point-set-topology

[sleek thicket]

All the proofs are magical

[limpid mural]

Well

To support this thesis just look at the cover of this book

http://math.univ-lille1.fr/~ramero/CoursAG.pdf

Lmao

[sleek thicket]

Omg I love this

This is my aesthetic

[limpid mural]

Ahahahah

[sleek thicket]

Brb learning French

I feel like I need to learn French to read EGA, just to say I did it

[limpid mural]

Yeah it was like my first though when I first saw the grimoire ahahahah

I just draw the cover in every blackboard just to spread the word

[sleek thicket]

I'll keep that in my back pocket

My goto blackboard signature is "that's yoneda babe"

[limpid mural]

Lmao

[rugged swan]

Haha

Oui

Grothendieck > all 😎

There's not english versions of EGA ?

[sleek thicket]

He asked for it not to be translated

[rugged swan]

Mh

[sleek thicket]

But he's dead and so people are flossing on his corpse by translating it

You can find it on github

[rugged swan]

That's suprising from Grothendieck

[sleek thicket]

It's a recent project so idk about quality

[rugged swan]

Why he refused ?

[sleek thicket]

No the translation on github

Tbh the point isn't to actually learn algebraic geometry from ega

It's the aesthetic

[rugged swan]

Lel

But idk why he'd refuse x)

[floral gust]

How would I start on part a?

[gentle osprey]

QuickMaffs:

[granite copper]

@floral gust By writing down what the topology on E x E actually is

[floral gust]

The topology on E x E should be the product topology that is generated by taking products of open sets in E which in our case suffices to be the sets such that p-1(b) would have an evenly covered neighbourhood

[limpid mural]

So fix a point in the diagonal and use the definition of cover map

To show that the set is open

Then you can show that Z / delta is open

[floral gust]

Um can you elaborate @limpid mural

The diagonal is in E x E and cover is for E

[limpid mural]

Yeah sure apply the definition on e

[gentle osprey]

QuickMaffs:

[floral gust]

Not sure how to proceed from this. @limpid mural

[gentle osprey]

QuickMaffs:

[sleek thicket]

@floral gust we're proving Δ is closed in Z

[floral gust]

Ah okay true true

Any idea however how to get started on it then @sleek thicket

[gritty widget]

@floral gust given any (e, e) in the diagonal, let b = p(e)

if p is a covering map, then you get disjoint nbds for all the f in E such that p(f) = b

out of these points, you take all the points except e

try that

[gritty widget]

Does anyone know what 'lub' stands for? Here's the example in book (metric topology if it helps): D(x,y) = lub{|x_i-y_i|}

or diamA = lub{d(a1,a2) | a1,a2 \inA}

I suppose sup? But there was max used in the book

Yeah that must be it.

Anyways, I'm not sure in one thing: I see a proof of two topologies being the same (euclidean and max). They say that: $$ \rho\left(x,y\right) \leq d\left(x,y\right) \leq \sqrt{n} \rho \left(x,y\right)$$ Where rho is max metric. They say that the first inequality shows that $B_d\left(x, \epsilon\right) \subset B_{\rho}\left(x, \delta\right) \forall x, \epsilon$. And I'm confused - shouldn't it be the other way around? If the distance euclidean is bigger than max, then for same epsilon the rho ball should be in d ball?

[gentle osprey]

Godel:

[limpid mural]

Least upper bound = lub

So yes sup

[gritty widget]

Or should I be thinking kinda this way: if rho metric is 'smaller' then for some epsilon more x's are in the rho-ball than d-ball?

Kinda makes sense but it's confusing

[limpid mural]

If rho <= d <= epsilon then if p is in the d ball then it is in the rho ball

Do u understand why?

Try to sketch the balls for the plane

[rugged swan]

hi guys, what's a "projective line" ?

I a book, the author is saying that if K is finite, projective line corresponds one to one with elements of KP^2

[gritty widget]

That's probably a typo

[rugged swan]

idk

it's in an exercice

[gritty widget]

I don't see it

Well, in general, the projective line over K (or KP^1) can be thought of as the set of lines through the origin in K^2

[rugged swan]

yes that's what I found on wikipedia

[gritty widget]

But KP^2 contains many copies of the projective line

so I really don't see how it can be in bijection with KP^1, especially if K is finite

[rugged swan]

so this can't be true since there is q+1 projective lines and q^2+q+1 elements in KP^2 where |K| = q

[gritty widget]

Ah but wait

[rugged swan]

the author is drunk

[gritty widget]

Did you mean KP^1 is in bijection with KP^2 ? Or the set of projective lines in the projective plane is in bijection with the set of points ?

[rugged swan]

"Show that, in KP^2, there's as many projective lines than elements"

[gritty widget]

Ah okay

okay this is true

[rugged swan]

I didn't understand the question then

[gritty widget]

No, I think you had phrased it correctly but I hadn't parsed it right

So let's see: What is a line in KP^2 ?

[rugged swan]

for me, elements of KP^2 are lines in K^3

[gritty widget]

yes

[rugged swan]

I don't see what could be a line in KP^2 since it doesn't have a vector space structure

[gritty widget]

Argh, unfortunately I cannot scan pictures, but basically you should think of it like this

[rugged swan]

oh ok, a line is a plane in K^3 ?

[gritty widget]

exactly

[rugged swan]

oh that makes sense then

[gritty widget]

in general, a subspace of dimension k in KP^n is the same as a subspace of dimension k+1 in K^{n+1}

(that is a definition)

So yeah, this is just duality

[rugged swan]

so in the sentence "projective line" means line in KP^2 ?

[gritty widget]

yes

but it is also geometrically a projective line in the previous sense

[rugged swan]

what do you mean by "geometrically" ?

[gritty widget]

this line corresponds to a plane P in K^3, and its points will correspond to the lines contained in P

[rugged swan]

yes a projective line is KP ?

oh ok

yes

thanks

[gritty widget]

all clear ?

[rugged swan]

yup

[gritty widget]

Do you see how to get the bijection ?

[rugged swan]

a plane is an equation ax+by+cz = 0 modulo colinear vectors to (a,b,c)

that's exactly RP^2

[gritty widget]

you mean RP^1 ?

[rugged swan]

the sets of planes

is RP^2

[gritty widget]

Ah yes

right

[rugged swan]

and that's duality since a plane is characterized by its orthogonal line

[gritty widget]

yup

likewise, you have a bijection between subspaces of dimension k and subspaces of dim n-k-1 in P^n

[rugged swan]

P^n = KP^n ?

[floral gust]

@gritty widget What do you mean take all the points except e? They lie in distinct sheets so how can they form a neighbourhood of e let alone an open one?

[sleek thicket]

I'll try to do the covering space thingy

Take an element (e, e) in the diagonal and let b = p(e). Let U be a canonical neighborhood of b. Then p^(-1)(U) is a disjoint union of open connected sets V_i which are all mapped homeomorphically into U by p. Let i_0 be such that V_{i_0} contains e. Then V_{i_0}×V_{i_0} is a neighborhood of (e, e) in E×E, and so (V_{i_0}×V_{i_0}) intersect Z is a neighborhood of (e, e) in Z

I claim that it is also contained in Δ. This amounts to showing that for any point (e, f) in V_{i_0}×V_{i_0} such that p(e) = p(f) we have e = f. This says exactly that p is an injection on V_{i_0}, which is true since it's a homeomorphism on there

@floral gust

[floral gust]

Lmao I just completed it too

[sleek thicket]

lol

[floral gust]

Thanks!

[sleek thicket]

I mean

It's good to do this

So it's probably for the best

[floral gust]

This was mine

[sleek thicket]

Is there a slick proof that Z is hausdorff?

