the orbit has at least 4 elements

if the orbit had 8 elements then you would end up with \ $|G_1| = |G_{1,2}| \cdot 8 \geq |G_{1,2}| \cdot 4$

[cloud walrus]

bee

[feral agate]

ohhh I see thanks

[warped fulcrum]

Aluffi

honestly I think it’s manageable, I’m currently going through it and I don’t have that strong of a background in algebra

If you don’t even yet know what a group or a homomorphism is, then certainly not, but if you do know the basics it’s quite a nice book (at least from what I’ve seen so far)

I did learn the basics from Artin, it’s also a nice book but a bit too long, did not complete it lol

[sinful mirage]

should these two be equivalent?

[agile burrow]

yes, but it's a little strange that they're using additive notation for the multiplicative group of C

I guess really the cocyle arises as the choice of a lift of a homomorphism to PGL to a map to GL

[sinful mirage]

how could one see equivalence?

for one direction this is the hint

but I fail to see why theta is a homomorphism

and for the other direction I have no idea

[sinful mirage]

it's me being too illiterate mathematically in notations. in the original text ti was denoted by $\mathbb{C}^{\times}$

[cloud walrus]

ProphetX

[fleet pelican]

yeah, it is the center of the linear group so it commutes...

[sinful mirage]

I just wrote ti as $C^{*}$

[cloud walrus]

ProphetX

[agile burrow]

It should just be a direct computation to show theta is a homomorphism

[sinful mirage]

yes. I tried,but don't see

let me show what I did. one sec.

[agile burrow]

Do you understand what the map GL(V) -> PGL(V) does?

[fleet pelican]

you are modding out by the center of a group, the center is always normal.

[sinful mirage]

not precisely

[sinful mirage]

i guess that's my confusion

[agile burrow]

Right, that's probably the first step

[sinful mirage]

$\theta(g_1g_2)=(\pi_V \circ T)(g_1 g_2)=\pi_V(T(g_1g_2))=\pi_V(\alpha^{-1}(g_1,g_2) T(g_1) T(g_2))$

[cloud walrus]

ProphetX

[sinful mirage]

and now i'm stuck cause idk what pi does

[agile burrow]

If V is a complex vector space, PGL(V) = GL(V) / C*. What this means concretely is that we're identifying all nonzero scalar multiples of a matrix

[fleet pelican]

Aka GL(V)/{cI:c in C, c not 0}

[agile burrow]

So in particular, pi takes an element of GL(V) to the equivalence class containing all nonzero scalar multiples of that element

[fleet pelican]

You will note that A(cI)B(dI)=AB(cdI)

[sinful mirage]

so in partcular we have that $\pi_V(\alpha^{-1}(g_1,g_2) T(g_1) T(g_2))=\pi_V(T(g_1) T(g_2))$

[cloud walrus]

ProphetX

[fleet pelican]

I should use E instead of I for legibility

[agile burrow]

That's right

Also, is there a reason you're using alpha^-1 instead of alpha lol

[sinful mirage]

T(gh) is alpha^{-1} t(g) t(h)

[agile burrow]

Oh, I see

[sinful mirage]

(alpha^-1 exists because alpha is nonzero)

[agile burrow]

Right

[sinful mirage]

(by assumption of alpha)

[agile burrow]

I guess it doesn't really matter which side you put alpha on since it'll always be invertible lol

But good, hopefully the forward direction is clear to you now

[sinful mirage]

and now the real question is: is pi_v a group homomorphism?

because if it is,we are done

[agile burrow]

Indeed it is. As memeke pointed out, the subgroup of scalar transformations lies in the center of GL(V), hence it is normal

[sinful mirage]

as PGL(V) is a group, i assume this is just the quotient morphism

[agile burrow]

That's right

[fleet pelican]

iirc it is exactly the center, but i dont recall the proof offhand

[agile burrow]

Yeah, it is the center. Maybe it falls out if you consider what commutes with elementary matrices

[sinful mirage]

(quick question before we finish this direction): does everything we said above hold for V infintei dimensional too?

or only for V finite dimensional

[agile burrow]

😬 idk about infinite dimensional stuff lol

[sinful mirage]

(ignoring continuity issues)

[agile burrow]

seems messy

[sinful mirage]

just algebraically

[agile burrow]

Maybe someone else can comment, I have no idea

[sinful mirage]

for finite dim case does 2=>1 too?

we showed 1=>2 so far

[agile burrow]

Yes, the other implication holds as well

The other direction is maybe a bit more involved, but I think we can work through it

So we're starting with a homomorphism G -> PGL(V), right?

[sinful mirage]

yes

[agile burrow]

The idea is that we want to try and lift this to a map G -> GL(V)

So what this means is that for each g in G, we should associate some scalar c(g) in C*

And then we can define our map G -> GL(V) by sending g to c(g) \theta(g)

[fleet pelican]

it is always the case that GL(V)->PGL(V)\equiv GL(V)/Z(V) is homo. There are many issues with I.D. this isn't one.

