point-set-topology

[marsh forge]

Heres a hint

It suffices to show that if I fix some point (x,y)

All other points have a path to (x,y)

You can choose a nice (x,y) if you want

That does seek

Seem

Like a nice choice

That is not a bad idea

Np

[eternal anvil]

so did I miss this or can you define an n-dimensional manifold as any set that is locally isometric to a subset of Rn

[marsh forge]

Locally homeomorphic

Or diffeo

If you wanna be fancy

[little hemlock]

About the first hint: is f(y) = d(x,y) continuous? If it is, then f(A) is bounded and attains its bounds and the conclusion (for the first part) follows.

[fervent citrus]

yes it is

[little hemlock]

I think I just wrote a proof of continuity. one sec

[fervent citrus]

it's even 1-lipschitz

[little hemlock]

Proof(?) Let $y \in X$, let $\epsilon > 0$ and take the open ball $V = (d(x,y) - \epsilon, d(x,y) + \epsilon) \subset \bR$. If $U = B(y, \epsilon/2)$, then we have that $f(U) \subset V$ by definition of $U$.

[gentle osprey]

kxrider:

[little hemlock]

Seems kinda trivial
I have no idea what 1-lipschitz means xd

[fervent citrus]

it means that for all $y$ and $z$ in $X$, you have $d(f(y), f(z))\leq d(y, z)$

[gentle osprey]

Tuong:

[fervent citrus]

which implies the (uniform) continuity of f

[little hemlock]

that would just follow from triangle inequality, right?

[fervent citrus]

Yes

for all y and z in X, you have
d(x, y)<=d(x, z)+d(y, z)
and
d(x, z)<=d(x, y)+d(y, z)

from which you can obtain |d(x, y)-d(x, z)|<=d(y, z)

[little hemlock]

i see. thank you!

[fervent citrus]

You're welcome

[fluid cloud]

Heylu

What's the natural metric on functions in [a,b]?

A question requires me to find d_1, d_2 and d_infty of certain functions

[marsh forge]

Where are they valued

[fluid cloud]

erm

[0,1]

Consider the function $f(t):=t$ and $g(t):=t^2$ for $t \in [0,1]$. Compute
$$ d_1 (f,g) \ d_2 (f,g) \ d_{\infty} (f,g) $$

[gentle osprey]

SolitaryWolf:

[fluid cloud]

ueruoer idk how to typeset on texit, but you get me right

i dont want the entire solution, i just wanna know what the metric is

[gritty widget]

d_2 is standard

[fluid cloud]

yeah idk the standard for functions

[gritty widget]

I gues d_1(x,y) is |x-y|

[fluid cloud]

sis-

please read my question, x and y are not real numbers

[marsh forge]

https://en.m.wikipedia.org/wiki/Norm_(mathematics)#p-norm

My best guess

[gritty widget]

idk my guess would be calculating distances of f(t) and g(t) for all t and taking supremum idk

[marsh forge]

These are the standard choices

Putting p=1,2,infty

[fervent citrus]

when working with bounded functions, d_infty is pretty natural

[marsh forge]

Whats your point lol

[fluid cloud]

i think godel might be right

[marsh forge]

These functions are trivially bounded

Godel told you dinfty

[fluid cloud]

idk how norm might help here tho

[marsh forge]

Norms induce metrics

[fluid cloud]

@fervent citrus idk what the natural is

[marsh forge]

The metric@induced for p=infty

[gritty widget]

yeah those 1 2 and infty are from p norm

[fervent citrus]

What's the natural metric on functions in [a,b]?
is pretty hard to answer when asked just like that

[marsh forge]

The p-metric induced by the p-norm

Is what you want

[gritty widget]

yeah the standard one is d_sup metric

[marsh forge]

He clarified later

[fluid cloud]

Honestly, I have no idea what I want, this is from my teacher's problem sheet; and-

let's just say-

he isnt that good...

[marsh forge]

Wolf pls trust me lol

[fervent citrus]

Big sad

[marsh forge]

I told you what you want

[gritty widget]

wolf pls trust me lol

[fluid cloud]

lmao

oki i trust u

[gritty widget]

yeah but those are functions, so you take p norm on all the points then take sup

[marsh forge]

Weird he didnt define them tho

Have you not read oart of the book or smth

[fluid cloud]

no book :(

he just blabbers stuff in lectures and thats it

yeah but those are functions, so you take p norm on all the points then take sup
@gritty widget that's exactly my concern, um, x isnt a vector right

Let p ≥ 1 be a real number. The p {\displaystyle p} p-norm (also called ℓ p {\displaystyle \ell {p}} {\displaystyle \ell {p}}-norm) of vector x = ( x 1 , … , x n ) {\displaystyle \mathbf {x} =(x{1},\ldots ,x{n})} {\displaystyle \mathbf {x} =(x_{1},\ldots ,x_{n})} is

‖ x ‖ p := ( ∑ i = 1 n | x i | p ) 1 / p . {\displaystyle \left\|\mathbf {x} \right\|_{p}:={\bigg (}\sum _{i=1}^{n}\left|x_{i}\right|^{p}{\bigg )}^{1/p}.} \left\|\mathbf {x} \right\|_{p}:={\bigg (}\sum _{i=1}^{n}\left|x_{i}\right|^{p}{\bigg )}^{1/p}.

um, @gentle osprey plis halp

[fervent citrus]

bad latex

[gritty widget]

you didnt put $$

?

[fluid cloud]

lmao

[gritty widget]

$ texit command $

[fluid cloud]

lemme just take a screenshot

[gritty widget]

texit doesnt work for free, you gotta give it the dollars you know what im sayin?

[fluid cloud]

BWAHAHAHHA

[marsh forge]

They specialize it to functions

W intregrsls

Later in the page

Scroll down

[fluid cloud]

oki hang on

:o

im dumb.

why dont i read the entire thing

lmfao

i think that's the one

$$ \int_{X} | f(x) -g(x) | ^p d \mu $$

[gentle osprey]

SolitaryWolf:

[fluid cloud]

okay, so X is [0,1], and whats mu?

[fervent citrus]

probably a measure but don't worry about it

[marsh forge]

Just integrate like normal

Ignore mu

[fluid cloud]

aite, thanks for your help!

[sleek thicket]

Let X be a (hausdorff, second countable?) topological space. Define an n-dimensional smooth structure on X to be a sheaf of rings F on X with local stalks such that (X, F) is locally isomorphic to (R^n, C^infty).

Let M be a smooth manifold. Let (S, ι) be a pair of a topological space S and an injective continuous map ι : S -> Μ. Let F be the sheaf of smooth functions on M and G the sheaf of continuous functions on S. Pullback gives us a sheaf map ι^* : ι^(-1) F -> G. Let H be the image of ι^*.

I want to say something like there exists a smooth structure on S making ι a smooth immersion if and only if H is a smooth structure on S (and in this case, ι is a smooth immersion). Does anybody have thoughts about this? Have I missed an obvious counterexample?

