point-set-topology

[uncut surge]

i'm sure they do in certain circles

[fading vale]

it sounds gendered

[uncut surge]

the world is big

[sleek thicket]

I do not fuck dudes

ugh

[frigid patrol]

Bro

[uncut surge]

no i mean no one's a bad person for any of this, but i think it's worthwhile to explore it

[frigid patrol]

I call my girl friends bro

[sleek thicket]

I would like to have a boyfriend and I hate covid but it's not the reason

I didn't have a boyfriend my first two years of college either

just a couple shitty dates

[fading vale]

pensive

[uncut surge]

keep pushing through fam

[fading vale]

i want bf but relationships are scary and hard

[uncut surge]

shitty dates are part of it

[sleek thicket]

yeah but lart I could just not

And then feel bad but over a long stretch of time

So I don't really notice it

[fading vale]

H^i(S^n/S^n-1) = H^i(S^n)^2 right

[sleek thicket]

umm

[fading vale]

uhh the quotient not set minus

[sleek thicket]

feels wrong lol

[fading vale]

S^n/S^n-1 = S^n wedge S^n right

[sleek thicket]

yeah but isn't that just two spheres

[uncut surge]

does S^n/S^{n-1} turn into the wedge product of two S^n?

[sleek thicket]

Wedge summed

[uncut surge]

yeah

[sleek thicket]

Yeah

[fading vale]

yea

so

its direct product

[sleek thicket]

Okay you're right

[fading vale]

lol

[sleek thicket]

Sorry mouth

[frigid patrol]

Direct product is

[fading vale]

H^i(S^n) oplus H^i(S^n)

[sleek thicket]

yeah, which is what you said

me too

yoooo

Spotify put misery business on my daily mix 2

sucks that it has like mysoginisitc lyrics

[uncut surge]

listen to later paramore

[sleek thicket]

why is all of the music I enjoy steeped in a sexist culture

ughhhh

[uncut surge]

it's more grown-up and cheerful

[sleek thicket]

emo music is sod ucking bad

[uncut surge]

well not cheerful, but more grown-up

[sleek thicket]

I don't want grown up I want whining

[uncut surge]

well, then you can listen to the song, resting assured in the knowledge that the artists have since then learnt from their faults and become better

and maybe that soothes the pain

[sleek thicket]

yeah

I am glad Hayley Williams is a good person

[fading vale]

remember when negative xp made basically a male version of the incel scott pilgrim song and it was literally the same thing but mildly worse

its uhhhhhhhhhhhh

i guess in the context of the first one it feels so forced to make a point and its like

[uncut surge]

lol i've never heard of this, but looking up the lyrics, this is peak 2016-"hi i'm a gamer and i got my politics from gamergate"

[fading vale]

just awkward

probably lartomato

[sleek thicket]

beuj

*bruh

[fading vale]

actually negative xp is kind of a sad person

i tuned into one of his streams

it was like

not irreverent edginess at all it was more like

jordan peterson lmfao

[uncut surge]

that spooks me

there should only be one of those

[fading vale]

gamer peterson gamer peterson

[uncut surge]

i feel like i keep derailing all the math channels

[fading vale]

Good.

[uncut surge]

but i have no questions about topology and geometry

it is all clear to me

[fading vale]

hurb

oh maybe i have to do double induction here

yeah thats probably it

hm still not quite apparent though

oh this still isnt that evident tho

hrm

ok nerd

anyway is there a way to show that 0 -> Z -> H^i(S^n)^2 -> H^i(S^n) -> 0 exact implies H^i(S^n) = Z

hm

oh maybe the splitting lemma?

ah yea i think it splits

[sleek thicket]

@limpid vault I started the derailment

Oops

@uncut surge meant you

[uncut surge]

curves scare me

[fading vale]

hmmm if we know that H^i(S^n) is finitely generated do we have H^i(S^n)^2 iso H^i(S^n) oplus Z implies H^i(S^n) iso Z?

[uncut surge]

it was a shared effort, @sleek thicket

[sleek thicket]

Curves are so weird fuck

[fading vale]

hurb

there is a problem to be solved here!!!

unless the joke is funny

then make the joke instead

anyway i guess more generally this is just

[uncut surge]

i think this follows from the structure theorem for finitely generated modules or so

[fading vale]

does $A \oplus B \cong A \oplus C$ imply $B \cong C$ given that $A, B, C$ are finitely generated

[gentle osprey]

Moth:

[fading vale]

idk that lartomato

[uncut surge]

yeah i had to google it, too

in your specific problem, you might be more lucky trying to make the isomorphism more explicit?

'cus usually these come from a LES or so

[elder yew]

isn't it 3 am for you ultra

do you guys even slaughter turkeys?

[fading vale]

mhm

[uncut surge]

oh yeah, homology groups are abelian i guess

[fading vale]

oh

i think i see

mhm

so we end up with like

$\bZ^{2n} \oplus \bZ_{q_1}^2 \oplus \ldots \oplus \bZ_{q_k}^2 \cong \bZ^{n + 1} \oplus \bZ_{q_1} \oplus \ldots \oplus \bZ_{q_k}$

[gentle osprey]

Moth:

[fading vale]

which only holds if the Z_q_i terms are all zero

and n = 1

[uncut surge]

yah that looks about right

[little hemlock]

im not understanding the underlined part.

how does the earlier part of the sentence imply that you can just replace fgy with y?

[uncut surge]

I guess it's implicit that k is continuous, right?

[little hemlock]

yea

[uncut surge]

Then the only way we could have $kfgy \neq ky$ is if one of them was 0 and the other one was 1, but since fgy and y are connected by a continuous path, that would result in a contradiction to continuity of k

[gentle osprey]

Lartomato:

[uncut surge]

Basically because then you'd get the map $k \circ h(y, -) : I \to {0, 1}$ assuming two different values, but there are no continuous nonconstant maps $I \to {0,1}$ by connectedness and all

[gentle osprey]

Lartomato:

[little hemlock]

ah okay i see, thanks

[fading vale]

jesus i still need to show that H^i(S^n) is zero for i neq n

this is surprisingly involved

mayer vietoris would be a lot easier : |

[uncut surge]

what are you even trying to do? building the homology of the sphere from some axioms or so?

[fading vale]

nah its just computing H^i(S^n) via an LES sequence and induction on n

but like only those

its a problem in hatcher

[uncut surge]

ah awright, yeah sounds like a bit of work

[fading vale]

(with coeff in Z)

oh god i just realized

its coeff in G

alksjflksdf

whatever

[frigid patrol]

Cohomology

[uncut surge]

friends of principal bundles, let us marvel at this definition of principal g-connections from wikipedia

I'm wondering, in the equivariance-condition 1), shouldn't one of the g's have an inverse? I think the transformation behaviour isn't quite right

oh no it's a pushforward and not a pullback

all is well

[gritty widget]

lie groups 😳

[red garden]

prove that Int(A') = closure(A)'

first side:

let x be in Int(A')

--> x is in the union of all open sets such that A is a subset

x is not in the intersection of the closed sets

x is in X-closure(A)

int(A') is a ssubset of closure(A)'

good?

[bleak helm]

What?

Do you mean open sets that are subsets of A?

[bleak helm]

Why does this follow?

[red garden]

yes

took compliments

[bleak helm]

Actually, it should be the union of all open sets that are subsets of A', not A.

