point-set-topology

[empty grove]

ahh I see

yeah I see a counterexample to being a lattice now and there is no tube in that case indeed

[swift fjord]

hmmm

wonder if there is one generally

[empty grove]

No there's not

That's what I just said

Example is take R³ and H is the xy plane but skewed to have some irrational slope along y so intersection of Z³ with H would be Zx0x0 which is lower dimensional so not a lattice and there's no tube and projection is also not discrete

[swift fjord]

nice

so probably the tube thing would work

I think

jsut gotta prove itnow

[empty grove]

ahh if there were points in L arbitrarily close to H, then consider their orthogonal projections onto H. These projections can be translated so they lie in a single parallelopiped in H, so by the same translations those original points could also be assumed to be projecting onto the same parallelopiped. Now this parallelopiped's epsilon neighborhood's closure is compact but contains infinitely many points of L so L can't be discrete

Used that the intersection is the same dimension as H to do the translations otherwise not everything will lie over a parallelopiped

[swift fjord]

not sure I see why you can always do that translation

but i'll think about it

[surreal estuary]

I am trying to learn some basic algebraic geometry and I'm struggling with understanding the Zariski topology and trying to find a proof.
There are the following two constructions:
$$\mathcal{V}(S) = \left{ a \in \mathbb{A}^n | f(a) = 0 \text{ for all } f(x) \in S \right}$$
where $S$ is a set of polynomials in $K[x_1, \dots, x_n]$, and
$$\mathcal{I}(W) = \left{ f \in K[x_1, \dots, x_n] | f(p) = 0 \text{ for all } p \in W \right}$$
where $W$ is a set of points in $\mathbb{A}^n$. Now the Zariski topology has the algebraic sets as closed sets, and I understand that part.
I was also able to prove some basic facts about $\mathcal{V}$ and $\mathcal{I}$, such as how they work with unions, intersections, subsets, etc.
Where I'm getting stuck is the following proposition:
The Zariski closure of a set $W \subset \mathbb{A}^n$ is exactly $\mathcal{V}(\mathcal{I}(W))$. I'm not sure how to even start on it.
I tried thinking of the closure as the intersection of all algebraic sets containing $W$, but got stuck and couldn't proceed.
Does anybody have a hint on how I can start to prove this?

[gentle osprey]

Anomalocaris

[empty grove]

You have to show that V(I(W)) is the smallest closed set containing W so it suffices to show that it is closed and that any closed set containing W contains it. The first one is by definition, for the second one, suppose there's a set S such that W is in V(S). What happens when you enlarge S? Can you compare S and I(W)?

[surreal estuary]

I'll try that now, I'm really slow at math stuff so I might not be able to get back for a long time depending on how hard I get stuck
It seems easier than what I was doing at least

[surreal estuary]

Nope, I have no idea what I'm doing here.

I wish the text would actually describe this stuff instead of defining it and saying the proofs are easy

[empty grove]

When you make S larger, V(S) becomes smaller. Try proving that S is contained in I(W)

This would solve the problem because it would say that all such S are in I(W), so therefore all such V(S) contain V(I(W))

Meaning V(I(W)) is the smallest closed set containing W

[surreal estuary]

And I don't know how to do it, because this is so unlike any topology I've ever worked with

[empty grove]

V(S) tells you how elements of S evaluate on W. I(W) is the set of all elements that evaluate that same way on W. Do you see it?

[surreal estuary]

V(S) tells you how elements of S evaluate on W
what do you mean about how element of S evaluate on W? I've never seen that definition before.

[empty grove]

Elements of S are polynomials

When you evaluate those polynomials on a point in W, you should get 0

Because you should get 0 for any point in V(S) which by assumption contains W

[surreal estuary]

I proved it I think. I don't really understand what it means yet, but I managed to show it

[broken belfry]

What's the euler characteristic of a pseudo-sphere?

[marsh forge]

what is a pseudo sphere

[broken belfry]

It's a shape with constant negative gaussian curvature

And I have been wondering what's its euler characteristic is

[marsh forge]

Weird I cannot find anything about its homotopy type

[hollow harbor]

Isn't it just homotopy equivalent to a cylinder

It's some cone

[marsh forge]

so like

that is my guess re: cylinder

but also when things get too analytic they can go wrong idk

it should have betti number 0 though

if its htpy eq to a cylinder

[sweet wing]

isnt this a hyperbolic space

you can very much form cursed quotients of H^n

[marsh forge]

I didn't see any mention of it being homotopy eq to a cylinder on wiki

there are a lot of cursed spaces with nice universal covers

maybe i missed something important

[sweet wing]

you should get spaces w/ negative gaussian curvature have a contractible universal cover so the only interesting betti number comes from H1(abelianize π1, which fully classifies your spaces up to isotopy)

[marsh forge]

I am not sure what you mean by this ari, a space can have contractible universal cover and many interesting homology groups

like a torus

Yeah but it doesn't determine your betti numbers

[sweet wing]

oh wait i was thinking of homotopy groups 🤦‍♀️

gg

[brisk horizon]

Categories can come with sheaves right

[marsh forge]

?

If a category has a good notion of covers you can define what it means for a presheaf to be a sheaf

[brisk horizon]

Ah okay

Is it to make something like the restriction map make sense?

[marsh forge]

the coverings tell you how to make sense of the sheaf gluing condition

i.e. if F is a presheaf and U is an object then covers of U should determine F(U)

[brisk horizon]

That make sense

[plain raven]

https://ncatlab.org/nlab/show/Grothendieck+topology @brisk horizon

[coral pivot]

I do not understand why the germ of f is phi. I have a rough visual of it(f should take values close to phi) but cannot see beyond that

[honest narwhal]

@coral pivot wait isn't it literally just like

Choose a ball

And uniqueness of analytic continuation?

[coral pivot]

Hmm I don’t see how that makes the germ phi tho

[true robin]

I may be misunderstanding, but doesn’t phi come equipped with a function and an open set, and since a is in that open set, the analytic continuation as they describe to any point in that open set is just the value of the function?

[coral pivot]

Oh lol I’m dumb yes I see it now

Like the continuation is going to be phi in the neighborhood of a phi is picked in

Ty dami and saketh, had a brain fart oof

[viral vector]

Why are Haken manifolds easier to deal with than other ones in 3 dimensional topology?

[brisk horizon]

@plain raven Grothendieck topology this is interesting

[tight agate]

:descent:

[tight agate]

[plain raven]

Cool! Do you have anything to say about descent?

[pseudo crane]

[plain raven]

nice

[tight agate]

[wet arrow]

hello! How do you derive this formula for the surface area of an ellipsoid?

[sweet wing]

#calculus but you integrate and approximate

[wet arrow]

ok thank you

[raw sedge]

"approximately equal to"

cursed

[swift fjord]

So I've been thinking about it and I don't understand what you're doing exactly. Are you translating them by points in L? And if so how do you guarantee they all lie in the same parallelepiped, and why can you assume they're all projecting onto the same one

[empty grove]

Pick a point x whose projection lies in a parallelopiped P. Then translations of P by points of L cap H should cover H. So you can translate proj(x) by some k in L cap H such that it lies in an arbitrary translate Q of P. Then proj(x+k) will lie in Q, and x+k is in L because x and k are

[swift fjord]

Ok yea that's the cocompactness of L cap H, but why does this imply there are infinitely many points in one of the parallelepipeds

[empty grove]

Not in one of them, but infinitely many points arbitrarily close to one. The assumption was that there are points arbitrarily close to H that are not in H (assuming p(L) is dense in p(R^n) is equivalent to this), and I just translated all of them to be over the same parallelopiped

That means that if you take a compact neighborhood of this parallelopiped then it contains infinitely many points which contradicts discreteness of L

[swift fjord]

Ok I think I get it. But i'm not sure I see where you actually use the fact that there are points arbitrarily close to H

[empty grove]

To get infinitely many points in a neighborhood of Q

[swift fjord]

But why would that not work if there was some minimal distance between points of L and H

[empty grove]

The same argument doesn't work. What I did was take x(n) to be a point of L at some distance between d ∈ (0,1/n] from H

And then all of these x(n) (translated appropriately) are in the closure of the radius 1 neighborhood of Q

[swift fjord]

But why would the translates still necessarily be arbitrarily close

Oh wait

k is in L cap H

Right

[empty grove]

They would be at the same distance from H as the original points because you're measuring distance orthogonally and translating along H

[swift fjord]

Yea alrighf

Gotcha

I think I totally get it now

Thanks!

[empty grove]

Np I learned a lot doing this lol

[swift fjord]

didn't actually need the tube lemma in the end lol

do you think the converse is true? i.e. if H cap L is not a lattice then d(H,L)=0?

