Right, you can generalize Gaussian integers to other quadratic forms ...

Right, I mean the one by Cox

I don't know the number theory behind this but there are good asymptotics on the number of ways to write $n$ as sum of $4k$ squares, which I learnt a bit about today

anyone have a handy chart of which conditions imply which for continuity, Lipschitz, differentiable, uniform continuity etc?

Lipschitz, please

you should figure that stuff out for yourself ;)

I could one day, but I still have assignments to do at the moment

Blah.

You're talking like a freshman, not like a grad student.

The heart of those lie in proving that for a certain subgroup $\Gamma$ of $\text{SL}_2(\Bbb Z)$, the cusp forms on $\Bbb H^2/\Gamma$ of weight $2k$ have Fourier coefficients of polynomial growth

which is a purely geometric fact to me imo

It's not for any particular assignment, it's just something I was curious about when reading the hypotheses of this problem

There are several faculty at UGA who do "analytic combinatorics," which is a form of number theory using some serious analysis @Balarka.

and I'm not very much like a graduate student at this point of my life tbh

right before finals

during

@TedShifrin I have vaguely heard of the circle method and so on, yeah. Dunno much

You can google Magyar, Lyall, Alex Rice, and others if you're curious, @Balarka.

Thanks, I'll forward this to my friend (who asked me to read a bit about a certain chapter in a certain set of notes on elliptic curves which discusses this and explain it to him)

So continuous => Lipschitz of course

Lol

well yes of course

Well, continuous => differentiable, too.

I just wasn't sure whether you could conclude $C^1$

from Lipschitz

@Balarka the algebra looks at the Hilbert class field of quadratic extensions.. the book I used gives a primitive element for those but in general to find such an element you need to use the j-invariant, which is a modular function, so $\operatorname{SL}_2(\Bbb Z)$ etc.

Nope

On compact intervals it's the opposite, $C^1$ implies Lipschitz

Lipschitz => a.e. differentiable

C^1 => locally Lipschitz

Lipschitz doesn't even imply differentiable, but Balarka just gave the deep theorem.

remembers the day when Balarka claimed to hate all analysis

oh that's really interesting

lipschitz is basically just C^1 for really cool people

It's very hard to get Lipschitz from uniform continuity. Adding hypothesis like slopes don't grow too fast at infinity doesn't help, you can have problem at 0 as well.

It's good to come up with examples of that sort

hmm ... or really aberrant people.

I think we've discussed some of this stuff in here before.

@TedShifrin I still don't really understand analysis

Hell, I don't understand anything.

ideally one should analyze all functions of the form $x^\alpha \sin(1/x^\beta)$ for appropriate choices of $\alpha$ and $\beta$, extending to $0$ whenever possible

i didn't do it but one should

those are a good source of examples

Yup.

what's that theorem that Lipschitz functions are secretly C^1-ish

rademacher is a.e. diff

is it due to whitney??

i should really know this

Which reminds me. I don't think you ever did my $\dfrac{x^\alpha y^\beta}{|x|^\gamma+|y|^\delta}$ problem :P

Ah no I didn't lol

I only know Rademacher, @Eric.

I don't even know your secret result.

@TedShifrin Thanks, this reminded me of something i read a while ago, I guess its a good time to read it again

Reminds me, this seemed like an interesting question but the OP doesn't seem keen on elaborating what they have in mind @Ted

(from here ihes.fr/~gromov/wp-content/uploads/2018/08/… )

In terms of Riemannian/hermitian geometry, @s.harp, the upper bound on the dimension of the automorphism group of an $n$-dimensional manifold is the dimension of $O(n)$ or $U(n)$, and you achieve that only for spheres/complex projective spaces. You can find neat stuff like this in Kobayashi's book on automorphism groups of ...

lol gromov

I saw that (and edited tags on it), @Balarka, but didn't want to mess with it.