[red garden]

yea i wanted to say that

but idk

u sure?

u implied that

if A is a subset of B

then A' is a subset of B';

that is not right tho rihgt?

[bleak helm]

Where did I imply that?

[red garden]

oh wtf

iom sorryi

its A' from the first place

omg omg yea sorry

yea yea

okay so im right

without that A

?

[bleak helm]

Which closed sets? And explain why this is true

[red garden]

demorgen

closed sets that contain A'

compliument of open is clsoe

im sorry im bad witrh math

im low iq

@bleak helm

[bleak helm]

Well, yes, that's true. But you need to specify which closed sets and which open sets you are talking about

And write it out in more detail

[red garden]

okay

thank you

[elder yew]

@gritty widget that exercise you showed me is in spivak's calculus on manifolds

(The prof didn't want you sharing bc we'll find out where they really get their exercises)

[gritty widget]

which

2-35

nice

the second part isn't in it though

i like the second part of that question

[elder yew]

Can somebody help me understand what this is

A punctured 3-sphere is the result of removing finitely many open
balls from S^3. A tangle (B, τ ) consists of a punctured 3-sphere B and
a properly embedded 1–manifold τ that has no closed components

[velvet finch]

@elder yew https://mathoverflow.net/questions/322274/manifold-with-no-closed-components

2
That's correct. – Najib Idrissi Feb 2 '19 at 11:29```

[nimble flower]

Suppose X is a topological space and G a non-trivial group

then is X ever homeomorphic to X/G?

[elder yew]

I meant overall what a tangle is

My friend says it's like a bridge braid box that can contain a maxima /minima on the inside

[velvet finch]

@nimble flower looks like this handout could have some info for your question http://math.stanford.edu/~conrad/diffgeomPage/handouts/qtmanifold.pdf

haven't looked at it in much depth though

[sleek thicket]

@nimble flower yes

Consider the action of the group C2 = {-1,1} on the circle S^1

By multiplication

The quotient space is also S^1

[fading vale]

hi shamrock : )

[sleek thicket]

hello!

Here's my proof: define $f : \mathbb{S}^1 \to \mathbb{S}^1$ by $f(z) = z^2$. Then $f(z) = f(w)$ iff $z = w$ or $z = -w$, and $f$ is a surjection between compact hausdorff spaces so it's a quotient map. Since $\mathbb{S}^1/C_2$ is the quotient space of $\mathbb{S}^1$ by the relation $z \sim w$ iff $z = w$ or $z = -w$, uniqueness of quotient spaces implies $\mathbb{S}^1/C_2 \cong \mathbb{S}^1$

ahhh

I thought I had this in my discord preamble

[elder yew]

Nice S^1

lol

there we go

[sleek thicket]

Nope

[elder yew]

still a bad S^1 floating around at the end

[sleek thicket]

I missed one

It's updating

[elder yew]

lol

[gentle osprey]

shamrock:

[sleek thicket]

all finite covering spaces of S^1 are iso to a covering of the form p(z) = z^n, so this actually gives a family of examples for all cyclic groups G

By taking products you get an example for all finite abelian groups G (products and quotients should work nice since we're dealing with compact hausdorff spaces)

oh you could also just have G act trivially lol. I'm gonna assume the asker wanted a faithful action of G

[nimble flower]

thanks people!

[sleek thicket]

simply exhibit a lie group isomorphism between SU(2) and SO(3)

okay sorry I just had to say it

so for a $3\times 3$ real matrix, $A \in \mathfrak{so}(3)$ iff $\tr A = 0$ and $A^T = - A$

[gentle osprey]

shamrock:

[sleek thicket]

for a $2 \times 2$ complex matrix, $A \in \mathfrak{su}(2)$ iff $\tr A = 0$ and $A^* = - A$

[gentle osprey]

shamrock:

[sleek thicket]

so clearly these conditions are pretty similar but I don't immediately see a map either way

[sleek thicket]

Sorry, I can't think of anything. It might actually be easier to show the lie groups are isomorphic lol

[woeful oasis]

they're not though...

There's a 2-1 map

which means the tangent spaces are isomorphic, but the groups themselves are not

[sleek thicket]

Fuck you're right lol

That's what I was thinking of, sorry

[woeful oasis]

Don't be sorry, do the Philipino Wine Dance

[sleek thicket]

something like SU(2) is the unit quaternions via an explicit map and then every unit quaternion acts by rotation on R^3?

I think that's the thing I was thinking of

Rotates by conjugation

(but this is clearly not an iso because a quaternion and its negative acts by conjugation the same way)

[woeful oasis]

ahhh that's the clever thing to remember I was looking for.

[sleek thicket]

which part?

[woeful oasis]

the negative acts the same

[sleek thicket]

Gotcha

But yeah this is gonna be a local diffeomorphism and a lie group hom

So it gives an iso on Lie algebras

[woeful oasis]

I'm looking at some basic identities again, I'm thinkin does this hinge on trig half angle identity?

I think that's part of what Rodrigues figured out that Hamilton didn't(?). I dunno I think this guy Altmann says that somewhere

[sleek thicket]

yes?

A smooth map induces a map on tangent spaces

in this case that map is an isomorphism

[woeful oasis]

You can get a lie algebra iso by looking at basis vectors tho

You can find it by looking at the matrix equations with coordinates

Considering each lie algebra as a matrix algebra, the relations defining each algebra. If you write these out in matrix form and compare each coordinate

So for so(3), $A^T = -A$ implies every diagonal entry is zero, and gives a correspondence between upper and lower diagonal entries

[gentle osprey]

Apopheniac:

[woeful oasis]

That leaves 3 coordinates,

Then, for su(2), A*=-A implies that the diagonal entries are imaginary

[gentle osprey]

Apopheniac:

[woeful oasis]

well that's wrong! i keep messing up negative signs

[neat beacon]

Just a curiosity:
If you have three points on a sphere you can make a triangle with them. Can it be / is it often considered to still be a triangle if it also happens to be a great circle? (or any other polygon for that matter)

[blissful dune]

i think it still is?

as long as the points are connected by geodesics, i think they still count as a triangle

not sure though

[neat beacon]

It follows then that a great circle is also every kind of polygon

Which is fun

If that's true

[gritty widget]

without knowing the definitions i'm tempted to say it's just like

a degenerate triangle

you know how a "degenerate triangle" in the plane is just a line

i.e. a geodesic

on the sphere your geodesics are great circles

so if you consider triangles to be collections of three geodesics intersecting nicely or whatever

this is the degen case

disclaimer: i don't know how one defines a triangle in non-R^2-spaces

basically what master chief said - they're a quick learner

[neat beacon]

Great circle segments

[gritty widget]

https://mathworld.wolfram.com/SphericalTriangle.html

maybe this is worth looking into

like if we're defining polygons on the sphere as collections of great circle segments (of which a great circle is still one) then a single great circle is as much of a polygon as a line is

[neat beacon]

Thanks

[gritty widget]

geometry 😌

[sweet wing]

imagine not having total angle of 0° in your polygon

cant relate tbh

[gritty widget]

[heady grove]

Anyone know how to go about sketching this surface

[gritty widget]

it's the graph of a certain function, that should make it easier

[feral copper]

Hi ! In the proof of Bianchi's second identity, there's something I don't quite understand... The following claim is unclear to me, how does one go prove it ?

