groups-rings-fields

[karmic moat]

as rings, CRT gives us the isomorphism $R/abR \cong R/aR \times R/bR$

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

(i think)

(maybe it's not true because aR + bR \ne R?)

but i guess if we consider the case where M = <x> (+) <y>, we have something like the submodules <x> and <y> are comaximal

[white oxide]

i'm not sure how to see this is true. if $p(x + (p^n)) = 0$ for all $x + (p^n) \in (R/(p^n))[p]$ then why is it not true that $p^{n - 1}(x + (p^n)) = 0$?

[cloud walrus]

okeyokay

[rocky cloak]

aR + bR = R, because a and b are coprime

[karmic moat]

oh wait yeah

oops thanks

oh maybe i should be considering something like M/aM and M/bM

either that or M/<x>M and M/<y>M, since <x> and <y> are comaximal

[rocky cloak]

If you think about R/ab = R/a x R/b, what are some nice elements in R/a x R/b?

[karmic moat]

a typical element would have the form (m mod a, n mod b) but i’m not sure of any nice elements

[rocky cloak]

Right, so what are the simplest elements of R you could pick for m and n?

[karmic moat]

0?

[rocky cloak]

0 is a good one, so maybe picking one of them to be 0 and another as something else simple

[karmic moat]

hmm

ab?

[rocky cloak]

Well, ab will just be 0 mod a and b.

[karmic moat]

yea

i guess anything thats not divisible by a or b

[rocky cloak]

So what is a simple element we could find in any ring

[karmic moat]

1

[rocky cloak]

1 is a good one

[white oxide]

yea how is this possible? say $p^{n -1} + (p^n)$ was a generator, nonzero by hypothesis. But then $r(p^{n - 1} + (p^n)) = pp^{n - 2}r + (p^n) = 0$

[cloud walrus]

okeyokay

[rocky cloak]

So an element like (1, 0), it corresponds to something in R/ab

[karmic moat]

mhm

[rocky cloak]

Does the notation (R/(p^n))[p] mean the submodule annihilated by p or something?

[white oxide]

yea

[rocky cloak]

If so p^n-2 is not in there

[karmic moat]

should you say (1,0) corresponds to 1*0 or 1+0 in R/ab

or nether

[rocky cloak]

It should correspond to an element that is 1 modulo a and 0 modulo b

[karmic moat]

ohh

1 mod a implies it has the form ra+1 and 0 mod b implies it has the form sb

wait maybe the first one isnt true

i think it is tho

[rocky cloak]

You're right

Now to finish it of ||multiply this element by x+y||

[white oxide]

oh so would the submodule annihilated by $p$ just be $p^i + (p^n)$ for $i \geq n - 1$

[cloud walrus]

okeyokay

[white oxide]

oh wait

this makes sense now

oops

[rocky cloak]

I mean it will be things of the form rp^n-1 + (p^n)

[white oxide]

right oops lol

thanks!

[white oxide]

does A being a cyclic R-module of order r in R just mean that it's generated by r?

[coral spindle]

Deeply messed up that they're using \varepsilon instead of \in...

This is a little weird to call it the 'order' but I believe they're saying that A is isomorphic to R/(r). It wouldn't be generated by r, no.

[white oxide]

$\varepsilon$

[cloud walrus]

okeyokay

[coral spindle]

It is generated by a single element, hence the name cyclic :)

[delicate orchid]

;3

[coral spindle]

:3c

[delicate orchid]

it's kind of insane how you can write every cyclic module as an ideal AND a quotient

quirked up to the max

[white oxide]

ah ok thx!

wtf is clifford theory

[delicate orchid]

study of how characters split when restricted to normal subgroups

@coral spindle I'm currently in the middle of setting the foundations to do clifford theory on fusion systems because I am a lunatic

[coral spindle]

Absolutely epic move…

I had to consider the Clifford theory of blocks recently. All I can say is I do not understand blocks still

[delicate orchid]

they're just funny connected components of the brauer graph :)

just work out the decomposition matrix ;3

[coral spindle]

UwU that’s famously such a cool and easy thing to do

I’ll bash it out asap

[delicate orchid]

wahhh wahh do it for C_2 at the prime 2 and then see if it's so hard nerd

[coral spindle]

Ok

Did it

What next pa?

[delicate orchid]

so now do it for any p-group at the prime p

and since all groups are p-groups we are done

[coral spindle]

[summer path]

wow i missed boy :3cing

[coral spindle]

Thank u sir I have seen the light

[summer path]

[coral spindle]

I am no longer boytjie

Wee has made me mantjie

Wew*

Lmao

[summer path]

cliffordjie

[barren sierra]

Rudin moment

[coral shale]

let $\varepsilon \varepsilon \bR^+$

[terse crystal]

Show ann(x+y)=abR

(Remember coprime ideals I and J we have I cap J=IJ

[jovial tundra]

,tex The number of elements not equal to their own inverse will always be even since every element $a \in S$ will have a corresponding inverse $a^{-1} \in S$. A group $G$ of order 4 should have three possible groups by the previous statement. One where there are 0 elements not equal to their own inverse, 2 elements not equal to their own inverse, and 4 elements not equal to their own inverse. However, 4 elements not equal to their own inverse is impossible since a group must contain the identity. Therefore, there are only two possible groups for 4 elements. The idea can be generalized to a group with n elements. There are then $\lfloor \frac{n+1}{2} \rfloor$.

[cloud walrus]

Fuzzy_Alpaca

[jovial tundra]

is this a true statement?

or are there any errors in my logic here

[delicate orchid]

C_3 only has one element equal to its own inverse

[jovial tundra]

ugh true

[delicate orchid]

What are you trying to do here

[jovial tundra]

trying to generalize it to n elements

[delicate orchid]

Generalise what

[jovial tundra]

Given n elements, how many possible groups are there (except for renaming the elements with different symbols)

[delicate orchid]

Good luck

It’s definitely not linear

Over 99% of the groups of order less than 2000 are precisely order 1024

[jovial tundra]

ok, i'll look up a paper. Thanks

[delicate orchid]

Might be 98%

Doing the number of abelian groups is much more doable if you know a few theorems

[crystal turtle]

Well, you mainly only need one theorem lol

[jovial tundra]

,tex The number of elements not equal to their own inverse will always be even since every element $a \in S$ will have a corresponding inverse $a^{-1} \in S$. For a group $G$ of order 4, the previous statement implies three possible scenarios: one where there are 0 elements not equal to their own inverse, one with 2 elements not equal to their own inverse, and one with 4 elements not equal to their own inverse. However, having 4 elements not equal to their own inverse is impossible, as a group must contain the identity element. Therefore, there are only two possible groups with 4 elements.

