point-set-topology

[empty grove]

great

[fading vale]

You only have to worry about the cohomology of any one ball here

because the balls are open, right

[flint pilot]

I'll fill in all the details when I write it up nicely.

yes they are open

[fading vale]

so their union is a finite union of open sets that covers itself obviously

and their intersection is trivial

by exactness we get isomorphisms from the direct sum of the cohomologies of the balls onto the cohomology of the union

so as long as you can compute H^k(B) you're good

[empty grove]

sorry I have to leave now

[fading vale]

cya

[flint pilot]

oh ok. Well thank you so much for your help! Let's see if @fading vale and I can work on part b. If you don't mind

[empty grove]

Np 🙂

[fading vale]

can you repost the problem please

its kinda hard to scroll up haha

[flint pilot]

yes. Here's part b: Let A = ${(x,y) \in \mathbb{R}^2: 1 < x^2 + y^2 < 4}$ Compute $H^k(A)$ for $k = 0,1,2,\dots$

[gentle osprey]

hype_db

[fading vale]

ah i see

mhm before we work on part (b) id like to say that theres a much simpler solution to part (a) using mayer-vietoris

its basically what i described before, the balls are open so their union is a union of open sets covering itself and we can apply mayer vietoris because its finite

[gentle osprey]

Moth In Shambles

[fading vale]

meaning the image of each map is contained in the kernel of the next

[flint pilot]

yeah that's going above my head. I'm sorry. That circle with lines through the middle i don't know, and we haven't even mentioned kernals in here. I appreciate the effort there, but I think I'm just going to go with the longer way lol

maybe once I take a topology course I'll get it better

[fading vale]

uh

wait

how did you define mayer vietoris

(the circle is the direct sum btw)

[flint pilot]

Let U and V be open sets in $R^n$ with U and V homologically trivial in all dimensions. Let $X = U \cup V$; suppose $A = U \cap V $ is non-empty. Then $H^0(X)$ is trivial and for $k \geq 0$ the space $H^{k + 1}$ is linearly isomorphic to the space $H^k(A)$.

[gentle osprey]

hype_db

[fading vale]

oh man okay haha

yeah nvm

[flint pilot]

so yeah it's ok. I'll just stick with what I have lol

let's do part b so I can be done with this problem. lol

[fading vale]

thats real wacky

[flint pilot]

I agree at least it's only one problem he asked about this stuff

[fading vale]

okay i havent really done much de rham stuff so i dont think i can help very much with the computation itself but as a hint: the cohomology should look like the cohomology of the circle

[flint pilot]

even cohomology wasn't defined in my book. I learned that word today

[fading vale]

[flint pilot]

so I really don't know how to do part b. It's ok though, I can wait for either Moldilocks to come back or someone else to jump in. I'll repost the question in case if anyone else wants to jump in

part b Let A = ${(x,y) \in \mathbb{R}^2: 1 < x^2 + y^2 < 4}$ Compute $H^k(A)$ for $k = 0,1,2,\dots$

[gentle osprey]

hype_db

[flint cove]

Do you have a visual understanding of what this space looks like?

[flint cove]

(also, is this integral or real cohomology? sorry if this was in the backlog)

[flint pilot]

I'd think it's a circle with radii between 1 and 4.

cohomology was a new word for me today, so I am not sure

[flint cove]

What tools do you have? Meyer-vietoris (i.e. Covering it with simple sets)? homotopy invariance (i.e. homotopy equivalences induce isomorphisms in cohomology)?

[gritty widget]

their class hasn't even defined homotopy

[flint pilot]

nope

[flint cove]

That's why I'm asking 😉

[flint pilot]

we were given a version of meyer-vietoris, but someone else earlier said it was a very bad version. We did part a of this problem from other principles that I understand even though it was longer

[flint cove]

Well, you can try applying whatever version of mayer-vietoris you have, if that's everything you have (C:
Do you have an idea of how one could cover this „circle“ (let's call it annulus instead) with „simpler“ sets?

[flint pilot]

I do not. I have no background in topology, and my instructor covered this on the last day of class without doing an example as he got stuck in the proof. It was really an unfortunate situation

[flint cove]

Mayer vietoris essentially tells you that H(U\cup V) is determined by H(U), H(V) and H(U\cap V).
So if U, V, and the intersection are „simple“ in such a way that you already know the cohomology for these sets, you should be able to calculate the cohomology of the whole space.

You can first answer out of „gut feeling“ how you can find such a cover (i.e. with sets that „feel conceptually simpler“ than the annulus)
Then you can try looking through your resources to find what spaces are „nice enough“ that the cohomology is already known.

[flint pilot]

I don't have a gut feeling. I don't even know what cohomology is. Like I said I have no background in Topology. I don't know what spaces have known cohomologies

so I still have no idea how to go about the problem

[flint cove]

Ignore the words „cohomology“ and „topology“, and try to think about it just visually – how can you cover this annulus with two simpler sets?

[flint pilot]

with circles? I don't know

[flint cove]

Ah, good idea, haven't thought about that

In this case we require open sets tho.

The answer would be that we take a set U that covers the left half plus a bit more, plus a set V that covers the right half and a bit more

[flint pilot]

ok so U would basically start at 1 and go to about halfway and V starts at that halfway point and then goes to 4

[flint cove]

Ah, wrong „right“, I think I should draw this

ah too stupid to rotate, my bad

[flint pilot]

it's ok. I can see it fine

[flint cove]

the point is that the sets U, V themselves have known cohomology because they're „essentially ℝ²“, and the intersection as well, just „essentially two portions that look like ℝ²“

But what the precise properties are that you need here depends on what your material covered.
Has the notion of „contractible“ been introduced?

[flint pilot]

it has not

[flint cove]

What book / lecture is this

[flint pilot]

Multivariable Analysis we're using Analysis on Manifolds by Munkres

[flint cove]

So it's de rham cohomology, right? I.e. closed differential forms up to difference of exact differential forms?

Perhaps the intent of the assignment was to explicitly calculate it, lol.

[flint pilot]

yeah he called it the DeRham group. H^k(A) = C^k(A)/E^k(A) where C^k(A) is the set of closed k forms and E^k(A) is the set of exact k forms

[gritty widget]

doesn't munkres have an entire chapter on de rham cohomology

[flint pilot]

and yes it is asking to calculate H^k(A) for k = 0,1,2,....

section 40 talks about it. It's a short section though, and my professor got stuck in one of the proofs so he didn't do an example or cover anything cause he got stuck. He also did not mention cohomology at all

[flint cove]

I assume it has been shown that H^*(ℝ^n) = (ℝ, 0, 0, …), right?

[flint pilot]

I can see if it's in the book. It wasn't covered in class

[flint cove]

I should've been clearer. with „explicitly calculate“ I mean „don't use any tools like mayer-vietoris and try to find the explicit classes one-forms/two-forms contributing to the cohomology“, which would be… perhaps possible, but way less systematic, and I don't see an obvious way how.

with MV it's clear how to arrive at that goal at least :>

[flint pilot]

well, let's just do it with MV. This is my last part of a problem for the entire semester, and I just want it to be done lol. I am hopeful I'll understand all of this better when I take an actual topology course, it's just tough right now with how my professor did it

[flint cove]

sure

So first of all we don't need to worry about H^k for k>2 since every such k-form must be zero.

Also, H^0 must be ℝ, since we only have one connected component (H^0 is the class of locally constant function up to difference of a constant function)

[flint pilot]

so that means it would be 1 since that is the dimension of R. And why is it 0 for everything larger than 2? If it's possible to explain with my limited background

[flint cove]

Because e.g. H3(X) = (closed 3-forms) / (exact 3 forms), but both things in this quotient are a 0-dimensional vector space

[flint pilot]

got it

[flint cove]

i.e. when there's no 3-form, there cannot be any „nontrivial“ one

[flint pilot]

that makes sense

[flint cove]

We also need that $H^k(A \sqcup B) = H^k(A) \oplus H^k(B)$, where $\sqcup$ denotes disjoint union

[gentle osprey]

lux

[flint cove]

Because recall that the intersection of our left half and right half is essentially two disjoint portions of ℝ^²

and that tells us that we can look at them separately

[flint pilot]

So this A and B correspond to the U and V you were talking about earlier?

[flint cove]

No, but to both halves of $U \cap V$, i.e. the top and bottom portion of the intersection

[gentle osprey]

lux

[flint cove]

they are also „nice“ as sets 🙂

[flint pilot]

oh ok. So U and V need to be defined in order to define A and B got it

and this is for the k = 1 case, correct?

