point-set-topology

[tough imp]

And apparently you can take some limit over like all n

To recover basically just a Taylor series

I guess this is actually for k[x]/(x^n)

Actually lmfao yes this is exactly the completion of k[x] at (x) which is k[[x]]

But yes, first order info

Sorry that was a bit of an aside

[bright acorn]

hm

that's nice

[tough imp]

I think the picture is more clear in the algebraic case here

IMO

And then you can just import that intuition to manifolds and just be like

“Okay Diff Geo and AG are really similar so it just kinda works out when we consider smooth functions not just algebraic ones”

[bright acorn]

It's just that I still didn't really study algebraic geometry

still taking my time to go through the commutative algebra bits

and I am more familiar with smooth manifolds

But I am pretty sure this will help me a lot when actually learning these concepts in the algebraic geometry setting

[tough imp]

This construction doesn’t require any fancy machinery

I moreso meant the dual numbers notion

[tough imp]

@bright acorn the language is maybe a bit different (premanifold), but Manifolds, Sheaves, and Cohomology by Torsten Wedhorn

Page 91-97

The specific thing with (I/I^2)-dual is on page 96

This makes an isomorphism between the derivations version of tangent vectors (which is shown to be equivalent to the more intuitive geometric notion in eg Intro to Smooth Manifolds by Lee, as well as earlier in this section of the book I mentioned) and the dual of I/I^2

[sharp frost]

Does anyone know what is meant by the "full cohomology algebra" here? I don't really know anything about this stuff but there seems to be a bunch of different flavours of cohomology and I can't find reference to "full cohomology" anywhere

[gritty widget]

they're talking about de rham cohomology probably, since it says "we realize cohomologies by classes of closed differential forms"

[sharp frost]

ahh cool thanks

[gritty widget]

and by "full cohomology" they just mean the direct sum of all the de rham cohomology groups

i think

that's usually what H*X means

[sleek thicket]

This is weird...

Why is this symmetric?

won't you get a sign change if ω1, ω2 are 1 forms?

[sharp frost]

yeah I thought about that too hahah

[sleek thicket]

Ah, "without odd dimensional cohomolgies"

Seems like they're using "a cohomology" to mean like "a cohomology class"?

Based on the parenthetical

[sharp frost]

what does it mean for a manifold to have no odd-dimensional cohomologies, does it just mean the nth cohomology group is trivial for odd n?

[sleek thicket]

That's how I would interpret this, yeah

This is still an incredibly odd description

Ah okay slightly less odd than I thought

there's no canonical fundamental class bc we have R coefficients and not Z coefficients

[sharp frost]

I've only seen De Rham cohomology defined as a group, not an algebra

[sleek thicket]

sure, you know the wedge product?

[sharp frost]

yeah

[sleek thicket]

So if you take all the groups of forms Ω^p(X) of varying degrees p and look at the direct sum, the wedge product is a multiplication on that vector space

Making it an associative algebra

[sharp frost]

ohhh yeah

[sleek thicket]

(it won't be commutative, generally there's a sign rule for swapping things. But if you only have even degree stuff it's commutative on the nose)

This multiplication satisfies $d(\omega \wedge \eta) = d\omega \wedge \eta \pm \omega \wedge d\eta$

[gentle osprey]

Chadrock

[sleek thicket]

And this formula allows you to define it on cohomology classes of closed differential forms instead of differential forms themselves

[sharp frost]

oh thats cool

[sleek thicket]

the point is just that algebra structure = wedge product

[sharp frost]

do you have a recommendation for a textbook that includes this stuff

I've been trying to use a differential geometry book by Isham but it's not powerful enough to keep up with the stuff I need for this research project

[gritty widget]

lee's introduction to smooth manifolds is very good

• comprehensive

[sharp frost]

ahh yes I've heard about that, thanks

[gritty widget]

i really like the approach of isham but you’re right it’s not very powerful at all

i wonder if there’s alternatives the follow the same approach

[versed junco]

There are two definitions: "An affine hyperplane is an affine subspace of codimension 1 in an affine space" and "A[n] [affine] hyperplane is an affine subspace of codimension 1 of \mathbb{R}^d". What's the difference between the two?

[honest terrace]

the first defines affine hyperplanes in any affine space, where the second only does it in R^d ?

[versed junco]

Does the second definition impose a redundant restriction?

[honest terrace]

its just more specific, since it restricts "affine space" in the definition to the case of R^d

but there's nothing redundant in it

[versed junco]

Isn't $\mathbb{R}^d$ merely an example of an affine space?

[gentle osprey]

JohnDark

[honest terrace]

yeah, it's just an example, the concept of affine space is much more general

but for someone who isn't familiar with, say, linear algebra with vector spaces over any field etc., the second definition will make more sense, ig ?

[versed junco]

Just these two definitions are different for me o.o

And the second one even seems wrong. It wouldn't be the case if it explicitly said an $\mathbb{R}^d$ hyperplane

[gentle osprey]

JohnDark

[versed junco]

I'm sorry for silly questions

But I do find it rather odd

[honest terrace]

I think that's clear from the context, but yes indeed it means an R^d affine hyperplane

[versed junco]

Thank you

I appreciate your help a lot

[gritty widget]

So I just figured out why open sets describe a topological surface. Took me a bit.

So if I understand correctly, one disc can't fully describe a sphere, you need 2 because the disc can't cover itself, is that right?

And you need 3 for a torus.

etc..

I'm leaning topology from point set topology, trying to understand why open disks/sets are foundational.

*learning

I'm learning it as well

I've just completed Cauchy sequences.

Yeah, the book is beginning to talk about metric spaces, but hasn't used it in a topological way yet.

how dist must be defined, etc..

is my intuition off, that you need 2 discs to cover a sphere, or no?

okay, I think I understand you there. I guess I need to study more to get a proper understanding

yeah, I feel like starting at point set topology is kinda backwards in curiculum.

It's foundational, but not really intuitive.

yeah, thanks for your input.

Yeah

ok

[frigid river]

Isn't it true that there cannot be any isolated points when d(x,y)=|x-y|?

Also, it seems so logical the no matter where and with which radius I centre my balls, I hit some number from R, but I have no idea how to formally prove this.

[river granite]

yes it is; center a ball of radius r at x and then try taking the average between x and x+r for instance

[river granite]

given any irrational number you can construct a sequence of rationals that converges to it

[frigid river]

What about (S,d) with $S=[0,1] \cup {3}$ though? Isn't 3 then isolated?

[gentle osprey]

veryhappyperson

[river granite]

ah sure, 3 is of course isolated then

[gentle osprey]

derivada.schwarziana

[frigid river]

Okay, and is this what I should write into my proof? Because, to be frank, I have no idea how to construct such a sequence. Would this mean that Q is complete in (Irrationals)?

[river granite]

no, it would mean that the closure of Q is R

since any set is a subset of it's closure, you just need to approximate the irrationals.

[gentle osprey]

derivada.schwarziana

[frigid river]

Okay, I am still confused about "given any irrational number you can construct a sequence of rationals that converges to it". If every irrational number is a limit point of Q, isn't this the definition of completeness?

[gentle osprey]

derivada.schwarziana

derivada.schwarziana

derivada.schwarziana

[frigid river]

Bruh, this makes sense, thank you xD

[river granite]

maybe the confusing bit was "converges", I meant converges in R

which isn't the same as convergence in Q or any proper subspace

[frigid river]

And this would mean Q is dense in R, right? Anyway, does density imply that there cannot be any isolated points?

Wait, $S=[0,1] \cup {3}$ should be dense in R, so I was talking stupid stuff again.

[gentle osprey]

veryhappyperson

[river granite]

no it shouldn't, it's closed in R so it's closure is itself

[river granite]

indeed Q is dense in R. As for the other question I'm pretty sure that's true it's not. For instance, (0,1)U{3} is dense in [0,1]U{3} and 3 is an isolated point

[frigid river]

Ahh

Okay, so I cannot use that for my argument either xD Sadly, I don't get yours. What does approximating the irrationals even mean, and how does this help with my ball proof?

[thin bramble]

can anyone help?

[frigid river]

@river granite , wasn't this sequence supposed to converge to an irrational ? I am so confused xD Even if it did, what would this prove?

[river granite]

well, if x is irrational then indeed it converges to an irrational. But note that even if x is irrational, each term of the sequence is rational since the floor function gives an integer.

[frigid river]

I get you now. The question remains though: What does this help me?

