point-set-topology

[unreal stratus]

Compactness is absolute like

It's a property of the subspace itself

[abstract furnace]

yea i think my brain is just rotting from thinking about metric spaces mb!!!

[gritty widget]

How do u guys remember the conditions for boundedness and connectedness

I keep forgetting them

Compactness too

[coarse night]

self descriptive

[unreal stratus]

I think doing enough problems / working w them should help you intuit them more and remember

[grim knot]

ah okay, it makes sense then. it confuses me because I thought that we needed an equality

and I didn't get that

Thank you!! @abstract furnace

[gritty widget]

Imho no but alr

[grim knot]

I know, I might have a lot of easy question, but I'm trying to make a bit of order in my head, and to understand the concept

I was doing this proof and I don't completely understand the last step. Why does the fact that every x in K has a neighbourhood imply that X\K is open?

[next crystal]

the second to last sentence?

[grim knot]

yep

[next crystal]

$p \in \bigcap_{j \in J} V_j \subseteq X \setminus K$

[gentle osprey]

michαel

[next crystal]

so we have found an open neighborhood of p contained in X\K, and since p was arbitrary X\K is open

[grim knot]

Easy question incoming. I am working with topological spaces, but is it always true that if we find open nbhd for every point, we always have open sets?

I'm kinda skeptical about it, this is why I asked

[next crystal]

yeah a subset U of X is open if for all x \in U there is an open set V such that x \in V \subseteq U

[grim knot]

okay, thanks for helping!

[next crystal]

its iff btw but thats the one that matters here

[woven badge]

Hey guys I have a maybe stupid question. Why is it that when we have some open set U of some space X, we can construct a continuous map f from X to [0,1] such that f^{-1} (0,1] is U. When the space is metric I choose the map which gives the distance from x to the complement of U. But how is this possible in the general case ?

[bitter smelt]

Why do you think this should be possible?

[unreal stratus]

Been trying to think of a counterexample lol

[coarse night]

Hmm take some irreducible set maybe

All maps to [0,1] will be constant

[woven badge]

Wow I misread what I was reading

One needs to add that there is an numerable open cover of X

[coarse night]

my example still works tho

[woven badge]

Yeah

That's what I don't get

[coarse night]

can you rather share the statement you are reading?

[woven badge]

Oh

I just got it

Nvm sorry

It was May's 'a concise course in algebraic topology' page 51, but I entirely misunderstood the stuff entirely my bad

[hidden crag]

i'm even more confused now, how does this relate to page 51 pf concise course

[empty grove]

Is that the section on neighborhood deformation retracts? I can see the similarity with that

[unreal stratus]

That's what I assumed

[tribal palm]

Fixed point theory?

[hoary breach]

Topology

[tribal palm]

oh alr, thank

hmm... seems my library doesn't have it? that's a first

wow yeah that seems to be the case

[gritty widget]

consider libgen

[tribal palm]

i hate reading things digitally, even on a tablet

.<

i have one of those e-ink tablets, but i just can't seem to bring myself to read on it, only write

so if i can't get my hands on a textbook physically, i simply can't use it

[limpid fern]

printers are a thing

[thorny agate]

What even is beta(f)?

like what is beta(f)(x)?

[empty grove]

That's what the exercise is asking you to define. Do you want the solution?

[cosmic socket]

That’s part of the problem

[thorny agate]

oh

I thought this was supposed to be something obvious from the definition

uhhh ok then I'll sit on it for longer

[umbral panther]

beta(f) is the unique map that makes the diagram commute. The main point is to prove that there is a map and that it is unique. The points in the last line are what make it a functor

[thorny agate]

ok so I have a guess. If $i_Y$ is the inclusion of $Y \to \beta(Y)$ then I guess the obvious choice is $\beta(f) = i_Y \circ f$ but then what if I have an element $x$ in $\beta(X)$ that isn't in $X$? Then $\beta(f)(x)$ isn't well defined right?

[gentle osprey]

spamakin

[thorny agate]

wait nvm dumb question I forgot that X -> beta(X) was an embedding and embedding meant bijection

I think that's right anyways?

so like if I have any element $x' \in \beta(X)$ I can find a unique $x \in X$ such that $i_X(x) = x'$

[gentle osprey]

spamakin

[coarse night]

write out the universal property of stone-cech compactification

[thorny agate]

since we're looking at the closure of the image of this map

there's a universal property?

[coarse night]

there is

[thorny agate]

oh

that would have been nice to know

[coarse night]

(I just looked it up kek)

[thorny agate]

I shall also look it up then

[tiny ridge]

Too much clutter. Take the 2D example Q(x) = (x+)^2 - (x-)^2. The gradient flowlines of Q are rectangular hyperbolas |x+|^2 |x-|^2 = const, which explains why those edges are pieces of the flowlines

Ignore if you already figured it out

[gritty widget]

yay ibsen is back

[tiny ridge]

sup

[urban zinc]

ayyy Ibsen! :D

where did you go

[tiny ridge]

had to write my msc thesis and i was procrastinating way too much here lmao

[urban zinc]

LOL

[tiny ridge]

done now

[balmy field]

W

[empty grove]

What did you write

[tiny ridge]

something something wavefronts

the title is “h principle for loose legendrian submanifolds”

[empty grove]

[tiny ridge]

speaking of wavefronts, heres something interesting. consider the differential equation dy/dt = 1/2(dy/dx)^2. introduce an auxiliary variable z and consider the solution y(z, t) = z^4/4 + t z^2/2 and x(z, t) = z^3/3 + t z. this y is a multivalued function of x, for a particular t.

the picture of the evolution looks as follows

in particular i think this equation does the following: given y(x, 0), treat it as a wavefront and evolve it according to huygens principle. y(x, t) is the new wavefront at time t

havent checked this but for many reasons i think this is true

now, @patent quarry: this is one third of the KPZ equation

the second term in 1D KPZ is d^2 y/dx^2. this makes it like the heat equation so my belief it is smoothens the shockwaves of cuspidal singularities when it occurs

the third term is gaussian noise, which means its everything above but random wiggly

a “random interface model” should be like a randomized version of wavefront evolution so that makes sense?

[fading vale]

Welcome back!

[tiny ridge]

the reason why im a bit excited by this (given the chain of reasoning above is true) is that the wavefront evolution is just Reeb flow for the unit cotangent bundle ST*R^2 = R^2 x S^1 with the contact structure cos(theta) dx - sin(theta) dy

[coarse night]

so you're done with the thesis

[tiny ridge]

some more dogwhistle: KPZ scales like 1:2:3. this is reminiscent of the cusp/twisted cubic (t, t^2, t^3) which appear as Legendrian submanifolds of the contact space

is there a connection?

[fading vale]

[tiny ridge]

https://www.desmos.com/calculator/nnxby2drar

Heres a plot of the above solution (slide s to evolve in time)

[fading vale]

This immediately crashed my firefox lmao

[tiny ridge]

Huh bizarre

[stable kite]

Does someone know an open reference on branched coverings?😇

[quiet summit]

someone help me. Pls give me hint

Let $\left{x_n\right}_{n=1}^{\infty}$ be a sequence which converges to $x$. Show that the subspace ${x} \cup\left{x_n\right.$ : $\left.n \in \mathbb{Z}^{+}\right}$is compact.

