point-set-topology

[pastel thistle]

XxY is homeo to X when Y has the trivial topology (also known as indiscrete topology ) ????????

I read it somewhere maybe is obvious and Im dumb

[honest terrace]

this can't even be true for cardinality reasons

[pastel thistle]

if the spaces are finite you cant even have a bijection

[honest terrace]

even if the space isn't finite

let Y = P(X)

Saying that you'd have a surjection X -> X x Y gives you in particular a surjection X -> Y = P(X)

this contradicts Cantor's theorem

[pastel thistle]

im trying to understand this

so this is wrong right ?

[honest terrace]

could you give the whole context ?

because some stuff isn't defined in your screen

[pastel thistle]

X is just the product of topological spaces where they are all indiscrete except for a countable number of those who are not indiscrete

the author says C is such countable set

[reef shore]

what is this trying to prove?

[honest terrace]

please screen the whole thing

[pastel thistle]

is quite long

[wanton marsh]

uncrop

[pastel thistle]

okey

give me a sec

this is what is trying to prove

[pastel thistle]

this is some preliminary stuff that helps the proof

[reef shore]

Yeah that line seems wrong, but I think first countability should be preserved under removing indiscrete factors

[honest terrace]

which line moldi ?

[reef shore]

the product of a given space with an indiscrete space is homeomorphic to the original, given space.

[pastel thistle]

yhea thats clearly wrong

[reef shore]

So they might be saying instead that if A, B are spaces and B is indiscrete then A x B is first countable iff A is

[honest terrace]

oh yes, then yeah what I said earlier is a counterexample

[reef shore]

which seems more reasonable

and all that is required for this

[pastel thistle]

maybe I cant still fix the proof if instead of saying homeomorphism I say that there is a open continious function

[reef shore]

does that prove first countability of the product from the first countability of the countable sub-product?

[pastel thistle]

sorry I mean to say, just a function that will preserve the first countability

[reef shore]

Right

[reef shore]

This seems easier to show

Because it is subsumed by the countable products case

[pastel thistle]

yes thats true

I will go that way, thank you all 🙂 😄

[reef shore]

np

[honest terrace]

[cerulean oriole]

Maybe their Kolmogorov quotients are homeomorphic?

[empty grove]

Seems reasonable

And kolmogorov quotient should preserve first countability

[cerulean oriole]

(x1, y1) (x2, y2) topologically indistinguishable iff all (U, V) st (x1, y1) in U x V, (x2, y2) in U x V
i.e. forall U, V st x1 in U and y1 in V, x2 in U and y2 in V.
In which case x1, x2 are indistinguishable because there exists a V.
Converse is obvious.

So K(X x Y) = K(X) x K(Y), so that's that.

[fluid adder]

Should I actually study Advanced Calculus before putting my hands on De Carmo's Differential geometry for curves and surfaces?

[hollow harbor]

what does "advanced calculus" mean

you should understand multivariable calculus

and probably some linear algebra, and knowing what an open set / closed set / continuous function is wouldn't hurt.

[gritty widget]

[gritty widget]

so like you have a topological space (X, T) and u wanna delete some points from it,

intuitively most natural thing to do is i think:

remove points from X and

new open sets will be same as before except without old points

if that doesnt work just take that as a basis

is there universal property for this

[reef shore]

sounds like the subspace topology

which does have a universal property

[gritty widget]

ah ofc

[gritty widget]

Can manifolds which are curved carry a vector space structure?

[gritty widget]

What I want to know is given a topological manifold with curvature, can I equip it with a vector space structure?

Oooh, that's unexpected.

say it ultra

Well, there are no global bijections for a general topological manifold to any R^d.

[wanton marsh]

[gritty widget]

The pullback would only exist locally.

also what's a topological manifold with curvature?

Um, I'm sorry?

[tight agate]

SSussy is talking about "actual" curvature

[gritty widget]

You know, the Riemann curvature tensor field?

That curvature.

ok, so you're explicitly referring to a smooth structure and riemannian metric

you should say so

Well, metric induced or not.

[honest terrace]

Are you looking for a topological vector space structure that has the same topology as the manifold's, or smth like this ? (In this case, the answer is no you can't)

[gritty widget]

No, an actual smooth manifold, which is curved.

What do you guys mean by a curved manifold?

Riemann curvature, right?

no, we're interpreting a topological manifold as a length space with respect to a metric inducing its topology

/s

I know that topological vector spaces exist.

What I wanted to know is that given a smooth manifold with curvature, can it also carry a vector space structure?

Well, the vector space structure isn't given.

What I want to know is could it be given one.

Let's say the manifold is the sphere.

With all the garnish.

How can the sphere be bijected to R^2?

There are only local bijections.

Can you give me an example?

[marsh forge]

do you believe that both R^2 and S^2 are of the same cardinality?

[gritty widget]

Yes?

[marsh forge]

then you believe there is a bijection between them

thats the definition of cardinality

[gritty widget]

Sure, but, can you give me an example?

[marsh forge]

i am honestly not sure there is an easy way to write one down lol

it will be inherently gross

[gritty widget]

Well, my intuition was that manifolds like R^d that has a vector space structure are flat.

[marsh forge]

the obvious thing to think of is like, remove a point, take the stereographic projection S^2->R^2 and then convince yourself you could find a place for the extra point if you wanted to

i think you are misunderstanding

a bijection does not know about any structure

it doesnt care if you have a vector space, a field, a topology, etc

its just sets

[gritty widget]

True.

Okay, cool, thanks!

[flint cove]

If a topological group acts on a topological space, is there any obvious way to relate the cohomology of the orbit space to the cohomology of the original space and the group?

For instance, if I quotient out an ℝ-action, I expect the cohomology to not change, because I would expect to get a deformation retract

and I was wondering whether that relates to ℝ being cohomologically trivial

Let's also assume that the action is free

[tight agate]

@flint cove puppe sequence

oh wait cohomology

lookup the serre spectral sequence

if the actions start getting weirder you will have to do EG x_G X instead of X/G

and you will get "Borel equivariant cohomology"

which is insanely complicated for super simple spaces

for example, if G acts on a point, then the associated cohomology is the group cohomology of G

[gentle osprey]

TTerra

[gritty widget]

i am in dire need of a sanity check. ping if you respond

(assume the flow is complete if needed)

[fathom cave]

i don't understand this. If we take wedge product of a n and n-1 covectors, we get a 2n-1 covector which is clearly not in the algebra. What am I missing?

[flint cove]

You're missing that it automatically becomes zero due to antisymmetry
Take say R² and wedge (e1 wedge e2) to anything else, you'll see that you only get terms containing (x wedge x)

[flint cove]

Okay, let's pretend I know how spectral sequences work for a second.
So the first thing here would be to notice that X onto X/G gives us a fiber bundle (or at least a fibration)? Does this still hold if I was non-free, say when I act with scalar multiplication on a vector bundle (since it includes the zero section)?

So if the fiber were cohomologically trivial, I would get $E_2^{p,q}= H^p(X/G, 0)\Rightarrow H^{p+q}(X)$

[gentle osprey]

lux

[flint cove]

But wouldn't that imply that E2 had only zero entries? How can that converge to the cohomology of X?

