The h Bar

@heather So recall that a map $T:X\to X$ of Banach spaces is a contraction if there is a $\lambda\in(0,1)$ such that $||T(x)-T(y)||\le\lambda ||x-y||$ for all $x,y\in X$.

That is, it strictly decreases the distance between points.

heather

right, yes, i remember now.

0celo7

@heather Do you know what the image of a map is?

heather

let me check my book

ah, right.

"if $\alpha : S\to T$ then $\alpha(S)$ will be called the image of $\alpha$."

0celo7

7:06 PM

@heather Ok, so take some $S\subset X$. What can you say about $T(S)$ in relation to $S$?

heather

well, T(S) is a contraction of S, the points will be closer together, right?

0celo7

@heather Exactly. So what about $T^2(S)=T(T(S))$ vs. $T(S)$ vs. $S$?

assume that $S$ has finite diameter/size

heather

well that will contract it even further

so if you have $T^n(S)$, as $n$ gets larger, the points in $S$ get closer together

0celo7

exactly

so if $L$ is the map you get by applying $T$ an infinite number of times, what should $T(S)$ be?

heather

well...one point, right?

a point in the center of all the points in the set, probably.

0celo7

7:20 PM

@heather Yeah, one point, but it might be outside of $S$

the first $T$ could shove all of $S$ to some other place in $X$

@heather So, if you know a little analysis, you can prove rigorously that $L(X)=\{x_0\}$, a single point.

@heather So what do you think $T(x_0)$ is?

heather

@0celo7 hmm....$x_0$?

(i'm expecting something tricky to happen =P)

0celo7

well, that is the trick

it's $x_0$

heather

that makes sense.

0celo7

So we found a point $x_0\in X$ such that $T(x_0)=x_0$

Such points are called fixed points

So, we have Banach's Fixed Point Theorem: If $T:X\to X$ is a contraction of Banach spaces, then there is precisely one fixed point of $T$ in $X$.

@heather So, can you solve my integral equation now?

You have the major result you need

The Dark Side

Hey hbar !! Maybe I wasn't around for sometime, so I'm out of loop. Discovered from the comments below this answer that Chris White is no longer around. Apparently he deleted his account? What happened here?

heather

7:31 PM

g(x) is just a point in that equation, right @0celo7?

so it would be a direct application of this theorem.

Bernardo Meurer

@TheDarkSide No one knows, and we shouldn't discuss it

PLEASE COME BACK CHRIS WHITE

8

The Dark Side

@BernardoMeurer I sniff something bad with "we shouldn't discuss this" !

Ahh. Search bar. I should've contemplated that earlier.

0celo7

@heather you first have to make sure that the integral operator is a contraction

and you have to select the correct Banach space

Bernardo Meurer

@TheDarkSide The mods go bananays if we speculate on this stuff

2

Hmm, my CPU design has gone rebel and refuses to do logic

The Dark Side

Yes, you are right. Just asked because I was out loop. Am diligently digging through Search bar results. Thanks :)

Bernardo Meurer

7:39 PM

@TheDarkSide No worries, good searching

heather

@0celo7 okay

The Dark Side

:)

2017

@BernardoMeurer "bananays"...that's quite an accurate description....lol :D

except that there is a spelling error :-P

Bernardo Meurer

@2017 There is absolutely no spelling error there

The only spelling error here is this proprietary malware you run

dmckee

@TheDarkSide There is nothing sinister that I'm aware of, but Chris didn't say in public why he was leaving and so we are left to guess. We have discouraged writing up speculation as if it was fact.

2017

7:45 PM

@BernardoMeurer "bananays" ?

Bernardo Meurer

@dmckee I have figured out VHDL

dmckee

Of which there has been quite a bit at times.

Bernardo Meurer

The secret is it's not meant to work. So if you just try really hard to break it then it works

dmckee

With a number of people claiming that Chris' departure proves their point. Even when those people are on opposite sides of some issue.

