point-set-topology

[gritty widget]

there exists a closed even-dimensional orientable manifold with non zero second cohomology which doesnt admit a symplectic form

[pearl holly]

Okay maybe I did a really bad job at explaining yesterday. Let me make it clearer now. You already got it and tterra already answered but I kind of feel bad for just leaving you confused yesterday so I will make this super clear now.

Like you said, if $f:X \rightarrow Y$ is a continuous map of spaces (let's assume that they are path connected, i.e the base point doesn't matter) then the functor $\pi_1$ must assign a morphism to a morphism, just like tterra said. Morphisms are in this case continuous maps between two (based) spaces. We will define this "assignment" $\pi_(X) \rightarrow \pi_1(Y)$ by simply mapping the morphism into it's induced homomorphism, i.e $\pi_1(f) = f_$ where $f_[p] = [fp]$. We want to show that $\pi_1(f) \pi_1(g) = \pi_1(fg)$, i.e $f_g_ = (fg)*$. But this is not hard to check since $fg_(p) = f_([gp]) = [fgp] = (fg)_[p]$. We also want to show that $\pi_1(1) = 1$, i.e $1_* = 1$. This is also simple because $1_*[p] = [1p] = [p]$, so the induced homomorphism of 1 maps everything to itself so it is the identity.

[frigid river]

wtf

[pearl holly]

I might have overdone, now it's all written out in baby steps lmao

[gritty widget]

f(X) -> Y

[pearl holly]

frick

[gentle osprey]

Tokidoki ✓

[gritty widget]

looks good to me

[pearl holly]

pi_(X)

[reef shore]

old notation of functions was literally f(X) \subset Y the f: X \to Y thing started around the same time as category theory

[gritty widget]

i already got you once i'm not gonna go twice

[reef shore]

holylocks

[gritty widget]

CP^2 # CP^2

i think

[frigid river]

godlylocks

[reef shore]

[frigid river]

[gritty widget]

god

ms paint ruined the transparency

[pearl holly]

CATLOVE???

[frigid river]

Imagine investing 100 dollars a year into emotes

[pearl holly]

Imagine having :catlove:

[gritty widget]

symplectic geometry is dope

why are u catyikesing me

it's true

[reef shore]

geometry

[gritty widget]

[pearl holly]

More like rigid topology

[gritty widget]

there's a reason symplectic is sometimes called "symplectic topology"

[reef shore]

I have never seen any geometry so all I know about geometry is charts and integrals

[gritty widget]

all symplectic manifolds of the same dimension are locally isomorphic

so there's really no local study of symplectic manifolds

it's all global

that's a cool fact i think

[reef shore]

so you are saying

It is just integration

[gritty widget]

it's easier to appreciate this fact if you've spent an ungodly amount of time doing riemannian geometry

where everything is local

eg curvature

integration

all of math is integration

[gritty widget]

i was under the false impression they were the same thing

oh no

i guess to me "geometry" implies the existence of local invariants, like in riemannian geometry

brain rot

[reef shore]

ultra is the funniest person on the server.

[gritty widget]

go back to the old pfp now

wait

[empty grove]

[gritty widget]

🥴

[hollow harbor]

[maiden oracle]

is it correct if i say that a free group on a set X is the same (in the sense that it's isomorphic) as the free product of $\abs{X}$ copies of Z ?

[gentle osprey]

chrisply

[empty grove]

Yes

[maiden oracle]

so free groups are just a special case of free products ?

[empty grove]

Yeah

[maiden oracle]

ok ty i just needed a sanity check since i'm not that comfortable with the material

[teal nova]

Can someone help me with this ?

Let be $X$ a compact space and ${C_n}{n \in \mathbb{N}}$ a subsequence of closed not empty subsets of $X$. If $C{n+1} \subset C_n$ for every $n \in \mathbb{N}$, show that $\bigcap_{n\in\mathbb{n}}C_n \not = \varnothing$.

[gentle osprey]

galois.theory

[flint cove]

Given that the notion of „closed space“ does not make sense in topology, you need to give a little more context

metric spaces?

[bleak helm]

Your terminology is a bit off, since that's just a sequence not a subsequence. But I think I understand. Are you familiar with the finite intersection property characterization of compactness?

[teal nova]

yes

[flint cove]

(ah, compact makes more sense)

[bleak helm]

Try to show your family of closed set satisfies the FIP

Then you can use that characterization

[teal nova]

I'm sorry I do not speak much English and I translated it with many errors

[bleak helm]

It's okay

Do you understand what I am suggesting you try though?

If you can show the collect ${ C_n: n \in \bN }$ satisfies the finite intersection property (and because the $C_n$'s are closed and the space is compact), then the intersection is nonempty.

[gentle osprey]

Lunasong the Supergay

[teal nova]

Oh! i've got it, thank you very much

[bleak helm]

👍

[flint cove]

In the context of complex manifolds, how precisely would one define a complex-valued differential form (i.e. some structure such that dz=dx+idy makes sense)?
I am remembering that instead of Ω(M) we can consider Ω(M,V) for arbitrary vector spaces (like e.g. a Lie algebra 𝖌), but I'm not quite sure how the construction would look like. I mean we probably want a DGA such that pullback becomes a morphism of DGAs.

My guess would be something like Ωⁿ(M, V) = ℝ-multilinear (n arguments) antisymmetric maps →V, and then one could define iω as (X_1, …, X_n) ↦ ω(iX_1, …, iX_n), where multiplication i in this context is a bundle homomorphism TM→TM

Motivation: I want to „understand“ complex integration literally in terms of of differential forms

Ah, okay, The pdf from Wilhelm Schlag does that in Chapter 6, but not in “general”, only over Riemann surfaces.
dz just seems to be the differential of the function z: (x,y)↦x+iy, U→ℂ for an open U in our MF

(https://gauss.math.yale.edu/~ws442/complex.pdf)

[maiden oracle]

looking at the material again i realized that i really dont understand what's going on at all. The free group on n generators is defined to be the free product of Z n times. And the general free Group over a set M is defined to be the free product of copies of Z indexed by M. Now that means that for finite sets M which have let's say m Elements there really is no difference between the free group over M and the free group on m generators, right ? I also don't see how the two definitions of a free group are compatible. One definition your words consist of elements in an arbitrary set M and in the other one your words are just integers concatenated ?? What would be an isomorphism between these two?

[swift fjord]

You just identify the generator (1) of each copy of Z with an element in the basis of the free group

You can think of each integer as a power of the generator

That's exactly your homomorphism. You send each element of M to a generator of a copy of Z then extend by the universal property

Then m^k will be sent to k for example

[maiden oracle]

well that actually sounds plausible

[swift fjord]

The key is that each 1 in each copy of Z is different

Since the underlying set of the free product is a disjoint union

[maiden oracle]

so since it is built upon a disjoint union it really shouldn't make a difference whether you take the free group over {1, ..., n} or another set M with the same cardinality, right?

[swift fjord]

Free groups over bases with the same cardinality are always isomorphic. The free product of some copies of Z though isn't exactly the same. It's almost purely notational though (Exponentiation becomes multiplication etc.)

So instead of i^k you have k*1_i =k_i
(Where 1_i, k_i are elements of the i-th copy of Z in the free product, and i is the i-th element of the free group over {1,2,..,n})

I mean if you wanna think of it being indexed over [n] when it's finite that might make it easier

I hope i'm making sense

[maiden oracle]

yeah i think it does make sense

thanks for your help. I think especially what the copies of Z mean made it click for me (and looking at it now it does seem kind of natural even)

[swift fjord]

Np

Well I wrote this so i'll just leave it here

CamScanner 07-14-2021 23.19.27

[swift fjord]

^

Although having something concrete is good too

[flint cove]

Tbh the universal property is one thing, but, for me personally the point where I got really comfortable with F(X) was when I learned about universal algebra and had to deal with term algebras all the time.

