yo

yo

Aight nerds

Oh HerrWarum wants to join us as well, @Balarka and @Mathein any objections to letting him in?

nope, none.

nope

@HerrWarum aight

Type something

@HerrWarum What brings you here?

hi

something

Here to talk mathematics!

Well, general chat about mathematics is done on the "Mathematics" room.

Okay so since none of you plebs know ring theory, as we go through Brown I'm gonna have to stop you sometimes and test your understanding of basic ring theory concepts

@Daminark only normies do ring theory

Daminark are you also on the math discord??

If you explain it satisfactorily, we can move on. Otherwise... we can still move on but yeah

Brown?

Yeah @HerrWarum

Nice, I'm KanExtension there

Oh, so Herr is from the math discord

Yup

What exactly is this room for?

And Herr: this is the general math chat

@Daminark I dare to say that I know some ring theory

This room is our study group for group cohomology

Also, going through Brown? Which book is that

Brown: Cohomology of Groups

@HerrWarum It started with the purpose of making a group study on something. That something is for the time being group cohomology

Oh cool, what are the prereqs

@Mathein oh no I meant that I only know the first few pages of Atiyah-Macdonald so every now and then I'll be like "ufcknw0tm8?"

Group cohomology was mentioned by my prof during the group theory course

After introducing short exact sequences

You probably wanna know the basics of algebraic topology

Oh god my copy of Brown that I rigged from libgen looks awful

I don't think Brown directly references alg top

How much of hatcher would you say suffices

@Daminark tbh you don't need a lot of ring theory. If you accept that modules over a noncommuative ring are a thing and that the tensor product is left adjoint to the hom functor, you'll be fine for the most part

There are some parts in my seminar on group cohomology which can be solved by nontrivial ring theory, not sure if they appear in Brown

hey someone give me a good copy of Brown bro

Aight, I'll email it

Well it'll come as Google Drive since it's > 25 MB

That's fine by me

Thanks, btw

No problem!

The book is "with 4 figures" haha

Well it seems too hard for me right now, so you guys carry on

@Daminark Atiyah-Macdonald considers commutative rings only. Group cohomology is one of the prime examples of non-commutative ring theory

Non-commutative rings are fake news but ok

(What's a good way to learn about those btw? A-M is my main ring theory book :V)

Lam is pretty good

Lam has a book "A First Course in Non-Commutative Rings" and a second book "Lectures on Modules and Rings". Both are highly recommended

@Daminark danks

No problem!

Anderson/Fuller covers a similar amount of topics in "Rings and Categories of Modules", but the approach is more categorical

Ted will stab me

In both group cohomology and also representation theory, all that you do is studying rings over certain non-commutative rings

so everyone who says non-commutative rings don't matter can shut up

Frick

(I'm drunk) I actually mean studying modules over certain non-commutative rings

"Rings over non-commutative rings" -Mathein

studying (ring) extensions of a given ring is a thing, so I don't feel too bad about that

Okay so this whole hom-tensor business

Is that saying that $\text{Hom}(X\otimes Y, Z) \cong \text{Hom}(X,\text{Hom}(Y,Z))$?

yeah basically

and also what is important is that if $X$ is $(S,R)$-Bimodule and $Y$ is left $R$-module, then $X\otimes_R Y$ is a left $S$ module

What's a bimodule?

I guess there's both an S and a R-action on X

Well yeah but is there some intercompatibility condition on the actions?

I think the point is S acts on the left and R on the right

Compatibility should be s(rm) = (sr)m

For rings $R$ and $S$ a $(R,S)$-Bimodule is an abelian group $M$ that is simultanously a left $R$ module and a right $S$ module such that $(sm)r=s(mr)$ for all $r \in R, s\in S, m \in M$

(kind of sniped)

why do i care about bimodules tho

Bimodules are a natural setting for considering functors between the categories $R-\mathbf{Mod}$ and $S-\mathbf{Mod}$

and in some sense, they are a natural setting for tensor products

I don't get it

if $M$ is a $(R,S)$-Bimodule and $N$ is $(S,T)$-Bimodule, then $M\otimes_S N$ is a (R,T)$-Bimodule

Bimodules include regular left and right modules by considering $(R,\Bbb Z)$ and $(\Bbb Z, R)$ modules

Okay, but all I am asking is what is inherently interesting about an (R, S)-bimodule structure?

