point-set-topology

[rugged swan]

here, we have A1 = [0,1/2] and A2 = [1/2,1]

(I'm trying to state a generalization)

[sleek thicket]

Yes in the finite case

Yes in the locally finite case, iirc

(meaning every point had a nbhd that intersects finitely many members of your cover)

No in general

Take any T0 space and a function out of it. It's continuous on each point set {x}, and these form a closed cover of your space

But in general they won't be continuous

Have you heard of the gluing lemma? @rugged swan

Suppose A, B cover X and A\B is contained in Int A, B\A is contained in Int B

[rugged swan]

nbhd ?

[sleek thicket]

Oh sorry neighborhood

Open set

[rugged swan]

ok

[sleek thicket]

Anyways under the conditions there, a piece wise continous function on A, B is continous, and the conditions hold if A, B are closed

[rugged swan]

B cover X means that X subset of B ?

[sleek thicket]

No

[rugged swan]

I didn't studied topology in general yet (only in metric spaces) and I'm learning algebraic topology x)

[sleek thicket]

A, B cover X mean that they're subsets of X whose union is X

You should probably learn more general top. before AT

I mean I skipped most of it

[rugged swan]

oh I didn't see the "," lel

[sleek thicket]

I didn't bother with all the separateness or tychonoff or anything

But you still need basic topology

[rugged swan]

I know how topology works but it's technical

[sleek thicket]

Yeah it is lol

But you need to know basic theorems in topology

Facts about connectedness/local connectedness/compactness/that stuff

The book Topology and Groupoids does general topology with a view towards AT

[rugged swan]

mh

I'll look at it

I have a topology cours in the second semester but I can wait till the course lel

[sleek thicket]

Sure

Just don't rush to get ahead of yourself

[rugged swan]

who wrote your book ?

[sleek thicket]

Ronald Brown

[rugged swan]

ok thx

oh it's a free book nice

[sleek thicket]

Yup

[rugged swan]

btw I didn't hear of the gluing lemma

[floral gust]

Is this correct for symmetry?

[rugged swan]

yes

[floral gust]

Idk why but a paper suggested that the desired homotopy is g(s,t) =h(s,1-t) so was kinda confused

[rugged swan]

this depends of your definition of homotopy

the idea is that you can transform continuously your first path to your second path

the "transform continously" is in your first variable here

because you ask for h to satisfy h(0,t) = y(t)

and h(1,t) = y'(t)

(by varying your first variable, you go from y to y')

in the paper you read, the author's definition of an homotopy was h(t,0) = y(t), h(t,1) = y'(t)

btw, it is more common that the "transform continuously" variable is the second

I think.

All papers I've read use the second variable definition

[sleek thicket]

@rugged swan the gluing lemma is what I stated above

[rugged swan]

oh ok

that's inuitive

but for what it is used for ?

[sleek thicket]

The thing you started out asking about

[floral gust]

Here is an example of the lemma

[sleek thicket]

^

The thing you originally asked about is false if I is infinite and follows from the gluing lemma when I is finite

[rugged swan]

ok ok

[floral gust]

What would f_hat will be in this case?

[marsh forge]

Lifting lemma who

Do you have the lifting criteria

To get f# from S1 to R

[floral gust]

no 😦

[rugged swan]

@sleek thicket w8

your lemma is weird

if I take A = ]0,+oo[

B = ]-oo,0]

and X = R

we have A\B = A subset of Int(A) bc A is open

but B\A = B isn't subset of Int(B)

bc Int(B) = ]-oo,0[

[sleek thicket]

Yes? That's means it doesn't apply

And that's good

Take f to be x on B and x+1 on A

[rugged swan]

I didn't understand what's the exact statement of your lemma then

I've read that if AUB = X and A\B subset of Int(A), then B\A subset of Int(B)

[sleek thicket]

No

The condition is that all three of those hold

And the result is that you can define continuous functions piece wise

[rugged swan]

oh ok

[sleek thicket]

Meaning that for any continuous f defined on A and g defined on B which agree A intersect B, there is a unique continuous function h defined on X which equals f on A and g on B

Does that make sense?

[rugged swan]

yep

[floral gust]

Is a path just a homotopy of points?

[rugged swan]

lol

you can see it as a function h(u,t) = y(t) yes

where h(_,t) is a path for fixed t

(this is a constant path)

[sleek thicket]

@floral gust yes, this is a nice way to think of it because it generalizes to higher homotopies

We can think of points in X as functions from a one point set into X

So a path is a homotopy between two of those functions, with chosen start and end

And then a path-homotopy is a homotopy between two of the paths with fixed start and end paths

And so on

[marsh forge]

Lmao

A hypothesis on my midterm

“Let X be a space where cover theory works”

[sleek thicket]

lmaooooo

modd

*mood

[sleek thicket]

locally path connected connected semilocally simply connected

Did I miss anything?

[marsh forge]

Sometimes you want to throw in hausdorff when doing cover stuff

But yeah that’s normally sufficient

For like universal cover etc

[sleek thicket]

I have a dumb q

Let B^n be the closed n ball

We have a map F : B^n -> R^(n+1){0} given by F(x1,...,xn) = (x1,...,xn, sqrt(1 - Σxi^2))

And a quotient map q : R^(n+1){0} -> P^n

Is the composition q ° F a quotient map?

Oh wait it's not surjective is it

For some reason I thought it would be but now I doubt

It surjects onto the copy of P^(n-1)

It is

Geometrically it's image is a paraboloid

Which intersects every line at least once

Let p : X -> Y be a quotient map and A a closed saturated subspace of X. When is Y the adjunction space of X and p(A)?

It holds in the case where A is the union of all nontrivial fibers

Which is what I nerd

*need

I figured out how to show that P^(n+1) is obtained by attaching an n+1 cell to P^n

By general results about pushout squares

Feelsgoodman

[floral gust]

Why do we denote loops based at point p in X by Ω(X,p) instead of Ω(p)? Doesnt the fact p is in X makes it much more useful to use the latter?

[marsh forge]

Uh

Consider A subset X both containing p

What’s Omega(p)?

