Mathematics - 2020-07-09 (page 1 of 2)

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).

Calvin Khor

1:10 AM

Hi, can someone ping me? Testing sound alert

user2103480

@CalvinKhor

Calvin Khor

dang I heard nothing. Thanks @user2103480

user2103480

You're welcome

Balarka Sen

@BalarkaSen This is the cheating way

@BalarkaSen Like this

Calvin Khor

Really? @CalvinKhor

Balarka Sen

1:13 AM

Naw you have to reply to some earlier message

Calvin Khor

@CalvinKhor oh I see

sick, tyty

Balarka Sen

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

Calvin Khor

Lmao

Balarka Sen

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

Calvin Khor

the numbers are increasing but the forward differences are weird

yeah ok that makes sense

Balarka Sen

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

Calvin Khor

lol

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

Balarka Sen

A

Calvin Khor

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

Balarka Sen

B

Calvin Khor

lol just 2?? Or did you stop

Balarka Sen

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

Calvin Khor

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

Balarka Sen

"- You can perform this action again in 5 seconds. - retry / edit / cancel"

That's what it said

Calvin Khor

rip dreams

Balarka Sen

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

Calvin Khor

lmao

i didn’t think they allowed chat bots

Balarka Sen

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? $\renewcommand{\Bbb}{\mathscr}$

Calvin Khor

$\mathbb R$

as my forefathers wrote

Balarka Sen

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

Calvin Khor

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

Balarka Sen

Haha rekt

Calvin Khor

RIP lol

Balarka Sen

1:28 AM

No blackboard bolds in my town

I renewcommanded \Bbb first time not \mathbb

Calvin Khor

Oh lol

Balarka Sen

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

$\Bbb{R}$, $\mathbb R$, $\mathscr R$

Calvin Khor

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

Balarka Sen

Yikes

How do I fix

Calvin Khor

lmao

Balarka Sen

1:30 AM

Damnit I forgot

Calvin Khor

Well you didn’t keep it in a temp var

i think its gone until we push the message off screen and refresh

Balarka Sen

Yeah

Shit

Calvin Khor

lol

Balarka Sen

$\renewcommand{\Bbb}{\mathscr}$ $\Bbb{R}$

$\renewcommand{\Bbb}{\mathbb}$ $\renewcommand{\mathbb}{\Bbb}$

$\Bbb R$

Ah but I changed both commands

Of course, I am an idiot

Calvin Khor

Yup hahahaha

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

Balarka Sen

1:32 AM

LMFAO

Calvin Khor

dont star these messages they’ll mess up chat forever

Balarka Sen

OH fuck

I am in deep trouble

Calvin Khor

Looooool

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

Balarka Sen

Yeah we can shift it to another room

@Alessandro Please shift these renewcommand messages to trash

Calvin Khor

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

Balarka Sen

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

Mike Miller

$\Bbb R$

Lmao

$\mathbb R$

Balarka Sen

Yeah I fucked both of these

Calvin Khor

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

Balarka Sen

Euler-Lagrange?

Calvin Khor

No, motion of perfect fluids

Balarka Sen

1:44 AM

Oh I dont know this. What's the eqn

Calvin Khor

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

Balarka Sen

Oh man

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

Calvin Khor

LOL

Mike Miller

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

Balarka Sen

@CalvinKhor ah ok

Calvin Khor

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

Balarka Sen

@MikeMiller I think it just gets fixed if the problematic messages are pushed away from the screen

thats a weird equation mate

Calvin Khor

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 $\operatorname{div} u = 0$

Balarka Sen

Hmmm

I buy this

Calvin Khor

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

Balarka Sen

yeah some Onsager conjecture thing

Calvin Khor

1:53 AM

Yeah that collection of notes

Balarka Sen

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

Calvin Khor

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

Balarka Sen

Oh you know that stuff

Calvin Khor

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

Balarka Sen

You gotta teach me

I don't understand the theorem

Calvin Khor

1:57 AM

the statement?

Balarka Sen

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

Calvin Khor

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

Balarka Sen

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

Calvin Khor

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

Balarka Sen

that rings right to me

Calvin Khor

2:00 AM

the price you pay is that C^2 blows up

Balarka Sen

yeah has to happen

Sonal_sqrt

hello there

Calvin Khor

I think the technical lemme is probably the way to reduce the error

lemma*

the rest of it is carefully tuning parameters

Sonal_sqrt

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

Balarka Sen

Yikes more PDE people

What is it with this chat today

Calvin Khor

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

Sonal_sqrt

Alright thanks

Chapter 2 of evans should do it

Balarka Sen

@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 $\Bbb R^n$.

Calvin Khor

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

Balarka Sen

Grumble I used bb

Calvin Khor

lol

Balarka Sen

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.

Mike Miller

@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!)

anakhro

who ruined \Bbb

Calvin Khor

Lol

Balarka Sen

:(

gulp

Calvin Khor

so it begins

you could renew\Bbb as \mathbf for now maybe

Mike Miller

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.

anakhro

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

Balarka Sen

@CalvinKhor Let's sit tight for a few more messages

We

can

push

them

out

Calvin Khor

is it due to vertical height or number of messages

Balarka Sen

of

sight

Oh still not fixed

WHAT

Calvin Khor

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

Balarka Sen

2:11 AM

$\Bbb$

Calvin Khor

aww didn’t compile lol

Balarka Sen

But we pushed em out man

Calvin Khor

Yay lol

also It did compile actually and now its horrible

Balarka Sen

$\Bbb$

HOW IS IT NOT FIXED

Calvin Khor

$\Bbb R$

Works for me

Balarka Sen

2:12 AM

Doesn't for me

Calvin Khor

would upload a screenshot but iPads hard enough to type on

Balarka Sen

Ah it does

$\Bbb{R}$

$\mathbb{R}$

Finally.

