advanced-algebra

[spice idol]

you can define it as some enriched category with one object so the localisation just becomes category theory again lol

[lone jacinth]

There definitely are people that study them, but I don't know what the relevant questions are.

And non-Noetherian rings show up in normal circumstances. The noncommutative polynomial ring for example, and other infinite dimensional algebras that show up in rep theory

[fierce steeple]

Bruh TIL this isn't noetherian

[spice idol]

i mean what can you expect, noncommutative rings to be nice? welcome to the real world, potato

[ornate atlas]

Yeah I guess I know that to be true enough, my non com course despite being about Noetherian rings did certainly look at some non Noetherian rings

[ornate atlas]

There is however a non commutative version of Hilberts basis theorem

It’s possibly a bit of a stretch to say its an exact generalisation but the vibes the same, if you have a ring S and a sub ring R, and some x in S, then provided that R+xR= R+Rx and the map from R<x> to S is surjective, then if R is left(right) Noetherian so is S

[spice idol]

R+xR = R+xR?

"if R is left Noetherian, so is R" lol

[ornate atlas]

Oops lol thanks

Yes obvious typos being that the x should be on the left and the right there and then the second one should be S

[spice idol]

👍

hm

i wonder how I could translate this into universal algebras

[acoustic kayak]

guys what are the different interpretations of a matrix's rank, in relation to the linear transformation it represents

[unborn rampart]

wrong channel, try #linear-algebra

[acoustic kayak]

Alr

[fierce steeple]

💀

[ornate atlas]

It’s actually a very nice result, it’s an easy ish test and something you can actually work with

[fierce steeple]

Sure hm

[round seal]

is my understanding at the bottom of the connecting homomorphism correct? He did not define it in class and ive been working on trying to chase around what it should be and this is what ive come up with

[near lantern]

I feel it should be mentioned that for any b such that g(b) = c there is a unique d such that beta(b) = h(d), and different choices of b must differ by some f(a) and then the corresponding d's differ by alpha(a).

[worldly zealot]

@lone jacinth https://arxiv.org/abs/2412.08904v1 did you see this lol

[fierce steeple]

Is this masterful clickbait

Lmao

[worldly zealot]

hm?

its just a very funny passive aggressive paper

see other remarks

[formal rock]

hardly passive

[worldly zealot]

xd

[lone jacinth]

Fei is very butthurt indeed. To some degree rightfully so I suppose, but it's not a good color

[lone jacinth]

If you want more "passive" aggressiveness by Fei
https://arxiv.org/abs/2409.12743

[worldly zealot]

😭

[fierce steeple]

I mean I clicked it on it cause of the "AI"

LOL

I guess I misunderstood but when I saw it I was like oh did AI give someone a 2 line proof lmao

[fierce steeple]

Lol i did wonder what should happen in this case

Like if someone else's work easily follows from yours

[lone jacinth]

And I guess what happens is they get a bunch of awards and praise for their ground breaking research and you write salty papers

[worldly zealot]

in any case it feels like tau-tilting theory and semistable torsion classes are completely different languages and for their purposes, the right language

so 🤷‍♀️

[fierce steeple]

LOL

[vague pawn]

Hi, regarding profinite groups, there is this exercise I find very nice. Would like some feedback on my solution. Is it correct? Is there a nicer way to explain things? Is there a more natural definition that I missed?

[near lantern]

Meanwhile me just being amused by AI in the title

[lone jacinth]

It looks good to me, though you might need to say something about continuity.

[vague pawn]

Yeah I haven't gotten to that yet

also why do we need G to be abelian in the last line of the question?

[lone jacinth]

I don't think you do

[vague pawn]

yeah its weird

[near lantern]

Are there any references studying, for a Borel subgroup B of a semisimple algebraic group G, the orbits of [b, b] (b being the Lie algebra of B) under the adjoint action of B?

Apparently this is a hard/wild question.

[spice idol]

I'm kinda struggling with this, I managed to prove that i and iii are equivalent, and that i and iii imply ii, but not really managing to prove that anything follows from ii

I've reduced it down to showing that, if I = (x1, ..., xn), then I * (xi) = (xi)

[weak lodge]

I think you can show (ii) implies (iii) by applying Nakayama's lemma to get r such that rA = 0 (in particular, rI = 0) and 1 - r ∈ I. Then take ||e := 1 - r||

[spice idol]

ah wait

I think I can use the local nature of flatness here

òf being zero I mean

lol

[lone jacinth]

If A is commutative you can just use Nakayama. In general I'm not sure the result is actually true

[spice idol]

it is an algebraic geometry book, so everything is commutative

I'm not sure how to apply Nakayama here though? I don't think the jacobson radical is guaranteed to contain I

wait no nvm that's not what we want here anyways

[lone jacinth]

Nakayamas lemma says if M is finitely generated and IM = M, then there is an i in I with im = m for all m in M

[vague pawn]

Not im=0 instead if im=m?

I might be misremembering

[lone jacinth]

No, though you might have phrased it as an element r congruent to 1 modulo I with
rm = 0

(So r = 1-i)

[vague pawn]

Yeah thats the version I know

checked on wikipedia

[spice idol]

is that easily seen to be equivalent to the jacobson radical definition?

[vague pawn]

See wikipediahttps://en.m.wikipedia.org/wiki/Nakayama%27s_lemma

[lone jacinth]

I would say this version is stronger because the Jacobson definition also works for non-commutative rings.

[spice idol]

right

[ornate atlas]

There’s about 15,000 statements of nakayma unfortunately

[spice idol]

hmm, I think I see how the jacobson version implies this version

[lone jacinth]

What you might try is to localize with respect to 1+I

That places I into the Jacobson radical at least

Then IM = M implies the localization of M is 0 which implies there is an element of 1+I annihilating M

[spice idol]

I would've never come up with that lol

[lone jacinth]

I mean, the idea is just if you want to use the Jacobson radical you need I to be in the Jacobson radical.

So by definition you 1+I should be units. How to make that happen? Just localize! From there the problem solves itself

[spice idol]

I think the main problem is I don't know enough about the jacobson radical

lol

[rose mirage]

the jacobson radical is the annoying bit of the ring everyone hates and all of the theorems involving it tell you when you can ignore it

[spice idol]

localisation is slowly dawning on me though

[lone jacinth]

The key is mostly that localization of commutative rings is nice.

I think the only two things to know about the Jacobson radical is that it's the intersection of all maximal left ideals and it's the set of x such that
1 - rx is a unit for all r.

[spice idol]

yeah that last bit was missing from my knowledge

I knew it sometime but forgot

[rose mirage]

I also like thinking of it as the annihilator for all simple modules

[fierce steeple]

Nakayama.

[rose mirage]

Naka ya mama

[spice idol]

right, as every simple module must be cyclic and isomorphic to a quotient by a maximal ideal, which is the annihilator of said module

[lone jacinth]

Everyone's favorite Japanese mathematician from the 1940s

[spice idol]

so their intersection is simply the intersection of all maximal ideals

[rose mirage]

very quick thinking monsieur _music

[spice idol]

as if to taunt me the next question is laughably easy using this

[rose mirage]

nakayama is that recent? ZAMN

[fierce steeple]

Is this a joke lol

[spice idol]

why thank you sir tbh

[fierce steeple]

I mean how old are schemes or even just "modern" classical AG bro

[fierce steeple]

Badly written question

[rose mirage]

I would've expected some mf to notice nakyama's lemma in like, 1432

[fierce steeple]

"Show that for every A module M, M is flat iff A is a field"

[rare walrus]

I would expect you need ring theoretic language to express it really

Like how would you otherwise express it

[rose mirage]

yeah but rings were not invented in the 1930s

[rare walrus]

The CRT can be expressed elementarily and that’s why it was known for so long, for example

[fierce steeple]

Still not too much earlier right

[spice idol]

this just in, every commutative fucking ring is a field

[lone jacinth]

"All A-modules are flat iff A is a field"

[rare walrus]

Rings were like, late 1800s?

[fierce steeple]

Dedekind et al right

[rose mirage]

yur that's what I thought

so I would've expected nakayama in like, the 1910s or 20s

[rare walrus]

Idk I feel like this still requires more ring theoretic language that didn’t exist for very long prior

Yeah fair point Wew

[fierce steeple]

I still don't like this wording

Smh

[rose mirage]

I think the wording is fine

[fierce steeple]

"A is a field iff all A-modules are flat"

Sounds better

Weirdly

I should say I don't actually mind like I am not being that pedantic lmao

Just I find it amusing when statements are linguistically slightly ambiguous or odd

[spice idol]

rather than continuing with these exercises I should probably go home and eat lmao

the uni library is a nice place to work though

[fierce steeple]

Nice

[spice idol]

no way, our small little STEM library has the book on quandles by mohammed elhamdadi

[weak lodge]

he is everywhere....

[spice idol]

istg

but that book is acc pretty cool

doesnt cover the cohomology as much as id like them to though >.>

also apparently cohomology is used in universal algebra???

to study extensions

i guess that really shows just how "grouplike" malcev varieties are

[spice idol]

i cant find anything abt it

huh

grargahrga i hate this

[sand python]

Would a class using a mix of linear algebra by shapiro and axler's linear algebra qualify as advanced algebra? or where would I as questions?

[spice idol]

#linear-algebra probably

[cloud karma]

A slight confusion I have. Let $\mu = x \lambda$ be a regular weight where $x$ is a reduced element in the Weyl group and $\lambda$ is dominant. My confusion is if we take $x w_0$. Is this reduced or not?

[broken turtle]

Delteto

[cloud karma]

Not necessarily right? If so then how do you make sure it is reduced?

[molten drum]

does finite support of (a_i) mean that only finitely many a_i's are zero or is suppor t refering to something else?

[fossil thicket]

it means finitely many a_i are non-zero.

[molten drum]

oh okay that makes sense. thank you

[limpid horizon]

Bruh why they doin all that

[spice idol]

I know

[near lantern]

when le local ring

[digital parcel]

van Gogh wrote Yoneda's lemma on the back of the Mona Lisa actually

[rose mirage]

ahh... ze artisztic beouty onle growz viv time...

