The h Bar

Slereah

1:01 PM

I tend to view $x$ as being the coordinate in $\Bbb R^n$ and $p$ as the manifold point itself

Sir Cumference

I'm seriously wondering why this is on the hot network

14

Q: Why don't astronomers use meters to measure astronomical distances?

In astronomy distances are generally expressed in non-metric units like: light-years, astronomical units (AU), parsecs, etc. Why don't they use meters (or multiples thereof) to measure distances, as these are the SI unit for distance? Since the meter is already used in particle physics to measure...

0celo7

@Slereah coordinates on $\Bbb R^n$ are $\xi$, obviously

ACuriousMind

@SirCumference Best thing is to stop wondering about HNQ :P

Slereah

@0celo7 $\xi$ are coordinates in the tangent bundle, sorry

or cotangent, maybe

0celo7

if you're using coordinates then you're a pleb, sorry

Slereah

1:07 PM

try to prove the existence of a normal neighbourhood without coordinates

0celo7

@ACuriousMind Yosida has a pretty neat construction of the completion of a NVS. If $X$ is the space, then $X'$ with the strong topology is a Banach space. Take the bidual with the double-strong topology, $X''$. There's an isometric embedding (not necessarily surjective) $J:X\to X''$. Then the completion of $X$ is the closure of $JX$ in $X''$.

ACuriousMind

I've seen that before.

Slereah

Man the construction of distributions in GR is so bloody awful

It's full of diagrams

0celo7

@ACuriousMind Sadly one cannot define the real numbers like this because it requires Hahn Banach :(

And, well, $\Bbb Q$ isn't a normed vector space

but anyhow

ACuriousMind

Yes, all the elegant notions of completion require that you have done the completion of $\mathbb{Q}$ "by hand" first.

Slereah

1:11 PM

isn't it?

It's not a complete one certainly

But you can put a norm on it

Oh I guess you need multiplication by $\Bbb R$ for a vector space

nvm

DanielSank

@SirCumference Can be.

It's particularly good as BBQ ribs or as part of a spicy dish in Chinese style.

ACuriousMind

@0celo7 Now, if you want at least a somewhat "powerful" completion functor that you can apply to $\mathbb{Q}$ to get $\mathbb{R}$ without being circular, then you can define a completion of topological groups without any reference to $\mathbb{R}$ and then complete $\mathbb{Q}$ by that.

Slereah

please @ACuriousMind, this is a channel to discuss pork recipes

ACuriousMind

It boils down to the usual construction by Cauchy sequences, though.

0celo7

Completion functor? Is there a categorical functional analysis that I should be aware of?

Completion of topological groups?

Whut are you talking about

DanielSank

1:21 PM

@ACuriousMind You boil pork? Gross.

I need forgiveness from you guys.

I argued with someone here last night.

I shouldn't have done it.

0celo7

I told you he was a trol

You ignored me

I was quite saddened

DanielSank

I didn't ignore you.

I knew you were right.

Forgive me.

0celo7

Then you ignored my advice!

Slereah

what did you argue about

ACuriousMind

@0celo7 Given a neighbourhood basis $G_i$ of the identity in a topological group $G$, you may define a Cauchy sequence $(g_i)$ by $\exists n : \forall i,j\geq n : g_i - g_j \in G_n$

0celo7

1:22 PM

You are forgiven, if you listen to a topology lecture

ACuriousMind

Then you can complete as the usual set of equivalence classes of Cauchy sequences.

0celo7

@ACuriousMind Yeah that's the usual notion of Cauchy sequence in topological groups

DanielSank

@Slereah I don't even know. The other guy was saying stuff about science being composed of an accepting orthodoxy etc. etc.

@0celo7 No.

0celo7

@ACuriousMind Hmm, I guess that works

@ACuriousMind What's hard to show is that the completion of an LCS is an LCS, Yosida omits the proof :o

ACuriousMind

The nice thing is that this construction becomes purely algebraic if the neighbourhood bases $G_i$ are subgroups, since then the completion is the projective limit of the $G/G_i$.

Alas, this construction gives you the p-adic numbers, not the reals.