If my brain isn't dead by now, this should mean that every point of the irrationals is adherent to the rationals, and then, @river granite ? I am sorry if I am bothering you with my questions xD

[river granite]

indeed that's what it means

[river granite]

I think you asked here whether it's true that any ball centered in a rational point must contain irrational points. It's true since Q is dense in R, and a way to prove that is by building rational sequences as we did

[frigid river]

This was my the original question. I have already forgotten how I got to the others stuff we were talking about, @river granite xD

I think my very first sentence is just wrong. I think I meant to say, "No matter where and with what radius I centre my balls in Q, I hit some numbers in Q, therefore there cannot exist any isolated points."

[river granite]

yes, that's true

[frigid river]

Okay, thanks again, but that is exactly what I have to prove -.-

[river granite]

this idea works, for instance

the whole thing about Q being dense in R is kind of unrelated, I thought you meant that balls centered in Q always have some irrational point in R

but since we're regarding (Q, | . |) as a subspace it suffices to find some rational point in any ball, other than the center of the ball of course

[frigid river]

I finally got it now, thank you, @river granite !

[thin bramble]

anyone? <@&286206848099549185>

[fading vale]

this map is linear right? cause its obviously linear in the first factor and also in the second?

im doing this problem, so here the image of the differential of psi is just the image of psi right?

[gentle osprey]

Moth

Moth

Moth

[fading vale]

I'm not sure where you need to explicitly work with a local paramaterization?

[obtuse meteor]

@sleek thicket
You thought you got me didn't you
https://estrogen.fun/i/qdy4.png

[sleek thicket]

I don't think this is linear

[obtuse meteor]

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

[sleek thicket]

I was very excited

hahaha

[obtuse meteor]

we did some riemannian metric stuff on friday

let me drop the worksheet

[sleek thicket]

anyways @fading vale why is the 2nd coordinate linear in y?

[obtuse meteor]

and it convinced me einstein good

[sleek thicket]

nice!

It is indispensable for riemannian stuff

[sleek thicket]

but creates some truly horrible expressions

Okay this is embarrassing as someone who just defended the honor of Einstein summation notation

But

I think raising/lowering index notation is so unnecessary and confusing

you're just using the iso with the dual space!

[obtuse meteor]

It’s actually based

[fading vale]

yeah its not i was being very silly

i am still confused about tau though

[gritty widget]

[sweet wing]

idts tbh makes things a lot nicer looking

[shut moat]

^

[gritty widget]

[long hornet]

Is it true that removing the closure of regular coordinate ball from a 2-manifold the same as removing a point?

If p is a point of B, and V is a neighborhood of cl(B) that is homeomorphic to the disk, then V - cl(B) is homeomorphic to V - {p}, but I am not sure how to extend this to a homeo. from M - cl(B) to M - {p}.

[sleek thicket]

Not sure about the topological case, yes in the smooth case

Should be homotopy equivalent in the topological category

[long hornet]

I see

What can make my argument fail is that there's no control on what happens on the boundary bd(V)

[sleek thicket]

in the smooth case you can take a collar neighborhood of the boundary of M\cl(B)

[long hornet]

I think this can be fixed by choosing a suitable neighborhood W inside V.

[sleek thicket]

I think

And in the topological case you can sort of homotope stuff out in the coordinate ball

[long hornet]

I know nothing about smooth manifolds, unfortunately

[sleek thicket]

ah okay

so a collar neighborhood is a neighborhood of the boundary which is the image of an embedding [0,1)×partial M -> M that sends {0}×partial M to partial M

this is for a smooth manifold M with boundary

But we can look at just a topological manifold with boundary and require that the map be a topological embedding

Ah it looks like this holds in the topological category also https://math.stackexchange.com/q/703009/408287

I think this gives that M\pt is homeomorphic to M\cl(B)

Or wait...I'm being very very silly

I'm not sure why I thought you needed a collar neighborhood

I think you can just do a homeomorphism in the coordinate ball

Containing cl(B)

Find a homeomorphism of R\closed unit ball and R\origin which is the identity outside of the closed ball of radius 2

[long hornet]

Weird, I thought this would be easy to prove.

[sleek thicket]

The collar neighborhood or your original claim?

[long hornet]

The former.

[sleek thicket]

I think your original claim is easy and I was confusing myself

Ah okay yeah it's definitely nontrivial

[long hornet]

Exactly!

[sleek thicket]

you should be able to do this with flows

set up the right ODE basically

[long hornet]

Wait, what are you talking about? The collar neighborhood?

[sleek thicket]

No, constructing the homeomorphism I described

Or maybe you can just write it down explicitly? It's possible I'm being silly

Oh yeah I think I'm being silly

[long hornet]

That's what I thought of: Suppose we have a coordinate ball B around a point p, and the following inclusions: B < cl(B) < W < cl(W) < V, where all these are coordinate balls, and we have a single homeomorphism f : V —> B(0; 3r) that induces a homeomorphism between cl(W) and cl(B(0; 2r)), and between cl(B) and cl(B(0; r)). Now there is a homeomorphism G sending cl(B(0; 2r)) - cl(B(0; r)) to cl(B(0; 2r)) - {0}, so we can use G to construct a homeomorphism g* : W - cl(B) —> W - {p}, and the nice thing is that g is equal to the identity on bd(W), because cl(W) is a disk. So we can extend g* to g : M - cl(B) —> M - {p}, by letting it equal the identity elsewhere, and this will be a homeomorphism.

[sleek thicket]

Oh, I was saying you can use flows to construct G

I agree with your argument

Er wait, no

I don't know what I'm saying

I should go to bed

[long hornet]

Well, I think the closed disk with a closed disk removed is homeomorphic to (0, 1] x [0, 1]

[sleek thicket]

yeah, you're right

I think I was getting hung up on stuff being smooth

When you just want continuous

[long hornet]

Oh, I see

[sleek thicket]

just like, if you do stuff piecewise then it won't be smooth on the boundary

But mostly i should just go to bed, I've made a bunch of silly mistakes in this conversation

[long hornet]

No problem!

Thank you though.

[warm hearth]

Hi

I have to solve this problem

I was able to find the equation of the tangent plane by plugging in the coordinates of M:

this is the equation: 3x+4y−6z−6=0

From here how can I prove that this plane cuts the surface after two lines? And how can I find the angle between these two lines?

[fading vale]

is intersection number mod 2 secretly homology with Z/2Z coefficients or something

[fair idol]

Can't you just substitute the plane equ into the hyperboloid and solve? If you get two solutions to the quadratic then you found the two lines I think

[cold vine]

Does anyone have a hint on showing that Hopf fibrs S3->S7 ->S4 and 7->15->8 are not null homotopic?

Exactly. Yeah I'll try that idea

[fading vale]

oops ill wait

unless the channel is free

[gritty widget]

[cold vine]

yeah its this

[fading vale]

moth doesnt understand how to take derivatives of multivalued functions part 10138409234

[cold vine]

go ahead ^^

[fading vale]

so for the derivative of the function psi here

if you compute the jacobian you end up with like ((1, d/dv (dg_v)^t w), (0, d/dw (dg_v)^t w)) i think

[cold vine]

yup ^^

[fading vale]

So I need to compute the derivative of this map at (y, 0)

i.e basically explicitly get the tangent space of the normal bundle

my idea was to do this explicitly w/ the jacobian and for that im pretty sure you end up with like

$(x_1, x_2) \mapsto (x_1, (\frac{\partial}{\partial y} d g_y^\star (x_2))$

[gentle osprey]

Moth

[fading vale]

but i have no idea how to work with this second term

like at all lol

im not even sure intuitively what this means? or like i kind of do, because at (y, 0) in the normal bundle we have Y x {0} and {y} x N_y Y (where N_y Y is the orthogonal complement of T_y Y) and we're like... varying in a neighborhood of y along {y} x N_y Y

maybe im being dumb and misunderstanding how the jacobian works in the context of computing the differential at a point

also that y is a typo it should be (x1, x2) |-> (x1, blah blah blah)

but texit wont let me change it

[fringe blaze]

can someone tell me how to show this property proves two metrics are equivalent: ∃a,b>0, ∀x,y∈X such that d1(x,y)≤ad2(x,y) and d2(x,y)≤bd1(x,y))

im reading its called strong metric equivalence but i cant find a proof for it

[honest terrace]

what's the definition of equivalent metrics you know ?

[fringe blaze]

two metrics that generate the same topology?