Tbh it’s much more common to see p dropped

And just use OmegaX

With basepoint being implicit

[floral gust]

How does one reads this diagrams

[marsh forge]

Uh the top and bottom

Are like the end results of homotopy

And the middle describes the homotopy

[marsh forge]

Hi

Does anyone here like, fully understand spectral sequences

I’ve been like bouncing them around, and I get the general concept

But the specifics of how the differentials/gradings/resultions work out

Has been elusive for me

@ me if you are willing to ELIU for me

Actually I’m just gonna write stuff

OK

So

what do we have

A bunch of Z-bi-graded pages E^r

W differentials

Such that $d^r: E^r_{p,q}\to E^r_{p-r,q+r-1}$

[gentle osprey]

MaxJ:

[marsh forge]

That is, they are diagonal bios

Bois*

Ok

So we want $E^r_{p,q}\cong H(E^r_{p,q})$

[gentle osprey]

MaxJ:

[marsh forge]

Where we define $H$ in the obvious way using the differentials (grading H doesn’t really make sense here I guess bc it would be a nasty formula with the bigrade a

[gentle osprey]

MaxJ:

[marsh forge]

If $E^r_{p,q}$ stabilizes we call it $E^\infty_{p,q}$

[gentle osprey]

MaxJ:

[marsh forge]

Jesus this is so much data

I should take like, a few hours this weekend

And write up a nice set of notes on this stuff

[midnight jewel]

is the plural of complex complices or complexes?

[vocal wharf]

is this a joke question

it's complexes

[sleek thicket]

This stuff sticks

*sucks

Like a year from now I'm going to have learned spectral sequences and think they're great

But they just look at annoying

[gritty widget]

Can someone help me

Side length of square is 2cm. I need to find the area of the shaded area

Vvithout calculus

[sleek thicket]

Go to #geometry-and-trigonometry

[gritty widget]

Can you do it

[midnight jewel]

is this a joke question
no, because

1. matrix - matrices and index - indices exists and are commonplace
2. I didn’t look up the etymology of complex to realize it was actually complexus in latin; I assumed it was complex all the way down

[marsh forge]

I have trouble learning them

Bc I don’t have any real reason to rn

No immediate applications or paper to write or anything

I feel like machinery needs to be used to be truly understood

[sleek thicket]

I'm planning to do sheaf and group cohomology this year (definitely sheaf, maybe not group) so I'll find a reason

[sleek thicket]

AG prof blew my mind

Apparently the cocycle condition is a shadow of a simplicial object

[dawn sable]

im taking a course on manifolds using Lee's smooth manifolds book next sem

what should i read beforehand

[dim meadow]

Lee's smooth manifolds

Kidding aside, probably spivak's calculus on manifolds

[dawn sable]

o I had thought that would a higher level than Lee's book

[dim meadow]

No

The opposite

[dawn sable]

okay fantastic

ty

oh he also has a series in differential geometry

[dim meadow]

Yeah the first book of that series is roughly equivalent to Lee's smooth manifolds

And Spivak's calculus on manifolds is basically a precursor to the first book of that series

[dawn sable]

i see

[floral gust]

Can somebody explain how this proof works? I know that a group isomorphism = bijective homomorphism. Since the function is continuous, it induced a homomorphism. How does the above proof convinces that it is also bijective?

[marsh forge]

That’s one equivalent def

[bitter yoke]

Functions that have an inverse are bijective

[marsh forge]

The much more natural definition of isomorphism

Or homeomorphism

Or whatever version of iso you want

Is that it’s invertible by a function of the same type

So the fact that the homomorphisms has a (full) inverse

Tells you that it’s an isomorphism

This is very important for understanding situations like this, the general term is that pi1 is functorial

And this is how we think about isomorphisms in categories

[floral gust]

So the proof should just require one to write that since the function is homeomorphic it admits an inverse, and since it is continuous it induces a homomorphism. Thus it induces an isomorphism. Would this be correct?

[marsh forge]

One last edition

Observe that that the induced homomorphism of identity is identity

And so the inversion property is preserved

Just getting bidirectional homomorphisms isn’t quite sufficient, and in nonfunctorial settings would actually be false

[floral gust]

How would I show the independence

[midnight jewel]

if you can show it’s an integer then showing it’s continuous would do it (haven’t thought through it tho)

like show the righthand side is continuous in t

and always an integer

[sleek thicket]

Showing it's an integer is just like when is e^a = e^b right?

Oh lol I read it as "independent of choice of lift"

[midnight jewel]

wouldn’t it depend on the covering map tho

[sleek thicket]

Which is harder

I assumed the covering map was the exponential

[midnight jewel]

I assume there’s some underlying assumption that the cover is like $\R \ni t \mapsto e^{2\pi i t} \in S^1 \subset \C$?

[gentle osprey]

Sascha Baer:

[sleek thicket]

Yeah that's what I assumed too

Because that gives a good definition of degree

[midnight jewel]

btw is that the same degree as in degree via homology?

[sleek thicket]

Yeah

I'm pretty sure

[midnight jewel]

or, well, ends up being the same

[sleek thicket]

I don't really know AT well

But I believe so

[midnight jewel]

I am really struggling with AT because I don’t have the time/energy to read up on the things I couldn’t follow in class

I’m like… following the general ideas but failing to understand a lot of the details

[sleek thicket]

@QuickMaffs we need a little bit more context

[midnight jewel]

and I just know it’s gonna come back to bite me

[sleek thicket]

That's where I'm at with AG right now

I need to do a bunch of really messy stuff this weekendb

That the prof was like "okay do this on your own"

Or else I'll get totally lost

[midnight jewel]

well this week I’m mostly preparing for a midterm in another subject so even less time

and diffgeo+TAing is eating up the rest of my time

[sleek thicket]

All of that is also me!

That wasn't a sentence lol

I need to grade rn

[midnight jewel]

also I disagree, four dimension better

[sleek thicket]

And understand the classification of surfaces

And so an algebra take home midterm by Wednesday

[sleek thicket]

And do a computability theory mditsrm tomorrow

[floral gust]

This is the full question

[sleek thicket]

*monday

Yeah so @floral gust can you show it's an integer?

[floral gust]

Yeah I have shown that it is an integer

[sleek thicket]

Okay, well let g(t) = f'(t+1) - f'(t)

Where f' is the lift

This is continuous on R

And is integer valued

Right?

[floral gust]

yep

[sleek thicket]

What can we say about g from that?