[gritty widget]

Hello, someone have any PDF about Algebrical Geometry?

[flint cove]

Vakil is a great resource: http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf

[gritty widget]

Thank you very much

[marsh forge]

converge is technical term

it doesnt have to give you the actual answer

[frigid river]

Could someone please explain to me what an induced homomorphism is? 0.0

[empty grove]

In the context of fundamental groups?

Whenever you have a continuous function (X, x) →(Y, y), you get an associated homomorphism π₁(X,x) → π₁(Y,y)

[frigid river]

and that's special about fundamental groups?

[empty grove]

No many other things have this kind of a property

Like continuous maps induce homomorphisms between homomology groups

Or chain maps

Or even simpler, group homomorphisms induce set maps, between their underlying sets

This inducing maps thing is studied in cat thy as functors

But idk if that's what you were trying to ask

[frigid river]

I think that's what I needed

[frigid river]

but that look like a very weird property

[empty grove]

Why is it weird

[frigid river]

how would that continuous function look like?

[empty grove]

It works for any continuous function

You get a group homomorphism

The group homomorphism is just applying the continuous function on the paths in X

[frigid river]

🤦

[empty grove]

ie you have an element of fundamental group of (X,x), say [f]

And say the continuous function is g

The image of this element under the induced map is [g ∘ f]

Soundness of the definition has to be checked

But this turns out to be a group homomorphism

That's included in the definition of maps between pointed spaces

[frigid river]

wait, just to avoid any further stupidity of mine: If there exist a continuous function between (X,x) \to (Y,y), then there must exist a homomorphism between their fundamental groups?

[empty grove]

Homomorphism

Not homeomorphism

[frigid river]

whoops, sorry

[empty grove]

And yes that is true

For example there's a continuous map from R to S^1

The corresponding induced map for the standard map will be the homomorphism from the trivial group to Z

[frigid river]

I'm sorry, I am not with you yet, I don't even get that function part.

Why does that take a basepoint?

(X,x) I mean

[empty grove]

That denotes a space X with a basepoint x

[frigid river]

Let me get back to you in 5 min

[empty grove]

[flint cove]

I know that „E ⇒ blah“ means that it starts to stagnate for some index, and the entries at that stagnation point should be blah, correct?

because then having a zero page would mean all the successive pages are zero as well after taking cohomology, so we're already stagnant

I feel like I have a very basic misunderstanding here

[frigid river]

I also feel like I am having a very basic misunderstanding here. Why does the basepoint of a fundamental group matter? I can't think of an example where the basepoint would make a difference.

pi_1(S^1,x)=pi_1(S^1,y) also for x \neq y, isn't it?

[flint cove]

Oh, no, I'm just stupid, I forgot that the 0th cohomology is nontrivial

then I have E₂ consisting of one column corresponding to H^p(B, ℤ)= H^p(B) and that stagnates when taking cohomology, because to the right and left everything is zero

likes spectral sequences now

[gritty widget]

one stupid thing you can do is take the disjoint union of a point and a circle

then the base point definitely matters

[frigid river]

Oh.

I didn't even think of stuff that isn't connected 0.0

[gritty widget]

for path connected spaces, the basepoint doesn't matter

any two fundamental groups of the space are isomorphic, and you can construct an isomorphism by choosing a path between the basepoints

it's in munkres

[frigid river]

Thank you ❤️

[gritty widget]

of course the natural question to ask is if connectedness is enough (instead of path connectedness)

that's an exercise

[frigid river]

[gritty widget]

just think of the usual connected-not-path-connected space

maybe with some modifications

[frigid river]

if I could only think of such an example 0.0

[gritty widget]

topologist sine curve

[frigid river]

OH

[pearl holly]

Now find a counter example that shows that the closure of a path connected subset doesn't need to be path connected

[gritty widget]

well, give me one

[pearl holly]

You already gave it

[gritty widget]

gimme another one

[pearl holly]

another one?

Well idk, a curve that is "infinetly dense" at its limit points somewhere but not as dense elsewhere, just like the sine curve thingy

[gritty widget]

[pearl holly]

But what other examples are there tho?

probably some really weird, non visual and abstract ones or something

[bronze lake]

It's a long shot, but would anyone happen to know where I might be able to find an english version of this paper? https://link.springer.com/content/pdf/10.1007/BF02950724.pdf
It's german title is "Zur Isotopie zweidimensionaler Flachen im R4", translating to "On the isotopy of two-dimensional surfaces in R4", published by Artin in 1925

[sweet wing]

i doubt it exists, but id say just force your way through, math german isnt too hard to read even with basically 0 backgroynd

google translate is useful :)

[gritty widget]

"just learn german"

[sweet wing]

yes

i agree

[gritty widget]

based

[sweet wing]

you get lots of hot guys

[gritty widget]

this is also a good reason

[bronze lake]

thank god it's only 3.5 pages

I just want to see the proof that the fundamental group of a spun knot in 4-space is isomorphic to the fundamental group of the knot that is being spun

[marsh forge]

Spun?

like rotated?

[flint cove]

I've started translating it and should be done in say half an hour
Gonna sound terribly awkward tho because it sounds awkward in German already

[bronze lake]

I would really appreciate that! It looks like most of the stuff I care about is on the latter half of the second page, where it talks about fundamental groups in the next to last paragraph. I more or less understand everything up to there

[frigid river]

@pearl holly are you here?

[pearl holly]

Yes

[frigid river]

Okay, let me post this then

[pearl holly]

okay good

Okay so the first thing is: what the heck does he mean by "depend only on the endpoints of the path h"?

Like if we assume that it is abelian, what should we arrive at?

Maybe that if $h_1$ and $h_0$ are two path with the same endpoint, then $\beta_{h_1}[f] = \beta_{h_0}[f]$

[gentle osprey]

Tokidoki ✓

[frigid river]

Sorry, I am still thinking about that "change of basepoint" definition you sent.

[pearl holly]

okay wait

this one?

[frigid river]

yep

[pearl holly]

It's basically conjugation like in group theory

since the h bar thing is kind of the inverse of h

[frigid river]

Yeah, I get that.

[pearl holly]

OKay good

[frigid river]

oh, so "dependent on only the endpoints of h" means that we could just pick any f and it would still be the same, I guess.

but that's visually very weird to me

[flint cove]

progress: mid third page

[pearl holly]

Hmm

[pearl holly]

Do you think this is right?

[frigid river]

i really don't know xD

[pearl holly]

Because I can pick two completely different paths h and h'. If beta is only dependent on the endpoints of these paths, and if the have the same endpoint, then b_h = b_h'. Right?

