point-set-topology

[shut moat]

what makes noether's theorem interesting to a mathematician btw

i suppose it's a different aspect of it from what physicists like about it

[gritty widget]

it's interesting because i actually have a clue what a momentum map does now

[shut moat]

nice

[gritty widget]

thinking of a symplectic manifold as a ""phase space"" and a symplectic group action

momentum maps find quantities conserved by the action

[quasi forum]

I saw no disagreement, so I consider that an "you got it." 😁

[gritty widget]

the term momentum map comes from the example of SO(3) acting on R^3 x R^3 (coordinates position, momentum)

the conserved quantity being angular momentum apparently

[shut moat]

oh nice you care about it for the same reasons then

[gritty widget]

and the momentum map comes out to be precisely that

(after making a million identifications)

another example being that of $S^1$ on $\bC$ by rotation. the momentum map comes out to be $\mu(z) = \pm |z|^2 + \mathrm{const.},$ which is definitely something that's preserved under the action

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

(plus-minus because the sign varies in the literature)

[shut moat]

why C instead of R^2?

[gritty widget]

writing things in complex coordinates is convenient

[shut moat]

fair

[gritty widget]

that example directly generalizes to the torus (S^1)^n acting on C^n

and that's interesting because then with the right choice of constant (any one works here), the zero level set of the momentum map is just the sphere

oh it might have to be |z|^2 / 2

[gritty widget]

and then if you quotient out by the symmetries of the group action ("symplectic reduction") you get complex projective space

[shut moat]

spooky

[gritty widget]

this is probably incorrectly worded so take it with a grain of salt

ah symplectic reduction is when you quotient out the zero level set by the restricted action

and that's about all i have to say

[shut moat]

ty :o

[gritty widget]

thank you for coming to my #events talk

now max can stop bugging me

[shut moat]

I've been googling the physical interpretations of this stuff and now my brain is melting out my ears

[gritty widget]

mathematically it's all somewhat straightforward

but physically?

something something symmetries and particles idk im not a physicist

[shut moat]

idk I saw a bunch of word spam like jacobi fields and homological

[gritty widget]

jacobi fields

[shut moat]

will understand in a few years I hope lmao

[gritty widget]

jacobi fields are a riemannian geometry thing

i wonder how they intersect into symplectic

[gritty widget]

arnold's classical mechanics book has a chapter on symplectic geometry that might have a good explanation or two

[shut moat]

oh oops jacobi and fields showed up in two disjoint places idk why i fused them

[shut moat]

yeah I've been meaning to give it a read, it seems rly good

[gritty widget]

the book by mcduff and salamon might also have some stuff

it's more on the mathematical side of things but it has a section or two devoted to building up to symplectic geometry from only classical mechanics

i havent actually taken a careful look at arnold's book but i really hope it mixes in physical explanations with mathematical ones

i wanna know what's actually happening not just how to move symbols around

[shut moat]

I haven't read much of it but I think it's not as rigorous as it could be at times

but ig it's a semi physics book so that's ok

[gritty widget]

rigor is for the weak

[shut moat]

like I was reading out of goldstein (a classmech book) and the derivation of euler lagrange felt fucked

[gritty widget]

just differentiate under the integral

[shut moat]

so I went and looked at arnold because it was supposedly a math book but it didn't spell out the analysis in detail

v sad

[shut moat]

yeah that's how it ends up working out but I wanted to understand what it actually was

and it ended up being a frechet derivative or something I think?

[gritty widget]

i took an optimization class that did this

i actually don't remember

i differentiated under the integral a lot in that course

[shut moat]

based

[gritty widget]

so it probably comes up

the joy when i learned in riemannian geometry that geodesics are just solutions to a variational problem

geodesic equations are just euler lagrange equations for the energy functional

i wonder if the jacobi equation J'' + R(J, c')c' = 0 can be interpreted as the euler lagrange equation(s) to something

anyways im just vomiting words at this point

[shut moat]

nice talking to ya, i should probably go back to preparing for the pile of midterms I have this week

[gritty widget]

good luck

petthecat

[shut moat]

ty, I'll need it

[bitter kestrel]

Hey sorry to like butt in but i just sort of had this thought that differential k-forms are like the dual vectors of singular k-chains. Does this actually work? Perhaps with the assumption that everything is nice and smooth? Could we then define the exterior derivative as like the dual operator of the boundary operator? Are there any nice applications of this, like direct arguments for Stoke's theorem or Poincaré duality or anything like that? I'd love to read some articles on this perspective if any exist, and also sorry if this is trivial or something its not really my main area of research 😅

[gritty widget]

check out chapters 17 and 18 of lee's intro to smooth manifolds

[gritty widget]

I want to show that two divisors are linearly equivalent. So I have to show that their difference is a principal divisor.

so I need to construct a meromorphic function with given roots and a given poles

I'm really not how to do this. If we were on C this would be easy, but on an a Riemann surface we won't have global coordinates

Here is the full question from Mirandas Algebraic Curves

[wanton marsh]

where does it say "riemann surface" in the question ?

[gritty widget]

good point

[obtuse meteor]

presumably you can do things like you would for manifolds

and define the meromorphic functions locally

and check it agrees on the chart

[gritty widget]

okay i think this might work

3p0 is is the intersectin divisor with z=0
the other one is the intersectin divisor for y=0

and two intersection divisors are linearly equivalent

[proud pewter]

Hello, i hope i don't intrerupt anyone. I come in a moment of despair because i know nothing about geometry, and i promised i would help a friend of mine with some collage-grade geometry. She is supposed to solve two exercises, one of them being trough carthesian and afinoid repers
i have some indications, hoping that would help
but atm i don't have any drafts of my friend's work so far.

[ivory dragon]

could you be a bit more specific?

im not sure what you mean by "carthesian and afinoid repers"

[proud pewter]

hmm

i don't either, hope i am not that problematic, but i am parallel and equidistant with mathematics. I just want to help xD i study modern applied languages so. Let me try to find a way to explain

i have some guides but i will try to translate them myself from romanian

[proud pewter]

Cartesian landmark and affinoid landmark i think this is the correct form

oh oops

Translation: Show that the ratio of points A B C and D is -1. (Circle of Apollonius)

i am willing to help you with whatever i am able to in order to make it easier for you, i am not looking for someone to solve these exercises, i am looking more to explain what my friend has to do in order to solve and understand them

[ivory dragon]

apologies, i'm unfamiliar with techniques of euclidean geometry; maybe someone else will be able to help

[proud pewter]

oh, it's alright

thanks for letting me know

hope tese notes can help for whoever stumbles upon this request

[sand void]

anyone got any ideas on how to approach the problem?

