The h Bar

Phase

00:00

Have you clicked the bookmark that says "start chatjax"

enumaris

a group homeomorphism only preserves the group operation

I have the extension enabled

Phase

What's a group operation w.r.t a vector space though

What does that mean

enumaris

where's the bookmark that says start...

ACuriousMind

@Phase Can we take a step back here? What exactly was the original question?

enumaris

a group and a vector space are different things...

Phase

00:01

@enumaris I never mentioned a group originally so why are you?

enumaris

a group is any collection of elements that have a group operation defined such that the group operation obeys certain rules

Phase

Yeah I get that

enumaris

wait, wasn't the question regarding group homeomorphisms

Phase

No, homomorphisms

enumaris

oh

lol

ok o.O

ACuriousMind

00:01

@Phase Well, homomorphisms of what?

0celo7

@Phase addition!

Phase

a vector space? idk :(

@0celo7 Right, ok

So.. a Homomorphism is just a mapping from a space of one type to another of the same? Vector space to a vector space etc?

enumaris

usually yeah, and it preserves the operations in that space in some way

0celo7

a linear map is just a special group homomorphism (ACM will kill me for doing this)

Phase

So all maps from a vector space to another are homomorphisms, and then all the properties of the map are tacked onto that?

enumaris

00:04

I click the little extension button but I don't see a "start" button...

0celo7

but I'm trying to get you to do an abstract nonsense proof of this

use the universal property of the quotient for this

enumaris

$f$

dang it

Phase

You're basically talking to a fuckin rock

:/

enumaris

not all mappings are homomorphisms

all homomorphisms are mappings

Phase

Between the same types of spaces? @enumaris

enumaris

00:06

only mappings which preserve the algebraic structure are homomorphisms

Phase

I was under the impression that a vector space has to have addition defined to be a vector space

enumaris

A vector space's definition is like 8 attributes long -.-

but yes, a method to add vectors is one of the definitions

0celo7

a vector space is just a module over a field

@ACuriousMind can u help him, I can't help

Phase

So there are mappings from a vector space to another vector space that aren't Homomorphisms @enumaris ? that seems counter to the definition

enumaris

then u'll just get people asking "what's a module?

"what's a field?"

ACuriousMind

00:07

I'm confused what exactly we're trying to explain here

0celo7

a module is just a vector space but over a ring instead of a field :P

enumaris

and round the ring we go lol

Yes, you can map a vector space to another vector space in such a way that the algebraic structure is destroyed

such a mapping is not a homomorphism

Phase

idgi

how can the structure be destroyed but preserved

meh

enumaris

oh, the new space has the properties of a vector space, but the mapping of individual vectors from one space to another might make it like A+B=C but A_new+B_new != C_new

0celo7

@Phase some people are saying stupid things

can you please tell @ACuriousMind and me what you want

enumaris

00:10

is that "some people" referring to me :(

Phase

idk my question was just about "how can I use the principle of extending a basis to prove rank nullity"

0celo7

@enumaris and me

Emilio Pisanty

@Phase there's several possible proofs

0celo7

@Phase can you just read the proof on Wiki and ask exactly what the problem is?

Emilio Pisanty

is there some specific scheme that you're trying to follow?

0celo7

00:11

@EmilioPisanty Typesetting on a scale from 1 to 10?

Emilio Pisanty

@0celo7 depends on the marking scheme

enumaris

eyes...bleeding...

Emilio Pisanty

if it was typeset by a five-year-old in the seventies, then I guess it gets a three out of ten

enumaris

please don't ever post such an atrocity again

Phase

Oh ok

I was getting confused because in my head it seemed trivial / tautological, saying that "the kernel basis can be extended to the domain spaces, and then by definition the kernel basis vectors vanish leaving just m many non vanishing"

but I only got up to "checking for redundancy" or whatever

So maybe this will be the thign

Emilio Pisanty

00:14

@Phase it really is close to that simple

0celo7

@EmilioPisanty eh, a 40 year old in '79?

