advanced-algebra

[spice idol]

if youre talking about nontransitive relations then sure ig tolerances are pretty useful to finite lattices

[verbal panther]

Are there weaker forms of invertibility

[spice idol]

in monoids youve got semiinverses

or smt

a, b st aba = a and bab = b

[verbal panther]

Interesting

Wdym

[spice idol]

if theres an element n such that an = b then n may not be unique

[verbal panther]

Is this it’s own property?

[cerulean cove]

I guess you could just describe it by saying the function of multiplying by a on the left is surjective, for all a, but idk if there’s any like specific name for it?

[eager hound]

You'd call it a-divisible

[cerulean cove]

Thank you!

[eager hound]

For every a there is not really a word for it

[verbal panther]

Would there also being an n so that na=b change the structure?

Or would those statements be the same

[verbal panther]

Seems kind of like a weaker version of this?

[eager hound]

Quasi quasi group

[foggy galleon]

can anyone explain what Grothendieck's spectral sequence is saying? In some texts this is buried under dozens of definitions, but I don't think it should be that complicated to state what's going on?

If I understand correctly, there should be maps d^(p, q) : R^pG o R^qF(A)-->R^(p+2)G o R^(q-1)F(A) (by "o" I mean composition), such that dd=0.

But then what's the relation with R^(p+q)(G o F)(A)? Should it be

ker d^(p, q)/im d^(p-2, q+1)=R^(p+q)(G o F)(A) ?

If so, doesn't this imply R^1G o F (A)=R^1(G o F)(A) ? Because R^pG o R^qF=0 if one of p or q is <0?

[zealous spruce]

the relation is that said spectral sequence R^p G o R^q F converges to R^(p+q) GoF, in particular there is a priori not the equality you speak of.
It basically amounts to "how a resolution of F and a resolution of G amounts to a resolution of G o F", and grothendieck spectral sequence is the answer.

As to how it function, you can see each page as a more precise approximation of what you're trying to get

the relation will depend on what properties you have, it can be what you wrote in some cases but will not most likely, you can check whatever source you want for the definition of the limit (i recall even wikipedia) and write out the isomorphism you get from the filtration

[foggy galleon]

so can it be stated precisely in simple terms or no?

[zealous spruce]

in full generality, i do not know such thing and it probably isn't possible

well i mean not better than what the definition would give

[foggy galleon]

I'm just asking if the definition is simple or not

Rotman gives 29 definitions before getting to Grothendieck's spectral sequence

[zealous spruce]

I think you should read them if you haven't

[foggy galleon]

why?

I'm just asking how economically can you state precisely Grothendieck's spectral sequence. I doubt you need 40 pages

[fierce steeple]

I mean this just sounds like a general question about spectral sequences tbh

[ornate atlas]

No ones ever accused Rotman of being too concise

[spice idol]

💔

[zealous spruce]

You can always read ncat or Wikipedia if you want more concise things, but you cannot expect such a general tool to have a more concise statement than "this spectral sequence converges"

[foggy galleon]

I mean yeah that's what I'm asking the definition of. I'm just taking Grothy's spectral sequence as a concrete example

[eager hound]

Im confused what you’re asking when you ask about the “definition” of a spectral sequence. But I think the useful statement that pops out is that the spectral sequence converges to R^( F \circ G ), I.e. point wise in (p,q) the differentials become zero, and R^(F \circ G) has a filtration with subquotients the “stable” entries of the sequence

Eg R^n(F \circ G) has a filtration with graded pieces \lim_{i} E^{p, n - p}_i, (the limit eventually stabilizes)

[golden osprey]

No idea if this is the right channel but I'll just ask here and someone tell me to ask somewhere else if needed. 
For reference this comes from the proof of Theorem 1 of the paper \emph{Complexity Issues in Bivariate Polynomial Factoring} by Bostan et al but I don't think that is super important.
This is likely just some basic power series stuff that is eluding me. 
I wonder if I should ask on math SE instead.

Let $k$ be any field and $F(x, y) \in k[x, y]$ a polynomial of total degree $d$.
We will assume that the degree of $F$ with respect to $y$ is also $d$ so that $F$ is monic in $y$.
Furthermore, we will assume that the resultant $\text{Res}_y(F, \partial_y F)(x)$ is such that $\text{Res}_y(F, \partial_y F)(0) \neq 0$.
Then say $F(x, y) = \prod_{i = 1}^s \mathcal{F}_{i}(x, y)$ is the factorization in to irreducibles of $F$ over $K[[x]]$.
Let $\phi_j(x)$ be any series solution of $\mathcal{F}_j(x, \phi_j(x)) = 0$.

First question: the paper states that since $\text{Res}_y(F, \partial_y F)(0) \neq 0$, we must have that $\phi_j(x) \in K[[x]]$ instead of $K((x))$.

Second question: the paper states that $\partial_x \phi_j(x) = \partial_x F / \partial_y F(x, \phi_j)$.

I don't see how either of these hold.

[broken turtle]

Spamakin🎷

[lone jacinth]

The second question is just the chain rule at least.

Like G(x) = F(x, phi(x)) is identically 0, so
0 = d/dx G(x) = d/dx F(x, phi(x)) + d/dy F(x, phi(x)) * d/dx phi(x)

Then you rearrange to isolate d/dx phi(x)

[golden osprey]

oh duh

[woeful crane]

I have a pair of matrices
$$X=\begin{pmatrix}
0 & 1 & 0 & 0 \
x & 0 & 0 &0 \
0 & 0 & 0 & y\
0 & 0 & z & 0
\end{pmatrix},$$
%
$$Y=\begin{pmatrix}
0 & 0 & 1 & 0 \
0 & 0 & 0 & u \
v & 0 & 0 & 0 \
0 & w & 0 & 0
\end{pmatrix}$$
where $x, y, z, u, v, w$ are non-zero complex numbers, and for... reasons, I know that I should be able to decompose $\mathbb{C}^4=U \oplus V$ where $U, V$ are 2-dimensional and each are invariant under $X$ and $Y$.

What is the most pain-free way to determine $U$ and $V$?

[broken turtle]

kr1staps

[woeful crane]

Here "determine" means provide bases as column matrices whose entries are (presumably) some combinations of x, y, ..., w

[lone jacinth]

Well, U and V would need to be sums of eigenspaces.

It looks like X should be pretty easy to diagonalize, then you can just see what Y does to the eigenspaces.

[woeful crane]

Thanks, I'll give it a go

[lone jacinth]

Is there a specific relationship between these numbers x, y, z, ...?

Seems to me you need one

[worldly zealot]

is there some generalization of ringel-hall algebras to infinite fields? like for a simple module (say, supported at one vertex of a quiver), computing the hall number of S, S, S (+) S is basically asking for injections from S into S (+) S, which is naturally P^1. so instead of counting you are looking at an appropriate variety – this seems tractable for someoen much smarter than me

[buoyant harbor]

please hellp

i have no idea what the lattice looks like

please

[spice idol]

well you can just start writing out all the meets and joins

the meet of N and M is N ∩ M, and the join is NM

[buoyant harbor]

im struggling with drawing the lattice

https://en.wikipedia.org/wiki/Zassenhaus_lemma

[spice idol]

well you start with A and X, and then little by little fill in the rest if the objects youre working with

[forest turtle]

you are looking at the lattice right here!

its the butterfly

[buoyant harbor]

wait, is the question exactly that?

but like C is x and D is Y???

huhh??

[spice idol]

yeah

[digital parcel]

they turned u into a lie algebra... wtf...

[spice idol]

i got frakked

[spice idol]

the point is, if you draw the lattice yourself (even with this reference) youll understand better whats happening in the proof

[buoyant harbor]

did they purposefully came with the butterfly lemma just to draw a butterfly>?

[spice idol]

its used in the proof of either Schreier refinement theorem or Jordan-Hölder

both are very important theorems

mathematicians just like to give silly names to things

[digital parcel]

snake lemma, horseshoe lemma

[spice idol]

the lemma that is not burnside's

(also known as burnside's lemma)

[spice idol]

(i think my favorite has to be the pentagonator though)

or this

[near lantern]

If R is a (finite) root system in a vector space V and W a vector subspace of V, is W ∩ R a root system (in whatever vector subspace of W it spans)?

[spice idol]

yes

at least, looking at the definitions from wikipedia i cant see any reason why not

[weak lodge]

yes, that's right

[foggy galleon]

Just asking the definition, I was just overwhelmed with all the abstraction. So in page 2 we have R^pG o R^qF(A) and maps d^(p, q) : R^pG o R^qF(A)-->R^(p+2)G o R^(q-1)F(A). Then the next page consists of E_3^(p, q)=ker d^(p, q)/im d^(p-2, q+1) and we have maps E_3^(p, q)-->E_3^(p+3, q-2) and we can form the fourth page E_4^(p, q), etc. Then for all large enough n E_n^(p, q)=R^(p+q)(F o G)(A), yes?

[eager hound]

No

You just get a filtration in the end

Whose subquotients are the E_n(p, q)

For n >> 0

To me (to everyone?) a spectral sequence always is basically the spectral sequence of a filtered chain complex

If you take appropriate resolutions Grothendieck’s spectral sequence reduces to this case

For a filtered complex it’s much more transparent what is going on: you’re taking a filtration of a chain complex and by allowing cocycles to have differential zero up to some element of the filtration and requiring that coboundaries come from some level of the filtration you are calculating successively better approximations to the cohomology of the complex

A prototypical example is the spectral sequence associated to the filtration of the chain complex C_(X) by the two step filtration with nontrivial piece C_(Z) for Z a closed subspace of X and the complexes those of simplicial chains

There you see that you can compute the cohomology of X in terms of the cohomology of Z and the relative cohomology of X wrt Z but the two pieces only give you a filtration on the answer

Then if you are lucky you can hope that the filtration splits for some reason

Or you only care about the rank of the answer so the filtration is fine

Etc.

[foggy galleon]

what does this mean then? Is this the filtration stuffs but implicit?

E_infty^(p,q) is lim_n E_n^(p,q) which stabilizes as you said, no?

