@yuggib Is Rudin's Fourier transform book the usual intro for harmonic analysis?

GR jargon is indeed annoyingly weird

I think I forgot what "past inextendible" means, actually

@BenNiehoff it just means the curve cannot be extended in the past direction any

it's the precise meaning of "can be extended" that is in question

it's not a statement about geodesic completeness, for example

I think

@BenNiehoff there's a smooth curve $\gamma_1$ such that $\gamma_1|_I=\gamma$ with $I$ the domain of $\gamma$

$\gamma_1$ should be timelike/null/whatever

@BenNiehoff Above I said geodesic but it works for any curve

so, if I just cut out a finite piece of Minkowski space, say between t=0 and t=1, is it past (future) inextendible?

will what be?

we're talking about extendability for curves, no?

say my manifold M is a cube cut out of Minkowski space

inextendability for manifolds means you cannot isometrically embed into a larger manifold

now consider the curve x=y=z=0

is it future extendible?

certainly not within M

It is, but inextendability does not mean geodesic completeness

so it's fine

ok

In Riemannian manifolds it does!

wait, it is, or isn't?

Your curve is inextendable

Any extension would do something weird and not be smooth

why should it make a difference whether we're in a Riemannian manifold?

cut a cube out of R^3 and ask the same question

I'm saying that geodesic inexendability implies completeness in Riemannian manifolds, but not in Lorentzian ones

It's a comment, not a retort to anything you said

you can't have a geodesic end on a singularity in Riemannian manifolds?

@BenNiehoff Singularity?

Not if it's complete, not

but it is inextendible, isn't it?

If all geodesics are defined on all of $\Bbb R$, then your manifold is complete. That's the Hopf-Rinow theorem.

the question is whether inextendibility implies completeness

I don't think it does

because a geodesic could run into a singularity at a finite distance

Oh, oh. Yeah

I need to read Choquet Bruhat one of these days

Maybe after I understand hyperbolic equations

@Slereah do you understand stings?

@BenNiehoff Here's an interesting question: Take $U\subset\Bbb R^n$ open, and $f:U\to\Bbb R^n$ injective. Is $f(U)$ open?

@0celo7 Hard to say

why, do u have new insight

Nopw

@0celo7 don't know exactly...

@BernardoMeurer one of my coworkers is a liberal type and for some reason she uses the black emojis (she's very white)

@0celo7 Name

I asked her if she got a bad sunburn or maybe radiation burns

@BernardoMeurer no

@BernardoMeurer I will tell you if you can convince rebecca to free herself from proprietary malware

@DanielSank not much just chillin' out

@ACuriousMind If $x = \Lambda x'$ and $\psi '(x') = S\psi(x)$, then why is there a $\Lambda$ in the first term after the third equation? imgur.com/a/hqB9o I suspect it's because $\partial_\mu$ acting on $\psi(\Lambda^\nu_\rho x^\rho)$ produces a $\Lambda^\nu_\mu$ because of the chain rule, but then we would have $\partial_\mu\psi(\Lambda x) = \Lambda^\nu_\mu\partial_\nu\psi(\Lambda x) = \Lambda^\nu_\mu\Lambda^\rho_\nu\partial_\rho\psi(\Lambda x)$ and so on, which does not make sense.

@Bass It's not $\psi'(x') = S\psi(x)$, it's $\psi'(x') = S\psi(\Lambda^{-1}x)$.

@ACuriousMind In the lecture script I'm reading, it is. Here's the whole paragraph: imgur.com/a/aB5CL

Oh, you are unusually supressing the prime on the $x'$. Well, that's a choice. Okay, but then, what's the question. $\psi'(x) = S\psi(\Lambda x)$ in your convention, and then it's chain rule as you said. Is what is confusing you that there's implicitly brackets, like $\partial_\mu (\psi(\Lambda x)) = \Lambda^\nu_\mu (\partial_\nu \psi) (\Lambda x)$?

Doesn't $\psi'(x') = S\psi(x)$ make sense? For scalars we have $\phi'(x') = \phi(x)$, so we just plug in a transformation in spinor space which depends on $\omega$.

@Bass Eh, I always get confused with the primes, don't listen to me ;)

@ACuriousMind Ah, those brackets $(\partial_\mu\psi)(\Lambda x)$ are basically saying "derivative of $\psi$, but the chain rule has already been used.."?

