point-set-topology

[pearl holly]

I speedrunned that part tho

[empty grove]

[tawdry widget]

Oh so you are saying that H_1 is the abelianization of π and the abelianization of free group of rank n equals Z^n?

I see thanks a lot

[pearl holly]

the second thing looks like the Euler characteristic, which is (-1)^n rank(H_n(X)) and you know that this is equal to verticies - edges + faces

[empty grove]

No faces in graphs

But yes that's the Euler char of the graph

[pearl holly]

ye you have components instead

[empty grove]

That's coincidental

Same definition applies

[pearl holly]

ye I know, I meant that like you take rank(H_0(X)) and move it to the other side

[empty grove]

Ah right

[shy moss]

What conditions must a collection of convex sets satisfy for its union to also be convex?

[drifting sundial]

does a sigma-compact space necessarily admit an exhaustion by compacts?

[maiden oracle]

I'm trying to understand the fundamental group of a configuration space and it's connections to braid groups Let $Conf_n(\mathbb{R}^2) = \product_{n=1}^n \mathbb{R}^2 \setminus {(x_1,\ldots,x_n) | x_i = x_j for some i \neq j}$ and let $UConf_n(\mathbb{R}^2)$ be the configuration space quotiented by the action of the symmetric group, so the order of the points here doesn't matter anymore. I basically drew what i mean in the picture. From the picture and intuitively the loops in the fundamental group should all be trivial since R^2 has no holes but wikipedia claims that this is isomorphic to the braid group on 2 strands. I could see how such a path could give rise to a braid by taking the strand f_i to be f_i(t) = (gamma_i(t),t) where t goes from 0 to 1 but the isomorphism in wikipedia just confuses me. I hope what i wrote is readable, putting this into latex would take me ages

[gritty widget]

for locally compact spaces it does

[drifting sundial]

what's a counterexample for non-locally compact?

[drifting sundial]

ohhh right there's Q

what a strange space

[gritty widget]

does anyone have a favourite recommendation for K-theory

algebraic K-theory

[gritty widget]

Q admits exhaustion by compact sets

[dusk heron]

Is there any book about complex manifolds, Kähler geometry, etc., which is similar in style to the books by John Lee (Introduction to Topological/Smooth/Riemannian Manifolds)?

[drifting sundial]

wait really

I thought all interiors are empty

[gritty widget]

not in the topology inherited from R

[drifting sundial]

can you give me one then

[gritty widget]

$U_n = (-n, n)\cap \mathbb{Q}$

[gentle osprey]

Moniker

[drifting sundial]

hmm but U_n is not compact?

[gritty widget]

its relatively compact

[drifting sundial]

ok now I'm confused

[gritty widget]

it's a relatively compact open set

[drifting sundial]

relatively compact as in the closure is compact?

[gritty widget]

yes

[drifting sundial]

youre taking closures within R right

[gritty widget]

no, in Q

[drifting sundial]

oh

[gritty widget]

why would I take closures in R if my space is Q

[drifting sundial]

that seems weird then, it doesn't seem to be compact

[gritty widget]

Q?

it isn't compact

[drifting sundial]

no the closure of U_n

the closure is [-n,n] cap Q right

[gritty widget]

right, it isn't

[drifting sundial]

so your example doesn't really hold right

[gritty widget]

yes

If there was an exhaustion by compact sets in $\mathbb{Q}$ then there would have to be a non-empty relatively compact open set, in particular, for some interval, say $(a, b)\cap \mathbb{Q}$, its closure would have to be compact it's never compact

[gentle osprey]

Moniker

[gritty widget]

because if we pick arbitrary irrational in (a, b) then for any sequence of rationals which converges to it in (a, b), we wouldn't be able to find a convergent subsequence (in Q)

I was trying to give another counter-example which is N with co-finite topology

[long hornet]

If X has an exhaustion by compact sets, and Y is a closed subspace, then Y also has this property, right?

[drifting sundial]

I guess so

[drifting sundial]

yeah, and also all the compact subsets of Q have empty interior

[gritty widget]

that's not true

[drifting sundial]

hmm?

[gritty widget]

{0} has empty interior

[drifting sundial]

oh sorry

I meant to say empty oops

[drifting sundial]

and because of this, the condition that each compact in the exhaustion lies in the interior of the next cannot be satisfied

@gritty widget that clears it up right?

[gritty widget]

why are you asking me, you asked the question

[drifting sundial]

I was confirming that you didn't have anything more to say

but thanks for the help anyway

[gritty widget]

ok

[pearl holly]

yo how long is a piece of string?

[tough imp]

Like 7

[maiden oracle]

in case anyone is interested in this. i finally figured out why the fundamental group is not just trivial. We typically say that a braid is the same as another braid if one can deform on into the other such that they do not intersect in the process(pretty intuitive) Now n distinct points of R^2 constitute a single point in the configuration space Conf_n(R^2)
The fact that these points are distinct is captured exactly in the definition of a configuration space, where we forbid to points to be equal. So a loop in the configuration space may be visualized how i did it, as n distinct loops where each is in our manifold(R^2 in the picture) but it is fatally wrong to then just take the topology of R^2 for each strand of the braid(or in this picture rather for each loop) individually and assume that this is the same topology that the one we have in the configuration space(that's what i wrongly assumed would happen) Instead if we take a loop in the configuration space and visualize it as n individual loops in R^2 then for every time t in [0,1] we must always have n distinct points in the configuration space. This implies then that you can't simply deform each of the n loops individually since when you combine these n distinct deformations into a single one it might very well happen that at a particular time t we may have the same point twice, but must instead find some deformation that respects this restriction, thus we can have pretty complicated fundamental groups and so braids aswell. wow im happy now

[tight agate]

complex geometry by Huybrechts

[tough imp]

@bitter yoke said he’d buy me this book

During the yellow book sale

And he never did

[gritty widget]

Around a piece of string long

[orchid forge]

i feel like toki is throwing shade at some reddit post

[bitter yoke]

Plssssssssssss

[tough imp]

You literally did

I can find the messages

@bitter yoke you can’t deny me

[bitter yoke]

Yea I offered

And then I asked for your address and you never gave it to me

So

[tough imp]

WTF NO YOU DIDNT

[bitter yoke]

I'm pretty sure this is what happened

Go find it I'm busy

[tough imp]

https://www.springer.com/us/book/9781461418085

Okay it was a different book

And you kinda did

I just missed the message

@bitter yoke sorry Zoph I was wrong :(

[bitter yoke]

Do u still want it

[tough imp]

Yeh

[bitter yoke]

What's ur address

[tough imp]

For real?

[bitter yoke]

??????? It was for real last time too

[tidal cedar]

This is actually really cool, I didn't know braid group pi_1 arose so naturally

[shy moss]

is the ideal I(V(S)) of an affine variety V(S) a subset of S?

[tough imp]

No

Let J be the ideal generated by S

Then I(V(J)) = rad(J)

[limpid leaf]

this is true right? theres some defn or theorme that says that the basis given by B_1 x B_2 generates the topology on T_1 x T_2 i tihnk...? Sorgenfrey is the lower limit topolgy btw

[true robin]

There is a catch though, a,b,c and d belong to Q in this set

[limpid leaf]

does (a, b) where a, b rational not generate the usual topology

[true robin]

The usual top is fine

[limpid leaf]

oh

[true robin]

But check the lower limit topology

[limpid leaf]

oh yeah the previous question was checking that hte lower limit topology on R2 was not generated by [a, b) x [c, d) where a b c d rational

i just wasnt thinking

okay so this is false because of that previous question

[true robin]

Yeah, but you aren’t done yet. You just showed that [c,d) for rational c and d is not a basis for the lower limit topology, but there could be a freak coincidence making the above set a basis

By above set I mean the (a,b)x[c,d) set

[limpid leaf]

I think this must not work because of the same failure as the last one which is a point like (sqrt(2), pi) or whatever

where it being irrational in both places means any basis element containing the point is not a subset of it

[true robin]

I think you have the right idea, but I don’t exactly understand your phrasing. You need an open set that is not a union of open sets in the given set. So what open set around (sqrt(2),pi) are you considering?

