point-set-topology

[abstract pagoda]

so we are trying to show intersection of nonempty open dense collection of sets is nonempty where all these sets belong to a compact hausdorff space

just like b4

So we just need to take two dense sets and show that their intersection is nonempty

in normal space

[gritty latch]

we need to show that the countable intersection of a bunch of dense open sets is nonempty

[abstract pagoda]

yea but that involves just showing two in the collection are nonempty

wait a second

low iq

[gritty latch]

gonna go eat

maybe we can just map to a metric space and just use the proof in the book lol

[abstract pagoda]

kek

actually stumpted

[plain raven]

this is a truly dope theorem and it is the foundation of like, so much of logic lmao

locally compact hausdorff spaces are also pretty good for the same reasons

reading Folland's real analysis book was pretty awesome, learned a lot of point set topology there

[gritty latch]

oops just needed to create a nested sequence of closed sets

[hollow harbor]

Wait baire category is relevant in logic

[plain raven]

obviously when we speak about "category" without qualifiers we are talking about Lusternk-Schnirelmann category smh

[abstract pagoda]

hold up

[hollow harbor]

So logicians dabble in two kinds of category theory

Wait, maybe 3

[abstract pagoda]

Where did you come to that conclusion

[gritty latch]

well notice that if you have a sequence of nested closed sets U_i then their intersection can't be empty

[abstract pagoda]

Yes

Each element of the sequence needs to be dense in Xthough

Wait

and open

[plain raven]

i think you're misunderstanding something celina

[abstract pagoda]

I am.

[plain raven]

the most basic kinds of forcing arguments in set theory and computability can be seen as baire category arguments, i'm not exactly sure how far i can push the analogy but like, a basic fact about Turing machines and other basic models of computation is that they perform a finite number of tasks in a finite amount of time. so if you write down the characteristic function of some set A\subset N on a Turing machine's oracle and ask it to carry out some computation, possibly querying the oracle to determine membership in A when it's needed, then at most it'll only ask finitely many questions of the oracle before terminating (if it does indeed terminate.) The fact that the computation only depends on finite information about this infinite set can be seen as saying that it depends in a continuous way on the infinite set, in the same way that giving finitely many decimals of an irrational real is like giving an open neighborhood that contains that real, and so continuous functions f : R -> R have the property that to compute the output out to finitely many decimal places you only need the input to finitely many decimal places.

[abstract pagoda]

Is the exercise to show that a countable intersection of open dense sets in a compact hausdorff space is nonempty?

[plain raven]

you can formalize this by topologizing P(N) by identifying sets with their characteristic functions in {0,1}^N, and then giving this set the product topology (where { 0,1} has the discrete topology)

[hollow harbor]

thanks for this, i'll try to read through it

[plain raven]

then this space is compact hausdorff (by Tychonoff, since {0,1} is obviously compact) and so baire category applies

[abstract pagoda]

How does this translate to showing intesection of nested closed sets is nonempty.

[hollow harbor]

I see, very clever

That's also really satisfying

[abstract pagoda]

Is there a dense set A in X such that its set of limit points is not equal to X?

I am confident that the answer is no because A dense in X means Every open set intersects A.

[coral pivot]

Yeah there is none

[abstract pagoda]

Beated it!

[coral pivot]

Often instead of that A is said to be dense in X if the closure of A is X

since these are equivalent

[abstract pagoda]

Yea I was tryna show they were equivalent

[pearl holly]

Okay so this is the exercise that I'm working with: let X be the quotient space of S² obtained by identifying the north and south poles to a single point. Put a cell complex structure on X and use this to compute the pi_1(X). So I am basically picking two antipodal points and pinching them together. I think that the fundamental group here is Z since if I pick a loop at the pinching point then this loop isn't homotopic to the constant loop etc etc. The presentation is <a, b | a=b^-1> = <a>. So I have a wedge of two circles and I need to attach a 2-cell according to ab because then by a theorem in my book the fundamental group will be <a, b | ab>. But I'm having trouble visualizing this cell complex and I can't see how that is X

Imagine there being a program that lets you play around and build your own cell complexes

That would be a very complex program

[gritty widget]

make one

[empty grove]

One cell structure I can think of is

one 0-cell

one 1-cell

one 2-cell, glued along ||a a'||

Where a is the 1-cell

But explaining why this works would be the genus g surface thing all over again

But yeah that structure obviously has fundamental group Z

[pearl holly]

Okay so the 0 cell and 1 cell combined form a circle?

[empty grove]

that's always true if you have 1 1 cell and 1 0 cell

[pearl holly]

oh okay lmao

and what is a'?

[empty grove]

The 1 cell

[pearl holly]

and why did you censor it?

[empty grove]

But in the opposite direction

[pearl holly]

rubiks cube notation be like

[empty grove]

I thought maybe you could try that part on your own 🤡

[pearl holly]

oh crap, I already saw it

[empty grove]

😵‍💫

[flint cove]

I hereby pronounce you dad of the week lmao

[gritty latch]

shit now I want to continue with my topology book into fundamental groups instead of functional analysis

[gritty widget]

why not do both

[gritty latch]

can't do both in one day but maybe I could alternate between the two

is there some overlap? i.e will knowing fundamental groups help with func

[hollow harbor]

pretty much no overlap whatsoever

[gritty widget]

awaiting a sully from ultraproduct

[hollow harbor]

[marsh forge]

I’m with ryc here

[hollow harbor]

ok, i obviously mean in an introductory course

operator K-theory or whatever the fuck doesn't count

[empty grove]

Second sully has been scored

[marsh forge]

But luckily

Fundamental groups are way cooler

Than functional analysis

[empty grove]

[hollow harbor]

nothing could be more false

[marsh forge]

cope

[gritty widget]

arguing personal tastes goes in #chill or #「ivory-tower-emeritus」 thx

[hollow harbor]

amalgamated "free" products. lol.

that's all fundamental groups are.

[marsh forge]

Only needs use that term

Nerds

[hollow harbor]

just a bunch of boring universal constructions, slapped on, over and over

[marsh forge]

They are pushouts

[gritty widget]

normed spaces. that's all functional analysis is. just a bunch of boring vector and metric space constructions, slapped on, over and over.

[flint cove]

I think topology basics are helpful tho.
I felt I could put stuff a little better in context by knowing about separation axioms (keyword kolmogorov quotient) or how to rigorously show stuff with nets and what first countability means

[hollow harbor]

*exciting

[flint cove]

Not strictly necessary but makes some things easier to navigate

[gritty widget]

*exciting

[hollow harbor]

universal constructions are never exciting

[gritty widget]

normed spaces are never exciting

[hollow harbor]

ok

fine

here's a legit reason

[marsh forge]

Idk why I’m pretending to be a fundamental group simp everything ryc is saying is correct

[gritty widget]

you know im going to copypaste what you send and change a few words right

[marsh forge]

Its a boring invariant

But it’s step one

[hollow harbor]

you study functional analysis and see all this stuff, then someone shows you how if you change your language just a little bit suddenly you've already learned a big portion of the theory of quantum mechanics!

isn't that cool?

[marsh forge]

no

That’s a downside

[hollow harbor]

it's not super exciting in a vacuum

[flint cove]

Or realizing that closed subspaces are precisely the ones you can quotient out in the category of normed spaces
Not an amazing fact but that still sorta calms my inner „why is this notion important??“ voice

[hollow harbor]

(except that you get to use choice a lot, which is very exciting)

[empty grove]

Rice is an applied mathematician

[gritty widget]

you study fundamental groups and see all this stuff, then someone shows you how if you change your language just a little bit suddenly you've already learned a big portion of the theory of algebraic topology!

sorry ryc

[flint cove]

also, this channel is noisy as hell atm I clearly picked the wrong time to try to contribute lmao

[empty grove]

Terra is just better

[hollow harbor]

that literally makes NO SENSE

[gritty widget]

exactly.

