point-set-topology

[obtuse meteor]

so is it sufficient to find like

[sleek thicket]

in the opposite of a pose with H the inclusion this says y contains H(xi)

[obtuse meteor]

an object A where A contains U and V

that is you have arrows

U <- A -> V

because that works with n = 1 in the definition

[sleek thicket]

yup

but also with an arrow y -> A

[obtuse meteor]

but then you also need y contains A

okay got it

[sleek thicket]

right

so I think I found a counterexample

[obtuse meteor]

what is our y here

[sleek thicket]

any nbhd of x

[obtuse meteor]

in the scheme case you're trying to work out

ok

a neighborhood of x which contains the distinguished opens U and V

we want to find a distinguished open which contains U and V and is contained in the neighborhood of x

ok this is fucked up

alternatively we can like

chain things together

but that scares me

for what should be obvious reasons

[sleek thicket]

lol same

[obtuse meteor]

1 is the only natural

if you come up with a counterexample for n = 1

we will try to meme together something for like n = 2 or n = 3

and if that doesn't work

we'll complain about stacks project

perfect plan

[tidal cedar]

LOL

[sleek thicket]

let $X = \mathbb{A}^2$ and $U = X \setminus {0}$. Then $D(x) \subseteq U$ and $D(y) \subseteq U$. If $D(x) \subseteq D(f) \subseteq U$ then $(x) \subseteq (f)$ so $f$ divides $x$, which means $f = x$ or $f = 1$ (up to units), so $D(f) = D(x)$ or $D(f) = X$, and in the second case $D(f) \not\subseteq U$. So if we have arrows $U \to D(f) \to D(x)$ in $\mathsf{Open}(X)^{op}$ then $D(f) \to D(x)$ is the identity. Same logic applies to $D(y)$, so if there was a chain like in (2) then we'd have $D(x) = D(y)$, which is false

[gentle osprey]

shamifold

[sleek thicket]

this should be a counterexample for any chain

basically if you take the plane minus a line sitting inside the plane minus a point, there's no intermediate distinguished opens at all

So the nice categorical lemma I proved isn't actually relevant here -_-

[obtuse meteor]

A2 over C?

[sleek thicket]

Yeah

Should work over any integral domain

[obtuse meteor]

¯_(ツ)_/¯

C is the only thing that matters right? ,,,,right

[sleek thicket]

Im just doing geometry okay

I'm picturing it over R

[obtuse meteor]

lol

[sleek thicket]

X is basically the smallest/simplest example of a non affine open subset of an affine scheme/variety

Which is why it's showing up here

[obtuse meteor]

hrrrmmm

okay I think I might agree with you one second let me check my brain

and look at stacks again

aha

sham

you're gonna laugh

but I think this works

U = U -> A <- V = V

[sleek thicket]

What's A here?

[obtuse meteor]

take it to be the intersection of U and V

[sleek thicket]

right

Wait uhh did I fuck up duality

[obtuse meteor]

rip moment

[sleek thicket]

So we have those maps in the opposite category

Yes

But the maps in stacks go the other way

[obtuse meteor]

no they don't

you have x0 = U x1 = U x2 = U cap V x3 = V x4 = V

n = 2

U <- U -> U cap V <- V -> V

right

[sleek thicket]

Oh lmfaooo

This is incredibly based

[obtuse meteor]

🧠

[sleek thicket]

And I see why my counterexample doesn't work

Okay this is mega sick ty faye

[obtuse meteor]

me not understanding how this helps in alg geo but being able to spot the dumb solution

[sleek thicket]

Tytyty

Lmfao

[obtuse meteor]

as a category theorist

[sleek thicket]

Well basically the structure sheaf is terrible but it's nice on a basis (and that basis is closed under intersections)

[obtuse meteor]

ah

[sleek thicket]

So I would rather compute a colimit over that basis than a general one

[obtuse meteor]

yeah I gotcha ^^

yes

[sleek thicket]

This is mega sick lmfao

[obtuse meteor]

I can make sense of structure sheaf on basis

[sleek thicket]

Thank youuuu

[obtuse meteor]

because like

localization

and formal inverse

[sleek thicket]

right exactly

[obtuse meteor]

structure sheaf elsewhere???

scary

not real

it can't hurt me

[sleek thicket]

So this lets you say prove the stalk at x is colim A_f over f in A\x

since f in A\x iff x in D(f)

We go from the stalk being computed over all nbhds of x to just the distinguished ones

[obtuse meteor]

oh super based yeah

[sleek thicket]

And then you can use commutative algebra to say that colimit of A_f is the localization at x

[obtuse meteor]

that's great

"and then you can use commutative algebra"

see you lost me there :P

[sleek thicket]

Yeah!! And this kind of cofinality argument with distinguished opens comes up a lot

Hahaha

[obtuse meteor]

sham

what are the odds

that in a poset

condition (2)

reduces to n = 1 or n = 2

[sleek thicket]

lol I'm not sure

So uhh like

[obtuse meteor]

feels like

it should

lmao

for spiritual reasons

[sleek thicket]

You proved that it holds if your poset has upper bounds

Right?

[obtuse meteor]

I can't explain why but it feals right

yeah

[sleek thicket]

Err like upper bounds under a fixed thing

[obtuse meteor]

sub-poset has upper bounds

slice category

has upper bounds

@gritty widget

wait lol

[sleek thicket]

Lmao

[obtuse meteor]

I didn't mean to do that sorry

[sleek thicket]

Okay so like the slice of the subcat over an object of the supercat

Weird

Anyways yeah

[obtuse meteor]

mhm

this is neato

wait no you don't even need upper bounds under a fixed thing

because if you have x -> z <- x'

and y -> x and y -> x'

you naturally have y -> z

[sleek thicket]

Ooh yes

[obtuse meteor]

transitivity

[sleek thicket]

Nice Nice Nice

[obtuse meteor]

exists

[sleek thicket]

I was getting confused by duality again lol

[obtuse meteor]

¯_(ツ)_/¯

happens

you need upper / lower (one of the two) under a fixed thing

and the other one you just need existence

I think is the point right?

like if you're trying to get x <- z -> x' then you need an explicit y -> z

otherwise you're frick-fracked

[sleek thicket]

Yes, I think so

If I understand

oh also we didn't even need to know that U cap V is a basis element here

You could take any basis element W contained in U cap V which contains x

[obtuse meteor]

oh yeah

true

[sleek thicket]

So in the general case this says you can compute stalks on a basis

[obtuse meteor]

super based

basis'd

[sleek thicket]

"Guy who calls all basis elements coordinate neighborhoods"

[obtuse meteor]

[sleek thicket]

Sheaf on a basis? You mean defining something in coordinates?

