groups-rings-fields

[twilit wraith]

its only pointset and homotopy

for the top part

[mint seal]

yeah passing 3 is usually the requirement, and often you can pick your 3 from a selection of a few more than that

I've heard of students who are exceptionally well prepared, entering a PhD program and taking quals right away. It's not common but yeah you can possibly skip some classes that way

[twilit wraith]

if you arent prepared for any of them you likely have to do six courses

to prepare

[mint seal]

I should also mention there are written quals, and then a separate oral qual, which is similar to a mini PhD thesis defense

[twilit wraith]

man i feel like i dont understand nilpotent groups much at all

[mint seal]

usually the oral qual is around the 3rd or 4th year

[mint seal]

me either. I kinda remember them coming up somewhere around Jordan Holder series decomposition stuff

[twilit wraith]

theres just a ton of problems i have to get done concerning them in the next two hours but i can barely do one

i shouldnt even say barely

strangely enough the math programs at top universities end up being kinda slowly paced

[merry harness]

group actions are so cools icl

[twilit wraith]

eh i just went close to home

i love math a lot but i love my friends and family more at the moment

though who knows maybe galois theory will surpass them next semester

just kidding

perfect reaction lol

[twilit wraith]

im still pretty stumped about the G/Z(G) nilpotent implies G is nilpotent problem

[velvet hull]

oh, what do you have right now?

[karmic moat]

if you want to dip your feet in geometry, you can look at Humphrey's Linear Algebraic Groups

All nilpotent groups are solvable and solvable groups are awesome

[velvet hull]

oh, im curious, what is there to know?

[twilit wraith]

honestly nothing

i thought itd be immediate from finding Z(G/Z(G)) somewhere in the upper central series of G but nope

i think i have to use like preimages of the projection map G to G/Z(G)

[karmic moat]

one result is that if G is a connected solvable algebraic group, then it's nilpotent if and only if G_s (the set of semisimple elements of G) is a subgroup

And then you can write G = G_s \times G_u, where G_s is the semisimple part and G_u is the nilpotent part

[velvet hull]

keep thinking about the upper central series

[twilit wraith]

i have been for a while

[velvet hull]

is Z(G/Z(G)) isomorphic to anything perhaps

[velvet hull]

that seems kind of random, is there any bigger theory about why nilpotent groups are relevant

[twilit wraith]

uhhh

nothing comes to mind right now but maybe

[karmic moat]

i think the importance is in the result that G = G_s \times G_u

[twilit wraith]

Z_2(G)/Z(G) i guess

[karmic moat]

there's also this https://terrytao.wordpress.com/tag/nilpotent-groups/

[twilit wraith]

im still stuck lowkey

ive not struggled with algebra this much until this weeks homework

not good

[velvet hull]

can you tell me what the upper central series of G/Z(G) looks like

[twilit wraith]

0th step is just Z(G)

1st step is Z(G/Z(G))

then $Z_2(G/Z(G))/Z(G/Z(G)) = Z((G/Z(G))/Z(G/Z(G)))$

[cloud walrus]

hiidostuff

[twilit wraith]

i mean this is just so complicated

[twilit wraith]

im guessing this isnt quite what u were looking for though

[tidal raft]

How do groups of Lie type relate to Lie groups?

[velvet hull]

I had to go through my notes, the proof was harder than I remember it being

but it boils down to proving that ||Z_i (G/Z(G)) = Z_i+1(G)/Z(G)||

[twilit wraith]

i see

i imagine this is by induction

i was just expecting this to much more elegant

[velvet hull]

theres an alternate proof that is slightly more general

I can send you my notes after youve proven it

[twilit wraith]

cool

[velvet hull]

I'm not aware of a nice proof lol

[twilit wraith]

its so late where i am i might just have to not turn in the homework tomorrow honestly

[velvet hull]

mainly because nilpotentcy is a hard thing to pass down from a quotient

[twilit wraith]

even harder when i barely know anything about it

and have 0 intuition

[velvet hull]

the only two normal subgroups that I know this actually works for is the center and the Frattini subgroup

[prisma ibex]

Finite groups of Lie type come from reductive groups over finite fields, reductive groups are the algebraic notion of compact Lie groups

Groups like SL_n(F_q) or Sp_2n(F_q) are good examples

[tidal raft]

Oh neat, so the idea is that we take some finite field, and apply the same constructions to arrive at the lie group to that finite field, and now we have a finite group with a correspondence to the Lie group?

[prisma ibex]

The correspondence is a bit more complicated but the classification of reductive groups over fields is very closely related to the classification of Lie groups/Lie algebras over the complex numbers for example

The classification is the same over algebraically closed fields and then you need a bit more to classify things over non algebraically closed fields

This is why for example the classification of Lie groups over the real numbers is a bit more complicated than it is over the complex numbers

It is likewise a bit more complicated over finite fields and this is where you see various twisted forms of reductive groups showing up among the finite groups of Lie type

[tidal raft]

oh wow, I didn't even think you could do such a thing as classifying all the lie groups you can embed in R or C. I will be looking into that.

[prisma ibex]

Well you can classify semi simple Lie algebras over C for example

This is the main classification result that classical Lie theory gives you

Then there is a tight relationship between this classification result for semi simple Lie algebras and semi simple Lie groups for example

But the point is that the same classification results work out identically for semi simple reductive algebraic groups over any algebraically closed field

[tidal raft]

I'm quite new to this stuff, as a sanity check, every Lie group has a corresponding canonical lie algebra, and you can map between them using the exponential map yes?

[prisma ibex]

Essentially yes, there are mild subtleties having to do with the center and with connected components but ignoring this yes that is the idea

Like several distinct Lie groups may have the same Lie algebra

[tidal raft]

like, a circle, and a slightly bigger circle?

[prisma ibex]

No no those would be the same up to isomorphism

[tidal raft]

oh different lie groups not up to isomorphism having the same lie algebra

does this happen in cases much like chairality?

[prisma ibex]

But for example R and S^1 have the same 1-dimensional Lie algebra

But these are clearly quite different as Lie groups

[tidal raft]

oh ok I understand

you're throwing away some information when you go to the lie algebra, namely the periodicity, and probably some other thing I can't think of as well

the lie algebra of R is just R as a field with the exponential map being the identity?

anyway, thanks for your patience and help it's been very helpful :)

[rocky cloak]

So with this definition it is immediate right.

The central series for G will just take one extra step to terminate than G/Z(G), otherwise they're identical.

[twilit wraith]

it does end up that they take one more step but it actually ends up being quite unclear

because the central series for G and G/Z(G) never actually meet

at first i also thought it was immediate but the proof was actually kinda horrible

[rocky cloak]

I don't know that you mean by meet.

I mean usually you think of it as subgroups Zi < G, but the quotients G/Zi are identical, just shifted by 1

[twilit wraith]

i dont think they are though

[rocky cloak]

Zi+1 is by definition the preimage of Z(G/Zi)

[twilit wraith]

the main thing to prove was that Z_i(G/Z(G)) = Z_i+1(G)/Z(G)

[rocky cloak]

Which is just the definition of Zi+1, yes?

[twilit wraith]

im not sure

[rocky cloak]

Zi+1(G) is the group such that
Zi+1(G)/Z(G) = Z(G/Z(G))

[twilit wraith]

i thought Z_i+1(G)/Z_i(G) = Z(G/Z_i(G)) was the definition

[rocky cloak]

Yeah, that's what I meant to write

[twilit wraith]

but isnt that not close to what im going for

[rocky cloak]

I mean, say G/Z(G) = H for notation
G/Zi+1(G) = (G/Zi(G)/Z(G/Zi(G)) = (H/Zi-1(H)/Z(H/Zi-1(H)) = H/Zi(H)

So G/Zi+1(G) = H/Zi(H)

So it really use comes down to of
H/N = K/N, then H=K.

[torpid veldt]

[twilit wraith]

im having a hard time following these equalities

but i am like really inept with nilpotent groups so far

[rocky cloak]

Like both Zi(H) and Zi+1(G) are just defined by a subgroup in
G/Zi(G) = H/Zi-1(H)

[rocky cloak]

So if Zi(G) = Zi-1(H), then Zi+1(G) = Zi(H)

That's all the equations above are saying

[twilit wraith]

oh i see

like just by the recursive definition of Zi+1(G) it has to be in G/Zi(G)

and same for Zi(H)

[thorn jay]

if you pull back Z(G/Z(G)) through the projection p : G -> G/Z(G) then you exactly get the second term in the upper central series of G

in general, if one has a normal subgroup N such that both N and G/N are nilpotent, then G is nilpotent. You do this by pulling back a central series of G/N through the projection p : G -> G/N, and combining that with a central series of N I was thinking of solvable groups

[rocky cloak]

This is not true.

Take G = S3 for example

[thorn jay]

no? I was thinking of solvable groups then lol

[rocky cloak]

It's true for solvable groups yeah

The problem is that even if
[G, G] = N, the fact that N is nilpotent tells you nothing about [G, N]

[thorn jay]

I see

of course

[rocky cloak]

If course if N was the center of G, then [G, Z(G)] = 1 and all is well

[thorn jay]

[swift tundra]

I am looking at the last part of 11 which is asking for a map A^m—>A^n injective with m strictly larger than n.