[empty grove]

just did the proof of tube lemma at the end essentially, but exploiting metric to allow using sequences

[empty grove]

Will have to think about that 😵‍💫

[swift fjord]

Same

[gentle osprey]

Micael

[prime shuttle]

Tbh i dunno where my question belongs, people said that it should be in differential geometry so i'm posting here. Sry if i'm wrong\
Hello, i'm trying to find the continuum limit of the formula of the laplacian on a grid, that means going from here:\
$\nabla^2 v(\overrightarrow{x})=\sum_{\mu=1}^d [-2v(\overrightarrow{x})+v(\overrightarrow{x}+a\hat{e}\mu)+v(\overrightarrow{x}-a\hat{e}\mu)]$\
To this (admitting an small a):\
$\nabla^2 v(\overrightarrow{x})=\sum_{\mu=1}^d \frac{\partial^2}{\partial x_{\mu}^2} v(\overrightarrow{x})$\
I kinda get the answer by doing the expansion in taylor series (centered in x) of $v(\overrightarrow{x}+a\hat{e}\mu)$ and $v(\overrightarrow{x}-a\hat{e}\mu)$. But then i have this: $\nabla^2 v(\overrightarrow{x})=\sum_{\mu=1}^d v''(\overrightarrow{x}) (a\hat{e}_\mu)^2$, how to manipulate this to get the final answer?

[gentle osprey]

Micael

[prime shuttle]

fixed, heh
i tried copying and pasting my last message and because of discord commands like _x_ things got messy

[gritty widget]

oh god what is that im geniunely scared to do that math

wait its a both

ok nvm

[jagged pivot]

If I wanted to find the Ehrhart polynomial of
$$ conv{ (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0),(1,1,0) } $$
I would have
$$ \frac{1}{2}t^3 + at^2 + bt +1. $$ I'm having a little trouble finding $a,b$ and was wondering if someone could point me towards a good direction

[gentle osprey]

beeswax

[jagged pivot]

Bc by Ehrhart's th'm, the leading coefficient of this is just the area, so I am like 90% confident my 1st and last terms are correct

[gritty widget]

Woof woof woof!

Maybe this will bring him back

[jagged pivot]

And if I use the 1st dilation and sub in t=1, which would set the polynomial =6, I get a+b = 9/2.

Or I'm just way way off

[bronze lake]

I can see that n is an upper bound for L (e.g., you can do it with n lines). If you can prove that n is also a lower bound, that it requires at least n lines, then the result immediately follows

[gritty widget]

Yeah this was my first thought

[bronze lake]

rip

wait the original n must be even

I'm assuming that condition isn't there for no reason, it might be to eliminate a strategy like this

oh I guess in that case you'd still be able to do n=4 with L=3

[gritty widget]

Ah, maybe I got it

Nvm, I don't have it. I was doing some bullshit thinking that was nonsense. I was thinking of trying to trim off the edges of a (n+1)x(n+1) square to make 4 nxn squares, then take their intersection to get a (n-1)x(n-1) square and do some kind of contradictjon, but what I was thinking of didn't make sense

Trim off the edges, like eg a [0, n+1]x[0, n+1] square becomes [0,n]x[0,n]

[gusty flame]

Some days ago I posted this question to the multivar channel, but it never got answered , so I'm trying here (please tell me if this doesn't fit here)

"Some weeks ago, I encountered a definition of an orientable manifold in the multivar book that i'm reading, the thing is, I can't connect this definition with my intuition of the properties that an orientable manifold will have, like, the fact that clockwise and anticlockwise movement is consistently defined, so I'm asking for any tips, hints, texts to read, or pretty much anything that can help me
This is the def"

So in short, I don't understand why imposing that the jacobian determinant must be positive will "induce" orientability (talking about a intuitive notion of orientability of course), how does it relate?

[gusty flame]

that makes a lot of sense, I really appreciate it

[swift fjord]

this feels like it has to be true

Feels to me like the best way to prove it would be to prove that H cap L is cocompact and therefore a lattice if the distance between H and L-L cap H is nonzero

but maybe that's not very productive

actually maybe using the torus could work

that might be it actually

actually i'm not sure if that helps a lot since the quotient can be a bit wack

[coral pawn]

What exactly am I supposed to do here?

What does it mean when it says consider polar coordinates as a coordinate chart

[hollow harbor]

The map $\phi : (r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ is a coordinate chart.

[gentle osprey]

ryc

[hollow harbor]

for R^2

[coral pawn]

Oh okay

[hollow harbor]

you have to figure out what the domain is that makes this a coordinate chart

(i.e., diffeomorphism from a subset of R^2 to a subset of your manifold, which is R^2)

[coral pawn]

So I guess I use R x [-pi,pi] - {0} as the domain

and R - {0} as the codomain

[hollow harbor]

coordinate charts are defined on an open domain

[coral pawn]

Oh yeah my bad

We definitely need R in the first slot

Because the radius can be as much as we want

[hollow harbor]

well

can it?

[coral pawn]

ignore 0 for now

[hollow harbor]

can the radius be negative?

[coral pawn]

Oh lmao my bad

We need positive radius

So R+

But we also need all angles between -pi and pi (including those two)

Wait

[hollow harbor]

well

you're choosing a maximal domain such that this is a coordinate chart

you might not be able to get everything in the range

(the point of this exercise is that this is a bad coordinate chart)

[coral pawn]

Not quite sure if I follow

[hollow harbor]

-pi and pi are the same angle

your coordinate chart won't be injective if you include those

[coral pawn]

Oh yeah my bad

So I guess it would be like [0,2pi)?

I don't really understand what the manifold in question is

It says the polar coordinates on R^2

But what does that mean?

Oh wait I think I see it

The manifold is R+ x [0,2pi)

With the subspace topology

and I can map this to R^2 - 0

homeomorphically

R^2 - 0 is open

So it it a manifold

Is this correct?

Or am I completely misunderstanding the question

[plain raven]

i think coordinate charts have to be between open subsets

[coral pawn]

Aren't these sets open?

[coral pawn]

This is open in the subspace topology

And R^2 - 0 is an open subset of R^2

[plain raven]

[0,2pi) isn't open in R

[coral pawn]

That's kind of what I was confused about

What set does it mean by "the polar coordinates on R^2"

[plain raven]

right lmao

ok

let me say it this way.

there is a well-defined function R^2 -> R^2 which sends the pair (r, \theta) to the point in the complex plane re^{i\theta}

right?

[gritty widget]

Yeah, I'm also confused what manifold they are talking about

[coral pawn]

Yes

[gritty widget]

What's the topology on the domain R2?

[coral pawn]

I think that's a part of the question

[plain raven]

the standard topology.

[coral pawn]

Like when it says "what domain do you have to use"

I think its asking us to pick a subset of R^2 which is enough to define every non-zero element as polar coordinates

Import the subspace topology on it

By definition, that subset will be open in the subset topology

And then somehow construct a map onto an open set of R^2 which is a homeo

[gritty widget]

Then, is it locally euclidean at 0?

[plain raven]

yes, both are locally euclidean at every point.
anyway call this function f.
There are some open sets U in the domain such that f restricted to U gives a diffeomorphism U \cong f(U)

[gritty widget]

But there is no open set around 0 such that a (polar) coordinate chart on it is injective right?

[plain raven]

and there are some open sets W in the codomain such that f has a right inverse (a section) on W, a map g : W-> R^2 such that g is a diffeomorphism onto its image and f(g(x,y)) = (x,y) for all x,y in W.

the problem is to find a maximal such W on which f has a section

a hint is that no W can possibly contain the origin

[coral pawn]

I'm not following this at all. What is the manifold you are considering?

[plain raven]

ok let me start over.

[coral pawn]

Like what is your definition of the "polar coordinate" manifold

[plain raven]

yeah.

let's just give the spaces two different names X and Y lol to talk about their functional role. Both X and Y are just R^2.
so X is the "manifold" space we are trying to give coordinate charts for, and Y is the space which is serving as the coordinate plane, i.e. we reduce points in X to elements of Y by means of some local homeomorphism from an open subset X to Y.

A chart for X is an open subset W of X together with a continuous map g : W -> Y which is a homeomorphism onto its image.
Ok. There is a function f : Y -> X which sends the pair (r,\theta) to re^{i\theta}.
A chart g : W -> Y is said to be polar if f \circ g = id_W.
The problem is to give a maximal open subset W of X that admits a polar chart g: W -> Y.

[coral pawn]

So R+ x (0,2\pi)?

[plain raven]

yep! that should work

[coral pawn]

Thanks for the help

It's a dumb question

Phrased really bad

[plain raven]

np i'm glad we figured it out in the end haha

[coral pawn]

Yeah

Thanks

[coral pawn]

If I have a function f: X --> R^n and for each i, the composition pi_i \circ f is continuous, then f is continuous

Is this true/well-known fact from analysis?

[hollow harbor]

it is false!

wait

no it's true i'm thinking of the wrong direction

is it?

[coral pawn]

Yeah I'm basically trying to prove that (r,\theta) maps to (rcos,rsin) and it's inverse are continuous

Without doing any calculations

[coral pawn]

is it?

[hollow harbor]

i thought i remembered a counterexample

[coral pawn]

pls end my suffering

[coral pawn]

What is it?