[shut moat]

what do you know about complete subspaces of R^2?

[gritty widget]

A good hint is to recall the definition of complete

and also consider (-1,1) as a subset of R and check if that is complete

or check that is is incomplete if you want a better hint

[gritty widget]

Does anyone have an idea for this?

I have a curve C and a line L in CP^2. Take a point p not on the line, we can define the projection of CP^2-p to the line which sends a point q, to the intersection of L and PQ (where PQ is the line joining p and q)

this map is holomorphic, and restricts to a holomorphic map C-p to L

how can we extend this map to be holomorphic on C ?

[sinful pecan]

is the answer just the intersection of L with the tangent line of C at p?

[quasi forum]

Is the subset D closed?

That may be the best hint I can give ya without just giving away the answer.

[gritty widget]

maybe but I don't see this? im not even sure how to properly extend the function?

ooh

this is just the only clever line to pick

[sinful pecan]

i just thought about q approaching p on C, which has the projection approaching the tangent line's intersection with L

[gritty widget]

okay well that helps alot

now i just need to check that this is infact holomorphic

[quasi forum]

The fact you two have the same profile picture makes tracking this conversation super confusing

[sinful pecan]

compact mode gang

[gritty widget]

what is nitro?

i saw someone offering to buy someone else nitro on a different server

[sinful pecan]

i think it lets you use emotes from particular servers anywhere. imagine paying for discord tho haha

[gritty widget]

imagine not using the default little green dude

[flint cove]

Is cech cohomology $\v{H}(X;A)$ a functor?
i.e., is there a well-defined pullback?

[gentle osprey]

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

[flint cove]

I'm having a hard time coming up with something natural, since covers can only be pulled back, hence functions from (thing derived from cover) to A can only be pushed forward

Wiki casually states that Cech is “naturally isomorphic” to De Rham (CW homotopy type), but does not in any place discuss how that is a functor.

[gritty widget]

likely a very small brain answer

but i would guess yes

if its naturally isomrphic to singular homology or deRham

a natural isomorphism is a map between functors

so I

i'm not sure how you could define a natural transformation without having functors

soz if this wasn't helpful

[flint cove]

eh, what can you do

… ¯_(ツ)_/¯

[tough imp]

Homology forms a functor

In particular, they’re delta functors, so I think they’re isomorphic as delta functors is what it means

[tough imp]

Pretty sure the functor is covariant

You form Čech cohomology by making a complex and taking cohomology right?

If you had a sheaf map F -> G you can just apply this on each of the things in the complex (taking massive products)

And then this gives you a map on cohomology

[flint cove]

On a fixed cover UU, yes, then direct limit

[tough imp]

Are you defining Čech cohomology as the direct limit?

Even if you are, you can do it on each cover element and form a map between direct systems

Then you get a map on the direct limits

[flint cove]

Yes.
The context of what I know is purely topological and does not use the word „sheaf“. Although I know the basics of that

[tough imp]

Uhhh

Well it probably works the same

[flint cove]

on a space X, we're taking the sheaf of continuous functions Γ(-,G)

so a sheaf map would be what in that case?

(let G=ℤ as an example)

let's fix f: X→Y first

[tough imp]

👀

You wanted a functor on X?

Oof

[flint cove]

Just be aware that my advisor already told me „you don't wanna do this“

[tough imp]

Yeah idk

[flint cove]

I just didn't want to shut you down 😄

[tough imp]

Sorry, I only know how to do Čech cohomology on sheaves

In this case it’s more like

You fix a soace X

And imagine you vary what functions you’re looking at

[flint cove]

And consider different sheaves over X, right?

[tough imp]

Which really means varying the sheaf

Yeah

[flint cove]

I see, that makes sense

[tough imp]

What you’re trying to do sounds like something scarier

It reminds me of the Leray Spectral Sequence type stuff

This relates Sheaf cohomology on a topological space with its pushforward

[flint cove]

fun fact
what I'm trying to prove eventually is injectivity of the pullback P(E)→X for a bundle E→X
multiple authors just refer to the Leray SS (or the Serre SS)

[tough imp]

Lol

[flint cove]

I'm currently seeing whether there are simpler arguments

[tough imp]

Well, I guess I’m on the right track

In my experience thinfs which are best done with a Spectral Sequence tend to be best done with Spectral Sequences

[flint cove]

Just out of curiosity, what is your intuitive image of the Leray SS that made your conceptual pattern matching trigger here?

[tough imp]

Which is a tautology, but really I think that when they’re used

it’s for a reason

Basically the leray spectral sequence relates cohomology of a sheaf on X with its push forward.

So like let’s use the continuous function one

[flint cove]

pushforward along what

[tough imp]

So if you had topological spaces X and Y

and a map f:X -> Y

You could define a sheaf on Y by saying that f_*F(U) = F^-1(U)

So like

In the continuous functions on X case

What you do is map to U, the continuous functions from f^-1(U) -> G

From what I saw you were doing earlier

Then you can look at cohomology of this (which is related to the Čech cohomology)

[flint cove]

This is that „sections functor is not exact, take derived functors“-stuff, isn't it?

[tough imp]

Right

So you end up getting a spectral sequence relating H^i(X,F) and H^j(Y,f_*F)

If you’ve seen the Grothendieck spectral sequence

You can construct it this way

You’re looking at the f_* and Gamma(X,-) functors (I guess really what you’re doing is observing that Gamma(Y,-)•f_* = Gamma(X,-) by construction)

Oh lol

I’m looking at the Wikipedia and the classical construction was for Čech stuff

[flint cove]

(btw, my long-term goal is to understand the core statement in the special case of the projective bundle. I def want to understand how SS work and what the intuition behind some of them is, but I will refrain from using that as far as possible to remain accessible to someone knowing only basic cohomology)

[tough imp]

¯_(ツ)_/¯

Idk geometry

I just know diagram and functor

[flint cove]

well then you know geometry

[drowsy trout]

Hey, could someone give me a few pointers about algebraic group $PGL_2$?
I'm working with Milne's etale cohomology.
Following Milne, we let $PGL_n$ be the representable functor s.t $PGL_n(S) = Aut(M_n(\mathcal{O}_S))$ where we consider automorphisms of $\mathcal{O}_S$-algebra.
Now, in the case where $S$ is the spectrum of a local ring $R$, Skolem-Noether theorem works and tells us that $PGL_n(S) = GL_n(R)/R^*$.
On general scheme, this only works locally, and so we only get an exact sequence:
$$1 \rightarrow \mathbb{G}_m \rightarrow GL_n \rightarrow PGL_n \rightarrow 1$$
According to Milne, this sequence is exact as a sequence of sheaves for the Zariski, Etale and Flat topology.