I feel like there probably is

Also the amount of notation in that proof is making my eyes burn

I don't think I believe that proof. You prove Δ contains Z intersect K, which is good enough, but they won't be equal in general

[floral gust]

Why is showing Δ contains Z intersect K good enough? Dont the subspace topology require that they must be equal to some open set

[sleek thicket]

A set is open if it contains a neighborhood of each of its points

More precisely, if you take the union of all the Z intersect K's you'll get Δ

[floral gust]

Oh yeah yeayh youre right

because I am considering for an arbitrary e

[sleek thicket]

Yup

Proof for closedness: take e≠f such that p(e) = p(f). Let U be a canonical neighborhood for p(e) which lifts to a neighborhood V of e and V' of f. Then (V×V') intersect Z will a neighborhood of (e, f) disjoint from Δ. This shows the complement is open, so Δ is closed

[floral gust]

I guess my first statement is false. (x,x) in U doesnt imply (x,x) in V_a but rather some V_i such that V_i in the collection. hence the first inclusion fails

[sleek thicket]

That sounds right. I didn't read closely

[floral gust]

Would your proof encounter any resistance if V=V'?

Becuase then I dont think it would be disjoint from Δ

@sleek thicket

[sleek thicket]

Yes, but that's impossible

[floral gust]

Oh nvm

Yep figured

[sleek thicket]

Np

[floral gust]

The point would end up lying in Delta due to the p being homeomorphic on the subset.

[sleek thicket]

Yup

[floral gust]

Is there a slick proof that Z is hausdorff?
This is true if Diagonal is closed which we can get by considering the point in the complement and considering the open neighbourhood of the point.

[sleek thicket]

Right, I meant as a substitute for proving the diagonal is closed

[limpid mural]

Do you know what does mean "generic perturbation of a smooth map" in differential geometry?

[frigid patrol]

Punch your function till its straight

[limpid mural]

Kay, ty

[gritty widget]

Lol my dad genericaly peturbated me when I was young

[gritty widget]

did it work?

[fleet rapids]

so If I have a manifold M

does the diagonal map just take x in M to (x,x)?

[gritty widget]

yes

you have anything and a diagonal, then x is taken to (x, x)

[rugged swan]

hey guys, if I'm quotienting RP^2 by the classes {[x,y,z]} when z=/= 0 and {[x,y,0] | (x,y) in R^2}, does it is homeomorphic to the complex projective line CP^1 ?

(and then to the 2-sphere)

[floral gust]

I am trying to show that the quotient map from S^2 to RP^2 is a covering map. Is this collection of sets would suffice as an even covering of the open neighbourhood?

[sleek thicket]

What about a point whose first coordinate is 0?

[floral gust]

): So I should rather consider x>0, x<0, y>0,y<0, z>0,z<0

[sleek thicket]

Those aren't disjoint though

[floral gust]

Any idea to salvage this or would I have to work a completely different approach? I kinda have written a lot on it so ):

[sleek thicket]

I think the first idea works, you just need to do cases on x and pick U properly

So like if x = [a : b : c] where a ≠ 0, then { [u : v : w] in P^2 | u ≠ 0 } is a neighborhood of x

And it's homeomorphic to R^2

What's the preimage of that neighborhood under the quotient map?

[floral gust]

{ [u : v : w] in S^2 | u ≠ 0 },{ -[u : v : w] in S^2 | u ≠ 0 }

@sleek thicket

[sleek thicket]

You don't want homogeneous coordinates there

But yeah

You've got the upper and lower hemispheres

Those are the connected components

And your quotient map restricts to a homeomorphism on them

So for a point whose first coordinate is nonzero, this works

Right?

[floral gust]

yeah...? So like do this for all three?

(as in just say wlog)

@sleek thicket

[sleek thicket]

Yup

But you need to do them separately

And be explicit about what U is

Sorry, I don't mean you can't wlog it

Just be clear that's what you're doing

[floral gust]

Yeah Yeah I got your point! Thanks. The reason why { [u : v : w] in P^2 | u ≠ 0 } is open is because its preimage is just S^2 take out the x axis right?

[sleek thicket]

Sure

[gritty widget]

If X x [0,1] is compact does X also have to be compact?

I feel like it has to but I'm not sure how to formaly justify that

[floral gust]

Take an open cover {U_i} of X. Then {U_i x [0,1]} is an open cover of X x [0,1]. Thus there exists a finite subcover say {U_1 x [0,1] ,.... U_n x [0,1]}. The collection {U_1,...,U_n} is hence a finite subcover of X.

Or just take the first projection map. They are surjective so the image is X. They are continuous so they preserve preserve compactness. Thus Im(proj) = X is compact.

[sleek thicket]

Also, X×{0} is a closed subset of a compact space, so it's compact, and it's homeomorphic to X

[floral gust]

{ [u : v : w] in S^2 | u ≠ 0 },{ -[u : v : w] in S^2 | u ≠ 0 } these are not disjoint right? So I should rather consider u>0 and u<0? Correct?

@sleek thicket

[sleek thicket]

Yeah lol I misread

The point is that the components of the preimage are the right thing

[floral gust]

Just one more thing. I am a bit uneasy about my understanding of a homotopy lift. Can you see if this proof makes sense (as in it uses the lifting properties correctly)?

@sleek thicket

[sleek thicket]

I'm very confused about a thing

Let Y = P^1

Then S(Y) = k[x, y] (this is the homogenous coordinate ring, I'm using definitions from Hartshorne)

Then K(Y) = k(x, y)

But K(A^1) = k(x)

However Y and A^1 are birational, because the open subset {[x, y] in Y | y ≠ 0} is iso to A^1

And Hartshorne claims two varieties are birational iff their function fields are iso (as k algebras)

Which is totally false here!

Can anybody see where I went wrong?

[sleek thicket]

<@&286206848099549185> if anybody can help with AG ¯_(ツ)_/¯

[sleek thicket]

Someone on another server helped

[clear jackal]

Hey I've been thinking about an equivalence relation on metric spaces. I don't know if it's already a well known thing but I've proven some results about it, I just want to know if it looks like anything someone here has heard of before:

[gentle osprey]

datorangeguy:

[clear jackal]

The main results I've been working at are on how else to characterize it (using strictly increasing functions to switch between metrics) and I've also been trying to see which further assumptions I can make to relate it to better known equivalencies

I also feel like category theory could help out a lot, but since I'm still learning the basics of categories, I'd rather see what is already out there on this relation

[light rock]

i don't see why category theory would help

but functions like these seem good

I don't think I've seen them before

in particular they are injective

idk sounds interesting to think about

[clear jackal]

well the function mapping between the metrics themselves is injective yes but that's different from the phi here, which is assumed already to be a bijection

[light rock]

you shouldn't need to think about bijections

seems to me like the natural thing to do is just consider maps satisfying this in one direction

and then maps doing both

[clear jackal]

I agree! I haven't been able to think of functions like that (not bijective, that is) over non-discrete metric spaces though. Many easy examples over discrete metric spaces.

You can think about it like mapping scalene triangles to scalene triangles, isosceles to isosceles, and equilateral to equilateral, which is much easier to show in discrete spaces

I've been pointed to shape theory by some of my professors, but the books I found at the library are way over my head :\

[light rock]

bijective functions on discrete spaces are trivial basically

well

idk

somehow these functions are preserving some kind of ordering

more than topological stuff

[clear jackal]

yes, in my current draft I've opted to call it "geometric order"

and I've conjectured on the further assumptions needed to show topological properties

[light rock]

have you proven any result about them?

[clear jackal]

for the bijections yes I've proven minor results

I've proven that $(M , d_1) \cong (M, d_2)$ if and only if there is some strictly increasing function $\sigma$ over the range of $d_1$ such that $d_2 = \sigma \circ d_1$ (function composition)

[gentle osprey]

datorangeguy:

[clear jackal]

and then I used that to prove the analogous result between distinct sets $(M , d)$ and $(N, \rho)$

But that function $\sigma$ from the former result is of particular interest. My conjecture is that if $\sigma$ and $\sigma^{-1}$ are continuous at 0, then $d_1$ and $d_2$ are equivalent metrics in the usual definition (same convergent sequences / open sets)

[gentle osprey]

datorangeguy:

datorangeguy:

[clear jackal]

I think I'm close to proving that but there's some minor details that I need to justify better

The main reason I'd like to consider bijections first is to get straight into the group of all such bijections over the same set, looking at all metrics on a set which have equivalent "ordering"

[dim meadow]

What do you mean continuous at 0?