[cloud walrus]

Fuzzy_Alpaca

[jovial tundra]

Ok for now, is that good? Before I go down some rabbit hole with this nonlinear stuff

Given 4 elements, how many possible groups are there (except for renaming the elements with different symbols)

[delicate orchid]

Yes there are only two groups of order 4, but I would say that your even argument only holds because the group is order 4

[tiny pagoda]

u just haven't discovered the last one

[jovial tundra]

,tex Ah I found an additional thing. If the order of G is even, there is at least one element x in G such that x \neq e and x=x^{-1}. Does that make it more concrete

[cloud walrus]

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

[jovial tundra]

,tex Ah I found an additional thing. If the order of G is even, there is at least one element x in G such that $x \neq e$ and $x=x^{-1}$. Does that make it more concrete

[cloud walrus]

Fuzzy_Alpaca

[terse crystal]

Your last statement is true. Just not sure how it will let you achieve your original goal of classifying groups of the same order

[jovial tundra]

,tex My final result is this: If $S$ is the set of all elements of G which are not equal to their own inverse, the number of elements not equal to their own inverse will always be even since every element $a \in S$ will have always have a corresponding inverse $a^{-1} \in S$. Conviently, $\lvert G \rvert$ is even, so the number of elements that are their own inverse is even. So a group of order 4 either has exactly one non-identity element that is its own inverse or all the elements are their own inverse since if the order of G is even, there is at least one element x in G such that $x \neq e$ and $x=x^{-1}$.

[cloud walrus]

Fuzzy_Alpaca

[jovial tundra]

Given 4 elements, how many possible groups are there (except for renaming the elements with different symbols)

[crystal turtle]

there are 2 groups of order 4

[jovial tundra]

Yeah, that's my final conclusion, but is the logic correct? (like the big chunk of text at the top)

[tawny cloak]

your logic here is correct in classifying the possible numbers of elements equal to their own inverses

but more broadly, you have an issue in your approach if you're trying to classify groups of a certain order up to isomorphism which is what you said

which is that the structure of a group carries a lot more information than just the number of elements of order 2 (and it just happens to be true in order 4 that the possible values for this correspond to the two possible groups)

classifying all of the finite groups up to isomorphism in general is an extremely difficult problem. I think the closest thing we have is the classification of finite simple groups, motivated by the Jordan-Holder theorem. However, there are very complicated ways potentially to put simple groups together so that doesn't classify all finite groups.

and there doesn't seem to be any cheatcode that'll tell you the number of groups of order n without understanding which groups exactly they are

however, since you seem interested in the number of elements equal to their own inverses, which are just solutions to x^2 = 1, I suggest generalizing that to the number of solutions to x^p = 1

personally I found a way to show that if G is a group of order n where p divides n, and k is the number of solutions to x^p = 1 in G, then ||k = p mod p(p-1)||

in the case p = 2 this is equivalent to your observation that the number of elements not equal to their own inverses must be even

[ivory trail]

in your latter case isn't k always 1 or am i missing something

[tawny cloak]

oh right I am stupid

of course

I edited it to make more sense

it just happened that the same reasoning could be used to get a similar condition in the other case but I was silly and didn't notice it's a degenerate case

[rocky cloak]

You may want to look at this

https://oeis.org/A000001

Btw, is there an oeis bot here like for wa?

[mighty kiln]

,w oeis a069420

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by arxetex

[rocky cloak]

That works i guess

,w oeis A1

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jagr2808

[cerulean trout]

Is it possible to deduce the weak nullstelensatz from the hilbert nullstelensatz

ok I'm sure it is but how?

[coral spindle]

It is, and it's a good exercise. Pick a good choice for the ideal in Hilbert's nullstellensatz.

[rotund aurora]

what choice are you referring to?

the implication is immediate

they are asking about nulstellensatz -->weak nulstellensatz, not the other way

[coral spindle]

Oh yeah mb Idk why I talked about choices at all

[summer path]

I also read it as weak nullst to nullst

[coral spindle]

I didn't read it that way, I just had the wrong idea in my head.

[delicate orchid]

here's my attempt at showing it:
The weak nullstellensatz claims that V(J) is non-empty for any proper ideal J of k[x_1,...,x_n] with k algebraically closed, because (V(J) is non-empty if and only if there is some point where all of the polynomials in J vanish by the weak nullstellenstatz)
So if we take a proper ideal J and apply the Nullstellensatz, I(V(J)) = \sqrt(J), if V(J) was empty then I(V(J)) would be k[x_1,...,x_n] but the \sqrt{J} cannot contain 1, as that would imply that 1 was in J contradicting the fact that J is proper, so V(J) cannot be empty, showing the weak nullstellensatz

I completely disagree that this implication is "immediate" as it took me 5 minutes of thinking

[jovial tundra]

Are there any errors in the proof for a group of order 4? So for now, I'm not trying to generalize it

I'm just trying to show that a group of order 4 only has two possible groups

[void cosmos]

to show that

R is a ring , if all R-modules are projective then R is semisimple

do i consider R as a module over itself , take N to be any submodule

and then consider 0 --> N --> R --> R/N

and this is split exact so N is a direct summand of R and im done?

cuz i have shown that every submodule admits a complement basically?

[delicate orchid]

I think this works - you know it splits because R/N is projective right

[void cosmos]

yes

[delicate orchid]

yurrr

[void cosmos]

i have the worst question

in existence

can i ask it

[rocky cloak]

What you have shown is that for a group of order 4, either all elements are their own inverse of exactly two of them are. And that is correct.

But you would still need to show that there is only one group where all elements are their own inverse, and only one where exactly two are.

[delicate orchid]

... yes? lol

[void cosmos]

when showing that all modules over a division ring is free

why cant i just say if A is a D-module ( or any module in general )

then A is isomoprhic to D^n/ker(phi) where ker(phi) is an ideal of D^n ( which is trivial idk )

i think i messed up the categories or idk

cuz like

this isomorphism is of modules right?

a module isomoprhism

[delicate orchid]

yes so ker(phi) would be a submodule of D^n as a D-module

[void cosmos]

which is not necessairly trivial right

unless A is a subring

[delicate orchid]

thinking

[void cosmos]

like unless A is just a subset of D

this is not trivial

cuz its not a division thing

[delicate orchid]

nah it wouldn't be

[void cosmos]

yea

so the proof is hard sadly

[delicate orchid]

do you have the result that tells you that division rings are semisimple

[void cosmos]

i think we just did it no? cuz all modules over divison rings are projective

[delicate orchid]

exactly, perhaps we can leverage that here?

[void cosmos]

idk how to hmmmmmmmmmmmmmmmmmmmmmmmmm

[delicate orchid]

same

[void cosmos]

lmfao jagr is typing

so the problem is donezo

[rocky cloak]

You can just reuse the proof that every vector space has a basis from linear algebra I guess

[void cosmos]

yea

zorn and stuff yeha

[delicate orchid]

does that work despite non-commutativity? epic

[void cosmos]

okay

[rocky cloak]

Yeah, the proof never uses anything about commutative.

[void cosmos]

yea

so i go like find the maximal linearly indepednant set

by zorn

and then take any element not in the span

take this set U that element

this is linearly dependant by maximlaity

and then write that element as a lienar combination of the elements in the maximal linearly independant set cuz i can divide now

and im done?

is that how it goes

i skipped linear algebra

[rocky cloak]

That is indeed how it goes

[void cosmos]

cool

tysm eren jeager

but wait

i will get deducted marks

if i do not shows

show

this is nonemtpy right?

