advanced-algebra

[plucky arch]

Epimorphism means precomposition is injective

[worldly zealot]

They should call the category of small categories Kitten

[limpid horizon]

f: X->Y is an epimorphism if hf = gf implies h=g

[plucky arch]

Yes, that’s a rephrasing of this

[limpid horizon]

The word injective doesnt make sense for all categories does it?

[plucky arch]

It does here

Postcomposition is just an ordinary function

As is precomposition

[limpid horizon]

The defn in class was f: X->Y just a morphism in a category

[plucky arch]

And?

The definition i supplied works for any category

[limpid horizon]

Im wondering y u said ordinary function

[plucky arch]

.

It’s a function between sets

Makes sense to ask whether or not it’s injective

[spice idol]

the function on hom-sets is injective

a morphism f : X -> Y defines functions f* : Hom(Y, Z) -> Hom(X, Z) and f_* : Hom(W, X) -> Hom(W, Y)

f* is injective iff f is epi and f_* is injective iff f is mono

in general this is functorial; fixing a Z yields the functors Hom(-, Z) and Hom(Z, -) from your (locally small) category to the category of sets

these are super super important, because they each faithfully encode all the information of your objects

(I.e. if you know that F = Hom(-, Z) for some Z, then this Z is uniquely determined up to isomorphism, and this is essentially the basis for universal properties)

[limpid horizon]

Thank you

[limpid horizon]

f* injective for all Z iff f is epi right

[plucky arch]

yes

[limpid horizon]

Is this because of yoneda lemma or something

[plucky arch]

yes

[spice idol]

yeah thats what i meant lol mb

[golden osprey]

If we consider the ring of symmetric polynomials in $n$ variables $\Lambda_n = \mathbb{Z}[x_1, \ldots, x_n]^{\mathfrak{S}_n}$, let $e_1, \ldots, e_n$ denote the elementary symmetric polynomials in $n$ variables. Then we in fact htat that $\Lambda_n = \mathbb{Z}[e_1, \ldots, e_n]$.

However, we also have somethign stronger, the $e_i$ are algebrically independent over $\mathbb{Z}$.

[broken turtle]

Spamakin🎷

[golden osprey]

Ok we can consider the inverse limit of all these rings of symmetric polynomials and we get the ring of symmetric functions

https://en.wikipedia.org/wiki/Ring_of_symmetric_functions it's constructed as a limit

So then we get that $\Lambda = \mathbb{Z}[e_1, e_2, \ldots ]$ so that $e_i$ is the degree $i$ elementary symmetric function in infinitely many variables.

[broken turtle]

Spamakin🎷

[golden osprey]

(Lambda is the ring of symmetric functions)

How does one talk about algebraic independence for infinitely many elements?

Cause like polynomials don't have infinitely many inputs yea?

So what's the right term to say that all the e_i are algebraically independent when talking about them as symmetric functions in the whole ring of symmetric functions, rather than specializing to the n-variate case

[plucky arch]

would it work to say every finite subset is algebraically independent?

[golden osprey]

Sure but my question is specifically is how do you formulate this for infinite sets, not finite subsets

[plucky arch]

i see

i was going off the linear independence def

[golden osprey]

Is this a thing that has a name?

[near lantern]

No fnitary relations

[near lantern]

It is in fact equivalent to this

There is some motivation to do it this way

A family of elements in a ring R indexed by X induces a unique morphism from the polcnomial ring in variables indexed by X, ℤ[X], to R. Call the elements algebraically independent if this morphism is injective.

The equivalence with all finite subsets comes from Z[X] = U of Z[all finite subsets]

which is "always" true in algebraic situations

[plucky arch]

yes that's the characterisation i know of in terms of free objects

i just wasn't quite sure what free objects in this case were

but e.g. this is exactly how it works for vector spaces

[plucky arch]

mfw filtered colimits commute with finite limits

[near lantern]

This makes sense since polynomial expressions are the only ones you can evaluate to see if they give a true relation

[near lantern]

No, I think it's commutes with the forgetful functor that matters here

[plucky arch]

hm ok

[near lantern]

I think "CRing is monadic over Set" is jargon for this property.

or something like that

[plucky arch]

well you need a finitary monad

i think

[spice idol]

injectivity is a property that doesn't depend on the finitaryness of the monad

as in; you always only have to check injectivity on substructures generate by two elements lol

[plucky arch]

Hm

I’d want to verify this for something non-finitary like the ultrafilter monad

[spice idol]

well, on the level of sets injectivity is measured by checking on all 2-element sets

and the substrutures generated by a pair certainly contains that pair

[cyan dagger]

Hey, is it true for $F$, $M$ left $G$ modules, $F$ free over $\mathbb Z[G]$, that $F \otimes M$ is projective as a $\mathbb Z[G]$ module (or equivalently that induced modules from the trivial group are projective?) I don't see how torsion in $A$ will not obstruct projectiveness

[broken turtle]

Brindille Connexe

[cyan dagger]

(here I am tensoring over Z of course)

[lone jacinth]

It's not true. Like you say, if M has torsion you will get into trouble.

[cyan dagger]

i see, thanks for confirming my doubt

[cyan dagger]

Hey i'm trying to do part 3. For me the isomorphism is obvious, as we can make the above exact sequence periodic and the isomorphism on cohomology follows. However I do not see how to get it with connecting homomorphism (I understand the decomposition of 2n short exact sequences, but I don't see how we can move onto 2n cohomology groups)

it's quite important for the rest of the exercise so I can't just skip it

[lone jacinth]

So you break the sequence into short exact sequences
K1 -> F0 -> Z
K2 -> F1 -> K1
...

Then the LES gives you
Ext^i(Fj, M) -> Ext^i(K[j+1], M) -> Ext^(i+1)(Kj, M) -> Ext^(i+1)(Fj, M)

As Ext(Fj, -) = 0 the connecting homomorphisms is an isomorphism
Ext^i(Kj, M) -> Ext^(i+1)(K[j-1], M)

So composing these you can get an isomorphism
Ext^i(K[2n], M) -> Ext^(i+2n)(K0, M)
Since K[2n] = K0 = Z, these are the cohomology groups

[cyan dagger]

thank you so much, I had found the argument with vanishing in the middle but didn't connect that the kernels on the side corresponded with the homology group. Thanks again!

[edgy pond]

i want to show if two representations are isomorphic over an algebraically closed field, they do not necessarily stay isomorphic in a subfield

so i take the algebraically closed field as C and the other field as Q. and want to find two isomorphic C[G] -modules that won't be isomorphic as Q[G] modules

are there any hints?

[woeful crane]

Is this even true?
Funtors preserve isomorphism, so if youre doing this via restriction of scalars functor, any isomorphoc CG moduñes stay iso over Q.

However, its possible the extension of scalars of two nonisomorphic QG modules become isomorphic as CG modules.

[edgy pond]

it is a question from my course to either prove or give counterexample

so i was trying to find counterexamples

[lone jacinth]

Are you saying you were asked to prove that if two reps are isomorphic over C then they are also isomorphic over Q?

If so you should try to prove it instead of finding counter examples

[edgy pond]

The question is "let V, W be G-reps over k. Suppose their extensions of scalars V' and W' are isomorphic as G-reps over \bar k. Are V and W isomorphic over k? and i was hoping to find a counterexample, though it is true it seems

oh proving is also easy

[lone jacinth]

This question is sightly different to what you asked before

[limpid horizon]

Are u k theory youtuber?

[woeful crane]

That's me.

[worldly zealot]

lol

i liked your vid with soergel

[edgy pond]

a \bar k [G] module homomorphism is automatically a k[G]-module homomorphism. so they are isomorphic over k too. Are there any mistakes with this argument?

[limpid horizon]

Bruuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuh

[lone jacinth]

Yes, you're arguing that given two isomorphic k-bar G modules their restriction of scalars to k are isomorphic, but the question is about the extension of scalars

Like if your given V and W as k[G] reps, and you let V' and W' be the extension of scalars, then if V' ~= W' are isomorphic k-bar[G] modules they are also isomorphic k[G] modules, but the question is showing that V and W are isomorphic

[edgy pond]

ah i see

i have no clue now then

i'd think this is not true anymore

[lone jacinth]

In pretty sure it's true, but it's more nontrivial to show

[spice idol]

excessive u's

[storm basin]

Yo. I’m a fan lol

[spice idol]

peak content about peak math

[limpid horizon]

No.

Stop that.

😚

[spice idol]

i will not stop uh
talking

[limpid horizon]

You have my permission to continue talking.

[lone jacinth]

If your field is infinite there is this slick argument:

Pick bases so that you can think of the representation V and W as group homomorphisms
G -> GL(n, k)

The extension of scalars is then just given by composing with GL(n, k) -> GL(n, k-bar).

As V' and W' are isomorphic, there exists an invertible matrix M over k-bar with
M phi_V'(g) M^-1 = phi_W'(g)
Which we can write as

M phi_V'(g) = phi_W'(g) M

Now write M = M1 e1 + M2 e2 + ... + Mm em for k-linearly independent elements of k-bar.

Then because of the linear independence and the fact that phi(g) has entries in k we must have
Mi phi_V'(g) = phi_W'(g) Mi
for all i.

Now consider the polynomial
det(M1 t1 + M2 t2 + ... Mm tm) in m variables. As this polynomial is nonzero on (e1, ..., em) it is not the zero polynomial, so if k is an infinite field there must exists values of ti in k for which it is nonzero. Letting a1, ..., am be such elements
M' = M1 a1 + ... + Mm am
Gives an isomorphism between V and W

Maybe there's a simple argument in the case k is finite... Idk

[woeful crane]

What do you mean by the right hand side M = M1 e1 + M2 e2 + ... + Mm ? The Mi are matrices valued in \bar{k} and the ei are elements of k?

[lone jacinth]

The Mi are matrices over k and the ei are in k-bar

[woeful crane]

Oh, I see. That's a really nice argument actually

[lone jacinth]

This same argument works for any algebra (and finite dimensional modules).

I guess if you're in the case of a finite group and char(k) relatively prime to |G| you can reduce to V and W being irreducible and then I guess it's not too hard to see that V' and W' can't be isomorphic if V and W are not

[woeful crane]

and kind of surprising tbh.

[woeful crane]

I was just about to ask.

Even infinite-dimensional k-algebras?

[lone jacinth]

Yeah, the algebra is hardly relevant to the proof, but you need the modules to be finite dimensional

It would just be reducing to the image of the algebra in End(V) anyway, which is finite dimensional

[woeful crane]

why can't the modules be infinite-dimensional? Is this some general fact about modules over algebras?

[lone jacinth]

Well, the argument uses that a homomorphism is given by a finite matrix and that the matrix is invertible iff the determinant is nonzero

So the argument wouldn't work in the infinite dimensional case

The statement is probably still true, idk

[woeful crane]

oh, ya, osrry I get that the argument only works for the fin dim case.

[woeful crane]

ah, I see, so you're not claimming it for the inf dim case

[undone idol]

There no way it’s k theory!!!

