Exactly so

So I'm guessing here that we look at $q((f-f(p)) + \mathfrak{m}^2)$

That works :)

Alright, and to see that this is a derivation

$D(fg) = q((fg - f(p)g(p)) + \mathfrak{m}^2)$

Do we want to play that $fg - f(p)g + fg(p) - f(p)g(p)$ game or something similar?

I was thinking that, yes

Because $fg(p) - f(p)g$ vanishes at $p$ so it's in $\mathfrak{m}^2$

Mhm

Er. It's in m, right?

Oh frick

Yeah it also just doesn't do what we want it to

How about -f(p)g + f(p)g

:GWChadTHONKery:

lol

So that's def in $\mathfrak{m}^2$ because it's just $0$

mhm

It gives you $q((f - f(p))g + f(p)(g - g(p)) + m^2)$

Is that what we want? It's not $g(p)$ is the problem

OTOH if you do -fg(p) + fg(p) you get f(p) as the problem

Yeh

Equate them?

idk

Maybe do it backwards. What's $f(p)D(g) + g(p)D(f)$?

$f(p)q(g - g(p) + m^2) + g(p)q(f -f(p) + m^2)$

$q(f(p)q - f(p)g(p) + g(q)f - g(p)f(p) + m^2)$

This looks bad

Where even is the $fg$ term

Maybe add and subtract both and see if the leftover terms die

Fuck I don't even know if that works

This is obvious if one of $f$ and $g$ are in $\mathfrak{m}$, right?

'Cuz $f(p) = g(p) = 0$

Maybe not.

say $f \in \mathfrak{m}$ and $g = g_0 + g'$ where $g_0 \in \mathfrak{m}$ and the other boi is a unit. $fg = fg_0 + fg'$

$q(fg + \mathfrak{m}^2) = q(fg_0 + fg' + \mathfrak{m}^2) = q(fg' + \mathfrak{m}^2)$

This was a mistake.

-Big Smoke

Ah, wait.

@Daminark We're all dummies. $q(fg - f(p)g(p) + \mathfrak{m}^2) = q((f - f(p) + f(p))(g - g(p) + g(p)) - f(p)g(p) + \mathfrak{m}^2) = q((f - f(p))(g - g(p)) + f(p)(g - g(p)) + g(p)(f - f(p)) + \mathfrak{m}^2)$

$f - f(p), g - g(p)$ are both in $\mathfrak{m}$. So that gets cucked because the product is in $\mathfrak{m}^2$

So it's $q(f(p)(g - g(p)) + g(p)(f - f(p)) + \mathfrak{m}^2)$

Which is indeed $f(p)Dg + g(p)Df$

I see

It was a lot of cancellation and chugging things in m^2

Also I'm gonna say a lot of stuff to push it off screen since it leaves the page

Okay so teach me now about varieties

lmao

Because I know that algebraic groups are supposed to also be varieties where you stick Zariski

But OK that's the proof. m/m^2 is the ultimate algebraist's definition of a tangent space and that's the thing you use in algebraic geometry

Alright, I'm gonna write a manifolds book and use that

A lot of manifold books do actually

It's not that bad

because

It keeps track of the 1st order Taylor series (because m^2 gets cucked) of functions on the manifold which vanishes at p

In general I'm gonna make everything as algebraic as humanly possible in that book. It'll be a manifolds book with not a single picture

And that's precisely what a tangent space should be doing

@Daminark I was initially going to say tough luck selling it but it'll probably gain a cult following

FUCK

all the French algebraists will read it

My response would've been "Rudin Principles of Mathematical Analysis"

Rudin is only good because of it's problems

I don't think anybody unironically self-studies from Rudin

Just ask for a real analysis book recommendation and everyone's gonna be like Rudin. Some will say Pugh as being more pictorial and with better exposition. Others are gonna be recommending this one called Abott because Rudin's hard

@Balarka you'd be quite wrong, a lot of people do I think

Really?

Hm

I did, I think Eric did

God.