[eager hound]

I don’t really understand the notation here, maybe you are in some situation where only one (p, q) gives a nonzero E^{p, q} on the last page?

Otherwise I don’t see how this could be true

[foggy galleon]

https://ncatlab.org/nlab/show/Grothendieck+spectral+sequence

[eager hound]

I think they must be using some fancy notation

But that map is not, say, an isomorphism.

[foggy galleon]

so for practical purposes H_bullet is just the disjoint union of the R^n(G o F)(A), no? Or is there some other relevant structure?

also for some reason they have changed to homology notation I think?

(this is from https://ncatlab.org/nlab/show/spectral+sequence)

and in the Grothendieck case the filtrations are finite no? As in you have ...subset F_(p-1)H_n subset F_p H_n subset... for each p<=n and for p=n this is R^n(G o F)(A)? Because in this case p, q should be nonnegative I think (or if one is negative then it should be zero)?

-# it's possible I have mixed up homology and cohomology notation, but ignore that

[near lantern]

Fun exercise (which also gets someone to check my work for free): what is the ring structure (under tensor product) of the Grothendieck group of finitely generated projective modules over a Dedekind domain R with ideal class group C?

Actually you can do the Grothendieck group of all finitely generated modules too. But in this case don't add relations for short exact sequences, only direct sums.

[foggy galleon]

so another question I had is to which extent does knowing the subquotients of this implicit filtration determine the R^n(G o F)(A)? I mean it's not even true for finite abelian groups that the subquotients of a filtration determine the group (eg. 0 subset 2 Z/4Z subset Z/4Z and 0 subset Z/2Z x {0} subset Z/2Z x Z/2Z). Isn't this disappointing?

[foggy galleon]

doesn't this boildown to computing a oplus b and a otimes b for a, b ideals of R? I believe they are R oplus ab and ab, respectively?

[toxic trout]

Here is lang's butterfly

[eager hound]

You can apply this in any abelian category, for some of them knowing the subquotients is very good (like vector spaces where it’s everything or modules over a PID where it’s pretty good)

A lot of times it’s useful because you can show that most of the subquotients are zero

That’s like 90% of the utility of spectral sequences

[foggy galleon]

nice

[near lantern]

Sure, and also the "organisational" task of seeing what ring that results in.

[foggy galleon]

Can Grothendieck SS be formulated as a spectral sequence of functors with natural transformations or something like that? I'm not sure when functor categories are abelian, so idk if this pov is useful

[lone jacinth]

You'd think it by the group ring ZC

I guess the only thing to show is that fractional ideals are isomorphic iff they are multiples of each other

[forest turtle]

something straight out of dr. doofenshmirtz’s evil lab

[fierce steeple]

Lol I would've just called this an error unless I am mistaken

[foggy galleon]

Edit it

[near lantern]

Nope, definitely not (assuming you mean to identify the basis C of ℤC with the corresponding rank-1 projective modules). Part of the classification theorem is that I (+) J = R (+) IJ, so R, I, J, IJ are not linearly independent. I think you are right that the fact that I (x) J = IJ means the group ring maps onto it and that you can express the ring as the quotient ℤC/(gh+1-g-h : g, h in C).

Which I now realise is even more interesting than my initial description...

[eager hound]

It does seem very wrong

[spice idol]

now i cant stop reading it in his voice

[vast storm]

so that's why there was a faint phineas and ferb smell

[forest turtle]

behold, the pentaganator-inator!!

[spice idol]

the pentagonatorator, an isomorphism relaxing the pentagonator

[limpid horizon]

that sounds interesting. what area of math is that?

something with number theory?

[near lantern]

I suppose it's a (1) question in module theory. Answering it requires knowing the (2) classification theorem for fg modules over a Dedekind domain and (3) familiarity with Dedekind domains to be able to compute certain tensor products. (2) and (3) are both primarily known by alg NT and comm alg people IG. (2) much more comm alg people than (3).

[lone jacinth]

So did you have a description more interesting than ZC/(gh-g-h+1)?

[near lantern]

Well, ||(gh-g-h+1) is the square of the augmentation ideal, so it's ℤC/I^2||. My original description was: ||if we map C to the Grothendieck group by sending the class [I] to e_[I] := I - R = I - 1 (ie formal difference of rank-1 modules I and R), then gh-g-h+1 = 0 becomes e_[I] e_[J] = 0. So you get ℤ ⨯ C with addition componentwise (using multiplication on the C factor, and as a ring C just squares to 0 (equivalently, the unitisation of the rng C with zero multiplication)||. Which of these is more interesting is up to you.

[lone jacinth]

Both pretty cool I guess

[near lantern]

I'm kind of surprised they're the same, honestly.

I/I^2 = G in general?

[lone jacinth]

You need G commutative at least

In general I guess it should be the abelianization of G

It would be H_1(G)

[near lantern]

Let phi: G -> I/I^2: g -> g-1. I is spanned by the range of phi. Also phi(gh) = gh-1 = g+h-2 = (g-1)+(h-1). So indeed phi: G/[G, G] -> I/I^2 is a group homomorphism. Its range spans, so it defines a surjective G/[G, G] (x) k -> I/I^2, where k is the base cring. Now why should it be injective...

[lone jacinth]

I guess the indirect rout is that group homology is the homology of K(G, 1) and first homology is abelianization of the fundamental group.

At least for k=Z that works

I guess you can also just explicitly construct an inverse.

As an abelian group I is just Z^(|G|-1) with basis g-1, so you may g-1 to g, and you can see this sends I^2 to 0.

[nocturne summit]

So idk where to go for this. Im guessing advanced algerbra because its collage algerbra. But I need help with somthing that may sound a little dumb to need help with

[digital parcel]

If the class is called "college algebra" try #prealg-and-algebra

[ornate atlas]

Feel free to ask theres no stupid questions, but I suspect youre looking for #prealg-and-algebra

[nocturne summit]

its called algerbra 3. i callit collage algerbra bc its a collage course. But ok. ty!

[ornate atlas]

Maybe #prealg-and-algebra is about to see its first LES

[spice idol]

LES?

[digital parcel]

long exact sequence

[ornate atlas]

Long exact sequence

[digital parcel]

wow second time you've copied me within the last 3 minutes

it's just so shameless

[spice idol]

omg universal delta functor or wtv

[ornate atlas]

Hey I hurt my wrist today typing is hard leave me be

[digital parcel]

damn

now im the asshole

[ornate atlas]

Yeah think about what you did

[nocturne summit]

lol

[digital parcel]

i am bro it hurts i feel awful

[spice idol]

im used to seeing SES but maybe i havent encountered LESs enough

[digital parcel]

just add a bunch of 0's (or whatever your zero object is) to the left and right and boom

[ornate atlas]

You take your SES, slap on some snakes, and boom, its long

[spice idol]

loooooonnngg

ig homology doesnt really pop up in UA

which is sad i wanna do hom alg

[ornate atlas]

You should find a way to start homotopical UA

Like condesed maths but for UA

[spice idol]

well its gotta be motivated by smt right

[digital parcel]

your motivation is a fields medal

just like scholze

[ornate atlas]

Motivated by wanting your name up there with Scholze

god damnit

[digital parcel]

sorry i'll add a delay to my messages

i'm being super insensitive

[spice idol]

cuz id ever win a fields medal

[spice idol]

im fucking wheezing at this LMAO

[ornate atlas]

I think if you made other people care about UA thatd be fields medal worthy

[spice idol]

you two are just on the same wavelength

[digital parcel]

i think nope gets credit for this one tho he mentioned condensed math

[ornate atlas]

Thank you, i need the appriciation

[digital parcel]

🫂

[spice idol]

maybe idk

it would be cool to lift current methods of UA to categorical settings like operads?

or algebraic theories

and i am making progress on categorizing the basics but im a little stuck as there is no obvious way to generalize the notion of a congruence relation

maybe the categorical notion? is that ever used anywhere??

[ornate atlas]

I mean, you have like a significant head start on any of this stuff compared to probably anyone else alive. I can’t imagine there’s anyone else who knows as much UA as you at your age so like, I think it’s incredibly conceivable that you could have a huge impact on the field if that’s what you wanted to do

[spice idol]

at what point does the math i do stop being low hanging fruit because its UA and its kinda dead

[digital parcel]

enumerative geometry (at least according to wikipedia) was dead for a while

but with motivic homotopy theory it's so prolific rn

[spice idol]

i can imagine UA is very popular under computer scientist people rn

with constraint satisfaction problems

[digital parcel]

boom

there u go

[spice idol]

though its still not enough rahhhh

everything
must be UA

[fierce steeple]

I wouldn't say it was dead without motivic homotopy theory and stuff. At least in the UK it seems a large proportion of AG I see is enumerative stuff

[weak lodge]

you mean like, congruence relations as coequalizers? i think borceux handbook of categorical algebra 2, chapter 3 discusses this

[fierce steeple]

For example because of connections to physics and stuff

[spice idol]

or internal equivalence relations

though this probably doesnt work if i want to extend to operads

coequalisers might work

[digital parcel]

Wikipedia loses again

[spice idol]

wait huh how to congruences correspond to coequalisers?

i see

that may be helpful, thank you

is there a natural way to turn these coequalisers into a lattice?

or poset, at least

[weak lodge]

haven't fully read this paper, but this might have what you're looking for? https://user.informatik.uni-bremen.de/porst/dvis/PORST_AlgebraicLattices_revfinAU.pdf

[spice idol]

yes, thank you very much

i suppose it is nice that this way we can define a quotient of X as a coequaliser c : X → E

and it happens for varieties that c : X → Y is a coequaliser iff the underlying function of c is surjective

[near lantern]

Wow, that's so simple I feel a little bad about missing it! Nicee.