[honest terrace]

Right.\
So try to show that if for every $(x, y) \in X$, there exists $a_{x, y}, b_{x, y} > 0$ s.t $a_{x, y} d_1(x, y) \leq d_2(x, y) \leq b_{x, y} d_1(x, y)$, then the topology generated by $d_1$ and $d_2$ are the same.\
Strong metric equivalence is even stronger than that (it says that the $a$ and $b$ are the same for any pair of $(x, y) \in X$).\
To show the two generated topologies are equivalent, try to show that every open ball of $(X, d_1)$ is included in some open ball of $(X, d_2)$, and reciprocally

[gentle osprey]

Shika-Blyat

[fringe blaze]

okay ill work with that thank you for your response

[west spindle]

aight so imma post this here since it's a computational geometry problem

what i THINK might work is first making the delaunay triangulation of P and then searching through its edges

but im not sure this actually gives what i need

and im not sure how to go about proving it either way

[gritty widget]

I guess there is an obvious O(n^3) algorithm using the hint, maybe it can be improved. (For each pair of points, draw a line between them, and consider all ways to "complete" the quadrilateral)

[west spindle]

the DT is my attempted improvement

[hazy siren]

hello

can someone help me with an example for path connected but not connected?

[gritty widget]

did you mean "connected but not path connected?"

[hazy siren]

i tried websearching and a few examples popped up but i didnt understand

nope i mean i=the other way

[gritty widget]

well, all path connected spaces are connected, so there are no examples

[hazy siren]

our teacher said this at the end of the class, and told us to look into it

idk, maybe i heard him wrong?

i was confused too

[gritty widget]

the canonical example is the topologist's sine curve

google that

[honest terrace]

sin(1/x^2) or smth like that, right ?

[gritty widget]

("canonical" in the sense of "this is the most common example")

[hazy siren]

ah alright thanks, ill look into that

so that ones an example for connected but not path connected?

[gritty widget]

yup

[hazy siren]

thanks

[gritty widget]

it's connected because it's the closure of a connected set, but not path connected, since joining (0, 0) to any point on the graph part by a continuous path will be impossible

[long hornet]

Is there any "practical" application of the fact that every topological manifold is metrizable?

[honest terrace]

all results on metric spaces immediately becomes true, isn't it enough ?

[gritty widget]

at least for smooth manifolds, i think some stuff to do with the tubular neighbourhood theorem involves choosing a metric

i've always blackboxed it so i don't know the specifics

but you can maybe look there

[long hornet]

I see

I wonder whether it's possible to ensure that, for example, all sufficiently small balls are actually regular coordinate balls, or something similar

[long hornet]

Unfortunately I don't know what a normal bundle really is.

[shut moat]

tangen't bundle

[gritty widget]

@west spindle if you confirm that your DT thing works, or if you find a solution in general, I will be curious to know 🙂

[west spindle]

i'll let you know

in the meantime

[west spindle]

<@&286206848099549185> see this, if any of y'all wouldn't mind.

[gritty widget]

all manifolds live in R^n for some n

easy

[gritty widget]

I think the DT thing fails (at least, without modifications). Eg consider the set of points: A=(0,0), B=(N, 1+eps), C=(N+eps,1), D=(-N, 1+eps), E=(-N-eps, 1) where N is large and epsilon is big. Then the optimal quadrilateral is the one that excludes the (0,0). But there exists the DT that consists of ABC ABD ADE

@west spindle

Lmao, "where N is large and epsilon is big", I meant "epsilon is small"

[west spindle]

hrm

okay, so DT won't work

shit

[plush needle]

@west spindle Maybe this can help, although I suppose your problem allows concave quadrilaterals as well? https://www.ics.uci.edu/~eppstein/pubs/EppOveRot-DCG-92.pdf

[wanton marsh]

they say time complexity of O(n³) ?

[gritty widget]

Looking at this pdf feels a bit close to giving up

But O(n^3) is basically easily achievable with the hint

And the easy O(n^3) algorithm is O(n^2) if the points are the vertices of a convex polygon, but I'm not sure if the original problem is reducible to this case

[west spindle]

okay so i have a new idea

if we know the min area quadrilateral among n-1 points, and we add another point, can we find the new min area quad in O(n)?

if yes, then we can get the O(n^2) total run time we want

[wanton marsh]

I think I have an O(n²log n) algorithm :(

[west spindle]

oh you do?

do share

i think i might get at least partial credit for that

[wanton marsh]

make a list of all the slope of the lines between pairs of points

and sort it

[west spindle]

aha. slopes

okay

[wanton marsh]

knowing the list allows you to keep a list of points sorted by "altitude" for all the slopes as you turn

[west spindle]

altitude?

[wanton marsh]

for a slope of 0 that would be sorted by y-coordinate

[west spindle]

oh

[wanton marsh]

and we want such a list for all the slopes

[west spindle]

okay i think i see what you're getting at

[wanton marsh]

because to get the minimal quadrilateral we want to find for each pair of points, a third point closest to them on top and a fourth point closest to them under

[west spindle]

for each slope take the points with the positive and negative altitudes closest to zero

[wanton marsh]

like the hint says

[west spindle]

the slope sorting thing came up in class actually

so it may not be far off

[wanton marsh]

if you can do the slope sorting in O(n²) you win

[west spindle]

a linear time sort would be quite the breakthrough

[gritty widget]

I'm not sure I follow how a list sorted by slopes is used for this

[wanton marsh]

start with a slope of 0 and have a list of points sorted by y-coordinates

have your list of pairs of points sorted by slope ready

as the slope goes up a little, you will get the first pair of point with a small positive slope

they will be consecutive points in your list of points

you look at them plus the one above them plus the one below them in the list and see how big the quadrilateral they make is

then you swap those two points in the list, and now you have the sorted list of points for a slope slightly higher than the slope of their line

and now you continue with the next pair of points in your list of pairs of points sorted by slope

and so on

[shadow charm]

isnt there a nice quadrilateral area formula dependent on diagonals?

maybe there's a way to think about it only in terms of the diagonals of the quadrilaterals?

just a thought

[wanton marsh]

this allows you to visit the n² pairs of points while for each of them, have a complete list of all the points sorted by the distance to the line between them

[honest terrace]

there is

[wanton marsh]

which if you do it naively is O(n³ log n)

[shadow charm]

n^2 log n no?

[wanton marsh]

no you if you do it naively and sort the list from scratch every time it's O(n³)

[shadow charm]

right

[honest terrace]

It's like $\frac{d1 \times d2}{2} \times \sin \theta $ where $d1$ and $d2$ are the length of the two diagonals and $\theta$ is the smallest angle formed by the two diagonals

[gentle osprey]

Shika-Blyat

[honest terrace]

iirc

[gritty widget]

You are not claiming that consecutive items in the sorted list create small quadrilaterals right?

[wanton marsh]

I am

[gritty widget]

Are you considering pairs of points with one of the points fixed?

[wanton marsh]

the hint says that any quadrilateral can be obtained by glueing two triangles

so you can just sort quadrilaterals by their diagonal (the convex ones have two ways to do this)

and then for a given diagonal where we want to glue triangles on

we just need the smallest triangles on each side

hence you have to sort the points

[gritty widget]

Yes, that I follow (and which I was saying there is an obvious O(n^3) algo, and O(n^2) algo if the points form the vertices of a convex polygon)

[wanton marsh]

by how far they are from the line

[gritty widget]

How do we sort the points (efficiently)?

[wanton marsh]

I explained that

when the slope you use to sort changes a little, you only need to do little changes to the list

you just swap to points P and Q when your slope goes through ... well the slope of the line through P and Q

[honest terrace]

.sort()

[wanton marsh]

you might have an annoyance when there are three or more points aligned

because for getting the smallest quadrilateral you need to pick a pair with the shortest base length for your diagonal

so you have to actually care about the order you pick them

[gritty widget]

Hm. So I have a sorted list of points (P1, Q1),(P2,Q2),..., (Pm, Qm), let's say the diagonal we are consifeting is (P2, Q2), what are you saying we do?

[wanton marsh]

say I have a sorted list of points according to their distance to a line of slope t - epsilon

AND a sorted list of pairs of points according to slope of their line

and I am visiting a pair of points (P,Q) of slope t

then I find P and Q in the list of points

if the list looks like ... X, P, Q, Y, .... I look at the quadrilateral made by the triangles XPQ and PQY

then I swap P and Q in the list of points

so I get ... X, Q, P, Y, ....

now my list of points is sorted according to their distance to a line of slope (t + epsilon)

and then I go to the next pair of points (P',Q') of slope t' > t

and I do it all again

[west spindle]

1/4 * product of diagonals * sin(angle between them)?

or something

wait if that's actually true

then i think we might have a breakthrough

holy shit

...wait

hm

[wanton marsh]

O(n⁴) algorithm ?

