point-set-topology

[obtuse meteor]

it's not though???

[desert bloom]

I think I might just ask the professor in person

on zoom

[obtuse meteor]

what happens to the topologies is given by like a certain universal property in a slice category

unironically this is the first thing I thought of I'm broken

[desert bloom]

I dont' know what that is...

[obtuse meteor]

no problem

I agree with slim that you should think of it as subsets

[desert bloom]

oh ok

I asked the professor on email and he told me to ask him on zoom, so I chickened out and asked here

yeah thanks

yeah

It does make sense if I don't think about topologies

Actually I think I don't really understand it, but I don't know what I don't know

so

I'll just ask the professor

he seems like a good guy, so I will probably be ok

[fading vale]

save urself

they r disjoint here anyway so introducing slice category stuff is unnecessary

[desert bloom]

This kind of feels like when I'm little and my parents are arguing

except this time the iq involved is significantly higher

[fading vale]

i cant marry faye we are both too gay for that

i guess depending on ur parents that might be accurate though

[obtuse meteor]

cringe

[fading vale]

currently having a gamer moment

it should be literally illegal to make me verify bijectivity

[obtuse meteor]

but wait am I reading "Moth | not male" incorrectly???

[fading vale]

im nb but im only really attracted to guys

or like idk masc leaning nbs

[obtuse meteor]

ahhh I gotcha

I'm attracted to the occasional nb

[fading vale]

one or two in the sea

[obtuse meteor]

but I get it now

yes

[fading vale]

also

so much pain

[obtuse meteor]

why pain?

[obtuse meteor]

please verify bijectivity of this parameterization of the genus g surface sitting in R^3 like this

have fUn

[gentle osprey]

Moth | not male

[obtuse meteor]

transport-simple??? This is a cringe terminology

[gentle osprey]

Moth | not male

[obtuse meteor]

oof

[fading vale]

its for fiber reasons

[obtuse meteor]

this looks like covering space theory

[fading vale]

it is

[obtuse meteor]

which makes sense

lol yeah

[fading vale]

the cover comes later

[obtuse meteor]

actions of groupoids

[fading vale]

[obtuse meteor]

~ covering spaces

[fading vale]

pain

[quasi forum]

Question: what does that upside down capital beta mean?

[fading vale]

ummmm

$\coprod$?

[gentle osprey]

Moth | not male

[quasi forum]

Yea. That thing

[fading vale]

disjoint union

[quasi forum]

Ohhh. That had me confused, but that makes sense. Ty

[fading vale]

np

[shut moat]

also that's an upside down pi

[fading vale]

yea xd

anyway i have to show that $\Phi(w)(z) = \Phi(w')(z')$ implies $u = u'$ and $z = z'$ where $w$ and $w'$ are paths from $b$ to $u, u' \in U$ and $z, z' \in \Phi(b)$

[gentle osprey]

Moth | not male

Moth | not male

[fading vale]

er

hm

[gentle osprey]

Moth | not male

Moth | not male

Moth | not male

[fading vale]

but i mean its not like an arbitrary functor from the fund. groupoid to set has to be injective or anything...

well...

hm

ugh no idea

[gentle osprey]

Moth | not male

Moth | not male

Moth | not male

[obtuse meteor]

have you ever considered that groupoids are fake?

[fading vale]

sadge

yeah idk how to show this

it doesnt even make sense to me tbh like why would Phi(w) be injective

[gritty widget]

I am working through some details of SU(2) being a double cover of SO(3). I am having trouble seeing that this map is actually smooth
So we have phi:SU(2) to GL(R,3) which sends X to the matrix Ad(X)
Its make sense that this should be smooth because X is being sent to the Ad(X) and the map Ad(X) is smooth since it Y\mapsto XYX^-1 is just polynomials(/maybe rational functions since we are taking the inverse) in each coordinate

But this is just saying that the image of an element under phi is a smooth map, this doesn't imply that phi itself is a smooth map

[sinful pecan]

\Pi(B) is a groupoid so isn't this true by def'n of functor?

[fading vale]

no

functors are not by defn injective but thats not what this is saying

w is a path so its a morphism in Pi(B)

Phi(w) is a morphism in SET i.e a function between sets

but that doesnt imply that Phi(w) is an injective function

[sinful pecan]

ya but w as a morphism in Pi(b) is iso, so you have an iso in Set too right?

[fading vale]

is it?

i dont think ive ever seen a defn of isomorphisms in Pi(B)

i

i guess it is?

[sinful pecan]

its a groupoid so every morphism is an iso

[fading vale]

yea i guess that makes sense

so nvm ur right

ok checks out

blegh ok next up surjectivity

[sinful pecan]

same argument right?

[fading vale]

i guess this works likewise, for x in p^-1(U), x is in Phi(u) for some u in U and then we have a path w from b to u which gives a path Phi(w): Phi(b) -> Phi(u), Phi(w) is surjective so there is a z in Phi(b) with phi_U,b(u, z) = Phi(w)(z) = x

yea more or less

ok now time to put the topology on X(Phi)

i assume whats going on here is the coarsest topology that the phi_U, b are homeomorphisms?

[sinful pecan]

it says its unique

[fading vale]

wouldnt it be generated by the subbasis with open sets bla bla bla

so that phi_U, b is both open and continuous for all U, b

which would be unique

[sinful pecan]

yes

[fading vale]

so yea we just have to verify that the phi_U, b and phi_V, c agree right?

like it does

agree might be the wrong terminology but u get what i mean

[sinful pecan]

yes gluing

[fading vale]

mhm

[gritty widget]

toe pology

[sleek thicket]

simp

[gritty widget]

SYMPLECTIC TUBULAR NEIGHBOURHOOD

If $N \subseteq (M, \omega)$ is a Lagrangian submanifold, then there is a neighbourhood $U$ of $N$ in $M$, a neighbourhood $V$ of the zero-section in $T^*N$, and a symplectomorphism $U \cong V$ taking $N$ identically onto the zero-section.

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

flonshed629835071465062410

[tight agate]

tinktonk

I need to compute the inverse of a connecting homomorphism

[gritty widget]

just diagram chase 4head

[tight agate]

and compose it with another connecting homomorphism

[gritty widget]

just chase it backwards 5head

[fading vale]

feeling pretty tubular rn

[sleek thicket]

Let $E \to X$ be a vector bundle

[gentle osprey]

Shamrock (not a furry)

[sleek thicket]

Ah nvm I think I see it

sick this makes sense

[tight agate]

lol

[sleek thicket]

😌

Needed to get a smooth nonnegative function which vanishes iff some sections are linearly dependent

But in my problem I know there's an embedding into a trivial bundle

So I can write things wrt a frame for the trivial bundle and then take a matrix of coefficients

Err I guess the matrix isn't square

So I need to take all the minors of the matrix and then take a smoothed maximum

But it basically works

[tight agate]

petthecatwiggle

[sleek thicket]

wym?

lol

I can see the face of god

well, the face of god up to homotopy

[gritty widget]

Let $F: {\mathbb{R}}^2 \to \mathbb{R}$ be a differentiable function and $ q $ a point such that $ F(q) = 0 $. If $ (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y})\rvert_{q} \neq 0 $, then there exists a neighbourhood N of point $ q $ in ${\mathbb{R}}^2 $ and a parametrized curve $ \alpha: (a, b) \to {\mathbb{R}}^2 $ such that $ { p \in N | F(p) = 0 } $ is a tray of curve $ \alpha $.