Strange, never before seen such notation. But seems to make sense.

Thx :)

@Bass Yes. $\partial_\mu (\psi(\Lambda x))$ means taking the derivative of the function $\psi \circ \Lambda$, then plugging $x$ into it. $(\partial_\mu \psi)(\Lambda x)$ means taking the derivative of $\psi$, then plugging $\Lambda x$ into it. That's how the chain rule works, after all!

Ah, of course. Now I feel dumb as always.

Sorry to have bothered you with chain rules :-D

@Bass no worries :)

Yo @ACuriousMind wtf does $d\bar{z}dz$ mean?

There are a bazillion posts on the internet about this and I still don't get it.

$dz = dx + idy$

Why is there 1/2?

or maybe the signs are the other way round.

@DanielSank It's a weirdo notation for $\mathrm{d}x\mathrm{d}y$ ;)

@ACuriousMind ok.

@DanielSank Sorry, there isn't.

Does it have something to do with coherent state path integration?

It should appear when you are integrating complex functions over subsets of the complex plane, but really, the complex character doesn't really matter, you're just integrating a function on $\mathbb{R}^2$.

@ACuriousMind ok

What's going on here

@0celo7 internets

@DanielSank Well, since you can label coherent states by complex numbers, it appears there, but it's not really linked to it

In what sense does it mean $dxdy$

If I compute naively I get $dx^2+dy^2$

^

@Bass $a\bar a=a_x^2+a_y^2$

there has to be something else going on here

So $dz d\bar{z} = (dx + i dy)(dx - i dy) = dx^2 + dy^2$

some complex geometry thing no doubt

@0celo7 In the sense I said - you're integrating a function in the plane

::raises arm:: I have a doubt

Whether you write that function as $f(x,y)$ or as $f(z,\bar{z})$ makes no difference to its integral

@ACuriousMind $dz d\bar z=dx^2+dy^2$ though

if we're doing physics math

And to me, a geometer, that's exactly what it should mean

$dz d\bar z$ is the Euclidean metric

But in terms of integral measures idk what's going on

Yeah. I don't think whoever wrote down that measure had anything more in mind that "just integrate over $z$ and $\bar{z}$".

Isn't it something like this: $dxdy = d\frac{z+\bar{z}}{2}d\frac{z-\bar{z}}{2i}=\frac{dzd\bar{z}}{2\pi i} + \frac{dz^2-d\bar{z}^2}{4\pi i}$

It's just a notation for integrating a complex function, I don't think there's something deeper behind it

Somehow the last term has to vanish.

@ACuriousMind proof?

@Bass Well, squares of infinitesimals vanish ;)

$f(x,y)$ vs. $f(z,\bar z)$ are quite different things

@ACuriousMind Ah yep. But then that's it, right?

@ACuriousMind Uh but $dzd\bar{z}$ is a square of infinitesimals too.

@0celo7 How. They're both functions on $\mathbb{R}^2$.

@Bass he was joking

Are they Grassmannian?

@Bass Well, they obviously only vanish when that gives the right answer!

@ACuriousMind In the same way that $f(z)$ and $f(x,y)$ are different

@0celo7 No. $f(z)$ and $f(x,y)$ are different because only depending on $z$ is a strong restriction (being holomorphic, if we're talking smooth functions on $\mathbb{R}^2$). A function $f(z,\bar{z})$ is just a rewriting of a generic function $f(x,y)$

@ACuriousMind Noted for future use. This term vanishes if it should :)

@ACuriousMind Is $z$ supposed to be $x+iy$ in $f(z,\bar z)$?

@0celo7 Yes

And $\bar{z} = x-\mathrm{i}y$, correspondingly

@ACuriousMind Ok suppose I take $f(x,y)=\chi_C$, where $C=[0,1]^2$

Then $\int_C f \,dA=1$, clearly

How does this work for a complex integral?

I'm not arguing I just have no clue what's going on

@0celo7 How would you define $\mathrm{d}z\mathrm{d}\bar{z}$ if not by that very integral?

That's what I'm asking you!

That's why I said it's just a silly notation for $\mathrm{d}x\mathrm{d}y$!