If S has a smooth structure such that ι is a smooth immersion, it has at most one, see ISM 5.18 and 5.32 (i guess with the sheaf definition this is up to isomorphism)

Here's why I'm thinking this: Let S be an immersed submanifold of M and suppose U is an open subset of S. If a function f : U -> R is smooth, we can cover U by {Ui} where Ui is am embedded submanifold of M. Then the restrictions fi of f to Ui are smooth functions on an embedded submanifold, so there is some Vi open in M containing Ui and a smooth function gi : Vi -> R such that fi = gi|Ui. Thus a smooth function on an immersed submanfold is locally the restriction of a smooth function on the big manifold. Conversely, such a function is obviously smooth

[silver nexus]

What are barycentric coordinates?

[dusty condor]

Baricentric coordinates are coordinates in a simplex in such a way that the coeficients sum 1. I mean, you have a simplex given by the vertices and you can describe the points in the simplex with affine combinations with coefficients whose sum equals 1

[silver nexus]

😮 Do you know where I could read more about these?

[dusty condor]

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

[pseudo crane]

Is delta here understood to be a member of X?

[tough hamlet]

no it's just a real number

[pseudo crane]

Hm

It seems like this wouodn’t make sense if the members of X are not real numbers

[tough hamlet]

generally if it gets used in an inequality

it's a real number

the metric maps into R

[pseudo crane]

That’s what I would assume

[tough hamlet]

it still makes perfect sense

[pseudo crane]

Oh

Ok

Yeah for some reason in the next problem I forgot that delta represented a distance

thanks

[tough hamlet]

np

[marsh forge]

@honest narwhal this doesnt belong here but i have the other channels muted

Can you give me a sketch of the uniformization thing

that you were talking about

[honest narwhal]

So I'm not terribly aware of the proofs of uniformization, and lol that super works here

[marsh forge]

this looks like some pretty heavy machinery

is this reasonable for a project

[honest narwhal]

It might be on the tough end, mostly just feels like a cool theorem. I think a lot of proofs pretty much boil down to constructing these things called "Green's functions"

[marsh forge]

This actually isn't too bad

i found a like 10 page paper

oh lmao this was overseen by Bhargava

maybe its an reu paper

[honest narwhal]

Which are you looking at?

[marsh forge]

Uh it was one of the first results but im going to dinner now

[honest narwhal]

Alright, well there may be other things, I mostly suggested it since it's an S-tier theorem which involves complex analysis and links to topology

[marsh forge]

Yeah im into it

[hybrid echo]

Hey nerds, so my prof was saying something, and I wanted to check to make sure I understand a few things properly

Is xsin(1/x) rectifiable using Cauchy-Crofton?

Because at least, my intuition is that the lines that intersect an infinite number of times would have measure 0, and therefore we can assign a length over any finite interval

[coarse kestrel]

@dim meadow this channel's description 👀

[dim meadow]

[celest flare]

Suppose I have a topology on X. Then can I say that the boundary of a set A is X - int(A) - ext(A)?

[west spindle]

sure

[gritty widget]

Hey, I need some help with this problem https://imgur.com/a/1PcxqBV

🥴

(Differential Geometry)

[sleek thicket]

I haven't been able to think about my manifolds hw for the last couple hours, so I'm going to do it in here

Let M be an embedded submanifold of R^n

I've shown that if y is some point in R^n and x a point in M minimizing |x-y|, then x - y is orthogonal to the tangent space T_x M

because the function f(p) = |p - y|^2 is smooth and minimized at x, and so its differential at x is the zero map, but it's differential is given by dotting with grad f(x) = 2(x - y)

Okay so define E : NM -> R^n by E(x, v) = x + v. This has a bijective differential at each point (x, 0), and so at each point there's a neighborhood of (x, 0) on which it's a diffeomorphism

We can assume that neighborhood is of the form V_δ(x) = { (x', v') in NM : |x - x'| < δ and |v'| < δ } by shrinking it

I want to shrink δ so that there's a unique closest point in M to anything in the union of the V_δ

The confusing this is that this isn't a local condition

I'm also confused about how to show existence of a closest point at all, let alone uniqueness

[glad badger]

I have a problem, I hope this is the right chan, don't know what "degree" this is on but anyway

I have n number of 3D points (x,y,z) any 3 of these form a plane

Based on A plane I want to convert all the 3D points to 2D, mainting ratios etc (I want to do 2D calculations on the 3D points). I hope this makes sense.

[bitter yoke]

This should just be projection onto your plane

[pliant radish]

I found the first general solution

https://media.discordapp.net/attachments/387449305885048852/682155743432278065/unknown.png?width=911&height=462

How can I find the next part?
for the next point?

please help : )

[gloomy plover]

Not sure if you're even asking in the right channel

[ivory dragon]

[definitely not]

[stable jackal]

$sphere: (x-1)^2+(y-2)^2+(z+1)^2=9 \ curve: l(t)=(1+t,2+t^2,3+3t^2), \quad t\inℝ$

how do I find the shortest distance from the sphere surface to the curve numerically?

[gentle osprey]

Squik:

[sweet wing]

you're basically minimising distance with constraints

[dire warren]

How would I find this form that pulls back to dtheta if it wasn’t given in advance?

<@&681259184582688842>

[keen cliff]

idk how to help for that 1 but also idk which channel dynamical systems goes in so I'll put it in here for now

I'm doin exercise 2.1.6 in brin & stuck (excuse the weird formatting idk why that happened)

i can show it's dense, Q²/Z² are all periodic so they're dense

i have no idea how to find a non recurrent point tho

[wanton marsh]

can you remind us of all the definitions

[keen cliff]

Ya

$x \in R(A) \iff x$ is a limit point of the sequence $x_n = A^n(x)$

[gentle osprey]

woog:

[wanton marsh]

is A given by a matrix in M2(Z) ?

[keen cliff]

A is a 2×2 integer matrix with determinant 1, and has eigenvalues non-roots of unity

so like, eigenvalues have |λ|≠1

there's one |λ|>1, and the other is 1/λ

[wanton marsh]

yeah

[keen cliff]

NW is non-wandering but that part is easy

i am just completely bamboozled on R(A) ≠ T²

I've been trying to get an argument that just like

"finds" a point that does the trick

my friend suggested to me suppose R(A) = T²

and I was like

hwhat

[wanton marsh]

the exercise about the circle is easier lol

[keen cliff]

i haven't done that one

or have i

feels like there's some sort of cantor set- like thing

[wanton marsh]

it's not exactly the same problem since you don't have the det=1 requirement

though

[keen cliff]

i was having very loose ideas about this thing last night tho

some sort of 2d cantor set thingy but couldn't put anything concrete enough together

is (T²,A) conjugate to anything that might be helpful

[wanton marsh]

I would try to pick a point in one of the eigenlines but

[keen cliff]

yeah like

[wanton marsh]

then a small ball around it corresponds to lots of small intervals on that line

and you have to make sure you avoid them ?

[keen cliff]

[wanton marsh]

wait am i just dumb

why am i not thinking about the eigenline that has a small eigenvalue

[keen cliff]

it'll go to 0

[wanton marsh]

every nonzero point on that line should work

[keen cliff]

SHIEEEET

well I'm dumb too

I've been working on this problem for ages

thanks!!!

[wanton marsh]

lol I was thinking about the circle exercise

where Em is a dilatation

so it automatically made me pick the other line

first

xD

[keen cliff]

fooo

k

[dire warren]

@keen cliff oh hell you r doing Brin and stuck

We shud make a new adv channel for dynamcial systems

[keen cliff]

Yeah I am @dire warren

[dire warren]

I beleive chapter 2 has that beast of a theroem

Van de waarden

Have fun haha

This book is wild

[keen cliff]

[dire warren]

The difficulty level varies so much between parts

Is this for a course or self study?

