12:00 AM
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"

12:01 AM
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

12:07 AM
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

12:10 AM
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 :)

12:12 AM
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

12:16 AM
@Daminark Hm?
rehi @Ted

rehi

guys, i have a question

hi @TedShifrin

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

$U$ is echelon form.

12:18 AM
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.

12:20 AM
@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
Ok

huh? Demonark

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

I'll ignore this.

12:22 AM
@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.

12:25 AM
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

12:27 AM
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

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

12:31 AM
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

12:34 AM
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 :)

12:37 AM
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

12:39 AM
@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

12:41 AM
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.

12:45 AM
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.

12:46 AM
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

12:48 AM
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.

12:51 AM
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.

12:55 AM
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.

12:56 AM
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.

12:57 AM
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

12:58 AM
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

12:59 AM
@MATH, this isn't right.

@Semiclassical is there a fundamental constant for ass regeneration?