point-set-topology

[quartz edge]

but i saw bachman uses similar befuddlyduddery

[sleek thicket]

sure, you can get a covector via the inner product on R^n

(covector = linear function from vectors into R)

[quartz edge]

ye

a forb

form

[sleek thicket]

Yip

1-form and covector mean the same thing

[quartz edge]

so is I_P a 2-form

[sleek thicket]

no

Well

[quartz edge]

it's not doing anything with webge prodcuts

[sleek thicket]

What is a form, to you?

[quartz edge]

product

[sleek thicket]

It's a symmetric bilinear form

but it's not an alternating form

[quartz edge]

aha

[sleek thicket]

Which is what eg differential forms are

[quartz edge]

ok i was trying to relate the two

[sleek thicket]

you can see this by symmetry

[quartz edge]

yeah

it doestt have antisymmetry

ok that is what i was missing

[grand plover]

Hi, could someone help me understand a topological thing?

I also posted the question here: https://artofproblemsolving.com/community/c7h2482712p20856932
but I'm still very confused, and I'm in a real hurry, so I'd appreciate your help.

Question:

If f(z) = 1 / p(z) where both are entire complex functions.

And we have a closed disk of radius R

I want to show that the function has a maximum value on the disk, but in my paper, I don't really have time to prove and explain it with compactness.

I was wonder why the following argument isn't valid:

If f(z) isn't bounded, then for some sequence in the disk, it must tend to infinity. Which means p(z) tends to 0 for that sequence

[gritty widget]

This is true, but why does the sequence have a limit?

[grand plover]

Mustn't it have a limit in D since a closed set has all its limits?

[gritty widget]

"a closed set has all it's limits"That is true, but sequences in a compact set do not need to be convergent, they just must have a convergent subsequence

[grand plover]

But if the sequence has a limit in D, mustn't all subsequences also converge to this limit?

[gritty widget]

But the sequence doesn't need to have a limit in D, eg 0.5, -0.5,0.5,-0.5,...

This has a convergent subsequence but it is not convergent itself

[grand plover]

But since D is closed, mustn't the sequence have a limit in D?

[gritty widget]

I said above that this is false

And you can see that the sequence I gave is not convergent

If the limit exists, then it is in D

But a limit can fail to exist in the first place

[grand plover]

I see

But if the limit doesn't exist, doesn't this also imply that no sequence will make f(z) go to infinity, and therefore f is bounded?

[gritty widget]

I'm not sure I understand your reasoning here

[grand plover]

If f is unbounded, on must be able to find some sequence in D for which f goes to infinity. Or am I mistaken?

[gritty widget]

This is true

[grand plover]

So mustn't this sequence also converge to some limit?

[gritty widget]

(btw, if you have access to some theorems from calculus, that I guess depend on compactness, in particular extreme value theorem, I think you can prove what you want that way as well)

You are just repeating yourself

[grand plover]

I was just unsure whether it was okay to simply cite the extreme value theorem without proving it. But to prove it, I need to show compactness, which would fill out a lot of my paper...

(My readers are high school teachers)

So I was just wondering whether there was an easier way to go about it

[gritty widget]

hey quickly does this definition work

I feel like it’s a bit sus but I can’t put my finger on it

[grand plover]

Ah, I think maybe I've misunderstood something here?:
If f of a sequence tends to infinity, then the sequence must converge to some limit.

[gritty widget]

This is still false

(unless you could prove p has at most 1 zero)

Er

[grand plover]

Oh, the assumption is the problem is that f has 0 zeros

[gritty widget]

Sorry, I guess that should be unless you could prove p is bounded way from 0 except at <=1 points

I think, or something like that

[grand plover]

If the assumption is that p is defined on the whole complex plane, and p has 0 roots. In the problem, I previously show that for some radius R, for all values outside this R, p is less than 1.
So I wanted to show that p is also bounded on the closed disk of radius R.

[gritty widget]

You mean f, right?

[grand plover]

Ah yes

(I am trying to show the fundamental theorem of algebra by contradictly assuming p of degree n>0 has no roots and showing that it leads to p being bounded on the whole plane)

[gritty widget]

If you are doing it with this liouville argument, I'm not sure how to get out of talking about compactness

[grand plover]

Hmm, yes that was what I was going for

Dang

Hm, not sure I myself understand yet, but it seems everyone agrees that my argument doesn't work. I will try with compactness then. Glad I won't hand in something wrong.
Thank you 🙂

Sorry, now it's bugging me a little: would f having at most roots validify the argument, or am I still just wrong?

at most 1*

the part about the sequences having to tend to some limit

[obtuse meteor]

btw @sleek thicket bc I need to complain

https://estrogen.fun/i/o3y5.png

Here's the syllabus for the manifolds class right

we have 5 weeks left in semester

and we haven't finished A (given we have done the multilinear algebra portion of B)

[fading vale]

monkaGiga

wait how many weeks have passed so far

[obtuse meteor]

we've just spent like tons of time juggling domains and codomains for charts

[fading vale]

[obtuse meteor]

how many weeks so far?

hm

we're on our 7th pset

so like

maybe 8?

[sleek thicket]

Wow

[fading vale]

huuuuuuuuuuurb

wait how r there only 5 weeks left in the semester

wtf

[tough imp]

This is a grad class right?

[obtuse meteor]

no

[tough imp]

wut

[obtuse meteor]

but last year they like

[tough imp]

you have manifolds for UGs?

[obtuse meteor]

literally did

de rham cohomology

and this year we are just vibing

defining the tangent bundle for submanifolds of R^d

[fading vale]

umich ends in mid april??

[obtuse meteor]

this weekish

yeah for exams

[fading vale]

o

hrm

[obtuse meteor]

last day of exams is April 31

[tough imp]

wait

wut

summer break starts in May?

[obtuse meteor]

ye

[tough imp]

when do u start school in the Fall?

[obtuse meteor]

september-ish?