[hollow harbor]

it's ALREADY algebraic topology

[gritty widget]

ok and

[hollow harbor]

functional analysis isn't already quantum mechanics!!!!!

[gritty widget]

[marsh forge]

Drank the rest of that redbull

Feel awful

[flint cove]

But I guess pointless verbosity is the only thing geometers are good at.
ducks and runs

[gritty widget]

we should stop arguing and let lux talk about actual mathematics (read: peace out lmao)

[hollow harbor]

GET IM JADE!

[marsh forge]

Is anyone but tterra a geometer

[hollow harbor]

tterra btfo

[flint cove]

sorry, my motivation is gone and I should do different stuff anyway I guess x)

[novel acorn]

I dabble in polytopes lmao

[gritty widget]

no need to apologize

although i was referring to a sully on ryc's message, this works too

[hollow harbor]

Ultra agrees with my message.

HA!

for very advanced topics *in one crank-tier subfield of functional analysis

[empty grove]

He's gonna gloat for weeks now

[hollow harbor]

modern banach space theory is a crank-tier subfield of functional analysis

[obtuse meteor]

I remember there being an interesting paper about this

[marsh forge]

Just the one

[hollow harbor]

wait, is this like

path spaces and calculus of variations or something

do they care about fundamental groups

if so, then that's not crank tier

i'm assuming this is the sort of thing you mean though: https://en.wikipedia.org/wiki/Kuiper's_theorem

[pearl holly]

Hmm now when I think of it, why is the fundamental group of the torus ZxZ and not the free product of Z with itself? The 1-skeleton (wedge sum of two circles) has a Z*Z as fundamental group. If we look at the torus, those two circles should determine the free group of the torus. You can make one loop around one circle and one loop around the other circle etc. so it feels like it should be Z*Z and not ZxZ. So something happens when we attach the "shell" (2-cell, rectangle) to the wedge sum of those circles but I can't see what.

[reef shore]

Yeah what happens when you attach the 2-cell is exactly what van kampen tells you

Imagine you have a 0-cell, a 1-cell, and a 2-cell glued along the 1-cell in the simplest way

just going around once

can you picture this space?

[pearl holly]

Like a blob?

[reef shore]

should be a disk

lol

[pearl holly]

So you first have a circle. Then your 2-cell is a disk so you take your boundary of this disk around that circle

[reef shore]

For attaching the 2-cell, I am taking the identity map S^1 → S^1 as the gluing map

where I identify the 1-skeleton with S^1 in the obvious way

[reef shore]

Yes so it is like taking a disk and a circle

and identifying the circle with the boundary of the disk

so you just get the disk again

[pearl holly]

okay so like a double layer of disks?

[reef shore]

no just a disk lol

[pearl holly]

Like the union of two identical disks?

Okay just a disk

[reef shore]

Like just taking a disk and gluing a string along its boundary in a circle

it doesn't introduce anything new into the space

[pearl holly]

oh okay yeah I'm with you now

[reef shore]

So the 1-skeleton here is the circle with group Z

but when you paste the disk going around once, you get the trivial group

because you are able to homotope all the loops on the 1-skeleton using this disk

[pearl holly]

yeah okay so you just cover up the hole right?

[reef shore]

Yep exactly

Now paste this disk on the same skeleton, but the gluing map is the constant map onto the basepoint

Do you see what happens?

[pearl holly]

So I get like an inflated balloon on 1 point on the circle?

[reef shore]

yep, do you see what this does to the fundamental group?

Hint: ||S^2 is simply connected||

[marsh forge]

Toki this is secretly that same exercise

About maps that extend over disks being nullhomotopic

[pearl holly]

It's still Z?

idk lmao

[reef shore]

Yep

[pearl holly]

Okay good

[reef shore]

Lol try to prove that

[marsh forge]

This is an easy exercise with van kampen

So is the torus one tbh

[reef shore]

But the point is, that when you glue along a loop, p, the disk allows you to homotope that loop to a point

which should be easy to see from Max's favourite theorem

Then van Kampen tells you that this is essentially the only loop that becomes nullhomotopic when you attach the 2-cel

(plus the stuff it generates ofc)

And in the torus, p is exactly aba'b'

So you get commutativity

[pearl holly]

Oh okay so aba'b' becomes nullhomotopic in the torus but not in the wedge sum of two circles?

[marsh forge]

Yes

[pearl holly]

and therefore we have to quotient it out

[marsh forge]

Yes

[pearl holly]

Hmm but I don't quite see how aba'b' becomes nullhomotpic in the torus but not in the wedge sum of two circles

Well I kind of see it

[reef shore]

Did you see the extremely well drawn images I sent yesterday

[marsh forge]

If you understand how to glue the disk for the torus it should be something you can picture

[reef shore]

If you want non van kampen intuition

[marsh forge]

Although

I have an easy picture for this

Toki

[reef shore]

Like a gave a homotopy from ab to ba

[marsh forge]

You know that a punctured torus

Is htpy eq to wedge of circles right

[pearl holly]

yeah I've seen an animation on this on youtube lmao

[marsh forge]

Okay

Take a torus

Pick a point

Draw a small circle around that point

We agree this is nullhomotopic right

[pearl holly]

yeah

[marsh forge]

Puncture the torus

[reef shore]

tinky see this I spent 10 mins drawing this for you

[marsh forge]

And in your minds eyes

[reef shore]

It is a homotopy from ab to ba

[marsh forge]

Picture where that little loop ends up

As you slowly retract onto the wedge of two circles

The answer is aba’b’

And thats why that loop is nullhomotopic in the torus

I.e. aba’b’ is nullhomotopic because small loops on the surface of the torus are

[pearl holly]

(wait moldi I will see them in a sec after max has wrote the stuff that he want to write)

[marsh forge]

There is no analogue of this

In the wedge of two circles

Okay I’m done

this makes it obvious that the purple loop will retract onto aba'b'

[reef shore]

is retract the right word here?

[marsh forge]

I mean to say that as we retract the punctured torus onto the wedge of two circles the purple loop will be brought to the boundary

puncture somewhere in the middle of that purple loop

and then convince yourself that all of my sketchy ignoring basepoints is irrelevant

[pearl holly]

okay wait wait so a circle is homotopy equivalent to a point?

[marsh forge]

uh

no

you know a circle is not homotopy equivalent to a point lol

[reef shore]

Why are homotopy equivalences useful toki?

Like what was the whole point

[pearl holly]

Okay wait so what's the argument here? I see that the purple circle deformation retracts to the loop aba'b' but then what?

[marsh forge]

well

you agree the purple loop is contractible here right

[reef shore]

The composition is giving a homotopy from aba'b' to a point

Just homotope the outline of the square to a single point (another way to think of what max is saying)

[marsh forge]

did you mean "the purple circle is homotopy equivalent to a point"?

because that is true within the torus

[reef shore]

aba'b' is exactly the boundary of the square

Wait by homotopy equivalent do you mean homotopic?

[pearl holly]

yeah lmao

[marsh forge]

i mean

most topologists and homotopy theorists

are very loose

w this language

[reef shore]

[marsh forge]

in fact i would just say "that circle is a point up to homotopy"

[pearl holly]

wait what does contractible mean here?