[obtuse meteor]

once again we are poasting

[sleek thicket]

If you couldn't tell from this and my tweet I'm trying to learn AG again lol

[obtuse meteor]

lol

I should try that again

¯_(ツ)_/¯

I feel like

I have been slowly absorbing

all the sheafy knowledge and stuff I'm going to need when I tackle the bear

and the algebra knowledge too

and eventually I'll actually be able to do it lmao

[tough imp]

AG good

[sleek thicket]

@tough imp bugs me about ag problems all the time so I feel like I should just learn it at this point

Oh lol

[obtuse meteor]

lol

[sleek thicket]

I was about to summon you

[tough imp]

I predicted

[obtuse meteor]

algebra is a huge weak point for me rn

[tough imp]

Chmonkey sense

Algebra ez

Things go brrrr

[obtuse meteor]

linear boisssss

[sleek thicket]

Hmm doubt

[tough imp]

Algebra actually ez when you have two good resource

1: commutative algebra book

2: @sleek thicket

[sleek thicket]

(2) Brendan Murphy

lmfaoooo

[obtuse meteor]

algebra easy when you cite commutative algebra book w/o proof

[sleek thicket]

That's the way to do it

[tough imp]

No the real based thing is

[obtuse meteor]

list of "commutative algebra things that are true"

[tough imp]

If u need result

[obtuse meteor]

"this reduces to problem in commutative algebra"

[tough imp]

See if you only need a little bit to prove

If so, learn it

If it requires a lot

Black box and promises your self you will learn it

[sleek thicket]

Student who reduces AG problems to the local case and then calls it a day

[tough imp]

It’s like taking out a loan

[obtuse meteor]

you won't

[sleek thicket]

no comm alg for me thanks

[obtuse meteor]

based

[tough imp]

Lmao

Once something is a statement about affines

He just says “follows by commutative algebra@

Doesn’t know what commutative algebra it follows from

[sleek thicket]

The student is chmonkey and "calls it a day" means "dm sham"

[obtuse meteor]

lsadjhfahs

[tough imp]

-_-

I try myself

[sleek thicket]

lol

[tough imp]

Then I come to you when stuck

Look I have held off asking you something

But now you have caused it

Do you know what regular sequence is

[sleek thicket]

Oh no

[tough imp]

Okay so let M be a module with a length n regular sequence in an ideal I

Can I embed M into some injective module N

Such that N/M has a length n - 1 regular sequence in a

Answer: I think no

But I don’t know how else to solve this ;(

Reason is I think if A is local and I = m, then any injective has depth 0 lol

[sleek thicket]

hmm

[tough imp]

So N/M would have depth 0

But I need this to do descending induction

[obtuse meteor]

sees commutative algebra

[sleek thicket]

Have u tried an example

[obtuse meteor]

flees

[tough imp]

Via LES 😎

[sleek thicket]

Lol

[tough imp]

I mean not really becuz

Wtf is arbitrary injective

It seems hard

[meager python]

”If $D(x) \subseteq D(f) \subseteq U$ then $(x) \subseteq (f)$ so $f$ divides $x$, which means $f = x$ or $f = 1$ (up to units)

[gentle osprey]

tolaria

[sleek thicket]

I'm thinning over Q

[meager python]

This isnt true btw

[tough imp]

Over Q?

[sleek thicket]

Oh yeah tolaria you're right, sorry

[tough imp]

Only module is 0

I mean only ideal

So like...

Regular sequence no exist

Haha

[sleek thicket]

It works out because I can replace work with the radical

Over Z Alex

With injectives being Q or Q/Z

[tough imp]

Oh

[sleek thicket]

The logic that D(x) = D(f) or D(f) = A^2 still holds, I think?

Since the radical of (f) contains the maximal ideal (x)

Err wait

No

That's wrong

hmm

(x) isn't maximal

I think the radical here should be the ideal generated by the squarefree part of f

Since we're in k[x, y]

So it should still work(?)

[obtuse meteor]

hey there's probably a way to construct a counterexample for n = 1

but we solved it using n = 2

so don't worry

:)

[sleek thicket]

It is still a counterexample I think, my logic was just off

But yeah "we" solved it for n = 2 haha

Thanks

[tough imp]

Wait so sham

Have you solved the sheaf on a base

For general category

[sleek thicket]

No it took too long

I stopped doing it

[tough imp]

Lol

Fair

[sleek thicket]

I see how to do it

[tough imp]

But have you now fully done Spec A?