2017

@BernardoMeurer Aaaaaaaaaaahhhhhh, I like malware so much !

Bernardo Meurer

7:46 PM

Chris' departure clearly shows that Free Software is the only way to go forward

dmckee

@BernardoMeurer You poor, poor thing. There are support groups to help you recover. Or so I'm told.

By a ... friend.

Bernardo Meurer

@dmckee My CPU has gone rogue too. I made some mistake on the design and it has literally taken over the FPGA dev board

It keeps printing things on these little 8-seg displays, we think it's trying to say something

It's even attacking the VGA buffer

dmckee

Help! Help! I've fallen and I can't get up!*, perhaps?

Bernardo Meurer

I wrote an autoimmune malware instead of a CPU :(

The Dark Side

@dmckee I totally understand and respect the mod-stance. Sorry, raised the topic since I only noticed now. No other intention from my side too.

dmckee

7:49 PM

@TheDarkSide I quite understand your shock. I'm still trying to figure it out myself. Alas, I have no way to contact him and ask.

Bernardo Meurer

I tried emailing him but he didn't respond. I didn't attempt further contact not to bother him

dmckee

I have my own guess, of course. But sauce for the goose and all that.

The Dark Side

:)

And ofc, "bananays" (whatever that means) :)

Bernardo Meurer

@TheDarkSide @dmckee Is clearly possessed by anger right now

The Dark Side

@BernardoMeurer Lets hope he doesn't use his mod-hammer !!

2017

7:51 PM

I don't think it will be difficult to locate Chris White on the Internet if one does some thorough googling.

dmckee

Bwahahahahahahah!

::dry washes hands::

Bernardo Meurer

@2017 It's not, it's more about respecting his right not to be bothered by us orphaned children

ACuriousMind

@dmckee Uh, your mad scientist is showing

2017

@BernardoMeurer That's a completely different point.

I'm just saying...

Bernardo Meurer

@ACuriousMind Bitch please, I just made an autoimmune CPU that has destroyed not one but TWO FPGA dev boards

The Dark Side

7:54 PM

@BernardoMeurer Great accomplishment! Which side of the "wall" are you?

Bernardo Meurer

@TheDarkSide Hm? Which wall?

The Dark Side

Ahh OK. Discovered from your profile that you are in Lisbon. Never mind then, the wall is not relevant then :P

Bernardo Meurer

:^)

0celo7

@heather Hint: show that $C[0,1]$, the space of continuous functions $[0,1]\to\Bbb R$ is a Banach space with the norm $||f||=\sup_{x\in[0,1]}|f(x)|$.

heather

@Bernardo is this guy the same, or different? (there are many chris whites brought up by the google)

Bernardo Meurer

7:58 PM

@heather Different

heather

@0celo7 i'm sorry, i really don't know enough to prove it. i think i get the idea of the general proof, but i don't know enough of the specifics.

Bernardo Meurer

CW was a handsome tall lion

astro.berkeley.edu/researcher-profile/3343634-chris-white

2017

Something like this? ^

The Dark Side

^ He beat me to this!

2017

This is the tallest lion I found

Bernardo Meurer

8:00 PM

Woah! How did you find such a personal picture of him?!

2017

So he must have gone hunting for his wife and children ;) He doesn't have any time left for physics :D

The Dark Side

Just imagine what he will feel reading this transcript if he stumbles onto this someday :D

0celo7

Pretty sure it's the female lions who hunt and that CW was too socially awkward for girls.