[spice idol]

lmao

[woeful crane]

I should probably change my name and pfp on here. I love the support gang, don't get me wrong, but it's prob annoying for others for discussions to get interrupted by people asking if it's me.

[ebon cove]

lol

[woeful crane]

oh, well looks like I don't have permission to change my name here anyways.

[spice idol]

idk, inf dim stuff is weird

id say try to deduce smt by looking at all finite dim submodules but thats probably not gonna help much if the algebra is annoying enough for there to not be finite dim submodules

[lone jacinth]

Could change your name to
K-theory [yes it's actually me]

[spice idol]

thats weird? maybe you need an active role for that or something

"John K-theory"

[limpid horizon]

Lol

[woeful crane]

for the inf dim case, what about taking an extension F/Q. Then I would guess that F[x] and Q[x] are not isomorphic as Q[x]-modules, but there extensions to \bar{Q} should be, right?

or am I tripp'n?

[spice idol]

how can you prove that \bar{Q} ⊗_Q F ≈ \bar{Q}?

or is that not needed for this case

[woeful crane]

Oh, ya I dont think that's true actually, so maybe my example doesnt work.

[spice idol]

i feel like there should be a name for algebras with this property

related to faithful flatness

[digital parcel]

U can dm modmail

The mods can change it for you

[digital parcel]

Yea active or higher

[lone jacinth]

Here's an infinite dim counter example @woeful crane @spice idol

https://mathoverflow.net/a/152083/157483

[spice idol]

oo funky

[undone idol]

no way its the jagr2808

[spice idol]

gahh I wanted to change my name to add [no, the other one] but i can't it'd be too many characters :despair:

there goes the fraktur font qwq

[limpid horizon]

Lol jagr name

[rose mirage]

That guy is cool

[digital parcel]

mick jagr

[lone jacinth]

I got them moves

Moo-oooo-oo moves like jagr

[spice idol]

no way you fuckin put 67 in there

out

banned

[digital parcel]

low swag levels detected... from your direction...

[spice idol]

i am in hell

[ebon cove]

this chat has gone down the drains

[spice idol]

lol

what are you verified in

[ebon cove]

oh I just noticed that, huh

must mean I'm legit 😄

[spice idol]

type shiii

i thought you were doing what seems to be a small naming trend too

[digital parcel]

<@&268886789983436800>

[limpid horizon]

"It is not possible for fewer than d elements to generate an ideal whose radical is m because then the dimension of R would be less than d." (context is x1, ... xd is a system of parameters and d = dim R)

can I have a hint for this? doesnt seem like it should be hard

do I need dim R < dim R/I?

[digital parcel]

this is essentially the content of the first part of Atiyah-Macdonald chapter 11

[limpid horizon]

Oh ok so its not like a trivial statement

[digital parcel]

it uses hilbert functions and the like. there might be a more straightforward proof but this is the one i know

[limpid horizon]

Ty

[wanton spoke]

if an ideal is generated by r elements then any prime minimal over it can’t have height bigger than r

[limpid horizon]

This is krulls principal ideal thm or smth?

[wanton spoke]

generalized version

I think krull is the case r=1

[mental escarp]

Did you find a book with exercises about systems of parameters?

[limpid horizon]

I did not

I mean ig matsumura probably has it

I should read a/m ch11 tho. Thats my plan to understand this stuff better

Or ch10, whatever it is

[edgy pond]

How can I find two irreducible representations V, W of a finite group G over a field with finite characteristic p > 0 such that their tensor product V \otimes W is not semisimple?

I found this answer online but I don't get why the tensor product is not semisimple

[wanton spoke]

little bit hard as question

[woeful crane]

Well, since you just need one example you can try writing this out explicitly. Take p=3 and compute V_2@V_2.

[wanton spoke]

over an algebraically closed field k of characteristic 3 the 3-dimensional simple V2 can be realized as Sym^2(k^2)
then V2 ⊗ V2 maps onto Sym^4(k^2) by: (f,g) ↦ f·g. Now Sym^4 is already not semisimple in char 3: it contains a 1-dimensional SL2-stable line spanned by the polynomial x^3 y − x y^3 (this is the standard invariant x^p y − x y^p at p=3) so you get a submodule isomorphic to the trivial representation inside Sym^4
the quotient Sym^4 / k·(x^3 y − x y^3) has dimension 4 and is simple
so Sym^4 is a nontrivial extension of that 4-dimensional simple by the trivial module hence not semisimple since Sym^4 is a quotient of V2 ⊗ V2
the tensor product V2 ⊗ V2 cannot be semisimple either

[limpid horizon]

JustSumGuy

Lol

[wanton spoke]

another example if you want
take k algebraically closed of characteristic 2 and G=S_3 and V the 2-dimensional simple kS_3 module

and compute V ⊗ V

[lone jacinth]

Don't even need k alg closed, could just take k = F2 if it simplifies the calculation

And it kind of is this example since SL2(F2) = S3

[wanton spoke]

the only reason I said algebraically closed was to make the Schur’s lemma feel automatic over F2 it’s still fine because your 2-dim module V is just the natural module for SL2(F2) ⊂ GL2(F2) and the centralizer of the whole group in End(V) is still only the scalars so (V ⊗ V)^G has dimension 1

so I agree you can just take k=F2

[lone jacinth]

Whatever floats the boat

[flint cloak]

whats the general approach to proving this statement?

i was thinking i could say V contains no irreducibles and show if that were the case then each of the subreps would be nontrivial and have dimension greater than zero and take more and more nontrivial subreps

[torn harbor]

isn't it basically immediate by semisimplicity

[flint cloak]

yeah

[torn harbor]

semisimple = direct sum of simple (hence irreducible) so just pick one of them

unless the way you're defining it is different

[flint cloak]

that hasnt been proven yet though

its the next problem

but i could just show it here

[torn harbor]

how did you define semisimple then

[torn harbor]

ah ok yeah I was gonna say

[flint cloak]

i guess i could just let dimV be n and then gather n+1 subreps in a direct sum from decomposing it that way

[torn harbor]

its a general fact for modules over a ring that semisimple implies direct sum of simple, I believe you prove it with zorn's lemma but u might be able to do it directly because finite dim

if there are no (nontrivial) subrepresentations, then its already simple, otherwise you can split and apply induction

(induct on dimension)

[flint cloak]

yeah from the definitions im working with it seems like the latter will be the way to go

we havent really talked about simple only semisimple

well we have the notion but not a definition

[torn harbor]

simple is the same as irreducible

[flint cloak]

yeah

ahhhhh

that makes sense with semisimple / completely reducible

[torn harbor]

yeah idk if you've seen this but you can view a representation of a group as a module over the group ring

and then everything fits in with the theory of semisimple/simple modules

which is quite nice

[flint cloak]

yeah unfortunately ive yet to get into those notions

i only know super basic ring/module theory

cool thanks blake

[lone jacinth]

So two key facts I would use here:

Since every representation has a finitely generated submodule, V has a finitely generated direct summand.

Any finitely generated representation has a maximal subrepresentation, hence an irreducible quotient (use Zorn's lemma)

[pastel agate]

Does anyone know of an intuitive explanation for why modules are decomposable iff there exist nontrivial idempotents?

I have the proof right in front of me but I really don't have any context other than that

It was useful for a particular homework problem but beyond that this proposition seems pretty odd to me

Why should I expect this to be the case were I defining things like direct sums etc

[wanton spoke]

an idempotent endomorphism e of M (meaning e∘e = e) is exactly a projection: it picks out a part of M and ignores the rest if M = A ⊕ B you get a nontrivial idempotent by projecting onto A along B applying the projection twice changes nothing so it’s idempotent
conversely if you have a nontrivial idempotent e then M splits as im(e) ⊕ ker(e): every m can be written as e(m) + (m − e(m)) the first term lies in im(e) and the second in ker(e) and the intersection is zero because anything in both would satisfy x = e(x) = 0 nontrivial just means neither piece is all of M or 0 so you really get a decomposition

[pastel agate]

Ah I see thinking of it as a projection makes things clear, thanks!

I only wish my instructor used this language but oh well

[wanton spoke]

dw I can be your new instructor

[hollow briar]

this is the definition i saw on universal algebra

and i don't really understand why i need both \mathscr F and F...?

why i can't just define F to be a family of finitary operations at once

without involving \mathscr F at all?

[hollow briar]

will it be not well-defined if i define like this?

[forest turtle]

you can, and it will be well-defined, but you want to be able to talk about and compare different algebras of the same type, so it is convenient to fix a set of function symbols and look at homomorphisms of algebras of the same type.

[spice idol]

\mathscrF is the so-called signature

F is just an interpretation of that signature

[spice idol]

i.e. \mathscrF is a collection of sets indexed by the nonnegative numbers (0, 1, 2, ...), and F assigns to each f in \mathscrF_n some function f^A : A^n -> A

an example might help: take \mathscrF to be { e, +, - } where e is 0-ary (so a constant), + is 2-ary, and - is also 2-ary.

Then we might take the algebra <G, F> over \mathscrF, where G is the underlying set of some group, e is taken to be the identity element, + to be the multiplication of that group, and - the operation taking (a, b) to ab^-1 in G.

Or another example, let X = { 0, 1 }, and let the interpretation of e be 0, and take + and - to be the operations sending a pair (a, b) to 1. This shows why we make a distinction between \mathscrF and F, as two operations in \matscrF may be the same function

in general, we need the distinction between \mathscrF and the actual operations on an algebra to define homomorphisms between algebras

[spice idol]

in conclusion: you can; sometimes these are useful, but if you want a notion of homomorphisms or products then you need \mathscrF to have a "canonical symbol" for each operation

it's like how you define a group to be a set with some operations that you generically assign symbols (for example e, * and ^-1). These generic symbols then form the \mathscrF of that theory

[hollow briar]

thanks, that was really helpful

i was just trying to understand by looking at a few more pages of the book and your examples

and now it seems i'm getting it

[spice idol]

youre using Burris and Sankappanavar right?

[hollow briar]

yes!

[spice idol]

nice! it took me a bit to wrap my head around at first too

ive noticed that a lot in UA, it kinda defines things you subconsciously use doing algebra, and getting your head to accept that new formal definition takes a bit of time

[spice idol]

based, though it just neglects to cover any categorical stuff

almost a third of the book is about applications of boolean algebras to universal algebra

which, don't get me wrong, is awesome, but i wouldnt cover it in a "first course"

[hollow briar]

maybe becuz categories are much general than UA? not sure though

[spice idol]

theres definitely connections of category theory to UA which i find important

for example, (co)limits and in particular direct and inverse limits

[hollow briar]

i think mathematicians always name their book without even thinking..

[spice idol]

lol so true

I would've also liked some basic stuff about tame congruence theory or commutator theory, however those may not even have existed yet at the time the book was written

we are long overdue for an explanation of those that isnt written as a research paper though...

anyhow that is far in the future for you for now, the book is really well written in that regard

[hollow briar]

thanks for help again

[spice idol]

nw! i think im the only active universal algebra guy here lol

[hollow briar]

lol tbh it's really hard to find someone really into UA

[spice idol]

lol you can tell me that again

[hollow briar]

i've tried to ask a few questions to my professor, but he said he isn't professional in that area and he didn't know either.