[sleek thicket]

I did it on a whiteboard

not typing sheaf on a base up

And yeah chm

[tough imp]

Pog

[sleek thicket]

The argument max made is perfectly clear now

[tough imp]

I don’t remember it at all lmfao

But I’m glad it works

[sleek thicket]

Well we sort of talked about how it works

So because D(f) = Spec A_f it suffices to check the sheaf condition on global covers

and you can reduce to finite many distinguished opens covering it by compactness

[tough imp]

Right but I mean like

[sleek thicket]

This is saying that $0 \to A \to \prod_{i=1}^n A_{f_i} \to \prod_{i=1}^n \prod_{j=1}^n A_{f_i f_j}$ is exact

[tough imp]

I know the argument now thanks to you

[gentle osprey]

shamifold

[sleek thicket]

Oh okay lol

Sorry

[tough imp]

I just meant I don’t remember max doing this

[sleek thicket]

Ahh yeah

[tough imp]

I recall he said like

“Look up sheaf on a base”

[sleek thicket]

He definitely did because I wouldn't have come up with this lmao

[tough imp]

Then I thought it kinda ended at that

[sleek thicket]

Nah I remember faithful flatness

[tough imp]

Fair

I only remember that from like

Umm

[sleek thicket]

And after you brought it up I remembered the chain homotopy

[tough imp]

Spring

Also I like

Barely knew what flat was

Actually

No I didn’t know

Period

[sleek thicket]

yeah I had just come back from AM pilling myself

[tough imp]

I didn’t learn that really till spring break

[sleek thicket]

so I had the definition of faithfully flat fresh in mind

[tough imp]

Lol

How did you find that wiki link

On that complex?

[sleek thicket]

Twitter

Check my qt of the one you liked

[tough imp]

Ohhh

[sleek thicket]

Definition: A scheme is a locally closed subset of projective n space over an algebraically closed field

[uncut geyser]

Im high on coffee, but I must ask just to make sure Im not thinking complete bs. the index of a singularity of a vector field is the degree of the gauss map at the boundary of a nonzero disk around it. should I be able to just use the degree of the vector field itself instead?

[slow tree]

I've never worked with hyperbolic geometry before, so can someone help me with drawing (displaying with a computer) an infinite tree in the Poincare disk like on the left side of the cover of this book

[frigid patrol]

woah whats this

that cover looks cool af

I love putting integers inside of disks

[marsh forge]

If you dig through Andrew Putmans Twitter you can find software that does it

[bright acorn]

So uhh

There's an institution here where I live that each summer offers tons of free courses in lots of areas of mathematics.

I'm think about either studying one entitled ''fundamental group and recovering spaces'' or Functional Analysis.

I have a certain idea of what algebraic topology is, because I've read some of it in Munkre's Topology book.

But I'm not entirely sure if I'm prepared to study it yet.

[marsh forge]

A real first course in AT would require a decent understating of groups/rings/modules and point set topology

[bright acorn]

I've got a good understanding of topology because I've read Munkres' book already, but I've also used it a lot through real, complex and functional analysis.

So I'm quite used to it

But I'm not sure if I know algebra enough

I've only read a bit of Dummit and Foote's book

mainly the first chapter dealing with groups

that's my main theoretical knowledge of algebra

[marsh forge]

Yeah I’d say you want at least group theory and linear algebra down well

But rings and modules really are better

You can’t use any textbook directly without understanding rings/modules

[bright acorn]

even tho I know what's a ring and a module, I don't know the main results in the theory of these two algebraic structures.

[marsh forge]

You don’t need a very deep understanding I suppose

Nothing like AM

[fading vale]

idt you need rings and modules if you're just doing intro stuff to like

pi_1 and covering theory

[marsh forge]

I don’t count that

I think a course needs to cover (co)homology to be considered a “real first class”

And you can do cohomology over fields only

But that’s not accessible via any textbook I know of

[fading vale]

yea but theyre talking about a summer program called "fundamental group and recovering spaces''

[marsh forge]

Oh

Lol I missed that

[bright acorn]

yeah

[marsh forge]

Yeah group theory is sufficient

[bright acorn]

it's only a 2 months course

Ok so uhh

[fading vale]

u probably want to know what a free group and a presentation is

[bright acorn]

the content that they cover is the following:

[marsh forge]

I can’t imagine they won’t cover that

[fading vale]

it could be covered in the class tho

max my full semester of AT didnt even reach covering theory

[marsh forge]

That’s just an incorrect name

At that point

They only wanted to discuss a subgroup of the ideas

[fading vale]

pain

we spent half the time on point set lol

it was a little silly tbh he said he was sad he had to cover so much but i feel like if thats the case he could have assigned some of it as reading

[bright acorn]

Homotopy, Fundamental Group, Examples and Applications: fundamental group of the sphere S^n, projective spaces, classical groups, Seifert-Van Kampen Theorem, Covering Spaces, the relations between covering and the fundamental group, universal covering, fundamental group of compact surfaces, orientation double cover, other topics.

https://impa.br/ensino/programas-de-formacao/mestrado-academico/disciplinas-mestrado-academico/grupo-fundamental-e-espaco-de-recobrimento/

It's in portuguese, so I made a free translation

[fading vale]

[marsh forge]

Talking about orientation before introducing homology is broke

But yeah you should be fine

Tbh the group theory involved in AT is usually very basic

[fading vale]

i just started that section in hatcher max lol

[bright acorn]

I guess that they do that because they introduce homology in a course on Manifolds

[marsh forge]

Broke

But makes sense ig

[bright acorn]

and this course on Manifolds is mainly taught before algebraic topology in general

here

they have a lecture on the subject

[marsh forge]

You might want to take that first if they expect you to be familiar w objects like manifolds

[bright acorn]

https://www.youtube.com/watch?v=Lv3zcmtJ_XA&t=3850s

[marsh forge]

But courses in manifolds tend to talk too much about smooth stuff no one cares about

(this is sarcastic only I don’t care)

[bright acorn]

I have some understading of manifolds mainly because of A Comprehensive Introduction to Differential Geometry

Vol 1

[marsh forge]

Lol

[bright acorn]

Didn't read the other books

[marsh forge]

That sounds fine then

[bright acorn]

why ''lol'' tho? 😳

is it a good book?

[fading vale]

idk which book that is

who's the author

[marsh forge]

It was directed at ultra’s hodge theory comment

[bright acorn]

Michael Spivak

[fading vale]

tbh half the humor on this server is "lol what if [statement] but formulated in [more complicated theory than traditionally used]"

which is based because it indirectly exposes me to random shit

[bright acorn]

btw would it sound dumb if I asked what is hodge theory really about? I mean, I have heard it is some way to study cohomology groups on smooth manifolds using PDEs. But other people have said it somehow makes a connection between lots of stuff in algebraic geometry and differential geometry. For some reason I also think it has lot to do with complex analysis so I'm not sure.