[limpid leaf]

Like just take any open set with all rational corners right? then it's in the basis

[true robin]

Also I think you’ll only need irrationality of the second coordinate

[limpid leaf]

yeah

i was just saying the other one failed for a pt like that

[true robin]

What is a rational corner?

[limpid leaf]

the corners of the rectangle given by (a,b) x [c, d) are rational for a,b,c,d

[true robin]

I see

So I suppose you understand how to finish the problem from here?

[limpid leaf]

it's actually a yes/no problem but yeah i see the soln is very similar to the previous one

thank u very much

[true robin]

Np

[gentle osprey]

deadpan2297

[tawdry widget]

Let X=(R^2)^n-{(x_1,…,x_n):x_i=x_j for some i,j} I think in order to see the connection between configuration space and braid group we can view a closed path I—> X/S_n as a path g from I to X such that f(1) is a permutation of f(0)=(e_1,…,e_n). Now let f_i= (P_i • g, id_I): I—>R^3 mapping t to (P_i(g(t)),t) where P_i is the i th projection from (R^2)^n to the i th R^2 clearly t—> (f_1(t),…,f_n(t)) is a braid.

[limpid leaf]

i do not know how to make the appropriate ball here

also i jsut want to clarify what i want to show

[abstract pagoda]

uh

[limpid leaf]

which is that for some $a \in A$ where $A$ is the 1/n set, I want to show that there is some $B_{\epsilon}(a) \in A$ such that $B_e(a) \cap A = {k}$

[tough imp]

Wha

[limpid leaf]

wha

is this the wrong place

[abstract pagoda]

No its good

[tough imp]

No it’s just like

[abstract pagoda]

i like

[tough imp]

Between 1/n

And 1/n+1 and 1/n-1

The distance is gonna be fixed

And you can just pick a number smaller

[limpid leaf]

ok thats what i thought but theres a piazza post that is fucking with my head

[tough imp]

What’s the dude saying kekw

[gentle osprey]

jesse

[abstract pagoda]

I that works

[limpid leaf]

is this correct *

[abstract pagoda]

That should work

[limpid leaf]

I just dont understand which set the ball is in

[tough imp]

It shouldn’t be in A

It’s not a set in anything

I mean it’s like

In the power set lol

But it should be a subset of R

[abstract pagoda]

You want to show that the intersection intersect R is equivalent to discrete topology

[limpid leaf]

its in P(A)?

[tough imp]

I mean

[limpid leaf]

subset R

[tough imp]

Technically yea

Yeah

[limpid leaf]

yes ok

[tough imp]

More importantly

It’s in the topology

[abstract pagoda]

Ya

[tough imp]

If you go by the “topology is a subset of P(X) such that blah blah blah”

[limpid leaf]

the subspace topology?

like T_{x | A}

[tough imp]

Uh

[abstract pagoda]

i never saw that notation tbh

[tough imp]

No

[limpid leaf]

i think its whats in Munkres idk

[tough imp]

I’m just saying that B_epsilon is an open set

That’s all

[limpid leaf]

yeah but in which topology

[tough imp]

The Euclidean topology

On R

[limpid leaf]

ok

[abstract pagoda]

no?

what

yea

wait

no

[tough imp]

What?

[abstract pagoda]

brb

[tough imp]

Lmfao okay

B_epsilon(a) is an open set in the Euclidean topology which is what you want

Since you’re considering your set of 1/n as a subspace of R under the Euclidean topology

[limpid leaf]

okay this piazza post is just making this more confusing

yes

this makes sense

[abstract pagoda]

B_e is a neighborhood of 1/n

[tough imp]

What does it say on Piazza

Yes

[abstract pagoda]

which is in R

[tough imp]

Yea…

[abstract pagoda]

Think of like ball in R

[tough imp]

Yes????

Is this addressed to Jesse or to me

Lol

[abstract pagoda]

i am developing occupational disease

[limpid leaf]

yes ok it is a ball in R

[abstract pagoda]

ya

[tough imp]

[limpid leaf]

and then the intersection with the epsilon ball around 1/n and A is always equal to the point

[abstract pagoda]

contradiction is easy

[tough imp]

Yeah

[limpid leaf]

why do i need a contradiction what

[tough imp]

Bro what contradiction?

Just pick epsilon small

[hollow harbor]

[limpid leaf]

theres a theorem that just says T is discrete iff for any a in P(X) then {a} \in T

i tihnk

[tough imp]

Yeah

Wel no

A in X

[limpid leaf]

yeah sorry a in X

[tough imp]

a in X

It’s cuz any set is a union of singletons

And unions of open r open

[limpid leaf]

yeah we proved it last pset

[tough imp]

Okay

Good good good

[limpid leaf]

what does this mean

[tough imp]

The point is that

The closest any point is to 1/n

Is either the distance to 1/n-1 or 1/n+1 (it will be the distance to 1/n+1)

So if you pick epsilon < that distance

[limpid leaf]

right

[tough imp]

Then the only thing in that ball is 1/n

[hollow harbor]

[abstract pagoda]

is it possible to show this is true just with a continuous map?

[tough imp]

Show what’s true

[hollow harbor]

Yes

[abstract pagoda]

A is discrete

[hollow harbor]

It is

You can make functions on (0, 1) which are continuous and have any values on these points

The restrictions to A are then continuous

[tough imp]

Hurb

[hollow harbor]

Hence all functions on A are continuous

[tough imp]

This is definitely gonna be harder

[hollow harbor]

And the topology is discrete

[abstract pagoda]

its simpiler fym

[tough imp]

You’re like gonna be taking bump functions and smoothing out sround each 1/n

Make the fuckinf function then

[hollow harbor]

Sure

[tough imp]

Write it down for me

[abstract pagoda]

constant

functions

work

[hollow harbor]

[tough imp]

????????

[abstract pagoda]

no?

[tough imp]

Are you fucking with me?

[limpid leaf]

meth

[hollow harbor]

How would that work

You need to show any function on A is continuous

[abstract pagoda]

ok contradiction maybe?

like assume every function from A to R is not continuous?

uh

no

hmm

so

[hollow harbor]

Ok, alternative idea

Take the map 1/x from (0, 1) to (1,infinity)

Its a homeomorphism

[abstract pagoda]

ok

[hollow harbor]

Then its restriction to A is a homeomorphism onto its image

[abstract pagoda]

x/1

[hollow harbor]

But the image is N

[abstract pagoda]

1/x

[hollow harbor]

😇

N is discrete obviously, now every point is in its own 1/2 ball

[abstract pagoda]

and thats it ig

dont u want every function from A to R though

thats just using tricks

i want to show it via every functions way

not homeomorphism

[jovial perch]

given that there exist a unique line passing through (x1,y1) and (x2,y2) for any two points (x1,y1) and (x2,y2), how would there be another polynomial defining C?

i would also appreciate an example of an affine plane and a line defined by two polynomials

for further definition

<@&286206848099549185>

[empty grove]

You could always square the polynomial

The roots won't change

Or multiply by a non zero scalar as 2.13 says

If k isn't alg closed you can multiply by any polynomial that doesn't have a root

[jovial perch]

oh, you mean like if the affine plane is A^2(R), then X^2+Y^2-1 is a algebraic curve on A^2(R) which is also represented by (X^2+Y^2)^2-1 ?

[empty grove]

No, by (x²+y²-1)²

Because a² + b² - 1 = 0 iff its square is 0

Since k is an integral domain

Oh wait in this case your thing also works lol

[empty grove]

Since this is true when you replace 0 with 1

Wait no, you get extra solutions

Because (a,b) such that a²+b² = -1 all become solutions

[jovial perch]

oh dang

[jovial perch]

yes that makes sense

so in essence any nth power of the lowest-degree polynomial where n is a positive integer represents the same line

[jovial perch]

wdym?