I don't really know how one computes ∇R, do I have to go in coordinates ?

Oh wait, nvm, it's the very definition of the covariant derivative ! Sorry ! :x

[gritty widget]

welcome to the confusing world of tensors

[feral copper]

Another topic now ! 😄 I bountied my MSE question to draw more attention, I'll cross-post over here in case someone happens to know the answer 😛

https://math.stackexchange.com/q/3907570/259363

[fading vale]

hn

if i knew any homotopy theory beyond pi_1 shit id try to help : (

[feral copper]

No worries 😉

The Ext discussion we had the other day was already helpful 😛

[fading vale]

[feral copper]

Homotopy theory is very interesting, if you have the time you should take a look ! 😉

[fading vale]

mhm

im still on ch 3 of hatcher

but im excited to check out ch 4

starting the section on cup products soon(ish)

[feral copper]

Are you following courses or just the book by yourself ?

And yeah cap/cup products and Poincaré duality is awesome too !

My favourite theorem is Hurewicz, that you'll see in chapter 4

[fading vale]

Just the book by myself

im in a course rn but its really slow and we're still doing like

van kampen

@feral copper we literally spent more time on point set than we did AT so far this semester : /

[feral copper]

Aaaah no way ! u_u

[fading vale]

c_c

its sad

hopefully the second sem is better

[willow spear]

If someone could lead me through a topological proof of this, that woudl be great!

Im really struggling on how to go about proving things in topology

[sleek thicket]

What have you tried?

Have you proven either direction of the iff?

[willow spear]

I haven't proven either direction

I just need some insight on how to approach soemthign liek this

[sleek thicket]

I think a lot of this is more about getting familiar with the mechanics of like, what is a topological space and what is a basis, not so much having special insight. Let's try and prove it together!

So first off, totally arbitrarily, let's do the τ1 = τ2 case

So assume τ1 = τ2. What do we want to prove?

[willow spear]

yo bet lets do this thanks

we want to prove that the two conditions are true

assuming that the two topologies are equal

Useful stuff: Union of basis elements = X

[sleek thicket]

Yeah okay so let's do (1) first

how do you prove a "for all" statement?

[willow spear]

I have no clue honestly

or wait

you write: "let b, the basis element, be any element in $B_1$

[gentle osprey]

DjokerNole

[sleek thicket]

Yup!

Sorry I had irl stuff

[willow spear]

lol its fine

[sleek thicket]

Then what's the next step?

[willow spear]

Then you do: let x be any elemnt in b, the basis element.

[sleek thicket]

Yup

Okay well now we hit a backwards E

How do you prove an existential statement?

[willow spear]

"such that b' is an element of $B_2$"

[gentle osprey]

DjokerNole

[willow spear]

' = prime

[sleek thicket]

so you need to find some b' yeah?

with the desired property

we don't have a lot of information about B2, which is actually a good thing! It means there's not a lot of ways to go wrong

What do we know about B2, and how can we use that to find an element b' with the desired property?

[willow spear]

The union of the elements of B_2, which is the union of b elements is equal to X

[sleek thicket]

What do you mean by "union of b elements"?

And that's true, but it's not sufficient here. The union of the collection {X} is equal to X, but b2 = {X} wouldn't satisfy property (1)

[willow spear]

Well we know that B_2 is a basis for T2 right?

[sleek thicket]

Yup

So what does that mean?

Like, what's the definition?

[willow spear]

oh B_2 is a subset of T2

?

[sleek thicket]

That is not the defintion

See my (non) example of B2 = {X}

[willow spear]

There exists a b_1 and b_2, which are elements of B, such that the intersection of b_1 and b_2 is teh union of elements of B_2

[sleek thicket]

Nope

That's still true for {X}

Check your textbook/notes

[willow spear]

Is it the fact that every element of X, is an elemet of teh Basis set, B_2?

[sleek thicket]

No

[willow spear]

x is an element of b' which is an element of B_2?

@sleek thicket yeah im kinda stuck here

[willow spear]

Is it just that teh topology, T2, is generated by the basis B_2?

[summer jolt]

Hi, I'm trying to prove that the following statement is equivalent to saying the map f is proper.

Does anyone know how to prove this?

X, Y, T are topological spaces with X and Y LCH

[tough imp]

What is LCH?

[tight agate]

is that not the definition of proper?

oh

and i also misread the question

I thought the map was the pullback of f along some map to Y

[tough imp]

I mean if you take T = a one point space this says f is closed

then a closed map of LCH spaces should be proper yeah?

Oh lol, I think the other direction is gonna be the harder one maybe

Brain no worky

Only can think of valuative criterion of properness for AG

[summer jolt]

@tight agate yeah, sorry. I should have been clearer. LCH = Locally compact Hausdorff and also we have that it is assumed that the projection map T x X -> T is closed.

[summer jolt]

I mean this is not obvious to me

[gritty widget]

when checkig if some function is continuous at some point x, is it enough to look at just small open neighborhoods of f(x) (preimages)? Or would I have to show the preimage is open for every open set contating f(x)?

(im thinking about some wlog argument, in this one problem the preimages of open sets are easy to deal with for epsilon small enough (metric space), but it's quite annoying and harder to check if epsilon is big enough)

[unkempt bolt]

If you want to check that a function f is continuous, you only need to check that preimages of "small" open sets are open, like you say. In you case of a metric space, only checking something like epsilons < 1, or < whatever number works for you, is enough.

[gritty widget]

But I was thinking - what if preimage of open epsilon =1000 ball isnt open? The definition says forall eps

[summer jolt]

@gritty widget whatever ball you have it will be a union of "small" balls

[gritty widget]

K ye true thx

But nono wiat

The thing is

Nvm

I mean even if its an union of a lot of small balls, I dont know if tje further small balls are open ubder preimage

Or do you mean like we dont care about further ones since they will not have the point

We wnnt to check continuity at

[unkempt bolt]

If you are checking continuity at a single point x, I believe it isn't necessarily true that preimages of open sets around f(x) are open. It may happen to be true, but it isn't guaranteed if the function is only continuous at x.

Continuity at a point x means that for any open set V containing f(x), you can find an open set U containing x which maps into V. In a metric space, you usually take V to be an epsilon ball around f(x), and if you want you can only check "small" values of epsilon.

[heady olive]

Hello,

I am trying to understand a section of a proof on a question in Do Carmo's differential geoemtry. The question states:

Prove that the inverse image of a regular value of a differentiable map:
$F: U \subset R^3 \rightarrow R^2$
is a regular curve in 𝑅3. Show the relationship between this proposition and the classical way of defining a regular curve as the intersection of two surfaces.
Slader has the solution (look at section b): \url{https://www.slader.com/textbook/9780132125895-differential-geometry-of-curves-and-surfaces/68/exercises/17/#}

I understand the first chunk of it, but I am getting lost here:

Since $c$ is a regular value of $F$ it follows that the 2 vectors $\nabla f_1(p)$ and $\nabla f_2(p)$ are linearly independent for every point $p \in F^{-1}(c)$

This is what I am not understanding. $c$ being a regular value, implies that the map $df_p: T_p R^3\rightarrow T_{f(x)}R^2$ is surjective. Why does that imply linear independence of those 2 vectors? It's not immediately obvious to me.

[gentle osprey]

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

[heady olive]

Oh jesus that did not format prettily... What do I do?