[marsh forge]

exactly what it always means

[reef shore]

nullhomotopic

[marsh forge]

homotopic to a point

[pearl holly]

Okay then I don't agree lmao, I can't see that

[marsh forge]

huh

just

do you believe any loop in R^2 is nullhomotopic?

because this is the exact same idea

[reef shore]

Take the boundary loop on this square. Do you agree that that is nullhomotopic?

[marsh forge]

that square is filled in

[pearl holly]

but doesn't this mean that any circle is contractible?

[marsh forge]

no

[reef shore]

Like just the square, don't identifty the stuff yet

When you make it into a torus you get non trivial loops

[marsh forge]

do you agree that every circle you can draw in R^2 is nullhomotopic in R^2?

[marsh forge]

you can do this specific loop in either order

[pearl holly]

yeah homotpic to the constant loop right?

[marsh forge]

yes

[pearl holly]

oh omg

lmao

[reef shore]

Yeah but easier to imagine on the square than on the torus

[pearl holly]

okay now I get it

[marsh forge]

okay

so

[reef shore]

and then compose that homotopy with the quotient map

[marsh forge]

the purple loop is homotopic to aba'b'

and the purple loop is homotopic to a point

by transivitivty...

[pearl holly]

yeah okay

so aba'b' is homotpic to the contant loop

[marsh forge]

(you can use the picture I drew with van kampen to compute the fundamental group of the torus exactly)

i.e., the first space is inside the purple loop

the second space is outside the purple loop

[pearl holly]

yeah I have actually done that lmao

[marsh forge]

:sully

[pearl holly]

oops

[marsh forge]

toki wtf

lol

[pearl holly]

what?

OH

lmao

wait

[marsh forge]

im so confused

[pearl holly]

okay sorry guys I'm really tired

[marsh forge]

how could you do that van kampen computation and also still not get what that image was lol

stop doing math when you are really tired lol

[pearl holly]

yeah idk what I'm doing

[reef shore]

Also toki here's what max was saying earlier: you pasted the disk along aba'b'. Then take the map S^1 → torus, mapping the circle to aba'b'. This extends to a map D^2 → torus, map D^2 to the 2-cell you just pasted. Then by that theorem, aba'b' is nullhomotopic

[marsh forge]

ezpz

[pearl holly]

ohh

[reef shore]

So when van kampen tells you that the loop you paste along becomes nullhomotopic it's not telling you anything new

[marsh forge]

fun fact

using cells like this to kill specific elements of homotopy groups

is super important in AT

[reef shore]

The new stuff is that it tells you that not much other stuff becomes nullhomotopic

[marsh forge]

and a fundamental part of many constructions

actually

toki

do you know about group presentations

[pearl holly]

yeah it's like <a | a=1>

[marsh forge]

great

Exercise

Let G be any group

build a CW complex (using the torus as a hint)

such that $\pi_1(X)=G$

[gentle osprey]

MaxJ (Glass Animals Arc)

[marsh forge]

You may assume G has a presentation with generators g_\alpha and relations R_\alpha

maybe i should use different indicies

(do this exercise when you are less tired

[pearl holly]

yeah crap, hatcher already proved this You take the wedge sum of S^2 g times and attach a 2-cell along the relations or something

[marsh forge]

well

the first part is wrong

and the second part is uselessly vague

so go actually prove it

lol

if you understood that proof

the torus question wouldve been immediate

[pearl holly]

Hmm okay. I will take a break first tho lmao

[marsh forge]

sure

[pearl holly]

And btw thank you guys for the help!

[novel acorn]

He didn't prove it lol
He just said it as fact

You take the wedge sum of 2g circles and attach the 2-cell on the path [a1, b1]...[ag,bg]

[marsh forge]

That is not the same statement lol

reread what i wrote

[novel acorn]

I though that was what he was referring to lol

[marsh forge]

it might be

but that only gives you very specific pi1s lol

[pearl holly]

Okay I can assume that G has a representation?

[marsh forge]

presentation

yes

[pearl holly]

oh sorry

[marsh forge]

representation of G is very different haha

[pearl holly]

Okay so let $G = \langle g_\alpha ; | ; R_\beta \rangle$. Now begin by having $\vee_\alpha S^1$. This has a fundamental group $F(a_\alpha)$ (free group with $a_\alpha$ as generators). But now we have a problem since the $R_\beta$'s must be nullhomotopic and now they aren't. So we do the exact same thing as we did with the torus, we attach $\beta$ 2-cells to our cell complex along the words $R_\beta$ because that makes them nullhomotopic

[gentle osprey]

Tokidoki ✓

[marsh forge]

yep!

the actual conclusion follows from van kampen

[pearl holly]

Thank you so much! Now I feel like I really understand this stuff

[novel acorn]

We can prove that the words Rbeta become nullhomotopic using van kampen?

[marsh forge]

yes

[novel acorn]

Orrr

I'll have to try that then

[flint cove]

No, but that nothing else becomes nullhomotopic I'd have guessed

[marsh forge]

no

well

yes

[empty grove]

Yeah I mean van kampen includes both things

[marsh forge]

you don't need van kampen to prove R_beta becomes nullhomotopic

[empty grove]

The proof that those do become nullhomotopic is easier

[marsh forge]

you use it to prove that each surgery only kills the normalizer of R_beta

[novel acorn]

Also don't mind me I'm gonna start asking questions here about this stuff as well soon lol
Good thing I even found a place where I can ask questions about this

[flint cove]

Nullhomotopy, more like nullhomodope-y

Glad to see some traffic about this stuff since it's kinda my weak spot

[novel acorn]

My friends would disagree

[trail tiger]

Skorokhod embedding from graphs to manifolds, anyone got a reference?

[gritty widget]

Is there a pseudo-generalization of https://en.wikipedia.org/wiki/Nash_embedding_theorem that applies to all geometries? I'm curious about the limitations of embedding manifolds in Euclidean space. What is the status of this field? I'm interested in universal representation learning/embeddings for machine learning.

In particular, I want universal angle-preserving embeddings.

related: https://sites.stat.washington.edu/mmp/geometry/reading-group17/html/samKslidesRMetric.pdf

[trail tiger]

Check out Greene's Isometric Embeddings of Riemannian and Pseudo-Riemannian Manifolds. It is too much to hope to get something totally general. For instance, you can't in general get Euclidean embeddings of symplectic surfaces preserving that 2-form data. I think for e.g. Lorentzian manifolds there are some embedding theorems but this is Google Scholar work right here.

[gritty widget]

thank you for your response. it is worth noting that the only operation i might really be interested in is cosine similarity, so it might be the case that all that needs to be preserved are the angles in a non-strict sense -- i.e. we want to be able to recover the angle information and essentially compute anything desirable via cosine similarity operations on this embedding space, but that does not necessarily mean that the mapping itself has to be strictly angle-perserving a priori

what i had in mind was basically the introductions of many extra dimensions, dummy variables, and data in order to possibly make this feasible

[trail tiger]

soo have you looked into lifts of polytopes in optimization? sounds roughly like what you're looking for

hmm but not quite, manifold learning and lifting is new territory tbh

[gentle osprey]

Sozin

[pearl holly]

Okay so I'm on this exercise: the mapping torus T_f of a map f: X -> X is the quotient of X x I obtained by identifying each point (x, 0) with (f(x), 0). In the case X = S¹ v S¹ with f basepoint preserving, compute a presentation for pi_1(T_f) in terms of the induced map f_*: pi_1(X) -> pi_1(X). Do the same when X = S¹xS¹. [One way to do this is to regard T_f as built from X v S¹ by attaching cells.]

