point-set-topology

[dim meadow]

What a strange thing to brag about

[alpine rain]

Hm not bragging actually

Sorry for my English

[dim meadow]

@marsh forge is the definition that you need a computable dense set with a computable metric defined on it? I guess you want the metric to take values in computable reals on inputs from your dense set

That would be my initial guess for a definition

[alpine rain]

They found the paper describing that one could exhaustively search for candidate in cantor space

[dim meadow]

Oh cantor space is something else

Computability theorists care about cantor space a lot

[alpine rain]

Then they thought it is topologically generalizable

[dim meadow]

And it's pretty well studied

Idk if what they are saying is anything useful, searching is a pretty basic thing in computability

What sort of thing are they searching for?

Also what paper are you referring to?

[alpine rain]

It is not a paper just website tbh

[dim meadow]

What website?

[tidal cedar]

Wait they were saying exhaustive search

Of a compact space

Regardless of cardinality

[dim meadow]

You can only care about cardinalites up to the continuum

If you want a countable dense subset

[alpine rain]

http://math.andrej.com/2007/09/28/seemingly-impossible-functional-programs/

[tidal cedar]

Unless you have additional info, you can't exhaustively search even a countably infinite search in finite time, right?!

[alpine rain]

Ye flona exactly.. it's meh

[tidal cedar]

Like you need additional info of SOMETHING

You can't just give any compact space and expect to search through all the elements

[dim meadow]

Yeah

You need to know what the finite subcover is

Which is not something that's computable

[alpine rain]

Indeed

[tidal cedar]

I can see like, orders, gradients, etc. allowing it maybe, but the general thing is impossible

Yeah

Is there a Turing degree which can?

[dim meadow]

Well idk if Turing degrees even can talk about this stuff

[tidal cedar]

Ahh

[dim meadow]

Because this stuff is big

[tidal cedar]

I was wondering if like computing a cover or something is halting, then adding an oracle for that allows for exhaustive search, IDK though

[alpine rain]

I get facepalm when cs ppl (or ofc any ppl) go like this, thinking of generalizing v specific thing

[dim meadow]

Idk actually, I feel like we are talking very vaguely now and I would need to sit down and make things solid before I would be able to say anything

[tidal cedar]

I'm really intuition-reliant with topology and even I basically think of uncountable spaces as my go-to example

Like uncountable as in underlying set, roughly "continuous" in the real sense IG

[dim meadow]

Okay here's me trying to make things solid

Open sets in cantor space correspond to sets of finite strings.

An open cover is going to correspond to a collection of sets of finite strings

[alpine rain]

Oh, that clarified things up

[dim meadow]

Let's say we are even only considering the computable open covers, that is we have a collection of indices of ce sets W_e which are sets of finite strings (I think this is a pretty good notion to consider)

[tidal cedar]

This is because cantor space in general is homeomorphic to a countably infinite product of some n-point discrete space, correct?

[dim meadow]

Well the basis elements are just the infinite strings that start with a given finite string

[alpine rain]

Start vs. Finitely many somewhere I guess

Product space just require finitely many iirc

[dim meadow]

Okay so now the problem is that you are asked, given a computable subset of the naturals that describes an open cover as I outlined above, to output a finite subset of that set which is a subcover

[marsh forge]

yeah something like this although I think "computable dense set" is kind of a misnomer

[dim meadow]

Now that I have described this I can actually talk about Turing degrees

[marsh forge]

bc most of the time you sort of assert that some subset is the computable dense set

or you base it on something that makes sense already

[alpine rain]

Woah talking abt Turing degrees

[dim meadow]

I'd imagine you can just have the naturals as your base if you want

That's what we do in computable structure theory lol

Okay so now an interesting question is "how hard is it to compute a finite subcover, given the setup I outlined above"

Another interesting question is "which indices describe open covers of cantor space"

[chrome dew]

interesting, sounds a lot like how I think about the 2-adic integers

[dim meadow]

Yeah they are topologically the same I think

[chrome dew]

naturals are dense, etc

yeah but what you're saying sounds like it's actually useful

[alpine rain]

Btw the cs person who thought this exhaustive search property would be generalizable on topological apaxw claimed that they are math major. Hmmm.

[dim meadow]

I feel like I'm having 2 discussions at once and maybe the streams are getting crossed

[chrome dew]

like what are you doing, like I want to steal the derivative and use it now to do some kind of gradient descent on what you're talking about

[dim meadow]

Haha

Yeah it turns out doing useful stuff is hard and they should try carrying it out instead of talking about it cascadar

Even describing what you want to do is hard tbh

[alpine rain]

Exactly

[chrome dew]

yeah, like seems as though you could just solve chess this way, this is not really realistic stuff here

[dim meadow]

I'm actually pretty interested in the Turing degrees of the stuff I described now, probably they are pretty terrible

I doubt they are even arithmetical

[alpine rain]

They also seem to think classical proof could easily be converted to constructive proof. Idk if this is true

[dim meadow]

Lol

[tidal cedar]

That is a GRAVE sin

[dim meadow]

Sounds like they've solved math

[alpine rain]

Lmao

[marsh forge]

constructify.py

[dim meadow]

We can all go home now

[tidal cedar]

Only heathens who assume LEM believe that

[dim meadow]

It's over

[tidal cedar]

The based and Coqpilled non-constructivists know that constructive proofs are not that easy

[alpine rain]

Idk, it looks silly when they worked out exhaustive search on naturals

And then think they could go further easily

Do mathematicians care about constructivism proofs?

[marsh forge]

constructive proofs are always nice

they normally allow for further exploration more easily

[alpine rain]

I see

Hm I should have said

Is there some mathematicians who do not care about constructive proofs?

[marsh forge]

it depends what you mean

i think very few mathematicians would not prefer a constructive proof given the option

as there is no downside

but lots of mathematicians don't try to make all their proofs constructive

in fact most don't

[alpine rain]

Oh ya, exactly

Latter is more of what I meant

[marsh forge]

there are also theorems that aren't true constructively

like theorems that aren't true if you remove LEM

[dim meadow]

Like LEM

[marsh forge]

i start every proof by contradiction

even if i find an explicit example

or a general construction

[tidal cedar]

burning hot take: LEM is intrinsically true, you not only have to not assume LEM but add ~LEM

I am entirely joking Ultra

[alpine rain]

Wait anticonstructivist existed

[marsh forge]

no

it was a joke

[alpine rain]

Oh

[tidal cedar]

Mathematics should be pure and unsullied of construction

[alpine rain]

Lmao

[tidal cedar]

These mathematical objects are not reachable by the mortal means of construction / description, only by the divine probing of some limited facets of their properties

I should make an anticonstructivist manifesto on my site or something

[alpine rain]

Great meme

[marsh forge]

to attempt to compute is to attempt to ascertain divinity

it is a sin

[tough imp]

Y’all are silly af

Just use the axioms whenever u feel like it who even cares all this is fake bruh

Except AG I saw a scheme b4

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

[sweet wing]

computationally nonstructive proofs are a pain

like sure there exists some object

but not being able to make it exist in code is

[frigid patrol]

cry

tfw your spaces are too big to compute anyway

[sweet wing]

or when LLL is taking hours and hours

at least it isnt years

[tight agate]