[gentle osprey]

Ji

[gritty widget]

a hint? use the definition of covering compactness and look at the open set containing the limit of the sequence

[quiet summit]

this is what i've come up so far

Since $\left{x_n\right}{n=1}^{\infty}$ converges to x then for every neighborhood $U$ of x there is an $N \in \mathbb{Z}^{+}$ such that $x_n \in U$ for all $n\leq N$ . Thus $\bigcup{\alpha_n \in N} U_{\alpha_n}$ covers ${x} \cup\left{x_n\right.$ : $\left.n \in \mathbb{Z}^{+}\right}$.

but i don't know how to proceed

[gentle osprey]

Ji

[tiny ridge]

did you mean n >= N

i botch inequalities all the time, but “all but finitely many” is easy to conceptualize

[quiet summit]

is this definition wrong?

this is from our prof

[empty grove]

It should be n >= N at the end

[tiny ridge]

according to your prof, 1 0 0 0 0 0 … converges to 1

which would be a bummer right?

[quiet summit]

yoooooo

im just using this module

[pulsar lagoon]

If I take the set [0, 1] and quotient it via 0~1 and x~y iff x = y and x, y =/= 0, 1

Is there a natural way of taking a metric on [0, 1] into [0, 1]/~

[unreal stratus]

Not really

[pulsar lagoon]

sadge

but not unexpected tbh

[coarse night]

use the metric on the circle

[unreal stratus]

I mean yeah you can give it the unique metric such that the homeo with the circle is an isometry

[novel acorn]

use the complex metric

[pulsar lagoon]

yeah but thats not a "natural" metric on it, cuz I was hoping for a metric on [0, 1] (like any metric)

[unreal stratus]

But then what's the point lol just work with the model of the circle you like

[pulsar lagoon]

I was just fooling around lol

[unreal stratus]

Fair o nuff

Ig this is part of why working w topological spaces is often nicer tho

Being able to do the quotients in the first place hrhe

[pulsar lagoon]

I just got to the quotient topology and this construction looked cool

[unreal stratus]

It is

[umbral panther]

Yeah, there’s a natural metric on a quotient. It might not be easy to write down, but it’s natural

In the case of just gluing two points on a space with a path metric, d2(x,y) = min(d(x,y), d(x,0)+d(1,y), d(x,1)+d(0,y))

[empty grove]

Idk how generally this would work, but you could do something like this
Let d' be the metric on [0,1]
Define d([x], [y]) = 0 whenever [x] = [y]
Otherwise, define d([x], [y]) as minimum over all paths from [x] to [y]. That is, take a minimum over all choices of finite sequences
[x_1] = [y_1], [x_2] = [y_2], ..., [x_n] = [y_n]
of the quantity
d'(x, x_1) + d'(y_1, x_2) + d'(y_2, x_3) + ... + d(y_n, y)

Seems to work in this case

[pulsar lagoon]

can I ask where this part comes from: min(d(x,y), d(x,0)+d(1,y), d(x,1)+d(0,y))

[umbral panther]

Does Moldilocks answer the question, provide more motivation?

[empty grove]

I can try

[pulsar lagoon]

yeah I think the shortest path thing makes sense

[empty grove]

I am taking the length of the smallest path from x to y as d(x, y)

But this path need not just go x → x_1 → x_2 → ... → x_n → y

It can jump from each x_i to a y_i in the same equivalence class without adding any length to the path

Infimum of all such paths should work as long as these infima are always non-zero for distinct points

I am guessing that a necessary and sufficient condition for that to happen would be for each equivalence class to be closed in the metric space

[tiny ridge]

yes

that should be enough

[pulsar lagoon]

Interesting

[empty grove]

https://math.stackexchange.com/questions/121023/why-are-quotient-metric-spaces-defined-this-way

[pulsar lagoon]

I will try to see this more

[empty grove]

Hell yeah

Seems like this is it

[umbral panther]

Another way to do it would be to say the function d2 with the largest values that is less than the original metric, but has d2(0,1)=0 and respects the triangle inequality

[pulsar lagoon]

based

[empty grove]

Why is the existence of such a function obvious?

[umbral panther]

If the maximum of two metrics is not a metric, then we’re in trouble. If it is a metric, then just take the maximum of all appropriate metrics

There is an issue about metrics vs pseudometrics. If you take [0,1] and glue all [0,1) to a point, the procedure says that 1 is distance 0 from the rest. Which is the right thing to do

[empty grove]

Not just 2, for this you would want the supremum of any collection of metrics to be a metric

[pulsar lagoon]

what does "supremum of a collection of metrics" mean?(sorry for the dumb questions)

[empty grove]

A metric is a function. You can take a supremum of a collection of functions pointwise

[empty grove]

This seems to be true but there's an even bigger issue

[pulsar lagoon]

ah I see

[empty grove]

How do you show the existence of even 1 metric satisfying this condition?

[umbral panther]

d(x,y) = 0 for all pairs

[empty grove]

assuming all metrics are bounded above by some function

[empty grove]

That's not a metric though

It needs to be non zero for distinct x, y

[umbral panther]

They’re bounded by the original metric

[empty grove]

Yes, I was pointing out that my own earlier statement was incomplete because I just said any collection of metrics

[umbral panther]

As in the example of collapsing an open set, you can’t ask that. If you take [0,1] and collapse [0,1), you have to collapse 1 as well

You can follow up and figure out what conditions on the equivalence relation should guarantee no further collapse

[empty grove]

Yeah sure, but my point is that your construction is useless even when you do have a condition on the equivalence relation

All it gives is a pseudometric, and to prove that it will be a metric in the nice case you would still need to make some explicit construction at some point

[empty grove]

It is not clear a priori, at least to me, that it will ever be non-zero

[swift mountain]

so cool

[empty grove]

[swift mountain]

sus

what do you mean by largest value?
sup over x and y of d2([x], [y])?

if thats the case I could define it to be the correct metric in some closed subspace

and then have fun elsewhere

[coarse night]

imposter metric

[swift mountain]

my highschool nickname

[empty grove]

so cool

[coarse night]

sniped

[umbral panther]

Largest over all pseudometrics satisfying some hypotheses

[swift mountain]

yeah whats largest?

like when is f_1 < f_2?

[umbral panther]

Read the backscroll

[empty grove]

pointwise supremum

[swift mountain]

cool beans

[swift mountain]

so what I wrote then?

[empty grove]

No clue what you wrote lmao

[swift mountain]

based

[empty grove]

[swift mountain]

yeah I understand now @umbral panther

[umbral panther]

Instead of taking the sup over pseudometrics, we could take the inf over metrics. Take the inf over metrics d_n which doesn’t collapse things to zero, but just to 1/n. And we could define dn by either an inf or a sup

(I think this works badly without compactness)

[swift mountain]

yeah the inf is the usual description

like generally when quotienting

first time seeing the sup

[void gazelle]

Hi, guys, is it true that if A,B\subset C, and B\subset C open, A' is the closure of A in C, then closure of A\cap B in A'\cap B is A'\cap B?