[flint cove]

The formula with the A and B holds in general tho, not only in our specific example, and for every k

[flint pilot]

ok got it

[flint cove]

If you care for the reason: I think it's possible to show it directly when plugging in the definitions and noting that a form over A \cup B is exact iff the restriction over A is exact as a form over A and the restriction to B is exact as a form over B. Note that the „hard part“ is the converse, i.e. showing that an exact form over A and an exact form over B „together“ make an exact form over A and B together. But this works in this context since A and B don't intersect anywhere.
(with closedness we have the same, but it's easier since it's a local property)

Soooo

We have $H^\ast(U) = H^\ast(V) = (\mathbb R, 0, 0, \ldots)$ and $H^\ast(U\cap V) = (\mathbb R\oplus \mathbb R, 0\oplus 0, \ldots)$

[gentle osprey]

lux

[flint cove]

And this is enough data to be „plugged in“ to mayer-vietoris

[flint pilot]

I will write down the reason to look at in the future. I don't like accepting things without knowing the reasons, but for right now I'm ok just going with it. So then, what is this H^*? It's not the same as H^k is it?

And that ending result, is that R^2, or is that expression meaning something different

[flint cove]

Yeah I feel ya, I don't like accepting such statements as well.

The * is just a placeholder, right. I could've also wrapped parentheses around it and write stuff like (H^k)_k = (…)

(Because H^k refers to a single cohomology group, whereas I wanted to write the values down simultaneously, so it's just syntactical shenannigans)

[flint pilot]

got it

[flint cove]

The end result should be H¹(X)=ℝ, H²(X)=0, if I'm not being really, really stupid

[flint pilot]

well, if you don't mind, could we go through to check that? I'm so sorry, just without seeing an example actually plugged into MV I think seeing it in action would help

so in my statement of MV, H^0(A) is trivial (we have it as R) and H^{k + 1}(A) is linearly isomorphic H^k(A). But what exactly becomes H^k(A)? And if they are linearly isomorphic, does that imply they would have the same result?

[flint cove]

Yes
I'm just verifying that I didn't do any stupid mistakes, I'm doubting my statement about H^0 a bit

let me think

[flint pilot]

ok no worries.

this makes me want to take a topology class really badly lol. Part of regrets not taking it this term

[empty grove]

This is the poincare lemma I mentioned earlier

[flint pilot]

I mean it was either Topology or Computational ODE and I research dynamical systems so I guess it made sense

oh hey welcome back @empty grove

[empty grove]

Hello

[flint pilot]

and yes, you mentioning Poincare really helped. I read the section so once he said that I was ok with it.

[empty grove]

Ah cool

[flint cove]

Ah, yes, I was stupid
It's not locally constant functions up to constant shift
It's all locally constant functions

[empty grove]

I'll catch up on some of the chat

[flint cove]

so I remembered correctly. But note that in that case H^0(A) is never 0 unless A is empty

H^0(A) is always ℝ^{number of connected components of A}

Which is what confuses me about the formulation of MV that you wrote, where it said that H^0 would become trivial.

[flint pilot]

here let me write out the entire thing again just to be safe. This is word for word from Munkres

[flint cove]

that would be a good idea

[flint pilot]

Let $U$ and $V$ be open sets in $\mathbb{R}^n$ with $U$ and $V$ homologically trivial in all dimensions. Let $X = U \cup V$; suppose $A = U \cap V$ is non-empty. Then, $H^0(X)$ is trivial, and for $k \geq 0$, the space $H^{k + 1}(X)$ is linearly isomorphic to $H^k(A)$

[gentle osprey]

hype_db

[flint pilot]

if you're curious, because it is easy to get the PDF online, it is on page 337 of the book

[empty grove]

What does homologically trivial in all dimensions mean? All cohomologies are zero dim?

[flint cove]

This is just wrong, unless Munkres has a very nonstandard way of calling the vector space H⁰ „trivial“

[flint pilot]

let me check the book. My professor didn't talk about it

[empty grove]

Yeah any non empty space would have non trivial 0th Cohomology

[flint pilot]

I don't know if this helps, an open set A of R^n is homologically trivia in all dimensions if it is star-convex.

[empty grove]

Oh cool

It does

It is all the sets on which poincare lemma applies

[flint cove]

Yep, so the poincare lemma justifies this naming

[empty grove]

Cohomologies being R, 0, 0, ...

[flint cove]

Still think it's very misleading to call the specific vector space H^0 trivial, but „cohomologically trivial“ is a totally valid term.

[flint pilot]

do you think they mean cohomologically trivial when they say trivial in this book?

[flint cove]

no, because the property „cohomologicall trivial“ applies to a topological space. H^0(A) is not a topological space.

[flint pilot]

got it

[flint cove]

okay, let's use this

[empty grove]

But it's still calling H^0(X) trivial when it's 1 dim

Which is weird

[flint cove]

We've established that H(U) = H(V) = (ℝ,0,0…), i.e. U and V are „cohomologically trivial“
Where U was the left and V the right portion

The theorem now says that $H^{k+1}(X) = H^k(U\cap V)$

(avoiding „A“ because that's what I called one of my intersection pieces)

[gentle osprey]

lux

[flint pilot]

Oh that's true. So I'm thinking of X as the one in the statement of the problem got it

[flint cove]

yop.
So for X it's the entries for U\cap V shifted one to the right
And on the 0th entry we have the „clearly trivial vector space“ ℝ 😄

[flint pilot]

yes lol

[empty grove]

0th Cohomology of intersection would be R² right?

[flint cove]

yep

[empty grove]

I see, cool

[flint pilot]

and that comes from the $H^*(U \cap V) = (R \oplus R \oplus 0 \oplus \dots \oplus 0)$ you had earlier right? At 2, that is $\mathbb{R}^2$?

[gentle osprey]

hype_db

[flint cove]

Every second oplus should be a comma, but yes

[flint pilot]

oh yes my bad

alright. Well that seems to make sense

[flint cove]

Although I'm not sure what you mean by „at 2“, since the second entry is 0 instead of ℝ²

[flint pilot]

I meant for k = 2. My bad

k = 1 *

I don't know why i said 2 lol

[flint cove]

Oh, no, the first entry is k=0

[flint pilot]

because that will give me H^1 ahhh

[flint cove]

Yep, so

\begin{align*}
(H^k(U\cap V)){k\geq 0} &= (\mathbb R^2, 0, \ldots)\
\implies \quad (H^k(X)){k\geq 0} &= (\mathbb R, \mathbb R^2, 0, \ldots)
\end{align*}

[gentle osprey]

lux

[flint cove]

Wait that looks wrong tho

[flint pilot]

is it though? The theorm gave you H^{k + 1} not H^{k}

[flint cove]

hence the values are shifted to the right, the first entry is just our H^0 we know anyway

i.e. H¹(X) = H⁰(U\cap V), H²(X) = H¹(U\cap V), etc

but that cannot be right, the first cohomology of the annulus must be ℝ as it retracts to a circle

[flint pilot]

and the theorem says it is trivial

here, do you think we should write out the whole proof? To see if it's all good? I can try to do it

[flint cove]

You can try that, I'll try to recover this proof from full mayer-vietoris

[empty grove]

Yeah you should get R, R, 0, 0, 0, ...

[fading vale]

its nonreduced cohomology i guess

[flint pilot]

We only need to worry about $k = 0,1$ as for $k \geq 2$, every $k$-form is $0$. This is because everything above this point will be in a $0$-dimensional vector space. Define $U$ as the left half of the region (plus a little more) and define $V$ as the right half of the region (plus a little more). Define $A$ and $B$ as the upper and lower portions, respectively, of $U \cap V$. We have that $H^k(A \cup B)$= H^k(A) \oplus H^k(B)$ because $A$ and $B$ are disjoint, and exact forms over $A$ and $B$ separately make an exact form over $A$ and $B$ together. So, we have that $H^(U) = H^(V) = (\mathbb{R},0,0,\dots)$ and $H^*(U \cap V) = (\mathbb{R} \oplus \mathbb{R},0 \oplus 0, \dots, 0 \oplus 0).$ Now, Mayer-Vietoris tells us that $H^{k + 1}(X) = H^k(U \cap V)$ and that $H^0(X)$ is homologically trivial, meaning $H^0(X) = \mathbb{R}$. Plugging in $k = 0$, one has $H^1(X) = \mathbb{R}^2$. Any other $k$'s give us a value greater than 2, meaning they are $0$. Therefore, $H^0(X) = \mathbb{R}$, $H^1(X) = \mathbb{R}^2$, $H^k(X) = 0$ if $k \geq 2$

[gentle osprey]

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

[flint cove]

From the long exact sequence we get an iso $H^k(U\cap V) \simeq H^{k+1}(X)$ only for $k\geq 1$

[gentle osprey]

lux

[fading vale]

Lux apparently this isn't the form of mayer-vietoris they were given

[flint cove]

I know

I just wanted to verify it

[fading vale]

oh

yeah

[empty grove]

the version he has is just for the specific case of contractible spaces pretty much

[flint cove]

Ah, so this is probably missing „is linearly isomorphic to Hk(A) for k≥1“

which is why we can't calculate $H^{0+1}(X)$ with that

[gentle osprey]

lux

[flint pilot]

In my book it says k >= 0 not 1

[empty grove]

for k = 1 it will be $\frac{H^0(U\cap V)}{\mathbb{R}}$

[gentle osprey]

Moldilocks

[flint cove]

That matches what I got from the sequence

[empty grove]

yeah I plugged it into the sequence too

[flint cove]

ALL HAIL THE SEQUENCE

[flint pilot]

so wait, is it right then? The way I typed it out earlier?