@gritty widget you might find this interesting: https://arxiv.org/abs/1906.08618

[sleek thicket]

@tight agate you end up getting the torus, not the klein bottle

I'm a little fried so I don't want to write out all the details rn

But basically the intuition is that we're acting on S^1 by a rotation

If you acted by complex conjugation you'd get a klein bottle

but rotations don't like...flip badly enough lol

[chilly rune]

Anyone know/ have an example of an explicit formula for a cellular approximation map to the diagonal map for some nice saces say RP^n T^n for small n say 2,3. Id like to understand how to get an explicit formuka for higher dimensional cases but am having trouble with the low dim ones. I'm currently interested in the cup product in an SO(n) chain complex, so inderstanding these diaginal maps is key. Would appreciate any comments/ideas

[tight agate]

@sleek thicket that makes perfect sense lmao, all the morphisms in your structure group were orientation preserving

[sleek thicket]

Yeah exactly

[tight agate]

For some reason it felt like -1 had negative determinant

[sleek thicket]

the next problem is about finding a bundle with total space the klein bottle

So I was very careful and did mayer vietoris

And realized that at one point you get x - y since conjugation is degree -1

But for the torus you'd get x+y since negation is degree 1

And that's what I'd screwed up before

[tight agate]

noice

your class sounds dope

they should have a class that does just bundles and characteristic classes over here

[sleek thicket]

It's pretty cool

We're doing fibration LES tomorrow pog

[gritty widget]

is the reason ad is notated ad with lowercase letters because it's like an infinitesimal version of Ad

[wanton marsh]

yes

[gritty widget]

how did this only occur to me just now

[sleek thicket]

you know why it's called the adjoint representation right?

If you're sufficiently galaxy brained you can frame it as a 2-adjunction of monoidal categories

||this is false I just want to make tterra afraid||

[tight agate]

once you do the fibration LES you can build up to the classical ASS very quickly

and stuff about killing homotopy groups

say you have an n-connected space X

Then Hurewicz lets you compute the n+1 homotopy group from cohomology

if pi_n+1(X) = G

then you can find a map X \rightarrow K(G,n+1)

such that it induces an iso on pi_{n+1}

and then take the homotopy fiber, F of that map

all the homotopy groups up to degree n+1 of F are 0

and all the higher homotopy groups of F are iso to the higher homotopy groups of X

as F is (n+1) connected, you can compute pi_{n+2} using Hurewicz

and repeat

lemme try drawing the diagram

[\begin{tikzcd}
{...} & {X_2} & {X_1} & {X_0} \
& {K_2} & {K_1} & {K_0}
\arrow[from=1-2, to=1-3]
\arrow[from=1-3, to=1-4]
\arrow[from=1-4, to=2-4]
\arrow[from=1-3, to=2-3]
\arrow[from=1-2, to=2-2]
\arrow[from=1-1, to=1-2]
\end{tikzcd}]

[gentle osprey]

Brofibration

[tight agate]

X_1 = F

the K_i are wedges of shifts of eilenberg maclane spectra

keep doing this

and apply pi_{*}

and you get

[\begin{tikzcd}
{...} & {\pi_*X_2} & {\pi_*X_1} & {\pi_*X_0} \
& {\pi_*K_2} & {\pi_K_1} & {\pi_{}K_0}
\arrow[from=1-2, to=1-3]
\arrow[from=1-3, to=1-4]
\arrow[from=1-4, to=2-4]
\arrow[from=1-3, to=2-3]
\arrow[from=1-2, to=2-2]
\arrow[from=1-1, to=1-2]
\arrow[dashed, from=2-4, to=1-3]
\arrow[dashed, from=2-3, to=1-2]
\end{tikzcd}]

[gentle osprey]

Brofibration

[tight agate]

and pi_*K_l is easy to compute

so you have a spectral sequence converging to the associated graded of pi_*(X)

and you get a description of the E2 page by applying cohomology to the first diagram

which is still hard to compute

[violet sonnet]

Hey i noticed some neat tikz pics here and there's not channel for latex help so i hope you dont mind i ask here :)

\begin{tikzpicture}
\newcommand \ra {0.9317451479533244};
\newcommand \rb {0.5190351276169526};
\newcommand \rc {0.9924344287178753};
\coordinate (A) at (0, 0);
\coordinate (B) at (0.448315760255185, -1.3797740347935905);
\coordinate (C) at (1.7949222969810386, -0.6933404582734711);
\draw (A) circle (\ra);
\draw (B) circle (\rb);
\draw (C) circle (\rc);
\draw (A) -- (B) -- (C) -- cycle;
\end{tikzpicture}

[gentle osprey]

jn3008

[violet sonnet]

im trying to label this diagram, should be simple

i want to draw a line from the midpoint of one of the circles to its boundary, I can calculate the coordinates by hand but i want to be able to generate this quickly for any coordiantes and radii i provide

something like the blue lines

[violet sonnet]

i figured it out, but does anyone know how to get the node text on the other side of the line?

[sweet wing]

the latex server (dm the bot like ~help or smt) is a better place to ask
idk what code you used to make the text but jus google "node text in opposite direction" or something

[violet sonnet]

thanks

but sometimes i can't find what im looking for on google

ill look for the latex server 👍

[alpine rain]

Is classical like LEM extra structure over constructivist math? E.g. when talking about topology? Got that response while arguing with this person who says that one can say compact when referring to computationally compact, and that is no extra structure

(I think) the person I'm arguing with is mentioning that being computationally compact lifts to being compact, so that's not much a point in differentiation

[rugged swan]

is there an example of a surjective morphism of sheaves f : F -> G which isn't surjective on sections ?

[tough imp]

Tons

Here’s a good example

Consider F and G on C where G(U) is the set of non-zero analytic functions on U and F(U) is the set of analytic functions on U

You have a map F -> G which on an open U sends a function f to e^f

This is not surjective on sections as not every analytic function has a logarithm, consider something like just the function f(z) = z on the unit disk

However locally every analytic function has a logarithm, you can construct it using a line integral i think?

[gritty widget]

Hello 🙂

I need help with algebraic topology it as applied to the blue brain project, a neuroscience study on the brain's (neocortex sample) capacity to form mathematical objects in up to 11 dimensions via neuronal connections.
https://interestingengineering.com/the-human-brain-is-capable-of-building-structures-with-up-to-11-dimensions

More specifically my friend is in need of understanding it within the boundaries of this project

[chrome dew]

I remember seeing some kind of e textbook of algebraic topology for neuroscientists a year or two ago but

maybe I hallucinated it

yeah first I heard of it was I was at an algebraic topology conference and was talking to a guy doing some kind of neuroscience stuff with the development of the brain

maybe 5+ years ago I forget when this was now

[gritty widget]

they turned mathematics into commutative algebra, and now they have the audacity to turn neuroscience into commutative algebra!?

😡

[chrome dew]

neurons are objects and axons are morphisms 😳

[rugged swan]

yeah thanks good example !