[hollow harbor]

lord if i know

this seems fine to me

i take epsilon > 0 and x, find delta_i such that when y is within delta_i of x we have pi i circ f(x) within epsilon of pi i circ f(y), and then i just take the min of the delta_is and use the fact that the distance from f(x) to f(y) is the square root of the sum over i of the squares of the distances from pi i circ f(x) to pi i circ f(y). this says f(x) is within sqrt(n)epsilon f(y). that's fine.

idk what this doubt i was having is...

[tight agate]

it is true

it should be iff

that's pretty much the definition of a product

if X and Y are top spaces, then X x Y is a space with maps to X and Y such that a map Z ---> X x Y is precisely the same data as a map from Z ---> X and Z ---> Y

[coral pawn]

What kind of a curve does y^2=-x^2+x+1 equation give?

Is it an ellipse?

[wanton marsh]

a hyperbola

[unreal stratus]

(isn't that an ellipse, in particular a circle?)

(x-1/2)^2 + y^2 = 5/4

[lean marten]

The xs will have a negative coefficient

[tawdry widget]

A circle

[flint cove]

Does the stone-cech compactification have anything to do with stone spaces?

Or, let's be more precise, the stone representation theorem

In stone-cech, the slogan is „points give rise to (distinct) principal ultrafilters, so let's treat every ultrafilter as a point“ if I understood that correctly

[grave maple]

https://en.wikipedia.org/wiki/Stone–Čech_compactification#Construction_using_ultrafilters

[coral pawn]

Any ideas on how to show smoothness?

[plain raven]

this seems like a pretty direct application of the theorem

[plain raven]

definitely, there's a lot of connections between them

the categories Set, Top, BoolAlg, and CompHaus are all woven together by a bunch of interesting adjunctions

for one thing, if you have a set X, and you regard it as a discrete topological space, then the stone cech compactification of X is given by looking at all ultrafilters of the powerset of X, so you're first sending X into P(X) in BoolAlg^op and then looking at the functor BoolAlg^op -> CH given by taking ultrafilters

so

and the category of CH spaces is monadic over Sets

[coral pawn]

The theorem only guarantees continuous differentiability right?

[plain raven]

well can you just post it

what is continuous differentiability

there are different versions of the implicit function i guess, the one i learned makes this trivial

[coral pawn]

What is the one you learned?

[plain raven]

superscripts are dimensions

i'm sure these are equivalent but i don't really remember the proofs off the top of my head. the references to 1.33 and 1.30 are about forming slice submanifolds (if U is a coordinate chart, then the subset of U with x1=0, ....xk=0 is a submanifold of dim n-k)

and like

using linear independence of tangent vectors to prove that functions define a coordinate chart locally

[coral pawn]

We haven't covered tangent vectors yet lmao

[plain raven]

groan. let me think

[haughty wave]

Around every point you can apply implicit function theorem to find a 2D chart

[plain raven]

ok. you definitely need to start by proving that the differential is surjective at every point in the fiber

that's not hard

[spiral silo]

Does anyone know if there's a general way to go from a topological space to a basis for continuous functions on that space?

Examples:
R, C -> Taylor series (or Laurent series)
[0, 1] or S^1 -> Fourier series
S^2 -> Spherical harmonics

That is, is there a general way to go from a description of a space (possibly its homotopy type expressed in some formal language) to a basis for the space of continuous functions over it?

(The C -> Taylor series example only gives holomorphic functions, not all continuous functions, and in the R, C -> Taylor series example there might be a finite radius of convergence, but I'm open to either a more expressive basis, or a restriction to "nice" enough functions for that basis to work. In fact I could have given the Hermite or the Laguerre polynomials as the basis for certain functions on R because those polynomials are orthogonal.)

[coral pawn]

I think I found a better way

We know that h(x,y)xy= 1 where (x,y) are in a neighborhood around (a,b) for some point (a,b,c) in the level set

And f(x,y,h(x,y))=0

Thus, x or y cannot be 0

So we get h(x,y)=1/xy

This explicit formula does not depend on the choice of (a,b,c) in the level set or the neighborhood around it

So we can simply glue this collection of h to a function from R^2 - (the coordinate axis)

and it will be h*(x,y)=1/xy

This is smooth on R^2 - (the coordinate axis)

@plain raven Do you think this works?

[haughty wave]

yeah, you just showed that φ(a, b, c) = (a, b) is a chart on the entire level set

[plain raven]

^

[lean marten]

Surely you can only construct a basis for continuous functions into a vector space

[marsh forge]

?

[eternal nimbus]

Dummy question, how do we define a line in finite fields?

[marsh forge]

One dimensional linear subspace

[eternal nimbus]

so there are no lines that don'( pass through the origin?

I'm basically trying to make sense of this, in projective geometry

[eternal nimbus]

(and the three points at infinty bit trips me out tho)

wait nvm i figure it out

[gritty widget]

does anyone have an idea what the "obvious matrices" here are? I'll type up the extra terms after posting this image but this is from page 248 of Lectures on Morse homology by Banyaga and Hurtubise

For the $Y_{r,s}$ matrices, we define $Y_{r,s}(z)$ to be the matrix in $\mathfrak{u}(n+k)$ with $z$ in the $(r,s)$ entry, and $-\overline{z}$ in the $(s,r)$ entry, and zeros everyhwere else.

[gentle osprey]

lime_soup

[gritty widget]

$U(n+k)\cdot x_0$ is the orbit under the adjoint action not just $U(n+k)$ multiplied by $x_0$.

and $N(U(n+k)\cdot x_0)$ is the normal bundle of $U(n+k)\cdot x_0$ so $N_{x_\sigma}(U(n+k)\cdot x_0)$ should be all the points normal to $U(n+k)\cdot x_0$ at $x_\sigma$?

[gentle osprey]

lime_soup

[lean marten]

@eternal nimbus hey what text are you using for projective geometry

I tried to learn some synthetic projective geometry last year but it didn’t fully click

[gentle osprey]

Hausdorff

[vast estuary]

I checked F(x,0) = k(h(x)) and F(x,1) = k'(h'(x))

but how do I show continuity of F?

[reef shore]

It's a composition of continuous functions

[vast estuary]

$H: x\mapsto H(x,t)$ is continuous. $\pi_2: (x,t) \mapsto t$ is continuous. $F = K \circ L$ where $L: (x,t) \mapsto (H(x,t), t)$. Why is $L$ continuous?

[gentle osprey]

Hausdorff

[vast estuary]

Is it just this?

[gentle osprey]

Hausdorff

[reef shore]

yes

[reef shore]

something like that, don't feel like verifying this lmao

but x,t maps to H(x,t) and x,t maps to t are both continuous

so their product is

The you compose with K

[tawdry widget]

kh ~kh’~k’h’

Or equivalently kh ~k’h~k’h’

[vast estuary]

what do you mean?

Any ideas for (b)? Here's what I'm thinking: Consider two paths $f: I\to Y$ and $g: I\to Y$ in $Y$. Join $f(1)$ and $g(0)$ by a path $h: I\to Y$, $h(0) = f(1)$, and $h(1) = g(0)$ (since $Y$ is path-connected, we can do this). Now we need to find continuous $F: I\times I \to Y$ such that $F(y,0) = f(y)$ and $F(y,1) = g(y)$ for all $y\in I$.

[gentle osprey]

Hausdorff

[vast estuary]

Umm. does this hold in particular for path-connected spaces or in general?

[gentle osprey]

Hausdorff

[vast estuary]

but I wonder how that helps our problem

Oh actually it looks like the solution is much simpler

[gentle osprey]

Hausdorff

[vast estuary]

Sorry for asking trivial questions but I've just started self-learning algebraic topology

I am wondering if we can concatenate homotopies, the way we concatenate paths? The definition seems immediate.

[reef shore]

yep

homotopy from f to g and from g to h concatenate to f to h

[gentle osprey]

Nobody

[vast estuary]

Yep thanks!

[gentle osprey]

Hausdorff

[vast estuary]

The latter looks cleaner and more convenient

and you can see how this generalizes to finite products of length (ns, ns-1, ns-2, ... so on)

[gentle osprey]

Hausdorff

[eternal nimbus]

It's really basic stuff, it's the undergrad Nigel nitchin lecture notes that you can find on his site

[reef shore]

@vast estuary if you have a homeomorphism f: X → Y, and A is a subset of X, then f|_A : A → f(A) is also a homeomorphism

Summarised as restrictions of homeomorphisms are homeomorphisms, with the understanding that you restrict both the domain and the codomain

[gentle osprey]

Hausdorff

[novel acorn]

Is p a homeomorphism tho?