[empty grove]

no, the book missed the quotient by R

for the H^1(X)

for bigger k it is correct

[flint pilot]

So H^0(X) = R, what does the quotient come out to for H^1(X)?

[empty grove]

H^0(U cap V) is R^2 as 2 path connected components

and we quotient it by a 1 dim subspace so we get R

everything upto isomorphisms

[flint pilot]

odd. SO the book's theorem is wrong. So other than the division by R, is the rest of how I wrote the proof ok? It almost looks like A and B play no roll

[empty grove]

A and B being the components of the intersection?

The intersection does play a role

[flint pilot]

it just looks like I mentioned them but then they didn't come up again later.

everything switched back to U and V

[empty grove]

We got $H^*(X) = (\mathbb{R}, \frac{H^0(U\cap V)}{\mathbb R}, H^1(U\cap V), H^2(U\cap V),...)$

Yeah the A and B were for notational convenience here

you use them to figure out the cohomologies of the intersection

[flint pilot]

oh ok. SO just stating them in the way I did would be ok?

[gentle osprey]

Moldilocks

[empty grove]

in which way?

[flint pilot]

in the long proof I wrote out above. I tried to write out the entire thing

[empty grove]

yeah you were able to write $H^k(U \cap V) = H^k(A) \oplus H^k(B)$

[gentle osprey]

Moldilocks

[empty grove]

thats why you were able to figure out the LHS

other than that yeah they didnt have a use

[flint pilot]

ah I see

alright. And then all I need to do is add the division by R in the last step, and I'm ok?

[empty grove]

yep

also its quotienting, usually wouldnt call it division lol

[flint pilot]

oh alright lol

wow so that question is done

[empty grove]

congrats

xD

[flint pilot]

Well, thank you guy so much! That was a very long time to solve it. I hope it goes well. Thanks again!

[empty grove]

Welcome 🙂

[pale sky]

Hi I was wondering how to start this question?

part a) was this

these are results from part a)

althoug i cannot say i am 100% confident about my results

here is our cup product from class

[sharp yoke]

im having trouble with b

my teacher said it might be easier to define the vector field by describing it on each chart, and then checking that these local definitions agree in the intersection of the domains of two charts

[shut moat]

if it helps you can think of the vector field on R^2 and restrict it to S^1

there's a standard tangent vector field you can then use

[sharp yoke]

whats the standard one

[gritty widget]

doesn't (x, y) -> (y, -x) work

you just need a vector field which gives a non-zero vector at each point of S^1

[sharp yoke]

yea except im trying to do it formally using the definition for the tangent bundle

where the tangent space at each point is like

[shut moat]

^yeah I was referring to this

[sharp yoke]

equivalence classes of $A_p$x$R^m$ where $A_p$ are the charts in your atlas containing the point and m is the dimension of the manifold

[gentle osprey]

『Urahairywizard』

[gritty widget]

sounds painful

[shut moat]

wait what

[sharp yoke]

yea its a super abstract definition its hard to work with

[shut moat]

i've never seen that defn before

[gritty widget]

just use part a!

[sharp yoke]

yea in Lee they use derivations which apparently only works for smooth manifolds

so our teacher gave us one that works for C^k manifolds as well

and they're equivalent in the case of smooth manifolds which we proved on our last p set

if you wanna look up the definition i think its in uh

[shut moat]

y∂_x - x∂_y should work then because S^1 is C^\infty

[sharp yoke]

"Tangent spaces as equivalence classes of n tuples"

[gritty widget]

just define cotangent space as a certain quotient in C^\infty(M) and take the dual, works for manifolds of any regularity

ah that's in lee

he gives a list of alternate definitions; that's one of them

[sharp yoke]

oh yea it is in lee

well a brief description of it is in lee

[sharp yoke]

ya i think im gonna use the derivation definition and use this one

i don't 'really understand the other definition too well

wait how do you show this forms a basis for the tangent space at a point

[gritty widget]

the tangent spaces are 1-dimensional so any non-zero element gives a basis

[sharp yoke]

ah of course

[dusk heron]

Consider a smooth and free $S^1$-action on $S^2\times S^1$. Is this action equivalent to the action where one just acts on the second factor?

[gentle osprey]

gustavn64

[dusk heron]

Or alternatively, is the orbit space always diffeomorphic to $S^2$?

[gentle osprey]

gustavn64

[summer jolt]

Hey I'm trying to compute the addition of points on a cubic curve using group law.
The curve is given by $$zy^2 = yz^2 +x^3 -x^2 z$$

Let $A=(0:1:1), B=(1:1:1)$, I want to compute $A+B$ using $O = (0:1:0)$ as the identity. To do this I found the equation of the line between them to be $y=z$. Then plugging this into the cubic I get the only points of intersections are $A$ and $B$. From here I'm not sure how to proceed.

[gentle osprey]

snypehype

[wanton marsh]

a line intersects a cubic curve in three points if you count them with multiplicity

[summer jolt]

So one of them must have multiplicity two?

[wanton marsh]

probably

[summer jolt]

hmm in that case I'm a bit unsure on how to compute A+B

Oh I just noted that $(0:1:1)$ is an inflection point, but not sure how that would help

[gentle osprey]

snypehype

[wanton marsh]

have you tried factoring zy²-yz²-x³+x²z after replacing y with z ?

[summer jolt]

Yes, I get $-x^3 + x^2y = 0$

[gentle osprey]

snypehype

[wanton marsh]

have you tried factoring that

into three linear factors

[summer jolt]

As I wrote in the question the only solns are A and B

{x = 0, y = y}, {x = y, y = y}

[wanton marsh]

.......

have you tried

factorizing

-x³+x²y

into

three

linear

factors

-x³+x²y = ? * ? * ?

you need to factor it completely if you want to know the multiplicities of the roots

[summer jolt]

Ok, I see. So you get $x^2(x-y)=0$. But I don't understand how knowing the multiplicity helps with computing the sum.

[gentle osprey]

snypehype

[wanton marsh]

well know that you knows x appears twice as a factor and (x-y) appears once

it tells you that the line intersects the cubic in A with multiplicity 2 and B with multiplicity 1

and so, A+A+B = 0

[summer jolt]

Ah ok, I see. That works because O is an inflection point right?

[wanton marsh]

no

that works always

[summer jolt]

Ah ok because I was looking at this theorem in my lecture notes

[wanton marsh]

are they not saying that O is the identity of the group ?

[summer jolt]

Not in this proposition no

but perhaps it's implied

[wanton marsh]

A is an inflection point if and only if A+A+A = 0

if and only if the tangent line at A intersects the elliptic curve at A with multiplicity 3

for any line, if it intersects the curve in three points A,B,C (not necessarily distinct) then A+B+C = 0

only this however doesn't completely determine a group law and that's why you have to arbitrarily choose an identity element to make a group

[summer jolt]

@wanton marsh ok thanks a lot for the explanation

the books on AG are so terse

I find it hard to make sense of these theorems sometimes

[wanton marsh]

and I said something wrong somewhere and I'm not sure where xd

but hmm yeah most of the time we pick O = an inflection point

it's not A+B+C = 0, but A+B+C is independent of the line chosen, I guess

so you were right about needing O to be an inflection point, sorry

[gentle osprey]

bdobba

[cursive flume]

how can a chart map into R*?