[gentle osprey]

Montessiel

[drowsy trout]

But I'm interested in a very specific case: I want to understand PGL_2 (and its cohomology) on quadratic orders.
So I'm trying to figure out what's happening there.

If I look at the long exact sequence, I get:
$GL_2(R) \rightarrow PGL_2(R) \rightarrow H^1(R,\mathbb{G}_m)$
with the last group (it's a group since the multiplicative group is abelian) being the Picard group

[gentle osprey]

Montessiel

[drowsy trout]

If I'm not mistaken, for quadratic orders (or at least for those that are integrally closed), this H1 is non other that the ideal class group, which may very well be non trivial

So $PGL_2(R)$ will not simply be a quotient $GL_2(R)/R^*$.

[gentle osprey]

Montessiel

[drowsy trout]

But is is Zariski-locally, so it shouldn't be that bad. Is there a way to describe the elements of $PGL_2(R)$ ?
The obstruction to surjectivity is a finite group so I'm also hoping that is might make my life easier, but so far I failed to find information on this in the litterature

[gentle osprey]

Montessiel

[split needle]

hey, what is the relative interior of a set and why do we need this? This comes up during convex analysis

[split needle]

alright I found a nice answer if anyone else is curious: https://math.stackexchange.com/a/2774285/243316

[tight agate]

Hatcher might have what you're looking for

section 4D does a version of the leray-hirsch theorem

and if you're doing this to obtain the splitting principle, then I think you can obtain a weaker result (than the usual splitting principle) by meming around with lambda rings

[flint cove]

Riight, Thanks for the reminder. I also found something else where I feel the injectivity is proven directly for compact spaces by induction over cover size.
I need to take some time and work through all variants here, haven't managed to do that yet.

Btw, does someone have a quick example of a space you see „in the wild“ that is not homotopy equivalent to something compact?

[tight agate]

just compact?

[flint cove]

Yes, all my mental examples detract to, say, a circle or a point or something like that

[sleek thicket]

My instincts say maybe the long line?

[flint cove]

I really need to expand my landscape of examples there :>

[tight agate]

you can very easily get weirder examples than this

like the p-adics

but that's still compact

[flint cove]

Was that R^2 with the lexicographic order topology?

[sleek thicket]

No, it's a weird construction involving the first uncountable ordinal

Basically you glue uncountably many intervals to eachother along their ends

[flint cove]

Let me look it up quickly. I should know this.

[sleek thicket]

I think the reason I'm thinking of it is that the homotopy groups all vanish but it's not homotopy equivalent to a CW complex

[tight agate]

oh wait does Q work?

[sleek thicket]

It's sort of too long for things about it to be detected by homotopy

oof

Oh maybe that's not so bad

Homotopies are constant

right? (edit: no, wrong)

[tight agate]

also just an infinite disjoint union of points should work

[sleek thicket]

Oh lol that is much simpler

[flint cove]

||that's what she said|| sorry, needed to get that out of my system

[tight agate]

lol

Z_p is compact, but does not have the homotopy type of a CW-complex

[flint cove]

Hmm, but there the connected components are compact, so all cohomology results for compact spaces should easily be transferable, right?

[sleek thicket]

Good for her

[tight agate]

sure ig

[flint cove]

Uh that sounds good

[tight agate]

Z_p is horrible as a top space

[sleek thicket]

You're horrible as a top space

[flint cove]

I'm trying to get a feel for when it can go wrong to show something for compact spaces, mumble something about „paracompactness“ and „similar argument“ and just continue, lol

[sleek thicket]

Probably depends on the argument ¯\_(ツ)_/¯

Although reducing from compact to paracompact seems weird to me

[flint cove]

Seems to me like many arguments for bundles work like this (although I probably forgot Hausdorffness)

[sleek thicket]

in my head the arguments for bundles don't really go through compactness, it's more about using paracompactness to reduce to the situation where you're working on a trivial bundle

Like usually (from my perspective) you don't even change/go local on the base, just the bundle

[flint cove]

Hm, I should retract that statement I guess.
But I feel like for some things, it's easier to convince me that / how an argument works by looking at compact spaces first, since often that case needs less technical fiddling

[sleek thicket]

Yeah, definitely

[flint cove]

Homotopy invariance of the pullback was one example

[quasi forum]

Why would a smaller collection be coarser than a larger one? I feel like it'd be the opposite 🤔

[ivory dragon]

think of the elements (open sets) of the topology

rather than the topology itself

[quasi forum]

So T_1 contains less open sets than T_2

[shut moat]

munkres had a nice analogy with pebbles or something

the open sets are the pebbles

if you crush the pebbles you create a finer topology

[quasi forum]

That analogy makes sense to me

[chrome dew]

yeah, crush up the open sets of T_1 to make smaller ones

[quasi forum]

So at it's most basic, we can compare the discrete and indiscrete topologies on X.
So since the indiscrete topology is just X and the empty set, we can 'crush' X to become every open set, which forms the discrete topology.

Wait, I think I went the opposite direction

There we go.

Is that line of thinking valid?

[chrome dew]

ok quick quiz, what's the finest and coarest topology you can put on a set?

[quasi forum]

The ones I just mentioned.
The coarsest would be indiscrete ${\emptyset,X}$, and the finest would be discrete: all open subsets of X.

[chrome dew]

sorry I just went to get something to drink and didn't read what you wrote at all lmao

funny coincidence

[gentle osprey]

dackid

[quasi forum]

Well, I think that means I was right :p

[chrome dew]

yep

[quasi forum]

So the way I am interpreting it is kind of like a microscope. At the end of the day, we are looking in the same space, but the greater the magnification, the more detail we get out of it, and so the finer the topology.

[viral atlas]

Yes!

[quasi forum]

Sweet. Tbh, going from metric spaces to topological spaces was not nearly as big a jump as I thought.

So far, this is pretty understandable

Thank you for the help. I'm definitely going to pick at this more tomorrow.
For instance, a question in my book wants me to show the intersection of the topologies T_1 and T_2 are also a topology, but the union may not be.

