point-set-topology

[vocal wharf]

anyways, i think its fine the only thing i think is weird is "since G_1 is open it must be an interior point" because to me it sounds like G_1 is an interior point (of something)

[shadow charm]

oh yeah i guess the wording could be better

but i meant 0 is an interior point of G0

[vocal wharf]

i would also write down the indexing set

[shadow charm]

im not sure what you mean

[vocal wharf]

consider an open cover ${G_i}_{i \in I}$

[gentle osprey]

Lochverstärker

[vocal wharf]

(i wouldnt use n as an index since to me it implies countable)

but this is just nitpicking at this point, the proof is fine

[shadow charm]

ah okay

i see what you mean

that's fair

[vocal wharf]

also the first sentence should be "Consider an open cover {G_n} of K"

[shadow charm]

i was mainly hesittant on the step where i say that G0 contains infinitely many elements of K so that finitely many are not in G0

right

[vocal wharf]

its fine if you know that 1/n converges to 0

[tough imp]

Well

Just because G0 contains infinitely many elements of K doesn’t mean that finitely many aren’t in G0

You can chop an infinite set up into two disjoint infinite subsets

What you really need to know is that for some N, and for all n > N, that 1/n is in G0

Then you can only possibly have 1/1,1/2,...,1/N not being in G0

[shadow charm]

yeah thanks, i discussed it with rokabe and he helped me polish out the problems

[red garden]

how can i talk about the figure 8

in topology terms

[gritty widget]

what do you mean "in topology terms"

are you looking for some kind of way to describe the figure 8 that isn't just "funny loop over itself?"

like a parametrization?

$\gamma \colon [-\pi,\pi]\to\bR^2$ given by $$\gamma(t)=(2\cos t,\sin 2t)$$ works

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

the image of this curve is a figure eight

[pastel linden]

its fundamental group is the free group on two letters iirc

Is that what you want?

[gritty widget]

sounds right

that moment when you recover a space from its fundamental group

[marsh forge]

that does not classify the figure eight

You can classify it up to homotopy as a K(F_2,1) though

@red garden it is also S^1 v S^1

[frigid patrol]

@gritty widget do you know about moment maps in symplectic geometry or equivariant morse theory?

[gritty widget]

i know like

one or two facts about moment maps in symplectic geometry

nothing about equivariant morse theory though

[silent folio]

Maybe I'm having a dumb moment but in the context of de Rham Cohomologies, $[H^\bullet _{dR} (M)]$ has a graded ring structure defined by

[gentle osprey]

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

[silent folio]

has a graded ring structure defined by $[[\omega_1]\cdot [\omega_2] \stackrel{def}{=}[\omega_1 \wedge \omega_2]]$ why is this well defined?

[gentle osprey]

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

[sleek thicket]

it sure does

Oh I missed the question lol

So to show it's well defined you need two things

(1) if you wedge two closed forms you get a closed form

(2) if you wedge a closed form with an exact form you get an exact form

Both should follow from the product rule d(ω ^ η) = dω ^ η ± ω ^ dη

[silent folio]

oooo

[sleek thicket]

The sign here is 1 if ω is an even degree form and -1 if ω is odd degree

Iirc

See if you can derive this!

Err, prove my conditions (1) and (2) from the product rule

Deriving the product rule is just annoying lol

[silent folio]

thank you sir shamrock

[sleek thicket]

Woog should knight me

I want to officially be Sir Shamrock

[silent folio]

maybe I shouldn't just sleep at this point but why are we using boundaries of k-cubes to integrate forms?

oh wait

Oh I think I understand it, is it related to the fact a 0-cube is a just a point, and then a 1-cube is a curve

then we can just pull-back the form along the image of the k-cube

[astral cedar]

I’m not really sure, but I’d say it’s mostly because it’s convenient

You could use simplices and it would still work

[sleek thicket]

I don't think what you're saying is standard

I just integrate forms along submanifolds ¯\_(ツ)_/¯

Is this for defining the exterior derivative? Iirc Arnold's mathematical methods(?) book defines the exterior derivative by using a kind of infinitesimal stokes theorem where you integrate along cubes

[astral cedar]

If I recall correctly rudin does this as well

Not that I would recommend anyone to study Stokes from Rudin

[shut moat]

yeah arnold's def is a limit of an integral over a parallelogram

$d\varphi(\mathbf{v}1, ..., \mathbf{v}{k+1}) = \lim_{h\to 0}\frac{1}{h^{k+1}}\int_{\partial P(\mathbf{v}1, ..., \mathbf{v}{k+1})}\varphi$

[gentle osprey]

~S^1

[shut moat]

which is p nice and suggestive of stoke's

although it caused hubbard to push the proof of exterior derivative identities to the appendix (hubbard's treatment was inspired by arnold)

[gritty widget]

[shut moat]

speaking of forms

is there any geometric intuition to be had for 2-forms in R^4?

[gentle osprey]

~S^1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[shut moat]

the phi here denotes that integrating it over a surface (with constant time) gives the flux of the magnetic field through the surface

[gentle osprey]

~S^1

[shut moat]

How do you think of this term, geometrically?

[cloud owl]

i might be waaaay off the mark here but it's just a tiny piece of the surface, right? a tangent plane?

[shut moat]

ig that's how you think of integrating most 2-forms?

but I'm trying to understand this in particular

like for instance if you have two vector fields v and w in $\mathbb{R}^3$, $W_{v}\wedge W_{w} = \Phi_{v\times w}$

[gentle osprey]

~S^1

[shut moat]

which is nice and easy to think of

[sleek thicket]

Re:Arnold. I found that a really cool intuition for what the exterior derivative is but I never worked out the details

It does justify me in thinking "the exterior derivative is the thing that makes stokes work" though

[tight agate]

I think Dami mentioned this at some point

[cedar pebble]

yea this is a good way to think about things

[tight agate]

the exterior derivative is the unique operator from k-forms to k+1-forms

[cedar pebble]

it's literally the only way I can remember the vector calculus integral theorems

[tight agate]

which commutes with pullbacks

[sleek thicket]

this is true. I was thinking about the fact that you can actually define it by a version of stokes

The thing approx posted above

[shut moat]

(shit there's a typo there mb lol, put an h in the arguments of P in the integrand)

[sleek thicket]

uniqueness is good but that formula is a little more concrete/actually defines the thing

[cold vine]

Does anyone know of a great place to read more about Van Kampen; I'm somewhat confused although we just covered it. I don't think our material covers it the best.