[keen cliff]

course

[dire warren]

Nice

Are they gonna cover the whole book?

[keen cliff]

theorem 2.8.1 van de waarden
exercise 2.8.1 prove theorem 2.8.1 using proposition 2.8.2

[dire warren]

Ppft

[keen cliff]

nah not the whole book but I'll probably end up continuing as self study cuz this stuff is hella cool

[dire warren]

😄

Indeed

If u ever want more dynamical systems reading I can recommend some

[keen cliff]

do u have any books that does things a little bit slower

i am KIND OF getting destroyed right now

[dire warren]

Katok hassellblatt’s first course in dynamics covers similar material at a slower pace

NOT the “intro to modern dynamical systems”

That one is kinda MonkaS

[keen cliff]

that one is very monkaS

i tried going thru the beginning of that during reading week

got stuck in the 0th section

[dire warren]

Hahaha

[honest narwhal]

Oh I have lecture notes on the Van der Waaerden section

[keen cliff]

@honest narwhal pass em

[honest narwhal]

@keen cliff

[wheat axle]

Someone help me with my geometry hw

[sleek thicket]

Let G be a lie group. A smooth representation of G is a finite dimensional \R-vector space V and a smooth homomorphism G -> GL(V). To each of these, we can associate an \R[G]-module V. When does an \R[G]-module give rise to a smooth representation of G?

Like you can just say that it's finite dimensional over R and the action map G -> GL(V) is smooth, but we can we characterize in a more direct way?

[viscid shadow]

are there 3d uniform tilings ?

[oak thistle]

Can anyone help me with cone and cylinder?

[gritty widget]

what kind

[oak thistle]

Uh

Problems on it

[marsh forge]

#geometry-and-trigonometry

[gritty widget]

they might be talking about the cone of a topological space or the mapping cylinder

I joke

[oak thistle]

I don't know topology

[gritty widget]

#geometry-and-trigonometry

[oak thistle]

Oh

But it's not pre University level

Equation of cone and stuff

This problem

[floral gust]

https://www.reddit.com/r/math/comments/cuv8me/give_me_any_space_and_ill_tell_you_why_s1_is/

[marsh forge]

this is a good take

[sweet wing]

This is topology. Our space are nice. Our sequences are exact. Our groups are abelian.

[midnight jewel]

This is a thread about math not set theory.

[west spindle]

can someone ELI5 what the fuck Spec(Z) is

[marsh forge]

Set of prime ideals

With a topology

[honest narwhal]

"This is a thread about math not set theory"
Hot damn someone got destroyed

[verbal prairie]

What's the significance of this theorem? Why are these 0,0 0,1 1,1 vectors chosen?

[dim meadow]

Fundamental group is abelian???

Wtf

[marsh forge]

Our spaces are simply connected

[dim meadow]

Maybe if you only study loop spaces and topological groups

[marsh forge]

Woke

Dont forget spheres

And homotopy spheres

[verbal prairie]

[small obsidian]

I agree I've never understood Spec(R)

[sleek thicket]

Suppose I start with an \R-algebra A and build up a manifold M such that A = C^infty(M). I can then define TA = C^infty(TM) (where TM is the tangent bundle). Is there a more direct way to characterize/build TA in terms of A?

A book I'm reading says you can do it but it would be too far out of the way

[peak narwhal]

Can half space (like half planes) work in three-dimensional (x,y,z) space? Do they just remain two-dimensional planes?

[midnight jewel]

yea, a half-space in general is just the set of all $x$ such that ${\lambda(x) \geq 0 \mid \lambda\colon \R^n \to \R \text{ linear}}$

[gentle osprey]

Sascha Baer:

[midnight jewel]

the boundary, where λ(x) = 0, will be an (n-1)-dimensional hyperplane

note that there’s the degenerate case λ(x) = 0 ∀x, which you may wanna exclude, or not

we didn’t exclude it in our definition because it allows you to say that a manifold with boundary has each point having a neighborhood homeomorphic to a half-space

[cursive nebula]

hey guys, what is S^2?

[delicate spire]

surface of a sphere

[cursive nebula]

AH!

i was suspecting that but wikipedia says "two-dimensional"

so i thought "two-dimensional" meant it was on a plane

[bitter yoke]

The surface of a sphere is two dimensional

[delicate spire]

topology people are weird

[cursive nebula]

now i'm not really sure i know what it means to be two-dimensional.

[bitter yoke]

I mean, the surface of the earth is locally flat

It looks flat to us, and so the surface is a two-dimensional object

[cursive nebula]

okay. ahh

i just started studying topology today, so these things are a little alien still

thank you for being patient

[bitter yoke]

You'll learn about manifolds soon where you can formalize this idea

[cursive nebula]

so i was thinking...
"two-dinensional" might mean "its points can all be described with two coordinates"

but i can't do that with the surface of a sphere

i could cut the sphere in half though and i could do it with the two halves

[floral gust]

Are you familiar with tangents?

[rugged swan]

You can locally do it

[cursive nebula]

yeah

i'm a math student

[floral gust]

You can define the dimension as the number of independent tangent vectors

[cursive nebula]

uhh

so, it's the surface of a sphere.
that number can't be more than 3

how do i prove it's 2?

well, it's at least two, of course

because i can picture a tangent PLANE in each point in the surface of a sphere

[floral gust]

Yes. For which the number of independent tangent vectors can be atmost 2.

[cursive nebula]

why 2 and not 3?

[midnight jewel]

because linear algebra dictates that a plane cannot have three independent vectors

[cursive nebula]

yeah, intuitively i dont see how there could be more

[floral gust]

number of independent tangent vectors = number of vectors in the basis of the tangent plane = 2 since a plane is spanned by two vectors namely the one in x direction and the one in y

[midnight jewel]

if you see the tangent space at a point as a vector space

[delicate spire]

latitude and longitude

[cursive nebula]

OH YEAH i know that!!

ahh alright

so. the surface of a sphere is locally a two-dimensional object

[midnight jewel]

ye. and a thing which is everywhere locally of the same dimension is called a manifold of that dimension (this is a very rough definition)

[floral gust]

Yes you can say that. To formalize this notion of dimension, you will study more in manifold theory

[midnight jewel]

(like, very rough)

[cursive nebula]

do you consider the tangent space for... each point of the surface?

[midnight jewel]

what you do often is consider charts, that is maps from a region in the space onto ℝⁿ for the right n

[cursive nebula]

because if you consider two points that arent polar opposites you get two different planes

ahhh regions. okay

[midnight jewel]

oh, I completely misunderstood your question

sorry

[floral gust]

It is the dimension of those planes which is our concern

[midnight jewel]

yes, each point has its own tangent space

there’s a notion of the tangent bundle, which is basically the “union” of all these

(but you also remember where you are on the surface)

[cursive nebula]

AH!! ok. ok.

[midnight jewel]

but yea here we just need the dimension

[cursive nebula]

so i was thinking of this tangent bundle and i wanted to say that it's R^3

but for a Single Point, it's a plane

[midnight jewel]

it’s not just the union, so it’s more complicated than jsut saying it’s ℝ³

[floral gust]

Dont... dive into these higher level topics like tangent bundle just now. They require a bit of prerequisite knowledge of manifold theory and this sort of abstraction so fast would leave you confused.

