point-set-topology

[orchid forge]

then the result is this guy

[dusk heron]

I agree that the two spaces in your first picture are homeomorphic. I guess the second space in your first picture is homotopy equivalent to the space in your second picture. But why, exactly?

[orchid forge]

you are just contracting this red arc here

[dusk heron]

Yes, but you have to contract the rest of the space with it in a nice way. I'm not sure exactly how to do this at the precise moment where the two points merge into one

[orchid forge]

i'm not sure exactly what the issue is

but it would maybe help to be more formal about it

you can apply van Kampen's theorem here, X = U \cup V where U is the sphere and V is the circle consisting of the line segment & the red arc i've drawn

or maybe just to help with intuition, any time you have a loop in this space it must pass through the line segment (since X \ the line is simply connected), but any loop through the line segment can be made to start and end at a given point on the sphere

for example, by moving it onto the red arc above

so contracting the red arc onto that point doesn't add or remove any loops

[dusk heron]

Yeah I agree that one can see it with van Kampen's theorem. I just want to be able to see why the following holds:

Like, what does the homotopy equivalence (the mappings between them) look like?

[orchid forge]

you can leave the entire sphere unchanged outside of a neighborhood of the circle

consisting of the line segment + the red arc

i'm not sure of a better way to describe it except just by sliding one point onto the other

you can keep track of any given point under this transformation

the map is a bijection except for at the red points, because it maps the entire red path to a point

i'm not sure what you're looking for, is it more formality? or what

[dusk heron]

This kind of explanation is exactly what I was looking for, thanks!

[orchid forge]

oh ok

[dusk heron]

And the "homotopy inverse" is the map which is the identity on the "surface" part, and maps the loop to the loop consisting of the segment between the two points together with the red line, I guess

Okay yeah, I'm convinced

[orchid forge]

it might be worthwhile to examine the proof that contracting a contractible subspace A in X is a homotopy equivalence when (X,A) is a CW pair

[dusk heron]

You mean that the quotient map $X\to X/A$ is a homotopy equivalence?

[gentle osprey]

gustavn64

[orchid forge]

yes

[dusk heron]

Ah yeah. I haven't really worked through any algebraic topology text, I've just picked up bits that I need here and there, but maybe I should take the time to do that

(To learn about small lemmas and stuff like that)

[plain raven]

unrelated but whenever somebody talks about important properties of CW pairs i just start getting angry at how hatcher introduces the concept of cofibration in chapter 2 but under a different name ("good pair", presumably because he just doesn't want to frighten readers with terminology) and under a nonstandard (but equivalent) definition

it was like another 6 months before i realized that good pairs and cofibrations were the same thing

i'm never going to let this go.

[obtuse meteor]

wait wtf

I'm also in this boat

tfw I just realized a good pair is a cofibration

malding that my algtop professor didn't say this in class

[plain raven]

yeah i'm looking it up and Hatcher defines a good pair is a "nonempty closed subspace A in X which is a deformation retract of a neighborhood in X"
This is basically equivalent to a cofibration. More precisely, if (X,A) is a good pair then the inclusion of A in X is a cofibration, and a cofibration is always a subspace inclusion whose image is the deformation retract of a neighborhood; if we work in a category of good topological spaces (compactly generated and weakly hausdorff) then a cofibration is closed

[shy moss]

if i have an affine variety how can i find its irreducible components?

[plain raven]

i don't know much about this subject but i think that this is more or less equivalent to primary ideal decomposition.

At the risk of saying something stupid

faye's reaction has given me validation, i plan to repeat this complaint in 1.8 months and possibly cause the same reaction in another person who stumbles into this channel at the time

[finite heath]

I am trying to prove that the interior of the closed ball in R^2 is itself

i defined an element (a, b) in closed ball

then there exists an open ball in the closed ball s.t.

(a, b) in open ball in closed ball

Now im stuck =p

[gentle lark]

Grab any interior point, then choose a ball centered at that point small enough for the radius of the interior point + the radius of the new ball to be smaller than the radius of the original ball. Then apply triangle inequality (I think?) to prove that ball works.

btw what's the etiquette here re: bumping questions?

[orchid forge]

what about the points on the boundary?

[finite heath]

Oh, I apologize

I meant to say

that the interior of the closed ball in R^2 is the open ball

My instruction to show this: "Give complete arguments based
on definitions, properties and characterization theorems"

My reasoning: by def. of interior, int(B closed) is the union of all open sets in B closed -- in the standard topology on R^2, open sets are open balls, hence the union of open sets of B closed is a union of open balls, which together form an open ball themselves ==> int(B closed) = B

@orchid forge am I way off?

[orchid forge]

open sets are unions of open balls, not necessarily open balls themselves

your first argument seemed to be more promising, a point is in the interior if you can fit an open ball around it, which is contained in the closed ball

[gritty widget]

would euclidean geometry go in here or one of the early university channels?

[tough imp]

Early university

Or no

Even the pre university

[gritty widget]

woops no i meant elementary diff geometry

[tough imp]

#geometry-and-trigonometry

Oh

[orchid forge]

[gritty widget]

here or #multivariable-calculus

[tough imp]

Kind of depends on

If you have the word manifold defined or not

[gritty widget]

Yeah, I wish we had manifolds n stuff but right now its Pressley's book, which doesn't really go that deep into the manifolds or riemannian stuff

so would that be multivar-calc then?

just ask it here no ones gonna get mad at you

[tough imp]

Yeh

[gritty widget]

posting in #multivariable-calculus will get it buried anyways

[tough imp]

Just don’t ask no SAS triangle questions

[gritty widget]

reminder

lmfao no I'm just trying to prove some stuff to do with curvature and I feel like I'm on uneven footing, let me get this cleaned up for looking over

[gentle osprey]

Manzareh

[orchid forge]

have you tried the pre university channel?

[tough imp]

[orchid forge]

(jk)

[gritty widget]

I feel like I'm skipping some steps here, or I goofed up since my final statement seems kinda messy I guess

[orchid forge]

I'm not sure what we're trying to show with all of this

reversing the orientation of a curve can only really do one of two things to a quantity involving the tangent vector right?

like intuitively it's either gonna fix it or reverse it

[gritty widget]

Right, but how would I go about showing that? I have a theorem that states that I know for certain that two curves with the same signed curvature are equivalent to isometry, which is similar to what you're saying if i'm catching your drift

which is to say if i really poke and prod it, it should look like the two are different by either a rotation or a translation, right?

[orchid forge]

they would be exactly the same curve

they're just different parametrizations

[gritty widget]

so the curvature would be the same, but the sign of the tangent and the sign of the normal would have to agree with that?

[orchid forge]

the curvature at each point, being intrinsic to the curve, would be the same

but the signed curvature depends on the orientation

so you have to think about it for a sec

[gritty widget]

so then they'd really only differ by a minus sign?

if they were different in anyway

cause kappa is a scalar quantity, so it really wouldn't have much flexibility either way, right? Am I understanding this?

[orchid forge]

that's what i'm saying, the conclusion should be either that the curvature is the same or negative

it should just be a computation

Like you did in the beginning

if $u = \pm s + c$, then $\frac{d^2}{ds^2} \gamma(u) = \gamma''(u) \cdot (u')^2 + \gamma'(u) \cdot (u'')$

[gentle osprey]

Kogasa

[orchid forge]

now $\gamma''(u)$ is just the scalar curvature at $u$, $u' = \pm 1$ so $(u')^2 = 1$, and $u'' = 0$

[gentle osprey]

Kogasa

[orchid forge]

so this is just $\kappa(u) = \kappa(\pm s + c)$

[gentle osprey]

Kogasa

[orchid forge]

or, intuitively, since the curvature tells you which way the tangent vector is turning: reversing the orientation means that you're now turning in the opposite direction, but also the tangent vector is pointing backwards

[gritty widget]

okay I think I got it now, thank you!

[orchid forge]

wait wait wait

i think i misread the problem

the curve $\gamma$ is supposed to be the signed curvature of some other curve $\tilde \gamma$?

[gentle osprey]

Kogasa

[orchid forge]

and you're trying to figure out how the curvature of $\tilde \gamma$ relates to the curvature of $\gamma$?

