point-set-topology

[delicate hollow]

not all of X

just the ones that map over

[sleek thicket]

Map into what?

Everything in X maps into Y

[delicate hollow]

i guess thats true

the preimage is the domain then?

well it cant be

its the domain that gets used

im not making sense atm lol

i guess the preimage is the domain

[sleek thicket]

So if you have a subset S of Y

The preimage of S under f is defined to be all the sets x in X such that f(x) in S

[delicate hollow]

yes

so the preimage of a set T are all the sets which get mapped to T

[sleek thicket]

All elements

So x in A means what?

[delicate hollow]

x in A where A = f-1(O) is the set of all elements in X which get mapped to O

[sleek thicket]

Right, so what does this mean

It means x is mapped into O, right?

[delicate hollow]

yes

[sleek thicket]

So what do you know about O?

[delicate hollow]

O is the image of the graph of f

no

O is an open neighborhood around points in the image of f

[sleek thicket]

We know exactly three things about O

It's open, contains f(x), and is disjoint from B

[delicate hollow]

yes exactly

oh

damn

disjoint from B

so its inverse image is also disjoint

from B

[sleek thicket]

That doesn't really make sense

[delicate hollow]

wait

[sleek thicket]

B is a subset of Y, the inverse image is a subset of X

The relevant fact here is that O contains f(x), by definition

[delicate hollow]

yes O contains f(x)

so x is in f-1(O)

[sleek thicket]

Yup

So A, B are open

Contain x, y

[delicate hollow]

ok

[sleek thicket]

What else do we need to check?

[delicate hollow]

that they are subsets of U

A x B is a subsret of U

[sleek thicket]

Yup

[delicate hollow]

but thats where my confusion still is

for A atleast

i dont really see how Ax B is U

i understand for B

but not for X

[sleek thicket]

Well take a point in A×B and try to show that that point is in U

[delicate hollow]

i guess its all about the visualization still

A is chosen to be the inverse image of an open set containing a part of the graph disjoint from B

[sleek thicket]

No

it's the inverse image of an open set in Y

The graph is a subset of X×Y

[delicate hollow]

oh ok

is it ok to think of the graph like this?

(A1 U A2 U...) x (B1 U B2 U...)

[sleek thicket]

I don't know what A1, A2, etc are

[delicate hollow]

just sets that comprise of the graph

so the preimage and image right?

is that what graph is?

obviously we were given a definition

is graph the cartesian product of the preimage and the image of f, or am I overismplifying?

ok

so if following definition

graph f = {x,f(x)| x in X}

im sorry, but I just dont easily see how the inverse image of O satisfies picking (x,y) in U

specifically the x part.

[sleek thicket]

You are oversimplifying

[delicate hollow]

ok thank god

[sleek thicket]

You don't include points like (x, f(x'))

Which would be in the cartesian product

[delicate hollow]

oh

for clarification again

why does f-1(O) work but not f-1(B)

[sleek thicket]

Because f(x) is in O and f(x) is not in B

So we still have something left to do in our proof, namely the claim that A×B <= U. Try to prove this. Take a point q = (a, b) in A×B, see if you can show q is in U (ie, that q is not in the graph)

[delicate hollow]

ig thats the hard part for me

proving q isnt in the graph

i thought choosing B and A was why

because we choose A and B to be subsets of components of U, so AxB is a subset of U would be my answer

but im not sure if thats rigorious enough

[sleek thicket]

Yeah so here's something very important to keep in mind

If S is a subset of the product X×Y

And we take the projection of S onto X and the projection of S onto Y

Taking the product of those two again will in general be bigger

Like we say with S = the parabola. The product was R×R^+, which is much bigger

[delicate hollow]

ok

so

to show A x B is a subset of U I need to do what?

[sleek thicket]

Take any point in the first set and show it's an element of the second

[delicate hollow]

wat

you dont mean show A is an element of B

[sleek thicket]

that is what being a subset means

No

[delicate hollow]

you mean show A is a subset of X?

and B is a subset of Y

[sleek thicket]

By the first set I meant A×B

[delicate hollow]

yes

[sleek thicket]

And by the second I meant U

[delicate hollow]

yes

but dont we choose A and B based on that criteria?

[sleek thicket]

No

We chose B to be a neighborhood of y disjoint from some neighborhood O of f(x)

We then chose A to be f^-1(O)

We still need to actually check that A×B is contained in U

[delicate hollow]

so showing that (x,y) is not in the graph pretty much

for all x,y in A x B

[sleek thicket]

Yes, but let's not use x, y

x, y are already the coordinates of our point p

[delicate hollow]

oh i get it

p are just all the points that satisfy it

but we are just trying to show that for any point chosen in the A x B it is in U

well

honestly not entirely sure

because

every p is in U by definition

and with every x,y in p we find an A and B that are open and contian them

[sleek thicket]

No

We start with a p and build A×B

But now we need to check something for all other points q in A×B

Namely that q is in U

[delicate hollow]

so showing that q is also not in the graph of f

[sleek thicket]

Yes

[delicate hollow]

yeah i honestly dont know then

it makes sense visually

[sleek thicket]

Try to use the definitions

q = (a, b)

You want to prove a negation, so assume for contradiction that q was in the graph

[delicate hollow]

q=(a,b) and a inA, b in B

[sleek thicket]

What does the tell you about a, b?

[delicate hollow]

suppose q is in the graph

then (a,b) is in the graph

for all b in B, it is doesnt touch graph

because disjoint in hausdroff

not sure for a though damnit

[sleek thicket]

no

It doesn't make sense for b in B to touch the graph

[delicate hollow]

yeah

b cant touch the graph

[sleek thicket]

Because B is a subset of Y and the graph is a subset of X×Y

I'm not saying it's false that b touches the graph

I'm saying it's not meaningful

you seem to be mixing up the image and the graph

[delicate hollow]

image is just a componenet of the graph

ok

so

if q = (a,b) is in graph (f)

im trying to make a contradiction here

i thought it was relevent though

no it wasnt you are right

[sleek thicket]

So what does it mean that b is in B?

