point-set-topology

[little shoal]

I believe it can be deduced from the results contained from page 210 to page 214 in Spivak’s “A comprehensive introduction to differential geometry”. I have vol 1, third edition.

[gritty widget]

how to define boundary of a manifold?

take a (topological) manifold and demand that the charts can be open subsets of {(x_1, ..., x_n) : x_n >= 0}. the boundary points are the points which have a chart taking the point to something with x_n = 0.

x_{n}>=0 means all x_i are >=0 or just the last point?

last

is there any intuition behind it?

draw a picture

let me give one

kind of shitty

how does x_{n}>=0 give rise to the 'boundary'?

[ivory dragon]

separates the stuff on the manifold from the stuff off it

[gritty widget]

manifolds with boundary are locally modeled on H^n, which has a "boundary"

[cedar pebble]

I mean the boundary has charts that look like this

[gritty widget]

anyways i don't feel like competing with other people's explanations so i'm going to defer to lee

[cedar pebble]

the charts here are an upper half space, the subspace of R^n where say the last coordinate is ≥0

[gritty widget]

lee ISM ch2 has a good explanation

[gritty widget]

so it's true that every manifold is a manifold with boundary,but not every manifold with boundary is a manifold,right?

[gritty widget]

ohhh this makes senes

[cedar pebble]

yes every manifold is trivially a manifold with boundary

[ivory dragon]

obviously im looking at the last coordinate here

[cedar pebble]

any manifold with nonempty boundary is not a manifold

but such a thing typically admits a stratification where the boundary is a manifold of dimension 1 less, and the manifold with boundary minus the boundary is a manifold

[gritty widget]

is a manifold without boundary just a manifold?

or a specific manifold

my prof defined a closed manifold using the notion of manifold without boundary but didnt define that

how to define manifold without boundary? or what's the difference between a manifold, and a manifold without boundary?

based on this a manifold without boundary is just a manifold,is this right?

[bitter yoke]

Yes

[gritty widget]

thanks!

every closed manifold is a compact manifold with boundary,but the other way doesn't work. does this seem right?

since manifold without boundary implies manifold with boundary

[gritty widget]

you mean without boundary?

In mathematics, a closed manifold is a manifold without boundary that is compact.

@gritty widget

[hallow bloom]

Are there any simple examples of paracompact spaces? Or some good example to compare a compact and a paracompact space to get some intuition about the definition of paracompactness?

[gritty widget]

inb4 "manifolds!"

[sleek thicket]

it's hard to find a space which isn't paracompact

haha owned i didn't say manifolds

[gritty widget]

i was going to

but i figured not great since that's usually in the defn

[sleek thicket]

Examples of paracompact spaces: all metric spaces, all cw complexes, all manifolds (as long as you include "second countable" in your definition of a manifold")

yeah haha

oh and: all compact spaces

all finite spaces

[gritty widget]

empty space

[sleek thicket]

tterra literally all of my examples include the empty space

hello rounin 🙂

I can give more concrete examples, but honestly if you name a space it's probably paracompact

[hallow bloom]

Then maybe a counter example would be better

I assume you'd say something that is not second countable

[sleek thicket]

Yup haha

Do you know what the long line is?

[hallow bloom]

nope

[sleek thicket]

Ah okay, the construction is sort of involved so I won't go over it

[hallow bloom]

I can google it, np

[sleek thicket]

ehh

I would prefer to give you an example which is easier to understand

[gritty widget]

all spaces are paracompact

[sleek thicket]

the construction of the long line involves ordinals

which are a little tricky if you haven't thought about them in depth before

[sleek thicket]

but I can't think of any 😢

[hallow bloom]

it's ok, np, just wondered maybe I'm just failing to find something simple but I guess there's no trivial example

[sleek thicket]

Yeah, the class of spaces I posted is pretty big

[gritty widget]

I think here is one: ||R with a negative infinity, with the topology where [-infty, a) is open||

[sleek thicket]

ah I think that works 8da

you don't even need negative infinity right?

[gritty widget]

Wait a second

[sleek thicket]

ono

[gritty widget]

Nvm I think I'm being a dumbass

No wait

[sleek thicket]

wait waht

I think R with open sets (-infinity, a) for each real a is a topology which is not paracompact

[gritty widget]

Ah ok nvm yeah I think it works. And yeah we don't even need negative infinity huh

[sleek thicket]

right yeah x < sup a iff x < a for some a

so the union of rays like (-infinity, a) is the ray (-infinity, sup a)

@hallow bloom did this example make sense?

actually 8da we can do this with N too right?

[gritty widget]

Ah yeah huh

[hallow bloom]

let me think about it

[bright acorn]

So guys

Has anyone here read a bit of Stacks Project?

[gritty widget]

chmonkey probably

[hallow bloom]

so what breaks here is that the refinement is not locally finite? e.g. for cover {(-\nfty,a): a \in \mathbb{R}}?

[bright acorn]

Is it a good way to learn algebraic geometry?

[gritty widget]

@tough imp

chmonkey is the resident algebraic geometer

sham too maybe

[bright acorn]

Lmao

That's cool

I've started studying a little bit more of algebra

More specifically commutative algebra

I am studying modules now

And I got interested in knowing like

How some of these concepts naturally arise from algebro-geometric contexts

Like local rings

And etc

I'd also like to ask

[tough imp]

I have read stacks

I think it’s good if you know what you want to learn

Not as like a textbook to just go in order and read

[bright acorn]

How much of commutative algebra is needed in order to start studying algebraic geometry?

Like

[tough imp]

Depends at what level

If you want to do schemes you need to be fluent with localization

[bright acorn]

I need to know the basics of commutative rings and modules over them, localization, noetherian rings

What else?

[tough imp]

And quickly you need to know dimension theory

The more you know the easier things go

[bright acorn]

I mean

[tough imp]

But no matter what, you’re gonna have to be prepared to be willing to prove commutative algebra on the fly

[bright acorn]

I am talking about more of a beginner's level algebraic geometry

Like algebraic curves over algebraic closed fields

[tough imp]

If you’re dealing with varieties you don’t need as much

[bright acorn]

Or more generally algebraic varieties as you put

[tough imp]

Oftentimes sources for that treat commutative algebra alongside it and develop what you need as you go along

If you have the equivalent of say, a year of undergrad algebra

And are comfortable with localization and maybe some dimension theory then you should be fine

[bright acorn]

Nice

[tough imp]

I’d just keep a commutative algebra book handy

[sleek thicket]

you're still gonna want some dimension theory

[tough imp]

So you can look things up as they come up

[sleek thicket]

for just varieties

also standard comm alg results like noether normalization

but some of this stuff can be picked up as you go

[tough imp]

That kinda falls into dimension theory

[sleek thicket]

yeah haha that's why I mentioned it

[tough imp]

Noether normalization at least

But yeah

I think it’s not too hard to treat deimsnion theory tho

[sleek thicket]

if a collection of open sets in this space is locally finite, it is finite. try to prove this and see why it breaks paracompactness

[tough imp]

I think Matsumura does most of what you’d need in jjst section 5

But idk how much of the prior stuff you need to understand it

[bright acorn]

Nice

I am reading Michael Atiyah's book on commutative algebra, doing the exercises and watching some really good youtube lectures on the subject

I was kind of curious about like

[tough imp]

Atiyah-MacDonald has like

[bright acorn]

what would be the next natural step

[tough imp]

More than what you need to get started for sure

In CA?

[bright acorn]

nope

I am more interested in algebraic geometry really

[tough imp]

Ah

[bright acorn]

but I knew that I needed to study commutative algebra

But like

[sleek thicket]

I am going to blindly recommend miles reid's undergrad alg geo because I like the accompanying comm alg book and I've heard good things

but I literally haven't read it so this is not a first hand account

[bright acorn]

What would be next step after reading atiyah's book to study algebraic geometry?