[gentle osprey]

Kogasa

[plain raven]

heading to the library right now to get a bunch of books on applications of spectral sequences in differential geometry

[orchid forge]

can we just double check that the problem is correctly stated

should it be, something like, "How does the scalar curvature of $\hat \gamma$ relate to $\gamma$" rather than "relate to that of $\gamma$"?

[gentle osprey]

Kogasa

[orchid forge]

because if not then my answer shouldn't have made any sense

[brittle lichen]

hi I am reading algebraic topology a first course and encounter some questions in the proof, why we need to define a new function $g$ and use another path to prove $g_y=q$ and thus $f_y=g_y=q$, but not using f and the original path directly?

[gentle osprey]

siv

[brittle lichen]

and here is the proof of proposition 1.8 (see (i)=>(iii)):

[lunar yoke]

I think this is just a choice by the author, you could show it in a different way

so since $\omega = pdx + qdy$, the definition of $f$ coincides with $f(x,y) = \int_{y_0}^y q(x_0,t) dt + \int_{x_0}^x p(t,y) dt$

[gentle osprey]

Phil P

[lunar yoke]

you could now use differentiation under the integral sign to show $f_y = q$

[gentle osprey]

Phil P

[lunar yoke]

instead the authoer chooses to use greens theorem in combination with a different path

[brittle lichen]

oh I thought there is a special reason that I cannot use the same path

thank you

[gritty widget]

Could anyone help me working out the following example?

The absurd is clear, the construcion of the exterior is not

[gentle osprey]

c squared

[rancid umbra]

does anybody have any tips/hints on how to show this result?

[gritty widget]

restrict to preimage of some such interval and apply extreme value theorem

[rancid umbra]

what is "some such interval" referring to?

[gritty widget]

(-\infty, c]

[rancid umbra]

uh. do you have a way to find a good interval (-\infty,c] to restrict to?

oh wait

[gritty widget]

preimage of

[rancid umbra]

yea i think just restricting it to a non-empty preimage of (-\infty, c] works

thought it was going to be more complicated than that, so i tossed that idea out first

[gritty widget]

oh yeah nonempty oops

[rancid umbra]

guaranteed at least one such preimage since X is non-empty

[gritty widget]

right

[fallow venture]

What would be the complementary of a cartesian product of 2 intervalls of R ?

Is there any way to use it in order to prove that two closed intervals in a cross product is also a close set ?

I know about unions of open sets, but cartesian products ... ?

[gritty widget]

Let me DM you a copy of the problem statement, hold on

[orchid forge]

ok so my solution was fine

we were just looking at how the curvature changes if we change between unit speed parametrizations

[bleak path]

Hi, got a couple of basic topology questions that I'm not 100% sure about

My course currently is dealing with continuity in the most basic sense, not even to the stage of homeomorphisms but past what you might expect from an analysis class, just for context on my understanding right now

One question I have is: is it possible to have a continuous surjection from [a, b] U [c, d] to [x, y]? What about the reverse

I am not sure about this at all, on one hand I can loosely imagine a function mapping a1 -> x1, c1 -> x2, a2 -> x3, ... . But I am not sure if this is a valid mapping

[gritty widget]

yep! I just wanted to be sure I wasn't missing anything, thank you again!!

[gritty widget]

first try proving that, for any two intervals [a, b] and [c, d] in R, there's a surjective continuous map [a, b] -> [c, d] (assuming a < b and c < d to get rid of trivial cases). try to imagine translating and scaling one interval to the other, and use this to come up with a good map.

for your question: if the intervals are disjoint then you just put your two maps together into a piecewise function, and if they're not, you got another such interval and you can apply the result directly. the converse won't be true if [a, b] and [c, d] are disjoint, since then their union is a "disconnected space," but as you'll see later, the continuous image of a connected space is connected (and [x, y] is connected). intuitively you won't be able to map [x, y] onto [a, b] u [c, d] without "cutting" [x, y] apart if the intervals [a, b] and [c, d] are disjoint

[bleak path]

Thanks for the tip on the first bit.

Re: the second bit; is the example of a mapping I've given sufficient for a piecewise mapping? Or is there a bit of logic I've missed. Dealing with how infinite the reals are is not something I'm completely familiar with yet so I'm not 100% sure if I'm right there. I'm a little worried about what happens at the border specifically, for example if the example sets were to be [0, 1] U [2, 3] to [4, 6], and we mapped [0, 1] to [4, 4.99999...), is that mapping valid? Because the second set is half open, and if it were to be closed it feels like there would be "missing" numbers (specifically those between 4.999.... and 5)

Intuitively it feels that I can't map [x, y] onto [a, b] U [c, d], and this was my gut feel as well. Is it correct to say that this mapping is not continuous because if for [x, x'], a subset of [x, y] and the mapping [x, x'] -> [a, b], there is a point close to x' but is not close to [a, b] and hence it is not a continuous function?

[fading vale]

What map [0, 1] -> [4, 5) are you thinking of exactly

[ivory dragon]

big if true

[tough imp]

Todd is a known troll. I calculated 5 -4.99… and it is 0.00…01

As everyone knows 5 = 4.99… is equivalent to 5 - 4.99… = 0 but 0.00…01 ≠ 0

[orchid forge]

That's not very Hausdorff of you

I feel attacked by this react

What radius ball separates 0 from 0.000...1

[gritty widget]

alright, i'm back once more with something I've done with torsion if no-one is doing something currently, I think I messed up though since the resulting equation seems too clean this time.

[hollow harbor]

Todd doesn't know about topologically indistinguishable points... 😩

[tough imp]

What the fuck is Hausdorff, my space is irreducible

[gentle osprey]

Manzareh

[swift fjord]

0.000....05 duh

[bleak path]

Not 100% sure, just napkin math and trying to work it out.

Thank you all for your help

[lunar yoke]

Consider the torus $\mathbb{R}^2/\mathbb{Z}^2$ and the loops $\alpha_1(t) = [(t,0)],\ \alpha_2(t) = [(0,t)],\ \alpha_3(t) = [(t,t)]$. For each $i=1,2,3$ I want to find a 2-fold covering of the torus so that the lift of the loop $\alpha_i$ is again a loop, whereas the lifts of the other two loops are not.

[gentle osprey]

Phil P

[lunar yoke]

By essentially guessing and checking I got the coverings for $i=1,2$ (total spaces $\mathbb{R}^2/\mathbb{Z} \times 2\mathbb{Z}$ and $\mathbb{R}^2/2\mathbb{Z}\times\mathbb{Z}$ with the obvious maps), but im stuck at $i=3$.

[gentle osprey]

Phil P

[lunar yoke]

There's gotta be some clever way to go about this. Anybody got any ideas?

[kind cedar]

What does it mean "As the zero set of finitely many polynomials, R is a closed subset of S ×S" ??? I don't see the relation.

[gentle osprey]

Phil P

Phil P

[gentle lark]

I find that justification a bit weird

As in, I'm not sure why it says "finitely many", because I think infinitely many would also mean R is a closed subset and, at any rate, it's just the one polynomial

Mabye there's something I'm misunderstanding

[lunar yoke]

what do you mean by "the one polynomial"?

you get one for each minor

[gentle lark]

y-tx

[lunar yoke]

???

[gentle lark]

I may be misunderstanding the definition of R there

[lunar yoke]

oh thats in the definition of R

no thats not the polynomial the author talks about

[gentle lark]

Ohhh the t is not fixed

Reading what the proposition is trying to prove would've probably helped me realize that. Sorry. 😅

[gentle lark]

Not sure what a minor is but I still don't get why you need them to be finitely many

[lunar yoke]

the whole proof relies on minors

a 2x2-minor is the determinant of a sub-2x2-matrix of the original matrix

[gentle lark]

Ahh ok ok

[lunar yoke]

vanishing of all 2x2 minors is equivalent to the two columns being linearnly dependent

[gentle lark]

But like, to justify why the intersection of the zero sets is closed, you don't really need to say there are finitely many of them, right?

It's easy to prove they're all closed, therefore the intersection is closed.

[lunar yoke]

yeah

[gentle lark]

That's why the "finitely many" part throws me off

[lunar yoke]

i dont think its weird considering we have finitely many minors

its a more precise description of R than if you would leave out the "finitely many"

[lunar yoke]

on an unrelated note I'd like to shamelessly bump my own question

[gentle lark]

Does the same construction as for the other two maps not work?

