point-set-topology

[gentle osprey]

Cogwheels of the mind
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[gritty widget]

lmao

love the latex thing

the partial would be a germ, so it takes in functions and spits out values in R

so, Im saying: take the partial derivative of the function x_*

the function is the "pushfoward", and the pushfoward function takes in the chart x as an input

[tawdry widget]

Partial over partial thing is an element of T_p M to me

x^j is a map from M to R.

Never mind let’s wait someone else to answer

[gritty widget]

ily

I hate this problem

[tawdry widget]

Which page are you reading?

[gritty widget]

page of what?

[tawdry widget]

Oh never mind I thought it came from the book you mentioned, differential geometry written by loring W. Tu

[gritty widget]

no

it comes from "Introduction to Manifolds"- Tu

I was looking at something

a theorem and lemma, or whatever

and then I was like, well thats cringe- he didnt prove it fully.

So I tried to prove what he just skimmed

and then it led me to here

[bitter yoke]

where is it in that book? I can maybe try to take a look

[gritty widget]

ill dm

i dont want to be too specific

[gritty widget]

Again, the problem I made isn't an exercise, it's just trying to fill in the gaps for something he said in the book.

It also would be nice to prove, like I said: the exterior derivative version simply has d(df)=0, by definition. The pushfoward is something else though, which can be proved to be connected to the differential- if you use a function/0-form (f). And this pushfoward doesnt have D(pushforward(f))=0 in its definition.

^what I mean by connection is that: $(f)_ {*, p} (X_p) = (df)_p(X_p)d/dt$

[gentle osprey]

TempMathAcc
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(You may edit your message to recompile.)

[pearl holly]

Okay so this is the exercise that I'm working with: Show that $H_0(X, A) = 0$ iff $A$ meets every path component of $X$. I think that I managed to prove the implication $H_0(X, A) = 0\implies A$ meets every path component, but I'm stuck on the converse. The long exact sequence tells me that
[\begin{tikzcd}
{H_0(A)} & {H_0(X)} & {H_0(X, A)} & 0
\arrow["{i_}", from=1-1, to=1-2]
\arrow["{j_}", from=1-2, to=1-3]
\arrow[from=1-3, to=1-4]
\end{tikzcd}]
and to show that $H_0(X, A) = 0$ I can try to prove that $i_*$ (homomorphism induced by the inclusion) is surjective. But I don't know how to do that. I also know that $H_0(X)$ is generated by the path components and the same goes with $H_0(A)$ but I don't know how to use this fact.

[gentle osprey]

Tokidoki ✓

[pearl holly]

there is probably something very stupid that I'm missing here lmao

[swift fjord]

You can represent relative homology as a direct sum of relative homologies of the path components

And what do you know about rel. Homology of a path connected space?

Relative to a nonempty subset

[pearl holly]

hmm I'm not sure

[swift fjord]

Specifically H_0

[pearl holly]

it's trivial right?

[swift fjord]

Yep

[pearl holly]

wait so X is path connected?

[swift fjord]

No

[pearl holly]

or is A path connected?

[swift fjord]

^

H_*(X,A) specifically

[pearl holly]

lmao I didn't even know that I could decompose it like that with relative homology

[swift fjord]

Did hatcher no show that?

Try proving it, it's not too hard

[pearl holly]

I don't remember it

yeah I will try to in a second after I understand this

[swift fjord]

It's like, pretty much the same proof

Just note what the relative set needs to be

[pearl holly]

okay but how am I decomposing H_(X, A)? Into path components in A or in X?

[swift fjord]

In X

[gentle osprey]

Nobody

[eternal nimbus]

anyone knows of good video lectures on projective geometry? bonus if it follows nigel hitchin book

[swift fjord]

Key point in the explicit way is the assumption that A meets every path component

[pearl holly]

hmm yeah but I don't see how I can use that fact in any way at all

[swift fjord]

so you know that $H_0(X)$ is free abelian and is generated by the classes of some element from every path component (In fact every element in the same path component is in the same class)

[gentle osprey]

ShiN

[swift fjord]

Similarly you can show that $H_0(A)$ is free abelian of the same rank, and you wanna show that $i_*$ is surjective, that is, it maps generators to generators

[gentle osprey]

ShiN

[flint cove]

Wait how, I thought H^0 was always of rank 1 in the path-connected case
Like H^0(X,A)=H^0(X/A) is still the homology of something connected, right?
Or do I get H^0(X/A,{A}) and that is something different?
(Just learning about relative homology)

[swift fjord]

it's not always the case that H_0(X,A)=H_0(X/A)

H_0(X,A) is always trivial in the path connected case

H_0(X) is always of rank 1 in the path connected case

[flint cove]

Hm, X/A might suddenly become connected for starters, right?

[swift fjord]

I'm not far enough in to know a solid counterexample but rotman said that this only really happens in 'nice' cases

[flint cove]

Good to know I have to be careful.

[pearl holly]

hmm I'm looking at $H_0(X, A) = \oplus_\alpha H_0(X_\alpha)$ where $X_\alpha$ are the path components of $X$ and to show that the left hand side is trivial, I have to show that every $H_0(X_\alpha)$ is trivial, which seems weird I guess

[gentle osprey]

Tokidoki ✓

[swift fjord]

nono

It decomposes into a sum of relative homologies still

[pearl holly]

oh lmao I misread your message then

so it should be \oplus H_0(X_\alpha, A)?

[flint cove]

Probably A\cap X_\alpha

[gentle osprey]

Nobody

[swift fjord]

yup

you gotta make sure the subspace is inside your space

note that $A\cap X_{\alpha}$ are exactly the path components of $A$ by assumption

[gentle osprey]

ShiN

[pearl holly]

yeah okay and then by a thing you mentioned before, this is trivial right?

Like homology relative to a path connected space is trivial

I think you said that but idk

[swift fjord]

yup

as long as the subset is nonempty

that's where the assumption comes in in my solution

[pearl holly]

bruh it feels like I missed like a whole fricking page in Hatcher lmao

he doesn't talk about these stuff

[swift fjord]

that sucks

[pearl holly]

okay so now I have to prove these fact lmao

1 implication turned into like 3 exercises

[swift fjord]

Try following the proof of the analogous fact for regular homology

[pearl holly]

yeah okay, will do that. Thank you all for the help!

[flint cove]

Look at tom Dieck if you have a bit of patience
For me the exposition to rel. Cohomology was much clearer there

[pearl holly]

cohomology

[flint cove]

I meant homology

For me homology only exists to introduce cohomology lmao

But in all seriousness, you have most of the long exact sequences in cohomology as well because you just build it on the opposite category, so lots of things map 1:1

[flint cove]

Lmao I even switched to superscripts without noticing

[swift fjord]

Yea that confused me

btw rotman proves what we talked about for CW complexes and subcomplexes

or more specifically that
$H_n(X,Y)\cong \tilde{H_n}(\sfrac{X}{Y})$

[gentle osprey]

ShiN

[pearl holly]

yeah Hatcher does prove that too but doesn't (X, Y) need to be a good pair for that?

[swift fjord]

idk it's in chapter 8 lmao

rotman lays all the groundwork for homology before exploring different theories

rn i'm in the chapter on excision

[pearl holly]

hatcher proved the invariance of dimension using the excision theorem lmao

that's kind of cool

[swift fjord]

owo what's that?

[pearl holly]

if nonempty open sets U subset R^m and V subset R^n are homeomorphic then m=n

that's what Hatcher writes

[swift fjord]

ah

cool

haven't done that yet

probably in this chapter then

since it's 'excision and uses'

[pearl holly]

man the proof of the excision theorem

[swift fjord]

about to prove the homology of the sphere and consequently brouwer's theorem too (Even though we proved it in chapter 0 just without knowing what homology is)

it does not seem fun

[pearl holly]

brouwer's theorem in all dimensions?

like the fixed point theorem?

[swift fjord]

yea, he said 'assuming we have such and such facts about homology functors, let's see why this is useful'

yes

[pearl holly]

ouff that seems kind of nice

Hatcher will be doing that later I think

[swift fjord]

I Know for compact sets (Or maybe it's for polyhedra, i'm not sure), the extremal points of a convex set span it (In the sense of convex combinations). Is this true for unbounded compact sets?

