point-set-topology

[pastel linden]

maybe I should take a proper course on topology before I make broad statements about how stupid this text is

[gritty widget]

my coffee is done, let me chug it and then write my thoughts

latter

ah if $i : \bR^m \to \bR^n$ is the inclusion i described and $f : \bR^n \to \bR^m$ is a homeomorphism, then $i \circ f : \bR^n \to \bR^n$ is an injective continuous map (being a composition of those), but it's certainly not open

[gentle osprey]

TTerra:

[gritty widget]

i know spivak wrote calc on manifolds when he was young

unsure about diffgeo series

[pastel linden]

I assumed he learned how to write by the time he wrote calculus

[gritty widget]

the calculus book is genuinely well written

iirc

no "stokes' theorem is trivial" going on in that one

[pastel linden]

alright here we go

never done anything with spivak and this already feels typical of him

[gritty widget]

not open = closed

[pastel linden]

shit did I write closed

let me fix that

corrected

[gritty widget]

ill check over it after i check my complex pset

slim you're up

[gentle osprey]

slimvesus:

[pastel linden]

oh shit, I think I meant to take the inverse of the homeomorphism

because then the inverse is injective and continuous and R^m is open, so then i[R^n] should be open

isn't $(\Psi \circ i^{-1})^{-1} : \mathbb{R}^m \to i[\mathbb{R}^n]$ valid? It's a continuous injection from $\mathbb{R}^m \subseteq \mathbb{R}^m$ into $i[\mathbb{R}^n]$ such that the image isn't open, which is a contradiction

[gentle osprey]

bacono:

[pastel linden]

it probably is, I just took the inverse of what I had before because I realized my mistake

sorry if I've been confusing you, I appreciate the help

[gritty widget]

~~bacono i wrote a proof some ways up if you wanna check 😉 ~~

[pastel linden]

yeah that works too, I think we had the same thing but yours is simpler in that you applied invariance in R^n instead of R^m

[nimble cipher]

Is this proof correct?

[quiet pilot]

Open in A does not imply open in X

[nimble cipher]

So I start from the open cover of X?

Then from then I just take the open cover of A by intersecting the open cover with A

[coarse kestrel]

No you go backward

Starting with an open cover of A, for each set U in A, there exists open set U’ in X such that U=A intersect U’

By definition

[nimble cipher]

Then I can build the open cover for X as these open sets U' right?

then I take a countable subcover. So when I intersect them with A, they are a countable subcover of A

[coarse kestrel]

Yes

[nimble cipher]

Thank you

[tight agate]

Does anyone have an intuitive explanation of what a flattening stratification does?

an equivalent thing in topology land might also help

[digital peak]

how does one prove that closure(interior(-)) is idempotent?

[sweet wing]

prob can use int(X) subset X subset closure(X) a few times to prove c(i(c(i(x)))) subset c(i(x)) and c(i(x)) subset c(i(c(i(x)))

[digital peak]

oh that's clever

$$iciX \subset ciX$$
$$ciciX \subset cciX = ciX$$
$$ciX \supset iX$$
$$iciX \supset iiX = iX$$
$$ciciX \supset ciX$$

[gentle osprey]

mniip:

[sweet wing]

hence conclude there are like 14 sets you can make from using closure and complement

[digital peak]

yeah that's what I was trying to do

except in coq

this probably has terrific generalizations

[gentle osprey]

Divya Ranjan:

Divya Ranjan:

[feral dragon]

Okay, so finally, 😰 Does this proof make any sense ?

[little hemlock]

looks good

[gritty widget]

i like the excited "contradiction!" at the end

I like to keep my math assignments not so dry too

[sweet wing]

i prefer presenting refined sols ngl feels neater

tho it looks super emotionless

[zealous glen]

just looking at which spheres can be quipped with a structure of a lie group, so far i have the trivial case of dim=0.. the unit circle case on the complex numbers with length 1, so dimension 1 but having a hard time showing for dim=3

anybody got any ideas?

and concluding why only these dimensions can be equipped with a Lie Group structure?

[zealous glen]

@gritty widget could you help here? I know you're good at topology 🙂

[tight agate]

it isnt a very easy problem to solve

you can conclude S2 isnt a Lie group if you know the hairy ball theorem

[zealous glen]

my main problem is finding the algebraic structure for the lie group of dim=3

for this

[tight agate]

do you have a guess?

[zealous glen]

nope, so any hint would be good :x

[tight agate]

it's a matrix group

or another hint would be think quaternions

[zealous glen]

number system of quaternions?

[tight agate]

yeah the quaternion algebra

[zealous glen]

Ah kk

[tight agate]

you can think of the quaternions as R4 with a special multiplication

[zealous glen]

and proving that these dimensions are the only dimensions is

[tight agate]

that's not very easy

at least the proof ive seen

[zealous glen]

any theorems off the top of ur head

that I can just look up without looking up the proof

[tight agate]

do you know about lie algebras?

[zealous glen]

ye

well

[tight agate]

and their connection with simply connected lie groups?

[zealous glen]

im studying it

yes

[tight agate]

oh then you might be able to do it yourself

the argument did involve a wee trick tho

[zealous glen]

mhm

[tight agate]

if you've never seen that sort of thing before

[zealous glen]

does this involve Hopf's Theorem?

[tight agate]

do you mean hopf invariant one?

[zealous glen]

yes i think so

homotopy invariant

trying to find a way to prove that the even dimensions of a sphere cannot be a lie group

but i iz stuck

[tight agate]

the idea to do it in general is show that you have a nonzero 3 form on the lie group

[zealous glen]

O

[tight agate]

for even dim spheres you can do it using hairy ball

[zealous glen]

i can do if the lie group is abelian or not

whats hairy ball

Oh

hedgehog theorem

kk

[tight agate]

no nonvanishing vector field on 2n sphere

[zealous glen]

yeah we have a different name of it here

[tight agate]

yeah that implies that tangent bundle is not parallelizable

[zealous glen]

right right

[tight agate]

but a lie group is parallelizable

so that gets rid of the 2n case

finding the 3-form is where the trick is

because once you get it then you know the 3rd cohomology is non-trivial

which implies that it has to be a 3-sphere

[zealous glen]

well we can just consider some map a(x,y,z)=<[x,y],z>

and changes sign swapping any of these values

so a is a 3-form

[tight agate]

noice

you have the right idea

[zealous glen]

and assuming this is nonabelian then we find some stuff that dont commute

mhm maybe i can do this, but i think im ganna be stuck on the rham cohomolgy

[tight agate]

wut why lol

you just did the hard bit

[zealous glen]

oh kek

[tight agate]

do you know the de rham cohomology of spheres?

[zealous glen]

not very well

[tight agate]

no i mean the cohomology groups

anyway, if you don't know the groups then they are not very hard to compute

[zealous glen]

sorry nope

i was just looking at the left-invariant apart

and trying to equate it to the right

of the diff-3-form

yeah im way too tired for this rn, hehe 6am ... so will look at it in the morning :x

but thank you!

[tight agate]

aight, it's a reasonably fun exercise to compute the cohomology of spheres

if youve never seen it before

[zealous glen]

mhm ok, ill look into it

im trying to complete all these problem sheets

and running out of time, but hopefully tomorrow if i finish some QFT stuff

[tight agate]

QFT eh

fancy

[zealous glen]

cry

[tight agate]

i will stick to reading about hairy balls

[zealous glen]

Haha

we can always swap

you do my qft assignments

ill dive into the hairy balls

[tight agate]

i know absolutely nothing about qft

[zealous glen]

yeah more than me bro

[tight agate]

you'll get a 1-sphere as your score

i hope to read about it sometime tho

[zealous glen]

cheeky

[tight agate]

they keep mentioning physics stuff in my diff geo class

but i have no idea on what theyre talking about

[zealous glen]

mhm like what?