So if I pick different endpoints, then the b_h and b_h' should be different

[frigid river]

yeah, sure

[pearl holly]

Okay so now we can at least start with something

let's start working with $\beta_{h_1}[f]$ and show that this is equal to $\beta_{h_0}[f]$ using the fact that the group is abelian

[gentle osprey]

Tokidoki ✓

[pearl holly]

let me try this on paper first

[gritty widget]

[frigid river]

$\beta_{h1}[f]=[h \centerdot f \centerdot \bar{h}]$ and that's just equal to $[f]$, since it's abelian, right? 0.0

[gentle osprey]

verysorrowfulperson ✓

[frigid river]

my coffee is still too hot for it to be consumed

[pearl holly]

ahhhh the neko server pinged me!

I leaved the server lmao

[frigid river]

leaved

you could also just mute it by right clicking it

[pearl holly]

well the concatenation is not abelan I think. We know that [f][g] = [g][f] tho

[frigid river]

huh, why isn't it?

[pearl holly]

because it's the fundamental group that is abelian

[frigid river]

the group operation of the fundamental group is concatenation

[pearl holly]

yeah but the group elements are [f], right?

[frigid river]

[pearl holly]

it's the equivalence classes that are in the group, not the loops itself

[frigid river]

yeah, I see :/

[pearl holly]

yeah we need some sort of relation between b_h_0 and b_h_1 I think

otherwise, we won't get anything I think

[pearl holly]

because how do we proceed from this?

that b_h1[f] = [h1 f h1 bar]

[pearl holly]

and all we know is that they have the same endpoint

[frigid river]

so they are homotopy equivalent, right?

[pearl holly]

yeah I think so

[frigid river]

so it's trivial xD

[pearl holly]

hmm

wait

wtf

something's wrong here...

[frigid river]

Okay.

[flint cove]

There you go

[pearl holly]

okay well they have the same start and endpoint, but that doesn't mean that there's a homotopy between them, right?

[frigid river]

no, it doesn't

I think

no, it really shouldn't

[pearl holly]

lux what a bro! Translating a whole paper, that's really nice of you!

yeah so it's not trivial, right?

[frigid river]

no, nothing here is trivial, for me at least

probably still is trivial for the solution manual authors

[pearl holly]

yeah I just thought about this

because if that was the case, then it actually would be trivial

without the need of the group being abelian

it just follows

[frigid river]

but if they have the same starting and end point, they obviously eliminate each other (h and h bar I mean), and weirdly this doesn't require "abelianness"

[flint cove]

@bronze lake feel free to ping me if some things are unclear; sometimes I don't quite understand what he means in German as well, it's kinda ancient regarding some parts of the nomenclature

[flint cove]

well it's clearly better than doing numerical analysis homework lmao

[pearl holly]

wait how do they eliminate each other?

[frigid river]

I am so confused by this, I think my visual intuition is completely leading my astray.

h and h bar are inverse paths, I thought.

[pearl holly]

well they are, h bar just moves in the other direction

so h bar = h(1-s)

[frigid river]

I think of it like vector addition

If I go h then f then \bar{h}, it's the same as going the green thingy, right?

so the whole thing is only dependent on f, and not h and h bar, right?

[pearl holly]

No I don't think so. By going h and then h bar you will go back to the starting point

[frigid river]

or am I just dead lost

[pearl holly]

Yo tterra is here

you're more than welcome to help us both

[gritty widget]

im eating dinner

[frigid river]

and we're suffering

maybe we could annoy ask Max

to at least explain what the exercise actually wants from us

[flint cove]

what was the question?

backlog got rather comprehensive

[frigid river]

and that's our definition for a basepointchangingthingy

[pearl holly]

Like is this what he means when he says "depend only on the endpoints of the path h"?

is that equivalent to that statement

[frigid river]

but I don't understand how it is dependent on h anyway

In my mind, they cancel each other.

[marsh forge]

whats the Q

im on a walk

and on my phone

so its time for some more bad typing

[pearl holly]

Yo max is here bois

[pearl holly]

this is the question

[marsh forge]

oh id need to think for this one I dont think its necessarily trivial

but yes its saying that the basepoint change homomorphism only depends on the basepoints

[frigid river]

but why don't the h and \bar{h} cancel each other?

okay, I asked a bad question apparently

[pearl holly]

Okay so we can show that if the group is abelian, then $\beta_{h_0}[f] = \beta_{h_1}[f]$ for two paths $h_0$ and $h_1$ that have the same ending point?

[gentle osprey]

Tokidoki ✓

[frigid river]

why, visually, isn't [h \centerdot f \centerdot \bar{h}]=[f]?

yeah, visualising them as vectors wasn't the smartest of ideas 0.0

~~just don't ask, please ~~

[pearl holly]

oh man

I think I know now

this will take a while to write down tho

[gritty widget]

[pearl holly]

okay hol up, let me think first

[frigid river]

share thy knowledge, please

[gritty widget]

i remember doing this exercise

[frigid river]

how long did you take? 0.0 3 minutes?

[pearl holly]

Okay I think that this is just a group theory thing

[gritty widget]

how did you get that figure?

[pearl holly]

we have that $\beta_{h_1}[f] = [h_1 f \bar{h_1}] = [h_1 f \bar{h_2} h_2 \bar{h_1}]$, right?

[gentle osprey]

Tokidoki ✓

[pearl holly]

since the h_2 and h_2 bar cancel

so then $\beta_{h_1}[f] = [h_1f \bar{h_2}] [h_2 \bar{h_1}]$

[gentle osprey]

Tokidoki ✓

[pearl holly]

so then, because the group is abelian, $[h_2 \bar{h_1}][h_1f \bar{h_2}] = [h_2 f \bar{h_2}]$

[gentle osprey]

Tokidoki ✓

[pearl holly]

then we have proved this

I think that this is right

[frigid river]

[pearl holly]

just manipulating the equation

[frigid river]

we need some referee for that proof

[pearl holly]

It's literally just manipulating the whole thing

but idk, could be wrong

[pearl holly]

this

[frigid river]

$\beta_{h}1[f]=\beta{h}_2[f]$, which we assumed would be equal to the task 0.0

[gentle osprey]

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

[pearl holly]

paths with same endpoint

[frigid river]

[pearl holly]

yeah, we have the other one too

I think that it should be kind of similar

idk tho

[frigid river]

whilst toki thinks, I will just write down this one.

[pearl holly]

wait

[pearl holly]

so you mean this part where I split it?

well a long time ago, we went through a proof with max and he showed me that you can split this thing like I did

yeah yeah I know sorry lmao

[frigid river]

and why can you split them up? 0.0

[pearl holly]

well crap, I don't really remember the argument that max gave

wait, I will scroll up

[frigid river]

and also, why is this? xD

I assume you can combine them due to the same argument.

[pearl holly]

oh no

he says that it follows from definition

[frigid river]

Ah great.

[pearl holly]

yes, here I am combining them and the step before it uses the fact that the group is abelian

or not the previous step, but the previous step combined with this step kind of

okay so how does that follow from definition?

[frigid river]

magic

don't technically all theorems follow from definitions?

[pearl holly]

aaahhhhh I shouldn't be doing all this stuff when I can't even remember basic definitions!

I should just go back to Munkres man

[frigid river]

no, stay here.

