point-set-topology

[marsh forge]

Wait

We want TM to be

Rank n

Wait that’s trivial

For a manifold

[dim meadow]

So the idea is that you can always make a frame locally

Because of the local triviality condition

[marsh forge]

Right

[dim meadow]

So you have a bunch of local frames

And you can't always paste them into a global frame

You have a global frame iff your bundle is trivial

[marsh forge]

Hmm ok

So a framed manifold is a manifold w trivial tangent bundle

That excludes all nonorientable manifolds

But also like

S^2

[dim meadow]

Yes

[marsh forge]

Which is weird to me

[dim meadow]

So one of the ways you define orientatability

Is in terms of local frames

[marsh forge]

Interesting

[dim meadow]

You want to be able to transition between the frames

With a positive determinant

[marsh forge]

Sorry what’s the form of this transition map?

[honest narwhal]

Please @ me next time there's a difftop discussion

[nimble jolt]

lol dami so thirsty

better re-add your helper role

[marsh forge]

So like we have maybe some path

[honest narwhal]

[marsh forge]

And we want to be able to transition our local frames

[dim meadow]

So I mean that a frame gives you a basis of your vector space in TM

[marsh forge]

Along this path?

[dim meadow]

At a point

And 2 frames give you 2 different basis

[marsh forge]

@honest narwhal would bott tu

Be a good source

To learn more about this

[honest narwhal]

I only really did the very beginning stuff in Bott-Tu but maybe

[dim meadow]

Read Lee lol

[honest narwhal]

Though that's more about characteristic classes

[dim meadow]

It's long af but very good

[nimble jolt]

Lee is where I learned all of this

[honest narwhal]

Vector bundles raw aren't too bad though

[marsh forge]

Which lee

[nimble jolt]

is careful about all the details which I think is important for a first encounter

smooth lee

[honest narwhal]

"Intro to Smooth Manifolds"

[marsh forge]

So wait

[nimble jolt]

you won't ever be as verbose as him when working with or discussing these htings in the future

[marsh forge]

I’m still a little under informed about the orientation picture

We get a local frame at each point

[nimble jolt]

but building the initial intuition kinda needs some verbosity by nature I think

[honest narwhal]

So do you know what an orientable vector bundle is?

[marsh forge]

No

[honest narwhal]

Okay so, let's say you have a vector bundle p:E->B

There's a local trivialization, right?

[marsh forge]

Ok

Yes

all tangent bundles are trivial

[honest narwhal]

Yeah on each of the open sets of a cover

[marsh forge]

No I mean

As a nickname meme

Anyway continue sorry

[honest narwhal]

Ohhhhh

Lel

But yeah so on the intersection of any two of these open sets you have a transition map

[marsh forge]

Ok

This is induced by the basis

But is noncanonical?

Unnatural maybe is the better term

[honest narwhal]

I mean it depends on your trivialization

[marsh forge]

Like I gues my point is in general we have two trivialization

Giving two bases

But in general a nonunique map between them

[nimble jolt]

ofc, but orientability is a statement about the EXISTENCE of a nice collection of trivialisations.

or equivalently a nice cover of local charts.

[marsh forge]

Ok ok

[honest narwhal]

And yeah you can talk about a reduction of the structure group

[marsh forge]

So we have these transitions

For any tangent (vector) bundle

Right

[honest narwhal]

Basically are we able to choose some trivialization such that the transition maps are all in some given subgroup H of GL_n(R)?

The vector bundle is orientable if you can reduce to GL^+

Namely if you can find some covering by charts such that the transition maps have positive determinant

[marsh forge]

Oh I like this

So we take any open cover or all open covers?

Which one?

[honest narwhal]

Any open cover

Like if there exists a trivialization such that blah

Then your vector bundle is orientable

[marsh forge]

Ok then if all the transitions can be chosen such that they are all of positive determinant

Then our manifold is orientable?

[honest narwhal]

Yeah for a specific open cover

[marsh forge]

For compact manifolds this reduces to finitely many choices

[dim meadow]

Yes

[honest narwhal]

This structure group business is nice because other reductions give different things. For example, I think I read that an O(n) structure basically is the same data as a Riemannian metric

[marsh forge]

So let’s say I cover S^2 by the two D2s

And let them overlap a lil

Oh wait sorry

Nevermind that was a bad example

But yes I get the picture

[honest narwhal]

Well so you can use the cover given by stereographic projection

[marsh forge]

Thank you everyone

[dim meadow]

Use the complements of 2 points

Cause stereographic projection

[marsh forge]

What does the proof that nonorientable manifolds have nontrivial tm

[honest narwhal]

Anyway yeah orientability of a manifold means the tangent bundle TM->M is an orientable vector bundle

[marsh forge]

Look like

[honest narwhal]

Well if the tangent bundle is trivial then it's obvious

The open cover can just be the whole manifold

[marsh forge]

Yes

Other direction tho?

[honest narwhal]

So you choose that open cover and the only transition map is the identity

(nonorientable => nontrivial TM) = (trivial TM => orientable)

[marsh forge]

Oh

Right

[honest narwhal]

Orientable manifolds need not have trivial tangent bundle

S^2 is the easiest counterexample

[marsh forge]

Yes that’s what started all of this

Someone said this contradicted Harry ball

Can you expand on that?

[honest narwhal]

Yup

Okay so hairy ball theorem says that there's no vector field on S^2 which vanishes nowhere

[marsh forge]

Yes

[honest narwhal]

If M has a trivial tangent bundle, then choose any nonzero v\in R^n

The map M->TM=M\times R^n given by x\mapsto (x,v) is a vector field

[marsh forge]

Oh I see

So framed manifolds are like

Super nice

Is a torus framed

[honest narwhal]

What's a framed manifold?

One with trivial TM?