[tight agate]

yang mills stuff

[zealous glen]

i mean QFT ⊂ Lie Groups ⊂ Topology ⊂ Diff Geom etc

[tight agate]

how is topology included within diff geom

[zealous glen]

tangent spaces?

and all that fancy stuff

oh

mbe otherway round

[tight agate]

geometry is sorta topology + extra structure right

yeah it should be the other way

[zealous glen]

yeah im tired

af

[tight agate]

do you know any yang mills stuff

[zealous glen]

is that shiz on feynman diagrams

because i think thats in the next 2 weeks

Lecture 08: Feynman Diagrams for the Yang-Mills field

yeah i have that in 2 weeks

[tight agate]

i dont think so

theyre probably related

but that's not what they were talking about in my class

[zealous glen]

semisimple lie algebra and an abelian lie algebra with compactess

mhm apparently is YM theory

which its based on SU(N)

mhm it seems quite interesting actually

[tight agate]

I have heard the word SU(N) guage theory thrown around, yes

but in the lecture they were looking at this functional

[zealous glen]

describing the behaviour of elementary patricles using non-abelian Lie groups and is the core unification of the EM force and Weak forces

[tight agate]

given by taking the inner product of the curvature with itself

[zealous glen]

mhm

[tight agate]

and integrate

(with the volume form)

and then they started looking at critical points of that

[zealous glen]

idk why

[tight agate]

and finally concluded that self adjoint connections are solutions to whatever

yeah idk

the whole thing was similar to Chern-Simons theory stuff

where the critical points of the functional turn out to be flat connections

[zealous glen]

mhm ok

interesting

i havent read about chern-simons theory

or mbe i have an just cant remember

[tight agate]

say you have a complex manifold

and a trivial bundle E over it

[zealous glen]

yeah

[tight agate]

then d (exterior derivative) is a connection

[zealous glen]

if i remember does this have some intergral with wedge product?

[tight agate]

look at the one paramter family d + tA

where A is a End(E) valued 1 -form

yeah and then the transgression is the chern simons form

A wedge dA + 2/3 A wedge A wedge A

I assume that's what youre talking about

[zealous glen]

yesyes

[tight agate]

yeah and you integrate that

[zealous glen]

oooo

[tight agate]

and then look at e^-k of the thing you get

[zealous glen]

so you take some path intergral over all SU(2) connections on the manifold

[tight agate]

yup yup

[zealous glen]

kewl

[tight agate]

and the critical points of the functional are flat connections

but yeah in general it's a path integral over End(E) valued 1 forms

the space of connections is an affine space over that

[zealous glen]

does this have properties of ricci flatness

or am i going in the wrong direction

[tight agate]

idk

by flat I just mean delta^2 = 0

[zealous glen]

the curvature vanishing right

[tight agate]

yup

[zealous glen]

Moduli space of flat connections on a riemann surface is naturally a complex manifold

so maybe it does exhibit some properties of ricci flatness

?

who knows

do a phd paper about it

yoour welcome

OOHH

maybe this comes into play with calabi-yau manifolds

[tight agate]

holy crap it's the jacobian

never thought of that before

flat U(1) connections

on a riemann surface

[zealous glen]

yeye

[tight agate]

if the mathoverflow post im reading is correct

[zealous glen]

link me broskarian

[tight agate]

https://mathoverflow.net/questions/313889/moduli-space-of-flat-connections-over-a-riemann-surface

[zealous glen]

mhmm

[tight agate]

you have supplied me with the hype I need to get through next week

[zealous glen]

ha! no problem

now i go for a smoke and I shleep or mbe some maff

[tight agate]

I need to write a history essay lmao fml

[zealous glen]

something cool to look at: The complex tangent bundle of the Riemann sphere

then corellate this with chern class

then calabi-yau manifolds

some cool stuff all coming in to play

wait wut history?

[tight agate]

im doing a history class

[zealous glen]

how u do history and maff

or its the history of math

[tight agate]

we need to do a bunch of 'general education' classes

[zealous glen]

:fivehead:

are u in uni?

[tight agate]

ye

[zealous glen]

"This is America"

[tight agate]

quite a lot of general ed classes actually

[zealous glen]

ah im from the UK we just study what we want

[tight agate]

sounds fun lol

[zealous glen]

mhm yesyes

oki i need a smoke but gl with the essay 😉

[tight agate]

cya

[zealous glen]

ciao bella

[feral dragon]

i like the excited "contradiction!" at the end
@gritty widget Well if you're someone like a logician that's the only thing in a proof to excite you

[burnt spruce]

Which scheme should be called 'The Ponzi Scheme'?

[meager python]

The scheme without closed points

[hard wind]

To show a space is connected, does it suffice to find a contradiction from supposing there is a pair of disjoint, non-empty, open subsets that cover the space, or must I suppose more generally there exist a collection of disjoint, non-empty, open subsets that cover the space?

I would think it does, since if I had A, B, C disjoint open subsets of a space X that covered X, B union C is itself open, so A and B U C do the trick.

[tight agate]

yup a pair works

a space is disconnected iff it is the union of two disjoint open sets

nonempty

[feral dragon]

How does my first drawing of Klein Bottle look like ?

[sweet wing]

art major chad

now help me draw H^3/<z->z+1,z->1/z>

[feral dragon]

haha I can be pretty decent at pencil drawing sometimes

now help me draw H^3/<z->z+1,z->1/z>
@sweet wing now help me understand what that creature is, first

[sweet wing]

it may just be a cusped torus

with a funny metric

[night parrot]

does anyone have any good resources for learning more about discrete differential geometry? i have a pdf from keenan crane's course that looks pretty good, but wondering if there are other books out there

[quiet pilot]

What's the main use of topology in algebraic geometry? I know we introduce the Zariski topology, but do we use any theorems from topology that helps us study it?

[tough imp]

It is t very heavy

If you know basic point-set, that’s most of what you end up using most of the time. Other stuff about codimension and stuff are sort of topological I guess but you wouldn’t talk about it in a general topology course

From what I gather AG doesn’t really need all that much of topology other than the notion of open covers since eventually you’ll abstract away from even topological spaces into Grothendieck topologies and stuff which mainly only retains the notion of open covers and the property of continuous maps we care about is that they pullback an open cover to an open cover

This isn’t to say topology plays no role, my professor published a paper recently which shows for some class of varieties the topological space actually determines the variety up to isomorphism which is pretty interesting

[quiet pilot]

Ah I see, I know some basic point-set and algebraic topology

The course I'm taking is called "Intro to algebraic geometry" and seems to only cover classical algebraic geometry so I don't think we'll cover schemes

[tough imp]

Oh okay, yeah you should be fine

You’ll have to learn new stuff since there’s topological properties you won’t cover in a normal point set class since they’re super weird

Things like an irreducible topological space, Noetherian topological space, etc

But that’s really easy to pick up

[quiet pilot]

Ah I heard about the first one in my commutative algebra course at least

Thanks for the answer, and thanks for the help with commutative algebra btw. Wrote the exam yesterday

[tough imp]

Uhhh, I don’t actually remember exactly what it was I helped you with but I’m glad I could help 😂😂

[quiet pilot]

Classification of prime ideals in Z[X], and a few other things 🙂

[tough imp]

Ahh, no worries haha

I worked through a classification of maximal ideals in Z[x] with someone once but I realize I made a slight error with it 😂

So if that was with you my bad OOF

[quiet pilot]

Yeah that was probably with me. The explanation made me look up a lot of things about ideals and quotients though and we covered it more extensively than the book did on a later lecture anyway

[tough imp]

Ah, okay. The part where I failed was showing that the maximal ideal has to have a prime number p. I think I said to intersect with Z, but you have to rule out the case that’s empty and in order to do that you have to extend the ideal to Q[x] I believe.