If I take the definition of the tangent bundle charst and star everything I do get this result

but I can't interpret this as being a chart

[uncut surge]

It's a bundle chart, right? And the codomain of such a map should be something like (open set in R^n) times some vector space

And (R^n)* is just some vector space -- not even that different from R^n, since they are canonically-ish isomorphic

I guess if you're just asking "how is this a manifold chart" -- in principle, the only thing you still need to do is identify (R^n)* with R^n via some diffeomorphism, then you can use this map to construct a map with codomain U' x R^n, and that's an open set in R^{2n}

@cursive flume

(If you don't know what a bundle chart is, ignore my first two lines)

[cursive flume]

how to prove r* is diffeomorphic to r @uncut surge ?

[uncut surge]

Do the expressions "dx_1, ... , dx_n" mean something to you? These are the standard basis of the dual space (R^n)*, so for a diffeomorphism between R^n and its dual, you could just take the mapping which maps e_1,...,e_n to dx_1,...,dx_n

[cursive flume]

why would this map be smooth and bijective?

I am familiar with dx^1,...dx^1n, they are the basis for the dual space

it's bijective because we map e_1 to dx^1 and so on so it's 1 to 1

but why smooth?

for it to be continuous would make senes because the dual basis is defined with the help of an inner product,which induces a toplogy on R* and it's the same as on R

but i can't see how the smooth structure would be the same aswell

[feral copper]

Hi, dumb question ! A handlebody decomposition with k1 1-handles, k2 2-handles etc gives a CW decomposition with k1 1-cells etc, right ?

By deformation retracting all the handles ?

[charred root]

Is there an inverse operation for the exterior derivative, basically a translation of Biot-Savart law and its relatives into the language of differential forms? So, if I have a, let’s say, 1-form $\omega=udx+vdy+wdz$, satisfying $d\omega=0$, how to find at least one solution to $d\varphi=\omega$?

[gentle osprey]

BIGfoot496

[uncut surge]

https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincaré_lemma The poincaré lemma gives a very explicit way to find such an "antiderivative" if you're on a starshaped domain

[charred root]

Thanks a lot! I’ll go and try to translate that Wiki text from Mathematical to English then😋

[pale sky]

hi was wondering if anyone could help me with this question

[sharp yoke]

having trouble with the reverse direction for a and also the example for part b

oh i just looked at lee for part b i can just take any submanifold that isn't properly embedded

[dense mica]

Hi guys, I really need help with this question :/
(my sketch:https://www.geogebra.org/classic/hgyxdyxu )

Let ABC be an acute, non-isosceles triangle with D is any point on segment BC.

Take E on the side AB and take F on the side AC such that ∠DEB = ∠DFC.

The lines DF, DE cut AB, AC at M, N, respectively.

Denote (I1), (I2) as the circumcircle of DEM, DFN.

Let (J1) be the circle that internal tangent to (I1) at D and also tangent to AB at K,

let (J2) be the circle that internal tangent to (I2) at D and also tangent to AC at H.

Denote P as the intersection of (I1) and (I2) that differs from D and also denote Q as the intersection of (J1) and (J2) that differs from D.

(a) Prove that these points D, P, Q are collinear.

(b) The circumcircle of triangle AEF cuts the circumcircle of triangle AHK and

cuts the line AQ at G and L (G, L differ from A).

Prove that the tangent line at D of the circumcircle of triangle DQG cuts the

line EF at some point that lies on the circumcircle of triangle DLG.

[viral atlas]

Better suited to #geometry-and-trigonometry @dense mica

[dense mica]

alright

[flint cove]

I think this rather belongs in #geometry-and-trigonometry.
Also, please don't add so much unneccessary whitespace, your message covers over half of my screen 😉

Oh how weird my discord didn't load the last two messages, only after I pressed enter o.O

[fading vale]

you must have a big monitor, this covers so much more than half of mine

[manic garden]

why are divisors so important?

is there an arithmetic version of the RR?

[tight agate]

divisors are the easiest sub-things of a scheme that one can study

higher codim sub things are hard

(see: Hodge theory)

[tough imp]

As in, we don’t really know anything about them

This is the answer Litt gave me lol

[cursive flume]

Let $P=T^*G$ be the cotangent bundle of a smooth manifold G. Let $\pi_1: P \to G, \pi_2: TP \to TG, \pi_3:TP \to P$ be the canonical projections defined on the bundles. Let $\alpha_q \in P, \omega_{\alpha_q} \in TP$ along the fiber $\alpha_q$. Define the maps $\theta_{\alpha_q}:T_{\alpha_q}P \to R, \theta_{\alpha_q}(\omega_{\alpha q})=\alpha_q(\pi_2 \omega_{\alpha_q}), \theta_{0}: \alpha_{q} \to \theta_{\alpha_q}$. Then $\theta_0$ is a smooth covector field and $\Omega_{0}:=-d \omega_{0}$ is a symplectic form on P.$

[gentle osprey]

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

[cursive flume]

trying to prove this theorem but struggling really hard

this is how my prof started the proof(this is still ok)

can someone help me understand the last line here please?

i can't really see how it follows

[dusk heron]

View $S^1$ as a subset of $\mathbb{R}^2\cong\mathbb{C}$, and view $S^3$ as a subset of $\mathbb{R}^4\cong\mathbb{C}^2$, and define an action of $S^1$ on $S^3$ by $z\cdot(w_1,w_2):=(zw_1,zw_2)$. Does there exist an orientation-reversing smooth involution $S^3\to S^3$ which commutes with this action?

[gentle osprey]

gustavn64

[dusk heron]

My guess is that the $S^1$-action is somehow connected with the orientation on $S^3$, forcing a smooth involution $S^3\to S^3$ to preserve orientation. But I am not sure how to justify this.

[gentle osprey]

gustavn64

[gritty widget]

If you have a metric space (X, d) and a subspace (E, d), and if a set is open in (X, d) then is it open in (E, d) also?

[strong heron]

In general, no. Open sets in $E$ will be of the form $G \cap E$ where $G$ is open in $X$. So, for example, if you take an open subset of $X$ that does not intersect $E$, it cannot be open in $E$.

[gentle osprey]

whysee

[gritty widget]

the empty set is open

[uncut surge]

So the big problem is that an arbitrary subset of X will of course not be a subset of E. The most reasonable thing you can ask for is "If a set A is open in X, then A intersected with E is an open set in E", and this is true

[strong heron]

Correct

[gritty widget]

oh i misinterpreted whysee's statement

lmao

[strong heron]

No, but your point about empty set is correct. I should've said non-empty in my statement.

I have almost a silly question, but if X and Y are two topological spaces and I have a continuous bijection f: X --> Y and a continuous map g: Y --> X such that gf=id, that implies X and Y are homeomorphic, right?

[tight agate]

yes

[strong heron]

Nice, it means I solved that problem. Yay!

[tight agate]

you might want to prove that tho

depending on the level of the class

[strong heron]

Yes, I can. I was just making sure.

gf=id and f surjective together implies f is invertible. So, g=f^{-1}, and hence g is a bijection too.

[empty grove]

How do you get fg = id?

[marble socket]

that's direct from gf = id and f being a bijection

[empty grove]

Right I see

[quasi forum]

Hey y'all I'm having a bit of trouble here. I need to show that A int. B is compact with the definition of compactness. But I'm really not sure where to start.

A U B was pretty easy, the intersection is the one I find difficult.

[fading vale]

have you seen that a closed subspace of a compact space is compact?

and that compact subspaces of a hausdorff space are closed?

[quasi forum]

Doesn't matter if we did. We have to use the definition of compactness here

[fading vale]

hmm

[quasi forum]

My thought is that we choose an open cover for A int B and then consider an open cover of B that contains the open cover of A int B

[fading vale]

yea if you could extend the open cover of A cap B

well @quasi forum if you can show that A and B are closed (idk if you guys have heine borel) then A cap B is closed in A (or B, whichever you choose) and then you can extend it very easily by throwing in A - (A cap B)

this is a cover of A

and now use compactness

[quasi forum]

But not an open cover of A

[fading vale]

oh it is tho

because A - (A cap B) is open

and it covers A - (A cap B) obviously

and then we take our open cover of A cap B and throw them together

to get an open cover of A

[quasi forum]

Why is A/(A int B) open?

[fading vale]

oh yeah you have to show that A cap B is closed

idk if you have heine borel

but you can show compact in metric space -> closed

[quasi forum]

Yes we do, that is one of the first things we learned

[fading vale]

right?

so A and B are closed and thus A cap B is closed

[quasi forum]

Yes

[fading vale]

and then X - (A cap B) is open

so A cap (X - (A cap B)) = A - (A cap B) is open in A

alternatively A cap B is closed in X so A cap A cap B = A cap B is closed in A

[quasi forum]

But A is closed, how are you getting openness out of this

[fading vale]

oh ur taking the complement of A cap B

which is open cause A cap B is closed, right?

