advanced-algebra

[lone jacinth]

You might start by describing maps
cone(id_A) -> B

[solar turret]

in ii) we are first localize B at A/p as A-module then defining ring structure on it, right?

[sly rune]

well, sort of? the relevant map is injective but when the question says “B/𝖇 integral over A/𝖆” you need to have in mind a map A/𝖆 → B/𝖇 for the definition to even make sense

[solar turret]

Yeah

[solar turret]

What about it?

[sly rune]

A/p?

a multiplicative subset in A is also a multiplicative subset in B

[solar turret]

Oh my bad

Thank you

[hard kite]

Let R be a Dedekind domain and I an R-ideal. Does there always exists an integral extension S of R such that the extesnion IS of I to S is principal? If I has finite order in the ideal class group of R then this is true. What about the general case?

[sacred sentinel]

The ideal class group is finite right?

Ah right this needs number field stuff, ignore me

[hard kite]

I think it is not true

If I becomes principal in an integral extension S then [S:R]I=0 in the class group i think

using the relative norm map

[waxen fractal]

in a variety (in the sense of universal algebra) let 1 be the final / trivial algebra. can we ever have a non-identity epimorphism 1 -> A for some algebra A

[lone jacinth]

Consider the identity on A and the map A -> 1 -> A.

These are only equal if A = 1

This is just true in any category really.

Any map 1 -> A is split mono, and split mono + epi = iso

[sleek zenith]

Is there a combinatorial interpretation for the concept of projective modules/resolutions and injective modules/resolutions?

What I mean by this is the following.

A free module over a ring A with generating set S can simply be thought of as the set of all possible strings formed by the symbols on S, subject to rules mimicking the axioms of an A-module.

It is free in the sense that the symbols in S satisfy no non-trivial relations.

A free resolution of a module M can be thought as giving a description of it in terms of:

1. Generators.
2. Relations.
3. Relations between relations.
4. Relations between relations between relations.
5. etc...

Are there similar interpretations for the concepts of projective modules/resolutions and injective modules/resolutions?

For instance, in what sense is a projective resolution giving a combinatorial description of a module?

Same goes for injective modules/resolutions (which seem much harder to interpret conceptually).

[ornate atlas]

Is that specific to free resolutions? Like isn’t this just the Dold Kan correspondence?

[sleek zenith]

What the fuck is the Dold Kan correspondence?

[past cove]

😭

[ornate atlas]

The categories of simplical R modules and non negatively graded chain complexes are equivalent

This is a surprising fact, but essentially just true by what you said lol

[ornate atlas]

https://math.uchicago.edu/~amathew/doldkan.pdf

this is a nice exposition

But yeah like I could be wrong (it’s late and I’ve not thought about this too much) but I think you may just have rediscovered the Dold-Kan correspondence, if that’s at all helpful

[sleek zenith]

Idk, this doesn't seem to help me much

Like

Let me put in this way

Algorithmically, what's the difference between a projective and a free resolution?

When you find a free resolution of a module, in a sense what you did is manage to describe the module using generators and relations using minimal information.

[past cove]

depending on the category free resolutions might not always be available

[ornate atlas]

As a possibly even more unhelpful answer, projective resolutions are the cofibrant objects in chain complexes, so they’re essentially like CW complexes. For the purposes of homological algebra, you can essentially just replace anything with a projective resolution like you would a CW complex in algtop, and these are nice maps that you understand pretty well

[sleek zenith]

So I guess a projective resolution must have some sort of redundancy in the algorithimic/combinatorial description of the module.

What redundancy is that?

[sleek zenith]

I am just thinking about modules here

Abelian groups even

[sleek zenith]

Hmmmm idk

Like

This seems to answer a conjecture I vaguely made up to myself.

About how chain complexes could be thought of as a categorification of the inclusion-exclusion principle.

[sleek zenith]

The inclusion-exclusion principle is almost like

[ornate atlas]

I’m not sure I have a great answer to this, but maybe as places that are good to look, Miller and Strumfels Combinatoral commutative algebra? They talk about minimal free resolutions and things like that, and @golden osprey may have something interesting to say about stuff like that (if not sorry for the ping), I know he does combinatorial algebra stuff

But yeah I think the main perspectives I have on why projective resolutions are nice are kinda homotopical and probably not that helpful here

[sleek zenith]

Intrinsically built up inside the notion of a simplicial complex.

Because they are build up from a iterative process via gluing of simpler pieces.

[ornate atlas]

But it does sound like what your talking about is at least somewhat related to the idea of simplical objects in a category, and that’s what dold Kan is about

[sleek zenith]

This seems to vaguely be saying that all chain complexes are somehow related to the much more down to earth chain complexes which appear from the combinatorial data describing the gluing of (n+1)-faces into the n-skeleton of a simplicial complex to build the (n+1)-skeleton.

[wanton spoke]

<@&268886789983436800>

[sleek zenith]

Namely, the simplicial chain complex which you use to build simplicial homology.

And which is very explicitly just describing the inclusion-exclusion principle.

[waxen fractal]

oh i forgot to reply to this. this makes sense

[golden osprey]

I got nothing sorry @sleek zenith

[waxen fractal]

r ○ s = 1 [s is split mono]
s ○ r ○ s = 1 ○ s
s ○ r = 1 [s is epic]
yea

[golden osprey]

The books I'm thinking of are
Cox Little and O'Shea, Using Algebraic Geometry
Bruns, Conca, Raicu, and Varbaro - Determinants, Grobner Bases, and Cohomology

But those books seem to exclusively focus on free resolutions

[sleek zenith]

Makes sense

[golden osprey]

Honestly may have better luck asking on MO

[sleek zenith]

Algorithmically speaking, free modules just seem much more intuitive.

[golden osprey]

If you don't find anything in those two texts

I'm surprised the second book doesn't seem to talk about projective resolutions at all

[hallow bone]

free modules are also just a great (hell, functorial) example of a projective object with an epi to your desired object

[golden osprey]

Lots of good info in that text usually

Alot of the classical enumerative combinatorics is really asking stuff about cohomology of various varieties so like

Alot

See: any intersection theory text

Probably

[sleek zenith]

HOMOLOGICAL ALGEBRA IS THE CATEGORIFICATION OF THE INCLUSION-EXCLUSION PRINCIPLE

Unironically

[golden osprey]

That sounds like a fun talk for a grad student seminar

[sleek zenith]

One day I want to write a book on homological algebra and algebraic topology which takes this seriously.

[ornate atlas]

I need to sleep so I can’t say too much more (and don’t have any definitive answer to say anyway), but the thing about objects, maps between them, maps between the maps (syzigies/homotopyies), maps between maps between maps etc is exactly what simplical objects capture, and this is closely related to chain complexes through that correspondence, so that may be helpful.

As for why projective resolutions etc are so nice, well it’s kinda what I said before. They always exist, and up to a certain weak level of equivalence you can just replace shit with projective resolutions and it’s fine, they’re like CW complexes but for homological algebra (in a precise sense, I mean they’re cofibrant in the projective model structure on Ch(R))

[sleek zenith]

That's literally what I had in mind in doing lol.

But first

I will write a blog post about it.

[ornate atlas]

I unfortunately can’t say anything particularly helpful, but maybe that provides something helpful to look at, if not idk best of luck lol, I’m interested to hear what you work out though

[golden osprey]

Ooo blog

[ornate atlas]

I saw you posting about the inclusion exclusion thing the other day

[golden osprey]

Can you send me your blog (DM if you want)

[sleek zenith]

Oh, I am still building it.

But I already have some posts in mind.

1. Exposition on Homological Algebra via examples. (How Homology Explains How Parts Create a Whole)

2. The Wigner Theorem in Relativistic Quantum Mechanics.

3. On the different notions of continuity in mathematical analysis.

4. How Symplectic Structures Model Classical Mechanical Systems, and More.

[golden osprey]

Are you rolling your own blog system

Cause if so

I have advice as someone who did that

Which is don't

[sleek zenith]

Lmao

I will use github

[golden osprey]

Smart

https://github.com/Spamakin/abel

This was dumb

You can see that it's been so long since I've updated it because I don't want to

So fucking stupid

[golden osprey]

This is all I learned

Which honestly

Valuable lesson

[cloud karma]

yeah I came to the same conclusion

[waxen fractal]

wait but i already did for a year now

[mystic rivet]

Hi guys i am new.

@mystic rivet

What is the area of the Red Triangle

[lone jacinth]

Depends a little on the ring I guess.

Like if you're working with a path algebra, then the protectives are kinda like free modules at a vertex and you can think about it as local generators and relations.

In general I think projective modules are a bit more abstract. A better way to think about it might be that if you take a free resolution you typically get a lot of contractable summands which then encodes redundant information. So it makes sense to trim of those summands, but then you're left with a projective resolution, since summands of free module need not be free.

Injective resolutions are in some way nicer and in some way more complicated I think. Nicer in the sense that you always have a minimal injective resolution. But more complicated in that injective modules are more complicated.

If you're over a commutative Noetherian ring then every injective module is a direct sum of injective envelopes of R/p for a prime p.

Breaking the injective envelope of M into such a sum you find an essential submodule of M isomorphic to a sum of R/p. This feels a bit like a generating set to me, and then further you are describing how the injectives get glued together so that's like relations, etc.

[distant harness]

That's not relevant in this channel. Try #geometry-and-trigonometry -- or perhaps #competition-math or #recreational-math, since it doesn't look like a run-of-the-mill geometry problem either.
You may want to state explicitly if ABCD is assumed to be a parallelogram, and whether the line to the right of "72" and "8" is guaranteed to be parallel to the on to the left of "79".

(It's a cute problem, but it still needs to go in the correct place).

[mystic rivet]

sorry, i didn't know

[distant harness]

Now you do. :-)

[waxen fractal]

can we have a variety of algebras V, A, B ∈ V, and a monoepic map f: A -> B with |A| = 2 and |B| > 2?

[jagr answered my question on this yesterday with the number here being 1 instead of 2]

(admittedly this might be way harder but also i was genuinely stumped for the "1" version and it ended up being kinda trivial so i might as well ask again)

{if anyone is curious, the answer for "3" is "yes"}

[lone jacinth]

What's an example in that case?

[waxen fractal]

im glad you asked, wasn't sure if ppl actually wanted to hear. yea, take the variety of bounded distributive lattices, and consider the chain 0 < x < 1 embedded in {0, x, ¬x, 1} (isomorphic to 2^2)

this works cuz complements are preserved by bounded distributive lattice morphisms

a complement of x is defined as ¬x such that x ∨ ¬x = 1 and x ∧ ¬x = 0

[sleek zenith]

Wdym by "contractable summands" and in what sense do these encode redundant information?

[lone jacinth]

The contractable summands will be (sums of) things off the form

0 -> F = F -> 0

They are redundant because they essentially introduce a generator and then also add that generator as a relation, so tell you nothing

(and contractable means homotopy equivalent to 0)

[sleek zenith]

OOOOOOOOOOOOH

:)

Very interesting

This is the kind of down to earth interpretation I was hoping for

Thx

[sleek zenith]

I am still trying to understand what you said about injective modules/resolutions more intuitively tho

Hmmmm

From what I understand

Taking the sheaf theoretical perspective

Injective resolutions appear when you are studying extension problems, right?

