The h Bar

user54412

12:06 AM

So I asked myself, "How can someone who makes such comments have a nonnegligible reputation?" It turns out much of that comes from asking a question with lots of pretty, irrelevant pictures to mask how little content there is.

user54412

Thankfully this strategy isn't quite as abused here as on some of the other sites.

ACuriousMind

@ChrisWhite ::sigh::

user218912

@ACuriousMind I'm preparing books to read in the fall.

user218912

gonna try and finish qft and gr (3rd attempt) in 70 days.

user218912

I already know the basics of both so it should be ok.

ACuriousMind

12:13 AM

@3750 I think you might vastly underestimate the depth of both subjects.

user218912

probably.

user218912

I don't mind though, I just want to do string theory already!

user218912

the depth will fill in later.

ACuriousMind

Yeah, it doesn't work that way.

user218912

then what should I do D:

ACuriousMind

12:16 AM

Forget about string theory and learn its prerequisites for their own sake.

user218912

fine.

Slereah

5

user218912

lol

Mike Miller

"forget about string theory [...]" :)

ACuriousMind

@Slereah that's not wrong!

Slereah

12:18 AM

Don't let Lumo hear you

Really I don't even think QFT and GR are enough to start on string theory

You need some CFT as well

ACuriousMind

@Slereah I said that already

Random facts from differential geometry and group theory are also pretty much required.

Slereah

yeah

Isn't there like

Virasoro stuff too

ACuriousMind

@Slereah That's basically CFT

Slereah

all that for a rubbish theory

ACuriousMind

One can reduce basic 2D CFT pretty much to the study of the Virasoro algebra

Mike Miller

12:23 AM

so what's a gravitational anomaly

ACuriousMind

Quantum anomalies happen when the quantum theory (read: the path integral) is not invariant under a symmetry of the classical field theory you started with

If the symmetry is just some random global symmetry, this is nothing to worry about, and actually not very interesting in itself.

Mike Miller

better question: do Witten's global gravitational anomalies exist?

ACuriousMind

But if the symmetry is a gauge symmetry (associated to a connection; is a local transformation), then such an anomaly just utterly destroys the quantum theory. Being gauge/gravitationally anomaly-free is a consistency requirement for every quantum theory.

12 hours ago, by ACuriousMind

Does anyone know of a refutation of this paper? I've been looking into it for two hours now and I don't know which formalism does which time ordering why anymore.

@MikeMiller ^I was wondering the same thing earlier today

What is your reason for doubting their existence?

Mike Miller

I don't doubt it. His abstract says they're the same as exotic 4-spheres, a well-known open problem.

I would be interested if there's a physical prediction of exotic spheres.

ACuriousMind

@MikeMiller It doesn't say 4-spheres

It just says spheres.

I think the gravitational anomalies discussed here are in dimensions 4k+2, k > 0.

So no 4-spheres for you

Hm, might not be 4k+2, that must've been a different paper, but regardless I think this is about higher dimensions

Slereah

12:39 AM

@ACuriousMind Does this make sense

From the uniformization theorem we have all metrics on $R^2$ conformal

So that means all metrics have the same eigenvalues

Mike Miller

@ACuriousMind Woah, my bad. He's interested in exotic diffeomorphisms, and those are only the same as exotic spheres in sufficiently large dimensions (4k+2, k>0 suffices).

Slereah

Well, they have different eigenvalues, but they are of the form $diag(f(x,y), f(x,y))$

Mike Miller

There are no exotic 4-spheres that come from exotic diffeomorphisms of a 3-sphere. So you'd like to know why there are exotic spheres in certain dimensions?

NeuroFuzzy

@Slereah I'm on panel two!!!!!

Woohooo!

Mike Miller

If gravitational anomalies in dimension $n$ are exotic diffeomorphisms of $S^n$ mod isotopy, then they're in bijection with exotic $(n+1)$-spheres. Milnor and Kervaire proved that the group of exotic $(4k-1)$-spheres is quite large.

Slereah

12:43 AM

I don't think it makes too much sense because that would mean that the eigenvalues of all real symmetric matrices with det > 0 have the same eigenvalues

Mike Miller

A reasonable conjecture is that there are exotic $n$-spheres for all sufficiently all sufficiently large $n$; let's say $n>126$.

