point-set-topology

[gritty widget]

I thought a metric had to be d:X cross X -> R ?

[fading vale]

yes and it defines a topology

which is a subset of the power set of X subject to certain conditions

i think you need to review your definitions here

[gritty widget]

ok

[fading vale]

and look at a couple more examples

[gritty widget]

ok

I guess I'm trying to make examples of simple sets that are topologies both with and without metrics in the natural numbers and my definititions are messed up.

[fading vale]

you need to understand that like

this is a weird way to think about it

you start with a set

you put a topology on the set

[gritty widget]

ok

ok, so like set {0,1}

becomes {0,1, {0,1}} as a topology

[fading vale]

it can

that is one possible topology to put on it

[gritty widget]

so how would one put a metric on {0, 1}?

ok

[fading vale]

the discrete metric given by d(x, y) = 1 if x =/= y, else d(x, y) = 0

[gritty widget]

how could one put a metric on {0, 1} then?

[fading vale]

would induce this topology

i mean as a two point space its not like it really matters because the only non-zero value the metric takes on is going to be d(0, 1) lol

[gritty widget]

ok, so as a set with ordered pairs for d, { 0, 1, (0, 1, 1), ( 0, 0, 0), ( 1, 1, 0) }?

[fading vale]

as long as its positive its fine

ok this doesnt make sense again

is (0, 1, 1) supposed to represent the metric and the output

[gritty widget]

yes

[fading vale]

this is not in the topology though

the topology is just a subset of the initial space {0, 1} that fulfills certain axioms

the metric induces a topology on {0, 1}

[gritty widget]

can't that be visualized as a set too?

the metric induced on a topology?

[fading vale]

yes

but (0, 1, 1)? is not in that set

also its not really a visualization lol its just what it is

what are you learning topology out of

[gritty widget]

ok, what would that set look like for {0, 1} and the metric d from before?

bert mendelsons into to topology

[fading vale]

this metric induces the discrete topology

where every subset is open

[gritty widget]

intro

[fading vale]

i really think u need to reread lol

and look at more examples

[gritty widget]

ok, I'm trying to find more examples, hard to search the internet for specifics when I mess up the wording, lol

[fading vale]

try munkres or another textbook maybe

(just grab a free pdf)

[gritty widget]

ok, munkres, great!

[woeful oasis]

https://unapologetic.wordpress.com/2007/11/09/topologies-as-categories/

[feral copper]

Hello ! Just trying to understand the notation here, what does the square brackets refer to ?

Hom(-,-) would be my first idea, but there's no reason why we could interchange the roles of X and Y in this case...

Like Hom(Z/2,Z) and Hom(Z,Z/2) are not the same for instance

Ooooooh that totally makes more sense now

Thanks !

However as I see it, it is defined for X=Y, what am I getting wrong ?

Like is it assuming we are taking X≅X×{basepoint} and Y≅{basepoint}×Y in the product X×Y ?

Thank you 🙂

[summer jolt]

Hi guys I'm trying to find a continuous surjection from the closed unit interval to the unit square. Is such a thing possible?

[cloud owl]

peano curve?

[elder yew]

I think it is possible

[rugged swan]

lol

[gritty widget]

it's absolutely possible

now find a homeomorphism 😉

[summer jolt]

now find a homeomorphism 😉
@gritty widget Wait there is even a homeomorphism????

How's that possible

[gritty widget]

(it's not possible, if the unit square is given the subspace topology)

[elder yew]

It's impossible for a 1 dimensional thing to be homeomorphic to a 2 dimensional thing

[gritty widget]

are we talking "unit square" as in the set {max(|x|, |y|) = 1} or the filled in square

[cloud owl]

who tf uses unit square for a hollow square

[gritty widget]

i just did i guess

i was thinking "unit circle" and that's basically a hollow square yeah?

so my mind said "unit square"

[cloud owl]

begone

[gritty widget]

[ivory dragon]

i was thinking "unit circle" and that's basically a hollow square yeah?
ah, the L^1 norm

[summer jolt]

I'm talking about [0,1]x[0,1]

[gritty widget]

haha my bad

[summer jolt]

peano curve?
@cloud owl hmm haven't heard of this before. I'll look it up

[feral copper]

https://www.youtube.com/watch?v=RBWwlfcftYM
Interesting, I had never actually seen it

[gritty widget]

[feral copper]

I have no clue how to compute this relative homology group... Would anyone have any advice please ? 🙂

$$H_4(S^2\times S^2,S^2\vee S^2)$$

[gentle osprey]

Matplotlib:

[fading vale]

you can use the LES from relative homology to get ... -> H_4(S^2 V S^2) -> H_4(S^2 x S^2) -> H_4(S^2 x S^2, S^2 V S^2) -> H_3(S^2 V S^2) -> ...

the first and last of these are zero so we by exactness we get an isomorphism H_4(S^2 x S^2) -> H_4(S^2 x S^2, S^2 V S^2)

@feral copper do you know how to compute H_4(S^2 x S^2)

it requires the kunneth theorem but the computation itself is fairly simple

[feral copper]

Oh yeah that was simple x)

Now I have two reasons to feel dumb xD

Thank you !

[fading vale]

np

: )

also dont feel dumb haha everyone messes up on stuff like this at some point

its normal

[feral copper]

Yeah ^^" ty ^^

In my case, none of the groups will have torsion, so I can use Kunneth's formula just like if I had a field, right ? (I need homology over Z to use Hurewicz's theorem)

[fading vale]

yeah

its just

[feral copper]

Thank you again ! 🙂

Tensor product in the middle though

[fading vale]

ah yeah oops

[feral copper]

Np ^^

[gentle osprey]

Moth:

[feral copper]

Tyvm !

[fading vale]

:D no problem

[feral copper]

Another question ! Can"t I just have H4≅Z by orientedness ?

[fading vale]

uhh possibly?

i dont know too much about orientability and stuff

but i think yes

assuming we know that S^2 x S^2 is orientable anyway

[feral copper]

A product of orientable is orientable I believe (I hope so ! XD)

[fading vale]

i think you're right

[feral copper]

Love that emoji

[fading vale]

it is very good yeah

[gritty widget]

flexing your nitro

[fading vale]

[fading vale]

nooooo

why did someone type here and then delete

TTerra ur mean

[gritty widget]

i want to make a really bad joke

about the exterior derivative satisfying d^2 = 0

[meager python]

What are the quintessential examples for cohomology in algebraic geometry that one should do

[cedar pebble]

If by cohomology you mean coherent cohomology then THE series of examples you should understand is the cohomology of line bundles/vector bundles on curves

Also cohomology of line bundles/vector bundles on P^n

But the curve case can keep you busy for a long long time and most more complicated examples reduce to a very good understanding of the curve situation

[meager python]

Exactly what I was looking for, thanks 🙂

[cloud owl]

hi. i'm really really bad at topology, ie. just starting out.
A function f : X → Y between two topological spaces is continuous if and only if f^-1(U) is open in X whenever U is open in Y.

1. are you allowed to just say U is open in X? wouldn't you have to define the topology U is open in?

2. why does the book suddenly take a hard left into category theory?

but can't there be multiple topologies on a space?