[marsh forge]

I feel like the explicit description is easier and more approachable than this convo makes it out to be

And the concept of generators and relations that arises from it is indispensable

[flint cove]

Oh, yeah, generators and relations might be the best approach to it.

well that's kind of a cok-y statement

[abstract pagoda]

Is that true? So each rule is just a homomorphism between free groups? What does is an example for D_n

[abstract pagoda]

I just googled the universal property of free groups. The normalizer of the subgroup generated by {a^n,b^2,baba} is the image of f?
So we let X be our set of letters and
g:X->G implies there is a unique group homomorphism
f:F(X)->G where f is an extension of g.
In this case g acts as the function that maps the relations to the freegroupp of generators of D_n?

What can you do with this information though?

[abstract pagoda]

Oh ok I thought that recontextualizing it might have atleast one other theorem attached to it

[marsh forge]

It is (in my opinion) pretty rare for a more abstract characterization to give more theorems than a concrete construction

[abstract pagoda]

What do you mean by categorical presentation?

[fading vale]

I do not think this message bleongs here

[gritty widget]

bleongs

[swift fjord]

I think you mean normal closure, not normaliser

[abstract pagoda]

What is the difference?

Adding elements that are fixed under conjugation to those elemtns?

[plain raven]

a categorical presentation of something would characterize it by a universal property which can be explained only in the language of category theory, i.e. using arrows, objects, composition, etc. For example the tensor product has a universal bilinear mapping property, the localization of a commutative ring at a multiplicative subset has a universal property. The characterization usually characterizes the object in question up to a unique isomorphism. On the other hand the existence of such an object might be given by a set-theoretic construction where one specifies what elements are in the set, what the addition and multiplication laws are, when two elements are considered equal, etc.

[swift fjord]

Normaliser N of H<G is the largest subgroup of G such that H is normal in N. Normal closure of H is the smallest normal subgroup containing H

Smallest and largest being w.r.t inclusion

[abstract pagoda]

Oh ok thats what I thought, thank you so much!

[abstract pagoda]

So the object D_n, dihedral groups of order n has a categorical presentation?

I dont remember exactly how to verify is something is a category

If I remember correctly a category is a class of Objects, Arrows such that there is an identity arrow and composition is allowed

So objects would be groups D_n, but im not sure what the arrows would be

[frigid patrol]

@cedar pebble Can you help me with this problem in Szamuely?

I know that the genus of Y has to be n/2 + 1

[cedar pebble]

Mhm

So what you have to do is 1) produce a transitive action of Dn on a set of 2n elements, then 2) compute the cycle decomposition when you go around a small loop around a puncture at a ramification point

Take say the right action of Dn on itself

If you compute the cycle decomposition you should see the branching you want

[frigid patrol]

@cedar pebble I hate this book

The genus of Y is now n-1

[cedar pebble]

yes

wait have you not been reading the version with the errata?

and yes this book has a lot of atrocious mistakes but it keeps you on your toes and it's good practice

[frigid patrol]

no

its terrible

I have been reading this book with the errata

Still spent several hours on this problem

on an incorrect problem

[frigid patrol]

Okay

I see that the commutator is just sigma^2

And this partitions the fiber into 4 sets of n/2 cycles

How do I contruct the cover now ?

@cedar pebble

Do I have to use some covering spaces = transitive pi-1 sets equivalence thing?

this

[cedar pebble]

so maybe an analogous situation that is clarifying that I took a whole course on

to specify a cover of P^1 minus points a_0,...,a_n you need to specify generators \gamma_0,...,\gamma_n with the condition \gamma_0...\gamma_n=1

[frigid patrol]

this is what the picture looks like around the 4 branch points

[frigid patrol]

This is kinda fun

[frigid patrol]

[covert inlet]

Hello, reading the following article (http://www.doiserbia.nb.rs/img/doi/0350-1302/2011/0350-13021103049I.pdf) I have trouble understanding lemma 2.3, only in the part that says that it is easy to show that the sets \mathcal{A} and \mathcal{B} it defines are closed. We work on the Vietoris topology over 2^X, and in the first case you could see that the set \mathcal{A} is the Vietoric of C and I think that C is closed, but I don't think this argument is correct. So I hope you can help me understand why those 2 sets are closed in the ordered arc. Thanks!

[rugged swan]

is it true that for a variety, every local ring is regular <=> every local ring at closed point is regular ?

[pearl holly]

Okay so this is the exercise that I am working with:

Show that for a space X, the following three conditions are equivalent:
(a) Every map S^1 -> X is homotopic to a constant map, with image a point.
(b) Every map S^1 -> X extends to a map D^2 -> X.
(c) pi_1(X, x_0) = for all x_0 in X.

I know that (a) and (c) are equivalent since Max explained to me that the elements in pi_1(X, x_0) could be regarded as homotopy classes of maps S^1 -> X instead of maps I -> X. Now I am stuck on proving that (b) and (c) are equivalent. Let [f] be in pi_1(X). Then f: S^1 -> X and this can by assumption be extended to a map g: D^2 -> X. But now what? What can I do with this map? I think that I am supposed to use a theorem that involves some sort of relation between a map and the extended version of that map, but I can't think of such a theorem. I can't either figure out a homotopy between g and a constant map right away. The book has not proved any results involving D^2 that will help me I think. The only theorem involving D^2 is the Brouwers fixed point theorem, but I doubt that it will help me here. Any hints?

[reef shore]

Are you sure that you know that (a) and (c) are equivalent

Because if you regard loops as circles, their homotopies would have to fix the base point, but in (a) there is no mention of fixing the base point

What you said only gives (c) → (a)

[pearl holly]

Hmm okay let me think over this again

[reef shore]

For what you said about D^2: If f is homotopic to f', and g is homotopic to g', and the compositions gf and g'f' make sense, then gf is homotopic to g'f'

[pearl holly]

Okay I will take that as a hint. I will try to fix (a) <=> (c) now lmao

[marsh forge]

Side note

I think this is the most important exercise in hatcher

[pearl holly]

I'm still kind of confused. Doesn't this work?: If pi_1(X, x_0) is trivial for all x_0 in X, then the loop f: I -> X is homotopic to c (the constant map). This loop can be based at every x since pi(X) is trivial for all x_0 in X. This f can be seen as a map from S^1 to X and so this map is also homotopic to c. Since x (the basepoint) can be whatever it wants to be, then every map S^1 -> X is also homotopic to c.

[marsh forge]

I think the logic is a little awkward here but it’s the correct idea

[reef shore]

Yeah (c) → (a) this works

[marsh forge]

The map S^1 -> X comes with a basepoint in the sense that we think of S^1 as equipped with a basepoint and the basepoint in X is determined by wherever the first one is sent

[reef shore]

The other one is a bit trickier because you might have nullhomotopies that don't fix the base point

[marsh forge]

Then of course your logic plays out nicely

[pearl holly]

omg I was supposed to show the other direction smh lmao

[reef shore]

btw don't try to do (a) → (c) directly

[pearl holly]

Hmm why not?

[marsh forge]

Hes saying that (b) might be helpful

Honestly b is like

Ugh

I could give a talk

Just on the importance of (b)

[pearl holly]

[marsh forge]

Ping me when you get it and I’ll send you a generalization of it for you to prove

[pearl holly]

Okay great, thanks! Btw, does a -> b, b-> c and c->a sound like a good idea in this case?

[reef shore]

Yep that is what I have in mind

[pearl holly]

I'm not that comfortable with these equivalent multiple things questions

[reef shore]

They need to form a strongly connected directed graph 😌

The idea is that implication is a transitive relation. So you prove a bunch of implications, you automatically prove their transitive closure

Which means all finite sequences of those implications are done

[marsh forge]

Ive seen some hilarious bad ones

With parts a-h or whatever

And proven in like the worlds most complicated graphs

The truly woke do like

We prove a->b and then that b+c->d and finally that…

[maiden oracle]

how would one describe homeomorphism vs homotopy equivalent geometrically/visually? For example it's always said that you can stretch and bend, but not tear or glue something with a homeomorphism. Can this be made more rigorous? For me it makes sense if you think of continuity in terms of neighborhoods. Since that just means that things that are supposed to be at somewhere the same place stay at the same place relative to each other (or formally it preserves the topology) But how would one, on a intuitive level argue that you can't fill out a shape (like S^1 is homotopy equivalent to S^1 x D^2 (a filled out torus) but not homeomorphic ) with homemorphisms ? I just don't see how this sort of "filling out" shapes violates the definition of continuity?