Is 0 in your metric space?

[light rock]

sigma:R->R

it's the map between the metrics

[dim meadow]

Oh hmm

Oh I see

This notion seems very restrictive honestly

I think sigma would have to be continuous everywhere

In order for both metrics to be metrics

Cause metrics are continuous

[light rock]

but no

[dim meadow]

Oh I see

[light rock]

it's very odd

[dim meadow]

If you have say a discrete case

[light rock]

yeah

[dim meadow]

Then you would be good

[light rock]

it's capturing the information about the metric which doesn't come from topology at all

[dim meadow]

This kind of sucks

[light rock]

it's very strange

which means it's probably a good relation to mod out by

who knows

[dim meadow]

Do you know any of the standard non Topological theory of metric spaces @clear jackal

Like bilipschitz maps, isometries

Metric maps

But okay, clearly any "nice" metric spaces will have only the continuous increasing maps R \to R

Suppose I imbeds in your space

Then every sigma must be continuous near 0

What else can you prove

Suppose a convergent sequence is in your space

Okay that's weird

You're right @light rock

This captures some weird ass data about your space

You can be discontinuous at 0 and it's totally fine

Even for a non discrete space

[light rock]

it's an odd question

I'm curious if you can prove results about this even

maybe not

because it captures such a weird structure

[clear jackal]

Yes firstly you must assume that $\sigma d $ has triangle inequality because you can not always produce a metric with a function that is merely increasing, take $ x^2 $on euclidean for example

[gentle osprey]

datorangeguy:

[clear jackal]

And of course for an arbitrary function $\sigma$ you must have that $\sigma(0)=0$ to try to confirm anything else

[gentle osprey]

datorangeguy:

[clear jackal]

But we're trying to define a function $\sigma :$ range of $d \to [0, \infty)$ such that $$\sigma d : (x,y) \mapsto \sigma(d(x,y))$$ is a metric on M

[gentle osprey]

datorangeguy:

[clear jackal]

MUCH looser than it would seem at first. All sorts of weird mappings work out. You can always define a discretization
$$\sigma(x)= \begin{cases} 0 & x=0 \ 1 & x \neq 0 \end{cases} $$

Which doesn't always yield an equivalent metric, since it maps any metric to the discrete metric

[gentle osprey]

datorangeguy:

[clear jackal]

But in this case it's not continuous at 0 if the domain is an open neighborhood of 0 relative to [0, infty) to begin with

But I'd assume this stuff is already developed nicely on which functions work over which metrics. I'm interested in the case where sigma is strictly increasing

Because it becomes much less arbitrary all of a sudden, a certain aspect of the fundamental geometric structures ends up being preserved

[clear jackal]

And it might not always have to produce an equivalent metric to preserve the geometry, in the discrete case it always does though, which lends nicely to graph theory. But I think maybe normed vector spaces themselves might have some operations that work without being continuous, but I can't be sure yet I'm already stumped in my original question because I don't know what it would be called to research further.

[zinc mortar]

Does anyone know how to do contour integrals for a complex expression from a -1 + i to 1 + i along y=x^2?

Just had an exam and it absolutely cucked me

Only seen 1 example or x^3 but have never seen any others

[small obsidian]

An anti-derivative of x² exists everywhere

So you can just do that
1/3 (1 + i)³ - 1/3 (-1 + i)³

[zinc mortar]

Not sure what you mean. I’m on the tram atm, I’ll write up the question in a few minutes 🙂

[small obsidian]

Kind of like a regular integral, you can just integrate and plug in the endpoints

Oh wait, the path is y = x², not the integrand

Sry misunderstood

[zinc mortar]

Yeah

If the path is along a circle or some line with real and complex coordinates it’s pretty easy

Only seen 1 example being listed along y=x....

Then got it in the exam, pretty toxic tbh

My intuition in the exam was just set γ(t) = t + t^2i from t element of [-1,1]which was similar to the example I saw

[small obsidian]

Parametrize the path.
z = t + t²i
dz = 1 + 2ti dt

Over (-1, 1) yeah

Then sub everything in, you have a real integral of t

[zinc mortar]

Wait shut

I wrote I wrong

Supposed to be x^2

along y = x^2

[small obsidian]

Oh, well same idea

Fixed

[zinc mortar]

hmm

I did that, it jsut got fucking messy lmao

i will try agin now

[small obsidian]

Yeah it might blow up a bit

That cos doesn't look too pretty

[zinc mortar]

na, i sorta bulshitted half an answer to get marks

the cos bit is easy

[small obsidian]

That's cos(t²) which is an impossible integral

[zinc mortar]

because coszdz fits well with cos(y(t))(y'(t)

[small obsidian]

Eeh but also zbar is there messing everything up

[small obsidian]

Oh right

cos has an antiderivative everywhere could have just done that oop

But zbar doesn't, so we needed to parametrize for it

[zinc mortar]

wait so you have to transform it again?

[small obsidian]

Oh no you're good, we did the work

You have to expand and integrate

[zinc mortar]

You reckon that’s the answer?

I checked wolfram

I crunched the numbers correctly

But if assumption is wrong

[sleek thicket]

I'm trying to prove stuff about hypersurfaces

I think I'm able to show that minimal nonzero prime ideals of a UFD are principal

And the ideal of a dimension n-1 irreducible subvariety of A^n is a minimal prime ideal

But why is it true for P^n

I mean that a dimension n-1 irreducible subvariety of P^n is the zero set of a single polynomial

It suffices to show that if Y is a variety and U, V an open cover of Y, and that dim U <= k and dim V <= k, then dim Y <= k

Take an affine open cover of P^n and use the contrapositive

Wait nvm lol

Apparently any nonempty open subset of an irreducible variety has the same dimension as that variety

According to my book

Book says the affine case is trivial, fun stuff

"we may take a chain in U of the form {(point of U) < line < plane < … < A^n} intersect U"

It's not obvious to me that a chain exists where all tje inclusions stay proper

I mean I get that U is BIG

I think wlog I can take U to be a distinguished open, since any open set contains a distinguished open (and dimension is monotone)

So let f(x1,...,xn) be a nonconstant polynomial in n variables. I want a chain of length n of irreducible subsets of D(f)

[gritty widget]

It is formal

It follows from the fact that a nonempty open subset of an irreducible space is dense

The hypothesis that the point belongs to U implies that the intersection of U with each element of the chain is a nonempty open subset of that element

Then if, for example, U inter F_i = U inter F_{i+1}, density imposes F_i = F_{i+1}, which contradicts the fact that it was a chain in the first place

And for the P^n thing, if you have an irreducible hypersurface, then consider its annihilator ideal in K[x_0, ..., x_n]. It is a homogeneous minimal prime ideal of K[x_0, ..., x_n], hence it is principal by the result you mentioned

For the proof of that result (minimal nonzero prime in a UFD => principal), consider a minimal nonzero prime ideal p in a UFD R. Then, there exists a nonzero f in p.
Since p is proper, f is noninvertible, hence it can be written as a product of irreducibles. Since p is a prime ideal, one of these irreducible elements, say g, belongs to p.
Then, we have (0) < (g) <= p. Since R is a UFD, (g) is prime, hence p = (g) by minimality.

Let (X,d) be a metric space and X' its discrete subspace. If X is compact, then is X' countable?

If Im not mistaken discrete space is compact only if its finite right? So X' also has to be finite?

Although I'm reading that 'Each closed subset of a compact space is compact.' so it could be infinite then and bigger than aleph_0?