[agile burrow]

Ok neat, thanks for the confirmation

[sinful mirage]

right

[agile burrow]

There's an extra layer of structure imposed on the map c because theta is a homomorphism

Oh wait, I think I have this wrong

Ok yeah let me back up for a second

[sinful mirage]

so what we have: $\theta:G \to PGL(V)$ homomorphism

[cloud walrus]

ProphetX

[agile burrow]

Right, we start with a homomorphism G -> PGL(V)

Let's try to lift this to a map T: G -> GL(V)

[sinful mirage]

We want a map $T:G \to GL(V)$, such that $T(g)T(h)=c(g,h) T(gh)$ for some $c(g,h):G \times G \to \mathbb{C}^{*}$ satisfying some condition

[cloud walrus]

ProphetX

[agile burrow]

Right

So in general, our lift T won't be a group homomorphism

[sinful mirage]

yes

[agile burrow]

However, T(gh) and T(g) T(h) both get mapped to theta(gh) under the projection to PGL(V), right?

What this means is that in GL(V), T(gh) and T(g) T(h) differ by some nonzero scalar multiple because they lie in the same equivalence class in PGL(V)

[sinful mirage]

supposing T satsfies the equality given

but first, we want to construct a map T

then show it satisfies the equality

[agile burrow]

Sure, but we can just define T by arbitrarily choosing a lift of theta

Like just choose some preimage of theta(g) in GL(V)

[sinful mirage]

can you please elaborate? I am not very familiar with the word 'lift'

[agile burrow]

Yeah, let me be more rigorous. So a lift of theta is a map T: G -> GL(V) such that pi o T = theta, where pi is the projection GL(V) -> PGL(V)

[sinful mirage]

right ok

so how we define this T?

using the data of theta

[agile burrow]

The naive thing you can do is just choose some preimage of theta(g) in GL(V) at random

and by construction, this gives you a lift

[sinful mirage]

(also wait one second)

there might be a mistake because of my notation

the cocycle condition should be addition or multiplication?

is this correct?

or should be * between them

[agile burrow]

It's easier to write it as multiplication here because it takes place in the multiplicative group of the field

I'd rewrite it with *

[sinful mirage]

but wait. alpha(g_2,g_3) is a complex nonzero number right

so alpha(g_1,g_2) is added or multiplied with another alpha?

[agile burrow]

Multiplied

Yeah, more generally one usually defines 2-cocycles as maps G -> A satisfying this condition, where A is some abelian group

In this case A is the multiplicative group of C, so the group operation is multiplication

[sinful mirage]

one second,let me get back to this. so if theta(g) in PGL(V), then preim theta(g) in G.

how does this tell me anything about T?

[agile burrow]

Oh, let me rephrase

So theta(g) lies in PGL(V), which is an equivalence class of elements of GL(V)

You can choose some representative of this equivalence class, which will lie in GL(V)

And just set T(g) to be this representative

Obviously this isn't well-defined, as different choices of representatives lead to different functions T

But bear with me for now

Now T(gh) and T(g) T(h) both lie in GL(V), but there's no reason they need to be equal, right?

[sinful mirage]

one second. let me try to write down formally what you wrote

to make sure i understand

$\theta:G \to PGL(V)$. Then $\theta(g) \in PGL(V)$, but as $PGL(V)=GL(V)/ \mathbb{C}^{*}$, we define a map $T:G \to GL(V)$ as $T(g) \in GL(V)$, such that what?

[cloud walrus]

ProphetX

[agile burrow]

We're choosing $T(g) \in GL(V)$ such that $\pi \circ T(g) = \theta(g)$

[cloud walrus]

walter

[sinful mirage]

no, that is clear

but T(g):=?

I don't understand what is the image of g under T

[agile burrow]

So we haven't really made a specific choice for T

The point is that theta(g) is a whole equivalence class of matrices, right?

[sinful mirage]

yes

[agile burrow]

Let me just pick out one of these matrices, and I declare that T(g) be equal to this matrix

And I make this choice for every g in G

[sinful mirage]

which one?doesn't matter?

i pick randomly one for each class?

[agile burrow]

Indeed, it (sort of) doesn't matter as we'll show later

Right now, different choices will lead to different definitions of T

but assume that we've fixed one such choice of T

[sinful mirage]

so now I have g many linear maps V->V, namely T(g)

[agile burrow]

That's right

We've just built some set theoretic function T: G -> GL(V)

And the only thing we know about it right now is that pi o T = theta

Ok, so now let's try to deduce some more structure/information about our map T

There's no reason why T(gh) should be equal to T(g) T(h) since we've just chosen representatives at random for theta(gh), theta(g), and theta(h)

But we can say something nontrivial about them - namely, $\pi \circ T(gh) = \theta(gh) = \theta(g) \theta(h) = (\pi \circ T(g)) (\pi \circ T(h))$

[cloud walrus]

walter

[sinful mirage]

right,this follows from theta being homomorphism

[agile burrow]

Exactly

In particular, T(gh) and T(g) T(h) lie in the same equivalence class after we pass to PGL(V). Explicitly, they differ from one another by a nonzero scalar

[sinful mirage]

true

[cloud walrus]

walter

[sinful mirage]

we could formallry write: $\pi(T(gh))=\pi(T(g)) \pi(T(h)) \iff [T(gh)]=[T(g)][T(h)]$, right?