It doesn't tell us very much

we know very little about B

[delicate hollow]

oh

q in graph means

q is in {(x,f(x))|x in X}

[sleek thicket]

Yes

So b = f(a)

[delicate hollow]

but we choose b to be not in that

[sleek thicket]

We can use this with the fact that a is in A and b is in B

[sleek thicket]

What do you mean by that

[delicate hollow]

b is still in B right

[sleek thicket]

b isn't in that because b is an element of Y and that set is a subset of X×Y

Yes, b is in B

[delicate hollow]

so

b=f(a)

im not entirely sure why this leads to a contradiction

it just wouldnt be possible though

when I thought about the problem in my head visually I cant really make sense of what it would mean if q was on the graph

oh

so f(a)=b

=> a = f-1(b)

[sleek thicket]

no

f may not have an inverse

[delicate hollow]

We know a is in f-1(O) and b is in B

[sleek thicket]

Yes

So if a is in f^-1(O)

And b = f(a)...

[delicate hollow]

b is in f(f-1(O))

b is in O?

[sleek thicket]

Yup

f(f^-1(O)) might not be O

[delicate hollow]

now i have to show O != B

oh

[sleek thicket]

But it's contained in O

[delicate hollow]

ok

oh true

why do I know O != B

oh wait

lol

i choose it to be disjoint

[sleek thicket]

Yeah

[delicate hollow]

so f(f-1(O)) not in B

[sleek thicket]

But it's not enough to be not equal

They need to be disjoint

If they're not equal that just says there's some element in one that's not in the other

That element might not be b

We need that any element in one is not in the other

[delicate hollow]

isnt it simpl argument

[sleek thicket]

I mean it's true that they're disjoint here

[delicate hollow]

f(f-1(O)) subset of O and O and B are disjoint

[sleek thicket]

I'm just pointing out O ≠ B is not suffisent

[delicate hollow]

so f(f-1(O)) and B are disjoint

[sleek thicket]

Since that's what you had been saying above

[delicate hollow]

yeah I can see that

[sleek thicket]

Yup, that works :)

[delicate hollow]

ok

i think I understand the x part better too

or why our choice of A works

because

p=(x,y) needs to be in X x Y - Graph f

f(x),y have disjoint open neighrborhoods O,B

f(x) in O

by continuious

x is in f-1(O)

the part I wasnt understanding

was the association between x in (x,f(x))

and x in p=(x,y)

I just thought of them as unrelated symbols n stuff

@sleek thicket thank you teacher, what classes do you teach

[sleek thicket]

I'm not a teacher

I work as a TA for some computer science courses

[delicate hollow]

oh

undergraduate?

graduate ta?

phd?

[gritty widget]

sham is a tenured professor

they just call their teaching duties TAing

[honest narwhal]

Okay so

If you give me an immersed Lie subgroup of a Lie group

[sleek thicket]

lmao

[honest narwhal]

Can it be dense?

[sleek thicket]

I'm an undergrad

Yes dami

Take an irrational curve on the torus

[delicate hollow]

cap

[honest narwhal]

Is that map a group hom? Also I forgot the adjective complex

[sleek thicket]

Yes

[honest narwhal]

Complex feels like it makes life difficult, an embedded complex submanifold should have real codim 2 so its complement is connected

Which makes me think/hope that even immersed can't be that bad

[sleek thicket]

hmm

[honest narwhal]

Also if the image is normal

[sleek thicket]

Normal is in normal subgroup?

[honest narwhal]

Yeah

[sleek thicket]

Okay so the setup is

G is a complex manifold with a group structure

Multiplication analytic and whatever

You have an immersed subgroup H of the same form

I think maybe my example still works?

[honest narwhal]

Yeah. Codimension 1

[sleek thicket]

Just take products

[honest narwhal]

R isn't complex

Or oh no

[sleek thicket]

So define γ : R -> T^2 by γ(t) = (e^(πi t) , e^(απ i t))

Where α is irrational

And then take γ×γ

[honest narwhal]

Yeah this destroys my idea. The ambient problem is

[sleek thicket]

Oh hmm this won't be dense

[honest narwhal]

If G is a simply connected Lie group with solvable Lie algebra

[sleek thicket]

The closure will be the diagonal I think

[honest narwhal]

Then it's diffeomorphic to a vector space

[sleek thicket]

Huh

[honest narwhal]

So now I can find an ideal in the Lie algebra of codim 1

Which corresponds to a Lie subgroup which is possibly immersed. I'm trying to say oh okay closure is a codim 1 closed subgroup

And then I can do an induction

[sleek thicket]

I see the concern

Yeah I think the complex case might be nice enough for this to work but I'm not sure, sorry

Oh wait maybe I see how to modify my example

Take two different irrational angles

Idk, I would look here

((e^iπt, e^iπαt), (e^iπs, e^iπβs)) on T^2×T^2

I would expect the closure to have real dimension 3 actually (all points of the form ((z, w), (z, u))) nvm I'm dumb

Which is weird

[honest narwhal]

@gritty widget any ideas?

[sleek thicket]

My thing won't work I think

I have to take real and imaginary parts so it's not analytic

Is there anything nice we can say about immersions?

In the complex case

Like, are they magically nice than real maps or something?

[honest narwhal]

Yeah ima just say we can find a codim 1 closed normal subgroup

Details are an exercise to the reader

[sleek thicket]

this seems like a big detail to sweep under the rug

wait a second...

Huh, I thought you could always quotient out by a normal subgroup

And get a lie group

Guess that's not true

See the torus example

[gritty widget]

laughing i just learnt about cellular approximation

cw is so much better then smooth sorry

[sleek thicket]

lmao do you see why max got mad at me now

also tbf I did tell you about cellular approximation!

[gritty widget]

yes but i didn't get it at all then

[marsh forge]

i dont even remember getting mad about this

but now im mad that you apparently said it 😠

[rotund thicket]

What happens if I take the connected sum of two tori along their inner circles?