Say you cover your space by some finite open sets.

Then you successively study the problem of extending a function that is initially defined only on union of the intersections of k+1 of these open sets to the union of the intersections of k of these.

Can you interpret injective resolutions more abstractly in this way too?

Like

I know that injective modules are those for which extension problems are always solvable.

Namely, given any inclusion X < Y of a smaller submodule X into a bigger module Y, and a homorphism f : X -> I into an injective module, then there's some extension F : Y -> I.

[sleek zenith]

So like

Can you interpret injective resolutions as sort of successfully trying to solve an extension problem?

Like

0 -> M -> I_1 -> I_2 -> I_3 -> ...

Where I_1 measures the "first level of obstruction" to extending a map f : X -> M to a map F : Y -> M, for X < Y. I_2 measures the "second level of obstruction" to extending a map f : X -> M to a map F : Y -> M and etc...

I am pretty sure there is

But like

I just don't know how to make this interpretation independent of the particular extension problem being studied.

Namely

It seems that the existence of an injective resolution is pretty much directly tied to the choice of a particular extension problem f : X -> M, which at first seems kind of arbitrary to me.

Dually, free and projective resolutions seem much more analogous to the topological problem of decomposing a space by smaller and smaller pieces which are then iteratively glued together.

[solar turret]

In theorem 5.21 i don't get how k' = k[\bar x ] ?

[past cove]

are you familiar with tensor products

[solar turret]

Yes

[past cove]

because this is just (B/m)[X] = B/m (x)_B B[X] = B[X]/mB[X] = B[X]/m[X] = k'

[solar turret]

But k' = B[X]/m'

[past cove]

Oh okay I misread

[near lantern]

B' is generated by B ∪ {x}, so the same remains true after killing m'. But the image of B and x in k' = B'/m' are B/m = k and \bar{x}.

[near lantern]

Fix a finite crystallographic root system with a base. For any dominant weight w, let D(w) = {w': w' is dominant and w-w' is a non-negative integer linear combination of simple roots}.
(i) For wi a fundemental weight, is D(wi) = {wi} always?
(ii) Is it always true for v in D(w) that there is a chain v = w0, w1, ..., wn = w with all wi's dominant and wi - w_{i-1} a simple root?
(iii) Under what circumstances can I assert D(w1) + D(w2) = D(w1+w2) (note that the ⊆ direction always holds)? If w1, w2 are both in the root lattice?

[olive schooner]

(i) is false. if you look at the C_2 root system, one of the fundamental weights is a_1 + a_2 (where a_1 and a_2 are the simple roots). So D(a_1 + a_2) also includes 0

(ii) is false. let w = omega_1 + omega_2 in type A_2 (where omega_i is ith fundamental weight). thus, omega_1 + omega_2 = a_1 + a_2 (a_i are simples). now consider the chain 0 = w_0, w_1, w_2 = w. then w_1 in your chain would have to be either a_1 or a_2 and neither are dominant

(iii) im not sure tbh, but restricting to the root lattice is definitely not strong enough. in type A_2, we have D(a_1 + a_2) = {0, a_1 + a_2} so D(a_1 + a_2) + D(a_1 + a_2) = {0, a_1 + a_2, 2(a_1 + a_2)}. but then D(2(a_1 + a_2)) contains 3omega_1 since

2(a_1 + a_2) - 3omega_1 = 2(omega_1 + omega_2) - 3omega-1 = 2omega_2 - omega_1 = a_2.

in fact, I wonder if (i) might even be type A specific. in type D_4, if you let the simples be a_1, ..., a_4 with a_2 the valence 3 vertex in the Dynkin diagram, then one of the fundamental weights is a_1 + 2a_2 + a_3 + a_4 so you also get 0 in D

[olive schooner]

for (i), an additional condition that you could add is for w_i to be minuscule (<w_i, alpha> <= 1 for all positive coroots) by Lem 30.2 + Prop 30.4 (Q_+ is the positive part of the root lattice) of these notes: https://ocw.mit.edu/courses/18-755-lie-groups-and-lie-algebras-ii-spring-2024/mit18_755_s24_lec_full.pdf

type A is special because all of the fundamental weights are minuscule

[near lantern]

Wow my brain was really off

Thanks!

[near lantern]

Ah this is the thing where you only take fundamental weights lying in the affine fundamental chamber / corresponding to the simple roots with coefficient 1 in the highest root?

Oh of course this is necessary and sufficient. I realised it was necessary for simple roots but didn't think to just treat other roots as roots instead of sums of simple roots.

OK, that's nice!

[near lantern]

Hmm so re (iii), the key feature is that the ability to subtract a2 and remain dominant "appears" for 2(a1+a2) without appearing for any of its "dominant summands" m omega1 + n omega2 (0 ≤ m, n ≤ 2). I suspect if we take w1, w2 with all even coefficients (i.e. in 2 WeightLattice) then the result might be true.

[near lantern]

Maybe (ii) is true if we increase the set allowed? E.g. there is a chain with differences positive roots, or elements of some even larger (but finite) set.

[solar turret]

Now i got it, thank you

[waxen fractal]

i got an answer from MSE answering this! the answer is yes

i can give details later if people are interested, rn my body is a walking corpse

[edgy pond]

how does one prove that sp4 is simple

[wheat meadow]

if you mean the complex lie algebra, then you just do the usual root decomposition stuff and see that it's of type C2

[edgy pond]

i see is there a more elementary way

without root decomposition

[wheat meadow]

not really -- root space decomposition is the key tool in understanding the structure of complex semisimple lie algebras

[last talon]

I mean you can just manually check what the ideals can be

[wheat meadow]

yeah but that is approximately as painful and way less principled

[last talon]

At the level where you’re asking about just sp4, I get the feeling that you’re supposed to do it that way so you can later see “oh, I can do it this much cleaner way now”

[edgy pond]

is showing that the standard representation of sp4 is simple easier?

[waxen fractal]

update: i also baited Martin Brandenburg into asking a related and interesting question apparently

[waxen fractal]

answer details: https://math.stackexchange.com/a/5133606/1370674

Brandenburg's new question: https://mathoverflow.net/q/510431/533379

[hushed bone]

Doxxed doxxed doxxed doxxed doxxed doxxed

[waxen fractal]

bro idc visit me irl if you want 🥀

im tryna build my personal brand, see?

btw mais is kinda a chad

wait does my github profile even dox me

oh i forgot it links to my personal website

welp i guess i should just like. view this as my 100% public account

lemme just go over the servers im in to make sure there's nothing too bad about this

freaky servers... nah im fine w that

gaming servers... hmmm idk if i really want that

[last talon]

G*mer

[wanton spoke]

send localization lol

[wheat meadow]

k[[x,y]]/(f,g)

[fallow zealot]

Can anyone help me understanding the definitions of pushouts and its properties? I've been reading Rotman's book on homological algebra for three hours straight and his definition is not clear for me

[rose mirage]

they're the colimit of a span

to actually be useful: the pushout of what exactly?

[fallow zealot]

He was only speaking about the pushout with sets with only three elements, and two of them are not comparable,

In modules

[rose mirage]

yeah so that's the colimit of a span, which I'll explain in more detail

[fallow zealot]

But he was also was speakingabout other cathegories, like abelian groups or topological spaces and so on

[fallow zealot]

Sorry, but what is your definition of span? I imagine that is not similar to that one from linear algebra

[rose mirage]

so the heuristic is that a pushout, like all colimits, are gluing along morphisms. What I mean by this is you quotient the coproduct by a particular equivalence relation which ensures that all of your morphisms agree

[rose mirage]

category of shape * <- * -> *

I find this article quite useful https://stacks.math.columbia.edu/tag/002U

[fierce steeple]

Almost completely unrelated

Beyond the meaning of the word span

[rose mirage]

colimits in almost every category you encounter will look like this, just instead of taking the straight equivalence relation you'd need a stronger condition, so for example in modules the pushout of, say A <- X -> B with maps f and g you'd take the submodule generated by relations f(m)-g(m) for m in the coproduct of A and B

[rose mirage]

this hopefully makes this "gluing" notion a bit clearer

[past cove]

you should take a look at the topological case of pushouts where they correspond to gluing spaces

[fallow zealot]

I saw it relates with Van Kampfem theorem

[past cove]

it's more basic than that

[fallow zealot]

Yes, thanks

[past cove]

gluing of cells in a CW complex is a pushout

[rose mirage]

yeah van kampen's theorem is a bit more complicated cause you're taking the pushout of non-abelian groups which is gross (although granted not that gross as the maps involved are just inclusions)

[past cove]

this is the picture I think you should have in your head imo

[fallow zealot]

Ok, thanks

[fallow zealot]

He used a similar equivalence relationship to state that every direct system in R modules has a colimit

[rose mirage]

oh god I hate the direct limit terminology. That's the colimit of a chain right

chat?????

[distant harness]

Yes, or more generally the colimit of a directed diagram.

[ornate atlas]

Direct and inverse limit are terrible names that need to die out

[fierce steeple]

Or a filtered diagram

[ornate atlas]

They seem to be dying but it needs to happen faster

[wheat meadow]

hot take i actually kind of like them

[fallow zealot]

Oh god, I'm kind lost to be honest. I never took an algebraic topology class, so the examples are making me confuse to be honest. I just have a notion of homology and cohomology by a course in differential topology that I did last semester

[fierce steeple]

I like them but probably just grown too acquainted lol

[rose mirage]

ur a chud

[ornate atlas]

This is why topologists shouldn’t be given rights

[fierce steeple]

Just like N and N^op indexed limits/colimits are very common

[rose mirage]

tbh this is more of a category theory construction than a topological one

[fierce steeple]

Here I am speaking more through AG aha

[rose mirage]

just compute a few examples to get a feel of how they work

[ornate atlas]

That’s worse, they don’t even know how to index co and contravarient functors

[wheat meadow]

but idk they're names that are very descriptive of what is happening

[fierce steeple]

What lol

[ornate atlas]

h^* for homology

[fierce steeple]

Idk what you mean lol

[wheat meadow]

every example i've seen in AG has been just H_n for homology and H^n for cohomology

[rose mirage]

I genuinely just write the star in superscript or subscript based on a coin toss I genuinely do not care for homology

[ornate atlas]

All the AG people I’ve spoken to denote homology like H^n and cohomology by H_n

[fierce steeple]

I have never seen this in my life

Lol

[last talon]

[wheat meadow]

other way no

[fierce steeple]

We always do cohomology by H^n

[ornate atlas]

Is Warwick just like some sort of backwards hell hole?

[rose mirage]

no shit sherlock?

[last talon]

Do you need to ask that?