[rocky cloak]

The empty set is linearly independent I guess

[last spoke]

So I'm struggling to deal with the group action part of finding the splitting field

I got the order easy enough, but then I think I got (\mathbb{Z}_2 \times \mathbb{Z}_6) when the answer should be (D_6)

[cloud walrus]

StarvinPig

[white oxide]

wait jagr srry if someone already asked but are u phd?

i know u teach right

[rocky cloak]

I'm a phd student yeah

[white oxide]

oh nice, hope i can get to ur point one day

[rocky cloak]

I'm sure you will

[white oxide]

what's your concentration?

or idk what to call it lol

[rocky cloak]

I do rep theory of finite dimensional algebras

Or just homological algebra more broadly I guess

[white oxide]

oh wow okay nice

[zinc sail]

Hi! I recently started learning group theory, and I'd like to know if my proof for the following problema is correct:

"Let $G$ be a group, and let $N$ be a cyclic subgroup of G such that $G/N$ is also cyclic. Prove that $G$ is generated by two elements."

Here is my proof:
Since $N$ is cyclic, there exists $n \in N$ such that $N = \langle n \rangle$. Similarly, there exists $g \in G$ such that $G/N = \langle gN \rangle$. Then, for any $a \in G$, we have that $a \in g^iN$ for some $i \in \mathbb{Z}$.
So $a = g^in^j$ for some $j \in \mathbb{Z}$, Therefore, $G = \langle g,n \rangle$.

Does this look correct?

[cloud walrus]

ucheo

[delicate orchid]

that looks perfect

[south patrol]

Yeah nice

[karmic moat]

u look perfect

[lethal dune]

[rotund aurora]

Ok fair

[sturdy mirage]

Can anyone help me see why a k-module M (k a field) must be isomorphic to some $\bigoplus_{i \in I} k_i$? I know modules are a sort of generalization of vector spaces, but ignoring that and just going by the abstract definition of modules... how would I see this?

[cloud walrus]

reking

[coral spindle]

It takes a fair bit of work to see that.

And absolutely you must look at more than just the definition.

Anyway, this is equivalent to every vector space having a basis. That’s what this is saying.

The same does not hold for modules in general.

[sturdy mirage]

well I understand that actually, that every vector space has a basis i am taking for granted

but what is the link between that and modules over fields, i guess is what i need to see

[delicate orchid]

not sure what you mean by "link"

[coral spindle]

Are you aware that a vector space is, by definition, a module over a field?

[delicate orchid]

modules over fields are exactly vector spaces

the slight generalisation is that of a "free module"

[sturdy mirage]

oh okay, no i wasn't aware of that definition of a vector space

[delicate orchid]

what is your definition of a vector space

[coral spindle]

whatever your favourite definition is, I suggest you try to convince yourself that they are equivalent.

[delicate orchid]

because that statement is true by the standard definition of a vector space

[sturdy mirage]

hmm maybe something that looks like $\bigoplus_{i \in I} k_i$. thats a vector space to me

[cloud walrus]

reking

[delicate orchid]

concerning

[coral spindle]

[delicate orchid]

have you done lin alg?

[sturdy mirage]

yes

[delicate orchid]

ok so you should know that a vector space is an abelian group with scalar multiplication (with the scalars in a field)
and a module is also just an abelian group with scalar multiplication (with the scalars in a ring)

[sturdy mirage]

well what i mentioned above is just the set, you need to take it with addition and scalar multipliication and some axioms i suppose...

[coral spindle]

It's most certainly not just the set! We do not use that notation for mere sets.

[delicate orchid]

well no that is the vector space but I was just making sure you knew the actual definition

tfw Set isn't abelian

[prime arch]

i feel like when they give you examples of vector spaces, they don't do it in the form \oplus_i k_i. so you probably think of vector spaces in more ways than just that direct sum

[coral spindle]

Maybe we should make this question clearer: do you know the definition of a vector space, as found in linear algebra?

[sturdy mirage]

not off the top of my head, but i have done linear algebra

is the point that the definition coincides exactly with the definition of modules?

(when the module is over a field)

[delicate orchid]

exactly, yes

[sturdy mirage]

i guess i will go through the proof of existence of a basis in a vector space and just insert the word module as i go along

[coral spindle]

I must underline the fact that this does not hold for modules in general.

[sturdy mirage]

oh i know that

thats maybe what confused me a bit

[white oxide]

just a little sanity check... Perm(G) is isomorphic to S_n where |n| = G, since we can just rename each element of G by {1, ..., n}

[delicate orchid]

yup

[white oxide]

why do i overthink things

thank you

[void cosmos]

consider {Z_p^i,phi_i,j} where phi_i,j Z_p^i --> Z_p^j is just the inclusion

why is the limit of this system divisible

[topaz solar]

Consider looking at it as subsets (subgroups even) of the circle group in C?

If I’m reading your thing right

(Which inclusion seems like an odd term imo)

[void cosmos]

oh u mean like

elements that are of order p^n for any n

right?

[white oxide]

just to clarify, saying that $a$ has two conjugates is basically saying that if $H = {gag^{-1} \mid g \in G}$ then $|H| = 2$

[cloud walrus]

okeyokay

[delicate orchid]

these are the prufer groups yes

[topaz solar]

Basically, I think that’s what you have

[delicate orchid]

I remember telling you about these when we were talking about the injective hull of Z_24

[topaz solar]

If I’m reading inclusion correctly

Ye

[void cosmos]

yea but

i am now trying to learn about that hahaha

[topaz solar]

Z(p^\infty) time

[void cosmos]

if u remember i told u did not understand the answer

[topaz solar]

Or smth

[void cosmos]

now i have to sadly

and im learnig about the prufer group

[delicate orchid]

ah ok, yeah the proof that they're divisible is kind of the same that they're their own injective hulls

[void cosmos]

i just wanna be able to show its the injective hull of Z_p

and thats it im done

[topaz solar]

I’d think showing it’s divisible would be easier but idk injective/projective hull stuff well so

[delicate orchid]

lemme recall the definition of a divisble group just so I don't have to hold it in my head
a group is divisible if and only if nG = G for all integers n

[void cosmos]

yea if its divisible as Z-module its injective

[delicate orchid]

the Z-mod action on the p-prufer group is given by n\omega = \omega^n right

[void cosmos]

idk

[delicate orchid]

has to be

[void cosmos]

i really dk anything abuot the prufer group and i cant find any resources

like literally

just an nlab and wikipedia

[delicate orchid]

this is just abelian groups atm

[void cosmos]

which are useless

[topaz solar]

It’s that direct limit

[void cosmos]

im not entirely comfy with direct limits

i saw that its Z[1/n]/nZ

[topaz solar]

Or, ya know, associate subgroup of the unit circle in C

[void cosmos]

as in like a+b/n ?

mod n?

[delicate orchid]

oh true fair point

duh

[topaz solar]

Yeah lmfao

[onyx trout]

Direct limits can be presented as a disjoint union of the underlying objects

[topaz solar]

Showing it’s divisible directly sounds way easier

[delicate orchid]

so now we check out our little z^n doodads, obviously G -> nG bijective if n is coprime to p but what about if it isn't?

[void cosmos]

what

[topaz solar]

Especially since I have 0 clue what an injective hull is

[void cosmos]

what bijective

[onyx trout]

Modulo the equivalence relation of eventually being equal if you follow successive morphisms

[delicate orchid]

I'm thinking outloud

[void cosmos]

ok ok

[topaz solar]

Colimits as induction

[void cosmos]

so for example what would Z_3^inf look like

[delicate orchid]

yeah, it's basically an element of the product where morphisms between elements behave as they should
one can see this directly from the diagram

[onyx trout]

So for a chain of inclusions here you get a sort of disjoint union of them as groups

[onyx trout]

What does this mean?