[marsh forge]

maybe I am wrong but afaik connect sum is usually only defined for a contractible loop

do you just mean identify the inner circles?

you get a “double torus” if so. you can visualize it by just thinking of two (hollow) donuts where the smaller donut fits perfectly as a ring inside the bigger donuts hole

[rotund thicket]

Hmmm, ok. I guess I don't really know the definition of a connected sum. My book is pretty hand wavy and pushed the definition to the end of the book

[marsh forge]

so usually connect sum is like

you take a nice filled in circle on the surface

cut out the filling

do this for both shapes

and then glue along the created border

[rotund thicket]

Ah I see, so my inner circle thing doesn't really make sense

[marsh forge]

yeah it makes sense to think about

just not to call it a connect sum

[rotund thicket]

because there is a well defined equivalence relation between the two circles

so that's why i was thinking about the quotient space of the union of the tori

why do you hollow out the circles you are connecting? how is the equivalence relation defined?

[marsh forge]

so as a topologist

my equivalence relations

are entirely visual

you can define it explicitly but i think visual approach is useful

anyway

connect sum is an operation used for closed manifolds

[rotund thicket]

welp

[marsh forge]

so you want to only allow connect sum in ways that give you a manifold out

when you put manifolds in

this requires the surgery i described

[rotund thicket]

i dont know anything about manifolds so ill just take your word

[marsh forge]

you can replace manifold with surface

if you know what the latter means

[rotund thicket]

not rigorously

[marsh forge]

i think it might be hard to understand connect sum rigorously without understanding surfaces generally

[rotund thicket]

hmm, i guess im being introduced to connect sum to build an intuition. rigorousness comes later

thanks, though

[warped flare]

the importance of connected sums is that every surface can be written as connected sums of tori or copies of the projective plane

so it makes sense to introduce it early when talking about surfaces

it also has simple descriptions in terms of the cut-and-paste polygons if you've seen those

[long hornet]

The proof I have seen of embedding a compact topological n-manifold proceeds by covering the manifold M by charts and then using partitions of unity to somehow combine these together to get an embedding (I don't quite remember the details). It was stated that the proof for noncompact manifolds is substantially harder, because/and it uses dimension theory.

[long hornet]

This was supposed to be a background for a question that I was going to ask, but I realized it was stupid, so.

[warped flare]

the proof in the noncompact case is similar

you cover your manifold with nice "partition of unity" charts, and then proceed to immerse them together

the point is that almost every map from an n dimensional manifold to a 2n+1 (or higher) is an immersion

(and this follows from the fact that almost every linear map from R^n to R^{2n+1} is an immersion)

then you massage your immersion to be an embedding

going down to 2n sucks though

[gritty widget]

@marsh forge is this advanced homotopy theory still visual to you?

Doing diff geom, I’m fine with picture a vector bund as a spikey ball and then making sure I don’t use any of the intuition that is specific to the low dimension

[obtuse meteor]

Oh god homology homework

,,,it's time to get algebraic

[marsh forge]

@gritty widget the parts that can be visual are

the parts that arent never could be to begin with

i like to think of algebra as first a way to formalize visual intuitiin

but secondly and more importantly

a way to extend it to contexts where visualization fails

[wide kayak]

"algebra without geometry is blind, geometry without algebra is dumb" - somebody really smart

[ivory dragon]

im just dumb

[fading vale]

"""algebra without geometry is blind, geometry without algebra is dumb" - somebody real smart" - me" -slimvesus

[tough imp]

I do algebra without geometry

Can confirm, feels blind

[fading vale]

god i love that pasta

[tough imp]

https://tenor.com/view/spongebob-squarepants-spongebob-nick-nickelodeon-patrick-gif-17195320

[sleek thicket]

I'm both

wait no I'm not a teenager anymore

[fading vale]

lol

nerd

[tough imp]

Lmao

I did that when I was 21

I was like

[sleek thicket]

Old

[tough imp]

“Two teens found dead”

Then remembered neither me or the other guy was a teen

[sleek thicket]

Lmfao

[fading vale]

shamrocks bones are creaking

[sleek thicket]

I unteened during quarantine so it doesn't count

Time has no passed

Since March 2020

[tough imp]

What’s March 2020

[fading vale]

a homophobic dogwhistle

dont listen to shamrock

[sleek thicket]

what

[tough imp]

What’s homophobic dogwhistle

[sleek thicket]

@gritty widget might know

[tough imp]

Is it a dogwhistle only gay dogs can hear?

[fading vale]

yes

[sleek thicket]

DO YOU HAVE SOMETHING AGAINST DOGSSSSS

see my status

[tough imp]

The channel in your status doesn’t exist

[fading vale]

see

[sleek thicket]

Oh whoops

[fading vale]

gay dogs only

[tough imp]

I only know #point-set-topology

[sleek thicket]

Fixed

[tough imp]

Oh I see

That belongs in #prealg-and-algebra

[sleek thicket]

#❓how-to-get-help

[tough imp]

#❓how-to-get-help

[fading vale]

[tough imp]

I have a topology question

[sleek thicket]

#hmmm-to-get-help

[fading vale]

anyway i spent like 4 hours playing system shock 2 so now im going to read dieck

[tough imp]

Why is donut = mug

Clearly donut is not mug

[sleek thicket]

This is not a research level question

[fading vale]

its because the roman god of donuts is actually the greek god of mugs

[sleek thicket]

Please take it to MSE

[tough imp]

Oh

[sleek thicket]

furthermore I am closing it as a duplicate

[tough imp]

I have voted to close

[fading vale]

MO overflow 10 years ago: explain the axioms of a topology
MO today:

[sleek thicket]

Yeah lmao old MO posts are wild

[tough imp]

Math Overflow overflow

I haven’t seen old MO posts

I mean I’ve seen old Gindi MO posts

[fading vale]

is it really?

[tough imp]

Did ppl ask questions like that on MO in the past?

[fading vale]

yes chmonkey

[sleek thicket]

That's actually what he got the fields medal for

[tough imp]

I will ask the first MO question to not be closed

[sleek thicket]

Showing the set of questions MO won't close is nonempty

Lol

[tough imp]

The attention it got

Tfw a MO question got enough attention for ultra to think it got too much

[fading vale]

MO is just middle school

[tough imp]

-_-

[sleek thicket]

Lmfao

[tough imp]

Was Peter a part of [redacted]

That would be funny, I think

Yes

[sleek thicket]

=

[tough imp]

I’ll do it again

No sham you did an equals sign

= vs _____

[sleek thicket]

=

[tight agate]

=

[sleek thicket]

Hello brofib, do you think scholze is overrated?

Possibly worthless?

