point-set-topology

[stable lance]

So that's a way, even though I don't know this result yet

[tight agate]

cool, I would also be interested in knowing if there's a covering space way of doing it

the way you get the result on genus is just by computing the canonical bundle of the curve

which is not hard if you know the canonical bundle of projective space

[stable lance]

I was trying to mimic the n=2 case, but no success yet

I'm thinking this time some non elementary result will enter

[burnt spruce]

@tight agate Something seems off about the approach via genus. E.g. the quadric gives you a CP^1 after projectivization, but the fundamental group of the cone is actually Z/2Z. I think the C^* bundle over the sphere is what introduces the loop.

[tight agate]

Oh no I think I misunderstood the problem, I thought you're trying to show that the fundamental group of the projective plane curve was nontrivial

yeah the argument does not apply to the cone

that's why I asked if you're quotienting out

[tight agate]

pi_1 takes products to products

So the fundamental group of the product is the product of the fundamental groups

[marsh forge]

If the fundamental group is G just take a product with a K(1/G, 1)

Then ur simply connected

[river plinth]

<@&286206848099549185>

[wind hull]

uh that works

[river plinth]

<@&268886789983436800>

[wind hull]

..

[river plinth]

@wind hull

[gritty widget]

Yes?

[river plinth]

help @wind hull

[wind hull]

uh i dont think we are supposed to ping for that

[river plinth]

he needs help

[gritty widget]

it's kind of you to try by your means to help Joshua

but you're not supposed to ping moderators for helping

[river plinth]

i cant im stuid

[keen cliff]

wrong channel. this isn't topology or geometry.

[gritty widget]

🤔

[river plinth]

oh my bad i thought moderators helped people

[wind hull]

im in geometry

[keen cliff]

or at least, not the level of geometry for this channel

[wind hull]

oh

which one then

[ivory dragon]

#geometry-and-trigonometry

[wind hull]

kk thanks

[ivory dragon]

as an aside @river plinth , please don't ping helpers for other people, and also don't ping it before 15 minutes have elapsed

as per the rules in #❓how-to-get-help

[wind hull]

did he get banned

lol

[ivory dragon]

i didnt do it

¯_(ツ)_/¯

[burnt spruce]

@ripe thistle @tight agate Yes, but its not the trivial bundle. The fundamental group would be Z in that case, but its only Z/2Z.

[proud maple]

Is this a place where I can ask algebraic geometry too? I have a question about closed point in a scheme. I always thought since a scheme is locally and affine scheme, the existence of closed point in the scheme is to be guaranteed. But my google search seems to indicate it isn’t a trivial question. May I know what is wrong with my naive answer here?

locally an affine*

I reckon it has something to do with likely erroneous understanding of mine that any affine local chart the point will always correspond to maximal ideal of each ring?

[honest narwhal]

In an affine chart you'll get a closed point, and if you're quasicompact you're fine

But in general you take an open cover, you want it to be closed wrt each element of the cover containing it, and that can be a problem in general

[proud maple]

@honest narwhal thanks!

[fallen canopy]

So there is this theorem that says “If $X$ is any space and $Y$ is a compact space, then the projection map $\pi_X : X \times Y \to X$ is closed”. This follows quickly from the tube lemma. However, the converse of this theorem is apparently true. I DO know that if we have the projection map $\pi_X$ is closed for any choice of X and Y compact, then the tube lemma holds. But this is as far as I have gotten; does the tube lemma being true imply $Y$ is compact? I don’t see any reason why it would, but I’m not sure what else to do.

[gentle osprey]

Michael Harp:

[lean swan]

$Y$ is compact if and only if $\pi_X$ is closed for all spaces $X$.

[gentle osprey]

leoli1:

[exotic badger]

Hi! can anyone help me? For i.) my idea is to show that any open ball with middle z in A_k lies in A_k. For that I'd have to take a y from the ball and show that dist(y,C) < 1/k right?

[bleak helm]

What is the exact definition of dist(x, C)? @exotic badger

[exotic badger]

$dist(x,C) = inf{d(x,f) : f \in C }$

[gentle osprey]

Cone:

[bleak helm]

Thanks. Ngl, you kind of saved my soul here. For some reason I had never seen it before but I recently read it in a paper, and could not find an explanation of it anywhere.

Okay, so you should take a point in A_k, and then find a ball around that point that is contained in A_k

[exotic badger]

the point is contained in A_k if dist(y,C) < 1/k right?

[bleak helm]

Yes

So let $x \in A_k$ then $dist(x, C) < 1/k$. Now consider $B(x, 1/k)$, can you show that $B(x, 1/k) \subseteq A_k$ ? (changed it to x because I prefer it, lol)

[gentle osprey]

Lunasong:

[exotic badger]

does that work?
$ dist(y,C) = inf{d(y,f) : f \in C}
\leq d(y,f)
\leq d(y,x) + d(x,f)
\leq \epsilon + 1/k $

[gentle osprey]

Cone:

[exotic badger]

and then send epsilon to zero

i took epsilon as the radius instead of 1/k

[bleak helm]

Wait,I think my radius is wrong. And you can't let ε approach 0, it has to be a fixed radius for your ball

Eating quickly then I'll think about it and @ you

[exotic badger]

ok np, enjoy!

[bleak helm]

Oh, duh: ε = 1/k - dist(x, C)

@exotic badger

[exotic badger]

oh yea, i remember this trick

thanks

[bleak helm]

Np

[exotic badger]

$ dist(y,C) = inf{d(y,f) : f \in C}
\leq d(y,f)
\leq d(y,x) + d(x,f)
\leq 1/k - dist(x,C) + 1/k
< 2/k - 1/k = 1/k $

i got this now

[bleak helm]

You should use \ before curly braces in latex

[gentle osprey]

Cone:

[exotic badger]

oh no d(y,x) is < 1/k - dist(x,C)

[bleak helm]

Oh, nvm, I misread badly

[exotic badger]

d(xf) is < 1/k

[bleak helm]

That's fine

Deleted to hide my shame :p

[exotic badger]

np!

u think it works?

[bleak helm]

Yes

[exotic badger]

alright, thanks!!

[bleak helm]

@exotic badger no that's wrong

Sorry, I'm not paying attention properly

-dist(x, C) > -1/k

And notice you have yet to use the fact that C is closed. There is more to this.

[exotic badger]

ok ill try again

thanks

[bleak helm]

@exotic badger btw, I don't think you need to use that C is closed

But you still need to fix what you did

[exotic badger]

on it!

[exotic badger]

hmm im not really getting anywhere

im stuck at
$ dist(y,C) = inf{d(y,f) : f \in C}
\leq d(y,f)
\leq d(y,x) + d(x,f)
\leq 1/k - dist(x,C) + d(x,f)$

[gentle osprey]

Cone:

[exotic badger]

also i noticed d(x,f) < 1/k doesnt have to hold

[lean swan]

Notice that $d(x,C)=\inf{d(x,z)\mid z\in C}\leq d(x,y)+\inf{d(y,z)\mid z\in C}$, so we have $|d(x,C)-d(y,C)|\leq d(x,y)$, therefore $d(-, C)$ is continuous

[gentle osprey]

leoli1:

[bleak helm]

I have $1/k > dist(x, C) \geq d(x, y) + dist(x, C) = d(x, y) + \inf {d(x, f): f \in C } = \inf {d(x, y) + d(x, f): f \in C } \geq \inf { d(y, f): f \in C } = dist(y, C)$

[gentle osprey]

Lunasong:

[bleak helm]

Only the steps where I take d(x, y) into the inf, and where I apply triangle inequality inside the inf needs to be justified, but I think you can do that.

[exotic badger]

thanks both! ill get to work

how did u conlcude $dist(x,C) \geq d(x,y) + dist(x,C)$?

[gentle osprey]

Cone:

[bleak helm]

Lol, idk what I wrote there, I am just going to take a picture of what I wrote down earlier when I did it, if I can find it

Okay, just ignore the 1/k > d(x, y) part.

It should have started with d(x, y) < ε = 1/k - dist(x, C), so 1/k > d(x, y) + dist(x, C) and then continue from there

@exotic badger

[exotic badger]

ok thanks

[nimble cipher]

Is the product topology on $\mathbb{R}^\omega$ path connected?

[gentle osprey]

emphatic_wax:

slimvesus:

[nimble cipher]

Ohh nice. and infinite product of continuous functions is also continuous?

[ivory dragon]

not necessarily

[nimble cipher]

ohhh I get this now

the component functions are continuous and hence the function is continuous because it's in the product topology

[gritty widget]

Yah

[nimble cipher]

Thanks!

