point-set-topology

[dim meadow]

and sierpinski space happens to work out but for dumb reasons

lol no

lots of constant maps

yeah

it's kind of a cool property tbh

[limpid mural]

Let $M, N$ be two manifolds such that their De Rham cohomology is finite dimensional. Let $\chi(M) = \sum_{i = 0}^{n}(-1)^i b^i(M)$ and $\chi(N) = \sum_{i = 0}^{n} (-1)^i b^i(N)$ their euler characteristic, where $b^i$ are the betti's number(aka dimension of the $i-$th cohomology space). I want to show that $\chi(M \times N) = \chi(M)\chi(N)$. Do you have any hint?

[gentle osprey]

emme:

[limpid mural]

I think this is the one where you use UCT and poincaré duality, emme
@gritty widget what is uct?

thank you

[dim meadow]

universal coefficients

idk if you have the assumptions for poincare duality

have you proven kunneth?

[limpid mural]

I can use the PD which in my course was introduced as isomorphism between H^k(M) and H_c^(n-k)(M)

where in H_c we consider only differential forms with compact support

have you proven kunneth?
@dim meadow Kunneth formula? Yes

Now that I look at it

the fact is just an application of kunneth formula lol

[dim meadow]

yeah

you just have to do some computation

[limpid mural]

yes

thank you guys

[dim meadow]

about what?

I don't think so

you don't have compactness

I don't know the proof of kunneth off hand

besides for taking the tensor product of the complexes

or whatever

for poincare you need compact orientable (not really cause you can do Z/2 orientable)

lol

yeah goodnight

[limpid mural]

Let $M$ be a compact orientable connected manifold, we know that this means that the de rham cohomology on $M$ is finite-dimensional. Can we say something about the dimension of cohomology of the open sets of $M$? I think that the cohomology should be finite-dimensional too.

[gentle osprey]

emme:

[marsh forge]

Do you mean connected open sets

Bc you can just take inf many epsilon balls

[limpid mural]

Yes connected one

you're right

[marsh forge]

Let me think abt this

[limpid mural]

sure

[marsh forge]

Hm. Take an open disk of dimension greater than 3. You should be able to embed a sequence of shrinking solid tori all connected whose cohomology would be infinite in degree 1

[limpid mural]

mh ok, then I make a further assumption. How about the open sets $M \setminus {p}$

[gentle osprey]

emme:

[marsh forge]

I believe that should work. Consider the homology of M rel a nbhd of p homeo to R^n this will be the same as M. And near x the change should be small, and then mayer vietoris can be used i believe

Im running errands so this is just a sketch

[limpid mural]

Yes using Mayer-vietoris I can show that the bettis number of $M\setminus {p}$ are the same of $M$ except for $b^n$

[gentle osprey]

emme:

[limpid mural]

but I used a fact about finite dimensional spaces

[marsh forge]

Oh you want inf dimensional? Im not going to conjecture abt that bc things go wrong and i dont know a lot abt them

[limpid mural]

That is: if I have an exact sequence $0 \to V_1 \overset{f_1}{\to} V_2 \to \ldots \to V_{k-1} \overset{f_{k-1}}{\to}V_k \to 0$ then $\sum_{i = 1}^{k} (-1)^i dim V_i = 0$

[gentle osprey]

emme:

[limpid mural]

I used applied this result to the mayer vietoris sequence

so I was assuming that $H^i(M \setminus {p})$ was finite dimensional

[gentle osprey]

emme:

[limpid mural]

for example taking your open sets, just call $N = M \setminus {p}$ and $V = $diffeo to an open ball of radius one

[gentle osprey]

emme:

[limpid mural]

We get that $N \cap V $ is homotopic equivalent to $\mathbb{S}^{n-1}$

[gentle osprey]

emme:

[limpid mural]

so $H^{i}(N \cap V)$ is isomorphic to $\mathbb{R}$ for $i = 0, n-1$ and trivial otherwise

[gentle osprey]

emme:

[limpid mural]

and $H^i(V) $ is isomorphic to $\mathbb{R}$ for $i = 0$ and trivial otherwise

[gentle osprey]

emme:

[limpid mural]

so to compunte $b^i(N)$ for let's say $i > 1$ I have the following sequence

[gentle osprey]

emme:

[limpid mural]

$0 \to H^i(M) \to H^i(N) \to 0$

[gentle osprey]

emme:

[limpid mural]

Without already knowing that $H^i(N)$ is finite dimensional I can't say that $dim(H^i(M)) = dim H^i(N)$

[gentle osprey]

emme:

[limpid mural]

right?

[sleek thicket]

@limpid mural i only skimmed the conversation above, but are you asking whether exactness of that sequence implies the two vector spaces are the same dimension?

Because if that sequence is exact then the map H^i(M) -> H^i(N) is an isomorphism

[limpid mural]

Yes

Nvm got it

😅

[sleek thicket]

Also I think max got scared off because he thought you were asking about infinite dimensional manifolds, not vector spaces lol

[limpid mural]

Lmao

[marsh forge]

Oh I just got busy hahaha

[unique wyvern]

So RxR is the set of all ordered pairs in R, how is this different to R2?

[uncut surge]

Not at all, that's usually how R^2 is defined

[unique wyvern]

Ok, I think its a product topology thing im getting confused with

Yeah its that, basically was getting tripped up over considering R2 via euclidean open sets then the the product metric on RxR

[unique wyvern]

https://gyazo.com/9f8219b502646a82bcda0eba2d4ed7a2

Is that supposed to say infinite

[midnight jewel]

yea definitely

finite intersection of open sets is open

[unique wyvern]

arbitrary union

so take complements to show?

[midnight jewel]

which would then imply that all sets are open and closed, which is obviously false in general

[unique wyvern]

Ok thanks

[midnight jewel]

just follow the hint

show that those sets are all open and that their intersection is S

[unique wyvern]

Oh yeah I've done this question

[midnight jewel]

then you’re done

[unique wyvern]

I was thinking about using it

no actually i did use it for the pointwise topology

showing that the point zero is closed

by saying its the intersection of of closed sets which i then proved were closed

[limpid mural]

Let $T = \mathbb{S}^1 \times \mathbb{S}^1$ the torus, $p \in T$, $M = T \times T$, $N_1 = T \times {p}$, $N_2 = {p} \times T$ and $X = M \setminus (N_1 \cup N_2)$

[gentle osprey]

emme:

[limpid mural]

I want to compute the De rham cohomology of the smooth manifold $X$

[gentle osprey]

emme:

[limpid mural]

So I try to study the Mayer-Vietoris sequence

with $U = M \setminus N_1$ and $V = M \setminus N_2$

[gentle osprey]

emme:

[marsh forge]

im not sure this approach makes sense how are you gluing those two

[limpid mural]

$M = U \cup V$ and $U \cap V = X$

[gentle osprey]

emme:

[marsh forge]

oh thats a setminus my bad

i was thinking quotient

okay sure

[limpid mural]

oh nono, thank god it's not the quotient

[marsh forge]

thats just a smash of the torus though for the record

but anyway

[limpid mural]

ok so $U$ should be diffeomorphic to $V$

[gentle osprey]

emme:

[limpid mural]

Now I don't know how to compute their de rham cohomology groups, I think an homotopy equivalence could be the way to compute them

But I suck at this, my thought is that $U$ should have the same homotopy of $T \times \mathbb{R}$

[gentle osprey]

emme:

[marsh forge]

why do you think this

So (T x T)-(T x p) is the same as T x (T-p)

[limpid mural]

Because I think about a torus minus a circle, which is another situation but seems similar

So (T x T)-(T x p) is the same as T x (T-p)
@marsh forge oh

[marsh forge]

Does that help you?

