point-set-topology

[sweet wing]

soooo

[sharp yoke]

🤨

im pretty sure ur not supposed to do it this way. Meaning ur not supposed to think of the 2n-gon embedded in R^2 and then say its closed and bounded and hence compact

i could be wrong

im definitely supposed to use this theorem

[sweet wing]

yup that's basically the proof

cuz well

you need to give a topology to the 2n-gon

and the only reasonable one is subspace of R^2

[sharp yoke]

im pretty sure ur not supposed to do it that way but

yea im 100% certain ur not supposed to do it that way

[sweet wing]

i dont rlly see other way rn lol

[sharp yoke]

you didn't even use that the polygon has an even number of sides

[gritty widget]

What do you mean you're not supposed to do it that way?

[sweet wing]

this is really so that the quotient makes sense

[sharp yoke]

oh true

that proof just doesn't seem at all connected to what i was reading in the section of this book

that asks this question

[sweet wing]

(see "surface" part of the thing)

occasionally you still do need to go back to point set stuff to prove things

(btw you still need to show it is a surface)

[sharp yoke]

yea i think ur proof works i just don't think it was the intended proof

[sweet wing]

this is why 2n

[sharp yoke]

why does 2n let you show its a surface

this book defines surface as a connected 2-manifold

[sweet wing]

yup

so you jus need to show at every point

you can find a open set homeomorphic to R^2

there are really only 3 points to check

[sharp yoke]

so basically ur saying if there's an odd number of edges there's going to be one edge that can't be paired up. Then any nbhd of a point on that edge can't be homeomorphic to R^2

which is why we need an even number of edges

[sweet wing]

yup!

[sharp yoke]

this still feels like it is completely unrelated to the proof it wants us to do. Why would they give us theorem 5.1 and this polygon naming scheme if we're not supposed to use it in the exercise

plus the case of odd edges would still work but instead you would just prove its a 2-manifold with boundary

it just doesn't feel like ur supposed to do it that way

i could be wrong tho

[sweet wing]

it's just so that like

whatever you are doing

actually makes sense

it's very common to have to fall back to basic point set to make sure everything you're doing makes sense

like

what if there is some naming scheme that doesnt result in a compact surface

then youll have to somehow be more precise in what is "allowed" and ahat sint

[sharp yoke]

that's exactly the proof i was trying to think about

you suppose for contradiction that the 2n-gon isn't compact (for some naming scheme)

then theorem 5.1 tells you its not homeomorphic to a sphere, connected sum of tori or connected sum of projective planes

and then ur somehow gonna produce a contradiction from there

[sweet wing]

uhhhhh

that feels it's going to be cyclic somewhere

[sweet wing]

to prove this classification theorem you kinda need that you get compact surfaces from identifying n-gons?

[sharp yoke]

yea i was just reading through that

i think you're right

im just gonna heine borel like a boss fuck math and fuck this p set

[sweet wing]

it's really jus a point set exercise to make sure nothing is wrong

[sharp yoke]

im going to be very proof sketchy about homeomorphisms on the boundary and hope i don't get points taken off

[sweet wing]

ik it is very meh but these kinda basic stuff to make sure everything is ok is really jus point set

[summer jolt]

If $f \in C^\infty(M)$ then is $X(f) \in C^\infty(M)$ for $X \in \mathfrak{X}(M)$?

[gentle osprey]

snypehype

[bitter yoke]

yes

[summer jolt]

That's because $X(f)(p) \in \mathbb{R}$ for $p \in M$? So we have $X(f) : M \rightarrow \mathbb{R}$?

[gentle osprey]

snypehype

[bitter yoke]

I mean that's how you define X(f) yes

but you would still need to show that its smooth

[uncut surge]

okay people i need some intuition

I've never really gotten used to working with complex forms, because to me it's really weird that you suddenly have to start caring about _anti_holomorphic things. Is there a good intuitive reason why, if we're generally only interested in holomorphic functions, and when we require all transition functions of a complex manifold to be holomorphic, why suddenly "antiholomorphic" forms like dx - i dy start showing up?

Would I be missing a lot if all my life i just worked with holomorphic forms and just ignore the complex-conjugated coordinate \bar{z}

okay i guess the dolbeaut operator absolutely requires antiholomorphic things to exist, but only if i've already admitted that i care about non-holomorphic sections in my bundle or something

[gritty widget]

can any symplectic kings or queens help me see why symplectomorphisms pull back hamiltonian vector fields?

if $f$ is a symplectomorphism and $X_H$ a hamiltonian vector field, i want to see that $f^*X_H=X_{H\circ f}$

[gentle osprey]

lime_soup

[gritty widget]

I know that $X_H$ is defined to be the unique vector field satisfying $\omega(X_H, -)=dH$. so $X_{H\circ f}$ vector field satisfying $\omega(X_{H\circ f}, -)=d(H\circ f)$

[gentle osprey]

lime_soup

[gritty widget]

But then $d(H \circ f) = f^dH=f^\omega(X_H,-)$

[gentle osprey]

lime_soup

[gritty widget]

But this gives $\omega(df X_H, df(-))$ which I can't see why this is equal to $\omega(X_{H\circ f},-)$

[gentle osprey]

lime_soup

[uncut surge]

First, we should talk about the pushforward of the Hamiltonian vector field, rather than the pullback, right?

That probably doesn'T make anything clearer yet but it's good to be precise lol

If it holds true that $\iota_{f_* X} \omega = (f^{-1})^* ( \iota_X (f^* \omega))$, where $f_$ and $f^$ are the pushforward and pullback respectively, then the statement is true fairly straightforwardly; here $\iota_X \omega$ denotes the interior product of the vector field $X$ with the form $\omega$

awww shit okay i forgot that a symplectomorphism is also a diffeomorphism

[gentle osprey]

Lartomato

[uncut surge]

Okay, the important thing is to show that $f^* X_H$ indeed fulfils the defining equation for the Hamiltonian vector field of $H \circ f$, that is a good starting point. Hence, consider $\iota_{f^* X_H} \omega$, you want to show that this is $d (H \circ f)$. I believe something like $ \iota_{f^* X_H} \omega = f^* \iota_X (f^{-1})^* \omega$ should follow very generally for all diffeomorphisms $f$, vector fields $X$ and forms $\omega$.

[gentle osprey]

Lartomato

[uncut surge]

Once you have this, use that f (and f^-1) are symplectomorphisms, use the definition of X_H as a Hamiltonian vector field, and use that the de Rham differential commutes with pullbacks

@gritty widget now i can ping you without breaking my tex

[frigid river]

I simplified my task and am now wanting to prove this first: "The set of isolated points, $A$, of $\mathbb{Z}$ is a dense subset of $\mathbb{Z}$." Could you please tell me whether my argument it correct or not? Since $X={x}$ is open for all $x\in \mathbb{Z}$, every $x \in \mathbb{Z}$ is an isolated point. Therefore, $A=\mathbb{Z}$. Hence, $\bar{A} = \mathbb{Z}$.

[gentle osprey]

veryhappyperson

[uncut surge]

I guess the topology on Z is the one you get from considering it as a subspace of the real numbers? Then yeah, this seems fine; you may want to make sure that you understand why a point x is isolated if {x} is open (and also why {x} is open)

[marsh forge]

Yeah you need to be sure 1) you're using the discrete topology or subspace topology (they coincide)

and 2) which of them you are using in order to prove you have open points

The original question makes me unsure this is the correct topology because I don't know why they'd phrase it that way

[frigid river]

Thank you. The original question was "Prove that the set of isolated points of a countable complete metric space X forms a dense subset of X." I was just trying to prove a simpler case first, because I had no idea how to approach this.