[gritty widget]

yeah

compact iff finite

note that if you have a finite union of closed sets, then the union is closed

but this may fail to be the case if the union is that of an infinite amount of sets

so you could take for example an infinite compact space (not a problem)

for example the unit circle

there is a bijection between [0, 1) and S^1

so that is bigger than aleph_0

but if you take an infinite discrete subspace of that thing, it won't be compact

and that won't be a problem because that is not closed

lmao i'm an idiot

?

suppose you had an infinite discrete subspace of X

Like I feel like X being metric has something to do with it

then you would have small balls around the points in your subspace

hmm

i don't know

i can't finish any thought today

no worries

is it true that subspace of a compact space is compact? Not always right?

something like taking very small balls around each point in X

Only if subspace is clsoed?

then there would be a finite subcover of that ball thing

but it couldn't exist because the discrete subspace would then be covered by finitely many balls

and yeah

the subspaces of compact spaces need not be compact

Ok yeah I see waht you mean, like, if X' is infinite theres a finite cover of it, but we can take a cover that each open set has only 1 element, but it cant be finite?

(0, 1) is a subspace of [0, 1]

but (0, 1) is not compact

any closed subspace of a compact space is compact

(try proving this, it's a nice exercise)

and if your space is hausdorff, then every compact set is closed

so a subspace of a compact hausdorff space is compact iff closed

Ok I think I understand, thanks

Also, why is [0,1] in lower limit topology not compact? I've seen an answer being that only countable intervals are compact in lower lim top, but I don't understand why. Any ideas? Or maybe other argument to why isn't it compact?

you can be a cunt and cover it like this: [0, 1/2), [1/2, 1/3), [1/3, 1/4), ...

that is an open cover for [0, 1]

but it has no finite subcovers

for if it did, then it would end at some n

no sorry

[0, 1-1/n)

this form

if it had a finite subcover, then you'd get at most [0, 1- 1/n) for some n

and that would be strictly smaller than the set it is supposed to cover

Ahh I see...

But for n>=2? (Just making sure)

because 0 is not open here right?

i guess yeah

start with n=2

for simplicity

idk what [0, 0) would actually be

most likely the empty set

oh yeah lmao I jsut thought that it will be 0

And the argument for [0,1] being compact in euclidean space is that (a,1] and [0,a) for some a are open?

so there wouldn't be that problem of the right side of the interval like in lower lim topo?

nvm, there are better justifications for that, anyways, thanks a lot.

[sleek thicket]

That's not an open cover, it never gets 1

[sweet wing]

Just need to union with [1,2) and thats a cover

[bitter yoke]

On the topic of coverings

Does anyone have a source for information on covering spaces? Things like regular coverings or universal coverings

Also my prof mentioned something about subgroups of the fundamental group being related to regular coverings?

[sleek thicket]

I'm liking the treatment in Lee's topological manifolds

I think Munkres does it too?

If you want to learn it the wrong way look at topology and groupoids, lol

And yeah, I think the exact result is that conjugacy classes of subgroups of the fundamental group correspond to homeomorphism classes of covers of your space (where the homeomorphisms have to respect the covering maps)

I'm missing a ton of conditions there

[floral gust]

What does it mean to identify a homomorphism. Like I know a continuous maps induces a homomorphism between fundamental groups. What does it mean to identify one?

[bitter yoke]

Just write down what the map is

[floral gust]

Considering its 2 marked (which for my prof means 2 paras), I doubt thats what he wants but thanks ):

[gritty widget]

@bitter yoke rotman's alg topo book

Is set ${ f \in C \left[0,1\right] : |f\left(t\right)| \leq 1}$ compact in topology with metric $d_{sup}$?

[gentle osprey]

Godel:

[gritty widget]

$d_{sup} \left(f,g\right) = sup{|f\left(t\right) - g\left(t\right)| : t \in \left[0,1\right]}$

[gentle osprey]

Godel:

[gritty widget]

Should be, but how do I show that for EVERY covering we can find a finite subcovering?

[honest narwhal]

So this is a metric space, meaning compactness = sequential compactness

Can you apply Arzela-Ascoli?

Err wait actually no this is an infinite dimensional Banach space

The unit ball can't be compact

@gritty widget consider the sequence x^n

Show it has no convergent subsequence

[sleek thicket]

Integration is continuous wrt this metric right? But the integrals of continuous functions on [0, 1] grow arbitrarily large

So you have a noncompact image under a continuous function

[valid lava]

Hi!
I was hoping to steal some of everyones favorite videos or online sources discussing reimannian geometry and topological spaces

[light rock]

which of them? @valid lava

those are two separate topics

[sleek thicket]

Let's think about homogenization and projective closure, as friends

[mild venture]

but what if i wanna think of them not as friends :c

[sleek thicket]

No I meant we (myself and anybody lurking in this channel) will be friends, and think about them together

As always, algebraic geometry is the enemy

[mild venture]

thats a funny way to spell laplace

[sleek thicket]

Laplace is the enemy?

[mild venture]

yes

if i had one wish it would be to go back in time and then kick laplace in the shin

like i could stay in the past after idc

[sleek thicket]

Nvm I just realized I can avoid thinking!!!!!

He'll yeah

I can just say "dimension lol"

[mild venture]

[sleek thicket]

wait no

I need to think a little bit about it :(

Is the closure of an irreducible subset irreducible?

I think so

Let Y be a subspace of X and U a nonempty open subset of cl(Y). Then Y intersect U is nonempty, since Y is dense in its closure. But Y intersect U is open in Y, and so it is dense in Y, i.e. it's closure in Y is Y. The closure of U in cl(Y) contains this, and so it is a closed set containing Y, and thus is all of cl(Y)

oh nvm I was getting confused because I thought x^2 + 1 was irreducible

lol

Okay, so why is the homogenization of an irreducible polynomial irreducible?

[dim meadow]

Hmm

[sleek thicket]

Suppose homogenization(f) = gh. Then dehomogenization(homogenization(f)) = dehomogenization(g) dehomogenization(h), so wlog the dehomogenization of g is a constant

so g is a power of x0

[dim meadow]

Oh that's fun

[sleek thicket]

But homogenization only adds in as many powers of x0 as needed

[dim meadow]

Yeah

So gg

[sleek thicket]

Neat

So this let's me just do dimension stuff

[dim meadow]

End all of your proofs with gg

Just saw a talk from kontsevich lel

Was nice

[sleek thicket]

Fun

[dim meadow]

I now kinda care about birational Geometry

[sleek thicket]

Lmao

I saw a talk from Christopher Hacon a while ago

Did not understand anything

It seemed very cool

Iirc it was on birational geometry but I got lost five minutes in

[dim meadow]

I keep not getting lost during talks

It's a weird feeling

Maybe I should go to harder talks

[sleek thicket]

That's very cool and good honestly

Goals

[dim meadow]

I think I just stopped paying attention to detail

[fleet trench]

quick question about the canonical one-form on the cotangent bundle: if we have a manifold $M$ and coordinates $q^i$ on it, then $\dd{q^i}$ should be a basis of one-forms; i.e. a basis of $T^\ast M$. but everything ive looked at says that the canonical one-form on $T^\ast M$, which lives in $T^\ast(T^\ast M)$ is written as $p_i \dd{q}^i$. how does that work?

[fleet trench]

wait, $(q^i, p_i)$ are the coordinates on $T^\ast M$. but since $\dd{q}$ is a one-form, shouldnt it be in $T^\ast M$ too?

