advanced-algebra

[spice idol]

i saw a proof for Tor using spectral sequences to abstract away all the messiness of the double complex, is that dualisable here?

[digital parcel]

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

[lone jacinth]

Yeah, the proof should be the same I think

[onyx imp]

ah okay, that is interesting, hopefully i will get to see something more explicit connecting the two soon

[spice idol]

Bosch's book on AG and CA mentioned it after a long ass proof lol

[spice idol]

can this in any way be extended to additive bifunctors? with some analogue of left/right exactness

[lone jacinth]

Some discussion here.
https://math.stackexchange.com/a/4586490/306319

But I think a key part of the step is that
Hom(-, I) is exact for I injective, so if your bifunctor doesn't satisfy that it's probably not true

[spice idol]

i see, that makes sense

[compact summit]

Does anyone have a nice intuition for direct/inverse limits that they can share?? I seem to often forget their construction and I generally have troubles working with them

For instance, its not immediately obvious to me that every abelian group is the direct limit of its finitely generated subgroups 🥲

[torn harbor]

injective (direct) limits are basically like generalized unions

the maps from one group to the next tell you how one group sits inside the next

a nice example is the group of all p^n-th roots of unity in the complex plane

which is an injective limit of cyclic group $\varinjlim_{n}C_{p^n}$

[broken turtle]

Blake

[torn harbor]

or as you said, every group is the direct limit of its finitely generated subgroups because every group is the union of its finitely generated subgroups

projective (inverse) limits are a bit more difficult to understand at least for groups.

In general if you have groups G_i, then in the injective limit (with a suitable directly family of maps) each element of each G_i will give you an element of the limit

conversely in the projective limit, each element of the limit will give you an element of each G_i

this is basically how I easily tell if something is a projective or injective limit

[compact summit]

Why would it be a union? Wouldn't it be better to use free products to be compatible with the group structure??

Thank you so much for your response Blake!! I found it very helpful!

[spice idol]

free products are very hard to work with

while using a quotient of disjoint unions an explicit construction can be obtained

[summer quest]

$\varprojlim_n\mathbb{Z}/n\mathbb{Z}=\widehat{\mathbb{Z}}=\prod_p\mathbb{Z}_p$

[broken turtle]

nGroupoid

[summer quest]

$\varprojlim_n\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}_p$

[broken turtle]

nGroupoid

[forest turtle]

i like to think of colimits as building an object by gluing together smaller parts and limits as reconstructing an object from its shadows.

[summer quest]

$\varinjlim_n\mathbb{Z}/n\mathbb{Z}=\mathbb{Q}/\mathbb{Z}=\bigoplus_p\mathbb{Z}[1/p]/\mathbb{Z}$

$\varinjlim_n\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}[1/p]/\mathbb{Z}$

[broken turtle]

nGroupoid

nGroupoid

[summer quest]

the examples of p-adic integers/profinite integers and the dual examples of Prufer groups are kind of the most basic and canonical examples to look at for both limits and colimits of Abelian groups in sequences like this

Q/Z has the nice interpretation in terms of all roots of unity in C*, similarly for Z[1/p]/Z and p-power roots of unit

[compact summit]

I guess it is easier conceptually to work with disjoint unions; but free products seem more natural to me here

[summer quest]

well you almost never want to consider these examples as just abstract groups

that's one issue with taking the free products approach

[compact summit]

I love thatttt!! Thank you :))

[compact summit]

Sorry, what do you mean by that??

[summer quest]

\hat{Z} and Z_p carry a natural topology since they are profinite groups

a lot of important structural features go out the window if you throw away this additional structure

[compact summit]

How does the structure go away??

I've become used to at this point that if you have a map X \to Y of topological spaces; then it should be considered continuous for instance. I haven't seen many situations where we don't unless its a question about that property (e.g. being continuous)

[summer quest]

well for example when one has a profinite group G like \hat{Z} one typically studies representations G->GL_n(F) with coefficients in some field F, and you get bad theorems unless you demand that these representations are continuous

if you know a bit of Galois theory then absolute Galois groups are maybe the most natural and interesting supply of profinite groups that nature hands you to study

\hat{Z} itself is the Galois group Gal(\bar{F_q}/F_q) for any finite field F_q

[compact summit]

I haven't seen examples where this goes bad before; I only just got introduced to Galois representations this morning and they are continuous?

[summer quest]

you want these to be continuous yes

[compact summit]

Is there a nice example you have in mind where this goes bad? 🥲

[summer quest]

like if you have a number field F and consider Galois representations Gal(\bar{F}/F)->GL_n(C) then continuity forces these to factor through a finite quotient Gal(E/F) for some finite Galois extension E/F, otherwise these things are completely out of control and pathological in general

[compact summit]

Right!

[summer quest]

another nice example to consider is if G=R/Z is the circle group and you consider the trivial representation M=R, then throwing away continuity you get Hom(G,M)=Hom(R/Z,R)=Hom(R/Q,R) which is huge and poorly behaved, whereas for continuous morphisms you get Hom_cont(G,M)=Hom_cont(R/Z,R)=0

[compact summit]

I guess I'm convinced that you generally want continuity. I'm interested in your claim that sometimes you don't want continuity!

OH, sorry, I didn't read properly!

[summer quest]

but yeah in the example of Galois representations you're meant to think of Gal(\bar{F}/F) as being built as a profinite group out of its finite quotients Gal(E/F), and the representation theory is meant to reflect that the representations of Gal(\bar{F}/F) are the representations of these finite quotients Gal(E/F) all bundled together

trying to do homological algebra with topological groups like this is classically kind of a headache, there are modern ways to avoid these headaches but not necessary to learn at first

[compact summit]

How do I verify these examples? Is it best to write out the direct limit from its construction as a quotient of a disjoint union, or should I verify the universal properties??

[summer quest]

I think in these cases it is easiest to write out the direct limit construction and identify the result explicitly

[compact summit]

Okayyyy :))

[summer quest]

both are manageable for examples like these

for absolute Galois groups in general part of the magic is that you have no a priori control on what finite groups are showing up as quotients, these are very highly mysterious groups in general

like already for Gal(\bar{Q}/Q) you really cannot explicitly name more than two elements of this group without doing some very non-canonical things involving the axiom of choice, nevertheless we can say a lot about the representation theory here

[compact summit]

I'm a little lost 😅 . Why do we want to control what finite groups show up??

[summer quest]

I mean you don't have to, but the point is that writing Gal(\bar{F}/F) as the projective limit of its finite quotients Gal(E/F) doesn't tell you very much if you don't somehow know a ton about what the tower of finite Galois extensions E/F looks like somehow

for finite fields you can understand what this tower looks like very explicitly

if you pass to the maximal Abelian quotient Gal(\bar{F}/F)^ab=Gal(F^ab/F) and so only consider finite Abelian Galois extensions then there is a lot more you can say

[compact summit]

Why is that? Are abelian Galois extensions easier to work with?

[summer quest]

well yes understanding them amounts to understanding 1-dimensional Galois representations only, and class field theory completely describes what is happening here

[compact summit]

Yesss, right!

[summer quest]

if you try to play a non-Abelian version of this game you already run into massive open problems for 2-dimensional Galois representations

[compact summit]

I guess you got me interested to learn more about Galois representations! 😂 . Do you have any recommendations to learn it??

[summer quest]

well the place to start (beyond basics of algebraic number theory) is to learn how class field theory works

Cassels-Frohlich has a fantastic book for this, as does Milne

[compact summit]

I've only had a first course in algebraic number theory 🥲 . Do you need more than that for class field theory?

[summer quest]

if you've had a first course you can start learning class field theory, you will spend some amount of time learning various supporting machinery like continuous group cohomology/Tate cohomology and just getting better working with local and global fields, but usually people just learn this stuff for the first time when learning class field theory anyways

class field theory is kind of the first really major big proof around algebraic number theory from some larger perspective

[compact summit]

From a google search looking for Cassels-Frohlich's book, I got a book on algebraic number theory. Is that one it??

https://math.arizona.edu/~cais/scans/Cassels-Frohlich-Algebraic_Number_Theory.pdf

[summer quest]

yeah

also worth checking out Milne https://www.jmilne.org/math/CourseNotes/CFT.pdf
https://www.jmilne.org/math/CourseNotes/ANT.pdf

the ANT notes are very nice for filling in any background, and then the CFT notes develop this story more or less completely

[compact summit]

Ooo, sounds like I need some homological algebra?? I'm still reading through (super slowly) Weibels book

[summer quest]

you learn it as you go

both Cassels-Frohlich and Milne have nice chapters on Lubin-Tate theory which gives you a very elementary and concrete proof of class field theory for all p-adic local fields, without any cohomological machinery

though the cohomological machinery gives a lot of further insight into the local situation, and then it's genuinely necessary to prove things in the global situation

[compact summit]

I'm a little worried that CFT is too far ahead for me; my ANT course was very basic in my opinion

[summer quest]

it takes time it's supposed to be a bit further ahead than any first course will prepare you for

you don't understand these sort of proofs all at once, you break them down into much more manageable pieces

if you're already reading through Weibel's book then group cohomology shows up sooner rather than later in that book, and applying this sort of abstract stuff to an actual goal like class field theory is how most people interested in number theory learn group cohomology in the first place

a lot of this machinery around Galois cohomology breaks this stuff down into manageable pieces, like all the actual computations reduce to group cohomology for finite groups

[compact summit]

Mmm, I haven't learnt group cohomology yet. I feel like I've hit a dead end with Weibel's section on spectral sequences...

[summer quest]

yeah I would recommend returning to that section later when it becomes more necessary

besides one typically learns spectral sequences through actual examples, not through an abstract treatment like Weibel

regardless you can get started with something like Lubin-Tate theory without any of this machinery and it's a beautiful story on its own

[compact summit]

I don't feel like I want to give up on it just yet. I'm motivated to learn it so that I can learn about Grothendieck's spectral sequence (I think specifically the Leray spectral sequence)

(I'm interested in all this for learning sheaf cohomology)

[summer quest]

generally one needs a rich enough supply of examples before being able to get the benefits of spectral sequences like this (like you should learn some basics of sheaf cohomology and learn the examples that you can compute by hand, before ever needing the various spectral sequences that help you with more complicated examples)

[compact summit]

Would that be do-able for what I've learnt so far from Weibel (chapters 1-3)?

I've learnt that sheaf cohomology is the right derived functor of the global sections; is it enough for concrete examples??

[summer quest]

I would skip straight to chapter 6 in Weibel and read this alongside the chapter on group cohomology/Tate cohomology in Milne. The two complement each other greatly

for sheaf cohomology I would recommend picking up places where this is actually used first, like in algebraic geometry from Hartshorne, or complex analytic geometry from most texts on Hodge theory

or in differential geometry from Bott-Tu

depends on your interests

[compact summit]

Thank you for your suggestion!! Is it for trying to learn class field theory, or sheaf cohomology??

