serious-discussion

[sharp mulch]

Is it matrix multiplication

[surreal sapphire]

it can be formulated as such at least partly

i think this is good, right?

[sharp mulch]

Matrix multiplication is a very well studied problem

Especially if you know things about your matrix

Like is it sparse

Or symmetric/hermitian

Etc..

[surreal sapphire]

ok, this will give me something to look into

i will read up on MPI, thanks 👍

[sharp mulch]

Also, things parallelize differently if you have shared or separate memory

Communication between separate memory pools takes a really long time

And things like matrix multiplication fall into a paradigm called "embarrassingly parallel" where there are no intermediate dependencies so if you do it in a shared memory (openmp) framework, you can essentially get linear speedup

[surreal sapphire]

i see

[sharp mulch]

Openmp is also definitely easier to do because it's almost entirely done via compiler directives

#pragma omp .....

You may also look into kokkos, which is a performance portability framework

The idea behind kokkos is that you rewrite your code in kokkos and then it handles all the parallelization

[bright hill]

CHESS STREAM

COME SEE SHURI GETTING ANNIHILATED IN OBSIDIAN

[violet wraith]

your chess skills are equivalent to carbonated fecal matter

[sharp mulch]

[neat frost]

@long matrix omg death!!

[long matrix]

he lost btw

[neat frost]

Only took like 15 episodes

[long matrix]

come to stream

[neat frost]

Nono I binge watch

[long matrix]

ok whatever, dont spoiler me

[neat frost]

Also I can’t unmute now, so that takes out all the fun

[long matrix]

i havent finished it.

[neat frost]

Have you finished the first season?

[long matrix]

yes, i did.

[neat frost]

So you’re still quite a bit ahead of me

So I will look up spoilers instead

For you <3

[fair mural]

what show

[neat frost]

“Quantum’s mom’s crazy adventures”

[long matrix]

psychopass

[bronze pelican]

quantum's mom

[fair mural]

[bronze pelican]

hi quantim

[fair mural]

hi

[rocky shuttle]

What are you binge watching?

Oh, psycho pass

[violet wraith]

I'm binge watching quantum's mom

[storm sage]

peter and the wolf is so fun

[violet wraith]

who the fuck are you

[bright hill]

Lmfaooooo

Ereh has been more active than you

[split island]

[white vale]

Hi, so the topic of my slogan is how to defend human rights
is my text for it good?

• Each voice is a shield for everyone's futures, a sword against everyone's foes -

thanks

if I can ask for a bit of help

[sly anvil]

It seems a bit wordy IMO

probably simplify it

[neat frost]

• Human rights, pretty cool imo -

[long matrix]

cat rights where

[fast moat]

https://upload.wikimedia.org/wikipedia/commons/thumb/7/7f/Visual_proof_mutually_touching_solids.svg/1024px-Visual_proof_mutually_touching_solids.svg.png

the coloring of a 3d map needs infinite colors

[bright hill]

Good morning slurp!

[neat frost]

Good morning DarQ!

[bright hill]

[neat lintel]

Imagine a new color

[fast moat]

wdym

[neat lintel]

Try to imagine a new color

[fast moat]

and?

[neat lintel]

Add see if you can find that color in that 3d map

[fast moat]

and?

[neat lintel]

Can you?

[fast moat]

why does this matter

[severe oasis]

@fast moat bc it does

[rocky shuttle]

@alpine kindle @bright hill I'm free now, tell me when you're ready for some calculus on manifolds

[alpine kindle]

dinner is being made

so

hm

about an hour?

[rocky shuttle]

Sure

[bright hill]

Sadly, i can't chime in today

[alpine kindle]

cool

[bright hill]

I promised neamesis we'd do rudin together

[rocky shuttle]

Np, DarQ, just finish Spivak

[rocky shuttle]

Oh, that's also fun

[bright hill]

I will

I'm doing algebra now tho lmao

Thanks for pinging me tho

[rocky shuttle]

[meager glade]

any online resources to prepare for accuplacer math college freshman

[zealous garden]

Calculus on manifolds sounds so fun

[silver meteor]

nice pfp

[severe oasis]

@true bough hi Deepy

[rocky shuttle]

still up for it?

[alpine kindle]

2 secs need to do laundry

[eager reef]

i think i'll join in a bit as well if that's alright for you

[rocky shuttle]

Sure

@bright hill was Rudin cancelled?

[eager reef]

yeah he had some questionable past tweets

[bright hill]

Apparently

[rocky shuttle]

[naive chasm]

RUDIN IS CRINGE

[bright hill]

No u

[alpine kindle]

@rocky shuttle are you still awake

[rocky shuttle]

jk, yeah I am

[alpine kindle]

it's diff form time

[rocky shuttle]

longest two seconds of my life

alright!

[alpine kindle]

so

recap

what's a differential form again

[bright hill]

Lmfao

[rocky shuttle]

A differential $k$-form is an alternating multilinear form $$\omega\colon\mathrm{T}_p{M}\times\dots\times\mathrm{T}_p{M}\to\mathbb{R}$$ with $k$ copies of the tangent space $\mathrm{T}_p{M}$ of a differentiable manifold $M$ at $p\in M$.

[alpine kindle]

texit is down

[rocky shuttle]

NO

NOOOOOOOOOOOO

[alpine kindle]

uh

[rocky shuttle]

https://mathb.in I guess

[alpine kindle]

yeah

[rocky shuttle]

So, how familiar are you with tangent spaces and classical vector analysis in general?

[alpine kindle]

not very familiar

[zealous garden]

Really?

[alpine kindle]

why are you doubting that

[zealous garden]

Your aka is the Exterior Algebra

[rocky shuttle]

btw why is the quadratic form missing from your aka

[zealous garden]

Because it's not a (general) Clifford Algebra

[rocky shuttle]

oh okay

[alpine kindle]

clifford algebras triggered chmonkey's ptsd

so i removed

[zealous garden]

Lmfao

[rocky shuttle]

anyways, perhaps we should restrict ourselves to integration of differential forms on submanifolds of R^n first, if you don't mind

[alpine kindle]

yeah

i mean that doesn't seem too specific

[rocky shuttle]

Okay, so is the alternating multilinear form thing clear to you?

[alpine kindle]

yeah i know what those are

like where f(x0,x1,...,xn,xn+1,...) = -f(x0,x1,...,xn+1,xn,...)

[rocky shuttle]

Yeah, and if you put two linearly dependent vectors, then it's zero

Alright, so let's assume that the k-dimensional manifold $M\subset\mathbb{R}^n$ can be parameterized by a singular k-cube $\varphi\colon[0,1]^k\to\mathbb{R}^n$.

[alpine kindle]

zan mathbin

[rocky shuttle]

How should I do it? Send the link everytime I tex something?

