The h Bar

DanielSank

00:42

I was just thinking: maybe we don't need to have a dedicated homework policy. I think the "homework" part throws people off, e.g. because they say "THIS IS NOT A HOMEWORK PROBLEM" or whatever.

Instead, maybe we should just clearly say that we only support questions with a conceptual core.

In other words, if the post asks how to solve any specific problem it's off topic.

Just get rid of the word "homework" entirely.

3

0celo7

@DanielSank YESSSSSS

@DanielSank I have some closed questions from back in the day.

That's exactly what I said.

DanielSank

@0celo7 Back when you were a n00b?

0celo7

Back when I was more of one

Pls upvote my question Mr. Google :)

Arnold is hard

DanielSank

@0celo7 What question?

0celo7

linked right above

of course, if you know the answer, feel free to write one

DanielSank

00:52

@0celo7 I upvoted it, but I'd recommend that you read it to yourself and clear up the language a bit. There are a few parts where it reads weirdly.

@0celo7 Nope. I don't know what an interior derivative is.

0celo7

I just read it...but I've been thinking about this for the better part of a day

Which parts?

DanielSank

> (I understand how to obtain the equation for the boundary of the homotopy of γ. gt is the flow of X.)

That part sounds strange.

Oh wait, is that dot after the gamma a period?

0celo7

It's a period

Should I switch the sentences?

DanielSank

@0celo7 Ha. Ok, that's why I never start a sentence with a math symbol >.<

0celo7

I'll switch them when I get a chance

NeuroFuzzy

00:56

@DanielSank I like this. Do you think this can actually happen? It seems like a pretty entrenched/old part of the site.

DanielSank

@NeuroFuzzy I'm writing a meta post right now. I hope to word it up to the standards of this site.

If we all approach this like true scientists I think it can happen.

Obviously, the word "homework" shouldn't really completely go away. If nothing else we should say "...and homework questions tend to lack the conceptual core..." or whatever.

0celo7

Yesssss

DS can save us

Wonder if homework questions will go down

Maybe we should try it and run an SLQ

DanielSank

@0celo7 What?

@0celo7 What's SLQ?

0celo7

What's that thing that you can do queries with

DanielSank

@0celo7 SQL

What do you want to check with an SQL query?

0celo7

01:02

we should change the language from homework to non-conceptual for a week

then run an SQL to find the number of closed questions

see if there's a statistically significant decrease

0celo7

01:13

@DanielSank I made some minor tweaks

DanielSank

01:30

@NeuroFuzzy Entrenched doesn't mean good.

dmckee

02:44

@0celo7 First of all I"m an experimenter. And secondly I'm not worried. I mean, how many theorists can tie actual knots in physical rope, anyway? No fair counting phenomenologists.

@NeuroFuzzy I've floated a similar notion once before. It got a kinda luke-warm receptions. But I would get behind it, and I think you might be able to sell John R. as well.

0celo7

I know how to tie a noose. Does that mean I'm an experimentalist?

dmckee

@0celo7 Hmmm ... might be. How do you feel about grease, wrenches and solder fumes?

0celo7

@dmckee I polished a crystal today and used a microscope

Does that count?

dmckee

Very promising. An acid test: how would you feel about being asked to conduct the Geiger and Marsden experiment (AKA the original Rutherford scattering experiment) the old fashion way?

user54412

@dmckee I was going to say I tied my shoes today, but that might not even be true...

dmckee

03:05

@ChrisWhite I developed some plantar fasciitis since I started teaching and have switched mostly to birks, so I only tie laces for the gym and for work boots now.

And I am looking every more the picture of the eccentrically casual physics professor.

DanielSank

04:01

@0celo7 are you around?

@NeuroFuzzy?

@dmckee I started getting that, and I actually manged to nip it in the bud by consciously not letting my feet pronate.

0celo7

04:16

@DanielSank yes

playing BF4, but yes

DanielSank

@0celo7 awesome.

I'm about to post my meta question about the homework policy.

0celo7

whatcha need?

DanielSank

You seemed interested.

0celo7

I am

DanielSank

It's a self-answered post.

I'd like your feedback so I can fix it up before everyone wakes up tomorrow.

You interested?

@0celo7 Uh... you there?

-_-

Ok, well, I'm going to post it. I hope you'll take a look.

0celo7

04:24

@DanielSank playing!