[summer quest]

the suggestion with Weibel and Milne is for class field theory, it seems like you could start learning this without too much more background

for sheaf cohomology you will probably need more background to get to computing examples where this machinery actually matters, even if the abstract definition makes some sense

like in algebraic geometry the first real sheaf cohomology computation one does is the cohomology of the line bundles O(n) on projective spaces P^d

this appears somewhere in Hartshorne or you can look this result up and there are probably about a hundred shorter self contained notes various students have written before on this

[compact summit]

I'm actually thinking of reading through Gathmann's notes on AG over my holidays instead of Hartshorne 🥲

[summer quest]

yeah there are loads of more or less interchangeable books on this subject, so long as they cover what you want to learn

[compact summit]

I guess Hartshorne is a little scary for me 😂

So my plan was really to go through Gathmann's notes first; and then come back to Hartshorne :))

He has a section on sheaf cohomology

[vague pawn]

I thought this was incomplete because it was only 190 pages

But its 2 pages per page

But I will try to get a physical copy

Thanks nG and lucas, your conversion was helpful to me too

[lone jacinth]

The main benefit of direct limits as opposed to arbitrary colimits is that you can work with unions instead of free products.

This is because the forgetful functor to set commutes with direct limits.

[compact summit]

OH, that's nice!! Thank you!

[torn harbor]

the free product is a generalized union, you basically take all of the groups and put them into a big bag and then you allow all possible multiplications with no relations

In some sense at least

I don't remember if you need to use the free product for the inductive limit of general groups

[plucky arch]

In this case I think a union would actually suffice

[torn harbor]

Yeah I think so

[plucky arch]

The main thing you want is the colimit

Using a quotient of a free product is one construction of the colimit

But the useful part is the universal property, and I think the union would satisfy that

[distant harness]

Well, not a plain union, but a union quotiented by making elements that map to each other by a morphims in the diagram equal.

(If all the maps in your directed system are actual set-theoretic inclusions, then yes, just the union).

[plucky arch]

Yeah that’s what I was considering

But you’re right that in the general case you want the quotienting

So long as all your maps are injective, though, your directed system should be isomorphic to one where they’re all genuine set-theoretic inclusions, I think…

[distant harness]

Yes, if you identify everything with its image in the direct limit.

[fierce steeple]

I realise now that I am very used to saying filtered colimit rather than direct limit, but not cofiltered rather than inverse lol

(I guess filtered colimits etc kinda appear more in some areas of math though idk – and "cofiltered" is a bit ugly)

[torn harbor]

I hate the term direct limit so that term is probably better anyway

[distant harness]

What's the concept behind "filtered" here?

[compact summit]

Why is that??

[torn harbor]

because it's not actually a limit

it's a colimit

in the language of category theory

[compact summit]

But the alternative language also calls it a limit, doesn't it??

[fierce steeple]

No, you could say directed or filtered colimit

[compact summit]

OH, by alternative, I mean inductive limit

[compact summit]

I like the name "directed colimit"!! Sounds nice!

[compact summit]

I guess the adjective "filtered" can also apply for inverse limits too?? Something like filtered limit? Or would people have no clue what I'm talking about haha?

[fierce steeple]

Hm I am trying to think of some illustrative examples. I guess in my mind the point is just that it's slightly more flexible and general than "directed" and all you need to prove the usual things you prove about directed colimits. But it is necessarily slightly hard to give very good examples, because given a filtered diagram D you can always find a cofinal map D' -> D where D' is directed (so that the colimit of any functor out of D can be computed after composing with D' -> D)

[fierce steeple]

That would be what is sometimes called a cofiltered limit

[compact summit]

I thought that filtered means a sequence of nested subobjects. But the nesting can be increasing or decreasing?

^So I would of thought by dualising, you just get filtered again

[distant harness]

I was mostly wondering about the wording -- it sounds like there's a verb "to filter" which describes something that has been done to the limit, and I'm unsure what the relevant sense of filtering something is.

[fierce steeple]

Ah it is more that the diagram itself is filtered

[distant harness]

Is it just by analogy to "a filter is closed under binary intersections"?

[fierce steeple]

Like especially if you think of the prototyping example of having an increasing sequence of subsets or something, then this gives stuff a filtration

[fierce steeple]

I wouldn't think in terms of filters in that sense, but more filtrations in this sense

Which is presumably older

[distant harness]

I have no good sense of "filtration" either, though.

By which I mean, I can look up the definition at Wikipedia, but I cannot find an intuition that would connect it to something like a coffee filter.

(Of course mathematics doesn't owe anyone to make etymological sense ...)

[fierce steeple]

Hm I guess I would think of like when you have some increasing sequence so you are gradually adding in bits in stages. Or maybe the reverse (I.e. decreasing sequences) is more evocative in removing bits bit by bit

But this is a good point lol. I am unsure who coined this terminology for example

But I think this may be the spirit at least

[past cove]

Maybe the idea is like in a coffee filter you extract some concentrate from some mush and similarly here since for any two elements you have an element lying over them you're somehow concentrating the elements into something "finer"

Tho also I just came up with this so idk lmfao

[plucky arch]

It’s a categorification of “directed set”

[distant harness]

It looks like there's at least sometimes a distinction between "filtered colimit" and "direct limit" in that the former doesn't require the index category to be posetal.

[spice idol]

its more general

i believe

[glacial tangle]

I am currently in a seminar and somebody claimed M in Mod(R) M projective implies M free if R is a principal ideal domain

For finitely generated it is clear what about the infinite case?

[spice idol]

yes

over a PID every submodule of a free module is free

uses choice though, of course

[woven loom]

A bit, and I think strictly so, but direct limits get a lot of milage anyway

[woven loom]

Iirc the keyword is it’s a result of dedekind?

[spice idol]

Hilton & Stammbach do not mention a source lol

[woven loom]

Average citation

I think he explicitly wrote it out for Z (before Nielsen-Schreier)?

[glacial tangle]

Thank you

[lone jacinth]

Fun fact: for a connected commutative noetherian ring every infinitely generated projective module is free.

So in particular this also holds for PIDs (though you can probably prove it more easily for PIDs)

[spice idol]

you with your connected rings

[ornate atlas]

I feel like it should be more immediately obvious to me why no idempotents iff spec(R) is connected

Maybe theres secretly some AG magic to do, and I think i see why idempotents get you a disconected space, but not the other way

[last talon]

Idempotents => disconnected is relatively easy, but the other direction is bleh

[ornate atlas]

Ok thats good enough for me lol

I dont need to think about AG anymore than I need to, and I can see this direction

[last talon]

The other direction is errr
Spec(A) = V(I) union V(J) with their intersection empty
So I + J = (1) and IJ is nilpotent
So there are e \in I, f \in J with e + f = 1, ef is nilpotent
Then you modify your choice of e, f so that e + f = 1 and ef = 0

[past cove]

just think disconnected => space is disjoint union of connected components => show each component is affine => Spec maps direct sums to disjoint unions (and vice versa) => R = (+)R_i => idempotents

[ornate atlas]

Yeah this looks midly more involved and not something I should just be able to see lol

I am ok with that

[last talon]

Spec maps direct sums to disjoint unions is not immediately easy on the level of locally ringed spaces?

[past cove]

I mean is it that difficult?

Tho okay point taken

but that diagram isn't exactly step by step there are details left out ofc

[last talon]

(I’m mostly saying that bc I tried that approach a couple weeks back when we had it set as an exercise and that was where I got stuck lol)

[spice idol]

clopen sets correspond to idempotents

in particular this implies one direction of Stone duality

[lone jacinth]

So one way you can prove it fairly directly is just from the definition of the Zariski topology and Chinese remainder theorem you have that
R modelo the nilradical decomposes as a product if spec(R) is disconnected.

Then you can use the idempotents lift modulo nil ideals (which is something that is true and useful also for non-commutative rings, so is something you should prove anyway).

The alternative root is to build up more AG machinery, so that you can show there is a section that is 1 on one component and 0 on another.

[limpid horizon]

I heard u dont like ag much? Ive only recently started to get a little into it

That question seems cool tho

Ag also seems real tricky lol

[ornate atlas]

The stuff I’ve done I haven’t loved, though I haven’t done a huge amount by any means. I also would learn more if the opportunity to take a scheme theoretic course showed up

But like it’s a huge field, there’s loads I won’t have seen, and what I have done is very much influenced by like 3 people who work on pretty much the same area, and I know other people who do AG feel similarly about that area lol

[limpid horizon]

Yeah im perusing the idea of applying to phd programs so im wanting to be more sure on what area i want to go deeper into

[lone jacinth]

There a sort of middle ground argument here as well more in the spirit of AG without too much machinary:

So let R be your ring, N is nilradical and you've shown that
R/N = R1 x R2

pick e1 a lift of (1, 0) and e2 = 1 - e1. Then we want to show that
R = R[1/e1] x R[1/e2]

We of course have a natural map R -> R[1/e1] x R[1/e2], let us show it is injective.

Say r maps to 0. That means there is an n such that
r e1^n = 0 and r e2^n = 0. Thus
r (e1^n + e2^n) = 0, but e1^n + e2^n is a unit (because it is a unit modulo the nilradical), so r is 0.

For surjectivity note that there is an n such that (e1e2)^n = 0

Let u be the inverse of e1^n+1 + e2^n+1

Then consider r = u e1^n. As e2^n r = 0, the image in R[1/e2] is 0. Notice also
e1^n+1 r = u e1^n * e1^n+1 = u e1^n (e1^n+1 + e2^n+1) = e1^n
means that
r/1 = 1/e1 in R[1/e1].

Similar arguments show that (0, 1/e2) is also in the image. Together this give you surjectivity.

[hoary urchin]

I mean if Spec A = U disjoint union V, the map A -> O(U) \times O(V) is directly an isomorphism by definition of a sheaf (surjective because you can glue, injective because you can glue uniquely)

unless I missed something and you want to avoid using the structure sheaf

[verbal panther]

What properties does an algebra need to have to be integrable?

[spice idol]

what is integrable hereM

?

[verbal panther]

i guess then a better question is is there notions of integrability for operations other than addition

[verbal panther]

like integration for the numbers but instead of infinitely adding infinitely small numbers it would be the operation outputting infinitely close elements to the input infinitely many times?

sorry if this is completely incomprehensible

[spice idol]

well its just not well defined at all

you have to choose what aspect of integration is important and then you generalise that concept

for example, a derivation on rings is a generalisation of the differential, which can be generalised to arbitrary categories using Beck-modules

but this takes very specific algebraic properties and generalises it which you might not care about

the first sign that this question is ill-posed is that differentiation doesnt even have a nice analogue for arbitrary rings besides "preimages of derivations"

[lone jacinth]

There is some work in showing that the structure sheaf is a sheaf I guess

[near lantern]

Yes, but we can make it concrete: we use that A -> A[1/a1] ⨯ ... ⨯ A[1/an] is injective with range exactly (a_i) such that a_i = a_j in A[1/(ai aj)], which I think is a useful AG-inspired pure algebra fact (proved by pure algebra as you did above) just like local-global principles. In this case e+f=1 so this fact applies and ef is nilpotent so A[1/(ef)] = 0; hence A = A[1/e] ⨯ A[1/f].

Another way to use AG for ideas but algebra for the proofs: given A, I, J as before, we expect that A is a product B ⨯ C with I = B ⨯ {0}, J = {0} ⨯ C, but in general a closed subset or its open complement doesn't have a unique defining ideal because V(I) = V(sqrt I), so I = B ⨯ {0}, J = {0} ⨯ C won't be true for the first I, J you choose. But if we replace I, J by their radicals we can be sure that I = B ⨯ nilrad(C) and J = nilrad(B) ⨯ C (and in general any choice of I, J will be B ⨯ (some nil ideal), (some nil ideal) ⨯ C). The point of all this is that in the product representation we should have e = (1-f', e'), f = (f', 1-e') with e', f' nilpotent. (Now AG is done and we can go back to algebra.)

