"The superscript "s" refers to "special"."

Also, @JohnRennie, damn, I went and did it without noticing, at physics.stackexchange.com/posts/270852/revisions. I just wanted to put it on the queue, not blast it to bits.

Gee more "special" algebras

How rude

@Slereah Wow

What kind of conservative garbage are you reading

it sounds fascistic

"Geometric theory of generalized functions with applications to general relativity"

It seems to be the big book on Colombeau algebras

@ACuriousMind Do you know what "complex current density $J^*(\omega)$" is?

"$\Omega \to \mathcal G^s(\Omega)$ is a fine sheaf of differential algebras on $\Bbb R^n$"

One day I should learn what a sheaf is

Also why is my electric field $E_0\exp \mathrm i\omega t$ allowed to be complex

That's not very physical

The measurement of that fied is the real part of that

@0celo7 I guess it's the Fourier transform of the density $J(x,t)$?

@ACuriousMind Perhaps, but where does $x$ go

It's perhaps supressed in your notation? I have no idea what you're reading and I've never done much classical EM

I'm reading physics

What does the symbol $\subset \subset $ mean

@0celo7 damn book?

@Slereah compactly contained.

@EmilioPisanty Not too bad. Writing the penultimate chapter now. It's hardly gonna be a stellar piece of work though. How's yours going?

Ticking along on chapter 6, but still missing some significant chunks

I'm pretty sick of it already

@EmilioPisanty Perhaps a little more acerbic an answer than I usually go for? Don't really see what else could be said about the question though...

@EmilioPisanty Really?! I'm just starting to enjoy it to be honest

but I said I'd submit by early September and I sort of have to stick to it

I'm at the point where I can just write stuff down without thinking every paragraph over for a week

@MarkMitchison naw, I'm sick of rehashing old shit

@EmilioPisanty yeah well there's that

@MarkMitchison (no, really, I really like it)

@MarkMitchison hear hear

@EmilioPisanty But we have to submit by late september in any case, right?

or is your situation different?

@EmilioPisanty why are you here if you need to write

this is the blackest of productivity black holes

@MarkMitchison formally, yeah, but that's sort of the timeline we drew up and I'd rather stick to it than draw out the pain another month

Plus my external is flying in from the states so the viva date is pretty hard-coded

@0celo7 cf. supra "I'm sick of rehashing old shit" ;-)

btw @MarkMitchison will you be in London this weekend?

Really looking forward to catching up

@EmilioPisanty External?

@0celo7 in the UK system you have one internal examiner and one external one.

Huh

So just two profs in total?

@0celo7 yup. Behind closed doors, your supervisor can't go, and you can't discuss your thesis with either examiner before the examination. Rather different to other places.

Hm

So I think that the generalized functions can be written as functions $f:\Bbb R^n \to \mathcal K$

But then the question is

What class of functions are they

Is it continuous, smooth, something???

ALTHOUGH

$\mathcal K$ isn't a field, apparently

Just a ring

Might be tough to build a Hilbert space out of it

Since there are non-0 elements without an inverse

Is the existence of inverses very important for Hilbert spaces

I'm guessing yes

@Slereah If you're only over a ring, you get the theory of modules, not of vector spaces, which is far less nice

yeah I'm guessing so

Hm

Why is everything so hard

Apparently there are several sets of generalized numbers

But $\mathcal K_e$ is ALSO not a field

Dang it

I guess I won't be revolutionizing QFT this week

there seems to be a paper about a field of generalized numbers, but I don't know if those are at all related to generalized functions

Generalized numbers have this weird thing where there's several things that are "zeros"

You have negligible numbers and strictly zero numbers and shit

"We present a differential algebra of generalized functions over a field of generalized scalars by means of several axioms in terms of general algebra and topology"

Aw yis

@EmilioPisanty crazy

@ACuriousMind In a nutshell, why do you get an i from geometric quantization

"This article is written mostly to satisgy the interest of mathematicians and scientists who do not necessarily belong to the Colombeau community"

Do you belong to the Colombeau community

Dunno

"In our approach the set of scalars (the constant functions) of our algebra of generalized functions forms an algebraically closed Cantor complete field, not a ring with zero divisors as in the original Colombeau theory"

Take that Colombeau

You hack

According to @ACuriousMind I'm having an identity crisis

Also what's a Cantor field

pokepedia.fr/Colombeau

That's not even nonsense

That sounds like a real thing

Same as Snarxiv

Good old Markov generator

@ACuriousMind Do physicists like BBS have any clue what they're talking about when they write about characteristic classes and such?