Hmm

I'd like to know more about how they are a natural setting for considering functors between R-Mod and S-Mod

That seems interesting

There's the Eilenberg-Watts theorem which says that every right-exact functor $R-\mathbf{Mod}\to S-\mathbf{Mod}$ is given by taking tensor products with some $(S,R)$-bimodule

that's what I was thinking of

Now we already have a homotopy of chain maps jfc

Oh, that is interesting

@Daminark Yup, chain homotopy is beautiful

Okay so let me guess here

If you were to take the standard chain complex for a topological space

Homotopic maps between space should induce homotopic maps between their singular complexes (now I just remembered the name)

That is true

Here's how I understand chain homotopy: self-promotion while we're at it

the key construction is the prism operator

Also, if you want to go all categorical, there's a $2$-category where the objects are given by some ring $R$, 1-morphisms between two rings $R$ and $S$ are given by ismorphism classes of $(R,S)$-Bimodules, composition of $1$-morphisms is given by tensor products and 2-morphisms is given by maps between $(R,S)$-bimodules which are left-$R$-linear and righ-$S$-linear

Oh no it's a picture

@MatheinBoulomenos Oh wow I actually like that

So 1-morphisms between R and S are basically right exact functors between R-Mod and S-Mod, you could say, by Eilenberg-Watts?

right exact functors, but yeah

Thanks

The Eilenberg-Watts theorem in particular says something about category equilvalences between $R-\mathbf{Mod}$ and $S-\mathbf{Mod}$

This feels like Morita theory

these category equivalences are also called Morita equivalences and they appear in algerbaic K-theory

yeah, was about to say that

Exciting

I think you can drop the "isomorphism classes of ..." if you consider Bicategories instead of 2-Categories, but tbh I don't know that much about higher category theory

Okay so if $f$ is a chain map, it induces a map on homology, which I'm guessing is $H(f)_n(a_n + B(C_n)) = f(a_n) + B(C_n')$

Where $a_n \in Z(C_n)$

(Do continue with your stuff, I'm mostly writing out thoughts here)

That's right

@Mathein I guess you could say 2-morphisms are "natural" to choose there if you think of 1-morphisms as functors R-Mod --> S-Mod and say they are "representable" by (R, S)-bimodules and demand Yoneda's lemma

All the stuff about tensor products of bimodules is not that relevant or group cohomology, but it does appear when talking about induced modules

(Being representable usually means F = hom(-, A), but in this case F = - \otimes A. Tensor and hom are dual ideas, anyhow, so that should make sense in an appropriate way)

@Balarka I'm not that happy with "representable" because we're not talking about a Hom-Functor, but a tensor product

oh you sniped me again

lmao

This room is a war zone

"copresentable" maybe

I guess.

And if $f\simeq g$, then we want to say that $f(a) + B(C') = g(a) + B(C')$. Well, we know that $f(a)-g(a) = d'h(a)$

($+hd(a)$ but $d(a) = 0$)

ya

And $d'h(a)\in B(C')$ so yeah

vsauce theme music but what is, $f - g = dh + hd$?

I'll go to sleep now. Bye @Balarka @Daminark

See you @Mathein!

Bye @Mathein

It's interesting to notice how $h$ shifts degree by 1

reed ma answr dawg

B... but it has pictures! whimpers

Nah really though I looked over it briefly and I can kinda buy what you're doing

The point is

if you use the cubular chain complex

which is $C_*(X)$ where $C_n(X)$ is the free abgrp generated by maps $I^n \to X$ from $n$-cubes

then clearly a homotopy between $f, g : I^n \to X$ gives rise to a map $F : I^n \times I = I^{n+1} \to X$

that's your chain htpy $C_n(X) \to C_{n+1}(X)$

So that plus what I said above about chain-homotopic maps inducing the same maps on homology shows that ordinary homology satisfies that homotopy axiom

ye

Nice. Lol these sessions are where I actually figure out how to do stuff instead of how to listen to other people do stuff

I love stuff like these

I feel like I learn more

Well, continuing on

This next bit is actually pretty nice. While it may not explicitly assume you know AT I can kinda see why it's good to know

If only so the terminology is less random

Yep

May as well be careful about details since I'll be black boxing a lot of the ringsy business

So $D_n(f) = d'f - (-1)^nfd$

I need to read this carefully too

Probably not going to be able to do it right now, but later today

Lol aight, well I'll be rattling this off now, and yeah I guess you can check it out whenever you can

I'll catch up for sure

For sure, I'm gonna be going quite slowly

So if $f\in \text{Hom}(C,C')_n$

Then we want to show that $D_n(f) \in \text{Hom}(C,C')_{n-1}$

So we know $f:C_* \to C_{*+n}'$

Then $fd:C_* \to C_{*+n-1}$

As does $d'f$, so that makes sense

Now, we want to say that $D^2 = 0$

$D_{n-1}(D_n(f)) = d'(d'f - (-1)^nfd) - (-1)^{n-1}(d'f - (-1)^nfd)d$

So this gives $(-1)^{n+1}d'fd + (-1)^nd'fd = 0$

0-cycles are those $f$ such that $d'f - fd = 0$, precisely the chain maps. 0-boundaries are those $f$ such that $f = d'h - hd$, which is a homotopy from $f$ to $0$

Trailed a bit but yeah (not just a bit)

But okay so if you define $(\Sigma^n C)_p = C_{p-n}$ and $\Sigma^n d = (-1)^n d$, then $H_n(\text{Hom}(C,C')) = [\Sigma^n C, C']$

This because an n-cycle is some $f$ such that $0 = d'f - (-1)^nfd = d'f - f\Sigma^n d$, and an n-boundary is some $f$ such that $f = d'h - (-1)^nhd = d'h - h\Sigma^n d$, so it's nullhomotopic in $\Sigma^n C$

And we say degree n maps are homotopic in that sense

IM BACK

Hai!

