advanced-algebra

[fierce steeple]

My point is that you do not just get the sequence 0 -> 0 -> R/I -> R/I -> 0

[hexed tangle]

so we are talking about the map f : I -> R and (f)/I = 0 -> R/I which is injective

[fierce steeple]

Once again

[inner kindle]

I = R•I. The submodule of R•I corresponding to the ideal I is I•I = I².

Thus you just need to find an example when I/I² -> R/I is not injective

[hexed tangle]

so we get the sequence 0 -> 0 -> R/I -> R/I^2 -> 0

[fierce steeple]

Well also the sequence you get is 0 -> I/I^2 -> R/I -> R/I. So if this is exact then I/I^2 = 0

[fierce steeple]

No

[hexed tangle]

...?

[fierce steeple]

Well like consider what no tea said

The functor sends M to M/IM

So if M = I what do you get

[hexed tangle]

dosen't (R/I)/I = R/I^2?

[fierce steeple]

No

Because what is I.(R/I)

[hexed tangle]

what's does . mean here?

[fierce steeple]

Like just I(R/I) if you want

As in same as in the notation IM

[hexed tangle]

that is (IR)/I = (I/I)=0

[fierce steeple]

Yeah

So R/I gives you R/I after applying this

And R is sent to R/I as well

Now we have said what I is sent to as well

[spice idol]

I is not a subset of R/I, so the syntax (R/I)/I does not even make sense

[hexed tangle]

I don't understant that

[fierce steeple]

I think it is almost okay cause we are viewing R/I as an R-module but yeah bad

[spice idol]

Hm, oh as the image of R/I wrt that functor lol

[fierce steeple]

Well cause we are doing (R/I)/(I R/I)

And that is just (R/I)/0 = R/I by what you said above

[hexed tangle]

ahha

so we get R/I -> R/I

[fierce steeple]

Yes

[hexed tangle]

that's an iso?

[fierce steeple]

Yup, just the identity

[hexed tangle]

ok, so how does this help=

it seems that we have shown that it is exact

[fierce steeple]

Well look at the sequence you get after applying your functor

What is it

[hexed tangle]

0 -> 0 -> R/I -> R/I -> 0

[fierce steeple]

We haven't

[fierce steeple]

It's not this

Again by this...

[hexed tangle]

how?

(I)/I = I/I = 0

[inner kindle]

.

[fierce steeple]

Lol are you reading what we have said

[hexed tangle]

is there someone here that would like to help me instead of making fun of me?

[past cove]

idk this notation is pretty shit ngl
The functor is M -> M (x)_R R/I

maybe that makes things clearer

so a priori you aren't just quotienting by I
You're tensoring with R/I

[hexed tangle]

I see

I didn't realize that

[rose mirage]

Not the R-torsion…. 🥀

[fierce steeple]

Sorry I mean it just feels like you are ignoring the poinys we have given

[hexed tangle]

I'm really not

[past cove]

yeah there's an identity that says that M/IM is canonically isomorphic to M (x)_R R/I

[hexed tangle]

So we have I (x)_R (R/I) = (IR)/(I^2) = I/I^2

[past cove]

yes exactly

[fierce steeple]

Or let the notation help

The fjncto5 sends M to M/IM

So you get I/II = I/I^2

[hexed tangle]

indeed

[past cove]

fjncto5

[hexed tangle]

0 -> I/I^2 -> R/I -> R/I - >0 so if I/I^2 != 0 then the sequence won't be exact

[past cove]

that last term ain't correct

[hushed bone]

Yes it is

[fierce steeple]

It is correct

[past cove]

bleh

okay I had it different in my head

[hushed bone]

Tensoring with R/I does nothing to I-torsion things

Where uh this version of I-torsion is where I literally kills the module

[fierce steeple]

Made up words

[past cove]

made up works that make sense dammit

[hushed bone]

Milky milk milk

[past cove]

I thought it was like R/I^2 and I used the third iso theorem incorrectly to conclude it lol

[hushed bone]

Good name

Lol

I mean actually there’s a big categorical reason to this

A map R -> S is an epimorphism of rings

If and only if S (x)_R S -> S the multiplication map is an isomorphism

If and only if the (co?)unit map of the push-pull adjunction is an isomorphism

Meaning for any S-module M, the map S (x)_R M -> M the multiplication map is an isomorphism

And cuz R -> R/I is surjective it’s clearly an epimorphism

Very overkill for this, but it kind of tells you the situation when you get S (x) S = S are basically generalizations of quotient maps = surjections

[past cove]

very cool explanation

I enjoy it almost as much as a good glass of milk

cow blood makes you stronger

[limpid horizon]

Goo goo gah gah baby wants his milky 🍼

[rose mirage]

PLEASEEEEEEE

[hushed bone]

https://tenor.com/view/dennis-prager-i-want-mommy-i-want-milk-i-want-to-be-held-i-want-to-be-comforted-gif-23575839

[limpid horizon]

Oh Hell Nah!!

[past cove]

I think it's well known that every person that drinks more than 1 glass of milk per day has a mommy kink going on

[fierce steeple]

What

[past cove]

I said my peace now only god can judge me

[fierce steeple]

Counit

[hushed bone]

I will never put to memory which is which

[fierce steeple]

I just remember like unit goes from identity always

[hushed bone]

And I love myself for it

[limpid horizon]

If i view the girl im dating as motherly and that makes me more attracted to her does that mean i have a mommy kink

[fierce steeple]

Like same for groups or any monad or whatecer

[past cove]

okay serious question do you actually need to ever use monads for anything

[fierce steeple]

Real

[fierce steeple]

Depends what you are doing

[hushed bone]

Monadic resolutions are very nice

And I think some silly stuff with monads can like, ensure that the colimit of groups, rings, modules, etc. has underlying set the colimit of sets

Which is cool I guess?

And it canonically induces the algebra structure on it

[fierce steeple]

In some parts of algebra and related things they are unavoidable

[past cove]

do they ever show up in like algebraic geometry?

which is a broad question

[fierce steeple]

I had in mind AG when i said that aha

[hushed bone]

A categorical construction? In algebraic geometry? No way

[fierce steeple]

For example "descent" type statements are often basically statements about functors being (co)monadic

[hexed tangle]

thanks guys

[hushed bone]

What

[past cove]

if another person tells me something in AG holds because left adjoints commute with colimits the rapture will be second thing humanity will have to worry about

[hushed bone]

[plucky arch]

they do though

[fierce steeple]

Isn't this usual like

[hushed bone]

I haven’t seen it

Tbh I actually haven’t seen descent properly

Like I know faithfully flat descent for specific ring theoretic properties

CM, Regular, Noetherian, etc etc etc

But this isn’t the big thing people call faithfully flat descent in terms of modules and descent data and stuff

[fierce steeple]

I mean this

[hushed bone]

Yeah I know lol

[fierce steeple]

Ye lol

[hushed bone]

But I don’t see how comonadicity says anything about that

I mean wtf is a comonadic functor

Blech

[past cove]

what the fuck does blech mean

[hushed bone]

Wouldn’t you like to know, weather they

[fierce steeple]

Well the point is like if you have a map X -> Y then you get pullback f^*: QCoh Y -> QCoh X and then you can view stuff in QCoh Y as modules over the comonad f_* f^* which unwinds to the usual statements

[hexed tangle]

Doesn't IM need to be contained in ker alpha for this to work?

[fierce steeple]

That holds automatically in this situation

[limpid horizon]

Because N is an R/I module so IM will map to 0 thru alpha i think

[fierce steeple]

Yes

[hexed tangle]

ah!

[limpid horizon]

Ya when i first read it i missed the fact that N is R/I mod

[hexed tangle]

thanks

[warm swan]

what

[hushed bone]

We didn’t need to know that brister

[warm swan]

fine I'll delete 🙄

you guys are lame

[past cove]

you showed weakness by deleting it

he literally never told you you had to remove that comment

and he'll think he's hot shit and push you around
You called the wrath of the chmonkey upon yourself buddy

[hushed bone]

It was a joke

[limpid horizon]

😢

[warm swan]

☹️

[hushed bone]

DMAN

I got the extra tall frown

:c

[limpid horizon]

grr

[past cove]

invisibelle you gotta be more confident

don't let a monkey chair hybrid push you around

[hushed bone]

Chair monkey hybrid*

[warm swan]

it's okay

:3

[warm swan]

what's the difference

[hushed bone]

I always say chair monkey hybrid

[fierce steeple]

Chmonkey

Oh lol nvm i have an answer

I was wondering like are there any good interesting examples where I = I^2 for non-Noetherian rings

But I think the classic examples would be stuff like A_inf

[spice idol]

Chmonkey or monkchair

[fierce steeple]

Hi enpeeace

[spice idol]

One sounds good while the other doesnt

[warm swan]

monkchair sounds way better

[spice idol]

Delusional..

[spice idol]

Heya

[limpid horizon]

hi enpeace

[spice idol]

You mean I^2 = I for all ideals?

[fierce steeple]

nah like

Well okay so in a noetherian ring i believe like I^2 = I means I is generated by an idempotent

[hushed bone]

I think if you take an infinite polynomial ring

You can then enforce a bunch of relations

[fierce steeple]

I think it's like if I is f.g. as an R-module and I^2 = I then by Nakayama's lemma there's x in the Jacobson radical with (1+x)I = 0 or smth like this

[hushed bone]

To get m=m^2

[fierce steeple]

Yeah

[hushed bone]

Like you say that x1 = x1x2

Or something like that

Not that actually

But you can do this and ensure each xi is in m^2

Other than that idk

[fierce steeple]

ye i think there are examples with perfectoid rings and stuff but ye

this came up to me cause it's like

if j: X -> Y is an closed immersion of schemes and you consider the derived pushforward j_*, when is this (derived) fully faithful

and the answer is like almost never because it implies the ideal sheaf is idempotent iirc

[astral ginkgo]

Where is the openness of the kernel being used (second to last line)? Isn't it just that if we map into a finite group G there are at most m^|G|=m maps?

istg david just says things sometimes

wait actually is it m^|G| or |G|^m

either way I think it's just m

no it's |G|^m since it's maps from S to G

is that generally m if |G| is finite and m is infinite?

actually no it'll probably be 2^m or smth

Ok so the kernel being open implies that if S is a generating set converging to 1 then its intersection with the complement of the kernel is finite, i.e. only finitely many elements in each map are getting mapped to a nontrivial element, so actually the amount of maps now becomes the set of finite subsets which has cardinality m

ok cool

actually it's a little more complicated. It's finite sets + a choice of elements in G but that doesn't add cardinality since G is finite

formally it's bounded above by FinSets(S) x Finite Sequences(G)

and finite sequences of elements of G is just aleph_0

[limpid horizon]

Ive been encountering lots of different ways that one can compute local cohomology. Is this definition with the direct limit system useful?