[tough imp]

wtf

[obtuse meteor]

like first day of september

[sleek thicket]

Uw starts really late Alex

[tough imp]

I mean I know but

[sleek thicket]

We had 1 day of september this year

[obtuse meteor]

anyway

the point is manifolds sad

and make me feel bad

[sleek thicket]

[median glade]

wait you have 4 months of vacation

[pastel linden]

this is literally like 90% of basic smooth manifold stuff though

[obtuse meteor]

fuck manifolds

[pastel linden]

it's important to have a bank of them to work with though

without regular submanifolds the theory would be garbage

[obtuse meteor]

sure but like

when doing these psets and stuff

I don't feel like I've really learned anything

or am any better at these manipulations

they're just very banal and boring and just take time to do

[pastel linden]

understandable, it definitely sucks ass in the beginning

but the cool parts easily make up for it

5 weeks is plenty of time to zoom through differential forms

[median glade]

a lot of math is building up a good repertoire of technical gadgets to prove some intuitive things formally 😛

at least that's been my impression in a few fields

[gritty widget]

the only decent results in an intro manifolds course are frobenius's theorem and stokes's theorem, everything else is just definitions and immediate corollaries

[median glade]

invariance of dimension is kind of cool

but also like one of those things, where you're like "wait, this isn't trivial?"

[pastel linden]

can't prove that in an intro manifolds course though

[obtuse meteor]

you can prove it for submanifolds right?

we just like

did that

[obtuse meteor]

this isn't really my impression of other fields tbh

[pastel linden]

the general proof requires retractions I believe

[gritty widget]

you can prove it for smooth things using some very basic linear algebra

topological manifolds? a lot harder

you can use de rham cohomology for that i think

[obtuse meteor]

yeah for smooth things it's immediate

bc from a diffeomorphism you'll get an invertible derivative

and you'll be like happy

[median glade]

do you use much analysis in diff manifolds?

[obtuse meteor]

mmm

the qualitative theorems a lot

and then like

existence of bump functions

[median glade]

existence of bump functions is topological no?

[obtuse meteor]

which we had to reprove in one dimension on the midterm and it's like why

sorry

rant

[median glade]

bump functions exist because of normality?

yeah yeah, Urysohn's lemma

[obtuse meteor]

can you do it that way?

neat

wait no that doesn't give you smoothness

you need smooth bumpy bois

[median glade]

oh, a smooth bump function

is there a smooth Urysohn's lemma?

lol

I'm asking, because I know calculus well, of course, but I'm wondering if it's worth a detour to analysis or if you can just get by for the most part

[obtuse meteor]

learn analysis

it's good for the soul

[median glade]

I mean, I read a book on measure theory once

so I know how like, Lebesgue integration works

idk, I find analysis kind of boring

[obtuse meteor]

I mean same but

you need a lot of the theorems

[median glade]

tru

[obtuse meteor]

just do topology instead

[median glade]

I've done topology heh

by that I mean, I read the first half of Munkres

[obtuse meteor]

munkres makes me sad imo

but maybe that's because I'm bad at books

[median glade]

well I'm not a math major

so I just self-study textbooks as a hobby

I feel like I learn way more through self-study than I can in a course

but I probably have a weird pacing

because courses will skip parts of books, and give a broader perspective usually

whereas reading the book gives you a bunch of details

[obtuse meteor]

I get bored by those kinds of details very quickly

[pastel linden]

how do you usually study coho

[obtuse meteor]

shitposting

tbh I probably learn more from wikipedia and nlab than other sources

classes primarily

but those two are big for me

[quartz edge]

did somebody just say urysohn's lemma

[fast minnow]

furrysohn's lemma

[sinful pecan]

read the references at the bottom

[cedar pebble]

it's fairly self-contained on a lot of pages

depending on what you're looking at

[sleek thicket]

step 1. know higher category theory

[gritty widget]

the nlab pages on topics surrounding symplectic geometry are surprisingly digestible

rather pleasing

[gritty widget]

yea some pages can be great

[marsh forge]

you can just like

ignore the word infinity

and the symbol

and most of its still true

@gritty widget

i was being serious haha

using nlab is the art of separating useless abstraction from what you need

almost by design

[river granite]

making sense out of abstract nonsense

[obtuse meteor]

reading nlab is learning why saying the words "L is adjoint to R" by themselves is a proof :^)

[sleek thicket]

Here's a neat thing I was thinking about: the only contractible compact topological manifold is the point (up to homeomorphism, the claim is significantly less interesting up to homotopy equivalence)

Curious to see what proofs of this other people can think of

(it's false that contractible manifolds are automatically Euclidean, see the whitehead manifold)

[gritty widget]

I am probably being a dumbass, but if I take X=[0,1] and a homotopy H(x, t)=(1-t)x, I thought this would imply [0,1] is contractible.

[sleek thicket]

In my terminology manifolds are by default without boundary

It's true that closed balls are counterexamples otherwise

[gritty widget]

Oh I see

Was your proof an AT proof?

[sleek thicket]

yup

used a bigger theorem than I would like

that's essentially my proof

oh huh

yeah

I was gonna say anything is Z/2Z orientable

So top dim cohomology with Z/2Z coefficients is nontrivial

But proving this is a little tricky imo

maybe not actually

like

Let's go with your proof

Orientation double cover isn't so bad

So if you're not orientable π1 is nontrivial by covering space theory

so say you're orientable

This means you can choose a consistent local orientation

but then you can glue together to a global homology class

still idk, feels like there should be a more direct way

I guess this provides some intuition

lol

what a mood

Nah it's okay haha

So for your proof

You can make it work without the Z/2Z coefficient stuff

if your not orientable you have a connected double cover

A point upstairs is a point of your manifold and a choice of local orientation at that point

does that intuitively make sense?

Nah I'm saying not orientable => not contractible

Because this was the hole in your original argument yeah?