[marsh forge]

I think that the only way to build this intuition is to see a lot of examples and understand the rigorous reasons

unless you are asking for a proof that S^1 x D^2 isn't homeomorphic to S^1

[maiden oracle]

no i know how to prove it

just remove one point et voila

two*

[marsh forge]

One suffices 🙂

[maiden oracle]

it does ?

they are both connected afterwards

[marsh forge]

you need a different contradiction

but it still works

donut with small bite

or a small air hole

[maiden oracle]

could one say that S^1 x S^1 is a deformation retract of torus with one point removed and then maybe something with fundamental groups ?

[marsh forge]

Oh its simpler than that

S^1-pt is contractible

whereas no matter what point you remove from S^1 x D^2

it won't be

that means the homotopy equivalence can't be a homeomorphism

anyway yeah working through a lot of examples

is how you get intuition for when two spaces are homotopy equivalent

and homeomorphic

[maiden oracle]

how do you see that? isn't that essentially through fundamental groups?

[marsh forge]

That's one method

a space X is contractible iff every map Y->X is contractible

so it suffices to find one nontrivial loop

no need for the full pi1

[maiden oracle]

hmm ye true

[pearl holly]

Okay I think that I finally proved it lmao. First $(a) \implies (b)$: first notice that the circle can be expressed as $e^{i v}$ where $v$ is some angle and that $D^2$ can be expressed as $ae^{iv}$ where $a \in I = [0, 1]$. Let $f: S^1 \to X$. Let $h_t: S^1 \to X$ be a homotopy so that $h_1 = f$ and $h_0$ is the constant map. Then $g: D^2 \to X$ defined by $g(te^{iv}) = h_t(e^{iv})$ is the desired extended map since the restriction to the circle is $g(1e^{iv})$ and when $t=1$, $h_1 = f$.

Lets do $(b) \implies (c)$. Let $[f] \in \pi_1(X)$. We will show that this is homotopic to the constant map. By assumption, this map extends to a map $g: D^2 \to X$. Since $D^2$ is path connected, there is a deformation retraction $h_t$ that "deformation retracts" to a single point $x_0$. Now $gh_t: D^2 \to X$ induced a homotopy $g \cong g(x_0)$. So $g|{S^1} h_t: S^1 \to X$ induces a homotopy between $g|{S^1} = f$ and a constant map.

and I already did c->a

[gentle osprey]

Tokidoki ✓

[pearl holly]

and composition and restrictions "preserve" continuity so the associated map is still continuous after all the manipulation

[marsh forge]

looks good to me even though you are breaking my heart with all the explicit maps and stuff ❤️

[pearl holly]

:catlove:

[marsh forge]

probably a good thing while you are getting used to it

i'm just lazy

[pearl holly]

what do you mean by explicit maps and stuff?

Is that something that I should ignore?

[gritty widget]

keep doing what you're doing, just for the sake of annoying max

[marsh forge]

oh i just mean

[gritty widget]

/s

[marsh forge]

my proof would go like this

[empty grove]

Quotient topology universal property?

[marsh forge]

actually no i'd probably do the same basic thing

it works well

I realized my proofs would only make sense if i could draw pictures lol

[gritty widget]

tikz

[empty grove]

A nullhomotopy is a continuous map from S^1 x [0,1] to this space such that S^1 x {1} is constant, so it factors through (S^1 x [0,1])/(S^1 x {1}) ≈ D²

[marsh forge]

Yes

Thats what I had in mind

But written like that idk if its simpler

I would ditch the e^ix stuff

I’d just say

[empty grove]

Yeah that part just makes me not wanna read it

[marsh forge]

Let S1 be a unit circle and u a vector, then the points of the disk D^2 can be written at tu for t in [0,1] blah blah

[pearl holly]

yeah that's true lmao

[pearl holly]

[marsh forge]

But toki here’s my better theorem

[empty grove]

You've seen quotient topology in munkres right?

[marsh forge]

Let f:X —> Y be continuous. Let CX denote Xx[0,1]/~ where ~ collapses the copy of X at time 1 to a point

[pearl holly]

i will let max write the thing out first before I ask my question

[marsh forge]

Show that f is contractible iff it extends to a map CX->Y over the inclusion at time 0

[empty grove]

Ok don't look at my argument now

[marsh forge]

Then explain why this generalized (b)

[pearl holly]

Hmm okay I will write this down and try it out

but what does "factors through" mean?

and the S^1/blablabla thing that you wrote moldi?

[marsh forge]

Ignore moldi anyway

No he was right

Just it’s a strong hint

[pearl holly]

Oh okay

[empty grove]

Yeah forget about factors through for now

[marsh forge]

Bonus points if you can explain this result using only vague geometric language

[abstract pagoda]

Wikipedia is the best

[pearl holly]

Okay I will look at the thing that moldi wrote earlier later when I've tried this "exercise" thing, but I have one more question

When solving this exercise, I used the argument that D^2 is path connected so that it deformation retracts to a single point. Why is path connectedness needed for this?

I understand it intuitively, but not formally

[abstract pagoda]

invariance points for fundemental group

so you can start with any basepoint

[empty grove]

Not all path connected spaces deformation retract to a point

[pearl holly]

and I just snatched this fact from Hatcher lmao

[empty grove]

S^1 doesn't

[abstract pagoda]

there is a basepoint change homtopy

because d2 is convex

[marsh forge]

The better way to see this might be to use the extension to construct an explicit null homotopy

[empty grove]

Wait did I misunderstand what you were saying

[pearl holly]

In Hatcher he says that "all path connected spaces deformation retract to a single point" while talking about a counter example that not all retracts are deformation retracts. But why is path connectedness needed?

This is my question. Ignore the thing that I wrote, I might have used a bad langauge there

[marsh forge]

Wait

[empty grove]

That is just wrong lol

[marsh forge]

Thats

[pearl holly]

Or something like that wait

[marsh forge]

Not true

Yeah hahaha

I was losing it for a second thank u moldi

I was like have I forgotten the defn of a def retract

[empty grove]

Because deformation retractions preserve fundamental group in particular

So that would mean S¹ has trivial fundamental group

[gritty widget]

it doesnt?

[pearl holly]

Okay here: "but a space that deformation retracts onto a single point must be path connected"

[marsh forge]

Yes

[pearl holly]

Yeah that is the part that I was reading. But why is path connectedness needed?

[marsh forge]

Exercise

Lmao I’m such a dick

But its a good exercise

[gritty widget]

[empty grove]

[pearl holly]

Yeah I thought about this while doing the other exercise and I think that it has something to do with that H: X x I is continuous. But I can't find a map from I to X that is continous

[empty grove]

Try to track points as you deformation retract the space to a point

Like just track the motion of one random point

[pearl holly]

yeah that becomes a path from the random point to the deformation retraction point

But how can you see it using only homotopy?

[marsh forge]

Wdym

[empty grove]

Homotopies are continuous in t

[abstract pagoda]

Use the change in basepoint homotopy

Its goated

[marsh forge]

Huh

[abstract pagoda]

The proposition in the book

[empty grove]

The family h_t actually represents a continuous map h from X x [0,1] to Y

[marsh forge]

Thats not super relevant this is straightforward

the maps f_t as hatcher writes them are continuous in t

So in particular

$t\mapsto f_t(x)$ is a path

[gentle osprey]

Karen

[marsh forge]

Thats ur hint

[empty grove]

Isn't that the full solution

[marsh forge]

no

I mean

Anything counts as a full solution

To a sufficiently lazy grader

But this wasn’t obvious to toki so I’m gonna make him write it out

[empty grove]

Fair lol

[marsh forge]

Pls delete

But yeah more point is just like

You dont need any lemma or anything

Its just definitions

[abstract pagoda]

Sorry for messing up the exercise

[marsh forge]

No ur fine

[pearl holly]

Okay I think that I see it now lmao. So H: X x I -> Y defined by H(x, t) = h_t(x) is continuous by definition. So f: I -> X defined by f(t) = h_t(x) for some x in X is continuous. f(0) is the identity so it equals x and f(1) is x_0 so we get a path there

[marsh forge]

Just to be extra pedantic how do I use this to get a path from arbitrary x_1 to x_2

[pearl holly]

Oh

Well if I have a path from x_1 to x_0, then I can have a path from x_0 to x_2. THis is a equivalence relation to there is a path from x_1 to x_2

or using the pasting lemma or something like that

to construct a path from x_1 to x_2

[marsh forge]

Just reverse the second path 🙂

[pearl holly]

Boom

[marsh forge]

This of course leads one to construct the fundamental groupoid

That’s a joke don’t worry about that it’s lame

[pearl holly]

Okay anyway, thank you all so much! I will tackle the "better theorem" sooner today since I have to go now

[empty grove]

Sooner today or later today?