[ornate atlas]

I mean yes but in other ways too

Mathematical ways

[wheat meadow]

yes, but also all of the ww people i've spoken to do it the correct way so idk what ur talking about

[ornate atlas]

I feel like I’m going insane

[fierce steeple]

Yeah I have interacted w Warwick ag lol

[ornate atlas]

I swear I’m not making this up

[last talon]

How?
Warwick doesn’t exist?

[fierce steeple]

No that's Bielefeld

[wheat meadow]

tbf warwick doesn't exist in the sense that there is no warwick at or near the university of warwick

[fierce steeple]

Sorry I mean Leamington Spa

[last talon]

Why not both?

[wheat meadow]

warwick doesn't locally exist

[rose mirage]

Shithole

[ornate atlas]

It’s in fucking Canley, leam is still 40 minutes away

[fierce steeple]

Literally who

[wheat meadow]

eh no leamington spa is actually pretty nice

[ornate atlas]

Exactly

[rose mirage]

I’ve been there and theres no soft play area

[last talon]

Nothing exists except Oxford, Cambridge, and the former east-west railway

[distant harness]

Warwick and Bielefeld are secretly the same place, linked by wormholes in the town square.

[ornate atlas]

Apparently Warwick castle is nice but I wouldn’t know because it’s fucking ages away

[wheat meadow]

relatedly all mathematicians are in oberwolfach, if people's website photos are to be believed

tho until i go there i will probably have to make do with a photo in qmul's building lol since there's a great view of the city from there

[ornate atlas]

Anyway yeah sorry that we took over here lol, I found pushouts confusing too at first, have you taken a look at like Reihls book? There’s a good few examples in there iirc

But yeah I just think it’s best to try to understand colimits well, and specialise to whatever shape of diagram

You’re right that SvK is just an example of a push out though, just a bit of an annoying one

[fallow zealot]

It's ok dude, the talk was really fun, but I'm only reading Rotman's book

I'll take a look on Reihls book

[ornate atlas]

Yeah It may honestly be worth just learning about colimits more generally if you’re comfortable with the idea of a category, it’s what i found to be the most helpful. Or like if you know what pullbacks/fibre products are, just think about the opposite of that (which is like half a joke, they are dual but you need to think about what this actually looks like)

[wheat meadow]

our first homological algebra lecture was a crash course on cat theory and the lecturer was basically like "you should all go and read category theory in context instead of trying to get it all from this 2 hour lecture, but that's a big book and i have to teach this at some point so this is how it's happening"

[ornate atlas]

I wish I had a homological algebra course

I just had to suffer through Weibel on my own during my UG thesis

(Though that was kinda entirely on me, bit if a side quest when I needed the 5 lemma)

[last talon]

why do Homalg more general than like module categories /silly

[ornate atlas]

Because it’s what gets me a PhD

[last talon]

Hmmmm
Have you considered:
better topic

/silly

[ornate atlas]

My mathematical career centres around triangles at varying levels of abstraction

[last talon]

Mine centres around lööps

(I am a fundamental groups merchant 🙃 )

[fallow zealot]

I'm doing electrical engineering, so my mathematical career has not started yet

[ornate atlas]

Ah yeah that seems like quite the tangent lol

[wheat meadow]

Primary school maths topic 1: shapes

This is a circle ⚪
This is a square ⬜
A triangle is a distinguished triple of objects and morphisms X -> Y -> Z - > X[1]

[last talon]

A square is a 4 sided triangle

[rose mirage]

a triangle is a size 6 subset of an incidence structure

[wheat meadow]

somehow i do not think ofsted will like the educational methodology at the Kan academy \silly

[last talon]

Start with infinity categories :3

[wheat meadow]

i mean all concepts are kan extensions so maybe those deserve two days

but yes

[fallow zealot]

Sorry to be silly, but can I ask something?
The pushout of a system like this, with Three elements that i <= j and i <=k is the quotient of $\oplus M_i \oplus M_j \oplus M_k$ quotiented by the submodule generated by $(-m, {\varphi_j}^i(m), 0) and (-m, 0, {\varphi_k}^i(m))$ for all m in $M_i$ ?

[broken turtle]

Phil

[past cove]

Yes the idea is that you glue Mj and Mk along the image of Mi in both

[last talon]

Yes
You can simplify it a bit, but that works

[fallow zealot]

Ok, I need to think because is not making sense to me, but thanks for confirming

[fallow zealot]

Using only two modules?

[last talon]

Yeah

[fallow zealot]

thanks

[fallow zealot]

$M_j \oplus M_k$ quotiented by the submodule generated by $({\varphi_j}^i(m), {\varphi_k}^i(m))$

[broken turtle]

Phil

[last talon]

I think there should be a negative on one of those

[fallow zealot]

why?

[hallow bone]

homological algebra is trivial

[hallow bone]

because you want to identify them

identifying a with b is quotienting by a - b

you want to identify phi^i_j(m) with phi^i_k(m), essentially, so you quotient by (phi^i_j(m), -phi^i_k(m))

[fallow zealot]

Ah, perfect, thanks @last talon and @hallow bone

[hallow bone]

logic is just knot theory

or something

idk i saw something about using knot theory for logic

RAHHHHH ribbon hopf algebras

quantum enveloping algebras

my beloved

[drowsy flame]

Tell me more I wanna know more

[hallow bone]

lowkey dont remember

let me see if i can find anything

okay a paper i found is ass

wtf

just nonsense

[primal grail]

in etingof one of the problems is to show that End_A(A) is iso to A^op but End_A(A) is defined to be all intertwining operators between A and A. is it safe to assume that these must be between the regular representation?

the usual method when defined as A-linear maps is to make the standard computation x mapsto x phi(1) = phi(x)

but it seems slightly difficult to generalize this for an arbitrary representation

[scarlet ermine]

are intertwining operators not just exactly A-linear maps, and your argument works? What is A here, just a ring? (maybe we should also be careful about left/rightness, and showing that composition is multiplication backwards to get the op carefully)

[primal grail]

it was defined as "commuting with the representation", i.e. phi(f(a)(v)) = g(a)(phi(v)) where the representations are given by f and g

so not explicitly commuting with everything in A

is this somehow equivalent to being standardly A-linear?

[scarlet ermine]

oh I understand your question now. Yes, A is the regular representation (if you are thinking of A as being kG or something). In order for End_A(M) to make sense, M is already equipped with an A-module structure (that is, M is a representation of A), and when we write End_A(A) we are considering the A in parentheses to be the abelian group underlying A, together with an A-module structure by multiplication coming from A, where A is a ring

assuming you mean A is a ring (like kG for example)

[primal grail]

associative algebra

but close enough

[scarlet ermine]

here you really get to blend the module language and representation theory language

associative algebras are rings

[primal grail]

yes

[scarlet ermine]

oh then we agree haha I wasn’t sure if you were stipulating anything

[primal grail]

yeah it's mostly been a few years and i just wanted to re-confirm, esp since End_R(M, N) or similar was not defined prior to the exercise as specifically being R-linear

[scarlet ermine]

maybe I’m confused what you mean, the R in the subscript means R linear

[primal grail]

it was all quite arbitrary here

and when revisiting i realized that it only works if you took the regular representation, and did not quite make sense when you took the definition of End_A(V) as is and any arbitrary representation

[scarlet ermine]

maybe this is a clarifying perspective:

representations = modules

If you are coming at this from a background in group representation theory, this is about being able to translate between

a rep, which is a group hom
G->Aut_k(V). You could also boost this up to a ring hom kG->End_k(V) by k-linearity.

and a kG-module V, which is a k vector space V together with a “scaling map”
kG x V -> V.

More generally, when you have a k-algebra A, the data of a left module is equivalent to either

a ring map A->End_k(M)

OR a scaling map A x M -> M (satisfying axioms like associativity)

so the “intertwining operator” is the perspective that we should consider A representations to be pairs (M,f), where M is a k-vector space and f is a ring hom A->End_k(M)

the A-linear perspective is that we should consider A representations to be just A-modules as usual. Now, you should check that under this correspondence, intertwining operators are exactly A-linear maps.

and then the regular representation business is saying that A as a representation is always taken to be the k-vector space A together with the map
f: A -> End_k(A)
sending “a” to multiply by “a” on the left. You should check under this correspondence, the regular representation is exacted the usual A-module structure on A by multiplication

[primal grail]

under this correspondence, intertwining operators are exactly A-linear maps
this is probably what i need to check

i.e. going from here to it is also A-linear

[scarlet ermine]

mhm

at the end of the day you are doing module theory, they are just using representation theory words, and it’s good to see that these are the same

I guess one thing that I think is confusing at first when you start thinking about this correspondence is that often people will write representations as pairs

(V, f), where f is the map from A->End_k(V) (or in group rep theory, f: G-> Aut_k(V)),

but when you think of V as an A-module instead, people don’t write the module as a pair (V, m) where m is the scaling map m: A x V -> V, even though it is an essential part of the data. People just leave it implicit and only write the symbol V to mean both the vector space/abelian group AND the data of the scaling map.

[primal grail]

oh when we mean A-linear we don't necessarily mean by the natural multiplication given by A, but rather by the module structure

[scarlet ermine]

well there is no “natural” multiplication necessarily, you need to get it from somewhere

like when they write a * v

they mean use the scaling/multiplication map A x V-> V coming from V being an A-module

I can’t make sense of a*v otherwise

[primal grail]

yeah i just meant when you also take V = A, so there is already a natural A-multiplication on top of whatever additional A-module structure you decide to add

[scarlet ermine]

right, but when we write A we do mean to give A the usual A-module structure which is multiplication

it would be very bad style to write A to mean the k-vector space A with a different scaling.

(at the end of the day, A has an underlying vector space k^n, and it would be bad style to write A when the only thing you were actually borrowing was the underlying vector space k^n rather than the “interesting bit” which is multiplication)

if you were only borrowing the vector space, just write k^n or invent a new letter like V

[primal grail]

right. i just took the "let V be a representation of A" plus all homomorphisms of representations V -> V to mean we can pick arbitrary representations for the left and right V. however, there should only ever be one representation considered here and the choice V = A implies the usual multiplication for End_A(A). if we instead said let V = A but the action is to multiply twice i.e. a * b = aab, then End_A(B) should not necessarily be A^op

[scarlet ermine]

yep!

