2:00 AM
@ACuriousMind Nothing---it was a corollary of learning a tiny little bit of the idea of Floer homology
which uses a perturbed version of the CR equations

@ACuriousMind Something interesting. Consider the set in $\Bbb R^3$ obtained by stacking two books in an "L" shape, such that the ends overlap.
Does that make sense?

Fduck, nvm, I'm just an idiot

Books = rectanges

These perturbed CR equations are, in that setting, the flow equations for the vector field defined by the gradient of some functional on loop space.

@0celo7 If I may offer a tip: if you're going to flag something, just flag it, and if you're not going to flag it, don't, but either way don't say that you're flagging something. It's pointless at best. (This I'd consider part of general chat etiquette, so it applies to everyone)

2:01 AM
Super cool stuff.
Extremely geometric

@0celo7 yes

but also very abstract
mystifying

@ACuriousMind Do you think this domain is Lipschitz?

@ACuriousMind Well to be fair, there was that time I was completely confused on degenerate gases

@0celo7 Reading the definition of Lipschitz domain on wikipedia just now---yes, I do think that.

2:04 AM
@Danu (Lipschitz boundary is one of those "technical conditions" I was mentioning.)

Or do you mean to take into account the actual pages
which would be terrible
I'm considering the books as planks.

No, I meant just solid rectangles.
A prof showed me this using two PDE books :)

Then it looks like almost a submanifold to me
if it didn't have corners
since submanifolds are locally graphs of smooth functions

@Danu It's a little hard to describe, but there's a point on the boundary that cannot be locally described by the graph of a Lipschitz function.
In fact, it cannot be described by any function.

this should probably (locally) be the graph of some (Lipschitz-)continuous function

2:05 AM
So what is your question, @SirCumference

@Slereah I realized what was wrong with my math, I'm just stupid

@0celo7 The corners?

No, not the corners. It's where the books meet.

the most important lesson of all

I was gonna ask why the comoving distance to the event horizon was $d_{eh}(t)\propto\frac{1}{a(t)k}$ while the proper distance was $d_{eh}(t)\propto\frac{1}{k}$, but then I realized that the comoving distance is equal to the scale factor times the proper distance...

2:07 AM
@SirCumference Why not get a standard book on Cosmology?

@0celo7 I have one

Which one?

@0celo7 The function $x\mapsto |x|$ seems to describe 90 degree angles pretty nicely (I'm thinking of the 1D picture)?

@Danu This is very much a 3D phenomenon. Let me take a picture

Are you restricting the function to take the x-y coordinates as input and the z as output?

2:08 AM
@0celo7 It's a class textbook, 21st century cosmology & astronomy or something like that
I'd need to see it to tell you

Please forgive the floral bedsheets, they are not mine
The part by the "m", @Danu

Ahh, I imagined the wrong shape -.-

Is that Dirac's QM book

Shankar

@ACuriousMind Indeed, me too.
I thought of the obvious extension of the L
i.e. no non-trivial information in the third direction
I maintain that that should probably be a pretty nice domain :P

2:12 AM
I meant "such that the ends overlap." to convey that shape :P

By "ends overlap" I thought you meant

@Danu Submanifolds are actually terrible

that the L was not disconnected at the corner

Any open set is a submanifold

@0celo7 bhahahaha :P

2:12 AM
But open sets can get bad

Open sets are the best

Not when you have to Sobolev embed

I love open sets
I also prefer them to be compact
if you catch my drift

Compact open sets?
Lucky bastard

:D
Compact manifolds or bust
no pathologies

2:14 AM
No?

A minimum of pathologies :)

Ever considered to broaden your horizon to stratified spaces or at least orbifolds? Compact manifolds are pretty boring, they don't give non-Abelian gauge groups or chiral matter :P

In fact, I prefer them to be almost complex as well

They don't give non-Abelian gauge groups?

Gotta get some Chern classes

2:15 AM
compact manifolds are pretty pathological in GR

I find that claim suspicious @ACuriousMind

@ACuriousMind Proof?

The sentence is missing an "when you compactify M-theory on them" ;)

Hmm.
Of course it is.