[quiet pilot]

Yeah, we did that and then used Gauss' lemma to show the ideal was principal and generated by the primitive polynomial of the polynomial with least degree in that case.

A nice proof now that I understand it haha

[tough imp]

Ah nice. I don’t remember if I actually did do that lmao

[delicate hollow]

What is definition of a separation? I can't find rigorous definition online and making my own doesnt seem to work well.
So far I have, two sets A,B in M are separated iff
1.A U B = M
2. A intersect B = {}
3. A,B Nonempty.

But I can think of examples that counter this.

[tough imp]

I think this is commonly referred to as a disconnection, and we require A and B open

This doesn’t always exist, and if one doesn’t exist then we call that space connected

[delicate hollow]

oh A and B have to be open and I think the fixes the problem

before I was having a problem where I was giving two sets that were touching

[ivory dragon]

yep, if you require A, B open this is precisely the usual definition of a disconnection

[delicate hollow]

🙂

[ivory dragon]

we say M is disconnected if there exist such A, B open [with AUB = M, A and B nonempty and disjoint]

and otherwise M is connected

[delicate hollow]

proving connected seems hard

[ivory dragon]

generally speaking, yes

[tough imp]

Generally you show it can’t be disconnected

[ivory dragon]

fortunately there is a useful theorem:

the image of a continuous function on a connected set is connected.

[delicate hollow]

yes

[tough imp]

I think like all my connectedness arguments are by contradiction.

[delicate hollow]

I was going to ask that question

[tough imp]

It’s just much easier to attack that way IMO

[ivory dragon]

yeah, certainly if you want to do it "directly" from the definition the usual way is

"suppose we have a disconnection {U, V} for this set S. then consider this other set T that we know is connected. then we can construct new sets out of {U, V} by... and this is a disconnection of T because... but that is a contradiction."

in some cases this can be done very "directly"

for example, proving the closure of a connected set is connected

[delicate hollow]

1. Say you want to prove there is no homeomorphism between two sets A,B then is it enough to show that A-S is connected and B-S is disconnected, for some subset S in A and B .
2. Is homeomorphic an equivalence relation?
   Two questions ive been thinking about

[ivory dragon]

homeomorphisms preserve connected components, yes

[tough imp]

Uh, you have to be careful right?

If you show this over all such S then you’re good

[ivory dragon]

and yes, homeomorphisms are an equivalence relation; see https://proofwiki.org/wiki/Homeomorphism_Relation_is_Equivalence

oh yeah good point @tough imp

homeomorphisms preserve number of connected components

so you'd have to show it for all such S

[tough imp]

You can like, assume the S’s in each set have the sane cardinality

[ivory dragon]

(i.e. that A has one connected component, while B has >1)

[tough imp]

But besides that it’s pretty general

[cursive glade]

You can extract the number of connected components from betti numbers of the space

[tough imp]

I’m trying to understand the characterization of morphisms into P^n from a scheme X as being equivalent to an invertible sheaf L on X and a set of n + 1 sections which generate L

I’m following Hartshorne, and on page 150 he says given a map phi: X -> P^n we get these sections s_i = phi^*(x_i) which generate phi^*(O(1))

I don’t understand what phi^*(x_i) is

For a second I thought it could be anything in the fiber of x_i... but that doesn’t even make sense!

[meager python]

Why doesnt that make sense?

[tough imp]

Umm, I think the ring map is going the other way actually

So these are the images of the x_i under the ring map on global sections?

I realized that given an invertible sheaf and sections which generate it you have to make a map into P^n and given the construction in Hartshorne this at least seems to be what’s going on?

[meager python]

Well if you have X into P^n then you have a map of sheaves

And x_i are elements of that sheaf

[tough imp]

Right, but this actually goes the other way

I thought it would be thinfs in the fiber

Aka stuff mapping to x_i

But the map goes the other direction so actually it should be the image of the x_i I think

I think this actually makes sense. Geometrically if we think of these as homogenous coordinates we want to define stuff by just saying “this is where this coordinate goes” which is what this does. But if you look at the actually construction of the map, since a homogenous coordinate isn’t well-defined, only ratios of them we define stuff by sending x_i/x_j to s_i/s_j

Which is really just the proper way to say we send x_i to s_i

Or something

[nimble cipher]

Is ${(p,q):p,q\in\mathbb{Q}}\cup{(p,q)\setminus K:p,q\in\mathbb{Q}}$ where $K={\frac{1}{n}:n\in\mathbb{N}}$, a countable basis for $\mathbb{R}_K$?

[gentle osprey]

emphatic_wax:

[digital peak]

So consider $[0,1]^{[0,1]}$ in product topology, what's a subset that's sequentially compact but not compact?

[gentle osprey]

mniip:

[digital peak]

I could probably construct something isomorphic to the long ray, but is there possibly a lower tech example?

actually now I doubt the ability to construct a long ray here...

[digital peak]

maybe if I took one of the connected components

nope not seq compact

[digital peak]

Let $\phi : [0, 1] \to \omega_1$ be a bijection, then
$$A := \left{ f \in [0, 1]^{[0,1]} \middle|\exists \alpha \in \omega_1 : \begin{array}{l}\forall \beta > \alpha, f(\phi^{-1}(\beta)) = 1 \ \forall \beta < \alpha, f(\phi^{-1}(\beta)) = 0\end{array} \right}$$
is homeomorphic to the long ray

[gentle osprey]

mniip:

[digital peak]

oops

swap 0 and 1

but it doesn't matter I guess

[digital peak]

would appreciate a simpler example still

[hard wind]

"Find a subspace of R^2 that is path connected but not locally path-connected at any of its points."
Is this even possible? I've been racking my brain for hours trying to think of how to modify the topologist's sine curve or something. Any hints?

Please ping me.

[gritty widget]

Do you mean the opposite? Locally path connected but not path connected? (I think path connected implies locally path connecredl

In that case, maybe like 3 topologists sine curves connected end to end

So that from each point there exists another point that you have to pass the fucky part of the topologists sine curve to get to

@espio

@hard wind

[coarse kestrel]

no, path connected does not imply locally path connected

[hard wind]

Nope, that was the exact wording.

Path connected but not locally path-connected at any of its points.

[coarse kestrel]

and also the problem wouldn't have made sense if you switch it since path connected at a point doesn't make sense

[hard wind]

Attaching those three topologist's sine curves wouldn't be path connected

Ye

[gritty widget]

Path connected means between any two points there is a path, and locally path connected at a point p means there is a neighborhood around p that is path connected, right?

[hard wind]

Yes

An open neighborhood specifically

[gritty widget]

Then doesn't path connected imply locally path connected?