Well I am generally dissatisfied

I just thought you might want an outline of a proof without the coefficient stuff, sorry

Kk

So say M is a manifold

Which is not orientable

If it's disconnected then it's not contractible

So say it's connected

Then we can construct a double cover M' of M

The points of M' will be pairs of a point in M and a local orientation of M at that point, you project onto M by forgetting the local orientation

It turns out that M' is connected iff M is not orientable, this isn't so hard to prove

So in our setup M' is connected

But the existence of a nontrivial connected double cover of M implies π1(M) ≠ 0

yeah?

Yeah :(

You need singular homology, orientability, orientable+compact => top dim homology is nontrivial, orientation double covers, and covering space theory

Yeah I guess so

[gritty widget]

$T_{id}(\mathrm{Diff}(M)) = \mathfrak{X}(M)$

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

[sleek thicket]

That feels right to me

Very right

[gritty widget]

i like it

[sleek thicket]

flows!!!

it's like

flows are infinitesimal group actions

sort of

[gritty widget]

so if you have a lie group action $G \to \mathrm{Diff}(M)$ and you take the differential at $e \in G$ you get the infinitesimal action

[sleek thicket]

yeah?

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[sleek thicket]

!!!

[gritty widget]

!!!

great minds think alike

infinite dimensional manifolds

[sleek thicket]

I do want to learn that stuff

Especially because I liked banach spaces

hmm

Think

[gritty widget]

🤔

im reading the nlab article on momentum maps

[sleek thicket]

thinking about bundles

[gritty widget]

why wasnt i reading nlab before?

[sleek thicket]

[gritty widget]

some of these pages are so clear and concise

[sleek thicket]

lmao

[gritty widget]

uh oh

[sleek thicket]

Very excited for this tterra arc

[gritty widget]

is this not the general case

[sleek thicket]

Gonna get category pilled

[gritty widget]

tterra arc this

tterra arc that

plz

im just in a really slow category theory arc

[sleek thicket]

Let me know when you want to learn lie groupoid/foliation stuff

[gritty widget]

foliations

foliations are nice

frobenius' theorem was my favorite result from intro manifolds

[sleek thicket]

It's super cool

[sweet wing]

@quasi forum can jus ask here

tho i have some test in like 14 mins lulz

[sleek thicket]

I can try to answer

[quasi forum]

I got everything except showing the inverse (d' is the metric of the domain) is continuous.

I could not figure it out for the life of me

[sleek thicket]

inverse of the identity?

what map are you trying to show is continuous, I mean

[quasi forum]

Well, we have two different metrics. We need to show that the distance metrics are a continuous mapping on the identity function.

[sleek thicket]

sure

so are you stuck on showing that id : (X, d') -> (X, d) is continuous?

[quasi forum]

Yea, that is precisely the issue

[sleek thicket]

so what does this mean concretely

that's a good starting place

hello ari

[sweet wing]

herro

i mean there is kinda a direct way

[quasi forum]

So although this may be a weird way to go about it, this is how I imagine what's going on

[sweet wing]

||show for each open ball U in (X,d) and each point x you can find a smaller open ball V in (X,d') which contains x and is in U||

and vice versa like how you would prove 2 basis generate the same topology

[sleek thicket]

ari i am trying to get them to solve the problem

not you

[quasi forum]

So it is very trivial if $d(x, y)\leq 1$ However, I am struggling for when it may be greater than 1

[gentle osprey]

dackid

[sleek thicket]

so before that

[quasi forum]

Ari, can you delete that please? I don't want the answer spoiled

[sleek thicket]

what does it mean for this function to be continuous?

concretely

[quasi forum]

Like formally?

[sleek thicket]

yup

just, what's the definition

[sweet wing]

i just gave the definition in munkres for 2 basis to be the same but sure

[sleek thicket]

always a good place to start 😛

okay sure lol but getting into the habit of checking the definition and working with it is a skill one needs to develop

[quasi forum]

Okay sure:
$f:X\to X$ is continuous if $\forall x\in X$ and any $\epsilon>0,\exists \delta>0$ so that:
$\forall x': d'(x, x')<\delta\to d(x, x')<\epsilon.$

[gentle osprey]

dackid

[sleek thicket]

sure

so take an arbitrary x in X, yeah?

[quasi forum]

Sure yeah

[sleek thicket]

then for some unspecified x' and delta, what does it mean to have d'(x,x') < delta?

[quasi forum]

It must be less than or equal to 1

[sleek thicket]

sure

so what im thinking is like

we can do cases

and get something more specific

so either (1) d(x,x') <= 1 or (2) d(x, x') > 1

then d'(x,x') < delta iff
(1) ...
(2) ...

[quasi forum]

Okay, let's look at greater than. 1 is not hard, it writes itself

That is case 1

Case 2 is what confuses me

[sleek thicket]

sure

so lets back up

in case (1) we have d'(x, x') < delta iff d(x,x') < delta, right?

[quasi forum]

Yea

[sleek thicket]

and in case (2) we have d'(x, x') < delta iff 1 < delta

yeah?

and you can handle case (1)

you agree with all that?

[quasi forum]

Yea, 1 must be less than delta

I agree

Since d'(x, x')=1

[sleek thicket]

sure

so

I'm trying to think of how to proceed without spoilers

[quasi forum]

Why not choose $\delta=\frac{1}{\epsilon}$

[gentle osprey]

dackid

[quasi forum]

And then we just choose $\delta=\min{\epsilon,\frac{1}{\epsilon}}$

[gentle osprey]

dackid

[sleek thicket]

I think that would work, and it's very clever!

[quasi forum]

Which unless my logic is off, that covers both cases

[sleek thicket]

I was just thinking delta = min(varepsilon, 1)

[quasi forum]

Oh, well that works too 😆

Okay, let me write it out and see if it looks okay. Afterwards, I want to add one follow up question to this if that is okay

[sleek thicket]

np, go ahead!

[quasi forum]

Wasn't much to write out here 😆

[sleek thicket]

nope lol

[quasi forum]

I'd have to justify it, but that wasn't too hard either

[sleek thicket]

so my intuition for this problem is

qualitative observation: on sufficiently small scales, balls look the same in either metric

[quartz edge]

wah, delta = 1/epsilon?!