[spice idol]

are you self learning it?

[spice idol]

gotta be then lmao
if i may ask, how did you find out about it?

[hollow briar]

for UA, yes. there were no specific courses that only cover UA in my college

[spice idol]

same 😔

[hollow briar]

i don't remember exactly though... maybe while studying category theory?

[spice idol]

hmm, interesting

categorically youre basically studying monads, when doing UA

(yk the monoid in the cat of endofunctors n all)

however universal algebra has certain nice advantages

namely, there is essentially no analogue for congruence theory in general so anything which relies on that doesnt work

uh, you'll learn what congruences are soon i believe

[empty citrus]

Im reading Jones’s publication on Von Neumann algebras for a research project. Should I take the plunge into a book like Takesaki or Pedersen if I have completed all the exercises?

[wanton spoke]

go Pedersen first and then Takesaki

[empty citrus]

Thank you I appreciate the help

[wanton spoke]

if you don’t have time just go Takesaki

is it modular theory your project ?

[empty citrus]

Yes it is mod theory on type 3 factors

I wasn’t sure if I should jump into takesaki right away considering it sort of is the authority on modular theory of von neumann algebras

[wanton spoke]

Takesaki: expect it to be dense but it’s exactly your topic
but keep Pedersen nearby for quick refreshers on C*-basics, weak topologies, approximate units, etc

[empty citrus]

I kinda figured that was how I should approach it but im still in undergrad so wanted to get a consensus

Thank you

[forest turtle]

how does that work?

btw, planning on picking up some UA soon too

[spice idol]

hmm its a little conplicated

very beautiful though

basicslly, an algebraic theory is a small category where every object is the product of some generic elemenr

it is well-know that each variety has an associated theory, and two theories are isomorphic iff their varieties are "isomorphic" (in a specific but very natural sense)

now, for a monad T on Set take the Eilenberg-Moore category C of T-algebras. Then if n denoted the set of n elements, we can associate a theory to this as follows: the objects are the natural numbers and hom(n, m) = hom_C(T(m), T(n))

essentially this is the same construction as the associated theory but purely categorically

now if T is finitary, then it happens to be totally decided by its theory (you can think if this as only containing finitary operations, so its decided by how its theory behaves on finite sets), and so youve got a correspondence between varieties of algebras and finitary monads on Set

[torn harbor]

Takesaki is a good reference but is going to be very difficult to read, I'd mainly use it to skim for results

[blissful adder]

hey :) i have a rep theory problem i'm stuck on

if V, V' are both $\mathbb{C}G$-modules, $v\in V$ and $v'\in V'$, i need to prove that $\Big(\frac{\dim V}{|G|}\sum_{g\in G} \overline{\chi_V(g)}g\Big)v=v$ and $\Big(\frac{\dim V}{|G|}\sum_{g\in G} \overline{\chi_V(g)}g\Big)v'=0$

[broken turtle]

blutac

[blissful adder]

my first idea was that this expression on the left (in the exercise they shortened it to e_V) looks really similar to the inner product for characters

$\langle\chi_i,\chi_j\rangle:=\frac{1}{|G|}\sum_{g\in G}\overline{\chi_i(g)}\chi_j(g)$

[broken turtle]

blutac

[blissful adder]

but i dont really know how to connect these ideas :')

any help would be appreciated

(the first part of the question asked me to show e_v is central, so that might help. but i cant see how it is)

[lone jacinth]

One useful trick is that an element e of CG is central iff
g e g^-1 = e

So try writing out what
g e_V g^-1 is

[spice idol]

facebook is gonna love this

[blissful adder]

will do. ty :)

[lone jacinth]

Maybe put the shitposting #chill or #1203471755449073774

[blissful adder]

'my category theorist friend wants to get high. i told him to add joint'

[spice idol]

okay bro

aside from the technically spam this isn't even an eigenvector

[lone jacinth]

And I'm guessing V and V' where supposed to be irreducible(?)

If so, some big further hints could be ||multiplication by e_V is a homomorphism|| and ||schurs lemma and what is the trace of this homomorphisms||

[blissful adder]

oh yes hehe sorry

V and V' are simple/irreducible otherwise their characters don't fit in the inner product nicely

[lone jacinth]

Well the first part should still be true in general, but the
e_V v' = 0 will require something more

[spice idol]

💔

[limpid horizon]

Lol

[spice idol]

man you've got shit taste

lmai

[blissful adder]

im not really sure how to write $g e_V g^{-1}$ in any way other than just $\frac{\dim V}{|G|}\sum_{x\in G} \overline{\chi_V(x)}gxg^{-1}$

[broken turtle]

blutac

[blissful adder]

but that doesnt really help me ;-; (because its basically doing nothing)

what did you mean by writing out what it is ?

[lone jacinth]

So notice here that as x ranges over all elements of g, gxg^-1 also ranges over all elements of G

[blissful adder]

yea

[lone jacinth]

This suggests you can simplify this a little by a change of variables

[blissful adder]

what, gx=y ?

[wanton spoke]

<@&268886789983436800>

[blissful adder]

i feel like gxg^{-1} just changes the location of my problem

[lone jacinth]

Well I was suggesting y = gxg^-1

[spice idol]

yes but the character has a very nice property

[lone jacinth]

How would you rewrite the expression as a sum over y?

[blissful adder]

$\frac{\dim V}{|G|}\sum_{y\in G}\overline{\chi(g^{-1}yg)}y$

[broken turtle]

blutac

[lone jacinth]

And what does
chi(g^-1 y g) equal?

[blissful adder]

just chi(y) no?

[lone jacinth]

Yup

[blissful adder]

but i dont understand what that achieves

i understand e_V = g e_V g^{-1}

so its just provin that really

[lone jacinth]

So then what this equals is
Sum chi(y)* y
which is exactly the definition of e_V

[blissful adder]

yea

sorry maybe my question is unclear

i have proven e_V is central

im trying to prove the next bit

[lone jacinth]

Which is the thing we set out to prove

[blissful adder]

sorry i think maybe i was unclear

[blissful adder]

when i said this

i meant 'i cant see how it is helpful'

i proved e_v is central without much difficulty

[lone jacinth]

If
e = g e g^-1
then
e g = g e

I.e. e commutes with g

[blissful adder]

i am stuck on showing V' is in the kernel of the action of e_V

not showing e_V is central, i did this

[lone jacinth]

Okay, you already had that part.

Then my next hint would be: multiplication by e_V is a function V' -> V'

Is there anything special we can say about this function?

[blissful adder]

i guess that it's CG-linear ?

[lone jacinth]

That's right

[blissful adder]

so it's a module hom

/ intertwiner

[lone jacinth]

And what do we know about homomorphism of simple modules?

[blissful adder]

OHHH

it's zero or iso

...

i cant believe i didnt spot that

[lone jacinth]

Yes, even more specifically it's given by multiplication by a complex number

[blissful adder]

oh yea

schurs lemma right?

ok perfect

thank u for the help i cant believe it was such a straightforward solution

i thought for sure i would have to invoke row orthogonality of character table

:')

i spoke too soon

e_V is an endomorphism V' to V'

not V to V'

so there's no direct sledgehammer that says they're not isomorphic

now i know e_V (v') = zv' for some complex number z

and e_V(v) = wv for some complex number w

[lone jacinth]

Yes, so now the next hint is that the trace of multiplication by z is
z dimV'

What is the trace of multiplication by e_V on V'?

[blissful adder]

let me think about that one haha

is it by chance the sum of eigenvalues of $\phi(v)=\sum_{g\in G}\overline{\chi_V(g)}g\cdot v$

[broken turtle]

blutac

[blissful adder]

which is $\phi(v)=\sum_{g\in G}\overline{\chi_V(g)}\chi_{V'}(g)$

[broken turtle]

blutac

[blissful adder]

i.e. the inner product ?

[lone jacinth]

Ding ding ding

[blissful adder]

okay nice

is that a property of trace?

linearity?

[lone jacinth]

Yes

[blissful adder]

okay makes sense

im forgetting some simpler results in my quest to pick up more advanced ones

alright !!!! thank u

all the pieces r together now

much appreciated <3

[vague pawn]

Let O be a one-dimensional noetherian integral domain. Why is every nonzero ideal of O not contained in almost all primes of O?

[spice idol]

because of primary decomposition essentially

[limpid horizon]

What is almost all?

[vague pawn]

except a finite amount

[spice idol]

simply, let I be a nonzero ideal. Then √I can be minimally written as the intersection of primes p1, ..., pn. Suppose then there is another prime ideal q with I ⊂ q. Then we have p1 ∩ ... ∩ pn = √I ⊂ q. However, q is prime, so there is some p_i such that p_i ⊂ q, and we get a sequence 0 ⊂ p_i ⊂ q. Now we may use that O is 1-dimensional: there can only exist chains of length 1, so we must have p_i = q (as p_i =/= 0 by assumption of I being nontrivial). Therefore, there are only finitely many primes laying over I: its associated primes

[vague pawn]

I see

[limpid horizon]

Nice

[vague pawn]

Then √I can be minimally written as the intersection of primes p1, ..., pn.
why are you doing this here with sqrt{I} and not just I?

[limpid horizon]

To consider all chains or smth?

[spice idol]

because an ideal can be written as an intersection of prime ideals iff it is a radical

[fierce steeple]

radical? Or is reduced synonymous

[spice idol]

uh yeah i mean radical

[fierce steeple]

Yeah I guess you can rephrase enpeace's argument as like: primary decompose says each ideal has a finite collection of primes minimal wrt containing it (the ass. primes) But of course the "minimal" is redundant by dim 1

[vague pawn]

we don't need minimal anyway

[fierce steeple]

Wdym lol

[vague pawn]

because we just want to show there are a finite number of exceptions

[fierce steeple]

Yes but I'm saying like

This is what the primary decomposition says

[limpid horizon]

Whats the intuition for it to even be “almost all”?

[fierce steeple]

Consider Z

This is saying any integer has only finitely many distinct prime factors

[lone jacinth]

O/J is 0-dimensional so artinian so has finitely many maximal ideals

[limpid horizon]

Yall are too good at math

[spice idol]

in general you intuit there to be an infinite number of prime ideals

so you can intuit geometrically ig

[fierce steeple]

This is also somewhat standard for some like

The theory of Dedekind domains uses a lot of this stuff

[vague pawn]

🙂

exactly where my question comes from

[fierce steeple]

So it appears in some alg NT for example

[limpid horizon]

Oh cool, sounds kinda nice

Im at the point where id need to sit with that question for a bit to work out understanding it

[fierce steeple]

(E.g. rings of integers of number fields)

[vague pawn]

I am currectly reading a bit about orders, which are examples of one dimensional noetherian domains

[fierce steeple]

Yee like dw about it just saying like we already have some prior exposure rather than just magicking it

[fierce steeple]

Noice, where from

[spice idol]

so its basically just comm alg

[fierce steeple]

Ye lol comm alg bit of alg nt ig

[spice idol]

orders like order theory

[vague pawn]

neukirch

[fierce steeple]

Unfortunately no

[spice idol]

*dies

[fierce steeple]

Lol

[digital parcel]

<@&268886789983436800>

[near lantern]

Well, if there are infinitely many things for something to possibly hold for and it holds for all but finitely many... it seems reasonable to say that.