[lunar yoke]

what would that look like?

[gentle lark]

My thinking is

You know how in the other two maps

You basically grabbed the one "perpendicular" to the one you wanted to preserve

And then you collapsed it "in two"

?

(I assume you did that)

Then grabbing the loop perpendicular to (t,t) (so (t,-t)) and collapsing it might work?

Maybe not

[lunar yoke]

im not really sure what you mean by "collapsed in two"

i didnt think too much about these two cases, as i said I just guessed and it worked xd

[gentle lark]

Like, foliating the torus by circles and mapping those circles by z -> z^2

If that makes sense

[lunar yoke]

yeah thats what i though of

[gentle lark]

Where if you foliate it vertically you get one of the maps, and horizontally the other

[lunar yoke]

never heard of foliating

[gentle lark]

I'm saying that if you do it diagonally (perpendicular to the loop you wanna preserve) that might work too

[lunar yoke]

but (z,w) -> (z^2,w) and (z,w) -> (z,w^2) were my initial guesses

[lunar yoke]

hmm ok

now i only need to figure out how to do that

[gentle lark]

A foliation is basically covering a manifold with smaller-dimensional submanifolds

These are foliated tori, for example

Hmm wait this might not work

It's a bit more involved

Hmm I think it might work but I'd have to write it down

(sorry if my random throw-shit-at-the-wall approach is not particularly helpful 😅 )

[lunar yoke]

nah its good actually

[gritty widget]

Any have a good introduction to stacks

[gentle lark]

Ok I've drawn some diagrams and I'm about 75% sure it does work

[lunar yoke]

could you show the diagrams

im more interested in the approach to this than the actual solution

[gentle lark]

I'm not sure that'd be helpful. But let me think if I can explain the idea instead, which should be helpful. 😛

Grab the loop you wanna preserve, then grab a loop "perpendicular" to it.

Make many copies of said perpendicular loop so that they fill the torus in the natural way

[gentle lark]

Like in this picture

[lunar yoke]

by translation?

[gentle lark]

Yes

If you then use the double cover z -> z^2 on the circle on each of those loops

You've got a double-cover of the torus (from the torus)

This is essentially what you did in the other two cases

[lunar yoke]

ohh now i think i get the picture

[gentle lark]

Now, if you go along the vertical loop, it will, intuitively, have a "component" in the foliating loops, and will thus get "carried around" when you lift it

Same for the horizontal loop

But the diagonal loop is entirely perpendicular

So it does not get carried along.

There's probably a very non-geometric, write-down-the-formulas-of-the-maps-and-check-they-do-what-you-want way of doing this rigorously

But that's the geometric intuition

[lunar yoke]

thanks for this, really nice

[gentle lark]

Now I'll bump my own question 😅

[gentle lark]

And also this question

[kind cedar]

As much as I loved your discussion, I still don't get the relationship between "being a zero set of polynomials" and "being closed"

[lunar yoke]

polynomials are continuous, so the pre-image of the closed set {0} is closed

[kind cedar]

ah, so!

thank you

and here, how can Xp act on f, if f belongs to another set of germs?

[gritty widget]

but X_p is acting on f \circ F

[kind cedar]

But Xp by definition is a map from Cp(N) to Reals, not from Cq(M) (q = F(p))

or am I missing something

[gritty widget]

f \circ F is in C_p(N)

[kind cedar]

I can't see how. F maps P from N to M, then is followed by f

[orchid forge]

and f maps from M to R

N --> M --> R

[kind cedar]

But Xp belongs to TpN, so it should act on functions from N --> R

Ah, I think I get it

[orchid forge]

yes, the composition f \circ F is a function N --> R

[kind cedar]

yeah, dumb question, ahah. Thank you.

[finite heath]

Hey guys

I am really stuck on this question

I don't know how to follow through with the reasoning.

What I have thus far.

@orchid forge any ideas =[?

[orchid forge]

I'm guessing they write this explicitly as a characterization theorem

A point $y$ is in the interior $B^\circ$ if and only if there exists an open set $U$ so that $y \in U \subset B$

[gentle osprey]

Kogasa

[finite heath]

Yes -- that is for the interior

My work thus far is for the closure

which is that

y is in closure of A iff any open set contiaining y intersects A

[orchid forge]

Yes

step 1 would be to figure out what the closure is, then step 2 would be to actually prove it

[finite heath]

step 1: ✅ cl(closed ball) = closed ball

[orchid forge]

ok, so take a point $y \in \bar B$ and prove that it's in the closure of $\bar B$

[finite heath]

step 2: i began with the theorem, but i am not sure how to go about proving such

[gentle osprey]

Kogasa

[orchid forge]

(every open neighborhood of $\bar B$ automatically contains $y$, since $y \in \bar B$)

[gentle osprey]

Kogasa

[orchid forge]

now take a point $y \in (\bar B)^c$ and prove it's not in the closure; there exists an open set $U \ni y$ which is disjoint from $\bar B$

[orchid forge]

this should say "any closed set" btw

[finite heath]

that's what i thought sir but

my textbook says =[

[orchid forge]

oh I see, every open neighborhood of y

it's correct, I was thinking about an open neighborhood of $\bar B$

[gentle osprey]

Kogasa

[finite heath]

oh gotcha

[gentle osprey]

Kogasa

[orchid forge]

note that the above is equivalent to saying "There is a closed neighborhood $U^c \supset \bar B$ which does not contain $y$"

[finite heath]

oh wait so -- do I prove that closure(closed Ball) = itself by first stating that OR do I work up to it?>

[gentle osprey]

Kogasa

[orchid forge]

isn't that what the problem is asking us to prove?

[finite heath]

yeah -- asking us to "find" it

[finite heath]

you reference the closure as if we define it first

so we do define it first?

[orchid forge]

we're not defining anything

this was a proof strategy

we can see intuitively that the closed ball is closed

so then we decide to prove this is the case

[finite heath]

so for an element in a set, every open neighborhood in that set contains the element?

[orchid forge]

if $A \subset B$ and $x \in A$ then $x \in B$

[gentle osprey]

Kogasa

[orchid forge]

this is just what a subset is

[finite heath]

but here

we have an open neighborhood of B \subset B and element in B, not the open neighborhood

=[

[orchid forge]

i mean

[gentle osprey]

Kogasa

Kogasa

Kogasa

[finite heath]

yes

but here we are working with open neighborhoods

[orchid forge]

there is no reason we have to

but anyway, every open neighborhood of $y$ intersects $\bar B$ too... at $y$

[gentle osprey]

Kogasa

[finite heath]

i am trying to branch off of the characterization theorem for closure sir (or ma'am)

[finite heath]

yes!

[orchid forge]

i feel like you can safely use the definition of closure lol

[finite heath]

oaky i like that too

closure of set is intersection of all closed sets containing said set

im sorry if i am being difficult

this is my first class taking proofs

i am so terrible at them

[orchid forge]

that's why they're making you do em, to practice

[finite heath]

but i feel like im not even getting better at them

like i understand more

and can understand more proofs when i get my hand held

but i am seemingly unable to come up with this stuff on my own

i spend 10+ hours studying this subject a week easily and i cannot even do this proof

even tho it's quite apparent that Cl(closed ball) = closed ball

[orchid forge]

maybe try doing more exercises on your own? and remember to show two sets are equal, the usual way is to show $A \subset B$ and $B \subset A$

[gentle osprey]

Kogasa

[orchid forge]

or, what I suggested was $A \subset B$ and $A^c \subset B^c$

[finite heath]

yeah ive seen this a lot

is that what we are essentially trying to do here?

cl(Bhat)=Bhat?

[orchid forge]

yes

[finite heath]

which way are we proving first?

[gentle osprey]

Kogasa

[orchid forge]

it doesn't really matter does it?

[finite heath]

I reckon not!

Would you be willing to voice call? I don't want to waste ur whole evening

[orchid forge]

I'm a bit busy, sorry

[finite heath]

Can I start the proof by saying like

W.T.S Cl(Closed Ball) is a subset of Closed Ball

Then there will be a second portion where I say

W.T.S Closed Ball is a subset of Cl(Closed Ball)

[orchid forge]

Sure. I would start by stating the thing you're actually trying to show though. "We claim $cl(\bar B) = \bar B$." Then go on to show both of those things, and then you're done

[gentle osprey]

Kogasa

[finite heath]

Oh okay

Sounds simple don't it

Hey kogasa

This direction was easy to prove I think

[little hemlock]

are you allowed to use that the closed ball is closed?

