@tb Thank you.

So, does anyone here speak Scots Gaelic?

Brian might he speaks all sorts of funny languages.

Oh, I thought you were just guessing based on his last name.

My god this story is facinating

@Cowbell he's interested in linguistics in general and in historic linguistics in particular. His specialty is Old Norse.

@tb No, I just know something about a lot of languages. And while Sc.Gael. is one about which I know something, it definitely isn't one that I can stumble through even with a dictionary.

@N3buchadnezzar How many deer testicles did you eat today?

You're Varg right?

@JonasTeuwen Quite a few I am sorry for that.

@BrianMScott Ciamar a tha sibh!

@BillDubuque It is not a question. I was just sharing.

@robjohn Ah, ok, so then I do have a long way to go to catch up.

@Gigili I guess it's because an exhibition is viewed as something static and not something that extends over time. If you say während it usually means that you're there all the time: Viel Spass während des Fussballmatchs, viel Spass während der Ferien, etc.

@tb Isn't that immer während?

@Gigili that's more like forever: Immerwährende Liebe.

@Cowbell That little bit I can puzzle out. Not too bad, thanks.

Today I'm unlucky with my follow-up comments :) I keep getting interrupted.

@tb Got it, thank you.

Hey

Hi, Ben

just got up theo

I just saw the election had a parachute candidate

good morning!

@tb Hey I just heard of this thing yesterday, the topology of compact convergence

it's in some chapter in munkres

@BenjaminLim yeah last minute

@BenjaminLim Hey, what time is it there?

9.14am @Gigili

Ah, okay.

@BrianMScott math. (How coincidental that the word for 'good' is 'math') Ok, I will stop bothering you now.

@Cowbell Very coincidental, but undeniably amusing!

@Cowbell so, have you got that exterior derivative thing figured out?

@BenjaminLim what about it?

@tb It looks very interesting :D :D

A new area I would like to look at

it's also very useful

perhaps it would help improve my analysis

area is maybe a bit exaggerated :)

area - you mean the thing that i talked about?

I'm not aware of a field called compact convergence :)

area was meant as....

@tb Well, I should be complete here. I'm just trying to learn the basic rules and definitions for use in complex analysis. I haven't really studied differential forms. (So I don't really know what the $\wedge$ product is, or what exterior derivatives are yet.)

@BillDubuque I will take your word for it. I know I have hinted, quite explicitly, at my age, so I assume you know that.

@ymar Thanks for accepting my answer!

@Cowbell I see. But you should pursue the computation to the end! It's very nice: you get the curl from this.

I have to get ready to head out to dinner. I will try to get back by 1:00 UTC.

@tb I don't doubt it. I'll probably try to pick up an actual book on the subject soon, so I don't have to fake it anymore.

@robjohn don't worry about it too much, enjoy your day!

ok guys, I have to get ready to go to class now, bye

bye.

later

@robjohn I may have missed the hint, but I now realize that I was confusing you with someone else who is very close to my age. Enjoy your dinner.

@Cowbell the nice thing is that if you take the exterior derivative of a function $f$ you get the gradient. If you take the one of a $1$-form, you get the curl, and if you take the one of a $2$-form you get the divergence...

So the exterior derivative generalizes all three of the operators ocurring in vector calculus, and it also gives the higher dimensional analogs.

The line integral in complex analysis also is an integral on forms, that's a very good way to look at it.

why is it clear that $L^2(\mathbb{T}^n)$ is mapped onto $l^2(\mathbb{Z^n})$ by the fourier transform?

@EricGregor what happens to the standard basis?

Do it in one dimension first

the basis on the former space?

yeah, the $\exp{(2\pi i \langle k, x\rangle)}$-thingies

i guess i can see it. i'm wondering why it's clear, though. after Folland shows those thingies are an orthonormal basis for $L^2$ he says this shows that the map is onto

he says that this is the content of the theorem

again. where are those $\exp$-things sent?

what's their image in $\ell^2$?

1s

ones what?

corresponding to the lattice point $\kappa$

err, sorry

Well, yes, the lattice points give you what in $\ell^2$?

it's 1 or zero depending on which lattice points

ok, maybe that's not it

i guess they get sent to their squares?

Don't guess, think!

You just need to spell it out... It's right there. (not the squares)

Let me tunnel from he other end: what's $\ell^2(\mathbb{Z}^n)$?

functions $f$ such that $\sum f^2 <\infty$

and over what do you sum?