[elder yew]

They haven't taken analysis yet

[tight agate]

moonbears that's kinda sorta what comm alg does

it replaced calculus

and analysis = calculus

[sleek thicket]

brofib attempting to give all the geometers in chat a rage induced heart attack

(algebraic geometers are algebraists so they're not included in what I said)

[tidal forge]

is the center of mass of a piecewise linear surface the weighted average of the centers of those pieces?

[chrome dew]

yes

[shut moat]

is there a nice explicit proof that the Mobius strip is not orientable?

Hubbard just said "lmao look at it"

I found these GATech notes online

it was also basically "lmao look at it"

[sleek thicket]

What's your definition of orientable again?

[shut moat]

an orientation is a continuous map from B(M) --> {-1,1} who's restriction to each tangent space is an orientation of the tangent space

the almost-frame-bundle thing

although Hubbard also introduced a list of ways to get orientations

with nonvanishing k-forms or normal vectors

also could it be done without charts? The remainder of the online stuff uses that, although I'm not super comfortable with it since Hubbard sticks to embedded manifolds and parametrizations (which are strictly global)

and hey terra lol

[gritty widget]

hello approximately circle

[sleek thicket]

Is your mobius strip compact or infinitely extended?

[shut moat]

compact I think

[sleek thicket]

how is it defined?

[shut moat]

it isn't rly "defined"

[sleek thicket]

My thinking is basically if you have an orientation then at each point you know which of the two boundary components is in the positive direction

[shut moat]

with this figure

[sleek thicket]

oof

[shut moat]

lol right?

[sleek thicket]

Alright so here's my high level argument

Choose an orientation

Let r : M -> S^1 be the retraction onto the center

[shut moat]

retraction meaning just projecting onto S^1?

[sleek thicket]

well there's this circle in the center right?

[shut moat]

yeah

[sleek thicket]

So we're sending a point of M to the point on that circle on the same line

it's a projection onto the circle

[shut moat]

gotcha

[sleek thicket]

for any point p in M, there's two boundary points of r^-1(r(p))

The orientation should actually allow us to choose a continuous function b : M -> 👌M such that r(b(p)) = r(p) and

[shut moat]

lmao

👌 M, I should use that from now on 😂

[sleek thicket]

so basically like, at each point you have a positive direction

[shut moat]

[wait nvm]

[sleek thicket]

Oops sorry I forgot about this

The idea is basically you can continuously measure the "distance" of a point in r^-1(p) from one of the boundary points and get a value in [0,1]

and then this gives a global homeomorphism M -> S^1 × I where the center of the mobius band maps to (z, 1/2)

and then you observe that you can cut out the center of a mobius bundle and get a cylinder, whereas if you cut out the center of a cylinder you get two connected components

Sorry I don't really know a good definition of the mobius band in R^3, here's how I usually think of it

[sleek thicket]

topology

i am thinking

let $Y = (I \times \R) \amalg (I \times \R)$ with inclusions $j_1,j_2 : I \times \R \to Y$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

I want a quotient map $r : Y \to \mathbb{T}^2$ with the following fibers

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

\begin{align*} r^{-1}(r(j_1(t,s))) &= {j_1(t,s+n) : n \in \Z} \quad \text{if } t\in {0,1} \ r^{-1}(r(j_2(t,s))) &= {j_2(t,s+n) : n \in \Z} \quad \text{if } t\in {0,1} \ r^{-1}(r(j_1(0, s))) &= {j_1(0, s + n) : n \in \Z} \cup {j_2(1, s + n) : n \in \Z} \ r^{-1}(r(j_1(1, s))) &= {j_1(1, s + n) : n \in \Z} \cup {j_2(0, s + n + 0.5) : n \in \Z} \end{align*}

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

I thought I saw how to do it by shearing

but I no longer see it

@gritty widget change your status from homogeneous to homogenius

[gritty widget]

no

[sleek thicket]

oh okay

[gritty widget]

not until i fully understand this proof in my symp geo class

only then can i do that

[sleek thicket]

okay anyways enough about simp geo or whatever

someone help me toruspill this problem

also to be clear, "having these fibers" is just a compact way to say "makes these identifications"

[sleek thicket]

hmm I think I have an idea

I can first glue I x R and I x R along (1, s) ~ (0, s + 0.5)

and reduce to looking at a quotient of [0,2] x R

hmm except this just moves other stuff over........

[sleek thicket]

What surface does < a, b, e | aeba¯¹e¯¹b¯¹ > present?

Feel like it should be the torus

I mean it's the torus unless I fucked up lol

[hazy stratus]

Oh the torus came up in like our last two problem sets

[hazy stratus]

What kind of notation is that? I haven't seen it before

It looks like a generated group

[sleek thicket]

It's supposed to haha. It's called a "polygonal presentation of a surface", and if there's only one relation then the fundamental group of the surface has that presentation

[hazy stratus]

Which seems kind of weird in this context

Ahh

[sleek thicket]

So here's an example

[hazy stratus]

Oh that's cool

It's a representation using symmetries?

[sleek thicket]

These are the presentations
<a, b|abb^-1a^-1>
<a, b|aba^-1b^-1>
<a, b|abab>
<a, b|abab^-1>

Presenting the sphere, torus, projective plane, and Klein bottle

[sleek thicket]

Not really. It's telling you how to glue the shape together

So you'd glue together two edges labeled "a"

And you might swap them or not depending on whether it's a or a^-1

[hazy stratus]

Like say, which points on two opposite lines of a square are glued together?

Looks kind of like origami instructions

[sleek thicket]

Yeah, that's exactly it!

It's not quite folding, more gluing

but the idea is you could literally take some paper and put it together

(maybe not for all of them, some surfaces can't be embedded in 3d space...)

[hazy stratus]

That's pretty cool 😄

[tall flame]

idk if this fits here but i ask

I have different weights for a neural network which attempts to return the mask of an object on a picture

the thing is, if it does pretty well on certain images, it takes background from others

I will keep training the nn, no worries. But

Can i somehow get the perfect mask combining the output of different models?

Like the intersection or something?

[sleek thicket]

i do not think this fits this channel. I mean I don't really care if you post it here but you'll probably get more help asking somewhere else

[tall flame]

where?

[sleek thicket]

Maybe on a cs/programming server? It's not really a math question

[tall flame]

it is about masks

in the end

a group of pixels

[frigid patrol]

Math babble babaabbbaab

[gritty widget]

why not just make the honorable names white at this point

[sleek thicket]

robble robble

[tall flame]

my ... idk how to say, last project for my degree

was a python script to identify any manifold

:)

https://gyazo.com/e1d6aac7fe48e83fa4ea4a72fbd698b2

this xD

[frigid patrol]

I think you meant genus

[tall flame]

ah

XD

[frigid patrol]

But that's cool

[tall flame]

maybe

i didnt made it on english

:P

[gritty widget]

ah yes

gender 1, 2, 3, etc.

