The h Bar

David Z

17:04

@ACuriousMind FYI this is not the sort of thing NAA flags should be used for, although I approved it so as not to throw you into another flag ban right away

Danu

@DavidZ Aren't we supposed to be in a test period??

I'm confused by this situation

David Z

@Danu For VLQ flags, yes. Other types of flags work as normal.

It was an NAA flag. I edited to clarify that.

Danu

Sorry, but it's not possible to flag something as "very low quality", AFAIK.

David Z

@Danu Then where do all those "very low quality" flags in the mod queue come from? ;-)

Danu

All one has is "spam", "rude", "not an answer" or "custom"

So NAA is the only possibility if you wanna signal low quality without insinuating things that aren't true.

I take NAA to mean "so bad that it's not an answer"

David Z

17:09

@Danu I would advise against that, at least on Physics. When something qualifies for an NAA flag, it's not because it is bad.

Danu

@DavidZ That's strange...

David Z

VLQ flags are only available on posts within the last week. Maybe that would explain it.

Danu

@DavidZ ^ on a post 2 hours old.

David Z

Which post?

I think there's also some criterion about votes

Danu

I haven't seen any other possibilities ever, as far as I can recall.

David Z

17:13

Well, that would be quite a coincidence, but the flag definitely does exist.

Danu

@DavidZ Ah, that could explain it!

I think that's terrible, though.

Seriously, upvoted answers should not be above the VLQ queue.

David Z

What do you mean, above the NAA queue?

Danu

Sorry, corrected.

David Z

I mean, what does it mean to be above a queue?

Danu

It's not possible to flag as low quality on upvoted answers.

0celo7

17:16

@Danu It does not use Morse theory

Well, one of the proofs does

Danu

Actually, the idea sketched in G&P doesn't sound bad.

0celo7

I don't think Ted was right

David Z

@Danu Ah, yeah, that's an unfortunate criterion, but I question how much of a problem it is. How many posts that deserve VLQ flags receive an upvote? (Rhetorical question, I'm not expecting you to count :-P)

0celo7

I've scoured the book and I cannot find the proof

Danu

@DavidZ Well, I know certain users...

0celo7

17:18

I will use el Google

ACuriousMind

@DavidZ Uh, okay. This is confusing because both NAA and VLQ go into the VLQ queue, so when non-mods handle the flags there's no feedback on the distinctions. The more I think about it the more this whole system needs a serious rework.

Also, how is it possible that I was able to flag that post, you marked the flag as helpful, and yet I'm now banned again?!

Danu

Thanks for wording it more eloquently

@ACuriousMind The site finally detected that you're trying to destroy it and turn it into a Physics MO ;)

0celo7

@ACuriousMind wow stop getting banned

ACuriousMind

@Danu As reasonable an explanation as any :D

David Z

Best guess, maybe you dropped below 10 flags in the past 7 days, but the last helpful one put you back over

DanielSank

17:33

@EmilioPisanty I was thinking Sevilla for the scenery more than the voice.

Emilio Pisanty

@DanielSank yeah, fair enough, maybe Buenos Aires doesn't quite make 'scenery'

not to say that it isn't an amazingly beautiful city

DanielSank

Although now that I think about it, Sevilla's scenery isn't all that amazing.

I would like to travel to Buenos Aires.

Emilio Pisanty

@DanielSank do

It's awesome

DanielSank

Also, thanks for explaining neotango vs. tango nuevo. I think perhaps those terms are commonly mixed nowadays.

Emilio Pisanty

@DanielSank could be, I'm not in touch with many Agentinians

btw I've got this on at the moment

Otros Aires - Otra Noche en La Viruta - Official Videoclip

►

DanielSank

17:35

In a previous phase of my life I went to the weekly milonga at Yale. Very fond memories. I also preferred dancing the neotango pieces over the more classical ones.

0celo7

what does it mean to "turn a function upside down"

DanielSank

The only exception is "A La Gran Muñeca", which is very fun for dancing.

