The h Bar

@Slereah I see. To me charts are literally R^n, but if that's your definition, then I agree. In any case, the answer to your question is still a yes; an open subset of R^n has trivial tangent bundle.

Slereah

Ah yes

I suspected so from the whole tangent bundle topology definition

Anonymous

@SwapnilDas What is that /a/ ?

Anonymous

|a| ?

Slereah

Since basically if it's a single chart you just have a single trivialization

Swapnil Das

Yup, my vertical line key doesn't work :P

Slereah

10:53 AM

So the bundle is just the trivial one

Balarka Sen

@SwapnilDas No, but look: $|v + w|^2 = |v|^2 + |w|^2 + 2v \cdot w$. $(|v| + |w|)^2 = |v|^2 + |w|^2 + 2|v||w|$. As $v \cdot w \leq |v||w|$ (Cauchy Schwarz), the former is always less or equal to the latter.

Slereah

Of course the hard part is proving whether or not manifolds that are not that are trivial or not

I guess if they're trivial I can just check if $M \times \Bbb R^n$ obeys the proper topology

Swapnil Das

@BalarkaSen Yup! There are two things that kept me hooked.

Cs Inequality in JEE material is a bit weird.

And another, what's that // // in CS inequality? @BalarkaSen

Balarka Sen

Cauchy-Schwarz is life. I don't know how JEE treats that material but it is life.

Swapnil Das

Yup!

Balarka Sen

10:56 AM

|| . || is the same as | . |

It means the norm of the vector, or the length.

Swapnil Das

then why two lines? Extension to n dimensions?

Anonymous

@BalarkaSen I didn't know that inequality has a name =P

Balarka Sen

Nah, just notation. || . || is generally used to denote "norm" for arbitrary vector spaces, which is a generalization of length.

But you don't need to know that probably.

Swapnil Das

Now

ACuriousMind

@BalarkaSen Indeed it is! I keep telling people the uncertainty principle is just CS but they want to keep talking about "wave-particle duality" :P

Anonymous

10:58 AM

@SwapnilDas Hmm, do you know cosine of any angle lies between -1 to +1 ?

Swapnil Das

yup, I used the same logic ;)

Anonymous

That is enough to understand the proof

Balarka Sen

@ACuriousMind Yeah!

I actually understood the uncertainty principle when I saw Hardy's uncertainty principle in Fourier theory.

Which is, of course, just Cauchy-Schwarz in the L^2 world.

Swapnil Das

@BalarkaSen Do I prove the reverse Triangle Inequality in the same way (using vectors)?

Anonymous

What is reverse triangle?

Anonymous

11:02 AM

so many names :D

Balarka Sen

Reverse triangle inequality is literally the consequence of triangle inequality.

Anonymous

Oh that difference of two vectors

Anonymous

It's the same thing

Swapnil Das

Could you explain how?

oh oh !

b replaced by -b?

Anonymous

Yeah

Swapnil Das

11:04 AM

lol.

@blue Could you guess from where I'm bringing this?

Anonymous

@SwapnilDas Bringing what?

Swapnil Das

This question.

Balarka Sen

Hmm. What does electric permittivity measure? I see that if the medium's $\varepsilon$ is greater, Coulomb attraction between two charged particles decrease. Does it measure how much resistance the medium creates against the electric field? That sounds completely contradictory to "permittivity" :P

Also this may be my first actual physics question in this chat :P

Swapnil Das

It tells you how much interaction does the medium permit ;)

Slereah

Oh no

The proofs of the triviality of the circle's tangent bundle I find are all from the perspective of an embedding

But non-Hausdorff manifolds aren't embeddedable at all!

John Rennie

11:09 AM

@BalarkaSen it tells you how much the medium is polarised by the electric field passing through it.

Balarka Sen

@SwapnilDas How so? From what I just said, it measures how much the medium resists to the interaction. In which case, $\varepsilon$ should be termed "electric resistance" and $1/\varepsilon$ should be the true electric permittivity. :P

John Rennie

If a medium is highly polarisible the electric field creates dipoles in the medium that oppose the applied field

Balarka Sen

@JohnRennie Hmm, what does polarised mean?