[pearl holly]

yeah I will lmao

let me just think

[frigid river]

Well, it's brackets and it's a group, so the laws of associativity must apply

[gritty widget]

[frigid river]

[gritty widget]

i am having an existential crisis

how the fuck does anything in geometry work

[frigid river]

[pearl holly]

are you kidding me

[frigid river]

[pearl holly]

BRUH

why am I doing this when I literally don't even know the definition?

[frigid river]

I remembered there being something about that property, but couldn't quite recall what it was xD

[gritty widget]

🤨

[pearl holly]

well anyway, that literally follows from definition. Should we try to do the other implication now?

[frigid river]

,ti

[gentle osprey]

The current time for veryhappyperson is 01:14 AM (CEST) on Sun, 27/06/2021.

[frigid river]

are you sure about that?

[pearl holly]

yeah same here

well idk

[frigid river]

the other seems harder, tbh.

[pearl holly]

I should honestly review all this before trying to do the exercises

wait let me think for a minute

[gritty widget]

skip reading just do the exercises xd

[frigid river]

don't worry, even max hasn't responded yet, so it can't be that simple

[pearl holly]

but Max was on a walk

just hol up a minute, I don't think that it will be that hard

[frigid river]

yeah, but he said he had to think about it

[pearl holly]

the only tricky part is to get the manipulation right

[frigid river]

I don't know how to begin this.

[pearl holly]

ok so we want to show that [f][g] = [g][f] right

and [f][g] = [fg]

so now we want to manipulate this thing to somehow get to the other side

so we need to add something to the equation

so what about the identity?

like e: [0, 1] -> the space?

oh wait

what???

hold on

right so [fg] = [e fg e bar] right?

oh there's now way

[frigid river]

what, no, why?

[pearl holly]

but it's the identity

oh no

[frigid river]

oh

[pearl holly]

we can just make it a constant loop

because now we at least have that [fg] = \beta_{e}[fg] right

so now if e ends at the same point like g, then we have that \beta_{e}[fg] = \beta_{g}[fg]

and that is just [gfg g bar] = [gf]

and so [fg] = [gf], right?=

[frigid river]

[pearl holly]

and the equality follows from the assumption that something something independent of h soemthing something

[pearl holly]

so from this

[frigid river]

I don't know, but I know that I will seek my bed now 0.0

[pearl holly]

okay lmao me too, but I think that we solved it

[frigid river]

you*

I didn't contribute anything.

[pearl holly]

it's just a lot of manipulation of the equations

nothing hard, just hard to see the manipulation thingys

[frigid river]

Okay, good night

[pearl holly]

and here e can be a constant loop that equals the end point of g

[pearl holly]

yes good night!

[gritty widget]

[pearl holly]

okay now I need to rest. Stardew valley time

[frigid river]

it's helpful getting the greenhouse before winter, so don't forget to plant fruit trees.

[pearl holly]

Well I’m on summer now so I think that it is fine

[frigid river]

trees take 28 days to grow though, and we really shouldn't have that conversation in here xD

[marsh forge]

oh

i think i figured it out on my walk

i just got back

[pearl holly]

[frigid river]

toki thinks he figured it out too

[pearl holly]

Might be wrong tho idk

Brb I’m in mobile and I hate mobile

[gritty widget]

math walks

[pearl holly]

🚶‍♂️ 🧠

[frigid river]

my brain just shuts off when walking

[marsh forge]

i just pictured stuff in my head

until i figured it out

[frigid river]

and how exactly did you figure it out? 0.0

[marsh forge]

Do you want hints or a solution

[empty plume]

What's the fundamental group of 2-torus?

Isn't that the abelianization of it?

[gritty widget]

[marsh forge]

slim

thats not the fg of the 2 torus

[marsh forge]

its a free group on four generators modulo a product of commutators

[gritty widget]

[fading vale]

I do not understand the argument in the corollary

(in this defn branch points are those with e_y > 1)

set of branch points sorry

im moreso confused about the claim in the proof

Like

why does the proposition imply that each point of Y has a punctured open set containing no branch points

how the argument made in the proof of corollary 3.2.4 follows from the proposition

Because it only has branch points at 0?

I see

and the fiber argument works similarly i guess

ok

that makes sense

thank you ultraproduct

write down explicitly that z |-> z^k has only 1 branch point at 0?

[gritty widget]

[fading vale]

or am i misinterpreting

I see

[gritty widget]

[gentle osprey]

slimvesus

slimvesus

slimvesus

[gritty widget]

[fading vale]

ultra if this explanation is done I think i understand the general notion of how to do the computation im just not sure how to show that in the points of C where z |-> z^k isnt a branch point in the usual defn its not a branch point in the defn we have here

like if we use the transition maps properly we get this square where the objects are all C, the one of the horizontal arrows maps z |-> z^k and its given by local behavior near y, the other horizontal maps z |-> z^d for some d and its given by local behavior of phi near y', and the vertical arrows are transition maps

so i guess i just have to show that if ur not a branch point of z |-> z^k in the usual sense then ur not a branch point of z |-> z^k in the sense defined above

ughhh this sounds like complex analysis i dont know

every day is moth vs """basic facts from complex analysis"""

i know

it should be like

$\begin{tikzcd}
\mathbb{C} \arrow[r, "z^k"] \arrow[d, "t_1"] & \mathbb{C} \arrow[d, "t_2"] \
\mathbb{C} \arrow[r, "z^d"] & \mathbb{C}
\end{tikzcd}$

pain

[gentle osprey]

Moth In Shambles
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fading vale]

$\begin{tikzcd}
\mathbb{C} \arrow[r, "z^k"] \arrow[d, "t_1"] & \mathbb{C} \arrow[d, "t_2"] \
\mathbb{C} \arrow[r, "z^d"] & \mathbb{C}
\end{tikzcd}$

AHHHHH

[gentle osprey]

Moth In Shambles

[fading vale]

holy shit finally

anyway z^k is top map between charts t_1 t_2 are transition maps z^d is local map and we know that for z in C nonzero corresponding to y' neq y in U_y cap U_y' we have z^d cause yeah

(you might have to shrink transition maps appropriately or something)

but how do we show that z nonzero implies d = 1

ofc z is not a branch point of z |-> z^k in the usual defn of branch points but idk how to show thats equivalent to the ramification index defn of branch points

Im not i hope slim suffers for 1000 years

what im trying to say is that like

i dont know how to show that branch points of z^k are branch points in the first place

under this defn

if they are then of course d = 1

so ur done

but idk how to show that z non-zero implies d = 1

i guess thats not rly what you asked me to do though

idk i probably just need to sit down and learn complex analysis in detail to get into the nitty gritty of a lot of htese constructions

ok what i mean is like

i know that z^k has no non-zero branch points under the usual defn of having to pick an argument

i dont know how to show that z^k has no non-zero branch points under this defn, i.e i dont know how to show that z^k locally looks like z |-> z for non-zero points

once you take charts that translate those non-zero points to 0

does that make more sense?

no like

under the defn of branch points im given

how do i know that 0 is the only branch point of z |-> z^k

Oh i may have been very unclear

Yeah the thing i showed earlier with the charts and the z |-> z^k

so points are branch points of phi if k > 1

thats my defn

yeah the proposition is a definition

it shows that every map has that z |-> z^k form given some charts

and then defines branch points as ones where k > 1

and im just not sure how to show that when X = Y = C and phi = z |-> z^k that points other than 0 arent branch points

this makes sense!