In general that's considered the gold standard for analysis books because it's got a lot of material and good problems

I never learnt real analysis so maybe my opinions don't count :3

:333

Aside from maybe Pugh I don't know any others aside from watered down books that are halfway between Spivak and Rudin

what about

STEIN SHAKARCHI

they are my spirit animals

Stein's totally different, it's what you read after you read Rudin

I'm talking Baby Rudin here

though i have never read the real analysis text

Big Rudin, it's less canonical

yeah i know

You've got SS, Rudin, Folland, Royden, blah blah

I like Stein complex from what I've seen in it

I retroactively like it

(Though I ended up doing Freitag because it had more stuff in it)

The guy with whom I took a reading course on Stein complex was quick to spot when the book goes downhill and said "This thing must be written by Shakarchi"

@Daminark I think Stein has a lot more fancy stuff than having a good foundational basics

All the Jensen formula, Weierstrass product formula stories, Phragmen-Lindelof, etc

Then a chapter on Fourier analysis

I mean Stein's got easily enough on foundations

I said Freitag had more meaning fancy stuff

Is that so

I found Stein to be super fancy

And that's volume 1

Oh yeah I didn't read Stein's bit about special functions

what I enjoyed the most in SS was the harmonic function theory

Second volume of Freitag was Riemann surfaces, several complex variables, abelian functions, and higher modular functions

I see

But yeah Freitag's got some of the stuff like modular forms I wanna learn (though I should see whether that's the best/most efficient way to do so)

Stein's quite good, and I did glance through it quite a lot

I think I only liked it because I took a reading course

I wouldnt have enjoyed it on my own

@Daminark Modular forms are holomorphic sections of line bundles on $\Bbb H^2/\Gamma$ where $\Gamma$ is a congruence subgroups :3

Topology topology topology

I wish I knew anything about those other than that one sentence

i have ze bookz by Iwaniec-Kowalski on analytic number theory that i once found

on a bookfair

the shiny AMS copy

Nice

it has a chapter on those

don't think i can read it tho

Oh yeah modular forms is as much a branch of number theory as it is complex analysis

tru

But yeah right now I'm feeling unproductive but sorta not having too much of a problem with my current juggling of subjects, it's sorta fun somehow to just take a random walk for a few weeks and let me lack of focus run wild

Oh, which reminds me

I was supposed to give you a paper to read

Tru

ams.org/journals/bull/1973-79-04/S0002-9904-1973-13293-8/…

But yeah I did email some algebra profs at my place like yo, what's your stance on pregaming classes?

I sorta mentioned how the group theory class was felt a bit slow, and generally my experience wasn't assisted much, by having known some going in

Ah good idea

But on the other hand, it feels like not knowing any ring/module theory (and to some degree Galois theory) is a bottleneck on my ability to do stuff I want

so what did they say?

No response yet, chances are since it's Christmas, people aren't checking their emails

That's true

To write it down before I forget: so the point of the paper is that topological tangent space (the curve thing, if you prefer) agrees with the algebraic tangent space ($(\mathfrak{m}/\mathfrak{m}^2)^*$) only if the manifold is smooth, but not if the manifold is a $C^k$ manifold for $k < \infty$.

Once it gets closer to class time I'll see what they say and then decide. Hopefully it'll give direction

Read the proof and teach me why that happens

Also wait hmm

@Daminark Yeah hopefully they'll have useful comments to make

So is the $C^{\infty}$ structure on some $C^k$ manifold unique?

Unique in which sense?

If you mean upto diffeomorphism, that's the exotic business and a hard theorem that it's "no"

But you could also mean upto compatibility

in which case the answer is no, but easily so

Okay, so you can have two non-diffeomorphic $C^{\infty}$ structures on a manifold that both agree with a $C^k$ structure. Even compatibility but we have diffeomorphism

So in that case, yeah it makes more sense to me

That the tangent spaces might get fucked

Well the topological tangent space does not get fucked

It's your algebra that gets fucked

i.e. algebra is useless

m/m^2 becomes infinite dimensional if you have a C^k structure

which is what the paper proves

Topology never gets fucked. It's always beautiful :)

(see Washington)

My response to Washington: no u

But really tho I mean, algebra sprung out of number theory to a large degree

Making it categorically good

my response in washington

Ok, I should get some sleep now

Lol aight, see you!