[void cosmos]

the prufer group

at 3

[delicate orchid]

the 3-prufer group

[onyx trout]

A direct limit is of a system

Oh

[delicate orchid]

i.e lim Z/3^nZ

[void cosmos]

so u would have like

elements of Z_3 and Z_9 and ...

[topaz solar]

Moamen, spit Z/3^nZ into the unit circle

[void cosmos]

with them being equivalent iff what?

whats the morphism

[delicate orchid]

the includion of Z_3 into Z_9

[void cosmos]

how is that

[delicate orchid]

n -> 3n

[void cosmos]

Z_3 is not a subset

of Z_9

[delicate orchid]

what

that's why it's an inclusion??

[void cosmos]

arent they diff equivalence classes

[delicate orchid]

do you not know about cyclic groups?

[topaz solar]

Moamen, literally just look at it as roots of unity on the unit circle

[void cosmos]

i do

[delicate orchid]

then Z_3 is obviously a subgroup of Z_9!

[delicate orchid]

anyway this is the way to go who gives a fuck about cyclic groups they're all in S^1 anyway

[topaz solar]

This literally makes it very obvious

[delicate orchid]

Z_p^\infty is the group of all points on the unit circle with order equal to a power of p

[void cosmos]

as in like elements with x^3^n = 1?

[delicate orchid]

correct

[topaz solar]

Yes

[void cosmos]

okay

[topaz solar]

For p=3 ofc

[void cosmos]

wait thats it?

thats the prufer group?

[delicate orchid]

ngl I can't yet see how p(Z/pZ^\infty) = Z/pZ^\infty quite yet, obvious if we replace p with something coprime to p

[void cosmos]

arent they diff sets

[topaz solar]

Now imagine how this is a limit of taking (3^n)-th roots of unity

[delicate orchid]

Z_9 = <g|g^9 = 1> then Z_3 = <g^3> idk what your confusion is

[topaz solar]

Well it still has order a power of a prime, since like x^p^n = 1, then (x^p)^(p^(n-1))

[delicate orchid]

well yes

but how do we get the elements of order p back

[topaz solar]

And that’s ur p operation

[void cosmos]

i thought like

as in Z/3Z and Z/9Z they are diff equivalence classes

[delicate orchid]

it's obvious it shunts all of the grade n elements to grade n+1 but

[topaz solar]

Because there’s also all the terms of order x^p^(n+1)

[void cosmos]

right [2] has different elements

in Z_9

than in Z_3

or idk

[topaz solar]

Because it’s infinite

[delicate orchid]

I haven't considered isomorphic objects "different" in literally 3 years

[topaz solar]

So this is passing down via Z/p^(n+1)Z -> Z/p^nZ

[delicate orchid]

I suggest getting into the same mindset (all categories are skeletal) - but your point isn't valid? An inclusion is an injective homomorphism

it need not be a subset

[delicate orchid]

no I'm not following, it's passing up isn't it?

oh no

no yup

[topaz solar]

I like to reserve inclusion for literal inclusion when it’s a literal subset, but if not then it’s just injection

[delicate orchid]

yup yup yup yup

[topaz solar]

It’s rotating around the disk

[delicate orchid]

you're raising to the power of p but that drops the order by a power of p

[void cosmos]

idk what ur saying hahaa

anyways

okay forget about this subset

[topaz solar]

Yep

[void cosmos]

now wdym

[topaz solar]

You go down in order, but whoops, N is infinite

So gg you get it back out

[void cosmos]

what 😄

[white oxide]

oh wait, this is because if we let G act on itself by conjugation then the orbit of a has two elements, so the index of the stabilizer of a is 2 so it's normal

[void cosmos]

can u

write out

with examples please

sorry im slow

[cloud walrus]

Dragonslayer Sharp

Dragonslayer Sharp

[void cosmos]

yea

[delicate orchid]

ceral2

[void cosmos]

n?

p^n

[topaz solar]

No because stacking powers

Try p^(n-1)

Since p * p^(n-1) = p^n

[void cosmos]

oh yeah lmfao

okay

[topaz solar]

Yep

It’s not injective

[void cosmos]

what

wdym injective

[topaz solar]

x -> x^p

[delicate orchid]

because it sends the order p elements to 1

[topaz solar]

But we still get the same thing back out

[cloud walrus]

Dragonslayer Sharp

[void cosmos]

idk just mod p lmfao

like just mod p it

or yeah i would love to divide by p ig

im lost honestly

[topaz solar]

Try multiplying by p

[void cosmos]

do u mean the other way around

[topaz solar]

No

[void cosmos]

from p^n --> p^(n+1) ?

[topaz solar]

You know how to go up, by what wew said

How do you go down, multiplying by p

Draw it out for 3^2 and 3^3

[void cosmos]

this is so confusing omfg

wdym multiply by p

isnt p 0 in mod p^2

[topaz solar]

bro what

[delicate orchid]

Is 3 0 mod 9?

Very concerning comment

[void cosmos]

oh

oh yeah

its lack of sleep ig

[topaz solar]

Bro I don’t know what to say to that

[void cosmos]

my bad

sorry guys happens

okay okay i get it

i switched things out

okay

[topaz solar]

Map Z mod 27 to Z mod 9

[void cosmos]

yea so ur saying u just go a mod 27 to 3a mod 9

yea

right?

[topaz solar]

Yeah, now use that to show the group is divisible

[void cosmos]

can u

describe the group

as the direct limit

with these morphisms?

or nvm

if its easier with

like x^p^n = 1

[topaz solar]

These are the oppose direction

[void cosmos]

yeah

Z_p^n --> Z_p^(n+1) is just the "inclusion"

ig

[topaz solar]

If you do this, it’s literally inclusion

Use the circle group subgroup and see it’s divisible

Then, use that intuition to show it for the direct limit

[delicate orchid]

Wrote out my solution and I’ll post after

[topaz solar]

Because I think you should use them at least once

As for injective stuff, I honestly don’t know why you’d use that

But my modules & homalg are sorely lacking

If you wanna try that way too, pass me a defn

[delicate orchid]

Hmmm injective modules le cohomology chungus

[topaz solar]

Yeah idk why you’d use this here

[delicate orchid]

Injective hulls are the largest essential extensions which SOUNDS important

[topaz solar]

I have no idea what essential means but it sounds essential

[delicate orchid]

F-essential subgroups!