[sleek thicket]

and continuity only requires you to look at sufficiently small balls

[quasi forum]

You no like?

[quartz edge]

what dark magick have i just stumbled upon

[sleek thicket]

lol

i think it's simpler as delta = min(epsilon, 1)

[quartz edge]

choose epsilon arbitrarily small

[quasi forum]

You can't

[sleek thicket]

well, you can

[quartz edge]

well you only care about local convergence

[quasi forum]

Epsilon is not up to you

[sleek thicket]

this might be a good exercise

dackid, try to prove this:

[quartz edge]

an upper bound on epsilon is up to you

[quasi forum]

Hold on, can I ask the question first

[sleek thicket]

sure

[quasi forum]

So our class isn't going too much into the topological spaces. We've talked about it here and there, but metrics has been the main goal.

How would we generalize this to topological spaces?

[sleek thicket]

you can't really

this is very much a problem about metric spaces

and distance

ariana's thing at the start is close

the topologies have a common "basis"

so they're identical

[quasi forum]

Can we do a continuity problem of this nature in respect to topological spaces?

[sleek thicket]

the issue is figuring out what an analogue of d' would be

[quasi forum]

Well, doesn't have to be d' exactly

I just want a problem regarding topological equivalence

[sleek thicket]

well for topological spaces there aren't really other finer kinds of equivalence

like

yeah

[quasi forum]

I basically just want to get my feet wet

[sleek thicket]

sure, do you know what a basis for a topology on a set is?

[quasi forum]

Not exactly

[sleek thicket]

hmm

[quasi forum]

Is that the collection of open sets?

[sleek thicket]

no, it's sort of like a smaller set of opens which generate the whole thing

so say $(X, \tau)$ is a topological space

[gentle osprey]

EShamrock -> BShamrock

[quasi forum]

It has been said

[sleek thicket]

?

a collection $\mathscr{B} \subseteq \tau$ of open sets is called a basis if (1) for every $x \in X$ there is an open set $U \in \mathscr{B}$ with $x \in U$ and (2) for any $U, V \in \mathscr{B}$ and any $x \in U \cap V$ there is an element $W \in \mathscr{B}$ with $x \in W$ and $W \subseteq U \cap V$

[gentle osprey]

EShamrock -> BShamrock

[sleek thicket]

(1) says that the basis covers X

[quasi forum]

Okay, hold on. I need a minute to digest

[sleek thicket]

(2) says you can cover any intersection of basis elements with another basis element

[quartz edge]

can i ask what eshamrock -> bshamrock is a reference to

[sleek thicket]

hmm, this may be to complicated

do you know what a bundle is?

[quartz edge]

i know what a tangent bundle on a manifold is

[quasi forum]

Hold on. Let me work with this please

[sleek thicket]

more generally a bundle is like a space which is locally a product

dackid, gristle is asking me a question. feel free to ignore it

so like think about the mobius strip

[quartz edge]

yea

[sleek thicket]

this locally looks like a circle times a line segment

[quasi forum]

Oh sorry. Thought you were referring to me here

[sleek thicket]

yeah?

[quartz edge]

oh ur doing that weird thing where u take products of spaces

i havent encountered that yet

[sleek thicket]

ah

then the definition of a bundle is unlikely to make sense

[quartz edge]

i see

[sleek thicket]

it's something which is locally a product

if G is a topological group then we can find a very special bundle EG -> BG

[quartz edge]

i will be ready for your name in a few months

[sleek thicket]

it's somehow "universal" with respect to bundles "involving" G

good luck!

oh actually here's a cool example

take your favorite manifold M

you have a tangent bundle TM

do you know what a "frame" is?

[quartz edge]

depends

related to moving frames?

[sleek thicket]

probably not

[quartz edge]

smh

[sleek thicket]

I don't know any physics if it's about that 😅

[quartz edge]

u high topologists and your advanced educations

wel

l

for example the frenet apparatus

would involve a moving frame

[quasi forum]

Okay, so how does this differentiate from the basic collection of open sets?

[sleek thicket]

oh look at the time, i need to finish my french homework by midnight

[quartz edge]

au revoir

[sleek thicket]

i have been threatened with an explanation of physics

[quartz edge]

it's just canonical bases for curves

[sleek thicket]

oh wait i looked it up

this isn't so bad

yeah this is exactly what i'm talking about

[quartz edge]

ah ok

[sleek thicket]

a frame is a smoothly varying basis

[ivory dragon]

easy example: one basis of the standard topology on ℝ is the collection of open intervals (a, b)

[quartz edge]

but for a manifold

M

yea

[sleek thicket]

right

not just a 1 dim manifold

[quartz edge]

right

[ivory dragon]

this differs from the collection of all open sets

[sleek thicket]

more precisely, it's a basis for the tangent space

[ivory dragon]

since, say, (0, 1) U (2, 3) is in this collection

[quartz edge]

ok T_P M

[ivory dragon]

but it isnt an open interval of the form (a, b)

[sleek thicket]

right

[quartz edge]

or TM rather

[sleek thicket]

so I'm gonna define a new space for you

[quasi forum]

Since $(0,\infty)$ does not cover $\mathbb{R}$

Would that be a workable example?

[sleek thicket]

let F(TM) be the set of all pairs of a point p in M and a basis for the tangent space T_p M of M at p

[gentle osprey]

dackid

[ivory dragon]

...an example of?

[quasi forum]

A non-basis

[quartz edge]

ok

[ivory dragon]

its not a basis for many reasons

[sleek thicket]

we can give this a topology

[ivory dragon]

for one, its a set, not a collection of sets

[sleek thicket]

if (U, F) is a chart on M then we have this subset F(U) of F(TM)

[ivory dragon]

but yes, your objection is another reason the set {(0, infinity)} is not a basis for the euclidean topology

[sleek thicket]

we're going to declare sets of this form to be a basis for our topology

[quartz edge]

oh jesus christ

i need to draw a picture

[sleek thicket]

hahaha

[quartz edge]

brb keep going

[sleek thicket]

okay so

what is this nonsense

well

here is my secret trick

let M = S^2

[quartz edge]

😮

[sleek thicket]

nobody can ever tell you otherwise!

ask them to think of another space

they cannot

[quartz edge]

well

[sleek thicket]

so just pretend

[quasi forum]

Hmm, okay.