[quasi forum]

Yes, but A is closed. How does takinh the intersection of an open and closed set imply it's open?

[fading vale]

no i mean A - (A cap B) is open in the relative topology on A

because X - A cap B is open, and so its intersection is open in the relative topology

like it looks like this

this is A - (A cap B)

maybe i said this in a confusing way

you only need your sets to be open in A

not necessarily open in X

[quasi forum]

I think it's worth pointing out our class mainly focuses on metric spaces. So topology/relative topology really doesn't mean jack to me yet

[fading vale]

ah hmm okay

can you give me your defn of compactness

@quasi forum

[quasi forum]

A is compact if for every open cover of A there exists a finite subcover

[fading vale]

right

so like

you need open sets in A

lemme post an image showing what i mean by this

like U cap A isnt open in X right? thinking of X as R^2 or whatever

and A as R x {0}

open sets in A are open intervals

and those intervals arent open in X

but they are open in A

and furthermore if i take an open ball in X and intersect it with A i get an open interval in A

does this idea make sense? like a set can be open in our line without being open in the plane?

[quasi forum]

I appreciate the help, but I'm not sure I am getting much from this. I think I need a mental break

[fading vale]

yeah thats fair

that happened a lot to me when learning topology id take a break and come back to something a few hours later and it went much much easier

[dusk heron]

Let $X$ be a connected topological space with nice enough properties to admit a universal cover. Assume that the fundamental group of $X$ is finite, so that the universal covering $\pi:\widetilde{X}\to X$ is finite-sheeted. If $f:X\to X$ is any (continuous) involution (i.e. $f^2$ is the identity on $X$), then the composition $f\circ\pi:\widetilde{X}\to X$ lifts to a continuous map $\widetilde{f}:\widetilde{X}\to\widetilde{X}$, satisfying $f\circ\pi=\pi\circ\widetilde{f}$. Does this map $\widetilde{f}$ have finite order, i.e. is there some $n\in\mathbb{Z}_{>0}$ such that $\widetilde{f}^n$ is the identity on $\widetilde{X}$?

[gentle osprey]

gustavn64

[dusk heron]

Since $\pi\circ\widetilde{f}^2=f\circ\pi\circ\widetilde{f}=f^2\circ\pi=\pi$, we at least know that $\widetilde{f}^2$ preserves fibers. Can it be shown to be a permutation on each fiber?

[gentle osprey]

gustavn64

[thin bramble]

Can anoyone help? Pick an arbitrary point D on the side BC of the triangle ABC, and consider the bisectors of the two angles at D and their intersections E,F with the sides AC and AB. Show that AD, BE, CF are concurrent (via Ceva's theorem).

[bleak helm]

@quasi forum You can do it similar to what Moth was doing, but using the compact subset defn(i.e. looking at open covers in X rather than showing it is compact as a subspace). Then you can just take the complement in X. Take an open cover $O$ of $A \cap B$. Note that since A and B are closed, $(A \cap B)^c$ is open. Add $(A \cap B)^c$ to $O$ then you get an open cover in X of both A and B. ||Since A is compact, you can find a finite subcover of A which is then also a cover of $(A \cap B)^c$. You can then remove $(A \cap B)^c$ from the cover to make it a subcover of $O$ again.||

[gentle osprey]

Lunasong the Supergay

[bleak helm]

I guess spoilers don't work in latex.

So you basically just add (A int B)^c and then remove it again.

Where ^c is the complement wrt X

Nvm, I overcomplicated, editing

You can do it similar to what Moth was doing, but using the compact subset defn(i.e. looking at open covers in X rather than showing it is compact as a subspace). Then you can just take the complement in X. Take an open cover $O$ of $A \cap B$. Note that since A and B are closed, $(A \cap B)^c$ is open. Add $(A \cap B)^c$ to $O$ then you get an open cover in X of both A and B. Since A is compact, you can find a finite subcover of A which is then also a cover of $(A \cap B)$. You can then remove $(A \cap B)^c$ from the cover to make it a subcover of $O$ again.

[gentle osprey]

Lunasong the Supergay

[quasi forum]

I think I get it now. A int B closed implies X/(A int B) is open. So the open cover of A int B and X/(A int B) is an open cover of A, or B. So that'll get the job done.

[fading vale]

right: more specifically once you intersect the X - (A cap B) with A (or B) to get an open subset of A rather than of X you will get an open cover of A

also you should probably use backslash rather than / for set minus because it might be confused with quotient spaces

or -

[quasi forum]

I never know which slash is the right one :p

[fading vale]

xd

i only know cause quotients ngl

[quasi forum]

How's this @fading vale ?

[honest terrace]

,texsp ||spoilers works||

[gentle osprey]

Shika-Blyat

[fading vale]

yea this works

[quasi forum]

Sweet!

[fading vale]

i think once u see the relative topology what im talking about will make more sense

but its the same concept

[quasi forum]

That'll be something I do in grad topology (fall semester)

[bleak helm]

Thanks!

I would rewrite {U, X\(A int B)} as U union {X\(A int B)}. And the finite subcover of A doesn't have to necessarily include X\(A int B), but as I write this, I realize you could just include it anyway.

[quasi forum]

The whole point of including X-(A int B) is to show that the rest of the open sets must cover (A int B)

[obtuse meteor]

surreal that they don't do relative topology in undergrad top???

[tough imp]

wtf is relative topology

[quasi forum]

@fading vale so quick question. This argument did not really use the notion of metric/hausdorff spaces. Why exactly does it fail if it is not a hausdorff space?

[fading vale]

if a space isnt hausdorff then a compact subspace isnt necessarily closed

for example, equip a space with the trivial topology

in general if you want to see why a Tn axiom is necessary just look at the trivial topology

[quasi forum]

Tn axiom?

Also, when you say the trivial topology , do you just mean the space and the empty set?

[fading vale]

yes but this is actually a bad example lol let me think of a good one

uh

two point space {x, y} and let the open sets be the empy set, {x}, and {x, y}

then {x} is compact but not closed in {x, y}

[fading vale]

uh separation axioms, hausdorff is T2, T1 means singletons are closed, things like that

dwai if u havent seen them most dont matter very much

[tough imp]

T1 does not mean singletons are closed, it means given two distinct points x,y there exists an open set containing x not containing y, and an open set containing y but not x

...

which is equivalent to points closed

[sharp yoke]

well said you gibbon

[lean marten]

@tough imp isn't that T2?

[tough imp]

nope

For T2 the open sets are disjoint

as in you have nbd U,V of x,y such that U\cap V is empty

[lean marten]

oh ok

i thought there just had to be one open set per pair

like in the sierpenski two point space

[tough imp]

that's T0 I think

Let me look it up so I'm not fucking it up

[feral bay]

yea

[lean marten]

ah

right

[feral bay]

we talked about that today once

i thought it was T₁

[tough imp]

I forgot what T1.5 was

I think varieties are T1.5

but not T2

[feral bay]

is there a T1.5?

[tough imp]

there's something about it

[feral bay]

isn't it just 2.5 and 3.5?

[tough imp]

maybe

[lean marten]

its 2.5 for urysohn spaces

[tough imp]

I swear that varieties satisfied soemthing slightly stronger than T1 tho

[lean marten]

and 3.5 for tychonoff

[tough imp]

Like the version of varieties that are embedded in affine spaces and whatnot

[feral bay]

maybe R something

[lean marten]

If it's embedded in affine space won't it be at least t4

[tough imp]

no, the topology is way fuck

[feral bay]

you mean zarsiki topology right?

[tough imp]

yeah

[lean marten]

oooooooooooooooh

[tough imp]

Whatever, I left varieties a long time ago

I deal with T0 spaces now

sober spaces gang

[feral bay]

lol

[lean marten]

nah continua are where its at

[tough imp]

does T0 imply Sober?

Right

Schemes are Sober

right, I just mean that Schemes don't fit anywhere on this separation axiom scale

they're sober, but the most you can get from that is T0

[lean marten]

Oh true

[feral bay]

my book has the proof T₃ is hereditary then just adds "why does this proof not work for T₄ spaces"

[tough imp]

imagine learning the separation axioms lmao

this meme brought to you by "I dropped my general topology course cuz it sucked ass" gang

[feral bay]

no wait, it was the product one

[lean marten]

idk urysohn's lemma was fun in my topology class

[tough imp]

Is that related to the partitions of unity or whatever

or like

open set inside a clsoed

exists a bump function blah blah

supported on

or whatever

[feral bay]

i learned the stone-weierstrass proof from nlab but then the prof asked me to prove some function is not continuous

and i froze up

btw, are absolute extensors useful?