??

ohhh okay

right, yeah, i remember now

[sweet wing]

consider the trivial topology and the discrete topology

any function to the trivial topology is continuous

2. why does the book suddenly take a hard left into category theory?
   also what book you should switch books

[cloud owl]

it's the one pinned in #book-recommendations

[sweet wing]

oml

pls dont use it

use like

munkres

for your point set intro

[cloud owl]

then why is it pinned lmao

[sweet wing]

what one person finds useful another may find absolutely garbage

i say this because idk why max seem to think(?) that it is a good intro
i think the category theory feels too forced, it seems more like "in theory we can teach like this"

same for me with like hatcher sure it is geometric but heck i cant learn from it

feel feel to just ask for book recommendation! mentioning your background and goals always helps

[cloud owl]

i saw no description lol

[sweet wing]

the book preface seems to imply so tho and chap 0 says the prelims are just a quick review of topology kinda implying you already know topology 🤔

i usually read preface to gauge if book is too hard for me lol

[summer jolt]

Can someone explain why the finite product of discrete spaces must be discrete. I have seen many explanations online but can't fully make sense of them. I know that if you a discrete spaces then their base must be singletons (and obviously they are open). But how do I show that all subsets of the product are open?

[sweet wing]

essentially you want to show some element (a_1,a_2,...,a_n) is open

[summer jolt]

@sweet wing Ok so an element like that would be open because the projection maps pn are continuous so their preimage is open?

[sweet wing]

yup

something along that lines

alternatively you know it's basis is products of open sets from the indiv topologies (cuz finite product)

and (a_1,a_2...,a_n) is a_1\times a_2\times...

[cloud owl]

it pains me to already have to ask for help for the first set of exercises but

i don't understand this

[fading vale]

its ok topology is hard

[cloud owl]

the first one definitely works

i think

but surely for the unions/intersections of any of the others they'd no longer look like

wait nvm i got it

i see the answer

i'm a fool

[fading vale]

haha

[cloud owl]

</rubberduck>

[fading vale]

yeah

[cloud owl]

pain

[uncut surge]

is (x) the surprise topology

that's a funny one

ah no, probably (ix), but i'm not sure why (x) couldn't work

[gritty widget]

What is (x)?

[uncut surge]

in kaisheng's screenshot

[gritty widget]

surprise topology

Oh lol. (Surprise topology? Surprise ideal)

[uncut surge]

oooh! i get it now, it's actually (i) that doesn't work

[cloud owl]

aaaaaa

how

[uncut surge]

are unions of intervals always intervals?

[cloud owl]

ah crap

[uncut surge]

that was a fun problem tho, you go through the whooole list and you're like "no, no, no, no, no" and then at the end you're hit with these really weird topologies that actually work

[night parrot]

im working my way through some beginner differential geometry stuff, and just got to the first / second fundamental forms. i wanted to check that my understanding here was correct

so in this equation, is ∇f(x0) a jacobian matrix? since the function that describes the surface, f(u, v), is a mapping from R2 to R3?

and a and b are just arbitrary 2D vectors?

[gritty widget]

yeah ∇f(x0) would be a 3x2 matrix, so the first fundamental form would be given by the 2x2 matrix ∇f(x0)^T ∇f(x0). as you say, a and b are vectors in R^2 (picture says this exactly)

[night parrot]

but i mean they can just be anything right?

and the product shown at the bottom is basically showing how the relative lengths of a and b change, with respect to the surface or something? just having some trouble building intuition here

or i guess, how the angle between them and lengths change, is how im interpretting it.. but not sure

[gritty widget]

and the product shown at the bottom is basically showing how the relative lengths of a and b change, with respect to the surface or something?
kinda, yeah. it's a little weird since this is only an approximation (one that gets better for a and b of smaller magnitude). it might help to draw a surface and a parametrization f, and then visualize ∇f(x0)a and ∇f(x0)b as tangent vectors to the surface at the point; then the inner product <∇f(x0)a, ∇f(x0)b> is how you compare them

if i had access to something to draw on i'd show you what i mean, but unfortunately i don't

maybe you can do a simple example, like with a parametrization of a sphere (or even simpler, a plane)

that might help you build some intuition

[night parrot]

thats a good idea, i have some ways i might be able to visualize a simple example

[gritty widget]

like everything in diffgeo should be tested against simple examples lol

I keep confusing covariance with contravariance, what's a way to distinguish them?

[fading vale]

dan

[gritty widget]

in the differential geometry context? it's either something to do with whether it takes in vector fields (contravariant) or covector fields (covariant), or with the position of indices (if you're a physicist)

[fading vale]

oh do you mean covariant vs contravariant functor

[cloud owl]

one has 'ntra' in it

[gritty widget]

differential geometry

i think lee's ISM covers this, mathacka

[fading vale]

covariant functors preserve the direction of arrows

contravariant functors reverse them

[gritty widget]

this is in the diffgeo context moth

not category theory

ISM?

[fading vale]

its probably related

[gritty widget]

diffgeo loves abusing terminology lul

ya maybe

mathacka: introduction to smooth manifolds

ok, thanks!

i actually use the appendix b of his riemannian manifolds book more, tbh, so i'll rec that too

[fading vale]

it seems like the same general concept of preserving vs inverting direction

[gritty widget]

oh sorry @gritty widget i got it wrong, covariant takes in vector fields, and contravariant takes in covector fields

fucking confusing

yeah

haha

well it's all in lee's books, i'd check those out

@fading vale there is actually some kind of connection, maybe. contravariant and covariant differ in how they transform with respect to changes of bases

i don't remember the specifics

but one goes one way and one goes the other way lol

the wiki page is clearly written for physicists but it's a good read

[fading vale]

im not sure if theres a genuine mathematical connection or if its just shared terminology (like "co" for flipped shit/contra for flipped maps)

[gritty widget]

i'd guess just shared terminology, but i'd love to see if there's a genuine connection

[fading vale]

someone categorify diff geo

[gritty widget]

""naturality"" already shows up quite a bit

it's prolly been done

[fading vale]

its kind of dumb that homology is covariant but uses chains while cohomology is contravariant but uses cochains

smh

[gritty widget]

that's gotta be confusing

contravariant eats covector fields, but covariant eats vector fields

physicists why?

[sleek thicket]

ISM does have a good explanation

[fading vale]

hello shamrock

[sleek thicket]

Hello

I started writing an explanation and then someone pinged me in a different server

[fading vale]

lol

[gritty widget]

hi shamrock

i looked into it more, the terminology does make sense

contravariant things transform with the inverse of the change of basis matrix

which makes a lot of sense and im surprised i didn't learn about it sooner

[elder yew]

The way I think of it is contravariant goes against the change

covariant goes with the change

contra = against

co = with

[sleek thicket]

@gritty widget that's what I was saying in my deleted post basically

I think it's okay terminology, it's just not really how I think about these things

[fading vale]

LMAO

@sleek thicket holy shit i went on the like wikia webpage or whatever and i got an ad for ISM

lmfao

targeted advertisement is amazing

[sleek thicket]

lmaoo

i get math textbook ads too

lots of hopf algebra stuff??