[empty grove]

yes

[empty grove]

That was wrong, ignore that

[jovial perch]

i see

idk what this is saying tho

the last sentence

[mint rose]

I would guess it just means f=0

As a polynomial

[swift fjord]

I think they mean why is it true

[jovial perch]

so f(x,y) = 0 for all (x,y) = 0 iff f(x,y) = 0? eh?

[mint rose]

No there's a difference

[swift fjord]

It's iff

[mint rose]

A polynomial evaluating to zero everywhere and being zero is not the same

For finite fields you can construct a polynomial which evaluates to zero everywhere by taking just the product of (x - a) for each a in F

[jovial perch]

so the polynomial is the zero polynomial whenever all "zero points" are zeroes of the polynomial

[mint rose]

In infinite you can't because the number of distinct zeros of a polynomial is inferior or equal to it's degree

And a polynomial is necessarily of finite degree

[mint rose]

The zero polynomial is just 0

[jovial perch]

yes, i'm aware

[mint rose]

Ah I think I'm reading wrong what you're writing

[jovial perch]

hm, but how would you translate that for each (x,y) = 0?

[mint rose]

As they did for F_2

Product for x and also for y and some those two up

[jovial perch]

wdym

[orchid forge]

the product of terms (x-a)(y-a) over all a in F

[mint rose]

The idea was (x-a)(x-b)... + (y-a)(y-b)... but that also works

[orchid forge]

yeah i just realized what you meant sorry

[mint rose]

Actually you could leave out the y terms and it would still be an example of a non-zero polynomial which evaluates as zero everywhere

[orchid forge]

that do be kinda true though

[pearl holly]

Is this some AG memery?

Or what is this lmao?

[empty grove]

Ye AG

[shy moss]

Let $S=\langle x_1^2-x_2^6,x_1x_2-x_2^3 \rangle \subset C[x_1,x_2]$, is it right that the zero locus is $V(S)=\lbrace (0,0),(1,1)\rbrace$?

[gentle osprey]

Or x1

[tough imp]

Jesus Christ

I think so

From the first relation you get x2 = cbrt(x1)

Then the second one says x1 = x1^4/3

Which is just to say that x1 is either 0 or 1

[true robin]

Missing (1,-1)

@shy moss

[wary furnace]

how do I find the geodesics of a 3d function? I know that I have to plug something into the euler-lagrange equation, but what exactly?

oh oops

this is the function I'm working with

[gritty widget]

what's a geodesic of a function

geodesic of its graph?

[wary furnace]

yeah

idk if this would help but I was able to describe it in cylindrical coords like this

[gritty widget]

figure out what the lagrangian of the length functional looks like in your preferred coordinate system (like the polars you got there), then plug in the components of your geodesic in your coordinates into the euler lagrange equations for that

iirc

you could also use the action functional (half the length functional and with squared lagrangian), that'd give you the same thing but with constant speed parametrization

it's been a while since i did this so i apologize if anything's off

[wary furnace]

thanks and np, I'm myself very new to all this

[tight agate]

lol tterra physician

[wary furnace]

would that lagrangian be something like this? I remember it from lagrange multipliers as F-λC
where F is the thing you wanna minimize and C the constraint
(oh also, the integral on the left is the length functional in cylindrical coords)

[gritty widget]

by lagrangian i mean the thing inside the length or action integrals

[wary furnace]

oh it's already inside the integral

yeah this is where I'm suck

yeah I'm really not sure what to do or how to incorporate the surface into this, do I add it or multiply it by the root thingy? or am I doing it wrong completely?

[shy moss]

thanks

[jovial perch]

the underlined confuses me

[gritty widget]

what is A^2(R)

[jovial perch]

the affine plane R^2

[gritty widget]

take theta = pi/4 to see that (0, 0) is in \Gamma i guess

[jovial perch]

hmmm

[jovial perch]

so it is also established that (x,y) = (rcosθ, rsinθ) = (0,0)?

[gritty widget]

don't know what you mean

they're saying that if (x, y) = (0, 0) then (x, y) is in Gamma and you can see that by the definition of Gamma by taking theta to be 1/4 in r^2 = cos 2theta

[jovial perch]

sure, that gives (r,θ) = (0,0)

but how is it related to (x,y) = (0,0) being in Gamma?

isn't (x,y) = (rcosθ, rsinθ) the missing link?

[kind cedar]

how is f_bar "clearly a bijection" ?

[gritty widget]

have you tried to prove it

[jovial perch]

:<

[gentle lark]

The two points that wind up at the same place are quotiented

[kind cedar]

I kinda don't get exactly what it means. They have an equivalence relation to them, but I still see the function as not being injective since it points 2 elements to the same element

or due to the quotient, those 2 elements are the same?

[gentle lark]

That

[kind cedar]

Ah, thanks, it makes sense

[jovial perch]

<@&286206848099549185>

[swift fjord]

note that the underlying set of I/~ is a set of equivalence classes, where each class is a singleton except the class of 0 and 1 which is the pair {0,1}, so in the eyes of I/~ [0] and [1] are the same object

[kind cedar]

If a continuous function f is also surjective, will it map open sets to open sets?

[cedar pebble]

No

[frigid patrol]

A continous function thats injective on R^n will map open sets to open sets

This is Brauwer's invariance of domaon

[pearl holly]

domaon

[plain raven]

Brauwer

[fading vale]

yamin absolutely ratioed

ratio'd?

[bleak helm]

Both, but the first is better

[gentle osprey]

亜城木 夢叶

[long coyote]

got it

[steel glen]

it would seem to be implied

above that statement they say

Observe that if we set x = rcos(theta) and y= rsin(theta)
so it looks like you're working with the same assumptions

[bright acorn]

One of the things that I was thinking is the following

If you fix a vector field $X \in \mathfrak{X}(M)$ on a manifold $M$, then the interior multiplication with respect to X
$$
\iota_{X} : \Omega^{\ast}(M) \rightarrow \Omega^{\ast}(M)
$$
Is a degree -1 antiderivation and in fact forms a chain complex, i.e
$$
\iota_{X} \circ \iota_{X} = 0
$$
These are some very nice properties, I wonder if there's some homology theory for manifolds that is based off the interior product?

[gentle osprey]

MisterSystem

[bright acorn]

I think people might not care too much about it

Because we first have to choose a vector field

But if there was some way to like

Get rid of this arbitrary choice but still work with properties of the interior multiplication

that would be kind of nice

[orchid forge]

what if you use a frame instead of a single vector field?

[fair idol]

Let F:M->N be a smooth map. How can it be shown that the components of F are constant off dFp is the zero map

[gritty widget]

what is dF_p in local coordinates in terms of F

[bright acorn]

I think that working with frames may be a really strong imposition.

[orchid forge]

say locally

or even just in R^n, using the standard frame

[bright acorn]

Because not every manifold admits a global frame

[orchid forge]

i'm pretty sure you recover something simple

[gritty widget]

I guess not because

You can also do the same with wedging with some form

And this gives you a crap cohomology

[coral pawn]

Is there a way to do b. that isn't painful?

[lunar yoke]

wouldnt it suffice to show that F is a homeomorphism?

[coral pawn]

But it isn't

[lunar yoke]

oh

[coral pawn]

0 and 2 both go to the same thing

[lunar yoke]

I though F(R) = F([0,2]) and then use compactness, continuity and that T^2 is hausdorff

ah

then maybe choose parametrization from S^1 with same image?

[coral pawn]

What do you mean?

[lunar yoke]

considering a different map altogether that just has the same image

like g : S^1 -> T^2 with g(S^1) = F(R)

[coral pawn]

What map might that be?

[lunar yoke]

maybe $g(e^{2\pi i \theta}) = [(2\theta, 1/2 + \theta)]$ would work

[gentle osprey]

Phil P

[lunar yoke]

basically the same as F just make it that actually 0 = 2

[coral pawn]

what does this do though?

[lunar yoke]

well if this g is an embedding then you get what you want, no?

