The h Bar

6:04 PM

@ACuriousMind the German restaurant lady is from Hberg

I need Hberg stories, dammit

The proof btw is that $$\int dx\ x^a \rho(x) = \frac{1}{2\pi} \int dx\ x^a \int dp \hat \rho (p) e^{-ipx} = \frac{1}{2\pi} \int dp \hat \rho (p) \int dx\ x^a e^{-ipx} $$

and using $\textbf{F}^{-1} x^a = i^a \delta^{(a)}(p)$

Since $\hat\rho(0) = 1$ and its derivative at $0$ is always 0 it follows

0celo7

@ACuriousMind what is meant by Coker on page 24?

ACuriousMind

That's standard notation for the cokernel

0celo7

I know, what is that

ACuriousMind

The target space modulo the image

0celo7

6:08 PM

Oh lord.

ACuriousMind

For instance, it's zero for a surjective map and the entire space for the zero map

Slereah

The coke kernel

0celo7

Well, I must say its a mystery to me why the kernel of the difference map is the first cohomology @ACuriousMind

Again on page 24

Er, zeroth

I haven't thought about its Coker yet.

ACuriousMind

@0celo7 The series is exact.

And you can add a zero to the left of $H^0$, i.e. $0\to H^0\to \mathbb{R}\oplus\mathbb{R}\to\mathbb{R}\oplus\mathbb{R}$.

0celo7

So?

ACuriousMind

6:22 PM

So by exactness of the sequence, $\ker(\delta) = \mathrm{im}(H^0\to\mathbb{R}\oplus\mathbb{R}) \cong H^0$ since the image of an injective map is the same as the source.

0celo7

Yes, I see that.

Now on to this Coker business

Target space modulo image...

Slereah

What's a "Taylor series with remainder"

0celo7

Well, by exactness, the image of the connecting homo is the kernel of delta

@Slereah Taylor's theorem.

Slereah

I suppose

But what class of function has a Taylor series with remainder

0celo7

So the image is one-dimensional, right @ACuriousMind

Slereah

6:31 PM

Analytics?

ACuriousMind

@Slereah I think it just refers to the case where to abort the series after the n-th term and write the rest into a remainder

Slereah

Or is it larger

0celo7

Of d*

Hmm, how do you actually get that Coker thing

This is confusing

ACuriousMind

@Slereah Well, every $C^k$ function has a Taylor "series" with k terms + remainder.

0celo7

@ACuriousMind how do I do this when rank nullify cannot be used?

Slereah

6:33 PM

Well it's written as $$f(x + a) = \sum_i^q \frac{x^i}{i!}f^{(i)}(a) + \frac{x^{q+1}}{(q+1)!}f^{(q+1)}(x + a)$$

Does that cover any smooth function

ACuriousMind

@0celo7 How do you do what?

0celo7

Get the Coker result

Finding out what it's dimension is and that it's the first cohomology

Slereah

Is this the coker result

ACuriousMind

@0celo7 The map to $H^1$ is surjective, so $H^1 = \mathbb{R}\oplus\mathbb{R}/\ker(\mathrm{d}^\ast)$. By exactness, $\ker(\mathrm{d}^\ast) = \mathrm{im}(\delta)$, which gives $\mathrm{coker}(\delta)$.

0celo7

Oh, the connecting homo is surjective?

ACuriousMind

6:36 PM

@Slereah lol

@0celo7 Again by exactness

0celo7

What the heck

ACuriousMind

If $A\to B\to 0$ is exact, then $A\to B$ is surjective. You are reading a book again that seems to suppose a rather larger familiarity with algebra :P

Slereah

Also Colombeau uses the notation $\mathcal O_M(\Bbb R^3, \Bbb C)$, which is apparently the class of smooth functions with derivatives growing slower than a power of $|x|$ at infinity

Which I've never seen before

0celo7

@ACuriousMind of course, that's trivial

Slereah

Apparently the sifting property of powers of delta is $$\int dx f(x) \delta_\varepsilon^m(x-a) \approx \frac{M[m,0]}{\varepsilon^{3(m-1)}} f(a) + \mathcal O(\varepsilon^{\mathfrak{some shit}})$$

0celo7

6:48 PM

I don't see where I have such an exact sequence

Slereah

Somewhat divergent

ACuriousMind

@0celo7 You do not see the $\mathbb{R}\oplus\mathbb{R}\to H^1\to 0$ right on the page?

