Mathematics - 2024-11-27

Ben Steffan

00:00

Serre's thesis is one of the most important papers in algebraic topology

IThinkHighlyOfEiligh

I know like 100 math history names, while the laymen only knows "Einstein"

Sine of the Time

Yeah, bro is overrated

Ben Steffan

even funnier, pretty much directly after his thesis Serre pivoted hard into algebra & algebraic geometry and became one of the leading mathematicians of that field instead

he single-handedly revolutionized algebraic topology and then said to himself "yeah that's enough" and went to help revolutionize algebraic geometry instead

IThinkHighlyOfEiligh

Serre was a real cerebrate

Ben Steffan

Serre is still alive, to the best of my knowledge

IThinkHighlyOfEiligh

00:02

O__O thx I didn't know

Dang! Does he still do math?

Ben Steffan

not sure, probably not much if any

guy's 98

IThinkHighlyOfEiligh

He did all the math already

Ben Steffan

pretty much

IThinkHighlyOfEiligh

Mathematicians live to be 100 sometimes, that's good news for us, friends

Ben Steffan

Vietoris published his last paper at the age of 103

IThinkHighlyOfEiligh

00:05

!!!

in his twin prime

Ben Steffan

he died at 110 and is to this day the oldest verified austrian man to ever live

IThinkHighlyOfEiligh

Well, once you crack cohomology, you pretty much are a genius about your health I guess

Claudio

@BenSteffan built different fr fr

Thorgott

the most incredible part is that he was still skiing well into his 90s

IThinkHighlyOfEiligh

Anyone uses Python for math projects, I wrote a MathJax formula parser that knows around 30% of the language on the KaTeX support page (KaTeX == MathJax for this purpose); I'm using KaTeX since quiver app is using it. Anyway, if you were wondering how to parse this complicated language I can help you. (used Lark parser-gen library).

I say 30% because there are a lot of features I didn't implement on purpose, for simplicity, and they are rarely seen operators etc.

So for example if you need to find all the variables in a formula or conver the formula to a variable-substitution-matching Regex, it's doable.

It's private code, so you'd have to ping me here.

copper.hat

02:52

MathJax is not that complicated to parse?

IThinkHighlyOfEiligh

04:47

@copper.hat I hope you're joking

My grammar is over 300 lines and that's just for 25% of it

Also, please define what would instead be hard to parse

That's just a single section covering some of the operators found in KaTeX

There's also the standard form Lark transformer and the variable subst one that have to be written

Or else the UX would be too buggy

But nevermind, I guess what I'm doing is nothing

$\varnothing$

If MathJax is so easy to parse, where the hell does there already exist a parser that does what I need it to?

katex.org/docs/supported.html

You're telling me that's easy to write a flawless parser for?

Maybe if you're on meth, but you'd be dead in a week from exhaustion

You sound like a layman, so I'm done with this. Oh yeah, the twin prime conjecture is obvious, it doesn't even need a proof, right? Because you're a layman it's obvious...

copper.hat

05:48

@IThinkHighlyOfEiligh I think you need to chill a little.

think_meaning_buildß

06:31

>_>

<_<

copper.hat

too subtle for me

think_meaning_buildß

looks left & right, nervously

Simple Cartoon / Animation: Man Looks Left & Right Nervously

►

copper.hat

06:55

:-) i seem to have a knack for triggering folks.

Soumik Mukherjee

@Thorgott what makes a scheme well behaved?

copper.hat

mother

leslie townes

any source of strong discipline, really

nickbros123

Why Dummit n Foote massive, but herstein smol

Soumik Mukherjee

@nickbros123 number of authors. 2>1

copper.hat

07:09

always wanted to submit a spoof article with the authors Dunning & Kruger

Soumik Mukherjee

@nickbros123 why not write Herstein with a capital H?

copper.hat

when i think of algebra, i am reminded of AE Housman's quote "Perfect understanding will sometimes almost extinguish pleasure."

nickbros123

07:33

@SoumikMukherjee cuz:

Soumik Mukherjee

07:48

I See, anti-capitalism✊🏼

3

think_meaning_buildß

LoL

copper.hat

just i.n. case

think_meaning_buildß

e. e. cummings

copper.hat

08:03

my memory is truly awful, we used to get slapped if we couldn't recite in elementary school. i developed a deep dislike of poetry as a result.

one of those teachers was Cillian Murphy's (the actor) grandfather.

think_meaning_buildß

08:15

yup, rote memorization was standard practice back in the day

copper.hat

it is also partly why i don't speak my supposedly native language.

i'm stuck trying to understand a straightforward combinatorics answer. its driving me crazy.

