point-set-topology

[gritty widget]

the current answer on the post doesn't address my actual question at all which is why i'm asking here

[marsh forge]

jesus christ

[gritty widget]

pain

[marsh forge]

why do you learn tbis

this*

[gritty widget]

geometry good

but do carmo's abuses of notation and consistent sweeping under the rug of technical details like this is irritating

unless i'm just being stupid

[long gorge]

Wait isn't this just a thing you can generally do, extending a vector field defined along a curve to the manifold

Unless I'm misunderstanding and there's some relevant context here

[gritty widget]

if the curve is an embedding you can

[long gorge]

Oh ok so the problem is that it isn't regular

[gritty widget]

yeah

[long gorge]

But assuming you can construct such a curve making it normal shouldn't be an issue with Gram-Schmidt right

Since that process doesn't change the definition of A(s) along the curve that you already have

[gritty widget]

$\int_{\mathbb{C}^n} \log|x_1x_3x5 \ldots x{2n - 1} - 1| \neq 0$ for $n \neq 1$

[gentle osprey]

Learner

[gritty widget]

I am trying to prove this result

[frigid river]

Could someone explain to me what's going on here?

p is some random polynomial (with its first coefficient being 1) that doesn't have any roots

[ivory dragon]

what about that dont you understand

[frigid river]

what this is

Does this end up being a circle?

[ivory dragon]

p evaluated at re^(2pi * i * s) divided by p evaluated at r

huh

[gritty widget]

It’s a complex polynomial but doesn’t have any roots?

[frigid river]

Well, Hatcher says the whole thing defines a loop in the unit circle based at 1.

[frigid river]

He's trying to prove that only constants can do that

[ivory dragon]

ah i see

intuitively x/|x| gives you xs "direction"

as a vector of length 1

this is a concept from linear algebra (normalization)

[frigid river]

Yeah, but what's the "x" in this case doing?

Does p(re^(2piis))/p(r) just end up being some random function, is there nothing special about it?

[ivory dragon]

i dont know hatchers argument

[pearl holly]

I mean, formally it is a loop, but I can’t understand what it is doing intuitively

[hazy nexus]

IIRC the intuition was wrapping up the polynomial p around the unit circle

I'm not sure I remember how that f_r accomplishes that, but it might also be too early for me

[pearl holly]

Okay so if I understand it correctly, the domain is [0,1] because the e^blabla “captures” every complex number so it’s sufficient to have a domain of [0,1]. Now f(0)=f(1)=1 and the function is continuous and therefore a loop??

And if it had a complex root then it would be a division of 0 so it wouldn’t be a loop?

[frigid river]

Yeah, it's assumed that this doesn't have any roots.

But what about the "r" though?

After that he says that f_r is a homotopy of loops based at 1.

and I don't know what that means

[pearl holly]

[frigid river]

Usually we say that we have a homotopy between 2 paths

So the function is a loop and a homotopy between all other loops based at 1? 0.0

[coral pivot]

the homotopy is when you vary r

the loop is when you vary s

[frigid river]

so is $r \in [0;1]$?

[gentle osprey]

veryfrustratedperson ✓✓

[strong heron]

May I ask why are you reading this proof without learning the definitions of paths and homotopy first?

[pearl holly]

I’m just curious about this question so that’s why I’m here lol

[frigid river]

I did learn the definitions of paths and homotopy first

[coral pivot]

well you dont need that

the idea is that you take any of the loop f_r, and you can see it is homotopic to f_0

(the homotopy arguement would be you are doing the loops f_{rt} with t\in [0,1])

[frigid river]

oh, okay, thanks.

[strong heron]

What do you understand by a homotopy?

[frigid river]

A continuous transformation of one function onto another

(Hatcher always keeps the endpoints fixed)

[strong heron]

Correct. Then f_r is a homotopy here between f_0 and f_1

And each f_r for r in [0,1] is a loop in the circle

Which is parametrised using the variable s

[frigid river]

Yeah, but what worried me is that Hatcher didn't limit the r in his proof

[sweet walrus]

If there a homotopy between f and a constant fn, f is said to be null homotopic. It might be useful to prove the equivalence of the following facts given h is a function: S_1 -> a space X
1.h is nullhomotopic
2.h is extendable to a function k:B_2 (disc) -> X

Couple it with the fact that z^n 'dominates' the value of f(x) polynomial with degree n you can show that under certain condition, which you can enforce every time, f(x) when restricted to S_1 is homotopic to the loop going n times around the circle where n is the degree of f.

But you have a contradiction! (Assuming you are proving ftoa)

[strong heron]

Yeah, you don't need the interval [0,1] strictly. A homotopy is just a transformation, it can be done over any interval. So, in Hatcher's argument, f_0 is homotopic to f_r for any r bigger than 0.

[sweet walrus]

Since the loop going n times around cant be nullhomotopic, assuming you know about covering spaces/fundamental group of circle

[strong heron]

So Hatcher goes f_0 ~ f_r ~ w_n and hence by transitivity, w_n is trivial in pi_1(S^1) which is Hatcher's contradiction

[hazy nexus]

Guys, r isn't a real number here, it's a complex number

The way you get a homotopy between $f_{r_1}$ and $f_{r_2}$, for complex numbers $r_1$ and $r_2$, is by composing any path from $r_1$ to $r_2$ in the function $f_r$

[gentle osprey]

Tormeson

[hazy nexus]

So if you define this path by $\alpha$, say, then the homotopy is given by $H(s,t) = f_{\alpha(t)}(s)$

[gentle osprey]

Tormeson

[hazy nexus]

modulo how you order the variables in a homotopy

[frigid patrol]

@fading vale https://youtu.be/8EI_jKhHhdA

[fading vale]

Based

[pearl holly]

PTYamin, more like PYTamin get it? YT? Like YouTube?

oh you have videos on Galois theory! I will watch them and like them and then I will subscribe to the channel

[sweet wing]

https://ncatlab.org/nlab/show/paracompact+Hausdorff+spaces+equivalently+admit+subordinate+partitions+of+unity

Since when did nlab contain these articles

[sweet wing]

(from lee ISM) isn't it sufficient to show that the maps aren't topological embeddings or am i missing something

[shut moat]

I think in principle you can have an image that is a submanifold even if the map itself is not a homeomorphism onto the image

[gritty widget]

loop around the circle twice xd

even better, the standard parametrization of the circle on [0, 2pi) is an injective immersion, but not a homeomorphism onto its image. the image is still an embedded submanifold

[sweet wing]

ah lol

lemme unstone myswelf

ohhhhh

righttt

cuz it says the image here

not the map

LOLL

ye thanks

[gritty widget]

https://cuhkmath.wordpress.com/2012/03/16/taylor-expansion-of-metric/

i just realized the author's name is kkk

yikes

well, yeah

it's just repeated differentiation

no different than an elementary calculus course

[sweet wing]

pain

and suffering

this is why i stopped doing physics

[gritty widget]

why would you ever need anything past like the second order term

what cursed shit requires you to expand the metric to the sixth fucking order term

you can already get some cool stuff with just the second order term

[sweet wing]

ikr

[gritty widget]

wtf is higher????????

[sweet wing]

thanks physics

apparently there is stuff in the opposite direction

idk why would one ever use those

[gritty widget]

[swift fjord]

Do physicists have an index fetish

I could mean anything

I see, it's like Schrodinger's cat. That's a physicist's favourite.metaphor

[pseudo crane]

what

that's true

[frigid river]

This isn't only true for the unit sphere, right?

