point-set-topology

[red garden]

OH WOW

we like

pair it with that x

and by assumption its in the Y_n

right?

[bleak helm]

Yeah

[red garden]

col

cool

is that it

[bleak helm]

So for every y ≠ x, there is a connected subspace K_xy. Their intersection is not empty since they contain x, and their union is X since x is in it, and for every y ≠x there is a K_xy in the union which contains it

That's it

[red garden]

yea yea got it

tysm

so

any hint for number 3?

i always struggle with showing like

specific shit like thjis

like showing somerthing is disjoint of something etc

any hints?

[bleak helm]

Sorry, writing a test soon

[placid thorn]

I wanna prove that any two intervals (a,b) and (c,d) are homeomorphic over the rationals

[chrome dew]

I think you can construct it by writing the equation of a line

[placid thorn]

really?

f(x) = c+(b-a)(x-a)/(d-c) will work?

I feel like it won't map rationals to rationals

[tepid depot]

i dont think this should work

[chrome dew]

why wouldn't it map a rational to a rational?

[placid thorn]

well...provided that a, b, c, d are rationals, then that big quotient sum soup will compute to a rational

[chrome dew]

yup

[tepid depot]

a=0, b=1, c = sqrt(2), d = sqrt(2) + 1. x = 1/2

[placid thorn]

hm...

[tepid depot]

there's no requirement the intervals have rational endpoints

that's why this is tricky

[placid thorn]

for my purposes, this isn't going to work

since I want $a, b, c, d$ to be reals, not rationals

[gentle osprey]

Corgwn

[placid thorn]

so that I can chop up the interval $(0,1)$ into a bunch of intervals bounded by irrationals

[gentle osprey]

Corgwn

[tepid depot]

i like this, that should work

[placid thorn]

ooooo I've got it

[tepid depot]

we can do it pretty simply without showing it's homeo to Q

[placid thorn]

what if I show that for any (a,b), there's a least rational and a greatest rational

[tepid depot]

there isn't

necessarily

[placid thorn]

fucc

[tepid depot]

actually yeah there will never be

anyway

[placid thorn]

right...

[tepid depot]

actually i think, we just want to show any rational interval is homeo to Q

i kinda think this is going to be something weird using choice

it has positive image

it's not injective, but it's getting there

i think

oh

nevermind

ok, i that works for (-1,1)

trying to get that to work for non-rational endpoints is more annoying

so i decided that's harder than it looks

works when you translate or scale by rationals

I kind of like your function, i just wanna see if there's a way to force it to take rationals to rationals on arbitrary endpoints

you see i knew it would be something insane

[placid thorn]

could I just use the fact that (a,b) \cap Q is gonna have a continuous bijection with Q?

[tepid depot]

ok now prove that

[placid thorn]

well I dunno how to attack that

[tepid depot]

lol

[placid thorn]

it feels true

[tepid depot]

yeah same

i want to do something weird with choice or zorn's lemma to prove such a thing exists

but i haven't worked out how that would go yet

i figured

yeah i was just thinking this actually

you enumerate Q

send the first point to something random in the interval

then define the next point in the way respecting the order of the previous points

you need a bit more

to make it dense

like put it half way between the points from before

that should be fine

ooooo yeah

god i think i did this at some point in my life

lol

if you back in forth in a way that respects order, you're guaranteed a bijection. this + monotonic means it's a homeo 🙂

i think either way back and forth is easier

i don't think a cantor set is possible, but i think the image might be dense without being onto

true

so in my initial approach, I went Q -> (a,b) cap Q

if I take something to x, something else to y

I'm forced to get a dense subset of (x,y)

because eventually i get the exact midpoint of that interval

and this applies to every interval

in my construction

but it's definitely not guaranteed to be onto

in fact it's definitely not since i'm basically just doing midpoints after i choose some initial points

[chrome dew]

yikes this got scary

[sleek thicket]

hey does anyone want to help me find the sign error in 4 pages of computations in stereographic coordinates?

would really appreciate it thx

[gritty widget]

make sure you squared the complex number right

[sleek thicket]

i did but I still have a conjugation and a sign

lmfao

this is for bundles

[gritty widget]

the analysis TA would write an equally long, condescending essay about how you don't know how to compute

sniped

[sleek thicket]

lol

I'm like

[gritty widget]

[sleek thicket]

60/40 on Lee fucking this up

I think the map he gives is orientation reversing

Which is sus

lmfaooo

I still haven't sent that regrade request

Didn't want to edit it down from RageMode

[summer jolt]

Maybe because I'm too tired but I don't see this

The definition I was give was this

I can see that the second term vanishes obviously but how do the other two vanish?

[gritty widget]

it says \nabla_X Z = 0 for all X

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

@summer jolt

[summer jolt]

Lol, I should go to sleep

[gritty widget]

[shut moat]

I've been having a bit of trouple proving that the wedge product is associative. Here's what I tried:

Let $\varphi_1 \in A^{k}(\mathbb{R}^n)$, $\varphi_2 \in A^{\ell}(\mathbb{R}^n)$, $\varphi_3\in A^{m}(\mathbb{R}^n)$

[gentle osprey]

~S^1

[sleek thicket]

Sorry, what's A?

Should that be Λ?

[shut moat]

$(\varphi_1 \wedge \varphi_2)(\mathbf{v}1, ..., \mathbf{v}{k + \ell}) = \sum_{\text{\shuffles}\ \sigma \in \text{Perm}(k, \ell)}\text{sgn}(\sigma)\varphi_1(\mathbf{v}{\sigma(1)}, ..., \mathbf{v}{\sigma(k)})\varphi_2(\mathbf{v}{\sigma(k+1)}, ... \mathbf{v}{\sigma(k+\ell)})$

[gentle osprey]

~S^1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[shut moat]

it's this book's notation for the vector space of k-forms on R^n

[sleek thicket]

Huh

[shut moat]

sry

[sleek thicket]

Never seen that before

[shut moat]

apparently it's A for alternating multilinear map? idk

[sleek thicket]

Oh that makes sense

Anyways, sorry for interrupting

Definition of wedge looks good

[shut moat]

so then throwing in the second wedge:

[sleek thicket]

Err what's Perm(k, l)? I would expect Perm(k+l)

But I might be misremembering how the wedge is defined

[shut moat]

oh sorry I fucked up the latex, it's supposed to be shuffles \sigma in Perm...

so the set of permutations where sigma(1) < .... < sigma(k) and sigma(k+1) <... < sigma(k+l) independently

[sleek thicket]

Gotcha

[bright acorn]

Is it always true that for a general topological space X, a point x is a limit of point of a certain subset S of X iff there exists a sequence of points in S such that the limit of that sequence is x?