[midnight jewel]

like you don’t just take the union of some physical planes floating in space

yea

[cursive nebula]

mhmh..

guys, can i ask you how you learnt this stuff?
do you know books you found very useful?

[midnight jewel]

these things will be defined in due time, and possibly quite soon even

[cursive nebula]

i have to study on my own because we're quarantined (coronavirus)

so, no classes for now

[midnight jewel]

I’m just taking classes. didn’t use any textbooks for diffgeo, which is where we covered manifolds

[floral gust]

Why do you have to study manifold theory just now? Take your patience with it. Learn what you are studying right now with patience and detail.
But if you must, start with Differential Geometry of Curves and spaces, move onto introduction to topological manifolds and then onto smooth manifolds

the latter two are by John Lee

[cursive nebula]

ahh i can think of many answers

one is that i have to take the exam in a couple of months

haha!

yes, thank you, i will check that out!

i will also probably be hanging around in this channel ;v;💞💞

our teacher let us download some solved exercises

so i know what kind of stuff will be on the exams, but....
i need theory

thank you for helping me :>
my questions will get progressively less and less dumb, and then... chef kiss diffgeo mastery.

[marsh forge]

Uh what

No

You dont need diffgeo

To do manifolds

[floral gust]

Yeah it’s not like a prerequisite but more like preparatory material to get you familiar with the ideas you’ll see much formally in manifolds. Like differential forms etc. But maybe that’s just me

[cursive nebula]

i dont really know what a manifold is, yet

this is a two semester course called "geometria 2". the first semester is about topology and algebraic topology and the second semester is called "curves and surfaces"!

we also have a "geometria 3" course in year 3, i don't know in which course we're going to cover manifolds, but i'm sure we will!

(i think differential geometry is in this course though!)

[marsh forge]

Youll cover it in the first

The second is diffgeo

[cursive flume]

when defining the standard topology, we have to define the soft ball, in which we basically use the "distance" or "norm"

however it is said that it we don't need norm to define that

do we need vector space structure?or we don't need anything more than a set on r^n to define the soft ball?

[vocal wharf]

what

[cursive flume]

we define soft ball as follows: they are d- duples, for which holds, that $ \sum_{i=1}^d (q_i-p_i)^2<r^2$

[gentle osprey]

Lina:

[cursive flume]

does this not imply the definition of norm?

[vocal wharf]

this just looks like the open ball with radius r around the point p in d-dimensional euclidean space?

i mean, generally you can give a topology to a metric space that way, but not every topology is generated by some metric

[cursive flume]

yes, I do understand this, but I'm watching Schuller's lectures on GR, but he says something fishy...

https://www.youtube.com/watch?v=7G4SqIboeig&list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_

min 22

he says we don't need norm

[vocal wharf]

well, if you forget the vectorspace structure on R^d and regard elements simply as d-tuples, you can still write this down and define open balls that way

[cursive flume]

so you don't need the definition of norm? just metric?

[vocal wharf]

i'm not quite sure what he's getting it

in general you do not need a metric to define a topology

but the standard topology on R^d comes from a metric

[sleek thicket]

Can someone explain why we care about lie brackets of vector fields to me? I have no idea what they represent

[cursive flume]

@vocal wharf i've found something. we don't need a metric to define a topology, but to define the standard topology we do need a metric, however a metric does not need the structure of vector space, just a metric space. a vector space which is normed is more special, because to every norm we can associate a metric, but metric can be defined independently, without the need of vector space structure, I think this is what Schuller means, do you think this is an ok explanation?

[vocal wharf]

seems good

also notice that only euclidean space has a standard topology though

[dire warren]

They don’t need a norm because all norms on fin dimensional vector spaces are equivalent in that they induce the same topology. So they just work with this given topology @cursive flume

[gentle osprey]

Celephinnor:

[versed pivot]

@rich fiber
That condition implies $\frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} \leq 2|x - y|$, is that right?
Since $(a-1)^2 \geq 0$ you have a simple inequailty $a + \frac{1}{a} \geq 2$ for $a > 0$
Taking $|x - y| < 1$ would contradict this

[gentle osprey]

Seoin:

[versed pivot]

I just swapped the roles of x and y and added the two inequalities

oh I see, I didn't read that part carefully enough

you can probably fix the argument if you replace x by a nearby rational but I would have to write it down to make sure

or maybe not, if f isn't continuous

yes maybe this is too simple, I missed the part that y is supposed to be rational

[chrome dew]

not sure if this is helpful or not, but pick some z in ]0,infty[ and since Q and R\Q are both dense in R we can find a limit of terms of x -> z and y->z. Then you'll end up with 1 <= 0

[dim meadow]

Something I was thinking about is maybe you can let U_n be the set of points st f(x) is < 2^-n or > 2^n. I think you can show that those sets are open and dense

And then the intersection is clearly empty

[chrome dew]

I assumed continuity, ignore me

[dim meadow]

Lol bad assumption

We do though

Wait maybe this is bad

No I can show it

You can show that around a rational or irrational point there is a small open neighborhood not including that point

I'll write something up

[dim meadow]

Okay got it

Given an irrational point x_0, suppose y is a rational point with |x_0-y| < (2^-n)f(x_0). Then f(y) >= f(x_0)/|x_0-y| > 2^n

Now given a rational point y_0, suppose x is an irrational point with |x-y_0| < 2^-n/f(y_0). Then f(x) <= f(y_0)|x-y_0| < 2^-n

This is a little off rn but I think you can make this work

There's still a bit of work to be done though

Yeah me too

Sure

[dim meadow]

I messed up a bit and made a correction

@rich fiber

Good

[gentle osprey]

Celephinnor:

[dim meadow]

Yeah juggling seems hard

What do you mean and?

[gentle osprey]

Celephinnor:

[dim meadow]

Welp

That's just empty

👀

Oh lol

Why is that open?

I wouldn't think it would be

But if it is that means the other one is closed

But I don't think it is

Oh wait

There's also the case of equality

But that doesn't seem so important

I don't agree with your close enough statement

Cause the problem is when you go closer, the radius of the ball shrinks

It's the juggling problem

I don't think that will do it

Okay here's something interesting

So you start off with the balls of radius 2^-n/f(y) about each irrational point

Then you intersect that with the balls of radius 2^-n f(x) about each rational point

This is an open dense set with the property that each point has f(x) < 2^-n or f(x) > 2^n

And then you're done I think

Yeah wtf

I think that does it

Fuck

Yeah this definitely does it

Wow that took me way too long

Cause it's an intersection of open sets

@chrome dew

We got it

What do you mean?

We are intersecting 2 open sets

👍

Baire is super applicable in weird ways

[chrome dew]

cool I'll check it out

[dim meadow]

What don't you get?

Say the two sets are U_n and U_n'

They do though

Okay look

U_n cap Q has the property

And U_n' cap R-Q has the property

So (U_n cap U_n') cap Q has the property

And (U_n cap U_n') cap R-Q has the property

(the property here meaning that all the elements either are < 2^-n or > 2^n)

Therefore A_n = U_n cap U_n' has the property

What's wrong with this?