0celo7

Oh

Now I do!

I thought you meant that was a general phenomenon

I'm having a bad day

You get those zeros because of the cohomology of R, right

Just like you get the other Rs because of the cohomology of R

ACuriousMind

yes

0celo7

Was homology or cohomology discovered first

Slereah

6:53 PM

homology first, otherwise it would be called cocohomology

0celo7

Shiggino

Slereah

And now the algebra $\mathcal G(\Bbb R^n, L(\Bbb D))$

The horror

0celo7

Oh god, why is this so hard

It's just the circle

ACuriousMind

@Slereah Maybe replace all the weird letters with pictures of kittens

Kitten-algebras can't be horrifying :)

Slereah

$😸(😻^n, L(😿))$

2

Things are looking up

I find it slightly odd that the class of "moderate" functions are functions like $f_\varepsilon \approx \varepsilon^{-N} f$

Most of those are the infinite part of the algebra

Nothing moderate about that!

amazon.fr/Geometric-Theory-Generalized-Functions-Applications/…

I should buy this

80 euros, tho

Durn it

John Rennie

7:25 PM

@NoahP Hi Noah. see How long was the universe radiation dominated?. I worked out that the radiation dominated era ended at about $a=0.00029$, which corresponds to $z=3446$. So I'd say you were pretty close :-)

ACuriousMind

@Slereah So the conut was discovered before the coconut?

Slereah

It was

Also : the coa

0celo7

Ha :D

ACuriousMind

Fascinating

0celo7

I guess nuts are not reflective modules

Noah P

7:42 PM

Hi @JohnRennie, sounds great!

That's all the questions now, polished off the rest of six with basically no problem

Well no problem at all actually

Thanks for all the help! I wouldn't have had a chance at doing them a week ago, but much happier with them now @JohnRennie

0celo7

7:59 PM

@ACuriousMind I think they assume basic familiarity with de Rham

What do they mean by "signed inclusion" on page 26

ACuriousMind

@0celo7 It's written directly underneath.

0celo7

@ACuriousMind Perhaps, but why the sign?

ACuriousMind

Because without the sign the sequence is not exact.

0celo7

How does the sequence of inclusions give that "signed inclusion"

@ACuriousMind That's obvious

But the sequence should just be the first one with the compact de Rham functor applied

ACuriousMind

It isn't

Just applying the functor gives $\Omega_c(U\sqcup V)$ in the middle. You have to use that that has an isomorphism to $\Omega_c(U)\oplus\Omega_c(V)$ and figure out what that does to the maps.

0celo7

8:07 PM

What does that even mean

where do you get a sign from, there's no orientation

do you have to repeat the computation done on the bottom of page 23?

hmm, no, I doubt that

@ACuriousMind Can you please be more precise

I'll try to fill in the details by myself

I'm not sure what I actually have to prove.

How do you get the reverse $\mapsto$, anyway

ACuriousMind

I have been precise. Applying the functor only gives you $\Omega_c(U\sqcup V)$. You have to describe the canonical isomorphism $\Omega_c(U\sqcup V)\to\Omega_c(U)\oplus\Omega_c(V)$ explicitly and see what concatenating that with the induced map $\Omega_c(U\cap V)\to\Omega_c(U\sqcup V)$ gives you.

0celo7

Describe the isomorphism explicitly? Is it not just $\omega\mapsto (\omega|U,\omega|V)$?

With inverse given by $(\alpha,\beta)\mapsto \alpha+\beta$

ACuriousMind

Wait

forget what I said

But why are you confused about this minus and not about the one appearing in the ordinary MV sequence?