(For what it's worth, I would not want to say "almost all primes" for "all but finitely many primes" for a ring with finitely many primes.)

[spice idol]

Is there a nice description of the kernel of the natural map Z(AxB) -> ZA x ZB? (where ZX is the free abelian group on the set X)

[rose mirage]

yeah I think there is. I've just worked this out explicitly and guessed the map you're after is [\sum_{(a, b) \in A \times B} n_{a,b}(a,b) = \sum_{a \in A}\sum_{b \in B} \mapsto \left(\sum_{a \in A}\sum_{b \in B} n_{a, b}a, \sum_{b \in B}\sum_{a \in A} n_{a,b} b\right)$, by freeness of $\mathbb{Z}[A]$ and $\mathbb{Z}[B]$ the latter tuple is $(0,0)$ if and only if $\sum_{b \in B} n_{a,b} = \sum_{a \in A} n_{a,b} = 0$. Double check this but I'm like 90% sure this is right

well that's good isn't it

one second please

just render it myself I cba with this robot bs!

[spice idol]

peak

[spice idol]

okay yeah

a nice generating set is acc what i need i am bad at wording lol

[rose mirage]

you want a big square of numbers where each row and column adds to 0!!!!

so you can take like

hmm no this is tricky

I was thinking you put a 1 anywhere and then a -1 anywhere in the same row/column but the sum of these need not lie in the kernel

it reminds me a bit of the classical 8 queens problem

[spice idol]

magic square ahh

[ornate atlas]

If I’m writing more than like 2 lines of latex I tend to do this, I couldn’t have survived in the pre linter days

I need someone to colour my brackets or I get confused :(

[rose mirage]

I write my latex into what is essentially notepad and half of the time I'm not even looking at the monitor I'm typing on

[ornate atlas]

https://tenor.com/view/jerry-meme-tom-and-jerry-meme-jerry-points-get-a-load-of-this-guy-this-guy-gif-11384860625931393733

[rose mirage]

you swine I will have your head for ts insubordination

[fierce steeple]

Based

Vim or emacs

[urban granite]

me: puts "echo "blah blah" >> document.tex"

[fierce steeple]

Based

sudo solve_maths_problem

[wanton spoke]

permission denied: cannot compute without --rigorous try: sudo apt install brain

[lusty galleon]

hi, trying to do this ring theory problem. i got to the fact that 2yx=2xy by computing (x-y)^3 and (x+y)^3 and setting them equal and cancelling, and then multiplying by x on the left / right and cancelling again, but now i dont know what to do, since i cant just cancel the 2's. please help 😭 this problem is beating me up real good

[fierce steeple]

sudo grok

[spice idol]

how are you getting a monomial of degree 2 lol

[lusty galleon]

are u asking for my work

[spice idol]

sure

[lusty galleon]

sorry for the mess

[lone jacinth]

This is a pretty hard problem, but here are some hints to one possible approach:

||Show that x^2 is in the center||
Subhint for showing this: ||x^2 is an idempotent, so x^2(1 - x^2) = 0, which means x^2 y (1 - x^2) is nilpotent||
Then
||Massage xy = x^3 y^3 to get yx||

[lusty galleon]

i haven’t learned what idempotent / nilpotent means yet 😔

nor what center means for rings (i know it for groups only)

[cloud karma]

These questions are either the easiest thing or the trickiest thing ever

[spice idol]

fun fact! there is no yet know standard deduction for these problems

[lone jacinth]

idempotent means
e^2 = e

Nilpotent means e^n = 0 for some n (in this case n=2)

[spice idol]

(proving that x^n = x => xy = yx)

[cloud karma]

is this true?

[spice idol]

with the right definition of standard deduction

[lone jacinth]

little latin lesson
Idem = self, potent = power, nil = zero

So power of something is itself/zero

[cloud karma]

Now I know why it's called nilpotent

Thanks jagr

[lusty galleon]

the version with x^2=x was so chill and now this version is so much worse

[cloud karma]

yeah many such cases

[spice idol]

wait till you get x^4 = x

[lone jacinth]

And I don't believe an approach with cancelling 2s should be able to work, because in R = Z/6 you cannot cancel 2s

[lusty galleon]

the thing is i’m unsure if that’s the intended way my prof wants me to do it (or herstein ig, the book i’m using), if i haven’t learned idempotent or wtv

[lone jacinth]

Though I guess you could break R into a product of a characteristic 3 ring and a characteristic 2 ring and then argue them separately

[cloud karma]

You can just say e^2=e without mentioning idempotent

[lusty galleon]

i suppose that’s true
how did u go about coming up with that approach to it?

[lone jacinth]

I guess it's kind of a trick that if you have an idempotent that is not central you automatically get nilpotent elements, which comes from thinking of breaking your ring into a matrix

And one thing you can notice immediately is that 0 is the only idempotent

I guess just using everything that has to do with powers of elements makes sense

But there isn't really a good way to come up with this I feel

[lone jacinth]

Maybe this approach makes more sense to you

As 2^3 = 2, R has characteristic at most 6, so breaks into a product of a characteristic 3 ring and characteristic 2 ring.

Then your argument shows the characteristic 3 ring is commutative.

For the characteristic 2 ring you might notice
(x^2 + x)^3 = 0, so x^2 = x

Which is a case you already covered

[lusty galleon]

ah, uh, maybe i should say i don’t know that much ring theory (it’s been 2 weeks)
i know about characteristic in a very vague sense, but idk about ring products at all

[lone jacinth]

Who is the person handing you these exercises instead of teaching you ring theory?

[digital parcel]

Maybe it’s a test to see if you’re cheating

[lone jacinth]

Slightly joking, but also don't feel this is a very productive exercise for a beginner

Like if you don't have any fancy tools, I'm not sure how much learning there is in trying random things and manipulating equations.

Like, okay there definitely is learning, but feels like it would be more frustrating than productive

[lusty galleon]

LOL blame herstein

he’s a weird mathematician and i mean that affectionately but also his book is so strange

the rest of the problem set for this week seems so reasonable too 🤔 like most of it was on ideals, and then some homomorphism stuff at the end, but this one is just thrown in there for some reason

[cerulean cove]

.

[hushed bone]

What does this mean actually. Like, we don’t know if x^n=x for all x implies xy = yx for arbitrary n?

[spice idol]

we dont know a standard elementary deduction. Like, only by "doing the algebra"

[jaunty sparrow]

Does the property hold for all n?

[spice idol]

yes, idr who but someone did prove it

[jaunty sparrow]

Ah

[lone jacinth]

I think you can even have n depend on x can't you?

[fierce steeple]

Is this Jacobson?

[spice idol]

yeah that sounds familiar

[fierce steeple]

I think that may have n depending on x but not sure

https://arxiv.org/abs/2310.05301

[spice idol]

omg!!

this is so recent lmao

[ornate atlas]

Everywhere I go, I see his name. Martin Brandenburg haunts me.

[fierce steeple]

Ig he's on stackoverflow a lot and iirc not in academia anymore so probably more able to do stuff like this which is cool

[ornate atlas]

That is actually very cool though, I’d seen special cases of that result before but I didn’t realise it held for all n

[fierce steeple]

I have seen his actual AG work be relevant to me which is cool

I mostly see lots of posts about locally ringed spaces

[ornate atlas]

Oh shit yeah he’s a software dev, I just assumed he was still in academia. It just seems that every stackexchange/overflow post I see he’s somewhere around there

And yeah I’ve come across a couple of his papers

[spice idol]

is he perhaps the one who made a paper about explicitly constructing limits in Sch and AffSch?

[fierce steeple]

Oh idk

Though for affine schemes that sounds a bit odd lol as it is easy

[spice idol]

yeah i guess i meant for schemes
apparently the proof that products distribute over coproducts in Sch does not require that coproducts distribute over products in CRing

[rose mirage]

products distribute over coproducts because I've yet to have been convinced otherwise

[spice idol]

one can hope

[unborn rampart]

What are the applications of quivers and path algebras? A quiver is a kind of graph right, so are they used in graph theory?

[jaunty sparrow]

One thing that comes to mind is bound quiver algebras, which are quotients of path algebras by "admissible ideals". A theorem of Gabriel shows that any finite-dimensional associative algebra is Morita equivalent to a bound quiver algebra.

i.e. their module categories are equivalent

The main use I have seen for quivers is studying the representations of finite-dimensional associative algebras.

[unborn rampart]

I see, thanks

[rose mirage]

they're a special case of category algebras

so you can do cohomology of quivers with them :zamnzaniel:

there's also interesting connections between quiver algebras being of finite type and real reflection groups

[unborn rampart]

interesting, thanks jagr

[rose mirage]

if u wish to look into ts more google "Gabriels theorem"

[limpid horizon]

Jager

[spice idol]

does this have any relation with singular (co)homology of the underlying graph?

[spice idol]

"algebras and representation theory" by erdmann is a good book covering this

[rose mirage]

I think the directedness of the paths involved cause them to be quite different but I'm far from an expert in this field

[spice idol]

guh
makes sense

[rose mirage]

I suppose this forgetful functor has an adjoint sending a (for now, simple) graph to a (for now, doubly laced) quiver

so you might be able to leverage something out of the quiver => graph direction

I don't actually know if that's an adjoint but I'd eat my thesis if it wasn't

[spice idol]

when im in an adjoint competition and my opponent is that functor:

[rose mirage]

yeah they're adjoints. EXERCISE: Construct the unit and counit of this adjunction and verify they satisfy the triangle equalities

[ornate atlas]

Stole the man’s name and research interests smh

[rose mirage]

I did some googling. Apparently the right adjoint here HAS a right adjoint. That's crazy talk

[spice idol]

oh nice a triple

Hom(A, -) ⊢ restriction of scalars ⊢ - ⊗ A

ahh

[rose mirage]

oh I see it now. If a directed graph is an ordered relation then the double right adjoint sends it to the graph with edges corresponding to "equivalent" (in the sense of a preorder) elements

[rose mirage]

yeah this is how they always look cause of Kan extenstion bullshit

[spice idol]

actually really cool

:0

[rose mirage]

but I came up with the first right adjoint all by myself

[spice idol]

im proud of you

[rose mirage]

my mummy says I'm very cleaver

I was about to ask if such a thing existed for directed hypergraphs and then I realised I am reinventing infinity category theory AGAIN

[spice idol]

🔥

aw darn

there you go again

with your silly higher category theory

[rose mirage]

like ok so a (for now, finite) directed hypergraph is a a pair of sets E, V equipped with a map s : E -> V and a map t: E -> P(V) where P is the power set functor. This is exactly the dual data to that of a 2-opetope. So I suppose it's not infinity category theory just 2-category theory

[spice idol]

well just continue the pattern

[rose mirage]

yeah I guess you could define a directed n-hypergraph as a sequence of sets V, E_1, ..., E_n together with maps s_{k} : E_{k+1} -> E_k and t_k : E_{k+1} -> P(E_k)

and these are again just backwards n-opetope. It's kind of nifty how if you replace the power set with just E_k you get backwards globular sets, just E_k^2 you get backwards simplical sets etc.

there's something weird going on!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[spice idol]

wdym?

