Oh it should be $\int^\infty_v ... $ ?

Where $v$ is the intial volume of the spherical shell.

@BalarkaSen Well, it kinda does (go to spherical coordinates, assume a spherically symmetric integrand, and then make an appropriate substitution v=v(r))

Buuuuut mostly not, and to the extent that it works it's as an abuse of notation.

@Semiclassical Then you should write it as $dr$ and not $dv$. That's what switching to spherical coordinates does.

I was think something else when I said from $0$.

It's still wrong, in any case.

Is $v$ correct ?

@A---B One does not simply write "integration blah to blah" when you're integrating volume forms.

$\int |E|^2 \,dV$ is a volume integral.

It's a triple integral.

^

hides quietly in corners

You can reduce it to a single integral, but you can't start there.

Hey @Ted.

hi @Balarka @Semiclassic

hi @BalarkaSen

After such condition is checked, we can then ponder about whether: $\lim_{n\to\infty}T_n\left(\sum_{i=1}^n\right)=\textrm{id}_I$

Spherical coordinates is definitely the way to go in this problem, anyways.

Well, I will think on it. I am throwing random variables now.

The integrand doesn't care what direction in space you are from the sphere, only how far you are.

@Semiclassical Since I don't know spherical coordinates what show I do ?

Hmm.

Try considering a thin shell, thickness $dr$ and radius $r>R$, of space outside of the sphere.

You can compute the electrostatic energy contained in this shell without using spherical coordinates. (Or much in the way of integration, for that matter.)

Yes I think I can.

Then you should consider how to add those contributions up.

Back later.

The above overgeneralised case might be nonsense. However a subset of it is actually found in the hilbert space of functions, where you can extract the coefficients of the sum by exploiting the fact that the inner product of two eigenfunctions in the basis gives dirac delta

and in the context of combintorics, the use of generating functions

hi chat

@Semiclassical Do you meant that I should consider those thin concentric shells ranging from 0 to infinity ?

I'll investigate that later, I need to get back to chemistry

In that question, is the sphere in question concentric with the original charge distribution? We seem to be assuming so ... @A---B @Semiclassic.

hi @Eric ... how was the exam?

I just finished it two minutes ago

And yet here you are ...

4 questions, each multiple parts, overall it was pretty good

@TedShifrin Yes it is.

OK, good, @A---B.

i didnt have time to finish one part of a question because I spent a bunch of the test being confused about a calculation until i realized i was working in an orthonormal basis

so that's a little sad

Hey everyone!

yo @Daminark how was your midterm

Not good

oh nvm I just saw your fb message

I was being an idiot on one of the problems

The comment I would add (if Semiclassic didn't say it) is that spherical symmetry tells you that you can reduce everything to a $dr$ integral, and so you basically want to know what value of $x$ means that the integral $\int_0^x \,dr$ is 90% of the integral $\int_0^\infty\,dr$.

Oh oh @idiocy.

The argument in hindsight is easy, but I didn't think to do that at the time

@Eric: I guess my brainwashing of you is not yet complete :D

Let $S = \{x \in X | x \mbox{ is locally connected } \}$, where $X$ is some topological space. Does anyone know if $S$ is generally open? If not, please don't furnish a counterexample; I would like to come up with my own.

How did both you guys have your midterms before noon? Isn't one of the classes in the afternoon?

Nope, my class is 9-10:30, Eric's is 10:30-12

Oh ... Neves likes early.

No time in between classes? That's convenient for students on campus.

I think undergrad classes have times that were determined 30 years ago or something

@user193319: You mean $S=\{x\in X: X \text{ is locally connected at }x\}$?

yeah he's always in his office doing math before 8

And there are 10 minutes, it's to 10:20

ohhh, thanks, Demonark. Still, big campus for just 10 minutes break.

@TedShifrin Yes, if that makes more sense. I just came up with the idea.

@Daminark he referred to your class as the little kiddos when trying to sort out the different piles of exams lol

Hello, nerds!

@TedShifrin Oh, that simplifies the problem. I think I can do rest of the problem my self.