[verbal panther]

is there any basis for operations outputting multiple elements?

so like if you had g(a,b,c) and f(d,e,f)={g,h} you could have g(n,f(x,y,z))

or would that even have any interesting behaviour

[spice idol]

thats just a pair of regular operations

[rose mirage]

it's giving operad but not quite

[spice idol]

this is just a clone brotato chip

i love clones <3

monoids but better

[rose mirage]

congratulations you've defined something so general it is everything and thus nothing. The height of your cold throne has left you unable to see the warm gardens of earth

[digital parcel]

Chills…

[spice idol]

"Thank you"

[vague pawn]

You should really thank him

[spice idol]

i could say the same about operads or categories

[spice idol]

i truly owe it to him

[limpid horizon]

Thank you Wew Lads Tbh

Wew Lads Tbh did indeed help me when i was a wee little boy learning some group theory for the first time in 2023

[digital parcel]

damn u are a server unc been here since 2023 damn

[limpid horizon]

Ikr

I joined in like september 23

[fierce steeple]

https://tenor.com/view/cat-scared-gif-27466237

[limpid horizon]

But then i got kicked cause i was trolling 💀

[digital parcel]

i got a 2 week ban for trolling

like 2 or 3 years ago lol

[limpid horizon]

Yup that was before i realized how genuinely valuable this server is 😭

[digital parcel]

originally perma then i was like "please" then they said "okay two weeks"

[limpid horizon]

I asked wew nicely to let me back in and he did 🥺

[spice idol]

2019 is crazy

as long as my account has existed

[digital parcel]

yea being here since 2019 is crazy

dont search my messages and sort by old to find when i joined that's irrelevant to this whole thing

[limpid horizon]

I was in first year undergrad at the time

[spice idol]

proud to say ive never trolled on discord 🙏 🙏

[limpid horizon]

Haha yea i couldnt see u doing that

[spice idol]

great way to get people to do anything is tell them they shouldnt do it under any circumstance

[digital parcel]

what can i say, i have a lot of trust and love for people

[vague pawn]

I am always trolling

even at this very moment

[limpid horizon]

wow

[rose mirage]

I dread to think when I first joined. Might've been 2021

[vague pawn]

I took a course in university "Trolling 101"

[limpid horizon]

Course code TROL101

[rose mirage]

TROL101 - Motivic Mackey Functors

[ornate atlas]

Im also a server unc technically but I was but a wee lad when I joined so Im not sure it counts

[spice idol]

ok boomer you aint fooling no one

[ornate atlas]

22 is young, I am youthful, I can still acurately be described as early 20s

[last talon]

Ancient
(Anyone older than me is ancient)

[spice idol]

WE ARE YOUNG (hey) THERES A FIRE IN YOUR SOUL

[limpid horizon]

Im 25 soon heh heh 🤡

I told a girl my age she said im pushing 30

[spice idol]

thats crazy

[limpid horizon]

Ikr

[fierce steeple]

Yeah moth had not even been born

Ig in 2019 I was starting to apply for uni stuff

[digital parcel]

In 2018 when i joined i had just started high school

[limpid horizon]

U started highschool in 2018?

Bruh i graduated hs in 2018

Lmao

[digital parcel]

Yea i can’t wait to graduate it

[fierce steeple]

Wait are you a high schooler anamono lol

Why did I not know this

[digital parcel]

No 😭

[worldly zealot]

7 year high schooler

[limpid horizon]

Lol

He had to be held back …

[digital parcel]

I’m in my last year of ug

[fierce steeple]

What is this then lol

Oh

[digital parcel]

A joke 💔

[fierce steeple]

Wait how many years is high school in US lol

[digital parcel]

4

[fierce steeple]

To me high school is like 7 years fr lol

[digital parcel]

Oh yeah I guess it’s diff in other countries

I think most of US it’s k-5 is elementary school, 6-8 middle, 9-12 high school

[vague pawn]

high school here is 3

[fierce steeple]

Year 7-13 is here cause no middle ting

[vague pawn]

10-12

[limpid horizon]

Where i grew up it was grade 8-12

[digital parcel]

Yea varies across US too

[fierce steeple]

Ok phew anamono is as old as I suspected then

at least educationwise

[digital parcel]

Yea

[digital parcel]

Who is that btw in the emote

Is that gothamchess

The chess YouTuber

[vague pawn]

yes

[digital parcel]

Lol

[limpid horizon]

It used to remind me of the nostalgia critic

[spice idol]

i started it in 2019

[limpid horizon]

😂

Bruh moment indeed

[digital parcel]

U just started ug right?

[spice idol]

yurr

[limpid horizon]

W

[spice idol]

finally experiencing REAL math

[digital parcel]

Intro to linear algebra

[spice idol]

introduction mathematicw

[fierce steeple]

Ah nice

[limpid horizon]

Lol

[fierce steeple]

Lol

I found these funny always. Like yeah I had to do well in hs exams etc and then arrive and we are taught stuff we have already done

But no I mean makes sense esp w people from diff education systems

[digital parcel]

In the US before undergrad we can get college credit via things like dual enrollment at a local community college or AP exams

Do u guys have that in the UK?

[spice idol]

no not "lol" i hate it i have to write down EVERY LITTLE THING im going insane

[fierce steeple]

Never heard of such a ting in uk

[digital parcel]

damn

[ornate atlas]

If it makes you feel any better I took a masters level analytic number theory course last semester and had to do this

I got marked down for knowing what the roots of unity were and stating “this follows because this is a group homomorphism”

Turns out that lecturer was notoriously pedantic and it resulted in me being more paranoid than I’d ever been before in my life

Pulled through and smashed it but fuck me it’s annoying having to justify every single minute step

[pastel shoal]

Also adds artificial difficulty to the exams if they are timed

[ornate atlas]

Thankfully that course was just coursework, I’d have went insane if I had to do an exam marked by that man

My friend currently is taking further complex analysis with him and that does have an exam, so I fear for him

[spice idol]

i had an amazing math teacher in hs who'd usually not mark small mistakes because he knew you understood the subject matter

granted, there wasnt a final exam in that subject (advanced math), so that probably contributed a bit

[ornate atlas]

I actually also got marked down in that class for denoting the cyclic group C_n, bro said that’s not notation he’s ever seen anywhere before and I should use Z/nZ

[spice idol]

what thats horrible

[pastel shoal]

I had a prof who keeps the maximum score slightly less than the sum total of the questions so making a couple silly mistakes is fine

[spice idol]

actually crazy

how have you never seen C_n before

[spice idol]

goated

[ornate atlas]

Yeah I wasn’t overly happy I think because it was 100% coursework they were just ridiculous in the marking to get a spread

I wised up after the first homework and included every imaginable detail and I did really well but fuck me it was stupid

I hated that course

Also because I just don’t care about analysis or number theory, and the guys notes were terrible, they were in literally the opposite of the logical order

[worldly zealot]

thats pretty dumb

artificial grade spreads

[ornate atlas]

It wasn’t entirely artificial because a good few people did just legitimately struggle, but I think there was a reasonable number of us doing pretty close to perfect and so they had to kinda separate the best of the best

Which kinda sucks but I mean it’s fixable, I think I ended the course with like an 86 or something (70 is an A)

I also lost marks by just entirely fucking up a question on character theory which really should’ve been the one I could do

[rose mirage]

write down the entire character table of S_5 (2 marks)

[rose mirage]

do you remember what the question was

[ornate atlas]

I don’t, but writing down the character table of some group was part of it

And my UG account is finally dead, can’t even check 🥀

[lone jacinth]

Pretty sure I created discord just because someone invited me to this server

[fierce steeple]

You created discord

Damn

[ornate atlas]

I joined to beg for help in my nat5s (GCSEs), I was definitely a child mass pinging in the help channels

[spice idol]

and jagr said, let there be discord

[rose mirage]

1 -1  1 -1  1  1 -1 
4  2  1  0 -1  0 -1
4 -2  1  0 -1  0  1
5  1 -1 -1  0  1  1
5 -1  1  1  0  1 -1
6  0  0  0  1 -2  0```

had to actually get a pen for ts one

[spice idol]

what if i dont want to

[rose mirage]

do S_4 then

S_4 is nice and easy

[spice idol]

right

id still have to get pen and paper

lol

[vague pawn]

W ANT

[rose mirage]

nah you don't I believe

the 2 dim is a pain ig

[spice idol]

conjugacy classes are ways to partition { 1,2,3,4 } right

[rose mirage]

nah actually it isn't I forgot S4 isn't almost simple

[rose mirage]

yur

[digital parcel]

been a while since i've done character tables lol

about a year

[spice idol]

soo 5?

i promise it did not take 30.minutes

[rose mirage]

4 3+1 2+2 2+1+1 1+1+1+1 yeepers

[spice idol]

wow im not stupidddd

right so you of course have the two dim 1 reps corresponding to the two C2 ≈ S_4 / A_4 reps

[rose mirage]

correct

there's another lift you can do

[spice idol]

ah from K4 right

S4/K4 ≈ S3

[rose mirage]

yur

then you know what the last two are :pauseChamp:

[spice idol]

ah i suppose all of these are lifted from S3 too lol

oh lemme guess thats the character of S3 acting on C^2 by the triangle symmetries?

dimension 2

ah right the schur orthogonality relations are useful here

man im slow my bad 🙏

[fierce steeple]

Mod 2

[rose mirage]

not really needed. The last two characters are just the standard rep and the standard rep times the sign

[spice idol]

standard rep?

[rose mirage]

what's the most obvious representation you can think of for a symmetric group

[spice idol]

just acting on C^4?

[rose mirage]

yur

unfortunately, not irreducible

the subspace spanned by <(1,0,0,0)+(0,1,0,0)+(0,0,1,0)+(0,0,0,1)> is fixed pointwise under this action, so you have to subtract off the trivial character

this is the "standard" rep, and the character values are always the number of fixed points of the permutation minus 1

[spice idol]

right i see

my head is not in the representation game right now

okay yeah filled it in

turned out not to be that hard lmao

thank you wew, my brain has been embiggened

[rose mirage]

Additional clues were like, you knew the last two reps both had to be 3 dimensional

[spice idol]

yes that i figured out, and i figured out the last row using the orthogonality relations

[rose mirage]

Then under the light assumption that one was the sign times the other you can use the fact that they add up to the regular character to deduce the values

[spice idol]

type shit

filling a character table is weirdly not unlike those silly math crossword thingies

[vapid axle]

For question 8 in showing that the p_i are primary, could I perhaps use induction on i?

