01:00 - 08:0008:00 - 23:00

1:00 AM
Considering a metric $ds^2=dt^2-(1/t)dx^2$ on $R^2$.
Why is this metric _not defined_ at $t=0$ instead of just being non-smooth (non-continuous) and non-degenerate at $t=0$?

Or in other words, what is the difference (regarding metrics) between _not defined_ in a point and _not continuous_ in a point? I assumed a diverging metric coefficient means that the metric is considered not continuous (in that point).

1:10 AM
Hi, can someone ping me? Testing sound alert

@CalvinKhor

dang I heard nothing. Thanks @user2103480

You're welcome

@BalarkaSen This is the cheating way
@BalarkaSen Like this

Really? @CalvinKhor

1:13 AM
Naw you have to reply to some earlier message

@CalvinKhor oh I see
sick, tyty

Copy message permalink and put a : in front of the numbers - yeah
There was a time when a bunch of us tried to ping future messages

Lmao

But like, not every number appears because the indexing is done network-wide

the numbers are increasing but the forward differences are weird
yeah ok that makes sense

1:17 AM
And you won't end up pinging messages in other chatrooms ofc
Yeah
Let's see
:54889417 ABC
Blah
Some fucker said something in that time span

lol
Next question then is whats the maximum number of messages I’m allowed to send per second

and who bans me if I spam in this chat .....hypothetically of course

lol just 2?? Or did you stop

1:20 AM
But the chat shows you to retry a message in 1 second I think
A
Yeah it just did
"Try posting the messages in 2 seconds" or something like this
A
B

hmm ok. So you’re gonna need insane coordination in a cyber cafe or something

"- You can perform this action again in 5 seconds. - retry / edit / cancel"
That's what it said

rip dreams

Oh did I tell you about the time we brought two chatbots and tried to loop them
So they keep spamming and the chat crashes

lmao
i didn’t think they allowed chat bots

1:22 AM
yeah there's a bunch of them out there
Ah there's also this
@CalvinKhor How do you denote real numbers when you write it on the blackboard? Like, what style do you use? $\renewcommand{\Bbb}{\mathscr}$

$\mathbb R$
as my forefathers wrote

Hm, I'm confused
$\renewcommand{\Bbb}{\mathscr}$ $\Bbb{R}$
$\Bbb{R}$
Oh
@CalvinKhor $\renewcommand{\mathbb}{\mathscr}$ Try again
Tell me what style it is again

$\text{I think this is not affected due to scope}\mathbb R$

Haha rekt

RIP lol

1:28 AM
No blackboard bolds in my town
I renewcommanded \Bbb first time not \mathbb

Oh lol

Let's fix this. $\renewcommand{\Bbb}{\Bbb}$ $\renewcommand{\mathbb}{\mathbb}$
$\Bbb{R}$, $\mathbb R$, $\mathscr R$

$\renewcommand{\sin}{\cos}$ please write $\cos^2$ in terms of sines

Yikes
How do I fix

lmao

1:30 AM
Damnit I forgot

Well you didn’t keep it in a temp var
i think its gone until we push the message off screen and refresh

Yeah
Shit

lol

$\renewcommand{\Bbb}{\mathscr}$ $\Bbb{R}$
$\renewcommand{\Bbb}{\mathbb}$ $\renewcommand{\mathbb}{\Bbb}$
$\Bbb R$
Ah but I changed both commands
Of course, I am an idiot

Yup hahahaha
$\newcommand{\BbbR}{|\!\text{R}}$ behold $\BbbR$

1:32 AM
LMFAO

dont star these messages they’ll mess up chat forever

OH fuck
I am in deep trouble

Looooool
i mean I guess someone in power can just delete the message if we really need it

Yeah we can shift it to another room
@Alessandro Please shift these renewcommand messages to trash

in the main site this isn’t possible anymore because they made the definitions apply in scope only

1:35 AM
haha yeah
Uhhh I guess its my responsibility to talk about something so these messages go away
What math are we upto
Oh but shouldnt talk math when the Bbb command is fucked

$\Bbb R$
Lmao
$\mathbb R$

Yeah I fucked both of these

Apparently the Euler equations formally pop out as geodesics on a certain infinite dimensional manifold

Euler-Lagrange?

