The h Bar

ACuriousMind

15:00

@0celo7 Okay. Let $\gamma_1,\gamma_2$ be the curves. Let $h$ be the chart in which $(h\circ\gamma_1)'(0) = (h\circ\gamma_2)'(0)$. Now, evaluate $(g\circ\gamma_i)'(0)$ by the chain rule by inserting $h^{-1}\circ h$ between $g$ and $\gamma_i$.

The desired result is that you get the Jacobian $D(g\circ h^{-1})$ acting on $(h\circ\gamma_1)'(0)$.

0celo7

Hey that's the same notation I used

wait what is $g$ now

ACuriousMind

Some other chart

0celo7

ah

ACuriousMind

Didn't want to use $h'$ because the prime is already differentiation

0celo7

right, right

@ACuriousMind and by linearity the result follows

I need to look up the abstract definition of the Jacobian...

ACuriousMind

15:07

@0celo7 Invertibility of the Jacobian, rather.

0celo7

@ACuriousMind uh, right

it has no kernel

ok, I know that, I think, but how do I prove it?

ACuriousMind

Hm, or, well, I guess linearity suffices

0celo7

@ACuriousMind no, it could have a kernel?

but that gives 0 either way

yeah, linear maps map 0 to 0

ACuriousMind

@0celo7 No, the Jacobian of diffeomorphisms is always invertible.

0celo7

@ACuriousMind hence the question mark, I'm not sure how to show that

ACuriousMind

15:09

It's the definition of a diffeomorphism.

0celo7

@ACuriousMind I know that

Well that doesn't make much sense, does it?

I know it but I don't know it :D

@ACuriousMind how is the Jacobian defined, abstractly

$\mathrm{d}f(v):=Df\cdot v$ or something?

$\mathrm{d}$ is the tangent map

ACuriousMind

@0celo7 Since we want to construct the tangent spaces here, we can't use "abstract" definitions. In this case, it's only defined for a differentiable map $f : \mathbb{R}^n\to\mathbb{R}^m$ as $(Df)_{ij} = \frac{\partial f^i}{\partial x^j}$

@0celo7 What the hell is the "tangent map" supposed to be, we currently don't even know that the tangent space is well-defined!

0celo7

@ACuriousMind ayy

calm down

@ACuriousMind ok but from that it's not clear to me that it has an inverse

ACuriousMind

@0celo7 It doesn't for arbitrary $f$.

0celo7

I mean for diffs

ACuriousMind

15:19

It's part of the definition.

0celo7

@ACuriousMind So a diff is $f:\mathbb{R}^n\to\mathbb{R}^n$

"A bijective differentiable mapping φ, whose inverse φ−1 is also differentiable is called a diffeomorphism."

That means we have a square matrix for the Jacobian

so maybe try to calculate the determinant

ACuriousMind

@0celo7 In this case, yes, and $g\circ h^{-1}$ is precisely such a thing

0celo7

so we take a volume form

no, this is getting too complicated

ACuriousMind

Somewhere in the definition of the charts there has to be stated that these transition functions have a Jacobian with full rank. I don't know what you want to do

0celo7

I don't know why the rank of the Jacobian is $n$

ACuriousMind

15:22

By definition!

There's nothing to show, the Jacobian of a transition function always has full rank, otherwise these things weren't charts

0celo7

After the fifth time you should see that I don't see that :(

why does it have full rank!?

ACuriousMind

@0celo7 $D(g\circ h^{-1}) D(h\circ g^{-1}) = 1$.

Thus it has an inverse.

0celo7

Ok you have to know that doesn't help me at all

just work with some diff $f$

Why is $Df$ invertible

ACuriousMind

Okay, if $f$ is a diffeomorphism, then it is invertible. $1 = D(\mathrm{Id}) = D(f\circ f^{-1}) = Df\cdot D(f^{-1})$, thus $Df$ is invertible.

0celo7

Ok, now we're getting somewhere. I don't understand the last equality.

ACuriousMind

15:27

Chain rule.

Write it out in components.

0celo7

this was an exercise in Straumann I never figured out, either

Chain rule? I really don't see this o.o

ACuriousMind

Oh, I'm sorry!

This is indeed a bit more complicated

0celo7

$$Df\cdot D(f^{-1})=\frac{\partial f^i}{\partial x^j}\frac{\partial (f^{-1})^j}{\partial x^k}$$

:/

wait, if I switch them...

hey isn't that the rule for...pushforwards/pullbacks/some star thing?

er, no, that won't work either

ACuriousMind

That the inverse of the Jacobian is the Jacobian of the inverse is precisely the inverse function theorem.

Sorry, one can't show this straightforward, it's something one has to know from analysis.