[crimson imp]

If $\nabla f$ were the zero vector, then the derivative would be a linear bijection between $\mathbb{R}^2$ and $\mathbb{R}^3$

[gentle osprey]

pmorelli

[tepid depot]

you can edit and the bot will rerender

[quartz rover]

im troubled by the first bit

'show that f''(x) =... '

i get that the force of gravity acting upon a particular point is the numerator (if you integrate it over the curve dx)

but how on earth do they get that this is f''???

[heady olive]

@crimson imp Why would that be a bijection and why would it being a bijection between those 2 spaces contradict any of the assumptions in the problem?
Like formally why does

Since $c$ is a regular value of $F$
Directly imply
it follows that the 2 vectors $\nabla f_1(p)$ and $\nabla f_2(p)$ are linearly independent for every point $p \in F^{-1}(c)$
Formally?

[gentle osprey]

Makogan

[heady olive]

(Lol I like that the bot is hispanizing questions)

[crimson imp]

As a linear map, in order to be injective, it's necessary and sufficient that its kernel be the zero vector

since it is surjective it would be a linear bijection (hence a linear isomorphism) between euclidean spaces with different dimensions

[heady olive]

hmmmm Imma sleep on this one, it's not clicking for some reason, thank you however

[heady olive]

I get it now, thank you

[cold vine]

So a contionuous map from compact to compact is closed?

Or is this using the fact that Compact -> Hausdorff is closed...

[honest narwhal]

@cold vine the latter is better

[cold vine]

Yeah I thought so too, but this (from a model solution) seems to imply the former which i dont think is right

But maybe im reading too much into it

[honest narwhal]

Yeah I think he meant to say p(S^n \times I)

Or something

[cold vine]

Yeah ok, thanks ^^

[honest narwhal]

Don't worry about it though, you've identified the correct reason

[cold vine]

nice!

[crimson imp]

Can someone help me with an exercise?

Let $D$ be the distribution on $\mathbb{R}^3$ given by $D(a,b,c)={(x,y,z),|,z-bx=0}$. How can one prove that two arbitrary points $P,Q$ of $\mathbb{R}^3$ can be connected by a path $\mathbb{\alpha}$ tangent to $D$, ie, $\alpha'(t)\in D(\alpha (t))$, for all $t$?

[gentle osprey]

pmorelli

[crimson imp]

The tools I have for distributions are the definition and Frobenius' Theorem. I can't use "differential forms"-like arguments

I actually managed to find a path dividing in cases but the last one lacks on smoothness 😦

[cold vine]

Hmmm. Am I right that I can say that H_0(D^n, D^n-{x}) is Z by path connectness of of D^n (Closed disc here)

[sleek thicket]

@crimson imp so tangent to $D$ should be some differential equation right? Like let $α(t) = (x(t), y(t), z(t))$ and we're requiring $z'(t) = y(t) x'(t)$ for all $t$

[gentle osprey]

Sh(amrock)

[sleek thicket]

So this completely determines z, up to a constant

And it should have unique (starting at P) solutions for all time as long as x, y are C^1? I think?

So like if $x(t) = (Q_1 - P_1)t + P_1$ and $y(t) = (Q_2 - P_2)t + P_2$ then $z(t) = \int (Q_2 - P_2)(Q_1 - P_1)t + P_2(Q_1 - P_1) dt = \frac{(Q_2 - P_2)(Q_1 - P_1)}{2} t^2 + P_2(Q_1 - P_1) t + P_3$

[gentle osprey]

Sh(amrock)

[sleek thicket]

But this won't give Q_3 at time 1 in general...

hmmm

Maybe you do need to use frobenius here

Iirc the proof of frobenius gives an algorithm for choosing your coordinate chart, so maybe it can give you global coordinates here?

[sleek thicket]

Hmm I don't think this distribution is involutive. $X = \frac{\partial}{\partial x} + y\frac{\partial}{\partial z}$ and $Y = \frac{\partial}{\partial y}$ are a global frame for it but $[Y, X] = \frac{\partial}{\partial z} \notin D$

[gentle osprey]

Sh(amrock)

[sleek thicket]

so frobenius doesn't apply...

What paths did you find pmorelli?

[crimson imp]

Oh, sorry I didn't mentioned that, the question was divided in two parts, the first one was actually showing it isn't involutive (which I did by finding the generators)

[sleek thicket]

oh lol np

I just figured I might check if we could use frobenius here

so $z(t) = \int_0^t x'(s) y(s) ds + P^3$ right?

[gentle osprey]

Sh(amrock)

[cold vine]

So I'm calculating H_p(D^n, D^n-{x}) with the closed disc. I'm doing it by taking the LES of the pair. Theres so many cases here with |x|=1, |x|<1 not to mention p>1, p=1, and then with different dimensions for n so I'm wondering wheter I'm missing an easier method?

[crimson imp]

The cases I divided were, letting P=(p1, p2, p3) and Q=(q1, q2, q3):

1. If p1=q1 and p3=q3 it is straightforward to construct a line segment that belongs to D
2. If p1≠q1, we can use the fact that the points (p1,b,p3) and (q1, b, q3) can be joined by a line segment contained in D, with b=q3-p3/q1-p1, and construct a curve using such line

[wanton marsh]

that's what I found too

[crimson imp]

As for the last case, which is p1≠q1 and p3≠q3, the path was the union of three line segments contained in D (given by cases 1 and 2)

But such union isn't differentiable, that's where I'm stucked

[sleek thicket]

I just got 2 lol

wait sorry I'm confused, wouldn't the last case be p1 = q1 and p3 ≠ q3?

[wanton marsh]

are you allowed to decrease your speed to 0

[sleek thicket]

Ooh good idea

I think that makes sense, you could like multiply the speed by a bump function when you turn

[crimson imp]

You're right, I type it wrong

[crimson imp]

Maybe, what do you mean by that?

[wanton marsh]

I mean alpha'(t) = 0

[sleek thicket]

The definition seems to allow it

[wanton marsh]

then you can continuously decrease your speed to 0 as you approach the turns, then your path can be C1

it can even be smooth with suitable bumps

[sleek thicket]

so like, $(\alpha \circ \gamma)'(t) = \gamma'(t)\alpha'(\gamma(t))$, right? So let $\gamma : \R \to \R$ be a smooth ($C^\infty$) curve with $\gamma(0) = 0$ and $\gamma(1) = 1$ and all derivatives of $\gamma$ vanishing at 1

[crimson imp]

But line segments have constant speed don't they?

[gentle osprey]

Sh(amrock)

[wanton marsh]

says who

[sleek thicket]

Pmorelli, you can reparameterize the domain to change the speed of α

[wanton marsh]

can you have alpha(t) = (0,t²,0) ?

[sleek thicket]

Without changing its range

[crimson imp]

Oh, I got it now

Maybe it can work then

Hmm in this case the entire curve on case 3 would be differentiable, regardless of the edges

This helps a lot

[sleek thicket]

Yeah, this is how you can like give a smooth curve R -> R^2 whose image is a square

Which is wacky

[crimson imp]

Wow, this is actually the first time I came into a conclusion like this

It is kinda strange but makes sense

Well, I'm certainly way further than I was at the beginning of this chat, I'll try to organize the infos that we talked and see if I can build up an argument. Thank you all!