[fading vale]

amazing

[sleek thicket]

actually i got an ad like that on tinder once

it's saved somewhere on my phone

[gritty widget]

"don't do guys, do math"

it was trying to tell you something

[sleek thicket]

springer graduate texts in homophobia

[fading vale]

want

[sleek thicket]

want

https://twitter.com/grassmannian/status/1290859768760815616?s=20

Brendan (@grassmannian)

are you fucking kidding me

Twitter · Aug 5, 2020, 3:58 AM

this is why we tweet out everything that happens to us

[tough imp]

wait wut

meme

[gritty widget]

736 followers, popular

[sleek thicket]

follow me

iam a clout chaser

[gritty widget]

no

[sleek thicket]

::(

[gritty widget]

my twitter account is cursed

[sleek thicket]

is it mostly furry porn or something?

[gritty widget]

[sleek thicket]

lol

[gritty widget]

Is geometry empirical or a priori?

[fading vale]

what

ask kant lol

[gritty widget]

That's why I'm asking 😭 idk if einstein refutes kant or not

[fading vale]

dan

the question you are asking does not have anything to do with geometry

[gritty widget]

I mean you guys do geometry so I assume you'd know

this is a philosophy question

[fading vale]

it has nothing to do with geometry

[gritty widget]

Alright my bad

[fading vale]

this question is basically just "does synthetic a priori justification exist" but thats a philosophical question lol

its not really about geometry

[gritty widget]

He thinks the opposite

That geometry is empirical

"Until some time ago, it could be regarded as possible that Kant’s system of a priori concepts and norms really could withstand the test of time. This was defensible as long as the content of later science held to be confirmed*) did not violate those norms. This case occurred indisputably only with the theory of relativity. However, if one does not want to assert that relativity theory goes against reason, one cannot retain the a priori concepts and norms of Kant’s system."

Well he says this

So apparently it has to do with relativity

[fading vale]

i dont think einstein is talking about geometry here

^

wow thank goodness i read deleuze

it was all for this moment

[gritty widget]

Deleuze
Oof

But thanks guys uwu

[gentle osprey]

Davide | Math student:

[gritty widget]

[cold vine]

So I have top.space. X and point x in S^n. Now since X×S^n retracts to X×{x} we see that Hn(X × S^n) is isomorphic to Hn(X×{x}) + Hn(X×S^n, X×{x}) (by splitting of the LES of the pair. We have shown this previously)(+ is the direct sum). Now since X×{x} is trivially isomorphic to X we see further that Hn(X × S^n) isomorphic to Hn(X) + Hn(X×S^n, X×{x}) Q.E.D.

Does this make sense . Seems a bit too easy so I'm thinking if I missed something

[fading vale]

@cold vine im not sure you can assert that the LES splits here so easily

or rather idk what argument ur using

maybe i have to review some stuff here ik that we get an injection H_n(X x {x}) -> H_n(X x S^n) but i dont remember the construction of the LES so idk if the map in the LES is injective

[cold vine]

So I've proven earlier that if A is a retract of X then the sequence splits and thus H(X) = H(A)+H(X,A)

and here I choose A = X×{x}, X=X×S^n

[fading vale]

yeah okay i think this works

the result is definitely true by mayer vietoris

[quiet pilot]

Is there some general method on how to describe certain curves in projective space P^2?

For example, a line in P^2 is defined as the zero set aX+bY+cZ=0, except the origin. A conic is gotten by ax^2+bXY+...+fZ^2=0

[elder yew]

yeah there are theorems on this

[quiet pilot]

Do we always just add in a Z and make things symmetric?

[cold vine]

@fading vale Alright thanks

How is this seen by Mayer-Vietoris

[fading vale]

@cold vine sorry i misremembered bc this problem is in hatcher

but theres a part 2 which uses mayer vietoris

[quiet pilot]

But ax+by is already homogeneous

[cold vine]

@fading vale Alright no probs ^^

[quiet pilot]

Ah of course

Thank you

I guess it becomes slightly messier for conics?

Ah right, thanks again

[cold vine]

So I'm doing an excision of a manifold X and a point x in X so now I have the nbd of x homeomorphic to R^n which is A, and I get from the excision of X-{x} and A that H(X,X-{x}) = H(A,A-{x}) =H(R^n, R^n-{*}) but now I should get that the last group is Z which I don't believe holds since that last group seems to be homology group of a two point space?

[fading vale]

use the LES of the relative homology to get H_n(R^n) -> H_n(R^n, R^n - *) -> H_n-1(S^n-1)

the first is 0 so we get an isomorphism of the 2nd onto the 3rd and the 3rd is just Z

[uncut surge]

Why do you think it has to do with homology of a two-point space?

[cold vine]

Ah yeah I'm wrong on that one. I was thinking we quotient everything but the point away to another point and then take the homology but thats def wrong.

Alright I see!

Thats very nice

[cold vine]

Hmm another question... So I have a space X constructed from three disks D1, D2, D3 with their boundaries identified. Now we also have subsets X1 and X2 which are made by identifying D1,D2 and D2,D3 respectively (and they are homeomorphic to S^n). So clearly X1 and X2 are closed but are they also open? I'm not quite clear on the quotient topology here. Is it done from the disjoint union of D1, D2, D3? I would presume so and thus they are also open.

[fading vale]

Im kind of unclear what you mean

by "we have subsets X1 and X2 made by identifying..."

do you just mean like X1 = D1 U D2 in D1 U D2 U D3/~

intuitively it looks like X1^C = int(D3)

[cold vine]

ahhh sorry. So X1 is D1 and D2 with boundaries identified

D1 is disk

[fading vale]

ah yeah

so i see what you mean by it being homeomorphic to S^n

[cold vine]

so now im doing excision on the three identified disk with the subsets of two identified disks

and thus i need that X1 and X2 are open in this space

[fading vale]

are you sure?

what excision are u trying to do specifically

[cold vine]

ahh sorry not excision buy Mayer-Vietoris

but*

[fading vale]

ah

[cold vine]

So I think X is formed from the disjoint union of Di

[fading vale]

um i dont think you need X1 and X2 to be open necessarily

you just need their interiors to cover X

[cold vine]

that is true

but if they are closed i think their interiors go to zerp, no?

so then i would need some other subspaces

but i would think that they are indeed open since inverse image of X1 is D1 U D2

and they are open if it is in disjoint union, I think?