[gritty widget]

[gritty widget]

topolog

[empty grove]

A donut may be topologued into a coffee cup

[frigid river]

How would I start here?

[empty grove]

Hint: an endomorphism of Z is uniquely decided by the image of 1

[fading vale]

Im kinda confused as to how the last sentence implies that the stabilizers are cyclic

[strong heron]

@empty grove' hint in other words: What are all the homomorphisms from the set of integers to itself?

[sweet wing]

it is locally z->z^e

but not globally

e = ramification index

[sweet wing]

i.e. there is a complex patch thatlooks like this

[fading vale]

Am I dumb

Why is this set open by definition

Oh I see

ok this makes sense

[gritty widget]

[sweet wing]

moth you need to draw more circles

[flint cove]

So here's one thing I don't quite understand about the serre spectral sequence
If I try to see what happens with the trivial fibration B→B, it reads

[gentle osprey]

lux

[flint cove]

So this gives us a 2-page that consists of only one column (or row?) filled with $H^p(B)$, and this already stagnates because codomain / domain of the respective arrows is 0

[gentle osprey]

lux

[flint cove]

But I read the arrow as "eventually, our page looks like $(H^{p+q}(B))_{p,q}$"

[gentle osprey]

lux

[flint cove]

which can't be true because we only have nonzero entries in one row / column and don't have constant entries in each codiagonal

[marsh forge]

You should read it as $\bigoplus_{p+q=n}E_\infty^{p,q}\to H^nB$

[gentle osprey]

The New Maxnormal

[marsh forge]

You have to worry about hidden extensions and stuff

bc you

are actually summing up an associated graded thing

not the real thing

im not sure where you would be learning about this that doesn't define convergence though

@flint cove

[summer jolt]

Does anyone know what is meant by "spectral topology" in the context here https://ncatlab.org/nlab/show/Gelfand+spectrum?

Ok thanks. Do you have any reference for this? because I can't seem to this definition anywhere

[fading vale]

If i have a cover p: Y -> X and two points x1 x2 in X with distinguished open sets U1 U2 that have non trivial intersection

does every section on U1 agree with some section on U2

on the intersection

[flint cove]

Fair, I guess I shuold bite the bullet and rigorously go through a text
I wanted to start off doing some „simple examples“ to convince myself that the whole machinery is worth it
Thanks for the helpful remarks.

[marsh forge]

Let me clarify

Convergence is the thing you need to actually pay attention to

feel free to completely ignore the setup of the spectral sequences

and the proofs

you just need to know what convergence actually means

I have a video lecture up on the Serre SS with some exercises at the end

if ur interested

[flint cove]

Okay, that helps to focus things
Vakil seemed to be an accessible resource at first, do you know of anything else?

[marsh forge]

McCleary is okay but has some typos

Hatcher's intro to the Serre SS sucks imo

[flint cove]

Found a copy of McCleary.
If you have a link to the video that'd be excellent!

[marsh forge]

https://www.youtube.com/watch?v=vsy8OY_0Er8

oh shit when did that get to 300 views lmao

[fading vale]

weed probably

[marsh forge]

I'll reveal the full computation of $\pi_*S$

[gentle osprey]

The New Maxnormal

[flint cove]

[marsh forge]

Im so torn

i want to give talks

but I am also

lazy

i should just do it

[flint cove]

Just do an over-eager announcement then you have to commit

works for me every time, now everybody knows that I have no clue about the Yoneda Lemma and I know a little more about it lmao

[marsh forge]

i already did

but i missed the first talk date

[void ingot]

I have a question for Compact Manifolds in general. Is the metric tensor bounded? So does $ C_1 \leq \sqrt{g} \leq C_2 $ hold? With C_1 > 0 and for a particulare choosen chart.

as function in general on the defined region with \sqrt{g} := \sqrt{ det g}

my professor said it is true, but the explanation was a bit off

[bold canopy]

Black hole go grr

[void ingot]

\sqrt{g}(p) is only defined on a region defined by a chart, why should scaling always work? In General the prof. said that the compact part was important

The Question is more about why this holds, because the prof. said it too me a week ago and since i was not able reproduce this argument. I want to show, that for a given chart part of an compact Manifold M (without boundary) there always exits a c_1,c_2 with

c_1 \leq \sqrt{g}(p) \leq c_2 \forall p \in F \subset M

where F is the subset defined by the chart.

[bold canopy]

If it goes to 0 its degenerate, if it blowd up then at the blowup point its bad news?

[void ingot]

yes thats the thing i want to show does not happen when it is on a compact manifold

He said i should look a the \wedge * Operator

but that does not help me in any way

even tho i only have Subset, where this specific Chart defines the metric tensor?

because this one can be relative open

ok, but i think about it, this seminar i am having right now is a bit hard. Thank you ^^

[pastel thistle]

people

why the first fundamental group is not conmutative

?

ups sorry im interrupting

no no just asking for the basic stuff

yes a wedge of circles I was thinking about that one

[cursive nebula]

hey guys, i have a question on differential geometry

there's an exercise our professor solved, and i don't understand the last passage,

so, the exersice is:

1. show that X is a differentiable manifold
2. show a base of the tangent space of X in point p_0

and there's the solution!

i understand everything except how he got a base (BIG CURSIVE B) of the tangent space in p_0

I think it might be... a simple linear algebra problem, but i've been thinking about it a lot and i don't get it, can anybody help? (please @ me if you reply!)

[cursive nebula]

I saw a past exam for this course, and one of the question was to do something similar for S^n, the n-dimensional sphere.

I can see that it could be seen as the pre-image of a regular point (1) of a smooth function in R^(n+1) (x1, x2, .... , xn+1 ) -> x_1^2 +.... + x_n+1^2
, the gradient of that function is gradf=(2x1, 2x2, ..., 2xn+1), and it's only zero in (0, ..,0) but that's okay because that point doesn't belong to the n-sphere.
So, that would make the sphere a smooth manifold of dimension n+1 -1 = n

The tangent space in a point x would be ... the set of the vectors that are orthogonal to the "radius vector" in that point, so i can describe that set TxS^n as the set of vectors y such as <gradf(x), y > = 0
BUT HOW DO I EXPRESS A BASE IN THOSE TERMS.... (like in the exercise my prof solved...)

(also can this be done GLOBALLY on a sphere?? because intuitively, I DO NOT THINK SO)

[rare storm]

anyone know how to do b) ?

[bright acorn]

Does a homeomorphism between topological spaces X and Y endowed with the Lebesgue measure send null sets to null sets?

I should ask this in analysis prolly

[haughty wave]

can't you consider like cantor function or something?

[void ingot]

Sadly i don't see the way to solve it. I even tried a different approach with the eigenvalues of the metric tensor. Because if it explodes at on point tge eigenvalues or at least one has to go to infty and the same with 0. But it does not work😞 i getting a litte depressed about it tbh, thats really bugging me long enough now.