One idea I had is to take sequences in the Integers indexed by the natural numbers. Then you can construct a map A^2–>A by defining the sequence in A to alternate between terms in the first and second component of A^2. The problem is that this looks like an isomorphism to me, but then 2 would equal 1 which is absurd. Is the function I defined not a homomorphism?

[rocky cloak]

So I'm not sure I understand your description exactly, but not that the map is supposed to be a homomorphism, not just a function

A homomorphism from A^2 can be described by where each of the two basis vectors is mapped

So maybe you can describe where your map sends those

[kind temple]

they are saying to send the pair
((x1,x2,…),(y1,y2,…)) to (x1,y1,x2,y2,…)

[rocky cloak]

Okay, so A = Z^N

Then sending
((0, 1, 0, ..), (0, 0, ...)) to (0, 0, 1, ...) is not a homomorphism

Because of e = (0, 1, 0, ...)
then f(e* (e, 0)) = f(e, 0), but e*f(e, 0) = 0

[kind temple]

right. this is a bijection, however

[swift tundra]

Hmmm well this shows it’s not a homomorphism of A-modules, but is it not a homomorphism of rings?

[rocky cloak]

It is a homomorphism of rings, but that's not really the question

[swift tundra]

Oh of course haha. Thank you!

[white oxide]

Could I have a hint for 8? The ideals are prime since quotienting by k[x_1, \dots, x_n] gives an integral domain. For the primary powers, I was thinking about proceeding by induction. Something like, assume that p = (x_1, ..., x_m)^k is primary for all k. Then p[x_{m + 1}^k] = (x_1, ..., x_{m + 1})^k is primary by Exercise 7 (I don't know if this equality holds and if it applies, but it's the first thing my mind went to).

[south patrol]

I would again consider the quotient, using the criterion that "primary" = "in the quotient, every zero divisor is nilpotent"

[white oxide]

Okay I'll try that ty

What is k[x_1, \dots, x_n]/(x_1, \dots, x_m)^k isomorphic to

[south patrol]

I don't think there is any easier way to write it

In particular as it is different to just quotienting by the x_i^k

But you can explicitly describe the zero divisors

[mental lake]

Show that every group G of order 12 can be written as a
semidirect product of Sylow subgroups (you will need an additional argument
to show that you cannot have ‘many’ 2- and 3-Sylow
subgroups at the same time).

how do you do this because if u want to use the 3rd sylow theorem you wouldnt get that the sylow group of order 2 nor 3 would be normal

and they need to be normal to split it up into a semidirect product

so what do i do

ping me if u can help

[velvet hull]

think about what might happen if there was no normal sylow subgroup

[mental lake]

then you dont get the semidirect product

[velvet hull]

dont look so far ahead yet

I claim that that is actually impossible, a group of order 12 always has a normal sylow subgroup

[mental lake]

but if you use the 3rd sylow theorem youd get that n2 | 3 and n2 = 1 mod 2

and for n3 | 4 and n3 = 1mod 3

so n2 is either 1 or 3

and n3 is either 1 or 4

and for it to be normal you want it to be 1

[proud vigil]

have you seen counting arguments before for proving that groups are not simple

[mental lake]

because then its unique

[velvet hull]

so let's assume none of them are normal

so n2 = 3 and n3 = 4

is there anything you can do with this information?

[mental lake]

then u have 4 sylow 3 sub groups

and 3 sylow 2 sub groups but idk what to do w that

[proud vigil]

distinct prime sized subgroups intersect trivially

since every element of one generates it

(Besides the identity)

so if there were 4 sylow 3 subgroups, how many distinct elements would there be

[mental lake]

8

?

@proud vigil

so what then

[delicate orchid]

You have 8 elements of order 3. There are assumed to be 3 Sylow 2-subgroups of order 4, thus you have at least 6 elements of order 2 or 4

See if you can find a contradiction from here

[mental lake]

so we cant form 12 elements if wed have 3 sylow 2 subgroups? unless it would be one sylow 2 subgroup

so wed have 8+3+1?

[delicate orchid]

yeah one Sylow 2-subgroup could theoretically work

in fact it does

[mental lake]

well isnt that the contradiction then, if wed have 3 subgroups it wouldnt be possible considering the elements that should sum up to 12

or is there something i overlooked

[delicate orchid]

yeah that's what I'm saying, one sylow 2 would work but 3 wouldn't

[mental lake]

oki

ty

[fiery dirge]

https://discord.com/channels/268882317391429632/1430590936903192749
If anyone can help regarding fields I'd be really thankful! I didn't want to post here since it's maybe a bit too lengthy for this channel and would steer the needed attention away from users that also need help.

[delicate orchid]

I'm kind of embarrassed I saw (C_2^2) \rtimes C_3 before C_12. Am I cooked?

[elfin wraith]

If it makes you feel any better I was just typing up a whole question asking why a sequence splits only to realise C_n was a chain complex, not the nth cyclic group

[delicate orchid]

the one downside to the C_n notation. I'm moving to just "n" now

(2+2):3 is way cleaner than C_2^2 \rtimes C_3 anyway

[elfin wraith]

Z/nZ - {ring structure}

Not all of it, just the multiplicative monoid part

sets arent that useful

[elfin wraith]

Turns out im still stuck even after learning to read. Ive proven the hint, thats fine, just first iso + exactness (I guess its all just exactness) so we know that C_n \cong K_n (+) Im delta. But now i feel like im just being dumb and I should basically be done, but im not seeing it

Are we just considering the kernel and image at each element of C_n? whats with the at most 2 non-zero terms part? Im guessing something to do with d^2=0? Also im not using that theyre free abelian which feels wrong

[delicate orchid]

I might be stupid cause I don't know what this means either

[astral ivy]

How do I read this? ord(g) = lcm(ord(g1​),ord(g2​),ord(g3​))??

order of element g is equal to lcm of orders of 3 other elements? it doesnt make sense

[elfin wraith]

This is part of the Hatcher experience

[delicate orchid]

I think the 2 non-zero terms part has to do with constructing a free resolution of the kernel as a subcomplex of the free complex (which you can always do with length 2 as it's coming from a chain complex of fere modules), then you interlace these to give the entire complex again (using the fact each term is free to ensure you do indeed get the same complex)?

[elfin wraith]

I would ignore it but he uses it for the universal coeffcicent theorem stuff

[elfin wraith]

[thorn jay]

this looks like a wrongly stated variant of the fact that a long exact sequence is just a bunch of SES woven together

but I'm not sure if that's related

[delicate orchid]

this is what I'm getting at I think

that's the picture I had in my head when I said "interlace"

[thorn jay]

(wrongly stated because it would be a decomposition into SES, not 0 -> A -> B -> 0

[delicate orchid]

this is where we'd use freeness I'd imagine

subgroup of free is free so the kernel is free

so 0 -> K -> K -> 0 is a free resolution

nah that doesn't seem like the right approach

[thorn jay]

lol

ah wait no I think I see what its doing

[elfin wraith]

Aha I forgot groups are nicer than rings

[thorn jay]

the hint here is helpful

[elfin wraith]

I would hope so

[thorn jay]

because of freeness you can write each C_n = im d_n (+) ker d_n

so take the subcomplexes 0 -> im d_n-1 -> ker d_n -> 0

[elfin wraith]

Why does this follow from freeness?

[thorn jay]

sequence 0 -> ker d_n -> C_n -> im d_n -> 0 splits

[thorn jay]

this is a subcomplex exactly because d_n d_n-1 = 0

wait no that's not right

[elfin wraith]

Is that just like rank nullity?

[thorn jay]

yeah I think?

maybe?

[delicate orchid]

trivial

it's Z^n yunc

yeah it's rank nullity nvm u got it

https://tenor.com/view/bad-ass-uh-oh-ndt-neil-neil-degrass-tyson-gif-14802463180959376587

[tardy hedge]

morgan freeman true

[elfin wraith]

Ok maybe this is dumb but Z is a PID so free = projective, we know projective iff it splits as ker + im is kinda how I got there lmao

not equal

[delicate orchid]

I was gonna say that for about 500 milliseconds

[elfin wraith]

but ya know implies

Ok lemme think then

[thorn jay]

this is the gist ig

i wrote a cochain complex ignore that, doing some cohom shenanigans

[tardy hedge]

FINGER REVEAL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[gray parcel]

Okay this might be a bit of a low-level question for this channel, but is there any group-theoretic significance for row equivalence? Like if I were to take the space of matrices of a certain dimension and then quotient by row equivalence, what do I get?

[thorn jay]

I guess it's nothing more than the cosets of an action on the set of matrices

take the group of permutations on the set of matrices generated by elementary row operations, then the orbits partitition the set of matrices, and you take that quotient

[tardy hedge]

what does row equivalence mean?

[gray parcel]

Equivalent up to row operations

Like row swaps, linear combinations of rows, and scalar multiplication of rows.

[thorn jay]

(equivalent to equal kernel when treated as a linear map apparently)

[fiery dirge]

what is the root field and how do I find it? For example:

$a = \sqrt{2 + \sqrt{5}}$. $a^4 - 4a^2 - 1 = 0$, can't be reduced in $Q$ since all of the roots are in the form $2 +- \sqrt{5}$ and $\sqrt{5}\notin Q$.