Swapnil Das

It isn't termed like that because perhaps we have electric resistance already existing lol.

John Rennie

@BalarkaSen If you have a neutral medium and you apply an electric field to it then any mobile positive charges will move in one direction and any mobile negative charges will move in the other direction.

This charge separation creates an electric dipole.

ACuriousMind

11:12 AM

@BalarkaSen I think it's more instructive to think in terms of the susceptibility $\chi = \epsilon/\epsilon_0 - 1$. The susceptibility tells you how strongly the material will be polarized by an electric field, i.e. the charges in it move a little bit in response to the electric field and form little dipoles, the greater $\chi$ (or $\epsilon$) is, the stronger this formation of dipoles is.

Balarka Sen

@JohnRennie Ah, yes, that is true.

John Rennie

Everything I say is true (for a limited and somewhat eccentric subset of meanings of the word true :-)

Swapnil Das

@ACuriousMind I'm really sorry to interrupt. Is the rumor that 'grad students usually lose their mastery over newtonian mech. true ?

Balarka Sen

@ACuriousMind Cute! I like that.

Anonymous

@BalarkaSen In a medium the net force between two charges is given by $F_o-F_m$ which is dented by $\frac{F_o}{\epsilon}$. $\epsilon$ is also called as the dielectric constant.

Anonymous

11:15 AM

That's another way to view it :)

Slereah

Wait, for branching spaces, I can just use the pullback bundle, right?

Balarka Sen

@JohnRennie Using the parenthesis as the smile of your smiley always disorients me up.

Slereah

$$f : \Bbb R \sqcup \Bbb R \to \Bbb R_Y$$

Anonymous

ACuriousMind

@SwapnilDas You need not feel sorry to interrupt, but I don't know what exactly that means. What is true is that one gets less "good" at arbitrary Newtonian mechanics exercises since one doesn't practice it any more, but I would strongly question whether "solving exercises" is the same as "mastery" over a physical subject.

John Rennie

11:16 AM

@BalarkaSen yes, but using two right parentheses looks wrong ...

user228700

@JohnRennie I've looked into it before and no, there isn't :'-(

Slereah

Wait no, wrong way around

Balarka Sen

@blue Yeah, I am aware of the dielectric constant.

John Rennie

@Kaumudi.H mug someone and steal their bicycle?

user228700

I wish :-/

John Rennie

11:18 AM

@Kaumudi.H how much longer are you stuck there for?

Anonymous

@BalarkaSen Relative permittivity is the same as dielectric constant.

user228700

Oh, not too long. I go back on Saturday.

Balarka Sen

@ACuriousMind Ok, so, if $\epsilon$ is larger, the dipole formation becomes stronger. Why does stronger dipole formation imply weaker Coulomb interaction between two charged particles?

@blue I know, I was just asking what all these really measure.

user228700

Another 11-hour journey back home but at least I'm going home!

user228700

I'll be home on Sunday morning.

Anonymous

11:20 AM

@BalarkaSen It is like a book keeping tool. F-F_m is represented by F/epsilon. It measures the the amount by which a medium is polarized

John Rennie

@Kaumudi.H are you doing anything to celebrate getting home, or is it straight back to work?

Balarka Sen

@blue So it seems.

Anonymous

@BalarkaSen The dipole inside the medium is developed in opposite direction to external field...

Balarka Sen

Ah.

Well, that makes sense.

Anonymous

11:22 AM

See this picture^

Balarka Sen

Right, I understand now.

Anonymous

@Kaumudi.H Got your rank?

user228700

@JohnRennie :-) No, no celebrations or anything, I'm afraid. Straight back to work it is!

user228700

I'm doing all the relaxation I can over here but the trouble is that there isn't much to do here :-/

John Rennie

@blue I think we should adopt the principle of not asking room members about their exam results. If they want to volunteer the results that's fine, but it's rude to ask.

user228700

11:24 AM

@blue Nope. For reasons that I don't wish to divulge, I'm not going to check my rank until BITSAT.

user228700

@JohnRennie I agree.