Originally I was going to try and understand how that comes about but now im not sure if its necessary to

Is that important for proving Thm A.7.1 there?

In any case, where can I find a proof for Thm A.7.1?

[hushed bone]

Yes absolutely

[hushed bone]

I think you can do it without?

[hushed bone]

Bruns Herzog

Stacks project

EGA

Does EGA have it? Maybe it’s in an SGA

[limpid horizon]

What is u_i here?

from bruns and herzog

Also, what is E^. (M)?

[limpid horizon]

i mean it looks like it should be that Ext thing thats used in defining depth but idk why it called u_i here

[limpid horizon]

injective resolution of M?

[hushed bone]

You need to go back and read more lol

This is related to minimal injective resolutions and bass numbers

But this also isn’t what you want anyway

To show it vanishes below the depth is not difficult, I think you’ve pretty much already done it from what you’ve already read

In fact this will tell you it doesn’t vanish at the depth

The difficult part is showing it won’t vanish at the dimension (and I guess you also need to see why it vanishes past that)

[plucky arch]

What’s the story/idea/motivation behind Tannakian reconstruction?

Or ig tannakian stuff generally

[vapid axle]

bro speedrunning higher algebra

[limpid horizon]

Lmao

Ya chmonkey is like slow down bucko

[potent plaza]

tensor braided categories i think

[plucky arch]

could you elaborate…?

[potent plaza]

not really honestly, but what i know is that you can just reconstruct the group from the tesnor braided category of the representaiton of the group

looking at the aut of the forgetful functor

i just went to a talk really but i am not so sure about like

the mechanics of it

check out pierre delignes paper "tensor categorie"

[plucky arch]

I’ve heard Deligne is associated with some scary math

[earnest osprey]

Hey, I read that a map f:X->Y between two indecomposable modules over a Nakayama algebra (so all indecomposable modules are uniserial) is irreducible iff it is a mono with simple cokernel or epi with simple kernel. Showing that any irreducible map is of this form is clear, but I'm unsure how to prove the converse. Does anyone know how it works?

[gusty void]

In a Nakayama algebra every indecomposable module is uniserial so its submodules line up in a single chain. Take a morphism f from X to Y that is injective and whose cokernel Y / X is simple. Because Y / X has no proper subfactors X is a maximal proper submodule of Y; there is nothing strictly between them in the chain. Now factor f as g followed by h, with g from X to some Z and h from Z to Y. The image of g is a submodule lying between X and Y. Maximality forces Z to be either X or Y. If Z is X then h is the original inclusion and splits, if Z is Y then g is the identity and splits. In every factorization one leg splits so f is irreducible. The dual argument (using quotients instead of submodules) shows that a surjective map with simple kernel is irreducible as well. Hence for indecomposable modules over a Nakayama algebra a morphism is irreducible exactly when it is either a monomorphism with simple cokernel or an epimorphism with simple kernel

[gusty void]

^

[earnest osprey]

Why is the image of g a submodule between X and Y? Isn't g:X->Z?

And why would Z need to be X or Y? I understand that the image of Z needs to be X or Y, and if it is X it's also clear that it splits. But I don't see how it splits if the image is Y

[gusty void]

Because f = h followed by g is injective g itself must be injective. That means g embeds X into Z, its image g(X) is just X sitting inside Z. Then h carries every element of Z (and therefore every element of g(X)) into Y. So you get the chain of submodules X = g(X) inside Z inside Y which is exactly what “g(X) lies between X and Y” means

[earnest osprey]

But h is not injective?

[gusty void]

Because Y is uniserial and X is a maximal proper submodule (Y ⁄ X is simple) there is no module strictly between X and Y. So in any factorization f = h g we must have Z equal to either X or Y. If Z equals X g is just theidentity on X and therefore splits. If Z equals Y h is the identity on Y; an identity map is an isomorphism and automatically has an inverse so h splits. In either case one leg splits making f irreducible

[earnest osprey]

We could just take Z = Y + Y

I agree that h(Z) must lie between X and Y

But there does not need to be an inclusion Z into Y

[gusty void]

When Z = Y we have f = h g with g : X -> Y injective. Assume h is not injective; then ker h is a nonzero submodule of the uniserial module Y. Because submodules of a uniserial module are totally ordered,ker h must either lie inside g(X) or contain g(X). But h acts injectively on g(X) (since h g = f and f is injective) so ker h has intersection g(X) = 0, forcing ker h = 0. Thus h is injective after all. For a finite length module like Y an injective endomorphism is automatically surjective hence an isomorphism which splits.

[earnest osprey]

Ker h is not a submodule of Y, but Z

[gusty void]

When Z equals Y h maps Y to itself,so its kernel is a submodule of Y. Since h g = f and f is injective h is injective on g(X) which means the kernel of h intersects g(X) only at 0. In a uniserial module that forces the kernel to be 0 so h is an isomorphism and therefore splits

[earnest osprey]

But Z does not equal Y.

The image of Z under h equals Y

Just take X->Y+Y->Y. This factors, and h is not injective

[gusty void]

If Z is not Y the maximal submodule property forces Z to be X (there is no middle option in a uniserial chain). In that case g : X -> Z is just the identity on X so g has an obvious inverse and splits. No matter which case occurs one leg of the factorization splits and f is irreducible

[earnest osprey]

????

Z is not a submodule of Y

h is NOT INJECTIVE

f = h g only implies that g is injective, not h

As I said multiple times, consider X -> Y + Y -> Y, which is of course injective, but Y + Y -> Y is not

[limpid horizon]

its gettin' heated in advanced-algebra

[gusty void]

Sigh

[earnest osprey]

Also, Z is not uniserial as it does not need to be indecomposable

As I said, just take Z as the direct sum of Y with Y

Then h is not injective

[gusty void]

We are working with Nakayama (uniserial) modules so every submodule of Y sits in one linear chain. Your example Y riect sum Y breaks this rule because Y direct sum Y is not uniserial so it is not a counter example. Take any factorisation f = h∘g. Because f is injective, g is injective. Let W be the image of h; W is an honest submodule of Y. The copy of X that g embeds ends up inside W, so we now have “X contained in W” and “W contained in Y”. Since Y∕X is simple X is a maximal proper submodule of Y hence W must be either X or Y. If W equals X then h lands inside X and g is essentially the identity on X which splits. If W equals Y, h is onto Y. Assume for contradiction that h is not injective; its kernel is then a non zero submodule of the uniserial Y. In a uniserial module any two submodules are comparable, so this kernel must either lie inside X or contain X. It cannot contain X because h acts like f on X and f is injective and it cannot lie inside X because that would force W to be X, not Y. Therefore the kernel must be zero making h an isomorphism which splits. Therefore in every possible factorisation one of the two maps splits so f is irreducible as claimed

[limpid horizon]

we need a certain Chmonkey to resolve this conflict

[earnest osprey]

Again, the kernel of h is not a submodule of Y

h:Z->Y

How would ker h be a submodule of Y?

[limpid horizon]

isnt the kernel always a submodule

dafok

oh

[earnest osprey]

Yes, but ker h is in Z

[limpid horizon]

submod of Z

lol

[earnest osprey]

Not in Y

Yeah

Plus, Z does not need to be uniserial. Only indecomposables are uniserial in Nakayama algebras, so Z = Y+Y is totally valid

[gusty void]

True... the kernel of h lies in Z, not Y. That is why we replace Z by W = h(Z), which is inside Y. From there the rest of the argument uses W not the original Z.

[earnest osprey]

What?

h(ker h) = 0

So yeah, of course h(ker h) = 0

How can you deny that X->Y+Y->Y is a valid factorization of X->Y?

[gusty void]

In the factorisation X -> Y direct sum Y -> Y you are picturing the second map h is just the projection onto one summand. That projection does split - the inclusion of Y into that same summand is a right inverse so h followed by that inclusion is the identity on Y. Because h already splits the definition of an irreducible map is satisfied

[unborn rampart]

Chmonkey isn't enough, we need Scholze and Mochizuki to settle this

[earnest osprey]

Yes, of course it splits

But all your arguments say that the projection would be injective

Are you trolling me or what

Considering ker h as a submodule of Im(h) 😭

[gusty void]

No, our injectivity conclusion relied on one extra fact: Y (and therefore W) is uniserial so any two non zero submodules must overlap. In the direct sum module Y direct sum Y that property fails so the projection can have a non zero kernel and still split. So our earlier argument never forces your projection to be injective it only forces injectivity inside the uniserial setting we started with

[earnest osprey]

W is the image of Z

Which is also uniserial for Z = Y+Y

W is either Y or X.

W is not isomorphic to Z

You literally say assume h:Z->Y is not injective, then it's kernel must be a submodule of Y

How do you view ker h as a submodule of Y?

[gusty void]

Yes, wth Z taken as Y direct sum Y the map h is the projection onto the first copy of Y so its image W equals Y. That sits in the “W = Y” branch of the argument. A projection from Y direct sum Y onto Y is a split surjection. The standard inclusion Y -> Y direct sum Y is a right invers so h already splits. We never needed h to be injective in this branch only to split and the projection satisfies that requirement

[earnest osprey]

Your "argument" would still show that the projection is injective

[gusty void]

That “kernel inside Y” line only applies in the special branch where Z actually is Y. In that case h is an endomorphism Y -> Y so its kernel is automatically a submodule of Y. When Z differs from Y (for example Z = Y direct sum Y) we never use that kernel argument, there g already splits and the proof is finished before we reach that step

[earnest osprey]

Z is not Y....

The image of Z is equal to Y

[gusty void]

"Only applies"

[earnest osprey]

Okay and what is Z is not Y?

So yes, if h is the identity then it is clear.

[gusty void]

If Z is not Y then W = h(Z) is a proper submodule of Y that still contains X. Because X is maximal W must equal X. Now h maps Z onto X and g embeds X into Z with the property that h ∘ g is the identity on X. That already gives a left inverse for g so g splits. Thus when Z != Y one leg (g) splits, when Z = Y the other leg (h) splits. In every case f is irreducible...