[fierce steeple]

Yeah

[near lantern]

An element of the Weyl group is neither reduced nor non-reduced. A word representing an element (by which I mean a sequence of simple reflections whose product is the element) can be reduced or non-reduced. If x is any element then l(x w_0) = l(w_0) - l(x) (this is a fact about finite Coxeter groups), but of course the product of two reduced words for x and w_0 will have length l(w_0) + l(x). It is true that there exists a reduced word for w_0 which starts with the reverse of a reduced word for x. If you use that reduced word then you can cancel x and that prefix, and what is left will be a reduced word for xw_0.

[plucky arch]

is there a ring-theoretic notion of "partition of unity"?

[fierce steeple]

Sometimes people use this term yes

[plucky arch]

I’ve noticed that this concept appears in a few different places

[fierce steeple]

https://math.stackexchange.com/q/4052910/932674

Though here it is almost literally a POU for the affine scheme lol

[plucky arch]

It comes up in the Chinese remainder theorem

And it comes up in Lagrange interpolation

[fierce steeple]

Hm I just feel this is the idea of having an orthogonal basis or smth

[plucky arch]

And it comes up for annihilating polynomials in linalg

Orthogonal wrt what

[fierce steeple]

Though this is a special case of CRT ig aha

[plucky arch]

[fierce steeple]

I mean vaguely orthogonal here like not a rigorous ting

As in just splitting up identity / a constant into like somehow "orthogonal" parts

To reduce smth global to more local

[lone jacinth]

You can translate the usual definition on C(X) into just ring theory at least:

A partition of unity is a set of elements fi such that for each maximal ideal m, only finitely many fi are not in m and
Sum fi + m = 1 + m

But if you require the partition to be finite, then it's just CRT

[fierce steeple]

With POU it is funny here in that like affine schemes are quasicompact. So the finiteness bit isn't that relevant

[plucky arch]

but it's not like partitions of unity on, say, a smooth manifold have to be "orthogonal", right?

they can have nontrivial overlaps

[lone jacinth]

Locally almost all of them are pairwise orthogonal.

So that's something

[fierce steeple]

I mean like they are all localised and vanish outside them

Like locally yeah you only care about finitely many

[digital parcel]

Coined by serre or something according to vakil

It’s a fun proof

[maiden pecan]

is this the abstract algebra spot? #advanced-algebra

[spice idol]

depends

for groups rings and fields youve got #groups-rings-fields

[plucky arch]

-# but if you want monoids or semigroups then here’s where it’s at

[spice idol]

or quandles/racks/spindles/shelves/BCK-algebras/cBCK-algebras/leibniz algebras/quasigroups/loops/boolean algebras/lattices/modules/etc

[limpid horizon]

breh.

[spice idol]

what, gotta make sure everyone feels included

[weak lodge]

people who study Leibniz algebras:

[spice idol]

oh my god there exist lie racks

this is amazing

uhhhh

[silver goblet]

what is a loop in this context
- a topologist

[spice idol]

quasigroup with identity element

basically, a magma where you've got left and right division, and an identity element

[silver goblet]

messed up

[spice idol]

lmao

[silver goblet]

im not convinced that anyone has ever used something here that is not a monoid nor a group

lol

[spice idol]

a magma is just nice shorthand for defining stuff

quasigroups are used in studying latin squares

and semigroups are important in computer science

[spice idol]

and apparently in cryptography

[silver goblet]

huh wild

[silver goblet]

idk i feel like they can be easily described without using a word that makes it seem important

lol

[spice idol]

i guess?

rather say magma than set equipped with binary operation

[foggy galleon]

[forest turtle]

prolix

had a grader known throughout our math department for his feedback. most of the time it was just the word prolix

[sly knot]

i accidentally clicked on this page instead of number theory but what the helly is magma i thought that was a volcanic rock 😭

[wise sedge]

A group but it doesn’t necessarily have associativity or inverses or an identity

[weak lodge]

equivalently, a set equipped with a binary operation (with no other assumptions)

[fierce steeple]

I prefer partial magmas

[vague pawn]

Here I don't really understand the notation $I_{G(L|K)} A_{L}$.

[broken turtle]

ExpertSqueeSQUEE

[vague pawn]

Here $A$ is a multiplicative G-module a $A_{K}$ is the fixed module of A under $G_{K}$ which is some subgroup of G.

[broken turtle]

ExpertSqueeSQUEE

[lone jacinth]

Anything specific you don't understand about it?

[vague pawn]

like why does this notation give this definition

[lone jacinth]

I mean, the I probably stands for ideal

[vague pawn]

and why the a^-1?

[lone jacinth]

If you mod out by
a^sigma a^-1
then
a^sigma = a

[vague pawn]

oh I see

this means sigma fixes a

ok thanks

[lone jacinth]

But yeah, as for the notation.

The ideal in the group ring ZG(L|K) generated by sigma-1 is the augmentation ideal.

Thinking of A as a ZG module this is IA, so the notation should make sense

[spice idol]

mfw being a unit is a local property

is every local property a direct consequence of some module being zero?

I.e., can it be directly proven from the fact that a module is 0 iff all its localisations by maximal ideals are 0

[lone jacinth]

I feel like most properties of modules can be reframed as some module vanishing

Actually I take that back. Finitely generated and other finiteness conditions are not, but they are also not local properties, so

[spice idol]

hm, I see

[spice idol]

this, for example, can be turned into the property of R/aR vanishing

and then using the exactness of localisation you get (R/aR) \tensor R_p = R_p / aR_p

[spice idol]

so we can use this

which I find awesome

and it proves that an element of the coordinate ring of X that is never zero on X must be a unit, as it is a unit in all localisations at primes (as those are the rings of regular functions at points of X)

[fierce steeple]

Maybe I am silly but why is this true?

I guess okay nvm I see what you mean

Too much schemes brainrot

It's the fact that the maximal ideal of the local ring is given by functions vanishing there lol

I interpreted this as like "vanishing as an element of the local ring" or smth which was silly

[spice idol]

lmao

I'm going through classical AG so I can motivate schemes for myself

it's great practice for my com alg too

schemes are really just manifolds but affine lmao

[fierce steeple]

Well not necessarily affine (usually not)

But ye like affine varieties subbed for R^n

Which makes it more interesting! As even the local structure varies

[spice idol]

I meant manifolds but instead of locally euclidian n-ball locally affine

[fierce steeple]

Yee sure

I just mean cause affine also has a meaning lol

Like R^n being affine n-space

[spice idol]

hehe

ball

[fierce steeple]

:cereal_2:

[frosty ember]

I was recently shown the following recipe for producing an algebraic group from a compact Lie group G: the space of G-finite vectors in L2(G) is an algebra over C, and in fact the group structure on G makes it a Hopf algebra. Then take the spectrum of this algebra.
does someone have a reference for this construction? it was claimed that the original group should be a compact subgroup of the C-points of this algebraic group, but this is quite unclear to me …

[near lantern]

I don't have any reference, but in the interest of understanding, do you know if L2(G) is to be considered with pointwise multplication or convolution? Because the latter seems more usual for this kind of argument (e.g., for a finite group, the group algebra with convolution as the multiplication is a Hopf algebra with the original group encoded in it as the group-like elements), but that would probably be non-commutative and non-unital, so IDK what its spectrum would be.

[lethal scarab]

I believe theres a character theory proof for this, but i wanted to find something avoiding that

Usually the approach would be to consider a non zero S_n - submodule of Alt^k(V), then look at some nonzero element and use some nice permutation (transpositions work very well usually i guess) to generate the entire module. But i cant seem to find a way to do this

Maybe another way to do this would be to show that every non-zero element's stabilizer is precisely the identity permutation?

[summer quest]

is this not just a funny restatement of classical theorems by Chevalley and others?

the category of compact Lie groups is equivalent to the category of R-anisotropic reductive groups over R whose connected components have R-rational points.

[void plank]

I have a bit of a silly question but given a chain map, we know the mapping cone can be used to describe the kernel/cokernel of the induced map in homology

But suppose I have chain maps p: C ---> D and τ: D --> C such that p \circ τ = id (I'm working over Ch(R-mod) here)

Since homology H_i is a functor between Ch(R-mod) and R-mod, doesn't p \circ τ = id immediately imply H_i(p) is surjective and H_i(τ) is injective? Or am I missing a subtlety here?

If this is true, I'm curious how I can use the mapping cone of p and τ to reproduce this

[lone jacinth]

Yes, so what this means is that the induced map
H_i(D) -> H_i(Cone(p))
is 0

[void plank]

Ahh ok, thanks!!

[wise sedge]

Are all Gelfand pairs (G,K) where G is a sporadic finite simple group known?

[limpid horizon]

There is a theorem that says R is Cohen Macaulay iff Rm is Cohen Macaulay. Later the book wants to compute H^imRm(Rm) to determine if Rm is CM, and so uses this theorem 3.5.6 to say H^imRm(Rm) = H^i(Rm (x)R C)

but I am confused because H^i(Rm (x)R C) means we are viewing Rm as an R-module but the theorem im assuming is interpreted as R is CM (as an R-module) iff Rm is CM (as an Rm-module)

hopefully my question makes sense, and hopefully someone can help me because im so over all of this

Another question: why is the Cohen-macaulay property only defined over local noetherian rings and fg modules

[worldly zealot]

it is defined over general noetherian rings - you check cohen-macaulayness at each localization

[limpid horizon]

ok

[limpid horizon]

wdym we check it at each localization

[fierce steeple]

Like you can define a Cohen-Macaulay ring/module as a ring/module whose localisations are Cohen-Macaulay

[fierce steeple]

Oh lol but now I am confused by this theorem you state

[limpid horizon]

Ok

[limpid horizon]

It was an exercise ill show u

[fierce steeple]

Kinda confused cause you say it's only defined over local rings but also above seem to use Cohen Macaulay for non-local rings

Or maybe you mean like R is an algebra over some local ring or smth

[worldly zealot]

bruns-herzog? i completely forgot what the asterisk business is supposed to be

[limpid horizon]

graded or something

[worldly zealot]

M_m is an R_m module

[limpid horizon]

ok thanks

its just too much and now since i dont care about going further into math i have such low motivation to do this stuff

its kind of an annoying position im in atm

im hoping once i meet with my supervisor again she can help me

[worldly zealot]

i think CM/gorensteinness is a little unintuitive and a slog without any geometric intuition of what they are telling you

[limpid horizon]

(c) is the important one right

R is CM as R-module iff Rm is CM as Rm-module is from (c)?