That second part is messing with my head. But, the more I think about it, the more I have an idea as to why. I think I just need sleep.

[viral atlas]

I think ||the union of open sets from T_1 and T_2 can possibly not be contained in either of them.||

[quasi forum]

I agree. I think the thing that would really break it is that the intersections of some of the open sets may not be in T_1 and T_2 anymore

[viral atlas]

No, I think all finite intersections would be contained, but unions of open sets would break down

Hmmm, or possibly yes

Maybe working with an explicit example would help here

[gritty widget]

[quasi forum]

For instance. If $T_1={\emptyset, \mathbb{R}, (3,5)}$, and $T_2={\emptyset,\mathbb{R}, (4,8)}$, so then by taking the intersection of (3,5) and (4,8), we get (4,5), which is not in $T_1\cup T_2$

[viral atlas]

Yepp, makes sense.

Same goes for union. (3,8) is neither in T_1 nor in T_2

[gentle osprey]

dackid

[quasi forum]

Precisely! :D

And then the intersection part of the problem should be clear as all the empty sets satisfy the 3 axioms for both T_1 and T_2.

[viral atlas]

Yep!

In fact

It works even for non-empty intersections

[quasi forum]

Absolutely!

[viral atlas]

This argument seems to be in the same spirit as proving intersection of vector subspaces or subgroups is again a vector subspace or subgroup

[quasi forum]

Part of the question also asks if it is true for an arbitrary intersecrion of topologies, and I think with this reasoning, the answer should be yes.

[viral atlas]

Both topologies at least agree on the set itself(X) and the empty set

[quasi forum]

Yes, that is definitely key. That'd be a problem if it didn't :p

[viral atlas]

Hmmmmmm, arbitrary intersections are beyond my understanding haha

[quasi forum]

I was gonna go to bed, but.... topology 😆

[viral atlas]

Are you using Munkres for topology?

Hahaha.

[quasi forum]

Is that a book?

[viral atlas]

Yes haha

[quasi forum]

Wilson A. Sutherland

[viral atlas]

Topology by Munkres is kinda the canon for learning point-set topology

Ohhh, I see

[shut moat]

they're just like finite intersections- $\bigcap_{\alpha \in A} X_{\alpha}$ is the set of points $\qty{x|\forall {\alpha \in A}; x\in X{\alpha}}$

[gentle osprey]

~S^1

[gritty widget]

use lee itm

[viral atlas]

It doesn't have point-set

[quasi forum]

In that case, it is definitely true!

[viral atlas]

Yeah, but it gets a bit weird with nested interval and things like that in R, right?

[gritty widget]

lee has all the point set anyone who isn't an analyst cares about

[viral atlas]

I'm an analyst

Don't you see my pfp and username

[gritty widget]

[viral atlas]

Anyway I'll look into ITM after I'm done with Pugh

Maybe I need some more mathematical maturity to parse the things he says

Like at the very beginning he talks about dimension of spheres and balls

And that didn't really get through me

[sleek thicket]

[viral atlas]

This probably illustrates that if we pick a topology, and a possibly finer topology, then chances are that intersection of open sets in the two topologies would not be an open set in either.

[quasi forum]

Would this be a reasonable way to go about this argument?

[viral atlas]

Uhhh

You should probably show how it agrees with the axioms haha

Like, write it out in a bit more detail

[gritty widget]

wat

[quasi forum]

But if $U\in T_1$ and $U\in T_2$, the fact it follows the axioms should be automatic

[gentle osprey]

dackid

[gritty widget]

you need to show that T_1 cap T_2 is a topology

[viral atlas]

Automatic how

[gritty widget]

what does that entail?

checking that

1. it contains the empty set, and the whole of X
2. it is closed under unions
3. it is closed under finite intersections

T_1 cap T_2 is a collection of subsets of X, equal to those subsets of X which are in both T_1 and T_2

[quasi forum]

Yea, I agree with that

[gritty widget]

so "the topological space axioms" don't apply to this particular set U you've chosen from T_1 cap T_2

[gritty widget]

you need to prove these three things about T_1 cap T_2

[quasi forum]

But that U is in T_1 and T_2 🤔

[gritty widget]

but how does this show any of 1 to 3?

[tough imp]

The axioms are on the set of opens

Not a particular U in T_1 or T_2

[quasi forum]

Better?

[tough imp]

Yeet

For verifying a topology is closed under finite intersections it suffices to look at just pairwise intersections btw

This might be helpful if you want to prove a particular thing is a topology

This follows by induction and associativity of the intersection

[quasi forum]

So by just looking at two sets instead?

[tough imp]

Yeah

If you don’t see why, I’d work it out

[quasi forum]

Oh okay. Does Yeet mean it looks good?

[tough imp]

Yeah

[quasi forum]

Sweet!

[viral atlas]

Isn't "yeet" some slang for throwing something away?

[tough imp]

Yeah

But also

Yeet

Idk

[viral atlas]

Lmao

[tough imp]

[quasi forum]

That's why I was confused 😆

[viral atlas]

[tough imp]

Tbh

I meant to say Yee

But said yeet instead

[viral atlas]

I figured that XD

[quasi forum]

Thanks for calling out my shit now, don't want to make that mistake later 😆

I definitely see where I went wrong. It's just taking me a minute to start thinking about collections of open sets instead of the sets themselves.

[viral atlas]

[fading vale]

uhh ok so consider like

[gentle osprey]

Moth

Moth

Moth

Moth

[fading vale]

what i need to do is show that this is a homeomorphism but im not really even sure where to begin tbh

like even injectivity and surjectivity idk how to do

actually wait

injectivity isnt hard

ok yea injectivity and surjectivity are fine

idk how to do continuity/openness

compact open topology hard

[feral copper]

Hi ! I don't understand Saveliev's proof of this statement... (from his book, "Lectures on the topology of 3-manifolds")... It seems to me that he is glossing over a lot of details, as can be seen from Hatcher's notes on 3-manifolds : https://pi.math.cornell.edu/~hatcher/3M/3M.pdf. What am I getting wrong ?

(also he is working in Top, and Hatcher in Diff, which may make things a lot different, but idk, Saveliev's proof is still a bit shady to me)

[cold vine]

I haven't used LES for a while; how do I form a LES from which I can split a sequence 0 -> H(X V Y) -> H(X xY) -> H(X smash Y) -> 0

assuming that X V Y is a strong deformation retract of it's nbd

[quasi forum]

Okay, so this was my first stab at a real topology problem. Do you think I handled it okay? If not, where did I go wrong?