[marsh forge]

Yes

Well

[honest narwhal]

Ah

I call that parallelizable

[marsh forge]

Specifically a framed manifold has a frame

Which is a specific choice of global trivialization

[honest narwhal]

I see

Anyway yeah the torus has trivial TM

More generally any Lie group

[marsh forge]

wait

I thought that

S^2 was a lie group

By rotations

[honest narwhal]

Nope

[marsh forge]

Is this not true

Oh shit

[honest narwhal]

SO(3) acts on S^2

By rotations

But SO(3) is RP^3

[marsh forge]

Wait so RP3 is S2/antipodes

[honest narwhal]

S^3

Oh wait

[marsh forge]

Right ok wait

Wait am I right

I always fuck up the indices

No it’s S3

[honest narwhal]

RP^n = S^n/antipodes = R^{n+1}\0 mod yeah

So yeah RP^3 is a quotient of S^3

[marsh forge]

So is there an obvious way

To associate a rotation with S3

I’ve always thought of a rotation of 3space being parametrized by the sphere

[honest narwhal]

Well, keep in mind RP^3 gives rotations of R^3

[marsh forge]

But I guess I never double checked this

Yeah I guess that’s not immediate to me

[honest narwhal]

That's not easy, you have to prove that RP^3 is diffeomorphic to SO(3), which is what acts in the obvious way

S^3 is a Lie group but that's just as unit quaternions

[marsh forge]

Oh

[honest narwhal]

Rather than some rotation group

[marsh forge]

Then I get it

I know the quaternion connectction

Right ok so rotations of 3 space

Are more complicated than I was thinking

I should play around w blender sometime

So let’s fix a point of S2

Maybe just on the positive x axis

Then every other point corresponds to two angles

But this isn’t enough to parameterize SO(3)

Which wants also a choice of axis

Right?

[honest narwhal]

Wait I think too many things are conflated here

In the case of SO(3), a single point corresponds to a rotation

Namely the rotation matrix

If you give me two points in SO(3), you compose the associated rotations to get a third and that's the multiplication

But rotations are just singletons here

In the case of S^2, even if you had a unique rotation given by two points, that wouldn't correspond to a third point (thus giving a multiplication)

Also are there only two rotations which send a point to another point on S^2?

Like, if you ask that a given point is fixed that's just choosing an axis so you have a lot of options

Does that make sense?

[marsh forge]

No sorry

I was thinking about like

What kind of information you need to specify a rotation of R3

I know a unit quat works

[nimble jolt]

you need to choose where to send the north pole, and then you are free to rotate about the axis through this new position of the north pole.

[marsh forge]

Is that injective?

Ok ok

@honest narwhal do we have a diffgeo class

[honest narwhal]

Third quarter grad topology/geometry is diffgeo

[marsh forge]

Oh lit

I wasn’t sure about 2nd quarter or 3rd quarter

What is 2nd

[honest narwhal]

2nd quarter is differential topology

Which is a superior subject tbh

Differential geometry is just bad

Though Amie Wilkinson is teaching diffgeo this year and she's supposed to be really good

Neves difftop is actually fucking amazing

[marsh forge]

Wait sorry

I mean difftop lol

So lee smooth is the standard reference?

I’ll try to go through it over the winter or something

[floral gust]

Are you guys in the same university?

[marsh forge]

Me and dami used to be

He graduated because he’s a Chad

[floral gust]

Is he a postgrad now?

[marsh forge]

Dami works for Goldman Sachs now

[dreamy smelt]

Quick question

Suppose I want to prove that f and g are one to one and onto if they're inverse functions.

If B is not f(A), do I use B or f(A) for the domain of g?

[vocal wharf]

what's A and B

[dreamy smelt]

A is the domain of f and the range of g

B is the range of f and thus the domain of g

[vocal wharf]

by range you mean codomain?

[dreamy smelt]

My book defined ranges; no codomains

[vocal wharf]

ok

if f(A) is not B, then f is not surjective

[dreamy smelt]

Though I know of their existence, I don't want to alter my definitions until I know them perfectly

[vocal wharf]

replace the word surjective with onto, if you prefer that word

[dreamy smelt]

Right. So if f is not surjective, which set do I use for the domain of g

[vocal wharf]

to show what?

[dreamy smelt]

That they cannot be inverses

[vocal wharf]

you want to show that there does not exist an inverse?

[dreamy smelt]

I want to prove that if f: A --> B and g: B --> A then both functions are one to one and onto.

I've taken care of one-to-one, but restricting B to a proper subset of f(A) creates confusion in choosing the domain of g for me.

[vocal wharf]

you want to show that if a function f has an inverse g, then it is one-to-one and onto?

[dreamy smelt]

Exactly

[vocal wharf]

if f: A -> B, then g is necessarily B -> A

if f(A) is not B, there is nothing to show, because it cannot be onto

[dreamy smelt]

That feels really loose to me

I've been tackling this problem with a proof by contradiction.

[vocal wharf]

there are 2 implications to show

onto and one-to-one => inverse exists

and

inverse exists => onto and one-to-one

which are you tackling

currently

[dreamy smelt]

Inverse exists

My book defines inverses by being able to compose both functions for an arbitrary element of each set and showing that the functions undo each other

So I was trying to restrict B somehow that I cannot pull the function back out from f.

*element back out

Though it's in the domain of g

[shadow ermine]

👀

[dreamy smelt]

I'll just show you so you can follow

[vocal wharf]

also, you might want to move this

this has nothing to do with topology

[dreamy smelt]

It's in my topology book lol

[vocal wharf]

lol indeed

[dreamy smelt]

It's just that the first chapter talks about sets and stuffs

[tough hamlet]

topology is just screwing around with weird sets

change my mind

[dreamy smelt]

Also it says P it should say f.

My bad

[shadow ermine]

hmm, you'd probably find this stuff at the beginning of an analysis textbook, or even abstract algebra textbook

[dreamy smelt]

Haha

[vocal wharf]

you would find this at the beginning of many books

[shadow ermine]

functions are essential in pretty much all branches of maths

[vocal wharf]

it's basics

[tough hamlet]

well being in a topology book doesn't really mean it's for the topology channel necessarily

[dreamy smelt]

So can I say that B is not f(A) and use f(A) as the domain or no

[vocal wharf]

tbh i would suggest not attempting a proof by contradiction

the domain of the inverse is B

[dreamy smelt]

Alrighty. I'll try another approach when I get sat back down.

Thanks mates

And sorry for asking in the wrong place lol

[sleek thicket]

What's the difference between diff geo and diff top?

