point-set-topology

[marsh forge]

Sorry are you saying that every map S^n->X is homotopic to a map S^n->X^n

Maybe up to me messing up the indexes

Because this is true

I think it should be X^n+1 actually

Anyway

[floral gust]

Yep kinda the same I think

[marsh forge]

Yeah this should be in hatcher

In his section on higher homotopy groups

X^{n+1} fully determines \pi_n(X^n)

[floral gust]

oh lemme check

Yeah that thing

[marsh forge]

If you find it

Lmk if my indexes are correct

[earnest epoch]

that result sounds familiar to me

because everything else has a trivial contribution

[marsh forge]

Yeah its pretty intuitive

Basically a map from Sn is trivial iff it extends to Sn+1

So its only the n+1 cells that matter

[marsh forge]

Ok time for another rant

So orientation is a continuous choice of generator for local homology

So if X,Y are oriented

[light rock]

orientation is a global choice of element in cohomology which passes to a generator in local cohomology

might as well go full algebraic

[marsh forge]

That means that we have elements x and neighborhoods U for each x such that H(X,X-U)->H(X,X-y)

[light rock]

and avoid continuous

not that that's wrong, cuz it isn't

[marsh forge]

I mean

I can’t just bust out new material for the pset lol

Anyway

So let’s think about X\times Y

Given continuous choices of generators for X,Y

Is there an obvious choice for X\times Y?

We have the kunneth map HxH->H(\times)

This gives us an element

But its not necessarily a generator?

Is it clear that the kunneth map sends generators to generators for the case of n,m manifolds?

It’s a map ZxZ->Z

Is it obviously surjective?

Wait ok

So for a connected n manifold X and a connected m manifold Y

[light rock]

check the error term in the kunneth formula maybe

[marsh forge]

By kunneth we have that the homology for both at 0 is Z

So we get Z\tensorZ at n+m

And we should get Z at H_0 with an iso

Duh

[light rock]

you can see that the Tor thing vanishes

I think

[marsh forge]

Something something flat something something

[light rock]

i mean

one term is (n) + (m-1)

the other is (n-1) + (m)

in both cases one of them is Z

and the other...

is something I dont remember

I know that this stuff is very well explained in hatcher

[bright rampart]

Can anyone recommend a book as interesting as this?

[marsh forge]

Why is this in topology and geometry lol

Anyway just read a math textbook

Theyre even cooler

[bright rampart]

Because it’s mostly topology

Anyway I found a bunch of books to check out by James Gleick and Rudy Rucker, not all topology but seem interesting

And GEB is a math book lol. It’s just not a general reference text. It’s a math philosophy book

[tough hamlet]

[bright rampart]

It’s speculation of mathematical conceptualization but I mean sure call it that I won’t argue with your definition. It’s fitting lol

I love your definition tho it’s fitting and funny

It talks a lot about interrelatedness of things, I consider that part of topology more broadly

[tough hamlet]

what do you think topology is

[bright rampart]

Where do I ask about this kind of stuff tho?

Sorry then

[marsh forge]

I had no idea topology had a colloquial meaning

[sweet wing]

given a point (x,y), can you find a nbhd that doesnt contain the diagonal? and try to deduce facts about x/the diagonal from this

[gritty widget]

is the product of two open sets open

[midnight jewel]

by definition of the product topology, yes

the product topology has as its basis all products of open sets (for finite products)

[gritty widget]

ah great thanks

[floral gust]

How can I argue π_1(X^1) is surjective to π_1(X)?
I have already proven that π_1(X^2) to π_1(X) is bijective so I assume I only have to show that π_1(X^1) to π_1(X^2) is surjective. X^n denotes the n-th skeleton of X which is a CW complex. Any help?

[light rock]

pi1 is generated by closed 1 cells mod some relations

is morally the answer

chase this through the definitions

[feral copper]

Hello ! I'm sneaking outta #category-theory to ask you : does anyone know of a good reference about knot theory, besides Rolfsen (which I've already read) ? 🙂

[sleek thicket]

Is it possible for the fundamental group of a wedge sum of two simply connected spaces not to be simply connected?

If the basepoints have neighborhoods which strong deformation retract onto those points, then no

But in general?

[floral gust]

I dont think its true in general, you need to apply Seifert Van Kampen to show the wedge is simply connected so if it fails to satisfy the criteria it wouldnt be simply connected I think.

I am positive I have seen a counterexample but cant recall atm

[narrow depot]

this question should only care about things up to weak homotopy equivalence unless you are defining wedge sum in a weird way. So the answer is yes, it is true.

[fleet rapids]

So Im looking at https://www.seas.upenn.edu/~jean/sheaves-cohomology.pdf and not sure why the pth homology module is a quotient module

like if the sequence elements (of the given chain complex) are just albelian groups, where do we get rings from for them to have module structure?

[fleet rapids]

ty

[vocal surge]

hey quick q

[bitter yoke]

Okay I'm still listening

[vocal surge]

lol

Want me to say rewrite it

[bitter yoke]

Just keep going

[vocal surge]

How do I show that space is not compact?

Like I know that I need a cover that emits no finite subcover

but idk how to produce that in this topology...

[bitter yoke]

Wait so your topology are all the finite sets, and all the subsets that contain infinity?

[vocal surge]

Its the integers union infinity

topology is finite set and anything union infinity is closed

and open is ur compliment is closed

so basically everything is open and closed

[bitter yoke]

Uh what? There are some sets that are only open or closed, but not both

[vocal surge]

Yes but finite intersections of open sets are open

[wanton marsh]

the closed sets are the finite sets or subsets containing infinity

iirc ?

[vocal surge]

yes

[bitter yoke]

Ah no, I'm wrong

[wanton marsh]

well then you only need to show that the space has no finite open cover

wait

that there is an open cover with no finite subcover

[vocal surge]

yes

that is the definition we have of compact

[wanton marsh]

or maybe it is compact

[vocal surge]

Someone in class said it wasn't

I'm struggling with this tbh

[wanton marsh]

"someone in class"

proof by authority

qed

[vocal surge]

Haha

Also

Isn't it that I need

[bitter yoke]

Hm

Oh

[vocal surge]

a open cover that does not emit a finite subcover?

[wanton marsh]

maybe you should show that the space is compact

[bitter yoke]

Each of the singleton sets form an open cover

[vocal surge]

Yes I thought that as well

[bitter yoke]

E.g., take {1},{0},{-1}

etc

[wanton marsh]

but {infinity} isn't open

[vocal surge]

but are they their own subcover

and yes infinity is not open @wanton marsh

[bitter yoke]

Hm yeah

Oh yeah

infinity causes a problem because you need to add all but finitely many to be able to make it open

Okay yeah I think this is compact now

[vocal surge]

How do I know EVERY open cover emits a finite subcover

when there's so many

[wanton marsh]

with a proof

infinity is a special point in that space that is different from all the others so maybe you should start investigating around there

[vocal surge]

Well anything union infinity is closed...

[marsh forge]

Depending on how you add infnity

You’re just talking about the 1pt compactification

[vocal surge]

Yeah in this topology

that's in the problem

[marsh forge]

The way to prove something is compact is to play the following game

Suppose someone gives you any open cover U

You want to construct for me a finite subcover

Without making any assumptions about U

If you can do this for an arbitrary cover you can do it for all of them

[vocal surge]

Can you do this for the singletons?

