point-set-topology

[gritty widget]

theyre not easy for a beginner

i can give up an example in a bit but in on a call

[midnight jewel]

(you can actually make texit have custom preambles)

[gritty widget]

a 1-form assigns to each point on your manifold a linear map from the tangent space to the reals (but smooth) i think

doesnt have to be smooth, just needs to be linear

its a functional

[wanton bone]

I've taken many of the relevant classes but I'm pretty rusty, especially with terminology I guess

OH @gritty widget what you just said made this paragraph im reading make so much more sense

[honest narwhal]

Yeah there's a formal way to do things with manifolds, namely smooth sections of the cotangent bundle

It's a good exercise to see why the standard dx business in R^n or open sets aligns with that

Keeping in mind that you can think of all tangent spaces as being the same

[gritty widget]

wait, why are they same

[wanton bone]

"We emphasize that for every tangent vector $v$, a 1-form $f$ defines a real number $f(v)$; and for each point $p$ in $R^3$, the resulting function $f_p: T_p(R^3) \to R$is linear."

[gentle osprey]

chrstfer:

[gritty widget]

(section on the cotangent bundle just means that for every point in your manifold, it assigns a cotangent vector)

[wanton bone]

so, for any point on the manifold there is a set of tangent vectors, $T_p(R^3)$, right? thats the tangent space?

[gentle osprey]

chrstfer:

[gritty widget]

yeah

yea

[wanton bone]

How does that relate to the cotangent bundle?

[gritty widget]

they are, hand wavingly, the derivatives of the curves passing through the point

i cannot spell today

[wanton bone]

the tangents are? or the cotangents?

[gritty widget]

the tangents

[wanton bone]

right yeah

[gritty widget]

that's intuitively what the cotangent bundle is, it's a larger manifold where you assign a cotangent space to every point on your manifold

[wanton bone]

can you define cotangent space?

[gritty widget]

the cotangent space is the dual space to the tangent space

^

[wanton bone]

ok

[gentle osprey]

grade a failure:

chrstfer:

[gritty widget]

exactly

[wanton bone]

ok

[gentle osprey]

chrstfer:

[wanton bone]

rather, i meant to say space of 1-forms over $\mathbb{R}$

[gentle osprey]

chrstfer:

[gritty widget]

no the 1-forms build points to linear functionals

[wanton bone]

OH

[gritty widget]

that are in the cotangent space

[gentle osprey]

chrstfer:

chrstfer:

chrstfer:

grade a failure:

chrstfer:

grade a failure:

chrstfer:

[gritty widget]

basically

[wanton bone]

basically? what am i missing?

[gritty widget]

no i mean basically ya thats correct

but again, the 1 form isnt from $T$ to $\mathbb{R}$, but from $U$ to a function $(T\to \mathbb{R})$

[gentle osprey]

grade a failure:

[wanton bone]

ok. So a differential form, then, is the 1-form that I can "plug in" a point to, to get a functional from a tangent vector $v \in T_p $ to a real number

[gentle osprey]

chrstfer:

[gritty widget]

yup

[wanton bone]

right, the 1-form operates on points to get functionals (which are elements of the cotangent space), and those functionals take us back to the field?

[gritty widget]

yes

points to functionals and those functionals go from the tangent space to the field

[wanton bone]

the mapping from an element of $U$ to the functional $(T_p(U) \to \mathbb{R}$, which is a mapping from tangents to scalars

[gentle osprey]

chrstfer:

grade a failure:

[gritty widget]

yea

[wanton bone]

I like that darkrifts, puts it in a nice programming context

[gritty widget]

what does the curry mean

[wanton bone]

though i guess programming got it from math so it would be the other way around

in programming its just a function with input x that returns another function which has x in its namespace (i think namespace is the right term)

[real notch]

It's a joke about currying @gritty widget

[wanton bone]

so like, x->(y-> x+y)

[gritty widget]

ya

[real notch]

$\lambda a. \lambda b. x$

[gentle osprey]

Darkrifts:

[real notch]

anyway, carry on

[gritty widget]

dont call my phone and talk like that

[wanton bone]

wait, darkrifts can you elaborate on the lambda calc there

specifically what is x

[real notch]

no, its a generic currying from a to a function which takes b to do x

[wanton bone]

right

ok

[real notch]

just assume it's some term where a and b are free

carry on

[wanton bone]

thanks

[gritty widget]

bro

do you do maths chrst?

[wanton bone]

i try

and chrs works better, im not a savior by any means

[real notch]

So, a 1-form associates each point p in an open set U with a point in the cotangent space T_p *(U), so what makes something a differential form? a smoothness or what?

[wanton bone]

what do you mean though, grade a?

darkrifts my book gives "Roughly speaking, a differential form on $R^3$ is an expression obtained by adding and multiplying real-valued functions and the differentials $dx_1$, $dx_2$, $dx_3$ of the natural coordinate functions of $R^3$."

[gentle osprey]

chrstfer:

[real notch]

that's not exactly what I was looking for, but hm

[gritty widget]

a 1 form is a differential form of degree 1

in general its more complicated

you need to know alternating forms

[gentle osprey]

chrstfer:

[wanton bone]

but this is all next chapter

[gritty widget]

theres more to it

the sign depends on the sign of the permutation

do you know permutations from group theory?

[wanton bone]

no

[gentle osprey]

chrstfer:

[gritty widget]

its the same thing just notation

[wanton bone]

right

but the first notation was confusing me deeper into the chapter i didnt quite pick up on it

[gritty widget]

we use < > notation

for functionals

[wanton bone]

so $\phi<v_p>$

?

oop, that didnt go right

[real notch]

or $\langle \phi_p, v \rangle$ maybe

[gentle osprey]

Darkrifts:

[wanton bone]

also that notation doesnt seem to be in my book, but it is reminiscent of c++ templates, which are functionals more or less

so is that the inner product of $\phi_p$ and $v$?

[gentle osprey]

chrstfer:

[wanton bone]

not sure how

[gritty widget]

exactly, we write $\langle df(p),v\rangle$

[gentle osprey]

grade a failure:

[real notch]

or somesuch since different functionals have mildly different conventions on where you put the p

[gritty widget]

this is because in hilbert spaces, linear functionals can be represented by elements of the space

google riesz representation theorem

[wanton bone]

ah, because the element df(p) is a function taking an input v, and can be thought of as a row vector like you said before

[gritty widget]

ya sort of

[wanton bone]

so vector inner products (dot products) work

[real notch]

So grade a, i'm rather poor at whatever-forms, but is there a distinct difference between differential forms and not so differential forms?