That's ultra's topic of the night

[tough imp]

Yes you did, you just deleted it because of your shame

Me and Sham bear witness

2 vs 1

[sleek thicket]

I think it's plausible that he's on discord tbh

Like

He literally named something anime

Of course he's extremely online

Wtf

can someone please homotope away the burned bits

(this is topology because they're homeomorphic to donuts)

[tough imp]

Topology is the study of donuts?

[sleek thicket]

I tried

[tough imp]

Should’ve just ate the burnt bit

[gritty widget]

homophobic dog whistle?

[sleek thicket]

What's homophobic, dogwhistle?

Nothing much, what's homophobic with you?

[gritty widget]

$\textbf{Definition:}$ A map $F: X \to X$ is called a $\textsl{homophobism}$ if $F(x) \neq x$ for all $x$.

[gentle osprey]

8da

[gritty widget]

$\textbf{Theorem}$ (Brouwer's homophobism theorem): There exist no continuous homophobisms of the closed ball.

[gentle osprey]

8da

[gritty widget]

heteromorphisms 🤮

[long hornet]

Isn't immersion only a thing for differentiable manifolds, though?
Also, the proof I learnt actually uses a copy of R^N (for some N dependent on the manifold) for each "partition of unity chart," so I realized the exact same method won't work because we will end up with an embedding in R^w.

[median glade]

yeah that works for compact manifolds only

[warped flare]

well yeah the process I described is for differentiable manifolds, completely missed you said topological

and you don't send each one to a different R in this process, you send each of them to the same R^{2n+1}

while being careful about it

[cloud owl]

d(x, y) = A, d(x, z) = B, d(y, z) = C

and d' = A', B', C'

A' = A/1+A

ok this may not be the smartest approach

uhhh A' = 1 - 1/(1+A)?

B' + C' = 2 - 1/(1+B) - 1/(1+C)

so if A' < B' + C' then 1 - 1/(1+A) < 2 - 1/(1+B) - 1/(1+C)

ie. 1/(1+B) + 1/(1+C) - 1/(1+A) < 1, for A, B, C > 0

[median glade]

Ok, I think I have a simple way of doing this

[cloud owl]

or, well, probs should be using >=

[median glade]

you have d(x, z) / (1 + d(x, z)) right

[cloud owl]

oh wait i think i have it

[median glade]

first use the triangle inequality on top

then, try and get the denominators to the right spot

for that, note that d(x, y) <= d(x, z)

[cloud owl]

wait no

[median glade]

and d(y, z) <= d(x, z)

so if I make the denominator smaller, I get something larger

[median glade]

I'm actually unsure if this is true

now that I think about it

[cloud owl]

you can probs wlog it

[median glade]

the triangle inequality gives the other direction

[cloud owl]

that's the opposite of what you want

[median glade]

my thinking is this, if you can show that d(x, y) >= d(x, z)

err, less

d(x, y) <= d(x, z)

well you can get here, right

the indirect path is greater

oh wait

oh yeah no

🤔

[cloud owl]

P: 1/(1+B) + 1/(1+C) < 1 + 1/(1+A)

A < B + C
1/A > 1/(B+C)
(1 + A)/A > (1 + B + C)/(B + C)
A/(1 + A) < (B + C)/(1 + B + C)

B/(1+B) + C/(1+C) = (B + C + 2BC)/(1+B)(1+C) = (B + C)/(1 + B + C + BC) + 2BC/(1 + B + C + BC) which is quite close but no cigar

P: 1/(1+B) + 1/(1+C) < 1 + 1/(1+A), given that A < B + C

1/(1+B) + 1/(1+C) = (2 + B + C)/(1 + B + C + BC) < (2 + B + C)/(1 + B + C)

1 + 1/(1+A) > 1 + 1/(1 + B + C) = (2 + B + C)/(1 + B + C)

i think i've done it?

fairly short too

[median glade]

don't you want to show that:

A / (1 + A) <= B / (1 + B) + C / (1 + C)

maybe I'm missing something in your argument

[sleek thicket]

Isn't it 1-1/(1+A) = A/(1+A)?

Not plus

Ah you negate both sides

And add 2

I think this works

[cloud owl]

A' = A/(1+A) = 1 - 1/(1+A), and similarly for B', C'
then P: A' < B' + C'
ie. P: 1 - 1/(1+A) < (1 - 1/(1+B)) + (1 - 1/(1+C))
ie. P: 1/(1+C) + 1/(1+B) < 1/(1+A) + 1

and then as before:

1/(1+B) + 1/(1+C) = (2 + B + C)/(1 + B + C + BC) < (2 + B + C)/(1 + B + C)

1 + 1/(1+A) > 1 + 1/(1 + B + C) = (2 + B + C)/(1 + B + C)

hurts the eyes but i think it works

brb learning texit

[sleek thicket]

I remember just multiplying everything out the last time someone asked about this

Couldn't find the post though

This seems like it works

[median glade]

ok yeah this looks to work

[sleek thicket]

Ah wait, I think I have a nice proof

$\frac{x}{1 + x} = \frac{1}{1+x^{-1}}$ is increasing in $x$

[gentle osprey]

s_n -> Shamrock weakly

[sleek thicket]

So $\frac{A}{1 + A} \leq \frac{B+C}{1+B+C} = \frac{B}{1+B+C} + \frac{C}{1+B+C} \leq \frac{B}{1+B} + \frac{C}{1+C}$

[gentle osprey]

s_n -> Shamrock weakly

[sleek thicket]

I think this works?

