@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

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

I'm confused by this situation

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

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

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

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

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"

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

@DavidZ That's strange...

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

@DavidZ ^ on a post 2 hours old.

Which post?

I think there's also some criterion about votes

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

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

@DavidZ Ah, that could explain it!

I think that's terrible, though.

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

What do you mean, above the NAA queue?

Sorry, corrected.

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

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

@Danu It does not use Morse theory

Well, one of the proofs does

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

I don't think Ted was right

@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)

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

@DavidZ Well, I know certain users...

I will use el Google

@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?!

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 ;)

@ACuriousMind wow stop getting banned

@Danu As reasonable an explanation as any :D

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

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

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

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

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

I would like to travel to Buenos Aires.

@DanielSank do

It's awesome

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

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

btw I've got this on at the moment

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.

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

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

@DanielSank Here, have an ñ

Gracias

De nada, caballero

You aware of Otros Aires?

No!

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

Also Plaza Francia

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

@DanielSank Bajofondo is also awesome

^ Truth

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

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

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

contorts hand to do right hand rule

how does this thing work?

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

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

lol

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

Protect Mr. X at all costs!

Also I hate how bad bibtex is :P

@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

@Danu do you have your summary on dropbox

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

thanks

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

My style?

For a summary, that's fine

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

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

or is what you're talking about

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

That is good

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

"the embedding"

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

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

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

@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?

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

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

@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?

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

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

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

@Danu Yes, they're all smooth.

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

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

@BalarkaSen Exactly

Mike's supervisor proved that!

@BalarkaSen No.

(for all dimensions $\geq 5$)

They do not prove it.

They give the idea.

You said that.

@Danu Yup :)

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

Fair enough.

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.

Good luck with your proof.

Want to hear it?

No.

Aww

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.

Oh

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

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.

C^1 and C^infty are not different.

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

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

Yes.

Ok.

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

Not $k=0$.

That's silly.

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

Non-smooth literally means that.

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

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."

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.

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

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

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

I don't understand it yet

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

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

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

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

Nice, tell me the page number.

123

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

You don't want a hint I assume?

No.

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

Good to know.

@BalarkaSen hmm, can you make that precise?

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

there are two theorems which are very similar

@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 $E$.

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

Well, that's what a generic section is.

Transverse to the zero section.

@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

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

@BalarkaSen No, the $k$ is arbitrary.

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.

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

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

Duh.

Duh?

Aka, "obviously". By transversality.

I don't understand your example

oh, nevermind

Duh :P

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

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

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

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

Yes.

Cool.

transversality of diffeomorphisms = diffeomorphisms are stable class?

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

My hint would have been exactly that.

I see.

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

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

Whoops, yes, transverse map I meant.

Sorry for wrong terminologies.

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

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

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

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

Oh no that's way too hard for me.

Huh?

It's a Čech-de Rham cohomological definition.

Not very pretty tbh.

It's an element of the Čech cohomology.

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

It is algebraic

Algebraic using the Čech-de Rham augmented double complex.

I see.

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

They are completely classified by the fundamental groups.

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?

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?

Yes. Yes it does. Interesting.

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.

@0celo7 Why's this true?

@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

Mm, I see.

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

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

Oh well.

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

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

I can't make sense out of the proof

sure, why not. i am not your boss.

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

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

OK, gotta go.

@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

@DavidZ yes!

@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)

@Danu Once you prove triagulability, sure.

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

@0celo7 That's true.

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.

@Danu You're going through G&P?

hey guys, slight thermodynamics derp

does raising pressure lower gibbs free energy

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

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)