Since I recall that Donaldson based his investigations of 4-manifolds on $SU(2)$ Yang-Mills theory
I'm pretty sure he <3 compact manifolds too
though clearly he thinks $\Bbb R^4$ is great too

2:16 AM
@Danu I actually proved the Myers-Serrin theorem on noncompact manifolds (without boundary)
It took some work

You're telling me you can't bootstrap the result on $\Bbb R^n$?

I'm assuming you googled what it means?

indeed
and I wanna say bump functions

And yes, you can. But there's work to be done!
@Danu Yes. But the hard part is proving that globally defined weak derivatives give rise to weak derivatives in charts
Maybe it is easy after all, but my advisor didn't seem to think so

Oh yes, I also do not have too much love for non-smooth things...

2:21 AM
And you also need to prove that there's a locally finite covering by charts diffeomorphic to open sets in which Sobolev approximation holds

Though some weird-ass $C^k$ stuff showed up in the gauge theory course

@ACuriousMind found a nice theorem in Lee that gives it
@ACuriousMind BTW, Lee's proof is exactly what I was describing. Good to know I learned something from that book.

So anyway

@Danu Smooth? This stuff isn't even continuous.
You're lucky it's measurable :D

What's a PDE solution that is locally unique but not globally unique?

2:24 AM
What does that mean?

@Slereah I already gave one example earlier, zero modes of the Laplacian (acting on p-forms). Locally unique by contractibility, globally not-unique if the corresponding Betti number is non-zero.

Locally unique? What?

(since by Hodge decomposition there's a bijection between cohomology classes and harmonic forms)

The Picard whatever theorem, @0celo7

If you have multiple global solutions you have multiple local solutions

2:26 AM
Picard Hasselhof theorem

I must be misunderstanding the question here

Hodge decomposition is another analysis result I'd like to understand

@Danu It involves Sobolev spaces, you might want to run away from it
Although de Rham has a proof of it without Sobolev

The Picard Lidelhof makes a big deal to specify that solutions are only unique on some interval
But I can't really think of an example where it isn't outside

Although he did say PDE, so I conclude he's confused.

2:30 AM
@Danu ..I just realized that doesn't work because the statement is for compact manifolds but the local neighbourhoods are non-compact

I still don't know what is being asked!

And of course 0celo7 is right that you can just restrict the global solutions to get local ones

@ACuriousMind Use elliptic regularity to get a solution in a small chart :^)

Well I'm guessing the global solutions are the same on that interval, but may differ outside
But I'm not sure what would be an example of that

2:32 AM
yes

Well, first of all, it has to be nonlinear.
@BernardoMeurer looks like the kid from iCarly
that's where I know him from

I do, that is true

@ACuriousMind What about compact manifolds with boundary? :P

That's a good question
Does it hold for those?

Geroch has a whole list of true/false spacetime topology question at the end of his paper
No proof, left as an exercize

2:40 AM
@Danu It doesn't hold; the partial integration needed to obtain the relation between $\Delta$ and $\mathrm{d}$ incurs an annoying boundary term that I don't see going away

@ACuriousMind It depends on how one is defining the forms.
In geometric analysis it is common to do things on $\mathrm{int}\, M$ and then say everything is uniformly continuous
Then your bump functions are only defined on compact subsets of the interior so the Laplacian is still essentially self-adjoint

@ACuriousMind You mean in the proof that $*\text{d}*$ is the adjoint of $\text{d}$?
(up to sign)

@Danu Nah, in the proof that $\Delta f = 0$ implies $\mathrm{d}f = 0$.

@ACuriousMind Well once you have the definintion in terms of d and its adjoint that's immediate

What?

2:43 AM
@Danu No, you have to integrate by parts

@Danu check yourself before you wreck yourself

I think the presentations we saw were different
It's the same problem anyways
Once you have $\Delta=dd^*+d^*d$
and that $d^*$ is the adjoint of $d$ wrt the L2 inner product
it's immediate
that latter statement goes wrong without Stokes
so I don't think I wrecked myself, @0celo7 :P

@Danu What is $\Delta$ if not $(d+d^*)^2$?

@0celo7 It's never that anyways
because that square expression doesn't make sense :P

Why not?
Do you dislike inhomogeneous forms
You might want to tell de Rham his book doesn't make sense.