@raw sedge after thinking about it I think you're right btw

I think that would work

[flint cove]

Is there a canonical way to make a topological space hausdorff?
For T₀ we can quotient out by the „indistinguishable points“ equivalence relation (i.e, the Kolmogorov Quotient), and for T₁ we can take the quotient which identifies connected components in the specialization preorder.

[reef shore]

Yes

[flint cove]

inb4 „collapse everything to a point“

[reef shore]

You quotient by a slightly more annoying relation, let me find

[reef shore]

youg get the same universal property as the kolmogorov quotient

but with different category

https://www.math.leidenuniv.nl/scripties/BachVanMunster.pdf

[flint cove]

with „the same“ you mean that the canonical morphim X → Haus(X) is a reflector, right?

[reef shore]

Yes

[flint cove]

kewl, did not expect that

[reef shore]

You essentially make the diagonal closed

by identifying a and b if (a,b) lies on the closure of the diagonal

and I think you need to take transitive closure of this since its not an equivalence relation necessarilly

[flint cove]

Without reading that my initial guess would be that we can identify points x, y if every neighborhood of one contains the other in its closure or something (the line with two origins being my intuition here; both origins would have to be identified) – but perhaps that's too weak

[reef shore]

[flint cove]

Oh right I always forget this „diagonal is closed“ classification

hold on writing abbout this one construction is enough for a bachelors thesis?

I should reevaluate the expectations I set for myself

[pearl holly]

Hmm can I assume that (S¹, A) where A is a finite set of points in S¹ is a good pair, i.e A is closed and there is an open set U containing A such that U deformation retracts onto A?

[reef shore]

Yes

There are only finitely many distances between pairs of A. You can take a distance small enough so that the points of A "don't interfere with each other"

In the sense that if you take a distance r smaller than half the distance between any pair of A, then you can take the radius r ball around each point of A as a neighbourhood that deformation retracts to that point

The union of all these r-balls is the good neighborhood

[pearl holly]

yeah okay I see. I got kind of confused since this thing depends on the topology and the exercise doesn't mention what topology it has

because now with this fact, this exercise is kind of trivial right?

I think you know which exercise I'm doing lmao

[reef shore]

S^1 always has the standard topology

[pearl holly]

oh lmao

[reef shore]

is it exercise 17?

[pearl holly]

yeah lmao

[reef shore]

Why is it trivial

[pearl holly]

wait maybe I'm dumb

[reef shore]

[pearl holly]

isn't the homology group of (S^1, A) trivial for n greater than or equal to 1?

[reef shore]

for n greater than 1 yes because it is a 1 dimensional complex

but first homology group is Z when A is empty for example

because then its just H_1(S^1)

[pearl holly]

oh lmfao I drew the wrong long thing on paper

long sequence of homology thing

[reef shore]

wrong long thing 😌

ok now your turn to help 🔫

[pearl holly]

how tf am I supposed to help you moldi????

[reef shore]

why is the subgroup of cycles also finitely generated?

Are subgroups of finitely generated abelian groups always finitely generated? I know that that's not true for non abelian groups or modules in general

[pearl holly]

Exercise to the reader.

no lmao I'm joking let me read it

[reef shore]

oh maybe structure theorem for finitely generated abelian groups will do it

[pearl holly]

hmm yeah I don't know, I think that I didn't read that part in depth

[reef shore]

lol I don't see how to use this since a subgroup need not be a direct sum of its projections

Subgroups of Z^n would be finitely generated because its a free module

So higher rank free modules can't inject into it because of determinant stuff

[pearl holly]

😵‍💫

[coral pivot]

This is true for modules in general yeah

[reef shore]

wait determinant only says no isomorphism

[reef shore]

submodule of finitely generated module is finitely generated?

[coral pivot]

Mhm

It’s bc of the commutavity

[pearl holly]

I found a stackexchange thing lmao

[reef shore]

Every homomorphic image of a finitely generated module is finitely generated. In general, submodules of finitely generated modules need not be finitely generated. As an example, consider the ring R = Z[X1, X2, ...] of all polynomials in countably many variables. R itself is a finitely generated R-module (with {1} as generating set). Consider the submodule K consisting of all those polynomials with zero constant term. Since every polynomial contains only finitely many terms whose coefficients are non-zero, the R-module K is not finitely generated.

wikipedia

[coral pivot]

Oh oof

[pearl holly]

apparently subgroups of an abelian finitely generated group is always finetly generated

[reef shore]

oh lets go

[pearl holly]

should I post the thing here or do you want to try to prove it yourself?

[reef shore]

I will try first

Thank you Toki 😌

[swift fjord]

Isn't that a consequence of the theorem of modules over a PID?

You can prove it with SNF

[reef shore]

Ye that's what I am trying to prove

[swift fjord]

Ah

[coral pivot]

Yeah ig it’s not true in general modules

[reef shore]

oh I see neat

[swift fjord]

Smith normal form is the way we proved it

[coral pivot]

Just black box it smh

[swift fjord]

I think that's standard

[reef shore]

Oh shit I think we proved this in a course too

[coral pivot]

Yeah ok that makes sense

Never had to work with non noetherian ring kek

[reef shore]

That's neat not worked with many noetherian modules though, haven't seen that theorem

[swift fjord]

Imagine being non noetherian

Kinda cring

[reef shore]

I see

Thanks everyone

[flint cove]

If your linear algebra knowledge does not increase linearly over time, you're doing something wrong.

(Lie- and Homological algebra plus modules over PIDs counts as linear algebra, right?)

I just proved that cl(A)×cl(B) = cl(AxB) and I feel stupid for how long that took.
But I'm glad I could write that down cleanly.

[abstract pagoda]

wait

[abstract pagoda]

cl like closure?

using what definition of closured to show it?

[reef shore]

complement is open?

are there other definitions in general topology

[abstract pagoda]

i think so

intersection of all closed sets contaaining A

[reef shore]

oh of closure

[abstract pagoda]

oh i didnt write it lol

[reef shore]

yeah smallest closed set containing A

[abstract pagoda]

A U limitpoints(A) is easiest definition for me to use

i always default to it

(A U A’) x (B U B’)= (A x B) U (A x B)’

[flint cove]

I used the „smallest closed set“ definition and the equivalent characterization that x not in cl(A) iff x can be separated from A by an open set

And then I wrote:
Recall that A×B⋂U×V = A⋂U × B⋂V, which implies
that the intersection of two boxes is empty
iff the intersection of one of their sides is.

If (x₁,x₂) can be separated from A×B via U×V, i.e. their intersection is empty,
hence one of A⋂U or B⋂V is empty.
If the former is empty, we have that x₁ not in cl(A),
implying in particular that (x₁,x₂) not in cl(A)×cl(B).
If the latter is empty, x₂ is not in cl(B), providing the same conclusion.

In summary, the outside of A×B is contained in the outside of cl(A)×cl(B) (which is closed!),
and so cl(A)×cl(B) is contained in cl(A×B).

On the other hand, cl(A)×cl(B) is a closed set containing A×B,
implying that cl(A×B) is contained in the former.

[reef shore]

Can't you say that cl(AxB) can't be smaller because each horizontal and vertical slice is homeomorphic to the projection and so the projections' closure is contained in this, and for the other inclusion prove that product of closed sets is closed (because the complement is union of A^c x Y and X x B^c)

[abstract pagoda]

Can yoy clarify first sentence using symbols. I think I know what you meant mr mold

[pearl holly]

mr mold

[abstract pagoda]

lux i like it it feels intuitive

[reef shore]

For any $(a,b)\in \overline A \times \overline B$, ${a}\times B \subseteq \overline{A\times B}$ by looking at each vertical slice $X \times {y}$, which is homeomorphic to just $X$, and then $(a,b)\in \overline{{a}\times B}$ by the same argument

[gentle osprey]

Coldilocks ✓

[flint cove]

Oh wait so you mean I should show that cl(A×B) is in cl(A)×X₂ and in X₁×cl(B), hence in the intersection?
I've thought about it but haven't seen why the closure of the slice is the slice of the closure

[reef shore]

each slice is homeomorphic to its projection on the correct axis

[flint cove]

but why

[reef shore]

so you can project, take closure, then invert projection

[reef shore]

product with terminal object

like by slice I mean slice of the whole product space

[abstract pagoda]

tbh examples would help

[pearl holly]

lux is literally writing latex without writing latex

[gentle osprey]

Coldilocks ✓

[abstract pagoda]

yes

agree

[reef shore]