[sleek thicket]

also don't let "undergrad" raise your hackles or whatever, it does serious AG

[bright acorn]

Any book recommendations besides hartshorne's algebraic geometry?

[tough imp]

You could just jump into schemes if you don’t care about understanding the abstract stuff, but I would not recommend that

Which is what I did

[sleek thicket]

> miles reid's undergrad alg geo

this is a book

[tough imp]

I’ve heard that it’s good as well

[sleek thicket]

@gritty widget people get weird about book names okay

[tough imp]

I think there’s like
A royal road to algebraic geometry

[sleek thicket]

ive told people the same but reversed about a book with graduate in the name

[bright acorn]

Why would I learn schemes before learning about algebraic and projective varieties?

[tough imp]

And Shafarevich

[sleek thicket]

yeah chm

who would do that

[tough imp]

Because you’re me

Idk haha

You shouldn’t do that

[sleek thicket]

I also liked Gathmann's notes

for varieties

https://homepages.warwick.ac.uk/staff/Miles.Reid/MA4A5/UAG.pdf
https://www.mathematik.uni-kl.de/~gathmann/de/alggeom.php

[bright acorn]

Oh

Undergraduate algebraic geometry looks short

[gritty widget]

shamrock accelerating everyone's ag arcs

[sleek thicket]

except his own 😔

shamrokc ag arc has been going for like 2 years at this point and it's not really showing much progress

[gritty widget]

algebraic geometry is on the "possible reading topics for the summer" list

[tough imp]

You could also ask a professor at ur uni

If you know any of them decently well

[sleek thicket]

the books I recommended might work for you as well

[tough imp]

Just be warned their advice could be bad, but I like prof’s advice more than randos most of the time

[sleek thicket]

although you would probably want to read some comm alg first, tterra

[gritty widget]

atiyah macdonald?

[tough imp]

Yeah TTerra CA is gûd

[sleek thicket]

miles reid's comm alg book is also based

AM is kind of dry and everything is in the exercises

[tough imp]

A-M makes you want to k-word yourself if you’re me

[bright acorn]

I'll take a look at Miles Reid's book

[tough imp]

So don’t do that

[sleek thicket]

I would suggest undergrad comm alg by miles reid

[bright acorn]

And also these lecture notes

[sleek thicket]

it gives some geometrical intuition too

[bright acorn]

I am not so used to studying using lecture notes

Most of them have few exercises from my experience

[sleek thicket]

chm might suggest matsumura but I think it's bad as a first exposure

[tough imp]

I actually wasn’t

[sleek thicket]

ah okay

[bright acorn]

And I learn a lot from doing a bunch of explicit computations and exercises

[tough imp]

Since I don’t think TTerra has the necessary algebra background??

[sleek thicket]

yeah haha

[gritty widget]

my algebra background is garbo, the course im in sucks balls lol

[tough imp]

If that’s it, maybe try doing the exercises of Hartshorne

[sleek thicket]

that's fine though, I think undergrad comm alg would be a good book at this point

[bright acorn]

Also

[tough imp]

Like I wouldn’t recommend Matsumura if you aren’t at a point you’re very comfortable with tensor products

[bright acorn]

What would be ''the next step after CA'' you were talking about chair monkey?

I got a bit curious

[tough imp]

Matsumura

Or Eisenbud

If you want to read like 800 pages lol

I’m just a Matsumura simp basically haha

It’s a super good book

[sleek thicket]

lmfao

from undergad comm alg

[bright acorn]

I mean

My idea was to like

take a topic

and focus mostly on it

for the rest of the year

And I was interested in doing it with algebraic geometry or algebraic topology

maybe eisenbud is going to be my new bible who knows

But yeah, I don't usually mind about really big books

[sleek thicket]

what the fuck, undergrad comm alg doesn't talk about tensor products

at all

I might need to stop recommending this book

[bright acorn]

I usually read slow and take a months to read the entire thing

[sleek thicket]

it does the snake lemma and primary decomposition

no tensor products

[bright acorn]

and it feels like a journey

[gritty widget]

tensor products

wow if only i learned that!!!!!!!!!

[sleek thicket]

okay well I would say this book is perfect if you fear tensor products

because

it doesn't do them

????

[gritty widget]

Tensors make me sad

[sleek thicket]

I mean I liked everything I read from it

[bright acorn]

I remember that I tried learning about tensor products a year ago. And this year I tried relearning it and did so much better 😄

Mainly because

Now I got a better feeling of what a universal property is

And how important it is to prove things relating to the tensor product

instead of explicitly giving a construction of it

Which is a pain

It feels so weird when you finally grasp a concept after so much time

[sleek thicket]

Definitely haha

[quartz edge]

broh wtf

this book is saying a vector field is parallel if its covariant derivative is zero along a curve on a surface

but then it says theres a unique parallel vector field for any curve

as long as the vector field lies in the tangent space at some point of the surface along the curve

i dont buy that

like take a plane as the surface and make a vector field of normals

all different lengths

have some zero vector somewhere

[delicate hollow]

ok

quick question

actually

ill give overview of proof

So im asked to prove f:Z->X and g:Z->Y are continuous iff h:Z->XxY is continuous

I decided to do (<=) first

Assuming h:Z->XxY is continuous. Since XxY with product topology has continuous projection maps p1:XxY->X and p2:XxY->Y. We can take an open set U in X and p1^-1(U) is open in XxY and subsequently h^-1(p1^-1(U)) is open in Z. So an open set in X is open in Z.
Similarly we can take an open set V in Y and p2^-1(V) is open in XxY and subsequently h^-1(p2^-1(V)) is open in Z. So an open set in V is open in Z.

Ill wait for approval on this part before I give the rest

[little hemlock]

looks fine for that direction. the point is basically that f = p1h, g = p2h, and the composition of continuous maps is continuous

[delicate hollow]

but p1 doesnt take X it takes XxY

My explanation for second implication is handwaving in that I pretty much let T be an open set in XxY, so T is Union of open Ai in X and Bi in Y. So f-1(Ai) and g-1(Bi) are open in Z

Also p1^-1(Ai) and p2^-1(Bi) are both open in X x Y.

So any open T is open composed of open sets which are open in Z, and since their projection mappings are continuous, they correspond to open sets in XxY, So sets in T are open in Z.

QED

Personally I dislike my explanation for the second direction because it isnt clear with notation

And Im sure it isnt perfect at communicating what I want.

[little hemlock]

but p1 doesnt take X it takes XxY
yes? Was there something wrong with what i wrote?

Yea, so the other direction is less straightforward, but to simplify things a bit, It suffices to check continuity on a basis. i.e. You only have to check continuity for open sets of the form U x V for U, V open in X and Y respectively. Again you'll want to write f = p1h, g = p2h and use continuity of this composition (along with the definition of p1, p2).

[quartz edge]

is it acceptable to consider a vector field to be a subset of a tangent bundle? or is this abusing notation/structure of bundles (considering the bundle is usually defined as a set containing all the points of a manifold paired individually with the tangent space at that point)

[bitter yoke]

I mean, you can consider the image of a vector field as a particular subset of the tangent bundle I guess

[gritty widget]

vector fields are injective functions M -> TM so why not

[quartz edge]

^ this is what i was thinking, though TM usually is like {(p, v) | p in M, v in T_p M}

[gritty widget]

right, so for each p instead of considering all v in T_pM, just consider one of em

that's a vector field

[quartz edge]

ye but the semantics r funky

in programming usually this is easier because you just can do elementOfTangentBundle[1] and get the actual vector

like

when you're dealing with typed data structures

ordered pairs etc

cadr if you want to be lispy about it

hear me out.

the correct term is apparently 'section' of a fiber/tangent bundle

but like

the terminology is harder to feel comfortable using idk

i need to read a lot more topology

[bitter yoke]

Idk, section to me feels like you're taking a slice of the tangent bundle so that's why it's called a section

[quartz edge]

in haskell i'd have written this as something like

selectorTermThingy :: (PointOnManifold, TangentSpace PointOnManifold) -> Vector (TangentSpace PointOnManifold)```

goofy but w/e

where i guess a tangent bundle would be

type TangentBundle = [(PointOnManifold, TangentSpace PointOnManifold)]```

just feels like i'm doing something wrong calling it a 'subset' of TM

cuz elements of TM are ordered pairs

[bitter yoke]

So are vector fields?