No, motion of perfect fluids

1:44 AM
Oh I dont know this. What's the eqn

In cartesian at least $\partial_t u+ (u\cdot \nabla)u = -\nabla p$

Oh man
Imagine if we replace $\nabla$ by $\Delta$

LOL

I don't even use TeX when I read chat so it doesn't bother me

@CalvinKhor ah ok

1:47 AM
im trying to get why this $\nabla p$ is a lagrange multiplier, I guess I formally see it but not sure how rigourous it can become

@MikeMiller I think it just gets fixed if the problematic messages are pushed away from the screen
thats a weird equation mate

The LHS is the transport of u along u itself
the RHS makes sure that $u$ remains measure preserving
oh I forgot to add that $\operatorname{div} u = 0$

Hmmm

Tao has written the equation with more diff geom notation but I’m not fluent enough to do it myself

yeah some Onsager conjecture thing

1:53 AM
Yeah that collection of notes

I read a bit of it while trying to understand Nash's isometric embedding theorem

i sat through a few mini classes of onsager conjecture type results, still not super clear
the calculations are quite heavy
nash is a tiny bit easier

Oh you know that stuff

i’ll dare to claim I know that stuff when I publish a paper on it

You gotta teach me
I don't understand the theorem

1:57 AM
the statement?

Yeah I sort of know the statement, but I don't know any analysis - you seem to be an analyst so maybe you can tell me what the hard part of the theorem is

i never formally learned diff geom so I kind of handwave through all the geometry setup, when i get to some equations i start paying more attention lol

I seem to remember you require some technical lemma but I can't remember exactly where. If you want to bear with me on the geometry I can maybe tell you exactly how much is clear to me

iirc you create a bunch of short embeddings together with a way to reduce the error in C^1

that rings right to me

2:00 AM
the price you pay is that C^2 blows up

yeah has to happen

hello there

I think the technical lemme is probably the way to reduce the error
lemma*
the rest of it is carefully tuning parameters

can some one recommend me a PDE book that contains analysis of equation of a vibrating string, heat equation, Laplace equation and their solutions

Yikes more PDE people
What is it with this chat today

2:01 AM
errr. If you see Evans, run the other way cuz u want something simpler @Sonal_sqrt that’s all i got for you
or maybe it is in Evans.
Not sure. Anyways Evans is a classical reference

Alright thanks
Chapter 2 of evans should do it

@CalvinKhor So here's how I conceptualize things. I have a Riemannian manifold $(M, g)$ and I want to isometrically embed in $\Bbb R^n$. We can Whitney embed $M$ inside a massive torus and extend the metric to the torus, so suffices to isometrically embed any metric on $T^m$ to $\Bbb R^n$.

What notes i have scribbled on this are in my office which are out of bounds atm @BalarkaSen lol but maybe i can peek at a paper somewhere

Grumble I used bb

lol

2:07 AM
Let Sym be the space of symmetric tensors on the torus, inside of which Riemannian metrics are the positive definite guys, a cone.

@Thorgott If addition is smooth then inversion is as well: since $+$ is transverse to 0 at (0,0), it's also transverse to 0 at $(a,-a)$ (the map $(+a, -a)$ identifies the map near (0,0) with the map near $(a,-a)$). Then you see that $\{(a, -a)\}$ is a smooth manifold. Similarly you check that projection onto each coordinate is smooth, and in particular inversion is a smooth map.
So your $\Bbb R^4$ is an abelian Lie group, hence isomorphic as a Lie group to the usual R^4. (The identity map in fact is clearly an isomorphism!)

who ruined \Bbb

Lol

:(
gulp

so it begins
you could renew\Bbb as \mathbf for now maybe

2:09 AM
In general any contractible Lie group is diffeomorphic to $\Bbb R^n$: use the KAN decomposition (which provides a global diffeomorphism to A x N, where A is Abelian and N is nilpotents), then use that the exponential map is a diffeomorphism for nilpotent Lie groups.