Yo Narcissus!

Hey! I have to read above

A friend is here for today and tomorrow, and then gone again for six months, so I am highly unavailable though :P.

He's catching me up on AG he's done for 6 months

(What little he can)

Oh basically I worked through the stuff you said about commutators making things into Lie algebras

And then talked to Balarka a bit about tangent spaces to manifolds, on one hand as derivations, and an another as the dual of the cotangent space

Now I'm wearing my number theory hat

@Daminark You betcha haha. It may still get said in the coming text I've yet to read, but there are many formalisms. The interesting thing is that all of the different ways to define a Lie algebra from an algebraic group never using legit derivatives (which we don't possess in the algebraic setting) give you the same Lie algebra you would obtain from the Lie group setting (where we take the identity fibre of the tangent bundle).

[Later: We can get a zariski tangent bundle in the scheme-theoretic setting]

@Daminark If you want to see all of this up to here done in the algebraic setting, see Humphreys - Linear Algebraic Groups - Section 5.

@Narcissus Downloading now

And that's actually pretty sick tbh

Like, so many different things just lead to the same result is pure witchcraft

Damn, the second edition is only on libgen in the format where 2 pages are side by side

Yo @Balarka

Eyo

Alright so, let $f\in \mathfrak{m}$

Where we have some $C^k$ structure on the manifold, and let's say we're at the point $p$

Define $o(f) = \sup \{\alpha: \lim_{x\to 0} \frac{f(x)}{|x|^{\alpha}} = 0\}$

Okay

(Basically, asking how high a degree of a "polynomial" still grows faster than $f$

And ofc this is something you'd define on the germ by picking any rep

Well, what if $f(x) = \exp(x)$

Am I allowed to write infinity?

Anyway, theorem. If $f\in I^2$, then $o(f) > k+1$ or is an integer ($k$ being the level of smoothness on our structure).

This didn't say but yeah it should be

Ok fine

Now the idea is that for $g\in I$ you use Taylor's theorem

$g = \sum_{i=1}^{k-1} a_ix^i + x^kr$ where $r$ is the germ of some continuous function at $p$

Mmkay

Apparently this implies you can write $f = \sum_{i=2}^k a_ix^i + x^{k+1}r$

My multiplying two things like that togather?

Seems fair.

Yeah, I haven't thought about the details but it's basically that

Anyway so if the $a_j$ are all 0, then $o(f) \ge k+1$

May not be an integer since it'll depend on $r$

True

But if some of the $a_i$'s remain, it'd be an integer

Exactly

But then what happens is that $\{|x|^{\sigma} : k < \sigma < k+1\}$ has to be linearly independent

Ah

Because otherwise some finite linear combination is gonna be rip in peace

I see

That's fun

And then wait rip in piece = order is illegal

Now, I say I don't know about this for sure because doing Taylor's theorem on a manifold feels... somewhat sketchy to me

So the point is having a $C^k$ with $k < \infty$ structure makes your germs only have a $(k-1)$-order Taylor series

The error term fucks you up

This is an interesting proof

@Daminark Well, do it on a chart :) This is an algebraic proof, actually. Here's the thing: $C^\infty_p(M)$ is isomorphic to the ring $\Bbb R\{x_1, \cdots, x_n\}$ of convergent Taylor series in $n$ variables at $\mathbf{0}$. Convergent means has a positive radius of curvature near the origin.

But, $C^k_p(M)$ is something worse.

It's isomorphic to the direct sum of a polynomial algebra with the ring $C^0_{\mathbf{0}}(\Bbb R^n)$ of continuous germs of functions on R^n at the origin.

Or at least, I think this is what it means

But that ring of continuous germs is totally bad and non-Noetherian

Approximately this is what is happening

EDIT: Radius of curvature? What the shit? I meant radius of convergence.

I love me some of them sweet radii of curvature