Integrals over $z$ should be contours

And integrals over $\bar z$ should be contours in the opposite direction

Yes, they are. But neither gives a natural notion for integration over both.

It is my firm opinion there's nothing to see here except confusing notation :P

Hmm? $$\int f\, dzd\bar z=\int_{C'}d\bar z \int_C dz\, f(z,\bar z)$$

@DanielSank The only thing that can save ACM right now is the observation that $dz d\bar z$ gives the Euclidean metric in terms of manifolds. If we define integration by integration wrt. the volume form of a metric, then it's true that $dz d\bar z$ gives the same integration measure as the Euclidean one, which is $dxdy$. Calling them by the same symbol is really dumb though, since there's a conceptual difference.

@ACuriousMind Also this makes no sense

@0celo7 It does if you don't insist on finding fault with it :P I really don't care for a rigorous discussion of integrals or measure theory

Writing $\int f\,dzd\bar z$ makes as much sense as $\int f(x,y)\, (dx^2+dy^2)$

@ACuriousMind I wasn't looking for anything rigorous, I know better

It's always strange seeing someone you know on Wiki

under an integral sign, what they really mean is $dz \wedge d\bar z$

probably needs a factor of $-i/2$ or something

then you should be able to satisfy your geometer's senses and still get $dx \, dy$

which is really $dx \wedge dy$, after all

@BenNiehoff indeed

Yep, that's the reason the last term here vanishes.

also, you can make sense of it if you hold your nose and say "the symbol dz under the integral sign doesn't actually mean anything except that we integrate with respect to z"

which I'm afraid is how most people see it

those people have to keep reminding themselves to put in Jacobian determinants when they change coordinates

as if it's just some rote step that's needed!

@ACuriousMind Hahaha, a new theme in discussions with @0celo7.

@BenNiehoff Yes yes

@BenNiehoff For a while I refused to write $dx$ in integrals.

...unless I was explicitly thinking about forms.

@DanielSank what did you write, then?

because the standard integral notation literally comes from Riemann sums

@BenNiehoff I let the set over which I was integrating speak for itself.

i.e. $$\int_{x=4}^6 x^3$$

$\int$ is an elongated S, for sum

Writing $dx$ adds zero information.

so, you are summing over the value of a function times an infinitesimal width of a rectangle

@BenNiehoff yes, I went to high school :-)

Nowadays I put the $dx$ in.

I'm glad I never had to grade your homeworks, that sounds atrocious

@BenNiehoff I only started doing that in college when I learned about what a differential form is.

Munkres writes $$\int f(x) dx$$ for an integral of a form and $$\int f(x) \, \mathrm{dx}$$ for a Reimann/Lebesgue/whatever integral.

@BenNiehoff saying $dx\wedge dy$ and $dxdy$ are the same is in...can't think of the word

like sincere

but the opposite

disingenuous

?

ah, yes

thanks

I'm talking about under an integral sign

@0celo7 @ACuriousMind why are you guys always video game characters?

a "double integral" is just the integral of a 2-form

oh boy

it's disingenuous because it should be the other way around

modulo concerns about orientability

the integral of a 2-form is a double integral

@DanielSank 'cause I like video games? :)

^

what's so hard to understand. Why are you a whatever you are @DanielSank

bush or plant or whatever

@DanielSank lol that's terrible

@BenNiehoff Put it this way: how are you going to define the integral of a $p$-form without reducing it to a multiple integral on $\Bbb R^p$?

I prefer to use upright d for forms, and if I ever write down "just an integral", I use italic d for that

It should always be clear from context what is meant...

sometimes you have a form whose coefficients are integrals ;)

ayy, that's clearly unworthy of consideration

as for integrating forms: I agree that you at least must pull back the form onto the place where it's being integrated

but I'm pretty sure you can then define an integral there, without having to walk all the way back to iterated integrals

I'm not talking about iterated integrals

fine then

While we are on the subject, I will say that the usual Fubini's theorem does not actually apply to Lebesgue integrals

which theorem is that?