2:46 AM
@Danu Ah, I see you meant the same as I did.

@ACuriousMind Exactly.

@Danu It does if you consider it as an operator on the full space of forms, not only on one degree.

I do indeed dislike using inhomogeneous forms
and de Rham is indeed dead
Which may or may not be due to repeated use of inhomogeneous forms

:P

2:48 AM
Are you guys pretty bored?

At least there are no pasteurized forms

I have a problem I've bee nstuck on for a while

it takes some time to explain

What's the subject?

2:50 AM
Almost complex manifolds
the twistor space construction

Urgh
I've never seen anything nice follow the word "twistor" :P

(not very closely related to Penrose's stuff)
no spinors
nothing like that
just good old geometry
It's apparently a generalization of Penrose's thing but meh

@ACuriousMind What math is "nice" except for point set topology

I repeat, no pathologies

2:53 AM
Ok, so what is the question

never mind---I'll explain it some other time
You guys don't seem too psyched plus it's late

Make it tie into G2 manifolds and ACM will go nuts

What is G2
Isn't that one of those shitty Lie groups nobody uses

@0celo7 It is related to that

Along with E8

2:56 AM
but ACM doesn't actually care about G2

Automorphism group of some 3-form or something

since in physics you don't actually need it much apparently

ACM was telling me about G2 structures the other day
and how it relates to harmonic spinors

He kept telling me G2 doesn't show up much :P

@ACuriousMind Well, what's the story?

2:58 AM
@Danu Well, it shows up but you don't need to know much about $G_2$ itself
@Slereah Everyone uses it in M-theory compactifications because for a 7-fold having holonomy contained in $G_2$ means it has a parallel spinor and hence the compactification preserves supersymmetry

It sounds super neat

Why does parallel spinor mean susy?

He said, disingenuously

@ACuriousMind Now that we have LaTeX, can you please explain that proof?

One nice thing about owning paper versions of GR books is that I can close the pdfs open on my computer

3:02 AM
@Danu If you set the field strengths of the higher gauge fields to zero ("fluxless compactification"), the supersymmetry variation of the gravitino is $\nabla\epsilon$ for the spinor $\epsilon$ that parametrizes the SUSY transformation.
Preservation of SUSY means the variation vanishes at least for some $\epsilon$, and "parallel spinor" just means exactly a spinor $\nabla \epsilon = 0$.

ok

Now, a spinor with $\nabla \epsilon = 0$ transforms trivially under holonomy, but if holonomy is the generic $\mathrm{SO}(n)$ there cannot be a non-zero spinor that does that

why not?

So for parallel spinors to exist, the holonomy must be reduced to a group under which the spinor representation of $\mathrm{SO}(n)$ splits into reps of which at least one is a singlet/trivial representation

3:06 AM
why?

@0celo7 If the spinor's covariant derivative vanishes, it is mapped to itself upon parallel transport along a loop.

Yes, I know what that means
But why does the nonexistence of a $G_2$ reduction imply there is never a parallel spinor?
I mean, what even is $\epsilon$? A spinor field? Why not just take $\epsilon=0$?

Being mapped to itself upon parallel transport means the holonomy group acts trivially on it, but the full $\mathrm{SO}(n)$ has no fixed non-zero spinors.

Ok, I did not know that latter fact.

@0celo7 It's just the statement that the spinor rep is irreducible.

3:09 AM
So if the full group is SO, by group theoretic reasoning, one should not be able to have $\nabla\epsilon=0$?

Exactly

Now, if one has $G_2$, how does one get $\epsilon$ such that $\nabla\epsilon=0$?
That's what I'm interested in

Geroch references a philosophy book
Apparently it discusses causality violations
I might have to get it just for the historical aspect

@0celo7 Ah, that's the annoying thing, there's little constructive about the properties of $G_2$-manifolds. You just know it exists because the spinor representation has a singlet in it. Given the spinor, you can construct the $G_2$-structure as the form with components $\bar\epsilon\Gamma_{\mu\nu\sigma}\epsilon$, but that's not an operation that's easy to invert in a way one could write down.