[cold vine]

In this LES of the pair (D1,S0) why can we expect the last group to be 0? Is it just in this case because it is H0(D1,S0) which I think is the reduced homology group H~(D1)=0 or is this more general result that stems from something else?

[tight agate]

A square can be given a smooth structure but it is still not a smooth submanifold of R2

[sleek thicket]

right

The trick we were talking about specifically involves having vanishing speed at some point

So you don't get an immersion

[tight agate]

yup

[wanton marsh]

I'm pretty sure you can have paths that don't have a vanishing speed but it would be a real bother to find one that does just what you want

[sleek thicket]

Yeah, probably

This problem seemed very undetermined

[tight agate]

No smooth structure makes is a submanifold tho

[sleek thicket]

oh I thought zef was talking about paths tangent to the distribution

[wanton marsh]

yes paths tangents to the distribution

instead of sharp angles you would have curves but then you would need to adjust stuff so that it works

[little hemlock]

#advanced-analysis is occupied so I am asking this here.

[honest narwhal]

So okay complete metric space

And density

What's the first theorem that comes to mind? Yeah BCT

[little hemlock]

yea ik, but I don't see why bct is needed. I have a "proof" that doesn't use it

[honest narwhal]

Oh lol

Shoot

[little hemlock]

it must be wrong somehwere, but here it is:

[honest narwhal]

I guess I instantly thought you were asking for a hint or smth

[little hemlock]

Let $x \in M$. Then $x \in E_k$ for some $k$. If $x$ is in the interior of $E_k$, we're done. Otherwise, $x$ belongs to the boundary of $E_k$ in which case every open ball containing $x$ intersects $E_k$.... Ah wait nvm it doesn't have to intersect the interior tho oof.

[gentle osprey]

kxrider

[honest narwhal]

F

[little hemlock]

I swear this kind of thing constantly happens when I ask here lmao. So hmmm, back to the drawing board

[sleek thicket]

Happens to me in here all the time lol

[crimson imp]

I thought about the parametrization stuff we discussed earlier, and in order to make the curve in case 3 differentiable, I can assume it is parametrized by arc lenght, can't I?

Then all of the edges would have unit speed and we could conclude it is a differentiable path that is contained in the distribution

[sleek thicket]

hmm this seems suspicious to me

[crimson imp]

actually I thought the same thing after writing lol

because there may be different vectors with the same norm

[sleek thicket]

So consider doing the same trick for a square

Like

Take a horizontal line and a vertical line

Use the same reparameterization-with-bump-function trick to get a smooth injective curve whose image is a corner

Like this $\lrcorner$

[gentle osprey]

Sh(amrock)

[sleek thicket]

If you parameterize by arc length I think the result won't be smooth

I'm pretty sure(?) arc length reparameterization will only give a smooth result as long as the curve has nowhere vanishing derivative

[crimson imp]

yeah, it actually is most likely not smooth 😓

[sleek thicket]

Okay yeah I think I see the issue

So define $s(t) = \int_0^t |\gamma'(u)| du$

For some curve γ

[gentle osprey]

Sh(amrock)

[sleek thicket]

If eg γ' is zero for a while, this will only be weakly increasing, and so won't have an inverse function

even if it is strictly increasing, its derivative s'(t) = |γ'(t)| might vanish

and so you can't use the (one dimensional) inverse function theorem

[crimson imp]

hmmm I think I got it

[sleek thicket]

This is making me a little concerned about something I did on my last homework lol...

[crimson imp]

The bump function trick seems more useful, it forces the derivative to be zero at the edges

[sleek thicket]

Yup

[crimson imp]

great, everything is clear now 🙂

thanks

[little hemlock]

okay back to this.

I have that M\D is not all of M since at least one of the E_n has non empty interior, but I am not really sure how that helps

[nimble jolt]

If not, then some closed epsilon ball B about a point in M is covered by the E_n, but disjoint from the int(E_n). I.e. B is a complete metric space that is written as a union of nowhere dense subsets, 🐻 says this cannot be the case.

@little hemlock

[little hemlock]

ohhh interesting, thanks!

[little hemlock]

so... my idea is this. A closed interval is of the second category, so you'd just have to show that removing the two endpoints still gives a second category set.

hmm

oh wait, if $I = \bigcup_{n=1}^\infty E_n$ with each $E_i$ nowhere dense then for any closed sub-interval $[a,b] \subset I$ you would have $[a,b] = \bigcup_{n=1}^\infty [a,b] \cap E_n$, contradicting completeness/bct for $[a,b]$.

[gentle osprey]

kxrider

[sleek thicket]

that makes sense to me

[little hemlock]

alright, nice

[cold vine]

So I have a map f:D^2->R^2 that is continuous and f(-z)=-f(z) for all z in S^1. Now there is a point z that maps to 0 in the disk. How do I go about proving this with Borsuk Ulam? I'm a bit stuck

We are working with the basic one athough we did prove some equivalences like that f:S^n-> R^n then there is x in S^n which maps to 0 which seems kind of similar here

sorry 😄

https://puu.sh/GT6o9/620d84073f.png

We also have sone that this is equal to Lusternik–Schnirelmann theorem

We also have that there does not exist an antipodal map S^n->S^n-1 and that there is no antipodal map D^n->S^n-1 which restrict to antipodal map on the bdry

[gentle osprey]

slimvesus

[cold vine]

hmmm ok Ill think about that one ^^thanks. yeah ive tried a couble of contradiction approaches but no breakthrough

[marsh forge]

if you ever see a problem of the form "there must exist z, f(z)=0"

try dividing

yeah

FWIW if you keep struggling I think this might be in hatcher

[cold vine]

that makes sense 😄

[marsh forge]

i think he proves borsak ulam in this form or something

but its been awhile

[fading vale]

99% sure he does

[marsh forge]

(there is also a similar proof of the fundamental thm of calc)

[cold vine]

ahh nice!

ok that worked, phew

[rugged swan]

hi, for a topological space X, is there a quasi-isomorphism beetween C_{*-1}(X) and C_*(Sigma X) ?

where Sigma X is the suspension of X

A quasi-isomorphism for degrees n > 0 I mean

[cold vine]

I'm trying to prove that homeomorphism D^n->D^n induces isomorphisms in Hp(D^n,D^n-{x}) -> Hp(D^n,D^n-{f(x)}). No clue where to start. I have calculated the homology groups in cases |x|<1 & |x|=1 but I don't see if they help here.

[marsh forge]

how much of a hint do you want @cold vine

[cold vine]

I'm so stuck I dont know where to begin so that's at least nice for now xD

[marsh forge]

okay the big idea is to use the LES of a pair

(also in case you aren't aware of this fact, a homeomorphism $f:X\to X$ induces a homeomorphism $f:X-a \to X-f(a)$)

[gentle osprey]

MaxJ

MaxJ

[cold vine]

okay nice. wasn't aware but that makes sense. Apparantly the pair isn't (D^n, D^n-{x}) which I used to calculate the groups

[marsh forge]

this is in fact the correct pair

i cannot say much more without giving it away, I think

[cold vine]

alright thanks!

[marsh forge]

lmk if you want another tiny hint that is more observation than hint

[cold vine]

ok thanks!

ahh so this seems to be 5-lemma with Hn(D^n-{x})->Hn(D^n-{f(x)}) induced by that fact you mentioned and the original homeomorphism

[marsh forge]

Yeah, my only other hint would have been that you in fact have two pairs, not one

[cold vine]

Yeah, true!