DanielSank

@0celo7 k

0celo7

killing the peoples

DanielSank

@0celo7 gross

@0celo7 I'd really appreciate feedback on this

0

Q: Should we rename the homework policy?

The homework policy is a constant source of confusion for new (and some times established) users. We see this confusion, for example, when users respond to closures based on the homework policy by defending their post with "This is not a homework problem", or similar. Some users have even been co...

discussion

0celo7

some times -> sometimes

> Some users have even been conditioned to claim a non-homework basis for their posts preemptively.

highlight, bold, italics

THIS SO MUCH

DanielSank

@0celo7 Too long?

0celo7

04:32

@DanielSank oh wow, that is long

DanielSank

@0celo7 Is it high quality long or this-is-longer-than-needed-to-make-the-point long?

0celo7

have not read

DanielSank

@0celo7 I compare the words of the help center, close reason, and canonical meta post with comments on why they're not only confusing but actually not even what we want.

I am afraid that if I don't make a complete case people will just say "no, it's fine, don't change anything".

"We talked about this before, etc."

Physics Meta

1

Q: Should we rename the homework policy?

The homework policy is a constant source of confusion for new (and sometimes established) users. We see this confusion, for example, when users respond to closures based on the homework policy by defending their post with "This is not a homework problem", or similar. Some users have even been con...

discussion

DanielSank

Woah... those come up automatically?

0celo7

04:38

yes....

Huy

04:50

wat

0celo7

06:52

@DanielSank

> If folks are really afraid we can do a few week trial period. Stack Exchange gives us awesome SQL tools so that we can measure the effects objectively.

+1

:)

DanielSank

07:11

@0celo7 Oh, yeah, totally used your suggestion. It's a great idea.

Ghost

am happy with how my question is going - in trms of answers and comments

Slereah

09:28

Know what annoys me?

The classical 2-body problem is still not solved

There's no analytic solution for $r(t)$

You'd think they would have figured it out in 400 years

Huy

11:35

no comment

JokelaTurbine

11:49

@Slereah Yeah, but Maybe Feynmans QED theory of Gravitation was right. (Messenger lections, Part 2 at 9:50) Maybe it works! Maybe there is other reason for the rotation of Earth. I have found a a theory for that, but I think it's not "mainstream physics" in a way that is accepted here. Though it is actually basic physics, thermodynamics, etc,

Slereah

12:31

what

Slereah

13:03

Is there a link between the path integral formulation and the Schrodinger wavefunctional formulation

The first one has $\langle F[\phi] \rangle = \int \mathcal{D}\phi F[\phi] e^{iS}$

While the second is $\langle F[\phi] \rangle = \int \mathcal{D}\phi F[\phi] \Psi[\phi] \Psi^*[\phi]$

Is there some relation between $e^{iS}$ and $\Psi\Psi^*$

ACuriousMind

@Slereah Should be the same for $\Psi$ the ground state, since the path integral computes the expectation values on the vacuum.

Slereah

Well not necessarily

You can compute the path integral between two field configurations on the boundaries

ACuriousMind

@Slereah So, what boundaries do you choose for your wavefunctional expectation value, and which ones do you choose for the path integral value?

Also, what the heck is $\mathcal{D}\phi$ supposed to mean without the exponential in front?

I'm very sure that measure doesn't exist.

The continuum limit relies on the Wick-rotated exponential to get a well-defined (Gaussian) measure on the fields even in the free case. You can't just write down $\mathcal{D}\phi$, if you go back to the limiting procedure that "defines" it, that limit just doesn't exist without the exponential

Just don't use wavefunctionals, they're completely useless :P

Slereah

@ACuriousMind Well the boundary values are for S, not for the path integral

And the wavefunctionals $\Psi$ are exponentials

at least all the ones I saw

ACuriousMind

@Slereah I don't know what that's supposed to mean.

Slereah

13:14

$e^{iS} = e^{i{\int_{\phi_1}^{\phi_2}\mathcal{L}}} $

I believe that is the basic notion for path integrals, no?

ACuriousMind

@Slereah The Lagrangian density is integrated over spacetime, not over field space. You can't give it fields as boundary values.

Slereah

Well the boundary conditions are that $\phi(x) = \phi_1$ for $t = t_1$

And same for $t_2$

ACuriousMind

No, that's a boundary condition for the path integral, not the integral in the action

Slereah

Hm

Let me check how the wavefunctionals do it then

ACuriousMind

$\mathcal{L}$ is integrated over spacetime, you could only give it spacetime points $x\in\mathbb{R}^4$ as boundary

Slereah

13:23

It's not easy finding schrodinger functional ressources

ACuriousMind

That's because they're frigging useless :P

Slereah

You're frigging useless

What is the vacuum expectation value boundary condition, by the way?