It's a problem because one of my audits is very far from another, and we're often held a few minutes after class in that audit. The professor in the other one understands and is alright with it

@Eric Lmao, nice

Do you understand my point, @A---B?

Demonark: At UGA we had 15 minutes, and I had students who were always late because their preceding teacher would always run over. I was never so inconsiderate.

@Eric: I used to refer to single-variable calculus as "baby calculus" in my multivariable class occasionally :P

luckily 75% of my classes are in the same building

But yeah we had 3 problems. First had 2 parts, one of which was to prove that $S^1\times \mathbb{R}$ is an algebraic submanifold (got), and the other was to show that its normal bundle was (didn't get it, since I didn't know the characterization of the normal bundle, we didn't talk about it much).

Yes @TedShifrin, to summries, you said that I can integrate along a line rather than over the volume. That is I need to find 90% decrement in energy along the line.

Second problem had part 1 to show that $S^n$ is not homotopic equivalent to $S^m$ if $n\ne m$, and part 2 was to show that anything homotopic equivalent to a contractible space is contractible

@A---B: Because $dV = r^2 dr d\sigma$, where $d\sigma$ is the area element of the unit sphere ... and what you're looking at is constant on spheres.

Demonark: Soooooo different from my exams. I guess I'm not surprised.

Third problem was to show that if you have a closed manifold $S$, that $S\times [0,1]$ is not contractible. That was the one that I should've gotten but for some reason didn't

@Ted In what respects do they differ?

@SimplyBeautifulArt actually, we have that $$\sum_{n=1}^{\infty} \frac{p^n \cos(nx)}{n!}=e^{p \cos(x)}\cos(p \sin(x))-1,$$ and it's useful to integrate with respect to $p$ from 0 to 1 and after that we immediately get the desired result.

The $[0,1]$ is a purple herring.

@TedShifrin So would you happen to know whether that set S is open?

Just show a compact manifold w/o boundary cannot be contractible. One-line proof.

Oh, Demonark, it's false when $S$ is $0$-dimensional, connected.

We assumed that $S$ had dimension positive, this was stated

Write down the definition of locally connected at a point, @user193319.

@TedShifrin That will become a double integral ?

But yeah, I was trying to use that $S\times [0,1]$ had a boundary, $S\times \{0,1\}$ and was trying to use that you couldn't retract to it

Actually, Demonark, I never discussed homotopies on anything but manifolds w/o boundary. Because if you cross a manifold with boundary with $[0,1]$ you now have a manifold with corners. Ugh.

$S \times I$ is contractible iff $S$ is.

$\iint ... dr d\sigma$ ?

I had to show that constant sectional curvature implies that the Ricci tensor satisfies $R(X, Y) = (n - 1)Kg(X, Y)$, and come up with an example to show the converse is false, prove some thing about Jacobi fields that was just a simple gauss lemma application

Not double. It's really triple, because $d\sigma$ is an element of surface area, so you're integrating $d\phi\,d\theta$ for that.

then there was a problem showing some properties of the height function for surfaces in $\mathbf{R}^{3}$ which was easy

@Waiting I used complex sine and Frullani's integral.

@Eric: I guess I don't know a counterexample off the top of my head.

@SimplyBeautifulArt I see. Indeed, it works.

then we had to prove that on a simply connected non positively curved manifold there is a unique geodesic connecting two points, then come up with a counter to show its false if it isnt simply connected

Oh, what about a surface?

I can do the last one :)

@TedShifrin I will work out on the problem. l

Good idea, @A---B.

and then come up with a nonpositively curved manifold with nontrivial fundamental group and no closed geodesics

@TedShifrin Thank you.

@Ted the converse of the ricci tensor thing is what I didn't have quite enough time for

@Eric: What if you just take any $2$-dimensional manifold?

Or did he specify higher dimension?

oh does any surface work

I guess I don't know what the converse means.