[sweet wing]

the true way to van kampen is via groupoids

[sleek thicket]

eh

I don't really agree

[sweet wing]

i find treating it via groupoid then specializing to π1 nicer tbh

[tight agate]

[sleek thicket]

Pushouts of groupoids are harder to compute and if you want to actually talk about multiple basepoints and still get useful info about π1 you need to do awkward things

like the computation of π1(S^1) with groupoids

you need to somehow take one basepoint in the whole space but two in the intersection

It's doable but kind of unclear

I think it's probably better to just see it with groups first

[marsh forge]

im sorry but i just cant believe anyone genuinely thinks this lol

[sweet wing]

rip

maybe my path in learning alg top is just messed up

[marsh forge]

i mean like

i dont understand what the advantage is lol

at all really

the pi1 case is easier to computer, more geometric, simpler to state, and requires less machinery

and honestly 99% of the time more useful

[little hemlock]

back to my question from the other day: $G$ is a finite group acting continuously and freely on a topological manifold $X$. The goal is to show that $X/G$ is a topological manifold, and right now I am trying to show that $X/G$ is locally euclidean. So far, I only have some intuitive guesses how charts may arise on $X/G$: if $(\varphi, U)$ is a chart in $X$ then I'm guessing I need to somehow define a chart $\varphi' : \pi(U) \to \bR^n$ which somehow in a well defined way uses $\varphi$ to map orbits to points in $\bR^n$, but i haven't had any luck so far.

[gentle osprey]

kxrider

[little hemlock]

I haven't needed freeness of the action yet, so I'm guessing that should come up somewhere

(pi : X to X/G is the quotient map, and we know from earlier that its open, so pi(U) is open)

[astral cedar]

https://math.stackexchange.com/questions/3172278/finite-group-action-on-hausdorff-space

This might help?

Once you have that the action is properly discontinuous, it should not be hard to show that the quotient is locally euclidean

[marsh forge]

^ what they said

you just need to show that you can choose any one of many disjoint open neighborhoods for lifting a small neighborhood of x in X/G

[little hemlock]

ugh, still kind of struggling with proving proper discontinuity. I just don't see any way to remove all of the possibly infinitely many points which are fixed inside of an open set by the action of G

[tight agate]

shrink the nbhd element by element

say you have an open set U

look at U and g_1U

if they intersect, make U smaller

hmm

wait a sec

yeah just do it for each element

and intersect them

it will still be open as it's a finite intersection

ignore what I wrote above, lemme rewrite it in a cleaner way

say we have a point p

for each g in G, there are disjoint open sets Up, Ugp separating p and gp

[little hemlock]

ohhh i think i see where this is going

[tight agate]

choose a nbhd of p inside Up such that g maps the nbhd into Ugp

(which you can as the function is cts)

now do this for all g

and intersect them

yeah that should do it

[little hemlock]

ok im kind of confused. What does intersecting them do here? If U is their intersection, wouldn't that just mean gU \subset Ugp for all g in G?

oh wait

gU \subset Ugp and Ugp is disjoint from the original Up which contains the intersection U.

[tight agate]

there is no U

let Up and Ugp be disjoint nbhds separating p and gp

[little hemlock]

U is just what i called the final intersection

[tight agate]

there is a nbhd Vg of p s.t. g Vg \subset Ugp

intersect all the Vg

intersect the intersection with Up

[little hemlock]

and then call that intersection "U."
U \subset Up and gU \subset Ugp for each g but Ugp is disjoint from Up and therefore gU is disjoint from U, right?

[tight agate]

yup

[little hemlock]

amazing. hopefully i've got it from here lol. thanks!

[tight agate]

[viral atlas]

Can someone explain why this function is discontinuous at x_0 by the topological definition of continuity?

[gritty widget]

topological definition of continuity being that preimages of opens are open?

[viral atlas]

I get that the open set (f(a),f(b)) has an open set (a,b) as its pre-image, implying continuity over that interval.

Yes, Terra.

[gritty widget]

is f(x_0) the part on the left or the right of the jump

i wanna draw a picture but i need to know that

i'm gonna show you an interval whose preimage isn't open

[viral atlas]

Let it be on the left.

[gritty widget]

[viral atlas]

Hmmmmmmmm

So having it on the left, the open interval ending on the left bit and starting on the right bit will both leave out x_0 in their pre-images?

[tight agate]

the open interval is f(x_0) +- epsilon

ye

[gritty widget]

so the picture is a little wonky

but

on the vertical axis i've shown you an interval I around f(x_0), yeah?

the preimage is just the stuff that gets mapped into there

but the preimage of this guy isn't an open set!

(this is not a rigorous argument, but i think you can come up with one from it )

[tight agate]

wdym

this is rigorous

[viral atlas]

Dumb question, but why is it not an open set? If I take the union of (b,x_0] and (x_0,c), I do get an open set?

[gritty widget]

sorry brofibration my analysis TA says rigorous = including absolutely every single step in mathematical symbols

[gritty widget]

no open set around x_0 is contained in it, despite it containing x_0 (concisely, x_0 is not an interior point)

since it's half-closed

(think about why)

[viral atlas]

-1 for not stating the Archimedean property

[tight agate]

you should turn in homework

[viral atlas]

Ohhh, that makes sense!

[tight agate]

with everything written in symbols

no words

[gritty widget]

lol

[tight agate]

just a wall of quantifiers

[gritty widget]

i'm still malding about that connectedness question

[little hemlock]

oh i think i saw that. I literally couldn't comprehend the criticisms lmao

[gritty widget]

every time i work on complex analysis or symplectic geometry, i think "yeah, that analysis class means nothing"

[tight agate]

how many points did you lose tho

[gritty widget]

1

[tight agate]

[gritty widget]

which actually makes me even angrier

[tight agate]

you're malding over 1 point?

[gritty widget]

yes.

[tight agate]

is it out of 2?