Phase

Yeah it really is

Emilio Pisanty

@0celo7 4/10

Phase

@EmilioPisanty this might be the one problem where I got it near instantly and overcomplicated it because I thought it must be weirder

It doesn't cover @0celo7 's proof though as far as I can tell, can you write it out? You dont need to explain it further after, I'll just give it a go at trying to understand it

Emilio Pisanty

@Phase the thing is, the argument you just laid out gives you $n-m$ candidate basis vectors for the image space (the other $m$ vectors having vanished)

but it doesn't prove that they're actually linearly independent

they are, but you need to prove it

Phase

00:16

But if extending gives the basis of V, then V is linearly independent, and regardless of how many basis vectors you discard you're still gonna have linear independence

doesn't that mean that they are

since a basis is linearly independent

Emilio Pisanty

@Phase the $u_1,\ldots, u_n$ are a basis in $V$

(following the WIkipedia notation)

but that doesn't immediately imply that the $T(u_1), \ldots, T(u_n)$ are linearly independent in $W$

no, wait, my notations are off

0celo7

@Phase You define a function $\tilde f:G/\ker f \to f(G)$ in the natural way. Now this is a homomorphism by the universal property, and it's surjective. It clearly doesn't have a kernel. So it's an isomorphism.

And that's rank-nullity.

Basically trivial.

Emilio Pisanty

@0celo7 please don't use moduli spaces, it'll just confuse him more.

0celo7

@EmilioPisanty I don't see what a moduli space has to do with any of this.

Phase

@EmilioPisanty If they aren't linearly independent under T, then say a basis pair existed such that under the transformation, they're no longer independent, you could add them to get zero and by definition the point that is mapped to that would lie in the Kernel, which it cant as it's already stated to be in the image space

Emilio Pisanty

00:19

@0celo7 $G/\ker f$ is a modulus space

or whatever you wanna call it

don't use it

0celo7

A quotient space?

Emilio Pisanty

yeah, whatever

don't use them

0celo7

No thanks

Everyone should know how to use them

Emilio Pisanty

@0celo7 yes, but you have to work up to them

0celo7

I recall @Phase discussing them at length with other users.

ACuriousMind

00:20

@0celo7 Do you remember how much you used to hate me pulling out the universal properties?

Emilio Pisanty

in this specific conversation, they'd do more harm than good, unless @Phase can confirm that they're OK with the concept

but in any case, Phase specifically requested a basis-extension proof

Phase

I'm.. Meh with them, I've had a brief conversation with balarka where he explained to me the notion of a quotient space and defining them with the equivalence relations, but I'm always up for learning new shit

0celo7

No, he also requested me to explain my proof.

Phase

@EmilioPisanty is my argument above correct?

Emilio Pisanty

@Phase close to

Phase

00:22

Fuuuuuuuuuuck

0celo7

@ACuriousMind No?

Emilio Pisanty

use the proper definition of linear independence

i.e. set an arbitrary linear combination to zero and show that the coefficients need to vanish

Phase

if they didn't it would lie in the kernel though, whats wrong with that approach?

Emilio Pisanty

@Phase it's a muddled argument so it's hard to be definitive

Lemme fix some notation: take a basis $u_1,\ldots, u_m\in V$ for the kernel of $T:V\to W$, and you extend it to a basis $u_1,\ldots, u_m,w_1,\ldots, w_n$ for $V$.

Phase

ok ill try writing it formally

Emilio Pisanty

00:24

You want to show that $T(w_1),\ldots, T(w_n)$ are a basis of $T(V)$

0celo7

@ACuriousMind You don't even need a universal property, just show directly that it's one.

But I think it's the unique map from the quotient, so that's why I used the universal dude.

Emilio Pisanty

To show that they're linearly independent, you take an arbitrary linear combination $\sum_{i=1}^n a_i T(w_i)$ and you set it to zero

$\sum_{i=1}^n a_i T(w_i)=0$

therefore $T(\sum_{i=1}^n a_i w_i)=0$

therefore $\sum_{i=1}^n a_i w_i$ is in the kernel of $T$

therefore the coefficients need to vanish, because the $w_i$ are complementary to $\ker(T)$

so you've proved that $a_i=0$

and that's the definition of linear independence

Phase

$T(u_1,...,u_m,w_1,...,w_n) = T(w_1,...,w_n)$ by definition of the bases u_1 to u_m lying in the Kernel, but now if you were to have it so that a point q $\in$ domain, expanded as a sum of at least two nonzero scalars and basis vectors, were to add to zero / be linearly dependent then you'd have $T(w_q,...w_{qn})$ = 0 for nonzero coefficients, so it necessarily must lie in the Kernel

I dont get why it doesnt work :(

Emilio Pisanty

@Phase what does $ T(w_1,...,w_n)$ mean?

