point-set-topology

[abstract saffron]

In French, Compact means quasi-compact and hausdorff

That's totally true

I'm blessed to know both worlds 😄

[void gazelle]

hi, guys, is it true that if Y\subset U is an irreducible closed subset of U, and U\subset X is open, then the closure of Y in X is irreducible?

[languid patrol]

The closure of an irreducible set is irreducible. This is a good exercise!

It doesn't depend on U at all really.

[void gazelle]

oh i see, Y\subset U is irreducible, then Y\subset X is irreducible, then closure of Y in X is irreducible

Thanks!

[tidal lynx]

is it true that the image of an injective continuous map R -> R is an interval

i’m dumbed that’s they’re for continuous functions in general

[vast estuary]

(cat on keyboard, sorry)

[tidal lynx]

by a combination of some evt ivt

[languid patrol]

It's just ivt.

As a subexercise: show that a connected subset of R containing two points contains all the points between them.

That's basically the whole problem.

[coarse night]

Hey that’s my job

[tidal lynx]

isn’t this like stronger ivt

well hmm there’s no mention of function

[languid patrol]

It depends on what you consider ivt to be. But since the exercise I gave is one line it is not that much stronger than ivt.

[tidal lynx]

I was talking about version 2 here, which sounds very familiar

[languid patrol]

Yes that would immediately solve your problem.

[tidal lynx]

as in the subexercise right

[languid patrol]

As in the question that you asked before

[tidal lynx]

oh uh

that’s true without the injective part right

[languid patrol]

Without the injective part it could be a point.

[tidal lynx]

oh and that’s continuous too yikes

can we say anything similar about smooth curves in the complex plane

i.e. injective continuously differentiable maps from a closed interval real [a, b] to C with non-vanishing derivative

[languid patrol]

It depends on what you mean by "anything similar"

Maybe the analogue is about f: D \to C holomorphic nonconstant. But the conditions are much stronger of course.

Where D is the open disc

[tidal lynx]

I see

[languid patrol]

For the complex numbers ivt says that "the image is connected" but there are a lot of paths in the plane, it's not clear what you'd want to claim about them.

[tidal lynx]

what is the point of identifying smooth curves with each other

I mean like this

Well tbh it says why here “defining curves independent of parameterization”, but I don’t understand the equivalence relation

“This can be done by considering equivalence classes of smooth curves with the same direction”

?? what does this mean

[pseudo coral]

does any1 have a copy of topology by Hocking, Young by chance

[languid patrol]

Yes

[pseudo coral]

could you help me on the details of theorem 4-24 on page 180 @languid patrol im a bit stuck on how to define h2,h3,h4

so from the figure, it seems as though h2 is only fixating on x1 but in what sense

im trying to prove pi_2(X,x_0) is abelian by brut force

seems like its a 4 stage process? continuously deforming one product into the other

assuming n=2, the text defines a homotopy h1:I^2 x I \to X via f(2x1,2x2/2-t) on 0<=x1<=1/2, 0<=x2<=1-1/2t and g(2x1-1,(2x2-t)/(2-t) on 1/2 <= x1<= 1, 1/2t <= x2<= 1

so many variables lol and so many constraints lol

[pseudo coral]

does h2 := f(2x_1,x_2) for x_1 \in [0,1/2] and := g(2x_1-1,x_2) for x_1 \in [1/2,1]?

[pseudo coral]

or is it f(2x1/2-t,x2) and g(2x1-t/2-t,x2)

[next crystal]

for part (b), I dont see how the closure of I is [1,3]. I understand why [1,3] is contained in the closure, but for x \in [0, 1), any basis element containing x must look like (m, n) where m<x and n\geq 1 (where m,n are in Z). Then (m,n) intersects I, so shouldnt the closure be [0,3]?

[unreal stratus]

I don't rly understand your reasoning

Hm

[next crystal]

im showing every basis element containing x intersects I

so x is in the closure

[unreal stratus]

What about (-2,1)

Oh waiy

[next crystal]

yeah that intersects I

[unreal stratus]

Sorry I misread

Yes they made a mistake since like

[1,3] doesn't even contain I

[next crystal]

okay thats what i thought

bc i got the same final answer for the boundary so it had to have been a typo

[obsidian spoke]

for this do we have to show that if U is open in X x Y then U is open in X' x Y'?

[novel acorn]

I'm confused as to how Neukirch managed to find two distinct elements of Gamma in that neighbourhood

Like the intersection is non zero but how does that imply the existence of two distinct elements in the intersection?

[nocturne basalt]

any nbhd of 0 contains some other element of Gamma

[novel acorn]

Why need this be true

Gamma is any discrete subgroup of R^n here

So like in R i can take Gamma = Z and any neighbourhood around 0 only need contain 0
It doesn't need to contain 1 or any other integer

[nocturne basalt]

well the contradiction comes after assuming it is not closed (the closure contains extra points)

[novel acorn]

Okay i mean yeah
But this only implies that any nbhd of x has a non-empty intersection with Γ
But why does that intersection have to have at least two distinct elements in it

[civic verge]

I am pleased to inform you that I have lost the partial of analysis in several variables, the professor will do the exam again, I am afraid I do not feel well for that, do you think it would be a good idea to talk to him to let me a range of exercises to prepare for the exam?

[deft bear]

So I'm working through Tverberg's proof of JCT, and he gets to "It is clear that Ni consists of 2 components". How would I show that?

[gritty widget]

wanna post the proof so people actually have any idea what you mean?

[deft bear]

I end up finding a different proof that does this step better, but here it is

[obsidian spoke]

Could someone explain how we can compare the topologies

[nocturne basalt]

I mean you just want to show inclusion right

I think you need to first consider the subspace topologies on X, X' and Y, Y'

[obsidian spoke]

is X/X' a quotient?

[nocturne basalt]

no I just mean and

changed it to commas

[obsidian spoke]

say X and X', Y and Y' are disjoint

then X x Y isn't even a subset of X' x Y'

so it cannot be open

[abstract saffron]

You don't need it be a subset

Basically you endow X' x Y' with subset topology. In that case, X x Y would be empty in the subset topology of X' x Y', which is indeed open

[obsidian spoke]

Ohhhhh

so X and X' are open sets in a top. space A with topologies T and T'?

[abstract saffron]

Yeah

This exercise is more about warping your mind around abstract definitions more

Just follow all the dot in definitions, you'll be fine

[obsidian spoke]

yeah this confused the hell out of me lol

[obsidian spoke]

so what I have is that a basis of topology on $X$ should look like ${E \cap X \mid E \in T}$ and on $X'$ it is ${E \cap X' \mid E \in T'}$, in $Y$ it is ${E \cap Y \mid E \in U}$ and finally on $Y'$ it is ${E \cap Y' \mid E \in U'}$ so the basis for the product topology on $X \times Y$ looks something like ${(A \cap X) \times (B \cap Y) \mid A \in T \text{ and } B \in U}$ and similarly on $X' \times Y'$ it is ${(A \cap X') \times (B \cap Y') \mid A \in T' \text{ and } B \in U'}$. So taking a basis element of $X \times Y$ we get $(A \cap X) \times (B \cap Y)$ but $A$ is also in $T'$ and $B$ is also in $U'$ hence $A \cap X'$ and $B \cap Y'$ are open in their respective topologies meaning that $(A \cap X') \times (B \cap Y')$ is open in $X' \times Y'$

Does this look right?

