point-set-topology

[delicate hollow]

ok

question

if I want to construct smallest open ball around a set

[barren sleet]

that's not well defined

in general

[delicate hollow]

would it be B = {x| d(x,A)<M}

[barren sleet]

there may not be a smallest open set

[delicate hollow]

if its bounded

there is

[barren sleet]

no

[delicate hollow]

?

[barren sleet]

what's the smallest open set around [0,1]

[delicate hollow]

ok

ive been beat

damnit

so I can just say that there is a ball

around a set

i dont really need it to be smallest );

[barren sleet]

yes

I see where you are going with that

[delicate hollow]

i glanced at the stack overflow answer he provided qq

i feel bad

[barren sleet]

just use def of boubdef

pick SOME ball that contains your set

doesn't need to be smallest

[delicate hollow]

ok

[barren sleet]

just needs to be bounded itself

[delicate hollow]

yeah the se gave entire answer

so like

if B in C where B is the bounded set and C is the open ball

then X/C is in X/B

[barren sleet]

indeed

by the way

[delicate hollow]

and the points in X/C are connected since X/C is connected

[barren sleet]

the problem never requires that the complement only has one component

it requires that it has only one unboundef connected component

[delicate hollow]

yea

all points in X/C make one unbounded connected component

[barren sleet]

this should not help. in general don't think of the points themselves as being connected. that is more like "path connected" which is stronger than connected

[barren sleet]

that set may not be a component of X\B

[delicate hollow]

wait

did I say thaht?

i didnt mean to say that

[barren sleet]

ok

[delicate hollow]

wait

why might X/C might not be a component

or atleast not a complete one

needs more elements maybe?

[barren sleet]

it might not be a complete one

[barren sleet]

exactly

[delicate hollow]

yea

[barren sleet]

that is tricky to work with though

[delicate hollow]

but

i got one thing

[barren sleet]

try to assume that there are 2 disjoint connected unbounded components and get a contradiction

[delicate hollow]

the needs more elements cant be in X/B - X/C

i can try that too

[barren sleet]

they can only be in tbag set

that set*

[delicate hollow]

yeah

woops

again

i meant to say that

i wonder if there is a more straightforward way to argue without contradiction that elements of component of X/C cant be in X/B - X/C

the two sets do form a seperation

[barren sleet]

uh

idk what you're saying

[delicate hollow]

ok

[barren sleet]

elements of X\C cannot be in X\B-X\C by definition

[delicate hollow]

yeah

but can they share componenets?

components

like there might be a connected set D that contains both of them

[barren sleet]

yeah

do you have a clear direction on this proof

[delicate hollow]

i might?

um

arguing by contradiction

wait

you were saying 2 disjoint connected unbounded components cant be a thing

so we have X/C is one

[barren sleet]

no

X\C is not oje

[delicate hollow]

ok

X/C is in one

is a subset of one

[barren sleet]

but there is a connected component containing X\C

exactly

that is one

now if D is another

[delicate hollow]

the other one would need to have points in C intersect B

[barren sleet]

what does it mean for D to be unbounded?

[delicate hollow]

that there is no open ball around it

[barren sleet]

no don't think of this. C intersecti B is fine

[barren sleet]

use this definition

[delicate hollow]

wait what do you mean be fine

[barren sleet]

like

[delicate hollow]

oh lol u gave me a visualization

ty

[barren sleet]

you said this as if it was the problem. but this is not the problem at all

[delicate hollow]

ok i see that

um

so problem with D being unbounded

is that it would need to be disjoint from X/C

and not contain points in B

[barren sleet]

yes

[delicate hollow]

but in doing this it cant really exist

it would need to be in X/C

[barren sleet]

can you turn this into a formal argument

[delicate hollow]

um sure

[barren sleet]

you're on the right track

when you are done I will tell you mine

[delicate hollow]

imma just type it

[barren sleet]

alright

[delicate hollow]

Let B be bounded in R2
Take C to be an open ball containing B
So X/C is in X/B
Since X/C is connected we have that X/C is a subset of a component of X/B.
X/C is an unbounded subset of a component of X/B
Suppose there is another set A that is an unbounded component of X/B disjoint from X/C.
Then There is no open ball containing A.
So A is in X/C, but this is a contradiction, so there is no other set A that is an unbounded component of X/B disjoint from X/C.
So X/C is the only unbounded component of X/B.
QED

Im going to be honest

[barren sleet]

some statements are not justified

[delicate hollow]

I am feeling a little insecure

with this statement

[barren sleet]

yes

here is my hint

[delicate hollow]

Then There is no open ball containing A.
So A is in X/C, but this is a contradiction, so there is no other set A that is an unbounded

[barren sleet]

it will be a lot easier if C is an open ball centered at the origin and you consider the radius of C as a revelat quantity

relevant quantity

[delicate hollow]

really?

[barren sleet]

yeah

then the complement of C is all points at least some distance from the origin. this will play along well with the definition of unbounded when doing your contraction

[delicate hollow]

ok

so like

C = {x in R2| d(0,x)<d(sup({p_x}),sup({p_y}))} where p are points in B and p_x,p_y are their x and y components.