[limpid horizon]

Using this we want to say H^i_mRm(Rm) = H^i(Rm (x)R C) = H^i(Cm) = H^i(C)m

But that is viewing Rm as an R-module not an Rm-module

Does my question make sense

[obsidian aurora]

anyone hear know basic algebra to plot points and make lines I am kind of Dum

[weak lodge]

#prealg-and-algebra

[obsidian aurora]

ty

[spice idol]

one way to make a property local (for Noetherian rings)

lmao

[lethal scarab]

would appreciate some help on this!

[spice idol]

you could try to find the character, and take the inner product and show that it is 1

like the
$$\frac{1}{|G|} \sum_{g \in G} \chi(g) \chi(g^{-1})$$

[broken turtle]

.enpeace_music

[astral maple]

Very random question: let V be an irreducible cuspidal representation of GL_2(F_p). Is a matrix representation of V (say, depending on its character) known?

[jaunty sparrow]

How does one go about finding all injective representations of a given quiver?

[lone jacinth]

If A is the path algebra of your quiver then every indecomposable injective is given by
D(eA)
for a trivial path e. And all injective modules are direct sums of indecomposable ones.

Or more explicitly in terms of the quiver: reverse the arrows, find the projective representations, reverse the arrows back by taking transpose of the linear map corresponding to an arrow.

[jaunty sparrow]

Yes, I see it now. Thanks

[unborn rampart]

If B is isomorphic to C, then A (x) B is isomorphic to A (x) C for any modules A, B and C, right? I think this follows from functoriality of the tensor product?

[spice idol]

also because its a universal property

I suppose a universal property guarantees functoriality

[unborn rampart]

I see I was curious about this, why it in principle isn't enough for (A ⊗ B) ⊗ C to be isomorphic to A ⊗ (B ⊗ C) to get generalized associativity. I think the point is that we need the substition property of equality: A ⊗ (some parenthesization) = A ⊗ (other parenthesization) because (some parenthesization) = (other parenthesization). And it turns out we have the substitution property for isomorphism too

I have no idea what "pentagonal" diagram he refers to

[spice idol]

that is weird

tensor products are functorial; isomorphic substitution should hold

maybe its something to do with canonical isomorphism?

[spice idol]

ohh the diagram must commute and that is not immediately obvious from simply the associativity isomorphism of the case of 3 modules

i see

thats really what it means to be associative up to canonical isomorphism: every way of turning two expressions into one another must be the same, that is, if we have two paths from one expression to another where every step is some canonical isomorphism (A ⊗ B) ⊗ C ≅ A ⊗ (B ⊗ C), then those paths must compose to the same isomorphism

[spice idol]

and it turns out that this diagram commuting is enough for all such paths to be the same

[lone jacinth]

I believe it is true that once you have the Pentagon, then all the higher associativity relations will hold

[unborn rampart]

ah, I see, and if you just use isomorphic substitution and (A⊗B) ⊗ C iso to A ⊗ (B ⊗ C) then you won't necessarily get a canonical isomorphism?

[spice idol]

well, in the case of tensor products you will, because this diagram is commutative for all modules

but in a general category with some associative bifunctor, this diagram might not commute

so that means there are two ways to "associate" two expressions which give different isomorphisms, which is not what you want of course

because if two expressions are the same up to associativity, you dont want there to be a choice of isomorphism between them

[unborn rampart]

that makes sense, thanks

[spice idol]

[mellow night]

This is a case of the pentagonator object in the category of modules no?

[limpid horizon]

If I subset J then I^n subset J^n right.

ty

[mellow night]

Yh my intuition also says so

[spice idol]

i believe i read somewhere that the pentagonator is an isomorphism that actually relaxes this diagram?

yes

[mellow night]

yeah it gave off pentagonator vibes

[spice idol]

this is simply called the pentagon identity

[mellow night]

Beautiful

[spice idol]

category theory is p cool

youve got associahedra

for any magma expression

[mellow night]

the best category theory concept ever is verdier duality imo

its not cat theory but it is based on cat theory

[spice idol]

dont know enough topology to know that sadly

[mellow night]

You can think of it as two versions of the same calculus on sheaves

[spice idol]

hmm

[spice idol]

not for this channel

#prealg-and-algebra

or maybe #precalculus ?

no actually I've got no clue where this should go

but not in this category that's for sure

[spice idol]

also it's not clear how this should continue

does it continue to 1.1 or to 0.11

[rose mirage]

1.1

[spice idol]

ah okay

[cloud wind]

0.2 increase

It's a cool question

[spice idol]

in that case it should be just 0.5 I guess

[rose mirage]

also how is it bracketed

[spice idol]

in the interesting way

[cloud wind]

True

[cloud wind]

Power is right associative

[spice idol]

power towers are conventionally
a1^(a2^(a3^...)))

[rose mirage]

nah

[spice idol]

wdym "nah"

[cloud wind]

If it wasn't right associative it wouldn't be solvable

[spice idol]

well, you'd still take the limit but it would just end up 0

[cloud wind]

Computer science

Is there a formula way to do this

[rose mirage]

the fuck does dequeue mean

[cloud wind]

Lmao

[spice idol]

and also why post it here

[cloud wind]

Because I have no friends

[spice idol]

I meant the channel

[cloud wind]

Oo

[spice idol]

remove from queue

[rose mirage]

what the hell is a queue

[spice idol]

bro has never programmed

I guess it means you've got a list

[cloud wind]

First in first out

Like a real life queue to get pokemon merch

[rose mirage]

ok so a stack

[cloud wind]

Last in first out

[spice idol]

not really, stack is pop of put on

this is more fucky wucky where you need access to both top and bottom

[rose mirage]

so.... a shift register?

[spice idol]

I guess yeah

[rose mirage]

ah so like the stack then

[cloud wind]

Fortran progarammer

[rose mirage]

to actually answer your questiom, this is an example of a discerte dynamical system on (if I'm understanding what dequeue means correctly) the power set of {1,2,...,100} so yes there is "a formula way" of doing it

I lied actually sorry, it might be on a much larger set

[spice idol]

the sets are ordered though

[rose mirage]

cause order matters

yeah

point is it's a discrete dynamical system

[spice idol]

lmao

[cloud wind]

I don't understand how but good to know

Ik very less math

Although math is a cooler subject than programming

And programming is one of the coolest

[rain scroll]

is this college algebra can i do or what this

lol

[limpid horizon]

Dis be advanced abstract algebra

[rain scroll]

oh n m lol

nvm*

[limpid horizon]

You probably want a pre university channel
Like precalculus or smth

[weak lodge]

#prealg-and-algebra #precalculus

[unborn rampart]

how does it being a discrete dynamical system imply that there is a formula (or a "formula way" whatever that means 😅 )? Isn't Collatz a kind of discrete dynamical system?

[rose mirage]

and collatz is written using a very simple formula

[unborn rampart]

okay, but I'm pretty sure conway's game of life is a discrete dynamical system. Checkmate

[rose mirage]

and it is described using a very simple formula

[spice idol]

lmao

[spice idol]

read the description

[unborn rampart]

I mean, the dequeue question also has a very simple formula, given in the question. It depends on what you mean by formula, but I interpret it as some closed-form thing, or atleast a way to compute it that is simpler than just brute forcing through the steps given in the question

[limpid horizon]

everytime i get off a plane i find it funny how they say "deplane"

I just noticed now that for this modified cech complex, the xi form a "system of parameters" for R

Why do we need that?

i dont think ive seen that being necessary for my purposes

[digital parcel]

reducing to a system of parameters is almost always nice

because the ideal generated by SOP is m-primary

and local cohomology is invariant under taking radicals

so if you want to compute things like the cech complex, koszul complex, etc, it's nice to use system of paremters

it gives you nice results about things like depth, length, dimension, etc

Serre's Local Algebra has a nice treatment of this iirc

it also gives nice characterizations of cohen-macaulay rings that i can't recall off the top of my head

[limpid horizon]

ok cool ty

[limpid horizon]

yes i know this fact, how does this connect to / is related to the ideal being m-primary

[digital parcel]

^

[limpid horizon]

Local cohom is invariant under taking radicals (so like H_I(M) = H_rad(I)(M)) but why is that related to I being m-primary

Cause you said “and”

[digital parcel]

m-primary means its radical is m

so computing local cohom of H_m is the same as computing the local cohom of H_(x1, ..., xd)

and thus you can explicitly write down terms in your koszul complex and your cech complex

[rain scroll]

Thanks mb I read it wrong so- 😭

[muted sierra]

In my infinite boredom, I'm skimming Wikipedia, and in the span of 10 minutes, I ran into two baffling things.

First, every compact Lie group is isomorphic to a matrix group. So I was like "Wait, is the spin group isomorphic to a matrix group?"

So, of course I jumped to the article on the spin group, and found that the Clifford algebra of a vector space with a definite quadratic form is naturally graded. And now that's even more confusing, because Cl(V, q) is a quotient of the tensor algebra TV by a nonhomogeneous ideal, so where in the hell does the grading come from?

Someone please unconfuse me.

I don't even care about the general case. Just the case where q is the square of the Euclidean norm is enough.

[jaunty cargo]

Is there any function which is R integrable and having infinite number of limit points?

[muted sierra]

#real-complex-analysis is over there. -->

[lone jacinth]

Notice that it says "graded as a vector space" not "graded as an algebra"

[scarlet ermine]

yeah I think it's a little dum to just grade as a vector space. I think the thing you actually want to think about is a filtration on Cl(V), where you say F_k Cl(V) are the elements who can be written as sums of products of at most k elts of V, and so F_k Cl(V) * F_l Cl(V) lands in Cl_{k+l} (V), and this is a filtered algebra. Then the associated graded of this algebra is the exterior algebra, and I think wikipedia is just lifting that grading to the vector space

[lone jacinth]

So the grading should just be things that are the product of k linearly independent things (and sums of those)

[muted sierra]

Okay, that filtration makes sense.