[earnest osprey]

An easy counterexample, Take Q the quiver 1->2->3. It is uniserial. Take X = P[3], Y=P[2], Z = P[1]. We then have X->Z->Y, h(Z) = Y and h is not injective

[earnest osprey]

No....

[gusty void]

Yes...

[earnest osprey]

h(Z) can be equal to Y without Y=Z

[earnest osprey]

Take this

g is not split

h is split, but not injective

h(Z) = Y but Z is not Y

[gusty void]

That triple is not a valid factorisation of the original map f: X -> Y. For h ∘ g to equal f h must act injectively on the copy of X that g embeds. In your quiver any surjection from P(1) to P(2) kills the bottom layer of P(1) – exactly where g sends X – so h ∘ g becomes zero... NOT f. To keep h ∘ g = f the kernel of h must avoid g(X) and in a uniserial module the only way a non zero kernel can avoid a non zero submodule is for the kernel to be zero. Thus h has to be injective after all

[earnest osprey]

😭 😭 😭

h is not injective

[gusty void]

3/10 rage bait

[earnest osprey]

dude are you trolling

Take Z = Y + Y

[gusty void]

YOU'RE TROLLING

[earnest osprey]

Okay so what you are saying is

Either Z = X or Y, correct?

[gusty void]

No

[earnest osprey]

What are you saying then?

[gusty void]

What collapses is h(Z): if that image equals X then g splits and if it equals Y then h splits so f is still irreducible

[earnest osprey]

Yes, the statement is of couse true

but h is not injective

You say: if h(Z)=Y, then h is injective. This would show that h is an isomorphism.

Which it does not have to be

[gusty void]

Can you display an explicit non zero element in ker h whose submodule has zero intersection with g(X) inside this uniserial module?

[earnest osprey]

Take g:X->Y+Y inclusion on the first factor. Take h:Y+Y-> projection to the first factor. ker h = second factor, which has zero intersection withb g(X).

[gusty void]

That fails because h ∘ g = 0, not the original map f so it is not a valid factorisation

[earnest osprey]

Oh sorry

I meant first factor of course

Projection

inclusion to first and projection to first.

[gusty void]

That still breaks the uniserial assumption that Y direct sum Y is not uniserial so the must overlap rule never applies and the kernel argument is irrelevant

[earnest osprey]

No it does not break the uniserial assumption

Only X and Y are uniserial

Z does not have to be uniserial.

[gusty void]

Could you show the chain of sub modules that makes Y direct sum Y uniserial? In a uniserial module every sub module must lie in a single total order

[earnest osprey]

Why would Y+Y be uniserial?

It doesnt have to be

Why would it need to be?

If Z is uniserial, sure it works

But Z does not need to be uniserial

[gusty void]

Our whole irreducibility claim is stated for maps between uniserial modules only.
Y direct sum Y is not uniserial so once you put that module into the factorisation you are working outside the hypotheses

[earnest osprey]

No, X and Y are uniserial

At no point Z is required to be uniserial

Neither in the definition of an irreducible morphism, nor in my question

[gusty void]

If Z is Y direct sum Y and h is the projection onto the first copy of Y then h is surjective and already has a right inverse (the standard inclusion of Y into that first summand). A map that has a right inverse is a split epimorphism so in the factorisation f = h ∘ g the “h splits” alternative of irreducibility is fulfilled. Injectivity of h was never required we only needed one leg of every factorisation to split and here h does

[earnest osprey]

Yes, it of course splits cuz this is the claim

But you "proved" that h is injective anyway

the claim is that h has to split. I give you an example of a split non-injective: "Yeah but it splits already so it's not a counterexampe"

[gusty void]

I showed h is injective only when Z is uniserial because the “any two non zero submodules overlap” step needs that property. If your Z is Y direct sum Y, Z is not uniserial so the injectivity step never fires we just note that h already splits and that is enough

[earnest osprey]

Okay but Z is not uniserial

[gusty void]

Ok?

[earnest osprey]

"we note that h already splits" yeah dude that has to be proven

[gusty void]

When Z is not uniserial I never claim h is injective... I only note that h already has a right inverse so it splits which is all the irreducibility condition requires

[earnest osprey]

So you are saying if Z is uniserial then its injective, and if its not uniserial then it splits (which you never proved and is essentially the whole statement of my question)

bruh

[gusty void]

If h(Z) hits all of Y then Z cannot be any larger than Y. Otherwise the pre image of X under h would give a submodule strictly between X and Z which would push forward to a submodule strictly between X and Y impossible because Y∕X is simple and Y is uniserial. So Z = Y and h is an onto endomorphism of Y; onto plus finite length forces h to be an isomorphism hence it splits

[earnest osprey]

What?

Again, just take Z = Y+Y

And yes, this splits, but where does your "argument" break down?

Z can be much larger than Y. This is the point. Either g splits or h splits, so Y is a direct summand of Z

And Z = Y is not always true

[gusty void]

Again, if Z = Y direct sum Y and h maps onto the first copy the pre image of X under h is a submodule lying strictly between X and Z which pushes forward to a submodule strictly between X and Y contradicting that Y∕X is simple so such a factorisation cannot realise f

[earnest osprey]

What?

Are you seriously just saying that X->Y+Y->Y does not factor f?

[gusty void]

It cannot because if h maps Z onto all of Y then by the “Y∕X simple” argument Z must actually be Y so choosing Z as Y direct sum Y breaks the required equality Z = Y

[earnest osprey]

Let us make it easy. Take the linear quiver with 3 vertices. Take X = rad(P[1]), Y=P[1]. Then x->(x,x), (u,v)->v definitely factors the inclusuion x->x

And Y/X is simple

Woah

magic

[gusty void]

That factorisation uses Z = X direct sum X and g : X -> X direct sum X given by x -> (x,x) already has the leftinverse (a,b) -> a so g splits hence this example still fits the criterion and does not contradict it

[earnest osprey]

no

it uses Z = Y+Y

Dude

X->Y+Y->Y being the inclusion is like

clearly true

[gusty void]

The diagonal map g(x) = (x,x) has a left inverse given by the projection (a,b) -> a so g already splits the example therefore still satisfies the irreducibility criterion

[earnest osprey]

Dude

Of course it splits cause thats what we are proving

I'm just saying your argument is wrong.

You keep saying that Z has to be Y and h needs to be injective

Which is wrong

The statement is that h is a split epi, so of course Z is not necc. Y

[gusty void]

????

[earnest osprey]

When I give you an example of a non-injective h you say yeah but that already splits, like all of them do

Z does not have to be Y

[gusty void]

I never said “Z must always be Y and h must be injective.” We've went over this like 5 times. The claim is that in any factorisation f = h g the image W = h(Z) is forced to be X or Y. If W = X, g splits. If W = Y, then (inside the uniserial chain) this forces Z = Y and a surjection Y -> Y is an isomorphism so h splits.
Injectivity of h is invoked only in that second branch it is not a blanket requirement.

[earnest osprey]

??????

You LITERALLY JUST SAID

If h(Z)=Y

THEN Z = Y

Let us make it easy. Take the linear quiver with 3 vertices. Take X = rad(P[1]), Y=P[1]. Z = Y+Y. Then g:X->Z, x->(x,x), h:Z->Y, (u,v)->v definitely factors the inclusuion x->x, X->Y

The cokernel is simpe

[gusty void]

You're taking “Z = Y” in the literal sense while the argument means “Z is isomorphic to Y once you identify Z with its image

[earnest osprey]

You cant identify Z with the image

Yeah of course Im(h)->Y is an isomorphism

But this does not mean that Z->Y is an isomorphism or even split.

[gusty void]

When ker h = 0 h gives an isomorphism Z ≅ Y so up to isomorphism Z “is” Y for the splitting argument

[earnest osprey]

Yes, but ker h is not zero

[earnest osprey]

!!!!

How can you keep ignoring multiple counterexamples

[limpid horizon]

Wtf

You guys are still at it

[earnest osprey]

i think hes trolling me

[limpid horizon]

maybe

[earnest osprey]

either hes trolling or always asking chatgpt what to reply

[gusty void]

Your diagonal map g already splits: define r : Y plus Y -> X by r(a,b) = a. Then r g sends x to r(x,x) = x so r g is the identity on X. Because one leg of the factorisation splits the irreducibility criterion is satisfied bro

[earnest osprey]

Yes of course it splits

[gusty void]

?!?!?!

[earnest osprey]

But where would your argument break down?

[gusty void]

Jesus

[earnest osprey]

Dude

Every fucking choice of h splits

You keep saying that h is an isomorphism

"If W=Z, then Y=Z"

Which I gave you numerous counterexamples to

Which is extremely obvious, since h is supposed to be a split epi, so Z = Y+M, so Z = Y is almost never true 💀

[gusty void]

The original claim is about maps between indecomposable (and hence uniserial) modules over a Nakayama algebra: such a map f is irreducible if and only if it is a monomorphism with simple cokernel or an epimorphism with simple kernel. To prove the "if" direction you must consider any factorization f = h ∘ g. If the image of h, say W is equal to X then g splits. If the image of h is equal to Y and you assume Z is uniserial then the kernel of h must intersect g(X) trivially (since f is injective) which is impossible in a uniserial module unless the kernel is zero so h is injective and being a surjection between finite length indecomposables is an isomorphism and splits. Nowhere did I ever claim that h is always injective or that Z must equal Y in all cases. In your example with Z = Y plus Y and h the projection g already splits via a left inverse so the irreducibility condition is still satisfied. The injectivity argument applies only when Z is uniserial and h(Z) = Y, otherwise you rely on whichever leg does split... I'm not going entertain this YAP sesh anymore because it doesn't seem like it's going to end up anywhere. Good luck!

[severe atlas]

It is pretty late, so excuse me if I mess up. I think you can split Z into indecomposables, denoted Z_i, choose i maximal such that the image of X->Z->Z_i->Y is maximal among these, and then you can show that one of the maps in the factorization X->Z_i->Y splits. I do agree, however, that ash's proof is very much incomplete, as the morphism does apriori not need to factor over an irreducible module, but with a bit of work, as I pointed out above, you can reduce to the case of Z being indecomposable.