[rose mirage]

It’s fun for a while but thank god theres automated algorithms for it

[spice idol]

yeah lol

although i wasnt implying they were fun

[rose mirage]

Another good one: C_p semidirect C_{p-1} (or if that’s to hard, D_n)

[spice idol]

C_p \rtimes C_p-1 is the holomorph of C_p right?

[rose mirage]

No clue what that means. It’s Aut(C_p) acting on C_p

[spice idol]

yes yes that

a similar construction comes up when defining extensions of quandles lol

[rose mirage]

The Frobenius group of order p(p-1) if you want

[rose mirage]

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

[spice idol]

crying

[vapid axle]

Why is this true? The claim is that $\mathfrak{p}_2 = \pi^{-1}(\overline{\mathfrak{q}_2}) \cap A$. But if $p \in \pi^{-1}(\overline{\mathfrak{q}_2}) \cap A$, we only have $\overline{p} \in \overline{\mathfrak{p}_2}$.

[broken turtle]

okeyokay

[vapid axle]

Nvm

[hard kite]

What is a good way to see that every principal artin local ring is a quotient of a dvr

Or is there a more direct way to see that such a ring has only finitely many ideals and that they are totally ordered by inclusion

[limpid horizon]

Where does that come up, is that relevant for like number theory?

[hard kite]

So I’ve been told T_T

But it’s more representation theory atp

I miss numbers

The context is some sort of theorem that says if two representations with conditions over a dvr R are equivalent over the fraction field then they are already equivalent sort of over R

[hard kite]

But it’s really bad and you have to either add a lot of conditions that are way to strong (so I’ve been told lol) or you have to do something else that is an extreme downgrade

Oh yea

That makes sense

[fierce steeple]

Also like artin rings always have only finitely many prime ideals

[hard kite]

Yes

[fierce steeple]

What I said was silly actually lol

The hard bit is showing they are totally ordered by inclusion lol. Then it should follow that all are just powers of the maximal

[hard kite]

I think you did it right

Every ideal will contain some power of the maximal ideal right

[fierce steeple]

I guess there is a funny point that like all ideals of such a ring should be principal and then the generator must lie in m^(k-1) \ m^k for some k

[hard kite]

And then you do that thing with the dimensions

[hard kite]

Oh wait

[fierce steeple]

Yeah I think that is good. What made me nervous is where being principal is used

[hard kite]

For the dimension being leq 1

[fierce steeple]

But I think maybe smth like this would make it ok

[fierce steeple]

Nice

And then yeah all should be good

[hard kite]

Yea

[fierce steeple]

Like if the ideal is (a) then a lies in m^k \ m^(k+1) for some k, so it is a nonzero element of the 1-dim A/m vector space m^(k+1)/m^k and hence spans this whole thing, so (a) = m^k

And ig m being principal is used when we say that that vector space is 1 dim as well

[hard kite]

Artin => noetherian => we have primary decomposition + every m-primary contains power of m so any ideal contains a power of m

[fierce steeple]

Or like, the radical of I is m (unless I = 0) and hence as I is fg we have m^k contained in I for some finite k

However I am sleepy so I may be being silly lol, sorry I should sleep

[hard kite]

You and me both

It is low key 4 in the morning for me

[fierce steeple]

Yeah 3 here

[silver goblet]

healthiest phd student sleep schedule

[spice idol]

real

[hard kite]

Not even a phd smh

I hope I can find one i realised im not in a optimal position

[silver goblet]

oh ngl i didnt see your message, my eyes were drawn to the doggo and i only read the message above 💀

i mean i went to sleep at a later time than i currently do in undergrad 💀

but i also had to wake up much later in undergrad so that nullifies my point

[silver goblet]

how so?

[hard kite]

don’t really have any connections + supervisor doenst really fwm atp

• my thesis is straying away from number theory which I have control over I guess

[limpid horizon]

R u in undergrad or doing a masters?

[hard kite]

Masters

[limpid horizon]

What happened with your supervisor?

[hard kite]

Nothing I just don’t know him that well and he is really busy

[limpid horizon]

Oh dang

[silver goblet]

is there a reason your thesis has to be NT?

[hard kite]

Because I like that a lot

Lole

[silver goblet]

fair enough lol

fwiw i think grad committees understand profs are busy and might not know their students super well

[hard kite]

I feel like if im trying to do nt then writing a thesis on that would put me in a better position

Sorry for being off topic algebra people

[fierce steeple]

Lol

Me editing a shared Overleaf at 2am

[ornate atlas]

2am is one thing, it’s when the edits happen at 7am

[fierce steeple]

Lol

[hard kite]

oh it is actually really simple. Let $A$ Artin local and principal and write $\mathfrak{m}=(\pi)$. Since $A$ is Artin, $\mathfrak{m} = \mathrm{Nil} (A)$ and thus $\pi$ is nilpotent. Let $I=(a)$ be a non-zero ideal of $A$. Then there is some maximal $n$ such that $\mathfrak{m}^n \supset I$. Write $a=u\pi^n$. Since $n$ is maximal $u$ has to be a unit.

[broken turtle]

1728

[fierce steeple]

oh lol

[hard kite]

Yea

I still wonder how to show that it’s even a quotient of a dvr though

I have never realy seen results like that

Like I know how to add ideals at the top but not at the bottom if that makes sense lol

[lone jacinth]

I think this might be a little tricky to prove. It's a special case of the Cohen structure theorem at least.

My thinking would be to construct the dvr inductively. If the maximal ideal is (p) you start with Z_p otherwise F[[x]] for the prime field F. Then you can sort of adjoin one element of the residue field at a time. If the element is transcendental it's easy, if it's seperable you can use Hensels lemma. If it's inseperable I'm not sure what you should do....

[hard kite]

Ah ok ill look into that some time

sounds tricky

[near lantern]

Maybe you take a separating transcendence basis (IDR if those exist for infinitely generated extensions).

[lone jacinth]

They don't exist in general (like consider the algebraic closure of Fp(t))

[lone jacinth]

There is a proof here of a non-commutative version
https://www2.math.ethz.ch/EMIS/journals/BAG/vol.46/no.1/14.html
(Theorem 5.2)

Though it's annoyingly fiddly

Things simplify a little by assuming commutative, but not that much

[spice idol]

noway, lattice theory!!

[lone jacinth]

It's basically classifying rings where the ideals are totally ordered

[spice idol]

this local coordinisation stuff is awesome

[muted sierra]

Are there non-Noetherian local rings (A,m) such that \bigcap_n m^n = 0?

[foggy galleon]

doesn't power series over a field in infinitely many variables work?

[muted sierra]

Mmm... How do you define that power series ring? I can think of two inequivalent definitions:
\begin{itemize}
\itemsep 0em
\item Ring of formal infinite sums $\sum_n f_n$, where each $f_n \in k[x_1, x_2, x_3, \dots]$ is a homogeneous polynomial of degree $n$.
\item Ring of formal infinite sums $\sum_J a_J x^J$, where $J$ ranges over the sequences $(j_1, j_2, j_3, \dots)$ with finitely many nonzero entries, $a_J \in k$ and $x^J = \prod_n {x_n}^{j_n}$.
\end{itemize}

[broken turtle]

Eduardo León

[muted sierra]

I guess at least in the first case it's guaranteed to work, though.

[foggy galleon]

I refered to the second (edit: actually wait yeah maybe it isn't so clear lol, but either interpretation should be fine)

is there some subtlety in the second case?

[muted sierra]

No, you're right. I was just being dense. The n-th power of the maximal ideal contains anything that has no terms of degree < n.

[lone jacinth]

Polynomial ring in infinitely many variables modulo m^2

[foggy galleon]

yeah note also that "local" is barely a restriction, since you can always localize

[muted sierra]

Thanks, you two!

[hushed bone]

You can make a ring Noetherian doing that tho

I don’t think you can do that “often” but it can happen

[verbal panther]

Can you have integration for structured with an identity and dense elements?

I’m sure dense is the wrong word there but I wasn’t sure how else to describe it

[long parcel]

You can

[verbal panther]

What can I google to learn more about that

[long parcel]

You can probably just copy paste your message into google

[hard kite]

Nice

[plucky arch]

i'm trying to understand how young diagrams give you irreducible representations of symmetric groups

i can see how the young symmetrisers are defined but their definition feels plucked from thin air

[plucky arch]

reading through fulton and harris

i do not understand the first 3 lines

this is lemma 4.25

[lone jacinth]

So c_lambda is an element of the group algebra A, and
V_lambda is the submodule Ac_lambda of left multiples of c.

Lemma 4.23(2) tells you that the elements of
c_lambda A c_lambda are all scalar multiples of c_lambda.

If c_lambda W contains Cc_lambda, then W contains
Ac_lambdaW which contains Ac_lambda C c_lambda = Ac_lambda = V.

[plucky arch]

i'm trying to follow

W contains A c_lambda W?

[lone jacinth]

Yes, W is an A module, so if you multiply by elements of A you stay in W

[plucky arch]

uuu

hm..

ok, i think i understand...

c_lambda = c_lambda w for some w in W

so A c_lambda = A c_lambda w = (A c_lambda) w, which is a subset of W

[lone jacinth]

Exactly

[plucky arch]

is this supposed to feel like symbol soup

[lone jacinth]

I mean, it is just a bunch of calculations I guess

There is some broader theory about the relationship between Sn-modules and GL(V)-modules by comparing how both act on $V^{\otimes n}$

[broken turtle]

jagr2808

[plucky arch]

i still feel no closer to understanding how young diagrams give reps of S_n

[lone jacinth]

Yeah, I don't really have a good idea for why young diagrams are there, or why they came up with that.

But it works

[scarlet prairie]

reviewing some notes from class, i think i copied this wrongly. can i check if the last line doesn't make sense right? just a chain complex won't induce a map on the homology grps.