[frigid river]

sooner today?

[pearl holly]

Later today I mean sorry

[frigid river]

[empty grove]

[frigid river]

I like Max's new name, btw.

fitting

[marsh forge]

I’m gonna turn your next question into an exercise if u dont fix that attitude

[tepid depot]

i’ve got a student who is trying to learn about the basics of manifolds. they have some mathematical maturity but not a lot of background beyond basic analysis. can anyone recommend a resource for exercises?

i’m looking for practice with the general definition of a manifold, tangent spaces, smooth functions on a manifold, and other basic things of that nature

[gritty widget]

spivak (the little one, not the massive 5 volume set )

well maybe only for the super basic manfold stuff

[tepid depot]

honestly that’s ok I honestly forgot about that book

although i would like something that would help someone understand general smooth manifolds, not just submanifolds of R^n

but i’ll look at it again and see if there’s anything in there I can recommend

[tepid depot]

hadn’t heard of this one, this is actually great since they’re learning this for a physics related project

[pearl holly]

Okay so I'm back now and it's quite late but I still want to give this a shot today. But what does "extends to a map ... over the inclusion at time 0 mean?

It is the "over the inclusion at time 0" thing that I don't understand

[marsh forge]

So you have X x [0,1]

Let $i_0(x)=(x,0)$

[gentle osprey]

Karen

[marsh forge]

Then you have a triangle

X —> Y

|
|

V

CX

And an extension is a map CX->Y

Making it commute

[pearl holly]

Okay I see. And the inclusion is from X to CX right?

[marsh forge]

Yes

[pearl holly]

Okay great! Thank you!

[marsh forge]

Hint: draw CX for some easy spaces X

[pearl holly]

Okay wait wait. What does it mean for a function to be contractible? Hatcher only defined what a contractible space is

Like homotopic to a constant map?

But that is nullhomotopic lmao

[marsh forge]

Nullhomotopic is the same thing

Both terms are common

[pearl holly]

Okay great!

[pearl holly]

Okay I think that I got something going on. This is very handway lmao but I think that this still works. So let's prove the implication $\implies$ first. Let $f$ be the map that is homotopic to the constant map, $c$. Then there is a homotopy $h_t: X \to Y$ such that $h_0 = f$ and $h_1 = c$. Now $CX = (X \times I)/ \sim$ where $\sim$ means "take the last $Xx{1}$ and map that to one point". If we for a second replace $X$ with $S^1$, then $S^1 \times I$ is a cylinder. Then $CS^1$ is a cylinder with a "sharp edge/point" at the top and that point is no longer a circle. Hence $CX = (X \times [0, 1)) \cup {a}$ where $a$ is the point at the top. Now define $$g(x, y) = \begin{cases} h_t(x) t \in [0, 1) \ g(a, 1) t=1\end{cases}.$$ This is a map from $CX$ to $Y$ and we see that $gi(x) = g(x, 0) = f(x)$ so the diagram commutes. This is the first implication.

The second one is very similar to the other argument I had. Let $f: X \to Y$ that extends to $g: CX \to Y$. If we again replace $CX$ with $CS^1$ then we see that $CS^1$ is a deformation retraction of a single point. Let $h_t$ denote this deformation. Then $gh_t: CX \to Y$ induced a homotopy between $g$ and a constant loop. Then restrict $g$ so that it becomes $f$ and then we have our desired nullhomotopy.

[gentle osprey]

Tokidoki ✓

[flint cove]

why not? the vol 1 exercises were where I first encountered some neat concepts like the hole in a hole in a hole etc

I mean it's a comprehensive introduction, so one naturally shouldn't be scared by the volume of it
Plus old typesetting and lots of space

[pearl holly]

I fricking hate latex

,ban TeXit

[gentle osprey]

You don't have the required permissions to ban members here!

Jul 16, 2021, 11:10 PM

[flint cove]

that's okay, latex hates you too @pearl holly

I mean regarding your question, I share your sentiment, the whole nullhomotopy and homotopy equivalence thing kinda confused me for a while as well

although I was too lazy to read backlog in order to provide more detailed insight :[

[pearl holly]

And this is indeed a generalized "version" of the previous exercise since, like I said before, S¹ X [0, 1]/~ is a sharp cylinder at the top and then if I project it down like the steorographic thing that you (Max) gave me as an exercise before (or if I simply "unbend" it like playdoo), then this is homeomorphic to D²

[marsh forge]

perfectly done!

[pearl holly]

Oh wow great! I thought that something there would be wrong lmao

Anyway, thank you so much for the help and for the exercises! I will go to sleep now since I'm kind of tired. Bye!

[gritty widget]

i think the big ones require a bit more background than little spivak (eg point set topology) but i could be wrong

i guess i wanted to specify i meant the little one, and not necessarily to disqualify the big one

[flint cove]

I still vividly remember learning a ton about DiffGeo from that first volume even though I didn't formally know about topology yet
But that was more interested, nonlinear reading as opposed to conscientious word-for-word study

And I was studying physics back then so my judgement was clearly pretty terrible

[abstract pagoda]

S1 is not a retract of D2 and if someone gave a retraction r:D2->S1 with r(x)=x/|x| they would be wrong because r^-1(r(B(0))) is not open in D2? I feel like im explaining it wrong.

Where B is a open ball around 0

[coral pivot]

i mean r(x) isnt even defined on D2, the origin that is

However if you do D2- {0} then this would be a retract

[abstract pagoda]

So in the argument all I need to say is that r is not well defined so it cant be continuous

[marsh forge]

r isnt defined* would be more correct

Well defined is something more specific

Usually

[abstract pagoda]

I thought well defined meant it is defined over the entire function

[marsh forge]

No

Also domain

[abstract pagoda]

So x=y means f(x)=f(y)

[marsh forge]

Yes

[abstract pagoda]

Sorry I meant domain

[marsh forge]

Thats different from what you said though

the issue is that you haven’t told me what r(0) even should be or might be

You literally haven’t defined it

[plain raven]

max can you summarize your talk on the importance of proving that a loop is nullhomotopic iff it extends along the inclusion S^1 -> D^2

what's the short version lol

[marsh forge]

Its a shockingly useful intuition in my experience, generalizes to cones, and is the first time a lot of people are introduced to extension problems that matter

It also sort of parallels the visual intuition of homology in an interesting way

Ie a generation is trivial if it’s a boundary

Generator* fuck autocorrect

Idk I just think it’s neat haha

[abstract pagoda]

generator of?

[marsh forge]

Pi n

[abstract pagoda]

How is S1 x I/ S1 x {1} a quotient topology?

I thought you need a map X->X/~ and a equivalence relation ~

Would it be S1 x I -> S1 x I/~ where a~b iff a=b?

[marsh forge]

The equivalence relation is that all points in S1x1 are the same

And all other points are unchanged

[abstract pagoda]

Oh ok

So

S1 x I -> S1 x I/~ where a~b iff a,b in S1 x {1}?

[marsh forge]

Or a=b

But yes

[abstract pagoda]

I have to add the or a=b part to include the "all other points unchanged"

[marsh forge]

This is a very common construction

Where you have a space C

X not C

And a subset A of X

You write X/A

For “X with A smushed to a single point”

[abstract pagoda]

It isnt proper to think of it as X-A U {*} where * is equivalence class of points in A?