Do you know another way of describing T-p

[limpid mural]

wedge of two circles if I aint wrong

[marsh forge]

you aint

[limpid mural]

So (T x T)-(T x p) is the same as T x (T-p)
@marsh forge how did you see that?

[marsh forge]

So what is T x T

what I do is I take a torus

and I replace every point on the first torus w another torus

if I then remove one of the tori

its the same thing as never having added it in the first place

so it suffices to remove the point before I take the cart. product

[limpid mural]

so legit

thank you for your help!

[marsh forge]

np

[knotty pasture]

Hy everyone

I need some exercises, better if with a solution, on de rahm's cohomology, solvable with mayer vietoris. Do you know any?

[marsh forge]

have you done all the exercises in bott tu

if not do all the exercises in bott tu

Here's a good one

The (unreduced) suspension of a space X is given by XxI/~

where the quotient collapses all points $(x,1)$ to a single point and all points $(x,0)$ to a single point

[gentle osprey]

MaxJ:

[marsh forge]

(two distinct points)

Compute the cohomology of Suspension(X) via mayer vietoris

in terms of the cohomology of X

@knotty pasture

[knotty pasture]

Mmm nice

I try to think about the exercise you proposed, however I still have to learn how to use this technique well

For example I have a quick doubt, if x is a manifolds, how could I prove that suspension (x) is a manifold?

always if it is

mmm

idk

[marsh forge]

no sorry i forgot that you need a manifold for derham

you can pretend its a manifold

the computation is entirely algebraic

[sleek thicket]

it's just ||the cohomology of X shifted one degree up|| right?

Because ||you can take U = suspension minus bottom point and V = suspension minus top point, both are contractible and the intersection deformation retracts onto X||

[knotty pasture]

Shifted one degree it means H(X)^k=H(susp(X))^(k+1)

[sleek thicket]

That's what I meant

Sorry, I didn't mean to spoil the problem

[limpid mural]

I computed the de rham cohomology of $T - {p}$ and $T \times (T \setminus {p})$

[gentle osprey]

emme:

[limpid mural]

where $T$ is the torus $\mathbb{S}^1 \times \mathbb{S}^1$

[gentle osprey]

emme:

[limpid mural]

For $T \setminus {p}$ we have $H^0(T \setminus {p}) \cong \mathbb{R}$, $H^1(T \setminus {p}) \cong \mathbb{R}^2$ and $H^2(T \setminus {p}) \cong 0$

right?

[gentle osprey]

emme:

[limpid mural]

$\setminus$ is the set difference

[gentle osprey]

emme:

[limpid mural]

Am I right?

higher cohomology groups are trivial

While for $T \times (T \setminus {p})$ we have $H^0(T \times (T \setminus {p})) \cong \mathbb{R}$, $H^1(T \times (T \setminus {p})) \cong \mathbb{R}^4$, $H^2(T \times (T \setminus {p})) \cong \mathbb{R}^5$, $H^3(T \times (T \setminus {p})) \cong \mathbb{R}^2$, $H^i(T \times (T \setminus {p})) \cong 0$ for $i \geq 4$

[gentle osprey]

emme:

[limpid mural]

I hope there is something wrong in this calculations lol

[knotty pasture]

That's what I meant
@sleek thicket Sorry but i have obtain that X(S(x))=X(x) where X(M) is eulero-poincare characterist. Is right?

Also I have a problem applying the result with X = S1

I obtain H1(S2)-H1(S1)-H2(S2)=0

S(S1)=S2 right?

[gentle osprey]

slimvesus:

[knotty pasture]

Ahahah xD

Please explain me

ok

I write what i do

[gentle osprey]

slimvesus:

[knotty pasture]

ok

I don't understand why it should to be true

Wait

I know only de rahm cohomology

defined for manifolds

But is isomorphic right?

[dim meadow]

de rham cohomology is isomorphic to real singular

[knotty pasture]

Unfortunately I did not follow a real algebraic topology course, I know de rahm cohomology only for the application to differential theopology

But if you want to show me how you did the exercise maybe I understand where I went wrong, however now I try to do a tex with my reasoning

[dim meadow]

alternating sum of dimension of de rham cohomology

[knotty pasture]

$\chi(M)=\sum_{n}(-1)^n b^n(M)$

[dim meadow]

that second ^n is bad

[limpid mural]

better $b^n(M)$

[gentle osprey]

emme:

Davide:

[dim meadow]

you can probably just do stuff with mayer vietoris

[knotty pasture]

However, in de rahm comology (
always if it makes sense) i have $\to H^1(S(X)) \to H^1(X)+ H^1(X)\to H^1(X)\to H^2(S(X)) \to H^2(X))+H^2(X)\to H^2(X)\to$

[gentle osprey]

Davide:

[knotty pasture]

Xx[0,1) Xx(0,1]

[gentle osprey]

slimvesus:

[knotty pasture]

Xx[0,1) Xx(0,1]
@knotty pasture ahahaha loool

But the quotient is not trivial here

ahaha

sorry

i contract Xx(0,1] to X

[gentle osprey]

slimvesus:

[knotty pasture]

i can contract $X\times(0,1]\backslash\tilda$ to (0,1]

[gentle osprey]

Davide:
Compile Error! Click the reaction for details. (You may edit your message)

[limpid mural]

lol

[knotty pasture]

then i have $\to H^1(S(X)) \to 0\to H^1(X)\to H^2(S(X)) \to 0\to H^2(X)\to$

[gentle osprey]

Davide:

[knotty pasture]

yes is a cone

[marsh forge]

@sleek thicket yes, in fact suspension is axiomatized in Eilenberg-Steenrod

[sleek thicket]

huh I didn't know that

I don't think it's mentioned in the axioms for homology in Hatcher

[marsh forge]

There should be something like it, its possible it follows from the ones present

The classification of cohomologies via spectra demands it

I think it follows from the ones in hatcher yeah

@sleek thicket Hatcher, exercise 3, 165

[limpid mural]

Max, I made the computations for the cohomology groups of $X = T \times T \setminus (T\times {p} \cup {p} \times T)$ where $T = \mathbb{S}^1 \times \mathbb{S}^1$ using Mayer vietoris on the 4-manifold $T \times T$ with $U = T \times T \setminus (T \times {p})$ and $V = T \times T \setminus ({p} \times T)$. The sequence doesnt seem to help me at all, there are too many non-zero terms. Do you see any better open set to find the cohomology groups?

[gentle osprey]

emme:

[marsh forge]

Let me think about it for a second

[sleek thicket]

Emme, can you use de rham's theorem? I think you can actually write down a pretty concrete delta complex structure

I might be wrong though

[marsh forge]

that seems hard to me it's a 4-manifold

[sleek thicket]

Well you know a structure for T in which p is a zero cell

So I think you can get one for T×T in which p×T and T×p are subcomplexes

Just like, using the delta complex structure on the product

Oh no nvm

That won't give you a delta complex structure on X

I'm being silly

[limpid mural]

I wish I knew so much math but this is my first exposure to everything in a differential geometry course, i don't know much about complexes. I just know basic de rham theory for smooth manifold(homotopy invariance, mayer vietoris, PD and kunneth formula)

[sleek thicket]

No worries, it didn't work anyways

[marsh forge]

Use the identification of T-p with S1vS1

as we talked about before

and use kunneth

[limpid mural]

Can't use that

[marsh forge]

and mayer vietoris

why not?

[limpid mural]

S^1 V S^1 is not a smooth manifold

[marsh forge]

oh right lol deRham

[limpid mural]

but I computed the cohomology of T - p

I posted it

wait

[marsh forge]

this is why you shouldn't learn deRham first

anyway

uh i have no idea how to do this smoothly all of my idea rely on htpy invariance

[limpid mural]

For $T \setminus {p}$ we have $H^0(T \setminus {p}) \cong \mathbb{R}$, $H^1(T \setminus {p}) \cong \mathbb{R}^2$ and $H^2(T \setminus {p}) \cong 0$
@limpid mural .

