advanced-algebra

[drowsy niche]

but this is essentially the main part of proving that a fd Lie alg over a char 0 field with an invertible derivation is nilpotent

[ornate atlas]

Its for sure a useful lemma to know, at least in all of the problems ive done

[hallow bone]

right, thats cool

you can also use it to prove that the Jordan decomposition of a derivation must be into derivations

[wheat meadow]

gradings where most components are 0 are the best kind of gradings lol

easier to understnad

[hallow bone]

true

[astral ginkgo]

At that point just grade by a finite group

[distant harness]

Sweep everything into grade 0.

[hallow bone]

hmm yes let's add structure by not adding structure

[analog schooner]

gradings sound very intuitive

touched up on like superalgebras and mentioned that they are Z2 graded algebras

where the superalgebra is composed of a direct summation of two subspaces

isnt a superalgebra also a vector space

wait but i thought an algebra over a field K or a K-algebra considered a superalgebra or is that not how it works

thought its just an algebra but with more structure

[hallow bone]

well it is

[hallow bone]

everything "over a field" has an underlying vector space

a grading can be seen as a generalization of polynomial rings

in a polynomial ring, everything can be written uniquely as a sum of elements with a certain degree (either simply degree or a multidegree in the case of multiple variables)

these will be monomials

graded rings/algebras generalize this

[analog schooner]

Interesting

Direct sum decomposition of subspaces is what algebra grading is right

I thought it was a direct sum decomposition of sub algebras

[distant harness]

It's part of it. It only becomes a "grading" when the multiplication operation respects the grades.

[analog schooner]

Does that mean grading is structure preserving

[hallow bone]

well that cant be it

the unit element can only be a part of a single component

[hallow bone]

It means that An Am ⊆ An+m for all n, m

and 1 ∈ A0

in particular, A0 is a subalgebra

[analog schooner]

So 1 component can be a subalgebra and the other is a subspace ?

[hallow bone]

sure

yeah

well the 0 component has to be a subalgebra

and the others cannot be

[analog schooner]

Is that how even graded algebras work

[hallow bone]

well you can do this over any commutative monoid Γ

[analog schooner]

So the odd graded algebra can’t be a subalgebra ?

[hallow bone]

you mean graded over Z/2Z?

[analog schooner]

Yes

[hallow bone]

then yes, because it is not the 0 component

[analog schooner]

This is for multiplying elements of degree l and k resulting in an element of degree l+k right

[hallow bone]

yes

[analog schooner]

Could algebraic extensions also generalize superalgebras

[analog schooner]

What algebraic structure would you make if it were to be a direct summation of two even graded algebras over Z2

These are of homogeneous elements of parity 0 right

[analog schooner]

Would that mean the superalgebra is unital by definition (I don’t think superalgebras are associative by definition)

[hushed bone]

Maybe this is because I worked through a treatment of the derived category in decent detail from Weibel (and also a bit from Fourier-Mukai Transforms in Algebraic Geometry), but when reviewing this stuff from Residues and Duality, I’m surprised at how such a complicated object (relatively speaking) is kinda… obvious?

Like it’s still kind of homological algebra like where if you know what to do the steps in doing that are relatively obvious and just kinda doing the only thing you can

It can just maybe be a pain to verify that this construction has all the properties you need and whatnot. And then there’s some silliness when trying to extend stuff out of a derived category with some sort of finiteness condition to one with less finiteness conditions, but I think these become really standard stuff

[elfin ice]

-# re-self-promote my message; as stated before, any help is welcome

[solar turret]

I am stuck at iv), i don't see how the hint helps me here? Okay 1 \otimes x = 0 in A\xi \otimes M

[lone jacinth]

Note that
A -> A\xi
a |-> a\xi
is an isomorphism of A-modules

And you know that
a (x) x = 0 iff ax = 0
because
A(x)M -> M
is an isomorphism

[solar turret]

I got it, thank you

[void plank]

Oooh very neat that the torsion functor is left exact

[hushed bone]

I mean it’s kinda clear

It’s literally a subset of the module it comes from so it’s gotta preserve injections

I guess exactness at the middle is kinda interesting

[void plank]

Yeah, I can see it in hindsight. It just wasn't something that occurred to me

[waxen fractal]

@hallow bone is there a name for when a quasigroup's divisions are identical to its multiplication, and also, the quasigroup is idempotent?

it's like if steiner quasigroups were non-commutative

[hallow bone]

involutive

[waxen fractal]

whoa

[hallow bone]

x(xy) = y = (yx)x

so an involutive idempotent quasigroup

[waxen fractal]

i found something interesting geometrically that is naturally described by them

not UAG tho sadly

[hallow bone]

oh?

[waxen fractal]

just like, geometry-geometry

[hallow bone]

alright lol

UA has plenty of geometrically-like stuff anyways

[waxen fractal]

i was thinking about interesting ways to generalize how a line graph is the unique connected 2-regular tree

like the one whose vertices are integers and whose edges are given by consecutive integers

so instead i decided to consider what happens if you freely try to tessellate solid triangles together

by "freely" i mean escaping the constraints of euclidean space, just like as an abstract structure

and what i came up with was, let the vertices be elements of the free involutive idempotent quasigroup on two generators, and let triangles be of the form {s, t, st} for all terms s, t

actually i wonder if i could get sagemath to make some drawings for me

i did some on my own via a whiteboard

[hallow bone]

sounds hyperbolic almost

[waxen fractal]

yea that's part of what im thinking

admittedly, a while ago, i did something similar, by generating the unique connected 3-regular tree via the cayley graph of the group presentation <a, b, c | a^2, b^2, c^2>

these structures, as topological spaces, should be homotopic, although i dont have a precise proof, just vibes

but it's interesting to me how the quasigroup creates the "filled-in" version, which might possibly be a manifold with boundary, while the other one isn't. although im not sure if it's a manifold because of what happens at sharp corners

[hallow bone]

well it shouldnt have a boundary because every triangle is surrounded by triangles

[waxen fractal]

that's true. so the only problematic behavior can occur at vertices

cuz we end up having infinitely many triangles all attached by the same vertex

[hallow bone]

right

at every vertex

[waxen fractal]

yea kinda mind-boggling

[waxen fractal]

ok this is too big to be useful for understanding it

https://sagecell.sagemath.org/?z=eJyVUUFugzAQvCPxh73ZVimCqKdI9NIn9OhayCJLask1lqFA-_raBtKU5pI9jWdnR-vZEd3QQwWczCQD8kVEmrSywcgF3DmoQRlw0pyRHtgxTcCXwaket1mxcEE7Bm1srMKtMd1qhFKtH6oqmHZ8qKYzgzKf-G9Co6ETg2coQZrTasALca_HeOXBS3HfHlsGPgI6ZjCx2-0-l9aiOdGN2Oli3JuGB6PsMitW7RL1Q_Vrmiav6sNq1SipXzqPcKac94Pzf8odWu1dKcnCUckV8Ub2DCwEL4-PPoC_uaDu0Yd7OWwrIm4jDmsLlp-dtO-U5VZ3Aw2r4Vz36hurQ1EUGbTqHF9PBfsB63OlkQ==&lang=sage&interacts=eJyLjgUAARUAuQ==

yea ok this is terrible for visualization

ill just make my hand-drawings

actually wait maybe the computer is trying to tell me i was wrong

i need more axioms

[hallow bone]

yeah you cant form tetrahedrons

because then it wont be a tree anymore

[waxen fractal]

yea i checked via sagemath, the homology is nontrivial, which is very much not intended

well glad to know i was wrong

this is what happens when i just declare things true on vibes cuz i checked a few cases 🥀