[fading vale]

i think ur right

[cold vine]

This is a bit hand wavy but i think it makes sense

[fading vale]

ur right i think

[cold vine]

nice ^^

thanks

[fading vale]

[cold vine]

Ahh it actually doesn't seem to work since if we take the inverse image of the projection of X1 we get not only D1 U D2 but also the boundary of D3 which definetely is not open 😅 Well thats rough

[slender plank]

Hello I need help with .last question

https://cdn.discordapp.com/attachments/371045929093955586/777543779066445855/20201115_153957.jpg

French is only used to introduce things

I need help for question 6)c

we need to show that there exist a theta that veriifies this inequality

[gritty widget]

Neat problem, maybe one idea is, if the zj all lie on the circle, then maybe you can prove it for any theta by a pigeonhole argument (for any line through the origin, a majority of the points must lie on one side), although I didn't do any calculations

[gritty widget]

(nvm it can't work for any theta)

[uncut surge]

I'm not sure I get it but from the setup of the question, it sounds like you need to integrate both sides of the inequality from b)

and pray to jeebus that you can somehow show that the left-hand-side is an integrable function in theta

And then I suppose some mean value theorem about integrals or so

yeah that feels about right

@slender plank ton fromage est pret

and yeah the mean value theorem i mean is $\int_a^b f(x) dx = f(c) (b - a)$ for some $c$

[gentle osprey]

Lartomato:

[uncut surge]

ah in fact you need to show that the integrand is continuous for that to hold

might be a bit tricky

oh no, you can just take the supremum of the integrand instead

[slender plank]

Ok thx Lartomato

[red garden]

yo

Let Y be a subspace of a metric space X, and let A be a subset of the metric space Y. Show that A is open as a subset of Y <--> it is the intersection of Y with a set which is open in X

suppose A is the intersection of Y with a set which is open in X

Y is open in Y

let x be in A --> we wish to show S_r(x) is contained in A for some r>0

x be in A --> x is in Y and x is in N for some open set N in X

we wish to show S_r(x) is contained in the intersection

i cant do shit

help

plz ping me if help

hint

no spoiler

plz

ty

[gritty widget]

It's not S_r(x) contained in the intersection that you want to show, it's S_r(x) cap Y

@mo2men

[red garden]

thats what i thought cuz i am af ucking geinus

okay i wrote it down in paper before i told u

[gritty widget]

Nice

[red garden]

can i tell u

the proof tho

im p sure its still wrong

suppose A is open

i claim A is the intersection of Y intersect S_r(x) for x in X and r>0

or wait

so for each x in A

i wish to show that A = Y S_r(the same x) ?

for some r>0?

@gritty widget

[gritty widget]

I guess no. But I'm not exactly able to follow what you're saying

[red garden]

omg i cant do it

idk what the fuck is the set open in X

idk

just am etric space where the subspace is a meetrics space on the same metric

just the set is subset

idk this is the ssubspace topology but i am not supposd to know that yet

[red garden]

ohfuck

why couldnt i think of it

[tight agate]

Let X be an almost complex manifold. Then using the almost complex structure, we have a direct sum decomposition of the (complexified) sheaf of k-forms on X: $A^k(X) = \bigoplus_{p+q = k}A^{p,q}(X)$. If $d: A^k(X) \rightarrow A^{k+1}(X)$ is the exterior derivative, then we can define $\partial := p_{p+1, q} \circ d: A^{p,q} \rightarrow A^{p+1,q}$ (by projecting onto the summand after applying the exterior derivative), and simimlarly we can define $\bar{\partial} := p_{p, q+1} \circ d: A^{p,q} \rightarrow A^{p,q+1}$.

[gentle osprey]

Brofibration:

[tight agate]

Now, locally we aways have $d = \partial + \bar{\partial}$.

on a small enough nbhd

[gentle osprey]

Brofibration:

[tight agate]

by picking coordinates

The book I'm reading says that the almost complex structure is integrable if and only if $d = \partial + \bar{\partial}$

[gentle osprey]

Brofibration:

[tight agate]

But another definition for an integrable complex structure is one that comes from a complex structure

That equality needs to fail while passing from local to global

So the question I have is is there an example where I can see that equation fail to hold globally

I'm assuming the problem arises while trying to change coordinates using a transition function that is not holomorphic

Is there some weird way in which the eigenspaces of i and -i of the almost complex structure switch around to prevent the equality from holding globally

I think that might be it as the transition functions are holomorphic if and only if they commute with the action of the almost complex structure, and if they commute with it then they preserve eigenspaces

but that's pretty handwavy

[meager python]

In Spec k[x,y,u,v]/(xy+ux^2+vy^2), how does one see that D(x,y) is not principal?

[knotty pasture]

Why if the structure group of a (k-1)-sphere bundle E can be reduced to O(k), we can form a RIemannian vector bundle E' of rank k whose unit sphere bundle S(E')=E?

[frozen phoenix]

how I would prove that if U is an open subset of R^n then the tangent space of U to a point p in U, is equal to R^n ?

My thoughts : Show that the directional derivatives are in one-to-one correspondence with the vectors, a vector v \in R^n produces a map taking appropriate smooth maps to reals, then we need to show that given a derivation can be realized as such a directional derivative

is this along the correct lines?

ok ty

[frozen phoenix]

is this the best/efficient way?

[quiet pilot]

How would I rigorously show that V(XT-YZ) \ {0} in k^4 has the origin in its closure?

k is infinite

Intuitively it feels like a function which is 0 on the whole V(XT-YZ) except possibly at a single point needs to have XT-YZ as a component

Nvm I think I got it. If f isn't 0 on origin then V(f)(intersected with V:=V(XT-YZ) is closed in V. So {0} is open in V, which is absurd as V(XT-YZ) is irreducible

[gritty widget]

any tips how to show lower limit topology doesnt have countable basis

[cloud owl]

i know not enough about things but iirc you show things aren't countable by assuming countability and then using a diagonal argument to find contradiction?

[gritty widget]

you do not know enough about things

[cloud owl]

ok

[gritty widget]

although you probably can prove that statement by seeking cardinality contradiction (at least tahts what I was trying to do), but it has nothing to do with diagonal argument

[cloud owl]

fine, cardinality contradiction, yeah

i've been incorrectly calling that sorta thing where you list them and find one that isn't in them a diagonalisation argument then

[gritty widget]

although I find your comment kinda weird, its not like finding that missing element is trivial in every case. Basically you said just prove it lmao

[cloud owl]

okay

sorry

[gritty widget]

lmao all good

like why wouldnt [q,p) work for q,p rationals

[uncut surge]

Aaah, I get it now. There wouldn't be a way to get something like [\pi, 5). Unions of closed sets can form open ends, but not closed ones.

And that is the precise problem, if you had a countable base, there would be at least one set of the form [x,\infty) that is not contained in the base for some real number x, but no union of half-open-intervals can give you this interval, unless one of them was equal to [x,\infty) to begin with

@gritty widget tricky stuff

Or, uh, hmm, maybe I'm lying. You could have something like [\pi,x) union [x, 5). But I think this is the right direction

regardless, one of the sets in the union has to have \pi as its lower end

it happens so often that i spam a channel with maththonk and then nobody talks to me anymore

F

[gritty widget]

@uncut surge I dont understand what you mean

[uncut surge]

AWRIGHT here we go

If you take your proposed topological basis, how would you form the set [\pi,4) by unions?

[gritty widget]

uhh decimal expansion left side\

[3.14159265358979..., 4) but I guess it will be just (pi,4)

[uncut surge]

Recall that you can only take unions, so I'm not sure how the decimal expansion will help you

Yeah, exactly

[gritty widget]

but what does it prove/discproeve

what definition are u using

[uncut surge]

the wikipedia one

[gritty widget]

ure saying that a basis = topology

?

[uncut surge]

https://en.wikipedia.org/wiki/Second-countable_space

[gritty widget]

so we should be able to produce from basis elements any open set and vice versa?