[cloud walrus]

ProphetX

[agile burrow]

Yes, that's right

[sinful mirage]

yes,now it is left to see cocycle condition si satisfied

[agile burrow]

Right, and to see this we'll need to use associativity of G

Namely, we'll want to consider that $T((gh)k) = T(g(hk))$

[cloud walrus]

walter

[agile burrow]

It might be easier to write it as T(gh) = alpha(g, h) T(g) T(h), I'm not sure

But it doesn't matter either way

[sinful mirage]

can we please continue this way as this is how we use it in physics?

[agile burrow]

Ok, sure

[sinful mirage]

(it would just replace alpha^{-1} with alpha but to make it consistent)

[agile burrow]

Fair enough

Yeah, so evaluating the left side we get $T((gh)k) = \alpha^{-1}(gh, k) T(gh) T(k) = \alpha^{-1}(gh, k) \alpha^{-1}(g, h) T(g) T(h) T(k)$

[cloud walrus]

walter

[agile burrow]

Evaluating the right side gives $T(g(hk)) = \alpha^{-1}(g, hk) T(g) T(hk) = \alpha^{-1}(g, hk) \alpha^{-1}(h, k) T(g) T(h) T(k)$

[cloud walrus]

walter

[agile burrow]

Spacing is hard :(

[sinful mirage]

so we conclude that $\alpha^{-1}(gh, k) \alpha^{-1}(g, h)= \alpha^{-1}(g, hk) \alpha^{-1}(h, k)$

[cloud walrus]

ProphetX

[agile burrow]

Right

And inverting everything gives the desired condition

[sinful mirage]

wdym niverting everything

this alpha^{-1} is the inverse of the complex number,not the inverse of the alpha map,right?

i'm confused a bit

[agile burrow]

Yeah, alpha^{-1} is the inverse of the complex number

So we can just take the reciprocal of both sides of the equation

[cloud walrus]

walter

[agile burrow]

Just like we wanted

[sinful mirage]

yeah,makes complete sense

[agile burrow]

Right, so this technically solves your problem but maybe there's one little discrepancy left

We made a lot of choices in constructing T. Hopefully this computation makes it clear that every choice of T leads to some 2-cocycle alpha

[sinful mirage]

so the outline is,we define a map T, then using that theta is a homomorphism, we infer somethnig about this map T, namely that T(g)T(h) and T(gh) differ by a complex nonzero multile. We now define a map alpha;GxG->C^{*} and from associativity of the operation, we obtain a condtion on alpha

[agile burrow]

That's right

[sinful mirage]

what choices did we make?

[agile burrow]

Well, we made a choice of representative in theta(g) for each g in G

[sinful mirage]

yes

[agile burrow]

And making different choices of representative would've led to different functions T, which would've led to different cocycles alpha

[sinful mirage]

but nothing changes if we choose a different representative, as when we map by pi, we still get same T'(g) T'(h)=some number T'(gh)

[agile burrow]

Yeah, so you always get a 2-cocycle but my point is that you might get a different cocycle if you have a different lift T

(This part doesn't matter for your problem, but I think it's interesting so I'm talking about it anyway)

[sinful mirage]

this seems problematic then

or well

how do I guarantee uniqueness of the lift then?

[agile burrow]

Well, you don't

[sinful mirage]

wait is this the stuff which was in the book that the lft is unique iff H^2(G,U(1))={e}?

[agile burrow]

YES

Yeah, so we want to put some notion of equivalence on lifts

[sinful mirage]

ok this would be really nice to clarify,because i am interested very much in why the second cohomology group parametrizes lifts

[agile burrow]

Sure, yeah

So suppose we chose some different lift, say L

[sinful mirage]

to be very concrete this means different representatives, right?

[agile burrow]

Yes

Again, we get that L(g) and T(g) are both mapped to theta(g) when we pass to PGL(V), so they differ by some scalar

We can write L(g) = f(g) T(g), where f(g) is in C*

Oh, and I guess the lift L will gives us a different 2-cocycle, say uhh beta

[sinful mirage]

let's say \beta

we used alpha

[agile burrow]

oh yeah

beta

that's good

[sinful mirage]

L(g) L(h)=beta(g,h) L(gh)

[agile burrow]

Perfect

So now we get $f(g) f(h) T(g) T(h) = \beta(g, h) f(gh) T(gh) = \beta(g, h) f(gh) \alpha^{-1}(g, h) T(g) T(h)$

[sinful mirage]

alpha^{-1} i think no?