[marsh forge]

what do you mean?

[frigid river]

"For every continuous map f:S^2 \to R^2",
Is this only true for S^2 (the unit sphere) or could this be any sphere?

[marsh forge]

you have to be careful about what you mean by antipodal but a similar result should be true of any space homeo to the unit sphere

[frigid river]

Okay, thanks.

[marsh forge]

in particular any such choice of homeo to the unit sphere gives a sufficient analogue of antipodal

[gritty widget]

[versed junco]

Is there a name for a multi-set M of points in some space S such that there is a function from real numbers to M? Or maybe something that is similar? I would like to consider such "paths".

[marsh forge]

can you explain a bit more?

[versed junco]

It is not a homework question. I was thinking about the paths drawn by rotations of points around axis. Then I was thinking about the images of addition and multiplication by a constant (as functions). I thought that it could be a useful concept for describing generating functions, comparing them etc. Intersections of paths are ordered (as are real numbers). This concept can be generalized to surfaces and manifolds.

As opposed to sets of points, paths can wrap around themselves (as the case with rotations).

Paths don't need to be connected (so we can consider special cases of paths (such as piercing several layers of paper with a needle)

I can also consider paths in natural numbers (primality, residues modulo different powers, etc)

I know that the word path in topology is already reserved, so I wonder if such concept already exists and if it does what its name is

[past barn]

help with b pls

idk what to do

[hollow harbor]

Did you find dF?

To investigate exactness of omega, you may want to consider integrating it around the unit circle starting and ending at (1, 0). If it were exact, what should this integral be?

[past barn]

idk i missed the lecture

:yikes:

this is for my homework lol.

6th lecture

also do u have any idea about a? @hollow harbor

like rly what i mean isi have to do e xterior derivative right

[hollow harbor]

Yeah

[past barn]

i tried doing that

[hollow harbor]

So for a you want to show the exterior derivative of omega is 0

[past barn]

and this is the eqt for it

?

yes?

or am i dumb

cause i missed that so wikipedia it is.

[hollow harbor]

I guess so. Is this some weird kind of cycle notation?

Odd.

[past barn]

but in this case wouldn't the (dw)_11

be

$\frac{xy}{(x^2 + y^2)^2}$

[gentle osprey]

LolLol

[past barn]

can u give me ur notation for it @hollow harbor

if that's oki c:

cuz i really just dont get wtf is going on here lol

[hollow harbor]

There should definitely be some minus signs in that definition wikipedia gave, funny.
You can look at the general formula for the exterior derivative and it's super confusing, but when you have a 1-form like f(x_1, ..., x_n) dx_k it just ends up being df wedge dx_k. So the derivative just falls on f (only if it's a 1-form!)

So in this case, you want to compute d(-y/(x^2+y^2)) (the x partial derivative dx, plus the y partial derivative dy) and then you wedge both pieces with dx.

Since dx wedge dx = 0, the dx part will go away.

So you really just need the y partial derivative times dy wedge dx.

[past barn]

what is y partial derivative tho

😢

[hollow harbor]

Similarly, you need to compute d(x/(x^2 + y^2)).

What do you mean

[past barn]

so that is just derivative with respect to x multiplied by dx right

[hollow harbor]

That's all that will matter (because the other part, the derivative w.r.t. y, will cancel once we get dy wedge dy).

Now be careful of the signs. One of these terms will give you dy wedge dx. The other will give you dx wedge dy.

When you flip dy and dx, you need to add a minus sign to the front.

They're "antisymmetric."

And you'll see that the two things cancel out once you flip one of the 2-forms.

[past barn]

wait I think I understand now

so is it just,

[hollow harbor]

So that wikipedia thing you saw has some implicit minus signs in it, that's what makes it confusing haha

[past barn]

d(x/(x^2 + y^2)) wedge dy?

[hollow harbor]

That's the second term

[past barn]

I compute that to be 0?

or nah

do I have to do the first terma swell

[hollow harbor]

The first term is d(-y/(x^2 + y^2)) dx

You need to show the sum of these is 0

[past barn]

I see...

[hollow harbor]

Since we're summing the terms at the start

And since d(w + v) = dw + dv where w and v are any forms at all.

[past barn]

alright tysm

i think i understand

again thank you

I gotta go cause its fathers day but

luv u ll

lol*

[drifting sundial]

is there any difference between this and simply considering functions R -> S x Z?

[past barn]

@hollow harbor i returned on my phone,sorry for pongg but idm if general formula is confusing i still wanna see it 😄

pls

I beg

[hollow harbor]

Oh I don't even remember it, do I?

Hmm

[past barn]

ah xd

i think I understand for one forms

$d(f(x)dx) = df \wedge dx$

[gentle osprey]

LolLol

[past barn]

well for a 1-form with 1 var and 1 differential

[hollow harbor]

Yeah

[past barn]

yey

[hollow harbor]

Is it that for f * higher order form too?

[past barn]

but how do u get df for multivariable

[hollow harbor]

I dont remember

[past barn]

f(x,y)dx

sorry I pretty much forgot calculus lmao

oh nvm

[hollow harbor]

Well for df you are kind of doing a chain rule. It's "df = (df/dx) dx + (df/dy) dy + ....". It's helpful to think of things as cancelling

[past barn]

just checked gogol

ye

[hollow harbor]

Even though they don't

[past barn]

thanks a lot for you help ❤️

pretty much caught me up on the whole lecture

😛

so exactness means: The k-form is the exterior product of a (k-1)-form

[hollow harbor]

Exterior derivative

Yeah

[past barn]

teah derivative

[hollow harbor]

Basically "exact" means "in the range of d"
And "closed" means "in the kernel of d"

[past barn]

my brain is infinitely differentiable everywhere mb

and closed means: The k-form's exterior derivative is 0

so like

[hollow harbor]

Yeah

[past barn]

xdx is closed

[hollow harbor]

Yep, we get dx wedge dx and that's 0

[past barn]

dx wedge dx = 0

yup!

alright thanks so much

i cant like contain my happiness

[hollow harbor]

It's also exact, it's the derivative of x^2 / 2 (which is a zero form)

[past barn]

instead of wasting 2 hours on a lecture I come here and talk to u for like 2 minutes

ez

[past barn]

oh ye

its like the omega

[hollow harbor]

Every exact form will be closed, because d^2 = 0 (try to prove that d^2(function) = 0, it's harder to check for higher forms though!)

[past barn]

hmm

d^2f = f''dx^2

and dx^2 is dx wedge dx

:perhaps:

idk if that last part is true

doe

so its

f'' * 0

which is 0

or im oversimplifying lol

[hollow harbor]

True if f is single variable, if f is multivariable it gets more complicated.

[past barn]

Yeah lemme have a go for dat

wait a minute

what would d^2f be :yikes:

wait

i think i figure out

df' * dx?

[hollow harbor]

First compute df

Then compute d of that

(it's a bunch of terms, each term is a 1-form which we know how to handle)

[past barn]

df = df/Dx dx + df/Dy dy

i use D to represent partial there

so d of that would be like

ahhh

df = df/Dx dx + df/Dy dy, first term: d^2f/Dx^2 dx^2 + d^2f/DxDy dxdy, second term is: d^2f/Dy^2 dy^2 + d^2f/DxDy dydx

so I think I figure out

it simplifies down too

d^2f/DxDy dxdy + d^2f/DxDy dydx

which is equal too

d^2f/DxDy dxdy - d^2f/DxDy dxdy

so it becomes 0

I think that is it :p

I wont ping u cause

I dont wanna keep hastling

but perhaps u will check this chat 😄

[hollow harbor]

I will be driving soon lol

For most of the day

So someone else will have to answer

But yeah, that looks right to me!