Yeah so look at the expression for x/(1+x) I posted above

It's a composition of f(y) = 1/y and f(x) = 1+1/y, both of which are decreasing functions

I think you need to be careful if stuff is zero

But that's an easy case anyway

Yup, exactly

[gentle osprey]

mirzathenarcissist

[cloud owl]

oh, wow, ok

that first thing with A/(1+A) =< (B+C)/(1+B+C) would make my first attempt work

[sleek thicket]

Try to manipulate x/(1+x)

[cloud owl]

nice

[sleek thicket]

The way kai did it is good but it doesn't give us enough info

Or actually wait

[cloud owl]

?

i mean i swear it just works

[sleek thicket]

1-1/(1+A) is increasing in A for the same reason

Sorry, I misspoke

I agree that your proof works

But I also think it's a little ugly

[cloud owl]

cannot dispute

[sleek thicket]

nope

lol

So kai's rewriting works too

Oh no, I'm not differentiating any of this

That would be ass

Just saying decreasing ° decreasing = increasing

fair, sorry

I think this also provides some intuition

like

we have a function F(x) = x/(1+x)

Which is strictly increasing

[cloud owl]

yeah, that's the best way

[sleek thicket]

We're sort of renormalizing the distance

Or something

Idk

It tells you that if x, y are closer than p, q in the original metric the same is true in the new metric

[cloud owl]

profile pic ig

[sleek thicket]

It was drawn to be a goblin wizard

But I get the shaman vibes

lol

@gritty widget

hahaha

[cloud owl]

um

who's lime soup

[sleek thicket]

It was drawn by someone who thought my twin fantasy pfp was "some kind of goblin wizard thing"

(lime soup)

[cloud owl]

ah

[sleek thicket]

So it's supposed to look like twin fantasy

It's a great album

[fading vale]

twin fantasy very good

both mirror to mirror and face to face

[flint cove]

Is there some theorem like „A morphism of short exact sequences of cochain complexes induces a morphism between the long exact sequences in cohomology“?

In other words, is the construction of the long exact sequence a functor from the category of short exact sequences of cochains?

Sorry if this is an uninformed question, my cohomology knowledge is quite lacking

[trail ibex]

Yes there is

Given a morphism of complex you get maps between the terms of the long exact sequences by applying cohomology. Most squares trivially commute by functoriality of the cohomology functor, the problematic ones are those with the coboundary map. For those you can go back to the explicit construction of the coboundary map via the snake lemma to check that they indeed commute

[flint cove]

Thanks, this does not sound too complicated. I should try and prove this myself.

[trail ibex]

Also maybe you'll be interested in the notion of delta-functor

[sleek thicket]

This is essentially the statement that the connecting homomorphism is natural

(I'm not saying anything new, it just might be useful in the future if you see someone say eg "by naturality of the connecting homomorphism")

[zealous bison]

howdy folks -- I'm trying to understand the difference between homology and cohomology from a physicist's perspective. In particular I'm trying to gain some intuition by the way the chern number is measured vs. computing simplicial homology

it looks like when you compute simplicial homology, you're counting generalized holes in this structure.

but when you compute the chern number of some map... to specialize to physics, the wavefunction as a function of k, you're looking at the way that patches in the domain cover the codomain.

like if I integrate over k in the brilluoin zone, how many times do I envelop the topological charge

is there something to this notion of comparing computing invariants over some manifold to computing invariants over some map between manifolds?

somehow something something dualization and looking at \delta_i(X)->G instead of \delta_i(X) and something something dualizing the boundary operator requiring you to find solutions to the boundary operator which looks like integration... something something

[fading vale]

i dont know two much about chern classes and bundle stuff but this sounds correct?

[zealous bison]

it feels like there's something I need to understand about studying \delta_i(X) vs studying the map.

[fading vale]

im not totally sure what you mean by that tbh

or like are you asking about this in terms of chern stuff or just normal cohomology

singular coho vs singular homology

[zealous bison]

let's say singular coho vs singular homology

idk I just have a smattering of different ideas from looking at solid state physics

[fading vale]

you want to understand the relationship between them for an orientable manifold?

[zealous bison]

ehhh I'm trying to find an intuitive understanding of the difference between studying homology vs cohomology

i understand one forms a group vs. a ring and cup product and all this...

but I still don't like... get what the difference in computing these things looks like

or what information I get

[fading vale]

in the context of manifolds you have poincare duality

ok well orientable n dim closed manifolds

H^k(X) iso H_n-k(X)

so they are quite strongly related

and you generally have hte universal coeff thm and stuff

i guess if by "difference in computing" you want like visual physics stuff i cant really help you there but the primary intuition i have for it is that cohomology is simply a stronger invariant because the cup product gives it a graded structure

and in terms of computing cohomology and homology are super closely related because of universal coeff thm

[gentle osprey]

Moth

[fading vale]

its a very algebraic relationship which makes sense since it comes from dualizing :p

sorry i cant be more helpful though 😔

[uncut surge]

(if you're doing cohomology with coefficients in a field, this basically just means that H^n is the dual of H_n iirc)

[fading vale]

yeah the Ext is trivial

[tight agate]

there is this and also for some reason cohomology seems to be the thing which generalizes well

[fading vale]

i wonder if those are related

cohomology gets the nice product because we have a very natural map from X -> X x X but the only ones from X x X -> X in the general case are projections which dont give very much information on the homology

[tight agate]

oh I'm not talking about the product structure generalizing

but yeah that's the reason why you have products on coho

the maps seem to be going the wrong way for homology

[fading vale]

ya

[tight agate]

however, in some nice cases homology and cohomology fit together to give you a single thing

and you can define products on homology

[fading vale]

hmm products on homology

i should learn about pontryagrin stuff

[tight agate]

look up tate cohomology if you're interested

[fading vale]

👀

[tight agate]

essentially, when G is a finite group, you get a norm map H_0(BG, Z) --> H^0(BG, Z)

[fading vale]

i havent seen group cohomology sadly

is there some nice chain complex associated with groups

[tight agate]

well what I wrote down is just singular cohomology

[fading vale]

oh

wait blind

classifying space

ok

[tight agate]

bar resolution

for any group G

there's a projective resolution of Z (with trivial G action)

$\mathbb{Z} \leftarrow \mathbb{Z}[G] \leftarrow \mathbb{Z}[G^2] \leftarrow ...$

[gentle osprey]

Brofibration

[tight agate]

and then for any G-module A, H^n(G, A) is defined to be the cohomology of the complex you obtain by applying Hom(-, A) to this resolution

[fading vale]

hrmmm

[tight agate]

this resolution is secretly the simplicial chain complex of the classifying space BG

[fading vale]

i think this makes sense?

whats G^n here?

[tight agate]

G x G x G ... x G

[fading vale]

oh

lol

okay

whats the map between the spaces?

projection of some kind?