[uncut surge]

Yeah

A basis of a topological space is defined as a collection of sets so that every open set is a union of elements from this collection

[gritty widget]

how would I make [\pi,4) from uncountable base tho

[uncut surge]

I guess just [x,y) for all real x and y

[gritty widget]

I guess the problem is that we cant do all [irrational, whatever) cuz of countablility?

[uncut surge]

Yes, exactly

The countability is what fails

[gritty widget]

yeah Ig thats right

[uncut surge]

So from understanding this example, we see that if we have an interval like [x,y), then we can only get this by unions of half-open intervals A_1,A_2, ... , if one of the intervals A_i already is of the shape [x, something)

So we can only get the closed end by unions if one of the intervals has the same closed end

[gritty widget]

yeah

[uncut surge]

this is a bit awkward to formulate but i hope you can sorta follow

[gritty widget]

yeah makes sense the proof is done

thx

[uncut surge]

naisu

[gritty widget]

also, is subspace of separable space separable?

with induced topology I guess

example (R\Q)^2 as subspace of R^2

probably not true

[sleek thicket]

It's true for R^2 or any metric space because separable is equivalent to second countable for metric spaces

I doubt it in general

[uncut surge]

Yeah apparently it's not true in general

[sleek thicket]

Can you spoiler your example?

I want to think about it

[gritty widget]

uhhh the problem im supposed to do is find out whether (R\Q)^2 is separable, was just my idea if thats true in general lol

[uncut surge]

|| it's not a counterexample, take (p + pi, q + pi) where q and p go through all rationals||

[sleek thicket]

Can you take like an uncountable discrete space an attach one generic point?

[gritty widget]

oh yeah wtf obviously larto

[sleek thicket]

ty ultra, moonbears suggested it and said you'd like it

I think my idea works but I'm not sure

[uncut surge]

oh i spoiled your whole problem then

[gritty widget]

how didnt I think of that, I was trying to show its not separable lol

[uncut surge]

whoops

[sleek thicket]

So let S be an uncountably set and define X = S union {γ}. Define a topology on S by saying a set is closed iff it's a subset of S or is all of X. Then the closure of {γ} is X, so X is seperable, but the subspace topology on S is that of an uncountable discrete space, which isn't separable

Does that look good?

[uncut surge]

that's smert

[sleek thicket]

ty

[gritty widget]

ye nice one makes sense to me

also, is C[0,1] d_sup separable? should be right? Im thinking just take lines y = q q rational

or y =px

[uncut surge]

With that first set you'd only be able to approximate the subspace of constant functions I'm pretty sure

[sleek thicket]

My intuition is not separable but I'd need to think about it

[gritty widget]

oh wait yeah knowing you can approx with polyniomials you can take just all polynomials?

those are countable

with rational coef

[uncut surge]

Yeah I think that's fine

[sleek thicket]

that makes sense

[gritty widget]

big brain time

when I learned that theorem I was thinking its the most useless one in analysis

since we didnt use it anywhere in that specific analysis course

[uncut surge]

it takes some time to appreciate some theorems, yeah

"wtf this doesn't help me calculate integrals AT ALL"

[fading vale]

that just makes it better i think

[sleek thicket]

integrals are good actually

[quaint agate]

Anyone here want to talk about knot theory?

[gritty widget]

@ moonbears

[quaint agate]

Who is moonbears

Oh wait

Lmao

they’re one of my friends

I didn’t even recognize the name for a sec

[solar cape]

Okay yall

I got an absolutely bonkers question

[gritty widget]

[solar cape]

do we know about nets here

[gritty widget]

i've encountered them once or twice

[solar cape]

I have the forwards direction (phi converges) done and it took like a page and a half and a lemma or two

but the backwards direction is tougher

i cant tell if im just being dumn and not seeing the easy way but yeah

thought I'd ask hhere

[nimble cipher]

Hi. How do I get the inverse of $p$? Alternatively, do I need its inverse or is there another way of computing for the lifting? By the way, $p$ is given by $p(x,t)=(t\cos 2\pi x,t\sin 2\pi x)$.

[gentle osprey]

emphatic_wax:

[elder yew]

Don't you invert coordinate by coordinate

?

[nimble cipher]

Yeah someone told me to do that. I will try that. Somehow, my multivariable calculus left me

[rugged swan]

Hello, I'm looking for an example which satisfies the homotopy extension property, thus an example of cofibration. Could someone show me one ? I don't find any on wiki or nlab.

[rugged swan]

hi again, I found my answer to my question.
I have another one question. I was reading the homotopical definition of the mapping cone on nlab : https://ncatlab.org/nlab/show/mapping+cone.
As I've understood the definition, for me, the mapping cone of a function f : X -> Y is the unique space Z with an embedding Y -> Z making the arrow f homotopically trivial.
The nlab page claims that the suspension is the mapping cone of f : X -> *. I don't agree with this because f is already homotopically trivial here, the mapping cone of f should be the point * too. What I didn't understand ?

[nimble cipher]

Can someone help me find 3 spaces X,Y,Z, covering maps q and r such that p is not? That is, when we omit the condition that r^-1(z) is not necessarily finite.

[slender plank]

Yo can someone help me for a fats exercise

So i send you a sum and you need to show that it implies that the polygone is regular

https://cdn.discordapp.com/attachments/576511214025310260/779090249530867712/20201119_220617.jpg

[nimble cipher]

Yeah I'm trying to do something with the circle

[slender plank]

So w is the number above ei2π/n

And we need to show that if the sum is true then the polygone is regular

Do anyone as an idea how to do it ? Apparently you have to use the general case of triangular equality

[nimble cipher]

So I remove one property in that so it does not become a covering map

?

that's sad haha

Okay. I will read through this paper. Algebraic topology is fun

Ohhh will also look into that

and find that stackexchange so I can have my peace too

I have seen the answer from the paper you linked. I think that works

[slender plank]

<@&286206848099549185>

[sweet wing]

how do you do a reu on AT without reading hatcher or a equivalent

[fading vale]

so the basic concept here right is like

(for context this is cohomology and psi is an element of H^1(X, G) where X is a 2 dimensional delta complex)

in the case G = Z2 if we pick any triangle either two of its faces are mapped to 1 or none of its faces are mapped to 1 by psi if psi is trivial under delta

so then we can like i guess mark all of the edges which get mapped to 1

and those all have 2 adjacent edges (meaning edges sharing a triangle) that are also marked

and then we draw a curve transversing all the edges and not any of the unmarked ones?

[summer jolt]

Hi I have to show that the commutator of lie derivatives and the lie bracket of vector fields are the same when acting on a (1,1) tensor.

I managed to show this

where tao is a (1,1) tensor but not really sure how to proceed next.

[gentle osprey]

slimvesus:

[sleek thicket]

for a vector field this is the jacobi identity, right?