[cloud walrus]

HMD

[vapid vale]

two matrices A,B of dimension mxn are row equivalent iff A = MB for invertible M. so this is the "orbit space" of GLm acting on your matrices by left multiplication

[gray parcel]

Nothing particularly interesting, then.

[thorn jay]

¯_(ツ)_/¯

[gray parcel]

Gotcha

[vapid vale]

i mean there are mildly interesting consequences, like the grassmanian Gr(r, n) is isomorphic to these orbit spaces at each fixed rank r

well not really a consequence just an interesting fact

[elfin wraith]

Ok yeah I had to rewrite this and think about it for a bit but I agree, thank you

[thorn jay]

ofc!!

[rapid cave]

are you asking what is $\bQ(\alpha)$?

[cloud walrus]

ExpertEsquieESQUIE

[fiery dirge]

I think that's just expansion of Q by alpha

[elfin wraith]

Yeah, its the smallest field containing Q and alpha

[fiery dirge]

my question is finding the 'root' field such that it contains the roots of polynomial which is a

[rapid cave]

so the splitting field of x^4-4x^2 - 1?

[fiery dirge]

yes sorry I don't know the term in english

so that will have maximum of 4 solutions and we need to expand one field to fit them

[rapid cave]

you already know the roots, the splitting field will contain all of them

[fiery dirge]

but I also need to find the degree of expansion [K : Q] where K is that splitting field

I already got that [Q(a) : Q] = 4

The tricky thing is I didn't study either Galois or Sylow so it's difficult

this course is very weird, it doesn't cover rings or fields completely but rather only does parts of them so I can't see the whole picture

[rapid cave]

so you know the roots are $\pm \sqrt{2 \pm \sqrt{5}}$, so the splitting field is $$K = \bQ(\sqrt{2 + \sqrt{5}}, \sqrt{2 - \sqrt{5}})$$

[cloud walrus]

ExpertEsquieESQUIE

[white oxide]

Hmmm.. so if f is a zero divisor in A = k[x_1, ..., x_n]/(x_1, ... x_m)^k, f belongs to r(Ann(g)) for some g in A, so f^n g in (x_1, ..., x_m)^k for some n...

[velvet hull]

by the result of question 7, it suffices to prove that (x1,...,xn)^k is primary in k[x1,...,xn]

but then that is straightforward

[white oxide]

sorry what lol

isn't that the question itself

oh nvm I see the difference

[velvet hull]

yep

[white oxide]

you're doing it for all the variables

[velvet hull]

yes

use 7i, 7iii and induction

[velvet hull]

(recall that ||powers of maximal ideals are always primary||)

[sacred wharf]

7i, -7i ?

[fiery dirge]

Yes that makes perfect sense

but what about [K : Q]? I am unfamiliar with this notation and don't know what happens, in case of [Q(a), Q] I was just looking at the highest degree of the minimal polynomial so that was straight forward

maybe [K : Q] = [K : Q(a)][Q(a) : Q] and we already know degree of expansion of Q(a) is 4 so we just need "how much K expands further in comparison to Q(a)"

and as you wrote K = Q(a, b) where a and b are the roots of minimal polynomial so maybe the answer is K expands by 2 in comparison to Q(a)? So the end result will be 2*4 = 8

[velvet hull]

If E/F is a field extension, then [E:F], called the degree of the extension, is the dimension of E as an F-vector space

[rapid cave]

[K:Q(a)] is indeed 2

Because K=Q(sqrt(2+sqrt(5)), i)

[vivid tiger]

(it should remind you of the index)

[vocal pebble]

Why? Both sides of the equation are numbers and that equality is possible. Take G = (Z/2Z)^3, then ord(1,1,1)=LCM(ord(1,0,0),ord(0,1,0),(0,0,1))

[tardy hedge]

the amount of reacts to this comment is not justified

[sacred wharf]

Fr it was a shitpost

[tardy hedge]

lmao

[sacred wharf]

U reacted the most tho

[tardy hedge]

hehe

[sacred wharf]

Definition of prime ideal is:

if AB, where A,B are ideals, belongs to P
A belongs to P
Or B belongs to P

?

[tough raven]

(And P is proper,) sure.

[elfin wraith]

Yes, and if your ring is commutative you can relax this to ab in P iff a is in P or b is in P, I.e. it holds at the level of elements

[sacred wharf]

let me try and prove it

[elfin wraith]

And just as a fun fact, if an ideal of a noncom ring satisfies the commutative definition, it’s said to be completely prime

[knotty frigate]

Lets assume we have a ring $(R,+,0, * ,1)$\
Now lets assume that we have $c:R->R$ such that:\
$c(c(x))=x$ and $c(x * y)=c(y) * c(x)$. Can we prove that c(0)=0 or even that c(0) is 0 or 1?\
I assumed that was the case by using the cancellation law for multiplication but thats not true for rings so there probably is a ring with this kind of involution that disproves this. But can we construct a finite one where c(0) isnt 0 and isnt 1?

[cloud walrus]

Pymamba

[rocky cloak]

So completely prime doesn't imply prime? That's an interesting choice of terminology...

[rocky cloak]

Let x=0 and y=c(0), then
c(0) = c(xy) = c(y) c(x) = 0 c(x) = 0

[elfin wraith]

I think it does for two sided ideals, but yeah probably not the best overall

[rocky cloak]

I guess more generally you can notice that c is an isomorphism of monoids (between R and R^op) and 0 is the unique absorbent element, so is preserved

[knotty frigate]

Oh yeah generalizations of the complex conjugate to rings has smth to do with opposite. wikipedia did mention that

[rocky cloak]

I mean, the relation
c(xy) = c(y) c(x)
reverses the order of multiplication

That's the relationship with the opposite ring

[knotty frigate]

1 * x = x * 1 = x
c(x) = c(x) * 1 = c(x) * c(1) => c(x)[1-c(1)]=0
c(x)=0 or c(1)=1.
So if we have one x in R such that c(x)=\=0

[rocky cloak]

I mean
1 = c(c(1)) = c( 1 * c(1)) = 1 * c(1) = c(1)

[knotty frigate]

oh yeah lmao

[sacred wharf]

how does one motivate intuitively why we look at F? (Artinian ring nil left ideal implies nilpotent

[rocky cloak]

More generally c will be a bijection, so only 0 can have c(0) = 0

[sacred wharf]

It feels like so many fo these proofs at the end of my 2nd course in algebra are just plucked from the sky

[knotty frigate]

oh yeah cuz its an involution

sry im really tired

[rocky cloak]

Well your goal is proving B is 0. And you want to create some chain of ideal related to that.

Now you know that BB = B and you also know that B annihilates simple modules. So B must be killing some parts of itself, but what is the smallest part that isn't killed. Well make a chain of the stuff that isn't killed and use the artinian property.

[sacred wharf]

Annihalte means BX = 0 right

[rocky cloak]

So I'm not saying this is something that would be easy to come up with for yourself, but it is a natural thing to consider

[sacred wharf]

in my first course in algebra and beginning of my 2nd it felt like i could come up with the proofs for a lot of the theorems on my own, now it just feels impossible

[rocky cloak]

Who knows how long it took whoever first made this proof to come up with it

Sometimes a lot of trial and error lies before the stroke of genius

[sacred wharf]

Why does B annihilate simple modules

[quiet pelican]

B is a subset of the Jacobson radical, which by definition is a subset of any maximal ideal
And simple modules are exactly the quotients of the ring by maximal ideals

[sacred wharf]

idk what any of that means

[rocky cloak]

So I don't know which definition you're working with, but a common definition for J is all z such that zS = 0 for all simple modules S

[sacred wharf]

oh shit sorry I should clarify, J is just any nil left ideal

[rocky cloak]

Allright, then notice if
zS is nonzero then
RzS = S
so for each s in S there is an r such that
rzs = s
But then
s = (rz)^n s = 0
contradiction

So a nil ideal annihilates all simple modules.

Anyway, the point was mainly that nil ideals annihilate a lot of stuff, but clearly if B^2 = B and B is nonzero then there's also stuff that isn't annihilated, so focus on that

[sacred wharf]

hm so z is the nil ideal and S is the simple module

[rocky cloak]

Well z is an element of the nil ideal

And yes S is a simple module

[errant quartz]

@wraith cargo The proof I've seen (for Sylow 1) goes like this (I'm writing this from memory to make it exceptionally clear where I might be making mistakes). Also, I'm writing it out step-by-step to more clearly explain where my confusion comes from.

Let $G$ be a group, and let $p^n | |G|$ for prime $p$ and positive integer $n$. Then there exists a Sylow $p$-subgroup of $G$, defined as a subgroup of $|G|$ with order $p^m$, with $m$ being the maximal integer such that $p^m | |G|$.