[gentle osprey]

emme:

[marsh forge]

Yes

thats true

Use kunneth to get

[limpid mural]

While for $T \times (T \setminus {p})$ we have $H^0(T \times (T \setminus {p})) \cong \mathbb{R}$, $H^1(T \times (T \setminus {p})) \cong \mathbb{R}^4$, $H^2(T \times (T \setminus {p})) \cong \mathbb{R}^5$, $H^3(T \times (T \setminus {p})) \cong \mathbb{R}^2$, $H^i(T \times (T \setminus {p})) \cong 0$ for $i \geq 4$

[marsh forge]

Tx(T-p)

[gentle osprey]

emme:

[marsh forge]

then finish it with mayer

that should work I think

[limpid mural]

I can't finish with mayer

I mean I can compute $H^0$ and $H^4$

[gentle osprey]

emme:

[marsh forge]

Man this is so much harder with deRham

[limpid mural]

but not the others, since there are too much non zero terms

[marsh forge]

i don't really know how to picture generators for it

[sleek thicket]

Emme can you post the sequence in full?

[limpid mural]

yes I can

[marsh forge]

but you need to think about the LES geometrically

[limpid mural]

Give me 2 mini

I tex it

[marsh forge]

and use that to compute

you can determine some of the maps this way (hopefully)

i unironically think learning deRham first rather than as an obscure computational tool is bad

[limpid mural]

First of all do you agree that $T \times T \setminus (T \times {p}) \cong T \times T \setminus ({p} \times T$? Where $\cong $ is diffeomorphism

[gentle osprey]

emme:

[marsh forge]

Yes

[limpid mural]

nice

I'll tex the sequence

rn

[marsh forge]

also you don't really need that

you need that they are htpy equivalent

which is trivial

[sleek thicket]

Yeah I just want to make sure we're all on the same page before trying to figure it out

[marsh forge]

sham lmk if you want to try to figure out a category theory thing thats been messing w me today

its kind of involved tho lol

[sleek thicket]

Oh sure

[limpid mural]

http://mathb.in/42314?key=13d66081ee415911641cab0ad89803f7fd9f12d8

@marsh forge @sleek thicket

do you agree?

[sleek thicket]

I think you probably don't want to just look at the dimension of everything

Like, you know generators for some of these groups I think

[marsh forge]

ok i have to say you did a lovely job

of latexing that LES

[limpid mural]

I did my best

[sleek thicket]

It is very nice

[marsh forge]

yeah this guy looks bad enough

that you need to actually use the map

rather than dimension spamming

often times you know more about the maps than just the kernel = image thing

[limpid mural]

So I need to find $\delta$ and stuff

[gentle osprey]

emme:

[marsh forge]

hopefully

not delta

[sleek thicket]

You might not need to go that far

[marsh forge]

i would try everything else first

[sleek thicket]

You can sometimes concluded that like, something is a surjection by looking at generators

So the next map has to be zero

Stuff like that

[limpid mural]

But To get the generators I need to actually compute the groups and not their isomorphism class, right?

[sleek thicket]

Right, exactly

What's your formulation of the kunneth formula?

Does it give you an explicit iso?

I'm thinking that we know a generator for S^1

So we can get ones for T and M

If kunneth is explicit enough

[limpid mural]

$H^k(M \times N) \cong \bigoplus_{p+q = k} H^p(M) \bigotimes H^q(N)$

[gentle osprey]

emme:

[sleek thicket]

Sure but do you know an iso between them explicitly?

[limpid mural]

proof use induction on open sets

and five's lemma

I don't see any morphism constructed

[sleek thicket]

Hmm maybe not then. I'm thinking of the singular cohomology case, where you prove some explicitly defined map is an iso

I haven't seen it for de rham

[limpid mural]

they just tensorize e mayer vietoris

well but I know this

[marsh forge]

btw can I ask where you got this problem

[limpid mural]

I know that the map $H^p(M) \otimes H^q(N) \to H^{p+q}(M \times N)$ descend from the map

[gentle osprey]

emme:

[sleek thicket]

Yeah that's sufficient

Is it the wedge product?

On representatives

[limpid mural]

yes

of the pullback of the projections

btw can I ask where you got this problem
@marsh forge sure it's an exercise my professor gave last week, you can find in english here

[sleek thicket]

Okay cool so you actually know generators for the cohomology groups of T and M

[limpid mural]

http://people.dm.unipi.it/martelli/didattica/matematica/2020/Manifolds.pdf page 242 exercise 8.7.5

[sleek thicket]

Emme do you see what I'm saying?

[limpid mural]

I am thinking

I guess If I know the generator and I know the maps of Mayer-vietoris then I know pretty much everthing

[sleek thicket]

Well in principle, but you probably want to avoid looking at δ

But yeah that's what I'm thinking rn

I am outside on a walk so I can't bash everything out

But I will when I'm at a whiteboard next

[limpid mural]

So spamming dimension(to avoid delta) + generators should be the right path

[sleek thicket]

That's my hope

So write down mayer vietoris but don't just make everything R^n

[marsh forge]

yeah finding like, a 0 map for instance is almost as good as finding a 0

and stuff like that

Just write down everything you can and pick at it for awhile and come back if it seems untenable

[limpid mural]

Yeah I can see there is a zero map but only on the first row, that make the first map of the second row injective, that's all that I know so far

So write down mayer vietoris but don't just make everything R^n
@sleek thicket I'll try that

[sleek thicket]

There's probably a better way to do this

[limpid mural]

Maybe some better open sets

some low dimensional trick

[sleek thicket]

I don't have a good sense of what this space looks like

I'm just visualizing it with T a circle lol

[marsh forge]

One trick I like to use emme

is I write like

$R_\gamma \times R_\eta \times R_\alpha$ instead of $R^3$

[gentle osprey]

MaxJ:

[marsh forge]

and keep track of what generators gamma alpha eta refer to geometrically

or i guess for deRham analytically

and of course it suffices to think about your maps on just those generators

[limpid mural]

thank you for the advice max

I am going to sleep

thanks for the help

I hope you'll enjoy the others problems on the book

[sleek thicket]

That book looks cool

[limpid mural]

it is, he did a geometric topology book too if someone is interest in low dimensional topology (http://people.dm.unipi.it/martelli/Geometric_topology.pdf )

[sleek thicket]

Okay so things are actually super nice for the 4-torus I think

We have four angle "coordinates" θ, γ, φ, ψ

These aren't actually coordinates, but they do give a global frame for T^4

Namely $\partial/\partial \theta, \partial/\partial \gamma, \partial/\partial \varphi, \partial/\partial \psi$

[gentle osprey]

shamrock:

[sleek thicket]

This gives a coframe, i.e. a smoothly varying basis dθ, dγ, dφ, dψ for the set of 1 forms

And these forms are locally the differential of a function (you can choose charts where angle functions exist for real) so they're closed

This gives a basis for the space of k forms, namely take all wedges of k of these together

and these are all actually closed

They'll descend to give a basis for the de rham groups

I saw this by kunneth but you can probably think about it more explicitly?

I'm not sure what a basis looks like for T^2 \{pt} though....

Well the 0th group is easy, the 1th is free of rank 2, and the 2+th is 0

hmm so maybe it's actually the same thing?

