Mathematics

Matt N.

7:00 PM

@t.b. I mean that as of know even though it will help I don't yet understand what I want to understand.

anon

I was surprised I was able to answer an algebraic topology question.

Matt N.

Sorry for pinging you, didn't expect you to show up in here as a consequence.

anon

Wait, that's BenjaLim, not MattN, who's studying AT.

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.

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.

7:08 PM

@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.

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.

@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.

@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

a particular state of an ocean with a discrete set of waves :-)

Matt N.

That is equally helpful as Jonas telling me that I should tell the lecturer that his book sucks. : )

t.b.

7:20 PM

@MattN. No, I was just answering your question why the Pontryagin dual of a compact group is discrete.

Matt N.

Ok : )

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.

Ew, this reminds me of my struggle to do the dual of a finite Abelian group.

t.b.

You already asked (at least) twice why the dual of $\mathbb{R}/\mathbb{Z}$ is $\mathbb{Z}$ :)

Matt N.

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.

7:38 PM