So for any set in the product

For any slice of it

You take its closure within that slice

ie just along that axis

[abstract pagoda]

isnt this only for vertical slices

[reef shore]

also for horizontal

Except they are homeomorphic to X

[abstract pagoda]

yes

[pearl holly]

pi_1 and pi_2 moment

[reef shore]

And closure of any slice is contained in the closure of the projection

[abstract pagoda]

you lost be when you said by looking at each vertical slice and then gave a horizontal one

[reef shore]

because closure is order preserving

I am learning from lux

Second factor is the horizontal axis

[abstract pagoda]

strange

[reef shore]

(jk )

[pearl holly]

wait I might have misread and accidentally 'ed a wrong message

[abstract pagoda]

so closure of a x B is closure of vertical slice which is contained in A x B and closure of horizontal slice A x b is contained in closure of A x B

[flint cove]

oh I got it the projection π₁ is a retraction for any slice inclusion hence that's an embedding
I feel stupid

[pearl holly]

fundamental group moment

[reef shore]

lol now I am confused

[abstract pagoda]

first entry is in horizontal direction

wait wtf

[reef shore]

I am saying that if (a,b) is in the product of closures, then a x B is in the closure of the product by taking closure along each correct axis, and then (a,b) in in the closure of a x B

[abstract pagoda]

why didnt i just write cl

[reef shore]

I don't use A x b

[abstract pagoda]

oh ok

[reef shore]

oh ye I see what you are saying now

I thought you were giving complete argument

[flint cove]

(Bonus points to whoever jumps in and interprets this as a matrix-vector-product)

[abstract pagoda]

nada

[reef shore]

[abstract pagoda]

toki

i have stopped reading at covering spaces chapter

i think imma pick up later today

[reef shore]

Yeah so cl(AxB) > cl(A) x cl(B), but right side is already a closed set contained AxB

So this must be an equality

[abstract pagoda]

also since we are in topology i have question about homology

[flint cove]

I don't quite get yet why {a}×B is in cl(A×B) when a is in cl(A) but that doesn't sound too hard. I'm gonna have to eat something first tho. 🙃

[reef shore]

fix a y in B and check (a,y) along the slice A x y

[abstract pagoda]

If you want a bunch of extra context here:#math-discussion message
My main question is just to get an understanding of the fundementals of the purpose of what homology is used for.
Mr. Max said that it is used to tell when cycles are boundaries for graphs.
Mr. Zoph told me that H_n=Z_n/B_n=ker(d)/im(d)
How are do cycles and kernel of differential maps relate, same question for the image.
Looking at a graph what is the differential supposed to represent. I think I know that homology isnt used just for graphs, so are there other concrete examples that can be used?

[marsh forge]

(Both me and zoph identify as men)

[pearl holly]

mr mold

[abstract pagoda]

is Mrs for married men?

[marsh forge]

No

Only women

[abstract pagoda]

[marsh forge]

I’m also not married

Okay okay

So

I’m stuck in a car for an hour with nothing to do

So maybe I can help w the confusion

So, maybe a good starting place is to ask how many computations of homology you’ve done directly (not using long exact sequences)

[abstract pagoda]

0

[marsh forge]

Okay that is a problem for sure

But I will try to handwave a bit

Everything I am about to say is true up to choosing signs (or, if you prefer, exactly true over Z/2)

Picture a triangle that is not filled in, and give it the usual delta complex structure

Can you picture this?

(Let’s ignore orientations for now)

[abstract pagoda]

i only read a little bit but is delta complex structure when you give the edges a direction?

[marsh forge]

It’s when you have an almost-triangulation of your space

[abstract pagoda]

so like you have a simplex and each edge has a start and end point

[marsh forge]

So the delta structure I am talking about

Is 3 verities

And three edges connecting them

In terms of the chain complex, this will be 3 generators in degree 0, 3 generators in degree 1, and then no others

Does that make sense?

[abstract pagoda]

ok

[marsh forge]

Okay so

A cycle is a chain that is sent to 0 by the differential

Label the edges a b and c

Then when you apply the differential to an edge

You get two verticies with signs

[abstract pagoda]

ok

[marsh forge]

If you apply the differential to all three edges

With the right signs

Everything cancels

So in effect

d(a+b+c)=0

Im again ignoring sign issues

[abstract pagoda]

oh wait

[marsh forge]

So the sum of all three edges forms a cycle

In fact, any graph theoretic cycle will have this property if you choose the signs correctly

(A cycle being something in the kernel of the differential)

[abstract pagoda]

Is this the reason why d^2 =0?

[marsh forge]

No

[abstract pagoda]

you said differential of an edge gives two vertexes with signs

[marsh forge]

Yes

[abstract pagoda]

so does it work linearly

[marsh forge]

So just throw a plus/minus infront of everything I say and it will be less of a lie

Okay so

H1 is the kernel of d1 over the image of d2

But there are no generators in degree 2

So the image of d2 is 0

So H1 has one generator and it is a+b+c

So H1=Z and we see one “hole”

Namely the unfilled I’m triangle

[abstract pagoda]

ok

[marsh forge]

Now suppose we fill the triangle in and name the filled in part X

[abstract pagoda]

im following

[marsh forge]

Then

d(X)=a+b+c

So H1 becomes 0

Because we filled in the hole

That is like the basic intuition

But in order to actually understand homology

You just have to do a ton of computations

[abstract pagoda]

oh wait

i think i get what the differential does sort of

and how it relates to boundaries

lemme draw something

[marsh forge]

Wait

Are abcde edges or vertices

[abstract pagoda]

rip i switched it up didnt i

ok remove e

and they will be vertxxies

[marsh forge]

You need to label vertices too

Sorry edges

[abstract pagoda]

ill label edges by just putting them next to eachother

[marsh forge]

Okay

[abstract pagoda]

so like d(ac)=a+c?

[marsh forge]

Up to signs yes

[abstract pagoda]

and d(X)=ad+ac+dc

[marsh forge]

Yes

[abstract pagoda]

So then H_1=kerd1/imgd2?

[marsh forge]

Yes

[abstract pagoda]

but what is d1 now?

[marsh forge]

Wdym

You have a basis (namely the 5 edges)

And you can compute d1 on each

So you know the whole map by linearity

I think I have done as much as I can without you just having to get your hands dirty with computations

I’d be happy to help walk you through those, however

[abstract pagoda]

ok ty ill do more reading you cleared up the initial confusion

[marsh forge]

The real intuition

Is that homology detects parts of your space that can bound something, but don’t

Based on the dimensions of this hole and the potential boundary, that’s where it shows up

[abstract pagoda]

oh ok

i want to read about the connection between homology and homotopy because that was mentioned in hatcher's intro the the chapter

[marsh forge]

It is very deep and very complicated but there are good theorems about it in hatcher for sure!

Arguably the connection between homology and homotopy roughly describes a lot of what I study to this day

[grave maple]

Quick question: if $f \times g$ is a quotient map, then $\pi_1 \circ f$ and $\pi_2 \circ g$ are quotient maps?

[gentle osprey]

echoone

[grave maple]

I believe quotient maps also map open sets to open sets.

I knew it. But the converse is not necessarily true at least in Top, right?

That is, if f and g are quotient then f x g is quotient.

https://math.stackexchange.com/questions/31697/when-is-the-product-of-two-quotient-maps-a-quotient-map

In Conv, it turns out to be true. But the definition of quotient maps in Conv is pretty gnarly.

At least, at first sight.

I wish there some some giant reference with all this info.

The best I've been able to find is some abstract nonsense like "in Conv, final episinks are preserved by pullbacks along arbitrary morphisms".

[versed junco]

There are regions of $R^n$ of different dimensionality. While (open) orthants like (+,-,-) etc are indeed important, how are $(0,+,-,+)$, $(+,0,-,-)$ called?

[gentle osprey]

JohnDark

[sleek thicket]

Are you asking if there's a term like "orthant" for them? If so I don't know it

it's an orthant of the 3d subspace where the first/second coordinate is zero

[swift fjord]

Given some set $K$ in $\mathbb R^n$, is there a nice way to characterise the points of $K$ lying on the boundary of its convex hull? In particular in my case $K$ is discrete

[gentle osprey]

ShiN

[gritty widget]

I think this is a common problem in computational geometry

Not to characterize it, but to compute it

[grave maple]

Wouldn't it just be the intersection of K with the boundary?