0celo7

8:23 PM

Well, they say you take the difference of the maps $\partial^*_0$ and $\partial^*_1$

ACuriousMind

They aren't saying it comes from directly applying the functor. They're saying "as before", implicitly referring to the construction of the MV sequence on p. 22

So the reason is indeed that the minus makes the sequence exact, and nothing else.

0celo7

Oh, well I'm confused because they don't have the 0s on page 26 on the sequence

@ACuriousMind hmm

I will trust your wise words because you are wise

@ACuriousMind Stupid question probably, but what is the "deep" reason why $\Omega$ is contravariant and $\Omega_c$ is covariant?

ACuriousMind

I'm not sure there's a "deep" reason. It's easy to restrict arbitrarily supported things, but restriction doesn't preserve compact support in general. However, it is easy to extend compactly supported things on open sets smoothly just by zero.

And as they say, you could also consider $\Omega_c$ contravariant for proper maps

0celo7

@ACuriousMind Yeah.

is it just more convenient to consider it as a covariant functor?

So, $\Omega_c$, option (b), is a functor from what to what?

$\Omega$ is from smooth manifolds and smooth maps to vector spaces and linear maps

ACuriousMind

@0celo7 Smooth manifolds and open embeddings to vector spaces and linear maps

0celo7

8:36 PM

Open embeddings?

ACuriousMind

Injective smooth maps whose image is open.

0celo7

@ACuriousMind Embedding is a bit stronger than that

You need it to be an immersion, too

Or is that automatic

ACuriousMind

Probably choose it to be an embedding just to be safe

It's not a "good" functor as such anyway

0celo7

The compact one?

ACuriousMind

Yep

0celo7

8:39 PM

Why not, and what do you mean?

Define good without me having to open up MacLane pls :)

ACuriousMind

For one, its cohomology is not homotopy-invariant

Which can be desirable if you want information about the space that's stronger than its homotopy type, but which makes it not a cohomology theory in the standard sense.

0celo7

What does that have to do with functoriality

ACuriousMind

Nothing

it's a functor alright

Just not a "nice" one

0celo7

I thought you meant it was not nice in a categorical sense

0celo7

8:58 PM

@ACuriousMind Is it in some way less nicer than $\Omega$ in a categorical sense?

Tried to find Spivak Vol. 1 in the library, found this instead

ACuriousMind

@0celo7 Yes. But my answer involves sheaf theory :P

0celo7

@ACuriousMind Ugh...lay it on me

0celo7

9:18 PM

@ACuriousMind Good rubber duck

ACuriousMind

@0celo7 The $\Omega^\bullet$ forms an acyclic resolution of the locally constant sheaf $\underline{\mathbb{R}}$ on a manifold. It therefore computes a rather "simple" object - the cohomology of the locally constant sheaf. $\Omega_c$ is not so simple.

Furthermore, you can pullback differential forms, which is essentially a morphism from the target sheaf to the direct image $f_\ast$. Now, for compact support, you'd need to do that for the "direct image with proper support" $f_!$, which is a far uglier functor because it doesn't have an adjoint in the standard category of sheaves.

0celo7

Makes sense

@ACuriousMind Is there a categorical reason why people use $\coprod$ for $\sqcup$

I feel like we've been over this?

ACuriousMind

Yes, the disjoint union is the coproduct.

0celo7

I should learn what that actually means

Or...did we not talk about that like a day ago?

ACuriousMind

I feel like we did :P

0celo7

9:27 PM

Do you write $\sqcup$ or $\coprod$

$A\sqcup B$ $A\cup B$ $A\coprod B$

ACuriousMind

I write $\sqcup$.

0celo7

but you're a categorist

ACuriousMind

Yes, but I find the inverted pi ugly and out of scale

0celo7

aha, so you admit to being a category theorist

ACuriousMind

You might better call me a category enthusiast :P

0celo7

9:33 PM

Then I am a geometry enthusiast

ACuriousMind

Yes, you are

0celo7

@ACuriousMind I feel guilty reading algebraic topology instead of geometry :/

But I want to learn it

To do geometry

@ACuriousMind Ok, just checking here...on page 27, I'm trying to see that image $\delta$ is 1-dimensional. Is that because knowing one of the components of $(-(j_U)_*\omega,(j_V)_*\omega)$ automatically fixes the other as they agree on $U\cap V$ and outside of that they are identically 0?

