The h Bar

@0celo7 I see you already got an answer on your normal bundle question but the answer is, again, using a fiberwise $\Bbb R^n \to \Bbb R^n$, by shrinking everything to the open unit ball, then using this to construct a global map. I agree there's a little subtlety involved in making that chart-wise right, that's why you have to arrange it to be identity near the zero section.

Once you're done, this is a bundle-morphism also a diffeo to image.

BTW, I'm ready for Poincare-Hopf, if so are you.

Bernard Meurer

@0celo7 Prove it

0celo7

1:05 AM

@BalarkaSen Sorry, eating dinner and then I have to pack for a trip

Balarka Sen

Ah, well.

0celo7

@BernardMeurer $\Bbb R$ is uncountable.

Balarka Sen

It's ok though, have fun on the trip.

0celo7

Thanks

Bernard Meurer

@0celo7 Rebecca?

0celo7

1:06 AM

Yes

Bernard Meurer

Say hi from me

0celo7

@BalarkaSen It should be noted that the proof I know of Poincare-Hopf uses Morse theory

GP's proof doesn't

Balarka Sen

All the better.

0celo7

@BalarkaSen the proof requires two concepts

the index of a vector field and the gauss mapping

let $v\in\Gamma(TM)$ have an isolated zero at $p\in M$, i.e. there is no other zero in some ball around $p$

Balarka Sen

Gauss map sends a point on your manifold to the normalized vector in the vector field at that point, right?

Given a vector field on the manifold, that is.

@0celo7 OK

0celo7

1:14 AM

Let $B$ be some ball in this region containing $p$, then $v$ has no zeros on $B$, and we can consider $\hat v=v/|v|$ on this ball

Balarka Sen

Yes.

0celo7

rather, on the boundary of the ball

a sphere $S$

Balarka Sen

That's a map $B - p \to S^n$, yup?

OK, boundary. Same things though.

0celo7

so we have a map $\hat v:S\to S^{n-1}$

Balarka Sen

Yeah, I think degree of that is index of $X$ on $p$.

0celo7

1:15 AM

yeah

I think you need $M$ to be orientable for this

Balarka Sen

P-H says sum of indices is Euler char? I forget.

0celo7

yeah

Balarka Sen

@0celo7 Um, I don't think you need to be orientable to define index around a ball.

0celo7

But first you have to show that the index sum does not depend on the vector field you choose

Balarka Sen

I mean you're looking at a chart, which you can orient.

I do believe orientability is required for P-H.

I mean, of course, otherwise I can take arbitrary orientations at arbitrary charts which doesn't make any sense. Yes, you are right.

0celo7

1:17 AM

there's some details in proving the index does not depend on the exact ball you choose

Balarka Sen

Right.

0celo7

and that it's invariant under certain homotopies which become important later

Balarka Sen

@0celo7 mhm

0celo7

Oh, $M$ is compact.

Balarka Sen

otherwise sum doesn't make sense

0celo7

1:18 AM

yeah

Balarka Sen

Compact > noncompact. :P

0celo7

also you need a tubular neighborhood later, and those are never nice for noncompacts

Balarka Sen

Believable.

small thing: is index of a vector field related to curl? dunno. it should be.

0celo7

let $X$ be a manifold with boundary, then the Gauss map sends the outward pointing normal on $\partial X$ to $S^{n-1}$, for each point in $\partial X$

@BalarkaSen I think so? It relates how the vector field wraps a sphere around itself.

Balarka Sen

OK.

@0celo7 Yes. So I think you can conclude that it's the same as curl of the vector field around $p$ by integrating along a ~~loop~~ sphere.

I'll think through this later.

0celo7

1:22 AM

So, given a vector field $v$ on a $\partial$-manifold $X$ with only isolated zeros, we have the following theorem: index of $v$ = degree of Gauss mapping on $\partial X$

Balarka Sen

Sounds a hell of a lot like Stokes' theorem.

0celo7

Nah, you want a sketch?

Balarka Sen

There is something deeper going on with that curl analogy, I bet.

@0celo7 If you want to sketch it, go ahead.

0celo7

Remove a ball around each isolated zero

Balarka Sen

Oh

That's a cobordism

0celo7

1:24 AM

Then $\hat v$ can be defined on the remaning manifold

@BalarkaSen yes, but we don't need that

Balarka Sen

I mean, at ends of a cobordism you have same degree

That's it.