Like the inclusion T^2{pt} -> T^2 is possibly an isomorphism on H^1, meaning the (classes of the restrictions of) dθ and dγ are a basis for H^1(T^2\{pt})

[marsh forge]

that map should def be iso

[sleek thicket]

The map S^1 wedge sum S^1 -> T^2 is an iso on simplicial H_1 by direct computation, so the same is true in singular, so T^2 \ {p} -> T^2 is an iso in singular H_1 by homotopy invariance, so T^2 \ {p} -> T^2 is an iso in singular H^1 with coefficients in \R since \R is a field. Thus its an iso on de rham H^1

Okay so what does H^2 and H^3 of T^2 × (T^2 \ {p}) look like?

Should still be wedges of stuff

But now some will be zero?

Right so in U = T^2 × (T^2 \ {p}) we have [dφ ^ dψ] = 0, but all others are nonzero

So the map H^1(T^4) -> H^1(U) (+) H^1(V) is the diagonal

The map H^2(T^4) -> H^2(U) (+) H^2(V) sends a basis vector b to (b, b) if it's one of the middle 4, sends it to (b, 0) if it's the first, and sends it to (0, b) if it's the last

And the map H^3(T^4) -> H^3(U) (+) H^3(V) acts on the first to basis vectors by b -> (b, 0) and on the last two by b -> (0, b)

All of these maps are injective I think

So if I did my computation right, the maps in this diagram should be

Which are all injective

This says the mayer vietoris sequence splits up into nice exact sequences

And the final map is an iso, so actually we get that $H^3(X) \to \R$ is an iso

[gentle osprey]

shamrock:

[sleek thicket]

So
\begin{align*}
H^0(X) &\cong \R \
H^1(X) &\cong \R^4 \
H^2(X) &\cong \R^4 \
H^3(X) &\cong \R \
H^i(X) &\cong 0 \quad \text{for } i > 3
\end{align*}

[gentle osprey]

shamrock:

[sleek thicket]

And you can get an explicit basis from what I wrote above if you want, you know that the maps from H^2(U) (+) H^2(V) are surjections

@limpid mural

[spark parcel]

hey topologists. I was wondering, you cannot have parrallel lines on the sphere? What about parrellel curves? I tried to do a graph theory proof on the torus using parrelel curves, but just realized now that parallel curves can't be assumed to be existent, which is what my (probably bad) proof did.

[marsh forge]

What do you mean by parallel

Non intersecting?

Thats easy

[spark parcel]

i defined it to be a one to one correspondence of points with the points in another curve, that are all equidistant. not sure if thats a good definition

i dont really know anything about topology but we were doing planar graphs in a course and had some questions about drawing them on the torus

[marsh forge]

Topology might not be what you want

But i think the answer is yes

To any reasonable interpretation of the question

[limpid mural]

@sleek thicket thank you very much!

[sleek thicket]

Np!

You really have to get into the details with these kinds of problems (sometimes)

I still think there's probably a better way to do this though...

Like maybe X is homotopy equivalent to a nicer space

[clear jackal]

Is the product of continuous functions in a topological group continuous?

From a topological group to itself I should say

[limpid mural]

yes, since the product is continuous

[clear jackal]

Oh ok in product topology on $A\times B$ the function $h : X \to A\times B$, $x\mapsto (f(x), g(x))$ is continuous if $f$ and $g$ are continuous?

[gentle osprey]

datorangeguy:

[limpid mural]

yes, try to show it using the definitions

[clear jackal]

Oh right finite intersection

[tardy meadow]

some notes I was looking at define a cover of $X$ as a set $\mathcal U \subseteq \mathcal P(X)$ such that: $$\bigcup_{U \in \mathcal U} U = X$$. This isn't standard, is it? More or less every other source I've seen has $$X \subseteq \bigcup_{U \in \mathcal U} U$$ instead

[link because it breaks the latex: (https://dec41.user.srcf.net/notes/IB_E/metric_and_topological_spaces.pdf)]

[gentle osprey]

George!:

[marsh forge]

The latter is fine

The former works if X is your total space

Yeah obviously equality will hold as defined

[tardy meadow]

it threw me off because later on they say that given a cover V of [0,1], you can cover {0} with a subset of V

[marsh forge]

You should be able to prove this

Yeah

[tardy meadow]

which isn't necessarily true with the former definition?

[marsh forge]

Thats an open cover in the context of subspaces

But if you think about the subspace topology

It works

[tardy meadow]

right, just struck me as weird, though it makes sense now (and ofc the opposite inclusion is obvious as the union of open sets), thanks again.

[wanton terrace]

Hey, not too sure if this is the right channel, but I'm wondering if anyone can help me with metric transformations? That is how something like a box transforms being moved through a metric

I'm just a bit confused on how to use a metric I guess. Trying to teach myself some GR

[marsh forge]

It would provably help if you were able to formulate a specific question

[small obsidian]

What book?

[marsh forge]

Or give an example

[wanton terrace]

Ok, given a square box, where in regular cartesian coordinates has lengths L, and given the schwatzchild metric, how does the shape of the box change when its center of mass is at some point (r,theta,phi) in the schwartschild metric?

Or if it better communicates my struggles, how does the shape of the box change as it falls into a black hole? Only accounting for the metric and not special relativity or tidal forces or anything

[marsh forge]

Oh you might get more help on the physics side of things

#old-network

[wanton terrace]

I may have answered my own question actually. I think the metric just describes a geodesic of objects, but doesnt affect the shape of the object itself

[limpid copper]

Not quite. What you are looking for is geodesic flow. Every point of your shape moves according to a geodesic. Of course it can be distorted.

The key fact here is that a metric (any connection) induces a path structure.

[wanton terrace]

@limpid copper that does make sense. Do you think I could get around this box problem by modeling each of the corners of the box as a massless particle and tracking their paths?

I'd have to account for stress forces later of course

[limpid copper]

Sure why not. Your image will be rather coarse tho. But if your goal is to visualize distortion due to curvature I would recommend doing more Riemannian geometry. There are very 'visual theorems' about different parts of the curvature tensor, for instance about size change or shear.

Solving all these geodesic equations can get ugly.

[wanton terrace]

I'll look into it thanks. Ultimately my goal is to see how wave functions transform near a black hole

[loud scarab]

Hello, quick question, from this definition, I have a question, lets say a point p in U is a regular point, then the image is a regular value of F, but, what if another point, q in U is a critical point but which has the same image, i would say the image is a critical value of F, but how can both statements be true at the same time?

[limpid copper]

point p in U is a regular point, then the image is a regular value of F This is not what's written in the definition. It says a value is regular iff it's not critical.

[loud scarab]

well

I still dont know what happens if two points that have the same image are different

p critical --> f(p) critical

q not critical --> f(q) not critical

f(p) = f(q)

[hollow inlet]

Blessed Theorem

[limpid copper]

q not critical --> f(q) not critical Wrong. Rephrasing your definition: q critical value of F iff there exists a critical point p s.t. q=F(p). Negation of both statements yields: q regular value (not critical) of F iff there is no critical point p s.t. q=F(p), i.e. all points in the preimage are regular.

[loud scarab]

oh okay got it now

sometimes capturing the true meaning of a definition is not apparent at first glance

[limpid copper]

I actually also came with a question. I am kind of desperate at this point. Please give it a shot.

[gentle osprey]

Axion:

[sleek thicket]

Do manifolds with boundary (embedded in R^n) have a tubular neighborhood?

[honest narwhal]

Compact?

I think the answer if you're not closed is "Yes but maybe the radius is epsilon(x) instead of epsilon"

[sleek thicket]

Oh yeah, in my brain the defintion of tubular neighborhood allows for varying radius

You can probably use the same proof as for the empty boundary case then

I'm thinking about how to show the de rham groups of a compact manifold with boundary are finite dimensional

The proof in Hatcher is pretty indirect

[honest narwhal]

Do you know the De Rham theorem?