[wanton marsh]

the singletons is not an open cover

[feral copper]

Hello 🙂 I'm asking this again because I would like to add one to my Xmas list ! 😄 Does anyone know of a good reference about knot theory, besides Rolfsen (which I've already read) ? 🙂

[wanton marsh]

no :(

[vocal surge]

Okay compliments of singletons are open so {A}'is an open cover of a lot so then there exists an arbitrary union of closed sets that suffices as a subcover?

since arbitrary union of closed sets is closed or something?

[wanton marsh]

no, complements of singletons aren't necessarily open

[marsh forge]

Thats not true

[wanton marsh]

and no about the arbitrary union of closed sets being closed

[feral copper]

Thanks :x

[vocal surge]

Can't I show via induction that {A} U {B} etc is finite and thus closed?

This is bad I'm really struggling with this

[wanton marsh]

oh wait

was {infinity} open ?

[vocal surge]

no

[wanton marsh]

uuuh closed ?

[vocal surge]

I guess

[wanton marsh]

oh yeah lol

[vocal surge]

It says anything union infinity is closed

[wanton marsh]

yes

sorry, yes, then, complements of singletons are open

[vocal surge]

So they can be like one big open cover rt?

or nah

[wanton marsh]

yes

and you should be able to come up with a finite subcover

from it

[vocal surge]

yeah

Does that suffice or am I missing something

[wanton marsh]

no because you want to do that for any possible open cover

not just the easiest one possible

after {the wholet set}

[vocal surge]

dang it lol

How do I make the statement more general

[marsh forge]

Let U be an open cover

Find a finite subcover

[vocal surge]

Okay one last question @wanton marsh is {A}U{B} open and closed

[wanton marsh]

what's A and B

[vocal surge]

singletons

[wanton marsh]

well it's closed

but it's not open if one of them is infinity

[vocal surge]

@wanton marsh How do I know that the union of anything with infinity is not open

closed =/= not open

[wanton marsh]

it depends

[vocal surge]

On what?

I have that A U B finitely many times can be an open cover and subcover of itself

but what about infinity

[wanton marsh]

it depends on what you union infinity with

I have no clue what you're talking about

[vocal surge]

same problem as before

[wanton marsh]

I have that A U B finitely many times can be an open cover and subcover of itself

in this

[vocal surge]

Its the integers union infinity
topology is finite set and anything union infinity is closed
and open is ur compliment is closed

[wanton marsh]

yes

[vocal surge]

A U infinity is open?

[wanton marsh]

what is A

[vocal surge]

A singleton

[wanton marsh]

is {1 ; infinity} open ?

[vocal surge]

I know its closed

[wanton marsh]

yes it's closed, but you were asking if it's open

[vocal surge]

yes

[wanton marsh]

so is it open ?

what is its complement ?

[vocal surge]

Z \ {A,infinity}

[wanton marsh]

is that finite ?

[vocal surge]

no

[wanton marsh]

does it contain infinity ?

[vocal surge]

no

ah it is open

[wanton marsh]

so then, is it closed ?

[vocal surge]

no

[wanton marsh]

right

so {1 ; infinity} is not open

[vocal surge]

But {1,2} is open

[wanton marsh]

yes

[vocal surge]

and it can be its own open cover and subcover

thus the space is compact

[wanton marsh]

what are you talking about

no

[vocal surge]

{1,2} is an open set

[wanton marsh]

yes

[vocal surge]

it is thus an open cover of {1,2}

[wanton marsh]

I guess ?

[vocal surge]

and we have shown that finite open covers can be their own sub cover

[wanton marsh]

well, {{1 ; 2}} is if you want to be precise

no

you need to show that any open cover of {1 ; 2 ; 3 ; ... ; infinity} has a finite subcover

if you change the bolded things you aren't doing anything

[vocal surge]

well {{1}} is an open cover

[wanton marsh]

of {1}

not of {1 ; 2; 3 ; ... : infinity}

do you know what the word cover means ?

[vocal surge]

yes it cover the entire set and possibly more

[wanton marsh]

well it can't cover more than the entire set

[vocal surge]

like in r (1,4) covers (2,3)

oh

[wanton marsh]

well we are working on the whole set

not on {1}

not on {1;2}

not on {1 ; infinity}

[vocal surge]

Well can't the whole set be written as one of those three things?

[wanton marsh]

no

[vocal surge]

Like the union of them

oh wait

We're trying to show the whole set is compact

[wanton marsh]

yes

[vocal surge]

not any arbitrary subset

[wanton marsh]

yes

[vocal surge]

Well the whole set is an open cover of itself

[wanton marsh]

yes that's the easiest cover you may need to find a finite subcover of

newsflash

if you could do that for any topological space

you wouldn't need to invent a new word to say "compact"

do you know what any in "for any cover" means

[vocal surge]

not really

I understand it for like (1,4)

but not this

[wanton marsh]

do you know what "forall x" and "there exists x" mean and why they are VERY different

[vocal surge]

for all x is -> any x can be chosen and "there exists x" -> I can give you an x that meets some condition

[wanton marsh]

those sound like the same to me

[vocal surge]

well like for all E there exists delta

[wanton marsh]

here you have come up with one specific open cover

with two in fact, with the earlier try

so you have shown that there exists an open cover such that you can find some finite subcover from it

but you want to show that forall open cover, you can find some finite subcover from it

[vocal surge]

I can't picture another open cover

[wanton marsh]

there are LOTS of them

[vocal surge]

So itself and all the arbitrary unions of its elements

[wanton marsh]

and you're doing something fundamentally wrong if you are just trying to come up with open covers of the space

[vocal surge]

Well I'm trying to find a finite sub cover that works for all open covers right

[wanton marsh]

no

that makes absolutely no sense

you want to show

FORALL open cover of the space, THERE EXISTS a finite subcover

which is very different from

THERE EXISTS a finite subcover such that FORALL open cover the subcover is a subcover of the cover (???)

[vocal surge]

Well then I have shown that for all open covers you hand me

I have a subcover

[wanton marsh]

when have you shown that ?

because yes it would be great

[vocal surge]

If I can make like a generic subcover formula

[wanton marsh]

and let me add

that when you have to prove a statement that starts with "forall"

you always

ALWAYS

start with "let U be an arbitrary open cover of the space"

just like you would write "let epsilon be positive" in a limit proof

[vocal surge]

yes then I give my generic formula for a subcover demonstrating that a subcover exists

[wanton marsh]

I'm not sure such a formula exists

[vocal surge]

😦

how do I show it then

[wanton marsh]

by logic

the world is not made of formulas

[vocal surge]

Well what are the different forms this open cover could have?

also tysm btw this is helping a lot

[bitter yoke]

There are a lot of them

But, like we've mentioned, you know that infinity must be in one of your open sets since its an open covering

Think about what open sets that include infinity look like

[vocal surge]

They are the compliment of a finite set not containing infinity

[bitter yoke]

Yep, now think about that

[vocal surge]

So the union of that for every z in Z is an open cover

[bitter yoke]

Uh what?

