Mathematics - 2020-05-10 (page 3 of 3)

@BalarkaSen my bad, that was my mistake

You want $J$ inside the unions of the rectangles if you're trying to define the Riemann integral of $\chi_J$.

Otherwise you haven't defined the integral.

Yeah, he's doing inner measure, not outer, which is a little strange. Still fixable, because $\partial J$ has volume $0$ anyway.

10:03 PM

@TedShifrin yes that's what i've been observing, i need to get some way to do it without fitting $J$ inside some larger rectangle

Just another exercise in tedium

I will have to write a Sherlock Holmes fanfic one day, "A Study In Tedium", because that's what my life has become

I would sandwich $J$ in between, then.

Why is there an extra $T_0\times T_0T_0$ in your domain, @Balarka? I'm losted.

Because there are two factors of $T_0$ appearing in $T^{(2)} \Bbb R^n = T(\Bbb R^n \times T_0 \Bbb R^n) = (\Bbb R^n \times T_0 \Bbb R^n) \times (T_0 \Bbb R^n \times T_0 T_0 \Bbb R^n)$, so if you take $T$ again...

@TedShifrin how so?

What is a space that locally looks like hyperbolic space called

10:06 PM

A hyperbolic manifold

Oh right

Oh, I guess that's why it grows as a power of $2$ ....

Agh

yeah

It is rather horrid, I admit

@Stupid: Cover $J$ with rectangles, and then cut off little corners of them afterward to get subrectangles that fit inside. I like covering with rectangles so that I can define the integral of the function by extending by $0$ to outside $J$.

OK, @Balarka, so according to my table, up to $n=4$ you have more dimensions than I do, at $n=5$ we tie, and for $n>6$ you shouldn't have enough dimensions. But your dimensions are using all the levels of tangent vector, and I'm using only highest order derivatives.

@TedShifrin How can I guarantee that way that these new subrectangles will still allow me to conclude that $\text{vol}(J)-\text{vol}(R_1\cup \cdots \cup R_j)<\varepsilon$

10:12 PM

Last question what does a space that locally looks like a pseudo man if old and globally looks like a hyperbolic man if old?

Because you can get $\text{vol} (\cup S_k - J)<2\varepsilon$ (by definition of Jordan and integrability) and then choose $\text{vol} (\cup S_k - \cup R_k)<\epsilon$.

@TedShifrin That checks out, yes.

Something might be off in the way I am interpreting the formula for $T^{(3)} f$ then maybe

So I'm confused that most of your dimensions are coming from low derivatives and only one slot is coming from highest derivative.

I’m study manifold today out of the book 📖

Yeah haha nuts

10:16 PM

@BalarkaSen 😂😂🥜🥜🥜

So my approach, @Balarka, everything is sitting there and I feed in $f$. For your approach, you have to pick different tangent vectors every time to get the various pieces of information.

I can't find a reference for whatever version of Picard-Lindelöf this is supposed to be, where should I look?

So the data we input are totally different.

Right, so for example I will fully recover $D^2 f$ if I feed in $v_1, u_2$ various coordinate vectors in $T_0 T_0 \Bbb R^n$ and $T_0 \Bbb R^n$

And then extract out a component

What you said makes sense.

I've never seen it as a 2D theorem, @Thorgott. Just any application of contraction mapping proves existence/uniqueness in the appropriate function space.

10:18 PM

contract the map 🗺

@Balarka: And I only have to feed in $f$ to recover all the necessary derivatives. You have to feed in various basis vectors with multiplicities.

@Thorgott wwwf.imperial.ac.uk/~svanstri/Files/de-4th.pdf Theorem 3.1 and 3.5

@Thorgott Check Milnor, "Morse theory", right before he starts talking about gradient flows. It's stated there I think.

Or what leaky suggested

🧮

what Leaky linked doesn't have something that applies to this case, unless I'm being stupid

I'll check Milnor

10:20 PM

♟♟♟♟♟♟♟♟anyone?