[wanton bone]

i think differential forms are n-forms but not all n-forms are differential forms

[real notch]

seeing as you said "a 1 form is a differential form of degree 1"

i'm clearly a bab at this stuff

[gritty widget]

i love u guys but i gtg

ill be back tmrw

[wanton bone]

alright, thanks very much for all your help

you just made this soooooooo much clearer

[gritty widget]

The pairing between the dual vector and the vector isn't an inner product

[sleek canyon]

It's physicist notation

You write duals as <v|

And vectors as |u>

Since they deal with reflexive spaces it's fine

[wanton bone]

Yeah, i got confused by the notation

[gritty widget]

i've seen the inner product between two vectors written as $\langle v,w\rangle$ and the dual vector/vector pairing (i.e. the dual vector acting on the vector) as $\left(\omega, v \right)$ sometimes

[gentle osprey]

GKZDragon:

[gritty widget]

where $\left(\omega, v\right) = \omega(v)$

[gentle osprey]

GKZDragon:

[gritty widget]

(which in the case of real vector spaces, gives a real number, and in the case of manifolds, gives a real function $f : M \to \mathbb{R}$ on the manifold)

[gentle osprey]

GKZDragon:

[sleek canyon]

I dont think anyone writes it like that

[real notch]

Notational conventions are weird, but you could also just use the <,> thing too

[gritty widget]

Yeah, but I don't think that should be used for the dual pairing since it's also used for inner products

[sleek canyon]

So? The space is usually reflexive

So it doesnt matter

[gritty widget]

it is?

[sleek canyon]

Hilbert spaces are

[gritty widget]

i think the question was considering general manifolds

[real notch]

Oh boy I better stop using my $\dv{x}$ notation since that's also used for fractions

[gentle osprey]

Darkrifts:

[sleek canyon]

Finite dimensional vector spaces are hilbert

[gritty widget]

hilbert spaces also require an inner product

[sleek canyon]

They correspond to a choice GLn/On

Who cares

When you fix coordinates you have a natural choice

And in riemannian manifolds which physicists deal with the inner product is given

[gritty widget]

what about manifolds without inner products?

[sleek canyon]

I just said so

[gritty widget]

you fix coordinates and assign them an inner product based on the coordinates?

[sleek canyon]

If you want

Thats how to interpret the pairing as an inner product

But the point is <,> is fine

[gritty widget]

doesn't that require a specific choice of coordinates still?

[sleek canyon]

Yes, thats what "you fix coordinates" means

But the result is independent of coordinates

[gritty widget]

wait, what?

I thought that there was no natural pairing between one-forms and vectors on a manifold

without an inner product

[sleek canyon]

It's not natural in the sense that its functorial

But going back and forth is natural

V**=V is natural

So you can just pretend it's symmetric

The pairing

[gritty widget]

not V* = V...

[sleek canyon]

YES

jesus

The point is, if someone uses notation it's usually justified

It's on you to figure out how

Instead of just saying it's wrong

[gritty widget]

okaay

which is why i prefer the (,) notation, since it's more justified than using inner product to denote something that isn't an inner product

[sleek canyon]

You're free to use whatever notation you want and everyone will keep telling you thats the wrong notation

[gritty widget]

i didn't say that the inner product notation is wrong, you can insert a one-form where a vector should be if they're equivalent, yes?

[sleek canyon]

But dont confuse other people who are learning

[gritty widget]

and that's the notation i've seen

telling someone who's learning that inner products and dual pairings are the same is far more confusing imo

[real notch]

[gritty widget]

and why are you even adding dirac notation into this?

like, the user literally said they were confused by the notation

and thought that functionals were inner producgts

y'all don't have to get on my ass because i'm using slightly different notation to stress the difference between dual spaces and inner products

@real notch related note: especially if it's the first time they're encountering it, d/dx notation for the derivative can be confusing, which is why most intro calculus textbooks make sure to explicitly mention that it isn't a fraction

or else you get stuff like this

[real notch]

that's the joke

because you're complaining about confusing <w, x> with dual

I think pear's entire thing is about how you can by fixing a coordinates

So uh

"When you fix coordinates you have a natural choice"

I know one of them mentioned Riesz somewhere in there

[sleek canyon]

Yeah the point was that you are usually in the riesz setting

Aka reflexive spaces

The question pretended to be about general banach manifolds

But it was actually about (usually riemannian) manifolds

Where the pairing's fine

Who cares this is dumb

[gritty plinth]

hi, can anyone help me with a basic continuity problem?
Let M = (M, d) be a metric space. Check that
d: M × M → R
itself a continuous function (where M × M is equipped with the product metric)

[bitter yoke]

What definition of continuity are you using

[gritty plinth]

Preimage of open is open

[bitter yoke]

Okay, then what are you stuck on

[gritty plinth]

I would have to prove all open sets in R have open preimage in MxM, no? A set is open if there exists a r neighborhood that is contained within the set. For open sets in R (each point has a ball within the set), how can I know whether the preimage is open? Kind of there, I'm stuck

[sleek canyon]

Its enough to consiser a basis of open sets

As you see from the usual definition of continuity

With epsilon and delta

[gritty plinth]

what do you mean by basis?

[sleek canyon]

It's enough to prove it for open sets of R which are intervals

Prove this fact first

[gritty plinth]

because all other sets are union of the intervals, right?

[sleek canyon]

Yes

Now it should be easier to check that preimages of intervals under d are open

[gritty plinth]

I'm blank. This is the first time I come across the product metric, too

[sleek canyon]

Well that's where d is defined

But you probably arent used to dealing with products

[gritty plinth]

since it's MxM, I guess a, b contained in M, d:(a, b) = (a^2+b^2)^(1/2), that is the product metric, right?

[sleek canyon]

That's one metric

A function AxB->C is continuous iff the composition with the projections is continuous

This fact might help

[gritty plinth]

the projections?

[sleek canyon]

The projections AxB -> A given by p(a,b)= a

And similarly for B

Alternatively, you can try to give a more geometric argument

Maybe thats better for you

Given an interval (x,y) and a pair (u,v) in your metric space, you need to find an open set around (u,v) that maps to the interval (x,y)

You can use the triangle inequality to build such an open set

Recall that if U is open in A and V is open in B then UxV is open in AxB

This is another simplification, rather than directly showing the preimage is open, you find an open set around each point in the preimage

[gritty plinth]

@sleek canyon We can have a y- and a x-neighborhood around the origin of M and intersect these sets. The result would be an open set U in M. UxU is mapped to the interval (x, y), and we can construct all open intervals this way. So for every open interval its preimage is open (it's a donut). Is this correct? Regardless, how would you use the triangle inequality?

[sleek canyon]

you have left out the most important part - the set must map to the interval in question

I mean, you say it does

but does it?