What is the meaning of $K$?

sectional curvature

So now it depends on $p$?

the converse is that there's a surface so that $\textrm{Ric}(X, Y) = (n - 1)K(X, Y) g(X, Y)$ but $K(X, Y)$ is not constant

now it depends on $(X, Y)$ everything is happening at fixed $p$ i think

I think "converse" is sufficiently vague that I would just use a surface as a counterexample. Ric is everywhere $K$, but $K$ isn't constant.

OK I'm less worried now since it seems like a number of people didn't get 2a since they forgot that maps to a higher dimensional sphere are nullhomotopic while the identity isn't. And a number didn't get 3 either

hmmmmm i'm satisfied with that

I don't like the one with $M\times [0,1]$ for the reason I stated, Demonark.

Are you working with the smooth category?

Um, diff top, @Balarka.

Despite frequent forays into continuity :P

Yeah, I mean I'm apparently not the only one who was tricked into trying to use that $S\times [0,1]$ had a boundary and you couldn't retract to it

@Ted can you think of a nonpositively curved manifold with $\pi_{1}(M) \neq 0$ and no closed geodesics?

lol. That's what I would do on 2a if I were to write that exercise; write in detail how to prove it for C^0 maps S^n --> S^m

Continuous, I think

Ah.

Hi @robjohn

@TedShifrin howdy!

Well, you're probably supposed to do a smooth approximation or something. I would do cellular approximation.

Or at least a baby version of it.

@Eric: I can give you one with LOTS of non-closed ones, but I don't have an answer yet.

We defined two spaces to be homotopy equivalent if there are continuous maps between them whose compositions are homotopic to the applicable identities

T^2 has lots of non-closed ones for sure, yeah?

But I don't know an example.

I'm unsure about my example, do conformal maps preserve the sign of sectional curvature?

The cylinder is easier, @Balarka. You mean the flat $T^2$, of course.

Yes, flat T^2.

@Eric: The sphere is conformally flat.

ah yes true

@Balarka We've already proven that any continuous map is homotopic to a smooth one anyway so if all smooth ones from $S^n \to S^m$ are nullhomotopic, so are continuous ones

hmm I was thinking you could take a plane with a handle that looks like a bit of flat $T^{2}$ and smooth out where they connect

@Daminark Yep, smooth approximation.

Also that continuous homotopy between smooth maps implied smooth homotopy

For smooth maps you see that the image is measure 0 for m > n.

*image

And pick a point away from the image.

then stereographically project

We didn't go into details on how you do the approximation, our professor was just like "Yeah, convolution and molification. If you know what that means, that's how you do it. If not, take it on faith"

You probably weren't supposed to explicitly do the smooth approximation, then, except stating it.

Well, we didn't need to explicitly state anything about the approximation. In class we already proved that it works and gives the homotopy results we want

So we could invoke those results as well

Oh, @Eric, how 'bout the pseudosphere?

That's interesting, @Ted.

One of my exercises in the notes is to use Clairaut to give the geodesics on the pseudosphere. I'm pretty sure none is closed.

goes to check notes

ah ok I'm pretty sure that works

Pseudosphere... Lol

That sounds interesting to say nothing else

$K=-1$ instead of $K=1$, Demonark. Hence "pseudo"

Hi Ted

Hmm I wonder if my original thought could be made to work

You said earlier that $K$ is the sectional curvature, which I'm guessing is not the curvature that's just doing the circle thing?

Your remark about the Killing form was pretty helpful for my way of thinking about the octonions! It simplified some stuff

My first thought was to take a flat torus $T^{2}$ in $\mathbf{R}^{4}$ and invert through a small ball centered at $p \in T^{2}$ and get like a plane thing with the handle I tried to describe earlier.. I wonder

Hi @Danu

Sorry, @Eric, I missed your original thought.

Now I have a kind of lame question: Is the group $S(U(1)\times U(2))$ (of block diagonal matrices inside $SU(3)$) isomorphic (as a group) to $U(2)$?

It should be, I hope

Demonark: It's just curvature of the surface.