There has to be a cleaner formulation qq

oh

gotta add something

[barren sleet]

don't try to figure out the radius of C

[delicate hollow]

because it wont contain that point

[barren sleet]

just call its radius M or something

and define it to be an open ball containing A

which exists by definition

of bounded

[delicate hollow]

ok

that is way easier lol

so C = {x in R2|d(0,x)<M} where M is the bound of B.

well

that isnt defined

I should write out definition for bounded first

[barren sleet]

yeah

yeah

[delicate hollow]

and then just use that M

[barren sleet]

there exists some M

[delicate hollow]

probably easier tofolllow

[barren sleet]

doesn't need to be a tight fit or anything

your space could be the unit ball and M could be 500 and nobody cares

[delicate hollow]

oh

ok

once i do that

justifying the contradiction is another part

so

i assume there is an unbounded component A of X/B

my guess is to make this all more rigorous

but when I say there is no open ball of A

[barren sleet]

if it's unbounded, what does that mean

[delicate hollow]

there is no open ball containing A

[barren sleet]

A is already a set, don't use that name twice

[delicate hollow]

it is?

[barren sleet]

isn't it

oh oo0s

[delicate hollow]

nah i avoided it

[barren sleet]

you used B

I thought it was A

ok yeah

[delicate hollow]

ok

we can use D tho

[barren sleet]

so A

is the name of the set

[delicate hollow]

ok so A

yea

unbouned

[barren sleet]

there being "no open ball"

means something better

it means

[delicate hollow]

complement of an open ball

[barren sleet]

for each finite value you pick

it contains points father than that value away from 0

but R^2\C already contains all points with d(x,0)

M

[delicate hollow]

ya

[barren sleet]

so if D really has some points as far away from the origin as you want

pick M to be your number

it just contain points with d(x,0)>M

[delicate hollow]

so its in X/C

[barren sleet]

but R^2\C contains all such points

[delicate hollow]

ya

[barren sleet]

so it's not disjoint from C

yeah

[delicate hollow]

that makes it way more rigorous

[barren sleet]

now I have to go but I will give you my explanation for connectedness

it helps to have an intuitive sense of open and closed

closed sets contain everything that is close to them in the metric space

or topological space (since the topolgies define a sense of "closeness")

open sets provide "padding" for all of their points

[delicate hollow]

contsunf?

contains?

[barren sleet]

sorry

[delicate hollow]

ah

[barren sleet]

I fixed it

[delicate hollow]

yea

because it includes limit points

[barren sleet]

ok read again now

[delicate hollow]

and the set itself

[barren sleet]

I changed it

yeah so if a set X is disconnected in the intruitive sense

[delicate hollow]

oh wait

i get it now lol

[barren sleet]

it means some set D and it's complement C are "far apart"

[delicate hollow]

i get the natural padding part too

[barren sleet]

with clopen and everything?

[delicate hollow]

ok

by far apart

you mean doesnt contain those limit points

so it cant be closed

why cant it be open?

[barren sleet]

I mean an intuitove sense of far apart

[delicate hollow]

wait

this is how im seeing it

[barren sleet]

wrong

[delicate hollow]

oh

i assumed far apart means not close together

close together means it contains its liimit points

far apart should mean it doesnt have its limit points

[barren sleet]

D is open because it is far enough from C that some neighborhood of every point avoids C (think intuitively, D and C have some intuitive distance)

so D is open

D is also closed because it contains everything close to it

because C is not close to it

[delicate hollow]

wait what is C now

[barren sleet]

^

[delicate hollow]

oh

[barren sleet]

C can't be stealing any limit points from D

[delicate hollow]

if it does that means its close to some points in D

[barren sleet]

that's how I make sense of clopen

[barren sleet]

yes

[delicate hollow]

wait

im reading why D cant be open

why it is*

oh i guess that makes sense intuitively

so like

D's goal is to have all points avoiding C and at the same time not being close to C

avoiding C means each of its points have an open neighborhood around them, when unioning gives an open set

[barren sleet]

yes

[delicate hollow]

and not being too close means not having any limit points in common with C which means D can't be closed because inorder to be closed it needs to contain all its limit points

[barren sleet]

yes

[delicate hollow]

is closed*

[barren sleet]

or also

think

D is open

that's all you need

cause then apply the same reasoning to C

[delicate hollow]

oh

[barren sleet]

and C being open gives you D is closed

so whichever of the arguments is easier

[delicate hollow]

a set is closed iff it contians all its limit points right?

and itself

because Cl(A) = A U A' where A' is set of limit points?

ok

ty for intuition

maybe someday ill make an animation

or a gif

[barren sleet]

yes

[delicate hollow]

seems like its easy to visualize euclidean ig

[barren sleet]

I'm just explaining intuition

[delicate hollow]

idk how notion expands to not R^n

its not possible to visualize otherwise

ig?

maybe there are a bunch of other ways

i think ill do a presentation on this seems simple and fun

[barren sleet]

in math conversations it's easy. get used to everything being rigorous

but sometimes

irs ok to not be

and at this time

it's just getting intuition for how our idea of connected relates to the topolgical def

[delicate hollow]

@barren sleet

What grade

[gritty widget]

[barren sleet]

wdym

like year of school

[delicate hollow]

ya

@gritty widget helps me get perspective of my compotition

wow

competition

[gritty widget]

[barren sleet]

I'm a junior in highschool

[delicate hollow]

really?

[barren sleet]

yah

[delicate hollow]

damn im doomed

[gritty widget]

buddy there are highschoolers in here doing way more advanced shit

who cares if some highschooler is doing things more advanced than u lol

[delicate hollow]

?

[barren sleet]

I have no right to tell this to you because I literally such shit about comparing myself to other people, but it's honestly pointless

there are highschoolers smarter than me like beyong my comprehension

[delicate hollow]

oh yeah

IMO people

[gritty widget]

you didn't make IMO????? you're doomed lol gl getting into grad school literally 0.0001% chance

[delicate hollow]

I have always had an understanding there are going to be people way ahead of the curve than me

[sleek thicket]

I am significantly more competent than all of you and you should quit math now

[gritty widget]

compact

[delicate hollow]

are you competition chadrock?