[hallow bone]

lmao

[waxen fractal]

aha i see what happened

i entirely forgot that (xy)(yx) is a term

in my head, i was only generating more terms off of edge-connected ones

ok tl; dr everyone can ignore everything i wrote in this channel today cuz it was based off of a fundamental error in understanding

[hallow bone]

F

[waxen fractal]

ok i fixed my idea

this is what i originally intended

https://sagecell.sagemath.org/?z=eJxNTsEKwjAMvQ_2D7k1xTGq6EXw5Cd4LGOULauF2ZZ26BQ_3nY4MYfw8vLyXmIHJ7iYmx9NZ9R4dgnRjFKymVXAnqxpeFlYerSD6igmsWzKYnABWjAWgrKacM-PZQGpMk-9pryKXa2D8lfkdaYirqJcP8NaeU-2R5k1UjTVci-3CbA3gw18-RVtl3_WrGyRs352fxEpX_X9QmNu6cwHYydMi6u7udHpJ_LE_j3qRzfhncJEcxvNi04HIUQFg9HLtBP8A3xQXDo=&lang=sage&interacts=eJyLjgUAARUAuQ==

im using polish notation to reduce vertex name sizes

anyway this time the rules are very simple

each edge freely generates a face with that edge on it

this is more analogous to my previous idea cut in half, cuz there is an explicit edge boundary this time, specifically along x-y

ok yea now that ive not fucked up, it's actually planar

hi mico

sorry for clogging chat with me talking to myself a little, yall are free to like, use this channel as intended if yall want

this seems to be the largest it'll go before sage gives up trying to get hyperbolic space to fit into euclidean space nicely

[last talon]

It scares me mishu :3

[hallow bone]

it constantly seems like it wants to close up on itself lol

[last talon]

mishu no know what maths mishu want do so mishu bored

[hallow bone]

i do have UAG notes im working on :3333333333333

[last talon]

UA no is option

[hallow bone]

fucK

[last talon]

Scooped

[hallow bone]

bbut theyre cool I promise

keith can vouch for me

[last talon]

You is have opinions on fun group theory for mishu?

[hallow bone]

well im sure you know lots more group theory than i do lol, but quantum groups are always fun

[last talon]

What is good book to learn quantum groups?

[hallow bone]

Kassel!

[waxen fractal]

why does it scare you

is it beeg

[last talon]

It beeg and not dwagon shaped

[waxen fractal]

ikr

[last talon]

mishu steal :3

[waxen fractal]

that's true although i think mico actually once tried ua before

[waxen fractal]

you should teach me ggt

[last talon]

:3

[hallow bone]

idk maybe theyd like uag lol

[waxen fractal]

tbh i have no idea what kind of math it's mostly into

[last talon]

mishu probably no would
It not very mishushaped

[last talon]

mishu mostly do all the groups like micotses

[waxen fractal]

that makes sense when i think about it

[last talon]

https://tenor.com/view/aileen-solo-farming-in-the-tower-the-top-dungeon-farmer-manhwa-chill-gif-7214407249400814117

[hallow bone]

sadge

[last talon]

this is mishu

[hallow bone]

eepy

its kinda interesting to see how did systems interact with like math

[waxen fractal]

ok so there's a natural action on the faces by the free monoid on 2 generators

[last talon]

Wawawawawawawawa
mishu mostly just get in way of doing maths because cudl is more fun :3

[hallow bone]

lowkey can get behind that

for me tho i dont have like full dissociation between alters or wtv i do notice that math just seems to be like

mostly an ever present knowledge? not like my other skills like drums or breaking or whatever

[hallow bone]

right

an easy way to get a 3 regular tree is the term algebra on one variable and three unary operations

[waxen fractal]

oh wait

you're kinda right

😭

well the 4 regular tree depending how you count

[hallow bone]

oh fuck

[waxen fractal]

anyway i wonder if there's a similar thing that the free group would act on

[hallow bone]

no you're right im stupid

wait what if theyre all involutative though

then it should work

so like the variety of Gsets with G the product C2 * C2 * C2

[waxen fractal]

yea that does work

[hallow bone]

peak

[waxen fractal]

im sorta trying to get a surface version of that

[hallow bone]

yeye i see

[waxen fractal]

i think you can take a really large colimit

[amber orbit]

How does Jacobson's "Lie Algebras" compare to other, more modern texts?

[flint cloak]

i havent really studied a ton of it but i did buy it because its one of those cheap dover publishing books, i think it gives an good treatment of lie algebras that you could find in most modern texts but i think you could probably get better takeaways from texts that will talk about lie groups and lie algebras

its very focused on lie algebras hence the name so it doesnt really cover much about the connections between lie groups and lie algebras and how to relate representations of lie algebras to representations of lie groups

lie algebras can be taught independently just fine, but a lot of what you learn can feel unmotivated without seeing how it gets applied to the representation theory of lie groups

that was my experience at least, because my first exposure was introduction to lie algebras by erdmann and wildon

i think brian hall's textbook gives a better picture

[solar turret]

I am stuck at vii)

I proved one direction, Supp(M/aM) \subset V( a + Ann(M) )

[limpid horizon]

maybe you can use vi) cause i think there is an isomorphism M/aM = A/a (x) M

[hallow bone]

should be

[solar turret]

But then I need A/a has to be finitely generated as A module

[limpid horizon]

it is

[solar turret]

Yeah

[hallow bone]

take the SES
0->a->A->A/a->0
tensor with M
a ⊗ M -> M -> A/a ⊗ M -> 0
is exact. Clearly the image of a ⊗ M in M is aM, so M/aM ≈ A/a ⊗ M

[hallow bone]

so whats the problem lol

[solar turret]

Let me try

I got it, thank you @limpid horizon @hallow bone

[hallow bone]

peak

[amber orbit]

I'm fine with a purely algebraic text, I have some familiarity with Lie groups.

[flint cloak]

if you are already familiar with lie groups i think youd definitely get a lot more out of brian halls book but jacobson is fine if you just wanna learn about lie algebras

the treatment is pretty much the same as most other texts on lie algebras

[near lantern]

Is there a standard category whose automorphism objects are the finite Weyl groups of all other types, similar to the category of finite sets for type A?

I mention https://pages.uoregon.edu/njp/fsb.pdf as relevant.

[solar turret]

I am stuck at the last part, any hint?

[undone idol]

I think it may be helpful to recall that

$(M \otimes_A B){\mathfrak {q}} \simeq (M \otimes_A B) \otimes_B B{\mathfrak {q}} \simeq M \otimes_A (B \otimes_B B_{\mathfrak{q}}) \simeq M \otimes_A B_{\mathfrak{q}}$

[broken turtle]

HChan

[undone idol]

for the middle isomorphism refer to exercise 2.15, p.27

[solar turret]

Any hint for part c?