Emilio Pisanty

@DanielSank Here, have an ñ

DanielSank

Gracias

Emilio Pisanty

De nada, caballero

You aware of Otros Aires?

DanielSank

17:37

No!

Emilio Pisanty

@DanielSank Yeah, you probably want to give them a listen

Also Plaza Francia

DanielSank

Doing that right now. They apparently have done some collaborative works with Bajofondo.

Emilio Pisanty

@DanielSank Bajofondo is also awesome

DanielSank

^ Truth

@EmilioPisanty I've posted this before, but since it's perhaps my favorite piece ever, I'll bring Nostalgias to your attention again.

Emilio Pisanty

@DanielSank Wasn't aware of it but it's kicking off really well

DanielSank

17:42

It'a amazing. See if you can keep track of which instrument is carrying the melody ;)

0celo7

contorts hand to do right hand rule

how does this thing work?

Danu

@0celo7 You know, just the way you figure out signs of intersection numbers ;)

0celo7

@Danu what do you think I'm doing this for :D

Danu

lol

Emilio Pisanty

Trivial-ish question: does the x in soft-x-ray laser go capitalized?

as in e.g. journals.aps.org/prl/abstract/10.1103/PhysRevLett.115.083901

bibtex is lowercasing everything and I'm not sure whether to protect the X

Danu

17:51

Protect Mr. X at all costs!

Also I hate how bad bibtex is :P

Emilio Pisanty

@Danu hear hear

except that there's hardly an alternative

Also, I'm really tickled by the synopsis, physics.aps.org/synopsis-for/10.1103/PhysRevLett.115.083901

> Synopsis: A Circularly Polarized Laser for All Labs

↑ sort of belied by just how incredibly complicated the experiment actually is

0celo7

@Danu do you have your summary on dropbox

Danu

@0celo7 Once I finish it I can send it, sure.

0celo7

thanks

Danu

It's not your style though. I do not stick to notions developed in the book at all times.

0celo7

17:56

My style?

For a summary, that's fine

What upsets me is when people do not use techniques from the book to do exercises

Danu

Also of course it's just the same as the stuff in the book---it's not of any independent interest.

0celo7

or is what you're talking about

Danu

I just have 2-3 remarks

@0celo7 I don't think it happened in the exercises, because I only did so few.

Also I use charts instead of those stupid parametrizations :P

0celo7

That is good

But they use the embedding explicitly in a few spots I think

Danu

"the embedding"

0celo7

17:59

What

Aha, I think I understand how one can use Poincare duals to compute intersection numbers.

But it's not how I would have proved it, and I want to understand if/how my proof can be saved

Danu

18:17

@0celo7 Understand as in have a detailed proof or as in know how it must work out?

Balarka Sen

@0celo7 I know a proof of that modulo a fact which probably involves Morse theory.

0celo7

@Danu I have a detailed proof.

Modulo a fact about tubular neighborhoods that will probably be resolved by "take it really small"

Since everything is compact tubular nieghborhoods are nice.

@Danu It was immediate to me how it must work out :P

Bott & Tu explain that, but they shirk the details.

@BalarkaSen Yeah?

Balarka Sen

I worked out how the idea would go myself once. It was exciting.

0celo7

Like two days ago you had never heard of my version of Poincare duality.

Balarka Sen

@0celo7 $M, N$ submanifolds of $W$ of complementary dimensions, with triangulations such that no. of pts in intersection is a subset of the vertex set. Then one can give a triangulation of $W$ extending given triangulations of $M$ and $N$ - that's the fact.

@0celo7 Why's the relevant?

0celo7

18:21

@BalarkaSen The existence of triangulations is so hard to prove for manifolds, my proof will likely be easier.

Danu

^if they're smooth it's no problem right? :P

0celo7

@Danu It's a folklore theorem, quite difficult to prove.

Balarka Sen

@Danu Yes, they're all smooth.