0celo7

That theorem in the curriculum for my analysis course

ok, but I didn't need invertibility of the Jacobian here, as far as I see

@ACuriousMind well, what about the rule $(f\circ g)_*=f_*\circ g_*$

isn't that like the chain rule?

ACuriousMind

15:35

Uh, that is exactly the same writte differently

0celo7

right but you don't need analysis

it should follow from the definitions of everything

let's go to maps $\phi,\psi$

function $f$, vector $v$

ACuriousMind

Halt!

You're not yet allowed to use vectors as in "tangent vectors", remember?

0celo7

then $\phi_* v(f)=v(f\circ \phi)$

where exactly did I need "the Jacobian has an inverse"

all I needed was linearity, right?

ACuriousMind

Yes, we already said that, but you kept asking me why the Jacobian should have full rank

0celo7

38 mins ago, by ACuriousMind

@0celo7 Okay. Let $\gamma_1,\gamma_2$ be the curves. Let $h$ be the chart in which $(h\circ\gamma_1)'(0) = (h\circ\gamma_2)'(0)$. Now, evaluate $(g\circ\gamma_i)'(0)$ by the chain rule by inserting $h^{-1}\circ h$ between $g$ and $\gamma_i$.

38 mins ago, by ACuriousMind

The desired result is that you get the Jacobian $D(g\circ h^{-1})$ acting on $(h\circ\gamma_1)'(0)$.

ok so now tangent vectors are well defined

ACuriousMind

15:39

Also, I'm now confused whether or not the inverse function theorem follows from the chain rule

0celo7

and I can use geometry

@ACuriousMind I'm planning to go talk to Freire and Denzler about some stuff, you want me to leave you alone and bug them about it?

damn, it's lunch time

they ain't gonna be there

from my future analysis text

ACuriousMind

Okay, I was confused for a moment, but the inverse function theorem indeed follows from the chain rule

0celo7

Ok, so I don't need it to prove the Jacobian has full rank?

ACuriousMind

I'm now wondering why so many people present it as if it was some arcane theorem that requires other tricks

0celo7

I use linearity to prove the geometry makes sense

then use geometry to prove the chain rule on manifolds

and then the Jacobian has full rank

then I show the pushforward is the Jacobian

I think Carroll does that in his GR book...also Wald

@ACuriousMind Does that outline make sense?

Danu

15:51

@ACuriousMind I suspect that proof hides the tricky parts in the implicit function theorem

ACuriousMind

@Danu Ah, you're right, the tricky part is not showing the Jacobian of the inverse is the inverse of the Jacobian, but the invertibility of the function in the first place.

0celo7

Right but diffs are invertible by definition

ACuriousMind

Yeah, we don't need the theorem here.

Danu

@ACuriousMind I'm pretty sure there's no easy way for the pair

15

Q: Better Proofs Than Rudin's For The Inverse And Implicit Function Theorems

I am finding Rudin's proofs of these theorems very non-intuitive and difficult to recall. I can understand and follow both as I work through them, but if you were to ask me a week later to prove one or the other, I couldn't do it. For instance, the use of a contraction mapping in the inverse fun...

math.bard.edu/belk/math461/InverseFunctionTheorem.pdf

ACuriousMind

@Danu lol, the accepted answer is useless because it doesn't provide a proof, it just hides the difficult part by calling it "technical details".

Danu

16:02

@ACuriousMind Which was my point :)

ACuriousMind

I really wished more people would provide what is written there as motivation for the theorem, though

Danu

This is something very good about Guillemin & Pollack's book on differential topology

It is made very clear what one tries to do

not just in the individual theorem, but the entire development of the theory

0celo7

what exactly is diff topology

Looking at G&P, it seems like Lee + Jost covers most if not all of it.

Danu

Topology in the smooth setting

I don't think anything covers this stuff anything like G&P

It's all about transversality

0celo7

Jost touches on it

Danu

16:10

...a concept that is pretty much ignored in Lee, as far as I can tell.

0celo7

What importance does transversality have?

Danu

...read the book and find out :)

> All physical maps are transversal.

Just one of many interesting statements :D

0celo7

What's a physical map?

Danu

A map that represents something physical

0celo7

sigh

Example and counterexample if you don't mind

Danu

16:12

Observables

are physical maps

The wave function is not :P

0celo7

Are all transversal maps physical?

ACuriousMind

@Danu That statement doesn't mean anything unless you state transversal to what :P

0celo7

Is that a theorem, a conjecture or just an observation?

Danu

@0celo7 A theorem/observation

0celo7

I love having too much RAM

:D

Danu

16:16

@ACuriousMind To some submanifolds of course :P

But you can pick any submanifold

so it's not really a restriction

The only "problem" with G&P is that they leave a LOT to the exercises

0celo7

Arnold does that too

And some I can't do using the material he covers, I have to use "western" geometry

Danu

What's that? Algebraic geometry? :P

0celo7

No

@ACuriousMind Wanna help me embed $\mathrm{SO}(3)$ in $\mathbb{R}^9$ :)?