Thanks alot

[marsh forge]

np

honestly i really like these types of problems, computations in (co)homology is more or less my favorite part of AT

[cold vine]

Yeah these are pretty fun. I just got super confused since it's been a while we had a similar excercise and we had a proof which proved something similar in a very low level detail and I assumed I had to invent something similar😅

But yeah, shows how nice these sequences and functoriality are

[nimble cipher]

Hi. How do I answer this problem from Gamelin-Greene: Show that any path in $S^n$ is homotopic with endpoints fixed to a polygonal path on Sn, where “polygonal” is now interpreted to mean that the path is formed from arcs lying on great circles of Sn.

[gentle osprey]

emphatic_wax

[nimble cipher]

<@&286206848099549185>

[surreal flint]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[tight agate]

What have you tried?

[nimble cipher]

I'm trying to use the open cover of Sn by eight open half-hemispheres

then I want to use it to partition the images of the subintervals of [0,1]

which each image is contained in some open half hemisphere then I want to construct a polygonal path from that but I'm not successful

[tight agate]

have you tried drawing any pictures?

[nimble cipher]

I drew the sphere and the covering

I don't know how to incorporate the path

[tight agate]

If I gave you two paths with the same endpoints in Rn, can you find a homotopy between them?

[nimble cipher]

Yeah. Straight line homotopy

[tight agate]

okay, now can you do a similar thing

?

[nimble cipher]

So, I define it piecewise?

[tight agate]

or really can you use that to construct homotopies between paths on Sn

[nimble cipher]

I did the problem before this which is every path in Rn is path homotopic to a polygonal path

[tight agate]

okay, use that

[nimble cipher]

So I just do exactly the same but in the sense of the unit sphere?

[tight agate]

Sn - pt is homeomorphic to Rn

[nimble cipher]

Ohh yeah. Thanks that's helpful!

[tight agate]

use the homeomorphism to transport everything to Rn, do whatever you need to do over there, and transport everything back

[nimble cipher]

Ohh okay. makes sense to me. Thank you!

[tough imp]

Is there an elementary proof that if A,B are finite type k-algebras, then the map Spec A x_k Spec B -> Spec B is an open map?

Vakil uses the result in chapter 9 but offputs the proof until like chapter 21 where you prove it using some thing about flat morphisms and Chevalley's theorem which is more than I'd like to be using

[sleek thicket]

We talked about this a little and I think that maybe this map is the restriction of a projection A^(n+m) -> A^m

And so you can reduce to proving it for that map

(maybe?)

[tough imp]

Consider the map Spec k[x,y] -> Spec k[x] given by the inclusion of k[x] into k[x,y]

Consider a distinguished open D(f(x,y)), is the image of this a distinguished open in Spec k[x]?

I feel as though it ought to be D(f(x,0)) but am unable to show it

Maybe this is just flat out wrong lol

[frosty sundial]

@tough imp f(x,y) = xy - 1 is a counterexample I think

[tough imp]

Yeah I don't think it's true anymore :(

But I am almost certain that it's still open

Since I think that k[x,y] is a flat k[x] algebra

[tight agate]

Yup that map is open

[tough imp]

So I'm just going to use Chevalley's theorem

In this specific case it isn't too bad

It's just an application of Chevalley's with going down

[tight agate]

The map Spec R[x] to Spec R is always open

[tough imp]

Oh?

How do you prove that?

[tight agate]

uuuuh

[tough imp]

I have to use the fact that k is a field

in order to do it the way I'm trying to

[tight agate]

okay, let's say we have a standard open D(f)

gimme a sec i need to write it out

crap this is harder than I thought

[gritty widget]

i believe in you

[tough imp]

I’m not sure if it’s true haha

For fields it is

[tight agate]

no it is true

[tough imp]

Did you prove it yourself at some point?

Or did you read it somewhere?

[tight agate]

no someone told me

trusted source

[tough imp]

That source? Albert Einstein

[tight agate]

We're looking at the map Spec R[x]_f to Spec R

We have the inclusion map p to Spec R for some point p

pullback

and we have Spec k(p) \otimes R[x]_f

[tough imp]

Sure

[tight agate]

So p is in the image iff that space is nonempty

[tough imp]

Yeah

[tight agate]

is that the same ring as (k(p)[x])_f

?

[tough imp]

Uhhhh...

This is a tensor product over R yeah?

Or hmmm

[tight agate]

yup

[tough imp]

Maybe you could show it by some universal property shenanigans

I feel like I might’ve shown this sort of thing for sometbing Hartshorne does in II.6

But I remember it being really confusing

And spending a few hours working it all out

[tight agate]

it looks like it's true

but if it is, then Spec(k(p)[x]_f) = empty set iff f is 0

which means that all the coefficients of f are in p

yeah so the complement of the image is closed

[tough imp]

What set is it?

Like V((f))?

Well no hat doesn’t make sense

[tight agate]

V (coefficients of f)

f isnt in R

[tough imp]

Right

Hmmm

Damn

So just showing that isomorphism is the only step

[tight agate]

yup

k(p)\otimes S = S_p/pS_p

S an R algebra

(R[x]_f)_p = (R_p[x])_f

So the ring we're looking at is (R_p[x]_f)/p(R_p[x])_f

which should be the same thing as (R_p[x]/pR_p[x])_f

which should be the same thing as ((R_p/pR_p)[x])_f

which is the same thing as k(p)[x]_f

[tight agate]

https://stacks.math.columbia.edu/tag/00FB

there ya go

that was the argument

[sleek thicket]

ttera moment

[gritty widget]

?

[sleek thicket]

[gritty widget]

[sleek thicket]

im doing RG again

I met with lee and he doesn't hate me for not showing up to class

[gritty widget]

😌

[sleek thicket]

yeah, he agreed to write me a letter of rec for REUs and also I realized I was less behind than I thought/the lecture recordings are not actually missing

perhaps I can actually learn RG

Probably dumb question

Is there an infinite discrete subgroup of O(n)?

[tepid depot]

i dont know the answer, but it’s not too hard to start thinking about n=1,2,3

[sleek thicket]

Yeah, that's true

[tepid depot]

and i think there are descriptions O(n) in higher dimensions that build on the lower dimensions, that could help you come up with something

[sleek thicket]

Yeah I mean O(n) should embed in O(n+1) by just adding a 1 in the bottom right corner of the matrix

And filling out with 0s

So it suffices to find one in O(2)

[tepid depot]

i’m guessing one isn’t going to exist in O(2)

everything in O(2) can be written as rotation composed with reflection right?

[sleek thicket]

And I think that will happen iff there's one in SO(2) ≈ S^1, since [O(2) : SO(2)] = 2

Yeah

[tepid depot]

right

well S^1 doesn’t have one i’m pretty sure

[tight agate]

S1 is compact

[tepid depot]

nice

[sleek thicket]

All O(n) are compact, that's why I asked originally

I wasn't sure if lie subgroups were automatically closed

I didn't think they were...

Maybe that's what I was missing?

[tepid depot]

if the set is topologically discrete, it’s closed right?

idk that could be wrong but it’s also probably true for lie groups

[sleek thicket]

Isn't {1/n : n in N} a discrete subset of [0,1]?