[fading vale]

oh? i dont know anything about hodge theory so i wouldnt know

this is traditionally where you explain the basic notions behind hodge theory to me in an accessible manner btw

[marsh forge]

and me

[fading vale]

there are no jokes on this server only opportunities for sucking knowledge from the marrow

[bright acorn]

I mean, I'm a self studying person so I got introduced to a lot of concepts in mathematics because of dumb memes lmao

first time I heard Grothendieck was because of something related to him being like a mage or something.

not sure

Anyway thanks

[fading vale]

so am i

all the high schoolers on this server are more or less

[bright acorn]

Still not sure if I join the course on functional analysis or that one on algebraic topology

[fading vale]

ok all the active ones

(mostly)

[bright acorn]

I mean, I guess functional analysis is going to be easier maybe

[fading vale]

is there fake math in deleuze?

[bright acorn]

The main reference for the course is Peter Lax's book.

[fading vale]

oh no

[bright acorn]

and from what I know, the content of the course may be something in this line of reasoning:

https://www.youtube.com/playlist?list=PLo4jXE-LdDTTIIIRwqK35CbFJieSJEcVR

It starts with pretty basic concepts

Basically a reminding of basic concepts you study in linear algebra.

That's how Peter Lax's books starts btw

You think so because I feel more confident studying it?

Yeah, I've read about 100 pages of Peter Lax's book and it looks promising enough.

[river granite]

iirc these Functional Analysis lectures follow Lax's book very closely

[bright acorn]

yup

[bright acorn]

Yeah, I use these as a way to make sure I understand what I just in the book.

[blissful dune]

aight so

I have a quick question

what is orientation

ok that was phrased poorly

more like, how come orientation is a thing?

If changing coordinate systems doesn't matter and preserves the end result, how does the orientation of basis vectors matter

[ivory dragon]

from what perspective are you looking at orientation from?

it sounds like youre talking about orientation of a vector space?

[gritty widget]

you care about orienting vector spaces because it ultimately ends up being involved in integrating forms on manifolds and making sure that's well-defined

[rugged swan]

For me, it is the ability of a manifold to preserve the left and the right. You can define this rigourosly by saying that locally, you can chose a canonical left and right (think of a 2-dimensional manifold, locally it is R^2 and you can chose left and right), this is formalized with a sheaf. The ability to well orient the whole manifold is measured by taking the 1st cohomology group of this sheaf (or less abstractly, you can take some local coodinates cover of your manifold and compute the Cech cohomology of this cover). If the cohomology group is trivial, it means that you can correctly glue one by one open sets of your cover "orientationwise".
For instance it does not work on a Mobius strip. Take a, cover of two simply connected open sets. You can locally choose a basis, thus an orientations on both of them but you can't glue them properly. If it is glued properly on one side it wouldn't be glued properly on the other side ! You can see it by figuring out that if you do one turn of your Mobius strip, your left becomes your right.

@blissful dune

[tough imp]

Is there an abstract criterion for when a sheaf of modules is (quasi)-coherent over a (Noetherian) scheme?

I’m thinking of something like the various criterion for affineness of a scheme, not just “show locally it’s M~ for modules M”

[silent oasis]

Let $M$ be an $n$-dimensional orientable manifold with boundary $\partial M$. Lefschetz duality gives isomorphisms $H_i(M) \cong H^{n-i}(M, \partial M)$ and $H_i(M, \partial M) \cong H^{n-i}(M)$. Over the rational numbers, using the universal coefficient theorem we obtain a perfect pairing $H_i(M) \otimes H_{n-i}(M) \to \mathbb{Q}$. Consider the long exact sequence associated to the inclusion $\partial M \hookrightarrow M$, more specifically the resulting arrows $p_k: H_k(M) \to H_k(M, \partial M)$. I have read in multiple places that the Lefschetz pairings induce perfect pairings on $\operatorname{im} p_i \otimes \operatorname{im} p_{n-i}$, but I have been unable to prove this. I would appreciate if somebody who had more knowledge of the inner workings of Lefschetz duality could help me out here.

[gentle osprey]

Erder

[silent oasis]

https://mathoverflow.net/questions/27429/how-does-the-lefschetz-poincare-dual-torsion-linking-pairing-on-manifolds-with-b

The comment made by Tom Goodwillie on this MO post gives a sketch of a proof which works as follows:

1. Restrict the perfect pairing $H_i(M) \otimes H_{n-i}(M, \partial M) \to \mathbb{Q}$ to a pairing $H_i(M) \otimes \operatorname{im} p_{n-i}$.
2. This pairing factors through $(H_i(M) / \operatorname{ker} p_i) \otimes \operatorname{im} p_{n-i}$, hence through $\operatorname{im} p_i \otimes \operatorname{im} p_{n-i}$.
3. The resulting pairing is perfect.

I got stuck showing 2.
Let's call the pairing $b$ and consider it as a bilinear map, then what I tried to show was that $b(x, y) = 0$ for all $x \in \operatorname{ker} p_i$ and $y = p_{n-i}(z) \in \operatorname{im}p_{n-i}$. I noticed that by exactness of the long exact sequence, one has $\operatorname{ker} p_i = \operatorname{im} q_i$, where $q_i: H_i(\partial M) \to H_i(M)$ is the inclusion map in the long exact sequence. Since the pairing on homology morally counts intersections of chains, it is sensible that $b$ should vanish on all pairs of the form $(q_i(w), p_{n-i}(z))$, as the former is represented by a chain which lies in the boundary of $M$, while the latter is represented by a chain in $M$ which avoids the boundary. However, I don't know if one can literally interpret the pairing on homology like that, let alone how one would prove this.