[scarlet ermine]

usually I've seen the spin group as the even elements of Cl(V) that are norm 1 and conjugate V back to itself, I guess this is supposed to somehow be the same?

[fierce steeple]

Woah Clifford algebras... I love

[muted sierra]

Wait, now I'm confused again. The definition of filtered algebra that I know uses a filtration by ideals, but those F_k Cl(V) mentioned above aren't really ideals of Cl(V). They're just vector subspaces.

[lone jacinth]

I think the usual definition is just that
Fm * Fn < F[m+n]
nothing about them being ideals

If they were ideals the filtered condition would be kinda moot since A*Fm < Fm < F[m+n]

[muted sierra]

Okay, then I just have to live with the fact that $F_0 \oplus F_1/F_0 \oplus F_2/F_1 \oplus \dots$ can only be identified with the algebra's underlying vector space, not with the algebra itself.

[broken turtle]

Eduardo León

[muted sierra]

I had only encountered descending filtrations by ideals before. Like, the ones you use to construct normal bundles in algebraic geometry.

[woven loom]

I mean I’ve seen ascending filtrations used that weren’t even subalgebras (in some generic sense like vector subspaces are subalgebras)

[muted sierra]

Wait, where the terms of the filtration weren't even vector subspaces? That's just crazy.

[inner ledge]

Symmetric positive definite matrices always have a Cholesky decompositon ?

[muted sierra]

#linear-algebra is over there. -->

But yes.

I mean, what could be a more explicit proof than the algorithm that computes one?

[granite yew]

The N grading on clifford algebras comes from the chevalley isomorphism to the exterior algebra, they are naturally isomorphic everywhere except char 2 (I hear that careful definitions can reclaim this isomorphism in char 2).

[muted sierra]

Algebra isomorphism of mere linear isomorphism?

Because the latter isn't terribly surprising.

[scarlet ermine]

it’s still just a linear isomorphism. In the exterior algebra, v^2=0 always, and this need not be true in the clifford algebra (unless your quadratic form is identically 0)

[granite yew]

Linear isomorphism.

That said, you can define the exterior product from within the clifford algebra by using this linear isomorphism

And then it's an algebra isomorphism

[muted sierra]

Mmm, can we think of this linear isomorphism as arising from a deformation process? I mean, in the construction of the Clifford algebra, we quotient $TV$ by the two-sided ideal generated by $v \times v - q(v)$ for every $v \in V$. Now let's replace the generators with $v \otimes v - t q(v)$ and make $t \to 0$.

[broken turtle]

Eduardo León

[mental escarp]

If I have a Lie algebra A and a ring ideal J in A is it true that A/J is also a Lie algebra?

[lone jacinth]

Is a ring ideal another word for ideal, or is it something different?

If you mod out an ideal you get a lie algebra at least

[mental escarp]

I am just worried that only certain ideals J would give Lie algebras but okay sweet thanks!

[granite yew]

Indeed, T(V)/<v^2-Q(v)> with Q=0

[granite yew]

I can't really recite the details very well, but this is my preferred reference

[muted sierra]

Thanks.

[spice idol]

Been getting familiar with local properties more and the fact that x in J(R) iff 1 - rx is a unit for all r is due to a local property!

(=>) Suppose 1 - rx is a unit for all r in R. Suppose then x notin m for some maximal ideal m. Consider the quotient R/m. Then r has a unit v, and as such we must have that 1 - vx + m= 1-1 + m = 0 is a unit in R/m, which of course is impossible, hence x must be in m.

(<=) Suppose that x in J(R). Then, for all r, rx in J(R). Passing to the localisation R_m; this is a local ring, so rx being a nonunit (as it is in J(R)) means that 1 - rx is a unit in R_m. But being a unit is a local property, so this means that 1 - rx must be a unit in R.

[spice idol]

do you mean a lie algebra ideal or an ideal of a ring structure, the commutator lie algebra of which being your lie algebra?

[mental escarp]

Say A is an associative algebra that happens to also be a Lie algebra.

I am asking if any arbitrary ring ideal J of the associative algebra A induces a quotient A/J with a Lie algebra structure

I am guessing by your question, it might be necessary to require the ring ideal to also satisfy some compatibility condition with the bracket?

[spice idol]

lie algebras are never associative I believe

like, they must be abelian then

oh wait you meant that your vector space carries both a ring and a lie algebra structure

like different operations

[mental escarp]

I mean A has three operations.

Yeah exactly

[spice idol]

then yeah of course the ring structure and lie structure must be related

couldn't tell you exactly though

because since we're only talking about ideals, that doesn't tell you much about the ring structure itself

[mental escarp]

So your saying the quotient A/J won’t necessarily be a Lie algebra unless J has a certain form?

[spice idol]

A sufficient condition, of course, is that we have an equality xy - yx = [x, y]

[spice idol]

I mean that the ring structure and lie algebra structure being unrelated basically guarantees there are ring ideals which arent ideals of the lie algebra and vice versa

[mental escarp]

I guess I didn’t know there was a notion of a Lie algebra ideal okay. And these must be the ideals that quotient to other Lie algebras

[spice idol]

like, take any k-algebra and consider the abelian lie algebra structure on it. Then most ideals of the lie structure (they are nothing more than the vector subspaces) wont be ideals of the ring structure

[spice idol]

I guess its why would you expect the ideals of the ring to have anything to do with the lie structure

[mental escarp]

Okay so also not all ring ideals of A will also be Lie algebra ideals. Thanks got it

[spice idol]

yis

but "ring ideals of A" is kinda nonsensical because a lie algebra does not inherently carry a ring structure

[mental escarp]

I have another Lie algebra question, say (L,<,> ) is a Lie algebra and L is a commutative algebra (L,+,*)

I know <x,x>=0 for any x in L. Say x is invertible. Does this imply that <x,x^-1>=0 or one? I feel like the bracket should give me this value but idk

[spice idol]

no, because the algebra and lie structures have nothing to do with eachother, unless you specify that

it's like putting two different group operations on the same set and expecting them to interact in some way

[mental escarp]

Ah okay I see thank you.

[mental escarp]

What if I had a Poisson bracket? In a Poisson bracket there is a Leibniz identity which does relate ring multiplication to the bracket

[spice idol]

[1, x] = [x^-1, x] x + x^-1[x, x] = [x^-1, x] x
=> [x^-1, x] = [1, x] x^-1

maybe you can prove its 0

idk

[mental escarp]

Ah okay I think I got the same thing dang. It could be the case that [1,x]=0 but I need to check. Thank you so much ! I needed a sanity check

[spice idol]

[1 • 1, x] = [1, x] + [1, x], so indeed

[fierce steeple]

I have a funny question. I hope Im right in thinking the "trivial Lie algebra" functor Mod_k -> Lie_k (notation hopefully clear, feel free to assume char k = 0) is the unique section of the forgetful functor Lie_k -> Mod_k? I.e. unique way of functorially endowing every vector space with a Lie bracket. I guess any section must send k to the trivial Lie algebra on k (no choice there) and since the forgetful Lie_k -> Mod_k is conservative and preserves filtered colimits and biproducts, the same must be true of any section, which forces the trivial (i.e. abelian) Lie algebra structure on every vector space

[quick sparrow]

hi girls, i’m in my representation theory arc

i missed the first month of lectures so i might be cooked already but uh yeah

[mental escarp]

that means <x,x^{-1}>=0 thanks!

[spice idol]

aligns with the intuition that [-, -] is some form of commutator

as inverses commute

[lone jacinth]

We talking groups, lie algebras, associative algebras, quivers, something else?

[quick sparrow]

the course is supposed to be an introduction to the representations theory of finite groups, compact Lie groups and complex Lie algebras

[lone jacinth]

Cool beans

[quick sparrow]

idk what lie algebras are

i should probably be reading right about now but i’ve way too much on my mind guh

[spice idol]

think of an associative algebra A, and the commutator [a, b] = ab - ba

that's what a Lie algebra tries to model

[lone jacinth]

Lie algebras is basically the structure you get by "taking the derivative" of a lie group.

[quick sparrow]

ohh lovely so like tangent space stuffs?

[spice idol]

question that just popped into my mind: is there such a thing like lie algebras but for algebraic groups/group schemes?

[lone jacinth]

There is such a thing yeah

[spice idol]

cool! what's it called?

[lone jacinth]

https://en.m.wikipedia.org/wiki/Linear_algebraic_group

"The Lie algebra of an algebraic group"

[lone jacinth]

Yes, given a lie group the multiplication induces a lie bracket on the tangent space at the identity.

[spice idol]

I see

thanks

[lone jacinth]

If you think of lie groups as just matrix groups and X(t) and Y(s) are two smooth maps from R to GLn that are the identity at 0, then

d/ds X(t) Y(s) X(t)^-1 = X(t) Y'(s) X(t)^-1
and
d/dt X(t) Y'(s) X(t)^-1 = X'(t) Y'(s) - Y'(s) X'(t)

So the lie bracket of matrix multiplication is just [x, y] = xy - yx.

This same thing you can do in general, taking a derivative of conjugation like this

[quick sparrow]

right

[unborn rampart]

I've always wondered why people study Lie algebras by themselves. I know it gives you information about Lie groups - every Lie algebra is the Lie algebra of a Lie group - but what importance do Lie algebras have in "pure" algebra?

[rare walrus]

One way they naturally occur is in Hochschild cohomology. There are some interesting open questions that Linckelmann is asking about the Lie algebra structure of HH^1 of blocks of finite groups, and also the whole algebra HH^* has a Lie bracket which remains relatively difficult to work with. Linckelmann has shown that there appears to be a connection between the Lie algebra structure here and the representations in the block of the finite group.