I always liked this characterization the most

[sleek thicket]

Hey, someone is currently using this channel

It's a little rude to interrupt

[bright acorn]

Hm sorry

[shut moat]

$$\text{sgn}(\sigma)\text{sgn}(\alpha)\varphi_1(\mathbf{v}{(\sigma \circ \alpha)(1)}, ...\mathbf{v}{(\sigma \circ \alpha)(k)})\varphi_2(\mathbf{v}{(\sigma \circ \alpha)(k+1)}, ..., \mathbf{v}{(\sigma \circ \alpha)(k + \ell )})\varphi_3(\mathbf{v}{\alpha(k+\ell + 1)}, ... \mathbf{v}{\alpha(k + \ell + m)})$$

wth

ok I'll just leave the sums implicit lol

[gentle osprey]

~S^1

[shut moat]

gdi

hopefully you get the gist of it

so at this point, I have on the LHS a sum over shuffles alpha in Perm(k+l, m) and a sum over shuffles sigma in Perm(k, l)

[sleek thicket]

ill be real with you

I would not do this

Both sides should be multilinear

In v, w, u or whatever

So you should be able to check on a basis

Which should be easier

This just seems inanely irritating

[shut moat]

oo expanding in a basis sounds nicer yea

i just thought I'd end up wrestling with permutation shit anyway

[sleek thicket]

Basis vector wedge together nicely

Oh also to be clear

[gritty widget]

for $\omega\in\Omega^n(M)$ and $\sigma\in\Omega^m(M)$, $$(\omega\wedge\sigma)(X_1,\hdots, X_{n+m}) =\frac{1}{m!n!}\sum_{\pi\in S_{n+m}} (\omega\otimes\sigma)(X_{\pi(1)},\hdots, X_{\pi(n)})$$

[gentle osprey]

spinsicle

[gritty widget]

if you want to go that way

[sleek thicket]

I'm saying a basis for the dual space

[shut moat]

because you expand each form in terms of the elementary forms, which requires a sum over increasing indices

[gritty widget]

sorry i'm missing a factor of sgn(\pi) in that

[sleek thicket]

Is this right? I think they don't have that normalization out front

You can see their definition up above

Dividing by 1/(m!n!) is a matter of convention

[gritty widget]

ya

[sleek thicket]

Maybe it works because they're summing only over shuffles

Idk

[gritty widget]

or you could do the alternating map thingy

[sleek thicket]

anyways approx

[gritty widget]

which is the same thing really

[sleek thicket]

I would suggest checking on a basis

For both the alternating maps and for the input vectors

[shut moat]

I'll give that a try, ty

[shut moat]

sry went and made tea

so this is fairly easy to prove then

if you expand in a basis you just need to prove that the wedge of elementary forms is associative

and those are determinants

and prods of determinants are determinants of prods which are associative

I think that should be the gist, although I need to actually write this out lol

[gritty widget]

[sleek thicket]

i agree with t

t

e

r

r

a

[placid thorn]

the interior of any one dimensional open interval (a, b) is just (a, b), right?

right, and any 1 dimensional set that you can describe $(a, b)$ for some $a$ and some $b$ is always open, yeah?

[gentle osprey]

Corgwn

[placid thorn]

also, $X - S = X \ S$, right?

[gentle osprey]

Corgwn

[placid thorn]

it was supposed to be a slash

just realized that it's equivalent to X-S, yea

[ivory dragon]

youre probably looking for $X \setminus S$

[gentle osprey]

Namington

[ivory dragon]

in which case yes, \setminus is the same thing as -

in terms of arithmetic of sets

[placid thorn]

yeah thanks @ivory dragon

[cedar pebble]

S left quotient by X

[gritty widget]

[placid thorn]

hmmm

am I on the right track to showing that uh

$\Bar{X-S} = X- I(S)$

[gentle osprey]

Corgwn

[placid thorn]

where $I$ is the interior of $S$

[gentle osprey]

Corgwn

[placid thorn]

by showing that any neighborhood of an arbitrary $x \in X -S$ $G$ intersects $X-S$ nontrivially?

[gentle osprey]

Corgwn

[placid thorn]

or at least getting to a point where I have that

leveraging that to show that $X- I(S)$.

[gentle osprey]

Corgwn

[little hemlock]

assuming G is an open neighborhood of x, the only thing that shows is that x is in the closure of X - S.

What you need to show is that x is in the closure of X - S if and only if x is in X and not in the interior of S

is this even true?

i am thinking cl(Q - (0, 1)) = R - (0,1) != Q - (0,1)

not sure what the assumptions were, but pretty sure this isn't true in R w/ its usual topology by above, unless im being dumb

[placid thorn]

really?

I saw written as fact in a lecture that for any X and subset S of X

X - int(S) = closure(X - S)

right

[little hemlock]

It just occurred to me that X is probably supposed to be the whole space, not a subset

[placid thorn]

yeah that's right

X is a topological space in fact

[placid thorn]

I think I've got a proof:

We will show that the closure of the set $X - S$ is equal to $X - I(S)$, where $I(S)$ denotes the interior of $S$. To do so, we will consider an arbitrary $x\in \Bar{X-S}$, and show that it is a member of $X- I(S)$. Consider that $x\in \Bar{X-S}$ is equivalent to the statement that $x$ is adherent to the set $X-S$. Thus, we know that $x\in X$ for an arbitrary neighborhood $G$ of $x$, $G$ intersects $X-S$ nontrivially. Thus, if we can show that $x$ is a member of $X$, and moreover that it is not a member of $I(S)$, it will be contained in $(X- I(S))$, and our proof will be complete. We know that the first statement is given, however, and so all we need to show is that $x$ is not a member of $I(S)$. \
Suppose that $x\in I(S)$. Then $x \in S$, and $S$ is a neighborhood of $x$. Thus, since all neighborhoods of $x$ intersect $X-S$, there is at least one element $a$ within $S \cap X - S$. Thus, suppose that $a \in S$. Then $a$ is not in $X-S$, since every element of $S$ is removed from $X-S$. Thus, we have wrought a contradiction, and as such $x\not\in I(S)$. QED.\ \

[gentle osprey]

Corgwn

[placid thorn]

"Let $X$ be an arbitrary topological space, and let $S_\alpha$ be a collection of $I$ subsets of $X$. We will prove that the union of all such $S_\alpha$ is a subset of the complement of the closure of all $S_\alpha$."

[gentle osprey]

Corgwn

[placid thorn]

I'm unsure of how to attack this one. Maybe show that the closure of a union of sets is equal to the union of a bunch of closure'd sets?

although that feels wrong

wait

I just need to show that for any given point $x \in S$, x is in the closure of $S$, right?