[gentle osprey]

Erder

[silent oasis]

corrigendum: Pairs of the form $(q_i(w), p_{n-i}(z))$

[gentle osprey]

Erder

[blissful dune]

Oh that makes sense thanks

[blissful dune]

Yeah I am

[quartz rover]

this paragraph is causing me great strife

the whole thing

P is a planar curve

or, well, a plane

what exactly are these tangent vectors

like what sort of object or action do they correspond to with respect to the plane P

and why does the component with the basis moving in the x direction get ascribed a coefficient dy

instead of dx

here's an illustration

it doesn't really help me with this point

[blissful dune]

wait hold on

can you clarify your question

wdym what are they

They're the derivatives of the map

[quartz rover]

ok let me step slowly

because my brain isnt latching on well to this construction

so we need a basis for the tangent space at a point p

[blissful dune]

yes

The basis is then defined as the derivatives of the map

[quartz rover]

the tangent space is just all arrows starting at the point p

the derivatives in the direction of each coordinate are a natural easy way to get a basis for the tangent space because they are localized to the point p

and they span the entire space

[blissful dune]

Yeah exactly

You just define those as the basis

[quartz rover]

ok so now

i have

wait wait dont jump ahead this is the part i am hating

i have the two objects

[gentle osprey]

kwadoo

[quartz rover]

and vice versa

now

[blissful dune]

ok

[quartz rover]

what exactly is this, first of all

[blissful dune]

that's the derivative of the point

[quartz rover]

in what sense

[blissful dune]

so if you have a map, a(t)

[quartz rover]

a map from a parameter t

[blissful dune]

a(t) = (x(t), y(t))

[quartz rover]

to a point on the plane

[blissful dune]

yeah

[quartz rover]

and you are just drawing

2 coordinate curves

[blissful dune]

at that specific point, (x, y)

[quartz rover]

with origin at p

one horizontal

one vertical

[blissful dune]

you take the derivative and define it as a vector

[quartz rover]

wait

you take the tangent vector

to the coordinate curves

[blissful dune]

yeah

[quartz rover]

ok

how does that relate

to the derivative of a function on the plane

at a point

[blissful dune]

ok so

the tangent vector to a map

a(t)

is

[quartz rover]

a'(t)

[blissful dune]

$$a'(t) = (x'(t), y'(t))$$

[gentle osprey]

Merry Chief

[blissful dune]

this is a vector

[quartz rover]

yep

ive done some basic diff geo

[blissful dune]

so the tangent vector @ t = t1

is

a'(t1)

[quartz rover]

yep

[blissful dune]

yeah

that's it

[quartz rover]

a'(p) = (x'(p), y'(p))

[blissful dune]

yeah

[quartz rover]

so

x and y are

coordinate functions

[blissful dune]

the tangent vector IS the derivative at that point

[quartz rover]

wait... processing

[blissful dune]

yeah

[quartz rover]

processing

processing

let me stare off at the wall for a few seconds brb

oh my god

[blissful dune]

do you get it lol

[quartz rover]

it was right there under my nose the whole time

[blissful dune]

yeah exactly lol

[quartz rover]

i was already doing this

[blissful dune]

yup

[quartz rover]

so you can like

ohhh frick

tangent space is kinda like phase space

[blissful dune]

I mean, if you want to think of it like that, sure?

[quartz rover]

like

im trying to consider how this works like

if you embed a curve in R2

the tangent space is only defined at points along the curve right

like theres no ambient notion of a tangent vector at all

[blissful dune]

yeah

[quartz rover]

ok

so lets see

[blissful dune]

it's only defined at points along the curve ig

[quartz rover]

what happens when u deform tangent curves

hmmmm

[blissful dune]

🤷‍♂️

the tangent vectors also transform

and become different vectors

[quartz rover]

yea the original curve is just kinda like the integral of the tangent curve

ok

ok ok

[blissful dune]

yeah i think you were just overthinking it

[quartz rover]

well see

i havent been properly introduced to this so

[blissful dune]

yeah that makes sense

[quartz rover]

hmm ok next piece

now what exactly are we doing here

dx * y'(p) + dy * x'(p)

why is this the way the coefficients are plugged in

[blissful dune]

ok so

I'm not entirely sure, but

wait no I think I might get it

I think it's because of how you write it out

like when you want to get dy/dx it affects the sign order

I think

[quartz rover]

like i see the picture and

it makes sense

[blissful dune]

honestly I'm not entirely sure on this one

[quartz rover]

and i see the derivative

and it makes sense symbolically

[blissful dune]

yeah i get what you mean

[quartz rover]

and i see plugging in the result of the symbolic execution to the visual graph

and it makes sense

[blissful dune]

yeah im also trying to figure this one out

[quartz rover]

but my god it does not make sense when you look at the coordinates and the curve together

[blissful dune]

🤷‍♂️

[quartz rover]

x'(p) gives a vector pointing horizontally

or wait

does it

[blissful dune]

idk just think about it for a couple hours and it'll make sense

[quartz rover]

NO

it does not

it's the gradient

[blissful dune]

yeah it is

it's the gradient

[quartz rover]

so it points normal to the x coordinate

[blissful dune]

OH

[quartz rover]

LOL

[blissful dune]

You're right!

[quartz rover]

that sucks so bad

[blissful dune]

lmao yeah you're right

thanks for that

[quartz rover]

holy hell i hate it

what even

why is x'(p) vertical

so... dy is how much y is changing proportionally to the x component of the gradient?

wait. recoup

what exactly is d(x+t, y)/dt

it's just a coordinate curve

well

a unit vector pointing in the positive direction

which is colinear with the coordinate curve

so that's

friick

that's... a horizontal vector wrt x and y

i.e. horizontal in R2

but it's vertical in the tangent space

oh good god

so really we can define a tangent space as a property of the ambient original space

we just cant ascribe any actual vector to it

unless a curve lies at that point

and indeed for curves which cross over themselves at points q

they will have multiple tangent vectors at the point q

oh shiiit

so theres no restrictions for how many vectors 'live' within the tangent space at a point

but I'M STILL CONFUSED. WHY is the coordinate curve for x vertical in the tangent space?

unit vector along the coordinate curve

= x'(p)

= <1,0>

ok the coordinate curve for x is not the same thing as the coordinate curve for dy

that much is now clear to me

differentiation spits out this like

thing

in tangent space

if we parametrize the x coordinate curve by t s.t. we get (x+t, y)

then the tangent vectors to each point along this curve

are given by dy

wat

this is nonsense

mathematicians are trolling

this is a no-fly zone for rational explanation

i think u just have to accept this point and move on without really deeply understanding the mechanics of this transformation

tterra please help

why on earth

when i differentiate the coordinate curve for x

do i get dy

why is dy a tangent vector to (x+t, y)

at constant x and y

[sleek thicket]

what's the question?