You are lucky Knuth is not dead yet, Balarka. Otherwise you'd become his latest haunt.

@CalvinKhor Let's sit tight for a few more messages
We
can
push
them
out

is it due to vertical height or number of messages

of
sight
Oh still not fixed
WHAT

$$\binom{\binom{\binom{\binom{a}{\binom{a}{\binom{a}{\binom{a}{\binom{a}{b}}}}}}{b}}{b}}{b}$$

2:11 AM
$\Bbb$

aww didn’t compile lol

But we pushed em out man

Yay lol
also It did compile actually and now its horrible

$\Bbb$
HOW IS IT NOT FIXED

$\Bbb R$
Works for me

2:12 AM
Doesn't for me

Ah it does
$\Bbb{R}$
$\mathbb{R}$

Finally.

Whew

2:13 AM
Requiem.

I shat myself for a solid minute

I’m gonna skim this arxiv.org/pdf/1609.03180.pdf i think it covers it...if it doesn’t apologies but again not at a computer lol @BalarkaSen

How did you ruin it?

Oooh Calvin is a PDE guy.
Need more of you around here.

@Knight I’m gonna reply u here because I can’t open the other chat for some reason this is pushing my iPad skills to the limit lol. Phone is an iPhone 7

2:18 AM
@CalvinKhor OK here is how the story goes. $\text{Sym}(M)$ be space of symmetric tensors. There's a map $C^\infty(M, \Bbb R^n) \to \text{Sym}(M)$ by pulling back the Riemannian metric on $\Bbb R^n$, where $n$ can be anything. The image of $\text{Imm}(M, \Bbb R^n)$ by this map is $\text{Good}(M)$, space of good metrics on $\text{Sym}(M)$ (these have to be Riemannian metrics because you're pulling back by immersions).

@anakhro i forgot what a ring is (not really but that’s close to what algebra i remember)

Are you sure they aren't great metrics
It's okay to forget definitions.
Don't be ashamed about forgetting things you don't regularly use.