In particular, for any sufficiently high power of e the second component will become actually 0 and we might be able to make it actually idempotent. Say e'^m = 0 and f'^n = 0. Maybe two ways to go from here:
(i) (e+f) = 1 ⇒ 1 = (e+f)^N = binomial sum and if N ≥ m+n-1 then every term has either e^m or f^n. So we can take new e = e^N + N e^{n-1} f + ... + NC(N-m) e^m f^{N-m} and new f = other terms. (This ought to be almost the same in spirit as showing sum of (commuting) nilpotents is nilpotents: ef = (0, (1-e')e') + ((1-f')f', 0). But IDK how to do it using that directly.)
(ii) e^m = ((1-f')^m, 0) is not quite (1, 0). But ((1-f')^m is invertible with inverse p(f') = (1+f'+f'^2+...+f'^{n-1})^m, which is the first component of p(f). So e^m p(f) ought to be an idempotent we can use, for sufficiently large m and n. I haven't thought about how to prove this yet.

[robust rain]

(posting here because #groups-rings-fields is taken)

I am trying to find the radical of $(XY-X^2, X^3-Y^3) \subset \mathbb{Q}[X,Y]$

[broken turtle]

swifteeee

[robust rain]

I have proved some things about radicals, namely:

But I am struggling to use these to find the radical of the above ideal

I'm very certain it is (X-Y), but I can't figure out how to get there fully

r(I) is notation for the radical of I

[vague pawn]

you can factor out X-Y from both generators

[robust rain]

I've tried going down this route, and I end up with (X-Y) (X^2 + XY + Y^2 - X)

and idk if that is eaiser to work with

[vague pawn]

(X^2 + XY + Y^2, X) = (X, Y^2) = (X) + (Y^2)

[robust rain]

I dont see the first equality sry

[vague pawn]

X^2 + XY + Y^2 = X(X+Y) + Y^2

like if you could write an element of (X^2 + XY + Y^2, X) as
f(X,Y) * (X^2 + XY + Y^2) + g(X,Y) * X then you can write it also as
f(X,Y) * Y^2 + (g(X,Y) + f(X,Y) * (X+Y)) * X

[robust rain]

Yes of course

I feel really silly

[pallid wraith]

I would prove directly that $(X-Y)^3$ belongs to the ideal, so $X - Y$ is in the radical

[broken turtle]

Adayah

[robust rain]

then r(X,Y^2) = r(r(X) + r(Y^2)) = r((X) + (Y)) = r(X,Y) but (X,Y) is prime in Q[X,Y]. All in all, r(XY-X^2, X^3-Y^3) = r((X-Y) \cap (X,Y)) = r(X-Y) \cap (X,Y) = (X-Y) as it is a subset of (X,Y^2)

okay nice

[robust rain]

thank you so much : )

[robust rain]

Maybe. I did want to use the results that were used as part of the question though

[wicked nova]

To show that a permutation p in Sn is a k-cycle, is it enough to show that exactly k points are moved around while the others stays stationary?

[torn harbor]

no

for example the permutation (12)(34) moves exactly 4 points but isn't a 4 cycle

[plucky arch]

Hm does additionally having order k work?

[torn harbor]

I believe so

[plucky arch]

Cool

[torn harbor]

since the order is the lcm of the sizes of the disjoint cycles that make it up

for a counterexample you would need to find a set of numbers with lcm = their sum

actually maybe possible?

thanks AI overview

[torn harbor]

actually false, consider (1 2 3 4 5 6)(7 8 9 10)(11 12) this permutation has order 12 and moves exactly 12 points

(because lcm(2,4,6)=2+4+6)

[plucky arch]

Rip

[vapid axle]

Can I have a hint for the backwards direction of ii please

I know that the left side is contained in the right side by part i

[past cove]

remember that you can write the radical as the intersection of primes lying over your ideal

[vapid axle]

Hm okay, so we want to show that if q is a prime ideal containing S_p(0) then it contains p

[past cove]

yes

[vapid axle]

I have no clue what to do Imma try localizing both sides

[past cove]

try a contradiction

[near lantern]

What's a linear algebraic representation (i.e. polynomial components) of GL_n on some vector space V and a vector subspace W of V such that the stabiliser of W is the subgroup B_n of upper-triangular matrices?

[torn harbor]

GL_n acts on n by n matrices* by matrix mult and the stabilzer of B_n is B_n I believe although that's a bit boring

[worldly zealot]

sort of crude but i think you can do like V = k^n (+) wedge1 k^n + wedge2 k^n +.... and let W = span {e_1, e_1 ^ e_2, e_1 ^ e_2 ^ e_3,...}

[woven loom]

I don’t quite think GL_n, B_n is a vector space

[near lantern]

Ah, OK. Basically Stab(W1 (+) W2) in V1 (+) V2 is Stab(W1) ∩ Stab(W2)? That's neat.

[torn harbor]

oh oops I mean Mn

[near lantern]

OK dang. That's also really neat. And even cooler: both use the same direct sum idea (the only difference is the exterior power step).

OMW to wonder why I didn't think of this...

[worldly zealot]

yeah, i was just constructing something where you are preserving the flag

[young forge]

hey uhh.. does anyone have resources on finite alternative loops.

I may have just made a cryptographic breakthrough. (Generalization of conjugacy based cryptography to nonassociative alternative loops). I'm trying it over finite field octonians but wanna know how to cryptanalyze it.

If my intuition is correct, this would actually get rid of most of the weaknesses of conjugacy based crypto.

Namely, there is no linear representation to exploit.

[wary elbow]

You can view this in terms of weight theory for reductive groups. GL_n is reductive and B_n is a Borel subgroup. Thus, any irrep of GL_n has a highest weight space, and its stabilizer is the Borel. These are always linear algebraic representations (I'm assuming you're working over the complex numbers).

[near lantern]

Ah, also neat. Although I guess the highest weight needs to have a positive component for all the fundamental weights (i.e. be "regular" / in the interior of the standard chamber) to avoid having a large stabiliser? (Which also means it's easy to find a low-dimensional rep with stabilisers other parabolics. )

[wary elbow]

Ah yes, that is correct. It needs to be a regular weight (otherwise the stabilizer will be a parabolic that is strictly larger than the Borel).

[near lantern]

Does anyone know a concise exposition classifying all closed subgroups between a Borel B of a connected reductive group G and G? I think I used to know the argument but forgot it.

[rare walrus]

So you mean classifying the parabolic subgroups of a reductive group? I think Malle & Testermann has it in their book

I wouldn't call it entirely concise though...

I think Digne & Michel's book on the representation theory of finite groups of Lie type (the 2nd edition) has a VERY concise coverage of parabolics and Levis

Yeah, in chapter 3.2

[worldly zealot]

would you recommend this book

[rare walrus]

Malle and Testermann?

It's...

I want to recommend it. It feels a bit weird

I think it's a good book but it is one of those books that assumes that you sort of already know a bit of what's going on

If you know vaguely what a reductive group or a group of Lie type is, then I think it's a good book

The caveat is that I don't know of any great books on these things

[ornate atlas]

The Hatcher/algtop problem I see

[wary elbow]

It's straightforward to classify them. If you fix a Borel B, the subgroups containing B are all parabolic, and they correspond to subsets of simple roots. In particular, there's only a finite number of them. This should be in any book on algebraic groups.

As far as introductory books on algebraic groups, there are many out there and they're kinda interchangeable (Humphreys, Springer, Borel, etc). Jantzen's book is the bible and is mandatory if you want to dig deep, but it's not a good introduction.

[near lantern]

I don't want to know that there are finitely many, I want an exact list. And it's not as simple as a sum-closed subset of negative roots because the bracket [ga, gb] ⊆ g(a+b) could still be 0.

[near lantern]

Thanks, I'll check it out.

[near lantern]

Ah, you said subset of simple roots. This is the part I can't reproduce.

[wary elbow]

This is an exact list. There is a bijective correspondence between the power set of the set of simple roots and parabolics containing a given Borel. This is section 8.4 of Springer's book on algebraic groups.

This result requires some work -- it's nontrivial.

The construction is easy though: a subset of simple roots determines a root subsystem. You obtain the parabolic by adjoining the corresponding negative root subgroups to your Borel. The hard part is showing that this is exactly the list of parabolics.

[near lantern]

I agree that's the hard part, and it's what I want a reference for. It is surely in most books on reductive algebraic groups, but not necessarily in one place where I can read just that.

[onyx imp]

The tensor product of an irreducible $\mathbb{C}G$-module with its dual contains the trivial $\mathbb{C}G$-module as a submodule.

[broken turtle]

somethingwrong

[onyx imp]

Anyone familiar with this know where I can find an easy proof of this?

the place where I got this from is a more general book on KG-algebra where K might have non-zero character so the proof is abit more involved but I feel that for the CG-case, there should be an easier proof

[rose mirage]

Does this work: If $U$ is the irreducible $\mathbb{C}[G]$-module with character $\chi$ then its dual has character $\overline{\chi}$, now $\langle 1, \chi\overline{\chi} \rangle = \langle \chi, \chi \rangle = 1$

[broken turtle]

FIELD WITH ONE ELEMENT LOVER

[plucky arch]

yeah that's a character-theoretic approach you can take

you can probably also just construct the trivial submodule directly

[onyx imp]

i am abit unfamiliar but the first equality just comes from expanding the definition of the inner product right?

[onyx imp]

I thought that maybe the trivial submodule would just be the sum of all simple tensors

[plucky arch]

so long as you're finite-dimensional

[plucky arch]

it's close to this

[onyx imp]

yep working on finite dimensional*

[plucky arch]

there's an element of $V^* \otimes V$ which corresponds to the identity map

[broken turtle]

Pseudo (Cat theory #1 Fan)

[rose mirage]

yeppers

[plucky arch]

it can be written as $\sum_i \vec e_i^* \otimes \vec e_i$ for any basis $\vec e_i$ of $V$, with $\vec e_i^*$ the corresponding dual basis

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

but this sum is basis-independent

perhaps the easiest way to do this problem, then, would be to use the isomorphism $U^* \otimes U \cong \text{Hom}(U, U)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[onyx imp]

hmm i dont follow this, what would the link be between the tensor product and the identity map

[plucky arch]

then finding a trivial submodule is equivalent to finding a nontrivial G-equivariant map

which the identity always is

[plucky arch]

are you familiar with the isomorphism between $V^* \otimes W$ and $\text{Hom}_{\text{Finite Rank} }(V, W)$?

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

this works for any vector spaces, even infinite-dimensional ones

[onyx imp]

ah okay

i am not as familiar as I should be, thanks alot I think i can fill in the details now

[plucky arch]

if either $V$ or $W$ is finite-dimensional, then of course $\text{Hom}_{\text{Finite Rank} }(V, W) = \text{Hom}(V, W)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

note that this isomorphism also works at the level of $\mathbb{C}[G]$-modules, not just vector spaces

[broken turtle]

Pseudo (Cat theory #1 Fan)

[lone jacinth]

A particularly short proof could be: as CG is semisimple having the trivial module C as a summand is the same as having a non-zero map to it.