[tepid depot]

ok you have a very good point

so im wrong lol

[sleek thicket]

Works in R^* too

Np lol

[tight agate]

crap

[sleek thicket]

I'm trying to prove things about "spherical space forms" which are riemannian manifolds with constant positive curvature, or equivalently quotients of S^n(R) by discrete subgroups of O(n+1)

So I would like to understand those subgroups

[tight agate]

okay but the topology must be symmetric

[sleek thicket]

wym symmetric?

I mean it's invariant under left translations

And inversion

[tight agate]

yeah I realized my statement doesn't make much sense

[sleek thicket]

So I guess my question is

Does S^1 have any infinite discrete subgroups?

[tepid depot]

ok no, here’s an idea

[sleek thicket]

oh wait can I use sequential compactness? Enumerate the points and pass to a convergent subsequence?

[tepid depot]

if there’s any irrational point in the group, it’s not discrete

[sleek thicket]

True

[tepid depot]

by irrational point i mean irrational angle

[sleek thicket]

Yup

[tepid depot]

ok so every point is rational if it’s discrete

[sleek thicket]

Oh yeah?

Why's that ultra?

[tepid depot]

you beat me to that by a little bit 🙂

haha it’s ok

[gritty widget]

ultra is too fast

[tepid depot]

but yeah definitely no infinite discrete subgroups here

[sleek thicket]

Why does not being dense imply there's a minimal rational angle?

Oh yeah

I think that makes sense? You could keep translating your small angles around

And get near anything

okay so does this generalize to SO(n)

Or O(n)

[tepid depot]

off the top of my head no, but also my feeling is it shouldn’t be possible because of what i initially said

very similar to this projection thought even if that doesn’t work

basically big O(n) should be built out of smaller ones

(probably)

what is O(n+1)/O(n)?

[sleek thicket]

I'm glad the answer to my question wasn't just "here is an obvious example"

I want to say it's a sphere?

Like O(n+1) acts transitively on S^n

And the stabilizer of the north pole is the set of rotations through that axis

does that make sense?

[tepid depot]

yeah that seems correct

so we can write O(n+1) = O(n) \oplus S^n right?

that seems helpful

[sleek thicket]

I don't really know what you mean by that

like it's not a product of the two (I think)

I think O(n+1) is an O(n) bundle over S^n?

[tepid depot]

as you can see my group theory is a bit shaky

[sleek thicket]

haha

I think that like, there's a spheres worth of copies of O(n) in O(n+1)

Is the right way to interpret this?

Or really an RP^n's worth since antipodal points have the same axis

[tepid depot]

that sounds correct

i’d believe it

[sleek thicket]

That's like a lot

So maybe we can try to project

Like arbitrarily

Maybe sards can tell us there's one that works (???)

[tepid depot]

ooo nice

[sleek thicket]

wait do you even need to invoke homogeneity

like

Oh the limit might be outside of the subgroup

[tight agate]

that is sorta what I tried to say but im confused

[sleek thicket]

Alright sick this rocks

hmm okay now the hard part lol

Tfw not all subgroups of SO(n) are cyclic

I think you can rotate along two disjoint planes in SO(4) so they can't be

[sleek thicket]

So I figured out a proof that any discrete subgroup of a compact lie group is finite

If $H$ is a discrete subgroup of $G$ then $H$ is countable (cover $G$ by countably many charts and use the fact that a discrete subset of $\R^n$ is countable)

[gentle osprey]

Shamrock (pejorative)

[sleek thicket]

Then $H$ has the structure of a $0$-dimensional lie group, and the inclusion $H \to G$ makes it an embedded Lie subgroup

[gentle osprey]

Shamrock (pejorative)

[sleek thicket]

But embedded subgroups are automatically closed

[sleek thicket]

I have confused myself

A lot

Consider eg $A_5$ as a subgroup of $SO(3)$, acting as the rotations of a tetrahedron

[gentle osprey]

Shamrock (pejorative)

[sleek thicket]

Nvm lol

[nimble jolt]

@sleek thicket what? is that argument really necessary? can't you simply take a sequence x(n) in H converging to some x in G, and then show that x(n+1)^-1x(n)->e contradicting discreteness?

[sleek thicket]

I think people said this earlier and I did not understand it lol

[nimble jolt]

do you understand it now? lol

[sleek thicket]

Yeah

we have a sequence of nontrivial elements

converging to the identity in H

[nimble jolt]

yep

[sleek thicket]

I think this is one of those nights where I kept doing math several hours past the point where I was still productive

[nimble jolt]

yeah fair

[jaunty nebula]

My goal is to point out that triangles PAD and PCD have the same angles. There's a trivial case where O is on the angle bisector and I don't really have to prove that, but if O moves around a bit and stays within the angle, then this is a bit more difficult.

I've tried to find a point which would allow me to use homothetic transformations to prove this, but that doesn't seem plausible.

[elder yew]

Can you try drawing a line between A and D, C, and B and try to play angle games?

[jaunty nebula]

I'll try that, thanks. My initial ideas was to use AC and BD to draw lines which intersect somewhere and then I'd have another angle besides P cornered one, but I couldn't find any use for it

Well, one of my initial ideas

[elder yew]

That was my initial thought too

but I didn't see it going anywhere

but with AD and BC you might be able to construct similar triangles

or whatever they're called

Yeah you can make triangles like

ABD

ACD

BCD

BAD

until something works?

IDK Euclidean geometry is not my strong suit

[jaunty nebula]

I remember thinking that geometry was easy and fun. I think university has taken the easy part out of it

[elder yew]

I never passed a euclidean geometry course

[sweet wing]

there is euclidean geometry in uni

[nimble jolt]

yeah @sweet wing some places bundle a bunch of geometry topics together and teach it as a relatively low level course.

[sweet wing]

ahhh

[nimble cipher]

Does anyone know how to answer these questions?

I can't think of a sequence. I feel like I need to define it recursively

I know the limit somehow should be such that the derivative is zero

[sleek thicket]

Any infinite set contains a sequence of distinct elements

Choose any initial element x1, then choose an element x2 distinct from x1, then choose x3 distinct from both x1 and x2,...
If at any point you can choose a next value then your set is equal to {x1,...,xn}, which is finite

[nimble cipher]

so the sequence can be arbitrary?

how does this show that there is a point in the preimage of {p} such that the derivative vanishes?

[sleek thicket]

well, what can you say about this sequence?

The hint says to look at its limit, right? That's not quite correct though, since an arbitrary sequence like this might not converge...

[nimble cipher]

Hmm I don't understand sorry

[tepid depot]

i thank shamrock is saying that you can find a convergent sequence in f^{-1}(p)

where all the points in the sequence are distinct

and that this may be helpful

[nimble cipher]

So I have to find that specific sequence iteratively?

[tepid depot]

well you just have to prove such a sequence exists

shamrock says: "well it's easy to find a sequence of distinct points, but that sequence may not have a limit"

is there anything you can do to fix that?

[nimble cipher]

isn't given since [0,1] is compact?

[tepid depot]

not every sequence in a compact set converges, but compactness does help

also you want to be using compactness of S^1 here

since f^{-1}(p) is a subset of S^1

[nimble cipher]

so what i need to do is to show that there is a convergent sequence in f^{-1}(p) with distinct terms right?

[tepid depot]

yes

[nimble cipher]

how do i use this to show that there is a point where the derivative vanishes?

does completeness help in showing the one about the sequence?