@TedShifrin True.

no way IN HELL my construction is tensorial, but I still am having hard time believing it doesn't determine r-jet equivalence at points.

either im not interpreting my calculation, which seems correct, properly or some dimensions are getting lost in the process you described; the input data being different than what a jet bundle eats

@geocalc33 time control?

Well, the standard way (and I gave Pig one reference I own) is looking at the span of all mixed partials, definitely not your way. Plus, that builds in the contact structure on the jet bundle.

@TedShifrin What???

oh wait... are you fixing $x$ and then just view it as an ODE for a function in $t$?

10:24 PM

Well, you want to know smooth dependence of the solution on $x$, but yes.

@TedShifrin Are you replying to my previous chat?

Two Stupids makes this ridiculous. There's another Stupid here.

@TedShifrin are you talking with StupidQuestionsInc?

I would politely request that someone change names.

But I am the read stupid here lol.

10:26 PM

The second Ted that appeared seems to have disappeared or changed his name — I know not which.

But I claim priority, having been here for seven years or more. Yikes.

I think the reason I cared about this construction in the first place is that I thought helps me define jets sheaf theoretically without bundle formalism in the first place.

Namely, for any bundle $p : E \to B$, define a sheaf $\mathscr{J}^k$ which eats an open subset $U$ of $B$, and spits out a tuple $(s_0, s_1, \cdots, s_k)$ consisting of a local section $s_0$ of $p$ over $U$, a local section $s_1$ of $Tp : TE \to TB$ over $TU$, etc etc a local section $s_k$ of $T^{(k)} p$ over $T^{(k)} U$ of $T^{(k)} p : T^{(k)} E \to T^{(k)} B$ ... such that the obvious diagram commutes

Lol btw will take me 30 more days to change name.

Ight

@Balarka: The other issue to think about is that $J^1(M,\Bbb R)$ also needs a slot for the value of the function. That's needed for the contact structure.

Guys are you talking about functional analysis?

Or ode?

@TedShifrin I should look into the geometry of the jet bundle pretty seriously, this is something I know nothing about. $J^1(M, \Bbb R)$ is $T^*M \times \Bbb R$ in a canonical way, right?

10:28 PM

So if you care about PDE or geometry, you definitely want that contact structure, @Balarka. Not that that is built into my viewpoint, etiher.

So it makes sense that you need this $\Bbb R$... odd dimensional

Should come from futzing with the solder form which gives the symplectic structure on $T^* M$

@LeakyNun what is time control

Well, and then you mappings pull back $dz - \sum p_j\, dx^j$ to $df-\sum f_j\,dx^j=0$.

Ah

Gotcha

I meant the big ones, @StupidQuestionsInc.

10:30 PM

Of course, I forget this tautological guy everytime

And then I chopped them to get your $R$'s.

@TedShifrin But how can we cut them that way so that it works for arbitrary $\varepsilon$

You are misunderstanding. The sets depend on $\varepsilon$.

@TedShifrin Yes I get that, my question is how can we achieve that so that their difference with the regular rectangles is less than epsilon

I'm still missing something. You're looking at the IVP $\frac{dF(x,t)}{dt}\vert_{t=0}=V(x),\,F(x,0)=x$, but if $x$ is fixed, this isn't even a differential equation (well, it is a degenerate one, but we would want to prescribe derivatives for some interval worth of $t$'s, no?).

10:33 PM

@geocalc33 it means, how much time each side has

yeah

I just realized that

10 min

@Thorgott Sorry, that should be $dF(x, t)/dt = V(F(x, t))$, period.

I'm saying that you cover $J$ with big rectangles and then remove tiny rectangles covering the boundary of $J$. When you remove those tiny rectangles, you're left with stuff on the inside and it can't be far off in volume from $J$ because the big ones were already not far off.

Your IVP shouldn't have $|_{t=0}$ in it.

The $|_{t = 0}$ condition was in the statement of the theorem, I want to cook it up. The ODE I wrote down will cook it up.

The $F(x,0)=x$ is the initial value.

You want all time derivatives.