[marsh forge]

No

That gives the wrong topology in general

Take the following example

Let D be the unit disk in R^2

And let X be the 1/2 radius disk

Then D/X is homeomorphic to D (exercise)

But D-X U * is not

At least with the normal topology

As a set the two are the same but the quotient specifies the topology, and if you write the union people will assume your mean the disjoint topological union

[abstract pagoda]

q:D->D/~ where a~b iff a=b or |a|,|b|<1/2, q is continuous and D/~=D/X
A is open in D/X iff q^-1(A) open in D

The goal is to have a map f:D/X->D that is continuous from here

And then im thinking I can show f^-1 is continuous

Im thinking I can make f a function that takes the equivalence class made from |a|,|b|<1/2 to the origin

And it takes all the other points and and somehow stretches it to the origin continuously

[marsh forge]

Thats the right idea

[abstract pagoda]

I was thinking f:D/X->D with f({*})=0 and f(x)=1/2(1+x) for x!={*}

Nah

I ran into a problem i thinks

I guess I should write it in terms of e^itheta

[marsh forge]

You can come back to this later haha

[abstract pagoda]

I think I get the idea

similar to how quotient groups take cosets and treat them as elements, quotient topologies take equivalence classes of points and treat them as open sets

It just threw me off seeing S1 x I/ S1 x {1} instead of S1 x I/~ and then seeing the definition of ~

Oh lol

I think I understand the proof from that review, thank you.

[plain raven]

hopefully this doesn't come across as too obnoxious, but both quotient groups and quotient topologies have universal properties which justify the construction

you can prove that if X is a topological space, and ~ is an equivalence relation on the underlying set of X, then for any continuous map f : X -> Y for another space Y, satisfying x \sim x' -> f(x) \sim f(x'), there is a unique factoring of f through the quotient map f : X -> X/\sim

Somewhat similar to Noether's first isomorphism theorem

[abstract pagoda]

How would you check x~x’ => f(x)~f(x’), Do you have to do it explicitly or is there a roundabout way. Continuing an example with homotopies if you have a homotopy between f,g:X->Y being
H:X x I -> Y with H(x,0)=f and H(x,1)=g. Is there is a relation that satisfies x~y => H(x,t) ~ H(y,t)?

The only thing I can think of is picking the relation to be x~y iff H(x,t)=H(y,t)

But then I dont know what X/~ would be or look like

Oh wait I made a mistake in asking

I cant think of a relation that makes that work

Oh you might have written it wrong

[marsh forge]

no

what he wrote is correct

if the map itself satisfies that condition

[abstract pagoda]

because x~y=>f(x)~f(y)?

[marsh forge]

that is a property that that map can satisfy but does not have to

what clerk is saying is that maps that do end up satisfying that

have the unique factorization described

[abstract pagoda]

Then why is the equivalence relation based on the underlying set because f(x) is in Y, not X?

Or is it that the equivalence relation cant be anything like a~b if a and b satisfy a property specific to X

[empty grove]

It should probably be x ~ x' → f(x) = f(x') right?

We don't have an equivalence relation on Y

[abstract pagoda]

Yea thats why I was confused

because if there was an equivalence relation on X it would have to be independent of elements of X and say something about the set in general

[marsh forge]

oh yes

im silly

sorry very tired

there is a more general statement where like

X and Y come equipped with equivalenve relations X/~ and Y/~'

[abstract pagoda]

Oh

[marsh forge]

and a map f: X -> Y "descends" to a map X/~ -> Y/~'

if the condition clerk wrote is satisfied

[abstract pagoda]

These universal property things are category theory related, right?

[marsh forge]

They are often formalized with category theory but often you don't really need any real category theory to discuss them

categories help us see how all universal properties can be seen as the same type of thing i guess

[pearl holly]

Hmm this looks awfully similar to the exercise I was doing yesterday and the solution that moldi and Max had in mind. What does this mean though? How do you know that h factors through the quotient?

wait let me check Munkres

Okay there's a theorem that says that if p:X->Y is a quotient map and g: X -> Z is contant on each fiber of p, then g induces a map f:Y -> Z such that fp = g. I guess that this solution uses this theorem

[gritty widget]

first isomorphism theorem for topological spaces

[pearl holly]

Is that actually a thing?

[gritty widget]

yeah i just made it

[pearl holly]

oh wow

fields medal where

[gritty widget]

already have one

[reef shore]

First isomorphism theorem is just the universal property of the quotient

[gritty widget]

🤔

[hazy nexus]

I mean, it pretty much is the first isomorphism theorem, lmao

[gritty widget]

😻

[pearl holly]

:catlove:

[plain raven]

yeah sorry all that was a mistake

i shouldn't have written ~ between elements of Y but rather equality =

[marsh forge]

unacceptable, two more strikes and you're banned

[flint cove]

What's your hand wavy summary of the baire ct?
To me it seems like it establishes thar small sets (nowhere dense) cannot be countably combined to something large (ie with nonempty interior)
And in vector spaces I guess this makes sense because open sets have to, like… stretch in all dimensions?
Not sure how good this mental model is.

(I guess my goal is to get a feel for what the role of baire is in certain proofs, but that gets into analysis territory)

[abstract pagoda]

There is this website called proof wiki and it shows you how to make the map explicitly, but the proof starts off as defining f:X/~->Y as a relation, I guess its harder to write it explicitly because you need to go from taking normal looking elements in X to elements in Y into taking equivalence classes formed by elements in X to elements in Y

In our case where h(s,1)=x we would just need to make the equivalence class from [x]=S1 x {1} in S1 x I / S1 x {1} map where f[x]=h(s,1)=x and then all the other elements in S1 x I / S1 x {1} have equivalence classes that look like their original elements in S1 x I so we can have f[(s,t)]=h(s,t).
Thats what I am thinking atleast

[plain raven]

i usually do it in the converse way. if a set is comeagre then it should be almost everything. so it states that the intersection of sets that are almost everything is nonempty. this gives nonconstructive existence proofs: if you want to prove there exists x such that P(x), decompose the predicate P into an infinite conjunction $\bigwedge_{n\in \mathbb{N}} P_n(x)$ and then prove each $P_n$ holds on an open dense set

[gentle osprey]

diligentClerk

[plain raven]

in classical algebraic geometry the analogue would be conditions where each $P_n$ holds away from the vanishing set of a polynomial (which is open and dense in the Zariski topology). Points that live in a large collection of open dense sets are "generic", and many important results in classical AG (I think to both Krull and Noether) asserted the existence of points sufficiently generic for whatever argument. Scheme theory just adjoins a formal generic point to the space which lives in every open dense set by definition, I guess the fact that this works shows that the arguments are really taking place in the opens of the sheaf and so the points are inessential.

[gentle osprey]

diligentClerk

[plain raven]

Don't take any of this too seriously though, I'm not really knowledgeable about these kinds of arguments

[flint cove]

Ah, that's a great perspective! I think I will start to look for things that „work almost everywhere“ and see what other, stronger properties I can get by taking conjunctions. I think this will be really helpful when I dive deeper into the proofs.

[plain raven]

in computability theory this perspective is very common. Computability theory is the study of the powerset of the natural numbers together with the Turing reducibility preorder, and the resulting poset you get when you quotient out by the preorder, and questions about the poset can often be phrased in terms of existence proofs for sets. for example, if you want to prove it's not linear, you show that there exist incomparable elements. if you want to prove it's dense you assume A, B are given and show there exists some C in between A and B. If you want to prove it doesn't have meets, you have to construct two elements that don't have a meet - actually the way this is done is you construct a pair of elements A, A' together with an ascending sequence {D_n} such that each D_n is less than both A, A', but any B which is below both A and A' lies below some D_n, so the {D_n} are cofinal among elements less than both A and A' - and this proves that A, A' can't have a greatest lower bound.

the reason this strategy works is that there are only countably many computer programs/Turing machines/inputs for those programs/outputs for those programs/etc. so the property "A is not computable" is the conjunction of "A is not computable by turing machine 1" "A is not computable by Turing machine 2"...

if you want construct A where A is not Turing reducible to B, you have to satisfy "Turing machine 1 does not carry out a reduction of A to B" and "Turing machine 2 does not carry out a reduction of A to B" and...

i guess i should be clear about what topology i'm talking about lol

the powerset of the natural numbers is 2^N so you can equip it with the product topology on {0,1}^N

which is homeomorphic to the Cantor set, it's compact, hausdorff and totally disconnected. very nice space

(done)

[flint cove]

That whole range of arguments kinda sounds like a countable version of compactness, or a local-global principle as long as you stay countable

(here I mean FO compactness)

[flint cove]

Does that coincide with compact-open? Because compact sets are precisely the finite ones in N, and that reminds me of the product topology generators

[plain raven]

Yeah, that's interesting. I wonder if what you're commenting on can be explained by the fact that both 2^\omega and the type space arising in the Compactness theorem for FOL are the Stone space of ultrafilters on a boolean algebra. So it's the same special type of compactness that arises from Stone's representation theorem, you see a lot of that in logic because of the ubiquity of Boolean algebras in logic

I don't know if it's the compact open topology. Your argument makes sense tho. Shouldn't be too hard to work out

[obsidian socket]

Wondering if someone knows what o(1) stands for in this this version for the Manin conjecture? The full paper is here: https://arxiv.org/pdf/math/0511041.pdf

[gritty widget]

some function that goes to zero as B goes to infinity, probably

just judging by the picture

[obsidian socket]

thanks! that does make sense

[flint cove]

the boundary is always closed, since it's the closure intersected with the complement's closure, correct?