[swift fjord]

Ofc, but that's not very useful if you don't know what the boundary of the convex hull looks like

Which is also a problem

Even in R2 i'm not seeing a straightforward way to determine whether a point is sitting on the boundary or not

[grave maple]

It's not straight forward. The boundary is a closed curve, so the Jordan curve theorem applies.

[swift fjord]

In my case it may be a bit simpler. I'm specifically looking at the convex hull of a lattice in the positive orthant. I was wondering if there was smth more general but maybe in this more special case there's smth you can say about the points on the boundary

For example I think it's true (Haven't tried proving it yet, might be smth well known) that every point on the boundary lies on some supporting hyperplane. In particular I think it's true that the supporting hyperplane will be the affine hyperplane orthogonal to the point

I'm not sure if the second part would generalise tho

You can prove that this shape is asymptotic to the axes, so intuitively this should be true since the angle between 2 lattice points on the boundary becomes duller as you keep going

[gritty widget]

Just want to double check, X is a compactifaction of Y if X is compact and Y is dense in X?

[swift fjord]

p(A,x) is the metric projection of $x$ onto $A$

[gentle osprey]

ShiN

[swift fjord]

Why do we have that
$$\langle y-p(A,x), x-p(A,x)\rangle >0$$?

[gentle osprey]

ShiN

[swift fjord]

Maybe I just don't understand how the hyperplane through p(A,x), orthogonal to x-p(A,x) is parametrised

Also doesn't $\overline{y} \notin (p(A,x),y]$ imply $\overline{y}=p(A,x)$ or am I missing smth??

[gentle osprey]

ShiN

[swift fjord]

Like, I understand this would be the case were E the orthogonal complement of x-p(A,x), but in general it would be an affine hyperplane no? So shouldn't the inequality be greater than some constant that is not necessarily 0?

[swift fjord]

why tho? that's my problem. In the general case E would be an affine hyperplane so shouldn't it be defined by an equality of the form $\langle \cdot, x-p(A,x)\rangle = \beta$?

[gentle osprey]

ShiN

[swift fjord]

Maybe they're implicitly shifting the setup such that p(A,x) is 0?

Even so weird that it's not even mentioned

[bright acorn]

Guys, just a quick question. But suppose that M is a symplectic manifold and N is an isotropic submanifold. Then, is it true that there always exist a lagrangian submanifold that contains N?

[brisk horizon]

is this the channel to ask question about algebraic geometry?

[tawdry widget]

Please share your question. Though I just started reading algebraic geometry, meeting varieties of questions is always nice.

[brisk horizon]

oh i just started reading Perrin's book. I was confused by this statement. "The affine algebraic subsets of a line is the line and finite sets."

Where the affine space is one dimensional

[tawdry widget]

In which page btw?

[brisk horizon]

page 9

[tawdry widget]

Oh I see so first the line k itself is an algebraic set (k=V({0}) )therefore we need to check the other half

That is when S doesn’t equal {0} then V(S) is a finite subset of k

Notice that V(S) equals V(<S>) where <S> is the ideal generated by elements of S

Since S doesn’t equal {0} therefore <S> is a non-zero ideal of k[x]

WLOG We assume that S=<S> is a non-zero ideal of k[x]. First if S doesn’t equal k[x] then since k[x] is noetherian, S is finitely generated. Let’s say S=Σ{k[x]P_i 1<=i <=n} then V(S) =V({P_i :1<=i <=n}

Therefore V(S) is the set of common roots of those P_i therefore a finite subset since any non-zero polynomial in k[x] has at most finite number of roots in k

If S=k[x] then V(S) is empty. Done

[cedar pebble]

For k algebraically closed the topology on MaxSpec(k[t]) (the set of maximal ideals of k[t], with the Zariski topology) is the cofinite topology.

this is more or less equivalent to what is above

[brisk horizon]

I'm not sure what affine algebraic subsets of a line mean. i know what an affine algebraic set defined by S means. Would an affine algebraic subset of S mean affine algeebraic set of subsets of S?

[tawdry widget]

Meaning an affine algebraic set in terms of k[x] which is also a subset of the affine algebraic set k=V({0})

[brisk horizon]

Ah, okay, I think i understand. k is an algebraic set because it is V({0}). Then the other algebraic sets are finite sets because polynomials have finite roots.

[tawdry widget]

Yes . And affine line / plane is just a name. Affine line is V({0}) in terms of k[x], affine plane is V({0}) in terms of k[x_1,x_2]

[brisk horizon]

Ah, that makes more sense.

In the Affine plane, other than the empty sets and the plane, "curves in the form V(F)" are algebraic sets. Curves here should mean algebraic curves right?

[tawdry widget]

Curve actually means that it’s dimension is 1

In the affine case V(S) is a curve in k^n means that the Krull dimension of k[x_1,...,x_n]/P=1 for any P which is a prime ideal that is minimal among prime ideals that contain <S>

[brisk horizon]

I vaguely remember Krull dimension

[tawdry widget]

Commutative algebra is needed I think, in order to read algebraic geometry

[bronze lake]

https://cdn.discordapp.com/attachments/708983945974513694/838779931818721310/20210503_101135.jpg

what are y'alls thoughts on this kind of final for an intro to topology class

went on for 3 pages of this

[ivory dragon]

very dumb

[gritty widget]

I like it

[versed pivot]

I wouldn't mind like 2 questions like that. But 3 pages?

[ivory dragon]

you have access to munkres too wtf

half of those are always examples in munkres

[bronze lake]

it gets into some really pathological examples

and munkres also doesn't cover all the properties of all these spaces

[sleek thicket]

Hate this

Really really bad

[empty grove]

Don't really mind it, looks like free marks

[gritty widget]

I thought it was easy points, but actually I don't know what lindelof means

Neither do I remember the metrization theorems

[empty grove]

Every open cover has countable subcover

[bleak helm]

This is big oof, imo

Could at least have asked you to prove it

Is this test real though?

Can't use a Ouija board lmaooo

[rancid umbra]

lol

[gritty widget]

Is oujia stuff real?

[bleak helm]

[brisk horizon]

If it can help you with this test, then it’s as real as it gonna get

[bleak helm]

I bet your next test won't tell you not to use one, so try it out

[obtuse meteor]

the exam for a good point-set topology class should be anything but point-set topology

just like how the class should be anything but point-set topology

[ivory dragon]

ive seen lines like that a lot

theyre intentionally ridiculous to draw attention to them

and if students are like

"okay i know i cant use [x] but can i use [y] as long as i dont post a question?" or things like that

it sort of preemptively answers them through hyperbole

"you cant use ANYTHING i didnt explicitly allow, including ridiculous stuff"

[bleak helm]

Ah, that makes sense but still amuses me. I've never seen it.

[tawdry widget]

Is there any book that’s specifically about Mandelbrot set? I want to find the reason of many properties about it.

[swift fjord]

This is the laziest exam I've ever seen

[novel acorn]

So I don't fully understand how this works
When Hatcher says we modified the sequence does he mean that we extended it?
Like after the 0 in the first sequence we added the Z?

And then because there's still a 0 before the Z, the reduced homology becomes the same as the normal one

[pearl holly]

Tbh I didn't get that part either but the fact that the reduced relative homology is the same as relative homology follows from definition. You have chains
$$\cdots \to C_n(X) \to \cdots \to \mathbb{Z} \to 0$$ and
$$\cdots \to C_n(A) \to \cdots \to \mathbb{Z} \to 0$$
then by definition, you define the relative reduced homology as the homology group of the chain complexes "divided" by each other. But since $\mathbb {Z}/\mathbb{Z}$ is trivial, you get the original chain for the non-reduced relative homology groups.

[gentle osprey]

Tokidoki ✓

[pearl holly]

maybe that doesn't help you at all lmao, but I didn't understand that part too so I just skipped it once I saw this

[novel acorn]

Lol
Yeah I guess you could look at it this way
I just dunno if you can divide it that way

[pearl holly]

why wouldn't you? That's the definition of relative homology

but maybe this is completely wrong idk

[swift fjord]

Nah you're right toki. The reduced chain complex for relative homology is the same as the regular one

[novel acorn]

So you can divide two chains that way

so that the Zs also factor to 0?

[swift fjord]

Yea

[bright acorn]

Guys, at the end of the semester I will have to give a talk on some topic in symplectic geometry.