[gentle osprey]

FamilyFriendly

[obsidian spoke]

I'm ngl I still dont quite understand whats happening here

[pseudo coral]

if any1 has a copy of Young Hocking Topology and would like to help me in finding the remaining homotopies to show pi_2(X) is abelian @ me plz

[feral copper]

Hatcher shows that \pi_n is abelian for n>=2
The proof is actually very visual (you use a diagram to 'construct' the homotopy between fg and gf)

[pseudo coral]

I need to construct the 4 homotopies explicitly though:/

is the second homotopy the following

[feral copper]

You could use Eckmann--Hilton too: https://stanford.edu/~sfh/homotopy.pdf

[pseudo coral]

$h_2(x_1,x_2,t)=f(2x_1,x_2)$ and $g(2x_1-1,x_2)$

[pseudo coral]

we haventlearned that yet :/

[gentle osprey]

MyMathYourMath

[pseudo coral]

wait but that doesnt use t shux

$h_2(x_1,x_2,t)=f(\fr)$

$h_2(x_1,x_2,t)=f(\frac{2x_1}{2-t},x_2)$ for $x_1 \in [0,1-\frac{1}{2}t]$

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

does this map 4-15 to 4-16 here:

it doesnt look like it :/

[obtuse meteor]

If I have a Tychonoff space B, I know that the R(B)-Module of sections of a vector bundle xi splits as S(xi) = Q + Q’. How can I show that dim Q is locally constant?

Calling F_b(Q) = {q(b) | q in Q}

I think I want to show that F_b(Q) intersect F_b(Q’) is zero. But I can’t quite figure out how to do this

(It also of course might not be true)

[nimble portal]

So just to be clear this is a chart from the graph of a function to some subset of the real line? Unless x & y are vectors

So I guess not always huh lol

[hidden crag]

phi does not map to the real line

[nimble portal]

Yeah

It would if our graph was on R x R right?

Not necessarily the entire real line but a subset of it at least

[hidden crag]

yes

[nimble portal]

Okay

So I understand the definition just fine

But I'm having trouble connecting it to the intuition for what I think a chart is supposed to do/be for/represent

[hidden crag]

try working this out for low dimensional examples

if you draw some cont. function R->R in a two dimensional coordinate system you will see why this looks like R

[nimble portal]

Like the point of a chart is that we can directly map a region on our manifold to Euclidean space, and using the coordinate functions of our chart we can get a rough idea of "where we're at" on the manifold right?

Like looking at a map of a city to find where you are

But this chart only gives us the first coordinate of where we're at

I'm assuming the use of the chart is to establish a local coordinate system

Oh wait I guess it does that if we're in R x R then once we know our local position x we can calculate our relative position on the manifold with (x, f(x))

Same idea for higher dimensions

[unreal stratus]

Hm why are you relating charts with these graphs

Or am I missing something

[nimble portal]

Lee's just giving an example of a chart

[unreal stratus]

Sure

[nimble portal]

On the graph of a function f: U -> R^k, U in R^n, which is a manifold

I'm switching books I like the way this guy talks

[gritty widget]

tu was recommended for a reason

[nimble portal]

People also recommended this book though

If we were in R^3 for example then U_i^+ would be the first octant?

Oh no lol not every coordinate has to be positive

What is B^n supposed to be here?

Just a subset of S^n? And u is an n-dim vector in S^n

Also doesn't U_i^+ cap S^n union U_i^- cap S^n partition S^n?

[obtuse meteor]

B^n is definitely the n dimensional unit ball without boundary

[nimble portal]

I see

Gah I'm confusing myself again with the stuff about dimensions and n-sphere vs n-ball

So the unit 2-sphere lives in R^3 but it's a 2D manifold

[nocturne basalt]

yes

[nimble portal]

Whereas the unit 2-ball lives in R^3 but it's a 3D manifold

[hidden crag]

no

[nimble portal]

fugg

[hidden crag]

D^2 is the disk in R^2

with boundary S^1

[nimble portal]

Right

[hidden crag]

i mean in these examples you can see it

take a point on the sphere S^2

cut out some part around it

flatten it a bit

[nimble portal]

I understand the idea intuitively

[hidden crag]

looks like a patch of R^2

[nimble portal]

Yep

[hidden crag]

hence why it's 2 dimensional

i also don't quite agree with what you wrote about charts earlier

[hidden crag]

here

[nimble portal]

Why?

[hidden crag]

but i am too tired to elaborate and i will type more about this tomorrow

you're not really navigating your mfd in that sense

[nimble portal]

Are there any simple YouTube videos or quick reads I can go through to more-properly understand the n-sphere and n-cube and whatever other important similar spaces there are?

Emphasis on quick read, just to understand where the concepts are different, not do much math with them 😭

[hidden crag]

...

[nimble portal]

no flame pls I know I really should just pick up a topology textbook

[hidden crag]

if you're not willing to properly learn about some of the most intuitive examples of manifolds idk

yes you should

[nimble portal]

What does "properly learn" mean?

Do exercises with them?

[unreal stratus]

Not really sure what you mean by "properly understand the n-sphere"

[nimble portal]

If that's what it takes to leave with a proper understanding then I'll do it I suppose, I just don't want to spend much time on it

[thorny agate]

Exercises, examples + counter examples, knowing the common tricks in proofs, etc

[nimble portal]

Like

[unreal stratus]

Well like you can go anywhere from the definition to like

idk

As hard as you want lol

[gritty widget]

oh you understand spheres? give me a positive-curvature metric on S^2 x S^2

[hidden crag]

compute the homotopy groups

good exercise

[nimble portal]

I just understand the 2-sphere as a ball in R^3 with the interior removed, just the surface

[unreal stratus]

That was correct except idk why you said ball instead of sphere

But yeah

[coarse night]

Cohomology ring

[nimble portal]

So is the 2-ball the same thing but with interior?

[swift fjord]

hint: use the adams spectral sequence

[unreal stratus]

Stfu nerds

[coarse night]

Küneth

[unreal stratus]

2-ball should be the disk in a plane i guess

[nimble portal]

I see

[unreal stratus]

Also I would'nt use the word interior here

[nimble portal]

You know what I mean though right?

[unreal stratus]

Like here you're using it in an informal sense I guess but yeah

[nimble portal]

Yeah

[unreal stratus]

I mean I would use words like insside or smth lol idk / volume but yes

[nimble portal]

I just have trouble because when I've considered sphere and ball equivalent all my life, as 3D objects ("3D circles" or whatever you'd call it in elementary school)

[hidden crag]

Spoiler?

[nimble portal]

Just like a bouncy ball or whatever 😭

[unreal stratus]

Just use homotopy mayer vietoris lol trivial

[unreal stratus]

Yeah fair enough

Just think in terms of the like local dimensions

If ur living on a 2-ball or a 2-sphere it's like being in the plane

[nimble portal]

Yep

[unreal stratus]

if ur small enuff

lol

[nimble portal]

I just need to shift to equating ball with neighborhood, when I visualize neighborhood I think of what's being referred to as a ball

Why is living on a 2-ball "like" being in the plane? Is it not literally being on a plane? I mean I guess you're still in R^3 but the ball is on the plane

So what do you call the object that I previously associated with sphere & ball? The..round shape...with a volume and surface...?