[empty grove]

Yes

[flint cove]

Thanks for answering the probably most stupid question I have asked here

[frigid patrol]

I'm having a hard time understanding why the transition functions for a vector bundle should be linear

Essentially this boils down to whether or not $\phi_U^{-1}\circ \phi_V(x,v)$ is linear in $v$

[gentle osprey]

PTYamin

[frigid patrol]

Which I dont really see why it should be

[reef shore]

on the same page

[gentle osprey]

Coldilocks

[frigid patrol]

ah ok

[reef shore]

yup

[obsidian socket]

thanks!

[flint cove]

All that might become conceptually clearer if you realize that a trivialization associated to U is actually an isomorphism in the category of U-vector bundles.
Then chart changes give an automorphism of the relevant product bundle in the category of U\cap V-bundles.

[lyric quartz]

Does anyone know if the following statement is true or false? Let $A$ be a simplicial abelian group. Consider $\pi_n(UA,0)$, where $U$ is the forgetful functor from Simplicial Abelian group to Simplicial Set. Let $x\in UA$ be a sum of degenerate element of $A$. Is it true that x is homotopic to $0$?

[gentle osprey]

ClearlyHalfAlive743

[keen urchin]

noob question here: Let F be a map from M to N, two topological spaces. N has the trivial topology {∅, N}. Is it true that F is always continuous?

I wonder because this seems to be written in this lecture (example b on the right), although I can't make sense of it.

[marsh forge]

Yes (exercise)

[hollow harbor]

what does it mean for a map to be continuous?

[keen urchin]

i see it now. since the preimage of the codomain must be the whole domain.. (right?) it was the preimage concept that confused me

[flint cove]

Judging by the picture, if for every open V in the codomain, the prelecturers head is open in the domain.

[hollow harbor]

Yes

[covert sail]

kind of a simple question but does anyone know if this is true for * all * complex polyhedra?

i think its a polyhedron in which a line that connects two vertices is contained in its interior

like convex as opposed to concave

like cubes and tetrahedrons i thikn

oh wait

does it imply that?

[covert sail]

just asking if thats true (im not really sure)

if all convex polyhedra are topologically equivalent

sorry if that sounds like a dumb quesiton btw

im supposed to write a passage but i dont know much ab the topic so am checking

wait really?

for all convex polyhedron or are there some exceptions

alright thanks, much appreciated!

[pearl holly]

Btw, nice pfp DN

S^1 homeomorphic to S^2 confirmed

[covert sail]

Yep

[gritty widget]

toki, i explained something similar to you a while ago

this is your time to shine

maybe

[pearl holly]

What?

I don't remember

[gritty widget]

convex stuff homeo to balls

[hollow harbor]

convex implies star shaped about a center point probably

right?

[pearl holly]

oh like bounded and convex subset is homeo to the disk

[hollow harbor]

if it's star shaped just normalize along rays

convex does imply star shaped around any interior point, yes

[flint cove]

Except for pointy balls like B[x,0] :3

[plain raven]

?

[swift fjord]

Locally compact hausdorff implies regular right?

[gritty widget]

yes

[swift fjord]

Wait I think I proved that actually

Lol

Ty

[abstract pagoda]

Is there a way to show singletons are closed in regular spaces without directly mentioning that it is hausdorff or going through the proof that singletons are closed in hausdorff spaces?

My idea is
Let X be a topological space and x in X. Suppose {x}^c is closed, then x and {x}^c admit disjoint open neighborhoods U,V respectively. Then {x}^c is open?

Yea my proof isnt right

yea i realized

its possible that {x} and {x}^c are neither open or closed

You can union those

Ah ok thats good proof

What do you mean this is weaker? I thought being regular was a stronger condition

Or are you talking about the condition of singletons being closed is weaker

Yea it isnt iff right, what js a counter example

Counter example for singletons being closed but not hausdorff

And standard topology?

[marsh forge]

Line with two origins is a common construction in topology

you can google it to see a defn

[gentle osprey]

Search results for line with two origins

From https://ncatlab.org/nlab/search?query=line+with+two+origins
Direct page found at: line with two origins

1 result where query appeared in title.

line with two origins

9 results where query appeared in body.

Cantor space
Introduction to Topology -- 1
line with two origins
long line
non-Hausdorff manifold
non-Hausdorff topological space
topological manifold
topological space
topology - contents

[abstract pagoda]

I remember going over line with two origins

But no formal construction

[gritty widget]

non hausdorff spaces

[abstract pagoda]

You just take the origin as the example for the two pointsv

And then they cannt be any disjoint neighbors

What is an example of a regular space that people think of as the standard example

Oh I guess R

[swift fjord]

I didn't read the whole conversation but singletons aren't necessarily closed in a regular space

Only if it's T3

Regular is being able to separate a point from a closed set not containing it

T3 is additionally being T1

That's how I was taught it

That's also equivalent to being regular and T2 which is how it's sometimes defined

[abstract pagoda]

Given a closed set and a point not in that set, there exist disjoint open neighborhoods of both the set and the point

[abstract pagoda]

I never formally learned seperation axioms, tbh my ungraduate topology is meh, there is no graduate topology so idk what that would entail

Only learned about Hausdorff and thas it

Im a noob and browse wikipedia and look things up when people post about them.

It looks like graduate topology isnt a thing for most schools and they are separated into topics classes

Yeup Im agree in ignorance. I am trying to learn AT through Allen Hatcher’s book but Im slowly losing a motivation for learning math as I progress through the topic. I for some reason thought that topology would have a lot more applications in physical world

It probably does but I havent learned a lot

[marsh forge]

Honestly

AT in particular

Doesn’t have that many

If you want some topology with applications

Fred Schiller has some good YouTube videos

[grave maple]

Topolgy usually serves as a foundation for other math subjects.

[obtuse meteor]

Persistent homology is screaming

[marsh forge]

Until someone explains to me why it’s useful in neuro

Besides just stating that as a fact

I’ll keep dunking

[obtuse meteor]

King

[pseudo crane]

is the burden of proof on the dunker or dunkee

[swift fjord]

not having any practical uses IS the motivation for learning it

[past pasture]

rather than an equivilance class?

pls ping me btw

[flint cove]

What is α, what is ξ, what is p? Notation is not that consistent with regards to all that
From guessing I would've said that ξ is not a point but a vector (the brackets will probably refer to a commutator, right?)

[past pasture]

@flint cove here's the page :P

remark 1.1.15 (ii) is where my question lies

[past pasture]

no it does not it refers to equivilance class

[flint cove]

Oh, I see.