[void cosmos]

i cant do it honestly

show that its divisible

or even p-divisible

im too tired

[topaz solar]

whuh

[delicate orchid]

l||et p^infty be the set of all roots of unity with order equal to some power of p. We wish to show this group is divisible, that is to say that np^infty = p^infty. For n coprime to p this is obvious as raising the p^kth roots of unity to n is a bijection for all k if n is coprime to p. So we reduce the case to showing pp^infty = p^infty. Raising p^kth roots of unity to the power of p results in a p^{k-1}th root of unity. This map is surjective, so pp^infty = p^infty. Thus p^infty is divisible.||

[delicate orchid]

different use of "essential" dw about it

[void cosmos]

a bigger module that has the property that all submodules of that module intersect nontrivially with the smaller module

[delicate orchid]

yeah like uhhh Q_8 and <-1>

not really like that at all

[void cosmos]

yea i think i see what ur saying now

yea

i got git

showing its p-divisible works

[topaz solar]

I see what ur saying
Bro peeked smh

[void cosmos]

i dont know why it works

with the direct limit thought

if the map is

inclusion

[topaz solar]

Anyhow, show it’s divisible with the direct limit anyway

[void cosmos]

and the direct limit is like

[topaz solar]

It is quite literally the same thing

[void cosmos]

the union of those and any two elements that differ by some power p are the same?

is that it?

no

or idk

that would be the inverse one

[delicate orchid]

you can include the group of units with order at least p^k into the group of units with order at least p^{k+1} by inclusion

[void cosmos]

with mulitplication with p

[delicate orchid]

and this time I do mean inclusion

literal subsets of complex numbers
hopefully with this you can construct an isomorphism

[void cosmos]

okay so the equivalence relation would be like

a ~ b iff a = b mod p^k for some k

or what

[topaz solar]

no homie look at the definition

[void cosmos]

of what

[topaz solar]

The limit

If f(a) = b, then a ~ b, where f is one of the maps in your system

[crystal turtle]

filtered colimits

[topaz solar]

That’s not symmetric, but you can extend that

It’s kinda just building something inductively

[crystal turtle]

transfinite composites

[white oxide]

is b just asking if we're given some stabilizer G_xi, then we can write it as yG_xjy^{-1} for some y in G and x_j in S?

[void cosmos]

so a ~ b iff a mod p^i = b mod p^j for any i,j j>i

[delicate orchid]

lets put some sort of grading on the complex numbers thing right
let p^\infty_k = {\omega \in S^1 : o(\omega) = p^m, m \leq k}
then our inclusions in the direct limit is basically saying that every element in p^\infty_k is the \emph{same} as that element in p^\infty_{k+1} - which is obvious when we embed p^\infty into S^1 as the maps are literally set inclusions!

[void cosmos]

im talking about the direct limit

[topaz solar]

yes

These two notions of Z/p^n Z and p^n th roots of unity in C are the same?

Like my guy they’re isomorphic

[void cosmos]

yea

[topaz solar]

So replace them by roots of unity

[void cosmos]

ok

[topaz solar]

And see that your inclusion on the Z quotients

[void cosmos]

yeah so ur like

[topaz solar]

Are the same as the inclusions in the roots of unity

It’s a square

It commutes

[void cosmos]

talking about the sets {z^p= 1} and {z^p^2=1} , ... and so on>?

[topaz solar]

that is what roots of unity are

So yes

[void cosmos]

yea and ur taking the disjoint union

of those

[topaz solar]

No??

[delicate orchid]

kind of

gimme a moment

wait is direct limit is colimit right

[topaz solar]

We could, but things are equivalent iff they are equal under an inclusion

Yes

[void cosmos]

so ur tkaing disjoint union of those?

[topaz solar]

How do you define the direct limit homie

[void cosmos]

okay fuck my definitions

give me

one

i trust u

lol

let A_i be abelian groups indexed by I

and phi_i,j be phi: A_i --> A_j be homomorphisms

and then consider the disjoint union of A_i

[delicate orchid]

I only got one for general direct limits lying around lol

[topaz solar]

Real

Now moamen

Use ur thinker for just a second

[delicate orchid]

I lost my spiel by copying and pasting this ffs

[void cosmos]

define a ~ b iff phi_i,j(a) = phi_k,j(b) for any i,k,j

then the direct limit is the set of all equivalence classes under this relation

thats it

[topaz solar]

What’s your equivalence relation if A_i is a subgroup of A_j for j>i

[void cosmos]

is that correct?

[topaz solar]

Yes

[void cosmos]

okay

[topaz solar]

It’s literal inclusion

So when are things gonna have the inclusion be equal

[void cosmos]

its like when u go back

ig

in a way

or wdym

[topaz solar]

Give an explicit characterization

[void cosmos]

arent then they equivalent if they are in the same group

[topaz solar]

Bro no

[delicate orchid]

now lets say \iota_k is the inclusion from the p_kth thing to the p_{k+1}th thing
so consider an element in the coproduct \coprod_{i=1}^\infty Z/p^iZ, call this element a = (a_1, a_2, ...)
this element can only be in the direct limit if \iota^{op}_k(a_k) = a_{k+1}

I think - sharp, thoughts?

[topaz solar]

I don’t like it

[delicate orchid]

I don't either

[topaz solar]

For one, that’s not how coproduct should be

[void cosmos]

if they are the same element ig

[topaz solar]

Direct sum is cofinitely zero

[topaz solar]

Yes

[void cosmos]

yea

[topaz solar]

So take a union

[delicate orchid]

agreed, something wack is occouring

[topaz solar]

That’s the inverse limit

But you swapped the arrows

[delicate orchid]

oh for FUCKS sake I hate this naming convention

[topaz solar]

I know

[void cosmos]

who

are u talking to

[topaz solar]

Wew?

[void cosmos]

okay

[topaz solar]

Bro?

There’s a reply feature used there

[delicate orchid]

I'm putting an ^op on the inclusion I don't care anymore

[topaz solar]

Still doesn’t work because coproduct don’t do that

Quotient the coproduct

[delicate orchid]

no.

[topaz solar]

That coproduct should be a disjoint union, so in set anyway

[delicate orchid]

anyway focus on moamen

[topaz solar]

Anyhow, moamen convince yourself that if we coherently replace each thing in the direct limit with an isomorphic thing and the squares commute, then we get an isomorphic direct limit

[void cosmos]

im trying

to

[topaz solar]

If you have that, then you automagically get the direct limit out of the unit circle one

[void cosmos]

write out the direct limit

with a mod p^j --> pa mod p^k

being the morphism

so i have the relation being pa mod p^j = pb mod p^j iff a ~ b

[topaz solar]

ORRRRR
Show it’s divisible by just using how multiplying by p shifts you down a step surjectively

[void cosmos]

i wanna show that this is the same as the p^k roots of unity for k inte

[topaz solar]

It is not the whole circle

[void cosmos]

yea b

mb

if u can help me show this

then both i dealt iwht the direct limit and i can show its divisble

[void cosmos]

this is how wew did it i think

and i would be eternally grateful

untill the end of the world

[topaz solar]

That’s basically what I did to convince wew

[crystal turtle]

Sorry, are you suggesting the direct limit should first be formed in Ab by taking disjoint union?

[topaz solar]

Yeah because concrete things are quite nice

[crystal turtle]

It's direct sum my friend

[delicate orchid]

same thing it's a coproduct

[topaz solar]

Real

It would be quotiented out anyhow

[void cosmos]

anyone? im super tired

and i wanna sleep and im so stupid now

[delicate orchid]

are you telling me that if.... a = a... then... a ~ a....

[void cosmos]

is htis correct

[topaz solar]

The inductive treatment of it via the morphisms won’t work for moamen atm

[void cosmos]

direct limits make sense i think but im just stupid now so sorry guys

its like what cat said

u can have ur own sense fo eventually getting closer now

of*

[topaz solar]

Can you not show Z/nZ is iso to the nth roots of unity?

[void cosmos]

with ur morphism of choice

[void cosmos]

i can

[delicate orchid]

then do that but more

[void cosmos]

okay wait

ohh

i think i got u now

if i think of the Z_ns as those circles

then it wont be multiplication by p now right?

it owuld be the inclusion

is that correct?