[coarse night]

Lol

Closed ball maybe

[nimble portal]

In R^4? Lol

[novel acorn]

man what is up with these convos about the semantics of calling something a ball vs. a sphere this is like the second one I've seen

[nimble portal]

The last one was also due to me

[nimble portal]

And this is why

[novel acorn]

it isn't a problem

just interesting

[nimble portal]

I finally get the example now at least lol

Thank you everyone

qeqiuwheqwuhem this makes so much sense

Wtf that's so cool

[nimble portal]

And is this right?

[novel acorn]

no
It doesn't include the case where x^i = 0

[nimble portal]

Yeah I wasn't sure how to put that

In my head it's like

You're missing the "boundary" connecting the two halves

[novel acorn]

you need too union this with S^{n-1}

[nimble portal]

Righttt

That makes sense

[gritty widget]

feather there's a name for spaces you can partition into two opens

[nimble portal]

Connected?

[novel acorn]

opposite of that

[nimble portal]

But I feel like the name connected implies the boundary is there

Yeah LMFAO okay

[gritty widget]

these are precisely disconnected spaces

S^n is pretty connected to me (n > 0)

[hidden crag]

n>0 mega pedant precaution

[gritty widget]

yeah lmao

[hidden crag]

[nimble portal]

Oh okay looking at the Wiki article I think I get it

[gritty widget]

with the recent crowd of people on this server i feel like i have to take huge precautions

[coarse night]

S^(-n) is connected

[novel acorn]

(for n<0)

[gritty widget]

yeah i had you in mind

[coarse night]

lol

[novel acorn]

there are two ppl here and idk who you're referring to

[nimble portal]

Why are there 2n+2 charts for S^n? You have two charts from U_^(+-) intsct S^n, where do the other 2n come from?

Oh wait

[gritty widget]

two charts for each x^i. there are n + 1 of them

[nimble portal]

You have two charts for every coordinate in S^n

Right

Wow that's so cool

[coarse night]

(you need only 2 charts tho, inefficiency )

[nimble portal]

Only?

[gritty widget]

stereographic projection

[coarse night]

different charts, not the one you are using

[novel acorn]

take S^n-{1} and S^n-{-1}

[nimble portal]

Oh okay lol

[coarse night]

||but you can't do it using 1 btw||

[gritty widget]

can you cover S^n by only one chart?

fuck

[coarse night]

fuck

[nimble portal]

I was thinking you'd need at least 3 in order to map the "boundary" connecting the +- U_i

[novel acorn]

there's probably some good picture in needham where he went and took like an orange and did some horrific things to it to show you can cover it with two charts

those poor eggplants...

[coarse night]

[languid patrol]

You just remove the weird stem bit on either side. It's not horrific. Stereographic projection is probably unpleasant to experience though.

[hidden crag]

stereographic projection has only ever been unpleasant

[gritty widget]

timo confirmed can't compoot

[hidden crag]

do not ask me for the derivative of something that isn't a polynomial

[coarse night]

but I like the connection with minimal number of contractible open covere with cup length

[languid patrol]

What is the derivative of log(e^{x^2 + 1})?

[novel acorn]

Ah needham

[unreal stratus]

Yeah

[coarse night]

[unreal stratus]

So based

Tbf S^0 does count

[hidden crag]

to show that not every suspension is simply connected maybe

[novel acorn]

Everything is a polynomial if you add enough terms

(Do not ask me about e^{1/x^2})

[coarse night]

[unreal stratus]

So true

[unreal stratus]

Yeah

[gritty widget]

do it

[unreal stratus]

Oh the cover by one chart?

Uhh well since S^n is homeomorphic to R^2 this is easy

[coarse night]

[gritty widget]

potato is a flat earther

[coarse night]

exposed

[unreal stratus]

Yes

Man configuration spaces are yummy

They should be used as the canonical proof of invariance of dimension

Jk

[nimble portal]

Oh boy

Do I consider R^(n+1) as R x R x ... x R (n+1) times or can I think of it as the vector space R^n?

I'm assuming it's just a topological space 😭

[coarse night]

both

[nimble portal]

I don't know how else to interpret "sending each point x in R^(n+1)\{0} to the subspace spanned by x"

In my head all I can think is

We take a point in R^2

And it spans a line

But idk why I can't extend that to taking a point in R^3 and spanning a plane

[coarse night]

think as identifying antipodal points on a sphere

[nimble portal]

Sure

[untold lily]

you won't live through the projective space example btw

this is where you drop this book

[coarse night]

also if you haven't seen projective spaces before, feel free to skip the example till you need it

[nimble portal]

LOL

two types of people

[untold lily]

you could skip it, but the issue is it ain't gettin easier man

[coarse night]

no it does

if you see in a different context

[untold lily]

feather hasn't seen any topology and the book hits you with "the fundamental group of a manifold is countable" like 10 pages after that example

it is over

[nimble portal]

Not any topology but not muchhh lol

[coarse night]

it's not Lee, is it?

[untold lily]

it is

[nimble portal]

not the guess

[coarse night]

thought it's tu

[nimble portal]

it was until I decided I like this book better

[coarse night]

it is just too lengthy

[nimble portal]

This I understand at least!

So uh

How do we go from

S^1 x S^1

to this picture

I mean I see the two S^1's

But they're perpendicular to one another, how did we get that?

Doesn't that depend on what "basis" you have for each S^1?

I'm not sure what exactly I'm trying to say there

BUt I'm thinking about it like R^2 = R x R

You can have a nonorthogonal basis and then you won't just get the x & y axes perpendicular to one another

[coarse night]

it's just one "representation" of S¹xS¹

[nimble portal]

What would I search up to find other representations?

[coarse night]

no

I don't mean representation in a mathematical way

[nimble portal]

I know

Other visualizations, if that word is better

[abstract saffron]

Study topo harder 😄

[coarse night]

ex like S¹ and ellipse are diffeomorphic

[abstract saffron]

Study it long enough, then you'll know all sorts of representation

[abstract saffron]

Think of coordinate system: You have a surface with two numbers for coordinates, and they both loop as circles

Now what does it look like? Well, the first number gives a circle

The second number gives another circle, but this new circle travels around the old circle

[nimble portal]

Yes

[abstract saffron]

As it moves, it shows a surface: that's a torus

[nimble portal]

I understand

But like

If the "new circle" was at an angle (not perpendicular to the old circle)

That's still a torus right?

[abstract saffron]

This particular picture does. But again, we're in topo world, everything is up to homeomorphisms.

[abstract saffron]

Yes

It can be any solid mess with a hole, and that will be counted as a torus

Or at least, the boundary of that mess

[nimble portal]

But since the angled new circle & the perpendicular new circle are homeomorphic then they're all essentially the same thing

[abstract saffron]

Exactly

[nimble portal]

That's awesome

[abstract saffron]

Welcome to real topology 😄

Now you get to play with the good stuff

[nimble portal]

Wtf is a precompact set now that's a word I've never seen

From Googling I think it's a set where the closure is compact?

[abstract saffron]

Ah, that's a French invention

[nimble portal]

not the french again

[abstract saffron]

Compact = Precompact + Hausdorff

Or is it quasi-compact?

[swift fjord]

That's quasicompact

Precompact is what feather said

[abstract saffron]

In French it's like précompact

[nimble portal]

Oh sweet

I think I see why

[abstract saffron]

I don't think I do 😄

[nimble portal]

Since it's second-countable, it has a countable basis

Then since it's a manifold, if we look at a specific region centered at point p with the corresponding chart, we can map onto R^n

[abstract saffron]

Is it though?