[quartz rover]

lemme latex it

bachman constructs the coordinate system of the tangent space to a plane at a point p as:

[gentle osprey]

kwadoo

[quartz rover]

where dx and dy are the components of a tangent vector

oh and

[gentle osprey]

kwadoo

[sleek thicket]

this notation makes no sense to me, sorry

[quartz rover]

same

[gritty widget]

let me read it

i had to get up

[sleek thicket]

I don't understand where t is coming in here

good morning ttera

[quartz rover]

they just shoehorn in a t

[sleek thicket]

Merry Christmas if that's your thing

[gritty widget]

that's tomorrow i think

i got a copy of irm for christmas 😌

merry christmas #point-set-topology

[quartz rover]

to get a parametrization for a coordinate curve

yea but not all are alive at a given time

or in use

or whatever u want to say

[sleek thicket]

Okay so how about this

Do you have a curve?

[quartz rover]

sure, take a parabola

[sleek thicket]

So you're not talking about the tangent space to the plane

But to the curve?

[quartz rover]

well, in general, to a point on the plane

i.e.

[gentle osprey]

kwadoo

[sleek thicket]

But then what is t lol

[quartz rover]

a parameter for the coordinate curve

for x or y

here's the source text

[sleek thicket]

Oh

This makes sense

[gritty widget]

ah, so the tangent space here is given by the tangent vectors to curves passing through that point. given a point p = (x, y), you want to come up with a "canonical" basis for T_pR^2. to do so, you want curves that go through (x, y) that give you things that look like e_1 and e_2, and (x+t,y) and (x,y+t) do the job

[quartz rover]

ye, right

let's see

yes, agreed

so now

why is dy paired with (x+t, y)

LMAO

[gritty widget]

[quartz rover]

like it checks out in the picture

but

WHY does it check out

cuz symbolically it confuses the hell outta me

[gritty widget]

what do you mean by paired with?

[quartz rover]

this

[gritty widget]

hmmmmmmmmmmmmmmmmmmm

[sleek thicket]

That feels like a typo? The dx <0,1> + dy <1,0> does

[quartz rover]

nope

[gritty widget]

ya it feels like it's a typo tbh

[quartz rover]

wait lemme think

are u serious

wut

[sleek thicket]

Read the next line

They say dx(<2,3>) = 2

And dy = 3

[quartz rover]

umm

[gritty widget]

[quartz rover]

are you serious

did they really make that big of an error

[sleek thicket]

Yeah haha

[gritty widget]

it happens

[quartz rover]

holy fuck that sucks

[sleek thicket]

Well it's a 2 symbol typo

[quartz rover]

i lost 2 whole days of thinking on this

[gritty widget]

:(

f

[sleek thicket]

lol my textbook next quarter is NOTORIOUS for typos and I'm scared

I've lost days on it before (with this book)

[quartz rover]

ok so

[gritty widget]

good luck.

[sleek thicket]

f u weibel

[quartz rover]

you guys have done differential forms before right like

[gritty widget]

i like to think i am comfortable with them

[quartz rover]

this tangent coordinate system is not complex

like

you're sure the transformation to the tangent space

or whatever you'd call it

can be thought of as

dx * d(x+t, y)/dt

and same for the other component?

[sleek thicket]

Every tangent vector in R^2 can be written uniquely in this form

[quartz rover]

ok good

i will scribble out the line in the book, thanks so much

[gritty widget]

sorry you had to spend so much time on the typo 😔

[quartz rover]

hehe

[gritty widget]

always sucks

[quartz rover]

i'll post a correction to it in my review after i'm done

[gritty widget]

does it have an errata?

[quartz rover]

idk but i usuallly do goodreads reviews

lemme search for one

[gritty widget]

just look up "author (book) errata"

[quartz rover]

couldn't find any

[gritty widget]

make one

[quartz rover]

will do cap'n

[long hornet]

I hope this question is suitable here: How transitive is the action of Homeo R on R?

I know it's at least doubly transitive (we can construct a linear function...)

Also, is Homeo R isomorphic to a familiar group?

[small phoenix]

it's probably not triply transitive -- i'm sorta guessing that means any two ordered triples are connected by a homeomorphism, yes? seems false since a homeomorphism should kinda-sorta preserve ordering.

for instance, i doubt there is a homemorphism which sends (0,1,2) to (0,2,1)

[uncut surge]

Yeah, that's clear since a homeomorphism of the reals needs to be strictly monotonous

But at least if you restrict to pairs (a_1,..., a_n) and (b_1, ... , b_n) which are both "equally" ordered, i.e. there is a permutation of the (a_1,..,a_n) which strictly orders both the a's and the b's simultaneously, then I think homeomorphisms act transitively on those, independent of n

You can show by constructing piecewise linear functions, for example. I guess that's as strong as it gets in terms of transitivity

[fading vale]

what exactly are the natural maps in this case?

[tough imp]

The natural ones?

[sleek thicket]

Can you remind me what | means moth? Is it H_n(M, M\{x})?

If so we have a map of pairs (M, M\B) -> (M, M{y}) given by the identity

[fading vale]

yea

[sleek thicket]

does that make sense then ?

[fading vale]

kind of?

do you know where i could see this done for RP2

[sleek thicket]

Which part?

Like showing it's not orientable?