What have you got

I see you're doing some witchery here

Kek, I mean this is just definition unpacking really

Wait this is black magic

What is this

How so?

i dont get it i dont get it i dont get it

Balarka breathe

$Hom_R(C, C')$ is a weird object

@Daminark breaths super fast

Dammit that wasn't the intention

I DONT GET HOM(C, C')

WHAT THE FFFF IS THIS

Lol I'm mostly contenting myself with a "the definition makes sense" thing for Hom(C,C'), and then trying to understand it using its properties

but what does it MEAN

Though really that's how I understand most things

its not hom in the category of chain complexes over R-Mod

wHaT is ThIs bLaCK MAGigck

Oh... Yeah it's defined differently

And the boundary map

OH MY GAH

Up until now I only watched the first 15 seconds

Where she was like "OH MY GAH"

But good lord the rest is worse

It

's a work of art

It really is

And it led to Pewdiepie's cowboy in the sky thing

Yup

I must say what cracks me up more than OH MY GAH is the entry of the yellow cat

HERRO EBURYNIAN

That spelling is actually spot on

I think Hom(C, C') is a manifestation of a double complex turned into a chain complex

Yeah that should be exactly it

Consider the lattice of points on the plane

with each point with $(i, j)$-coordinates corresponding to $Hom(C_i, C_j)$

and there are vertical and horizontal differentials going

$d_{ij} : Hom(C_i, C_j) \to Hom(C_i, C_{j - 1})$

$d'_{ij} : Hom(C_i, C_j) \to Hom(C_{i-1}, C_j)$

This gives you a double complex, yeah?

And somehow you still made it a picture... I'm impressed

And yeah that'd do it, under what I expect the def of a double complex to be

Not yet, let's see if it works out

Oh yeah double complex is exactly what you think it is

Formally it's an object of the category $\mathrm{Ch}_\bullet (\mathrm{Ch}_\bullet (R))$

aka a chain complex of chain complexes

So yeah there's this construction which turns a double complex over R into one single chain complex over R

ncatlab.org/nlab/show/total+complex

Ya boi $Hom(C, C')$ should be exactly the product total complex of this dude

Ah, I see

Lol you unironically linked ncatlab, this is an interesting day

hahah

Aight now that we got that figured out, the rest should make sense

And now back to the world of chain maps (phew)

So you define homotopy equivalence and weak equivalence in the "well duhhh" way

Now, a homotopy equivalence is a weak equivalence, never got around to proving that for spaces but it should be pretty easy

Right

$H_n(Hom(C, C')) = [C, C']_n$ is actually genuinely interesting

It means homotopy theory is homology theory for chain complexes

So $f'f\simeq id_C$ and $ff' \simeq id_{C'}$

I guess look at $ff'f$

Well, you need to observe that $H(f')$ and $H(f)$ are inverse to each other, no?

Oh yeah $H(ff') = H(id)$

But that's just $H(f)H(f')$

ya

So in particular, $H(f')$ can't have a kernel

Since $H(id) = id$ on the right domains

mhm

Oh wait I'm dumb

Yeah okay so $H(f')H(f) = id$ means $H(f')$ is surjective as well

That's right

And then do it in all directions

I have no idea why that took me so long to process

's okay my man

Ah proposition 0.3 is serious business

It actually proves it

I'm still hung up on Hom(C, C')

I don't get it

This is blax m0gic

I mean I think what you said above is about right, and when I'm having difficulty understanding something through its definition I've found it helpful to understand it based on its properties

I started taking that approach when I was trying to understand compactness and it just wasn't working

but the property $H_n(Hom(C, C')) = [\Sigma^n C, C']$ is bl000x magx bro\

But once I sorta proved all the theorems around it, that kinda helped me get an "intuitive" idea of what the thing really was. That kind of intuition sorta helped me in algebra type things as well

it's enconding homotopy type using homology type

it's like [X, K(G, n)] = H^n(X; G) and the Dold-Thom stuff

there's some fuckery going on deep down

This is magic, and that magic gives insight as to what the nature of Hom(C,C') is, as an avenue of encoding homology with homotopy, with its partner being suspension

but why should it?