[rose mirage]

cause like, an opetope with two edges is a globe, with three edges is a simplex etc. etc.

doesn't hold for cubical sets due to woke

[spice idol]

what the fuck is an opetope

woke nonsense

[rose mirage]

they're just lil guys

do NOT ask how you compose them

[spice idol]

how do you compose them

[rose mirage]

very carefully

[spice idol]

how do neither nlab nor wikipedia have an actual explanation of what they are

[rose mirage]

chapter 7 of https://arxiv.org/pdf/math/0305049

any they probably don't explain them because they are THE most convoluted model for infinity categories imo

[olive schooner]

along the representation theory line, there's been a lot of activity in using them to study cluster algebras via cluster categories if you're interested in that route. if you look up BGP reflection functors and tilting theory, cluster categories give an environment to study the cluster algebraic analogues

and some things about cluster algebras have representation theoretic proofs that don't have known combinatorial proofs

[rose mirage]

hey I applied to do a PhD on cluster algebras

[spice idol]

yeah i can imagine
but i also can somewhat see how theyd apply

[unborn rampart]

Interesting, thanks it's funny how half the sources on the wiki page for tilting theory is by people from my uni

[spice idol]

where does the name tilting come from lmao

[lone jacinth]

Tilting induces a linear map on the dimension vectors / Grothendieck group which "skews" or "tilts" the dimension vectors

[lone jacinth]

I mean Idun Reiten and Sverre Smalø where very instrumental in developing tilting theory

(Idun is the R in both APR tilting and HRS tilting, with Sverre being the S)

[spice idol]

i see!

[lone jacinth]

I'm not 100% on the etymology, but this is something that makes sense at least

This is from Brenner--Butler 1980, I think they were the first to call it tilting

[spice idol]

"and because the word tilt inflects easily"

[ornate atlas]

Im not sure if this is better for here or #alg-top-geo-top but its basically just a homological algebra issue.

I have this problem, and so I just need to show that like f_*delta = delta f_*, so just a bit of diagram chasing. The issue is that im realising im not exactly sure (or have forgotten) how delta actually acts on elements

[last talon]

Draw out the diagram at the level of complexes and chase the only way for an element to get between the two places you want

[ornate atlas]

Wdym

[last talon]

So your LES of pairs comes from the SES of complexes 0 -> C(A) -> C(X) -> C(X)/C(A) -> 0
Draw out two “layers” of that SES (namely, the n and n-1 layers) and figure out how to get an element from C_n(X)/C_n(A) to C_{n-1}(A)

[ornate atlas]

I just realised I litterally have a homalg book next to me that tells me exactly how its defined

[spice idol]

well, the LES is functorial no? so if you can show that a morphism of pairs induces a morphism of SES's of complexes

[last talon]

This is kinda asking you to prove the LES is functorial in this specific case, so I assume it’s saying you can’t use the general result

[spice idol]

meow

well getting your hands dirty is good once in a while

[lone jacinth]

Functors from Top to Abelian groups is an abelian category. Snake lemma / the existence of the LES holds in abelian categories.

Conclusion the connecting homomorphisms is a morphism in that category aka a natural transformation

[prisma gull]

What is H_n(X,A) in this case?

[spice idol]

the homology of the chain complex C_*(X) / C_*(A)

[prisma gull]

Okay

It's actually equivalent to show, H_n is homological delta functor

[spice idol]

i guess but thats not very useful lol

you just need to show that the homology LES is functorial

[prisma gull]

I did this last year, my approach was morphism of using exact sequences.

But I don't remember details

[prisma gull]

Let's assume that G->G' surjection of fibrant simplicial groups, then does kernel of this simplicial natural transformation also be fibrant?

[wanton spoke]

yes

the kernel is a simplicial group and every simplicial group is a Kan complex hence fibrant

@summer quest is a kind of kan complex too

[summer quest]

so structured, yet so truncated

[wanton spoke]

yes

kan n-type lol

[prisma gull]

I know but I want to prove it

I'm trying now

Um...

[prisma gull]

This will help me thanks!

I saw that n-groupoid is a kind of kan complex on higher category theory and homotopy from Cisinski, last week, lol

But I only know brief explanations, since I'm beginner of that book

[cloud karma]

just straight to the point (Enpeace)

based

[spice idol]

took the words right out of my mouth

quite literally, i might add

plagiarist you cited me tyvm :3

[prisma gull]

Even Weibel book said fibrant with Kan conditions, to make sure that reader check details from Kan-complex

I think it was clear argument from him

[prisma gull]

I figure out why it's true

Simplicial group always be kan complex as a set

Good

Thank you

I proved it

[hushed bone]

Idk what this discussion is about honestly, but all I wanna say is I’m convinced that it is impossible to write an accurate homological algebra book

Every source I’ve ever come across on it has some weird random errors, a map is written wrong, there’s maybe typos, some definitions are flipped or something

I feel like homological algebra is something you need to really just get down and dirty with to really understand, and it’s because of all this random crap when you try and write it all down you end up with weird errors

I guess caveat:
stacks project is probably mostly devoid of errors and whatever Grothendieck wrote probably is too

[wanton spoke]

honestly even Grothendieck isn’t immune to mistakes
some of his texts have gaps, typos, or places where details are left to the reader, and later people had to clarify or fix things
he’s insanely influential, but that doesn’t mean everything he wrote is perfectly error-free

[hushed bone]

Yeah I mean, this is totally true, but I think it’s on a different level than most things

EGA is, honestly, pretty complete. I feel the stacks project is sort of even more complete, but EGA has fairly few errors for the massive thing it is, and there are errata in later volumes

There’s still errors for sure, like Zariski-locality of projective modules which makes a proof for something involving infinitesimal lifting incomplete, but it’s leagues better than a lot of other things on homological algebra haha

SGA and FGA are sort of different because those are of a different nature and aren’t supposed to really be textbooks

[wanton spoke]

yeah that matches my experience
EGA is impressively solid and Stacks wins on completeness mostly because it’s updated

[midnight mirage]

57 is a prime number btw

we somehow call it as the Grothendieck prime

for some reason

[ornate atlas]

Can someone please just double check im not being stupid, if I have a split exact sequence of abelian groups, applying Hom(-,G) (for G an abelian group) still gives a split exact sequence right? Like all there is to check is that its exact at the right after dualising, and that B*\cong A*\oplusC* right? But then the second thing just comes from Hom(-,G) preserving the direct sum, and then exactness follows instantly from the fact that it tells you your maps are just inclusion and projection?

[spice idol]

yes, split epis are preserved by any functor

ig you need left exactness to prove the first map remains injective and exact nvm

[lone jacinth]

Fun fact: a functor is additive (F(f+g) = F(f) + F(g)) iff F preserves split exact sequences

[spice idol]

since addition is given by a split sequence?

[lone jacinth]

I guess that's a way to put it

[ornate atlas]

Yeah I just wanted to double check I wasn’t being stupid because there was like nothing to it

And all the other problems took fucking ages lol

[spice idol]

homological algebra do be like that

[ornate atlas]

To be fair the prior problem was an easy diagram chase, I’m just stupid

[plucky arch]

Diagram chasing is certainly an interesting proof technique

[spice idol]

it's fun

[ornate atlas]

It wasn’t even the diagram chasing that was the issue I was chasing properly, I just defined the action of a functor on morphisms the wrong way around and was then confused why nothing worked lol

[spice idol]

nice pasttime

[ornate atlas]

And every book I checked to see where I was being dumb said “this is routine but long”

[limpid horizon]

lol

[limpid horizon]

Basic Hom alg by Osborne seem to go thru the nitty details of that diagram chase stuff

[ornate atlas]

Oh nice that’s good to know.I’m generally ok with the arguments now, like for this problem I did know the right things to check and things to use, I just set it up wrong, but it’s good to have another source for when I’m stuck

[limpid horizon]

Yeah i think its the friendliest hom alg book tbh

[ornate atlas]

My first pass with a lot of this was Weibel so anything is friendlier

[near lantern]

This may have been asked before (including by me possibly) but what's the best resource on (triangulated and/or) derived categories for building intuition?

[worldly zealot]

https://arxiv.org/abs/1610.09640 ive used sections of this, recall it being useful

[near lantern]

Thanks!

[wise sedge]

What is the second cohomology of the alternating group on n points acting irreducibly on $\mathbb{Z}^{n-1}$?

[broken turtle]

Mecejide

[storm basin]

I haven’t read this specific book but I kinda feel that writing the details of many diagram chases makes the book genuinely harder to understand than just saying “you should do it urself”

It’s just a wall of text that is really hard to interpret without just doing the proof yourself anyway

Hom alg is one of the fields where I feel leaving it as an exercise is actually good pedagogy

[storm basin]

As an example of a book where I don’t like when it does this, some of the stuff in Hatcher

I like most of Hatcher but a lot of the hom alg stuff is just holy wall of text that would be 10x more readable if he gave some hints & drew the hints on the diagrams and said do it urself

[ornate atlas]

I don’t know that I agree with “it’s not comprehensible as a long paragraph”, it’s just that it’s often long and beyond knowing what elements to pick and where, there aren’t many ideas.

But like fundamentally all you’re doing is picking an element somewhere and following it around a diagram, there’s no reason that would be hard to follow

[gaunt solar]

Hey guys I have a banger i am on it for 2 months now please help

[storm basin]

The only even remotely moral way to like typeset a diagram chase in a textbook is to just redraw the diagram with labeled elements after every 1-2 lines of the proof. If you don’t want to do that (which is fine imo) it is 10x better to just leave it as an exercise.

The first time I read the diagram chases in Hatcher I was like wtf is going on and crashed out for like a day cause I didn’t know how a diagram chase is supposed to work

It’s just not something that reads well in book format

[lone jacinth]

Or just draw the diagram and have elements labeled x1, x2, x3, ... by the relevant objects.

[storm basin]

Maybe my ideal version of pedagogy for something like that is to do this once to show how to do it and leave the rest as an exercise

[storm basin]

Yes but if it’s just diagram -> wall of text below it’s still really hard to read if it’s your first time encountering stuff

[lone jacinth]

Maybe followed by a paragraph like there is an x2 mapping to x1 because [insert property]

[storm basin]

You need multiple diagrams interspersed throughout the text to simulate the flow of a blackboard lecture

[lone jacinth]

Idk, it's not that hard to move your eyes up slightly to look at the diagram

[storm basin]

Yeah but it’s really distracting on first read

Especially if there are multiple elements on the same diagram it takes mental strain to clearly see each step

Anyway it’s not that big a deal but I think the best pedagogy is to be as clear as possible the first time and leave literally all the rest as exercises

I was crashing out on Hatcher’s proofs for so long until eventually I realized that best practice for learning homological alg is just to ignore the authors proofs completely and write your own

[lone jacinth]

I guess trying to prove things for yourself is generally a good tip in a lot of cases

Some math is best enjoyed in the comfort of your own head

[ornate atlas]

Yeah like, if I’m reading a proof like that I genuinely just trace around with my finger where elements are in the the original diagram, it’s fine

I’m also not saying that all the proofs should be included either, I think leaving most of them as exercises is fine because as I said they’re generally routine. The slight issue I have is that a lot of books tend to just leave them all, and they can be tricky if you’ve not seen any of them done before. I think Rotman strikes quite a nice balance in that regard

[limpid horizon]

Profound.