[nimble portal]

Isn't this what second-countable means

[abstract saffron]

Some authors add more conditions, like second countability or paracompactness

It depends on the book

[gritty widget]

what are you talking about

[abstract saffron]

Strictly speaking, a manifold can be defined as something locally homeomorphic to R^n

Without any addition conditions

[untold lily]

I mean... it's pretty common to have second-countability as a condition?

[unreal stratus]

Not to mention Hausdorffness lol

[abstract saffron]

I have seen enough examples to never make assumptions 😄

[coarse night]

ppl work with non hausdroff as well

[abstract saffron]

If only there's one definition to agree upon

[gritty widget]

people work with nonhausdorff/paracompact/whatever. feather is working with hausdorff second countable

you are talking to feather

[coarse night]

apparently they show up in foliation theory

[abstract saffron]

Yes, the real beasts

[nimble portal]

Then we can take open sets in R^n as the basis for its topology? Something like that?

[untold lily]

yeah but when someone says the manifold has a countable basis since it is second countable, it seems weird to challenge that

[nimble portal]

And if we take the union of those all over our manifold we'll get a countable basis for the whole thing

[abstract saffron]

Well, that won't work, but dw, you have second countability in this case I think

[nimble portal]

And they're precompact since open intervals in R are precompact

[coarse night]

no

[nimble portal]

Blegh this is where my lack of topology knowledge becomes painfully apparent huh

[abstract saffron]

Basically second countability is like going to Walmart: it gives you almost everything you want

You have a countable basis for the whole space

[coarse night]

unless you don't consider (0, inf) as interval

[nimble portal]

o

But inf isn't in R

[gritty widget]

...

[nimble portal]

💀

im gonna cry

[unreal stratus]

Walmart has never given me anything

[untold lily]

I mean this is just not possible

[nimble portal]

never felt so dumb in my life

😭

[coarse night]

😑

[abstract saffron]

(0, inf) doesn't include inf, you know that? 😄

[nimble portal]

BAEWBYWHEIUQWHE

IM SO DUMMY HEHEHEHE

[abstract saffron]

that's why we have parenthesis

[untold lily]

you are trying to tackle a book that requires at least a 100 pages of topology as a prereq

[nimble portal]

Wait okay sorry the union of finite open sets in R

[untold lily]

with barely any

you can't do this

[nimble portal]

is a countable basis for R

Watch me lems

[gritty widget]

we're all watching you feather and not many of us are confident

[nimble portal]

I should just Google and double-check before I speak

[abstract saffron]

Well, yeah, feather be feather. Get used to it. We tried, it didn't work

[nimble portal]

I'm not confident either

What Megu said!

I enjoy this

[gritty widget]

we dont

[untold lily]

I studied a fair bit of topology before trying to get into manifolds and I still struggled hard with the topology

[nimble portal]

Then don't respond :p

[untold lily]

it just won't even be fun

[nimble portal]

I'm having fun

[untold lily]

aight

[abstract saffron]

I studied a shit ton, and I still struggle 😄 it never stops

[coarse night]

try Lee ITM first then

[unreal stratus]

Mood

[abstract saffron]

Not Lee, noooo. We had this convo already

[nimble portal]

The only reason I want to study this in the first place is so I can do fun calculus shenanigans :3

[coarse night]

lol agree

[gritty widget]

can't you do fun calculus shenanigans... in calculus?

[nimble portal]

I have for the past three years

[coarse night]

but itm,ism,irm is a series that almost covers everything you nedd

[nimble portal]

This is just harder fun calculus shenanigans ^_^ new stuff for me to learn

[gritty widget]

you are months away from getting to any calculus shenanigans if you are doing this

[unreal stratus]

Wait like what

[nimble portal]

Learning about differential forms and generalized Stokes thm and so on :o

[gritty widget]

you can do that on R^n you know

[nimble portal]

Yep

[coarse night]

long journey

[nimble portal]

I don't mind

[abstract saffron]

Not like I'm confident either, but I'm actually curious how far feather will go. I'm also learning knot theory myself with a pinch of (algebraic) topology, intuition, imagination, and sheer willpower

So I completely understand

[nimble portal]

If I give up, it'll be of boredom, not because of difficulty

If I'm interested in it I'll try my hardest to learn it, difficulty be damned

[coarse night]

it takes time for the subject to grow and gain maturity

[abstract saffron]

Well, not doing anything doesn't make it shorter 🤷

[nimble portal]

I'm gonna do some more reading and see if I can justify this lemma to myself first before I read the proof

[coarse night]

doing some point set topology certainly does

[abstract saffron]

Kids like to bang their heads against the walls first to make sure there's no shorter way.

[nimble portal]

[abstract saffron]

Sometimes I wish I still had the same naivety

I learned topo because no one told me it was supposed to be hard

[coarse night]

lol

i don't even care about point set anymore

[abstract saffron]

And it wasn't! I understood up to Hilbert's fifth problem in a week

Still cannot replicate that speed anymore

[coarse night]

godspeed

[abstract saffron]

Oops, Hilbert's 10th is the one about function. That's number theory in disguise, another story

[abstract saffron]

It's actually not so bad. Just follow the definitions and make sure you understand them all concretely

[nimble portal]

Yep

That's what I'm doing

I forgot coordinate ball was an already defined term, I really need to start writing this stuff down

[coarse night]

also give some time for the concepts to grow.

[nimble portal]

Yeah

[coarse night]

Like don't expect to solve every single exercise in first read through

[abstract saffron]

Don't. Keep going back to the definitions. It's like learning vocab for a new language: writing them down and never look at it doesn't help. But searching for it the 4th time, and you will remember it.

[nimble portal]

Alright

[abstract saffron]

At that point your brain will be tired of going back that it will just remember stuff to save time

best by test

[nimble portal]

Love the psychology there force my brain to learn out of sheer laziness

[unreal stratus]

Generalsied stokes theorem is a let down

[abstract saffron]

I'll be surprised if she pulls that off we never know

[coarse night]

Also if you can survive through Bredon then 2nd chapter is a nice run through of all the important concepts

[abstract saffron]

We have a name for that, no? Not just generalised stokes

[nimble portal]

I think it's awesome, Moth helped me with the intuition for when I first learned the Kelvin-Stokes theorem a couple years ago

It's so cool that the idea extends to other spaces

[unreal stratus]

wdym lol

[abstract saffron]

When you have a hammer, everything is a nail

[nimble portal]

Dude

[coarse night]

lol

[nimble portal]

You're like

[unreal stratus]

oh I mean idk the proof ends up being pretty dry lol

[nimble portal]

Confucius reborn?