I have heard from my friends who do p-group stuff that there is supposedly a way to classify p-groups through Lie algebras over F_p, but I don't know anything about this so I will leave it there.

I am trying to remember a name, give me a second...

Yes there we go

And also something something they're inherently interesting blah blah etc. Cushing, Stagg, and Stewart recently discovered a large number of simple F_2 Lie algebras which is in itself very curious I think.

[lone jacinth]

The classification of semisimple lie algebras and the connection to Dynkin diagrams is pretty interesting

[rare walrus]

(Btw it's precisely that this classification fails in F_2 that the paper above is interesting)

[spice idol]

of course it fails in F_2

[lone jacinth]

They have interesting representation theory and lots of connections with physics and quantum algebra

[spice idol]

i particularly enjoy Lie algebras because they are a particularly nice example of algebraic structures where congruences correspond exactly to substructures

[unborn rampart]

Thanks for the examples my reading speed drops to 2 WPM when I'm in this channel for some reason

[rose mirage]

Linckelmann mentioned

[limpid horizon]

it needs to be a short exact sequence of complexes right (lol almost 2 months later reply)

[rare walrus]

You could probably triangulate me from this if you didn’t already know

[spice idol]

derived functors are about left/right exact functors, and in particular measuring how much they fail to be resp. right/left exact

[rose mirage]

call ts triangle distinguished the way I'm translating over to ur location

[limpid horizon]

oh yea i was thinking of the general thing where a ses of complexes gives a les in homology but that theorem is just used in proving the les in derived functors i think

[fierce steeple]

Fun times

[fierce steeple]

This is v important in general

I am interested in this boytjiw now as someone who has thought a decent bit about hochschild stuff but not reps

Collaboration time

[limpid horizon]

Yeah i like that theorem

I enjoyed learning about hom algebra and i got depressed once i had to learn combinatorial algebra

[fierce steeple]

What is combjnatorial algebra tbh

I know a friend doing some

[limpid horizon]

I dont know, im learning about “stanley reisner” theory. Correspondence between simplicial complexes and quotient of polynomial ring by squarefree monomial ideal

The nonzero monomials in the quotient correspond to faces in the complex sort of things

@ornate atlas knows most about it in this server probably

From what ive seen at least

[rare walrus]

Hom alg is like 80% bookkeeping it's frankly exhausting

[rare walrus]

I'm afraid it's not something I personally know about (yet -- I am interested!) but you should definitely talk to Marcus Linckelmann if you're interested. He has high hopes

[fierce steeple]

Aha nice

I think a friend is interested in something quite close to this

[fierce steeple]

There is also the area of deformation theory which is now known to be controlled (in large part) by generalised Lie algebras. I work with lie algs for this reason

[fierce steeple]

One thing that is interesting in this though is that there should be a lot more structure than a Lie algebra structure

Well for starters like it should be a dg Lie algebra but much more than that

[rare walrus]

Marcus was saying it's a graded Lie algebra

I don't really know much more

But that already gives us a lot ofc

[fierce steeple]

Ye

[limpid horizon]

Ye.

[rare walrus]

Ye.

[spice idol]

graded stuff is funky

[weak lodge]

t's not pure algebra, but another motivation for Lie algebras sans Lie groups is to define weight systems with connections to finite-type/Vasilliev invariants in knot theory; see chapter 6 of this book

[spice idol]

mm thats why ive seen lie algebras come up with quandles before

[weak lodge]

also the leibniz algebra - lie rack correspondence

[spice idol]

yeah that one is what i was thinking of

lie rack is such a weird ass thing to say lmaoo

[weak lodge]

real

[spice idol]

i love all the connections in math

[limpid horizon]

ts 🥀

[spice idol]

ive been thinking on how I can somehow generalize the notion of regular map (i.e. locally "nice") to universal algebra. For commutative rings this would be locally rational, but in arbitrary structures theres no such thing, sadly

however, if your notion of algebraic geometry is nice enough, there is some notion of "product" of congruences by which "prime congruences" can be defined, i wonder if something can be done with that

[ornate atlas]

I know a bit but not a huge amount, there’s someone else here who seems to know a bit more but I can’t remember who, they pop up sometimes

That is essentially the basic idea though, you can often turn hard combinatorial questions into problems about relatively easy comalg questions (or well not easy, but like problems with more angles of attack)

At least at the basic level there’s that anyway, you then get into Hilbert schemes of points and stuff and I know much less about that stuff

[spice idol]

schemes? consider me enticed

[near lantern]

This sounds very interesting; can you elaborate (especially: what kind of "deformation theory" do you mean)?

[near lantern]

I think the reason the word "locally" works in comm alg is that you can define A_a to be an "open subspace" of A and {A_a1, ..., A_an} to cover A iff (a1, ..., an) = A and this actually behaves like it makes sense. This suggests to get a notion of "locally" for another signature with category of algebras C one should look for Grothendieck topologies on C^op.

[spice idol]

hmm, i have heard of those

rn what my intuition would be is to look for nice-enough-but-not-too-nice notions of functions on some set A

like, if A carries an algebraic structure the obvious notion is that of terms or polynomial operations but fields show that this isnt really enough

maybe theres some way to classify whenever the notion of "locally polynomial" is too restrictive

oo thats a cool idea

[spice idol]

to show two extremes: polynomials for sets are exactly the identity and constant functions, and the polynomials (hell, even just the terms) for any primal algebra are every operation on that set

[scarlet ermine]

here's how I think about why to study Lie algebras. In a linear algebra/ring theory course, you spend a lot of time on the structure theorem for PIDs, particularly with the application in mind for the k[T]-module structures on a finite dimensional vector space V... just a fancy way of saying you study a single operator T: V->V, but you already get lots of worthwhile results, like the jordan/rational canonical forms.

A finite dimensional k[T]-module is the same data as a finite dimensional representation of the 1-dimensional Lie algebra/k, which I'll call kT, so this is a morphism of Lie algebras kT -> gl(V), which is again picking out a single operator.

The study of Lie algebras then generalizes this study from 1 operator to several operators (the generators of your Lie algebra).. In the same way that we get a lot of mileage in algebra/geometry by recognizing group actions, I see Lie algebra representations as recognizing that some collection of operators on a vector space have formal properties in how they commute. So, it's maybe worthwhile see what has to be true about "realizations" of operators given some data about how they commute (so studying the representation theory of a certain presented Lie algebra).

For example, in Hodge theory (sorry I guess this isn't strictly algebra, but this is the example I have off the top of my head), you get operators on the cohomology ring (the Lefschetz operator, it's dual, and the counting operator). If you know Lie theory, you might wonder how these commute with each other, and then you would notice that these satisfy the same relations as the usual generators of sl2, so cohomology is an sl2 rep, and we can then automatically apply that structure theory to define primitive cohomology

[fierce steeple]

By deformation theory I mean studying problems of the following type: I have an object x defined over a ring B and a map A -> B of rings, and I want to find some y over A that reduces to B. Like e.g. x might be a B-module N and you want to find an A-module M such that M (x)_A B = N. Or you can play a similar game with schemes or representations etc. Usually the focus is on "infinitesimal" deformations e.g. surjective maps of Artinian rings, and e.g. you may be interested in going from F_p to Z/p^2 to Z/p^3,... with the hope of lifting to Z_p.

You can then formalise this sort of problem (i.e. in terms of functors satisfying a couple of properties) but what's interesting is that in many cases the problem of lifting is somehow related to differential graded Lie algebra (or dgla). It turns out that if you generalise these functors enough to define them on more general (homotopy-theory-flavoured) rings (and value them in spaces rather than sets) then it becomes literally true that dglas correspond to these deformation problems (this was proven independently by Pridham and Lurie), at least in char 0

But yeah tl;dr the point is you can study these often quite geometric lifting problems by studying something more algebraic

Like it is kinda amazing that an entire problem of lifting some object to any ring is encapsulated by a single algebraic object, about which you can ask questions (e.g. is this Lie algebra abelian?)

[scarlet ermine]

just saw a talk a little bit about this last week! it seems very cool

[fierce steeple]

Oh wow nice lol

[fierce steeple]

Was there anything thjs was about in particular?

Or like a general ting

[limpid horizon]

For I an injective module, in the proof of something bruns and herzog say let I = E(k). Would anyone know what this “E(k)” means?

[spice idol]

what is the context?

[limpid horizon]

Ill send screenshot one sec

Oh R is noetherian local ring

So im thinking k probably means residue field idunno

Cause it later says assume I = E(R/p) where p is not maximal ideal

[spice idol]

inhective hull, apparently

ah, the largest essential extension

it is the "universal" injective module for M, in some sense

[lone jacinth]

For a commutative noetherian ring any injective module is the direct sum of modules of the form
E(R/p)
the injective hull of R modulo a prime ideal.

Presumably here R is local, so k is the residue field

So E(k) means E(R/m)

[spice idol]

are the x_j fixed here?

[limpid horizon]

I think they are a system of parameters for R so they generate an m-primary ideal . I dont know the significance of this atm

So yea i think they fixed

[scarlet ermine]

it was about whether central degree two elements of the homotopy lie algebra all come from embedded deformations

[worldly zealot]

does this condition on injectives imply noetherian? i think i remember hearing that but i cant verify

[molten vale]

Can the inverse limit of commutative rings with totally bounded krull dimension be notherian with infinite krull dimension?

[fierce steeple]

Hmm interesting

[fierce steeple]

By totally bounded do you mean there is a universal bound for the dimensions of all of those rings?

[molten vale]

Also, if the inverse limit of commutagive rings R_i is noetherian, is the inverse limit of ideals (of R_i) precisely the ideals of the inverse limit?

[molten vale]

yeah, i asked a prof a while ago and they said they suspect it to be hard for the inverse limit to have Krull dimension more than N+1, if it is noetherian and if the bound is N.

[fierce steeple]

Yeahh that is my intuition

(With the N+1 case have easy examples like e.g. Z_p being limit of Z/p^n)

[molten vale]

I dont have intuition beside these examples since idk what the inverse limit being noetherian tells us. If the second statement about ideals corresponding to inverse limit of ideals is true (given that the inverse limit if noetherian), it might be much more approachable.