I've never thought of a covering space as a homeomorphism so this seems a little strange

[flint cove]

Thanks, that explains a lot l
In particular I thought every such compactification would work with Ultrafilters; thale fact that it is specific to discrete spaces makes the connection clear

[flint cove]

When you say „over sets“, are you talking about specific properties related to the adjoint pair forgetful |- discrete top?
I only know that a monad is an endofunctor, I'm not sure what this has to to with that full subcategory CHaus

[vast estuary]

It is, when restricted to V_alpha

[novel acorn]

Oh I missed that line lol

sorry

[swift fjord]

time for a new question i'm struggling with. Half thinking out loud here but so far it's stumping me

So as always we have a lattice $L$ in $\mathbb R^n$, and we consider its klein polyhedron, that is, $K=\text{Conv}(C \cap L - {0})$ where $C$ is the positive orthant. We then define the sail as $S = \partial K$. Now, I want to know if the lattice points on the sail always span the lattice

[gentle osprey]

ShiN

[swift fjord]

intuitively, I wanna say of course

in 2d it's easily provable

I tried tackling the problem with induction but reducing the dimensionality always requires some quotient or projection, which the boundary does not play nice with

it's led me to ask some questions about how the sail of a sublattice relates to the sail of a lattice

I also know that the extremal points of the klein polyhedron (Which are all in L) span (in the sense of a convex hull) the klein polyhedron, but I don't think that helps me much

Once you fix a lattice there's not a lot of things i've found you can say about the lattice points on the sail

If there is a counterexample it has to be in at least R^3, where it's much harder to calculate the sail in general

Something else that may help is a sufficient and necessary condition on lattice vectors to be a basis for the lattice

but the arguments i've been able to come up with are a bit circular

the reason I feel like induction might work is that if the lattice satisfies some condition (namely that the intersection with all hyperplanes where one coordinate is 0 is trivial), then every lattice point on the sail should be able to be completed to a basis for L

since i'm pretty sure it's primitive (That is, there is no $l$ such that $\frac 1 l v \in L$). Although I haven't proven that yet

[gentle osprey]

ShiN

[obtuse meteor]

I anti-enjoy thinking about a proof of the excision axiom

[swift fjord]

same

[novel acorn]

That 4 page Hatcher proof scarred me

[fading vale]

Can an affine curve over an arbitrary base field not be finite for whatever reason?

[marsh forge]

That is a classic proof for any good lecturer to give a sketchy proof for

[fading vale]

the correct way to prove excision is to simply dare people to prove you wrong

[marsh forge]

Proof by “find a counterexample”

[fading vale]

[novel acorn]

The pit that the Riemann hypothesis is currently lying in

[fading vale]

I wonder if DVRs can never be finitely generated over any field

thats kind of fucked up

[ivory dragon]

dvrs are a fake algebraic structure

prove me wrong

theyre so weird

[fading vale]

So true!

I think the argument being used here is that finite morphisms of curves are surjective because non constant morphisms Y -> X correspond to inclusions

but like

for Y -> X to be non constant you need Y to be infinite

and i dunno why an arbitrary affine curve has to be infinite

[plain raven]

yeah this is really interesting... this is a bit much to get into right now but i'd be happy to get into it some other thing. suffice it to say that one can reconstruct the category of compact hausdorff spaces from the endofunctor on Sets that results from composing the adjunction T : Sets -> CH -> Sets, together with certain distinguished natural transformations id_Sets -> T and T\circ T -> T. this is a result of E. G. Manes

I think it was probably Lawvere who first pointed out that using these "Monads" on the category of sets, one could give like, a somewhat categorical treatment of universal algebra, basically the endofunctor T might send a set X to the set TX of all terms in the language of group theory with coefficients in X, modded out by the equivalence relations that occur in the free group. then X is a group iff there is an interpretation map TX -> X sending each word in the formal language to the thing it's supposed to denote in X. this interpretation map has to satisfy certain laws, of course

but this gives a general schema for talking about categories "algebraic over sets", once you understand the basic idea it's not hard to prove that groups, rings, R-modules for a given ring, etc are all "algebraic over sets" in the monadic sense

and a category of algebras over a monad has certain nice properties, for example the forgetful functor Alg -> Sets creates limits, so that whenever you have a diagram D in Alg, D has a limit and the limit is computed exactly the same as it is wrt the underlying sets

so it was really shocking when Manes pointed out that CompHaus is monadic over Sets, because in what sense is topology "algebraic"? essentially the map sending each ultrafilter on a CH space X to its point of convergence satisfies similar associativity and unit laws to algebraic term evaluation

very cool

but yeah you can think of a monadic functor as a special kind of forgetful functor with a left adjoint that has better free-forgetful properties than an arbitrary adjunction. i think probably like, the usual hom-tensor adjunction on R-mod is not monadic, for a counterexample, i mean i'm just guessing here tbh but that's my intuition, a monadic adjunction should look more "Free-forgetful" than that

manes' book on universal algebra via monads is pretty readable, modulo some ancient notation and composition of morphisms from left to right lol

[obtuse meteor]

the beginning of my second course in algtop seems to start with it

tomorrow at 9am

me listening to someone do subdivision

painful

[plain raven]

is subdivision a functor SSet -> SSet
if so is there a nice way to describe it in some abstract-nonsense way

[marsh forge]

I think barycentric subdivision is such a functor but it’s not like particularly nice afaik

Like for example for certain theorems you need to apply it twice

[plain raven]

bs

[raw sedge]

is the map Set -> CHaus here the Stone-Cech compactification of the discrete space on the set?

[plain raven]

yep

and the other direction is the forgetful functor

[gentle osprey]

Keith Elliott Peterson

[tight agate]

pi_{q+1}(M) = maps(S^{q+1}, M) = maps(S^q, maps(S^1, M))

by the smash product hom adjunction

maps(X smash S^1, Y) = maps (X, maps(S^1, Y))

yes

you can write down the maps pretty easily

yup

yeah it's true for any triple of spaces

idk why I wrote it down for just S^1

yeah yeah

nice spaces

point-set problems

[coral pawn]

Is there a good description for the maximal atlas defined by the identity on the real numbers?

It should contain all linear functions on R

But is that it?

[bitter yoke]

I'm not really sure what you're asking. Isn't this maximal atlas just the usual structure of R?

[coral pawn]

Yeah

Like take the chart (R, id)

It is contained in some maximal smooth atlas (because it is tautologically smooth)

What do elements of this look like

I think it should be linear functions on open sets of R

(locally linear?)

But that would imply globally linear

So its just the linear functions on R

[bitter yoke]

What do elements of the maximal atlas look like?

Any diffeomorphism between open sets of R works right?

[coral pawn]

I don't think so

x maps to x^{1/3}

That's a counterexample right?

[lean marten]

Yes

I believe its an exercise in Bott and Tu

[coral pawn]

Do you know what the result is?

Like what is a good description of the standard maximal atlas on R

[lean marten]

Ummmmm

[bitter yoke]

A counterexample to what? maybe I'm misunderstanding something here

[lean marten]

Oh thats just with identities as the transition functions

[bitter yoke]

I feel like something like x maps to x^3 + x should also be in the maximal atlas

[lean marten]

Keep in in mind $x\mapsto x^{1/3}$ is not smooth

[gentle osprey]

Oatman

[coral pawn]

Zoph you know how every smooth atlas is contained in a maximal one right?

[bitter yoke]

Right I get that

[coral pawn]

So take (R, id)

[bitter yoke]

You're asking what elements of the maximal atlas look like

[coral pawn]

That's a valid atlas

Yeah

[bitter yoke]

I get that, I'm just not sure what meant by x -> x^1/3 is a counterexample

[coral pawn]

Oh wait my bad

[bitter yoke]

Yeah, like Oatman said, this isn't smooth

[coral pawn]

Okay

[bitter yoke]

I think the example I gave of x maps to x^3 + x is okay though

its smooth and has a smooth inverse because its derivative doesn't vanish

[coral pawn]

I think your answer that its the set of all diffeomorphisms is kind of a hack though

Because that's like the definition

[lean marten]

I think the point is it kind of doesn't matter

We generally only care about the explicitly given generators of the maximal atlas

[bitter yoke]

That's true, I mean you could phrase it as the set of all infinitely differentiable bijective functions with infinitely differentiable inverse from one open set to another

[lean marten]

That one exists is just nesseccary to move on to the meat of the subject

[coral pawn]

Lmao maybe I should write that on my hw

[lean marten]

Oh is this a homework question lol

[coral pawn]

"You're not asking the right questions"

And get administratively dropped from the class

[bitter yoke]

Yeah I agree this is a weird question to ask

But I don't really know of any better answer than what I said

[coral pawn]

Fair enough

I'm just gonna say your answer

And maybe that it includes linear functions cause those are particularly nice

[alpine bolt]

(x,y) is a coordinate but what are the coordinates of $(m_2-m_1)/1+m_1 m_2)$?

[gentle osprey]

bunkermush

[coral pawn]

Show that it is... hyperplanes

What does this mean?

[coral pivot]

the stereographic projection is defined using two charts right

[coral pawn]

Yeah

[coral pivot]

this is asking if the transition map is smooth

[coral pawn]

I did that part

I don't know what it is referring to when it says the projection onto the corrdinate hyperplanes

I can't think of any chart from the sphere to R^n which is not a translated stereographic projection

So I don't know what charts it is referring to

[coral pivot]

hyperplane is the subspace that is like, one of the coords is 0. ig they want you to basically forget one of the coordinates of your sphere and use those as your charts?