[coarse night]

agree

[nimble portal]

Everything you spit sounds cool as fuck (Megu)

[abstract saffron]

Nah. Try living in France for a few years, you'll see how it affects you

[quick bough]

how does one show that any interval I cannot be homemorphic to S^1 "using an argument about connectedness of appropriate subspaces"

any hints

[gritty widget]

remove a point

[quick bough]

yeah, ik that it is homeomorphic to S^1 without the north pole

[coarse night]

homology

jk

[quick bough]

but like would i be "using an argument about connectedness of appropriate subspaces"

[gritty widget]

if f: X -> Y is a homeomorphism and S is a subspace of X, then X\S is homeomorphic to Y\f(S)

[untold lily]

remove a point, check the connectivity of your space after

[quick bough]

shouldn't that still be connected

[abstract saffron]

Actually I'm curious about 3) 😄

[quick bough]

it's even path-connected

[untold lily]

? for which one

[coarse night]

same right? remove 2

[untold lily]

this is meant to be a hint not the full damn answer

[abstract saffron]

The easiest way is domain invariance, but we don't have that tool here, But yeah, removing points is a cool trick

[untold lily]

you remove points

[abstract saffron]

2 is lame, let's go with 3

[coarse night]

remove entire S¹ and claim by invariance of domain

[abstract saffron]

Go with fundamental group instead because it's cool

Well, that's basically removing points, but as a grad student

[coarse night]

true

[abstract saffron]

Yes feather? We're all waiting

[nimble portal]

I'm just trying to type it out here while I go back and forth between Google and the book

[abstract saffron]

Maybe try writing them down once. Not because you'll look at them again, but for the sake of writing.

Idk, it helps for me. Especially at the beginning

[fickle hamlet]

If A is a subspace of X and B is a subspace of Y , then the product topology on A × B is the same as the topology A × B inherits as a subspace of X ×Y. I don't really understand the difference between the two. Is this correct? product topology on AxB is {UxV| U in subspace topology A of X, V in subspace topology B of Y} and topology AxB inherits as a subspace of XxY is {AxB intersect with PxQ| PxQ is open in topology of XxY}?

[abstract saffron]

One is $(U \cap X) \times (V \cap Y)$, and the other is $(U \times V) \cap (X \times Y)$.

[gentle osprey]

Megumi_Tadokoro

[unreal stratus]

Thi is correct

They are the same pictue

[abstract saffron]

It depends on you taking product first, or subspace topo first. In this case it's the same, but it's not trivial

[unreal stratus]

I assume this is a typo lol

I would say it is pretty trivial they are the same

like

$(U \cap A) \times (V \cap B) = (U \times V) \cap (A \times B)$ straight up

[gentle osprey]

potato

[abstract saffron]

Oh yes, should be A and B

[unreal stratus]

I think

So they have the same standard basis

[abstract saffron]

I mean, everything is nice in finite case.

It's just the infinite case that's worth the rigour.

[unreal stratus]

Wdym finite case

Oh then sure

It's just you said it wasn't trivial lol

[abstract saffron]

Well, it can still be an exercise 😄

And in infinite case, it's not for me

I'm not sure where it holds anymore

I guess it holds even in general case? Product topology is nice enough for these kinds of ops

[fickle hamlet]

i dont get what u mean by infinite case

[abstract saffron]

When you have a family of spaces $(X_i)_{i \in I}$

[gentle osprey]

Megumi_Tadokoro

[abstract saffron]

If I is countable, sure. But if it's not, well....

That's where category theorists will shout at us about their importance

Also, there's this whole mess about Axiom of Choice

Just, stay away from it whenever you can, for your own sanity

[fickle hamlet]

oh ok but this is just for two its just in the first chapter of the topology book

[abstract saffron]

Then yes, it's trivial

Ahhh, point-set. Good ol' days...

Actually, I just notice

(see ⁠get-advanced-access to use this channel) point-set topology (topological spaces), algebraic topology (homotopy/homology/etc.), geometric topology, anime
Why anime?

[long prairie]

why not

[nimble portal]

Okay so if we have a manifold M^n (dimension n) then by definition (second-countability) there's a countable basis for the topology on M

Now look at a neighborhood U in M, and consider the chart phi: U -> U', where U' is a subset of R^n

R^n naturally has the Euclidean topology, for which n-balls form a countable base. The closure of an n-ball is the (n-1)-sphere, and the union of these two is closed and bounded, so by the Heine-Borel theorem it's compact. Hence n-balls are precompact

U' comes with the subspace topology, of which precompact n-balls form a countable base. Then, taking phi inverse of these precompact n-balls, we have a countable base for U

So if you do this for every neighborhood U in M, and take the union, then we'd get a countable base for M

[abstract saffron]

Eww, weebs

[nimble portal]

Pretty sure I'm missing a bunch of details but is that the right idea? lol

Still coming to grips with what a basis for a topology is

[abstract saffron]

It took me a good week. Take it easy.

[nimble portal]

How about this?

I think that's the best I can do, guess I'll just go read the proof now

[abstract saffron]

It's a sketch, but I can't find anything wrong, hmmm

I mean, there are details left out, but that's the trivial part.

[nimble portal]

Wtf isn't this what I said

I don't understand why we need that last part about V in U_i being precompact though

Oh wait

[abstract saffron]

I mean, you have to show the preimage forms a basis. That's why we have this V thing

[nimble portal]

I don't think that's why

[abstract saffron]

Also you haven't shown precompactness, but that should follow easily(?), or I'm hallucinating

[nimble portal]

The sentence before the V thing shows that the preimage does form a countable basis (when we take the unions)

Yes, that's what I was about to say

The V thing is to show that our countable basis of coordinate balls is precompact

[abstract saffron]

I mean, oh c'mon, you have all the nice maps in the world

That's nuisance, you got the good grip. Dw, just be careful next time.

[nimble portal]

I haven't properly read that last bit yet

Let me try to show to myself why the coordinate balls are precompact

[obtuse meteor]

Megumi I don’t really think you’re helping much

[obtuse meteor]

A coordinate ball is not necessarily precompact

[abstract saffron]

Yes, but we're in finite dimensional case

[obtuse meteor]

Consider a coordinate ball given by a homeomorphism of (-1,1) onto R

(Even then Megumi)

[abstract saffron]

Ok, I'm hallucinating obviously

[nimble portal]

I see

[obtuse meteor]

Also @nimble portal I came here to say do the RP^n example. It is arguably the most important early example of a manifold

[nimble portal]

Yeah it's the one Tu mentioned too

I'll come back to it once I've gotten to play around with other topology stuff for a bit

[obtuse meteor]

It will also build your intuition for what the theory of manifolds is allowing you to do

[bitter smelt]

Also it comes up constantly in the book later. It's a nice example of a nonboring manifold

[abstract saffron]

That will bend your mind quite a lot actually

[nimble portal]

How is this a coordinate ball?

[obtuse meteor]

It’s not super easy to cook up an explicit embedding or RP^n into R^m. But showing it is a manifold allows you to still work with it

[nimble portal]

The radius of the ball is infinite

[obtuse meteor]

Left side is a 1D ball. Right side is our whole manifold. We have one chart

Radius of left hand ball is 1 :)

[nimble portal]

Yes

Hm

Does this have to do with the manifold being Hausdorff

[abstract saffron]

I doubt

[obtuse meteor]

No

[nimble portal]

Hm

[unreal stratus]

Manifold

[nimble portal]

So then how does Lee conclude that the open balls are precompact

[obtuse meteor]

It is true that any coordinate ball contains a coordinate ball which is precompact

[nimble portal]

I feel like you need Hausdorff since that means we've got a finite cover right?

[obtuse meteor]

I would think about this and try to prove it

[obtuse meteor]

Hausdorff has nothing to do with having a finite cover here

[nimble portal]

Blegh

[obtuse meteor]

Hint: do it this way, and use that this is true in R^n

[nimble portal]

Okay wait pause one second

[obtuse meteor]

Also use that you have a homeomorphism in your hands

[obtuse meteor]

In this example, the homeomorphism restricted to (-1/2,1/2) will have a precompact image

[nimble portal]

Oh

Well because we're in R^n we can just write our not-precompact coordinate ball as a union of precompact coordinate balls right?