[umbral panther]

No. What if B=C? Then A’ cap B is A’

[coarse night]

You mean subset instead of in?

[void gazelle]

oh, sorry,

Yes

[empty grove]

Yes. This follows from the the fact that closed subsets in the subspace topology are the closed subsets of the larger space intersected with the subspace

Wait

I assume you meant to ask for the closure of A cap B in B

Rather than in A' cap B

The answer is the same in any case I suppose

And I don't think you need B to be open

[swift mountain]

wont A'\cap B introduce new limit points then?

I mean closure of A\cap B

oh wait

nvm,

A\cap B\subset A

[void gazelle]

nah, i want to see closure of A cap B in A' cap B

[empty grove]

Hmm should be correct for the same reason then yeah

Wait B needs to be open

[umbral panther]

Why does B need to be open?

[swift mountain]

A = (0,1) and B =[1,2]

[empty grove]

My reasoning was incomplete, still trying to see where we need it to be open

But we need it because there are counterexamples otherwise and it seems to my intuition that that is resolved when B is open lol

[void gazelle]

when i want to prove this, X\subset C closed, such that X containing A cap B, then X may be strictly smaller than A', and X cap A' cap B containing A cap B. is there any way to see this is not true

[echo oyster]

Is the lower limit topology locally connected?

[coarse night]

what are the compact sets in lower limit topology?

[coarse night]

ig you don't need that if A ⊂ B

[echo oyster]

I’ve not read compactness yet

[coarse night]

oh sorry, your question was locally connected, not compact. mb mb

so first, is [a, b) connected in lower limit topology

[echo oyster]

no, it can be written as union of [a,c) & [c,b)

[coarse night]

so can you see why it's not locally connected?

try to recall the definition of locally connectedness and basic open sets in lower limit topology

[echo oyster]

Locally connectedness is, for any nbd U of x , a connected set V can be found for x belongs to V subset of U

Oh I get it

We can’t find any connected V

Thank you

[swift mountain]

Brother @ me if you figure it out

[empty grove]

Got busy lmao will maybe get back to it later

[swift mountain]

Coolcool

[coarse night]

is it the problem of the closure?

[swift mountain]

It's about why B needs to be open

[coarse night]

ah ok

[tribal palm]

flipping through munkres… it seems very pedagogical but also lame in a way that i can’t quite place

[novel acorn]

It's dry

[empty grove]

Yes. Take a point x of A' cap B. You want to show that x is in the closure of A cap B. If x is in A, you are done. Otherwise, you can show that it is a limit point of A cap B and hence in the closure of A cap B. Take any open neighbourhood (N cap A' cap B) of x in A' cap B where N is open in C, then N cap B is an open nbhd of x in C so there is a point y of A in it, so this nbhd of x has a point of A in it.

@swift mountain

You have to use the fact that N cap B is open in C

[swift mountain]

Oh based

[void gazelle]

Thank you guys!

[swift mountain]

Yeah I think intuitively it makes sense and we just had to write it down to make sense of it

Thanks @empty grove very cool

[empty grove]

[urban zinc]

Why is A required to be nonempty here?

[tiny ridge]

there are no maps to the empty set

C_n(empty) is uh

the zero vector space?

i mean the zero abelian group

[urban zinc]

Oh I see, ty!

[umbral panther]

You can allow A to be empty if you define the reduced homology of the empty set correctly. H_0 = 0 and H_-1 = Z
Wrong first try: H_0 = Z and H_-1 = Z

[tiny ridge]

deranged

[urban zinc]

is this because H_n(X/A, A/A) is isomorphic to Htilde_n(X/A, A/A) which is isomorphic to Htilde_n(X/A) since Htilde_n(A/A) = 0 and we can draw out the long exact sequence of reduced homology groups

[urban zinc]

also, does the group H1(R, R \ {0}) count how many (oriented) times a 1-chain crosses zero?

[tiny ridge]

yup

likewise for H_n(M, M \ pt), M n dim

[urban zinc]

ty!! :)

[tiny ridge]

@urban zinc let M be an n dim manifold. consider, for any open set U of M, the group O(U) := H_n(M, M \ U). For an open subset V of U, there is a natural "restriction" map O(U) -> O(V). Prove that there is an exact sequence 0 -> O(U \cup V) -> O(U) (+) O(V) -> O(U \cap V) -> 0.

[urban zinc]

Oh no

[tiny ridge]

(Such a thing, ie an O that satisfies the properties above, is known as a "sheaf")

[urban zinc]

Terrifying

[gritty widget]

sheaves are easy it's just the same thing as continuous functions

[tiny ridge]

agree with TTeppa

not scary

[urban zinc]

so an element of O(U \cup V) is an n-chain in M whose boundary is contained in M \ (U \cup V), mod out by relative boundaries of n+1-chains

so for the first map we want to take the n-chain and split into subcomponents where the boundary is contained in M \ U and M \ V, respectively

maybe

[tiny ridge]

M-U is bigger than M-(U \cup V), notice

[urban zinc]

Oh wait yeah

Okay just send [c] to ([c], [c])?

[tiny ridge]

Yep

[urban zinc]

And then send ([c], [d]) to.... [c] - [d]?

I don't know if that one works

Hm

It's the only thing I can think of though

[tiny ridge]

Yes

Absitively

[urban zinc]

ayyyyy :D

okay and then... I need to figure out why that one is surjective

[tiny ridge]

That's all.

[urban zinc]

So I want to take an n-chain whose boundary is in M \ (U \cap V) and break it into an n-chain whose boundary is in M \ U and an n-chain whose boundary is in M \ V

[tiny ridge]

Right.

[urban zinc]

This feels like barycentric division

[tiny ridge]

Yep

[urban zinc]

Hmmm but I only know that for open covers

Oh wait does (M \ U)^o \cup (M \ V)^o = M \ (U \cap V)

[tiny ridge]

(M \ U) \cup (M \ V) = M \ (U \cap V) believe it or not

[urban zinc]

That is logical lol

[tiny ridge]

Ol'e de'Moivre figured it all out

[urban zinc]

Hm but M \ U and M \ V aren't open

Does everything still work

[tiny ridge]

Oh, fair point.

Probably requires a little fiddling

[urban zinc]

I see I see

That's cool though

[tribal palm]

going to try to find a copy of dugundji's topology, looks lit

[urban zinc]

wtf how

[queen prism]

hey i remember this

[unreal stratus]

So in fact you can prove stronger things than this lol

[queen prism]

there's another result you can prove that includes R and C plus the quaternions I think

using some topological wizardry that i can't imagine

[unreal stratus]

Basically if R^n can be the structure of division algebra over R then S^n can be endowed with the structure of a group for example

More generally yes you can check when S^n is an H-space

[urban zinc]

magical

[unreal stratus]

Which means it has an operation which is unital up to homotopy etc

then H-space structures on S^n correspond to elements of some homotopy groups of spheres

Using the so called Hopf Invariant

[urban zinc]

this is so cool

[unreal stratus]

I wrote part of my bach thesis on it aha

but yes the more general problem here is Hopf Invariant One which is quite deep and was proven by Adams - it gives as a corollary that S^1, S^0, S^3 and S^7 are the only spheres with H space structure, so that R^2, R, R^4 and R^8 are the only finite dim vector spaces over R which admit a division algebra structure (without requiring even associativity)

The easiest proof uses K-theory but the original proof was very long and used detailed calculations with spectral sequences etc

Very cool result :)

[urban zinc]

Also does anyone have any favorite applications of the Euler characteristic?

