I thought that mean tensor algebras.

They're honestly just a vector space with a quadratic form.

@BalarkaSen It does not.

Nothing fancy. But all the things that come with it are wonderful.

OK, I believe you. I stand corrected.

I need to stop skipping a head in Boothby, he probably does

I'm reasonably certain one who wants to do mathematics (as opposed to physics) learns about other cases of these after sufficiently understanding $\Omega^*(M)$.

I don't see anything in the table of contents explicitly stating it, but he does have a very high-level-low-level conversation about differential geometry.

I'm not familiar with $\Omega$

It's rather Greek to me.

It's the space of $k$-forms on $M$.

Oh, the k-linear forms? Or all the non-linear ones, too?

differential forms.

Okay. That's one of the chapters I skipped past.

That aint a good chapter to skip.

It's not so much skipped.

What do you refer to when you say "other cases of these", PVAL?

It's that I'm working backwards from the result I want to understand to the premises for it.

It's worth knowing what differential forms are. That's good stuff.

And this has some tree-like structure that doesn't reach every foundation

I'm doing a reread of the whole book from beginning to end.

I'll hit it eventually.

No skips, this time

@BalarkaSen Ya nevermind about that.

$H_1({\rm\Bbb RP^2};\Bbb Z_3)=0$, right?

What's \rm?

$\rm$

Makes it $\rm\Bbb RP^2$ instead of $\Bbb RP^2$ (turn on MathJax). Just a typography thing.

@Akiva It shouldn't be very hard to compute.

Huh? It's Z/2.

no

Hey no fair you changed your coefficient there

I made a typo the first time

I meant Z3

Then duh yes it's 0. Try computing it.

There are probably a million ways to do it, but universal coefficient theorem is a general tool to settle these sort of things, which you don't know yet.

In fact you can do it geometrically quite easily (take the generating loop, wrap it thrice around. That's 0 since coeff is Z/3. But wrapping thrice is same as wrapping twice (= 0) and then wrapping once again. So generating loop = 0 in homology)

Yeah, I just wanted to be sure

Hello, I have just a question about notation,

And, of course, Mayer–Vietoris works fine with every choice of coefficients

Should I right f follow P if $\lim_{x \to y} |f(x)-f(y)| \le |x-y|$ or $\lim_{x \to y} |f(x)-f(y)| \le \lim_{x \to y} |x-y|$ ?

write *

(lol sorry)

Sure.

What do you mean by "follow P"?

A function which verifies a property P

Ah

It can be any property

Choice of coefficients most of the time is not a problem in computation of homology groups. If you know what it is in $\Bbb Z$ coefficient, you know what it is in all $\Bbb Z$-module coefficients.

I guess it depends on what your property P is supposed to be, then. @Shadock

@Akiva So, what are you learning?

And in general what is the better one?

@BalarkaSen History

Nah, I mean in terms of math.

@Shadock The second one means $\lim_{x\to y}|f(x)-f(y)|\le0$, which is true iff $f$ is continuous at $y$. I feel like this is not what you meant to say.

I think you meant to say that, $|f(x)-f(y)|\le|x-y|$ is true for all $x$ in some neighborhood of $y$

which neither of the two things you wrote mean

(The first one doesn't really mean anything, actually; I'm pretty sure the symbols can't be put together like that)

('cause on the left-hand side, $x$ is a dummy variable, and on the right, it's not)

@BalarkaSen Beginning of real analysis, but I haven't really started yet

Where are you learning from?

Baby Rudin, I think it's called

Nice. Real analysis is nice.

Just in the second chapter

and I have a history paper I need to write

Well It's because I have a friend who sent me a mail saying let $f$ and $g$ defined on $\mathbb{R}$ to $\mathbb{R}$, we say $f$ and $g$ are closed to each other iff $$\forall (x,y) \in \mathbb{R}^2 , \lim_{x \to y} |f(x)g(x)-f(y)g(y)| \le |x-y|$$. Are sin and cos closed to each other. Well I am in trouble with the notation and he is sleeping now and I'd like to know if i'm the only one in trouble

There's a cool problem in ch 2 Rudin. It asks if there are any perfect sets in $\Bbb R - \Bbb Q$.

You might want to look into that. It's a hard problem, but I know of a slick solution.

@BalarkaSen Didn't read that, but IIRC yes? I thought about it (or something similar) a while back

I also know of a solution using hyperbolic geometry ;)

@Shadock Yeah, I don't think that's the right notation

@AkivaWeinberger Yes. In fact there's a cantor set. In fact there cannot be anything but a cantor set.

Ok thanks, i will wait to answer him so ^^

See you

I think it means that there is an $\epsilon>0$ such that, for all $x$ where $0<|x-y|<\epsilon$, the inequality holds

That is,

the inequality holds for all $x$ in some neighborhood of $y$

…Although, now that I look at it again, that doesn't make sense either

I guess too, but he is a thug when he use mathematics notations lol

because of the $\forall(x,y)\in\Bbb R^2$ in front

I'm really not sure

@BalarkaSen Tell me about whatever you want.

Well I gonna sleep bye !

@BalarkaSen Yeahh

@BalarkaSen Set of everything of the form:

$$0.a_10a_200a_3000a_40000\dots$$

where $a_i\in\{1,2\}$ for all $i$

Ugh, but yes, that's a solution.

Set of "Liouville numbers" or whatever.

Or,

same exact thing but continued fractions

$[a_0;a_1,a_2,\dots]$, $a_i\in\{1,2\}$

@MikeMiller OK, so first about the coordinate free definition of volume form. We choose a smoothly varying orthonormal basis $\{e_i\}$ for each $T_pM$. Then define $\omega(e_1, \cdots, e_n)= 1$.

@BalarkaSen Just to check: Where is your smoothly varying basis defined?

Everywhere?

I don't know this stuff but "everywhere" seems wrong

You asked to prove (1) that's smoothly varying (2) that's well defined (3) same as $\sqrt{|g|} dx_1 \wedge \cdots \wedge dx_n$.

@MikeMiller I mean I have a bunch of vector fields which at each tangent space $T_pM$ form an orthonormal basis.

@BalarkaSen OK. You cannot do that globally, that would imply the manifold is parallelizable.,

Ah, good point.

How do you define a smoothly varying basis on a sphere? Choosing one of the bases would give a nonzero vector field

I mean, you cannot even have one globally nonvanishing vector field on $S^2$.

There we go

@AkivaWeinberger: This is not your conversation. :)

Fair enough.

But I'm not wrong, right?

No.

@BalarkaSen: But you can do it locally. Since there was the subtletly with doing it globally, maybe you can add that as a thing to do: see why you can do it locally.

(Locally = in some open neighborhood of $p$.)

I can choose a chart. Inside that it's $\Bbb R^n$, which is parallelizable.

and gram schmidt etc etc. Ok.