[alpine kindle]

ig
only for bits i don't get

i get this bit

[rocky shuttle]

You can also just copy my texts into mathbin

[alpine kindle]

yh

[rocky shuttle]

That's faster

[alpine kindle]

yep

ok

[rocky shuttle]

So, for such a manifold $M$ (with the parameterization $\varphi$) the integral of a $k$-form field $x\mapsto\omega(x)$ defined on a domain in $\mathbb{R}^n$ containing $M$ is defined by $$\int_{M}\omega=\int_{[0,1]^k}\omega(\varphi(\xi^1,\dots,\xi^k))[\frac{\partial}{\partial{x^1}},\dots,\frac{\partial}{\partial{x^k}}]\mathrm{d}{\xi^1}\dots\mathrm{d}{\xi^k}.$$

[alpine kindle]

issue

oh

[rocky shuttle]

pdv thing

and dd

[alpine kindle]

dd still not working

[rocky shuttle]

Why do you have to be down TeXiT 😔

How can you put plain text in discord?

[alpine kindle]

gotta be honest i have no idea what's going on in that integral

[rocky shuttle]

Yeah, we're going to discuss this now. So, I wanted to start with the objective

[alpine kindle]

it seems to be basis dependent

[rocky shuttle]

Yeah, but it turns out to be not basis dependent!

[zealous garden]

Man I wish texit was working

[alpine kindle]

(also isn't x |→ ω(x) just ω)

[rocky shuttle]

It's a differential form field, which assigns each point on the ambient manifold a differential form $\omega(p)\colon\mathrm{T}_p{M}\times\dots\times\mathrm{T}_p{M}\to\mathbb{R}$.

Usually, the term field is omitted though

[alpine kindle]

oh i see

wait shouldn't that be omega(p)

[rocky shuttle]

Hmm, yeah, I need to be consistent with the notation

[zealous garden]

I thought differential form was the term for the field, and an individual was simply a form

[rocky shuttle]

Yeah, that's common in the literature,

I like to distinguish between differential forms (alternating multilinear forms, where the vector spaces are tangent spaces of some manifold) and differential form fields though

So, to understand the above formula, maybe we can look at line / path integrals in R^n first

[alpine kindle]

so omega(x) is a linear map from \bigwedge^k T_x M to F

[rocky shuttle]

Yeah, multilinear and alternating

[alpine kindle]

?

with bigwedge^k those turn into just linear i think

[zealous garden]

yes

[alpine kindle]

cool

[zealous garden]

I think that's the point

[alpine kindle]

what's the sinplest nontrivial case

that usually helps

[rocky shuttle]

Yeah, so, for path integrals, our manifold is a curve (1-dimensional) in 3-dimensional space. Given a coordinate chart $x$, The space of differential 1-forms $\Omega^1(\mathbb{R}^3)$ is spanned by $\binom{3}{1}=3$ basis vectors: $dx^1$, $dx^2$, $dx^3$.

So, we can write a 1-form $\alpha$ as $\alpha=\alpha_1 dx^1+\alpha_2 dx^2+\alpha dx^3$.

[alpine kindle]

so dx^n is π_n basically?

[rocky shuttle]

Basically yes

[zealous garden]

what's pi_n?

[alpine kindle]

nth projection morphism

[zealous garden]

[rocky shuttle]

More formally, the tangent space is spanned by the coordinate frame $\frac{\partial}{\partial{x^1}},\dots,\frac{\partial}{\partial{x^n}}$ and $dx^1,\dots,dx^n$ are the dual basis vectors to them

[zealous garden]

nice and simple

[alpine kindle]

coordinate frame?

is that just a basis

[rocky shuttle]

Given an $n$-dimensional manifold $M$ and a chart $x\colon M\to\mathbb{R}^n$, the $i$-th coordinate basis vector is given by the tangent vector $\frac{\partial}{\partial{x^i}}\in\mathrm{T}p{M}$ corresponding to the derivation $$\frac{\partial}{\partial{x^i}}f(p)=\lim{t\to 0}\frac{f(x^{-1}(x(p)+te_i))-f(p)}{t}.$$

A particular basis formed by a coordinate chart

[muted wedge]

zanarkand sory to interrupt but what the f

[rocky shuttle]

what?

[muted wedge]

what are we talking about

given that

i dont speak latex

[zealous garden]

calculus on manifolds, read what's been said before now

[muted wedge]

a

wh

ok

[zealous garden]

start here

[alpine kindle]

wait hold on isn't x the identity function

[rocky shuttle]

in R^n, yes

[alpine kindle]

what

[rocky shuttle]

Are you familiar with the definition of a manifold?

[neat frost]

Ugh math

[alpine kindle]

i thought that was just a notational shortcut

[bright hill]

I wish I could partake

[alpine kindle]

yes

[neat frost]

Wanna do algorithms with me?

[bright hill]

I have to sleep now tho

[neat frost]

what???

[zealous garden]

gn (dar)^3

[rocky shuttle]

Every point has a neighborhood that is homeomorphic to some open subset of R^n

[alpine kindle]

yeah

[rocky shuttle]

A chart x at p is such a homeomorphism

[bright hill]

I like this one lmfaoo

[rocky shuttle]

gn kyuu

[alpine kindle]

ok sorry it just wasn't clear what x was

nvm

I'm blind i didn't see you defined it earlier

[rocky shuttle]

Okay, let's look at a simple example

[alpine kindle]

yes

[rocky shuttle]

Let $M$ be a curve in $\mathbb{R}^3$ parameterized by $\varphi\colon[a,b]\to\mathbb{R}^n$.

[alpine kindle]

oh wait so is \pdv{}{x^i} a vector field

[rocky shuttle]

Yeah

[alpine kindle]

cool

[rocky shuttle]

And $\alpha=\alpha_1 dx^1+\alpha_2 dx^2+\alpha_3 dx^3$ be the 1-form we want to integrate over $M$.

(I need to find out how to disable formatting in discord)

[alpine kindle]

can you not use the hidden backslashes pls

it's not loading in mathbin

[rocky shuttle]

Does it work now? (still with the hidden backslashes)

[alpine kindle]

not really

[rocky shuttle]

Hmm okay

[alpine kindle]

has a bunch of underscores

[rocky shuttle]

I removed them

So, the integral would be

$$\int_{M}\alpha=\int_{a}^{b}\left\lbrack\alpha_1(\varphi(t))dx^1(\frac{\partial{\varphi}}{\partial{t}})+\alpha_2(\varphi(t))dx^2(\frac{\partial{\varphi}}{\partial{t}})+\alpha_3(\varphi(t))dx^3(\frac{\partial{\varphi}}{\partial{t}})\right\rbrack dt.$$

[alpine kindle]

ooh that is

spicy

[rocky shuttle]

Or short

[alpine kindle]

how do you get that

[rocky shuttle]

$$\int_{M}\alpha=\int_{a}^{b}F(\varphi(t))\cdot\dot{\varphi}(t)dt$$

[alpine kindle]

that's worse

[rocky shuttle]

where $F=(\alpha_1,\alpha_2,\alpha_3)$ is a vector field with the components of alpha

[zealous garden]

the notation and terminology is different, but I recognize everything

[alpine kindle]

the second one is much harder to get

[zealous garden]

I think it stems from the fact that the dx^i are covector fields

[rocky shuttle]

Hmm

[zealous garden]

and in finite dimensions a covector is isomorphic to a dot product with a fixed vector

[rocky shuttle]

Yeah, so this is the usual definition of a path integral in classical vector analysis that can be defined without knowing about differential forms

[alpine kindle]

ok

can we go back to the nicer long version

[rocky shuttle]

lmao

[zealous garden]

nicer?