1. Let $X$ be the set of subsets of $G$ with size $p^m$. Then we define a group action on $X$ by $g \cdot A = gA$ for $A \in X$. Note that the size of $X$ is ${|G| \choose p^m}$.
2. Note that $p^m$ therefore does not divide $|X|$. This is because all factors of $p^m$ are factored out of $|G|$ during this "n choose k" operation, and the unique prime factorization of $|G|$ prevents $p^m$ from being a factor of the size of this new set.
3. Since the orbits of members of $X$ partition $X$ and $p^m \nmid |X|$, there must be at least one member of $X$ with an orbit whose size is not divisible by $p^m$. Let us denote this set by $A$.
4. The stabilizer of $A$ is a subgroup whose order is $\frac{|G|}{|O(A)|}$, where $O(A)$ is the orbit of $A$. Since $p^m \nmid |O(A)|$ but $p^m | |G|$, it is clear that $p^m | |stab(A)|$. But $|stab(A)| \leq p^m$ because $|stab(A)| \subseteq A$, so $|stab(A)|=p^m$, and thus $stab(A)$ is a Sylow $p$-subgroup of $G$.

Off the top of my head, I don't know where the idea to focus on divisibility comes from. I know this will vaguely classify the size of different orbits, but I don't know why a subgroup would come out of this process, or why that subgroup would specifically be the stabilizer of a member of $X$. In other words, the jump from 1 to 2 and then 3 to 4 is confusing (though I understand how to get from 2 to 3).

[cloud walrus]

muffin hamster

[noble nexus]

this is the unhinged proof

its simultaneously genius and awful

the idea of acting on all subsets of size p^m is just so crazy

[errant quartz]

Upon further reflection, the jump from 3 to 4 makes more sense to me now

But caring about divisibility feels like it comes out of left field

And to some extent the idea of using a group action here feels mysterious to me

[noble nexus]

I think divisibility is actually very natural, though I'll admit at first it seems weird

because basically the defining feature of a p-group is the divisibility of its order

acting on subsets is definitely a bit mysterious

but all proofs of sylow (to my knowledge) rely on the fundamental fact that a p-group acting on a set with number of elements not divisible by p must have a fixed point

which is imo the really important takeaway

[quiet pelican]

Group actions are how you prove a lot of stuff in group theory
The specific group action is definitely an interesting choice though
Although it kinda makes sense that a “sufficiently nice” p^m size subset should have orbit of size |G|/p^m (and so stabiliser of size p^m)
So you check one exists

[thorn jay]

mathematicians love redundancy

don't know what to pick? just pick all of them!

[sacred wharf]

to prove first first sylow, using induction is my fav proof

[sacred wharf]

axiom of choice moment

[rocky cloak]

If we're just taking about the first theorem you can prove it using the class equation and induction:

|G| = |Z(G)| + sum [G:C(x)]

If |Z(G)| is a multiple of p, then it has an element x of order p (use for example classification of finite abelian groups). Then G/(x) is a smaller group, so our induction applies.

If |Z(G)| is not divisible by p, then one of the [G:C(x)] must not be either, and then C(x) is a smaller group with the same power of p dividing it, so you can apply induction.

[errant quartz]

I might think more about this proof since it seems more intuitive to me

[sacred wharf]

yesi think u can also go and use this method to prove that it has a group of all orders of powers of p iirc

[noble nexus]

Yeah this is nice, but also the proof of the class equation uses the p-group action result

[rocky cloak]

I wouldn't say the proof of the class equation uses the p-group result, or anything about p-groups.

But both are just about counting sizes of orbits for a group action.

[thorn jay]

this is so clean!

[rocky cloak]

Like the proof of the class equation is just, what are the orbit of G acting on itself by conjugation + orbit stabilizer

While the p-group result is: orbits of a p-group action is either trivial or a multiple of p

[quiet pelican]

The proof you gave needs Cauchy, which is the same sort of “pick a random action that happens to work” bs
But I’m not sure you can avoid that so
Actually no isn’t there an easier proof in the abelian case (inb4 classification of fg abelian groups)

[rocky cloak]

They proof only needs Cauchy for abelian groups, which you can prove without any group actions

Through for example the classification of finite abelian groups

Though that result is more complicated to prove

[noble nexus]

yes that's what I mean

[quiet pelican]

Called it
I’m not sure this is that much more satisfying than Cauchy

[noble nexus]

I guess it's not specific to p-groups but it's morally the same

[rocky cloak]

You can prove it much easier actually.

Say A is abelian, pick an element x. If the order of x is a multiple of p you can make an element of order p. Otherwise A/(x) is smaller, so induction

[sacred wharf]

I guess u teach algebra course at ntnu

[thorn jay]

shadow wizard finite group gang
we love casting induction

[rocky cloak]

I have done so, I'm not currently doing it

So anyway, no random group action, though the class equation itself is sort of a "random" group action (not that random though)

[quiet pelican]

Class equation is conjugation which is like
Arguably the most natural action

[sacred wharf]

hv u taught sheddow ?

[sacred wharf]

Acting on cosets >>

[rocky cloak]

Well I don't know sheddows irl identity, so idk

I'm guessing no just from the time frame...

But hard to say

[sacred wharf]

Damn you guys have probably seen each other irl without talking but spoken many times here

[thorn jay]

the conjugation action has more interesting orbits

[sacred wharf]

Yeah ik i was just playing devils advocate

My teacher thinks its so important that after teaching orbit stabilizer and stuff she reproves the exact same stuff for conjugation action

[quiet pelican]

Death penalty

[rocky cloak]

Proving Sylow 2nd and 3rd will definitely require some group action stuff, but I guess here the actions are quite natural.

Acting on G/P and acting on the conjugates of P by conjugation

[sacred wharf]

Ccl(K) and u pick any other group of same order and have that act on ccl(k)

Very neat because u can use this proof to trivially get third sylow

[rocky cloak]

What's ccl?

[sacred wharf]

Set of all gKg^-1

Conjugacy class of K

[rocky cloak]

Makes sense

I'm not sure I'm following the proof you're suggesting though

[sacred wharf]

So

The idea is you pick 2 group of order p^m

And show they must be conjugate

First group is K

Next is H

You have H act on ccl(K)

And then u can use orbit stabilizer

And idea is you show H is in ccl(K)

[rocky cloak]

Well, I'm still not sure I follow

[sacred wharf]

So

I Ccl_H(P) I = I H : N_G(P) n H I
Where P is in ccl(K) right

[rocky cloak]

Yeah that seems right

[sacred wharf]

Yeah u r right im skipping some details which i typically remember as i reach this stage
so one detail is of course ccl(K) not divisible by p

To see this just reuse orbit stabilizer

[rocky cloak]

I see, so you get a conjugate of K such that HP = PH, which must mean H is in P by maximality

Nice.
The proof I know for the 2nd goes like
If H acts on G/K by left multiplication, then the fixed points of the action are exactly gK such that
g^-1 H g is a subgroup of K.

Now if H is a p-group and G/K is relatively prime to p, the action must have a fixed point.

[sacred wharf]

Fixed point is Gx = x right

Neat

[lofty storm]

Hi, im trying to use the distinct conjugacy class sizes to show that they cannot add up to a specific number in S4, but im having difficulty proving that the conjugacy classes are restricted by the order of the permutations? could anybody possibly help?

[wraith cargo]

Wdym "restricted by the order of the permutation"?

Do you mean like bounded by the order of the permutation?

[lofty storm]

yeah

like 3 cycles all being a conjugacy class

Do i have to use the class equation or smth?

[wraith cargo]

Uh you can

But hold on I'm thinking

[rapid cave]

conjugacy classes in Sn are all permutations with the same cycle-structure

so all 3-cycles form a conjugacy class

[lofty storm]

oh yeah i just found a proof in the lecture notes about it

thanks for your help tho

[rocky cloak]

There are two distinct conjugacy classes of elements of order 2 though

[lofty storm]

Two permutations in Sn are conjugates if and only if they have
the same cycle type.

this is the proof i found in the notes

[vivid tiger]

So, the way to remember is to know that conjugating a permutation is just permuting the numbers

Like it's easy to see in notation

[lofty storm]

Does anyone have any good books / yt playlists to learn group rings and fields my lecturer is really bad

[karmic moat]

borcherds has good youtube videos

dummit foote is a standard textbook

[vivid tiger]

sigma = (2 3 4 1)
tau = (3 4 1 2)
so look at tau sigma tau^{-1}
where does 1 go? to 4. where does 2 go? etc. you will get (4 1 2 3)

The point is like

It's a basis transformation

(i'm assuming you know linear algebra)

[lofty storm]

thanks i'll have a look at them

[karmic moat]

gathmann's commutative algebra notes are nice https://agag-gathmann.math.rptu.de/class/commalg-2013/commalg-2013.pdf

it has a more geometric flavor

[vivid tiger]

Actually you can be more precise about this I think. You have a homomorphism from S_n to GL(R^{n})

given by permuting the axes

So now, we literally have basis transformation

[delicate orchid]

HMM... INDEED....

[vivid tiger]

Yeah I noticed even when I first learned it that it was like it, but now I realize it's literally it

[karmic moat]

wew have you heard of F_1 geometry

[delicate orchid]

changed my name for ts joke I hope it's appreciated

[karmic moat]

aka "absolute algebraic geometry"

[delicate orchid]

I have but I like F_1 from the combinatorial standpoint

[karmic moat]

woah

what does that look like

[delicate orchid]

Schur-Weyl duality on steroids

[wraith cargo]

This exists?????