[gentle osprey]

Corgwn

[tough imp]

What’s the difference between a group scheme and an algebraic group?

[tight agate]

I think algebraic groups are group objects in finite type schemes over a field

while group schemes are just group objects in scheme

[tough imp]

Ohhh

Makes sense

They say algebraic groups are group objects of varieties

As like the classical thing where you end up getting all the matrix groups as algebraic groups and stuff

[tight agate]

yup

definitions might vary a bit between sources tho

[tough imp]

Yeah

Sucks tbh

Afaik there’s not a great source of algebraic groups in the context of schemes

There’s milne’s thing but it only looks at closed points

[tight agate]

I was about to say milne lmao

[tough imp]

Idk I talked to a grad student and he said he didn’t like it for ignoring the non-closed points

There was a topics course on it last year

I did not go

As I was a babby learning what a sheaf was

[tight agate]

might make sense to learn some of the classical stuff first idk

[tough imp]

Then I’d need to learn the other classical stuff

[tight agate]

[tough imp]

Fml

Do you understand divisors?

II.6 exercises are hell and a grad student told me to skip them

But I still don’t get them at all

[tight agate]

depends on what you mean by understand

[tough imp]

¯_(ツ)_/¯

I mean

[tight agate]

lemme look at 2.6

[tough imp]

I know definitions

And that’s it

Pretty much

Idk how you use them, what they’re useful for

Etc

Someone told me to check silverman’s arithmetic of elliptic curves

[tight agate]

mumford

[tough imp]

Mumford?

[tight agate]

curves on algebraic surface

excellent book

[tough imp]

Is it

[tight agate]

lots of stuff on divisors

[tough imp]

Schemes

[tight agate]

yes

[tough imp]

Thank god

God I’m such a ducking poser

[tight agate]

poset

heh

[tough imp]

“Is it the abstract one cuz Idk the classical stuff”

😔

Guy who can’t solve a quadratic over |R and asks for everything in ring theoretic terms

[tight agate]

anyway, one of the reasons for divisors is that they're subschemes that you can actually say something about

[tough imp]

😔

Unless it’s a Cartier one...

[tight agate]

it's very hard to study higher codim stuff

[tough imp]

Thsts what Litt told me

[tight agate]

which is why hodge theory is a big deal

[tough imp]

As to “why we care”

And said yeah hodge conjecture is only known for codim 1 basically

But it doesnt really give me an idea on how to use them

I should have kept up last year

When we did curve stuff and riemann roch and stuff lol

😔

[tight agate]

I mean what do you want to use them for?

[tough imp]

Ugh I need to understand differentials too

Well

[tight agate]

divisors are an object of study

[tough imp]

They get brought up a ton in seminars I go to

Blowups

[tight agate]

yes

[tough imp]

And I’m doing curve shit like

We’re trying to show that M_g-bar is a blah blah blah

When you talk about curves I’m pretty sure you start talking about divisors really quickly

[tight agate]

yes

[tough imp]

So

[tight agate]

mumfords book studies deformations of curves

[tough imp]

Basically I just want to get a better feel for it so I can kinda understand wtf is going on

Haha

[tight agate]

say you have a surface F

and a connected scheme X

then a family of curves on F parameterized by F is a relative effective cartier divisor on Fx X

[tough imp]

🥴

Yeah you see

The latter half of that sounds like gobblygook to me

Hahaha

[tight agate]

and the problem in mumford is to find a universal deformation family

cartier divisor you know

[tough imp]

Yeah but relatively effective

[tight agate]

you know effective too

relative is just "transport the notion to morphisms"

[tough imp]

Was that the stuff in the image of the nice map

Or was that like a principal one

Effective should be the corresponding thing to like the degree of each codim 1 thing being >= 0 right?

[tight agate]

effective means the thing is locally given by f = 0 for some nonzero divisor f

nonzerodivisor*

[tough imp]

Wait wut

Okay so I know a Cartier divisor like is

Blah blah on a cover elements of blah fucking blah such that

I should reread II.6 tbh

But it was so fucking painful to get through oh my god

[tight agate]

that is what I wrote, but I said what the blah blahs are

[tough imp]

If it’s locally 0

Why isn’t it globally 0 then haha

[tight agate]

?

oh

by f = 0 I mean the vanishing of f

like some equation = 0

like x = 0 in the affine plane

[tough imp]

I mena

Sure but idk what that means over a random ass scheme but

I looked it up and it was what I thought

The point was you don’t need to go to fractions

It’s just like in the actual structure sheaf

Locally

[tight agate]

it is an ideal sheaf, yes

there's also the cocycle description

[tough imp]

Right that’s what I meant

Like locally it’s elements in the total ring of fractions of Gamma(U_i,O_X)

Such that on intersections the ratio is actually in Gamm(U_i\cap U_j,O_X)

And then effective is that there’s a cover where they’re just in Gamma(U_i,O_X)

[tight agate]

oh wow, there's a lot of starred problems in 2.6

haha

[tough imp]

Half

And I literally can’t make sense of most of them because they are like

Let Y be the blah blah (Chspter I section 7) and I’m like

-_- ok so idk what that is as a scheme

I think I don’t know algebraic geometry

I know scheme theory

Yes

[tight agate]

well at least youll be doing moduli of curves this quarter

so that's algebraic geometry

might even learn some GIT in the process

[tough imp]

Ehhhhhh

Debatable how much of the moduli of curves bit I’ll actually understand details of haha

If I can understand stacks and how they work, get comfortable working with them, and have high level understanding of wtf the moduli stuff is

So I could explain to a friend “oky so basically...”

I’ll take it as a W

[tight agate]

GIT is pretty cool

[sleek thicket]

you cannot

[tight agate]

lol nice username

[sleek thicket]

im gonna block you the next time you try to schemepill me

ty

[tight agate]

you are a fiber

I am a fibration

[tough imp]

Sham

i literally told Jessie that

If I can learn stacks and prestack, be comfortable with that, be able to work with them

And know enough of the moduli stuff to explain to you what a moduli space is I’d be happy

It’s too late

You’re already in my grasp

[tight agate]

you don't need to get into stack land to talk about moduli spaces

[tough imp]

Sure

But we are

So ¯_(ツ)_/¯

I’ll learn the hilbert scheme... sometime

[tight agate]

there's simpler ones

like grassmanianns

[tough imp]

Right

[tight agate]

the hilbert scheme is even a sub thing of a grassmannian

[sleek thicket]

I am a fan of the grassmannian

[tight agate]

yes, grassmannian good

except when I had to prove that it is a smooth mfld

[sleek thicket]

lol

I had a very very nasty problem

Alex will probably get it this year too

[tough imp]

Uh oh

[sleek thicket]

It's about like, when can every element of the kth exterior power of V be written as a wedge of elements of V

In terms of k and n = dim V

and the idea is basically dimension counting

[tough imp]

So wait

Is this like saying

When is everything a simple tensor

[sleek thicket]

like you embed the grassmannian in the projectivization of Λ^k V and voila

[tough imp]

But wedge

[sleek thicket]

Yes

but you need to know that map is like, full rank or whatever

Or even better show it's a closed embedding

[tough imp]

Chogpamp

Closed embedding?