Slereah

Which I don't think is true

ACuriousMind

@MikeMiller My translation of pg. 207f.: It's not that the anomalies are directly related to the spheres. It's that certain anomalies are related to certain "gravitational instantons" (in this case certain manifolds) and physically, we only need to include such an instanton causing an anomaly in our path integral if there is a corresponding anti-instanton than when connectedly summed with the old one gives the usual $S^n$.

Now the Poincaré conjecture tells us that the instanton is actually already topologically a sphere - and the question is now where this gives us a way to construct the anti-instanton

For n>4, Witten knows that it's an exotic sphere associated to some exotic diffeomorphism, and just taking the inverse diffeomorphism, he gets the anti-instanton. For n=4, the lack of knowledge about exotic 4-spheres simply makes it impossible to say anything about the (non-)existence of such anomalous things.

Mike Miller

OK, sorry, so a gravitational instanton is a manifold that connect sums to the standard sphere?

ACuriousMind

@MikeMiller At least, those are the physically relevant ones, yes.

There might be more instantons in general, but Witten gives a physical argument before that why in general only instantons that sum with an anti-instanton to the trivial configuration are physically of relevance.

@Slereah I think you're misapplying the uniformization theorem again.

A metric on $\mathbb{R}^2$ need not be conformal to the flat metric - it could as well be conformal to the hyperbolic metric on the unit disk.

Mike Miller

12:59 AM

@ACuriousMind OK, but like you say "manifolds that sum to $S^n$" are the same as exotic spheres, so your gravitational anomalies really are the same as exotic spheres, then.

A nice equivalent condition is that a metric on $\Bbb R^2$ is conformally equivalent to the plane iff every bounded harmonic function is constant

Slereah

@ACuriousMind Yeah I think I will never get it :p

But

ACuriousMind

@MikeMiller No, because not every anomaly is guaranteed to come from such an instanton. Instantons cause anomalies, but to my knowledge there is no guarantee that all anomalies come from instantons (Witten seems to agree, check the beginning of section III)

Slereah

If it is conformal to the flat metric

Does what I say apply

ACuriousMind

It's the gravitational instantons which are the same as exotic spheres.

Mike Miller

Sorry, sorry, my bad.

I'm playing rocket league so I'm a little slow right now.

So what's the claim that you're skeptical of?

ACuriousMind

1:06 AM

@MikeMiller Ah, well, it's not me who is sceptical, but I found this paper that claims Alvarez-Gaumé's and Witten's computation of gravitational anomalies made a subtle but rather elementary error and that the proper physical quantities instead of what they computed are non-anomalous. The reaction in the physics community to that paper seems to be largely non-existent, which puzzles me even more.

Trying to actually check whether there is an error or not has led me down a rabbit hole of related things but so far not brought me closer to deciding whether that paper doubting Witten is nonsense or not.

And now it's 3am and I should better go to bed :)

Mike Miller

@ACuriousMind OK, gotcha. Funny coincidence.

vzn

1:42 AM

@Slereah lol didnt know 0celo7 modelled for cartoons!

@ACuriousMind maybe email witten? ;) & hey we are always looking for new special guest speakers & you dont seem interested :P

Slereah

65537-gon

In geometry, a 65537-gon is a polygon with 65537 sides. The sum of the interior angles of any non-self-intersecting 65537-gon is 23592600°. == Regular 65537-gon == The area of a regular 65537-gon is (with t = edge length) A = 65537 4 t 2 cot ⁡ π 65537 {\displaystyle A={\frac {65537}{4}}t^{2}\cot {\frac {\pi }{65537}}} A whole regular 65537-gon...

The illustration is just a circle

Those cheeky buggers

vzn

← walked into caltech bookstore & bought feynman bio book, phd comic dvd, and caltech tshirt today =D

Slereah

10:31 AM

How do Wilson loops work

$$w_C = \mathcal P e^{\oint_C A^\mu dx_\mu}$$

But then

How do you determine $A_\mu$

I thought the point was that we could solve the gauge field without the gauge equation

Is there an equation for Wilson loops to solve

ACuriousMind

10:43 AM

@Slereah ?

@Slereah ??

Slereah

Yeah I dunno

Trying to find a good introductions to wilson loops

ACuriousMind

What do you mean "solve the gauge field without the gauge equation"?