So basically I'm taking something that looks like a cylinder and then I fold it like a torus specified by the map f. I guess that the last part in [] is a hint and I assume that I should add 2 cells because then I can just use a theorem in my book. In the case where X = S¹ v S¹ and f(x) = x, I get a "double stacked" donut with a 1-skeleton consisting of the wedge sum of 3 circles. Then I just need to attach a 2-cell by abc a^(-1) b^(-1) c^(-1) I think depending on how you label the generators. That expression will be my relation. But when f is something else other than f(x) = x is where I'm stuck. I don't see what the relations will be then (btw I got it now. There's two relations involved and you use f_*(a^-1) instead of a^-1)

[craggy elbow]

Just for part(c) I thought using the geodesics from part (a), (b) and letting the functions of $\phi=0$\and then rearranging to find $\theta$

[gentle osprey]

Sozin

[novel acorn]

Think about the fact that he wants you to observe the basepoint preserving map f, and think about how that relates to the fundamental group of S¹ v S¹
(Btw these are just some first thoughts I had I'll try to work something out and get back to you)

[pearl holly]

yeah I got it now but thank you!

crap, don't read the last sentence in my post

[novel acorn]

But like you're quick with these problems gotta give you that
Already on no. 11 lol

[pearl holly]

lmao but I skip some of them tho

[gritty widget]

nooooo nooooo you have to do all of them every single one nooooooooooooooooooooooooooooooooooooooooooooooooo

[pearl holly]

I wouldn't consider myself to be quick... I was stuck on the same page for like 3 days...

about the gluing stuff on donut thing

[novel acorn]

I'm currently thinking about no. 3 in the section before this one (since I didn't do most of those)

Maybe a walk will spark an idea

At first I thought that no. 2 would help
But I was mistaken

no. 2 is an easy and fun one tho

[pearl holly]

In chapter 1?

or which chapter are you referring to?

[novel acorn]

1,1

[pearl holly]

no lmao

[novel acorn]

For a path-connected space X, show that π1(X) is abelian iff all basepoint-change
homomorphisms βh depend only on the endpoints of the path h.

[pearl holly]

ah yeah that one. I did it with another person before

[novel acorn]

2. was easy
   Now this is the real challenge

[pearl holly]

I can't really give a hint I think. It's more like manipulating things until you get what you want I guess, just like 2

I did that at least

[novel acorn]

Hm well I did 2. by ||Showing that if I take any two h's that are homotopic to one another I'll get the same βh from them thus showing that it only depends on the homotopy equivalence||

At least I hope that that's a valid proof

[pearl holly]

yeah that sounds right but I could be wrong lmao

[marsh forge]

okay i did 10min of research

time to goof off

[pearl holly]

research

[novel acorn]

The mathematician's life

[pearl holly]

like research research?

[marsh forge]

Im reading a paper that I am trying to generalize and prove an interesting result about

so yes

research research

anyway

[pearl holly]

[novel acorn]

Something
topology related?

[pearl holly]

that's really cool

[marsh forge]

what problem are you working on

[novel acorn]

For a path-connected space X, show that π1(X) is abelian iff all basepoint-change
homomorphisms βh depend only on the endpoints of the path h.

[marsh forge]

oh i figured this out the other day for toki didnt i

[novel acorn]

That thing above in spoilers is for another earlier problem

Show that the change-of-basepoint homomorphism βh depends only on the homo-
topy class of h.

[marsh forge]

did i ever actually explain the answer

lol

[pearl holly]

No lmao but I think that I figured it out at the end

with veryhappy

[marsh forge]

nice!

i think this exercise mostly requires like a few clever ideas

so i dont want to give hints

[novel acorn]

I have come up with an idea
I will introduce new paths at the end points and show that if I move the endpoints along these new paths βh will change

[marsh forge]

you need to explain your idea better lol

i have no clue what you mean

[novel acorn]

Like we have the path h

[marsh forge]

im not even sure what you are proving

[novel acorn]

Ehh it's like from an earlier proposition gimme a sec

[pearl holly]

I don't think you need to use any sort of theorem or lemma or whatever for this one

But maybe you could idk

[novel acorn]

Yeah probably not
I'm just showing what h is

[pearl holly]

oops sorry lmao

[novel acorn]

Hm I'm gonna give this one a think while I walk

be back in a bit

[marsh forge]

That’s how I solved it

Unironically I was on a walk

This is one of those problems where like

There are only so many things you can try

Bc the result is so general

[novel acorn]

I will update if I come up with something on my walk

[marsh forge]

Nice

I’ll go back to research I guess

[novel acorn]

I AM ||yet to get|| a breakthrough

Although I have come up with some ideas

such as the fact that the "iff" says that I have to first prove that if it's abelian it only depends on the endpoints (although I feel like this is already by definition) and that if it only depends on the endpoints it has to be abelian

now about both I'm unsure about how to do right now

[gritty widget]

Since no one tries to help you, I’ll try

[craggy elbow]

@gritty widget thank you

Would you like to see my workings for the previous parts?

[gritty widget]

Yes, I can

[craggy elbow]

Yes you would like to?

[gritty widget]

I’d like to see your works, has any problem

[gritty widget]

That’s alright

[fair idol]

Can somebody explain to me the significance of the Riemann mapping theorem?

[marsh forge]

Solving qual problems

[obtuse meteor]

😎

[novel acorn]

OK I will try to do this problem again tomorrow

||Not that I've given it much though since my walk lol||

[plain raven]

given geometric objects you can classify them with different levels of roughness, from very fine grained equivalence relations to more coarse ones. so biholomorphism is a very strong equivalence relation. arguably the "classification problem" is one of the central problems of any branch of geometry - classification of surfaces, classification of varieties up to birational equivalence. two biholomorphic domains or riemann surfaces or whatever are essentially the same from the perspective of complex analysis, because functions or differential forms or whatever can be transported along the biholomorphism. then, below that you'd have the weaker notion of diffeomorphism, and weaker still of homeomorphism, and lastly of homotopy equivalence, which is a very coarse relation.

so from the perspective of complex analysis if you want to classify things up to biholomorphism you should start by classifying them up to homotopy. of course if you have two domains with distinct fundamental group, they cannot be homotopy equivalent, and thus not homeomorphic. this is like, one of the weakest algebraic invariants in the book, second only to just counting the number of connected components.

so ok, let's restrict our attention to domains in C which are connected and have trivial fundamental group and try and classify those. intuitively you'd try and figure out which are homeomorphic, which then which are diffeomorphic, and then last which ones are biholomorphic, and you'd have finer and finer partitions each time you asked the equivalence relation to respect more structure

what's absolutely shocking is that this approach turns out to be unnecessary. it turns out there are two, and only two, simply connected domains which are open subsets of the complex plane, up to biholomorphism: C and the unit disk. Moreover, the equivalence class of C only contains C itself, and the equivalence class of the unit disk contains all other simply connected open subsets of C.

[fair idol]

Thank you for the really helpful response. I see that the Riemann mapping theorem is about classification of open and connected subsets. This is very neat thank you so much.

There is a side of me which is a tad confused about why this is relevant to a complex analysis course. I understand biholomorphism is a within the realm of complex analysis theory yet this feels like a purely topological/geometric result.

[plain raven]

there are aspect of complex analysis that are more 'analytic', like Montel's theorem, and there are aspects of it which are fundamentally geometric. lots of stuff under the general theory of conformal mappings falls into that latter category

[novel acorn]

@marsh forge I have a question
How do you check if your proof is rigorous enough
I've been doing some other problems from Hatcher but I'm never really sure if my proof is rigorous enough
So how might you deal with this lingering feeling that it isn't complete

[marsh forge]

Uh

You eventually get a feel for when you are right

One thing you can do

Is like

Read your solution

And ask

“If pressed could I justify this”

And if you don’t know how to answer it

Maybe expand a little

[novel acorn]

Hm interesting idea
I'll try that

Thanks for the advice

very impressive
Are they all in Inter-Universal Teichmuller theory
So that only you and Mochizuki know they're correct

[fair idol]

I've noticed a bit of that. I appreciate your help explaining this.