Whew

2:13 AM

Requiem.

Balarka Sen

I shat myself for a solid minute

Calvin Khor

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

NoName

How did you ruin it?

anakhro

Oooh Calvin is a PDE guy.

Need more of you around here.

Calvin Khor

@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

Balarka Sen

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).

Calvin Khor

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

anakhro

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.

Calvin Khor

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

Mike Miller

Oh who cares it's readable either way

Calvin Khor

@MikeMiller if you’re on about $

$\mathbb R$ then refresh it should work

Balarka Sen

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 $\varepsilon > 0$

Mike Miller

@CalvinKhor I don't even have the TeX thing available, I just read it without compiling.

Balarka Sen

I don't exactly remember what the content of the hard part is.

Calvin Khor

@MikeMiller ahh ok sure

Balarka Sen

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.

Calvin Khor

I think that’s prop 4.1 of the arxiv paper i linked (up to change of notation)

but you need the quantitative bounds

Balarka Sen

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

Calvin Khor

Sure

i can believe it

anakhro

trivial analysis

the softest analysis

Cashmere analysis

Calvin Khor

lol

Balarka Sen

@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 $F : \Bbb R^2 \to \Bbb R^2$, $(x, y) \mapsto (\eta(x) \cos(cy)/c, \eta(x) \sin(cy)/c)$

Calvin Khor

yes things like this

Balarka Sen

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

Calvin Khor

yeah sure

Balarka Sen

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

Calvin Khor

Kik

lol

Balarka Sen

lmao

I dont know how to prove Gromov-Rokhlin

Calvin Khor

never heard of it, and I didn’t think you could compartmentalise that part of the theorem as a separate result

Balarka Sen

2:38 AM

free maps are mysterious to me

yeah Gromov has tried time and again to rewrite Nash's proof

Calvin Khor

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

Balarka Sen

I am trying to look around for a proof of Gromov-Rokhlin

Calvin Khor

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

Balarka Sen

Lmao

Oh come on Leray's stuff has nice applications

Calvin Khor

Yes i use Leray’s work all the time

Balarka Sen

2:42 AM

I doubt he phrased his spectral sequences in terms of pure homological algebra

That was Serre

Calvin Khor

Oh that work

Balarka Sen

math.ucla.edu/~hmkhang24/expository/nash.pdf

Ch 5

Pure analysis, looks completely foreign to me

Calvin Khor

Haha trust you man im just not gonna get any of it

Oo lol

“Who at least knows what a vector field is”

Balarka Sen

That's why they do this stuff on the torus instead, to do Fourier analysis

something man

Calvin Khor

Fourier analysis is just doing cool shit with characters (not that I know what a character is)

Balarka Sen

2:46 AM

lmao

i dont know either viewpoints, analytic or rep stuff

Calvin Khor

hahaha

This doesn’t look easier than the arxiv paper i linked

Balarka Sen

Haha

Calvin Khor

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

Balarka Sen

@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

Mike Miller

Yeah, the KAN thing takes more work than I'd like

Balarka Sen

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

Mike Miller

Most references only prove that the KAN map is a homeomorphism (== continuous bijection)

Hah you're linking to Khang's notes. Good guy

Balarka Sen

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

Mike Miller

Does it? I didn't know that

Balarka Sen

I actually don't know how

It feels intuitive

Yeah I should compute this

not constant negative I just meant negative

or nonpositive

Balarka Sen

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

Mike Miller

3:57 AM

@BalarkaSen core.ac.uk/download/pdf/82428733.pdf

In partic 2.4 seems to say this is not a fruitful line of attack

Balarka Sen

@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 $G/K$

Mike Miller

nilpotent lie groups are contractible

K = 1

Balarka Sen

Oh because KAN, N = Nilpotent

Forgetting what words mean

Mike Miller

Yeah, I really think the fruitful thing to do is figure out why N has diffeomorphic exponential map

(I don't know why)

Balarka Sen

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

Mike Miller

4:15 AM

@BalarkaSen But not by a left-invariant metric I guess?

Balarka Sen

Yeah your example seems to say that

just meant i have no approaches left (punpun)

Mike Miller

I bet it's really negative scalar curvature

that you find

Balarka Sen

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

Mike Miller

but think of your career!

Balarka Sen

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

Mike Miller

4:30 AM

@BalarkaSen It's the integral over $\text{Gr}(2,n)$ of $\text{sec}$

Balarka Sen

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

Mike Miller

no idea lmao

Balarka Sen

yeah whatever man

Anonymous

5:26 AM

What choice of topology makes [0, 1) homeomorphic to S^1 as mentioned in this answer (math.stackexchange.com/a/495940)?

Leaky Nun

just transport the topology along the bijection

Anonymous

@LeakyNun Umm...what does that mean?

Anonymous

Do you have some reference?

Anonymous

I understand the idea of quotient topology though

Leaky Nun

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

Transport of structure

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...

Anonymous

5:38 AM

@LeakyNun Yes, I get that [0, 1) and $S^1$ are in bijection as sets

Anonymous

@LeakyNun I see, I never heard of this concept

Leaky Nun

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"

Anonymous

Oh, as homeomorphisms are continuous (pre-images of open sets are open) it makes sense to define the topology like that on [0, 1)

Anonymous

Interesting!

Sebastiano

7:17 AM

@Croissant Thank you very much again. I hope that my question not is closed again @CalvinKhor Best regards.

Manish Mittal

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

Q: Integral $ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $

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...