[gritty widget]

no

10

for the whole question

[tight agate]

[dim meadow]

[gritty widget]

i'm malding because it's such a stupid fucking thing to lose a mark on

it's literally nothing

but he still did it

he could have just posted an ascii middle finger for the feedback and it'd be the same

[dim meadow]

lol just get to a level where you don't care about grades

[gritty widget]

already am

[tight agate]

lol

[dim meadow]

nice

[tight agate]

don't lmao

unless you dont want to go to grad school

[dim meadow]

I mean in grad school

[gritty widget]

i don't care about grades because they're already good

i literally got a 90 on this problem set

this is unnecessary anger

the other marks i lost were because i couldn't be assed to write down a parametrization of the topologist's sine circle and a fully formal and rigorous proof that it's not locally path connected

not even worth the effort

(yes, the TA wanted us to do that)

[viral atlas]

Okay, I gave some more thought to this definition. If I have an open ball centred at x_0, the corresponding image is the union of two sets which are not open. Am I right in thinking so?

[tight agate]

are you trying to show that the preimage is not open?

[viral atlas]

Uh yes

[tight agate]

right, so you want to show that there is no open ball in R around x contained in the union of those two intervals

[viral atlas]

Yes.

[tight agate]

okay, so say you have an open ball of radius r around x_0

why is it not contained in the union of those two intervals

oh wait it's just one interval

not two

[viral atlas]

I'm actually not entirely convinced about that Since x_0-r lies in the first interval, and x_0+r lies in the second, I feel like I have an open ball contained in the union.

(Referring to Terra's drawing)

[tight agate]

if you look at tterra's drawing

there's a gap

between x_0 and the other interval

points in that gap are still in the open ball

remember we are looking at the open ball in R

not in the subspace topology

[sleek thicket]

I submitted a regrade request on a hw I got 43/44 on last week

[tight agate]

ignore the subspace topology remark

[sleek thicket]

they changed my grade 2 days after I got it back and the new deduction was blatantly wrong ("you didn't fix x, this is showing uniform continuity" when I explicitly fixed the variable)

I was mad and felt justified regrade requesting 1 point

[tight agate]

the set we're looking at is (a, x_0] union (b,c)

[gritty widget]

brb emailing my TA

[tight agate]

where b and x_0 are some distance away

so any open ball containing x_0

will contain some points between b and x_0

[viral atlas]

OH, I got it!!!!

[tight agate]

so not in the set

[viral atlas]

I had a brain-fart moment when I couldn't process the gap, I thought we had intervals (b,x_0] and (x_0,c). Turns out the second open interval has some gap, so its something like (c,d). So any open ball centred at x_0 is not in the preimage.

[tight agate]

yes

[viral atlas]

[tight agate]

[viral atlas]

Thanks both of you!

[gritty widget]

[viral atlas]

The discontinuity at x=1 is explained by the fact that f(1) is not in any open set in the range?

[little hemlock]

Take a small open interval around f(1). It’s preimage is a point (unioned with stuff that comes from the right side of the graph)

That’s how I would think of it anyway

[viral atlas]

Thanks!

[viral atlas]

To claim that a subset of R which has a finite complement is open, can I claim that it is the finite union of open intervals in R such that these finitely many deleted points are excluded?

[sweet wing]

yup

alternatively singletons are close and finite union of close sets are closed

[viral atlas]

Good point, thanks!!

[shadow charm]

when considering the metric space Q with d(p,q) = |p-q|, am I allowed to considerr open neighborhoods with non rational radius?

say for example the neighborhood $N_{\sqrt{2}}(q)$ of q

[gentle osprey]

Little Narwhal

[tough imp]

yes

[shadow charm]

alright i dont know i didnt feel sure

cause im trying to show that the set 2 < q^2 < 3 is closed and bounded in Q and I made use of such neighborhoods for that so I was a little unsure

so now to show that it isnt compact

[tough imp]

consider that metric spaces can be defined on sets which aren't subsets of R

so it wouldn't make sense that you'd have to only look at rational radii

[shadow charm]

for the compactness part, is considering an infinite cover that approaches a point between sqrt2 and sqrt3 from both sides the right way to go?

nvm

it isnt

and i just realised i messed up the first part too reeeeee

[viral atlas]

Is a certain subset A of X said to be open because it is included in the topological space (X,T) for some topology T, or is the topology T a collection of open sets? I'm kinda confused about which property is more primitive here-the open-ness of a set, or the definition of a topology on a set.

[shadow charm]

i think it is the first

[viral atlas]

Okay, thanks!

[shadow charm]

id wait for a second opinion though

[tough imp]

This set is closed because the complement could be written as (-infinity, sqrt(2)) \cap Q union (sqrt(3), infinity)\cap Q which is open in Q

[shadow charm]

first im trying to see if i can fix what i messed up

i just didnt consider negative q but i think it should be an easy fix

[tough imp]

oh wait me too lol

but yeah easy fixx

[tough imp]

definition of topology on a set

A subset is open with respect to a certain topology

it can't be the first because then every set is open as it's included in the in discrete topology

[viral atlas]

Aaah, makes sense.

Thanks !

[tough imp]

Chmonkey

[shadow charm]

aight fixed, back to compactness

i think i thought of a good cover in the meantime

$G_i = (-\sqrt{3} + \frac{1}{i}, -\sqrt{2}-\frac{1}{i}) \cup (\sqrt{2} + \frac{1}{i}, \sqrt{3} - \frac{1}{i})$ and take the union $\bigcup_{i=1}^{\infty} G_i$

[gentle osprey]

Little Narwhal

[shadow charm]

this should work right?

[tough imp]

obviously you need to intersect with Q but I trhink so

[shadow charm]

aight and the final step was to say if it is also open an it seems pretty clear to me that it is

can always construct subset neighborhoods

[tough imp]

idk how much topology you know but

The topology you have is the metric topology induced by the normal metric on R but on your subset

this coincides with the subspace topology

so the fact that your sets look like "open set in R intersect your set"

tells you it's open

[shadow charm]

oh yeah ofc

we showed that theorem a little earlier

that it's open relative to a sub metric space Y iff it is an open set in X intersected with Y

[viral atlas]

To claim T is the discrete topology, it should be sufficient to demonstrate that every singleton subset of X is in T; can I take pairwise intersections of two infinite sets precisely in such a way that their intersection yields a singleton? It seems doable in concrete settings, but can I do such a thing for arbitrary(possibly uncountable) infinite sets?