[cedar pebble]

this depends entirely on the audience

some proofs fly with some audiences and are considered mere sketches by others

the vast majority of mathematical proofs are not fully rigorous but are still fine for most purposes

[ivory dragon]

thm: $A \vDash A$

prf: $A \qed$

[gentle osprey]

Namington

[empty grove]

Think of how you'd explain your solution to a class of students (who ask questions)

[empty grove]

Shouldn't the proof also be A ⊨ A

[unreal stratus]

me, a physics student, realising how bad that would be as a criterion for whether smth is rigorous

[empty grove]

It's not supposed to be a criterion lol

[unreal stratus]

yeah sorry lol

[empty grove]

But it allows you to find mistakes often

[native raptor]

i keep watching lectures on handle decompositions and i'm still not sure what's going on

[unreal stratus]

Indeed

[native raptor]

is there anyone who'd be willing to go over stuff with me?

[empty grove]

Like even if you can't imagine what the counter questions would be like, just putting an explanation into words does a lot

[native raptor]

smh i got confused with handlebodies

[livid drift]

anybody up for reading spivak part-I in vc

[pearl holly]

okay so I'm on this exercise and I just can't understand how the fundamental group involves no quotients at all. Isn't this space homotopy equivalent to an orientable surface of infinite genus? Because then the fundamental group will be <a_1, b_1... | [a_1, b_1] ... > but this group involves quotients. But idk, infinity is weird lmao

[hollow harbor]

Does that stuff only work for compact surfaces?

Idk, just asking

The orientable surface of genus g stuff

[pearl holly]

Hmm idk tbh. All I know is that it works for orientable surfaces of genus g lmao

[plain raven]

the quotient of the free group on A by an equivalence relation that is generated by identifying elements in A will still be free

like

if you take the free group on { g1, g2,g3} and quotient out by g1g2 = e

this is still free

[pearl holly]

wait so <a_1, b_1... | [a_1, b_1] ... > becomes free?

where [a_1, b_1] ... denotes the product of the commutators

[plain raven]

I didn't claim that. i was just saying that under certain cases a quotient of a free group will still be free.

but let me think about this case

[novel acorn]

But here you're quotienting over the commutators so you're just doing the abelianization of X right?
So you just have to show that the abelianization is free

[coral pivot]

I imagine this won’t work because you do the genus g surface by looking at it as a quitiont of some polygon

You cannot do that here as the surface you have isn’t compact

[pearl holly]

Hmm yeah that might be the case

But the 1-skeleton is still made out of the wedge sum of infinitely many circles and then I just apply 2-cells to get my surface right? Because then the theorem still applies

[coral pivot]

Let me think about the difference

[novel acorn]

Yes and applying the 2-cell makes the path by which you applied it nullhomotopic i.e. you quotient it out. So you're left with the abelianization of the group of the wedge sum\
It's the same thing that happens to the torus. The group of $S^1\vee S^1$ is $\bZ*\bZ$, but when you apply the 2-cell onto it it abelianizes and you end up with $\bZ \times \bZ$

[gentle osprey]

Irony Incarnate

[coral pivot]

But that ain’t free

[pearl holly]

Yeah exactly but when you quotient it out it doesn’t mean that it is free

[novel acorn]

Something something infinity
But I dunno
I'll have to think about it more

[reef shore]

This sentence doesn't make sense

Groups don't have infinitary multiplication

[pearl holly]

oh okay so I can't use this then?

[reef shore]

Nope

[pearl holly]

ah okay I see. So it's like finite intersection thing with open sets?

[reef shore]

Or rather, the gluing of the 2-cell(s) is not done like that here

Try to figure out the cell structure first

[pearl holly]

wait why the cell structure tho? I won't be able to use the theorem anyway right?

[reef shore]

Van Kampen theorem is still true tho

You can't use the one for finite genus orientable surfaces

[pearl holly]

lmao if pi_1(X) = pi_1(X) * pi_1(X) then what is pi_1(X)? Does it need to be trivial or can it be a free group with infinetly many generators?

[reef shore]

can be the latter

[pearl holly]

lmao but this is probably so wrong

Like you pick your A and B so that the intersection is just a cone and doesn't contain any holes, so it is contractible. The A and B have the same fundamental group as the big space X. So then pi_1(X) = pi_1(X) * pi_1(X) and pi_1(X) can't be trivial

yes slime

[pearl holly]

this is what I mean, just for clarification

I don't know

[reef shore]

Isn't the intersection here an annulus

which deformation retracts to S^1

[marsh forge]

this is a neat problem

the only "solution" i can think of is pretty sketchy

[pearl holly]

hmm idk. Isn't it a filled-in cylinder?

[marsh forge]

no

[pearl holly]

frick

[marsh forge]

wait is it

no its not

surfaces of genus g are never filled in

[pearl holly]

ah yeah frick I forgot

how the heck am I supposed to use van Kampen then?

am I even supposed to use it?

[obtuse meteor]

Van kampen can be applied when the intersection is nontrivial

It just takes more work

[pearl holly]

yeah but my approach here was to find a intersection that is trivial because then I will not have to quotient out by anything and so I will get a free group

but maybe this just doesn't work

[obtuse meteor]

Perhaps the first thing is to prove what the generators are

And you might want to do it for something that’s only one sided infinite first

And use what you learn to do the two sided infinite

[marsh forge]

hm i did some digging and outside of my sketchy solution everything seems pretty involved

[obtuse meteor]

It’s a very strange problem. I guess it emphasizes why a lot of our tools work best for compact spaces…

[marsh forge]

It's a bit annoying because Hatcher does not specify how to construct this space. It's obvious what it should be, but if Hatcher gave the CW structure explicitly it would not be as bad

Actually, I guess this doesn't have a CW structure

The best proof I can find requires some nontrivial lemma || https://math.stackexchange.com/questions/1082073/fundamental-group-of-surface-with-infinite-genus-is-free-on-infinite-generators ||

@pearl holly my suggestions would be that you should explain a few facts that amount to my sketchy proof, and then read the above. This is a cute exercise but not essential.\

1. Explain why the generators for the fundamental group come from the "1-skeleton" of this space

2. Explain why no finite word on these generators is contractible

That basically proves the fact, but the explanations might be a bit sketchy

[pearl holly]

Okay great thank you!

[marsh forge]

I guess an alternative would be to prove that A and B are both free on infinitely many generators and apply van kampen

this might be doable

Actually yeah I think my idea combined with this second idea is a real proof

[obtuse meteor]

That was around what I was thinking

This lemma is a cool thing to keep in my back pocket though

[marsh forge]

that lemma is cool

You can really read it as like

pi_1 preserves very very nice directed colimits

[pearl holly]

the lemma that starts with "if a space X is the union of a directed set of ..."?

[marsh forge]

yes

[pearl holly]

I don’t even understand that

[sullen ether]

Fast topology question, if we have Topological space X, and lets say it isnt connected. Are connected components unique? I mean if we try to write X as union of connected components is this representation unique?

thats enough 😄 Thanks 😄

[sullen ether]

If H is open subgroup of topological group does H neccesseraly contains identity component of G?

[reef shore]

What do you mean by component? Connected component?

It contains the identity element

It doesn't necessarily contain the connected component containing identity

[empty grove]

oof true

[sullen ether]

Okay open subgroups are closed I get that. Now if $H$ is that open subgroup and $G_0$ identity component. Than $H\cap G_0$ is open. How ?