[alpine kindle]

yes

nicer

[zealous garden]

You're quite the character

[alpine kindle]

it makes more sense

[rocky shuttle]

okay, let's look at the integrand

[alpine kindle]

of what representation

[rocky shuttle]

The whole idea is to flatten / pullback the curve M to the interval [a,b]

[alpine kindle]

hm yes

I've heard of pullbacks before

no idea what they are

[zealous garden]

like parameterizations (?)

[alpine kindle]

sth related to categorical limits iirc

[zealous garden]

I've seen the definition

[rocky shuttle]

I don't know the relation to pullbacks in category theory

I'm not used to explaining things by text only

[alpine kindle]

sorry

[zealous garden]

f(y(x)) = g(x), g is a pullback of f

that as an example

of the categorical notion

[alpine kindle]

hm

[rocky shuttle]

So, our differential form $\alpha$ is defined on the curve $M$

[alpine kindle]

yes

[rocky shuttle]

The pullback $\varphi^*\alpha$ is defined on the interval $[a,b]$

[eager reef]

that relation isn't relevant for your cause rn anyway

[rocky shuttle]

Okay

[alpine kindle]

how is phi*alpha defined

[rocky shuttle]

I'm getting to it

[alpine kindle]

ok

cool

[rocky shuttle]

So $\alpha$ at the point $p\in M$ takes a tangent vector $v\in\mathrm{T}_p{M}$ and maps it to a real number $\alpha(p)[v]$.

[alpine kindle]

yes

[rocky shuttle]

The pullback $\varphi^\alpha$ at $t\in[a,b]$ takes a tangent vector $u\in\mathrm{T}_t{[a,b]}\simeq\mathbb{R}$ and maps it to $$(\varphi^\alpha)(t)[u]=\alpha(\varphi(t))[D\varphi(t)[u]].$$

[vivid halo]

texbot my beloved

[rocky shuttle]

$D\varphi(t)[u]$ is the differential / pushforward of $\varphi$ at $t$ in the direction of $u$.

[alpine kindle]

ok yes I'm imagining this equation and it makes sense

[rocky shuttle]

So, essentially, the integral of $\alpha$ over $M$ is given by the integral of its pullback over the interval

$$\int_{M}\alpha=\int_{[a,b]}\varphi^*\alpha$$

[alpine kindle]

so

ok

phi is from R to R³

[rocky shuttle]

And yeah, the very first integral I showed you was just the pullback

[rocky shuttle]

for a curve, yes

[alpine kindle]

yeah

let's say phi is from R^k to R^n

[rocky shuttle]

In general, we have singular k-cubes, that is, $\varphi\colon[0,1]^k\to\mathbb{R}^n$.

[alpine kindle]

yeah

that

[rocky shuttle]

Then $$\int{M}\omega=\int{[0,1]^k}\varphi^*\omega$$

[alpine kindle]

wait what if our curve is unbounded

[rocky shuttle]

For a k-form omega

[alpine kindle]

would you just go over all R^k instead of [0,1]^k

[rocky shuttle]

So far, we have been talking about manifolds that can be globally parameterized by a singular k-cube

In general, it's not possible

[alpine kindle]

Dphi is a function [0,1]^k to the set of linear operators on R^n right

[rocky shuttle]

To the set of linear operators from R^k to R^n

[alpine kindle]

yes ok

ok yeah that makes sense

takes it to the corresponding tangent vector at phi(t)

[rocky shuttle]

Yeah

[alpine kindle]

so how do we deal with boundaries

or is that already done

[zealous garden]

don't the boundaries deal with themselves

[rocky shuttle]

wdym?

[alpine kindle]

like how do you take the derivative of phi at the boundary

[zealous garden]

I mean between two adjacent k-cells in the manifold, along their shared boundary, between orientation and measure the boundary values vanish

[alpine kindle]

of the k-cube

[zealous garden]

is this the part where you cheat because bla bla smooth

[rocky shuttle]

For the integral over the k-cube, the values on the boundary do not matter since the boundary is of measure zero

[zealous garden]

see that guys, measure zero

I was right

this is just a more general claim than I made

[rocky shuttle]

really?

[alpine kindle]

ok so

[rocky shuttle]

Anyways, my exposition was not very structured, but is integration of differential k-forms over singular k-cubes clear now?

[alpine kindle]

what does singular mean

[rocky shuttle]

I think singular here comes from https://en.wikipedia.org/wiki/Singular_homology

[long matrix]

u could give us a talk on differential forms

[rocky shuttle]

galwah first

[long matrix]

i have a long week ahead of me, but maybe the weekend

[rocky shuttle]

I mean I could've also done a vc

And do some doodles, but I forgot my pen tablet in the office

[alpine kindle]

i couldn't have vced idt

[rocky shuttle]

Np

[alpine kindle]

so zan

where does the exterior derivative come in

[zealous garden]

[rocky shuttle]

It's hard to structure all of this...

[alpine kindle]

the previous bit was easy to follow

[rocky shuttle]

Alright, maybe we can continue with integration over chains first?

[alpine kindle]

[rocky shuttle]

So, a chain is just a formal sum of singular k-cubes

[alpine kindle]

ok

[rocky shuttle]

(free Abelian group generated by the singular k-cubes)

[alpine kindle]

oh

that was unexpected

[vivid halo]

this is an awfully large group

[alpine kindle]

how do you define this group

[vivid halo]

when comparing this to differential forms it's fine to use R coefficients, so if you like you can instead think about the R-vector space with basis given by singular k-cubes

[alpine kindle]

?

[zealous garden]

what the fuck

[alpine kindle]

what

[vivid halo]

I mean it's infinite dimensional but that's fine

[eager reef]

lmao nG

[vivid halo]

elements are finite linear combinations of singular k-cubes

[rocky shuttle]

yeah... but aren't the coefficients in Z?

afaik

[vivid halo]

sure but when you compare singular cohomology to de Rham cohomology it's with coefficients in R

[rocky shuttle]

oh okay, I didn't know that

[alpine kindle]

can someone explain what a formal sum is

[rocky shuttle]

[vivid halo]

oh it's just like

a linear combination

[alpine kindle]

how are you even adding k-cubes

to make a group

[vivid halo]

you're just adding them formally

[rocky shuttle]

I mean, a polynomial is also just a formal sum

[vivid halo]

yeah you're just like

treating the k-cubes as variables if you like

[zealous garden]

this means that you add them

and leave it at that

the sum is the sum

[alpine kindle]

yeah but a polynomial isn't a group is it

[zealous garden]

right?