[west spindle]

hkshfhffg fuck

yeah

[shadow charm]

im personally stumped my paper is full of dots and lines, and no improvements

[west spindle]

WHAT A BREAKTHROUGH! [digs straight down]

[shadow charm]

this is probably completely wrong but could the minimal quad be made from 4 consecutive points on a minimal spanning tree?

i dont have any justification for this it just seems so visually and maybe it can help idk

nvm found a counterexample by mesing around

[west spindle]

minimal Euclidean spanning tree?

[shut moat]

oh no sham left :(

[west spindle]

there's a part b to the min area quad problem by the way

but i think i can solve it

kinda?

part b is to count the number of convex quads with vertices from P in O(n^3 log(n))

which i believe can be done with the slope-sorting method @wanton marsh presented

orrr wait maybe not. shit

[tough imp]

He’ll be back

[shy moss]

hi

the fundamental group is defined to be te equivalence classes of loops, but is possible to define a group on arbitrary paths, not necesary loops?

[shut moat]

l think if you fix two distinct endpoints then you get the fundamental groupoid

[gentle osprey]

Search results for fundamental groupoid

From https://ncatlab.org/nlab/search?query=fundamental+groupoid
Direct page found at: fundamental groupoid

3 results where query appeared in title.

fundamental groupoid
fundamental groupoid of a cubical set and the cubical nerve of a groupoid
simplicial fundamental groupoid

116 results where query appeared in body. (Page 1/11)

2-groupoid of Lie 2-algebra valued forms
2009 August changes
2009 March changes
Atiyah Lie groupoid
Calculus of fractions and homotopy theory
Categories and Sheaves
Chern-Weil theory in Smooth∞Grpd
D-topological infinity-groupoid
David Michael Roberts
Galois cohomology
Galois theory

[fading vale]

here tau is just summation lol and N(S^2) is the normal bundle

intuitively should this have any critical values?

i know for sure that its regular at S^2 x {0} and i dont see why it would ever not be?

like its not like N_(u, 1) S^2 should be any smaller at S^2 x {1} or whatever

so i Think we can literally just take charts of R that translate a neighborhood by n? cant we?

and then tau is regular on S^2 x {n}?

[rugged swan]

no, tau take a point x of S^2 and a normal vector y at this point and "translates" x to x+y

you should see something no regular

to help you, such an irregular point is called a focal point

[empty grove]

You dont fix endpoints, the fundamental groupoid is the set of all paths (upto path homotopy) in a topological space, along with a partial binary operator which is the concatenation of path given that their respective endpoints match up

[fading vale]

yeah i guess im just gonna have to hunker down and compute the differential

lmao

sigh

i guess identifying it w/ S^2 \times R will make it a bit easier

@honest narwhal so re: the thingy we talked about yesterday (thanks for that btw it worked out)

the thing with the 2nd derivatives and such

derivative of psi and all

so im working with the case here where i need to find the derivative of psi at an arbitrary point (y, v) in the normal bundle

so here we get like:
1, 0
d^2 g_y^t (v), dg_y

and if we wanna find what this derivative does to a point (x1, x2) its like, (x1, d^2 g_y^t(v) * x1 + dg_y^t(x2)), right?

we have to fix a v in N_y Y and evaluate the hessian at that v

here im taking it specifically for the sphere... so i guess i have to find a g about y in S^2 that maps an open neighborhood of that to zero

im guessing this is where the focal point stuff comes in because somethings omething stereographic projection

[rugged swan]

tho you don't need to use local parametrization to compute the differential of tau

you can see your normal bundle as a submanifold of R^4

then it's easy to compute tau, it's the sum lol

it's linear

[fading vale]

yea its diffeomorphic to S^2 x R

[rugged swan]

did you succeed the tau exercise ?

[fading vale]

maybe im being silly but then i dont really see where it would fail to be surjective

[rugged swan]

this

oh

maybe try (x,y) -> x-y from R²->R

you'll see

[fading vale]

find the focal points of that?

[rugged swan]

uh no it doesn't work lol

for me the derivative of tau is (h,k) -> h+k

[fading vale]

yeah for sure

tau is linear after all

[rugged swan]

oh yes, and (h,k) is in the tangent space of S^2xR

thus in Ker(df)xR where f(x,y,z) = x²+y²+z²-1

[fading vale]

um is the domain of d tau here ker(df_y) x R?

[rugged swan]

yes

[fading vale]

oh i think i kind of see, before when i was working with a general Y and (y, 0) i had to compute the differential of psi to determine the tangent space of N(Y)

but in the case of S^2 I can just use that N(S^2) is diffeo to S^2 x R

so the tangent spaces are diffeomorphic as well

so i dont have to compute d psi to find the domain of d tau

[rugged swan]

yeah

because you know N(S^2)

[fading vale]

right

mhm

ok thanks haha i think i get it better now

[rugged swan]

btw I still don't see where there would be a non regular point for me but I think it's (x,-x)

a focal point is a point f such that there's x and y in S²xR s.t. tau(x,y) = f

[fading vale]

mhm

[rugged swan]

(I've given the bad definition before)

[fading vale]

maybe im being dumb but i still cannot see a point where this fails lol

im not sure how (x, -x) works here since x is in S^2 and -x in R

[rugged swan]

yeah I don't see why too, I think it's true that it's not regular on (x,-x) bc it's written in my course xD

if you have the answer I'm aware, it's really weird

[gentle osprey]

PAHUS

[cold vine]

I have projection to $B =[0,1]\times{0}$ from $E={(x,y) | \leq x, 0\leq y, x+y \leq 1}$. This is clearly a fibration since projections are fibrations. Any clue on how to show that this is not a fiber bundle ?

[gentle osprey]

PAHUS

[obtuse meteor]

this is very false? unless you mean homotopy not rel {0, 1}

yeah

plus you can't take a group structure on it

what you said would imply that fundamental groupoids are all contractible

which would be very bad

[gilded shell]

Anyone know what a Gdelta set is?

https://cdn.discordapp.com/attachments/486841085210132490/834210473812819968/unknown.png

[gritty widget]

G_delta means it's a countable intersection of open sets

[gilded shell]

Cool thanks. Why G? Why delta?

[gritty widget]

it's a german language thing, i think

The notation originated in German with G for Gebiet (German: area, or neighbourhood) meaning open set in this case and δ for Durchschnitt (German: intersection).

^from wikipedia

[gilded shell]

That is so weird

Thanks

God damn Germans...

[shut moat]

ah ok

[fading vale]

hmm do you think by this they meant that its like

((0, 0, 1), -1)

agh this is so confusing

@gritty widget do u have like any idea how to do this

ok like part 1 is fine.

the critical values are so confusing to me

namely i dont feel like they should exist

i know for a fact d tau is surjective at (y, 0) and like

i dont understand why moving up would matter

at all

[west spindle]

im still wondering how to do that min area quad problem from yday tbh

O(n^2) seems so out of reach oof

[gritty widget]

Did you understand zeps n^2 log n solution?

[west spindle]

yeah

but the problem said O(n^2), so...