0celo7

Yes, because of the boundary lemma

Balarka Sen

@0celo7 Did you not mean "sum of indices of $v$", instead of "index of $v$"?

0celo7

or whatever GP calls it

Balarka Sen

Right.

0celo7

1:25 AM

@BalarkaSen In my personal notes, I just call the index sum the "overall index"

Balarka Sen

OK, good terminology.

Sure. This is done.

But I am betting that curl analogy is not useless. This is the analogue of Stokes' theorem.

0celo7

So that shows on a boundary manifold, the index is does not depend on the vector field

but we are far from done

Balarka Sen

("swirly around boundary is the same as local swirlies inside")

0celo7

you now technically have to deal with nondegenerate zeros

Balarka Sen

Nondegenerate zeroes?

0celo7

1:27 AM

we require that all zeros of $\hat v$ are regular values

Balarka Sen

Ah, yes.

But I can homotope a bit

0celo7

exactly

Balarka Sen

By transversality theorem, I can make $0$ a regular value

0celo7

but for now we have to work with nondegenerate zeros

Take your compact manifold, and Whitney embed it

Then take a closed tubular neighborhood of constant thickness

Balarka Sen

Go on.

0celo7

1:29 AM

the closed part there is why compact is so important

in general noncompact manifolds do not have closed tubular neighborhoods

Balarka Sen

i agree, carry on

0celo7

Then the index sum equals the degree of the Gauss map of the boundary of the tubular neighborhood

This proof is hard

It uses Riemannian geometry

Balarka Sen

Hmm, OK. What have we accomplished till now?

0celo7

We've shown that the index sum on a compact manifold does not depend on the vector field

So if we can find a vector field that computes the Euler characteristic, we're done

Balarka Sen

@0celo7 Wait, wait, how did we do that?

We showed degree of Gauss map on bd is index.

Then we said this Whitney & tubular nbhd thing

0celo7

1:32 AM

@BalarkaSen Yeah, and Gauss map does not depend on the vector

Balarka Sen

OK. True. But I am missing out the relevance of the last tubular nbhd thing

0celo7

We showed that the index depends only on the manifold itself

@BalarkaSen Because the compact manifold does not have a boundary

We fatten it in $\Bbb R^{2n+1}$, and take the Gauss map on that

Balarka Sen

Ah. Hah.

Understood.

So, only remaining part is to construct a vector field w/ index = euler char.

0celo7

Yeah

That's where Morse theory comes in

Balarka Sen

Coolio.

0celo7

1:35 AM

You can find a Morse function whose gradient has that property

Balarka Sen

I see.

That's nice.

0celo7

It's probably an exercise in Hirsch

Balarka Sen

I have to study this proof carefully. Thanks for the intro.

0celo7

Milnor doesn't prove all of it, of course

And I'm not sure where to find the proof of the closed tubular neighborhood thing

It's an exercise in Milnor, and GP has an almost proof of it

Balarka Sen

Seems like good stuff I could ponder on.

0celo7

1:39 AM

Oh, i did have a proof in mind at one point

You need some exercises in GP

Balarka Sen

Kindly don't reveal; I'd want to think about it when I get there.

0celo7

That was the roughest of sketches anyway

Balarka Sen

heh.

Alright, I gotta go. Pack well.

0celo7

Oh shit

Thanks for reminding me

Balarka Sen

:P

0celo7

3:06 AM

@ACuriousMind If the fundamental group of a manifold has a torsion element, what does that imply for the universal cover?

lucas

3:33 AM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Thanks! But I cannot learn new language because I have lost my memory ability (I sometimes forget my friends name). And to learn a new language you must have a great memory. BTW, thank you for your goodwill to help me! Don't care me so much. I will leave the site for a long time after I earn fanatic badge (3 days later). It was too hard for me to visit the site for 100 consecutive day and I don't want to lose it for 3 days. Thank you very much because of all your helps! Good luck!

John Rennie

@dmckee None at all, and the chances I would actually buy one are near zero - it looks like a yuppy toy to me. But it caused enough amusement (bemusement?) to be worth posting :-)

Bernard Meurer

4:56 AM

@JohnRennie What's a cyclotron?