[gentle osprey]

Vanhousen

[gritty widget]

What I've tried: I assumed without loss of generality that $\frac{\partial F}{\partial x}\rvert_{q} \neq 0$

[gentle osprey]

Vanhousen

[gritty widget]

Then I have tried to write a definition of a limit point in $q = (x_1, y_1)$ and I got this $\lim_{h \to 0}{\frac{F(x_1 + h, y_1)}{h}} = p \neq 0 $

[gentle osprey]

Vanhousen

[gritty widget]

But I can't seem to find a neighborhood of q such that all points in it have a value of zero

And I'm out of ideas how I shall move forward

[shy blaze]

I have been trying to get an idea of how the fundamental group and the first homology group of a space are connected, and I read that the first homology group is the abelianization of the fundamental group. I was just wondering, how can you sort of visualize the elements of the first homology group?

They would have to be loops that are not commutative I guess

are 2 loops around different punctures in the punctured plane commutative? I would think not right?

[gritty widget]

Maybe someone has a better intution than me but I tend not to think about the actual elements in a homology group

The intutive facts you get are from looking at the dimension of the homology group in certain coefficents

[shy blaze]

I guess the only case Ill be looking at is the punctured disk, maybe theres some intuition on that specific case / similar cases?

[gritty widget]

Well the punctured disk pi_1(X)=H_1(X)=Z

[shy blaze]

well, not the singly punctured disk, the n-punctured disk haha

[gritty widget]

If map of lie algebras is an isomorphism can we conclude that the there is a surjective map between their lie groups\

[gritty widget]

at least if the lie groups are simply connected (there's actually an isomorphism then) but im not sure about the general question

ah yes thank you

I'm trying to work through the details of SU(2) being a double cover of SO(3)

[fading vale]

(here B is path connected locally path connected and semilocally simply connected)

at the end here, why does continuity of the transition maps follow from this?

[marsh forge]

the preceeding statement is false, so continuity is a trivial consequence

[fading vale]

monkaS

actually

[marsh forge]

idk what it is about that writing style

that makes me really not want to read it

[fading vale]

its just really fucking terse

also tom dieck is ESL

[marsh forge]

its more the symbols i think

just looks like symbol salad

[fading vale]

sigh

this part is kinda lame

constructing inverse functors or whatever is zzzzz

[fading vale]

ummmm

ok

so this is weird

lets say $p \colon E \to B$ is a cover

[gentle osprey]

Moth | not male

Moth | not male

Moth | not male

[fading vale]

F_b is the fiber of b btw

so alternatively a functor X was defined here

[gentle osprey]

Moth | not male

Moth | not male

Moth | not male

Moth | not male

Moth | not male

[fading vale]

(w/ a map to E)

i know that E is the disjoint union of the fibers

but it shouldnt be the disjoint union of the path components of the fibers

and i dont see what the homeomorphism is here or how it can possibly be the identity

bc... path components vs points

[flint cove]

Is there any reason why the classifying spaces for (say real) n-plan bundles is the classifying space of O(n), and not of GL(n)?
Or are these two homotopy equivalent and one of them is easier to deal with?

(side note: I know nothing about the homotopy theory involved, I only know classifying spaces exist some times and how to construct the grassmannians)

[cedar pebble]

they are homotopy equivalent: GL(n) deform retracts onto O(n)

[flint cove]

That's what I thought. Is the deformation retract given by some Gram-Schmidt-orthogonalization?

Like, I would've guessed that you Take A = (a1, …, an) in GL(n) and gradually transform an into something orthogonal to span(a1, a2, …, an-1) etc, but I have no clue as to why that would be a retract

At least it preserves the flag associated to A

Ah wait, I would just need a homotopy, I already have the embedding O(n)→Gl(n,ℝ), there's not really much left to prove except for constructing that process formally as h: [0,1]×Gl(n, ℝ)→Gl(n,ℝ) and showing it's continuous

[cedar pebble]

yup, this is the deform retract

[sleek thicket]

imo it's fucked up

O(n) being a deformation retract of GL(n) is something which has never sat right with me

the next time we design math I'm changing it

[obtuse meteor]

it seems right?

what is bad about it?

I just want someone to appreciate how I used tikz for my algtop homework
https://estrogen.fun/i/qysj.png
https://estrogen.fun/i/z0lo.png
https://estrogen.fun/i/jpep.png

if anyone can guess what kind of question this is: you are a real gamer

[marsh forge]

if you use quiver

you can get rid of the arrows

oh wait maybe you want them i was thinking you were drawing universal covers but thats bc im dead tired

ignore me

[gritty widget]

Ah, I recall doing problems like this in kindergarten

It is "connect the numbered dots", right?

[flint cove]

I have no idea how to use this in an algtop exercise, but that's a cayley graph of D_6 = <x,y | x², y³, (xy)²>.
I'm really not sure what all the „loops“ should represent; they don't seem to have too much in common.

[obtuse meteor]

I want them bc I'm calculating generating sets for a kernel of a surjective group homomorphism h : F_n -> G

G in this case is S_3

and n = 2, picking generators a, b for F_2 the map sends h(a) to an order 2 element and h(b) to an order 3 element

You basically use some maffs to draw the correct universal cover with edges labeled and colored

then pick a spanning tree to get a generating set

very cool ^^

Yeah that's right. The reason this comes up is because if h(a) = x and h(b) = y then these are the relations you get from it being S_3

Also I used straight up tikz, not quiver or tikzcd because I'm a masochist

[flint cove]

Wait, so is the basic idea here that the kernel as a subgroup of a free group must be free?

I never did algtop formally so I may just know too little about F2

[obtuse meteor]

ye and you want to compute the generating set for the kernel

and you do that with covering space theory

[flint cove]

What does that mean?

I know what a simply connected space is and that in some instances there may be a notion of „universal covering“, but that's about it

[obtuse meteor]

do you know much about covering spaces?

ah cool I can explain it if you'd like!!!

there's a lot of cool stuff here

[flint cove]

sure

[obtuse meteor]

do you know the defn of a covering space?

[flint cove]

Surjection, locally the preimage decomposes into disjoint union of the image?

[obtuse meteor]

yep!

[flint cove]

then yes

[obtuse meteor]

And you know that they enjoy the homotopy lifting property?

[flint cove]

ah, yeah, I have heard that

[obtuse meteor]

how we'll use it here is the following fact

if you have a covering space p : Y -> X

and a path gamma in X, then picking a point y lying over gamma(0) there is a unique lift of gamma to Y which starts at y

furthermore if you start with two homotopic paths gamma, gamma' downstairs, you lift to homotopic paths upstairs

[flint cove]

Do you wanna introduce the monodromy action of π1 on the fibers?