[unreal stratus]

More general than Euler characteristic but have you seen the Lefshetz fixed point theorem

The proof uses a sort of generalisation of Euler characteristic

Which is v cool

[gritty widget]

lefschetz theory is good

[unreal stratus]

Says you

[urban zinc]

Not yet

About to

In uh

Nine more one-hour videos

[unreal stratus]

Lol nice

Which series

[urban zinc]

https://www.youtube.com/playlist?list=PLpRLWqLFLVTCL15U6N3o35g4uhMSBVA2b

[unreal stratus]

Ah he's so cool

[urban zinc]

Wait do you know him lol

[unreal stratus]

Nah I just watched some of his videos like a year ago

[urban zinc]

He's handsome too ngl

[unreal stratus]

When I was at the stage you are at now basically

Yes

I liked the historical bits

Nicely contextualised stuff

[urban zinc]

yea he's great

[tiny ridge]

Does it? It is not clear to me why S^n being an H-space implies R^n+1 admits a division algebra structure.

The division algebra result is quite classical, you do not need topology for that IIRC

[tiny ridge]

RP^2 is not boundary of a compact 3-manifold with boundary.

[unreal stratus]

Yes agreed, I'm just saying there are many topological results in this direction

[unreal stratus]

Well I had in mind more just the restriction on the "n"

In that it means we can't get any more spheres with H-space structure beyond the ones that obviously have one

[tiny ridge]

Let M(n, R) be the space of n x n real matrices. Call a subspace V of M(n, R) totally nonsingular if every nonzero element of V is an invertible matrix.

The maximal dimension of totally nonsingular subspaces is 1 + #(linearly independent vector fields on S^n-1)

[unreal stratus]

rho(n) moment

Yeah I want to read the proof of like the linear result using Clifford algebras

like representations thereof

[tiny ridge]

The above is a cute exercise but I forget how to do this

The maximal ones are the Clifford algebras as you say

[grizzled ibex]

Have you guys read any book about top alg that doesn't have this "tradition"/"appeal" to combinatorics?

[unreal stratus]

Well I assume you mean linearly independent linear vector fields right

Proving that it coincides with the general case is much harder ofc xd

[tiny ridge]

I don't think I mean that

[unreal stratus]

Well like

[tiny ridge]

The other direction is harder, but at least there's a bound

[unreal stratus]

Hm I meant like

[tiny ridge]

I agree that one of the direction is easier

[unreal stratus]

Do you mean subspace V in the linear sense?

[tiny ridge]

Yes

[unreal stratus]

Okay like

Well yours fairly easily coincides with 1 + # of linear linearly independent vector fields right

[tiny ridge]

Why?

[unreal stratus]

Hm if I'm not mistaken the families of totally non-singular things almost exactly correspond to linearly inndependent linear vector fields

[tiny ridge]

Yeah but why

[unreal stratus]

But I can't remember how lol

[tiny ridge]

It's a nice proof but I forget how to do it

Oh

[unreal stratus]

But anyway the point is that like iirc, that was already known like before Adams' result

[tiny ridge]

Consider the map (V \ 0) x S^n-1 -> S^n-1, f(A, x) = Ax/|Ax|

[unreal stratus]

Yeah I think basically you can take an orthogonal basis of your subspace of M_n or smth

Something roughly like that lol but maybe I'm smoking

[tiny ridge]

This gives a map V \ 0 -> Diff(S^n-1), or equivalently S(V) -> Diff(S^n-1). Take its derivative

[unreal stratus]

Oh okay lol rip

[tiny ridge]

You get a map R^n -> X(S^n-1)

This is injective because meh

f(I + tA, x) = (I + tA)/|I + tA| * x/|x|

What is this as t -> 0

lol

Well I want to take the derivative of the above at t = 0

[unreal stratus]

Okay, my bad, I confused what you said with something else

Sorry

[tiny ridge]

nw, I didn't entirely follow what you wrote

Tell me the derivative of (I + tA)/|I + tA| x/|x|

Is it the component of Ax orthogonal to the sphere

Geometric approach

[unreal stratus]

Okay sorry so what I had in mind was the original place (afaik) where hurwitz radon numbers appeared, like number of orthogonal skewsymmetric anticommuting real matrices

which corresponds to sets of linear orthogonal vector fields

[tiny ridge]

So to argue injectivity of the derivative we just need to ensure Ax is never parallel to x for any A in the totally nonsingular subspace

This is because if Ax = a x, then (a I - A) x = 0. But I, A are both in my totally nonsingular subspace so is there linear comb... oh, I may not be there

If I is there this proves n - 1 <= #(lin indep vector fields on S^n-1)

Dunno how to nudge this further

[tiny ridge]

This is because Cliff(n, R) is actually a maximal nonsingular subspace of M(n, R), that's all.

[unreal stratus]

Oh nice lol

[tiny ridge]

Um

I mean Radon-Hurwitz number is the maximal m such that Cliff(m, R) admits a faithful rep in R^n

I think

And then that's the maximal nonsingular sub

Eg: In M(4, R) sits H which is the span of the four (++++) signature Dirac matrices 1, I, J, K with I^2 = J^2 = K^2 = -1, IJ = -JI, etc

That's maximally nonsingular

(S^3 = SU(2) has 4-1 = 3 lin indep vector fields)

This is the numerological reason for the apperance of Radon-Hurwitz, but those numbers are nothing special IIRC

[mental rose]

Yeah this is purely algebraic I believe

[unreal stratus]

Yeah i think like

nlab has an article doing the entire proof lol

I'm gonna try to learn the proof of that actually for fun hm

[coarse night]

what's the easiest way to see if M is a n-manifold with boundary, then H_n(M)=0

assuming nothing about orientability

(let's say compact)

[tiny ridge]

delete boundary

homology unchanged

noncompact Poincare duality

[coarse night]

yeah but how to see deleting boundary doesn't change homology? without going into cellular homology

[tiny ridge]

boundary admits collar nbhd

youre a fan of Lee he should tell you that

[coarse night]

yeah ik that

[tiny ridge]

excise

[coarse night]

hatcher showed it for non smooth ones well

[tiny ridge]

even easier than excise

deformation retract an open collar to a closed collar

M heq to M \ dM

[coarse night]

yeah that works, cool

but in the excise argument, where are we excising from?