If you're worried that that might be gauge variant - it's not, the exponential is the holonomy around $C$, which is a gauge-invariant quantity.

Slereah

Well from what I have gathered

Apparently you can find the value of the field from the wilson loops

Or is that wrong

ACuriousMind

If you are able to calculate arbitrarily small Wilson loops, that gives you a way to calculate the field strength, not the gauge field (I think the holonomy<->field strength thing is the Ambrose-Singer theorem or something)

Slereah

But then, how do I calculate the Wilson loops?

The formula given implies that you use the gauge field, but then if you have the gauge field, you already know what the field strength is

ACuriousMind

10:48 AM

Well, it really depends on what you are given.

Slereah

What can I be given

ACuriousMind

For example, in lattice theory, you usually don't have "the gauge field", but an assignment of group elements to edges of the lattice (which should the thought of as the integration of $A$ along the edges)

From that data, calculating Wilson loops is easy - just multiply the group element belong to the edges in the loop together in succession, but you don't really have a mechanism to extract values for $A$.

Slereah

that is fine, but

Why would you be given the group elements on the edges of the lattice?

Is that the boundary conditions?

ACuriousMind

@Slereah With "edge" I mean any line connecting two neighbouring lattice sites.

It's not boundary condition, it's the field configuration (that you probably generated by some Monte-Carlo scheme since you're probably trying to evaluate a path integral over all gauge field configurations)

Slereah

Oh right

ACuriousMind

10:53 AM

Perhaps this conversation can go better if you tell me why you're looking at Wilson loops? ;)

Slereah

Well I'm trying to solve a field

Classical one

Equation is a bit tough

Geodesic equation, not so much

Wondering if there is some way I can use this to help

ACuriousMind

What have Wilson loops to do with either of that?

Slereah

Well that is why I am investigating

To learn about them

ACuriousMind

But I didn't hear anything about a gauge field anywhere there

Or is your classical field a gauge field?

Slereah

Well I don't mind all that much

I used a scalar field, could be an EM field :p

ACuriousMind

10:56 AM

But a scalar field doesn't have Wilson loops

Wilson loops occur only in gauge theories

Slereah

Hence why I said an EM field

The point is solving a field on it

ACuriousMind

I don't think Wilson loops help you in any way classically

They are interesting in the quantum theory because they often are order parameters for phase transitions and because it's sometimes better to use them directly as the dynamical variables, especially at strong coupling.

FrancescoS

Hi guys! physics.stackexchange.com/questions/265802/… anyone?? :)

Bass

@ACuriousMind Is T-product the time-ordered product? And what's the T* product?

ACuriousMind

11:12 AM

@Bass Both are time-ordered products but they differ with their behaviour w.r.t. the time-derivative - the T* one is that which the path integral computes and commutes with time derivatives, while the T-product (the standard time-ordering) picks up certain equal time-commutators when you try to pull a time derivative past it.

For polynomials in fields, these do not differ, since fields usually have an equal time CCR that makes the commutator vanish.

But if you have derivatives in the operators you're considering, one can get non-zero corrections.

Bass

11:26 AM

@ACuriousMind Like this $[\partial_t,[\phi(x), \pi(y)]]=[\partial_t,i\delta^{(3)}(x-y)]=0$? What's an example of a field where this commutator does not vanish?

ACuriousMind

@Bass That doesn't make sense, you can't take the commutator of a time derivative with a field - the time derivative is not an operator. What I meant is that if you examine $\langle T\partial_0\phi(x)\phi(y)\rangle$ and $\partial_{0,x}\langle T\phi(x)\phi(y)\rangle$ then you find the two quantities differ by $\langle \delta(x^0 - y^0)[\phi(x),\phi(y)]\rangle$.

I'm starting to think that this paper claiming Gaumé and Witten are wrong is nonsense though because after reading through everything carefully, I can't really say what exactly they think the error is, and I think they misunderstood how the argument about the anomaly is supposed to work.

Well, back to reading Witten, then...

Bass

11:43 AM

@ACuriousMind In QM you can: $[x,p]=[x,-i\partial_x]=i$, no? There the spatial derivative is an operator, no?