[sleek thicket]

no idea but I just finished a milestone in the problem I'm working on. I've shown that any projective hypersurfaces is the homogeneous zero sets of a single polynomial! Now I just need to show they have affine complement (d-uple embed so they it's a hyperplane) and prove that affine+projective=just a point (requires me to read about complete varieties >.>)

[sleek thicket]

Okay, I finished grading so I guess I'll try that first step

Let H = Z(f) be hypersurface in P^n

So f is a homogenous polynomial (say of degree d) in x0, x1,...,xn

I want to show P^n\H is affine

Well, let N = (n + d choose d) - 1 and let σ : P^n -> P^N be the Veronese embedding

f has an expansion f(x) = Σ_{|α| = d} c_α x^α, where α is a multi index and x_α a monomial of degree d

And be definition of the Veronese embedding, σ(Z(f)) = { [x^α | α a multi-index of degree d] : x in P^n, f(x) = 0 }

Which is the intersection of im σ with Z(Σ_{|α| = d} c_α z_α)

The point being that we've turned this weird hypersurface H into the intersection of a Veronese variety in P^N with a hyperplane H = Z(linear equation)

Since f is nonzero, there is some β such that c_β. Then we have a projective change of coordinates which sends z_β to Σ_{|α| = d} c_α z_α. The image of H' under the inverse of this is a hyperplane Z(z_β)

The complement of this is affine (duh) and so the complement of H' is as well

So P^n\H is mapped isomorphically into the closed subset σ(P^n)\H' of P^N\H'. Since the latter thing is affine, σ(P^n)\H' is as well, and so P^n\H is

[sleek thicket]

Then if Y is any closed subset of P^n disjoint from H, it is also a closed subset of P^n\H, and so is both an affine and projective variety

If I knew how complete varieties worked, I would be able to conclude Γ(Y) is f.d. over k, and so it's just a bunch of points

But I don't :(

My definition of what it means for Y to be complete is that it is separated and for any variety X, the projection π_Y : X×Y -> Y is closed

[urban panther]

i have a simple question that will probably be 2-3 lines of proof but i can't seem to make the mental jump : how can i prove that a locally compact hausdorff space is regular? I don't want to consider compactification, which is the answer on mathoverflow to the question. For some reason I'm having a mental block on the other way. It's from munkres' book! Thanks so much in advance!

[light rock]

do it first for compact hausdorff

now to separate a point and a closed set work with an open neighborhood of the point with compact closure

[floral gust]

What would the homotopy look like between different members of F*G

I have this although I am a bit doubtful whether the concatenation should be done over t or s

[light rock]

it doesn't matter which one you pick

[floral gust]

I mean say I pick t_1. Then what would the homotopy conditions look like? Like H(0,t1,t2) = f(t1,t2) ; H(1,t1,t2) = g(t1,t2); H(s, boundary I) = x_0. Correct?

@light rock

[light rock]

yeah

[floral gust]

The lemma requires the existence of a homotopy between alpha and beta but I ended up proving it without. Can anybody point out the flaw?

[floral gust]

<@&286206848099549185>

Or was it just needed to describe the path?

[wicked trout]

Yeah i think you only used the homotopy where you used “h” to denote the path

[next eagle]

Suppose I got some wavy surface defined by some two argument function, how to define distance between points A, B on this surface, assuming I want to "walk from A to B on this surface".

Is there some kind of method similar to rectification on the curve for more dimensions?

[small obsidian]

You're probably going to resort to a line integral

Let me see what I can pull up

[next eagle]

For example:
f(x,y) = cos(x)*sin(y)+1
distance from A(1,2) to B(5,3)

[light rock]

the distance is the inf of length of curves connecting them

[floral gust]

Use the arclength distance after equipping the space with the standard euclidean distance

[light rock]

what do you mean @floral gust

[gentle osprey]

QuickMaffs:

[floral gust]

where y(t),x(t) are parametrization of the curve connecting the two points.

[next eagle]

Thanks @floral gust

[light rock]

yes that's the length of a curve

but what's the standard euclidean distance?

[wanton marsh]

if you have to solve for the shortest curve from A to B that's going to be pretty difficult

[light rock]

there need not exist one

in general

the distance is the inf over all curves

[wanton marsh]

well there is a "distance of a shortest curve"

[light rock]

there doesn't need to be a shortest curve

[wanton marsh]

for well-behaved functions on R² ?

say, C²

[dim meadow]

Take R^2 minus a point as an example of where the inf is never achieved

[light rock]

yeah if they're globally defined you're fine

but otherwise no

[floral gust]

Say I have a continuous function from f:X -> Y. Consider the compactification of X denoted as X'. Does there exist a continuous extension of f ?

[light rock]

no

take f:R->R

identity

compactify the first R

even if Y is already compact, still no

[floral gust]

Oh okay. But why we dont have a continuous extension in this case?

[light rock]

the limit as x->-infty is -infty

the limit as x -> infty is +infty

they're different and neither exists in R in the first place

[floral gust]

Oh I meant that a continuous extension in the sense that the codomain will also change (preferably to the compactification of the previous codomain)

[light rock]

what do you mean by compactification then?

is there some canonical one you're thinking of?

one point compactification of a metric space?

even then, still no

map R -> S^1 = (-1,1]/~ by sending (-infty, infty) to (-1/2, 1/2)

the limits are different and they exist in S^1 already

so when you compactify you dont get a map

[floral gust]

Ah okay thank you thank you

[signal sleet]

Hey my university was slack and didn't offer any differential geometry, but I'm trying to turn a set of coordinates in latitude and longitude (on the earth) into coordinates x & y relative to some central location. Any suggestions on where to look, or ideas? The points are relatively close together so I was just going to assume the earth is locally flat, but I'm not sure on how to do the conversion and I feel like it'd be simple for somebody in here

[floral gust]

Earth is globally flat

[chrome dew]

putting the globe in global

well as long as the points are relatively near each other and you're not too far away from the equator you can just directly use the latitude and longitude as your x and y coordinates directly

I guess it depends, we can get more serious if you have more information about like what sort of things you want to use your map for if it's really that critical

[remote grotto]

Let M be an n-dimensional manifold and let omega be a k-form on M. Let N be a an oriented submanifold of M diffeomorphic to S^k and let i : N->M be the inclusion.

Suppose $$\int_N i^*\omega = 0.$$

[gentle osprey]

ubergoliath:

[remote grotto]

Show that $$\text d\omega = 0.$$

[gentle osprey]

ubergoliath:

[remote grotto]

ok so i think i solved this problem but i cant see where i assumed N to be a k-sphere, could someone let me know if my proof is invalid and where i would have to invoke the fact that N is a k-sphere?

my sol: if the integral of a form vanishes on the manifold then the form must be exact, so $$i^*\omega = \text d\tau$$ for some (k-1)-form tau

[gentle osprey]

ubergoliath:

[remote grotto]

then $$i^*(\text d\omega) = 0,$$ and since i is injective, it follows d omega = 0?

[gentle osprey]

ubergoliath:

[remote grotto]

wait do injective maps between manifolds induce injective pull-backs

i think this part might be wrong

[gritty widget]

So I have this problem which I think I solved but kinda not sure about one thing: Let X = [0,1] x [0,1], Y = [-1,0] x [-1,0] $\subset \mathbb{R}^2. f: \mathbb{R}^2 \to \mathbb{R}^2, f\left(x,y\right) = \left(-x,-y\right)$. Given lexicographical order on X and euclidean topology on Y, a) determine whether or not $f:X\to Y$ or $f:Y \to X$ is continuous. b) same thing but let X and Y have no borders.

[gentle osprey]

Godel:

[gritty widget]

I think for a) none of those are continuous but not sure: I'll send my drawing

First drawing shows that this set is closed in euclidean topology but isn't in lex order topology. On the right this will be open in lex order, but it won't be in euclidean. Is it right?

[dim meadow]

Huh, this is actually interesting

I wanna say that that set on the left is closed in both Topologies

It is, yeah

[gritty widget]

no, their complement in lex is also the upper border

above it

right?

[dim meadow]

Oh okay

You're probably right

[gritty widget]

cause if you take any point from top above it, any neighborhood will interesct the bottom closed set

[dim meadow]

I don't agree with that

[gritty widget]

Why

[dim meadow]

Okayz I agree with it

Sorry

Sure, that set isn't closed in lex

And yeah the second is clear

[gritty widget]

Didn't really think about it yet what happens when there are no borders, but just wanted to make sure about those, thanks

[dim meadow]

Cause the projection map isn't open

[gritty widget]

Wait, sorry, @dim meadow what do you mean by that?

[dim meadow]

Projection maps are open

[gritty widget]

But like projection on Y?

[dim meadow]

On X

You get 2 points

[gritty widget]

nooo

sorry didnt write cleaner, but all the lines between the one I drew are also there

[dim meadow]

Oh

Anyway you can just do a single line segment

Doesn't matter

Also you would project to a closed interval in your picture

So

[gritty widget]

ahh yeah true, like, would be getting the first pic interval kinda?