2017

@BernardoMeurer Some thorough googling :D

uryga

8:22 PM

hi, anyone here good with Clifford algebra?

damn ok this might be better as a question on the actual board, not in chat

ACuriousMind

@uryga Depends on what you mean by that, just go ahead and ask what you actually want to ask

uryga

in clifford algebra, there's Bivectors. People keep referring to them as "oriented areas", but then draw them with a spinny thing inside: https://sureshemre.files.wordpress.com/2013/06/ga_basis_of_3d_multivector-svg.png
And i'm trying to get some intuition on why that happens and what a bivector represents

(poorly phrased, i know)

ACuriousMind

@uryga If you have two vectors $v,w$, then $v\wedge w$ represents the parallelogram they span. The "spinny thing" tells you the orientation of that parallelogram - it indicates the direction in which you'd have to rotate $v$ through that plane to get to $w$.

uryga

why is it defined that way though? what's that good for

a parallelogram full of twistiness

also, how would that work for cubes like the one in the picture?

okay nevermind, i think i found an okay resource on this. Thanks though!

AccidentalFourierTransform

8:44 PM

so there is this question with a bounty that is about to expire, and it has no answers...

is it unethical/discouraged to post an empty answer so as to get the bounty, and offer the bounty back?

ACuriousMind

@AccidentalFourierTransform Yes, that is discouraged - you'd be posting not an answer and it would be deleted as such.

The possibility that the bounty is lost if the question is not answered is part of what you should ponder before you offer a bounty.

AccidentalFourierTransform

oh well

ok then

Emilio Pisanty

9:07 PM

df?

gscholar is off its meds

0celo7

Lol, I can have an undermounted SMG on this rifle

ACuriousMind

@EmilioPisanty That or they have email in the kingdom of the dead now

0celo7

@heather The proof goes something like this: First consider $B[0,1]$, the set of bounded functions $[0,1]\to\Bbb R$. You show that this is a Banach space in the norm $||f||=\sup_{x\in[0,1]}|f(x)|$. You then show $C[0,1]\subset B[0,1]$ is closed. This is the usual proof that uniform convergence preserves continuity.

AccidentalFourierTransform

wait Turing died!?

fuck 2016

ACuriousMind

@AccidentalFourierTransform ?

heather

9:13 PM

uh...ya turing died...

ACuriousMind

He's been dead for much longer.

heather

he died in 1954

Emilio Pisanty

@ACuriousMind I'm split between finding it completely ridiculous and being miffed that they list an email at a completely unrelated university

heather

63 years ago this june

Emilio Pisanty

It shoulda been Manchester U or none at all

@ACuriousMind @heather I'm pretty sure AFT's aware of that

0celo7

9:14 PM

@heather You also have to prove that a closed linear subspace of a Banach space is a Banach space. That I'm sure you can do already

Emilio Pisanty

@0celo7 heather does Banach spaces and uniform convergence now? goodness me.

whizzkids'll overtake us all

heather

@EmilioPisanty that would be because of 0celo =) it's super interesting.

Emilio Pisanty

@heather yeah. just... don't listen to the sirens' call.

you know the Feynman thing about math and physics and so on

heather

actually, i don't.

0celo7

@EmilioPisanty Maybe you're right

@heather How about first showing that a Cauchy sequence is bounded?

heather

9:19 PM

@0celo7 sure.

0celo7

Or even simpler: show that a convergent sequence is bounded

I'll show you how do to it for convergent ones, you do it for Cauchy

heather

okay.

0celo7

Ok, so we have $x_n\to x$ in a normed vector space $X$

Can you tell me again what that means?

heather

well, a vector space is a set of vectors that is closed under addition and multiplication by scalars, and a norm is just a way of finding the distance between two elements.

a normed vector space is one which has a consistent rule that follows the three rules for norms for finding distances between any of the vectors in it.

0celo7

Good, but I meant $x_n\to x$ specifically

@heather Actually, the norm $||x||$ gives the length of $x$. The distance is $||x-y||$

heather

9:24 PM

oh, $x$ is the limit of the sequence $x_n$ where $x_n$ is of course within $X$

0celo7

@heather What does the limit mean exactly?