ACuriousMind

@Bass Yes and no. It is an operator if you choose the position representation $L^2(\mathbb{R}^3,\mathrm{d}x)$.

But, in quantum field theory, the field operators themselves depend on spacetime, and derivatives act on the fields, not on the space of states.

Bass

@ACuriousMind of course. Thanks!

ACuriousMind

The QM analogue to try ing to write $[\partial_x,\phi]$ is trying to write $[\partial_t,x(t)]$ in the Heisenberg picture.

Bass

12:54 PM

Ah I see, both the time derivative and the T product are operators on the space of operators (fields) which act on spaces. So the commutator is something like $[\partial_t, T]$, right?

ACuriousMind

@Bass Well...I wouldn't write it like that

Bass

@ACuriousMind Why is it $\langle T\partial_0\dots\rangle$ but then $\partial_{0,x}\langle\dots$? Why the $,x$ in the subscript of $\partial$?

ACuriousMind

@Bass I should've written it as $\langle T (\partial_0\phi)(x)\phi(y)\rangle$. The $,x$ is there to denote on which of the variables $x,y$ the derivative acts.

You could already write $\partial_{0,x}$ inside the brackets, but there it seems kinda superfluous.

Bass

Ah yes.

@ACuriousMind Isn't $p$ an operator too in the momentum representation? In the position representation, $x$ is an operator too: $\hat x: \psi(x)\mapsto x\psi(x)$

ACuriousMind

@Bass I'm not quite sure what you mean. Both position and momentum are operators - but multiplication with the variable $x$ or differentiation or something like that is but a representation of these operators on a particular space.

Bass

1:09 PM

I see. p is always an operator, but the derivative only in the position rep.

0celo7

Good el morning

@ACuriousMind So, when we talk about Ehresmann connections on G bundles, there is a globally defined connection form

But when we talk about them on vector bundles, we can only define it in a bundle chart?

Is that right

ACuriousMind

Not exactly

The globally defined connected form live on the bundle. If you want to get a connection form living on the base space, you also only get a locally defined one

0celo7

@ACuriousMind Ambrose-Singer is something like $\mathfrak{hol}_p\cong\{\Omega_p(u,v)\mid u,v\in\mathcal H_p\}$

Something like that...

fucking TeX

bah

ACuriousMind

You're missing a backslash before to opening curly bracket.

0celo7

@ACuriousMind So how is the globally defined connection form defined on a vector bundle

I could look it up in Kobayashi (Abrose Singer)

Because I have to hunt down that other thing anyway

I thought there was an Ambrose Singer for Riemannian geometry...

And it was completely different

ACuriousMind

1:24 PM

@0celo7 Dunno, never saw it

0celo7

o I was thinking of Ambrose Hicks

Where did I read about it tho

@ACuriousMind ???

you're not a very good geometer..

ACuriousMind

I'm not a "geometer" at all!

0celo7

@ACuriousMind kek

@ACuriousMind bullshit

Your profile says you're interested in geometry

You don't know anything besides principal bundles and even then you don't know the proofs

I bet you can't even prove that connections exist!!!

Theorem. Every principal G bundle admits a connection.

ACuriousMind

@0celo7 No, I'm interested in geometrical aspects of physics, not "geometry"

0celo7

Prove it please.

Hint: Partition of unity black magic.

@ACuriousMind What does that even mean?

ACuriousMind

1:32 PM

@0celo7 Set $\omega=0$. Q.E.D :P

0celo7

@ACuriousMind ...huh?

you're either trolling or a genius.

ACuriousMind

I guess I was trolling because what I thought doesn't work :P

0celo7

@ACuriousMind I don't think the connection form can be the zero map...

How does $\omega(A^*)=A$ work

ACuriousMind

I was thinking of the local connection form.

But if it is zero in one trivialization it's not zero in the next.

0celo7

@ACuriousMind damn physicist

ACuriousMind

1:37 PM

@0celo7 Yes, that's what I am.

0celo7

My brother's internet is so slow

@ACuriousMind You're welcome.

Bass

@0celo7 from what book is that?

0celo7

@Bass The only book that matters.

If you could save two books after the apocalypse, what would they be @ACuriousMind @Bass

ACuriousMind

Something on farming and probably something about basic machinery :P

0celo7

Make it three.