[tight agate]

you might be able to guess it yourself

it's not a projection

but very similar to the differential you see in usual cohomology

[fading vale]

hm

[tight agate]

think ugly alternating sum

[fading vale]

im guessing the first map sends it to the coeff?

like sum a_i[g_i] |-> sum a_i

[tight agate]

yup

[fading vale]

and then on the second its...sum a_i(g_i1, g_i2) |-> sum a_i(g_i1 - g_i2)?

or something?

wait thonk

[tight agate]

not quite

[fading vale]

right we need d^2 = 0

[tight agate]

ignore the a_i for now

if I give you (g_0, g_1)

what's a nice way to get an elt of Z[G]

which looks like a differential

(use the group operation)

oh wait I think you got it right

[fading vale]

[tight agate]

no wait you did not

gimme a sec I need to write it down

[fading vale]

is there some nice relationship between Z[G^2] and Z[G][G]

tfw group rings

[gritty widget]

Hi guys

[tight agate]

it should be (g_1, .., g_n) mapsto g_1(g_2, ..., g_n) + \sum (-1)^i (g_1, ..., g_{i}g_{i+1}, ..., g_n) + (-1)^n(g_1, ..., g_{n-1})

up to sign

[fading vale]

dear god

[marsh forge]

i think its just deletion

according to wiki

unless im misreading

[gritty widget]

You guys are talking about tensors?

[fading vale]

[tight agate]

okay a 0-cochain with values in A sould take a to (g mapsto ga-a)

[fading vale]

we are talking about group cohomology

[gritty widget]

Cool

[fading vale]

hello tterra

:tterralurking: when

[gritty widget]

hi, moth

[fading vale]

[gritty widget]

You guys like Riemann geometry?

yeah

basically only me

on this s erver

[fading vale]

xd

shamrock moment

im fairly certain riemannian geo gave sham a split personality

[gritty widget]

@gritty widget want talk about tensors?

the greatest crime uoft has committed is not offering a second riemannian geometry course

[gritty widget]

i'm in a lecture right now so i can't, sorry

[fading vale]

is that even greater than the intl tuition

[gritty widget]

No problem

[tight agate]

oh yeah nvm it's just deletion

rip Im sorry moth

[gritty widget]

I love tensors

symplectic geometry is just

oh there's this interesting thing you want to learn?
better learn all of this homology and cohomology first

[fading vale]

oh like alternating sum of the deletions

[tight agate]

yes

[fading vale]

ok

that makes sense i guess

its cringe but it makes sense

[gritty widget]

I use it all times when I calcule something in Modern Physics

[fading vale]

i think we should do some kind of revolution against alternating sums of deletions

i am tired of them and their indices

[gritty widget]

Interesting

[marsh forge]

they are nice

[tight agate]

but ones you dualize and start looking at cochains

you need to multiply

[fading vale]

hrm

ok that was interesting

[tight agate]

I need to read the first chapter from the book on simping again

[fading vale]

the book on simping

[tight agate]

goerss and jardine

[fading vale]

[tight agate]

I was confusing face maps and degeneracies

I think

[marsh forge]

the map you describe is the chain differential for the simplicial bar com-lex

complex

[tight agate]

yes

oh wait both of them work

[marsh forge]

join me in mod2 land

[tight agate]

they give isomorphic things

[marsh forge]

up to chain htpy?

or on the nose

[tight agate]

probably this, lemme try writing it out

yeah it might just be an iso on cohomology

both of them are projective resolutions

so both work

there should be a way to write down an iso of chain complexes too

by using a change of basis

something like (g_1, g_2, ..., g_n) ---> ( g_1, g_1g_2, ...., g_1g_2g_3..g_n)

[sleek thicket]

This is true

[fading vale]

im kind of confused as to what quotient is actually being described here

is it supposed to be the 4 point space {1, -1, a, b} with varnothing, {a}, {b}, {a, b}, X open

i think {a, b, 1} and {a, b, -1} should also be open, no?

oh wait they are lol

[sleek thicket]

lol

This is like the quasicircle right

Or pseudocircle

[fading vale]

yeah

i was so for a second because i was like "u need 1 and -1... why arent they in there"

[little hemlock]

does anyone have any idea what this hint is talking about?

[gritty widget]

reduce it to the case where you can check it on coordinate vector fields

there has to be a better way lol

[dusk heron]

This might be a long shot (and tell me if this is off-topic for this channel), but does anybody know if there is a Swedish word for "orbifold"? The Swedish word for "manifold" is "mångfald", and the word for "orbit" is "bana", so I am thinking it should be "banfald".

[gritty widget]

@little hemlock if you know cartan's formula you might be able to use that, instead of doing messy computation

[little hemlock]

im not sure if I know cartan's formula

[gritty widget]

$$\mathcal{L}_X \omega = di_X\omega + i_Xd\omega$$ for $\omega$ a differential form and $X$ a vector field

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

(L is lie derivative and i_X is contraction)

[little hemlock]

ah, nope

[shut moat]

prove it then use it

[little hemlock]

i could get pretty far without using the hint, but I'm trying to understand why you can do this. Is it basically just because at each p, the k form is linear in tangent vectors?

[little hemlock]

idek what a lie derivative is

[gritty widget]

u pullback the form along the flow and differentiate

generally, anything that eats in vector fields and spits out smooth functions, which is a homomorphism of modules over the ring of smooth functions, actually only depends on the arguments pointwise

examples being differential forms

if $T\colon\mathfrak{X}(M)^{k+1} \to C^\infty(M)$ is the right-hand side of the equation you want to prove, then you'd want to check that for all vector fields $X_0,\dots,X_k$ and smooth functions $f_0,\dots,f_k$, you have $$T(f_0X_0,\dots,f_kX_k) = f_0\cdots f_k \cdot T(X_0,\dots,X_k)$$

(\mathfrak X (M) means the set of smooth vector fields on M)

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

for the sake of preserving your sanity you can probably just show that it's multilinear (in the module homomorphism sense) in the first argument, and then exploit antisymmetry of some kind

[gritty widget]

let me find a reference for this fact

[dusk heron]

John Lee's book

[gritty widget]

yeah lmao

i was gonna go for that

[gritty widget]

multilinear over C^\infty(M) just means this

as well as respects addition

woops

[little hemlock]

right, okay, that kind of makes sense

[gritty widget]

i kind of butchered the explanation

but it's all in lee

[little hemlock]