[sleek thicket]

Especially you tterra

I am not your competition. You and your competition are like ants to me

[delicate hollow]

damn

comparing myself isnt pointless

[barren sleet]

to me it is

[delicate hollow]

staying doomed is a good thing though

[barren sleet]

comparing to me

[delicate hollow]

well look at this

my goal

or one of them

do math research at a high level

[barren sleet]

same as me

[delicate hollow]

I might be able to achieve my goal

but not to the most specific degree

[barren sleet]

whenever that thought enters my mind my first thought is "I'm never gonna be good enough every9ne else is so much better"

[delicate hollow]

well yea

i know thats true

[barren sleet]

do

I honestly don't know how it works

[gritty widget]

the key is to just think you're better than anyone unironically

[delicate hollow]

Ive only been the best at a couple of menial things in my life

[barren sleet]

and I'm not at that stage anyway

[sleek thicket]

I think this is a thorny and complicated issue, but my main take is this

[barren sleet]

ow

[gritty widget]

@​Moderators

[delicate hollow]

I dont get it

[barren sleet]

ok everyone shun me until tomorrow

I need to do homework

[sleek thicket]

Remove advanced role

Add studying role

[barren sleet]

good idea

[gritty widget]

,ban ethan

[gentle osprey]

This may only be done by a moderator!

Apr 16, 2021, 3:43 AM

[sleek thicket]

Delete discord

[barren sleet]

thank you

[sleek thicket]

,ban ethan

[gentle osprey]

This may only be done by a moderator!

Apr 16, 2021, 3:43 AM

[delicate hollow]

oh we can go to #discussion

[barren sleet]

,ban mniip

[gentle osprey]

This may only be done by a moderator!

Apr 16, 2021, 3:43 AM

[gritty widget]

oh since #advanced-analysis is being used i kind of want to ask my integral question here

[delicate hollow]

wait

we cant

[gentle osprey]

No selfroles matching moderator.
See ,selfroles --list for the list of valid selfroles.

Apr 16, 2021, 3:43 AM

[sleek thicket]

Wait was kxrider typing

I thought I saw someone

Do they have a real question

[delicate hollow]

probably

[sleek thicket]

okay whatever ask your stupid integral here

[gritty widget]

nvm

i want to bash my head into the wall some more

i don't think enough blood has pooled

[delicate hollow]

i want to also

i want to get corona vaccine

so that my blood can clot in my spine

[gritty widget]

if you're dead you can't get corona

[delicate hollow]

and ill die with acute pain

[gritty widget]

coronavirus vaccine

[delicate hollow]

true

[gritty widget]

[sleek thicket]

tterra brain damage arc???

[gritty widget]

you mean my life?

[delicate hollow]

i mean

[sleek thicket]

tterra exsanguination arc???

[delicate hollow]

lets be honest

sometimes there is no solution qq

[sleek thicket]

I think this conversation should change

[delicate hollow]

what is the question ttera

[sleek thicket]

And not continue

[gritty widget]

i agree, it should.

someone teach me about poisson cohomology so i dont have to read

[tough imp]

@​Moderators

[sleek thicket]

I will do this

[delicate hollow]

how did you avoid ping?

[sleek thicket]

Poisson cohomology is a form of sheaf cohomology

[tough imp]

wdym

[delicate hollow]

how did ytou not ping mods

dont try and bait me noob

[sleek thicket]

Sheaf cohomology is the derived functor of the inclusion of category of sheaves into the category of presheaves

[tough imp]

Ohhhhhh right

​

[tough imp]

?????????????????????????????

[delicate hollow]

lol

[tough imp]

No it's a cohomology theory of Fish

[sleek thicket]

Chmonkey, please do not interrupt my incredibly correct explanation

Derived functors are when you take F and put a dot over it

[tough imp]

but...

that's...

the wrong...

whaaaaaaa

[sleek thicket]

chmonkey

please stop interrupting me

[tough imp]

[sleek thicket]

I am trying to inform tterra

[delicate hollow]

@gritty widget imma pm u but im too lazy to use pointer

[sleek thicket]

About sheaf cohomology

[tough imp]

Don't listen to Sham, he's a sham

Sheaf cohomology is the derived functor of the inclusion of O_X-modules into Sheaves

[sleek thicket]

This is so obvious

[gritty widget]

actuall

y

eh nvm

im not going to start

[sleek thicket]

Oh no was that actually helpful

wait

What are you not starting

[tough imp]

oh cochain cohomology of its Chevalley-Eilenberg algebra, that makes sense

[gritty widget]

i was going to go on a rant about how yesterday i was thinking about poisson structures and how, like, there ought to be some way to measure how much poisson vector fields fail to be hamiltonian

but

id literally be the only person understanding it

[tough imp]

Hamiltonian like in graph theory?>

🙂

[sleek thicket]

Awww

[gritty widget]

anyways it turns out that poisson cohomology actually does measure this

[sleek thicket]

I mean you're right

[tough imp]

Just like in graph theory

[sleek thicket]

It's cool that you figured this out

nice

[tough imp]

Is TTerra actually just UGDG

[gritty widget]

the actual definition of poisson cohomology is pain

but at least the first degree is easy to understand

[sleek thicket]

How are you only realizing this now chm

[tough imp]

I thought he was just meming

but the memes never ended

[sleek thicket]

[tough imp]

and now he says something like hamiltonian and poisson cohomology

so now I know he is serious about his DG memes

[sleek thicket]

What's the first degree of poisson cohomology

[tough imp]

okay...