@rich fiber

Lol

You're just inexperienced

Eventually you'll be able to deal with these kinds of arguments fairly easily

👍

[dim meadow]

👀

@gritty widget what do you mean?

Do the problem above

I'm writing some hw up right now

Please don't give me a problem

Yeah that was top tier

Yeah that shit is gross

The hint is baire category theorem

[dire warren]

@gritty widget try challenge problem 152

[dim meadow]

👀

@gritty widget R+ is R

Lol

Do you not know this?

(0, \infty) is homeomorphic to R

Locally compact hausdorff

[dire warren]

It’s a topological invariant

[dim meadow]

Locally compact, hausdorff

[dire warren]

The locally attaches to compact

It’s two properties

Hausdorff, and locally compact

Yeah

[wooden scarab]

I'm looking for a classification of universal covering spaces of closed 2 manifolds. It should be in some intro topology textbook I assume?

[honest narwhal]

So we can assume orientable

Because take the orientable double cover

[dim meadow]

For orientable we have S^2, R^2 and H^2

[honest narwhal]

Yup, this is called the uniformization theorem

[wooden scarab]

Where H^2 is a disk?

[dim meadow]

Hyperbolic plane (or disk)

[honest narwhal]

Actually is it true that any orientable surface is a Riemann surface?

[wooden scarab]

I think so *(for manifolds)

[honest narwhal]

Like if so then yeah what you said is an immediate corollary of uniformization

[dim meadow]

Classification of surfaces says yeah

[honest narwhal]

What's the classification of non-compact surfaces?

Or wait

[dim meadow]

[honest narwhal]

He said closed

Lol nvm

[dim meadow]

Yeah

[honest narwhal]

Yeah lol uniformization ez

[dim meadow]

For non compact there is a classification

[honest narwhal]

Well it's not ez but yeah

[dim meadow]

Yeah

You need ends and shit

[wooden scarab]

how about compact with boundary?

[sleek thicket]

Is the following variant of Whitney's approximation theorem true?

Let M, N be smooth manifolds and f : M -> N a homeomorphism. Then f is homotopic to a diffeomorphism

I'm guessing not

[edgy crescent]

I wouldnt call this a variant of whitney but it's true

nevermind

it is not

this is troubling

because continuous maps are homotopic to smooth ones

and homotopies may be replaced by diffeotopies when they exist

but thats not enough

[sleek thicket]

What are diffeotopies?

[edgy crescent]

smooth homotopies

[sleek thicket]

Oh you're saying homotopies between smooth maps, not the homotopy you get from Whitney approximation

Can you find a counterexample?

[edgy crescent]

yes

no but google can https://math.stackexchange.com/questions/1068805/is-every-self-homeomorphism-homotopic-to-a-diffeomorphism

[sleek thicket]

Oh lol I should've googled it first

[gritty widget]

anyone here familiar with basics of differential form

seitani ur familiar right?

[edgy crescent]

yeah my cousin was a differential form

rude

im much cuter

[gritty widget]

lol serious tho

i wanna ask a q

and idk if it belong to multivariable or this channel

[edgy crescent]

ask

[gritty widget]

so here

part c, is it closed?

[edgy crescent]

just compute dw

[gritty widget]

I got dx1^dx2+...+dx_{n-1}^dx_n

maybe i should find a counterexample

[edgy crescent]

well that's not zero right

?

[gritty widget]

yea, but just wondering if some operations could make it 0

[edgy crescent]

they're always linearly independent vectors in Alt^2 R^n

[gritty widget]

if not then ok ill try to find a counterexample

[edgy crescent]

find a basis for Alt^k R^n

it will help you be more comfortable with this stuff

[gritty widget]

we didn't use basis to define differential forms though...

we just define k-forms as summation of index

[edgy crescent]

yes, but they form a vector space

and you want to show that this sum is not zero

it can't be if the vectors are linearly independent

[gritty widget]

in this case what are the vectors?

[edgy crescent]

dx1^dx2, dx2^dx3, etc

[gritty widget]

ok, ill try, thanks

[edgy crescent]

as an example so you can sanity check, a basis for Alt^3 R^4 is given by
dx1^dx2^dx3,
dx1^dx2^dx4,
dx1^dx3^dx4,
dx2^dx3^dx4

[gritty widget]

wait im not even sure if d(x1 dx2) is dx1 dx2

im new to this stuff sorry

[edgy crescent]

it is but go make sure

do carmo's differential forms book is good

[gritty widget]

ah ok

im using calculus and analysis in euclidean space jerry shurman

[edgy crescent]

I see

you can google lecture notes for these things

[gritty widget]

i heard calculus on manifolds is good

[edgy crescent]

it's good to compare several sources

when you're confused

[gritty widget]

ok

oh wait...

d(x1 dx2) is not dx1 dx2

i gotta do product rule...

man

this sucks

[edgy crescent]

it is dx1 dx2

d isn't derivative

although it kinda behaves like it

[gritty widget]

according to book its supposed to be d(w^v) = dw ^ v + (-1)^k w ^ dv

where in my case x1 is a 0-form

[edgy crescent]

okay that's a valid way of computing it

but d(dw) = 0

always

[gritty widget]

yea that ik

[edgy crescent]

so when you have something of the form f dx_I

then you have an explicit formula

https://en.wikipedia.org/wiki/Differential_form#The_exterior_derivative

given by the partials of f

[gritty widget]

oh

that i think were learning later

lol

so we wont be able to use it this hw

[edgy crescent]

you can deduce it immediately from what you said

you have x1 dx2

[gritty widget]

but yea, i think then d( x1 ^ dx2) is not dx1 dx2 because of product rule for d operator

[edgy crescent]

so you go dx1^dx2 + x2^ddx2

= dx1^dx2

[gritty widget]

oh right

[edgy crescent]

so you can always just apply d to the scalar

[gritty widget]

ohhh

interesting

so its just a corollary of product rule for 1-forms i guess?

[edgy crescent]

ye

that product rule is for general forms

not 1-forms

[gritty widget]

w/bout k-forms

ah ok

[edgy crescent]

it's general for d(v^w)

[gritty widget]

where v is scalar

0-form

[edgy crescent]

v and w are any forms

[gritty widget]

d(v^w) = (dv)^w when v is any form??