[vivid halo]

polynomials form an Abelian group! you can add them

[alpine kindle]

yeah but a polynomial on its own isn't a group

[eager reef]

a polynomial is an example of a formal linear combination here

[zealous garden]

it spans one doesn't it

[alpine kindle]

oh wait

was zan saying that chains form an abelian group

[vivid halo]

there's nothing deep going on here

[rocky shuttle]

yeah

[alpine kindle]

oh

i misinterpreted

[zealous garden]

you can count each k-cube

[vivid halo]

if you have a set S you can form the free Abelian group Z[S] whose elements are formal Z-linear combinations of elements in S

so expressions like a_1s_1+...+a_ns_n

you don't need any addition on S for this to make sense

[alpine kindle]

i thought this was saying that a formal sum of k-cubes is an abelian group in and of itself

[rocky shuttle]

no, it's an element of the free Abelian group

[alpine kindle]

yeah i get it now

[rocky shuttle]

Anyways, the integral over a chain is then obviously defined as the sum of the integral over each singular k-cube with multiplicities

[alpine kindle]

yes

wait

you can't use the same diff form field over all of them

[vivid halo]

you can

if you have a single k-form \omega and you know how to integrate it over a single k-cube, then you know how to integrate it over a linear combination of k-cubes

[alpine kindle]

how do you integrate a 1-form field over a surface

[rocky shuttle]

the diff form fields are defined in the ambient space

[alpine kindle]

for example

[vivid halo]

you don't

[alpine kindle]

oh is this derivatives

[vivid halo]

you integrate k-forms along k-dimensional spaces

[zealous garden]

wait that's a hard limit?

[vivid halo]

it doesn't really make sense to integrate if the dimensions don't match up

[alpine kindle]

oh

so

[zealous garden]

why does it not?

[alpine kindle]

you can't integrate over arbitrary chains?

[zealous garden]

what about integrating surface flux?

[vivid halo]

what about it

I mean maybe what you're referring to is how like in R^3 there's some way you can conflate 1-forms and 2-forms

[zealous garden]

no

that solves what I'm talking about in R^3

what I'm talking about is before the conflation

[alpine kindle]

nG

[zealous garden]

integrating a 1-form over a surface

a very basic problem (for vectors)

[vivid halo]

[alpine kindle]

what i thought was implied earlier is that you can integrate over any chain

[vivid halo]

I don't think this is strictly true

what is true is that you can integrate any k-form over k-chains

[zealous garden]

so this is a sharp distinction between GA and Forms then

[alpine kindle]

you can't integrate over [0,1]^k + [0,1]^(k+1) tho

[vivid halo]

yes this is one reason why GA is bad

[vivid halo]

right

this doesn't make sense for forms

[zealous garden]

How is this a negative?

[rocky shuttle]

integrate what?

[alpine kindle]

idk

[vivid halo]

like you can't integrate a k-form or a (k+1)-form on these domains

[alpine kindle]

nvm

[vivid halo]

anyways the cleanest definitions come from integrating k-forms over k-dimensional regions

[alpine kindle]

so what is the basis for the set of k-chains

[rocky shuttle]

all the singular k-cubes

[alpine kindle]

all of them how

[vivid halo]

literally just all of them

this is a very infinite dimensional space

[alpine kindle]

is it just like

anything homeomorphic to a k-cube

[vivid halo]

it's quite literally like

for every continuous map \sigma:[0,1]^k->X

you get a new basis element

[alpine kindle]

oh ok

all of them in that sense

[vivid halo]

the main point is this just gets you talk about expressions like a_1\sigma_1+...+a_n\sigma_n

call this space C_k(X), this huge vector space spanned by k-chains

the boundary of a k-cube is a formal linear combination of (k-1)-cubes, so you get a linear map C_k(X)->C_k-1(X)

since we're in a very infinite dimensional setting it's like

kinda hopeless to actually compute very concretely with this

[wintry heron]

thats actually my favourite saying

[vivid halo]

this is sort of the problem with introducing singular homology first, it's not very computable but it has nice formal properties

[alpine kindle]

so this is where cohomology comes in

[rocky shuttle]

and exterior derivatives

[vivid halo]

well so you can define the singular homology H_k(X,R) in terms of these C_k(X)'s and this boundary map between them

[alpine kindle]

?

what is a homology

[vivid halo]

the definition is like

[alpine kindle]

like

[vivid halo]

you have these two boundary maps d_k:C_k(X)->C_k-1(X) and d_k+1:C_k+1(X)->C_k(X)

the homology is the kernel of one modulo the image of the other

H_k(X,R)=ker(d_k)/im(d_k+1)

the immediate problem is that both ker and im are infinite dimensional so this isn't really computable

[alpine kindle]

if the sequence is exact you get the trivial group

[vivid halo]

but it does work as a definition and you can show this has nice properties

[alpine kindle]

?

[vivid halo]

right yeah

[alpine kindle]

idk when that would happen

[vivid halo]

so it's measuring the failure of the sequence C_k+1(X)->C_k(X)->C_k-1(X) is exact at C_k

[blazing pawn]

when things are boring and sad

[vivid halo]

well so what would it mean if it's exact there it's like

somehow this is telling you about k-dimensional holes in your space

[alpine kindle]

hm

is part of this like
unsolved

[vivid halo]

nah this stuff is all very well understood

[neat lintel]

What is the (topological) Chiral Homology?

[vivid halo]

the way things are defined with singular homology makes things sound impossible to compute, but for instance if you fix a triangulation of your space there is another definition that only involves finite dimensional vector spaces and gives you the same answer

[stable patrol]

“Muh topological arms are so continuital!” Smh do probability problems instead

[bronze wedge]

you kind of actually need topology to do probability

[vivid halo]

the tradeoff is like, okay great now you have a theory that involves finite dimensional objects but now it's not so clear how this depends on the choice of triangulation, it doesn't work for topological manifolds that don't admit triangulation, maybe you're working with a space you don't understand well enough to fix a good triangulation, etc

[stable patrol]

So you say I’ve been doing topology all this evening?

[alpine kindle]

hm

[vivid halo]

the same kind of tradeoff happens for de Rham cohomology also: the spaces of differential forms there are also infinite dimensional since they involve spaces of smooth functions

[alpine kindle]

so how does our boundary homology relate to the exterior derivative

[stable patrol]

So math not only has smooth brains, but smooth functions?