The key thing here is that d/Dx and d/Dy commute (that means df/DxDy = df/DyDx)

[past barn]

yup thats what i did

since they're not actuaally differential forms i think

cause they're partial

cause u notice i did DxDy rather than DyDx

which was a detail I made sure to account for 😛

this can also be proven with logic not reallyp roven but conjectured

cause here take this for example,

x + y

we do x then y

it willgo to 0

both ways

x^2 + y, 2x then y, derviative goes to 0

xy^2, y^2, 2y, 2xy, 2y

it shows no signs of changing

[vast estuary]

"But sets of the second and third types are not open in the larger space..." - I understand this, but what is the relevance of this comment?

[bleak helm]

I think they're just trying to emphasize you get open sets that aren't open in the original space

[vast estuary]

but that's weird because the two topologies are the same - that's what they conclude

[bleak helm]

Well, that's not part of the reason

The reason is because those sets form a basis for the order topology

[vast estuary]

Yeah, both topologies have the same basis and hence are the same

So both topologies have the same open sets

[vast estuary]

Contradicting this

[bleak helm]

How does it contradict that?

[vast estuary]

Oh yeah sorry, my bad

Got it

[bleak helm]

Y under the order topology and Y under the subspace topology have the same open sets. But it also includes sets not open in R

Okay

[sweet wing]

can someone help me sanity check my sol to ISM prob 6.1:

(lee ISM 6.1: prove that for a smooth map F:M->N where dim M<dim N, F(M) has measure 0 using the fact that images of measure 0 sets under smooth maps remains as measure 0)
isn't M measure 0 in MxR^k and define F':MxR^k->N by F'(m,x)=F(m)
then doesn't it follow immediately that F'(M,0) has measure 0 in N since we can just piece together a covering of N, seems super overkill for lee to use sards if this works 🤔

[past barn]

What is the equation to work out the hodge star operator?

yes

liek, what is it

$\star \omega$

[gentle osprey]

LolLol

[past barn]

what would dis mean?

or is that incorrect way

of using it

for a k-form omega

me gusta where you go

[gritty widget]

you can probably find this on wikipedia or in any analysis on manifolds text

https://en.m.wikipedia.org/wiki/Hodge_star_operator

@past barn

[past barn]

@gritty widget ive looked at that page but I still dont really understand how to compute it

:/

the first problem I have is, where does n come from, and what would it's value be

the second problem I have is, what does the $\beta_{1}^{\star}$

[gentle osprey]

LolLol

[past barn]

mea

n

[sweet wing]

truedat

smt liddat

yea

[past barn]

if I have a 1-form

would the hodge star operator of this be,

[gritty widget]

yes

for fun, compute *d*df

[pearl holly]

Okay I learned about these planes that you glue together to get a cool thing while reading about quotient maps. You can formalise the construction of a donut from a sheet of paper with quotient maps. (You take a "sheet of paper", fold it to get a cylinder and then you fold the endpoints togehter) But can't you do it with homeomorphisms too? What "visual" differences are there between quotient maps and homeomorphisms when they act on things?

[obtuse meteor]

Quotient maps can glue things

Homeomorphisms cannot

[pearl holly]

Oh lmao

okay and why is that formally using the definitions?

[honest terrace]

quotients are about changing the spaces (and after doing that you get a map from the initial space to the new space, the quotient map), homeomorphisms are maps between spaces that are "topologically the same"

the point of homeomorphisms is that they don't change anything about the topology, so they're not exactly suited to glue stuff, because usually that will change the topological structure

[tawdry valve]

also a homeomorphism, as a function between sets, is a bijection, so you can't glue things because that would require identification

[pearl holly]

okay and what is a identification?

[tawdry valve]

oh I mean just a function being non-injective

like

for f: X->Y, if f(a)=f(b), then you might say that f identifies a and b

like for example, if you wanted to make a cylinder out of [0,1]^2

you would want to maybe glue the top and bottom sides of [0,1]^2

"gluing" in this context is just making them be the same thing. you replace each points (x,0) and (x,1) with a single equivalence class

and the projection map is the map pi: [0,1]^2 -> [0,1]^2/~ that glues (aka identifies) the top and bottom

I think shika was making a good point about homeomorphisms too

homeomorphisms=isomorphisms in the category of topological spaces

from the topologist's perspective, if two spaces are homeomorphic they are really just relabelings. You might call a point blue, I might call a point purple, it's just names

[pearl holly]

Right okay, I get it now! In this case, gluing the [0, 1]^2 to a cylinder with the equivalence classes would require a non bijective function so it can't be a homeomorphism. But why is the definition of a quotient map like it is? So p^-1(U) is open iff U is open. How does that help with the "gluing" process?

[marsh forge]

Its backwards

the way its presented is

what you want to do is like

take some space

glue it in some way to get a new "space"

but whats the actual topology on this new space?

the topology you give it is exactly the condition in a quotient map

idk why people insist on presenting it in that way

basically the "definition" of a quotient map there is telling you how the topological data of the unglued space descends to give topological data for the glued one

[pearl holly]

ahhh okay. So a quotient map is a way to transcribe/give data about the topological structure of the unglued space and the glued space?

So maybe it's a way to "deliver" data about the glued space?

[marsh forge]

thats a good way of thinking about it yeah

[pearl holly]

Hmm but it almost feels more natural to define a quotient map to be p(U) is open iff U is open. But with this, we can't really say what the open sets in the glued space are and therefore it's "better" to define it using p^-1(U) is open iff U is open because then we have the information about the open sets in the glued space. Is this a good way to think about this?

[tawdry valve]

For me, the thing that motivates the quotient topology is that it is the finest (most open sets) topology that makes the projection map continuous

[verbal wraith]

from the point of view of semi-lattices quotients are what we usually call subobjects and subobjects are downwards closed sets

it's just a smaller collection of closed sets that's closed under intersections

[pearl holly]

oh boy, here we go again

[verbal wraith]

I mean you seem to be having trouble understanding quotients in topology

[pearl holly]

Sorry but I don't really know what a semilattice is. I really appreciate the effort, but I don't really get it yet

[verbal wraith]

just think about what happens to P(X) when you take a set-quotient

[gritty widget]

the former construction might not be closed under intersections. let $p(U)$ and $p(V)$ be open sets. then $p(U \cap V) \subset p(U) \cap p(V)$, but the other inclusion may not hold.

oh no my tex

[honest terrace]

yeah this ^

oh no my queen

[pearl holly]

yeah I actually thought about that but then you deleted your post and then I started to worry that I was wrong lol

[gentle osprey]

TTerra

[gritty widget]

yeah i deleted it since i wanted to get on my computer and type a better answer lol

[honest terrace]

Half of Rosen's "oh no my queen" moments

[gritty widget]

the thing is, $p^{-1}(U \cap V) = p^{-1}(U) \cap p^{-1}(V)$, so this isn't a problem in defining a topology

[gentle osprey]

TTerra

[pearl holly]

yeah now I get it lol

[gritty widget]

some might say "pre-images behave nicer with respect to intersection"

[pearl holly]

okay I get it now. Thank you all so much, it really helped me to understand the intuition behind this!