[dim meadow]

An easier notion of continuity to use for this might be the preserves limits definition

ie sequential continuity

[gritty plinth]

The intersection of the neighborhood with distance less than y with the one with distance less than x is open and by the distance formula it is mapped to the open set that goes from x to y

The problem said MxM->R though, also I'm not adding as you said but using the product metric so I'm not sure how this applies here

[gritty widget]

Bro

Use sequential continuity

[dim meadow]

Lmao

Yeah sequential continuity still seems like the best idea

[sleek canyon]

@gritty plinth I don't get what your neighborhoods are doing

but they really don't seem to work

you start with a point (a,b) which maps to z in the interval (x,y)

and you want a neighborhood of (a,b) which still maps to the interval (x,y)

[gritty plinth]

oh right, I messed up really bad with the neighborhoods thing. I'll rewrite it with what I had in mind. We can have a x- and y-neighborhood from the MxM origin and take the complement of the x-neighborhood with the y-neighborhood and take the complement of that with the limit points (or just the circle with radius x; picture a 'open' donut). Call this set S, it should be open. S is mapped to the interval (x, y) as the distance from 0 to the limit of the x neighborhood is x, and to the limit of the y neighborhood is y. We can construct all open intervals this way. So for every open interval its preimage is open. Does this make sense?

[sleek canyon]

that's not gonna work

it doesn't make sense to try to build a neighborhood of the form U x U

you want a neighborhood of the form U x V where U is close to a and V is close to b of the pair (a,b)

how close? close enough, depending on how far x and y are from d(a,b)

[ember maple]

If I remember right, UxV can be used as basis.

[sleek canyon]

yeah

that's why we can do this

[gritty plinth]

I'm really lost lol

[sleek canyon]

think about it like this

let's say d(a,b) = p

right? for some p in (x,y)

say (p-e, p+e) for simplicity, whatever

now surely you can find a neighborhood of a such that for any A in it you have d(A,b) ~= p

and similarly for b

right?

can you do it for both now?

the hint was the triangle inequality

[gritty plinth]

nope, still no idea...

[sleek canyon]

you'll have to think about it

alternatively you can drop this definition of continuity and do it by sequences instead

[gritty plinth]

(p-e, p+e) what do you exactly mean with this?

(smol pp, big pp)

[sleek canyon]

forget that part

the point is you have some wiggle room inside (x,y)

cuz it's open

the important part is the rest

[dim meadow]

Lol Jaco, you've been at this for hours

[sleek canyon]

lmao

[frigid patrol]

d(a,b)

[gritty plinth]

this is how I see it, yet I can't see where the triangle inequality can tell me this is an open set

[sleek canyon]

why is it a torus?

doesn't make any sense

you can move b slightly, and you can move a slightly

the wiggle room is like a ball

[gritty plinth]

if the distance is less than x, its outside the interval

[sleek canyon]

the distance between a and b

not the distance from the origin

whatever the origin means

there's no origin

[gritty plinth]

Yes but there are many a and b's

[sleek canyon]

no you fix one

that's the point

you fix one pair (a,b) and find a ball around it

that goes into (x,y)

[gritty plinth]

you can find a ball within the torus, no? I drawed a torus because of how the distance is defined so the distance goes radially... The distance between a and b is supposed to be the distance from the pair to the center of the picture

[sleek canyon]

no

the picture doesn't make any sense

it doesn't represent d(a,b) in any way

note that d(a,a) = 0

[gritty plinth]

d(a,a) is not 0 in the product metric

it's sqrt(2)*a, no?

[sleek canyon]

WHAT

oh okay

I think you misunderstood

d: MxM -> R is the distance function in M

it's not a "product metric", however you wanna define one

it's just the usual metric

[gritty plinth]

what is the usual metric?

[sleek canyon]

d(a,b)

the distance between a and b

defined by the metric space

the product metric would live in MxM x MxM

[gritty plinth]

and the problem says the MxM has a metric called the "product metric"

[sleek canyon]

okay sure

I've been just using the topological space MxM

but you can give it an explicit metric that generates it

if you want

but I think that's confusing you, more than anything else

the product metric is a function MxM x MxM -> R

[gritty plinth]

so...

[sleek canyon]

I can't do this anymore

you're tearing me apart lisa

[gritty plinth]

yeah, I'm sorry, thanks for helping me though

[sleek canyon]

maybe leave it for a while

and give it a fresh look tomorrow

I could just tell you the answer but I don't think that's instructive

I think the geometric argument is good for you to figure out on your own

[gritty plinth]

👌 sure

[lucid turret]

Hello. Can I bother you with simple homology stuffs? Let X=R^2, p point in X.
Is it true that Hn(X, X\p) = Z if n=2 and 0 otherwise?

or am I going mad

[lucid turret]

follows from LES ?
...->H2(X\p) -> H2X -> H2(X, X\p) -> H1(X\p) -> H1(X) -> ...

[sleek canyon]

ye

[lucid turret]

Thank you

[plain panther]

Oh lordy here we go. Okay so, to begin with, when it comes to topology and geometry, I have barely any background in it.
Right now I'm on set for a project to construct a Hexagonal Sphere (Might not be the proper term, but seem gif below)

Quite frankly, I'm really lost. I've found tons of resources from stack overflow (https://stackoverflow.com/questions/46777626/mathematically-producing-sphere-shaped-hexagonal-grid) to some bits of code, but nothing that's truly helped me understood it.
If anyone could help clarify some of these concepts, I'd be really grateful. To have a bit of an insight on what I'm trying to do exactly, I'm trying to program (In Lua) a function that takes a number representing the (rough) radius of the solid, as well as the number of tiles and then generating such an object. Sadly I've been really lost when it comes to the actual assembly of it. I'm not even sure if it's physically possible without semi deformed shapes.

https://upload.wikimedia.org/wikipedia/commons/e/e7/Géode_V_3_1_duale.gif

Sadly, I'll have to look back later, but I appreciate any kind of advice. I'm completely stumped lmfao. (Also if there's a better channel for this, apologies!)

[marsh forge]

Does it need to be strictly hexagonal? The one you posted is not

Also, researching sphere tilings should be what you want. In theory radius should be irrelevant but there are probably restrictions on the number of polygons

[sleek canyon]

read the stackoverflow thread

it's an answered question

[coarse kestrel]

Is any region of the complex plane homeomorphic to the whole complex plane?

[sleek canyon]

yes

a ball

[coarse kestrel]

What do you mean by a ball?

[sleek canyon]

https://en.wikipedia.org/wiki/Ball_(mathematics)

[coarse kestrel]

Yeah I know what a ball is

Do you mean any region is homeomorphic to a ball

And a ball is homeomorphic to the complex plane?