[vivid halo]

the relation is like

[delicate mulch]

please stop disrupting the conversation

[vivid halo]

we defined homology H_k(X,R), cohomology H^k(X,R) is dual to this

[alpine kindle]

ah

[vivid halo]

you play the same game in reverse, you have cochain groups C^k(X)=C_k(X)* (dual space)

[blazing pawn]

counting down the seconds until this somehow becomes nG explaining sheaf cohomology

[vivid halo]

and coboundary maps going the other way

[alpine kindle]

so

[stable patrol]

I’m too interested in smooth functions

[alpine kindle]

what's a cochain

[vivid halo]

a cochain here is a functional on chains

[alpine kindle]

hm

[vivid halo]

so an element of the dual space C_k(X)*

that is a linear function C_k(X)->R

the main example of such a linear function is integrating a k-form \omega

you get a linear function sending k-chains to the integral of \omega over that k-chain

[alpine kindle]

interesting

are there other elements of C_k(X)*

[vivid halo]

sure, but we only care about what happens on cohomology in the end

the claim is that we've produced a map H^k_dR(X)->H^k(X,R) that sends the class of a k-form \omega to this linear function defined by integration, and this map is an isomorphism

[alpine kindle]

hold on

you've just added like 2 bits of terminology there

[vivid halo]

ahh oops

H^k(X,R) you do the same thing as with homology

kernel modulo image

[alpine kindle]

yes

i get that

[vivid halo]

for the boundary maps on cochains

H^k_dR(X) you also do the same kernel modulo image thing, now with the exterior derivative on differential forms

[alpine kindle]

so it's a subgroup

?

because of Stokes' theorem

[alpine kindle]

or something to that effect

[vivid halo]

a priori these two groups H^k(X,R) and H^k_dR(X) have nothing to do with each other

we've defined a map from one to the other using integration

[alpine kindle]

oh i see

[vivid halo]

H^k_dR(X) is the kernel of d:\Omega^k->\Omega^k+1, modulo the image of d:\Omega^k-1->\Omega^k, so elements are equivalence classes of k-forms

[alpine kindle]

ohh

[vivid halo]

so these live in very different spaces

so it's sort of a miracle that you end up with the same cohomology theory in the end

[alpine kindle]

what does this equivalence relation look like intuitively

or is it not intuitive

[vivid halo]

no no it's intuitive it's like

two k-forms \omega_1 and \omega_2 are equivalent if they differ by an "exact k-1 form", that is there exists a k-1 form \alpha such that \omega_1=\omega_2+d\alpha

[alpine kindle]

hm

[vivid halo]

I mean this is what it means to take the quotient by im(d)

[alpine kindle]

yeah

[vivid halo]

it might not be easy to verify this in practice!

like how do you actually find this \alpha, idk

[rocky shuttle]

H^0_dR(X) are precisely the locally constant functions on the manifold

[alpine kindle]

how does one visualise the exterior derivative

above df = Df

for f a function

[rocky shuttle]

and the homology group corresponds to the connected components of the manifold

[vivid halo]

it's kinda hard to visualize

as you've noted for functions it's essentially the derivative

[alpine kindle]

yeah

[vivid halo]

this is sort of dodging the question but idk that I usually go about trying to visualize this in general

[alpine kindle]

ok

fair enough

I'll think of it in terms of the diagram

[vivid halo]

the other answer is like, you can say what d\omega is if you can say how to integrate it

and then yeah the relation between the exterior derivative and the boundary map is exactly Stokes's theorem

[alpine kindle]

wait what is the coboundary

[vivid halo]

for singular cohomology we have the coboundary maps C^k(X)->C^k+1(X)

that are dual to the boundary maps on C_k

[rocky shuttle]

I'm going to sleep, thanks all for the discussion!

[zealous garden]

so what's so bad about being able to integrate 2 dimensional quantities on 3 dimensional manifolds?

[vivid halo]

I don't get how you're defining it lol

[zealous garden]

simple, the same way we defined it with reals way back in the beginning

[vivid halo]

[zealous garden]

you have a value for the function over a sufficiently small area

and you multiply it by the sufficiently small area

[vivid halo]

okay

idk I've never run into a situation where I was like "dang if only differential forms could do that"

like can you give me an actual example

where you would use this

[zealous garden]

that's not what I asked, you said it was a bad thing

why

[vivid halo]

[zealous garden]

I mean I can come up with an example where k and m don't match, but not with 2 and 3 like I said

[vivid halo]

I mean one thing that's bad about this is that it's very clearly not compatible with cohomology

unless you want a bunch of things to be 0

yet another L for GA!

[zealous garden]

No

L for me

I can't respond on a topic I don't know

[mortal igloo]

Is the GA joke about how elements of a GA can be non-homogeneous and that has no physical meaning?

[zealous garden]

in classical mechanics the faraday field is non-homogeneous

[vivid halo]

[neat lintel]

https://twitter.com/saturnalreturn/status/1596303742328242176?s=46&t=kJ_aYJQrY4miI8YeNG-1zw

a small leaf (@saturnalreturn)

bro are you doing ok? all the sonnets youve been submitting in sonnetchat have a tone of sombre resignation

Likes

10750

Retweets

1295

Twitter · Nov 26, 2022, 12:44 AM

[zealous garden]

idk ng, it seems to play out the same way in cohomology

[quick pasture]

Anyone here good at chem

[sharp mulch]

There is a chemistry server linked in #old-network

[zealous garden]

worst case scenario we just agree to be as weak as differential forms and then we can do the cohomology stuff

[vivid halo]

again I don't think I've ever run into a situation where this is necessary

I'd love to hear of a situation where this actually matters

[zealous garden]

so that's why you guys only integrate k-forms on k-dimensional spaces

to do anything else requires a metric

[vivid halo]

right, and GA is sort of a sneaky way to try to avoid this

[zealous garden]

What? Metric is in the name

I mean, I see what you're saying, why someone could be misled

[mortal igloo]

You need a quadratic form to construct a Clifford algebra lol

$\operatorname{Cl}(V)\coloneqq T(V)/(v\otimes v-q(v))$

[jovial ember]

I hate Clifford algebras I hate Clifford algebras I hate Clifford algebras

[fathom swallow]

Icy001

[jovial ember]

My Lie groups class did spin geometry

And these dudes were like the bane of my existence

Classify all of them over R

[zealous garden]

yes

[mortal igloo]

Something you can learn from that class is to say that the spin group is the double cover of the orthogonal group over and over

[jovial ember]

Yeah

[fleet oyster]

^ is this a bot shilling NFTs

[leaden torrent]

yes

banned

ty for the report

as an fyi: do not invest in anything, crypto-related or otherwise, that you found on discord.

and i mean absolutely anything.

even if it isnt a scam (which is very very unlikely), have you seen the intelligence of the average discord user? you really expect these people to have good financial advice?