[vocal surge]

Well I need to cover the whole set rt?

[bitter yoke]

Yes, that's what an open cover is

[vocal surge]

The compliments are open

So the union of all the compliments is an open cover

[bitter yoke]

The complements of what

[vocal surge]

The singletons

[bitter yoke]

Sure

How does this help you

[vocal surge]

Idk I need to show every open cover has a finite sub cover

Maybe if I can show that all the types of open covers have a finite sub cover?

All we did in class was do one proof where we prove a lack of compactness, not this

[bitter yoke]

Just think about infinity

[vocal surge]

okay

it is closed

[bitter yoke]

As in

Think about the open set in your open cover that contains infinity

[vocal surge]

There's a lot of them

[bitter yoke]

Not necessarily

[vocal surge]

What open cover is this?

The union of the compliments?

[bitter yoke]

Any open cover

[vocal surge]

I'm not sure what you want me to notice

[bitter yoke]

You've already mentioned it before

"They are the compliment of a finite set not containing infinity"

Your open cover must include at least one of these sets

[vocal surge]

But this is not a finite subcover is it?

Its an open sub cover

[bitter yoke]

No you need to think more

[vocal surge]

Ah if A'UB' are our complimements -> AUB is in the open cover

[bitter yoke]

I have no idea what this means

[vocal surge]

The open cover is the union of compliments of a finite sets not containing infinity

But in their union the "holes" are filled in

and the union of those holes is finite

[bitter yoke]

Yeah I still don't really know what you're saying

But

It's possible your open cover only has one open set that contains infinity

[vocal surge]

idk what I suggested doesn't work

idek

[floral gust]

Take any open cover of the space. Since ∞ is in the space and the union of sets in an open cover contains all points in a space, we have that there exists atleast one open set in the cover such that it will contain the point ∞. The only open sets (call it O∞) that contain ∞ are of the form { complement of a finite} U {∞}. Thus the only remaining points to be covered in the space are those that weren't covered by O∞ which are finite in number (Can you reason why? From this and the fact that the open cover covers the whole space can you complete the proof?)

@vocal surge

[vocal surge]

Are they finite because they are the arbitrary union of singletons?

[floral gust]

O∞ covers a complement of finite set and the point ∞. In other words, O∞ covers every point in the space except points in some finite set.

[vocal surge]

okay so that's why what remains is finite

[floral gust]

Indeed! Can you complete the proof by using the fact that what remains is just some finite set and my open cover contains the whole space?

(what remains to show is for you to extract finitary amount of open sets from the original arbitrary open cover. One of the open sets should be O∞ as we proved earlier. What else should go in the finite subcover/ finitary amount of open sets that cover the space?)

[vocal surge]

Well we still need the remaining points rt?

so then the open cover is made up of the union of finite things so any subcover is finite?

or maybe not, sorry this is hard for me to conceptualize idk

[floral gust]

Try to write it out.

Remember open cover is an arbitrary collection of open sets. From this, you will extract a finite subcover.

[gritty widget]

Let $f: Y \to Y$ be a continuous function of compact space Y. Show that there exists a nonempty closed subset $A \subset Y$ such that $f\left(A\right) = A.$

Any ideas? A and f(A) compact

[bitter yoke]

This isn't true

[gritty widget]

Wait what xD

[bitter yoke]

It's not hard to think of a counterexample

[gritty widget]

What's your counterexample?

There's no way that problems wrong

[bitter yoke]

Constant functions are always continuous

[gritty widget]

Umm

But you can take A to be a point

[bitter yoke]

So you meant to write f^{-1}(A) then?

Oh

I didn't see your edit

[gritty widget]

I didnt edit anything

[bitter yoke]

Shit jk

[gritty widget]

The thing I edited was added 'of'

[bitter yoke]

I thought it said f(A) = Y

My bad

[gritty widget]

Ahh ok I see what you meant now

[gentle osprey]

Godel:

[dim meadow]

Yeah this isn't true

Take the discrete Topology on a finite set

Oh hmm

Yeah this is just not true

Take three points {a, b, c}, map a and b to c and map c to a

@gritty widget am I missing something?

[gritty widget]

I mean, you want to show that there exists a mapping such taht its true @dim meadow

[dim meadow]

Oh

Lol

That's not what you said smh

[bitter yoke]

Oh what

[gritty widget]

Wait wha

[dim meadow]

Yeah this makes no sense

👀

[gritty widget]

why

[dim meadow]

Cause take id

Can you restate the question?

[gritty widget]

But we don't know what the f really is

cant say its id

[bitter yoke]

?????

[dim meadow]

So is the question that you're given f, and you want to show there is some A so that f(A)=A?

Cause that's not true

[gritty widget]

Why isn't that true

Given what we know in the problem

[dim meadow]

I gave an example lol

Of a space and a map

[gritty widget]

What's your example again

[dim meadow]

Take three points {a, b, c}, map a and b to c and map c to a

Nothing gets mapped to itself

[bitter yoke]

Can't you take all of Y

as A?

[gritty widget]

@dim meadow in this example take A = {a,c} then

[dim meadow]

No

Oh you're right

Hmm

Maybe this is correct

[gritty widget]

yes it is its a problem I seen

[bitter yoke]

A = {a,c} wouldn't work, since f({a,c}) = {b,c}

[dim meadow]

Ah well

[gritty widget]

a ghets mapped to c and c to a

so it does work

[bitter yoke]

oh you're right

[gritty widget]

And it's supposed to be a short answer question

And using the fact its compact

the 'all spaces are compact' guy should know

[dim meadow]

Oof

[vocal surge]

hey I've got another quick question but I can wait for this guy

[gritty widget]

Just ask, I'll repost it tomorrow if I don't come up with an answer by then

[vocal surge]

okay

If i take f : I -> R where I is an open interval in R, to be differentiable at x and that f'(x)> 0 show that there exists an open interval around f'(x) not containing x where f'(t) is greater than 0

My attempt is take lim as t -> x of f(t)-f(x)/t-x = lim t-> x f'(t)

then there exists delta st B(x,delta) \ {x} where f' cts and then fix an open interval v where t-x>0 in the delta ball

then v intersect I \ {x} is good rt?

[versed pivot]

@vocal surge
at first you said i take f : I -> R where I is an open interval in R, to be differentiable at x
then in the proof you said where f' cts
are you assuming that f' is continuous or not?

[vocal surge]

f' need not be continuous

see x^2*sin(1/x)

I need that it is continuous around the point x but not necessarily at that point and then I can find delta s.t. there is an interval of that open interval that contains another point where f'>0

[gritty widget]

how would i go about proving the trace operation on tensors is well defined? \begin{equation}\text{tr}: T^{l+1}_{k+1}(V) \to T^l_k (V) \end{equation}

[gentle osprey]

brug moment:

[gritty widget]

i've come up with a few ideas but they all seem stingy

the way the book defined it was for $Q \in T^l_k (V)$ \begin{equation}\text{tr}Q(\omega_1, \hdots, \omega_k, v^1, \hdots v^l) = \text{tr}Q(\omega_1, \hdots, \omega_k, \cdot, v^1, \hdots v^l, \cdot) \in \text{End}(V)\end{equation}

[gentle osprey]

brug moment:

[gritty widget]

idk i've probably got the wrong idea

[gritty widget]

<@&286206848099549185>

[proud locust]

can anyone help me lead me to the right direction?