[marsh forge]

Ah I see

[frigid river]

Honestly, I have still no idea how to approach this xD It doesn't seem like I can transfer my simple argument to this problem.

[gritty widget]

ah thank you this makes sense

i abusing notation and just writting an empty entry

so i didn't take into account the extra factor from the interior product

[candid mulch]

The definition for a derivation involves a linear map from an algebra to itself. Apparently there's a more general definition from an algebra to a bimodule over the algebra.

Now, I'm confused since when defining tangent spaces via derivations, we take a 'derivation' at some point $x$ as $D: C^\infty(M)\to\mathbb{R}$, but this doesn't seem to be covered by the definitions above (or it might be and I'm just big dumb). What am I missing?

[gentle osprey]

tanninStains

[candid mulch]

I guess I'm just confused on what can constitute a derivation. Is it enough for the domain and target to be algebras over the same field as long as the Leibniz condition is satisfied? Is that more than enough?

[uncut surge]

The tangent space definition at a point is a bit of an outlier here, since it works by evaluating at a specific point; so D(fg) = f(x) D(g) + D(f) g(x); you of course can't do that in a general algebra; you think of this kind of derivation D(f) as "first you 'derive' f in some way, and then you evaluate at x"

I'm wondering if there's some point of view to rescue this...

Yeah, I think the correct notion here would be the algebraic notion of a derivation of modules, i.e. if you have an algebra $A$ and an $A$-module $M$, then a function $f : A \to M$ is a derivation if $f(ab) = a \cdot_M f(b) + f(a) \cdot_M b$, where $\cdot_M$ denotes the action of $A$ on the module $M$. In your setting, the algebra would be $A = C^\infty(M)$, the module is $M = \mathbb{R}$ and the module structure of a function $f \in C^\infty(M)$ on a real number $r \in \mathbb{R}$ is evaluation at the point $x$ of the manifold, so $f \cdot_M r := f(x) \cdot r$

[gentle osprey]

Lartomato

[uncut surge]

This is not a super common perspective from my experience, but it's the correct one here at least 🤷

@candid mulch

[frigid river]

Sorry to ask again, but could someone please give me an example for a subset of a metric space, which closure has an empty interior? I am having a hard time comprehending such things.

bruh

[honest terrace]

a less trivial example:

[frigid river]

and the union of multiple {} is then of first category, right?

[honest terrace]

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

there's definitely simpler examples

like {a} for any a in R

it's closed so it's its own closure (speaking about the cantor set, but that's also true for {a} lol)

[candid mulch]

Hm, it's weird to think of R as a C^infty(M)-module... but I think I see it. Thanks 🙂

[frigid river]

That makes sense xD Thank you. But the union of such sets is of first category, right? So does that mean [0,1] is of first category, for example?

[honest terrace]

I'm sorry i'm not very familiar with the english maths vocabulary, what does being of first category means here ? @frigid river

[uncut surge]

Yeah, you'd think it would go the other way round 😄 But that's exactly what an evaluation map does to a function!

[frigid river]

"A subset is of the first category if it is the countable union of nowhere dense subsets."

[uncut surge]

https://en.wikipedia.org/wiki/Meagre_set This is an alternative name for "of first category"

(find your own language in the sidebar)

[frigid river]

Ohh 😦

What would be? Nothing in R?

Is there a meagre subset of R?

Is that the only one?

[honest terrace]

cantor set is nowhere dense, so indeed it is a countable union of nowhere dense subsets

[honest terrace]

thanks !

[frigid river]

Wait, is "nowhere dense subset" equivalent to "countable union of nowhere dense dubsets"?

[honest terrace]

no

here i'm just using the implication

(since if A is a nowhere dense subset, then the union U A is countable (only one element) and well, every subset of the union is a nowhere dense subset)

[frigid river]

Okay, thank you. I think I have at least some intuition now.

[gritty widget]

for example Q is a countable union of nowhere dense subsets but is dense everywhere in R

[frigid river]

Just to be sure, "nowhere dense subset" implies "countable union of nowhere dense subsets", right?

[honest terrace]

yes

but the conserve is false, as the counter example of carla shows

[frigid river]

Does this imply Q has an empty interior?

[frigid river]

If I remove a non-isolated point, x, from a countable metric space, X, must {x} necessarily be closed in X?

Or is is it just not open?

$X\setminus{x}$

[gentle osprey]

veryhappyperson

[frigid river]

Now I am confused. Isn't an isolated point a "finite point set" that is open?

"Anyway finite point sets are always closed in metric spaces"

So an isolated point is open as well as closed? I am sorry if I misinterpret everything xD

That is weird.

Okay, thanks. I can finish up my proof then, I think.

[marsh forge]

(sets that are both closed and open are what we call (unions of) connected components of a topological space)

[frigid river]

@gritty widget One last question (I hope xD): Does this mean the complement of the set is open and closed as well?

[marsh forge]

yes

[frigid river]

Okay, thanks, again.

[tough imp]

What

Is this actually true?

Any clopen set is a union of connected components?

[marsh forge]

yes

its not so hard to prove

the disconnecting open sets are immediate

[tough imp]

It isn’t true the other direction right?

[marsh forge]

@tough imp it's always true that they are closed and the converse is true if there are finitely many connected components

[tough imp]

Yeah

But not in general

[marsh forge]

no

luckily all good spaces

have only like 3 components at most

[fading vale]

is my proof incorrect

or did i like implicitly call upon p(z) and r(z) having no common zeroes somewhere

[frigid river]

I hope Moth isn't mad because of my interfering with his question, but is this finally correct now? And if it is, why only for "countable" metric spaces?

I take everything back, give me 5 min to rethink

xD

[marsh forge]

side note

im gonna start ominously putting rigorous terms in quotes

just to keep people on their toes

The function f can clearly be seen to be injective and "continuous"

[frigid river]

I meant to write x, not U_x.

[sleek thicket]

.pin

[frigid river]

{x}

I never said that though. I meant that it is not open, therefore it is only closed, not open as well.

Why not? If {x} is closed, its compliment must be open, does it not?

xD Nothing, just wanted to let the reader know that I knew xD

Why? Does Baire only apply when a set is only open, not closed as well?

And to finish this up, why does this only work for a countable metric space?

My book only states "Let ${U_n}{n=1}^{\infty}$ be a sequence of dense open subsets of a complete metric space $X$. Then $\bigcap{n=1}^{\infty} U_n$ is also dense in $X$. It doesn't say anything about ti only applying for countable union, except I am missing something again.

bruh

Okay, thank you, yet again.

[gentle osprey]

veryhappyperson

[long coyote]

Let $\alpha:I\to S^2$ be injective. Show that $\alpha(I)$ has an empty interior.

[gentle osprey]

亜城木 夢叶

[sleek thicket]

Is I an open or closed interval @long coyote?

[long coyote]

I is closed,

[sleek thicket]

Sure, so α is automatically a homeomorphism onto its image

Right?

[long coyote]

yes

[sleek thicket]

So suppose the image had nonempty interior

Then it would contain a ball

(something homeomorphic to the disk in R^2)

We could pull that back along α to get a subset of I homeomorphic to a disk

Try to prove this is impossible

[long coyote]

i see

[tawdry widget]

α is not surjective therefore we can assume that it maps I to S \ {s} which is homeomrophic to R^2

Using that fact that I is compact we can get that α(I) is a measure-zero set

Where any open ball isn’t

[sleek thicket]

How do you get that?