Danu

9:52 PM

@ACuriousMind You can scale it, relative to the $\sqcup$

(use the scalerel package) :D

0celo7

@ACuriousMind Is this the first isomorphism theorem?

ACuriousMind

@0celo7 I think you mean the correct thing but you are expressing it strangely

0celo7

@ACuriousMind How would you express it?

ACuriousMind

@Danu It's still an inverted pi then. It just looks drunk to me, not like an operator :D

Danu

Is it exactly an inverted $\Pi$?

ACuriousMind

10:01 PM

@0celo7 The image of $\delta$ in $H(U)\oplus H(V)$ is the elements of the form $(-x,x)$ in $\mathbb{R}\oplus\mathbb{R}= H(U)\oplus H(V)$ since $\int (j_U)_\ast \omega = \int (j_V)_\ast \omega$

@0celo7 yes

@Danu I think so, yes

$\prod \coprod$

0celo7

But is $\prod$ exactly a $\Pi$

$\sum$ and $\Sigma$ are not the same

ACuriousMind

No, okay, it's an inverted product sign

Still I find it just doesn't look like it's a proper operation

0celo7

@ACuriousMind Oh my god all this algebra

ACuriousMind

And it's difficult to see at which end it's open in some fonts and sizes

0celo7

@ACuriousMind Ahhh, an injective function is an isomorphism onto its image?

ACuriousMind

10:34 PM

@0celo7 Yes, if bijective morphisms are isomorphisms

0celo7

---?

ACuriousMind

I feel I've seen this question before but I can't find it

@0celo7 It would have separated my messages had you not interrupted :P

0celo7

@ACuriousMind This is probably stupid, but what do they mean by an "orientation-preserving trivialization" on the top of p. 30

Charts are not maps between $\Bbb R^n$, so what does it mean for one to preserve orientation

ACuriousMind

You know, I also only mostly figure these things out from context instead of knowing them.

0celo7

What does that mean

Are you subtly telling me to figure it out myself

Oh, it means that $(U_\alpha,\phi_\alpha)$ is consistently oriented with $[M]$.

I think?

user54412

10:41 PM

@ACuriousMind There are a lot of questions asking about going from the spectrum to the hamiltonian, e.g. physics.stackexchange.com/questions/13480/…

0celo7

That's how Lee seems to define it, let's go with that.

ACuriousMind

@ChrisWhite Well...going from the spectrum is a bit different than going from an explicit solution in position space, but that might be what I thought of.

0celo7

@ACuriousMind Also, the Haar measure is derived and shown to be unique in Lee.

ACuriousMind

@0celo7 So?

0celo7

@ACuriousMind Just letting you know!

Qmechanic

10:45 PM

-1

Q: Incorporating gravity

What is (currently) seeming to be the best way to combine general relativity and quantum mechanics? I know string theory doesn't really have any evidence, and I've read about loop quantum gravity, but I was wondering what other options were, and which one was the best. Bonus: What experimenta...

primarily opinion-based?

user54412

@Qmechanic would have said yes, but the edit just now makes it borderline enough that I'm not swayed to vote

user218912

@ChrisWhite I'm exploring different areas of physics to see what I like, do you have idea how I can get into computational numerical relativity and relativistic hydrodynamics?

user218912

in terms of books and other resources.

ACuriousMind

@Qmechanic not opinion-based, but it will attract opinion-based answers since I believe there is no objective answer currently known. I don't think that makes the question off-topic, but the answers could use some more scrutiny than on average.

user54412

@3750 Hmmm. NR is new enough that there aren't many books on it. The only one I've heard of is this but I haven't read it since I learned the topic before it was written. At least it was written by two very big names in the field. And it's blue :p

user218912

11:01 PM

@ChrisWhite alright, what about relativistic hydrodynamics and also computer simulations/coding?