[orchid forge]

i would imagine not

[little hemlock]

"by defintion, the closed ball contains its boundary" is not far from making that assumption

[supple locust]

If we have a covering from complex plane minus few points to complex plane minus few points, given by a complex polynomial why does the deck transformation have to be analytic on the whole complex plane?

[coral pawn]

I came up with the vector field X which is just the pushforward of d/dx by sigma_N^{-1} at all points except the north pole, where I defined it to be 0

How would I show that this vector field is smooth at N?

[gritty widget]

Are d, d' metric corresponding to equivalent norms?

[orchid forge]

pass to a chart for S^2 containing the point at infinity

[supple locust]

should I be asking algebraic topology questions in this channel?

[gritty widget]

Yes

[gritty widget]

Have any of you people tried to learn qft

Kind of feels like something the “modern” mathematician should know

[supple locust]

What is a strategy to find all covering spaces of something like a klein bottle?

Finding all subgroups of its fundamental group seem really tough task

[uncut surge]

Given a smooth manifold $M$ and a sufficiently nice subspace $N \subset M$, define the space $\Omega^k(M,N)$ of differential forms $\omega \in \Omega^k(M)$ with the property that, for any $p \in N$, the form $\omega$ is flat at $p$, meaning all its derivatives at $p$ vanish in any choice of local coordinates (equivalently, its infinity-jet vanishes at $p$). I have reason to believe that this behaves like a kind of relative de Rham cohomology (i.e. I think it probably fulfills Eilenberg-Steenrod axioms). Anybody got some thoughts on that, or a good reason for why that could be the case?

[gentle osprey]

Lartomato

[uncut surge]

e.g. the associated cohomologies for (R^n, {0}) and (S^n, {x}) where x any point of the n-sphere seem to give the correct relative cohomologies, and that's just kinda cool to me

[long hornet]

Hmmm..

What happens if we embed Omega^k(N) in Omega^k(M) somehow (bump functions?) and then consider the quotient?

I think this is somehow related

[fading vale]

this is relative derham cohomology i think yea

theres another equivalent one thats like Omega^n(M, N) = Omega^n(M) oplus Omega^n-1(N) i think

sheafy proof that this works out https://mathoverflow.net/questions/122872/de-rham-model-for-relative-cohomology?rq=1

[uncut surge]

oooooh thanks to both of you

haha this is exactly what i've been doing but like a caveman

[shy moss]

can a 3-sphere be obtained as a 3-simplex identifying oposite faces?

[amber tapir]

Can someone help me start this, I’ve been stumped for so long

[hollow harbor]

the set of all pairs of straight lines is countable.

what is the set of points which are equidistant from two fixed, given straight lines?

[fickle marsh]

What can we say about a function from RxR to RxR that has continuous coordinate functions

In particular with std. topology

But also in general

[empty grove]

Continuous coordinate functions implies continuity

[fickle marsh]

But isn't that only for like A -> X x Y where A subsets X

[celest flare]

Locus of points equidistant from any pair of lines in a countable set of lines is a union of countable set of lines S. Union of all vertices of countable set of equilateral triangles is a countable set V. Cardinality of set of all lines in R^2 is cardinality of continuum, hence there exists a line m in R^2 not in S.

Now, as intersection of two distinct lines is either a single point or a null set, m intersection any line in S is a single point or a null set. Hence, union of m intersection all lines in S is atmost countable set A. Cardinality of set of all points of m is continuum, cardinality of A is at most countable, cardinality of V is countable. Hence, there exists a point R on m which is not in A and is not a vertex of any equilateral triangle in the countable collection. That point R is not equidistant from any pair of lines and is not a vertex of any equilateral triangle.

[empty grove]

No I might be misinterpreting what you are saying, is that some theorem you've seen?

The product topology is defined in a way such that a function into the product is continuous iff all its coordinate functions are

[fickle marsh]

Yeah that's in the munkres book

They do a whole section about continuous every variable sep. not implying continuous

They only talk about a set A -> XxY

But I'm asking about RxR -> RxR

It may be a result from a second course in real analysis I'm not sure really

[empty grove]

Weird

Not it's not too hard to prove

The thing I'm mentioning

[fickle marsh]

Yeah

Well

But it's not true in general

Like f XxY -> A could be continuous in each coordinate

But

F not continuous

Hmm

Okay maybe I should just cite the book

[empty grove]

I think coordinate functions being continuous is stronger than continuity along each axis

Because coordinate functions being continuous also tells you that the function's restrictions on say horizontal lines vary continuously

[fickle marsh]

So my question is what if A is a product space R already

It's not really clear by these two statements

[empty grove]

Oh yeah domain is continuous then not much you can say

I'm doing this on the codomain

[fickle marsh]

But what about when co domain is domain

RxR to RxR

[empty grove]

You can still do the same, there are 2 coordinate functions, both RxR → R

[fickle marsh]

Same as in it will be continuous?

Or that it won't?

[empty grove]

It would be, assuming that by coordinate functions you mean these

[fickle marsh]

I'm thinking like ex. F((x,y)) = (x + y^2, y)

Can I say oh well this is continuous since it's continuous in both coordinates

[empty grove]

Yes

Think of this as still A → RxR where A just happens to be RxR itself

[fickle marsh]

Quirky

Thanks

[charred root]

Guys, I might be way over my head here, but. Today in uni on analytical mechanics we were told about the Poisson bracket of two smooth functions on a differentiable manifold. I instantly thought: this reeks of a Lie algebra. For finite-dimensional Lie algebras there’s always a corresponding Lie group. Is there one for the algebra of smooth functions? What does it look like? I can’t wrap my head around it

Or should this question go to #groups-rings-fields?

[plain raven]

no i think this is the right channel

[gritty widget]

this is the right channel.

[gritty widget]

Poisson stuff comes from vector fields

As far as I know there is no “poisson group” that induces whatever poisson bracket in some way

[charred root]

Like, usually you make a Lie group out of a Lie algebra by introducing some sort of an exponential operator that obeys Baker-Campbell-Hausdorff. But heck if I know how to do that if instead of matrices you have functions and instead of the commutator you have the Poisson bracket

[gritty widget]

I have the weil algebra W(g)

so its the tensor of the Symmetric algebra on g with the exterior algebra on g

and we grade it by an element in Lambda^p has degree p

an element in S^q has degree 2q

seemingly its obvioust that the zero degree of this is then the base field of the lie algebra

can anyone see why this is the case?

[gritty widget]

in particular it would be enough if the degree zero terms of the symmetric and exterior algebra for a vector space were just the base field

[quasi forum]

Okay, so the question is you have two top. Spaces X and Y, where Y is finer than X.
What can one space being comnected imply about the other.

So I am pretty sure Y connected means X is connected, but not the other way around.

However, I am having a hard time thinking about how to actually prove this.

[orchid forge]

remember a separation is just a pair of open sets with a certain property

and "Y finer than X" means that open sets of X are open in Y

so if you have a separation of one, is that also a separation of the other? if not, can you come up with an example?

[quasi forum]

Hmmm.
Well, if we take any two open sets in X, we know that they are not a separation in Y.
However, what if this is because there is a set in Y, but not X, that intersects them both?

[orchid forge]

you mean what if they're no longer disjoint in Y?

presumably X and Y are just different topologies on the same set

[quasi forum]

Wait, I am being silly.
If we take two open sets A,B in X, we know they are not a separation of Y.
But since we are acting on the same set, then clearly A and B also can't be disjoint in X.

So yes. Y connected implies X connected

But the opposite is not necessarily true. Because we can find two open sets that are not in X, so we cannot deduce if they form a separation based on the topology of X.

[orchid forge]

not immediately. but you won't know for sure until you find a counterexample

[quasi forum]

Easy. R and R\{0} both with the standard topology

[orchid forge]

needs to be the same set

otherwise finer/coarser doesn't make sense

[quasi forum]

Shoot true.

I got excited

[orchid forge]

you're right to think simple though

[quasi forum]

Hmmm, I am not so sure

[orchid forge]

there are some topologies we can always define on any set

[quasi forum]

Discrete/indiscrete

The Discrete space is always connected, but the indiscrete space is not connected.