Sorry for the handholding im requesting here, but I am starting with a nowhere vanishing $m$-form $\omega$. In local coordinates $u^1, \dots, u^m$, we may write $\omega = f du^1 \wedge \cdots \wedge du^m$. Now, if $v^1 ,\dots, v^m$ is another coordinate system, then we have that $$\omega(\frac{\partial}{\partial v^1}, \dots, \frac{\partial}{\partial v^m}) = f\det\left(\frac{\partial u^i}{\partial v^j}\right).$$

[gentle osprey]

kxrider

[little hemlock]

ok i thought i understood what nowhere vanishing is supposed to mean, but now i don't think i do

[gritty widget]

nowhere vanishing means at every point p, the top-degree alternating tensor on T_pM obtained by evaluating the form at p is non-zero (i.e., it sends a basis of T_pM to a non-zero number)

in particular

a non-zero top-degree alternating tensor on a vector space determines an orientation

and then asking that a differential form on a manifold do this is the same as patching the orientations together smoothly :petthecat:

[trail ibex]

I'm a noob in diff geo so maybe my question is trivial. Say I have a map T^m -> T^n between higher dimensional tori with coordinates given by Laurent polynomials. Is there an explicit condition on the coefficients so that the map preserves the Haar measure ?

[onyx crow]

Here K^(q) is the q-skeleton of K. In (b), is it sufficient to show that |K^(2)| is path connected? Cuz \iota_2 is a homeomorphism to its image? I don't see why simplicial approximation is needed.

Wait nvm. It seems \iota_2 being a homeomorphism to its image is irrelevant.

[obtuse meteor]

geometry gtfo. Proof by "long exact sequence" is my new best friend

[fading vale]

literally so true

[sleek thicket]

This is so based

[sleek thicket]

This doesn't sound trivial. I feel like you can determine whether it preserves the Haar measure by checking that it pulls back the riemannian volume form to itself?

That form can be written down explicitly and you can try to compute the pullback by an arbitrary Laurent polynomial

[obtuse meteor]

hmmm

is implicitly choosing isos

of maps

fine

I believe it's fine

like

if I'm talking about the induced map

H_1(S^1 x {1}) -> H_1(T)

I can just say we can choose isomorphisms to Z and Z^2

so that this is k -> (k, 0)

and then be on my way

(there's another problem where we calculate the induced map formally)

[tough imp]

Idk, this sounds possibly not trivial

Like... it probably works out and maybe you can do some weird naturality argument

But a prior I don’t see why any arbitrary map would be able to be turned into a map of the form k -> (k,0) so you’d need to examine the original map a bit

[fading vale]

okay so idk if you guys have seen that computation of pi_1(S^1) with van kampen for grpoids

but its like this:

so anyway im doing this exercise that says to show that if X is path connected and Y = X x I/X x bdry(I) pi_1(Y) = Z

and we can cover Y by two cones of X x [0, 0.5] and X x [0.5, 1]

so we end up with another case where we are taking van kampen of grpoids with 2 contractible spaces and an intersection equal to X U {*} for X path connected

is there a way we can like. generalize this argument somehow? or something broad we can say about these kinds of situations?

[obtuse meteor]

well it's not arbitrary. It's the induced map of the inclusion

[tough imp]

Right, but that’s why you’d need to analyze it a bit to make sure it works

You’d need some way to verify that with the iso morphisms to Z and Z^2 it ends up that way

I mean you could just assume it does but ¯_(ツ)_/¯

[obtuse meteor]

"it is clear"

[fading vale]

just write "so true" in parentheses

better yet draw the guy

[obtuse meteor]

in that problem I'll probably end up writing something like <a + im(partial_2)> -> <a + im(partial_2), b + im(partial_2)> is the induced map on homology in the problem on that

and then say something like

"if we choose the canonical isomorphism from a free abelian group to ZZ^n we see that this is equivalently a map Z -> Z^2 of the following form"

[obtuse meteor]

this would be so cringe christ

[shut moat]

sanity check- To show that the one point compactification of R is homeomorphic to S^1, I can just embed R into S^1 and invoke uniqueness up to homeomorphism, right? (Pls don't give away too much, this is hw)

[obtuse meteor]

this sounds good yeah

if you've proved 1 point compactifications are unique up to homeomorphism

[shut moat]

great, ty!

[urban yarrow]

tom Dieck omg

[fading vale]

lmao yeah

yeah.

[sleek thicket]

Faye I thought about your torus thing and decided it's almost but not quite formal

You're welcome

ah wait no I think I've decided it is formal, nice

@obtuse meteor do you know the splitting lemma?

[obtuse meteor]

I don't know it by name

[sleek thicket]

It's very very handy!

Especially in the homological stuff you're doing right now

Say you have a short exact sequence 0 -> A -> B -> C -> 0

(note: sequence of abelian groups or modules, this doesn't work for nonabelian groups)

The following are then equivalent:

1. The map A -> B has a left inverse
2. The map B -> C has a right inverse
3. There's an iso B -> A (+) C turning the map A -> B into the inclusion a |-> (a, 0) and the map B -> C into the projection (a, c) |-> c

So if the injection/surjection "split" then you actually have a direct sum

It's a really great exercise

(and it's up there in usefulness with like the 1st isomorphism theorem)

This is relevant to your problem because the topological map S^1 -> T has a left inverse, so the map on homology does to, so by the splitting lemma H_1(S^1) -> H_1(T) like an inclusion Z -> Z (+) M for some M = coker (H_1(S^1) -> H_1(T)). Since H_1(T) = Z (+) M is free abelian of rank 2 you can show necessarily M = Z

[obtuse meteor]

Oh that one

yes I know it

this is clever

[tough imp]

Idk if you know this but

Brendan is kinda smart

That’s why we have an emote called braindan in Chmonkey math which NO ONE CAN SHIW BECAUSE NO ONE HAS NITRO

[sleek thicket]

Doing algebraic topology while being much better at algebra than topology has taught me that you can grind absurd info out of things without actually thinking about the maps too hard

[marsh forge]

this is in my opinion the entire point!

topology is hard

[sleek thicket]

.pin

[sleek thicket]

wtf I swear we used to have more pins in here

Like WHY IS ONE DONUT EQUALS ONE COFFEE CUP or something

[marsh forge]

thats there

im looking at it

[cloud owl]

can confirm

[sleek thicket]

Wtf

[marsh forge]

sad

[tough imp]

Don’t tread on me, lol

[tepid tulip]

Would anyone here happen to understand why the Laplace-beltrami operator of a function is a bilinear form on T_pM (X) T_pM? I have been having a lot of trouble understanding this, especially since the normal Laplace operator in R^n is simply some value in R.