Easy lol just show the ring map is surjective

[sleek thicket]

anyways we had to do this with the charts on the grassmannian

[tough imp]

Uh oh

[sleek thicket]

Since we didn't know eg it was a quotient by a group action

Or a quotient at all

And the charts are very very very bad

[tough imp]

Triple uh oh

[sleek thicket]

It was one of the worst problems in 54x

[tough imp]

Fuck

What chapter?

[sleek thicket]

Which is saying a lot lol

Not sure

[tough imp]

We’re doing like

3,4 then to 7

To cover Lie groups

So maybe by the time we get it we’ll know it’s a quotient...

😖

[tight agate]

the plucker embedding is good

you get actual equations

[tough imp]

Ugh

Stop

That part of II.5

Sucks

[sleek thicket]

Huh

That's an interesting order chm

But lie groups are so fucking cool

[tough imp]

The ones where you do bullshit by like probing deep into Projective space on the affine parts

[sleek thicket]

I love them

[tough imp]

Blech

[sleek thicket]

Also yeah brof it is

But like

[tough imp]

Yeah well if you love them so much why don’t you marry one

[sleek thicket]

Also really annoying

To construct

With the tools we had

[tight agate]

it is annoying to do it

but once you have it

it's great

[tough imp]

Which ones the plucker?

[tight agate]

span(v1, ...., v_k) to wedge

[tough imp]

Like I swear I first told sham about this

And then he saw it later

And then learned it for actual

[sleek thicket]

no

I think I saw it first in aluffi

[tough imp]

Oh

There’s an exercise

[sleek thicket]

Possibly with you

Yeah

[tough imp]

A really like nasty one

Yes

I asked u for help

And we were like

Wtf

[sleek thicket]

Hmm okay this sounds familiar now

Oh wait uhh

Actually

I went to a preseminar talk

[tough imp]

VIII.4.8

[sleek thicket]

It was by Jake, the postdoc I did the AT reading course with in spring

[tough imp]

This was like a 2D case

[sleek thicket]

And he defined the plucker embedded

[tough imp]

Huh

[sleek thicket]

He does a lot of combinatorics/ag with the grassmannian

I don't understand any of it

But he was jealous when I got @grassmannian

[tough imp]

Wait dude wtf Aluffi does some like weak version of Artin Rees which gets you

For A Noetherian I•\cap I^n = \cap I^n

Which is a lot of why Artin-Rees has mattered for me besides showing stuff about the I-adic topology of a submodule being induced by the I-adic topology of the module itself

Maybe this is secretly the Matsumura proof of Artin-Rees since it isn’t the one in Lang

[cursive flume]

does anyone know a reference where the representation theory of galilei,lorentz,poincare,diffeomorphism group is being done in detail, such that a physicist can follow?

[obtuse meteor]

I must contract spaces

also CW complex shit is going to make me a sad girl

[obtuse meteor]

This comment inspired by showing that S^∞ is contractible

[tidal cedar]

[obtuse meteor]

What?

The prof gave a hint so it shouldn’t be that bad

I have I think figured it out

But displaying a particular homotopy is always a bit painful

[tidal cedar]

S^infty is like, sequences of reals with some metric

I saw an analysis-ish proof that was painful

I guess maybe like cell complex approach could be easier?

Like, Hilbert space and showing all functions can be homotopied into a constant function was kinda wack

[obtuse meteor]

It’s just union of all spheres with standard embedding

And take topology to be U is open if and only if U intersect S^n is open for all n

[placid thorn]

what does it mean for a base to "admit" a countable basis?

[gritty widget]

probably "has a countable subset that's also a base"

[placid thorn]

can I show it within the context of the thingie I'm working on atm?

[gritty widget]

sure

[placid thorn]

So, the lower limit topology on $R$, right?

[gentle osprey]

Corgwn

[placid thorn]

we have another base $B$, which is the set of $[a,b)$ such that $a < b$ and a and b are rational numbers

[gentle osprey]

Corgwn

[placid thorn]

the lower limit topology on $R$ is the same, I suppose, but without the restriction that $a < b$ and $a$ and $b$ are rationals

[gentle osprey]

Corgwn

[tidal cedar]

Contractibility and cell complex structure of S^infty are exercises in chapter 0 of Hatcher

I remember going like upon first seeing that

[placid thorn]

What I’m trying to show is that the lower limit topology can’t admit a countable basis

but again

I'm not sure what that means

you mean, the size of the topology is uncountable?

Its a collection of subsets of a given space that define a topology over a space

I think I get it

[obtuse meteor]

It seems it's just like not too bad

being careful with the details and checking things are continuous

is the brunt of it

but I basically just yelled "pasting lemma" and "polynomials are continuous"

when I constructed the homotopies

and didn't look back

:)

[obtuse meteor]

ok fuck CW complexes

the product of CW complexes is not equipped with the product topology

[placid thorn]

if I have a closed set $A$ such that $S\subseteq A$, $S$ is a neighborhood of $A$, correct?

[gentle osprey]

Corgwn

[pastel linden]

a neighborhood of the set would enclose it, not be enclosed

[placid thorn]

Consider an arbitrary closed set $A$ which contains $S$. we, consider an element of the intersection of all closed sets containing $S$, $x$. We know that $x$ will be in the intersection of this set if $x$ is any given set closed set containing $S$. Let $S'$ be any such closed set such that $S \subseteq S'$, then. We will show that $x\in S'$ implies that $x \in \Bar{S}$. It will be enough to show that $x\in S'$ implies that $x$ is adherent to $S$, that is, every neighborhood $N$ of $x$ intersects $S$ nontrivially. Thus, suppose that for all neighborhoods of $x$, including $N$, $N \cap S$ is the null set. We know that since $S \subseteq S'$, however,

[gentle osprey]

Corgwn

[placid thorn]

I'm trying to work this one out.

[tough imp]

People talk about neighborhoods of closed sets as well for open sets containing a closed set

This for example makes it easy to state the theorem like

[placid thorn]

I don't know where to go with this proof though.