[quartz edge]

mmh, yes

[sleek thicket]

gristle your are actually getting to something interesting here

Do you know anything about dependent types?

[quartz edge]

hardly, i used to

[sleek thicket]

ah okay

Well the tangent bundle is the type family (x : M) -> T_x M

the total space (the actual space you're thinking of) is the dependent pair type Σ_x T_x M

So an element is a point in M and a point in the tangent space

[gritty widget]

UGCT!!!!!

[sleek thicket]

GO THE FUCK TO BED

[quartz edge]

UGCT!,!,!

[sleek thicket]

TOP TERRA

anyways gristle

A section of the bundle is a dependent function in Π_x T_x M

So to each point x it assigns a tangent vector

Hott makes a precise analogy between type families and fibrations

[quartz edge]

ah

[sleek thicket]

(the tangent bundle is a fiber bundle, so in particular a fibration)

Warning: I didn't actually read the previous discussion

[quartz edge]

relevant tho

ur very articulate shamrock

[sleek thicket]

Thank you I haven't been drinking very much

[quartz edge]

what's the sigma thing

[sleek thicket]

It's the dependent pair type

So say you have a type family

What is this

Well fix a base type A

Then a type family on A is a function A -> Type

One example is lists

If A = Type then List : A -> Type

This is like, parametric types

But there's also stuff like vectors of a fixed length

[quartz edge]

ye i saw those in idris

[sleek thicket]

We have a type family Vect : N -> Type where Vect n is the length n vectors of ints

nice!

And then finally, if A = M is a manifold we have a type family T : M -> Type of tangent spaces

Does this make sense?

[quartz edge]

one sec

yep

[sleek thicket]

So if you have a type family P : A -> Type, we can consider pairs (a, x) where a : A and x : P a. The type of the second coordinate depends on the type of the first, so this isn't just a product type. It's the dependent pair type, Σ_{a : A} P a

BTW, I'm actually TAing a course on PL theory/dependent types/coq right now :)

[quartz edge]

the 'type families' remind me of type constructors

[sleek thicket]

Yeah, that's when the base type A is the universe of all types

Like the list example

[quartz edge]

aha

[sleek thicket]

But here the base can be something more interesting, like the type of natural numbers

(like in the vector example)

[quartz edge]

ye

[sleek thicket]

Anyways, dependent pair types

Are what the tangent bundle is

[quartz edge]

oooooo

i get this now ok

yep that's very straightforward now that the notation is clear

awesome

[sleek thicket]

yeah!!!

Dependent types r cool

[quartz edge]

i would like to learn coq proper

maybe in a few months or so i will devote some time to this

[wide kayak]

can you recommend an introduction to this type stuff you're talking about?

is that homotopy type theory?

[sleek thicket]

Yup

[flint cove]

yooo what kinda backlog did I miss out on here
Bundles w/ Haskell and dependent types, that's something I didn't expect

[quartz edge]

just realized this text specified very explicitly 'a vector field X on M has X(P) in T_P M for every P in M'

smh

[gritty widget]

[cursive flume]

how can one prove that the exterior derivative of an n-form can be written as such in a chart?

[flint cove]

You use einstein summation, right?

[cursive flume]

yes

[flint cove]

The underlying property used here is that for smooth functions $f \in C^\infty(U)$, we have $d(f dx^I) = (\partial_i f) dx^i \wedge dx^I$, when the $x_i$ are local coordinates. Is your question why that holds?

[gentle osprey]

lux

[cursive flume]

yes exactly

why that holds

[flint cove]

Well, how have you defined the exterior derivative?

Because I know that as one possible definition

[cursive flume]

the axiomatic defintion, sec

[dusk heron]

Consider a Hermitian manifold $M$ with complex structure $J$, and let $\omega,\eta$ be $1$-forms such that $J\omega=\eta$. Is there then a complex coordinate $\zeta$ and a real function $u$ such that $\omega+i\eta=e^u \dd{\zeta}$? This is what I guess a paper uses, but I'm not sure how to see this.

I am not very familiar with complex manifolds in general.

[flint cove]

In that case I would give the hint that $f dx^I = f \wedge dx^I$

[gentle osprey]

lux

[flint cove]

since wedging with a 0 form is literally just multiplication

[gentle osprey]

gustavn64

[cursive flume]

this follows from 4. axiom?

[flint cove]

What you want follows to part from axiom 4, yes. The fact that wedging a 0-form is multiplication follows from the definition of the wedge product.

Although I'm not sure why we need the lie derivative in axiom 2 instead of just saying df(X) = X(f)

but I may forget something stupid, I'm not that solid in the foundations

[frigid river]

If B is dense in A, does this mean that B must be open in A?

[gritty widget]

are the rational numbers an open subset of the real numbers?

[gritty widget]

(the answer is "no")

[cosmic socket]

Isn’t it the case that B must never be open. Because some neighborhood of B contains some point in A.

Or is there some counter example

If B and A aren’t equal

[gritty widget]

R\{0} is open and dense in R

[cosmic socket]

Nvm it should be B can never be closed

[gritty widget]

see pins in #advanced-analysis

[cosmic socket]

Lol

Amazing

[cursive flume]

related to my question yesterday,I managed to prove it for holonomic basis,i.e. the coordinate induced basis on the vector fields

does anyone have ideas how to extend the proof for non-coordinate induced basis?

trying to rpove equivalence of these 2 definitions

[gentle osprey]

Not A Horse

[winter rose]

I was asked to show that if ${\tau_i }$ is a family of topologies on X, then \cap \tau_i$ is a topology on X, then $\cap \tau_i$ is also a topology on X. The correct solution is to let $U_k \in \cap \tau_i$, for all k. Then argue that $U_k \in \tau_i$, for each i and k. Then conclude that $\cup U_k \in \cap \tau_i$. But my question was whether this is necessary. After all, if we start by saying "let $U_k$ be in $\cap \tau_i$, for all k, wouldn't the union over all k of $U_k$ be in $\cap \tau_i$ right away?

[gritty widget]

you didnt close your first $

[winter rose]

whoooops

[gentle osprey]

kirafa
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winter rose]

I guess what I'm saying is that I don't see why we would need to use the fact that the each of the individuals tau's are topologies on X

[hard wind]

I'm not exactly sure how to show this

[winter rose]

isn't it true that if $A_i \in B$, for all i, then $\cup A_i \in B$?

[gentle osprey]

kirafa

[winter rose]

sorry, i meant to write $A_i \subseteq B$

[gentle osprey]

kirafa

[winter rose]

really? Can't we just specialize to the case where $A_i = U_k$ and $B = \cap \tau_i$?

[gentle osprey]

kirafa

[winter rose]

oh i think i'm starting to see the problem

we assumed that $U_k \subseteq \cap \tau_i$, but we're tying to show that $\cup U_k \in \cap \tau_i$ right?