[coral pawn]

No

Wait

[lunar yoke]

I though if you have a manifold M and an embedding f: M -> N then f(M) is a submanifold of N

or am i just completely on the wrong track? xd

and was thinking that embedding = homeomorphism that is a smooth submersion

I also just recently learned about all of this, so maybe i got things mixed up

[coral pawn]

What you said might be true, but we haven't defined it that way

And we haven't proved it either

Why does my professor have to make the hw so sucky?

Literally all the hw problems are working out examples

No proving general results

[lunar yoke]

better than no examples at all i guess

you get general results in lecture prob

[coral pawn]

No I would take no examples at all over this

[coral pawn]

Just the theorems

[lunar yoke]

well what is your definition of embedded submanifold then

is your course based on some book?

[coral pawn]

I think I'm gonna go to sleep

This shit is melting my brain

[lunar yoke]

ok gn, sorry if i confused you

[coral pawn]

Oh it's not your fault lol

You didn't confuse me at all

It's just the problems in this class don't usually have a slick way to do them

And I find that really annoying

[gritty widget]

This will make you so good though

[gentle osprey]

Brian485

[mint rose]

Not really a concrete math question, but if I want to go in the general direction of topology+algebra+geometry and I had to choose between graph theory and functional analysis course, which will be better suited (although I'm aware they're both different domains of maths)?

[ivory dragon]

probably functional analysis

it has random connections to topics in AG in particular

and graph theory feels easier to pick up on the fly

since its (at the intro level) feels more like a collection of various techniques and facts, not structured like a typical math course

[green quartz]

Hi. I am trying to show that the map $f : \mathbb{R}^{n+1} - {0} \to S^n$ given by $f(x) = \frac{x}{||x||}$ is either closed or open. Can someone give me a hint?

[empty grove]

do you mean x/|x|?

[green quartz]

Yes, sorry.

[gentle osprey]

11de784a

[empty grove]

consider asymptotic stuff

and consider open balls

conclusion would be ||open, not closed||

[green quartz]

I am sorry if it's elementary, but what do you mean by asymptotic stuff?

[mint rose]

That's a good point

[empty grove]

I was being vague so you can figure out what it means assuming you know what asymptotic means

[green quartz]

Oh, okay. Thanks!

[wise walrus]

Can someone push me in the right direction here? I'm trying to prove that every basis for a second countable topological space has a countable subcollection that is also a basis for the same topology.

I know topology goes against a lot of intuitions sometimes, but I just can't fully wrap my head around picking out countable subcollections that still form a basis for the same topology. I know my notion of set "size" doesn't matter because we're working with countably infinite sets, but I just dont know what get's removed y'know?

[lunar yoke]

you know there exists a countable basis. This basis essentially tells you which sets from the given basis you want to choose

[wise walrus]

I just can't figure out how to generally pick something out of the basis and still have the basis be present in all sets of the topology like it needs to be

like, would this proof imply that the countable basis always has some "extra elements"?

[ornate lark]

you know a countable basis exists. play around with the idea of the intersections between the sets in the countable basis and the uncountable basis you started with

[wise walrus]

but the topology is second countable, so doesn't that mean that we're starting with a countable basis and finding the also countable subcollection?

[lunar yoke]

no, second countable just means that there exists some countable basis

other bases might not me countable

generally if you take all open subsets for example, this will be a basis, but might not be countable

[wise walrus]

oh, so there exists at least one countable basis, and this is generally the subbases we're looking for to be the subcollection of the much more prevalent non-countable subbases? That makes a lot more sense.

I'll have to somehow formalize the idea that an uncountable basis will always have "more elements" (in an infinite way, idk) than the countable counterpart, is that the right direction?

then move onto intersection stuff

[lunar yoke]

i was tallking about bases in general, not necessarily subbases

im not sure what you mean by formalizing that an uncountable basis has more elements

You can approach the problem like this

Let $X$ be a topological space and $(C_i)_{i \in I}$ a basis of $X$, where $I$ is some index set.

[gentle osprey]

Phil P

[lunar yoke]

this basis need not be countable

[wise walrus]

yeah, makes sense

[lunar yoke]

If $X$ is second countable, we also have a countable basis $(B_n)_{n \in \mathbb{N}}$ of $X$

[gentle osprey]

Phil P

[lunar yoke]

Now, we need to find a countable subcollection $(C_j)_{j \in J}$ for $J \subseteq I$ countable which is still a basis

[gentle osprey]

Phil P

[wise walrus]

okay yeah, that all makes sense

[lunar yoke]

Since countable unions of countable sets are countable, it suffices to find for every $B_n$ a countable collection of $C_k$'s whose union is $B_n$

[gentle osprey]

Phil P

[long coyote]

Let $f,g:\mathbf{I}\to\mathbf{I}\times\mathbf{I}$ be continuous; let $f(0)=(a,0)$ and $f(1)=(b,1)$, and let $g(0)=(0,c)$ and $g(1)=(1,d)$ for some $a,b,c,d\in\mathbf{I}$. Show that $f(s)=g(t)$ for some $s,t\in\mathbf{I}$.

[gentle osprey]

亜城木 夢叶

[empty grove]

Do you know the Jordan curve theorem?

@long coyote

or could you tell what tools you are allowed to use?

[long coyote]

i don't know the Jordan curve theorem, currently, i have the brouwer fixed point theorem

[long coyote]

thinking to come up a map that maps IxI to IxI, so I have two disks

[bold canopy]

Consider (x, y) -> midpoint(f(x), g(y))

And use brouwer

(might have gotten order backwards)

Actually, that might not work oops lol

[plain raven]

wait does this imply brouwer as well?

in two dimensions

i feel like this can be proven somehow with just the intermediate value theorem but idk how.

[bold canopy]

You can use a similar proof to brouwer to do this problem, idk if theres a simpler way

For (x, y) in I^2, look at f(x), g(y), assume no fixed point, then you can connect with line and project to boundary, you get a map I^2 to S^1

[lunar yoke]

If we have a meeting point for some f and g, would it somehow be possible to say that given a homotopy from f to h, h must also have a meeting point with g?

[bold canopy]

You also have the inclusion of boundary into I^2, so you have a sequence
Circle to disk to circle
This composes to give something homotopic to identity map (so should send Z to Z of the pi1 of the first and last circle) but this is impossible because disk is simply connected

[lunar yoke]

im imagining that this point continuously wanders around during the continuous deformation of f to h

Im really not familiar with homotopy stuff yet

[plain raven]

I continuously wander around too, could be ADHD

[lunar yoke]

but if this works, you could start by looking at the straight lines connecting two pairs of starting points, which clearly have an intersection, and then deform the lines to f and g respectively

[orchid forge]

this is all you need. pick some real number x between c and d, then by IVT there exists t in (0,1) with g(t) = x, and again by IVT there exists s in (a,b) (or (b,a)) with f(s) = x

[plain raven]

dope. i suspected it was true just because it was an essentially 'one dimensional' problem but i didn't know how to formalize it

[orchid forge]

actually my argument doesn't really make sense

i was thinking about projections of f and g onto one of the two copies of I

instead of just f and g

[plain raven]

Damn!!!!!

[orchid forge]

my next choice would be to homotope both curves to straight lines

which someone already suggested

i think you're right though, you can use IVT

[plain raven]

i haven't read the existing proofs but the way i'm thinking about trying to solve this is by using a hairy-ball theorem style argument. like, if f(s) never equals g(t) then the map (s,t) -> f(s)-g(t) defines a map IxI -> R^2 -{(0,0)}, which i think restricted to the boundary of I x I probably has nontrivial winding number, which is a contradiction

but i haven't thought through it

yeah. i think this argument should work.

[ornate lark]

I^2\im(f) is disconnected, if g is continuous and doesnt hit the im(f) then it cannot ever reach (1,d)

but yeah thats an ivt argument

[coral pawn]

Will it?

For c, is it sufficient to say that f^{-1}(-1) has codimension 1, so has dimension 2, and d/dx, d/dy will be the basis elements?

[orchid forge]

(the image of) d/dx and d/dy maybe

under the induced map

but considering this surface is naturally embedded in R^3, you can describe the induced tangent vectors in terms of d/dx, d/dy, d/dz

which might be what they're going for

[coral pawn]

Ah I see. So they want me to consider the inclusion map and compute the pushforwards of those?