[delicate orchid]

take any formula you want for GL(n, q) and put q = 1 in and it'll be the formula for S_n

[karmic moat]

the future is now

[delicate orchid]

specifically I like that the Sylow structures of GL(n, p^k) at p are just S_{p^k}

[wraith cargo]

This is really cool

[delicate orchid]

I'm trying to find the source I used to have that made it really clear one sec

[thorn jay]

I remember you told me about this a while ago

ts is crazy

[delicate orchid]

can't find it :(

[karmic moat]

did you have fun getting my hopes up

[sacred wharf]

Dont use it

4/10

Tbh i just use my lecturers notes now instead

[thorn jay]

the Hall-Senior genus is such a random condition lmao

[delicate orchid]

I can remember the $S_n$ side, the Sylow $p$-subgroup of $S_n$ is given by $\prod_{k=1}^{\nu_p(n)} C_p^{\wr a_k}$ where $n = \sum_{k=1}^{\nu_p(n)} a_kp^k$. The $GL(n, q)$ looks similar but with $S_n$s in the wreaths

[cloud walrus]

FIELD WITH ONE ELEMENT LOVER

[sacred wharf]

I recommend just trying 2-3 books reading a subchapter each and going on what u like

[delicate orchid]

and some random bullshit at the start of the product . Point is that it all vanishes when q = 1
and you're left with just the big product of wreath powers

[sacred wharf]

What field has one element i guess just zero

[karmic moat]

#book-recommendations message

[delicate orchid]

https://ncatlab.org/nlab/show/field+with+one+element

[karmic moat]

i have no idea what a wreath is tbh ive seen it before and got kinda scared

seen it in passing

[delicate orchid]

en passant

[thorn jay]

wew do you know about this / have worked with this?

[karmic moat]

too many damn products to keep track of these days

[thorn jay]

wreath is interesting cuz apparently every extension of G by H is contained in some wreath product involving G and H

[delicate orchid]

wreath powers of C_p have a nice interpretation as the automorphisms of a rooted p-ary tree that preserves the height of each node

https://mathoverflow.net/questions/164808/sylow-3-subgroups-of-symmetric-group

see the second answer for a case where p = 3

[karmic moat]

oh interesting

[delicate orchid]

can't say I have 💔

but Hall's name has activated my neurons

[thorn jay]

https://groupprops.subwiki.org/wiki/Groups_having_the_same_Hall-Senior_genus

[karmic moat]

i wonder if wreath products show up in geometry

[delicate orchid]

I'm reading ts now lol

[thorn jay]

wikipedia doesn't even have a page

[karmic moat]

well i guess there's my answer

[delicate orchid]

the whole isoclinic thing has never really made sense to me

[thorn jay]

like the lattice isomorphism thing I can, like, see somewhat

[delicate orchid]

wreath products are easy if you know semidirects - do you know semidirects?

[karmic moat]

yeah

[thorn jay]

the lion concerns himself with lattices of normal subgroups

[delicate orchid]

I mean the lattice of normal subgroups is probably the most intuitive graph I've heard associated to a group

[karmic moat]

they show up in levi decompositions

well actually i guess they show up everywhere

the borel is the semidirect of the unipotent and the maximal torus

[delicate orchid]

the wreath product $G \wr H$ of two finite groups (along with the data of a map $H \rightarrow S_n$ for some $n$) is simply $G^n \rtimes H$ with $H$ permuting the factors of $G$ in $G^n$

[rocky cloak]

No one has ever seen the field with one element, but it leaves traces of its existence.

[cloud walrus]

FIELD WITH ONE ELEMENT LOVER

[delicate orchid]

for example, D_8 is isomorphic to C_2 \wr C_2 (or S_2 \wr S_2 if you prefer)

[karmic moat]

oh interesting

not that bad then

[noble nexus]

wreath products are honestly some of the most intuitive examples of semidirect products

also the greatest group https://en.wikipedia.org/wiki/Lamplighter_group is a wreath product

[karmic moat]

does the S_n here come from, say, Cayley's theorem

[delicate orchid]

like you've got the involution graph, the Brown lattice etc. etc. all really weird things

[delicate orchid]

it can do but it doesn't have to. Your map doesn't even have to be injective

if you pick the trivial map H -> 1 -> S_n you just get G^n x H

[karmic moat]

I see

within the last hour i learned two new things then

wreath product and that the ppl who light streetlamps are called lamplighters

[delicate orchid]

In the category of groups, straight up lightin' it... and by "it" lets just say.... my lamps...

[karmic moat]

BOOOOOO 🍅

[delicate orchid]

BOOOOOOOOOOOOOOOOO

[karmic moat]

virgin "-1" vs chad +18 answer

[thorn jay]

where's this even used? It seems cool though

[thorn jay]

struggling to see how that isn't a research question

like it could very well come up in research to want the sylow-3 subgroups of S_n

[karmic moat]

yeah

[delicate orchid]

I think it's moreso that it's not too hard to figure out the structure in general

[thorn jay]

but why do the work if you can see if the work's been done for you

[karmic moat]

real shit. that's why i google the solutions to every exercise after 5 seconds of thinking

just kidding

[thorn jay]

people who use LLMs for their homework:

[delicate orchid]

I went insaneballs on the combinatorics of this Sylow subgroup back in my first year phd

good times

[thorn jay]

why would you ever want to touch combinatorics 💔

[delicate orchid]

algebra is combinatorics you coward

[thorn jay]

category theory sometimes has a combinatorics-y flavor which I enjoy ig

[delicate orchid]

it's..... yummerss......

[thorn jay]

[elfin wraith]

This is just…

[karmic moat]

didn't richard feynman say something like "if you can't explain something to a ≤10 year old, then you dont understand it"

≤10 because i dont know the exact number

[elfin wraith]

Your index is like half the length of the equation

[karmic moat]

do you think he would change his mind if he saw this kind of shit

[delicate orchid]

I worked on that formula for 14 hours straight from 1pm to 3am and started violently shaking when it worked

good times

[thorn jay]

richard feynman is famous for his wise words and takes

[karmic moat]

true. i also hear einstein treated his wife with grace and love

[delicate orchid]

I could explain this. It follows directly from the structure of iterated wreaths along with Mackey's irreducibility criterion (which is taught in all good first schools I presume)

[karmic moat]

oh well duh

i mean you dont have to explain it to me i mean it's right there i could see it clear as day

[quiet pelican]

You know I don’t think you could explain QM to a <= 10 year old (barring very extreme exceptions)
Like you can give a vibe which is pseudoaccurate
But that’s not explaining

[rocky cloak]

The trick is knowing the right 10yos

[thorn jay]

thisis a stupid fuckign quote anyways lol

[noble nexus]

its morally true

but of course 10 year old is not really accurate in general

[karmic moat]

read it and weep

[elfin wraith]

It’s possibly true in physics where you can, if you dumb shit down sufficiently far, relate it to something the kid can picture

But like, how are you going to explain natural transformations to a 10 year old

[delicate orchid]

draw a homotopy

[karmic moat]

tbh sometimes i have to look up the definition of natural transformation

[delicate orchid]

it's a square yunc 💔

[noble nexus]

for abstract math I would say if you can explain it to a curious first year undergrad

[thorn jay]

the quote is bad because a different notion of "explain" and "understand" is used than the one most people would use

[noble nexus]

that's the hack for boosting your understanding btw

[karmic moat]

oh yeah it is a square i forgot it's a square

[noble nexus]

find some undergrads and start yapping

[quiet pelican]

I can explain groups to a 5 year old if you let me say symmetry enough /hj

[delicate orchid]

SIKE! It's actually a 2-disk

[elfin wraith]

I think I could explain groups to anyone who understands multiplication

So maybe not 5 year olds but like 7 year olds

I honestly don’t really know what kids know when

[quiet pelican]

No its /nat\alpha FGfAB

[rocky cloak]

I can explain most things to a 5 year old, just some of the things might take me around 20 years to explain

[delicate orchid]

qwertyuiop to you too!

[elfin wraith]

I can explain anything I know to a 5 year old, provided they’re not required to understand it

[karmic moat]

azerty layout users won't get this joke

[delicate orchid]

it actually pmo that people do higher categories simplically when the cubic and globular interpretations are much easier

pmo...

[thorn jay]

yes but

but

simplices

<3

[delicate orchid]

azerty layout users don't get many things, like "invited to parties"

[karmic moat]

maybe what feynman should've added was that the ≤10 year old should engage in a discourse with you relevant to what you're explaining

i thought a lot of the nordic countries use azerty

[thorn jay]

the <= 10 year old should not wish to be able to use the information and reason with it in a meaningful way

[delicate orchid]

maybe what feynman should've done instead is not be a HACK FRAUD

[delicate orchid]

u ever been to a party in Reykjavik? didn't think so

[karmic moat]

just got back from one last night actually

[thorn jay]

have YOU?