[gentle osprey]

kirafa

[winter rose]

oh right

i'll try it again and see where the reasoning fails - thank you for your help

oh, i see what you mean now

i see the mistake i made

they're elements, not subsets

i was getting it confused because the elements themselves are sets

ok, i see it clearly now, thanks again

[flint cove]

So what is your specific question? I'm not sure what you mean by „non-coordinate induced basis“

[uncut surge]

Probably means how to do it globally when you don't have a coordinate basis available, i.e. you're not in a single chart

in that case: partition of unity time my d00d

[flint cove]

I would be confused if we needed such heavy machinery for an equality like that

especially since that is determined purely locally

[frigid river]

I have a short, dumb question: Q is dense in R, right?

At least that would intuitively make sense to me.

Okay, thanks. Does this also imply Q is open?

Since $closure(Q) \neq Q$?

[gentle osprey]

veryhappyperson

[gloomy plover]

Not closed does NOT generally imply open

[frigid river]

😦

Ahh, okay.

[gloomy plover]

Q has empty interior in R

Q is doing a more impressive job in my eyes

[chrome dew]

in fact you can make R from Q by cauchy sequences

[wanton marsh]

Q is closed in itself

[chrome dew]

not exactly

there are other completions of Q depending on the choice of absolute value

at least Ostrowski's theorem shows these are the only possibilities

yeah it's not too difficult either and is actually a bit more general

you can relax it to something called a generalized absolute value and even then it still holds

[gritty widget]

I quite like that name, "OstrowskI"

[frigid river]

May I ask another stupid question? A dense subset, A, of a metric space, S, can only be closed if S=A. Is this correct?

[gritty widget]

yes

see #advanced-analysis pins

[frigid river]

Thanks.

[sweet wing]

wait wdym

[gritty widget]

[sweet wing]

oh lmao

lmfaoLOL

[gritty widget]

yes

[sweet wing]

we have a legend here

[cursive flume]

what I mean is that I used that $X=\partial_i$

[gentle osprey]

ProphetX

[cursive flume]

what if I would use $X^{i} \partial_{i}$

[gentle osprey]

ProphetX

[cursive flume]

the issue is then that the computation gets so tedious I get lost 😦

[flint cove]

Yeah, I figured. I don't know the solution, and I cannot concentrate today
it's a good exercise tho, perhaps I find time to think about it 🙂
this „omit then commutator“ formula is actually one of the things I never really looked at in my DiffGeo class

shame on me I guess

[cursive flume]

yeah no worries,let me know if you have any idea how to do it 😅 thanks! 😄

[summer jolt]

I was looking at the solution to the following question:

Now intuitively, I would think that the map from the cone to X is the same as the homotopy map so that means they are equivalent. Am I correct in my deduction?

Since I assume that the equivalence class [z,0] will be sent to a single point in X

[flint cove]

note that the explicit definition is additive because every operation used distributes
And if I see it correctly the you can show that it also commutes with the C^\infty(M)-multiplication

(as does every proper differential form, since its values are defined pointwise)

[cursive flume]

yes showing how it commutes with c^infty multiplication is what is truggle with

diff forms for sure do cause they're c^infty multilinear

but why wedge

[flint cove]

Ah okay
I proved it for n=1 only

What do you mean by „why wedge“

[cursive flume]

sorry

i menat exterior

[flint cove]

I'm confused because the question misses a verb, I know what wedge means :)

We can't wedge vector fields like we can multiply them with functions so that doesn't type check for me

[cursive flume]

i meant if i proved the equivalence on exterior derivative for partial_i partila_j

to prove for x^i x^j partial_i partial_j

that would mean i could pull out x^i x^j

and apply proof from before right?

(soryr for sloppy formulation,im physicist)

[flint cove]

Well our d\omega is then a C^\infty(U)-linear combination of the d\omega-values at the basis vector fields \partial_i (for each argument), so yes

[fading vale]

@summer jolt yeah, a map on the quotient is equivalent to a map on S^1 x I thats omega on S^1 x 0 and constant on S^1 x 1

[cursive flume]

right so what I have to prove is multilinearity of d omega

[fading vale]

(Sorry about interrupting its a short aside so i think its not too intrusive :p)

[flint cove]

I don't fully grasp how to solve it yet but it's a powerful way of resucing questions of d\omega to questions of the values at the basis vector fields

[cursive flume]

we know that it maps forms to higher forms but we gotta check multilinearity

thats not an axiom is it?

[flint cove]

Well technically we would have to check that d\omega is actually such a differential form as claimed in the codomain

Which may be half of the work here, but is nontrivial

*as claimed by saying that the codomain is \Omega^{foo}

[cursive flume]

yes exactly

[flint cove]

I'm too drunk to want to look up the wedge formula I'm afraid.
I just remember that it was a bit complicated.

[cursive flume]

it's ok dw,i'll try to check it,it's for certain a good exercise 😅

i didn't find the whole proof in any dm book so far yet

[flint cove]

I feel like there must be an elegant way tho

[cursive flume]

yeah there must be

computing/brute forcing is hell

also another trivila question mb but i can't see

how can one prove uniqueness of such a map? even if we prove equivalence of these 2 definitions

[uncut surge]

The uniqueness from the axioms follows once you know that you can write every differential form as a finite linear combination of terms $f dg_1 \wedge \dots \wedge dg_k$

[gentle osprey]

Lartomato

[uncut surge]

Because then: The product rule axiom implies that you can calculate the exterior derivative by calculating the exterior derivative of all the factors f, dg_1,... ,dg_n; the axiom that d^2 = 0 tells you that the exterior derivative of the dg_1,...,dg_n is zero; and the exterior derivative of f follows from the last axiom

The fact that you can write every differential form in this way can, for example, be proven with partitions of unity over coordinate sets, if you understand that locally every differential form is something like f dx_1 \wedge ... \wedge dx_n

[cursive flume]

how can one prove that? can you give a reference for the proof with partitions of unity please?

[pale sky]

was is the relation between orientability of a surface and torsion?

[sleek thicket]

do you know about the orientation double cover?

hm, I'm not sure this is exactly right but it's certainly relevant

basically if you have a manifold M you can construct a double cover it whose points are pairs of a point in M and a local orientation at that point

this is connected iff M is not orientable

this does connect orientability -> covering spaces -> pi1(X) but it's not quite right

I think for compact manifolds you can do some stuff with Z-orientability as defined in Hatcher

aha! see Corollary 3.28 in hatcher

@pale sky

[pale sky]

no whats that?

[sleek thicket]

this

but I think this statement from hatcher is more relevant

[pale sky]

ah thank you

ahhahaha

[sleek thicket]

haha

[pale sky]

exactly what i was looking for

[sleek thicket]

nice!

[pale sky]

tyty ❤️

[sleek thicket]

That section talks about orientation in a lot of depth

would recommend reading it

[gritty widget]

what about orientation and torsion of a connection in a vector bundle over M

[tawdry widget]

I have a question related too. I don’t understand the relationship between two definitions of orientation. One is defined in differential geometry, each x from M define an orientation on T_x M . The another one is defined in algebraic topology, like each x from M choose a generator of H_n( M, M \ {x},Z) or something. Those two definitions, are they equivalent? How?

Never mind I need to learn algebraic topology. Someone told me that Hn(M, M \ {x}, Z) is isomorphic to Z . I don’t know that.😂

[shut moat]

how exactly do you motivate the Urysohn lemma and its proof?

so far I've seen it used in the proof that all compact manifolds can be embedded in R^n

but like, I wouldn't have known to ask that question about normal spaces

if you know what I mean

[ivory dragon]

i think munkres literally just says "thisll be needed for the urysohn metrization theorem"

and doesnt elaborate more

even though he gives a full chapter

i agree, im just explaining the munkres approach

[shut moat]

walking backwards from the urysohn metrization theorem feels reasonable

also lol are point-set-topologists an actual thing?

[gritty widget]

Wait, you guys think urysohn lemma is unmotivated?