[rose mirage]

The exceptional types should just be the delooping of the corresponding Coxeter groups, embedded into FinSet via their canonical group action as real reflection groups. In the notation of the paper you linked one can most likely define FS_D by a similar process of embedding a unique involution on-top of FinSet as D_n is isomorphic to C_2^{n-1} \rtimes S_n, similar to how B_n is Z_2 wreath S_n but I admit I haven’t worked out the details - the involution in the paper seems to corresponds to the diagonal C_2 in B_n but the semidirect product that forms D_n isn’t as simple so the precise action of this involution on the finite sets isn’t as clear to me

[stuck canopy]

Hey guys, i understand that the "If condition (i) (and therefore (ii)) holds..." statement should be true but how do i prove it?
This guy here says it's an easy consequence of (i) so i think im missing something obvious
The (satisying (7)) => (forming a basis) implication is clear but i cant see how to do the other implication

For info, "B as above" means N-graded k-algebra where k is a field and "h.s.o.p." is homogeneous system of parameters, as in homogeneous, algebraically idependant elements a1,a2,...,an€B such that B is a finitely generated k[a1,a2,...,an]-module

this is how far i got

[limpid horizon]

simple examples work if you just try stuff out

[void plank]

Does anyone here have a reference for learning more about Koszul-Tate resolutions? I'm specifically interested in ones for group algebras, but I can't find a good resource for this

[robust rain]

To check that I am understanding correctly:

The sheaf $\underline{{p}}$, which I will denote $p$ from now on, is the sheaf such that for any open $U \subset X$, $p(U) = {p}$.

[broken turtle]

swifteeee

[robust rain]

That is, to any open set, I am associating the set {p}

Hom(p, F) is the sheaf on X given by associating to any open set U, the set of morphisms from the restrictions of sheaves. Hom(p, F)(U) = Mor(p|U, F|U)

And a morphism $\varphi : p|_U F|_U$ is given by the following data: to any open $V \subset U$, I get a morphism of sets $\varphi(V) : p|_U(V) \to F|_U(V)$, that commutes with restriction

[broken turtle]

swifteeee

[robust rain]

So this is just a map {p} -> F(V)

Ts is like an onion twin ✌️ tf is going on 🥀

long story short: I don't know where to go. I am completely overwhelmed by the amount of data in this one sentence. Help.

[past cove]

This is not the correct definition of a constant sheaf

The constant sheaf are all continuous maps from U to {p}
I think in this case it's the same because they're always the same map

But you've gotta be careful in general

[hallow bone]

its the sheafification of the constant presheaf

[past cove]

Yes

[hallow bone]

in this case its the same

[robust rain]

the set {p}(U) is cts maps from U to {p}?

aren't singletons final in Top

so there is one map, sending each element of U to p

[past cove]

In this case what you wrote is correct but for other bigger sets you have to look at the continuous maps

[robust rain]

okay

[past cove]

So I just said that so you don't get the wrong intuition from a special case

[robust rain]

I see, thank you

I think where I'm struggling is to describe a map from Hom({p}, F)(U) -> F(U)

[past cove]

Well what does Hom({p},F)(U) look like

[robust rain]

A section of Hom({p}, F)(U) is a sheaf morphism from {p}|U to F|U, where | denotes restriction.

I.e. to any open V \subset U, I get maps {p}(V) -> F(V)

that play nice with restriction

[lone jacinth]

And what is {p}|U?

[past cove]

Well notice that maps {p}(V) -> F(V) are in bijection with the elements of F(V)

[robust rain]

Oh I see my problem

[robust rain]

I thought this wouldn't work, because it wouldn't 'play nicely' with restriction. Take f : {p}(V) -> F(V), say f(p) = v. Then if we take a subset W of V such that v is not in W, then it wouldn't restrict appropriately, but of course my arrow is the wrong way

I'm not explaining myself too well, but I see now

Okay let me try to write this up

[past cove]

If you want a concrete answer the way you do it is you look at the global sections of your morphism and map it to it's image

And the inverse is given by taking a section s in F(U) and defining the morphism by phi(V): {p} -> F(V) where p is mapped to the image of s in F(V) by the restriction map and then this glues by the fact that we're working with sheaves

[robust rain]

Writing these things out still feels really awkward to me, so I would appreciate any comments on clarity/brevity/notation

[robust rain]

This is the following data: for any open subset V of U, p|_U(V) = p(V) = {p}

this whole paradigm of doing ag better become useful soon

[past cove]

I think you should clarify what f is and that you're taking global sections of something in Mor(p_V, F_V)

[robust rain]

Could you please elaborate on 'clarify that you're taking global sections of something in Mor(p_V, F_V)?

[past cove]

Well I mean a morphism between sheaves is a collection of maps for each open

So the idea is to look at the map for the biggest open (this is U since that's what we're restricting to)

And looking at the image of that map

[robust rain]

I'm fairly sure I know what you mean, but I'm kind of struggling to put it into words.

Perhaps I need some air

Thank you for your help Irony

[void plank]

Let R be a ring and suppose it is freely and finitely generated over a subring S.

Apparently this implies that R is integral over S, so for any r in R, it is the root of a monic polynomial with coeffs in S.

Under what additional conditions is it the case that the constant term of this monic polynomial is non-zero?

[fierce steeple]

This is false unless you are using "freely and finitely generated over a subring" in a nonstandard way

Since this should just mean it is a polynomial ring over S

I assume you mean it is a finite free as an S-module for S a subring

[void plank]

Yes, sorry

I do mean R = S^n for some finite n

I should add that R is commutative here.

Ultimately, I'm trying to prove (or disprove) that if M is a finitely generated R-module, its torsion submodule as an R-module is the same as its torsion submodule as an S-module

Its clear that if m in M is an S-torsion element, it is an R-torsion element.

If m is an R-torsion element, then I was trying to use that R is integral over S to show that some r that annihilates m is the root of a monic polynomial with coeffs in S. But unfortunately, I'm not sure if I can say the constant term of this polynomial is non-zero

If I can, then I can conclude m is an S-torsion element (the constant term is the element in S I need)

[lone jacinth]

R being an integral domain is a simple condition at least

[void plank]

Why is that?

[fierce steeple]

Divide the poly by x

[lone jacinth]

If x * f(x) = 0, then either x or f(x) is 0

[fierce steeple]

Idk if there is much nicer stuff you can say

Idk if you want a hint for the question you are ultimately doing though

[void plank]

Is what I'm trying to prove even true in general?

[fierce steeple]

No

One issue is like this module structure isn't massively helpful

Like let's take S to be a field k

Then the hypothesis holds for any finite dim commutative k algebra R

[void plank]

I in fact have that in my specific situation

[fierce steeple]

Well now there are.simple counterexamples I mean

[void plank]

In my application, R is a polynomial ring with coeffs in k

ahh

[lone jacinth]

Since you're talking about torsion elements at all, presumably R is an integral domain. So then there's no issue

[fierce steeple]

Like k x k

[fierce steeple]

I'm confused

Then it isn't fd over k

[lone jacinth]

Even in general you would probably define torsion element as annihilated by a regular element, and then again there's no issue

[void plank]

Yeah you're right, I specifically have a quotient of a polynomial ring

[fierce steeple]

To be fair that is implied already by being integral over k

[void plank]

So to summarize, you're saying that if R is a finite dim k-algebra and M is a fg module over R, then torsion_R(M) = torsion_S(M) (where M is interpreted as an S-module on the RHS)?