[burnt spruce]

Ohh nice. and infinite product of continuous functions is also continuous?
@nimble cipher As a simple counter example, take the infinite product of $f(x) = x$ with itself on the domain $[0,1]$.

[gritty widget]

Huh? But for a basis set in the codomain U1 x U2 x ... x Un then inverse image is U1 \cap U2 \cap ... \cap Un, open, am I missing something...?

@burnt spruce

[glacial portal]

@nimble cipher As a simple counter example, take the infinite product of $f(x) = x$ with itself on the domain $[0,1]$.
@burnt spruce

How this function looks?

[burnt spruce]

Huh? But for a basis set in the codomain U1 x U2 x ... x Un then inverse image is U1 \cap U2 \cap ... \cap Un, open, am I missing something...?
@gritty widget No I think we are just talking past eachother -- the product in the sense of the map into the product is continuous. Maybe I was confused about what the poster meant, I thought they were asking something like, we know that if $f,g$ are continuous then $fg$ is continuous. So what about $f_1 f_2 \ldots = \pi_{i = 1}^{\infty} f_i$.

@burnt spruce

How this function looks?
@glacial portal The product of $x$ with itself infinitely many times (pointwise, not categorical) is $0$ on $[0,1)$ and $1$ at $1$.

[glacial portal]

@glacial portal The product of $x$ with itself infinitely many times (pointwise, not categorical) is $0$ on $[0,1)$ and $1$ at $1$.
@burnt spruce

Aaah, it's true, hahaha

I was thinking of R

[meager python]

Stupid question, X = Spec k[x, y, z] what is O_X(U) for U = X \ V(x,y)?

[meager python]

That’s what I’m thinking based on ”geometrical picture” since U = D(x) \cup D(y) but I kinda don’t see it algebraically 😦

[cedar pebble]

I mean it’s all localization

You can do the localization element-wise

[meager python]

But can we use O_X(D(f)) = R_f twice or smth?

[cedar pebble]

Essentially yes

[meager python]

Like if D(x) and D(y) would be disjoint then the sheaf of the union would be the product

[cedar pebble]

An easier way is to note how localization at distinguished opens behaves under union

Well so the point is that

$S^{-1}R=\varinjlim R[f^{-1}]$

[gentle osprey]

nGroupoid:

[cedar pebble]

Taken over elements f of your multiplicative subset S

[meager python]

I guess maybe looking at ker(prod O_X(U_i) —> O_X(U_ij))?

The intersection would be on the elements over z

[cedar pebble]

Yes! That’s essentially the limit appearing here

[meager python]

Hence why we don’t get two copies of z but only one

[cedar pebble]

I think that is right?

[meager python]

So we identify the product terms on the right which agree on the intersection, otherwise they get distinct copies

Outside their intersection

If that makes sense

[cedar pebble]

Yea I see what you are saying

[meager python]

Thanks

[gritty widget]

what is a holomorphic vector bundle

[crisp sparrow]

https://en.wikipedia.org/wiki/Holomorphic_vector_bundle

[gritty widget]

complex manifolds

[fading vale]

when ur defining a homology theory, why does going from reduced/absolute to relative homology mean you change the third axiom from wedge sums to disjoint unions

like i guess i get that its more general because you can just take all the basepoints, call the set of all of them A, and then $(\coprod X_\alpha, A) = \bigvee X_\alpha$

texit please

idk maybe im just being dumb and overthinking this

[small phoenix]

is it clear that you can't keep the 3rd axiom as written in unreduced homology? it's just false since h_0 counts connected components.

[fading vale]

oh yeah i think that clears it up

ty

i was being dumb

[marsh forge]

Maybe a side note: you should think about reduced (co)homology as being the correct notion for based spaces which is justified by the isomorphism $\tilde{H}(X_+)=H(X)$

[gentle osprey]

MaxJ:

[marsh forge]

In this context wedge is the correct notion of coproduct

and we want to preseve them

@fading vale

[fading vale]

ahhh

ok that makes sense

ty

[thick ferry]

I'm trying to give an example of a measurable space $(\Omega,\Sigma)$ a function $f:\Omega \to \mathbb{R}$ such that $f$ is not $\Sigma$-measurable but both $|f|$ and $f^2$ are. I was thinking of letting $\Omega = [0,1]$ letting $\Sigma$ be the smallest algebra containing all open sets in $[0,1]$ and defining $f$ so that $f(x) = 1$ if $x\in A$ and $f(x) = -1$ if $x\notin A$ where $A$ is some non measurable subset of $\Omega$. The problem is I cant find such a set $A$, does it even exist?

[gentle osprey]

Kraft Macaroni:

[floral gust]

For what a non-measurable subset could be, check out https://en.wikipedia.org/wiki/Non-measurable_set

[thick ferry]

Thanks mate, out of interest I'm struggling to visualize what a measurable set actually is. Is it a matter of just getting used to the definition and not bothering whether you can visualize it or not?

Also doesn't the Cantor set have measure zero, does that mean it's not measurable?

[floral gust]

Yes, I misremembered the definition of non-measurable

[marsh forge]

two questions

the first is

is Spec Z/p^n homeomorphic to Spec Z_p where the latter is the p-adics?

the second depends on that?

my rational here is that both should be the sierpinski space?

[tight agate]

Are you asking if lim Spec Z/p^n = spec Zp?

[marsh forge]

no maybe im just being dense bc im trying to help someone with a pset for a class im not in

but the prime ideals of Z/p^n should be 0,p and likewise for Z_p

right?

so they should be homeomorphic as spaces

[tight agate]

sounds alright I think

[marsh forge]

equally potentially dumb question

if i have two rings R and S with prime ideals

then i keep seeing that the prime ideals of RxS are of the form PxS and RxP

for P a prime ideal in the other

but the 0 ideal is also prime but not of that form right?

[tight agate]

0 is not a prime ideal in thr product

the*

(r,0) (0,s) = (0,0)

[marsh forge]

oh shit

thank you im so dumb

i forgot this was an integral domain thing

[tight agate]

yeah the proof of that fact that the prime ideals of RxS are of that form also uses the same idea

[fading vale]

is (b) even true

like if n = 2 and k = 1 wouldnt taking out a circle in S^2 just partition it into two disks

with trivial homology

@marsh forge it is your time to shine max

wait nvm

lol

in that case i would be 2 - 1 - 1 = 0 so checks out

ok i think i see the intuition now

[small phoenix]

oh yeah. schoenflies theorem.

actually, that's weaker. schonflies says that the components are homeomorphic to disks.

the funny thing is schoenflies doesn't generalize to higher dimensions because of funny, pathological embeddings like alexander's horned sphere.

and homology can't detect such pathologies.

[gentle osprey]

emphatic_wax:

[cold vine]

Hi! I'm kind of stuck. I'm supposed to prove that if I have a topological group with normal subgroup such that the quotient group and the normal subgroups are connected that the whole group is connected.

I know that projection is open map and if I assume for contradiction that there is an open & closed set then I should maybe get some contradiction, but I can't see how.

[cold vine]

w.l.o.g H is in X and thus all the cosets should probably be in X?

at least each coset is connected as well

ah yeah i see, i was going about it backwards

thanks a lot that helps quite a bit!

do you have any clue if it would be harder with the characterization of connectedness that there is no subset that is both open and closed, which was what i first tried to do

ah yeah makes sense!

[zealous glen]

if we have an atlas of M, and find additional charts which are compatible with all the other charts in the Atlas. It says any two such additional charts are also compatible with each other (in the same atlas).

am i thinking that this is trivial as axioms just still holds? as being in the same atlas

or is there another way to think of this?

like... the sets will still be open, hence intersection is open... and transition maps are still smooth

[gritty widget]

so you're asking if chart 1 and chart 2 are each compatible with some atlas A, then chart 1 and chart 2 are compatible with eachother? this is true and it's not difficult to prove, but there's more to it than what you wrote

[zealous glen]

yeah thats what im asking

just wrote the proof, is this gucci or needs more work?

[gritty widget]

i'm assuming the explicit setup is you have two charts $(U, \phi)$ and $(V ,\psi)$ each compatible with the atlas ${(U_\alpha, \phi_\alpha)$ on some locally euclidean space, and you're wanting to prove that $(U, \phi)$ and $(V, \psi)$ are compatible with eachother, meaning that the transition maps between $\phi$ and $\psi$ are $C^\infty$ (i.e. the maps $\phi \circ \psi^{-1}$ and $\psi \circ \phi^{-1}$ are $C^\infty$)

[gentle osprey]

TTerra:

[gritty widget]

in which case i have no idea how this proof shows that

unless we have some different meaning of compatible in mind

proof in spoilers in case you don't want to be spoiled

[zealous glen]

ah this is the book im studying from

[gritty widget]

nice

[zealous glen]

but i didnt want to cheat huehue

[gritty widget]

i guess for a hint i can say: you want to rewrite the transition map $\phi \circ \psi^{-1}$ as a composition of two maps you know are smooth (e.g. by being compatible with the atlas you're given, and the fact that it's an atlas is important)

[gentle osprey]

TTerra:

[gritty widget]

you know the fact it's an atlas is important because tu discusses that chart compatibility is not transitive

so try using that

[zealous glen]

O

of course

i think i was just making it harder for myself

tyty

[stuck kraken]

is the tangent space of a manifold a vector space?