the implication is the complete lack of knowledge of the theory lol

Oh who cares it's readable either way

@MikeMiller if you’re on about \mathbb R$then refresh it should work 2:21 AM It suffices to prove for every Riemannian metric$g$on$M$,$g \in \text{Good}(M)$to prove Nash's theorem. Indeed, take a Whitney embedding$f : M \to \Bbb R^n$, rescale so that$g' := f^* g_{\text{Euc}} < g$. Then$g - g'$is a Riemannian metric on$M$, hence good, and thus$g - g' = i^* g_{\text{Euc}}$where$i : M \to \Bbb R^m$is some immersion Then$f \oplus i : M \to \Bbb R^n \oplus \Bbb R^m$is the embedding such that$g = (f \oplus i)^* g_{\text{Euc}}$The easy half of Nash's theorem is Nash's approximation theorem, which says every Riemannian metric on$M$is approximately good. That is, for any metric$g$there's a symmetric tensor$h$for which$g + \varepsilon h$is good for all$\varepsilon > 0$@CalvinKhor I don't even have the TeX thing available, I just read it without compiling. I don't exactly remember what the content of the hard part is. @MikeMiller ahh ok sure Ah, here's the point. The easy half says (a little more than)$\text{Good}(M) \subset \text{Sym}(M)$is dense. It is NOT open, you can perturb a metric which comes from an immersion to a Euclidean space very little so that it no longer comes from an immersion. I think that’s prop 4.1 of the arxiv paper i linked (up to change of notation) but you need the quantitative bounds 2:28 AM Yeah I'm just trying to recall where exactly analysis enters. Nash's approximation theorem is almost trivial I think I don't think I need do any analysis to prove that Sure i can believe it trivial analysis the softest analysis Cashmere analysis lol @CalvinKhor Here's the heart of Nash's approximation theorem, I think. Suppose you wanted to approximate the quadratic differential$\eta(x) dy^2$on$\Bbb R^2$by good metrics. Consider the quickspiral map$F : \Bbb R^2 \to \Bbb R^2$,$(x, y) \mapsto (\eta(x) \cos(cy)/c, \eta(x) \sin(cy)/c)$yes things like this 2:32 AM$F^*(dx^2 + dy^2) = \eta'(x)^2/c^2 dx^2 + \eta(x)^2 dy^2$and let$c \to \infty$assuming$\eta$has bounded derivative Yeah And so it's just termwise approximation of the metric by quickspirals yeah sure I guess this is how it enters: Call a map$f : M \to \Bbb R^n$free if$\partial_i f_p, \partial_i \partial_j f_p$are all linearly independent vectors. Of course this forces$n \geq m(m+1)/2 + m$or something Let's call image of$\text{Free}(M, \Bbb R^n) \to \text{Sym}(M)$,$f \mapsto f^* g_{\text{Euc}}$to be$\text{Strong}(M)$, the strong metrics. The strong ones out of the good ones are stable, in the sense that if$g$is a strong metric and$h$is any$C^\infty$-small symmetric tensor then$g + h$is good. This is the Gromov-Rokhlin lemma Gromov-Rokhlin + Nash approximation proves Nash I believe. So the analysis must be in Gromov-Rokhlin Kik lol lmao I dont know how to prove Gromov-Rokhlin never heard of it, and I didn’t think you could compartmentalise that part of the theorem as a separate result 2:38 AM free maps are mysterious to me yeah Gromov has tried time and again to rewrite Nash's proof So....there’s a way to do it without quantitative estimates? does it also get the result that you can C^0 embed into an arbitrarily small ball I am trying to look around for a proof of Gromov-Rokhlin kk i need to learn enough algebra so that if i end up drafted for war i can pull a Leray and pivot to something with no applications Lmao Oh come on Leray's stuff has nice applications Yes i use Leray’s work all the time 2:42 AM I doubt he phrased his spectral sequences in terms of pure homological algebra That was Serre Oh that work Ch 5 Pure analysis, looks completely foreign to me Haha trust you man im just not gonna get any of it Oo lol “Who at least knows what a vector field is” That's why they do this stuff on the torus instead, to do Fourier analysis something man Fourier analysis is just doing cool shit with characters (not that I know what a character is) 2:46 AM lmao i dont know either viewpoints, analytic or rep stuff hahaha This doesn’t look easier than the arxiv paper i linked Haha Also this doesn’t get the C^1 result right maybe I’m skimming it wrong anyway i need to prep lunch, cya non-PDE dudes @MikeMiller Also thanks for the comment. Should have noticed multiplication by$-1$is automatically smooth; also nice that you dont have to use the full power of uniqueness of analytic structure on Lie groups Just suffices to note abelian Lie group Yeah, the KAN thing takes more work than I'd like 2:52 AM Cya @Calvin. Oh yeah I don't really know the$C^1$detail I know regularity is somehow lost when you do h-principles but I dont know how that happens in this case Most references only prove that the KAN map is a homeomorphism (== continuous bijection) Hah you're linking to Khang's notes. Good guy Oh you know him? Oh he's from UCLA But a Lie group has a maximal compact, so if you're contractible that has to be trivial, which