It's $\Bbb R^2/\Bbb Z^2$

Well, they take C mod the lattice

Now give $\Bbb R^2$ the usual complex structure by identifying with $\Bbb C$

I'm wondering which one is better to think about

You can't really use the fact that product of complex manifolds is a complex manifold here though

$S^1$ are real 1 dimensional manifolds.

"Actually I didn't need that last part"

I don't understand what you mean by the last part.

The product of manifolds

I remembered that I didn't need it

ah ok

hello

Actually the $\mathbb{C}/\Gamma$ stuff isn't too bad

It's cool with me, not sure why you don't like it

I just anticipated it to be annoying but it's actually okay

i need to prove that R is not totally bounded, with the usual metric. so i let the converse be true, and let R be contained in an epsilon net {x1,x2,...,xn}. Then i can say that this epsilon net is contained in a ball of radius 2n(epsilon)+1 about any xi in the epsilon net, but clearly this ball does not contain xi+2n(epsilon)+3. so R is not totally bounded, is this okay?

More generally, totally bounded sets are bounded

yes, in R^n

In a metric space generally

but is my proof ok?

oh

Gimme an $\epsilon$ net $\{x_1,\ldots,x_n\}$

Then the union of $\epsilon$ balls around those points is bounded

but R is not, hence follows

Just choose a point $x_0$, let's say $x_n$ is farthest away from it, and find a ball that's $2(\epsilon + d(x_1,x_n))$

Yeah basically

i wrote very informally above, but will that proof work?

You have the idea down

ty :)

So if you just translate that to formal language you're a-ok

@Balarka god that music video...

cute catz

Complex manifolds? Say what?

I see Balarka's broken his sleep again.

Alright so removable singularities in a Riemann surface will also work by taking a limit, I imagine

@Ted yeah I'm joining Balarka (for now at least) in doing Forster

Huh?

Oh, for a day?

Hopefully for longer

@Daminark Hm?

rehi @Ted

rehi

guys, i have a question

ask away

hi @TedShifrin

can i calculate the row echelon of a matrix using only the LU decomposition?

$U$ is echelon form.

If $X$ is a Riemann surface and $U$ an open subset, $f$ is holomorphic on $U\setminus\{a\}$ and bounded on a neighborhood of $a$, you can extend to a holomorphic function

rehi NV

Oh, Riemann extension theorem.

@Daminark Right, so how do you plan to do that?

Taking a limit you can set $f$ on a deleted neighborhood of $a$ as is, and $f(a) = c$ where $c$ is whatever the limit is

But you have to check it's holomorphic

That's a continuous extension at best

how do you know there is a limit?

actually no, if i calculate the gaussian elimination of a matrix, the matrix is different form the U

@tazdingo: Echelon form is far from unique.

But $U$ is an echelon form.

@Ted also good question

I mean in the case of $\mathbb{C}$, you'd use $(z-a)^2f(z)$ and $0$ on $a$

Oh you know the story

huh? Demonark

@TedShifrin That's differentiable, the thing Daminark defined right now

I'll ignore this.

@TedShifrin thank you!

If I recall correctly there's another way to prove the Riemann extension theorem by using the Cauchy integral theorem.

i hope one day i'll be able to discuss topics with u guys, not just ask questions

you're welcome, taz

Taking smaller and smaller keyhole contour, etc

I'm not convinced, Balarka.

About what I just said or Daminark's proof idea?

I'm not following Demonark at all.

@Daminark Good exercise: Explain your proof to Ted.

@Ted so let $g(z) = f(z)(z-a)^2$ for $z\ne a$ and $g(a) = 0$

All I'm assuming is that $f$ is holo and bounded in a punctured neighborhood of $a$?

Yup

Hmm. OK.

But yeah at $a$ you have $\frac{g(z)-g(a)}{z-a}$

That's $f(z)(z-a)$, which tends to $0$ since $f$ is bounded nearby

OK

So now $g$ is holomorphic, and $g(a) = 0$, but $g'(a) = 0$ as well, so the Taylor series for $g$ starts from $(z-a)^2$

So $g(z) = c_0(z-a)^2 + \ldots$

Uh huh

But then away from $a$, you can write $f(z) = g(z)/(z-a)^2$

$c_0 + c_1(z-a) + \ldots$

But then define $\tilde{f}$ to be that Taylor series

OK.

That's a holomorphic extension of $f$

I guess I prefer my usual proof, because it's less ad hoc.