[sleek canyon]

yes

in fact U c C is homeo to C iff it's simply connected

ball's the stuff inside

aka an open disk

[coarse kestrel]

http://en.wikipedia.org/wiki/Riemann_mapping_theorem

Are you talking about this theorem?

[sleek canyon]

this is much harder

but sure, this implies the result quickly

[plain panther]

@sleek canyon yes, I saw it was answered but I was hoping for a more clarified answer as I'd like to understand more about it.

[gentle osprey]

mathbb:

[sharp dove]

I see now, we can check for periodicity by checking if there is some $c>0$ such that $\psi(x,y,z, c)=\psi(x,y,z,0) + 2\pi(n,m,k) $ for some $n,m,k\in\mathbb{Z}$. If this were true, we must have $k=0, 2\pi n=c\cos(z), 2\pi m=c\sin(z)$. Now for any point with $z\neq 1$, we must have $\tan(z)= \frac{m}{n}$. Clearly the tangent of any $z\neq 1 $ is not necessarily rational so the flow is not periodic. However, I’m still wondering how I can get an explicit flow map $T^3\times \mathbb{R}\rightarrow T^3$. We could try $\pi^{-1}\circ\psi\circ\pi$, but $\pi^{-1} $ is only defined locally. Would I have to get a collection of these on a finite cover of the torus and patch them together with a partition of unity?

[gentle osprey]

mathbb:

[dim meadow]

Given a sufficiently nice topological space X (second countable, connected, hausdorff) we define a notion of boundary as follows. The n-boundary of X is the set of all points which do not have a neighborhood around them homeomorphic to R^n. It is pretty easy to verify that the n-boundary of X is closed. Suppose the complement of the n-boundary is a n-manifold and the n-boundary is a n-1 manifold. Is X a n-manifold with boundary?

Any ideas would be appreciated

@ivory dragon any thoughts?

[ivory dragon]

So the key here would be showing whether X is a (topological) manifold

Being a topological manifold with boundary does not imply being a topological manifold, which is somewhat confusing

But the other way around is true

If something's a topological manifold, then it must be a topological manifold with boundary

If you can show that X is a topological manifold, it should follow that it's an n-manifold with boundary

Can you see why?

[dim meadow]

Um the point here is that it can have boundary

Any connected manifold with boundary should satisfy my conditions

I know about manifolds with boundary lol

[ivory dragon]

Then what's the problem?

[dim meadow]

So this construction is designed to generalize the idea of taking the boundary

[ivory dragon]

Yes

[dim meadow]

With the goal of asking the question "is a connected space a manifold with boundary iff it's boundary is a n-1 manifold and it's interior is a n manifold"

I should probably think about this a bit more when I'm more rested lol

There is a nonzero chance that I'm being stupid

[sleek canyon]

it's not

take R^2 and add a line R passing through the origin

the complement of the n-boundary is R^2 \ {0}, a manifold. the n-boundary is R, a manifold.

the union is not a manifold with boundary

the intrinsic problems in the definition are pretty evident from this

the condition of being locally a half space is quite strong

compared to this boundary condition

[frigid patrol]

I like that example

[dim meadow]

@sleek canyon thanks that's very helpful

[zinc tinsel]

Uhh

this is a physics kinda qution

but i will ask away here since it's still differential geo

What is an intuitive explanation of differential forms formulation of Maxwell's equations?

[wraith bluff]

oh wow

right up my alley

The formulation of Maxwell’s equations with differential forms consists of the following two equations:

F=dA

and

d ⋆ F = J

Though to be perfectly honest

I'm not sure what you mean by intuitive

If you don’t understand how to get from these to the four Maxwell equations in vector calculus notation,

then you need to learn more differential geometry, since that’s a very basic exercise in transforming differential forms in index-free notation to index notation,

then applying the definition of

A = Aμ

and

J=Jμ

omg

i assume I don't need to correct myself

If you’re looking for something else, let me know specifically what sort of intuition you’re looking for and I’ll try to provide that. @zinc tinsel

[zinc tinsel]

:)

To me the most important two equations relate the time derivative of the electric field to a magnetic field and the time derivative of the magnetic field to the electric field.

[wraith bluff]

So...

F=dA?

That's what the equation says.

[zinc tinsel]

I don't recognise the notation.

[wraith bluff]

A is a one-form (the potential one-form A= A_μ=⟨ϕ/c, A⟩ where ϕ is the scalar potential and A is the vector potential, and d is the exterior derivative. F=dA ⟹ dF=0, which is the two Maxwell equations relating time derivatives of one field to the curl of the other. (To see this, just remember that in three (plus one) dimensions, the exterior derivative is basically the curl of the vector potential (plus a constant times the time derivative of the scalar potential).

@zinc tinsel

[zinc tinsel]

:D

Ty

[wraith bluff]

If you need any further explanation, ask in the physics channel. I don't want to be banned by @honest narwhal

I've seen Hank linger around in that place.

[zinc tinsel]

Hank?

[wraith bluff]

Ye, some physicist who isn't too spectacular at math but could likely tolerate that question.

[zinc tinsel]

Um

I have another question

if you don't mind

[wraith bluff]

shoot.

[zinc tinsel]

What is a layman's explanation of why simplicial complexes are used in topology?

[wraith bluff]

The answer is that simplicial complexes offer a combinatorial model for a large class of topological spaces (including all topological manifolds.)

That is to say, that these offer discrete things to calculate for a lot of spaces we care about (up to homeomorphism.)

This can sometimes be a great theoretical advantage at well (for example, Riemann-Hurwitz and the Lefschetz fixed point theorem become near trivial in this setting.)

Ultimately, CW complexes were introduced as alternatives to some of the clunkier models used in simplicial complexes

RP^3 is a good example of a gross triangulation, and indeed the CW-structure is much simpler.

These offer a nice halfway point, where the structure is looser in its requirement, but still offers a lot of theoretical advantages for proving theorems, but it is still hard to speak of an algorithm to get a result out.

For example, if you want to form the nerve of a covering, it would be laborious to do so for CW-complexes (and basically pointless since simplices work just fine there.)

This is true for making nice arguments, but also true if you want to program an algorithm to compute some topological invariants.

Do you understand this now?

I tried to put it into laymans terms as well as I could

[zinc tinsel]

Yes ty

[wraith bluff]

Might I ask you something @zinc tinsel

[zinc tinsel]

suress

[wraith bluff]

From what I have seen, you have been struggling with set point topology notation.

These questions you're asking imply you have made some sort of a quantum leap in your studies.

[zinc tinsel]

aha two people use this account.