[lone jacinth]

A ring is Noetherian iff the direct sum of injective modules is injective.

So yes. Though I think E(R/p) might still be the only indecomposable injective modules for a commutative ring R, not sure...

[lone jacinth]

How about this:

Let R = k[x1, x2, ...] be the polynomial ring in infinitely many variables and J = (x1, x2, ...).

Then R/J^n is zero-dimensional for all n.

The inverse limit should be some kind of power series ring thingy. Which I'm guessing should have infinite dimension...

Indeed the inverse limit should be exactly the power series ring
k[[x1, ...]]
and
(x1) < (x1, x2) < ...
is an infinite sequence of prime ideals

Shoot, you also said Noetherian

I don't want to think about Noetherian rings with infinite Krull dimension

[spice idol]

its unnatural and should not exist

🙂‍↕️

[molten vale]

yeah this example was why I added noetherian

although i did the inv limit of k[x_1,..,x_n]/(x_1,...,x_n)^n cuz i didn't think of R/J^n being 0 dimensional

[fierce steeple]

Why not

🔥

[limpid horizon]

Potato the true algebraist

[fierce steeple]

I am baby

[digital parcel]

Smart ass baby damn

Was hartshorne ur Goodnight Moon

[sonic skiff]

A,b,c >1 is false btw only abc>1 is given and a,b,c are all positive.

[hushed bone]

This would belong more in #real-complex-analysis

This channel is for like, Lie algebras, commutative ring theory, homological algebra, homotopical algebra, etc

[sonic skiff]

I see

[limpid horizon]

Lol

[low obsidian]

Got assigned a hw problem to classify all irreducible representations of D4, but we haven't dont characters yet. Any ideas how to proceed?

[lone jacinth]

Say D4 = (s, r)

r must be sendt to a matrix of order 4. These are diagonalizable, so you may assume it's diagonal.

Then s needs to be sendt to a matrix of order 2 with srs = r^-1

[vague pawn]

Also sum of squares can help

It means there is 4 1-dim and 1 2-dim irreducible representations

[lone jacinth]

You can get that without characters?

Anyway, another useful thing could be that determinant and trace are preserved under conjugation, so r and r^-1 must have the same of those

[vague pawn]

I think so

[lone jacinth]

I mean, you can prove it from the group algebra being semisimple I guess

[vague pawn]

Yes thats how I remember it

[low obsidian]

How can you show this

[low obsidian]

Makes sense, im stuck on why we would be forced to be dim 2 or 1

[lone jacinth]

So my suggestion is essentially to classify all the representations, then just notice which are irreducible.

So you may think about decomposing the space into the eigenspaces of r.

[fierce steeple]

Naur im baby

But thank

[soft parcel]

Is there information missing here?

[last talon]

Why do you think there is?

[soft parcel]

It just seems like very little information to go off of

I can conclude some things under slightly stronger assumptions, so idk

[ornate atlas]

Hmm, this feels like some segre embedding stuff that we did in my alggeo class, but I remember that proof essentially being just a bunch of linear algebra and we showed it was something to do with the rank of the matrix of the coefficients I think?

I don’t know if that method could be extended to generic R modules (I’d be surprised) but maybe that’s somewhere to start?

FWIW I don’t know, that does seem hard in general but I haven’t thought about this at all really, so

[last talon]

Yeah I’d be surprised if the answer isn’t “iff c_11 c_22 = c_12 c_21”, but I don’t even know how to prove that such elements are pure tensors if we’re not in a UFD
Actually I don’t even think that’s true outside a UFD
Like if we took c_11 = 2, c_12 = 1 + sqrt(-5), c_21 = 1 - sqrt(-5), c_22 = 3 in Z[sqrt(-5)]

[lone jacinth]

I'm not sure there's missing information, but it's not so clear what the question is asking for.

Like conditions on cij that make it true for arbitrary M and N maybe?

[soft parcel]

I think the question is supposed to be an "it is a pure tensor only if ..." type question

But seeing as this is a commalg course and no algebraic geometry has been covered, I think whoever wrote this exercise probably just forgot to add some conditions or something

I will just skip it

[plucky arch]

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

this is an absolutely wild statement to be independent of ZFC

[digital parcel]

Yea I remember my prof talking abt that in my alg course

[spice idol]

ive done a little digging and found that the cohomology for this is actually a specific version of something called monad cohomology. Namely, we take the canonical resolution of a comonad, take any abelian group object Y, and consider Hom(-, Y) : C → Ab. This defines a cosimplicial object in Ab, so we can simply take its homology

[neat nexus]

How do we prove that for a ring, if all its right ideals are projective then the ring is right hereditary (submodule of any projective right module is also projective) ?

[candid igloo]

Let R be an integral domain with infinitely many elements, of which only finitely many are irreducible.
Suppose every non-unit has an irreducible factor. Show that R has infinitely many units.

[spice idol]

refrain from double posting

#groups-rings-fields is the better place for this anyways

[limpid horizon]

The dreaded... Double Post.

[limpid horizon]

Would anybody know how one would prove this?

Or just where to even start really

I will first review what the *local is

[limpid horizon]

I feel like it might not be that hard really but i havent actually tried going thru it yet

[woeful crane]

I've noticed something weird, but it takes a little set-up. I'm not actually interested in reps of finite groups, but this is a good "test case".

Let $H$ be a subgroup of a finite group $G$ and let $\phi:\mathbb{Z}/2\to \text{Aut}(G)$ such that $\phi(1)(H)=H$. I will write $G^+:=G\rtimes_\phi \mathbb{Z}/2$ and likewise for $H^+$, which is a subgroup of $G^+$.

If an irrep $\pi$ of $G$ is not isomorphic to $\pi\circ \phi(1)$, then there is unique irreducible representation $\tilde{\pi}$ of $G^+$ such that $\text{Res}_{G}^{G^+}(\tilde{\pi})\cong \pi \oplus (\pi\circ \phi(1))$. Otherwise, there are exactly two non-isomorphic irreps $\pi_1, \pi_2$ of $G^+$ such that $\text{Res}(\pi_i)\cong \pi$.

Suppose now that $\pi$ is a complex irrep of $H$ which is isomorphic to $\pi\circ \phi(1)$, and let $\pi_1$ be any one of its two irreducible extensions to $H^+$. The irreducible composition factors of $I_{H^+}^{G^+}(\pi_1)$ are exactly the extensions of the irreducible composition factors of $I_H^G(\pi)$.

Claim: If $\sigma, \tau$ are irreducible composition factors of $I_{H^+}^{G^+}(\pi_1)$ such that $\text{Res}^{G^+}_G(\sigma)\cong \text{Res}^{G^+}_G(\tau)$, then $\sigma\cong \tau$.

[broken turtle]

kr1staps

[woeful crane]

To put this another way, if $\rho$ is and irreducible composition factor of $I_H^G(\pi)$, then $I_{H^+}^{G^+}(\pi_1)$ only contains one of the extensions $\rho$ to $G^+$.

[broken turtle]

kr1staps

[limpid horizon]

Im having a hard time interpreting M-sequences

an element of R being M-regular is good but then say x2 being M/x1M-regular gives me pause

i guess a key point is that x2 may not be M-regular right

[hushed bone]

It’ll be true if R is local and M is finite

[limpid horizon]

For x2 to be M-regular?

[hushed bone]

If it’s M/x1M regular yeah it’ll be M-regular

[limpid horizon]

Why is that?

is this also related to cohen-macaulay stuff somehow?

[fossil thicket]

it boils down to some cute lil application of nakayama

idk if it's true without noetherian assumptions, so me assuming R is noe local

[limpid horizon]

For (left or right) derived functors, given M->N we have R^iM->R^iN. Do we get that map from first making a ses using M->N ?

[fierce steeple]

Well this is what Weibel calls the horseshoe lemma, like given a map between things you can find a map of injective resolutions

Like this is how it is a functor

[limpid horizon]

Yea what i saw so far from that is, if you have a ses and have an injective resolution of first and third entry, then there is an injective resolution of the middle that is also a direct sum of the two other resolutions

And also true for projective stuff

[rose mirage]

filling in the horseshoe

[digital parcel]

my prof once had us prove the horseshoe lemma in a homework but he didn't call it that

but i remembered reading about it somewhere so i titled that problem "Horseshoe Lemma" and he left a comment saying "i told you not to use outside resources"

[limpid horizon]

Thats why i was asking do we set up an ses from M->N and then do horseshoe lemma thing?

Like just ker M image

[digital parcel]

do you need horseshoe lemma to show that M -> N induces a map R^iM -> R^iN?

i thought it was just some homotopy lifting argument

[limpid horizon]

Yea im just wondering where do we get that map

Derived “functors” so like how is it a functor type shi

[digital parcel]

you can probably find a proof of this in any homological algebra textbook

[fierce steeple]

True lol I guess in my mind horseshoe was basically this lifting thing lol

Mb

[digital parcel]

lift lift lift

[limpid horizon]

I saw lemma of if u have map M->M and two projective resolutions of M then u can connect them

Does that work with any M->N

[digital parcel]

yeah that's basically what i mean by homotopy lifting argument

you have a map of modules M -> N and take projective resolutions P and Q of both

your map M -> N lifts to maps between P_i -> Q_i

[fierce steeple]

Kinda funny doing all the stuff w existence of resolutions and then it usually only gets used for modules or smth in a first course, where there are canonical things

But lots of maths is like this ig

[digital parcel]

i would guess authors assume if you're reading about it then you've seen it in the context of modules

[limpid horizon]

And does the map given from that way coincide with the maps between derived functors you get when you extend a ses M N L to an ses of derived functors

[digital parcel]

and just handwave any of the nitpicky details that you prove/see in modules

[fierce steeple]

Ig you mean LES

[limpid horizon]

Yes mb

[fierce steeple]

But yes by construction

Since the LES comes from a SES of complexes

[limpid horizon]

Riiight its like the same thing

[digital parcel]

in the same way that homology and cohomology are functors

[fierce steeple]

And the (non-boundary) maps are just those induced by applying (co)homology

[digital parcel]

you can just think of these as generalizations of them

[limpid horizon]

Yea, ig the core problem was i wasnt totally sure how to interpret homology as a functor

[digital parcel]

well actually not generalizations it is homology and cohomology

[limpid horizon]

Yeah

[digital parcel]

if you want a concrete computation I think Hatcher has this

in the context of simplicial homology or singular homology whichever he uses at that point

more formal treatments you can find in any hom alg textbook

[limpid horizon]

Yea cause not all homology comes from injective or projective resolutions kind of thing right

[digital parcel]

weibel, rotman, etc

[limpid horizon]

I only really been studying that because i needed to know Ext and local cohom and those are both derived functors

[limpid horizon]

Yea i mean, simplicial homology has nothing to do with resolutions right ..?