John Rennie

5:35 AM

Cyclotron

A cyclotron is a type of particle accelerator invented by Ernest O. Lawrence in 1932 in which charged particles accelerate outwards from the centre along a spiral path. The particles are held to a spiral trajectory by a static magnetic field and accelerated by a rapidly varying (radio frequency) electric field. Lawrence was awarded the 1939 Nobel prize in physics for this invention. Cyclotrons were the most powerful particle accelerator technology until the 1950s when they were superseded by the synchrotron, and are still used to produce particle beams in physics and nuclear medicine. The largest...

All the modern big accelerators are synchotrons (or linear), but cyclotrons are simpler to build and operate and I think they are still used where the beam energy doesn't have to be high.

DanielSank

6:38 AM

I'm trying to find a room in which I explained a thing to CuriousOne.

Anybody know how to do that?

@ACuriousMind I disagree with that so much it hurts.

Please do not state your motivation for doing science as if it represents the entire scientific community.

So much physics has happened precisely because the scientists involved had a very specific application in mind.

Nuclear fission, semiconductor junctions, the entire field of quantum computing...

This paper and all the science it represents happened because I was trying to perfect a very specific application.

...I hope someone reads these messages...

Secret

7:11 AM

^ journals.aps.org/pra/abstract/10.1103/PhysRevA.91.042312 multilevel quantum systems vs entangled qubits?

peterh

7:39 AM

@DanielSank I regularly search for my nick on the google. The chat logs are indexed there.

@DanielSank I've read. I think the main problem is that the practical usability of most new developments seems very far away.

@DanielSank And it has a collective demoralizing effect.

@DanielSank But: if we see, how are we living differently as a century before, the change is extreme. I think, the hardcore theorists moved always parallel with the practice. Einstein has got the Nobel prize for the photoelectric effect, and not for the GR.

Balarka Sen

7:57 AM

@DanielSank I do not think ACM meant "nobody does science to apply it in the real world". Rather, I'd interpret it as "people do not care about science just for it to be applied. Aka, stuff which doesn't have any potential applications doesn't mean it's useless"

But that's just me.

peterh

Of course nobody knows what will be once useful.

David Z

8:29 AM

@DanielSank hm, well if you look at your chat profile you can see some information that might be useful. Rooms you own, recent messages you've posted, recent replies to you, etc. (I'm not sure how much of this stuff is visible to you and how much is mod-only.)

DanielSank

8:51 AM

@DavidZ Thanks for the tip. Don't seem to be able to find what I'm looking for.

Secret

Daniel, have you heard about multilevel quantum systems (see journal link above). What are the advantages and disadvantages of using them instead of qubits?

DanielSank

@peterh I've chatted a lot. Not sure if I can find a specific conversation via Google. Will try.

@peterh Well, in my experience, most experimental labs have some applications always in mind.

@peterh Huh? Demoralizing? You're saying that doing science with applications in mind is demoralizing?

@ACuriousMind I think this idea that the artist doesn't art to make money sort of ignores the whole thing about living creatures propagating their genetic material, etc.

@Secret I have no idea.

Probably similar to using tri-level classical logic.

But that's a guess.

peterh

9:08 AM

@DanielSank No! I only suspect, doing science with the knowledge that the applications are far away, it could demoralizing.

@DanielSank Google search does this task very smartly, maybe you can search for keywords in it.

peterh

9:21 AM

@DanielSank You can search also like this: danielsank site:chat.stackexchange.com <- this will limit the results to the chat.

DanielSank

@peterh Ahhhh nice.

peterh

@DanielSank It is quite funny to know, that everybody can search for you in this way.

Secret

9:57 AM

This is what happens when you make $(-1)^2=-1$

Secret

10:07 AM

So basically the change is that, while for addition, the structure is identical to that of the integers, for the multiplication structure, you end up with an ideal generated by the element -1

That is, one you fall into the ideal $\langle -1 \rangle$, you have no way out except going via the long route of keep adding 1 s to get back to the positive integers

(Sorry there are two typos in the x table, two of the 6q should be 6)

John Duffield

12:31 PM

@DanielSank : Easy peasy.

bolbteppa

12:45 PM

@JohnDuffield Why do you think Griffiths calls $S = I \omega$ spin angular momentum for a rigid body with orbital angular momentum $\vec{L} = \vec{r} \times \vec{p}$?

John Duffield

12:56 PM

@bolbteppa : I don't know. Maybe it's because he thinks the electron is a point particle?