Emilio Pisanty

@heather the (possibly misattributed) quote goes "physics is to mathematics as sex is to masturbation"

heather

what the sequence converges too or approaches @0celo7

@EmilioPisanty strange quote, but okay

Emilio Pisanty

seriously, y'all gonna flag Feynman?

AccidentalFourierTransform

who's the prude that flagged that?

0celo7

9:25 PM

@heather the actual definition please

Emilio Pisanty

@0celo7 that's boring. go for 'every bounded sequence has a convergent subsequence'.

AccidentalFourierTransform

> Physics is like sex: sure, it may give some practical results, but that's not why we do it.

0celo7

@EmilioPisanty Not true in general Banach spaces.

Get out of here :P

@heather I want you to write it out, from memory. This isn't for me

Emilio Pisanty

@0celo7 a.k.a. the "general Banach spaces are boring" theorem =P

what is it you actually need for that one?

0celo7

@EmilioPisanty Now, it can be shown that holds iff the space is finite dimensional. That's very interesting.

Emilio Pisanty

9:28 PM

Obviously completeness is not enough

heather

okay.

0celo7

sequential compactness $\iff$ local compactness $\iff$ finite dimensional (For normed vector spaces)

Emilio Pisanty

@0celo7 only if, as well?

goodness

@0celo7 this only for Banach spaces, then?

I'm pretty sure there was a separate characterization

essentially giving the property a name

Baire's theorem, possibly?

0celo7

I think it works in general.

I'll go even further and say it works in any TVS

ACuriousMind

@EmilioPisanty Consider the unit ball - it's only compact in finite dimensions since otherwise a basis of unit vectors forms a non-converging bounded sequence.

Emilio Pisanty

9:31 PM

@0celo7 how can it work in general?

I'm pretty sure you can formulate the property in spaces where dimension doesn't make sense

0celo7

@ACuriousMind The problem is that outside of Hilbert spaces you can't get a basis of unit vectors

It requires some trickery for general normed vector spaces, and pure genius for TVS

ACuriousMind

@0celo7 Yeah, but the Hilbert spaces are the nice ones - if it doesn't work in them, no reason to expect it to work in any Banach space

0celo7

@EmilioPisanty In a general topological vector space I mean

Emilio Pisanty

@0celo7 what does "bounded" mean in a general TVS?

the natural setting for that theorem is over a metric space

heather

for all "small values" greater than zero, there exists a set $N$ in the set of natural numbers such that the sequence $n$ is greater than or equal to $N$ where the length of the sequence minus the limit of the sequence is less than the "small value".

0celo7

9:35 PM

@EmilioPisanty Bounded for a general TVS means that $U\subset X$ is absorbed by any neighborhood of $0$

i.e. for each nbhd $V$ of $0$ there is a constant $\alpha>0$ such that $\alpha U\subset V$

Emilio Pisanty

this is what I meant, I think

Sequentially compact space

In mathematics, a topological space is sequentially compact if every infinite sequence has a convergent subsequence. For general topological spaces, the notions of compactness and sequential compactness are not equivalent; they are, however, equivalent for metric spaces. A metric space X is sequentially compact if every sequence has a convergent subsequence which converges to a point in X. == Examples and properties == The space of all real numbers with the standard topology is not sequentially compact; the sequence (sn = n) for all natural numbers n is a sequence that has no convergent subsequence...

0celo7

@EmilioPisanty What exactly is your question? I know a very nice proof that "unit ball is compact" $\implies$ "$X$ is finite dimensional"

For a normed vector space of course.

You can convince yourself that the compactness of the unit ball is equivalent to the Bolzano-Weierstrass property for normed vector spaces

Emilio Pisanty

@0celo7 that's easy via its contrapositive

0celo7

Oh?

You were surprised when I said it above!

Emilio Pisanty

@0celo7 I thought better of it

0celo7

9:40 PM

Ok, how does the contrapositive do it?

Emilio Pisanty

@ACuriousMind's example springs to mind pretty quick

0celo7

@EmilioPisanty In a general normed vector space you can't use Gram-Schmidt to get an orthonormal basis.