[orchid forge]

as long as there's at least 2 points in the space, that is true

backwards though

[quasi forum]

Oh, whoops. Indiscrete is just X and the empty set

Still taking time to realize which is which :p

[orchid forge]

they're both trivial though

"indiscrete" as in "does not separate any points"

like "discrete" means "separates every point"

my professor called them both "stupid topologies"

or y'know

finest topology / coarsest topology

would probably be good

[coral pivot]

I Call it codiscrete topology

[orchid forge]

i was gonna suggest something categorical but

i don't think we ever talk about the category of topologies on a fixed space

[coral pivot]

Like, every map out of a discrete space is continous

[orchid forge]

it's really just a poset

[coral pivot]

Every map into a codiscrete space is continous

It’s kinda dual

[orchid forge]

yeah it makes sense i think

[bleak helm]

codiscrete reminds me of cofinite / cocountable, so it sounds like the complement

Big sully

[coral pivot]

Hmm true ig

[orchid forge]

well, it is the topology generated by the complement of the discrete topology

[quasi forum]

Btw, thx for the help Kogasa

[quasi forum]

So I think I am pretty close to solving this. But I am having a hard time seeing how the connectedness of the fibers help here.

f is supposed to be p

[orchid forge]

$A \cup B$ a separation of X?

[gentle osprey]

Kogasa

[quasi forum]

Yes, I literally just noticed that :p

[orchid forge]

do you know that f(A) and f(B) are even open?

[quasi forum]

No we do not

However, we can show that A is just a union of fibers, and so is B

[orchid forge]

that would be the same thing

f(A) and f(B) would be open if A and B are saturated with respect to p

which is to say they are a union of fibers

[quasi forum]

Ah, that is the key ingredient yes.

So then we just need to find a way to say that a union of connected fibers is connected.

[orchid forge]

again A and B aren't necessarily unions of fibers

that would make them saturated

but here's the thing

if A and B are saturated, then f(A) and f(B) are open, and that would make $f(A) \cup f(B)$ a separation of $Y$, which is impossible

[gentle osprey]

Kogasa

[orchid forge]

if, say, A is not saturated, then there is $a \in A$ so that $f^{-1}(f(a)) \not\subset A$

[gentle osprey]

Kogasa

[orchid forge]

if the fibers $f^{-1}(f(a)) \subset X$ are connected, and $A \cup B$ is a separation if $X$, what can we say from here

[gentle osprey]

Kogasa

[quasi forum]

That f^-1(f(a)) intersects with B (in some compacity)

And that would be mean A and B are not separations since f^-1((f(a)) is connected.

[orchid forge]

if you have a separation of X into two disjoint pieces, any connected subspace must live entirely in one of the two pieces

[quasi forum]

Yes, but we just found one that is in both. Which is the contradiction

[orchid forge]

a contradiction even

[quasi forum]

That makes perfect sense. Thank you

[plain raven]

i think this is worth doing if you are looking at the forgetful functor Top -> Sets and you want to look at its fibers. i think echoone can say more about this but even though the fibers are just poset categories they have some interesting properties. For example, if X is any set and you have a family of maps f_alpha : X -> Y_alpha into a family of topological spaces (even if this family is a proper class) there's a unique coarsest topology on X (object above it in the fiber category) that makes all these maps continuous. If you have a map of sets f : X -> Y and a topology on Y you can pull it back along f to get a topology on X by taking preimages. Stuff like that

So although poset categories on their own are uninteresting a big bundle of them together can be interesting.

also the discrete and indiscrete topology functors Set -> Top are the left and right adjoints respectively to the forgetful functor Top -> Sets

no. right and left adjoints respectively?

hmm

no i was right the first time lol

[orchid forge]

i can see all this, i think the adjunction is probably the thing i was looking for (a good categorical description of these topologies)

[quasi forum]

Wait, so why are we saying f(A) and f(B) is a separation here?

Nah, I got it.

[orchid forge]

ok good

i was writing it out but it's not like an instant observation

you have to kind of go through it

[quasi forum]

By the definition of A and B, f(A) and f(B) are disjoint.
Moreover, since p is a quotient map, the union is Y.

So f(A) and f(B) are a separation, but this can't happen since Y is connected.

[orchid forge]

f(A) and f(B) need not be disjoint for any quotient map f. but if they're not disjoint, then there's some $x \in f(A) \cap f(B)$

[gentle osprey]

Kogasa

[orchid forge]

the fiber over x is connected, so it must be contained entirely in either A or B

but that's a contradiction

[quasi forum]

Ah, I see. That makes sense, yes

[orchid forge]

so f(A) and f(B) are disjoint, they cover Y, and you can show they must be open (i.e., A and B are saturated)

the last thing is to argue that neither of them are empty

then we would have a separation of Y, which we know is impossible

[quasi forum]

Well if one is empty, then the other, say f(A), would just be Y

[orchid forge]

which means f^{-1}(f(A)) = X, but as we said A is saturated, f^{-1}(f(A)) = A

[quasi forum]

How is this? I am not entirely sure if this does it

Not sure this works. If it did, then I think this same logic would imply X\Y is connected (which is not true)

[quasi forum]

I was trying to show that since there is no open and closed set in X, there is no open and closed set in YUA or YUB.
But again, this feels problematic since we can make the same type of argument for X\Y, but we know for a fact there are clopen sets (the separation) in X\Y

So I do not believe this works. I am also not too sure how to fix it

[quasi forum]

I see my issue. Unfortunately, this doesn't do a damn thing 😭

[vast estuary]

Hi, is this the right place for differential geometry?

[gritty widget]

yes

[gentle osprey]

Hausdorff

[vast estuary]

How do I do this?

[gritty widget]

consider something like $$s \mapsto \int_a^s |c'(t)|,dt$$ i believe

[gentle osprey]

TTerra

[vast estuary]

this is the phi?

[gritty widget]

idk

[vast estuary]

hmm

[gritty widget]

[gentle osprey]

Hausdorff

[gritty widget]

yea

[gentle osprey]

Hausdorff

[gritty widget]

yea so take the inverse of phi and you're good

[vast estuary]

hmm i don't really understand it still

could you explain some more? @gritty widget

[gritty widget]

the reparametrization you want is the inverse of the function you have here

note that c \circ \phi might not even make sense for the phi you wrote here

[vast estuary]

True

but then, the inverse also doesn't make sense, unless the range of phi is properly restricted

[gritty widget]

idk what you mean

well maybe i do

but it's not a terribly important detail

[vast estuary]

okay yeah i understand it now

[gritty widget]

good vibe

[vast estuary]

quick question, does such a reparametrization always exist?

problem being, in certain cases, the function we have defined may not be invertible

[gritty widget]

you need some regularity assumption

"c' \neq 0 except on a finite set of points" should also work

[vast estuary]

okay yeah i am assuming c' > 0 indeed

[gentle osprey]

Hausdorff

[vast estuary]

Here's the complete proof

[gritty widget]

looks good

maybe a few words on why f is a bijection would be good, but you got the important ideas

hausdorff DG arc

[gentle osprey]

Hausdorff

[vast estuary]

Could you check surjectivity?

[orchid forge]

f is differentiable with positive derivative everywhere right?

[gentle osprey]

Hausdorff

[orchid forge]

so it's strictly increasing, f(a) = 0 and f(b) = L

[jagged pivot]

For the d- dimenstional cube, what is another way to write $$[0,1]^d?$$

[gentle osprey]

beeswax

[empty grove]

What do you mean by another way

What kinda form do you want

[jagged pivot]

I am trying to prove that
$$ conv({0,1}^d) = [0,1]^d, $$ and I'm just trying to see what the elements in each set look like

[gentle osprey]

beeswax

beeswax

[empty grove]

conv means convex hull?

[jagged pivot]

Yes

[empty grove]

Yeah you can do this directly with what you have. I don't know if there's a more efficient way to say I^d than that

[jagged pivot]

How would I argue that if some vector v is in [0,1]^d, then it's also part of the convex hull?

[empty grove]

A ⊂ conv(A) for any set A

||consider the constant line segment||

[jagged pivot]

Wait what do you mean constant line segment

[empty grove]

Hmm what is the definition you have of conv(A)

Because usually this contains A by definition

Are you inductively constructing it by including line segments?