[agile burrow]

Sure, maybe it's an inverse

Yeah, it's an inverse

[cloud walrus]

walter

[agile burrow]

And rewriting yields $\beta(g, h) = f(gh)^{-1} f(g) f(h) \alpha(g, h)$

[cloud walrus]

walter

[sinful mirage]

right

[agile burrow]

Hmm, I'm maybe forgetting what happens now

I think the point is to show that $f(gh)^{-1} f(g) f(h)$ is a coboundary

[cloud walrus]

walter

[sinful mirage]

what is the definition of a coboundary?

i am unsure but thiseems very related tolamba here

we have almost same formula

actually,it is thesame

[agile burrow]

Yeah, we're very close

Oh perfect

Ok yeah

So the point is that lambda here is a coboundary

I'm a little confused though

[sinful mirage]

the book didn't define coboundary

[sinful mirage]

in our case f= lambda, beta=w', alpha=w

we have the concrete exact same formula

i just don't know what we use it for

[agile burrow]

Hm yeah I'm missing like one piece lol

But I guess for your purposes, this suffices

[sinful mirage]

we should rewrite $f(gh)=\beta^{-1}(g,h) \alpha(g,h) f(g) f(h)$ and we have the same formula exactly

[cloud walrus]

ProphetX

[agile burrow]

The point is that two different lifts of the same projective representation lead to two cocycles which differ by such a function lambda

Conversely, a cocycle gives us a projective representation, and you can probably convince yourself that equivalent cocycles gives you equivalent projective representations

So we get a bijection between projective representations and H^2

Isn't that neat?

[sinful mirage]

did we not show this?

[agile burrow]

Yeah, that's what we showed

[agile burrow]

Ok I've resolved this

Yeah, so formally a 2-coboundary is defined to be a map phi: G x G -> A such that there exists a function f: G -> A for which phi(g, h) = f(g) + f(h) - f(gh)

I'm ignoring some more technical stuff

But the point is that our function f that we had above gives us this exactly

[sinful mirage]

yeah,i've read that you need to use tor/ext functors,and to get here you use a bar resolutioin

(whatever those words mean),for now i don't mind/care about those details per se

[agile burrow]

Right, yeah this is all group cohomology stuff

But yeah, no need to concern yourself with that stuff right now if you're working on other things

I've only briefly glanced at connections between group cohomology and projective representations so it was really fun to work through this

[sinful mirage]

but wait,can we elaboraet on this please?

[sinful mirage]

so i'm at the point we stopped:here

how to continue from here?

[agile burrow]

Yeah, so our starting point was that we were assuming we had two different lifts L and T of the same projective representation theta: G -> PGL(V)

And from there we deduced that the associated cocycles alpha and beta are equivalent in H^2

[sinful mirage]

how?

[sinful mirage]

we concluded this

[agile burrow]

Yeah, and that's precisely the equality that holds in the definition here, right?

[sinful mirage]

yes,lambda=f, beta=w',alpha=w

[agile burrow]

Yeah, so the cocycles alpha and beta become equivalent in H^2

[sinful mirage]

ohh you mean this

$[\alpha]=[\beta]$ for the equivalence relation defined in the picture

[cloud walrus]

ProphetX

[agile burrow]

Yes, that's right

Sorry, I should've clarified

On the other hand, every cocycle gives us a projective representation, and equivalent cocycles in H^2 give us the same projective representations

[sinful mirage]

so we showed: two different lifts give riise to two different cocycles, which become the same in the cohomology group

[agile burrow]

Yes

[sinful mirage]

every cocycle gives us a projective representation: this is 1=>2

no?

[agile burrow]

Right

[sinful mirage]

now the last part i don't se

[agile burrow]

Ok, we can work that out explicitly

[sinful mirage]

equivalent cocycles in H^2 give same proj rep

[agile burrow]

So if we have equivalent cocycles alpha and beta

We can write $\lambda(gh) = \alpha(g, h) \beta(g, h)^{-1} \lambda(g) \lambda(h)$

[cloud walrus]

walter

[sinful mirage]

right

[agile burrow]

Let's rewrite this as $\lambda(gh) \beta(g, h) = \lambda(g) \lambda(h) \alpha(g, h)$

[cloud walrus]

walter

[agile burrow]

Oh, hm

Let's see

Ok so maybe I was lying a little bit

I don't know if there's a bijection between H^2 and projective reps

Because I'm not sure if it's the case that every cocycle is realized by some map T: G -> GL(V)

Our starting point for 1 -> 2 is that there exists a map T such that T(g) T(h) = alpha(g, h) T(gh), but I don't know if it's the case that for any cocycle such a map exists

[sinful mirage]

I believe it is true but the proof is too technical for me to relate it to our language

[agile burrow]

Hmm ok I'll look into it

[sinful mirage]

is this related? or comletely dfferent

[agile burrow]

Oh, sure so this is related

Yeah the classical result on H^2(G, A) is that it classifies group extensions 0 -> A -> E -> G -> 1

[sinful mirage]

there's full proof of this,the bijection is written out and also showed that is both inj and surj

[agile burrow]

Right, yeah it's nice how explicit this is

[sinful mirage]

how is this related to proj. reps?