[obtuse meteor]

yeah so that's like the first thing you do

pi1 acts on the fibers

but you actually get something even better

when the image of pi1(Y) in pi(X) is normal

pi(X) / p_ast(pi1(Y)) acts on the fibers

and it acts in the fibers in a way that's compatible with a bijection between pi(X) / p_ast(pi1(Y)) and a particular fiber

compatible in the sense that it's the same as the action of pi1(X) / p_ast(pi1(Y) on itself

[flint cove]

wait… so essentially curves which are pushed down from Y to X have trivial monodromy action? or what is that saying

[obtuse meteor]

yes

this should be clear

a curve pushed down from Y to X

is exactly a curve in X that gets lifted to a loop in Y

[flint cove]

Hm… let's say we have a double cover S^1 → S^1, and let γ be a loop on the left. this is a double loop on the right. However, that acts on the fiber as addition by two.
So this doesn't seem to be correct

(at least the initial statement that these curves have trivial monodromy action)

[obtuse meteor]

addition by two on the left is the same as addition by 0

for that double cover

non?

[flint cove]

Oh, right, the fiber is a two point set, and we're coming around again

[obtuse meteor]

mhm

[flint cove]

for some reason I forget that other covers of the circle exist and I equate everything with ℝ

[obtuse meteor]

lol it be like that

[flint cove]

even if I start the sentence knowing better, lol

[obtuse meteor]

so now the magic comes around

and what it essentially tells us is that

so well first we need this bijection

[flint cove]

(pun intended I guess)

[obtuse meteor]

fix a basepoint y in Y, define phi(H[gamma]) as follows, lift gamma to a path in Y starting at y, and take its right endpoint

exercise: verify this gives a well-defined bijection from the right cosets of H = p_ast(pi1(Y, y)) with p^{-1}(p(y))

so how you do one of the problems like this where you're trying to find a generating set is

you know that the number of sheets in a cover (aka the number of vertices in your graph) is the index of H in pi1(X)

[flint cove]

Wait, let's recap. what is H, what is φ, …?

Notation is lost a bit in the backlog.

[obtuse meteor]

H = p_ast(pi1(Y, y))

phi is a map we're defining now

from right cosets of H to p^{-1}(x)

[flint cove]

right

[obtuse meteor]

which depends on a basepoint y in Y lying over x

following up to this point?

[flint cove]

Yes, writing stuff down, give me a sec

it's 4AM over here and I had two beers

[obtuse meteor]

lol

[flint cove]

I can assume that the lift is a aleft inverse to the pushdown, right? so elements π_*(δ) of H are lifted to δ (up to based homotopy)

because then left-multiplying with π_*(δ) does literally nothing to the right endpoint, as with δ we stop where we started

[obtuse meteor]

hrmmm

I think yes?

[flint cove]

I mean yes we can, as „being a lift“ is defined as „something in the preimage projected to our curve in X“ and the lifting property ensures precisely that this lift is unique

[obtuse meteor]

ye

[flint cove]

Wait can I assume that our cover is (path-)connected?

[obtuse meteor]

yeah

it's a theorem (tm)

that for nice base spaces

you can get a path-connected locally path-connected covering space

which has the correct fundamental group downstairs

this is the big theorem which makes this work

How I envision it is like

[flint cove]

isn't it locally path-connected already if it's guaranteed to be path-connected?

[obtuse meteor]

nah

there are path-connected spaces that aren't locally path-connected

[flint cove]

uh dear god

[obtuse meteor]

although being a covering space over a locally path-connected space means you're guaranteed to be locally path-connected

welcome to topology

the reason I said nice base spaces

is the real condition on your base space is

path connected, locally path connected, semi-locally simply connected

but it doesn't matter bc any reasonable space is this

in particular this is true of any CW complex

[flint cove]

Yeah, okay, I should not research counter-examples, in AlgTop you ignore them anyway

[obtuse meteor]

yes

okay so like

how I envision it

is you use this big theorem to build your covering space which has ker(h) as your fundamental group

you use the index result to draw the number of vertices, since it will be the index, which is |F_n/ker(h)| = |G| by first isomorphism theorem

[flint cove]

The surjectivity is clear as we can just take the lift of the pushdown of the path to the point y

[obtuse meteor]

ye

we then like, try and fill in edges

and how we fill in edges is

we label each of our vertices by elements of G

and the upshot of like using this bijection and understanding how it interacts with the monodromy action

is it turns out that if you're lifting an edge alpha in your base space to find out how to glue it into the top space

[flint cove]

Oh that is interesting

[obtuse meteor]

then if you glue the left edge to y_g

then you have to glue the right edge to

y_{gh(alpha)}

this allows you to understand exactly what graph this covering space is

(knowing that it already exists, so you can carry out this process without much worry)

and then you use the result about fundamental groups of graphs

to find a generating set for the fundamental group of this covering space

projecting it down will get you a generating set for ker(h)

it's a pretty awesome and powerful technique

[flint cove]

huh

That's a lot to swallow, but from a high-level this seems really intriguing

like, I always imagined presentations of groups to be the hardcore term-manipulating way of verifying properties, but this feels a lot more geometric

[obtuse meteor]

yeah this is intensely geometric

[flint cove]

and finding generators of things is rarely something you have a canonical method for

[obtuse meteor]

This is also like

totally an algorithm you could program

idk if it's what sagemath or stuff actually uses to compute certain generating sets

but like

[flint cove]

not sure, might be permutation group stuff (schreier sims)

[obtuse meteor]

¯_(ツ)_/¯

[flint cove]

(Not that I would understand what that does in particular, lol)

[obtuse meteor]

but yeah working through this result on homework I was like

:o

and then

[flint cove]

dark magic

[obtuse meteor]

very powerful tools

algtop feels broken

[fading vale]

anytime ur doing anything with path connectivity and you want to check counterexamples just look at the topologists sine curve

wait

BRUH

i forgot that the fibers were discrete so the path components can be identified w/ the space

asdflkasdflk

[obtuse meteor]

wdym?

[marsh forge]

i assume moth is talking about a finite cover

[sleek thicket]

moth i regret to inform you that the fibers of the topologists sine curve are not discrete

i think you should be able to prove {0} x [0,1] is not discrete with some calculus (e.g. the intermediate value theorem) so I don't blame you for not seeing it

[obtuse meteor]

🧠

@flint cove for a counterexample

clearly path-connected

but not locally path-connected at (0, 0)

[flint cove]

ah, because there is a neighborhood basis that shrinks in a sort of „independent“ direction to where the connecting paths move

like, the ε-nbhds will go concentric while the paths go fucking everywhere and out of each neighborhood a zillion times

makes sense

[sleek thicket]

I would simply connect the dots

Literal children can do it

[obtuse meteor]

🧠

[flint cove]

Hm, perhaps an example simpler to visualize might be a ladder with increasingly tiny step sizes
(i.e., union of [0,1] \times {2^{-n}} for all n plus vertical lines at x=0,1)
That surely is path-connected, but at (1/2, 0)it's not locally path-connected, right?

[obtuse meteor]

Yeah good intuition :)

[cobalt seal]

Hi there,
I'm working through Einstein notation/early tensor stuff and am currently working through a question involving the below term:

I'm fairly new to this stuff, so it's entirely possible I've made a mistake, but am I correct in reaching a kronecker delta as below?

[gritty widget]

you combined an x and an x' into a kronecker delta, doesn't seem right

that said i don't know that the x with one up index and one down index even is so

[cobalt seal]

we were taught this relationship. Is that not correct?