@honest narwhal

Currently taking a sequence on "the topology and geometry of manifolds"

Using ITM in 1st quarter and ISM in 2nd/3rd

[wide umbra]

hey guys idk if this is the right place to ask but what are some of ur undergrad book recommendations for geometry of the euclidean space through vectors

[dim meadow]

With Geometry you have stuff like geodesics and angles

@sleek thicket

[sleek thicket]

Ah yeah

I think Lee said something about that

How you really need more structure

[dim meadow]

Yeah

[sleek thicket]

Diff top seems cooler

[dim meadow]

I agree

If I ever have to talk about the law of cosines I'm immediately bowing out lol

quits math PhD

[honest narwhal]

Yeah differential geometry is bad

Differential topology is good

[sleek thicket]

I think we're just doing that

This year

But if I took the special topics geometry stuff next year it's something to keep in mind

[marsh forge]

Wait

So if lee is diffgeo

What’s the best text on difftop

[sleek thicket]

By that definition wouldn't Lee be difftop? Smooth manifolds just involves a smooth structure I think?

I haven't read it yet

But I thought diff geo would be more like intro to reimannian manifolds

[honest narwhal]

Lee is neither difftop nor diffgeo tbh

It's like

The definitions surrounding smooth manifold stuff and basic mechanics

To me the actual topology is like

Let's say transversality and intersection theory

[bleak crescent]

would anybody be able to provide any hints as to these three? i'm able to come up with bijections, but coming up with ones that are continuous is proving to be trickier

[sleek thicket]

What are your bijections?

@bleak crescent

For the first one, think about where you could get the extra point to add to [0, 1) to get [0,1]

[bleak crescent]

my (i think non-continuous) bijection for A -> B would be

well, i restricted it to a bijection from [0,1) to [0,1] first

if x = 1/n for some positive integer n greater than 1, then f(x) = 1 / (n - 1).
otherwise, let f(x) = x.

[sleek thicket]

Ah yeah, that's not continous

[bleak crescent]

yeah hmm

[sleek thicket]

I guess my hint is that on that restriction, it's just the inclusion map

But since it's surjective, you need to pull the 1 from somewhere else

[bleak crescent]

oh... interesting

did not think about that

[sleek thicket]

I think there is a continous bijection from A to [0,1] actually

Challenge problem, lol

[bleak crescent]

guess i'll think about if this is continuous lol

If x > 1, then f(x) = x.
If x in [0,1), then f(x) = x. (It's just the inclusion!)
If x = -1, then f(x) = 1.
If x < -1, then f(x) = x + 1.

[sleek thicket]

Yeah, that's exactly the map I came up with

[bleak crescent]

it... looks cts hm

[sleek thicket]

It is

[bleak crescent]

given each restriction is continuous

(thanks! i will think about 2nd part now haha)

[sleek thicket]

Because there's some space separating each set in the union A, a function out of A is continous iff each restriction is

And each restriction is just a polynomial

[bleak crescent]

Ahhh, yeah

i just wanted to double-check, but: each of these individual singleton sets or half-open intervals are "open" in the subspace topology of A, right?

[sleek thicket]

Yup

[bleak crescent]

because {-1} = (-1.5, -0.5) intersect A

and same argument with half-open intervals

[sleek thicket]

That's what I mean by "some space separating them"

[bleak crescent]

Ahh

[sleek thicket]

Yeah

Have you learned about connectedness?

[bleak crescent]

And then the restrictions (how i listed them) are either opens themselves or arbitrary unions of opens

yep

[sleek thicket]

Yeah, so the components of A are the sets in the union

[bleak crescent]

ah, right

[sleek thicket]

Because they're also open (this means A is "locally connected"), A has the disjoint union topology wrt them

I don't know how much you know, that might not make sense

Also I've convinced myself that there's a continuous bijection A -> [0,1], but it would be bad to write down

[bleak crescent]

haha i'm thinking about the continuous bijection B -> A, and i can see why it's harder than the former question

[sleek thicket]

No, I think I unconvinced myself again

I'm too tired for this

Yeah, B -> A is trickier

Okay, got it

[bleak crescent]

is it fair to say that there's going to be a restriction [0,1]? i'm guessing i can't separate

case 1: x = 1
case 2: x in [0,1)

because i want a continuous function

[sleek thicket]

Your logic is right

Knowing the function on [0,1) will determinate its value at 1

Map two points onto the endpoints, so you just need to get a map A -> (0,1). Map a single point onto 0.5 to bisect the interval. On the left half, write (0, 0.5) = union_n [1/(n+1), 1/n) and map countably many (but not all of!) your half open intervals in A into this segment. In the other segment, repeat this process

This is for a continuous bijection A -> [0,1]

It's probably not the clearest, lol

[bleak crescent]

ooooooooooooooooooh

interesting

[sleek thicket]

Yeah, after your get B -> A I'd recommend thinking about this

[bleak crescent]

got it!

(well, after i show the non-homeomorphic part haha)

sec, shower aaa

[bleak crescent]

@sleek thicket is it fair to say that, because [0,1] is connected, then f([0,1]) must be connected if f is continuous?

so then the image of [0,1] must be entirely contained in one of [0,1), [2,3), [4,5), etc.

[rugged swan]

f(K) is compact if K compact and the image is T2

[west spindle]

you don't need that last condition do you

[rugged swan]

Idk

In a book I've seen someone showing T2 to prove that the image of some continuons equation was compact

I didn't understand why lol

[marsh forge]

Take an open cover of f(K). Preimage of open is open. It pulls back to an open cover of K, which has a finite sub cover. Then the images of the finite subcover cover f(K).

I do not think hausdorff matters here

[midnight jewel]

you need hausdorff if you wanna argue about relations between closedness and compactness

but the image of a compact set is always compact under continuous maps

[marsh forge]

im hijacking this channel

to do pset problems

hope thats ok w everyone

Ok let X be path connected

we have some 1-cycle c

that is, $c=\sum k_i\sigma_i$ for some simplicies $\sigma_i$

[gentle osprey]

MaxJ:

[marsh forge]

and in particular $\del(c)=0$.

