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7:00 PM
@ACuriousMind Okay, cool. I'm out of this now. Putting 0celo7 on ignore forever.
 
Oh no :(
What will I do now
I won't have anyone to talk to about JEE
 
@0celo7 You might also want to disengage now
 
let's talk about something else! um...*::grabs nearest book::*
oh, hey, i had a legitimate question about this.
 
@heather Ok, so contractions in Banach spaces
 
@0celo7 okay.
 
7:02 PM
You went on a 30 minute errand that lasted all night
 
Clearly had something to do with relativity, then
 
i came back after an hour-ish, if you'll check the transcript
but yes, i do apologize
 
Oh, I was probably playing MGSV
Gotta grind up my R&D level for that sweet automatic tranq pistol
 
i figured it was that or a lab report =)
 
no lab report this week thank god
 
7:03 PM
@heather If this doesn't seem like an attack then sorry, I disagree chat.stackexchange.com/transcript/message/36126227#36126227
 
oh, real quick: could whoever starred that comment I made saying "JEE = physics problems" unstar it?
 
@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.
 
right, yes, i remember now.
 
@heather Do you know what the image of a map is?
 
let me check my book
ah, right.
"if $\alpha : S\to T$ then $\alpha(S)$ will be called the image of $\alpha$."
 
7:06 PM
@heather Ok, so take some $S\subset X$. What can you say about $T(S)$ in relation to $S$?
 
well, T(S) is a contraction of S, the points will be closer together, right?
 
@heather Exactly. So what about $T^2(S)=T(T(S))$ vs. $T(S)$ vs. $S$?
assume that $S$ has finite diameter/size
 
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
 
exactly
so if $L$ is the map you get by applying $T$ an infinite number of times, what should $T(S)$ be?
 
well...one point, right?
a point in the center of all the points in the set, probably.
 
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?
 
@0celo7 hmm....$x_0$?
(i'm expecting something tricky to happen =P)
 
well, that is the trick
it's $x_0$
 
that makes sense.
 
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
 
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?
 
7:31 PM
g(x) is just a point in that equation, right @0celo7?
so it would be a direct application of this theorem.
 
@TheDarkSide No one knows, and we shouldn't discuss it
PLEASE COME BACK CHRIS WHITE
8
 
@BernardoMeurer I sniff something bad with "we shouldn't discuss this" !
Ahh. Search bar. I should've contemplated that earlier.
 
@heather you first have to make sure that the integral operator is a contraction
and you have to select the correct Banach space
 
@TheDarkSide The mods go bananays if we speculate on this stuff
2
Hmm, my CPU design has gone rebel and refuses to do logic
 
Yes, you are right. Just asked because I was out loop. Am diligently digging through Search bar results. Thanks :)
 
7:39 PM
@TheDarkSide No worries, good searching
 
@0celo7 okay
 
@BernardoMeurer "bananays"...that's quite an accurate description....lol :D
except that there is a spelling error :-P
 
@2017 There is absolutely no spelling error there
The only spelling error here is this proprietary malware you run
 
@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.
 
7:45 PM
@BernardoMeurer "bananays" ?
 
@dmckee I have figured out VHDL
 
Of which there has been quite a bit at times.
 
The secret is it's not meant to work. So if you just try really hard to break it then it works
 
With a number of people claiming that Chris' departure proves their point. Even when those people are on opposite sides of some issue.
 
@BernardoMeurer Aaaaaaaaaaahhhhhh, I like malware so much !
 
7:46 PM
Chris' departure clearly shows that Free Software is the only way to go forward
 
@BernardoMeurer You poor, poor thing. There are support groups to help you recover. Or so I'm told.
By a ... friend.
 
@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
 
Help! Help! I've fallen and I can't get up!*, perhaps?
 
I wrote an autoimmune malware instead of a CPU :(
 
@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.
 
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.
 
I tried emailing him but he didn't respond. I didn't attempt further contact not to bother him
 
I have my own guess, of course. But sauce for the goose and all that.
 
:)
And ofc, "bananays" (whatever that means) :)
 
@TheDarkSide @dmckee Is clearly possessed by anger right now
 
@BernardoMeurer Lets hope he doesn't use his mod-hammer !!
 
7:51 PM
I don't think it will be difficult to locate Chris White on the Internet if one does some thorough googling.
 
Bwahahahahahahah!
::dry washes hands::
 
@2017 It's not, it's more about respecting his right not to be bothered by us orphaned children
 
@dmckee Uh, your mad scientist is showing
 
@BernardoMeurer That's a completely different point.
I'm just saying...
 
@ACuriousMind Bitch please, I just made an autoimmune CPU that has destroyed not one but TWO FPGA dev boards
 
7:54 PM
@BernardoMeurer Great accomplishment! Which side of the "wall" are you?
 