Anonymous

@JohnRennie Is that rude? Sorry. We were all discussing the possible colleges we might join in the other chat room....so I thought it's natural. Anyhow it's @Kaumudi.H's wish....

John Rennie

@blue if someone has done badly then asking them directly will probably embarrass them.

user228700

@blue It's only understandable. It's common for literally anybody to ask the rank of other people over here.

Anonymous

@JohnRennie We have all done badly =P

Anonymous

11:27 AM

Anyhow

Anonymous

I'll not ask

Balarka Sen

JEE is too hard.

2hard4me

user228700

@blue It doesn't apply to me alone. I'm not a sensitive bird, I can tell you my rank if you want (if I decide to check it after all) but it is a bit rude. It puts one on the spot and yes, a little embarrassment does ensue.

Anonymous

@Kaumudi.H Lol, Yashas bugged me to reveal my marks in the other room :P Okay okay fine. I don't know since when asking rank is considered rude. :/ You are free not to tell. I thought it is common to discuss among friends.

John Rennie

eBay has just mailed me to congratulate me on buying my 1000th item. I'm not sure that's an achievement to be proud of :-)

user228700

11:29 AM

@JohnRennie Omg :-o ...

John Rennie

@Kaumudi.H they weren't all laptops :-)

Balarka Sen

I think there's too much ego and embarrassment involved with the ranks of various exams in India, partly because it's almost a life-and-death matter. I think this is bad for health, but oh well - such is life.

user228700

@blue It's always been a bit inconsiderate, I feel; it won't seem that way only because it's common practice here for even strangers to ask.

0celouvskyopoulos

Hello

user228700

@JohnRennie So u say ._.

0celouvskyopoulos

11:32 AM

@Kaumudi.H wot

Anonymous

@Kaumudi.H Sorry. I'll never ask your marks again. :) Sorry!

user228700

@0celouvskyopoulos Wot? right back at ya.

John Rennie

@Kaumudi.H I take it that's a sceptical smiley? :-)

user228700

@blue Lol. No need to apologise. I'm not comfortable revealing my marks over here, that's all. Besides, like I said before, I haven't even checked yet and I won't either. Not until BITSAT anyway.

Anonymous

Anyhow, I'm sorry. I didn't think it would become such a serious situation. My test didn't go as well as expected too.

user228700

11:34 AM

@JohnRennie Sort of, yes.

user228700

@blue Serious situation? :-P It's no serious situation. Ah, I see. If this is any encouragement, one of my friends who is now in IITM didn't do well in Main either but he prepared well for Advanced and did much better :-)

Anonymous

@Kaumudi.H Right. Thanks :)

John Rennie

Is the Advanced exam just for the IIT? Or does sitting it and getting a good, but not quite good enough, result benefit you in applying to other colleges?

Anonymous

@JohnRennie Only for IITs

Anonymous

And IISC

Balarka Sen

11:38 AM

Does IISc have an individual entrance exam of it's own?

I have 0 idea about all these.

user228700

@BalarkaSen Yes. KVPY.

John Rennie

Is it that big a deal to go to an IIT? I guess in terms of money, i.e. well paid jobs, it is but if you just want to do science are the IITs still best?

Balarka Sen

@Kaumudi I see.

macco

Hej

user228700

@JohnRennie IITs don't offer particularly great courses in pure Science; it's an institution for Engineering.

0celouvskyopoulos

11:41 AM

Bah

Anonymous

@JohnRennie IITs aren't science institutions. They are more of engineering colleges. But yes, they are the best in Engineering.

John Rennie

@Kaumudi.H but then I guess every Indian parent wants their children to be engineers ...

Balarka Sen

Does one take KVPY after grade 12?