[jagged pivot]

Yeah, the set of points is contained in the convex hull. And for the proof, I am just showing that each set is a subset of the other

[empty grove]

Right so this is by definition then right?

[jagged pivot]

Oh, so [0,1]^d is the d dimensional cube.
S = {0,1}^d is the set of 01-vectors. The exercise wants me to prove that the convex hull of S forms a d dimensional cube

[empty grove]

Ohh mb I didn't notice the curly on the left

So you can show, in order, that given any convex set containing {0,1}^d, it must contain
the vertices of the cube
the edges
the faces
the whole volume

[jagged pivot]

Hmm

And correct me if I'm wrong, this should prove subset both ways?

[tawdry widget]

Induction

[empty grove]

Not in the way I phrased it at least, because the convex set I took could very well be R^d itself. I did not mention any smallness constraints

But the other way is easier, because [0,1]^d is already a convex set containing {0,1}^d

So must contain the smallest such set ie the convex hull

[tawdry widget]

(x_1, …, x_n) =(1-x_n)(x_1 ,…, x_(n-1) ,0) + x_n (x_1, …, x_(n-1) ,1)

And the statement is trivial when n=1

[empty grove]

That seems to be (1,...,1,x_n)

Ah I see

Nvm

[jagged pivot]

With this one, you are saying $${0,1}^d \in [0,1]^d \implies {0,1}^d \in conv({0,1}^d )$$?

[gentle osprey]

beeswax

Moldilocks ✓

[empty grove]

@jagged pivot

conv A gives the smallest convex set containing A, so any convex set containing A contains conv A

[grand salmon]

I gotta look into this a bit more.. I'm mostly doing the visual geometry part of higherdimensional objects...

As in: how to fold a tesseract in the most efficient way from any 3D net or how you could visualize a rubik's tesseract... for a game
reason being that I just do this as a hobby.. been studying 4D objects for roughly 2.5 years

[empty grove]

What do you mean by efficiency of the folding?

[orchid forge]

Like a cube can be drawn as a series of adjacent squares on a paper, cut out and folded up?

Never thought about doing that in 4D, but not sure what efficiency would mean either

Minimal number of folds?

[empty grove]

That feels like it should be the same

For any net

[orchid forge]

Yeah lol

[empty grove]

Because it's the number of edges in a spanning tree of the graph you get

With faces as vertices and edges between adjacent faces

[orchid forge]

Not necessarily but up to some trivial reductions

[empty grove]

Wait is it not?

[orchid forge]

I mean if you're just saying the number is well defined, then it may be

[empty grove]

At least should be a special case of a spanning tree I think

[empty grove]

Yeah and I'm saying that that's because a net should correspond to spanning tree of the graph

And folds should be the no of edges in it

[orchid forge]

Yeah

[empty grove]

And no of edges in a spanning tree are the same for any spanning tree

[orchid forge]

I was thinking you can get a new net which adds folds but only in kind of trivial ways

Which would correspond to splitting an edge a --> b into a --> c --> b for example

[empty grove]

I don't think that works, you are changing the entire tree

You would end up with 2 occurences of c if you just did that, and that means you'd have a cycle

[orchid forge]

c is intended to be a new node here

[empty grove]

Oh are you taking nets with overlappings allowed?

[grand salmon]

exactly

[orchid forge]

I think I have to be

[empty grove]

I see

[grand salmon]

the difference between 2D -> 3D and 3D -> 4D is the complexity of folding it... as in: what would be the universal first move for folding a tesseract?

[empty grove]

There should be a way to reduce to the no overlaps case somehow

[orchid forge]

Yeah that's what I meant by up to some trivial reductions

[empty grove]

Universal first move? 😵‍💫

[grand salmon]

Having a net like this:

makes it very easy

[empty grove]

Don't get what universal first move means here

Surely you can fold in any order?

[grand salmon]

you move the inner cube in the 4th dimension, which would make it appear 2 times larger... like this:
(Step 1)

[orchid forge]

Turtle

[long hornet]

(unrelated: what's the name of the software?)

[grand salmon]

I'm calling one step: all actions that brings one edge (2D-3D) / one face (3D-4D) from one starting point to another without moving the entire object.

For example:
⬛1️⃣⬛⬛
1️⃣🟦1️⃣🟦
⬛1️⃣⬛⬛
in 2D -> 3D this means moving at most 2 connected vertices .. one edge.. per step.

1: fold the outer squares up
this leaves us with something like this:
⬛🔽⬛⬛
▶️🟦◀️2️⃣
⬛🔼⬛⬛ where the arrows are faces that are set up like this: |[]|¯

the second step would be folding the last edge over to the left square's edge that is folded upwards, like closing a lid.

[grand salmon]

blender

[empty grove]

So you're counting those 4 foldings as 1 step?

Or 4 steps

[grand salmon]

1 step

as long as an adjacent square isn't folded, it remains one step... wait.. I'll just do it in blender.

[empty grove]

I see

[grand salmon]

so: starting point, step 0

[empty grove]

So I'd like to claim that the first step should always be to fold up the end squares ie the squares that only connect to one other square

[grand salmon]

step 1

[empty grove]

And the most efficient net should be the next with the most branching

[grand salmon]

step 2

yes

[empty grove]

Let me see if I can prove that

[grand salmon]

should be the case

[empty grove]

Yeah clearly the net you sent can be folded in 2 steps

The one for the tesseract

[grand salmon]

exactly

[empty grove]

Do you want to just find the most efficient net or the most efficient folding given any net?

[grand salmon]

the latter

[empty grove]

I see

[grand salmon]

sort of a folding formula.. how to fold any net without having to break your head for about 10 minutes

[empty grove]

Lol it should start from the outermost cube and move inwards

[empty grove]

Outermost in this sense

[grand salmon]

and maybe even an actual formula for how you could calculate the minimum amount of steps you need for folding a tesseract

[empty grove]

Like leaves in graphs

[grand salmon]

yes

[empty grove]

Number of steps should be minimum depth of the tree starting from any node

[grand salmon]

This is also sort of connected to the "Homeomorphically irreducible trees of size n=8" then.. right?

cuz there are certain nets that are basically the same.. just moved around a bit

[grand salmon]

yeah

[empty grove]

There are less of those than there are nets

Because they require branching at each internal node

[grand salmon]

hm.. u right

wait... then we'd have to limit the connections per node to 4 for 2D-3D... and 6 for 3D-4D

wait no.. same thing... still the same thing

[empty grove]

Those are limited automatically

Because each face has at most 6 adjacent faces

[grand salmon]

oh wait... then it's not homeomorphically irreducible... nope unrelated... since you can put faces of a cube together like this:
🟦🟦⬛
🟦⬛⬛
🟦🟦🟦 <- that isn't irreducible

[empty grove]

Where are you pointing bro

[grand salmon]

I'm confusing myself a bit here..

no no no... I didn't mean to imply that

[empty grove]

I know I'm just kidding

[grand salmon]

there we go

though... that would be the net with the most branched net.. right?

[empty grove]

Damn

Hmm no it is the least i think lol

Only 2 leaves

That's the least possible

[grand salmon]

2️⃣1️⃣⬛
1️⃣⬛⬛
🟦1️⃣2️⃣... wait .. you're right

or... would it be like this then?
2️⃣3️⃣⬛
1️⃣⬛⬛
🟦1️⃣2️⃣

since the folding of one part would affect the starting points of the right edge on 3?

[empty grove]

Yes

[grand salmon]

I might have to rethink the idea of a step...

one step = moving all edges / faces that are unaffected by another edge / face that is being moved..

so:
the green squares are those that are moved along while another face moves...
the red squares are the faces that are in their correct places.
🟩⬛⬛⬛
🟩1️⃣⬛⬛
⬛🟥1️⃣🟩

🟩⬛⬛⬛
2️⃣🟥⬛⬛
⬛🟥🟥2️⃣

3️⃣⬛⬛⬛
🟥🟥⬛⬛
⬛🟥🟥🟥

done

wait a minute...

have I gotten the net wrong this entire time?