[gritty widget]

yes

[stuck kraken]

ty

[burnt spruce]

@stuck kraken Tangent space at a point

shakes fist pedantically

(You could also have the vector space of all smooth vector fields.)

[stuck kraken]

ok hold on

so the tangent space is just the vector space tangential to a point on a manifold

what do you call the set of all tangent spaces

[gritty widget]

tangent bundle!

well

close enough :^)

[stuck kraken]

is the tangent bundle a vector space?

[gritty widget]

not quite

how could you possibly add two tangent vectors to a manifold living in different tangent spaces?

[stuck kraken]

i mean that was what i was asking i suppose, whether there is a sensible definition of addition in this sorta context

so if it's not a vector space, what kind of object is it (or is it just a thing in its own right)

[gritty widget]

i can say it's (it being the tangent bundle) something known as a vector bundle, but i don't actually know enough about those to say anything meaningful

[stuck kraken]

fair enough

[marsh forge]

Its a manifold

(Which is easy to prove via the defn of local triviality)

[stuck kraken]

also, can someone give some sort of intuition for why this should be true?

[coarse kestrel]

Can someone give me a hint for this

Here's the definition of a neighborhood system

I was thinking maybe we can use the collection of N(x) as a basis then generate a topology, but that doesn't quite work as if we take R^n and the set of closed balls with positive radius, then this collection satisfy the hypothesis and the corresponding topology is the usual topology, but the topology generated by these closed balls is not the usual topology

[nimble cipher]

Hi. How do I show that $[0,1]\cap\mathbb{Q}$ is not locally compact?

[gentle osprey]

emphatic_wax:

[tight agate]

Yo can anyone help me out with this: Let M be a 4-dim smooth manifold. Let $\sigma \in A^2_{\mathbb{C}}(M)$ be a closed form such that $\sigma \wedge \sigma = 0$, and $\sigma \wedge \bar{\sigma}$ is nowhere vanishing. Show that there is a unique complex structure I on M such that $\sigma$ is a holomorphic 2-form on M.

[gentle osprey]

Brofibration:

[tight agate]

I assume we can use the 2-form to define a map from TM to TM*

and the kernel would work out to be the eigenspace of -i oncee we define I

And is the intuition for an almost complex structure just defining the action of i on each of the tangent spaces (making them complex vector spaces) in a compatible way?

[tidal forge]

How do you show that the complex projective space is compact?
What are some common ways to prove compactness in general? The definition doesn't seem like a very useful way to prove the compactness...

[gritty widget]

the quotient of any compact space is compact

write projective space as the quotient of a certain compact set and you're good

@tidal forge

"the quotient of any compact space is compact" basically falls right out of the definition of the quotient topology

one or two line proof lol

[tidal forge]

Is $\mathbb{C}^n \setminus {0}$ compact?

[gentle osprey]

shifubear:

[gritty widget]

no

but that's not the only set of which complex projective space is a quotient

can you think of any other ones?

the one i have in mind is ||the sphere||

[zealous glen]

why can a graph have the condition of continuous and still be a smooth manifold

even though the graph can have corners etc?

like |x|

[gritty widget]

because the graph of any function can be given a smooth structure

it might not be a "smooth" manifold in the sense you'd think

but after all, a smooth manifold is just a set with some maps you call smooth

however

it will not be what is called a smooth submanifold of R^2

i.e. its smooth structure is not "inherited" from R^2

the atlas is just the projection to the domain of the function whose graph you're considering

so while the graph has a smooth structure, it's not the one "coming from R^2"

[zealous glen]

ah so it may represented by (x,y) on some space its a smooth manifold in R^n but not in R^2

[gritty widget]

a particularly "offensive" example is the square. it's homeomorphic to the circle, so it can be given a smooth structure, but it's sure as hell not going to be a smooth submanifold of R^2

not sure what you mean by that

[zealous glen]

i meant the exact example u just gave in a sense

just badly worded

[gritty widget]

well if it's badly worded i'm not gonna know what you mean

it's still a subset of R^2, and it's a smooth manifold, but it's just not a smooth submanifold of R^2

as for what a submanifold entails, the idea is that it's a subset, which is also a smooth manifold, whose smooth structure is "inherited" from the larger manifold

[zealous glen]

ahhhh ok

[gritty widget]

it's actually not that hard to prove that the graph of |x| isn't a smooth submanifold of R^2

or at least intuitively think about it

[zealous glen]

but why can we just define a continuous function being a smooth manifold?

[gritty widget]

the function itself isn't a manifold, its graph will be

[zealous glen]

why is it true to be smooth on R^n but not a smooth submanifold of R^2

yeah a graph of a continuous function

[gritty widget]

why is it true to be smooth on R^n but not R^2
no clue what this means

the graph of a function will be a smooth manifold because you can give it a smooth structure

[zealous glen]

how can you just give it a smooth structure?

[gritty widget]

you can pick a particularly awful function and get something that doesn't look like it'll be "smooth" in the normal sense of the word, but you're still getting a smooth manifold by definition

as follows

if $f : \bR \to \bR$ is continuous and $\Gamma_f = {(x,f(x)) : x \in \bR}$ is the graph of $f$, then, if $\pi : \Gamma_f \to \bR$ is the projection $\pi : (x, f(x)) \mapsto x$, the set ${(\Gamma_f, \pi)}$ is an atlas on $\Gamma_f$. by adjoining to this collection all of the charts compatible with $(\Gamma_f, \pi)$, you get a smooth structure on $\Gamma_f$ making it a $1$-dimensional smooth manifold

[gentle osprey]

TTerra:

[zealous glen]

so any topological manifold where a homeomorphism exists can be considered as a smooth manifold?

[gritty widget]

"where a homeomorphism exists" ?

[zealous glen]

which there exists

[gritty widget]

you're gonna have to be a bit more precise

but i think i know what you mean

any manifold exists homemoprhism manifold can be considered smooth maniofld exist?

any topological space that's homeomorphic to a smooth manifold is itself a smooth manifold. this is easy to prove (pull back the charts etc. etc.)

and is actually a more general case of the construction given here

is that what you mean?

(and that's the max reaction amount)

i remember doing a ton of exercises on this stuff way back when i was first learning manifolds

fun stuff

||👻||

[zealous glen]

Let M be a topological manifold of dimension $n$. Suppose that there exists a homeomorphism

$\phi \colon M \longrightarrow \mathbb{R}^n$

defined for all $x \in M$. That is, $\phi$ is a continuous bijection with a continuous inverse. Thus, it would be appropriate to call $\phi$ a chart on $M$ and, it seems to me, we could define an atlas $\mathcal{A}$ on $M$ consisting of the single chart $\phi$,

$\mathcal{A} := {(\phi, M)}$

Using this atlas, if my reasoning is correct, one can consider $M$ to be a smooth manifold. Therefore, any topological manifold $M$ for which there exists a homeomorphism $\phi \colon M \longrightarrow \mathbb{R}^n$ can actually be considered as a smooth manifold.

[gentle osprey]

!R. ¬:

[zealous glen]

is that line of reasoning correct?

[gritty widget]

protip: \varphi

this is correct, and a special case of what i mentioned earlier

however!

you may want to be careful and say that A induces a smooth structure on the manifold

(if you want to be pedantic)

since we're on the topic of smooth structures and atlases and what not

[zealous glen]

mhm ok, and how can one make this much more general?

[gritty widget]

let $M$ be a smooth manifold and let $N$ be a topological space for which there is a homeomorphism $h : M \to N$. if ${(U_\alpha,\varphi_\alpha)}$ is an atlas on $M$, consider ${ (h(U_\alpha), \varphi_\alpha \circ h^{-1}) }$

[gentle osprey]

TTerra:

[gritty widget]

this is more general because $\bR^n$ is covered by a single chart $(\bR^n, id)$

[gentle osprey]

TTerra:

[zealous glen]

ok, thank you.