Phase

That any vector under T is just equivalent to the components along the basis vectors not in the Kernel

Sticking to wiki notation

Emilio Pisanty

00:29

@Phase where does Wikipedia have $T$ acting on more than one vector at a time?

it's a terrible practice

0celo7

wot

Phase

It's components in my case

0celo7

@Phase link?

Phase

the components correspond to the bases of the same name

shit idea i know

Emilio Pisanty

@Phase the $w_i$ are not components, they're vectors

Phase

00:30

but I cant edit now

Ik ik

but its too late to edit

:(

Emilio Pisanty

whatever you're writing, if I was marking it, it would come back bleeding red ink, I should think

Phase

I'll give it one more go

0celo7

they have T acting on a sum of stuff.

Phase

Let $U_i$ and $W_i$ be bases, with respective values $u_i$ and $w_i$ for components. the set of vectors $U_i$ span the kernel, and $W_i$ is obtained by extending the basis. By definition of the kernel, $T(u_1,....,u_n,w_1,....,w_m)$ reduces to $T(w_1,....,w_m)$, and if this basis is linearly independent then the only time $T(w_1,....,w_m) = 0$ is when all the values $w_i$ are equal to zero. If this wasn't the case, then a non-zero point in the domain space spanned by U and W, would be mapped..

to zero and therefore lie in the Kernel, which by previous argument and definition it doesn't

Emilio Pisanty

@Phase $T(w_1,...,w_n)$ is meaningless until and unless you define it

it's not standard notation

(you shouldn't define it, but that's another matter)

Phase

00:35

I guess because I'm not sure how to use the notation to communicate properly, I figured losing generality and talking about vectors would suffice

I feel like even when I have the right answer I can't communicate it

It's incredibly frustrating

Fuck it my entire argument boils down to the fact that if a vector in $V$ lies within the span of $W$, then by definition it's not in the kernel, so no linear combination of nonzero basis vectors in the transformed basis $T(W)$ could add to zero

Does that work

please tell me it works

Emilio Pisanty

@Phase almost

there is a vector in the span of the $w_i$ that is in the kernel of $T$

the zero vector

enumaris

oh

latex is finally working

hurray

Phase

edited to add nonzero

I kept that in last time but I forgot it in the recent one

:(

edited again because points could include p and -p

Emilio Pisanty

@Phase the general idea is OK

but the implementation needs a ton of work

you need to be talking a whole lot more about linear combinations

Phase

The maths lecturer I speak to tells me my notation is shit too

Emilio Pisanty

00:41

@Phase so fix it

Phase

Idk how, it's hard to rewire what's natural

Emilio Pisanty

@Phase yes, it is

you still need to do it

Phase

true

Maybe I should keep reading Artin and Abbott and just see if it gets better the further I get through

Anyway, thanks for the chat and help, just chatting in general @0celo7 @ACuriousMind @EmilioPisanty I'm gonna head to bed, have a good night

Rumplestillskin

01:30

Hello all! A quick sanity check. In the classical two body problem. With regards the system energy $ E = v^2/2 + U(x)$, where $U(x)$ is the potential energy of the system. In the limit $U(x) \rightarrow 0$, $v^2 \rightarrow 2E$??

Bernardo Meurer

@Phase Abbot is pretty good

0celo7

@BernardoMeurer drinking bourbon in your honor

Bernardo Meurer

@0celo7 Facetime me dawg

I'm drinking beer in bed

0celo7

I'm in the kitchen

how did you get beer

Bernardo Meurer

I bribed a polish homeless dude

That's also how I get cigarettes

0celo7

01:32

...you smoke now?