[empty grove]

Lmao

The 3 vertex should go left

[grand salmon]

yeah

[empty grove]

But ye folding looks correct

[grand salmon]

there we go

okay nice

[empty grove]

Wait no this doesn't work either

[grand salmon]

... yep

[empty grove]

The 2 2 squares overlap

[grand salmon]

dangit

[empty grove]

oof

[grand salmon]

now

finally

[empty grove]

Sorry lol

[grand salmon]

np .. thx for checking it

[empty grove]

But I think the 3 vertex and the right 2 vertex coincide

[grand salmon]

hm

[empty grove]

[grand salmon]

dammit u right XD

[empty grove]

Always

[grand salmon]

🟩🟩1️⃣⬛⬛
⬛⬛🟥1️⃣🟩

🟩2️⃣🟥⬛⬛
⬛⬛🟥🟥2️⃣

3️⃣🟥🟥⬛⬛
⬛⬛🟥🟥🟥

[empty grove]

Looks good

[grand salmon]

this should be right.. yeah

[grand salmon]

Alright I'm putting wayyyyy too much time into a system that isn't actually useful here.... so i'll just scrap that idea entirely T_T.. if you wanna reread that.. feel free to :/

I got to the point where I wanted to call this:
⬛🟦⬛⬛
🟦🟦🟦🟦
⬛🟦⬛⬛ 1x1-(2)-1x2-(2):3 .... so naahhh let's not

I really dislike the hyperfocus I get from my ADD sometimes... because I just focus on things that are unnecessarily pointless... trying to find a meaning or formula for everything

[pearl holly]

#topology-and-chemistry

[empty grove]

chmeomistry

[grand salmon]

I was bored... so I started building a 5D .. object

that's just a cube, eh?

not exactly

roughly a look into it

making the object wasn't the problem... it was just... adding the faces in was a pain

[pearl holly]

yo what program is this?

[grand salmon]

Blender

[pearl holly]

oh okay I see. Thanks!

[long hornet]

Suppose A is a closed subset of a manifold (say) M. Is there a relation between whether M retracts onto A, and the homotopy type of M - A?

[finite heath]

and x obviously lies in A, right?

[lunar yoke]

no, x need not lie in A

for example x=1 is a limit point of A = [0,1) in X = R

[gritty widget]

this is more a question about intuition on general cases not something specific

when you see some statement about connected lie groups

what kind of obstructions will you have preventing the same statement being true for non connected lie groups

[plain raven]

i wanna read the book Cohomology of Sheaves by Iversen.

it focuses mostly on the treatment of Poincare duality by means of sheaves
and tries to show how basic arguments in analysis can be seen from the point of sheaves.
I think this is meant to be like a book that transitions you from a decent general knowledge of homological algebra and some idea of applications to the point where you're ready to study derived categories, perverse sheaves and applications to geometry
i am really flaky and so i need somebody to buddy up with to help me read this lol

so lmk if you would wanna discuss it with me or read it with me.

[gritty widget]

I have been trying to read Equivariant sheaves and functors by bernstein but feel i am missing stuff. If is another person trying to read this with you i would like to try join, but If I was the only one I think I am a bit too slow to be a good partner

[orchid forge]

I am interested in reading the book but I'm currently struggling to find time just to exist

[pearl holly]

is the cup product a homomorphism?

when I try checking it on the chain level it doesn't seem to add up because of distributivity

[empty grove]

homomorphism from what to what

[pearl holly]

wait wtf

I mean like $(\varphi \smile \psi)( \sigma_1 + \sigma_2) = (\varphi \smile \psi)(\sigma_1) + (\varphi \smile \psi)(\sigma_2)$

no wait

[empty grove]

where's psi on the right side

[gritty widget]

smiley

[gentle osprey]

Tokidoki ✓

[pearl holly]

ehhh

why is it still loading?

[empty grove]

$\phi\smile\psi$ is a homomorphim to $G$, thats what you have written

[gentle osprey]

Moldilocks ✓

[empty grove]

but that is different from cup product being a homomorphism

[pearl holly]

$0 \smile 0$

[gentle osprey]

Tokidoki ✓

[empty grove]

that would have to be from H^m x H^n → H^(m+n) I suppose

but this won't be a homomorphism

instead of being linear it is bilinear (thats what distributivity of multiplication is)

and so this is really a homomorphism between

[gentle osprey]

Moldilocks ✓

[orchid forge]

$\bigcup \smile \bigcup$

[gentle osprey]

Kogasa

[pearl holly]

ye right. I mean something like this

[empty grove]

right so phi cup psi is a homomorphism from C_{m+n} to G

otherwise the cup product wouldnt give you something in H^(m+n)(C; G)

because the elements in this are equivalence classes of homomorphisms

[pearl holly]

yee okay I see. But is that true?

[gritty widget]

is there any nice result about

the classifying space of non-connected lie groups

[empty grove]

are you doubting me that hard did I say something blatantly wrong in the last call

[pearl holly]

no wait what?

sorry I'm really sleepy

I don't know what's going on

[empty grove]

im confused

I am saying it is a homomorphism

[pearl holly]

okay but how do you prove that? I don't see how that is true

[empty grove]

it was like a linear combination of homomorphisms right?

with coefficients in Z?

[pearl holly]

yee

[empty grove]

that should also be a homomorphism by chmommutative algebruh

[pearl holly]

when I try to do it I get something different from (\varphi \smile \psi)(\sigma_1 + \sigma_2) because of distrubitivity

I'm too lazy to write the whole thing out

like it's (\varphi \smile \psi)(\sigma_1 + \sigma_2) but plus some terms

[empty grove]

I am too lazy to answer your question

let me look at hatcher once lol

oh wait its not a linear comb of homomorphisms

uh

isnt it defined by linear extension of a map on the basis

all such sigma form a basis of the group C_(k+l)(X)

He is defining a map on the basis and then extending it linearly

using universal property of free ab groups

This will definitely be a homomorphism

[pearl holly]

okay so like $(\sigma \smile \psi)(\sigma_1) + (\varphi \smile \psi)(\sigma_2) = \varphi(\sigma_{1[v_0, \cdots, v_k]} + \sigma_{2[v_k, \cdots, v_{k+l}]})\psi(\sigma_{1[v_0, \cdots, v_k]} + \sigma_{2[v_k, \cdots, v_{k+l}]})$ and now $\varphi$ and $\psi$ are homomorphism and now when you distribute stuff you don't get $(\psi \smile \psi)(\sigma_1 + \sigma_2)$

[gentle osprey]

Tokidoki ✓

[pearl holly]

oh

lmao

okay so Hatcher just defines how the cup product "works" on basis elements?

[empty grove]

yes

[pearl holly]

OOOH

lmao

yee okay I completely forgot how C^n looked like

I should sleep. Thank you so much!

[empty grove]

[pearl holly]

but it's still really early

imma take a nap isntead

[empty grove]

I support your decision

[orchid forge]

@plain raven i think this book is like the exact kind of material i need right now, so if you really want to read it with a buddy i am willing to do it

the copy i have literally appears to be printed off a typewriter though which is cool

have you looked at kashiwara & schapira at all?

[gritty widget]

If you want to make a seperate discord or something i would join

[plain raven]

Toki the cup product is a homomorphism on the tensor product

[empty grove]

I was there first

[plain raven]

sheaves on manifolds or the cat one

[plain raven]

toki i know hatcher gets a lot of hate around here but that said

I HATE how hatcher introduces the cup product.

[empty grove]

really sounded like you were gonna praise it there

[plain raven]

The cup product is NOT the most fundamental product. The most fundamental product is the CROSS product which he only defines 10 - 20 pages later in TERMS of the cup product

makes me so mad

or made me mad later once i understood it

[pearl holly]

cross product linear algebra moment

[plain raven]

i won't give you a huge rant right now

but

here's the short version

[gritty widget]

[plain raven]

hmmm trying to come up with the short version...

I've told the same stories enough that i'm sure people are realizing that i have some idiosyncratic axes to grind or whatever, so, y'know, take what i say with a grain of salt.

[orchid forge]

sheaves on manifolds

[plain raven]

ah yeah. ok i mean i vaguely know that one but i couldn't tell you off the top of my head if it's something i want to get into rn, let me look into it and get back to you

[orchid forge]

good book, some overlap, might be worth checking out

[finite heath]

may i ask a little question

[gritty widget]

you can ask a big question if you want to

[finite heath]

:]

what's the purpose of this big N here

why we have to say that n greater or equal to it?