[past pasture]

kinda cringe notation ik xd

well not really but ehh kinda cringe

since ec's are denoted with []

ping me again cause I can see u typing alot

[flint cove]

@past pasture the short answer is that since our target space is ℝ^m, we can identify tangent vectors in T_φα(p) ℝ^m with actual elements (or “points”) of ℝ^m

there, the author implicitly considers ℝ^m to be a manifold with only one atlas, namely the identity

[past pasture]

oh wait im dumb lol

i thoguht it was referencing

[flint cove]

And so for equivalence classes [β,ξ] in T_whatever ℝ^m, where β in {β}=B is said atlas, there is only one reasonable representative

[past pasture]

[flint cove]

I mean it's slight abuse of notation if I see that correctly

because when the author writes ξ they mean [β,ξ]_whatever, it's just that φ_β = id_ℝ^m

[past pasture]

ohh I see

thanks C:

[flint cove]

yw :3

[pearl holly]

lux is writing in latex without even needing latex

[gritty widget]

That definition gave me cancer

There's no sane reason for writing it like that, surely

[plain raven]

differential geometry is notorious for dense notation, i don't think this equation is really out of place with what you'd see in any diff geo book

[hollow harbor]

Tterra in shambles.

[gritty widget]

this is garbage writing

[pearl holly]

Let me get this straight: a two dimensional CW complex can be a disk. A three dimensional CW complex could "look like the mapping cylinder"? So like a cylinder glued to the disk. Is this right? And attaching a 2-cell is basically gluing something homeomorphic to the 2D ball?

[plain raven]

this is what differential geometry is, to me

tokidoki can you explain more about what you mean? A disk is definitely a 2D CW complex but i don't understand the bit after that. when we talk about attaching a 2-cell we mean taking an existing complex X, taking a map phi: S^1 -> X and forming the pushout of phi along the inclusion S^1 -> D^2

[gritty widget]

kolar michor slovak

[plain raven]

yeah it's a really interesting book i'm just having trouble getting too much intuition from it

and it's a bit dense lol

i've read warner and that took me ... a long time

this one is maybe even more challenging

[pearl holly]

oh sorry for not responding. I mean that when we attach a 2 cell, intuitively, we just glue something homeomorphic to a 2 dimensional ball into the existing complex. Is this right?

[plain raven]

yeah, along its boundary

[gritty widget]

[plain raven]

if X is the old complex and X' is the new complex

then set theoretically X' is the disjoint union of X with the interior of the disk

because the gluing is along the edge of the disk, the circle

hatcher's appendix on CW complexes is pretty good, Strom's modern classical homotopy theory has a pretty good intro to CW complexes

[pearl holly]

Okay so let's say that I want to glue a cylinder to a space. Then I have to glue the boundary of that cylinder to the space. But this seems weird since if I take the whole boundary and glue on to the space, then I will have no cylider glued there at all

So if I want to glue a cylinder, I only want to glue along the circle below the cylinder

Like the lower part of the cylinder

But hatcher says that I need to take the whole boundary

yeah I need to think over this again lmao

[plain raven]

are you talking about a solid cylinder

or a hollow cylinder

you're gluing on an n dimensional thing along its n-1-dimensional boundary

can you be more specific about what you mean by a cylinder

[pearl holly]

Just a solid cylinder that is filled in. So the boundary of that will be a hollow cylinder

[plain raven]

gotcha. yeah if you only glue along the boundary, it won't have the effect of identifying points inside the cylinder with points inside the complex. that's hard to visualize because there's no good way to realize this in 3D space, you'd have to work in 4D so that the interior of the cylinder is disjoint from the complex

so when you say this

But this seems weird since if I take the whole boundary and glue on to the space, then I will have no cylider glued there at all

i think i disagree with that, the interior of the cylinder will still be there

[pearl holly]

yeah okay I see

You know what, let me draw a picture brb

So now when I try to map the sphere to the blue region, the whole boundary would need to be glued to the blue space and then the whole sphere just collapses

no wait, shouldn't psi go from S² to X¹?

But that's even weirder

[plain raven]

you can totally define a projection from the boundary S^2 onto that flat surface

it'll collapse down the sphere but not the interior

there will just be a 3d blob with the property that if you start at the bottom and move to the top you'll end up where you started.

kind of like if you took a solid donut and glued it to the plane along a cross sectional disk

except whenever you reach the edge of the donut you collapse back to the plane

these are not really good ones to get intuition for lol

you want to start with like

the case where the boundary of the cell is already kind of laid out for you

for example take three points

x, y, z

now if you have a copy of D^1 (the unit interval) its boundary is S^0, a pair of points; there are natural homeomorphisms S^0 \cong {x,y}, S^0 \cong {y,z} and S^0 \cong {x,z}

if you were to glue three copies of D^1 along each of those attachment maps, you'd have a graph with three nodes and three edges which was homeomorphic to a triangle

You could also have the attachment map S^0 -> {x} which sends both points to x, then gluing D^1 along that attachment map would give you a closed loop, an edge which starts and ends at the same point

Now if you take the triangle T we just built

then if you have a copy of D^2, you can glue it to the triangle by choosing a homeomorphism between S^1 and the triangle T. then the gluing of D^2 to the complex along that map would be a solid triangle, because you'd be using the solid disk to fill in the interior

If you took two copies of D^2 and glued them both to the triangle along the same homeomorphisms, then you'd have two solid disks which are disjoint on their interiors but which agree on their boundaries, S^1 = T= S^1

this would be homeomorphic to a sphere, S^2

now you could glue a solid ball D^3 into this sphere along the attachment map S^2 -> S^2 and so on

[pearl holly]

ahhh yes okay

now I see

Thank you so much for clarifying and for taking your time! I really appreciate it!

[plain raven]

np good luck, enjoy topologizing

[bronze lake]

okay this might be the stupidest question asked in this channel

but what's the name of that series of books by spivak that this is from

(might not be spivak, but I'm like 90% sure)

differential geometry, there it is

I got it

[gritty widget]

[gritty widget]

Are those the covers of A Comprehensive Introduction to Differential Geometry?

[flint cove]

Looks like it (C:

[strong shard]

yeah I've never seen them in print in person but apparently those are the covers

as far as i can tell from internet pictures there's not even the title anywhere on the cover, just the picture

[bronze lake]

my uni library copy has actual titles on the cover I think

I'll have to check them out again

[strong shard]

that would make more sense than these pictures lol, unless he titled volume 5 "All the way with Gauss-Bonnet"

[cedar pebble]

pain

[fading vale]

[gentle osprey]

Ultragirlboss

Ultragirlboss

Ultragirlboss

Ultragirlboss

[fading vale]

So honestly this is kinda hard to think about because I havent seen an explicit description of the fiber product of schemes in the non affine case

or of the diagonal map

[cedar pebble]

yea these things are always funky

[fading vale]

Im not really even sure what the diagonal map is?

Like

[cedar pebble]

so at the very least it's going to be induced by universal property

so now we just have to think how this works out

[fading vale]

It says its induced by the identity on Y

But

The only thing i really know about the fiber product here is that its the pullback of Y -> X by itself

[cedar pebble]

mhm! :^)

that's how you form the diagonal map

[fading vale]

well like

$\begin{tikzcd}
Y \times_X Y \ar[r] \ar[d] & Y \ar[d] \
Y \ar[r] & X
\end{tikzcd}$

[cedar pebble]

the simplest case is when you're taking the diagonal \Delta:X->XxX where the pullback is taken along the two canonical morphisms X->Spec(Z)

[gentle osprey]

Ultragirlboss

[fading vale]

If this is all i know about the fiber product here how do i get an "identity map" in both coordinates?

I dont even know that Y x_X Y has coordinates

[cedar pebble]

it does, think about what fiber product means (e.g. if you did this in the category of sets)

[fading vale]

Well that would be the pullback of the two maps right

Explicitly its like points of the product that commute

{(x, y) | f(x) = g(y)}

is this true of schemes too??

[fading vale]

Nvm i have done some soul searching...

Szamuely is dumb

anyway i get it now

szamuely gamered me

But this makes sense

[hollow harbor]

Think about how confusing it is to read on compact mode.

Think about how confusing it is to people who aren't me and therefore are not phenomenologically aware that I am not the one typing.

[fading vale]

Yet...

Landian ultra arc?

[hollow harbor]

At least I can trick Moth.

[fading vale]

No u cant

[hollow harbor]

Revenge for outing my Nikita prank.

[fading vale]

You literally did this last time pretending to be nikita and i found u out

immediately

[hollow harbor]

[mossy ermine]

i am soul

[fading vale]

Hello Gabriel

Knight IV

cat mustache puzzle

[mossy ermine]

can u change my name to moth whiteknight

[fading vale]

Wait that is from 3.