[agile burrow]

Ok so first let me make an observation

[sinful mirage]

so what I am sure about,but can't prove: a projective representation <=> U(1) central extension

[agile burrow]

We had that a lift of a projective representation theta: G -> PGL(V) gives us a map T: G -> GL(V) such that T(g) T(h) = alpha(g, h) T(gh)

[sinful mirage]

in this case, H^2(G,U(1)) would classify U(1) central extensions, would classify proj reps

[sinful mirage]

this is the missing piece (accepting that proof)

[sinful mirage]

yes

[agile burrow]

Now if alpha(g, h) = 1 for all g, h in G (i.e. if alpha is the trivial cocycle) then the lift T is actually a group homomorphism

What this means is that if the cocycle associated to a lift is trivial, then our projective representation lifts to an actual linear representation

[sinful mirage]

this is true

is this true also for cocycles equivalent to the trivial cocycle?

or not necessarily

[agile burrow]

Yes, it's true if you're equivalent to the trivial cocycle as well

(I think)

Now in general, a cocycle won't let you lift to a linear representation. However, it's always the case that you can lift to a linear representation of a different group H

And this group H will end up being a central extension of G

[sinful mirage]

why?

[agile burrow]

Why can we lift?

[sinful mirage]

yes,to a different group

[agile burrow]

I will tell you

So explicitly, if theta: G -> PGL(V) is our projective representation, we can construct a group H = {(g, A) in G x GL(V) : theta(g) = pi(A)}

There's an obvious surjection from H onto G by projecting onto the first factor

[sinful mirage]

right

but one second

what group structure does H get?

[sinful mirage]

this is a set.

[agile burrow]

H is a subgroup of the product group G x GL(V)

so you just have componentwise multiplication

[sinful mirage]

ok yes

[agile burrow]

The kernel of this surjection is the subgroup {(e, lambda I)} where lambda is any nonzero scalar in C

And hopefully it's clear that this is contained in the center of H

[sinful mirage]

sure

[agile burrow]

Finally, we get a linear representation H -> GL(V) given by projecting onto the second factor

and so we have a central extension 0 -> ker -> H -> G -> 1

And we've lifted the projective rep G -> PGL(V) to a linear rep H -> GL(V)

In this case, we mean a lift in the sense that the composition H -> GL(V) -> PGL(V) agrees with H -> G -> GL(V)

[sinful mirage]

pr(H((g1g2,A1A2))=A1A2=pr(H(g1,A1))pr(H(g2,A2)), this is why it's a group homomorphism?

[agile burrow]

That's right

[sinful mirage]

what's the diff between 0 and 1?

[agile burrow]

Oh, I just use 0 on the left since the kernel is abelian and 1 on the right since G need not be abelian

In both cases they refer to the trivial group

It doesn't really matter, if anything it's bad notation on my part

[sinful mirage]

here ker=?

[agile burrow]

ker is the kernel of the map H -> G given by projection onto the first factor

[agile burrow]

this thing

[sinful mirage]

ohhh

ker->H this -> is inclusion?

[agile burrow]

Yes

[sinful mirage]

so im(inclusion)=ker so this is exact

yeah makes sense

[agile burrow]

[sinful mirage]

thanks a lot!

one last question if you don't mind: do you agree that proj reps <=> U(1) central extensions?

or proj reps <=> central extensions?

we showed that from a projective rep,we can obtain a central extension by H

is it true that from a central extension H,we can obtain a proj rep?

[agile burrow]

I'd have to think on it a little bit more. I want to say yes

[sinful mirage]

my confusion here is. are proj reps central extensions, or U(1) central extensions?

[agile burrow]

Yeah, I'm not too clear on it either

are these like unitary projective reps?

[sinful mirage]

yes

[agile burrow]

Ahh ok

So I think if everything is unitary then every time we have a map into the multiplicative group of the field, I think we can just replace it with U(1)

because the point is that in P(U), everything differs by a scalar but since everything is unitary, these scalars have to have norm 1

[sinful mirage]

this seems slightly related,but does not relate central extensions to proj res

but my guess is this: claim: proj reps <=> H extensions. unitary proj reps <=>U(1) extensions

this makes sense to me

but i can't prove

[agile burrow]

Sure it does, the extensions k* x G are central extensions of G

I think it's saying the same thing we were saying before

Ok, so we've shown that projective reps give central C* extensions

[agile burrow]

And if you believe this, every unitary proj rep gives a central U(1) extension

[agile burrow]

I guess I forgot to clarify here that the kernel is isomorphic to C*

Conversely, if we have a central extension 1 -> C* -> H -> G -> 1 and a linear rep H -> GL(V)

[sinful mirage]

why?

we showed they give H extensions

[agile burrow]