[chrome dew]

well you just haven't defined what it is, it's not standard notation, x could be anything

[cobalt seal]

Right, sorry. x^i is the coordinate of a general point of a manifold, and x'^i is the same coordinate in a separate coordinate system

[gritty widget]

can you post some context @cobalt seal

also can anyone help me interpret this

i think here that they define the corresponding atlas on $M$ by taking charts $\varphi: U\to V$ and defining $U_0=\varphi^{-1}((\bR^n\times{0})\cap V)$ and $\varphi_0$ to be the restriction of $\varphi$ to $U_0$

[gentle osprey]

spinsicle

[gritty widget]

specifically i'm not sure why $D(\eta\circ\varphi^{-1})$ needs to have that form

[gentle osprey]

spinsicle

[fading vale]

i hate u

anyway i was talking about another thing

equivalence of TRA_B and COV_B

and it relied on the functors being discrete for it to work

[fading vale]

no it was a non sequitur but i had a revelation

also finished a tom dieck section

😌

[quick wing]

Can anyone tell me why a dominant morphism from an integral scheme X to the projective line determines a rational function on X?

I have been stuck on this for ages and I think it should be simple

It is from Fulton's intersection theory if that is relevant, from 1.6 where is showing the alternative definition for rational equivalence of cycles

[trail ibex]

Do you see why a dominant morphism to A^1 determines a regular function ?

Then just patch things together I guess

[long hornet]

Can a proper closed subset of R^n be homeomorphic to R^n? I feel like this should be easy (at least, assuming invariance of domain)

[gritty widget]

at least this is false for R right? this proper closed subset must be connected, so it must be an interval, say [a,infty). but i can remove a from this set and it is still connected, but we cannot remove a point from R and it is still connected

[long hornet]

Yes

Using the same reasoning, C (our closed set) must be connected, so V = R^n - C must also be connected

[gritty widget]

oh, i guess you can also argue that every point of R^n is locally euclidean, but this is not so for a proper closed set

[long hornet]

Makes sense!

[gritty widget]

wait a second... is this circular lol

[long hornet]

Ah, I think this reduces to proving ball \cap C is not homeomorphic to R^n..

[gritty widget]

oh, is it really just a direct implication from invariance of domain? let f be a homeomorphism C->R^n. then f^{-1} satisfies the hypothesis of invariance of domain. it follows that C is open in R^n, which is a contradiction (grader: 0/10 for not addressing C=emptyset case)

[long hornet]

Is that what the invariance of domain says?

It is!

Apparently I only knew about invariance of dimension.

[flint cove]

Hope it's okay to ping @marsh forge for a delayed question about the SerreSS-talk.
Is there any setting where the idea of taking cohomology groups with values in another cohomology group naturally pops up? This seems like a really strange idea to me (who only knows basic singular cohomology up to the künneth formula).
I always imagined cohomology as assigning some number to a nontrivial chain in a way that reflects its nontriviality, like a winding number around a hole or an element of ℤ/2ℤ to measure something 2-graded or whatever.
But matching that metaphor to what we had as definitions in the E2 page would be like „I assign this chain the element flips pages in large book of cohomology classes x, where x is a canonical generator of H*(ℂP^∞). Of course, the natural choice!“, which doesn't really give me a hint of what's actually going on.

[marsh forge]

not an exact answer

but taking coho w coefficients in homotopy groups

is a key ingredient in obstruction theory

i think that you should think abt my univ coeff remark

namely that taking coho w coefficients in another coho

is up to torsion

the tensor of the two cohos

[flint cove]

Hm, so basically just „extension of scalars“ of \otimes ℤ to \otimes new fancy coho-group

I think I can accept that as a formal device.

[long coyote]

Let X be T2 and suppose that K⊂X is compact. what is X/K

[gritty widget]

are you asking what properties the complement of K has?

e.g. compact sets in T2 spaces are closed, so it'll be open

[sleek thicket]

or the quotient?

[gritty widget]

that's why i'm asking lol

[long coyote]

its quotient

[gritty widget]

just wanted to be sure

[long coyote]

is X/K T2 as well

[gritty widget]

i think it'll be

not entirely sure

the quotient is T2 iff the equivalence rln is closed in X x X, maybe you can look there not true unless the quotient mapping is open

[flint cove]

For Pairs of distinct points in X/K we have two scenarios: either both are not K, i.e. of the form {x1}, {x2} for x1, x2 in X\K, or precisely one is K

[sleek thicket]

The equivalence relation will be Δ union K×K

[flint cove]

So I'd say the question is whether K can be separated from {x}

[sleek thicket]

which is a union of two closed sets

Δ is closed by hausdorffness

[gritty widget]

very nice

[sleek thicket]

(Δ = { (x, y) in X×X : x=y } is the diagonal)

@long coyote the relevant facts here are that a space is T2 iff it's diagonal is closed in the product and that compact sets are closed in a T2 space

So try to compute the diagonal of X/K

oh wait maybe there's a concern here

in that products of quotients maps may not be quotient maps

It's something like this anyway

[flint cove]

Which is the case since open sets around K in the quotients are in bijection with open sets containing K in X. So we can separate x from k for every k in K by means of Uk around x, Vk around k. Picking a finite subcover Vi1, …, Vin gives a cover of K which is disjoint from the Ui1 cap … cap Uin, separating K from x

[long coyote]

thank you

[sleek thicket]

I'm having trouble proving this

I think you additionally need the quotient map to be open

[sleek thicket]

I don't think my proof is valid, sorry

check out what lux is doing

[flint cove]

So for the first case I mentioned, to separate {x1} from {x2} in the quotient it would be sufficient to have open sets separating x1, x2 which lie entirely in the complement of K, so they're corresponding to open sets in the quotient.
But taking open U1, U2 separating x1, x2, we can just take U1 \setminus K and U2 \setminus K – since x1, x2 not in K, U1, U2 still contain x1, x2, and since in T2 compacta are closed, subtracting a closed set from an open one remains open, ta-daa

(Huh, I never realized that in T2 spaces you can strengthen the separation to compact sets, learned something new)

[gritty widget]

you are correct, sorry

tfw forgetting how to do math

[sleek thicket]

you are a complete loser because you forget irrelevant gentop fact #10169420

[gritty widget]

[tough imp]

You are a loser because you forgot the 5 adjectives you apply to a scheme map for this statement to be true

[sleek thicket]

"I'm chmonkey I do algebraic geometry"

[tough imp]

[flint cove]

hrr

[tough imp]

If I don't constantly remind ppl I do algebraic geometry then how will I remember to do algebraic geometry?

[shut moat]

why do we define the quotient topology using a quotient map? The equivalence relation definition feels much more intuitive

[sleek thicket]

what definitions are you thinking of specifically?

[obtuse meteor]

Are these not the same???

[shut moat]

idk maybe I haven't read into it enough lmao

but like one is the coareset topology with respect to which a map is a quotient map

and the other is basically the same but you're actively treating the codomain as an equiv class thing, which would be equivalent to a map X -->X/~

[sleek thicket]

i mean the equivalence relation one only works for a specific set

the quotient map business is essentially trying to detect when a surjection p : X -> Y is secretly X -> X/~ for some equivalence relation

i.e.\ when ie there an equivalence relation and a homeomorphism $h : Y \to X/\sim$ such that

$$
\begin{tikzcd}
& X \arrow[dl, twoheadrightarrow, "p"] \arrow[dr, twoheadrightarrow] \
Y \arrow[rr, "h"] & & X/\sim
\end{tikzcd}
$$
commutes

[obtuse meteor]

Yeah I usually think of the second one

[sleek thicket]

oof

[obtuse meteor]

lol

[gentle osprey]

SHAMROCK YELLING EMOJI

[shut moat]

I see

Ty!