[wraith bluff]

oh xD

[real notch]

"hey what's a perfect set"
"hey hows simplical complexes work"

[bitter yoke]

yeah woah that confused me too

[zinc tinsel]

xD

Gn

:DDDDDDD

[gritty widget]

lol darkrifts

[zinc tinsel]

In #help-9

plz

wrong channel

[wraith bluff]

?

xD

you haven't sent a message in that channel

[real notch]

sorry im about as intelligent as a poptart, but if you have a k-form at p, you define a linear functional from the k-th exterior power of the tangent space at p to the scalar field?
like, for a k-form $\beta$, you get something like $\beta: U \ni p \mapsto (\bigwedge^k T_p(U) \to K)$ or something?

[gentle osprey]

Darkrifts:

[real notch]

so each p maps to a linear functional from the wedge power boy?

[wraith bluff]

nice tex skills.

[tidal cedar]

https://youtu.be/VvCytJvd4H0

This is REALLY basic topology, but I'm actually very surprised that Matt Parker didn't solve it immediately since he specifically wrote in one of his books about how genus 1 surfaces could have more planar graphs (like the utilities graph) that wouldn't be possible on a genus 0 surface.

[umbral surge]

probably playing it up for the camera

[small obsidian]

Agreed, any of them would have immediately been tipped off by the mug

[wraith bluff]

lmao

[midnight jewel]

well, vi hart didn’t play along… but also didn’t make the cut for that video

(she released it as a standalone video a bit later)

(ruining a perfectly fine donut in the process)

[vocal wharf]

drawing K_7 on a torus is part of a class i was TA-ing

my colleague knitted a torus and stitched the K_7 on it

to visualize it for the students

[frigid patrol]

Lol @dim meadow

[ember maple]

"Algebraic topology (also known as homotopy theory) is a flourishing branch of modern mathematics."

I didn't know algebraic topology = homotopy theory

[dim meadow]

It does not

[honest narwhal]

"Algebraic topology = homological algebra + pictures" -Serre

[dim meadow]

"although a lot of recent model theory has centered around fields"

I'm paraphrasing

From hodges

Let me see if I can find the quote

[marsh forge]

Does anyone have a recommendation for a good introductory text to algebraic geometry? Something that would help me get the flavor of the field?

[sleek canyon]

gathmann's notes

or shafarevich

[honest narwhal]

EGA

[dim meadow]

SGA

[sleek canyon]

FAC

[cedar pebble]

if you really want a panorama that goes too fast check out Holme's A Royal Road to Algebraic Geometry

otherwise a good beginner textbook is Eisenbud and Harris Geometry of Schemes

[sleek canyon]

monkagiga

a bit hard for a beginner

[wraith bluff]

I mean

lmao i'm braindead

[sleek canyon]

this isn't a math person

remember

[wraith bluff]

Wait wait what

did a whole branch of topology disappear into thin air

[honest narwhal]

Basically that the description makes no mention of geometric topology

A lot of these descriptions are outdated

[sleek canyon]

hey dami

the channels suck

change them

xd

[honest narwhal]

They seem fine tbh

[sleek canyon]

yeah i'm jesting

[honest narwhal]

Though you suck and I should change you

[sleek canyon]

it was but a ruse

[honest narwhal]

:0

What do you guys think of the new description?

Gottem

[wraith bluff]

@gritty widget

Something to entertain yourself with.

[sonic niche]

Why is a finite CW complex compact?

Hatcher explains on page 5 how a CW complex can be constructed inductively by attaching n-cells i.e. open n-dimensional disks.

[wraith bluff]

hatcher

[sonic niche]

On page 520 in the appendix he writes "A finite CW complex, ... , is compact since attaching a single cell preserves compactness."

Now my question: why is this obvious? An open disk is not compact, so how can I see that sticking two together is?

[sleek canyon]

what the fuck are you babysnake? @wraith bluff

[wraith bluff]

No.

[sleek canyon]

then why the fuck are you posting that

[wraith bluff]

:)

[sleek canyon]

fuck off mate

honestly

you're very annoying

[wraith bluff]

Tilted

[sleek canyon]

@sonic niche attaching a cell is the same as taking a union with a closed disk

with the boundary already being in the set

[wraith bluff]

lol

You are attaching closed discs in a CW complex (In the notation of hatcher D^n is the closed n-disc cf. page XII). Each closed disc is compact.

[sleek canyon]

ye that's the algebraic analogue of this geometric notion

[honest narwhal]

For real though, probably something like "Point-set, algebraic, and geometric topology. Differential and complex algebraic geometry" sounds good?

[sleek canyon]

why the fuck are you repeating what I just said @wraith bluff

sounds good

oh my god

[wraith bluff]

"The CW compex is formed by taking a quotient of a compact space (the finite union of compacts is compact). Taking a quotient preserves compactness (Since the quotient space is the image of the original space under the projection map, and continuous maps preserve compactness)."

[sleek canyon]

can you just ban this guy @honest narwhal

[wraith bluff]

sh

:)

[sonic niche]

time to go back to melbourne

We need to find another server @wraith bluff

[sleek canyon]

@honest narwhal pls

[wraith bluff]

let's go to numbers &co @sonic niche

[tough hamlet]

lmao

[wraith bluff]

I have all day.

All the time in the world.

Don't have any qualifications.

Got sacked

I've been blacklisted by the society of "janitorial contributions to society"

I can't even leave my room!

[honest narwhal]

Take this discussion in particular somewhere else, and Cat God, your situation is going to be reviewed but do not continue trolling here

[wraith bluff]

Where do I troll?!?!?!?!?!

Proper math is dead

[restive mural]

hi, I'm new here

I'm having trouble defining a geometric notion of point sets

does that sound like it fits in this channel?

if yes, I'd like to elaborate

[wraith bluff]

geometry

[restive mural]

but I haven't studied much topology so please bear with me

[wraith bluff]

#discrete-math

[restive mural]

oh sorry I skipped a word

a geometroc notion of convergence of point sets

I mean points in a space with a defined metric

I don't know enough topology to know the exact necessities, so let's just go with normal, euclidic space

and the point sets change continuously, because I think that is probably easier

so if you have a 2D disc, with boundary that is shrinking with "time"

I can google that but I can also explain where I'm at with my own reasoning

long story short, I think I have figured out a way to define convergence for most sets

but it breaks down for some pathological ones

like every irrational number but not rational ones

for example

and certain special infinite dimensional fractals

(I think)

if you take the set of irrational numbers

and this set does not change over time

then I still don't know a way how this function can converge to the set of irrational numbers

which is what I want

does that make sense?

my method of convergence that I made up is the following

the function F assigns a set to each real number, I'm interested in the limit of F(x) x->y

first, I take the "insides" (not sure of the correct word) of the sets at each x and look at the closure at y

except, I think "closure" is not quite right

which coordinates can you go to arbitrarily close to, in a continious path

the path must never be outside F(x), and the path must end with an x coordinate of y

the result will always be the closure of the limit that I want it to have

so the inside is always correct, but the boundary might not be

[small obsidian]

This relates to completeness unless I'm misunderstanding?