[delicate orchid]

no further questions at this time

[thorn jay]

bro is closing his office

[quiet pelican]

(Ignore the claimed error, editor doesn’t realise \nat induces displaymode because my commutative diagram shortcuts are all designed to be auto-displaymode, it compiles fine)

[delicate orchid]

anyway back to the field with one element, specific representation theory over it

[delicate orchid]

https://tenor.com/view/old-man-its-peak-its-way-too-peak-its-honestly-more-than-peak-meme-gif-14944785001351870503

[karmic moat]

can you send the macro please

[delicate orchid]

very epic but I like customising my squares in quiver.io

[karmic moat]

^

[elfin wraith]

https://tikzcd.yichuanshen.de on top

Fuck quiver

[thorn jay]

I just type it directly into latex

you all are WEAK

[elfin wraith]

Me and all my homies hate quiver

[quiet pelican]

Yeah so do I but if I’m livetexing, really common stuff like SESs and naturality squares get inline shortcuts

[delicate orchid]

really common stuff like SESs
I wish I was a topologist 💔

[karmic moat]

what's wrong with quiver quiver is awesome

[elfin wraith]

TeX me right now the LES in cellular homology

[karmic moat]

the thing is that even if i set up all these things i end up just forgetting about them

[elfin wraith]

It’s a bloated rip off hack version of the one I linked

[quiet pelican]

It takes longer than a 5 or 6 argument command

[thorn jay]

I don't even know cellular homology

[quiet pelican]

And oh yeah I forgot about \doublefunc which I have in my cat theory notes for writing parallel pairs

[rocky cloak]

I mostly just type them directly too, feels like a hazzle opening up some secondary program, and then suddenly I want the arrows to bend in some way then I would need to read the unreadable thing quiver produces.

[elfin wraith]

What are they even teaching these students in their first sem of UG anymore

[karmic moat]

:(

[delicate orchid]

I have a macro \marhrm that just replaces it with \mathrm automatically because I make that typo so often

[karmic moat]

W

[quiet pelican]

I’ve considered doing this for \codt (\cdot)

[karmic moat]

i have a macro \cat that's just \mathrm

[delicate orchid]

this mogs quiver cause the curved arrows just work

[karmic moat]

in case i ever decide to change how i want my categories to look

[delicate orchid]

I'm switching

[thorn jay]

programming 1 💔

[delicate orchid]

I don't use mathrm for categories lol

I use "whatever I damn well please". A little mathfrak here a little textbf here...

[karmic moat]

wew the kinda guy to use sans serif...

[elfin wraith]

I genuinely don’t like quiver, I’m 99% sure the one I linked came out first and quiver copied, but it’s just more bloated and shit

[elfin wraith]

If I had my way the use of mathfrak would be punishable by death

[karmic moat]

i think every other day i change my mind between whether i want sheaves to be mathcal or mathscr

[delicate orchid]

$\mathfrak{S}$wag

[karmic moat]

but i always leave O to be mathscr O

[cloud walrus]

FIELD WITH ONE ELEMENT LOVER

[karmic moat]

because it looks cool

mathcal O looks stupid

[delicate orchid]

NO IT DOESN'T

[karmic moat]

yes it does

[delicate orchid]

unhinged take

[thorn jay]

it looks so stupid

[karmic moat]

it's like a loser version of mathscr O

[thorn jay]

Die.

[delicate orchid]

I'm crashing out rn kicking and stomping my feet

[elfin wraith]

My friend genuinely messaged me the other day tweaking wondering wtf the group \mathfrak{S}_n is because it’s literally not an S

[thorn jay]

You should be happy you're not saying this in #algebraic-geometry

[delicate orchid]

most recent thesis draft has FIFTY \mathcal{O}s and I love them all equally

[thorn jay]

mathcal has got to be my fac

fav

[karmic moat]

$\mathcal O$ looks so dumb. $\mathscr O$ looks so awesome

[cloud walrus]

anamono

[karmic moat]

wait

[delicate orchid]

wait til you see the oh no no no no look at the top of his $\mathfrak{A}_n$

[karmic moat]

wait no sorry i got them mixed up

mathcal O is awesome

[delicate orchid]

wrong one

[karmic moat]

nonono sorry i got them wrong \mathcal O is better

[cloud walrus]

FIELD WITH ONE ELEMENT LOVER

[quiet pelican]

But $\mathfrak{sl}_2$ how else do I do my Lie algebras

[cloud walrus]

mico

[karmic moat]

issuing a public apology soon

[thorn jay]

lowercase frak is alright

[karmic moat]

Lie(SL_2)

[noble nexus]

yeah mathcal O is pretty awesome

[thorn jay]

that's actually crazy

[noble nexus]

so is mathscr H

my beloved

[thorn jay]

$\mathscr{F}$

[cloud walrus]

.𝖊𝖓𝖕𝖊𝖆𝖈𝖊_𝖒𝖚𝖘𝖎𝖈

[delicate orchid]

... it has 1829 \mathcal{F}s....

[noble nexus]

classic as well

[karmic moat]

yea that one is awesome

[quiet pelican]

Death penalty

[thorn jay]

what is your ass needing all those \mathcal{F}'s for

[elfin wraith]

I would rather we used hiragana than mathfrak

[noble nexus]

I switched to the less slanted mathscr and I think I like it more

[delicate orchid]

check my discord status

[thorn jay]

type shit

[noble nexus]

except for mathscr M which looks like a droopy dog

[elfin wraith]

I’d rather just fucking guess what you mean than mathfrak, because that’s already what I’m doing with that dumbass font

[karmic moat]

$\mathcal M$

[cloud walrus]

anamono

[thorn jay]

writing Beck-modules over an algebra using \mathcal{F}
couldn't be happier, it looks like I'm doing something important like sheaves

[karmic moat]

i have my files set up with a very nice font i like

\usepackage{mathpazo}
\usepackage[libertine,cmintegrals,cmbraces,vvarbb]{newtxmath}
\DeclareMathAlphabet{\mathcal}{OMS}{cmsy}{m}{n}

so awesome

that last line is specifically so that i can have the nice \mathcal O

because the one that comes with libertine doens't have a curl it just looks like an O which is stupid

wait maybe it does i forgot

anyway it looks stupid

[elfin wraith]

The only thing I do to my latex is make the margins not awful, computer modern is a great font and doesn’t need to be changed

[quiet pelican]

Tbh if you make me read maths with thinner margins now it feels claustrophobic

Especially in pdf form

[karmic moat]

tbh i prefer thinner margins

[thorn jay]

you should see commutator theory by Ralph Meckenzie

[karmic moat]

okay i guess for casual things like if im typing up solutions to a book im reading i'll use wide margins

[thorn jay]

I'd say about half of the page is white space

[karmic moat]

but for papers i like thin margins

[thorn jay]

😭

[karmic moat]

wait does thin margins mean more space from the sides or the opposite

damn im stupid as hell today

[elfin wraith]

There’s a prof from my UG who doesn’t use like a header gap

[thorn jay]

I assume the former?

[elfin wraith]

The body of the text literally just continues from the very very top of the page

[karmic moat]

oh okay yeah that's what i thought too

[thorn jay]

huh??

[karmic moat]

LMFAO

[thorn jay]

psychopathy

[rocky cloak]

The margins are the spaces, so thin margins mean text close to edge

[karmic moat]

oh okay

[elfin wraith]

Her documents look awful, I always assumed it was just her lecture notes but nope her actual published papers do it too

I think she just gets away with it because she’s unbelievably cracked

[thorn jay]

but Nope

[karmic moat]

i like text further from the edge if im reading a paper or something, closer to the edge if im typing up solutions

[karmic moat]

that's crazy

[karmic moat]

lmfao

[elfin wraith]

Our number theory notes had the exact same pages defining a ring interspersed in them randomly like 3 or 4 times

Lovely woman, amazing mathematician, cannot use a computer at all

[karmic moat]

well i mean you gotta go out of your way to do that right?

i thought by default latex has some margins

[elfin wraith]

It has massive margins

This has always been my confusion

[thorn jay]

maybe this is why we're getting programming classes...

[quiet pelican]

I think we should have lines be 1 character wide as standard

[karmic moat]

w
t
f

[quiet pelican]

All margin, no content

[noble nexus]

default side margins are way too large I find

[thorn jay]

imagine commutative diagrams like this

[karmic moat]

do equations have to be one symbol per line too

[quiet pelican]

Of course

[karmic moat]

awesome

[thorn jay]

aren't certain specific margins required sometimes?

[karmic moat]

i remember seeing some paper a year ago or so where the equations were left-aligned, they had an indent but were left-aligned

[quiet pelican]

I heard you like the align environment so I aligned everything

[karmic moat]

tbh i lowkey liked it

[thorn jay]

I remember that I used to write word documents with everything centered

[elfin wraith]

They actually are, there’s like genuine typographical reasons why to do with the optimum amount of text on a line and how computers do stuff smaller and so they picked that tradeoff

I hate it though I’d rather have more text per line and reasonable margins

[karmic moat]

the euler math font sucks

it's the one vakil uses

[thorn jay]

it looks too much like the typewriter font of old math textbooks / articles

giving me ptsd trauma flashbacks

[karmic moat]

yeah my issue is that it's hard to distinguish from normal text

[noble nexus]

like this font?