[tall flame]

si quieres te hablo en español despojo social

:)

[gritty widget]

gender 0

[tall flame]

better now? :)

[gritty widget]

that is pretty neat, though

[chrome dew]

implying there are countably many genders

bigot!!

[gritty widget]

[sleek thicket]

I'm thinking about this as a polygonal presentation

If I restrict to 0 <= s <= 1 on both copies

I get this

Does that seem right?

s is the horizontal coordinate in these pictures, I guess

and t is upside down...

But besides that

[marsh forge]

what space is this

[sleek thicket]

Certain quotient of (I × R) disjoint union (I × R)

It's supposed to be the torus

I'm not seeing it

[obtuse meteor]

The beast returns

https://estrogen.fun/i/l2u5.png

[red garden]

question number 8

[astral cedar]

I am pretty sure that the first one is homeomorphic to the real line with 0 removed

Yeah, the projection on the x axis is a homeomorphism with R\{0}

The second one is connected, but it takes a bit of work

You can look up the topologist’s sine

[little hemlock]

Think: the closure of any connected space is connected

Showing it’s not path connected does take some work tho

[sleek thicket]

This is a reason to prefer connectedness to path connectedness

it's how I justify the definition to people who are learning (even though I think path connectedness is more intuitive)

[fading vale]

i am so confused

wtf is p_0 here

a map from I -> S^1?

that makes this into a path?

ok nvm i found it its the exponential function from I -> S^1

[sleek thicket]

you're welcome

@fading vale can you help me with my topology problem

[fading vale]

i will try

sure

[sleek thicket]

This

I think I can have a proof via polygonal presentation stuff but I can't get the details right, and also I would like an explicit map

[fading vale]

oh jeez

[sleek thicket]

It should just be like

Write down some fucky exponential map lol

Here's an equivalent thing

That might be easier, idk if I missed something

[fading vale]

maybe im dumb but arent the first and third/fourth lines contradictory here since the first should imply that r^-1(r(j_1(0, s))) = {j_1(0, s + n) : n in Z}

[sleek thicket]

I may have made a typo, let me check

oh shoot first two should be \notin

[fading vale]

ok that makes more sense lol

[sleek thicket]

Let U, V be the circle minus (distinct) points. Let C1, C2 be the components of the intersection of U and V. Let π : T^2 -> S^1 be the projection onto the first component. Find homomorphisms φ : π^-1(U) -> U × S^1 and ψ : π^-1(V) -> V × S^1 over our base such that φ(x, z) = ψ(x, z) if x in C1 and φ(x, z) = ψ(x, -z) if x in C2

here's the original problem I want to solve

[fading vale]

[sleek thicket]

It's tricky

Don't you want to bundlepill yourself

This is for your own good

("this" being you doing my homework for me)

[fading vale]

i take the bundlepill in 12 chapters

[sleek thicket]

This problem is so annoying lol

[fading vale]

yea its kinda hhhhhhh

[sleek thicket]

I can show it's abstractly iso early

*easily

[fading vale]

meanwhile i am doing cool based homotopy

[sleek thicket]

Mayer vietoris + classification of surfaces

but I need compatibility with the projection

[fading vale]

what class is this for?

[sleek thicket]

also I'm also doing homotopy theory in this class rn. I saw a proof that π3(S^2) is infinite cyclic generated by the hopf fibration!!!!

This is for bundles

This is like the first time I've ever worked with higher homotopy groups

And we like, proved the LES from a fibration

It was awesome

I was supposed to read that in the book and understand the proof today and then I didn't lol

[fading vale]

oh i think im doing stuff like that soon

or in the next few chapters

ch3 is covering theory, ch4 elementary htpy theory, ch 5 is entirely about fibrations and cofibrations

[sleek thicket]

Niiice

Yeah so like

The book doesn't even define fibration I think

Iirc Lee said the name in class

But it proves fiber bundles are fibration and then deduces the les from that

[fading vale]

unfair that uni students just get to casually take classes under ppl like lee

hhhh

[sleek thicket]

lol

look I had to do this whole email exchange

arguing that I should be in the class

two years ago

[fading vale]

based

[sleek thicket]

He sent a very long list of requirements

and I was like

Except I had to say I never learned metric spaces. Didn't end up seeing them in a class until a 1 day review at the start of grad analysis

The same quarter I was taking a course on riemannian manifolds

[fading vale]

as it should be

[remote locust]

Hello

I read chapters 2, 3 of munkres and started reading chapter 4

Would that be enough to start exploring algebraic topology?

[gritty widget]

munkres has a little intro to algebraic topology that you could try reading

part 2 of his topology book

iirc it doesn't depend on anything from chapters 5-8

[remote locust]

Cool

Thanks

Any nice book to start algebraic topology

[gritty widget]

hatcher

[remote locust]

Thanks

[limpid leaf]

Morning TTerra

[gritty widget]

good morning, jesse

[limpid leaf]

knew I'd find you here

[gritty widget]

of course

[gritty widget]

S^1 x S^1 with fundamental group Z x Z

[cloud owl]

circle group T?

no?

oh wait i misread

f

[cedar pebble]

or you can just see it "intuitively" by actually doing the computation

I mean I guess the intuitive reason for this is that SU(2) is the universal cover of SO(3); SU(2) is homeomorphic to S^3, while SO(3) is homeomorphic to RP^3 (which has the same fundamental group as that of RP^2)

[shut moat]

np ty again sham!

[sleek thicket]

I think it's just kind of awkward to talk about the mobius band as a subset of R^3

[gentle osprey]

(T*(Terra), -dτ)

(T*(Terra), -dτ)

[gritty widget]

need this as a lemma to something else

maybe need connectedness on G? i saw the theorem somewhere else with a connectedness assumption, but here we're not making it

i'm not entirely sure it will even be smooth, since for any t there could be another b(t) taking m to gamma(t) (kind of like a "jump" in the curve in G?). then a(t) and b(t) would differ by an element in the stabilizer of m. maybe that's important?

unfortunately airing out my thoughts here has not given me any new ideas, so any insight or help is appreciated

ah i think i got it

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

😌

[tight agate]

petthecat

[gritty widget]

that was a lot simpler than i was expecting

[sleek thicket]

simpler version of the thing I was asking yesterday

what's a quotient map $r : [-1,1] \times \R \to \mathbb{T}^2$ such that $r^{-1}(r(t, s)) = {(t,s+n) : n \in \Z}$ if $-1 < t < 1$ and $r^{-1}(r(1,s)) = {(1, s+n) : n \in \Z} \cup {(-1, s+n+0.5) : n \in \Z}$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

you can mod out the second coordinate and start looking for a map $\tilde{r} : [-1,1] \times \mathbb{S}^1 \to \mathbb{T}^2$ such that $\tilde{r}^{-1}(\tilde{r}(t, z)) = {(t,z)}$ if $t \in (-1,1)$ and $\tilde{r}^{-1}(\tilde{r}(1,z)) = {(1,z), (0, -z)}$

so it's like folding up a cylinder to a torus

but we twist the bottom by a half rotation

I guess

hmm so like

continuously rotate the second coordinate

as you go down

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

oh lol this is a homotopy

between f(z) = z and f(z) = -z

I think

hmm kind of

$\tilde{r}(t, z) = (e^{(t+1)\pi i}, e^{i (\pi (t+1)/2)} z)$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

this seems plausible?