There are certainly compact sets that have positive measure

Even ones with empty interior

(eg take a fat cantor set in R and look at its product with itself)

[tawdry widget]

Sorry it is a mistake it isn’t because I is compact. The set of rational numbers in I is {x_1,x_2,...} any positive real number ε we can find a open neighborhood I_n of x_n in I such that α(I_n) is contained in B(x_n, ε/(2^n)

[sleek thicket]

what?

Are you saying compact => measure zero?

[tawdry widget]

Not any more

It is not true as you said

[sleek thicket]

Right so I think the issue with this proof is that the I_n might not cover I

But I'm having trouble understanding it

[tawdry widget]

Dead end ... I am completely lost😂

Btw how to illustrate that it’s impossible for I to contain a open subset homeomorphic to an open disk in R^2...

[sleek thicket]

connectedness

If you remove a point from a connected open subset of I you'll get at least two connected components

Wait no sorry

If U is a connected open subset of I then it must contain some point x distinct from the endpoints of I. Removing such a point disconnects U

But removing any point from a disk will leave it connected

For higher dimensional versions of this question ("invariance of dimension") you need something stronger like homology

[tawdry widget]

I see. Thanks I got it.

[obtuse meteor]

Statement: Computing the cellular boundary maps action on a 2-cell is easy--just read off the coefficients from the exponents of the word you glue along. You can't do this for >=3-cells

Question: But what if instead of a free group--you had something better to glue along???
Question: Am I about to reinvent operads?

[sleek thicket]

Tfw you like Faye's tweet and then you see it on discord

[obtuse meteor]

figured someone here might know the answer?

[marsh forge]

im not sure i understand the question

[manic garden]

1. what 2) no

[sleek thicket]

This seems like a very strong reaction

[obtuse meteor]

do you understand the statement?

[marsh forge]

let me revisit in the morning but i dont see an immediate connection to operads

[obtuse meteor]

mhm

as for rephrasing the question

Question: Combinatorially we describe gluing data of a 2-cell onto the 1-skeleton by specifying a word (aka an element of a free group) in the 1-cells. Is there a corresponding type of structure that gives the gluing data of an n-cell onto the (n - 1)-skeleton, and can you get a similar result about computing cellular homology

[manic garden]

well, the word structure comes from the fact that an attaching map for a 2-cell is a map from S^1, i.e. an element of pi_1, so part of some non-abelian group

but looking at pi_2 and above is abelian, so "words" in this sense don't really mean much

[obtuse meteor]

sure

but I don't think the gluing data is given by just looking at the coefficients in pi_2 right?

or am I missing something?

[manic garden]

uh, sort of

that's pretty much the crux of obstruction theory

[obtuse meteor]

:o neat!

[sleek thicket]

Here's something interesting: the image of a continuous injective map I -> R^2 might not have measure zero!

https://en.m.wikipedia.org/wiki/Osgood_curve

[gritty widget]

lebesgue measure 🤢

[shut moat]

real mathematicians use the counting measure on everything

[sleek thicket]

You are all degenerates

Tterra especially

Lebesgue measure is neat

[hazy siren]

can someone help define total boundedness in metric spaces?

[gritty widget]

go on wikipedia lol

[sleek thicket]

mean

[gritty widget]

a metric space is totally bounded if for every r > 0 there's a finite covering by balls of radius r

[hazy siren]

ok this definition makes a lot more sense than wikipedias

thanks

[gritty widget]

that was the definition from wikipedia

[hazy siren]

i found the epsilon net def wth

eh nvm just saw it

[quasi forum]

I'm having a real tough time with this one. I could really use some help understanding this

[marble socket]

(try to look at the contrapositive of the condition)

[quasi forum]

If the arbitrary intersection is empty, then there exists a finite subset where the finite intersection is also empty.

[marble socket]

yep!

[quasi forum]

Ohhhh, then if that's the case, the complement is the space itself!

[marble socket]

exactly! and complements of closed subsets are open subsets!

[quasi forum]

So that arbitrary union is an open cover of X

Wow, that's really straightforward.

[marble socket]

hehe

[quasi forum]

I need help on another one, but let me wrap my head around this one first real quick.

Thank you very much!

[marble socket]

[quasi forum]

This is the next one. I also ran into a similar issue where I wasn't really sure where to start.

So hold on. Does X not being compact imply it is either not bounded or not closed?
I know the converse is true, but I don't think this implication is true

So how is that hint useful?

Wait... X is a subset of the real numbers

Heine Borel is at play here

[marble socket]

yep

[quasi forum]

We haven't proved that direction in class yet, so it's time to take things for granted!

[marble socket]

hehe okie

so say X was unbounded, can you give me an unbounded function from X --> R?

[quasi forum]

The most trivial would be f(x)=x

[marble socket]

is that continuous?

[quasi forum]

Well, $|x-c|<\delta \Rightarrow |x-c|<\epsilon$ where $\epsilon=\delta$

[gentle osprey]

dackid

[marble socket]

yep!

so we're done with this case!

what if X was not closed? This means that there is a limit point p of X which is not in X

[quasi forum]

Yea, I agree with that

[marble socket]

so try to find a an unbounded function from X --> R

[quasi forum]

Then what about $\frac{1}{x-p}$

[gentle osprey]

dackid

[marble socket]

yep this works!

[quasi forum]

Proving continuity might be a little trickier

[marble socket]

this is continuous on R\{p} so continuous on any subset of R\{p}

[quasi forum]

You mind if I try the epsilon delta way? I could use the practice

[marble socket]

yea okie

[quasi forum]

Okay, I am a little stuck

I am not sure how to deal with that x-p

[marble socket]

so we know that p and c are not the same

[quasi forum]

Absolutely

[marble socket]

but if you let delta be huge, then x might get very near p

so first lets assume delta is small enough, we'll assume delta <= |c-p|/2 this will guarantee that x doesn't not go very near p

[quasi forum]

Okay, I think I follow

[marble socket]

what can you say about |x-p| now

[quasi forum]

Would it be smaller than |c-p|/2?

[marble socket]

yep

(if you want to see that, look at |p-c| <= |x-p| + |x-c| <= |x-p| + |p-c|/2)

[marble socket]

oh no, its greater

we pick x near c, to make x away from p

In any case, we have |x-p| >= |p-c|/2

[quasi forum]

I'm sorry, I'm not sure I see how we got there

Ohhhh, wait a sec. I see it now

[marble socket]

draw a picture plot p and c and draw x near |p-c|/2 distance of c

[marble socket]

using this, we get |f(x) - f(c)| <= (2/|p-c|^2)* |x-c|

[quasi forum]

Yep! I see it now

[marble socket]

so i can take delta = epsilon * |p-c|^2/2

but recall we wanted delta to be smaller than |p-c|/2

else our calculations would be wrong

so final delta should be min(|p-c|/2, epsilon*|p-c|^2/2)

[quasi forum]

Wait, why the minimum? I understand the epsilon term

I'm just not sure why that isn't sufficient here

[marble socket]

sorry i was afk

so the given epsilon could be huge

[quasi forum]

This is true

[marble socket]

what if the epsilon given to you was 2/|p-c|

[quasi forum]

Then delta would just be |p-c| (if we just look at the epsilon term in the mininum).