[floral gust]

Yeah I can see that g must be constant everywhere but a proof doesnt come to mind

[sleek thicket]

Prove that the following is equivalent to connectedness of X:

All functions from X to a discrete space are constant

[midnight jewel]

continuous

[sleek thicket]

all functions are continuous fam

[midnight jewel]

you’re continuous

[sleek thicket]

I won't even look at discontinuous functions

Analysis is bullshit because it has functions other than rational functions

[floral gust]

And R is X in my situation?

[sleek thicket]

Complex analysis is on thin fucking ice

Yes it is

But the general result is good to know

And easier to prove because you don't need to worry about the specifics

[floral gust]

Okay perfect. So I have that connectedness of R <=> continuous function from R to discrete (in this case integers) is constant => deg_f(t) = deg_f(k) and thus independent

[sleek thicket]

Yeah, exactly

[floral gust]

Thanks!

[sleek thicket]

Np

[floral gust]

btw damn

Thats a very slick proof

[sleek thicket]

The connectedness thing?

Oh wait you can actually do it more easily on R

Just write down "intermediate value theorem lol"

Which is what all connectedness arguments are in their bones

[floral gust]

Btw how would I argue that deg_f(t) induces the discrete topology on Z

[sleek thicket]

?

Z already has the discrete topology

[floral gust]

We dont know that a priori?

[sleek thicket]

Well like

g is a function into the real line

And is continuous

Right?

Because it's integer valued, it induces a function R -> Z

And by the "characteristic property of the subspace topology", this function is continuous as a function into the subspace Z of R

So really the point is that the discrete and subspace topologies on Z coincide

[midnight jewel]

which follows from the fact that you can easily draw some open intervals around each element in ℤ

so each point is open

[floral gust]

Sorry, I still dont see how it follows. What I am given is that deg_f(t) is a map from (R, t_std) to (Z, t_unknown). Once I show that t_unknown = t_discrete I can argue that deg_f(t) is continuous and it is a map to a discrete space thus it must be constant.

Now I understand the point that if deg_f(t) is a map from R to the "subspace" Z of R then I can just say that it maps to (Z, t_sub) = (Z,t_discrete).
So my question is, what allows me to just say that Z is a subspace of R (that is it is endowed with subspace topology) and not, lets say, indiscrete topology?

@sleek thicket

[sleek thicket]

No that's not what you're given

You have a function deg

Forget about it

Define g : R -> R by g(t) = f(t+1) - f(t)

[floral gust]

Ah okay

[sleek thicket]

Yeah

Does that make sense?

[floral gust]

We know that f is a map from R_std to R_std so g is also a map from R_std to R_std. However, the domain is just Z so it maps to the subspace Z. Which has the discrete topology. Correct?

[sleek thicket]

Yup

That's exactly what I'm saying

[floral gust]

Thanks! And apologies for the constant bother

[floral gust]

Also how would I go about for this one

@sleek thicket

[sleek thicket]

So like we know f(t+1) = f(t) + n for some n

Right?

And g(s+1) = g(s) + k

Then g(f(t + 1)) = g(f(t) + n) = g(f(t) + n-1 + 1) = g(f(t) + n - 1) + k

Do induction

You'll get g(f(t)) + nk

@floral gust

[floral gust]

But how do I know that h, the lift of gf, has the same form as g'f'

@sleek thicket

[sleek thicket]

Oh yeah lol

I'm stupid

Didn't you prove uniqueness?

Oh no you didn't

And shouldn't

Okay so really what we want is that deg f is indepent of lift

Okay so like

[floral gust]

This is all I have and the only context I have are the path lifting theorems

[sleek thicket]

Your definition of degree is bad

You see why?

You don't know that deg f is independent of choice of f hat

[floral gust]

But should that matter?

[sleek thicket]

Well it means deg f isn't a function of f

It's a function of f along with a choice of lift

And so g' ° f' will lift g°f

Just paste the two diagrams together

That specify f' and g' are lifts

Okay so check this out

Let f' and f'' lift f

Consider the function s(t) = f'(t) - f''(t)

What can you say about s?

Hint: same trick as before

[floral gust]

AHHHHHH

LMFAO

[sleek thicket]

You see my concern and solution?

Any two lifts differ by a constant integer value

But that gets killed off in the degree

[floral gust]

Yes, concern being it is possible that upon changing the lift, the degree might change. And the concern is resolved by proving it doesnt

[sleek thicket]

Yup

So you get to choose the lift for g°f

And by putting the diagrams for g and f together, the composition of their lifts lifts their composition

[floral gust]

Okay so I will have that h'(t+1) - h(t) = deg(g o f ) = g'(f'(t+1) ) - g'f'(t) and the remaining I can handle

THanks!

[sleek thicket]

I think you're missing a prime but yeah

What class is this for?

I think we're doing a similar pace

[floral gust]

Topology

[sleek thicket]

Except mine is like two weeks earlier because we start super late

Intro?

Because if so, nice

[floral gust]

Yeah

[sleek thicket]

This is good to cover

And some intro top courses are very bad

[floral gust]

Although the prof tries to make the classes fast paced so we covered almost all pointset in first two weeks

[sleek thicket]

So like

Good

lol

[floral gust]

Ye....... Book recommednations?

I am trying to balance it with Lee's Introduction to topological manifolds but that gets advanced real quick

[sleek thicket]

That's what I'm using /shrug

I learned originally from Topology and Groupoids

Which is weird

People like Munkres

[floral gust]

Munkres has a bad interface 😦

Like I dont like the formatting of text

which makes it idk bad for me

[sleek thicket]

Oof, sorry

I get that

[floral gust]

What are you guys covering atm?

and which course is yours?

[fading vale]

munkres formatting is bad?

[sleek thicket]

Lee's

Topology of Manifolds

Next two quarters smooth stuff

[floral gust]

"physics ≥ math" is false*

Ah nice

[fading vale]

@floral gust long story short i was gifted nitro on the condition that i put "physics > math" in my name

so i did this

doing "physics >= math" would break the terms

[floral gust]

):

@sleek thicket Is this your first topology course?

[sleek thicket]

Technically yes? I studied stuff on my own though

It's supposed to be approachable for grad students whose undergrads didn't have a topology class

[floral gust]

So wait are you a grad student?

[sleek thicket]

No but the undergrad topology course is a joke

[floral gust]

Ye... honestly...

Like thanks for these pathological examples I will never ever use.....