Four.

user116211

1:44 PM

What would you want to save @0celo?

0celo7

Kobayashi vols 1, 2, Lee SM, Helgason

Need to keep geometry alive!

@Bass Kobayashi Nomizu vol 1

Bass

@0celo7 is it good? Looks comprehensive..

user116211

2:00 PM

@0celo7: It seems too costly. I would go with EMIs to buy it ;/

0celo7

2:41 PM

@Bass Very comprehensive. Notoriously difficult.

Don't read unless you're a geometry junkie and want Bourbaki-style math.

Bass

Isn't it a bit old? Much has happened since the 60's..

0celo7

It's foundational material.

And you need to know a lot about Lie groups before reading it.

Bass

@0celo7 Uhuh.

0celo7

But it is the most comprehensive reference on connections on principal bundles out there.

Everything else skims details or assumes you've read KN.

yuggib

@0celo7 is "let's" acceptable in formal writing or is it better to use "let us"?

0celo7

2:50 PM

@yuggib That's tricky because "us" is already informal.

But as a general rule, contractions are never formal.

yuggib

@0celo7 so you would not use the construction let us at all?

it seems pretty common in scientific literature...

0celo7

@yuggib In formal writing I would not use first person.

@yuggib I have not done any of that...

I'm not the best person to ask.

yuggib

substitute "formal" with "scientific"

0celo7

I have no experience with that

Danu

I'm taking a lecture which is very elementary and covers the physically relevant stuff in about 200 pages.

yuggib

2:52 PM

I mean, you have read a lot of mathy books

0celo7

I guess go to your favorite books and ctrl F "let's"

Danu

"let us", you may find.

"let's", definitely not (in most)

0celo7

let us is what is in Jost

(I had Jost open on my desktop already)

Danu

@Bass ^^^^^

0celo7

@yuggib Lee uses let us over let's

@Danu hmm?

Danu

2:55 PM

On principal bundles

(but they're math notes)

0celo7

@Danu Kobayashi Nomizu doesn't cover Yang Mills

@Bass ^

yuggib

@0celo7 I see

0celo7

But if you ask Mike Miller something about G bundles and he doesn't know right away, he'll probably check KN

So that might tell you something.

@yuggib Lee is a gold standard for "nice" writing

Jost isn't

But he's a director of a Max Planck institute so he might have some authority

@yuggib Weinberg uses let us over let's

Danu

My course does :D

0celo7

hmm?

Danu

2:59 PM

@0celo7 a topologist I talked to said that Jost is well-known for having serious mistakes in his papers haha

0celo7

@Danu His books are well-regarded though

Danu

(but the right ideas)

0celo7

@yuggib so it's fine to use let us

yuggib

I will

0celo7

Even Kolar et al. use let us

And they're crazy prceise

unlike me

I do not know how to spell that word

@Danu ever heard of a "gauge natural bundle"?

Danu

3:05 PM

No

0celo7

I've heard rumors this has connections to physics.

yuggib

@0celo7 I've seen people doing $p$-adic and even adelic quantum mechanics

that does not mean they have any connection to physics whatsoever

Danu

lol

Bass

3:20 PM

@Danu ^^^^^ ?

Danu

@Bass I thought you were asking about principal bundles; only later did I see it was just 0celo7 talking to himself ;)

Bass

@Danu Are you texing the notes on principal bundles?

You seem to be texing everything :)

Danu

The prof is doing it himself

mathematik.uni-muenchen.de/~hamilton/gaugetheory.php

0celo7

> Prior knowledge: Linear algebra and calculus.

Sounds like KN

But then they define manifolds via pseudogroups of transformations and have zero pictures

Translation: algebra at the level of Lang and calculus at the level of Papa Rudin

Danu

This is really elementary.

(no pictures, though)

0celo7

3:32 PM

@Danu Do you consider KN elementary?

Danu

Never read it

0celo7

@Danu Suppose we have two metrics (on a metric space) which determine the same topology. If one is complete, is the other too?

exclude the trivial case in which the metric space is compact

ACuriousMind

@0celo7 No.

0celo7

@ACuriousMind can you provide an example please?