[gritty widget]

[versed junco]

@drifting sundial What would Z represent in f:R->SxZ? Current multiplicity of point on the path? If so, it is a viable model.

[marsh forge]

You want to define the open sets of the image (post glue) based on the original (pre glue)

oh whoops nvm

yes tterra has a better answer

very sleepy shouldnt be mathing

[hazy nexus]

Oh hey, it's the lattice guy

My hero

He's an expert in his field

Of making all of topology incomprehensible

[gritty widget]

[tight agate]

[hazy nexus]

I dream for the day we get Urysohn's lemma in terms of upper semi-lattices or whatever

[honest terrace]

[verbal wraith]

I mean, I just actually know topology... SMH, you should really lose your sully privileges for sullying people who know what they're talking about. This is like stack exchange all over again.

[hazy nexus]

Please formulate Urysohn's lemma in terms of semi-lattices

[gritty widget]

i don't think there was ever doubt that you know topology. you just present it from a different viewpoint that most people here don't share or find pedagogical to people learning it for the first time

[empty grove]

Imagine where society would be if y'all just accepted his opinions

We'd have so many more godly copy pastas

[empty grove]

But seriously tho do you have any resources where one could study this?

Because that would be better than just bits and pieces from you over text without knowing prerequisites

[honest terrace]

this

that doesn't sound uninteresting tbh

[empty grove]

Ye, I'm actually kinda interested in seeing if these kinda definitions would make some parts of topology easier

[honest terrace]

it's just meme-tier when you answer with this stuff to someone asking something about elementary topology

but yeah same, I'm also interested

[cerulean oriole]

+++++++++++++++

I can't even say whether this is good or bad because I don't know half the terminology.

[drifting sundial]

yep

[modern osprey]

holy hell

[drifting sundial]

I think you could even interpret this as an actual topological path. in the case of rotations, choose some basepoint like (0,1) and identify all of its copies to get a quotient.

[past barn]

alright!

[gritty widget]

it should come out to something familiar

[past barn]

I will try in a few hours because

i Have fricking school

in a bit

[gritty widget]

[past barn]

😢

(im bri'ish)

[gritty widget]

just for simplicity do it for n = 3

you've already done part of the computation anyways

[past barn]

alr

lol the person in my server mad at me

this is

d*df

now star of that

wait

is that like

$\star d\star df= \frac{\partial^2f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2f}{\partial z^2}$

uhhhh

UHHHHHHHHHHHHHHH

[gentle osprey]

LolLol

[gritty widget]

[past barn]

there we go

[gritty widget]

correct

[past barn]

yay!

[gritty widget]

this is the laplacian!

[past barn]

Oh yeah

pog

I understand hodge star opertaor

interesting

[gritty widget]

laplacian is $\Delta f$, hessian is $\nabla^2 f$

[gentle osprey]

TTerra

[past barn]

my electromag book...

HAS LIED TO ME!!!

[gritty widget]

lol

[past barn]

delta!

alright

i use

that

now what would

$\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star d\star df$

be

[gentle osprey]

LolLol

[past barn]

coool thing is i realized u can rewrite physics equations with this

cause maxwells equations use a lot of those operators

which can be written in termsof exterior algebra operations :p

[viral atlas]

Please avoid spamming in topic channels. 😅

[vast estuary]

Why is the part in blue true?

[true robin]

If Vn is not dense, its closure is not X, so the compliment of Vn closure is a nonempty open set. hence we have an x0 and that x0 has a neighborhood around it that is not in Vn closure

[gentle osprey]

Hausdorff

[true robin]

Yes it is. If |x|<r then |x+x_0-x_0|<r so x+x_0 is not in Vn. If it is the equality that is bothering you, we could take the closed ball of radius r/2 instead.

[vast estuary]

Thanks!

[vast estuary]

Could I get some hint on how to get (3)?

[gentle osprey]

Hausdorff

[woeful oasis]

I think it's just a matter of applying $\Lambda$ to $x_i$, and since $\Lambda$ is continuous you get $\Lambda x_i \to y$.
The rest seems like a matter of bookkeeping

[gentle osprey]

Apopheniac

[vast estuary]

Wait what do they mean by THE open unit balls

Are they assuming that they are centered at origin by default?

[vast estuary]

Okay assuming that the unit balls are centered at origin, I have a proof. Let me type it out

[marsh forge]

if you see an alg topologist using coordinates (which they rarely do) its probably based on the most obvious choice of imbedding

[obsidian socket]

hi, I've been trying to understand weil divisors, and am wondering if someone could help or point me in the right direction to a source which doesn't require knowledge of schemes etc

[cerulean oriole]

(I don't know.)
Does this definitely need the Baire theorem?
It seems so much simpler somehow.

[gentle osprey]

Hausdorff

[vast estuary]

Looks good? Let me know if you see any errors

[vast estuary]

I realized I wanted to post this in #advanced-analysis but here it is now, oops! Hope it's not too off topic

[empty grove]

@verbal wraith

[marsh forge]

In the spirit of not just sullying but explaining the sully, no one is sullying you because you are wrong. they are sullying you because you keep trying to shoehorn a POV that people don't respond to. If you acted like this on stackexchange I have no doubt people would be annoyed as well.

[verbal wraith]

Im writing something up currently but realistically I'm not close to done.

[gritty widget]

new copypasta?

lessons in why it's important to learn math before category theory

[marsh forge]

i mean you can actually phrase a lot of point set very elegantly in categorical ways

but thats not related to the semilattice business

in some ways the category theoretic stuff can even be expressed without a fore-knowledge of category theory

but I am hesitant to try it out on people who are already confused by the way their source or teacher presents stuff

[past barn]

@gritty widget hi, you is pro at exterior algebra thingies, cans you explain interior product owo

[gritty widget]

it's relatively simple

[past barn]

gud gud

[gritty widget]

if you have a k-form on a vector space and a vector v, you slot v into the first argument of your form and get a (k - 1)-form

[past barn]

ohhh

that's ez

so for example, a 1-form

$xdx$

[gentle osprey]

LolLol

[past barn]

idk why i did latex

to write that lol

then

wait

what would that be

owo

[gritty widget]

if $\omega$ is a one form and $X$ is a vector field, then the interior product of $\omega$ by $X$ would just be $\omega(X)$

[gentle osprey]

TTerra

[past barn]

yeah but if

$\omega (x) = xdx$

[gentle osprey]

LolLol

[past barn]

what would interior product be

cause I think it

XdX

[gritty widget]

let me be a little more precise for a moment

[past barn]

but that still have differential

[gritty widget]

i told you the linear algebra version

now let me tell you the differential forms version

[past barn]

o ogi

i feel like this gonna be big

😨

[gritty widget]

linear algebra: if $\omega\colon V^k \to \bR$ is a $k$-form (so alternating and multilinear), and $v \in V$, then the interior product of $\omega$ by $v$ is the $(k-1)$-form defined by $$(\iota_v\omega)(w_1,\dots,e_{k-1}) = \omega(v,w_1,\dots,w_k)$$

[gentle osprey]

TTerra

[past barn]

[gritty widget]

to get the version for differential forms on manifolds (e.g., open subsets of euclidean space), replace "form" by "differential form" and "vector" by "vector field"

roughly

[past barn]

can u do example

[gritty widget]