[gritty widget]

prolly a good exercise to work out the details on that one

it's not terrible; you know what to check

[tidal forge]

Is the sphere homeomorphic to the projective space?

or do i need to take some quotient of the sphere

[gritty widget]

projective space is a quotient of the sphere

which answers your question of whether or not projective space is compact

if you're used to representing projective space as a quotient of C^n \ {0} (or whatever) and not comfortable with that statement, it shouldn't be terrible to come up with a homeomorphism between the two quotients

(projective space is obtained by identifying antipodal points on the sphere)

[tidal forge]

ohhhhh ok i vaguely recall seeing that definition before.

[gritty widget]

now that i've muted all of the other channels i can camp this one

[tidal forge]

haha you're the best

[gritty widget]

this is also like the only advanced channel i can post in, since of all of the advanced channel topics the only one i've had a full course in is topology

[coarse kestrel]

Is there any point in memorizing these, and how can you memorize these more easily

Ah makes sense ok

[ivory dragon]

they should be fairly intuitive though

if you think about them

[coarse kestrel]

The thing is that it doesn't feel very intuitive to me

[gritty widget]

visualize subsets of R^3

[nimble cipher]

Assume that a point in the space has a neighborhood contained in a compact set. Choose a sequence of rationals in the neighborhood converging to an irrational...
@gritty widget Thank you!

[versed geode]

Just try drawing some examples; it feels pretty instinctual after some practice

[gritty widget]

imagine two balls touching

[coarse kestrel]

How is quasi-component not just the components

[coarse kestrel]

Is there an example of a space where the quasi components are not the components

Oh I think I see where my reasoning went wrong

So a space $X$ is connected iff every continuous map $d$ from $X$ to a space $D$ with discrete topology is constant. Then I thought if the quasi-components are the equivalent classes of the relation $x\sim y$ if $d(x)=d(y)$ for all $d$, then if $Y$ is a quasi-component then $d|_Y$ will be constant so $Y$ will be connected. But this is false since not all maps $d:Y\to D$ has a continuous extension to $X\to D$

[gentle osprey]

Whoever:

[gritty widget]

bredon

[sleek thicket]

let M be a compact, oriented, connected riemannian manifold with boundary. Suppose X is a vector field on M and that $\int_M (X f) d V_g = 0$ for all $f \in C^\infty(M)$. Is $X$ necessarily zero?

[gentle osprey]

shamrock:

[sleek thicket]

I don't think it should matter that M is riemannian, feels like this should be true for any volume form (or false in this special case lol)

lol nvm I'm dumb the problem I'm trying to solve is much simpler than this

[gritty widget]

what is this from if you don't mind me asking?

[sleek thicket]

trying to show that if two harmonic functions agree on the boundary (and the boundary is nonempty) they agree everywhere

Wlog u = 0 on the boundary is harmonic. I was trying to use green's identity
$\int_M p \nabla q dV_g = \int_{\partial M} p N q dV_{\hat{g}} - \int_M \langle \grad p, \grad q\rangle dV_g$

[gentle osprey]

shamrock:

[sleek thicket]

If you just take p = q = u you get that $\int_M |\grad u|^2 dV_g = 0$, so necessarily $\grad u = 0$, but I was being dumb and taking q = u but letting p be arbitrary

[gentle osprey]

shamrock:

[sleek thicket]

that gives you $\int_M (\nabla u)p dV_g = 0$ for any smooth function p : M -> R

[gentle osprey]

shamrock:

[gritty widget]

i see

this is just a bit beyond me, but it looks like an interesting problem, so i'll keep it in mind (currently learning RG)

thanks

[sleek thicket]

sure

the original thing or the one about harmonic functions?

[gritty widget]

harmonic stuff

[sleek thicket]

I am also currently learning riemannian geo

in week 2.5 of a course

these are exercises from ch2 of lee's introduction to riemannian manifolds

we aren't gonna talk about them in class but I wanted to learn the material

[meager python]

I’m a bit confused. Classical algebraic varieties seen as ringed spaces are not locally ringed spaces because the pullbacks do not send maximal ideals to maximal ideals, right?

We lack the generic points?

I’m confused that one calls say fg reduced k-algebras seen as schemes also affine varieties

[sleek thicket]

you can build a scheme out of an affine variety

and this "build out of" is a (full and faithful iirc) functor

[tough imp]

it is

it's called t

in Hartshorne

[uncut surge]

@sleek thicket Let $M = S^1$ and $X = \partial_\phi$, $\phi$ being the usual coordinate on the circle. Choose the volume form $d \phi$. Then $\int_{S^1} Xf d \phi = \int_0^{2 \pi} f'(\phi) d\phi = f(2 \pi) - f(0) = 0$ for all functions $f \in C^\infty(S^1)$

[gentle osprey]

Lartomato:

[uncut surge]

Does that make sense? I think that makes sense

[sleek thicket]

@meager python see proposition 2.6 in hartshorne

[gritty widget]

why is there an algebraic geometry in this chat smh

[tough imp]

I think in section 4 he shows the image is just blah blah reduced finite type separated k-schemes

Because this is where AG goes

unironically

[meager python]

Is there an alg geom chat?

[sleek thicket]

that sounds right lartomato, I need to take a second to think about it

no tolaria you're in the right place

nice username btw 🙂

[gritty widget]

well, either here, or abstract-algebra chat

[tough imp]

It's just here

[uncut surge]

naisu

[sleek thicket]

i mean ive asked questions about ag in algebra

and answered them

[tough imp]

Yeah but

I am of the opinion AG goes here

commutative algebra that is secretly AG goes in algebra

[sleek thicket]

anyways @meager python does that answer make sense? we have a way of embedding the category of k-varities into the category of k-schemes and you don't get any new morphisms by doing this or anything

[tough imp]

varities

varititties

[sleek thicket]

the construction is essentially "add in all the generic points"

[meager python]

Is variety in this case a lrs?

Or just a rs

Before adding generic points

[sleek thicket]

I think it should be a locally ringed space but I'm not totally sure

you have a sheaf of regular functions and the stalk at any point should be(?) the same as the stalk of the structure sheaf of the corresponding scheme at the corresponding (closed) point

that might be wrong though...

[meager python]

But normal mspec doesnt send maximal ideals to maximal no?

Unless we add the generic points

[tough imp]

You’re never gonna get that even when adding in generic points I think

[sleek thicket]

so if I understand right you're worried that a morphism of varities might not be local on stalks?

(also the result from hartshorne proves the claim I made about the regular functions/structure sheaf)

wait i screwed up the word varieties again didn't I

I should go to bed and stop thinking about harmonic functions

[meager python]

Well yes dont we need locality on stalks?

[sleek thicket]

we do

err hang on

there's two things here

[meager python]

Otherwise schemes are not op ring etc

[sleek thicket]

(1) are varieties local ringed spaces with the sheaf of regular functions
(2) are morphisms of varieties morphisms of locally ringed spaces or just ringed spaces

[meager python]

Well for 1 to make sense we need 2 also no?

[sleek thicket]

I thought you were asking about (2) but you really meant (1)

yup

so you're asking if $\mathcal{O}_{X, p}$ is local, where $X$ is a variety

[gentle osprey]

shamrock:

[meager python]

Well that it is, since its A(X)_p

[sleek thicket]

that's what I was going to say lol

just trying to find the reference in hartshorne to make sure since it's been a while since I did varieties

so what's the issue?

O_{X, p} is the stalk at p of the sheaf of regular functions

[meager python]

So are the morphisms local? Ie maximal ideals are sent to maximal?

[sleek thicket]

right, so that's (2)

so say you have a morphism $f : X \to Y$ of affine varieties and a point $P \in X$. You want to say $f_P : \mathcal{O}{Y, f(P)} \to \mathcal{O}{X, P}$ satisfies $f_P(m_{f(P)}) \subseteq m_P$

did I get all that right?

[meager python]

Yes

[gentle osprey]

shamrock:

[sleek thicket]

no but I fixed it lol

[meager python]

Something like that

Inverse of a maximal ideal should be mapped into a maximal ideal

[sleek thicket]

okay so $f$ is dual to $\varphi : A(Y) \to A(X)$, we want to check that $\varphi_P : A(Y)_{\varphi^{-1}(m)} \to A(X)_m$ is a map of local rings for any maximal ideal $m$ of $A(X)$

[gentle osprey]

shamrock:

[sleek thicket]

does that sound right?