[fading vale]

.pin

[sleek thicket]

say we have a continuous map $j : X \to Y$. is the statement that $j$ is a cofibration the same as saying that

$\begin{tikzcd} X \arrow[r, "i_0"] \arrow[d, "j"] & X \times I \arrow[d, "j \times \mathrm{id}"] \ Y \arrow[r, "i_0"] & Y \times I \end{tikzcd}$

is a pushout square?

[gentle osprey]

Sham aced their analysis final

[sleek thicket]

where i0(x) = (x,0)

ah no i see

we don't assert uniqueness

[sleek thicket]

similarly this isn't quite a cokernel type deal

[desert bloom]

Am I hallucinating, or does this mean that N hat has the discrete topology

[bleak helm]

You can't think of an infinite set that doesn't contain infinity?

[desert bloom]

I can

probably

wait I'm confused

[bleak helm]

Well, then that set is not closed and its complement is not open

So it's not the discrete topology

[desert bloom]

oh ok

[ivory dragon]

"infinite set" and "set that contains ∞" are different things

for example, {1, ∞} contains ∞ but is not infinite

meanwhile, ℕ is infinite but does not contain ∞

of course, you can have both at once

{1, 3, 5, 7, 9...} U {∞} for example

[desert bloom]

If I define this infinite set without infinity U, then I define V= (N\U) union {infty}, then surely, Nhat without V is open right?

[ivory dragon]

i.e. the set of odd naturals and infinity

[desert bloom]

but isn't Nhat without V just U?

so U open?

[ivory dragon]

uh

i think U in your case is just ℕ

if i understand you correctly

which is indeed open, as the complement of the closed set {∞}

[bleak helm]

An infinite set without infinity is not closed according to the defn. This is not the same as being open necessarily.

[wanton marsh]

if the topology were discrete then every subset of Nhat would be closed

[desert bloom]

but V is definitely closed right?

and U is just any infinite set tbh

[ivory dragon]

im not totally sure what U is though

oh

[desert bloom]

{2,4,6,8, 10 ... }

anything

But then I just proved that literally anything can be in the topology

[ivory dragon]

in any case, V is closed since it contains ∞

[desert bloom]

it doesn't though?

[desert bloom]

oh wait

[bleak helm]

Oh, you are right. Any set that doesn't contain infinity is open since its complement contains it and is closed. But those arent the only open sets.

[desert bloom]

yeah, I defined it wrong I think

but then, for any finite set F, If I define Nhat without F, then it is also in the topology

but then, that means that any set with infinity is also in the topology

since I could just arbitrarily union them together

so is that not the discrete topology?

[bleak helm]

Why is any set with infinity open? Not true.

Any set without infinity is open.

[desert bloom]

because N-hat / F has infinities in them

[bleak helm]

No

[desert bloom]

If F itself doesn't contain infinity

[bleak helm]

Yes

[desert bloom]

so F is closed

[bleak helm]

No wtf

[desert bloom]

but complement of close sets is open

its an or statement

it can either be a finite set, or contain infinity or both

so a finite set without infinity is definitely closed

[bleak helm]

Oh yeah said F is finite, I apologize.

[desert bloom]

then I've basically just found a case for every set to be in the topology

which is strange

[wanton marsh]

no

[bleak helm]

No you have

haven't

[wanton marsh]

N is not closed in Nhat

because it is not finite

and it doesn't contain infinity

[bleak helm]

The open sets are the cofinite sets and the sets that don't contain infinity. That's what you have shown.

[wanton marsh]

{infinity} is not open in Nhat

[desert bloom]

I'm gonna think about it over breakfast and come back

I think I just have a bad case of dum

[desert bloom]

So let me try to just start fresh

so a set is closed if it is finite, contains {infty} or both

[ivory dragon]

right

[desert bloom]

then the set of cofinite topologies with {infty}, any set without infinity, or cofinite without infinity is open?

[ivory dragon]

a set is open if it either is cofinite or does not contain infinity (or both)

oops typo

so the non-open sets are those that contain infinity and are not cofinite

[desert bloom]

so the power set of the naturals is in the open set?

just the naturals, without the hat

[ivory dragon]

uh, ℕ is open (since it does not contain infinity) but im not sure why you said power set

oh you mean

any element of the power set of ℕ is open

[desert bloom]

yeah

whooops

[ivory dragon]

yes, those are some of the open sets.

[desert bloom]

but then

if I'm reading it right, I can union anything in the topology and get another open set?

then if any elements of the power set is in the open set, and cofinite sets with infinity are in there, then the union will be any elements of the power set with an infinity?

I think I need to think harder probably

[ivory dragon]

the union of a cofinite set with another set is cofinite

[desert bloom]

Oh

right

[ivory dragon]

so yes, unions of open sets will be open

(you can approach this more formally by arguing by cases)

a set is open if it either is cofinite or does not contain infinity (or both)
so consider what happens in each case:

• the union of two sets without infinity
• the union of a set without infinity with a cofinite set
• the union of two cofinite sets

now replace "two" with "infinitely many"

the argument should be the same

[desert bloom]

oh right

thanks so much

sometimes my stupidity frightens me

[gritty widget]

The sum of two nilpotent ideals in a lie algebra is nilpotent

this makes sense because for n large enough (I+J)^n will only contain elements in I^k J^k'

but I don't see how to actually show this inclusion

[cold vine]

I have freely null-homotopic map f:(X,x)->(Y,y) and now I should have that f*:pi_n(X)->pi_n(Y) is trivial. How should I go about this?

[gritty widget]

Do you know how f* is defined

I think writing that out explicitly should be a good hint

[cold vine]

Is it this definition?

[gentle osprey]

Bany, um futuro phideas

[gritty widget]

Hmmm

Interesting