[delicate orchid]

.

sorry I put my headphones down and they hit my keyboard

[topaz solar]

I find it very funny to think you sat them down because of this

[topaz solar]

This is what we have been saying

[void cosmos]

yea mb

i got it now

and now

with the inclusion being the morphisms

a ~ b iff a = b anyways

right?

[topaz solar]

Yes

[void cosmos]

so it would just be the disjoint union

[topaz solar]

So union em

[void cosmos]

of all the

p^ns rootsh of unity

so thats it literally

[topaz solar]

Well don’t disjoint Union

[void cosmos]

like x is in this iff x^p^k = 1 for any k

[void cosmos]

normal union

mb

[topaz solar]

They should be very much so overlapping

[void cosmos]

yea

cuz they are all

subsets

anyways

lol

[topaz solar]

One

Word

Every

Message

[void cosmos]

yea mb adhd

phew well i hope ur grateful u dealt with one of the greatest minds in math history

thanks guys

[white oxide]

could i get a hint for c) pls

[topaz solar]

What’s G_x

[delicate orchid]

stabliser for x perhaps?

[topaz solar]

I meant for them to write it, but welp consider N is a kernel of some sort

[delicate orchid]

oh shit sorry lmfao

[terse crystal]

Stabilizer
Consider G/G_x-> Bij(S,S)

[topaz solar]

Ur the group guy, I was the one interning your domain

[topaz solar]

“”Hint””

[terse crystal]

I know p-Prüfer group is divisible obviously. But how do we prove the limit is p-Prüfer in the first place… why it can’t be larger…
Like an element (x_1,x_2,…) such that image of x_i under inclusion Z/p^iZ->Z/p^(i+1)Z is x_ i+1
How do we know that when n large enough it’s constant…

[delicate orchid]

upon the witnessing

[topaz solar]

Well, for one, it’s inclusion so…

[delicate orchid]

^

once it hits anything it's constant forever

[topaz solar]

It’s kinda constant by definition

Literally that whole sequence is the same element

[delicate orchid]

this is what seemed off to me in my little doohicky up above

[topaz solar]

That’s not just if the map is literal inclusion too btw

[delicate orchid]

hmm

[topaz solar]

Since x_i ~ x_{i+1}

[delicate orchid]

I really don't like the fact that it's constant forever

I could believe that once it wasn't the identity it was constant forever

I might just be skill issuing here

[terse crystal]

Oh nvm… I view Z/p^iZ as a subgroup of p-Prufer group , and inclusion Z/p^iZ-> Z/p^(i+1)Z as inclusion of two subgroups of p-Prufer group in the first place…

[delicate orchid]

like don't get me wrong it absolutely is a constant sequence by the definition of a limit lmfao

[topaz solar]

We have $\phi_i: A_i \to \lim_{\rightarrow} A_i$ and $\phi_j(\iota_{ij}x_i) = \phi_i(x_i)$

[cloud walrus]

Dragonslayer Sharp

[delicate orchid]

but then how do we get anything other than pth roots of unity?

almost like we need a bicomplex going on or something

[topaz solar]

Well, we get the p^n+1th too

[delicate orchid]

no we don't. Inclusions preserve order

[topaz solar]

Because inclusion isn’t necessarily surjective

[delicate orchid]

exactly

that's my problem

[topaz solar]

??

[delicate orchid]

what is the representation of a p^2th root of unity in the direct limit

[topaz solar]

an element from the p^2th set passed up through the inclusions

[delicate orchid]

we cannot place anything in the x_1 slot as then the entire sequence is equal to a pth root of unity

[topaz solar]

They’re not sequences

[delicate orchid]

you know what I mean

tuple

[topaz solar]

So there isn’t an x_1 slot

[delicate orchid]

it's a quotient of the coproduct

[topaz solar]

So it’s not sequences

[delicate orchid]

this is so much easier when it's over an arbitary category one second

take the direct limit over (N, \leq)

consider the funny triangles

right ok yeah I got it now

[topaz solar]

Coproduct isn’t functions

[delicate orchid]

you do literally just start at the p^2th root of unity and keep going up

[topaz solar]

Coproduct is like

Direct sum

[delicate orchid]

ok you don't have to patronise me

I'm very biproduct pilled from working in R-mod for 2387465 years

[topaz solar]

I’m just concerned why we have sequences

Biproduct’d

[delicate orchid]

tbh there's a good chance i've got direct and inverse limits the wrong way around in my head again

[topaz solar]

Infinite coproduct not being biproduct

[topaz solar]

Classic naming conventions never being confusing

[topaz solar]

It’s \bigoplus A_i not \prod

(And further quotiented)

[delicate orchid]

happy to know it's not just me fucking this up

there's just something about limits man I can never get a good intuition for them

[topaz solar]

If you need anything further, try applying ||b) to the fact that N < G_x||

[delicate orchid]

ok, lies

I get limits but not colimits intuitively

[topaz solar]

Tf is that emote

[topaz solar]

Colimits are just inductive-y constructions

[delicate orchid]

thanks!

[topaz solar]

Limits are screwy

[delicate orchid]

limits are completions of rings but more so

[topaz solar]

Based

[delicate orchid]

like they're so easy to think about

[topaz solar]

I think colimits are nicer

[delicate orchid]

I realise now that my general category definition I posted before was for an inverse limit not a direct limit

[topaz solar]

Names

[crystal turtle]

inverse limits and direct limits can go to hell

[delicate orchid]

NOOOOO MY WHOLESOME SEGAL COMPLETION THEOREM NOOOOOOOOOOO

[topaz solar]

Call em limits or colimits

[delicate orchid]

hmmm... maybe

[topaz solar]

With appropriate adjectives

[delicate orchid]

I'll call em inverse limits and colimits

[terse crystal]

So what was it… the person who asked said limit, I assumed he meant limit as in category theory…

[delicate orchid]

it was a colimit

[topaz solar]

Yes

[delicate orchid]

I think I'm going to cry

[topaz solar]

Direct (co)limit

[terse crystal]

Then it’s just p-adic integers?

[topaz solar]

No

[delicate orchid]

that's the limit

[crystal turtle]

that's just a limit in the dual category

[topaz solar]

Direct limits are colimits

[crystal turtle]

p-adics is limit with projection maps

[delicate orchid]

WOW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[topaz solar]

Inverse limits are limits

[terse crystal]

Oh yeah…

[topaz solar]

Wew & cog actively getting beat by names

[delicate orchid]

I'm at my fucking limit

[topaz solar]

Colimiting

Ultraproducts are colimits

[delicate orchid]

welp guess I'm never doing model theory

[topaz solar]

Well, they’re directed colimits across a bunch of products

[crystal turtle]

based

[delicate orchid]

I need some direction in my life

[topaz solar]

Same

[crystal turtle]

same

[terse crystal]

I am dumb… yeah the cone Z/p^iZ to p-Prufer group is universal…

[delicate orchid]

galactic, even

[crystal turtle]

galactic properties

[uncut girder]

Limit, colimit. Direct limit, inverse limit.

Limit = inverse limit. Colimit = direct limit.

[coral shale]

ok lets be real, how many math terms are unnecessary

over half i bet...