forces your manifold to be diffeomorphic to$\Bbb R^n$I think the proof goes like,$G/K$admits a constant negative curvature metric and is simply connected so Cartan-Hadamard gives what you want Completely smooth proof Does it? I didn't know that I actually don't know how It feels intuitive Yeah I should compute this not constant negative I just meant negative or nonpositive 3:16 AM I guess I should compute the curvature of a left-invariant metric on$G$and then use this formula to get it on$G/K$I'll$K$-average it out on$G$first, so the metric actually descends to$G/K$. And the point must be that fibers are positively curved, because on$K$this guy is now bi-invariant I have no idea how to compute curvature of a left-invariant metric 3:57 AM In partic 2.4 seems to say this is not a fruitful line of attack @MikeMiller Oh, interesting. But are you sure those lines of positive Ricci curvatures aren't along the cosets of$K$? Then they get killed in$G/K$nilpotent lie groups are contractible K = 1 Oh because KAN, N = Nilpotent Forgetting what words mean Yeah, I really think the fruitful thing to do is figure out why N has diffeomorphic exponential map (I don't know why) But google tells me G/K is indeed nonpositively curved I don't know why at all Feels like I should be able to compute Confused. I guess I'll give up 4:15 AM @BalarkaSen But not by a left-invariant metric I guess? Yeah your example seems to say that just meant i have no approaches left (punpun) I bet it's really negative scalar curvature that you find Everything should be like SL_n(R)/O(n) = H^n man The curvy stuff should be inside the maximal guy and when you crush it you only have curving out I don't get it but also im blatantly announcing how naive my mathematical thinking process is lmao but think of your career! LMFAO Yeah I should say by deep computation I have convinced myself it should be true But I think something is wrong because I am using Ad-invariance of the little ad "ad invariance of the little ad" is the catchiest meme line i have made @MikeMiller how do you think of the scalar curvature dude i have no mental picture for it Ricci sort of makes sense because its average sectional in a direction, and basically says how fast a geodesic ball grows in that specific direction? what is scalar wikipedia says it's the simplest curvature invariant of a Riemannian manifold. even wikipedia understands this better than me 4:30 AM @BalarkaSen It's the integral over$\text{Gr}(2,n)$of$\text{sec}$hmm, ah so i guess how fast the geodesic ball grows in ALL directions in average of course, that's what tr Ric should mean no idea lmao yeah whatever man 5:26 AM What choice of topology makes [0, 1) homeomorphic to S^1 as mentioned in this answer (math.stackexchange.com/a/495940)? just transport the topology along the bijection @LeakyNun Umm...what does that mean? Do you have some reference? I understand the idea of quotient topology though do you know that [0,1) bijects with S^1? then that bijection allows you to view them as "the same set" so you can transport the topology on S^1 to a topology [0,1) then they will be homeomorphic In mathematics, particularly in universal algebra and category theory, transport of structure refers to the process whereby a mathematical object acquires a new structure and its canonical definitions, as a result of being isomorphic to (or otherwise identified with) another object with a pre-existing structure. Definitions by transport of structure are regarded as canonical. Since mathematical structures are often defined in reference to an underlying space, many examples of transport of structure involve spaces and mappings between them. For example, if V... @LeakyNun Yes, I get that [0, 1) and$S^1$are in bijection as sets @LeakyNun I see, I never heard of this concept then let$f:[0,1) \to S^1$be the bijection you can say that$U \subseteq [0,1)$is open iff$f(U)$is open this is how you transport a topology they're essentially "the same set" 5:40 AM Oh, as homeomorphisms are continuous (pre-images of open sets are open) it makes sense to define the topology like that on [0, 1) Interesting! 2 hours later… 7:17 AM @Croissant Thank you very much again. I hope that my question not is closed again @CalvinKhor Best regards. Hi, I'm new here and not sure if this is the right place , but still, ... Does anyone know book from which this question has been taken ? 16 I have this difficult integral to solve. $$\dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx}$$ Now my approach is this: split$(\sin x)^{\sqrt 2 + 1}$and$(\sin x)^{\sqrt 2 - 1}$as$(\sin x)^{\sqrt 2}.(\sin x)$and$(\sin x)^{\sqrt 2 - 2}.(\sin x...

I have been told it's from a book...

7:41 AM
Is there any kind of book like number theory for physicsits? I do know a little complex analysis
But would love to learn number theory

7:59 AM
@MoreAnonymous
I know one book
elementary number theory
by

01:00 - 08:0008:00 - 23:00