[sleek thicket]

np! hope this helped

[shut moat]

so like, for some common examples I've seen informally in a knot theory seminar

you have like a square

and you identify the sides of the square to produce a cylinder or mobius strip or projective plane or something

[sleek thicket]

right

[shut moat]

would different "gluings" corrospond to different choices of p?

[sleek thicket]

Sure, and the codomain might change

if you want to think of the quotient as giving you a copy of the torus or something, that's saying there's a quotient map p : square -> torus

yeah?

and then maybe you glue the edges differently and get a quotient map square -> projective plane

[shut moat]

right, cool

ok this is much cooler than the previous point set stuff

[sleek thicket]

quotient stuff is awesome haha

very powerful and also finnicky

[ivory dragon]

quotient stuff is good because it annoys coq people

[sleek thicket]

Namington is a leanhead confirmed

[shut moat]

mniip harassed me with his coq one day in VC 😔

[summer jolt]

Can someone explain what is meant by the union of two algebraic varieties?

[ivory dragon]

set-theoretic union

what are you confused about?

[obtuse meteor]

[obtuse meteor]

Yeah quotient stuff is good for your health. Point-set is generally garbage but quotients allow you to construct neat spaces quickly and efficiently

inb4 cw complexes "approaching me" meme

[flint cove]

You Refer to the „Rest“ in „Top is CompHauss and Rest“, right?

[summer jolt]

So the definition I was given was the following: $V(I) \subset C^n$ consists of all points $P \in C^n$ s.t. $f(P) = 0 \forall f \in I$ where $I$ is an ideal in $C[x_1,...,x_n]$. Now I am unsure how to interpret this geometrically. I assume that by the union of two varieties we mean $V(I) \bigcup V(J)$ for some ideals $I,J$. However, what does their union look like? Like what is a typical element in the union?

[gentle osprey]

snypehype

[flint cove]

It consists of points that vanish either on a p in I, or a q in J.
If you think hard enough you might find a ring-theoretic combination of I and J such that all elements in there vanish on every point on V(I) cup V(J) 🙂

[summer jolt]

@flint cove ok thanks I'll give it a go

[flint cove]

If you want a specific example you can take ||I = (x1), J = (x2) for the plane ℂ². What's V(I), V(J), and as what vanishing set can you describe their union?||

[hard wind]

Brain hurty

I'm not sure exactly what facts about path-connected and simply-connected spaces i need to be taking for this proof

[flint cove]

My mind would go to existence of unique lift and that homotopic maps have homotopic lifts

[hard wind]

good idea, thank you -- we recently discussed unique lifting so that is probably right

so in this case we are lifting paths?

[flint cove]

Yes, I think the key point should be to notice that ||all loops in B are contractible, hence must have contractible lift ||

A similar argument: first show that ||π1(B) acts (at least in this case) transitively on each fiber||, and then you can use that ||π1(B) it's trivial, hence each fiber must be a singleton, giving injectivity ||

[hard wind]

Thanks for the spoilers, that was quite thoughtful

[obtuse meteor]

an alternative approach

the identity B -> B lifts to a map B -> E since Id_ast(pi1(B, x)) = pi1(B, x) = 0 is a subset of p_ast(pi1(E, xtilde))

more generally this shows something even better

[cedar pebble]

Faye out here enjoying AT more and more

[obtuse meteor]

and that B is path connected (and hopefully locally path connected???)

we might not need the locally path connected

let's check

hm, we'd need locally path-connected as well

this is an L

well so there's a really nice result

that if the base space is path connected and locally path connected, then if pi1(B, x) = p_ast(pi1(E, xtilde)) then you get homeomorphism by applying lifting properties via meme

do we believe in simply connected but not locally path-connected spaces @cedar pebble

And yes I am very much enjoying AT

[cedar pebble]

yea there are examples

e.g. the comb space

but all examples are going to be like

[obtuse meteor]

but do we believe in them nG

[cedar pebble]

really shitty

[obtuse meteor]

I expect there to be garbage examples but

[obtuse meteor]

do we care about them lol

yeah this was brought up earlier

[cedar pebble]

I mean sometimes people care about them, e.g. people that study shape theory

[obtuse meteor]

it was the first thing I thought of

sad times

[cedar pebble]

but as long as you're doing relatively tame things then no we don't care about them

[obtuse meteor]

good

[cedar pebble]

I mean they are an artifact of the category of topological spaces being bad

they will never show up in more algebraic approaches to topology

[obtuse meteor]

in showing continuity of this lifting guy

you need local path connectedness of the left hand space

really garbage

sad

@cedar pebble is there a theory of killing generators of your fundamental group by inserting a wavy sin(1/t) kinda boi

[cedar pebble]

[obtuse meteor]

this feels like a thing that should exist

like that's what that does right???

[obtuse meteor]

that's why this is simply connected

[cedar pebble]

I mean it should exist

[obtuse meteor]

you can't actually complete the loop

bc you inserted a bad boi

hmm

[cedar pebble]

but is this really something that should exist?

[obtuse meteor]

can you detect these using the techniques of algebraic topology at all?

[cedar pebble]

I mean any time you work with dreadful spaces like this you need to use something like shape theory, the usual techniques from AT start to fall apart a bit

[obtuse meteor]

idk what the fuck shape theory is lol

[tough imp]

triangle is a shape

[cedar pebble]

shape theory is sort of a toolbox for actually working with really pathological spaces like this

[obtuse meteor]

like okay here's an example of what should be a nice (if dumb) invariant

this thing is simply connected or whatever but is not an interval

bc I can take more than three points out of this

and keep it connected whereas I can't do that with interval boi

at least one should be able to by removing from the line at 0 non?

[cedar pebble]

usual homotopy theory is to shape theory as singular cohomology is to something like Cech hypercohomology

[obtuse meteor]

IDK wtf cech cohomology is in the first place

Shape theory????

SHAPESSS

[cedar pebble]

Cech methods are like

[obtuse meteor]

:)

https://estrogen.fun/i/xiz6.png

AHA

so you can

[cedar pebble]

so usually you compute something like cohomology by finding a nice open cover and then computing with that

[obtuse meteor]

you can detect this algebraic topologically

[cedar pebble]

Cech cohomology is sort of like

[obtuse meteor]

I mean

maybe not if you're a dirty algebraist

but

[cedar pebble]

you take the limit over all possible open covers ordered by refinement

[obtuse meteor]

tf

get back to me when I learn cohomology

we're starting homology on monday

am excited

[cedar pebble]

yay yay

exciting!