[hard wind]

Errr, no

[coarse kestrel]

no no no

that's not the definition

locally path connected means for every point p and open set U containing p, there is a smaller open set V that is path connected

[hard wind]

Sorry, yeah thats the real definition

[gritty widget]

But your definition implies mine, and my definition implies yours

[hard wind]

Take the topologist's sine curve and connect the two ends (0,0) and (1,0) with a curve

[coarse kestrel]

your definition doesn't imply mine

[hard wind]

Actually not quite

I meant not quite at mathbath

Thats an example of path connected but not locally path connected yeah

[coarse kestrel]

yeah at every point

[hard wind]

But i need every point

Which honestly, makes no sense to me

I dont see how thats possible

At least not as a subspace of R^2

[coarse kestrel]

i was thinking of maybe take $\br{\bigcup_{x\in\bQ}\brc{x}\times\bR}\cup(\bR\times\brc{0})$

[gentle osprey]

Whoever:

[coarse kestrel]

But the Rx{0} part doesn't work

[hard wind]

Interesting idea, I had thought about comb spaces earlier

Yeah... it's hard

I couldnt seem to find a comb space that works

[coarse kestrel]

Oh I found the answer on stackexchange

[gritty widget]

Hm, I guess I fail at topology

[hard wind]

Damn, impressive whoever

[coarse kestrel]

welp

[hard wind]

I could not find it on stackexchange

[coarse kestrel]

do you want it?

[hard wind]

Can i have a hint first

[coarse kestrel]

the answer kind of used my idea

except it's

connecting (0,0) to (1,p) for every rational p in [0,1]

and you try to do something else to make the space not locally path connected at (0,0)

[hard wind]

hm

oh ok i get it

now i must mess with (0,0) yeah

[digital peak]

hmm

subspace of [0,1]^[0,1] in product topology whose sequential closure is not sequential closed

[nimble jolt]

Pasting here so not lost in the clutter of chill:

Consider the indicator functions of an algebra of sets (e.g. the opens). If a sequence of these indicators converges pointwise, then both the liminf and limsup converge to this same limit, which is the indicator of the set A that is either the union of cofinite intersections or intersection of cofinite unions we discussed earlier (these must coincide in fact). This already shows that if you start with the opens, a convergent sequence can only give you something at most two steps up the Borel hierarchy.

On the other hand, as each step of the Borel hierarchy forms an algebra, you can always write an arbitrary union or intersection of sets in one step as a monotonic union or intersection. This shows that a convergent sequence can get you anything in the next step up the Borel hierarchy.

Paragraph 2 shows you get get arbitary finite step Borel sets by repeatedly taking sequential closures, Paragraph 1 shows you can't get all of these with only one sequential closure operation. Hence the sequential closure is not sequentially closed.

[meager python]

Why is there no channel for algebraic geometry alone?

[honest narwhal]

Algebraic geometry \subset algebra \cup topology and geometry

[ancient zenith]

if you are however looking for algebraic geometry discussion, Ravi Vakil runs a math discord (I kid you not) called the algebraic geometry syndicate

that's the only reason I have the (CA) in my name; they wanted us to indicate our locations

[honest narwhal]

@ancient zenith one thing that's worth noting is that discord has a way to have a nickname on a server separate from your global username

So if you'd rather not use your real name and location here you can change your overall user to something more anonymous and then set a nickname on AGS that complies with what they want

[ancient zenith]

oh my god

you're telling me I've been unnecessarily doxxing myself this whole time

rest in rip me

[gritty widget]

but u need to solve a challenge problem to get nickname changing privilege, better get solving quick!

[sweet wing]

ripppp

[red garden]

Algebraic geometry \subset algebra \cup topology and geometry
@honest narwhal algebra is a subset of topology and geometry?

[honest narwhal]

Huh

[rugged swan]

Algebraic geometry = Algebra tensor Geometry

[meager python]

Dynamical systems why? When theres analysis

Category theory? You mean foundations? Etc

[gritty widget]

I claimed that the set $A$ defined by
$$A={ (x,y,z)\in\mathbb{R}^3 \ \lvert \ x+y+z\in\mathbb{Q}}$$
has interior point as $\varnothing$ and closure as $\mathbb{R}^3$? But how to show that? I have shown that set $A$ is a neither open nor closed set in $\mathbb{R}^3$.

[gentle osprey]

weilam06:

[tight agate]

Proving the closure is R3 should be easy, as A contains Q^3 (and Q is dense in R)

To show that the interior is empty, fix (x,y,z) in A, for any epsilon ball in A, pick an irrrational delta such that (x + delta, y, z) is contained in the epsilon ball.

[gritty widget]

So we pick $\delta=\dfrac{\varepsilon}{2}$ for irrational $\varepsilon$ and $\delta=\dfrac{\varepsilon}{\sqrt{2}}$ for rational $\varepsilon$ to show its interior?

[gentle osprey]

weilam06:

[tight agate]

yeah those work

[gritty widget]

Nice

Proving the closure is R3 should be easy, as A contains Q^3 (and Q is dense in R)
@tight agate But I still don't get the idea since set $A\subseteq \mathbb{Q}$.

[tight agate]

no, A contains (sqrt(2), -sqrt(2), 0), which is not in Q^3

Q^3 is contained in A

So Q^3 closure is contained in A closure

And what is Q^3 closure?

[gritty widget]

R^3

[tight agate]

so?

[gritty widget]

We can conclude that cl A=R^3!

Thanks

I get it now

[tight agate]

yeah A is a lot bigger than Q^3

Q^3 is countable

A is not

[gritty widget]

I understand now, so formally written, Q^3 ⊆ A ⊆ R^3 => R^3=cl Q^3 ⊆ cl A ⊆ cl R^3=R^3 yields the result

[tight agate]

yup

[little robin]

Hey I'm looking for an academic journal that contains a proof of the Hyperplane Separation Theorem. Does anyone know where a good place to look would be?

[tough imp]

Why do you need a journal?

I think you can find proofs just out on the web

[little robin]

Dumb reasons. XD
Basically I gotta get 12 citable items for a survey research paper

Web proofs =/= citable items

[tough imp]

gross

[little robin]

ikr

[tough imp]

Good luck lol, don't think many people consider this theorem "journal-worthy"

maybe try some computational geometry shit or something idk, some applied geometry journal

[little robin]

Yeah, but I can't get ahold of books in time because I kinda blew this off. XD

Yeah, it's my own fault and I'll suffer for it. Thanks anyway

Thank you so much

I'll look into it.

[summer jolt]

Hi, I am trying to prove that the real projective plane is NOT homeomorphic to a circle. However, I am a bit lost. I'd like to argue in terms of connected components but not sure.

[gritty widget]

aren't they both connected?

how overkill do you want your solution to be

the universal cover of RP^2 is S^2 and the universal cover of S^1 is R

another maybe overkill solution is that RP^2 has fundamental group Z / 2Z and S^1 has fundamental group Z

but connectedness is not the topological invariant that fails here

[coarse kestrel]

idk i wouldn't say that's an overkill solution

[gritty widget]

im blanking on a non-AT topological invariant that breaks here

maybe manifolds are too nice

true

@summer jolt what level is this?

line goes that way or line goes this way

[summer jolt]

@summer jolt what level is this?
@gritty widget undergrad

The circle is oriented
@gritty widget we haven't covered orientation yet

but connectedness is not the topological invariant that fails here
@gritty widget I was thinking that if I remove two points on the circle then I form two connected components and RP2 you don't? But I don't see the picture

[gritty widget]

everything i mentioned is undergrad level to me, i meant if this is a gen topology class or an AT class lol sounds right

i like that argument

maybe it'll help to look at RP^2 as something like this, or maybe as the sphere with antipodal points identified

(if you want a picture)

[summer jolt]

i like that argument
@gritty widget the only thing I am unsure is what would be the number of connected components in RP2 if I remove two points.
Thanks for the pic, but I really don't get why it represents RP2. I am used to thinking that RP2 like a half sphere with antipodal points but even that is hard to grasp.