[marble socket]

and then x could get very close to p, as p was a limit point

and our calculation was based of the rough estimate the our |x-c| <= |p-c|/2

so this wouldn't apply and f could be huge

[quasi forum]

Ah, I see. So the minimum accounts for both cases

[marble socket]

yep

[quasi forum]

Ah okay. I see now. Thank you for bearing with me, I'm not sure why I am struggling so much on this one

And now we know the function is continuous and unbounded, so we can conclude the proof

[marble socket]

yep!

or like i said, you could use the fact that if you have 2 real valued functions f and g which are continuous at a, and g(a) != 0, then f/g is continuous at a

the proof is pretty much what we just did.

[quasi forum]

Oh... well, that works too :p

[shy blaze]

does anyone know here whether Christoffel symbols are constant or dependent on the position in a bundle?

I think they should be dependent on the position in a bundle, but I am not sure

[uncut surge]

Yeah they're functions of the position

[shy blaze]

thank you very much!

[uncut surge]

Wasn't much, but no problem 😄

[shy blaze]

heh

the notation was just a bit confusing sometimes

especially with the einstein summation convention in the mix as well

[chrome dew]

they're basically the components of the partial derivatives of your basis vectors, at least understanding that the question becomes obvious

[shy blaze]

and the book Im reading uses differential operators as vectors as well, which confused me at first but makes sense

[uncut surge]

Yeah exactly, what merosity said

[shy blaze]

yeah its related to vertical and horizontal subspaces and covariant differentiation

[uncut surge]

The perspective of "tangent vectors = differential operators" is very common (fundamental even) to differential geometry, so it's not all that strange; tho yeah, it takes getting used to

[shy blaze]

I can type out how I understand it and you guys can see if its correct maybe

so as far as I understand, the vertical subspace of a principal bundle is very simple, and is just tangent to the fiber, so d/dg_i are the basis vectors, then for the horizontal subspace the basis is of the form