0celo7

1:24 PM

With the usual direct limit topology I assume?

Is it well known/easy to show that group operations are continuous in the direct limit topology?

I guess they're continuous on all of the subgroups

So by the universal property it's true

DanielSank

This isn't physics. I'm offended. ;-P

(Is that joke old yet?)

0celo7

very

DanielSank

k

0celo7

@ACuriousMind where would one need the completion of a topological group?

ACuriousMind

@DanielSank Say Maxwell's equations 20 times and work out three variants of the twin paradox and the Science Orthodoxy shall absolve you of your sins.

0celo7

1:27 PM

Most groups I know of are compact already (are compact groups complete?)

oh...Lorentz/Poincare group

Slereah

Also $\Bbb R^n$

ACuriousMind

@0celo7 We did it in algebra. You want to do it of a ring, usually - the completion of a ring is often easier to study than the ring itself but retains some of its properties

Slereah

Very non-compact

DanielSank

@ACuriousMind I'd rather be excommunicated.

ACuriousMind

@DanielSank The writ declaring you a heretic will be announced shortly, in that case.

0celo7

1:28 PM

:(

poor Daniel

DanielSank

@ACuriousMind Can I bribe the pope with something?

0celo7

I should learn some algebra

@DanielSank the pope wants no mead

ACuriousMind

@DanielSank If you don't know already, then you can't.

DanielSank

::Slips Pope $20::

The Dark Side

Not sure why "The" meta post is getting downvoted. Irrespective of what transpired, "Should discussions on topics which disturb long time users of the chat room be banned?" is a perfectly valid question for Physics meta.

I don't think we should make this question 2017 (i.e. the user 2017) specific.

ACuriousMind

1:35 PM

@TheDarkSide Imo, they made the question specific by quoting specific messages from this particular incident.

So it's no surprise people vote on it based on their opinion about this incident and not on the general question

The Dark Side

@ACuriousMind Yes that is true. Also it was anger driven. But away from this context, it is perfectly valid policy question. No?

0celo7

The meta question seems to be more of a hit piece directed at ACM and myself.

DanielSank

It was a poorly posed meta post, even if the underlying question is good.

Many meta posts suffer for this reason.

ACuriousMind

@TheDarkSide Yes, it is.

The Dark Side

@DanielSank I agree.

ACuriousMind

1:37 PM

Most people don't vote on the "underlying policy question" though. They vote based on what they see.

DanielSank

^ conjecture

0celo7

@ACuriousMind Hey, in shog's post, what is my deleted comment?

Slereah

Hm

The Dark Side

@0celo7 That's not unusual. AFAIK you are the king of deleted posts :P

Slereah

I suspect that if a path is a sting, it will only be homotopically equivalent to other stings

The Dark Side

1:40 PM

Hahaha ^

0celo7

@Slereah nope

Slereah

do you have a counterexample

The Dark Side

All existing posts are counterexamples. Well played :P

Slereah

well, I guess it's more

0celo7

Yeah. Take $\Bbb R^2$ Minkowski space. Look at the line going from $(0,0)$ to $(0,1)$

That's a sting if you go back and forth

Slereah

1:42 PM

It can't be homotopically equivalent to a smooth curve

(timelike)

0celo7

It you bulge it out a bit while mantaining the kink at the top, it's a timelike loop with two corners

ACuriousMind

@DanielSank Yep. One based on observing voting habits for a while though ;)

Slereah

I think trying to smooth it it will cease to be timelike

ACuriousMind

@0celo7 Nothing that would make you look good.

0celo7

@ACuriousMind what does that mean?

ACuriousMind

1:44 PM

@0celo7 That means it is in your best interest if you don't ask me to disclose it.

0celo7

@ACuriousMind Huh? I already said it, what could be so bad?

The Dark Side

:: Imagines ::

ACuriousMind

In fact, I think it won't do any good to repost it in any case. Just drop it.

0celo7

Strange reaction.

The Dark Side

Yes, if it is context specific, it may escalate bitterness. Responsible mod behavior.

0celo7

1:46 PM

I'm not bitter

I'm just curious what I said

I wonder if I have a key logger

The Dark Side

Freak question, why does SE give a diamond to mods, why not enclose their names within two $\vert$ to make a pun?