[gritty widget]

i strongly believe it will again be one connected component

im writing something down to convince myself lol

think of it like this maybe: if you remove two points from RP^2, how many points do you remove from the hemisphere with antipodal points identified? could you possibly disconnect it?

that argument might be a little sketchy

[summer jolt]

Yeah you're right it would still be connected.

[gritty widget]

and then RP^2 minus those points is the image under the quotient map of your hemisphere minus some points, which is still a connected domain

i like what slimvesus is saying more than what im saying lol

listen to them

[summer jolt]

RP2 - point = mobius band
@gritty widget oh wow! I didn't know this.

[gritty widget]

like i said, just nuke it with universal coverings

or throw a grenade at it with the fundamental group

i don't actually know a lot about the projective plane

you're the expert here

[summer jolt]

Isn't that the circle again?

[gritty widget]

so many ways to do this 😌

[summer jolt]

Hmm so you saying that if I remove a point from RP2 you get a circle?

Ah!

Ok I am getting to understand the reasoning now!

Yes, I can imagine. That makes sense.

[gritty widget]

Here I have a question. I am asking to write down a single subset of $\mathbb{R}^2$ that satisfies all conditions: \
(i) $A$ is connected and not pathwise-connected.\
(ii) $A$ contains the set $K={(x,y)\in\mathbb{R}^2 \ \lvert \ x\in[0,1], y=0 }$ as a subset.\
As the example that I know, the set $P=Q\cup R$, with $Q={(x,y)\in\mathbb{R}^2 \ \lvert \ y\in[-1,1], x=0 }$ and $R=\left{(x,y)\in\mathbb{R}^2 \ \lvert \ x\in[0,1], y=\sin\dfrac{1}{x} \right}$ is the most famous example to show that connected does not imply pathwise-connected in higher dimension. Thus I modified a bit for my answer, that is, take set $A=K\cup G$, where set $K$ is given on the question, and\
$G=\left{(x,y)\in\mathbb{R}^2 \ \lvert \ y\in[0,1], x=\dfrac{1}{2}\sin\dfrac{1}{y}+\dfrac{1}{2} \right}$\
Am I in the right track?

[gentle osprey]

weilam06:

[sweet wing]

i) you mean not path connected?

oh do people use pairwise connected for path connected or is it smt else

[gritty widget]

i) you mean not path connected?
@sweet wing Oops I meant pathwise-connected. It seems I have a typo.

[sweet wing]

icic

but yh the idea is that

just put the sine curve in the correct location

such that Q aligns with K

[gritty widget]

Yeap I wonder if I get the set $G$ wrong because I have to adjust the graph

[gentle osprey]

weilam06:

[gritty widget]

If there are other example that satisfies the condition, it would be better

[nimble cipher]

Hi. Why is the subspace topology here discrete? Note X is a metrizable space.

[stable lance]

'A has no limit points' implies that every subset of A is closed in A

[nimble cipher]

Ohh that makes sense, which means that every subset is also open

[summer jolt]

Question: what do you get gluing the opposite sides of 2n-polygon? Square gives the torus, I think the hexagon also gives the torus, the octagon gives the double torus (I looked this up). However, I'm failing to identify the pattern (is there any?)

[gritty widget]

https://math.stackexchange.com/questions/3145232/n-gons-gluing-diagram-proof talks about this

there is a pattern

[summer jolt]

Interesting!

So you just glued tori together

[gritty widget]

maybe you can visualize the octagon case like this:

use a line to split the octagon into two halves. if we collapse this middle line to a point while doing the gluing, we'll get two squares joined at a point. each square can then have its sides glued together like you would to get a torus (as pictured). then you have the double torus

(i really hope i didn't identify the sides wrong lmao)

it's been a while since i did this so please correct me if my square-to-torus gluing is wrong

[sharp yoke]

having trouble proving the map is injective

its easy to visualize but im having trouble writing it out formally

[gritty widget]

I think the important thing here is that the interior of A is nonempty

Otherwise eg a line segment in R2 through the origin does not give rise to an injective p

iirc there is some tricky linear algebra argument about compact convex sets with nonempty interior satisfying some nice properties that are relevant here, but there's a good chance I'm misremembering/talking nonsense

[nimble jolt]

@sharp yoke

If x is a point of the boundary of A with maximal distance from 0 along a given ray L from 0, I claim it is the unique boundary point of A on this ray. To see this, consider apply the convexity condition to an epsilon ball about 0 and x. This gives you a narrow cone of points contained in A, and all points of the ray L other than x are interior to this cone and hence not boundary points of A.

[gritty widget]

but why is there an epsilon ball around x?

i think proving this is tricky

ah, i am a dumbass

i guess i am probably mixing this up with something else

[nimble jolt]

epsilon ball around 0, not x @gritty widget

picture x as the point (0,1) in the plane for instance, and you are putting a tiny ball about (0,0) and drawing line segments between every point in this ball and (0,1), forming essentially a cone.

[gritty widget]

yeah i see it, i was being a dumbass

[sharp yoke]

thank you

[summer jolt]

maybe you can visualize the octagon case like this:

use a line to split the octagon into two halves. if we collapse this middle line to a point while doing the gluing, we'll get two squares joined at a point. each square can then have its sides glued together like you would to get a torus (as pictured). then you have the double torus
@gritty widget yeah that's exactly how I thought about it!

[feral dragon]

@oak skiff You're mapping to a set with the same topology (an open set) or something else ?

Why don't you send that problem so that it'll be easier to explain with an example.

[feral dragon]

@oak skiff mostly it's obvious, but are you aware of "The Rules for constructing continuous functions" ?

ahh,I guess you construct continuous functions by proving that they're continuous over specific topologies

So you know these and have you practiced examples on this ? @oak skiff

btw what book is your course referring to ?

ahh Munkres shouldn't create any confusions tbh. But you should be looking at Chapter 2 Section 18 i believe.

That's fine, good luck with that though. Hope you get the confusions cleared up

[quiet pilot]

Do you even need that X1 and X2 are algebraic here?

I'm not sure if I'm missing anything because the equality seems to follow straight from the definition

[tough imp]

I don’t think so

In response to do they need to be Algebraic

[quiet pilot]

Ok that's good to hear

[gritty widget]

Hi ! Just to be sure : A manifold is simply connected if it is arcwise connected and has the trivial fundamental group (i.e. for all x in M, all loop from x to x are homotopic to the identity) ? (I hope this is the place to ask, I'm new here, just tell me if it is not 🙂 )

that sounds like the right definition of simple connectedness yes

you only need that the fundamental group at some point is trivial, since you have path connectedness

you only need that the fundamental group at some point is trivial, since you have path connectedness
@gritty widget I didn't get that, I think I am lacking something

[elder yew]

If it's path connected, then it suffices to check that the fundamental group is trivial at a single point because you can always use that change of base theorem in Hatcher

Showing that the fundamental group at a base-point contained in the same path connected set are isomorphic

@gritty widget does that make sense?