But
Hom(V*(x)V, C) = Hom(V, Hom(V*, C)) = Hom(V, V)

[rose mirage]

he's smooth wit it...

[lone jacinth]

(One Hom is over CG and one over C)

[rose mirage]

this is essentially my argument but less woke

[lone jacinth]

I guess in the end what is shortest comes down to what you've already established about / how you define tensor, dual, etc

[near lantern]

For V finite-dimensional (certainly true if G is finite), V (⨯) V* is isomorphic to Hom(V, V) as a G-representation, and span{id} is a copy of the trivial G-representation inside it. This works in the modular setting as well.

[onyx imp]

thanks all, i managed to get quite abit more reading in and got here

I am abit confused though, the whole point of my reading was that maybe the following statement held specially for C but it seems to hold for any algebraically closed field over any character:
For simple FG modules, V \otimes V* will contain the trivial module

[lone jacinth]

Don't need algebraically closed either

[onyx imp]

a

[onyx imp]

ah okay thank you, i think for my purposes, i am only looking at algebraically closed fields, maybe i will get to take a look at this in the future

[onyx imp]

oh and by divide, its just mean contains a copy

ah i have a couple typos here

Suppose the trivial module does not divide $M^* \otimes M$, by contrapositive of 2.2 $p\mid \dim M$ and thus by proposition 2.3 $M\oplus M $ divides $M\otimes M^* \otimes M$. But from what I just know, is that $M^* \otimes M$ will always contain the trivial module if $M$ is simple. So this means that $M \oplus M$ will never divide $M\otimes M^* \otimes M$ if $M$ is simple?

[broken turtle]

somethingwrong

[lone jacinth]

It seems you're picture uses the notation divides to mean has a direct summand, not just a submodule

[onyx imp]

oh yes your right but I thought that it meant the same thing?

ah that would only be true over something like C right?

[lone jacinth]

In the semisimple case it does. So if p doesn't divide the order of your group

[onyx imp]

so this would be right as long as p doesn't divide order of the group, is that correct?

[lone jacinth]

So you're saying that if the trivial module doesn't divide, then some stuff follows.

But since the trivial module does divide it you can't really conclude anything

[onyx imp]

ah okay, the argument doesnt make sense at all in any case then actually right

[lone jacinth]

But if you're over C, then p=0.

So then I guess 2.3 just gives you what you want

(I don't know if it's implicit that p>1 in 2.3)

[onyx imp]

hmm okay thank you, i think I have direction on what to look at now, thanks as always for the help

[fierce steeple]

Funny message out of context

[spice idol]

no pp 😔

[lone jacinth]

0, my favorite prime number

[dusky bone]

so wait, I've been thinking, with GCH-less ZFC, 2^X = 2^Y but X != Y
is the vector space F^X isomorphic to F^Y?

X and Y are cardinals

[dusky bone]

The answer seems to be yes because their basis's have the same size but I still need to check

[fierce steeple]

lol i meant more it sounds like C has char p

[lone jacinth]

Well it does

[nimble orchid]

Is there anyone here who can explain me what is Clifford algebra?

[ornate atlas]

Its like a normal algebra but big and red

[rose mirage]

if M is an R-module and Q is a quadratic form M -> R t's a quotient of the tensor algebra T(V) by the relations v (x) v - Q(v)(1 (x) 1 )

essentially defining how you should "multiply vectors" corresponding to the quadratic form

[nimble orchid]

In a popular book by Roger Penrose "the road to reality" I read that it is related to rotations in R^n and quaternions, but I cannot understand the link

[rose mirage]

the quaternions are a specific case of a clifford algebra over R, as are the complex numbers

specifically I believe you take the quadratic form Q : V -> R to send a vector (v_1, v_2) to v_1^2-v_2^2 for the complex numbers and (v_1, v_2, v_3, v_4) to v_1^2-v_2^2-...-v_4^2 for the quaternions

but I'd need to double check that

[nimble orchid]

Ok, ok, thank you! I will check by myself, don' t worry!

[rose mirage]

If I am wrong then it's definitely some combination of +s and -s

just not the one I wrote

[nimble orchid]

Another question Is It correct to say that exterior product Is the generalization of cross product?

[rose mirage]

yeah

although I'd prefer to just go all the way to "lie algebras are a generalisation of the cross product"

[plucky arch]

In what way

[distant harness]

R^3 with the cross product is one of the main examples of a Lie algebra.

[rose mirage]

the cross product satisfies the jacobi identity and thus gives rise to a lie algebra structure on R^3

as tropo said it's a main example of a strictly non-associative lie algebra

[potent plaza]

iirc its actually one of two non-abelian 3 dimensional lie algebras

real

or nvm u need also the condition that the derived subalgebra is of dimension 1 for this "classification" to be true

[rose mirage]

didn't know that, that's pretty cool

[potent plaza]

i mis-wrote that

this classifes real dim 3 liealgebras with L'=L

the other being sl(2,R)

if you were to take this over C, youd have only 1 such lie algebra (sl(2,C))

you know the two (sl(2,R) and the cross-product) are not the same by noting that

in sl(2,R) there is an element such that the adjoint representation is diagonal

while you can never have that with the cross-product lie

alg

[atomic acorn]

yeah for clarity so(3) is the cross product one

[fierce steeple]

Glad you said this

[potent plaza]

yeah you only have one if you consider over C right

[fierce steeple]

Lol I meant just cause I like Lie algebras away from R and C

[potent plaza]

right 😄

[onyx imp]

Let $F$ be a field and $G$ a finite group. The group algebra $FG$ is called semisimple if every $FG$-module is a semisimple module, i.e. decomposable into a direct sum of simple $FG$-modules.

[broken turtle]

somethingwrong

[onyx imp]

im trying to find something which explains why the decomposiiton is unique up to isomorphism/order

[ornate atlas]

This is just Maschkes theorem and the Artin Wedderburn theorem right?

Well perhaps just AW, maschke just tells you when FG is semi simple

[onyx imp]

i am not sure of the exact names, i know the former is Maschkes theorem. i chatgpt it but i know i will need the result but the proof chatgpt gave is abit involved

any idea of a book which goes through it in abit more detail?

[ornate atlas]

There’s a pretty good proof on Wikipedia, that’s probably the only modern proof I know where to find (this was an exercise in my noncom class)

There is a proof in like Hungerford but that goes a bit of a different way about it

[onyx imp]

i am not familiar with what an artinian ring is actually

[pastel shoal]

You don't need AW. Take two decompositions and restrict the identity map to the simple parts and use definition of simple. Serre's Linear representations of finite groups should have a proof

[onyx imp]

I will take a look, thanks both

[ornate atlas]

Ah nice, my representation theory is borderline nonexistent, I’ve just seen this as an application of AW

[pastel shoal]

Yeah I just looked it up, AW seems to be a lot more general

[lone jacinth]

Say you have a decomposition into simple modules
M = Sum S1 (+) Sum S2 (+) ...
where the Si are distinct simple modules.

Then Hom(Si, M) = Sum End(Si)
which is a vector space, so has a well defined dimension.

So the decomposition is unique

There's also the stronger
Azumaya-Krull-Remack-Schmidt theorem: if a module is the direct sum of modules with local endomorphism ring, then the decomposition into indecomposable modules is unique.

And the endomorphism ring of a simple module is a division ring, hence local

[cyan dagger]

heyy! I'm trying to solve this question. I understand why the composite passes to the quotient (mimicking the second order terms vanishing) , but I'm not quite sure how to show that the kernel of such map is the image of the first.

For me, the natural choice for the first map would be to send a dervation d to the vector (d(pi(xi)))_i , where pi denotes the surjection

however, i am struggling a bit to show that double composition vanishes. If I restrict myself to one summand, given a differential d, exactness should yield $d(\pi(x)) \cdot \pi(\tfrac{\partial f}{\partial x}) = 0$

[broken turtle]

Brindille Connexe

[cyan dagger]

for a f in the ideal I

however, after trying various manipulations with leibniz rule, I can't quite get this result. If someone has an idea to give me (I may have the wrong first map), that would be amazing!

[lone jacinth]

d(f) is exactly equal to Sum_i d(xi) df/dxi though right. And since f=0 in A, d(f)=0

if you don't already know this then it's enough to prove it for f a monomial (because of linearity) and then you can do induction on degree

[onyx imp]

I have been reading more abit more about ordinary representation theory when $\operatorname{char}(F)\nmid |G|$. Is the representation ring for the positive case exactly the same as for $\mathbb{C}$? I.e. take the free $\mathbb{Z}$-module over basis elements given by isomorphism classes of simple $FG$-modules where addition is essentially the direct sum of $\mathbb{C}G$-modules and multiplication is given by the tensor product. Is there any notable difference within the ordinary case that I should be aware?

I also recalled my professor talking about the representation ring over finite field $F$ being not very interesting and so when we talk about positive characteristic, we specifically mean an algebraically closed field. Is this still true for just within the ordinary case?

[broken turtle]

somethingwrong

[cyan dagger]

thank you!

it makes sense in a differential geometry point of view, but i didnt it held for algebraic derivatives

[lone jacinth]

The definition of the representation ring is the same yes.

I'm not sure why it would be less interesting over finite fields. As long as your field is a splitting field for your group the representation ring will be more or less exactly the same over the algebraic closure.

[onyx imp]

okay maybe i will look at this splitting thing, thank you

[hard kite]

Working in an abelian cat with enough injectives. If I have an object $A$ a resolution $A \to R^\bullet$ and an injective resolution $A \to I^\bullet$ you get choose an induced map (unqiue up to homotopy)$R^\bullet \to I^\bullet$ making the triangle commute. Can this be done naturally? What I mean is that if I had $A \to B$, a resolution $B \to S^\bullet$ and an injective resolution $B \to J^\bullet$ together with maps $R^\bullet \to S^\bullet$ and $I^\bullet \to J^\bullet$ can I choose $R^\bullet \to I^\bullet$ and $S^\bullet \to J^\bullet$ such that everythign commutes?

This must be very confusing to read ill draw it

[broken turtle]

1728

[hard kite]

I kind of want to say that you can view $I^\bullet \to J^\bullet$ as an injective resolution of $A \to B$

[broken turtle]

1728

[hard kite]

yea that works

it is an injective complex since it is bounded

[near lantern]

Off the top of my head, by all the uniqueness up to homotopy results, the map is (unique and therefore) natural if you view it as a map in the homotopy category. Whether it can be done natural without identifying homotopic maps, I don't know.

[hard kite]

Yea

I thibk it can

The idea I had is that you consider the category of complexes. Then you can view A -> B as an element of that category and (A -> B) -> (R->S)^* is a resolution of A->B. Since I^p -> J^p is a bounded complex in which every object is injective it is an injective complex. So (A -> B) -> (I -> J)^* is an injective resolution

And then you get an induced map (R->S)^* -> (I->J)^* making the whole thing commute

[lone jacinth]

When you say injective complex you mean injective object in the category of complexes?

If so you also need the complexes to be exact, which their not.

But if you're allowed to choose I and J aswell then it can definitely be done

[plucky arch]

So I’m trying to understand how the lie bracket gets defined for a Lie group

Say I have two tangent vectors X and Y at the identity

I can make them vector fields by setting the value at g to be g . X and g . Y right

Then… is the lie bracket the commutator of these two vector fields…?