ACuriousMind

Trying to find it

I'm beginning to think that one does not actually talk about "boundary conditions" in any precise sense for the path integral. In the QM case, one integrates over the conditional Wiener measure over the space of paths with fixed endpoints - there is no "boundary condition" because the measure one is integrating against is not defined on other paths to begin with.

Slereah

Well the boundary conditions are the fixed endpoints

Also I forgot that wavefunctionals are only for the wavefunction at a time $t$

So a full propagator would have also $e^{i\hat{H}}$

ACuriousMind

@Slereah Yeah, but it doesn't really make sense to talk about that because changing the boundary conditions changes the measure. It's not like $\int_{x_0}^{x_1} f(x)\mathrm{d}x$ where you can change $x_0,x_1$ at will without changing the measure.

Slereah

13:34

I guess that would be the link between the path integral and the wavefunctionals

But aren't there path integrals with specific boundary conditions, like in non-equilibrium processes

ACuriousMind

@Slereah Yes! One can separate out the initial condition from the path integral in that Schwinger representation. Let me look it up

Slereah

From my days of doing non-relativistic path integrals boundary conditions were pretty important

ACuriousMind

Not saying they aren't, I'm saying they are of a different nature since they actually change the measure. Since we're mostly doing only formal manipulations of the integral though, it's probably fine to ignore that, mostly.

Slereah

Hm

I should buy a fancy path integral book

Maybe go back to Demichev

ACuriousMind

:( Everything I can find writes states like $\lvert \phi_1 \rangle$ as if it was completely obvious that a classical field configuration specifies a quantum state.

the only way I see how that could be done is saying it is the state the operator $\phi_1$ generates from the vacuum.

But then, it is not really obvious why $\langle \phi_2 \vert U(t) \vert \phi_1 \rangle$ should be the path integral with boundaries $\phi_1$ and $\phi_2$. It would be the case for the QM path integral and position eigenstates, but how should one define the $\phi_i$ so that they are the QFt analogon of position eigenstates? Are they eigenstates of the operator $\phi(x,t_i)$ at every $x$?

Ah, I think that's it.

Slereah

13:55

Eigenstates of $\phi$ are coherent states

ACuriousMind

So?

Oh, they're not orthogonal...

Wait, if $\phi$ is a real field, they should be, it's a Hermitian operator after all.

...what do you mean by "coherent state" in a QFT?

Slereah

14:08

States of minimal uncertainty?

ACuriousMind

For which operator?

BTW, here Lubos says that the "infinite time evolution" suppressed all states but the ground state, so the path integral without boundary conditions evaluates stuff on the vacuum by construction. I'm inclined to agree since this is very similar to an argument in the non-path integral derivation of the LSZ formalism.

dmckee

@DanielSank I've no idea where to even start on that. Is the a class of specialist to see for advice or help in that kind of matter. My podiatrist is rather old fashioned, I think.

ACuriousMind

He also says that $\lvert \psi_i \rangle$ is supposed to be essentially a delta function for the field configuration $\psi_i$. That is, it is kinda ill-defined.

::sigh::

0celo7

@ACuriousMind someone answered

ACuriousMind

QFT is a mess.

0celo7

14:17

That's why you should do GR, which is mathematically well-defined

ACuriousMind

@0celo7 Looks alright to me.

0celo7

GR or the answer?

ACuriousMind

@0celo7 The answer

0celo7

14:32

what you got against GR

@ACuriousMind I don't understand the Fubini's theorem part

@ACuriousMind is the $\mathrm{d}s$ being moved past the differentials in $\omega$?

@ACuriousMind also why does $\int_{H\partial\gamma}\omega=(-1)^{k-1}\int_0^1\int_{\partial\gamma}i_X\omega\,‌\mathrm{d}s$

ignore the "also why does"

ACuriousMind

14:50

@0celo7 Somewhat. Fubini's theorem states that one may exchange the order of integration, and that's what's being done. Instead of integrating first over $[0,1]$ and then over $\gamma$ we now integrate first over $\gamma$ and then over $[0,1]$.

0celo7

I get that

So why the sign change?

ACuriousMind

@0celo7 Because of forms.