[gusty void]

Here, cleaned it up: Let f : X -> Y be a map between indecomposable (hence uniserial) modules over a Nakayama algebra. (=>) If f is irreducible it is either injective or surjective but not an iso; in the injective case Y/X cannot have a proper sub module (else f would factor and split) so Y/X is simple and dually a surjective irreducible map has simple kernel. (<=) Assume the injective case with Y/X simple (the surjective case is dual). Factor f as X —g-> Z —h-> Y. First break Z into indecomposable summands and keep the one whose image under h is largest; call this image W. Because g is injective X sits inside W and since Y/X is simple and Y is uniserial W is forced to be either X or all of Y. If W = X then h lands in X and h g = id_ X so g has a left inverse and splits. If W = Y then h is onto Y; if its kernel were non zero it would have to intersect g(X) in a uniserial module contradicting the injectivity of f so the kernel is zero, h is an isomorphism, and h splits. Thus in every factorisation one leg splits so f is irreducible. Therefore a morphism between indecomposable Nakayama modules is irreducible exactly when it is a monomorphism with simple cokernel or an epimorphism with simple kernel.

[severe atlas]

Not exactly, when choosing i such that the image of X->Z->Z_i->Y is maximal, there is apriori no reason that this map is equal to f (but this does not matter in the end). You still need to see that this composition is injective, it's image is X and that showing that this being split is enough to conclude. If Z is indecomposable, it is trivial and I do not believe that Max doubted that. Reducing to the above mentioned case however does take some work.

I can work it out tomorrow, I'm too tired now

[gusty void]

Break Z into indecomposable pieces Z_1 + … + Z_t and pick Z_j whose image W = h(Z_j) is largest among the h(Z_i). Because g is injective the map X -> Z_j -> Y is still injective so its image sits inside W. Now Y∕X is simple so W is either X or Y. If W = X, then h restricted to Z_j lands in X and the composite X -> Z_j -> X is the identity giving a left inverse for that piece of g, hence g splits and so does the whole factorisation. If W = Y, uniseriality forces ker h|{Z_j} to be zero so h|{Z_j} is an isomorphism onto Y and splits. Thus after reducing to the single indecomposable summand Z_j, one leg of the factorisation splits

[gusty void]

Alright 😊

[severe atlas]

X->Z_i->Y is not injective because g is injective, you really need the maximality of the image for that. Moreover, that in this case h is an isomorphism, or more precisely injective, has nothing to do with Y being uniserial, but now since Z is assumed to be idecomposable, it is uniserial.

But I think if you work it out in detail, it is easy to see

[gusty void]

Choosing the summand with maximal image ensures g(X) maps into W non trivially so the composite X->Z_i->Y stays injective with Z_i indecomposable (hence uniserial) any non zero kernel would force overlap with g(X) so the kernel is zero and h|_{Z_i} is an isomorphism

Oops

Thays the one

[west tangle]

Hello, I have been struggling to understand what an F-Algebra and Initial Algebra are (For context, my interest is as a Computer Scientist studying type theory)

I have been reading the attached definition from wikipedia: https://en.wikipedia.org/wiki/F-algebra

So just walking through the definition: A is the carrier set, which we could say is the "data". In Agda (or Peano Numbers), the terms of Nat would be zero, suc(zero), suc(suc(zero)), etc. So this would be the carrier set for some algebra about natural numbers.

And a (alpha) is defined as "F(A) -> A". I am really lost on what this is supposed to be. It is a functor, from A to some other categorical object, that is then mapped back into A? Expanding it into something like type notation, I am interpreting it as this: "A -> B -> A"

I believe that, in other definitions, alpha is supposed to be a set of fucntions Af(t1,t2,t3,t4,...) -> A

I read this as a set of all the functions that take in terms of A and return a term back that is in A (closed functions). If this is a correct alternative definition, then I am really confused as the what or how "F(A) -> A" connects.

[west tangle]

Yes. After reading: https://bartoszmilewski.com/2017/02/28/f-algebras/

Specifically, the attached part. It is clear that F is supposed to be some sort of Mapping from A that captures the idea of functions on terms of A.

Maybe I just don't know what a Functor is. It is a mapping from A into a different object, but I just dont see how that constructs the functions on A.

Oh. I get it.

I was interpreting the functor as mapping elements of A to elements of a different categorical object. I think I didn't know what a functor was. It takes in A, the object, not the elements, and maps it to an object with elements for each defined function. The mapping is not element to element, but object to object. I think this makes sense.

[spice idol]

It's helpful to see category theory as a layer of abstraction that does away with elements

We only care about the objects and their relationships, rather than the actual elements of these objects

[severe atlas]

No, choosing i such that the image is maximal ensures that X->Z_i->Y has image equal to X inside of Y, since the sum of the images of X->Z_j->Y contains X, and we have that Y is uniserial, so one of them has to contain X. But it can't be bigger than X, since then it would have to be Y, but l(X) < l(Y), so we can't have a surjection X->Y, so it has to be X. Thus, it is an isomorphism (since a surjective endomorphism of f.g. modules is injective), and so X->Z_i->Y is injective, so X->Z_i is injective.

[rare walrus]

I know that this says 'algebra' in it, but this isn't really the channel to ask this in unless you want a group-theoretic example or something. You are likely to get better responses in #category-theory from people who actually know relevant facts.

If you are looking for a genuinely algebraic example, then the simplest possible one is that an F-algebra for the functor F : Set -> Set given by F(A) = A x A is precisely what it means to be a magma, which is arguably the simplest possible kind of algebraic structure. If we also impose some additional requirements, then we can encode what it means for a magma to actually be a semigroup, and so on.

[young escarp]

F-algebra in cat theory and algebra means very different things lol. In Algebra it’d be an “associative unital algebra over the field F”

[young escarp]

Algebraic F-algebras often are those given by polynomial endofunctors.

A × A is a polynomial (of degree 2) :)

[rose mirage]

if only there was some catchy term for, say, monoids in this so called "category of endofunctors". What a world that would be....

[spice idol]

Oh I know!! Algebraic theories!!

[rare walrus]

Banned phrase

[limpid horizon]

hahaha

[spice idol]

😭

The innocence of that question

[fierce steeple]

Triple

[rose mirage]

ohhh baby a tripleee

[fierce steeple]

Real

[spice idol]

MOM GET THE CAMERA

[hexed tangle]

Why is x_n integral over k[x_1,...,x_(n-1)]/I' in this proof?

[echo summit]

Is this solution reasonable?

[hushed bone]

I think it works, but I didn’t quite follow the logic

I think you can make it clearer what you’re doing by saying something like “one of (I,f) or (I,g) has infinitely many minimal primes over it”

Because I guess the point is that each prime minimal over I is minimal over either (I,f) or (I,g), so one of them has infinitely many

I re-read it and get what you’re doing, but I think it can be clearer is my input

I think I’d format it like this,

If I < p is a minimal prime, then as fg in p, either f in p or g in p, in which case (I,f) < p or (I,g) < p, and p is a minimal prime over these ideals as if it were not then p is also not minimal over I. Then, the subset of primes minimal over I is a subset of the union of primes minimal over (I,f) and (I,g), so one of those sets must be infinite and we contradict I’s maximally

[broken turtle]

Sleepybear

[hushed bone]

The cardinality of A doesn’t matter, you can just keep picking some elements not in the previous ideal

This might use some dependent choice or something I guess

[fierce steeple]

Another thing you can do: any ideal in a Noetherian ring is fg (often this is the definition...), so pick a finite set of generators and you can write each of those as a finite linear combination of elements in ur original generating set

But yeah I mean Noetherian has many equivalent definitions so there are lots of ways to do this - another would be to consider the set of ideals which are generated by finitely many elements of your generating set and by the Noetherian property you know this set has a maximal ideal, which is necessarily all of I

[hushed bone]

I think this is better

Because it removes the Noetherian hypothesis

I fg implies among any generating set you can find a finite subset that still generates it

[fierce steeple]

ah yeah good point ye

it's just like i know this is smth you can do with vector spaces lol

[gusty void]

Pick Z_i so W = h(Z_i) is largest. g is mono so h g(X) = X but W can still be Y. example: Z = Y+Y, g(x) = (x,0), h(u,v) = u. Then W = Y, l(X) < l(Y) holds, yet no X ->> Y surjection appears. if W = Y ker h|_{Z_i} would overlap g(X) in a uniserial module, impossible, so ker = 0 and h splits. if W = X h lands in X and g splits. either way one leg splits, QED

[hexed tangle]

in this lemma, why can one choose a basis for m/m^2 among the elements of m?

is it meant that the span of the basis chosen is isomorphic to m/m^2?

[fierce steeple]

Pick elements of m whose images in m/m^2 form a basis

[hexed tangle]

thanks

I have a thing where I take things a bit too literlally 🙂

[marsh fiber]

I have not taken a basic course in Representation theory, but know about it from Advanced Linear Algebra course i credited

Can someone explain Kazhdan property(T) in simple terms for a group (topological group or topological vector space too) and why is it important to study them with some examples ?

[echo summit]

I am working on an exercise which asks me to show that the ring k[x^2,xy,y^2]\cong k[u,v,w]/(uv-w^2) is not isomorphic to a polynomial ring. What is the best strategy for this? My idea to start was that this ring has trancendence degree 2, so if it were a polynomial ring it would be isomorphic to some k[z,w]. From here maybe we can do something with grading? These rings are both graded by N\times N (with N containing 0), but I'm not sure if this really works, since you would need to show that regardless of the grading there is no isomorphism.

[feral rune]

Perhaps it's not a UFD?

[echo summit]

good call, thanks

[feral rune]

Yeah, actually the relation uv=w^2 does it I think. Because w is irreducible, and so are u and v.

[echo summit]

yep

[mental escarp]

Is there a procedure for going from one basis of a module to another basis of the module? I want the coefficients of the new basis in terms of the old coefficients.

[fierce steeple]

It sorta depends what you mean but you should be able to do the same thing as for vector spaces

But with vector spaces you would usually do this by comparing to another basis anyway - what I mean is there's not really gonna be a magic way to do this

(as "change of basis" matrices require you to already have a basis and then write stuff in terms of that basis, lol)

[short vine]

suppose v: A->B is a surjective homomorphism, what is a necessary/sufficient condition for a surjective homomorphism w: A -> B so that there exists an automorphism h of A such that v compose h = w?

[fierce steeple]

Idk what you mean by conditions here

Conditions for such a w to exist? Well it always does by taking any h. So do you want to know when two giben surjective homomorphisms A -> B are related in this way?

[short vine]

yeah the latter

[severe atlas]

Yes, of course the image of h can contain X properly, but the point is that image of "the new" g can't be larger than X, which is not true in general (e.g.,the image of 2Z->Z, 2x->x contains 2Z properly). Moreover, you really need that the modules are finitely generated, otherwise X->Z_j->Y might no longer be injective. You keep ignoring these things, and essentially omly proving the very easy part. It is not directly clear why you can reduce to the indecomposable situation (i.e., why is the new map still injective, has image X, and so on, which I explained a couple of times now).