[lone jacinth]

Should probably say, "if the cohomology is always 0"

[scarlet prairie]

ah yes that makes sense

thank you

[harsh coyote]

can i ask question in here?

[spice idol]

no, you're not allowed to ask questions until you've written 20 books and published 1000 papers

[vague pawn]

what ✨enpeace ✨ means is that you are welcome

[harsh coyote]

Thank you to say so!

[spice idol]

but seriously don't ask to ask in a chat about asking questions

[harsh coyote]

ah i will

i'll keep that in mind

[ornate atlas]

Whats your question?

[harsh coyote]

I'm learning representation therory in undergraduate lectures
It's hard to understand because of things like wedge, tensor.. etc
It is better I just memorize(without understand) the proof and go to graduate school to understand it?

[spice idol]

I don't think memorizing will help much

[ornate atlas]

Yeah im guessing youll have an exam and youll have to actually engage with those ideas

What is it about the wedge and tensor products youre struggling with?

[hearty timber]

Sorry if this is a stupid question, but what precisely does the problem here mean by "act"?

[lone jacinth]

Presumably making it into a lie algebra module.

But it's a little vague. Like you could just give it a trivial structure, so presumably whoever made the exercise has something natural in mind

[hearty timber]

The rest of the problem goes like this

[hearty timber]

Hmm, I'll try to see if I can find something

[lone jacinth]

I mean, sp(V) is a subalgebra of gl(V) right? So it naturally acts on V, and then you get the usual action on V(x)V which you must then show induces something well defined in the exterior power

That would be my interpretation at least

[hearty timber]

What is V(x)V?

[lone jacinth]

The tensor product of V with itself

For two lie algebra modules V and W the lie algebra acts on the tensor product V(x)W by
g(v (x) w) = gv(x)w + v(x)gw

[harsh coyote]

Actually, I'm a freshman, so i dont know any algebraic structure corresponding to tensor..
I've been trying to understand for a month, but I don't think I've even started

[past cove]

What book are you learning from?

Also is the rep theory class an undergraduate one or graduate?

[ornate atlas]

Theres definitely ways we can help, but if you just dont have the background for the course can you maybe come back to it later?

[harsh coyote]

i learning on undergraduate class(Abstract Algebra 2)
Representation Theory: A First Course(Springer, 1991)

[lone jacinth]

What do you mean by "algebraic structure corresponding to tensor" exactly? Like what are you trying to understand?

[harsh coyote]

I thought that 'it would be easy to understand if there was an example'.

[ornate atlas]

Have you taken linear algebra?

The tensor product is a bit of a weird one, its nice and its useful, but I dont think its super intuitive

[harsh coyote]

I studied alone so i maybe weak at linear algebra

[ornate atlas]

Do you have to take representation theory? I would reccomend understanding LA first

[lone jacinth]

Okay, so you're just asking for an example of the tensor product of two things?

I guess you can try to see why
Z/2 (x) Z/3 is 0

[ornate atlas]

The chapter on the tensor product in "A primer on Algebraic D-modules" is really nice IMHO

But if you dont really know much algebra that might still be hard to follow (Hence why im wondering if you can perhaps just take this course later)

[harsh coyote]

Things like this come out

I should study other things more than representation for this exam...

Thank you for trying to help me

Thanks for give me your time

[spice idol]

I do advice you to get a strong background in LA before doing rep theory

[hearty timber]

Is there a reason why you chose this action and not g(v (x) w) = gv (x) gw?

[spice idol]

without LA, nothing in rep theory will feel motivated or make sense

[spice idol]

works nice with the characters

there isn't a canonical kG-module structure on the tensor product anyways, as kG is not a commutative ring

canonical is not the right word here

you can't expect a similar kG-module structure on the tensor product as for the R-module structure on the tensor product of R-modules, where R is commutative, because kG is not commutative

[lone jacinth]

It comes from the derivative of the corresponding lie group action

[hearty timber]

This seems so random to me. Do you have any book recommendation that could help make things clear for me?

[lone jacinth]

Idk. They don't talk about actions in whatever you're taking pictures from?

[hearty timber]

Also, we have an action of g on V (x) V, but don't we want it to act on A^k(V) (x) A^l(V)?

[hearty timber]

No mention of it at all

[lone jacinth]

I assumed you defined the exterior power as
V (x) V / (v(x)w + w(x)v)
so then you would need to check that it gives rise to an action on the quotient.

[hearty timber]

ohh

I thought we defined it as 2-forms on V

[lone jacinth]

Another reason for not defining it this way is that this is in fact not a lie algebra action

[lone jacinth]

I mean, maybe you do. Lots of equivalent definitions

[hearty timber]

That's equivalent??

[hearty timber]

I can't believe this form was never mentioned in my LA book if this is equivalent... It's so much simpler to work with

[lone jacinth]

I guess technically you would need to do
V^* (x) V^* for it to be equivalent

[hearty timber]

And then the action of g in Sp(2) on an element x in V^* would be by precomposition?

[fierce steeple]

In char not 2 i guess

Actually I did wonder this lol. To me it should always have like v (x) v = 0. Is this the most common convention or am i tripping

It's just funny to me cause writing the relations v (x) v is also more concise

[lone jacinth]

Yeah, I guess v(x)v, would be better.

Point was really just as a quotient of V(x)V

[fierce steeple]

For example this is the "most natural" definition in that, for example, it is needed if you want exterior algebra to be compatible with base change (since this holds over Z), at least for free modules

[fierce steeple]

Oh yeah I should say like i guess the char 2 thing was not so much an actual issue but it did make me think again about this

[lone jacinth]

Yes, though you need to throw in a negative sign to get an action

So
(gf)(v) = -f(gv)
for f in V*

[hearty timber]

Why does the sign matter?

[lone jacinth]

You want
[g, h]f = ghf - hgf

If you write it out you see that the sign is wrong.

Morally what's happening is that taking the dual turns a left action into a right action, so you have to do something to make it back into a left action.

And flipping the sign does that because
[g, h] = - [h, g]

[hearty timber]

Ahh, I see. I wasn't thinking about the bracket

I was just concentrating on understanding the action itself

But that makes sense

[hearty timber]

So in reality, is "show that a Lie algebra L acts on a Lie algebra K" just supposed to be interpreted as "show that there is a Lie algebra homomorphism L -> gl(K)"?

[plucky arch]

I think I have an idea for how I would tell this story now

It took me a while but understanding how Specht modules are made helped

[lone jacinth]

An action of a Lie algebra L on a vector space V is the same data as a lie algebra map L -> gl(V).

And I guess it would make sense to just ignore the lie algebra structure on K here, not sure what the context is for L acting on K...

[hearty timber]

Oh yeah

[hearty timber]

Ok, now I'm totally on board with what is happening

Thanks a bunch!!

[hearty timber]

the email I just got

[lone jacinth]

Well, that does explain it

[scarlet prairie]

this was copied down so there might be mistakes in the picture itself. I understand mostly how hom_R(-,W) is a contravariant functor allows us to construct the complex of homs. I don't understand why we are allowed to jsut forget about V and throw in a 0 at the start/say that d_1^* is injective. Can i also check whether we used the fact that hom_R(-,W) is left exact here at all? Left exact specifically still requires a SES to start with right?

[lone jacinth]

So they're not claiming that d1* is injective.

As they say "the complex is usually not exact".

The only thing required to be a complex is that the differentials compose to 0. That's why it's also fine to add in 0, because anything composed with 0 is 0.

So far they have not used that Hom is left exact, but you will need that for showing that
ker d1* is equal to Hom(V, W)

[scarlet prairie]

ah okay right cause this is jsut a cochain, we don't have exactness at hom_R(P_0,W).

[lone jacinth]

In general you won't necessarily have exactness anywhere

[scarlet prairie]

for the left exact contravariant functors, it means we have this kind of thing: 0->X->Y->Z->0 is exact means f(Z)->f(Y)->f(X)->0 is exact?

[lone jacinth]

That would be right exact

[scarlet prairie]

ah oops

hmm okay I do have the part your talking about for ker d1* equal to hom(V,W) on my next page.

but over here it only has the form X->Y->Z->0, is that enough to use left exactness?

[lone jacinth]

A sequence like
X -> Y -> Z -> 0
can be broken into
0 -> K -> Y -> Z -> 0
and X -> K -> 0

So you can always go to short exact sequences if you like

[scarlet prairie]

ah using the sequences you have there, i get 0->f(Z)->f(Y)->f(K) where f is left-exact contravariant. by just contravariant, i also get the map f(K)->f(X). Then i compose the maps between f(Y),f(K) and f(X) to get 0->f(Z)->f(Y)->f(X). Note that exactnmess as Y is preserved because composition preserves the kernel

thanks alot for the help. im convinced but i do find it abit weird why we choose to state the full condition as having a SES 0 -> X -> Y-> Z -> 0, is this specifc case of left exact special?

[lone jacinth]

I think that's just to make it easier to talk about left exactness, right exactness and exactness simultaneously

[scarlet prairie]

ah that makes sense, thanks again for the help

[lone jacinth]

I think you'll also find people defining left exactness as preserving left exact sequences

[vague pawn]

if $k[x_1,...,x_n]$ has a graded structure given by homogeneous polynomials, after localization by $x_n$, does the localized ring have a graded structure? It makes sense to me that such a graded structure would be $$\bigoplus_{d \in \bZ} T_d$$ where $T_d$ is the set of homogeneous elements of degree $d$ (with 0 maybe?). Am I right?

[broken turtle]

ExpertEsquieESQUIE

[lone jacinth]

It is graded yes.

The degree of f/g is degf - degg

So for example x1/xn has degree 0

[vague pawn]

I realized what confused me while typing my problem

thanks jagr

[lone jacinth]

The rubber duck method

[vague pawn]

that is how its called?