In some sense, it was secretly built to do so. This feels like BS but I've found that this is the type of statement that retroactively shapes how I think of the object in question

I don't think your POV explains anything about this object at all

It's like accepting the object

as it is

There's some deep homological algebra at play here that I do not understand

I mean, the definition to me feels like it's essentially trying to formalize the notion of, I'm trying to find an object that does ___. I guess the fact that you even can link homotopy and homology doesn't strike me that strongly because I don't associate homology using triangulations much, if anything Dold-Thom is somehow what I've got in mind.

the total homology of $Hom(C_\bullet, C'_\bullet)$ spits out $[\Sigma^\bullet C, C']$. What in the world

But I think the magic is less likely to be encoded in the algebra and more in the topology

@Daminark Yeah I don't think your understanding of homology is at all correct when you motivate yourself to think in terms of [X, B^nG] or in terms of Dold-Thom

Like, the fact that homology defined this way links up nicely with the geometric interpretation of homology feels like a heavy fact

That these things hold strikes you as surprising only when you see both sides of the story

homology is not meant to encode homotopical info

Like, by construction

You were spoiled by J P May when he told you this surprise as a definition and robbed you of the amusement of seeing your intuition break down

It's a good experiment on his part, but I think it has a lot of side effects, like this

Well, what do you think is the conceptual link between E-M spaces and cohomology? Like, in a sense that's satisfactory to you, why ought they match up?

I have no fucking clue!

And lol yeah, if anything he replaced that intuition with a sort of formal one that makes this stuff feel alright

Hmm

Anything grasp on Dold-Thom?

Well, I guess I know why [X, K(G, 1)] = H^1(X; G) should hold

Like I really do think that the formal bit, which is really how I'm processing all this, works itself out in some nice way. Like, when I was writing that stuff out, in my mind was "This is true because, well of course it is"

What do you mean? Your definition of cohomology is [X, K(G, n)].

If it's true to you, it's tautologically so

To the rest of us, it comes as a surprise

No I'm talking about the above stuff on H(C,C')

Oh

What I'm saying is that nothing within the formal workings, if you just process it within that framework, is particularly impressive, it's borderline contentless in that world. The fact that it spits out a statement that's surprising topologically must therefore be a result of how the topology allows you to define formal stuff in this way

That's the vibe I'm getting from here, and it sorta explains why I'm just like "Ya this is chill"

shurg

Fuck

shrug

Well, that's fair, I guess. Tbh it seems to me that it's a super dry approach to learning mathematics but hey, different people have different ideas I guess

Eg that's why I asked Mathein why I should care for (R, S)-bimodules earlier today

and he gave an answer which was, well, pretty fantastic

even if it was algebraic

Things must have meanings, regardless of algebraicity or geometricity

Why even define this meaningless object called Hom(C, C')? People didn't know the algebra would pan out the way it does before doing the computation.

I suggested the double complex, which is slightly more natural. Still unclear why it should have anything to do with [C, C']

Well, I actually looked at Hom(C,C') as, alright well chain maps and their homotopy is degree 0, can we "parametrize" stuff by degree? Ah, we can, let's see what comes out of it?

But for what it's worth, until I read your thing above, the definition of homotopy to me felt entirely unmotivated

I was just like, well alright this feels like you've found a clever invariant but sure let's roll with it

Hmm

But once you did the prism business I was like aight, so this is why that stuff is homotopy.

I see

That's why I said I think the surprise lives there

Like I don't fully get the connection because my understanding of triangles is limited

(That statement sounds horrible now that I've said it)

lmao

No I mean it's fine

I just think my understanding is very much meaning-associated

I can feel Ted's smack across chats

I won't tell on you

Ok, I shall get back to this later. Need to study physics today

Keep on reading, meanwhile

Lol, I imagine you'll like this more than chemistry?

topological fuckery >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> homological algebra >>>>>>>>>>> physics > chemistry

Though I will say organic chem is what helped Araske understand symmetry groups

Ah yes

all the optical isomers

n' shit

He immediately looked at the tetrahedron and the notion of parity in that context convinced him that its symmetry group is A_4

ah i see

Which it is. I only ever became convinced of this when we did orbit-stabilizer to get that it has size 12, then pushed it into S_4

At which point it had to be A_4 because that's the only subgroup of index 2

Otherwise, parity and tetrahedra would seem to be disjoint to me

So idk chem will help for geometry (tbf, physics will do that much more)

i think it's in general the other way around :3

we help them

not they help us

fuck them

Kek, true

crystallographic groups do relate to organic chemistry and solid state, though, yep

Hmm

BALEETED

But do you agree?

Absitively

;novathonk

@BalarkaSen Thank you

What was beleeted :O.

VERY SEcReT Info

:O