[west spindle]

draw the square

[wanton marsh]

that question is amazing when you look at it

[sweet wing]

p sure this shouldn’t be here anyways

[floral gust]

Is such a CW complex correct? (I mean the mathematical redaction more importantly)

[gritty widget]

Is the set of integers with p-adic metric compact?

[marsh forge]

@floral gust yeah

Not sure what redaction means here

[floral gust]

I was just confused about the middle figure because it didnt seem to resonate with the usual picture of a klein bottle

[marsh forge]

Well its not a klein bottle yet

[floral gust]

Yeah yeah but like for a torus it makes sense

[marsh forge]

?

[floral gust]

As in the middle figure makes intuitive sense if used for the torus 1-skeleton

[marsh forge]

Torus and klein have identical 1-skeletons

You create both by gluing the two cell in a specific way

[floral gust]

Oh okay okay then! Thanks.

[gritty widget]

Do you guys know any good websites to practice GEOMETRY proofs?

[floral gust]

@gritty widget Note that every compact metric space is complete. Thus not-complete implies not compact given metric. Consider the geomteric series 1+p+p² ... . What can you say about its convergence given the p-adic metric?

[gritty widget]

Haven't learned about complete spaces yet

[sweet wing]

Metric spaces are hausdorff, compact spaces are closed there

[marsh forge]

@honest narwhal this is why we need to change the name

[proud locust]

@gritty widget it's on my geometry hw though

[floral gust]

How does one sees the last line?

Is this by noting that S^n{x} ~ R^n which is contractible?

[marsh forge]

Any map going thru 0 is 0

[floral gust]

But here no map is going through 0? no?

(assuming by going through 0 you mean f(0) )

[marsh forge]

No

I mean 0

Like the trivial group

Its denoted 0

[floral gust]

Yeah but I am asking how did they know that H_n(S^n -{x} ) is a trivial group

[marsh forge]

Think about what S^n minus a point is

[floral gust]

R^n-1 which is contractible?

[marsh forge]

Yep

[floral gust]

Thanks!

Secondly can you also tell how to conclude the last equality where the degree is defined as the generator of the image under the isomoprhism with the integers

Seems like i have to argue f(deg(g) ) = f(1) * deg(g), but I cant really find a justification for this

[marsh forge]

The justification is that f is a homomorphism

[floral gust]

Can you elaborate a bit more? From that all I am seeing is that f(deg(g) ) = f(1 * deg(g) ) = f(1) * f(deg(g)) = deg(f) * f(deg(g))

[marsh forge]

Let f: Z->Z. Then f(n)=nf(1)

Its a good exercise

So i wont prove it unless you ask

[floral gust]

Is $d_n$ supposed to be $$ d_n = \sum_k d_n(e^n_k) $$

[gentle osprey]

QuickMaffs:

[floral gust]

<@&286206848099549185>

[marsh forge]

Its supposed to be cellular boundary

[floral gust]

I am aware of that but like the book only provides a formula for d_n(e^n_k) and does not say anything about d_n. So from above I assumed that d_n just represents the sum.

[marsh forge]

Well in that para they are only taking in on e_k^n’s right?

Anyway because you’re dealing w free commutative groups

You get the unique Z-linear extensions of d_n’s values on the basis elements

[floral gust]

Yes. So in the case we have two e^k_n, what would happen?

[marsh forge]

You’d sum bc its linear

[plain panther]

So, I'm trying to create a very, very truncated isocohedron in code (specifically in Lua, if you're curious). I fonud a post that really addresses a lot of this and I'm particularly interested in this solution: https://stackoverflow.com/a/47455940

However it doesn't really go into the math of how to do such operations, can anyone clarify? Nor can I truly find anything that extensively talks about this.

[plain panther]

<@&286206848099549185>

[plain panther]

RIP lmao.

[marsh forge]

Sorry these channels tend to only get answers when the ppl who frequent them find the question interesting

[plain panther]

Yea that's definitely understandable. I'll just have to keep looking and hoping.

As well as trying to solve in independently.

I'm really stuck on angles though....

Does anyone know if all vertices (or centers of the shapes) are located on a sphere?

Yea, unfortuneately it's less for the intrigue of mathematics and more for the practicality.

[gritty widget]

I've been told to come here. I'm trying to implement a bezier curve in some regard into a C++ program to be used to solve a BVP. I've found 2 algorithms for doing this. The first was too symbolic, but would probably work, but then I'd be focusing on simplifying equations, etc. The second I've found seems the most numerical, and it's the approach I am trying to take, but there seems to be a hole in the algorithm of some sort.

Let $B_{n,i}$ denote the $i$-th $n$-degree Bersntein polynomial s.t $B(t)=\sum_{i=0}^nB_{n,i}(t)P_i$, where $P_i$ denotes a control point.

The paper goes on to redefine the bezier curve as $\sum_{i=0}^n B_{n,i}(t)(\frac{i}{n},f(\frac{i}{n}))$.

Moving on to the beginning of the algorithm. Let $(P,Q)$ be a pair of unknowns and construct this Bezier curve spawned by the control points array $A={(a,y_a),(p,q),(b,y_b)}$, where $f(a)=y_a$ and $f(b)=y_b$ are the boundary values of the differential equation. It says we can compute the values of $p$ and $q$ respectively by the following two equations, included in the image.

[gentle osprey]

Dodsy:

[gritty widget]

and here is an example where a numerical answer is given to the first part of the algoritm

I'm not exactly sure how they got the point (0.56344,1.68501)

I've tried mostly elementary things, and don't get anywhere near these values for (p,q)

[random slate]

Now the waiting game begins :P but there should be someone here who can help out

[marsh forge]

Depends who is around, not many people come to this channel

[gritty widget]

I don’t mind waiting! I’ll tinker with it :) thanks again guys

[gritty widget]

Real stupid Q

Does homotopy between curves cares for orientation?

[marsh forge]

Not really, but the question is a bit vague

For example you can easily have a based homotopy between two curves of opposite orientation

But if you apply a homotopy to an oriented curve theres an obvious induced orientation

[gritty widget]

Here is my problem

I got a text that said that the integral of a form is the same on the following curves

Because they are homotopic

But aren't they curves with different orientation?

If so, shouldn't integral of formes cares for orientation?

[sweet wing]

,w parametric (sinx, cos^3x)

[gentle osprey]

Display results online and refine query

Requested by Ariana#0640

[sweet wing]

should be somewhat clear why this is homotopic to γ1 i believe?

[gritty widget]

That is clear but doesn't it change the orientation of the curve?

[sweet wing]

Yea it does
Homotopy doesnt necessarily preserve orientation

[marsh forge]

wait do they not both go counter clockwise am I dumb

oh

no they dont

[sweet wing]

γ1 is +ve and γ2 is -ve i believe

[gritty widget]

Ok, so this was the point of Q

The orientation is not a factor when evaluating a form on homotopic curves

?