Anonymous

@BalarkaSen Nope

Balarka Sen

@JohnRennie A surprisingly large number does.

Anonymous

11:42 AM

In 11 th and 12 th

user228700

best is certainly arguable but best in terms of placements after college, yes.

user228700

@blue What? No. If enrolled in a course of pure Science, one can certainly sit for KVPY even after 12th.

Anonymous

@Kaumudi.H Oh, that is a different story

user228700

@JohnRennie Most, yes. Although, the situation is improving now.

Balarka Sen

interesting. I wonder if I want to take that exam.

Anonymous

11:44 AM

@BalarkaSen It has physics and chem too :P

Balarka Sen

yeah, meh

i don't

John Rennie

@macco Swedish for hello?

@BalarkaSen where would you go to do a maths degree?

user228700

Ooh, @Balarka: Why don't u try CMI?

Anonymous

ISI and CMI are far better for maths

Balarka Sen

I don't really know, I'm a clueless guy. I plan for CMI and ISI.

@Kaumudi That's what I have been suggested.

the entrance exam scares me a little

user228700

11:46 AM

Right. One of my friends is studying there now.

Anonymous

Or go abroad and join MIT

Anonymous

They'll take you

user228700

But she's the only girl in her batch :-/

Anonymous

You're talented :)

John Rennie

@BalarkaSen Chennai Mathematical Institute?

Balarka Sen

11:46 AM

i don't want to go outside

user228700

Yes, yes. If I were @balarka, I wouldn't be worrying about the exam itself.

Balarka Sen

@JohnRennie Yeah

Anonymous

I don't even know half the maths you do @BalarkaSen

user228700

^ Neither do I.

user228700

Much more than just half, I'm sure.

John Rennie

11:47 AM

@BalarkaSen and ISI = Indian Statistical Institute?

Balarka Sen

Yeppers

macco

@JohnRennie Umm yes, its just aesthetically more pleasing than the English version

John Rennie

Whatever did we do in the days before Google? :-)

user228700

@JohnRennie Libraries! Geez, I cannot imagine living at that time!

Balarka Sen

@blue @Kaumudi I don't think that's a problem. I just do what I do; you are both great what you do. I probably can't do half the problems in JEE

I am terrified about exams because I am not good at writing them

John Rennie

11:48 AM

@Kaumudi.H it was hard when I was a child before electricity was invented :-)

macco

Does anyone know why people think this is a duplicate? physics.stackexchange.com/q/327960/140964

user228700

@BalarkaSen This particular exam won't be too bad for you, I'm sure. It's just like you said: 10 questions and 3 hours or something like that.

Balarka Sen

i also value being able to do mathematics more than "knowing math". I am personally trying to get good at the former.

oh, interesting.

John Rennie

@macco it does look pretty similar to your question. In both cases the questions ask why the emission and absorption don't cancel out.

macco

But mine is about how it works in the Sun, which is different

user228700

11:51 AM

@BalarkaSen Look it up, man. I think you have an excellent chance of making it. One of my friends-who doesn't do nearly as much Math as you-made it last year.

John Rennie

@macco yes, that's a fair point, and the answer is related to the temperature distribution in the Sun so that is specific to your question.

user228700

@JohnRennie It's funny how sometimes you insist that you aren't old and sometimes, you say things like this :-P

John Rennie

@Kaumudi.H in my efforts to entertain I regard the truth an optional extra :-)

user228700

Haha, OK :-)

John Rennie

I don't think this is a duplicate because the answer to this question is specific to the structure of the Sun and the Solar atmosphere. It's true that the proposed duplicate discusses the underlying physics involved, but that alone is not enough to answer this question. — John Rennie 16 secs ago

user228700

11:55 AM

I think I'll walk to town to buy some coffee now--something to kill time and it's also been like, 4 days since the last time I had a cup of coffee.

macco

@JohnRennie thanks

John Rennie

that's a good way to relax too. And you could get some decent food :-)
I'm also killing time as I'm bored with programming. I think I'll also head into town.