[sleek thicket]

s p h e r e

[quartz edge]

fuckin good enough

[sleek thicket]

right??

so

coordinate charts

idk fucking

hemispheres

yeah?

good enough charts

[quartz edge]

sure

[sleek thicket]

so what happens on a hemisphere

well we have coordinates

just project down

yeah?

[quartz edge]

yea sure

[sleek thicket]

so we have a canonical basis

at each point

[quartz edge]

im feeling quizzical

[quasi forum]

@ivory dragon so were you trying to say $(0,1)\cup(2,3)$ is not a basis?

[gentle osprey]

dackid

[quartz edge]

about that

[quasi forum]

Because I agree there

[sleek thicket]

well

think about the disk

down below

what's the tangent space to the disk?

this is not a trick question

[quartz edge]

its the disk

[sleek thicket]

hmm, not exactly

sorry I mean like the tangent space at a point

[quasi forum]

Correct me if I am wrong, but this sounds a lot like compactness

[ivory dragon]

okay we need to go back a bit

[quartz edge]

its just

[ivory dragon]

a basis isnt an individual set

a basis is a subset of your set of open sets

[quartz edge]

a local linearization of the disk at that point

[sleek thicket]

sure

it's a plane yeah?

[quartz edge]

yep

[ivory dragon]

i gave ONE example of a basis for the euclidean topology on ℝ

[sleek thicket]

spanned by the vectors like

(1,0,0) and (0,1,0)

[ivory dragon]

the collection of open intervals of the form (a, b)

(0, 1) U (2, 3) is not in this particular basis

[sleek thicket]

similarly if you're living up on the sphere

[quasi forum]

Yea I agree. What is a non-example? I think that would help

[sleek thicket]

in this little hemisphere

[quartz edge]

its another plane

[sleek thicket]

you can draw arrows pointing forward and to the left

sure it's a plane but like

[quartz edge]

its rotated

[sleek thicket]

we can write down a basis

on this chart

[quartz edge]

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[quasi forum]

Wait sorry, why is that?

[sleek thicket]

something like uhh

[ivory dragon]

since it isnt an open interval of the form (a, b)

[quasi forum]

Ohhhh!

[ivory dragon]

theres no way to write it as (a, b)

[sleek thicket]

take the inclusion of the disk onto the sphere D -> R^3

[ivory dragon]

the point is that

[sleek thicket]

take the jacobian

[ivory dragon]

if we take a basis for a topology

and allow arbitrary unions

[sleek thicket]

the columns of the jacobian will be a basis for the tangent space

[ivory dragon]

we get all open sets in the topological space

[quartz edge]

ok yes

[sleek thicket]

point is

[ivory dragon]

so while not every open set is in the basis

every open set can be written as a UNION of the basis

[sleek thicket]

all the other bases are now simple

take your canonical basis (v,w) at p

[ivory dragon]

(0, 1) U (2, 3) can be written as the union of basis elements (0, 1) and (2, 3)

[quasi forum]

Ohhh, I see

[sleek thicket]

then other bases are just (Av, Aw) for a linear transformation A in GL(2)

[ivory dragon]

(0, infinity) can be written as the union of basis elements (0, 1) U (0, 2) U (0, 3) U (0, 4) U (0, 5) U ...

[quartz edge]

gl(2)

[sleek thicket]

sorry not quite

[ivory dragon]

[note that we're allowing infinite unions]

[quartz edge]

linear group 2?

[sleek thicket]

yup!

[quartz edge]

havent learned it, googling

[quasi forum]

Which is still technically in the basis

[quartz edge]

ive taken no algebra yet

[sleek thicket]

if A = {{a,b}, {c,d}} then we get a new basis (av + bw, cv + dw)

[quartz edge]

yea agreed

[sleek thicket]

it's not quite matrix vector multiplication

its like we multiplied the whole basis by a matrix

anyways

these are what other bases look like

@quasi forum you are not entitled to either my help or this channel

another discussion started when you stopped posting

I stopped responding since namington seemed to be helping

[quartz edge]

where tf are we going on this wild ride

[sleek thicket]

sorry haha we've gone kind of far

so the point is

if you have a chart U for M

(with some fixed coordinates)

T(U) is easy to describe

[quartz edge]

very

[sleek thicket]

it's just a point in U and a matrix

so topologically

F(U) is homeomorphic to U x GL(2)

[quartz edge]

this smells a lot like a bastardized e-math version of what ive been doing in differential geometry lately

[sleek thicket]

lol

e-math?

anyways

poitn is

[quartz edge]

yea ppl that know shit i dont know from far off lands

[sleek thicket]

F(M) looks nice locally

[quartz edge]

[quasi forum]

Well, thanks I guess. I got some help at least.

[sleek thicket]

that's me

[quartz edge]

ye it do

[sleek thicket]

dackid, I helped with your original problem and then continued to talk to you

you asked me to come up with a problem off the cuff

I thought of something and then you stopped responding for 4 minutes

[quasi forum]

It's okay. I'm moving on

[sleek thicket]

so @quartz edge F(M) is sort of weird and abstract globally

[quasi forum]

Like I said, I got the help I initially wanted

[sleek thicket]

but in coordinates it's nice

this is called "differential topology"

sometimes it's the other way around though

so anyways point is

we have this group GL(2)

and we have this space which is locally like a product of GL(2) and our manifold

[quartz edge]

F takes a tangent bundle on M and maps that to tangent basis "bundles"

[sleek thicket]

right!

exactly!!!