True!

@digital peak gabe has made a request

[mossy ermine]

no i read bernstein instead

idk i didnt feel the proofs were very geometrically motivated

a lot of times it was like

haha silly coincidence

so im reading some more stuff 2 make it less like that

just geometric rep stuff in general

i think bb localization is supposed 2 be intepreted as

looking at the nilpotent cone and springer resolution

and going from one to another

no his lecture notes on d modules

do u do these memes too

functional analysis is rep theory.

yea the only characteristic that matters

fuck all ye number theorists

pls do something fruitful with ur life

ooh what with

[tight agate]

Sometimes you need char p results to prove a char zero result tho

[mossy ermine]

are you proving another 2839474 big things

[bitter yoke]

yep gabe

[mossy ermine]

logic ppl really be on a different plane of existence huh

[bitter yoke]

Gabe says fuck number theorists while begging me to read rational points on varieties w me pls explain this

[mossy ermine]

rational points over C

also i am reading why are you not reading

im literally reading moonen rn

nvm forgot u dont know ag

😔

someone told this to me too

[tight agate]

Time to move to physics

[mossy ermine]

do everything over a alg closed field and then galois descend

[bitter yoke]

Are you actually

Should I catch up

[mossy ermine]

nah i just started

the book is incomplete though

but theres enough

[bitter yoke]

What about ur research tho

[mossy ermine]

to warrant a reading

idc i do what i want!!!!!

[bitter yoke]

yep gabe

[mossy ermine]

rep

semisimple

category O

yea

it is very cool ok!!!!!!!

category O is literally just fin dim

[frigid patrol]

[west brook]

I just discovered the first examples of unicursal uniform polytopes with the same point group as the Leech lattice.

[gritty widget]

[past pasture]

what is the bottom equation meant to represent geometrically?

Or is it just saying that, if that derivative exists for all (a_1,...a_m) in N_0^m then it's smooth?

[pearl holly]

First part

Second part

How does Hatcher know that N is a normal subgroup?

[coral pivot]

When they say normal subgroup generated by all the loops, they dont actually mean the subgroup generated by these elements

but the smallest normal subgroup that contains these

[pearl holly]

bruh

[coral pivot]

like, a subgroup generated by given elements is the smallest subgroup containing said elements

same for normal subgroup generated by given elements

its the smallest normal subgroup that contain those

[pearl holly]

So N is the smallest normal subgroup that contains that loop?

(Just to make sure lmao)

[empty grove]

You should also prove that a smallest such subgroup exists while you're at it

[coral pivot]

seems like they are talking about more than 1 loop, so would be the one containing all those

but yeah

[pearl holly]

oh yeah lmao, alpha varies

Okay thank you so much!

[swift fjord]

Why would you not just say normal closure

[pearl holly]

:bruh:

[empty grove]

[pearl holly]

I don't even know where to start lmao

How do I show that something even exists?

Well there's always a smallest and largest normal subgroup

So that one has to be in between these or something idk

[coral pivot]

hint is ||intersection of normal subgroups is also normal||

[pearl holly]

hgmm okay let me think

[empty grove]

Too direct a hint lol

[pearl holly]

Not for me lmao

[coral pivot]

oof

[pearl holly]

oh okay now I see lmao

[empty grove]

[coral pivot]

my bad ig

[swift fjord]

You can also construct such a group explicitly

And show it must be contained in every other normal subgroup containing the set

[pearl holly]

I think you can use that fact to show that A_5 is simple or something too right?

oops sorry

[swift fjord]

The proof that An is simple is mostly just a lot of casework iirc

[pearl holly]

the fact that johnDirichletseries mentioned I mean

no wait I'm thinking of something different

Oh yeah, A_5 is the only non trivial proper normal subgroup to S_5

[marsh forge]

Construct it

[pearl holly]

[marsh forge]

Or prove it cant not exist

That covers basically every case

[empty grove]

Does zorn's lemma fall into the second category

[marsh forge]

Zorns counts as a construction

[empty grove]

[swift fjord]

Ok then, construct a hamel basis for R over Q please

[empty grove]

Constructivists in shambles

[marsh forge]

Apply Zorns

Done

[swift fjord]

Touché

[hallow swan]

this convention is similar to that of adjoining elements to a field right

[empty grove]

Yes

[pearl holly]

Okay so the van Kampen theorem says under nice conditions that pi_1(X) = the free product of the "components"/N where N = ker(phi), phi being the surjective map from the free product of "components" to pi_1(X). Hatcher now says N is "the normal subgroup generated by the image of the map pi_1(A \cap B) -> pi_1(A)". I guess that in both cases, the normal subgroups are equal. How?

[marsh forge]

Can you post the passage, I can’t imagine hatcher states this without proof

[pearl holly]

Yeah okay sure, wait a sec

This is the theorem along with some info about the maps at the top

wait a moment

[marsh forge]

Oh

Yeah

He explains it

[pearl holly]

[marsh forge]

Do you have a question about the explanation

[pearl holly]

I don't see the expiation

[marsh forge]

Oh lol you’re reading the theorem

Keep turning pages

There will be a proof lol

[pearl holly]

The next proposition in my book is the one that I'm reading now lol, there's no other theorem or anything in between

[empty grove]

Hatcher does proofs later

Examples and such first

Just wait for it lol

Also physical copy of Hatcher

[pearl holly]

yeah but this isn't a example. I've already read the proof of the van Kampen theorem and seen the examples too. It's like Hatcher turns to the applications of this now

[marsh forge]

Okay first

Kampen

I assumed it was a typo the first time

[pearl holly]

Oops sorry

[marsh forge]

You’re okay I just don’t want you to get laughed at for it in person lol

Second I’d have to find my hatcher pdf but there should be a proof of 1.20

[pearl holly]

yeah I've read the proof of 1.20 (which in my version is the van Kampen theorem) but there's no proof that the kernel is the image of the map pi_1(A \cap B) -> pi_1(A)

Maybe I somehow missed it idk

[marsh forge]

Thats like

The entire statement

Lol

Although I’m not sure where you are getting that statement @pearl holly the description of N is given elsewhere

Your description is not correct

Yeah wait that’s very wrong where are you getting that

[pearl holly]

This: kernel is the image of the map pi_1(A \cap B) -> pi_1(A)?

[pearl holly]

This?

Just type "yes" moldi lmao

[empty grove]

wrong opinion toki

[marsh forge]

Yes that is wrong the kernel in van kampen is not the image of the intersection

The intersection can contain nontrivial homotopy classes

[pearl holly]

This is what I am reading now

no crap sorry, I don't mean the image

[fading vale]

1.20 might be the trivial intersection version?

[marsh forge]

Its not

[fading vale]

Oh

[marsh forge]

I’m confused can you post much more

No reason to crop so much

[pearl holly]

Okay sorry let me retake

[marsh forge]

Oh

No

I can explain it

This follows from van kampen

Here is the intuition for van kampen

If I take the free product of pi1A and pi1B

Then I am counting some loops twice

Namely those in A cap B

So what we do is we take a quotient to force these double counted loops to be the same

[pearl holly]

[marsh forge]

Then if B is contractible

And I force every loop of pi1A that is contained in pi1AcapB to be the same as the loop in pi1B

That forced them all to be contractible

Hence the kernel is exactly as described

[pearl holly]

ohhhh

Okay now I see

Thank you so much!

[frigid river]

I've been following some lecture and I'm a little confused 0.0
If I have three groups, G_1, G_2 and H, and both G_1 and G_2 are homomorphic to H, does this necessarily mean that G_1*G_2 is homomorphic to H, where * is the free product?

In hindsight, this is probably more of an algebra question but it's from a topology lecture, so it can't be that far off

[wanton marsh]

homomorphic ?

[pearl holly]

So basically, if you have homomorphisms f_1: G_1 to H and f_2: G_2 to H, then this extends to a map f from their free product to H. But why is f(g_1 g_2) = f_1(g_1) f_2(g_2)?

(I'm working with veryhappy here btw)

[frigid river]

wait, that isn't a word? 0.0

english hard