[coral pawn]

But those aren't charts

Because the range won't be open

For example, in if we take the S^2 in R^3 and project it onto the xy plane, we would get the closure of the unit circle in R^2

Unless...

[coral pivot]

hmm

[coral pawn]

We remove the part that intersects with the xy plane

[coral pivot]

maybe they mean like

pick a nbd around a point and project it down to hyperplane

theres some nbd that is homeo to this projection

and these are your charts?

[coral pawn]

since it uses "the" instead of "a"

I don't think we can choose a nbhd

[coral pivot]

I mean it will be the same smooth structure regardless right

[gritty widget]

on $S^1$ you have an atlas given by the four charts
\begin{itemize}
\item ${(x,y)\in S^1:y>0} \to (-1,1)$, $(x,y)\mapsto x$,
\item ${(x,y)\in S^1:y<0} \to (-1,1)$, $(x,y)\mapsto x$,
\item ${(x,y)\in S^1:x>0} \to (-1,1)$, $(x,y)\mapsto y$,
\item ${(x,y)\in S^1:x<0} \to (-1,1)$, $(x,y)\mapsto y$.
\end{itemize}
the maximal atlas determined by this one is the smooth structure on $S^1$ they're referring to

[gentle osprey]

TTerra

[gritty widget]

generalize to arbitrary n appropriately

[coral pawn]

Yeah I think that's what I realized in S^2 and R^3

Thanks

Is there a way to do this problem without working with the cases where the north pole is or isn't in the covers separately?

[swift fjord]

is there a natural way to extend the metric space of compact nonempty sets given by the hausdorff distance into a topological space of all closed subsets?

actually in general

is there a way to take a 'metric space' where the distance can be infinite and turn it into an actual metric space

[empty grove]

Discrete metric

Oh wait extend I'm guessing that will enforce pseudometric at the very least

Does "set of all (U, f_U) where f_U is a diffeomorphism U → V ⊂ ℝ" not work?

[swift fjord]

the metric does distinguish points, the problem is some sets can be infinitely far away

I don't care if it's entirely a metric space but I want it to generate a topology

[empty grove]

oof I see, closed sets

[swift fjord]

yea

just noncompact/unbounded generally

[empty grove]

The usual basis doesn't work?

Balls of radius r

All finite radii

[swift fjord]

hmmm

that could work

I thought of that too

I just don't know if it'll fully model the relationship between the sets

[empty grove]

What do you mean by that

Do you want the nonempty compact subsets to form a subspace of this

They should be

[swift fjord]

I think they will

[empty grove]

Yeah

[swift fjord]

idk honestly, i'm just taking shots in the dark here

would this be metrisable?

it'll at least be sc hausdorff I think right

[empty grove]

Yeah

[swift fjord]

I think it should be metrisable according to urysohn

like it should be T3 cuz it sort of has an underlying metric space structure

[empty grove]

Seems legit

[swift fjord]

where's the cat in that emote

[empty grove]

[swift fjord]

lmao

ok turns out metrics with infinity is like, very legit

and also has very nice properties

like breaking down into a disjoint union of regular metric spaces

which are called 'galaxies' lmao

also my bad it's not sc

only fc

[empty grove]

oh right

[hollow harbor]

Unironically great terminology

[empty grove]

Analysis channel is above this one rice

[swift fjord]

lmao

actually you can also always bound your metric by composing it with some increasing nonnegative function with a finite limit

but I feel like this would be less interesting

feels like you're losing information to conform to the usual definition of a metric

[reef shore]

Would you retain triangle inequality?

[swift fjord]

right

it needs to be subadditive too

[reef shore]

Yeah

[hollow harbor]

metric spaces are analysis, analysis channel is above this one shin

[reef shore]

[unreal stratus]

they're just a special case of this channel though /s

[hazy nexus]

I feel like you should be able to do this via the homeomorphism [0,∞] -> [0,1]

[true garden]

Hello. How can one intuitively find open sets of a torus as a quotient space of a rectangle?

[reef shore]

open sets of the rectangle which have the property that whenever they contain a boundary point they also contain the point opposite to it

ie open subsets saturated wrt the quotient map

which is the definition of the quotient topology

[true garden]

So ...?

[eternal nimbus]

easy trivia time just to see if i'm right. Projective transformations are defined by linear transformations with 0 kernel. So if

$T:V ->W, T(v)=w$

such a linear transformation, then the projective transformation is simply

$\tau:P(V) ->P(W), \tau([v])=[T(v)]=[w]$

Where [v] are homogeneous coordinates

[gentle osprey]

PCR_Anibal

[eternal nimbus]

(i give up on the spacing, sorry if channel is busy, there are not "geometry 2" x) )

[empty grove]

The open sets of the torus are the images of the saturated open sets under the quotient map

[true garden]

Ok. But, how to visualize them? I know the open sets of a rectangle and know the quotient map, but I cannot find their images on torus.

[empty grove]

oh for visualisation it's better to view it as a subspace of R³

It has the subspace topology

Around each point the small neighborhoods just look like small open disks on the surface

[hollow harbor]

One open subset of the unit square [0,1]^2 is a semicircle centered at the point (1/2, 0). This semicircle has boundary, part of the bottom edge of the square.

This is not an open subset of the torus. If we wanted it to be an open subset of the torus, we'd need it to spill over the top edge of the square.

Like Moldi said, mathematically, this means that whenever we have a point on one side of the square in the open set, we need to have the corresponding point on the other side.

And having that point necessarily means having a blob around that point.

[empty grove]

The images aren't that hard to visualize actually, like you know how to roll up the square to form the torus? Just draw whatever saturated open subset of the square, then roll it up and that drawn thing is the open subset of the torus

[hollow harbor]

So you can think of open subsets like balls that slide around the square, but when they go out one side, they come back in the other (sometimes being split into what looks like two pieces - actually, it's one piece because of the quotient stitching it back together)

[true garden]

I did that, but I could not find them.

[swift fjord]

Roll up a piece of paper into a torus, draw an open set on it then unroll

[empty grove]

Very hard to do given the curvature of the torus

[swift fjord]

It can be donr

[hollow harbor]

not living in R^4

[swift fjord]

Ok maybe you.need more flexible paper

[true garden]

You mean in practice, or in mind?

[hollow harbor]

i will open up MS paint

and draw a picture

[empty grove]

#sponsored

[swift fjord]

Good enoigh

[empty grove]

Nice wristband bro

[swift fjord]

Wristbands are just tori anyways

[hollow harbor]

enjoy this gorgeous rendition of an open set on a torus

the light red interior is the open set

the boundary isn't included

[swift fjord]

Why is the boundary included

[hollow harbor]

but! it does include the bits of the boundary of the square that it looks like it should include

yes

[empty grove]

Reminds me of the sistene chapel

[hollow harbor]

i just didn't want to waste time drawing that

[swift fjord]

Everyone knows solid lines mean closed

And dotted lines mean open

[hollow harbor]

u try using ms paint splines

[empty grove]

You stop shilling for ms paint

[swift fjord]

I'd rather not tyvm

[hollow harbor]

anyway i could also have drawn one in the corner, and then it would go into all 4 corners

[true garden]

You mean I should first find the saturated subsets of the rectangle and then find the images of them?

[hollow harbor]

idk what saturated even means

[empty grove]

Yes

[swift fjord]

It means f^-1(f(A))=A

[empty grove]

Saturated means things going to the same image are all in the set

[arctic plover]

subsets of Rectangles for what?

[empty grove]

Or all not

[hollow harbor]

to me it's just "find the sets in the quotient space which pull back to open sets in the rectangle

[true garden]

Ok. What are the saturated subsets of the rectangle?

[hollow harbor]

well, the quotient map pulls back points on the boundary to two points

so you need open blobs around both of those

[empty grove]

Whenever a set contains a point in the boundary it should contain the point opposite to it

[hollow harbor]

not boundary but

you know

[empty grove]

That would be saturated

You know what open is

[true garden]

And what are the images?

[flint cove]

But in that case a T-algebra would consist of a map (underlying set of stone-chech of the set X)→ X… how can that be interpreted? it seems to me like a somewhat arbitrary thing to look at, to play the devil's advocate here

[empty grove]

Roll it around 😌

Like can't really give a better description

Just roll the square like you do to get the torus

And see where the stuff you've drawn ends up

[true garden]

In mind?

[empty grove]

Yeah

[true garden]

I did that, but I could not because I was confused.

[empty grove]

Try to do it for ryc's drawing

The set he drew is open saturated

[plain raven]

it sends each ultrafilter to its unique point of convergence

[empty grove]

Just see how it behaves under roll

[plain raven]

it's the evaluation map that sends the filter to its unique limit point

[flint cove]

This is a very natural idea to me given that classical algebras are about „construction“, ie taking things of the form (a₁,a₂,…) to things in your structure; iirc this is the prototypical example of (polynomial functor)X→X, right?
I've never seen this applied in any topological context tho

[plain raven]

i'm so stoned man

[true garden]

Ok. Thanks. Let me try aging for nth time.