[sleek thicket]

Jesus christ this is

Very stupid

[tough imp]

different DG than what DG means to me

[sleek thicket]

not gonna lie to you

chm

[tough imp]

True

[sleek thicket]

Do some real geometry...Spec \R

[tough imp]

lmfao

No no remember that a real projective variety

is a technical term

[gritty widget]

basically euclidean geo right???????

[tough imp]

it really is like Euclidean geometry then

[sleek thicket]

Tterra what is the first degree of poisson cohomology

[tough imp]

How to solve every geometry problem in 8th grad

draw as many triangles as possible

stare at it until it becomes obviuos

[sleek thicket]

This is called a "simplicial set"

[tough imp]

which is equal to its 1skeleton

what is a do Carmo

[gritty widget]

i have not

i wonder if my uni gives me a nice copy

let me check

it does not

[sleek thicket]

Wait

A random pdf I found claims poisson cohomology is the same as de rham cohomology

Ie singular cohomology with real coefficients

How are all the cohomology theories the same

It's fucked

[gritty widget]

at least in the case that the poisson structure comes from a symplectic form this is obvious but i don't see it

link

[sleek thicket]

oh sorry this was for symplectic manifolds I can't fucking read

[gritty widget]

anyways

there is still some checking to be done

do not fret

[sleek thicket]

Sorry

[gritty widget]

you can have stupid poisson structures

like

{f, g} = 0 for all f, g

[sleek thicket]

I looked at the definition though and it seems very similar to lie algebra cohomology

Which ig makes sense

Since a poisson structure is like a lie bracket on the smooth functions with an extra condition, right?

[gritty widget]

right

that condition being that fixing one argument gives you a derivation

ie vector field

[sleek thicket]

Right

[gritty widget]

so you can ask what vector fields arise as such

which is what lead me to poisson cohomology (or to looking it up lol)

i will look into lie alg cohomology

[sleek thicket]

Oh so is the first group like all vector fields/Hamiltonian ones?

Or some subset of all vector fields maybe?

[gritty widget]

some special subset

[sleek thicket]

That feels cohomological

Closed/exact

[gritty widget]

so alternatively, a poisson structure is a special bivector field, and then you call a vector field X Poisson if L_X(bivector field) = 0

[sleek thicket]

Right

[gritty widget]

then its {poisson v fields}/{Hamiltonian v fields}

[sleek thicket]

Ooh

[gritty widget]

which was very pleasing to me

[sleek thicket]

Wait

What's L_X? Don't bivector fields eat like pairs of covectors?

So like it can't be the interior multiplication

[gritty widget]

lie derivative

oops

[sleek thicket]

Ahhh

Gotcha

That makes sense

[gritty widget]

defined as like, time derivative of pushforward wrt flow or some shit

i dont remember lol

[sleek thicket]

Yeah ik how lie derivatives work

Just take the lie bracket and then use the appropriate product rules 4head

It is so fucked up that [X, Y] = L_X Y

What's going on there

[gritty widget]

ikr

like what

[sleek thicket]

[gritty widget]

proof: insane computation

[sleek thicket]

Hahaha

sad but true

[gritty widget]

i had it on an assignment

pain

[sharp yoke]

@sleek thicket thx i think i wrote an 85% coherent proof

[sleek thicket]

Nice!

I can't believe I missed that detail

[gritty widget]

let $(M, \omega)$ be a symplectic manifold. we say a vector field $X$ is symplectic if the Lie derivative $L_X\omega = 0$, and Hamiltonian if $\iota_X\omega$ is exact. by cartan's formula $$L_X\omega = d\iota_X\omega + \iota_Xd\omega = d\iota_X\omega,$$ so $X$ is symplectic iff $\iota_X\omega$ is closed. cohomology.

[gentle osprey]

TTerra

[sleek thicket]

ah that makes sense

You mixed up the name of the formula though

It's cartans *magic * formula

Because every time you see it

You go

[gritty widget]

[sleek thicket]

"what the fuck is this magic"

[gritty widget]

it just works lol

cohomology.

[sleek thicket]

cohomology.

[gritty widget]

i might bring up poisson cohomology during my presentation

[sleek thicket]

That would be neat

[gritty widget]

if i can find anything about it that isn't just the definition then i absolutely will lol

[sleek thicket]

You could do a quick detour into lie algebra cohomology

Lock the doors

This is now a homological algebra talk

[gritty widget]

im finding that literally everything i find to present about is just a generalization of some shit from symplectic

so

[sleek thicket]

hahaha

[gritty widget]

i'll feel weird including too many details

nothing im talking about is new

[sleek thicket]

Do you have interesting non symplectic examples at leaat?

[gritty widget]

i do

i think

[sleek thicket]

Nice

[gritty widget]

ah yeah

i wrote down a little bit about torii being poisson manifolds

just as a cute example

[sleek thicket]

Oh neat

[gritty widget]

but since those are lie groups i can probably go more in depth with that example!!

[sleek thicket]

Those are not symplectic

Because of compact

Wait no that doesn't make sense lol

The even dimensional cohomology groups of a torus are nontrivial

But because torii can have odd dimension works

There

[gritty widget]

[sleek thicket]

Neat!

Tterra did you know that $\binom{n}{k} = \binom{n}{n-k}$?

[gentle osprey]

Chadrock

[gritty widget]

uh

no

i didnt

mindblowing

[sleek thicket]

Oh okay, here's an easy proof

By the kunneth formula, $\dim H_k(\mathbb{T}^n, \R) = \dim H^k(\mathbb{T}^n, \R) = \binom{n}{k}$

[gentle osprey]

Chadrock

[gritty widget]

[sleek thicket]

By poincare duality and orientability of the torus, $H^k(n\mathbb{T}^n, \R) \cong H_{n-k}(\mathbb{T}^n, \R)$

easy!