(assuming you know the lie derivative and lie bracket coincide for vector fields)

I think we can say $(\mathscr{L}_X \omega)(V) = X(\omega(V)) - \omega(\mathscr{L}_X(V))$, so $(\mathscr{L}_X \omega)(\mathscr{L}_Y \omega))(V) = X((\mathscr{L}_Y \omega)(V)) - (\mathscr{L}_Y \omega)(\mathscr{L}_X(V)) = X(Y(\omega(V))) - X(\omega(\mathscr{L}_Y(V))) - Y(\omega(\mathscr{L}_X(V))) + \omega(\mathscr{L}_Y(\mathscr{L}_X(V)))$

[gentle osprey]

shamrock:

[sleek thicket]

then $(\mathscr{L}_X \omega)(\mathscr{L}_Y \omega))(V) - (\mathscr{L}_Y \omega)(\mathscr{L}_X \omega))(V) = X, Y + \omega(\mathscr{L}_Y(\mathscr{L}_X(V)) - \mathscr{L}_X(\mathscr{L}_Y(V)))$

[gentle osprey]

shamrock:

[sleek thicket]

hmm i dont see it

oh wait lol

that last bit is $(\mathscr{L}{[X, Y]}\omega)(V)$ by the first computation I did and the fact that $\mathscr{L}{[X,Y]}(V) = \mathscr{L}{X}(\mathscr{L}{Y}(V)) - \mathscr{L}{Y}(\mathscr{L}{X}(V))$

[gentle osprey]

shamrock:

[sleek thicket]

so this shows it for 1 forms, just by how the lie derivative of a contraction works

then to check it on $(1,1)$-forms it suffices to check it on simple tensors $f V \otimes \omega = f \otimes V \otimes \omega$

[gentle osprey]

shamrock:

[sleek thicket]

also note that it holds for functions because the lie derivative of a function is just applying the vector field to that function

so now assume $\mathscr{L}_[X, Y] T = \mathscr{L}_X \mathscr{L}_Y T - \mathscr{L}_Y \mathscr{L}X T$ and $\mathscr{L}[X, Y] S = \mathscr{L}_X \mathscr{L}Y S - \mathscr{L}Y \mathscr{L}X S$ for tensors $T, S$. Then $\mathscr{L}{[X, Y]}(T \otimes S) = (\mathscr{L}{[X,Y]} T) \otimes S + T \otimes \mathscr{L}{[X, Y]}(S) = (\mathscr{L}_X \mathscr{L}_Y T) \otimes S - (\mathscr{L}_Y \mathscr{L}_X T) \otimes S + T \otimes (\mathscr{L}_X \mathscr{L}_Y S) - T \otimes (\mathscr{L}_Y \mathscr{L}_X S)$

[gentle osprey]

shamrock:

[gritty widget]

[sleek thicket]

on the other hand $\mathscr{L}_X (\mathscr{L}_Y (T \otimes S)) = \mathscr{L}_X(\mathscr{L}_Y T \otimes S + T \otimes \mathscr{L}_Y S) = (\mathscr{L}_X (\mathscr{L}_Y T)) \otimes S + (\mathscr{L}_Y T) \otimes (\mathscr{L}_X S) + (\mathscr{L}_X T) \otimes (\mathscr{L}_Y S) + T \otimes (\mathscr{L}_X (\mathscr{L}_Y S))$

[gentle osprey]

shamrock:

[sleek thicket]

from this we see $\mathscr{L}_X (\mathscr{L}_Y (T \otimes S)) - \mathscr{L}_Y (\mathscr{L}_X (T \otimes S)) = (\mathscr{L}_X (\mathscr{L}_Y T)) \otimes S + T \otimes (\mathscr{L}_X (\mathscr{L}_Y S)) - (\mathscr{L}_Y (\mathscr{L}_X T)) \otimes S - T \otimes (\mathscr{L}_Y (\mathscr{L}_X S))$

[gentle osprey]

shamrock:

[sleek thicket]

and that's what we had up above

so this bit was proving that your property is true for two tensors it's true for their tensor product, it only used the product rule for the lie derivative

it holds by definition on 0-tensors (functions)

it holds by the jacobi identity on vectors

and you can extend it to covectors via the identity $\mathscr{L}_X (\omega(V)) = (\mathscr{L}_X \omega)(V) + \omega(\mathscr{L}_X V)$

[gentle osprey]

shamrock:

[sleek thicket]

any tensor of arbitrary rank is a sum of tensor products of (0,0)-tensors, (0, 1)-tensors, and (1,0)-tensors, and the lie derivative is additive, so this establishes it for tensors of arbitrary rank

@summer jolt

anyone in this thread smoke weed

[elder yew]

no, but I love to fuck with people that are high

[chrome dew]

'grassman'ian 😎

Moichizuki is so straightedge

[sleek thicket]

i just felt awkward dumping tensor bullshit and then complete silence

[gritty widget]

lol

just never deal with mixed / contravariant tensors

join the covariant-only club

cartan only holds for lie derivatives of forms tho right?

[sleek thicket]

I didn't use cartan?

lol

i think i had this as an exercise in my smooth manifolds course

like

it's not really so bad

right

then you use the formula $\mathscr{L}_X (\omega(V)) = (\mathscr{L}_X \omega)(V) + \omega(\mathscr{L}_X(V))$

[gentle osprey]

shamrock:

[sleek thicket]

to get the case of a 1 form

then you argue that it's preserved under tensor products

both of those last two things are very very easy

just lots of symbols

you just do the product rule or w/e when you need

it's really not so bad, you just follow your nose

i mean cartan only does the 1 form case

which is not so bad

it is easy lol

people danning are wrong

this isn't DG

@gritty widget I don't really see how you do (1,1)-tensors without doing (p,q)-tensors

like

because there's no metric

this is diff top

[gritty widget]

well

now this is just being pedantic

[sleek thicket]

this is all just about smooth structures on manifolds!

no jommetry

not quite slim

it's a sum of those

and you mean \otimes lol

yeah, $\sum_{i=1}^\infty f X \otimes \omega$

right

so... show that it's preserved under tensor products?

[gentle osprey]

shamrock:

[sleek thicket]

yeah exactly

that gives you the (p,q)-case lol

yes

well you have three terms here anyways

$f \otimes X \otimes \omega$

[gentle osprey]

shamrock:

[sleek thicket]

like you kind of need to induct, but I think if you're doing this level of stuff it's okay to say "it holds for tensor products of 2 things so it holds for tensor products of n things"

yeah?

yeah so like

if you look at my proof again

there's two sections

(1) establish it for 1 forms

(2) do it for tensor products

yup

it's an easy problem

just like

lots of symbols

exactly

there is no cleverness needed

I think cartan will actually be annoying

versus the formula I used

?

$\mathscr{L}_X(\omega(Y)) = (\mathscr{L}_X \omega)(Y) + \omega(\mathscr{L}_X Y)$

i mean which formula are you asking about

there's like 3 formulas that are relevant here

the one I posted is just a product rule but for applying a covector to a vector

[gentle osprey]

shamrock:

[sleek thicket]

then there's a product rule for the tensor product

the statement we were trying to prove overall is that $[\mathscr{L}_X, \mathscr{L}Y] = \mathscr{L}{[X, Y]}$

I don't really get your question

like

the lie derivative is a thing that we care about

this is saying it satisfies a good property

i.e. it's a derivation

Pretty much

[gentle osprey]

shamrock:

[sleek thicket]

that sending a vector field to its lie derivative is a lie algebra representation

a lie derivative takes (p,q)-tensors to (p,q)-tensors

you look at how the tensor changes as you flow it along a vector field

[elder yew]

^_^

[gritty widget]

"fisherman derivative"

my RG prof called it that

[sleek thicket]

awww

i like that

[elder yew]

If I recall correctly, lie derivatives are the same thing as lie brackets

If things are nice enough

[sleek thicket]

@elder yew for vector fields

right

[gritty widget]

u might need C^\infty for that moonbears but also all manifolds are C^\infty

[sleek thicket]

it's like a directional derivative

but we need more structure

than just a single vector

we need an entire vector field so we get a flow

[elder yew]

can you define flow?