[tiny ridge]

dunno said the first thing that came to mind

might not be correct

[coarse night]

[tiny ridge]

second thing is

solve homology exercise problems putting close to epsilon thought

throw the kitchen sink at random

itll work out

@coarse night learn k theory and tell me how the k theory proof of atiyah singer goes

homology is done you passed AT

time to move on

[coarse night]

yeah I was doing random things and suddenly remembered I needed his result

I did start K theory (vector bundle one) from hatcher

[tiny ridge]

patrick shanahan has a book

[coarse night]

How much k theory do I need for this

[tiny ridge]

where he proves AS index theorem by K theory

idfk man

i don’t understand that proof

[coarse night]

Lol

so much to read so little time

[tiny ridge]

you have one more year than me

use it

oh no

youre already a phd

[coarse night]

ye

[tiny ridge]

F

[coarse night]

[tiny ridge]

decided what youre specializing on?

[coarse night]

I'll break rijul's record

[tiny ridge]

this could be goodbye as you go on to become an algebraic geometer

[tiny ridge]

did he take 7

[coarse night]

~8 maybe

[tiny ridge]

damn

[coarse night]

he's done now

think mohit took 6-7

[tiny ridge]

yeah saw a notice with his thesis defense

mohit switched around a lot

he was doing ggt but then lol

you know what happens

[coarse night]

he did htt no?

[tiny ridge]

his masters thesis was ggt

idk about this htt arc

[coarse night]

ok lol

I don't want hardcore AG, that'll take ages to pick up

[tiny ridge]

what is softcore ag

[coarse night]

idk learn some AG move on

[tiny ridge]

rep theory lmfao

[coarse night]

fuck no

[tiny ridge]

arvind only algebraic geometer left at this point

and he does rep theory

[coarse night]

yeah, holla doing stacks

[tiny ridge]

rofl

that could be fun

[coarse night]

idk what mirror symmetry does, he calls himself algebraic geometer

[coarse night]

ye sounds interesting

[tiny ridge]

rep theory

[coarse night]

something something conformal blocks

[tiny ridge]

yeah rep theory

[coarse night]

no one does symplectic here

[tiny ridge]

nah

[coarse night]

just saw on pinned msgs

[coarse night]

unable to find, which one exactly

[tiny ridge]

https://link.springer.com/book/10.1007/BFb0068264

[abstract saffron]

If you don't, there's no way I'll stand a chance 😄

[molten oyster]

Imma start dw

Also ppl do symplectic here

[coarse night]

I didn’t mean the server

[novel acorn]

that reminds me I need to read Arnol'd to learn symplectic stuff

[rough skiff]

Consider honology with coefficients in a commutative unital ring R.
Does it's additive structure the same as taking the coefficients in the additive group R?

[coarse night]

yes

[empty grove]

Do you have a separate definition for homology with coefficients in a ring?

Separate from the one for coefficients in an abelian group

[rough skiff]

Just taking everything as R module

[empty grove]

Yeah the scalar multiplication is never used for the homology computation so nothing changes

It is only used to define the scalar multiplication on the homology

[obtuse tundra]

Definition question: for X a space that is sequentially compact does that mean that any convergent sequence of terms in X has a limit point in X?

[paper wedge]

yes

no

a space is sequentially compact if every sequence has a converging subsequence

think of bolzano-weirestrass but without the bounded condition

[obtuse tundra]

Ah

[paper wedge]

what ur saying is X is closed

[queen prism]

bolzano-weierstrass my beloved
too bad it's a myth

[obtuse tundra]

I keep getting closed and seq comp mixed up ffs

[paper wedge]

they are similar

so ur doing okay

can something be compact but not closed?

what about the other way around

[cosmic socket]

R is closed

[paper wedge]

yup

[obtuse tundra]

R is closed in R, yeh

Oh R is also compact

[paper wedge]

u sure about that

[obtuse tundra]

oh wait I'm thinking B-W

which has boundedness

[paper wedge]

yea

and btw

sequentially compact != compact alwaays

only in metric spaces

in general they are not equivalent

[obtuse tundra]

yeyeye

compact implies seq comp

and the reverse holds iff the space is metrisable

[paper wedge]

so here is a fun exercise for you

think of any fucked up topological space

and produce counter examples for all sorts of things

[empty grove]

Only for sequential or first countable spaces

Not in general

[paper wedge]

a set that is closed and bounded but not compact

[empty grove]

The stone cech compactification of ℤ is compact but not sequentially compact

[paper wedge]

for example

[coarse night]

reposting

[paper wedge]

^ yea this is something to bear in mind wires

alot of good stuff in metric space topology just doesn't work

[obtuse tundra]

That's a funny quote enough I might just remember it lol

[marsh forge]

Poincaré predicting the theory of infinity groupoids

[queen prism]

is it soon or never

[obtuse tundra]

Oh hold on

[paper wedge]

the fact that algebraic topology was invented first before point-set is very cool

[coarse night]

also why is named as :specz:

[obtuse tundra]

If I say X is a compact space, I need to specify what it's compact in, shouldn't I

[paper wedge]

no

compact means heine-borel compact

open covers

thats just the default

[obtuse tundra]

Yeah ik

[paper wedge]

sequentially compactn implying that is just a nice bonus

u cant have that

in general

[marsh forge]

it’s an extremely cursed meme

[obtuse tundra]

But to define a cover, do I need subsets of X

[coarse night]

lol

[obtuse tundra]

or subsets of the space X is in

(okay Elon Musk's gonna come at me with that)

[paper wedge]

an OPEN covering of a space X is just a collectino of open sets such that the union of those is the space

[empty grove]

It will not matter because the open subsets of a subspace are those of the larger space (intersected with the subspace)

[obtuse tundra]

Yh so any union of open subsets of X will be itself a subset of X

So the definition's nigh-on moot

[marsh forge]

?

[paper wedge]

idk

what u mean

black mirror new season is out btw

[obtuse tundra]

Can an open cover of X be a strict superset of X?

[paper wedge]

an open cover

of X

is not a set

nor a subset

its a collection of open subsets of X such that if u take their union you get X

[obtuse tundra]

Okay, the union of the sets in an open cover then

[empty grove]

It's union has to contain X

Its*

[obtuse tundra]

Yh that's what I'm thinking

[marsh forge]

Usually we require that an open cover consist of subsets of X, possible just X itself. Sometimes authors are a little sloppy and cover subspaces by open sets of the ambient space, but this can be remedied as that open cover becomes a normal open cover by intersecting all the open sets in the cover w the subspace

[paper wedge]

isnt that conventional

[empty grove]

But if you take an open (in X) covering you're fine

[obtuse tundra]

I'm just wondering how the convention came to be that "X is contained in (union)" was used for this

(ig this also includes calling it a "cover")

[empty grove]

Idk lul

[paper wedge]

i think it is

[marsh forge]

Well that’s the whole point hahaha

You want every part of X to be covered

Hence the term cover

[paper wedge]

yea so its like any element of your space is atleast in one of the sets in your open cover

any point*

[obtuse tundra]

Ah right igu

See, my notes use something like $X \subset \union U_i$

[gentle osprey]

waes_wires
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[paper wedge]

then

moldilocks

is righ

for ur case

[empty grove]

[marsh forge]

It’s very common in say real analysis where basically everything is a subset of R^n

[obtuse tundra]

(been a while since using latex for set notation ngl)

[marsh forge]

To allow open covers’ component open sets to not be subsets of X

[empty grove]

It's \bigcup not \union

[obtuse tundra]

lol bigcup

But yh igu

[coarse night]

\bigtorus

[obtuse tundra]

Btw, is it that a space is complete? if every Cauchy sequence is convergent or...?