Sounds like a question for Mother Meta?

ACuriousMind

@TheDarkSide ...because that'd be a mathy pun and the SE network deals with many other subject areas?

The Dark Side

@ACuriousMind Yess. Imagine what English / Sociology people would make out of it.

Dang. You just cost me some rep points :P

0celo7

I don't get it

The Dark Side

^ Which field, 0celot?

(He/she does more Maths here than anyone else.)

0celo7

1:54 PM

The |mod|

The Dark Side

Of course not JEE level :P

Slereah

modulo, @0celo7

ACuriousMind

@Slereah modulus

Slereah

w/e

so what is the chat session about

ACuriousMind

@Slereah Good question. Is there something we should put on the agenda?

The Dark Side

1:56 PM

@Slereah Anything goes as always, perhaps the new stipulation is - it should not offend long term users of hbar :P

Slereah

well no, the chat session is very much so the opposite of anything goes

ACuriousMind

@TheDarkSide We've done something where we've streamlined the chat session by putting up an agenda and not just talking about anything

Alas, I don't really have anything for today's agenda.

The Dark Side

OK I see. Not very regular.

ACuriousMind

If someone wants to discuss any of the recent meta posts that'd be fine, but I don't think there's much that still needs to be said and wasn't said on meta

The Dark Side

@ACuriousMind Theory of turbulence maybe? There has been some "development" apparently, although nothing on the theory side. More of numerical stuff.

Dang!

physics.aps.org/articles/v10/25

OK just the link.

or

journals.aps.org/prl/pdf/10.1103/PhysRevLett.118.114501

Love to hear from the experts how much of a development does this amount too.

Slereah

2:01 PM

paywall

The Dark Side

And, if this is PRL worthy at all?

ACuriousMind

@TheDarkSide "Talk about physics" is the default if there's nothing else on the agenda, anyway ;)

Slereah

Is it?

You mostly talk about math

0celo7

@ACuriousMind aha

ACuriousMind

@Slereah Not during chat sessions

0celo7

2:02 PM

I was wondering what "absolute mod" was supposed to be :P

The Dark Side

@Slereah Hahaha ....

@0celo7 Community managers, e.g. Shog9? OK bad joke.

Slereah

let's check arxiv for new GR papers

There's so many papers about random spacetimes in alternative GR models

it's the background noise of GR papers

The Dark Side

@Slereah One more data point in favor of the fact that this room is too GR dominated.

Which is not bad though.

Slereah

it is best

GR is great

The Dark Side

GR8 ?

Slereah

2:07 PM

yes

0celo7

Lol, this theorem is so stupid

The Dark Side

What does it say about the theory of turbulence?

0celo7

Something is dense in the bidual with the weak* topology, but the bidual is defined with the strong topology

Physics Meta

1

Q: Why do I get the "Bad question warning" when first posting here?

I was going to post a question and I got the "bad question ban" warning. I have searched my previous posts and it turns out I have never posted anything on this SE - could one of the moderators please check if there are any deleted posts that I simply forgot about? If there are no past posts - wh...

support

0celo7

.-.

Secret

2:10 PM

0

Q: Escaping a pair of orbiting black holes through the saddle

Imagine a pair of black holes in orbit around each other. I'm wondering how this would distort their mutual event horizons. In fact would it be possible to "break" the event horizons by reducing the escape velocity below the speed of light at the point in between the two holes. If my mental mode...

science-based physics gravity orbital-mechanics black-holes

That's a physics question, not a worldbuilding question

The Dark Side

Quoting from the Physics article:

A different tack that has become increasingly popular [2–5], and the one that Schatz and co-workers follow, is to view fluid flow as a dynamical system, whose possible states lie in a high-dimensional space [6]. In this framework, the time evolution of a flow’s velocity field is represented by a trajectory through state space, in which each point corresponds to a solution to the Navier-Stokes equations.

Wonder what you GR people say about that?

Slereah

I don't think there's a GR equivalent of Navier Stokes

The Dark Side

Wow Leonhard Euler is in our chat room!