[gentle osprey]

MaxJ:
Compile Error! Click the reaction for details. (You may edit your message)

[marsh forge]

I want to show that there is some map $S^1\to X$

[gentle osprey]

MaxJ:

[marsh forge]

such that when I look at the induced map $f_*:H_1(S^1)\to H_1(X)$

[gentle osprey]

MaxJ:

[marsh forge]

ie $\mathbb{Z} \to H_1(X)$

[gentle osprey]

MaxJ:

[marsh forge]

there is a generator of $s \in \mathbb{Z}$

[gentle osprey]

MaxJ:

[marsh forge]

such that $f_*(s)=c$

[gentle osprey]

MaxJ:

[marsh forge]

Ok whats the idea. Well, let's think about what $\del(c)=0$ really means

[gentle osprey]

MaxJ:
Compile Error! Click the reaction for details. (You may edit your message)

[marsh forge]

well if it were over Z/2Z then this would literally just mean we have a loop

but we aren't working over Z/2Z

but if all the coefficients are 1/0 this is the same thing

Lets say I have something simple like a triangle

with sides e1 e2 e3

then I get something like e1-e2+e3 as the oriented loop around

this has a clear map S^1->triangle representing it

how did I get it? well I subdivided by circle and then chose to follow the simplicies between the verticies in the specified direction

Ok in particular a map from $\Delta^1$ is an oriented map $I\to X$

[gentle osprey]

MaxJ:

[marsh forge]

So if $\sigma_1,sigma_2$ are 1-simplicies in $X$, can we in gener compose them?

[gentle osprey]

MaxJ:

[marsh forge]

Yes, if their orientations match up, but not really if they dont

like $a\to b \leftarrow c$

[gentle osprey]

MaxJ:

[marsh forge]

that's not a path

maybe think about this form another angle

given $f:X\to Y$ what exactly is $f_*$

[gentle osprey]

MaxJ:

[marsh forge]

Right ok

so up to homotopy

f takes simplicies to simplicies

in particular we want to look at a map from S^1 to X

we partition S^1 in the obvious way to have "enough" simplicies

then we map S^1->X and match up simplicies

wait not exactly

we're going to want to send the standard delta structure on S^1

to our exact linear combintation

this is basically a pointed loop, but it shouldn't matter where we send the "vertex" as long as its really a loop

in particular we can just look at a map $I\to X$ and prove that if the linear combination is a cycle

[gentle osprey]

MaxJ:

[marsh forge]

then we can descend to a map $S^1\to X$

[gentle osprey]

MaxJ:

[marsh forge]

so how do I want to naively make this "path"

start at a vertex

we have a cycle, so in particular, we have at least one path leaving and entering the vertex

otherwise the values would not ~cancel~

so choose one thats leaving it

we get to the next vertex

now its possible we again have ~many~ choices that we can make

clearly the last one could be arbitrary

can this one?

ok so theres no backtracking that has to be done

why?

because the coefficient ie either negative or positive

but we can always collect these coefficients

so we in particular only have one direction to travel along any vertex.

so clearly the next choice can be arbitrary too

[marsh forge]

i think this idea works

bc you can just keep going

and you either end at the starting point

or "leave" it too many times

...not sure why I need path connected though

[bleak crescent]

hmm

[marsh forge]

maybe once I work out the details this will be clear

[marsh forge]

Ok time to think about another problem

bc ive been very unproductive today

[cunning temple]

Anyone know a good resource for a free particle on the surface of a sphere? I want to do simulations given initial theta, phi and their time derivatives.

[verbal epoch]

What type of material should I have down to start intruductory topology? So far, I've learned Linear Algebra, Number Theory, Graph Theory, and some elementary abstract algebra. Is that enough to start with metric or some intro topology?

[sleek thicket]

It helps to know some analysis

Because a lot of intro topology (esp. metric stuff) is generalizing open balls in R^n

[verbal epoch]

What type of analysis? I've done some analytical calculus course but nothing like real analysis

[sleek thicket]

Real analysis

Preferably real analysis in R^n

I mean there's only so far you can go with that before you need to talk about topology more seriously

But having thought about basic topology of R^n already helped me a lot

When learning real topology

Like, if you know the definition of connected/path connected/compact and like have solved a problem with heine borel, you'll be good

[honest narwhal]

Alright AG is geometry so I'll migrate my ranting here

[gentle osprey]

Daminark:

Daminark:

Daminark:

[gentle osprey]

Daminark:

[wooden scarab]

Do people compute cohomology of spaces from the definition in terms of a cochain-complex or do you usually compute homology and use the various dualities?

It’s not really obvious to me how to ex compute the cohomology of a delta-complex

[deep light]

I'm somewhat stuck on this differential geometry exercise

We're working with the tautological line bundle tau of CP^1

We have the standard charts on U0 and U1, which are

$$ U_0 = { [a, b] \in \mathbb{CP}^1 \mid a \neq 0 }$$

[gentle osprey]

Ridder:

[deep light]

And similarily for U_1

Local frames are given by

$$\mathbf{e}_0(z) = e_1 + z e_2 \text{ and } \mathbf{e}_1(z) = z e_1 + e_2$$

[gentle osprey]

Ridder:

[deep light]

Where e_1 and e_2 are standard basis of C^2

So the usual stuff

Now the question is:

Obviously we can place the tautological line bundle inside CP^1 x C^2

Quite by definition

On CP^1 x C^2 we have the canonical connection

Which should then induce a connection on the tautological line bundle

And the exercise is to show that it's given locally by the 1-forms

$$ A_0(z) = \frac{\overline{z}}{1+|z|^2} \mathrm dz $$

[gentle osprey]

Ridder:

[deep light]

Well on U_0 at least

I've tried the following:

Instead of taking any section and any vector field, just take e_0 and partial/partialz for now

Calculating the connection in the trivial bundle C^2 by definition it becomes 0

By definition of the flat connection

Hence projecting onto the tautological line bundle it stays 0

And well that's kinda a thing because that would mean the connection is 0 on tau as well, which I have to show it isn't

The full question for anyone interested (section b)

[deep light]

<@&286206848099549185>

If there's anyone around knowing some differential geometry 😄

[deep light]

This is the one question I'm having trouble with, so I'd love some help

And it looks so awesome because it is leading up to calculate the first Chern class of the tautological bundle

And thus proving that the tautological bundle is not trivial 😮

But I need some help here

[deep light]

¯_(ツ)_/¯

[zealous bison]

Anyone have a suggestion for an intro to De Rham cohomology for a dumb dumb physicist?