[tough imp]

In a “paracompact Hausdorff space (I think?) given a closed set S, and a neighborhood U there exists a function identically 1 on S and with support contained in U”

Paracompact says like

Any open cover hs a locally finite refinements

I think????

All I know is “it gives partitions of unity or sometbinf”

In the context of manifolds

Lol

[obtuse meteor]

lol

point set scares me

and verifying point set facts

scares me a lot

[tough imp]

Ah

Yeah I dunno, I like being able to call somethig neigborhoods

After all what is a closed set but a really big point

Ignore the fact not all points are closed

[obtuse meteor]

a compact set is a big point

closed maybe not

[tough imp]

That’s cuz u aren’t woke enough

[azure kindle]

Is this a true statement?

$$ A' \subseteq A \iff A^c \text{ is open }$$

[gentle osprey]

Melvin

[azure kindle]

I cant find a proof for this

[gritty widget]

what's A'

[azure kindle]

The set of limit points

got the notation from wikipedia

[gritty widget]

isn't this just the definition of a closed set

[azure kindle]

yes

But I wanna see if theres a proof for it

[gritty widget]

whether or not there's a proof that's more than "it's by definition" depends on what your definitions of open, closed sets are

point set is cursed

[tidal cedar]

I have no intuition for spaces that aren't "nice" and I really need to read Counterexamples in Topology at some point

[obtuse meteor]

There is no such intuition

[azure kindle]

ok It makes sense

[shut moat]

this is why any topological space worth looking at is a manifold

[tidal cedar]

wait actually no that's not a take that differs significantly from my current takes

[gritty widget]

the only spaces worth looking at are locales

[shut moat]

ok pointless topology looks cool

vaguely related: I was surprised to learn today that you can think about integrating differential forms over more general subsets of R^n than just manifolds

is there any such generalization for abstract manifolds [well, I guess spaces lol]? :o

[gritty widget]

fuck it the only spaces worth looking at are sites

@shut moat https://ncatlab.org/nlab/show/Stokes+theorem

[shut moat]

oh ik you can generalize stoke's theorem to manifolds-with-corners, I meant like something even more general

[gritty widget]

https://arxiv.org/pdf/math/9310231.pdf

[shut moat]

owo

[gritty widget]

also you can't integrate on general (topological) spaces if that's what you're referring to

you need a volume form

[shut moat]

yeah general topological spaces seems too broad a class to even attempt to define it on

[gritty widget]

integrate on the long line

[sleek thicket]

based

[shut moat]

hot

[gritty widget]

what's a bad topological space

hmmm

[sleek thicket]

R

S^2

[gritty widget]

manifolds \neq bad

[shut moat]

[sleek thicket]

uhh wait how about a point

[gritty widget]

vacuously hausdorff

based

[sleek thicket]

>has a distinguished element
based

[ivory dragon]

({}, {{}}) is a pretty great topological space

[sleek thicket]

What do you mean by (,)?

[ivory dragon]

topological spaces are notated (S, T)

where S is a set and T is a topology on S

[sleek thicket]

I'm meming about sets

[ivory dragon]

[gritty widget]

you're a set

[ivory dragon]

well you can interpret that the usual way

for ordered pairs

[sleek thicket]

yes

[ivory dragon]

which would give {{{}}, {{}, {{}}}}

truly the most comprehensible notation

[obtuse meteor]

Remember that all of mathematics is just empty sets smushed together in fancy ways

[placid thorn]

this is going to sound very silly

but if I have a topology $T$ over $R$, and a subset of $R$ $S = (0, \infty)$

[gentle osprey]

Corgwn

[placid thorn]

how do I show that no sequence of $S$ $s_n$ converges to $a$?

[gentle osprey]

Corgwn

[placid thorn]

I've been doing so much topology and there are so many definitions swimming around my head that I don't remember

[obtuse meteor]

You’d show that there’s some open neighborhood of a which intersects with the s_n only finitely many times

That’s a strong way to do it

But you can also do something like exhibit a neighborhood so that no matter how far our you are in the sequence there’s a term which is not in the neighborhood

[placid thorn]

So, in this topology, every neighborhood of 0 are uncountable sets containing 0, N, such that R - N is countable.

So, if I have a sequence over the real positive real numbers, s_n

[west brook]

Using https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Markov.pdf and some polytope geometry, I was able to figure out that the Hadamard conjecture is true iff an n-simplex can be vertex-inscribed in a (n/2-1)-rectified n-simplex for all positive singly even n. Can it?

[placid thorn]

okay...last problem in this damned set

I need to show that the standard topology on R^{2n} =~ C^n is finer than the Zariski topology on C^n

[gritty widget]

Can someone help me through this? The product of two smooth varieties is smooth.

It is enough to show that if X is smooth at p and Y is smooth at q then X\times Y is smooth at (p,q).

If X is smooth at p, then from the definition of smooth we get a (U,\phi, f_1,..f_n-d) as in the defintion. Simillary for Y at q we have a (V,\psi,g_1,...,g_b).

Now U\times V is certainly an open subvaritey of X\times Y. And \phi\times\psi is a map from U\times V to some closed set, but I cant see why the closed set must be Z(f_1,..f_a,g_1,..g_b) and I can't see why the matrix still has full rank

using this definition for smooth

[gritty widget]

that is a fun problem

[placid thorn]

is it

I have no idea how to attack it

[gritty widget]

how much have you seen of the Zariski topology?

[placid thorn]

I'm still trying to wrap my head around what it is

[gritty widget]

so what way is your definition phrased

[quick wing]

If $X$ is an integral scheme, and $f$ is a rational function on $X$, does $f$ determine a map from $X$ to $\mathbb{P}^{1}$?

[gentle osprey]

Grothendieck

[gritty widget]

so the tricky thing is with the zarski topology you should think in terms of closed sets

so try and do this problem by showing that every closed set in the zariski topology is a closed set in the standard topology

[placid thorn]

this is what I'm looking at btw

what does the wiggly equals mean?

are we just looking at R^{2n} with the standard topology in this case?

or... R^2 =~ C^n?

[gritty widget]

that challenge problem better not be what i think it is

[placid thorn]

the thing with 14 unique subsets?

why

is it an evil problem

okay cool

epic

[gritty widget]

in general its going to mean "isomorphism"

[placid thorn]

very cool Ciprian very cool

[gritty widget]

so in topology it will mean homeomorphic, in group theory isomorphic as groups, in ring theory isomorphic as rings

[placid thorn]

hm...

so I think I've got my head wrapped around the zariski topology

its the collection of points where a finite collection of complex functions are 0.