The union of all infinite subsets of X should be X, and removing a single element from an infinite set still gives an infinite set.

So I could take an arbitrary proper infinite subset A of X, remove an arbitrary element x from it to produce a new set A', then x is in an element of (X-A') which is itself an infinite subset. Now (X-A')\cap A is the singleton {x}. T being a topology, it is closed under finite intersections and hence we have {x}\in T for every x\in X. Thus T is the discrete topology.

[tough imp]

X\A' might not be infinite

[viral atlas]

How do I fix this?

[marble socket]

just don't take an arbitrary A

[tough imp]

You need to use choice for something like this, but

Take a countable subset

then split that up into two sets, the even and odd ones

you can throw x into this countable subset too, and just throw it into whichever one you like for example

I don't know of other ways to show that you can put x in an infinite subset with infinite complement

[viral atlas]

Oh, this works!

[tough imp]

To grab a countable subset you definitely need choice

if you want a proof that one exists, you can use well-ordering

I don't know the proof that it's necessry, i looked this up earlier this year

[pastel linden]

couldn't you also just take two infinite sets containing the same point?

take an infinite set and choose a point in it

[viral atlas]

That was my initial idea

[pastel linden]

then add that to another infinite set and take their intersection

[viral atlas]

But I thought it wasn't acceptable

Oh

Is it legitimate?

[tough imp]

Uh

There's no reason that those two's intersection is only x

[pastel linden]

oh wait yeah

sorry 2 am brain is not good at topology

[tough imp]

I'm actually not sure if you need choice to show that you have an infinite set with infinite complement

[viral atlas]

Can we not define these sets to be such a way that their intersection is guaranteed to be a singleton?

[tough imp]

but if you want to do it grabbing a countable subset you can't

lmfao

without choice you can't even show the existence of a proper infinite subset

[viral atlas]

[tough imp]

so you're going to ultimately need to do something janky with choice

[marble socket]

X -{a} ?

[tough imp]

but grabbing a countable subset is not too bad

https://math.stackexchange.com/questions/2774459/what-do-i-have-to-suppose-to-say-an-infinite-set-has-a-proper-infinite-subset-an?noredirect=1&lq=1

oh wait

nvm I didn't read the entire thing

LOL

but actually this is perfec

this is exactly what we wanted

proper ifninite subset with infinite complement

needs choice

[viral atlas]

Can I create arbitrary partitions of infinite sets?

[tough imp]

wdym?

[viral atlas]

Like, take infinite subsets A and B of X such that A\cap B=\emptyset such that they both don't include some x\in X, and A\cup B\cup{x}=X?

[tough imp]

Yeah do what was said earlier

take a countable subset Y of X

wplit up Y into Y_1 and Y_2

these are disjoint

Now call Z = (Y_1 U Y_2)^c

now just remove x from Z, Y_1, Y_2

Now you can do

Y_1, Y_2 U Z

both are infinite

x isn't in either

and they're disjoint

and then Y_1 U (Y_2 U Z) U {x} = X

sorry, I didn't say what Y_1 and Y_2 are, just like even and odd

[viral atlas]

Okay, makes sense. Can I claim all this without any additional assumptions(choice, etc.) which I should quote in my proof?

[tough imp]

the only thing you need is choice

In order to grab the countable subset

You could do this with say, well-ordering

you can enumerate X as element x_alpha

where alpha are ordinals, and you can just take all of the x_alpha where alpha is a natural number for example

I think choice also tells you that any cardinal contains subsets of all the smaller sizes inside it

so you can abstractly grab a countable subset

Tbh

I wouldn't worry about this

I'd just say "take a countable subset"

and I think any reasonable person will accept that you can do this, unless your prof cares about choice

but the ppl who care about choice are pretty few and far between, and even those who do probably accept it and use it in a context like this

[viral atlas]

Nw about that, I'm studying on my own at the moment. I don't even think my uni offers topology at the undergrad level. I just wanted to know what I'm assuming here in order to prove the result. Thanks a ton!

[tough imp]

I mean I think it's something anyone would believe at first mention pretty much

Like if I said

Let x_1 be an arbitrary element of X

then let x_n be an element in X\{x_1,...,x_n-1}

[viral atlas]

Yeah, it sounds intuitively reasonable to me.

[tough imp]

then take the set of all x_n

you just need choice to grab all those elements I think, but this is pretty reasonable I think

[viral atlas]

Agreed.

[tough imp]

Oh, I guess the only thing to mention is you want countably infinite lol

since countable includes finite (for most people's definition)

[viral atlas]

Oh yeah, I'll write it down as countably infinite. Thanks again!

[tough imp]

[viral atlas]

[sleek thicket]

[pseudo crane]

[sweet wing]

[gritty widget]

[marble socket]

[pastel linden]

[shut moat]

[red garden]

product of totally disconnected spaces is totally disconnected proof

help?

[swift hemlock]

to separate two points in the product, just separate each of the coordinates/components, then take the product of those open sets to get a separation in the product

[little hemlock]

@tight agate yo sorry for the ping, but about the whole X/G is a manifold thing. Do you happen to know where freeness of the G-action is used? I haven't needed it at all which is worrying

[sleek thicket]

I can show you an example where it fails for nonfree G

if that would help

[little hemlock]

sure

[tight agate]

if the action wasnt free then we couldnt have separated g and gp

[little hemlock]

ohh yea because gp = p is possible for g!=e in a non-free action

[sleek thicket]

consider GL(n, R) acting on R^n by multiplication. any two nonzero vectors can be swapped by the action of an invertible matrix, so the orbits are R^n \ {0} and {0}. The quotient will be the sierpinski space

I guess this isn't so good because you're only looking at finite groups, not topological groups with proper actions

[little hemlock]

thanks anyway sham

[sleek thicket]

in my head the issue is that points can get stuck close together

you can get like singularities in your space, or nonhausdorffness

[tight agate]

there's plenty of natural examples if you're thinking about (non-finite) topological groups

the action of Sl(2,Z) on the upper half plane

comes up when you're trying to construct the moduli space of elliptic curves

[sleek thicket]

yee, I'm actually having trouble thinking of a finite one whose quotient isn't a manifold though

I guess {-1,1} on R

this is still a manifold with boundary though so it still feels a little too nice

[tight agate]

permutation rep of S_n?