[karmic moat]

bc it doesn't have that slight slant like usual

yeah

[noble nexus]

it is a tad offputting

I don't mind it though

[karmic moat]

i've gotten used to it

[thorn jay]

oh man, I did not think how much difference a font and better margins would make

[sacred wharf]

Woke paper

[thorn jay]

I just set the l and r margins to 2cm

[noble nexus]

p sure my notes would be like twice as long with default margins

let me actually check

[karmic moat]

my thesis was 6in x 9.5in

in that one i actually used euler

[noble nexus]

lol +25 pages

[karmic moat]

in my solutions document it's just 1in on each side

[thorn jay]

this is the way to go forward

[karmic moat]

too much space above

[thorn jay]

zero margins

[noble nexus]

I just want some way to make continuous page latex documents

but it doesn't work with pdf apparently

without doing jank solutions

[thorn jay]

yummy

[elfin wraith]

Perfect

[karmic moat]

looks good

actually when i print out papers i usually scale it smaller a bit so i have more room in the margins to write stuff

[thorn jay]

I love how it just cuts off the proposition at the bottom

[karmic moat]

it's a sign that it's probably not important

[thorn jay]

no def not

[white oxide]

What is a good way to see this highlighted line? Since (-1)^a 5^b raised to the 2^{l - 2} power is equal to 5^{b 2^{l - 2}}, it suffices to show that 2^l divides 5^{b 2^{l - 2}} - 1. I was thinking about writing 5^{b 2^{l - 2}} - 1 = (2^2 + 1)^{b 2^{l - 2}} - 1 and using the binomial formula, but I don't know if this is going too crazy

[thorn jay]

oh right this is an algebra channel

[tardy hedge]

bruh moment

[white oxide]

Lol I would post this in elementary number theory but that place is crickets

got aired in #elementary-number-theory 😭

[tardy hedge]

lol

[thorn jay]

I'm gonna crash out

overleaf won't compile my paper anymore

it's not even a big paper

[noble nexus]

overleaf

[thorn jay]

💔 its convenient usually

[thorn jay]

seems like some fermats little theorem stuff

omg I've written a paper with 150 pages

[rocky cloak]

The preceeding paragraph claims 5 has order 2^l-2

It says "from (2)", but idk what (2) is.
Edit: actually it's just the line directly above

[tardy hedge]

it looks lovely Enpeace

Truly a sight to behold

[thorn jay]

this ONE trick turns a 25 page paper into a 150 page paper

[karmic moat]

not enough whitespace, my eyes are tired from going back and forth

[thorn jay]

you just want to be able to hold the down arrow

one character per page

[karmic moat]

Yeah

I should be able to read it with a zoetrope

But like a huge zoetrope

[tidal schooner]

you're writing a paper on universal algebraic geometry? are you going to share it here once it's finished? 👀

[thorn jay]

well its basically already finished, I just have to gather the courage to go to a prof at my uni to ask for endorsement 🙏

[sacred wharf]

bro is trying to speedrun fields medal

[sacred wharf]

did u solve a new theorem or smth ?

[thorn jay]

no but you need endorsement to post on arXiv

[karmic moat]

vixra

[thorn jay]

why would I ever want to post on vixra, im doing actual math here

[kind temple]

first paper i clicked on from vixra

https://vixra.org/pdf/2507.0100v1.pdf

[delicate orchid]

I’ll endorse you….. for a price!… Your soul!!!!

[rocky cloak]

For me it was enough to have a university email. But I guess it depends on your affiliation

[lusty marlin]

I think in specific categories endorsement is needed and a university email is insufficient.

[rocky cloak]

From ArXiV

arXiv may give some people automatic endorsements based on subject area, topic, previous submissions, and academic affiliation. In most cases, automatic endorsement is given to authors from known academic institutions and research facilities. arXiv submitters are therefore encouraged to associate an institutional email address

[lusty marlin]

Yes, but there was someone on the advanced number theory channel not too long ago who required endorsement despite having an institutional affiliation (in particular one that I know has been given automatic endorsement by arxiv in other instances).

[rocky cloak]

I'm guessing number theory is a particularly fruitful place for crankery

[lusty marlin]

Probably, yeah

[vivid tiger]

i wanna see vixra postings that are legit!

[elfin wraith]

Inb4 RH is solved for 15 years before anyone notices the proof on vixra

[shrewd sandal]

Im reading some rep theory notes, and they write in an example that given a transitive group action of $G$ on $X$, and $x_0\in X$, then $G/\mathrm{Stab}_G(x_0)\cong X$ naturally as $G$-sets.

The isomorphism comes from the orbit-stabilizer theorem, yet I don't understand where the naturality comes in: in particular this is a natural isomorphism between what two functors?
Specifically what is the functor on the left, where there seems to be a choice of $x_0\in X$?

[cloud walrus]

Poopoopeepeepants

[rocky cloak]

I'm assuming they just mean naturally in a more informal sense.

G-sets can be realized as functors, but then of course saying "naturally" and "as G-sets" would mean the same thing, so probably not what was meant.

[shrewd sandal]

God fucking dammit

Why cant people mean what they say

Thank you though

[rocky cloak]

Actually, I suppose you could make a category of pointed transitive G-sets. Where the objects are transitive G-sets with a chosen point and the morphisms are homomorphisms that preserve the base point.

Then you get to functors to the category of G-sets, taking (X, x0) to X or G/stab(x0) respectively

It is a natural isomorphism between these, so probably that's what they meant

[shrewd sandal]

Ah okie

I will forgive them then

Thank you!

[delicate orchid]

I do not forgive them because that is silly

[rocky cloak]

The main takeaway no matter the interpretation should just be that this isomorphism isn't some weird add hoc thing, but just the obvious map g |-> gx0

[shrewd sandal]

Youre silly

Youre just jealous that i finally know what a character is

[rocky cloak]

Turns out learning rep theory really builds character, ey?

[knotty badger]

This is the naturality statement for orbit-stabiliiser that I'm familiar with

you view these as functors on the action groupoid associated to the action

[tardy hedge]

hey, why is C fg as a k-algebra?

[delicate orchid]

now THIS is what I'm talkin about

lets get some kinda uhhh elmendorf reconstruction in ts jawn

[knotty badger]

there are further ways you could define this naturality as well

since the act of turning a group action into its action groupoid is essentially a grothendieck construction

so you get naturality that way

[rocky cloak]

It's generated by t1, ..., tn and 1/f

[tardy hedge]

Oh, yea i think im getting confused on fg as a module vs fg as an algebra

ig i forgot u can multiply generators together in the ring sense there too

[noble nexus]

unless the book specifically is using and mentioning category theory you shouldn't take the term naturally too seriously

It may or may not actually be a natural transformation, and usually its not the point the author is trying to make

[thorn jay]

this is crazy

NO text

[thorn jay]

i think it depends on the primary category, lol
maybe the secundary ones too

[tough raven]

IDK what you're talking about.

To my beloved Evaggelia Leleki for all the good memories.

[thorn jay]

okay sorry

ONE text

[thorn jay]

dangit.. i sold it for like two money just a couple days ago

[chilly ocean]

interesting! Thanks.

[knotty badger]

[tough raven]

Higher categorical money

[white oxide]

Hm, is it something like, assume that (x_1, ..., x_m)^k is primary for all k, then we want to show that (x_1, ... x_m)^k[x_{m + 1}^k] is primary in (x_1, .... x_n)^k or something

I'm just thinking about why it suffices to show that (x_1, ..., x_n)^k is primary

If p = (x_1, \dots, x_m) is primary, then p[x_{m + 1}] is primary...

But we need powers of these ideals

Is $(x_1, \dots, x_m)^k [x_1] [x_2] \dots [x_m] = (x_1, \dots, x_m)^{k + 1}$?

[cloud walrus]

okeyokay

[white oxide]

No I guess not since that would imply (x_1, \dots, x_m)^k is contained in (x_1 \dots, x_m)^{k + 1}

[covert cliff]

I dont know much rep theory but I have seen SU(2) representations on polynomials in 2 variables before. Does anyone know the motivation behind this? It seems quite arbitrary to look there (at least from the outside)

[delicate orchid]

what else we gonna act on? monomials? 😹

[chilly ocean]

Can anyone suggest me a book on GROUP THEORY?

[vapid vale]

well the finite subgroups of SU(2) have been well understood for quite sometime, since there is a double cover of it onto SO(3). so you get cyclic, binary dihedral, and binary symmetry groups of the platonic solids (tetrahedral, octahedral, icosahedral).

in the late 1800s invariant theory was considered a very important field, essentially asking questions about the polynomials (the symmetric algebra) on a space invariant to a certain linear transformation (better put, a group representation).

felix klein studied this in the case of SU(2,C) and C^2. what he noted was that for the polynomial ring C[u,v] (the symmetric algebra over C^2) and a subgroup G, the invariant ring could be generated by three polynomials x(u,v), y(u,v), and z(u,v) subject to a single relation f(x,y,z)=0. you can graph this, and there is a singularity at the origin – these are called the kleinian singularities, associated to these subgroups.

with algebraic geometry came tools to resolve (or smooth out) singularities, and patrick du val studied these minimal resolutions (so they are also called du val singularities). i believe he was the person who discovered that the minimal resolution for these singularities replaces the origin with a bunch of P^1's which intersect in a pattern corresponding to the ADE dynkin diagrams.

there is much more interesting stuff where this comes from (the mckay correspondence)

[karmic moat]

#book-recommendations message

[rocky cloak]

So SU(2) canonically acts on C^2 (that's just how SU(2) is defined), so then you also get an action on functions defined in C^2. And polynomials are nice functions

[chilly ocean]

Yep

[covert cliff]

Oh I see, and since polynomials have just been studied forever its the natural thing to gravitate to?