[gentle osprey]

MotionMath

[sullen ether]

hmm okay but what about my problem from up there. How is $G_0\subset H$? Is it true that $G_0\cap H$ has to be either empty either whole $G_0$? and why?

[gentle osprey]

MotionMath

[sullen ether]

My problem is to prove that $G_0$ is contained in open subgroup $H$ of $G$.

[gentle osprey]

MotionMath

[sullen ether]

I dont know how to do it except it has something to do with the fact that $H$ is also closed. Obviously both $G_0$ and $H$ containts identity so intersection is not empty. But I dont know how to prove that $G_0\subset H$

[gentle osprey]

MotionMath

[sullen ether]

Identity component, so it is biggest connected set containing identity

anyone ?

Why is it open

hahaha

ohhhh

Thanks

that solves my problem xD

Thanks a lot I forgot to look at it like that completly xDDDD

[fading vale]

ok time to assblast some setup for this question:

[gentle osprey]

kyhoji, ho!

[fading vale]

I hate you

$\begin{tikzcd}
X \times_S S' \ar[rr, "\text{etale}"] \ar[dd, bend right] & & S' \ar[dd, "\varphi"] \
X_i \ar[u, "i"] \ar[rru, \text{etale}" \
X \ar[rr, "\text{etale}"] & & S
\end{tikzcd}$

HereX_i is a connected component of X x_S S'

Texit pls

So much pain

[gentle osprey]

kyhoji, ho!
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fading vale]

Fuck offfff

Anyway basically what im trying to ask is that if i have a finite etale cover X -> S i can base change to another finite etale cover X x_S S' -> S'

and these correspond to subgroups U and U_S' of pi_1(S) and pi_1(S') (imagine i picked ab asepoint lol)

but i also have a finite etale cover X_i -> S' given by restricting to a connected component of the base change

corresponding to some subgroup U_i of pi_1(S')

whats the relationship between U_S' and all the U_i?

Like can we express it with internal direct sum or something?

is it more complicated? is there an easy way to describe this at all?

Maybe this belongs in #math-discussion

@cedar pebble anyway when you see this itd be neat if you had a reference/answer

[cedar pebble]

No ❤️

(Reading)

[fading vale]

yea its np

[cedar pebble]

Lmfao did you just give up on the diagram

[fading vale]

Yes.

Feeling:

oh i was literally missing a bracket

ok whatever it doesnt matter

[cedar pebble]

Yea so internal direct sum isn’t the right word for it but it’s like

In the category of finite etale covers you’re doing a coproduct decomposition (and a finite etale cover is connected iff it’s coproduct irreducible)

[fading vale]

nGroupoid: "no i wont explain"
also nGroupoid:

[cedar pebble]

When you take your finite index subgroup U of etale pi_1 the way you’re getting the action is taking the regular representation and then projecting to the quotient

[fading vale]

[cedar pebble]

Regular meaning, if you have a group G there is a natural action of G on itself right?

[fading vale]

Yea i follow

[cedar pebble]

So if you have some decomposition of this projection of the regular representation, can we conclude anything about the subgroups that defined this?

[fading vale]

Uh I'm not really sure

My first instinct is profinite completion...?

[cedar pebble]

No need

[fading vale]

cause we have all these finite quotients

I see

[cedar pebble]

I mean okay you can use the fact that these quotients are finite

But you can try to think about this for like

A finite group G

You have U in G yielding G->Aut(G/U) and U_1,…,U_n yielding the same kinds of representations

Say the first one is the coproduct of all these smaller ones

[fading vale]

Right

[cedar pebble]

Try to translate this into something a little more explicit about these subgroups. I think this just amounts to writing out what it means for a finite permutation rep to be a coproduct and so on

[fading vale]

[cedar pebble]

(It’s also kinda fine to just say oh U and these subgroups are related in this way iff the corresponding regular reps decompose like this, that’s a very rep theoretic way to think about it)

[fading vale]

Yeah i dunno any rep theory at all

Im trying to think about this because of like

the characterization of when a subgroup contains the kernel of the induced morphism on etale pi_1

Is there like a source on this i could read

Or something idk i just know literally zero rep theory

[cedar pebble]

You don’t need rep theory. I just mean like, write out what the coproduct of finite permutation representations is

(Hint: disjoint union of sets, with the induced action)

I think stacks 57.3 has some of this spelled out

[fading vale]

does coproduct of reps mean like of the underlying sets

such that the actions are compatible or whatever

[cedar pebble]

57.3.9 is the key I think

Yea

[fading vale]

so i have that pi_1(S')/U = pi_1(S')/U_1 cup .... cup pi_1(S')/U_n

[cedar pebble]

Yup as disjoint unions

[fading vale]

I see

[cedar pebble]

Then you take the induced action

So I think we said we want each of the U_i’s to correspond to connected covers

[fading vale]

yeah

[cedar pebble]

What can we say about the action on each of these quotients?

[fading vale]

Uh it stabilizes the basepoint? or something

[cedar pebble]

U corresponds to a connected cover iff the action on pi_1/U is transitive

[fading vale]

Oh wait yes

Duh

[cedar pebble]

So what you’re doing in general is decomposing the representation into transitive representations

[fading vale]

Right that makes sense

[cedar pebble]

Makes sense? This feels like a satisfactory characterization to me since this is a group theoretic statement

[fading vale]

Uh well i was hoping there was a relationship between U and the U_i

like a more direct one

[cedar pebble]

Hmm

I mean you can unfold this a bit further but it’s kinda like

[fading vale]

cause see the condition for X' -> S' corresponding to U' subset pi_1(S') containing the kernel is that theres an X -> S and connected component X_i (corresponding to U_i subset pi_1(S')) of X x_S S' -> S'

with a map X_i -> X' corresponding to an inclusion U_i in U'

[cedar pebble]

Sure you can start by saying that all the U_i’s are in U

[fading vale]

But I am trying to find a way to think about why its an inclusion specifically of a connected component

and like get some intuition

Right

[cedar pebble]

Yea so I think to actually write down the condition that relates these as subgroups you kinda have to write down this thing involving the regular rep anyways

Right like you really do have to do these things directly, you can’t work directly with U as a subgroup

Since it is typically some crazy subgroup of some crazy profinite group, you can’t really write down elements to see things directly

So you have to work with the finite quotients, and then the condition amounts to this orbit decomposition of the regular rep

[fading vale]

[cedar pebble]

[fading vale]

so we have a map pi_1(S')/U_i -> pi_1(S')/U'

Oh

is a choice of component sort of like a choice of base point

in S'

[cedar pebble]

Is the direction wrong? Taking quotients should reverse the direction

[fading vale]

Oh

Szamuely moment

[cedar pebble]

And yes the choice of a connected component amounts to decomposing the finite set into a bunch of group orbits

And picking (a point in) one of the orbits

[cedar pebble]

If the action is transitive you only have one orbit

Yea I think Szamuely has this inclusion wrong lol

[fading vale]

Hate

Malevolence

[cedar pebble]

Like smaller covers correspond to larger subgroups right?