I’m okayish with simplicial Homology... the singular stuff I don’t really understand but would like to

[marsh forge]

I think the idea is kinda straightforward

Any lecture notes online

Should probably work

[zealous bison]

ugh maybe I'll just crank through hatcher

[sleek thicket]

Can someone give me a pointer for this problem?
Suppose h : S^1 -> S^1 is a homeomorphism. I want to extend it to a homeomorphism H : S^1 × I -> S^1 × I such that H(x, 0) = (h(x), 0) and H(x, 1) = (x, 1)

Essentially H is a homotopy from h to id, but with the extra constraint that it's bijective if we add in t

[marsh forge]

Uh

Maybe I’m being dense but can you take the canonical factorization

The mapping cylinder

[sleek thicket]

You are probably not being dense

But I don't understand the mapping cylinder

[marsh forge]

Let me think for a sec

[sleek thicket]

The class this is for has not gotten to homotopy yet, so I might be doing the original problem entirely wrong

I'm not sure if this is even true, my homework problem is to show that if f : 👌D -> 👌D' is a map between closed n, m cells and p, q are interior points of D, D', there is an extension F : D -> D' of f sending p to q

[marsh forge]

Yeah I was pausing

Bc I’m not sure that every automorphism is homotopies to id

Actually I think it’s false

[sleek thicket]

Oh nice

Cool cool cool

[marsh forge]

Take a finite topological space w discrete topology

Any permutation should be an homeo

But I don’t think this is homotopy eq to id

[sleek thicket]

Hold up: isn't a map of degree not equal to 1 noninjective?

Something like that

For the circle specifically

[marsh forge]

Yeah I believe that should be true

[sleek thicket]

I can write down a homotopy

[marsh forge]

Sorry I’m in the middle of a thick green out

[sleek thicket]

For each x in S^1, go along an arc counterclockwise from h(x) to x

Nah you're good

Maybe that's not continuous?

[marsh forge]

I wouldn’t be surprised if all homeos of the circle are homotopic to id

[sleek thicket]

I think the degree thing proves it, right?

"all homeos of the circle are homotopic to id" iff "all homeos of the circle have degree 1"

And if maps of higher degree are noninjective, that's gotta be true

[marsh forge]

Any map of degree > 1 induced an non invertible Map of homology

So in particular it’s not a homeo

[sleek thicket]

Cool

[marsh forge]

Is that how your class defined degree btw?

It’s normally defined by the map on homology

[sleek thicket]

No, we're like 3 weeks away from degrees and don't do homology

[marsh forge]

Oh ok

[sleek thicket]

Which I'm not happy worth

*with

[marsh forge]

Interesting, I guess you can do it for S1 with pi1

But that’s kinda inelegant bc it’s hard to generalize

[sleek thicket]

Yeah that's how it's defined

[marsh forge]

You have to look at pi_n(S^n) in general

Yo generalize that

[sleek thicket]

Makes sense

[marsh forge]

But higher htpy groups are usually intoruced after homology

[sleek thicket]

That's not good enough is the problem

Because it also needs to be bijective

[marsh forge]

I’m a little confused w your approach to the problem

But I haven’t thought about it much

What are you thinking?

[sleek thicket]

The cell problem or the homotopy thing?

[marsh forge]

Cell problem

[sleek thicket]

Wlog they're closed n balls

It suffices to show there's a automorphism of an n balls fixes the boundary and sends p to 0, for any p

Because we can first do that, then do like F(x) = |x| f(x)

And then do it in reverse on the codomain

Does that make sense?

Well I can define a homeomorphism from a closed n ball to itself which sends p to 0 and permutes the boundary

[marsh forge]

Tbh it would suck to write down

[sleek thicket]

Yeah it would

[marsh forge]

But there’s an obvious homeo that does that

[sleek thicket]

Wait that homeomorphism?

Lee does it for me

I have an explicit construction

[marsh forge]

Sorry you have a construction for the homeo sending p to 0

For all p?

[sleek thicket]

For any p

But it can fuck up the boundary

[sleek thicket]

Really all I need to do is (for any p) construct a homeomorphism B^n -> B^n which sends p to 0 and is the identity on the boundary

This should be easy

[marsh forge]

I mean you can describe such a thing

It’s not hard to see it should exist

Just like stretch the ball

[sleek thicket]

Won't that move the boundary around?

The obvious one Im visualizing will

Like what I'm thinking is translate so p is at the origin

Then normalize the length of each point on the boundary

And do the same for the rest of the points

[versed pivot]

you want f : B -> B so that if |x| = 0 then f is the constant map with value p and if |x| = 1 then f is the identity map on the boundary

you can write down a simple such f by deforming linearly between these two maps as |x| varies from 0 to 1

so f(x) = (1 - |x|)p + x

@sleek thicket

[sleek thicket]

Is that a homeomorphism?

I did come up with that map

But it wasn't obvious to me that it's injective

[versed pivot]

hmm

[sleek thicket]

or surjective lol

[versed pivot]

it's easy to visualize what it does so I'm sure it is

but idk how to prove it

[sleek thicket]

yeah for someone who's taking two graduate level geometry classes I sure am bad at visualizing things

lol

[versed pivot]

it sends line segments from 0 to the boundary to line segments from p to the boundary

right?

should be possible to turn that into a proof

[sleek thicket]

That should be (1-|x|)p + |x|x

Right?

Otherwise you'll be too big or smth

[versed pivot]

errr

I'm thinking like this

you want the line segment between 0 and x/|x| to go to the line segment from p to x/|x|

so you get (1 - |x|) p + |x| x/|x|

then the |x|/|x| cancels

[sleek thicket]

Okay yeah I think I'm being dumb

I appreciate the help but I've been doing math since 11am continuously

So I'm going to take a look at this tomorrow morning

[versed pivot]

👌

[obtuse raven]

Could anyone explain to me what the use/motivation is behind subspace topology?

[midnight jewel]

well uh… if you have a space, and you have a subspace, and you wanna talk about open sets in the subspace…

like you have to define it in some way and the way it’s defined seems extremely natural to me, what do you not understand?

[obtuse raven]

I understand the idea but I don't really see a point to it

If the inheritance property were to be consistent I would. But if inheritance is not always the case I don't really see a point.

I'm writing an essay right now and I am starting a chapter about subspaces but I'm having a hard time introducing it

[sleek thicket]

Inheritance?

[obtuse raven]

Something I read about where a subspace inherits property from it's motherspace (Probably wrong terminology but you get what I'm saying right?)