[gritty widget]

Yes I get the idea and the intuition

but I did not see that theorem

i'll check it

[elder yew]

It's chapter 1

I forget which theorem it is exactly, but the proof isn't too difficult

Just compose paths cleverly

[gritty widget]

ok but basicly it says that if there exist a continuous path between two path and if the fundamental group at one of them is trivial, then it is also trivial for the second one ?

if you have a path from a to b, then the fundamental groups are isomorphic by the following map:
-take a loop at b
-map to a loop at a by starting at a, following the path to b
-do your loop at b
-go back to a using the path in reverse

in alg lecture rn, can't go super in depth

it should be in literally any topology book

(yeah yeah, homotopy class of loops, i get it)

[tough imp]

Just use the path a to b (and b to a) to turn a loop at a into a loop at b

And a loop at b to a loop at a

Conclude these processes are inverses

There’s only one way to do this that you should be able to think of

I guess verifying this plays nice with the operation or whatever becomes more of a pain but once you have the map you can just verify details

[gritty widget]

Ok so there is a way to map a loop at a with a loop at b but i don't see how this help us to conclude that the fundamental groups are isomorphic

And why the fact that they are isomorphic implies that if one is fundamental, the other one has to be

one is fundamental

if one fundamental group is trivial*

sorry i'm a little tired 😅

why if one is trivial the other is
well if two groups are isomorphic and one has only one element, so does the other

oh yes make sense

chmonkey described why the map gives you an isomorphism of pi_1(X, a) and pi_1(X, b)

you can find the details in e.g. munkres or hatcher or on wikipedia probably (or just do it lol)

ok, thanks a lot for your patience 🙂

[tough imp]

If $\mathscr{E}$ is a locally free $\mathcal{O}_X$ module of finite rank, I want to show that
$\mathscr{H}\text{om},(\mathscr{E},\mathcal{F})\cong \check{\mathscr{E}}\otimes \mathcal{F}$ for any $\mathcal{O}_X$ module $\mathcal{F}$

[gentle osprey]

Chmonkey:

[tough imp]

On an open set $U$ for which $\mathscr{E}$ is isomorphic to a free $\mathcal{O}_X$ module, I was able to produce an explicit isomorphism, and was (with trouble) able to show that these agreed on intersections to give a global isomorphism

[gentle osprey]

Chmonkey:

[tough imp]

Is there a way to define such a map globally however, and then to simply check that it's an isomorphism locally? If so I'd much rather do that, but it's really not clear how you can get a map either direction

[tight agate]

We can define the following map from the presheaf that you get when you take the tensor product over each open set

[tough imp]

I figured it out ;w;

[tight agate]

oh okay nvm

[tough imp]

I literally couldn't think of a map Hom_A(M,N) to Hom(M,A) (x) N

for like an hour and a half

Err, I guess it's the other way around

[tight agate]

yeah the map I have goes the other way

[tough imp]

You send like

f(m) (x) n to f(m)n right?

[tight agate]

yup

[tough imp]

I had this like exact idea, but thought for some reason I had to pick an arbitrary n to use

forgetting you get given an n to use

😔

and sat around for an hour and a half mulling about how tf I can do this

[tight agate]

but this is just a map from the presheaf right

[tough imp]

Right, but you can show it's a local iso

[tight agate]

you need to sheafify after that

[tough imp]

so it doesn't matter

[tight agate]

yup

[tough imp]

You don't even need to localize

[tight agate]

or just on stalks

[tough imp]

err stalks

the issue is

to know that sheaf hom plays nice with stalks you need like

the first thing to finitely presented or something

at least that's what I saw on Stacks??

[tight agate]

hmm

[tough imp]

And I don't want to deal with that shit right now haha

The idea is sheaf-hom_{O_x}(F,G)_x is just Hom_{O_X,x}(F_x,G_x) right?

But I think this fails in general

[tight agate]

I think youre fine in this case as they're locally free

[tough imp]

Yeah, but basically

I don't want to prove that equality

XD

I mean I guess I should, but also it's equally easy to just go to a trivializing cover

to get the map we defined is an iso

but to prove the equality of the stalks above is more work and I'll just... save that for later

[brittle widget]

Omg my geometry lecturer just wrote f holomorphic from compact Riemann surface X to CP^1, d for degree of f, F' for # faces of a triangulation T of X, F for # faces on f[T], and then reasoned F' = dF XD

[elder yew]

Sounds like an analyst

That's the kind of Prof. I can get along with

[sleek thicket]

ughhhh my brain is too small for riemannian geometry

hello ttera

[gritty widget]

hello shamrock

i was typing a witty response to what epicguy posted

but then i realized i had read their post wrong

[sleek thicket]

tfw

Do you want to help me think about riemannian geometry?

My brain is fried

[gritty widget]

it's 4 am for me so the chances i'll be able to contribute anything are infinitesimal

but ok

[sleek thicket]

kk

Let G be a lie group

With a bi invariant metric g

And lie algebra \g

[gritty widget]

[sleek thicket]

I'm supposed to show <ad(X) Y, Z> = - <Y, ad(X) Z> for all X, Y, Z in \g

[gritty widget]

kinda looks like the "connection is compatible with the metric" condition but equal to zero with some terms shuffled

[sleek thicket]

I tried to do this as follows:
Let γ be the one parameter subgroup with γ'(0) = X. Define vector fields V, W along γ by V(t) = d(L_γ(t))_e(Ad(γ(t)) Y) and W(t) = d(L_γ(t))_e(Ad(γ(t)) Z). Then d/dt <V(t), W(t)> = <D_t V, W> + <V, D_t W> by compatibility with the metric

By left invariance of the metric <V(t), W(t)> = <Ad(γ(t)) Y, Ad(γ(t)) Z>, and by bi-invariance this is constant

(bi invariance is equivalent to the inner product on \g being invariant under the adjoint representation of G)

okay so the derivative of a constant is 0

So it suffices to show (D_t V)(0) = ad(X) Y

(and by symmetry the same thing but for W)

This feels plausible to me because ad(X) = d(Ad_e)(X) = (Ad ° γ)'(0)

But I haven't gotten any further than that

wtf is D_t

I hate it

[gritty widget]

V(t) = d(L_γ(t))_e(Ad(γ(t)) Y)
should it be Ad(γ'(t))Y inside? or am i missing something about lie groups

[sleek thicket]

Nope

Big Ad is different than little ad

[gritty widget]

in what way?

[sleek thicket]

Ad : G -> GL(\g)
ad : \g -> \gl(\g)

Ad(x) is the differential of the conjugation by x map : G -> G at the identity

ad(X)(Y) = [X, Y] but also ad is the differential of Ad at the identity

[gritty widget]

i see

give me a few minutes to digest everything you wrote

[sleek thicket]

okay well one more thing: if we define $A_x = d(L_x)e(Ad(x) Y)$ then $A$ extends $V$ so $(D_t V)(0) = \nabla{\gamma'(0)} A = \nabla_X A$

[gentle osprey]

shamrock:

[gritty widget]

hmmm

just write it in coordinates

[sleek thicket]

fucc u

[gritty widget]

lemme try it for a little bit and see if i come up with something

[sleek thicket]

Sure, I'm mostly just trying to sound off the problem again and see if I notice anything new

[gritty widget]

the only technique i know of to make calculating covariant derivatives easier is writing things in parallel vector fields but i don't really see how that'd make this easier lol

i'm still thinking on it tho

[sleek thicket]

I think I maybe figured it out but I'm confused by something

So like

If you instead define V(t) = d(L_γ(t))_e(Ad(γ(t)^-1) Y) then I think V(t) = d(R_γ(t))_e(Y)

Because in G if you conjugate by γ(t)^-1 and then multiply by γ(t) on the left, that's the same as multiplying by γ(t) on the right

Actually I think I see it now

ty for the help

[gritty widget]

does working with this new definition of V(t) and W(t) solve it?

also it would be V(t) = d(R_γ(t))_e(Y), can confirm

there's a proof of this in do carmo

[sleek thicket]

I think so

I mean, I actually changed it to d(R_gamma(t))_e(Ad(gamma(t)) Y)

but things work out right I think

So with this change V has extension the left invariant vector field corresponding to Y

hmm no this does not immediately solve my problems

but it is insight

and I should go to sleep

[gritty widget]

i guess i can give an uneducated hint

[sleek thicket]

Sure

That's what I asked for :P

[gritty widget]

it's in the section before covariant derivatives in do carmo so you shouldn't need those

[sleek thicket]

What is?