They write so imprecisely, it makes me think they don't

I get that vibe from a lot of PSE posts about string theory

@0celo7 How am I supposed to know that?

@ACuriousMind You know everything

^

@ACuriousMind What color is my underwear?

Silly question from a man who knows everything.

@BernardMeurer potato

@ACuriousMind Depending on the state of decay of said potato, yes

@ACuriousMind What's the color of mine

@0celo7 Trick question, you're not wearing any

lol

Hello everyone.

I have a thing to say.

@ACuriousMind I'm not a slut

Oh god

When you grow up and review scientific papers, try to remember that the point of science is to discover things and explain them.

As opposed to what

What has your jimmies rustled

@DanielSank I thought the point of science was to make books with complicated titles

That's mathematics

@ACuriousMind I'm just so frustrated by physics math

This may sound trivial, and you may think I am joking.

I am not. This is important and I hope to the heavens and the powers of the universe that everyone here will remember this when they review a paper that delivers new information matter-of-factly.

We see papers for review all the time which essentially say "You might think quantum systems are actually classical. If so, you would analyze it like this. In this work, we do an experiment and show that the classical thinking doesn't work by analyzing the data in a new way".

@ACuriousMind Let $E\to M$ be a vector bundle. How do I see that there is an exact sequence $0\to TM\to (TE)_M\to E\to 0$?

We also get papers rejected for being not exciting enough (literally got that on a referee report) despite the fact that the paper delivers a solution to a decade-old and very important problem.

IN other words, we didn't sex it up enough, so they rejected it.

Hilariously, both referees said the paper was very clear, and then the second referee rejected the paper for being "too obvious".

That's like reading a really good math proof and saying "don't publish this in a book because it makes the theorem too obvious".

@DanielSank Add some germans words too it

Thanks for listening.

That's what I did in philosophy essays

"Our set-theoretical framework is the usual ZFC axioms in set theory with the axiom $2^{\mathfrak c} = \mathfrak{c}^+$, where $\mathfrak{c} = \text{card}(\Bbb R)$"

CH rules

I still don't know what "too obvious" is supposed to mean.

@ACuriousMind That would mean $(TE)_M\cong TM\oplus E$

I don't see that

If no one's done it before, it can't have been that obvious

@ACuriousMind Well, obviously I don't either as I can't contact the referee.

It seems like $TE$ should contain information about tangents along the fibers of $E$

@DanielSank And those referees can lick my fat Brazilian... Nevermind, don't want to get banned again

But somehow that's captured by $TM$?

@ACuriousMind A world famous leader of quantum information publicly declared the problem as "the skeleton in the closet of our field". I solved this problem, and the referee says it's too obvious.

So yeah, that's my world right now.

Plot twist: the reviewer is that leader

BS like this is a big part of why I left academia.

@DanielSank Gimme his address, I'll beat him up with my gang

@DanielSank You fool

@BernardMeurer Nah, it's cool.

(We're not very effective. I'm the only member)

You'll change your mind after working for the private sector

@Slereah What?

@Slereah ...super happy so far...

@ACuriousMind Let's look at the dimensions for a moment

@DanielSank Well, find someone who will publish it, then demolish the people who rejected you when it's praised for what it is.

I'm not :(

The dimension of $E$ is $n+k$ assuming it's a $k$-bundle

@Slereah Still doing research but my job doesn't depend on publishing.

So the tangent bundle has dimension $2n+2k$

Why is publishing meaningful when you have Arxiv? Isn't the whole point "getting it out there"?

But in $TM\oplus E$ you get $3n+k$

Something isn't right there

@ACuriousMind Oh yeah, super funny additional note: the paper I just reviewed cites the paper which was rejected for being too obvious.

@BernardMeurer Peer review is very, very important.

However, peer review should be more than 2 people :)

@ACuriousMind In other words, we're already being cited, despite having done an "obvious" experiment :P

@ACuriousMind How to demolish?

@DanielSank Get a turing award

or a Nobel

@ACuriousMind I'm trying to show that the normal bundle of $M$ in $E$ (viewed as the zero section) is $E$ itself

@DanielSank That's...funny if it weren't enraging

or a medal from someone

...intuitively that seems obvious

now that I think about it

@DanielSank Hmmmm...win some award, complain about it in an acceptance speech? You're right, there doesn't seem to be a suitable medium for that

@ACuriousMind It's encouraging!

@ACuriousMind My plan is basically to go around the community and loudly voice my opinion that we need to value clear useful results more and dramatic bullshit less.