[sleek thicket]

hmm I was thinking that would just give you R but maybe not

oh no it's weirder

because of like (x,x)

yeah C2 acting on R^2 by swapping will be fucked up

or maybe not so much, I think it's like a reflection along y = x

so you get the upper half plane

but in higher dimensions it's more fucked up

[tight agate]

S_3 on R^3?

what is the irrep decomp

[sleek thicket]

yeah, I think the quotient won't be a manifold with boundary

there's the sign rep on a 1 dim subspace I think

[tight agate]

you have (x,x,x)

[sleek thicket]

right

oh right sorry not sign but trivial

[tight agate]

is the complement an irrep?

[sleek thicket]

I believe so

so we have p : R^3 -> R^3 given by p(x,y,z) = (x+y+z,x+y+z,x+y+z)/3

this is a projection onto the 1d rep

what's the kernel?

so it's everything where x + y + z = 0

this is invariant

[tight agate]

yup

[sleek thicket]

basis given by (1,-1,0) and (1,0,-1)

oh right so like

S_3 -> C2

then we act on R^2 by swapping

that's the other irrep

[tight agate]

okay the orbits are super messed up

[sleek thicket]

good haha

oh right when you want to deal with the braid groups this comes up

I've definitely thought about this before

the nth braid group Bn is pi_n of the quotient of R^n \ {bad points} by S_n

[silent folio]

Hey can anyone help me with trying to prove that the stereographic projection of n spheres S^n is a smooth manifold of dimension n?

so I've realised we can cover S^n by two charts, I've taken out the poles of both these, but I'm a little stuck on how to define our coordinate maps φ

[sleek thicket]

so you have two sets

presumably $U = \mathbb{S}^n \setminus {(0,\ldots,0,1)}$ and $V = \mathbb{S}^n \setminus {(0,\ldots,0,-1)}$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

you're not sure how to define $\varphi : U \to \R^n$ and $\psi : V \to \R^n$

yeah?

[gentle osprey]

Shamrock -> E -> B

[silent folio]

$U_{N} = \mathbb{S}^n \setminus {(-1,\ldots,0,0)}$ and $U_{S} = \mathbb{S}^n \setminus {(1,\ldots,0,0)}$

[gentle osprey]

𝓐eteer

[sleek thicket]

ah okay sorry

other coordinate

[silent folio]

yah, np

[sleek thicket]

so you're stuck defining the maps from these to R^n?

[silent folio]

yeah I'm having a brain mush moment

[sleek thicket]

that's fair lol

[silent folio]

how do I define our φ to R^n

[sleek thicket]

do you have a geometric sense of what this map does?

[silent folio]

kind of im guessing that

ok so for the 1-dimensional sphere, the circle it kind of cuts the circle and pulls it out right

and maps it to R

[sleek thicket]

sure

this is sort of a bad case imo

[silent folio]

and this is now mapping to some plane R^n?

[sleek thicket]

like when I think of the circle I think of parameterizing it by (cos(t), sin(t))

and then you map the circle minus a point onto (0,2 pi)

and then pull that out to R

I think R^2 is better

S^2

[silent folio]

yeah yeah, I did this in two ways, i used 4 charts and then showed I could do it for 2

now I'm trying to do the S^n... R^n+1 case

so for our S^1 we had that we could define our circle by x, sqrt(1-x^2) right

[sleek thicket]

sure

[silent folio]

how do we define our coordinate maps

for S^N?

[sleek thicket]

I think it's better to think about S^2

so like

has anyone ever explained to you the idea of stereographic projection?

for the sphere

[silent folio]

Well not really. But I think I watched this dude

and he kinda explained it

😄

[sleek thicket]

hahaha

right I think I remember that video

something about pythagorean triples

so here is the idea

[silent folio]

actually is was about quaternions

[sleek thicket]

oh

[silent folio]

but yeah

[sleek thicket]

so think about the sphere

minus its top point

[silent folio]

but yeah I understand we are intersecting different points of the circle/sphere and finding where that point lies on our R space

[sleek thicket]

and think about the xy plane

right!

so we draw a line

connecting e.g. the north pole to our point

and that intersects the plane z = 0 at a unique point

[silent folio]

when we say our point

any point?

[sleek thicket]

yup!

so we want to define phi(p)

take p

connect it to the point you're projecting away from

look at where that hits the plane

[silent folio]

O

erm

can we see it as

you know the point we are picking on our sphere

ahh I think I understand

rubber duck method is helping

[sleek thicket]

np lol

[silent folio]

right right, so how can we formulate an equation for this cheeky line

wait our line going through our point P and touching our North pole

is the line going through the sphere?

right

do I have to use $d(a, b) = \arccos\left(\frac{a \cdot b}{1}\right)$ where a and b are points on our n-sphere?

trying to find the line that maps these points using the stereographic proj.

[gentle osprey]

𝓐eteer

[sleek thicket]

uhhh

you can parameterize a line through two points in R^n

tp + (1-t)q

and then solve for when the last coordinate is 0

or the first one in your case

[silent folio]

oh my

what ist i doing

I just differentiated arccos to find the gradient

and just got into confusion

I forgot about this

$L_N (λ) = (−1, 0, . . . , 0) + λ (1 + x_0, x_1,\ldots , x_n)$

[gentle osprey]

𝓐eteer

[silent folio]

i got dis

Ok, I think we gucci now lol thank you sir shamrock

[tough imp]

⚔️ ☘️

There’s no knight emoji :(

[silent folio]

☘️

[tough imp]

@sleek thicket important

[sleek thicket]

It's catching on

[tough imp]

☘️

[red garden]

let x and y be points --> x=(x_1,x_2,....,x_n) and y =(y_1,y_2,...,y_n) , let X_i be seperated by A_i U B_i where x_i is in A_i and y_i is in B_i ( each X is totally disconnected )

is this right? @swift hemlock

[proud ridge]

exterior derivative of a p-form is just the sum of the partials right?