[quartz edge]

ok ok

[gritty widget]

need principal fibre bundles

[sleek thicket]

it's called the frame bundle

(of the tangent bundle)

[quartz edge]

aww sheeit

[sleek thicket]

so things like this

spaces whihc are locally a product U x G and have a "nice" action of G on them globally

are called "principle bundles"

the action in our case is that thing with like (av + bw, cv + dw)

[quartz edge]

im not gonna lie this is dope and i wanna run through a text on things like this

[sleek thicket]

hahahaha

I've been doing this stuff this quarter

[coral gale]

talking about bundles does this proof look right?

[sleek thicket]

so the kicker is

[quartz edge]

is that red diff topo book good?

[sleek thicket]

EG -> BG is sort of a "universal principal bundle"

it's a principle bundle

[quartz edge]

guillemin pollack

hmm

interesting

in what way is it universal

[sleek thicket]

and another other principal bundle on say M comes from pulling back along a "unique" continuous map M -> BG

[quartz edge]

agh pullbacks

im almost there

[sleek thicket]

"unique" is in scare quotes because it's only unique up to homotopy

haha

pullback is a very overused word

it means a lot of things

[quartz edge]

saw it in bachman, didnt finish yet

[sleek thicket]

i haven't read it, sorry

[quartz edge]

got u

[gritty widget]

best book on fibre bundles is steenrod

[shut moat]

bachman A geometric approach to diff forms?

[quartz edge]

yes

i need to finish tu and go on to bott tu

[shut moat]

seems cool ty

[quartz edge]

but people keep yelling at me to read lee instead

[gritty widget]

tu is a watered down version of lee

[quartz edge]

so they do say

[sleek thicket]

Lee is good

The bundles book I've been using is a Lee book but it's not public yet

[quartz edge]

i might even nab lee for riemannian instead of do carmo though i already have the latter

[sleek thicket]

and I'm not sure I'd recommend it

[quartz edge]

do carmo's RG is just like

hard as fuck to read lol

[sleek thicket]

lol my brain saw "nab lee" as "nlab" and got very concerned

[quartz edge]

LOL

same, somehow

[gritty widget]

do carmo is a waste of time

[sleek thicket]

I liked IRM

[quartz edge]

yeah?

like

[quasi forum]

Good luck. They've been on this roller coaster ride for a hot minute. Enjoy the ride!

[quartz edge]

i considered do carmo for the euclidean diff geo text but

went with shifrin instead and it is way nicer

despite being lesser known

i guess maybe the recommendations i got for do carmo must have come from students using him as course texts

i want off mr bones wild ride

[coral gale]

well to be fair the conversation started before i asked my question

[sleek thicket]

Weren't you just complaining about people interrupting you dackid?

Sorry yeti, I really do have homework I've been avoiding

[coral gale]

and nobody is obligated to answer anybody's questions

[sleek thicket]

Otherwise I would look at it

[coral gale]

no no problem lmao

[quartz edge]

do ur homework now

NOW

[quasi forum]

I'm certainly not interrupting anything rn. I was just stating facts.
And yes, I was frustrated, but frustration doesn't get anywhere inthe longterm, so it's all good.

I do also plan to ask for help in the future, so it would be dumb to be butthurt and break those ties over something silly.

In case it was not clear, I do appreciate the help you have given me. 😁

[quasi forum]

Man I really don't like when channels are left in a state like this...

[viral atlas]

I suppose you could help me out, then.

Is the unit interval [0,1] homeomorphic to R?

[gritty widget]

if we remove 0 or 1 from [0,1] we get a connected space

if we remove anything from R we get a disconnected space

@viral atlas

[viral atlas]

Aahhhh

Thanks!

[quasi forum]

So does the fact that connectedness isn't preserved mean it is not a homeomorphism?

[shut moat]

yeah- connectedness is a topological property

[quasi forum]

Oh wow, that's a very neat consequence

[shut moat]

yeah it's pretty slick

proving that two spaces aren't homeomorphic is much harder than proving that they are, I think, so it's v nice to have extra topological properties like these which are easier to check

[quasi forum]

So if I may, what is connectedness exactly?

I know that'll come up later in the semster. But ya know, it isn't later yet :p

[shut moat]

a topological space X is disconnected if there exist two disjoint open sets who's union is X. And it's connected if it's not disconnected

[median glade]

Another version:
A space is connected iff every continuous function to {0, 1} is constant

intuitively: there's no way to paint the space with 2 different colors continuously

[quasi forum]

Oh, so as soon as you remove a point a from R, you have the intervals $(-\infty,a)$ and $(a, \infty)$ which is definitely disjoint

So that's why you say it's disconnected

[gentle osprey]

dackid

[gritty widget]

a space X is connected iff the only clopen sets are X and \varnothing

best definition

[quasi forum]

Hmm okay. I think I understand

[viral atlas]

Why does a circle stay connected if we remove a point from it?

[quasi forum]

So a common question from real analysis is to show that $(a, b)≈\mathbb{R}$. So can this idea of connectedness be enough to show that's true for any a and b?

[gentle osprey]

dackid

[viral atlas]

You'll probably need an explicit bijection?

Pugh explains it very neatly with a picture

[quasi forum]

Since normally it can be tough to find a bijection between each interval

?

[viral atlas]

Map the points on (a,b) to a semicircle with radius |b-a|, centred at (a+b)/2 through vertical lines from (a,b). Then create projections from the centre of the semicircle to the entire line.

[sleek thicket]

The circle minus a point is an interval

[viral atlas]

[sleek thicket]

Think about parameterizing via sines and cosines

Or stereographic projection

[viral atlas]

Aaahhhh

Okay

That's genius hahaha

[quasi forum]

Alright. This radial projection thing is interesting, but a bit above my understanding. Thank you for the help

[viral atlas]

It's pretty easy

Let me share the picture

[median glade]

Or you define a circle as an interval with endpoints identified

[quasi forum]

It's also 3 am, which has a very significant factor 😆

[sleek thicket]

I agree but I also think that's not a great thing to say after someone says they're confused

it can be a little discouraging

[viral atlas]

Sorry, I didn't mean it that way

[sleek thicket]

It's okay, I'm not a mod or anything

[viral atlas]

I'm pretty sure I wouldn't have understood it the way I explained it either

The picture is pivotal

That's what I meant

[quasi forum]

The picture intrigues me

[viral atlas]

So you see

[sleek thicket]

good picture

[quasi forum]

Oh, that is not at all where my mind was going

[sleek thicket]

Project the interval onto a semicircle

Project the semicircle onto R

You get a nice rational function

[viral atlas]

You can map points on the interval to points on the semicircle through those vertical lines

Then take lines from centre of the semicircle to R

[quasi forum]

So if I may, how exactly are we showing that the interval can map to the semicircle?