yeah I think this is right pogchamp

or actually $\tilde{r}(t, z) = (e^{(\pi/2 + (t+1)\pi)i}, e^{i (\pi (t+1)/2)} z)$ but still

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

no....

just $\tilde{r}(t, z) = (e^{(\pi/2 + t\pi)i}, e^{i (\pi (t+1)/2)} z)$

[gentle osprey]

Shamrock -> E -> B

[tepid depot]

counter example: f(x,y) = x?

am i missing something here?

if it's properly a polynomial in both variables I should have a cute proof of this

[tight agate]

C is the Y-axis

[tepid depot]

oh duh

thanks

[gritty widget]

why does it look like a smiley face

[tepid depot]

$\omega \cup \omega$

[gentle osprey]

doubledual

[tepid depot]

i guess the cup product cup symbol is different

[marsh forge]

$\smile$

[gentle osprey]

doja max

[gritty widget]

$[\omega]\smile[\eta]$

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

nice

[sweet wing]

[tough imp]

what the fuck is happening in this proof?

Granted I'm tired as shit, but they never took an arbitrary neighborhood of p as far I could tell and I have no idea why they look at V_0\cap D

[gritty widget]

[tough imp]

I mean i got that part

[rugged swan]

If F is a sheaf of rings over Y, phi : X -> Y and x in X, do the local rings phi^-1(F)_x and F_phi(x) are isomorphic ?

[tough imp]

I mean as far as I can tell like... what it wants to literally say is p is just on the fucking boundary of H

but I don't see what the hell he's doing in the proof

[tough imp]

I think so. Write it out in terms of colimits and both should be the colimit of thigns F(U) for U a nbd of phi(x), you might have to do some "colimits commute" thing but I think once you commute them one of them just kind of does nothing

[gritty widget]

an arbitrary neighbourhood of p will always contain a boundary chart by taking intersections ("shrinking")

[rugged swan]

yeah for me it is true

so I don't understand what it means to be a "local homomorphism"

[tough imp]

Local homomorphism is a property of map of local rings

if you have a map phi:(R,m) -> (S,n) where m,n are the maximal ideals

[rugged swan]

it preserves the maximal ideals ?

[tough imp]

Then phi(m) < n

[rugged swan]

yeah ok

[tough imp]

or equivalently phi^-1(n) = m

[rugged swan]

makes sense then

ty

[tough imp]

but at the end you're still essentially using that p looks like something (x^1,...,x^n-1,0) and that nbds of that in M (aka R^n) all have poitns with x^n < 0

right?

Like this proof seems stupid overly complicated idk why he does any of this but it might be me being no-sleep mode and literally just like "bruh once you go local this is literally just in R^n"

[gritty widget]

sounds right

lee likes to be technical with these kinds of arguments

i agree with you lol it basically is just "go local, R^n, qed"

just draw a picture

[tough imp]

This is AG brain

Also you say be technical

but he's still implicitly associating it with R^n by taking coordinates like that lol

[gritty widget]

maybe technical wasn't the right word

[marsh forge]

we should have a separate channel for arguments involving coordinates

[gritty widget]

lee likes to spell things out

[tough imp]

He's being a dummy

is what it is

just

R^n

qed

[gritty widget]

analysis class is starting, suffering time

[sweet wing]

[spring relic]

So I just want to make sure I'm understanding this definition right

So in this definition we're sayiing U is a subset of X and T_f = {m in P(U) s.t. X - U is either finite or is X}

Just want to be sure I'm getting this right

[marsh forge]

yeah

its also commonly called the cofinite topology

[spring relic]

k, also when we say let T be a collection of some set X does that mean T has subsets of X?

cuz now that I'm read the definition again I'm interpreting the definition differently

I think Ima move to the questions section for this

[marsh forge]

huh

it says "Let T be the collection of subsets of X such that..."

its clear that every element of T is a subset of X

[spring relic]

@marsh forge I think this edit I made makes it more explicit

[marsh forge]

"subsets U of X" is perfectly clear

and very common writing-style

Let Y be the collection of all elements/covers/supersets/etc y of X with property....

"all subsets U where U subset X" is very painful to read lol

[spring relic]

idk, I kinda like it. The previous phrase makes it sound like T_f is the collection of all possible subsets you can make from X

[marsh forge]

I dont know if I am just used to it because this way of writing is incredibly common

but I completely disagree lol

"subsets U of X such that..." makes it completely clear you're taking a subset of P(X) with some property

[marsh forge]

like just objectively this is bad writing style

its okay to find another way to write it

but that is clunky and bad

i guess my point is that i dont see an argument for any real ambiguity and this kind of phrasing is super common so you should get used to it haha

[spring relic]

k will do. I just wanted to make sure I'm getting the definition down correctly in my head. thx btw

[pure sigil]

So I'm familiar with the concept of weak/weak* convergence. One of my profs today was talking about weak and weak* topologies. I was wondering if anyone had a good explanation for it?

Ok so it's the topology that defines the notion of convergence in that sense? (I'm sorry I'm applied math and not super familiar with topology as a whole)

Oh ok that does make much more sense!

Thank you so much - I really appreciate it!!

[little hemlock]

Here, G is a finite group.

Does anyone have a hint possibly? I've been stuck trying to prove X/G is Hausdorff for an eternity

oh yea, and the action is continuous: i.e each of the maps T_g : X to X which send x to g.x are continuous

[gritty widget]

(ignore what i just deleted, misread!)

[tight agate]

prove that every point x has a neighborhood U such that G.x \cap U = x

[little hemlock]

The only potentially helpful thing I've found is that sets of the form $\bigcup_{g \in G} T_g(U)$ are open in $X/G$ when $U$ is open in $X$

[gentle osprey]

kxrider

[tight agate]

essentially show that the quotient map is open

[gritty widget]

welcome to the piss role channel bro

[tight agate]

once you have this, there's an obvious way to define charts on X/G

[tight agate]

you see, I don't get the whole piss color joke

perhaps I'm too hydrated to understand

[gritty widget]

what, are you using light mode?

tfw you can't even read honorable names on lightmode

[tight agate]

if your pee isnt clear, then death is near

[little hemlock]

i have $$\pi^{-1}(\pi(U)) = \pi^{-1}{G.x : x \in U} = \bigcup_{g \in G} T_g(U)$$
when $U$ is open in $X$ and $\pi$ is the quotient map $X \to X/G$ which proves that $\pi$ is open, but haven't been able to see why there exists a nbhd $U$ of $x$ such that $G.x \cap U = {x}$.

[gentle osprey]

kxrider

[sleek thicket]

can you do it for a single element of G?

@little hemlock

like in my head, we want to use finiteness of G to keep cutting down some neighborhood

removing one element from the intersection at a time

does that make sense?

[little hemlock]

yea, hmm. Let $U$ be a neighborhood of $x$. You can remove $G.x \setminus {x}$ from $U$ since $G.x \setminus {x}$ is closed (a finite union of closed singletons).