[marsh forge]

dm him and ask lol

well the theorem is about fixed-endpoint htpy

[gritty widget]

In that case I can always find an homotopy that changes direction to the curves

[marsh forge]

But like, this isn't a fixed endpoint htpy if we parametize both by [0,2pi]

that's not true abes

[sweet wing]

For R^2 yes(since the fundamental group is trivial)

[marsh forge]

that's also not true ariana

pi1 is about loops

try homotoping [0,1] to a point without moving the endpoints

or try reversing the orientation

Can you post where this Q is coming from abes

it feels like something is missing

because these curves have different basepoints

[sweet wing]

o rite they have different base points

[gritty widget]

I'll try and do my best at translating

We're considering exercise 2 and I'm given that form

[marsh forge]

which problem

[gritty widget]

Or vector field as u prefere

[marsh forge]

form

i prefer form lol

anyway it's just asking you to evaluate right?

[gritty widget]

To show that it is not conservative (exact) he integrates on the curve (cos(t), sin(t))

As it holds a value different from 0 it cannot be exact

[marsh forge]

ok

[gritty widget]

At this point I'm asked to find the value of such form when integrating on the second curve (sin(t), cos³(t))

[marsh forge]

t in [0,2pi] yes

[gritty widget]

At this point he concludes because he already know the value on the first curve and their are homotopic

While I agree with everything I do not understand why the change of direction does not hold any consequence

I thought the curve orientation was important when evaluating form integral

Is it not important here because I have a closed curve?

[marsh forge]

I think that might be the trick here re: closed curve

[gritty widget]

(he has already shown that the form is closed in the domain)

[marsh forge]

but importantly your prof is playing fast and lose

bc the theorem is only true for fixed pt htpy

and this isn't one w/o reparam

you need to 'rotate' the first curve or the second curve

or you can just change the start/end point by choosing a different interval

[gritty widget]

Ok, it clarify many things as it was my problem

[marsh forge]

bc its periodic

yeah my guess is this:

[gritty widget]

While I agree that the results has to be true I think I was misguided by his "losiness"

[marsh forge]

1: it's impossible to reverse the orientation unless the curve is closed

2: so its probably the only time it doesn't matter

[gritty widget]

Thanks bro, appreciated

[marsh forge]

np

[marsh forge]

tensor is left adjoint to hom i need this tattooed

[unkempt shard]

hello?

[marsh forge]

Hi

[gritty widget]

Given the following surface why I can't just find the singular point as those that are zeroes for the Gradient?

[fleet rapids]

So whats the effect on changing the base of the log used to define topolgical entropy in terms of what its used for?

(if we define it this way, that is)

Like do we just pick log base e or base 2 and call it a day for using this concept ^^^?

[floral gust]

Can somebody check if this is correct?

[pine pasture]

This chat dead? I'm trying to understand geodesics

[marsh forge]

Whoops i meant to respond to the thing abb

Abv

@floral gust your answer is correct

So i assume the technique was lol

[floral gust]

Well I knew the answers beforehand xD

[marsh forge]

Lmao well

[vocal surge]

Given top spaces x and y I have a continuous fnx f that maps $\mathbb{X} \to \mathbb{Y}$ and $n \to x_{n}$ converges to l on $\mathbb{X}$. I need to show $n \to f(x_{n})$ converges in $\mathbb{Y}$

I was thinking maybe path connectedness?

or something with sub sequences

[floral gust]

Is the function continuous?

[vocal surge]

ah yes

[gentle osprey]

OPz qt:

[floral gust]

What definition of continuity do you have?

[vocal surge]

uniform and point

[floral gust]

Is it not the definition of point continuity that given f, if x_n → x then f(x_n) → f(x) ?

[vocal surge]

yeah in a way yeah

we did it with epsilons and deltas

so I thought that only held in R plus I kind of haven't used point continuity in a while

[floral gust]

Can somebody tell me how can I go formally from {2ne | n in Z} to 2Z?

[marsh forge]

Consider e^i-1 formally as a symbol

And or just write down the isomorphism

[gritty widget]

Hi i was working on a proof which im not sure can work: Show R^2/A where A is countable, is path connected. I tried induction depending on cardinality of A, but not sure if that makes sense. Does it follow from induction, if we show the step, that it Will work for countable infinite set?

I talked to one person and his answer was that induction doesnt imply sth works for infinitely many numbers, which I think I disagree with

[rugged swan]

induction on cardinality of A ?

A is countable not finite

it doesn't make sense

[gritty widget]

At most countable then

Aleph 0 max

[rugged swan]

"works for finite case => works for countably infinite case" is false

[gritty widget]

Yep thats what I thought is wrong

What about transfinite induction tho

[rugged swan]

I think that transfinite induction is just the finite induction + the implication "works for finite case => works for countably infinite case"

in fact, the implication is formally true because true => true but unformally, it's not because you know how to do the finite case that you know of to do the countably case

you don't need any induction to prove the result btw

[rugged swan]

you didn't read the whole conversation...

[gritty widget]

Is there any way my argument could work though?

With trans induction

And using lemma that if there exists a path then there exists another path that differs every where but start and finish

[dim meadow]

Something shitty you can do is say there are uncountably many paths which do not intersect between any 2 points

Then you're done pretty instantly

And this is an obvious fact

@gritty widget

[gritty widget]

Not when you take out uncountable set

[dim meadow]

That's the point lol

You're only taking out a countable number of points

[gritty widget]

But thats a shitty way

[dim meadow]

So you're only fucking up a countable number of paths

Actually I like this way

It isn't shitty

[gritty widget]

Probably wrong Channel, what Im not sure is how to trans induction

[dim meadow]

Lol

Prove it for all successor ordinals, show that if it's true for all ordinals less than a limit ordinal it is true for the limit ordinal

Tbh just use what I said, it's easier

And it's probably the intended solution

[gritty widget]

No

[dim meadow]

Smh

[gritty widget]

I know intended solution

Its you put a Line between 2 points

And show there are too many paths

[dim meadow]

That's what I said lol

[gritty widget]

Ye

Its not my hw, just thinking

[dim meadow]

Oh you just want to prove this using transfinite induction

For yourself

[gritty widget]

Ye

[dim meadow]

Cool

Tbh this doesn't seem like the sort of thing you would want to prove with transfinite induction

Like the finite case for all n shouldn't imply the countable case

[gritty widget]

Ill try and if I fail Ill go to a set theory professor

[dim meadow]

Transfinite induction isn't hard, it just doesn't apply here

[gritty widget]

Not with that attitude

[marsh forge]

Wait

That doesn’t make any sense

Transfinite induction doesn’t work here because we don’t even care about 99% of the ordinals

And its patently untrue for any uncountable ordinal

Anyway liquids proof is the correct one

[tough mirage]

"correct proof"

[marsh forge]

Its a well known result with a well know best proof that has stood the test of time

If someone thinks of something cuter Ill rescind it

[gritty widget]

did i prove this right?

[gritty widget]

<@&286206848099549185> been > 15 mins

[sweet wing]

what is the problem

oh

well that looks ok

[marsh forge]

Looks fine from a brief glance

[fleet trench]

have any of yall used Bredon's book and/or have thoughts on it?

