Conversation started Jul 19, 2012 at 18:46.
anon
anon
Jul 19, 2012 18:46
@MattN Is going to be furious.
t.b.
t.b.
@MattN. Assuming that you have normalized Haar measure on $G$, the orthogonality relations tell you that $$\int_G \gamma(g)\,dg = \begin{cases} 1 & \text{if } \gamma = 0 \text{ is the trivial character}\\ 0 & \text{otherwise.}\end{cases}$$ But that integral is nothing but the Fourier transform of the constant function $f(g) = 1$, so $\hat{f}(0) = 1$ and $\hat{f}(\gamma) = 0$ otherwise.
But this tells you that $\{0\}$ is an open set of $\Gamma$, so $\Gamma$ is discrete.
Matt N.
Matt N.
Teddy!!! : ) mwah
t.b.
t.b.
Hi, Matt, mwah, back :)
@anon phew
Matt N.
Matt N.
Meh I was planning to not sit in this chat room for the rest of today. : )
t.b.
t.b.
I'm not going to stay very long...
Matt N.
Matt N.
Jul 19, 2012 18:53
I assumed.
t.b.
t.b.
But I have a few minutes.
Matt N.
Matt N.
Man, I don't know what a Haar measure is. Been meaning to learn about it but nasty homework sheets etc have been keeping me from learning anything I wanted to learn : )
Ok, thanks, let's leave it at that for now. I'll read about Haar measure and stuff.
anon
anon
just know it's a (left or right) translation-invariant measure on certain topological groups unique up to normalization.
Matt N.
Matt N.
It would take too long otherwise.
t.b.
t.b.
@anon exactly. Locally compact is what you need.
@MattN. On the circle it's just the usual angular measure (the same as Lebesgue measure on the interval)
Michael Boratko
Michael Boratko
Jul 19, 2012 18:58
@anon so $dx/dt=A$ removes the $A^2$ term completely, should I reply to my own question with that as an answer or would you like to?
Matt N.
Matt N.
Ok. But I need some time with this. I feel too pressured into understanding it because you don't have time now.
anon
anon
@MichaelBoratko when was the last time you checked your question/inbox ;)
Matt N.
Matt N.
@t.b. Can I just take this comment and think about it?
Michael Boratko
Michael Boratko
@anon Very good! Thanks again!
anon
anon
no, thinking is not allowed, you know that Matt
t.b.
t.b.
Jul 19, 2012 18:59
heh :)
@MattN. what do you mean? sure, that's why I wrote it...
Matt N.
Matt N.
@t.b. I mean that as of know even though it will help I don't yet understand what I want to understand.
anon
anon
I was surprised I was able to answer an algebraic topology question.
Matt N.
Matt N.
Sorry for pinging you, didn't expect you to show up in here as a consequence.
anon
anon
Wait, that's BenjaLim, not MattN, who's studying AT.
t.b.
t.b.
@HenningMakholm You don't need very much for that. Hahn-Banach is plenty enough (and with HB you can prove a version of Banach-Tarski without further choice).
It's a bit more straightforward with the ultrafilter lemma, though.
Matt N.
Matt N.
Jul 19, 2012 19:05
Oh. Sort of "of course". The sum in the inner product of finite Abelian groups becomes an integral on compact groups.
@t.b. One point open implies the whole set has discrete topology? I don't believe.
t.b.
t.b.
@MattN. sure, $x \mapsto \gamma x$ is a homeomorphism $\Gamma \to \Gamma$, so every point $\{\gamma\}$ is open.
Maybe I should have said that the Fourier transform of an integrable function is continuous and $1$ is integrable because $G$ is compact.
Matt N.
Matt N.
No it's ok. I'm just a bit tired. Did some 9 hours of concentrated work (Atiyah-MacDonald).
So I'm a bit bonged out.
@t.b. Ok, now I believe you. I didn't realise your argument involved this map.
t.b.
t.b.
@MattN. A group's topology is determined by its neighborhoods of the neutral element, just as a vector space's topology is determined by the zero-neighborhoods. For precisely this reason: homogeneity.
Matt N.
Matt N.
@t.b. Oh yes, I remember that from the FA notes. Thank you.
@t.b. Hm, this tells me that the dual $\Gamma$ has the discrete topology. But it doesn't tell me that $\Gamma$ looks like $\mathbb Z$.
But perhaps it does and I'm just tired.
I think I'll understand this by tomorrow.
And I think this is going to be an exam question: explain what a Fourier series is.
anon
anon
a particular state of an ocean with a discrete set of waves :-)
Matt N.
Matt N.
Jul 19, 2012 19:20
That is equally helpful as Jonas telling me that I should tell the lecturer that his book sucks. : )
t.b.
t.b.
@MattN. No, I was just answering your question why the Pontryagin dual of a compact group is discrete.
Matt N.
Matt N.
Ok : )
anon
anon
A continuous homomorphism $\Bbb R/\Bbb Z\to \Bbb C$ lifts to a periodic homomorphism $\Bbb R\to \Bbb C$; now we have Cauchy's functional equation but with boundary conditions $f(0)=f(1)=1$...
the functional equ has exponentials $x\mapsto e^{ax}$ as solutions (requiring $0\mapsto 1$), and the periodicity condition tells us $e^{a}=1\implies a\in2\pi i\Bbb Z\cong \Bbb Z$
Matt N.
Matt N.
Ew, this reminds me of my struggle to do the dual of a finite Abelian group.
t.b.
t.b.
You already asked (at least) twice why the dual of $\mathbb{R}/\mathbb{Z}$ is $\mathbb{Z}$ :)
Matt N.
Matt N.
Jul 19, 2012 19:26
I think I asked at least three times: twice in chat and once on main : )
@t.b. Hehe, yeah the second link is exactly what I meant with struggle.
Something I didn't understand properly otherwise I'd be able to remember.
@anon Thank you : )
@t.b. Ok, I'm going to bed soon. Is everything alright over there? How is algebraic topology doing?
t.b.
t.b.
@MattN. More or less, yes. But that one non-semilocally simply connected space I know is one nasty beast...
Matt N.
Matt N.
@t.b. Just punch it : ) Or let me do it, I'd love to.
t.b.
t.b.
Oh, no need to punch it, but thanks for the offer.
Matt N.
Matt N.
Jul 19, 2012 19:41
: )
Come on, say goodbye/good night, I feel so guilty because you're here.
t.b.
t.b.
Good night! No need to feel guilty.
Matt N.
Matt N.
Good night! It was nice to see you.
t.b.
t.b.
See you soon!
 
Conversation ended Jul 19, 2012 at 19:45.