[orchid forge]

"the sequence is eventually contained in U"

is exactly the same as "there exists an N so that when n >= N, x_n is in U"

[empty grove]

It's saying that after a certain point (which is N), all terms of the sequence lie in U

[finite heath]

what is the purpose of saying that tho, can't they just say that the sequence is contained in U?

[plain raven]

fuck. ok you guys got me

it's going to be a long post.

[orchid forge]

it's not. it's eventually contained in U

[empty grove]

We don't want absolutely all the terms in U because that is too strong a condition (everything in the sequence would be arbitrarily close to x, intuitively)

[gritty widget]

i want to see diligentClerk's holyfuck.txt of hatcher

[empty grove]

For example the sequence 1/n converges to 0 as it should, but if you said that the whole sequence should be in U, then it wouldn't

Because you would be able to take U = (-1/3,1/3) and there are terms of the sequence outside

[plain raven]

should i start a thread

[empty grove]

What we are saying is that the terms of the sequence are eventually always within U

[empty grove]

Yes

Tag mods

[plain raven]

ah nvm i don't think i have that power

[hollow harbor]

[plain raven]

dear mods can i start a brief thread?

[empty grove]

@unborn lotus I do not fear you

[empty grove]

Brief he says

[plain raven]

I am a pathological liar.

Damn irrelevant tangent but there used to be this pathological liar character on SNL played by Jon Lovitz and like, i could recommend you check it out but it's so damn unfunny it's not worth doing.

[empty grove]

Is everyone hammered today

[orchid forge]

it is a thursday

[plain raven]

THIRSTY THURSDAY

as we used to say at my undergrad

i went to state school.

[hollow harbor]

😵‍💫

[orchid forge]

I want to read the post

[plain raven]

yeah idk i'm trying to find a compromise between saying what i want to say and saying it in a reasonable amount of time and space.

i can't keep you on the hook that long.

[orchid forge]

you could always put it in a pastebin

[plain raven]

Yeah

ok let's say i'll write it now and uhh don't expect it in the next fifteen minutes

just go away and come back later lol

i'll ping the people who are here rn

[orchid forge]

oh, don't feel pressured to spend too much time on it but if you do, i will read it at least

[hollow harbor]

I HATE how hatcher introduces the cup product.

[empty grove]

[gritty widget]

i love hatcher

only because it makes people mad

[plain raven]

Hatcher is my best friend. We like to hang out

[lunar yoke]

so should you actually not read hatcher as your first exposure to algtop?

[gritty widget]

take hatcher on a date

nice and romantic, get some food or something

[lunar yoke]

if so, what would you guys recommend instead

[empty grove]

Play skyrim

[lunar yoke]

dont even have a mouse to play games kek

[plain raven]

phil it's not the worst. i would say like, skip chapter 0, really take him seriously when he says it's meant to be superficial. if you try to work out the details of what he's saying it's honestly quite hard to read. Chapter 1 is absolutely fine. I don't remember having any problems. Chapter 2 is ok, i mean i struggled but homology is just pretty hard.

[hollow harbor]

Rotman

[plain raven]

Chapter 3 i just bailed and never looked back.

Rotman i think is probably a really good intro.

[lunar yoke]

ok thx

[plain raven]

I haven't read Rotman's book but i've heard a lot of good shit about it.

[lunar yoke]

ill check it out then

[empty grove]

Where'd you learn chmology then

If you did

[gritty widget]

hatcher did once reply to one of my emails

[empty grove]

I'll write to him too

I don't wanna lose to you

[gritty widget]

I tried to ask him for his latex template

[plain raven]

if you're asking my personal opinion I really look the book by Spanier, that's what i ended up settling on, it's extremely formal tho whereas hatcher is more handwavy. I don't think the formalism impairs the geometric content tho. Not early on at least. later it becomes a struggle but like... i think lots of people have trouble finding geometric meaning in cohomology

I would definitely recommend the first 3 chapters of Spanier as an introduction to algebraic topology.

It's dense tho so like, it's ok to pass through it a bit quickly.

However Spanier does have its own problems.

[empty grove]

You hearing this toki

[pearl holly]

yeeeee I be listening to every word here

[lunar yoke]

well i guess theres pros and cons to every book

also depends on personal taste

[plain raven]

The book was meant to be a kind of compromise/balance between a textbook and a reference and i just don't think it succeeds. Rather it evolves slowly from textbook to reference. The chapters on homology and cohomology are truly encyclopedic.

He also strives to work in like, kind of ridiculous generality at times, considering all the possible arguments of all the functors. For example like

there's a notion of 'cohomology with coefficients in an abelian group G' which is based on maps from the space X into G

and it's a theorem that when G is a ring, the cohomology inherits a ring structure

but instead of saying that, spanier says:

If G, G', and G'' are Abelian groups and mu : G\otimes G' -> G'' is a bilinear map, then for all n, m there are induced maps H^n(X;G) \otimes H^m(X;G') -> H^{n+m}(X, G'')

[gritty widget]

[plain raven]

I just don't think that level of generality is really necessary. You're almost always concerned with the case G = G' = G'' = R.

I will say that like, outside of the homology/cohomology chapters it's readable.

Chapter 7 on homotopy theory is quite good. The early chapters are quite good.

Also like

[gritty widget]

i really enjoy being told all this fancy math from a lovely smiley dog

[plain raven]

the formalism is really visually intimidating, like when you open the page you're like "Jesus christ". But that just means it's not skimmable. If you read it line by line and think about the constructions they make a lot of sense.

He just prefers to give explicit line by line coordinate based homotopies in some cases, especially early on.

But later on like

he doesn't do that, you're expected to get the concept of homotopy by then lol

matter of fact there are a shitload of unproven statements whose proofs are left as exercises

and that's not even counting the actual exercises at the end of each chapter

which i've never done as i had quite enough of a workout from the text itself

[empty grove]

If clerk isn't lovely smiley dog irl I'll be disappointed

[plain raven]

I am a dog.

[sullen lynx]

hello, i have a question about topology, should i post it here or in math help?

[plain raven]

ask the question and if it's a bad place for it we'll redirect you elsewhere

[sullen lynx]

I just started to learn topology at uni and I have a very basic question, yet i can't find any answer in my lectures.

Let's say I have a set X and a set Y and I know if they are open/closed/not open and not closed/open and closed. What will the cartesian product X x Y be like?

[orchid forge]

you need to define a notion of topology for cartesian products

[sullen lynx]

i am learning about topology in R^n

[orchid forge]

the natural one for a cartesian product of two sets is called the "product topology," and is given by a basis consisting of sets U x V, where U is open in X and V is open in Y

so for example, an open set in R^2 is a union of elements like (a,b) x (c,d), since the intervals (a,b) are the basic open sets of R

[sullen lynx]

i see, but what if the sets are not finite? like R^+ x R ?

[orchid forge]

the sets can be anything, finite or not

the product topology is still generated by products of open sets

there is a subtlety when you take infinite products

like R x R x ...

[sullen lynx]

ok ,so if i have R (open and closed) and [0,1] (closed set), is R x [0,1] closed ?

[empty grove]

Yes

In R²

[orchid forge]

given that products of open sets are open, you can show that products of closed sets are closed too

[sullen lynx]

and if i have X (open ) and Y (closed), then X x Y is neither closed, nor open? (X is in R, Y is in R)

[orchid forge]

it's hard to say in general. for example if 0 is the empty set, R x 0 = 0 is both open and closed. if U = (a,b) and V = [c,d], then U x V is neither open nor closed

[sullen lynx]

oh i see

thank you very much, that was very helpful

[gentle osprey]

Kogasa

[orchid forge]

So usually, in RxR, an open x closed is neither open nor closed, with the only exceptions occurring when one of those sets is R or empty. This holds when R is replaced by any connected space

[sullen lynx]

i have an exercise like : Find out if R+ x R is closed or open or both; how would you tackle it ?
I thought about taking the complement of R+ , so R- , which is open. If R- is open, then R+ is closed. So R+(closed) x R (open and closed) gives us a closed set?... I'm not sure if that's correct...

[orchid forge]

Well as we said a product of closed sets is closed

[sullen lynx]

but R is closed and open, so i am confused