[brittle socket]

[vestal solstice]

crypto?? isnt that the ice type pokemon

(please do not ban me i am not an nft bot)

[neat lintel]

Ima nft bot

[fair mural]

cope

[autumn hornet]

If i drop a Apple the Apple falls down

[zealous garden]

no

[ember fog]

I just wanna vent that uni calculus is obliterating me. I already started my course with little confidence in my math skills because my foundations suck and I get anxious around math. Now my class is pulverising the little foundations I did have

I got by in high school but I wasn't a stellar or olympiad student

I'm trying to assess my situation and from what I can tell I'm ill equipped with math study skills itself

I can see how the problems are different in high school vs now as well. It's much more rigorous now and less "mechanical"

[woven whale]

olympiad math has nothing to do with uni math

[ember fog]

People are using them as a metric for our exam difficulty "even the Olympiad kids found it hard"

[woven whale]

lmao that's stupid

[cinder zephyr]

Honestly the issue with most calc courses is that the teachers don't put in the effort to make good lectures

The content isn't even that bad

It's just poor profs / TAs

[ember fog]

I feel that, they make lectures on how to solve very small individual pieces but exams and hw are just a palimpsest of materials in a small amount of time. I guess they could argue they don't want to be handholding us?

[sharp mulch]

As a grad student who has taught calculus, a lot of people struggle because of poor foundations

[ember fog]

That seems to be my problem

[sharp mulch]

You can use khan academy to brush up

[ember fog]

Khan academy doesn't work for me unfortunately but I do like textbooks and professor Leonard

[sharp mulch]

Oh ok as long as you have the resources

[cinder zephyr]

It's a combination of both

I've seen some awful calc profs / TAs

Lack of foundation is an inevitable issue considering how poorly we pay elementary/middle/high school profs

[deep mango]

Yeah a lot of students experience this learning curve of a transition from mechanical thinking to more creative and problem-solvingy thinking. It's a tough hurdle to cross and the best things you can do are talk to a lot of people (this is a good space for that, also study groups and such with your classmates) and practice. it does start to feel natural and instinctive! but there's no way to avoid the time it takes to get that way.

[cinder zephyr]

Poor foundations do not make a profs life easier tho, I agree with that part

[cinder zephyr]

The study group part is so important

Cause very likely you aren't the only one struggling

Also take advantage of office hours (meaning start your HW early so that you have the opportunity to go to OH)

[deep mango]

And I hear that it's so awkward to ask people if they want to get together and work on stuff especially if you feel behind, but literally every time i do this the person is like "omg yes im so glad someone else asked"

Lol

[ember fog]

I really want to join a study group but y'all the institute at my uni makes us sign with ink on paper that we will not be sharing any part of our homework and will not ask for any form of help from another person or calculator

And also we can't share course materials

Even the module screenshots

[tight comet]

your institute is dumb

[cinder zephyr]

That's so fucking dumb lol

[broken egret]

Hello

[rocky shuttle]

Hi

[broken egret]

How are u

[rocky shuttle]

Fine, what about you?

[broken egret]

Great

[rocky shuttle]

Welcome to the math cat server

[zealous garden]

category theory

[rocky shuttle]

not excluding that

[tight comet]

none of this is (not) abstract nonsense

[rocky shuttle]

it's general abstract nonsense

[neat lintel]

My algebra book has this exact diagram

(which makes sense)

[dense musk]

could anyone whos good with graph theory here send me a dm pls

[lament flicker]

[lament flicker]

what theorem is that actually? it kinds reminds me of the 1st isomorphism theorem

[long matrix]

the... 1st isomorphism theorem

[eager reef]

[lament flicker]

wow

[sharp mulch]

Lol

[long matrix]

otherwise better known as discussion's greatest moment

[lament flicker]

hey don't blame me I still haven't taken a full algebra course

[mossy oriole]

Discussion question: The real functions of a variable that express the displacement (velocity or acceleration) of a body as time passes defined on the interval "A", A included in R, are all these functions continuous?

For example, if f is not continuous at x=a because , in this case, f(a) is different than lim x-> a+ , we would be saying that "the body teleported" since the displacement.

(assume that the function is the sign function, the domain is all R positive and zero) The image of 0 is 0, but the image of values ​​greater than zero start to be 1. This means that the body has moved 1 meter (or 1 inch, the units don't matter). So we would be indicating that the function represents an impossible phenomenon (that the body has teleported).

Another case to analyze:

If we assume that all functions are continuous we could now ask what happens with differentiability.

We know that continuity is a necessary condition, but it is not enough that f is continuous to say that it is differentiable.

In many cases, there are graphs that represent velocity or displacement (lines with slopes 0, other lines with positive and other negative slopes). These graphs can come from a function defined by parts, where each part is a line. The function is continuous but not differentiable. It is not differentiable because it is not differentiable at some points, those points where each part of the graph begins and ends (if we study lateral derivatives we see with the naked eye that they are different, and if we do it analytically we arrive at different derivatives too)

[burnt dune]

:eeyes

[uncut tusk]

hey quick question

how do you give someone kuddos on this discord? There's a bot command that gives someone some points

someone just helped me out in a big way and I want to recognize them 🙂

[sharp mulch]

I don't think there is one

[split island]

Yeah, there isn't a mechanism in place for rep

[uncut tusk]

ah okay, thanks anyways!

Ah there is! I found the command t!rep @<username>

[sharp mulch]

Oh the tatsu thing

That is global and not server specific

[uncut tusk]

oh I gotcha

[storm sage]

Yeah the Helpful role is the only award for helping out

[rocky shuttle]

Do you have to be a Helper to get the Helpful role?

[hollow jackal]

no

[naive bolt]

why do we say mathematician and not mathologist?

[rancid meadow]

Because we are considered artists not scientists

Or something idk thats the pretentious answer

[naive bolt]

that is pretty pretentions

but also based thanks

[sharp mulch]

Physicists aren't physologists

[naive bolt]

dam youre right

[sharp mulch]

Chemists

[naive bolt]

now I need to look up grammer of the suffix

or just think harder lol yeah

[rancid meadow]

Im sure this is partially historical too

like being "a person who does math" has been around for a lot longer than some other fields

[scarlet jasper]

Probably similar to how physician comes from physiology, mathematics to mathematician

[swift sinew]

I think this is close to the truth. -ist means practicer rather than researcher and it’s likely the case that math was initially seen as a tool rather than a field of study

[naive bolt]

mathematician seems pretty unique though, I can't think of any other fields where a practicer or expert is an "atician"

[sharp mulch]

Aesthetician

[naive bolt]

quit disproving by counterexample 梦境倒流

[swift sinew]

Statistician

[naive bolt]

god damnit

[static loom]

I think the -atic- part shouldn't be taken along for the ride since it's part of 'mathematics'

[sharp mulch]

Lol

[rancid meadow]

im a swagician

[static loom]

Physician ends in -ian, wouldn't make sense to call them Physaticians

[naive bolt]

ah true yeah

[swift sinew]

It could be a relic from Latin’s mathematici

[wooden flax]

mathist

[neat lintel]

Is MIT 18.01 a good introduction to calculus?