[gritty widget]

you might want to be a bit more precise with that statement

partials of....?

you want to get a (p + 1)-form

[silent folio]

why are the inverse functions for the coordinate maps for the n dimensional sphere like this

are we trying to map our points of the form x/(1-x_0) to S_n \ {(1,0,...,0)} ? and if so how does this work by taking the differential of our sphere?

[proud ridge]

Sorry I'll rephrase. The exterior derivative is a (p+1)-form, whose entries are the partial derivatives of the coefficient function with respect to each coordinate basis, wedged with the coordinate basis you take the derivative with respect to

[silent folio]

ooo this is defined for our y>0 and y<0 (upper and lower hemispheres) respectively and the 2y_1... 2y_n gives us our curvature gradient of the sphere right

[pastel linden]

is this not just a stereographic map?

my topology is not great but it looks like that

I guess the standard atlas is more straightforward

[silent folio]

yes it is

[shut moat]

sry if I'm interrupting (the person seems to have left), but I was wondering how useful the theory of differential forms is in solving ODEs

first order ODEs seem to be equivalent to finding integral curves of 1-forms, but is that change in perspective actually useful?

and can this be generalized to higher order and/or nonlinear ODEs?

[gritty widget]

i guess i have wondered something like this too, or maybe, what does PDE on manifolds look like? bunch of differential form crap?

probably something i don't actually want to think about

finding integral submanifolds of distributions on a manifold is the same thing as solving a PDE iirc

frobenius' theorem gives you necessary and sufficient conditions for when you can do that

[shut moat]

I think there was also some generalization of laplace's equation using *d*d as an upgraded laplacian (I really don't know what I'm talking about here so could very well be wrong )

[gritty widget]

spooky

frobenius' theorem has an equivalent in terms of differential forms

maybe you can look there?

there's probably a connection between PDEs and differential forms that you can dig up

ISM chapter 19

[shut moat]

man lee does all the things lmao

[sleek thicket]

It's a good book

[gritty widget]

just wait until you see kobayashi-nomizu

[sleek thicket]

Also not all pde in diff geo is forms

There is also riemannian stuff

Or wait no that's an ode isn't it

Just higher order

[gritty widget]

[sleek thicket]

I was thinking about geodesics and parallel transport

[gritty widget]

those are ODEs

hmmmmmm

but yeah there's probably some PDEs lurking in riemannian geometry too

[shut moat]

that seems more like an application of ODEs to manifold memes than the other way around

[gritty widget]

frobenius theorem is exactly application of manifold to DE memes

well "exactly"

iirc you can say some meaningful things

[sleek thicket]

@nimble jolt can probably say more, I think his work is either pde or closely related to pdes

[gritty widget]

what i'm thinking is

ODEs are to PDEs as finding integral curves is to frobenius' theorem
frobenius' theorem has meaningful applications to PDE theory
frobenius' theorem has a differential form equivalent
thus, differential forms apply to PDEs

yes gomez say more this is something i like

[shut moat]

during my digging online, I looked for a characterization of curves defined by $\int_C F_i dx^i = 0$ (which is basically a DE problem right?) And it lead to this random obscure book called the Geometry of Vector Fields

[gentle osprey]

~S^1

[shut moat]

and the families of curves are apparently called "non holonomic manifolds"

[gritty widget]

interesting

le holonomy groups have arrived

[shut moat]

also it defines the curvature of a vector field which was strange

was a fun find but the book is a bit dry so I probably won't properly read it lmao

[gritty widget]

https://issuu.com/mohammeddeaibes4/docs/_yu._aminov__the_geometry_of_vector

this?

[shut moat]

yeah

[gritty widget]

ugh i hate this book viewer

seems neat

[shut moat]

here's a pdf if you like

[gritty widget]

🚔

[shut moat]

[gritty widget]

ah yes, each component \xi_i of \tombstone

[swift hemlock]

yeah, just take the product of the sets A_i and the product of the sets B_i to get a separation of x and y

[red garden]

so im right?

[swift hemlock]

well you have to separate x and y to prove this^

[swift hemlock]

so just add this and you're done

[red garden]

.

so yea the space here would then

for x and y distinct points we say X = Product(A_i U B_i)

where the coordinates of x and y are seperated by each A_i and B_i

is this right

so for every two points

we found a seperation

[swift hemlock]

no, X = Product(A_i) U Product(B_i)

where x is in Product(A_i) and y is in Product(B_i)

and both sets are open in X by definition of the product topology

[red garden]

yea yea

fisrst problem ever done

first* in topology

tysm

[swift hemlock]

good luck with the rest haha

[red garden]

ty

[obtuse meteor]

anyone wanna help me show more smooth compatibility

[sleek thicket]

Sure

[obtuse meteor]

the right answer was no sham

the right answer was no :((((

[sleek thicket]

oh no

[tidal cedar]

[sleek thicket]

I am always happy to subject myself to awful manifolds computations for a friend

[obtuse meteor]

I just don't wanna do manifolds ever again lmao

right now I need to show a fun thingy

So I did this problem right

https://estrogen.fun/i/tb8v.png

By very nasty construction

@tough imp is familiar

:)

arctan time

and now I am doing this problem

and I am just here to complain

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

[sleek thicket]

https://tenor.com/view/तीरचलाना-दिलपेवार-उदास-दिलटूटना-arrow-in-the-heart-gif-14975254

Oh yeah I helped chm with that problem a couple weeks ago lol

well I will not get in the way of complaining

[obtuse meteor]

This part ii

I am confused by

because like

the maps I've computed

can have derivative 0

and that is

very not good

[sleek thicket]

[obtuse meteor]

for being diffeomorphisms,,,,,

[sleek thicket]

Concerning

(also I love that, I found out about it from this tweet https://twitter.com/chagnie/status/1355996191763222531?s=19)

MyNameIs Sándor (@chagnie)

@grassmannian @Category_Fury https://t.co/QmZJjPqEkX

▶ Play video

Twitter · Jan 31, 2021, 9:47 PM

[obtuse meteor]

yeah that's a good one

I should download

[sleek thicket]

(this is my algebra prof from last year who's an algebraic geometer)