And I think I will try to talk about Floer Homology

I would like to ask someone who is more experienced in symplectic geometry / topology some references on Floer Homology. Some of its applications in low dim topology, surgery theory, its influence on the Arnold's Conjecture, how it is related to morse theory and etc.

It seems such a powerful tool, I just don't know exactly about what exactly to talk about it.

Some ideas and references would be good

[plain raven]

reduced homology is neat

[marsh forge]

one might call it... based

[gentle osprey]

Nobody

[cedar pebble]

what do you mean by this?

like what do you mean by "correct"

if we grant that HZ represents ordinary cohomology with Z coefficients then this is an immediate consequence of S-duality

[bright acorn]

😳

[gritty widget]

https://link.springer.com/book/10.1007/978-1-4471-5496-9

you know where to find a copy

[bright acorn]

Yeah, I have checked this one out already. Just wanted to know if anyone here had some more interesting references and stuff

thanks again TTerra

[reef shore]

This diagram commutes, so we can identify the 2 groups at the top with the 2 at the bottom which are isomorphic to Z. Then the map f* at the top becomes multiplication by an integer. Wouldn't it be the same integer as the f* at the bottom by commutativity of the diagram?

[reef shore]

Is it important that the arrows aren't going directly from the bottom groups to the ones at the top? But I think that in a square
A → B
↓ ↓
C → D
if the 2 vertical maps are isomorphisms then this square commutes iff the square with these 2 replaced with their inverses commutes

ah the middle line doesn't exactly split the square into 2 squares with that property

[sweet wing]

isit jus me or does the puppe sequence intuitively feel like it's in the wrong direction

[sweet wing]

my intuition is like

a "(co)fibre sequence" is "approximately like 0->F->X->Y->0 is exact except we dont have the 0-> and ->0 actually" so Hom(A,-) or Hom(-,A) should be left exact

[cedar pebble]

Depends on if you’re talking about the exact puppe sequence or the coexact puppe sequence

In some sense the exact puppe sequence goes the wrong direction yes

Since you’re taking the fibration dual to the mapping cone

It’s just that in practice the coexact one is harder to use

[sweet wing]

hm the coexact puppe sequence is also "flipped" unless im directionally impaired again right cuz you start with
X -> Y -> C(f)
and get
... -> [C(f),B] -> [Y,B] -> [X,B]

[cedar pebble]

Ah okay I see what you mean

[sweet wing]

perhaps this is why cohomology is more "natural" whole homology you need to like wedge then take hom

i always thought of the puppe as "derived functors" and i just realized it is flipped lol

[cedar pebble]

Yea so I mean cohomology is a lot more natural in the sense that you’re just taking homs

Homology requires the smash product and that’s a pain

[sweet wing]

yeah

[cedar pebble]

These two things are monoidally dual, but setting this up is definitely a pain in its own right

[sweet wing]

oh huh

will see that someday i guess

[cedar pebble]

So I mean smash product gives Ho(Spectra) monoidal structure, and you get a smash/hom adjunction as you would with the usual tensor product for example

So any time you have cohomology you can dualize in this sense to get homology

[sweet wing]

ahhhh right thats true

[cedar pebble]

The problem is just that setting up a good smash product of spectra is a really nontrivial ordeal; it doesn’t work for sequential spectra for instance

[sweet wing]

yea

[pearl holly]

Okay so Hatcher wants to prove that the homomorphism induced by the inclusion $i_*: H_k(X^n) \to H_k(X)$ is an iso if $k < n$. He writes "From the long exact sequence of the pair $(X, X^n)$ it suffices to show $H_k(X, X^n) = 0$ for $k \leq n$. Since $H_k(X, X^n) \cong \tilde{H_k}(X/X^n)$, this reduces the problem to showing $\tilde{H_k}(X) = 0$ for $k \leq n$ if the $n$-skeleton of $X$ is a point". I don't get the last part. Why is that sufficient? If the $n$-skeleton is just a point, shouldn't it just be the 0-skeleton or at least a subset of it?

[gentle osprey]

Tokidoki ✓

[pearl holly]

oh and $X$ is a CW complex and $X^n$ is the $n$-skeleton of it

[gentle osprey]

Tokidoki ✓

[tawdry widget]

I think that the n-skeleton of X being equal to a point means that any cell of X has dimension greater than n, so X is not necessarily a 0-skeleton

({*}=X^0=X^1=…=X^n<=X^(n+1)<=X^(n+2)<=…)

Sorry , any cell except one cell of dimension 0 that maps to {*} in this case * is the image of X^n under v:X—>X/X^n

[pearl holly]

oh okay I see. But I still don't see why showing that tilde H_k(X) = 0 is sufficient for k <= n if the n-skeleton of X is a point

[tawdry widget]

I don’t get it. X/X^n is a CW complex whose n-skeleton is a point so if you proved that tilde H_k(Y)=0 where Y is any CW complex whose n-skeleton is a point then you have your original statement: H_k(X^n) is isomorphic to H_k(X) using long exact sequence

[pearl holly]

oohh lmao yeah I see now. Thank you so much for the help!

oohhh I thought that Hatcher used the same X in both lines smh

[obsidian socket]

Does anyone know how to prove the following plucker equation? $\kappa ^* =3d(d-2)-6\delta -8\kappa$

[gentle osprey]

Felix_

[obsidian socket]

I can prove it for the non-singular case when $\kappa^*=3d(d-2)$, but am a bit stuck on the singular case

[gentle osprey]

Felix_

[bright acorn]

I want to show that given a symplectic vector space $(V, \Omega)$ an a collection of lagrangian subspaces $W_{1}, \cdots, W_{k}$ of $V$, then there's a lagrangian subspace $L$ such that $L \cap W_{i} = {0}$, $\forall i \in {1, \cdots, k}$.
\
\
I have already proven, using the fact that every lagrangian subspace admits a lagrangian symplectic complement, that there always exists an isotropic subspace, let's call it $L' \subset V$ such that this property is satisfied.
\
\
My idea then, is to consider the collection,
\
\
$\mathcal{C} = { L' \subset V , \vert , \text{L' is isotropic and} , L' \cap W_{i} = 0, \forall i = 1, \cdots, k , }$
\
\
As we have seen, this set is non-empty.
\
\
And I want to use Zorn's lemma somehow to prove that this set indeed has a maximal element, and that this maximal element is in fact a lagrangian subspace that satisfies the property I am looking for.

[gentle osprey]

MisterSystem

[bright acorn]

The problem is that I am bit stuck and I am not able to use Zorn's Lemma to prove this

The professor gave a hint to use reduction spaces and all

But yeah, I am still not able to figure out if this is the way to go or not

This idea came to me basically because lagrangian subspaces are maximal isotropic, and maybe by finding a maximal element in C I may be able to somehow deduce that it is lagrangian too.

[bold canopy]

does a dimensional argument work?

think about the space of all lagrangian subspaces of V; the condition of nontrivial intersection cuts out some positive codimension spaces, so you'll always be able to find another lagrangian subspace that has trivial intersection with any finite collection?

[bright acorn]

Oh yeah, you are thinking about a proof using the lagrangian grassmanian

[bold canopy]

ya

[bright acorn]

I know one that uses some topological properties of the lagrangian grassmanian

But I want to use something more elementary

https://w3.impa.br/~henrique/teach/symp21_1.pdf

look at the hint the professor gave to this bonus problem

It must be easier than this lmao

[bold canopy]

unless im crazy; the hint (using problem 2) is exactly proving that the nontrivial intersection property cuts out positive codimension in the lagrangian grassmanian

seems kinda natural; to me, using zorn's lemma would be more complicated lol

[bright acorn]

yeah

I will try to use this then

And write down the details

[abstract pagoda]

Why is every function in hom(X,S) continuous for X any topology and S trivial topology?

with S being a two point set

All this means is given any function f, I need to show f^-1(S) is open in X

and f^-1 empty is open in X

[true robin]

f^-1(S)=X

and for completeness sake, f^-1(empty set) is empty

[abstract pagoda]

but what if you have a function that only maps to one point in S

oh lol

[true robin]

Do you understand now?

[abstract pagoda]

yes i didnt think enough

thanks

[true robin]

Np

[marsh forge]

@abstract pagoda if you havent yet, as an exercise prove that if X has the discrete topology and Y is any space, then every function X->Y is cont.