[restive mural]

so I subtract the boundary of the result

the next step is to take the boundaries of each set and if they are not all empty, I repeat the steps

and add the inside of that result to the endresult

so if you have shrinking circles that have a boundary of one half but not the other

and they shrink to a point in the limit where the radius is 0.5

then the first step would add a circle without boundary with radius 0.5

the second step would add a boundary on one half, but not the end points

the third step will add the end points, if they are there

then the procedure bottoms out and gives the result that I want

but if I try this with the unchanging set of irrational numbers

then it never bottoms out

because the boundary of the boundary of the set of irrational numbers is the set itself

and I see no way to fix that

this is probably a bit hard to follow, I'm sorry

I can make some drawings

and I will google the gromov hausdorff metric

I haven't completely groked the gromov hausdorf metric yet, there are a lot of terms that I don't understand

but I get the feeling that it does not deal with boundaries, is that correct?

in other words the metric can't help me tell the difference between a disk with boundary vs no boundary

I made an illustration

https://gyazo.com/6b8ea4ba0ee6cadf4780afef8ff18557

pictured is a 2d example

or a 1d, rather

the sets are lines in 1d space

and the function assigns a line to each real number x

grey and red are the points belonging to the sets, red being a boundary point

on the right is a naive convergence

the naive convergence is just the convergence of each coordinate

in the upper example, some points will be in the set, until the line has shrunken too much and then it is never in the set again, so the convergence of this coordinate is "not in the set"

this convergence has the benefit that the negative of the limit set is equal to the limit of the negative sets

this is a property that I definitely want to keep

BUT

it cannot tell if boundaries are supposed to be in the limit

in the upper example the boundaries are correctly included, but in the lower example they are missing

[sleek canyon]

looks impossible to understand

can you define your convergence concisely?

[restive mural]

the naive convergence that I described right now, yes

the more elaborate one needs pictures I guess

[sleek canyon]

I don't think you know what you're trying to define

but a better question is, why?

what's the goal here

[restive mural]

I want to put a definition to my intuition

[sleek canyon]

what's the intuition

[restive mural]

if you take a geometric set, and you take cross sections of it

[sleek canyon]

what's geometric set

[restive mural]

just any shape

[sleek canyon]

uh

what's a cross section

[restive mural]

like a 3d cube

and you take the intersection with a plane

[sleek canyon]

ok

[restive mural]

that would be a cross section

now you can move this plane "up and down", right?

[sleek canyon]

sure

[restive mural]

if you let the location converge to some position

then you have a convergence of cross sections

[sleek canyon]

ok

[restive mural]

and the limit, for almost all shapes, should just be the cross section of the shape at the point you're converging to

[sleek canyon]

is that what you're trying to say?

[restive mural]

of course, the shape may just be something completely different at point, but usually not

I'm failing to define the convergence of the cross sections

[sleek canyon]

it's just the union of sequential limits

for each sequence of points

[restive mural]

can you explain that in different words?

I may know what you mean but I'm not sure

[sleek canyon]

if you're taking cross sections through the point x

and you're going to take the limit of the cross sections as x -> y

the only sensible way to define it as the set which contains all limits p(x1), p(x2), p(x3), ... where xn -> y and p is any point in the cross section at x1

[restive mural]

I think I know what you mean

but wouldn't that limit always be closed?

[sleek canyon]

no

more importantly, the limit is not going to be the cross section at y, even under very generous conditions

for example even if your space is a manifold

it's going to fail wherever the associated projection map isn't regular

[restive mural]

I don't understand that

[sleek canyon]

I can imagine

[restive mural]

for know I'm interested in euclidian space if that helps

[sleek canyon]

manifolds are locally euclidean

[restive mural]

if I can solve that I can think about extensions

[sleek canyon]

they are the nicest possible subsets of euclidean space

for these kinds of things

and even those aren't gonna work

but they work almost everywhere

[restive mural]

what if I require global euclidianess?

[sleek canyon]

global euclideaness means R^n

R^n, of course, works

[restive mural]

good

I still don't understand why the limit with your definition isn't always closed

in R^n, that is

let me check that I understood what you said

first of all, you are assuming a sequence of cross sections

not a continuity

[sleek canyon]

no

it's the same thing

if you want you can take paths

instead of sequences

[restive mural]

ok

[sleek canyon]

it's the same

[restive mural]

let's say a sequence because it's easier to describe

[sleek canyon]

I mean

[restive mural]

you take a point of the first cross section, then a point of the second and so on

[sleek canyon]

it's closed inside your space

[restive mural]

the coordinates of the points may converge or they may not

if you take all possible sequences that converge

and take the unit of all their limits

that would be the limit of the cross sections

is that correct?

[sleek canyon]

ye

[restive mural]

I don't understand how that is not always closed

I thought about this definition and it does not do what I want

[sleek canyon]

what do you want

[restive mural]

if the cube is borderless, then the limit of the cross sections should also be borderless

[sleek canyon]

it is

[restive mural]

but with your definition it has a border

[sleek canyon]

you take the limit in your space

not in the ambient space

[restive mural]

how do you define the space?

[sleek canyon]

that's why I'm saying it's not closed in the ambient space

what

the space is your "geometric set"

I'm telling you that a reasonable definition should be a manifold

[restive mural]

ah, but what if I don't know what the geometric set is

imagine you only have the cross sections

[sleek canyon]

then you can't define things on it

[restive mural]

you don't know what the rest of the shape looks like

[sleek canyon]

this is the only sensible way to define the limit

[restive mural]

I'm dissatisfied with that...

[sleek canyon]

too bad

that's how you do it

[restive mural]

can you explain why there can be no other way

[sleek canyon]

well

that's a hard question

[restive mural]

let me put it this way

the series 0,1,0,1... does not converge

...with the normal definition of convergence

but there are extensions that let it converge to .5

[sleek canyon]

i mean it wholly depends on what exactly you're trying to do

if you're going to be dealing with geometric properties, this is the definition you want

[restive mural]

and these extended definitions give the correct answer to all "normally" converging series

so I think there often is not just the one true way

[sleek canyon]

you would need to have a better characterization of the kinds of objects you're dealing with

and the kinds of problems you want to solve

as long as it's just a toy problem, with no significance, you can do whatever you want

and you can come up with a lot of definitions that might satisfy you but won't do anything

definitions in math are in service of proving theorems

they come after someone realizes that those are the right objects that get you places

and that give you powerful results

I don't think I have a good reason for why this is the canonical definition other than "this is how it works in geometry"

but I do know that every mathematician you ask would tell you the same thing

[restive mural]

it is a toy problem, I'm not a mathematician

but finding a definition that has real value and gives powerful results is not my problem right now

my problem is that I can't find any definition that satisfies me at all

it's possible that there is just no way to do it

[lavish dirge]

@restive mural What if you rigorously defined "satisfaction"? Not the idea, but what you are seeking.