[orchid forge]

yes

[coral pawn]

Won't we then have to know what the charts on f^{-1}(-1) are?

The regular level set theorem gives us that f^{-1}(-1) with the subspace topology of R^3 has a smooth structure, but it doesn't say anything about what the smooth structure is other than the fact that the inclusion should be smooth.

[orchid forge]

we don't need anything about a smooth structure here

[coral pawn]

What do you mean?

How else would you compute the pushforward of the inclusion map?

[orchid forge]

i see what you mean, the fact that the regular level set theorem gives an embedding should fix a unique smooth structure on the level set

but you can solve the problem without thinking so technically, by thinking about the tangent space of the level set as a literal subset of the tangent space to R^3

defined by some equation

[coral pawn]

Ah I see

So I can use the orthogonal vectors thing

[orchid forge]

in particular, if z^2 - x^2 - y^2 = -1, then z dz - x dx - y dy = 0, so the tangent space to the level set is the subspace of T_p(R^3) consisting of the solutions to that equation

the orthogonal complement of (x, y, -z) in the standard basis, yes

[coral pawn]

Do you mean (1,0,0)?

[orchid forge]

that should be the case at p=(1,0,0) yes

[coral pawn]

Seems too handwavy to me

[orchid forge]

you can state and prove it precisely

as long as it's clear what you're actually doing

[coral pawn]

Could you show me how to do it?

[orchid forge]

well you first have to decide on your basis for T_pM. you said d/dx and d/dy before, but what precisely do you mean if you're not regarding T_pM as a subset of T_pR^3

[coral pawn]

We would need charts to precisely define those

[orchid forge]

(also i think you meant d/dy and d/dz, right?)

[gritty widget]

i know a groupoid is a category where all the morphism are isomorphisms

i know this is a natual generalisation from the point of view of

[coral pawn]

Does it make a difference?

[orchid forge]

if we want to describe a basis in terms of the pushforward into R^3, yes

d/dx is the normal vector at p=(1,0,0)

[gritty widget]

a group is a category with one object and only isomorphism

is there a geometric picture for groupoids?

is there a notion where groupoids can act on a space

[coral pawn]

Wait

I think I can just state what the basis is and give some explanation without proof

Because it simply states "give a basis for..."

[orchid forge]

yes, but "give a basis" kind of implies some comprehensible description of what the basis is

we know there is a basis

[coral pawn]

We know it would be (0,1,0) and (0,0,1)

And these correspond to d/dy and d/dz

at the point (0,0,1)

[orchid forge]

inside of T_pR^3, yes

since an embedding induces an injective map between tangent spaces, you can just choose the appropriate two vectors in T_pR^3 and pull them back (uniquely) to T_pM

[coral pawn]

Computing the pullback would require us to know the charts

I think I can just leave them in T_pR^3

[orchid forge]

there's nothing to compute, and the charts are given by the embedding into R^3

that's what i was saying before, you can think about all of these things as being literal subsets of R^3 and T_pR^3

without loss of generality

so I would say that "d/dy and d/dz [regarded as elements of T_p(f^{-1}(-1))] form a basis" is perfectly fine

with the proof being that they satisfy the equation defining the tangent space to a level surface

(and span the whole thing)

[coral pawn]

We haven't proved anything like this

About the tangent space satisfying any equation

I know what you mean by that

Like the tangent space is the space of all vectors orthogonal to the surface

But we haven't proved anything like that

We proved it for the sphere I think

But not for surfaces in R^n in general

[orchid forge]

if f : R^n --> R, the tangent space to R^n at f^{-1}(a) decomposes into ker df and its orthogonal complement

[coral pawn]

Which is what really annoys me about the hw in this class. We do all these examples but we almost never prove general results. If we did, then a lot of the problems in the hw would become much less work

[orchid forge]

the subspace ker df consists of directions in which f is constant

so ker df is the tangent space of f^{-1}(a) at p

really, the pushforward under the inclusion into R^n

you can prove this relatively easily with the picture in mind

[coral pawn]

Again, I believe everything you say, but we can't really use this stuff without proof in the hw

And proving all this stuff would make the hw much longer

For b, won't there just be 2 different types of level sets?

One for f^{-1}(0) and one for everything else?

[orchid forge]

f^{-1}(a) is disconnected for a>0

[coral pawn]

Ah I see

Can you help me out with another problem?

[orchid forge]

maybe, but i'm sure someone can if not me

[coral pawn]

What is the best way to do b?

[orchid forge]

first of all, do you see how R^2/~ is a torus, and what the subspace F(R) looks like both in R^2 and on the torus?

[coral pawn]

We know R/~ is the circle and constructing a torus is done by placing a circle at each point on the boundary of another circle circle

So that's why the R/~ x R/~ is the torus

I think F(R) looks like stripes?

Like two stripes on the torus

[orchid forge]

it looks like two stripes on the fundamental domain [0,1] x [0,1]

in particular the line segments (0, 1/2) --> (1, 1) and (0,0) --> (1, 1/2)

[shy moss]

let G be a grupoid then an G-action on a sapce X is a functor G->Aut(X) where Aut(X) is the category with object X and morfism the automorfism of X.

[orchid forge]

if you glue the horizontal edges of [0,1] x [0,1], you get a cylinder, with the boundary circles coming from the vertical edges

then glue the vertical edges (now circles) to form a torus

what happens to those stripes?

[coral pawn]

By horizontal edges do you mean the top and bottom or the sides?

[orchid forge]

i meant the top and bottom of [0,1] x [0,1], but you can glue opposite sides in either order

[coral pawn]

Can you tell me what happens to those stripes?

I'm having trouble keeping track of those stripes when I move things around

wait

I see

[orchid forge]

the bottom edge goes to one circle on the torus, and the left edge goes to the other circle. the path f(R) passes over the bottom edge twice and over the vertical edge once before intersecting itself

so it turns into a curve winding around the torus

[coral pawn]

Yeah

How does this help in answering the problems though?

We would still have to choose charts and stuff on this

[orchid forge]

if you have the picture in mind, it is intuitively clear what F(R) is and why F is a smooth immersion

then you just need to translate that into the formal proof

[coral pawn]

That's what I was asking about haha

Is there a way to do this without coming up with charts are doing all the verification about them being smoothly compatible?

[shy moss]

you mean something like that?

[orchid forge]

you should use the fact that T^2 comes from a quotient

you can think about F as being a map R --> R^2 followed by the projection R^2 --> T^2

like say F = pi \circ G. then to prove F is an immersion, you just need to argue that dG is injective everywhere, and d pi is injective on im dG

[coral pawn]

Uhhhh

I think we may be using different definitions

[orchid forge]

what do you mean

[orchid forge]

i didn't even say "embedding"

a (smooth) immersion is a (smooth) map which induces an injection on the tangent spaces

[coral pawn]

Yes

[orchid forge]

you have a map R --> T^2, to show it's an immersion you just need to show that there is no point where dF(dx) = 0

[coral pawn]

Yeah so we showed F is a smooth immersion in a

[orchid forge]

but T^2 is a quotient R^2 / ~, so we can understand the tangent space of T^2 by starting with R^2 and applying the map induced by the projection

[coral pawn]

Now we have to show that Im(F) is an embedded manifold

[orchid forge]

okay, i didn't realize you were only asking about b

[coral pawn]

Sorry I probably should have made it more clear

[orchid forge]

for b it's the same idea, use the fact that T^2 is a quotient of R^2

consider the subset [1/2, 3/2] x [0, 2]

on which f(R) looks like the diagonal

projecting to the torus gives the entire space f(R)

and the projection is one-to-one everywhere except along the edges

the only points of the line which are identified are the endpoints

[coral pawn]

I'm not entirely sure I follow

Aren't you just coming up with charts?