[lone jacinth]

That's when you try to explain your problem to a rubber duck. The act of explaining helps you solve the problem

[vague pawn]

I see

that is so true

[lone jacinth]

And today I got to be your 🦆

[vague pawn]

you are my

[digital parcel]

usually i explain it in my shower to my shampoo bottle

sometimes my conditioner bottle gets to listen too

[digital parcel]

that is the most beautiful thing i've ever seen

love is real

[vague pawn]

double duckies, double explaining, double understanding

[lone jacinth]

Don't get to use the rubber duck method much anymore, they became so busy after they got married

[vague pawn]

||maybe you will get a third duckie soon ||

[lone jacinth]

Eggcellent

[worldly zealot]

i used to have imaginary friends that i would explain stuff to

[digital parcel]

did they get married too?

[worldly zealot]

they became complacent and incurious

[digital parcel]

oh

[ornate kindle]

Should one of the alphas inside the brackets there be negative?

Like

[broken turtle]

NotABot

[ornate kindle]

I'm like 99% sure that's a typo, and have marked the book accordingly. I'll erase it if anyone tells me I'm wrong

[spice idol]

"we are trying for a baby!"
Thank you for telling me you guys are doing it raw every night

[plucky arch]

Schur functors are a pretty cool concept

[limpid horizon]

i barely understand any math at all atp lol

[ornate kindle]

The more math I learn the more I understand how much math I've yet to understand

[limpid horizon]

Unfortunately

[wary elbow]

Yes, one of them should be negative. Without the negative that bracket is zero.

[ornate kindle]

Yeah that was what first gave me pause. Thanks for confirming!

[lime python]

This is from the paper https://arxiv.org/pdf/2504.11841. Is the permutation resolution of $M_2$, $k \rightarrow kG \rightarrow M_2$. Why does $M_1$ and $M_3$ being permutation modules mean that the maximal permutation dimension of an indecomposable is $1$?

[broken turtle]

Funky_Funktor

[lime python]

Sorry I forgot to mention that $M_2$ is the $C_3$ modules of dimesnion $2$

[broken turtle]

Funky_Funktor

[hard kite]

Does anyone know some good sources to learn about representation theory of groups over rings not just fields? Specifically local rings, dedekind domains things like that?

Also looking for a source to learn from about the Grothendieck group of module categories

[lone jacinth]

M1 and M3 has dimension 0, and M2 has dimension 1, so then the maximal dimension is 1

[worldly zealot]

gonna meet richard stanley next week :3
i am only mildly familiar with his work so if anyone has burning interesting questions i may pick one and ask

(sorry to crosspost, i just only check certain channels so i am assuming others do the same)

[ornate atlas]

Of Stanley-Reisner fame?

[worldly zealot]

the very one

[limpid horizon]

breh

i didnt know he was still alive lowk

but ig the combinatorial algebra stuff was only in 70s

[ornate atlas]

Yeah apparently him and Hochster are still kicking

good for them

[spice idol]

having to use the fact that right derived functors are universal delta functors to show two cohomology theories are equal

hom alg is kinda cool

[ornate atlas]

Ok so ive (with a lot of enpeaces help) done a), but im pretty instantly stuck on b).

We know that each $C_n $looks like $\mathbb{Z}^n \oplus \mathbb{Z}/p_1^{a_1}\mathbb{Z} \oplus\cdots\oplus \mathbb{Z}/p_n^{a_n}\mathbb{Z}$ and weve shown in a) that we can write each of these as in terms of the of the image of the previous plus the kernel and theres at most 2 of these which are not zero, sure. But im not really sure what the map $L_{n+1}\to K_n$ looks like, theres plenty of information there but im not sure how to bring it together really. Im guessing I want to start by choosing a basis for all the $C_n$ or something? Idk im a bit lost

[broken turtle]

Nope

[spice idol]

splitting into 0 → Z → 0 is crazy

[ornate atlas]

Its certainly a sequence that exists

[spice idol]

"youre doing great buddy, contributing so much!"

C_n is free

so you can immediately throw away all the Z/p^aZ

[ornate atlas]

FML I forgot theyre also free

Ok yeah I see it now

[spice idol]

nice!

reading the problems correctly is the hardest part of math

[ornate atlas]

I think I do anyway, let me go try to write it out, but I think i see it

I think ive got that but ive not done any matrix stuff

Of course you dont need to, linear maps and matrices are as good as each other, but im worried im missing something

Like am I being dumb or is it just that any homomorphism of Z is just multiplication?

I suspect it is the first thing but I dont see why (other than 22 years of evidence to suggest thats generally the case)

No actually I dont think thats right

Idk im going home now, tired and hungry

[spice idol]

Hom(Z, -) represents the identity functor

[lone jacinth]

The key idea that makes this work is that subgroups of free abelian groups are also free

[spice idol]

(so yes)

[lone jacinth]

And in fact a similar thing is true for any hereditary ring

[ornate atlas]

I keep flip flopping between thinking this is trivial and not really getting it

I definitely don’t see how to use the row reduction hint though

[lone jacinth]

Ah, I thought you were thinking about part a.

Part b is Smith normal form, which i guess requires slightly more than just standard row reduction

[spice idol]

right i was just thinking about that

[ornate atlas]

Part a) is fine I’ve got that, but I feel like part b) I think I see but I don’t actually

[summer quest]

yeah this is a Smith normal form problem

[lone jacinth]

I mean it's just the proof of the classification of finitely generated abelian groups

[summer quest]

yeah lol

[lone jacinth]

So I wouldn't call it trivial, but perhaps something you've vaugely seen before

[ornate atlas]

I’ve definitely seen it before, but then I guess I am right and it’s trivial because i just started my solution with the theorem

Like that combined with part a) tells you the C_n look like either Z^2 or Z? Then just look at what homomorphisms of Z are?

[lone jacinth]

Cn are just arbitrary fg free abelian groups

[ornate atlas]

So they’re any Z^n?

[lone jacinth]

Yes

Not necessarily same n as in Cn xD

[ornate atlas]

I thought part a) saying there’s at most 2 non zero terms in the sequence gave us n is at most 2

[ornate atlas]

Yeah lol

[lone jacinth]

It splits into complexes of the form
L -> K
So of the form
Z^n -> Z^m
in the fg case

(and if you follow the hint you can assume the maps are injective)

[ornate atlas]

Ok yeah sure

That makes more sense as to why I’m doing matrix stuff

I’ll think about this fresh tomorrow and maybe not be struggling as much

But I do think I see it now

[spice idol]

I am so frustrated right now
the category of Beck-modules of an algebraic structure A is this close to being a category of modules

:(

the only issue is that A might be infinite, so the base ring might not be a Beck module in a natural way

[onyx imp]

was asked to prove the horseshoe lemma as homework, can i check that this is the statement of the lemma? just using a different name?

[lone jacinth]

This is indeed the horseshoe lemma

Guess not everyone enjoys naming diagram lemmas after the shape of the diagram

[onyx imp]

alright thank you

[digital parcel]

“Simultaneous resolution” is so lame

What do they call snake lemma

[lone jacinth]

"long exact sequence of kernels and cokernels lemma"

Taking kernels is left exact
A wonderful functor in fact
But then to the right
Oh what a sight
The cokernel, it's finishing act

[digital parcel]

Bravo

[limpid horizon]

wow.

[golden osprey]

is there a way to check if a sequence of polynomials forms a regular sequence via grobner bases?

idek what text I would look at for this stuff

[past cove]

Maybe eisenbud has stuff about this

[near lantern]

IDK whether there's a more "direct" or "optimal" method, but I think there is definitely a way, since f1, ..., fr is regular iff (0) : f1 = (0), (f1) : f2 = (f1), ..., (f1, ..., f(r-1)) : fr = (f1, ..., f(r-1)) (and maybe also (f1, ..., fr) ≠ (1) depending on your definitions), and colon ideals and ideal equality can be computed using Gröbner basis-based algorithms (c.f. Cox Little O'Shea (end of) section 4.4 for colon ideals).

[golden osprey]

Ah I see

Makes sense yea, colon ideals didn't come to my mind

[onyx imp]

If I have a two homomorphism of two SES and either splits, then the connecting homomorphism in the snake lemma is the zero map.

[onyx imp]

can i check whether this sounds correct?

[lone jacinth]

I guess you mean you have one homomorphism, but yes correct.

[wise sedge]

Is PSL(2,R) a finitely presented group for R the ring of integers of any global field?

[summer quest]

I don't believe so, consider PSL_2(O_F) for F=Q(\sqrt{-d}) with d different from 1, 2, 3, 7, 11

[wise sedge]

Oh wow

That is not the answer I was expecting

[near lantern]

A commutative domain is Prüfer iff binary ideal sum distributes over binary ideal intersection. In which ones does infinitary ideal sum distribute over (still binary) intersection?

(Answers for general commutative rings also welcome.)

[lone jacinth]

If the ring is Noetherian every infinite sum can be reduced to a finite sum, so that would work.

But I guess that just means the Noetherian case is uninteresting

[near lantern]

Dedekind domains win again

[near lantern]

Glad I didn't stipulate that anything is Noetherian then lol

OTOH at least the answer is precisely known in the Noetherian case.

[lone jacinth]

Actually, doesn't infinitary follow from finitary?

Like
if x is in I \cap (Sum J), then x is in some fininte sum Sum' J, so in Sum I\cap J

[spice idol]

How do injective objects "look like", in general? Super vague question but I vaguely know an example of an injective abelian group but that's basically where it stops. How do they look like for module categories, for example?

[lone jacinth]

So Hom(R, Q/Z) is an injective cogenerator for ModR. So every injective looks like a direct summand of products of copies of that.

Not sure how enlightening that is...

There's also a good analogy with divisible groups.

Notice that an abelian group being divisible can be framed as every homomorphism from
nZ to A
lifts to a homomorphism from Z to A (i.e. you can divide by n).

In general a module M is injective iff every homomorphism from an ideal I to M lifts to a map from R to M. So you can sort of "divide" by I.

[spice idol]

ah I see

[lone jacinth]

For a commutative noetherian ring all injective modules are direct sums of E(R/p) where p is a prime ideal and E is injective envelope.

And for artinian rings, they are direct sums of E(S) for simple modules S

[spice idol]

sadly I have doubts the ring I'm implicitly working with is going to be Noetherian lmao

[lone jacinth]

What's the ring?