[gentle osprey]

kxrider

[sleek thicket]

hmm I think this might actually be the wrong condition

[little hemlock]

i was about to say, im not really sure how this helps tho

[sleek thicket]

yeah so you want a nbhd such that $gU \cap U = \varnothing$ for $g \neq e$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

the image of a connected such U under the quotient map should be evenly covered

how does that sound

[little hemlock]

hmm, so hausdorffness is supposed to come out of this somehow?

[sleek thicket]

oh lol I was misremembering the statement you were trying to prove

I think this should imply Hausdorffness

but what I was saying would be for showing X -> X/G is a covering map

(I think I see how Hausdorffness pops out of this though)

[little hemlock]

i desperately want to remove the set ${s \in U : G.s \cap V \neq \varnothing }$ from $U$, but there's no way its finite.

[gentle osprey]

kxrider

[little hemlock]

taking U and V to be disjoint open sets in X containing elements x and y respectively

[small phoenix]

for Hausdorffness: open sets in X/G correspond under \pi to G-invariant open sets in X. I think given x and y in X, and disjoint open sets U_x and U_y, if you push U_y by G, you get only finitely many intersections with U_x. you can trim both U_x and the orbit of U_y (exploiting Hausdorffness of X and finiteness of G) to make neighborhood of V_x which is disjoint from the orbit of V_y.

[meager python]

Anyone’s still taking Jarod Alpers class here?

[sleek thicket]

@tough imp is

[tough imp]

Yee

[broken robin]

Hi guys, I have problems to show the claim, can someone give me a hint to proves that N is open in W_k? 😔😔😔😔

[zealous glen]

Is the kernel of a homeomorphism, the pre-image of the identity?

Getting a little confused with tangent spaces of regular level sets

[gritty widget]

is the kernel of a homeomorphism the preimage of the identity
the word you're looking for is homomorphism

tangent spaces of regular level sets
if your level set is given by a function F, then the tangent space to the level set at x is given by ker (dF_x), where dF_x is the differential of F at x

there isn't much more to it

[zealous glen]

ah right I think i understand. So for our level set... our push forward defines some tangent space basis (from one manifold to another), and then we can use the kernel to find the tangent space at some point in our original manifold ?

[gritty widget]

the level set needs to be given by the preimage of a "regular value" to be a manifold, which is to say the differential of the defining function is surjective at every point of the level set

[zealous glen]

right that makes sense. So with that being said we can define our ker(dF_x) = TpM?

M being our original manifold

and p being in that manifold

such that F(p) = c and c is our regular value

[shadow charm]

Could someone give me an example of a sequence of non compact sets where this fails?

[marsh forge]

(0,1/n)

n->infty

[shadow charm]

hmm right that's empty since they're open

[marsh forge]

(0,1/n] also works

[shadow charm]

what about an example with closed non compact sets?

[marsh forge]

i would just google that one

[tepid depot]

[n,\infty)

n -> infinity again

[marsh forge]

oh nice duh

i was like

must be closed but not bounded

no useful examples of that

(its early be nice)

[shadow charm]

right okay thanks tons

also what's this new role i see

advanced helper?

[marsh forge]

not super new

[tepid depot]

yeah i bribed the mods with my proof the riemann hypothesis

[shadow charm]

[summer jolt]

Can someone explain what is meant by "extendible" sections on a vector bundle?

[gritty widget]

presumably it means something like: if $E \to M$ is a vector bundle over $M$, $S \subset M$ is a submanifold, and $s$ is a section of $E$ defined on $S$, then you can extend $s$ to a section of $E$ defined on all of $M$, or maybe on an open subset of $S$ in $M$

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

the most familiar example being when s is a vector field (i.e. E = TM)

i can't imagine it meaning much else lol

[summer jolt]

Here is the relevant section from my notes, but I'm struggling to interpret what it means

[gritty widget]

ah, sections along curves

so you can have a section of E defined along gamma, right?

and you want to ask whether any given section along gamma is actually the restriction of a section defined in a neighbourhood of gamma

this isn't always possible!

it's possible if gamma is an embedding iirc

it might help if you interpret this in terms of vector fields, for some visual intuition

To make the definition precise: let $\sigma$ be a section of $E$ defined along $\gamma$, i.e. a map $\sigma\colon [0,1] \to E$ such that $\pi\circ\sigma=\gamma$. We say $\sigma$ is extendible if there exists an $s \in \Gamma(E)$ such that on $[0, 1]$, $\sigma = s \circ \gamma$.

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

i just reworded what the pic said

yeah the phrasing in the pic is kinda odd imo

lee defines this for tensor bundles like this

screenshots of lee = proof by authority

[summer jolt]

Ok just to clarify: a section $s$ along $\gamma$ is ${p \in \gamma(t) | s(p)}$ right?

[gentle osprey]

snypehype46

[gritty widget]

i'm not sure what this means

the section is a set?

[summer jolt]

Yeah sorry I mean the the image of the section is as above right?

[gritty widget]

well swap the left and the right in your set lol

but yeah, kind of

eh

nah, you should be precise

a section along a curve gamma always has the same domain of definition of gamma, so you should be writing s(t)

the "along a curve" part is very important

To make the distinction clearer: a section of $E$ along $\gamma$ is a map $\sigma\colon[0,1]\to E$ such that $\pi(\sigma(t))=\gamma(t)$ for each $t \in [0,1]$. It's basically a curve in $E$ defined "along" $\gamma$. Without the descriptor "along $\gamma$," a section of $E$ is a map $s\colon M \to E$ with $\pi(s(p)) = p$ for all $p \in M$.

[gentle osprey]

(T*(Terra), -dτ)

[summer jolt]

I see this makes sense

[gritty widget]

also this

[summer jolt]

So for the extendible part we are just saying that curve in the yellow and green paths is the same then the section is extendible right?

Ah ok yeah, that is what I meant the picture was just to give me an intuition. Ok I think I get the definition, but I'm still confused about why "extendible"?

[gentle osprey]

slimvesus

slimvesus

slimvesus

[summer jolt]

Thanks that makes much more sense than my lecture notes

[summer jolt]

Yeah I also get the definition of my notes now.

[zealous glen]

Hey can anyone help me figure out the intuition behind local flows on manifolds?

is it do with mapping an integral curve or have I misinterpreted the text 😄

[gritty widget]

intuition behind local flows
package all of the information about a vector field and its integral curves into one function

basically

solving an ODE

the theorem about the existence of flows is in fact a direct analogue of a theorem from ODEs you might be familiar with

let me see if i can find it

eeeeeeeeeeeeeehhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[zealous glen]

OOOOO

so essentially we are mapping our integral curve which has some interval into our vector field

and calculating its velocity etc

since its velocity vector is in the tangent space of our manifold

Ty sir

when we talk about vector fields being over a manifold are we basically saying embedded onto the manifold?

or is this just another way to define a tangent bundle

[gritty widget]

no clue what "mapping into our vector field" means. a flow is a function $F(p, t)$ associated to a vector field $X$ on a manifold $M$ such that for each $p \in M$, the curve $t \mapsto F(p, t)$ is the integral curve of $X_p$ (and it also satisfies the property $F(F_s(p),t)=F(p,s+t)$)

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

and F(p, 0) = p, for all p

[zealous glen]

by mapping i mean taking every point and redefining onto our vector field

[gritty widget]

i still don't know what that means

you should check out chapter 9 of lee's smooth manifolds book

[zealous glen]

doesn't the integral curve map some interval J to the manifold M?