d/dx_mu + Gamma_{mu, i, j}g_j d/dg_i
``` (with einstein summation over i and j), and the Christoffel symbols sort of correct for the curvature of the space (so in the direction of your fiber)

which then depends on the position in your bundle (makes sense), and the manifold and fiber themselves

is that about correct?

[uncut surge]

So the g_i are the fibre-coordinates I suppose

[shy blaze]

yes

yeah a principal bundle over a manifold M and a fiber/structure group G with coordinates x and g respectively

[uncut surge]

I think your understanding might be right, tho I always forget the precise formulas with which the horizontal subspace is defined from a connection

[shy blaze]

yeah Ive been having a lot of trouble understanding connections and parallel transport in a rigorous way, but I think I finally got it

it seemed like a bit of a chicken and egg scenario

where the connection and the horizontal subspace seem to be dependent on eachother

[uncut surge]

Though what is a bit weird: If you had the trivial bundle and the trivial connection, your horizontal subspace should basically just be spanned by the d/dg_i

or wait, no

that's the vertical one

[shy blaze]

no by the d/dx_i

yeah

[uncut surge]

yah

No then this is probably good, yeah; the christoffel symbols correct exactly for the curvature

[shy blaze]

worst part was that I had a lot of trouble finding a good definition of a connection, especially in the book I was mostly using :/

[uncut surge]

Haha there's like fifty

[shy blaze]

yeah

from what I understand its mostly just a function that describes how local coordinates of different neighborhoods in a bundle relate

[uncut surge]

But the one which is like "an operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamm(E)$ which is $C^\infty(M)$-linear in the first argument and Leibniz-ey in the second one is probably appropriate here

[gentle osprey]

Lartomato
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[uncut surge]

That's a good intuitive view I'd say, yeah

[shy blaze]

yeah, though this is for my thesis, and I'm getting kind of side-tracked with this topic, so I might just keep the exact definition of a connection kind of open haha

but after like a week of struggling I finally understand what I need to I think

[uncut surge]

Ah, it's good to write down one, and then appeal to your favourite diffgeo book to say "connections have a bunch of equivalent definitions, which give rise to the following stuff i need" or so

[shy blaze]

Ill be mostly describing what it is along the covariant derivative since it's an example of one I think, and thats the one I mostly need

yeah I referred to a book for a more precise definition

[uncut surge]

But yeah, that's like 90% of the work in diffgeo, getting a good feeling for all the objects that float around, so don't feel bothered too much if this takes you some time

[shy blaze]

yeah...

[uncut surge]

Diffgeo is less "super cool theorems about super cool things" and more just a good useful framework to view geometry through

[shy blaze]

it was a lot harder than I expected

[uncut surge]

So understanding the definitions well is an effort in itself

[shy blaze]

and this is mostly for a physics part of my thesis, where I feel like Im allowed to be a bit less rigorous at least haha

[uncut surge]

Yah I know how the physicists are

They might even be happy if you leave the details in the dark a little

[shy blaze]

I usually prefer math, but for stuff like this, where the math behind it is very involved...

[manic garden]

there are someee cool theorems (though half of them are "this analytic object ends up really nkt caring about the metric at all")

[uncut surge]

Which one are you thinking of fam

[manic garden]

gauss-bonnet is a standard one

[uncut surge]

Tho I won't deny that there are some cool magic tricks there, most of them are something like "we define this object in charts and then show that no matter which charts we used, it's a globally well-defined boio"

Which is cool in its own right, but rather than revealing something magical, it just kinda tells you that the framework of differential geometry is a good one to work with

But yeah gauss-bonnet is a fair one

It's not something i've personally ever seen treated in a diffgeo course weirdly enough

[manic garden]

i mean, similar things can be said for alg geo

but there are sick theorems in alggeo

[shy blaze]

just to be sure, are the \omega_{\mu, a}^b in this math stackexchange answer in fact the christoffel symbols?: https://math.stackexchange.com/questions/2337456/explicit-derivation-of-the-horizontal-subspace-connection-and-covariant-derivat

they are usually denoted with Gamma I guess, but they are in the place where I thought the christoffel symbols would be

[uncut surge]

Oh yeah they are

[shy blaze]

whew

[uncut surge]

This is the cooler way of thinking about Christoffel symbols; you might have heard that they are not globally-well defined/not a tensor; but they are components of the so-called "connection 1-form", which is well-defined globally

Christoffel symbols don't transform well, but christoffel symbols times 1-forms transform well

That's not super important for you but it puts the mind of differential geometers worldwide at ease

[shy blaze]

yeah I kind of skipped the part of the 1-form, cause I wasnt really sure what it was haha

[uncut surge]

since we don't have to work with weird objects which only make sense in local coordinates

[shy blaze]

the book I used kind of dropped it out of nowhere

the connection 1-form did seem like a more general thing than the christoffel symbols though

[uncut surge]

It's the same data, just packaged in a globally-well-defined way

[shy blaze]

hmmm

[uncut surge]

It's like if you have a vector field "X" and you choose local coordinates "X = X_i \partial_i", then the collection of functions X_i across all the different charts does not stitch together to a globally-well-defined function X_i, because its transformation behaviour across different charts sucks

But together with the \partial_i it starts transforming well -- and stitches together to a globally-well-defined vector field, namely X

But that, too, is something difficult to get used to, since "usual" mathematics just takes place in Euclidean space, where the concept of "transformation behaviour" is never something that matters

[shy blaze]

as in, the field X is transformed when "moving around" the bundle?

[uncut surge]

In a sense, yes

[shy blaze]

I mean, when moving around a sphere or a cone youll need it too

[uncut surge]

Or when moving between two different overlapping charts

[shy blaze]

yeah I guess thats the most important part

but then what is the 1-form?

is it just a collection of those transformations?

[uncut surge]

Yeah, written as $\Gamma_{ab}^c dx_a \otimes \partial^c$ or so in local coordinates

[gentle osprey]

Lartomato

[uncut surge]

It's also a bit weird since it's a tangent-bundle-valued form, lol

[shy blaze]

but where is the structure on that collection?

just composition or something?

I wasnt sure what to make of it

[uncut surge]

What kind of structure are you looking for?

[shy blaze]

well just any structure on the 1-form

since its not just a collection I imagine right

[uncut surge]

Oh, hm, so that it must fulfil some properties or so, like how the christoffel symbols fulfil some identities

Yeah I think this is where my spontaneous knowledge ends

I'm nooot sure if these things will be super important to you quite yet anyway; it's probably just as healthy to think about connections in terms of christoffel symbols for now

[shy blaze]

heh, I'll be avoiding it anyway when writing, but thanks very much for your help and confirmations

[uncut surge]

Yeee maybe that's good for now. These are the things that I even sometimes lose my head in, even though I've worked with this stuff for a couple years now

But good luck on your thesis in any case!

[shy blaze]

thanks!

[steep chasm]

There is a one to one correspondence between levi civita connection on M and the globally defined o(n) valued connection one forms

[uncut surge]

oh yah that was one perspective yes

Tho idk how much easier it makes it on a diffgeo-newbie to think of "o(n)-valued 1-forms" lol; but matter of taste

[shy blaze]

heh

makes it a lot scarier IMO

[steep chasm]

The christoffel symbols are thereby coordinate patch components of the connection 1-form

With values in the tangent bundle

Which can be regarded as a associated bundle to the O(n) principal bundle over the riemannian manifold

How u can imagine it is that the connection one form in this case describes how we move between tangent spaces (which are the fibers of this bundle which can be seen as just the O(n) group at each point ( a little imprecise cause the fibers are actual O(n) torsors))and we can globally define it

If we now look at a coordinate patch on the manifold we can also regard this connection one form locally as the christoffel symbols

Also u can view the “ way we move between tangent spaces” by saying the connection one form measures how a vector field for example devitates from being horizontal to the fibers

Aka if the vectorfield is horizontal the form vanishes aka is 0

So the covariant derivative d+Γ acts on horizontal vectorfield X through d(X) + Γ(X) = d(X)

[summer jolt]

I'm trying to understand the proof that the fundamental group of homotopic classes of paths is a group. I'm trying to intuitively make sense of the proof of the existence of inverses.

In this diagram, what exactly is happening?

I can see that we start with $[\alpha \circ - \alpha]$ and end up with the constant path.

[gentle osprey]

snypehype

[hazy siren]

hello

so compactness implies closeness and boundedness

however the converse is not necessarily true for non euclidian, or anything other than Rn and Cn

i know this

however cant come up with an example for the latter statement

[gritty widget]