$\mathbb{Z}^n$

where do the functions live?

i don't understand your question

Just ignore...

@EricGregor yes, so, you have functions on $\mathbb{Z}^n$.

that are square summable.

ok, so here is what i was trying to say earlier. the exp guys get sent to the sum of the lattice points in their argument

for example

@EricGregor - I mean a math theorem with extreme long formula that could impress people...

oh, it just gets mapped to itself, right?

No.

since $\left<E_\kappa,E_j\right>E_j=0$ unless $\kappa=j$

sorry i'm being stupid

i'm reading schramm's real analysis and he defines a linear order as a binary relation with trichotomy. no mention is made of transitivity. then he says let {S_a : a \in A} be a collection of sets and suppose A is linearly ordered. define the lim inf and lim sup of {S_a}

No problem, I just want you to say it. It's right in front of your nose.

(as an exercise)

i'm thinking he made a mistake in the definition because i don't see how one could define lim inf and lim sup on a relation that's only trichotomous

@EricGregor so you fix $\kappa$ and you let the $j$ run through the lattice points, right?

For $j = \kappa$ you get $1$ and for the other $j$'s you get $0$. That's what you just said.

yes

so it seems to me $E_\kappa$ should map to $E_\kappa$. what am i missing?

So this $e_\kappa = (\langle E_\kappa, E_j\rangle )_{j \in \mathbb{Z}^n}$ is a sequence indexed by the lattice points $j \in \mathbb{Z}$. In other words a function $\mathbb{Z} \to \mathbb{C}$.

this this :)

@AbstractionOfMe I think that you're right: I see no way to avoid adding transitivity as a requirement.

And every function $f: \mathbb{Z}^n \to \mathbb{C}$ can be written as $\sum_{\kappa \in \mathbb{Z}^n} f(\kappa) e_\kappa$

@EricGregor Now we're talking about $\ell^2(\mathbb{Z}^n)$ and for $\ell^2$, the $(e_{\kappa})_{\kappa \in \mathbb{Z}^n}$ are an...

orthonormal basis

exactly.

So you start with the orthonormal basis $(E_{\kappa})$ and end up with an orthonormal basis $(e_\kappa)$.

wasn't that what i said originally??

well, not originally. but at some point :)

No, you mentioned that the $E_\kappa$ are an orthonormal basis, and you said something that they were sent to themselves.

what are they sent to?

Again: the orthonormal basis $(E_{\kappa}) \subset L^2$ is sent to the orthonormal basis $(e_\kappa) \subset \ell^2$.

in both cases isn't it just $e^{2\pi i \kappa x}$?

folland uses the same symbol even

i.e. he doesn't distinguish b/w upper and lower cases as you're doing

The former are lattice-periodic functions on $\mathbb{R}^n$ (or functions on $\mathbb{T}^n$) if you want, while the latter are functions on the lattice $\mathbb{Z}^n$

That's two completely different beasts, in my opinion.

@tb, one question. if a function on $\mathbb{Z}^2$ can be written $\sum_\kappa f(\kappa)e_\kappa$, to use your notation, what is the argument of such a function?

$f: \mathbb{Z}^n \to \mathbb{C}$.

is it buried in the basis?

the $x$ in $e^{2\pi i \kappa x}$?

Yes, the basis is a function on the lattice: $e_{\kappa}(j) = 1$ if $j = \kappa$ and $e_{\kappa}(j) = 0$ otherwise.

So if you plug in a $j$ from the lattice, you'll simply get $f(j) = \sum_{\kappa} f(\kappa) e_{\kappa}(j)$

ok, i think i fully understand what you are saying now

i was thinking that the lattice was just the index of the sum, not also the domain of the fourier series

i don't know if folland's exposition is too terse or if i'm just dense. probably more the latter

I guess Folland assumes some acquaintance with the stuff. Don't have doubts in yourself, have doubts in the book, or at least if the book suits you.

i actually like folland most of the time

@ClarkKent are you referring to "silent-pings"?

@EricGregor then try a bit longer, more. Which one is it?

I would just suffer through. Suffering usually leads to nice things 8-).

First you have to break down completely to make things good.

At least...

I only leafed through the abstract harmonic tome.

@tb try what a bit further?

oh, folland, you mean

i don't mind hitting my head on the wall

Folland is Excellent.

the good people here help me get bandaged up

I have his QFT, Real analysis and Abstract Harmonic Analysis.

The Best

@ClarkKent I'm not aware of anything else than "Teddy Bear", tuberculosis and this