How do you do it?

But I've seen proofs like this before.

Laurent series.

Ah, that makes sense

The proof Daminark did (and the one I know) is doing a Laurent series proof in a restricted case I guess

I guess I always think of Laurent series as a description of holomorphic functions near it's poles

isolated singularities, not poles

more powerful because of annuli, etc.

And transferring it to the Riemann surface case, find a punctured neighborhood of $a$ on which $f$ is bounded and which lies in the domain of a single chart around $a$

Demonark: It's the usual manifold game. Anything local you prove locally and it's just the usual theorem.

@TedShifrin Ah, OK.

I guess it does describe holomorphic functions near essential singularities as well

Especially :)

Yeah, alright

My definition of essential singularity is that the negative Laurent series has infinitely many nonzero terms.

Right. I guess that's from the fact that if it had a Laurent series which stops at $1/z^k$, look at $z^kf(z)$ and that's holomorphic, making $z = 0$ a pole.

Yeah when working through the book which will not be named, we defined things like so as well, if I remember right

My definition of pole is finite negative Laurent series with some nonzero term.

Though my physics TA defined it based on the limiting behavior

@TedShifrin haha

And I don't do residues with those horrid limits, either.

So removable singularity is if the function has a finite limit there, pole is infinite limit, and essential is no limit

That's a theorem, Demonark, to me.

@Daminark Same in my reading course

In my typical computational vein, I start with the Laurent series.

Yeah at some point soon I'm gonna finish the last bit of chapter 2 of Stein (I think I stopped at this reflection thingy) and get to chapter 3 where it does that stuff more

btw @Daminark check DC, i have a dank music video for you

I still dislike Stein because he downplays/delays the log until the end.

Doesn't he do it in chapter 3?

Or is that treatment incomplete?

He waits till meromorphic functions I think

It's ridiculously late. I taught it earlier, but I was pissed.

i leave for a couple hrs n u guys racked up 910 new messages

Shows you should never come back, Faust.

i was doing math!

That's great! But why come back?

Hmm, I mean I guess I basically skipped chapter 1 so that's why I didn't feel as late :P

fiance needs car to drive to work.

that doesn't mean you need to be here :P

i was at the school libary

plus i figured out a really hard question on th drive back

Doing hard math while driving is illegal.

really?

Almost as bad as doing hard meth.

lol i doubt it

never done meth but i hear it rekts you pretty good

I've never smoked a cigarette, but I did try dating a guy who was a meth addict, and it's NOT good ... at all.

i have smoked some cigars and i have an old root pipe

i smoke tobacco out of once a month or so

but its not really the same thing as cigarettes

Cigars probably do more damage than cigarettes, but people pretend not.

diffrence is smokers smoke 24 cigarettes a day and i smoke 2 or 3 cigars a year

pls, amphetamines not methamphetamines

2 or 3 a year probably won't kill you.

lol

awesome picture

Erdös

Certainly alot safer to have some Adderal or Dexidrine than meth

I don't know how to do the Hungarian version of ö

at least its made by chemists in a lab

I don't want anywhere near any of this.

eh its not that bad my doctor says im supposed to take it everyday but i try and avoid taking mine cause it makes me stupider but it defiantly helps me do my hw. used to drive him nutz cause i wouldn't take it when i used to work.

I can see that you might need something, Faust :)

that being said i only take 15-20mgs qand most people with what i have take like 60

to 120

how can I find the formula of this function? :S

you can't, Twink, and neither can I.

I'm trying to graph it, I found a formula that looks like the first part

hmm :(

Why're you trying?

Yeah, the first part could be a parabola.

well at least one that looks like it and coincide with the values at those points

I used the formula of a circle

Whoever did that probably did it in a graphics program and dragged bezier curves. That's what I would do.

It's not part of a circle.

It does not have constant curvature, man.

can that be done in latex?

I guarantee you this was drawn in something like Illustrator.

its clearly 3 functions

No.

Huh? @Faust

well it looks like 3 functions

well, it's at least two

defined on diffrent parts of the domain

I got this using latex

lets say there are points a and b in an f(x), if we know that there is a point between those two called c, how could we find c?

well its def at least 2

Yeah, but the original curve does not have constant curvature.

what my teacher said was f(c) = (f(a), f(b))

Why are we doing this, Twink?

thats confusing me cause its not making sense to me

@MATH, this isn't right.

@Semiclassical is there a fundamental constant for ass regeneration?