with what metric?

[obtuse meteor]

this is the most boring example though. An example with a complete space is harder

for metric spaces the better condition than closednesss and boundedness is completeness and totally bounded

[gritty widget]

pick your favorite infinite-dimensional banach space

[obtuse meteor]

there's a sequence example

with ell^1 or whatever

but I'm slow

[obtuse meteor]

ball go brrrrrrrrrr

here's a good example of what TTerra was talking about

https://estrogen.fun/i/gjrj.png

[summer jolt]

Yes, I understand that I think. What I'm struggling is how one "slowly" deforms the path into the constant map visually. That is what the diagram is supposed to do.

[obtuse meteor]

the diagram is a way of subdividiing I x I in order to make the homotopy more obvious (these diagrams appear in a lot of AT textbooks)

Hm? I don't think that's that hard? It's not sequentially compact

[obtuse meteor]

I guess the idea is like. First go all the way along alpha and then all the way back along -alpha
Then go 0.9 along alpha then 0.9 back along -alpha, and kinda like "unfold" it this way

[gritty widget]

for example, the closed unit ball in C^0([0, 1]), the space of continuous functions on [0, 1], with the sup metric, is not compact. the sequence {x^n} has no convergent subsequence (why?)

oops im late

the questioner is long gone

[obtuse meteor]

I'm going to go do my linguistics lecture? I think it's good to see why both conditions are necessary, not just completeness?

totally bounded though

[summer jolt]

My understanding is the following: you do $\alpha \circ -\alpha$ and then "do" the constant map. And slowly you increase the speed of $\alpha \circ -\alpha$ and so you have "more time" to do the constant map. Then eventually you'll be at a point where you're spending very little time on $\alpha \circ -\alpha$ and most of the time on constant map. Would this be the correct way of thinking about it?

[gentle osprey]

snypehype

[obtuse meteor]

hmm

something feels weird about this

yeah

if you were just doing it faster

you could do this for any loop

and that would be bad

[summer jolt]

but what does it mean to do "less" of it?

[obtuse meteor]

this is a clever example! Thanks for sharing

[summer jolt]

So is this what you mean by "less"?

[obtuse meteor]

mm

we're only doing this at one endpoint

so instead the right half should get smaller

but not the left half :)

(this is required if we want a path homotopy)

[summer jolt]

@gritty widget oh I see this makes sense. So the speed in the case is unchanged

Ok thanks I think I get it. I'll try to parse the equation and relate it to what you said later.

[obtuse meteor]

the diagram is slightly different than your proof slimvesus

or no it's the same, I misread

so if you draw a horizontal line in the diagram

then the first section of the line is "wait at x_0", the second section is "do alpha for this time", the third section is "do -alpha for this time" and the fourth section is "wait at x_0"

[summer jolt]

@obtuse meteor Ah brilliant! that makes it easier to understand

[long coyote]

how do you use lebesgue number lemma when proving S2 is simply connected

[obtuse meteor]

So the goal of this proof is you have a loop gamma in S^2 based at x_0

you want to homotope gamma so that it's not surjective, aka it's in S^2 \ {x} for some x != x_0

because S^2 \ {x} is just homeomorphic to the plane, and everything is trivial there

The intuitive picture is you draw a small open ball B around some point x, and then whenever gamma passes through that ball B, you homotope gamma inside of B so that gamma traverses still inside of B without touching x

The way the lebesgue number lemma is used is that you know the preimage of B under gamma is open, so it is composed of finitely many intervals (a_i, b_i)

the preimage of x is closed in [0, 1], so it's compact. The lebesgue number lemma tells you that this compact set is contained in only finitely many of the (a_i, b_i).

and then on each interval [a_i, b_i] which contains this compact set you just nudge gamma a little bit in order to miss x to make your homotopy

the key point is that there are finitely many such intervals, so you can do all of these homotopies no problem

Nice fact: This same proof method shows that for a manifold M with dim M >= 2, there is an inclusion pi_1(M, x_0) into pi_1(M \ {x}, x_0) for any two points x != x_0

[long coyote]

👍

[obtuse meteor]

Even nicer fact: I think this same proof method shows that for a manifold M with dim M > n there is an inclusion pi_n(M, x_0) into pi_n(M \ {x}, x_0)

max seems like the relevant expert to check me on this :P

[sleek thicket]

That isn't true. Consider M = T^2 and n = 1. Then M \ a point is homotopy equivalent to the wedge of two circles, so pi_n(M, x_0) = Z^2 while pi_n(M \ {x}, x_0) is a free group of rank 2. There is no inclusion of Z^2 into a free group of rank 2 by nielsien-schrier (I think a similar thing was on your midterm review faye)

[marsh forge]

I think the key issue here is that you can construct (it is nonunique) a nice map from $\Omega M\hookrightarrow \Omega(M-x)$. The issue is twofold: you must make a choice for each map and the map is not injective yet. Of course, obviously before you actually remove x it is clear all possible choices are homotopy equivalent and the places where injectivity goes wrong share a homotopy class by construction (in M), so maybe this will work out. Unfortunatley, after you remove the point you remove some possible homotopies so choices are not unique even up to homotopy and depending on your choices you may have also broken homotopy-injectivity. Sham gives an example that shows that even if you try to make your nonunique choices carefully its not quite good enough

@obtuse meteor

Okay after like a million edits i think this is fine now lol

The right way to think about the S^2 example might be that S^2 is exceptional in that you can remove a point and it turns out this point wasn’t necessary for doing any homotopies because you could always go around the other way

[sleek thicket]

what's Omega?

I would read it as diff. forms but that doesn't makes sense from context

[marsh forge]

$\Omega X:= \operatorname{Maps}_*(S^1,X)$ is a common notation from alg top sorry

[gentle osprey]

MaxJ

[sleek thicket]

ah okay sorry

right right I have seen this

[marsh forge]

It’s adjoint to $\Sigma X := S^1\wedge X$

[gentle osprey]

MaxJ

[sleek thicket]

but when talking about manifolds and homotopy invariants it's not my first guess for omega 😛

[marsh forge]

at least you knew from context i wouldn’t bring up differential forms hahaha

[sleek thicket]

this is interesting tho

[marsh forge]

I think its a convincing reason to see that like, homotopy groups are very sensitive to most changes to a space despite the fact that we often think about them as very weak invariants

[obtuse meteor]

maybe I made a mistake. Instead we want a surjection pi_1(M \ {x}, x_0) to pi_1(M, x_0). Is this true?

[sleek thicket]

ah yeah this is true in the example I gave

and the map is going the right way

[obtuse meteor]

this way there are no choices to make

you just take the inclusion

and it's surjective via the argument I made

This is of course less useful, because understanding quotients is harder than understanding subgroups (at least imo)

hmm, is there anything really nice you can draw about this?

[marsh forge]

I decided that was too pedantic of a message

anyway

I think it is worthwhile to think about the kernel here. Take Sham’s example.

Fix a point

make it a disk

the kernel of the pi_1 map is the commutators

and it turns out the loop represented by the boundary of this disk

is that commutator

In general the obstruction to that being an isomorphism should have to do with the associated small disk which you can always draw in a manifold

I’m being a bit cheeky but we can either use path conjugation or or we can move the basepoint to the edge of the disk for this to be actually true

So the kernel of the pi_1 morphisms somehow tells you how much extra connectivity the point you removed was accounting for

(this works well w my discussion of the S^2 example because in degree 1 removing a point has no kernel and therefore that point accounted for no connectivity)

[obtuse meteor]

yeah this makes sense ^^

unfortunately ofc you can't provide such simple descriptions of what you lose in higher dimensional cases

[marsh forge]

yeah i have no rigorous backing for this really

but like

to me it seems geometry only ever really tells you anything about one homotopy group

[sleek thicket]

@fading vale related to my thing about covres

not directly applicable to your problem

but interesting still

[fading vale]

thinkies

[sleek thicket]

good problem

I don't think you should try it yet

but also, what do I know

I think the number of sheets of the cover will be the degree of f

but if you talk about degrees later

think about it ig

idk I'm not your boss

[fading vale]

capitalist shamrock arc

[sleek thicket]

lol

[gritty widget]

[sleek thicket]

i am not now nor have I ever been a member of the american capitalist party

[gritty widget]

shamrock

im typing up a really long response to my prof's "do part a more carefully"

it's not me insulting him

[sleek thicket]

[fading vale]

malevolence

[gritty widget]

i want to

i really want to

im just summarizing all the things we discussed

including book page references

[sleek thicket]

[fading vale]

book page references

this is so demented

idk if thats what tterra means

but im pretty sure this prof like

makes u cite result by result from the book

i hope so

[gritty widget]

i require immense clarification because the things in the problem im asked to look at don't even have definitions in the book

[sleek thicket]

I would simply take a better course?

[gritty widget]

hahahaahahahahahahahahahahaahahhahahaa

[honest narwhal]