Emilio Pisanty

@0celo7 so this is for a general normed VS?

you can't use Gram-Schmidt

but surely you can start with an infinite linearly independent sequence

0celo7

Yes, $X$ is a general normed vector space, not necessarily Banach.

Emilio Pisanty

and build a bounded sequence where all vectors are $d$ away from each other

0celo7

9:43 PM

@EmilioPisanty That's the usual proof, but I'm saying I know a very neat one

That proof gets pretty messy (the one you outlined), it contains many $\epsilon$s

@EmilioPisanty For a pre-Hilbert space, you take an orthonormal set $\{e_i\}$. If it's infinite-dimensional, this gives you a sequence. But $||e_i-e_j||=\sqrt 2$ for $i\ne j$, so the sequence has no Cauchy subsequences. Thus no subsequence can converge.

Emilio Pisanty

@0celo7 yeah, ok, for a normed space I can see that's probably true

@0celo7 yeah, that's what I meant earlier

0celo7

10:16 PM

@ACuriousMind In a TVS, is a translated open set absorbed by the original open set?

I.e. for $G\subset X$ open, consider $G_x=G+x$. Is there $\alpha>0$ such that $G_x\subset \alpha G$?

Oh, no, not true. But what if $G$ is bounded?

bolbteppa

10:55 PM

Why isn't Schwarzschild easy :(

Ben Niehoff

what's not easy about it?

bolbteppa

Remembering the Christoffel symbols, remembering the field equations :(

Takes ages to get them

Can't remember them

AccidentalFourierTransform

BTW if you guys use Mathematica, you might find this useful:

2

A: Compute covariant derivative in Mathematica

Here is some code I wrote a while back: ClearAll["Global`*"] SetAttributes[Rs, Constant] $Assumptions = Rs > 0; Coordinates = {t, r, \[Theta], \[Phi]}; dim = Length[Coordinates]; MetricTensorLL = {{(1 - Rs/r), 0, 0, 0}, {0, -(1 - Rs/r)^-1, 0, 0}, {0, 0, -r^2, 0}, {0, 0, 0, -r^2 Sin[\[Theta]]^...

Ben Niehoff

ew, I hate Christoffels

I use spin connections

bolbteppa

Even the $R_{\mu \nu} = \frac{\partial }{\partial x^{\nu}} \frac{\partial}{\partial x^{\mu}} \ln (\sqrt{-g}) - \frac{\partial \Gamma^{\sigma}_{\mu \nu}}{\partial x^{\sigma}} + \Gamma^{\sigma}_{\nu \alpha} \Gamma^{\alpha}_{\mu \sigma} - [ \frac{\partial}{\partial x^{\alpha}} \ln ( \sqrt{-g} ) ] \Gamma^{\alpha}_{\mu \nu}$ simplification doesn't save that much effort

and doing it with tetrads takes even longer

ACuriousMind

10:57 PM

@BenNiehoff Not surprising considering you're the SUGRA type ;)

Ben Niehoff

no it doesn't

first of all, since you know there is spherical symmetry, why bother writing out spherical coordinates?

bolbteppa

Last time I wrote it up it turned into an A4 page!

Ben Niehoff

ooh, one A4 page! scary

AccidentalFourierTransform

lol

bolbteppa

In mocha-chino latte land, I want the Christoffels and EFE by immediate thought :(

heather

10:59 PM

i've taken multiple pages on simple problems like multiplying matrices or whatever because i space it out and i make idiotic errors (if that makes you feel better)

Ben Niehoff

when I took my GR course in grad school, I made it a point of pride to do each (multi-part) problem on a single page :P

so, e.g., 6 problems = 6 pages

ACuriousMind

@BenNiehoff Was that a training in elegance or in small handwriting? ;P

Ben Niehoff

there was another student who was literally turning in 30-page homeworks

training in elegance

bolbteppa

I bet it was scribbled too

Ben Niehoff

oh, no, he had very neat handwriting

but was pedantic

heather

11:01 PM

also, my handwriting is terrible...i've taught myself to cross my sevens and z's so that i don't mix them up with twos or whatever, and so on. it's bad.