(if k = 0, we define the interior product to be 0)

[past barn]

lol

oh

i see

so like can U do with

vector field = (x,y) and differential form is xydx + xydy

[gritty widget]

if our vector field is $X$, and the form is $\omega$, then it's the $0$-form (a function) given by $xy,dx(X) + xy,dy(X).$ since $X(x, y) = (x, y)$, $dx(X) = x$ and $dy(X) = y$, so the interior product ends up being the function $\bR^2\to\bR$ given by $$(\iota_X\omega)(x, y) = xy(x + y)$$

[gentle osprey]

TTerra

[past barn]

wait wa

im notfamiliar with

dx(X)

lol

[gritty widget]

differential 1-forms eat vector fields and give functions

(one way of looking at them)

(also my favorite since it keeps the notation simple and computations cleaner)

[past barn]

:o

I see, thanks for da help again sir

C;

[gritty widget]

[past barn]

sadge

my brain does not understand

applying dif form to vector field

stinky brain

:C

like how u compute

dx(x,y)

=

x

hows

😢

[honest terrace]

learning math is precisely learning ct, nothing else is real math

[woeful oasis]

$dx:T_p M \to \mathbb{R}$ takes as input tangent vectors. This form in particular satisfies
$dx(\partial_x) = 1$

[gentle osprey]

Apopheniac

[empty grove]

Ty

[gritty widget]

[woeful oasis]

Perhaps "orbit"? Or "the orbit of a point (or points) under the action of a 1-parameter subgroup of a Lie group"
"Orbit" means the points you get after applying the action of the group. "1-parameter subgroup" meaning you'd like to move the points according to a curve, not a surface etc.

[fathom cave]

can you recommend me a few sources on graded algebra @gritty widget

[gritty widget]

idk open up a commalg book

i literally don't know lol

[fathom cave]

i looked up in gallian and dummits book
didnt find it

[gritty widget]

did you check atiyah macdonald?

[fathom cave]

Nope

alright i will look into it

[gritty widget]

that's the canonical commalg book afaict

(i'm not an algebraist)

[tacit sleet]

matsumura chapter 13

atiyah--macdonald chapter 10

[woeful oasis]

Eisenbud does a lot of graded stuff throughout

[gritty widget]

would anyone like to do a reading group on hatchers spectral sequences over the summer?

[frozen otter]

These are three persistence diagrams produced by different datasets that represent the same underlying phenomenon. The points themselves are in a 512-dimensional space, but my computer doesn’t let me go that high. I know that these points lie roughly on a sphere in 512D, and the main question I’m interested in is how they are distributed on that sphere.

The 0D part of the diagram tells me that there are multiple connected components (usually 2, but potentially up to 4 in one of the plots) and no higher level structures (which is expected if they are distributed around a 512D sphere). I'm not sure what (if any) meaning the large gap in the 1D and 2D PH along the line x = y means.

Has anyone seen things like this before / have advice about interpreting these diagrams?

[severe stump]

I have a topology question: In the moore plane (https://en.wikipedia.org/wiki/Moore_plane), do there exist disjoint open sets U and V such that U contains all (x, 0) points where x is rational, while V contains all (x, 0) points where x is irrational? I suspect the answer is no, but I don't know how to prove that.
(I'm self-studying, and this was an exercise, and I don't know how to even start this.)

[pearl holly]

Okay so this is the theorem that I'm reading now: Let X be a space. Then X is locally compact Hausdorff if and only if there exists a space Y satisfying the following conditions:
(1) X is a subspace of Y.
(2) The set Y-X consists of a single point
(3) Y is a compact Hausdorff space.
After the proof, the author writes "if X is not compact, then the point of Y-X is a limit point of X, so that the closure of X is Y". I don't understand how that follows. Why is the point of Y-X a limit point of X when X is specifically not compact?

okay so if that point were not a limit point, then X would be closed in Y since it would contains all its limit point. But then it is a closed subspace of a compact space so it must be compact, a contradiction I guess

[tepid depot]

Seems correct, though I'd hope there's a nicer way

I think it's a little more clear by contrapositive. If the closure of X is not Y, it must be X. So X is a closed subset if a compact Hausdorff space, and thus compact.

[pearl holly]

yeah that's true. Thank you!

[gritty widget]

petition to go back in time and tell old geometry authors that \langle and \rangle are a thing and to not use < and >

[gritty widget]

If $Y={P_1,P_2}\subset \mathbb{A}^2$ how do i find $I(Y)$. Am I right in thinking its $\langle (x-p_{1,x})(x-p_{2,x}), (y-p_{1,y})(y-p_{2,y}),(x-p_{1,x})(y-p_{1,y}),(y-p_{1,y})(x-p_{1,x})\rangle$

[gentle osprey]

lime_soup

[gritty widget]

And is this how I should be thinking of this question

[frigid patrol]

This channel is super cool

https://www.youtube.com/channel/UCQp_QaILaVzfa5K_yDJP12A

I just wanted to share it

It animates some cool homotopies and covering spaces

[gritty widget]

Does anyone know of a good way of book keeping this

say I have r points in affine n-space

all distinct

I want to write down generators for the ideal

call the kth point P_k=(p_{k,1},p_{k,2},...,p_{k,n})

[true robin]

By the ideal do you mean the ideal corresponding to the n points? In that case the ideal is generated by elements of the form: (x_i-p{k,i})(x_j-p{l,j}) where i,j,k,l range over all possibilities

[gritty widget]

yes

is this enough?

i would have thought we needed all the coordinates

for example if we have 3 points in A^3

by this (x1-p11)(x2-p22) is a generator

but this does need not vanish on P3

[true robin]

Oh wait it should be intersection of ideals not products, damn. Well in that case i can’t see a nice way to get the generators.

Sorry I got a bit mixed up.

[true robin]

Ah wait nevermind, since all the maximal ideals are coprime the intersection is the product. So instead of a product of 2 generators, take a product of all r generators. Concretely: a_1a_2…a_r where each a_i is of the form x_l-p{i,l} for some l should be a generating set (when you take all possibilities for l)

[gritty widget]

okay thanks

[severe stump]

No answers? TT_TT

[reef shore]

Hint: if such a V existed, writing V as a union of B_i = {i} union B((i,r_i),r_i) for each irrational i (any other basis elements in V can be removed WLOG), then any rational q will be a limit point of {(i,r_i)} in the standard topology of R^2.

Actually I'm not sure if that will prove anything, nvm

[severe stump]

Yeah. I can construct disjoint open sets U and V such that U contains (0, 0) and V contains all other points on the X axis. If your method of proof worked, it would also probably contradict this counterexample.

[reef shore]

right

I feel like you could do something by induction. Take the basic set in U containing (0,0). That would have some finite radius which will allow you to choose an irrational close enough to 0 that its basic set in V has its radius bounded by 1/2. Then take a rational close enough so that its basic set in U is bounded by r/4 and so on. This should be a convergent sequence in R, and its limit should have its basic set bounded by every r/2^n

[severe stump]

This kind of construction doesn't work, I think.

For any distjoint sets A and B that are subsets of the X axis, if A and B are countable, then a construction similar to that results in open sets U and V such that U contains A and V contains B.

[reef shore]

Mine uses density of both the sets

[severe stump]

Basically, order the points in A and B somehow, and for each of the point, choose a circle that's small enough that it doesn't overlap the previous points in the order.