[fierce steeple]

I'm saying that that is false

Take R = k x k and M = R

[lone jacinth]

Well depends what Torsion_R means

[fierce steeple]

Sure

[void plank]

like m in M is in Torsion_R if there exists an r in R such that rm = 0

[lone jacinth]

I mean, if that's your definition that's fine. But it's a stupid definition that isn't even a submodule

[void plank]

Ah sure, I guess I need regular element in general like you were saying

My situation is basically this.

R = k[x,y], S = k[x^b, y^b], so R = S^(b^2) as an S-module (I will need to consider quotients of these rings later).

I have a map M: R^2 --> R given by the 1x2 matrix [f g], where f and are g are in R.

I have an induced map M': S^(2b^2) --> S^(b^2). I need to consider Torsion(coker M) vs Torsion(coker M')

It wasn't clear to me how these are related, but I'm skeptical there is a relationship based on what you two said

Thanks for the help!

[lone jacinth]

So I guess these maps are a bit of a distraction.

What you're saying is your have an R module and you take restriction of scalars to S.

As R is a domain an integral over S, the two torsion modules should be the same

[void plank]

Ah ok, and is the proof obvious?

[lone jacinth]

I mean you already wrote it earlier

If rm = 0 take the minimal polynomial of r, then the constant coefficient also annihilates m

[void plank]

Ah ok. And if I then quotient S and R by a common ideal, the proof breaks since R is no longer necessarily an integral domain

[lone jacinth]

At the very least it would depend what torsion should mean for this new ring

[void plank]

Thank you two once again for your help!

[near lantern]

I think torsion is best considered associated with a mulitplicative subset of a ring (at least if it's commutative).

Or more generally a filter of ideals, but in any case not just a function of the ring and module.

[lone jacinth]

Could define it with respect to an element in a flat module, as the kernel of the map M = M(x)R -> M(x)F.

Then you get it with respect to a multiplicative closed set by picking 1 in S^-1 R

I wonder if this gives you more than multiplicative sets in the commutative setting...

Another nice way is do define it as the torsion part of a torsion pair. So for example fix a module Q and call a module torsion if it has no map to Q.

If Q is the field of fractions of an integral domain you get the usual notion. Not sure if there is a nice cannonical choice associated with a multiplicative closed set for this

Okay, I think you can pick Q to be an injective cogeneratot for S^-1R modules

[hexed tangle]

This definition seems rather circular and I don't know how to interpret it

[lone jacinth]

What part are you struggling with?

[hexed tangle]

all of it

[lone jacinth]

If you're thinking of the differential then a more verbose description would be

$(df)n \coloneqq d_B \circ f_n - (-1)^n f{n+1} \circ d_A$

[broken turtle]

jagr2808

[hexed tangle]

what are the f_n's?

also what are the quantifiers?

[lone jacinth]

Okay, forget that equation and let me step back.

Let C be the Homcomplex we're constructing.

Then Cn = Prod_i Hom(Ai, Bi+n). That is an element f in Cn is a tuple of f_i : Ai -> Bi+1.

Now we need a differential d: Cn -> Cn+1. It should take this tuple f to a new tuple g = df.

Then g_i is defined as
d_B o fi - (-1)^n f_i+1 o d_A

A more abstract perspective could be that Cn consists of homomorphisms of graded abelian groups A[n] -> B (i.e. we forget the differentials) and then the differential d: Cn -> Cn+1 is checking if these homomorphism commute with the differential. (So df = 0 iff f is a homomorphism if chain complexes)

[hexed tangle]

are there more than one f in \underline(Hom)(A,B)_n?

ah so the differential is a map of of tuples of maps

[lone jacinth]

Underline-Hom(A, B)_n is a product. So an element f in it is an element of a product, i.e. a tuple

[hexed tangle]

indeed

[lone jacinth]

So all understood?

[hexed tangle]

yes I think so

thank you

I don't understand the definition of P^(i) in

[lone jacinth]

The idea is you start with P^(i) = A, then modify it one differential at a time to replace the objects with objects in A_0

[void plank]

I have a probably dumb thing I wanted to check, but suppose I'm working with chain complexes of vector spaces. Let f: A --> B be a chain map.

We know the homology of the mapping cone (by taking the usual LES) helps us determine if the induced map in homology f* is surjective/injective at various degrees.

But in the vector space setting (and I think any hereditary category?), it can be shown MC(f) is homotopy equivalent to ker(f) (+) coker(f) (up to some degree shift I'm being lazy about). So surely I can just analyze the homology of ker(f) (+) coker(f) to get the same information about f*?

[lone jacinth]

You say "just", but I'm not sure why this would be any easier

[fierce steeple]

Well I guess it depends what info you want about f

[void plank]

In my application, my f is in fact surjective. So I only need to analyze ker(f), and I know the boundary maps on this complex are induced from those on A

That's why I was hoping this would help

[fierce steeple]

I guess it is unclear to me what you want to know about f

[lone jacinth]

If you have a short exact sequence of chain complexes you get a LES in homology.

So if you just want to know the dimension of the homology of the cone, then it's the same as for ker(f)

[void plank]

It's specific to my application, but I just want information about whether certain homology groups of the mapping cone vanish and to relate that to properties of the boundary maps on A and B (their kernels and ranks)

[hexed tangle]

I don't understand how the limit comes into play

[lone jacinth]

The limit is then a complex where every object has been replaced by something in A0

Like P^(n) has replaced every object for i <= n

Then in the limit as n goes to infinity, every object is replaced

[hexed tangle]

I see

[ornate atlas]

Why isnt this nilpotent? This is the argument presented in the solutions, but Im not sure if thats what I get? Like H is a cartan subalgebra, so its abelian, so [A,A] = \oplus_i L^1_{alpha_i} right? But then going to A^2, dont we end up taking like [L_alpha,[L_alpha,L_alpha]]\subset L_{3alpha} and get the same argument as with L^(n)?

Im slightly just confused about why the same arugment doesnt apply to both the derived and lower central series, I guess im misunderstanding the argument that its soluble too (which is possible, the solutions mention heights which afaik we didnt cover this year, this is an old past paper)

[last talon]

So we get
[a, [a, a]] contains [h, [h, L_a]] = h^2(a)L_a

Now there’s some h with h(a) =/= 0

But with [[a, a], [a, a]] we can’t do a similar thing, since neither side contains H

[ornate atlas]

Im not sure I follow

[wheat meadow]

explaining it is annoying to phrase correctly because i have forgotten all of lie algebras, but if you just work things through with sl_n (where this is the upper triangular traceless matrices) it becomes pretty clear why it's solvable but not nilpotent

[ornate atlas]

Ok this is probably wise, just work with sl_n as a toy is the mantra of lie algebras lol

[robust rain]

Just as a toy example. Let F be a sheaf on X, equipped with the discrete topology. Let U \subset X, where U = {p1, p2, p3}. If f is a section of U, let [f]_p denote the germ at p.

Let f, g be sections of U. Then does ([f]_p1, [f]_p2, [g]_p3) consist of compatible germs?

I can pick open subsets of U, namely U1 = {p1, p2}, U2 = {p3} and f in F(U1), g in F(U2), such that the germ of any element of U1 is the germ of f?

By f in F(U1) i mean the image of f in F(U) under the appropriate restriction map

I'm fairly sure this is correct looking at the equivalent definition.