[spice idol]

Looking at Beck-modules of some arbitrary algebraic structure. The ring is.. you can get a construction lol but I haven't played around with it much yet

[limpid horizon]

ma boi wat da FRICK is a beck module

[lone jacinth]

Allright, so you have some abelian category, which happens to be equivalent to the module category of a ring?

[digital parcel]

Rotmans intro to hom alg has stuff on injective stuff if u want

[limpid horizon]

u been on this beck module grind

[spice idol]

only a full subcategory

[limpid horizon]

actually u did tell me the defn once

[lone jacinth]

Well then the injective objects could be totally different

[spice idol]

arghh

they do exist because the resulting category is Grothendieck

(says the nlab page at least)

[lone jacinth]

I believe it

[spice idol]

the main problem is that for every element of the algebra you've got an idempotent, and 1 should be "the sum of all those idempotents" which of course isn't defined when the algebra is infinite

[rose mirage]

can you not phrase this as a coproduct over the ideals (viewed as R-modules) generated by those idempotents being isomorphic to the ring itself?

that's normally the reason why we care about that condition

[spice idol]

yeah that's indeed equivalent

you want every module to be isomorphic to the direct sum of all e_x(M)

[spice idol]

that is probably implied by this

yeah because free module stuff

[rose mirage]

yeah the projections would factor through the free module right

[spice idol]

ye exactly

algebra is so easy, just say "because free object" and you're done

[rose mirage]

well at least you can now actually state your problem

[ornate atlas]

Big day for field theorists

[spice idol]

this just in, the category of fields is monadic over Set

[rose mirage]

such a dog shit object it's unreal

[ornate atlas]

Real, doing Galois theory is making me go insane

[rose mirage]

ignore my username

[spice idol]

it's alright because the field with one element isn't real

[rose mirage]

the field with one element is not a field it is a infinitude of spheres glued together

[ornate atlas]

The last 3 problem sheets have all essentially been “write a as a Q linear combination of b!”

[spice idol]

god

[ornate atlas]

Thanks that clears it all up

[lone jacinth]

I guess this is what you would call a unital module over a ring with enough idempotents

[rose mirage]

I'm only being slightly facetious when I say that

there's deep connections between F_1 and the sphere spectrum

[ornate atlas]

I have heard your yap on it before and I remember it being vaguely reasonable

[rose mirage]

that makes one of us

[ornate atlas]

Vaguely is doing a lot of heavy lifting

[spice idol]

hmm

I see

that would make things nicer, I wonder what kind of impact that would have on f.e. Ext modules

if all goes well then the cohomology of algebraic structures should just be certain Ext-modules so, if they behave nicely in rings with enough idempotents and an equivalent condition exists that can be checked purely using the algebraic structure, would be great

just thinking outloud

[summer quest]

it kind of is though!

[spice idol]

idk I haven't seen it with my own eyes

so it must not exist

[vapid axle]

Why do they write x in B? Shouldn't x be in A, since A is integrally closed?

[limpid horizon]

Thats a cool result

I need to read that chapter

[lone jacinth]

A integrally closed means that the only integral elements in the field of fractions are those in A.

You can still have integral elements in B.

For example Z is integrally closed and Z[sqrt(2)] is an integral extension

[limpid horizon]

The only elements in the field of fractions that satisfy a polynomial over A, are elements of A? Is that what that means?

[lone jacinth]

monic polynomial

[limpid horizon]

Why do we need monic? I did know this once upon a time i think

[lone jacinth]

Well, 1/2 satisfies 2x - 1.

I guess more to the point you want
A[x] to be generated by 1, x, ..., x^n-1. So you need the leading coefficient to be a unit in order to write x^n as a combination of them

[worthy grotto]

Yo guys I am having trouble understanding metrices

Specifically with multiplying when finding Cofactor of matrices

[near lantern]

I would suggest finding a concrete and specific question that encapsulates your confusion, then asking it in #linear-algebra.

[finite marsh]

hello! would someone please suggest books / resources / lectures for a first course in ring and module theory?

[urban granite]

any abstract algebra books should contain them. For abstract algebra book I suggest dummit and foote.

(as written in the channel description it's not the channel for intro groups,rings,modules, fields stuff, these should go to #groups-rings-fields)

[compact summit]

Does anyone happen to know how this family of subobjects comes about?? How are the subobjects defined explicitly?

(For reference, this is from Weibel's book on homological algebra; the chapter on spectral sequences)

My guess is this: Since $E_{p,q}^r$ is a subquotient of $E_{p,q}^a$ (as argued by Weibel), there is an epi $\pi_{p,q}^r: E_{p,q}^a \to E_{p,q}^r$. So we can define $$Z_{p,q}^r := (\pi_{p,q}^r)^{-1}(\ker(\partial_{p,q}^r)), \qquad B_{p,q}^r := (\pi_{p,q}^r)^{-1}(\text{im}(\partial_{p + r, q - r + 1}^r))$$

[broken turtle]

Lucas

[compact summit]

As another reference, I found the section on Wikipedia (https://en.wikipedia.org/wiki/Spectral_sequence) "Interpretation as a filtration of cycles and boundaries" to also be confusing

Ideally, I want the constructions to work in any abelian category. So I'd rather not make any definitions based on elements!

[urban granite]

[wary elbow]

These objects are part of the definition of a spectral sequence. B is the image of d and Z is the kernel. See Definition 5.2.1.

[summer quest]

all of this stuff works without needing to talk about elements

although by Freyd Mitchell this doesn't actually matter

if you want to prove general diagram chase results which apply to any Abelian category, either you can appeal to Freyd-Mitchell and then argue in terms of elements, or you can argue with generalized elements. I don't think I have encountered any situations where insisting on avoiding elements actually clarifies anything (it certainly doesn't lead to more general theorems)

as an aside, if you're learning spectral sequences for the first time it almost never helps to stare at these kinds of formulas, you pretty quickly understand where they actually come from if you learn how to actually compute things with spectral sequences before learning abstract results about them

the actual idea that leads to spectral sequences (say of a filtration) is not mysterious at all, it's exactly the same idea as how short exact sequences induce long exact sequences in (co)homology

You can view a 1-step filtration as a short exact sequence and this gives you a long exact sequence in (co)homology. You can view a 2-step filtration as two short exact sequences spliced together, and it's a good exercise to work out how the corresponding long exact sequences end up being spliced together as a result. If you do the same exercise for a 3-step filtration you quickly notice the pattern and you have rediscovered how spectral sequences work

[inner ledge]

What does classification, characterization of something in mathematics means? Context: " If you know a little abstract algebra, the Jordan canonical form is also of interest in the sense that it completely classifies the conjugacy classes of matrices over the complex numbers (and some other fields as well), and is a special case of a more general phenomenon regarding module homomorphisms."

[digital parcel]

I just read it to mean like "one can recover all the information of conjugacy classes [...] from the Jordan canonical form"

[scarlet ermine]

to classify something is to put it in bijection with something you understand better. Maybe that’s a finite set, or maybe it’s countable, or maybe there’s moduli (you have some parameters that vary in something like the reals or complex numbers).

For conjugacy classes of complex matrices, you can put them in bijection with jordan normal form (up to reordering of jordan blocks)

jordan normal form matrices are much easier to understand—it’s not hard to write down what all of them look like

and then when it comes to properties of complex matrices that only depend on conjugacy class, you can use this classification to make your life much easier. For example, calculating the derivative of the determinant as a function from complex matrices to the complex numbers

[compact summit]

I don't understand 🥲 . There is no mention of B and Z in Definition 5.2.1 in my book. I don't get the feeling that these subobjects come with the definition of a spectral sequence because Weibel seems to suggest that it follows from each E^{r+1} being a subquotient of the previous term E^r

[compact summit]

I've been told to be careful about that; as I'm a beginner, I'm not exactly sure what the boundaries are when just merely working in R-mod

For instance, I might falsely assume that every abelian category has enough projectives because R-mod does? 😅

[summer quest]

there are none

[compact summit]

Are there some nice examples of computing with spectral sequences I can look at? I'm getting stuck on the general theory for now!

[summer quest]

oh well okay sure

but I just mean as far as basic diagram chasing is concerned this is not an issue whatsoever

you can always talk about elements rather than generalized elements without any worry. Questions about projectives/injectives is not an issue of elements

[compact summit]

I guess so, but what about in making definitions? If a definition rests on elements, then it might be "hidden" that you're using them in a proof

[summer quest]

playing around with the Serre spectral sequence (for cohomology of topological spaces) or the Hochschild-Serre spectral sequence (for group cohomology) are both good things to play with, there are loads of very simple accessible examples

[compact summit]

I don't have much algebraic topology background; my course in my school only got up to the Mayer Vietoris sequence in Hatcher. So I don't know much about the cup product for instance

[summer quest]

I mean if you have a particular Abelian category in mind that you have fixed where you know that you cannot talk about elements, then ideally you should avoid this sort of indirection

[compact summit]

^I'm assuming that its important to view cohomology of topological spaces as rings??

[summer quest]

if you're proving general results about general Abelian categories this doesn't matter

[summer quest]

it is important but plenty of spectral sequence examples don't require this, and plenty of examples are simple enough that you can just look up what that ring structure is

[compact summit]

I'm mostly interested in sheaves I guess as my main example of an abelian category without elements :))

[summer quest]

sure but then it's not so hard to talk about generalized elements either

[compact summit]

Is there an example say in Weibel, or youtube that you have in mind??

[summer quest]

here is a very nice example to consider, the exposition is especially nice https://people.maths.ox.ac.uk/tanj/notes/sssguide.pdf

[compact summit]

Thank youuu :))

[summer quest]

if you have sheaf cohomology in mind then one spectral sequence that will appear over and over again is the Grothendieck spectral sequence

[compact summit]

That's what I want to build up to!! But unfortunately, its one of the last things Weibel discusses in the chapter ;((

[summer quest]

that and the Leray spectral sequence

[compact summit]

Unfortunately, I still feel a little lost on my original question though 🥲

[summer quest]

it will get easier once you go through the guided examples I posted above

[compact summit]

Okayy, I'll have a read now :))

[woeful crane]

Ive been reading through "Users guide to spectral sequences" in my live streams every morning. Made it through some background, but also reached the conclusion I just need to see some more computations "in action".