[viral atlas]

They map points on semicircle to points on the entire real line

[gritty widget]

you use invarients to prove things aren't homeomorphic

[quasi forum]

I definitely get the intuition

[viral atlas]

I'm just happy with the intuition, probably someone else could give the formal idea here.

[quasi forum]

Ohhh okay, since $-1\leq \cos(\theta)\leq 1$

[gentle osprey]

dackid

[sleek thicket]

I generally use the strongest invariant (the homeomorphism class). I'm built different tho

[quasi forum]

No, I mean that is true regardless of what theta you choose

That is the bounds of cosine

Oh, I'm used to thinking about angles in theta :p

That is all

[coral gale]

[quasi forum]

You said we can map (a, b) to (-1,1) and the way we can do that is to let f(x)=cos(x)

[gritty widget]

take $\tan:(\pi/2,\pi/2)\to\mathbb{R}$

[gentle osprey]

spinsicle

[sleek thicket]

Spin, please don't introduce this example. It's going to complicate the other bijection being defined

In terms of projection

The thing up above has nothing to do with angles or trigonometry

[quasi forum]

Isn't that the example that shows (-1,1) is uncountable

[sleek thicket]

That is one bijection

[sleek thicket]

But the one here is different

You don't use any sines or cosines or anything

[quasi forum]

Okay, I guess I am not following

[sleek thicket]

Sure, so there's two steps

Let S be the semicircle

S = {(x, y) : y < 0 and x^2 + y^2 = 1}

[quasi forum]

Okay, I am good with that.

[sleek thicket]

we have a map (a, b) -> S and a map S -> (0,1)

The second map is easier to write down

It's the vertical lines in the picture

Just project out the x coordinate

So so yeah if p = (x, y) has angle θ we'll send p to cos(θ)

But imo that's more complicated then just f(x, y) = x

[quasi forum]

Okay fair, that was just the first thing I thought about

[sleek thicket]

Sorry, I didn't understand what you were saying. It makes more sense now

but the first map is trickier

[quasi forum]

I wasn't thinking in the world of projections. It's a different mindset than I'm used to

[sleek thicket]

The second map is like this

So the second map is S -> R

Oh right it's also important to note the thing about cosine would be about the the map S -> (-1,1) (but it is better to think about it as just giving you the x coordinate)

Not between that and (a, b)

[quasi forum]

Hmm, okay

[sleek thicket]

(-1,1) and (a, b) are homeomorphic for simpler reasons

Just translate and scale

so the really tricky bit about all this is the connection between S and R

S and (-1,1) is just (x, y) |-> x

And the intervals are just similar shapes

But S and R is the clever bit

[quasi forum]

What does that notation mean

The line with the arrow

[sleek thicket]

Ah sorry

It's the function which takes (x, y) to x

[quasi forum]

Ah okay! Gotcha

[sleek thicket]

What do you mean mirza?

Oh i missed the word rigorous lol

I think so

So here's what the picture is depicting dackid

Take a point on R

Actually sorry one thing

We're going to think of R as sitting below S

[quasi forum]

Okay, I can do that

[sleek thicket]

So like R × {-1}

sure

So take a point on that line

[quasi forum]

So hold on.

[sleek thicket]

mhm

[quasi forum]

Why the direct product?

Not sure where that came in

[sleek thicket]

It's the set of points (t,-1) with t in R

[quasi forum]

Oh, silly me. Alright, go ahead

[sleek thicket]

No worries

so we have this line

Sitting tangent to the semicircle

Given a point on the line we can connect it to the origin

By which I mean

Draw the line segment between them

this intersects the semicircle at exactly one point

we'll send the point on the line to that intersection point

Conversely, given a point on the semicircle you can connect it to the origin and look at the intersection with the line

and this gives an inverse map

[quasi forum]

So we are referring to drawing a point from an element in R to the origin?

[sleek thicket]

yup

[quasi forum]

Okay, I follow

[sleek thicket]

That's the slanted lines in this pick

That all meet at the origin

This is a bijection between the semicircle and R. If you write it all out and figure out what the intersection points are you get some rational function (so in particular everything is continuous)

[quasi forum]

Oh, that does give you an inverse map indeed. And therefore f is a bijection.

[sleek thicket]

This is really cool imo because there's no need for trig!

[quasi forum]

Projections are a really neat tool. They take some time to get used to, but they are quite nifty

[sleek thicket]

Yup! This kind of projection away from a point is very useful in algebraic geometry

[quasi forum]

What is algebraic geometry exactly?

[sleek thicket]

At its heart it studies shapes that are defined by some number of polynomial equations

So like conic seconds are some examples of shapes like this

Or the curve y^2 = x^3

It uses the tools of abstract algebra to analyze these shapes

It is a very complicated field of math that has branched off a lot in many different directions though

[quasi forum]

So how does algebraic geometry look at shapes like that? Moreover, what does it do that elementary algebra/geometry cannot?

At least for y^2=x^3 I mean.

Are abstract algebra and algebraic topology related? Ya know, apart from the word algebra?