ACuriousMind

@0celo7 I thought you didn't like counterexamples :P

Take $d(x,y) := \lvert \arctan(x)-\arctan(y)\rvert$ on $\mathbb{R}$. The topology induced on $\mathbb{R}$ is the standard one, but the metric is not complete.

0celo7

3:44 PM

Oh god what kind of counterexample is that

@ACuriousMind I need to make sense of Hopf-Rinow given Ozeki-Nomizu

Riemannian geometry that you wouldn't understand

ACuriousMind

@0celo7 The one I saw in one of my courses :P

0celo7

@ACuriousMind why is it not complete :P

ACuriousMind

@0celo7 The sequence $(n)_{n\in\mathbb{N}}$ is Cauchy w.r.t. it, but doesn't converge in the standard topology.

0celo7

@ACuriousMind I leave the proof to the reader and believe you

@ACuriousMind Time to introduce my 5yo nephew to geometry

He's of age

Bass

@Danu Cool!

ACuriousMind

3:51 PM

@0celo7 I got the proof right here (it's 1.1b)).

@0celo7 of what age?

0celo7

@ACuriousMind the age at which geometry becomes accessible

We shall start with Lee...

@ACuriousMind Suppose $E\to M$ is a vector bundle with structure group $G$. Let $G\to P\to M$ be the associated principal bundle. IF $E$ has a bundle metric $g$, how is this structure represented on $P$?

ACuriousMind

@0celo7 What's a bundle metric?

0celo7

@ACuriousMind I used to look up to you. It's a smooth assignment of an inner product to each fiber.

ACuriousMind

I don't think that's a structure that makes sense on principal bundles

0celo7

@ACuriousMind It makes sense on vector bundles.

ACuriousMind

4:01 PM

@0celo7 so?

user218912

4:14 PM

@0celo7 do you still read on your ipad?

0celo7

@ACuriousMind hmm?

@3750 yes

user218912

I was considering buying one for reading.

user218912

do you like yours?

John Rennie

4:32 PM

@3750 I use a tablet for reading novels, and it's very good. I hardly read paperbacks these days. But it's rubbish for science/maths books because it's too slow to flick backwards and forwards.

user218912

@JohnRennie I was worried about that.

John Rennie

If you can work steadily though a maths book I guess the tablet would work, but you're like me you have to keep flicking back to find out why you can't understand the current chapter. Tablets are rubbish for that.

Slereah

WAIT

I think I got it

Finding the eigenvalues of the metric will not work because that can only do it up to linear coordinate change

I need to find a generic coordinate transform to get the conformally flatmetric

Tho I guess diagonalizing it might help to do that, still

Check those sweet eigenvalues

knzhou

5:43 PM

Youll have to take that claim up with prof micheal bender from whom i learned this method/ideas — user122066 2 mins ago

You just cant sum an infinite series term by term by term. Infinity isnt a sensical well defined concept. Thats where summing ny hand runs into trouble and why the infinite sum of integers is NOT divergent — user122066 47 secs ago

Every damn time. I try to explain something to somebody who looks a bit more confused than usual, and then this stuff pops out.

Where does this crap come from?

Slereah

6:44 PM

But you don't sum an infinite term

That's what the regularizer is for

dmckee

8:15 PM

@knzhou It takes a special kind of person to persevere through the minutia of rigorous treatments of infinite summations in a autodidactic education.

Most of us need a mentor to keep on us about that.

But without understanding that stuff (or at least recalling the time when you did understand it) you can get into this kind of trouble right very quickly.

Personally I've had to recover that knowledge every damn time I've gone on to deepen my analytic understanding.

Danu

9:22 PM

@ACuriousMind There might be some big news coming... More in about a week.

ACuriousMind

@Danu That sounds...ominous :)

Danu

You best believe

Mike Miller

danu is pregnant

20

Danu

You clearly don't belong here, stranger

Steven Stewart-Gallus

Given vacuum energy shouldn't there be a tiny correction to Bose-Einstein statistics saying that the number of particles below a certain energy is zero?

ACuriousMind

9:36 PM

@MikeMiller The question is - will the baby be a mathematician or a physicist?

3

Danu

^^

Mike Miller

I think there are more pressing questions.

John Duffield

@Danu : interesting. I'll look out for that. And give comment as appropriate.

Danu

@JohnDuffield I think you'll love it ;)

ACuriousMind

That sounds even more ominous.