[gritty widget]

the proof of the thing you wanna show

[sleek thicket]

oh uh

[gritty widget]

(the first thing)

[sleek thicket]

I'm confused

The thing about ad(X) being skew adjoint?

[gritty widget]

ya

[sleek thicket]

Maybe I'm fucking this up then

It's in the section on the Levi-Civita connection in Lee

But there's more to the problem

[gritty widget]

lemme take a peek at lee

[sleek thicket]

So maybe this first bit doesn't use it

hmmm

[gritty widget]

i might look at the problem more tomorrow

[gritty widget]

i don't have a lot of homework this week so i have some spare time to mess around with

oh nice copy

[sleek thicket]

It's on my homework so I'm going to think about it more for sure lol

Yee

I have the trio

[gritty widget]

i am jealous

both of my classes concerning smooth manifolds so far have completely ignored lie groups

[sleek thicket]

Oof

[gritty widget]

and my rg prof is speedrunning do carmo, any%

[sleek thicket]

My course last year didn't do a lot with them

like

The important stuff if at the end of ISM

But do to covid we didn't get there

and so they did it in prelim prep seminars over the summer

But I was doing other stuff

[gritty widget]

ism chapter 19

[sleek thicket]

I read it on my own but it's not the same

Oh 19?

We did that

We did cover the closed subgroup theorem

foliations are cool

[gritty widget]

idk what the lie group exponential is so i don't understand lee's hint

is it related to the geodesic exponential

where you go along a geodesic for time t = 1

[sleek thicket]

Well uj

That's this problem lol

See part c

if you have a bi invariant metric they're the same

[gritty widget]

oh

[sleek thicket]

but it's pretty simple

Same idea I mean

You know about like, integral curves?

Did you know that left invariant vector fields on a lie group are complete?

[gritty widget]

that's a nice fact

i vaguely recall reading about it

[sleek thicket]

Pretty cool

[gritty widget]

i kinda wish my courses would go into this stuff, seems like i have to go read it on my own

[sleek thicket]

well like, the exponential of a vector X in the lie algebra is γ(1), where γ is the integral curve with γ(0) = e and γ'(0) = X

[gritty widget]

oh so it's basically the same thing but the integral curve isn't a geodesic

lol

[sleek thicket]

So like how the exponential on a riemannian manifold works but with integral curves instead of geodesic

Yeah

This is an integral curve of the left invariant vector field which is X at the identity

and which starts at 0

[gritty widget]

and exists for all time 😌

[sleek thicket]

Yee

[gritty widget]

so you don't have to worry about scaling issues to define gamma(1)

[sleek thicket]

And then conversely γ(t) = exp(t X)

And γ is a group homomorphism R -> G

[gritty widget]

👀

[sleek thicket]

Yee

This is like

"one parameter subgroup"

idk I'm too tired for this

[gritty widget]

maybe i'll read the lee lie group stuff over my break

[sleek thicket]

It's good!!

I wish I understood it better

Me and a friend of mine are thinking of reading a book on Lie groups/lie algebras

But time...

[gritty widget]

how much stuff from algebra actually comes into play? i'm taking algebra right now so it'd be nice to have the two (algebra, smooth manifolds) really mix together

(for lie groups)

[sleek thicket]

Idk

I assume rep theory of lie groups is different

But on the other hand, there's some stuff which is the same

[gritty widget]

algebra has just been "symmetric group goes brrr" and i want to see a little more

[sleek thicket]

oh okay lol I was gonna ask if you'd seen rep theory for finite groups

So like, I was thinking about this on my own the other day

[gritty widget]

oh god no lmao

[sleek thicket]

There's this theorem called maschkes theorem in finite group rep theory

It's like the first theorem in that

And it involves averaging something over your group

[gritty widget]

the most rep theory i've seen is skimming the gelfand naimark duality stuff from my func anal book, and it's not like i actually read it

[sleek thicket]

And the proof works the exact same for compact lie groups!!

[gritty widget]

oh neat

[sleek thicket]

You just need to assume your representation is smooth and take an integral to average

anyways, I think there's definitely bits that are algebraicy and bits that aren't, but I haven't studied it yet so idk

My uni is doing a completely algebraic course on lie algebras right now and I dropped it

Lecturer was terrible and I want some geometry

[gritty widget]

geometry > algebra

tbh

anyone who says "but algebra = geometry" is smoking crack

good thing it's 5 am and the people who know more mathematics than me can't come get angry at me for that

anyways i'm going to head off to sleep now

[sleek thicket]

lmao

geometry >= algebra

But also algebra >= geometry

[gritty widget]

order is not antisymmetric

😌

[sleek thicket]

we're only in a pre-order though so algebra might not be geometry

haha

[gritty widget]

if i notice anything when (if) i look at the problem tomorrow i'll post it here

good luck on the problem (and rest of hw)

[sleek thicket]

ty

I have two weeks so I'm not worried

Benefits of being in post-quals courses lol

[quiet pilot]

I'm working in k[X1,X2,X3,X4] over some unknown field, not necessarily algebraically closed. I want to write the intersection (X1,X2) \bigcap (X3,X4) more explicitly. My guess is (X1X3,X1X4,X2X3,X2X4), but I'm not sure how to prove the inclusion to the right

[sleek thicket]

@gritty widget you were right about not needing covariant derivatives

any bilinear function T : V x W -> U between fd vector spaces satisfies $dT_{(p, q)}(v, w) = T(p, w) + T(v, q)$

[gentle osprey]

shamrock:

[nimble flower]

is there some easy way of showing a function from a compact space to a compact space is continuous with looking at the open sets?

[gritty widget]

Maybe the closed graph theorem?

[pastel linden]

if the domain and range are metric spaces I believe the function is uniformly continuous

[little hemlock]

i interpreted their question as: "is there a quick-shortcut for testing continuity of functions between compact spaces"

[tight agate]

The fastest way I know is to just claim that it is continuous

[little hemlock]

i believe a continuous map between compact spaces is closed. I can't think of a sufficient condition easier than the usual definition of continuity tho

[nimble flower]

i interpreted their question as: "is there a quick-shortcut for testing continuity of functions between compact spaces"
@little hemlock yeah - that's correct

hmm maybe the closed graph theorem

[gritty widget]

Oh, I misread the original question, I thought it said "without" open sets, not "with" open sets

[coarse kestrel]

what is a visualization of the one point compactification of a countable set with the discrete topology?

[gritty widget]

i guess i have never used this space before, but why not just Z+infinity at the end?

[coarse kestrel]

i meant like a visualization of it

like how the one point compactification of R is visualized as a circle

i guess the one point compactification of Z is just the subset of the circle that corresponds to {∞} U Z

xD

[gritty widget]

I guess that works

[river granite]

a countable set with the discrete topology?
isn't that already Hausdorff (if it has ≥2 elements) and compact?

[gritty widget]

How is it compact?

I guess countable to me means infinite countable (maybe some people use it to mean finite?)