🤯

[bitter yoke]

Maybe one explanation is that algebraic geometry associates a ring of functions to such shapes. So for example, you have continuous functions from R to R, and more generally, on a manifold, you have a (vector space) of functions from your manifold to R. It turns out that you can use this vector space to study your manifold. Algebraic geometry tries to do a similar thing with "algebraic" shapes by giving them a ring of algebraic (polynomial) functions

[quasi forum]

I barely understood what you said Zopherus. But hey, I did ask and I appreciate the involved answer 😁

If there is anything I've learned today, it's that I need to spend some more time in these channels. You guys talk about some cool stuff.

[bitter yoke]

What exactly don't you understand? I could maybe explain a bit more in detail

[quasi forum]

So you're gonna have to break it down a bit.
I do not know what either rings or manifolds are.

[bitter yoke]

Just think about things like toruses or spheres for example

We can consider continuous functions from the sphere to R, and use this set of continuous functions to study in the sphere in certain ways

[quasi forum]

Okay yeah. I follow

What do you mean by locally?

So is it actually the addition and multiplication operator? Or is it some other binary operator that just has those names?

[bitter yoke]

Basically, algebraic geometry tries to do the same thing, but instead of smooth things like spheres or toruses, we have to study algebraic sets like y^2 = x^3

So we replace continuous function with certain "algebraic" functions

[quasi forum]

What is so special about this y^2=x^3 🤔

[bitter yoke]

nothing really

[quasi forum]

Scalar multiplication and addition right?

[bitter yoke]

the easiest example that's not smooth I guess

[quasi forum]

Oh really!? Well, that does make sense, since multiplying a matrix by a matrix gives back another matrix

So you said it had to be monoid under multiplication. What does that mean exactly?

Oh okay. But matrix multiplication has an inverse yeah?

True true

So invertibility is not a requirement

Oh, well that makes sense. Invertibility is in the name.

[sacred gust]

wait mirza, are you still going through D&F or have you switched to lang

[sleek thicket]

It's a very very useful counterexample

[sacred gust]

[sleek thicket]

Useful to keep in your back pocket when doing ring theory stuff

Sure, in that they both serve as counterexamples to things

[quasi forum]

Dually noted. Well, it's 4 am here and I still haven't slept yet. I'd like to continue, but my body does not want to. Thank you for all the insight.

[sleek thicket]

Here are my top three counterexamples in ring theory:

k[x, y]/(y^2 - x^3)

k[x1, x2,...]

an infinite product of F2's

These rings are very bad it rules

No! And this is a very important distinction

Power series can have infinitely many terms

We're looking at polynomials which can only have finitely many terms individually

But the set of variables they can draw from is infinite

So like you can have x1, x1+x2, x1+x2+x3,...

Power series rings are denoted k[[x]]

That's for one variable

or like k[[x, y, z]]

Hmm, I'm not sure what you have in mind

lol

It might

¯\_(ツ)_/¯

AoC is important in ring theory

Specifically zorn's lemma

[sleek thicket]

Actually if you're looking for set theory this ring is closer

Err

[sleek thicket]

This one

By infinite I meant countably infinite

[viral atlas]

Shamrock talk when

[sleek thicket]

The structure of this ring is deeply related to the stone cech compactification/set of ultrafilters on N

[sleek thicket]

Sham talk on Wednesday

But it's for a class

[viral atlas]

Oh, goodluck!

[sleek thicket]

I'm not gonna hand out the zoom code

Thanks!

[viral atlas]

Hahaha, understandable. Hope you can give a talk here when you have the time.

[sleek thicket]

Yeah, I was going to this summer but I had a covid scare and freaked out

couldn't focus on prepping or giving the talk

[viral atlas]

:(

[sleek thicket]

It turned out okay!

Just was scary

[viral atlas]

That's nice.

[viral atlas]

It must've been, I guess. Had you given any talks before?

[sleek thicket]

Nope

This is going to be my first one

But I've worked as a TA a lot and there's some overlap

[viral atlas]

Aaah, that makes sense.

[sleek thicket]

just the information density/goals are a little different

[viral atlas]

Fair. What's your audience this time?

[sleek thicket]

Grad homological algebra course

Other students in the class are also giving talks

[viral atlas]

[sleek thicket]

it's not so intimidating, I've seen them give talks over the past couple days :P

It feels like a very safe environment

[viral atlas]

That's nice.

[sleek thicket]

I just hope people walk away with something

[viral atlas]

I'm sure they will!

Goodluck, once again.

[sleek thicket]

Ty! I'm going to sleep now so I can wake up for my classmates' presentation on frobenius categories 💤

[viral atlas]

Hahaha, yeah, it must be super late for you rn.

Goodnight!

[gritty widget]

if $(M, \omega)$ is a symplectic manifold and $J$ is an almost complex structure which is compatible, i.e. such that $g(X,Y) := \omega(X,J(Y))$ is a riemannian metric, is there anything interesting we can say about the levi-civita connection on the riemannian manifold $(M, g)$?

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

would play around with this but a lecture's starting, so putting this here in case anyone has any comments

only thing i can think of doing off the top of my head is writing the condition that $\nabla$ be metric, i.e. $$X g(Y,Z) = g(\nabla_XY,Z) + g(Y,\nabla_X Z)$$ in terms of $\omega, J$

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[quasi forum]

So question: can a topological space have a finite number of elements in it?

[sleek thicket]

yup

One interesting example is the sierpinski space S = {0,1} with the topology where {0} is open but {1} is not

[quasi forum]

Hmm, okay. So follow up question. How would we show in a finite space X that any finite subset is closed?

[cloud owl]

watch this: {{}, {X}}

[sleek thicket]

This is untrue

[cloud owl]

boom, i'm a regular topological

[marsh forge]

another side note

Peter May has an entire book about the theory of finite topological spaces

it's pretty neat

not particularly useful

but neat nonetheless

[shut moat]

@ terra you could probably ask hankel in the physics server, he knows a bunch about symplectic geo (he's an insufferable asshole though)

[quasi forum]

Is it true if you have a finite metric space?

[gritty widget]

my symplectic geometry prof is having office hours after this lecture so i might just ask her

but thanks

[sleek thicket]

yes, but finite metric spaces are way less interesting than finite topological spaces

They're all discrete