[gritty widget]

show this to discrete math students

[sleek thicket]

someday

[gritty widget]

does this end up being circular or is it actually a genuine proof

[gentle osprey]

Chadrock

[sleek thicket]

I think it's a genuine proof

The computation of homology could end up sus

But I don't think so

[gritty widget]

handwave it enough and it's fine

[sleek thicket]

Kunneth formula says this

Oops

[gritty widget]

[sleek thicket]

Okay whatever I can't find a pmg

*png

[gritty widget]

[sleek thicket]

It says if you have like finite CW complexes and you're working over a field or whatever

The kth cohomology of their product is the direct sum of the tensors of the ath cohomology and bth cohomology where a+b = n

So you keep pulling off circles

you can do an inductive argument

And get that it's n choose k

Probably not circular

[gritty widget]

i see

meh even if it were circular you should still give it to intro discrete math students

make them shit

[sleek thicket]

lol

Did you give the proof with the weird duality thing on your algebra hw

[gritty widget]

i am back to wondering if there is a link between poisson cohomology and lie algebra cohomology

https://en.wikipedia.org/wiki/Poisson_algebra

[sleek thicket]

I mean yeah the definition of a poisson structure only involves the algebra of smooth functions

Ooh these are allowed to be noncommutative though

[gritty widget]

let me find a definition of poisson cohomology

[sleek thicket]

🐟

[gritty widget]

clearly

[sleek thicket]

How are your fish puns coming along

[gritty widget]

i have not nearly included enough in my presentation

[sleek thicket]

https://tenor.com/view/russia-bear-truck-theft-fish-gif-3468320

[gritty widget]

here is a more down-to-earth definition

P the bivector field

and [P, ] being the fucky bracket on multivector fields that i dont understand

[sleek thicket]

This is more down to earth

[gritty widget]

i have written "Is there a link between Lie algebra cohomology and Poisson cohomology?" down in my notes

i do not think i will pursue it any more tonight

but ty for bringing that up

[sharp yoke]

how do you justify that f is differentiable here?

oh wait i just realized how to do it as a posted it

nvm

[sleek thicket]

Nice!

[plush needle]

If I am given two geodesics $γ_1$ and $γ_2$ and a distance function $d$ that gives me the distance between any two points on the geodesics, what conclusions can I draw about the geodesics from the information given by $d$?

[gentle osprey]

Roenbaeck

[plush needle]

(this is a more general question than one I posted earlier about "boxes", but essentially the same problem)

I'm also interested in any constraints under which it would be possible to say much more. Like, "if we assume the curvature is uniform, then..."

[plush needle]

I have closed form expressions for $d(γ_1(t), γ_2(t))$ in a set of spaces, as a clarification.

[gentle osprey]

Roenbaeck

[silent oasis]

Not sure whether this room or the category theory room is more suitable for this question. Let $k$ be some field, $V$ a $k$-vector space. In group scheme theory, people work with the functor $\operatorname{Aut}(V)$ from the category of $k$-algebras to $\mathbf{Grp}$ which assigns to a $k$-algebra $R$ the group $\operatorname{Aut}_R(V \otimes_k R)$, where $\operatorname{Aut}_R$ denotes the $R$-linear automorphisms. What does this functor do on morphisms?

[gentle osprey]

harraq

[silent oasis]

If we restrict to the finite-dimensional case, for a morphism $f: R \to S$ of $k$-algebras, $\operatorname{Aut}(V)(f)$ goes from $GL_n(V \otimes_k R, R)$ to $GL_n(V \otimes_k S, S)$, where $n$ is the dimension of $V$, and can be defined by pushing forward the coefficients. One could extend this definition to the case when $V$ is not necessarily finite-dimensional using choice to construct a basis $\lbrace v_i | , i \in I \rbrace$, observing that $V \otimes_k T$ is a free $T$-module for any $k$-algebra $T$, hence any automorphism is given by an "infinite matrix" in a sense that can be made precise, and pushing forward the coefficients of that matrix as in the finite-dimensional case. However, it remains to show that this preserves invertibility and is overall a bit cumbersome.

I feel like there should be a more conceptual, diagrammatic way to define how $\operatorname{Aut}(V)$ acts on morphisms. Can anyone help me out here?

[gentle osprey]

harraq

[tawdry widget]

? How is Aut(V)(f) defined? Where f is a morphism from R1 to R2. f isn’t necessarily isomorphism.

[silent oasis]

this is exactly my question.

I'll illustrate the action of $\operatorname{Aut}(V)$ on morphisms in the case that $V$ is finite dimensional. In that case, $V \otimes R$ is a finitely generated free $R$-module and any $R$-linear automorphism of $V \otimes R$ is given by an invertible square matrix with entries in $R$. Let $f: R_1 \to R_2$ be a ring homomorphism. Then $f$ preserves units. If $M$ is an invertible square matrix with entries in $R_1$, then applying $f$ to every matrix entry yields a square matrix with entries in $R_2$. Moreover, the resulting matrix is invertible, because its determinant is a unit in $R_2$. I hope I could illustrate the finite dimensional case. Note that this relies on a choice of basis, which is something that you generally want to avoid

[gentle osprey]

harraq

[timber junco]

i need help in learning how to factorize

[tepid nest]

@timber junco This isn't the correct channel

[timber junco]

ahhhhhhhhhhhhhhhhhhhhhhhhhh

i thought the quadratic formula was geometry based?