[subtle plaza]

I know of two natural ways to define the bracket for the Lie algebra of a Lie group. One is as you describe: given vectors x and y in the tangent space at the identity, extend them to be left-invariant vector fields X and Y on all of G by pushing forward by left-multiplication. Then their commutator [X, Y] is again left-invariant (one should check this) and the value at the identity is what we define [x, y] to be. I don't really like this definition as I think it obscures what the connection to study homomorphisms between Lie groups actually is.

An alternative way is as follows. Given an element g in G, conjugation by g defines a smooth map F_g: G -> G which takes the identity to itself. Therefore, the differential of this map at the identity, denoted Ad_g, is a linear map from T_eG to itself. One can check that Ad behaves well with respect to composing the F_g maps, so we get a homomorphism Ad: G -> GL(T_eG) taking g to Ad_g. By differentiating Ad at the identity, one gets a map T_eG -> End(T_eG) which takes a vector x to an endomorphism ad(x). One can define the bracket by [x, y] := ad(x)(y). A more expository derivation of this is described in Fulton and Harris' book I think (as well as a description of why these two definitions are equivalent).

[plucky arch]

i've seen the conjugation definition before but i always got confused by the double differentiation

[subtle plaza]

Fair, I guess I tried to reconcile this by observing that we are treating GL(T_eG) as a Lie group and thinking hard about why Ad is a smooth map so I can differentiate again and what this means intuitively in terms of infinitesimal conjugation

[plucky arch]

yeah why is Ad a smooth map

[wary elbow]

It's the composition of left and right multiplication by g and g^-1, respectively. Multiplication in the group is always smooth.

[fierce steeple]

Ig this is more about Ad(g) rather than Ad itself

[plucky arch]

yeah exactly

i can totally believe Ad(g) is smooth

but that's different from Ad itself being smooth

[fierce steeple]

Should come from considering the map G x G -> G sending (g,h) |-> ghg^-1 locally, like ig this is a submersion

Actually this is essentially the approach here I guess https://math.stackexchange.com/questions/4049481/adjoint-map-textadg-rightarrow-textaut-mathfrakg-is-smooth

[hard kite]

yea i was being kind of confused

i fixed it out though

because in my situation i had a ses of resolutions and of injective resolutions

and its easy to check that an ses of injectives is an injective complex

and then everyhting works out

[mild imp]

Anyone know of any abstract algebra tutors?

[near lantern]

Ahhhh so one has to use the tangent bundle like this instead of just tangent spaces.

I was musing over this recently; it's nice that the answer just appeared in front of me.

[plucky arch]

What’s a good proof of the hook length formula

[worldly zealot]

i don’t think there’s any thats not annoying combinatorics and sum manipulation

[plucky arch]

Nothing I could show my students?

[worldly zealot]

yeah iirc a lot of the combinatorial identities you can use for rep theory are really nontrivial combinatorial facts

and have involved proofs

wikipedia shows a probabilistic pseudo-proof so maybe you can do that but also i felt like i didkt lose anything when my professor told us to trust the combinatorics

[fierce steeple]

Tbh I remember doing this in like first year undergrad or before, didn't think it was too bad?

Maybe it was the probabilistic method

[plucky arch]

How would you explain it

[fierce steeple]

Actually sorry I more meant to reply to hk

[worldly zealot]

was it like, taught?

or an exercise

[fierce steeple]

Actually n o I think I got it confused with smth else. My bad

Definitely not first year lol

I think it was the special case for Catalan numbers that we did

[worldly zealot]

oh

[plucky arch]

i see

This is where my lack of understanding of combi holds me back

[worldly zealot]

i don’t even think it’s an “understanding” thing

like it’s not conceptually deep it’s just very involved

[plucky arch]

Do you mean combinatorics

[worldly zealot]

the proof

[plucky arch]

hm…

in my experience this usually means there’s something I’m missing about the proof

[near lantern]

I feel like I remembering seeing a proof in some slides online that seemed pretty motivated, but IDR anything about it.

[ornate atlas]

I mean when it comes to combi I think it just often is the case that there’s nothing to do but suck it up and count, which can be a bit of a painful process but not all that conceptually hard

I don’t know this formula or the proof so I couldn’t comment on that, but just as a general point about combinatorial things

[plucky arch]

hm…

I’m not used to that

I guess like

I’m very happy doing big calculations in physics

Not as much for maths

[ornate atlas]

Yeah, I don’t really know what to say to that beyond that’s just how combinatorics tends to be. Not to say there aren’t very clever arguments there, but often times you just have to consider cases and get to counting

[worldly zealot]

https://sites.math.washington.edu/~billey/classes/561.fall.2019/past.articles/glass.ng.2004.pdf i found this which calls itself simple , its 4 pages and uses a complex analysis theorem

[ornate atlas]

That looks gross

[worldly zealot]

lol

[fierce steeple]

Ig more just a generating function trick right

Rather than really complex analysis

(As one would hope lol)

[worldly zealot]

yeah

i was clickbaiting sry

[plucky arch]

I’ve often found it hard to “get” combinatorics

[drowsy niche]

this really depends on what they/you have seen. my personal favorite proof uses some facts about symmetric functions/rep theory of lie algebras that you may or may not have seen.

[ornate atlas]

Same here, counting is hard

[plucky arch]

To me it’s always felt very disconnected and hard to make sense of

[worldly zealot]

i like algebraic combinatorics

[drowsy niche]

https://mathweb.ucsd.edu/~ssam/old/22W-202B/notes.pdf#subsection.5.2
roughly follows my preferred proof

[ornate atlas]

I don’t know much in general but the combinatorial comalg I’ve done is pretty sweet

[plucky arch]

Like it very rarely felt like i was ever “getting better” at combi

[drowsy niche]

this is kind of the appeal of combinatorics imo

[plucky arch]

the… appeal?

[drowsy niche]

yes, some people like doing difficult things asap

[worldly zealot]

oh yeah. met richard stanley on monday it was cool

[drowsy niche]

which is easier to do in combi than other areas

[ornate atlas]

I feel the same, but I also wouldn’t say I’ve really put enough effort into getting better at it to know if that’s a general truth or not

[ornate atlas]

Oh yeah you mentioned that! Guys a great author very jealous

I briefly spoke to David Eisenbud on the stairs last week though

[plucky arch]

This I don’t quite understand

[fierce steeple]

V cool

[ornate atlas]

Guy is a great lecturer, very engaging

And from the very brief chat I had with him about getting lost on the way to the talk, lovely too

[lone jacinth]

I got an email from Eisenbud the other day. Felt a little starstruck.

[ornate atlas]

Oh shit! Very nice haha

This is a very nerdy game of mathematical one ups manship happening here

[lone jacinth]

It was just journal formalities, not an actual conversation, but I can feel a little starstruck if I want

[fierce steeple]

Ok I mean I zoom called [prof] recently

[ornate atlas]

Oh if I ever got an email from Eisenbud I’d be milking that

[plucky arch]

It’s fun being an outsider and having no clue who eisenbud is

[ornate atlas]

He posed an interesting question at that talk I could’ve possibly worked on as my dissertation but nah

[ornate atlas]

Amazing author, huge in alggeo and comalg, head of the MSRI for years

[fierce steeple]

Well-known and respected mathematician in algebraic geometry / commutative algebra who also wrote one of the most-used comm alg texts

[ornate atlas]

Also appeared in that one very famous video about hagoromo chalk

[plucky arch]

Makes sense i wouldn’t know anything about him

[fierce steeple]

TIL that Maclane was one of Eisenbud's advisors

[plucky arch]

I’m quite far from AG and comalg

[ornate atlas]

https://youtu.be/PhNUjg9X4g8?si=2n5zddxZ-la0WIvk

Crazy that I remember watching this when it came out, before knowing anything about maths, only to end up being a big fan of the chalk and meeting the guy in the video

[fierce steeple]

Lol v cool

[lone jacinth]

He has many appearances on numberphile
https://youtube.com/playlist?list=PLt5AfwLFPxWK45hX6QOhtri2cZwwX8G-5&si=3utEnTUl8sfR3RVd

[ornate atlas]

I did always wonder “it can’t be that much better right?” And honestly they had a point, only chalk worth using

[fierce steeple]

I would say w conferences you should fairly quickly meet a lot of your heroes

Indeed I have after just a couple confs

[plucky arch]

Hm then probably none of the videos I’ve watched

[ornate atlas]

I do think he’s still worth knowing about just because he’s a very good writer and orator

[plucky arch]

really…?

[ornate atlas]

I will make it there, one day

[fierce steeple]

Just turn up for free food

(When that is available)

[ornate atlas]

There was actually a really cool conference over summer that I really wanted to go to but unfortunately… not a PhD student…

[ornate atlas]

I think so yeah

[fierce steeple]

I think often masters students can go to stuff but ye depends

[plucky arch]

this would be good for me if I didn’t suck at my job

[fierce steeple]

Naur I am sure ur great

[ornate atlas]

I was at that time just unemployed lol, and it was ran by people from my uni I knew quite well “the ring theoretic aspects of Lie Theory”, seemed very cool

[fierce steeple]

It's when like

You calculate the truth by using rings

[plucky arch]

given my record so far, not especially

[ornate atlas]

Yeah it was a summer school on the Boolean ring

[fierce steeple]

Hug I have belief

[fierce steeple]

Nice

[plucky arch]

im trying to keep some amount of belief

[fierce steeple]

(This actually reminds me of smth I meant to look up w lie algs lol thanks)

[spice idol]

boolean rings mentioned

[ornate atlas]

Boolean rings plural?

I’ve only ever met like F_2 (using join and meet)

[spice idol]

well you know, x^2 = x so boolean ring = boolean rings

[spice idol]

take any set of propositions closed under finite OR, AND, and NOT. This defines a Boolean ring

[unborn rampart]

I don't think it's great for your self-confidence to constantly remind everyone how you suck at math tbh

[spice idol]

or, equivalently, the powerset of a set forms a Boolean ring

[ornate atlas]

Interesting, I hadn’t tthought about that

[spice idol]

(more strongly a complete atomic Boolean algebra)

[spice idol]

there is an equivalence of Lawvere theories of Boolean rings and Boolean algebras

from this lense, Stone duality is just a nice case of the spectrum of a ring being a contravariant functor

[ornate atlas]

Those are a lot of words that I’ve seen but don’t understand

[spice idol]

(actually there are a couple lenses as to why one might expect Stone duality to exist)

(one of them being universal algebraic geometry)

[plucky arch]

This is about me sucking at my job actually

[spice idol]

(but thats not a conversation we're ready to have yet)

[unborn rampart]

What's your job?

[plucky arch]

Being a physics phd student

[unborn rampart]

Hmm, I doubt you can draw the conclusion that you're not good enough to be a phd student already, you only started this fall, right?