[woeful crane]

Might start reading this kne tomorrow.

Someone also suggestes Bott and Tu, so that might be the next step after that. Though, I think Im ultimately more interested in seeing it put to work in group cohomology

[near lantern]

The Freyd-Mitchell Embedding Theorem says that every small abelian category is isomorphic to a full exact subcategory of R-Mod for some (unital) ring R. That means that you can assume that all objects of A are R-modules (but not that all R-modules are objects of A), that morphisms of A are R-linear maps (and all R-linear maps between objects of A are morphisms of A), and that the objects of A are closed under (the zero module,) finite direct sums, kernels and cokernels of maps, and anything else you can build out of that, such as images. (It follows that the direct sum, kernel, cokernel (and hence all finite limits/colimits, such as pullbacks) of stuff in A is the same as the same construction for R-modules.)

Those are the boundaries of what the embedding theorem lets you assume, if you want to know that technically.

[compact summit]

But, when proving say the snake lemma using the embedding, you construct the connecting morphism based on elements and it's not so clear how it's defined in any abelian category

My concern is that I only know it exists

I guess it's enough for a proof, but there are times where knowing how it's explicitly defined is nice!

I suppose that my concerns developed after reading several mathstackexchange posts about it from Martin Brandenburg

Here are some:

• https://math.stackexchange.com/questions/4145820/can-homological-algebra-be-done-in-rmod-without-loss-of-generality?rq=1
• https://math.stackexchange.com/questions/361351/applications-of-mitchells-embedding-theorem
• https://mathoverflow.net/questions/32173/mitchells-embedding-theorem

[spice idol]

oh! you're K-Theory??

[woeful crane]

You can definitely do it just in terms of universal properties of kernels/cokernels, that is, without elements
https://math.stackexchange.com/questions/3135969/snake-lemma-without-elements-exactness

[woeful crane]

But for me, "explicit" means there's some elements that generate the objects in question and we write down exactly what happens on each generator.

[woeful crane]

yup, it's me, ya boi

[spice idol]

cinema

[clear ermine]

is this the right place to post this?

[spice idol]

this seems like contest math lmao

also: where did the \nu come from

[torn harbor]

that's the p-valuation

https://en.wikipedia.org/wiki/P-adic_valuation

[spice idol]

I see

[torn harbor]

ok wikipedia embed thanks for filling my entire screen

[spice idol]

wikipedia embeds are a nightmare

[spice idol]

case in point: "let n = 2025"

[merry nebula]

lol

[spice idol]

returning to this: suppose 1 = a1 e_x1 + ... + an e_xn and that A is infinite. Then there must exist a y ∈ A not equal to the xi. Then:
1 * e_y = (a1 e_x1 + ... + an e_xn) e_y
= a1 e_x1 e_y + ... + an e_xn e_y
= a1 * 0 + ... + an * 0 = 0
which is a contradiction.

[compact summit]

I'm still really lost 🥲

I guess seeing some examples using the Serre spectral sequence was nice; it clears up for me their use, at least in a vague way for now

But I'm not sure where the cycles and boundaries come in to all this

I'm even lost about why Weibel introduced them in the first place. They give a nice general definition of E^\infty page, but is it used at all?? Aren't most spectral sequences that pop up bounded and so E^\infty is just the page at which things stablise?

[spice idol]

in general, given a set E of orthogonal idempotents (meaning ef = 0 for e≠f ∈ E), then R is the direct sum ⊕_e∈E Re only if E is finite

[silver goblet]

love your videos!

[rose mirage]

It’s oh so very woke

[woven loom]

Surround links with <> to hide embeds, like https://en.wikipedia.org/wiki/P-adic_valuation

[distant harness]

You should also be able to dismiss the embed yourself after posting it. (On in-browser Discord, there's a cross icon in the upper right corner that's only visible when you mouse over the embed. On mobile there should be a "remove embed" item in the context menu for the message).

[compact summit]

This result seems to just follow by functorality of homology...?

I'm not sure why the 5 lemma is needed 😅

(This is from Weibel's homological algebra book)

[limpid horizon]

who is it?

[spice idol]

man this problem is hard

i dont think i could ever solve this

[digital parcel]

#prealg-and-algebra

[chilly mist]

Ok

Thanks

[limpid horizon]

This aint no ordinary algebra … this is… ADVANCED ALGEBRA.

[limpid horizon]

Bro probably was graphing linear functions at 4 years old

[broken turtle]

mawzi

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

mawzi

[clear ermine]

oh wait

i think i get it 👍

[cyan dagger]

hey, I'm a little stuck trying to understand why the inverse map works here. I perfectly understand how ZG is a free ZH module, and any morphism of ZH module needs only to be defined on a choice of left cosets. The first map makes a lot of sense, but I don't understand how the given inverse map is compatible with the tensor on ZH

What's bugging me is that if I write g as hy for a left coset representative y, then $hy \otimes m = y \otimes hm$

[broken turtle]

Brindille Connexe

[cyan dagger]

but then the map gives $xy^{-1}h^{-1}m = x y^{-1} hm$

[broken turtle]

Brindille Connexe

[lone jacinth]

This is not true.

The correct relation would be
yh (x) m = y (x) hm

[cyan dagger]

oh, I was unaware the tensor coefficients were "passing by the middle"

I guess that comes from me working with tensor on commutative rings before, thanks a lot

there is something else that is bothering me: I came up with an inverse that sends a true tensor yh (x) m to the ZH linear map that sends y to hm, and is 0 on the other cosets. Unless I am incorrect again, this is the inverse, but i don't see why it coincides with the given inverse there

[lone jacinth]

So from writing
yh (x) m
it seems you're saying y is a representative for the left coset yH.

But the homomorphisms ZG -> M are determined by their values on right cosets. So maybe what you meant to describe was that it sendt y^-1 to hm, and is 0 on the other cosets.

Or do you pick a coset representative z such that
y = h'z^-1, and then consider the map that is h'^-1 h m on z^-1 and 0 on other cosets...?

Either way I think you might run into trouble

[lone jacinth]

But yeah, I think the inverse given in your image is wrong.

Maybe they meant to say that it should be 0 unless xg^- was in H...

[cyan dagger]

yeah because otherwise the double composition gives you new tensor terms that don't seem to cancel each other

I think I need some time to think about what you told me to define an appropriate inverse, many thanks for your insight!

[cyan dagger]

my best guess was that due to tensor relations they would die, but I can't see it

[lone jacinth]

The main idea anyway is that if
gi is a set of representatives for G/H, then
ZG = Sum ZH gi^-1 = Sum gi ZH
So a homomorphism is exactly determined by where gi^-1 is sendt and any tensor decomposes as
Sum gi (x) mi

So in both cases what you get is isomorphic to M^n determined by these mi and f(gi^-1) respectively.

[lone jacinth]

I mean, I guess just computing it in simple cases like
G = C2, H=1 shows that it doesn't work

[cyan dagger]

yeah..

[cyan dagger]

sorry if this is trivial but why can you have the first equality without assuming H normal?

ok sorry im bad

not enough sleep

[lone jacinth]

So (gH)^-1 = Hg^-1

[cyan dagger]

again thanks a lot, and sorry for asking somewhat simple stuff

[lone jacinth]

No problem

It is quite fiddly with inverses and left and right cosets. So even though there aren't really complicated ideas in play it's easy to get wrong

[onyx imp]

my professor proved this in class and then made the remark that since Hom(D,-) is a left exact covariant functor, we obtained the right covariant derived functor ext(D,-)

i understand how it follows from the theorem that the nth ext group is covariant functor

but i don't see how we got right exact

[spice idol]

such a long exact sequence is always associated to right derived functors

it follows from the fact that a SES of cochain complexes gives a long exact sequence of cohomology

[lone jacinth]

What do you mean? Where are you getting right exact?

[onyx imp]

would right covariant derived functor mean right exact?

[spice idol]

no, right exact means that a functor preserves exactness of sequences of the form A → B → C → 0

[onyx imp]

what would right mean then?

[spice idol]

have you heard of right derived functors before?

[onyx imp]

nope im not sure hwat derived means too actually

[spice idol]

the idea of them is that you extend the exact sequence 0 -> FA → FB → FC as "naturally" as possible

(if 0 → A → B → C → 0 is exact and F is left exact)

[onyx imp]

ah okay so. if F is left exact covariant, then given SES 0 ->A->B->C->0 we have
0->FA->FB-FC is left exact, F' is said to be right derived if i can make this

0->FA->FB->FC->(F' continues something here?)

[spice idol]

well thats a property but not exactly what defines it

[lone jacinth]

Yes, so the definition of right derived is usually just taken to mean apply F to an injective resolution and take cohomology.

But the result is that you extend 0 -> FA -> FB -> FC to the right, hence the name

[spice idol]

(there is a universal property here but this relates to a general object called delta functors)

[lone jacinth]

Similarly a left derived functor is when you extend the sequence to the left (this time given by projective resolution)

[onyx imp]

ah okay thanks for clarifying, i get a sense of it now, i think i will leave the details to next time since my prof only made a small remark on it

[onyx imp]

actually, in the proof of the above theorem, we started with a projective resolution of D

[spice idol]

they are definitely worth looking into, very important objects

[onyx imp]

i know there is a way to define the ext grp in terms of injective resolutions instead, is there anything to do with that?

[spice idol]

how do you even define Ext without injective resolutions

[lone jacinth]

Yes, it's a bit of a magical fact that the derived functor of
Hom(D, -) and Hom(-, C) coincides

[spice idol]

oh lol

[digital parcel]

With a projective resolution

[lone jacinth]

What is a projective resolution if not an injective resolution in the opposite category

[digital parcel]

Oh they already said that

Damn