[meager python]

Yeah

[sleek thicket]

well $\varphi(\varphi^{-1}(m)) \subseteq m$ in $A(Y)$, the same should be true in the local rings

[gentle osprey]

shamrock:

[meager python]

For every point m in X

[sleek thicket]

yup

so like, if q = f(p) we already have a relationship between the maximal ideal representing q and the maximal ideal representing p

right?

namely m_q = varphi^{-1}(m_p)

I think i got all the backwardness right

[meager python]

Image of maximal doesnt need to be maximal

[sleek thicket]

the preimage doesn't need to be either

[meager python]

Neither does the inverse

[sleek thicket]

yup

actually here's what makes this all work: if $A, B$ are finitely generated $k$-algebras and $\varphi : A \to B$ is a morphism of $k$-algebras, then $\varphi^{-1}(\mathfrak{m})$ is maximal whenever $\mathfrak{m}$ is maximal in $B$. The proof is that you have an injective $k$-algebra map $A/\varphi^{-1}(\mathfrak{m}) \to B/\mathfrak{m}$. Because $B/\mathfrak{m}$ is a field which is finitely generated over $k$, it's finite over $k$, and so finite over $A/\varphi^{-1}(\mathfrak{m})$. Thus you have an integral extension $A/\varphi^{-1}(\mathfrak{m}) \subseteq B/\mathfrak{m}$, and as $B/\mathfrak{m}$ is a field this forces $A/\varphi^{-1}(\mathfrak{m})$ to be a field, and thus $\varphi^{-1}(\mathfrak{m})$ to be maximal

@tough imp can you check this

[gentle osprey]

shamrock:

[sleek thicket]

so @meager python in general you don't get that the preimage of a maximal ideal is a maximal ideal, but varieties are nice enough that it happens

[tough imp]

This seems right to me

[meager python]

Ah ok

[sleek thicket]

huh

i didn't know this

[tough imp]

You just leverage finite extension for ID means one is a field IFF the other is

[meager python]

I’ll have to read through that

[sleek thicket]

the stuff I was saying earlier seems sus now

it uses some commutative algebra

@tough imp also a sneaky application of zariski's lemma

[meager python]

I basically ignored a lot of ag on classical varieties

[sleek thicket]

same lol

then schemes wrecked me

so I figure I should learn it eventually

like before I try for schemes again

[meager python]

Now I audit a course on classical old school ag with varieties and I’m trying to translate some stuff into scheme world

But try to see what goes ”wrong” etc

[tough imp]

Schemes didn’t quite wreck you, the exorbitant amount of time schemes take to learn wrecked you

[meager python]

I learned schemes first time in first year from liu, i passed the course but didnt really understand

[sleek thicket]

dang

That's pretty impressive

[meager python]

Well not really, I had lots of time on my hands lol

[sleek thicket]

haha fair enough

chmonkey and I took an AG course in our second year and it was brutal

[meager python]

But liu did sheafs really poorly

[sleek thicket]

I dropped out before cohomology because I didn't understand schemes well enough

[meager python]

So I was handwaving a lot

Like I didn’t really understand the basics well enough

Yeah seems like zariski’s lemma is key there

So the thing that confused me was that they dont mention they need to be morphisms of locally rs but as you write i guess that follows

[sleek thicket]

who?

[meager python]

Its actually quite instructive to go through the classical theory

[sleek thicket]

I've sworn of AG for the moment. I'm learning riemannian geo now and hopefully I'll hate that enough to push myself back to AG

[meager python]

Nice, are you second year?

[sleek thicket]

3rd year

[meager python]

Sounds like a very good idea to get some breadth

[feral dragon]

Geometers, tell me the difference between differential geometry, algebraic geometry, topology, differential topology and algebraic topology.

[sleek thicket]

can you give some context

Like

Do you know what any of those fields are, and if so what's your understanding of them?

[bleak helm]

I've sworn of AG for the moment. I'm learning riemannian geo now and hopefully I'll hate that enough to push myself back to AG
😂 😂 😂

[feral dragon]

Not much tbh. I'm kinda confused. I do know about topology and differential geometry (as in its relation with physics). But I don't know anything about AG, AT or Diff. Topology. I actually have been trying to study topology and differential geometry but I kinda see some similarities between other things of AG or AT but I don't know if I need them. So mostly I'm asking this to know what does each field deal with and if one is developed over another. So that I know where I should start from. (Most of what I am saying is purely related in the context of theoretical physics).

[violet sonnet]

im a bit stuck on how to begin these questions

im not even sure if i have correctly defined the sets in i)

i have that $f(D(x_0, r)) = {f(x) \in X | d(x, x_0) \leq r}$, and $D(f(x_0), r) = {x \in X | d(x, f(x_0)) \leq r}$

[gentle osprey]

jn3008:

[violet sonnet]

i tried to post in questions but got no reply

please tag me if you reply ! thanks in advance

[sweet wing]

@feral dragon
differential geometry -> basically calculus on manifolds
algebraic geometry -> linear algebra, but not really linear
topology -> basically set theory change my mind
differential topology -> differential geometry, but global (i.e. not looking at sufficiently small neighbourhoods)
algebraic topology -> either homotopy or (co)homology tbh but basically instead of having compactness hausdorff-ness simplyconnectedness you assign an algebraic object to sufficiently nice topological spaces

where should i start from
do point set first

[feral dragon]

Ahh.. thanks for that Ariana that kinda clears my doubt about those things. And Topology is just set theory, huh ? I have been learning stuff like compactness, connectedness, and well everything started from Set Theory but I thought Topology is something related to Geometry 😬

[sweet wing]

the answers are half serious but like

point set topology feels a lot like set theory

the boundary between geometry and topology is super fuzzy but i would say local stuff is more geometric and global stuff is more topological

compactness connectness etc. is pretty analysis-like

@violet sonnet suppose $x\in D\left(x_0,r\right)$, then what is the distance from $f(x)$ to $f\left(x_0\right)$

[gentle osprey]

ariana:

[coarse kestrel]

This depends on axiom of choice right?

For the definition of Phi

[versed geode]

Yes, I think

[coarse kestrel]

DC?

axiom of dependent choice?

looks like countable choice for some reason

hmm

[gentle osprey]

CityHunter:

[gritty widget]

I don't think you need DC for the second part

[coarse kestrel]

Why is this the generalization of subsequence

[fading vale]

is this from bredon

[coarse kestrel]

And not “a subnet of a net $\mu:D\to X$ is the composition $\mu\circ h$ where $h:D’\to D$ is a function with $x\geq y\implies h(x)\geq h(y)$”

Yeah

[gentle osprey]

Whoever:

[fading vale]

are you asking why nets generalize the concept of sequences?

[coarse kestrel]

No

[fading vale]

oh i see

[coarse kestrel]

Why is that the definition of subnet and not mine

[fading vale]

uh i think your condition is stronger right

but its not necessary for the subnet stuff to work out

oh wait no

@coarse kestrel uh i think what happens if you drop the condition and just use monotonicity is you lose some properties of sequences and subsequences that you want to generalize

like just keep going and i think you'll see an example

[coarse kestrel]

Ok

How do you lose properties by requiring a stronger statement

[honest narwhal]

Oh yeah this is Bredon's font 100%

[gritty widget]

How do you lose properties by requiring a stronger statement
mu won't be able to repeat its values

[coarse kestrel]

But the normal subsequences don’t repeat values either

[gritty widget]

thats y the generalization

[coarse kestrel]

That still doesn’t answer why it’s the generalization

But my question is already answered

Also I just realized Bredon said it too

It’s not that it will lose properties

It’s that even this weaker version will preserve all the properties of subsequences

So no need for the stronger one

[violet sonnet]

@sweet wing if x in D then d(f(x), f(x0))= d(x,x0) this is because f is distance preserving, then what do I do?

[sweet wing]

yup

so if x0 in B(x,r)

is d(x0) in B( d(x), r)?

[marsh forge]

why does bredon even go through nets

no one cares about nets

[coarse kestrel]

[hard wind]

I believe for the product topology the closure is R^omega, and in the box topology the closure is R^inf

Is that correct?

[honest narwhal]

His first chapter is a very general "All the point-set you need to know" book, so in particular it's got stuff that people in e.g. analysis might care about

Also it's how he proves Tychonoff in general

[marsh forge]

do people in analysis care about spaces that arent second countable

[honest narwhal]

Once you put in the weak topology anything can happen

[marsh forge]

who knew

[burnt spruce]

An important example is discussed here: https://mathoverflow.net/questions/157218/is-the-space-of-tempered-distribution-second-countable

[tidal forge]

To prove an atlas defines a smooth structure on a manifold, do i just prove that two arbitrary charts in the atlas are compatible to each other?

[small obsidian]

Yeah pretty much

[tidal forge]

ok thanks 🙂

also, what is the cartesian product of two charts?

is it a mapping from two inputs to two outputs?