[obtuse meteor]

it's simplex hours

oh god this homework in algtop this week

is super duper spicy

no hints but

http://www.math.lsa.umich.edu/~jchw/2021Math592Material/Homework7-Math592-W2021.pdf

[gritty widget]

[obtuse meteor]

Labeling hard lol

[gritty widget]

idgi

i thought its the due date

[obtuse meteor]

The due date is 12 March

She just mistyped

[gritty widget]

ah, i am a dumbass

[cedar pebble]

🥺

[pastel linden]

awwww

professor who care

[flint cove]

Isn't a less scary way of formulating that to say “cech cohomology is looking at covers that are fine enough“?
But tbf I haven't rigorously gone through it yet; still need to write that down for the bundle classification at some point

[cedar pebble]

Yes in nice enough cases you can just look at fine enough covers

Once you go to fine enough covers the colimit stabilizes and you win

For really pathological cases this doesn’t happen, and moreover you need to consider hypercovers instead of just covers

[obtuse meteor]

Imagine putting this at the end of a super long homework assignment

TLDR “this section is completely optional”

[cedar pebble]

practice self care! Also practice these 18 problems uwu

[sleek thicket]

fucking owned

Wait I have a better idea

Keep the title "well-being"

And the 2 questions

But add in a bunch of really awful tricky problems

not optional but still under "wellbeing"

[cedar pebble]

[sleek thicket]

"are you okay? Are you sleeping enough? Calculate the integral of sin(x)^3."

[obtuse meteor]

Lol

[flint cove]

Like Konfuzius said, if you have a choice, you can well-order your being every time

[sleek thicket]

"hey make sure to drink enough water and prove that (L^q)^* = L^p in this problem with parts a-f"

[cedar pebble]

Go take a walk! Speaking of walking, answer these five questions about the walking quiver and its category of representations

[flint cove]

Afer too many assignments like that I would start associating a healthy lifestyle with the darkest areas of mathematics and avoid it at all costs

[obtuse meteor]

Ohno

[sleek thicket]

Shamrock isn't reflexive but Shamrock** is

reflexive topological space

discuss

it's a space homeomorphic to C(C(X)) with the compact open topology

[flint cove]

Wait, is it a space X homeo to C(C(X)), or is it an arbitrary space that is homeo to C(C(X)) for some X, or is it a space X such that the embedding into C(C(X)) is a homeomorphism?

[sleek thicket]

should be last, you're right

For eg normed vector spaces you can have V** ≈ V but not via the natural map

I think there is no such X though lol

I can't think of one

[marsh forge]

C(\emptyset) has an element tho

unless im being dumb

I guess both interpretations are interesting

[sleek thicket]

well the double dual of the concept is interesting

yesss my prof just believed me when I said I forgot to turn in my homework on time

[marsh forge]

naturally

[sleek thicket]

I legit did

finished it days in advance

[gritty widget]

oh

[sleek thicket]

And then didn't realize it was due at 1:30pm instead of midnight

[marsh forge]

i feel like 95% of profs respond well to that for me

[sleek thicket]

yeah I mean one would hope

But the late policy on the syllabus is super strict

so /shrug

[marsh forge]

on one hand i feel like strict late policy on surface but nice in email is a good strat

but i know some people hate emailing profs for like

[sleek thicket]

eh it creates a lot of stress for the student

[marsh forge]

personal or cultural reasons

[sleek thicket]

Well there's that but also like

[marsh forge]

yeah see it creates very little stress for me so i have to empathize here

[sleek thicket]

You might just not think about it

Or preemptively assume they'll reject it

[marsh forge]

yeah

thats what i mean

[sleek thicket]

hmm I think I might be able to ignore all the diff top on the last homework

Err

Diff geo

I did the diff top stuff

There's 2 topology-y bundles problems and 3 diff geo ones

And I have 208/210 on the homework so far

and need above an 80% to get a 4.0 in the class

I should probably learn this stuff

[meager python]

So it is dominant map between integral schemes gives a field extension k(t) = k(P^1) -> k(X).

[alpine bolt]

I created a function F: W --> R^1

W is a neighborhood of a point p, and p is in the image of g

Am I trying to show F(0)=0 or F(p)=0?

[rugged swan]

you can prove that the tangent space at (0,0) isn't 1 dimensional

[trail ibex]

If you remove the self intersection point there are 3 connected components

whereas removing a point from an open interval only gives 2 connected components

[gritty widget]

Oh?

I thought it is 2 connected components

Ah sorry, I was being a dumbass. You obviously mean locally

[uncut surge]

Ye I was confused too for a moment lol, but the argument is for a small neighbourhood of the point I guess

[trail ibex]

Yes

[jolly ice]

Ayy I have to prove that the set of all condensation points in R^K is perfect

[thin bramble]

Can anyone help? Prove that translations and halfturns map any line {x+tv : t in R} to a parallel line (same direction v)

[jolly ice]

Bro I already asked a question :/

Yell if x is in R^k

Then a translation is just adding (t,0,0,0...)

So just prove that it would preserve parallelness

Wait say that if x goes to x' the distance is a constant

Thus parallel

[thin bramble]

Thanks. I'll see what I can do from here

[jolly ice]

You have to show that is some way the distance is a constant

Best way is to use the map itself to show it since you already know what the map is

[summer jolt]

I was asked to prove that the intersection of two algebraic sets is an algebraic set. So an algebraic set is given by $V(I) = {p \in C^n | f(p) =0 ,\forall f \in I}$ where $I$ is an ideal in $C[x_1,..,x_n]$.

The way I approached it was to assumed that given two ideals $I$ and $J$ then $V(I) \cap V(J) = {p \in C^n | f(p) =0 ,\forall f \in I \text{ and } J}$. So then I thought that this could be given by $V(I) \cap V(J) = V(I+J) = {p \in C^n | f(p) + g(p) =0 ,\forall f \in I \text{ and } \forall g \in J}$. Since, if $p \in V(I)$ and $p \in V(J)$ then $f(p)=g(p)=0$ so $f(p) + g(p) = 0$. Hence it would follow that $p \in V(I+J)$ as $I+J$ is an ideal.
However, I'm having trouble proving the converse direction, that is given $p \in V(I+J)$, then $p \in V(I)$ and $p \in V(J)$.

Any help would be appreciated.

[gentle osprey]

snypehype

[jolly ice]

Isnt this abstract algebra?

[summer jolt]

I mean yeah or algebraic geometry?

Should I post it there?

[jolly ice]

I think you might find better success in algebra?

[summer jolt]

Ok I'll post it there.

[thin bramble]

So I was given a hint to use $\tau_a (x) = x+a$ and $\sigma_a(x) = 2a-x$

[gentle osprey]

KillerWhale2498

[tough imp]

Algebraic Geometry stuff goes in here haha

If it’s in the form of pure algebra it goes there, but if it’s AG stuff it goes here is usually@how it goes. Tbf tho it’s kinda a grey area and it ends up in both

[sleek thicket]

I've posted very geometric stuff there and very algebraic stuff here

My metric is "which channel is open"

[jolly ice]

I have a question on some perfect sets

In E (_ R^k all condensation points of E make a perfect set

So E is a subset

Actually wait nvm solved

[jolly ice]

If anyone wants to help me check the proof pls come to math voice

[candid mulch]

Is (a, b) paracompact?

[sleek thicket]

yes

It's homeomorphic to R, so you might as well ask if R is paracompact

[candid mulch]

Awesome, thanks

paracompactness feels much weaker than compactness

[fervent citrus]

every metric space is paracompact

[candid mulch]

lol that came up a few minutes into the lecture once I posed the question

[flint cove]

definitely

This question only makes sense if taking the interior actually produces a cover, which is not the case in that example

Keep in mind that this imagery heavily uses that D is embedded into ℝ²

if you just take D as a metric space and ignore the surroundings the open balls around the radius one points are actually “half open” in the sense of the imagery

[jolly ice]

Couldn't you increase the radius of the closed ball by a size of epsilon and use that to create a finite subcover and then prove that the subcover must correspond to a sub cover of the closed balls

Since epsilon can be arbitrarily small all you have to prove such an epsilon exists

Yes but if you do so for an arbitrarily small epsilon you have to prove that the difference is insignificant

[flint cove]

I'm not clear about what you are referring to (what do you wanna prove / disprove?)