[steady lance]

I think having to write up diagram chases explicitly in my first algebraic topology/homological algebra class was one of the main reasons I decided not to go into that side of algebra lol

fun to write on a blackboard but awful to try to write on paper formally

[spice idol]

I think I've found it? My situation is slightly different and i genuinely forgot that this was the case on multiple occasions working in ts problem:
A, B and AxB are pointed sets, and the corresponding distinguished element (denoted x( gets identified with the 0, i.e. ZA gets modulo'd by Zx and so on. Now take the set
{ (a, x) + (x, b) - (a, b) | a in A, b in B }
I'm like 99.9% sure this generates the kernel of \hat{Z}AxB -> \hat{Z}A x \hat{Z}B (where \hat{Z}X stands for ZX with x identified with 0)

[edgy pond]

I want to prove that if V is a complex faithful representation of G then every irreducible representation appears as a summand of Sym^n(V) for some n. Now, I know that these appear as a summand in the n-fold tensor product of V with itself. Is there a way to conclude the same must be true for Sym^n(V) using this fact?

I can try to compute the inner product of an irreducible character with the a character of Sym^n(V) but the latter does not have a nice expression for large n

but the same approach with the tensor product works nicely, so it'd nice if I can deduce this from the tensor product case

[spice idol]

this works: decompose any sum into two parts where one has x in the first entry and the other x in the second entry, then it follows that if it gets mapped to 0 in \hat{Z}A x \hat{Z}B that both of the components have to be 0, so we are done

[near lantern]

TLDR the discussion but is it generated by ||{ (a1, b1) - (a1, b2) - (a2, b1) + (a2, b2) : a1, a2 in A, b1, b2 in B }||?

[near lantern]

Emacs

[spice idol]

yes

lol

but my situation is slightly modified and i didnt say so cuz im stupid lol, but A, B are pointed sets and you actually consider the groups ZX/xZ where x is the distinguished element

[spice idol]

in which case this is the answer

[near lantern]

One thing that I found convincing is that the category of modules of any finite-dimensional algebra is equivalent to the category of representations of a quiver with relations (as in KnightWatch's answer). The construction basically takes (IIRC) the vertices to be the projective representations, the number of arrows to be the dimnsion of hom spaces between the projective representations, and the relations to be the relations between those homomorphisms. So the quiver is genuinely a diagram of the (projective) representations. But you can do this for other representation theory settings, and I believe this is actually useful e.g. it come up in category O representations of a semisimple Lie algebra (I think??).

Anyway I'll stop replying to old messages. Not sure how I started.

[worldly zealot]

is this being assigned as a problem? i dont know any character theoretic way to show it

[edgy pond]

yes, and there are two hints:

1. estimate the inner product of an irreducible character with the character of Sym^n(V) as n gets big
2. Show that the action of G on the set of maximal ideals (x_1 - a_1, ... , x_d - a_d) of the symmetric algebra of V is faithful and deduce that the regular representation is a quotient of the symmetric algebra representation

here x1, ...., xd forms a basis of V

[edgy pond]

I am trying to solve the following:
Let G be a finite group and V be an irreducible complex representation of G. Let H be a normal subgroup of G then V restricted to H is isomorphic to m-copies of an irreducible representation W of H, as H-representations.

My approach is to take an irrep W of H contained in the restriction of V to H, then gW is H-stable, hence an H-representation. It is isomorphic to W and is also an irreducible H-representation. Now, $U = \bigoplus_{g \in G/H} g W$ is a G-stable subspace of V, hence it must be equal to V. This then proves the claim. The only problem I have here is whether U is actually a direct sum. Because it is not clear to me why $W \cap gW$ is trivial for every $g \in G/H$. Can anyone explain why this would be the case?

[broken turtle]

pink_panther

[near lantern]

I don't think this is true. Let G be S_3 and V the 2-dimensional irrep and H the subgroup of order 3, then V is the direct sum of two distinct irreps of H.

(This shows up in your answer in the part where you assume W is isomorphic to gW as a H-rep. This is not true (but it is still irreducible).)

[near lantern]

To answer your question, recall that W and gW are irreducible H-reps. So if the intersection is non-trivial, then - check that it is H-stable - what can it be?

But you have to show more than this in case there are more than two summands.

[buoyant fox]

What is this diagram supposed to be? my professor mentioned that its how you happen upon the definition of convolution for multiplication in the ring R[x], but the convolution operating doesnt make the diagram commute, so Im not sure whats going on

[foggy galleon]

Call the interrogation map H. I think you want f(n)g(m)=H(n+m)?

it's hard to tell

[buoyant fox]

yeah thats what would make the diagram commute

but thats not convluiton

convolution would be H(n+m)=sum over all i,j such that i+j = n+m (f(i)(g(j))

which is almost always going to be larger than that H which makes the diagram commute

[foggy galleon]

mmh I see. yeah idk maybe that's just a typo

[balmy pollen]

The regular representation of a group is equiv to just taking the group ring as a module over itself right

[spice idol]

yes that's why it carries so much information

[balmy pollen]

wahoo ty

[spice idol]

if you take complex representations then the group ring is semisimple hence why you can decompose it as a direct sum of all its simple modules (i.e. irreducible representations), so this view is nice

[scarlet ermine]

I’m more familiar with mutations of exceptional collections than quivers (as in the context of cluster algebras), is there any relation?

I was hoping maybe that this could be something like if you mutate a full strong exceptional collection (which gives you an equivalence to the derived category of quiver representations) this would still be strong and also give a different quiver, which is the quiver mutation… but that’s definitely not true (need not be strong, and also mutation of exceptional collection is not an involution, this is only a braid group action in general… but maybe you happen to get the same quiver?).

or I guess you could get a quiver out of an exceptional collection whether or not it is strong, just won’t get an equivalence

[regal bone]

Question Do I have to simply the Sigma? Or can I jsut put it in the caculator?

[spice idol]

advanced algebra 💔

[spice idol]

we should make this a sticker here

[regal bone]

same shit different font

[spice idol]

ts is not tuff bro

in any case #discrete-math prolly better for this q

[urban granite]

it's very funny that pre-uni people thinks what they do is too advanced to post it in #advanced-algebra when they do not know what a group is

[spice idol]

are sums #prealg-and-algebra ?

#precalculus ?

[regal bone]

mb chat i just sent it in a popular chat to ge a quick response

All Math Help are full

[spice idol]

well categories exist for a reason

[regal bone]

I apologies for my existence great math king! You will see my exit now at once!

[spice idol]

okay man

no need to be passive aggressive about it

[regal bone]

https://tenor.com/view/hii-cat-canycat-gif-11976152362355570375

Bye bye now my math superior!

[worldly zealot]

cringe

[spice idol]

yeah

ts crazy

i almost care

[regal bone]

https://tenor.com/view/hii-cat-canycat-gif-11976152362355570375

[maiden wadi]

goes to Burger King
“can I have a Big Mac”
we don’t serve that here
wow you guys r so elitist. The McDonald’s was busy so I came here

[regal bone]

Whats the closest thing to a big Mac
Burger
ok i will take a burger
Also I do admit and apologise for posting in the wrong channel 🙂

[spice idol]

it's alr lol no one's gonna hold it against you

just not being rude would be nice

[maiden wadi]

That’s fair ngl sorry for the toxic message

I let the urge to be funny on the internet take priority over not being an asshole that’s mb

[spice idol]

being funny on the internet is a top priority tho

[woven plinth]

Ts so advanced lil bro

[forest turtle]

https://tenor.com/view/ah-shit-here-we-go-again-ah-shit-cj-gta-gta-san-andreas-gif-13933485

[ebon cove]

take that to #category-theory

[limpid horizon]

Who the FUCK is that

[foggy galleon]

the isotropy type tells you about fixed points

the automorphism group of H is PSL_2(R). If you quotient by discrete subgroups of PSL_2(R) you obtain geometric spaces. Eg. modular curves. You can obtain all compact Riemann surfaces of genus g>1 in this manner, by the uniformization theorem

these modular transformations also come up when trying to classify elliptic curves over C

any elliptic curve over C is of the form C/L where L is a lattice (this is not obvious, but it essentially comes from the surjectivity of the j function). So you have to ask when two lattices gives the same answer. You can choose a representative Z+tau Z where tau lies in the upper half plane, and tau is chosen modulo PSL_2(Z)

also you get more refined information if you consider other types of moduli problems. Eg: cyclic subgroups, a point of order N, or a pair of points with torsion data. These are related to the modular groups Gamma_0(N), Gamma_1(N) and Gamma(N), and the space H/G is the "moduli space"

[limpid horizon]

I’m trying to exhibit that a set of prime ideals are associated primes of R/I by showing which element it annihilates. The ideal I is a squarefree monomial ideal plus some extra linear form generators. By computations on mac2 i think i know which elements will be the ones annihilated by the supposed associated prime, but i just need to show it algebraically now i guess

So if im claiming “ab” is the element the associated prime P annihilates,
then I need to show

ann(ab) = P

Or since we are in a quotient R/I thats the same as the colon ideal

(I : ab)

And computing these colons im getting stuck on

If I is a monomial ideal then there are formulas for this but the extra linear form generators is making it weird

R is a polynomial ring over a field

[storm basin]

I know this is algebra but this is also the isometry group of the hyperbolic plane so it’s important to understand it for DG reasons

[abstract dagger]

im pretty embarassed to question this but can someone explains me what are quotient spaces by relation operators 😭

i dont get the books explanations

some authors also call it canonical proyections

[worldly zealot]

i haven't heard the term relation operator in math

can you show what the book is referring to

[spice idol]

what book?

[abstract dagger]

i don't actually know who is the author or where is from, my teacher sent me only the quotient space chapter bc i asked him smth abt it

[wanton spoke]

I think he means equivalence relation

[abstract dagger]

oh and also i wanted to say an equivalence relation

exactly

[abstract dagger]

like its a 3 pages pdf

with only that chapter

[wanton spoke]

quotient spaces are just what you do when you’re too lazy to distinguish points anymore

[worldly zealot]

(this is better for #linear-algebra ) but with that definition 6.1, suppose V is R^2 and W is the span of (1, 0), so the x-axis. then if any two points (x1, y1) and (x2,y2) with y1 = y2, their difference will be in W. so (x1, y1) is equivalent to (x2,y2). so basically anything with the same y-coordinate is indistinguishable i.e. in the same equivalence class. then the later lemmas and observations are building up to the idea that if you treat every equivalence class as a point, you get a valid vector space

which they call the quotient space

[steady lance]

I don't think there is a standardized quicker way to do this in general tbh

which is why m2 is so nice

but if someone else knows of one pls @ me too in the response. Would be useful

[limpid horizon]

Ive been wondering then - how does m2 do the computation?