[sleek thicket]

I mean like I appreciate weird spaces

But like

Do we need to teach all of the separateness axioms in depth?

I just don't really get the priorities

[floral gust]

And those neverending prefixes.....

[sleek thicket]

Idk, I like Lee's approach of "everything is nice but also you need to know the conditions"

Like people should leave a course understanding locally path connected vs locally connected and why they aren't implied by the global versions

But that's not the point

If you know what I mean

[floral gust]

Yeah exactly like the focus is so much on irrelevant stuff like imo topology should be sort of built only as far it is useful and required for algebraic or maybe maybe geomteric topology

If I would never see Lindelof in algebraic, dont teach it. Maybe just include it as a small text box. Devoting a full chapter and a week doesnt make sense

[midnight jewel]

Like thanks for these pathological examples I will never ever use.....
I like pathological examples

they help me actually understand what can go wrong

[gentle osprey]

oprah winfrey:

oprah winfrey:

[gritty widget]

sorry just trying to figure out if the above is the right transformation rule for the gammas

because it doesn't seem like they transform like (1,2) tensors

and this is on some smooth mfd with a connection blablabla

[chrome dew]

that looks roughly right, I scratched out a little derivation but I wasn't careful since I'm tired

they aren't tensors, so it's to be expected

that's why they call them Christoffel symbols not Christoffel tensors

[rugged swan]

hi, do you have any good references for non euclidean geometry ?

[gritty widget]

what sort of?

i had a nice book on hyperbolic geometry

but it's just one type of non-euclidean

idk if it's what you want

[rugged swan]

I need to do a course for undergratuate students, I'm thinking about Galois theory because I know this domain but I thought also about non euclidean geometry that I don't know, maybe a little introduction would be great

[bitter yoke]

http://moussong.web.elte.hu/bsm/neg_notes_19.pdf

[rugged swan]

hm thx

non euclidean geometry is really related to projective geometry ?

I didn't know that

[gritty widget]

well if it's geometry but not euclidean then it's pretty much non-euclidean geometry

[bitter yoke]

Yeah

Projective geometry is just a type of non-euclidean geometry

[rugged swan]

ok, I wanted to do a historical approach like introducing the euclidean geometry axioms and look at what could we do if we change the 4th axiom

[gritty widget]

i think we used this https://www.springer.com/gp/book/9781852339340

doesn't require much pre-knowledge on anything

and it starts with stuff like what happens if you poke the parallel postulate

[floral gust]

How can I show this without using homotopic maps have the same degree

[midnight jewel]

I mean just lift a point up and argue that at least one of its preimages is contained in the image of the lifted map

if it has degree ≠ 0 then it’ll cover either [0,1] or [-1,0] so you can find a preimage in one of those two

[floral gust]

Oh I am not given [0,1] or [-1,0]. I am just given the lifts from R to R

@midnight jewel

[midnight jewel]

[0,1] and [-1,0] are in fact subsets of ℝ

I’m treating 0 as the starting point of the path in ℝ here, you can also just like, name it x or whatever

in which case you’d cover either all of [x, x+1] or all of [x-1, x]

the importance is just htat if the degree is nonzero, then you’ll have an entire preimage of S¹ in the image of f̂

[floral gust]

As in degree ≠ 0 => π(f'(R))=S?

[midnight jewel]

yea

that shows it’s surjective because π∘f̂ = f

[sleek thicket]

f ° π

But the point is the image of f°π and f are the same

[midnight jewel]

I’m saying you take a point x in S¹. its preimage under π is a set of the form {y + n | n ∈ ℤ} for some y. Using exercise (b) and the intermediate value theorem, you can now show that if the degree is not 0, one of those must be in the image of f̂, which will show that x is in the image of f

[dim vigil]

Is the length of the vector projection of a onto b always less than or equal to the length of a?

It makes sense but I’m not 100% sure

[floral gust]

A hypotenuse is always either equal or greater than its base by Pythagoras theorem

[dim vigil]

Great thank you

[floral gust]

Was this what you meant?

@midnight jewel

[dim vigil]

I’m not sure, that looks foreign to me.

[lyric bane]

i found this interesting thing and i cannot understand why it happens
we know that pi = 180 degree
cos(180) = cos(pi)
but if you do
cos(180*180) it is not equal to cos(pi * pi)
why does this happens

[dim vigil]

But you answered my question. It was what I had suspected, just needed a little reassurance haha

[sleek thicket]

@floral gust yeah that looks good

And sounds like what the other person was outlining

Although I really hate that you have k + n and k isn't an integer

[floral gust]

):

[sleek thicket]

It's just a little uhhhh

terrible

No offense

[floral gust]

Any ideas for the converse as well? (I really hate my TA for introducing degree theory when we havent even done lifting properly)

[sleek thicket]

What's the converse?

And also this is more intuitive imo

[floral gust]

Show there exists a degree 0 surjective map from S1 to S1

[sleek thicket]

So like

Here's my intuition

Do you know the fundamental group of the circle?

[floral gust]

Yeah

[sleek thicket]

I'm not going to use a result about it

Okay

Well the degree is measuring what power of the generator you are

Considering a map S^1 -> S^1 as a loop

This breaks a little because you can move around the basepoint

But do you see what I mean?

[floral gust]

Yeah I see your point

[sleek thicket]

So let's think about how you can have a loop which is nullhomotopic but goes around the circle

Well

Why did we need homotopy to define the fundamental group in the first place?

[floral gust]

Yeah so basically finding a surjective nullhomoptic function right?

[sleek thicket]

Yup

What were the two things we needed from homotopy to make π(X) into a group?

[floral gust]

concatenation?

[sleek thicket]

Right so concatenation of paths isn't a group operation

But it is on homotopy classes of paths

What changes?

[floral gust]

The endpoints ?

The fact that for two paths to be homotopic they must have the same endpoints?

@sleek thicket

[sleek thicket]

So what I'm thinking is inverses

Take a loop that goes around the circle once

That's not nullhomotopic

But if we concatenate with its inverse it is

And the image will still hit all of those points

Does that make sense?

[floral gust]

Yes it does.

Actually thats kinda what I had in mind to consider a piecewise function in which the first half is exponenital in forward, and the second half is exponential in backward

[sleek thicket]

Yeah, I think that's the same thing

[midnight jewel]

@floral gust the image you posted is pretty much exactlt what I said with some details filled in, yea

[floral gust]

Oh yeah that's what I wanted to ask! Thanks!