[jolly ice]

Prove that the enlarged balls subcover corresponds to the finite sub cover of closed balls

[flint cove]

Not sure what you mean by „correspond“. That they are equal? Because that they are not

[jolly ice]

Bijection

Because what I am proposing is a bijection between infinite covers

Yea That's what I want to prove

Can you reference it?

Isn't that a non question then?

[cosmic beacon]

lmao

[jolly ice]

Oh ok nvm I was trying to look at another question

Isn't that one obvious tho? Since the radius does not converge and compact spaces have to be bounded I believe

Wait nvm I shouldn't say that

[prisma seal]

it's not obvious imo, because it would hold with open balls, and so you might expect it to work with closed balls

[jolly ice]

It's non obvious but follows from compact sets

Oh I was referring to them being the same radius

No I think it's false because it follows from the being the same radius

Infinite covers with sets of the same radius cannot be compact

[prisma seal]

there is a pretty vacuous counterexample; take your favourite infinite compact metric space, and cover it with closed balls of radius 0 (all the points)

no finite subcover

yeah haha I did say vacuous

wait fr

Yeah I think the analysis course I took remarked that

[jolly ice]

Haha was that at waterloo @prisma seal

[prisma seal]

yes I'm a waterloo student!

[jolly ice]

I am in the same servers as you. You have answered my questions multiple times :)

[prisma seal]

that sounds right

Are you done with this question?

Ok awesome

I have a differential geometry question; what is the intuition/geometric meaning of the Gauss-Bonnet formula? This is how it was presented in the notes my course is using

On an assignment, I'm being asked to use this to find an area, but I don't actually understand what this is saying

[marsh forge]

my god thats painful to read

[prisma seal]

max this course has no lectures

just these notes

[marsh forge]

lmfao

[prisma seal]

I can honestly say that I'm gonna include it in my will to have someone project these notes at my funeral

everyone will understand what happened

[jolly ice]

Ngl these notes are making me excited for waterloo

What year is this?

[prisma seal]

this is a third year differential geometry course, you can do it in second year

[marsh forge]

yeah my suggestion nicholas

[prisma seal]

this course either provides no intuition or does this

[marsh forge]

is to take that much better explanation

and then

try to see

how what your given

js a special case

so you can use it for hw

[prisma seal]

there is another statement later that's this

[alpine bolt]

How do I construct a neighborhood out of X?

[flint cove]

do you have a plan on how to proceed?
Not sure what kind of neighborhood you are trying to look for.

[alpine bolt]

tbh im not sure either im just trying to use this definition to arrive @ contradiction

[flint cove]

oh, so you probably defined a manifold as a submanifold of ℝ^n

Did you prove a fact like „M is a submanifold iff around every point there is a local diffeomorphism to ℝ^n“?

[alpine bolt]

um

i did not

[flint cove]

like, not you in particular, but the material to go to
As in, is it something you are „allowed“ to use 🙂

Because the intuition is that locally, it's not one dimensional, because the lines are not „horizontally isolated“

[alpine bolt]

what does the word "locally" mean in the mathematical sense?

[flint cove]

„something holds locally“ usually means „for every point there is a neighborhood satisfying something“
e.g. „Locally path-connected“ means „around every point there is a path-connected neighborhood“

And something is a smooth 1-manifold if and only if „locally“, i.e. for every point around some neighborhood, it is diffeomorphic to an open interval

So one way to disprove that this is a 1D Manifold would be to pick any point, wave your hands, and say „Hey, this doesn't look like an open interval, no matter how small I make the neighborhood!“

and if you prefer rigor to hand-waving you can use your favourite topological property that holds in the open interval but not in the space X you're given :>

[sleek thicket]

That's not what a locally path connected space is

If it was path connected would imply locally path connected

Locally has multiple meanings, one is "happens in a neighborhood of each point" and one is "happens in arbitrarily small neighborhoods of each point"

locally path connected is of the second type

[flint cove]

ah gosh, sorry for spreading lies, should've looked that up, thanks for correcting

[sleek thicket]

it's okay, I am constantly bearing false witness

[flint cove]

in my defense I wanted to write „has a neighborhood base of path-connected points“ but didn't want to explain what a nbhd base is

[alpine bolt]

maybe i should just find when [x y]=0 and just find the jacobian at that 0 point

[marsh forge]

@sleek thicket i think locally almost always means arbitrarily do you have a counterexample?

[sleek thicket]

"a section of a sheaf is zero if it is locally zero"

"a manifold is locally Euclidean"

I mean it's true in both cases that this happens in arbitrarily small neighborhoods

But what's being said is "happens on a cover"

Idk when I hear local my first instinct is "can be checked on a cover"

[marsh forge]

I think my point is like

if the two notions differ

its always the one o saod

said*

[sleek thicket]

Sure

[sleek thicket]

Sometimes diff geo is good

My bundles class started doing more diff geo type things

Lots of truly horrible computations

And we ran out of the prepared stuff in the book so there was only lectures

I couldn't follow it at all

But now I'm doing the homework and like

I can just do computations

And it works

I don't need to have a deep fundamental understanding

[gritty widget]

post computations

[sleek thicket]

Let E be a rank k vector bundle

Everything is in the smooth category

Say we have nonvanishing k form Ω on E

This gives a reduction to SL(k, T) or SL(k, C)

Oh and say we have a connection \nabla on E

Then $\nabla$ is compatible with the $SL$-structure iff $\nabla \Omega = 0$

[gentle osprey]

!!Shamrock!!**

[sleek thicket]

Compatible meaning this

[gritty widget]

"not yet written"

share the book

[sleek thicket]

oh yeah lol

No

I would get in trouble

[gritty widget]

[sleek thicket]

I need that rec letter

Why do you think I got into fields

[shut moat]

graduate then share

[marsh forge]

how will he know u leaked it

[sleek thicket]

He might not know I leaked it but it's a pretty small class

It's also like a very incomplete draft copy

I'll leak it when it's published and more complete

[flint cove]

what's ISM?

ah

[gritty widget]

it has made me wonder. there were some draft lecture notes/books that my professors in college wrote that we used in their class. they are still unpublished, after 5 or 6 years. i guess it takes a long time to flesh out/publish something

[cosmic beacon]

Would the set of all sequences that start with 0,1 be open in this metric space?

[gritty widget]

as in, individually? (the set of all sequences that start with 0, and the set of all sequences that start with 1, are indiviudally open sets?)

[cosmic beacon]

no, the set of all sequences that go (0,1,something,something,etc.)

[gritty widget]

but every sequence starts with either 0 or 1 right?

so you are talking about the whole space?

ohh

i see

[cosmic beacon]

for example, the sequence (0,1,0,1,0,1,...) would be in my set. so would (0,1,1,1,1,1,1) or (0,1,0,0,0,0), etc.

It's supposed to be

but I cant express it as one is the thing

The "lowest" sequence is (0,1,0,0,0,0,...) which has a norm of 0.25. the "highest" sequence is (0,1,1,1,1,...), which is whatever the binary number 0.111111111.... is in decimal (0.49999999?). So, it's essentially the interval [0.25,0.5)?