[short vine]

the kernels must be isomorphic

But is it if and only if?

[hushed bone]

Idk what basis means, but over for a finite module over a Noetherian local ring a minimal generating set is related to any other one via an invertible matrix, in the exact same way it is for vector spaves

And I think the Noetherian case in general follows from this via localization

[old willow]

For context, alpha is in GL_2(Q)^+ and T_alpha is a hecke operator.

Does anyone get why S_k(Gamma(1)) is an invariant subspace?

[cerulean cove]

ask this on number theory channel

[fierce steeple]

Well it's equivalent to there being an automorphism of A that takes one kernel to the other

[short vine]

But this data of kernel isomorphism is not sufficient right

[short vine]

Like supposed v,w: G2 -> G1 have isomorphic kernels, does this mean that there exists some automorphism h:G2 -> G2 such that vh=w?

[fierce steeple]

No, e.g. Z has no non-trivial automorphisms but the ideals 3Z and 2Z are isomorphic as modules

[short vine]

Sorry forgot to mention G2,G1 are finite abelian p groups(the same p)

[molten vale]

If we have rings $Z(A)\subseteq B\subseteq A\subseteq U$ and we have $C_U(C_U(A))=A$, is it true that $C_U(C_U(B))=C_A(C_A(B))$? I suspected that this is false but I could not find a counterexample.

[broken turtle]

qwertytrewq

[fierce steeple]

Oh lol I think I assumed we were talking about rings aha sorry

[woven loom]

Well, it’s surjective, so you’re taking a quotient

You get something like 0 -> ker -> A -> B -> 0

When does it fail (if you don’t know necessary conditions)

[short vine]

First we basically have two exact sequences, and now we need the automorphism to restrict to an automorphism between the two kernels

im actually wondering when will it succeed

it's two exact sequences with morphisms in between, B -> B as identity map and ker -> ker as some map that exists. I'm suspecting that this isn't enough to induce an automorphism of A

[rough vale]

in octonions, we have 1, and 7 imaginary units, call them i,j,k,1',i',j',k'. Where prime on imaginary units are just from the cayley dickson consutuction as you would guess, so i'=(0,i). Say I define $$Hx^\perp:=span{\mathbb{R}}(x,ix,jx,kx) \text{ and } Hx:=span{\mathbb{R}}(1'x,i'x,j'x,k'x).$$ Is there an octonion $o$ so that $o H_x^\perp = H_x$ for all octonions $x$?

[broken turtle]

Horse Face

[vapid axle]

I'm a little bit confused about this proposition. I know that S^{-1}M can be viewed as an S^{-1}A module, and I suppose S^{-1}A and M as well, but here the tensor product is subscripted with an A instead of an S^{-1}A. Is the A a typo, and it should be replaced with S^{-1}A?

[past cove]

There isn't any typo

I'm not totally sure what's like confusing you but the idea is that the tensor product extends the scalars

so you go from your A-module M to an S^-1A module called S^-1M

cuz like M isn't a priori an S^-1A module

[fierce steeple]

oh lol i didn't realise it was you, irony

hello

[past cove]

yeah I've become the milk enby

hello how are you potato

oh arki and I will be reading some notes about K-theory this summer if you want to you can join us
They're focused on the (oo,1)-cat way to look at things

so you'd probably like it hehe

I can Dm you the link if you want

[fierce steeple]

Maybe some extra context might help? In general if you're given an $A$-module $B$ and a map of rings $A -> B$, then you can form the extension of scalars $B \otimes_A M$. Now to form this, you view $B$ as an $A$-module via the map, but you can then give $B \otimes_A M$ the structure of a $B$-module in the `obvious' way i.e. $b.(b' \otimes m) = bb' \otimes m$. This is a sense the best way to turn $M$ into a $B$-module: if you're given any $B$-module $N$, then you can view it as an $A$-module by restricting scalars along $A \to B$, and then any map $M \to N$ of $A$-modules corresponds in a unique way to a map $B \otimes_A M \to N$ of $B$-modules.

[broken turtle]

Prismatic Potato

[fierce steeple]

Then in this proposition the point is that $S^{-1} M$ is an $S^{-1}A$-module and there's an `obvious' map $M \to S^{-1} M$ of $A$-modules, which extends along $A \to S^{-1} A$ to give you the map $S^{-1} A \otimes_A M \to S^{-1} M$ of the proposition

[broken turtle]

Prismatic Potato

[fierce steeple]

Ooh yeah that sounds great, what were you planning to follow? I have been meaning to get better at K-theory lol

[fierce steeple]

very cool

[past cove]

hold on I'll dm you

[fierce steeple]

coolio

[plucky arch]

K theory sounds very cool

Unfortunately I do not know enough math to even start it :(

[woven loom]

Well, option A: grab a book and trial by fire

I mean, how else do you determine “knowing enough”

[digital parcel]

what's option b

also where did the K in K theory even come from

i know the grothendieck group is K_0 but where did that come from too

since i've seen some ppl just denote it w G

[woven loom]

Finding someone who knows what they’re doing and following a more guided approach

[plucky arch]

I mean I’ve heard it’s very hard algebra

[woven loom]

It sure ain’t easy

[plucky arch]

Algebra is my weakest area of math so

[void hare]

are R[x] and R[y, y^2/2!, y^3/3!, …] isomorphic as graded R-algebras if |x| = |y| and R contains Q?

this should just be x^n sending to y^n/n! right?

the latter is also called the divided power algebra i think

[woven loom]

id say its my weakest as well but tbh hard to tell with my general lacking

but just cracking open Weibel's book and seeing what you can or cannot do might be doable

[woven loom]

well, I'd sorta expect f(x^n) and f(x)^n to be nicely related which isn't ideal with that /n! term, but maybe I'm missing something about what notion of iso

[void hare]

so, you expect it not to be an isomorphism?

does the map i give not preserve the multiplicative structure

[woven loom]

well, consider like

f(x^2) should be f(x)^2 right?

[void hare]

yeah

[woven loom]

but wouldnt we get f(x^n) = f(x)^n / n! as a result?

[void hare]

we don’t, right?

we get f(x^n) = f(x)^n?

[woven loom]

well f(x) = y right?

[void hare]

i guess

[woven loom]

but f(x^2) is y^2/2

which looks kinda like f(x)^2 /2

[void hare]

yeah

[woven loom]

but not f(x)^2

so it doesnt look like a ring homomorphism if y * y is y^2

[void hare]

yeah, that’s true, i see

[spice idol]

Can't you just send x to y?

[void hare]

isn’t an iso

[woven loom]

me when R[x] is iso to R[y]

[woven loom]

why not

[spice idol]

Why not

Jinx

[void hare]

doesn’t send generators to generators?

[woven loom]

what do you mean it doesnt

[spice idol]

You're assuming Q is included in R right

[woven loom]

does y not generate that?

[void hare]

oh

right

yikes…

[woven loom]

remember, not all homomorphism send every choice of generator to every list of generators

because like, isomorphism maps between 2d vector spaces will only give you generators in the output, not necessarily map to a particular basis

so you dont need to send the list of generators {1, x, x^2, ...} to the list {1, y, y^2/2, ...}

[void hare]

yeah, ik that, i just forgot that Q is contained in R

or well, it wouldn’t be true if Q isn’t in R, right?

[woven loom]

well, it didnt respect multiplication so its not even a homomorphism

[void hare]

yeah

[woven loom]

the map x ~> y isnt an iso if you dont have Q yeah

not that it makes too much sense outside char 0 too

[void hare]

yeah, thank you

[woven loom]

anyhow shouldve been the biggest red flag fr

[plucky arch]

general lacking...?

[woven loom]

as in, general lack of skill in analysis, topo/geo things, logic

[plucky arch]

Huh, it doesn’t appear that way to me

[trim hinge]

I have a quiver $Q$ and relations $I$ such that $\frac{kQ}{I}$ has global dimension 2 (this quiver is actually a subspace configuration poset but I'm not familiar with the language of poset rep theory). I thus have an 'extended' Tits/Euler form $q$ that eats dimension vectors and spits out a number that is telling me the value of dimEnd - dimExt + dimExt^2.

I have a bunch of representations of this quiver of interest coming from some other construction. These representations are associated to elements of a root system with form $p$, and all these elements are real roots, in particular have 'p = 1' (its really p=2 for reasons).

All of the resulting $kQ$ reps I have constructed from $'p = 1'$ roots have had $q = 1$. So I'm trying to pin down the 'root system' associated to q to try and make some correspondence, or atleast an implication like "if p = 1 over here, then the constructed thing will have q = 1".

My problem is that the Gram matrix of q has positive off diagonal entries and it is my understanding that that means it cannot have come from a generalised cartan matrix and so I'm not sure that there is even a Kac-Moody algebra kicking around.

But there IS some structure hiding in q in the sense that if I just blindly enumerate a bunch of dim vects with q = 1, I'm only seeing the valid configurations that im independently constructing. Also the only q <= 0 guys showing up in the enumeration are lattice points in the cone generated by two minimal imaginary 'roots', i.e. two guys in the 2 dimensional kernel of the Gram matrix.

I guess my question is then if anyone knows or can signpost me to info on like structure results for these 'pseudo root systems' if thats even what I'm looking at. Sorry that it's all very vague, it would just be too long to go into the details. I can elaborate if people have any ideas, thanks

[broken turtle]

ΣAC
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thorny flax]

I don't think you are in the correct section. Try #groups-rings-fields

Anyways, if |x|=oo and you falsely assume |gxg^(-1)|=n<oo then by what you proved we have |gxg^(-1)|=|g||x||g^(-1)|=|g||g^(-1)||x|=|gg^(-1)||x|=|e||x|=|x|=oo so contradiction. @gritty knoll

[balmy pulsar]

disregard my last message, sorry

[past cove]

K-theory

[limpid horizon]

What is the motivation for finding generators of ideals “up to radical”?

[hexed tangle]

my commutative algebra exam was today... got the top grade 😄

[rare walrus]

Congrats!

[hexed tangle]

thanks!

[ornate atlas]

Nice one!

[hexed tangle]

Of course I had this Tshirt on

[unborn rampart]

Does it say "it makes people cry"?

[limpid horizon]

to justify that M is cohen macaulay iff H^i_m(M) = 0 for i < dim M, do i just need to justify the fact that depth_m M = min{i | H^i_m(M) neq 0}?