[gritty widget]

Can you explain the idea behind it? I still don't understand it 😞

[west spindle]

basically for every segment connecting two points, we drop perpendiculars from all the other points onto its line

and look for the shortest ones above and below

[gritty widget]

Yeah, and I believe he is saying we can find the closest points to this line segment (or rather a list of points listed by their distance), and when we iterate to the next set of two points, we can find the new closest points without recomputing everything. But I don't see how that is possible

[gritty widget]

I think of this as the trivial O(n^3) algorithm (when we iterate over all points to find the closest ones). (And we can optimize this to O(n^2) if we assume that the points form the vertices of a convex polygon, in which case intuitively it suffices to check the points adjacent to P and Q). But I don't follow how zef is optimizing this from n^3 to n^2 log n

In general, I don't see how the slopes "play nice" with the distance of points to the line segment PQ, or that a list of points sorted by distances to a line segment can be easily updated. I am fairly sure I can think of examples where iterating to the next slope causes O(n) swaps to the list of points sorted by distance

[rugged swan]

The derivative should be more precisely (h,k) -> h +kx for me, if you differentiate in S^2xR. But it's still wierd because h is orthogonal to x

[soft cradle]

I have an idea, but haven't proved it. Translate the points so that they are all in the first quadrant. Then sort the points such that $(x_i,y_i) < (x_j,y_j)$ if either $x_i < x_j$ or $x_i = x_j$ and $y_i < y_j$ in $O(n \log n)$. Arrange the points in a grid. Points with the same $x-coordinate$ lie in the same column and those with the same $y-coordinate$ lie in the same row. Now, for each point in the sorted list, consider the minimal quadrilateral with that point as the leftmost and bottom-most point. 
$$ $$
It's likely that we would need to consider points only on a certain number of adjacent gridlines which could be bounded by the sum of the number of points in those adjacent vertical or horizontal gridlines and that is $O(n)$. Showing this would probably need to consider a bunch of cases depending on if some of the other 3 points lie on the same horizontal or vertical gridline or not. Also having the the parallel gridlines should make area comparisons somewhat easier.
$$ $$
If that's possible, we would be done in $O(n log n) + O(n^2)$

[gentle osprey]

andreO

[west spindle]

this feels really hacky

and it also feels like its disregarding the hint i was given, which was to consider a quad as two triangles attached to a diagonal

[gritty widget]

Oh, maybe I'm starting to see zefs solution now

[wanton marsh]

we aren't sorting an arbitrary list. There are constraints on the slopes that pairs of points can have.

so it's not completely ruled out that sorting in O(n²) is possible

[plush needle]

If you project onto the diagonal, and are only interested in the closest points, and given that you have to calculate the projections anyway. Would it not suffice to keep the two points with the the shortest distance in memory instead of producing a list and sorting it?

[gritty widget]

My understanding is that: you have a list of points sorted by signed distance to a line PQ. Then the idea is that if you were to shift PQ without changing the slope, then the list of points is still sorted. And iterating to the next line segment from PQ is sort of like shifting the line segment, and changing the slope by epsilon. Is that a more or less accurate understanding?

[wanton marsh]

yeah

[west spindle]

we are rotating not shifting

oh

hrm

[wanton marsh]

well the next line of closest slope can jump pretty far

but since the sorted list doesn't change if you translate the line that's not a problem

it's only rotating the line (so changing the slope little by little) that will change the order

[west spindle]

see this point is where im not rly sure

how are you suggesting a sort in O(n^2)

[wanton marsh]

I am not

[gritty widget]

Do you need hard O(n^2) or is eg amortized O(n^2) okay too? (Maybe you can throw a ton of data structures at the problem to keep run times low)

[west spindle]

hard O(n^2)

[wanton marsh]

I am just saying that just because sorting an arbitrary list has a lower bound of O(n log n)

doesn't mean that for our case O(n² log n) is a lower bound for sorting the pairs of points

[west spindle]

oh ok i think i see your point

im not sure i get your and 8da's point about rotating

[gritty widget]

Yeah, eg maybe you can do merge sort

For each point, sweep over the other points such that the line segments you create are already sorted by slope

[west spindle]

i am assuming that we have some blackbox sort function we dont care about the innards of that runs in n lon(n)

n log(n)

wait did you say sweep

[gritty widget]

Then this creates a list of n line segments that are already sorted. Then merging this should take n^2 time

[wanton marsh]

isn't that already O(n * n log n)

[gritty widget]

Actually I'm not sure how to do the sweep such that you iterate over points in a way that they are already sorted

[gritty widget]

It would be n n log n if you manually sorted each list, yeah, but I guess I was thinking of some way such that it "automatically" gives you a sorted list

[wanton marsh]

also, merging n lists of size n takes O(n² log n) time I think

[west spindle]

n sorted lists?

[wanton marsh]

yeah n sorted lists

[west spindle]

fuck. this is making my head hurt

[gritty widget]

Ah, yeah, I am making too many mistakes/thinking too wishfully 🙂 I don't have an n^2 algorithm for merging n lists

[west spindle]

bottom line is i think i might be able to get pity points if i present the "sort slopes in O(n^2 log(n))" algo

so its not ENTIRELY hopeless

[wanton marsh]

in practice the difference between O(n²) and O(n² log n) is not that crippling

compared to a O(n³) algorithm

(it might also be a project euler problem but I can't really remember the number and a lazy search didn't bring up anything)

[plush needle]

It's certain that concave quadrilaterals are supposed to be included?

[wanton marsh]

I think so

[west spindle]

yes

i can assure you they are

[honest terrace]

Also I realize I should've kept this secret for some reason so I'll delete the msg

[wanton marsh]

I guess you have to submit it as a project euler problem and wait for people to solve it and maybe one of them will post a O(n²) solution

[west spindle]

lmao

[gritty widget]

I guess it is a question for ann how cheaty it would be to look at the algorithm from the book in your screenshot

[west spindle]

i dont have a book, all i have is a pdf with a bunch of problems our teacher gave us :P

[west spindle]

ah but is the smallest-area triangle necessarily contained in the smallest-area quad?

[gritty widget]

I just mean roenback posted a screenshot, and that book/algo for theorem 2 probably contains a great idea for solving the quadrilateral problem

[west spindle]

oh

i dont think itd be cheaty at all

[plush needle]

https://www.cs.princeton.edu/~chazelle/pubs/PowerDuality.pdf

(page 82)

[west spindle]

okay i'll look through this once ive had some food bc all this math shit is making me kinda hungry

[cold vine]

Im missing something very obvious but why does the definition of Fiber bundle give us the homeomorphism to the fiber of the singleton set? Since the definition only talks about homeomorphisms for the open sets and singletons are not open sets

[empty grove]

Restrict the homeomorphism phi to π^{-1}({p}). The restriction of a homeomorphism is a homeomorphism onto its image.

[cold vine]

Ahh true! Thank you!

[fading vale]

i figured it out btw

you had to explicitly write out the diffeo between N(S^2) and S^2 x R and basically use it as a local paramaterization of each point so like

if g: S^2 x R -> N(S^2) is the diffeo you have to compute d(tau circ g)

[cold vine]

Hmmm I have shown that the projection from $E={(x,y) | \leq x, 0\leq y, x+y \leq 1}$ to $B =[0,1]\times{0}$ is not a fiber bundle, but how can I show that it is a fibration. I think I can use the fact that B is a retraction and I believe that projections are fibrations, right? Am I in the right direction?

[gentle osprey]

PAHUS

[rugged swan]

yeah ok legit then. I've computed the differential wrongly. Btw someone explained me that if you restrict tau to (x,-x) it's constant equals to 0 then you can see that the Kernel of df is non trivial by differentiating with resepct of x

[fading vale]

yea

the critical values are just 0 btw

the points are S^2 x {-1}

you end up getting like (1 + t)v + t'w or something and v and the tangent plane span R^3 so you need (1 + t)v to be 0

i.e t = -1

[empty grove]

Use the fact that the projection from the square to the interval is a fibration, and that any initial condition on the triangle can also be viewed as an initial condition on the square. Lift homotopy to the square, then choose appropriate retraction from square to triangle to compose the homotopy with

A bit wordy but I hope that makes sense. If I'm not wrong there will be only 1 choice for the retraction to the triangle

[cold vine]

Ah I think I see this , thanks!

[feral copper]

Hello ! Struggling with visualization a bit here... Working in dimension four 🙂
So I managed to see that $$\natural^k(S^1\times D^3)$$
is a 1-handlebody. Now, I want to see a handle decomposition of $$\natural^k(S^2\times D^2),$$
what could it be ? It may be very obvious but I can't find it...

[gentle osprey]

Matplotlib

[feral copper]

The thing I want to do is trying to use the Laudenbach-Poénaru paper (http://www.numdam.org/article/BSMF_1972__100__337_0.pdf). I understood the paper and the outline of the proof, but I've read elsewhere that this paper means that the attaching of the 3-handles can only be done in a unique way, and I don't understand how this is implied by the results proven...

[gritty widget]

$\natural$

[gentle osprey]

Σ, liestinche

[gritty widget]

What it mean?

[feral copper]

Boundary connected sum

For M natural N you glue M and N along discs on their boundary

It is such that bdy(M natural N)=bdy(M) connected sum bdy(N)

[bright acorn]

When my professor was discussing non-isoled singularities, branch points to be more exact, he briefly mentioned how we can consider a group called the monodromy group in order to properly study how does a certain "function" like the complex logarithm fails to have a unique value if you run around a point.