[digital parcel]

for modules they should

[limpid horizon]

Oh, why?

[digital parcel]

umm wait actually lemme think about that

okay to be precise

homology is not a functor from the category of modules but from the category of chain complexes (or something like that)

[past cove]

as long as you have an acyclic resolution you can use it to calculate homology

[digital parcel]

in this case they come from projective/injective resolutions

[limpid horizon]

Acyclic meaning exact? I forget that defn

[past cove]

so you should be able to say something like if you category has enough projective/injectives everything works out

[past cove]

acyclic means 0 homology for i>0

[limpid horizon]

Do u mind spelling out more what u mean here

[digital parcel]

resolutions in abelian categories are the same as resolutions of modules

not necessarily projective/injective resolutions but just resolutions

when you're computing singular cohomology you're taking a chain complex of abelian groups

if memory serves me right

[limpid horizon]

A resolution of M is just 0 -> M -> M0 -> M1 etc exact?

[spice idol]

i think usually it ends with M

[digital parcel]

oh oops yeah

so not resolutions

[digital parcel]

you can write right resolutions like that

[spice idol]

right right

coresolution

[limpid horizon]

I got kinda lost on what we’re talking abt now tbh

[spice idol]

yes, you are correct

though a lot of homology comes from derived functors

group cohomology, for example

[limpid horizon]

Ya thats why i liked learning about derived functors they seemed nice / useful

[fierce steeple]

Real

[digital parcel]

this convo is much more fun than my stupid lie groups homework which has been sitting on my second monitor for a while

bvlah blah regular map blahblah

[limpid horizon]

Lol

[digital parcel]

this map is regular blah blah blah

[limpid horizon]

Anytime something is “regular” its boring tbh

[spice idol]

lie algebras are cool

oh its just lie groups

[digital parcel]

yeah just lie groups

[fierce steeple]

Simplicial homology can be expressed as sheaf cohomology which is a derived functor tbf

Lel

[spice idol]

hey dont you talk shit about regular maps of vatieties qwq

[spice idol]

its also a monadic cohomology

[digital parcel]

regular maps are useful

[fierce steeple]

Sure

[digital parcel]

i just dont like these routine computations

but anyway it must be done to get a good grasp of it

[limpid horizon]

Idk what im thinking of is regular sequences in algebra and thats kind of boring

[digital parcel]

"regular sequences" and "boring" in the same sentence is crazy

[limpid horizon]

Ok ik it is interesting i just havent thought of it enough

Ok but like listen the definition of it is not interesting

But the fact it can detect other cool stuff is interesting

I havent studied properly yet how it can detect homological things

[digital parcel]

the first one that comes to mind is that you use it to define depth

under some loose conditions i forgot

[near lantern]

Regular? For Lie groups?

[digital parcel]

you can use depth to prove things like existence of hilbert polynomials in two (or more) variables

[digital parcel]

we just started so it's moreso proving properties of the underlying manifolds

regular as in the manifold sense

[near lantern]

Oh

[near lantern]

Just smooth group homomorphism?

[digital parcel]

we don't assume smooth

we're working over C^k, real analytic, or complex analytic

[near lantern]

So regular means C^{whatever the manifolds are}?

[digital parcel]

basically

or real anaytic or complex analytic

[near lantern]

Right

[digital parcel]

wait no sorry

regular functions are C^k or whatever

regular maps pull back regular functions to regular functions

[ornate kindle]

Question about the phrase "symmetric part" in the following

[broken turtle]

NotABot

[ornate kindle]

On a previous page it talks about an r with

[broken turtle]

NotABot

[ornate kindle]

Though in that context r is more firmly defined.

So my question is, is it correct that "symmetric part" is referring to the r_12 + r_21? Or would it mean some component within r that is symmetric

[limpid horizon]

why is that an issue

pretty sure R is just noetherian

[hushed bone]

Then (I^t + Ann(M))M = I^tM = M < mM

[limpid horizon]

So, M = mM would be bad?

[hushed bone]

Nakayama ain’t it

[limpid horizon]

yeah was just reviewing it rn. Would this need R to be local so that jacobson radical is just m?

[hushed bone]

Hmmm

True

Oh

Well this is bad because Supp M = V(Ann M) so if m contains Ann M then M_m can’t be zero

But when you localize this says any m containing I^t + Ann M has M_m = 0

[limpid horizon]

Not sure abt this part

[hushed bone]

I literally argued it above

[limpid horizon]

This shows that Mm = 0? Im not sure why

[hushed bone]

This says M_m is 0

[limpid horizon]

What does?

[hushed bone]

That equality lol

[limpid horizon]

Uhh ok ik since Supp M = V(Ann(M)) then if we had I^t + Ann(M) < m then Mm ≠ 0, i dont know where Mm = 0 is coming from

[hushed bone]

Localize this equality and apply Nakayama

Now M_m is 0

You said it needs to be local so just localize and now you get to use Nakayama

[cloud karma]

In the context of affine weyl groups, let y = x w_0 t^(w_0 lambda) where here lambda is dominant and x is an element of the weyl group. I'm trying to show length(y)=length(x)+length(w_0 t^(w^0 lambda)). I know I'm supposed to use the length formula but I'm not sure how to deal with it

[limpid horizon]

Localize the modules over m?

[hushed bone]

Yuh

[limpid horizon]

If M = N then Ms = Ns for a localization?

[fierce steeple]

Localisation at s is a functor and in particular isos induce isos

[limpid horizon]

Yes i thought it was smth like that

[spice idol]

is being isomorphic local?

[fierce steeple]

Of course as stated it seems a little different aha

[fierce steeple]

Yes or no depending on what you mean aha

[limpid horizon]

Functors in general preserve isomorphisms right

Because they preserve identity maps and diagrams

[fierce steeple]

"Being an isomorphism is a local property of a map" is what I would say

[spice idol]

f : N → M being an isomorphism i can imagine is local

[fierce steeple]

Yes

[spice idol]

yippee

ofc you can just use the exact sequence
0 → ker f → N → M → coker f → 0
and then exactness of localisation

anyways sorry to interject

[spice idol]

ye that was his point

[fierce steeple]

Yeah and nice short exercise

[summer quest]

it should be referring to the Sym^2 component yes

[limpid horizon]

But also this doesnt help me concretely with what the map between things are

Like ik now how you get the map, ok do resolutions, fill in the diagram, apply functor, and then map on homology comes from chain map

But like …

In particular, the map M->M just multiplying by some x in R induces the same map Hm^i(M) -> Hm^i(M) on local cohomology somehow

Just multiply by x

[fierce steeple]

For example for this you can use just one resolution of M and then map it to itself via multiplication by x in each degree

Also for modules and stuff it is often possible to use like very canonical resolutions for these kinds of arguments, like any R-module M has the resolution using free R-module on M etc

[limpid horizon]

How does that help?

[limpid horizon]

So the map that just multiplies by x does the same on the map of derived functors for any of them

So like Ext would be the same

[fierce steeple]

Like you just reason using that resolution

[fierce steeple]

Ye

[limpid horizon]

But a map I1 -> I1' thats just multiply by x means after applying functor the map FI1 -> FI1' is also just multiply by x...?

for local cohomology i guess its just that because when u apply the I-torsion functor, maps dont change its just restricted

[limpid horizon]

Oh, we have M = mM and we can localize these R-modules over m to get Rm-modules Mm = mMm, then apply nakayama to say Mm = 0 which is the contradiction

sorry just want to make sure i really understand this properly

[limpid horizon]

Of course, if you have a maximal M-sequence x1, x2, ... xn, then M/x1M has a maximal M/x1M-sequence x2, ... xn right

[fierce steeple]

Depends on the functor

But for many things this is the case

[limpid horizon]

So you saying Ye here wasn’t totally true?

Because like you said now it depends on the functor

[fierce steeple]

Ok I guess I didn't rly understand what you meant by this lol

But it does depend on the functor ye

[limpid horizon]

Ya mb ik i was being a little careless

With describing it

[fierce steeple]

But a lot of these functors behave very well with such things ye

[wanton talon]

Whats algebra

[limpid horizon]

its like x+ 5 = 3 solve for x type shi

[wanton talon]

Ahh it’s easy

[limpid horizon]

ye

[spice idol]

it's like when you've got thingies and arrows between thingies

[vague pawn]

thats #category-theory

[spice idol]

and when youre doing geometry and the arrows turn around then youre doing algebra

[vague pawn]

we are not like them

[wanton talon]

But algebraic geometry its hard i think

[vague pawn]

sure is

[spice idol]

no super easy, barely an inconvenience

I mean sheaves and schemes, who doesn't understand them

[ornate atlas]

It’s incredible it took people so long to simply realise that points are prime ideals

Plainly obvious to any small somewhat educated child

[spice idol]

exactly, like its such an obvious step!

[limpid horizon]

I mean duh, any shmuck knows that

[obsidian aurora]

help a brother

[jaunty sparrow]

#prealg-and-algebra

[woven loom]

Suppose we have some function $w:{(n,m)\mid n> m} \to F_\infty$ where for any $n>m$ we have that $w_{n,m} \in F_m \leq F_\infty$ in the obvious usual embedding.