[hexed tangle]

it does

[scarlet ermine]

maybe one motivation is the Nullstellensatz—if you only care about finding equations f1,…,fi in k[x1,…,xn] to cut out a Zariski closed subset V(I) set theoretically, then the Nullstellensatz says (when k is alg closed) that V(I) = V(f1,…,fi) iff radical(I) = radical(f1,…,fi)

[limpid horizon]

well I guess we also need the fact that H^i_m(M) is nonzero at i = dim M, and then we are good

[balmy pollen]

can i have an example of a commutative algebra where krull and gk dim differ

[hushed bone]

What’s gk dim

[hushed bone]

Do you mean gl dim?

If you do, then for a Noetherian ring (of finite Krull dimension let’s just say) the global dimension being finite => gl dim = Krull dim which is equivalent to being a regular ring

So the only way these can disagree is if you’re not a regular ring in which case the gl dim is infinite and the Krull dimension is finite

Assumptions throughout is Noetherian and finite Krull dimension

[ornate atlas]

Gelfand kirilov dimension no?

[hushed bone]

Tf is that

[ornate atlas]

Something something D modules

[ornate atlas]

I brushed against it in my noncom class but never actually looked into it

Something about Hilbert polynomials and Bernstein’s inequality

[past cove]

If you meant gelfand kirilov this can only be true if your algebra is non finitely generated

So idk there are probably some pathological examples of that

[balmy pollen]

Yeah I’m just having trouble thinking of one

[balmy pollen]

Nah, gelfand kirilov

Like growth rate stuff

[astral ginkgo]

Can anyone help me with the following statement: There are only finitely many maps from a group G into a finite group H with a given kernel.

[hushed bone]

Let K be the kernel, then either G/K is infinite in which case no such maps exist as G/K needs to inject into H. So now if G/K is finite the maps G -> H with kernel K are in bijection with injections G/K -> H which is finite by cardinality

Obviously if K isn’t normal so you can’t form G/K the number is 0

[astral ginkgo]

it;s always normal, it's a kernel

[rose mirage]

K is a kernel it's gonna be normal

[astral ginkgo]

right, I was being stupid thanks chmonkey

[hushed bone]

As stated K could start off not normal

And then the number of these maps is 0

[rose mirage]

if K wasn't the kernel of the map your injection argument fails

[hushed bone]

Oh my god, if there are no maps G -> H with K as a kernel then K could be not normal, and then my argument doesn’t work cuz I can’t guarantee I can form G/K. But if that’s the case you already proved the result

[rose mirage]

also you started with "let K be the kernel"

[hushed bone]

That’s all I was saying

Whatever I meant let K be a subgroup

[rose mirage]

thanks for letting me farm my very active points for the day

[hushed bone]

Sybau

[rose mirage]

sYbau

[undone idol]

if you are talking about primary decompositions, then it's to do with decomposing subsets of the Zariski topology into a union of irreducible sets, which is related to Nullstellensatz as joesph said. alg geo stuff

[vapid axle]

Does anybody know why $\text{im } \overline{\nu} \subseteq \ker d$? If $\overline{\nu}(m) \in \text{im } \overline{\nu}$, then we have $\mu' (f' (\mu^{-1}(m))) = f(m)$ (and consequently, $f'(\mu^{-1}(m)) = 0 \in \text{Coker } f'$). I'm just having trouble showing that $\mu^{-1}(m)$ is well-defined

[broken turtle]

okeyokay

[vapid axle]

Nvm this doesn't work if \overline{\nu}(m) \neq 0

[cobalt sonnet]

I am looking at this proof of how non degeneracy of a pairing (defined in terms of having a copairing) is equivalent to the space being finite dimensional and and map between space and dual being injective
i am stuck in the converse part if the functionals are linearly independent does it follow that there is a dual basis?

[vapid axle]

Anyone?

[limpid horizon]

I cant see why they are the same if I and J have the same radical (im pretty sure it meant to say radical there, not nilradical)

If I^nx = 0 and rad(I) = rad(J), J^mx = 0 for some m?

[past cove]

If they've got the same radical you can show there's an n such that J^n subset I

And the other way around

For some different positive integer

Tho I think this is a thing that only works in noetherian rings (i.e. you need some f.g. assumption on ideals)

So I hope you're in a noetherian ring lawl

[digital parcel]

afaik local cohomology is always over noetherian rings (but dont quote me on this)

at least in the case of those notes which i've seen before

[limpid horizon]

Yeah i was wondering if there needed to be extra conditions

In the notes im reading it didnt mention that

[limpid horizon]

Oh?

[digital parcel]

those notes look familiar can you send them to me

i can't remember where i've seen them before

[limpid horizon]

Just says R is comm ring

http://homepages.math.uic.edu/~bshipley/huneke.pdf

[digital parcel]

p 2

well it should be noetherian at least

[limpid horizon]

Oo

Okay

[digital parcel]

i can't imagine huneke doing anything over non-noetherian rings

[limpid horizon]

Thank u

[digital parcel]

i'll have to think about this tho

hochster defines it over noetherian rings (as the direct limit of some ext modules)

[limpid horizon]

yea, i havent learned that defn yet

[digital parcel]

https://mathoverflow.net/questions/209337/local-cohomology-and-radical-of-ideal see here

[jolly carbon]

I came up with this question while thinking about the construction of real numbers via formal decimals. Does the usual decimal representation define an embedding of the field R of real numbers into the quotient ring Z[[1/10]] := Z[[x]] / (1 - 10x)? I think it is true that 0.999... = 1 in Z[[1/10]]

[limpid horizon]

if J is fg by like {j1, j2, j3} then something in J^n has components that are like j1^n, j2^n, j3^n and some other combination of the j's. The j1^n j2^n j3^n are all in I if n is big enough, but what do I do with the other ji components

[hushed bone]

Local Cohomology is unbelievably ass over non-Noetherian rings and without finiteness hypotheses

You can define it but it sucks

[thorny flax]

I understand that there is a group action of the symmetry group $S_4$ on the cube in $\mathbb{R}^3$.

How can I compute the number of orbits of $X_e \times X_i$ for each $i\in {f,v,d,t,c, e}$ where:

$X_e$ is the set of edges of the cube

$X_f$ is the set of faces of the cube

$X_t$ is the set of the two tetrahydron of the cube

$X_d$ is the set of diagonals of the cube

$X_c$ is the center of the cube

$X_v$ is the set of vertices of the cube

[broken turtle]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorny flax]

I am asking this in this section because I want to compute $c(\mathbb{C} X_e, \mathbb{C} X_i)$ for the $i$'s above.

The intertwining number of the quasiregular representations of the sets above.

[broken turtle]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[rose mirage]

quasiregular 💔 intertwining number 💔 S_4 acts transitively on all of these assuming I'm guessing what a tetrahydron is correctly.

much easier way to do this would to just write the characters down and take inner products

[fierce steeple]

Is it okay over non-Noetherian assuming like the ideal is fp or smth

Or are the issues more serious

I guess some dimension theory kinda breaks so oof

[lone jacinth]

1 - 10x is a unit in Z[[x]], so that quotient is just the trivial ring.

[teal depot]

Hi! I am currently trying to learn about schur functions for understanding a paper, I would love to know what books I could refer to understand it.

[hushed bone]

I don’t really know tbh, I just know shit breaks down fast

But like, the interaction with depth I think immediately fails

A key part in showing depth 0 <==> Gamma_m is nonzero requires something with associated primes

[jolly carbon]

Thank you

[vapid axle]

Someone save me!

[woeful crane]

Given $x''\in \text{Im}(\bar{\nu})$, we can take the "$x$" that appears in the screenshot you posted such that $x\in \ker(f)$. Thus (taking the corresponding $y'$ from the proof) we have
$0=f(x)=u'(y')$, and since the proof defined $d(x'')$ to me the image of $y'=0$ in the cokernel, this shows that $d(x'')=0$.

[broken turtle]

kr1staps

[limpid horizon]

should it say right derived there

[rose mirage]

those are the rightest looking functors I've ever seen

but alas, F is contavariant, so these are indeed left derived functors

I think?

[limpid horizon]

i dont really know what the terminology is referring to tbh

[rose mirage]

nah they are

why the hell they're denoted with an R is beyond me

[limpid horizon]

the first sentence talks about right derived

so theres some typos here ?

what is the "right" vs "left" derived refering to

[digital parcel]

If you have a right exact functor then you get a left derived functor which adds a bunch of stuff to the left

[limpid horizon]

so if u have a right exact functor it doesnt make sense to say it has right derived functor right

[digital parcel]

It measures how far your functor is from being left exact

[rose mirage]

well no it does it's just a boring one (the zero functor)

[limpid horizon]

ok i think that makes more sense

[rose mirage]

ngl I had to google if Ho(C^op) was Ho(C)^op

[rough vale]

say a polynomial in x_1,...,x_4 is invariant under SO(2) for example, like P(x_1,y_1, x_2,y_2)=P(r(x_1,y_1),r(x_2,y_2)) where r in SO(2), then is it some product or sum of x_1^2+y_1^2, x_2^2+y_2^2, <(x_1,y_1),(x_2,y_2)>. Only norm and inner product are the fundamental building blocks of such polynomials. Is this true? if so, what is this theorem called?

[rose mirage]

this theorem almost certainly does not have a name but this sounds a lot like the type of problem Invariant theory is designed to solve. You're asking for the degree 4 part of the invariant ring of the diagonal embedding of SO(2) -> SO(2) x SO(2) (viewed as a subgroup of GL_4(R), I presume) acting on R[x1, y1, x2, y2] as you described

if you're correct (which I do beleive) then you've essentially just found the primary invariants for this invariant ring

[rough vale]

any idea where i can find such a theorem?

[rose mirage]

I really wish I could remember but it's been years since I learnt this lemme try and think

[rough vale]

tyty

[limpid horizon]

this modified cech complex that can be used to compute local cohomology is what i actually need to learn, but i am not totally sure what background is needed to learn this and justify that it does give local cohomology

do i need to learn koszul complex too?

[fierce steeple]

Except the 0th one

Well I like to phrase in terms of total derived functors and then ur saying RF = F which is pretty

[fierce steeple]

(I should say like I said this as a joke)

[rose mirage]

ok for infinite groups I've completely forgotten how it works. But it kind of makes sense where this would appear right? O(2) is the largest group of matrices that preserves the inner product (in particular, the norm), so the invariant polynomials would correspond to norms and inner products. Try this for U(2) and see if the primary invariants look like the polynomials you've found but with complex conjugates in the approparite places

that'll be interesting to see and I think it works

[rough vale]

i don't understand this, how would one deduce that those are the fundamental building blocks of such polynomials?