Also this philosophy paper :

3:18 AM
Likewise, given the $G_2$-structure as a 3-form, you know there's an associated metric for which $G_2$ is the holonomy but no formula for the metric in terms of the $G_2$-structure is known

I'm going to sleep. Bye!

ahahah

@Danu night

3:19 AM
@Danu bye

:^)

@Slereah Stop
I will send it to you

I """bought it"""
too late
fucking hell
28 pages
"Although Einstein was led to general covariance by physical considerations, it was pointed out as early as 1917 by Kretschmann and concurred in by Einstein that this principle is devoid of physical content, and any theory whatever can be formulated in a generally covariant form"
who even needs covariance
"Both Kretschmann and Einstein thought that a generally covariant formulation of Newtonian mechanics would be extremely complicated; it was realized later, however, that this was not at all the case, and that Newton's theory of gravitation could be put into a form very similar to that of Einstein."
I think covariant Newton is done in MTW
Flat space with curved time
Oh man
that paper has a huge literature section on philosophy books regarding causality in physics

3:36 AM
I think I need to get a driver's license
It's useful to have an ID like that

what other id do you have?

@Slereah Eh? Isn't one of the big problem people have been having with various modified Newtonian theories (as alternative to dark matter) that they haven't been able to construct them in a covariant manner?
And if so, how does that fit with this line?

Well, as said
MTW has a covariant version of Newtonian gravity

@Slereah Right, but you need a covariant theory in which the power law changes at large distances.

do you want the reference the paper gives for that claim?

3:40 AM
@Slereah Not really. I've no time to follow up, I'm just wondering how these two notion get together.

@Slereah You understand Hindi ? :)

No
I also don't understand Serbian, and yet

@Slereah Where are you from ?

France

3:57 AM
@Slereah I see. I find it weird to listen to songs in languages I don't know :P

👌🏻
and of course, the enchanting tunes of russian music

@Slereah Didn't really like the tune of the Russian music (which you linked) XD Anyway, tastes vary! :)
@Slereah This is better :)

4:17 AM
You can tell that paper is pre-70's because it won't even consider tachyons

5:05 AM
perhaps they might be better at QFT also

Come and test your knowledge of spacetime topology!

Not being able to answer that image above is the primary reason I am reading munkres. Without topology, I just cannot make sense of nontrivial spacetimes in GR
and then, I end up going off a tangent to infinite sets. They are pretty awesome though

Quite unfortunate that there is no proof for those assertions
I'll have to try work it out

btw I have a small question for you. Is the following flow chart representing some causal structure of a CTC?

5:21 AM
whaaat
there's no jumps in GR
Paths are continuous

Ok, think those arcs as: identifying the pair of events shown.
For example, The following simpler case below will be a CTC

That is the Deutsch Politzer spacetime
Kinda

I see, and it seemed to be the simplest type of CTC geometry

5:37 AM
Well it has a topology change
So it is gonna be shit
The simplest CTC geometry is the time cylinder
$S^1_t \times \Bbb R_x$

Right

It is mostly trivial and well behaved

what's going on
@Slereah piecewise smooth, even

I'd even say smooth
Although
Quantum paths are only holder continuous

proof?

5:42 AM
It's in Demichev

what's that?

Path integral book
there's a proof that smooth paths are of measure 0

I did that in my analysis class

(wrt the wiener measure)
(heheheh)
(wiener)
"Theorem 1.1 (Wiener’s theorem). The Wiener path integral is equal to zero over both the set of
discontinuous and the set of differentiable trajectories. In more precise mathematical terms, the set of
discontinuous as well as the set of differentiable functions have a zeroWiener measure."

define the measure?

5:52 AM
I'm not sure how to define it rigorously but it's the path integral measure
$$\int \prod e^{\int \dot{x}^2 dt} dx$$
Of course $\int \dot x^2$ isn't well defined since most paths aren't differentiable
Gotta do it via some limit process

Can you please define it rigorously
I am curious
you have the book
what even is a quantum path?
@ACuriousMind The spinor representation?
Again, what is $\epsilon$?
If it's a spinor field, then it will depend on the global analytical/topological data of the manifold, no?

Well, the book that defines it rigorously isn't Demichev, I think
It's Jaffe
Lemme see