You need to take a completion product measure and the proof is a little different

honestly, I can't bloody remember theorems that are named after people...I wish this convention had never been adopted

@BenNiehoff $\int_{A\times B}f(x,y) \,d(\mu\times \nu)(x,y)=\int_A(\int_B f(x,y)\, d\nu(y))d\mu(x)$

under mild assumptions

turns out it's not true that $\mathcal L^1\times\mathcal L^1=\mathcal L^2$ for Lebesgue measure

and it doesn't apply to Lebesgue integrals, because the inner integral might not exist?

@BenNiehoff oh, if the inner one exists then both are infinite

Maybe that's Tonelli's theorem...

there are two very similar ones

But no, that's not the issue

The issue is that if you product Lebesgue measure on $\Bbb R^1$ with itself, you don't get Lebesgue measure on $\Bbb R^2$.

You miss some null sets, and have to add those back in

I'm not sure I know what that means

Which part?

all I remember about Lebesgue integrals is that they slice horizontally rather than vertically

oh :P

I've never studied measure theory

Maybe this conversation is for another person then

@ACuriousMind perhaps?

@0celo7 No, it's not.

You're wrong.

It's lean and avoids needless repetition.

For indefinite integrals I used to write e.g. $$\int_x f(x)$$

ok, then what's $\int x^2 y^3$?

@BenNiehoff You forgot the bounds and/or symbol under the integral symbol.

$\int_x$ actually means something...it's similar to $\int^x$

Basically, the point is that $dx$ is not the same kind of thing as $f(x)$, so it shouldn't be multiplying $f(x)$. The meaning of $dx$ is better associated to the integral symbol itself, IMHO.

They are now proposing coupling between dark photons and axions

and $\int^x$ means "do a definite integral where the upper bound is x, and I don't care what the lower bound is"

@BenNiehoff Yeah, I can't TeX what I want, unfortunately. I used to write the integration variable underneath the $\int$.

oh well

I'm wrong?

@0celo7 Yeah.

surprised?

probably you want \underset

$$\underset x$$

@DanielSank You do know I write my integrals as $\int f$, right?

Did not know that.

But if you insert $x$, put the damn $dx$.

@0celo7 No.

yeah, the dx is right and proper

$f(x) dx$ doesn't make any sense unless you're thinking about forms.

I mean, it's a fine formal notation, I guess.

But it's a bit inconsistent.

$dx$ means Lebesgue measure. What if you have other measure floating around?

What if you're working in a Banach space of measures?

You don't multiply functions by measures.

gtg

No one is saying that.

@BenNiehoff The really proper thing is $\int_D f(x)\, \mu(dx)$

Or $d\mu(x)$

Whichever floats your boat

I sometimes write $\int_D f \, \mathrm{vol}$

But many times there will be at least two measures so you have to specify which one you want

@BenNiehoff For Riemannian geometry I like $\int_D f\, dv$

Circumverting the uncertainty principle by dumping the uncertainty into the observable we don't care about

actually, I write the Einstein-Hilbert action as $\int_M \star R$

$dv_g$ if I have multiple metrics

@BenNiehoff multiple things wrong with that

such as?

First is that actions should not be over the whole manifold

You compute actions on (pre-)compact subsets

fine

And you're missing the GHY term :P

oh, well I don't care about the boundary

who is Y?

York

ah, ok

Other than that, it's a very noble way of writing the action

so I mostly approve

@Secret scientists quoting rolling stone song lyrics to explain their work. have seen it all now, interesting/ notable article/ advance, thx for sharing :)

Too bad we cannot do a similar circumvention for entropy though, cause every microstate has nonzero expectation value of energy, meaning that some energy and/or mass will end up there and reducing efficiency

We can, however, exploit entropy to our advantage in the context of self assembly

Btw the journal article itself suggested it is some kind of squeezed state

@0celo7 welll...., sure they are

I mean, $dxdy$ is not a well defined thing but that's what you mean when you use it inside an integral

Jeez, are you still arguing about integral notation? :P

@ACuriousMind I invoke the "I am never wrong" theorem so we can be done with it

I love you all!

@Secret saw you musing on the simulation/ holographic universe hypothesis, did you see this one? there is another ambitious experiment also at fermilab that got a lot of press, but basically a null result so far

@peterh be careful some people can be offended or delude themselves if when you say that

Don't say you love anyone

In fact don't say much

Argh

Read this answer, still have no idea what in the world "amount of substance" is supposed to represent