[obtuse meteor]

That’s true but I don’t believe you really even need that

But sure that’ll be fine

[nimble portal]

So just to remind myself what this is all about

I said coordinate balls are precompact

But that's wrong, as demonstrated by your counterexample

[obtuse meteor]

If you just grab yourself an open cover of precompact coord balls you’ll be happy. Bc then you can use second countability to get a countable basis of such guys

[nimble portal]

I think I just don't get the counterexample? LOL

Like (-1, 1) is a precompact coordinate ball

[obtuse meteor]

Yes. And there is a homeomorphism onto R

So R is a coordinate ball

[nimble portal]

How does using a homeomorphism to stretch that onto R change that

[obtuse meteor]

Ah ah ah

No it isn’t

Being precompact depends on what the total space is

[nimble portal]

How can R be a coordinate ball? Isn't the point of coordinate balls that the radius is unity??

[obtuse meteor]

Because closure depends on what the total space is

You’re misunderstanding the definition of a coordinate ball

It consists of a few pieces of data

A coordinate ball in a manifold M is a pair (U, phi) where phi is a homeomorphism phi : U -> B_r(p) for some r > 0 and p in R^n

In my example. The manifold is R, the coordinate ball is R, the map phi is phi : R -> (-1,1)

[nimble portal]

OH

The coordinate ball is supposed to just be U though, not (U, phi) right?

[obtuse meteor]

It’s the pair

To be specific

[nimble portal]

Okay

I see I see

[obtuse meteor]

What a coordinate ball is essentially is a piece of your manifold that looks like a ball in R^n

[nimble portal]

After we apply phi

Right

Idk why I just thought they were unit balls

I was totally misunderstanding the lemma then

BUt I don't think I'll have to hcange too much

Okay okay so

Proof sketch is something like

Consider a chart phi on U in M

Then because M is locally Euclidean, we can write phi(U) as a countable union of n-balls balls in R^n

These n-balls are precompact

Then we take phi inverse to write U as the union of the preimages of the n-balls

But then by definition of being a preimage of an n-ball, these subsets are coordinate balls

And they'd form a countable basis for U, and then you take the union over the manifold to get a countable basis of coordinate balls for the manifold

Now did I word it right?

I see why a coordinate ball is not necessarily precompact though

But now I just need to show that in this case these coordinate balls are precompact

Hm

Couldn't you just consider the closure of each coordinate ball to be the preimg under phi of the closure of the n-ball

[obtuse meteor]

This is like. Quite close

Though there is an error

[nimble portal]

Fuckkk

Where is the error and what am I missing?

[obtuse meteor]

Hint: it doesn’t work with the counterexample I gave

[nimble portal]

That's what I was working on lol

[obtuse meteor]

Not necessarily no?

[nimble portal]

How can an n-ball not be precompact?

Just take an n-ball of radius > 1

[obtuse meteor]

I think you need to take a step back and stop thinking so formally. Also it helps break to things up into pieces

[obtuse meteor]

Do you understand the counterexample

[nimble portal]

Wasn't the counterexample to show that coordinate balls need not be precompact

[obtuse meteor]

Yes

[nimble portal]

I'm saying that n-balls are precompact

[obtuse meteor]

Precompact depends on the total space

Your total space is phi(U) not R^n

I agree n-balls are precompact in R^n

In phi(U) maybe not

[nimble portal]

Ah

Fuck

I see

[obtuse meteor]

Here are the two steps I would do

1. show that for every point p in M there is a coordinate ball in M which contains p
2. show that you can take this open cover and make it into a countable sub cover using second countability (nontrivial. Hard topology exercise for a beginner)

[nimble portal]

Okay

And I technically haven't even done 1 yet since I assumed that the n-balls would be precompact in phi(U)

[obtuse meteor]

Yes

[nimble portal]

Well

[obtuse meteor]

There is a very quick proof of 1

[nimble portal]

phi(U) is homeomorphic to R^n

[obtuse meteor]

Maybe 1’ is show it without the precompact condition

[obtuse meteor]

Er it depends on your defn of a chart

[nimble portal]

Lee's definition has it homeomorphic

[obtuse meteor]

Some ppl require image to be R^n and some ppl say any open subset of R^n

Then sure the argument you gave above works then I have no issue with it anymore

[nimble portal]

Does two spaces being homeomorphic mean their subsets are homeomorphic too?

[obtuse meteor]

Which subsets?

[nimble portal]

Like what I'm thinking is

[languid patrol]

A homeomorphism restricted to any subspace is also a homeomorphism onto its image.

[nimble portal]

because phi(U) is a subset of R^n

[obtuse meteor]

If you’re asking if h : X -> Y is a homeomorphism and A is a subset of X then h(A) is homeomorphism to A via h then yes

Feather I do beg of you read a topology book at least a bit

[nimble portal]

And the basis for R^n is the union of n-balls

[obtuse meteor]

I thought you said before it was R^n

[nimble portal]

uh

I think I was wrong

💀

Let me check instead of trying to remember

Sorry...

Ohhh I see where I was misreading

[obtuse meteor]

I retract this statement lol

[nimble portal]

Whoops phi(U) is in R^n so it's fine then right?

Oh

[obtuse meteor]

No consider the counterexample I gave earlier

[obtuse meteor]

^

[nimble portal]

Yeah

[obtuse meteor]

Total space is Utilde

[nimble portal]

Yes

Well in that case don't we just fix it by saying that phi(U) isn't precompact?

Since phi(U) is still in R^n then it inherits the subspace topology so we can still write it as the union of n-balls right?

So even though phi(U) might not be precompact, each n-ball still is

Like with (0, inf) in your example

So then we just take the inverse

[obtuse meteor]

No?

(-1,1) is an n ball

It is the union of 1 1-ball

But that 1-ball is not precompact if (-1,1) is the total space

[nimble portal]

I'm making the same mistake every time, I see it now LOL

Wait okay then

Why don't you write phi(U) as a union of balls in phi(U)

[abstract saffron]

Wow, y'all are really dedicated, I must say

[nimble portal]

Which is fine since phi(U) inherits the subspace topology

These should be precompact in phi(U)?

[obtuse meteor]

(-1,1) is a ball in (-1,1)

[nimble portal]

Agh how do I use the right words

Balls that are proper subsets of phi(U)

[obtuse meteor]

(0,1) is not precompact there

[nimble portal]

I'm gonna cry

The only other thing I can think is like

phi(U) is in R^n

[obtuse meteor]

Why not try this?

[nimble portal]

Isn't that what I'm doing? I'm just stuck on the coordinate ball part, since that hinges on us being able to write phi(U) as a union of unit balls

[nimble portal]

But why do we need precompactness in phi(U) anyways

The domain might not be a coordinate ball but it can be written as the union of coordinate balls no?

[stark fog]

I think it has to do with the existence a subordinate continuou partition of unity for the open cover {U_i} of the manifold

dunno

[gritty widget]

...

[stark fog]

sorry if I'm wrong

[nimble portal]

I'm sorry Faye maybe I should just come back to it later

I feel like I'm missing the point and you're just talking to a wall 😭

[obtuse meteor]

Here’s a correct solution to part 1:

Take your chart phi : U -> Utilde. Let p be in U. By openness there is an r>0 so that B_r(phi(p)) lies in Utilde.