0celo7

wha

pls show

ACuriousMind

You'll have to write out the form in it's components and use Fubini for the coefficient functions and then reassemble the form to see where these signs come from

I'm not gonna do that, these calculations are both tedious and unenlightening

I've learned not to think too much about signs unless the result seems wrong :P

0celo7

but in Fubini there are no sign changes

so why would there be sign changes here

ACuriousMind

14:58

@0celo7 Because the antisymmetry properties of forms always induce such gradings, haven't you seen these annoying $(-1)^k$ pop up before?

0celo7

Dude but in Fubini's theorem there are no signs!!!

ACuriousMind

I know

0celo7

Then where do they come from?

ACuriousMind

@0celo7 You're moving the $\mathrm{d}s$ through $k$ of $\mathrm{d}x^i$.

0celo7

@ACuriousMind that's what I said above!

but why!

ACuriousMind

15:04

Because, when you write out the l.h.s., you must write those behind the $\mathrm{d}s$, but on the r.h.s., they should be written in front.

0celo7

Fubini's theorem makes no mention of that

ACuriousMind

@0celo7 Because it's not a property of Fubini's theorem!

0celo7

and there's no wedge!

ACuriousMind

Fubini tells you we may exchange the order of integration

It doesn't tell you how to move forms around.

0celo7

@ACuriousMind so is there a typo?

and should there be a wedge there?

ACuriousMind

15:06

What is happening is this: The integration of forms over chains is ultimately defined in terms of an integration over $\mathbb{R}^k$. The usage of Fubini is just to be able to integrate over that $\mathbb{R}^k$ prior to integrating over the time.

0celo7

dude

should there be a wedge

and should the order of the things be reversed on the rhs

ACuriousMind

Not as it is written

Because of the $\int_0^1$, the $\mathrm{d}s$ there is just an integral delimiter, not a 1-form.

0celo7

then why is there a sign dammit

look, I obviously don't understand something here

ACuriousMind

Because if you do it right you have to either put everythign as functions or everything as forms but the user did not show you the derivation because it probably one page of shoving indices around.

0celo7

so please show me instead of saying it's because of forms

ACuriousMind

15:09

@0celo7 I guess what I'm saying is that the origin of this factor is completely irrelevant to understanding the idea and method of this proof.

0celo7

I don't see what there is to show! if you write all the integrals down and use Fubini, there are no signs!

:/

@ACuriousMind well I obviously don't understand something basic

@ACuriousMind could you please show it for $k=1$ or something

ACuriousMind

@0celo7: $\int_0^1 \int_\gamma \alpha \mathrm{d}s = \int_{I\times\gamma}\alpha\wedge\mathrm{d}s = (-1)^k\int_{\gamma\times I}\mathrm{d}s \wedge \alpha = (-1)^k\int_\gamma \int_0^1 \alpha \mathrm{d}s$.

0celo7

wtf

ACuriousMind

Where in the first and fourth term, $\mathrm{d}s$ is just an integral delimiter and in the second and third, it's a 1-form

0celo7

I give up

ACuriousMind

15:19

It's a terribly confusing notational shell game.

0celo7

that's not helpful at all

ok that's more helpful

alright, next question

how did he get $\mathrm{d}F^{i_j}$?

ACuriousMind

@0celo7 By, uh, definition? $\mathrm{d}f = \frac{\partial f}{\partial x^i} \mathrm{d}x^i$, right?

0celo7

yes

I get the first equality

the second one, not so much

ACuriousMind

@0celo7 For the first summand, chain rule after observing that that $x(t)$ is given by the flow and that the derivative of the flow is the vector field itself.

0celo7

but $F$ is $g^t\gamma(x)$

where does $\gamma(x)$ go

also what is the index on $\mathrm{d}F$ supposed to mean

christ, this problem...

ACuriousMind

15:30

@0celo7 That's "defined" in the equation right above

0celo7

$\mathrm{d}(F^*y^{i_j})$?

ACuriousMind

@0celo7 Which is $F^*\mathrm{d}y^{i_j}$, yes.

0celo7

:(

I still don't get it

like any of it

fml

ACuriousMind

I think $\mathrm{d}F^{i_j}$ is a row of the Jacobian matrix of $F$.

But I'm not exactly usre, I hate it when exterior derivatives and differentials mix.

0celo7

so where does the $\gamma$ go?