[tepid nest]

No, these are the advanced channels

Just take it to #prealg-and-algebra

[timber junco]

sorry i'm in AP thats why i put in advanced

[weak idol]

we have a curve r(t).

r(0) = (2019,-1)
r'(0) = (0,1)
the curvature of the curve is fixed to be 1/t^2+5t+6

r(1) = ?

im sorry if this doesnt count as advanced, but i asked in #calculus and i didnt get any guidance

[weak idol]

rate of change of the unit tangent wrt the arc length

[gentle osprey]

slimvesus

[delicate hollow]

typo?

show connected by not path connected meant to say, show connected but not path connected?

[sweet wing]

probably

[delicate hollow]

☑️

quick question

for proving connected

can you argue this

suppose set is open

and then suppose that the set contains all its limit points

show some contradiction in that?

[sweet wing]

all sets are open in itself

recall how we define connectedness

[delicate hollow]

i know how we define it

its no clopen sets

the problem is to prove topologist sine curve is connected but not path connected

i was thinking of doing argument by assuming a seperation, then showing it doesnt exist.
then proceeding to show it isnt path connected, which i honestly don't know off top of my head

[sweet wing]

more accurately there are only 2 clopen sets

the empty set and the entire space

[delicate hollow]

ya

yea

um

def is

graph of your given f(x) unioned with verticle line from -1 to 1 at origin

and subspace topology on r2

so yeah the closure of graph

its continuous

so um

open sets get mapped to open sets

specifically

open in R^2 is open in (0,1]

and (0,1] is connected

wait uh

i can just say connectedness is perserved by continuity

and then just move on to arguing not path connected

but ig that might be harder

[long hornet]

Let p, q be two points in a connected topological manifold. Shamrock essentially proved that there is a homeomorphism f sending q to p and is equal to the identity outside a prescribed open set containing the path (I think? At any rate, this is possible), but I still wanted to find a coordinate ball V containing p, q. I think the following argument should work, but I feel like I am missing something.
Pick a coordinate ball B around p, and let p' =/= p be in B. Now assume we have f so that it fixes p and sends p' to q, and let B* = f(B), which is also a coordinate ball around p. Now the point is that f(p') = q, so q is in B*.
To make f fix p, we find a neighborhood of p that is disjoint from a neighborhood of the path C = f([0, 1]). This is possible because {p} and C are disjoint closed sets, and M is normal.

[worthy canopy]

how do i parametrize a curve if i know its curvature (as a function itself), a point on the curve, and a tangent vector at that same point?

[tough imp]

this is not what a continuous function does, it pulls back open sets to open sets btw

[worthy canopy]

how do i parametrize a curve if i know its curvature (as a function itself), a point on the curve, and a tangent vector at that same point?

[delicate hollow]

@tough imp h

yea

if f:X->Y continuous

then open in Y implies f^-1(Y) is open in X

say X was connected

then Y is also connected by continuity

or wait

i think it needs to be surjective

i dont remember

[tough imp]

the image of a connected set is connected

so it would need to be surjective

else every set is continuous by just taking a map from the one point set haha

[honest terrace]

if f: X -> Y is continuous, then f(X) is connected

so Y is connected is connected if f is surjective indeed

[tough imp]

why you copying what I'm saying smh smh you gotta change your name to Chika-Blyat now

[honest terrace]

i'm just too slow

[tough imp]

I'm sorry :(

it's okay

you can stay Shika-Blyat

[honest terrace]

fortunately every connected space is a connected space !

[tough imp]

Ultra you're wrong

I care about spaces not up to homeomorphism

[honest terrace]

[supple locust]

If I have a continuous function from unit disc to circle i.e. D -> S1 , is there always an x in S1 which is fixed under this function?

[barren sleet]

yes but idk how to show it without brouwer fixed point theorem

[supple locust]

but don't we have D -> D map in brouwer fixed point theorem ? I dont see how this follows from that

[gritty widget]

D is the closed unit disk? compose with the inclusion map S^1 -> D if you really want it to be D -> D

this makes no difference, a map into S^1 is practically the same thing as a map into D

[barren sleet]

if its continuous from D to S1 its continuous from D to D, just consider the original function to have codomain D

oh yeah terra said it better

composing with inclusion map

[gritty widget]

hm

[supple locust]

hmm. for some reason I thought that brouwer required the map to be surjective

[barren sleet]

nope

it works super well when its not surjective actually

[supple locust]

yea

I just checked the statement again

[barren sleet]

special cases of brouwer become much easier if you actually require the range to be "small" enough

[gritty widget]

this is now reminding me of a cute complex analysis problem i did earlier

[barren sleet]

idk complex analysis

and im not in a rush to learn

considering what else i dont know

[gritty widget]

find a holomorphic mapping of the open unit disk into itself with no fixed points. can you pick it to be biholomorphic?

[barren sleet]

idek what holomorphic means

[gritty widget]

everywhere complex differentiable

which is magically the same as complex analytic

[barren sleet]

yeah i have heard of thta

that seems wild

[gritty widget]

it is!

[barren sleet]

but idk enough complex analtsis

[gritty widget]

follows from cauchy's theorem

[barren sleet]

i could not eyeball a function and twll you whether its complex differentiable or not

[gritty widget]

if i show you the cauchy riemann equations

then you will be able to

[barren sleet]

which cauchy's theorem

ok i have a question

[gritty widget]

cauchy's theorem is "f is holomorphic only if f dz is closed" (converse is called "morera's theorem" in my notes)

which gives you cauchys integral formula which gives you complex analycity

[barren sleet]

if you have the open unit disk D and a smaller disk within it that has opposite points at some bounary point of D and the origin

so like

radius 1/2

and they would share a boundary point with each other

if the disks were closed

if you shrink the unit disk into that disk

there are no fixed points

[tight agate]

open disk --> upper half plane --> shift ---> back to disk?