[small obsidian]

That's an odd notion haha

[tidal forge]

yeah i'm trying to define a smooth structure on a product of two smooth manifolds

[small obsidian]

Since a chart comes with a set and a function

[tidal forge]

hmmm ok i guess that makes sense

[small obsidian]

Stupid source doesn't say, but it's probably smooth when each are smooth

[tidal forge]

lol

i think that's what im trying to prove

how do you show that an atlas is maximal?

[gritty widget]

it's sufficient to just have a smooth atlas to get a maximal one

you usually don't prove that a given atlas is maximal

since, given an atlas, you can just add to it of the charts on the manifold compatible with that atlas, which gives you a maximal atlas

[tidal forge]

ohhh interesting

[gritty widget]

it'd really suck if we had to check an atlas we come up with is maximal every time we want to construct a smooth manifold

[tidal forge]

yeah that's what i was thinking

lol

[river granite]

I think one uses Zorn to prove a maximal atlas exists in general?

[gritty widget]

https://math.stackexchange.com/questions/66554/is-zorns-lemma-required-to-prove-the-existence-of-a-maximal-atlas-on-a-manifold this says it isn't

but that also looks cumbersome, so...

[river granite]

oh, so it can be done but it's not necessary

zorn's lemma is op

[gritty widget]

all vector spaces have bases 😌

[coarse kestrel]

Oh you think all vector spaces have a basis? Write down a basis for the vector of space of real numbers over rationals then

[burnt spruce]

a basis for the vector of space of real numbers over rationals

[dim meadow]

Which vitali set?

[coarse kestrel]

[marsh forge]

thou shalt not have false vitali sets before me

[feral dragon]

So I've been spending hours in lectures for Topological manifolds and stuff related and there's a point when the Lecturer claims : Your whole life until now(possibly in physics) has been spent in charts This was kinda an interesting remark for me, the lectures were amazing though, very clearly explained bundles, fibre-bundles, charts, Atlas etc

[gritty widget]

it's physics don't expect it to make sense

probably something along the lines of "we are always using coordinates implicitly" idk

[feral dragon]

it's physics don't expect it to make sense
@gritty widget Yeah! Thats what he wanted to say. That everything that we've been doing in physics has been in one way or other related to charts.

[sleek thicket]

I think it's the latter

That was my understanding of it

[gritty widget]

"you undergrads"

[sleek thicket]

Implicitly you've been working in different coordinate charts

But you didn't have the language for it

eg "this transforms according to the following change of coordinates"

Is really expressing something about a global object

But you only see the local incarnations and how they relate to one another

also sorry to interrupt but I have a question

I was doing the following exercise: let $(M, g)$ be a riemannian 3 manifold. We have an isomorphism $\beta : TM \to \Lambda^2 T^* M$ given by $β(X) = X \lrcorner dV_g$. Define $\mathrm{curl} : \mathfrak{X}(M) \to \mathfrak{X}(M)$ by $\mathrm{curl}(X) = \beta^{-1}(d(X^\flat))$. Then $\mathrm{curl}(X) = (\ast d(X^\flat))^\sharp$

[gentle osprey]

shamrock:

[sleek thicket]

I said that for any 2-form $\omega = d(X^\flat)$ we have the equality $\omega = \beta((\ast \omega)^\sharp)$, and proved this by saying it's local and then choose an orthonormal frame and just bashing everything out

[gentle osprey]

shamrock:

[sleek thicket]

but this seems like a really shitty proof lol

Can anyone see a way to do this at a higher level?

For the previous problem about expressing div in terms of the hodge star I was able to just use the nice high level characterization of *

[tidal forge]

So composition of smooth functions is smooth, but the converse is not true right?

[sweet wing]

take any ugly bijective function and compose with its inverse

[tidal forge]

aaaa perfect

thank you thank you ❤️

[heady grove]

Anyone got any ideas

[graceful sequoia]

@heady grove For part a, you need to find $ \int_{0}^{4 \pi} || \gamma^{'} (t) || dt $

[gentle osprey]

Sup?:

[coarse kestrel]

I don't quite get this proof

here are the theorems

and a closed map is a map that maps closed set to closed set

So I think 8.3 holds if we can prove that {y} is closed in Y, which isn't always true

[gritty widget]

But should {y} have to be closed, and not compact?

Which it certainly is

[marsh forge]

What's the preimage of {y}

[coarse kestrel]

Xx{y}

[marsh forge]

is this compact?

[coarse kestrel]

It should be but idk

[marsh forge]

Is the product of compact spaces compact?

[coarse kestrel]

It should be but that’s the next theorem

[marsh forge]

Oh theres a more direct way

[coarse kestrel]

I guess this is homeomorphic to X which is compact

[marsh forge]

give me a space

yeah haha

[coarse kestrel]

xD

[marsh forge]

Okay

[coarse kestrel]

Ok

[marsh forge]

so Xx{y} is compact

the projection is closed

[coarse kestrel]

Alright

Yeah

[marsh forge]

what do we get?

[gritty widget]

Even, you don't need general product of compact spaces is compact

[coarse kestrel]

And now we get the projection is proper

[marsh forge]

yes

gj

[coarse kestrel]

Alright

[heady grove]

@graceful sequoia what about b

[coarse kestrel]

That problem should be #calculus ngl

[hard wind]

I did the first part, proving convergence in the product topology.

But frankly I'm not seeing how to show it doesnt converge in the box topology

[tight agate]

any product of open balls is open in the box topology

can you find a sequece of radii (r_i)i such that for all natural numbers N, there exists m > N such that x_m = (1/m, 1/m+1, ...) is not in the product of balls (i.e. for all m we have r_i < 1/m+i for some i )?

So basically pick a sequence that decreases a lot faster than 1/m

[hard wind]

oh, i am so stupid

thank you

[fallen canopy]

so i know that U-K is open (k is closed in a hausdorff space), and i know that this has a basis of pre-compact opens, but i cant make any progress - any advice? do i want to cover K with a cover of pre-compact opens?

[coarse kestrel]

I'm pretty sure they mean $\forall a\in A,f_\alpha(a)\to f(a)$ right?

[gentle osprey]

Whoever:

[summer jolt]

Hi guys how can I construct a homeomorphism from D2 x S1 (solid torus) to this subspace

[nimble jolt]

Map D2 to the circle in the y=0 plane, then rotate by the angular coord of S1 about z-axis.

[sleek thicket]

I s2g I am never going to remember all of these coordinate formulas and transformation rules for connections

there's just so many terrible formulas

What the fuck is a semicolon in a subscript, that should be illegal

[gritty widget]

[sleek thicket]

im actually so mad

Irate

[gritty widget]

it's some cursed tensor notation iirc

[sleek thicket]

It's somehow denoting the like, direction in which you take a derivative I think

That's my understanding

Or like

Those indices eat a vector field and you differentiate in the direction of that vector field

[gritty widget]

🤮

good thing im using do carmo

good luck

[sleek thicket]

That's exactly what I was looking at lmfao

Post I made in another server at the same time

Does do carmo do this better somehow?

[gritty widget]

nah do carmo doesn't go into this kinda stuff at all lol

do carmo only talks about connections in TM (no arbitrary vector bundle stuff)

how do i put it

the only connection to do carmo is the levi-civita one

his chapter on connections is also like 5 pages long

[sleek thicket]

oh lol

Yeah that wouldn't work for me

Lee is also teaching a course on vector bundles next quarter and it'll use connections on more general bundles

[summer jolt]

Map D2 to the circle in the y=0 plane, then rotate by the angular coord of S1 about z-axis.
@nimble jolt thanks that was helpful. I tried to think about an explicit map but failed. Any idea?

[coarse kestrel]

Why is X completely regular

[coarse kestrel]

@limpid vault I don't quite see how

[gusty yew]

Anyone know how to do 3.5.1, I don't really know how any of this works

[coarse kestrel]

@limpid vault what's precompact

ah

[fervent citrus]

thought that was relatively compact

[coarse kestrel]

https://en.wikipedia.org/wiki/Relatively_compact_subspace

Ok apparently they're the same?

I didn't find the word precompact on google for some reason

[versed geode]

tomato tomato 🤷‍♂️

[coarse kestrel]

But it did redirect me to relatively compact

[fervent citrus]

looks like "precompact" is a thing for uniform spaces

[coarse kestrel]

topology truly hurts me

[versed geode]

then why do you do it

[coarse kestrel]

all these terminologies

because it is fun

minus the terminologies part

[versed geode]

whoever is a masochist confirmed

Although I do see what you mean, after getting past and fully understanding all the tedious definitions, topology is really cool; it can become very intuitive

[fervent citrus]

ah there, found it
a uniform space X is precompact when for all entourage V of X, you can cover X with finitely many subsets of X that are V-small

tbh you don't really need to memorize all the defs, you just need to know where to find them when you need them

[versed geode]

What is entourage?