[charred root]

There is a simple procedure for calculating a knot group. Is there a reverse procedure? Can you take a group with a Wirtinger presentation and find some knot with this group?

[sweet wing]

abstractly knowing the fundamental group of any hyperbolic 3-manifold uniquely determines it up to isotopy so yes

as for explicitly drawing out the knot tho im not too sure :p

[austere furnace]

@empty grove 👀

[empty grove]

bro im your TA not theirs

shut up and learn 🔫

[safe nest]

hey what surface/manifold embedding is this emoji?

my guess was https://en.wikipedia.org/wiki/Boy's_surface , but I dont think it has the right genus

[thin jewel]

Can anyone help me understand this?

Is this like the Serge Embedding?

[quartz edge]

obligatory

[fading vale]

this is cute

i wonder who drew it

[obtuse meteor]

there should be more things like this

[feral bay]

https://twitter.com/omnisucker/status/927521141463199744 found the original

omnisucker (@omnisucker)

rkgk

Retweets

2237

Likes

5167

Twitter · Nov 6, 2017, 1:00 PM

[zinc hearth]

guys I'm not sure f is continuous if it's graph looks like this :/

[grizzled ibex]

If i have two topologic spaces X,Y and two continue functions f,g X->Y how do i prove that C = {x in X : f(x) = g(x)} is closed in relation to X?

sorry if it's trivial

i thought that it's a subset from id(x) then objectively it's closed

considering f(g^(1)) = x as a subset from Id(x)

[feral bay]

one of those two spaces needs to be T₂ i think

[grizzled ibex]

Y is t1, hausdorff

is it necessary?

[empty grove]

Yes

[grizzled ibex]

why?

[empty grove]

Have you seen Y is Hausdorff iff diagonal in Y x Y is closed?

[grizzled ibex]

i missed that i think

[empty grove]

You can take X → Y x Y given by (f,g)

And use the above result

[grizzled ibex]

hmm

ty

[fathom cave]

thats a beautiful drawing

[cold vine]

Trying to calculate pi_n(RPn,RPn-1): I have formed the LES of the pair and then since pi_n(RP_n) = Z and pi_n-1(RP^n) = 0 so I have LES: Z -> pi_n(RP^n, RP^n-1) -> Z -> 0 so the last map is surjective. What else do I need for the calculation?

[quasi forum]

Hey y'all, I'm really struggling with number 5. I'm not quite sure I understand where compactness comes in to play here

I know f is uniformly continuous, but how does that help? 🤔

Not in this class. We need to stay away from contradictions as much as possible

I'm interested

[rancid umbra]

honestly, excercise 4 looks useful here. given x in X, x would never be in the image of f

there are no fixed points. f(x)\neq x for all x

ah nvm

thought excercise four was saying something diff.

[gritty widget]

analysis profs are deranged

[tight agate]

one of my analysis profs was very pro contradiction

he liked to write contradiction even when it wasnt

he said something like "there might some sensitive souls who will claim that what I'm doing is not proof by contradiction, but I don't care"

[quasi forum]

Nah, my prof rocks!

https://tenor.com/view/rock-and-roll-headbang-rock-party-gif-12474624

Anyways, back to the problem

I was wondering if there was a way to make use of compactness in this problem

[thin bramble]

[quasi forum]

Isn't x a fixed point if x=f(x)?

How the heck would it be false?

[quasi forum]

I see you read what I said earlier ultra

Nice try though

[tight agate]

sadge

[gritty widget]

you have to reprove every theorem you use, but without using a proof by contradiction for them

[quasi forum]

Well it is in metric spaces

[rancid umbra]

what is a minima in an arbitrary metric space?

[quasi forum]

Anyways slim, I'm confused on your hint. I'm straight lost

I mean sure, f(x)=x

Well, the function we actually care about is d(x, f(x))

Since X is compact, the function attains it's bounds by the extreme value theorem

And... since x is not equal to f(x), that infinum cannot be 0

So the epsilon is the mimimum of d(x, f(x))

Do I need to prove this claim? 🤔

Hmm, I feel like I've done this before. 🤔

But I'm struggling to do it for some reason

Yes

What?

So we are looking at the metric d(x, y) where y=f(x)

But I'm not sure if that helps 🤔

[rancid umbra]

no consider the function g: X --> X x X, g(x) = (x, f(x))

[quasi forum]

Actually, I'm not sure how that's a composition. That's what confuses me

Ah, okay.

Then take the distance of that

[rancid umbra]

right

[quasi forum]

And I think we've shown in the last that compositions of continuous maps are continuous

[rancid umbra]

great. that finishes the proof

[quasi forum]

Yep, all I need to do is equip X x X with the max metric so it's really easy to see g(x) is continuous

So you told me earlier slim that the statement in the question may not be true if X is not compact. Why is that?

I'm not sure I understand why that's a counter example

[rancid umbra]

got an interesting question thats been bugging me for the last couple of days, so I might as well ask it here to see if anybody has any input. Below is the initial question. The more general case is what I am interested in.

Let d: R^2 x R^2 --> R be the usual metric on R^2. Suppose X is a subset of the plane such that when d is restricted to X x X, it induces the discrete topology on X. what is the maximum number of elements X can have?

if by |R| you mean uncountably many, then no

it is countable, but its not infinite

its a really geometric statement

all points are not a distance one from each other

then its not the discrete metric

[gentle osprey]

slimvesus

[rancid umbra]

its saying that all the points x,y in X are a distance one from each other, or they are a distance 0, in which case x and y are the same point, when measured with the usual metric

[gentle osprey]

slimvesus

[rancid umbra]

im talking about the metric d(x,y) = 1 if x\neq y and d(x,y)=0 if x=y. is that not the discrete metric?

but yes this is also what i was thinking

[honest terrace]

but that's not the only one that induces the discrete topology

for instance replace 1 with any nonzero nonnegative constant and that still works

[rancid umbra]

okay, then my fault for not being more precise

[honest terrace]

(yeah, I'm just talking about constants to be sure to preserve triangular inequality slim)

[rancid umbra]

but, i have been struggling to come up with a proof that it actually is n+1 in R^n

i immediately thought of doing that, but this proof doesn't really lend it self to induction

ah strong induction

hmm. but don't you have to set it up like this: suppose X subset R^n+1 has more than n+2 points?

unless im not seeing something.

so, once you have two points x_1, x_2, you can essentially look at the points equidistant from both x_1 and x_2 an n-dimensional subspace of R^n+1

right typo

right. wlog, you can assume its not affine, right?

[wanton timber]

You can be a little more direct by taking the average of the n+1 points in R^n, then looking at the orthogonal line to R^n in R^{n+1} through the average point. The distance from the average point to the n+1 points will be less than one, so by continuity you can find two points on that line that are distance one from the n+1 points (all points on this line will be equidistant from all n+1 points)

That at least gives you the lower bound of n+1. For an upper bound, I would try to use the fact that the convex hull of n+1 points in general position is n dimensional, so n+2 points must have one point as not a vertex of their convex hull in R^n.

[rancid umbra]

so, the orthogonal line is a function from R^n to R^{n+1}?

ok.

right

im going to have to read more about what a convex hull is, but thanks for the input guys. this was helpful

[wanton timber]

Having thought about it a little more, what I said about convex hulls is not correct

[rancid umbra]

oof

[wanton timber]

I remembered a square has four vertices in R^2...

[rancid umbra]

so the problem is still trying to get an upper bound...

[wanton timber]

Yeah, you might still want to try to argue that the convex hull of n points that are all equidistant from each other must be at least n-1 dimensional, but the argument is not obvious to me at the moment

Actually, in line with my lower bound argument, the line through the average of the n+1 points in R^{n+1} should be ALL of the equidistant points to the n+1. Only two points on the line will be distance 1 to the n+1 points, but they will be distance 2 from each other, so you can only include 1 of them.

[rancid umbra]

this is what i was noticing with lower dimensional cases, but i dont know enough geometry? algebra? (not really sure what this subject is called) to formalize what i was noticing

[wanton timber]

The points equidistant from 2 chosen points is a hyperplane (it's the hyperplane perpendicular to the vector from one point to the other, based at the average of the two points)

Thus the points equidistant to more than two points are intersections of hyperplanes

You should be able to leverage that to show that it's one dimensional

I have other things I need to work on right now, but hopefully that gives you a direction to ponder

[grizzled ibex]

anyone?