[shut moat]

do you know a motivation? :o

[gritty widget]

I don't know what you would take as a good motivation, but it feels apparently "right"

And I think there are moments where you have wondered, "what is a condition under which I can have a continuous function that distinguishes these sets?" And urysohn comes to the rescue

I guess so

[shut moat]

well ty

[fierce glen]

I was trying to find proof for this in my textbook, but didn't find any; does anyone have proof for this: \
For any oriented link $S^{3}$ is a closure $\hat{\xi}$ of some beginning $\xi$ to a braid group $B_{n}$, for some $n$ and the oriented links $\hat{\xi}$ and $\hat{\eta}$ are equivalent if $\xi$ and $\eta$ differ by a sequence of Markov moves of: \
\begin{itemize}
\item[i{,}-] Changing an element of $B_{n}$ to a conjugate element in the group.
\item[ii{,}-] Changing $\xi \in B_{n}$ to $i_{n}(\xi) , \sigma_{n}^{\pm 1} \in B_{n , + , 1}~\text{for}~i_{n} : B_{n} \to B_{n , + , 1}$, such that it's the inclusion.
\end{itemize}

[gentle osprey]

tryme

[fierce glen]

The book just blankly says that hence if you define $w : B_{n} \to \bZ$ it will be a homomorphism defined by $w(\sigma_{i}) = 1$...

[gentle osprey]

tryme

[lavish dove]

Heya, diff.geo problem here I believe, on the side of geodesics

I am trying to find a way to compute the distance between two points on a sphere along the sphere (context : two points on the earth)

I've searched up on the true circle method but I conjectured another method but it's not working out, could anyone let me where the gap is in my thinking

I took my two coordinates (in form of longitude & latitude), converted them into cartesian form

Expressed those two points as position vectors, then found angle between those two points using the dot product.

After that I used arclength formula for circles

still getting ~1000KM off for some reason

[gritty widget]

without knowing your computations all we can say is "check your computations"

[lavish dove]

have checked numerous times but still not sure where the problem lies

😔

the two position vectors both have length 6378 so they should be correct

true distance should be 5500km

not sure what I'm doing wrong.

I just tried the geogebra function to find the lengths

There is some fault in the vectors because my calculated length is not wrong

Maybe it's converting from longitude and latitude into spherical coordinates, do you have any idea about that?

oh my goodness, I was right, I was converting the angles wrong, I'm gonna go have a cry, thank you champ

[quasi forum]

So intuitively, why does it matter that compactness requires every cover to have a finite subcover?

This is what we started with in terms of motivation

How'd it jump from there to here

I get the idea of finite subcovers. I'm just trying to wonder why it needs to be true for every possible cover

I guess that's what is throwing me off

Well, I guess why does the provisional definition imply that it needs to be true for every cover?

Because isn't a family of open balls just one cover?

Here it does

So is the family of all open balls all the open covers of A?

Okay, that's what I was getting at. The book primarily focuses on metric spaces, so I'm not too concerned about the general sense just yet.

Is this kind of a similar idea to how the balls of a metric topology form a basis? So although there may be other open sets, just considering the balls is sufficient enough.

Okay beautiful. I just wanted to make sure. Thank you for sticking through it with me. I wasn't explaining my thought process too well :p

Yea, I agree with that

It is very easy to show the second claim is always true.

A is it's own cover

Not very interesting

True. But there are simplistic covers that don't tell us much

And it doesn't necessarily need to be true that the family of covers in inside A, just that A is in the family of covers

So as you mentioned, the topological space is a perfect example

Q for instance

Now is Q compact?

If not, what would be the thing that makes it not work?

Oh, a subset is closed iff it is compact

I forgot about that

What was it?

Oh, so it's specific to R?

Probably because of completeness I am guessing

[chrome dew]

as a hint, you can use an open cover for R that has no finite subcover to do the same thing as for Q to prove it's not compact, try drawing a picture and try covering it up

[quasi forum]

Wait, I said Q

Silly me

[quasi forum]

Ah, (-n, n)!

That union of sets is R, but there is no finite sub cover of the union of (-n, n) that contains R

[chrome dew]

yeah looks like it works to me, I had a different one in mind

[quasi forum]

What was yours?

[chrome dew]

cover it with (n-1,n+1)

if you remove any single one, you end up with a hole

[quasi forum]

Ah, yeah that works too!

[chrome dew]

well, maybe a few more than 1 needs to be removed but you get the picture lol

[quasi forum]

I sure do. Thank you you two

[chrome dew]

are the p-adics compact? 😛

[quasi forum]

Are we referring to $\mathbb{Q}_p$? If so, I don't know much about this extension

[gentle osprey]

dackid

[quasi forum]

But if we are referring to the p-adic absolute values, absolutely not.

The union of (0,n) spans R_+ which contains all the possible values of p-adic distance in Q. But once you remove one, you lose part of that set

So (0,n) does not have a finite subcover for the p-adic distance

[chrome dew]

ah, what are the open sets?

like, in particular how do you know the specific radii corresponds to open sets of an open cover?

[quasi forum]

Okay sure. P-adic distance is a metric. So we can just take balls around the point 0, given by $B_n(0)$

[gentle osprey]

dackid

[quasi forum]

So $\mathbb{Q}\subseteq\bigcup_{n\in \N} (0,n)$

[gentle osprey]

dackid

[quasi forum]

Ayy, I did it right.

And the p-adic distance maps Q into Q

So I think it just ends up being an iteration of the same question 🤔

Actually I have a class on compactness today. Let me do that so I can clear things up. I think I see how to solve the problem, I just need to be careful

[frigid river]

"Prove that the set of isolated points of a countable complete metric space X forms a dense subset of X." What is an isolated point? A point which has no adherent points maybe?

Or is an isolated point just a set with just one element in it?

[gritty widget]

x is isolated if it's the only point in some open set

[frigid river]

Wait, how is that supposed to be open then?

(0,0)?

Since it is the only point, the set must be a single point set, right? But how can such a set ever be open?

[gritty widget]

consider an example, X = {0, 1, 1/2, 1/3, 1/4, ...} with the subspace topology inherited from R. 0 is not isolated, but all other points are

e.g., 1 is isolated, since {1} is open in X. indeed, {1} = X \cap (1/2,3/2)

or in any discrete space, all points are isolated!

[frigid river]

I still don't get how {1} can be open in any metric space? If I recall correctly, a set is open if interior of the set is equal to the set. And a point is an interior point if I can draw a ball centered at that point with some radius bigger than 0 and still be within that set. How does this work for {1}?

I am sorry if I am asking stupid questions, but I just don't get it.

[quasi forum]

Since you are dealing with metric spaces, are you familiar with the discrete metric?

[frigid river]

That was the metric where the distance between two points was either 0 or 1, right?

[quasi forum]

Yep. And it is only 0 if they are equal

Choose a point $x\in X$. Then $B_r(x)={x}$ if $r\leq 1$.

[gentle osprey]

dackid

[quasi forum]

Since this is true for every point in X, every point is an isolated point.

[frigid river]

Thanks, that I get now. So a "single point set" is open in the discrete space. This does not apply for the "usual" metric space though, right?

[quasi forum]

Recall $B_r(x):={y\in X| d(x, y)<r}$

[gentle osprey]

dackid

[quasi forum]

Nope, it does not

[frigid river]

So an isolated point can only exist in a discrete metric? Because how is such a set then supposed to be open?

[quasi forum]

I'll take that back because it seems you are trying to prove something like that :p

Not necessarily no

It is just not generally true for all metric spaces

[gritty widget]

btw the example i gave is something you should keep in mind for your problem.

[ivory dragon]

its just that connected sets have no isolated points.

R is connected, hence no isolated points

[frigid river]

Okay, thanks guys, I will try the problem now.