@Daminark what is the circle thing

In general, if I have a transitive group action $G\curvearrowright M$ with stabilizer DIFFEOmorphic but not ISOmorphic (as a group) to $H$ then I cannot say $M\cong G/H$ right?

I think the idea is that the curvature is 1/R where R is the radius of the largest approximating circle

I think the answer to your first question is no.

I care because I have the above situation for $G_2\curvearrowright \widetilde{\operatorname{Gr}}_2(\Bbb R^7)$, and I'm not sure $S(U(1)\times U(2))$ is (group) isomorphic to $U(2)$ but I know I should get out $G_2/U(2)$

The one that Soug talked about at the end of 207

@TedShifrin Is that to me?

Yes, @Danu.

And ah @Ted, alright

damnit TeX

(this is for that quadric in $\Bbb P^6$ again, Ted :P)

@Daminark that's for curves

I mean, I guessed that you just replace circle with 2-sphere in the case of surfaces, no?

sectional curvature is intrinsic to your manifold, but curvature is not an intrinsic property of curves

Intrinsic meaning, not dependent on the ambient space?

Right.

I guess I'm ignoring Demonark and Eric and thinking about Danu's question.

I approve

@Daminark you actually replace the circle with the sphere when you go from planar curves to space curves

cause you the frenet frame grows up and gets an upgrade cause of the presence of torsion

I'll just roll with that for now, I imagine this will all make more sense in the bootcamp

but yeah @Daminark curves don't really have an intrinsic geometry, they're all locally the same

There's an issue with $S(U(1)\times U(n)) \cong U(n)$, @Danu. I don't believe you can do it.

@TedShifrin Yeah, I guess it's related to the issue of $SU(n)\times U(1)\not\cong U(n)$

Right. There are double-cover issues.

@Daminark last year at least Schlag took the curve stuff as prior knowledge, which not everyone had and we moved straight on to surfaces, I'd recommend reading some stuff on it before you get there in Ted's notes or do Carmo or something of that nature

I wish I knew more about this :P

But the paper I have says

Ah. I mean Soug did a bit with us (from Buck... angery reacts only)

But yeah I'll check it out then

What else did he sometimes assume?

@Danu: Of course they have the same dimension. But so what?

@TedShifrin Exactly!!

(that's the end of the proof)

Where's @arctictern when we need him? :P

uh I mean nothing else really, complex analysis/group theory/riemann surfaces started from the absolute bottom, so did probability @Daminark

of course you guys aren't doing the group theory/ riemann surfaces stuff though so idk what'll happen

@Eric: Re your earlier idea — it's basically impossible to decide what geodesics on such a thing would be, and whether there are any closed ones is very subtle.

Guys I have a small problem in measure theory

I guess it might be a good idea to find out how ODEs actually work for dynamics, I didn't learn that well at all in 208

Let $\hat{e_{\phi}} = (-\sin \phi, \cos \phi, 0)$. Show that $\nabla \cdot \hat{e_{\phi}} = 0$ in spherical polar coordinates

@Lozansky: Look up the vector operators in spherical coordinates.

I have

But I don't end up with $0$ lol

I only do this stuff with differential forms.

Do it again :P

I don't have it memorized.

OK let's see

Anybody has access to folland real analysis?

@TedShifrin This is what I thought, but was to embarrassed to say out loud :)

@Ted yeah this is what made me nervous about it, closed geodesics existing or not strikes me as crazy hard in general

@Zee I do, though my exposure is limited, just so you know

It's just a small clarification daminark

@Daminark I guess, I know absolutely 0 about dynamics, one of the people in my cohort of people teaching you guys apparently knows a bit so he'll probably take the reins on that front

So am looking at exercise 20a on page 32

@Eric: There are all sorts of conjectures even for surfaces re such things (cf. Zoll Conjecture).

But I don't get how you can find such an outer measure since by prop 1.13b on the previous page, the two measures equal

We have also the picture:

@MaryStar: Most of us are busy thinking about other things right now.