[lone jacinth]

Yes. In general whenever U is a disjoint union of open sets you don't need to check any compatibility between germs in coming from different sets

[robust rain]

Thank you : )

[hallow bone]

Notice that [H, [A, A]] is already simply [A, A], since each L_α is a generalized eigenspace, so L_α = [H, L_α] as α =/= 0

Thus
[A, A] ⊃ [A, [A, A]] ⊃ [H, [A, A]] = [A, A]

im a lil tired so i may be misinterpreting why youre confused

[ornate atlas]

I think the issue is the first line, but I’ll give this another go tomorrow with like sl_n and see if I can work out exactly where I’m getting confused

Because I’m pretty sure what I said about it being soluble is correct, I just need to work out why it doesn’t then carry over to nilpotent, but I suspect I’m just saying something silly

[hallow bone]

well its not nilpotent because for every L_α (with α =/= 0) there is an h ∈ H such that it acts invertibly on L_α

the H is important, because without it the algebra would be nilpotent (for example because its [A, A], and derived algebras of soluble algebras are nilpotent)

[clever nacelle]

what did u try to prove it

yea i would say that exactness « spreads » information in a way

wdym the rest?

[past cove]

the way it's stated is kind obtuse but the two cases are "the morphism is injective" and "the morphism is surjective"

so the way you prove it is to do a diagram chase

going around the rest of the diagram to show it's injective/surjective

[clever nacelle]

oh hn i can’t even read it 🥀

and it’s 3 am so i can’t even read AT ALL 😭

[queen tide]

https://tenor.com/view/ishowspeed-translate-gif-12052961640477791272

[hushed bone]

Bro it looks you held the pencil overhand with your whole fist

Are you right handed?

Ok well then I don’t think the fact you use your left hand matters much lol

[ornate atlas]

How does one make it to homological algebra before primary school 🥀

[last talon]

I knew homalg before primary school, it’s like
The first thing your parents should teach you

[hushed bone]

I learned the alphabet through homological algebra

Like M, I was like oh that’s like modules

[robust rain]

R is for R-Module

[lone jacinth]

A is for Algebra
B for BG 🎵
C is for chain complex
D for dg 🎵

E is for Eilenberg
And F is free 🎵

G is for group
H for homology 🎵

[last talon]

A is for abelian group
B is for bar resolution
C is for chain complex
D is for differential
E is for essential extension
F is for free module
G is for group ring
H is for homology
I is for injective module
J is for Jacobson radical
K is for killing form
L is for Lie algebra
M is for module
N is for Noetherian
O is for octahedral axiom
P is for projective module
Q is for Quillen
R is for radical
S is for short exact sequence
T is for torsion
U is for uniform module
V is for variety
W is for Weyl algebra
X is for exact sequence
Y is for young tableau
Z is for zig-zag lemma

[leaden ridge]

Amazing.

[lone jacinth]

X for eXact
Y for yoneda
Z for ... the integers

[last talon]

lol we came up with Z at the same time

[last talon]

Changed E to essential Extension and X to exact sequence
I prefer mine for Y (down with the categories!)

[lone jacinth]

Yoneda meaning yoneda extensions of course, not yoneda lemma

[hushed bone]

X is for Twitter

Z is for Zentrum

[lone jacinth]

B for bar resolution

[last talon]

Done 🙃

Q and Z are the ones I’m not happy with now

Maybe V too

[lone jacinth]

O is a little boring too

[last talon]

Yeah fair

[lone jacinth]

Z for zig-zag lemma

[robust rain]

I was thinking A for A-module, O for O_X-module...

T for Tate Module?

okay maybe only I find this joke funny

[last talon]

D for dwagon module :3

[robust rain]

O for orbit-stabiliser

[silk bolt]

Just joined, can i put questions in here or is this just for discussion and i should put my questions in regular maths help?

[last talon]

You can put them here as long as they’re (abstract) algebra questions beyond a standard first algebra series (questions from a first algebra series belong in #groups-rings-fields )

[hushed bone]

To add onto the answer above, you should not post these level of questions in the help channels cuz it won’t get seen by someone who can help you, most likely

[silk bolt]

thank you, having a quick look it seems im definiutely not at the abstract alegbra level yet thanks tho

[void plank]

O for octahedral axiom?

[last talon]

Added

[fierce steeple]

Down with the octahedral axiom

[distant harness]

Outer isomorphisms, then?

[lone jacinth]

Just rename it to the second isomorphism theorem and all is well

[void plank]

I have a question about this exercise in Weibel

Let P ---> A, Q ---> B be Cartan-Eilenberg resolutions. Its clear that when taking the vertical filtration, doing the vertical homology H_q(Tot(P_(i,*) (x) Q_(j,*))) computes Tor_q(A_i, B_j).

My first question: is the Tot^(+) in the second result just denoting the fact that when I take horizontal homology, I'm just doing something like taking the pth homology of Tor_q(A_0, B_0) <--- Tor_q(A_1, B_0) (+) Tor_q(A_0, B_1) <--- ....?

Second, I know that for CE resolutions, taking horizontal homology gives a projective resolution of the homology of the base complex at a particular column. But is it the case that P (x) Q is itself a CE resolution of A (x) B?

[void plank]

Or at least is it obvious that when I do the horizontal differential along Tot(P_(*,k) (x) Q_(*,l)), it factors as the direct sum of terms like H_k(A) (x) H_l(B), where k+l = q?

[robust rain]

Can someone please verify this solution? I'm slightly suspicious because it felt too easy. Comments on notation/clarity/whatever else appreciated : )

(implicitly I am assuming that F, G are (pre)sheaves on a set X, and U is an open subset of X)

[hushed bone]

This is not how I would have done it, but it’s pretty swag

As in it’s better

[robust rain]

I can't exactly take credit because the hint is to consider the diagram, but thank you haha

[hallow bone]

lol

ok bud

[void plank]

Nvm, when taking horizontal homology, I can use flatness + kunneth to distribute the homology over the tensor product. Then the columns are tensor products of resolutions of the homology groups of the base complexes by a property of CE resolutions, and taking vertical homology computes Tor of these

[hallow bone]

people who take classes on them usually don't understand it pretty well

that's why they take the classes :>

the fuck is a neurotropic

"the math channel" 💔

[hexed tangle]

I don't see how they conclude that that P^(n) -> P^(n-1) is a quasi-isomorphism

[hushed bone]

You only need to worry about the homologies being isomorphic in degree n and n+1. The fact the kernels of the chain map in degree n+1 and n are isomorphic (via the natural map between them) is a general categorical fact due to it being a pullback square.

Because the chain map is surjective, and the next map is identity, it’s obvious that H_n(P^(n)) surjects onto H_n(P^(n-1)), the question is whether it’s injective or not. Suppose a class x goes to 0 via this map, then x is mapped to something by the chain map to something in the image of P^(n-1)_n+1. Pull this back to an element y in P^(n-1)_n+1, and since the chain map is surjective in degree n+1 as well this pulls back to something z in P^(n)_n+1. Then the image of z maps to the same thing x does via the chain map, meaning it differs by something in the kernel. This means x = im(z) + x’ for some x’ in ker(P^(n)_n -> P^(n-1)_n). But because the kernels are isomorphic, x’ is the image of some z’ in P^(n)_n+1, so z - z’ maps to x.