[flint cove]

Your tangent space – call it V – consists of all vectors (v₁, v₂, v₃, v₄) satisfying 2v₁-v₂+v₃-v₄ = 0.
There's many possible bases here, but one possible base of that in particular would be ((1,0,0,2), (0,1,0,-1), (0,0,1,1)): they are three linear independent vectors satisfying above equation, so they form a basis of V.

And in terms of your manifold, ∂/∂x would be your first unit vector, ∂/∂y your second, and so on

[cursive nebula]

ohhh thank you so much @flint cove !! i got it now

IT REALLY WAS A QUICK LINEAR ALGEBRA THING

i just emailed my teacher about it and now i feel really embarrassed!

this would also be a valid base, right?

you just need any three linear independent vectors

[gritty widget]

differential geometry is just linear algebra parametrized by smooth things

i have heard of them a few times

but i don't know much

(anything, really)

yael karshon namedropped them a few times

what is it called

NEW SPACES

[cursive nebula]

RKrtejtklrltjk

BABE WAKE UP

i can't believe i woke up today, had a full meltdown over baby linear algebra and emailed my teacher about it and now i have to be embarrassed

guys stop saying the truth my mind is weak

[gritty widget]

one time i was confused about something for like 3-4 days and had absolutely no clue how to justify it

eventually went to the prof's office hours and it was literally a one liner

[cursive nebula]

that's okay ;w ; ❤️

it means you'll never forget it lol

[gritty widget]

the pain is etched into my soul

[cursive nebula]

best way to retain memory on any subject is embarrassment and pain

[gritty widget]

it was something about lie group orbits

and their tangent spaces

[cursive nebula]

ohh

so it was already somewhat advanced

[gritty widget]

i don't remember the exact issue but the one liner was basically like "the map sends the orbit to itself so it sends the tangent to the orbit to itself" and i was like fuck

so simple

[cursive nebula]

i failed an exam once bc i couldnt prove gramm schmidt's orthonormalisation algorithm in linear algebra like 6 whole years ago and now it lives in my brain rent free

i know it better than the lyrics of whatever i listened to at 14

honestly, a teacher's job should be making students understand why something is really obvious
but then students will surprise you on how not obvious something you think is obvious is

i had a calc 2 teacher with absolutely NO EMPATHY like.
she couldnt get HOW someone might not find ALL OF MATHS obvious
she was like "obviously, you're all playing stupid because there's no way you don't get this, i can't wrap my head around it!"
and it was really terrible to experience, but also hilarious

"what do you mean you don't get it? you're lying, right?"NO PROF

[lean marten]

I guess every problem is intractable until it's trivial?

Sorry should have been more specific

I mean that pretty much everything seems impossible till you understand it

and then it seems trivial

[bright acorn]

Does someone have an interesting example of a non connected manifold with varying dimension on each connected component?

[gritty widget]

tu's manifolds book (intro) allows for such manifolds. there's an MSE answer out there by him that explains why

if you can find that, then he should give an example

[bright acorn]

Oh cool, I don't particularly remember that.

[gritty widget]

it's a really small technical thing

never really matters

but it's a thing

if you can find it pls link it i actually want to see

or it might be on MO idr

[bright acorn]

You mean this?

I couldn't even find his account on MSE

[gritty widget]

ill see if i can find it when i get out of the shower

i swear ive read a post by him that explains it

[bright acorn]

FOUND IT

[gritty widget]

ya

[bright acorn]

This is actually a really cool example

But prolly too hard to describe with the current tools we have available at the course I am taking

[gritty widget]

group actions weird

is the moral of the story

wanna know a fun fact

every closed subset of R^n is the fixed point set of some smooth R-action

ok what hypothesis did i miss

i use fun sarcastically

i hate this fact

just goes to show how nice compact / proper lie groups actions are

imagine your fixed point set being some stupid shit like the cantor set

[bold canopy]

How to draw this smooth action if i want 3 points in R^2 as my fixed point set?

? Is it supposed to be >=0 function? How do you prevent additional maxima?

[gritty widget]

let f be a >= 0 smooth function which is zero at exactly those points and take the flow of f/(1+f) * (a coord vector field)

the zeroes of this vector field are exactly your points, and the boundedness of the coefficient ensures the flow is globally defined

(so is actually an R action and not anything local)

i read this on an MSE post a while ago and haven't forgotten it

more generally if your manifold admits a nowhere vanishing smooth vector field then you might be able to do the same thing

[bold canopy]

oh right what I was confused about when I was trying to draw what it should be was that I was only drawing minima or maxima type fixed points (for the 3 points in R^2), which can never work by some topological argument (like, 1+1-1 != 0, -1-1+1 != 0), but in this example f/(1+f) * (coord vector field), all the fixed points are hyperbolic, and 0+0+0 = 0

(?)

maybe the topological stuff I was saying is nonsense; I'm going to believe something like that in my head anyways

[pearl holly]

Hmm I read a small portion of the image that MisterSystem sent and I thought that the theorem in invariance of dimension seems kind of cool. How does one prove it? I can only find PDF's online that I either don't understand (there's too much background knowledge that they don't cover) or proofs that only prove it for small values for the dimension (where n = 0 and n=1). Maybe this proof actually is pretty hard to prove so this might be a little bit far fetched since I don't know a lot, but it would be cool to see a proof of this.

[summer jolt]

Hey, I've seen two different definitions of weak topology on a Banach space X, one says that it is the smallest topology that make linear functionals in X* continuous and the other talks about the topology induced by seminorms. I suppose that the two are the same, but what is the intuition behind the weak topology and why should these two definitions be equivalent?

[hollow harbor]

In some weird sense, the weak topology should be thought of as the topology of pointwise convergence. What do I mean by this? Well, a sequence x_n in X converges to a point x in X weakly if for every linear functional f in X*, we have that the sequence f(x_n) converges to f(x) as a sequence of real/complex numbers.
So what's really going on here is that the x's are acting like functions, and the f's are acting like the points that we're evaluating the functions on. Let's switch the notation: x_n(f) converges to x(f) for each f. Where I've just flipped the positions k
of f and the x's for notation purposes, the meaning hasn't changed. Now you can see that it's like a sequence of functions, x_n, is converging pointwise at every input f to a function x.
For me, understanding the weak topology in terms of how it captures pointwise convergence is really important. It's also how I actually use the weak topology in practice: to prove things using weak topologies, I almost always use weak (pointwise) convergence.

Ok, now why is this equivalent to the other two definitions you mentioned? It's actually kind of a bridge between the two imo.

First of all, we're demanding that each linear functional still satisfy sequential continuity (f(x_n) converges to f(x) when x_n converges to x: this was always true, but now it's our only restriction). Maybe you can try working out why it's true that the kind of convergence we've defined is precisely the kind given by the smallest topology that makes all the linear functionals continuous. Even if it's not totally immediate, now we can see that we're mostly just trading between usual continuity and sequential continuity.