[reef shore]

oh

[severe stump]

This only works for countable sets, and breaks down on the irrationals.

So I suspect that there isn't similar construction for the rationals and irrationals

@reef shore I'm actually not sure I understand your construction.

Are you trying to prove by contradiction that the two open sets can't exist?

[reef shore]

yeah thats what I was trying

I am taking alternating between taking rationals and irrationals. Lets say the basic set containing (0,0) has radius r. Then pick any irrational x with |x-0| < r, so the basic set containing this must have radius < r/2, then pick a rational at distance < r/2 from this, and that must have radius < r/4 and so on

[severe stump]

Oh. Wait. Lemme think.

That might work because that produces a series of numbers that must converge to something either rational or irrational.

Probably.

[reef shore]

yeah that's what I am thinking but I am failing to see why this doesnt prove that the construction doesnt work for countable dense sets

[severe stump]

And this convergence thing is what's missing from the countable sets thing.

[reef shore]

oh I think I see it

right

yeah that limit point will be close enough to everything in the sequence to put a bound of 0 on it

[severe stump]

Lemme try to formulate this properly...

[reef shore]

[severe stump]

(loading icon here)

So... start with (0, 0) is in the circle with radius r. Now we look at an irrational number between 0 and r/2.

Actually, let's just use an irrational number between 0 and r/4. So (x, 0). The radius of the circle there must be no more than r/2.

[reef shore]

I think choosing one between 0 and r also gives a bound of r/2

[severe stump]

sure lol,

Let's say the new radius is s. (s < r/2).

Pick a rational number between x and x-s/4. The circle there has radius no more than s/2.

Repeating this results in a series of ever-shrinking circles. And since we switch between going left and right, and the steps keeps getting shorter, the tangent point of the circles must converge to something.

Something something contradiction. Good enough for me lol

[reef shore]

lmao

You don't even need to alternate

[severe stump]

There are a few more annoying details that I'm not gonna bother dealing with.

Oh right, doesn't even need to alternate since it's bounded by a geometric sum.

[reef shore]

yep

[severe stump]

something something analysis is annoying

thx!

[reef shore]

[pine kernel]

Can anyone point me towards resources to help me tile a sphere with polygons?
Basically I want to do something similar to spatial partitioning but in a spherical domain. I am looking for the simplest way to achieve this but the internet indicates me(a math noob) I should give up. The only divisions I think I can do right now is dividing sphere surface into n wedge surfaces but its not the best division for my purposes.
Any suggestions?

[gritty widget]

Looks like B and D might be homeomorphic

But how draw the homeomorphism

[reef shore]

They're not, B is shaped like a cross and D is a single curve

[gritty widget]

proof by picture

[reef shore]

oh wait

[gritty widget]

B is not cross

[reef shore]

yeah I was reading A for some reason

Projection should be a homeomorphism

45 degree projection

of D onto B

[gritty widget]

(0,0) of B goes to where in D ?

[reef shore]

(1,1)

(0,c) should go to wherever the line x+c=y intersects xy=1

(c,0) wherever x-c=y does the same

[gritty widget]

I still not get the projection function ....

[reef shore]

Or I think easier way would be to translate D so that (1,1) is on (0,0)

Then D looks like some stuff in the second quadrant and some stuff in the fourth

project the stuff in the second quadrant onto the y axis and the stuff in the fourth quadrant onto the x axis

[gritty widget]

How there any stuff in 2nd and 4th quadrant ? D is a curve in first quadrant...

[reef shore]

I am translating D

like this

[gritty widget]

Cool

And then draw projection function somehow.......

On to the axises

[reef shore]

Yeah break it apart into the 2 pieces, projection on each thing is a homeomorphism, then use pasting lemma to say that the pasted function is continuous

ok even easier solution lol

B is homeomorphic to R because just bend the positive y axis onto the negative x axis

so (x,0) -> (x,0) and (0,y) -> (-y,0)

and D is homeomorphic to the positive x axis by projection

positive x axis is homeomorphic to R

[gritty widget]

just appeal to classification of connected topological 1-manifolds smh

the solution i had in mind was basically the same thing

[reef shore]

I came up with the same solution as teppa after only 3 tries

I can be a geometer now

[gritty widget]

one's the graph of a continuous function on (0, \infty) so it's homeomorphic to that, and one's homeomorphic to R by fixing one half and rotating the other. and R and (0, \infty) are homeomorphic

your idea is geometric and nice though

[reef shore]

[gritty widget]

invite link dm'd

[reef shore]

[gritty widget]

I was new......

What is pasting lemma...

it pastes

i used it the other day

What is "classification of connected topological 1-manifolds"

it's exactly what it sounds like

list out all connected topological manifolds of dimension 1 up to homeomorphism

[reef shore]

If you have 2 continuous functions defined on 2 closed sets U and V respectively and they agree on the intersection, you can paste them to get a continuous function on the union

[gritty widget]

(i.e., give a list of connected topological manifolds of dimension 1 such that no two are homeomorphic, and all (previous adjectives) 1-manifolds are homeomorphic to one on the list)

for connected 1-manifolds the distinct four are [0, 1], [0, 1), (0, 1), and S^1

[pseudo crane]

I like how you wrote (previous adjectives) instead of "connected"

[gritty widget]

repeating myself gets annoying

[pseudo crane]

true,true

[honest terrace]

what is there except n circle glued together , R, the long line and other longer stuff in the same spirit of the long line (except if you enforce countable basis, in which case the long line and longer stuff doesn't count as manifolds ?) ?

[gritty widget]

you can do it for the empty boundary ones using some simple riemannian geometry

[honest terrace]

oh wait

you can do edges of a cube too, for example, ig 🤔

[gritty widget]

i'm assuming second countability

and hasudroff

[reef shore]

edges of a cube dont give a 1-manifold

[gritty widget]

yeah the circles glued together don't either

[reef shore]

3 edges meet at a point

[honest terrace]

oh and then this point can't have a neibrd homeo to R ?

yeah right okay

ok so what is there except R and a closed segment

[reef shore]

tteppa already said the 4

[honest terrace]

and a circle

I missed it where he said it

[reef shore]

.

[honest terrace]

oh right

then I had all of them except [0,1)

and other stuff

that aren't manifolds

:pepega:

[reef shore]

3/4th a geometer, and a little bit of something else

[gritty widget]

pick a riemannian metric g on M. consider a point of M and a maximal unit speed geodesic emanating from the point. you can prove that this has to be onto using a connectedness argument. it's also not hard to see that the geodesic is a local isometry. if it's injective, then M is isometric to the open interval on which the geodesic is defined. if it's not injective, then it's periodic and hence isometric to a circle

roughly

[reef shore]

I agree with ultra here.

[honest terrace]

I agree with Moldilocks here

[gritty widget]

I agree with myself here.

i wonder if you can modify this argument for the manifolds with nonempty boundary

probably

[swift fjord]

I have a question about the standard proof of Brouwer's Fixed point theorem using homology, with the statement of the theorem as follows:

Suppose $f:D^n \rightarrow D^n$ is continuous, then it has a fixed point.

The proof i'm talking about is the one where you suppose by contradiction, then construct a retract onto $S^{n-1}$ by considering the intersection of the line spanned by $x,f(x)$ with $S^{n-1}$. My question is, it seems that $f$ must be continuous for this retract to be continuous, but I don't see exactly why this function must be continuous.