For non-smooth, there are manifolds which are not triangulable.

@0celo7 You mean Bott-Tu's proof :P

Danu

@BalarkaSen Exactly

Mike's supervisor proved that!

0celo7

@BalarkaSen No.

Danu

18:23

(for all dimensions $\geq 5$)

0celo7

They do not prove it.

Balarka Sen

They give the idea.

You said that.

@Danu Yup :)

0celo7

I should have been clearer. They give an idea, but it's not the one I used.

Balarka Sen

Fair enough.

0celo7

I'm still wondering how to convert their idea into something usable.

But the proof will involve working out co-orientations, i.e. orientations via normal bundles. And there might be some signs in there I'm not seeing.

Balarka Sen

18:25

Good luck with your proof.

0celo7

Want to hear it?

Balarka Sen

No.

0celo7

Aww

Balarka Sen

Mike told me a curious intuition on the fact that there always exists a triangulation on a smooth manifold extending a triangulation on a given manifold. Essentially, "space of triangulations on the disk" is contractible, whatever that means.

Danu

Oh

Balarka Sen

18:29

I plan to understand all that someday, but need to study Morse theory before that.

0celo7

Stupid question, but isn't triangulation a topological property? Why does it depend on smoothness?

Any $C^1$ manifold can be made $C^\infty$, so triangulate that.

Balarka Sen

C^1 and C^infty are not different.

It's C^0 and C^1 which are

0celo7

Or is the thing that there are nontriangulable $C^0$ manifolds?

Balarka Sen

Yes.

0celo7

Ok.

When you say "non-smooth" I think of $C^k$, $k<\infty$.

Not $k=0$.

Balarka Sen

18:31

That's silly.

0celo7

I agree, you should say "C^0" when you mean it.

Balarka Sen

Non-smooth literally means that.

And not including $0$ as a number $< \infty$ is not a good convention.

0celo7

What?

By non-smooth above you meant specifically $k=0$.

While what you said is strictly true, it's confusing.

You could be much more precise by just saying "topological manifolds."

Balarka Sen

I said "There exists non-smooth manifolds which are not triangulable". With your definition of non-smooth (C^k for k < \infty), that's still fine.

trilolil

Hello everyone, very quick and easy question regarding waves accross materials

someone hits the rod with a hammer and a wave propagates accross it:
http://www.imgur.com/0mGs3sI

on the second line, nex to the arrow you see $F=\Delta p - S ....$
I think both terms should be switched, because I think right now the force F will always be negative, and I am not sure whether that makes any sense. Could anybody concur?

0celo7

18:36

@BalarkaSen Did you say you are or are not familiar with Euler classes?

Balarka Sen

Nope.

It's got something to do with the Poincare dual of intersection of a generic section of a bundle with the zero section, IIRC

0celo7

That's not how Bott & Tu defines it, but they state that and the proof is quite hard

I don't understand it yet

Balarka Sen

But I haven't actually proved that generic section make sense.

0celo7

@BalarkaSen You can make the section transversal to the zero section

Balarka Sen

It's not guaranteed to be a section anymore, after making it transverse. There's more work than genericity of transversality involved.

0celo7

18:40

@BalarkaSen Correct. That's why Bott & Tu has a proof that you can make it transversal and still be a section.

Balarka Sen

Nice, tell me the page number.

0celo7

123

Balarka Sen

Excellent, I'll keep in mind not to stumble onto that page by accident until I prove it by myself.

0celo7

You don't want a hint I assume?

Balarka Sen

No.

0celo7

18:44

You should be able to do it with tools from GP, they don't use anything crazy in the proof

Balarka Sen

Good to know.

0celo7

@BalarkaSen hmm, can you make that precise?

Balarka Sen

Maybe I'll think a bit about it right now.

0celo7

there are two theorems which are very similar

Balarka Sen

@0celo7 $p : E \to M$ be my bundle (assume $M$ is compact). Then pick a generic section $s : M \to E$, whose image is a submanifold (it's an injective immersion). Look at it's intersection with the zero section in $E$. That's a homology class in $H_*(M)$.