How many groups can arise as quotients of groups of the form
$$G_w = \big\langle (e_i\mid i<\omega)\mid \forall i>j. e_j^{e_i} = w_{i,j}(e_1,\dots,e_j)\big\rangle$$

(For some choice of word sequences w)

[broken turtle]

Sharp, the Inevitable

[spice idol]

what does e_j^{e_i} mean?

[woven loom]

As in, conjugation

[spice idol]

i see

totally forgot that notation existed ngl

[woven loom]

But the idea is things that have many concentric subgroups that you can’t escape by conjugation from outside

Without being quite normal

(Necessarily, anyway, you could just permute the generators and be grand and run into no issues, or otherwise have the words be freely independent and generate F_n or wtv)

[woven loom]

I was hoping these would be nice enough to make an argument work but they don’t quite seem to be

[mental escarp]

given a real vector space V and a real number $\lambda$ is the tensor product $V^{\bigotimes \lambda}$ well defined?

[broken turtle]

HausdorffT1

[fierce steeple]

I don't think so and honestly have no idea what this would mean

[spice idol]

honestly that would imply real numbered bases then lol

[mental escarp]

Dang okay thanks. I saw something like this in the context of “tensor densities” but maybe that is a specific geometric construction.

[fierce steeple]

I may also be wrong

Do you have a reference or smth? Dw if not lol

[woven loom]

You can have weird dimensions in op alg things

[spice idol]

[north stream]

Hi fellas, I was wondering something about general rings in positive characteristic. Given a ring R with char R=p a prime number, given x in R nonzero, if we take the ring S=R[t]/(t^p-u) then the module of Kähler differentials of S over R is zero iff there is already an element u such that u^p=x.

This is obviously true if R is a field, but I was wondering some general case without even asuming that R is noetherian.
One side is obvious since if there's already a root for t^p-u then S is a product due to the chinese remainder theorem. The other side I was trying to figure out if it makes sense to asume the Kahler diferentials is something like S/(f'(t)) which is non-zero since f'(t)=0.

[sage totem]

just, MATH.

longin of PROUNOUNCE.

[fierce steeple]

It seems some letters have been swapped here, but ignoring that I feel this may be true for general R - I'll have a go myself now lol

[fierce steeple]

actually you can just read this off here

(Here the set up is you have R -> S -> T)

[fierce steeple]

The Chinese Remainder theorem shouldn't apply though. Since p = 0 in R, if t^p - u has a root b then t^p - u = (t - b)^p and your polynomial ring is just R[t]/t^p up to iso, not R x .... x R

More concretely like the 'differentiation' map $R[t] \to R[t]$ of $R$-modules descends to a non-zero $R$-linear derivation $S \to S$ since the derivative of $t^p - u$ vanishes. This shows that $\Omega^1_{S/R} \ne 0$ by the universal property

[broken turtle]

Prismatic Potato

[fierce steeple]

I guess you could also consider residue fields of R and reduce to the case of a field

[north stream]

Damn, I just looked at this and realized that I don't know what I was thinking. Thanks a lot!, this makes a lot of things clearer.

[fierce steeple]

No problem. I guess this is just a very funny thing where it would work were we not in char p with an inseparable poly lol

[cloud karma]

Just want to make sure. Say $\Phi^{+}$ are the positive roots of a root system and $\Phi^{-}$ otherwise. If I take the sum $\sum_{\alpha \in \Phi^{+}} |(\lambda,w_0 \alpha)+1|$, since $w_0 \Phi^{+} \in \Phi^{-}$ we can sum over the negative roots so we get $\sum_{\alpha \in \Phi^{-}} |(\lambda,\alpha)+1|.$ My question was is this right? If this is right then can we also just sum over the positive roots instead?

[broken turtle]

Delteto

[near lantern]

This is right but you can't just sum over the positive roots because |(l, a) + 1| ≠ |(l, -a) + 1|. You could replace the sum over negative roots by a sum over positive roots of |(l, a) - 1|.

[near lantern]

This is true in general. For any commutative ring R of characteristic p, Omega_{S/R} is a free S-module with basis dt since f'(t) = 0.

[cloud karma]

I see thanks for checking my sanity

I've been trying to compute a certain length of an element in the affine weyl group. So say you have the term x w_0 X^(w_0 lambda) where X here is representing the translation part and lambda here is dominant. I'm trying to show that ell(x w_0 X^(w_0 lambda)) = ell(x)+ell(w_0 X^(w_0 lambda))

the length of the big term is $\sum_{\alpha \in \Phi^+} |(\lambda,w_0 \alpha)+\chi(x w_0 \alpha)|$ where chi here outputs 1 when you hit a negative root and 0 otherwise

[broken turtle]

Delteto

[cloud karma]

We can split the sum for cases when this chi is either 0 or 1

I tried subtracting this sum with ell(w_0 X^(w_0 lambda)) to hopefully show that indeed I get ell(x)

I ended up doing so and getting some cancellation and at the end what I got is $\sum_{\beta \in \Omega^+} |(\lambda,\beta)| - \sum_{\beta \in \Omega^+} |(\lambda,\beta)+1|$ where $\Omega^+$ here denotes the set of all $beta \in R^{-}$ such that $x \beta$ is in $R^+$

[broken turtle]

Delteto

[cloud karma]

now since lambda is dominant and beta is in R^-, (lambda,beta) <=0 and from there I got partially what I got

when (lambda,beta) < -1 I think the terms simplify to 1 and you get ell(x) but otherwise it's a mess

[lusty juniper]

hello. I am new to abstract mathematics, so my question might be misinformed. regarding lie group lie algebra correspondence, I wanted to ask if lie algebras act similarly to the jacobian as a derivative. the surface level similarities feel obvious, tangent space to a manifold, lie algebra's to my current understanding study linear fields

the jacobian is a linear field

u get where I am coming from

oh but to clarify I dont think lie algebras study the rate of change, just that they can be used analogously as a means to study the manifolds/functions

[void plank]

If you have two chain maps p and tau such that p \circ tau = 0, this implies H_i(p) \circ H_i(tau) = 0

Based on this alone, is there really anything I can conclude about the injectivity or surjectivity of H_i(p) and H_i(tau)? I tried analyzing the homology of mapping cone for p and tau and didn't get anything useful...

[fierce steeple]

Not really cause one of them could be 0 and the other anything

[void plank]

Yeah, I had the same conclusion as well

[fierce steeple]

Or maybe more dramatically like you could take the complexes to just be in degree 0

And then you just have pi = 0 which doesn't let you conclude much

:(

[void plank]

In my practical situation, my complexes and chain maps do have a certain form.

But they're not specific enough for me to conclude anything more definitive about them based on the p \circ tau = 0 condition sadly

[fierce steeple]

Ah sure

[void plank]

What is weird is that I keep numerically finding that H_1(Tau) is surjective for my specific p, tau, and complexes (still obeying the condition that p \circ tau = 0)

But I guess that's just a consequence of the situation I'm dealing with, rather than something to do with general hom alg

[fierce steeple]

Hm then this forces H_1(p) = 0 right

[void plank]

Yeah it does by the property of epis in additive cats. I was hoping to prove in generality that something about p \circ tau = 0 in hom alg implies this instead of relying on numerics for my specific chain complexes and chain maps, but I guess that's out of the question

[weak lodge]

Sure, in a certain sense. By the correspondence, real/complex representations of a simply connected Lie group correspond one-to-one (via differentiation) to real/complex representations of its Lie algebra. This turns the geometric problem of classifying the former, which can be daunting, into the linear-algebraic problem of classifying the latter, which is often much simpler

[lusty juniper]

I see, thanks for the response!

[barren obsidian]

hello

@here Could I link my hypercomplex preprint for review and possible colaboration?

it is in osf

[spice idol]

whats it about

[barren obsidian]

it is about new methods for generating the multiplication table of hypercomplex numbers from patterns I observed I made pythons script for it and also probable a formula for zero pair divisors count for any hypercomplex dimensions, it got the results aligned with recent preprints

[frosty ember]

There seem to be two definitions of the Lie bracket:

• On a general manifold, the Lie bracket [X,Y] of two vector fields X and Y is defined by [X,Y]f = X(Yf)-Y(Xf), which one can check defines a vector field
• On a Lie group G, we can define [X,Y] as the "quadratic" term in the expansion log(exp(X)exp(Y)) = X+Y + 1/2*[X,Y] + ...
  Is there a quick way to see that these give the same answer?

I think I am asking for a proof of some easy case of the Baker–Campbell–Hausdorff formula

[jaunty sparrow]

You can find a proof of this result in "Symmetry, Representations, and Invariants" by Goodman and Wallach

[void plank]

Let f: C ---> D be a chain map of chain complexes of finite dimensional vector spaces

I can take the dual of the entire diagram, which gives the cochain map f*: D* ---> C*

On the other hand, since the boundary maps obey partial_i \circ partial_(i-1) = 0 = partial_(i-1)^T \circ partial_i^T, could we not also have a cochain map g: C* ----> D*?

If so, what's the relationship between f* and g?

[fierce steeple]

Not sure what you mean, like how are you defining g even if our complexes are in degree 0?

[near lantern]

Is there an analogue of Schur--Weyl duality for other semisimple/reductive groups in place of GL_n with their Weyl group in place of S_n?

[void plank]

Actually, disregard my question. My chain map in my application has a lot more specific structure where several of the vertical maps are identical. So my original question is irrelevant

[soft parcel]

I feel like my approach is wrong here. Could anyone steer me in the right direction?

[fierce steeple]

I'm not really sure what you are doing in the first line

[soft parcel]

If M is faithfully flat, then the exactness of the sequence gives N = 0

[fierce steeple]

Why should it be exact?

[soft parcel]

It's 0 everywhere

[fierce steeple]

I mean like what is your original exact sequence

[soft parcel]

0 -> 0 -> N -> 0 -> 0

[fierce steeple]

That isn't exact

[soft parcel]

I know

[fierce steeple]

?

[soft parcel]

But if M is faithfully flat, then the latter sequence gives that this would have to be exact too

[fierce steeple]

Oh wait now I see what you mean lol

That is cool

Sorry I am silly

[soft parcel]

no worries

happens

[fierce steeple]

Just I would make it more clear what you are doing