[obtuse meteor]

so like

the inverse of an angle function

is just e^(it)

right

and so

my issue is

[sleek thicket]

right

[obtuse meteor]

the standard structure on S^1

is just projection to the first or second coordinate

[sleek thicket]

yup

[obtuse meteor]

so you get

t -> cos(t) and t -> sin(t)

and um

these

[sleek thicket]

The domain should fix it

[obtuse meteor]

are not good functions

the domain should fix it

[sleek thicket]

I think

[obtuse meteor]

but the domain has to cover the whole circle right

[sleek thicket]

Nope

You're just working on the overlap

[obtuse meteor]

not the whole circle

ahhhhhhhh

[sleek thicket]

Which is a subset of a quarter circle

[obtuse meteor]

I see

hm

yeah

not a subset of a quarter circle

it's a subset of a half circle for us

[sleek thicket]

oh right sorry

[obtuse meteor]

Bc the angle function can be on any open subset of S^1

ye

[sleek thicket]

Yeah I was getting confused because there's four charts

[obtuse meteor]

ye

hmmm

I'm still not convinced

that domain will fix this

because like

-sin(t) and cos(t)

the derivatives

at least one of those is gonna go "badoosh zero"

somewhere

[sleek thicket]

hmm this is true

[obtuse meteor]

and when it goes "badoosh zero"

my diffeomorphism goes

:(

[sleek thicket]

okay so concreteness time

Angle function is 0 to 2π on U = S^1 \ {1}

[obtuse meteor]

ah I am not smart

I see now

[sleek thicket]

oh okay

[obtuse meteor]

ignore dumb comments

[sleek thicket]

I was still thinking lol

last year I just said they have smooth inverses

[obtuse meteor]

which chart you're in

will be like

[sleek thicket]

Which makes sense to me

[obtuse meteor]

the opposite of the one which makes them be zero

and you will be happy

yeah I'm going to invoke IFT here

[sleek thicket]

Oh right!

[obtuse meteor]

is why I want this

[sleek thicket]

yeah so like

[obtuse meteor]

bc I know I have a cts inverse

[sleek thicket]

Yeah this makes sense

Yeah I just assumed I could say that any branch of arcsin or arccos is smooth

since they're like, definable by an integral of a smooth function or whatever

[obtuse meteor]

ye

[sleek thicket]

The key to avoiding technical details is just to claim things are true

😌

[obtuse meteor]

🧠

analysis TA where are you

[sleek thicket]

lmfao

My holy war continues

I went to oh yesterday

And the TA gave a false argument when a student asked for help (problem was to say a certain map is surjective, they showed it has dense image and used continuity)

And then on the next problem was like "sorry I didn't actually look at this problem yet can anyone else help"

I have been a ta and done worse but I'm still salty lol

[obtuse meteor]

rip

I don't look at problems ahead of time either

but at least I know how to solve them all so

[sleek thicket]

yeah haha

This was like

A super technical argument

You need to do a weird induction thing

and then the ta was like

"what are you doing, this is super easy"

anger

[obtuse meteor]

anger time

now I am just needing to show that like

theta(U) will always be open

so that this is actually a chart

and things work

,,,I do not want to show this

[sleek thicket]

rip

uhh what would I say here

hm

Ah okay i think I see

I know why this all feels so familiar

It's because I was literally saying these things to chm a couple weeks ago

[obtuse meteor]

lol

[sleek thicket]

He just dropped manifolds

Somehow like all the uw core courses have gone berserk

this year

Every single one is faster and more stressful than it was last year, and they were already pretty high up there

[obtuse meteor]

oof

a bunch of my friends have dropped manifolds too

our manifolds is not that bad

oh god I just realized something disgusting

so like

for a general angle function

theta(U) might not be bounded

proof

split S^1 into countably many open subsets by just making intervals of smaller and smaller size, say 1/2 of the circle, then 1/4, then an 1/8 etc

use the standard angle function + 2pi * k

where k is the part you're in

this is continuous if you shrink the parts by a tiny bit so that they're like disconnected and can't interact

this construction is actually so cursed holy fuck

[sleek thicket]

makes sense to me

[obtuse meteor]

brb becoming a number theorist

god gave us the naturals

all else is fake

[gritty widget]

[obtuse meteor]

what?

[red garden]

out of toopic

cna u use topology in number theory

or maaybe not yet

HAHA jk

[sleek thicket]

lmfao

[red garden]

number theory is not about numbers

[obtuse meteor]

topology is heavily used in noombers theory

slimvesus is just a memer

[sleek thicket]

I agree with everybody here

Number theory *is* very important in topology

[obtuse meteor]

i just threw up

brb

[sleek thicket]

How are you supposed to prove that an irrational-angled curve on the torus is dense without using Dirichlet approximation???

You're just hating me for being so correct

[obtuse meteor]

[sleek thicket]

Im not even meming!

[obtuse meteor]

imagine having angles

[sleek thicket]

This is the only way I know to do it!!!

[obtuse meteor]

this is geometry

not topology

[sleek thicket]

it's lie groups

[obtuse meteor]

You should have just said a line in the plane with this loop de doo property

[sleek thicket]

wiat what

I don't know what you're talking about here

[obtuse meteor]

torus is just C/lattice

[red garden]

if two fields are connected

[obtuse meteor]

so it's equivalent to asking you be dense in C when you add in equivalencies with these lines

[red garden]

does this mean there must be a 'functor'

from this cat to this cat

[obtuse meteor]

which IDK feels more number theoretic to me

usually means that. Not always

that is one of the ways

[sleek thicket]

oh yeah true Faye

[red garden]

not sure

[sleek thicket]

in my head it's like

look at the 1 param subgroups of the torus

some of them are embedded

The others are dense

You might want to prove this

and figure out why it's true

hmm actually I may be lying

not sure the others are embedded

[obtuse meteor]

I don't know lie groups

[sleek thicket]

I think they are? Might have self intersection

[obtuse meteor]

and clearly I don't know anything about manifolds

[sleek thicket]

Lie groups are so cool I love them

[obtuse meteor]

I just know glue

and glue and glue

[sleek thicket]

They're like me fave thing in diff top/geo

maybe this server should do a lie group reading course

That would be fun

I want to learn more about them

like the rep theory

[tight agate]

rep theory is just K theory

so learn that

[sleek thicket]

I intend to

We're currently doing classifying spaces in my bundles course

So I should be able to start learning Bott periodicity stuff soon

[tight agate]

I need to present on some equivariant K-theory stuff in a week

and I don't know any

[sleek thicket]

hmm I am having a conundrum. I want to do math but I'm hungry. And I can't go get food because I need to do math

[obtuse meteor]

I need to prove that S^1 has the right fundamental group tomorrow

[sleek thicket]

how can I possibly resolve this

[obtuse meteor]

speedrun time

[sleek thicket]

Exciting!