[gritty widget]

[grizzled ibex]

one sec

if i take a neighbourhood filter at a point x generically, how i guarantee that for a element of this filter(Let's say A) x doesn't reside in the closure from this element A

i mean, how the elements from this filter really defines the general notion of neighbourhood and interior in topology??

sorry if i'm missing something important

@gritty widget any idea to explain this for me?

[gritty widget]

You are wondering how to recover the topology on a space X from the set of all neighborhood filters? I haven't thought about it carefully, but maybe you can do something somewhat indirect, like charqcterizing the continuous functions on X (into what space?) in terms of the neighborhood filters and recovering the topology from those continuous functions

But maybe not 🙂 I am just bullshitting thoughts

[cold vine]

Trying to calculate for real proj spaces $\pi_n (R P^n, R P^{n-1} )$ I formed the LES and then noticed $\pi_n (RP^n)=Z$ and $\pi_{n-1}(RP^n) = 0$ so I have in the end $\to Z \to \pi_n (RP^n, RP^{n-1}) \to Z \to 0$. I don't think the LES gives me the answer directly, but maybe something with the homom. for orthogonal groups $O(n)/O(1) \times O(n-1) \cong RP^{n-1}$ or some fibration or covering for the $RP^n$ but I'm not seeing it.

[gentle osprey]

PAHUS

[cold vine]

Still thinking of this one if any one has ideas for this

[grizzled ibex]

guys, a open set in a topology can be both open and closed right?

[obtuse meteor]

yes

[grizzled ibex]

is it relevant?

[cold vine]

Yes! Trivial example being the empty set since it's complement is the whole space which is open

[grizzled ibex]

if i construct a topology based on filters i get elements that are open and closed... i just noticed that

[ivory dragon]

if you construct a topology at all you get sets that are both open and closed

since the empty set and whole set are both open and closed

[grizzled ibex]

*all elements

[ivory dragon]

ah.

[grizzled ibex]

is it weird?

all elements being both open and closed

is it somewhat relevant?

[ivory dragon]

see https://math.stackexchange.com/questions/3151030/non-trivial-topology-where-only-open-sets-are-closed/3151066#3151066

[grizzled ibex]

it's talking about partition

i see

ty all

[ivory dragon]

im pretty sure thinking of these as topologies is typically the wrong perspective

[grizzled ibex]

@ivory dragon how?

[gritty widget]

Nice, this is a neat topology that I hadn't thought about really before

[grizzled ibex]

@gritty widget have u seen the topology induced by filters?

it has the same propety

[gritty widget]

Mm not really, my level of experience with filters is about at the level of being aware of them

[grizzled ibex]

i see

i recommend it

[gritty widget]

What do you mean by the topology induced by (a neighborhood) filter? You don't just mean the topology that created those neighborhoods in the first place?

[grizzled ibex]

since i was studying axiomatic set theory i was about to ask myself how partitions would work in a topology

we get something nice as showed

@gritty widget neighborhood filter actually

im kinda lazy to write LaTex but it's this

[gritty widget]

What's the squiggly with subscript at the end?

[grizzled ibex]

Vx is the centered filter

a filter that converges to x

A is a element from every convergent filter to x from A

that means that those points are adherent points from A (?)

we get that the closure from A is equal to A then A is closed... but also open since it's a element that satisfies the axioms

@gritty widget does it make sense? we can define adherent points with convergent filters

[gritty widget]

It has been a while since I have touched filters 🙂 and I am a bit occupied right now, so I am not able to follow atm

[grizzled ibex]

sorry

[gritty widget]

No no

[obtuse meteor]

What’s the open mapping? Without context we can’t tell you how ^^

[gritty widget]

well it is strictly monotone and continuous, i think you could use that

why not just use that f(x) is an elementary function

all elementary functions are continuous

[obtuse meteor]

Hmm

You can just use point set definitions

From metric space topology

[grizzled ibex]

what does it imply to say that the only element in the boundary from a set S is Ø? (besides the fact that any other elements different from Ø in S is in it's interior)

sorry if my question is bad written

[obtuse meteor]

@thorny flare so this problem for one depends a lot on your definitions of log and exp. However a standard definition of log is as the integral from 1 to x of 1/t dt.

with this definition

a standard argument can guarantee that log is cts

[grizzled ibex]

@obtuse meteor any light for me?

[obtuse meteor]

from metric space stuff. I.e. show that for all x in (0, infty) there is a e > 0 there's a de > 0 so that if |x - y| < de then |log(x) - log(y)| < e

as for showing that log is open

there's a standard result you should prove

which is that for a continuous strictly monotonic function f : I -> R, where I is an open interval, it's an open mapping and a bijection

The main property for this is that f(connected set (for example an open ball)) = connected set

and then connected sets in R are intervals

and you can argue that the connected set you get out has to be an open interval

bc it can't have a max/min (argument here)

[obtuse meteor]

I don't know what "an element different from empty set in S" means, since presumably S doesn't have empty set as an element

this will suffice to show open-ness, because reasons

[thorny flare]

yeah, I know that part

[grizzled ibex]

nvm

if Ø is in it's boundary S is a clopen set

[obtuse meteor]

What partial S = 0 tells you is that int(S) = S = closure(S)

yeah

[obtuse meteor]

*is its boundary

[grizzled ibex]

yeah

sorry

[obtuse meteor]

no problem just making sure

👍

[grizzled ibex]

🥳 🥳 🥳

[cursive flume]

is anyone familiar with the proof that smooth manifolds admit a symplectic form and has like 5-10 minutes?

I'd have some questions related to it,but i'd have to introduce some strange notations so if you don't have time/patience no worries

[gritty widget]

the proof that smooth manifolds admit a symplectic form

uh

what's your favorite odd-dimensional manifold?

[cursive flume]

no i mean 2n dimensional smooth manifolds sry

[gritty widget]

still not true

S^4 doesn't admit a symplectic structure

[cursive flume]

hm what ok i might get the statement wrong

[gritty widget]

(since its second derham cohomology is trivial + it's compact)

[cursive flume]

this is the precise statement

[gritty widget]

oh cotangent bundles

[cursive flume]

would you have 5-10mins to go over the proof?

but the notation is very yikes sry in advance

mathphys notation

[gritty widget]

ok

[cursive flume]

here it's a bit strange he writes stars for the charts,but it's ok since r^n* is diffeo to r^n

[gritty widget]

this looks like a foreign language

[cursive flume]

he proves it in charts

then he just denotes

and here he lost me

but i'm open to any proof if there is other, I just don't get this at all

like why does the cotangent bundle have a canonical symplectic two form

[gritty widget]

if you're open to another proof take a peek at prop 22.11

yeah this is just terrible to look at

i strongly recommend looking at another source.

[cursive flume]

they are the cotangent/tangent bundles

like TQ=P

TTQ=TP

XD

ah lel

[gritty widget]

yes

Let $M$ be a smooth manifold. Let $(q, \alpha) \in T^*M$. Define $\tau_{(q, \alpha)}\colon T_{(q, \alpha)}T^*M\to\bR$ by $$\tau_{(q, \alpha)}(V) := \alpha(d\pi(V)).$$ If $(x^1, \dots, x^n)$ are local coordinates for $M$, and $(x^1,\dots,x^n,y_1,\dots,y_n)$ are the associated coordinates for $TM$, then you can show that $$\tau = y_i,dx^i.$$ So $-d\tau = dx^i \wedge dy_i$, which is a symplectic form. (It's just the canonical symplectic form on $\bR^{2n}$ pulled back to $TM$, where $n = \dim M$.)

[cursive flume]

my prof is trying to prove that the liouville form is a form

[gentle osprey]

TTerra

[cursive flume]

thats half of the profo

[cursive flume]

this is all for that

just to prove theta is smooth

[gritty widget]

or

to prove it's smooth

write it in coordinates like me

no hassle

this is also the proof in lee i think

he uses xi instead of y but whatever

[cursive flume]

how can you show the tau equality?

also in order to write d tau like that you need to be in darboux coordinates right?

my prof said we can't use that until we prove that darboux coordinates do always exist

[gritty widget]

[gritty widget]

the point of the canonical form on the cotangent bundle is that every coordinate chart for the cotangent bundle coming from M is a darboux coordinate chart

when i wrote d tau

i literally just used the definition of d

[cursive flume]

ww w wat

this is enlightening

my prof did a whole course on why darboux charts exist

[cursive flume]

i'll note this

[gritty widget]

yeah it's a neat fact

anyways

[cursive flume]

ok the only point i don't see is the tau

[gritty widget]

you can prove this by just writing out tau as a linear combination of the dx^i's and dy_i's, and solving for the coefficients

discord took like 20 seconds to send that for some reason

[cursive flume]

how to use teh definition of tau you gave to compute what you wrote?