Have you tried being at a school where the TA for analysis isn't trash?

[gritty widget]

[honest narwhal]

@sleek thicket you know for a second I thought your theorem was wrong

And then I remembered Sard

[sleek thicket]

lol

[honest narwhal]

And I was like 😍

[sleek thicket]

Sard's good

[honest narwhal]

Sard is actually an S-tier theorem

[sleek thicket]

,ban slimvesus

[gentle osprey]

This may only be done by a moderator!

Apr 9, 2021, 2:21 AM

[gritty widget]

sard? more like fart

jk sard is based

[sleek thicket]

,bt

[honest narwhal]

shamrock it's fine no need to ban slimvesus

We all start as children

[sleek thicket]

I have been angered

[honest narwhal]

And we grow up

slimvesus is but a toddler

[gritty widget]

hehehe im still typing out this post

[sleek thicket]

Please

Slim why isn't it working

Slim please

[obtuse meteor]

I used sard this week

it was way overkill

[sleek thicket]

Delete the goddamnserver

Nice Faye!!!!

[obtuse meteor]

but it's funny to overkill things sometimes

[sleek thicket]

I love that theorem

[obtuse meteor]

I used sards

to prove that polynomials always have a value

[sleek thicket]

Bro you're scaring tterra stop

[obtuse meteor]

which is not repeated

lmao

[sleek thicket]

lmfao

alright Faye

That is

[gritty widget]

what is a measurable function M -> R??????????

[sleek thicket]

Beyond even my tastes

[honest narwhal]

Definitely on an orientable manfiold you can integrate against a volume form

And that defines a measure

But I'll say this

[sleek thicket]

What is measure :S

[gritty widget]

dami you lack context

[honest narwhal]

It's easier to define measure zero than it is to define measure

Just cap with charts 🙃

[obtuse meteor]

baby: derivative of a degree d polynomial has at most d - 1 roots, so there is one value which does the right thing
woke: sard's theorem

[honest narwhal]

Uncertainty principle man

The only microscopes good enough for these sets cost more than the department's budget can afford

So we call it measure zero

[obtuse meteor]

really smashing butterflies with hammers today

😌

[honest narwhal]

Hah that's a new version of that expression

I usually say nuke a potato from orbit

[obtuse meteor]

unrelated: sham I'm wondering how long it'll take before I have tons of cringe in my mentions. I already had someone whose twitter name is "ontos of intellect" quote tweet me

[sleek thicket]

Aw smashing butteflies makes me feel sad

oh it's coming Faye

I'm used the turn off reply feature more aggressively

It is peaceful

[honest narwhal]

ur face is 0 dimensional

^

Not even vacuous in my case

[obtuse meteor]

just got this one

https://estrogen.fun/i/ij9m.png

[sleek thicket]

lol

[obtuse meteor]

[sleek thicket]

lmao

[honest narwhal]

Sounds like something someone for whom the statement is vacuous would say but ok

[sleek thicket]

Twitter is bad

tterra has the right idea

Just look at filth and never interact with people

[obtuse meteor]

I mostly interact with people I top

so there's no problems

[honest narwhal]

All the fundamental math needed to understand nature at any practical level was mature by the early 20th century. There has been no progress in pure math useful to the public because there is none left to make

[honest narwhal]

More like

[sleek thicket]

We're just making things worse imo

[obtuse meteor]

math is done pack it up

[honest narwhal]

All the fundamental math needed to understand nature at any practical level was mature by the early 20th century. There has been no progress in pure math useful to the public because there is none left to make

[obtuse meteor]

hm?

[sleek thicket]

Math

since the early 20th century

[honest narwhal]

Math is creating undergraduate category theorists and not giving any benefit so

I buy it

[sleek thicket]

It's regressing

(i am meming)

[obtuse meteor]

reject modernity: a standard theory of groups
embrace tradition: a set of equations with substitution rules and a permutation group / general linear group

[sleek thicket]

schoomers btfo

[honest narwhal]

Idk if I buy that so much as

The stuff is more advanced so it's harder to appreciate

[obtuse meteor]

we haven't had time to look back imo

emily riehl has a model of a future where like

homotopy type theory is taught to undergrads

in 2100

[honest narwhal]

Like I can take 10 big results between 1900 and 1921

And a lot of undergrads/early grad students could understand their significance

[obtuse meteor]

and infinity-category theory is done as an advanced undergrad course in the language of HoTT

[honest narwhal]

Arithmetic QUE? Not as much

Langlands is apparently booming/about to boom rn

[obtuse meteor]

mhm

[honest narwhal]

Poincare conjecture

Which fits more largely in Thurston's geometrization

slim maybe, maybe not. But definitely the sentiment that less happened in 21st century compared to 20th def didn't exist 😛

[obtuse meteor]

I wana get into dynamics at some level but I feel dumdum

[honest narwhal]

I am hoping to learn a lot of dynamics over the coming year

[obtuse meteor]

my issue is I can't read books

even when there is one

[honest narwhal]

I spent approximately 4 weeks of my life trying to learn dynamics

[obtuse meteor]

I need to talk about maths with professors and stuff

[honest narwhal]

About 4-5 years ago

And yeah I used Brin and Stuck

[obtuse meteor]

and hear someone talk about something

to actually understand

[honest narwhal]

I guess my thing is

I found Brin and Stuck non-trivial but kinda doable as a rising third year undergrad

Now I'd be a rising third year grad student

So it's probably not nearly as hard lol

I'm prob gonna use two books

Brin and Stuck + Einsiedler and Ward

[obtuse meteor]

how do you people like

learn out of books???

maybe I'm just not mathematically mature enough yet?

but I can't keep myself motivated to read a book

and do exercises

compared to like talking about math with profs and stuff

[honest narwhal]

Yeah usually I prefer lectures tbh but I think the best thing is to not do it totally independently

Either do it under prof guidance or as part of a reading group

At minimum have some external source of accountability to force you to make progress

And optimally you have someone to discuss the math with regularly

[obtuse meteor]

mmm

maybe I should ask my like letter writer

[honest narwhal]

Idk I honestly found Brin and Stuck to be okay

[obtuse meteor]

if she can help me get into it

she works in complex dynamics

[honest narwhal]

Like it's hard for sure

At the time I read it, it was easily the hardest math I attempted to read

[obtuse meteor]

and I'm taking complex analysis next semester

[honest narwhal]

But as a second/third year undergrad I was able to do stuff

So I feel like now it'd be very doable

Since I'm much more experienced and since a lot of it wouldn't be a first pass anymore

That is tru

I mean dynamics has just one example which matters and it's the hyperbolic toral automorphism right?

jk

[gentle osprey]

slimvesus#9286's Avatar

Click here to view the image.

[honest narwhal]

I will say at the time I never really understood the point of the solenoid shit

B&S has a list of weird examples early on

Solenoid and horseshoe

[sleek thicket]

@obtuse meteor I don't think it's a maturity thing, at least not necessarily

I can't focus on lecture and so it just doesn't work for me /shrug

[honest narwhal]

HTA and circle rotation I can appreciate for sure

[obtuse meteor]

hmm I see

[obtuse meteor]

wdym by this?

[sleek thicket]

I mean I read books out of necessity, I guess?

Like if the choices are to read books or watch a lecture

As my main way of learning a thing

The lecture will do almost nothing for me

[obtuse meteor]

huh

I'm like opposite

[honest narwhal]

^

I'm very big on lectures lol

I think it helps that if I can get myself to focus

Then my memory is crazy fucking good

[obtuse meteor]

yeah I almost never forget like

statements of theorems

or things we've done in class

[honest narwhal]

I remember one time I was actually focused in class

[obtuse meteor]

but I can't read them and remember them as easily at all

[honest narwhal]

And a few hours later I basically recounted the lecture perfectly

[obtuse meteor]

I always have to refer back 10 times

[honest narwhal]

To someone who missed class

[obtuse meteor]

when I'm reading

[honest narwhal]

Except one detail in a proof

[sleek thicket]

Oh yeah and like

[honest narwhal]

Now I'm usually not focused but then I'm just fucked whether it's lecture or reading

[sleek thicket]

I did two reading courses on algebra before ever taking an algebra class

[honest narwhal]

I guess book is nice since you can consult later in that case

[sleek thicket]

And just found out it works really well for me

[obtuse meteor]

I just haven't taken an algebra class

[honest narwhal]

But yeah when my focus is good it's good

[obtuse meteor]

and am doing algtop

it's working out fine

[sleek thicket]

It's my strongest subject/the thing I have the strongest knowledge base in

Well my point isn't really about algebra lol

[obtuse meteor]

¯_(ツ)_/¯

wtf is /shop lmao

I'm doing algebra and complex analysis next semester

probably nothing else

bc I value my sanity

[honest narwhal]

Complex analysis is cute

[obtuse meteor]

I wish I could like

take the next course in algtop next semester

but I really need complex anal

and algebra

so it doesn't make sense lol

it's 596 so it's the qualifying exam level of complex analysis

let me looksee

https://estrogen.fun/i/ft41.png

the algtop class also probably uses higher algebra than I know

for a second course

ye

https://estrogen.fun/i/sq33.png

I also don't have any difftop background

[sleek thicket]

Both seem cool

I would go for complex but also maybe I wouldn't

[obtuse meteor]

hmmm yeah

idk

I'm caught in the weird area of things only being offered in fall/winter now I think