[dim meadow]

Yeah

Kinda

[gritty widget]

K didn't know projections are open

good to know thx

[dim meadow]

Try proving it lol

It's not very hard

[fleet rapids]

So why is defining a knot as the image of an embedding into the three sphere the same definition if we replace S^3 with R^3?

I heard it being something about S^3 being compact but I don't see how that helps

[light rock]

S1 is compact

[honest narwhal]

Not just being compact, but being the one-point compactification of R^3

[light rock]

and this yeah

[honest narwhal]

Yeah I guess compactness of S^1 is important since you wanna guarantee that an embedding misses a point of S^3

[fleet rapids]

oh I get it now

ok so its analogous to the stereographic projection (where we add a point to C), but instead with R^3

thanks!

can we generalize this to S^n \sim \mathbb{R}^n U ${\infty}$ ?

[honest narwhal]

Yeah

[fleet rapids]

dang Latex, trying to make brackets for set notation

set containing infinity

[honest narwhal]

You need to write \{

[fleet rapids]

thanks! 😄

there we go

[honest narwhal]

Don't use ~

Write \sim

Also write \mathbb{R} for the real numbers

Oh your problem was a bunch of this is outside the tex

[gentle osprey]

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

[honest narwhal]

$S^n \cong \mathbb{R}^n \cup {\infty}$

[gentle osprey]

Daminark:

[fleet rapids]

Should we use the block S for the n-sphere?

or is that nonstandard?

[honest narwhal]

I don't use it myself but it's a thing people do

[light rock]

block S is the sphere spectrum

S^n is fine

[fleet rapids]

I'm sure ive seen block S for the nsphere

[dim meadow]

I use it sometimes

[midnight jewel]

I’ve gone back and forth on it

currently I don’t use it

I find BB S to be annoying to write

[fleet rapids]

ok I have a problem

how is the Alexander horned sphere a 2-sphere

what is the generalized notion of a 2-sphere that we have in mind? Is it just any embedding of $\mathbb{S}^2$ into $\mathbb{E}^3$ ?

[gentle osprey]

Stephen ™:

[fleet rapids]

So if I'm right about that then I guess the JCT holds for any imbedding of $\mathbb{S}^1$ into $\mathbb{E}^2$ but not (in an analogous sense of bounding an n-ball) for 2-spheres as defined above?

[gentle osprey]

Stephen ™:

[floral gust]

Can somebody critique this proof (and if it is correct maybe also explain why the fact it is an isomorphism implies they are homotopic.

[floral gust]

<@&286206848099549185>

[remote grotto]

yo quick question, is the pullback of a local diffeomorphism between smooth manifolds an isomorphism on the exterior powers of cotangent spaces?

[light rock]

locally sure

globally I dont think so

[remote grotto]

oh er, yea but like on each tangent space right

actually here's the real question i have

if i have a k-form omega on R^n, and if there is a k-sphere S such that the pullback of the form to the k-sphere (using the inclusion i:S->R^n) is exact, then does that mean that omega is locally exact?

[light rock]

no, take a radial form

wait

[remote grotto]

oh er

hm

[light rock]

is the radial form closed?

I mean

all forms on R^n are closed

[remote grotto]

wait huh?

top forms?

[light rock]

R^n is contractible

hang on

[remote grotto]

that means all closed forms are exact

[light rock]

all closed forms are exact I mean

yes

ugh

im weird today

so being locally exact is being exact

[remote grotto]

oh right

wait ok this probably doesnt work then

wait the original statement i have to prove is:

the given condition is equivalent to: for every N diffeo to S^k, the pullback i*omega is exact

[light rock]

do you have the de rham theorem available?

[remote grotto]

(afaik)

what does it state?

[light rock]

it identifies de rham cohomology with singular cohomology

[remote grotto]

oh, nope we're "not allowed to use any results concerning singular cohomology"

wait i want to show is that for a k-form omega in R^n, if the pullback to any k-sphere is exact, then the k-form is closed

[light rock]

yes

[remote grotto]

this statement must be true right

[light rock]

it is

[remote grotto]

given that it is a special case of what i have to prove

[light rock]

I can see it immediately from the de rham theorem

but not directly

[remote grotto]

oh hm i see

what i was hoping to take advantage of was that the pullback commutes with exterior derivative

i want to somehow take the pullback jiomega from the k-sphere to R^n by some map j:R^n->k-sphere and relate it to omega

[gritty widget]

How does one tell that a graph is simply connected in a topological sense? Seems to me it would have to be a tree but I cant seem to find any information

[honest narwhal]

So are you familiar with the statement that if you take a CW complex and mod out by a contractible subcomplex, that's a homotopy equivalence?

@gritty widget

[gritty widget]

It sounds familiar

[honest narwhal]

So obviously a tree is contractible

And if you give me a (connected) graph, well there's a maximal subtree

And if you contract that to a point, you'll get a wedge of k circles, where k is the number of edges not in this tree

[gritty widget]

What is the criteria for an edge to be contractible, is it just that it doesnt become a multigraph?

[honest narwhal]

I mean an edge is just [0,1]

[gritty widget]

Ok, but am I allowed to retract loops then if I can have a multigraph

[honest narwhal]

So if you contract an edge, you may get a multigraph, this no longer becomes an operation on graphs but on topological spaces

So there's no problem

The point is that a tree is a contractible topological space, and the maximal subtree T of a graph G is a subcomplex

So we can make sense of the topological space G/T

And it turns out that this is a wedge of k circles where k is the number of edges outside of T

[gritty widget]

Oh right so if I have a 4-cycle

I can get a tree with 3 edges so I get 1 edge outside

[honest narwhal]

And if you crush the 3 edges to a point you get a circle

[gritty widget]

but if I contract edges 1 after 1 I get a point with a handle

yeah I guess a point with a handle is exactly a circle lol

[honest narwhal]

Yup 😛

[gritty widget]

This was very helpful

[honest narwhal]

But yeah the idea is that this is simply connected iff k = 0

i.e. if G is already a tree

[gritty widget]

So if I have a finite topological space does it make sense to make the points vertices and if there exists a continuous function from v to u then there is an edge from v to u?

Straight from v to u I mean not passing through another element

[honest narwhal]

I haven't thought about them this way, I know there's some business with finite spaces and posets

And if you give me a poset you can create its graph

This may or may not be isomorphic to what you describe

[gritty widget]

what I have is 4 elements, which are path connected in a certain way

My idea is to show that they are not simply connected by making a graph

so basically any loops would have to be made from some combination of the paths that are possible so they are the only thing to consider I think?

[velvet dagger]

is it correct that there are 29 topologies on {a,b,c}

[remote grotto]

yes

but only 9 inequivalent ones

wait ugh i actually cant solve htis

i feel like i tried everything

[remote grotto]

help :'(

[lament thistle]

Just ordered Munkres from Amazon, gonna have some fun for the next few months 🙂

[dim meadow]

Lol nah

[remote grotto]

is that "nah" to helping :'(

[light rock]

you could look at the proof of the de rham theorem anyway

maybe you can extract the result

without actually going through singular cohomology

[remote grotto]

what solution did you have in mind with de rham theorem?