You're in jack's RG right?

[sleek thicket]

yup

[gritty widget]

[sleek thicket]

I can define flow or I can try to give you a nice picture

and I think the picture is easier first

so like

[gritty widget]

define it and make sure to avoid all the domain issues too

[sleek thicket]

say you have a vector field

that's an integral curve of X

the flow bundles up all the integral curves together

[elder yew]

Looks like someone wants to do research in RG

[sleek thicket]

lol me?

I changed my name to this because of a tweeet where I said I hate RG

I legit don't think I want to continue with it after this class, it has been very annoying lol

but geometrically flow is like, take a point and follow the path of the vector field

yup

[elder yew]

(Peterson > Lee)

[sleek thicket]

yee, i get to see his unreleased bundles book next quarter

hype

[elder yew]

That's awesome though

[sleek thicket]

Yeah he's great!

[elder yew]

I liked RG mainly because I liked covariant derivatives so much

and thinking about manifolds

[sleek thicket]

also more cynically it'll be nice to have him to write me a letter of rec

manifolds are good

but I like manifolds without metric more 😛

[elder yew]

I don't think that's cynical to want good letters

From reputable mathematicians

I have relatively uknown mathematicians as my letter writers

so I'm a little concerned

[sleek thicket]

sheafs+manifolds are good

@elder yew I'm definitely aware of like, letters of rec being important and am trying to optimize for them

but otoh, it can feel a little grimy

[elder yew]

Do you have any research project?

Like do you think you can get something with jack in the spring?

[sleek thicket]

yeah I did that knot theory reu this summer

[elder yew]

Like a reading course

[sleek thicket]

am currently doing this weird bad physicsy project, it might turn into weird good deformation quantization stuff next quarter

jack is retiring this year 😦

[elder yew]

Oh what a shame

[sleek thicket]

I don't think he wants to do a reading course or research

but he will have had me for two years in small grad courses, and I made myself known

like after 1 quarter with him he was willing to write me my reu letter

[elder yew]

That's great to hear!

Man UW sounds like they have caring profs lol

[sleek thicket]

I've liked them a lot

[elder yew]

I'm going to a talk by Wilson on Monday

he's presenting at my research mentors seminar

[sleek thicket]

oh nice

He was willing to let me skip into the ug measure theory course freshman year, even though the honors analysis sequence I was taking didn't technically fit the prereqs

but I had a time conflict

[elder yew]

Yeah. He was going to let me into his topics PDE course but it cost too much

[sleek thicket]

oh that sucks :/

[elder yew]

It would've been 4k up front

whereas if I wait another year a course would be closer to 1.5k

[sleek thicket]

well if you end up taking a pde course next year I might see you! depending on whether I recover from my anti-RG sentiment

[elder yew]

It depends if I get into UW or not. I think I have a good chance

[sleek thicket]

I hope it works out for you

lol flows

[elder yew]

My letters and research are strong, my gpa is a little poopy

[sleek thicket]

honestly like

i hated ODEs so much

thought they were boring and poopy

but flows are so good

very cool

[elder yew]

I hated PDEs so much at first, and fourier

[sleek thicket]

also geodesics and parallel transport and stuff

[elder yew]

Cuz I thought it was just plug n chug

But now I think the methods are very good

[sleek thicket]

I haven't really done anything with them yet, except for like figuring out some ad hoc stuff on manifolds hw when it was necessary

but they seem interesting

and useful

[elder yew]

If I get in shammy, can you share toro's exams w/ me?

[sleek thicket]

yeah sure

[elder yew]

I wanna prepare for the quals

[sleek thicket]

I can send you the one we took last friday rn

[elder yew]

Sometimes they take exam questions from the prof that taught it prior

oh sick

I'll tuck that away

Will they let you take quals?

[sleek thicket]

ehh

chmonkey wanted to this year and they didn't let him

I don't really want to, it seems unpleasant

[elder yew]

The first problem is exactly a problem in stein and shakarchi

[sleek thicket]

i found it in there during the test lol

[elder yew]

I did that problem and made sure I had it down for my class. Thought it would be a great test question

Didn't show up

[sleek thicket]

I wish I had done it before the test, it was an optional problem :/

there are like 1-3 of those after every lecture

i mean yes

it is literally just solving an ODE

huh

I guess the picture just helps me a lot more than the formula

like

I understand how they relate and all

and I'm gonna use the definition of integral curve if I want to prove anything

Yeah, ik what you mean

ugh this reminds me I gotta do riemannian geo homework

I'm supposed to compute the sectional curvatures of $SU(2)$ under a family of metrics

this convo made me realize the approach I had thought of doesn't work lol

oh word?

[gentle osprey]

shamrock:

[sleek thicket]

is there a nice way to do it im missing?

lol

the issue is that like

there's a nice formula in the bi invariant case

this was like a hw problem

for me

but the metrics I have are only left invariant

so like, covariant derivatives aren't nice

r i p

mood

I'm not sure I agree

like, what other topics?

maybe I don't understand what you mean by accessible. I'm finding RG less hard than AG and slightly more hard than AT

like i'm at the point where most of the stuff I'm doing is like, obtuse and hard lol

yeah that's a mood

idk I felt like that with algebra a lot

(and then I woulld eventually get used to it and develop intuition)

I mean dummit & foote and basic comm alg

I'm thinking of the algebra course I took last year

like when I first learned group theory I found it really unintuitive

and when I first learned rep theory I couldn't figure out how it all fit together

because im bad at physics and don't find it interesting

anyways slim diff geo is really annoying

i agree

i didn't mean to be like, combatitive

I am the Riemannian Geometry Lover, so

I am cursed to be with it

despite my wishes

Yes

okay goodbye I'm gonna play magic the gathering with some friends

Good luck managing the pain of curvature :)

[elder yew]

I need magic to gather with my friends

Cuz I need to summon them from hell

[coarse kestrel]

So every complete metric space is the continuous image of the cantor set

Does this mean the cardinality of metric spaces is always not greater than the cardinality of R?

yeah the cardinality of complete metric spaces

so that's why it doesn't quite make sense to me

by a continuous image of a cantor set, the continuous function should be surjective right?

so that means the cardinality of the cantor set is always greater than or equal to that of any complete metric space

[little hemlock]

cantor set is compact tho

and R is complete but not compact

i.e. there is something fishy about that theorem.

the only theorem online i see about this is that every compact metric space is the continuous image of the cantor set. maybe u mixed up completeness and compactness?