Balarka Sen

You should watch a good movie. Best way to kill time.

The most productive way to procrastinate, for me.

Slereah

no, you should watch a bad movie

user228700

@JohnRennie Wokay :-) Yes, hopefully I will find something to eat as well! Have a nice evening, then.

John Rennie

11:57 AM

@macco though you've got a good answer, so it doesn't really matter if the question is closed, though I understand it's a bit annoying.

user228700

@BalarkaSen Actually, I did. I watched A New Hope this afternoon.

Balarka Sen

@Slereah That too

Is that Star Wars

user228700

Yep.

Balarka Sen

Cool. I haven't watched them but I have heard good things about them.

I still think you should watch some mind melting movies tho

everyone should

user228700

@BalarkaSen I don't have a good connection to the internet so there's no way for me to watch movies that aren't already on my laptop :-/

Balarka Sen

11:59 AM

Ahh.

I carry around Tarkovsky's Solyaris, Stalker, Nostalghia and Sacrifice with me all the time.

(Those are not mind-melting)

AccidentalFourierTransform

Airplane! - best movie ever

Balarka Sen

i need a week's break from life so i can go on a John Carpenter streak

Secret

Big data needs big theories: It is true that neural networks help crawl heaps of patterns in a phenomenon, but we cannot just rely on pattern crawling, we need to understand what drives those patterns in the first place

Yashas

@Kaumudi.H So you are not registering for JEE Advanced? You can get your parents check your rank and register for you :/

without telling you the results

0celouvskyopoulos

12:22 PM

@BalarkaSen ...

@Kaumudi.H Balarka does know a lot

sadly no analysis!

@JohnRennie I meant Giggs the rapper.

Slereah

1:04 PM

For the bundle to be trivial, is it sufficient to just check for a global bundle chart

0celouvskyopoulo7

yes

Slereah

Easy enough for $\Bbb R^n$ then

What is the proof for the triviality of the circle bundle though, without referring to any embedding

0celouvskyopoulo7

Any fiber bundle over $\Bbb R^n$ is trivial.

ACuriousMind

@Slereah "the circle bundle"? Do you mean the tangent bundle of the circle?

Slereah

yes

$T S^1$

0celouvskyopoulo7

1:09 PM

I'm not sure you can do it without an embedding. I have a full proof of it if you want.

I mean

Slereah

Well I'd like a good feeling of a general proof for proving triviality

ACuriousMind

@Slereah Then compute its transition functions in one particular trivializing atlas, show they are trivial/cohomologous to the trivial ones.

Slereah

But since non-Hausdorff manifolds aren't embedabble, that's not a very transferable proof

0celouvskyopoulo7

@Slereah Why don't you read an actual book on the subject

Slereah

I read this : math.uh.edu/~minru/Riemann08/vd.pdf

pretty good

but no actual proof on examples

ACuriousMind

1:11 PM

@Slereah Indeed, if you cannot embed you need to show triviality by some other means - existence of non-vanishing section, triviality of transition functions, that sort of thing

0celouvskyopoulo7

Did you check Lee?

Slereah

I did download it

I'll give it a look

@ACuriousMind Does showing it in a single chart work, or do you need an argument for arbitrary charts

ACuriousMind

I don't understand the question

Slereah

I guess that the property of manifold charts makes it so that if it works in one chart, it will work in all charts

(hopefully)

ACuriousMind

If your bundle would turn out to be trivial in some choices of charts but not in others, something is rather messed up :P

Slereah

1:17 PM

You know me, I have messed up manifolds

Also is there something like $M = M' / \sim \to TM = TM' / \sim$

ACuriousMind

No, I mean then your notion of bundle is messed up.