[ivory dragon]

it doesnt really work in R either

consider [0, 1)

well theres a relationship in that "closed" means "complement of open"

but the closedness of a given set has no relation to whether that set itself is open

[frigid river]

So an isolated point of a metric space is the only point in an open subset of the metric space, or am I getting this wrong again?

It doesn't state it in my book -.-

[ivory dragon]

a point x is isolated if {x} is open

so yes, if there exists an open subset of the space which contains only x

[frigid river]

Okay, thanks again.

[quasi forum]

Remember how I said I had a class on compactness today? Well, the zoom servers decided to shut down 5 minutes before it started -_-

[frigid river]

Nice.

I am sorry to annoy you again but could you give me an example for a metric space that is countable and complete?

It just says: "Prove that the set of isolated points of a countable complete metric space X forms a dense subset of X."

So I guess it may be finite?

Yes.

Does such a thing also exist for d(x,y)=|x-y|? I am still having troubles dealing with the discrete metric.

Thank you. I didn't even now Z was complete, honestly.

Well, now I know that all of them must converge, since it is complete xD

But proofs are hard -.-

[quasi forum]

Topology is a proofs course 🤷

[frigid river]

I like reading proofs. Writing them is another thing xD

[quasi forum]

I don't think the class cares too much about your ability to read proofs 😆

Try to prove this. If you have a cauchy sequence in $(\Z,|\cdot|)$, then it is eventually constant.

[gentle osprey]

dackid

[quasi forum]

That's the key piece to what slimvesus mentioned earlier

[frigid river]

Thanks, give me a day or two, I might have figured it out by then xD I will focus on the problem I sent though first, since the next Chapter is called "The real line" and I think that involves the proof of R being complete. I might be able to copy something then.

$\lim_{m,n \to \infty} d(x_n,x_m)=0$?

[gentle osprey]

veryhappyperson

[quasi forum]

Gonna wanna epsilon this to see what's going on

[frigid river]

So, for each $\epsilon >0$, there exists an $N$ such that $d(x_n,x_m) < \epsilon$ for all $n,m \geq N$.

[gentle osprey]

veryhappyperson

[frigid river]

The metric could be anything from $[0,\infty)$

[gentle osprey]

veryhappyperson

[quasi forum]

If this is true for any epsilon, what must d(x_n,x_m) be when epsilon is, say, 1/2?

[frigid river]

@gritty widget 11

@quasi forum d(x_n,x_m)<1/2

[quasi forum]

Sure, but what must d(x_n,x_m) be equal to?

[frigid river]

so 0

Well because there are no decimals in Z xD But I don't think that is what you want to hear.

[quasi forum]

It's a good deduction though

[frigid river]

Agreed.

x_n=x_m then

[quasi forum]

For all n, m>=N

don't forget that tiny detail

[frigid river]

That is is cauchy, because the above mentioned limit is 0

[quasi forum]

That was the original assumption

[frigid river]

If it is constant, I cannot just assume it converges, right?

well yes I can.

[quasi forum]

Remember, this was what we were trying to do originally

[quasi forum]

Sure can! Because the term it converges to is in the sequence (x_n)

And it is already assumed that $(x_n)\subseteq \Z$

[gentle osprey]

dackid

[frigid river]

Okay, but why did we conclude it must be constant again?

[quasi forum]

Because for $\epsilon \leq 1$ there is an N so that $d(x_n,x_m)=0$ for all $n,m\geq N$

[gentle osprey]

dackid

[frigid river]

I think I get it. Because of the definition, I can pick epsilon < 1, and then it must be constant, right?

[quasi forum]

For sure. The epsilon just gives you an explicit understanding of what is happening

[frigid river]

Okay, much thanks again.

[quasi forum]

Now the final take away is every Cauchy sequence is eventually constant, and so it converges

What does that tell us about the space $(\Z,|\cdot|)$

[gentle osprey]

dackid

[frigid river]

Only in a complete space though, right?

[quasi forum]

That is the definition of a complete space

A space is complete if every Cauchy sequence is convergent

[frigid river]

Yes.

Ahh, you were talking about Z

[quasi forum]

So yeah, $(\Z,|\cdot|)$ is a complete space

[gentle osprey]

dackid

[quasi forum]

Sorry if I butted in slimvesus. I learned this stuff recently so I kinda got a bit excited that I actually knew how to help with the problem. :P

[frigid river]

Doesn't this then also imply that Z is closed in R, since Z is complete?

[quasi forum]

Well, is it?

[frigid river]

-.-

[quasi forum]

Also, how do you know $\R$ is complete? Wasn't that thing you said you were all gonna prove later on.

[gentle osprey]

dackid

[frigid river]

Yes, but it was stated in this chapter, and they said they were going to proof it later xD

[quasi forum]

Seems like quite the assumption to me

Also yes, I am going to be that guy 😎

[frigid river]

And if I may bother you one last time (for today), for "Prove that the set of isolated points of a countable complete metric space X forms a dense subset of X.", do I go the usual way of trying to proof Closure(set of isolated points)=X or should I try that differently?

[quasi forum]

Sounds like a solid plan

[frigid river]

Okay, thanks.

[summer jolt]

Hey I'm learning about One-Parameter Groups of Transformations on smooth manifolds. I have a question about "infinitesimal generators" of such maps. My book mentions this

My question is the infinitesimal generation of a transformation uniquely defined?

Because it seems to me that the condition requires to fix the value of the vector field at t=0 but not elsewhere

[gritty widget]

X is uniquely defined, although i'm not sure i understand your question. the vector field isn't a function of t so i'm not sure what you mean by "fix the value of the vector field at t = 0."

[summer jolt]

what I meant was that the tangent vector that we "pick" at x is the tangent vector to the curve passing through that point

However, I don't understand how that condition alone uniquely defines the vector field.

[gritty widget]

given a one-parameter group of diffeomorphisms $\varphi\colon M \times \mathbb{R} \to M$, and a point $x\in M$, write $\varphi_x(t) = \varphi(x,t)$. then $t\mapsto \varphi_x(t)$ is a smooth curve in $M$ passing through $x$ at $t = 0$, so it defines a tangent vector, call it $X_x$, in $T_xM$, obtained by differentiating the curve $t\mapsto \varphi_x(t)$ at $t=0$

[gentle osprey]

Buncho Wolves

[bitter yoke]

I think that your confusion might be that X is defined by setting X_x = (\varphi_x)'_0 for all x, not just doing it for a single x in M

[summer jolt]

Ah I see! Yes, thanks that makes sense now.

BTW any idea why is this called the "infinitesimal generator"?

[gritty widget]

from lee's ISM, page 214-215: The term “infinitesimal generator” comes from the following picture: in a smooth chart, a good approximation to an integral curve can be obtained by composing many small straight-line motions, with the direction and length of each motion determined by the value of the vector field at the point arrived at in the previous step. Intuitively, one can think of a flow as a sequence of infinitely many infinitesimally small linear steps.

[sleek thicket]

didn't you need to do like symplectic geo or sometihng

[gritty widget]

i am doing symplectic geo!