Something grobner basis i guess

[ornate atlas]

That’s generally the answer to these kinds of things

[limpid horizon]

My colleague suggested i look at this theorem instead

Im trying to find regular sequences for a cohen macaulay simplicial complex

If its CM then max regular sequence <-> its a s.o.p

[steady lance]

Do u do linkage?

[limpid horizon]

Wdym, like the link operation on simplicial complex?

[steady lance]

Like linkage theory of ideals

Guess not haha

But that stuff requires similar considerations to what you’re asking about

[limpid horizon]

haha yeah i have heard of that stuff

i looked at it a while ago and it was a bit hard for me to parse at the time

[steady lance]

It’s pretty cool! Worth looking into imo

[limpid horizon]

Oh I know you hahaha

[steady lance]

Yooo

[limpid horizon]

Tall dude

Lol

Not sure if u remember me we just talked briefly

[steady lance]

Oh yeah! Wait does ur name start with a k?

[limpid horizon]

It does haha

[steady lance]

Ah sick small world!

Hope all is going well!

[limpid horizon]

Thanks, likewise

[spice idol]

hm i wonder what "kiand123"'s name is

[steady lance]

https://tenor.com/view/winnie-the-pooh-think-think-gif-15324290528311488474

[limpid horizon]

Lol

[subtle smelt]

How well known is the following?
A functor F is fully faithful iff there exists any natural isomorphism
Hom(A,B) -> Hom(FA,FB)
In case where F admits a left adjoint G, it equivalently suffices to show that there exists any natural isomorphism 1 => GF.

If you know to expect it, then its a relatively simple Yoneda-type argument of seeing that all the universal elements must be isos which combine to an automorphism of F.
But I haven't seen it mentioned before in literature.

Kind of makes a big difference when you have proofs like:
f^* : D^b(Y) \to D^b(X) is fully faithful if f_* O_X = O_Y (everything is derived).
Hom(f^* A, f^B ) = Hom(A, f_ f^* B) = Hom(A, B \otimes f_* O_X) = Hom(A,B)
the second iso. being the (derived) projection formula.
(you really dont want to explicitely check that the last two isomorphisms can be expressed as the unit and some additonal isomorphisms)

[fierce steeple]

Idk I feel it is worth checking anyway here and like projection formula as you used it is saying that f_* f^*(-) is O_Y-linear and so the map should be the unit without too much difficulty, like just unwinding what the maps are

[subtle smelt]

It isnt that bad in this case, but quickly escalates if you throw in like 10 natural isos.
It has to do with the lax monoidal structure being used to define the projection formula map in the first place, in my notes I had
"But this follows from noting that, up to canonical monoidal unit isomorphisms the maps comes from
[
\phi \mapsto \varepsilon \circ f^*(m \circ (\eta \otimes 1) \circ \phi )
]
where we then have the identity
[
\varepsilon \circ f^m \circ (f^\eta \otimes 1) = 1
]
as follows from the definition of (m) as induced by the adjunction and the triangle inequalities."

[broken turtle]

Kerr

[wanton spoke]

once you demand a bivariant natural isomorphism Hom(A,B) ≅ Hom(FA,FB) it is rigid in the Yoneda sense
concretely set η_A to be the image of id_A under Hom(A,A) → Hom(FA,FA); naturality forces α(f) = F(f) ∘ η_A = η_B ∘ F(f) so η is a natural automorphism of F and the given α is just the usual map f ↦ F(f) twisted by η in particular, since each η_B is invertible you recover that f ↦ F(f) is bijective on every Hom-set so F is fully faithful With an adjoint G ⊣ F any natural isomorphism Id → GF gives Hom(A,B) ≅ Hom(A,GFB) and composing with the adjunction gives Hom(A,B) ≅ Hom(FA,FB) hence full faithfulness by the same rigidity this is just a rephrasing of the standard “right adjoint fully faithful iff GF → Id is an iso” (equivalently Id → GF after inverting)
for the derived pullback argument the point is: as soon as your chain of isomorphisms is natural in both variables you don’t need to identify it with the canonical unit or maps existence of that natural Hom-iso already forces full faithfulness

[subtle smelt]

I know the proof, but the way I even figured out was by looking for a counter-example lol

"we see that this map is an isomorphism by just making up a different one and checking it" just sounded absurd to me back then, and somehow I never encountered this rigidiy of bifunctors before. Hence asking how well known it was

[wanton spoke]

totally fair if you haven’t internalized "bifunctor naturality is brutally rigid" it does sound like cheating

[subtle smelt]

What are your favourite tricks from the "bifunctor naturality is brutally rigid" world

[wanton spoke]

evaluate at the identity: given any bivariant natural gadget involving Hom(A,B) look at what it does to id_A or id_B and you usually extract a canonical natural transformation (or automorphism) that controls everything

[spice idol]

bivariant natural gadget is a great term

[subtle smelt]

Me anytime the opportunity arises form now:

[spice idol]

<@&268886789983436800>

peak work

what are prereqs for learning about triangulated categories?

[scarlet ermine]

to learn about the derived category you don’t need as much as I thought I would have needed. If you feel comfortable with homological algebra on the level of Ext/Tor, you could pick up Gelfand and Manin and start maybe in chapter 2. It’s not the awesome-ist book imo, and maybe it’d be good to pick up some category theory from maybe Riehl alongside GM chapter 2.

I only got comfortable working with triangulated categories afterward when trying to read Bayer’s a tour to stability (https://www.math.utah.edu/dc/dc-lecture-notes.pdf) in case you like something with alg geo applications

one exercise that I struggled with but made me feel way better about the machinery was proving that this definition of a bounded t-structure (i think due to Bridgeland) is equivalent to the definition in BBD

[dark fable]

Just did this exercise and I’m wondering if this is useful beyond just applying it for the case of complex conjugate (which is shown by taking the dual rep anyway)

[scarlet ermine]

I guess C’s automorphism group is just Z/2Z so there’s no other use here. I bet the problem is looking forward to doing representations over different fields, where you could have big interesting Galois groups

[lone jacinth]

It could be useful, but it's relyant on the galois group being easy to compute.

Like say Chi(g) = sqrt(5), then you know there is an automorphism of C that takes this to -sqrt(5), so there should be a character with Chi(g) = -sqrt(5). But calculating exactly what this automorphism does with everything else can be hard.

I guess it gives you some partial knowledge of a new character anyway.

[lone jacinth]

The automorphism group of C is uncountable

Any automorphism of a subfield of C extends to an automorphism of C

[scarlet ermine]

oops! thanks sorry

(ahhh i was remembering that R only has the identity field automorphism, which I feel like is bizarre)

[lone jacinth]

I guess it depends what you want to learn about.

If you know the Yoneda lemma and basic diagram chasing you can read the axioms and start proving diagram lemmas and some simple properties.

For examples the homotopy category is easy to construct if you just know what a module is. To get the derived category from there you need some version of localization of categories.

If you want to work with enhancements and stuff you might start to need some homotopy theory / homotopical algebra.

[blissful field]

Can an algebra be defined over a module in the same way as over a vector space?

[spice idol]

yes

[fierce steeple]

You'll want to be over a commutative ring so that the tensor product makes sense

[frozen gazelle]

now im wondering a question regarding continued fraction and SL(2,\mathbb{Z}) can be asked here or by number theory

[near lantern]

I will guess number theory. In general if your first channel tried doesn't get responses for a day to a few days, you can try another channel (or at least that's what I do). I've also heard of cross-posting if you warn that it's cross-posted.

[frozen gazelle]

i see then i have already posted my q in advanced number theory

[dark fable]

This does seem very practical then, didn’t know you could extend any automorphism of a subfield of C. I haven’t seen that proof in the course tho

[lone jacinth]

It's pretty much just Zorn's lemma.

True for any normal extension

[dark fable]

I recall some reps with roots, like for A_5 the reps on the isocahedron (?)

And yeah makes sense

thanks

[lone jacinth]

Yeah the 2nd and 3rd character here are related by such an automorphism
https://people.maths.bris.ac.uk/~matyd/GroupNames/1/A5.html

[distant minnow]

Hey, I’m looking for someone to help each other learn homological algebra. I’m a complete beginner. If you’re interested, DM me.

[limpid horizon]

@solar turret 😲 your clone

[spice idol]

oh i gen thought that was notknow

[limpid horizon]

hahahahha

[urban granite]

lmao

[granite yew]

This is one of the few things I don't really know in popular GA. I can tell you GA and GA-like methods tend to be some of the better options for posing, animating, and interpolation. Having advantages in speed, memory footprint, and numerical stability (depending on implementation).

CGA specifically I know is utilized in robotic control and robotic vision, but I'll be real and tell you I have no idea how it is used

Screw theory is about translations and rotations, and screw motions (translating along a line while also rotating around the line)

If I'm not mistaken, a line combined with an orientation, telling you how to translate and rotate, and a proportion between translation and rotation, is a screw

[golden osprey]

You'd be better off probably just asking questions here

And it'd help if you said what text you were thinking of working through

[distant minnow]

I am using the introductory book of rotman about homological algebra + I wanna do this bcs I am not supposed to learn homological algebra officially yet, so I am looking for sm1 so we give each other the motivation and respect a schedule.

[ornate atlas]

https://etale.site/teaching/s23-128/math-128-s23-lecture-notes.pdf

there’s this set of notes which is probably a horrible way to be introduced to homalg but I’d like to work through at some point. If I were to do so though it wouldn’t be for another couple of months yet

[solar turret]

sometimes me too

[limpid horizon]

Notknow VS. Nasmil_337

[drowsy compass]

Are there uncountably large tarski monsters

[storm basin]

surely if all ur proper subgroups are finite u cant be uncountable

idk how to prove this though tbf

"Every Tarski monster group is finitely generated. In fact it is generated by every two non-commuting elements." - Wikipedia, so no it cant be uncountable

(every finitely generated group is obviously countable)

[last talon]

Quick proof: take the subgroup generated by a countable number of distinct elements

This is a countable subgroup

[storm basin]

Nice.

Obvious in hindsight but I like it

[near lantern]

Let F: D_A -> D_B be a functor between trangulatd categories, both equipped with t-structures with hearts A, B respectively. If F is exact and maps A into B, is its restriction to a functor from A to B n exact functor of abelian categories?

[limpid horizon]

Hearts? Lol

[delicate swan]

Is there any concrete description of the span of all k-tensors of the form v^{\otimes k} (I mean as v varies in \mathbb{R}^n)?

I know that for k=2 you get all PSD matrices

So my question really takes place in k\geq 3

[near lantern]

I believe this is Sym^k V and this is proved in the course of any proof of Schur-Weyl duality.