[coarse kestrel]

I meant compact

Oops

[tight agate]

Anyone know a good reference on spectra?

topology spectra

I have been looking at some of Lurie's books

I was curious if there was a more 'classical' reference

in the orginal setting

sans quasicategories

unless the originial setting involved that

orginial*

ffs

original

nah I just wanted to see how they first came up

just as representing objects of cohomology theories I assume?

that sounds reasonable

http://www.math.uchicago.edu/~may/PAPERS/history.pdf

that article says that spectra were first introduced in the context of spanier-whitehead duality

ahahahaha

[summer jolt]

Thanks a lot

[gentle osprey]

weilam06:

[nimble flower]

if i have a cts map, t -> f_t, would the map t -> f_t^-1 be cts as well?

suppose the f_t are homeomorphisms

[gritty widget]

what's the topology on your function space

[nimble flower]

nvm actually - this isnt a fruitful direction

thanks though

[fading vale]

maybe this is a silly question but im a little confused by this

i get that the snake lemma gives us Z_n-1 -> B_n-1 -> ker delta -> Z_n -> B_n

well

the duals of the B and Z but like

yea

so is a composition with the quotient from ker delta to H^n(C; G) just like

implicit here

and this is actually Z_n-1 -> B_n-1 -> ker delta -> H^n(C; G) -> Z_n -> B_n

or am i misunderstanding something

🤔 i guess technically this is just hom alg but its from hatcher so wutever

wait this doesnt make sense lol

its that the maps from ker delta_n into Z_n vanish on im delta n+1 probably right?

so we can pass to the quotient?

[gritty widget]

@gritty widget

i saw that

ah, you caught me

[exotic root]

If X is an affine variety in A^n, is X open or closed in the projective closure? Is X open or closed in P^n?

Like if phi is the identification of A^n with D(x_0) in P^n then the Zariski closure of phi(X) in P^n should be the projective closure

But im not seeing where the stuff are open or closed

[tight agate]

Well if it is closed then it is it's own closure

So it is affine and projective

If you're working over a field

then global sections is a finite k algebra

which implies that the variety is a point/points

as it is affine

[exotic root]

But does that imply that X is open?

[tight agate]

If it is not a point/points

[exotic root]

X is X^pclos \cap D_+(x_0) in P^n

[tight agate]

yup

[exotic root]

So in its own closure X should be closed no?

Open

[tight agate]

?

You're taking the closure in Pn

[exotic root]

Yeah

[tight agate]

not in D(x0)

and by the argument above, if X is closed in it's projective closure then it is a collection of points

Oh you're asking if it's open

yeah it is

in its projective closure

[exotic root]

In both its closure and in P^n?

[tight agate]

in it's projective closure

wether it's open or not in Pn depends on X

If you take X = An, and embed X as D(x0), then it is open in Pn

[exotic root]

https://math.stackexchange.com/questions/496670/algebraic-geometric-interpretation-of-the-projective-closure-of-an-affine-variet?rq=1

They say Y can be identified with an open set in P^n

Thats whats confusing me I think

[tight agate]

It can be identified with an open subset of a projective variety

I don't think it can be identified with an open subset of projective space

Take a single point

(spec k)

[exotic root]

Hmm

[tight agate]

it isnt an open subset of Pn due to dimension reasons

[exotic root]

Opens in projective have high dimension no?

[tight agate]

same dimension as the whole space

[exotic root]

Being dense

[tight agate]

and dim Pn = n > 0

[exotic root]

Right

[tight agate]

but spec k is 0 dim

But yeah I assume they wanted to say it can be identified with an open subset of a projective variety

not projective space

[exotic root]

I see

So they meant open in its own closure

Thanks for clearing the confusion!

[fading vale]

hm

o

[gritty widget]

yo so I'm not sure about one thing: given lexicogrpahic square I want to find the interior and closure of (0,1) x {0}. so interior is empty and the closure should be (0,1] x {0} sum [0,1) x {1}. Not sure how to argue why (1,1) is in the closure. Would saying that (1,1) isn't a successor of any element be enough, so when taking a nbhd of that point we can't go just from the top without crossing any points on (0,1)x{0}?

not sure how to show it formally

[feral copper]

Hi (again) ! I asked a question about a week ago on MSE, and it didn't find any answers, so I thought I may ask it here ! It's about computing a homotopy group of a wedge of CW spaces : https://math.stackexchange.com/q/3907570/259363

[exotic root]

What am I missing? Why does it follow from the orange that Z x_B A -> Z is closed?

[cunning coral]

Just a quick sort of really informal concept check I need.

So in the hyperbolic plane, if you have a line m that contains the limiting parallel to a line l, m and l cannot share a common perpendicular. And I'm trying to understand a basis of why. My though process is that since m is asymptotic to l, then the distances between l and m is always increasing as you slide along a given direction of l. So you can't find two perpendiculars to l of equal length, so you cannot construct a perpendicular to both l and m.

Better way to say it, you can't drop equal length perpendiculars from m to l, so there is no common perpendicular to them l and m

[woeful oasis]

It is easy to picture in the poincare disc model

[cunning coral]

Poincare is the one with diameters and arcs of circles yeah?

I don't remember the models names

[woeful oasis]

The plane is good too, I just think sometimes the ball is easier to look at because it's compact

and you have just circles to worry about, instead of some circles, some lines

[cunning coral]

But yeah, I understand the visualization, I just wanted to make sure my word explanation made sense

[woeful oasis]

i don't understand "contains limiting parallel". You mean they're lines, but asymptotic?

[cunning coral]

Since they are asymptotic you can't construct the situation you need to get to to construct a common perpendicular

Yeah, in the hyperbolic plane you end up with some angles allowing the rays from a given point to be parallel, and other angles beyond that won't be parallel

[woeful oasis]

If two circles are tangent to one another and have common perpendicular, then they'd have the same center and hence be the same

[cunning coral]

And the limiting parallel is the one such that any angles beyond that will cross, so they have to be asymptotic

[sleek thicket]

shomework (sham homework) time

[fading vale]

👀

[sleek thicket]

Let $G$ be a lie group with a bi invariant metric $g$

[gentle osprey]

shamrock:

[sleek thicket]

I need to show $\mathrm{sec}(X, Y) = \frac{1}{4}|[X,Y]|^2$

[gentle osprey]

shamrock:

[sleek thicket]

When $X, Y$ are orthonormal left invariant vector fields

[gentle osprey]

shamrock:

[sleek thicket]

I know from a previous exercise that $R(X, Y)Z = \frac{1}{4} [Z, [X, Y]]$ for any left invariant vector fields X, Y, Z

[gentle osprey]

shamrock:

[sleek thicket]

Since $X, Y$ are orthonormal, $\mathrm{sec}(X, Y) = Rm(X, Y, Y, X) = \langle R(X, Y)Y, X\rangle = \frac{1}{4} \langle [Y, [X, Y]], X\rangle$

[gentle osprey]

shamrock:

[sleek thicket]

so for some reason we have $\langle [Y, [X, Y]], X\rangle = \langle [X, Y], [X, Y] \rangle$

[gentle osprey]

shamrock:

[quaint agate]

Hey Shamrock

[sleek thicket]

hello

[quaint agate]

I’m trying to find someone to talk about knot theory with

xd

[sleek thicket]

sorry, I'm not a huge fan of knot theory. I did a project on it this summer and came out not super interested

[quaint agate]

Ah

ye