Bernardo Meurer

Well, I couldn't drink so I found an alternative

But now that I can drink I don't smoke

I just need one or the other

0celo7

do math as an alternative

no time for that BS

Bernardo Meurer

Math just makes me sad because I suck at it, lol

0celo7

@BernardoMeurer you need to get obsessed with something that isn't a drug

Bernardo Meurer

Says who

0celo7

01:37

K

Bernardo Meurer

I'm joking

it would be better

But idk, I'm lazy

ACuriousMind

@0celo7 ::lights cig, opens beer:: No, he doesn't.

0celo7

@ACuriousMind Oh god, you smoke?

Bernardo Meurer

@ACuriousMind I want to sit down in a porch and smoke a cigar while having a beer with you

While we watch out kids wrestle

ACuriousMind

@0celo7 There are very few vices in this world I don't indulge in :P

Bernardo Meurer

01:45

@ACuriousMind This is why I like you

0celo7

Smoking is objectively bad.

Bernardo Meurer

So is drinking

0celo7

Nah

Bernardo Meurer

But morning coffee with a cigarette is damn good

0celo7

A glass of red wine a day is good

@BernardoMeurer do you have a mouse?

Bernardo Meurer

01:52

@0celo7 A shitty one

0celo7

@BernardoMeurer I don't have any mouse with me, nvm

I could try to get Bob's magic mouse to work

Bernardo Meurer

Magic mouse is cancer

It's so bad

0celo7

he likes his

he's such an apple fanboy

@ACuriousMind Oh, I'm going to get a tablet soon-ish

a new one with a pen

ACuriousMind

nice

0celo7

@ACuriousMind Any idea what condition you need for convergence to define a topology?

ACuriousMind

01:57

@0celo7 Convergence in what sense? That of filters?

0celo7

@ACuriousMind Sequences.

Filters? What are you, a logician?

inb4 nets are filters

ACuriousMind

@0celo7 So, what data is given?

0celo7

@ACuriousMind I have the cone of metrics $\mathscr M$ (well, a subcone of properly decaying ones) and I want to give it a topology using weighted Sobolev spaces such that the ADM mass becomes continuous.

Er, I guess that's just a metric topology

nvm

ACuriousMind

If you know for every sequence whether it converges and if so, to what, then I think you can define a topology by saying that a set is open if it contains infinitely many elements of every sequence converging to a point in it

0celo7

@ACuriousMind That's actually not always true, i.e. there can be multiple topologies with that property.

ACuriousMind

02:02

@0celo7 There can be multiple topologies inducing the same convergences, but saying a set is open iff it has that property should define a unique topology, no?

0celo7

I actually just want a metric topology given by the sum of some tensor norms

@ACuriousMind Well no, because all of those topologies have the space open sets in the sense of convergence.

But there might be more actually open sets.

Oh, I see what you're saying.

ACuriousMind

Yeah, but my definition selects the coarsest of these topologies, right?

0celo7

Yeah, hmm.

Yeah I was thinking about the converse problem

"Can one characterize a topology by its convergent sequences?"

enumaris

getting all mathy up in here

ACuriousMind

I admit my claim to being a physicist doesn't look especially strong after this conversation ;)

0celo7

02:10

It's weaker than $\mathscr D'(\Omega)$

and if you get the (bad) joke you're not a physicist either

ACuriousMind

Ouch

Bernardo Meurer

@ACuriousMind We should play CIV V one day

0celo7

@ACuriousMind did you get it :^)

ACuriousMind

@0celo7 Yes :|

enumaris

lol

There should probably be a threshold of amount of pure math you're allowed to do as a physicist. If you go over the threshold, you lose ur physicist creds...

Bernardo Meurer

02:12

@enumaris Don't worry about it, physicists can't do maths

0celo7

I think once you know that $\mathscr D'(\Omega)$ has the weak* topology your claim to physics is on shaky ground.

ACuriousMind

@BernardoMeurer Sure, why not

Bernardo Meurer

@ACuriousMind Only if you play as brazil and I as germany

to strengthen our diplomatic relations

ACuriousMind

Heh, deal

enumaris

ouch

0celo7

02:13

@BernardoMeurer why have you never asked me

Bernardo Meurer

@0celo7 You play Civ V?

I always knew Bajoran played

0celo7

no, but I could get it

there's a Steam sale

Bernardo Meurer

Then get it

We can play together

There's a fun setup which is three humans in a team, against everyone else being a bot in another team with domination victory only

It's intense

Get the expansion packs!