[delicate orchid]

Way under half

Weird collisions like this are always noticed because they’re so obnoxious and annoying

[white oxide]

dumb question. for the right direction, we have p divides G and every element an order of a power of p, how does that imply that G is of order p^j for some j \geq 0? what if we have G = p^jm for some m which doesn't divide p and j nonnegative?

[topaz solar]

m won’t divide p because p prime, but anyhow, see how if a prime q divides G then there’s an element of order q

[white oxide]

yeah

[topaz solar]

So if q is the order of an element

It’s a power of p

Because G is a p-group

[white oxide]

i see

i think i understand

[rain acorn]

Why the restriction (underlined with red) on multiplication?

[onyx trout]

A vector space is over a field of scalars

[rain acorn]

This is to define a vector space on a field extension to prepare for the study of constructibility

[onyx trout]

Here L is a space and K are the scalars

Akin to how we can multiply a vector in R^2 by a real number

But not by another vector

Does this help at all

[rain acorn]

Sorry for late response, thank you for the help

[rocky cloak]

Isn't the whole point of direct limits that you can construct them in Set (i.e. using the disjoint union).

The construction with direct sums works for any colimit, so then restricting to direct limits is kinda pointless.

[onyx trout]

Me??

[terse crystal]

Yeah… I don’t know why some people require colimit defined when index set is direct too, when the general case can also be constructed…

[rocky cloak]

I mean you can construct it either way, but the construction for direct limits is much easier to deal with an has much nicer properties like being exact.

[terse crystal]

What is the general result about colimit we have?
(Like do we have taking colimit is right exact or something? in what categories?)

(I thought I saw taking colimit is right exact in small abelian categories, something like that, not sure…

[delicate orchid]

Taking colimits is always right exact I’m pretty sure

Where it makes sense to talk about exactness of course

[rocky cloak]

Yeah colimits commute with colimits, because it's a left adjoint. And that works in any category.

In particular it's right exact.

[terse crystal]

Oh thank you both. I am dumb, I forgot adjoint things F -| G right where G is like constant , G(B)(i)=B G(B)(f)=id B thing right

[terse crystal]

Wait then what is special for direct limit ? You said exact, taking direct limit preserve monomorphism?

[rocky cloak]

Yes, in Mod R anyway

Like each element in the direct limit is actually mapped to by something in one of the objects in your system. And something is mapped to 0 in the direct limit if it is mapped to 0 by some function in your system

So you have a really explicit description of what the elements in the direct limit are.

Where as for a general colimit, you can think of it as sums of things coming from your objects, but it's pretty tricky to tell when two such sums are "the same".

[terse crystal]

Yeah, I see

Oh like some arrows having same source and target

I remember i saw counter example of colimit case once , coequalizer used

[rocky cloak]

Yeah, the cokernel is not exact

[terse crystal]

Thank you. No wonder there is a name of it

I just hope people in the future can call it direct colimit or something 😂

[rocky cloak]

The real question is why care about direct limits, when filtered colimits have all the same nice properties but are slightly more general. And also have a less confusing name

[delicate orchid]

Are direct limits explicitly on the funny chain category?

I thought they were literally just colimits but with a different name

[terse crystal]

I see, filtered , interesting

[rocky cloak]

Super annoyingly some people use "direct limit" as a synonym for colimit. Which is just the stupidest naming convention ever

[chilly ocean]

Maybe somebody can help me with this:
There are math overflow comments that the representation theory of the exterior algebra on a ≥3 dim vectorspace is wild, since it is a quotient of a pathalgebra with 3 loops. I cannot find any treatment at all of when quotients of pathalgebras are wild -- only of when pathalgebras themselves are wild. Does anybody have a reference?

[delicate orchid]

If a pathalgebra contains the rose with 2 petals it’s wild right?

So uhhh where did I read that fact

Something to do with simultaneous diagonalisation of two matrices I’ll have a look for it

[chilly ocean]

I think if there is even one loop in the pathalgebra it is wild. But I am a differential geoemtry phd student and not an algebraist.

[delicate orchid]

The delooping groupoid of the trivial group definitely shouldn’t have wild representation theory I don’t think

You could be right though I’m no expert myself

@rocky cloak thoughts?

[delicate orchid]

No this should definitely be the case - there’s only one parameter to change in the entire rep here

[chilly ocean]

here is what I meant about a single loop making it wild
https://www.math.uni-bielefeld.de/~sek/kau/leit6.pdf
Theorem. If Q is a connected quiver which is neither a Dynkin quiver nor a Euclidean quiver, then Q is strictly wild

[rocky cloak]

Ringel computes a few minimal examples of representation wild algebras
https://link.springer.com/chapter/10.1007/bfb0081230

If you can show that your algebra has one of these as a quotient, then it will be wild

[peak root]

Wait. Aren't direct limits the same as filtered colimits, where the index category is filtered or directed. I think that's why it's named so.

[chilly ocean]

Thanks, that sounds like a great reference

[rocky cloak]

direct limit = indexing category is directed system
filtered colimit = indexing category is filtered category

[chilly ocean]

If it has one of them as sub-algebra it is also wild, correct?

[rocky cloak]

No, I don't believe so. Not without further requirements anyway

[chilly ocean]

Alright, thanks. I'm sure I can reconstruct one of the algebras in this paper as a quotient of an exterior algebra, just need to think so more. So its very helpful 🙂

[rocky cloak]

The algebra he calls (a) is a quotient of the exterior algebra.

[chilly ocean]

Oh really, I was looking at the others because I didn't understand what M_2 is supposed to be

[rocky cloak]

It's just everything of degree two

[chilly ocean]

That means the quotient is the algebra with 3 generators and all products are zero?

[rocky cloak]

Exactly

[chilly ocean]

Wow, it feels like that should appear as a quotient of next to everything

[delicate orchid]

indeed! tame representation type is very rare

finite even more so

[rocky cloak]

I think its pretty easy to see that this is wild actually.
If (V, A, B) is a represenation of k<a, b> (A and B matrices acting on V)
Then just let X, Y and Z act by upper triangular matrices with I, A and B in the upper right corner and 0 everywhere lese

This will then give a representation embedding of k<a, b> into k[x,y,z]/M_2

should be fully afithful and everything

[chilly ocean]

I'm guessing it works like this: Suppose you have a decomposition of the rep. of k[x,y,z]/M_2 (wich is a rep on V \oplus V), project the decomposition to the first V factor to get a decomposition of V. I can't see why the result is invariant under A, B

[crystal turtle]

Oh is it? I don't actually use them that much lol, I had only just seen it written with direct sum

[rocky cloak]

An argument like this shows that if Phi is any idempotent xyz-linear map on V^2, then it's just given by an ab-linear map on V

So in particular if this is a projection down to a summand, then V also has a nontrivial summand

@chilly ocean

[chilly ocean]

Very nice, thanks 🙂

I was looking at certain families of spherical harmonics, it was a surprise when googling told me that the classification problem I encountered is a classical impossible problem xD

[rocky cloak]

Yeah, there is suprisingly little complication needed before problems become wild

[old spire]

Why are solvable Lie groups and algebras natural to consider?
What's the intuition behind them?
They seem to be closed under various things: https://en.wikipedia.org/wiki/Solvable_group#Properties

They show up in Levi's Decomposition theorem, but they don't seem intuitive.