Despite the evidence of things like the Einstein-de Haas effect, and despite things like the Poynting vector of the electron.

bolbteppa

How exactly does E-de H demonstrate spin? I mean what is the underlying reason spin arises in that case?

John Duffield

@bolbteppa : by virtue of conservation of angular momentum. See Wikipedia: "the Einstein–de Haas effect demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics."

"Considering Ampère's hypothesis that magnetism is caused by the microscopic circular motions of electric charges, the authors proposed a design to test Lorentz's theory that the rotating particles are electrons. The aim of the experiment was to measure the torque generated by a reversal of the magnetisation of an iron cylinder."

Note though that the electron isn't like some spinning ball. It isn't some "spinning charge". Feynman said "If you look at the energy flow, you find that it just circulates around and around". The electron is the electron because of this.

bolbteppa

Hmm, so looking at the picture, an electric current flies through the wire, a moving electric current generates a magnetic field rotating around the wire right? If the wire was straight it would just rotate around one axis, but since it's coiled the magnetic field rotates around 2 axes, right? Thus two rotational parameters in the magnetic field are used & so spin naturally arises?

peterh

1:13 PM

@JohnDuffield Sorry the silly remark :-) But on my little understand, yes, the electron is thought a point-like particle, with undefined location (more exactly, the wavefunction abs square is the probability distribution of its location).

@JohnDuffield And, if I upload beautiful pictures about the visualized wave functions, I get a lot of rep. :-) But I don't do this often.

John Duffield

@bolbteppa : forget about the current in the wire. You reverse the magnetisation of the cylinder and it turns.

bolbteppa

I don't really even understand the experiment, so current goes through the coil and because the magnetic field rotates around the wire it interacts with the block thing which reacts to the magnetic field and rotates in response to it?

John Duffield

@peterh : it isn't. The electron isn't some speck that has a field. The electron's field is what it is.

peterh

@JohnDuffield The electrons EM field? Next to it, it has a lot of other quantum numbers as well, isospin and hypercharge and similars, it has also a rest mass.

John Duffield

@bolbteppa : reversing the magnetisation in the cylinder changes the angular momentum of the electrons. Conservation of angular momentum means the cylinder turns.

@peterh : yes. The electron isn't some speck that has an electromagnetic field. It is electromagnetic field. If you prefer to think in terms of the electron field, the same applies there. It's quantum field theory, not quantum point-particle theory.

peterh

1:22 PM

@JohnDuffield And, it can be described as excitations of the electron field.

@JohnDuffield Ok, but it is also isospin/hypercharge field, too. Or not?

John Duffield

@peterh : I'm sorry, I don't know what you mean by an isospin/hypercharge field.

peterh

@JohnDuffield The quantum numbers what mixes to produce the weak interaction and the EM interaction on low energies.

@JohnDuffield The e has quantum numbers associated with them: the weak isospin and the hypercharge. In particle data tables, these quantum numbers are next to its electric charge.

John Duffield

@peterh : IMHO it's best to understand charge before moving on to hypercharge. Cut a paper strip like this, form it into a Möbius strip, then take a look at the Wikipedia spinor article, and the atomic orbitals article: "electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves".

Standing wave, standing field.

peterh

1:41 PM

@JohnDuffield Ok, I check the page. I suspect, you are thinking on an analogy, as if the charge would be same preserved topological phenomenon.

@JohnDuffield I think I already know what is the wave function and the bound quantum state. Now I am reading the spinor article.

John Duffield

The strip of paper represents the electromagnetic four-potential of a 511KeV photon. See depictions of an electromagnetic wave. The sinusoidal "electric wave" is the spatial derivative of it, whilst the orthogonal "magnetic wave" is the time derivative of it. See this section of the Wikipedia electromagnetic radiation article. But note that it's an electromagnetic wave.

There aren't really two orthogonal waves moving in tandem.

I have to go I'm afraid. Bye.

0celo7

2:59 PM

@yuggib my sunglasses are made in Italy

Danu

3:09 PM

What has the h bar come to...

0celo7

3:20 PM

@Danu hey

peterh

@JohnDuffield If an electron-left transforms to an electron-right, it is like a 360 ${}^\circ$ rotation of a spinor?

2

Danu

Priceless

peterh

@Danu What?

Danu

You and John Duffield are good conversation partners

also hey @0celo7

0celo7

3:40 PM

@Danu If my fundamental group has a torsion element, does that mean anything special for the universal cover?