[light rock]

it's an element which integrates to zero under all embeddings of S^k

this is precisely a cochain which is 0 on closed chains

so it's 0 in cohomology

[remote grotto]

o

wait

what about this

by stoke's theorem, the integral over the k-sphere is the same as an integral of d omega over the interior (the ball) of that sphere

if a smooth (k+1)-form over R^n is not identically zero, then there exists a (k+1)-plane on which the pull-back of the form is also not identically zero

(i havent proven this yet, but im just brainstorming)

this pullback is a top form on the (k+1)-plane; if it does not identically vanish, that means there is a ball over which it is a top form times a strictly positive or strictly negative smooth function

therefore its integral over that ball should be nonzero

but it vanishes by the given statement?

something like this idk

i think i have enough to figure this out now, thanks

[remote grotto]

ayy this works

[sweet wing]

in munkres top pg 256 when he is showing that for every regular space, every open covering having a refinement that is a countable locally finite covering has a refinement that is locally finite open covering, he seems to assume that this collection of open sets intersecting finitely many elements of a locally finite closed covering(B) is countably locally finite, tho that doesnt seem true, like say B is just a [0,1] in the subspace [0,1) with lowerlimit topology
so how can (3) be used if the starting set isnt countably locally finite
p.s. (3) i believe munkres is referring to for a normal space, a countable locally finite covering have a refinement that is a locally finite closed covering

[sweet wing]

?

empty file uh

[remote grotto]

[sweet wing]

whoos nvm somehow i misread the lemma, it meant 'for a regular space every open covering etc.' so thats already a open covering and by assumption (2) is already true

[remote grotto]

:o

[gritty widget]

So a preorder gives me a directed graph, but does the connectedness of the graph tell me anything about the connectedness of the space

[gritty widget]

What space

Yes an order relation induces a graph but your sentence has no space

[marsh forge]

@honest narwhal the space

[honest narwhal]

T H E S P A C E

@gritty widget it's an inside joke don't worry. And @gritty widget there is a natural way to get a topological space out of a pre-order

Specifically you let U_x = {y≤x}

And call that a basis

[marsh forge]

What’s U_x?

Open ball?

Oh wait nevermind I read that super wrong

[gritty widget]

I just disassociated at the bakery

[little radish]

Quick dumb question, is a hilbert space necessarily linear?

[gritty widget]

What does linear mean

Yes it's a vector space

Yes the norm comes from an inner product

How can the norm come from an inner product if the space isnt linear to define an inner product

[gritty widget]

Ok so if I have a specialization order then the corresponding graph has the same connectivity as the space?

[little radish]

Thank you levy

yeah that makes sense

[digital nova]

Quick question about immersions and submersions

I know that a composition of immersions is an immersion, same with submersions

Let's say we had an immersion from R^10 to R^20, then a submersion from R^20 to R^5. Can I say that the composition R^10 to R^5 is a submersion?

[remote grotto]

@digital nova i dont think so

because the composition of the push forwards might not be surjective

like consider the inclusion map from R^10 -> R^20: (x1,...,x10) -> (x1,...x10,0,...,0)

[remote grotto]

followed by the map R^20 -> R^5 that projects to the last 5 entries

being linear, the derivatives can be identified with the maps themselves

but the composition is not surjective and so is not a submersion

[digital nova]

thanks!

[remote grotto]

np

sorry it took me so long to finish saying that sentence

[digital nova]

oh no problem, I had to catch the bus so didn't see it until recently

[remote grotto]

got distracted lol

ah ok

[gritty widget]

i am having a hard time to grasp the definition of a topological space. It is just a set of subsets closed under union and intersection. I mean what is the motivation for that? I can't make any sense out of it.

[bitter yoke]

Think about analysis and R

And open balls in R

[floral gust]

Is the topology on S^0 the discrete topology in 0th homotopy group?

@gritty widget A nice way to think about it is what points are close and what points are not close and just let go of any notion of "closer" (which is made rigour in metric spaces). If y is close to x, we say that y lies in the neighbourhood of x. By considering points that live close to x, we develop a neighbourhood of x. Now lets see why would we want their collection to be closed under infinite union and finite intersection.
Well if I have a bunch of collection of points that are close to x then the union of all of them is also close to x. Thus, the infinite union is also a neighbourhood.
Now lets see why it falls for intersections. If I have a bunch of points that are close to x, and bunch of points that are close to x, then as long as the intersection is non-empty, the intersections contain points that are close to x (if the intersection is empty, we get the empty set which is also considered to be a neighbourhood, maybe just formally, but I dont have an argument for that rn). Thus we see the finite intersection is a neighbourhood.
But if you continue this infinitely, there can be that you would just end up with the point itself and it itself is not necessarily its neighbourhood. Thus infinite intersection are not always open.
Not 100% sure if all this is consistent, but I use this to intuit the concept.

[warm hedge]

when a set is open and close at the same time ?

[marsh forge]

Clopen sets are connected components

Well unions of connected components but we can just think about it as being “generated” by connected components

[remote grotto]

@gritty widget yea, it might seem a little arbitrary at first, because for example, open sets in R^n have lots of properties, so why specifically those? i think the book "modern differential geometry for physicists" by chris j isham gives pretty good motivation for the definition of a topological space based on even more fundamental ideas about the "neighborhood of a point"

essentially the point of a topological space is to give a way to compare qualitatively e.g. how far A is from B as compared to how far A is from C

a "metric space" is one concrete realization of this idea, but in isham's book, he attempts to motivate that topological spaces are sort of the most general notion you can use to make such comparisons

(note that in order to compare how far points are from one another, you dont actually need to know the distance between them, just like you dont need to know the number of atoms in the universe and the number of atoms in the earth to make the comparison that the number of atoms in the earth is smaller)

and perhaps what lies at the crux of all of this is that we would often like to talk about things like convergence of sequences and continuity in more general settings than R^n

but in order to make these definitions, we never really need to know how far things are from one another

in the usual definition of convergence in R^n, for example, "for every epsilon > 0, there exists a natural number N such that |x_n - x| < epsilon for all n > N"

the only reason we calculate the distance between x_n and x is to compare it to epsilon

but in the most general situation, we might not actually care about what that distance is

just comparisons between those distances

[midnight jewel]

Algtopo question time!
In this exercise, we assume we have to use Mayer-Vietoris to compute H(C) in terms of H(X) and H(Y). The issue is how to actually split C into two spaces.
Our first idea was to take just the cone over Y (including Y) as one set, and X as the other set, so that Y is their intersection; but we noticed that the interiors of those sets would be X \ cl(Y) and the cone over the interior of Y (relative to X); these interiors don’t cover C so the assumptions of Mayer-Vietoris aren’t satisfied

also even if it still worked we can’t really see how that helps because without knowing anything about H(X) and H(Y) I can’t tell a whole lot from that exact sequence

[marsh forge]

Put A = X and B = CY with intersection homotopy equivalent to Y

Like A is “X and a little bit of the cone (contractible to X)” and B is the cone up to that same overlap

[floral gust]

I am not sure how the bijectivity holds true for n>1. Intuitively, pi_0 tells you the number of connected components. Thus bijectivity means that the number of connected components of X and X^n are same for all n>0. But this is simply false demonstrated by following counterexample:

[marsh forge]

You last cell addition doesn’t make much sense

You have to glue D^2 to X^1

But can you describe your gluing map to me for your X^2?

(The idea for this proof is that X^0 establishes a bunch of points that could be connected and X^1 fully determines which are connected)

Then you have to glue higher cells along the existing skeletons, which doesn’t let you connect or add anything

Add any new connected components*

[rapid ridge]

When you say gluing, do you mean the whole boundary "must" be mapped to some point in the previous skeleton?

[marsh forge]

The fully rigorous way of saying it is

Given X^i we build X^{i+1} by pushouts on D^{i+1} <- S^{i} -> X^i

So yes, you have to define a map S^i into X^i

And then “glue” the disk along that

[floral gust]

Can you answer why this doesnt constitute as a counterexample, where the boundary of each disc is mapped to a point in the previous skeleton?

@marsh forge

[marsh forge]

Uh

What the fuck is it

Lol

The first diagram was clear enough but i really have no idea

What thats supposed to be

[floral gust]

Nvm the attachment for green wouldnt be continuous.

Lemme just try the proof then ):

[marsh forge]

I think you are misunderstanding the construction if a CW complex

[floral gust]

I tried to attach disc of increasing dimensions at every step so as to connect the two lines each time by sending the boundary to a point in the previous skeleton

[marsh forge]

You have to glue the full boundary

You have to send the full boundary to a point

Or i mean you can do other stuff

But you cant just glue one point of the boundary to a point and call it a day

[floral gust]

Yep yep, I was sending the full boudary to a point using the constant map. But it wouldnt be continuous for the green disc since it some point of the boundary must get mapped to the left blue line and some to the previous purple circle

[marsh forge]

Not sure I get what you’re saying

Here’s a quick sketch for you

The only cell with disconnected boundary is the 1-cell

If you use a continuous function

You send connected things to connected things

So a 1-cell boundary gluing can connect things

Because the boundary could a prior be disconnected

But when you glue higher cells, you have to glue all of their boundary to something already connected

So in particular they can’t “add” connections

[floral gust]

Ohhh, I was trying to argue that given two points in X^(n), if they are connected in X then theyre connected in X^(n), and thus the number of connected components is the same. But the proof seems to be turning very complicated, because to build a path in X^(n) seemed very difficult

But your proof seems very clear. Thanks!