I mean to me it just looks like a chart expansion since tau is a linear map from the tangent bundle of the cotangent bundle of M to R

but im confused a bit

if it's in the a section on tangent bundle,it should be a vector field

i'm missing smth

[gritty widget]

i'm typing something

long

let's say $\tau = a_i,dx^i + b^i,dy_i$, where here the $a_i$'s and $b^i$'s are functions defined on (open subsets of $T^*M$), and the $dx^i$'s and $dy_i$'s are the $1$-forms on $T^*M$ coming from the local coordinates $(x^1,\dots,x^n,y_1,\dots,y_n)$. then, $$a_i = \tau\left(\frac{\partial}{\partial x^i}\right), \qquad b^i = \tau\left(\frac{\partial}{\partial y^i}\right).$$ these are functions defined on $T^M$, so to compute them, take $(q, \alpha) \in T^M$. then
\begin{align}
a_i(q,\alpha)&=\tau_{(q, \alpha)}\left(\left.\frac{\partial}{\partial x^i}\right|{(q,\alpha)}\right) \
&= \alpha\left(d\pi\left(\left.\frac{\partial}{\partial x^i}\right|{(q,\alpha)}\right)\right)\
&=\alpha\left(\left.\frac{\partial}{\partial x^i}\right|_{q}\right)\
&=y_i(q,\alpha).
\end{align}
so $a_i = y_i$. can you compute $b^i$?

[cursive flume]

i cant really see this cause tau: TT*M->R

[gritty widget]

ugh

[cursive flume]

this seems to be a vector field on T*M

[gritty widget]

let me reformat this crap

also there's a slight notation abuse that i would like to bring to light

[gentle osprey]

TTerra

[gritty widget]

note

[cursive flume]

right

[cursive flume]

but one-forms can be written also as multilinear maps from TM->R

ok right

I can't see why do we have x^...n, y_...n as local coordinates for TM

if we have a manifold its tangent bundle contains 2n many coordinates

[gentle osprey]

slimvesus

[cursive flume]

1. being coordinates of points, 2) being coordinates of velocities

are the b_i-s coordinates of velocities?

sorry, i mean y_1,..y_n here

[cursive flume]

so TM has n many coordinates as base points, n many as coordinates from velocities

[gritty widget]

oh that should say coordinates for T^*M

discord killed my *'s

[cursive flume]

so n many for base points,n many ofr one forms

right?

de y_1,..,y_n are coordinates of one forms?

[gritty widget]

yes

y_i(one-form) = ith coefficient of one-form written with respect to dx^1, ..., dx^n

[cursive flume]

but x_1,...,x_n are coordinates of base points,why can we expand tau in terms of them?

the y part makes sense

the other parts of the computation are ok

i just don't see the first line

[gritty widget]

$x^i$ is defined on (an open subset of) $T^*M$ by defining $x^i(q, \alpha) := x^i(q)$. (so it would probably make more sense to write something like $\tilde{x}^i$ for the coordinate on $T^*M$)

[gentle osprey]

TTerra

[cursive flume]

and neither why b^i should be zero

[gritty widget]

\begin{align*}
b^i(q, \alpha) &= \tau_{(q, \alpha)}\left(\left.\frac{\partial}{\partial y_i}\right|{(q, \alpha)}\right) \
&= \alpha\left(d\pi\left(\left.\frac{\partial}{\partial y^i}\right|{(q,\alpha)}\right)\right) \
&= \alpha(0) \
&= 0.
\end{align*}

[gentle osprey]

TTerra

[gritty widget]

in jargon, d/dy^i is "vertical"

it gets killed by dπ

and that shows that tau has the required form (heh)

does that make sense?

[cursive flume]

ok now everything's clear except the last statement here

with the pullback

yes

[gritty widget]

ah

if i think of my coordinate chart as a mapping $\varphi$ from an open set $U \subset TM$ to an open set $\bR^{2n}$ (where $n= \dim M$), then $\varphi^*\omega_{std}=-d\tau$ on $U$, where $\omega_{std}$ means the standard symplectic form on $\bR^{2n}$

which is exactly what it means for phi to be a darboux chart

[cursive flume]

lol wait

[gentle osprey]

TTerra

[cursive flume]

this is very similar to something i've seen

is this similar to that? he defines the coordinates on TM by pulling back from TR^n

[gritty widget]

my coordinate chart phi is exactly this

(the (x^1, ..., x^n, y_1, ..., y_n))

[cursive flume]

now i see

thanks a lot!

[gritty widget]

[cursive flume]

are you following Ana da Silva's book btw on your lecture? @gritty widget

[gritty widget]

my class ended lol

[cursive flume]

ahh

[gritty widget]

but we did not follow it (or any book)

a homeomorphism will take distinct open sets to distinct open sets and conversely, so the spaces ought to have the same number of distinct open sets

[thorny flare]

right

ok thx

[gentle osprey]

lime_soup

[ivory dragon]

like, youre saying the image of your function is all isolated points?

if so, then... yes but not for the reason youre saying

singletons in the euclidean topology are not open

but that is NOT the same thing as being closed

(well, it is since R is connected in the euclidean topology, but "morally" not)

oh unless you mean

maybe im misinterpreting slightly

hold on

let me revise

yeah i was totally misunderstanding what you were asking

the answer is "no"

the sets you look at when checking continuity

can be ANY subset of the codomain

they dont have to be subsets of the image of the function

so for example

consider the function [f(x) = \begin{cases}1&x\in \bQ \ 0&x\not \in \bQ\end{cases}]

[gentle osprey]

Namington

consider the function \[f(x) = \begin{cases}1&x\in \bQ \\ 0&x\not \in \bQ\end{cases}\]

[ivory dragon]

this does map x to "isolated real numbers"

but its not continuous

because the preimage of (-0.5, 0.5) is not open

the preimage of (-0.5, 0.5), or equivalently the preimage of {0}, is R \ Q

which is not an open subset of R

in other words, it seems you were assuming the set in "preimages of open sets are open" was a subset of the image

but it can be a subset of the codomain

(-0.5, 0.5) is not a subset of the image {0, 1} of f

but it is a subset of ℝ

and so its preimage should be open if f is continuous

but it isnt.

does that make sense?

[thorny flare]

yeah I think so

so

if I have 1 if x in U and 0 if x not in U where U is clopen and (proper nonempty) subset of X and f maps X to R

then its cont?

cus then i'd just demonstrate all closed sets have closed preimages

[gritty widget]

i know the pushforward of a smooth form will be smooth

but if we have $\alpha=f^*\beta$ with $\alpha$ and $f$ smooth can we conclude that $\beta$ is smooth?

[gentle osprey]

lime_soup

[obtuse meteor]

no

take f : {*} -> R

pick any form beta on R

f^* beta will be smooth

and f is smooth

[gritty widget]

damn

the situation i have is i have a pr:C^n-0 to CP^n-1

i know that alpha is smooth

and alpha=pr^* beta

i've been trying to use that pr is a submersion but haven't gotten it

submersions admit local sections

what book is this

thank you

lee ism

if s is a local section of pr, then s^*alpha = s^*pr^*beta = (pr \circ s)^*beta = beta, so beta is smooth

very useful result

big thank you

actually can you clairfy

one direction of the proof is just linear algebra, and the other is the submersion normal form theorem

ok

i can try

(pr \circ s)^*beta = beta,

is just because

pr \circ s is the identiy

right, on an open subset of the codomain of pr

but i was lazy and omitted the restriction

yes thank you

i was fine with the restriction being omitted

but i was a little sus

in whether we had an identity on the map of manifolds

or only after taking the tangent map

[uncut surge]

let us say, hypothetically, i wrote this sentence

"Nonzero classes in singular homology characterize closed surfaces which do not bound a volume."

would you debate me on this

or does it spark joy

maybe a weird question but i'm wondering if this captures the correct understanding of singular homology in a snappy, intuitive slogan

[fading vale]

i think it basically captures the visual intuition for homology

if i was trying to explain it to people in a handwavy way this is how id describe it visually

[uncut surge]

vewwy gud uwu

[obtuse meteor]

there is also some like

literal way this is true

like you can view homology classes as maps in from special kinds of closed surfaces

and if they're nonzero those maps can't be extended

It's in the proof of Hurewicz in 2.A

of Hatcher