@DanielSank Well...*that* they cite you is encouraging. That they might get published while you are not, not so much, right?

"The only role of the GCH is to guarantee the uniqueness of the field of scalars $\hat{ \Bbb C}$"

Don't want too many of those

@DanielSank How can you be dramatic in a scientific paper?

@DanielSank People will loudly agree with you and then return to valuing dramatic bullshit more.

@ACuriousMind Well... they won't get published, at least not in the form the paper was in when I reviewed it.

...but that's not quite the point.

@ACuriousMind Well, one thing that gives me hope is that I am finally being sent papers to review!

They can value what they want, but I will continue to voice my opinion and exercise it in reviews.

what is considered exciting in your field, these days

Is it the quantum version of DOOM

When does that come

Wait a moment

FML that's not how you add dimensions in Whitney sums

@ACuriousMind Ok I could really use some help...

@Slereah Several things:

1. Application of quantum computing to new problems, such as molecular simulation, physics model simulation (i.e. running an digital simulation of a Bose-Hubbard model)

@DanielSank I wish you success in that :)

What exactly is the restriction of a bundle to a submanifold of the base @ACuriousMind

2. Doing pathetically simple experiments and showing that a new flavor of analysis predicts the outcome. This typically comes with statements like "You might have though classical mechanics would work here, but surprise! it doesn't as shown by our new flavor of quantum analysis".

Ah, the base dimension changes

3. Substantial increases in experimental capability. These are my favorite.

@ACuriousMind Thank you.

What the hell is $(TE)_M$

$TE$ is a vector bundle with base $E$

Anyway, I should go to bed. I won't be here till Monday, presumably, so don't get banned or burn the chat down in my absense! ;)

$E$ in turn has base $M$, so

@ACuriousMind Thanks for ignoring me!

Love you

@ACuriousMind See ya! Have fun at Wacken!

So $(TE)_M$ is a vector bundle over $M$

@DanielSank Which paper are you talking about?

Hmm

The molecular simulation one?

I think $TE$ is an $(n+k)$-bundle

Oh no, wait, that was already published

so that would make $(TE)_M$ an $(n+k)$-bundle over an $n$-base.

$TM$ is an $n$-bundle

$E$ is a $k$-bundle, so $TM\oplus E$ is an $(n+k)$-bundle

@MarkMitchison This one: arxiv.org/pdf/1606.05721v1.pdf

OK, guessed as much

(Since DAniel is first author)

You can spot an experimentalist paper

All those authors

@DanielSank You wouldn't happen to know about dielectrics, would you?

and linear response theory

Did you know @0celo7

You can derive the dielectric constant from QFT

I doubt that

You can

via the GORDON DECOMPOSITION

It's magic

Why did I even bother asking

DS hates me

@Slereah I've heard of that

@MarkMitchison Yeah that's the one that got the "this is obvious" rejection.

@BernardMeurer When linking to arxiv entires, link to the abstract page, not the pdf.

@DanielSank That's all I had in my history :p

But noted :)

@BernardMeurer Change the /pdf/ to /abs/

Ah, nice!

"We denote by $\mathcal E(\Omega)$ the differential ring of the $\mathcal C^\infty$ functions from $\Omega$ to $\Bbb C$"

Why

Isn't $\mathcal C^\infty$ already good enough

Eww, to $\Bbb C$?

@DanielSank I'm surprised, I thought beyond-RWA effects were very much "on-topic" at the moment

@Slereah "all bump $n$-forms on a connected manifold are cohomologous"

Welp, time to prove that

what's a good greek letter

$\varsigma$

$\Xi$

Looks like a sperm

@MarkMitchison looks like a harmonica

$\varpi$

@0celo7 Are we playing Pictionary?

No, we are playing "I should not have taken a 5 day break from reading this highly technical manuscript because I am now confused beyond all hope of recovery"

I am fairly sure that quoted thing is wrong.

it should be "all bump $n$-forms with unit integral"

Because the integral of the difference is $0$

Although...that does not imply cohomologous

Apparently non-archimedean fields have "residue rings", which sounds very gross

integrating to zero is certainly not enough to be exact, is it?

"For every infinite cardinal there is unique, up to isomorphism, non-standard extension $^* \Bbb R$ of $\Bbb R$ such that $\text{card}(^*\Bbb R) = \kappa^+$"

Hm

I wonder if that means that this set of generalized numbers is isomorphic to the hyperreals

Altho that is assuming the GCH

@MarkMitchison Well maybe, but apparently it's still too obvious.