Hey @TedShifrin :)

Thanks for the comments on my question btw

hi contradictory @Perturb :)

lmao

People who don't know we know each other from chat might think I was a little rude, but oh well.

Familiarity breeds contempt?

Speaking of contempt! Heya, @Fargle.

Hey @Ted :)

@TedShifrin Well I hope not, I didn't find it rude in the slightest

Also I understand what you said about cellular homology now, like computing the cellular homology of $\mathbb{CP}^n$ is ridiculously easy using it

I don't know any other way to do it (well, I can do deRham cohomology using some fancy stuff about symmetric spaces).

I really need to learn deRham cohomology once I get furhter in diff geom

The beautiful result to which I refer is that on a (locally) symmetric space $G/H$, $G$-invariant differential forms are automatically closed, hence represent cohomology. So you just have to use some Lie theory to find invariant forms.

More is, in fact, true. The cohomology is just given by the exterior algebra of invariant forms.

Noted, I'll definitely look at that result again in the near future

Very powerful.

@TedShifrin Wait, really?

One of my friends applied for a programming position at a company and actually got a call for an interview. Can companies choose what age they are willing to hire or are there laws on those things?

Thus the cohomology is entirely a representation theoretic phenomenon?

@user193319 Doesn't this sound a lot like regularity of a topological space?

Yup @MikeM, for symmetric spaces (at least with $G$ compact).

I actually proved this result in my thesis because it was crucial.

But it's classical.

Can you sketch the idea to me?

Oh, over $\Bbb R$, @MikeM. I'm not doing torsion, of course.

Right, neither was I

Thinking about (g/h)^H

It is slightly technical. Do you know what a symmetric Lie algebra pair is?

No but I can work it out.

From context, I mean

Looks like it's the infinitesimal data of the "symmetric" part if symmetric space

So you have $\mathfrak g = \mathfrak h \oplus \mathfrak p$.

Yup

You have an involution of $\mathfrak g$ which leaves $\mathfrak h$ fixed and is the negative identity on $\mathfrak p$.

Right

This will tell you that certain structure constants $c^\gamma_{\alpha\beta} = 0$, where $\alpha,\beta,\gamma$ are in the $\mathfrak p$ range.

(Unless I'm messing up.)

Now dualize and think about the Maurer-Cartan forms $\omega^\alpha,\omega^\mu$. What we just said about structure constants says that $d\omega^\alpha \equiv 0\pmod {\omega^\mu}$.

Hmm. Am I supposed to see the conclusion now?

Any form on $G/H$ is cohomologous to an invariant form (standard argument you've given here before). But any invariant form must be a polynomial with constant coefficients in the $\omega_\alpha$, and this computation shows it must be closed.

Also, is $\omega$ a 1-form?

Aha.

I don't see why "zero mod curvature" implies zero

These are $1$-forms, but they generate all the invariant forms.

I caught after your last comment. (I guess this is a representation theory theorem.)

No, not mod curvature. It's mod the stuff in the $\mathfrak h$ range.

That tells you you get closed forms on $G/H$.

I guess I didn't understand notation - the point is $\omega^\mu$ is in the h-direction, I guess

Right.

Oh sorry, I didn't make that clear.

I'd need to see the details before I could explain this to anyone but I think I see the idea.

So, anyhow, any invariant form is closed. But that also tells you that the only exact invariant forms are $0$. So the complex of invariant forms gives the deRham cohomology.

Exactly

Nice!

Griffiths once gave an argument for this that was totally wrong ... But geniuses often make silly mistakes.

That explains why I never knew examples with differential! I assumed it was low dim trickery.

Ah, glad I could help you ... my second contribution :)

I'm no genius but I found earlier that while copying a proof into my document I also copied a grocery list

ROFL

That isn't necessary and sufficient. :)

This is a good way to show a homogeneous space can't be locally symmetric. My favorite way.

You didn't make it into the paper acknowledgements since I ended up excising Gray, I think. But it would be a lie to not acknowledge you in the thesis, where acknowledgements (to me) representat people who made it possible.

I am not asking for public acknowledgment. :)

But it feels like I've known you an eternity :)

hi everyone

Hi @Leaky

Slightly longer than the project has existed, yes.

Um, slightly by 3 years or so.

pshhh 3 years is nothing, la revolucion is immortal

Ah, Eric the philosopher arrives.

@TedShifrin my source says that if I use a wrongly scaled $\mathrm dx$ then I get $\hat{\hat{f}}(x) = rf(-x)$ for $r>0$, but I'm wondering whether that should be $f(-rx)$ instead

apparently, la revolucion is a mexican restaurant in North Carolina.

You're talking about Fourier transforms, Leaky?

yes

@CaptainAmerica16 the dogs will subsume any symbol to reproduce the alienating machine

I have no idea what "wrongly scaled" means.

@TedShifrin let's say $x=ty$ for some positive constant $t$

I have no idea what this means. I thought you meant factors of $\sqrt{2\pi}$.

this is very strange, the change of variables formula should indeed give you $rf(-x)$

@ÉricoMeloSilva That got more real than I intended it to.

I have no idea why there's a negative sign.

oh that's because the fourier inversion formula and the fourier transform formula differ by a negative sign

$\hat f(s) = \int_{t \in \Bbb R} f(t) \exp(-2\pi i s t) \ \mathrm dt$

$f(t) = \int_{s \in \Bbb R} \hat f(s) \exp(2\pi i s t) \ \mathrm ds$

I haven't thought about this in 30 years.

Lololol

hmm, I wonder what's the Pontryagin dual of $\Bbb R \times \Bbb R_d$ where $\Bbb R_d$ is $\Bbb R$ with the discrete topology

it looks messy enough

Question: Is there a simpler way to express this group? $\langle x_i,y\mid x_i^2=yx_i=x_iy=x_{i+1}\rangle$

well your group is abelian and bigenerated

the first generator has order 2

Hello chatr

how do I write on latex without installing anything

in fact it is monogenerated

in fact it is just $C_2$

I have small papper to write with math symbols

@LeakyNun Hi

hi

do you know a good page for latex work ?