Also - what are the "extended definitions" like?

All of the ones you are using.

And how are you defining "geometric"?

When I think "geometry", I think of this: https://en.wikipedia.org/wiki/Klein_geometry

Maybe you don't!

[gritty widget]

what kind of prerequisites should one have before doing algebraic topology

[ember maple]

I think some of topology course or book insert a little at at its end. So I assume as long as knowing algebra and topology it's sufficient for intro

[sleek canyon]

ye

basic algebra, basic topology, that's fine

[ember maple]

Like I saw Seifert-van Kampen at the end of lee's topological manifold book

Hmm, lee's top manifold is intro topology book right?

[sleek canyon]

ehhh

not really

it's a weird book

I haven't read it

topological manifolds aren't such a core topic

compared to just point set topology, differentiable manifolds

and algebraic topology

[ember maple]

Well the book read like top intro book but for people interested in manifold theory potentially interested into diff geo. Though maybe that's just my lack of experience

[sleek canyon]

lee's smooth manifolds is the usual intro to diff geo

[ember maple]

Yeah that's the sequel

[sleek canyon]

I wouldn't call it a sequel

well

I dunno

[ember maple]

People keep calling it lee's trilogy for people going to specialize into dg or so from what I remember

[sleek canyon]

weird

I mean I don't really know what the geometers read

[ember maple]

Or manifold theory. Sorry maybe I misremember some things so nevermind

[sleek canyon]

dg is manifold theory

mostly

[lavish dirge]

Is Croom's book good for that?

I've seen the table of contents

But I haven't read the book

[sleek canyon]

for what

basic topology?

[lavish dirge]

It goes from "no topology background" to "introductions to category theory and algebraic topology"

[sleek canyon]

munkres is what most people use

but there's a lot of books

[lavish dirge]

Algebraic topology prereuisites

[sleek canyon]

I like sims and bredon

but I do recommend munkres

first

for the algebra any standard algebra book will do

D&F, rotman, artin, herstein, pinter

[restive mural]

@lavish dirge sorry, I was sleeping

rigorously defining what I'm seeking is problematic, I wonder if it can be done without already knowing the answer

imagine you're a mathematitian from a long time ago

and you're thinking about what happens to the sequence 1, 1/2, 1/4, 1/8...

you think the answer should be 0, but how can you define the result?

it's getting ever closer to 0, but technically, it's also getting closer to -1

I think you can't fully define what you're looking for without having the definition in the first place

[west spindle]

what are you talking about rn

[restive mural]

but the intuition is still important

[west spindle]

limits? or what

[restive mural]

ah, yes I was talking about convergence earlier

I'm looking for a definition of convergence for certain objects

[west spindle]

what objects

[restive mural]

I would gladly talk about it, but ivy wrote several replies that I also want to answer, hmm

[west spindle]

i mean... i'm a bit out of the loop on what you're talking about

[restive mural]

• this is a placeholder that I will fill have filled with answers to ivy -
  "Also - what are the "extended definitions" like?"
  -well, the easiest definition for convergence here is to take the limit of each coordinate by itself
  as you move the cross section, the status of a coordinate of the plane might be 0 1 1 0 1 0 0 0... and then 0 forever (1 = belongs to the set, 0 = does not belong) so the limit if that coordinate is 0
  do that for every point and you have a limit
  but that limit does not give the desired answer for boundaries
  I have a more elabore definition that gives the desired answer in most cases but it breaks for some pathological examples
  I tried to explain it earlier but I'm afraid it was impossible to understand, if you want I can try again with drawings
  "And how are you defining "geometric""
  -I'm afraid I don't understand many terms in that wiki article, it certainly possible that you assossiate geometric with something very different than I do
  when I said geometric, I think what I meant was differentiable manifolds
  but let's use just R^n in the beginning
  -end of placeholder-

imagine you have a 3D cube, in 3D space

you can take cross sections of the cube by intersecting it with a plane

if you move the plane up or down, the cross sections might change, depending on the angle

now if you let the position of the cross section converge to some location

then you get a series, or a continuum of cross sections

[west spindle]

uhh

[restive mural]

I am looking for a definition for the convergence of these cross sections

[west spindle]

what

now if you let the position of the cross section converge to some location

i don't understand what you mean by this

if you move the plane up or down, the cross sections might change, depending on the angle

or this, for that matter

what angle

[restive mural]

ok, imagine you have a cone

the tip is at the origin, and it's pointing in z direction

[west spindle]

ok...

[restive mural]

the plane for the first cross section is the x,y plane

now you move the plane up in the z direction

the cross section at first was a point

[west spindle]

the cross-section turns from a point into a circle

[restive mural]

and becomes growing discs

yes

[west spindle]

ok

so

[restive mural]

you keep moving the plane up to the position z=1

in the limit z=1, what does the limit of the cross sections look like?

[west spindle]

i'm struggling to understand the motivation behind this entire process

if there even is any

what is this all for

[gritty widget]

i think your example with 0 1 0 1 0 1 was better

than this geometric one

he had a problem with the limit of this sequence being 1/2 or non existent

iirc

[restive mural]

no, not really

[gritty widget]

alright ill see myself out again lol

[restive mural]

I used that example in a philosophical sense

I was arguing that there is often more than one useful definition for convergence for the same domain

[gritty widget]

well ya

but as reflexive mentioned, we work with the definitions most useful in the given context

[restive mural]

you're not the first to ask about the motivation of all this, ann

that's interesting

I just think it would be neat to have a concept for convergence of this kind of geometric objects

[gritty widget]

like sure, you can drop the positivity requirement for say a set function being a measure

[restive mural]

not sure how else to explain the motivation

[gritty widget]

but when youre working in intro measure theory, you keep it since its useful

and you call possible negative measures that

signed or negative or whatever measure

[restive mural]

of course, when you think about cross sections of the object of interest, instead of worrying of the limit as the cross sections approach a certain point, you could just look at what the cross section at that point is

but what if you send out the plane to infinity

then you can't do that

[gritty widget]

well, thats where continuity comes in

[restive mural]

is that enough as a motivation?

