12:00 AM
@DanielSank I learned it from a children's cartoon. Ni hao Kai-Lan
Quite instructive

@BernardMeurer I find it amusing that in Chinese "ni" means "you". So "nihao" is like "ni how" which is like "how are you?".

Hmm, that is interesting!

hao means how?

@DanielSank Is my QC ready yet?

@0celo7 I don't know, but it sounds the same.
@BernardMeurer No.

12:02 AM
That's it. I'm calling google's customer support
I want to be able to break RSA-2048 keys in under 30 minutes

12:15 AM
Do you want to hack Bob?

12:56 AM
@0celo7 What's bigger $\mathbb{N}$ or $\mathbb{R}$ ?

$\Bbb R$, assuming your notion of "bigger" is reasonable

Define reasonable

"there are more things in $\Bbb R$ than in $\Bbb N$"

@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.

@0celo7 Prove it

1:05 AM
@BalarkaSen Sorry, eating dinner and then I have to pack for a trip

Ah, well.

@BernardMeurer $\Bbb R$ is uncountable.

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

Thanks

@0celo7 Rebecca?

1:06 AM
Yes

Say hi from me

@BalarkaSen It should be noted that the proof I know of Poincare-Hopf uses Morse theory
GP's proof doesn't

All the better.

@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 $p$

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

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

Yes.

rather, on the boundary of the ball
a sphere $S$

That's a map $B - p \to S^n$, yup?
OK, boundary. Same things though.

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

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

1:15 AM
yeah
I think you need $M$ to be orientable for this

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

yeah

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

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

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.

1:17 AM
there's some details in proving the index does not depend on the exact ball you choose

Right.

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

@0celo7 mhm

Oh, $M$ is compact.

otherwise sum doesn't make sense

1:18 AM
yeah

Compact > noncompact. :P

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

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

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.

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.

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 $\partial X$

Sounds a hell of a lot like Stokes' theorem.

Nah, you want a sketch?

There is something deeper going on with that curl analogy, I bet.
@0celo7 If you want to sketch it, go ahead.

Remove a ball around each isolated zero

Oh
That's a cobordism

1:24 AM
Then $\hat v$ can be defined on the remaning manifold
@BalarkaSen yes, but we don't need that

I mean, at ends of a cobordism you have same degree
That's it.

Yes, because of the boundary lemma

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

or whatever GP calls it

Right.

1:25 AM
@BalarkaSen In my personal notes, I just call the index sum the "overall index"

OK, good terminology.
Sure. This is done.
But I am betting that curl analogy is not useless. This is the analogue of Stokes' theorem.

So that shows on a boundary manifold, the index is does not depend on the vector field
but we are far from done

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

you now technically have to deal with nondegenerate zeros

Nondegenerate zeroes?

1:27 AM
we require that all zeros of $\hat v$ are regular values

Ah, yes.
But I can homotope a bit

exactly

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

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

Go on.

1:29 AM
the closed part there is why compact is so important
in general noncompact manifolds do not have closed tubular neighborhoods

i agree, carry on

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

Hmm, OK. What have we accomplished till now?

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

@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

1:32 AM
@BalarkaSen Yeah, and Gauss map does not depend on the vector

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

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

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

Yeah
That's where Morse theory comes in

Coolio.

1:35 AM
You can find a Morse function whose gradient has that property

I see.
That's nice.

It's probably an exercise in Hirsch

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

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

Seems like good stuff I could ponder on.

1:39 AM
Oh, i did have a proof in mind at one point
You need some exercises in GP

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

That was the roughest of sketches anyway

heh.
Alright, I gotta go. Pack well.

Oh shit
Thanks for reminding me

1 hour later…
3:06 AM
@ACuriousMind If the fundamental group of a manifold has a torsion element, what does that imply for the universal cover?

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!

@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 :-)

1 hour later…
4:56 AM
@JohnRennie What's a cyclotron?

5:35 AM
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.

1 hour later…
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...

7:11 AM
^ journals.aps.org/pra/abstract/10.1103/PhysRevA.91.042312 multilevel quantum systems vs entangled qubits?

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.

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.

Of course nobody knows what will be once useful.

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.)

8:51 AM
@DavidZ Thanks for the tip. Don't seem to be able to find what I'm looking for.

@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.

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.

9:21 AM
@DanielSank You can search also like this: danielsank site:chat.stackexchange.com <- this will limit the results to the chat.

@peterh Ahhhh nice.

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

9:57 AM
This is what happens when you make $(-1)^2=-1$

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)