[covert cliff]

Is there any specfic reason for polynomials? It seems also algebraically motivated since polynomials are pretty algebraic from what I know

[rocky cloak]

Essentially the motivation of algebraic geometry: polynomials can be reasoned with purely algebraically and work over arbitrary fields, and still contain a lot of geometric information about their domain

[covert cliff]

I see, thank you

[delicate orchid]

not sure if this is related but the canonical way of viewing the nth symmetric power of the standard representation of a subgroup of GL(m, k) is via an action on the nth degree symmetric polynomials over k in m variables

no clue if this works for infinite fields but it does for finite

[covert cliff]

This is similar to the case of vector spaces it seems

[elfin wraith]

I just got the question, are there non trivial field extensions of C? And yes, you can take like the embedding of C into its field of rational functions, but are there any other “more interesting” examples?

[noble nexus]

I think if you are looking for finite dimension reps (which you are since SU(2) is compact) then it's somewhat natural to look at polynomials since they are kinda the most natural form of finite dimensional function spaces

together with using the action on C^2

idk if there's like a deeper reason though

[covert cliff]

I see, its just "we could"

[noble nexus]

That's how it seems to me but perhaps there's some better way to think about it

[covert cliff]

Ok, thank you

[prisma ibex]

only transcendental extensions since C is already algebraically closed

[rapid cave]

Found this https://math.stackexchange.com/questions/1466273/is-it-possible-to-construct-a-field-larger-than-the-complex-numbers

[chilly ocean]

is this done using field extension

[rocky cloak]

The algebraic closure of C(t) is isomorphic to C. So you can have fun embedding from C into itself, making it an infinite degree extension of itself

[lusty marlin]

Any field that contains another field is called an extension of the other field

[prisma ibex]

this is an absolutely schizophrenic way to think about \bar{C(t)} lmao

[vapid vale]

well historically it would have been natural. hilbert showed that these invariant polynomial rings are finitely generated k-algebras which is quite nice and perhaps justifies it after the fact

[lusty marlin]

How does one prove this first fact?

[elfin wraith]

Oh god lmao that makes sense to me but is equally upsetting

[elfin wraith]

They have the same cardinality, so I think they have to be isomorphic

[prisma ibex]

worth mentioning that the algebraic closure of C((t)) is something quite explicit

[noble nexus]

probably same cardinality basis over Q or some nonsense like that

[rocky cloak]

So you can prove that an algebraically closed field is determined by characteristic and transcendence degree.

And for uncountable cardinality transcendence degree is just equal to the cardinality of the field

[prisma ibex]

$\overline{\mathbb{C}((t))}=\mathbb{C}{{t}}=\bigcup_{n>0}\mathbb{C}((t^{1/n}))$

[cloud walrus]

nGroupoid

[lusty marlin]

No, ℝ and ℂ are not isomorphic as fields.

[elfin wraith]

R isn’t algebraically closed

[prisma ibex]

mama mia...

[rocky cloak]

First you pick a transcendence basis, then use that embeddings into algebraically closed fields can be extended along algebraic extensions

[delicate orchid]

That middle notation is making me froth at the mouth

[prisma ibex]

it's pretty standard notation

[lusty marlin]

Hmm I get it now

[karmic moat]

in a good way or bad way

[delicate orchid]

Im aware

[lusty marlin]

Yeah
Thanks

[delicate orchid]

Take one look at the texit post and let me know what you think

[karmic moat]

with you i can't tell

i've never seen it before so i rely on others opinions to tell me how i should feel about it

[elfin wraith]

To be clear btw, I was meaning what jagr said, I’m not completely dumb lmao

[rocky cloak]

And I guess a big family of interesting example is take any irreducible variety and take the field of fractions of its coordinate ring

[lusty marlin]

Ah

[elfin wraith]

Ah yeah I see that, but yeah they all vaguely look like just taking a field of fractions

I kinda expected that I suppose, I wasn’t sure what else it could be but maybe there’s something weird and whacky out there

[prisma ibex]

these are the "global" examples, there are loads of "local" examples that generalize the usual definitions of local fields also, just as every (1-)local field in the usual sense is either a finite extension of R or Q_p or F_p((t)), you get examples of (2-)local fields by considering finite extensions of e.g. R((t)) or Q_p((t)) or F_p((t_1))((t_2)) and so on

[rocky cloak]

I mean everything is some algebraic extension of some field of rational functions

[elfin wraith]

I don’t really know anything about those, I know they’re a number theory thing but that’s about it

[elfin wraith]

This is true

[rocky cloak]

They can still be pretty wacky I think

[prisma ibex]

local fields are what you get from completions of global fields at places (absolute values)

global fields are finite extensions of either Q or F_p(t), so local fields are finite extensions of either R or Q_p or F_p((t))

there is a more abstract definition of both local and global fields in terms of valuations but also this really short list suffices

(global fields are fields of fractions of Dedekind domains for which every nonzero ideal has finite index, and these have the property that all of the corresponding local fields arising from completion are locally compact for the topology induced by the absolute value)

[covert cliff]

I see, my lack of algebra knowledge is gate keeping me as usual

[alpine plank]

I guess this kind of nonsense is why we should think about field extensions and not just fields

[rocky cloak]

Either that or stay in the more sane world of number fields, and not this weird analytic beast that is C

[alpine plank]

number fields my beloved

[sacred wharf]

If its not in z its not a number

[tardy hedge]

what is a number type shi 😝

[sacred wharf]

Wdym shi

[rocky cloak]

Number 2 without the t

[sacred wharf]

wo

[tardy hedge]

type shi

[sacred wharf]

I havent done much fields

[alpine plank]

I did a philosophy course and we learnt what the pythagoreans conception of numbers was, lowkey they were straight up insane. either that or the text we followed was really bad

[tardy hedge]

maybe u could learn Galois theory now

[sacred wharf]

Wanted to be only offered even years here

[tardy hedge]

its so annoying when its like that

[sacred wharf]

Fr i got infinite groups instead which i am not taking

[tardy hedge]

sad

[sacred wharf]

Next year im mainly taking applied courses, this might be last terms of pure math for me ever

[tardy hedge]

wow

[thorn jay]

why?

[sacred wharf]

Next year is my final year of my degree

Since i do 3 year

[tardy hedge]

bro wants a job or smth

bro really out here wanting work 😂

[sacred wharf]

Cant end up like canadian youth fr

[tardy hedge]

the fact thats not even a joke

[sacred wharf]

From westernU which i think is solid uni

[tardy hedge]

yup same with my friend rn. Masters in eng from Uoft

[elfin wraith]

SWE is particularly fucked tbf

[tardy hedge]

yeah

[thorn jay]

swe?

[tardy hedge]

i never really wanted to do swe. took some comp sci in undergrad but wasnt really a fan

[elfin wraith]

After the massive hiring crazy during the pandemic there’s just not that many positions

[tardy hedge]

i took like 4 comp sci courses

[thorn jay]

o software engineering

all the companies want is ai people now smh

[elfin wraith]

I enjoy it enough, I’d take a SWE role probably

[tardy hedge]

me robot beep boop

[elfin wraith]

This is also true, but honestly for SWE it’s just how heavily they hired through Covid

But it’s certainly part of it

[thorn jay]

i enjoy software engineering but not the uni course

[elfin wraith]

Yeah quantum programming (which was like 50% monoidal category theory) is the only CS course I took

I kinda wish I took DSA though

[sacred wharf]

Crazy uoft is top in the world

[tardy hedge]

Yeah

he has been applying to stuff in toronto

all rejections so far

not even like interview

[sacred wharf]

Oversaturation, if people didnt rush into it it probably would have ordinary opps

[sacred wharf]

Are u sure there is no like cv issues or smth ?

[elfin wraith]

I think Canada is also just generally fucked jobs wise

[tardy hedge]

i dont think so man, he's like one of the most well rounded ppl i know

experience at amazon as swe too

and other jobs

[elfin wraith]

[tardy hedge]

yeah Canada feels especially cooked rn

but i know its pretty bad everywhere too

i mean canada just doesnt have an economy so like

something like huge percentage of canada economy is real estate lol

[sacred wharf]

Canadian engineers dream of working for american companies

[tardy hedge]

He’s having issues too of being hesitant to move cause he’s in a good relationship and the girl is in toronto

He was considering a Phd at ubc in vancouver

[sacred wharf]

Gotta hope carney fixes it

U seem like a carney fan

[tardy hedge]

Not really

Im not really a fan of any of them

[sacred wharf]

Canada went from being the envy of the world to horrid so quickly

[tardy hedge]

Lol real

[elfin wraith]

When was Canada the envy was the world lol

[sacred wharf]

2016

[elfin wraith]

Hm

[sacred wharf]

half my family moved to canada just for it to get screwed

well tbf one of the alternatives was the UK, at least they now live in a first world country

[tardy hedge]

Lol

Canada was always just that country nobody cares about

Its kinda nice that way

[sacred wharf]

nah it's the place everyone wanted to move to until a few years ago

[tardy hedge]

Yeah

[elfin wraith]

?

[sacred wharf]

apparently they make it easy to naturalize etc. if you speak french and english? i thought of moving a while back

[tardy hedge]

Not too sure of the specifics but we had too much immigration recently so the govt is trying to cut down from what im aware

I think for student visas specifically

[sacred wharf]

damn we be hated