and there's an issue of if I have the money to do a 4th year and stuff

and if I need to fast-track graduating and applying to grad school

in which case I need complex analysis

to have a good app

lol

mmm

that sounds like

a bad decision

considering how busy I've been this semester

and that I have to work as a grader next semester

(again money)

[sleek thicket]

Don't overload yourself

[obtuse meteor]

yeh

[strong heron]

When we use the van Kampen theorem, in the amalgamation product, are the original relations of the participating groups taken into account?

https://cdn.discordapp.com/attachments/749229218537144413/829781077199028244/unknown.png

Like, if in this picture a Mobius band is attached along its boundary to the meridian circle gamma, what would be the fundamental group of the resulting space?

Is it not just free on two generators?

[sleek thicket]

Yes

for your second question I would need to think about it, sorry

[strong heron]

Can you let me know when you have the time?

[sleek thicket]

I think by van kampen it would have presentation <a, b, c|aba^-1b^-1, c^2 a^-1>

So we can extend the torus and the mobius band a little to get open sets which cover your space and are pretty much the same

The mobius band's fundamental is generated by the loop c that goes around its central circle (you might need to change basepoint but I don't think it's a huge deal here)

And c^2 is the boundary loop of the mobius circle

the torus is generated by two commuting loops, the loop a called γ in your picture and a horizontal loop b

The intersection is (actually deformation retracts onto) the circle γ in your pic, and so is generated by γ. When looking at the groups, the generator of this will include as the element a in the fundamental group of the torus and c^2 for the mobius band

This is how I get the relation c^2 a^-1 = e

So this group can also be presented by <b, c|bc^2b^-1c^-2>

This contains a subgroup <b, c^2> which is not free so it cannot be a free group (subgroups of free groups are free)

[strong heron]

That's right. I also got this only. But then I got confused primarily because of this... When you calculate the fundamental group of the 2 holed torus

Then as the standard octagon, you get the presentation <a,b,c,d | aba^-1b^-1cdc^-1d^-1>, right?

[sleek thicket]

I think that's right?

I can check

Yup, that's right

[strong heron]

But if you take it is the connected sum of two 1-holed tori and then use van Kampen, you seem to get <a,b,c,d | aba^-1b^-1=cdc^-1d^-1=1>

How are these two groups isomorphic is what is confusing me.

[sleek thicket]

I think I know what you did wrong

So that second group is the free product of π1(T^2) and π1(T^2)

Right?

[strong heron]

Yes, supposed to be

[sleek thicket]

So if you had two torii touching at a point you could do van kampen to get this fundamental group

But this doesn't work with T^2 # T^2

What open subsets are you using?

Presumably they should be homotopy equivalent to T^2

[strong heron]

Oh, no. They're homotopic to torus-disk

[sleek thicket]

right

Exactly

[strong heron]

Ah, I see. So they are totally free on two generators!

[sleek thicket]

Yes!

Because torus \ disk is homeomorphic to torus \ point is homotopy equivalent to a bouquet of two circles

And you could track generators and stuff

But I think it would be ugly

I remember being very confused that the connected sum and wedge sum weren't homotopy equivalent when talking my first class on this stuff

[strong heron]

Yes, yes. So it's basically (Z*Z)*(Z*Z) where the amalgamation is over Z again? In the connected sum case.

[sleek thicket]

Yes, but the relations added by amalgamation aren't obvious to me

Oh wait!

I think I see it!!!

Okay think of two squares

[strong heron]

Hmm, the boundary of the removed disk is equal to the boundary of the torus, right? So I think we will get the same result as in the octagon case. Because the octagon can be subdivided by a line into two 1-holed tori with disks removed.

[sleek thicket]

With their centers cut out

And the edges identified to be torii

Then the loop in the center looks like aba^-1b^-1

[strong heron]

Yes, that's what I was saying too..

[sleek thicket]

Ah sorry haha

I just got excited because I saw the geometry of it

[strong heron]

Yeah, you take the hole at a vertex and then stretch it open, do the same for the other square and then join them to get an octagon!

Okay, great! This had been stumping me since morning that why are those relations being added. Many thanks!

[sleek thicket]

What book are you looking at for this stuff?

[strong heron]

Mostly Hatcher

[sleek thicket]

ah nice

hello ethan

[barren sleet]

yes hello?

[sleek thicket]

oh, I just saw you were typing

[barren sleet]

did you see that i solved my thing from earlier

[sleek thicket]

felt polite

[barren sleet]

oh

[sleek thicket]

nice!

I did not

[barren sleet]

i was gonna ask a qn about hatcher but shanged my minf

[flint flicker]

Hello i am trying to see what the fundamental group of $X$ is, where $X$ is the union of $S^1$ and $C$. $C$ is the plain square of side length $\sqrt2$ and centered at (0,0).

[gentle osprey]

kh

[obtuse meteor]

this seems homotopy equivalent to a wedge of two circles

but I would have to draw it

do you have a schematic of what this looks like in the plane?

[flint flicker]

it is a circle and the square in it

[obtuse meteor]

oh sqrt(2) centered at (0, 0)

one sec

so it's like a square

with the circumscribed circle?

[obtuse meteor]

yeah

probably the best way to approach this

is to treat it as a graph

do you have the result on computing the fundamental group of a graph?

[flint flicker]

No not yet

[obtuse meteor]

hmm

what about contracting subcomplexes of a cw complex?

[flint flicker]

YeS

yes*

[obtuse meteor]

So the idea is like

repeatedly contract edges

until you end up with a wedge of circles

[flint flicker]

so i will get the full circle ?

[obtuse meteor]

mmm

no you shouldn't

you should visually go through this contraction process

contract one edge at a time

to make sure you don't miss anything

[marsh forge]

Is that square filled in

you should be careful

nothing in the original statement makes me think it would or wouldn't be

but the answer changes

[obtuse meteor]

ah yeah that is important

[flint flicker]

Yes it is filled in

the second part of the question is what happens if it is not filled in

[marsh forge]

Okay so my advice

everywhere you see somehting like

[obtuse meteor]

then what we're doing here is if it is not filled in

[marsh forge]

-------

fuck

one sec

$----$

[gentle osprey]

MaxJ

[marsh forge]

there we go

a line between two verticies

contract it

and redraw

one at a time if you want to be careful

and eventually you should get some wedge of circles

Oh wait its filled in shoot

okay sorry

first step is deal with the square

Do you know how you might do that?

[flint flicker]

what i tried so far is to use Van Kampen

but i was not able to go far

maybe try and contract the square?

[marsh forge]

Yes

that last idea

is a good one

do that and redraw

[flint flicker]

this is what i got

[marsh forge]

Hmmm

Not quite

One thing to keep in mind

Everything that was connected before you contract the square

has to be connected after

Here's a procedural hint

What is your main goal? Contracting the square to a point

So start by drawing the point

Then you've finished w the square

But now you have to ask "How are all the lines that used to touch the square connected to this point"

Well, there is only one way to do it

(it might help to try to animate this in your head if you can)

(visualize the square shrinking and how this pulls on everything else)

[flint flicker]

my circle will become a 8?

[marsh forge]

Not quite

You have 4 little line segments touching the square

[flint flicker]

ah yes like a 4 leaf

[marsh forge]

Yep!

Do you know pi_1 of this

if not you will need van kampen but its not too bad

[flint flicker]

No i will try using van kampen

Thank you for the help

Your tips were very helpfull

[marsh forge]

Np!

Eventually after some practice

you'll look at these things and see the answer instantly hahaha

[flint flicker]

I hope so

[marsh forge]

something that might help build the skill

is to try to visualize homotopies without drawing them first

and like watch it happen in your brain and see if you can accuratley do it before you do it in the procedural way

might cause some headaches haha

[flint flicker]

haha thanks for the help!

i got the same figure after contracting each line in the empty square

it must be wrong

[marsh forge]

Hm?

Your end result should be a 4-leaf clover

But if the square was not filled in, it should be a 5-leaf I believe

(you can only contract 3/4 of the sides before you don’t have a contractible side)

[flint flicker]

this is how i am seing it

[marsh forge]

in 2->3 you identify two sides of the triangle but you arent allowed to do this

remember all you are doing is contracting one side

the other two can bend or whatever

[rotund thicket]

isn't this like taking the quotient space where the corners of the circumscribed square lying on a circle are equal?

[marsh forge]

not if it isnt filled in no

if it is filled in this is exactly what is happening yes

[rotund thicket]

why doesnt this happen when its not filled?

isnt this S^1 \ ~

[marsh forge]

you can’t preserve homotopy type by contracting a non contractible subspace

[rotund thicket]

uhhh

ill take your word then lol

[marsh forge]

(if your method worked then the filled in version would be htpy equivalent to the not filled im which is false)

[flint flicker]

how can i contract in the second step without merging the two lines

[marsh forge]

ill draw

[flint flicker]

ill need to retract 2 lines at the same time?

[rotund thicket]

but if this was less than 4 points then we wouldn't need to have the circle filled? (sorry for budging in with my question)