[edgy crescent]

no

but if you have a form w = f dx

[gritty widget]

when v is 0 form that is true though right

[edgy crescent]

not really

w has to be a pure form

with constant scalar

something like dx1^dx3

not x^2 dx^1^dx^3

more generally w has to be closed

the precise useful statement is in the wiki page i linked

[gritty widget]

oh exterior product

man

we just call it different names

its just product of a k-form and an l-form

yea i got that bit

im thinking though, maybe d(f ^ dx1 ^ ... ^ dxk) = df (dx1 ^ dx2 ^ ... dx^n) is a special case if you keep on applying product rule, but i dont know

because at least we seen that d(f^dx) = df ^ dx

[edgy crescent]

yes it is

that's the case in the image above

im going to play videogames if you need me yell bloody mary 3 times to your mom

[gritty widget]

ok lol

go have fun

so i got an answer

and if u can approve thats good

so since w= dx1dx2+...+dxn-1dxn

i let y(t) = (t, t, ..., t): [0,1]->R^n

so that integral of w over y is n-1 which is not 0

for n>1

[edgy crescent]

your loop should be closed

closed forms integrate to 0 under (contractible) closed loops

[gritty widget]

y should always be closed?

to be able to integrate

[edgy crescent]

you can always integrate but that doesnt give you information

the result is that closed forms integrate to 0 over contractible closed loops

[gritty widget]

but here the definition of closed is dw = 0

[edgy crescent]

oh then why are you integrating?

oh you're integrating dw aren't you?

not w

I see that confused me

yeah a way to show dw is not zero is to integrate it

but that's a bit overkill

[gritty widget]

yea

[edgy crescent]

linear algebra shows it's nonzero immediately

[gritty widget]

sorry didnt clarify

i guess about lin alg ill ask prof tomorrow

and pt d), i just say since w not closed, then w not exact

[edgy crescent]

yea

[gritty widget]

ok thanks

[chrome schooner]

yo

help

lol

hello ;=;

i just need help from smart math guy

[gritty widget]

hey i think i got what you mean by vector space now

so for a k-form, let all the possible indices dx_{I} be the coordinates right

and then the dx1 ^ dx2 + ... + dx_{n-1} ^ dx_{n} is something like a bunch of 1's and 0's on the nxn matrix

wild guess actually

or are they like vectors (1,1,0,...,0), (0,1,1,0,...,0),...,(0,...,0,1,1) that are linearly independent?

i think thats what i meant

@edgy crescent if u can confirm true/not true

[sleek thicket]

I'm taking my first course on smooth manifolds and we've just gotten to flows of vector fields. My book has an appendix with proofs of existence/uniqueness of solutions to linear ODEs, and those made sense, but the theorem about the solution smoothly depending on initial conditions seems really hard. Is it important for me to understand it right now?

[edgy crescent]

@gritty widget not exactly true

your formula would imply that dx1^dx2, dx1^dx3 and dx2^dx3 are linearly dependent

the vector space of alternating k-forms in R^n has dimension n choose k

where a basis is given by the dxJ where J is an increasing subset of [n]

[gritty widget]

oh

so the dx objects are the basis of the vector space which is the V^k(A) all k-forms

?

V as in the upside down V

but are they known as the standard basis just like the standard basis for a linear transformation?@edgy crescent

[edgy crescent]

yes

[gritty widget]

Ok thanks

[pseudo crane]

Smells like differential geometry in here

[midnight jewel]

fields whose name ends in y are the best since you can use the descender of the y to underline the title

[prime prism]

Can I just make sure: the intersection of two simplicial complexes need not be a simplex

In particular it can be a disjoint union of simplicies

(e.g. Two tori joined in two spots)

[gritty widget]

Are two circles with the same orientation concatenated at a point homotopic to a single circle?

I'm thinking no but I don't know how to prove it

like the loop concatenation $fg$ where $f(t)=(\cos(2\pi t), \sin(2 \pi t))$ and $g(t)=(2-\cos(2\pi t), -\sin(2 \pi t))$

[gentle osprey]

zd:

[gritty widget]

Assume a homotopy (by defn. continuous) exists from $fg$ to $f$, so if $C$ is an open (closed) set in $\mathbb{R}^2$, the inverse image $f^{-1}(C)$ in $[0,1] \times [0,1]$ is open (closed)... something something contradiction?

[gentle osprey]

zd:

[midnight jewel]

Are two circles with the same orientation concatenated at a point homotopic to a single circle?
are we talking in an ambient space? like, in ℝ² or sth?

[gritty widget]

Yeah

[midnight jewel]

if you have no requirement for the homotopy to be injective (for fixed t), then you can just contract everything to a point and then expand into a circle

seing as ℝ² is contractible

[gritty widget]

True, I was trying to understand/learn about fundamental groups of R^3-{some knot} so what about in that case

[midnight jewel]

or just contract one of the loops into the basepoint

so you can pull things apart if you don’t require the basepoint to remain fixed

ah wait no it even works generally

you can turn it into a big loop

[gritty widget]

I see what you're saying yeah

[midnight jewel]

by just moving the loop a bit away from the basepoint when it first returns to it

[gritty widget]

cause that loop you're shrinking can contract to a point

and u just have the big boi left

[midnight jewel]

in your case with the knot it would depend whether the loop goes around the knot or not

[gritty widget]

ye ye

I have a much clearer image of this now ty tho

Didn't understand that before

[gritty widget]

sorry I didn't connect the loops at an initial basepoint but in this case they're homotopic to that setup anyway
with the stuff on the bottom, pic 1 to 2 is concatenation of the loops, 2 to 3 is performing what I'm pretty sure is a valid homotopy right? (That's my question ;-; )

Space is supposed to be R^3\ {trefoil}

[fading vale]

So I'm trying to understand a proof in bredon that, given a hausdorff space X the following conditions are equivalent:
-X is locally compact
-X is a locally closed subspace of a compact hausdorff space
-X is a locally closed subspace of a locally compact hausdorff space

for showing (1) implies (2) he says that since X is locally compact its an open subspace of its one point compactification which ofc is a compact hausdorff space which makes sense to me

but i dont get why this implies X is locally closed

does local compactness imply locally closed?

[honest narwhal]

@fading vale Open subsets are locally closed

Open subset of a closed subset

The whole space is closed

boom

[fading vale]

owait

._.

i was being dum

[honest narwhal]

lmfoa

[fading vale]

ok tbf he defined locally closed literally the line above this and then went abt proving this

oh

@honest narwhal yea he didnt define it as being open in its closure

[honest narwhal]

What was his def again?

[fading vale]

every a in A, there exists a neighborhood U with a in U, A cup U is closed in U

this still works tho i was being dumb and not thinking

[dire warren]

I’m having trouble understanding the characteristic embedding described in Milnor’s lectures on the h cobordism theorem. Why are hyperbolic sine and cosine chosen?

<@&681259184582688842>

Screenshots for context

[gritty widget]

I honestly have no clue if this is relevant but maybe https://en.wikipedia.org/wiki/Integral_curve#Generalization_to_differentiable_manifolds has some insight as to the motivation of that last sentence in the section you're talking about? Again no clue lol

from hatcher: why is $ir \cong \mathbb{1}$? Is Hatcher using $a \cong b$ as meaning "$b$ is a restriction of $a$"?

also apparently I don't know how to write those symbols in latex oof off to google

It is clear to me that ri is the identity though, just include A, A=im(r) and r=r^2, done.

