Mathematics - 2020-04-21 (page 2 of 2)

« first day (3548 days earlier) ← previous day next day → last day (1769 days later) »

Sha Vuklia

22:28

guys, say we have a closed curve in the plane, part of which is given by $(x,x\sin 1/x)$ for $x>0$, and $(x,0)$ for $x\leq 0$

could you then still say this is a manifold in the plane?

like, would the locally euclidean property go wrong at $(0,0)$?

Thorgott

I think that's not locally connected at $(0,0)$

Leaky Nun

22:48

@ShaVuklia this is a manifold because this is a graph of a continuous function

but "manifold in the plane" is ambiguous

is that to be interpreted as submanifold?

in that case it would be at best a $C^0$ or $C^1$ submanifold

Sha Vuklia

ye sry, I meant embedded 2-manifold

Leaky Nun

C0 or C1, no higher

Thorgott

oh, I misread

Leaky Nun

probably C0

Thorgott

there's an $x$ factor in front of the $\sin$

Leaky Nun

22:53

The X Factor

The X Factor is a television music competition franchise created by British producer Simon Cowell and his company SYCOtv. It originated in the United Kingdom, where it was devised as a replacement for Pop Idol (2001–2003), and has been adapted in various countries. The "X Factor" of the title refers to the undefinable "something" that makes for star quality.The prize is usually a recording contract, in addition to the publicity that appearance in the later stages of the show itself generates, not only for the winner but also for other highly ranked contestants. Unlike Idols, where the judges only...

Sha Vuklia

hm, the reason I'm asking is

in Lee, Green's theorem is formulated for compact regular domains

(that is, embedded 2-manifolds)

and my book states stuff in complex analysis for jordan curves

I was wondering if they always give a regular domain

since my book uses Green

topologicalorientablesurface

Let $F$ be a field. Let $D\subseteq F$ be a subring of $F$ with the multiplicative identity. Put $D'=\{$ $ab^{-1}$ $:$ $a,b\in D$ , $b\neq 0$ $\}$. Then $D'$ is a subfield of $F$.

Is this true?

Leaky Nun

@topologicalorientablesurface have you tried verifying the properties?

topologicalorientablesurface

yeah, just stuff in showing that every non-zero is a unit

Leaky Nun

have you tried $ba^{-1}$

topologicalorientablesurface

22:57

Thats what I thought, but $a^{-1}$ may not be in $D$, right?

Leaky Nun

but $a$ is

read the question again

Sha Vuklia

I'm confused. why do some versions of Green's theorem talk about continuous curves

while a continous curve might not give a smooth manifold?

how would we then know how to differentiate on it?

oh wait, they do assume it's piecewise smooth

but then my questions stays

Balarka Sen

You can integrate a form on a piecewise smooth curve. Green's theorem in general works on compact embedded 2-manifolds (well, 2-disks, but this can be amended) with corners - the corner points are the points where the curve isn't differentiable.

Sha Vuklia

ah, so the interior is an embedded submanifold so that's fine, and if we integrate on the piecewise smooth curve, than we consider it as a 1-dim "piecewise" submanifold?

my main question is how do we know that such piecewise smooth curve gives us an embedded 2-manifold

Balarka Sen

Jordan-Brouwer separation theorem.

Sha Vuklia

23:10

ye I was aware I needed that

but

I still didn't see why it's a 2-dim manifold

like sure, there is an interior, and the topological boundary is the smooth curve

because the example with (x,sin1/x)

oh wait, that's not an example

Balarka Sen

The topological interior is the disk, so interior points are manifold points. The topological boundary is a smoothly embedded submanifold of R^2 (if you want to forget about the piecewise smooth garbage), so locally you can find neighborhoods which are diffeomorphic to the upper half plane.

That makes it a 2-manifold with boundary

Sha Vuklia

uhmm, why is the boundary smoothly embedded? because it's the image of a smooth curve?

Balarka Sen

That's what a smooth Jordan curve is?

A smooth embedding S^1 -> R^2

Sha Vuklia

oh, the definition I was working with was just that it's the image of a smooth (closed) curve

Balarka Sen

You want injective

But sure that's not different from what I am saying

Sha Vuklia

23:19

but..

as part of the curve

so that's not an immersion

Balarka Sen

Ah fine I see what you are saying. Yes, that's correct.

Sha Vuklia

oh oops, i thought I was completely off

Charlie

If I have a curve $\mathcal{C}$ in the complex plane, is there a notation that means "the region bounded by $\mathcal{C}$". So if I wanted to talk about a point $z_0$ inside it I could write $z_0\in$"notation for area bounded by $\mathcal{C}$".

to be clear I mean a closed curve

Balarka Sen

@ShaVukila Nah, the problem is your definition of a smooth curve is a bit iffy. Consider $\Bbb R \to \Bbb R^2$, $t \mapsto (t^2, t^3)$. According to your definition this is a smooth curve, the image is $x^3 = y^2$ though which is not a smooth submanifold.

It's a cusp

Sha Vuklia

sure

the thing is

the first version of Green's thm that I learned about, talked about piecewise smooth curves

now I learned a more general version in Lee I think (phrased for compact regular domains)

so now I'm wondering

is this first version a restriction of this other version, or is it in fact a different formulation?

Balarka Sen

23:32

I don't really know the statements of these versions in detail. Here is Green's theorem for me, and every other statement is certainly a special case of this: Let $M \subset \Bbb R^2$ be a $C^1$-manifold-with-corners, and $\omega$ be a $1$-form defined on a neighborhood of $M$. Then $\int_{\partial M} \omega = \int_M d\omega$

where $\int_{\partial M} \omega$ is defined by choosing $C^1$-parametrizations $\gamma_i$ for each boundary component of $M$, and computing $\int_0^1 \gamma_i^* \omega$, summing it over $i$.

The parametrizations has to be consistent with the orientation of $M$ (induced form the canonical orientation on $\Bbb R^2$) in a particular way but you can figure that bit out.

Are you happy with this, @Sha? :P

Sha Vuklia

I'm guessing that a piecewise-differentiable curve will yield a C^1 manifold with corners

have to look up the definition of a manifold with corners

Balarka Sen

That's exactly right

Sha Vuklia

okay, then at least I am reassured in that sense

Balarka Sen

I'm requiring piecewise $C^1$ actually, piecewise differentiable has issues because derivative isn't continuous, and I don't know how to integrate bad discontinuous functions.

You can of course assume bullshit less regularity like Lipschitz but who gives a shit, really

Sha Vuklia

ye no, that's fine

Balarka Sen

23:37

Right, so manifold with corners are very precise objects and that should satisfy your curiosity. One can do Stokes on them in general.

I think Lee talks about them? Don't exactly remember

Sha Vuklia

oh does he?

because

it's exactly Lee's formulation of Green's theorem that has disappointed me a bit:p

Balarka Sen

Damn

Sha Vuklia

let me see

ahh

ye, he treats it

Balarka Sen

Nice!

Sha Vuklia

along with Stokes, so that's good

Balarka Sen

23:39

Well if you have no other question I will head to bed :P

It's 5 AM here

Sha Vuklia

haha, I should do the same

almost 2 AM here

thanks a lot!

and nights

Balarka Sen

No problem, good questions. Night!

copper.hat

23:56

is there an easy way to delete a chat room that you created?

« first day (3548 days earlier) ← previous day next day → last day (1769 days later) »

all rooms

Transcript for

Apr20

Apr '2021

Apr22

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$