10:35 PM

Because of course if $dF(x, t)/dt = V(F(x, t))$, $F(x, 0) = 0$ then $dF(x, t)/dt |_{t = 0} = V(x)$ as well!

yeah, that makes sense

I usually start out with the queen walk followed by the pawn feign to the left and then strike with the king

Euler pentagonal number theorem proof

imagine trying to play higher dimensional chess

who in the right mind will even play higher dimensional chess

10:39 PM

3d chess?

Play torus chess

that's too difficult

oh wait

@Balarka youtube.com/watch?v=xbSaNnHZ-Kk

i see now

trve

10:41 PM

Fanks

npno

np

npPOpPnp

ppap

no, it's ppab

does the first p stand for president

principally polarized abelian variety

10:44 PM

rofl

oops, v not b

rip

that was a literal typo

@geocalc33 increment?

it's variety pronounced in spanish

I'm sure that's what you meant @TedShifrin

10:45 PM

@LeakyNun 30 seconds

Ok, let $M$ be a smooth manifold and $X$ a smooth vector field on $M$. Then $V\colon=\varphi_{\ast}X\colon\mathbb{R}^n\rightarrow T\mathbb{R}^n,\,x\mapsto d\varphi_{\varphi^{-1}(x)}(X(\varphi^{-1}(x)))$ is a smooth vector field on $\mathbb{R}^n$. We identify $T_x\mathbb{R}^n\cong\mathbb{R}^n$ canonically for all $x\in\mathbb{R}^n$ and think of $V$ as a map $\mathbb{R}^n\rightarrow\mathbb{R}^n$.
For each $x\in\mathbb{R}^n$, the IVP $y^{\prime}=V(y),\,y(0)=x$ has a unique solution $f_x\colon(-\varepsilon_x,\varepsilon_x)\rightarrow\mathbb{R}^n$ for some $\varepsilon_x>0$. Now we want these to…

@geocalc33 add me on lichess.org as ChEsSn0oBz

There's a running joke amongst the rest of the countrymen that people from Bengal switch V and B during pronunciation (this is kind of true)

@BalarkaSen and why is it a joke?

That's Vengali for you

10:48 PM

@LeakyNun There's a pretty famous institute in Kolkata called the Bose institute, they have a cyclotron, but it doesn't work

Do you know why?

why?

Because V x B = 0

is this some theoretical physics meme

that I don't understand

Lorentz force

nvm

added @LeakyNun

10:53 PM

@geocalc33 challenged

Anyone play chess.com?

@StupidKid ChEsSn0oBz

I'm level 1500 apparently

Kenny lau?

@StupidKid yeah

11:02 PM

It appears Milnor argues by compactness, but covers with open neighborhoods of the $x\in\mathbb{R}^n$, because apparently smooth dependence on the initial condition is already guaranteed, for some reason

I used to play chess there

with random indian

and they quit in middle of game

@StupidKid add me

or someone who just waits for 10 min

@LeakyNun I need to make new account

I have new account now

hi, demonic

Alessandro Codenotti

Hi

11:06 PM

added u liu

I rarely play chess nowadays

accepted

I may suck at playin it lol.

Good Game XD.

gg

I forgot all the stragety I learned and started to place randomly.

How long have u been playin chess?

a year

but I've been playing Chinese chess for longer

11:15 PM

@geocalc33 What's ur chess account name?

zetaspace37

May be I played for weeks if I sum all the time spend on chess. And it was long time ago.

@geocalc33 You can add me. DeAlembert

Just made a new account.

@stupidkid okay

I think I will start playing chess.

no result found of zetaspace37

@StupidKid: That should be D'Alembert :P

11:26 PM

ah, this is what I want mathoverflow.net/questions/18976/…

@TedShifrin But I can't enable ' in name lol.

Bye guys I got work to do.

Ah, my apologies :P

@TedShifrin Oh, that makes sense

will see how i can prove it formally

Even if I take the smooth dependence on the initial value for granted, I still don't see why this guarantees we can simultaneously solve the equations for all initial values in some small ball.

ODEs are scary