So, for example, 0 exists in the Zariski topology over $C$

[gentle osprey]

Corgwn

[placid thorn]

cause you can think of all the complex functions going through the origin and voila

and the entirety of the complex plane $C$ also lives in the Zariski topology over $C$

[gentle osprey]

Corgwn

[placid thorn]

cause you could just take the complex line along the x axis or whatever

and it's 0 at every point

am I understanding things correctly?

[tough imp]

I posted this in AGS, but I'll post it here as well.
I know that if you are trying to show a functor F:(Sch/S)^op -> Sets is representable it suffices to show it's a Zariski sheaf and that there is a covering by representable open subfunctors. I've heard however that once you show F is a Zariski sheaf one may assume that the base scheme S is affine.

It isn't completely clear to me what it means to "assume S is affine", perhaps this means that for an open affine U of S you consider the functor (Sch/U)^op -> Set by sending X -> U to F(X -> U -> S).

Secondly, I don't see how even if you were to do so how this could show representability of F, presumably this gives you a Zariski open covering of F by open subfunctors but even if you showed these functors on (Sch/U)^op were representable they can't in their current form be open subfunctors since they aren't even functors from (Sch/S)^op

[gritty widget]

where or what is AGS?

[tough imp]

Algebraic Geometry Syndicate, it's like Ravi Vakil's server or something

[placid thorn]

what characterizes the closed sets on $C^n$ in the standard topology?

[gentle osprey]

Corgwn

[placid thorn]

we just treat the complex elements of that like regular variables, right, and just euclideanistically calculate their distance?

[tough imp]

Just pretend it's R^2n

[gritty widget]

euclideanistically

[placid thorn]

yeah that's what I'm saying

"Consider the complex plane of numbers in $n$ dimensions, $C^n$. We will show that the standard topology over $C^n$ is finer than the Zariski topology on $C^n$. To do so, we will show that if a set is open with respect to the standard topology in $C^n$, it is open with respect to the Zariski topology. It is suitable to show the contrapositive of this statement for the purposes of this proof, IE, that if a set $a$ is closed over the Zariski topology, it is closed in the standard topology. Thus, let $A$ be an arbitrary set closed in the Zariski topology over $C^n$. Thus, there exist a collection of $m$, finite polynomial functions in $n$ variables $F_m$ such that for any $f \in F_m$, and for any $a\in A$, $f(a) = 0$. Thus, to show that $A$ is closed in the standard topology, we must show that $A = \Bar{A}$ over the standard topology. That is, for any $a \in A$, $a$ is adherent to $A$, which is to say that for an arbitrary open set of $a$, $U$, $U\cup A$ contains at least one element, $b$."

[gentle osprey]

Corgwn

[placid thorn]

am I on the right track here?

I'm drawing a blank on what I should do next

[dim meadow]

Think about basis elements

[placid thorn]

of which basis?

the standard basis?

[dim meadow]

Also your contrapositive seems incorrect

[placid thorn]

oh, right, because not being open doesn't imply that you're closed

[dim meadow]

Yeah it is in fact very incorrect

If A and B are 2 topologies on X, then a set being open in A also being open in B is equivalent to a set being closed in A also being closed in B

[placid thorn]

but "Every set that is closed as per the second topology, is also closed as per the first"

[dim meadow]

Yeah, or just basis elements of zariski being open in standard

Doesn't the standard topology have more opens than zariski, not less?

Yeah it was rhetorical lol

No worries though

The standard topology being finer means that there are more opens in standard than zariski but you wrote the opposite @placid thorn

[placid thorn]

did I then

[dim meadow]

Yeah

Okay anyway you want to show that if a set is closed in zariski it is closed in standard

What are the closed sets in zariski?

They are the zero sets of some set of polynomials. In other words they are the intersection $\bigcap\limits_{i\in S} p_i^{-1}(0)$

[placid thorn]

right

[gentle osprey]

Liquid

[dim meadow]

Lol there we go

[placid thorn]

that should prove that the zariski topology is fine than the standard topology, no?

*finer

[dim meadow]

Well we know arbitrary intersection of closed is closed

[placid thorn]

finer means that if it's open in Zariski, it's open in Standard

https://topospaces.subwiki.org/wiki/Finer_topology

[dim meadow]

So you just need to show zero sets of polynomials are closed

[placid thorn]

according to this, the two conditions are equivalent

[dim meadow]

You can switch open and closed, yeah

By just taking complements

Yeah

(incidentally when I said look at the basis earlier, the basic closed sets for zariski are just the zero sets of a single polynomial. Once you show that the basic closed sets are also closed in the finer topology then you are done)

[gritty widget]

really?

I don't mean this argumentatively but do you know of a book that uses them the "wrong" way around

just very curious to see

actually this is a good time to ask this question

a fine topology has more open sets

so a fine topology has less continous functions

so then when we are defining stuff like the quoitent toplogy

we pick the finest topology that makes the quoitent map continous

because if we picked the coarses topology you'd just get the trivial topology

so in general defining topologys in terms of making a map continous you always pick the finest topology where it is still continous

so say we want to define a topology on on a set U and we have a map f:Uto V

ah okay

so in the domain the the topology that makes sense is the coarsest

in the iamge the finiest

is there something categorical going on here?

retweet

[dim meadow]

How do you define quotients in category theory? Look at the diagram and check if it checks out

(ie look at the universal property)

[obtuse meteor]

hrrmm

so

explicitly defining homeomorphisms

makes me very sad

namely

https://estrogen.fun/i/ncql.png

this does give a homeomorphism D^n/boundary to S^n I'm nearly certain

but showing bijectivity will be very fun

[tight agate]

Is the problem just show D^n/bdry is homeomorphic to Sn?

because if so, there should be an easier way of doing it

[sleek thicket]

wait why?

Writing down an explicit map seems like the easiest way to me?