[gritty widget]

he explains very precisely what flows are, and proves their existence and uniqueness theorem with painful detail

yes

[zealous glen]

at different intervals t

I'm trying to think what they look like in terms of a picture, but haven't got a clue

is there a picture of it in the book?

[gritty widget]

probably

[zealous glen]

got it

thanks for the help!

[gritty widget]

maybe to be a bit more precise, if $X$ is a vector field on a manifold $M$, then for each $p \in M$ we get an integral curve $\gamma_p$, defined on an interval $(a_p,b_p)$ around $0$ (note that the interval may depend on $p$), so that $\gamma_p(0) = 0$ and $\gamma_p'(0)=X_p$. a local flow is then a function $F$, such that $F(p,t) = \gamma_p(t)$ whenever it makes sense. (making sense of it is the annoying part of getting local flows from vector fields)

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

so if you fix p, then F(p, t) is as a function of t the integral curve of X through p

[zealous glen]

is this correct?

[gritty widget]

still not sure what that means

[zealous glen]

this makes a lot of sense now

[zealous glen]

ok can you tell me what we mean by over a manifold?

[gritty widget]

i might have left out a condition or two this is very technical and annoying

yeah shamrock i knew you were gonna blast me with some edge case #3495873459384

this is why i rec lee

[sleek thicket]

lol

I was gonna say flows are very visual

If you have a vector field eg

you can put a point down somewhere

And follow the arrows

[zealous glen]

o

[sleek thicket]

Like, if you think of those arrows as being something like the current in a body of water

idk I don't do physics

But you're literally just being pulled along

By the arrows and the direction they point

[gritty widget]

hence, "flow"

[sleek thicket]

go with the flow dude

[gritty widget]

lee's flow chapter is good until you get the existence and uniqueness proof

then you taste blood

[sleek thicket]

🤢

I think I may have read it but I'm not sure

[zealous glen]

mhm okay thank you guys

[gritty widget]

i had the proof on an assignment once

[sleek thicket]

My class hit it at the start of covid

Like the week I moved down to Cali

[gritty widget]

perfect reason to blackbox it

[sleek thicket]

So I was more focused on uhhh

Everything else

Playing minecraft all day

[gritty widget]

@sleek thicket https://www.math.toronto.edu/laithy/367/367-ps4.pdf

go to the last page

this instructor was a little sadistic with his assignments

[shut moat]

this sounds like integral curves but for manifolds

[gritty widget]

well...

it is

[shut moat]

o nice

I thought flow was smth else

[gritty widget]

flow just packages all of the integral curves together

[shut moat]

pog

[gritty widget]

i asked him why he didn't just say "riemannian metric" and he said "i didn't want people to look it up"

[sleek thicket]

this sucks lol

[gritty widget]

yeah that course was genuinely painful

although it was the summer so the ridiculous homework was actually manageable since lower course load

[cloud owl]

what a mood

[shut moat]

so, a while back, hubbard made an interesting comment about the volumes of complex manifolds:

the comment is the bottom paragraph but I left in the rest for context

so I recently got to the actual exercise explaining the reason this happens:

Now, I managed to prove this

and the magic trick had to do with $a + bi \sim \begin{pmatrix}a & -b \ b & a\end{pmatrix}$, $\det z = |z|^2$

[gentle osprey]

~S^1

[shut moat]

and it became clear that $\det{D\gamma ^T D\gamma}$ is always a perfect square with the same reasoning

[gentle osprey]

~S^1

[shut moat]

but despite all this, I still don't have a geometric feel for why having a natural orientation means the volume of the complex manifold is always this natural to compute

like, I get why it works symbolically, but is there more to it?

[little hemlock]

what condition of the implicit function theorem requires that df/dy != 0?

the way it was explained in class, the jacobian of f should have maximal rank, but it doesn't specify that particular components should be nonzero
--yet, its necessary in order to conclude the last part with u'(x).

[shut moat]

The implicit function theorem guarantees some implicit function when the jacobian has maximal rank, but you need the extra condition to guarantee that y here can be chosen as the implicit function

it should be clear why this is the case if you look at the result

this'll shit its pants if the denominator is zero

[little hemlock]

take the statement in spivak for example. Its a little different than the "maximal rank" formulation i learned. But what part of it is different so that g(x) gets to automatically be differentiable?

err, i guess he checks that a certain submatrix of the jacobian is nonsingular instead. And in my case, that submatrix would just be df/dy?

[shut moat]

yeah exactly

the generalization of the fraction above is like -[D implicit variables]^{-1}[D active variables]

where the [D ] is ofc the jacobian

this comes from a similar chain rule argument

[shut moat]

don't the U_1OK and U_2OK also have to be connected in order for this to work

[shrewd moth]

@gritty widget can you help me with a proof

[gritty widget]

if you have a question just post it

[shrewd moth]

okie

im confused on this solution

in yellow

[gritty widget]

this isn't topology

#proofs-and-logic

[shrewd moth]

im not understanding it but is there another way to restate it

[tight agate]

Namington is helping you in the other channel

[shrewd moth]

can you help on proofs and logic please

[gritty widget]

bruh

[shrewd moth]

okie

[tight agate]

in the algebra channel

[shrewd moth]

i cant find the channel

[gritty widget]

#groups-rings-fields

[shrewd moth]

ty

[little hemlock]

The quotient map $\pi : \bR^{n+1} \to \bR \mathbb P^n$ sends an open set $U$ to the set of lines through the origin which intersect $U$ somewhere. How would you show that $\pi$ is an open map?

[gentle osprey]

kxrider

[tight agate]

I think you were working on something similar earlier?

anyway, in general if you have a topological group G acting on M

let pi: M ---> M/G be the quotient map

[little hemlock]

well instead of trying to prove that X/G is a manifold, im taking a break to prove that projective space is a manifold.

[tight agate]

let U be open

in M

then what is pi^-1(pi(U))?

[little hemlock]

G.U

[tight agate]

right

which is the union over g, gU

right?

[little hemlock]

yup. So G.U is open and pi(U) is open.

[tight agate]

right

oh and you want R^n - {0} btw

[little hemlock]

so i need a topological group acting on R^n+1 whose orbit set is RP^n

[tight agate]

minus the origin

[little hemlock]

ah yea, right

i see, so R-{0} acts on R^n+1 - {0} by scalar multiplication

[tight agate]

R^*, yes

and projective space is the quotient

[little hemlock]

nice, thanks

[tight agate]

another construction is the quotient of S^n by a Z/2 action

[cedar pebble]

same construction :^)

[tight agate]

(it is sort of the same construction but whatever)

[little hemlock]

in what sense is that the same?