[gritty widget]

Anyone omline

How do i go about seeking help

[versed pivot]

you ask a question

[gritty widget]

Last time i did that i got sent to #rules

And i cant read

So i wasnt sure

[sleek thicket]

Okay so like

Because I can't understand fiber products

I'm gonna think about applying them

Instead of the construction

In my class we looked how if you map Spec k -> A^1 by sending the point of Spec k to some point a and map A^1 -> A^1 by squaring, and then take the fiber product, you get Spec k[t]/(t^2 - a)

This is easy because we're working with affine varieties

But in the case a = 0, this is actually false. The fiber product is just Spec k because we're working with varieties and the reduction of k[t]/(t^2) is k

So it's either two points or a point with fuzz

Does this generalize in the obvious way?

Suppose you have a point a of Y and a morphism f : X -> Y. Define g : Spec k -> Y to pick out a. Is the fiber product of X and Spec k the fiber of f over a?

Suppose everything's affine, X = Spec B and Y = Spec A, so f makes B an A algebra (let φ be the map A -> B)

So a is really a maximal ideal m of A

And really f takes a maximal ideal n of B to its preimage under φ

Then f^(-1)(m) = { n maximal ideal of B : φ^(-1)(n) = m}

And the fiber product of X with Spec k is the spectrum of (the reduced ring of) B (×)_A k

Okay so if we choose presentations B = k[x1,...,xi]/R and A = k[y1,...,yj]/S, then as A algebras, B = k[x1,...,xi,y1,...,yj]/(R + S + (y1 - φ(y1),...,yj - φ(yj)))

So like I guess B (×)_A k is k[x1,...,xi]/(φ(y1),...,φ(yj))

Ugh I need to write this out more formally, I'm not confident in that

[remote grotto]

yo does an embedding of one manifold in another have to be a smooth map

like an embedding of one smooth manifold in another

[light rock]

yeah every map in the context of a smooth manifold is assumed to be smooth

[remote grotto]

damnit

ok i have a problem that says that there's an injective, closed immersion between smooth manifolds f:M->N

and i have to show f must be an embedding

so it's not enough to show that f is a homeomorphism onto its image?

i also have to show that the inverse of f on its image is smooth?

(by the way, does that happen to follow immediately from anything :p)

[light rock]

ah

okay homeo onto its image is enough

that's kind of weird

[remote grotto]

apparently this is some kind of well known thing or smth, at least according to this post: https://math.stackexchange.com/questions/1950646/when-is-a-smooth-homeomorphism-a-diffeomorphism

[light rock]

because we want stuff like x -> x^3 to be a closed embedding

[remote grotto]

wait why is homeo onto its image enough?

[light rock]

for the definition

we just say that's what closed embeddings are

we don't require diffeo to its image

[remote grotto]

oh, but according to this post diffeo follows

[light rock]

no it doesn't

[remote grotto]

oh ok

[light rock]

ah I see

the question asker is wrong

[remote grotto]

i wasnt in class when they defined precisely what an embedding was

[light rock]

the answers clears it up: there are immersion homeos that are not diffeos

[remote grotto]

ah gotcha

but they're still embeddings right

[light rock]

an embedding is an immersion that's homeo to its image

yes

[remote grotto]

gotcha, thanks

[remote grotto]

hm can anyone explain to me why the bottom line on page 3 is true

https://schapos.people.uic.edu/MATH549_Fall2015_files/Survey Charlotte.pdf

it says Z = g^{-1}(0)

but g is not even defined on all of Y

actually ig my question is

if Z is a submanifold of Y then why does there exist a smooth map g : Y -> R^(y-z) such that g^{-1}(0) = Z?

[light rock]

the chart has been picked around a point "a" in the image

so that 0 corresponds to the point you want

[remote grotto]

in the image of what?

a chart around a point "a" in the image of f?

[light rock]

yes

it is implied here that (0, ..., 0) corresponds to a

as well as (c1, ..., cz, 0, ..., 0) corresponds to Z

[remote grotto]

wait so "a" is chosen in Z intersect f(X)?

[light rock]

tes

[remote grotto]

ok but why does that mean g : Y -> R^(y-z)

the domain of g is not Y right

since c_i is a map from U to R for some U subset of Y

[light rock]

hmm

not sure exactly what they need

but you may extend it to the whole of Y

by extending it by 0

and smoothing out

[remote grotto]

huh

[light rock]

otherwise just work locally

it might be enough

[remote grotto]

but locally you dont get g^{-1}(0) = Z

which is critical

[light rock]

you get a piece of Z

[remote grotto]

yea

[light rock]

and you can piece together this argument for different points of Z

[remote grotto]

but that's not enough i think

hm

[light rock]

actually the rest of the argument is also very confusing

this isn't very well written

maybe try finding another source

[remote grotto]

at least it's better than the proof of the same theorem here: https://schapos.people.uic.edu/MATH549_Fall2015_files/transversality_handout_Greenblatt.pdf

[light rock]

for the results

which one is it

[remote grotto]

"construct a map g:Y->R^(y-z) such that g^{-1}(0) = Z"

[light rock]

ah

[remote grotto]

section 2

ok i guess if we work locally, we get another submanifold

[light rock]

i mean this is the idea of the other proof

you can just work locally

[remote grotto]

but the problem is idk if transversality still holds

[light rock]

transversality is a local property

you check it at points

[remote grotto]

right but it also depends on the manifold

right?