[fervent citrus]

an element of the uniform structure

[gentle osprey]

notwhale:

[dire rapids]

I think there is a way to do it

[gentle osprey]

notwhale:

[dire rapids]

Like maybe we should consider maps $z\to w$

[gentle osprey]

notwhale:

notwhale:

notwhale:

notwhale:

notwhale:

[dire rapids]

sorry if my messages are messed and unreadable

[sleek thicket]

What's topological charge?

I couldn't figure that out from your initial message

[gentle osprey]

notwhale:

[sleek thicket]

ah, gotcha

[dire rapids]

I would love any comments or just names of appropriate literature, where they consider this question (or give some route of it)

[sleek thicket]

key words would be degree theory

you can associate a homotopy invariant integer to any continuous map between two compact connected smooth manifolds of the same dimension called its degree, and for maps S^n -> S^n these classify when two maps are homotopic. There are also purely topological ways to do this for spheres specifically

Algebraic Topology by Hatcher does degree theory for the spheres, also other books probably

[summer jolt]

On a cantor space how can one form a basis of clopen sets?

[haughty anvil]

Would this be the right channel to ask about this?

[ivory dragon]

no, try #geometry-and-trigonometry or one of the questions-x rooms

[haughty anvil]

Okay.

[earnest epoch]

in regards to what was mentioned earlier. Pre-compact and relatively compact are not the same, but they are equivalent in complete metric spaces.

[gritty widget]

Sounds kinda dumb to me

"pre compact" totally just sounds like the set you have before you take a closure and it becomes compact

And it is also the definition that I am familiar with

[sleek thicket]

probably a terminology conflict where some people use it to mean something more specific than others

Like compact vs compact hausdorff vs quasicompact

[little hemlock]

why is $\ell^p$ not homeomorphic to $\ell^\infty$?

[gentle osprey]

kxrider:

[gritty widget]

one is separable and one is not

[little hemlock]

Ah okay

[gritty widget]

assuming 1 <= p < infinity at least

i'm unsure about what happens when p < 1 not even a norm lmao

i should know this

[gritty widget]

Terra did u recently add color to ur pfp? Or have I just been hallucinating that ur pfp was black and white

[untold rune]

yeah looks like the pride flag

[gritty widget]

i did add color

and it is indeed the pride flag

i want to change it so there's a bit more of the rainbow showing

[modern trail]

What if X and Y are not Hausdorff spaces?

i mean then sequence f(x_n) can have two limits

or theorem still holds (i guess it holds, since hausdorff space does not change defn of limit)

[river granite]

yeah, it holds regardless of what the spaces X and Y actually are

just read the proof

[modern trail]

ye

just like wondered if hausdorff space somehow changes it

but it does not

[river granite]

well, one direction doesn't use it at all and Hausdorff doesn't guarantee metrizable (not even first-countable, which is enough to prove the converse as Munkres notes), so yeah it's not very relevant here

[river granite]

I just found this site https://topology.jdabbs.com/

[cursive flume]

I have a question regarding GR diffgeo. some authors define minkowski spacetime as a four dim manifold and some as an affine space, however I can't see why are they equivalent

I found this, but don't understand how the metric structure turns the mfd into an affine space

[cursive flume]

it seems to boil down to the question of proving affine space=> manifold,but not the other way around

but idk how to prove that

[sweet wing]

it isnt rlly an affine space in the non-sr case isit?

[gritty widget]

If X is the projective space (seen as a prevariety) $P^1$ over an algebraically closed field k and we have the open cover by two copies of $A^1$, call them $X_1, X_2$ that glue together on $A^1 \ {0}$ by $x \mapsto 1/x$. I want to show directly that $\mathcal{O}_X(X) = k$. The definition in Gathmanns notes say that $\mathcal{O}X(U)$ is the set of functions $\varphi \colon U \rightarrow k$ which have the pullback by the maps $pi_i \colon X_i \rightarrow X$ in $\mathcal{O}{X_i}(X)$ on the patches. See 5.4 in https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2019/alggeom-2019-c5.pdf. I realize this is about showing that regular functions on each patch are polynomials $p(x)$ and that somehow we end up with $p(x) = p(1/x)$ implying that $p$ must be constant, this seems to hold over where you glue. But how do we show that this must hold globally $\mathcal{O}_X(X)$?

Somehow with this cumbersome notation I can't really show it, but I understand the proof in say Hartshorne

[gentle osprey]

Görtz:

[gritty widget]

Should I rephrase the question or add more explanations?

[snow gull]

need some help. I would say that conectedness in T' implies conectedness in T but the other way around is not necessarily true. Am I correct? Am i missing something?

[quiet pilot]

Seems correct

[snow gull]

But how do I prove my 2nd "claim"?. For the first one it makes sense that if every subset of X in T' is not open and closed at the same time (since X is connected by assumption) then none of the subsets of X in T will be closed and open at the sime time since every possible subset in T is also in T' because it is contained in it so it must be connected. The other way around i can't seem to justify it properly. Is my 1st justification even good though?

[quiet pilot]

I think you are correct, but you can even do it with definition

Simply show X not connected in T implies X not connected in T'

If X not connected in T we have X=A U B where A,B are in T, but then they are also in T' so X not connected in T'

For the other way around just find a counterexample

[snow gull]

Im sorry but I don't quite understand how showing X not connected in T => X not connected in T' is equivalent to my statement (X connected in T' => X connect in T).

[gritty widget]

contrapositive

[snow gull]

oh I see cuz the direction of the implication reverses

For the counterexample am I correct if I choose X=|R and T=usual topology and T'= left open topology.
In T |R is connected since u can't split it in 2 open subsets(A,B) such that AUB=|R since u would always be missing the point between the intervals, but in T' it can be disconnected if you chose A=]-infty,a] and B=]a,+infty] and |R=AUB. Am I correct or did I fail somewhere?

[quiet pilot]

Showing R is connected isn't that simple, A and B doesn't have to be intervals

But your example works if you just let T be the trivial topology instead

[snow gull]

trivial=usual?

so open subsets are of the form ]a,b[?

[quiet pilot]

Trivial topology is the topology where the only open sets are X and empty set

[snow gull]

oh I see, ty

[quiet pilot]

Are you using Munkres btw?

[snow gull]

yap

[quiet pilot]

Ah that's a nice book

You can usually find counterexample to these type of things by considering the coarsest topology i.e trivial, vs the finest i.e the discrete topology

[snow gull]

I see, should have thought about those, since they are simpler

[quiet pilot]

No need to get bogged down in technical proof of showing R is connected

[snow gull]

so just to make sure just need to change T to trivial? T'=left open works out and my example is not wrong?

[quiet pilot]

Yeah because every set is connected in the trivial topology

But R is not connected in left-open one as you noted

[snow gull]

Nice I was just worried about the intervals cuz in one we have infinity closed and in the other we have infinity open

is that not a problem?

[quiet pilot]

Oh wait

Infinity is not a point in R

Just let B =(a,infty)

It is open in left-open topology

All sets of form (a,b) are

[snow gull]

so the set [a,infinity] is open in left open? how?

[quiet pilot]

You get the left-open topology by letting the intervals of form (a,b) and (a,b] be basis elements.

[snow gull]

oh I see

[quiet pilot]

You can actually just let (a,b] be basis elements I believe

[snow gull]

the ( means open

that's why i confused it

[quiet pilot]

Because you can write intervals (a,b) as a union of intervals (a,c]

so the set [a,infinity] is open in left open? how?
@snow gull Should be (a,infty)

[snow gull]

how tho? wouldn't u always be missing b?

[quiet pilot]

b is not in (a,b)

[snow gull]

guess im retarded

[quiet pilot]

Nah takes a while to get a good grip on topology

[snow gull]

I guess my problem is with the definition of the left open topology, is it true that (a,b] and (a,b) are both open in left open

if yes then it all makes sense

[quiet pilot]

Yes they are both open

[snow gull]

thx a lot

[quiet pilot]

Np

As I mentioned above you can let the intervals of form (a,b] be the basis for left-open. It follows that the left-open is finer than regular topology on R by the comparison test in one of the early chapters

Because in every basis elt (a,c) of R, you can fit a basis elt (a,a/2+c/2] of left-open topology

[snow gull]

I see, makes sense, thx

[gritty widget]

linearly independent in the space of vector fields along gamma, right? since it doesn't really make sense to say you have 2n vector fields which are linearly independent at each point of an n-dimensional manifold

(do carmo riemannian geometry, page 112)

i'm used to reading "linearly independent vector fields" as "vector fields whose values are linearly independent in each tangent space" so i just wanna be sure i'm not going crazy

nvm i see it, worked some details and it makes sense

[modern trail]

won't f(0) in the second part just be an empty set

being open

ah no

i am dumb

nvm

[river granite]

Is Do Carmo's RG good? I liked his book on curves and surfaces and read most of it, but didn't use it as a main text (I followed some course notes instead).