This says that the class of x in H_n(P^(n)) is zero, and so the map on homologies in degree n is injective and thus bijective

You can do another similar element-wise argument to show that the homologies in degree n+1 are isomorphic which I leave to you

[hexed tangle]

is there a categorical argument that does not rely on element-wise arguments?

[ornate atlas]

There’s this marvellous thing called the Fryed-Mitchel embedding theorem

There will be an argument just using kernels and cokernels etc because it is true of abelian categories but I suspect it’ll be pretty painful

[hexed tangle]

Ahha

Also, I seem to be unable to react to messages with emojis?

[lone jacinth]

A -> B
C -> D
is pullback iff
0 -> A -> B(+)C -> D
is exact.
As B -> D is epi you get that
0 -> A -> B(+)C -> D -> 0
is exact, so the square is also pushout.

Pullbacks induce isomorphisms on kernels, pushouts on cokernels.

So in degree n+1 the kernels are the same, and the images coming from n+2 are the same, boom boom.

In degree n use that you can define homology as the kernel of the map
cok(d) -> P_n-1
and it's just the dual of before

Alternative approach:

Let K be the kernel of P^(n) -> P^(n-1)

From the square being pullback the differential K_n+1 -> K_n is an isomorphism, so the homology of K is 0 in degree n and n+1.

Then take the long exact sequence in homology

[queen tide]

i wonder if theres a way to automatically diagram chase

like with prolog

where you give constraints and it tries to prove things like that

[hushed bone]

Okay I wanted to try and do something like this but failed to see the kernel as a complex and LES

[ornate atlas]

Jagr with an incredibly slick proof as usual

[hushed bone]

Nope with the jagr glaze as usual

[lone jacinth]

The LES argument requires proving that first though. Which then ends up more complicated than the first one

[ornate atlas]

Youll write a good proof one day chmonkey, I promise

[hushed bone]

Proving what first

[lone jacinth]

The existence of LES in homology

[hushed bone]

Oh I guess but

I think you’d have that by the time you’re doing this thing

This is like something to setup the derived category

[lone jacinth]

Yeah true

[hushed bone]

Using projectives

[robust rain]

Let F be a presheaf of sets on X, let F^sh denote its sheafification. Let V \subset U. Is the restriction map F^sh(U) -> F^sh(V) given by essentially 'forgetting' all the germs of points not in V?

As, in, we send (f_p), p \in U to (f_p), p \in V? Is this even a morphism? It feels as if I am not specifying where some elements are sent?

[last talon]

That’s a morphism of (sets, groups, rings) fairly trivially
It’s just the projection morphism of the product

[robust rain]

Wait I am so silly

[last talon]

And yeah this should all be pretty tautological because we’re essentially saying “well if something works, it must be this. Does this work?”

[robust rain]

Yes of course (x,y) \mapsto x is a function

Thank you mico

[lone jacinth]

If you think of compatible germs as a function from U to the germs, then the restriction map is literaly restriction

[robust rain]

Oh this is neat. And the sheafification of a sheaf is the same datum as the sheaf itself because we showed earlier that "sections of a sheaf F over U" = "compatible germs"

[robust rain]

Do you mean as a function from F(U) to the germs?

[lone jacinth]

No

A set of compatible germs (f_p) is a function on U taking p to f_p

[robust rain]

I think I see

[lone jacinth]

Like the idea is if you let
X' be the union of all germs. Then you have a map
X' -> X
that just takes a germ to whatever point it is a germ over.

Then for a sheaf F, sections of F are exactly sections of this map and the restriction maps are restrictions.

[robust rain]

I need to ponder for a moment

[hexed tangle]

if 0 -> A -> B(+)C -> D -> 0 is exact isn't the cone on the morphisms is acyclic and hence they are quasi?

[vague pawn]

any hints for showing a localization of a UFD is a UFD?

[lone jacinth]

I mean. That's true yes

[lone jacinth]

A factorization of a/s is just a factorization of a times the unit 1/s

[hushed bone]

Lmfao

I think this is not that simple????

Oh no, I think what im thinking of is showing a regular ring is a UFD

At least in Matsumura I think you prove first that if a localization with respect to some multiplicative set with some special property is a UFD, the original ring is a UFD or some shit

[hexed tangle]

how does B -> D being epi imply that A -> B (+) C is a mono?

[lone jacinth]

It does not. That's unrelated.

If
0 -> A -> B(+)C -> D
is exact. Then
A -> B(+)C is mono, independently of whether B -> D is epi

But if B -> D is epi, then also B(+)C -> D is epi

[hexed tangle]

then how do you get that 0 -> A -> B(+)C -> D -> 0 is exact from B -> D being epi?

[lone jacinth]

0 -> X -> Y -> Z -> 0
being exact is equivalent to
Y -> Z being epi and
0 -> X -> Y -> Z
being exact

[lone jacinth]

Here you have a pullback square with P^(n)_n -> P^(n-1)_n epi

The point is that this square is also pushout

[hushed bone]

I’m pushout

[hexed tangle]

How do you argue that
"As B -> D is epi you get that
0 -> A -> B(+)C -> D -> 0
is exact, so the square is also pushout."
how is 0 -> A -> B (+) C known to be exact?

[lone jacinth]

The square is pullback

That's how P^(n)_n+1 is defined

[hushed bone]

You need a D at the end

[hexed tangle]

ah ok I see

[hushed bone]

And that’s the definition of a pullback via a formula

Or a construction I guess

[void plank]

How can I do moves in triangulated categories like jagr

[lone jacinth]

Just shoot for the stars, if it feels right then aim for the heart of a t-structure

[hushed bone]

Well if you miss the stars you’ll still be… wait it’s supposed to be shoot for the moon and if you miss…

[lone jacinth]

Shoot for the stars, if you miss you'll land among the vast suffocating emptiness of space

[robust rain]

you know in kung fu panda when po goes to the top of the mountain with master shifu? thats my plan for getting good

[lone jacinth]

The secret ingredient is ||that there is no secret ingredient||

[ornate atlas]

My plan is basically to think about them for the next 4 years, suffer a lot, and maybe become wise somewhere along the way

[hushed bone]

About what

[ornate atlas]

Maths

triangulated categories, likely among other things one would think

[urban granite]

I like stable ∞-categories

[hushed bone]

I don’t

[last talon]

Have you considered studying AC :3

[hushed bone]

What’s ac

[ornate atlas]

Axiom-o choice

[last talon]

axiom of choice Atiyah conjecture

[urban granite]

autistic cohomology

[hushed bone]

Idk what that is

[last talon]

Basically a whole cluster of conjectures about what values l^2 betti numbers can take

[hushed bone]

On what

[robust rain]

Aeanvalue Cheorem

[hushed bone]

What’s l^2

[last talon]

#1504882709569212616 message

[hushed bone]

What is C^2_n

[hushed bone]

Too much

Sorry

[last talon]

It no is simples

[obsidian marsh]

Peak

[robust rain]

Does 'natural' in this case mean 'unique'? Both of these objects are limits/colimits, so is the question asking me to verify that each side satisfies the others universal property?

[robust rain]

Can someone verify that this is the correct definition of the colimit? Vakil simply tells me to flip some arrows but I get confused sometimes about which arrows should be flipped and which arrows shouldn't be.