Could someone give me a better explanation of this?

[elder yew]

e^2pi i = e^0

This is a no no

[bright acorn]

hm

[elder yew]

Since it's periodic, the log function is an inverse to the exponential

So you have to cut off the angle to not go all the way around

otherwise you get multi-valued

[bright acorn]

Yes

But how is it related to monodromy groups?

Or what is it exactly?

[elder yew]

There's the monodromy theorem

You can try looking that up

[bright acorn]

oh

I will give a look

thanks

[cold vine]

I have a $E={[0,1]\times{0})\cup({0}\times[0,1])$ and the projection to the unit interval on the x axis denoted $B$. Now I have shown that this is hom eqv, and fibers are contractible, but I am trying to show that this is not a fibration. I have tried to come up with some homotopies in B that can't be lifted but I can't seem to think of one. Any ideas?

[gentle osprey]

PAHUS

[cold vine]

I was thinking about having a homotopy between identity and constant map to $(1,0)$ from either $B$ or ${0} \times I$ but itseems that these are both liftable

[gentle osprey]

PAHUS

[flint cove]

In May's „concise“, is this a typo, or are there meant to be two stars?
If yes, is this done to convey the difference to the direct sum, i.e. finitely supported coefficient sequences?

[cold vine]

bump

[rugged swan]

I guess the point (1,0) when projected in B is homotopic to (0,1), but this isn't true in E with respect to the fibers

@cold vine

the homotopy H(.,t) = (0,t) can't be lifted

bcp it would give something like h : [0,1] -> E where h(0) = (1,0) but p(h(x)) = (0,x)

then h(x) = (0,x) for x>0

[pale sky]

So i had this question asking me to prove this

i was wondering if my follow up answer is correct

I think the answer here must be true because of what i state right?

but it seems too straightforward

[obtuse meteor]

reversing the diagram like this

you don't have the same properties of the map

[pale sky]

hmm but it says the sequence is exact

so dont i still have the same properties?

[obtuse meteor]

Z^beta and T have swapped positions

[gritty widget]

Hi Everyone! I'm looking for an example for a covering number on S is smaller than the packing number on T, despite T being subset of S. Any hints?

[obtuse meteor]

so exactness tells you different things

[cold vine]

Thanks so much! Couldn't figure that one out although I did think of the homotopy

[empty grove]

Which book are you reading from btw?

[cold vine]

I have materials which are loosely based on Tom Dieck, but I own Hatcher which has been a help.

[pale sky]

oh this time i get Z^b + T right?

[cold vine]

Also checked out Munkres when relevant

[obtuse meteor]

oh @pale sky I see

you you didn't use anything about

Z^b and T right?

the statement is false

in general

it's equivalent to the sequence splitting

by the splitting lemma

[pale sky]

wait sorry i dont see why its false

isnt the + commutative?

[obtuse meteor]

like

your proof of the statement in part (a)

you said it didn't use any properties of Z^b and T

it's not true in general that if you have an exact sequence 0 -> A -> B -> C -> 0

that B = A + C

so your proof of part (a)

is wrong

[pale sky]

ah ok do u see where it is wrong? i have examined it for a bit and not sure where

[pale sky]

but hatcher says this

[obtuse meteor]

yes, when the sequence splits

[pale sky]

isnt that what i prove here?

[obtuse meteor]

you can't prove this splits without using properties of T and Z^b

[pale sky]

in a)

[obtuse meteor]

let me take a closer look

I think your proof is wrong in that

H/Im m_1

can't be identified with H/T

because T is not a submodule of H

[pale sky]

isnt it though? since this diagram has to be for homomoprhism of modules right

[obtuse meteor]

consider also like. An exact sequence 0 -> Z -> Z -> Z/5Z -> 0

where the first map is x5

and the second map is quotient

Z is not Z + Z/5Z

your proof would imply this

because Z/im (x5) is not Z/Z = 0 even though im(x5) is isomorphic to Z

[pale sky]

sorry im not sure what this means

[pale sky]

"where the first map is x5" and "and the second map is quotient"

[obtuse meteor]

so like

consider maps 0 -> Z -f-> Z -g-> Z/5Z -> 0

f(x) = 5 * x

g(x) = x mod 5

[pale sky]

ah i see

tyty

hmm for a)

is there a way to rectify this proof

or is it no where near what i need?

also im still abt confused about why T cannot be identitief with im m1 since m1 is a module homomorphism

[obtuse meteor]

so like

T and im m1 are isomorphic

that doesn't mean H/T and H/im m1 are isomorphic

in fact H/T doesn't even make sense

consider

5Z and 6Z are isomorphic as groups right

but Z/5Z and Z/6Z

are not isomorphic

5Z and 6Z are not isomorphic in the slice over Z

[pale sky]

ah i see thanks!

[empty grove]

Thanks

[fading vale]

diff top notation is...

dense

[gentle osprey]

Moth

[fading vale]

5 lines of this hurts my head

[gritty widget]

[sweet wing]

[dusk heron]

Let's say we have a smooth function $f:\mathbb{R}^3\to\mathbb{R}$ which goes to a constant value at infinity, i.e. $f(x)\to C$ as $|x|\to\infty$, for some $C\in\mathbb{R}$. Let's now say that we want to find a diffeomorphism $F:\mathbb{R}^3\to\mathbb{B}$, where $\mathbb{B}\subseteq\mathbb{R}^3$ is the open unit ball, such that $f\circ F^{-1}:\mathbb{B}\to\mathbb{R}$ extends smoothly to a function on the closed unit ball $\overline{\mathbb{B}}$ (which necessarily will have to equal $C$ on $\partial\mathbb{B}$). What kind of decay do we need on the derivatives of $f$ in order to be able to do this?

[gentle osprey]

gustavn64

[remote beacon]

Guys what's the scariest topology?
Zariski topology

[hazy siren]

is the R^2/QxQ plane connected and path connected?

[gritty widget]

Nice problem 🙂 the first thing that comes to my mind to try to show that if is not connected is to write down an equation in x and y like p(x,y)=0 that has no rational solutions

Then we can take p>0 and p<0 as open sets

Sadly I am not number theoretic enough to know how to write down such a p

[marble socket]

you can take a look at this

[gritty widget]

Oh, wow, it's connected?

[marble socket]

path connected 😛

yep ❤️

[gritty widget]

Ah, I was wondering what was going on since I intuitively agree that this space is path connected. This should be "only rational solutions" instead of "no rational solutions". Which sounds false

[marble socket]

(just to confirm, that's set difference, right?)

[honest terrace]

set difference is \, not /, no ?

[viral atlas]

[ivory dragon]

$\setminus$

[gentle osprey]

Namington

$\setminus$

[ivory dragon]

/ is usually a quotient

some authors dont care about your directionormative notation though

and will do whatever

[lucid turret]

Let (X,x_0) be a pointed topological space. There is a category with objects real numbers and morphisms x -> y being continuous maps f: [x,y] -> X with f(x)=f(y)=x_0 (i.e. a loop but the interval doesn't have to be [0,1]). Morphism composition of arrows f: x -> y, g: y -> z is the loop composition but now on an interval [x,z].

Please has anyone ever seen this somewhere except for here and now?

[flint cove]

No, but I haven't seen much math in the grand scheme of things

where did it come up?

Not necessarily, if you interpret [b,a] to be empty when b>a

because then no non-identity arrow is invertible

[sweet wing]

I've seem tom dieck briefly mention that we could do this but thats it

[flint pilot]

Hi everyone. This question came up in a Multivariable Analysis class, but I was told it's much more involved in Algebraic Topology, so they suggested I ask here. Thanks in advance! a.) Let $A = \cup_{j = 1}^{m} B_j$ where $B_1,\dots,B_m$ are non-empty open balls in $\mathbb{R}^n$, pairwise disjoint ($B_j \cap B_k = \emptyset$ if $j \neq k$). Prove $dim H^{0}(A) = m - 1 $ and $H^k(A) = {0}$ if $k \geq 1$. b.) Let $A = {(x,y) \in \mathbb{R}^2: 1 < x^2 + y^2 < 4}$. Compute $H^k(A)$ for $k = 0,1,2,\dots$

[gentle osprey]

hype_db

[empty grove]

Shouldn't dim H⁰(A) = m since there are m path components?

Wait this is de Rham Cohomology right?

Or singular homology?

[flint pilot]

@empty grove neither of those words sound familiar. I'm checking my book to see if they are stated.