[orchid forge]

I guess, but there are natural charts associated to a quotient

if you want a fundamentally different argument, you can use the regular level set theorem again

you can describe im f explicitly in terms of particular coordinates for T^2 = S^1 x S^1

im f is a curve that winds around one circle twice as it winds around the other once, so if (s,t) are (properly chosen) radial coordinates for S^1 x S^1, im f is the curve s = 2t

hence it is the level set g^{-1}(0) of the function T^2 --> R given by g(s,t) = s-2t

but if you want to use properties of F, i think the intention is to understand how to prove things by passing to a quotient

[coral pawn]

@orchid forge Will F(R) be an open subset of T^2?

[orchid forge]

no

F(R) is that curve winding around the torus twice

topologically it is just S^1

[orchid forge]

use the coordinates induced by R^2, namely at each point you have d/dx and d/dy which are both never 0, and df sends d/dx --> d/dx + (1/2)d/dy, which is never zero

oh that's still for a

but once you have an injective immersion of a compact space, it is automatically an embedding

so, take f and restrict to [0,2], now it's injective except at the endpoints

so it induces an injective map S^1 --> T^2, and it is an immersion because f is (and df is the same map at the endpoints)

this is an embedding S^1 --> T^2 whose image is im(f)

and what i was getting at by suggesting you look at the rectangle [1/2, 3/2] x [0, 1], this is a compact domain on which f attains all its values

for the other approach (level set theorem) the radial coordinates you want are just the x and y coordinates of R^2, taken mod Z

[coral pawn]

What even is this problem?

What does a mean?

[bright acorn]

$A \in \mathbb{R}^{n^{2}}$ is just a general square matrix and we want to compute the differential of the map $F : R^{n^{2}} \rightarrow \text{Sym}(n)$ at such a square matrix $A$.

[gentle osprey]

MisterSystem

[bright acorn]

Or are you asking about what does equation a mean?

that is indeed the differential of the map F at the matrix A in the direction of X.

And you are supposed to compute that

[coral pawn]

How is the differential computed using the paths?

[bitter yoke]

Do you know how the tangent space can be defined as an equivalence relation on the set of paths through a point?

[coral pawn]

Yeah

[coral pawn]

We have this

[bitter yoke]

yeah so you can also define the differential using this definition

[coral pawn]

Yeah

What I don't see is why does it suffice to check what the differential is on the paths A + Xt?

[bitter yoke]

Because all other paths are equivalent to one of this form

[coral pawn]

Oh okay

That kind of makes sense

[bitter yoke]

I mean, in R^n, the tangent space is canonically R^n, so you only need to consider the straight lines through the points

[coral pawn]

Because of the linear approximation

[bitter yoke]

To get the whole tangent space

[coral pawn]

So the computation looks something like this

Let $\gamma$ be the curve $A + tX$. Then $dF_A([\gamma])={f\mapsto \frac{d}{dt}|_{t=0}(f(F(\gamma(t))))}$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

this equals ${f\mapsto \frac{d}{dt}_{t=0}f(A^TA+t(A^TX+X^TA)+t^2X^TX)}$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

How is one supposed to make sense of this?

[orchid forge]

if you just compute the usual derivative lim_{t \to 0} (F(A + tX) - F(A))/t you will end up with A^TX + X^TA

the fact that A + tX is actually a path in R^(n x n) is why we needed to identify R^(n x n) with its tangent space, but this is just "a path through A with initial direction vector X"

so this derivative says that the map dF sends the tangent vector X to A^T X + X^T A

remember that the differential / pushforward of a map is just "the induced map on tangent vectors," which is intuitive: tangent vectors v in T_pM act on functions h : M --> R^n; given a function f : M --> N, to get a functional acting on g : N --> R^n, you precompose with f to get (g \circ f) : M --> R^n then apply v

it doesn't matter if you define the functionals as "derivations of germs" or as "derivative along curves" nothing else really changes

[coral pawn]

But that's not what we're computing. We need to compute lim_{t \to 0} f(F(A + tX) - f(F(A)))/t

We would've been able to apply the chain rule, but the thing inside f is a matrix

[orchid forge]

Given a functional (tangent vector) X, dF(X) is a functional (symmetric matrix) that sends f --> dF(X)f = X(f \circ F) = X(f(A^TA))

now what I mean by X(f(A^TA)) is "the derivative of f(A^TA) in the direction of X"

by the definition of X as a linear functional

and we can write this in terms of a curve with initial velocity X, as $\frac{d}{dt}\big|_{t=0} f((A +tX)^T(A+tX))$

[gentle osprey]

Kogasa

[coral pawn]

Okay, but how do you compute this derivative?

You'd need to compute $\frac{d}{dt}_{t=0}f(A^TA+t(A^TX+X^TA)+t^2X^TX)}$

[gentle osprey]

Finitely Many Bananas
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[orchid forge]

you don't need to compute it, you just need to recognize it as "the derivative (at t=0) of f : Sym --> R along a curve with initial velocity A^TX + X^TA and initial position A^TA = F(A)"

which is precisely what A^TX + X^TA is, as an element of T_(F(A))(Sym); a functional acting on functions f : Sym --> R

generalizing a bit, this tells us that dF|_A can be computed by just asking what F does to paths with initial vector X and initial position A

more precisely, we just need to compute (d/dt) (F(A + tX) - F(A))/t, which is why i had written this initially

for a sanity check, it should be clear that whatever dF does to X should really not depend on a choice of arbitrary function g : Sym --> R

and the induced map dF sends tangent vectors to tangent vectors, so given a curve L(t) through A with velocity v, it ought to send v to the velocity of the curve F \circ L(t)

all I did was show that this is compatible with the definition of tangent vectors as functionals

[spare hinge]

Hello maths people ! I am having trouble in showing that the Riemann surface $S:={(z,w)\in\mathbb{C}^2;w^2=z(z-1)(z-\lambda)}, \lambda\neq 0,1$ can be compactified uniquely, has anyonee ever done such things or has any good reference for this?

[gentle osprey]

Abriel

[spare hinge]

I am currently trying to find charts close to $p_{\infty}:=(\infty,\infty)\in \hat{\mathbb{C}}^2$ but this is painful and I feeel there is a simpler way

[gentle osprey]

Abriel

[orchid forge]

it may help to work in $\mathbb{CP}^2$ with coordinates $[z, w, \zeta]$ instead, with the zero set of $w^2\zeta - z(z-\zeta)(z - \zeta\lambda)$

[gentle osprey]

Kogasa

[orchid forge]

with $\zeta = 1$ we have the original equation, so each point of my zero set $[z, w, \zeta] = [\frac{z}{\zeta}, \frac{w}{\zeta}, 1]$ corresponds to a point $(\frac{z}{\zeta}, \frac{w}{\zeta}) \in S$

[gentle osprey]

Kogasa

[spare hinge]

I tried this and tried looking then for the intersection with ${\zeta=1}$ but then which set gives $p_{\infty}$?

[orchid forge]

provided that $\zeta \ne 0$

[gentle osprey]

Kogasa

Abriel

[orchid forge]

the only points in the projective zero set that don't correspond to a point in S are the ones with $\zeta = 0$, but that makes the equation just $-z^3 = 0$

[gentle osprey]

Kogasa

[orchid forge]

so it's the point $[0, 1, 0] \in \mathbb{CP}^2$

[gentle osprey]

Kogasa

[spare hinge]

It makes sense, I didn't really digged a lot into this method because this is not in the lecture

I don't reeally know what the professor's waiting for

Thank you for describing this method !

[orchid forge]

no problem

[spare hinge]

Actually this is what I wanted to do from the beggining but since this is not in the lecture, I refrained myself from doing it haha

[orchid forge]

you can probably do it directly, but i'd have to think about it

[uncut surge]

Let $M$ be a smooth manifold. I have these shitty ass diagonal sets $M^q_k := {(x_1,\dots,x_q) \in M^q : |{x_1,\dots,x_q}| \leq k}$, which are, by definition, the points in $M^q$ which have no more than $k$ different points of $M$ in their coordinates (if $k = 1$ or $k = q - 1$, these are called the thin and fat diagonal, respectively, but I'm also interested in the in-between cases). These aren't manifolds anymore in general, and I'm generally not aware of them admitting any nice structure other than being metric spaces as a subset of a metric space. This is very vague, but does anyone have anything to say about these things? Is there some nice literature where they're treated, or some natural setting to put them into?