[hallow bone]

essentially, yes

alternatively you can show that the morphism induced by their universal properties is an isomorphism

[robust rain]

thanks : )

[hallow bone]

all of them

all the arrows should be flipped, because youre working in the opposite category

[robust rain]

wait i think F(m) is going in the wrong direction

[distant harness]

It doesn't matter for the concept whether you flip m and F(m); at worst that just corresponds to taking I^op instead of I.

(At best, to drawing the mirror image of the same commuting triangles, in which case the actual content of the definition is exactly the same).

[near lantern]

Natural means in the categorical sense. But the unique isomorphisms between things with the same universal property are natural, so that is a way to obtain naturality.

[robust rain]

Wait I've just had a thought. I don't think this is good enough. \

Suppose $\ker(\mathcal{F}_p \to \mathcal{G}_p)$ satisfies the universal property of $(\ker(\mathcal{F} \to \mathcal{G}))_p$. As the stalk is a colimit, this induces a morphism $(\ker(\mathcal{F} \to \mathcal{G}))_p \to \ker(\mathcal{F}_p \to \mathcal{G}_p)$\

Now suppose that $(\ker(\mathcal{F} \to \mathcal{G}))_p$ satsifies the universal property of $\ker(\mathcal{F}_p \to \mathcal{G}_p)$ As the kernel is a limit, this induces a map $(\ker(\mathcal{F} \to \mathcal{G}))_p \to \ker(\mathcal{F}_p \to \mathcal{G}_p)$.\

Originally, I thought one of the morphisms was going in the other direction, then I could conclude they are isomorphic because limits/colimits are unique up to unique isomorphism, so they have to compose to the identity. But, now I'm not so sure. \

I've already shown the first part, that there is a unique morphism $(\ker(\mathcal{F} \to \mathcal{G}))_p \to \ker(\mathcal{F}_p \to \mathcal{G}_p)$, but I'm not sure how to proceed. \

Any advice?

[broken turtle]

swifteeee

[robust rain]

I suppose I can try to show that the induced morphism is an isomorphism

and we're assuming that F is a sheaf taking values in abelian groups, so maybe this won't be too painful

[lone jacinth]

Are you doing set valued sheaves or a more general category?

[robust rain]

sheaf of abelian groups

or O_X modules

but I'm doing abelian groups for ease

[lone jacinth]

Yeah, then you can probably just show explicitly that it's injective and surjective

[robust rain]

I'm going to have lunch then give this a go

[robust rain]

I'm just thinking, so far I've only shows the existence of a morphism using universal properties, and there is no way I can prove injectivity/surjectivity using just existence? Maybe I need to explicitly constrcut a morphism using the characterisation of limits/colimits we have for categories that are set with extra structure, in terms of some quotient of the (co)product. hm

[past cove]

in general categories this doesn't hold

the fact that limits and colimits commute isn't a true statement generally

so you have to use the properties of the categories you're given

so you can't do this "purely abstractly"

[robust rain]

this is true for abelian categories in general right? (that limits in abelian categories commute, not that sheaves with values in an abelian cateogry form a category in which limits commute)

[lone jacinth]

No, not true in general.

An abelian category where direct limits commute with kernels is called AB5
https://en.wikipedia.org/wiki/AB5_category

[rose mirage]

I like how AB4 categories aren’t mentioned in this page

[robust rain]

oh curious. this grothendieck guy seems pretty good at maths

[hallow bone]

they're not real

[ornate atlas]

Its also interesting that it doesnt just redirect to the abelian category page where they discuss this anyway

[digital parcel]

what an odd stub article

since all this is already said in the abelian category page anyway

[ornate atlas]

Bro took inspiration and decided to repeat what I said 🥀

[digital parcel]

it's meta if you think about it

you're like the abelian category article and im like the ab5 category article

[last talon]

90% of the actual content of the article was written by the original person to create the article in 2010

[digital parcel]

i totally didnt just make up this excuse

[near lantern]

Perhaps even metabelian?

[robust rain]

Am I being stupid. Is there an easier way of doing this

Or do I have to just thug it out

[lone jacinth]

I mean, this doesn't seem that hard.

It's the induced morphism essentially by definition.

Well definedness and injectivity should be immediate and there's not much to showing surjectivity either.

[robust rain]

Okay, maybe I am just being a weakling and should just shut up and write

After thinking for 2 minutes it turns out that basically everything is immediate. And all of a sudden after spending a solid 3 hours on this exercise, all of it feels trivial and it feels like I haven't done anything haha

thank you all for the help

[hallow bone]

youve done lots! familiarised yourself with the concepts more

[robust rain]

No of course, I know, I'm being dramatic

[hallow bone]

i feel you haha

[robust rain]

this is the first time i've really experienced a 'grothendieck proof' I think. at each step we're unravelling definition, or doing some obvious diagram chase, and in the end we have really done something nontrivial

I feel like a nut in water rn

[lone jacinth]

Typical day in the life of math

[robust rain]

Also what the hell, for sheaves of abelian groups, kernel, cokernel, mono, epi and exactness are all stalk-local

I'm slightly stunned

I'm flabbergasted in fact

[robust rain]

I hated stalks ten days ago and now I am a huge fan and this makes me happy

[hushed bone]

It’s kinda the point

[lone jacinth]

You can cover a lot with them at least

Eventually you'll reach the limit though

[subtle smelt]

It is good to know that they are all fancy variations of "exactness is stalk local" using the properties of abelian categories

IMO the more amazing part is how many complicated things are usually determined by "coherence condition + pointwise condition", in case of morphisms of sheaves you want to know the existence of some morphisms but then a lot reduces to just looking at all the points.

[robust rain]

In Vakil we show that kernels, images etc. are stalk-local first and then use this to deduce that exactness is stalk-local. Does this mean that we can show that exactness is stalk-local first then deduce everything else?

[subtle smelt]

I'd have to think about it, but no thats the normal way

Preserving exactness is equivalent to preserving short exact sequences... which in particular means preserving kernels and cokernels

So any reasonabe proof would end up proving these two cases first

[lone jacinth]

You can make reasonable proofs that just proves exactness without splitting into cases

[subtle smelt]

What's the idea?

[lone jacinth]

I mean just consider a germ in the kernel and show it's in the image

I guess from there it would depend how exactly you're defining image of sheaves

But I'm not sure how it would help to split into cases

[hallow bone]

local global principles!!

[last talon]

I prefer local to local principles

[robust rain]

I got really giddy in my commalg module when we proved that being the zero module is local so it shouldnt be a surprise that i got so giddy at this

haha

[hallow bone]

its a very nice result

[last talon]

Prove that AG is local

[hallow bone]

and basically all the results you mentioned can be proven from that

[hallow bone]

every geometric property is local, AG is about geometry

hence AG is local

[robust rain]

I was actually wondering if there is a difference between study of local properties and geoemtric properties

[last talon]

Not every geometric property is local :3

[robust rain]

I don't intuitively why geometry should be local i suppose

[hallow bone]

such as?

[last talon]

A curve being a geodesic

(In the metric geometry sense)

[hallow bone]

(metric geometry is not geometry)

[last talon]

I say in the metric geometry sense because metric geometers require a geodesic to be globally length minimising and Riemannian geometers usually only want locally

[last talon]

(It is)

[hallow bone]

(but geometry is when topoi)

[last talon]

(Least brainrot algebraic geometer)

[ornate atlas]

I mean have you seen algebraic geometry when you dont do all that nonsense?