Balarka Sen

9:42 PM

wink

John Duffield

@Danu : please do give me a hint as to why. Because all too often, when there's "big news coming", I end up disappointed.

Danu

@JohnDuffield Neh, just ignore it.

Ted Shifrin

They really will allow just anyone in these rooms ...

John Duffield

@Danu : meh, so there's no big news coming?

Danu

by the way guys

OMG ALL THE TEAMS ARE SO BAD IN THE EURO CUP

Let's just give it to Iceland already

ACuriousMind

9:45 PM

...why are we being invaded by the math chat? oO (I, for one, welcome our new mathematician overlords)

Balarka Sen

because of the big news, 'course

we are here to congratulate danu

ACuriousMind

@Danu OMG I DONT CARE ABOUT FOOTBALL ;P

Ted Shifrin

There's way too much math babble in here, at any rate.

John Duffield

@Qmechanic : does "reverting to our standard procedures" mean I'm OK to talk physics here now? Free speech in science and all that?

Danu

@ACuriousMind I think they're trying to return the favor after my gradual shift :P

bolbteppa

9:59 PM

Ok more fun, so you can do manifolds with pseudogroups, apparently more general than sheaves too, yay... "In any case, the type of geometry embodied in a particular flavor of manifold is controlled by a particular groupoid of transformations which preserves whatever geometric features one is interested in; cf. Felix Klein’s Erlanger Programm." I thought Erlangden did not even specify every geometry, but the most general form of manifold theory bases itself off this?

Danu

He specified the whole setup/idea: Geometry should be viewed as the study of properties invariant under certain transformations

In this sense, that setup is an obvious continuation of his idea

bolbteppa

But you can't actually get every geometry from that point of view right? "General Riemannian geometry falls outside the boundaries of the program."

Danu

I guess that's the point of introducing groupoids: It'll accomodate this

bolbteppa

If so, that's pretty cool

Danu

(I don't know if it is---it just sounds like that from that paragraph)

bolbteppa

10:08 PM

I'd love to read some book going through geometry as the Erlangden program from a modern perspective then (assuming it's true) showing why you need to extend it to groupoids to do differentiable/complex manifolds, where the book also does all of Euclid better and Hilbert too, making mnemonics so you can memorize every proposition in Euclid, is that too much to ask? :(

ACuriousMind

Do football fans in other countries also drive around and honk their cars' horns after their national team has won a match or is that a particular German idiocy?

Danu

Yes, and also set off fireworks :P

John Duffield

The latter. It's one of many.

Danu

(ambulance noises now...)

ACuriousMind

@Danu I live next to a rather busy street. I can't hear the fireworks over the horns :P

bolbteppa

10:17 PM

"The Erlangen program (in German, Erlanger Programm ) is a project, begun by Felix Klein at Erlangen in the 19th century (Klein 1872), to study geometry via symmetry groups of “geometric shapes”, hence from the perspective of group theory. The idea is to take the elementary building blocks of geometry to be not just Euclidean spaces but more generally homogeneous spaces G/H.

These have then also been called Klein geometries . When it was proposed, the Erlangen program served to unify various different kinds of geometry, discovered and studied at that time, into a common framwork. On the other hand, many kinds of geometries without global symmetries are not Klein geometries, notably Riemannian geometry is (in general) not.

But a (pseudo) Riemannian manifold is locally (tangentially) modeled on Euclidean space (Minkowski spacetime) and this local model space is a Klein geometry. The generalization of Klein geometry to such local situations is Cartan geometry. While many types of geometries (such as Riemannian geometry) are not in general Klein geometries, they are locally like Klein geometries. This generalization of Klein geometry is known as Cartan geometry.

In physics terminology this corresponds to “locally gauging” the symmetry group. Aspects of Klein geometry may be generalized from groups to groupoids and even categories or ∞\infty-groupoids. See at higher Klein geometry. Higher Klein geometry is the generalization of Klein geometry from traditional (differential) geometry to higher geometry: where Klein geometry is about (Lie) groups and their quotients, higher Klein geometry is about (smooth) ∞-groups and their ∞-quotients.

No idea why KN decided to use pseudo-groups if pseudo-groups are about some insane generalization so :p

Danu

10:48 PM

Neat-o.

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