[gentle osprey]

Lartomato

[gritty widget]

shitty ass diagonal sets

[uncut surge]

SAD! sets

I've been trying to stay professional

Maybe I'm overengineering, I really only want to show dumb statements like "if M is contractible, so is M^q_k for all q and k", but I probably don't need any funky topological properties for that if I try to explicitly construct a homotopy

[gritty widget]

What’s M^q

[shy moss]

i think is just the product of M q-times

if someone X is path connected, local path connected and pi_1(X) is finite, then is X simply connected?

[gritty widget]

RP^2 is path connected and locally path connected (it's a manifold), but its pi_1 is Z/2Z

[orchid forge]

You can build more examples with this idea, glue a disk to a circle with some torsion (here two twists) and then glue those

[uncut surge]

yeah exactly

[gritty widget]

Wow they are weird sets

[uncut surge]

lmao

They're not that bad, it's easy to get some intuition if you think about $\mathbb{R}^3_1$ and $\mathbb{R}^3_2$ and try to visualize those; in other words, the diagonal line $(x,x,x)$ in $\mathbb{R}^3$ and the set of elements $(x,x,z), (x,y,y), (x,y,x)$ in $\mathbb{R}^3$. With the second one you end up seeing that this can'T be a smooth manifold anymore, but it's the intersection of three planes (intersecting in precisely the diagonal line (x,x,x))

[gentle osprey]

Lartomato

[uncut surge]

I think they're cool and interesting but also annoying because it feels like there should be literature on them that I can't find

[gritty widget]

Yeah for R it’s not too bad

But even Tori get confusing

[surreal ocean]

I’m learning about bases of a topological space. I read that for some set $X$, $\beta$ is a base for a topology on $X$ iff the intersection of each pair of members of $\beta$ is a union of members of $\beta$. My question regards the case where the intersection of any two members of $\beta$ is in $\beta$. Can the intersection of sets be considered a union of itself, hence satisfying the condition?

[gentle osprey]

Justinflys

[surreal ocean]

ok thx just wanted to verify that

[orchid forge]

You need to stipulate that the intersection is in the basis

Otherwise it doesn't satisfy the condition

In general bases need not be closed under finite intersections, but the topology they generate will be of course. The intersection may be the union of no less than infinitely many basis elements, for example the basis for R^2 generated by open balls where A, B are distinct, intersecting balls

[orchid forge]

Are you familiar with Schubert varieties

[uncut surge]

Sort of, do you think they're related?

@orchid forge I sort of see a very rough similarity between them and my problem, but not too sure

[orchid forge]

just a hunch, I'm not sure either

[orchid forge]

if M=R, a point in your space with precisely k different coordinates corresponds to a choice of point from each a bunch of orthogonal lines, which is equivalent to a choice of (k-1)-plane

[finite heath]

Which sets are closed sets in the finite complement topology on a topological space X?

Would you guys say my work for this question is sufficient?

It does not say prove

[gritty widget]

you're missing a set

[finite heath]

oh, the empty set is open in the finite complement topology

so X \ empty set (or X) is finite and closed as well

?

and empty set is not closed

nor finite

@gritty widget im new =[

[gritty widget]

remember a set in the finite complement topology is open if its complement is finite or if it's empty

[gritty widget]

closed, yes, but not necessarily finite

[finite heath]

Hmmm

let me think about it

be back in 20 minutes sir.

I gtg rq

[gritty widget]

your statement that "a set is open in the FCT if its complement is finite" is almost true. you need to include the empty set here

[finite heath]

Right

i updated it to

FCT: topologyt whose open sets are the empty set :D and every set in R w/ a finite complement

[finite heath]

True --> empty set is open so we know X, it's complement, is closed... but finite? we do not know

[gritty widget]

correct

[finite heath]

How does that look, sir

[empty grove]

It kinda looks like you haven't proved that these are the only closed sets

If you had a set S, neither finite not the whole set, can it be closed?

[coral pawn]

How would I show d?

[orchid forge]

if M is a level set of f : A --> B, then the tangent space T_pM is (isomorphic to) ker df (at p)

we have a formula for dF, so what is ker dF at A=I

[coral pawn]

I don't think we have proved that

[orchid forge]

not much to do but prove it then

it's not a tricky proof or anything, it's pretty much just following the definitions

[coral pawn]

Is there any other way to do d?

[orchid forge]

how else do you intend to characterize the tangent space to a level set

[coral pawn]

Is this result true in general?

[orchid forge]

yes

under the conditions of the regular level set theorem

[coral pawn]

Ah okay

Idk I don't think this is what the professor wants us to do

[orchid forge]

I would be genuinely surprised if there was a different intent

it's not even really a theorem, just an observation

if f : A --> B and M = f^{-1}(q), then a priori it lives inside A, and T_pM is a subspace of T_pA

we know that the tangent space can be characterized by the derivative along a curve

so you consider all the curves in M which go through p

f sends all these curves to q

[coral pawn]

So you're saying that the map T_p(M) --> T_p(A) --> T_f(p)(B) is the zero map?

[orchid forge]

yes

[finite heath]

May I ask a lil question 🙂

if not i wait

[coral pawn]

go ahead

[orchid forge]

conversely, everything in ker dF also lives in T_pM

because that would mean there's a curve along which f is constant

hence equal to q

[finite heath]

is (a, b] open or closed in the lower limit topology?

[coral pawn]

How do you show that though?

Like can you walk me through the proof of it

[orchid forge]

use the definition of dF in terms of curves

[coral pawn]

Using the derivation definition of tangent space if possible

I'm not very comfortable with the curve definition

[orchid forge]

to define a tangent vector as a derivation of smooth germs is an abstraction of the idea of "partial derivative in a direction"

a vector v is the derivation "partial derivative in the direction v"

differentiation along a curve is just a parameterized version of this

where now v is just the initial velocity of some curve

you can always find a curve through a point with a given initial velocity (do it locally in R^n then transfer it onto the manifold with the chart)

so they're really the same thing, it would be good to become familiar with both definitions

[orchid forge]

it could be open, closed, both, or neither. how would you normally show it's open?

[coral pawn]

I see it!

[finite heath]

if (a, b] is comprised of a union of basis elements [x, y) it is open

[orchid forge]

right, so what's the issue? (a,b] contains b, so if we covered it with basis elements [x, y), one of them would have to contain b

but if b is in [x, y), then so are points to the right of b

[finite heath]

i don't think the union of 2 basis elements [x, y) and [x2, y2) can exactly equal (a, b]

so it is closed

[orchid forge]

it doesn't have to be a union of 2 elements, or even finitely many

it can be a union of infinitely many basis elements

also, "open" and "closed" are not mutually exclusive / inverse

you can have sets that are both open and closed

[finite heath]

o word

okay

so idk how to prove it's open, closed, both, or neither then :\

or not even prove but

intuitively tell

[orchid forge]

if you want to see if (a,b] is open, try covering it with basic open sets

(basic open sets which are contained in (a,b])

[finite heath]

let's say

for every x in A = (0, 1] there's a B = [c, d) s.t. x in B subset of A

let's say x is 1

no [c, d) that can contain 1 and be in (0, 1]

[orchid forge]

right

[finite heath]

so it is closed right?

that means it's closed

[orchid forge]

that means it's not open

[finite heath]

or does that just mean it's not open

oh word

[orchid forge]

you can't write it as a union of basic open sets

because any such union would also contain points that are outside it

[finite heath]

true

now, to tell if it is closed or not

[orchid forge]

to see if it's closed or not, you just have to do the same thing to the complement

[finite heath]

oh right!

[orchid forge]

remember closed means "the complement is open"

[finite heath]

and vice versa yes?

[orchid forge]

yes, that is the definition of closed

[finite heath]

but, let's say it's closed and not open

that breaks the definition

because then the complement of said closed set cannot be open

[orchid forge]

why?