Danu

3:57 PM

If it's pure torsion then you have a finitely-sheeted covering of course

peterh

@JohnDuffield In the Higgs lagrangian, the kinetical term + energy minimization causes that the Higgs field is the same in the whole Universe?

Danu

@peterh the universe is not described by classical field theory---this (constant field) can never happen for any field

peterh

@Danu Yes, but it is for the other part of the Lagrangian, the well-known mexican hat potential. But it has also another part, which seem to me as space-derivates of the field: theory.sinp.msu.ru/comphep_old/tutorial/node106.html Somewhere I've found, it is named kinetical term.

Danu

@peterh I've got to go mow my lawn :P Sorry, bye!

peterh

@Danu If I understand the formula well, then the only possibility to nullify the lagrangian, if both the kinetical term and the mexican hat potential is zero. What means, the squared sum of the Higgs components must be the Higgs VeV (to null out the mexican hat part), and these components must be the same everywhere in the universe (to zero out the kinetical term).

@Danu Good night :-)

0celo7

4:15 PM

@Danu not pure torsion

And why dos pure torsion imply finitely sheeted?

Does pure torsion imply a finite group?

Qmechanic

1

Q: Energy of tides: how does the change of spin of a body affect another body through gravity?

There are a few posts here at PSE that explain how the energy of tides comes at the expense of the Moon/Earth angular momentum. There is something that escapes me: how can angular momentum or KE be affected by what happens on Earth? How does the Moon know there are tides on Earth? its pull , whi...

Duplicate?

Danu

5:12 PM

@0celo7 Try it out :P

@peterh You're talking about classical things---on the classical level the Higgs field $\phi$ is indeed constant in a vacuum. I just pointed out that the classical level is not reality.

(@0celo7 in fact what I said was wrong---I had the case of a single generator in mind)

Burnside's problem

The Burnside problem, posed by William Burnside in 1902 and one of the oldest and most influential questions in group theory, asks whether a finitely generated group in which every element has finite order must necessarily be a finite group. In plain language, if by looking at individual elements of a group we suspect that the whole group is finite, must it indeed be true? The problem has many variants (see bounded and restricted below) that differ in the additional conditions imposed on the orders of the group elements. == Brief history == Initial work pointed towards the affirmative answer. For...

Bernard Meurer

6:56 PM

Hello fellow children

Emilio Pisanty

8:01 PM

@dmckee You pinged?

I imagine it's about this one but all I see is this.

ACuriousMind

11:11 PM

@DanielSank I read most things that are directed at me ;) I'll concede that my statement there contains hyperbole, what I wanted to express is that science, like art, is thought of as an end in itself, not merely as means to an end.

@DanielSank What "whole thing"? The external biological/sociological/historical/psychological reasons for a human's behaviour are completely disjoint from the internal narrative of that human for why they do things. Additionally, human society as we know it has existed for far too short a time to make it probable that any specific human behaviour is the result of selection under evolutionary pressure (most mutations are actually neutral with respect to fitness to begin with)

Balarka Sen

11:32 PM

@0celo7 Fundamental group corresponds bijectively to preimage of a point by the covering map.

Pure torsion implies finite, so...

Ah, Danu mislead. Lots of infinite pure torsion groups. $\Bbb Z/2 \oplus \Bbb Z/2 \oplus \cdots$.

OK, now I see Danu did mention that he mislead.

Balarka Sen

11:46 PM

I googled, and it seems it's open whether there is a pure-torsion finitely presented infinite group, see e.g. here. This means it is open too whether or not what Danu was trying to say is true even for compact manifolds, which is what I was thinking of.

Interesting.

apparently it's called the Burnside's problem.

ACuriousMind

7 hours ago, by Danu

Burnside's problem

The Burnside problem, posed by William Burnside in 1902 and one of the oldest and most influential questions in group theory, asks whether a finitely generated group in which every element has finite order must necessarily be a finite group. In plain language, if by looking at individual elements of a group we suspect that the whole group is finite, must it indeed be true? The problem has many variants (see bounded and restricted below) that differ in the additional conditions imposed on the orders of the group elements. == Brief history == Initial work pointed towards the affirmative answer. For...

Are you somehow selectively not reading what Danu wrote?

Balarka Sen

Darn.

Nope, I missed all of that.

Probably a result of not sleeping for 3 days.

Heh, but it's interesting to see how both of our trail of thoughts agreed.

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