Ben Niehoff

haha

I have another friend whose d, 2, alpha, and partial derivative all look the same

heather

oh, that's me as well.

AccidentalFourierTransform

@heather I only use roman numerals

heather

@AccidentalFourierTransform how do you live

ACuriousMind

My handwriting in school was terrible. It's become much better since I have to write things that I might actually want to re-read later :P

AccidentalFourierTransform

11:02 PM

$\alpha\propto da$

Ben Niehoff

oh, I forgot propto symbols

heather

it's terrible - people will ask to copy notes from me when they've been gone, and they'll have to ask me to transcribe words - and sometimes i don't even know what i wrote until i stare at it for a solid five minutes.

AccidentalFourierTransform

why do people write anyway?

latex for everything

Ben Niehoff

but the biggest thing, I think, is lowercase xi...if someone has a neat xi, then you know they care about their handwriting :P

heather

because our brains are bad and not everyone has a computer

$\xi$

oh, that one

my epsilon, e, and xi probably would all look the same

ACuriousMind

11:03 PM

@AccidentalFourierTransform Are you apt enough at typing and LateX that you can live-TeX a technical lecture?

dmckee

^ That.

heather

^that too, i'm not that proficient at LaTeX

AccidentalFourierTransform

idk, perhaps

I never ever take notes anyway

heather

i can type pretty fast, but probably not fast enough, as well.

Ben Niehoff

especially with the right macros

heather

11:04 PM

@AccidentalFourierTransform uh...

Ben Niehoff

but I, too, never take notes

AccidentalFourierTransform

I really dont understand why people take notes, sincerely

Ben Niehoff

I understand it, but it generally doesn't work for me

ACuriousMind

@AccidentalFourierTransform I find that I retain things I've written with my own hand much better.

Ben Niehoff

that much is certainly true!

ACuriousMind

11:05 PM

I also understand that that's not the case for everyone

Ben Niehoff

but, if I'm writing, I'm not listening fully

AccidentalFourierTransform

well in my case, it is much more useful to spend my time paying attention to the lecture

bolbteppa

It works if you use videos because you can pause and go at your own pace instead of rushing to keep up in real time

ACuriousMind

@BenNiehoff Ah, I've heard that complaint - luckily I'm good enough at multi-tasking that writing doesn't really take that much of my attention

AccidentalFourierTransform

how many cores do you have?

heather

11:06 PM

if it's a class in say social studies, i generally take less notes - if i need specific dates or whatever, i can always check google and tests aren't very hard in middle school. but in math, I basically always take notes.

Ben Niehoff

I've heard that studies show that most people who think they are good at multi-tasking...aren't ;)

AccidentalFourierTransform

@BenNiehoff but the duck is already dead

sorry, wrong room, wrong person

ACuriousMind

@AccidentalFourierTransform Ah, yes. Multithreading is a wonderful thing :)

Ben Niehoff

honestly, though, I can't even stand to sit through a lecture anymore

thankfully I don't have to

ACuriousMind

@BenNiehoff If I noticed that I was missing bits of the lecture because I was focused on writing I'd stop taking notes - but that's never happened. The things that distract me during a lecture are usually human-shaped and sit next to me :P

Ben Niehoff

11:08 PM

humans can be very distracting

heather

i hate most school classes because they are so slow. if i learned it on my own, or if i learned it just to write a paper, i'd probably remember enough for the future and be done much more quickly.

i expect things will become quicker in highschool, though.

Ben Niehoff

it's definitely true that you retain things that you actually use

dmckee

@heather Good luck with that.