Eg I think I have heard "enumerable" as finite or cardinality of integers

[river granite]

aaaaah yes you're right. then you can take e.g. an open cover of singletons and it fails catastrophically

for an infinite set

[coarse kestrel]

oh yeah by countable i meant infinite countable

So in bijection with the set of naturals

[gritty widget]

all sets are finite

just as all holomorphic functions are constant

[river granite]

t. wildberger

[tight agate]

I'm doing a problem which is asking me to interpret $\int_{\mathbf{P^n}}c_n(TP^n)$

[gentle osprey]

Brofibration:

[tight agate]

where c_n(TP^n) is the top Chern class of the tangent bundle of projective space

I computed that it is (n+1).Vol(P^n)

and I know that this is somehow supposed to correspond to the number of times a generic vector field on Pn vanishes

but I'm unable to make the connection

no pun intended

[tight agate]

I found an answer online which used Hirzebruch Riemann Roch and the Hodge decomposition lol

and the Borel Serre identity

there's gotta be a more low-tech way of doing this

[fathom radish]

https://cdn.discordapp.com/attachments/641733014711435264/773985263507472404/unknown.png

given I have the radius, the major axis is the blue line, and the minor axis is the red line, how do I find the shortest distance from the major axis for the curve of the circle

I was trying to make radial sanding blocks myself because they are expensive and I ran into this problem

im taking euclidean geometry next semester but for right now I only have 11th grader education on it

[small obsidian]

Sorry, your question isn't clear. Shortest distance between what and what?

[fathom radish]

oh sorry. Im trying to find how far apart the blue line and the edge of the circle are only on the y axis.

[small obsidian]

What do you know about the circle? Length of the blue line? Radius?

Angle from center to edge of blue line?

[fathom radish]

the radius, major axis (blue), and minor axis (minor) are all given. I am essentially setting it up by superimposing an ellipse on a circle where the radius is know

another thought is to find the velocity of y via partial derivative given that the edge of the ellipse is half of one period. but im also not sure if a formula for the perimeter of an ellipse based upon just the major and minor axis exist (it probably does).

and then maybe integrate the partial of y to get the distance from the major axis?

im not sure. Im just using my knowledge of calc because thats what I know. there is probably a much better way to do it

[small obsidian]

"Im trying to find how far apart the blue line and the edge of the circle are only on the y axis."
I took this to mean you're trying to find the length of the red line. But you have that as a given?

[fathom radish]

I only have the maximum though

I need to find it at every point

[small obsidian]

Oh haha I see

That would be
√[r² - x²] - height of the blue line

Height of the blue is just r - red

[fathom radish]

ah ty

[coarse kestrel]

I don’t see how the locally finite collection implies the closure is the union of the same U’s

[gritty widget]

local finiteness lets you commute union and closure, you can see a proof here https://math.stackexchange.com/questions/2623619/closure-of-union-of-locally-finite-collection-of-subsets-equal-to-union-of-closu/2623626

does bredon not mention it?

it's like one of the reasons you'd care about locally finite collections ig

[coarse kestrel]

No it did not

Goddamnit

[gritty widget]

😔

the proof in the mse post i linked is good imo

it's good to keep in mind, it comes up sometimes when you work with partitions of unity

[coarse kestrel]

Yeah

Thanks

That proof on mse is good

[sharp yoke]

whoever ur learning topology as well

[coarse kestrel]

[gritty widget]

Damn, I cannot understand this. I am not sure if this is "Whoever, you as well are learning topology" or "Whoever, you are learning topology as well"

[coarse kestrel]

No idea

But I’ll take that as a compliment

[sharp yoke]

apparently you can do the forward direction using that the identity function is a homeomorphism if you assume the topologies are equivalent

[sharp yoke]

when you assume continuous i can get the upper bound easily but i can't get the lower bound

ok i guess my process is wrong because doing the inverse does nothing for me

how do you use continuity to find the bound

[elder yew]

Is that true independent of dimension?

Even in infinite dimensions?

My fcnl background

Is a little lacking

I only know finite stuff well

lol

Really?

It's that basic?

No, it's ok

My background is fucking wonky

I deserve it

Ahh I see

That is fair

As you can tell, I was thrown into advanced topics courses without a whole lot of background

Ahh, bounded if and only if continuous

I knew it was true in finite dimensions

I didn't know it was true in infinite

Bounded if and only if continuous

Yeah

The proof is the same as the finite version

I only took an applied functional analysis class, where we were interested in Boundary Value Problems

and using things like compact operators

[uncut geyser]

bounded if and only if lipschitz if and only if continuous at the origin if and only if takes bounded sets to bouned sets yada yada

[elder yew]

To get results

Never took a pure functional

[gritty widget]

just read pedersen 😌

[uncut geyser]

isnt it like

bounded means |T(x)-T(y)| = |T(x-y)| <= C|x-y|

or am I tripping

[gritty widget]

this is correct

[hard wind]

Let p and q be quotient maps, and r be a map. Let p = q o r. Is r continuous? Is r quotient?

[gritty widget]

let q be a constant map lol

[hard wind]

Well, q might not be a constant map.

It could be any quotient map.

[gritty widget]

well then the answer is "not necessarily"

i guess you could then ask what conditions on q (or p) you need?

[hard wind]

Okay, thank you

What if p and r were quotient? Could we say anything about q?

[gritty widget]

q would be a quotient map

falls right out of the definition of one

well it's not hard to see

[gentle osprey]

TTerra:

[hard wind]

Makes sense.

Thank you

[sharp yoke]

@limpid vault thx. Also to address what you said before this was just one part of "walk-through" problem for proving that all norms on Rn are equivalent

i only posted that one part which is probably why it seemed weird

unfortunately that p set took me quite a while so now im gonna have to brew some coffee as i start my physics problem set also due tomorrow

[sleek thicket]

bruh

I cannot deal with this level of indices

[gritty widget]

[sleek thicket]

the proof is just as bad

[gritty widget]

lmao

this is an abomination

imagine writing it out with all the sums

[sleek thicket]

well less bad because it's just a solid block of notation and not א1 indices

Dear god

I mean it shows how good Einstein summation notation is

[dim meadow]

@sleek thicket you are like little baby

[sleek thicket]

I am very much a little baby

Riemannian geometry pushes the limits of my baby brain

[dim meadow]

Oh I meant with the indices

[tight agate]

Does anyone know if there's a pdf of FGA explained floating around?

managed to find a djvu file but those are always horrible

[sharp yoke]

what book is that from

[gritty widget]

the book shamrock posted is lee's "introduction to riemannian manifolds"

[tight agate]

they're pretty neat

a lot of cool theorems

[sleek thicket]

they're cool but as i said, my brain is too small

lots of ugly notation and computation

but the structure/geometry is really nice & good when you're able to see past that

[gritty widget]

how is studying riemannian manifolds?
if you can deal with the occasional terrible computation, it's super nice

take all the nice stuff about smooth manifolds, and then add some geometry, and now drawing funny warped squares and curvy lines becomes an actual proof technique

(/s)

[tight agate]

there's going to be a class on Fukaya cats over here next quarter

im not sure if I should do it as I know pretty much nothing about symplectic stuff

other than the definition

or the Kaehler case

[meager python]

@tight agate you have some of the chapters in draft versions here:

https://ncatlab.org/nlab/show/FGA+explained

[tight agate]

yeah I looked at that, some of the links seem to be broken

but thanks!

oh wait nvm that linked to another page and I could access some lecture notes

thanks!

Here's the link if anyone else cares:

http://indico.ictp.it/event/a0255/other-view?view=ictptimetable

[sleek thicket]

There's a course on "Canonical metrics in Kahler geometry" here in spring and I haven't decided whether I want to take it

I've heard bad things about the prof and the course announcement thingy says "Prerequisite: good handling of multi-variable calculus and mathematical maturity"

Which is very concerning to me

[sleek thicket]

Right lol