[sleek thicket]

Well yeah, it depends on the property

E.g. compactness is not always inherited

But we can't even ask the question "does a subset S of X have the topological property P" without fixing a topology on S

[obtuse raven]

Of course

But what confuses me is that if inheritance isn't consistent what's the point? I might be missing something, I am no expert

[sleek thicket]

The motivation isn't whether properties are inherited

Actually, let me back up

The unit sphere should be a space

Yeah?

[obtuse raven]

Yeah sure

[sleek thicket]

What's the topology on the unit sphere?

[obtuse raven]

You mean with B(x)

[sleek thicket]

I'm not sure what that means

[obtuse raven]

the open ball thingy

[sleek thicket]

There are lots of open ball thingys

I mean the set { x in R^3 : |x| = 1 }

[obtuse raven]

isn't it just all elements of the sphere

if with topology you mean tau

[sleek thicket]

A topology on X isn't a collection of elements of X

It's a collection of subsets of X

[obtuse raven]

Yes

[sleek thicket]

So what's the topology on the sphere?

[obtuse raven]

well the collection of subsets of said sphere

[sleek thicket]

Not all the subsets though

If it had all the subsets then the sphere would be discrete

[obtuse raven]

oh so you mean it's just a collection of subsets of said sphere

[sleek thicket]

Yup

That's what a topology (on a set) is

Satisfying some axioms

[obtuse raven]

Yeah I know the three you're referring to

[sleek thicket]

So, what's the topology on the sphere?

[obtuse raven]

What? I just told you, a collection of subsets contained within the sphere

[sleek thicket]

Yes but which subsets

[obtuse raven]

subsets for which the distance to the center is less then 1

[sleek thicket]

All such subsets?

What if I take just a point set {x} where x is a point on the sphere

[obtuse raven]

Well that wouldn't be an open set

[sleek thicket]

Yes, exactly

[obtuse raven]

since it's on the border

[sleek thicket]

So what are the open sets?

[obtuse raven]

Informally you could say any subset that isn't on the 'border'

or for which you can make a epsilon-ball

[sleek thicket]

What do you mean by the "border"?

By sphere I just mean the shell, not the whole ball

[bleak crescent]

(i might ask a quick question in a bit when he's finished haha)

[sleek thicket]

I guess my point is that sometimes subsets of a topological space can be interesting spaces in their own right

E.g.

the sphere

We get a topology on it automatically from the topology on R^3

[honest narwhal]

See here's the problem. I think we can agree on the statement "More is better", right? Well, that means the higher dimension the sphere, the more interesting it is

But S^{infty} is contractible

And points aren't interesting

[sleek thicket]

I can't tell how serious you're being

[honest narwhal]

Now you know how it feels

[sleek thicket]

But I want "high dimension better" as a nick

pls

[honest narwhal]

Milnor wants to know your location

[sleek thicket]

Ty

Low dimensional topologists btfo

[obtuse raven]

Oh so you're point is that it can be handy if a topology is carried over to another

[sleek thicket]

Well that's the whole point

It gives us a way to carry the topology over

To a subset

[obtuse raven]

Okay now I see

You're doing god's work my friend

[sleek thicket]

This is actually the second time today lol

[obtuse raven]

Thanks for your patience

[sleek thicket]

I spent all day explaining basic analysis and topology to my cousin

It was very very fun

@bleak crescent

shoot

@honest narwhal dumb question

What is S^infty?

Is it the obvious colimit?

Can we describe the topology more concretely?

oh is it a subspace of the countable product of R with itself?

I think that makes sense

[honest narwhal]

I'd be worried about the norm on such a space

I think you can take the unit ball in a Hilbert space?

(Anyway I should get to sleep soon so see you!)

[bleak crescent]

um, so that problem where we’re supposed to construct a continuous bijection from
B = ... {-2} U {-1} U [0, 1] U [2, 3) U [4, 5) ... to
A = ... {-2} U {-1} U [0, 1) U [2, 3) U [4, 5) ... , where A and B are subspaces of R

is it fair to say that uh

Because [0, 1] is connected, then f([0,1]) would be entirely contained in a connected space because f is continuous?
but f is a bijection, so f([0,1]) would have to be in one of [0, 1), [2, 3), ...

[bleak crescent]

oops, should include @sleek thicket haha

[marsh forge]

@sleek thicket s_infty has the following structure: the colimit is taken over inclusions

So an open set of s infinity is any U such that U\cap S^k is open

For all k

[sleek thicket]

cool, that's what I meant by the obvious one

Oh but you're saying because they're inclusions it's nice

Of course

@bleak crescent that reasoning is valid (you might want to justify it a little bit more)

Note that we could also have image contained in {-1}

[marsh forge]

Always takes me a sec when I have to actually consider element-wise constructions lol

Like real subsets instead of injections

[sleek thicket]

Same

I like just saying "it's a colimit"

But it's good to actually understand what the thing is a little more concretely

[marsh forge]

What would the totally abstract condition be? Take U in the colimit, then U is open if we take the image of the (composed) inclusion, intersect, pull back and then that set is open?

Yuck

What is the actual weak topology of a colimit now that I think about it

You’d just want all the universal maps to be continuous I guess

I think that thinking about it like that would lead you to the specific weak topology here

[bleak crescent]

@sleek thicket image f([0,1]) contained in {-1} would be impossible because f is a bijection, right?

[limpid mural]

Let $\varphi : \mathbb{S}^1 \to \mathbb{R}^3$, $\varphi(t) = ( x(t), y(t), z(t))$ be a smooth embedding such that $z'(t) = y(t)x'(t) \forall t \in \mathbb{S}^1$. Let $\pi:\mathbb{R}^3 \to \mathbb{R}^2$ the projection on the $(x,z) -$plane. Suppose that for $t = 0$ we have that $(x'(0), z'(0)) = (0,0)$, I want to show that if $x(t)$ is A Morse function then the map $\pi \circ \varphi $ is locally like $t \mapsto (t^2, t^3)$ around $t = 0$.