[rancid meadow]

yes

[deep mango]

MIT is a dodgy institution but that course is good 👍

[torn willow]

Why is MIT dodgy

[rancid meadow]

bottom tier

weird students

lots of their funding comes from making bombs to explode in other countries

[wooden flax]

MIT is dogshit

[bronze pelican]

oh

when i was a highschooler

MIT was the pinnacle

good to know i didnt miss much

[bronze pelican]

in terms of reputation

[true bough]

[wooden flax]

@bronze pelican I was being sarcastic

[bronze pelican]

Ryc wasnt

[wooden flax]

Huh

[still pond]

hi

[neat lintel]

Hi

[vestal solstice]

tony stark moment

[neat lintel]

Unfortunately weapons have to be made.

Human race isn't a Kardashev type I civilisation.

[deep mango]

Another MIT simp joins the masses.

[neat lintel]

@deep mango I think the most interesting things I've seen come out of universities in Japan and Nordish countries

Japan seldom brags though.

MIT just likes to brag constantly.

[deep mango]

I agree with that

I mean I'm mostly thinking from a perspective of their undergrad program, the people I know who went and the policies I know they have

It's obviously good but it is put on an unreasonable pedestal

[neat lintel]

It's also rather elitist.

A lot of them want to work at FAANG etc.

[deep mango]

I would suggest that I don't care where research comes from so much as who it comes from (in terms of research groups and academic networks). I don't feel like MIT is connected in any way to any of the fields that I interact with, which is odd, because they have faculty working in those fields.

[neat lintel]

I don't like FAANG companies at all.

[deep mango]

To me that says something about the school.

Yes I mean, I can't say a ton because I went to Berkeley which is a silicon valley feeder and now I'm at NYU which is a wall street feeder

[neat lintel]

I'm aussie our universities operate very differently.

[deep mango]

But I didn't feel like berkeley was elitist at all. Stanford gets that mantle

Ive heard

[neat lintel]

There isn't really any "curriculum" most of your learning forces you to be absolutely independent.

You are given due dates, tutorial material, lectures

If you show up or not they don't care.

It's a bit like British education system

Harsher and less emphasis on getting rolling pissed as a nit

Also you aren't allowed to pad your course with useless shit like in America.

[neat lintel]

Plenty of lazy dead beats though.

Oh I've read my fair share of what happened when the rich kids were all incredibly dumb and had a smile on my face.

[deep mango]

Lol

[deep mango]

Yea you can definitely do that

[neat lintel]

And it was written in black and white by the FBI's internal offices (Rich kids dumb)

[deep mango]

Here

[neat lintel]

Even though the cheating they did, said they were super smart.

Getting a car because you passed Precalc lame.

[deep mango]

[neat lintel]

Whats your field of research?

I couldn't be bothered with a PhD...

Also it doesn't pay well...

I'll stick to Software and Computer engineering.

[bright hill]

Wtf, people go to mit to go to FAANG?

[neat lintel]

Some do...

Stuff FAANG 35+ hours a week to be a code droid

[neat frost]

Famazon
Amazon
Amazon
Nozama
Gamazon

[neat lintel]

And any creative or technical work you do on company time they can appropriate.

Amazon has literally stolen from people who worked on company systems at work

And they found out about "their project"

and stole it.

FAANG companies are parasites.

[bright hill]

Reminds me of how Disney owns the right to SOO much porn and rule 34 stuff

[deep mango]

Mathematical fluid dynamics

[bright hill]

I don't understand people that go to top schools just to have a regular ass job in industry

[bronze wedge]

What else are you supposed to do with a bachelor's lmao

[bright hill]

Grad school?

[sharp mulch]

Some people need to get jobs for financial reasons

[bronze wedge]

Well yeah but there are lots of people who don't want to go to grad school

[neat lintel]

I'm more an applied math / physics / engineering type.

[bright hill]

Yeah, but like

[bronze wedge]

Don't group physics and engineering with applied math 😤

[bright hill]

Couldn't they have gotten the same thing with a school that's less expensive?

[bronze wedge]

Probably

[tight comet]

it's not always a choice

[midnight prairie]

Anyone good at physics here?

[sharp mulch]

#❓how-to-get-help

[tight comet]

no one

[neat lintel]

Really good engineers actually understand the math.

Unlike a lot of them that say "I took differential equations and have no idea what I learned"

[sharp mulch]

Are people who want jobs not allowed to get high quality educations

[deep mango]

Also important to remember that a lot of people think they want to do research before undergrad because they have no idea what they're talking about decide they don't once they actually do it

[neat lintel]

Maybe I am just referring to the old benchmark for engineers, many gears ago back in the 50's where it was a very high benchmark for engineers at the time.

[bronze wedge]

Engineers proving things: take Taylor series and set everything but the first term to zero

[neat lintel]

@bronze wedge I like the theoretical things I just care about their physical application.

Unless I can "do" something with it, I don't want to know.

Taylor series is used a lot in CS

For proving Algorithmic complexity.

Infinite series and closed / open form equations for function growth.

[tight comet]

(Don't tell the computer scientists complexity is just analysis under the hood)

[neat lintel]

It is...

[midnight prairie]

Not only diff equations but the whole course lmao

[neat lintel]

Doesn't bother me, spade is spade, shovel is a shovel.

[zealous garden]

Time to apply nonstandard analysis to complexity

[midnight prairie]

do we use volumetric density for 3 dimensional objects?

[neat lintel]

Volumetric density? What in the context of material / atomic weight?

and space between the atoms?

[midnight prairie]

It's to find centre of mass

Mass/volume

[sharp mulch]

Physics server

Linked in #old-network

[midnight prairie]

it's dead 💀

[sharp mulch]

X for doubt

[neat lintel]

Oh it's for fluid dynamics.

[midnight prairie]

nooo

I'm asking for objects

[bright hill]

I thought ivies and such didn't necessarily have a higher educational quality

[neat lintel]

Objects? In what way, density refers to the amount of stuff in a given volume

[bright hill]

It was mostly just prestige

[sharp mulch]

They probably have a better education experience

[bright hill]

And networking

[sharp mulch]

Like smaller class sizes

[neat lintel]

Higher the density more mass there is in a smaller volume

[midnight prairie]

Objects as in cube

cuboid

[tight comet]

They also don't necessarily have trashy education quality, fwiw.

[bright hill]

I mean, obviously

[neat lintel]

@midnight prairie Same thing, density is volume in a given mass.

E.g a block of lead is heavier than say a block of steel for the same volume

Volumetric mass density is mass per fluid unit volume.