[finite heath]

because closed means "the complement is open"

oh

[orchid forge]

yes. so a set which is closed and not open, has open complement which is not closed

[finite heath]

im dumbo

[coral pawn]

Kogasa how would you show the reverse inclusion?

[orchid forge]

Remember $f : A --> B$ and $M = f^{-1}(q)$. Suppose $v \in \mathrm{ker} df$. Take a curve $\gamma(t)$ through $p := f^{-1}(q)$ with initial velocity $\gamma'(0) = v$. Then $f(\gamma(t))$ is a curve through $q$ with initial velocity 0.

[gentle osprey]

Kogasa

[coral pawn]

Yeah, I'm guessing we need to use the chain rule now?

But can we do that?

[orchid forge]

hmm

actually i think i want to do it another way

Knowing that $T_pA \cong T_pM \oplus T_pM^\perp$, we can just show that any $v \in T_pM^\perp$ must have $df(v) \ne 0$

[gentle osprey]

Kogasa

[coral pawn]

You're using all these results that I need to prove haha

[orchid forge]

this is just linear algebra

a vector space decomposes into the direct sum of a linear subspace and its orthogonal complement

so "kernel" and "everything but the kernel"

[coral pawn]

Yeah okay

[orchid forge]

I'd recommend just looking up a cleaner proof, lol. I want to say if f(gamma) is a curve with initial velocity 0, it's homotopic to a curve that is 0 on some small time interval (-epsilon, epsilon), then this pulls back to a curve lying in M

[coral pawn]

Don't wanna use "homotopic" in my proof haha

That's for next semester when we do algebraic topology

People on the server have hyped it up as being super nice

[orchid forge]

it is great

oh wait, it's just a consequence of the dimension

[coral pawn]

How?

[orchid forge]

remember by assumption (regular level set) df is a surjection T_pM --> T_pN, so its kernel has dimension dim M - dim N

[coral pawn]

Yeah

But we don't know the dimension of O(n)

So I don't think we can use that

[orchid forge]

but we know the dimension of f^{-1}(q)

is also dim M - dim N

[coral pawn]

Can you explain this to me again?

We have that the kernel has dimension n(n-1)/2

[orchid forge]

I'm not talking about a specific level set

i thought you wanted to know why the tangent space to a level set is ker df

once we know that, the problem is simple: dF = A^TX + X^TA, so at A=I we have dF = X + X^T, and ker dF is the subspace consisting of X such that X + X^T = 0

now $T_pM \subset ker dF$ because as we saw, any curve in M maps to a point, so the differential the 0 map when restricted to $T_pM$

[gentle osprey]

Kogasa

[orchid forge]

and by dimension counting the reverse inclusion holds

[coral pawn]

Got it

So we have to use the think in the level set theorem about codimension

Thanks for the help!

[orchid forge]

no problem, hopefully this makes some sense

[coral pawn]

It did

[uncut surge]

@orchid forge somebody upvoted my post on fat diagonals on stackexchange lmao, one year after i posted it

was that you

in either case, i am now satisfied with my shitty diagonal sets; if U is contractible (or a finite disjoint union of contractible open sets) then all the diagonal sets U^q_k are contractible (or finite disjoint unions of contractible open sets)

constructing deformation retractions explicitly was so kawaii

why is math not always so nice

[shy moss]

let X be an affine variety and I(X) his ideal, how can is how that V(I(X))=X?

[empty grove]

A variety is by definition V(J) for some ideal J

So you need to show that V ∘ I ∘ V = V

Try from here, it becomes an exercise in language and set theory

[shy moss]

thanks

[orchid forge]

nope

[pearl holly]

So when I have Hom(Q, G), Q being the rationals and G a group, I know that an element in this is a homomorphism from Q to G. But should I like see Q as a group under addition or under multiplication?

[orchid forge]

Q isn't a group under multiplication since 0 doesn't have an inverse

[pearl holly]

oh yeah lmao

yeah okay lmfao. Thanks btw!

[quasi forum]

I feel silly for asking this, but I am struggling to figure out how to approach this. May I please have some assistance?

On part a)

[orchid forge]

Consider the basis for R consisting of balls (of radius epsilon, centered at x_0)

what would the preimage of such a ball look like?

[empty grove]

Suppose you are checking continuity at (x,y). Informally suppose x is close to x', y to y', then can you relate d(x,y) to d(x',y') using properties of the metric?

If you don't want to use bases and such

[orchid forge]

oh yeah that's a better argument

[quasi forum]

Well, you could use the max metric on XxX

[empty grove]

Yeah this is all under the assumption that you know the product metric

[quasi forum]

So no then. Hmmm

[empty grove]

All the usual ways of defining a product metric are equivalent so you can use the max metric

You are checking continuity, so both the domain and codomain must be metric spaces(/top spaces)

So the first thing you should think about when given this statement is what the metrics are

[quasi forum]

So the metric on R is the Euclidean metric, and the metric on X x X has the product metric

[empty grove]

Yes

[quasi forum]

I hate to say it, but I'm not sure

[empty grove]

With the understanding that product metric is not unique, and you can pick any one that satisfies the conditions

(if you have not seen this, then use the definition of the metric on the product that you have seen)

[quasi forum]

I think my brain just needs a break.

[kind cedar]

Considering the projetive space $\mathbb{R}P^n$, to prove it is second-countable I just need to show the projective map $\pi$ is an open map. But $\pi$ is an open map if given an open U in $\mathbb{R}^{n+1}$, the set of all equivalent points to points in U is also open. But this set includes the origin, and doesn't seem to be open at that specific point. Where am I doing something wrong?

[gentle osprey]

Necrowizard

[kind cedar]

Ah, I was missing that argument! Thank you

[kind cedar]

Okay, now I have questions that are even more basic and I feel stupid, but I need to ask.
I'm required to show that the following sets admit no differentiable manifold structure (no atlas?) compatible with the standard topology.
a) [0,1) in the real line
b) A cross + in R^2
c) the long line from the Munkres book.

I honestly have no idea what I'm supposed to do here. Demonstrate that no charts are possible? On case a) I just wrote "no open set contains 0 so it's not Hausdorff", but I'm not sure this is enough...

on b) I'm simply stuck and on c) I didn't understand the concept of the long line

[hollow harbor]

What do you mean no open set contains 0

[0, r) is open in [0, 1) for any r less than 1

However you're right to focus on that

Because the point is that no neighborhood of 0 is homeomorphic to a unit ball in R

So it might be good to try and prove that (that a half-open interval cannot be homeomorphic to an open one)

That shows there's no chart at 0

Similarly, can you identify a point of the + where there's no neighborhood of the point that could possibly look like a single interval or disk?

[kind cedar]

a semi-open set is open?

[hollow harbor]

Well, it's not open in R

But its open in [0, 1)

[0, 1) is its own space

[kind cedar]

ah, I see. so any open balls centered on 0 will not contain negative points, but those aren't in the space from the beginning so no contradiction, i guess?

[kind cedar]

I still don't understand how to prove the half-open interval cannot be homeomorphic to an open one

[hollow harbor]

Here's an idea. What happens if you remove a point from an open interval?

[kind cedar]

the problem is obviously on the (0,0) but can't I just chart it with Id to R2 ?

[bold canopy]

If you had a map [0,1) -> (0,1) that is a homeomorphism, where would you map 0?

[hollow harbor]

Charts have to be homeomorphic maps with open sets in some R^n.

[kind cedar]

It becomes a closed one? Or a disconnected, not sure

[hollow harbor]

The + in R^2 is not open

[hollow harbor]

Disconnected is right

Is it true for [0, 1) that any point I remove disconnects it?

[kind cedar]

that's a good reasoning

[kind cedar]

removing the zero doesn't disconnect it. However since I'm not well educated in math, I don't know any thing involving disconnection. I'll look in the book to see if there is something about it

[hollow harbor]

Yeah. This turns out to be a way to prove they're not homeomorphic.

There might be an easier way