@Slereah So?

Secret

for starters spacetime is not really a fluid, it does not really flow in GR

The Dark Side

Not something that I have mastered/know much about, but is this something like gauge-gravity duality applied to fluid flow?

Slereah

2:15 PM

what

The Dark Side

We seem to have lost @ACuriousMind, or atleast his interest!

@Slereah What what?

@Slereah I mean writing a higher dimensional version of a (3+1) D problem, and then using your artillery to try to crack it?

Sorry if it sounds too naive. Like I said, nothing appreciable in terms of what I know here.

Slereah

2:41 PM

@0celo7 Penrose doesn't define trips properly

He forgets to specify that the time orientation has to remain the same between trips

but it is assumed implicitely in the proof

since the images of the trip segments in the tangent space are in the two different light cones

Sir Cumference

Are all the images on the site broken for you guys, or is it just me?

ACuriousMind

It's just you.

Sir Cumference

K

Run like hell

I was wondering if you see the text latexed in this chat, I see the $ $ symbols and the text in the middle, is there a way to see it properly or do you just type it like this for clarity?

ACuriousMind

@Runlikehell Look at the link to MathJax in the upper right corner of the chat.

-1

Q: Two vectors A and b are such that (A+B)=c and A ²+B ²=c ² then find the angle be two vectors

. Umaer Manzoor from Kashmir

vectors

???

Slereah

2:57 PM

seems pretty clear to me

The Dark Side

@ACuriousMind Migrate to Maths maybe?

Slereah

or music

ACuriousMind

...click on the link, people.

The Dark Side

Ahh yes.

ACuriousMind

I'm not confused about the title, but about that giant picture with musical notes on it

The Dark Side

2:58 PM

Hahaha ...

Slereah

Apparently it is Umaer Manzoor from Kashmir, @ACuriousMind

The Dark Side

Puzzling.SE

Emilio Pisanty

@ACuriousMind good going on the timing

We will now hold you to that standard

The Dark Side

(It is a puzzle what the title has to do with the picture.) => Puzzling.SE

Emilio Pisanty

@ACuriousMind Bonus points if you can associate a favicon with the bookmarklet

0celo7

3:01 PM

@ACuriousMind ok I can take the rep class after all

He said there won't be homework so that's good

ACuriousMind

@EmilioPisanty :P

@EmilioPisanty hmmmm?

Emilio Pisanty

Terry Bollinger

Is it chat time, or has Daylight Saving Time messed me up again?

Slereah

it's always chat time

Come and chat

The Dark Side

^ That

ACuriousMind

3:07 PM

@TerryBollinger If you're looking for the chat session, you're an hour early, so, yeah, DST has messed it up again

Emilio Pisanty

though, given the sudden surge of people coming in, you might not be alone

--is it DST in the US already?-- so it seems

Terry Bollinger

Yep, they snuck it in last weekend. What's bad is my auto reminders seem messed up too.

ACuriousMind

@EmilioPisanty Three dashes for strikethrough, not two ;)

Emilio Pisanty

@ACuriousMind oh well

you win some you lose some

The Dark Side

~~Test~~

^ Ah. Learnt something new.

Secret

3:11 PM

I think I can find a cover that has no finite subcover. Consider open n-balls such that their size decreases as they approach the boundary of $D(0,1)$ Poincaré Hyperbolic Disk fashion. Then this cover is going to have no finite subcover since you cannot take any subcollection of it that is finite and still covers all of $D(0,1)$, making it noncompact

I am not sure how to formulate that though, I obviously need a sequence that decreases to zero for the radii of each n-ball in the cover, but the extra degrees of freedom in higher dimensions $dim X > 2$ makes it difficult to specify the direc…

Slereah

Here's a cover that has no finite subcover : $\forall n \in \Bbb N, (n, n+2)$

Perfect cover of $\Bbb R$

Secret

It is, since all the open sets are effectively stacked end to end and cover $\Bbb{R}$, thus taking any one of them off and suddenly $\Bbb{R}$ is no longer covered

Emilio Pisanty

@Slereah is $\forall a,b \in \Bbb R, (a,b)$ more concise?