[rose mirage]

show that they span the invariant ring

which is now firmly a YOU task

[rough vale]

actually I am not really familiar with any of this which is why I was wondering if there is a source I can find this in and learn the preliminary material

I have 0 background in invariant theory

[rose mirage]

I have essentially 0 background. I only know of the specialised "Reflection Groups and Invariant Theory"

[cobalt sonnet]

Anyone?

[thorny flax]

I don't know how to do it.
Could you please help me?

[tight marsh]

Hey, I'm trying to work through a paper: "On an easy transition from operator dynamics
to generating functionals by Clifford algebras"

(35) and (36) have totally lost me. What exactly is a clifford monomial?

Also, the 2nd and third lines of (35) these bra/kets seem to come out of nowhere. I can only assume they pop up due to some sort of inner product relation?
But why specifically |0> and |p>?

In the paper the partial derivatives are duals of j's which are elements of a vector space. P() is some ordering.

If anyone thinks they can help please let me know!

[vapid axle]

Wait what

Isn't this just the definition of d lol

What is y'?

[woeful crane]

Yes, after all, we're trying to prove that something is in the kernel of d, so we'll probably have to use the definition of d.

[vapid axle]

So why is y' in the image of f'?

[woeful crane]

I showed that y' = 0, so f'(0) = 0 = y'.

[vapid axle]

Wait by your logic, wouldn't it follow that d is the zero map?

Sorry I'm confused

I guess I'm confused why $x \in \ker f$. If $\overline{v}(m) \in \ker f''$, then $\overline{v}(m) = v(m)$, so here we take $x = m$. But how do we know that $f(m) = 0$

[broken turtle]

okeyokay

[vapid axle]

Can somebody sanity check me

Never mind I'm a dumbass

m is in the kernel of f 🤦‍♂️

thanks @woeful crane

[limpid horizon]

Hey, im not sure exactly how to interpret the direct sum in the C^k

Something like R_x1x2 is the localization of R wrt to the multiplicative closed subset generated by x1x2 right

also, what is a "system of elements"? Is that a technical term?

[digital parcel]

likely means system of parameters

given a local Noetherian ring $(R, \mathfrak m)$ of dimension $d$, a system of parameters is a set of elements $x_1, \dots, x_d$ such that $(x_1, \dots, x_d)$ is $\mathfrak m$-primary

[broken turtle]

anamono

[digital parcel]

yeah just the localization of R at the element x_1x_2

[limpid horizon]

thanks dude you are new but been helpful to me lol

[digital parcel]

glad to help

anyway the reason they can "fix a system of parameters" is because every local Noetheran ring admits a system of parameters

i think Atiyah-Macdonald has some stuff on this

Bruns-Herzog as well

[limpid horizon]

yeah right now my goal is to understand hochsters formula and reisners criterion from combinatorial algebra

its using a lot of commutative algebra that i need to catch up on

[digital parcel]

ah yeah nice

[limpid horizon]

sometimes its hard to tell what i should focus on whats important

[digital parcel]

bruns-herzog's called Cohen-Macaulay Rings is a good reference for this then

although it's boring as hell to read

[limpid horizon]

yeah

I have a copy my supervisor gave me but its back home lol

im in toronto now maybe uoft has it in the library but idk

[digital parcel]

https://agorism.dev/book/math/commalg/cohen-macaulay-rings_bruns-herzog.pdf

this is hopefully legal seeing as i just pulled it from a google search

[limpid horizon]

yea

nothing beats physical books for me tho haha

[digital parcel]

same

[limpid horizon]

what are we indexing over in the direct sum

[digital parcel]

all possible choices of k integers satisfying that

so like you could have 1 < 2 < ... < k

or 1 < 3 < 4 < ... < i_k

and whatnot

[limpid horizon]

so C^0 is = R_x1 (+) R_x2 (+) ... (+) R_xn?

[digital parcel]

I think C^0 is just R

C^1 is R_x1 + ... + R_xn

in C^k, there should be k terms in the localization of each summand

[limpid horizon]

oh cause it starts i_1 not i_0

[digital parcel]

ye

oh wow they use \mathbb{C} for the cech complex lol

[limpid horizon]

we're considering the C^k as R-modules and these are R-module maps right

[digital parcel]

should be yeah

[ornate atlas]

That’s an understatement

Its a nice book in so far as it has everything you might want it to, not so nice in terms of being a riveting read

[vapid axle]

Let $S^{-1}A$ the ring of fractions of $A$. Why is $S^{-1}A$ flat? Let $0 \to M' \to M \to M'' \to 0$ be any exact sequence of $A$-modules. The claim is that the tensored sequence
[0 \to M' \otimes S^{-1}A \to M \otimes S^{-1}A \to M'' \otimes S^{-1}A \to 0] is exact. Since $M \otimes S^{-1}A \cong S^{-1}A \otimes M \cong S^{-1}M$, I'm guessing the idea is to show that
[0 \to S^{-1}M' \to S^{-1}M \to S^{-1}M'' \to 0] is exact. There is a proposition in Atiyah-Macdonald stating that $S^{-1}M' \to S^{-1}M \to S^{-1}M''$ is exact at $S^{-1}M$, but nothing about the injectivity of the first arrow or surjectivity of the second arrow (unless $M'$ is a submodule of $M$. So how does this make sense?

[broken turtle]

okeyokay

[last talon]

Applying the proposition with 0, M’, M, we get that M’ -> M is injective, and applying it with M, M", 0, we get that M -> M" is surjective

[vapid axle]

Oh right lol

Thank you

[fierce steeple]

This is a general thing that to prove exactness it's enough to show it preserves exactness of these little things A -> B -> C

Indeed that is probably the better definition

[signal moon]

What is the notation $\Gamma\otimes_\mathbb{Z}\mathbb{R}$ on the last line? I know that this gives an $\mathbb{R}$-vector space of dimension n, but formally what does it mean?

[broken turtle]

person2709505

[shy cargo]

have you seen a tensor product before?

[signal moon]

Nope

I don't think I'll be able to figure out what that is (in a timely manner) either, so since someone has explained what it means in this context I'll leave it as a black box for now. Thanks anyways though

[shy cargo]

Okay so on vector spaces the tensor product of two vector spaces V and W over a field k is the set of ordered pairs (v,w) of vectors in V and W with the rule that (kv,w)=(v,kw), so we can factor scalars out of either coordinate

[signal moon]

That "rule" means we're quotienting the set of ordered pairs out by some equivalence relation I guess?

[shy cargo]

Then for modules, or vector spaces over rings, we just take this same construction. If M and N are both R-modules, $M\otimes_R N$ is ordered pairs $(m,n)\in M\times N$ where we've chucked in the equivalence $(rm,n)=(m,rn)=r(m,n)$

[broken turtle]

lance

[shy cargo]

Yep, it's formally M\times N/these equivalences

[shy cargo]

And then finally coming down to earth for your case, just remember every abelian group is a Z module and you're done

[signal moon]

I think I understand

[manic wasp]

i think you also need (a+b,c) equiv (a,c) + (b,c) and the same in the second coordinate

[shy cargo]

Oh true that’s not in fact a property of what I said

[ornate atlas]

commutative rings…

[spice idol]

Every ring is commutative

R-algebras need not be though

[fierce steeple]

Lmao I find that take amusing

[spice idol]

Hey it's essentially a back door

Just view noncommutative rings as Z-algebras

[fierce steeple]

Yee it's just funny cause hm

Cause of AG etc I usually think of A-algebra as being synonymous with a map A -> B but it is funny that what you said also is a kinda common thing

[spice idol]

I like group algebras and I wanna include them somehow

[rare walrus]

You mean group rings

Idk man this take gets posted here a lot

but like

...people already say rings

so

hmm yes let's make a new standard that everyone should follow. this will solve all problems with the old standard, since everyone will simply start using the new one!

[spice idol]

I hope you realize how non-serious I am about all this

[rare walrus]

https://xkcd.com/927/

xkcd: Standards

Fortunately, the charging one has been solved now that we've all standardized on mini-USB. Or is it micro-USB? Shit.

[spice idol]

the xkcd I was thinking about too lol

Notation and conventions are a mess but hey at least it isn't as bad as in music

Cue the tantacrul vid

[near lantern]

This is literally the opposite of the convention I am used to lool

[spice idol]

That makes sense I guess?

Kind of

to get the correspondence R-algebras <-> ring homs from R

[near lantern]

No wait

I'm used to rings being not necessarily commutative

and algebras being commutative associative unital, associative unital, or (just a bilinear multiplication) depending on context.

[ornate atlas]

This is the correct way (because it’s also what I’m used to)

[spice idol]

Commutative algebra is interesting qwq

But yes hur durr I hate interesting math I will never go out of my comfort zone

Stealing this meme though

[rare walrus]

Stealing this

[fierce steeple]

Lmao

[fierce steeple]

Wdym by this tho lol

Idk like "all rings are commutative" is a fine convention in some contexts but I don't think many would say rings should actually be redefined

[spice idol]

We should redefine rings such that the multiplication isn't even associative tbh

[fierce steeple]

Actually lol this is the thing with like

Normed division algebras are (in my experience) not assumed associative

[spice idol]

Unit quasigroup

Or is it just a multiplicative inverse, not division

[ornate atlas]

So glad my 2 minutes in my phones markup was worth it

[spice idol]

https://tenor.com/view/fire-writing-gif-24533171

[limpid horizon]

I dont understand the dk maps

[limpid horizon]

also these direct sum of localizations are like within what ambient structure?

is this like the "external" direct sum or something?

so something like Rx1x2 (+) Rx1x3 (+) Rx2x3 is the same as Rx1x2 x Rx1x3 x Rx2x3?

so would elements of that look like (r1/(x1x2)^m1, r2/(x1x3)^m2, r3/(x2x3)^m3)?