[gentle osprey]

ShiN

[gritty widget]

[coral pivot]

Uh do you mean why the retract is continuous?

[swift fjord]

yes

more, how do you show that it is continuous

well if it wasn't continuous you couldn't call it a retraction

[coral pivot]

Try writing it out explicitly ig

[gritty widget]

have you tried finding a formula for it

sniped me

[honest terrace]

it's in the name, retract, ct means continuous here

[swift fjord]

lol

[reef shore]

Haven't proven that it's a retract yet

[gritty widget]

[swift fjord]

I'm so bad with analytical geometryt

but in essence parametrising it should be the same in any dimension right

[coral pivot]

I mean it’s AT you just picture it and say it’s obvious

[swift fjord]

clearly

[gritty widget]

x + t(f(x) - x)

[swift fjord]

yea that's what I thought

ok you know what, I think i'll skip proving that it's continuous

I mean it's easy to see

[gritty widget]

a true topologist

[reef shore]

[honest terrace]

soo that's how we're supposed to do topology

[gritty widget]

anyways just find t that makes |x + t(f(x) - x)|^2 = 1 and you have th map

[honest terrace]

wow that makes everything so much easier

[gritty widget]

it should be straightforward

[reef shore]

Therefore no need to do it

[pseudo crane]

i think nothing in ag has ever been proven

people just assume someone before them made it rigorous

[gritty widget]

@swift fjord here's an actual idea. $|x + t(f(x) - x)|^2 = 1$ if and only if $$|f(x) - x|^2t^2 + 2\langle x, f(x) - x \rangle t + |x|^2 - 1 = 0.$$ the leading coefficient $|f(x) - x|^2$ is always non-zero by assumption. by the quadratic formula, to show that the root $t$ (and thus the point on $S^{n-1}$ obtained by the line from $x$ through $f(x)$) depends continuously on $x$, it suffices to show that $$ \langle x, f(x) - x \rangle^2 - |f(x) - x|^2(|x|^2 - 1) > 0 $$ when $|x| < 1$

[gentle osprey]

TTerra

[gritty widget]

(or >= 0, whatever)

maybe it's geometrically obvious that this thing should have a root and that this is a silly thing to check

but if you actually wanted to be sure of continuity without writing down an explicit formula for the retract, this would probably work with minimal pain

tl;dr the quadratic formula ensures continuity

moldi pls one up me with a real argument

ok fine having a root is geometrically obvious, the quadratic formula ensures continuity, now apply homology qed

[coral pivot]

i mean that seem quick enough (the root part follows from cauchy swartz)

[empty grove]

You could also say that given any epsilon neighborhood on S^(n-1) around the image of x under this function, contains the image of some delta neighborhood of x. For this, you can find a delta_1 small enough such that delta_1 balls around x and f(x) are such that any line going through both in the right order lands in that epsilon neighborhood (visually obvious, can probably prove) and then you can take a delta small enough so that the delta ball around x lands inside the delta_1 ball around f(x)

Then take min(delta, delta_1)

[gritty widget]

g&p has a proof of continuity iirc

or am i thinking of a different book

can't believe these geezers copied me

[swift fjord]

Maybe i'm dumb but I don't see where this argument uses the continuity of $f$

[gentle osprey]

ShiN

[coral pivot]

Last line

[swift fjord]

oh wait

I didn't see that line lol

ty

[gritty widget]

you need to be like

always looking

[swift fjord]

lmao

thank you though that's much nicer than finding an explicit formula

[gritty widget]

Hyperbolic geometry

[gritty widget]

I was trying to show in $\mathbb{H}$ -Poincaré half-plane model- that hypercycles defined for a given line $l$ and a given $a$ in $\mathbb{R}$ as $$H(l)={z \in \mathbb{H} : d_{\mathbb{H}}(z,l)=a } $$
are not lines.

To prove something I tryed to argue that given that $Mob(\mathbb{H})$ acts transitvely on the lines of $\mathbb{H}$ I can find the hypercycles of the imaginary axis and via the $\gamma \in Mob(\mathbb{H})$ that maps my line in the imaginary axis i can state $$H(l)=\gamma^{-1}(H(\gamma(l))$$

Where I'm using the fact that a Möbius transformations preserves distances. At this point I defined [already here i have doubts] $d_{\mathbb{H}}(z,l)$ for a given $z \in \mathbb{H}$ and a line $l$ as the inf of the distances $d_{\mathbb{H}}(z,w)$ with $w \in l$.

Given this and the general assumption $z=a+bi$ and $w=ki$ i'm left with trying to calculate the inf of $$arccosh(1+\frac{|a+(b-k)i|^{2}}{2bk})$$.

Is any of this correct? There is any "distance from a given point to a given line" standard formula in hyperbolic geometry? Thanks for any help and hint.

[gentle osprey]

Stephen

[west spindle]

@gritty widget idk how much it helps, but hypercycles based on the imaginary axis in the half-plane model are rays of the form y = mx with m real and nonzero

[gritty widget]

But how do you know it? @west spindle

[west spindle]

do you want the honest answer or an attempt at a mathematically sound answer?

[gritty widget]

The second one

[west spindle]

alright

[gritty widget]

To you think deriving the expression that i found might lead to something?

[west spindle]

so you have $\min_{k>0} \cosh^{-1}\paren{1 + \frac{|a + (b-k)i|^2}{2kb}}$

[gritty widget]

Yes imho this is the general distance

[gentle osprey]

Ann

[gritty widget]

And i defined distance as the inf

[west spindle]

that's the formula for the distance between two points in the upper halfplane model

[gritty widget]

Not sure this is the standard in hyerpolic geometry

Yes it is

[west spindle]

we might want to focus on minimizing $\frac{|a + (b-k)i|^2}{2kb}$ instead since the function $x \mapsto \cosh^{-1}(1+x)$ is monotone

[gentle osprey]

Ann

[west spindle]

so we have $\frac{a^2 + (b-k)^2}{2kb} = \frac{a^2 + b^2}{2b} \cdot \frac{1}{k} - 1 + \frac{1}{2b} \cdot k$

[gentle osprey]

Ann

[west spindle]

this is just algebra. do i need to go into detail here?

[gritty widget]

Ok the monotone thing makes it clean

[west spindle]

yes it does

[gritty widget]

I hope i would be able to end from here

[west spindle]

you can take the derivative of this wrt k and find the minimum that way

[gritty widget]

Hoping it gives the wanted results

[west spindle]

i believe you'll find that it is a function of (b/a)

[gritty widget]

b/a are fixed it's fine

[west spindle]

yes i know

it'll just prove my point that equidistants based on the line x=0 are rays at non-right angles to the absolute

for that matter, equidistants centered around other lines are circular arcs which intersect the absolute at non-right angles.

[gritty widget]

I feel kind of weird about this

with some small conditions we can make a statement like

almost all smooth functions are morse functions

but its still a pain to check if a random function is actually a morse function

is there must be some nice trick like, if we have a family of smooth functions parameterized by some continuous thing then at there are only finitely many non morse functions in the family

this statement is obviously not true

but it just feels so weird that they are everywhere but hard to check

[gritty widget]

Im not sure about the definition of distance from a point to a line but it should agree with what im doing

As lines are geodesic

[pearl holly]

So this is the exercise that I'm working with: if f: X_1 -> X_2 is a homeomorphism of locally compact Hausdorff spaces, show f extends to a homeomorphism of their one-point compactifications. What does it mean for a homeomorphism to extend to a different homeomorphism?