@TedShifrin for any epsilon and every lipschitz function i can find a C^1 function so that our lipschitz guy is equal to the C^1 guy outside a set of measure less than epsilon

Oh, so it's Lusin-esque in flavor.

@ÉricoMeloSilva that seems like Radamacher + Lusin

yeah

i was thinking of a harder result due to whitney that just happened to be next to this one in federer

how much of federer did you read

be honest

Oh, Whitney extension is the ultimate amazing theorem.

yeah thats fantastic

@BalarkaSen everything ive read in federer ive read bc i was sent there by something else

I learned Rademacher and Whitney first year of grad school in Pugh's dynamical systems course. Fascinating course.

@ÉricoMeloSilva HOW MUCH

LOL, he ain't answering.

idk i cant quantify it i just read whatever bits and bobs i need at a time

it's a reference book on my desk i dont read the damn thing

you madman

I have only been able to read like a little bit of basic measure theory, the first chapter or smth

its so dense

honestly it's just not made for reading

Nope.

And people complain about Kobayashi/Nomizu.

way more readable

still hard to read

Lots of undergrads complain that my books are too hard to read, or just too hard. Meh.

on similar vein i had to read hyperbolic dynamics from Katok-Hasselblatt some time ago

pain

ur multi book seems fine 2 me

i think there are no better multi books tbh

Well, you're never going to complain about my writing being too hard. We're talking other sorts of students.

most undergrads are cancerous so their opinions dont matter

say i, an undergrad

ill light a cig so that i can be cancerous in the literal sense

undergraduates need to be beaten with the stick of consideration and understanding

i agree up until "stick"

@TedShifrin that book made me feel intoxicated when I was an undergraduate, not in a good way

It's tough for grad students. I don't know too many undergrads who would attempt it.

I was an undergraduate for a long time

@Balarka solidarity

@ÍgjøgnumMeg ah ok

remind me what the Hilbert class field is again? the highest extension which has abelian Galois group, in the sense that the Galois group is Gal(Kbar/K)^ab?

It should be unramified too

Oh ok I don't remember what that means

I think grad students need to sit in comfy chairs and be given food and drink during class

something something every prime ideal of O_K lifts to distinct prime ideals in O_E

a prime $\mathfrak{p}$ decomposes into a product of primes $\prod_i\mathfrak{P}_i^{e_i}$ in such a way that none of the $e_i$ is greater than 1

Understood

@TedShifrin Hi Ted

@ÍgjøgnumMeg Hi champ

hi @Jacksoja

Am sorry about my comment yesterday

it's also unramified at the infinite primes (i.e. a real embedding of your number field extends to a real embedding

Hey @Jacksoja

it was meant as young generations use abbriviations not yours haha

LOL, what comment?

oh

that one

but that problem , i did not solve but found a solution ^^

it is cantor problem

but the creation of that set is so smart

lol that was where my pun was gonna be

I was talking about the Bertrand paradox, I believe it's called

"can't or won't"

I still do not get that pun

is it the name?

can'tor

yeah i see that

idk it was meant to be a hint

it was funnier in my head

but i would not have guessed that either

One cannot run the risk

haha might be funny for some, am not funny myself so i cant tell :D

Is there any other way to do that (other than cardinal arithmetic)?

the set is T = { s in S such that s is not an element of f(s)

so clever!

Well, I led you almost to it, @Jacksoja.

@ÍgjøgnumMeg So $\Bbb Q(i)/\Bbb Q$ is ramified at $2$, right? Because $2 = i(1 - i)^2$ and $1 - i$ is a prime in $\Bbb Z[i]$

I know but is it hard to come up with such thing @TedShifrin you led me in a good way but I did give up early

Right, so $(2) = (1-i)^2$

There are no unramified extensions of $\Bbb Q$

Oh interesting

Oh I see that's true

in proving the gcd( a,b) =c has this form c= ma+nb

$\Bbb P^1$ has no nontrivial covering spaces :P

quick Galois theory skills

how do we "guess " that form from the beginning ? @TedShifrin

lol

You want to make up the set so that $T=f(t)$ leads immediately to a contradiction.