[sleek thicket]

Yeah

That's what I meant by "Hatcher's proof"

He proves that the singular homology groups are finitely generated

From which it's easy

With de rham

[carmine thistle]

Hi
Can someone provide intuition on the conditions under which there is a unique local solution to Hamilton's equations on a symplectic manifold and the conditions under which it can be uniquely extended

For instance, we know that if grad H is Lipschitz, then there must be a local solution

And if the energy level curves are compact then there is unique extension

[brittle thicket]

Could anyone help me with the following:

Show that a polynomial f(z) with complex coeff can always be extended to a continuous map \tilde(f) from S^2 to S^2. Show that the degree of this map equals the degree of f as a polynomial. Show also that the local degree of \tilde(f) at a root of f is the multiplicity of the root.

I don't understand how to do this

I understand that z^k has degree k if it's considered from S^1 to S^1

But what happens with S^2?

[wanton marsh]

here S² is the Riemann sphere, C u { infinity }

topologically it's a sphere

[reef ether]

@brittle thicket are you familiar with homotopy?

[honest narwhal]

Pretty sure he does know that. And @brittle thicket yeah so you just extend to f(infinity) = infinity

As for degree stuff

Hmm

I guess f(z) = z is the identity so degree 1

And then the idea would be that deg(fg) = deg(f) + deg(g) where f and g are polynomials

And in principle you could also do that locally

[brittle thicket]

Yes but my issue is that I'm not working over S^1

Hatcher does exactly that computation in the case of z^k:S^1->S^1

So what do I do here?

S^2 is locally C so I can view z^k as a stretch in the plane?

Which should be homotopic to a rotation

Which is itself homotopic to the identity

I think I get the idea but I'm lost in the details

The other issue comes when I have a full polynomial

Not just a z^k

The degree of \sum^n a_iz^i is n

But deg f + deg g = deg f+g

Right?

So z^k + z has degree what, z+1?

How do I justify this locally?

Wait nvm

Hatcher doesn't say anywhere that deg f+g = deg f+deg g

So that's a non-issue

[marsh forge]

all polynomials of degree d should generated the map with class d

[brittle thicket]

What do you mean?

What's the map with class d?

[marsh forge]

You have a map S^2 to S^2

pi_2(S^2)=Z

so every such map has an integral rep

[brittle thicket]

Oh we haven't seen higher homotopy groups

That's Hopf right

Pi_n(S^n) = Z

Sadly we haven't seen this in class

[marsh forge]

oh i mean idk who proved it

[brittle thicket]

And I don't see how I could use it

[marsh forge]

How does hatcher define degree here

and local degree

[brittle thicket]

Okay so the degree is the integer defining the induced map f_*

It's a map from Z to Z so it has to be of the form da

For some integer d

[marsh forge]

Induced where

In homology?

[brittle thicket]

Yes

You look at a map f:S^2->S^2

[marsh forge]

Ok then just translate everything I said from htpy to homology lol

degree of polynomial should give which map it is

[brittle thicket]

That's precisely what I don't understand lol

In the text Hatcher does the following:

He takes z^k

[marsh forge]

So for the S^1 maps, as z gets very large all polynomials are basically z^n

[brittle thicket]

And sees it as a function S^1 to S^1

[marsh forge]

This should be the same idea

[brittle thicket]

Then what he does is he gives a homotopy argument

[marsh forge]

Let me think about a nice homological approach

[brittle thicket]

Namely, locally there are only finitely many points x_i in S^1 such that f^-1(y) = {x_1,...,x_k}

And near every point, f is a local homeomorphism

Then he homotopes that to a rotation

Which since it is a homeo and is homotopic to the identity, has degree 1

Then he just sums it

Surely I can't just copy paste this argument can I?

Or does it work just the same when working over S^2 (seen as the 1-pt compactification of C)

[marsh forge]

what page in hatcher is this

Oh wait this is immediate from some hatcher results

You want to use prop 2.30

Put y=0

then the fiber over 0 has some k\leq n elements

Then the degree of the map is the sum of all their local degrees

Therefore the problem quickly reduced to showing that the local degree is multiplicity

[brittle thicket]

Oh wow

So that's really what it boils down to ok

And this comes from the fact that there are finitely many roots

And that locally, those are homeomorphisms?

[marsh forge]

not just finitely many

but (with multiplicity) the degree = number of roots

I'm not sure what you mean by locally a homeo

we actually need these local degrees not to be 1

if multiplicity is higher

Oh duh

Okay so the linear map (x-x_i) has degree 1 because its a local homeo

Oh wait hm

Ok think about z^n. For any nonzero point we have n distinct roots, all of which will have degree 1

so that gives z^n with degree n near any point

Then (z-z_i) has local degree 1

by composition with z^n

(z-z_i)^n has degree n

and locally we can't distinguish p(z)(z-z_i)^n from (z-z_i)^n as long as p does not have root (z_i)

that should do it

you might need to fill in some of my boldly asserted lemmas there

[brittle thicket]

Lmao I'm so lost

Why are you composing here?

I don't get what this has to do with multiplicity of root = local degree

Maybe it's something along the lines of

If z_i is a root with multiplicity k then p(z) = q(z)(z-z_i)^k?

[marsh forge]

Ok so

if you compose degree maps

you get the product of the degrees

thats in hatcher

so to show that (z-\alpha)^n has degree n

it suffices to show (z-\alpha) has degree 1 near alpha

and z^n has degree n

because then we just compose

and multiply

[brittle thicket]

Okay that makes sense

Let's back track a little

Oh I see

At a root

I only have one point in the preimage

Wait no

The preimage of zero is all the roots

And their local degrees are all one

Wait no that's not what you said

Ah I see this solves the first point

But how do I deal with showing that the multiplicity of the root equals the local degree?

[light veldt]

Can a fractal embedded in an n-dimensional euclidean space have a hausdorff dimension larger than n?

[fervent citrus]

i don't think so

[midnight jewel]

I would be very surprised too but I can’t immediately find anything confirming it online

[fervent citrus]

the hausorff dimension is increasing and i believe you can show that the hausodrff dimension of R^n is n with some effrt

[midnight jewel]

the second bit is definitely true. by increasing you mean that if U⊆V then dim(U) ≤ dim(V)?

that would do it yea

[marsh forge]

@brittle thicket Thats the statement i proved hahaha

[marsh forge]

Multiplicity is given by terms of the form (z-\alpha)^np(z) for some irrelevant p(z) term

[gentle osprey]

welperooni:

[wicked mirage]

ignore it, got it

[brittle thicket]

OH

it makes a lot more sense now hahahahaha

why is p(z) here irrelevant to the discussion?

[marsh forge]

Uh thats something you probably have to prove

but the idea is that near a root

the root term dominates

there is probably a way to make this rigorous but I lost interest at that point lol

so in local homology we should basically not be able to differentiate

[cobalt lily]

ultra basic question- is the manifold the flat plane or the curved plane?

additionally, if anyone recalls hearing something early on in their math career that made manifolds click, would love to hear it. they sorta makes sense but not quite enough to answer questions about them

[uncut surge]

In that depiction, the manifold is the curved space, where the flat plane is the tangent plane to the manifold at the point x

I don't think there's a better philosophy for manifolds than "things which locally look like R^n, but can have a more complicated topology globally"

And to understand most basic diffgeo concepts, understanding this "locally manifolds are just R^n" picture is sufficient. At least to understand how you can do analysis on manifolds, i.e. differentiating

[cobalt lily]

ha! that makes a ton of sense now for several reasons, thanks so much!

[bold canopy]

is there an unknotting algorithm for 2-knots?