good night

think_meaning_buildß

cya

nickbros123

09:21

theres the following question: are there non empty metric spaces such that, for every possible superspace, it is open relative to that superspace

here superspace as in, we "identify" the space as living in the superspaces? like R kinda lives in C, we can make C kinda live in something bigger, maybe a larger set whose subset is C

i suppose we can just keep going like that, no?

and I suppose the larger metric must be so that it must restrict to what ever metric is in our initial set

Soumik Mukherjee

09:35

@nickbros123 what if you begin with a complete metric space?

nickbros123

that guarantees closed ness not open ness, i suppose

Soumik Mukherjee

So if you require it to be open as well then that makes the superspace disconnected

nickbros123

which superspace, though

Soumik Mukherjee

The superspace that is the completion of any superspace you're having your metric space in

nickbros123

I can always make a bigger superspace, no? The question alludes to having some property being true for every superspace

Soumik Mukherjee

09:52

So this doesn't hold for connected metric spaces

psie

10:18

Let $F:\mathbb R\to\mathbb R$ be increasing. Denote by $F(x+)$ the right-hand limit. It seems obvious, but how can I show $G(x)=F(x+)$ is right-continuous? We have $$\lim_{x\to c^+}G(x)=\lim_{x\to c^+}\lim_{t\to x^+}F(t).$$I'm not really sure where to go next.

psie

10:39

The definition of right-continuous is $F(a)=F(a+)$ for all $a\in \mathbb R$. So...ehm...I guess $G$ is the definition of a right-continuous function.

psie

11:05

Another basic question. The sets $E_r=(x,x+r]\subset\mathbb R$, do they "shrink nicely" to $x$? According to the definition of "shrink nicely", they have to

1) $E_r\subset B(r,x)$ for each $r$,

2) there is a constant $\alpha>0$, independent of $r$, such that $m(E_r)>\alpha m(B(r,x))$.

But $E_r$ is not a subset of $B(r,x)=(x-r,x+r)$, or?

Thorgott

11:19

@SoumikMukherjee in this context, well-behaved should mean locally noetherian

the case of non-locally noetherian schemes becomes a bit more complicated

an $A$-module $M$ canonically induces a sheaf $\tilde{M}$ on $\mathrm{Spec}(A)$ given on basic opens by $\tilde{M}(D(f))=M_f$

this is the local model for a quasi-coherent module on a general scheme

now in the case of locally noetherian schemes, this is the local model for a coherent sheaf if we additionally assume that $M$ is finitely presented (equivalently, finitely generated)

in the non-locally noetherian case, coherence is not quite the natural generalization of being finitely presented and/or being finitely generated, but I think the locally noetherian case is all you need for intuition

and, in general, the reason coherence is sometimes preferred over the alternative of being locally of finite presentation is that it always yields an abelian category, which the latter doesn't

Jakobian

@nickbros123 is $X\times \{0\}$ open in $X\times [0, 1]$

It isn't

So the answer is "no"

Thorgott

@psie it is indeed not

Jakobian

@psie huh?

@psie I feel like here it'd be easier to prove this by an epsilon-delta argument

psie

11:51

@Jakobian how? $x$ is not really a point, only $c$ is. I can only state some definitions. Given $\epsilon>0$, we want to show there exists $\delta>0$ such that $0<x-c<\delta\implies G(x)-G(c)<\epsilon$ (I've already shown to myself it is increasing, so we can remove the absolute value signs). We have $G(x)=F(x+)$, so there exists $\delta'>0$ such that $0<t-x<\delta'\implies F(t)-F(x)<\epsilon'$.

@Thorgott that's strange, as it is claimed to be in the proof of Theorem 3.23 in Folland. Perhaps I'm interpreting the definition of "shrink nicely" incorrectly? Maybe the $x$ in $E_r=(x,x+r]$ doesn't have to be the same as the center of the ball it should be contained in.

Thorgott

no, it's just a minor inaccuracy

Jakobian

@psie look at what you wrote and write it again

psie

@Jakobian everything or just some parts of it?

Jakobian

Some parts. Last one

psie

Hmm.

Jakobian

12:05

It'd be great to put some quantifiers in too

Soumik Mukherjee

@Thorgott thanks a lot, I need sometime to understand everything that you wrote

psie

12:25

@Jakobian Here's a new formulation. We want to show $G(a)=G(a+)$ for all $a\in\mathbb R$. This is the same as $F(a+)=G(a+)$ for all $a\in\mathbb R$ by definition of $G$. Now, $F(a+)$ means $\forall \epsilon_1>0$, $\exists \delta_1>0$ such that $0<x-a<\delta_1\implies F(x)-F(a)<\epsilon_1$ and likewise we have $0<x-a<\delta_2\implies G(x)-G(a)<\epsilon_2$.