0celo7

Maybe then @ACuriousMind will finally teach me German

ACuriousMind

I'm sorry I tend to have full weekends :P

Bernardo Meurer

02:17

@ACuriousMind Really? Busy with what?

ACuriousMind

@BernardoMeurer RPG sessions, boardgaming sessions and drinking/parties, mostly

Bernardo Meurer

@ACuriousMind Tsc, and you don't even invite me!

ACuriousMind

If I invited you and you didn't show up I'd be terribly hurt. You willing to pay to fly to Germany? :P

Bernardo Meurer

@ACuriousMind You have a fancy job now, fly me over Christmas :D

ACuriousMind

Ain't starting till January

Bernardo Meurer

02:21

Balls

2018

I'll bitch about it

and you're mean too! You never even added me on the bookface

ACuriousMind

Only people I've met in RL there, toldya :P

Bernardo Meurer

@ACuriousMind ::complains::

user400188

I'm having a bit of trouble asking a question at the moment.

Bernardo Meurer

@user400188 Ask it specifically and as unclearly as possible to @ACuriousMind

He will do his best to help you in length

user400188

The derivation (workings) take up more than a page, so I am finding it hard to include the context of the question. Also, the question is more like 3 questions. As I am having trouble with the interpretation, the math, and the formulation.

0celo7

02:27

what is the subject

user400188

Its a derivation from an introductory chapter to QFT

Its not real QFT yet however. And nothing I havent already covered before

0celo7

you'd better ask

before you get rekt

I can hear ACM fuming

user400188

Hmm, Ill give it a go

Bernardo Meurer

@user400188 Remember, write as abstrusely as possible

And don't use chatjax

ACuriousMind

Guys...

Bernardo Meurer

02:29

Ask @ACuriousMind specifically

@ACuriousMind Best way to get revenge

0celo7

ACM will make you feel stupid

be warned

ACuriousMind

don't confuse hapless visitors, please :P

user400188

Well, I guess Ill ask it in bits, since its too long for one question:

First bit:

$\int_{-1}^{1}d \cos(\theta)$. How do I interpret this integral? I assumed originally that $d cos(\theta)= d/d\theta cos(\theta) \d \theta$. So the integral is actually zero.

The text seems to think that it is equal to 2.

0celo7

it's $\cos(1)-\cos(-1)$

really? 2?

user400188

since cos is symmetric that should be zero

0celo7

02:33

Welp, I can't do it either

ACuriousMind

Yeah, I concur that integral is zero

user400188

Ill show the full equation: it might help

Bernardo Meurer

0celo7

Isn't that the usual spherical coordinates thingie tho?

Bernardo Meurer

I broke my fonts

(Look on top)

ACuriousMind

02:34

@BernardoMeurer It appears the aliens are trying to contact you

user400188

Yes its from spherical coordinates

$\begin{align}\label{JeansCubeIntegratedExpectationEnergy}
E(\omega)&=\big{<}\int_{}^{\omega}E_nd^3\vec{n}\big{>}\\
&=\int_{}^{\omega}<E_n>d^3\vec{n}\\
&=\int_{}^{\omega}d^3\vec{n}<E_n>\\
&=\int_{-1}^{1}d \cos(\theta)\int_{0}^{2\pi}d\phi\int_{0}^{\omega}d|\vec{n}||\vec{n}|^2<E_n>
\end{align}$

Bernardo Meurer

@ACuriousMind Yeah, I've deciphered that they are asking me if I want them to force you to befriend me

I replied yes

so get rekd

ACuriousMind

Wait, the boundaries there are weird

0celo7

^

user400188

yeah exactly

Its what my friend called a physicists equation. A bit wishy-washy skipping in the details.

0celo7

02:36

the sine should be integrated from $0$ to $2\pi$, causing a...something for the cosine

uh

ACuriousMind

I think $\int_{-1}^1 \mathrm{d}\cos(\theta)$ means integrating from where $\cos$ is $-1$ to where $\cos$ is $1$, which is then of course $1 -(-1) = 2$, but that's a really weird way to write it

0celo7

@ACuriousMind It's the standard thing though ;_;

ACuriousMind

it's weirdo notation, though

0celo7

@user400188 more like physicist BS.