[gritty widget]

@river granite i've read up to chapter 4 (on 5 rn) and i think it's good. it doesn't require a ton of theory to jump into, but you should be comfortable with notions from smooth manifold stuff since the intro chapter he gives doesn't really cover everything you need for the text (even though he says it does lol). explanations are usually pretty clear, and he gets right to the point with things

exercises are good, but a lot of them come with hints that are basically "fill in the details to the solution" hints or just full on solutions

also notation abuse is rampant

[river granite]

so it's similar to the curves and surfaces one in those last two aspects lol

[gritty widget]

the exercises cover a ton of material

[river granite]

I know some smooth manifolds so I guess I'll check it out

[gritty widget]

i know the later chapters go into fundamental group and covering space stuff, so it might be good to know that too

the chapters are short but he really does just get right to the point. there isn't a lot of time spent covering extra semi-related stuff (cough lee cough) - probably my favorite part of the book

[river granite]

the chapters are short but he really does just get right to the point
that's my favorite kind of book lmao

[gritty widget]

although sometimes he leaves out steps in a computation and you'll be staring at it going "what the fuck??"

at least i did a few times

ricci curvature ptsd

i guess that's my short review of the book

while i haven't read them yet, i've heard that the later chapters are really good

[loud scarab]

How do I get the x,y,z in terms of u's and v's so that I could calculate the Jacobian derivatives? When the equations are given to me this way I don't know how to get an expression for the partials of x and y with respect to u and v

is there a way to parametrize this?

[gritty widget]

If $X_1$ and $X_2$ are affine varieties then I have shown that the disjoint union $X_1 \sqcup X_2$ has coordinate ring/global sections isomorphic $A(X_1) \times A(X_2)$. How do I show that it is in fact an affine variety and not only a pre-variety?

[gentle osprey]

Görtz:

[gritty widget]

$A(X_1) = k[x_1, \ldots, x_n]/I, A(X_2) = k[y_1, \ldots, y_m]/J$ then $A(X_1 \sqcup X_2) = A(X_1) \times A(X_2)$ is also finitely generated but how do I conclude that $X_1 \sqcup X_2$ is isomorphic to an affine variety?

[gentle osprey]

Görtz:

[gritty widget]

I know that if $X, Y$ are affine varieties then they are isomorphic if and only if their coordinate rings are isomorphic. But if $X$ is a pre-variety (not necessarily affine) and $Y$ an affine variety then I can't conclude that if they have isomorphic coordinate rings then they are isomorphic, can I?

[gentle osprey]

Görtz:

[sweet wing]

@loud scarab you dont necessarily need x,y in terms of u,v to compute the jacobian

[loud scarab]

but how do i get the derivatives of x and y with respect to u and v as functions of u and v only? @sweet wing

[sweet wing]

If you have $\frac{\partial u}{\partial x}$, what does it tell you about $\frac{\partial x}{\partial u}$

[gentle osprey]

ariana:

[loud scarab]

I use the fact that the jacobian of the inverse function is the inverse matrix and try to go from there?

[sweet wing]

yesh

[pastel linden]

In this paragraph, Spivak is establishing that invariance of domain forces the neighborhoods of points needed to be homeomorphic to R^n to be open, but I don't really understand the necessity of this theorem to prove that

if we have a neighborhood of p homeomorphic to R^n, can't we just take an open subset U of R^n, and by the continuity of the homeomorphism we know that f^-1(U) is open in the neighborhood of p and so it's an open neighborhood?

e.g. the neighborhood contains an open set containing p, and so the definition refers to open neighborhoods

I'm probably just confusing definitions, he just sort of jumps directly into things

[pastel linden]

also if we're going by this stuff then there are a lot of manifolds that he just doesn't consider manifolds (e.g. SU(n))

[pastel linden]

i realize now that I'm confusing what he wrote

this is his definition of a manifold M

he has a bit where he essentially just wrote the proof I wrote determining that it doesn't matter whether we use neighborhoods or open neighborhoods since every neighborhood contains an open neighborhood with the same property

but then he says that the neighborhood "must" be open, which is the confusing part

could you explain why invariance of domain prohibits this? if openness is a topological property then I don't see the issue, since f(U) would just be closed which is true

oh wait I'm dumb, it wouldn't be both closed and open and so there is no such homeomorphism

never mind still confused

the SU(n) example was more a reference that invariance of domain only applies to subsets of R^n, but I guess if you define a metric such that the set is isomorphic to some subset of R^n we can still consider it a manifold

oh, I see

and then invariance of domain just makes it so those neighborhoods happen to coincide exclusively with open neighborhhods

[nimble flower]

Is the image of a disconnected set under a continuous function, disconnected?

[gritty widget]

zero function on [0,1] cup [2,3]

any constant function will always have a connected image

[nimble flower]

darn

wanted to use that "fact" in a proof

[gritty widget]

homeomorphism though

[pastel linden]

you do have that if the function is not continuous then the image of a connected set is not connected

[gritty widget]

squishing disjoint blobs together gives you one blob

[nimble flower]

sadly i dont have a homeomorphism 😦
Just a continuous bijection

[gritty widget]

is it from a compact space to a hausdorff space 👀

[pastel linden]

continuous bijection is homeomorphism in hausdorff I thought

[nimble flower]

it's from a discrete, closed subset of R^n to R^n

i'm trying to show there cant be an uncountable subset. so i suppose it exists, then i get a bijection with R^n. since it has the discrete topology i get a continuous bijection

[pastel linden]

i have spent like 30 minutes staring at this and I have no fucking clue where this argument comes from

where does he get the existence of the one-to-one continuous map from

[gritty widget]

(x_1,...,x_m) -> (x_1, ..., x_m, 0, ..., 0)

[pastel linden]

is that not an open subset of R^n

wait I'm being dumb

definitely not open

[gritty widget]

prove it 😌

[pastel linden]

take an open ball around literally any point?

although I don't see how this negates the possibility of them being homeomorphic unless I'm missing some really obvious fact

[gritty widget]

invariance of domain

i think cant use that unless n = m lmao

:(

why am i not seeing this lol

[pastel linden]

I know, this seems very non-trivial despite spivak pretending it is

[gritty widget]

maybe if i drink this coffee the idea will come to me

i will report back

[nimble flower]

if A is a countable subset in R^n with the induced topology, that induced topology should be the discrete topology, correct?

[gritty widget]

no, take Q in R, for example

singletons are not open

[nimble flower]

ah true

fk topology

not as bad as analysis - but things that i want to be true arent true lol

[pastel linden]

but knowing that one specific map is not open doesn't imply that there can't exist any homeomorphism

I assume there's something with invariance of domain but I don't see how to apply it here

[gentle osprey]

slimvesus:

[pastel linden]

oh I see

so R^n is homeomorphic to the R^n subspace of R^m. Assume R^n is homeomorphic to R^m. Then if the R^n subspace is homeomorphic to R^n, it's also homeomorphic to R^m by composition. Then since the R^n subspace has the same ambient space as R^m, we can use invariance of domain to show that there is no way they can be homeomorphic which leads to a contradiction

[gritty widget]

just spivak being spivak

[pastel linden]

no you helped a lot, let me tex something proper up and then I'll post it if it's nonsense

[gritty widget]

post it anyways lol

im curious

[pastel linden]

thank you spivak

[gritty widget]

just... define a manifold with dimension in the definition?? you lose no generality

ok, maybe you could get some stupid shit like a line disjoint union sphere

but like

who cares lol

[pastel linden]

I think tu's intro just says "we hold back on proving that topological dimension is well defined until chapter 7" when manifolds is chapter 2

[gritty widget]

i think there's a definition out there that allows that lol

[pastel linden]

https://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/

tao goes over why the proof is funky in this post a bit

[gritty widget]

tu allows stupid stuff like sphere disjoint union a line to be a topological manifold

[pastel linden]

well so does spivak with metric space nonsense

[gritty widget]

it literally never comes up in his book (intro to manifolds) though

[pastel linden]

for some reason he doesn't consider non-metrizable spaces and whatnot

[gritty widget]

[pastel linden]

which I guess makes sense from an intro pedagogy thing

[gritty widget]

all manifolds are metrizable so who cares 😌