@TheDarkSide Hm? Which wall?
 
Ahh OK. Discovered from your profile that you are in Lisbon. Never mind then, the wall is not relevant then :P
 
@heather Hint: show that $C[0,1]$, the space of continuous functions $[0,1]\to\Bbb R$ is a Banach space with the norm $||f||=\sup_{x\in[0,1]}|f(x)|$.
 
@Bernardo is this guy the same, or different? (there are many chris whites brought up by the google)
 
7:58 PM
@heather Different
 
@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.
 
Something like this? ^
 
^ He beat me to this!
 
This is the tallest lion I found
 
8:00 PM
Woah! How did you find such a personal picture of him?!
 
So he must have gone hunting for his wife and children ;) He doesn't have any time left for physics :D
 
Just imagine what he will feel reading this transcript if he stumbles onto this someday :D
 
Pretty sure it's the female lions who hunt and that CW was too socially awkward for girls.
 
@BernardoMeurer Some thorough googling :D
 
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
 
@uryga Depends on what you mean by that, just go ahead and ask what you actually want to ask
 
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)
 
@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 $w$.
 
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!
 
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?
 
@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.
 
oh well
ok then
 
9:07 PM
df?
gscholar is off its meds
 
Lol, I can have an undermounted SMG on this rifle
 
@EmilioPisanty That or they have email in the kingdom of the dead now
 
@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.
 
wait Turing died!?
fuck 2016
 
@AccidentalFourierTransform ?
 
9:13 PM
uh...ya turing died...
 
He's been dead for much longer.
 
he died in 1954
 
@ACuriousMind I'm split between finding it completely ridiculous and being miffed that they list an email at a completely unrelated university
 
63 years ago this june
 
It shoulda been Manchester U or none at all
@ACuriousMind @heather I'm pretty sure AFT's aware of that
 
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
 
@0celo7 heather does Banach spaces and uniform convergence now? goodness me.
whizzkids'll overtake us all
 
@EmilioPisanty that would be because of 0celo =) it's super interesting.
 
@heather yeah. just... don't listen to the sirens' call.
you know the Feynman thing about math and physics and so on
 
actually, i don't.
 
@EmilioPisanty Maybe you're right
@heather How about first showing that a Cauchy sequence is bounded?
 
9:19 PM
@0celo7 sure.
 
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
 
okay.
 
Ok, so we have $x_n\to x$ in a normed vector space $X$
Can you tell me again what that means?
 
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.
 
Good, but I meant $x_n\to x$ specifically
@heather Actually, the norm $||x||$ gives the length of $x$. The distance is $||x-y||$
 
9:24 PM
oh, $x$ is the limit of the sequence $x_n$ where $x_n$ is of course within $X$
 
@heather What does the limit mean exactly?
 
@heather the (possibly misattributed) quote goes "physics is to mathematics as sex is to masturbation"
 
what the sequence converges too or approaches @0celo7
@EmilioPisanty strange quote, but okay
 
seriously, y'all gonna flag Feynman?
 
who's the prude that flagged that?
 
9:25 PM
@heather the actual definition please
 
@0celo7 that's boring. go for 'every bounded sequence has a convergent subsequence'.
 
> Physics is like sex: sure, it may give some practical results, but that's not why we do it.
 
@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
 
@0celo7 a.k.a. the "general Banach spaces are boring" theorem =P
what is it you actually need for that one?
 
@EmilioPisanty Now, it can be shown that holds iff the space is finite dimensional. That's very interesting.
 
9:28 PM
Obviously completeness is not enough
 
okay.
 
sequential compactness $\iff$ local compactness $\iff$ finite dimensional (For normed vector spaces)
 
@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?
 
I think it works in general.
I'll go even further and say it works in any TVS
 
@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.
 
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
 
@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
 
@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
 
@EmilioPisanty In a general topological vector space I mean
 
@0celo7 what does "bounded" mean in a general TVS?
the natural setting for that theorem is over a metric space
 
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".
 
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 $\alpha U\subset V$
 
this is what I meant, I think
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...
 
@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
 
@0celo7 that's easy via its contrapositive
 
Oh?
You were surprised when I said it above!
 
@0celo7 I thought better of it
 
9:40 PM
Ok, how does the contrapositive do it?
 
@ACuriousMind's example springs to mind pretty quick
 
@EmilioPisanty In a general normed vector space you can't use Gram-Schmidt to get an orthonormal basis.
 
@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
 
Yes, $X$ is a general normed vector space, not necessarily Banach.
 
and build a bounded sequence where all vectors are $d$ away from each other
 
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.
 
@0celo7 yeah, ok, for a normed space I can see that's probably true
@0celo7 yeah, that's what I meant earlier
 
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?
 
10:55 PM
Why isn't Schwarzschild easy :(
 
what's not easy about it?
 