[honest narwhal]

I read chapter 1 of it to figure out point-set for real and I'm gonna soon do a crash course on algebraic topology, quite possibly using it

[fleet trench]

sick, i’m thinking of checking it out over break to see how far i can get with it

[gritty widget]

is the n-th de rahm cohomology isomorphic to $\mathbb{R}^{\text{#number of connected sets}}$

[gentle osprey]

mop:

[sonic hill]

No that's the 0th deRham cohomology isn't it

[gritty widget]

sorry 0-th ty

[remote grotto]

anyone wanna help me figure out a problem about indices of vector fields before my hw is due in one hour

[gritty widget]

naive question from someone who knows basically no topology: It feels like the side of topology that's the generalization of metric spaces is very different from the side of topology that's about stuff like manifolds. In practice, are these areas studied relatively separately but they get lumped together under the label of topology, or are they actually really similar if you have a more nuanced understanding of the subject?

[sonic hill]

They’re similar, topology removes the metric and studying metrics in manifolds goes by the name of Riemannian geometry

[gritty widget]

ah cool! thank you so much

[spice oracle]

I have a question

Let $S$ be a scheme, and let $G$ be a smooth affine group $S$-scheme. When we define the classifying stack $\mathcal{B}G:= [S/G]$ (with by definition trivial $G$-action). Is it a classifying stack in the sense that principal $G$-bundles over an $S$-scheme $X$ correspond to homomorphisms from the associated stack of $X$ to $\mathcal{B}G$? I imagine one proves this using some sort of $2$-yoneda lemma, but I don't know a precise statement.

[gentle osprey]

Rose:

[cedar pebble]

yes

the moduli property is essentially in terms of the Yoneda lemma for stacks, although you can also see this in terms of the functor of points for BG

[eternal crown]

Okay this is really basic question so feel like a complete dumbass for asking it

What would the area of the bottom right hand corner be, if we know the distance away from the top left hand corner?

E.g. say we know that the total length of the diagonal is sqrt(2)*a, but the line only intersects at 1, what is cotton hand corner area?

Oh also with assumption that length and width is a and a, and the area is a^2

[gritty widget]

is the solid line perpendicular to the dotted one?

[eternal crown]

Yes

[gritty widget]

ok so

struggling to write this in words

but the leg lengths of the triangle are a-something

can you find that something?

would you agree the bottom right hand corner triangle forms another 45 45 90 triangle?

[eternal crown]

Yes

It's a bad drawing sorry

[gritty widget]

the goal is an area in terms of a right? (just making sure)

[eternal crown]

Yup

Need way to find area depending on this distance from top left corner, which can change. Context of this is basically I'm modelling forest fires using cellular automata, and the solid diagonal line is a fire front

I need area because I want to work out the area burnt / total area

[gritty widget]

wait so what is the independent variable that we are picking

the length of the dotted line in the bottom right hand triangle?
ie, the distance from the bottom right hand corner to the intersection point

[eternal crown]

yes

that can change

but I don't know how the area that it encloses would change

[gritty widget]

i think i see how to do it, lemme find some paper to draw it

[eternal crown]

thanks :)

[gritty widget]

well it seems like...

the total area and diagonal length doesnt matter

just the distance you pick from the corner to the intersection

because i could make the whole square huge without affecting the area

so i dont think it will depend on your a

the area purely depends on the distance choice

see how the area of the triangle is not dependent on the area of the square?

as long as i pick the same x, they have the same area

also #geometry-and-trigonometry next time for questions like these

[eternal crown]

thanks @gritty widget

[gritty widget]

although... upon reviewing your question you also wanted a proportion of area to the total square

which is dependent on the total area of course

and that should be easy to make

which should be this i think

ignore the bottom i was just testing one number to reassure myself a little

so i think that should give you the proportion of area burned to total area for any choice of x and a

well wait not every choice but every geometrically possible choice

$p=\frac{x^2}{4a^2}$ for $a>0$ and $x \leq \frac{\sqrt{2}a}{2}$ where $a$ is the side length of the square, $x$ is the distance from the intersection to the corner, and $p$ is the percent burned

[gentle osprey]

Botnuke:

[gritty widget]

@eternal crown does that seem right?

[eternal crown]

I'm checking it out but it looks good, thanks so much @gritty widget

this is actually the equation that the paper I am working from used:

so the numerator is the total area - area burnt, and the denominator is the total area

and they get like 0.83, in the case that x = (sqrt(2)a - 1)

for working out A

the burnt out area in the corner

wait I'm still trying to work out what you've done...

[gritty widget]

i used 45 45 90 triangle properties

[eternal crown]

oh yeah

I'm really bad at basic geometry 😳

what would happen if x > sqrt(2)a/2 ?

would probably be then easier to just work out the burnt area

and use that to figure out the unburnt bit

cool cool cool

this is going to be a pain in the ass to code

[gritty widget]

it would become this area

[eternal crown]

yeah

[gritty widget]

and not sure it would hold anymore

[eternal crown]

yeah I think it would be quite tricky making a formula that holds for all values of x

[gritty widget]

you could make it a piecewise function

[eternal crown]

I could just have some condition that checks and then does things differently depending on whether x <= sqrt(2)a/2 as you mentioned above

[gritty widget]

yea and calculate the unshaded area (in the pic) instead if x is greater

and just subtract it from the total area

[eternal crown]

do you know whats going on in formula (5) that I linked @gritty widget ?

[gritty widget]

no, need context

[eternal crown]

the ((sqrt(2) -1)^2) * a^2) bit specifically

it is to work out the ratio of burnt area / total area

so the bit in the bracket in the numerator is the unburnt area, i.e. area A

that you were working out

[gritty widget]

if we have this, is the shaded area the burnt part?

or the unburnt

[eternal crown]

the unburnt bit

[gritty widget]

ohhh

ok flip around all instances of “burnt” ive said with “unburnt”

didnt understand the context

im still not sure whats going on in that formula, was a choice for x already made?

the proportion i made earlier will be a formula for the unburnt part btw, not burnt

burnt will be 1-p

[eternal crown]

yes the choice for x was already made

x = a

[tough hamlet]

why here

[stuck kraken]

@tough hamlet this is geometry obviously 😄

[gritty widget]

yep on the standard topology

nothing wrong here

[eternal crown]

i thought this was geometry 😭

[stuck kraken]

#geometry-and-trigonometry lol

[gentle osprey]

All functions are continuous:

[sonic hill]

How are you doing f+g

pointwise addition? Why would the values stay in the sphere

[floral gust]

Oh it is supposed to be f(x) ≠ -g(x)

[sonic hill]

Well the homotopy can be quite explicit then

[floral gust]

Yeah I know, I was thinking about in terms of degrees

[sonic hill]

for each $x\in X$ you know that the segment from $f(x)$ to $g(x)$ doesn't pass through the center so you can take [h_t(x):=\frac{(1-t)f(x)+tg(x)}{|(1-t)f(x)+tg(x)|}]

[floral gust]

That whether can we say something about the individuals knowing the addition

[gentle osprey]

Icy001:

[sonic hill]

Well two things

First the naive addition doesn't lie on the sphere

so you have to normalize it

Luckily, you can normalize it because $f(x)+g(x)\neq 0$

[gentle osprey]

Icy001:

[sonic hill]

Secondly, what you get is just the $t=\frac12$ phase of the homotopy

[gentle osprey]

Icy001:

[sonic hill]

Personally I don't see a plausible argument with degrees

[floral gust]

No I mean not the naive addition, but by giving the complex representation to the points in S and then use the addition over the complex plane. Would that work?