@ACuriousMind how would you write it?

ACuriousMind

or I can't think straight at 3:30 am, either of those :P

user400188

02:37

@ACuriousMind I saw it that way too. @0celo7 I've never heard of it been standard.

0celo7

@user400188 see my profile for some books to read if you want to abandon physics

@ACuriousMind Ah

it's perfectly consistent notation

we're just stupid

let $\cos\theta=\varkappa$

no

Bernardo Meurer

@user400188 Become a type theorist

ACuriousMind

It's $\int_0^{\pi}\sin(\theta) \mathrm{d}\theta$, I think?

0celo7

yeah I was right

$\int_{-1}^1 d\varkappa = 1-(-1)=2$

holy shit we can't integrate

@ACuriousMind from 0 to 2pi

ACuriousMind

No, from 0 to $2\pi$ would be 0

0celo7

02:39

lmao

Bernardo Meurer

lol

Chub and Tuck is not proud

0celo7

@ACuriousMind You're right, but I'm also right

ACuriousMind

Yes

Still not a good use of notation, imo

0celo7

$$\int_{-1}^1 d\cos\theta=\int_{-1}^1 d\varkappa =1-(-1)=2.$$

@ACuriousMind it's completely standard

user400188

@ACuriousMind I think that should be between $-\pi$ and $\pi$.

ACuriousMind

02:43

@user400188 No, integrating over a full period of $2\pi$ is always zero

0celo7

I agree with that

ACuriousMind

We have $\int^1_{-1}\mathrm{d}\cos(\theta) = \int^{\arccos(1)}_{\arccos(-1)} -\sin(\theta)\mathrm{d}\theta = -\int^0_{\pi}\sin(\theta)\mathrm{d}\theta = \int^\pi_0\sin(\theta)\mathrm{d}\theta$.

user400188

Oh your right. I was thinking of cos for some reason (and between $\pm\pi/2$).

@ACuriousMind ah thats perfectly clear now

enumaris

why have we devolved into trying to integrate cosines...

erm sines I mean

0celo7

lack of intelligence

ACuriousMind

02:53

@user400188 Alrighty then, what's next?

0celo7

oh right, 3 parts

ACuriousMind

@enumaris You'd rather have some more topology? :P

enumaris

are those my only choices lol

0celo7

Hmm.

enumaris

Can't we discuss some physics, like, why is loop quantum gravity so much better than string theory?

0celo7

02:54

@enumaris Oh, we can discuss barriers in elliptic PDE

I'm trying to find a modern argument for a Schoen and Yau barrier argument from 1979

enumaris

(Any Sheldon's in the house?)

ACuriousMind

I won't discuss that because it's not true :)

0celo7

I need to solve a PDE with specified asymptotics and it's a complete pain in the ass

enumaris

XD

0celo7

I think I need to solve Dirichlet problems on balls and send the radius to $\infty$, then get a $C^2$ convergent subsequence using Schauder and Arzela-Ascoli

but their calculations take like 5 pages and that's ridiculous in 2017

@BernardoMeurer 4 new Lil Uzi tracks!

user400188

02:57

Ill give the next two parts at once

This is actauly the draft of the question I was going to ask on the main site. I hope your all ok with if its a bit anbiguous at the moment:

0celo7

also a Jaden Smith record

it's #4 on iTunes; smh

user400188

The following is a derivation from the textbook "Quantum Field Theory and the Standard Model" by Matthew D. Schwartz.

The context of the derivation is that a few photons are trapped inside a Jeans Cube. The expectation value of finding a standing wave with particular energy $E_n=\frac{2\pi}{L}\hbar|\vec{n}|$, confined to a cavity with length $L$, is given by $<E_n>$.

$\begin{align}
<E_n>&=\frac{\hbar\omega_n}{e^{-\hbar\omega_n\beta}-1}
\end{align}$

I am comfortable with $<E_n>$'s derivation, what I have trouble with comes next:

0celo7

you broke chat

user400188

yeah I noticed. Not sure why becuase it rendered on the "ask a question" page.

maybe its the label?

Transcript for