Remembering the Christoffel symbols, remembering the field equations :(
Takes ages to get them
Can't remember them
 
BTW if you guys use Mathematica, you might find this useful:
2
A: Compute covariant derivative in Mathematica

AccidentalFourierTransformHere 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]]^...

 
ew, I hate Christoffels
I use spin connections
 
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
 
10:57 PM
@BenNiehoff Not surprising considering you're the SUGRA type ;)
 
no it doesn't
first of all, since you know there is spherical symmetry, why bother writing out spherical coordinates?
 
Last time I wrote it up it turned into an A4 page!
 
ooh, one A4 page! scary
 
In mocha-chino latte land, I want the Christoffels and EFE by immediate thought :(
 
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)
 
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
 
@BenNiehoff Was that a training in elegance or in small handwriting? ;P
 
there was another student who was literally turning in 30-page homeworks
training in elegance
 
I bet it was scribbled too
 
oh, no, he had very neat handwriting
but was pedantic
 
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.
 
haha
I have another friend whose d, 2, alpha, and partial derivative all look the same
 
oh, that's me as well.
 
@heather I only use roman numerals
 
@AccidentalFourierTransform how do you live
 
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
 
11:02 PM
$\alpha\propto da$
 
oh, I forgot propto symbols
 
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.
 
why do people write anyway?
latex for everything
 
but the biggest thing, I think, is lowercase xi...if someone has a neat xi, then you know they care about their handwriting :P
 
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
 
11:03 PM
@AccidentalFourierTransform Are you apt enough at typing and LateX that you can live-TeX a technical lecture?
 
^ That.
 
^that too, i'm not that proficient at LaTeX
 
idk, perhaps
I never ever take notes anyway
 
i can type pretty fast, but probably not fast enough, as well.
 
especially with the right macros
 
11:04 PM
@AccidentalFourierTransform uh...
 
but I, too, never take notes
 
I really dont understand why people take notes, sincerely
 
I understand it, but it generally doesn't work for me
 
@AccidentalFourierTransform I find that I retain things I've written with my own hand much better.
 
that much is certainly true!
 
11:05 PM
I also understand that that's not the case for everyone
 
but, if I'm writing, I'm not listening fully
 
well in my case, it is much more useful to spend my time paying attention to the lecture
 
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
 
@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
 
how many cores do you have?
 
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.
 
I've heard that studies show that most people who think they are good at multi-tasking...aren't ;)
 
@BenNiehoff but the duck is already dead
sorry, wrong room, wrong person
 
@AccidentalFourierTransform Ah, yes. Multithreading is a wonderful thing :)
 
honestly, though, I can't even stand to sit through a lecture anymore
thankfully I don't have to
 
@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
 
11:08 PM
humans can be very distracting
 
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.
 
it's definitely true that you retain things that you actually use
 
@heather Good luck with that.
 
oh, yeah, you'll probably never attend a lecture you don't think is moving too slowly ;)
 
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.
 
11:12 PM
@dmckee it won't?
 
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.
 
hmm.
 
@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.
 
the student's name? Albert Einstein.
 
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.
 
11:16 PM
^ ::chuckles::
That's a slick move.
Assuming you want an honest count.
 
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.
 
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.
 
nice story
I'll tell it as mine from now on
 
I mentioned that to my high school CS teacher when I saw her next and she laughed. He'd been her masters advisor.
 
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 :\
 
@BenNiehoff Just say vierbein, or, better, vielbein :P
Though in one dimension you should say einbein
 
I have occasionally seen einbein, zweibein, fünfbein, and elfbein
 
Yeah. But vielbein is just "many-legs", so it covers everything but the one-dimensional case where you just have a single leg
 
I'm pretty sure I've even seen ones in theories with multiple time dimensions jokingly referred to as zuvielbein
 
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?
 
@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
 
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
 
that is exactly how I feel about them :-P
 
and people have this nasty habit of writing them as e_\mu^a, which just forces you to write twice as many indices compared to using coordinate bases
which of course just turns people off
no-one wants RSI from writing so many indices
 
you really loathe indices, dont you
 
11:51 PM
yep :)
 
Ben fuck-indices Niehoff
DeWitt was a great guy
 
isn't he still?
 
anyway, we usually just call them "frames" in SUGRA, rather than inventing fancy mysterious Greek or German names
and the concept is pretty simple
every (finite-dimensional) vector space has a basis
the tangent space over a point is a vector space
so, choose a basis at every tangent space in a continuous way in some open patch of your manifold
thoses basis vectors are a frame
and can be used as a basis for tensors as well, of course
if you have a metric tensor, you can express it in your frame basis
usually it's convenient to choose a basis which is orthonormal
then your metric in that basis is just \delta_{ab}
 
You might either like this answer of mine or choose to add your own.
 

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