[empty grove]

Difference is idk singular homology lol

[flint pilot]

the set of closed k forms C^k(A) divided by the set of exact k forms E^k(A). We define H^k(A) = C^k(A)/E^k(A)

[empty grove]

Yeah that's de Rham Cohomology

Isn't singular homology abelian groups and de Rham Cohomology R vector spaces?

oh damn didn't know that was a thing

[empty grove]

But I think you'll have this since locally constant smooth functions on A are essentially m tuples

[frosty sundial]

You can take singular cohomology with coefficients in any abelian group A, though you usually want it to be a rinf

If you use the ring Z you get the usual singular cohomology

[empty grove]

I see

[frosty sundial]

In your definition, you probably look at integer linear combinations of chains

Err, cochains

[empty grove]

Right

[frosty sundial]

You can replace “integer linear combination” with “real linear combination”

And that’s (co)homology with real coefficients

[gentle osprey]

slimvesus

[frosty sundial]

That’s not true slim

No

See the universal coefficient theorem

[empty grove]

And the higher cohomologies will be dim 0 because H^k(A) will be the direct sum of H^k (B_i) which are all 0 as B_i are contractible

[flint pilot]

well, if I can prove it is actually equal to m, I'll be ok. My professor is known to have typos lol.

[empty grove]

Yep

[flint pilot]

yes I remember that

connected components? That sounds new to me.

[empty grove]

Connected component = each B_i here

[flint pilot]

the two theorems in the section given are the Homotopy Equivalence Theorem and Mayer-Victoris Special Case. But ok, I think I see what you mean. There are m B_i's, so m connected components. That implies dimension m?

[empty grove]

You can use homotopy equivalence here

A is homotopy equivalent to m points

Just contract each ball

[flint pilot]

Ok hold on, I'm sorry my topology is very weak since I haven't taken a class in it. My professor also got stuck during the proof of the theorem, so he never did an example using it. When you say contract each ball, how exactly would I do that? And how would that allow us to conclude A is homotopy equivalent to m points?

He didn't define homotopy. In my book, it says that if two functions satisfy the theorem, then they are homotopy equivalences. And I did not know that fact, but I don't see why it wouldn't be true.

[empty grove]

Yeah check the definition of homotopy

[flint pilot]

that's what I asked my instructor and he didn't have an answer lol. It's theorem 40.1 in the Munkres Analysis on Manifolds book. It's easy to get a PDF and it's a long theorem, so is it possible to look at it there?

[empty grove]

Have you computed the cohomologies for a single ball?

[flint pilot]

nope. I really hate that my instructor covered this section. He did it on the last day of class and he got stuck in the middle of a proof and then he gives us a homework problem on it

luckily this is the ONLY problem on it. So I'm just hoping to get through it with some sort of an understanding and then when I take Topology I'm hoping to learn more.

agreed. I really don't like it... especially on the last day of class

[empty grove]

Try to think about this case first, m=1

Then it's a matter of taking direct sums

[flint pilot]

then its derivative is 0?

[empty grove]

No he just said def of locally constant lol

What does the function look like, can you think of any examples?

[flint pilot]

so I feel that would mean C^k(A) = 0 meaning H^k(A) = 0.

[empty grove]

Why would C^0(A) = 0?

[flint pilot]

I didn't mean to. I thought our only assumption was m = 1? Nothing about k yet.

ohhh that was my bad I missed it

so yes then it would simply H^0(A) = 0. But we want it to be m (or m - 1 according to my instructor)

[empty grove]

Why would it be 0?

[flint pilot]

wouldn't C^0(A) = 0? Meaning C^0(A)/E^0(A) = 0 regardless of E^0(A)?

oh that's true. I don't have a reason to think that. my bad

yes that's my bad

it implies f = c (where c is constant)

well if ALL of the derivatives have to be 0 I'd think we could.

because f' = 0 implies f = c as we said. That's true for all f.

[empty grove]

By all the derivatives do you mean derivative at all points?

[flint pilot]

oh, well you said it's not always true, so I guess I'm incorrect then

[empty grove]

Consider the case where your function is 0 on (0,1) and 1 on (2,3), not defined elsewhere

Derivative is 0 everywhere, so what property of the space allows this kind of thing?

[flint pilot]

I'm sorry, I'm really not sure. I feel it would always be true just because of how derivatives are defined

[empty grove]

See this example for a function in R

[flint pilot]

oh, so you're saying certain functions can be defined in such a way where non-constant functions can have derivatives equal to 0

so in C^0(A) there could be up to m functions that are not constant, but have derivative 0. There could be less, but that's the most they could have right?

or am I jumping too far?

Alright forget what I said then.

ok so on B_1 there must be a function with derivative zero. With what you said, now I'm not sure it has to be constant. I could define a function like the one above on B_1 why couldn't I?

Why not exactly how the one above was defined? Let B_1 = (0,1) U (2,3)

oh that's true. So on B_1 such a thing can't be defined. It has to be constant. Even if it has jumps, those jumps must be constant

[empty grove]

What do you mean by "jumps have to be constant"?

[flint pilot]

like a step function. A sequence of steps.

[empty grove]

But then it won't be a constant function right?

And you would get non differentiability where those jumps happen

[flint pilot]

yeah I see what you mean

so is that where the union of open balls comes in? We union the open neighborhoods where the function is constant?

ok sure. So there's 1 such neighborhood, B_1

[empty grove]

Wait slim

2 definitions of local constancy being used

That's why it's confusing

One is df = 0 and the other is there exists some neighborhood of constancy

@flint pilot you've seen the first definition right?

[flint pilot]

yes the df = 0.

[empty grove]

Right so take a path from x to y in the ball

[flint pilot]

that path has to be constant right?

[empty grove]

Which is a function from [0,1] to ball

No x and y are any 2 points on the ball

Path has nothing to do with f right now

[flint pilot]

ok

[empty grove]

Now given a function f on B_1, you can get a function on [0,1] from this

By defining g(a) = f(whatever a maps to under the path)

[flint pilot]

sure that's fine

[empty grove]

Notice that if f is locally constant, g would be too

Chain rule

So now you have a locally constant function on [0,1]

[flint pilot]

ok that's fine

[empty grove]

Can you say anything about locally constant functions on intervals?

[flint pilot]

they have derivative 0 in this case

they map to locally constant functions?

like @empty grove said g(a) = f(whatever it maps to) they have to be constant

I'm sorry I don't think I'm seeing what you guys are getting at

[empty grove]

So idea is I'm trying to reduce locally constant on a ball to locally constant on an interval

We want to prove that locally constant on ball means constant on the ball

Do you know that for intervals in R, locally constant implies constant?

You can use mean value theorem to say this

[flint pilot]

They definitely defined it at some point, I just don't remember exactly where. It has to be earlier in the book

let me see if I can find it

oh I just searched from the PDF and it didn't come up. Maybe they don't define it

[empty grove]

Yeah we need path the be differentiable at least

Right so if you have a differentiable path connecting x and y in whatever space, and a locally constant function on that space, you end up with a locally constant function on the path, and because on intervals locally constant implies constant, you get that the function must be constant on this path

So the function would have the same value at x and y, in particular

If path is constant then x = y right

oh you're saying we need to handle it separately

Right, it'll be fine but will just make things unnecessarily complicated

[flint pilot]

I think I see that. Just to make sure I'm still following because we've been going at this for a while. For the case m = 1, we have that A = B_1. We want to show dim H^0(A) = 1. By assumption, we are considering locally constant functions and want to show that they are constant on the entire ball B_1. Any two points connected by a path must be equal since we have a locally constant function on the path. Because on intervals locally constant implies constant, the two points are equal. Therefore, the function is constant on B_1. This means C^0(A) = 1 since we have 1 function that is constant with derivative 0. So all that is left is to show E^0(A) = 1 so that H^0(A) = 1. Is that a summary so far?

[empty grove]

dim C^0(A) = 1

[flint pilot]

yes my bad, I forgot to write that

[empty grove]

Right

That is correct

So now think about m=2

You have a locally constant function on the union of 2 balls

[flint pilot]

well wait what about E^0(A)?

for m = 1

[empty grove]

E^0(A) is always the trivial space

Because you don't have any -1 forms to differentiate

[flint pilot]

ah ok.

so it is 1 as well

[empty grove]

It has dimension 0

[flint pilot]

oh? so 1/0?