If an element of X\A goes through ir, it will end up in A, so clearly it's not equal to the identity, so I'm guessing that Hatcher means something about restrictions with that symbol right

and no it's not included in the "standard notations" section even 😂

[coral pivot]

hatcher uses $\cong$ to mean homotopy

[gentle osprey]

JohnDoeSmith:

[coral pivot]

i.e maps ir and the identity are homotopic

its the first paragraph in that page

[gritty widget]

tfw I can't read 🤦

yeah just found it lmao

thanks 😂 it's a fun time

[coral pivot]

lol np

[gritty widget]

ree

[gentle osprey]

zd:

[gritty widget]

[marsh forge]

huh

[gritty widget]

I can't figure out how to get mathbb{1} to work here

[chrome dew]

there is no mathbb 1

$\mathds{1}$

[gentle osprey]

Merosity:
Compile Error! Click the reaction for details. (You may edit your message)

[chrome dew]

I guess this package isn't default

$\usepackage{ dsfont }
\mathds{1}$

[gentle osprey]

Merosity:
Compile Error! Click the reaction for details. (You may edit your message)

[chrome dew]

oh well I tried

[gritty widget]

oof

how 2 prove homotopy type equivalence is transitive?
Suppose spaces $X \cong Y$ and $Y \cong Z$. So homotopy equivalences $f_1$ and $g_1$ exist between $X$ and $Y$, and $f_2$ and $g_2$ between $Y$ and $Z$ likewise. Again by definition these equivalences have homotopies $F$, $G_1$, $G_2$, and $H$ such that $F$ is between $g_1f_1$ and $\text{id}_X$, $G_1$ is between $f_1g_1$ and $\text{id}_Y$, $G_2$ is between $g_2f_2$ and $\text{id}_Y$, and $H$ is between $f_2g_2$ and $\text{id}_Z$. Is this the right way to set it up? Cause it's a lot ;-;

My immediate thought is to try to find homotopies between $f_2f_1g_1g_2$ and $\text{id}_Z$, and $g_1g_2f_2f_1$ and $\text{id}_X$ but idk how

[gentle osprey]

zd:

zd:

[gritty widget]

if anyone has a hint toss it up so I could keep trying

[marsh forge]

Use the original homotopies to build a new one

for the composites

its not so bad

[gritty widget]

I was thinking I might need to find a homotopy between $g_1g_2f_2f_1$ and $g_1f_1$ since then transitivity of homotopy gives that the former is homotopic to $\text{id}_X$ as desired, but I have no clue how to construct one from the other ones

[gentle osprey]

zd:

[gritty widget]

If it's taking me this long to work through a trivial part of a paragraph in chapter 0 of Hatcher like this should I just give up on pursuing this at all

Will I just never be good enough to do things this quickly

[coral pivot]

dw im starting hatchers too

it takes a while, its intro to an entirely new field

[gritty widget]

Good to know I'm not alone lol

Do I need to show that compositions of homotopic functions are homotopic? as in going down the route of trying to show $g_2f_2f_1 \cong f_1$, and then composing both sides with $g_1$?

[gentle osprey]

zd:

[gritty widget]

In turn showing that composition in the other direction of homotopic functions yields two also homotopic functions would prove this completely right, cause we know $g_2f_2 \cong \text{id}_Y$, compose to get $g_2f_2f_1 \cong \text{id}_Yf_1=f_1$, again the other direction $g_1g_2f_2f_1 \cong g_1f_1 \cong \text{id}_X$, and done by those two lemmas?

[gentle osprey]

zd:

[gritty widget]

Proving this would also show that you can multiply equations defining homotopy equivalences together too
like $a \cong b$ and $c \cong d$ implies $ac \cong bd$, cause $ac \cong bc$ and $bc \cong bd$ by the lemma

[gentle osprey]

zd:

[gritty widget]

So it would be nice for it to be true, is there another way to do this proof of homotopy type equivalence being transitive without proving this too? cause Hatcher says proving homotopy type is an equivalence relation is "easy" soooo

[rugged raptor]

Isn't a torus the same thing as a knitted sweater?

Since you can entwine it around itself to make it into an arbitrarily long string, and the just knit with that string.

[cursive flume]

Schuller in his lectures makes the definiton of "manifold philosophy", which basically means: we do physics on manifolds, represent it in charts, then associate the chart properties with the real world properties, continuity works- for example, but then he goes into "ill defined properties", which can be chart dependent, if we do the physics on the whole manifold. can someone give me such an example?- we want physics to be chart independent

I was thinking of differentiability,if we do it on the whole manifold, without making restrictions, but we can solve it with restrctions (work on a more rigorous structure, namely a differentiable manifold), does anyone know such a "property", which can "not be fixed" by "adding more structure"?

[sweet wing]

Assuming some conditions holds, this is true
is literally how most theorems are

[floral gust]

I was thinking of differentiability,if we do it on the whole manifold, without making restrictions, but we can solve it with restrctions (work on a more rigorous structure, namely a differentiable manifold), does anyone know such a "property", which can "not be fixed" by "adding more structure"?
@cursive flume I don’t know think so because you can literally introduce the property as a fix. As in “we now consider the manifolds that will satisfy the given the property.” And the 0 dim manifold would most likely satisfy this due to some sort of vacuity. Interesting is then is to ask what properties ca;you introduce and still not run out of the base cases.

[cursive flume]

@floral gust yes, this was my question, what property can be defined such

[floral gust]

Take any self-contradicting property

[wheat brook]

@cursive flume Consider the sphere. We cannot map it to 1 chart. If we do physics on the whole sphere, we will always run into pole somewhere. The fix is to switch locally to a different chart.

[cursive nebula]

hey guus, can i ask a topology question?

ahh i did!!

NICE

thabk you!!

i need it for another longer proof

i will still check them

i dont know if it's the SMALLEST topology that you can generate with those sets

wait. it is

yeah,

/rotate

the exercise im trying to solve is on the base chapter

[cursive nebula]

i need this result (tau3 is a topology) to prove that thing

oh thanj you!

see i tried using the axioms, intuitively i can see it's reasonable for tau 3 to be a topology

but idk how to write it formally

,rotate

[cursive nebula]

this is the exercise i have to do

i have to prove that B is a base for a topology that is more fine than the topology is

so this is what i thought

,rotate

[cursive nebula]

if tau3 is a topology, i can easily prove that B is a base for tau3

so thats why i asked if tau3 was a topology

[gritty widget]

Hello there, I would ask you for some recommendations about some nice Topological subject to talk about for a group of undergrad students, please?

[midnight jewel]

what level are we talking? like is the target audience people who have zero experience with topology yet or ones who have already done some (or maybe even a lot of) basic topology stuff?

“undergrad” could mean anything from no experience to having taken two years of topology classes

[cursive nebula]

hey guys,

i'm trying to solve problem 25, but i don't know how to go about points ii) and iii)

(what does relative topology mean?)

,rotate

[cursive nebula]

this is the idea i had for points ii) and iii) but i don't know if i'm on the righr track (it would help if i had That Definition)

i can hit the second line in Each Point with open sets
(and they would generate the discrete topology)
but i cant do it with the first

is that it?

[midnight jewel]

relative topology is also called the subspace topology

if you have $X \subset Y$ and you have a topology $\mathcal{T}$ on $Y$ then you can define a topology on $X$ by simply saying the open sets in $X$ are all sets of the form $X \cap O$ for $O \in \mathcal{T}$

[gentle osprey]

Sascha Baer:

[midnight jewel]

ie you intersect the open sets of Y with X to get open sets in X

[cursive nebula]

ahh thank you so much!

okay, so
those lines are subspaces of R^2

i can easily prove ii) bc i know that topology includes every singleton so it's discrete

[topaz shuttle]

Hi there does anyone know anything about the topology of complex numbers?

[midnight jewel]

I mean it’s just the standard topology on ℝ²
if you have specific questions you should

[twin aurora]

I'm trying to work out a proof of the following theorem from basic algebraic topology, that the deck transformation group of a covering space is isomorphic to the fundamental group of the base space mod the fundamental group of the covering space. does anyone have any hints to drop/resources that i could check out to be pointed in the right direction?