[paper wedge]

what

space

[obtuse tundra]

Say X is a topological space

[paper wedge]

are u asking if compactness implies completness?

[paper wedge]

ok

X is a topological space

[obtuse tundra]

Nah I'm just trying go through definitions

What's it mean for X to be complete

[paper wedge]

it is as you said

[obtuse tundra]

ah cool

[umbral panther]

That’s the only definition I know. But that’s a definition for a metric space, not a topology

[obtuse tundra]

yh that's what I was thinking

Cauchy's difficult to define without using a metric

[paper wedge]

yes

it only makes sense if u have a metric

me go jim now

@obtuse tundra try to prove in a metric space that compactness implies completness

[obtuse tundra]

oo have fun

[paper wedge]

try to argue like u would do with riezs lemma or something

constructing a sequence that can never have a subsequence converging

god luck

[tribal palm]

i'm just through some topology things without having done any topology, so i'm probably way off with this, but: is the standard topology of a metric space (X,d) the topology generated by every open ball around every element of X ?

[paper wedge]

yea

metric topology

not standard

standard means ?

😄

[tribal palm]

what?

[marsh forge]

Yes

[tribal palm]

also ø and X is in the topology generated by this set of open balls, since they can be written as a union of resp. no empty balls and all empty balls around some element in X ?

[paper wedge]

this is anywhere

by definition

but that implies that they can be written as a union of open balls

( if they are in the topology generated by those )

but not since that then this

[obtuse tundra]

Yeah (X, d) the metric is said to "induce a topology"

[marsh forge]

I when you say “topology generated by” you are implicitly guaranteeing the axioms of a topology are satisfied

So even if the open balls didn’t give you X and the empty set

You’d have to add them in anyway

(It is true that it works out regardless, just a semantic note)

[tribal palm]

hmm

[obtuse tundra]

(hehe hehe hehe, "X, d")

[abstract saffron]

Even for one metric there can be many topologies tho

[obtuse tundra]

(so childish of me but I still find it humorous)

[unreal stratus]

Wait what lol

[marsh forge]

???

[abstract saffron]

Did I say something stupid? 😄

[marsh forge]

I don’t know what you meant

[tribal palm]

as in not taking all radii of the open balls?

[unreal stratus]

Me neither

[obtuse tundra]

Take X = R, and the discrete topology

[marsh forge]

Only one of these is at all related to the metric

[unreal stratus]

Like metric spaces already have a notion of open sets

And you just let the corresponding topological space have the same open sets

[obtuse tundra]

I'm trying to get at what Megumi meant

[marsh forge]

It is true that there are potentially multiple topologies for which all the open balls are open

But that’s not really fair to say that they are all coming from the metric

[unreal stratus]

Yeah lol

[obtuse tundra]

^

[queen prism]

are you talking about how you can often impose many different metrics on one set?

[obtuse tundra]

er...

[abstract saffron]

You can say it's convention and all, but I'm still annoyed if it's not explicitly stated

[marsh forge]

???

[queen prism]

ok I think I see now

[obtuse tundra]

Okay, but given a metric, it induces a single topology

[queen prism]

maybe

[obtuse tundra]

We can have two metrics induce the same topology

[queen prism]

right

[obtuse tundra]

A metric can't induce multiple topologies

In R^n, for instance, the Manhattan distance, the Euclidean distance and the infinity-distance l induce the same topology

(in R^n, this we call the standard topology)

[unreal stratus]

Indeed any metric induced by a norm

[obtuse tundra]

ah yeah, you can go from that too

[tribal palm]

do {B(x, r) | x in X and r ≥ 0} and {B(x,r) | x in X and r in N (the positive integers)} generate the same topology?

[unreal stratus]

Mo

No

[tribal palm]

i must admit i'm not even sure if the second here generates a topology at all

[unreal stratus]

Topology is sensitive to the small stuff

[obtuse tundra]

No, as one of them is a subset of the other

[unreal stratus]

You can always generate a topology from a set lol

[tribal palm]

waa

maybe i should just start reading

[unreal stratus]

Just intersection of all topologies containing the set

[obtuse tundra]

Yeah, a topology is just a collection of subsets satisfying some criterja

whose elements we call open

[unreal stratus]

Since the power set is a topology that is not an empty intersection

[marsh forge]

I always forget that that’s actually well defined bc there’s no set theoretic issues

[unreal stratus]

I am learn p completed K theory

[marsh forge]

(For example, if one tries to define algebraic closure the same way you’d run into problems)

[unreal stratus]

Oof any tips

[marsh forge]

Nice

Which K theory

[tribal palm]

i only stumbled on the intersection of the empty family of sets for the first time last week and had a slight existential crisis

[unreal stratus]

Okay so I mean E theory at ht 1

[marsh forge]

Ah okay sure

So KU

[unreal stratus]

but might read up on CHT for context idk I imagine useful

Well lurie notes look hot

[marsh forge]

Personally I don’t recommend starting w luries notes

They are great for a second pass

But something like the orange book is probably better for an intro

[unreal stratus]

Fair is there anything you'd recommend instead?

[marsh forge]

Ravenels orange book is really good

A little slow if you already know what spectra are but you can just skip

[unreal stratus]

Is that nilpotence and periodicity

[marsh forge]

Yeah

You won’t learn all about like Morava E theories

But you’ll learn like, the big picture of CHT and you’ll be much better prepared to go further

[umbral panther]

Don’t allow r=0, or you get discrete topology

If you allow positive real r or positive integer r, they do generate the same topology

But this is tricky. Normally you don’t want to allow arbitrary generation. Normally you say look at a basis. The topology generated by a basis is intuitive. If they don’t form a basis, the topology they generate may get out of hand and be surprising, as in this case

(I was assuming X was the real numbers. If it’s a small circle, the open set for r=1 is the whole set and the topology is boring)

[marsh forge]

I’d also recommend reading the original Nilpotence II paper

Nilpotence I is also good but very technical

[tribal palm]

hmm ok i don't understand this, but i really don't know what i'm talking about either, i'll come back when i've worked through some chapters of my book

[red yoke]

There's a thing called a subbasis

[marsh forge]

If B(a;0) the ball of radius 0 around a is open, then this set is just {a}

As a result, for every point in the space, the set containing just that point is open

[red yoke]

Where instead of taking arbitrary unions you can also take finite intersections

[marsh forge]

Since open sets are closed under union, you get that every subset is open

[red yoke]

Usually this is what is meant by a family of sets generating a topology

[tribal palm]

ohhh right, as i was writing i thought fsr that B(x,0) = ø

[marsh forge]

Ah yeah

[tribal palm]

which is obviously not the case

[marsh forge]

Nah

[tribal palm]

but yeah we'd still have the same problem

[abstract saffron]

I was thinking about two things at the time, and apparently it was not a good idea. One was a chart induces a metric on the surface, and we might have different charts for same region, but topology should be the same. The other was multiple topologies on R^2, like half-plane topology.