[marsh forge]

Also important

Path connected =/= connected

[floral gust]

Yep yep that I know

[marsh forge]

All reasonable CW complexes are path connected (and they are all obviously locally path connected)

[floral gust]

Over here to apply Van Kampen, they need to make sure U is open. But U being a disc is not open. Am I missing something here? (If the suggestion is U is a disc with boundary removed then it is not homeomorphic to S^{n-1})

[marsh forge]

Also more generally

As long as you’re doing “reasonable” things

In practice we ignore most of the technicals of taking open vs closed neighborhoods

Esp in the CW setting

[light rock]

yeah technically you have to pick an open set that's slightly bigger than the closed one you want

but you can always pick a homotopy equivalent one

even one that deformation retracts to your closed thing

[marsh forge]

I mean “always” is like kinda sketchy but yes this is the idea

[light rock]

yeah it's in that weird point where they're too useful not to mention but too unimportant and technical that they have a dumb name

[marsh forge]

May et al refer to them as NDR pairs

Actually NDR pair might be stronger

Anyway they are all technical any no one sane ever does anything that aren’t these

[floral gust]

In an exact sequence what is the map 0 -> A ? Is this f(0) = 0_A where 0_A is the identity element of A?

[rugged swan]

You have only one morphism from 0 to A

So it's inambigous

The map is effectively the trivial map

[dim meadow]

Lmao

[remote grotto]

homologous

[proven raven]

I dunno if this is the right place to ask but What do the hyperbolic functions have to do with the regular trig functions? Do they just share some properties?

I just had a lecture on them and it just went over the basic definition

[light rock]

it's the same but for the hyperbola

so under a change of coordinates they're the same functions

[midnight jewel]

I’d say the main relation between them comes from complex analysis

the hyperbolic ones are really just the regular ones along the imaginary axis

(up to like a factor of i)

I think it was sinh(x) = i*sin(ix) and sth similar for cos?

[light rock]

yeah that's the change of coordinates

[midnight jewel]

and because of that they:

1. share a ton of properties
2. actually have a similar geometric interpretation too (replace unit circle x²+y² = 1 with unit hyperbola x²-y² = 1 and make analogous definitions)

but yea it’s more complex analysis than topology&geometry I’d say

though they do show up in geometry ofc

[marsh forge]

Complex analysis is basically topology

[midnight jewel]

that seems a bit reductive

[marsh forge]

All things are basically topology

[light rock]

surely multiplying by i isn't complex analysis?

[dim meadow]

I think it's a little nicer than just a change of coordinates lol

(although it is exactly a change of coordinates)

Hyperbolic Geometry is very nice

[next eagle]

Hi, maybe is here anyone who can give me advice how to understand general theory of relativity. I mean, what mathematics background should I know to learn this. I read that differential geometry, pseudoriemann manifold, metric tensor are necessary, but is it all? I will be grateful if someone list me whole relevant (or at least a big part) mathematician stuff for general relativity.

[light rock]

what's your objective?

like what do you wanna do with it?

[next eagle]

I'm just curious about that.

[light rock]

you should look for books for physicists

cuz mathematicians do the theory in a much deeper way

get a book on general relativity

see what the prereqs are

and go from there

[sweet wing]

https://physics.stackexchange.com/questions/15002/mathematically-oriented-treatment-of-general-relativity
similar thread

[chrome dew]

woah

Ron Maimon's answer is awesome

basically explains why computer science proves physicist's index notation is superior to math notation

[marsh forge]

...

[light rock]

what?

the fuck?

[sweet wing]

...

[next eagle]

Can we define curve given by some f: R -> R as a 1-dimensional Riemann's manifold?

[floral gust]

basically explains why computer science proves physicist's index notation is superior to math notation
So whats next? "Tensor is something that behaves as a tensor" > Mathematical description fo tensor?

[light rock]

@next eagle you need to put a metric on it

but there's a natural one to put on the graph of a function

[next eagle]

@light rock arclength as a metric?

[light rock]

yeah

[rugged quiver]

Is it true that pullback of connection on vector bundle is contravariant? I mean $(f\circ g)^*\nabla=g^f^\nabla$

[gentle osprey]

Gigrise:

[random slate]

Here, have a fun thing from a project I'm working on

[chrome dew]

all parabolas are ellipses now? I can't keep up with this server meme

[light rock]

it's well known that all conics are the same under change of coordinates

idk if conic sections is appropriate for the >advanced mathematics section

@rugged quiver sanity check the domain and range of each f*, g*

and you'll see

[random slate]

Well, it's from projective geometry

[dim meadow]

It's a nice picture tbh

[random slate]

It's for my final project for my Projective Geometry class.

I'm basically going over the cross ratio, harmonic sequences, how they help you create a coordinate grid in perspective, then looking at what curves look like on that grid, going into homogeneous coordinates and how to find where a curve intersects the line at infinity, and ending with Bezout's theorem.

In the form of a Youtube video.

Here was what the construction of the perspective grid looked like.

[rugged quiver]

@light rock yeah actually it has to be contravariant ! ty

[fleet rapids]

so what determines where things intersect the line at infinity?

[random slate]

I could have sworn "projective geometry" used to be in the channel description before somebody turned it into "Liquid Echo Chamber".

@fleet rapids You basically have to turn your polynomial equation in x,y into a homogeneous equation in x,y,z, then set z=0 and see what possible (x,y) pairs work

So y = x² becomes yz = x² (you multiply each term by enough z's so that each term has the same degree in this case 2)

Set z=0 and now you have the equation y*0 = x², so y can be anything but x must be zero. So, the parabola y = x² intersects the line at infinity at the point at infinity in the direction of (0,1)

[fleet rapids]

hmm ok that's neat

[random slate]

It really is

[fleet rapids]

I have to do more projective geo at some point

[random slate]

There's more reason behind it than that obviously

I can send you the video once I'm done

[fleet rapids]

that would be appreciated, ty

[random slate]

Mind if I PM you so I remember

[fleet rapids]

sure

I've seen projective geometry also come up in my current area of focus too (algebraic topology) so all the more reason to study it

[chrome dew]

You mentioned you're going over the cross ratio, what do you use that for?

I know it's invariant under mobius transformations, and I think has some sort of geometric interpretation, but I'm unaware

[random slate]

So the cross ratio is invariant under projective transformations yeah

And if the cross-ratio comes out to -1 in particular, the four points form what's called a harmonic set

A particularly useful example is that if three points are equally spaced, then they form a harmonic set with the point at infinity

So what that means is that if a point in your picture is supposed to represent the point at infinity, then a harmonic set will be what equally-spaced points look like "in perspective"

So that's how to accomplish realistic foreshortening

[fleet rapids]

The context is homotopy theory in the picture below, guessing the bar means its an adjunction space right?

Ok I think I got it now

[marsh forge]

Huh

Its a quotirnt

Yeah thats the odea

[fleet rapids]

Where all of S^1x1 is identified right?

[marsh forge]

Yeah

[fleet rapids]

Ty!

[midnight jewel]

yea, though graphically I would say the bottom of that cylinder is 0 and the top 1 ^^ not that it matters

just more conventional

[fleet rapids]

Ah ok good to know

[marsh forge]

I dont really think theres a convention to drawing pictures loo

[floral gust]

My professor introduced us to a theorem that says if I take a homotopy class in a space X, I can find a representative whose image is entirely contain in X^n where X^n is a skeleton of X. Any idea what it is called? Or how can I look it up for more info?

[marsh forge]

Uh