Ahh ok

@Ted I guess we don't really have many ways to find things satisfying particular geometric/topological properties in general unless our spaces have loads of symmetries

the fact that so many simple sounding things in geometry are hard makes me excited

@TedShifrin I wrote down the definition of local connectedness, and I really couldn't get anywhere.

I've never been Riemannian-ly inclined, as you know, but there are astoundingly hard simple questions.

@Daminark actually I meant prop 1.13a

I can't think about it right now, @user193319.

Yeah one question that really excited me a lot when I first heard it is to show that closed $C^{2}$ surfaces (maybe with some more assumptions about convexity or something) have at least two umbilical points.

$$\nabla \cdot \hat{e_{\phi}} = \nabla \cdot (-\sin \phi, \cos \phi, 0) = \frac{1}{r^2} \frac{\partial}{\partial r}(-r^2 \sin \phi)+\frac{1}{r \sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \cos \phi)=\frac{-2\sin \phi}{r} + \frac{\cos \theta \cos \phi}{r\sin\theta}= \frac{\cos \theta \cos \phi-2\sin\phi \sin\theta}{r\sin\theta} $$

Something's afoot

@Danu: I still contend that there's a square root issue with your question.

@TedShifrin Yeah, definitely think it doesn't work.

But I was thinking about whether the homogeneous space still works

I haven't been thinking about that one.

and I am trying to convince myself that only the diffeomorphism matters

Shouldn't it, morally? Since the base won't be a group in any case

I don't know what $G/H$ means when $H$ isn't a subgroup.

So as long as I can smoothly identify its points with cosets $gH$, it should be OK

Come on folks, no one has folland and 5 minutes?

But I have something diffeomorphic to $H$, so I actually form the cosets using $H$

not caring that the identification with $H$ is bad algebraically

you know what I mean?

That's what I'm hoping is the case, anyways.

@Zee This is a bad way to get help.

What's the proper way?

That formula is wrong, @Lozansky. I'm assuming your $\phi$ is my $\theta$.

@Zee I've been looking at it for some time now and reading around. That kind of attitude is very much not the way to go about things, though

@TedShifrin Yeah $\phi$ is the azimuthal angle

@Zee You wait until either someone expresses interest by themselves in your question, or you first spend some weeks/months coming here and building a bond with people.

Or however you spell it

i

Ah

Wait, @Lozansky. I think you mean the angle in the $xy$-plane.

Right

Your formula is for the wrong $\phi$ :P

How do you mean?

Alright, am just impatient since I gotta leave for work soon

@Lozansky: So you have to add three terms, differentiating the coefficients of $e_r$, $e_\theta$, $e_\phi$.

But $e_{\phi} = 0$, no?

@Zee Then that's too bad. You'll just have to be patient...

I'm inclined to say this may be a typo but I'm not sure

If not then you'd use that $E\subset \bigcup_n A_n$, then $\mu^*(E) \le \sum_n \overline{\mu}(A_n)$

In the formula I have, the only term in the divergence that involves the coefficient of $e_\phi$ is that you take $\partial f/\partial\phi$ where $f$ is the coefficient of $e_\phi$. That is $1$ in your case, right?

And @Zee this may be a bit harsh, I'm really sorry if so, but you are the one asking for help in this context, so I don't know if you can be the impatient one here

I didn't realize we were a fast-order math joint now.

shrugs @Ted

@Lozansky: Am I done?

Hey chat

How's everyone?

Hi @Alessandro. Ah, @user193319 has a good question for you I haven't thought about.

Hey @Alessandro!

I'm going to stop working for now @Ted, thanks for the help. Kotschick finally decided to actually do this research tutorial i.e. weekly meeting for his students to discuss math.

Excellent, @Danu. I'll ponder a bit later.

Tomorrow's the first meeting. Then on Sunday I'm going to Cortona, Italy, for a week of learning about Kähler-Einstein metrics.

Pretty psyched!

Oh cool.

Hi Mike, bye Mike

Don't forget to plan for my visit, too :D

I have, don't worry

@TedShifrin Wait.. in spherical polar coordinates, $\hat{e_{\phi}} = (0,0,1)$? Does that make sense?