Showing it makes the right identifications is annoying but it's just algebra. It's a quotient map because it's a surjection between compact hausdorff spaces

[obtuse meteor]

Oh yeah it's not like

hrm

I have done all the theory part of it

I just need to show this is a bijection when you descend to the quotient

[sleek thicket]

sorry Faye I agree it's annoying

[obtuse meteor]

and I don't want to do the algebra

[sleek thicket]

I just don't think there's a simpler way

[obtuse meteor]

there probably isn't

just agree it is "geometrically" obvious brendan

[sleek thicket]

lol

[tight agate]

you can probably mess around with one point compactifications

[sleek thicket]

🧠

[obtuse meteor]

I also like

don't have a good intuition for how to write down a "nice" map for these purposes

like I know the above map works

[sleek thicket]

hmm

[obtuse meteor]

but showing it's bijective seems more work than finding a nicer map :P

[sleek thicket]

To me it seems like stereographic projection

The map you wrote down looks similar

[sleek thicket]

You might be able to describe it via protection onto a disk I think

[obtuse meteor]

absolute hate

[sleek thicket]

Like S^n{south pole} -> open disk

[tight agate]

stereo + shrink gets you sphere - pt to interior

ye

[sleek thicket]

Right

[obtuse meteor]

yeah

this is probably the inverse to that process

[sleek thicket]

And then you just notice it extends to the boundary

[obtuse meteor]

I can see how it looks similar having seen some stereo before

[sleek thicket]

Probably

[obtuse meteor]

but that doesn't mean

I want to do the algebra

[sleek thicket]

I spent the last couple days thinking way to hard about stereographic projection

[tight agate]

both the sphere and disk/bdry are one point compactifications

[sleek thicket]

Had a hw problem involving a map S^2 -> CP^1 and I just couldn't get the details right

[obtuse meteor]

"this is a bijection, as demonstrated by this geogebra file when n = 2"

[sleek thicket]

turns out the map is orientation reversing

Prof wrote down the wrong thing

[obtuse meteor]

oof

[sleek thicket]

and so the problem is false

[obtuse meteor]

I might like

show it to my prof

and see if she has any suggestions for a better map

or if this is like

a good one

and I should just chug

[sleek thicket]

And this seems like the nicest map you can get

[obtuse meteor]

brendan that's like telling me santa isn't real

[sleek thicket]

tfw

[obtuse meteor]

I am bad at algebra

hmmm

I could demonstrate an inverse probably via stereographic bois or something

[sleek thicket]

Demonstrating an inverse seems annoying

Just because you're mapping into a quotient

[obtuse meteor]

yeah

sadness

[astral cedar]

Could you try to do it by showing Dn/boundary~[0,1]^n/boundary and then using some trigonometry?

Although, I am not sure how trivial the trigonometry should be

[sleek thicket]

I don't really understand what you have in mind

When you say using trigonometry

[astral cedar]

Something generalizing the identification of [0,1]/{0,1} to S1 via t->(cos t, sin t)

[sleek thicket]

Oh, I see

Via spherical coordinates

[astral cedar]

Yep

[sleek thicket]

That would definitely work, yeah

formula in n dimensions is kind of messy though

[astral cedar]

Yeah, I vaguely remember

[shut moat]

So Hubbard usually avoids anything to do with the language of general topology and bundles, but in defining an orientation on manifolds, it does this:

and after some googing, B(M) seems to coincide with the frame bundle in more general smooth manifold stuff

So my question is, is this definition actually equivalent to that frame bundle? can you give it the topology inherited from R^n(k+1)?

[sleek thicket]

it should be

so like

i am going to say words and you may not understand them

say $M$ is some abstract smooth manifold and you have a smooth embedding $F : M \to \R^n$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

then you get an injective bundle map $dF : TM \to T\R^n$ over $F$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

hmm maybe this is not what I mean to say

I don't see abstractly why the frame bundle has the topology inherited from R^n, but it should be true

ah I think I see why this is true

so you can find an injective bundle map $\iota : TM \to M \times \R^N$ for some large $N$

by embedding M into euclidean space as above

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

this gives some set function $F(TM) \to F(M \times \R^N) = M \times GL(\R, N)$

which is injective

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

so the question is why is this an embedding

injective+locally an embedding should imply it's an embedding, I think

so you can work in trivializations of the tangent bundle

ie choose coordinates on M and see how this map looks

wait no

this is wrong, you don't get a map F(TM) -> F(M \times \R^N)

hmm

you do get a map F(TM) -> M \times (\R^N)^(dim M)

I think I would need to write this out more carefully

I am 99% sure this map will be an embedding

[shut moat]

is this bundle map \iota not dF?

[sleek thicket]

not exactly

so there's two reasons

we need to make the identification of T\R^N with R^N \times R^N and we need to restrict to the subset F(M)

(sorry for the double use of F, should probably call the map M -> R^N like G or smth)

but it's basically dG

[shut moat]

right ok that makes sense

not that I understand all of the words But from what I do understand, this seems to make sense

ty again Shamrock

[sleek thicket]

ill post after I write out the proof

sorry for the rambling lol

[shut moat]

not at all

[sleek thicket]

Alright so

Say you have an injective vector bundle homomorphisms i : E1 -> E2 over M

This is automatically an embedding

you can check its an embedding locally, and locally its trivial (heh)

Locally you should have a left inverse

Proof: let p be arbitrary. We argue that i has a left inverse on a nbhd of p. Choose a local frame X1,...,Xn for E1 on a nbhd U <= M of p. Then i(X1),...,i(Xn) are still linearly independent at p, so we can shrink to some V <= U around p and complete to a local frame i(X1),...,i(Xn),Y1,...,Yk for E2. Then on the local trivializations over V given by these frames the map E1 -> E2 is the inclusion R^n -> R^(n+k) on fibers

In the smooth case this also proves you get a smooth embedding

okay so this induces an injective function j : F(E1) -> E2^n, where E2^n is the whitney sum of E2 with itself n times (this means that on each fiber you just replace the vector space with n copies of the vector space direct summed together)

We want to know that j is also an embedding

Well we can factor it as F(E1) -> E1^n -> E2^n

The second map is an injective vector bundle homomorphism, and so an embedding

The first is the inclusion of a subspace (at least, depending on your construction of the frame bundle)

If you construct the frame bundle by abstractly taking the disjoint union of the bases for the fibers and then defining local trivializations (and so then taking the topology induced by these trivializations) then the map F(E1) -> E1^n is injective and locally looks like the inclusion of GL(n, R) into M(n, R) on appropriate trivializations of each, and thus is an embedding

@shut moat sorry for the long reply. Basically F(TM) -> M×(R^n)^k can be factored as F(TM) - > (TM)^k -> M×(R^n)^k; the first is an embedding by definition and the second is an injective vector bundle homomorphism, and so an embedding for linear algebra reasons

So the construction in your book should be the same as the abstract version

[gritty widget]

[obtuse meteor]

does anyone have tips for understanding / working with CW complexes?

[rugged swan]

idk if it can help you but thnik CW complexes as fibrant and cofibrant objects are a good way to think of them imo

I don't know a lot about cw complexes

[gritty widget]

can anyone help me with this?

[rugged swan]

if f(P) =/= 0 can't you take U = D(f) ?