So U” should be larger

[fading vale]

Yeah no what you're saying makes sense like the map

yeah

[cedar pebble]

Szamuely moment

[fading vale]

Isnt this entire proof irrecoverable then lmfao

or maybe the map X_i -> X' is supposed to be a quotient not an inclusion

Maybe ur supposed to assume surjectivity

of X_i -> X'

[cedar pebble]

That doesn’t sound right

[fading vale]

[cedar pebble]

God this is so dumb I’m pretty sure that proof is ruined

Just read the section in stacks lmfao

[fading vale]

Hate this book

[cedar pebble]

Yea this section is like

[fading vale]

what section is it

[cedar pebble]

Honestly trash

[fading vale]

in stacks i mean

[cedar pebble]

57.3 on Galois categories in stacks

[fading vale]

I see

[cedar pebble]

And just like

The surrounding sections

[fading vale]

:malevolence:

[cedar pebble]

I think like 57.3 through 57.15 is a good replacement

[fading vale]

is the claim of the proposition itself wrong

than X_i -> X' implies it contains the kernel

[cedar pebble]

Hmmm

I think it’s maybe okay just the proof has a mistake?

[fading vale]

does it work if u specify surjectivity

[cedar pebble]

Idk read the stacks sections and compare, they prove basically all the same things and those proofs are fine

[fading vale]

[cedar pebble]

Honestly just like don’t read this section in Szamuely it’s so bad

Return for the survey of advanced results though

[fading vale]

i already read this i was just reviewing

[cedar pebble]

Nice

[fading vale]

monkaS

I think hes implicitly assuming surjectivity

[cedar pebble]

OH

Ooh okay yea I see what’s going on

Okay yea the proof is fine now I think it’s just really poorly communicated what is being assumed about this map

[fading vale]

Hmm

@cedar pebble as a followup question: if i identify X with pi_1(S)/U does U correspond to automorphisms of the fiber functor that are trivial on X or something like that?

[sullen ether]

https://prnt.sc/1fd8rh1

[swift fjord]

when constructing a CW complex inductively, at each step the topology on the space is the quotient topology of the disjoint union topology of the space with the unit disks right?

[marsh forge]

Yes

[plain raven]

hey fam

i'm a couple beers deep

just read about the steenrod squares today.

think i'm starting to understand what's going on there

[marsh forge]

Very cool

Axiomatic or constructive?

[plain raven]

they stated the axioms first and then gave a construction but this book dodges uniqueness

it just proves existence

but they will revisit them later to give invariants associated to manifold embeddings so i'm looking forward to seeing tat

i understand they also have something to do with Stiefel-Whitney classes?

[summer jolt]

Any recommendations on introductory readings about etale spaces and sheaves?

[plain raven]

the book Sheaves in Geometry and Logic starts with a chapter on category theory and a chapter on sheaf theory. Lectures on Riemann surfaces by Otto Forster has a good introduction to sheaves with an immediate application to dealing with multi-valued functions in complex analysis, such as the integral of the natural logarithm. The book "Topologie Algebrique et Theorie des Faisceaux" by Godement introduces sheaves, etale spaces and sheaf cohomology. (Also a good introduction to spectral sequences, I might add.)

[summer jolt]

Ah thanks that Mac Lane book looks quite good!

[fading vale]

are you looking for something with a more general categorical perspective or something more rooted in the algebraic geometry

[plain raven]

i haven't read the bredon book. it looks interesting tho i'm curious about cech homology and iirc bredon addresses it there. i think he also talks about borel-moore homology?

[fading vale]

Bredon

I read parts of his intro AT the writing style was not my thing

[plain raven]

btw mac lane's book doesn't address cohomology at all, which is the reason sheaves were invented, and also later the reason topos theory was invented. kind of a glaring omission imo but yeah it's a good place to just learn the technical basis of sheaves and etale spaces

at least better than hartshorne

[fading vale]

is etale space here supposed to mean like finite etale cover over a scheme or the general definition of local homeomorphism in the slice category or whatever

[plain raven]

probably the latter, as sheaves correspond to those

[fading vale]

Yeah thats what it sounds like just wanted to make sure though

[novel acorn]

Um I just want to check here
Borsuk-Ulam still holds for the torus
👀
At least my proof says it does

I'm unsure however about the correctness of my proof so I thought I'd check if I got the correct result

[fading vale]

It does not

[novel acorn]

Then I think I know why

It was the step I was unsure about

So good to know that my assumption was wrong

[fading vale]

To be fair im not 100% clear on what you mean by borsuk ulam for the torus

are you asking if maps from T^2 -> R^2 has to map 2 antipodal points to the same point in R^2?

[novel acorn]

That any mapping $S^1\times S^1 \to \bR^2$ has to have some two points such that $f(x,y)=f(-x,-y)$

[gentle osprey]

Irony Incarnate

[fading vale]

Take the projection onto the plane

of the usual embedding

[novel acorn]

Ohhhhhh

damn

I was constantly thinking about the wrong thing

damn it

I was visualizing that $f(x,y)=f(x,-y)$ for some reason

[gentle osprey]

Irony Incarnate

[novel acorn]

I got about 60% of those words but still very intriguing

[summer jolt]

Categorical perspective. I don't know much algebraic geometry. I came across them in the idea of geometric morphisms and wanted to understand the basics.

[flint cove]

„Every map f: S¹→S² is homotopic to a map g that misses the north pole N“. How would we prove such a thing, given that terrible things like space filling curves exist?

I mean if the preimage were only finite then I would've guessed try to restrict to a nbhd of said preimages and move them away from N one by one

but we only know f⁻¹(N) is closed

Or can we guarantee that a point other than the north pole exists that has only finite preimage?

that's kind of what I want to prove 😉

or at least I want to know how Borcherds' remark along this way rigorously works

Hm, not sure, never looked at the van kampen proof that closely
I guess we could use celllular approximation, but I never looked at that proof either

sec, gotta find that video. Watched it a few months ago and suddenly the question popped back into my head

Ah, nevermind, he seems to adress it in the video

https://youtu.be/WL9BFRyWMUI?list=PL8yHsr3EFj52yxQGxQoxwOtjIEtxE2BWx&t=887 for anyone interested

Okay, his argument is a bit weird and I am irritated
He says to cover S² with four disks, and consider f to be [0,1]→S¹.
Now, „because [0,1] is contractible“, he claims there is a finite decomposition 0<a₁<…<a_n<1 such that theh image [a_i, a_i+1] is contained in a disk respectively

I am not sure what contractibility has to do with that at all

This sounds a bit like taking the preimage of each disk and using compactness

more precisely, we can construct an open cover by considering the preimage f⁻¹(D_i) of each disk which is a union of all open intervals (a,b) contained in said preimage
Then these intervals form a cover of [0,1] and each have the property that their image is bounded in a disk.
Then there must exist a finite subcover of such intervals, which we can refine to a closed cover of the form [0,a₁], [a₁,a₂], …, [a_n,1] (note that refinement does not change the property of being confined to one of the disks in the image)

So that should work, I can go to sleep now.

(the rest of the argument he gives is that f restricted to [a_i, a_i+1] is homotopic to the geodesic path from f(a_i) to f(a_i+1), which of course works fine)

[hollow harbor]

Borchie moment.

Yes it's compactness

He probably just said the wrong word

[gentle osprey]

craft

[jagged ocean]

hmm, equivalently I'm really asking why definitions 1 and 3 are equivalent on this wikipedia page: https://en.wikipedia.org/wiki/Fundamental_vector_field#Definition

[strong heron]

Hehe.

Absolutely, he meant compactness. This was pointed out in the comments on the video also. Also, I think fewer than 4 disks (like 2) may also be used and the argument can still go through.

[novel acorn]

Why did he last upload two months ago

[hollow harbor]

Borchie busy

I do think he'll be teaching this fall

but

it's not the fall

[meager python]

Why is it not ”weird” that pi^et_1 of a point is not necessarily trivial? Except when we look at algebraically closed fields? Why doesn’t this ”contradict” the relation to pi_1 in topology?

[topaz plover]

I've been told that I only need to study point-set for like 30 minutes to an hour before studying algebraic topology; is that advisable, given that most schools will offer point-set courses as prerequisites for algebraic topology?