A C* extension of G is a short exact sequence 1 -> C* -> H -> G -> 1

A U(1) extension would be a short exact sequence 1 -> U(1) -> H -> G -> 1

[agile burrow]

But given this, observe that the linear rep takes C* as a subgroup of H to C* as a subgroup of GL(V), identifying C* with the nonzero multiples of the identity matrix

Then the fact that G = H / C* implies that the linear rep H -> GL(V) induces a homomorphism G -> PGL(V), which is precisely a projective rep

Explicitly in the last map, if you have an element g in G, you choose a preimage of g in H, map it to GL(V) via the linear rep, and then pass down to PGL(V)

Maybe you want to check that this is well-defined but that isn't hard

So this tells us how to get a projective rep from a central C* extension (or a unitary proj rep from a central U(1) extension)

And I think these constructions should be inverses of one another

[sinful mirage]

yeah,makes sense. I'll try to work out this

but first,i'll type up in a TeX file everything we discussed. Thanks for the help

[agile burrow]

happy to help

[sinful mirage]

(in case you're a SE person and want free karma, feel free to answer https://math.stackexchange.com/questions/4671037/equivalence-of-definitions-of-projective-representations?noredirect=1#comment9872664_4671037)

[agile burrow]

it was fun to think about

lol maybe i'll get around to it eventually, i usually stick to discord

[sinful mirage]

(in case you are interested in physical applications of this pure math stuff)

this is the theorem I am aiming towards,but going in small steps

[agile burrow]

WOAH this seems neat

Yeah, I don't know any of the physics applications of projective reps, but I came across it a few times when I was reading about rep theory and some quantum mechanics stuff

[sinful mirage]

so we in physics basically want to classify projective unitary reps of the symmetry group G in a hilbert space H

[slate tide]

I saw this the other day, it was wild, but 2-cocyles are really common in physics for some reason and I guess this is another reason

[sinful mirage]

(this later turns out to be more involved than theorem 4.8)-> the real classifciation is as follows: proj unitary reps <=>unitary reps of the universal central extension of the Lie group (this is a cnetral extensions through which every other central extensions factors through uniquely)

[sinful mirage]

but theorem 4.8 gives us a classification for certain G,which satisfy that they are simply connected and H^2(g,R)=0

[agile burrow]

universal central extensions are cool

[sinful mirage]

I know nothing about them but there's this perfect thesis written by a mathematician https://www.math.ru.nl/~landsman/Nesta.pdf

language is not on the level which we discussed/used, it's much more elegant/abstract

but i'm slowly getting there

[agile burrow]

You got it king

[sinful mirage]

in what context did you see them?

[slate tide]

a string theory talk I went to, but don't ask me any details, I just went to absorb and not to understand lol

But there may have been some 2-groups floating around somewhere, and I think the main thing was gauge groups

[sinful mirage]

@agile burrow I think why the confusion with + and * appeared for me haha! if we consider projective unitary reprsentations, then the cocycles will not take values in C*, rather U(1), and hence we will have cocycle relations such as e^{itheta(g,h)}e^{...}=e^{}e^{}

and then indeed the thetas add

[agile burrow]

ahhh

[sinful mirage]

lel

[agile burrow]

ok yeah that makes a lot of sense lol

thanks for clarifying that

[sinful mirage]

yeah i was confused because i've read the statement in a mathematiics book

that's why PGL

but I use it for PU 😄

[sinful mirage]

@agile burrow we discussed that a lift in general exists for a central extension of G

in case G is a Lie group, do you perhaps have intuiton why G should be replaced by its universal cover?

cf

[agile burrow]

Ahh, I'm afraid I don't know all that much about Lie groups so I don't think I can provide any insight for this

[sinful mirage]

for our discussion G being algebraic(discrete) our result is true though right?

a discrete group is isomorphic to its universal cover

[agile burrow]

Yeah, this all works well for discrete groups

[fleet pelican]

You build a universal covering group as a central extension of a lie group, it is built so that the projection is a covering homomorphism. It's really more of a lie algebra thing though tbh.

[sinful mirage]

Right but we just showed that a lift exists for a central extension of the group

Is the central extension of the universal cover also a central extension of the group?

I think not

[fleet pelican]

The universal cover is built via central extension.

[sinful mirage]

But not by a C* extension

Rather by a pi(G) extension,I think

[fleet pelican]

It is really better to think of the extension in the context of the algebra.

[sinful mirage]

Not necessarily

[indigo ridge]

can someone help me understand the statement "and therefore any subgroup of order 3 is cyclic" what is the justification for this?

[solar shore]

are the two proofs similar?

[solar shore]

i think its talking about how any group with order 3 is cyclic

[indigo ridge]

yeah no kidding

[solar shore]

every group with prime order is cyclic

[indigo ridge]

sure but whats the justification?