[gritty widget]

Had to proof that if f extends to a continuous map $E^{i+1}\to X$, then f represents the zero class in $\pi_{i}(X; x0)$

[gentle osprey]

Stephen:
Compile Error! Click the reaction for details. (You may edit your message)

[gritty widget]

My attempt: Every element in $\pi{i}(X,x{0})$ it's a class of a map $(S^{i},) \to (X,x{0})$. Via the extension $\tilde f$ of $f$ we would have that the same class is produced via $\tilde f\circ i:S^{i}\to E^{i+1} \to X$. But now we recall that $E^{i+1}$ is contractible hence an element in $\pi{i}(X,x{0})$ that comes as image of an element in $\pi{i}(E^{i+1},)$ as to be $0$.

[gentle osprey]

Stephen:

[gritty widget]

is this ok?

[marsh forge]

Whats is E here

Oh that makes sense

[meager python]

Is this channel for algebraic geometry also?

[chrome dew]

yep

[meager python]

Do i need to learn stacks to get into moduli spaces of algebraic curves?

[bitter yoke]

I don't know a ton about moduli spaces but the answer is probably yes

If you only really want a vague, unrigorous idea of what moduli spaces are, you can work only with schemes

But when you try to actually construct some moduli spaces of curves, you run into problems

Like, it's probably not clear to you why this is a problem, but sometimes your algebraic curves will have non trivial automorphisms which means the naive way you'd want your moduli space to be can't be a scheme

And I think there are ways to fix this solely by working with schemes, but stacks are generally the more natural answer to fixing these problems @meager python

[meager python]

Thanks. I only vaugely know what moduli spaces are. I kinda lack the understanding exactly why you need the more general machinery of stacks

[bitter yoke]

https://ncatlab.org/nlab/show/moduli+space#because

Is basically one reason why things fail to work

Okay, this example is probably a bit complicated. I think there's an easier example using rational curves but

[vocal wharf]

the easiest way would be to construct the open sets in S^n, i think

eh, nvm that is probably not what you want

actually constructing the cones in R^{n+1} seems like a drag though

[uncut surge]

Alternatively, if you really want to stay in R^{n+1}, choose any open neighbourhood U of one of the lines that does not intersect the other line, and then by Hausdorffness of R^{n+1}, for every point of the other line you can find an open neighbourhood disjoint to U

oh wait that's too dumb

no no

[vocal wharf]

my general idea would be to construct open neighbourhoods of the preimages in S^n

and make sure that they are saturated

[uncut surge]

yeah that works too

[marsh forge]

just think of RP^n as a Z/2 quotient of S^n. Take two points in RP^n, lift to S^n, take neighborhoods giving hausdorfness in S^n of all 4 preimages, and then go back downstairs

Oh, well open sets of lines are going to correspond to antipodal open sets on S^n

Theres no real difference you're just hiding it

But I mean, it's not hard to take these cones on lines using like, degree-wise arugments

like a basic open set of lines will just differ by some small \theta, and these \theta can be chosen so that the cones dont overlap

well they will overlap at 0 but thats removed

yeah as stated I think any topologist would be fine but maybe you'll wanna work out the details

[ebon mortar]

I am trying to use a lemma to make a geometric argument about whether a given curve is asymptotic with respect to our surface M. The lemma says "Let M be a surface in R3, let a:I->M be a curve, and let U be the restriction to a of the unit normal vector field on M. Then a is an asymptotic curve iff a'' is tangent to M" (a=alpha)
Can someone explain what the bold part means visually or geometrically?
My teacher didn't explain restrictions very well.

[meager python]

Ok I kinda see the automorphism problem in regards to why coarse/fine moduli spaces are not enough. But it’s not obvious why generalizing to stacks solves that problem. Which is the gist of stacks im guessing

[bitter yoke]

Well, to answer that question you need to figure out what a stack is lmao

Which I don't really understand very well either but

[meager python]

Yeah precisely my point with my last sentence :p

[bitter yoke]

I guess an analogy is that people realized that working with varieties wasn't enough and we needed schemes

[meager python]

Digging into orbifolds explains the importance of the automorphism group and deligne-mumford stacks into the necessity of capturing more information

[bitter yoke]

Yeah, that's essentially the connection

iirc there are also a couple other problems that stacks solve but I'd have to look at some notes to remember

[ebon mortar]

Is it true to state that if acceleration of a curve alpha in a surface M is zero, the it is both normal and tangent to M?

[sleek thicket]

I am having trouble with some stuff with frobenius's theorem

[gentle osprey]

shamrock:

shamrock:

shamrock:

[sleek thicket]

okay wait actually things might be good

[gentle osprey]

shamrock:

[sleek thicket]

and the algorithm in ISM involves first finding commuting vector fields and then using their flows to find a flat chart

[gentle osprey]

shamrock:

[sleek thicket]

and so those coordinates should give a flat chart

ty for helping shamrock

hmm those coordinates don't look so hot anymore, and I erased the computation on my whiteboard

oof

[gentle osprey]

shamrock:

[sleek thicket]

and said iso send V and W to coordinate vector fields

okay, now I have a first order linear system of pdes

$f : U \to \R$ has to satisfy
\begin{align*}
z \frac{\partial f}{\partial x} &= x \frac{\partial f}{\partial z} \
z \frac{\partial f}{\partial y} &= y \frac{\partial f}{\partial z}
\end{align*}

[gentle osprey]

shamrock:

[sleek thicket]

I do not know how to solve pdes, halp

okay this is in lee somewhere for sure, I think

oh I also need that $\frac{\partial f}{\partial z} \neq 0$ everywhere

[gentle osprey]

shamrock:

[gentle osprey]

shamrock:

[sleek thicket]

so like, I reduced to finding a flat chart for the distribution $\operatorname{span}(\partial/\partial x, \partial/\partial y)$ on $\R^2 \times (0, \infty)$

[gentle osprey]

shamrock:

[sleek thicket]

this does sound easier

is $\R^2 \times (0,\infty)$ a cube?

[gentle osprey]

shamrock:

[sleek thicket]

wait can't I just take z' = ln(z)?

but ln(x^2 + y^2 + z^2) isn't a solution to my original differential equation...

okay yeah this works

but I am

confused

guess I fucked up setting up the PDE?

oh no, $\phi(x,y,z) = (\ln(x), \ln(y), \ln(x^2 + y^2 + z^2))$ isn't surjective

😦

[gentle osprey]

shamrock:

[sleek thicket]

I googled my exercise and found a seemingly incorrect answer on mse

https://math.stackexchange.com/q/3081078/408287

I don't think the flows of either of those frames exist for all time

Which means you don't get a global flat chart

Unless I'm misunderstanding something

i can't ask on there though because someone in my class got called out for posting homework questions on mse already lol

[meager python]

Does Spec(Z) being final in the category of schemes have any geometrical meaning?

[stark field]

I need a counterexample for two closed sets whose sum is not closed

[limpid mural]

what do you mean with sum?

[gritty widget]

I assume set of all sums x+y where x in X, y in Y

[stark field]

$A+B = {a + b | a \in A, b \in B}$

[gentle osprey]

Navix:

[fervent citrus]

Z and πZ in R maybe

[gritty widget]

yeah

[stark field]

any reason u chose pi? or is any irrational number plausible?

[gritty widget]

yeah

[fervent citrus]

any irrational works

[willow spear]

https://cdn.discordapp.com/attachments/630686968141053952/716050489455673454/Screenshot_45.png

Someone guid eme

Just getting started with Topology

[strange folio]

So what it's saying is prove it for two things first

[willow spear]

ike how would you even prove something,iek this

[strange folio]

e.g. let U and V be open sets. Prove that U intersect V is open.

[willow spear]

ok

open means within T right?

[strange folio]

yeah, open just means in T.

[willow spear]

ok

How would u show that they interesect

thats part that i don;t understand

[strange folio]

What are your axioms for T?