Slereah

@BalarkaSen said something along those lines

@ACuriousMind Well you know, I'm pretty sure that's true, but with math

You never know

Sometimes you get an infinite dimensional manifold with tentacles that's a counterexample to an obvious thing

The long line is pretty bad for that really

It is the worst manifold

I don't even know what the tangent bundle of the long line is supposed to look like

It's not trivial, but

what is it gonna be shaped like

It has that thing where all functions on it have to be $0$ eventually

Continuous, anyway

Yashas

1:36 PM

Simulation Suggests 68 Percent of the Universe May Not Actually Exist

Slereah

Does the long bundle even have a non-trivial structure group or is the non-triviality a weird reason

Jim

@Yashas that's a weird title. If the stuff doesn't exist, then it isn't part of the universe, and so once again 0% of the universe doesn't exist

Yashas

The title should've been "Simulation suggests 68 % of the universe of our most widely accepted model may not actually exist in reality"

ACuriousMind

@Slereah What does "have a structure group" mean. If the bundle is trivial then it has a reduction to trivial structure group, but the structure group is not uniquely defined.

Slereah

Well it's not trivial

But it feels weird

ACuriousMind

1:40 PM

Then it does not have a reduction to trivial structure group

Slereah

Since it's basically just a big $\Bbb R$

Jim

@Yashas dark energy?

Slereah

What "shape" does it have

Yashas

yea

Jim

naturally

what theory are they supporting this time?

Slereah

1:42 PM

@Jim The universe was made of asparagus the whole time

Yashas

They claim that everything fits with the data if you don't make simplifying assumptions

I am not good enough to understand that article really well. I'm still a newbie who is struggling with SR :P

Well, Indian textbooks are incomplete.

Jim

@Slereah that sounds like the description of a dream I once had

Yashas

Textbooks are like crashcourses

Jim

@Yashas that's a bold statement

Yashas

The researchers set up a computer simulation of how the universe formed, based on its large-scale structure. That structure apparently takes the form of "foam," where galaxies are found on the thin walls of each bubble, but large pockets in the middle are mostly devoid of both normal and dark matter.

The team simulated how gravity would affect matter in this structure and found that, rather than the universe expanding in a smooth, uniform manner, different parts of it would expand at different rates. Importantly, though, the overall average rate of expansion is still consistent with observ…

ACuriousMind

1:45 PM

@Slereah How do you know it's not trivial?

Slereah

The big argument for it seems to be that there's no continuous function on the long line that is non-zero

Which I guess translates to no non-vanishing vector fields

ACuriousMind

But...a vector field is not a function $L \to \mathbb{R}$, it's a function $L\to TL$

Jim

The linked article lacks the details I'd need to determine how valid the work is in showing the unnecessity of dark energy

Slereah

I guess the argument translates fine enough?

Like the vector field is just gonna be $f(x) \partial_x$

I mean, don't ask me, I can't topology enough to prove it, but

ams.org/journals/proc/1969-023-02/S0002-9939-1969-0246318-X/…

rob

@Yashas Actual paper: arxiv.org/abs/1607.08797

ACuriousMind

1:50 PM

Also, I do not believe you that all functions on the long line must be zero

Jim

@rob reading that now. The first thing that stands out to me is that they haven't actually done the work to see if it agrees with CMB observations yet. Just standard candle observations

ACuriousMind

What about the constant function $L\to\mathbb{R}, x\mapsto 2$?

It's obviously continuous, and it's obviously non-zero everywhere

Slereah

Is it obviously continuous

Physics Guy

Hello

Jim

hello

Physics Guy

1:51 PM

How are you, Jim?

Slereah

The issue seems to be with open sets that are bigger than $\Bbb R$

Jim

I'm well, Physics Guy

And you?

ACuriousMind

@Slereah Preimages of open sets are either $L$ or the empty set.

Physics Guy

Yeah, fine, thanks. I'm struggling with String theory right now.

String theory is great.

Slereah

8

A: Tangent bundle of the long line

The long line has the property that there is no continuous self map $f:L\to L$ such that $f(x)>x$ (or $f(x)<x$) for all $x\in L$. Indeed, $f^n(0)$ is an increasing sequence, hence converges to some $x$, which is a fixed point. So if there were an everywhere nonzero tangent vector field (for some ...