[primal grail]

ok

that clarifies a lot

[scarlet ermine]

(small tidbit but your example won’t work because a * b = aab won’t be a representation in general because it won’t have distributivity, since (a + c)*b wouldn’t equal a*b + c*b unless we are in characteristic 2… you can still find other sorts of examples in general though or be happy with characteristic 2 haha)

[primal grail]

ah

true

yeah it was mostly if we pick any other non-standard rep, the statement is not necessarily true

[scarlet ermine]

right, maybe if A=kG tensor, do any rep of smaller dimension than kG, and then direct sum a bunch of trivial representations or something until we get the right dimension

[primal grail]

related: i feel like there should be some natural isomorphism here as well but idk what to choose as a 2nd functor. A \to A^op is an easy first functor but im struggling to come up with a good way to functorialize maps f : X \to Y to maps End_X(X) \to End_Y(Y)

it seems fairly doomed because even if you consider having 3 things, X -> X, X -> Y, Y -> Y, i don't think you can use any 2 to get at the other

but phi -> phi(1)^op does feel fairly natural, so hopefully there is something somewhere

oh ig A -> A^op also fails as a functor when you try to expand f^op of a product, it seems hard to do

[scarlet ermine]

oh wait sorry I wasn't thinking, your op is the opposite ring not an opposite category

hmm

I think ^op is a fine functor actually, where f^op is actually exactly the same as f

because f(x *_op y) = f(yx) = f(y)f(x) = f(x) *_op f(y)

[primal grail]

oh right

when i did it in my head i missed bringing the *_op outside

and had it as normal *

but yeah, still struggling to see anything for End_X(X)

but it's 2am

maybe sleeping and a fresh mind will clear things up

[scarlet ermine]

wellllll if you believe the problem you showed that End_X(X) is isomorphic to X

and in fact you have a fixed isomorphism X->End_X(X) which sends x to multiply on the left by x

and so this is functorial in the sense that given an f:X->Y, and a phi in End_X(X), you get a psi in End_Y(Y) by just recognizing phi is mult on left by some x, then apply f to x, and then set psi to be mult on left by f(x)

but this is maybe a little silly

yeah maybe I need to sleep too lol

oh you can make End_{(-)}(-) into a functor Ring -> Ring by defining for f:R->S a ring hom, getting

End_R(R) -> End_S(S)
(phi: R->R) -> (phi (x)_R S: R (x)_R S -> R (x)_R S )

where you are giving S the structure of an R-module by f (i.e., extension of scalars along f). So if you wanted, you could see that this functor is naturally isomorphic to (-)^op, if you wanted to see this as some sort of natural isomorphism

but this will boil down to just pushing some definitions around, I think probably isn't very super interesting

[scarlet ermine]

Rings -> AdditiveCategories

sending a ring R to R-Mod. Then, there is a functor
AddCats -> Rings

which sends an additive category to the endomorphism ring of the identity functor. This is indeed a functor because given a functor C -> D between additive cats C,D, we can hit a natural transformation of 1_C with the given functor.

I claim that the composition of these two functors is what we want, since for C = R-Mod, the identity functor is Hom_R(R,-), and (Z-enriched) Yoneda tells us the endomorphism ring of this functor is exactly the ring Hom_R(R,R) = End_R(R)~~~

where I’m hiding the tensor because the way that Ring->AddCat was functorial was that given R->S, we get a functor R-Mod -> S-Mod by extension of scalars/tensor

hmm nlab claims the center is not functorial in general?? https://ncatlab.org/nlab/show/center+of+an+additive+category

[hallow bone]

do you mean functorial?

yes, that is true

center is only invariant under isomorphism, but not functorial

[scarlet ermine]

oh right if F:C->D is the functor, then i was only getting a natural transformation F=>F not 1_D=>1_D

[primal grail]

oh this comes from R (x)_R M = M i see

[outer lion]

So when it comes to injective resolutions, I kinda get how to "compute" them, you can just take injective hulls of quotients.

What about projective resolutions? Like if you hand me a module over a ring, how do I find a projective module that surjects onto it?

[eager hound]

that’s way easier

You take the generators

That’s a map from a free module

Then you resolve the relations

and so on

[outer lion]

oh, and a free module is projective. right. So all you need to do to compute this in practice is find generators and relations?

[eager hound]

well on most rings there will be higher relations as well

So you will have to resolve the kernel of the map from the relations free module to the generators one

[hushed bone]

nHail, projectives are way way way easier

In fact you can even make functorial resolutions (tho these are huge)

[eager hound]

on like a regular local ring I think you can even do this on a computer

[hushed bone]

By taking the free module generated on M itself, so it has a basis element for each element of M

And it sends the basis element associated to m to the actual element m

Take the kernel of that, and do this again

[outer lion]

Interesting. I suppose this is what happens when I show up to the injective week and not the projective week of class

[hushed bone]

If you do this you get an actual functor so maps M -> N lift to maps on the resolutions without even needing to mod out by chain homotopy

[outer lion]

yeah that would be huge but it does gurantee existence I suppose

okay, thanks

[hushed bone]

You can always take a generating set so if you’re Noetherian and start fg the resolution is always by finite free modules

Which is theoretically very very useful

Injectives are way more mysterious, and they have to be huge

If you’re a local ring and you have a fg injective module the ring is automatically Artinian local

Which is like n-steps away from being a field

[wheat meadow]

and ofc if you have a PID you have divisible iff injective which is at least some meaningful information

but yea over general ring injective modules are hard to say much about other than their homological properties

[outer lion]

Also, do you know of a good source that talks about Yoneda's construction of Ext^1 and the equivalence with the derived functor definition? I'm realizing the course notes on it are kinda cryptic

[hushed bone]

Also you don’t have to take injective hulls

[outer lion]

okay these kinda made sense to me, just mirroring the "divisible hull" construction in groups

[hushed bone]

Lowkey… work it out?

Or check stacks I guess

I think it might be tough to understand how to work it out

[outer lion]

stackoverflow or stacks project?

[hushed bone]

Latter

[eager hound]

I think the best way to think of an ext class is as a nontrivial map from Q \to N[1]

[hushed bone]

Ermmm what about by ext^2 class?

[outer lion]

yeah having f.g. local seems strong intuitively, a lot of the elements in your ring are close to being units somehow

[eager hound]

And from this it’s pretty easy to see the Yoneda version is equivalent

[eager hound]

I wonder what those could be…

the world may never know

[hushed bone]

TTEG u r so mean 2 me, its like your goal in life is to make me feel stupid

R u happy now? Do you wish I just thought wow I am no smarter than the monkeys in the zoo?

[eager hound]

you are smarter than the chairs at the zoo though

[hushed bone]

That’s true…

[eager hound]

that’s the benefit of crossbreeding a chair and a monkey

[hushed bone]

Actually when they took the specific chair and specific monkey for me to crossbreed the avg intelligence of both groups went down

[sleek zenith]

@eager hound I asked a similar question a few days ago and proposed an interpretion of injective resolutions in terms of trying to iteratively solving an extension problem, is this a good way of thinking about them?

[sleek zenith]

.

[lone jacinth]

Page 82 and onward here
https://opperman.folk.ntnu.no/HomAlg.pdf

[outer lion]

TY, I'll give it a look

[torpid rune]

Why are left-derived functor are named "left"-derived functors when it is from a right exact covariant functor btw? Is it because when you construct it, you lowk process to the left of the projective resolution? Or is it just one of those naming quirks that just stuck for no reason?

[hushed bone]

They go to the left

[lone jacinth]

Left derived functors measures failure to be left exact.

You don't have to apply the construction to right exact functors, it's just a little nicer to think about failure of left exactness when you don't also have a failure of right exactness to deal with

Right exactness just tells you that
L^0 F = F

[foggy galleon]

what is P^*-->J*(P*) ? Like it's a complex of complexes, no (for each j, a resolution of P^j)? But then what is the cohomology of Gamma(X, J*(P*)), shouldn't it be a complex? But it should equal H^p(X, F). Or do they mean the total complex or something

[lone jacinth]

I would think
P^* -> J^(P^)
is a quasi iso to a complex of acyclics. Which you could construct from making a complex of complexes and taking total complex if you wanted to.

[foggy galleon]

so just to be sure. We think of J*(P*) as a complex, yeah? So that J*(P*) otimes J*(G) is also a complex and J*(P*) otimes J*(G)-->J* is a quasi iso to a complex of acyclics?

ty

[torpid rune]

this lowk made sense

[fierce steeple]

Or possibly trivial

[ornate atlas]

Why is this the case exactly?

I suspect im just forgetting something basic from LA

[subtle plaza]

This comes down to simultaneous diagonalization: if you have a set of diagonalizable operators which commute then there exists a change of basis such that they all become diagonal

I guess I'm just restating what's in the text lol, but that's the key phrase to look up if you want to review a proof. If I remember correctly, you can do this for two operators by restricting one operator to the eigenspaces of the other and work from there or something like that

[ornate atlas]

No this is helpful lol, ive seen that before, I just didnt realise thats what was happening, but yeah makes sense, thanks!

This entire section is just very notation heavy, taking a while to wrap my head around

[hallow bone]

yeah, by commutativity an eigenspace of A will always be B-invariant

so you probably diagonalize B in that eigenspace
do this for all eigenspaces of A and youve got a simultaneous diagonalization

but im not sure how to generalize this to arbitrary sets of commuting operators

[subtle plaza]

I guess if V is finite-dimensional then so is End(V) so you can restrict an infinite set to finitely many elements that have the same span

[hallow bone]

right

[hallow bone]

of course this can be continued inductively taking instead the simultaneous eigenspace of A1, A2, ..., An

[plucky arch]

I think you can do this for arbitrary sets via induction on dimension

So you’d have the statement P(n) be “if S is a set of commuting diagonalisable operators in End(V), where V is at most n-dimensional, then it can be simultaneously diagonalised”

P(1) is true essentially by definition since every operator is diagonal

Suppose P(k) is true. Unless every operator in S is a scalar operator, there is at least one operator with at least two eigenvalues. Call this operator L

Then you can break up V into the eigenspaces of L, which are all proper subspaces

Moreover you can show each eigenspace is stable under S

Then you just use the inductive hypothesis on each eigenspace to simultaneously diagonalise S in that eigenspace

And that should do it?

[hallow bone]

that works too

cleaner than what I did lol

[lone jacinth]

I guess your argument covers finitely many operators on arbitrary space while pseudo covers arbitrary many operators on finite dimensional space

Whereas the arbitrary many operators on arbitrary space statement isn't true

[hallow bone]

that makes sense yeah

arbitrary spaces are not well behaved

[ornate atlas]

Infinity ruining all the fun as per usual

[hallow bone]

what do they mean by semisimple elements here?

just that ad(h) is diagonalisable?

[ornate atlas]

Yeah basically, that in the Jordan decomp the nilpotent part is 0

[hallow bone]

reading lie algebra stuff has taught me i lack some lin alg fundamentals

lol

[ornate atlas]

Yeah I’m definitely a bit rusty lol

I need to remember how to actually calculate the JCF of a matrix before this exam, I thought I was free

[hallow bone]

[distant harness]

Concretely: We can take space of all real sequences, and look at the family of operators that each kill all coordinates except one. Each of these operators is separately diagonalizable, and they clearly commute, but the only vectors that are eigenvectors for all of them are the ones with singleton support, and those don't span the entire space.

[hallow bone]

good example

[plucky arch]

Am I correct in thinking that for a finite dimensional vector space over an algebraically closed field, a linear operator is semi simple iff it is diagonalisable?