On the other hand, there's the topology induced by seminorms. Well now each seminorm just comes from a linear functional. Check that weak convergence as we've defined it is the same thing as convergence in each seminorm, and vice versa.

Maybe for a more direct path between the two definitions: you need the seminorm for the functional f in order to make the functional f continuous, so we need all those seminorms. This is just by writing down what it means for f to be continuous. But we also need the SMALLEST topology, so we shouldn't add any more beyond the topology these seminorms generate.

[summer jolt]

@cobalt turretC thanks a lot for your detailed answer! That made things a lot clearer

[tight agate]

@limpid vault it might be related to the topology on an infinite symmetric product

https://en.wikipedia.org/wiki/Symmetric_product_(topology)#Infinite_symmetric_product

I dont think so

it's a free commutative monoid on X

according to the wikipedia page

there's probably a nice way to topologize the groupification

[gritty widget]

Has anyone an idea what the $C^{\infty}_{loc}$-topology is?

[gentle osprey]

lime_soup

[tight agate]

on what?

[gritty widget]

for the full context im trying to understand the "space of broken trajectories"

from morse theory

This is from Audin's morse theory, they prove directly that this is compact. I saw on some stack exchange post that this is the compactifction in the C^{\infty}_{loc} topology

[tight agate]

it should be some kind of mapping space

did not pay attention to the topology

maybe try looking up gromov compactness

[gritty widget]

Okay thanks

[gritty widget]

[quartz edge]

Is the pullback notation evoking the hodge star operator?

or is that a notational overlap

[gritty widget]

they're completely different things

notational overlap i guess

[quartz edge]

hm, pure unadulterated sadness

[gritty widget]

[quartz edge]

I was kind of hoping that like

pullback of g on a k form omega = g(hodgestar(omega))

by some weird composition of n-k-forms in vector val'd functions

[gritty widget]

hmm

i don't think i know of any identity similar to that

https://en.wikipedia.org/wiki/Exterior_calculus_identities maybe you can find something here

[sleek thicket]

hodgestar...

[gritty widget]

shamrock activated

[quartz edge]

hodgerock shamstar

[misty mango]

can someone help withthis ?

[gritty widget]

have you sketched the set

[void ingot]

Professional

[gritty widget]

also haris don't multipost

#help-8

[misty mango]

yes I did

@gritty widget it's a disc with an interval (1,2)

?

[reef shore]

Can you see what the accumulation points and the boundary points are, from the sketch?

Also it should be the interval [1,2], not (1,2)

[misty mango]

hmm

what is the accumulation points?

those that are inside?

[reef shore]

They are points in R^2 such that you can find a point of A arbitrarily close to them (ie in any neighbourhood)

[misty mango]

so it must be the first answer then?

or what

wait

the boundary points are not defined

[reef shore]

Why are they not?

[misty mango]

like those on the boundary

is not in A

[reef shore]

Or do you mean you don't know the definition?

[misty mango]

also that I guess

[reef shore]

[misty mango]

hmm

[reef shore]

Boundary of A is A closure minus A interior

[misty mango]

ahhhh

so like a kind of limit

limit point

[reef shore]

kind of, but interior could also contain limit points so you remove those

[misty mango]

the answer is C then

[reef shore]

yes

[misty mango]

but the boundary is not defined

[reef shore]

It is

[misty mango]

how?

when we have strictly less

than

[reef shore]

A closure and A interior are defined

boundary is the set difference

[misty mango]

ahhhhhhhhhhhhhhhhhhhhhh

topology is confusing me

are you good at topology?

[reef shore]

I know some

[misty mango]

I got some practise exercises until august

but we won't get the answers, only for practise

also quadrature

like if you know numerical analysis

[reef shore]

I see, you can post them here if you get stuck on any

[misty mango]

I got 1 look

I got 3 questions I have been looking at for quite long

also on more in topologty

my guess here is number 2

because the neigborhood is in A

right?

[reef shore]

yeah that would imply that A has non empty interior

[misty mango]

exactly

and this one has a degree of 8

since we have 9 points

am I not right?

[reef shore]

it seems right.

[misty mango]

my guess is B here

since the inverse is self-mapping as well?

in the open interval

[cedar pebble]

why are you taking a multiple choice exam on topology

[gritty widget]

cohomology of RP^n multiple choice

[brave stag]

guys
i need to use the Alexander Subbasis Theorem to prove that the space 2^𝑋 is compact for every 𝑋.
and uh, i just need a lil help to know where to start 🤡

[flint cove]

Hmm… start with the definition of a subbasis?

It's virtually impossible to say something helpful here without knowing your context and which definitions you are comfortable with

[summer jolt]

I'm confused about the statement that says the weak-* topology is weaker than the weak topology. Here I assume the weak topology is defined on X and *-weak topology is defined on X *, but how can we compare topologies on two different topological spaces?

Also why is the weak-* topology weaker?

[wanton marsh]

isn't the dual of X also a Banach space, and so also has a weak topology ?

[summer jolt]

@wanton marsh is that true in general? I don't really know much functional analysis at all, but I thought that depended on the underlying field

[nimble jolt]

@summer jolt Elements of X* pair with both elements of X and elements of X** in a bilinear way. The weak* top on X* is that which makes all of the former cts, and the weak top on X* is that which makes all of the latter cts. The general embedding of X into X** is what gives you that the weak top is stronger (you are requiring more functionals to be continuous, i.e. you need more open sets).

Of course it immediately pops out from this that X reflexive gives you equality of these topologies.

[summer jolt]

@nimble jolt ok so in this case the book is talking the weak topology on X* not on X?

[nimble jolt]

Yep, because as you say it does not make sense to compare topologies on different spaces.

[summer jolt]

I see thanks a lot!

[nimble jolt]

no worries

[pearl holly]

Okay so this is the exercise that I'm working with. Here's what I (and veryhappyperson) thought:

So every homomorphism from Z to Z is of the form f(x) = ax for some integer a. This is because f(x) = f(1+1+1+ x amount of times) = f(1) + f(1) + x amount of times = ax where f(1) = a. But now what? You have to somehow "transform" this f(x) to be a homomorphism from pi_1(S^1) to pi_1(S^1). But how do you do that? To "change" the domain and range of this function, we need the variable x to be an element in pi_1(S^1), so something of the form [f]. But what does "a" correspond to in pi_1(S^1)? I assume that it must be a map from S^1 to S^1 because then I'm done, but I don't know how to prove it.

I also tried "conjugation", so hfh^-1 where h is the isomorphism from Z to pi_1(S^1) and where f: Z -> Z is a homomorphism. This will be a homomorphism from pi_1(S^1) to pi_1(S^1) but this doesn't give me a lot I think.

Then I tried to "replicate" the argument that every homomorphism from Z to Z is of the form f(x) = ax but with homomorphisms from pi_1(S^1) to pi_1(S^1), but it didn't go well because I only got "concatenation" in my expression instead of composition. So how would one do this?

[marsh forge]

Im confused

What do you mean change

pi_1 S^1 is Z

they are the same thing

[pearl holly]

Hmm yes but you still need to kind of "transfer" f(x) = ax so that it is "written in the language of" pi_1(S^1)

[marsh forge]

?