[delicate orchid]

do you want specifically lie groups or will an exposition about solvable groups be ok

cause I don't know enough about solvable lie groups to comment

[old spire]

Ideally, their relevance to Lie groups would be better.
Wikipedia provides some motivation with solving polynomial equations (in characteristic 0) and Galois theory.

[delicate orchid]

yes that's the classical motivation for them, I'll let someone else with more lie group knowledge answer instead

[onyx trout]

The original motivation was the Galois theory stuff

For Lie groups / algebras the decomposition here & the associated exact sequence can be seen as showing solvable ideals measure obstructions to being semisimple

[delicate orchid]

oh that's cool

[onyx trout]

By Cartan's criterion we can also look at the Killing form as measuring this

And in some sense semisimple / solvable are "opposite"

I don't know if there's any great intuition for this in terms of the derived series

Lie groups were originally considered to try and extend Galois theory to differential equations, if I remember my history

So I believe this was the first motivation for working with these

However it was quickly realized that Galois theory doesn't bring much to actually difficult differential equations

[white oxide]

i'm a little confused - so given a group K, we say that aKa^{-1} for a in G is a conjugate of K when it's of that form, but we say that H and K are conjugate if H = aKa^{-1}?

[delicate orchid]

H and K are conjugate if and only if.... H is a conjugate of K...

I don't mean to be rude but that's fairly obvious

[summer path]

I thought you were asking more advanced questions a few days ago

As in not group theory

[delicate orchid]

hey group theory can be advanced.....

[onyx trout]

Good to solidify your basics

[summer path]

Or maybe all the blue names have just blended together

[white oxide]

yeah i think i'm mixing up the definition of what it means to a conjugate of K and when H and K are conjugate

are they the same thing

[delicate orchid]

just because it's basically solved and a dead field doesn't mean it's not advanced...

[delicate orchid]

like this is such a minor difference in language I'm surprised you're getting tripped up on it this late in

[white oxide]

it's aight i'll just look it up somewhere else

[delicate orchid]

if H is conjugate to K, then H = aKa^-1 for some a

like, by defintiion

of both things

[summer path]

That's not what I meant but I thought okay was the one asking about modules

[delicate orchid]

that could be moamen?

[summer path]

Oh

[old spire]

It seems that Cartan's criterion might follow from the Levi decomposition theorem. That, or the other way around.

[summer path]

Blue names

[delicate orchid]

blue name [derogatory]

[summer path]

If I shitpost enough I'll get a blue name too

[white oxide]

yeah but here in the pic we say that hKh^{-1} is conjugate to K but it's not equal to K

[delicate orchid]

I didn't look at the pic tbh

and yes...? what's the problem?

K doesn't have to be normal in H?

[white oxide]

but i thought a conjugate of K had to be equal to K by definition?

[delicate orchid]

??????????????????????????????

[white oxide]

bruh

[delicate orchid]

post your definition of conjugate

there could be a conflict of language going on here

[old spire]

It seems that "solvable" should have a geometric motivation that isn't just its historical origin in Galois theory. I suppose the Levi Decomposition should suffice. I don't know what I'm asking for.

[white oxide]

Let H be a subgroup of G, acting on the set of subgroups of G under conjugation. For h in H, we say that hKh^{-1} is a conjugate of K (they're isomorphic but not necessarily equal)

[old spire]

Levi Decomposition reminds me of the Witt decomposition of a non-degenerate quadratic form.

[delicate orchid]

well there we go

[white oxide]

but they're isomorphic and not necessarily equal?????

[delicate orchid]

yes?

[white oxide]

are we considering them up to isomorphism then?

[delicate orchid]

there are three copies of C_4 in Q_8 all conjugate

[white oxide]

because fundamentally there's a difference between isomorphism and actually being equal

[delicate orchid]

we're doing the opposite? we're saying they're not equal

[old spire]

Because you've got these nice factors ("simple" ones maybe?) and then this giant horrible factor with "here be dragons" written all over it. In the Witt decomposition case, it's the "anistropic" factor, and in the Levi decomposition it's the solvable subalgebra.

[delicate orchid]

I have absolutely no idea where on earth you got the idea that conjugates are equal

perhaps phrasing it as

"two subgroups are conjugate if they're in the same orbit given by the action of conjugation by some larger group"

makes it clearer?

[white oxide]

yeah sure thanks

[onyx trout]

It uses Jordan decomposition which can be seen as a speciation / variant of Levi's

[onyx trout]

What did you do to me!!!

[onyx trout]

Though they are distinct

[delicate orchid]

smote

[white oxide]

could i get a hint for this question pls

[delicate orchid]

think about the elements of H that are fixed by conjugation from G

[white oxide]

yeah that's what i was considering earlier, here I let G act on H by conjugation but that gives H is congruent to H_0 mod p where H_0 = {h in H | gh = hg for all g in G}...

and H_0 is exactly the intersection of H with the center of G i believe

wait i'm stupid is that it

cuz |H_0| \geq 1

[analog tree]

Was searching around, didn't really find anything specific - has any study of the direct limit of the cayley dickson algebras been done? The most I could find were some stackexchange threads that culminated in "rather than looking at the direct limit of CD, just find the minimal CD algebra that contains your the two elements that are being operated"

[old spire]

"speciation"?

[white oxide]

oh wait no this wouldn't work bc they could both have remainder 1 when divided by p

fuckkkk

[onyx trout]

Sorry this is a nonsense term here

I meant as in Levi decomposition can be seen as a generalisation of Jordan decomposition

[old spire]

So Levi => Jordan => Cartan's criterion?

I don't see how Levi => Jordan.

You'd have to see an element g of a Lie group G as somehow giving rise to a Lie group in its own right, which you then decompose.

[onyx trout]

The Jordan decomposition of a linear operator has a case for (the Lie algebra of) endomorphisms of a vector space

Sorry I was a little vague I do not believe it directly implies this

[old spire]

Spiritually, they look similar. Just look at the terminology.

[onyx trout]

Yes

Nilpotent algebras are all solvable

[old spire]

Hmmmm is one the decategorification of the other? 🧐

[onyx trout]

You might actually be able to recover Jordan decomp from Levi via the adjoint representation

Sorry I just got back from lunch and am a little distracted so those last couple messages are a little nonsensical

[chilly ocean]

I have a problem

L commutative field and K a subfield of L with K≠L s.t. for every a in L\K, there is b in L\K with a+b in K and ab in K. If 1+1≠0 show there is x in L\K with x^2 in K and L={a+bx| a,b in K}

I supposed that for every x in L\K, x^2 in L\K

So x+x^2 in K and x^3 in K

If 1+x in K then x(x+1)=a with in K so x=a(1+x)^-1 so x in K false

So 1+x in L\K for every x in L\K

[topaz solar]

|H| divides |G|

And |H| > 1

[chilly ocean]

If 1 in K: 1+x=b with b in K so x=b-1 but because K subfield then x in K false

So 1 in L\K

x+1 in K for every x in L\K so x(x+1) in L\K because x in L\K

[topaz solar]

Are you sure 1+x is in K

It just guarantees some b for x+b in K

(And xb in K at the same time)

[chilly ocean]

Oh sorry

x in L\K implies x+1 in L\K

Yeah I was wrong

I need to rethink

2x in K

[topaz solar]

Think how you’d find i in C

@chilly ocean take x in L\K, and let’s say the guaranteed term is y in L\K

What’s (x+y)(xy)

[chilly ocean]

Thats in K