[plucky arch]

I started last fall

[unborn rampart]

I haven't done a phd, so I don't know what the experience is like, but I'm sure it's a whole lot of failing and struggling, until you eventually figure it out

[fierce steeple]

Tell me about it

[unborn rampart]

Anyways, my point is that when you constantly tell yourself (and everyone around you) that you suck, that's what you'll believe, regardless of whether it's true. You can struggle with math without actually sucking at math

You shouldn't make "sucking at math" into your personality

[fierce steeple]

Instead, you should tell other people they suck at math

[plucky arch]

This is the main place I talk about math so it naturally comes up here

[spice idol]

they call me the projector

📽️

[plucky arch]

I do suck at algebra at least

[spice idol]

means its.only up from here

[plucky arch]

-# im aware this will be used against me in the future but idc

[unborn rampart]

Well... I'm pretty sure I suck at math more than you, and I don't go around announcing it. It "naturally" comes up because you bring it up. Tbh it seems like a defense mechanism, like if you fail it's okay because you already know you suck. Sorry, I'm not trying to be rude btw, but I think you have some math-trauma it's worth working through

[plucky arch]

Sure but there’s lots of empirical evidence for me sucking at algebra

[lone jacinth]

Seem you figured out the rep theory of Sn pretty quick and well. Doesn't seem like sucking to me

In fact working through and understanding things that are hard is probably a better quality than magically having intuition for things from the get go

[unborn rampart]

Again, if that's what you want to believe, that's what you believe, but it's your choice to make "sucking at algebra" part of your identity. If your only criteria is that there are people who know more than you then pretty much everyone sucks

[rose mirage]

What if you actually do though

[plucky arch]

I just read Fulton and Harris and spent a few hours trying to understand the argument

[plucky arch]

My criteria is an inability to have intuition for algebra is all

[unborn rampart]

Yep, that's completely normal

I don't think many people have a natural intuition for algebra

[soft parcel]

Hey guys, if I want to figure out what the nilradical of a certain Lie algebra is, what would be possible things I might want to have a look at?

[plucky arch]

#advanced-algebra message

I got stuck on this basic thing

[limpid horizon]

You’re probably just on a downer, maybe just take a break and dont dwell on it

I know how you feel though sometimes i feel like a complete idiot

[unborn rampart]

But like, it's pointless for me to try to convince you otherwise. I'm just trying to say that if you stop saying that you suck, you'll eventually stop believing it

[ornate atlas]

<@&268886789983436800>

[limpid horizon]

Why tag moderators

[distant harness]

Scambot attack, dead now.

[limpid horizon]

o ok

[near lantern]

I saw [prof] in the hallway less than 12 hours ago.

[near lantern]

(more strongly a Boolean algebra, and nothing but Boolean algebras)

[near lantern]

It still shouldn't be part of one's identity.

[near lantern]

FYI I've tried to read the classification of reps of Sn at least 4 times (or 10? who knows) over 2-3 years and I still don't feel like I "really" understand it or have intuition for it. I've assumed it will be a matter of time, attempts and seeing it "in the wild" and made my peace with not "really" understanding it for the indefinite future.

And I'm literally trying to do research in representation theory.

[drowsy compass]

yeah, it's the sort of proof that you need to play with to really understand

[ornate kindle]

I've got a homework problem I'm really stuck on

Prove or find a counterexample, if the center of a finite group $G$ is cyclic, then $G$ has a faithful irreducible representation.

[broken turtle]

NotABot

[ornate kindle]

Any hints/suggestions?

[last talon]

Immediate gut feeling: false because how the fuck do you prove this otherwise 🙃

[ornate kindle]

I showed earlier in the question that G having a faithful irrep implies the center is cyclic, but I've been spinning my wheels on this for like 3 days

[ornate kindle]

Lmao thats practically where I'm at

.

This is the whole Q btw

Working on (c)

Any chance I have the converse statement wrong?

[last talon]

C_2 x C_2 works

[ornate kindle]

Doesn't C_2 x C_2 have a non-cyclic center?

[last talon]

mishu read hypothesis challenge

[ornate atlas]

I spent a good our on a problem yesterday wondering why something was true only to realise it was the assumption of the theorem I was trying to prove

[ornate kindle]

lol, happens

[rose mirage]

I know how to do the converse but not this way round

pmo

[ornate kindle]

Like faithful irrep exists => Z(G) cyclic?

[rose mirage]

yeah

[ornate kindle]

Yeah that was part (b) of the Q

[rose mirage]

ok so we just need a group with a non-cyclic centre

I still feel like this should be true

[ornate kindle]

How would finding a group like that help?

Like, (b) gives us that a group with a non-cyclic center won't have a faithful irrep

[rose mirage]

yes and we're now assuming that G does have a cyclic centre

[ornate kindle]

But we're trying to show that having no faithful irrep means the center isn't cyclic, or the center being cyclic gives the existence of a faithful irrep

[ornate kindle]

Or finding a counterexample to this

[rose mirage]

quickly looking for a counter example now but my gut tells me it should be true

[ornate kindle]

Same

[ornate atlas]

This question really ramps up in difficulty huh

[ornate kindle]

Yeah

Well this is an undergrad/grad split course, and this is the grad student part of the question lol

[ornate atlas]

That makes sense lol

[rose mirage]

I've been properly nerd sniped by this

[ornate kindle]

Blugh I'm really out of practice with proper group theory too, so I'm having trouble trying to even start looking for a counterexample

[rose mirage]

is it an inductive thing where you can assume the centre is trivial and then inflate back up through the central series

[ornate kindle]

Maybe.......

[ornate atlas]

Yeah I havent got a clue, but the fact that you dont know is what tells me this is very tough

[ornate kindle]

I'm looking through stuff on characters, since that's come up a bunch on this assignment

[rose mirage]

the kernel of a rep is exactly the kernel of the character, all normal subgroups arise as the kernel of a character. If every character is non-faithful then every character arises as an inflation from a quotient.

the issue is they can be different quotients

ok nope got a counter-example

I think

[ornate kindle]

Don't tell me what it is

[rose mirage]

ok I won't

I will say that the order is quite small and it's findable

[ornate kindle]

I really really really wanna know, but this is homework so I should at least try to find it myself lol

[rose mirage]

fair enough

[ornate kindle]

Coolio, I'll look around 👍

[rose mirage]

if you want helping hints lmk

[ornate atlas]

Time to crawl through groupwiki I guess lol

[ornate kindle]

groupwiki?

[last talon]

What is the order?
(I’ve checked the standard groups and constructs and all seem useless)

(Granted, these are “what can I compute in my head” but like)

Like I don’t think anything below order 16 should work
Although there might be one of order 12

[distant harness]

If you find it (groupprops), please let us know where -- the URL I get from searching has been dead for weeks.

[rose mirage]

it's not a standard group, it's order 18

[last talon]

Ah bleh I think I know what it should be

Yeah I think I know

[rose mirage]

raised eyebrow emoji.... go on....

[last talon]

Yup

[rose mirage]

well don't leave me in suspense - what's the presentation?!? what's the name!??!

[last talon]

||https://people.maths.bris.ac.uk/~matyd/GroupNames/1/C3sS3.html||

[rose mirage]

WOW

[distant harness]

18 sounds a lot like ||the wreath product of C3 by S2||

[last talon]

Like I knew the group then I spent a bit looking for the char table to check

[rose mirage]

lemme think about if these are iso

[rose mirage]

the humble method of little groups:

[rose mirage]

I don't think they are iso unfortunately

I'll check to see if this is another example

[distant harness]

No, and my proposal does have a faithful irreducible representation, so it falls apart there too.

[rose mirage]

ah, thanks for saving me the trouble

I think the ingredients that indicate a counter example are a large commutator subgroup and a large abelian subgroup

which are requirements at odds with each other, hence why counter exampels aren't common

[ornate atlas]

https://groupprops.subwiki.org/wiki/Main_Page doesn’t this work?

[ornate atlas]

I was thinking of groupprops, just forgot the name

[distant harness]

I get just 403 Forbidden there.

[ornate atlas]

Seems to work fine for me, hmm

[rose mirage]

there are 80 counter examples of order <= 200

[last talon]

It works for me

[ornate atlas]

Groupprops becomes the one website in the world to only work in the UK

[rose mirage]

another brexit benefit

[digital parcel]

Guys i dont want to ruin the fun but i'm in the us and it works :(

[ornate atlas]

I will say, groupprops is probably better and a great resource. But the name is just nowhere near as cool as DaRT

[digital parcel]

what's that

[ornate atlas]

Database of Ring Theory

[digital parcel]

Lfg didn't even know about that

[ornate atlas]

Searchable database for anything rings, it’s a nice way to look for counterexamples and stuff

[digital parcel]

AG has fanography

https://www.fanography.info

[ornate atlas]

Yeah but

[rose mirage]

groupprops does not have the table of marks for SD_{2^n} it's USELESS

[digital parcel]

wtf

[ornate atlas]

Someone suggested me an MSc thesis computing a bunch of 3-folds and just like, I’m good

[digital parcel]

"right-not-left"

[rose mirage]

computations are cool I like doing them

[digital parcel]

computations are the best way to learn something

and pretty satisfying to do

well

the process isn't always satisfying

but sometimes the result is

[ornate atlas]

I don’t disagree, the project I want to do will probably just be computing a bunch of cohomology, I just find alggeo uniquely miserable

My friend from UG is currently working on exactly the same problem of working out new fano 3-folds and shit just seems painful

[ornate atlas]

At least that kinda flavour of AG, it’s a big field that I don’t know that well, and all my exposure to it has been coloured by people who do 3folds and birational geometry

[digital parcel]

I know I like AG i'm just still finding what exactly I like

[distant harness]

Hmm, it does work for me on mobile, just not from my laptop (either Firefox or Chromium), though the two devices are on the same wifi.

[ornate atlas]

I’m on my phone just now, let me try on my laptop

[digital parcel]

I'm on my laptop

[ornate atlas]

Yeah works fine in safari and chrome on my laptop

This seems to be a very unusual issue on your end lol

[digital parcel]

Maybe tropo got banned from groupprops

[ornate atlas]

Hardware ban from groupprops would be an interesting one

[ornate atlas]

Im still keen to learn more, it seems like theres a lot of cool ideas in there. I just havent found anything that really blew me away yet, Ive just not loved its vibe. But like I also dont know what a sheaf is so im very far from being able to really pass judgment on the subject

[digital parcel]

I started liking ag because I have a friend who is doing his phd in it

And he would always tell me these cool things and he helped me with chapter 1 of hartshorne

He basically showed me all the cool stuff in ag through those exercises

[ornate atlas]

I saw recently that universal enveloping algebras of lie algebras correspond in some way to coordinate rings which is kinda cool I guess

[digital parcel]

Oh interesting I didn't know about that

We learned about universal enveloping algebras I think last week in my lie groups class

[ornate atlas]

I still dont know much about them, but my non-com class had us do some proofs with them

Im guessing they showed up there because that prof is interested in the intersections of AG and noncom ring theory, particularly through universal enveloping algebras lol

Im taking Lie Algebras next sem so ill maybe know what their deal is a little more than just cool rings soon

[digital parcel]

When I think universal enveloping algebra I just think "set [x,y] = x \otimes y - y \otimes x"

where [-,-] is the lie bracket and \otimes is in the tensor algebra (since it's just a quotient of the tensor algebra)

[spice idol]

a sheaf is the correct way to do AG

[digital parcel]

There's another way to define U(g) from some basis

It also has a nice universal property

[wary elbow]

Among other things, enveloping algebras are a way to apply ring theory to representations of Lie algebras

Humphreys has a really nice book on this

You can think of them as a noncommutative deformation of the ring of functions on the dual Lie algebra

[digital parcel]

Oh interesting

[spice idol]

its a particularly nice example of an adjunction induced by a term-reduct (when you make a new variety of algebras by taking some set of terms (in this case the underlying vector space and the term c(x, y) = xy - yx) and making that the new operations

[digital parcel]

Interesting

Oh yeah I watched this last night and somewhere like 3/4ths the way through they mentioned UA

https://youtu.be/vmcbm5FxRJE?si=-DpXVL6qMlprMOis

Apparently grothendieck was interested in it for a while

Awesome talk too but he kept reading the french part in french and so I had to keep pausing and translating

[spice idol]

omgmgomg

[young forge]

Alright first does anyone know any GAP libraries for:

1. Cryptanalysis of algebreic asymetric crypto schemes
2. Representation of and computation in finite exceptional groups of lie type? (Part of this cryptanalysis invovles doing stuff in the exceptional group G(p) for a large prime p.(that's the automorphism group of octonions over a finite field)
3. Algebra with octonions over finite fields(note split and non split octonions are isomorphic i think over finite fields).

uhh does anyone here know anything about this

[distant harness]

... you'll find out by seeing if you get an answer to your question.