[abstract pagoda]

any preimage of any set in Y, open or not, is going to be open in X

because every subset of X open in discrete topology

[marsh forge]

yep!

[abstract pagoda]

i was asking this because i read article about sierpenski topology

tbh idk why it matters, but they said something about category theory that i didnt get to

[marsh forge]

it is a category theory thing

you have basically shown that the forgetful functor of Top has both a left and right adjoint

[abstract pagoda]

also mr max ty for helping me understand homology earlier

i was reading a bunch of intuitional srticles about it but havent done many exercises, so i think its time for me to hit the books

[surreal flax]

is a homotopy literally just a path between paths, because in cubical type theory, thats essentially how they lay it out, to allow the construction of higher cubes?

especially with the univalence axiom

[empty grove]

iirc that's only true if the domain of the functions between which the homotopy lies (domain of domain of homotopy ) is compactly generated

[surreal flax]

like this definition of a homotopy in homotopy type theory is just a path between paths

eg, these proofs are equivalent

[empty grove]

Do you know anything about the spaces

[surreal flax]

i'm coming at this from a type theory direction, so my way of thinking about spaces is just a type or like a domain

like simplfied

i dont know much maths :/

[empty grove]

Idk type theory so can't help

[surreal flax]

:/ welp iz all good

this definition on the hott wikipedia page should explain to you the deal with type theory

it explained to me the base idea in homotopy theory

[marsh forge]

No

That is an example of a homotopy

but the general notion is far larger

[marsh forge]

I am not sure what you meant by this tbh

[empty grove]

X x Y → Z is naturally isomorphic to X → Z^Y when all spaces are compactly generated

Or when Y is locally compact I think?

[marsh forge]

Sure but that isn't a path between paths

A homotopy very much can be thought of as a path between maps

or rather a path within a mapping space

[empty grove]

But then aren't you setting X = Y = I?

oh

[marsh forge]

I don't think they ever mentioned doing that

[empty grove]

Lol

[marsh forge]

Not all homotopies are fundamental groups 🙂

[empty grove]

Wait so in path between paths Y is locally compact already so it works

[empty grove]

Wait did I assume that

[marsh forge]

Well, I thought maybe you forgot that we often care about the case where only Y is I

not X

[empty grove]

Ah yeah

But when only Y is I

Then it's not really a path between functions right?

You get X → Z^I

[marsh forge]

You can think of a homotopy between f and g as a path from f to g in the mapping space between X and Y

[empty grove]

But I → Z^X need not be nice?

[marsh forge]

Up to potentially replacing all spaces with nicer ones

[empty grove]

as in compactly generated?

Or is this something else

[marsh forge]

I am not sure what hypotheses you need in general, but replacing everything with a weakly equivalent CW complex should suffice

at the very least intuitively one should think of homotopies and paths in a mapping space as the same thing

[empty grove]

Yeah I see

[marsh forge]

(And in fact this generalizes to the notion of right and left homotopy in a model category)

@surreal flax btw if that was all unhelpful feel free to ping me

(In type theory of groupoids, calling a homotopy a path between paths makes sense in light of the fundamental groupoid)

[surreal flax]

ohh

[empty grove]

c.g. hausdorff spaces also Cartesian closed

Idk what k-ify means

I see, that's neat

[plain raven]

heyy

yo

mm nvm i was going to ask a question but i think it's ill formed. will come back later once i've thought this through

[swift fjord]

Given some point $z \in \mathbb R^n$, what would you call the hyperplane $z + z^{\perp}$

[gentle osprey]

ShiN

[swift fjord]

I wanna call it 'tangent' but then the question is 'tangent to what'

I guess yea

I see people talking about a tangent hyperplane at a boundary point of a convex set

and I think this is what they mean

but i'm not sure

and I can't find a proper definition

no idea

would this have to be the case if M is piecewise linear?

Can you give a counterexample?

oh well I guess it wouldn't have to be at the vertices

but what about not at the vertices

right

[versed pivot]

Maybe they mean supporting hyperplane? As in supporting hyperplane theorem

[swift fjord]

they are talking about situations in which the supporting hyperplane coincides with the tangent hyperplane

I don't think there's a general formula for the tangent hyperplane in this case then

[obsidian socket]

anyone know if the 28 bitangents of a quartic includes tangents to hyperflexes?

[raw sedge]

okay, so n dimensional real projective space is effectively defined by identifying antipodal points in the n-sphere. Am I correct with the intuition that thisw is equivalent to identifying the antipodal points in the (n-1 dimensional) boundary of the closed n-ball?

so, for example, I'm imagining it in the case of n = 2 by identifying the closed 2-ball with the lower half of the 2-sphere, and ignoring half the points here in some sense accounts for identifying them with their antipodal points (which are in the closed 2-ball), and then the remaining identifications are just the identifications of antipodal boundary points

how? the circle is the 1-sphere, and the closed 1-ball is a closed interval

wouldn't both quotients here form the circle again?

I'm confused

so the closed 1-ball is a 1-manifold with boundary

the boundary does consist of 2 discrete points

and identifying them, we get a circle

isn't this exactly what we want?

[a,b]/a~b is a figure 8?

ohhh

okay

that's not intuitive to me

S^1 doesn't have a boundary

we're not identifying two antipodal points

we're identifying all antipodal points with each other

[raw sedge]

I see how that would make an 8 shape

alright, thanks

I needed a sanity check for that

[gritty widget]

Or you're both insane

[raw sedge]

what if every mathematician is insane and no one else can tell because we're incomprehensible

[rough locust]

Hey so I'm trying to work through the rigid definition of a manifold, and I had a quick notation question: does C^0 mean the set of continuous functions?

Honestly just super confused trying to understand this beast of a definition lmao

thank you

any suggestions for getting intuition into the concept of a coordinate neighborhood?

The definition here has no motivation and I'm kinda confused

ah ok! Though why does a one-to-one function that maps the given subset to an open set function this way?

sorry I think I phrased that poorly.

I mean I don't quite see how the definition captures that idea tbh

A set $U \subseteq M$ together with a one-to-one function $f: U -> \mathcal{R}^n$ such that $f(U)$ is open.

[gentle osprey]

Benm

[rough locust]

Where M is the set (manifold) in question

[bronze lake]

The motivation for it being a neighborhood and a sort of 'local' condition on the manifold is that you can rarely find a function that works properly on the entire manifold

for a sphere S^2, for example, you could define a function on the upper half and lower half to R^2

[rough locust]

None explicitly stated lol

Just a handout my prof gave me

Later on he defines “c0 consistent” coordinate neighborhoods

Not explicitly

Ah ok that helps! Thanks

Ya lost me there ngl lol

Oh nvmind

Since they’re just in R^n? Lol

[bronze lake]

exactly

[rough locust]

Oh wait a second, I think I finally get the idea behind whole definition! So basically we show that our manifold is made up of subsets that are isomorphic to open parts of R^n?

And we want to make sure each point on the manifold is mapped to one point on R^n

Ah oops lol

Ah ok!

Talking through this makes it so much clearer! Thanks so much!

Yeah ngl before this my idea of a manifold was just “some surface that looks relatively flat” lmao

[brisk horizon]

I’m not scare of sheaves and schemes anymore

[gritty widget]

I want to try to become more aware of sheaves and schemes (without getting engulfed and becoming a goomer). I heard the red book is a nice intro, and I have been thinking of trying to read it at some point

[brisk horizon]

The rising sea of algebraic geometry is a lifesaver

It throughly motivate every new object.

[swift fjord]

Suppose you have a, i've heard it call 'Euclidean complex', it's a complex of polytopes that behaves like a simplical complex (i.e. the faces of each polytope in it are also in it, each pair intersects in a lower dimensional face). Is there a construction similar to simplical homology you can do for this kind of object? And if not why not

Or what would you have to require

Trying to think how you'd construct the boundary operator in this case

[solemn warren]

I've made a short video trying to give some intuition about smooth manifolds. I will appreciate feedback if someone is interested in watching it. https://www.youtube.com/watch?v=_DxwjSBA0dQ

[plain raven]

sheaves are ... good

i am going to slowly go through this server and one by one pronounce my final judgement on every subject of math

|| spoiler alert: everything is good. math is beautiful||

[plain raven]

this is a great question shin. idk the answer but i believe you have to have some kind of theory of polytopes to deal with poincare duality, poincare duality is a bit hard to express for simplicial complexes, it's easier to express for more general polytopes. take this with a grain of salt, i may be misinformed

so something like this must be somewhat developed

[swift fjord]

Interesting

I'll look into it but I may need some more experience with AT stuff

Thanks!