I seem to remember hearing something about how semisimplicity was a generalisation of diagonalisability

Not sure how accurate that is

[lone jacinth]

You also don't need findim for this

[plucky arch]

Hm

Maybe I need to review what diagonalisable actually means

[lone jacinth]

An operator is semisimple iff it is diagonalizable over the algebraic closure

[wheat meadow]

u using erdmann and wildon as well? i have come to profoundly dislike that book over the course of my lie algebras module lol

[lone jacinth]

There existing a basis of eigenvectors is a good one

[plucky arch]

Like I know what it means for a matrix to be diagonalisable, but not an operator

I see

[ornate atlas]

I think so lol, these are just the lecture notes but I think it’s what they’re based on

[wheat meadow]

makes sense

[ornate atlas]

The lecture notes aren’t fantastic either I don’t think

[plucky arch]

Is the statement unchanged in infinite dimensions?

[hallow bone]

does this have to do with the fact that simple k[X]-modules correspond to simple extensions of k?

[plucky arch]

Uh idk what that means sorry

[fierce steeple]

# linear-algebra jk

[wheat meadow]

i mean it's fundamentally a choice of content thing -- talking about just lie algebras and just their elementary theory is making out one of the most important and interconnected areas of modern algebra to be something niche, isolated, that we just classify for fun and do nothing with

[ornate atlas]

Yeah micoi shared their notes and they looked a bit better because they skipped a lot of proofs to have time to talk about some applications, and we have done like every proof in quite a lot of detail, and for the most part I’ve not found them to be particularly enlightening

[wheat meadow]

Also having toby gee teach the lie algebras course feels like... a massive waste lol. If ur gonna make him teach at least make him teach something interesting and fun

[distant harness]

I think for an infinite-dimensional topological vector spaces you get different concepts depending on whether you ask for a Hamel or Schauder basis.

[fierce steeple]

Well here I assume we were talking about vector spaces and not topological vector spaces

None of this should depend on R

[wheat meadow]

they had ana caraiani teaching number theory this year and it sounds like the course was much more interesting this time around as a result

[fierce steeple]

Is this the nt course which only does quadratic extensions lol

[hallow bone]

A k-vector space equipped with an operator is nothing more than a k[X]-module. Simple R-modules are always of the form R/m where m is a maximal ideal, so simple k[X]-modules are of the form k[X]/m where m is maximal, i.e. possibly degenerate simple field extensions k < k(a) where LMT by a is the operator.

[distant harness]

Yeah, but physics does a lot of simultaneous diagonalization on Hilbert spaces, so the difference may matter to Pseudo.

[fierce steeple]

Sure ye

[ornate atlas]

I learned number theory and commutative algebra from a (very lovely and incredibly clever) prof who started both courses by telling us about how much she hated number theory and comalg lol

Neither of them were great courses

[fierce steeple]

This is so funny lol

What was her field lol

[wheat meadow]

historically it was just like... pure elementary nt with Pell's eq and reciprocity and stuff it was honestly quite a sad course. I took it for free points ofc but it didn't really justify a third year module

[fierce steeple]

Lol sure

[wheat meadow]

alg nt was much more interesting because the course was written by george boxer so he made sure to include quite a lot of interesting and challenging stuff that got kept in at least until the year i took it

[ornate atlas]

Funky noncom algebra stuff, like set theoretic solutions to Yang Baxter

[fierce steeple]

Curious

[hallow bone]

braces

[ornate atlas]

Yes

[hallow bone]

weird stuff

[astral ginkgo]

Based

[ornate atlas]

Just wondering about showing that [x,[x,y]] (and [y,[x,y]] but itll be the same) are in M_a here. Am I right in saying that [x,y]\in L_0 = C_L(H) = H, so [x,y] = h\in H? I think this works but just to double check

Then just like [x,h] = -[h,x] = -alpha(h)x

(but im mainly checking the justification that [x,y] is in the cartan subalgebra)

[wheat meadow]

yeep

[ornate atlas]

Cool cool, that took me way too long so I wanted to make sure I wasn’t being dumb lol

[wheat meadow]

[L_a, L_b] is always inside L_a+b

which is useful if you know either that a+b = 0 or that a+b is not a root

[ornate atlas]

Yeah I was missing that for so long, then I remembered the last exercise I did was proving it

[wheat meadow]

ive been grinding past papers so hard lol ive done like 5 of them and i still haven't done any of the recent ones

turns out you can just google a lot of different uni's lie algebras exams and they're basically all the same

[ornate atlas]

This is good to know because I don’t have so many examples here and I think I need some extra practice (I mean I’m only just learning the course now, I ignored it all term)

[wheat meadow]

tho i've done "I and R/I solvable => R solvable" like 4 times now it's a really popular question lmao

[ornate atlas]

The thing that confuses/worries me about this is how much linear algebra I can assume

[wheat meadow]

i generally work under the principle that anything from linear algebra can be assumed

[ornate atlas]

Vs what I actually have to show, like the way products of (strictly) upper triangular matrices work or whatever

[wheat meadow]

which is consistent with the mark schemes i've seen from my uni and seems reasonable

[ornate atlas]

Yeah I assume anything I use would be fine but I’ll be upset if I get a random “why is this 0, prove it”

[wheat meadow]

i think just saying things like "if x and y are upper triangular then [x,y] is strictly upper triangular" is perfectly fine

they know you know how to show things like that

maybe if it was a bit more sophisticated there might be a mark for showing it, but worrying about that will cost more time than its worth

well at least if you're not at cam and therefore don't have to worry about stacking alphas lol

[ornate atlas]

Everything about that uni is nonsense lol

Yeah there’s a reason I never applied to part III lol

(Besides probably not getting in)

[urban birch]

Does anyone know how an Intro course to algebra at the graduate level would compare to a textbook like “Abstract Algebra by hungerford”? I just got into a masters program for math and I just am trying to plan some courses

[ornate atlas]

Hard to say generally, do you have a syllabus or anything?

[past cove]

I think something like Dummit and Foote is more standard

[ornate atlas]

D&F at masters level?

[urban birch]

I don’t have a syllabus unfortunately but I’ll try seeing if I can get one

[ornate atlas]

Surely you’d want something a bit faster moving if you’re doing a masters level intro

[past cove]

when I took graduate algebra in the US we used Dummit and Foote

tho it wasn't just like

all the stuff in dummit and foote it was more

idk how to frame this

we covered stuff like Sylow

er

a lot of ring theory ig

field theory

and some commutative algebra

but I think this is more a consequence of having really weak UG algebra courses in the US

[ornate atlas]

Yeah I guess, the concept of having an intro to algebra at a masters level just doesn’t really make sense to me to begin with so idk lol

[past cove]

well idk like

I don't think it's super standard to cover sylow stuff in UG on any level of depth

same with stuff like modules over PIDs and galois theory

we also covered stuff like localization

[urban birch]

My masters program “abstract algebra 1l detail is this

Arithmetic of the integers, unique factorization and modular arithmetic;
group theory including normal subgroups, factor groups, cyclic groups,
permutations, homomorphisms, the isomorphism theorems, abelian
groups and p-groups.

[past cove]

and the class was really fucking fast

[past cove]

uh are you sure this is a masters class and not a UG class

[urban birch]

It’s from university of Toledo, I feel like my undergrad classes covered these topics so that’s why I’m like wondering if its equivalent

And the next course in the sequence is “Ring theory including integral domains, field of quotients,
homomorphisms, ideals, Euclidean domains, polynomial rings, vector
spaces, roots of polynomials and field extensions.”
Which I feel like I covered relatively deep in a number theory class but not as sure

[last talon]

Having done a quick glance at that master’s, I’m not sure it’s worth the time of anyone who’s done a decent undergrad (ie, any undergrad that goes a few courses beyond the standard analysis/algebra series)

[urban birch]

Thank you for the input. I am a bit worried about the rigor of it. I applied because it was online and I still work full time.
My other option is csu Chico math masters which looks better at least

[last talon]

I think Toledo ends up covering more on average, although I’d get someone else’s input too

[torpid rune]

Is there a relationship between some construct of A and C and Tor_i(A,C)?
Like, there is one for the Ext functor, which is the Baer sum. But is there anything similar to that for Tor?

[fierce steeple]

Wdym by C here? Not sure what the question is

[wheat meadow]

there still isn't really a general "classification of X" object interpretation of Tor tho tbh

like the best i could give you is that if you have a presentation of $A$ then Tor classifies which relations in that presentation become trivial in $A \otimes B$

[broken turtle]

Dirichlet

[wheat meadow]

but that's certainly nothing like Ext^1 classifying extneions

[torpid rune]

Oh like A, C are R-modules.
and there is a way of interpretation of how Ext_1(A,C) measures how much when:
0 -> A -> B -> C -> 0 be an exact sequence that fails to be split exact. and you can extend it to Ext_i(A,C) as equivalent classes of exact sequence 0 -> A -> B_1 -> B_2 -> ... -> B_i -> C -> 0

Just curious if there is a similar thing with Tor

[torpid rune]

wym by presentation? Could you make that clearer for me?

[lone jacinth]

The closest I can think of would be how for a Noetherian ring

Hom_Z(Tor_i(A, C), Q/Z) = Ext^i(A, Hom_Z(C, Q/Z))

[Actually Noetherian shouldn't be necessary, just true in general]

[torpid rune]

is that an adjoint?

[lone jacinth]

It comes from the hom tensor adjunction sure

[torpid rune]

nice

[analog pumice]

Is anyone start solving Goldbach's conjecture

Every even number greater than 2 can be written as the sum of two prime numbers. Goldbach's conjecture said

[analog pumice]

I have another problem

About reimann hypothesis

[ornate atlas]

Please keep these channels on topic

[eager hound]

he might be using advanced algebra to prove the Riemann hypothesis

actually if there is a proof I would think it probably does use it

@alpine dirge please ban nope

[topaz oyster]

If you think about it, all modern math is advanced algebra

[eager hound]

they gave tox honorable what is the world coming to

[topaz oyster]

I was demodded for fighting for what I believe in, the right to dm green names slurs.

[eager hound]

the number of slur sayers on this discord is pretty high

[topaz oyster]

And the number of slur receivers is pretty low

Makes you think

Anyways, algebra

What the fuck is a semigroup

[hallow bone]

conjectures do not usually speak

maybe you should refrain from the gas station pizza next time

[distant harness]

They do, and it hurts them a lot that you don't listen.

[hallow bone]

something that should really be completed to a monoid

[hallow bone]

whuh ohhh so those are the voices i hear every day

[ornate atlas]

It’s true, I should’ve thought before setting maths back decades

[topaz oyster]

And then what?

I've been seeing the little turds more lately, and idk the general theory.

I assume, despite having a lot of parallel nomenclature, semigroup theory is significantly worse than ring theory?