[fading vale]

yea kinda

like i understand why we have maps from H(M | B) -> H(M | y) it makes sense that theyre both Z but its confusing to me concretely how this relates to the choice of generator

is this supposed to be like we have a ball around each point x such that all the y in B have the same choice of generator?

all 1 or all -1?

[sleek thicket]

Yeah exactly

That's the idea

[fading vale]

if ur map from H(M | B) -> H(M | y) is always the identity then i dont get how else the local consistency condition can be fulfilled

oh

ok hahahaha

[sleek thicket]

Well it's not the identity

[long hornet]

That makes sense but I can't pin down the reason why. Is it because R has the order topology?

[fading vale]

hmm

what is it then

train wifi moment 😭

[sleek thicket]

It's induced by the identity on the space M

But it's not the identity map of pairs

Sorry moth I lost connection haha

[fading vale]

i did too lol

so its the map from (M, M \ B) -> (M, M \ y) that acts as the identity on the first and i guess the embedding on the second?

[uncut surge]

A homeomorphism on the reals is in particular a bijective continuous map on the reals; if it was not monotonous, then by continuity, we can use the intermediate value property to find two elements x,y, mapping to the same f(x) = f(y)

[fading vale]

and it induces a map on the relative homology?

[sleek thicket]

Yup!

[fading vale]

does this map end up as the identity on homology?

cuz if it doesnt then i think the local consistency condition is more complicated than all the same choice of generator, isnt it?

ugh i dont have paper i have to imagine the diagram

[uncut surge]

I don't know much about the order topology but it might very well be related; I'm guessing that equipping a space with an order topology makes it possible to prove intermediate value theorems on it

sry 4 mixed conversations don't get distracted bbygirls

[sleek thicket]

Moth I don't know what you mean by the identity on homology

[fading vale]

like the identity map from Z to Z lol

as groups

[sleek thicket]

It's a map between two different groups tho

[fading vale]

slfj i mean like

it sends the 1 of the first infinite cyclic group to the 1 of the 2nd infinite cyclic group

[sleek thicket]

But there's not a unique 1

[fading vale]

oh ugh i see yea

we can pick 1 and -1 arbitrarily

[sleek thicket]

The whole thing about orientation is that infinite cyclic groups have a nonunique generator

[fading vale]

yea

[sleek thicket]

So heres what you can say

We have maps Hn(M|x) -> Hn(M|y)

For x, y in B

[fading vale]

mhm

[sleek thicket]

We have a given choice of generator on these groups

[fading vale]

right

[sleek thicket]

If you use that generator to identify each with Z then it's the identity

But I think this is more confusing than the original condition lol

[fading vale]

i kind of get it i think

okay mhm

i think i just need to see an example here cuz its not super clear to me how this all relates to the topology of any given space

like i dont get why exactly we cant do this for RP2

or why we can for S^2

[sleek thicket]

good question, lemme think about it

I tend to think of orientation in terms of smooth structures which isn't useful here

[fading vale]

yea

the way im thinking abt it rn is super like

im just imagining arrows at each point sjflksf

[sleek thicket]

Yeah that's called a tangent space :P

[fading vale]

mhm

it kind of works if you think about it like

for each point you have a sphere with arrows pointing outwards at each point

and then rotating will keep arrows pointing outwards while reflecting will cause them to point inwards

[sleek thicket]

oh yeah so you're actually onto something here lol

If M is an oriented riemannian manifold and M' a hypersurface in M, and we have a normal vector N to M' then we get an orientation on M'

[fading vale]

idk what that means but sounds cool

[sleek thicket]

so somehow the orientation on the sphere comes from the fact that we can choose a smoothly varying normal vector

[fading vale]

is the hypersurface here like

an S^n-1 in S^n

[sleek thicket]

Yup

[fading vale]

mhm

[sleek thicket]

Codimension 1 submanifold

[fading vale]

right

this makes some intuitive sense actually

like no matter how you lay out your arrows on RP^2 you can expect to see a problem where like if you pick a point there should be a ball around it where half are pointing "out" and half "in" i guess

you kind of need some notion of direction though ugh like

hmmmmmm

[small phoenix]

locally, you can orient any manifold as it's locally euclidean. the problem is extending local orientations over the whole manifold.

[fading vale]

yeah i understand the general idea its just hard for me to see how local consistency relates to the topology of the manifold

like i can kind of see why RP^2 cant be oriented if i fudge everything and think abt arrows but

hh

itd be really nice to see an example of an orientation put on like idk the sphere

in terms of the local orientations as generators of homology and such

tterra moment

[small phoenix]

i think you can see that there is a chain of open sets which sorta returns to itself, but in the journey, the local consistency requirement violates the global condition.

maybe think about the mobius strip.

[fading vale]

im not sure what you mean by a chain of open sets here

[sleek thicket]

(you can phrase the whole local orientations+local consistency business in terms of a sheaf)

[fading vale]

oh god

[sleek thicket]

Don't worry about it lol

[small phoenix]

yeah, the orientation sheaf.

[fading vale]

algebraic geometry arrives from the future

[sleek thicket]

I do not think it would be helpful to introduce the orientation sheaf here

[gritty widget]

tterra and not knowing AT name a more iconic duo 🐺

[sleek thicket]

Okay moth how about this

Take a generator β of H_2(S^2)

You should be able to cook up an orientation from this

[fading vale]

is our ball around any point S^2 - x?

[sleek thicket]

There's a map H_2(S^2) -> H_2(S^2|x) for each x yeah?

What ball?

[fading vale]

for the local consistency condition dont you need an open ball B within a neighborhood of each point

[sleek thicket]

Yeah but we need to define the orientation at each point first

[fading vale]

and the nthe natural maps from H_2(S^2 | B) -> H_2(S^2 | x)

um ok

[sleek thicket]

well like, we can't check consistency if we don't have generators at each point to begin with

Yeah?

[fading vale]

right

that makes sense

[sleek thicket]

So we're gonna take a global generator of the homology

[fading vale]

mhm

[sleek thicket]

and then restrict to each point, sort of

Like we have a map Hn(S^2) -> Hn(S^2|x) right?