This is where I'm stuck. I'm tempted to write $G(x)-G(a)$ in terms of right-hand limits of $F(t)$ as $t\to x^+$ and $t\to a^+$, but that doesn't lead anywhere.

Jakobian

@psie after you wrote "$F(a+)$ means..." I got lost

it surely should look different

in particular there shouldn't be any $G(a)$ or $F(a)$

psie

ok, let me try again, I think I understand

yeah $F(a+)$ by itself means nothing :)

Jakobian

No it does mean something, it means a certain number

but you didn't write the definition of that number correctly

psie

12:42

@Jakobian ok, so just change the occurrences of $F(a),G(a)$ to $F(a+),G(a+)$ above. In particular, we have $G(x)-G(a+)<\epsilon_2$. And now $G(x)=\lim_{t\to x^+}F(t)$, and so here somehow I'd like to argue that for $t$ close enough to $x$ (from above), $F(t)-G(a+)<\epsilon_2$ too. I don't know what to conclude here though.

Jakobian

@psie sure, but I'd still like there to be some sort of absolute value on the right side of implication

Sha Vuklia

What is the function here that we apply the implicit function theorem on? It should go from an open in $\mathbb R^3$ to $\mathbb R^2$ I believe

Initially all we have is a parametrization $r(s,t)=X_1(s,t)e_1+X_2(s,t)e_2+X_3(s,t)e_3$

Jakobian

or, at least, some kind of $0\leq ...$

Sha Vuklia

Or maybe it should start from an open in $\mathbb R^4$, since we get two balls of dimension 2

Jakobian

now the first term can get as small as we want, so the second term, $|G(x)-G(a)|$ is to be made as small as possible

Sha Vuklia

12:48

oh I see I think

Jakobian

and for that we recall the definition: $G(x) = F(x+)$ and $G(a) = F(a+)$

Sha Vuklia

$(s,t,x_1,x_2,x_3)\mapsto (x_1-X_1(s,t),x_2-X_2(s,t),x_3-X_3(s,t))$ probably

psie

@Jakobian indeed, but if we write out the definition, we have two limits tending to two different points

Jakobian

I guess you can always choose $x$ to be a continuity point of $F$, so that $G(x) = F(x)$

so you have $|F(a+)-F(x)|$

so you just need $0 < x-a < \min(\delta_1, \delta_2)$ and $x$ is a continuity point of $F$

psie

ok

Jakobian

12:52

then $|G(a+)-G(a)|\leq 2\varepsilon$, and since $\varepsilon$ was arbitrary, $G(a) = G(a+)$

Sha Vuklia

nvm, I'm still confused

Jakobian

because monotone functions have only countably many points of discontinuity

psie

indeed, that's a nice property

@Jakobian so...is $G(a) = G(a+)$ only for almost every $a$?

since what you've shown only works at points of continuity of $F$

Jakobian

ruminate on this

psie

@Jakobian I've ruminated :D we have $0 < x-a < \min(\delta_1, \delta_2)$, so we can find an $x$ in that interval that is a point of continuity

If there was no such $x$ in that interval, we'd have uncountably many points of discontinuity

nickbros123

13:04

@Jakobian here are u taking coordinate wise addition, or something like $\sqrt{d_1^2+d_2^2}$

Jakobian

@nickbros123 what are you asking me

nickbros123

the metric on the product

i think coordinate wise addition itself works

Jakobian

the product topology $X\times Y$ can be obtained by various metrics

there is no one canonical choice

the metric $d((x_1, y_1), (x_2, y_2)) = (d_X(x_1, x_2)^p + d_Y(y_1, y_2)^p)^{1/p}$ for any $1\leq p < \infty$ for example, induces product topology on $X\times Y$

nickbros123

right, yeah

Jakobian

or put it differently, $d((x_1, y_1), (x_2, y_2)) = \|(d_X(x_1, x_2), d_Y(y_1, y_2))\|_p$ where $1\leq p\leq \infty$ and $\|\cdot \|_p$ is the $\ell^p$-norm induces product topology on $X\times Y$

in the language of metric spaces, those all metrics are topologically equivalent

they induce the same open sets

nickbros123

13:11

makes sense

Jakobian

that's why, when it comes to products, you usually don't specify a metric, in this case the topological view is the one that's more beneficial

this is one thing where doing things only with metric spaces obscures the view

Astyx