[plain raven]

hmm i'm not sure how helpful this is but like

there must be a category of polytopes

whose only maps are like, injective linear face inclusions

i only bring this up because this is kind of the gluing data for a polyhedron

[swift fjord]

Wdym by 'there must be a category'

[plain raven]

i just mean there is a category whose objects are polytopes and whose maps are linear face inclusions. that's all. this category has an inclusion functor into Top and there is perhaps some kind of nice systematic categorical way of gluing these things together along the face maps in a way that would give a general notion of polytope

[swift fjord]

Ah ok gotcha

[shy moss]

here, why $\overline{\pi}:P/G \rightarrow B$ is homeomorphism?

[gentle osprey]

Or x1

[empty grove]

You can check injectivity, surjectivity, continuity and openness each on their own. Which one are you getting stuck on?

[shy moss]

continuity

[empty grove]

P/G has the quotient topology. The defining property of the quotient topology is that whenever we have a continuous map f from P to B which maps each orbit to a single element (or more generally if x ~ y → f(x) = f(y)) then f factors uniquely through the quotient map

And the maps in the factoring are of course continuous

So you can try proving this from just the definition of the quotient topology

If you need a hint to get started with this, just try writing the definition of the quotient topology. The condition about respecting orbits translates directly to inverse images of open sets of B being the inverse images of certain open sets of P/G under their respective maps. You should be able to just apply the definition from here.

[shy moss]

x ~ y mean Gx=Gy?

[empty grove]

Yes in this particular case

In general you can quotient by any equivalence relation so I was just mentioning that

And this property holds for P/~ in that case

[shy moss]

thanks

[shy moss]

here, how can $\sigma$ induce the identity map on the orbit space if the orbit spaces $P/G$ and $P/Q$ not need to be equal?

[gentle osprey]

Or x1

[marsh forge]

a morphism of principal bundles is always an isomorphism

[flint cove]

Is there a common name for the category of CW structures and cellular maps?

The category of „filtered spaces“ would probably be denoted Top(ℕ) following the Top(n)-convention, right?

[marsh forge]

Oh you mean like, objects are filtrations of CW complexes?

Cellular filtrations

[flint cove]

Yes

[marsh forge]

Are you only allowing the obvious filtration by n skeletons

[flint cove]

I want to have language and notation in order to clearly distinguish between „(relative) CW complex“ and „Pair equipped with a CW structure“

[marsh forge]

I am not sure I understand your meaning

I guess I myself am actually not clear on the distinction here

[flint cove]

I'm a bit confused here as well (sorta new to the topic). I mean actually there's three kinds of „objects“ here:

• (X, (A, X₀,X₁, …), (φ_α)_α) where the φ are attachhment maps, the Xᵢ arise inductively from the attachment pushouts, and X is the colimit of the Xᵢ.
  Since the Xᵢ are redundant data we may just write (A, X, (φ_α)_α).
• General Filtered spaces (A, X₀, X₁, …)
• Pairs of spaces (X, A)

The first thing we could call „Pair with CW structure“
It gives rise to a filtered space in the obvious way, and arbitrary filtered spaces give rise to pairs in the obvious way

I just want notation to be able to say „these two steps are forgetful functors“

[marsh forge]

What is A here

[flint cove]

A filtration which is of the isotype of the skeletal filtration of a CW structure could be called „CW filtration“
A pair which is of the isotype of the pair associated to a CW structure is called „CW complex“
At least that's my understanding, but I haven't seen it written down in a rigorous way like that (looked in tom Dieck and Hatcher)

A is an arbitrary topological space you fix to begin with, the thing to which you glue the 0-cells

[marsh forge]

Ah okay, right, A is what makes it a relative complex I guess

[flint cove]

Ye, sorry for the potential confusion

[marsh forge]

So when people say CW complex they almost always assume A=\emptyset

[flint cove]

Yep

[marsh forge]

Okay yes so I think the right way to think about this is like

or notate it

any time X is a CW complex, it has an implicit n-skeleton, and people rarely refer to attaching maps unless you are doing a like, computation that relies on them

So a CW complex is automatically filtered, and this filtration is cellular

The filtration on a space need not be by n-skeletons nor even cellular

So like, the category of CW complexes and the category of CW complexes along with their skeleta is the same category

(assuming all maps are cellular)

A pair of CW complexes is just a kind of different thing, although it does filter the original CW complex

Maybe im all over the place let me summarize my thoughts

1. A CW complex is the same thing as a CW complex filtered by its n-skeleta since this filtration is unique, and when maps are ceullar they play nicely with this specific filtration

2. When X is a space we normally want our filtrations to be cellular, and a filtered CW complex might have a filtration that differs from its skeletal filtration

3. A pair of CW complexes can be thought of as a two-stage cellular filtration

I think this language should capture all of the stuff you were referencing idk if im helping @flint cove

[marsh forge]

(A space X that is homeomorphic to a CW complex might have several CW structures but usually when we call X a CW complex we fix some CW structure on it)

[flint cove]

Well, I'm not quite sure whether people actually mean a CW complex to be „X such that a cellular filtration exists“ or „X equipped with a cellular filtration“
I can never really tell

[marsh forge]

Well again, there are cellular filtrations that are not the skeleton

[flint cove]

I mean with everything homology-related it doesn't matter in the end, but still

[marsh forge]

when someone says let X be a CW complex they mean let X be a space X equipped with a CW structure

i.e. X along with instructions for how to build X cellularly

normally we don't actually care about how we build X so often times the actualy attaching maps or whatever are omitted

in the same way that we refer to a group by the name of its underlying set

[marsh forge]

the big exception is in doing certain computations, for example the homology of RPn

[flint cove]

Okay, I have to read more attentively, tom Dieck in fact does what you say 🙂

[marsh forge]

its amusing to me that tom Dieck has become the go-to book around here

[flint cove]

Although he does not quite care about the precise attachment maps, just that Xₙ is obtained from Xₙ₋₁ by attaching n-cells

[marsh forge]

ew

thats bad notation

(X,A) should be reserved for CW pairs

[flint cove]

in what way? do I need to look out?

[marsh forge]

tbh ive never ever seen A\neq \emptyset in the wild

[flint cove]

hmmm

[marsh forge]

but like if you see people talk about pairs of spaces

(X,A) is like the standardish notation

theres a lot of uses of cellular approximation of maps and spaces about when people talk about this stuff too

[flint cove]

I don't quite see how this would be confusing tho, if (X,A) is a relative CW complex, (X, A) is also an object of Top(2) by forgetting the CW structure

[flint cove]

You're definitely helpful, but I guess I have to wait a while until the thoughts have settled a bit 🙂

so thank you

[brisk horizon]

What’s a cotopology

[marsh forge]

You normally want your filtration to be cellular

and A need not be Cellular

in the relative version

i.e. you want X to be a real CW complex and A a sub-CW-complex

[flint cove]

Well what would go wrong? I'm not that familliar with the technical details yet
I essentially need all this to define (co-)homology relative to A

[marsh forge]

Well just let A be not a CW complex

[limpid mural]

Hello

do you know where I can find a proof of Stone-Weierstrass theorem?

Any book that I can read?

[viral vector]

Rudin's Principle's of Mathematical Analysis (chapter 7) or Folland's Real Analysis (chapter 4) both have proofs

[limpid mural]

danke!

[gilded coyote]

Hello!, What you think about this proposition?: "given two topological spaces $(X, \mathcal T_X)$, $(Y, \mathcal T_Y)$ and a topology in $X \times Y$. The preimage, via projection, of an open set, is always open"

don't think it's true, but I couldn't find a counterexample, so maybe it is(?

[gentle osprey]

Polux12

[raw sedge]

I'll need clarification on what you mean by "and a topology in X cross Y". You mean X cross Y considered in absolutely any topology? Or do you mean considered in the product topology?

This is certainly true in the product topology.

[marsh forge]

(And it is certainly false if I get to pick any topology)

[gilded coyote]

At first, it refers to any topology, I have verified that it is true for the product topology, but with other topologies it is not so clear to me that it is false

[raw sedge]

Let X = Y be a discrete two point space, and consider X cross Y in the trivial topology.