[willow spear]

1. T includes null set and the set X

2. T contains the union of any subset of X

3. T conatins the intersection of any subsets of X

Definntion of a topology on X

[strange folio]

I think you're not being quite precise enough, as axiom 2 and 3 should look more different from each other

[willow spear]

wait what

[little hemlock]

You have that for any $U, V \in \tau$, $U\cap V \in \tau$. You want to show that for any finite collection of elements $U_i \in \tau$, the finite intersection $ \bigcap_{i=1}^n U_i \in \tau$

[gentle osprey]

kxrider:

[willow spear]

so how would show that mathematically

[little hemlock]

do you know how to write a proof by induction?

[willow spear]

No not really

I mean I know what it means

[vocal wharf]

you are probably missing serious prereqs for topology, then

[little hemlock]

i mean, topology without tears is a pretty gentle introduction. Do they not talk about induction at all somewhere in the book?

[willow spear]

Yeha thats the book i'm using

[vocal wharf]

this is an actual book problem?

this is so weird

why are there quotation marks around mathematical induction

[willow spear]

so how would u use math induction

couase they don;t talk about that in the book

[little hemlock]

i can guide you though this, but lets move to #proofs-and-logic

[willow spear]

Yeha epic thanks

[steep kestrel]

How can I define in a "simple" way the Ricci Scalar?

[willow spear]

Any of you guys know good latex to text convertors

[gritty widget]

let Y be a countable discrete space and X=\{1/n: n\in\bN\}. i'm struggling to prove that X and Y don't have the same homotopy type

all i can think about is compactness

[frosty sundial]

so, in my mind, X is a countable discrete space

it's countable

and the subspace topology inhereted from R is the discrete topology

so X and Y are homeomorphic

[sleek thicket]

i second what buncho said. You might be thinking of X union {0}? That's not discrete (it's countable and compact)

[gritty widget]

oh yea thanks you're right I mistyped

[sleek thicket]

I'm kind of surprised ig, it seems like homotopy equivalence shouldn't be able to distinguish these two?

I think they both have homology which is 0 in degree > 0 and free on countably infinitely many generators in degree 0

[gritty widget]

hold up there's a hint let me find it

[sleek thicket]

All paths into either have to be constant by connectedness

[sleek thicket]

Ahh I think that makes sense

So X is compact, right?

[gritty widget]

ya

[sleek thicket]

What does that tell us about a potential homotopy equivalence f : X -> Y?

What can we say about the image of any continuous function X -> Y?

[gritty widget]

it's compact?

[sleek thicket]

Yup!

What are the compact subsets of Y?

[gritty widget]

everything in Y?

wait idk

[sleek thicket]

No, Y itself is not compact

So let S be a subset of Y

What's the subspace topology on S look like?

E.g. what are the open sets?

[gritty widget]

it's discrete

[sleek thicket]

Yeah

So when is a discrete space compact?

[gritty widget]

if it's finite?

[sleek thicket]

Exactly

So what does that tell us about maps X -> Y?

[gritty widget]

oh

all but finitely many points in X get mapped to a single point in Y

wait

how

[sleek thicket]

?

Didn't you just prove it?

Here's a proof: let y = f(0). Then f^(-1)(y) is a neighborhood of 0, so it contains all but finitely many elements of X

Here's a proof: let y = f(0). Then f^(-1)(y) is a neighborhood of 0, so it contains all but finitely many elements of X

You don't really need to think about compactness ig

You don't really need to think about compactness ig

[gritty widget]

ok i think i get that

ty

[wanton marsh]

speaking of contractible spaces

is this space contractible

a double comb

there is a line segment for each rational slope

[dim meadow]

Lol

[gritty widget]

https://topospaces.subwiki.org/wiki/Double_comb_space

[wanton marsh]

you're spoiling all the fun

[pseudo crane]

Is this supposed to be p^tilda?

I haven't seen p tilda defined like this anywhere else

[gritty widget]

Yeah looks like a typo @pseudo crane

[pseudo crane]

Thanks

[lyric quartz]

Here, how do we know that we can just take $V_s$ to be the closure mapped to a unique $A_\alpha$? Thanks!

[gentle osprey]

ClearlyHalfAlive743:

[gritty widget]

i think it's this https://en.wikipedia.org/wiki/Lebesgue's_number_lemma

[gentle osprey]

TTerra:

[gritty widget]

well, that's not really what he's doing, but this should also work

[gritty widget]

is this a typo?

because contractable doesn't imply convex

[frosty sundial]

but convex implies contractible

which is what's relevant

[gritty widget]

yes

the wording makes it seem like contractible implies connectedness, and hence convex implies connectedness

idk maybe i can't read

[frosty sundial]

your interpretation wouldn't even make sense if it were true

mart

[gritty widget]

wait so is the question asking me to prove convex implies connected?

[frosty sundial]

if A implies B and A implies C, you can't make any connection between B and C

no, it's asking you to prove contractible sets are connected

and then deduce that convex sets are connected

because convex ==> contractible ==> connected

[gritty widget]

oh right ok sorry lol

[low arch]

I realize this is introductory Topology, but could anyone help with this?

[vocal wharf]

it's not really topology at all, but the flaw is that it assumes that every element is related to at least one other element

[dim meadow]

#proofs-and-logic

[low arch]

@dim meadow Thanks, the problem is from the Munkres Topology lol

[dim meadow]

probably from chapter 1

[vocal wharf]

(consider the empty relation on a nonempty set, which is both symmetric and transitive but not reflexive)

[low arch]

@dim meadow Yeah, it is.

@vocal wharf Thank you

[lyric quartz]

Here, it says $f_i$ is path obtained by restricting $f$ to $[s_{i-1}, s_i ]$. What does this mean? We can't just say $f_i: [0,1] \rightarrow X$ as map where everything is 0 except on $[s_{i-1}, s_i]$, since that wouldn't be continuous, and we can't reparameterize since that would defeat the purpose of proof. Thanks!

[gentle osprey]

ClearlyHalfAlive743:

[sleek thicket]

Why can't you reparameterize?

If $f_i$ is the map $f_i(t) = f((s_i - s_{i-1}) t + s_{i-1})$ then $f$ is homotopic to $f_1 \bullet (f_2 \bullet (\dots (f_{m-1} \bullet f_m) \dots))$

@lyric quartz

[lyric quartz]

You mean $f_i(t) = f((s_i - s_{i-1}t + (s_{i-1})$?

[gentle osprey]

ClearlyHalfAlive743:

[lyric quartz]

@sleek thicket

[sleek thicket]

Yes except you screwed up the parentheses a little

[lyric quartz]

Yeah, true.

[gentle osprey]

shamrock:

[sleek thicket]

I'm not sure what you meant when you said it would defeat the purpose of the proof

[lyric quartz]

it just didn't seem right lol

[sleek thicket]

I mean all the stuff about reparameterization is bookeeping. All that it's saying is that you can cut up the path

The author (Hatcher) isn't the most formal lol

[lyric quartz]

Yeah, having trouble reading because I'm someone who needs everything spelled out lol.

[honest narwhal]

So I've been thinking of reading this one book by Tammo tom Dieck

Which seems more formal than Hatcher

(It's written by an algebraic topologist rather than geometric topologist lol)

[lyric quartz]

I would probably be lost anyway, but it does look nice lol

[honest narwhal]

Yeah I mean, if you prefer things spelled out look into other books. Hatcher's kinda notorious for being wishy washy

Rotman I think is the best easy book on algebraic topology

[lyric quartz]

Thanks for the rec, I'll check it out!

[meager python]

By knowing all the fibres of a scheme over (p) for p prime, can we ”construct” the scheme?

[sleek thicket]

I'm not really sure how to interpret your question

Like, what would a counterexample look like? Two nonisomorphic schemes with isomorphic fibers over each (p)?

Okay maybe I do know how to interpret your question

What about Spec Z versus a big disjoint union of {(0)} and {(p)} for p prime?