[light rock]

only locally

it just depends on the tangent space at a point

in the domain and image

[remote grotto]

like if f : X->Y is a smooth map and f is transverse to some submanifold K of Y then f need not be transverse to Y

[light rock]

yes

but working locally is fine

[remote grotto]

but what if the open neighborhood we pick has a wildly different tangent space

[light rock]

it doesn't

the tangent space is a local construction

it's the same on any neighborhood around the point

[remote grotto]

ok true

ok sure

now i agree that if f : X->Y is transverse to Z a submanifold of Y then for any open neighborhood U subset of Y of a point p in Z, if we let the open neihgborhood V of p in Z be V = U intersect Z, then f is transverse to V

i guess then we can get that f^{-1}(V) is a submanifold of X

but why does gluing together the preimages f^{-1}(V1) ... f^{-1}(Vk) result in a submanifold of X

is there some generic submanifold construction for the union of two submanifolds

i dont even know of a topology that is naturally equipped on the union of two topological spaces

[light rock]

your set is f^-1(Z)

and you are taking open sets on it

and showing they are manifolds

[remote grotto]

how does a collection of manifolds whose union is a set become a manifold

[light rock]

a topological space which is covered by manifolds is a manifold

[remote grotto]

why is f^{-1}(Z) a topological space

[light rock]

it's a subspace of X

[remote grotto]

oh right

man we never mentioned in class that a top space covered by manifolds is a manifold

but i hope it's obvious enough

that i can just write one line about charts covering every point

[light rock]

well, there's a way of writing this in one line...

it's not very satisfying though

[remote grotto]

lmfao

F = f o g isnt even true

[light rock]

you can actually glue the local maps into a global map

[remote grotto]

F should be g o f

[light rock]

well okay but I mean the first part

you can glue local maps into a global map using partitions of unity

[remote grotto]

ah sure

is the partition of unity even needed?

[light rock]

maybe not even

[remote grotto]

er yea it is

[light rock]

you can get away with somewhat less probably

but it's the standard way

[remote grotto]

in order to make it agree on a neighborhood i think

[light rock]

it's the cleanest way of doing it

[remote grotto]

ok yea i see

i dont know if the manifolds are paracompact tho

(which iirc is the condition needed for the existence of a partition of unity)

[light rock]

they are

even if they are not you can build a partition of unity

only it's not made up of countably many things

but it's still locally finite

so it works the same

[remote grotto]

wait huh manifolds need not be paracompact right

[light rock]

they usually are

[remote grotto]

also according to https://ncatlab.org/nlab/show/paracompact+Hausdorff+spaces+equivalently+admit+subordinate+partitions+of+unity, a partition of unity may not be guaranteed if not paracompact

[light rock]

I mean

second countable + hausdorff are the usual conditions

and they imply paracompactness

[remote grotto]

the manifolds at least in the context im studying them are second countable + hausdorff

[light rock]

yes

[remote grotto]

oh does it

but dont there exist second countable hausdorff spaces that are not metrizable

and hence not paracompact

[light rock]

manifolds are metrizable

because they are locally euclidean

[remote grotto]

dont they have to be regular or smth tho

[light rock]

they are

another way: embed your manifold M into R^n, take the induced metric

[remote grotto]

oh right

damn

i never realized that lmao

wow i've been living my whole life thinking there existed non-paracompact manifolds

[light rock]

usually when people say that they mean they drop second countability

and it's done, rarely

[remote grotto]

i thought paracompact implied second-countable

oh er nvm

misread

anyway, thanks

[light rock]

yw

[remote grotto]

oh wait also while you're here, do u happen to know how i can show that if M is a submanifold of R^n then for any point p in R^n, the point q in M closest to p satisfies that p-q is normal to every tangent vector to M at q

i dont really see how to do it in general

in the special case where M is like the preimage under some function of a point or something i think i can do it

nvm think i got it

[light rock]

yeah if it's not then you kinda flow the other way

aka the non perpendicular direction

and get closer

[remote grotto]

wait wtf

in the definition of transversality it's not a direct sum??

[light rock]

because of dimension concerns, it is a direct sum

(dim k) + (dim n-k) = dim n
implies the sum is direct anyway

you shouldn't take at face value the comments/questions on these websites, only the answers

[remote grotto]

oh also another question

if Z is a submanifold of Y why does there exist a chart neighborhood V of Y such that V intersect Z is a chart neighborhood of Z?

oh wait nvm i see

[light rock]

yeah these details take a bit of getting used to

you can actually define a manifold by this property

[remote grotto]

oh wait irght

[light rock]

as a subset of R^n

[remote grotto]

that follows from the definition of submanfiold that we're using

[light rock]

such that opens intersect it as charts

ye

[remote grotto]

wait actually idk if it does

[light rock]

it might depending on what it is

[remote grotto]

for an intrinsic definition of submanifold

if N is a smooth manifold and M is a subset of N equipped with the subspace topology then every open set in M is of the form M intersect V for some open V in N

is it also true that every chart domain in M is of the form M intersect V for some chart domain V in N

[light rock]

yeah

[remote grotto]

why's this true?

[light rock]

you take a small enough open of N such that it's a chart

and then you keep making it smaller until the intersection with M is a chart

[remote grotto]

hm i see

well i dont see it rigorously

[light rock]

take an open set of M that is a chart

[remote grotto]

yes

[light rock]

since it's open in the subset topology it comes from an open V of N

[remote grotto]

yup

[light rock]

intersect this V with a chart open for N

now this is a chart open for N which intersects to a chart open for M

[remote grotto]

ojhhh

gotcha thanks

ok yea i dont think im gonna be able to write the proof that preimage of a submanifold transverse to a function is a submanifold of the domain

too many small things to write that im not able to keep track of

even after all of this i cant show that a point is a regular value

fukcing hell

wait

is the rank of the push forward of the projection R^n -> R^(n-k) to the last n-k coordinates equal to n-k on any open neighborhood of the domain

wait ugh

rank(AB) = rank(B) right

[light rock]

the rank is n-k yeah

it's the n-k identity matrix and the rest zeroes

[remote grotto]

yea

[remote grotto]

wait

im pretty sure that it's actually not a direct sum

i think it's just a union

because to be a direct sum we would have dim(Y) = dim(Z) + dim(X)

which is not true; only dim(Y) < dim(Z) + dim(X)

[next eagle]

for metrics d: X x X -> [0; inf) set X contains come n-th dimensional points right?

[sleek thicket]

@next eagle what?

[random slate]

So I know that the notation ∂S can be used to describe the boundary of a set S. Is there a sort of "inverse" operator to ∂ that takes a boundary and gives the region bounded by it (assuming that a notion of "outside" and "inside" has been defined)?

I guess KIND OF like $\partial^{-1} C$ for some closed curve $C$?