$\rho'(x)+x\rho(x)=0$

Well, actually, more "argument" is needed. That only shows $K(t)$ has no unramified extensions - that's true, right?

$K$ alg closed

@TedShifrin neat ! but I was talking about this problem " in proving the gcd( a,b) =c has this form c= ma+nb "

@Balarka you can actually just look at the discriminant of your number field

Oh.

Wonder if it's possible to get only people who don't know algebra on the generals committee @ÉricoMeloSilva

@TedShifrin first we want to show that the GCD do exist for any two integers and why directly go to that form ?

Now $\Bbb Q = \Bbb F_1(t)$ so that is enough argument :3

You think about the division algorithm

@RyanUnger dont u have to get grilled on algebra tho

You see that the Euclidean algorithm gives you the gcd.

$\Bbb F_1$ lol

@ÉricoMeloSilva the guy I talked to said all of Lang was fair game

sounds scary

right right

from EA we do get that form

he said he was asked about finite gp reps a lot

that's the assumption ive been going off of

well is it possible to just fail any Galois theory questions and still pass

yes, @Jacksoja, you see it's the smallest positive integer of that form.

"pass with disgrace"

well ive heard they're pretty forgiving especially if it's way outside ur wheelhouse

I don't even see a first year algebra course offered

@ÍgjøgnumMeg looks scary

I don't think the graduate courses are all up yet are they

there isnt one i think

There doesn't appear to be topology either

i think usually they only have like intro diff geo, intro pde, intro alg top, maybe intro something else

well the way I had it @TedShifrin was i start with a set M = all the intergers of this form an+bm such that m,n are integers , then i show that this set is not empty, and has positive intergers in it

but they usually have like maybe 4 intro courses total per year

Hmm well we can find some algebra ppl

Are you good at it?

I'm going through Atiyah-Macdonald right now

@RyanUnger "Count the number of irreducible representations of $S_n$ in $\text{GL}_m(\Bbb F_p)$"

@TedShifrin but then the argument of EA was the key

In theory I once knew how to do that

I won't be at your funeral

i used to be pretty ok, i havent done like any algebra recently but im hoping i can get up to speed quickly or maybe ill just sit in on whatever undergrad algebra they have

You're on a different continent, that's ok

such results are easy to use but tuff to see where they came from in a axiomatic way @TedShifrin IMO ofcourse !

Lol their undergrad algebra is AG

lol wtf

do you guys forget some earliar courses in math ?

and then revisit them again ?

registrar.princeton.edu/course-offerings/…

@ÉricoMeloSilva seems scary already

when i do too much algebra without analysis , i forget many calculus / analysis concepts

is that normal?

now there is this one registrar.princeton.edu/course-offerings/…

@ÍgjøgnumMeg Prime powers are primary ideals in Dedekind domain, or no?

but it's a 300 level

Maybe the numbers don't mean what I think they do

im probably gonna take the intro geo + pde for my special topics prep and then de lellis class and idk if ill do anything else in fall

i think 500 is grad and everything lower is undergrad?

idk we dont register till we get there

Yeah that's what I thought

So the 300 level is Artin

and the 400 is AG

where's the regular graduate algebra?

idk if they have one lol

Riemannian geometry and PDE is the only thing reasonable for the two topics

GMT could be a disaster

yeah

Fefferman is teaching a course on smooth functions lol

@Jacksoja: You know me well enough to know that I am not fond of axiomatics.

I'll want to take that too

My course load is too much

Do we have to take tests?

Princeton doesn't believe in standard first year grad courses ... or at least they didn't in the past.

oh yeah they usually have a functional intro and a riemann surfaces intro too @Ryan but i think it says somewhere on the dept page that they only have like 4 - 6 grad intro classes per year and they miss a lot of the basic subjects