0celo7

Hatcher is blue

user218912

well

user218912

Idk if I'll ever read hatcher because I'm afraid of real math now.

user54412

@3750 For relativistic hydro, the most cited book is probably this. But it's the sort of book that's more like a long expository research article -- useful for practitioners not particularly engaging. The field isn't new, but it's niche enough that apparently no one really writes books on it.

0celo7

that's not a good thing to be afraid of

how are you going to take analysis and geometry

user218912

11:03 PM

I'm backing off for now.

user218912

I'll take them in 3rd year for my degree requirements.

user218912

no more pure math for me for a while.

0celo7

anyone want to take bets on how long that lasts

user218912

yesterday, by 3750

@0celo7 my mind changes every few hours so don't mind me.

user218912

@ChrisWhite Ok, I'll take a look at that one.

user218912

11:05 PM

@ChrisWhite what about scientific programming?

0celo7

@ChrisWhite Shame I didn't know about that when I liked GR.

user54412

@3750 In general, with both these subjects I would be cautious about jumping in too fast. They're the sort of thing that requires a lot of background. Consider that most physics curricula don't require fluid dynamics, but this is necessary for both subjects (most effort in NR these days isn't about vacuum, it's about neutron stars and such).

user218912

@ChrisWhite oh you need fluid dynamics?

user218912

I know some stuff

user54412

@3750 Learn C.

2

user218912

11:06 PM

alright, why not c++ or fortran?

user218912

some other astrophysics guy told me that.

ACuriousMind

Those pointers still haunt me in my sleep ;)

0celo7

@ACuriousMind Pointers?

ACuriousMind

@3750 Oh, KyleKanos would tell you to learn Fortran were he still here, I'm sure :D

user54412

@ACuriousMind But he's not, so I'm getting out my soapbox

ACuriousMind

11:07 PM

@0celo7 C has pointers. They take a bit getting used to

0celo7

That question QM linked is gone.

user218912

@ChrisWhite which language is preferable for doing simulations like youtube.com/watch?v=8C_dnP2fvxk

user218912

which looks ULTRA COOL

user54412

@3750 Fortan died before even I was born. No one ought to use it. Compilers aren't written for it anymore (the few that exist are just Fortran-to-C translators on top of C compilers). It's not faster than C. It promotes bad habits. It's one selling point was that equations look natural in it, but this is compared to deader languages from the 50s. Equations are more natural in C than Fortran.

0celo7

@ACuriousMind Wait, is dR cohomology only $C^\infty$ homotopy invariant?

user54412

11:11 PM

@3750 C for the simulation. The visualization afterward can be done in any number of languages or visualization tools.

ACuriousMind

@0celo7 That's a strange question

0celo7

@ChrisWhite Everything in the nuke business is in Fortran...which probably is for the same reason that banks use COBOL.

user218912

@ChrisWhite got it, thanks.

0celo7

@ACuriousMind How come?

ACuriousMind

@0celo7 How would a non-$C^\infty$-map define a map on the differential forms in the first place?

0celo7

11:13 PM

@ACuriousMind Excellent question. So does the dR cohomology depend on the smooth structure?

user54412

C++ has its uses, but I would not recommend it as a starter language. The only way to start with it is to strip it down to C. If you jump right in with its abstract features you'll miss out on understanding imperative programming, which is necessary for scientific computation (though is disdained by modern and hip CS types).

ACuriousMind

@0celo7 No, since by deRham's theorem it is isomorphic to singular cohomology.

0celo7

@ACuriousMind Ah, right. I technically do not know about that yet. I'm assuming it will be proved later.

user54412

@3750 You're going to Toronto?

user218912

@ChrisWhite yeah.

user54412

11:15 PM

They're pretty big at NR, especially at CITA, so you'll probably have some exposure.

0celo7

Hmm, it's not in the index.

@ACuriousMind What would it be called besides the de Rham theorem?