Slereah

Well it is also a cover

Although that one isn't even countable

Emilio Pisanty

I guess it's ambiguous w.r.t. what happens when $a>b$

Slereah

3:15 PM

I think that's just the empty set then

Secret

@EmilioPisanty yeah left < right by convention of writing intervals

Slereah

or can be defined to be without troubles, anyway

Emilio Pisanty

anyways, it is objectively more natural

Slereah

I mean, by definition

it is the empty set

Since $(a, b) = \{x \in \Bbb R | x > a \wedge x < b\}$

If $b < a$ that is the empty set

Secret

Terry Bollinger

3:25 PM

I'm on mobile -- does anyone happen have the MathJax activation link easily available?

Secret

The infinite cover I am talking about earlier, not sure how to write it down in set notation though

Slereah

that's not a cover

I can see parts not covered!

Secret

I know, it is extremely hard to draw all the circles

Hmm, I think I can find a more straightforward example by picking one of the many hyperbolic tilings, that should work

Slereah

Tilings aren't good covers because tiles are closed sets

ACuriousMind

I'm afraid I've never tried to activate MathJax on mobile

Secret

3:31 PM

oops

DanielSank

@ACuriousMind hmm, I ought to try that.

For some reason, my chrome bookmarks don't show up on mobile.

Slereah

$$🤠^🤠(🤠)_🤠$$

Terry Bollinger

I'll rejoin later on laptop, thanks all...

Slereah

$$^🤠_🤠\overset{🤠}{🤠}^🤠_🤠$$

DanielSank

@Slereah wat?

Slereah

3:38 PM

many cowboys

Secret

$$🤠^{🤠^{🤠^{🤠^{🤠^{🤠^{🤠}}}}}}$$

Slereah

$$\overset{🤠}{\overset{🤠}{🤠}}$$

DanielSank

Why are there cowboys?

vzn

@TheDarkSide cool/ thx for sharing, youre interested in fluid dynamics? this reminds me of applying machine learning to physics which there are some increasing cases of lately, am collecting diverse instances, suspect it will work for some previously intractable fluid dynamics problems.

Slereah

$$\overset{🤠}{\overset{🤠}{\overset{🤠}{🤠}}}$$

The unicode consortium thought it was very important that there were cowboys

I'm guessing that the unicode consortion gets a lot more money for adding emojis than for adding obscure central asian scripts

So they do a lot of emojis these days

0celo7

3:42 PM

wtf is going on here

ACuriousMind

Apparently we are invaded by cowboys.

Slereah

$$\overset{🤠}{\overset{🤠}{\overset{🤠}{🤠}}}^{🤠^{🤠^🤠}}_{🤠_{🤠_🤠}}$$

they stretch to infinity

DanielSank

Btw, if you choose an emoji in Google chat on desktop, it won't necessarily look the same on another OS, I.e. mobile.

Slereah

I am aware

especially the gun emoji

0celo7

@ACuriousMind Do you know the theorem that says the unit ball of a Banach space $X$ is dense in the unit ball of $X''$ in the weak* topology?

Slereah

3:44 PM

Most platforms vs iOS

ACuriousMind

@0celo7 That's Banach-Alaoglu, no?

Ah, no, it isn't

0celo7

No, that's a compactness result

@ACuriousMind What I have is the following Lemma: Given $X$ Banach and $x''\in X''$, for $\epsilon>0$ and $f_1,\dotsc, f_n\in X'$, then $\exists x\in X$ such that $||x||\le ||x''||+\epsilon$ and $f_j(x)=x''(f_j)$ for all $j$.

So that means the weak* seminorms can't distinguish $x$ and $x''$, which is good

But the estimate $||x||\le ||x''||+\epsilon$ is going the wrong way

ACuriousMind

@Slereah Kinda weird that facebook has the gun pointing at the user :P

0celo7

Because if $||x''||=1$ then all I get is $||x||\le 1+\epsilon$ which does no good

Slereah

Well, slightly to the left of the user

ACuriousMind

3:50 PM

Okay, depends on how fat the user is and how close they sit to the screen

Emilio Pisanty