Look at it's dual in $H^*(M)$.

I think that's the Euler class of $E$.

0celo7

18:49

Yeah, in the smooth/de Rham case you have to demand that the intersection is transversal so it's a sumanifold of $E$.

But that's what I'm having a lot of trouble with

Balarka Sen

Well, that's what a generic section is.

Transverse to the zero section.

0celo7

@BalarkaSen Let $\pi :E\to M$ be an oriented $k$-bundle and let $s$ be a section with a finite number of zeros. (This always exists.) Then the Euler class of $E$ is the Poincare dual of the zeros of $s$.

@BalarkaSen I hadn't heard that terminology before

Balarka Sen

@0celo7 OK, so $E$ has the same dimension as $M$ here.

0celo7

@BalarkaSen No, the $k$ is arbitrary.

Balarka Sen

What? Take a line bundle on $M$. A generic section of that has a whole codimension 1 submanifold of zeroes.

That's hardly ever a finite set.

trilolil

18:53

I know some of you have read my post just above, plz feel free to share your knowledge and give me an answer

0celo7

@BalarkaSen Yeah you're right, the bundle rank is the manifold dimension.

Balarka Sen

Duh.

0celo7

Duh?

Balarka Sen

Aka, "obviously". By transversality.

0celo7

I don't understand your example

oh, nevermind

Balarka Sen

18:55

Duh :P

0celo7

If I tell you what I saw it would spoil the proof you want to write

Balarka Sen

Yes, it would. So preferably you shouldn't spoil.

0celo7

The thing with homotopy of fibers earlier relates to the result you quoted.

Balarka Sen

@0celo7 I have an idea on how to prove my thing.

$p : E \to M$ by my bundle and $s$ be a section ($M$ is compact). $p \circ s$ is identity. Homotope $s$ to a transverse section $s'$. $p \circ s'$ is then homotopic to $p \circ s$. I can choose the homotopy to be arbitrarily small, hence make $p \circ s'$ a diffeomorphism of $M$ (transversality of diffeomorphisms). Call this $f$. Look at $s' \circ f^{-1}$: that's the desired section though isn't it?

'Cause $p \circ s' \circ f^{-1} = f \circ f^{-1} = \text{id}$.

0celo7

19:11

Yes.

Balarka Sen

Cool.

0celo7

transversality of diffeomorphisms = diffeomorphisms are stable class?

Balarka Sen

Oh, yes, sorry. Stablity of diffeomorphisms, I meant.

0celo7

My hint would have been exactly that.

Balarka Sen

I see.

0celo7

19:16

@BalarkaSen Your $s'$ is not a section though.

Just a map $M\to E$, right?

Balarka Sen

Whoops, yes, transverse map I meant.

Sorry for wrong terminologies.

0celo7

I wish you hadn't mentioned that Euler class result.

I was happy to ignore it but now I want to understand it :)

Balarka Sen

heh. what's B-T's defn of Euler class?

0celo7

It takes roughly a page to define. See page 117.

Balarka Sen

Oh no that's way too hard for me.

0celo7

19:20

Huh?

It's a Čech-de Rham cohomological definition.

Not very pretty tbh.

It's an element of the Čech cohomology.

Balarka Sen

It looks algebraic to me; I haven't studied Cech-de Rham.

0celo7

It is algebraic

Algebraic using the Čech-de Rham augmented double complex.

Balarka Sen

I see.

0celo7

@BalarkaSen Very interesting problem: classification of connected Abelian Lie groups.

They are completely classified by the fundamental groups.

Balarka Sen

Hmm, that's interesting.

Wait, no, I don't buy that. What do you mean, "completely classified"?

Isomorphic fundamental groups imply Lie group isomorphic, in particular, diffeomorphic?