ACuriousMind

I know no other name for it

0celo7

It has to be in here.

It's in Lee, if all else fails.

user218912

@ChrisWhite yeah, if I learn enough I'll see if I can audit their graduate course on n-body simulations. because it has no prerequisites except familiarity with a programming language and basic physics.

user54412

n-body is a very popular choice for a first scientific code

user54412

11:19 PM

it's actually one of the few types of scientific code that uses a nontrivial data structure (a type of tree), if you're into that sort of thing

user54412

I guess biologists might have something fancy with BLAST, but that's about it afaik

user218912

I have no idea how to begin though, after I learn some C.

user218912

how do I apply C to physics?

user218912

is there a book on that?

user54412

that's... pretty broad

user218912

11:21 PM

like a book that can show you examples of scientific code (physics related).

user218912

to get started only .

user54412

once you're comfortable with the language, pick a problem and look up the associated algorithm

user218912

oh I see.

user54412

usually it boils down to updating an array of numbers every time step

user54412

once that works, pick a different problem and try to code it's algorithm

user218912

11:22 PM

@ChrisWhite also nice blue homepage background and blue shirt. :)

user54412

at some point you'll see patterns that are hard to explain a priori

user218912

@ChrisWhite what do you mean by this?

user54412

like, you'll get into the habit of seeing an equation and knowing what should be represented in the code and how it should be processed to get a numerical answer

0celo7

Were they on drugs when they made this??

user54412

is it a time evolution or seeking a steady state? does it deal with discrete or continuous things? are the interactions local in space? in fourier space? in phase space? are there conserved quantities? -- all of these questions hint at what type of code should be written

dmckee

11:27 PM

@3750 You start by learning c well. And then you start learning some numerical analysis and at the same time start learning how to abstract physical descriptions down to tractable mathematical models.

user218912

ok.

dmckee

Each of the three parts requires you to write a bunch of code and make a bunch of mistakes (but don't worry about the mistakes: they'll come naturally) in order to really get it.

The first hundred thousand lines are the hard ones. After than things get easier.

Until you start programing in a "grown-up" environment with collaboration and check-in requirements and long term planning and so on. All very necessary, but they dial the pain level back up.

Do you already know how to program in any language, or are you starting fresh?

user218912

fresh basically

user54412

collaborative coding: how to clobber others' code before they clobber yours

dmckee

Then don't just start on the bible. You'll want a more gentle introduction.

user218912

11:31 PM

which bible?

user218912

idk any C books

dmckee

For someone who already understnads programming The C Programming Language is a great book. But it's hell on beginners.

user54412

you know I never read that book

dmckee

@3750 The book that the authors of the language wrote. Called "K&R" and "the bible". It's very short and very dense, but if you are already a programmer it tells you exactly what you need to know about c.

0celo7

@ChrisWhite You're clearly a fraud

user218912

11:33 PM

alright then book recommendations please?

dmckee

I don't know. C was my second "serious" language and I never needed anything other than K&R.

user54412

@0celo7 more like an apostate at worst

ACuriousMind

I think it highly depends on the setting whether being a fraud or an apostate is worse...

dmckee

I know lots of people who never read it. Most everyone who learned C first. People who came to C by way of C++ and so on.

knzhou

Young person chiming in here; I think I'm at least ~10 years younger than both of you. I don't know any physics majors my age using C. Higher-level languages like Python, R, Mathematica, etc. are much more common.

0celo7

11:34 PM

@ChrisWhite I'll never read MTW, so I can sympathize.

knzhou

Do you anticipate a transition from C/C++ to one of those soon-ish?

0celo7

@knzhou They're 15 years apart :P

dmckee

@knzhou And they are the right thing to start most projects in. But you need to know what is going on under the hood and you need to have a low-level systems programming language available to you.

0celo7

In fact, @dmckee is 12.