[hallow bone]

smt smt greens relations

like you want some way to control how close it is to a) lattice b) group c) ring

i guess what you do with them just depends on what you need

[topaz oyster]

I see

[hallow bone]

this is a local example of general UA stuff

varieties of algebras can have very different "feels" to them, even if theyre all globally the same

for example, BCK-algebras feel ring-like, racks/quandles feel set-like, etc

the way of controlling this is called Malcev conditions

[spark ravine]

a hausdorff quasisemigroup

[hallow bone]

the shit is a quasisemigroup

[topaz oyster]

Jokes on you my semigroup is only semicontinuous

[astral ginkgo]

It's a joke on how "quasicompact" in AG means compact in normal people terms and "compact" means "quasicompact and hausdorff"

[hallow bone]

oh lol ty

[topaz oyster]

do you have any sources on greens relations btw?

[hallow bone]

i was looking around for it but got distracted lol

[topaz oyster]

understandable lol

I kinda suspect that the greens relations will be just... completely intractable on the semigroup I'm interested in

[hallow bone]

oh, how so?

you never know lol

[topaz oyster]

I mean I explained the motivation for algebra on beta-N to you the other day

it turns out you badly lose commutativity when you extend the semigroup structure over non-principal ideals lol

[hallow bone]

right

sounds horrible, hell yeah

[topaz oyster]

one of the worst things is we don't even know if it has non-idempotent elements of finite order 😭

[drowsy niche]

over a (semi)hereditary ring, is every projective module a direct sum of finitely generated projective modules?

[primal grail]

if A is an associative algebra, I a left ideal, and V a representation with an isomorphism to A/I (afaict this should be an isomorphism of vector spaces), is it true that they are also isomorphic as A-modules, given the standard A action on A/I (which i assume is a vector space quotient)?

ig if we upgrade the isomorphism to an isomorphism of representations i think it's pretty free?

i'm pretty sure the correct way forward is the image of 1 in A/I should generate V via the representation, but to do so i really need the action to commute with the isomorphism which i have not brought myself to upgrading just yet

[scarlet ermine]

No, they mean an isomorphism of A-modules. It is not true that V isomorphic to A/I as a vector space implies V is cyclic, since this only tracks dimension which is way too coarse

[primal grail]

ok excellent

that makes it free and i'm not missing anything

[scarlet ermine]

yup image of 1 is exactly the thing to study

[lyric rapids]

Quick question because I want to be sure about something. Given a R-Mod M. And W is a multiplicative set. What exactly is W^(-1)M?

My guess is that since. We defined R-Mod M as an abelian group M, with a homomorphism of R -> Hom(M,M) (the set of all group automorphism of M) similarly we define W(-1)M as the same abelian group M as before, and a homorphism of W^(-1) -> Hom(M,M). Such that this diagrams over here commute?

[hallow bone]

either you can define it as the tensor product W^-1R ⊗_R M

[lyric rapids]

I get that but that feel weird since it talk about W^(-1)M then prove it is isomorphic to the tensor product of M (x)_R W^(-1)R

[hallow bone]

or you can define it as the set { m/w | m ∈ M, w ∈ W } subject to the relation m/w ~ m'/w' iff there is a t ∈ W such that tw' ⋅ m = tw ⋅ m'

[lyric rapids]

Hmmm

Okay

That is fair

[hallow bone]

this should be defined

??

no?

[lyric rapids]

The book doesn't have it

Which is why i was reasonablely confused

But again this is a note from like a course years ago.

[hallow bone]

right

[lyric rapids]

Hmm now im curious if my definition is the same as that definition.

Maybe not

[hallow bone]

no

because youre doing nothing to the module structure

[lyric rapids]

Ohhh

Okay okay

[lone jacinth]

For a hereditary ring every projective module is a direct sum of ideals, so I guess you would need to investigate some non-noetherian hereditary rings.

The main example I know would be path algebras of quivers where the ideals should break down as direct sums of principal ideals.

[analog pumice]

1+2+3+4+5+6+7+8+9+......♾️=?

[hushed bone]

Very advanced algebra indeed

[analog pumice]

Huh

[wanton spoke]

[hushed bone]

U in the wrong channel little bro try #chill instead

[analog pumice]

👀oo why

[ornate atlas]

Please stop shit posting in here I’ve asked you before

[analog pumice]

Ooo

[nimble sierra]

man…

[analog pumice]

Shuttt...

[kindred ore]

How can we extend the Butterworth coefficient function so that the function a in the picture can have any real argument n?
Here I have implemented an analytical extension of the coefficient index, but not the order:
https://www.desmos.com/calculator/pyoo9y0xti?lang=ru

[astral ginkgo]

How is this relevant to algebra?

[kindred ore]

Maybe not, but most likely it is a real-complex analysis

[hallow bone]

then why didnt you just go to that channel 💔

[kindred ore]

Are you okay? I sent a message there over half an hour ago

[gilded remnant]

,tex could anyone explain what the derived tensor product in the base term does for showing $HH(R) \simeq R \otimes_{R \otimes_{\mathbb{Z}} R^op} R$

[broken turtle]

snow

[gilded remnant]

i’ve posted my full question here
https://math.stackexchange.com/questions/5133748/what-does-the-derived-tensor-product-do-in-the-base-term-for-texthhr-sime

[fierce steeple]

Maybe I should answer it there but basically there are two definitions of HH: one where you don't derive the tensor product over Z and one where you do. I would say nowadays you always derive the tensor product over Z

[near lantern]

Faithfully Flat Descent

[bitter vault]

Is this where we post cubic formula?

[distant harness]

No.

(Unless you're ready to discuss it from a Galois theory perspective, I suppose).

[primal grail]

logically this should be sound, right? here q-weyl means yx = qxy. to prove a basis for an algebra i can represent it with a functionally identical basis that somehow pushes the scalars through to obtain scalars = 0. it seems cheap though, but otherwise i don't quite see how to show that coeffs are zero with just the relations yx = qxy and xx^-1 = 1

i.e. even if the representation was somehow not injective it should still work

[warm owl]

I'm trying to look at the representation theory of $SU(3)$, and I don't understand how to arrive at the Gell-Mann matrices as a basis of $\mathfrak{su}(3)$. What I do get is that $\mathfrak{u}(n)_{\mathbb{C}} = \mathfrak{u}(n) \oplus i\mathfrak{u}(n) = \mathfrak{gl}(n, \mathbb{C})$.

[broken turtle]

Vanitas Daemon

[warm owl]

The condition that I need $\det U = 1$ for $M \in SU(3)$ tells me I want traceless matrices for the basis, and $U^{\dagger}U = 1 = UU^{\dagger}$ corresponds to the condition that the matrices in $\mathfrak{su}(n)$ must satisfy $X^{\dagger} = -X$.

[broken turtle]

Vanitas Daemon

[warm owl]

But then

Does this uniquely determine a choice of basis?

[warm owl]

Also, aren't I technically assuming simply-connectedness of SU(3) here? I know that the special unitary groups are in fact simply connected, but (1) I don't know or understand why, and (2), from what I recall, the proof to show that a Lie group G is simply connected is to prove the map exp: g -> G is surjective.

[near lantern]

Surjectitivity of exp should hold for a connected Lie group, not just simply connected.

[near lantern]

The Lie algebra of U(n) should be {X ∈ gl_n(ℂ) : X* = -X}. The Lie algebra of Su(n) should be {X ∈ gl_n(ℂ) : X* = -X and tr(X) = 1}. (The first condition exponentiates into e^X e^X* = e^X* e^X = 1, while the second exponentiates into det(e^X) = 1.)

[near lantern]

Not uniquely, but it is a well-defined vector space and it is well-defined which sets of matrices are and aren't a basis for it. You can use any basis for the vector space (or not use a basis at all if you reason differently) and presumably these Gell-Mann matrices are a popular choice because they make something easier to work out.

[narrow kraken]

I can't figure out how to show the hint here. I defined the left R-module structure on D by (d \otimes f(t) (x) = d x f(pi), where pi is a fixed transcendental element of D. I let N be a finitely generated R-submodule of D^n generated by (v^(1), v^(2), ... v^(m)) in D^n, where we have v^(i) = (v_1i, v_2i, ...v_ni). Then, let E := k(pi), which is a commutative subfield of D, and we have that for N is contained in sum from i=1 to n of D W_i, where W_i := span_E {v_i1, v_i2, ..v_im}, each of which are finite dimensional E-vector spaces. I wanted to argue via dimensionality or something but idk what to do now

also, is is true that a simple ring is isomorphic to a matrix ring over a skew field iff if it is Artinian? i thought we only know one direction, i.e artinian + simple -> matrix ring. actually I guess it's obvious M_n (D) is Artinian + simple..

[narrow kraken]

i understand why proving that the hint holds works though: ftsoc if we assume R \cong M_n (D') for some skew field D', we would have {R-mod} ~ {D'-mod} by Morita equivalence. Since D' is a skew field, all D'-modules are free, so trivially every D'-module contains a free submodule. So if we construct a specific R-module M such that for every finite n, M^n contains no nonzero free R-submodules, these module categories can't possible be equivalent, so we arrive at a contradiction

[lone jacinth]

At least for the n=1 case you have that d in D is annihilated by

pi d^-1 (x) 1 - d^-1 (x) t

So R cannot embedd into D.

I'm thinking maybe you can generalize this but killing of one coefficient at a time

Like in D^2 to annihilate (d1, d2)

You first do
pi d1^-1 ....
Which gives you
(0, pi d1^-1 d2 - d1^-1 d2 pi)
If this is 0 you're done otherwise you can annihilate the second term.

Taking product of those gives
(Product of nonzero things) (x) 1 + (bunch of stuff tensor powers of t)

[drowsy niche]

note that it suffices to prove this for countably generated projective modules by Kaplansky's theorem on projective modules.

if P = <x_1, x_2, ...> we want to construct f.g. direct summands P_n containing x_1, ..., x_n so that if P_{n+1} = P_n (+) Q_n we can decompose P into f.g. projectives as P = (+)_{n ≥ 0} Q_n (where we let P_0 = 0).

for any f.g. direct summand P_n of P we want to show that there exists an f.g. direct summand P_{n+1} containing the generating set of P_n and x_{n+1}.

in general, we can show that any finite subset of a projective module over a semihereditary ring is contained in an f.g. direct summand of said projective module: let F = P (+) P' be free and note that any finite subset S of P lies in an f.g. free direct summand F' of F. let p be the projection of F onto P' restricted to F' so that im(p) is an f.g. submodule of P' hence projective, thus ker p is an f.g. direct summand of F' containing S. writing F = ker p (+) C (+) F'', the projection F -> ker p restricts to the identity on ker p c P, so ker p is a direct summand of P containing S.

(sorry for the double ping lol)

apparently this result is originally due to Bergman according to this paper https://home.fau.edu/wmcgove1/web/Papers/mpr.pdf

[drowsy niche]

wrt an earlier failed proof: over a semihereditary ring are f.g. submodules of flat modules projective?

certainly if we remove f.g. this is false

since we can just take Z as our ring and Q as a flat module which is not projective over Z

[drowsy niche]

I imagine this statement is false too but idk a counterexample

[near lantern]

How different is this from the definition?