[sleek thicket]

also i like jjk

[gritty widget]

i figured out what to do my homework on

im going to fill in some of the details on the computations my prof did

since i didnt get them the first time around, and theyre kind of related to my presentation

[sleek thicket]

sounds "fun"!

if i could plug myself into diff geo computations all day long I would 😌

ohhhhhhhhhh

:himb: is from the end scene

[gritty widget]

yes

[sleek thicket]

i was wondering when it happened

in the show

nice

[gritty widget]

i think the dancing in the ending is based off of one of the manga extra chapters that came out around the time the anime started airing

[sleek thicket]

i like the characters a lot so far

aww and his uniform is cute

little red hood

[gritty widget]

[sleek thicket]

im getting naruto vibes

just overall

the trio

the teacher

[long coyote]

Let $\alpha:I\to S^2$ be a path. Supposeαis not surjective. Show that $\alpha$ is path homotopic to an injective $\beta:I\to S^2$

[gentle osprey]

亜城木 夢叶

[gritty widget]

maybe a starting point is by noting that any path on S^2 that misses a point becomes a path in the plane

via stereographic projection from any point on S^2 not hit by alpha

[tough imp]

Writes \alpha, inserts a real alpha, writes \alpha again

🤔🤔

Good idea Buncho

[gritty widget]

buncho wolves is a tterra alt

[long coyote]

🤔

[sleek thicket]

Harder version of this question: prove the same thing but for the surjective case

[tough imp]

Let X be an affine variety and Y a constructible subset. How could I show that Y contains an open set which is dense in the closure of Y?

[sleek thicket]

Pretty much yeah

[elder viper]

https://cdn.discordapp.com/attachments/497227343337881640/828852600026824714/image0.png

need help with this

[sleek thicket]

There's a typo in the exact sequence

The coefficient rings 🥪 and 🍔 are swapped in a couple places

You should email your prof and let them know

anyways, this is a pretty easy application of the universal coefficient theorem for cohomology. Check out Hatcher's book for more info

[tardy meadow]

https://mathoverflow.net/a/296605 - may be being a bit stupid but I can't see where \overline \Omega = \R is actually used in the proof

doesn't seem to impact the proof at all to just cut it out, since you're mainly working on the complement of Omega

[uncut surge]

There's a lot of things I don't get there; why can we assume in the first place that \Omega is a finite union of open intervals

ah fuck okay it's just annoying notation nevermind

" Since for every x∈X∩(a,b) there is a sequence xi→x, xi≠x such that f(n)(xi)=0" This sentence after formula (3), maybe this is where density is used? I've not quite yet understood where that sequence is coming from

Ah, okay, from X cap E_n having nonempty interior in X

i conclude that i have no idea where density is used

[marsh forge]

I also cannot figure it out lol

As a side note

Yeah nvm I don't see why its relevant lol

[tardy meadow]

ye thanks, thought I was missing something obvious. agree there's a lot of clunk

[sweet wing]

yea it's just not needed lol

[tardy meadow]

don't get why they did this cos you can split up intervals to make it a countable union eveni f it can be written as a finite one [in fact I don't even think the fact it can be a countable union matters either]

[sweet wing]

it doesnt idk wtf is notation there lol

[gritty widget]

i'm convinced that everyone who uses MSE or MO is literally insane

[cloud owl]

what was the q

[dim radish]

this

[tardy meadow]

i think they read the quantifiers the wrong way around

[gritty widget]

someone posts literally anything
HEY THIS DOESN'T BELONG HERE THIS IS A TRIVIAL QUESTION GET OUT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[tardy meadow]

lmao where's this

[gritty widget]

https://math.stackexchange.com/questions/1323787/what-is-the-relation-between-c-infty-linear-and-tensorial

[tardy meadow]

i love mathematicians

[uncut surge]

oh my god i love how angry he is

[tardy meadow]

OP ran over his cat and this is his revenge

[uncut surge]

i don't know why i made this

[tough imp]

Still wondering about this. Someone save me

[tough imp]

Wtf not using pinning bot?

Mod aboos

[ivory dragon]

this is why i dont stackexchange

i dont want to get yelled at for poor pedagogy standards in an answer i gave 4 fucking years ago

[tough imp]

Wait

Holy fuck

Jesus Christ people are insufferable

[urban yarrow]

https://blogs.ethz.ch/kowalski/2016/02/14/is-mathoverflow-insane/

[sweet wing]

lmao

[uncut surge]

that blog looks really extremely cringe to me

"haha, having fun with math things? gamifying the concept of explaining math to others? absurd. to me, math is suffering"

[tough imp]

Kowalski, analysis

[marsh forge]

honestly

[uncut surge]

i can't help but read that paragraph in a heavy french accent

[marsh forge]

that last sentence

kinda woke

[uncut surge]

like, this post, is just so garbage

mathoverflow is a platform on the internet; every platform on the internet needs some sort of guidelines and rules, if it wants to make the experience for the user nice; therefore, it may be worthwhile for the people who care about this platform to sometimes debate about these guidelines

idk sorry i'm cringing

[marsh forge]

okay but like

MO def is toxic with the "is this a good fit" stuff

and it is rarely constructive

and MO is like dying as a result of it lol

[uncut surge]

as in, you think they take it too seriously? maybe, i'm not involved enough i suppose

[tardy meadow]

MO says it's about research maths but isn't it more >= graduate-level maths generally? The question I linked is nowhere near research level lol

[uncut surge]

https://mathoverflow.net/questions/231078/recent-observation-of-gravitational-waves but asking "should this question about gravitational waves be on MO?" seems like a very reasonable thing to do

[uncut surge]

i've always taken it as "if you're a post-graduate level math-person, and you're wondering about a question which feels very specific or difficult to you, then MO is an okay fit"

[marsh forge]

This is not even the case

[tardy meadow]

yeah this is my understanding too

[marsh forge]

its often like a pissing competition

[uncut surge]

of course you can't ever perfectly know if your super sophisticated question can be solved with bebby's first arithmetic, or the other way round

[marsh forge]

about what is "good enough" for MO

which leads to very few people ever feeling comfortable posting a Q there instead of MSE

[cloud owl]

no true scotsman etc.

[marsh forge]

Certainly there are "graduate level" questions that would be laughed off of MO

[uncut surge]

i do admit, i always find MO scary to post in, but i've never had any personal bad experiences there

[tardy meadow]

Yeah I find MSE scary enough I dread to think in a few years where my questions might warrant posting to MO :p

[uncut surge]

i find that if you explain where you're coming from, or why you feel like your question is complicated to you, then you usually get understanding answers

that may not always be the case tho i'm sure; like all communities, it's made up of people, and math people are really unpleasant sometimes

[ivory dragon]

if you havent worked on a problem for at least 60 hours and cross referenced 10 different textbooks and skimmed 300 pages of arxiv

youre not allowed to post it on MO

thems the breaks

[sleek thicket]

.pin

[ivory dragon]

the end goal is to make it so that only literal open problems can be asked on MO

and drive humanity's mathematics knowledge forward through the incentive structure that is MO reputation

[sleek thicket]

The end goal is to define all of mathematics as at the level of a fist year calculus class

[ivory dragon]

someone put RH on mathoverflow and bounty it

ill reply that it seems at the level of a first year calculus course

[gritty widget]

that is so funny

[marsh forge]

We will eventually understand that all of known mathematics is "first year calculus" and that open problems are simply an infinitesimally fattened version of first year calculus

[viral atlas]

[tight agate]

are all specializations modeled by DVRs?

i.e. if I have a scheme X

and a point x which specializes to y

then is there a DVR R and a map Spec R --> X such that the generic point gets mapped to x

and the maximal ideal gets mapped to y

hmm okay the stacks projects says that it is true if we assume that X is locally noetherian

[broken robin]

Hi everyone, can somebody help me with scheme theory? I have an elementary doubt

[frozen otter]

This is much more on-topic here than it was in #category-theory, so I’m going to follow up on the conversation I had a couple days ago on equivariant convolutional neural networks here. An introductory reading list for equivariant DL
@marsh forge @limpid vault @gritty widget

Practical Papers:
Group Equivariant Convolutional Neural Networks https://arxiv.org/abs/1602.07576
Probabilistic symmetries and invariant neural networks https://arxiv.org/abs/2004.05154
A Group-Theoretic Framework for Data Augmentation https://arxiv.org/abs/1907.10905
Gauge Equivariant Convolutional Networks and the Icosahedral CNN: https://arxiv.org/abs/1902.04615
Generalizing Convolutional Neural Networks for Equivariance to Lie Groups on Arbitrary Continuous Data: https://arxiv.org/abs/2002.12880