[empty grove]

Yeah dim 1 space quotiented by dim 0 space

Gives you dim 1 space

[flint pilot]

oh ok. I didn't know that. That explains it

[empty grove]

Can you justify why C^0 has dim 1?

[flint pilot]

my thinking was because we only have 1 ball, B_1. It's constant on all of this ball, so it is 1 dimension.

[empty grove]

Yeah but how do you get from constant on the ball to dimension = 1?

[flint pilot]

that's true. I don't know

[empty grove]

What is C^0(A) as a vector space?

[flint pilot]

the set of closed k forms on A

[empty grove]

Right but can you tell me what the set is exactly?

Like it's the constant functions, but how many of those are there

[flint pilot]

maybe like C^0(A) = {w s.t. dw = 0} Something like that?

[empty grove]

Well more precisely it will be {f_c s.t. c is a real number, f_c is the constant function corresponding to it}

[flint pilot]

ah ok yes that is more precise.

now, I feel like there could be a lot of those functions. I mean, there are infinitely many constants.

[empty grove]

ie you have exactly one constant function corresponding to each real, and this correspondence is a vector space isomorphism

[flint pilot]

yes each real has a constant function corresponding to it

[empty grove]

Yeah and do you see that it's an isomorphism between the real vector space R and C^0(A)?

[flint pilot]

yes that's actually very clear. It makes perfect sense

each real constant has a corresponding constant function

[empty grove]

Nice, and R is 1 dimensional as a vector space over R

Yeah

[flint pilot]

exactly yes

wow so that's m = 1! I see it now!

[empty grove]

Yep

Do you know how k>0 will work for m=1?

[flint pilot]

and it'll be the same thing for m = 2 I believe. Are there any other ambiguities? And as of right now, no, but I think i have an idea. If k >= 1 (not 0 right?) then there aren't any functions that will come out with derivative 0 since we no longer have constant functions. So C^k(A) = 0. Is that at least an idea? Or is there much more to it?

[empty grove]

Nope, there will be forms on A which will have derivative 0

[flint pilot]

then I'm not sure

[empty grove]

For example x dx

Try finding the higher cohomologies for just a point then

Instead of a ball

[flint pilot]

what do you mean by higher cohomologies? That word wasn't mentioned in my book. I know that was an hour ago that I mentioned that so I thought I'd remind you.

[empty grove]

k>0 lol

[flint pilot]

ok fair lol

but doesn't it say k >= 1? that's in the statement of the problem

prove H^k(A) = {0} if k >= 1

[empty grove]

Yeah but C^k(A) doesn't have to have dimension 0 for that

C^k is actually infinite dimensional, but turns out that all closed k forms on a ball are exact when k ≥ 1

[empty grove]

I'm really not sure how you're supposed to do this without having seen any homotopy, but this is called poincare lemma

[flint pilot]

ah and he skipped the section on Poincare's Lemma. Lovely....

[empty grove]

ok damn

[flint pilot]

well here's the thing. It's the section before this one in the book. So let's go with that fact, and I'll read it myself lol

[empty grove]

Yeah cool

[flint pilot]

my goodness this is literally my last problem of the entire semester and my professor was like, "let's make it really tough." lol...

[empty grove]

I mean it wouldn't be so tough if all this stuff wasn't skipped

Or done without prerequisites rather

[flint pilot]

Yeah. It really stinks. But, at least it's the last problem. Ok, so C^k(A) = E^k(A) meaning C^k(A)/E^k(A) = 1 since they are the same right? But wouldn't that mean H^k(A) = 1? It doesn't mention dimensions on this part.

[empty grove]

C^k/E^k would be the trivial space

These are vector spaces not numbers

[flint pilot]

oh, so that becomes 0 then

[empty grove]

Yep

[flint pilot]

well there we have it

[empty grove]

so with this we know that dim H^0(B_i) = 1 and dim H^k(B_i) = 0 for k > 0

[flint pilot]

and for m = 1

[empty grove]

Yeah I just put B_i instead of A

[flint pilot]

yes that's fine

[empty grove]

So we can move to m=2

So for k = 0, again you have the space of all locally constant functions

But on 2 balls

What can you say about such functions

[flint pilot]

before I type it all out, does the same argument work? Because that's what I'd say.

[empty grove]

Nope

You may not have a path between any pair of points

[flint pilot]

then how can we get the fact that it is constant everywhere?

I mean that was the entire basis of our argument

[empty grove]

we cant

but you can use that argument partially

in the sense that whenever x and y are in the same ball, the function would take the same value on them

because then you can connect them by a path

[flint pilot]

Ok so what about this: on B_1 and B_2 separately, we can connect any two points with a path. Now on the area where B_1 intersects B_2, we have the assumption that their intersection is empty if j \neq k. So would that allow us to disregard that problem? Then the union of those balls is still open

[empty grove]

B_1 and B_2 are given to be non intersecting

all the balls are pairwise disjoint

so you wont be able to connect points in different balls by a path

[flint pilot]

Ok, so how can we handle that?

[empty grove]

and notice that a function can in fact take one value on one ball and another constant value on the other ball, while having derivative 0

[flint pilot]

that is true

[empty grove]

There will not be any relation between the values

ie each value can be arbitrary

so C^0(A) is exactly the set of functions which take a value c_1 on B_1 and c_2 on B_2, where c_1 and c_2 are arbitrary reals

[flint pilot]

yes ok that makes sense

[empty grove]

what do you think is the dimension of this space?

[flint pilot]

I'd think 2 just because it's taking on 2 values

[empty grove]

right, the isomorphism to R^2 is that the above function maps to (c_1,c_2)

[flint pilot]

oh yes exactly! Back to the isomorphism argument. So yes this works

[empty grove]

and as always, E^0(A) = trivial space, so H^0(A) = C^0(A)

[flint pilot]

and it is equal to 2

[empty grove]

yep

so can you try and handle the case for arbitrary m, k=0?

[flint pilot]

I think I can do it. It will follow this same structure from m = 2. I think I'm good on that part!

[empty grove]

great

so for k>0

You are familiar with direct sums of vector spaces right?

[flint pilot]

like the direct sum they cover in introductory abstract algebra?

[empty grove]

yeah

[flint pilot]

then I've done it, but my abstract algebra is very rough. So it'll take some reminders as I do it

[empty grove]

Ah ok

[flint pilot]

I research applied math lol. PDE/ODE dynamical systems

[empty grove]

oh i see

So notice what we had was C^0(B_1 union B_2) = direct sum of C^0(B_1) and C^0(B_2)

[flint pilot]

yeah off the top of my head I don't remember how I'd calculate that direct sum. I may remember if I see it done, it's possible.

[fading vale]

singular homology with coefficients in A gives you modules over A so they coincide if you take a field

[empty grove]

just think of it as putting both things together in a tuple

right

[fading vale]

ofc Z-modules are abelian groups so in that case you get the abelian group stuff

[empty grove]

yeah

[flint pilot]

ok so taking an element in B_1 and a corresponding element in B_2 and adding them. Simple enough

[empty grove]

yep

and the reason was that the behaviour of a form on one ball doesnt have any effect on its behaviour on the other one

because the balls are disjoint

[flint pilot]

yes

[fading vale]

are you computing the de rham cohomology of a disjoint union of balls? have you seen contractibility?

[empty grove]

yes but no background in topology

[fading vale]

thats... so weird

[flint pilot]

@fading vale nope. The reason we've gone on so long is because my professor didn't teach us or require the necessary background. This is from an Analysis class. Never even defined homotopy

[empty grove]

so you get similar behaviour for the higher forms, that if you have a closed form on B_1 and a closed form on B_2, putting them together (ie defining a single form by combining their domains) you get a closed form on the union

[flint pilot]

so then the same argument would work. C^k(A) = E^k(A) meaning they are both the trivial space, so they have dimension 0

[empty grove]

and conversely given a closed form on the union, you can restrict it to either ball to get closed forms on each ball

[gritty widget]

talking about de rham cohomology
haven't even defined homotopy

my condolences

[empty grove]

yep, same argument works for E^k, and so all higher cohomologies are the same

are 0*

[flint pilot]

got it. Wow so that's part a of the problem lol.

[empty grove]

yeah lol

[fading vale]

you said you were given mayer vietoris right?

[empty grove]

can you do the same for m>2?

[flint pilot]

@fading vale it's in the section yes.

@empty grove you mean on part b?

[empty grove]

no, so far i was talking about m = 2

but see that the same argument works for all m

[flint pilot]

yes, it follows based on what you said. And for above. I'm good on that part