ACuriousMind

15:41

@0celo7 Nowhere. Note that the vector field $X$ has to be evaluated at a point on the manifold, not on $\sigma$, so you have $X\circ\gamma$ as a vector field that takes the $x$ coordinate from $\sigma$ instead. (One doesn't need to use the chain rule, sorry for that)

But the user suppressed the notaiton of the point at which the vector field is evaluated since it is there anyway in the $F(t,x)$.

0celo7

ok

uh, what about the second term in $\mathrm{d}F$

sorry about this, I'm just really confused by what's going on here

ACuriousMind

@0celo7 Another notational shell game. The sum over partial derivatives there is $\mathrm{d}F^{i_j}\rvert_{t = \text{const}}$, i.e. the derivative is only taken on the spatial coordinates forsome constant time. But at constant $t$, $F$ is $g^t\gamma$, which is $H_t$ by that user's definition.

This reminds me why I didn't like my diffgeo lecture. Half the steps are just juggling one notation into another.

0celo7

@ACuriousMind ::sigh:: well I still don't see how that's $H^*_t\mathrm{d}y$

ACuriousMind

15:56

@0celo7 Well, because you took the $i_j$-th component of $F$ as a function into $M$. Doing $\mathrm{d}y^{i_j} \circ \mathrm{d}f$ selects the $i_j$ component of $f$.

0celo7

> $\mathrm{d}y\circ\mathrm{d}f$

what is this fucked up notation

Slereah

@ACuriousMind for $\phi$ and $\pi$

ACuriousMind

@0celo7 Concatenation, and the definition of $f^*\mathrm{d}y$.

0celo7

was rhetorical

ACuriousMind

@0celo7 How was I supposed to know that? :P

0celo7

15:58

::shrug::

I'm just really tired of this problem

this is so much harder than the standard proof

ACuriousMind

@0celo7 Then, uh, stop doing it?

0celo7

@ACuriousMind I can't read books in which there are problems I've started to solve but can't finish

ACuriousMind

Y'know, I suspect that since Arnold is of the Russian school, he intended you to have geometric intuition to see all those formulae as self evident and not prove them.

0celo7

which formulae

ACuriousMind

Because the actual formal proof is really getting ugly.

@0celo7 All those you don't see how to prove/why they hold.

0celo7

16:00

great

ACuriousMind

I don't have that kind of intuition, either, that's why I hate stuff written by the Russian school :P

0celo7

hmm

I'm thinking I should talk to Freire about this

he recommended the book

least he can do is help me with a problem

ACuriousMind

Plot twist: He doesn't like the book either and just wanted to see how long you try to work through it to please him.

0celo7

...

please no

these exercises are horrible

well, let's see what John Ma writes back

and what Alex says

thanks for your time

Slereah

Why is one of them a dog

0celo7

16:05

good question

ask tumblr

ACuriousMind

@Slereah You're not wondering about the dinosaur in the background?!

0celo7

hehe

Slereah

All old people are dinosaurs

0celo7

interesting point

Emilio Pisanty

No, Alpha, that's not what I meant

ACuriousMind

16:10

lol

Well, it is a line from Lyman to Alpha :D

Emilio Pisanty

@ACuriousMind Yeah, I can't really complain, can I?

0celo7

haha

Slereah

Now try doing the transition

(by plane)

DanielSank

@dmckee The podiatrist is the correct specialist. If yours doesn't know about this problem maybe get a referral to another?

0celo7

now we can add feet to the list of topics dicussed here

dmckee

18:49

@0celo7 Once your feet start hurting you become willing to discuss them anytime and anywhere. Take care of them, you use them a lot.

Slereah

19:16

6

Q: How is foliation of manifolds' theory useful in General Relativity?

I am interested on getting some hints on how Foliations Theory of Manifolds can be used fruitfully on General Relativity. I just started my Ph.D on Mathematics this semester focusing on studying Holomorphic foliations on projective manifolds. I just came across some texts which related the more g...

general-relativity differential-geometry resource-recommendations

$\int_{\partial V} stuff$

The most treacherous term

ACuriousMind

@Slereah Use \text, you savage!

2

Slereah

$\int_{\partial V} \\ text$

I wanted to make a joke but I forget how to escape slashes

ACuriousMind

I think you can't escape them, but there's some command to print a backslash, like \backslash or something

Danu

Hi all

ACuriousMind

Hi Danu

Danu

19:27

Is requiring that $\Delta t$ (and $\Delta x$) transform in the same way as $t$ and $x$ strong enough to imply linearity of the Lorentz transformation for a boost along $x$?