Like in LES of a pair there's a map into the relative homology group

[fading vale]

do we let the local homology at x be the image of beta under the map induced by (S^2, null) -> (S^2, S^2 - x)?

ok thank you wifi very cool!

there we go

right yea

[sleek thicket]

Yeah

We do

[fading vale]

woke

[sleek thicket]

These are both infinite cyclic groups tho!

[fading vale]

mhm

and its an isomorphism

so betas image has to generate Hn(S^2 | x) right

[sleek thicket]

Right

so the idea is now that if you take a ball around any point the induced maps Hn(S^2|x) -> Hn(S^2|y) will preserve these generators

Since they're all coming from a global generator β

[fading vale]

i think that makes sense

like if you pull out the sequence the induced map on homology gets u a commutative diagram

between the two LES

yea this makes sense i feel

[fading vale]

i gtg now for a couple of minutes but ty! i think i get it much better now

[sleek thicket]

So this commutes by functorality

[fading vale]

yes!

like that

[fading vale]

hmm

lemma 3.27 is used to prove thm 3.26

it says that part (c) follows immediately from part b of the lemma but im not rly sure how

i get that we get isomorphisms H_i(A; R) -> H_i(M; R) but im not sure how we're supposed to know that H_i(A; R) = 0

wait nvm u can let A = M

oops

what does the sum of sections and scalar multiply of section actually mean here

i guess i kind of see it

if we have f : x |-> a_x and g : x |-> b_x the sum is just the normal sum of two functions right?

and likewise w/ scalars

[quartz rover]

this book just kind of says random things and doesn't justify them at all

if we think evaluating a 1-form $w$ on a tangent vector $v$ as a dot product of the two, the book says this is equivalent to projecting $v$ onto a line $l$ which has a basis vector $\frac{w}{|w|^2}$

[gentle osprey]

kwadoo

[quartz rover]

and it says that using that particular basis vector makes the projection ( or the 1-form, rather) coordinate independent

how can this possibly be

if i dot w and v i get

[gentle osprey]

kwadoo

[quartz rover]

if we represent a vector along the line l as

[gentle osprey]

kwadoo

[quartz rover]

and i try to get w * v to relate to this parametrization of l

well first getting a scalar out of the line i get

[gentle osprey]

kwadoo

[quartz rover]

now what does it mean to project v onto l

hmm I guess we want to find t for the projection

$w \cdot v = |w| \cdot \text{comp}_l v$

[gentle osprey]

kwadoo

[quartz rover]

if we take $|l(t)|=\text{comp}_l v$ then we get...

[gentle osprey]

kwadoo

[quartz rover]

$w \cdot v = t$

[gentle osprey]

kwadoo

[quartz rover]

o_o

ok i guess that qualifies as coordinate independence lol

but what a wacky way to arrive at that

[fading vale]

these are the worst words in the english language

[frigid patrol]

Hatcher

[tight agate]

Pretty neat article:

https://www.ams.org/journals/bull/2007-44-04/S0273-0979-07-01178-0/S0273-0979-07-01178-0.pdf

[uncut surge]

@fading vale but it divides a polygon up into triangles by forming new triangles with the barycenter, subly

i think the name is fair (even tho i also don't like that you call it "barycenter" in english)

[chilly rune]

Anyone got a good method for computing degree of the followong map CP^1 to CP^1 where in affine coords we send z to f(z)/g(z) where f and g are polynomials of degree n and m respecitvely

Doing it by local degree seems quite tedious

Was wondering wether you could somehow use the zero dim cohomologies

And then poincare duality

The other way sorry

0 dim homologies

[sleek thicket]

unrelated but this is adorable https://twitter.com/AudreyRosevear/status/1342483539376353283?s=19 (check the replies too, Hatcher actually gave her mom a gift idea)

Audrey🏳️‍⚧️ (@AudreyRosevear)

My mom emailed Allen Hatcher for gift ideas 😭😭😭

Likes

185

Twitter · Dec 25, 2020, 2:53 PM

Also computing local homology doesn't seem that bad to me? I think you can compute it at the zeroes of g since then you're sort of mapping into the plane

It's been a while so that might be wrong lol

[gritty widget]

pain

🥖

[fading vale]

[gritty widget]

first displayed equation is making me thonk a little nvm that's what i get for not reading every part of the long-ass chapter 2

but whatever

[gritty widget]

ok nvm that wasn't that painful

😌

[sleek thicket]

lol

[uncut surge]

the only thing to remember about the stereographic projection is that it exists, nicely

and the picture ofc

[sleek thicket]

I disagreeeee

[uncut surge]

what else tho

OH i guess if you're into conformal things then you might wanna work with it all the time

[sweet wing]

[sweet wing]

yes

[red garden]

any finite T_1 space is discrete

proof: let X be a finite T_1 space

let A be a subset of X we wish to show that A is open in X

to do this we can show that A' is closed

now A' is finite --> A' = {a_1,a_2,...,a_n}

decompose A' into the singletons a_is

each one of them is closed ( as this is a T_1 space )

A' is the finite union of closed sets --> A' is closed

--> A is open

is this right

[long hornet]

I don't understand the A' step and don't see why it's necessary. We already have A as a union of finitely many closed sets (singletons), so it is closed. Since all sets are closed, all sets are open.

[red garden]

uh oh

okay

yea

yea sorry

tysm

[red garden]

is there anything between

quotient spaces and measure theory

[vocal wharf]

what

[marsh forge]

what

[cloud owl]

watt💡

[red garden]

like if i know measure theory

do i see quotient spaces in another way

will they be better understood

idk i was talking to the prof about the book not having quotient spaces

and he just went quotient spaces are importnat but they will be better understood if you learn measure theory

and i was just wondering what could be there

does any1? if any

[thin bramble]

@gritty widget how many math courses are you taking? Because you are also in the complex variables channel.

[gritty widget]

but i only shitpost in there

anyways, four