[wise sedge]

Up to isomorphism, there is exactly one split octonion algebra over any field. It just so happens that over finite fields no non-split octonion algebras exist.

[ornate kindle]

I fucking got it

God freaking dammit, I wasted so much time on this

But in class today my prof suggested using Sage to find a counterexample and it took me like 45 minutes

30 minutes to figure out how to build a checking function, 15 minutes to run small groups from the Wikipedia page through said checker until I found one

I want to throw up. Jesus christ

Well.... at least I got some practice with Sage it of it lmao

[rose mirage]

flexing my magma code that did this in 30s

[ornate kindle]

Thats pretty cool

I was using an online interpreter and have little knowledge of what languages like Sage actually have available, so I'm sure my code had tons of room for improvement

[ornate atlas]

I also want to learn to write sage well, I really only turn to it when I’m already desperate so I just kinda end up beating it into shape

[ornate kindle]

Lmao yeah, I might try using it more on the next couple assignments for this class

[ornate atlas]

Yeah my honours algebra course had a couple of submissions where we had to use it to make conjectures, and a couple of problems that would just be miserable without it (count the number of homs or whatever) but that was a while ago and even then I was kinda just making it work

People who are good with CAS seem to really make it work for them though, so I think it is worth learning

[digital parcel]

I had a class where we had to do some stuff in macaulay2 for hw

[fierce steeple]

Macaulay2 is fun

[young forge]

Yup. So if you try the normal caley dickenson construction, it ives you something isomorphic to the split ones

it seems

[digital parcel]

yea it's fun

sometimes it's frustrating but tbh that's just skill issue

[wise sedge]

You have access to Magma?

[rose mirage]

Nope but I do have access to the online calculator

[wise sedge]

RIP

[blazing schooner]

can someone help me with this linear algebra proof with hermitian matrices - i am getting lost with the steps because i am confused on how to multiply the matrices

im doing the forward case of A being normal implying HK = KH

my professor was gone so i missed quite a bit of lectures

[torn harbor]

right so you if you factor out the constants you just have to multiply $(A+A^)(A-A^)$

[broken turtle]

Blake

[torn harbor]

and show you get the same thing as if you multiply it in the other order

since the constants just commute with everything they don't really matter

[onyx imp]

im having trouble finding out what a C-algebra is?

[past cove]

an associative algebra over C

so it's a ring (possibly without unit) that also has a C-module structure

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

[onyx imp]

just for context, the funny $a$ is just some type of ring with additive group that is free abelian and $a_C$ is the extension of scalar i.e. $C \otimes_Z a$.

[broken turtle]

somethingwrong

[onyx imp]

so i need to treat the tensor product as a ring, then the C-module structure is just how C acts on C?

[past cove]

the C-module structure is just the natural action on C^n as in any vector space

[past cove]

but yes you give the tensor product a ring structure (just multiply componentwise) and you're set

[onyx imp]

hmm i think im not too familiar with taking tensor products with rings. from what I understand if i take the tensor product between a (R,S)-bimodule and an S-module, then the tensor product is naturally an R-module.

what is it that allows us to give the tensor product a ring structure here?

[past cove]

because both R and S are rings

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

[onyx imp]

1. Every ring is z-algebra including C.
2. by that wikipedia page, the tensor product is a C-module, which can naturally be made into a ring
3. Finally, a ring which is a C-module, is a C-algebra where the map is given naturally by phi(c)=c.1

would these sound right to you?

[past cove]

Yeah this is correct

[onyx imp]

ah okay, thanks for the help

hmm actually, by the wikipedia page i only have that the tensor is a Z-module right?

oh nvm let me think abit more

[near lantern]

It depends on what tensor product you take. If A, B are ℂ-modules then you can take A (⨯)_ℤ B or A (⨯)_ℂ B. The former is a ℤ-module with two different ℂ-module structures. The latter is a quotient of the former where the two ℂ-module structures are forced to coincide, so it has a single ℂ-module structure.

[onyx imp]

yep i figured out which was the one i needed

[near lantern]

This is hacky (because I didn't get around to checking out the references) but I think this can also be done using the Bruhat decomposition (or more precisely the (B,N)-pair theory about how double cosets multiply), since a subgroup of G containing B is the same as a set of double cosets closed under multiplication (which is multi-valued on double cosets) and inversion (which is just given by inversion in W).

Oh, that's exactly how Digne-Michel does it.

[fierce steeple]

Chad

[young forge]

would trying to generalize projective geometry's eliptic curve group law to a noncommutative field of scalars go here?

i'm trying to define eliptic curves over the quaternions

[spice idol]

i think that goes into noncomm alg geo territory

[fierce steeple]

Idk noncomm alg geo isn't rly like this usually lol

[spice idol]

yeah i guess

universal algebraic geometry then but that wont be of much help.im sure

[young forge]

what the fuck happens when we try to generalize the notion of a projective curve to a nonabelian ring

[fresh surge]

What the fuck is ts first day of school bro

[hushed bone]

Idk but it ain’t advanced algebra lil bro. #prealg-and-algebra

[young forge]

would AAG key exchange work over the octonions and would it be vunerable to normal linear-group attacks on aag?

[worldly zealot]

michael artin has a paper on noncommutative projective schemes

[young forge]

if so, i also have an idea to do AAG or something over the automorphism group of the (finite field split) octionions. Represented as conjugation (like, but nonassociativity) actions. Which provides a distinct way to represent elements of G2 disticnt from the typical(easily broken) matrix representations

tbh what group theoretic cryptography needs is good nonlinear representations of like... exceptional groups of lie type. Like we have for cyclic groups now with multiplicitve/eliptic curve groups

This is actually a canidate for that(but needs more research)

further research is also needed into using F4(automorphisms of octonion projetve plane) as well

but i don't understand that as well :3

[young forge]

link pls

[worldly zealot]

https://www.sciencedirect.com/science/article/pii/S0001870884710875

[young forge]

thank youuu

[languid canopy]

Hello everyone, I need your help. My new master's advisor asked me to start working and told me to read this:
algebraic definition of the explosion of an ideal.
Could someone please help me with some information and a little explanation?

[mental escarp]

Can you tell if he was asking you to learn the definition of a ring ideal?

[ornate atlas]

Do you mean blow up by any chance? That’s an algebraic geometry term

[lone jacinth]

Explosion would be a funny mistranslation then

[ornate atlas]

In that case it refers to blowing up in the sense of a balloon rather than blowing up like a bomb. I’m not aware of the explosion of an ideal

[spice idol]

FIRE IN THE HOLE

[pastel shoal]

I heard the term actually comes from blowing up a photo (zooming in, we take it for granted now but you actually had to do a chemical process to created a zoomed in version of a photo back then)

[ornate atlas]

Ah you might be right actually, I think I was told that in my class. I think the visual I have in my head of the blow up at a point just looks balloon like enough that I tend to think of it as that haha

[pastel shoal]

There's also the film Blow-Up (1966) where a major plot point is that a photographer discovers a murder after blowing up one of his photos

I never really got that movie but cinephiles seem to like it

[jaunty sparrow]

That was a movie I had to watch for a film course for my required humanities credits

[jaunty sparrow]

It was quite strange.

[pastel shoal]

Yeah but it did have some really beautiful shots, I loved the ending sequence with the ||mimes playing tennis||

[jaunty sparrow]

Oh of course. I meant strange in a good way lol

[languid canopy]

Yes, the professor told me to read the book on “Steps to Commutative Algebra.” I only read up to chapter 3, and then he told me to study that definition, since it's faster to study with a goal in mind.

I was searching and I only found concepts in algebraic geometry as mentioned, but then I looked at a definition of rings that says

Expansion of A_{X} by the ideal m_{x}

[mental escarp]

I looked in the subject index of that book and don’t see anything in terms of expansion. Maybe he isn’t using a standard term for something cause I’ve never heard of that term. Maybe you could share a screen shot of a part of that book you’re confused about. But this might belong in the other thread on rings for all I know.

[languid canopy]

"I suggest you work on the following problem:
Understand the algebraic definition of the explosion of an ideal. From there, give a justified description of what the explosion of a point in C^2 and in C^n is like. You can assume that the point is the origin. Subsequently, of a point on a curve. For example, the explosion of the origin on the curve y^2 - x^2 - x^3 = 0"

by my advisor

[mental escarp]

Is he referring to the concept of a blow up? A blow up is a concept in algebraic geometry. I think he’s talking about that. That’s the formal term I know. You should ask this in the algebraic geometry thread

[languid canopy]

Yes!

He deals with the theory of singularity in algebra.

[mental escarp]

Nice I like singularity theory too!

[languid canopy]

Because I was researching the topic and it goes into depth on schema theory, but he's not asking me to go that far into the subject matter.

I guess I'm in the right place. Yesterday I was reviewing my Dummit book and couldn't find anything about the explosion of an ideal. I was wondering if it could be the definition or the theorem. I don't recognize that definiti

"Blown up"

That's what the professor is referring to, I suppose.

[mental escarp]

I think you wanna take your ideal J=<x,y,z> in k[x,y,z] and consider something called the Rees algebra R(J). I think this is the algebraic description of a blow up. But I’m not so sure

[languid canopy]

I'm looking through several books to see what I can find, but it's difficult; not everything can be found easily because of the terminology.

[ornate atlas]

Yeah I thought that would be the case, just a funny mistranslation

[languid canopy]

Yes, thank you for taking the time to look it up and give me the information.

I've already been able to confirm a few things. Thank you very much. I'll have to start working on this issue. If I need anything, I'll be asking on the channel.

[languid canopy]

Yes my english is bad

[ornate atlas]

Your English seems fine! It’s a perfectly reasonable mistake, the term has like 3 or 4 different meanings in English

[spice idol]

#linear-algebra

[tight kiln]

Thanks mate

[haughty raft]

I liked that movie too 🙂 Didn't expect to see it in Advanced Algebra though!

[pastel shoal]

I watched that movie around the same time I was learning about varieties for the first time and I always remember it when somebody mentions blow ups

[void plank]

I've been trying to look up information on differential graded algebras and for some reason, I can't find proofs that the Leibniz rule implies the multiplication descends to a well defined operation in (co)homology

Is it obvious that this is the case?