[shut moat]

I think you might've flipped it? You want to show that it's closed under arbitrary unions, but finite intersections

oh wait nvm misread ignore me lmao

[quasi forum]

I took care of both possibilities for the union 😁

[quasi forum]

True true!

Other than that?

[marsh forge]

what part do you dislike

[feral copper]

Hi ! Starting from "there exists a homeomorphism carrying D1 to the upper hemisphere"

I mean, what is the homeo defined on ? What is the codomain ? And why does it exist ?

[obtuse meteor]

god these algtop problems are spicy

hmm

D1 not D2?

oh D_1 not D^1

I can read

[feral copper]

This is more like geometric topology actually, which I like a lot less than AT ! But yeah, it's spicy 🙂

[obtuse meteor]

I was talking about my midterm this week and the review actually :P

[feral copper]

Oh I'm sorry ! :x

[obtuse meteor]

If anyone wants to share in my suffering

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

[gritty widget]

based

[sleek thicket]

the converse of practice problem 1. is true for finite covering maps, which is neat

[shut moat]

cool diagrams tho

[obtuse meteor]

I mean Sham part of that is the content of 2!

[sleek thicket]

oh lmfao

nice

i did not read further than 1

because my friend was texting me

4(b) literally always catches me off guard lol

and there's a simple proof too!

just doesn't stick in my brain

[obtuse meteor]

I never think about CP^n so

I should like

consider this

[gritty widget]

it's just the symplectic reduction of the standard torus action on C^n

[obtuse meteor]

,,,,,,,,,,,,,

terra

[gritty widget]

Ah, I knew @obtuse meteor has babababa in profile status, is this about composing loops and homotopy stuff?

[sleek thicket]

tterra moment

lmfao

[obtuse meteor]

no 8da

it's this meme

tfw I have this on a keybind

super+r lesbiab

[sleek thicket]

yeah i would reccomend thinking about the CP^n thing

if only because you will facepalm

[obtuse meteor]

hmm

think I got it?

CP^0 and S^1

S^1 can't cover a point

big brain proof is

Z is not a subgroup of 0

🧠

[sleek thicket]

🧠

[obtuse meteor]

small brain proof is any open nbhd in S^1 contains lots of points

Not sure if the prof wants me to think in general

but

[sleek thicket]

what about other n though

[obtuse meteor]

yes

[sleek thicket]

oh fair

[obtuse meteor]

hmm

IDK for other n

[sleek thicket]

i will spoil it because i was to and it's a meme

[obtuse meteor]

okie

[sleek thicket]

go to evenly covered open sets

then

some open subset of CP^n is homeomrophic to some open subset of S^(2n+1)

[obtuse meteor]

sure

[sleek thicket]

CP^n is 2n-dimensional and S^(2n+1) is (2n+1)-dimensional

[obtuse meteor]

ah indeed

[sleek thicket]

so invariance of dimension for manifolds says this is impossible

(to justify invariance of dimension if yall haven't done that, take charts and look at the homology of the space minus a point)

[obtuse meteor]

psh

formality

this is algtop

we don't do formality in this class

me: <draws a cover>
me: hmmm, looks regular to me
me: <draws another cover>
me: not regular. Bingo!

related fun problem

22 is nice

and I solved it :o

[sleek thicket]

nice!

wait X is just like

a wedge of two circles

right?

[obtuse meteor]

mhm

[sleek thicket]

ah, so equivalently

is there a normal subgroup N in Z * Z and a normal subgroup K in N with K not normal in Z * Z

[obtuse meteor]

yeah but doing it

with that

is cringe

do it with geometry

fucking algebra nerds

[sleek thicket]

this happens in D8

which has a presentaiton with two generators

so you can pull back

[obtuse meteor]

I hate you

[sleek thicket]

tada

:^)

[obtuse meteor]

why

why would you do this

;-;

you wanna know the chad answer?

[sleek thicket]

cool problem faye

ty for that

sure

[obtuse meteor]

ok make a cover like this

two vertices

a transposition

for one of the loops

and identity loops for the other one

now make a cover

[sleek thicket]

I don't understand what you mean by a transposition for one the loops

[obtuse meteor]

ummm

[sleek thicket]

so we have two vertices

[obtuse meteor]

I should draw one sec

paint time

[sleek thicket]

lol

[obtuse meteor]

https://estrogen.fun/i/k8yd.png

this is clearly regular

bc just like

rotate

https://estrogen.fun/i/lzn8.png

and it's regularly covered by this guy

via flipping horizontally

but that guy doesn't regularly cover the wedge of two circles

because you have a loop on the left and right vertices but not on the center vertices

so there's no graph automorphism taking them there

[sleek thicket]

I think that makes sense (to me, I'm not doubting that your construction works)

[obtuse meteor]

ye

this is better than groups

imo

[sleek thicket]

you have a right to your opinion

the whole point of math is to turn things into algebra so you can do algebra on them

[obtuse meteor]

no

[sleek thicket]

[obtuse meteor]

false

[sleek thicket]

look it's not called topological topology

if you were supposed to do topology

it wouldn't be called algebraic topology

now would it

:^)

anyways these problems are cool

and absolutely not helpful for me for studying for my analysis final, which is what I should be doing

[obtuse meteor]

the point is to solve algebraic problems

with topology

🧠

[sleek thicket]

my prof didn't give us review problems she just said "do old prelims"

lol

[obtuse meteor]

ascending algebraic topology

[sleek thicket]

honestly it's fucked up that the universal covering space is only uniqueish

and similarly the algebraic closure

automorphisms should be illegal imo

(im looking at 12b)

[obtuse meteor]

lol

[tight agate]

this would kill a lot of bundle theory tho

[sleek thicket]

sacrifices must be made

it's time for a new name anyway

ah, algebraic topology 😌

faye is not allowed to criticize my proof now

[tight agate]

catwiggle

[obtuse meteor]

I really don't wanna do this problem lol

[sleek thicket]

rip

[obtuse meteor]

fuck group theory

all my homies hate group theory

[sleek thicket]

by the contrapositive

i am not faye's homie

lmao i thought this looked similar to something i put on my algebra problem sets

http://untyped.me/problem_sets/winter3.pdf practice problem 1

and

[obtuse meteor]

,,,shammy

sham you're my homie

[sleek thicket]

practice problem 2 is the example of non-transitivity of regularity of a cover

that I gave

wait wtf is problem 20

this seems incorrect

ah no

im being dumb

this makes sense

[gritty widget]