LOL

bye

G'night, @MikeM.

Hey @Mike!

No, @Lozansky.

@Alessandro Doing alright, how about you?

Where are you getting the formula you're using?

@TedShifrin which question?

Well I have $\hat{e_{\phi}} = (-\sin \phi, \cos \phi, 0)$

Is the set of points at which a topological space is locally connected an open set, @Alessandro?

@Daminark I got a cold, but apart from that it's all good

Right, @Lozansky. You write your vector field as $a e_r+ b e_\theta + c e_\phi$ and the formula is in terms of derivatives of $a,b,c$.

@Alessandro: I hope yours doesn't stay around over a month, as mine did.

@Daminark well I thought you forgot about me so that's why I asked others. Anyway, thanks for the answer

@TedShifrin So $a=b=0, c=1$?

Yes, @Lozansky.

Huh, yeah well it's trivial

In that case

(It's clear on geometric grounds that the divergence of $e_\phi$ is $0$, by the way. Think about the meaning of divergence.)

@TedShifrin Locally connected means that every point has a connected nbhd or a connected local basis? Anyway I'd guess this is true but I'll think about it

Yeah, how much it diverges? :D

I asked @user193319 what the definition was, but he didn't give it. For me, I believe it's that any nbhd $U$ of the point contains a connected neighborhood of the point.

Anyhow, @Lozansky, when you look up a formula, make sure you understand what everything means :P

@TedShifrin Thanks for the help :)

You're welcome.

Prof. @Ted and everyone \o

Heya @Studentmath :)

How you?

How is it going?

I am good, second weekend in a row getting to work on math, not taken for granted :)

Most of your compatriots probably look forward to something different :)

Haha, well I look for that too. It's usually sleep I tend to trade for it, not going out :) Or well, try to

I've been stuck with some analysis question, I posted on the forums too. Think I will try and give it another try tomorrow

I don't usually browse main that much outside of a few specific areas.

Differential geometry, etc.?

Generally.

What kind of analysis?

Is it bad if ears are buzzing after a fall?

Functional analysis. Should be, I think, rather a simple question. A pre-question to Banach spaces, about linear transformation in $l^2$

Yikes, Demonark.

Could be a concussion.

It's gone now but just wondering, is that something to see a doctor about or just wear it off?

Did you faint from lack of sleep/food?

Tripped on the stairs and landed hard on my backside

Did you bang your head? Probably should go to the student health services and be checked.

How long ago?

@Studentmath woo functional analysis!

@Ted I did not bang my head, just backside, and this happened approximately 10 minutes ago

And yeah that's probably a good idea

Leg still hurts so hopefully nothing got busted

Better to let them check you out. And you're paying for student health services.

For sure go and get an Xray ...

Better get a friend to go with you.

Yeah, I agree, thanks

Let me know later, please.

@Daminark these things can kick off a little bit later, even if you are feeling just fine now, so @Ted's advice about getting a friend with you is true. Better safe than sorry.

@Daminark don't waste your time, concussions can not be treated

@Zee well they can get worse..

I used to play sports, even when you get a bad concussion, nothing the docs can do for you

But they still would send everyone to the doctor, wouldn't they? ;)

Either way, I still think better safe than sorry.

Yes, please, @Studentmath.

Oops, please, Demonark.

Lunchtime for this Bonzo. See you all later.

Actually, I think being sorry is better, safety is dangerous

@Ted do you still have that link of a web page inspired by "counterexamples in topology" where you can select some properties and get a space with those?

bon appetit

(I think you gave it to me a while ago but it might have been someone else)

Oh, buon appetito!

Lolol @zee, surely. But yeah see you @Ted, and I'll check up

@Daminark btw I figured out the confusion prop 1.13a does not apply to all subsets

Ah

Hi all

@TedShifrin and @TedShifrin Sorry I didn't provide a definition. Here it is: A space $X$ is said to be locally connected at $x$ if for every neighborhood $U$ of $x$, there is a connected neighborhood $V$ of $x$ contained in $U$.

i feel meh today