[lone jacinth]

It might qualify as obvious I guess.

(f + d(f'))(g + d(g')) - fg = fd(g') + d(f')g + d(f')d(g')

fd(g') + 0 = fd(g') + d(f)g' = d(fg') and similarly for d(f')g while
d(f')d(g') = d(f'd(g')) for the same reason.

(I probably have the sign wrong, but 0 = -0)

Note I'm using that d(f) = d(g) = 0

[void plank]

Oh awesome, thanks!!

[thick oriole]

thoughts on notes from the underground by aluffi directly after ladr

[unborn rampart]

This is probably the wrong channel, but yeah, notes from the underground is great, and doesn't have a lot of prerequisites. I think you can read most of it without even knowing lin alg, as long as you have some familiarity with proofs

[thick oriole]

would it connect well with ladr?

[unborn rampart]

Um, I'm not sure what that means. There's no special connection between LADR and aluffi, it doesn't matter what lin alg book you read

I recommend asking in #book-recommendations instead

[thick oriole]

i meant coming straight from linalg, i wanna understand stuff from ladr better but maybe any algebra book would help

and yeah sure ill sstop talking abt it here

[near lantern]

Is it true that for (W, S) a Coxeter system (Coxeter group with the choice of simple reflections), for any w, w' in W there exist reduced expressions w = s1...skt1...tl and w' = tl...t1u1...um such that s1...sku1...um is a reduced expression for ww'?

Oh, it suffices to prove if l(ww') < l(w) + l(w') then there exists s such that w, resp. w', has a reduced expression ending in, resp. beginning with, s; because then we can induct on l (the required number of cancellations).

We can always do it if we allow the tj to be non-simple reflections.

For my algorithm to work it suffices that if w in W and s, s' simple are such that sw, ws' are reduced but sws' is not (so w^{-1}sw = s'), then for every prefix w' = s1...sl of every reduced expression w = s1...sk, the reflection w'^{-1}sw' is also simple. Conversely by considering the product (sw')(w''s') (where w'' := s(l+1)...sk), this is necessary.

[near lantern]

But if this were true, s1ss1 is simple ⇒ it is equal to s ⇒ s2s1ss1s2 = s2ss2 being simple would have to be equal to s, etc. In other words, no distinct simple reflections would be conjugate and the centraliser C_W(s) would be the parabolic subgroup generated by the simple reflections commuting with s. The former fact is very false (but interestingly the latter fact is true for at least type A_n where I can prove it geometrically).

[near lantern]

Re the latter fact we have the very interesting https://web.ma.utexas.edu/users/allcock/research/centralizers11.pdf which says something like it (the centraliser is generated by its reflections, if not by its simple reflections) is true in a Coxeter group with Dynkin diagram without loops but can pick up a free group factor otherwise.

[fierce steeple]

Just read the Dostoevsky

[thick oriole]

What

[near lantern]

@spice idol do you know about free division rings?

Do you know if they are well-understood in the literature*, rather.

[lone jacinth]

What are free division rings?

[near lantern]

Well.

The definition I read here starts with all expressions (call it E) in n (non-commuting) variables with operations 0, 1, +, ⨯, -, ^{-1}. Then for any choice of values for the n variables in any ring R (call this a "specialisation in R"), you define evaluation of expressions in the obvious way, getting a function ev_a(e) for a a specialisation and e an expression, which is not defined for all expressions.

Now define an expression to be "sensible" if it can be evaluated in some specialisation (this rules out 0^{-1} for example) and two sensible expressions to be equivalent if they have the same value in all specialisations where they can both be evaluated (there is always at least one specialisation where they both can be). The equivalence classes of sensible expressions form a division ring called the free division ring on the starting n variables.

(Caveat: the paper says to use all specialisations in a fixed large division ring of characteristic 0 but I'm hoping it's the same to use all specialisations in all rings.)

This has the universal property that any specialisation f of the variables in any division ring of characteristic 0 extends to a subring R (containing the non-commutative polynomial ring obviously) which is "closed under inverses" in that whenever a in R and f(a) ≠ 0, a^{-1} exists in R. (I'm not sure I like this universal property.)

[spice idol]

i see, cant say ive heard of this

seems very model theoretic

[near lantern]

OK actually they do cite Skew-field constructions by Cohn which I might check out, but I thought I'd ask first if this is all easier than it looks.

[spice idol]

many sources seem to talk about random matrices lol

[near lantern]

Exactly!

[spice idol]

well this is kinda how Id do it after a bit of prodding probably

the reason it is so messy is because of the inclusion of partial functions which SUCK

[near lantern]

In the sense that this seems like a pretty hard free object to use explicitly. Given an expression and specialisation, to tell whether I can evaluate I need to check all equivalent expressions for division-by-0 errors. I suspect there is a "simplest" equivalent expression (i.e. some kind of normal form), but it probably won't be sufficient to check that because there could be a specialisation where the normal form fails to evaluate but a more complicated expression doesn't.

[near lantern]

Makes sense.

[spice idol]

well the problem lies in the noncommutativity

for "free fields" you can leisurely take the set of quotients p/q that "make sense"

[lone jacinth]

Well I can tell you what it would be on a set with 0 or 1 elements at least

[near lantern]

And e.g. it would be nice if I can just say that it is a localisation of the non-commutative polynomial ring (i.e. specialisation extends to whole thing iff non-zero on all polynomials, in which case it extends uniquely). But the concrete description is complicated enough that this seems hard (at least for something I'm just reading on the side out of curiosity ).

[spice idol]

youve got stuff involving Ore conditions right

[near lantern]

I think you're right that it's the partial-ness more than the non-commutativity. Free groups aren't nearly this complicated.

[spice idol]

well in free groups you can just invert everything

here youve got the fear that simplifying an expression might suddenly make a bad expression appear in the right/left denominator

[near lantern]

Yes, you can concretely describe localisations (i.e., you can make something with a relatively simple description as a set of expressions with the desired universal property) if the Ore property holds. I'm not sure the non-commutative polynomial ring has it though.

IIRC Cohn has done a lot on the topic of localisations and embedding rings in division rings though. Maybe I should read something of his.

[lone jacinth]

It does not. In fact it has more than one non-isomorphic skew field of fractions

[spice idol]

this is why noncomm rings do not admit an algebraic geometry 💔

[near lantern]

I'm going to guess this "free division ring" is one of them (the "most general" one? but I would expect different localisations to be incomparable...).

[near lantern]

WDYM?

[spice idol]

havent proved it yet but im like 99% sure

[near lantern]

What do you need exactly in order to admit an algebraic geometry?

I've mused a little on this before and IG you need a notion of "open cover" such that enough local-global principles hold to do anything useful.

[spice idol]

well my definition is a functor to a generalisation of the category of ringed spaces to the analogue in some arbitrary variety V. This functor must be a section of the global sections functor, have enough open immersions, and the base space must be an object like a spectrum

that is what makes an ""algebraic"" geometry

[near lantern]

sorry functor from what to what

[spice idol]

from a variety V to the category of V-spaces (objects are sheaves with values in V, and morphisms are analogous to morphisms of ringed spaces)

[near lantern]

Oh like a Spec functor from V to V'd spaces?

Right.

[spice idol]

"object like a spectrum" means the spectrum of some coherent condition (which is a bit technical to define right now)

[near lantern]

owo

[spice idol]

(as in i may be taking up enough space here already)

[fierce steeple]

Woah homotopy theory

[spice idol]

im not sure its that kind of spectrum 💔

[fierce steeple]

V cool tho

[spice idol]

now hopefully i will have the motivation to continue this

"feet"

[vague pawn]

I didn't you are this type of guy

[spice idol]

thats the name of that fuckass penguin emoji

[worldly zealot]

movie called happy feet

[last talon]

How difficult is it to show that Out(F_2) is isomorphic to GL_2(Z)?
(I’m trying to tell if this is a “you should take this on faith” or a “you should prove this”)

[spice idol]

yeah the naming is just funnt

[spice idol]

what is Out here?

oh outer isomorphism group of the free group on two generators

[worldly zealot]

i imagine the hardest part is inventing the right presentation of GL_2(Z)

choosing

[spice idol]

just compare sizes duh

if GL_2(Z) and Out(F_2) have the same size they must be isomorphic

[last talon]

It’s easy to get a map in the -> direction
I’m not sure how to either show it’s inj/surj or to produce a map in the <- direction

[spice idol]

[a, b; c, d] sends x to x^a y^c and y to x^b y^d right?

thats a homomorphism cuz free and prob also an automorpjism

[last talon]

It’s not immediate that that’s an automorphism

[spice idol]

get to proving

[torn harbor]

u just need to crunch the inverse

[spice idol]

yis

[torn harbor]

then ur good

[last talon]

?

[spice idol]

or just show it is a monoid action

i.e. a functor

[last talon]

Oh wait I think I see?
But I’m not sure that works

[torn harbor]

its obv a homomorphism so you just need to show its invertible

[spice idol]

then as functors preserve isomorphisms you have a group action

[torn harbor]

and for that you just apply the inverse transformation

then if you have an arbitrary automorphism, its gonna send x and y to some words and I think if you just conjugate by the right thing you get it into this form

which proves the result

[last talon]

I think it should be -c

And -d

Then it works out

[rose mirage]

there's gotta be some incredibly stupid way to do this

[torn harbor]

yeah I think ur right with the -c -d

[last talon]

(I just computed what f(A)f(A^-1)(x) should be and that’s how it works)

[torn harbor]

actually this seems maybe a bit harder than I thought

[rose mirage]

like it's the coproduct (in Grp) of Z with itself so for every pair of maps Z -> F_2 we get exactly one automorphism and any automorphism yields a map Z -> F_2 (restrict to one of the generators)

[torn harbor]

but I assume there's some way

well you don't get an automorphism for every pair of maps is the tricky part

[rose mirage]

true they can be endomorphisms

[torn harbor]

also looking specifically for outer automorphisms

[spice idol]

they are always endomorphism

smh

[rose mirage]

u know what I meant

[spice idol]

i am braindead unfortunately

[subtle plaza]

I like to think of this topologically. Identify F_2 as the fundamental group of a wedge of two circles, which I denote by X. Then Out(F_2) is the group is unbased self-homotopy equivalences of X (since X is a K(F_2, 1)) and the map to GL(2, Z) just takes a homotopy class of maps to its action on H_1(X).

This map is surjective because given any element A of GL(2, Z), the matrix induces a linear map on R^2 which commutes with the action of Z^2. Thus, the map descends to the 2-torus T^2 and fixes the image of the origin, hence it induces a map on the once punctured torus which is homotopy equivalent to X. Then by construction the induced action on H_1 is represented by the matrix A.

The injectivity is not as clear to me, but I know a mapping class group trick which works here. Suppose a representative map f induces the identity on H_1. If I work on the once-punctured torus again, then let a and b be simple closed curves which intersect once. Then f(a) is isotopic to a and f(b) is isotopic to b, so we can modify f by a homotopy to fix a and b pointwise. Cutting the torus along these curves, we get an induced map on the punctured disk, which is homotopic to the identity (this last part follows from the Alexander trick) so f was homotopic to the identity to begin with.

[last talon]

Ah that’s a cool application of the Alexander trick
Like it feels fairly standard but also I wouldn’t’ve thought of it here