[gritty widget]

cayley transform is so useful

[barren sleet]

but i cant tell if mine is holomorphic cause i just have no clue how to work with that

[tight agate]

holeomorphism

[barren sleet]

whats a holeomorphism

google doesnt knpw

[tight agate]

it isnt a thing

it will be a thing

[barren sleet]

oh

[tight agate]

someday

[barren sleet]

@gritty widget what is wrong with my answer (unless its magically right)

[barren sleet]

is it hard to explain what this means in more detail? (hard = requires some complex analysis info to explain)

[tight agate]

it isnt hard if you're willing to accept that there is a holomorphic map from the unit disk to the upper half plane with a holomorphic inverse

[barren sleet]

hmm

[tight agate]

but if you're willing to accept that

then map to the upper half plane

shift the plane

and map back

[barren sleet]

omg

that is smarty pants

thats a bijection as well

mine (if it even works) is only an injection

[barren sleet]

and i am willing to accept that

i am trying to think of one rn

[tight agate]

this is probably the intended solution

they teach you about those maps in a complex analysis class

[barren sleet]

im confused about why its the half plane

cant you just as well shift the whole plane

[tight agate]

this is where the holomorphic-ness comes in

there is no holomorphic bijection between the complex plane and the unit disk

[barren sleet]

:(

[tight agate]

but there is one between the upper half plane and the disk

weird eh

[barren sleet]

thats so weird

like

..

i feel like i can think of one

must not be holomorphic

[tight agate]

it's called liouville's theorem

any bounded holomorphic function on the plane is constant

so if it maps into the disk

it's bounded

and therefore constant

[barren sleet]

whats wrong with this?: biject [0,1) with [0,infinity) in a smooth way that has a smooth inverse in R. then send eveythign in the disk to the number in the same direction but the abs value scaled appropriately according to the bijection?

[tight agate]

it wont be holomorphic

[barren sleet]

is complex analysis generally harder than real analysis (at the introductory level, i know each can like get as hard as you want)

[barren sleet]

:(

[tight agate]

real and complex differentiability are very different things

[barren sleet]

damn it

i thought smoothness was sure to be enough

[tight agate]

say I have a function from R^2 ---> R^2

and say it's differentiable at some point

then you have the corresponding jacobian matrix

if it's complex differentiable, the jacobian commutes with multiplication by i

[barren sleet]

am i misunderstandign something

or how can it not

oh

do you mean treating C as R^2?

[tight agate]

sure

[barren sleet]

ok

so in the real scenario multiplying by i is multiplying by the matrix:

(0 1)
(-1 0)

[tight agate]

so you want it to commute with a 90 degree counterclockwise rotation

[barren sleet]

or something

[tight agate]

ye

[barren sleet]

that is cool

[barren sleet]

what about this

[ivory dragon]

engineering students take complex analysis so

[tight agate]

so if the jacobian is $\begin{pmatrix}
a & b\
c & d
\end{pmatrix}$

[barren sleet]

oh....

[ivory dragon]

im shitposting, they take a non-rigorous computational version

[gentle osprey]

Brotivic t-structure

[barren sleet]

so like

[ivory dragon]

and even then, only certain disciplines (electrical) do

[barren sleet]

complex analysis\the interestinf stuff

[ivory dragon]

i mean it still covers the important theory

[barren sleet]

it needs to be symmetric right?

antisymmetric?

[tight agate]

Then we're demanding that $\begin{pmatrix}
-c & -d\
a & b
\end{pmatrix}$ is equal to

[barren sleet]

oh i see

[gentle osprey]

Brotivic t-structure

[barren sleet]

its a strong condition

a=-d and b=-c

[tight agate]

ye something like that

[barren sleet]

but wait

a=d as well

so a and d are zero

b=-c

omg

that makes perfect sense

the derivative is literallt just like a multiple of i

actually

that seems very restrictive

[tight agate]

no

nothing is zero

[barren sleet]

:(

[tight agate]

lemme write it out

[gentle osprey]

Brotivic t-structure

Brotivic t-structure

[barren sleet]

isnt it -1 of that

[barren sleet]

and -1 of that

[tight agate]

up to a -1

so a = d and c = -b

[barren sleet]

oh yeah idk how i got a=-d

that makes more sense

becaise then it acts like a general complex number

which is what i ex[ect

expect

[tight agate]

these are called the Cauchy-Riemann equations

[barren sleet]

but the othe way got me a multiple of i

[tight agate]

im pretty sure I did the calculation right

ye it's a = d and c = -b

[barren sleet]

yeah

[tight agate]

so you can already see that the jacobian of a holomorphic function is a lot more special than the jacobian of a real diffable function

[barren sleet]

yeah

well like

in this interpretation

R^2->R^2 is like a 1d function

and we already know for real differentiable functions, the jacobian is just [c] so it can be thought of as c=f'(x)

and this rwsult seems like the same thing

the jacobian is like the matrix equivalent if a general complex #

c

[tight agate]

right

[honest narwhal]

Ooh complex analysis

[honest terrace]

Hey quick question. What would be the prerequisites to start diving into algebraic topology, on the topology side ?
I've recently opened Hatcher but I was already kind of lost while reading the part on cell complexes in the chapter 0.
Or maybe is there a more noob friendly intro to alg top than Hatcher ?