ACuriousMind

@knzhou All the people doing simulations I know are using C, no matter their age. It depends highly on what the application is, I think

user54412

11:36 PM

@knzhou Not for hardcore scientific programming. Certainly 99% of all code written is not in that category, but that 1% just doesn't work unless you are close to the hardware.

dmckee

@knzhou I've noodled around with python from time to time. I'd like to have time to really dig in, but...

user54412

I didn't know anyone in undergrad who did this sort of programming. It's the sort of thing done deep in the basements by pale and undernourished grad students.

knzhou

Well, we're pretty pale and undernourished undergrads. :P

dmckee

They get plenty of calories. It's just that there aren't many vitamins in vending machines.

0celo7

@ChrisWhite We had a fat, pale grad student

knzhou

11:38 PM

I did put in the time to learn C, but I was more productive with Python after just a couple days with it.

Plus there are new Python-to-C things. Like, you can use ROOT in Python now!

...uh, not that one would ever want to.

But I'm banking on Python being at least somewhat feasible by the time I'm in grad...

user54412

I've certainly written more self-contained programs in Python than C. But I've hastened the heat death of the universe far more with C, and isn't that what counts?

ACuriousMind

I remember running one of your Python scripts for that self-evaluation meta post

dmckee

@knzhou It really depends on what you're doing. If you're not doing hard-core numerical work then it'll be fine. If you are doing heavy crunching you'll still want to get away from the hidden abstractions and close to the metal.

user218912

@ChrisWhite found another cambridge mathematical physics book, do you know it? amazon.com/…

dmckee

Or may they'll final come through on automated numeric fine-tuning. Right after fusion, and flying cars.

Slereah

11:41 PM

@ACuriousMind Is it true that a one particle state for a QFT Foch state is just $L^2(\Bbb R^3)$

dmckee

Speaking of old school programming stuff: I worked a problem from Programming Puzzles & Code Golf the other day. In yacc:

user54412

@3750 no, but it looks like it covers a lot. Section 6.1.4 could be the title of my thesis.

ACuriousMind

@Slereah Hmmm. Well, the one-particle states are spanned by $a^\dagger(\vec p)\lvert 0\rangle$. Since $a^\dagger(\vec p)\mapsto \exp(\mathrm{i}\vec x \cdot \vec p)$ maps that to the Fourier "basis" of $L^2(\mathbb{R}^3)$, I guess that is true.

dmckee

4

A: Count of "a"s and "b"s must be equal. Did you get it computer?

Bison/YACC 60 (or 29) bytes (Well, the compilation for a YACC program is a couple of steps so might want to include some for that. See below for details.) %% l:c'\n'; c:'a''b'|'a'c'b'; %% yylex(){return getchar();} The function should be fairly obvious if you know to interpret it in terms of ...

Slereah

So... Do we have like

$\bigoplus L^2(\Bbb R^3)^{\otimes n} = L^2(\mathcal S)$

0celo7

11:48 PM

@ACuriousMind I think one can use Whitehead to approximate a continuous homotopy equivalence by a smooth one, and obtain an isomorphism of de Rham groups that way.

ACuriousMind

@Slereah More like $\bigoplus_{i\in\mathbb{N}}L^2(\mathbb{R}^3)^{\otimes i} = L^2(S)$ since the l.h.s. is what defines the full Fock space.

I'm not sure how true that is rigorously, though

Well

It depends on what the "=" means, I guess

Since all separables Hilbert spaces are isomorphic, it's a bit pointless to talk about Hilbert spaces being equal/isomorphic without further structure that needs to be preserved

Slereah

$A = B \eq \forall a \in A, a \in B$

0celo7

Does $h\circ f=id$ and $g\circ f=id$ imply $h=g$?

Slereah

Wait, is it even $\bigoplus_{i\in\mathbb{N}}L^2(\mathbb{R}^3)^{\otimes i}$

ACuriousMind

No, there's a symmetrization missing

Slereah

11:54 PM

After all the proper scalar Fock space is $$\bigoplus_{i\in\mathbb{N}}L^2_{Sym}(\mathbb{R}^3)^{\otimes i}$$

Colombeau uses that representation to do the QFT, I wonder if it's necessary