Ben Niehoff

oh, yeah, you'll probably never attend a lecture you don't think is moving too slowly ;)

dmckee

I had a few professors in college who moved so fast I needed all my concentration to follow the lecture. And a higher fraction in grad school.

heather

11:12 PM

@dmckee it won't?

dmckee

But at that level a lot of teaching is just building the framework, and the learning happens when you go to apply it to non-trivial problems.

So screaming fast lectures are not needed: the right material is more important.

@heather Well, I don't know anything about the high school you're going to attend, but you seem to be pretty quick. If you're faster than the majority of your peers then the lectures will still be slow.

heather

hmm.

ACuriousMind

@BenNiehoff That reminds me of our first semester theoretical physics course. The prof had the impression we were uncomfortable with the lecture so suddenly, out of nowhere, he pointed at a random student and told him to come down here

The prof was a nice guy, but he could have a rather intimidating vibe, and everyone started wondering what he'd do to the poor student.

AccidentalFourierTransform

the student's name? Albert Einstein.

ACuriousMind

Then he turned his back to us and just asked everyone who thought the lecture was going too quickly to raise their hand. The student was just down there to count the hands.

dmckee

11:16 PM

^ ::chuckles::

That's a slick move.

Assuming you want an honest count.

AccidentalFourierTransform

lol

ACuriousMind

Almost everyone raised their hand.

Nervous laughter ensues, prof goes on a tangent how this is "no laughing matter", more nervous laughing

Then he asks who thinks the lecture is going too slowly.

A single student raised her hand. She never lived that one down.

@dmckee Yeah, that was the intent. Though up until he announced it you could feel the suspense in the air as to what was gonna happen.

dmckee

I had Marvin Marcus (a narrow gauge celebrity in math circles) for a course in college that was essentially algebraic structures and he covered ground like you wouldn't believe.

AccidentalFourierTransform

nice story

I'll tell it as mine from now on

dmckee

I mentioned that to my high school CS teacher when I saw her next and she laughed. He'd been her masters advisor.

Ben Niehoff

11:32 PM

@ACuriousMind haha...my own PhD advisor was at times possibly over-concerned...he would often interrupt himself in lectures to ask "Questions? Comments? Concerns?" if he thought people weren't asking enough questions

P.S. the word "tetrad" makes me ill, I think. Also the way they are usually taught :\

ACuriousMind

@BenNiehoff Just say vierbein, or, better, vielbein :P

Though in one dimension you should say einbein

Ben Niehoff

I have occasionally seen einbein, zweibein, fünfbein, and elfbein

ACuriousMind

Yeah. But vielbein is just "many-legs", so it covers everything but the one-dimensional case where you just have a single leg

Ben Niehoff

I'm pretty sure I've even seen ones in theories with multiple time dimensions jokingly referred to as zuvielbein

ACuriousMind

lol

0celo7

11:38 PM

@heather See if you can follow this: Let $x_n\to x$ in a normed vector space $X$. There exists an $N\in\Bbb N$ such that for $n\ge N$, $||x_n-x||<1$. This implies that $||x_n||< 1+||x||$ for $n\ge N$. So for any $n$, we must have $$||x_n||\le \max\{1+||x||,||x_1||,||x_2||,\dotsc, ||x_{N-1}||\}:=A<\infty.$$ Thus $||x_n||\le A$ for all $n\in\Bbb N$, so $(x_n)$ is bounded.

I think Bajoran first told me that proof

@dmckee Is narrow gauge a math field?

@AccidentalFourierTransform What?

Unless every course you've taken comes straight from a book, how are you going to recall things later?

AccidentalFourierTransform

@BenNiehoff where are they properly taught? I have a very basic knowledge of them

@0celo7 I dont need to recall things later

I only need to know where to find them when I need them

wikipedia is usually good enough

Ben Niehoff

I don't know any place where moving frames are properly taught that uses the word "tetrad"

but anyway, they are often taught as though they are something dark and mysterious and seldom used