[urban granite]

do metric geometers use the word "topoi"?

[last talon]

It’s all bs 🙃

[near lantern]

How about being simply connected?

[last talon]

That too

But I’d say that’s topological not geometric

[vague pawn]

let A = k[x,y,z]/(x^2+y^2+z^2) where k is a field (char(k) is not 2). Then A can be checked to be integrally closed in its fraction field (This is similar to calculating the ring of integers in quadratic fields).
What happens when k is not a field? is it still integrally closed?

[hallow bone]

same thing duh

[digital parcel]

I think you need to have that 2 is invertible

and probably k itself should be integrally closed

[last talon]

[vague pawn]

another question. If A is integrally closed, is A[x] integrally closed?

how do I show this?

[past cove]

idt it's entirely straightforward to prove

[vague pawn]

idt = I don't think?

[past cove]

yeah I mean it's not hard but you need to have some preliminary results set up

https://stacks.math.columbia.edu/tag/030A

[vague pawn]

I will just take it for granted for now xD

[fierce steeple]

In short, like: if K = Frac(A) then note K[x] is integrally closed, so it's enough to show that A[x] is integrally closed in K[x]. In other words we need to show that given some polynomial f = a_n x^n + ... + a_0 where the a_i are in K, if f satisfies some monic polynomial g over A[x] then each coefficient is in A. But by induction on n and using our hypothesis on A, it's enough to show that a_n is integral over K, which should just follow from staring at g(f)=0

[steep pivot]

<@&268886789983436800>

[vague pawn]

Let A,B be two k-algebras and domains that are isomorphic as vector spaces.
If A is integrally closed then is B integrally closed?

[lone jacinth]

Isomorphic as vector spaces tells you basically nothing. Any finitely generated k-algebra will be a countable (or finite) dimensional vector space

As an example compare k[t] with k[t^2, t^3]

[vague pawn]

I see

[lone jacinth]

That they're isomorphic as rings for example

[vague pawn]

Not sure I have that in my case

Looking at
k[w,x,y,z]/(wz-xy) and k[x,y,z,w]/(x^2+y^2+z^2)

[past cove]

which one do you know is normal?

[vague pawn]

The sum of squares one

Also I would like just a hint

I need to check if the isomorphism I have in mind is an isomorphism of rings

[lone jacinth]

If k has square root of -1 it looks like you can maybe do a change of variables

[vague pawn]

Its alg closed yeah

[lone jacinth]

I didn't check, but looks like it could be possible

[vague pawn]

I will try

[lone jacinth]

Btw do you have a reason you want to use one to prove the normality of the other as opposed to just proving normality

[vague pawn]

Thats the point of the exercise I think

[digital parcel]

are they even isomorphic? diagonalizing wz-xy should give you a sum of squares in four variables, not three

[vague pawn]

Yeah i meant 4 squares

[digital parcel]

this should just give you the answer then lol oops

[vague pawn]

Already finished this exercise

[rapid cedar]

There’s no way around groking these out it seems?

[obsidian marsh]

Bro go to #prealg-and-algebra or #competition-math

[rapid cedar]

Alright alright

[bright swallow]

For an n-dimensional symmetric real matrix is there any well known tight lowerbound for the smallest eigen value?

[obsidian marsh]

This goes to #linear-algebra or sth

Idk

[ornate atlas]

Dont try to move people if you dont even know what question theyre asking

[obsidian marsh]

Okay

[distant harness]

What do you know about the matrix that you'd hope to get that lower bound from? Merely the fact that it's symmetric will not give you any.

[bright swallow]

Sorry I meant, I have an n-dimensional symmetric matrix. And it's too difficult to calculate the eigenvalues by solving the characteristic polynomial. So is there a way I can get a lower bound for the eigenvalues

[shy cargo]

beyond standard ones like the matrix norm idts. Consider just like a diagonal matrix with your favourite arbitrarily large or arbitrarily small values on the diagonal

[distant harness]

Also do you mean absolutely smallest eigenvalue or would large negative eigenvalues qualify too?

(Not that I have a good answer in either case).

[bright swallow]

negative eigenvalue qualify too

[foggy galleon]

shouldn't there be an inequality with some norm or something?

[lone jacinth]

Not really a bound, but iterating
x |-> Ax / |x|
should converge to an eigenvector for the largest eigenvalue.

And the smallest for A would be the largest for A^-1.

This is in absolute value though. Negatives might be annoying

I guess you might also just solve the characteristic polynomial with newtons method or whatever

[foggy galleon]

this is probably interesting to you https://en.wikipedia.org/wiki/Spectral_radius#Symmetric_matrices

if you have a norm on the vector space, that induces a matrix norm on nxn matrices that's compatible with the vector norm in the sense that |Ax|<=|A| · |x|. Then if x is an eigenvector |lambda x|=|A x| <=|A| · |x|. Assuming x!=0, you get |lambda|<=|A|

and in the case of the l^2 norm you get equality with the maximum of the absolute values of the eigenvalues

for other eigenvalues you have this

[neat oxide]

WHY THE HELPER TAG BEING PINGED

[obsidian marsh]

because someone needs help?

[neat oxide]

No they are not following the rule of 15 minutes

[obsidian marsh]

they probably dont know the rule

or are just impatient

[round seal]

gershgorin + the fact that they’re real might get you pretty far

[hushed bone]

advanced-algebra is immune to typos

[distant harness]

Somewhere a moderator is thinking, "oh, a challenge ...".

[drowsy niche]

it's known that trace 0 matrices over any PID must be additive commutators, but what do we know about when determinant 1 matrices must be multiplicative commutators?

this doesn't even hold over all fields (consider 2 by 2 matrices over F_2)

[drowsy niche]

it is true however over algebraically closed fields

[eager hound]

It holds over F_q if q> 3 though

[eager hound]

Actually it holds for any field with more than 3 elements

You can just prove it by hand by writing elementary matrices as commutators of elementary matrices

[drowsy niche]

wouldn't this only give us that every det 1 matrix is a product of commutators

[grave pumice]

is the Kahler differential $\Omega_{k[[x]]/k}$ what one would expect ($k[[x]] dx$,) or does the fact that $k[[x]]$ is not finitely generated cause this to fall apart?

[broken turtle]

lexi

[hushed bone]

https://mathoverflow.net/questions/21189/the-structure-of-the-module-of-kähler-differentials-of-rx-over-r

tl;dr

Hard

[grave pumice]

that makes sense; the 'sensible' description of the module of kahler differentials of k[x] depends on a derivation being determined by where it sends x

but k[[x]] isn't finitely generated

[eager hound]

Oh I thought you wanted to know if Sl_n(K) was perfect, not if every element was literally a commutator

That fails even for Sl_2(R) so I don’t know what kind of statement you’re hoping for

[drowsy niche]

mostly hoping that it would at least hold for SL_n(K) when n > 2 and K is not char 2

and indeed this is stated in the literature (see for example the intro of https://www.jstor.org/stable/1993409) but I can't find a proof

[signal tide]

they proved theorem 1 in that paper
you don't have the full paper ?

[glass hearth]

why does jstor watermark the pdfs you download with your ip and date and time

[wheat meadow]

probably to discourage redistribution