"my algebra problem sets"

chad

[sleek thicket]

lol

it fizzled this year bc pandemic

i didn't want to push them

(or commit to organizing stuff for them really)

[gritty widget]

[sleek thicket]

[obtuse meteor]

20 seems hard

like

wat

[sleek thicket]

i think you can do it purely group theoretically

[obtuse meteor]

I guess one part is easy

[sleek thicket]

but i might have to

do topology

[gritty widget]

Sham, you know what this means, it is time for you to leave #groups-rings-fields

[sleek thicket]

shock

[obtuse meteor]

20b is false

[sleek thicket]

someone said A&M earlier and I didn't recognize the book title lol

[obtuse meteor]

20a is

[sleek thicket]

wait wym 20b is false?

oh like there's no covers

[obtuse meteor]

yeah

[sleek thicket]

not that it's ill posed

[obtuse meteor]

ye

[sleek thicket]

yee 20(b) is false by easy group theory

[obtuse meteor]

yes

20a is

sad

[sleek thicket]

20(a) is false by hard group theory (I think)

[obtuse meteor]

lmao

hmm

maybe you can exploit Schreier index

[sleek thicket]

ah I think I see a proof

by topology(!)

[obtuse meteor]

:o

[sleek thicket]

do you want to hear it?

or think more on your own

[obtuse meteor]

thonk

[sleek thicket]

thonk

ah wait I think my proof is wrong, rip

:c

this is hard

oh wait no

my original proof works i was just being silly lol

i went local on S^1 v S^1 for no particular reason

just because I felt like it

okay, do you want to hear my Correct, Actually proof for 20(a)?

(proof that there is no such cover)

[obtuse meteor]

I want to think

[sleek thicket]

👍

[obtuse meteor]

so like

problem should be the center point

intuitively what I believe is that like

a neighborhood of the torus if you take a point out is still path-connected or has some # of path components or something

but if you take the center point you get 4 components

wondering

if something similar happenns

for homotopy equivalents for torus

[sleek thicket]

that's pretty much my proof, yeah

look local away from singularity => 2-manifold homeomorphic to 1 manifold => problems because of removing a point after going local

[obtuse meteor]

hmm

but the problem is

S^1 x S^1

homotopy equivalents

[sleek thicket]

yup

[obtuse meteor]

can be weird

[sleek thicket]

can be fucked up

[obtuse meteor]

so 2-manifold homeo to 1 manifold

isn't the thing

[sleek thicket]

Oh wait this is immediate from Nielsen schrier

You even said something like that above lol

Z×Z is abelian, so if it were a free group it would be free on one generator, which it just isn't

[obtuse meteor]

me saying the fact

Shamrock being like "noooooooooooo"

me being like "thonking"

[sleek thicket]

sorry :(

[obtuse meteor]

it ok :P

it happens

I didn't realize at all

more mad at myself for not realizing this

[sleek thicket]

I didn't realize you had a proof when you said this

and yeah i don't think there will be a nice topology answer

probably

hmm or perhaps...

[obtuse meteor]

hmmm

21 seems

scary

I don't want to compute that fundamental group

[sleek thicket]

lmao that definition of X is a meme

[obtuse meteor]

and I definitely don't want to compute its conjugacy classes

should it be a clear space

[sleek thicket]

im not talking about the quotient

but the one point compactification of R^3 is a space you know

[obtuse meteor]

S^3?

[sleek thicket]

yup!

[obtuse meteor]

hmmm

this action

on S^3

is bad

[sleek thicket]

yes

[obtuse meteor]

it's like

[sleek thicket]

very bad

[obtuse meteor]

the RP^3 action

but

not

at one point

why

would you

[sleek thicket]

but fucked up

[obtuse meteor]

do that

[sleek thicket]

lmao

[sleek thicket]

actually im not sure it is like RP^3 that much

on the complement

because we're taking the antipode in R^3 and not in S^3 <= R^4

[obtuse meteor]

oh yeah lol

oof

hmm

maybe there's a way to give a nice CW complex structure on it

I guess I'll find out at office hours

[sleek thicket]

are they at 4:30?

[obtuse meteor]

7:00 ET

so 15 minutes ago

[sleek thicket]

gotcha

[obtuse meteor]

but I'm in class rn

[sleek thicket]

good luck!

[obtuse meteor]

and then they're for 2 hours

so

fun

exam tomorrow

very scared

me when someone hands me an orbit space:

[sleek thicket]

my analysis class's exam ends at 6:30pm pacific

[obtuse meteor]

do analysis!!!

[sleek thicket]

but i got permission to take it with another student tomorrow at 9am

[obtuse meteor]

I see

have fun!

[sleek thicket]

bc of my talk

yeah haha it started at 2:30

if i were going to do it today i would be mega fucked

srtarting 2 hours late

[obtuse meteor]

oop

[sleek thicket]

ah wait

this problem

is very very sneaky

[honest narwhal]

Ooh we doing problems now

[sleek thicket]

it looks asymetric

like

infinity being fixed

but!

0 is also fixed!!!

[obtuse meteor]

yes and ???

[sleek thicket]

still thonking on that

[obtuse meteor]

oh

[sleek thicket]

i think maybe if you take X \ 0 and X \ infty

[obtuse meteor]

I already realized that but have no idea if it helps

[honest narwhal]

What's the problem exactly?

[sleek thicket]

so these will descend to the cover

and you may be able to svkt

[obtuse meteor]

we're looking at my midterm review

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

SVKT,,,,,

[sleek thicket]

they should be R^3/weird action

[obtuse meteor]

scary words Sham

scary words

[sleek thicket]

it is a good theorem 😛

@honest narwhal problem 21

[cold vine]

I'm trying to show that (X smash Y) smash Z is isomorphic to X smash Y smash Z when X,Y,Z are locally compact hausdorff. I can't figure out the mapping, any help?

[sleek thicket]

yeah I think this makes sense

is that second space X smash (Y smash Z)?

[honest narwhal]

Oh this looks like RP^3 but it's not lol

[sleek thicket]

yeah haha

so it's like

take R^3 and mod out by x ~ -x

glue two of those together

along x |-> 1/x

so it's RP^3 but(!) first you identify positive and negative numbers

which is very weird

err

wait

not 1/x

that doesn't make sense

was thinking 1 dimension lower....

[obtuse meteor]

I am going to office hours

I will update yall :P

[sleek thicket]

see ya!