[gentle osprey]

emme:

[limpid mural]

Now

By Morse lemma we have a chart around 0 s.t. X(t) looks like $t^2$ and by the relation $z' = yx'$ we have that $z''(0) = y(0)x''(0)$. Now suppose that $y(0) != 0$. The book I am following says that If we choose the line parallel to the x axis, passing through the singularity as the first coordinate axis and the line with slope $y(0) = \frac{z"(0)}{x"(0)}$ ( also passing through the singularity) as the second coordinate axis around the singularity, we get a local coordinate system in the (x,z) plane such that the projection is given by $t \mapsto (t^2, f(t))$ where $f$ is such that $f(0) = f'(0) = f"(0)=0$

[gentle osprey]

emme:

[limpid mural]

My question is: why $f"(0) = 0$? The change of coordinates in the plane seems to be linear, so I don't understand why we get a function $f$ such that $f"(0) = 0$

[gentle osprey]

emme:

[sleek thicket]

@bleak crescent yes but you need to consider that case

You're trying to show such and f can't exist

[bleak crescent]

yeah

right now, i'm thinking more about how i'd construct that function period haha

[sleek thicket]

Go back to what you showed earlier

Wait hang on

What was the originating problem?

*original

I don't think f is a bijection, is it?

Maybe I forgot how it worked

[bleak crescent]

one sec

the continuous bijection from B to A haha

[sleek thicket]

Ah okay

So we're pretty much done

The image of [0, 1] is compact and connected

And it sits inside one of the [a, b)

What are the compact connected subsets of the real line?

@bleak crescent

You can use this to determine f([a, b))

[bleak crescent]

hmm

all closed, bounded intervals or one-point sets, right?

hmm, so all f([a,b]) should also sit inside other [c,d)....

[verbal epoch]

Book recommendations for self-studying intoductory topology?

[marsh forge]

Munkres

Or Janich

[sleek thicket]

@bleak crescent so is the image open in the subspace topology on A?

[bleak crescent]

could you theoretically

map

[0, 1] to [0, 0.5) and

map [2, 3) to [0.5, 1)?

and map [4, 5) to [2, 3)

and map [6, 7) to [4, 5)

and so on?

@sleek thicket

[sleek thicket]

No, that first bit couldn't work

[bleak crescent]

ah

[sleek thicket]

[0,0.5) isn't closed

But

That's not enough

[bleak crescent]

ah, right

[sleek thicket]

You could map [0,1] to [0,0.5]

And [4,5) to (0.5, 0.75]

And [2, 3) to [2,3)

And [6,7) to [4,5)

And so on

That's a continuous bijection between the two

Unless I'm missing something

[bleak crescent]

how would a map from [4, 5) to (0.5, 0.75] work?

oh

[sleek thicket]

Flip it

Bop it

[bleak crescent]

reverse it?

yeah

[sleek thicket]

Twist it

Squish it

[bleak crescent]

oooooh

[sleek thicket]

No I'm kidding

Just reverse and scale

You see what I mean though?

[bleak crescent]

yeah

[sleek thicket]

Then the image of those two will be [0, 0.75)

So we're back where we started

And we can keep filling it out with every other interval

This will define a continuous bijection from B to A

[bleak crescent]

wait, i don’t think i fully understand—what happens to (0.75, 1)

[sleek thicket]

You map [8, 9) to (1 - 1/4, 1 - 1/8]

And then [10,11) to [6,7)

[bleak crescent]

oh so you’re constantly

getting smaller and smaller left half open intervals

[sleek thicket]

Yup

[bleak crescent]

to “fill out” but never quite reach 1

[sleek thicket]

Yup

Index the half open intervals by positive naturals

[bleak crescent]

powers of 2

[sleek thicket]

Send the (2i-1)th interval to (1 - 2^(-i), 1 - 2^(-i-1)]

Send the 2nth interval to [n,n+1)

And send [0,1] to [0,1/2]

[bleak crescent]

yeah

[sleek thicket]

And leave each discrete point where it is

[bleak crescent]

aaaaaaa that’s clever

[sleek thicket]

Do you see what I mean?

Continuous bijection

I think I came up with this the last time you posted it

So back to the problem

We have [0,1] mapping to [a, b], which is contained in some half open interval

Right?

Is that set [a, b] open in A?

[bleak crescent]

i want to say no because [a, b] isn’t open in R

[sleek thicket]

That's not quite enough though

Since e.g. [a, b] is open in itself

[bleak crescent]

but each [a, b] in A would be the intersection of a closed interval in R with A, right?

and the only sets open and closed in R are R or empty set

[sleek thicket]

That's still not enough

{-1} is the intersection of A with both [-1.5,-0.5] and (-1.5,-0.5)

[bleak crescent]

oh oops

[sleek thicket]

Remember that closed and open aren't opposites

[bleak crescent]

so [a, b] is necessarily the intersection of A and either (c, b] or [c, b] in R, where c <= a, because [a, b] is contained in a half-open interval.
but neither (c, b] nor [c, b] is open in R

and [a, b] would have to be the intersection of A and a set open in R to be open in the subspace topology of A

[bleak crescent]

@sleek thicket wrote it out!! and thought about it more hahaha

thanks for the help

[sleek thicket]

Nice!

Sorry, I forgot about this lol

But good job

I'll take a look if you post it

[bleak crescent]

i just had to write about it on paper haha

[sleek thicket]

Ah okay no worries

[bleak crescent]

i think my prof wants to start algebraic topology tomorrow, so this was a way to segue into it for him

(though i’m not quite sure how yet)

[sleek thicket]

That would be really fast

What are you doing tomorrow?

[bleak crescent]

i think we start... defining homotopies? @sleek thicket

we're doing sections 51 and 52 of munkres this week, i believe

[sleek thicket]

Oh gosh

[bleak crescent]

so homotopy and stuff and then fundamental group

we just finished urysohn metrization theorem

[sleek thicket]

oh okay

Sorry I thought you were just doing like subspaces

For some reason

I'm pretty jealous ngl

[bleak crescent]

ahh, nah, this was a review problem haha

[sleek thicket]

My course is doing the classification of surfaces

Which is neat

But it relies on the existence of a triangulation of everh surface

Which we don't prove

[bleak crescent]

ooooooh

[sleek thicket]

And we don't get to homotopy for like 2 more weeks

[bleak crescent]

my prof is probably skipping what you're covering this week and next week haha

[sleek thicket]

Yeah, probably. We're focused on manifolds

It's taught by the manifold guy

And the next two quarters are on smooth manifolds

[bleak crescent]

oooooooooooooh

[sleek thicket]

Oh sorry by "the manifolds guy" I mean Jack Lee

Author of introduction to topological/smooth manifolds

[bleak crescent]

oooo that's really cool haha

hopefully great teacher, too

[sleek thicket]

He's pretty great

[bleak crescent]

i've only had like one or two profs who were relatively famous, and they ended up being really disorganized, unfortunately, hahaha

glad you don't have the same experience so far

[gritty widget]

What do you guys recommend to learn topology? Is munkres fine?