V = \mathbb{R}^n

I forgot how to actually do it but I think explicitly, you can induct on n. You want to express symmetrise(v1 (x) ... (x) vn) as a linear combination of v^n's when you know the same is possible for n-1.

[delicate swan]

Thank you very much🙏

I found it in stack exchange

[delicate swan]

I didn’t actually

😪

[scarlet ermine]

this has been alluded to me but I’ve never proved it, so I’ll try now! If A->B->C->A[1] is an exact triangle with all A,B,C in a heart, then 0->A->B->C->0 is an exact sequence in the abelian category sense—you can check exactness in the middle by seeing that after hitting it with Hom(W,-), you get

Hom(W,A)->Hom(W,B)->Hom(W,C)

exact because Hom(W,-) is cohomological. So, now we just need to see the A->B is injective and B->C surjective, so we check the kernels and cokernels are 0.

Recall that the kernel of A->B is a shift a truncation of the cone (which is C), but that truncation will be 0 because it’ll be in the wrong shift.

Similarly the cokernel of B->C is a the other truncation of the cone (which is A[1]), but this is again 0 for the same reason.

so an exact F mapping a heart inside a heart will be exact functor restricted to the heart in the abelian category sense because the exact sequences are exactly the triangles that happen to be in the heart

[subtle smelt]

Yeah. Just compose it with the (0th) homology

Provided your exact functor maps objects in the heart to the heart. Although with comparison with derived functors this shouldn't be surprising

If your objects are concentrated in degree 0, then h^-1(FC) = 0 and h^1(FA)=0. Furthermore h^0(FX)=FX under this assumption. This yields you an exact sequence
0 \to FA \to FB \to FC \to 0

[spice idol]

Are the machinery and results around Stone duality used in the theory of profinite groups?

[young forge]

Does anyone have any.. anything that has the following properties:

A large(order ~2^64-2^512), maximally "wild" (for the other propertues), quasigroup that is not a semigroup such that:

The quasigroup operation is not "ringlike". The natural representaton of the structure is not a linear combinaton of basis elements over any reasonable ring.

The quasigroup has many interesting, highly nontrival endomorphisms/automorphisms that can be represented and sampled from efficiently. Other then these the magma has a structure that seems random.

The quasigroup has some sort of O(polylog(|M|)) way to represent elements, and a polytime algorithm for the magma operaton on these bitstrings. And this is not in terms of generators/relations. Given a subquasigroup generated by <a, b>, and an element k, it should be hard to determine if k is in <a, b>

There is some vaguely coherent way to construct M

I get magma and quasigroup mixed up.

such a quasigroup almost certainly exists. The issue is constructing it.

in case your wondering why i'd be remoted interested in such a fucking weird structure, cryptography

If something with these properties is constructed, I already have a DH analogue for it that should be quantum resistant

Based on the following problem:

given some element k, and a list of tuples of (a_i, phi(a_i)), such that k is guaranteed to be in the subquasigroup generated by the a_i(substitute in a_0, a_1...), compute phi(k).

for some endomorphism phi

or a slightly modified version, given (w_i(a_i), phi(a_i)), where w_i(a_i) is some product of a single instance of a_i along with some godawful nonassociative product of elements randomly sampled from Ker(phi),(we don't know phi or it's kernal), compute phi(k).

basically, the idea is you have some set of generators of M(our quasigroup), a_i. And an endomorphism phi. publish (a_i, phi(a_i)), Someone who wants to send an e lement generates an element from <a_i>, and then uses the labeling to generate the same element from <phi(a_i)>, send the element w generated from <a_i> the other person with phi can apply it to w to get phi(w), which is the shared secret

sorry if this is totally incoherent also i dont wanna clog up this channel so feel free to dm me or make a thread ig

[midnight mirage]

🔯

Love the way how algebra look so cool

[past cove]

Grothendieck and Verdier were famously Talmudic scholars in their local Yeshiva when they were inspired to define derived categories

[fierce steeple]

What funny about this post? Straight facts

[digital parcel]

sometimes the truth is funny

[summer quest]

very hard to even define perverse sheaves without the notion of t-structure in the first place!

[lone jacinth]

Is the connection that you have some derived equivalence and the perverse sheaves come from the image of the standard t-structure? Or what's the history there?

[wanton spoke]

people didn’t invent t-structures first and then get perverse sheaves it was more like: we keep seeing this class of constructible complexes that behaves the way an abelian category should but the derived category won’t let us talk about kernels and cokernels or exact sequences
the fixes like: it’s just a complex concentrated in certain usual degrees or take some extension-closure miss the point because perversity is measured by stalk and costalk conditions not by ordinary cohomological degrees and you still don’t get a real abelian category
so they found that the perverse t-structure is the best way to fix the problem

[near lantern]

I'm not sure if this makes sense but: Consider the derived tensor product on D(R-Mod), R a commutative ring. Cohomology is a functor from D(R-Mod) to (R-Mod)^{Z}, which should be monoidal at least for R a field. But to make this true, do I have to make either of the (braidings of the) monoidal products supercommutative?

[lone jacinth]

Does the definition of monoidal functor say anything about the braiding?

Can you not have monoidal functors between categories unless they are braided monoidal?

[near lantern]

IDK that's why I'm not sure if the question maKes sense

I'm just thinking that cup products are supercommutative but IG that's not about the monoidal structure. It's an algebra wrt the monoidal structure and commutative wrt the supercommutative braiding.

[lone jacinth]

I mean there is something called "braided monoidal functor" / "symmetric monoidal functor" and cohomology probably is that too

[near lantern]

Yes lol. Maybe it's braiding-preserving if you use non-super on both sides or super on both sides.

[summer quest]

well so the actual motivation for why you would construct this sort of thing is to salvage Poincare duality in the presence of singularities

in some sense the construction works the same as how the usual category of constructible sheaves is recovered as the heart of the standard t-structure on the constructible derived category

in the constructible setting you are working relative to a fixed stratification, and you are gluing the standard t-structure from the standard t-structures on categories of locally constant sheaves on each strata

the perverse t-structure is constructed the same way, but with different shifts depending on the relative codimensions of strata, so that locally constant sheaves are replaces by suitable shifts of such

there are various different choices of shift which you can take here, corresponding to a notion of "perversity"

Poincare duality takes the form of Verdier duality and this exchanges the heart of the t-structure with this perversity with the heart of the t-structure with the dual of this perversity

the heart of the t-structure with middle perversity is then Verdier self-dual and Poincare duality is restored in this sense

[summer quest]

this is a little subtle and it depends

but yes there are plenty of situations where the "naive" functor you write down for cohomology ends up landing in supervector spaces rather than vector spaces, and you have to modify the braiding in order to land in vector spaces again

this tends to happen when constructing categories of motives (e.g. over finite fields) in the usual way involving algebraic correspondences as Homs, in this sort of setting you can modify the braiding if you assume certain standard conjectures about algebraic cycles (you need the Kunneth correspondences to decompose in the right way so that you can modify the Koszul sign accordingly)

another interesting situation where you naturally encounter this issue of Koszul sign and supervector spaces is when you consider the category of Z-graded complexes of (super)vector spaces

here there are two gradings, one is the internal Z-grading of complexes, one is the external Z-grading

you can then shear this bigrading

that shear functor is a monoidal equivalence, but not a symmetric monoidal equivalence when working with vector spaces

if you do the same shear but also change parity depending on degree when working with supervector spaces, this becomes a symmetric monoidal equivalence

when you restrict this back to the even supergraded objects you get this picture

[summer quest]

this should answer what you are wondering about here

this shearing functor is rather interesting and shows up implicitly in a lot of geometric Langlands and derived versions of Tannakian reconstruction theorems and so on

one funny upshot of this is that you get way more fiber functors in the derived setting, since you can always take any fiber functor and precompose it with iterates of shearing

[wanton spoke]

you really developed what I said briefly lol

[worldly zealot]

does anyone know how to compute, say, the symmetric power of a group representation on GAP and decompose it into irreducibles

[low carbon]

idk but there is a manual online that may be of some help

[warm relic]

if u think ai can make movies or is intelligent join now

i just made a better movie than the ai with trial equipment

[median halo]

@warm relic this channel is for mathematical discussion relevant to the channel topic. keep off-topic comments in #discussion or #chill

[verbal panther]

Any examples of commutative non associative structures?

[ornate atlas]

A non associative lie bracket over a field of characteristic 2 works

[urban granite]

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

[verbal panther]

Cool

[spice idol]

most commutative magmas

[ornate atlas]

And of course we all have a rich list of these to hand lol

[spice idol]

speak for yourself, i definitely do

min{x, y} + 1 on the naturals, in any case

[ornate atlas]

Yeah I’m joking a bit, I’m sure I could come up with a bunch no problem it’s just a funny way to phrase it

I think like f(x,y) =nx+ny on N would work?

[spice idol]

yes for n ≠ 1 or 0 of course

[ornate atlas]

Yeah sure lol

Probably a lot of metrics would work as well right?

[spice idol]

probably any associative operation on N followed by a +1

[ornate atlas]

True lol

[fierce steeple]

Lol

[ornate atlas]

This is really more #advanced-number-theory. Youll maybe want to ask a more specific question if youd like some help though!

[eager hound]

There is supposed to be a correspondence between certain (algebraic) automorphic forms and algebraic varieties, which works something like: automorphic form f for a group G <-> compatible family of galois representations \rho_{f, \ell} with values in \hat{G} <-> family of \ell-adic Galois representations H^i(X; Q_l) for all \ell prime. <- X up to some fine equivalence relation

For elliptic curves this says that weight two classical modular cuspidal eigenforms with rational fourier coefficients correspond bijectively with elliptic curves over Q up to isogeny. And there are generalizations for larger number fields.

[mental escarp]

Is the zariski closure of a line segment the corresponding line?

[eager hound]

You mean over the real numbers or something?

Whatever statement you're looking for (if true) will follow from the fact that a polynomial in one variable which vanishes at infinitely many points must be zero.

[mental escarp]

No I mean in any affine space

[eager hound]

then what does a line segment mean

What do you mean?

[mental escarp]

Uhh okay I guess I did mean in Rn for the notion of an interval in an embedded line. But ya that property about when a single variable poly vanishes on infinitely many points probably answers whatever question I meant.

[eager hound]

okay well I cannot explain the entire theory of modular forms on discord

you should read something about them

Gamma_0(N) is the subgroup of Sl_2(Z) consisting of matrices which are congruent to an upper triangular matrix modulo N. Associated to such a subgroup there is a space of modular cuspforms of weight 2 called S_2(\Gamma_0(N)) (some functions on the upper half plane with nice properties), and the Mellin transform of a cuspidal eigenform gives some completed L-function (just like the mellin transform of \theta gives the completed zeta function) and this is supposed to match the L-function of the corresponding galois representation. To learn more about things you'd have to read about the theory.

[hard kite]

Let $R$ be a Dedekind domain and $I$ an integral $R$-ideal. Does there exist an isomorphism $I^{-1}/R \cong R/I$ as $R$-modules? This is true when $R$ is a PID.