Like take a scheme vs the disjoint union of its fibers over p

That might not have the same topology

@meager python idk if this makes sense

But it seems like it answers your question in the negative

So I think concretely this will be Spec Z vs the disjoint union of Spec Q and Spec Z_(p) for each prime number p

[meager python]

Do you really mean Spec Q? @sleek thicket

[sleek thicket]

I think so. I calculated that that was the fiber of Spec Z over (0) but I might have fucked up

[meager python]

Z_p being Z/p?

[sleek thicket]

No, Z localized wrt to (p)

[meager python]

This is a stupid question just trying to get a feeling for what information we miss from looking at the fibres over (p)

[sleek thicket]

Sorry no

It should be Fp

I said that originally but I thought it was wrong

You miss how the fibers fit together

Like, think about the corresponding statement with topological spaces

[meager python]

I wonder what the geometric meaning of Spec Z being final in schemes is

[sleek thicket]

I think I have another counterexample, maybe?

What are the fibers of Spec Z_(p) over Z?

Is it Spec Fp? I think so

Because you end up with the fibered product of Spec Z_(p) and κ((p)) over Spec Z, but κ((p)) is Fp and so you're looking Spec (Z_(p) (×)_Z Fp). Tensoring with a localization is the same as localizing, and localizing Fp wrt (p) as a Z-algebra just gives Fp, so the fiber should be Spec Fp

That's the same as the fiber of Spec Fp over (p)

So that's a counterexample with two one-point spaces

Spec Fp vs Spec Z_(p)

[meager python]

Both have the same fibres? (I need to check that on paper cant see that in my head)

[sleek thicket]

I think so

No I am a fool

Z_(p) has a generic point that I was ignoring

The original example of Spec Z vs the disjoint union of Spec Q and Spec Fp for all p should work though

I should probably add a disclaimer that I'm bad at AG

[warm mirage]

I am trying to show that the convex envelope of a set of points of an affine space is the set of barycenters of these points affected with non-negative coefficients
Any ideas?

[fervent citrus]

the set of barycenters is convex so it contains the convex envelope of your set of points, and the convex envelope of your set of points contains your set of points and is convex so it also contains the barycenters

[warm mirage]

What I tried is I took two points P and Q barycenters of two systems of points with non negative coefficients
The segment PQ is formed of the barycenters of {(P,1-α);(Q,α)}, where 0<=α<=1, but why are these barycenters, barycenters of points in the original set?

And in the inverse statement I can't see why it should contain the barycenters, if you can elaborate please

[fervent citrus]

for "contains segments implies contains any barycenter of any finite number of points with non negative coeffs", it's just induction

[warm mirage]

Ok I see that

For the first implication I can't see why that is true

I should use the associativity of the barycenter operation

But as I said if I take P Q barycenters of two systems of points with non negative coefficients, the segment PQ is formed of the barycenters G of {(P,1-α);(Q;α)} where 0<=α<=1
Why these points G are barycenters points of the original set?

[fervent citrus]

which original set are you talking about ?

[warm mirage]

The set of points given in the affine space

[fervent citrus]

why would you think it would contain the segments ?

[warm mirage]

It doesn't have to, there's just points, the envelope contains the segments

I mean the set of barycenters, why does it contain segments?

These segments are formed of barycenters of the endpoints with coefficients 1-α α
Why these barycenters are in the set of barycenters of the points

[fervent citrus]

segments are made of barycenters

[warm mirage]

Yes

[fervent citrus]

so the set of barycenters contain the segments

[warm mirage]

This is what I don't get

The segments we're talking about are the segments with endpoints barycenters

Why the barycenters of the barycenters are in the set of barycenters

To show that I should show that these barycenters of two barycenters are barycenters of systems of point in the set of points considered with non negative coefficients

[fervent citrus]

you should be able to brutally rewrite a barycenter of barycenters as a barycenter of your original points

[warm mirage]

Yes the associativity property, but I can't rewrite that

Because G=bary{(P;1-a);(Q;a)} should be written as G=bary{(A1;a1);...(Ak;ak);(B1;b1);...;(Bm;bm)}

P is a barycenter, so it is affected of the coefficient Σci, and Q is a barycenter so it is affected of the coefficient Σdi

It should be that Σci=λ(1-α) and Σdi=λα to use the associativity, right?

[fervent citrus]

idk what kind of associativity principle you're trying to invoke

all of this should just be a verification

[warm mirage]

This is the associativity principle

Thanks for the help
I'll think about what you said

[lyric quartz]

here, why can we assume that we can do a small perturbation, and how can we assume that there are at least three rows? Thanks.

My guess with perturbation is that it's something to do with $A_\alpha$ being open, so you have some wiggle room or something. And for three rows, maybe if you have less than three rows, you get something trivial? But I'm not sure.

[gentle osprey]

ClearlyHalfAlive743:

[gentle osprey]

slimvesus:

[lyric quartz]

The proof says you can make a perturbation.

So the grids misalign.

Open cover of $X$.

[gentle osprey]

ClearlyHalfAlive743:

[lyric quartz]

this seems also relevant:

[stark field]

$$\text{dist}(K,A) = \inf{\text{dist}(x,A), x \in K} = \inf{d(x,y) : x \in K, y \in A}$$

How can I prove that if A and K are closed but not compact, disk(K,A) is not necessarily > 0?

[gentle osprey]

Navix:

[fervent citrus]

you can find an explicit example of two disjoint closed sets A and K such that dist(A,K)=0

for example in R², the graph with equation y=1/x and the line with equation y=0

[stark field]

@fervent citrus Is y=1/x closed because it isnt convergent? In R it would be open I think.

[fervent citrus]

the set {(x,y) in R² | y=1/x} is closed in R²

[sweet wing]

in general y=f(x) is closed for f continuous

[stark field]

thanks @fervent citrus @sweet wing

[fervent citrus]

you're welcome

[atomic totem]

Hi, I've just been introduced to the Minkowski distance metric and i just have a quick question, i understand that when p = 1 and 2, it corresponds with the Manhattan and Euclidean distance respectively and the Minkowski distance metric is mostly used when p = 1 and 2. How i do know which p to use?

[round flint]

What do you mean

[round flint]

Consider the

Topology generated by the basis of closed intervals with rational endpoints

Then show then convergent sequences in standard topology converges in this topology

Topology on R

[fervent citrus]

I think you can show this topology is finer than the standard one

[round flint]

Yes

I'm trying to show $(a, b)$ \subset $[a, b]$

[gentle osprey]

Mr. M:
Compile Error! Click the reaction for details. (You may edit your message)

[wanton marsh]

ooooh closed intervals

I was wondering why it wasn't completely trivial

[round flint]

I'm trying to show

for any open neighborhood of x in $T_1$ there is an open neighborhood of x in $T$ contained in it

[gentle osprey]

Mr. M:

[wanton marsh]

which one is T1 ?

[round flint]

The topology generated by the basis of closed intervals

with rational endpoints

[wanton marsh]

is {0} an open in T1 ?

[round flint]

yes

[wanton marsh]

does the sequence 1, 1/2, 1/3, 1/4, ... converge to 0 in T1 ?

[round flint]

yes

wait no

yes

[fervent citrus]

for all reals a and b such that a<b, you can check that the open interval with endpoint ms a and b is the union of all the closed intervals with rational endpoints it contains

so an open set in the standard topology is a fortiori also open in your topology

[round flint]

Yes

I've shown that already

[fervent citrus]

so you're done

[round flint]

And concluded (a, b) is open in T_1

How?

[wanton marsh]

I thought you had to show converges in T => converges in T1

you would be done if you had to show converges in T1 => converges in T

it also would be nice if you were not trying to prove something false

[round flint]

yes

I think the example you asked about before