[tepid depot]

you may recall $\lim_{x \to a} g(x) = L$ if and only if every sequence $x_n \to a$ of distinct points not equal to $a$ has $g(x_n) \to L$

[gentle osprey]

doubledual

[nimble cipher]

Ohh yeah. I apply this to the derivative?

[tepid depot]

yeah exactly

you already know it's differentiable, so you just gotta evaluate the limit for some sequence with that property

where g is the difference quotient

[tepid depot]

what do you know about sequences in compact metric* spaces?

[nimble cipher]

every sequence has a convergence sequence?

for metric spaces?

[tepid depot]

you're not stating that correctly

i think you know the result

[nimble cipher]

every convergent sequence*

[tepid depot]

every convergent sequence has a convergent sequence?

[nimble cipher]

has a convergent subsequence

sorry for the confusion

[tepid depot]

yeah that's it 😄

ok with that fact you should be able to do it

[nimble cipher]

Nice thank you!!

so for the last bit

[nimble cipher]

i don't get this one

f(x+h)-f(x)/h

is this what you're referring to?

then I let h go to 0?

[tepid depot]

it's easier to let g(x) = (f(x)-f(a))/(x-a)

and find a sequence converging to a

you could do it either way though

[nimble cipher]

Ohhh okay I understand and the limit should be 0, i.e., L=0

[tepid depot]

also i want to make sure it's clear, the theorem is "in compact metric spaces, every sequence has a convergent subsequence"

[nimble cipher]

Thank you!

Thank you @tepid depot and to @sleek thicket too

[tepid depot]

anyone know if there’s something like a fubini’s theorem for product manifolds? something along the lines of $\int_{M \times N} \omega = \int_M \int_N \omega$. Main thing i’m not sure how to think about is what the differential form on the right side should mean exactly

[gentle osprey]

doubledual

[tepid depot]

assuming something like this is possible

[tight agate]

probably works

the charts on the product are just products of charts

so you can do fubini locally

[sleek thicket]

hmm

I also don't know how to interpret the statement

I think you can maybe say something about the integral of like π1*ω ^ π2*η where π1, π2 are projection maps?

Seems relevant https://mathoverflow.net/q/350952/136523

I asked about it on Twitter and someone pointed me here

[gritty widget]

I asked about it on Twitter

[sleek thicket]

Also possibly this: https://math.stackexchange.com/q/3895777/408287

[sleek thicket]

well it seems Twitter was more helpful for me than discord was for doubledual :^)

[gritty widget]

[tepid depot]

wait whats your twitter?

(in b4 i follow you already and didn’t realize)

[sleek thicket]

@grassmannian

[tepid depot]

nice handle

now you get to see my face i guess lol

also that banner lmao

yeah like 50% of the people i follow on math twitter follow you so it’s a miracle i didnt follow you already

[sleek thicket]

I am very active on there

We have one mutual: Daniel Litt's private account

[sleek thicket]

@tepid depot I found the right version of fubini

[tepid depot]

nice

[sleek thicket]

It's in Amann-Escher Analysis III Chapter XII.2

Unfortunately in German

[tepid depot]

another pdf for my collection

never mind

[sleek thicket]

I think it's still readable enough

I can't really focus on it rn but it seemed simple

[tepid depot]

i will give it a go lol

[sleek thicket]

You can like partially apply a form

[tepid depot]

thanks a lot!

[quiet pilot]

Is it true in general that $$D(g) \subset V$$ has the same dimension as V? Where V is an affine variety

[gentle osprey]

AoiKunie

[meager python]

Dimension is local on irreducible components

At least in the noetherian case

Depending on your definition of variety you should have that k(U) = k(V) for U zariski open in V and so dim U = dim V.

[nimble cipher]

Hi. Can someone help me in proving this?

[tepid depot]

@nimble cipher do you have a good intuitive understanding of what's going on here?

there's a lot of terminology being introduced at once so it'd make sense if it's hard to parse

you have some z in the image, and s in f^{-1}(z). If d(s)=1, that means the curve is going counterclockwise at s. if d(s) =0, that means the curve is going clockwise at s

so adding up all the d(s) should tell you how many times you went around the circle counterclockwise

and that should correspond to \tilde{f}(1)-\tilde{f}(0) since that's the lift of f

this isn't a proof or anything but you should try and make sure that all makes sense before going at it

[nimble cipher]

I get the intuition but I don't know how to make it rigorous.

[tepid depot]

alright this might help get you going then

[nimble cipher]

I know d(s) represent how the loop is passing the point corresponding to the regular value

[tepid depot]

without loss of generality you can take z = 1, just by rotating the image of your curve

you can check nothing relevant will change

this makes it a little easier, because points in the preimage of 1 are always integers

for the lift

second tip: there are finitely many points in the preimage (of f or the lift)

i haven't worked this out, but i think with this induction should be viable

that or some kind of telescoping sum trick

[nimble cipher]

okay. I will try to work on this a bit. Thanks for the ideas.

Is it that I should work from the difference of the endpoint of the lift?

[tepid depot]

yeah

i would label all the points in the preimage 0 = s_0 < s_1 < ... < s_n = 1

think about those points for the lift, and how that relates to d

[nimble cipher]

do these points correspond to the degree of the the map?

[tepid depot]

you're showing $\sum_{i=0}^n d(s_i) = \tilde{f}(1) - \tilde{f}(0)$

[gentle osprey]

doubledual

[tepid depot]

and this is the degree of the map

it's a possible definition

[nimble cipher]

Ohhh okay okay thank you

but how do i know what the value of the lift at the endpoints?

[tepid depot]

well $\tilde{f}(1)$ ad $\tilde{f}(0)$ both map to 1 right?

[gentle osprey]

doubledual

[nimble cipher]

yes

[tepid depot]

since you send them through e^{i 2 \pi x}$

so they're definitely both integers

and similarly $\tilde{f}(s_i)$ is always an integer

[gentle osprey]

doubledual

[nimble cipher]

Okay this works thank you so much!

[gentle osprey]

Brofibration

[tight agate]

it turned up during a lecture

with no name

M and N are A-modules

[tepid depot]

@nimble cipher looking at this carefully, i think there might be a slight misstatement in the problem. if 0 and 1 are both in the preimage of z, you wouldn't want to include both of them in the sum, since they both correspond to the start/end of the loop

when you get a little more formal, a closed loop is a map from S^1 instead of a map from [0,1] with equal endpoints

and i also imagine that the loop is supposed to be C^1 at the endpoints of the loop

something maybe to check with your prof about

[nimble cipher]

Oh so the problem defines the function wrong?

[tepid depot]

well it's a bit of a technicality

if z = 1, then you put d(0) and d(1) in your sum

and that's double counting that point essentially

but if z != 1, you dont have this problem

[nimble cipher]

I think when my prof defined a loop to us is that it's a path with the same starting and end point

[tepid depot]

right

a loop is really a map with domain S^1

if you define it that way, you don't have this issue

your prof wanted to keep it a little simpler i think

but that slightly messes up the formula in this special case

if the domain was S^1 the formula would be right

[nimble cipher]

so i can't take z=1?

[tepid depot]

the formula is wrong if z=1, it's right otherwise

like let's do a simple example

f(t) = e^{i2pi t}

the lift is $\tilde{f}(t) = t$