[plain raven]

you could do this with polynomial functors or you could form the free term algebra in the language

[flint cove]

But a non-principal ultrafilter in βℕ does not have a limit in ℕ, so how can we build a map → ℕ here

[plain raven]

it's not a compact hausdorff space so it doesn't have an algebra structure 😦

if X is a set, then a T-algebra structure on X is exactly the same as a compact hausdorff topology on X

[flint cove]

Hmm… but what if we take βℕ→ℕ to be constant ↦0?

[plain raven]

this is what i mean when i say that working in the category of sets we "recover" the CH spaces

[flint cove]

what CH-structurc should that induce

[plain raven]

ah yeah

[flint cove]

Or is this a bialgebra thing because we want it to be compatible with the embedding X→βX

[plain raven]

sorry it's not enough space to explain all this stuff so i just dropped a bunch of details. the T-algebra structure map has to satisfy certain diagrammatic laws. for any set X whatsoever there's a canonical map eta : X -> TX (sending x to its principal ultrafilter) and also another map mu : T(T(X)) -> T(X) which is a bit harder to describe. in order to say that a map h: TX ->X is a T-algebra it needs to satisfy certain coherence conditions wrt eta and mu

namely

$$h\circ \mu = h\circ T(h)$$
$$ h\circ \eta = \operatorname{id}_X$$

[flint cove]

Oh right we don't hay “arbitrary T-algebra“ bbut “monadic T-algebra“ (not sure if the right nomenclature)

[gentle osprey]

diligentClerk

[flint cove]

no need to be verbose, I know what a monad is, I just forgot we care about this here

[plain raven]

yeah idk the nomenclature either it's a shitshow

i blame the CS people tbh

[flint cove]

in how far

[plain raven]

we never should have called an arbitrary map TX -> X an "algebra"

that's undeserving of the name algebra

it is so minimal in its structure

that it's not even worth giving a special name to

[flint cove]

I guess I'd believe you more if I knew more examples, I've seen TX→X rarely pop up
…but I suspect that's just because I'm not trained to seeing such monadic structures in non-abstract-nonsense settings

So you see it kinda as undeserving as calling an adjunction a „generalized galois connection“, a reflective subcat a „generalized closure operator“, and a cat a „generalized poset“? is that the right ballpark?

oh right I see now how we get an η: X→βX (inclusion) and a μ: ββX→βX (as all ultrafilters in βX are principal, right?)

[plain raven]

i think calling cats "generalized posets" and adjunctions "generalized galois connections" is fine if you're teaching category theory and want to start with the easiest examples first
but everything you named has a lot of structure to it
a map TX -> X is just not a lot of structure, i'm just complaining because math already has overloaded the word "algebra" pretty heavily and maps of the form TX -> X are totally ubiquitous, you could call everything a functor algebra at that rate

i don't remember exactly what mu does.

my brain shuts down around the third iterated powerset

[flint cove]

Ah, I see. So the wording it's sorta-correct but does not hit the nail on the head so to say

[flint cove]

that's a good way to avoid set-theoretic issues lmao

[plain raven]

it happens in topos theory too tho

this is a nice segue

in any topos you have a contravariant powerset functor which sends f : X -> Y to f^{-1} : PY -> PX

and it turns out that

well

this gives an adjunction between E and E^op

so the composition PP : E -> E is a monad

so that's pretty cool

[flint cove]

That seems pretty obvious on first glance

[plain raven]

it turns out that the algebras of the monad PP are exactly powersets, i.e. sets of the form PX
and the only maps PX -> PY that are compatible with the PP-algebra structure are the ones of the form f^{-1} : Y -> X

[flint cove]

like, toposes are a generalization of sets generalize the notion of „subset“ if i understood stuff correctly

[plain raven]

so the opposite category of E can be realized as the category of algebras of this monad, up to equivalence (replacing X with PX and f :Y-> X with f^{-1} : PX -> PY)

[flint cove]

okay, that seems less obvious

[plain raven]

which is wild to me

and as a consequence, like i said before, monadic forgetful functors have really good properties, you can lift limits along them (they "create" limits)

[flint cove]

Okay I think we're wildly off track considering this is #point-set-topology but I guess so far nobody has complained yet

[plain raven]

toposes are for sheaf cohomology
sheaf cohomology is geometry

lmao

[flint cove]

¯_(ツ)_/¯

hot take: any advanced channel should find a way to talk about monads

[plain raven]

but yeah so like
the forgetful functor E^op -> E creates limits, so that means if you have any diagram in E^op, you can project it down into E, compute the limit there and lift it back up
but a limit in E^op is a colimit in E

[pearl holly]

Is it fine if I ask a basic question real quick? 👉 👈

[plain raven]

it follows that if E has all limits (or all finite limits) then E^op has all colimits (or all finite colimits)

that's the end of my rant go ahead

toposes have all finite limits as an axiom, so this proves they also have finite colimits necessarily

[pearl holly]

So Hatcher says this: We can decompose the Klein bottle K as the union of two Möbius bands A and B glued together by a homeomorphism between their boundary circles. Then A, B, and A intersect B are homotopy equivalent to circles..." I don't really see how A intersect B can be homotopy equivalent to a circle

I found this picture online because I was struggling with the visualization and from this is seems that A intersect B should be empty

(this pic is not in Hatcher btw)

[flint cove]

Because A intersect B is the two dashed lines in the left half

and if they are precisely symmetrical then the right boundary point of the lower dashed line is glued to the left boundary point of the upper dashed line

because you glue the sides in reverse order

[pearl holly]

aaahh yeah okay I see. Thank you so much!

[flint cove]

lmfao discord knows sed-style s/pat/subs/ expressions

[flint cove]

oh NOW I actually got it, because we're in a topos setting we don't leave the category when going to the power set

that's where E^op comes into play

[plain raven]

yeah this is all internal, $A\mapsto \Omega^A$

[gentle osprey]

diligentClerk

[flint cove]

schweet

[plain raven]

sorry lol

[flint cove]

what a to-positive feature

[plain raven]

it's so weird

i was just thinking today about how grothendieck topologies on a category can be expressed as a kind of closure operator

and just like a galois correspondence is a special case of an adjunction, a closure operator is a special kind of monad

i don't think i fully understand what it means to take a 'monadic' pov towards these things tho

which is why i was thinking about it

[swift fjord]

@empty grove I've been thinking about the converse of the problem we talked about a few days ago (In fact a slight generalisation of it) and I think I (almost) have it, I just need a bit of help with one last thing. So the generalised version is that given some linear transformation $T:\mathbb R^n \rightarrow \mathbb R^k$, and some lattice $L \leq \mathbb R^n$, then $\text{ker} T=H$ is rational iff $T(L)$ is a lattice in $\mathbb R^k$. So i'm trying to prove that if $T(L)$ is a lattice, then $H$ is rational. I'm trying to prove this by the contrapositive, so i'm assuming that $H$ is not rational, and I wanna prove that $T(L)$ is indiscrete. This boils down to proving that the distance between $L - H \cap L$ and $H$ is 0 (That is, there are arbitrarily close points to $H$ from $L-L \cap H$). $$\$$
So suppose $H$ is not rational, then $H\cap L$ is free abelian of rank $< \text{dim}H=k$, then let $v_1,\ldots,v_k$ be a basis for $H \cap L$, and let $w_1,\ldots,w_l$ be a basis for the orthogonal complement of $\text{span}(v_1,\ldots,v_k)$ in $H$. Then note that $\text{span}(w_1,\ldots,w_l)=W$ is totally irrational with respect to $L$ (That is, $L \cap W = {0}$). Previously i've proven that the lattice is asymptotic to a totally irrational space (I've technically only proven this for a hyperplane, but I think the proof is similar in this case). $$\$$
Here's the problem, intuitively since $W$ is orthogonal to $H \cap L$, it seems to me that the points of $L$ arbitrarily close to $W$ should be coming from $L - L \cap H$, but I'm not sure how to justify this.

Wow sorry for the super long post

[gentle osprey]

ShiN

[flint cove]

I've thought about whether the „specialization preorder“ endofunctor gives us a monad but it isn't since the identity X→SpPr(X) is not continuous (the open sets on the right are the subsets of X upwards-closed wrt x₁≤x₂ iff cl(x₁) sub cl(x₂), and as a counterexample to continuity take a closed point in some space that is not isolated)

hm, can't we topologize a preorder as well by taking the upper sets to be closed? perhaps it works then idk

[swift fjord]

My bad I meant that rank(H cap L)=k, not dimH.

[flint cove]

(My motivation here would be that if it worked, I would wanna see what the „algebras“ SpPr(X)→X would look like – in the alexandroff case that would have to be the identity since SpPr(X) should carry the same topology iirc, but non-alexandroff cases would interest me)

posets does not sound right as the cardinality of indistinguishable elements per equivalence class matters

(think indiscrete of different cardinalities)

same poset (as the preorder quotient), mirorring the fact that they have the same T0-quotient (point)

[empty grove]

@swift fjord take a neighborhood of 0 in H, that doesn't contain any other points of L (discreteness of L in H) and then W x that neighborhood should work as a neighborhood of W not containing any non zero points of L cap H?