Definitely not a good idea

[umbral panther]

Oh, yeah, you’re right.

[tribal palm]

pff

i'll go eat my pasta now

B(x,r) = { y in X | d(x,y) < r }

[umbral panther]

Right, we’re talking about open balls. The open zero ball is empty

[marsh forge]

Lmao I’m the dumb one

Sorry about that

[tribal palm]

we’re all the dumb one

but yeah B(x,r) is always be open in (X,d)

[tiny ridge]

i havent spent a lot of time with it though. its the original atiyah proof

[trail charm]

is this direction right? i tried to follow Lee's proof for the finite case (bottom of second image)

(also lee’s proof has wrong notation, in his errata he says the k’s should be replaced with n’s)

[paper wedge]

yes

[trail charm]

im a genuus

genius*

[obtuse tundra]

Is it valid to claim that
"A topological space X is said to be connected if for all open subsets A in X, the set X\A is not open."?

(in effect I've tried to take the opposite of the definition of disconnected-ness)

[gritty widget]

indiscrete topology

[brittle rapids]

maybe proper nontrivial

[obtuse tundra]

shoot you're right

Well

Rather, take the discrete topology

But yh it falls apart fairly easily

[gritty widget]

no, indiscrete is correct

discrete is every subset is open

indiscrete is only the space and the empty set are open

any indiscrete topology is connected but the condition you wrote does not hold

[obtuse tundra]

Okay, but in the discrete sense, my proposed definition would make any top.space not connected

Which isn't correct

[gritty widget]

what is "in the discrete sense"?

[obtuse tundra]

I mean, with a discrete topology

Wait, "any indiscrete topology is connected"?

[gritty widget]

yes

by the usual definition

[obtuse tundra]

Oh right

[gritty widget]

but not by your definition

that's what im trying to say

[obtuse tundra]

Because if you could disconnect it, into say U and V, then either U or V is empty, contradiction, thus the space is connected

igu

[gritty widget]

you sort of just repeated the definition of connectedness

but yeah the argument is essentially trivial

[brittle rapids]

once you fix that, i think the statement should work

[obtuse tundra]

"fix it" how

[brittle rapids]

​

[obtuse tundra]

Ah

right then

[queen prism]

the only sets that are both open and closed are X and {}

[marsh forge]

Not to be pedantic but you should exclude the empty set and X

Oh someone already said it

[trail charm]

ok so if i understand it correctly, stone-cech compactification is the closure of the image of the function which maps from the ambient space to the set of continuous functions on [0,1]

but why exactly do we care about it

"set of continuous functions on [0,1]" is the wrong way to put it

but yk what im saying

more like function from the set of continuous functions to [0,1]

like sure it's got some cool properties like [0,1]^C is compact, its a functor on completely regular spaces, etc

[marsh forge]

i think it also has a universal property

i forget precisely

https://math.stackexchange.com/questions/4346580/is-stone-Čech-compactification-the-only-one-with-universal-property

[trail charm]

a bounded real value map on a completely regular map can be extended to a map from its s-c compactification to the reals

so the importance of the s-c compactification comes from the fact that any compactification can be factored via a s-c compactification map?

[marsh forge]

thats one reason yeah

its sort of like a "free compactification"

[trail charm]

because of the universal property?

[marsh forge]

yeah

[trail charm]

gotcha

it's also a bit of a headache to think about

a mapping from a space to a set of functions from a set of functions to [0,1]

😵‍💫

[marsh forge]

honestly this is a good example of the difference between concept and model

somehow I think the intuition from s-c doesn't come from the specific model you've been given

and it might be more helpful to think about it in terms of this universal property

[trail charm]

i dont have a model unfortunately, just the construction and some properties

[marsh forge]

thats what i mean by model sorry

[trail charm]

oh gotcha

[marsh forge]

like, there are plenty of ways to build a space homeomorphic to that model of stone cech i think

but the real intuition comes from sort of working with them

rather than thinking about it using this one specific lens

[trail charm]

yeah if X is completely regular, its stone-cech is compact hausdorff and there's an embedding from X to beta(X)

so that's kinda nice i suppose

unfortunately it looks like bredon doesnt use stone cech in the book, just includes it as a sort of fun fact

[marsh forge]

yeah it doesn't come up a lot in AT

[trail charm]

looks like it comes up much more in analysis

it's interesting but i don't think i'm gonna do the stone cech exercises here (like proving universal property and whatnot)

good to know for now i suppose

[marsh forge]

yeah i think thats fine

whats next for you in bredon

[trail charm]

probably review the rest of chapter 1 (i.e., paracompact, quotient, homotopy, top groups, baire category theorem)

then start on chapter 2 (differentiable manifolds)

i sparsely read chapter 1 throughout the last 2 months so im just going through and reviewing it all

the real question is what's next for you maxj

hmm?

oh i got very active again

[marsh forge]

I am currently doing a deep dive into the Burklund-Hahn-Senger paper Galois Reconstruction of Real Artin Tate motives

i think thats the name anyway

[trail charm]

oh wow yeah man i love that stuff (the only word i know in there is galois)

abstract looks interesting

i am really looking forward to delving deeper into category theory sooner or later

[thorny agate]

Speaking of Stone Cech

I am back asking the same question I asked a couple days ago (got stuck and only now coming back)

ok so I'm familiar with universal property of this compactification

however I'm not sure how to construct what beta(f) would be

[marsh forge]

I think you just use the universal property, unless you are asking for an explicit construction?

[thorny agate]

yea explicit if possible

[marsh forge]

you'd need to pick a model for it then

[thorny agate]

I was about to say I can argue existance using universal property

wdym pick a model

[marsh forge]

like, how do you construct the s-c compactification

[thorny agate]

oh

tada

from my notes (but from Bredon rly)

yea so I have this and I'm not sure how to use this to get something explicit

[trail charm]

bredon!

[thorny agate]

⭐ indeed ⭐ (I'm dying)

[marsh forge]

okay honestly i am a bit not sober tired

[thorny agate]

wish that were me

[marsh forge]

but

[thorny agate]

all g

[marsh forge]

My suggestion would be

try to figure out what the map X->beta(X) is

should be a "literally only thing you can do" type situation

and then define beta(f) to be f on the image of X and then find some clever way to extend this to the rest

[trail charm]

bredon gives an explicit X -> beta(X) map in thm 11.10

[marsh forge]

ah yeah then i would just look at the lemmas in the chapter

[thorny agate]

yea I gave said map here

[marsh forge]

Oh duh

Hm

I think the idea should be that for every point in the closure

you define betaf(x) to be the limit of a sequence approaching x

[thorny agate]

tried looking elsewhere in chapter + in the lemmas used in the proof that the definition is well defined but no luck

[marsh forge]

like every point in the closure is the limit of points in the image of X

[thorny agate]

oh

[marsh forge]

and you want to show the choice doesn't matter i think

[thorny agate]

as in choice of sequence?