[solar shore]

if a group G has order p, then lagrange's theorem says that any subgroup, say <a> for a in G must divide p, so |<a>| must be 1 or p

if it was 1 then its the trivial subgroup, so |<a>| is p

then |G| = p = |<a>|, then G = <a>

hence g is cyclic

[indigo ridge]

ahh makes sense thanks\

[solar shore]

np

[charred crescent]

are the number of left and right cosets equal in general?

[carmine fossil]

Yes

[lusty marlin]

There exists a very 'natural' bijection from the set of left cosets to the set of right cosets of a given subgroup of a group.

[shell brook]

i love this "in case you are interested in physical applications of pure math stuff" [screenshot of pure math theorem]

[chilly ocean]

Im trying to find all intermediate fields of extension $\mathbb{Q}( \sqrt[4]{2} ) | \mathbb{Q}$. I guess one way would be to find all the intermediate extensions of Q(p), where p is the minimial polynomial X^4 - 4 by galois correspondence and then only look at those which are contained in our given field. But this approach seems horrible. Is there a quick and elegant way to determine all intermediate fields? (I believe theresnonly one, but im not sure how to show it)

[cloud walrus]

matcool473
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sullen island]

how do i see that the dual of an injective R-module is projective/dual of a projective R-module is injective? where R is not necessarily commutative (i am not even sure if these statements are true, ive googled but to no avail)

[fleet pelican]

you can add some of the roots to create other roots.

[fleet pelican]

there are two intermediate fields

[fleet pelican]

Second is just special case of first. Obviously Z/nZ is cyclic, generated by 1.

[frigid lark]

P is projective if every exact sequence of the form $$0\to A \to B \to P \to 0$$ splits. Now apply the $Hom(_,R)$ functor (a contravariant functor) to the diagram above to get the exact sequence: $$0\to Hom(P,R) \to Hom(B,R) \to Hom(A,R) \to 0 $$ which you can argue that splits.

[cloud walrus]

Parrot Tea

[frigid lark]

Thx

[rustic crown]

dual of projective is not necessarily injective tho

like for pids R, injective is equivalent to divisible, which is not true even for free R-modules (unless it is 0)

[frigid lark]

Oh rip

I just thought that would work

[rustic crown]

yea that only checks very specific exact sequences

not every module is dual of something for eg

as dualizing kills torsion (for integral domains)

[sullen island]

hmmm

but this is true if i take the "standard" duality functor? as in duals of vector spaces

Hom(-, k) where k is a field

[rustic crown]

well vector spaces are weird

because projective, injective and free are all the same things >.<

[sullen island]

im having difficulty seeing this even for vector spaces: i can see that Hom(-, k) is exact, but why does that imply for a projective (resp injective) module M, Hom(M, k) is injective (resp projective) ?

[rustic crown]

because all vector spaces are projective/injective/free

[sullen island]

hmm i see

[rustic crown]

yea, projective and injective are duals in the sense of opposite category and not the dualizing functor

projective in R-mod is equivalent to injective in (R-mod)^op

[sullen island]

how do i prove that? 😮

[rustic crown]

basically this proof lmao

but instead of duals you did (-)^op

[frigid lark]

Is (R-mod)^op just the modules over the opposite ring?

[rustic crown]

nuuuuu

far from it

[frigid lark]

Oh ok

I was concerned

[rustic crown]

left R^op-mods are just right R-mods

so for commutative rings, there is literally no difference in the two categories

but (R-mod) and (R-mod)^op are very different

[sullen island]

to show Hom(P, k) is injective, i need to show that if i take arbitrary A, B in R^{op}-mod,
$$0 \to Hom(P, k) \to A \to B \to 0$$ splits

[cloud walrus]

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

[rustic crown]

[sullen island]

argh

[elder wave]

[sullen island]

latex hates me

[rustic crown]

there's no Hom(-, k) here >.<

[frigid lark]

Just do it in Google docs

[elder wave]

wait det i'm a bit confused

(R-mod)^op is not a category of modules right

[rustic crown]

in what sense

yea it's not directly some R'-mod

[elder wave]

yeah that's what i meant

[rustic crown]

but is still abelian so we can still define projective/injective

[elder wave]

but still abelian so you have an embedding

ah yeah that's what i was wondering

ty

[sullen island]

sorry i dont quite understand, im trying to prove your remark on projective iff injective in R-mod^op, so dont i take the duality to the opposite category?

[rustic crown]

okie what i mean is we're reversing arrows using (-)^op and not hom(-, k)... i didn't even define any k here

[sullen island]

oh i see...

[rustic crown]

so show P is injective in the opposite category, start with any monomorphism i : A --> B in (R-mod)^op and a map A --> P.
this is equivalent to having an epimorphism i^op : B --> A and a map f : P --> A in R-mod, since P is projective you can lift this map to P --> B and in the opposite category this gives us an extension B --> P

[pastel cliff]

ive moved on to JCF now and im confus

how do we know to choose those bases for RCF and JCF

i guess the process is outlined in the second lil paragraph

actually slightly different question, in the first part are we not decomposing straight into the invariant factors?

oh it's just the monic polynomials that make up each invariant factor

ok real question : is JCF the same as RCF but under a different basis?

i.e. different only by row operations on each block

[thorn delta]

Yes

[topaz spruce]

Hello guys I need ur help

[hallow kiln]

#advanced-analysis .