[gentle osprey]

DMAshura:

[honest narwhal]

In principle, by the Jordan curve theorem a simple closed curve splits R^2 into two connected components

Each component has the curve as its boundary, and only one of them is bounded

So you could say okay choose that guy given a simple closed curve

But it's never usually thought of as an operator really

It's just like, let D be the region it bounds

[random slate]

Would it be weird to write it as $\partial^{-1} C$ in my notes? XD

[gentle osprey]

DMAshura:

[honest narwhal]

Probably

[random slate]

Darn

. . . I vote we make this notation standard :V

[honest narwhal]

Lmao

If that were doable then the term "Electromotive force" wouldn't be a thing anymore in physics

[random slate]

Yeah, and all the textbooks would use τ 😛

But hey I just enjoy trying to improve on notation 😛

[sleek thicket]

I'm gonna think about an easier thing from office hours than locally representable bullshit

My prof claimed that if Z/2Z acts freely on a scheme X, there may not be a quotient scheme of X by that action

So like let's do an easy example

X = A^1\{0} (= Spec k[x, y]/(xy - 1))

So as a topological space, X is cofinite plus a generic point

And like let's say k algebraically closed

Let C2 = { ε, ι }, where ε is the identity and ι^2 = 1

Then C2 acts on X as a space by ε•x = x and ι•x = -x

Oh if we think of X as an algebraic group then C2 is the subgroup { 1, -1 } acting on X by multiplication

We can't do ι•x = 1/x because then it's not a free action

And the action of ι is a linear polynomial so its an iso on A^1, and it restricts to an iso on X

So morally the quotient X/C2 should be like an open ray?

Like maybe it's just A^1???

Oh no the structure sheaf has to be weird is the thing?

So let's let Y be the quotient

Y represents a certain functor

The thing which sends Z to { φ : X -> Z | φ(x) = φ(-x) }

I need to poke Y by choosing a good Z

[gritty widget]

Looks like it's still X

with quotient map $x \mapsto x^2$

[gentle osprey]

trex:

[sleek thicket]

Omfg I'm dumb

Something still feels off but I'm not sure what it is

Oh I think I know my propylene

*problem

I'm visualizing k = R instead of k = C

Geometrically

[gritty widget]

idk, the invariants of k[x, x^{-1}] under C2 is k[x^2, x^{-2}], which is precisely what you want

ah

that makes sense lol

[sleek thicket]

Literally every time I do this stuff

I get confused about that

It's been like 4 months since I started doing AG and yet

Okay, time to think about other free C2 actions

There's one on P^1

Which sends z to 1/z, aka swaps the homogenous coordinates

[gritty widget]

heh I can relate, I remember I spent a lot of time on a question during my Riemann surface exam before realizing that a connected complex curve minus some points is still connected, because it's not topologically a curve

[sleek thicket]

Yessss

It's so unnecessarily confusing

Like it's really obvious if you think for a second

This is still interesting though, since the map z -> z^2 gets weird at 0

Ohhh but I excluded 0

So it's unbad

[gritty widget]

yess

[sleek thicket]

Lol

[gritty widget]

life does work out

The one on P^1 is not free; it fixes [1:1]

[sleek thicket]

Oh, good point

Right and that makes sense when I think about it as z -> 1/z

It'll also fix [1:-1]

(assume char k = 0 I guess. Actually assume k = C)

What about z -> -1/z though? I'm not sure if i can make sense of that in projective coords

Wait no

z = i

Uhhh z -> -1/(conj z)?

Oh but conjugation isn't like a rational function

Idk if I can make algebraic sense of that

[gritty widget]

well yeah, if you work with Z/2, better not work in char=2

[sleek thicket]

Yeah good point

[gritty widget]

and yeah, conjugation is rational

but not complex rational

[sleek thicket]

Yeah that's what I Mena

*mean

Since I'm working with k = C

[gritty widget]

yeah, this is not C-polynomial

[sleek thicket]

Maybe it works though?

Oh no

It won't be a morphism

Because of what I was saying

If C2 acts by inversion, what does P^1/C2 look like?

We identify z and 1/z

[gritty widget]

so via -1/\bar z ?

[sleek thicket]

So [x:y] and [y:x]

[gritty widget]

or just 1/z ?

ah okay

well it's not a scheme

[sleek thicket]

Well that's what I'm looking for tbh

[gritty widget]

but probably like P^1

[sleek thicket]

But I was told to look for a free action with no quotient

I'll think about why it's not a scheme for rn

[gritty widget]

I think the classical counter-example is pretty complicated

[sleek thicket]

Classical counterexample as in complex analytic with no complex analytic quotient?

Or as in the standard scheme example?

[gritty widget]

standard scheme example

[sleek thicket]

I guess I'll think about weirder schemes then

[gritty widget]

but maybe I'm confusing this with another counterexample

[sleek thicket]

I'm not sure if he actually wanted me to find it or was just motivating why we would maybe move to more complicated categories of spaces

But it's good to think about anyways

[gritty widget]

certainly

[sleek thicket]

I think I figured out part of why things weren't working

Like actions of the form z -> p(z)/q(z)

A fixed point of this map is a root of p(z) - zq(z), which exists b.c. we're all algebraically closed here

This isn't 100% formal but still

[chrome heart]

is this the place for computational geoemetry?

[gritty widget]

How did he take the derivative of gamma here?

idk how this part works at all lol

thanks in advance 👌

you're most welcome

<@&286206848099549185> ^

[gritty widget]

rip

[sleek thicket]

rip

[gritty widget]

was my question too vague or something

[wicked thorn]

wow I haven't seen a green chalkboard in years

[gritty widget]

imagine using a whiteboard or using a dry wiping thing on your chalkboard

[fleet trench]

im not really sure how to interpret all the decorations on the gammas, but it looks like your prof is talking about normal coordinates, where the components of the metric tensor are just the kronecker delta and the gammas disappear at the point; if you look up "normal coordinates" you might be able to find an explanation that makes more sense to you

oh wait is that first subscript like telling you what parameterization the christoffel comes from

[gritty widget]

This is legit killing me

Does anyone know how he got that derivative above

[limpid mural]

Do you have clear why there exist a chart where the components of the transition function looks like alpha^ + cristhoffel?

[gritty widget]

i mean it's a chart you can do anything

[gentle osprey]

i made macarons:

[gritty widget]

i wrote the question up more clearly

[limpid mural]

P is fixed?