[tight agate]

Hurewicz

[obtuse meteor]

by 8pm tomorrow

No like for an alg top homework

we don't have that yet

[sleek thicket]

What proof are you doing?

[obtuse meteor]

we can use homotopy lifting property

w/o proof

[sleek thicket]

Ah nice

[obtuse meteor]

so I need to like

[sleek thicket]

So kind of covering spaces

[obtuse meteor]

prove it's a covering map

[tight agate]

prove S1 is a K(Z,1)

[sleek thicket]

right

[tight agate]

EZ

[obtuse meteor]

(aka handwave coordinates and hope the alg top prof doesn't care)

🧠

[tight agate]

the covering map isnt bad at all

[obtuse meteor]

the covering map is easy

[sleek thicket]

Faye I would simply cite the well known result that π1(S^1) = Z

[obtuse meteor]

angle functions are bad

[cedar pebble]

You can always use the complex analysis proof of this computation

[obtuse meteor]

manifolds course: construct with arctan
alg top course: It is clear that angle functions exist

[cedar pebble]

[sleek thicket]

do yall know what the first proof I saw was

[tight agate]

Hurewicz

[sleek thicket]

nope

Memier

[cedar pebble]

Pushout of groupoids

[sleek thicket]

Yup

Not even kidding

I wrote a term paper on it and everything

[tight agate]

do you mean van kampen?

[sleek thicket]

Without knowing any covering space stuff or degree theory

[cedar pebble]

Groupoid van Kampen

[sleek thicket]

Yeah brofib

[tight agate]

how's that memeir

[cedar pebble]

Yea that paper is nice

[sleek thicket]

it is absolutely memier

[obtuse meteor]

the fact that sham learned alg top

from topology and groupoids

is such a meme

[sleek thicket]

sham hasn't actually learned any alg top

I've done two shitty independent studies

[obtuse meteor]

[sleek thicket]

And learned the material really poorly in each

[cedar pebble]

Wow mood

[tight agate]

Brofib learned alg top by talking to his friends

[sleek thicket]

next year I should probably take the alg top class

but I already sort of kind of know the material

the eternal struggle of "nail down your foundations" vs "do cool stuff"

[obtuse meteor]

how relatable (tm)

[cedar pebble]

Me learning something for the first time: not doing enough exercises, rushing things, not learning everything properly
Me learning something for the second time: I already know this, shutting off brain

[sleek thicket]

I did a Hatcher reading course in spring but uh covid

And also I jumped into the 2nd quarter with cohomology

Spent like 2 weeks speed running Hatcher homology chapter

[tight agate]

do a seminar, you will be forced to learn the foundational stuff so that you don't look like an idiot when someone asks you a question

[sleek thicket]

X

[obtuse meteor]

holy shit stop it stop it

[sleek thicket]

I feel like I would learn stuff at a shallow level if I were trying to also keep up with a seminar

[cedar pebble]

[obtuse meteor]

I enjoyed the cat theory seminars

despite not knowing enough

but that's ultimate ugct

going to cat theory seminars

[sleek thicket]

I went to a couple ag seminars and was very lost

which is why I need to someday grind Hartshorne

[cedar pebble]

It’s midnight and I haven’t started the rep theory homework due tomorrow oops

[obtuse meteor]

nG bad

[sleek thicket]

Honestly I would prefer to do that than reus this summer

But I want $5K

[cedar pebble]

Also need to prepare stuff for meeting tomorrow supposed to have talk slides to show Daniel

[tight agate]

this is why the west coast is better, it isnt midnight yet

more time to do rep theory homework

[sleek thicket]

I would simply complete my rep theory homework

:smug:

[cedar pebble]

Yea I’m gonna

[sleek thicket]

we can crowd source it

Post and we'll all write solutions

Just don't check for quality

[cedar pebble]

I haven’t looked at it yet hold on

[sleek thicket]

I need to rewrite a really ugly argument because I realized I was misremembering how fubini works

it will be much better now

but I don't want to think about it

[cedar pebble]

okay so it's like

not great not terrible

[sleek thicket]

oh ive done the last one at least

and 3

Yeah the first one scared me because "quiver"

But this looks pretty reasonable

[cedar pebble]

1 is kinda annoying but not bad, 2 is easy, 3 is easy but annoying, 4 is annoying

yea not bad

last week's homework was really annoying I didn't do it

[sleek thicket]

I feel like I've done 2 but also I'm blanking on what the radical of a module is

So maybe not

[cedar pebble]

intersection of all maximal submodules

[sleek thicket]

ah yeah okay I've done 2

[cedar pebble]

yea so that's not bad it's just checking definitions

[sleek thicket]

I have run out of homework

So I need to like

Work on reu apps or something

blech

[obtuse meteor]

hopefully I get into the REU I'm working on stuff for rn

tfw the REU emails you and is like

"so you know that application you filled out?"

"turns out we need you to read these papers and write this response as well by say hmm, Wednesday next week, preferably Monday"

'it's """optional"""'

literal hell

only like 6-15 spots too according to the prof

so I'm like

AHHHHHHHHH

[cedar pebble]

[obtuse meteor]

@sleek thicket you know the REU I'm talking about and anyone who follows me on twitter saw me post about this :P

me: applying to an REU in noomber theory / probability
me: so yo wtfffff is a martingale

[sleek thicket]

OOF

i mean

good

but OOF

[obtuse meteor]

yeah

[cedar pebble]

oh hey 1a is cute and then the other parts are easy

computed the Jacobson radical by literally just classifying all the maximal ideals and then computing the intersection

[tough imp]

I think I did this one by myself