Now Vtilde = B_(r/2)(phi(p)) has compact closure in Utilde (since its closure is the closed ball)

Then let V be the preimage of Vtilde

Then phi|V : V -> Vtilde is a precompact coordinate ball about p

Work smarter not harder

@nimble portal maybe read this and see if it makes sense

[nimble portal]

How is V a coordinate ball?

oh

my fucking

god

WHY DID I KEEP GOING BACK TO n-BALLS MANNNNNNNNNNNNN

Idk why I kept thinking coordinate ball means phi(U) has to be an n-ball argh

[obtuse meteor]

I mean phi(V) is an n-ball

Since Vtilde is

[nimble portal]

Yeah

I get it

How does is Vtilde precompact?

Doesn't its closure just have radius r/2

Never mind I see what you're doing

I was trying to do the same thing too but got stuck in the wording since I wasn't focusing on a specific point

I was trying to do it this way but got stuck trying to word this part

But I understand now I think

Every B_r is inside B_r', so each B_r is precompact in B_r', and since we can cover Utilde with B_r's then we can cover it with B_r's, which will then be precompact in Utilde

Yeah I understand it all, this is what I was trying to do too, just struggled with putting it into words

I just still kinda don't get what this part is for

Okay so the coordinate balls have compact closure (this follows from just taking the preimage of all of the B_r', of which the preimage of B_r lie in)

So they're precompact in each U_i

[languid patrol]

I assume you’re trying to show that M has a countable cover by precompact open sets?

[nimble portal]

But what's the bit about M being Hausdorff about?

[stark fog]

compact subspaces of Hausdorff spaces are closed

[nimble portal]

I was trying to prove that to myself, I understand the idea behind it all except for the very last bit in Lee's proof

[languid patrol]

Okay so this is just a last little technical detail which would be familiar from point set topology.

[novel acorn]

I mean this last part is showing that the basis that you chose is precompact in M (since before you only showed it's precompact in Ui)

[nimble portal]

Right

I just don't understand where being Hausdorff ties into that

And I'm not getting it from SubGui's comment either 😭

[novel acorn]

Every compact subset of a hausdorff space is closed iirc

[languid patrol]

You have shown that there is a countable open cover U_i with subsets V_i \subset U_i which have compact closure in U_i, but the closure in M might be larger. However a compact subset of an open set is compact within the whole space, and in a Hausdorff space a compact set is closed, so the closure of V_i in M is contained within its closure in U_i, whence they are equal.

[novel acorn]

OK so I assume you understand why V is closed in M?

[stark fog]

also, I think that is a lemma (actually theorem)

Let X be a Hausdorff space, then X is locally compact if and only if given x in X and U a neighborhood of x, there is a neighborhood V of x such that x in V subset U and the closure of V is compact.

In other words, it has pre-compact neighborhoods V subset U for each x in X and neighborhood U of x

[novel acorn]

ok there are too many ppl trying to explain stuff here so I'll leave

[stark fog]

[nimble portal]

"However a compact subset of an open set is compact within the whole space"
So each V_i with its closure (which is compact) is compact within the whole manifold M
"And in a Hausdorff space a compact set is closed"
So each V_i with its closure is compact on M

Ahh but that means each V_i is precompact in M

I see I see

You know

maybe

just maybe

you guys have convinced me to brush up on my point-set topology

,

[languid patrol]

Yes but the V_i aren’t compact, their closures (call them K_i) within U_i are compact.

[nimble portal]

Yes

That's what I meant whoops yeah thank you

How long would it take to go over all the prerequisite psTop?

[untold lily]

lol

[nimble portal]

[untold lily]

wasn't so fun?

[nimble portal]

Nope

It was very fun

[stark fog]

you could study it from the initial chapters of Lee, Introduction to Top. Manifolds

I think it would not take so long assuming you have already seen it before

[nimble portal]

Yeah, I have

[languid patrol]

It’s not necessary strictly speaking to understand point set topology to do calculations/get the gist of what’s going on with manifolds if you want to see them as subsets of R^n, but if you want to do manifolds from a modern perspective all of the technical details will be lost on you if you’re not fluent in point set topology. Little flourishes like this one that would be second nature if you’d read a bit more might be very confusing.

[nimble portal]

Just not the stuff about connectedness & Hausdorff spaces

Yeah I would have never come up with that last bit on my own

I did get the gist of what was going on too for the rest of the proof

Just couldn't put it into words

[languid patrol]

Yeah which is not so much of a problem in itself, it just might not be clear reading the proof which parts are purely technical and what is the heart of the proof. And you might get very confused about some very small technical part of the proof which is basically “soft mathematics”

[novel acorn]

not too long
But I feel like that's the wrong question to ask

You should never rush learning math

[nimble portal]

I agree

I'm just an impatient little rat

The only thing I don't immediately see is that compact sets are closed in Hausdorff spaces. I think if I'd properly studied the point-set first too then I would've never had trouble putting my thoughts into words and we never would have spent so long on this in the first place

Also just better at paying attention to definitions

[novel acorn]

point set is like the technical prerequisites to be able to do cool topology

me when separation axioms

Ohh but this reminded me how some of my friends shit themselves when they took diff geo cuz first class first thing that the prof did was prove that every manifold was paracompact so everyone got scared it would be a super technical class lmao

[nimble portal]

Speak of the devil

That's what I'm about to get to

LOL

Having trouble with the definition of locally finite

[languid patrol]

Depending on what you’re trying to do it might be nice to do something like d carmo’s curves and surfaces

Where you do a lot of practical things with 1 and 2 dimensional manifolds in 3 dimensional space using explicit parameterizations but don’t have to deal with all of the baggage of the modern formalism of a manifold

[novel acorn]

Take some nbhd of x called A
Then locally finite means that if you have a cover indexed by some I, i.e. {U_i}_{i\in I}, then A cap U_i is non-empty only for finitely many i's

[nimble portal]

Can I think of paracompactness like a mesh? In the sense of like finite element analysis & CAD lol

[languid patrol]

In the end many undergrads/grad students read an entire manifolds textbook but find themselves unable to compute integrals of differential forms/curvature/etc. explicitly.

[novel acorn]

uhhh I wouldn't say so?

I don't think there's a very visual way to see stuff like compactness/related stuff at least I don't have one

[nimble portal]

Nope I get it now

[novel acorn]

Generally of all the topological properties I think that compactness is one of the more difficult ones for people to get

[ocean narwhal]

compact set = thin

[nimble portal]

The idea is that we can refine our cover as much as we want and every point will still only intersect finitely many sets in our cover (given a large enough neighborhood)

[languid patrol]

In the end what paracompactness gives you is that if you take an arbitrary collection of open balls covering the space, then this can be refined where each point lies in only finitely many at a time. This allows you to construct functions on open balls and average them together to get functions defined on the whole manifold. The averaging wouldn’t make sense if there was a point in infinitely many of the open balls

[nimble portal]

So we can make our open cover as fine as we want

[nimble portal]

Oh!

[languid patrol]

[0,1] = thin (0, 1) = thicc apparently

[ocean narwhal]

lol I should have said precompact

[nimble portal]

Wait a second

Is the reason we do this to define functions on manifolds?