[bitter yoke]

<@&681259184582688842>

[marsh forge]

Each subgroup correspondens to a cover

Use thiis

Im@so drink bj

[dawn geyser]

hi everyone! i'm working with double factorials (as in 9!! = 9 x 7 x 5 x 3 x 1, not (9!)!)) and trying to use them for a proof of n-spheres but i'm really struggling with a lot of the notation

[delicate spire]

hey whatchu need help with?

[dawn geyser]

im working with this right now

<@&681259184582688842>

im basically tryna get to here by combining the recurrences but idk know how to

[delicate spire]

Sn and Vn are surface area/volume of unit sphere?

[dawn geyser]

i think? i don't actually know what the S is meant to represent

[delicate spire]

ok yea like for a 3d sphere it has S2 surface area V3 volume

I found the proof I think you want on wikipedia O_o

looks like you need 1 more recurrence formula and it's easy to get the double factorial bits

[dawn geyser]

i got it from wikipedia yeah

im just trying to understand the steps and its not really explaining it that well to me

[delicate spire]

ok, I can't really follow the integration bits, but this part makes sense to me

what part is tripping you up?

[dawn geyser]

this is my first time doing anything involving n-spheres, so we haven't learnt a lot of the theory for it.

so basically all of it

[ivory dragon]

if $V_{n+1} = \frac{S_n}{n+1}$, then $V_{n+2} = \frac{S_{n+1}}{n+2} = \frac{2\pi V_n}{n+2}$

surely you follow at least this much?

[gentle osprey]

Namington:

[ivory dragon]

so one can use this to show inductively the two identities below

[dawn geyser]

yeah yeah i follow that

[ivory dragon]

i'll show the first one, the next case follows. let $k = 0$, then $V_{2k} = V_{0} = 1$ by definition, so clearly $V_{0} = \frac{\pi^0}{0!} = \frac{1}{1}$.

[gentle osprey]

Namington:

[ivory dragon]

now we can assume that this holds for some $k$ and wish to show it for $k+1$

[gentle osprey]

Namington:

[ivory dragon]

$V_{2(k+1)} = V_{2k+2} = 2\pi \frac{V_{2k}}{2k+2}$

[gentle osprey]

Namington:

[ivory dragon]

due to the identity shown earlier

and we know that $V_{2k} = \frac{\pi^k}{k!}$ by the inductive hypoth

[gentle osprey]

Namington:

[ivory dragon]

conclusion follows.

[dawn geyser]

ok!! thanks so much for that

i'm just gonna reread that all real quick

[ivory dragon]

ie

$V_{2k+2} = 2\pi \frac{V_{2k}}{2k+2} = 2\pi \frac{\pi^k / k!}{2(k+1)} = \pi \frac{\pi^k}{k!(k+1)} = \frac{\pi^{k+1}}{(k+1)!}$

[gentle osprey]

Namington:

[ivory dragon]

in case you couldnt finish the algebra yourself

this completes the induction

anyway a similar process holds for V_(2k+1)

this is very standard "first proofs class" induction, no real tricks

[dawn geyser]

i'm in 12th grade, doing like four unit maths. it's an extension question and we've barely covered any thing involving n-dimensions, so i'm not really able to do these sorts of proofs yet

thanks for helping so much though

[ivory dragon]

the gist behind induction is that

we prove something for a "base case"

in this case, our base case was when k = 0

and then we show that "k holds" implies "k+1 holds"

by a "domino effect", this shows it for all integers greater than the base case

since we've proven that k = 0 is true

and then k = 0 is true implies k = 1 is true

and then k = 1 is true implies k = 2 is true

and so on

this is called the "inductive step"

[dawn geyser]

okay yep, i get that

[ivory dragon]

and the fact that we're allowed to assume is the "inductive hypothesis"

if you show both the base case and inductive step, you have a complete proof by induction

thats the main "mechanics" of the above proof, everything else is just algebra

[dawn geyser]

so how could i use this an example of an application of double factorials? can i use it to find the n-dimension volume?

[ivory dragon]

well, if you want to get it into double factorial form, you have to show $\frac{\pi^k}{k!} = \frac{(2\pi)^k}{(2k)!!}$

[gentle osprey]

Namington:

[ivory dragon]

i omitted that from my proof

this isnt particularly hard to show, however

in any case, yes, this is how you calculate the n-volume of an n-sphere.

[dawn geyser]

where n=2k?

[ivory dragon]

yes

i.e. even n

for odd n you use the below formula

same gist either way

er well i should clarify

[dawn geyser]

yeah thats right, thank you for that. i think i've got it now.

[ivory dragon]

this is the volume of an n-sphere

of radius 1

for radius x you have to multiply by x^n

[dawn geyser]

so R is radius 1?

[ivory dragon]

(also technically "volume of the n-ball" is a more mathematically correct term but who cares)

the formulas you screenshotted assumed a unit n-sphere

i.e. radius 1 (R = 1)

if you change the radius, just multiply by R^n

[dawn geyser]

ok can you give me a practice question of volume of the n-ball with a value for k?

[ivory dragon]

i mean, you can make random numbers and just plug them in essentially

calculate the volume of the 5-dimensional ball of radius 7

or of the 12-dimensional ball of radius 2.1

use this to rederive the formulas for the area of a circle and the volume of a sphere

etc

[dawn geyser]

units^5 right?

[ivory dragon]

yes

(mathematicians generally don't bother to write units, so if you're looking into this and see units being disregarded, dont worry - it's just notation abuse)

[dawn geyser]

this kind of stuff is crazy, idk how you guys do this for a living honestly

[ivory dragon]

i mean, these methods are fairly routine

ignore the conceptual "woah n-dimensional sphere thats so craaazy" and just see where the definitions take you

alternatively, stop viewing "dimension" as something that's necessarily tied to spatial dimensions

we just use "dimension" to mean "an axis on which some data can vary"

if theres 3 dimensions or less, its usually convenient to represent that data as a location/position

that's the motivation for the cartesian plane, for example

but really, computers are regularly doing 1000-dimensional statistics whenever they do machine learning stuff

its nothing crazy, just maybe a bit hard to visualize

they're, of course, not asking "geometric" questions about this 1000-dimensional statistics (theyre not gonna ask what the "volume" is), but the idea of an n-sphere is just generalizing a geometric connection from 2/3 dimensions to higher dimensions

[dawn geyser]

yeah i think its the visualisation that i struggle with but your explanation cleared up with the axis yeah

just functions with another axis

its hard to think of but that makes it easier

oh by the way, what does it mean by this?

The dimension of n-sphere is n, and must not be confused with the dimension (n + 1) of the Euclidean space in which it is naturally embedded. An n-sphere is the surface or boundary of an (n + 1)-dimensional ball.

[ivory dragon]

ok so uh

this is a notational thing

and the reason we choose this notation is due to some fairly sophisticated topology

well, not particularly sophisticated but

basically a "2-sphere" is a 3d sphere

a "3-sphere" is a 4d sphere

etc

the reason for this is that a "sphere" is actually the boundary of a ball

specifically, the 2-sphere is the boundary of the 3-ball

and so on

[dawn geyser]

ok yeah thought so

[ivory dragon]

since its just the boundary, it has a lower dimension

since for example