[sleek thicket]

pretend everything is finite, or at least finite dimensional

induction is your friend

I might be able to say more specific things in a more concrete context

[marsh forge]

other than that i would just say they are nice spaces that are useful for computation

[marsh forge]

this depends on a choice of model structure

i.e. weak vs genuine

[gritty widget]

I think I might be a bit confused about the wording of the question, its this there is some Q where f does not vanish, or f does not vanish on all the Q

[shut moat]

[sleek thicket]

note that (TM)^k isn't the product space

The fiber over a point p is k tuples of tangent vectors at p

So it's like { (p, v1,...,vk) : v1,...,vk in T_pM }

[shut moat]

right

the notation for some of these bundles always feels a bit weird when I wiki surf

[sleek thicket]

Yeah haha

I would write it $(TM)^{\oplus k}$ usually

[gentle osprey]

Shamrock -> E -> B

[marsh forge]

sham i am sorry to interrupt but desperate can u check my question in ivory when u get a chance i will love you forever

[shut moat]

like tensor bundles/form bundles are denoted as exterior powers/tensor powers of TM when that's really shorthand for the bundle formed by the disjoint union of exterior/tensor powers of the tangent spaces right? it's not like an actual operation between bundles

[sleek thicket]

Oh sure I'll check uw's library

Does sci hub not have it?

Well approx you're right and you're wrong

It is the bundle formed by doing something on all tangent spaces

But this is an operation between bundles

do you know what a functor is?

[shut moat]

I vaguely know it's a category thing, but that's it

[rugged swan]

what is your definition of a curve ? A 1 dimensional algebraic set over C ? @gritty widget

[sleek thicket]

So here a functor is like an "operation on vector spaces"

For any finite dimensional vector space V you give me, I give you back a finite dimensional vector space F(V)

[gritty widget]

quasi-projective variety whos irreducible components are all one-dimesional

[sleek thicket]

And for any linear map f : V -> W I also give a map F(f) : F(V) -> F(W)

[gritty widget]

but here we assume its irreducible so, its a quasi-projective variety of dimesion 1 over an algebraically closed field

[sleek thicket]

We also need the rules F(id_V) = id_F(V) and F(g°f) = F(g) ° F(f)

Some examples are F(V) = kth exterior power of V

Or kth tensor power of V

[tight agate]

Is the problem just asking you to show that if the germ of a section is nonzero, then it is nonzero in a nbhd? @gritty widget

[sleek thicket]

if you have a functor like this, and it satisfies one extra condition I'll get to in a sec, you get a functor from the category of vector bundles on X to the category of vector bundles on X

Given by applying F to all the fibers and then taking a disjoint union

So it's true that the notation Λ^k (T^*M) is a little weird, but we really are applying an operation to the bundle T^* M

Namely this lift of the exterior power functor

The extra condition is that for any finite dimensional vector spaces V, W, the function Hom(V, W) -> Hom(F(V), F(W)) taking f to F(f) is continuous (the Hom sets are finite dimensional vector spaces, so they have a natural topology)

anyways point is Λ^k (T^* M) is an abuse of notation but not as bad as you might think! Λ^k E makes sense for any vector bundle E

[shut moat]

what's Hom here?

[sleek thicket]

Hom(V, W) is the set of linear maps from V to W

(which is itself a finite dimensional vector space, and thus a topological space)

[shut moat]

ah ok, that's kinda cool then

so a functor is like a big brain homomorphism

[tight agate]

Because if the question is this, then you do not need any of those additional assumptions. It works in any locally ringed space and follows pretty much from the definition of the stalk at a point.

[sleek thicket]

Yeah exactly!

It's a homomorphism where the algebraic structure is a category instead of eg a group or a ring

it just feels kind of different because of how big categories are

[shut moat]

that's rly cool!

[sleek thicket]

I was thinking about this a lot recently

It turns out that there's higher categorical stuff going on here

which I will not get into

but basically this process of lifting a functor to bundles has a lot of nice structure to it

[gritty widget]

I think its asking to show that if a germ in non zero, then we can find a neighborhood where it does not vanish anywhere on that neighborhood except possibly the point p

but maybe i have read the question wrong

[tight agate]

oh ok I think you got it right

I misread it

[gritty widget]

I think it is worded slighly badly

I would take "f(x) is non zero for all x in R" to mean that f is not the zero function,
and would say "for all x in R f(x) is not zero" to mean f does not vanish on R

[rugged swan]

that's what I meant but it doesn't work if f(P) = 0

you don't have P in your set

is it a real curve ? not a complex one ?

[tight agate]

In any LRS the set of points x such that f(x) \neq 0 is open

but yeah that's not quite what the question is asking

[rugged swan]

yeah but U should be a neightbourhood of P

[gritty widget]

A curve over a general field that is algebraicly closed

[gritty widget]

Are you saying the question is not true if we interpret it as find a neighborhood U such that f does not vanish on U\p?

[rugged swan]

no but the idea to take the set of f non-zeros doesn't work

[obtuse meteor]

grrr

I might have not done algebra correctly

or

my idea for showing this is a surjection is wrong

and I am very sad

[gritty widget]

Since f is non zero there must be some open set V containing p where f does not vanish everywhere

then is there some nice propety of open sets in an irreducible 1 dim variety

[rugged swan]

it works in analytic geometry, for holomorphic functions, then it works for your example by GAGA principle 🤓

(joke)

[tight agate]

yeah for 1 dim over alg closed field the topology is just the cofinite topology

just take the open set where f(Q) not equal to zero, and take the union with P

[rugged swan]

oh yeah that's what I was about to say lol

[gritty widget]

P need not be open?

[rugged swan]

btw why for dimension 1 variety the topology is the cofinite one in general

[tight agate]

the complement is still finite

so the set is open

wait am I bullshitting

[rugged swan]

I mean it's non obvious for, like, self intersecting curves

[gritty widget]

wait what

[rugged swan]

I think it is true

[tight agate]

yeah the complement is still finite

[gritty widget]

if every point was open then the topology would be (haussdorff or discrete?)

[tight agate]

no

a point is not open

the complement of a point is inifnite

but an open set union a point is open

as the complement is finite

[gritty widget]

oh

king

very cool

[tight agate]

yea the topology on 1dim irred over alg closed field is funny

any two curves are homeomorphic

[gritty widget]

quick question

[tight agate]

take any bijection and it works lmao

[gritty widget]

oh okay that answers my question

[tight agate]

wait wut

[gritty widget]

I was going to say this is clearly not the same as the affine line

so how do they both have the same topology

[tight agate]

oh okay lol

[gritty widget]

does this self intersecting count as a irredicible curve of dim 1?

[tight agate]

yes

[gritty widget]

whack

[tight agate]

the zariski topology is just too coarse to see it

[gritty widget]

that is ridiculous

very cool

[tight agate]

to see that the analytic topology can see that it is "not" irreducible, take the completion at the intersection

[sleek thicket]

r i p