[cold vine]

That's the one I'm trying to ultimately show but through this definition https://puu.sh/Hq6vc/2718a25796.png

[sleek thicket]

ahh gotcha

i regret to inform you that this looks very annoying and I do not want to think about it, sorry pahus

[honest narwhal]

Somehow it feels like the problem is basically a matter of, if a loop goes around 0 or infinity something funny is afoot

[cold vine]

Haha xD my feelings exatly, no probs dude

[sleek thicket]

so my thinking is

this space is covered by two copies of R^3/~

where we glue positives to negatives

and the intersection is (R^3\{0})/~, which is homeomorphic to RP^2 which has fundamental group Z/2Z since it's the quotient of the manifold R^3\{0} by a very nice (free etc) group action of Z/2Z which is R x RP^2

so you can seifert van kampen maybe

what's the space R^3/~ though? I'm not sure

[wanton marsh]

R x projective plane perhaps

[sleek thicket]

haha

oh you mean for R^3/~?

[wanton marsh]

for (R^3 \ {0} ) / ~

[sleek thicket]

yeah, I agree

so for R^3 / ~, can you just contract it all down?

like

[wanton marsh]

yes

[sleek thicket]

upstairs you can homotope the identity to the origin by H(x,t) = tx

and this descends

t(-x) = -tx

so yeah

it should be a simply connected space

by svkt

@obtuse meteor focus on your office hours/don't reply but I think the answer to 21 is that Y is simply connected, so there's exactly one iso class of connected covers

this seems like a bad question

like it's a trick question, right?

[marsh forge]

the screenshotted statement (a) is false but I wouldn't call it a trick?

a simply connected space is it's own universal cover and there can't be any intermediate ones

[sleek thicket]

eh, i guess

it's just like

"true but only in the case where X is contractible"

[marsh forge]

i would not say its a very interesting question

or that anyone would get it wrong for an interesting reason

[gritty widget]

I'm trying to understand this https://math.stackexchange.com/questions/251646/splitting-of-the-tangent-bundle-of-a-vector-bundle/2276379

but it seems that its much easier to actually explicitly define such a map

but now because what ive done is easier i assume its wrong

*i assume what i have done is wrong

[sleek thicket]

@obtuse meteor how was oh?

[obtuse meteor]

good!

[sleek thicket]

Nice!

[gritty widget]

do any symplectic kings or queens know how to see that the canonical form on the cotangent bundle agrees with the canonical from from the spillting of the tanget space of the cotangent bundle

[obtuse meteor]

https://estrogen.fun/i/mh76.png

@marsh forge turns out talking to you is useful for alg top midterm review :P

[marsh forge]

hahahah

[obtuse meteor]

(this was on the midterm review packet)

midterm is tomorrow :o

[sleek thicket]

Good luck!

[obtuse meteor]

cursed times

https://estrogen.fun/i/qi2z.png

so one could approach this problem with algebra

https://estrogen.fun/i/l9tp.png

but I think

it's better to do it by like

just drawing covers

and being like "these are the only regular ones by (handwave)"

[marsh forge]

idk of a way to make that rigorous off the top of my head

idk it feels like if u r intentionally avoiding the algebra in AT ur missing the point to me hahaha

[sleek thicket]

Oh I thought about that one

I couldn't do it with algebra

I think I showed there are 4 covers

[obtuse meteor]

exploit the fact that like it's given by graph automorphisms

and those have symmetry requirements

[sleek thicket]

But it seems like the prof wants explicit covers

[marsh forge]

well if you know how many there are

its not that hard to like draw them all given sufficient time

i guess

[obtuse meteor]

I mean I know they're given by subgroups of rank 4

but it's just like

how do I compute the normal subgroups of F_2 of rank 4

I don't know algebra

[sleek thicket]

Wdym rank 4?

[obtuse meteor]

they have index 3

by schreier index formula

they have rank 4

[sleek thicket]

Ahh okay

I do not know that formula lol

[obtuse meteor]

I have it in my notes

[sleek thicket]

I computed it by saying F -> F/K is a surjective group homomorphism onto a group of order 3 with kernel K

yeah?

But F/K ≈ Z/3Z

Since there's only one group of order 3

[obtuse meteor]

ah

[sleek thicket]

So it suffices to determine the possible kernels of a surjection F -> Z/3Z

[obtuse meteor]

that makes sense

but makes me sad

yeah

[sleek thicket]

And maps out of F are easy

sorry :(

[obtuse meteor]

it is

"how did not see" hours

[sleek thicket]

You've been practicing for a while!

[obtuse meteor]

wdym?

also I hate questions like this lol

https://estrogen.fun/i/e619.png

[sleek thicket]

When one does math for a long time one's brain turns into cornflakes that have been soaking in milk for 3 hours

yeah it's annoying

Like

I love these questions as long as they're not graded lol

[obtuse meteor]

yeah but I took an hour thirty minute break sham lol

[sleek thicket]

Fair

[gritty widget]

can we consider S^0 a symplectic manifold?

[sleek thicket]

huh, weird

I guess so yeah

Take the unique 2 form on it

It's nondegenerate because everything is degenerate lol

[obtuse meteor]

Jenny told me
"PLEASE COMPUTE"
https://estrogen.fun/i/238z.png

[gritty widget]

ah nevermind

symplectic 0-manifold

[gritty widget]

i relaise this was stupid

do you by chance now how to see that the canonical form on the cotangent bundle

is the same as the canonical form on T_pM \times T_pM^*

oh $\phi^\omega=\phi^ d\lambda=\lambda(d\phi)

[little hemlock]

so, im trying to compute the de rham cohomology of the circle. We have $\Omega^0(S^1) \xrightarrow d \Omega^1(S^1) \xrightarrow d 0$ so i pretty much just need to compute the image of $d: \Omega^0 \to \Omega^1$.

[gentle osprey]

kxrider

[little hemlock]

I believe this map is supposed to be surjective, but I'm not really sure how to prove it.

[sleek thicket]

If this map were surjective the 1th de rham cohomology would be zero

right kxrider?

(ty for the backup tterra 💪)

[little hemlock]

i suppose. is that bad?