ACuriousMind

No.

0celo7

D:

Not even a hint?

Should I be trying to embed $S^3/\mathbb{Z}_2$?

ACuriousMind

16:28

Can't give you a hint because I've never done that. That's a really boring and useless task.

Danu

I don't understand---$SO(3)$ is naturally a submanifold of $\mathbb{R}^9$ already

It's a matrix algebra

ACuriousMind

Oh, lol, you're right

Danu

$\operatorname{Mat}(n)\cong \mathbb{R}^{n\times n}$

ACuriousMind

...did you really type \Reals there? :D

0celo7

Danu

16:32

That's a shortcut I defined in my own TeXstudio :p

0celo7

I think he wants me to find the $\{f_{i}\}$...

ACuriousMind

It's completely useless to do that explicitly

0celo7

@Danu I know

Danu

So why are you even asking? :P

0celo7

Because I think he wants me to find the set of functions!

And I have no clue how to go about it.

Danu

16:34

Write down a matrix

the mapping to $\mathbb{R}^{n\times n}$ is realllly obvious

0celo7

Yes

Danu

So you have the map.

You should know the properties that determine $SO(n)$

Write them down in terms of your map

Check that these equations determine a smooth submanifold

0celo7

So we're on the same page, the map is just stacking the rows/columns of the matrix into a column vector.

ACuriousMind

77

Q: Can I add a baby as a co-author of a scientific paper?

Can I put the name of my baby as one of the co-authors of a scientific paper? I know it sounds disturbing, but it's a way of mine to protest against co-authors that haven't made any contribution (they haven't even read it or are part of the research area) to a paper, but they are part of the res...

publications authorship

oO

Danu

It led me to some interesting links :D

The one about "Stronzo Bestiale" made me laugh :D

ACuriousMind

16:47

@Danu I guess it shows you sometimes have to be an asshole...

0celo7

@Danu Can you please verify that's the map you had in mind

Danu

17:07

@0celo7 Just try it out yourself---I don't feel like acting as your personal tutor is a good idea (for either of us!)

Danu

17:18

Does anyone see a clear reason why $\partial^\mu B_{\mu\nu}=0$ for $B$ a 2-form only represents three constraints (as opposed to four)?

bolbteppa

@Danu you can solve for one of the constants and express it in terms of the rest

Since that expression is equal to zero

The personal tutor thing is interesting to watch haha

@Danu there is a fantastic power serious proof of those theorems if you're willing to go to old books, explains where the modern proofs come from

Danu

17:37

@bolbteppa Hmm?

What theorems?

@bolbteppa Ya mean $\sum_\nu \partial^\mu B_{\mu\nu}=0$?

0celo7

@ACuriousMind Did you ever figure out what Arnold meant by "first integral"?

bolbteppa

First integral has a well-defined meaning...

@Danu in Goursat's Course of Analysis he proves the implicit function theorem using power series (for one function, but it generalizes), and in the same course he proves the existence theorem of ode's using the same method, and then re-proves the ode's theorem using a naturally modified form of the natural power serious proof, successive approximations, so it's only natural to be able to re-prove those theorems using successive approximations,

but at least to me it only becomes intuitive when you see the history

Emilio Pisanty

17:58

Any mods around?

Huh.

I'll just drop this here and hope they catch it

@DavidZ, @ManishEarth, @Qmechanic

Someone should approach this guy

physics.stackexchange.com/users/96994/…

Or girl

And tell them to take it easy on the tag edits.

Bastian Treichler

18:14

@ACuriousMind I hope my comment makes the question physics.stackexchange.com/questions/215574 clear. Would be great if you could provide me an example for one concrete $W(J, \lambda)$

Anubhav Goel

18:41

how do we close a question on physics. stack exchange which I asked

0celo7

@AnubhavGoel you have a delete button on the OP

Anubhav Goel

19:09

will it delete my question or make it out of conversation

Physics Meta

19:55

0

Q: Raise in reputation bar for protection of questions

Sometimes questions are protected to prevent low-key answers. Is this applicable to comments as well? And I want to say that sometimes users with >10 rep also make bad responses. There are plenty of such cases faced by us editors of late answers and other questions. So I request the bar for prot...

discussion feature-request

ManishEarth

20:31

@EmilioPisanty done

Emilio Pisanty

20:52

@ManishEarth Cool

0celo7

@RobertBryant Yes, that bears out my impression, though since the authors are physicists, it is a little hard to figure out what they mean... — Igor Rivin Mar 25 '14 at 11:34

2

Emilio Pisanty

22:35

@0celo7 Them's fighting words.

Transcript for