[flint cove]

(anyway, should be the same construction)

[swift fjord]

you're saying that for any nbhd of 0 that doesn't contain points of L cap H, W x that nbhd won't contain nonzero points of L cap H, but should still contain points of L because of what I said?

Do you think it would matter if I took some random complement instead of the orthogonal complement?

[empty grove]

Yeah it should contain points of L assuming what you said

[empty grove]

hmm I don't think so

[swift fjord]

I don't think so either but it's hard to visualise

since very quickly we go past 3 dimensions

lmao

[empty grove]

Lol yeah

But shouldn't be hard to prove algebraically

That neighbourhood x W is open because both these sets are, and it only intersects H on (that neighborhood x 0)

So is disjoint from H cap L

[swift fjord]

wait why is W open

[empty grove]

oh wait no whoops

[swift fjord]

no but you could take a nbhd of 0 inside span(v_1,...,v_k)

and then do $$\bigcup_{x \in W}x + N$$ where $N$ is that nbhd

[gentle osprey]

ShiN

[empty grove]

Yeah isn't that span just H

[swift fjord]

we're assuming that H cap L is of lower rank than dimH

[empty grove]

Yes that is what I was doing lol

oh shit yeah H cap L

Not H

[swift fjord]

yeye

the contradiction comes from the fact that a subspace that is not rational will contain some completely irrational space

[empty grove]

Looks like a sheared cylinder, must be open

[swift fjord]

yeye

not every sheared since it's orthogonal

even*

[empty grove]

Yeah true

[swift fjord]

alright cool, thanks!

finally done with something at least

now onto the hard questions

[empty grove]

oof

[swift fjord]

very oof indeed

[eternal nimbus]

Well, after so much hard work on that, i can throw an easy question on the mix?

don't this statements...contradict each other?

I though to post in in linear algebra, but dunno if be being a projective space it changes anything

[hollow harbor]

Surely they mean linearly dependent? The line above literally says c'' + a'' + b'' = 0

unless i'm going crazy, or this is all different in projective space

it's weird

they also say afterwards that the 3 points lie in a 2d subspace of V

which should make them dependent

[swift fjord]

wait this doesn't actually work since this way we're staying inside of H, but the points in L will have to be outside of H by assumption, so this doesn't actually help us. and if we take a random open ball in R^n around 0 that doesn't contain points of L cap H, i'm still not sure I see why W + that ball will also not contain any

[empty grove]

We can take the subspace H

And if W is not arbitrarily close to L - 0 in this subspace

It is not in the original

[swift fjord]

wait yeah you're right

but then i'm still not sure i'm convinced why H doesn't intersect W + ball in H

[empty grove]

H does

H cap L doesn't

Because H cap L = span(L) cap L
W + ball in spanL cap H cap L ⊂ that ball cap L = 0

Oh we take that ball in span L

I must have written down H earlier

Because I confused notation

The cylinder is in H

[swift fjord]

not span L

span H cap L

span L is just R^n

[empty grove]

oof yes

[swift fjord]

but i'm not sure if we take the ball in H or in span H cap L

[empty grove]

So many caps 🧢

[swift fjord]

lmao

[empty grove]

I'm taking it in span of H cap L

[eternal nimbus]

alright, figure it out, the idea is that they're linearly independt 1 to 1, a from b, b from c, c from a, but all together you have that one is a linear combination of the others, so they are in a subspace of dimension 2

[empty grove]

And then adding W should give cylinder in H which is open in H and it's intersection with span H cap L should be that ball which doesn't intersect H cap L

And other parts of cylinder certainly can't intersect H cap L

[swift fjord]

i

i'm not sure if it's necessarily a cylinder. What if span H cap L is of lower dimension than W

[empty grove]

Yeah I mean that's just an analogy

But the point is that its intersection with span H cap L should be just that ball again

[swift fjord]

ok wait yea it doesn't matter

[swift fjord]

this feels like it should be so trivial but i'm not sure how to show it algebraically

[empty grove]

Because they don't lie in span (H cap L)

[swift fjord]

oh wait

yea I got it

[eternal nimbus]

So, a bit of clarification question if anyone could help me. Consider the projective plane and isomorphism $P^2(F_{2) \simeq F^2_{2} \cup P^1(F_{2)$ If i understand correctly, this is like seing it as the plane + a line at infinity right? But...how do i know which/how many points are there?

[gentle osprey]

PCR_Anibal

[bitter yoke]

What are you asking? Are you asking how many points are at infinity and which ones they are?

@eternal nimbus

[eternal nimbus]

Yes 🙂 and the method. I have the answer sheet but it only says "3" without explaining what or how x)

[bitter yoke]

I mean it depends on what the field is

The idea is that P²(F) has elements that look like [x:y:z] right?

By scaling z appropriately, you can make this become [x:y:1] or [x:y:0] depending on whether or not z is non zero

The elements that look like [x:y:1] are a copy of F^2_2

And the elements that look like [x:y:0] are the points at infinity

[eternal nimbus]

F is finite field of 2 elements, sorry

[tepid depot]

So given that F has two elements, I would write out all the possible points in homogeneous coordinates

[eternal nimbus]

So my points at infinity would would be [0,0,0] [0,1,0] [1,0,0] ?

[tepid depot]

First point isn't valid

Other two are correct, you're close

[eternal nimbus]

[1;1,0]

[tepid depot]

Yee

That's all 3

[eternal nimbus]

right, i completely forgot about (1,1) being there so i was like "uh..answer was three and thats the last posibility" x)

Right, thank you both!

[eternal nimbus]

...dimension of the empty set -1? WHAAAAT

[plain raven]

the dimension of the projective space of lines through the origin of the vector space V is dim V - 1

setting V = the zero vector space, dim V = 0 and so the dimension of the empty projective space is -1

😉

you see this in singular homology too, reduced homology is just the homology when you include the set of maps from the empty simplex into the chain complex (the -1 dim simplex)

[eternal nimbus]

ohhh that makes sense thank you

[lean marten]

Its also how you define the base case for inductive dimensions

[plain raven]

what's the inductive step

[lean marten]

You say that a space X has inductive dimension less than or equal to n if for each open U in X its boundary has dimension less than or equal to n-1

So a singleton has dimension 0

A line segment has dimension 1 because all boundaries of open sets are discrete or empty

ect .

[marsh forge]

Don't you have to assert that empty set has dimension -1 here anyway

Oh i didnt read up enough, whoops

[coral pawn]

Do the polar coordinates and Cartesian coordinates determine the same smooth structure on R^2?

I'm pretty sure they don't, but I can't find a counterexample

Ping me

[supple locust]

how does one deduce that $\pi _1 (X \times Y)\cong \pi _1 (X) \times \pi _1 (Y)$ using Van Kampen's theorem?

[gentle osprey]

bert

[supple locust]

A Concise Course in Algebraic Topology-J. P. May says that it is immediate from universal property of products

[empty grove]

Ye van kampen would be weird to use here

Since no pushouts are involved

Why do you want to use van kampen

[supple locust]

because of this note basically. I want to see how this works. I know the simple minded proof of this fact.

[empty grove]

Wait van kampen and universal property of products are completely different things

[supple locust]

yea van kampen is all about coproducts (colimits in fact)

[empty grove]

Yes And I don't see any way of using that here but there's a way to use the universal property of products

So you can construct some natural homomorphisms both ways

And they will be inverses

And to construct them you use the universal properties of both product groups and product spaces

[supple locust]

I see so it has nothing to do with van kampen. We use that pi _1 is covariant functor and universal property of products that gives us one way morphism

I got the other way morphism as well

[empty grove]

Great now you just need to show that both compositions of these 2 are identity

[supple locust]

nice then the uniqueness from the universal property will give the compositions are id

[tawdry widget]

Any way unless X and Y are path-connected you still need to specify their base points respectively…

[empty grove]

This should be true for fundamental groupoids too I think

Which would allow omitting base points

[supple locust]

Yes I think. Because otherwise the functor pi _1 wouldn't make sense

[lean marten]

@coral pawn Polar coordinates aren’t bijective in R^2 because of the point (0,0)

[plain raven]

it wouldn't be easier to omit the basepoint here i think

because you would just need to prove the lemma about path homotopy here for an arbitrary pair of points in the space rather than ... for an arbitrary but fixed basepoint in the space

idk

[alpine bolt]

Isn't the first function just (x,y)? It doesn't matter what the values of z are since the function doesn't even include them.

[pearl holly]

so Hatcher has just derived this exact sequence

and now he gives an example. Where does the sequence 0->Z_2 -> H_1(Z) -> Z -> 0 come from?

Where does Z_2 come from?

Hatcher writes a 2 above the arrow that goes from the integers to the integers and I assume that it means a map f(x) = 2x because 1-g_*(x) = x- -x = 2x. But then he also writes a 0 in the map in the right which I don't understand. Shouldn't it also be a 2 instead of a 0?