[gritty widget]

and you cant always look at the cross section at that point

[restive mural]

what? no

[gritty widget]

since limits dont require your thing to be defined in that point

[restive mural]

oh, can't

sorry yes, exactly

that's what I'm saying

[wanton bone]

is the wedge product of something with itself (df^df) always zero, or is that only true for the coordinate differentials (dx^dx)?

oh you

nah you're right and i should have done that immediately

i need more practice working with these algebraically anyway

and yes, they are

[mystic ridge]

what is some intuition behind the fact that knots aren't a thing in 4D space?

is there some kind of formal proof or intuition behind that fact?

Not the best at imaging 4 dimensional space

[wanton bone]

@gritty widget wouldnt that also be an extra degree of freedom to make a knot?

[mystic ridge]

🤔 Would S^2 then not give more degrees of freedom for making knots in 5D space?

it does for most

[wanton bone]

What are S^n?

[mystic ridge]

circles, spheres and hyperspheres

[wanton bone]

ok

[wanton bone]

yeah

[mystic ridge]

fire

[west spindle]

iirc you need to have codim 2 for knots

as in, for interesting knots

[gritty widget]

guys

Could someone explain me why R^n is hausdorff please?

[west spindle]

it's a metric space

all metric spaces are hausdorff

[gritty widget]

yes, i know the explanation for R

u can find 2 disjoint intervals for 2 diff points

for R^n is the same but with balls?

[west spindle]

balls

yes

[sleek canyon]

balls yes

kappa

[west spindle]

the two points can be housed off from each other by balls of radii smaller than half the distance between them

haha get it

[gritty widget]

so to explain it, i need to talk about what is a metric space?

[west spindle]

wait how are you talking about hausdorff without knowing what a metric space is lmfao

[gritty widget]

i know what a metric space is

[west spindle]

i mean

then you can prove all metric spaces are hausdorff

and then just apply that to R^n and be done

[gritty widget]

but again, metric spaces appear

[west spindle]

i mean they don't HAVE to appear if you don't want them to ig

[gritty widget]

isnt there a way to explain why R^n is hausdorff without mentioning the metric?

okey

[west spindle]

^

[gritty widget]

mmm okey, thanks

[west spindle]

oh i mean like i guess you could define the topology by viewing R^n as a product space

with R endowed with the order topology

and then you get a base consisting of open boxes

same shit tbh

[gritty widget]

for this i need to talk about the product topology 😛

is okey, i will talk about the metric just a bit

ty

[sleek canyon]

if you wanna do it with products then prove that X hausdorff, Y hausdorff implies XxY hausdorff

it's just as easy

[gritty widget]

i have a nother quest

question*

https://gyazo.com/086313f311f2c10c3559cdb03d494bb7

could u help me to draw that on a square form?

https://gyazo.com/a3c18948ae2f85ab769e7ce7771e8671

like that

if possible

[gusty fractal]

Is there a way to implement game engine that uses Lobachevsky's geometry in its core?

To simplify, lets say the space is 2d

[gritty widget]

#computing-software

[gusty fractal]

Perhaps using Poincare's half-plane model

My question is more related to geometry

More like how can one calculate shapes' deformations

[gritty widget]

Maybe ask there too just in case

[gusty fractal]

Ok

[gritty widget]

guys

e

every metric space is hausdorff?

yes, right?

and a hausdorff space has always a metric?

okey

so... I was about to say that Rn is hausdorff cuz u can always find 2 open sets with radius less than the half distance between both points

but that is pointless if every space with metric is hausdorff

what does that mean uwu

@gritty widget

why

😦

if i said something wrong i will appreciate ur help

ok ok

and... S1 (the sphere on a plane)

is the quotient of [0,1]x[0,1], right?

whats the diff from quotient topology and product topology?

[bitter yoke]

No, S1 is the quotient of [0,1]

[gritty widget]

ah

so what is a quotint space?

cuz for me S1 has dimension 2

and [0,1] dimension 1

[bitter yoke]

S1 is only the points on the boundary of the unit circle

[gritty widget]

ye

[bitter yoke]

E.g. it's only the points exactly distance 1 away from the origin in R2

S1 is a 1-manifold so

[gritty widget]

so S1 is the interval [0,1] but bent?

[bitter yoke]

A quotient space intuitively is where you identify different points together

In this case, S1 is the interval [0,1] but you say that 0 and 1 are really the same point

Which makes sense if you think about taking the interval and connecting the endpoints

[gritty widget]

so S1 is a circle or a disc?

circle, right?

[bitter yoke]

it is a circle

[gritty widget]

and the torus? why is it another quotient?

i know the torus is 2 circles

[bitter yoke]

Right so there are two ways to write the torus

You can write it as S1 x S1 as the product of two quotient topologies

Or you can write the torus as the quotient of [0,1] x [0,1]

[gritty widget]

so S1 is the quotint space of [0,1] where 0 and 1 are the same point? thats the definition?

[bitter yoke]

That's one way you can define it yes

[gritty widget]

and another one more precisly? could u give me it?

[bitter yoke]

You can define it as a subspace of R2

[gritty widget]

yes but i have to explain S1 as a quotient space

[bitter yoke]

It's really probably best if you read a textbook or some lecture notes on this stuff

[gritty widget]

i am reading the Jhon Lee book, and Massey

but i dont find an example for S1

[bitter yoke]

example 3.22 in Lee's book

is literally S1

But regardless, it's helpful to construct these examples yourself

You know what the definition of a quotient space is, you know what the identification has to be because I told you, so you should construct it and see how the quotient works

[gritty widget]

sure

!

[cedar pebble]

I can think of at least three ways to construct S^1 as a quotient

the first two are basically the same and the third is actually a suspension

For the first two, we could consider R/Z or we could consider [0,1]/{0,1}

for the third we could take the 0-sphere S^0 which is just two points, take the product with an interval I, and then quotient by gluing the respective endpoints of the two intervals

[lucid turret]

S^1 < S^2 < torus < noncommutative torus

ordered by how interesting they are

[west spindle]

noncommutative torus?

[lucid turret]

yes check wiki it's some cool C*-algebra stuff

imo operator algebras are pretty entertaining and deserve attention. they're robust and have tons and tons of structure

[cedar pebble]

Gonna have to disagree with this ordering. There are a lot of ways in which S^2 is way more interesting and complicated than the torus...

[visual grove]

my schools math department has a dedicated section called operator algebra

[honest narwhal]

Homotopy groups of the torus: ez
Homotopy groups of S^2: open

[sleek canyon]

yeah the torus is pretty much as complicated as S1 which is quite simple