[sonic hill]

by stereographic projection?

One point maps to infinity under such a projection and there's nothing stopping f or g from hitting that point so you'll have a problem

[floral gust]

No not via stereographic but by representing points in S by (cos theta, sin theta)

[sonic hill]

Oh I was thinking of $S^2$ this whole time lol

[gentle osprey]

Icy001:

[sonic hill]

well I mean the homotopy works for arbitrary dimensional spheres

but if you're thinking about $S^1$ then well...

[gentle osprey]

Icy001:

[floral gust]

Yeah, for the remaining, I can also just argue that they are contractible

And since [X,Y] is trivial if Y is contractible thus f and g are homotopic

[sonic hill]

Spheres aren't contractible

[floral gust]

for n >1?

[sonic hill]

yep

The infinite dimensional sphere is contractible which is what you might be thinking of

[floral gust]

Oh god.....

I guess I have to rely on homotopy then

[sonic hill]

yep

[floral gust]

I was thinking of impressing my TA by using some roundabout approach ):

[sonic hill]

:(

Spheres missing a point are contractible though

[floral gust]

Btw why is a sphere not contractible for n>1?

Oh I see why

A loop can be shrunk

but a n-sphere cannot be shrunk inside n-sphere right?

[sonic hill]

ya

[floral gust]

I suppose I cant even use the fact that f+g is nullhomotopic ? (Or do your objections above also apply to S1 instead of just S2?)

[sonic hill]

For S1 I guess you can do that

but I'd express it as complex number multiplication

to avoid confusion

wait a sec

Negative of an angle isn't the same as the antipodal point

So when you rephrase it in terms of angles

I mean let's define new functions

$f(x)=e^{ia(x)}$ and $g(x)=e^{ib(x)}$

[floral gust]

No I mean I would define in terms of pi + theta no?

[gentle osprey]

Icy001:

[sonic hill]

$f(x)\neq -g(x)\iff \bR/2\pi\bZ\ni a(x)-b(x)\neq\pi$

[gentle osprey]

Icy001:

[sonic hill]

Yea so $a(x)-b(x)$ is nullhomotopic so $a\simeq b$ and so $f\simeq g$

[gentle osprey]

Icy001:

[sonic hill]

ok simplified would be

$f(x)/g(x)$ misses the point $-1$ on the unit circle

so $f(x)/g(x)$ is nullhomotopic

[gentle osprey]

Icy001:

Icy001:

[sonic hill]

Are you sure $\mathbb S$ is $S^1$ though?

[gentle osprey]

Icy001:

[floral gust]

No I mean I have to do for Sn but I was interested in employing degrees

So just interested in S1

[sonic hill]

Hmm yeah

There is no continuous surjective map from $\bR^n$ to $S^n$ as far as I know

[gentle osprey]

Icy001:

[sonic hill]

for $n>1$

[gentle osprey]

Icy001:

[floral gust]

Not even space filling curves?

Because I remember in the proof for showing that fundamental group of S2 is trivial, he said that usually the contraction works because a loop would miss a point but there exists loops as well which would fill all of S2

[sonic hill]

hm

ok this is both false and not what I wanted to say

ok what we need is a continuous map $\mu\colon S^n\times S^n\to S^n$ such that ${\mu(x,-x)\mid x\in S^n}$ is just one point

[gentle osprey]

Icy001:

[sonic hill]

oh and that no other pair of points map to that point

[floral gust]

what is both false ?

[sonic hill]

what I said

[floral gust]

Oh okay

[sonic hill]

``There is no continuous surjective map from $\bR^n$ to $S^n$ as far as I know
''

[gentle osprey]

Icy001:

[floral gust]

Is u the measure?

[sonic hill]

No it's just some map for which subtraction of angles was an example of on $S^1$

[gentle osprey]

Icy001:

[sonic hill]

and I'm trying to find the minimal generalization of that that allows you to use your method

[floral gust]

Ah I see I see

[sonic hill]

We also need:

$x\mapsto \mu(f(x),g(x))$ constant implies $f\simeq g$

[gentle osprey]

Icy001:

[sonic hill]

I feel like this is another way to state the axioms of an H-space

https://en.wikipedia.org/wiki/H-space

[floral gust]

Oh.... NICE! I will check it out

Thanks for the reference

[gritty widget]

so I read that the wedge product on the grassman algebra is the linear continuation of the $\wedge: \Omega^n(M) \times \Omega^m(M) \to \Omega^{n+m}(M)$

[gentle osprey]

mop:

[gritty widget]

what is linear continuation

I can’t seem to find it anywhere

[gritty widget]

np

[remote grotto]

lol did u see that in frederic schuller's geometric anatomy of theoretical physics

if a map f is defined on some set V equipped with an addition (not necessarily closed), if the span of V is some other space U which inherits the addition then u can define f on U by defining f(v + w) = f(v) + f(w) and f(av) = af(v) for v,w in V and a in the underlying field

this is not always well-defined though, so you have to check that it doesnt lead to any contradictions

[gritty widget]

Hi can someone help me prove this statement: Let X be a connected metric space. For any two closed disjoint sets $A,B \subset X, \exists C \neq \emptyset, C = \text{closed }, $ such that $C \cap A = \emptyset = C \cap B $.

[gentle osprey]

Godel:

[fleet trench]

and singletons are closed in metric spaces

[gritty widget]

ohh yeah

lmao

I dont know why I didnt remember singletons being closed thx

Also in this one I'm not quite sure: Show that if in metric space $\left(X,d\right)$ for any open ball $B_d \left(x,r\right)$ satisfied is $\overline{B_d \left(x,r\right)} = {y: d\left(x,y\right) \leq r }$ then every ball in this space is connected.

[gentle osprey]

Godel:

[gritty widget]

Like, isn't it obvious balls in metric spaces are path connected?

if thats not the way, any tips on how to show it?

hmmm

yeah thats true

Differential geometry: I need to show that the general form of a surface of revolution, given by $\vec{X}(u,v) = (f(v)\cos(u), f(v)\sin(u), h(v)) $is regular only if $f(v) \neq 0, f'(v) \neq 0)$ or $f(v) \neq 0, h'(v) \neq 0).$ Is it enough to calculate $\vec{X}_u \times \vec{X}_v = (f(v)h'(v)\cos(u), -f(v)h'(v)\sin(u), f(v)f'(v)(\sin^2(u) - \cos^2(u))$ and then state that $\vec{X}_u \times \vec{X}_v \neq 0$ only when $f(v) \neq 0, f'(v) \neq 0)$ or when $f(v) \neq 0, h'(v) \neq 0)$? Or should I show more, if this is even an appropriate first step at all?