So if it's a cube it's still going to be the same, then again it would be dependent on the properties of whatever you are putting in that liquid as well

[midnight prairie]

I never stated how the object is made

or the material for it

[neat lintel]

So just general volume

[midnight prairie]

Yep

[neat lintel]

density = mass / volume

Re-arrage the equation d = v* m/v dv = m/d

so v = m/d mass / density

[midnight prairie]

No

VD= M/Volume

[neat lintel]

Rho is the symbol you were looking for

[midnight prairie]

YES

[neat lintel]

@midnight prairie That what you needed?

[midnight prairie]

Isn't rho also used in electricity

[winter python]

@neat lintel hello

[midnight prairie]

@neat lintel is impulse I= F . delta T

Or change in momentum= F . Delta T

Oh wait fuck both the are the same things

[neat lintel]

bruh

[long matrix]

@flat harbor where r u 😢

[deep mango]

Literally 3:30 am blo time

[wooden flax]

Lol

[deep mango]

Common sense and courtesy

[wooden flax]

should be up then

@flat harbor

[long matrix]

no i wont leave my favourite people alone

its 330 tho wut?

[deep mango]

Dummy

[long matrix]

+8 only? damn

[cold lantern]

For Christmas I can wish for many books from Springer. Which ones would you recommend?

[wooden flax]

#book-recommendations

[storm sage]

why the sully...

[wooden flax]

Asking for random books?

[cold lantern]

Asked it there.

[wooden flax]

noice

[storm sage]

I feel like this is a perfectly valid discussion topic lol

[cold lantern]

Basically yes. Especially books for undergrad meaning from say second year to masters sixth year. I'm looking for the absolute classics almost anyone would recommend.

[wooden flax]

any subjects?

what do youi know already

[jovial ember]

Absolute classics? Every Bourbaki book 🤡

[cold lantern]

Yes, because (1) I can filter and (2) I can make a big wishlist.

[wooden flax]

EGA

[jovial ember]

Not a Springer book

Dummy

[wooden flax]

@jovial ember might unironically learn french for ega

[jovial ember]

You don’t need to learn French

You just read EGA

And learn how to read French math by reading French math

[wooden flax]

dang

[jovial ember]

I’m completely serious

[wooden flax]

tried reading hartshorne but the exercises were killing me with the little exposition

might try that then

[cold lantern]

This is how I've learned English and how I'm currently learning French maths.

[jovial ember]

Reading EGA isn’t really gonna fix that

If you want a companion to Hartshorne try Mumford’s red book or Bosch Algebraic Geometry and Commutative Algebra

[wooden flax]

ch1 was fine, ch2 was where im having problems

will check out mumford

[jovial ember]

EGA is not easier than Hartshorne and the idea that it is because it doesn’t have exercises and that the proofs are complete are by ppl who haven’t tried reading it

There’s kinda no getting around just struggling for a while

[wooden flax]

how long did you spend on problems before giving up

[jovial ember]

Like days

[wooden flax]

oh

[jovial ember]

I would regularly spend over like 4 hours on one

[long matrix]

chm is where the biggest problems at

[wooden flax]

i need to just put my head down and try harder then

[long matrix]

[wooden flax]

[jovial ember]

Unironically yes, sorry

I mean

Try some other things too like

They’ll have some useful stuff that really should’ve been in Hartshorne

But at a certain point you just have to do hard math

And struggle

[wooden flax]

yea i agree

[jovial ember]

But anyway, EGA doesn’t solve the problem

Trust me

[long matrix]

chm wheres the m gone in your name

[jovial ember]

I gave it to Shamrock

[wooden flax]

:(

do you know him irl

[jovial ember]

Yeah

[wooden flax]

cool

[jovial ember]

We went to university together

[wooden flax]

oh thats nice

[jovial ember]

He’s the reason I started doing algebra

And AG basically

[wooden flax]

how did he convince you lmao

[jovial ember]

Well he started an algebra study group I joined

Before this he helped me when I was starting off in my analysis class cuz I didn’t know how to prove shit

Then at the end of the year we decided to study something, him, me, and another friend

I think he wanted to do homological algebra actually?

But the other friend wanted to do AG so I just went along with it

I hated AG at first btw

[wooden flax]

Lmfao

Homo alg would have been boring for you ptobably

if youre not seeing applications

[jovial ember]

No wait he wanted to do AT

I had already seen homological algebra

That group was very weird

The second quarter had module theory and ring theory and Gali’s theory

And I ended up like, proving the snake lemma with Sham

[bright hill]

Chmonkey lore

[jovial ember]

And we even defined Ext

Lol

[red kelp]

@hollow sundial you didnt help me btw my friend showed me the steps taken instead of giving me math lore from wikipedia

[jovial ember]

Lmfao

[sharp mulch]

What does this mean

[red kelp]

i asked for 1 step

and he gave me math lore

[long matrix]

@hollow sundial not doing what hes paid to do

[red kelp]

got me more confused than before

[sharp mulch]

Does riemann use he him pronouns

[red kelp]

but my friend sent the work so now i understand it

[jovial ember]

I went through and read it

[dusty wagon]

Who cares his name is male

[sharp mulch]

?

[jovial ember]

Lmfao

[long matrix]

tldr?

[jovial ember]

I’m pretty sure Riemann is he/him

[sharp mulch]

What does this mean

[jovial ember]

that guy is mad cuz bad

And the question was factoring

And Riemann said it didn’t disappear, it just was factored

[red kelp]

ye i knew that

i just wanted to see more steps to make it clear for myself

[jovial ember]

Bro there’s literally no step to write there

You factor

[red kelp]

how not

[jovial ember]

That’s a single operation

[long matrix]

ax + ay = a(x+y) ?

[jovial ember]

Yes

[long matrix]

rip

[jovial ember]

There’s no intermediary step

[red kelp]

idk what my friend showed m then but its correct still

[long matrix]

ax + ay = ay + ax = a(y + x) = a(x+y)

[rocky shuttle]

math lore: https://en.wikipedia.org/wiki/Distributive_property

[jovial ember]

[red kelp]

ill send that to everyone that asks for simple math steps and compliment myself

[long matrix]

waw, history right here

anyways, ax + ay = a(x+y) is indeed a property

that you kindof take for granted

you can verify it with some actual numbers if u want.

[red kelp]

its all good ty tho

i understand it now

[hollow sundial]

@red kelp why are you still talking about me. You're just being rude or obsessed with me

[long matrix]

@hollow sundial come to discussion more often

[red kelp]

i thought the convo ended

[sharp mulch]

riemann do you use he/him pronouns

[long matrix]

helping ppl >.>

[hollow sundial]

Says conversation ended. Still proceeded to send 10 messages about me

[hollow sundial]

Yes

[hollow sundial]

Man they really teach lore in all the math classes.

[bright hill]

You see, the conversation with did end

But they wanted a conversation with the manager

[neat frost]

Change your name DarK.

[bright hill]

Why?

[neat frost]

Because it’s not funny.

[bright hill]

Are you aggravated?