I mean that's what you said but like

[languid patrol]

It’s to be able to construct a lot of functions on manifolds

Yep

[nimble portal]

We just showed that we can cover M with precompact coordinate balls

By gluing together the precompact coordinate balls covering the domains of every chart

So if we have a function on a manifold

We can consider the function restricted to a specific subset

Then compose that with our chart (so phi(f)) to understand how the function behaves in a space we're familiar with, one rich with structure

And because our manifold is paracompact (which we're about to see the proof of) then we can do this within smaller and smaller domains and glue them all together to get an accurate picture of the entire function on the manifold

[languid patrol]

Eh, sort of. You don’t need paracompactness to glue the functions together if you already know they come from a global function.

[nimble portal]

It's the locally finite part of paracompactness that's important to what you were saying

[languid patrol]

What it helps you do is locally describe functions around a bunch of points and stitch them together into a function which is globally defined, smooth, and looks like a given function near a chosen set of points.

[nimble portal]

Otherwise we could do what I was saying with just any refinement of the cover, we don't need local finiteness for that

[nimble portal]

:o

[languid patrol]

Yes exactly

Paracompactness allows you to construct global functions with desired local behavior, going in the opposite direction is trivial.

[nimble portal]

I'm assuming this is part of what gives the manifolds here their "differentiable structure"

We can have a function on a manifold, and a definition of continuity thanks to being a topological space, but no definition of differentiability

[languid patrol]

No, they would have it even without paracompactness! But it would be really hard to prove anything about smooth functions/differential forms on them

The differential structure just comes from the charts.

[nimble portal]

But if we look locally we can glue together finitely many functions to smoothly approximate the global function around a point, which we can then do fun stuff with

[languid patrol]

Yep! It’s important in many technical arguments to know that non-constant smooth functions actually exist and paracompactness is a nice technical property that will allow us to do that.

[nimble portal]

Wow that's so freaking cool

Do we use Hadamard's lemma to show those functions exist

[languid patrol]

Not really, the key tool that’s used is something called a partition of unity. They are quite obnoxious to construct but once you do they are a very powerful tool.

[nimble portal]

I see

[languid patrol]

Basically you cover your space by U_i’s and V_i’s such that the V_i’s are disjoint, and construct functions f_i such that f_i = 1 on V_i, f_i = 0 outside of U_i, and the sum of all the f_i is the constant function 1

[nimble portal]

Oh

[languid patrol]

Then you can take whatever smooth functions g_i you want on the U_i and average them by \sum_i f_i g_i to get a function g which is smooth and looks like g_i on V_i

The sums are well defined because of paracompactness

[nimble portal]

I think I get it

f_i g_i is only going to be nonzero where U_i and V_i intersect right?

In U_i

So where V_i overlaps U_i

[languid patrol]

V_i is contained in U_i

[nimble portal]

Oh I see

[languid patrol]

But the U_i intersect each other inside U_i but outside the V_i

And only finitely many at a time

So only finitely many terms are nonzero at any given point

[nimble portal]

Right

I see

That's enough math for me today lol

Thank you so much

Mind if I ask a question or 2 about Lee's topological manifolds before I go?

[languid patrol]

Sure

[nimble portal]

So I definitely want to cover Chapters 2 & 4

Chapter 3 is important if I want to understand projective space

[languid patrol]

I think it’s all important

[nimble portal]

234 or the entire book? 😭

[languid patrol]

2,3,4 are all really important background

I’ve never read the book

[nimble portal]

Okay

pain

[languid patrol]

But judging by the table of contents it all seems like essential background. It’s probably the driest part of the book but it will make the rest easier

[nimble portal]

Sigh

Is it too late to just go back in time and swap majors to math

This is infinitely more interesting than engineering

Blegh sorry thank you again, have a good evening friend

[lean knot]

Let t and T be two Hausdorff topologies on the same space X, if a sequence (x_i) converges to x in t and y in T, does that mean x=y?

I think so, by generate a new topology through the union of the old ones, but I'm not sure if there is any flaw in my reckening

ok, it should be a net (x_i)

[umbral panther]

Not if you don’t assume there’s any relation between the topologies. Let one space be [0,1] and the other just swap 0 and 1. They’re just names. Then 1/n converges to 0 in one topology and 1 in the other

If you take the two and generate a third, it’s (0,1) plus two isolated points

[lean knot]

@umbral panther then how can he be sure there are such U and V here?

[umbral panther]

These aren’t unrelated topologies. Each one is finer than the last

[lean knot]

oh, yeah, I neglected that. Thanks a lot 😘

[gritty widget]

can someone help me with this problem

the continuous image of a dense set is dense in the image of the mapping

i can't read this, sorry

np

you should try typing up your argument

it'll be more likely to get feedback

Z+2piZ is dense in -1 , 1
x<y and a in Z
x<a+2pia<y
x-2pia<a<y-2pia
abs(x-2pia)<abs(a)<abs(y-2pia)
-1<cos(abs(y-2pia))<cos(abs(a))<cos(abs(x-2pia))<1
abs(a)=N
then cos(n) dense in -1 , 1

[nimble portal]

Is projective space as simple as the definition implies...?

Equivalence classes of lines passing through the origin in R^1

with the quotient topology

[gritty widget]

in R^1 this is trivial and boring

there is only one line through the origin in R^1

[nimble portal]

Yes

I'm not sure why it's any more interesting in R^2 or higher though

If I have my visualization right it's just a fat asterisk on the plane O.o a bunch of lines criss-crossing through the origin

[gritty widget]

that's like asking why people don't only focus on zero-dimensional manifolds

[nimble portal]

I don't know why Tt, enlighten me

I mean Faye mentioned earlier that projective space is a good early example of an interesting manifold

Because you can't embed it in R^n

[gritty widget]

0-dimensional manifolds are the absolutely least interesting examples of manifolds you could possibly think of

there's literally nothing to be done with them

[nimble portal]

OKay sure

I think it'd be a cool exercise to prove that projective space is a manifold (and define the charts)

[gritty widget]

it is a very important exercise

it is also a worked example in quite a few books

one you should understand well

[nimble portal]

No fun if you don't try to work it out yourself

[gritty widget]

based on what i've seen i think it'll be quite difficult

good luck

[nimble portal]

🙃

Let's see

How can it even be Hausdorff 💀

Let me look at the definition again

[gritty widget]

well, to even speak of it being hausdorff, you need to have a topology

so you should start there

[nimble portal]

Well yeah there's the quotient topology

Just trying to understand that

So subsets of P^n are open if the preimg is open in R^(n+1)

[gritty widget]

subsets of P^n

[nimble portal]

Thank you

[coarse night]

think of qoutienting a sphere instead of R^n+1

[nimble portal]

So I need to show that given two distinct points in P^n, there's are neighborhoods around each that are disjoint?

[gritty widget]

yes

given two different points x and y in P^n, find open neighborhoods U and V of each, respectively, which are disjoint

[nimble portal]

And "neighborhoods around each" means that each point is an element of some set in the quotient topology

[gritty widget]

i just mean that x is in U and y is in V

[nimble portal]

I know

[nimble portal]

I was trying to translate this

[gritty widget]

i did it for you

[nimble portal]

Yes sorry I just want to clarify that "open neighborhoods" means "sets in the quotient topology that contain x & y, respectively"

[gritty widget]

unless you have another topology on P^n in mind, that's all it could mean

[nimble portal]

I did not I just wanted to make sure I wasn't misunderstanding

Okay cool

I feel like I'm completely misunderstanding what projective space is supposed to be

Or rather look like (visualizations for P^1 and P^2)

Hm

Can't you like

Ok so take two distinct elements in P^n, [x] and [y]