[bitter yoke]

It's fine

[chilly silo]

I used Munkres and liked it

My class went a little fast, but if you're learning on your own and just covering the point set part of the book, you'll be good

[gritty widget]

Ive tried reading few books and they all had different approaches and definitions

[chilly silo]

different definitions

do you just mean notational nuances

or do you mean, like

wholly different notions

[gritty widget]

I mean, some definitions are equivalentl but stated differently

[chilly silo]

ok sure

there's not much of a difference there.

[gritty widget]

My course doesnt really use much of the limit definitions (hard to give example now, Im just starting) , which I found a lot of in few books

[chilly silo]

who uses limits in topology lmao

[gritty widget]

Kuratowski

[chilly silo]

only time I used limits in my gen top course was for certain metrics

you mean the textbook or the embedding theorem?

[gritty widget]

Kuratowskis text book

[chilly silo]

ah

well, I've never read it

seems too analytic

[gritty widget]

It really is

[bitter yoke]

What is it called

Nets or something

[marsh forge]

Limit points are a big part of topology definitions

You don’t need a metric, and they characterize closed sets

Nets are a different thing tho

[sleek thicket]

Limits are good

If you're just doing stuff in second (or first I guess?) countable hausdorff spaces

wowee I wonder why people like manifolds

[marsh forge]

Wait

I could be wrong or my point set rusty

but afaik limit points are completely general

[sleek thicket]

Yeah I meant genuine limits

Of sequences

[marsh forge]

Oh ok

Yeah then you need nets

[sleek thicket]

I'm thinking about covering spaces

Let p : X' -> X and q : Y' -> Y be covering maps

And let x, y be basepoints for X, Y

Consider the subspace Z = p¯¹(x) × Y' union X' × p¯¹(y) of X'×Y'

Is Z a covering space of X wedge Y?

Let r : Z -> wedge be the product of the maps p and q

Where wedge = {x}×Y union X×{y}

I guess Z = (p, q)¯¹(wedge)?

I'm just thinking out loud but if anybody wants to help that would be grand

Okay so I think it's definitely connected

I'm fine assuming X and Y are path connected, in which case that's easy

And r is continuous and surgeons

*surjective

So take a point (a, b) in the wedge

If it's the point (x, y) then take canonical neighborhoods U and V for x and y wrt p and q

Then U×{y} union {x}×V is the intersection of the wedge with U×V, and it's preimage under r is p¯¹(U)×q¯¹(y) union p¯¹(x)×q¯¹(V)

p¯¹(U) is the disjoint union of U_i for which p restricts to a homeomorphism U_i -> U

Similarly V_j for V

So this preimage is the disjoint union of the U_i × q¯¹(y) unioned with the disjoint union of the p¯¹(x) × V_j

Oh and each point of p¯¹(x) lies within one of the U_i

So really the index set for the i's is the fiber of x

[sleek thicket]

lol im fucked

talked to AG professor today

turns out we're going to try and speedrun schemes and cohomology next quarter

lemme just uhhhh

read two chapters of hartshorne

in 10 weeks

[floral gust]

Is this correct?

[sleek thicket]

Yup, looks right

Suppose p : X -> Y is a covering map between nice spaces

Don't assume it's a universal cover though

[honest narwhal]

Too late

[sleek thicket]

/(

:(

Let y0 be a basepoint for Y

We get an action of Aut(p) on X

And also on the fundamental groupoid of X with basepoints p^(-1)(y0)

And this is like a nice action

In that it acts by functors

Is there a way to get an action of Aut(p) on π(X)? I'm worried about noncanonical choices or w/e but I think it just works

If you contract the groupoid to a point

If so, more interesting question: can we interpret the semidirect product of π(X) and Aut(p) topologically?

In some nice way?

@honest narwhal you responded so you have to help

also @marsh forge you do algebraic topology

this is that

[honest narwhal]

My AT is really bad lmao

[sleek thicket]

Yeah me too

Mainly because I've never actually learned it

Just read the dumb groupoids book on my own

[honest narwhal]

Ronnie Brown wants to know your location

[sleek thicket]

I mean it's one of my favorite books

But also

Complete trash

[sleek thicket]

Hey uhhh

Let's attach B^(n+1) to P^n

For fun

So my thought is

P^(n+1) has an affine open like [1:...] in it

Wait sorry I mean it has a chart, wrong class

I want to map the interior of B^(n+1) in there

So like send it to the hyperplane at x=1 in R^(n+2)

And then quotient down to P^(n+1)

And handle the boundary separately

So

How do I handle the boundary?

I feel like I just use the quotient map S^(n+1) -> P^n

But won't the embedding of the interior make that fucked up?

Maybe I should do an example?

But I think I need to jump up to 4 dimensions

In order to attach B^3 to P^2

Maybe I'm thinking about this backwards

Or too forwards

There's an adjunction space P^n cup_p B^(n+1)

Where p : S^n -> P^n is the quotient map

Can I abstractly show this is homeomorphic to P^(n+1)?

[gritty widget]

what do you mean by "abstractly"?

[sleek thicket]

I mean without plumbing along a bunch of coordinates and quotient maps

[rugged swan]

let $X$ and $Y$ be topological spaces. Let $(A_i){i \in I}$ be a family of closed subsets of X satisfying $\bigcup \limits{i\in I} A_i = X$. Let $(f_i)$ be a family of continuous function from $X$ to $Y$. If I define $g : x \in A_i \mapsto f_i(x)$ ($g : X \rightarrow Y$) for all $i \in I$. Is it true that if $g$ is well-defined, then $g$ is continuous ?

[gentle osprey]

Zak:

[rugged swan]

I'm asking this question because of the fact that it is not easy to proove that, for instance, the composition of two paths is a path

it is obvious but the rigorous proof isn't easy

(or I don't see an easier proof, idk)

(I'm saying that to prove this is continuous isn't easy)