[sleek thicket]

Well sort of

The point of homology and cohomology is to detect n dimensional holes

(I'm simplifying, but this was the origin historically and is still a good motivation)

The circle has a 1 dimensional hole in it

so we should expect H^1(S^1) to be nontrivial

Anyways, it's not true that d is surjective

I can tell you the dimension of H^1(S^ 1) if you end up wanting to check your answer

[gritty widget]

just use mayer vietoris lol

(do not)

[sleek thicket]

Lol I thought you were serious

And was gonna say

"they might not know it"

[gritty widget]

i'm concerned that you've ever taken anything i've ever said seriously

[sleek thicket]

Idk why I'm having this conversation with a homophobe

also here's another perspective: you may not have seen this yet, but de rham cohomology is homotopy invariant, so we have an iso H^1(R^2 \ {0}) ≈ H^1(S^1). Then if H^1(S^1) = 0 we would have H^1(R^2 \ {0}) = 0. This says that closed iff exact for vector fields in the plane. Does that last claim sound right to you @little hemlock ?

[gritty widget]

closed vector field

don't you mean symplectic and hamiltonian

[little hemlock]

This says that closed iff exact for vector fields in the plane.
even though you've removed the origin?

[sleek thicket]

Yeah, I'm saying that this is a reason the cohomology group shouldn't vanish

Because it would imply this statement about vector fields on the punctured plane (the statement "closed iff exact")

[gritty widget]

are there any other ways we can prove/see \pi_k(S^n)=0 for k<n other then freudenthal

@ max

[little hemlock]

ah, i see what you mean. my intuition isn't exactly great. Not sure if its obvious that closed iff exact is false for vector fields on the punctured plane

[sleek thicket]

Whitney approximation+sard's theorem

[gritty widget]

interesting

[sleek thicket]

I mean I can prove it rn

[gritty widget]

please that would be really cool

[sleek thicket]

Any map is homotopic to a smooth map and the basepoints will line up and all

yeah?

[gritty widget]

mhm

[sleek thicket]

So it suffices to show smooth maps S^k -> S^n are homotopic to the constant map

sard's theorem implies that the image of such a smooth map is measure zero in S^n

Sard's theorem implies this about any smooth map from a lower dimensional manifold to a higher dimensional one

yeah?

[gritty widget]

ok yes

[sleek thicket]

And then it's easy

Just remove a point outside the image and you factor the map through a contractible space

[gritty widget]

wow ok i did not think of that very nice

thank you

[sleek thicket]

Yeah, it's neat!

Here's another similar proof

https://en.m.wikipedia.org/wiki/Cellular_approximation_theorem

[gritty widget]

that is neat

[sleek thicket]

You have cw structures on sphere where there's a single 0 cell and a single n cell

a cellular map S^k -> S^n will be constant

Any map is homotopic to a cellular map

Similar vibe as the smooth argument

[sleek thicket]

Sorry, I don't think it's obvious

But it's something that I think most people see in a vector calculus class

Sorry for assuming

Basically you can make issues happen by having stuff blow up at the origin

1/z on the complex plane minus the origin

Is closed but not exact

By which I mean, if you let f(x, y) = 1/(x+iy) = p(x, y) + i q(x, y) then the differential form p(x, y) dx + q(x, y) dy is closed but not exact

Closedness is pretty much the cauchy riemann equations, if you've seen any complex analysis

But it's not exact since the line integral around the unit circle is nonzero

[tight agate]

Hurewicz thm

[gritty widget]

i like the sard way that is pretty cool

[sleek thicket]

Yeah!!!

Smooth = good

Max was very down on this argument when I said it to him lmao

[obtuse meteor]

smooth bad

cellular good

[little hemlock]

ah yea, i think i remember. There are path independent vector fields on a non-simply connected domain which are not conservative, or something like that.

[gritty widget]

category of smooth manifolds is amazing wtf you talking about

[sleek thicket]

I would not go this far

The category is kind of ass lol

But the objects are very good

[sleek thicket]

Hmm, not exactly. Iirc that's the definition of conservative (but I may be wrong)

[gritty widget]

yes i know joking but damn that proof is just so satisfying

[little hemlock]

g is conservative if there is f such that df = g, i.e. g is exact, right?

[sleek thicket]

Ah I saw it defined by being path independent. They're equivalent on any smooth manifold

(even with boundary!)

But you may have df = 0 and f not path independent

On a non simply connected domain

Like the circle

Which says exactly thst H^1(S^1) = 0!

[sleek thicket]

Sard's theorem is so good haha

It's like

"Oh no, how do I know this works??"

"well if you choose a random option it'll fail with probability 0"

[shut moat]

$\frac{xdy - ydx}{(x^2+y^2)}$ is closed but not exact on the punctured plane

[sleek thicket]

"so there's gotta be some way to make it work"

[gentle osprey]

~S^1

[sleek thicket]

yes, this is the 1/z form I defined above

[shut moat]

oop I missed that

[sleek thicket]

Err, it's not quite actually

There's like a multiplication by -1

But that doesn't really change anything, same idea

[gritty widget]

society if 1/z dz integrated to 0 on a loop around the origin

[sleek thicket]

Np haha. I can only remember the explicit generator of H^1(S^1) if I think of it as 1/z

The formula is too awkward

[shut moat]

It's easier to remember it as the flux form of the vector field $\frac{\vec{\bm{x}}}{|\bm{x}|^n}$

[gentle osprey]

~S^1

[shut moat]

imo

[sleek thicket]

wtf is a flux form

go away

[shut moat]

what was that funny interior thing you showed me a while back?

[sleek thicket]

ah okay this isn't actually bad, that's just 1/(z conjugate)

[shut moat]

stick it into volume form

[sleek thicket]

Hmm, not sure what you mean

[shut moat]

like let $\mathbf{F}$ be a vector field on $\mathbb{R}^{n}$. Its flux form is the $n-1$ form given by $\Phi_{\mathbf{F}}(\mathbf{v}1, \dots, \mathbf{v}{n-1}) = \det[\mathbf{F}, \mathbf{v}1, \dots,\mathbf{v}{n-1}]$

[gentle osprey]

~S^1

[sleek thicket]

ah gotcha

Interior multiplication

This is F |_ dV yeah

Nice!

[shut moat]

so in coordinates the variants of the closed but not exact form would be $$\frac{1}{(x_1^2 + \dots + x_n^2)^{\frac{n}{2}}}\sum_{i=1}^n (-1)^{i-1}x_i dx_{1}\wedge\dots \wedge \hat{dx_i}\wedge \dots \wedge dx_n$$

[gentle osprey]

~S^1

[sleek thicket]

Yup

[shut moat]

that hat is 2 smol

[sleek thicket]

Ah I think I might decide this is obvious

Yes, this is obvious