Cool

Now if I could get somebody to help with it, that would be great

Biggest brain

[tough hamlet]

[gritty widget]

Let $f: X \to Y$ be a quotient map. Let $Y$ be a connected space and assume that for every $y \in Y f^{-1} \left(y\right)$ is connected. Does it imply X is connected?

[gentle osprey]

Godel:

[dire grail]

what do you mean by a quotient map

a surjection with the induced topology?

[dire grail]

so yes

the proof is very visual

do it by contradiction

here's an outline https://math.stackexchange.com/questions/302059/how-to-prove-this-result-involving-the-quotient-maps-and-connectedness

[gentle osprey]

All functions are continuous:

[gritty widget]

@floral gust thanks, actually thats what I tried to do but got stuck in the half

[floral gust]

Yw!

[gritty widget]

@floral gust wait, actually Im not sure about one thing: you said that for each y either f^-1 (y) \subset A or B since each fiber is conencted. Why is that? What does it being connected imply here?

[floral gust]

Say a set is connected. Is it possible for it to lie inside two disjoint sets?

[gritty widget]

ahh okay

[gentle osprey]

All functions are continuous:

[gritty widget]

I often think that f^-1(y) is just one point not sure why, gotcha

well if its one point its just obvious so I werent sure why was connected needed lol

[floral gust]

Blame analysis classes tbh

[hazy forum]

Idk where to ask this but does anyone know what a mirror manifold is? My friend was talking about mirror symmetry and how with mirror manifolds the symplectic and complex structures are switched

[gritty widget]

that's a term used in string theory

special calabi-yau manifolds which we do IIA and IIB on

mirror symmetry is the relationship between IIA and IIB

@hazy forum

[hazy forum]

Yee

[gritty widget]

[thorny flame]

I was thinking of this problem recently and I wanted to hear your guys thoughts on it. What is the best way to tile a region on (cover) the surface of a sphere with circles. That is if you had a 3 steradian section what is the optimal layout to cover a region on the sphere with the least amount of circles. I think the solution on just a plane region is to tile the region with hexagons then circumscribe the hexagons with circles. But not sure how that would translate over to curved geometries like a sphere. Any tip/hints/resources are helpful!

[hazy forum]

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

You're correct about the plane one!

[thorny flame]

Hmm thank you, I've seen circle packing before and it's generalization to higher dimensions although with packing it doesn't allow for overlap and doesn't cover the full region (correct me if I'm wrong). Is there an analog which covers the region with the least amount of circles allowing for overlap?

I'm looking into voronoi tesselations on surfaces right now but can't seem to make a great connection between the partitions and circles of equal diameter

[lost isle]

For continuity, there's two definitions usually. A metric one and one with open sets. They're equivalent.

This boils down to the metric topology is generated by open balls so it suffices to check continuity on the open balls which is the definition of continuity.

Why is the equivalence of definitions important/what problems does it solve? It makes one's mind set up connections between metric space concepts and topology concepts so that one can work with both better. In particular abstract results about topology can walk through the bridge to give us results about metric spaces. But they get stronger mostly.

My question is, what are some examples where this is striking?

One common example that is given is say we have a function from $\mathbb{R}^2$ to $\mathbb{R}$. And we are considering $\mathbb{R}^2$ once with the Euclidean metric $d_E$ and once with the taxicab metric $d_T$. It's claimed that this function is continuous on $(\mathbb{R}^2, d_E)$ if and only if this function is continuous on $(\mathbb{R}^2, d_T)$ BECAUSE $d_E$ and $d_T$ generate the same topology. However, this doesn't make a lot of sense to me. Can someone walk me through this example in detail? And what are some other examples where the equivalence of the metric and open set definitions of continuity are striking and can be used to solve interesting problems?

[tough hamlet]

what do you mean by striking

[floral gust]

Given space (X x Y,d) can I find a metric d' on X and d'' on Y such that the product of metric topologies induced by d' and d'' is equal to (X x Y, d)?

[dim meadow]

No

Not necessarily

[floral gust]

Can somebody explain how were these boundary maps obtained?

In particular I dont see why l0 and u0 should have opposing signs

<@&286206848099549185>

[midnight jewel]

think about this: if you had a closed circle (starting from a base point and returning to it), then you’d want its boundary to be 0 right? so the sign of the starting and endpoint must be opposite in order to cancel out

ultimately it’s just a definition, the signs of the faces alternate when you take the boundary

but what I just said is why

in this example here also you want $\partial u_1 + \partial l_1 = 0$. and you can only get that reasonably if you treat start and endpoints differently

[gentle osprey]

Sascha Baer:

[midnight jewel]

@floral gust

[floral gust]

@midnight jewel But why would I want it to be 0? Isnt the criteria that the d_n+1 d_n = 0 and not d_n= 0? Secondly, we were made to use the cellular boundary formula in which we consider the degrees of the maps. Would be very helpful if you can explain in that context

[midnight jewel]

what is the boundary of a closed circle?

intuitively

(it has to be some 0-dimensional set)

(well, a 0-chain I mean)

(not techncially a set)

as I see it, a circle oughn’t have a boundary or something has gone wrong

[floral gust]

No, its more like the degree of the map going from the boundary of the disc attached to the cells it was being attached ( quotient the remaining space to a point)

And boundary of a circle, shouldnt it be like two points ?

[midnight jewel]

which two?

[floral gust]

l0 u0

[midnight jewel]

why those

[floral gust]

because they were used as our 0-cells

[midnight jewel]

think more topologically. the boundary of a complex is defined in such a way that it’s always the thing that you’d intuitively want it to be. those two points clearly lie in the interior of that circle, so they can’t be boundary points

the boundary isn’t just the skeleton of one dimension lower

it’s only “exposed” bits of it, so to speak

you know, like a boundary

[floral gust]

well no points are being exposed

so an emptyset?

[midnight jewel]

yes, the boundary is 0
I wouldn’t call it the empty set per se because complexes aren’t really sets (they’re functions into your space)

well, actually no, sorry, that’s for Δ-complexes I think CW-complexes you do see as sets

[floral gust]

Oh okay then might be stupid but for clarification why i couldnt have the first one to be u0 +l0 and the second to be -u0-l0?

[midnight jewel]

I’m confusing formalisms here

well, sure, but what really distinguishes the two? you can slide around one to get the other and suddenly the signs changed?

but each has a clearly defined “front” and “back”

so you just say the front has positive sign and the back negative sign and it all works out

[wanton marsh]

how are the boundary maps defined ?

[floral gust]

Oh okay, lastly, for the last complex, why did the signs changed. Using just one orientation, both of them should yield u1 +l1, no? Or is it the same boundary argument?

[midnight jewel]

point is the boundary is just defined to be alternating in sign because then everything works

what do you mean with the signs changed?

[floral gust]

like d2u1 = -d2 l1?

[wanton marsh]

I would say that if we were talking about simplicial complexes yeah

[midnight jewel]

for the top one, orientation is one way round, for the bottom one it’s the other way round

like, it literally is, for one it goes clockwise for the other counterclockwise, just based on how the arrows are drawn

[floral gust]

How so?

[midnight jewel]

CW complexes are definitely the kind I’m least familiar with, we did most of algtopo with Δ-complexes

[wanton marsh]

same