[obtuse meteor]

I’d say more than 30 minutes to an hour, but the general sentiment is correct

[novel acorn]

Also for algebraic you do need abstract algebra

and I dunno if that generally comes before or after point-set

[topaz plover]

Yeah, I have that.

[obtuse meteor]

Strong claim

[hollow harbor]

[novel acorn]

You don't know the pain of reading Hatcher before learning AA

[hollow harbor]

homological algebra isn't algebra apparently

[obtuse meteor]

(I’m memeing but also not entirely joking)

[hollow harbor]

no but i kind of get the idea

it's like

different

[novel acorn]

Then just like read the first part of Munkres
That'll get you up to (very detailed) speed

[obtuse meteor]

You can just like

[hollow harbor]

and you can learn how to do it without learning a bunch of algebra

[obtuse meteor]

Learn the algebra you need as you’re doing AT

In my experience as long as you have basics “I know what a group and group homomorphisms are”

[novel acorn]

Well I mean you still need to understand how groups work and stuff like that

[hollow harbor]

i mean

sort of?

[novel acorn]

I assume that most AT textbooks assume you understand what a homomorphism is

[topaz plover]

How much algebra do you need? I know up to galois theory, but I don't know much commutative algebra and I don't know any homological algebra.

[novel acorn]

Yeah that's more than enough

I should actually read up on Galois theory tbh
Never got around to that

[pearl holly]

Monarch I know up to galois too and I’m getting through AT so you should be fine, but I’m still a noob so I shouldn’t be answering lmao

[obtuse meteor]

Knowing galois correspondence before galois covering space correspondence is bad and not good for the soul 😌

[topaz plover]

I only know galois correspondence.

[hollow harbor]

of course

you need to learn like

2 hours of group theory

and you need to know some linear algebra

[novel acorn]

well

maybe a bit more

but I dunno

[hollow harbor]

i suppose so

i think if you know like

axler linear algebra

you can learn the group theory / module theory you need for an intro alg top course super quickly

but if it's your first time doing algebra then it's a totally different ball game i agree

[novel acorn]

Well I mean I had no real contact with abstract algebra so I spent a bit more time learning everything about groups, rings, fields bla bla

[bold canopy]

when I was first learning groups, the only way I knew how to "understand" a group was by drawing its cayley graph and staring at it, which didnt really help that much
for ages people would parrot "understand a group by its actions" and that meant nothing to me
just a random memory

[novel acorn]

Learning about them through cayley graphs is pretty specific ngl

[marsh forge]

fun fact

the first time i truly understood cayley graphs

was the first time i took mushrooms in 2nd year

[novel acorn]

Math on shrooms
That feels like a life changing experience

[marsh forge]

i felt like i understood a lot at the time

turns out i was just pretty high

[novel acorn]

You finally get how to do that one Hatcher problem

[marsh forge]

the infinite one? I do think i know how to do it

[pearl holly]

ight time to get some mushoorms and grind all the exercises

[novel acorn]

Just any one that has been bugging you for a while

[marsh forge]

i am long past thinking about hatcher problems hahaha

why do authors before the existence of spectra insist on using machinery not available to spectra

also

someone needs to write a second course in alg top book

(me)

and introduce spectra and model categories properly

or maybe by that time Lurie and his acolytes will have fully cleansed this earth of the concept of a 1-category

[swift fjord]

What module theory would you need besides knowing what the free Z module is?

[hollow harbor]

I've never actually needed any real module theory for the alg top I've studied.

But I'm guessing it's useful for homological algebra stuff?

maybe not

[marsh forge]

You dont need much intense module theory

You should know the basic ideas and defns

[swift fjord]

ok ye makes sense

[fading vale]

ok the bespoke answer is that they should be learned simultaneously in the setting of riemann surfaces

[shy moss]

hi

why the classifying space of $\mathbb{Z}$ is $S^1$?

[gentle osprey]

Or x1

[cedar pebble]

BZ is a connected space with the property that its fundamental group is isomorphic to Z and all higher homotopy groups are trivial. But S^1 is such a space too, so they are equivalent

[marsh forge]

It might help for you to clarify the question as to what part of the claim confuses your

[hallow swan]

in figure 13.1, isn't (1) not satisfied?

all of the points outside of the two circles are not in any basis elemetns

same with 13.2

[gritty widget]

the basis consists of all such sets, not just the ones in each figure

[hallow swan]

wym by that

[hallow swan]

ohhh

so there's like inf of them

not 3

lol

[gritty widget]

yeah lol

[hallow swan]

thanks

[gritty widget]

it's just showing you what (2) in the definition of a basis means

using this particular basis

[hallow swan]

i see

and to be clear, X is the plane or R^2 ?

it's not really formally defined

[gritty widget]

i guess in this particular example it'd be the plane

it says so in the caption

[hallow swan]

yea

in 13.2 for the rectangles

can't the intersection of two rectangles be a line segment

and the line segment can't contain a basis element

wait it says interior nvm

[gritty widget]

[hallow swan]

also whenever munkres is checking the union case for a topological space he says "let us take an indexed family"

but "indexed family" sounds like it has to be a countable union

when it can be uncountable too?

[gritty widget]

indeed it can

[hallow swan]

yea so "indexed family" is bad terminology then?

or am i misinterpreting

[gritty widget]

why is it bad terminology?

you can always add "(un)countable" before

[bleak helm]

An indexed family just means a family of the form ${U_a: a \in A}$ where each set is indexed by some element of A. But A can be finite, countable or uncountable so that will determine the size of your family

[gentle osprey]

Lunasong the Supergay

[bleak helm]

It doesn't specifically mean indexed by the natural numbers if that's what you are thinking

[pearl holly]

Man I remember reading that part from Munkres

[hallow swan]

yeah that's how i interpreted

[hallow swan]

the analogy at the bottom

"the collection of open sets has been enlarged"

sure, you have a greater number of open sets

but the set corresponding to a collection of pebbles which was smashed no longer exists

like {1, 2, 3} turned into {1} and {2, 3}

[true robin]

Unions of pebbles are open as well

[hallow swan]

oh true

that's what a topology is

nice

wow the discrete metric induces the discrete topology

who would've guessed

[plain raven]

that's a crazy coincidence

[novel acorn]

Munkres is comfy ngl

[gritty widget]

my mind is blown

[unreal stratus]

What about the Euclidean metric?

Which topology does that induce? 😢

[coral pivot]

Euclidean topology

[unreal stratus]

Woahh

lmao

[swift fjord]

Guess what metric induces the trivial topology

[empty grove]

Finally, trivial metric

[wicked mirage]

you mean, imaginary numbers?

[novel acorn]

Those actually aren't real hehe
Geometric algebra corrupted me

[gritty widget]

How can I start to study topology? And what previous knowledge I need

[reef shore]

No previous knowledge needed but knowing about metric spaces can help with some intuition/motivation. Munkres is the standard text

[gritty widget]

Thanks

[hazy nexus]

If you don't know about metric spaces, knowing how to draw blobs can help

[zealous glen]

Is there any difference between an endomorphism T on V and saying that that V is T-invariant. The paper I am reading keeps saying T-invarient and I don't want to misunderstand the author.

[marsh forge]

Uh

Is V a subspace of some larger vector space

[hollow harbor]

well endomorphism usually means T is a map from V to V (V is the whole vector space). you usually say "W is T-invariant" for a subspace W if T(W) is a subset of W.

[marsh forge]

^

[hollow harbor]

so "V is T-invariant" means the same thing as "The restriction of T to V may be seen as an endomorphism of V after restricting the codomain"