[drowsy niche]

wdym? for semihereditary rings we have that fg submodules of projective modules are projective

another characterization is that semihereditary rings are precisely coherent rings of weak dimension at most 1

I tried to use this + Lazard's thm to show that an f.g. submodule of a flat module is f.p. and flat hence projective

but that argument doesn't work afaict

[near lantern]

Right, thanks. I forgot the exact definition.

[lone jacinth]

Let V be an infinite dimensional vector space and let R be End(V).

Then R is von Neumann regular so it's semi-hereditary and every R-module is flat.

Let I be the ideal of maps with finite dimensional image, then R/I is cyclic and flat, but not projective.

[drowsy niche]

oh right ty. this example was in lam iirc

ty (lam)

[lone jacinth]

If you have weak dimension <= 1 then submodules of flat modules are flat, so it really comes down to flat module that is fg, but not finitely presented

[narrow kraken]

oh i see, so we actually need to show that R doesn't embed into D^n for any n. for some reason I got confused and thought showing R does not embed into D suffices 🐭

yea I'm not sure how to explicitly construct an annihilator r for every (v1, v2, ..vn) in D^n

oh nvm, the D-dimension of R = D \otimes k(t) is dim_k k(t) = infinite, while the D-dimension of D^n is n, so obviously it can not embed.

[distant harness]

<@&268886789983436800> mrbeast

[drowsy niche]

what does this mean

can anybody just say the magic words "mrbeast" to get a message banished

[past cove]

there's a scam going around about like mrbeast promotoing some crypto thing

and trying to get people to use some app that scams them?

[drowsy niche]

wait that's a scam?

[past cove]

yeah I was surprised too

[drowsy niche]

my riches as a maths student have been wasted 😔

[maiden wadi]

hello everyone i am mr beast if u go to scam.crypto.com and send me $100 then i will give you $100000

[last talon]

Ok here’s 100 dollars
-# just don’t check them enough to realise that they’re Zimbabwean dollars from back when they were completely worthless

[maiden wadi]

thank u for subscribing to mr breast, i have told chandelier to wire 100,000 to your account

[bronze timber]

"even a non algebraist one"

hmmm

[primal grail]

https://mathoverflow.net/a/25828 i don't understand this answer. i'm fairly certain we are trying to study the free algebra k<x, y>/<xy - yx - h> (perhaps the variables are swapped from the normal presentation), but the writing of X = e^x and Y = e^y confuses me. if we treat this as a normal exponentiation of some free variable (whatever that means), then XY should be e^{x + y} which should not have this qXY relation. if we treat X as a map e -> e^x we do get something, i.e. YX = e -> e^{xy} = e -> e^{yx + h} but this e^h factor doesn't really seem to play nice with composition of functions

[last talon]

Prove that XY = e^{x + y} (hint: it’s not true since x and y don’t commute)

(The first interpretation is right)

[primal grail]

i guess since i don't know what X = e^x means here i'm just lost

do i just pretend it's finite dimensional and do the standard e^matrix thing

[last talon]

I’d probably interpret it as an element of the non-commutative formal power series ring in 2 variables

[last talon]

This should get you the right answer for reasons that aren’t fundamentally wrong though

[primal grail]

or wait, maybe we write e^x = 1 + x + x^2/2 + ... and multiply?

[last talon]

Yeah

[primal grail]

i see

[last talon]

(And use the [x, y] = h relation)

It’s very messy
But it’ll get you what you want

[primal grail]

ok that makes more sense

i guess i will make this computation in 3 hours after my social engagement in 4 minutes

[primal grail]

i think this is right so far but yeah it is incredibly messy

and i just need to bash out the finishing touches

but also maybe i made a mistake somewhere bc the infinite sum with binomial coefficients has a bunch of 0 terms

[primal grail]

hmm i appear to be missing a n!

[primal grail]

i guess my question from here still remains. why does the text go through the effort of considering linear operators on t^a k[a][t, t^{-1}]? if the method i used works, shouldn't we be able to do something similar with k[u, v]?

the following definition for faithful representations seems to suggest that there would be problems if it was not injective, but even if two elements correspond to the same action, shouldn't any algebraic relation formed by the coefficients still hold esp if the action somehow corresponds exactly with the chosen basis

[limpid horizon]

This problem doesnt seem to be too bad but :

0 -> N -> M -> L -> 0 ses of R-modules over a local ring

If depth M < depth L, depth N = depth M?

So depth N <= depth M is clear

[hushed bone]

Use local cohomology

[limpid horizon]

Oh really

Ok

[hushed bone]

Wait depth M < depth L

Sorry wait is that a question you’re asking

Or something you’re trying to prove

[limpid horizon]

Trying to prove, i saw it in a paper

[hushed bone]

Okay yeah use local cohomology

Or use this

https://stacks.math.columbia.edu/tag/0AVY

[limpid horizon]

Thanks ill try

[hushed bone]

47.11.2

Look at those inequalities and you’ll see that the equality you want is forced

By combining two inequalities

But with local coho it’s also easy

Depth is the first nonvanishing degree for local coho at m

So for all degrees < depth L you have H^i_m(N) = H^i_m(M)

[limpid horizon]

Just the first one right?

[hushed bone]

This gives you only that depth M >= depth N

The third one gives you the other

[limpid horizon]

Ok, I thought we could get depth N <= depth M through the injective map

Oh thats the same inequality lol?

[hushed bone]

Sure

Actually is that true

I’m not sure that’s actually always true

For any injective map

[hushed bone]

Anyway this tells you that the first time they are nonzero is the same cuz H^i_m(M) is gonn be nonzero in a degree lower than depth L

And so it happens at a degree where they’re equal

[limpid horizon]

Ok, im just not sure how you got this equality

[hushed bone]

The LES

[limpid horizon]

Oo

[hushed bone]

The groups on either sides of the maps are local coho groups of L

Which are 0 by the depth assumption

I mean except for degree 0 where the group on the left is just 0 always lol

[limpid horizon]

Ok cool yeah

Thx

I like this sort of algebra

Depth cohen macaulay stuff

[round seal]

If i want to show that a noetherian local ring of dim 1 has a MCM module, then the trick is to just consider R/p for some minimal prime p right

this is what works in kx,y at least

[hushed bone]

Yeah

To be CM it just needs to not be depth 0

Which would require zero divisors

[narrow kraken]

is there a nice proof of this identity for quaternion algebras? we have (a,b)_F (a,c)_F = (a,bc)_F in Br(F), where (a,b)_F = F<i,j>/(i^2 - a, j^2 - b, ij = -ji)? My proof was just giving an explicit homomorphism on generators (a,b)_F \otimes (a,c)_F \cong M_2 ((a,bc)_F) = (a,bc)_F \otimes M_2 (F), and showing it respects the algebra relations/commutativity, since and since any homomorphism of CSAs is an isomorphism, we have (a,b)_F (a,c)_F = M_2 ((a,bc)_F) ~ (a,bc)_F in Br(F).

I tried to show that (a,b)_F (a,c)_F (a,bc)_F = 1 in Br(F), which is equivalent since every F-quaternion algebra is 2-torsion in Br(F), using the fact that (a,b)_F splits iff. b = Nm{E/F} (z) for some z in E = F[sqrt a], but this only tells us whether or not (a,b)_F is 1, i.e, if its not 1, we can't deduce anything about (a,b)_F from this field norm right?

[proven turret]

I’m trying to learn universal algebra and I don’t understand this condition.

What does f(a,c) = f(a,d) <-> f(b,c) = f(b,d) have to do with commutativity?

Why not something like f(b,a) = f(a,b) <-> true?

[limpid horizon]

Enpeace Approaching

[hallow bone]

"abelian" should not be read as "commutative"

f(b, a) = f(b, a) is possible, but this is a pretty boring condition that ends up not being all that useful

what are you using for this?

iirc the situation simplifies if you're in Malcev variety (variety where every pair of congruences commute, i.e. has a malcev term m):
An algebra A is abelian iff for all operations f with arity n, we have:
m(f(x1, ..., xn), f(y1, ..., yn), f(z1, ..., zn) = f(m(x1, y1, z1), ..., m(xn, yn ,zn))
i.e., m is a homomorphism from A^3 to A

(something similar can be done for a congruence modular variety too but the proof is very hard and uninituitive, lol)

[hallow bone]

it then turns out that A is essentially a module over a ring, anyways, lol, so in some sense this case is a little boring

[hallow bone]

what resources are you using? i dont think ive seen this before

[proven turret]

What does it mean for m to take multiple arguments? I thought it had to act on an element from A and send it to some other algebra B

(I’m reading this: https://www.math.uwaterloo.ca/~rdwillar/documents/Publications/torino.pdf)

[hidden shore]

Oops

Oh wait for an algebra

Nvm

[hallow bone]

lol

in the case of groups, abelian and commutative coincide

[hallow bone]

m is a 3-ary operation

i.e. a function from A^3 to A

idk what you mean by this question

you know what the product of algebras is right

[hallow bone]

why are you learning UA with this lol theyre not even course notes or anything

i recommend smt like Burris and Sankappanavar if youve never seen UA before

[proven turret]

Idk google lol. I don’t know crap about UA but I’m trying to describe a type system with it because category theory was a bit (a lot?) too general

[proven turret]

I assume it’s something you get when you take the Cartesian product of the two sets of the algebras but idk anything about its structure

[proven turret]

Ok thanks I think I understand this. I don’t understand its consequences but I’m sure that will come with doing exercises

[hallow bone]

okay lol if you dont know what the product of algebras is i really suggest stopping tryna read about abelian congruences and starting from the beginning

[hallow bone]

breh there are no exercises in that paper

[proven turret]

lol I just meant in general
Yeah I think that paper is for people who have way more background knowledge but don’t know UA in particular

[hallow bone]

^

abelian algebras for modular varieties are actually a particular instance of stuff related to the commutator, might you be interested in their general theory

ralph McKenzie has a book on it

its.. alright

not the best

[hallow bone]

Say I have a semisimple Lie subalgebra L of gl(V), where V is a fd complex vector space. The abstract Jordan decomposition of some x in L is x = d + n for some d, n in L such that [d, n] = 0, ad(d) : L -> L is diagonalisable and ad(n) : L -> L is nilpotent. This is furthermore unique.

The book I'm reading claims that the abstract Jordan decomposition agrees with the usual Jordan decomposition, but implicitly assumes that in the usual Jordan decomposition of x, we already have d, n in L (if that is the case, the result is not super hard to prove)

I'm just not sure if that is true at all

(I'm reading Karin Erdmann and Mark J. Wildon's Introduction to Lie Algebras)

[hallow bone]

I got as far as showing that if x = d + n is the usual Jordan decomposition, then ad(d) and ad(n) restrict to derivations on L, and so by semisimplicity can be represented uniquely as elements d' and n' in L. Thus, since ad(x) = ad(d) + ad(n) = ad(d') + ad(n') is the Jordan decomposition of ad(x) in gl(L), x = d' + n' is the abstract Jordan decomposition in L.

But how do I know that d = d' and n = n'?