[lean marten]

What topology is being put on $X\times Y$?

[gentle osprey]

Oatman

[raw sedge]

read the messages following the question

[lean marten]

Lol

In that case just take the indiscrete topology on $X\times Y$

[gentle osprey]

Oatman

[lean marten]

And assume the spaces are bigger than a single point

[raw sedge]

@gilded coyote By the way, you may like to verify that the product topology has exactly and only the open sets necessary for this condition to hold.

This corresponds to the fact that a map into X cross Y in the product topology is continuous exactly when the maps resulting from projecting into X and Y are continuous, which corresponds to the intuition that movement in X cross Y is continuous precisely if your corresponding points in X and Y move continuously.

[marsh forge]

As an aside, the underlying set of a topological space tells us basically nothing about the topology, so one should never expect that a topological property should hold for any topology you could put on a set

[raw sedge]

this is why the product topology is the "right" topology on X cross Y

[gilded coyote]

Thank you all, now many things make sense!

[ripe lodge]

so uh I hope this goes here, but I'm curious how off the mark this meme is (like how much truth there is to it):

specifically the claim in the first panel

[sweet wing]

i mean it's true isnt it

[reef shore]

we have an algorithm to prove or disprove any statement about euclidean geometry

but is it realistically efficient

[sweet wing]

does it even matter

why would i want to stare at a bunch of lines

idk what alg is it but would jus gröbner basis the IMO problems

[reef shore]

an open problem would be a statement whose provability we don't know about

[sweet wing]

but open problems could be uninteresting

[reef shore]

yes so we have uninteresting open problems in that case

[sweet wing]

so what is the most interesting uninteresting problem

||ari cranking moment||

[swift fjord]

Collatz Conjecture

[bronze lake]

algebraic geometry is when you map algebraic statements to euclidean statements to solve them

that's basically how they solved fermat's last theorem with elliptic curves 😎

[pearl holly]

Okay so let me get this straight: the cellular homology of some CW complex is always the same no matter what "cell structure" you give the space right? Because cellular homology is the same as singular homology and singular homology is always the same right?

I just need this to be verified

[empty grove]

That seems legit 😌

[pearl holly]

yeah okay and so the Euler characteristic is well defined right?

[empty grove]

Give me an hour

[pearl holly]

Like it's the same no matter what cell structure you give it

the euler characteristic is just the alternating sum of the rank of the homology groups

or well, it turns out to be that way

[empty grove]

oh wait found the definition of Euler Characteristic

Yeah it should be independent of cell structure

[pearl holly]

that's actually really cute lmao

[empty grove]

[marsh forge]

yes to all of these questions

[gentle osprey]

ShiN

ShiN

ShiN

ShiN

[swift fjord]

Any thoughts would be appreciated

This is all in R^n of course

[swift fjord]

This is supposed to be $p(L)\cong \sfrac{L}{H\cap L}$

[gentle osprey]

ShiN

[swift fjord]

Thinking maybe smth with 3rd iso? Since that works for topological groups

but idk how that would help

[empty grove]

Are lattices discrete subgroups of R^n?

In that case you might be able to do something with the tube lemma

[swift fjord]

lattices are discrete cocompact subgroups of R^n

[empty grove]

You can reduce to a case where H is compact because L should be finitely generated (?) so it should repeat after some finite distance or something idk how to say this

[swift fjord]

or equivalently isomorphic images of Z^n

[swift fjord]

Wdym H is compact, it's a subspace

L is free on n elements

[empty grove]

Reduce to the case as in to produce a tube around H that doesn't intersect any points of L outside H, you can look at some finite part of H and produce a tube around it using tube lemma and compactness and that tube should work globally

And if you have a tube around H then you have a tube around p(H) such that p(L cap H) is a singleton in it

[swift fjord]

p(H) is always 0 tho

[empty grove]

Which would prove that it is discrete

[swift fjord]

you want p(L) to be discrete

[empty grove]

Yes

But you get an interval around it as well which contains no points of p(L)

[swift fjord]

ahh

IC

interesting

[empty grove]

I'm just not sure about the part where I'm applying the tube lemma because I'm just relying on intuition about lattices

[swift fjord]

I don't see where we use the assumption that H cap L is a lattice here tho

that's supposed to be a necessary condition

[empty grove]

Yeah idk then I can't even think of a situation where it isn't because I cannot visualise the definition of a lattice that you gave in generality

[swift fjord]

A lattice is just

take Z^n

and apply some linear isomorphism

[empty grove]

Yeah that's what I'm imagining

But if H is a vector subspace then it feels like intersection with H should always be a lattice?

[swift fjord]

You can engineer some line/plane such that the intersection with the lattice is not closed under addition

[empty grove]

Oh is it because H need not pass through origin

[swift fjord]

no we're talking about subspaces

Nah, take for example span(v) where v has linearly independent coordinates over Q

this should be a subspace that doesn't produce a lattice when intersected with Z^n

[empty grove]

Isn't the intersection just {0} because all coordinates can't be rational simultaneously for some cv

oh wait

[swift fjord]

shit you might be right

I can't remember the exact example rn

but believe me there's subspaces where that wouldn't work.

If a subspace is not rational then it will not be closed in the quotient R^n/L

[empty grove]

Lol anyway all I can show is the path, traversing it is your own journey

[swift fjord]

that I can tell you

I'm still not convinced, because this is supposed to be the key thing

[empty grove]

Yeah I suspect that either what I said is nonsense or reducing H to the compact case is using it somehow

[swift fjord]

I'll think about this direction

thanks

Why can't you fuckin pin comments for yourself on discord

Maybe uh

i'm not sure that the tube will be sent to an interval by p

[empty grove]

Isn't a tube defined as H x some interval

[swift fjord]

yea

[empty grove]

[swift fjord]

Ok I think I see why it would be sent to an interval

then I still don't get what is going wrong here

actually

I think

I think maybe you can't put a tube around H

at least not always

like either you can prove that you can't ever put a tube around H such that there are no lattice points in it

or you can prove that H must be rational

rational meaning H cap L is a lttice

[empty grove]

Yeah that's the part where I don't know how anything works

[swift fjord]

I think maybe H being rational can imply that the lattice is not asymptotic to H

but I need to think about it a bit

lol

cuz for example were H to intersect the lattice trivially then we could prove that it was asymptotic to it

[empty grove]

Wait if the intersection is a lattice

Then it's a subgroup of L

[swift fjord]

yea

[empty grove]

So you can take its cosets and the quotient should also be a lattice maybe?

In which case its cosets should be discretely placed maybe idk

[swift fjord]

the quotient needn't be discrete I think

[empty grove]

wack

[swift fjord]

Actualyl

I think the quotient of 2 lattices is always finite (But can be arbitrarily large)

but also I think I was incorrect in saying that L/Hcap L \cong K

yea def incorrect

even as groups

K has more elements

[swift fjord]

This is cuz you can characterise the index by the number of lattice points inside a fundamental domain of H cap L

and this will always be finite

when you take a compact fundamental domain

specifically a parallelepiped

[empty grove]

Isn't Z²/Z inside of R² infinite?

[swift fjord]

hol up

ok yea sorry

this only works if both lattices are freely generated by the same number of elements

my b

otherwise you can characterise it by the ratio of the covolumes

but this breaks down if one is lower dimensional

[empty grove]

That makes sense

[swift fjord]

since that would be covolume 0

which ig could be taken as infinite index

so yea I think in this case the index would be countably infinite

but i'm not sure if the quotient group need be discrete

I haven't played around with the quotient groups much tbh

beyond finite index subgroups

Ok I think i'll keep thinking about this tmrw tho

[empty grove]

Can't you say that the intersection of L with H will be finitely generated torsion free hence isomorphic to some Z^m hence a lattice

[marsh forge]

Anyone want to take a crack at a qual problem i can't think of a good solution to?

[swift fjord]

wdym. That's the assumption

[empty grove]

I can't imagine the intersection not being a lattice lol

Like I'm thinking of counterexamples

[swift fjord]

oh

[empty grove]

Ah being a subgroup isomorphic to Z^m is not enough since there are dense subgroups of R isomorphic to Z², but here we should also inherit discreteness from the whole space?

Is that enough for being a lattice

[swift fjord]

yes being discrete and isomorphic to Z^n is enough

but n needs to be the same as dimH

to be a lattice in H