[fast minnow]

Might’ve helpful to check the distance between (0, 1, 0, 0, ...) and (0, 1, 1, 1, ...)

[cosmic beacon]

given the binary number 0.001111111..., the distance between the two is that number in decimal. So basically almost (but not quite) 0.25, right?

[sleek thicket]

lmao link

or dm me or whatever

[cosmic beacon]

idrk what that tells me

yea, 1/8+1/16+1/32+1/64+...

ohh is it?

I forget it lol

👍

[fast minnow]

Another way to think about it “what is the closest point to (0, 1, 0, 0, ...) which does not start with 0, 1)”

[cosmic beacon]

0,0,1,1,1,1,1,1

[fast minnow]

Right, so what is that distance? Then make a guess for what the ball should be

[cosmic beacon]

the thing is, the norm of (0,0,1,1,1,1,1,1) is the same as the norm of (0, 1, 0, 0, ...)

so how can I have a ball which includes one but not the other?

[fast minnow]

You have to use the given metric to calculate distances

[cosmic beacon]

or, that's what I mean, the distance from any point to (0,0,1,1,1,1,1,1,...) is the same as the distance from that point to (0,1,0,0,0,0,...)

[fast minnow]

I don’t think that’s true, like (0, 1, 0, ...) is distance 0 from itself but it will have positive distance from the other one

[cosmic beacon]

well, compare it to the sequence 0 repeating. The first one is [\sum_{n=1}^\infty\left(\frac12\right)^n-\frac14-\frac12=\frac14.]
The second is also trivially $\frac14$.

[gentle osprey]

Isaiah

[cosmic beacon]

wait, I think I see actually, ignore that

[chrome dew]

nice ultrametric space you got there

[sleek thicket]

oof

[cosmic beacon]

alright would it be the ball of radius 1/8 centered at 0,1,1,0,0,0,0,0,0,...?

[fast minnow]

If you mean (0, 1, 0, 0, ...) as your center then it should be something like that yeah

I think the radius might be 1/2 or something though

[chrome dew]

either sequence is a center, the radius of the "open" ball should be 1/4 in this case

but all balls are clopen so by "open" I mean with strict inequality not equality like this: d(x,(0,1,...)) < 1/4

[cosmic beacon]

how did you get that 1/4?

[chrome dew]

maybe I'm wrong about the radius actually I'm computing it off

we were talking about something similar a while back haha but yeah it's nt exactly thes ame one

[cosmic beacon]

i cannot gain any intuition for what sequences are "close together"

ok, i'm pretty sure the ball would be radius 1/4, but I wouldn't even know how to prove it. Mind you, I'm supposed to show all sets containing all sequence starting with some fixed entries can be expressed as an open set, and I can't get even one set

yep

yep

[gentle osprey]

slimvesus

[cosmic beacon]

yes

[gentle osprey]

slimvesus

slimvesus

[cosmic beacon]

im an idiot lmao

yes of course

[flint cove]

Does any compact manifold admit a cover of contractible sets where every pairwise intersection is the disjoint union of contractible sets?

[cosmic beacon]

yes

[flint cove]

TIL it's called a „sufficient open cover”

[cosmic beacon]

yep

I can take unions of balls

I dont need to show its a ball in its own right

I just need to show its an open set

I see

Well, (0,0,1,0,0,0,...) has a distance of 1/4 away from (0,1,0,0,0,...), a problem.

oh wait

no thats wrong

its 1/4 + 1/8

so we pick a different midpoint in the set

but picking that midpoint is the problem as well

yes, I believe there is as well, but constructing the open balls is a PITA. The exact question is asking me to prove that the metric topology defined above is equivalent to the topology on the same set formed by taking all such sets containing sequences with fixed entries as a basis

do you just alternate the terms after the fixed terms?

[chrome dew]

did you try making the center (0,1,1,1,1,1,...) maybe?

or that doesn't work

[cosmic beacon]

im thinking maybe you pick the sequence thats farthest away from (0,1,0,0,0,0) and (0,1,1,1,1)

[cosmic beacon]

yea that doesnt work

lmao can you give me a hint

without giving it away

well, how did you figure out the radius if you dont even know the midpoint to pick?

or did you fully figure it out

you mean it'll be the same radius regardless of choice of midpoint?

but then what about thids

[gentle osprey]

slimvesus

[cosmic beacon]

jeez

is it just the union of the two endpoints with radius 1/4?

what point would be in the ball of radius 1/4 centered at (0,1,0,0,0,...) that shouldn't be there?

Like if I take the ball of radius 1/4 centered at (0,1,0,0,0,0,...) and the ball of radius 1/4 centered at (0,1,1,1,1,1,...) wouldn't htat give me what I need?

wasn't (0,1,1,1,1,1,1,1,...) the only one that wasn't included in the ball of radius 1/4 centered at (0,1,0,0,0,0,...)?

yea it is

lol np

these suck lmao

I'm pretty sure what I described works

lol ty for the help

[gentle osprey]

slimvesus

[cosmic beacon]

Lol I'll take your word for it, I'm too sick of this to be curious right now unfortunately

I'll take a look later

[fast minnow]

That sounds like it should work

Basically the metric tells you that if two things are within say 1/4, then they agree at the first 2 coordinates

[cosmic beacon]

yep

B(b_1,...,b_k) is the set with fixed first k entries

[fading vale]

wtf does this mean

[gentle osprey]

many such cases

many such cases

[fading vale]

or wait

[gentle osprey]

many such cases

many such cases

many such cases

[fading vale]

i guess that makes sense

[gentle osprey]

Ted

[viral atlas]

My visualisation is restricted to a continuous function from bR\to\bR, I'm having a hard time finding a counterexample

Like

Yeah, I need help here, I can't prove/disprove this

Aaahhh, okay, that makes sense.

[fading vale]

other counterexamples usually come from putting two metrics/topologies on the same space and taking the identity

like uh say Q' is the rationals with the discrete metric

and take the identity Q' -> Q

or you could do R' and R

mhm

[viral atlas]

I might have to think a bit more about this, but thanks for the help. :)

[fading vale]

(if it helps the geometric intuition here is that your domain has "finer" balls than your codomain)

[viral atlas]

In the discrete metric case?

[fading vale]

yes

[sleek thicket]

This is a good thing

Because there are no such counterexamples

By invariance of domain

[fading vale]

hi shamrock

[sleek thicket]

Let me try to think of an easy proof in the 1d case

hi moth!

[viral atlas]

Nice, at least one thing I got right lmao.

[sleek thicket]

lol

ah yeah

[fading vale]

yea monotonicity stuff

[viral atlas]

I'm trying to think more about what Moth said.

[sleek thicket]

Does that make sense Ted?

the increasing/decreasing thing

[fading vale]

the way i think about it Ted is that going from the discrete metric to the normal metric basically loses information right

[sleek thicket]

I don't think so?

[viral atlas]

Hmmmm, yeah, that does make sense

[sleek thicket]

Just add a gap

[fading vale]

so you cant go backwards

because that information is "lost"

[sleek thicket]

f(x) = x + sgn(x)

Bijection is continuous iff strictly increasing

[fading vale]

slimvesus moment

[sleek thicket]

:slime:

[fading vale]

🟢

[sleek thicket]

good emote