[hallow bone]

https://math.stackexchange.com/questions/487384/does-the-abstract-jordan-decomposition-agree-with-the-usual-jordan-decomposition

i found this

its definitely not trivial, lol

[primal grail]

this should be enough to characterize finite dimensional irreducible representations of the weyl algebra in characteristic p, right? we just give x and y for the JCF and then to get any particular representation you just look at what space y fixes and how x acts on that space, to find suitable change of basis matrix and everything should just work, yeah?

[primal grail]

ig the reason im worried is bc i think you can technically add any upper triangular matrix equal along the diagonals to x and the yx - xy = 1 relation should still hold, but that kind of screws with the nice basis but somehow y is not changing

[near lantern]

enpeace Lie algebras arc

[hallow bone]

thought itd be nice to have a grasp on the rep theory of lie algebras before getting my hands dirty with quantum enveloping algebras

lie algebras are pretty fun

[wheat meadow]

the proof of that in that book is busted yes

this is acknowledged on the errata page

[hallow bone]

okay lol that gives me some peace of mind

ive found a much cooler proof of Weyl's theorem with cohomology anyways

the interesting parts of which seem to generalize very nicely to general Hopf algebras

[wheat meadow]

👀

this won't be useful for my final on monday but i am interested

[wheat meadow]

(anyway for reference the correct proof is in humphrey's book iirc)

[hallow bone]

basically, it can be shown that H^i(L, M) ≈ Ext^i(k, M) where k is given the trivial L-module structure
Suppose L is semisimple, and M is a simple module. Let c be the corresponding Casimir operator. Then it is central, and acts invertibly on M, and acts as 0 on k as homomorphisms. As Ext^i(-, -) is a bifunctor, this means that c acts on H^i(L, M) both invertibly and 0, i.e. H^i(L, M) = 0. Thus, using induction and the cohomological LES, H^i(L, M) = 0 for any finite dimensional L-module.

Now, Ext^0(M, N) ≈ Hom_L(M, N) = Hom_k(M, N)^L, and this can be extended to a homomorphism Ext^i(M, N) ≈ H^i(L, Hom_k(M, N)), so by the above we have that Ext^i vanishes forall finite-dimensional L-modules!

[hallow bone]

yeah

[hallow bone]

absolutely goated bio btw i love it

[wheat meadow]

hehe :3

[wheat meadow]

ah neat this makes sense -- and H^1(L, hom(M,N)) vanishing pretty much immediately gives weyl right

[hallow bone]

yis, thats the second paragraph

since Ext classifies extensions

[wheat meadow]

ye

[hallow bone]

honomogical algebra is peakest

cant wait till i get to do homalg of quantum groups

[summer quest]

lol yeah needless to say this is a good idea

[hallow bone]

i think im gettin it now though, its not too hard

[summer quest]

yeah it's not so terrible

but yeah the quantum enveloping algebra is nicely understood in terms of deforming the usual category of representations

[hallow bone]

after rep theory of groups in char 0 probably the nicest nontrivial rep theory

[hallow bone]

yeah, ive worked with the sl_2(C) case and it really is just the usual rep theory of sl_2(C) but with a bunch of q's added everywhere

[summer quest]

for generic q the quantum story is very much like the classical story in char 0, for q a root of unity you end up with something with much more closely resembles the representation theory of groups in char p

[hallow bone]

i figured

modular representations of groups...
scary..

[summer quest]

it's not so bad, just different

[hallow bone]

its number theory and number theory scares me

[summer quest]

it's not so much number theory

like it's nice to compare how semisimple Lie groups/Lie algebras over C and over R are classified compared to the analogous situation over finite fields

it's basically the same classification and the kind of twisting that happens when you go from C to R is the same kind of twisting that happens when you go from \bar{F_q} to F_q

[hallow bone]

twisting?

[summer quest]

yes when you work over fields which are not algebraically closed the classification will reduce to the classification over the algebraic closure plus understanding the possible Galois twists

[hallow bone]

:0 similar to cocycles?

[summer quest]

yes these twists are classified by Galois cohomology

if you fix some object \bar{X} defined over \bar{F} then H^1(Gal(\bar{F}/F),Aut(\bar{X}))={objects X defined over F isomorphic to \bar{X} over \bar{F}}

when you do this for something like Lie groups or Lie algebras over R you get inner twists plus outer twists that come from actions of Gal(C/R) on the Dynkin diagram

this is why these things over R are classified in terms of Satake/Vogan diagrams rather than Dynkin diagrams

[hallow bone]

yurrr

[summer quest]

the same thing works over any field with Satake-Tits diagrams

[nimble orchid]

Hi, I' ve a basic question. If you have a space equipped with a non definite positive metric, such as that of special relativity with signature (-1 +1+1+1), what is the algebraic structure of this space? I mean, is there a sort of generalized scalar product? Is It a sort of extended Vector space, hilbert space? Think about the Minkowski space. I come from a physics background, and I feel that so often things are introduced without reasoning about the deep structure

[hallow bone]

i guess here is where the reals are convenient given that we actually understand Gal(C/R)

lol

as opposed to the absolute galois group of Q

[summer quest]

look into symmetric bilinear forms/quadratic forms and their signatures

[hallow bone]

i see

[hallow bone]

tits...

[summer quest]

yeah same with finite fields, although even for something like Q you don't actually have to understand this group explicitly to end up with classification results

or at least you can isolate the parts of the classification which involve some kind of arithmetic like this

like over Q the Galois action you are having to write down necessarily factors through some finite quotient of Gal(\bar{Q}/Q) so you end up just talking about some number field F/Q over which things split

[hallow bone]

right and we do understand these somewhat in nice cases right

[nimble orchid]

Is there a way to relax the definition of Hilbert space, such as L^2 in a such a way the scalar product Is not such as <f,f> >= 0?

[summer quest]

right and depending on what situation you are talking about you don't actually need to understand all of this

like for the semisimple case in general you can have arbitrarily many simple components permuted by Galois so that's sort of hopeless to completely classify

but as soon as your Dynkin diagram is connected there are not so many symmetries that can actually appear

in fact there are only involutions plus a few exceptional triality cases

so over Q at worst you will only see quadratic extensions and then possibly few cubic/sextic extensions in the triality case

this is why the simple classification over finite fields is so closely analogous to the classification over R: in both cases almost everything involves involutions on the diagrams

and you only end up with a few extra cases of triality which of course are not possible for R since Z/3Z and S_3 are not Galois groups over R

[hallow bone]

yes

where can i go into the details of this?

[summer quest]

Knapp's "Representation theory of semisimple groups (an overview based on examples)" is a really good reference for everything over R, with loads of examples as you would imagine

various texts about reductive algebraic groups will cover the analogous story over arbitrary fields in various levels of generality and detail

[hallow bone]

ah yeah of course its algebraic groups

thank you, another one to add to the pile of stuff im interested in

[summer quest]

Knapp's book is really beautiful I have a copy of it on my desk and use it constantly as a reference

[hallow bone]

https://www.degruyterbrill.com/document/doi/10.1515/9781400883974/html

what the fuck 💔

304 euros for ebook

[summer quest]

🏴‍☠️

[hallow bone]

okay it is 700 pages long but still

[vagrant tapir]

Do you guys do local fields or does this fall into another channel?

[hallow bone]

either this or #advanced-number-theory , probably

[vagrant tapir]

ty

[ornate atlas]

Unfortunatley due to discord TOS we cant allow discussion of piracy etc.

[hushed bone]

I am planning to take a ship full of my maties into the Mariana Trench and plunder some booty from any poor fool unlucky enough to be sailing by

[ornate atlas]

The mod team approves this message

[hallow bone]

Say I have a Hopf algebra H. In nice cases (like when it is generated by primitive or grouplike elements, or in the U_q case), the elements of Hom_H(k, Hom_k(M, N)) correspond to those of Hom_H(M, N). Is this always the case though? If so, its definitely not obvious to me

[lone jacinth]

So I guess you're making Hom_k(M, N) into an H-module by restricting the H(x)H^op structure along antipode and comultiplication (?)

So then Hom_H(k, Hom(M, N)) would be those f for which
h . f = counit(h) f
So
h(1) f (S(h(2)) m) = counit(h) f(m)
in sweedler notation.

If f is H-linear, then
h(1) f (S(h(2)) m) = h(1) S(h(2)) f(m)
and h(1) S(h(2)) = counit(h), so that works.

But I guess the converse is trickier...

[hallow bone]

yeah, the converse is where im stuck

working out the U_q(sl(2)) case by hand essentially uses the fact that the comultiplication is essentially that of the UEA of a lie algebra except deformed by a group-like element so everything works out nicely

but not in a way that i can see generalizing arbitrarily

maybe if youre a braided Hopf algebra you can do something?

[lone jacinth]

Alright, it should be true:

Say f has
h(1) f (S(h(2)) m) = h(1) S(h(2)) f(m)

Then also
h(1) f (S(h(2)) h(3) m) = h(1) S(h(2)) f( h(3) m)

The lhs is
h(1) f ( counit(h(2)) m) = h(1)counit(h(2)) f(m) = h f(m)

And the right hand side equals
counit(h(1)) f ( h(2) m) =
f ( counit(h(1)) h(2) m) = f(hm)

[hallow bone]

ohhh of course

[hallow bone]

thank you!

So it follows from here that Ext^n(M, N) is indeed isomorphic to the n-th Hopf algebra cohomology of Hom_k(M, N)

[lone jacinth]

I feel if this wasn't the case, then Hopf must have defined his algebras wrong

[grave pumice]

what's the motivation for caring about the cohomology of the kahler differential?

[hallow bone]

that is true

[grave pumice]

how do you repeat the construction when the kahler differential is no longer an R-algebra, unlike the starting ring?

[ornate atlas]

I should really learn more about Hopf algebras than their string diagram definition

[hallow bone]

similar to usual de rham, the higher degree differentials are defined using exterior powers

[grave pumice]

hmm i don't know exterior powers

ig i should learn

[hallow bone]

yeah lol

important construction

[hallow bone]

yesss

they are awesome

its the smallest structure needed for duality and a monoidal fibre functor

[grave pumice]

so the exterior algebra is $T(V)/({u \otimes u}_{u \in V})$?

[broken turtle]

lexi

[hallow bone]

yes

[grave pumice]

so you take the tensor algebra and make it alternating

hmm

ok ykw maybe i just havent seen enough motivation for this

and it will seem more natural

later

[hallow bone]

you probably should see de rham first

and get familiar with that

[grave pumice]

got it

so is this motivated by de rham

[hallow bone]

are you using a textbook? if not then i also recommend that lol

[grave pumice]

im using stacks project

[hallow bone]

why else would they be called differential forms

[hallow bone]

thats a terrible idea lol

[grave pumice]

ok

[hallow bone]

im sorry lol but stacks is not optimized for learning

its a reference

[past cove]

You will read Peter Mays more concise book oooohhhh 👻 👻 👻 👻

[hallow bone]

big collection of facts as opposed to something that introduces concepts in a logic and gradual manner

[hallow bone]

grrr Kassel is the best