if n = 3, im trying to understand what d2: C^2 -> C^3 map would look like

[digital parcel]

say $n = 3$ and my variables are $x_1 = x, x_2 = y, x_3 = z$. Then we have
[ C^1 = R_x \oplus R_y \oplus R_z \quad\text{and}\quad C^2 = R_{xy} \oplus R_{xz} \oplus R_{yz} ]
as an example, let's just look at the maps $\alpha$ going from $R_x$. the map $\alpha : R_x \to R_{xy}$ is localization at $y$, since ${1} \subset {1, \hat{2}}$

the map $\alpha : R_x \to R_{xz}$ is localization at $z$ since ${1} \subset {1, \hat{3}}$

the map $\alpha : R_x \to R_{yz}$ is the zero map

hold on lemme check this actually

or something like this

[broken turtle]

anamono

[limpid horizon]

thanks, i havent really seen localization of a localization before so im trying to understand what that would be like

[digital parcel]

so then when you take the differential you just apply the correct sign change

$R_x$ consists of elements of the form $r/x^n$, so $(R_x)_y$ consists of elements of the form $(r/x^n)/y^m$, where $r/x^n \in R_x$

[broken turtle]

anamono

[digital parcel]

another way to say this is that $R_x \cong R[x^{-1}]$ and so $(R_x)_y \cong R[x^{-1}][y^{-1}] = R[x^{-1}, y^{-1}]$

[broken turtle]

anamono

[digital parcel]

that's just in this case where we're localizing at elements

the more general case is this (exercise 3.3 of Atiyah-Macdonald)

you can use universal property of localization to prove this

or, if you prefer, chasing the definition of localization

if i were vakil i'd suggest you to do both, especially the one you don't prefer

[digital parcel]

basically all the map \alpha is doing here is the following:
0. note that when mapping from a component of C^k to C^{k+1}, you're either adding an element to the localization or not

1. if you're doing the former, then all this map is doing is localizing at that element
2. if you're doing the latter, then your map is zero

so like R_x \to R_{xy} "adds" the element y, so your map \alpha is just localization

but R_x \to R_{yz} doesn't add any element (because x isn't being inverted in R_{yz}) so your map is just zero

[digital parcel]

direct sum of R-modules (since the localization of an R-module is still an R-module)

[tight marsh]

Bumping because I still dont understand anything here

[limpid horizon]

Ya so this is just direct product right

[digital parcel]

Technically theyre not the same

Direct product is the actual product in the category of R-modules, direct sum is the coproduct

[limpid horizon]

In this case are they the same?

[spice idol]

They only differ in the infinite case

Coproduct seems to be used more often though?

I suppose a sum is nicer to work with than a tuple

In universal algebra, coproducts behave unpredictably, direct sums are really smt special

[limpid horizon]

u loveee direct sums dont you enpeace 🥰

[digital parcel]

Yeah forgot about dimension my b

[limpid horizon]

nathaniel boutta drop some heat

[analog abyss]

Not sure if this is the right forum, as my question is kinda analytic, but: in a functional analysis book I'm studying, they define the notion of a unitary representation, and demand continuity of the representation $\pi$ in the sense that for each vector $v$ in your Hilbert space, the map $g\mapsto \pi_g(v)$ is continuous in $g$. Why is this the ``correct'' notion of continuity to demand? For instance, why not demand that the map $(g, v) \rightarrow \pi_g(v)$ is continuous on $G\times \mathcal{H}$, or that the map $G\rightarrow B(\mathcal{H})$ is continuous with respect to the topology induced by the operator norm on $B(\mathcal{H})$? Is it secretly equivalent to one of these?

[broken turtle]

Nathaniel Kingsbury-Neuschotz

[analog abyss]

No, just a possibly stupid question lol

[fierce steeple]

Iirc this says G -> B(H) continuous for the strong operator topology on B(H)

But then you may wonder why one wants that lol

[analog abyss]

(If I should post in an analysis channel b/c continuity, or a number theory channel b/c unitary representation theory, lmk. I posted here since I think of it as a representation theory question, and I think of representation theory as "more" algebraic than analytic)

[analog abyss]

I think I saw this in a stackexange post

[fierce steeple]

But also this is like a coarser topology so you should only be able to get more reps

[analog abyss]

But as you said, why that topology on B(H)? (I have yet to see this topology in said functional analysis book lol)

[analog abyss]

Ah. So it's like the weakest continuity condition we can impose and get a meaningful theory, kinda?

[fierce steeple]

I think so. Honestly I am not an expert at all but just remembered that fact. Nlab has an example

But yeah it is sort of funny as often one is careful when setting this up stuff so it is "continuous enough" but here you have to weaken that

[fierce steeple]

I think this is such an important example that it is motivating though

[analog abyss]

Oh! Yeah I think it is! Regular representations can be used to do Fourier like things!

[fierce steeple]

Well also just this is one of the easiest representations I guess aha so if that goes wrong then you have a problem

(But again sorry I cannot be too helpful - I'm sure others can say more)

[analog abyss]

And it contains loads of irreps (in the case of a finite group G, ALL of the irreps), so it's important

[analog abyss]

Thank you for saying what you've said!

(I'm now gonna wait and see if e.g. one of the Langlands supergeniuses in the server like Aphelli or nGroupoid decides to say something and blow me away with cool math)

[spice idol]

Just jealous we don't get such nice constructions in UA qwq

I guess that's what we get for treading the land of generality

[fierce steeple]

Idk this feels like generality in two different directions

[spice idol]

Wdym

[fierce steeple]

I guess it's just like you can have very general things where coproducts make sense that aren't universal algebra

But there will be things in the other way too ig

[spice idol]

Coproducts make sense in any category

[fierce steeple]

Sure when they exist ye

[spice idol]

I guess, direct and inverse limits exist in any variety but they're not used often

In what I've seen at least

I saw them used in a proof that every algebraic lattice is isomorphic to the congruence lattice of some algebra

[spice idol]

[dense ore]

repeating from some other channel: memes and shitposting goes in #chill

[near lantern]

Yeah IG maps into infinitely many modules just don't come up that often (except maybe directed inverse limits?).

[near lantern]

The obvious notion of continuity is joint continuity as you mentioned. Under your hypotheses, this turns out to be equivalent to the weaker condition given (which is that G → U(H) ⊆ Maps(H → H) is continuous "pointwise", i.e., continuous wrt the product topology on the latter).

I think it would have been more reasonable (or at least pedagogically sound) to at least mention this fact near the definition, but oh well.

[near lantern]

There's actually a whole zoo of topologies on B(H) used in functional analysis. IIRC among the most common ones we have weak* ⊆ weak ⊆ strong ⊆ norm (where a ⊆ b means a is coarser than b).

(All I know about this comes from the Wikipedia page though - I'm no expert.)

[analog abyss]

Huh, I guess that given $\epsilon>0$ and a choice of $g$ and $v$, one can find a $U$ containing $g$ such that for $g'\in U$ one has $g'\cdot v$ within $\epsilon/2$ of $g\cdot v$. Since each individual $g'$ is assumed to induce an isometry of $H$, so long as $v'$ stays in an $\epsilon/2$ ball about $v$ one gets that $g'\cdot v'$ stays within an $\epsilon/2$ neighborhood of $g'\cdot v$, and overall $g'\cdot v'$ stays within $\epsilon$ of $g\cdot v$.

[broken turtle]

Nathaniel Kingsbury-Neuschotz

[analog abyss]

(While continuity in each argument isn't typically enough to get joint continuity, the assumption of unitarity gives us a quantitative bound on the continuity in v that's uniform across choices of g!)

Cool!

[cerulean hill]

why is it clear that a discrete valuation ring must be integrally closed?

This is from Milne's book, where he defines:

[analog abyss]

I believe the given conditions immediately imply that A is a UFD, and it is known that every UFD is integrally closed

[hushed bone]

Yeah this follows from the fact that it’s a PID

Like, not anything in the (a), (b), (c) portion

[cerulean hill]

okok i see

[forest turtle]

what is the motivation for calling G an extension of Q by N? In what sense are we extending Q?

[spice idol]

We are extending Q in the sense that to each element of Q we add more information (namely, the closet of N in G corresponding to that element)

[plucky arch]

ooh maybe this is some kind of, like, fiber bundle

a bundle over Q with fiber N?

[spice idol]

Youd need a topology for that

Right?

[plucky arch]

not necessarily right

bundles are a pretty generic concept

[spice idol]

But it's right that with the projection mapping p : G -> Q, the fibers are the cosets of N

[plucky arch]

idk i guess this just looked formally similar to, like, $S^1 \to S^3 \to S^2$ for the hopf fibration

[broken turtle]

Pseudonium

[plucky arch]

instead it's $N \to G \to Q$

[broken turtle]

Pseudonium

[plucky arch]

the fiber is $N$, the total space is $G$, the base space is $Q$

[broken turtle]

Pseudonium

[forest turtle]

there is a note about how some people refer to it as extending N by Q. how would you interpret that?

[plucky arch]

idk im not really an algebraist

but i like thinking about it like a fiber bundle

[spice idol]

I like that interpretation less

But I guess it's extending "the other way"

[forest turtle]

why? it feels like it goes in the direction of the arrows at least

that isn't like

[spice idol]

Extending stuff is basically always against the direction of the arrows

Unless you're doing a tower of inclusions

[forest turtle]

is there a term for going with the direction of the arrows?

[spice idol]

Covariantly?

It's a different kind of extension

[forest turtle]

its not like, retracting?

[spice idol]

Like you have two types of extensions: an injection B → E or a surjection E → B

[forest turtle]

right right

[spice idol]

For groups, a group extension can be seen as both

But I like the second type more

[forest turtle]

why?

[forest turtle]

idk enough about fiber bundles to make a meaningful connection, but maybe someday

[subtle plaza]

Maybe belongs in alg-top channel, but one can think of a group extension this way. If the groups are discrete, one can construct a K(G, 1) (this is a space whose fundamental group is G and whose higher homotopy groups are trivial). Similarly, one can construct a K(Q, 1) and a continuous map from the K(G, 1) to the K(Q, 1) which induces the map pi on fundamental groups. After homotoping things to make this a fibration, the long exact sequence of homotopy groups implies that the fiber is a K(N, 1)

[plucky arch]

@.@

srry i don't know much algtop :(

[forest turtle]

i wish i had personal pins from each server

[subtle plaza]

The point is that this short exact sequence of groups can be realized as coming from a fibration

[plucky arch]

:D

that sounds cool

[spice idol]

Wouldn't it be to K(Q, 1)?

[subtle plaza]

Sorry, yes. Edited to fix

[spice idol]

Since N is the fiber

Thingy

[spice idol]

Why I like the second type more?

[forest turtle]

yea

[spice idol]

Because I like going against arrows in a diagram

I don't know really

In universal algebra at least, a quotient algebra tells you much more about an extension than a subalgebra