[marsh forge]

Okay so

Let X_1* be the one-pt

then you have X_1 includes into $X_1^*$

[gentle osprey]

MaxJ (Sockoist) ✓
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[marsh forge]

thats ugly

okay anyway

you want to show that there is a homeomorphism

form X_1* to X_2*

such that

when you restrict the map to the subspace X_1

you get the original homeo

does that make sense?

so like a homemorphism g

such that g(x)=f(x) whenever x is in X_1

so for all but one point in this specific instance

[pearl holly]

hmm but how can I get the original? So the restriction will be like f: X_1 -> X_2*, but this can't be a bijection since X_2(star) is "one point" bigger than X_2?

[marsh forge]

oh

[cedar pebble]

you get the original when you restrict to the original space

[marsh forge]

yeah

so you want a homeo g between their one-pts

that restricts to a homeo f between the original spaces

[cedar pebble]

so the property that this morphism "extends" is a certain factorization property for the canonical inclusion X->X*

[marsh forge]

(this isn't exactly a normal extension problem since the target changes)

[pearl holly]

okay so I am restricting both the domain and range?

[marsh forge]

you are restricting the domain

but the range gets restricted automatically

[pearl holly]

ahh

[marsh forge]

it's not asking for the restriction to be a homeomorphism of X_1* to X_2

or X_1 to X_2*

sorry

[pearl holly]

yeah okay I see I think. So the image will be X_2?

[marsh forge]

yes

[pearl holly]

Okay great, thank you so much!

[gritty widget]

any good differential topologists here?

If I have a perfect morse function

you don't need to worry about the gradient flow lines to compute the homology

its just done from the indices of the critical poitns

ping dami for morse theory help

will they mind?

probably not. the only people i can think of who probably know morse theory past the first 3 sections of milnor are dami and brofibration

@honest narwhal i would like to summon you

[honest narwhal]

What's up

Lol I don't actually know shit in Morse theory but I'm down to struggle with you guys a little bit

[gritty widget]

Im working through the calculation of the cohomology of the complex grassmannians

it actually ends up being pretty slick at the end

[honest narwhal]

oo

[gritty widget]

the cohomology groups are

[honest narwhal]

Fam

You can see that people are doing things in here already

[gritty widget]

free groups of size counting schumber sells

[honest narwhal]

Don't interrupt

[gritty widget]

I like that it was

any help?

then huge wall lmao

[honest narwhal]

I'm deleting it lol

[gritty widget]

okay but what i am confused about is

Sorry, tought here you could ask

With more liberty

[honest narwhal]

From now on if you see a conversation happening

Don't frickin blow it up lol

Wait your turn

Got it?

[gritty widget]

there is a section that goes through calculating the gradient flows of the morse function

but

we also show that it is a "perfect morse function"

[honest narwhal]

What's perfect here?

[gritty widget]

and by this we mean all the critical points have even index

[honest narwhal]

Ah

[gritty widget]

so then we you make the homology complex, the odd degree things will vanish

(this tells you why CPn does have cohomoloy in odd degrees)

[honest narwhal]

Oh I guess that makes sense in hindsight

[gritty widget]

doesnt

but then i don't see why they calculate the gradient flows

maybe its just there for completion?

[honest narwhal]

Maybe? What are you working out of?

[gritty widget]

Lectures on Morse Homology

(I might give a talk on this after I figure it out if anyone would be interested)

yes

i would

[honest narwhal]

Very much down. But yeah lemme read that bit

[gritty widget]

i read the first few sections of milnor and loved it so i'd be down

guys for hyperbolic Geometry questions here is my best spot?

Yeah I got it, I'm waiting for the signal

fire away

I will paste as is a latex one okay?

I was trying to show in $\mathbb{H}$ -Poincaré half-plane model- that hypercycles defined for a given line $l$ and a given $a$ in $\mathbb{R}$ as $$H(l)={z \in \mathbb{H} : d_{\mathbb{H}}(z,l)=a } $$
are not lines.

To prove something I tried to argue that given that $Mob(\mathbb{H})$ acts transitvely on the lines of $\mathbb{H}$ I can find the hypercycles of the imaginary axis and via the $\gamma \in Mob(\mathbb{H})$ that maps my line in the imaginary axis i can state $$H(l)=\gamma^{-1}(H(\gamma(l))$$

Where I'm using the fact that a Möbius transformations preserves distances. At this point I defined [already here i have doubts] $d_{\mathbb{H}}(z,l)$ for a given $z \in \mathbb{H}$ and a line $l$ as the inf in k of the distances $d_{\mathbb{H}}(z,w)$ with $w \in l$.

Given this and the general assumption $z=a+bi$ and $w=ki$ i'm left with trying to calculate the inf of $$arccosh(1+\frac{|a+(b-k)i|^{2}}{2bk})$$

EDITED FROM HERE TO SHOW MORE WORK:

given that arccosh is a monotone function i need to find the inf of $$\frac{a^{2}+{b}^{2}}{2bk}+\frac{k}{2b}-1$$

taking the derivative I'm looking to the zeroes of:

$$\frac{-2a^{2}-2b^{2}+k^{2}b}{2bk^{2}}$$

solving and remembering that i need $k>0$ i get $k=(\frac{2(a^{2}+b^{2})}{b})^{\frac{1}{2}}$

looking now for the $a+bi$ such that for a given $p$ we have:
$$ d_{\mathbb{H}}(a+bi,(\frac{2(a^{2}+b^{2})}{b})^{\frac{1}{2}}i)=p $$

expanding this i got an expression of $z,z^{*}$ that is not a standard form for a line in the plane and it should end the exercise.

Is any of this correct? There is any "distance from a given point to a given line" standard formula in hyperbolic geometry? Thanks for any help and hint.

[gentle osprey]

Stephen

[gritty widget]

If anyone knows any reference for this problem

In book or any hint i would very much enjoy. Can't seem to find any answer

did you ask this already?

Yes

did you try asking on stack exchange?

Yes

can you link to your post?

https://math.stackexchange.com/questions/4180918/hypercycle-are-not-lines?noredirect=1#comment8666170_4180918

Thank you for helping

[honest narwhal]

I'm back fuckin finally

So Banyaga and Hurtubise?

@gritty widget

[gritty widget]

sorry

i am too slow to get this humour

oh

the authors

yes lmao

[honest narwhal]

Hahaha

Which page ish?

Is it chapter 8? The bit in chapter 1 isn't giving much deets lol

Ah 8.6?

Yeah so in the next section he does just state that since the Morse function is perfect you just need to count cells in order to compute homology

I'm guessing the point of computing gradient flow was some mix of completeness and possibly to understand the unstable manifolds (which is a task that one can care about from a purely dynamical systems pov, not just in service of algebraic topology)

@gritty widget that's my guess

[gritty widget]

opps sorry was getting food

yeah i think you are right

this probably belongs in baby linear algebra but

permutation matrices are in U(n) because they are similar to the identity matrix

[honest narwhal]

Uh

No matrices are similar to the identity matrix

Rows of an ONB are just the standard basis but in some different order

Which is still an ONB of C^n

@gritty widget

[gritty widget]

ah yes

thanks

[gritty widget]

cf

[gritty widget]

identity matrix in shambles

what is the most appropriate laugh react

[gritty widget]

this one

[pastel thistle]

hey need help with a question