0celo7

19:26

Namely, $\pi_1(G)=\Bbb Z^k$ for some $0\le k\le \mathrm{dim}(G)=n$, and $G\cong \Bbb R^{n-k}\times (S^1)^{\times k}$.

Does that make sense?

Balarka Sen

Yes. Yes it does. Interesting.

0celo7

The proof is really cool, IMO

You can break up the group into a bunch of 1-dimensional factors so $G=A_1\times\cdots\times A_n$.

By the classification theorem, each $A_i$ must be either $\Bbb R$ or $S^1$.

The fundamental group fixes each one.

Balarka Sen

@0celo7 Why's this true?

0celo7

@BalarkaSen You need the theorem: Let $G,H$ be connected Lie groups. For every Lie group homo $F:G\to H$, $F$ is surjective iff $F_*:\mathfrak g\to\mathfrak h$ is an isomorphism.

Now, we consider the Lie group homomorphism $\exp:\mathfrak g\to G$, where $\mathfrak g$ is an additive Abelian Lie group.

This is a homomorphism because of BCH and because $\mathfrak g$ is an Abelian Lie algebra.

Clearly $\exp_*=\mathrm{id}_\mathfrak{g}$ is an isomorphism, so $\exp$ is surjective.

Then $G=\exp \mathfrak g=\exp\bigoplus \mathfrak g_i=\prod \exp\mathfrak g_i=\prod A_i$.

where $\mathfrak g_i$ are 1-D subspaces of the Lie algebra

Balarka Sen

Mm, I see.

I don't understand much of that, admittedly, but cool.

0celo7

19:38

It's an exercise in Lee

One of the more interesting bits of that book

@BalarkaSen Bott & Tu are taking the Čech cohomology of a good cover with values in a covariant functor

It's starting to sound like nLab

Balarka Sen

Oh well.

Good thing I am not the one reading Bott-Tu.

0celo7

will you forgive me if I don't understand this Euler class generic section thing

I can't make sense out of the proof

Balarka Sen

sure, why not. i am not your boss.

0celo7

my one idea was to use homotopy invariance of cohomology but compact vertical cohomology is not homotopy invariant.

Balarka Sen

vertical cohomology? sounds like bott-tu has turned your life upside down.

OK, gotta go.

0celo7

19:44

@BalarkaSen Yes, the compact vertical cohomology of a vector bundle $\pi :E\to M$ is the cohomology of the double complex $C^\bullet(\pi^{-1}\mathfrak U,\Omega^\bullet_{cv})$, where $\mathfrak U$ is a good cover of the base and $\Omega^\bullet_{cv}$ is the compact vertical diff form presheaf

i.e. forms that have compact support along each fiber.

@dmckee Whatcha teaching this semester

DanielSank

20:16

@DavidZ yes!

Danu

20:44

@0celo7 the proof of equality of different definitions of Euler characteristics outlined in G&P, 3.7 seems pretty much entirely satisfactory (once you complete the final step from going to # of simplices to Betti numbers)

0celo7

@Danu Once you prove triagulability, sure.

But it might be easier to just learn Morse theory instead :P

Danu

@0celo7 That's true.

0celo7

Or is it triangulizability?

@Danu So that's four proofs.

Poincare duality, Morse theory, characteristic classes, triangulations.

Oh, there's a fifth via pseudodifferential operators my advisor mentioned.

@Danu But the Poincare duality is the most satisfying because it works for any intersection number, and the Euler characteristic is a happy coincidence.

Balarka Sen

@Danu You're going through G&P?

user507974

hey guys, slight thermodynamics derp

does raising pressure lower gibbs free energy

0celo7

20:54

@BalarkaSen Someone proved RH, we need you to read the proof and check it.

user507974

wait, obviously no it doesnt

but one thing is bugging me, if we raise the Gibbs free energy we make a state less stable, so if we increase the pressure (and by extension Gibbs) then shouldnt that make the other state more relatively stable (lower temperature change needed to elicit phase transition)

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