[tight agate]

take the sphere in R^n+1

The R* action on R^n+1 is the antipodal action on S^n

[little hemlock]

oh you just mean both quotients are the projective space?

[tight agate]

no, I mean it's the "same construction" in some sense

by restricting spaces

like a line in R^n+1 will intersect the unit sphere at 2 points

and the antipodal action identifies those

perform the quotient on R^n+1 and identify the entire line to a point

or work on S^n and identify those two points to a point

[little hemlock]

right, i see

[tight agate]

and even though it is the same definition, it has its merits

it lets you see the cell structure on RP^n immediately

also, it was useful while computing the de rham cohomology of RP^n

[viral atlas]

Is a spherical surface 2-dimensional because a point on a its surface can be specified by two parameters(angles with two of the 3 axes), while in a solid sphere, you can also vary the distance from the origin?

[tight agate]

yes

[viral atlas]

[tight agate]

[marsh forge]

is it true that a point on a 2-manifold can always be specified by two data points wrt the origin?

otherwise i think its wrong to say this is “why” a sphere is 2dim

[viral atlas]

Why is it 2-dimensional, then? I'm still kinda unconvinced about my reasoning tbh.

[tight agate]

you can do it, "locally"

[marsh forge]

It is two dimensional because if you take a point and zoom in on it

it locally looks like R^2

[viral atlas]

Oh

[marsh forge]

its important that you can do this for any point

this is also why a circle is 1dim

[viral atlas]

What is different about a solid sphere? When you zoom into a point on/in it, does it look like R^3?

[marsh forge]

and a ball (filled in) is 3dim

yeah

[viral atlas]

Oh

[marsh forge]

okay small techbicality

theres a notion of manifold and “manifold with boundary”

so a solid sphere has a boundary that is 2dim if you remove the interior

but the interior is 3dim

[viral atlas]

Owww

So the interior of such a shape has a +1 dimension relative to the dimension of its boundary?(Interior points of circles seem to be like R^2 on zooming in, and for spheres you get R^3)

[marsh forge]

the boundary is always dim-1

[viral atlas]

That's what I meant, sorry for the weird phrasing haha.

[viral atlas]

Is this true every time, or does this fail under certain conditions?

I can't think of anything which violates this idea

Just wondering if there are some pathologies

[marsh forge]

its true by definition but this follows from a rigorous definition of manifold with boundary

[viral atlas]

I see.

[marsh forge]

namely the interior has to be locally R^n

and small neighborhoods of the boundary points have to look like R^n cut in half

[viral atlas]

[marsh forge]

its easiest to see with a circle

filled in circle

disk

so inside the disk

it looks like R^2

but on the border

[viral atlas]

It looks like a line, so R^1

[marsh forge]

not quite

[viral atlas]

[marsh forge]

if i take a neighborhood of the arrow point

im also gonna capture some of the inside

[viral atlas]

Yes

[sinful pecan]

note the closed disk

[marsh forge]

so if you "unbend" the neighborhood of a border point

it looks like this

i can draw it if you want

[viral atlas]

Thanks, that would be nice.

I think I see the distinction but not entirely sure.

[marsh forge]

art is not my forte

also pretend i colored in the whole disk on the left

does this help

im cutting out the blue neighborhood

and unbending it

it looks like a 2-ball cut in half (a 2-ball happens to be a disk)

[viral atlas]

Aah, okay

Yes, this makes sense.

[viral atlas]

Owwwww, so you have to unbend the neighbourhood every time?

Unbending a neighbourhood on the sphere's surface would...

Oh yeah

That gives a hemisphere!

But how do you talk about "unbending" more formally?

[vocal wharf]

there exists a homeomorphism

bijective, continuous map with continuous inverse

[viral atlas]

Oh, okay. So you kinda map the points under some transformation preserving some properties?

[vocal wharf]

yeah

you map the thing you cut out to the half circle thing

[viral atlas]

(Please bear with my bare knowledge, I don't know what continuous maps and inverses are)

[vocal wharf]

continuity ensure that "close things stay close"

[viral atlas]

How? Physically, this feels like bonking the cut out portion, but I can't make sense of how to do that formally.

[viral atlas]

Oh, that makes sense.

[tepid depot]

@viral atlas between this question and what you asked in the analysis channel, I think learning some point set topology at this point would be very helpful.

these notes are really good: https://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf

idk if you're in a class and dont have time for this or anything

but if you do have time I highly recommend it

[viral atlas]

I started with Lee's ITM, I've been told it teaches point-set top with a manifolds tilt.

Oh, these are Hatcher's notes! XD

I'll take a look, I like his way of presenting things. Thanks!

[tepid depot]

just looking briefly at the lee table of contents, chapters 2-4 are the same content

but lee is taking a lot longer to get through it

so the difficulty density is probably lower, but idk if that's a super great thing in this case. I feel like hatcher's notes are actually really good and not that difficult

[viral atlas]

They look good! The idea explained above has already been explained by an example right in the beginning. :p

[tepid depot]

if you're interested in the content in the later chapters of lee, I think you could jump straight to chapter 5 after reading hatcher's notes

[viral atlas]

Great, thankyou so much.

To claim that a non-empty finite set(say, A) in R is not open, is it sufficient to say that no open interval could be a subset of A(for every such open interval is either empty, or has infinitely many elements), and hence none of the points in A is contained in an open interval which is also a subset of A?

[marsh forge]

yes

every point in an open set should be inside of some basis element

[viral atlas]

What is a basis element? 😅

[marsh forge]

oh

it captures the idea of a “basic open set”

on R the collection of all open intervals forma a basis

and all open sets are unions of intervals

[viral atlas]

Owwww, I see.

What are some other examples of basis elements(in probably a different setting)?

[marsh forge]

in any metric space the sets

B(x,delta):={y | d(x,y)<delta}

forms a basis

where x,delta are allowed to vary over the whole space and all of R+ respectively

[viral atlas]

Okay, so this is a generalisation of the open interval idea to all metric spaces.

[marsh forge]

but it works for all topological spaces

[viral atlas]

[marsh forge]

but if youre only familiar w metric stuff

[viral atlas]

I'm actually not familiar with any other topological spaces

[marsh forge]

this is the easiest way to think about it

[viral atlas]

Bar metric ones

[frigid patrol]

anybody know what equivariant morse theory is about?

[tight agate]

probably morse theory + group action

Atiyah and Bott wrote a pretty big paper on it I believe

[frigid patrol]

do you know what moment map means in this context?

[tight agate]

you should probably ask tterra

[shadow charm]

im just reposting a q i asked in a question channel since it fits better here: I just need someone to confirm that my proof is good enough and that it isnt lacking rigor cause i feel like it kinda does

where at then end it should read $G_0 \cup {H_n}$

[gentle osprey]

Little Narwhal

[shadow charm]

oh and also i wrote G1 in a few places and meant G0 oops

[vocal wharf]

your finite subcover is (technically) {G_0} \cup {H_n} right?

[shadow charm]

it exactly is yes