[tough imp]

It comes from the exact sequence in the diagram

You have the exact sequence
0 -> Z -2> Z -> H_1(Z) -> Z -0> Z

Once you quotient the second Z by the image of the first Z you end up with

0 -> Z_2 -> H_1(Z) -> Z -0> Z

Then since the last map is the 0-map the kernel is all of Z

Then since the image of H_1(Z) is the kernel of the next map, that map H_1(Z) -> Z is surjective

This gives you
0 -> Z_2 -> H_1(Z) -> Z -> 0

Does that make sense?

[pearl holly]

okay wait, what does Z -2 mean?

[tough imp]

Sorry

That just mean

The map defined by •2

I put it inside -> to be the name of the map haha

-2> meant multiply by 2

Similarly -0>

Is the 0 map

[hollow harbor]

Minus 2 is greater than Z.

[pearl holly]

hmm I don't really understand this step. Why can you just quotient out by the image?

[tough imp]

Because of exactness

So the map Z -> H1(Z) leads to an injective map Z/kernel -> H1(Z)

But the kernel is the same as the image of that last map

And the image is 2Z

[pearl holly]

yeah okay I see I see. But why is the last arrow the 0 map? Shouldn't it also be just multiplying by 2?

[tough imp]

Idk that’s what’s in the image you sent

I don’t actually know AT

[pearl holly]

so like here, the last map is 1-g_*. So it is defined exactly like the left most map, 1-g_*. But even though they are the same, one map is multiplying by two and the other one is just the 0 map lmao

[tough imp]

They’re between different homotopy groups tho

¯_(ツ)_/¯

[pearl holly]

yeah I guess

[tough imp]

Like the way you identify H0(S1) with Z and H1(S1) with Z is different

So I guess g_* acts differently

If 1 - g_* is 0 that means g_* acts as identity on the 0-th homotopy group

So maybe try to see why that’s the case?

[pearl holly]

yeah okay. Thank you so much for the help, I appreciate it!

[tough imp]

THEY CALL ME MISTER HOMOLOGICAL COMMUTATIVE ALGEBRA

[empty grove]

Chmister commutative algebra

[empty grove]

Didn't get your question. You're asked to find the image of U1 under psi1 and psi1 inverse

[tough imp]

You want to find the set, the fact that z > 0 and the constraint that x^2 + y^2 + z^2 = 1 means the image won’t just be all pairs (x,y). For example (2,0) will not be there

Tfw Moldilocks hahahaha

[swift fjord]

THEY CALL ME MISTER HOMOLOGICAL ALGEBRA

[pearl holly]

wait wtf, how do you "calculate" the homology H_1(M, \partial M) with M the Möbius strip using the cellular boundary formula?

I'm looking back at old examples and now I just don't understand them

So Hatcher writes this: In the simplest case of the degree 2 map S¹→S¹, z →z², this says that the Möbius band does not retract into its boundary circle. So here we would have an exact sequence: $$0 \to H_1(\partial M) \to H_1(M) \to H_1(M, \partial M) \to 0$$ that splits and so $H_1(M) = H_1(\partial M) \oplus H_1(M, \partial M)$ and I know that $H_1(M, \partial M) = \mathbb{Z}_2$ since if you contract the boundary to a point you get the real projective plane. But now I don't see how the degree 2 map is being used here

[gentle osprey]

Tokidoki ✓

[pearl holly]

So basically, how do you know that H_1(M, partial M) is Z_2 using the cellular boundary formula?

I'm constantly having massive brain farts

Does it even make sense to "calculate" relative homology with the cellular boundary formula?

ahhh this is just so frustrating

[empty grove]

Can't see how you'd do that

Maybe first compute absolute homology then do long exact memes

First step could use cellular homology

But why do you want to use cellular homology for this?

[pearl holly]

because hatcher writes something about the degree two map from S¹ to S¹ and I assume that it has something to do with cellular homology

so if you look at page 148, the diagram there says that H_n(M_f, S^n) is Z_m

and I can swear that I wrote something down on my paper that clarified this but I can't find it

and now I don't understand this

[empty grove]

Degree don't have anything to do with cellular homology

Degree 2 map from S¹ to S¹ means the map which goes around the circle twice

Degree of f: S^n to S^n is f*(1) where f* is the induced map on the nth homology and 1 is the identity n-simplex on S^n

[pearl holly]

but the cellular boundary formula tho?

it has degrees in it

[empty grove]

Doesn't mean that it's used anytime degree is used

[pearl holly]

I don't know what that is

okay moldi wait I will send a pic real quick

[rough locust]

He so I'm having some trouble understanding what this question is asking: "Prove the identification of points at opposite ends of diameters on the boundary of the circular disk D^2 defines a 2-manifold". What does this mean by "identification of points at opposite ends of diameters"? Best I could come up with is that we're working on D^2/~ where x ~ y iff x and y are opposite boundary points, but that doesn't make much sense

[pearl holly]

so start reading at "for a somewhat more subtle...". How does Hatcher know that the homology group of that big thing is Z_m?

[empty grove]

x ~ y iff x = y or x and y are diametrically opposite

[rough locust]

But then isn't it basically just the top half of the unit disk?

[empty grove]

You're not pasting the stuff on the interior

Only the end points of the diameters

[rough locust]

Oh so like it's a circle with the boundary only on the top half?

[empty grove]

Well it won't be that exactly

Because if you exit from the bottom you should re enter from the top

It's a very weird thing which I don't think can be visualised in R³

but idk

[rough locust]

ok so now I'm even more confused, cause the assignment's hint tells me to re-write it as the closed unit disk {(x,y) in R^2 | r <= 1}

oh wait

[empty grove]

I must be wrong then

[rough locust]

"with opposite boundary points identified" lol

nah you're right

I missed that part

[pearl holly]

yeah I guess I see that

[empty grove]

Now use exactness

This short exact sequence is part of the long exact sequence for the pair M_f, S^n

[pearl holly]

yeah okay I see lmao

[gentle osprey]

Nobody

[pearl holly]

so you can restrict the last map so that you can replace H_n(M_f, S^n) with Z/mZ right?

[rough locust]

quick question: does "diametrically opposite" mean they are both on the boundary and are opposite of each other, or just opposite to each other?

[empty grove]

I used it as the former, idk what it usually is actually

[rough locust]

ah ok. thanks

[pearl holly]

wait is the map H_n(M_f) to H_n(M_f, S^n) surjective?

[empty grove]

A → B → 0 being exact at B means that the first map is surjective

[rough locust]

thank you!

[pearl holly]

ah yeah of course

man

I can't do this shit

okay wait now, why is the map multiplication by m?

[empty grove]

I'm guessing you can deformation retract onto the codomain part?

[pearl holly]

okay wait. So this mapping cylinder just deformation retracts into one of its ends. One end goes around itself m times and this will be the generator of H_n(M_f). Now the map H_n(S^n) -> H_n(M_f) is induced by the inclusion and so it takes a generator to a generator. So if you go around one time in S^n you go around m times in M_f right?

[empty grove]

Would it deformation retract onto its domain part?

I don't think it will, at least not when f isn't injective

[empty grove]

The codomain part is just a circle. Even though "it goes around itself m times", the generator will still be the simplex that goes around once

But the important part is that going around the domain circle once is the same as going around the codomain circle m times

That's why the map is multiplication by m

[pearl holly]

yeah okay I see now. Thank you both for the help!

I keep getting these brain farts man

I understood this before but then I go back and boom I don't get it

[eternal nimbus]

Anyone can correct this?

So, i have $E=\tau(S \setminus {0})$ where $\tau$ is the canonical projection (takes a vector s in S and sends it to the projective point [s]), S is a non-empty set of V (vector space) and <S> is the vector subspace generated by S. I have to prove, given the first equality, the second one (and i'm lazy to type it in latex lol)

My idea seems a bit...complicated? and i'm not sure it really holds
I basically says that P(W) is the set of 1 dimensional subspaces of V, so i rewrite P(W) as the union of this subspaces, say that the condition E in P(w) is equal to the condition that every vector s of S is in the union of this. but the union of subspaces is a subspace so this intersection of unions of vector subspaces has to be span(E)....which honestly said like that seems like i'm sputting nonsense?

[gentle osprey]

PCR_Anibal

[bright acorn]

Given $I, J \subset \mathbb{R}$ countable sets such that $|I| \leq |J|$, does there always exist $f \in C^{0}(\mathbb{R})$ homeomorphism such that $f(I) \subset J$? A stronger result, which is the one I am in fact more interested in, would be to assert if there exists an orientation preserving diffeomorphism that does this.

[gentle osprey]

MisterSystem

[honest terrace]

(What does orientation preserving means here ? )

[bright acorn]

it just means the jacobian of this map being positive

in this case it just means monotonic I guess lmao

[honest terrace]

Also do you allow infinite sets to be considered countable ?

[bright acorn]

yeah, in fact the case of I and J having the same cardinality as N is really interesting

if we consider I and J finite