[limpid mural]

Anyway p should be fixed, so the gammas are just real numbers and do not depend on alpha^i

[gritty widget]

@limpid mural why does the 1/2 go tho

[limpid mural]

You are deriving a product alpha^j alpha^k

The notation is a little confusing, the j in. The gamma Is not fixed while the j in the deltas is fixed

I guess

[gritty widget]

ugh

i still don't get it lol

oh wait

oh fuck this is trippy

[limpid mural]

Write down the sum and derive with respect j fixed

Let's say the j in the gammas is now an h

[gritty widget]

$\pm \Gamma^i_{mk} \cdot (\delta^m_j \alpha^k + \alpha^m \delta^k_j)$

[gentle osprey]

i made macarons:

[gritty widget]

$\Gamma^i_{jk} \alpha^k + \Gamma^i_{mj} \alpha^m$

[gentle osprey]

i made macarons:

[gritty widget]

but the lower indices are symmetric bc the alphas commutes

$\Gamma^i_{jk} \alpha^k + \Gamma^i_{jm} \alpha^m = 2 \Gamma^i_{mj} \alpha^m$

because change of indices

[gentle osprey]

i made macarons:

[gritty widget]

@limpid mural does that look right?

[limpid mural]

Just one question

Can you write down $\Gamma_{(ij)}$ I mean the symmetrization

[gentle osprey]

emme:

[gritty widget]

the symmetrisation is implied from $\alpha^i \cdot \alpha^k = \alpha^k \cdot \alpha^i$ right?

[gentle osprey]

i made macarons:

[limpid mural]

Wellz the point is that at the beginning you are not deriving $\Gamma_{mk}\alpha^m \alpha^k$ but $\Gamma_{(mk)} \alpha^m \alpha^k$. now I don't remember exactly what is $\Gamma_{(mk)}$ and the symmetry of Christoffel's symbol should be a consequence of the torsion free of the connection

[gentle osprey]

emme:

[limpid mural]

So or you are assuming that the connection is torsion free or I don't know if the trick with the lower indexes is right

[gritty widget]

$\Gamma^i_{mk} (p) \cdot \alpha^m \cdot \alpha^i = \Gamma^i_{mk}(p)\alpha^i \cdot \alpha^m$

[gentle osprey]

i made macarons:

[gritty widget]

i could write the symmetrisation brackets

and i'm pretty sure you can do that with the lower indices

ok ty this helped a lot 👌

[limpid mural]

If you are sure then the proof is right, I mean it must be that way It's my fault If I can't see why

[gritty widget]

Are there some criterions which say when topological space X is metrizable or when it has a countable basis?

smirnoff's theorem if i remember correctly

https://en.wikipedia.org/wiki/Nagata–Smirnov_metrization_theorem

yup

[raven sonnet]

if I had euclidean geometry questions, would this be the channel to ask them in, rather than #geometry-and-trigonometry ?

I'm only unsure because it's not pre-university, but not exactly advanced mathematics imo

[fleet rapids]

So like, does this make sense to ya'll? I'm not really sure how you map bijective (they said it's homeo) to [1,2] from the unit interval when the function acts like an identity function on R, typo?

This is how they define the set Y

[gritty widget]

yes they are!

it should be x+1 for y=0

[gritty widget]

does this proof work?

$T^i_{jk} = \Gamma^i_{(jk)}$

[gentle osprey]

i made macarons:

[gritty widget]

torsion tensor

[gritty widget]

<@&286206848099549185>

[gritty widget]

$\prod_{n \in \mathbb N} [0,1] \subset \prod_{n \in \mathbb N} \mathbb R$

[gentle osprey]

Godel:

[gritty widget]

Is this open or closed?

It looks closed So I wrote its open during exam but not sure

[radiant basalt]

@gritty widget

if box, it's closed

[gritty widget]

Why not in product?

[radiant basalt]

wait

i might be talking nonsense

yeah it's closed

if a point is outside your set X, then one component is outside [0,1], so you can find an open ball around it, then take a product of it with infinitely many R's and you have an open set outside X containing your point

so the complement is open

it's not open because you can't make an open set that's completely inside it, because infinitely many components of the open set will be R

[pseudo pecan]

Anyone here know anything about shimura varieties?

[light rock]

what do you want to know about them @pseudo pecan

[gritty widget]

can anyone confirm this proof?

[chilly silo]

<@&286206848099549185>

:(

[gritty widget]

rip geometry

[chilly silo]

well

[chilly silo]

I figured out my issue!

[gritty widget]

<@&286206848099549185> plz above ^

[wicked thorn]

guess I should start studying topology so this channel gets a little love

[chilly silo]

do it

topology is so cool

learn point set, then algebraic

and then become a category theorist

[gritty widget]

reminder that category theory is merely a tool for the superhuman topologists and algebraists

[sleek thicket]

me irl

[limpid mural]

So tru

[sleek thicket]

I accidentally created one of these

It might be a bad idea to use aluffi for a first course

Because this is what happens

[limpid mural]

Get them back on algebraic geometry b4 it's 2l8

[sleek thicket]

That's a good idea. I should use vakil as our text and just skip the actual algebra

You can learn ring theory on the way

And groups are easy to pick up in the context of algebraic groups

Actually the person I'm thinking of wants to do AG in grad school

And knows basically 0 AG

Isn't even good at basic ring theory

[limpid mural]

Lmao

How bout complex analysis then, try to introduce to it

[sleek thicket]

I am considering linking a term paper he wrote, but that would be a little too far

The first paragraph includes the term "sheafs"

[limpid mural]

To Him*

[sleek thicket]

Oh he took complex analysis with me and did well afaik

He's not like bad at math

Just likes the fancy seeming things

Our prof for that class is actually cited in Hartshorne

[limpid mural]

Yeah I see, well I hope he will survive lol

[sleek thicket]

Got his start doing complex geometry and now does like harmonic analysis

Yeah hopefully

[limpid mural]

I have a prof who did a PhD in AG

And now he does only Calculus of variations

Lol

[sleek thicket]

I feel like AG is gonna break me before I even get to riemann roch or w/e

It's hard and confusing

[limpid mural]

Yeah, I find it hard too

I need to be more strong on commutative algebra to actually understand what the fuck is going on

[sleek thicket]

Mooooood

Gonna bunker down in December

Over break

[limpid mural]

It's kind of magic

[sleek thicket]

And do as many problems in AM as possible

[limpid mural]

Black magic

[sleek thicket]

That's what I always say!