[pearl holly]

yeah idk lmao

I'm really tired sorry

But how would you use f(x) = ax in this scenario? How do I proceed from here?

[marsh forge]

I think the issue might be that your thing with "ax" only makes sense if you view Z as a Z module

It might be better to write: every function Z->Z is of the form f(1)=n which determines every other value

But maybe this isn't helpful

okay heres a better way of thinking of this

We know $\pi_1 S^1\cong \bZ$ and that every homomorphism $f:\bZ\to \bZ$ is of the form $f(x)=nx$ for some $n\in \bZ$.

[gentle osprey]

The New Maxnormal

[marsh forge]

Now why do we know that first thing

How do we compare $\pi_1 S^1$ and $\bZ$?

[gentle osprey]

The New Maxnormal

[pearl holly]

Well we show that psi: Z -> pi_1S^1 defined by psi(n) = (cos2pi n, sin2pi n) is a isomorphism

Is that what you mean?

[marsh forge]

Well

can you put that into like

human words

what does psi(n) look like

[pearl holly]

it's a circle. So if we have loops we can "concatenate" them to get a loop of a winding number which is a sum of the other winding numbers

[marsh forge]

okay so

psi(n) is a loop living in S^1

in the dumbest words possible

what does this look like

[pearl holly]

Well it's a circle? A loop?

[marsh forge]

is psi(n) just one loop?

[pearl holly]

So it has winding number n

[marsh forge]

If I wanted to walk around the circle in the path described by psi(n) what would i do?

[pearl holly]

So it goes around itself n times I think

[marsh forge]

yes!

okay

so to the number n in Z

we associate the path "go around n times"

Also I messed with some variable names let me restate stuff

every function Z to Z is of the form f(n)=mn

Okay

so in the "language" of pi_1 S^1 what does this mean

if "n" is go around n times

[pearl holly]

well it means go around mn times maybe?

[marsh forge]

Great

now, this is completely defined by what happens for f(1)

or f(go around once)=go around m times

can you think of a map S^1->S^1 that would induce this?

[pearl holly]

Hmm well not really. I'm not quite sure what you mean by induce

[marsh forge]

You need a map $f:S^1\to S^1$ such that $f_*(1)=n$

[gentle osprey]

The New Maxnormal

[marsh forge]

The map $f$ induces the map $f_*$

[gentle osprey]

The New Maxnormal

[pearl holly]

Well maybe a map that maps (cos 2pi, sin 2pi) to (cos 2pi (n-1), sin 2pi (n-1))?

[marsh forge]

okay hahah side bar

we are putting on our topologist hats now

please never write down another continuous function like that

just describe it

pure vibes

[pearl holly]

okay lmao, so like a thing that goes around the circle n-1 times

[marsh forge]

n-1?

[pearl holly]

because then when I compose it with "going around 1", I get "going around n-1 + 1 = n"

[marsh forge]

you're composing-composing here

not path composing

[pearl holly]

Oh

[marsh forge]

So you have this map from the interval to S^1 describing going around once

and then you wrap that circle around another circle n times

so now your path goes around how many times?

[pearl holly]

n times?

[marsh forge]

yes

so you dont need to adjust by subtracting 1

[pearl holly]

ah yes okay I see

But how would I write it using math? Because now f_a[w_n] = [w_{na}] where w_n is a loop that goes around itself n times is an expression for all homomorphisms from pi_1S^1 to itself, right?

[marsh forge]

what weve already said counts you dont need symbols for something to be math

But if you want to be careful

you'd just do something like

Well I mean yeah you've basically already done it

unless you feel like you need to prove that

$\omega_n\circ f\simeq \omega_{nm}$

[gentle osprey]

The New Maxnormal

[marsh forge]

but you've already shown that all that matters here is winding numbers

and we already counted it

so its all formal and fine

[pearl holly]

yeah I know that it counts and it is a great way to think about it, but I got to this point once and I didn't know how to proceed from this so I just threw away the whole idea

[marsh forge]

?

i really dont know what you mean

[pearl holly]

yeah sorry idk, I should sleep and examine what I've done and then come back once I can express myself clearly lmao

[marsh forge]

if anything what you are saying sounds like you got so caught up in the symbols you lost sight of the geometry that gives you the answer

[pearl holly]

yeah that might also be true

But anyway, thank you so much for the help!

[gritty widget]

Someone here have any pdf about Riemann geometry?

[gritty widget]

i have plenty. give me 20 minutes or so and ill DM you a bunch

don't want to share them here

[gritty widget]

Alright

[gritty widget]

Hm ok

(pirated content)

Has virus?

lol

no

the mods don't want pirated content being posted on the server

Hm

I see

[gritty widget]

@gritty widget is it cool if i send them now

it might be a bit spammy

dont want to do it out of nowhere

@gritty widget In preference send it on my private

Because I’m gonna outta home

[hollow harbor]

Hey TTerra? Respect the rules please 🙂

[frigid patrol]

Nerds

[regal sonnet]

Is Coordinate Transformation & Homeomorphism the same thing, or what's the Difference?

[marsh forge]

not every homeomorphism can be expressed as a change of coordinates

[regal sonnet]

How is Coordinate Transformation different?(specifically, I want to know about coordinate Transformation of 3-d space to Hyperbolic space).

[marsh forge]

Again, not every topological space has a useful notion of coordinates

homeomorphism as a term is just vastly more general

[regal sonnet]

oh, ok

thanks

[tight folio]

can a fundamental group on a single connected component have an involution without being trivial?

[woeful oasis]

2 dimensional real projective space has fundamental group Z/2Z. In this case the only nontrivial group element has order 2.

[flint sinew]

sorry if this question is too simple for an advanced topic channel 😅 I'm only a math minor who's studying all kinds of maths as a light hobby

if I would define manifolds like this:

1. topological manifold is a special case of topological space
2. to classify as a d-dimensional topological manifold, each point in the topological space's set must have on open set in the topology containing said point and an infinite(/arbitrary?) number of (continuous?) functions that map that point to a point in a d-dimensional coordinate system

is this definition sufficient? does it leave something critical out? is it susceptible to misunderstandings?

[woeful oasis]

instead of "infinite/arbitrary number of functions", rather there exists an open set and there exists a function on that open set which maps the open set homeomorphically to an open subset of Euclidean d-space.

There end up being a bunch of functions, but the definition asserts the existence of at least one very special kind of function.

[flint sinew]

and that function must take any open set from the topology and be able to turn it into an open subset of Euclidean d-space in order for the top. space to be classified as a manifold?

does the function need to be bijective as well?

[woeful oasis]

Yes, with continuous inverse

[flint sinew]

in a lecture series I'm watching right now, the lecturer tends to talk about these functions as "maps". is this just a convention of topology or does it carry some special nuance?

[woeful oasis]

If you think of continuous in the forward direction as telling you no ripping/tearing, the continuity in the other direction you could think of as telling you there's also no squishing/collapsing. That way, for example, you can make sure you have a consistent notion of dimension