[gritty widget]

rotman

[honest terrace]

ooh

that looks a lot more approachable as a beginner

thanks a lot

[gritty widget]

i remember briefly reading hatcher when i wanted to learn algebraic topology. it was weird

[uncut surge]

it's a pretty book but somehow i also never liked the style

[gritty widget]

i liked it, however some exercises are a bit too vaguely stated

The following problem makes me wonder something

Let $$f:\mathbb{R}\rightarrow \mathbb{R} $$ continuous and $$ \lim_{x\rightarrow +\infty }f(x)=\lim_{x\rightarrow -\infty }f(x)=0 $$. Prove f is uniformly continuous.

[honest terrace]

add a _ after the \lim

so that you have \lim_ instead of lim

so that the "x -> +oo" goes under the lim

[gritty widget]

yea thanks i just copied it from #calculus lol

oops

[gentle osprey]

Carla_

[gritty widget]

If we extended f to the extended real line, it would still be a continuous function i believe. However idk if uniform continuity makes sense in that new space. Could this be somehow used to show f is uniformly continuous?

[honest terrace]

so uniform continuity may make sense in the new space

Just see it as a metric space

[gritty widget]

but it is not a metric space

what is the distance between -infinity and 0 for example?

[honest terrace]

it is metrizable

[gritty widget]

right

its equivalent to [0,1] in some sense

but idk if that equivalence can be used to transfer uniform continuity

cuz the metric might not be compatible with the original metric in it's interior

[gritty widget]

Hm, I wonder if you can do the trick where you swap the metric d with d(x, y)/(1+d(x, y)), and then define d(x,infinity)=1. I know this generated the same topology, but not sure if the metric is similar enough that uniform continuity transfers over

[barren sleet]

if it forms the same topology how could continuity be ruined

oh

uniform

[tough imp]

Could you do something like, let a be a number such that on (-infinity, -a) U (a,infinity) |f(x)| is < 1

Then on [-a,a] since this is compact f is uniformly continuous, and then use the fact f is bounded on the complement to show its uniformly continuous there as well or something?

That’s my first instinct at least

[gritty widget]

Yea ofc, but i'm curious if could use the extended real line somehow instead

[tough imp]

Ahhhh okay

Then I have absolutely no idea

[bright acorn]

Ok so

I want to have some intuition on the following thing

[tough imp]

[bright acorn]

Suppose you have an n-dimensional C^{\infty} manifold and you pick a point p in M.

We can see that the set of all real valued C^{\infty} functions f : M -> IR that vanish at p forms an ideal on the ring of C^{\infty} functions from M to IR

and you can use this ideal

call it I

To define the tangent space of M at the point p as the dual of the quotient I/I^{2}

But I really don't quite get the intuition behind this definition

you can define the tangent space of a manifold at a point using curves or derivations

and it's all fine

But I really don't get this one

[obtuse meteor]

I guess the intuition is

you're ignoring higher order information

which is what a tangent vector should be

only the first order information a la taylor series

[bright acorn]

how is it tho?

[obtuse meteor]

I mean I/I^2 fairly clearly reads as "you have I which is things vanishing at zero, view these as smol bois, throw away all smol bois * small bois which is I^2 and call those 0"

so you have only first order smol bois

[tough imp]

http://www.math.utah.edu/~ptrapa/math-library/humphreys/ Humphreys-Linear-algebraic-groups-1998.pdf#page517

This is not for manifolds I will preface

This is for varieties, and their tangent space, but I assure you the analogy is very similar

On page 38

the author goes through essentially this

You can identify a local chunk like being in A^n (this is like taking a chart to be in R^n)

And then you can define the tangent space via partial derivatives the way we usually think of it

From there they show that if you are embedded in affine space that the analogous I/I^2 is isomorphic to this tangent space

The analogy for manifolds is practically exactly the same

I’d give a manifolds reference if I could, but I don’t know of one

This helped me swallow the I/I^2 pill since you see an explicit isomorphism between what you intuitively view as the tangent space and I/I^2

[bright acorn]

I see

Yeah from what I have heard

this definition is usually more useful when dealing with algebraic varieties

But it's a nice little bonus to know that you can also define the tangent space of a manifold at a point this way

Thanks for recommending a reference for the subject

[tough imp]

Give me a few minutes, I have a book which might have an explanation for this

It’s a manifolds book done using locally ringed spaces

So if anything I know of would have this sort of algebraic description, it would probably be this one

[bright acorn]

That's kinda new

[tough imp]

Also for what it’s worth, do you know what the dual numbers is

K[x]/(x^2)

This is pretty much what Faye was sayinr, and also how I think of I/I^2 sort of

[bright acorn]

Not really :/

[tough imp]

So elements of this look like

a + bx

And then multiplication operates like

(a + bx)(c + dx) = ac + (ad + bc)x

So you’re considering essentially the first order information

The way to think of it is like

a is a point

And the value b is like this infinitesimal information, sometimes people use epsilon instead of x

Think of x as being some number thats sooooo small that when it’s squared it becomes 0

But it still exists

So the idea here is that like a + bx is kind of like the point a

(Say on the real line)

And b is this tiny little vector at a

So say, 1 + 2x is like the point 1 with this vector pointing two to the right from 1

So this captures tiny infinitesimal data which is kinda what you want in the tangent space

But now let’s consider if we instead looked at

(x)/(x^2)

Okay, this is a special case of I/I^2

Well an element of (x)/(x^2) looks only like

bx

Essentially the a = 0

So now all we’re left with is the vector

Aka the tangent vector

Also if we looked only at polynomial functions on R, then (x) is exactly the ideal of functions vanishing at 0

[bright acorn]

oooh

Ok I see it

So the intuition is literally considering information up to a first order approximation?

[tough imp]

Basically

You could consider (x)/(x^3)

To capture 2nd order info too