[pastel cliff]

and this basis only exists when it's over an algebraically closed field right

[topaz spruce]

Thanks

[hallow kiln]

yes, by definition polynomials over non-algebraically closed fields need not split into linear factors.

[thorn delta]

i think if the irreducible factors of the char poly happen to be linear, then this is enough for the existence of a JNF

but yea, you would get this automatically by just assuming algebraically closed

[pastel cliff]

JNF = JCF?

[south patrol]

Jordan normal/canonical form

Tbh I usually use JNF too lol

[rustic crown]

JCF is less C than RCF :p

[hallow kiln]

yes of course.

[rustic crown]

(cause in RCF you have a C way to order the individual blocks based on the invariant factors, but for JCF there is no natural way to order the blocks with different eigenvalues)

so like RCF of a matrix is unique, while JCF of a matrix is only defined up to permutation of blocks with different eigenvalues

ofc for bwocks of the same eigenvalue you can order them on the size

[south patrol]

RCF doesn't need cringe alg closure

[rustic crown]

i mean you can talk about JCF of only triangularizable operators

this word is so hard to say

trianguliozaidhafldif

i always end up saying it trianguralizable >.<

and then it sounds so weird

oh maybe owofied vewsion wouwd be easiew to say since w = w = w
twianguwawizabwe

whut did i say lol

[pastel cliff]

owoified

when finding JCF, can you reduce T before finding c_T?

[rustic crown]

whut does it mean to reduce T?

[pastel cliff]

like row reduce a matrix

[rustic crown]

oh

probably not

you can only replace it by a similar matrix

so like you can diagonalize/triangularize it if you want

row-reduction might change the eigenvalues

[pastel cliff]

doing this

do i have to do id * x - T and then reduce

[rustic crown]

nah nobody wants you do to that much work

find the char and min poly

for smol examples they are sufficient to deduce the JCF

[pastel cliff]

char technically not defined at this point

[rustic crown]

c_T >.<

[pastel cliff]

oh i meant characteristic polynomial by that

[rustic crown]

:nani:

[plucky flicker]

Hi! How can I prove that for irreducible characters the higher Frobenius-Schur indicator is an integer? Any hint would be appreciated

[pastel cliff]

is this a shortcut to charateristic polynomial then?

[rustic crown]

whut

[pastel cliff]

or like

how would you do this example

[rustic crown]

i was saying char poly and min poly tell you a lot about the JCF (in smol dimensions), so always useful to compute them quickly

[rustic crown]

pretty much like they did

[pastel cliff]

my question is how they got c_T(x) so immediately then

[rustic crown]

oh cause they asked wolfy

[pastel cliff]

[rustic crown]

there is a description of the char poly using the sums of principal minors

but asking wolfy is faster

or directly computing is also pretty much same

[rustic crown]

but this is only useful to like compute the coefficient of t^n-1 lmao

(because it's just trace)

[agile burrow]

I found an MO post on this: https://mathoverflow.net/questions/107256/sum-g-gk-frobenius-schur-indicators-s-n-invariants-in-freeassx-i-center-o

I'm almost certain there ought to be an easier way than viewing higher Frobenius-Schur indicators as Adams operations and using the connections to Newton polynomials, but I don't know of it lol

[plucky flicker]

thank you!

[hot lake]

iirc if we're talking about representations of finite groups, there is a fairly elementary way by interpreting morphisms of representation from V to V as G-invariant subspaces of V² (not sure about that one I'm a little foggy about the details) then using Maschke's theorem to say that this can only have dimension exactly 1. Whether it falls into the symmetric or antisymmetric part or whether there are 2 of them because V secretly splits over the complex numbers, tells you if V was real, quaternionic or complex.

[plucky flicker]

G is finite and we are working over the field of complex numbers

[hot lake]

you can also tie the thing into the existence of G-invariant symmetric or hermitian (?) bilinear forms on V

ah wait I only know about the k=2 case

[plucky flicker]

we proved the k=2 case

i've tried induction on k

[fervent rock]

hi i have a question

Lets say i have group G and centralizers Z(G). Is it right to say that Inn(G) ismorphic to G/Z(G)?

[thorn delta]

if by "centralizers" you mean the center, then yes

[plucky flicker]

Yes, take the homomorphism G --> Inn(G) which sends g to the automorphism of conjugating with g. This is surjective and the kernel is Z(G)

[fervent rock]

So then from that i have another question

If i am looking at D_5, its Z(D_5) = {e} (the identity)

so Inn(D_5) ismorphic to D_5/{e}

which is basically just saying Inn(D_5) isomorphic to D_5

[plucky flicker]

ye

[fervent rock]

So from that, pretty much, I can just say Inn(D_5) = D_5?

or is there any witchcraft I gotta do