ML Theory:
On the Generalization of Equivariance and Convolution in Neural Networks to the Action of Compact Groups https://arxiv.org/abs/1802.03690
A General Theory of Equivariant CNNs on Homogeneous Spaces https://arxiv.org/abs/1811.02017
Theoretical Aspects of Group Equivariant Neural Networks https://arxiv.org/abs/2004.05154
Intertwiners between Induced Representations https://arxiv.org/abs/1803.10743
On the Benefits of Invariance in Neural Networks https://arxiv.org/abs/2005.00178

Transformers:
SE(3)-Transformers: 3D Roto-Translation Equivariant Attention Networks https://arxiv.org/abs/2006.10503
LieTransformer: Equivariant self-attention for Lie Groups https://arxiv.org/abs/2012.10885
Group Equivariant Stand-Alone Self-Attention For Vision https://arxiv.org/abs/2010.00977

Prior convo: #category-theory message

[marsh forge]

oh boy a lot of reasing

[manic garden]

show us your doubts

[manic garden]

i think in general they're modeled by henselisations of local rings

but over the separable closure it aligns with a dvr

or maybe the other way

[frozen otter]

It’s definitely not necessary to read it all, but I had this reading list lying around and I figured I might as well be thorough

[frozen otter]

The first and last paper under “theory” are the most interesting ones, IMO.

[marsh forge]

Hahaha no its a good thing, thank you

[tawdry widget]

Since we can find an affine subset U defined by a noetherian ring A containing y (x also since y belongs to the closure of {x}) therefore we can assume that X is an noetherian affine scheme.

I remember that local noetherian rings are DVRs

So could we just take R to be the localization of A/p at the point q/p

[tough imp]

Local Noetherian rings are not DVRs

You need to enforce more conditions like being a domain and having principal maximal ideal or being integrallt closed and dim 1

[tawdry widget]

Oh then forget what I said😂,I would like to know the answer either

[tough imp]

I mean I agree

Definitely restrict to looking at affine Noetherian schemes

But idk what to do from there

[tawdry widget]

Me neither...

[sleek thicket]

a noetherian valuation ring is a dvr right? bc finitely generated ideals are principal, so it's a PID?

[tawdry widget]

Yes DVR implies PID

[sleek thicket]

this is true

i do not see how it is relevant to my question

I'm asking about valuation rings in general

[tawdry widget]

Btw Hartshorne seems difficult to me. Lots of people say that it shouldn’t be, but I still need to ask :Is there any easier materials to read? What books do you read?

[sleek thicket]

it is very difficult

I read hartshorne and now I don't know algebraic geometry

[tawdry widget]

Finally! I thought that something was wrong about me, about my brain.

[sleek thicket]

no absolutely not

I have cried in the math lounge at my school doing hartshorne problems

@tough imp is a hartshorne master and he will tell you it is a very hard and frustrating book

[tawdry widget]

I see. Any other recommendations?

[tough imp]

Lmao

I am@not a master

I am simply a victim

Anyway Hartshorne very hard

Makes me want to cry sometimes but instead I scream into my pillow

If you’re still on chapter II maybe check out Algebraic Geometry I by Görtz and Wedhorn

It doesn’t touch cohomology but is pretty thorough on Schemes

It’s also a lot more functorial which is eventually the true path and super useful

But it could be a bit too abstract if you’re starting out, but it does work it’s way up to that

Besides that there’s this other book I haven’t personally used but

A book by Bosch, I think it might just be called Algebraic Geometry

It seems pretty good

[tawdry widget]

Thanks I will check it out.

I can’t afford to be stuck any longer

[tough imp]

What are you stuck on

Everything is a perfectly valid answer

[tawdry widget]

😂😂

[tough imp]

This was a real question tho

If it’s like II.5 or II.6 for example those were insanely hard sections for me

I think II.6 is pretty much everyone’s hardest section

[remote beacon]

Anyone wanna explain to me what a "bundle" is?

Like vector or fiber bundle

[tough imp]

It’s a topological space living over your space (meaning it has a map to the space, call this space X) such that this map (think of it as a projection) looks like the projection map X x F locally

So depending on what kind of bundle you talk about what F is here is a bit different

But let’s just say F is a topological space right now, maybe the real line then you have this map X x F -> X which just projects the first coordinate

Think of X as like some blob, then X x F looks like this blob but like as a cylinder above it and the projection map just squashes it

This isn’t the greatest photo, I couldn’t find a good one, but look at the top left

The base space X is like that circle on the bottom, then X x F sits above it, it looks like a bunch of copies of X stacked on top of each other right? You have a real line sitting over every point of X if F is the real line

So a bundle is something which locally looks like this, we call this kind of a bundle the “trivial bundle” because it’s well... the simplest one

So now more generally we’ll say that B is a bundle over X if we have this map f:B -> X such that

There’s an open cover of X by sets U_i such that

f^-1(U_i) is homeomorphic to U_i x F in such a way that this triangle commutes

The fact that this triangle commutes says that doing the map f from f^-1(U_i) actually behaves like the projection from U_i x F to U_i

So over U_i, B looks like the trivial bundle over U_i

The different words “vector, fiber, etc” basically tell you what F is

If it’s a vector bundle then F is a (real) vector space

For a fiber bundle I think there’s no conditions

[remote beacon]

So the mapping from $f^{-1}(U_i \to U_i)$ is what?

[gentle osprey]

KingArthur

[tough imp]

It’s just f

[remote beacon]

Because the other thing is basically projection

[tough imp]

Restricted to f^-1(U_i)

[remote beacon]

ooh

yeah

[tough imp]

But the other thing is

When you stipulate a certain kind of bundle you might also want these identifications of f^-1(U_i) to U_i x F to play nice

So have you seen a manifold?

Like a smooth manifold or w/e

[remote beacon]

I'm in intro topology and sadly no

[tough imp]

Ah okay well

It’s basically saying it looks kinda like R^n locally

[remote beacon]

It's basically R6N up close

R^n

[tough imp]

with respect to the differential structure of R^n

And what this means is like

Given a point on your manifold it might live in these two patches

Which youve identified with some open subset of R^n

You want to make sense of differentiable functions near x

So identifying a nbd of x with R^n we can kinda do that right?

We know what a differentiable function is on R^n

The problem is what if you had a function defined near x and under one identification to R^n it was differentiable

And for a different one it wasn’t?

We don’t want that, so we enforce that all these various identifications of patches of our manifold to open sets in R^n sort of “smoothly vary”

Aka it respects the differential structure on R^n

This is just fancy talk to say that no matter what identification you use, something will always be either differentiable, or not differentiable

So the reason I brought that up is let’s say you wanted a vector bundle B

You’ve identified lots of patches f^-1(U_i) with U_i x F

But maybe we want those to be sort of compatible in some sense

So I think you might want to enforce some extra conditions on the nature of those homeomorphisms, but i don’t quite remember lol

Point is though, if you zoom in on X, a whatever bundle over X is something with a map to X which looks like the projection X x F -> X

[sharp yoke]

i think you can do this with induction 🤔

base case is easy

and then i think a 2n-gon is the connected sum of a 2(n-1)-gon and a 2-gon

not sure tho

don't even know how to write this formally

[sweet wing]

you dont need induction

think simple

maybe if you want some motivation what if it wasnt compact what does that tell you about the polygon ...

solution: ||it's a quotient of a compact surface||

[sharp yoke]

ur saying that any 2n-gon is compact

and then the quotient of a compact surface is compact?

[sweet wing]

essentially

yea

rmb your 2n-gon has it's boundary as well

and it is a subspace of R^2