@Slereah Hey, one of the profs at LMU is giving a seminar on foliations this semester ^^

ACuriousMind

@Danu Uh, what? For that to become answerable, how did you define a "Lorentz transformation" in this context?

Danu

@ACuriousMind $t'=\gamma(t-\beta x)$ you know, the good ol' one from intro to physics...

Also, I of course meant any linear combinations of times/$x$-coordinates of specific events rather than just $t_2-t_1$ and such.

It should be enough, right?

Slereah

@Danu Oh boy

you can cut up them spacetimes like sausages now

ACuriousMind

@Danu Uh, how is that not linear by definition?

Slereah

You know

I always wonder

Can you always foliate spacetime into achronal spacelike slices

Danu

19:35

@ACuriousMind Yeah :P

I know

hehe

Slereah

Obviously not all of them will be Cauchy surfaces

But otherwise is it always possible

Danu

Physically speaking, it means that one should be able to take arbitrary origins for coordinate systems, doesn't it?

ACuriousMind

@Danu No, the taking of arbitrary origins is encapsulated by the translations allowed by Poincare symmetry, but not in the Lorentz symmetry.

Danu

@ACuriousMind Think about it though, really

Also I think your comment (while correct) doesn't really imply anything about this.

ACuriousMind

::thinks about it though, really::

I don't understand what you mean.

Danu

19:38

Maybe a better way to say what I mean is

Assuming translation invariance, one derives linearity of the boost transformation

Is that correct or not?

ACuriousMind

@Danu You defined the boost transformation to be linear up there, I don't know what you want to derive

Danu

@ACuriousMind I want to justify it

I think it's not nice to postulate it to be linear.

I don't like having it as an assumption.

I don't wanna assume the $\gamma(t-\beta x)$ that I wrote up there, of course :P

ACuriousMind

@Danu Why not? How else could a group act on the affine vector space of spacetime if not by linear operators?

Danu

@ACuriousMind Right, affine vector space

That is what I mean by translation invariance (for the origin)

ACuriousMind

@Danu Well, you might also just say vector space, since the Lorentz part of the Poincare symmetry has nothing to do with the translations.

I really don't think that translation invariance is in any way related to the Lorentz transformations being linear.

Danu

19:43

Well

Picking the origin different exactly sends

$t\mapsto (t-t_0)$ and such

ACuriousMind

Just fix the origin. Still, how else are transformations gonna act if not by linear operators?

Danu

I think it's very reasonable to postulate that the transformation must be the same for all choices of origin

This (at least almost---I'm not sure if you get all linear combinations this way but probably you do) should imply linearity

@ACuriousMind I think you're just thinking in a different language.

ACuriousMind

Very possible, since I'm not really seeing what you want to do. What do you know about the Lorentz transformation if you don't want to know that they're linear?

Danu

I want to motivate linearity. You know that $t'=f(t,x)$.

Trying to formalize my statement I do get stuck, so I guess you're right.

I'd still like to motivate it somehow

ACuriousMind

16

Q: Proving that interval preserving transformations are linear

In almost all proofs I've seen of the Lorentz transformations one starts on the assumption that the required transformations are linear. I'm wondering if there is a way to prove the linearity: Prove that any spacetime transformation $\left(y^0,y^1,y^2,y^3\right)\leftrightarrow \left(x^0,x^1,x^2,...

special-relativity spacetime differential-geometry metric-tensor coordinate-systems

Looked at this?

Danu

19:53

@ACuriousMind Lol, I even upvoted Qmechanic's answer

Sigh, memory is a difficult thing.

That's a nice proof

0celo7

For your second question, note that the pullback is calculated by (if $F = (F^1, \cdots, F^m)$) $F^* dy^{i_j} = dF^{i_j}$. @0celo7 — John Ma 19 mins ago

Well that's not very helpful...

ACuriousMind

@0celo7 That is the same thing I said here :P

0celo7

I know...that wasn't helpful either...

ACuriousMind

@Danu Heh, I've sometimes read a question and thought "well, that's a nice one, never thought about this" and then realize I upvoted both it and some answers....

Danu

@ACuriousMind Use \mathrm, you barbarian!!!

(okay, actually not true)

ACuriousMind

19:59

@Danu Well, not for actual words which are meant to be text.

Danu

(but felt like saying something)

@ACuriousMind Yes, yes, I've read the relevant TeX - LaTeX post too :P

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