[marsh forge]

like i recall that the s-c compactification as some description as a space of formal limits

yeah exactly

this probably combines the concrete construction as well as regularity

[thorny agate]

that has the same flavor as the proof that the map X -> beta(X) is an embedding

so probably right track

I'll try it

this class is gonna kill me next sem

[trail charm]

ping me if u solve it

im curious but need to sleep

https://www.ams.org/journals/tran/2002-354-09/S0002-9947-02-03008-8/S0002-9947-02-03008-8.pdf

section 2 looks promising

tho it uses filters and ultrafilters

[thorny agate]

idk what those are

[trail charm]

X is completely regular, so it has a s-c compactification i_X : X -> βX

by universal property, f : X -> Y extends uniquely to f’ : βX -> Y

but Y is also completely regular so it also has a s-c compactification i_Y : Y -> βY

so the composition X -> βX -> Y -> βY gives the desired result?

uniqueness of the diagram follows from uniqueness of universal property

so ig let βf = i_Y \circ f’?

smth like this

im in bed so im too lazy to tex it out

even if it’s wrong it makes a ton of sense in my head and that’s enough to get me to go to sleep

[copper raven]

I need help understanding the following claim from this video: The guy on the video claims that using the property of covering spaces we can map a sufficiently small neighbourhood of y_0 into an evenly covered set U

I don't understand how he claims that

[tiny ridge]

look up the lebesgue number lemma

you cover X by a collection of open sets evenly covered by the covering map, and then pull it back by F

then finitely refine it by compactness of K x I

then you can choose however fine a covering you need

[empty grove]

What is K?

Or is this in the context of the video

[tiny ridge]

Oh Y

[empty grove]

Y isn't given to be compact though

[tiny ridge]

Ok fine

Use compactness of I

[empty grove]

And does the Lebesgue number lemma require metric spaces

[tiny ridge]

you just need to refine vertically

I is a metric space

[empty grove]

Oh ok lol

[empty grove]

You can also take the image of y_0, find an evenly covered neighbourhood U of that, and then the preimage of U is an open neighbourhood of y_0 that maps into an evenly covered nbhd

[copper raven]

Ah yes

But image under what @empty grove

F I guess?

[empty grove]

Yes

Maybe image of (y_0, 0), not sure what the notation is

[copper raven]

Yeah I thought the same, but he was like "we're constructing a function" and didn't write F there so I thought what was it

More like (\pi_1(y_0), 0) xD

[empty grove]

The function being constructed is probably F tilde

[midnight parrot]

What is the last one. I've never heard of "anime" in the context of topology

[tiny ridge]

https://cdn.discordapp.com/attachments/1021163726822449283/1114963028199354448/D5F71BBD-EE2B-4D88-8964-5BFD540B33AD.jpg

Still you doubt, atheist?

[empty grove]

Oh they misspelled anima

From condensed math

[midnight parrot]

ahaha

[coarse night]

anime is the new big math thing

[tiny ridge]

tbh 90% of homotopy theorists are weebs

its the norm

[coarse night]

you aren't upto date

[empty grove]

I went to a homotopy theory conference and was chatting with the speakers and one of them asked what I did over the weekend there and I said I played Minecraft and they got very interested

They were all very puzzled that I play Minecraft at my age and what it is all about

[tiny ridge]

great convo

[empty grove]

Spent 30 mins answering their questions about the game

[tiny ridge]

they should get into it

would vibe well with the subject

or a better use of their time

one of the two

[coarse night]

should have said you watched anime during the weekend

[empty grove]

Why would I lie

[coarse night]

to get them interested

and suggest them to watch boku no pico

[empty grove]

Bruh one of them got worried because she said that her son plays Minecraft and she was expecting that he'll grow out of it soon but now that she's heard that I "still play Minecraft??!!" she lost all hope

[empty grove]

No clue what you're on about

[coarse night]

Minecraft ain't that bad tbh

[unreal stratus]

Lol

[tribal palm]

hmm ok so i had a look at previous exams in my uni's topology course and it looks weirdly... easy?

https://www.uio.no/studier/emner/matnat/math/MAT3500/oppgaver/MAT3500-4500_2012.pdf

[trail charm]

can someone check this when they get a chance :)

[trail charm]

this is original problem

[tiny ridge]

only Stone can Cech this

[drifting jungle]

To apply universal property of s-c compactification the codomain of your function must be compact Hausdorff

Y may not be compact, so instead you first compose f:X to Y with i_Y:Y to betaY and then apply the universal property.

[trail charm]

ohhh gotcha

ty

like this?

,rotate

[empty grove]

Ye

[trail charm]

woohoo

@thorny agate see above

[trail charm]

i wonder if minecraft would be a good tool for visualizing topology

[empty grove]

Good luck visualising ℝP² in Minecraft

[trail charm]

i can barely visualize it in my head

[feral copper]

Hey! I'm taking two (smooth, closed, connected, and hell, oriented even) surfaces S and S' in a (smooth, closed, connected and oriented) 4-manifold X, whose intersection form I denote as Q. I let n=Q(S,S'). Does the fact that there are two ambiant isotopic copies F~S and F'~S' such that F intersects F' in exactly |n| points have a name?

[thorny agate]

I guess I was hoping for something more constructive/ explicit than "apply universal property" since I had a explicit definition for way the S-C compactification is done. I might just be being overly pedantic

[trail charm]

oh gotcha

[feral copper]

Because I also like playing a bit, you could do it actually using command blocks. Remember that RP² is nothing but a disc with antipodal boundary points identified in pairs. Now, take two copies of a square (one is the mirror of the other), and when you reach the edge of the first, teleport to the edge of the second

[trail charm]

oh i just noticed that lol

my bad

[thorny agate]

Said exact definition but again I'm being overly pedantic probably lol

Nah nah you're fine

[umbral panther]

That’s not true, right? Do you mean a name for pairs that satisfy this?
(It’s not true for two reasons. One is dimension 4. The other is fundamental group)

[feral copper]

Uhm I guess I can ask that X is simply-connected
Is it really not true? I thought I read that somewhere a while ago...

It's asking about cancelling pairs of intersections with opposite signs, and I assume if X is simply-connected that's always feasible?

[empty grove]

Here comes the nerd jesus christ

jk lol

Idk if I would call that visualisation

[feral copper]

Yeah well, it won't embed in 3-space so that's about as good as it gets if you don't want this weird immersion (that I never understood really xD)

[empty grove]

Oh yeah that immersion

[feral copper]

Like this one

[empty grove]

I guess I would consider that visualisation because you can see the entire space from a third person pov

as opposed to the earlier thing that you suggested

I have never seen that immersion drawn though

[feral copper]

Try plotting an algebraic curve on both, and you'll see that the disk one is very good 🙂

[tiny ridge]

This is not even true in 2D

[empty grove]

hmm

[feral copper]

You mean for curves on surfaces?

[tiny ridge]

Yes

[feral copper]

There's no symmetric intersection form tho

But I guess mod 2 it is

[tiny ridge]

Why should that be important

[feral copper]

Oh and also, surfaces with genus are not 1-connected!